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15400	https://www.khanacademy.org/math/grade-7-math-tx/xa876d090ec748f45:one-variable-equations-and-inequalities	One-variable equations and inequalities \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. 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Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. 7th grade math (TX TEKS)11 units · 111 skillsUnit 1 Number and operationsUnit 2 One-variable equations and inequalitiesUnit 3 Geometric equationsUnit 4 Proportional reasoning with ratios and ratesUnit 5 Graphs and two-variable equationsUnit 6 SimilarityUnit 7 ProbabilityUnit 8 Circles and composite figuresUnit 9 Volume and surface areaUnit 10 Data representationsUnit 11 Teacher resources Course challenge Test your knowledge of the skills in this course.Start Course challenge Math 7th grade math (TX TEKS) Unit 2: One-variable equations and inequalities 2,000 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test Testing solutions to equations Model two-step equations Two-step equation visual models Two-step equations Graph two-step equations One-variable equations and inequalities: Quiz 1 Two-step equations with decimals and fractions Find the mistake: two-step equations Interpret two-step equation word problems Two-step equations word problems One-variable equations and inequalities: Quiz 2 Testing solutions to inequalities Model two-step inequalities Two-step inequality visual models Two-step inequalities Two-step inequality word problems Write two-step word problems One-variable equations and inequalities: Quiz 3 Discount, markup, and commission word problems Compare discounts Scale a budget Variable expressions with exponents Simple and compound interest One-variable equations and inequalities: Quiz 4 One-variable equations and inequalities: Unit test About this unit Let's solve one-variable equations and inequalities. We'll model and solve them visually, then explore how to work backwards through the order of operations to solve them algebraically. We'll also reason with variables to model financial and other real-world contexts, such as finding simple and compound interest. Unit guides are here! Power up your classroom with engaging strategies, tools, and activities from Khan Academy’s learning experts. PDF Solutions to equations Learn Testing solutions to equations (Opens a modal) Practice Up next for you:Testing solutions to equationsGet 5 of 7 questions to level up! Start Not started Two-step equation intro Learn Modeling two step equations (Opens a modal) Intro to two-step equations (Opens a modal) Two-step equations intuition (Opens a modal) Worked example: two-step equations (Opens a modal) Graphing two-step equations (Opens a modal) Practice Model two-step equationsGet 3 of 4 questions to level up! Practice Not started Two-step equation visual modelsGet 3 of 4 questions to level up! Practice Not started Two-step equationsGet 5 of 7 questions to level up! Practice Not started Graph two-step equationsGet 3 of 4 questions to level up! Practice Not started Quiz 1 Level up on the above skills and collect up to 400 Mastery points Start quiz Two-step equations with decimals and fractions Learn Two-step equations with decimals and fractions (Opens a modal) Two-step equations with decimals and fractions (Opens a modal) Find the mistake: two-step equations (Opens a modal) Practice Two-step equations with decimals and fractionsGet 5 of 7 questions to level up! Practice Not started Find the mistake: two-step equationsGet 3 of 4 questions to level up! Practice Not started Two-step equation word problems Learn Two-step equation word problem: garden (Opens a modal) Two-step equation word problem: oranges (Opens a modal) Practice Interpret two-step equation word problemsGet 3 of 4 questions to level up! Practice Not started Two-step equations word problemsGet 3 of 4 questions to level up! Practice Not started Quiz 2 Level up on the above skills and collect up to 320 Mastery points Start quiz Solutions to inequalities Learn Testing solutions to inequalities (Opens a modal) Practice Testing solutions to inequalitiesGet 3 of 4 questions to level up! Practice Not started Two-step inequalities Learn Modeling two step inequalities (Opens a modal) Modeling inequalities with negative coefficients (Opens a modal) Two-step inequalities (Opens a modal) Practice Model two-step inequalitiesGet 3 of 4 questions to level up! Practice Not started Two-step inequality visual modelsGet 3 of 4 questions to level up! Practice Not started Two-step inequalitiesGet 5 of 7 questions to level up! Practice Not started Two-step inequalities and equations word problems Learn Two-step inequality word problem: apples (Opens a modal) Two-step inequality word problem: R&B (Opens a modal) Writing two step word problems (Opens a modal) Practice Two-step inequality word problemsGet 3 of 4 questions to level up! Practice Not started Write two-step word problemsGet 3 of 4 questions to level up! Practice Not started Quiz 3 Level up on the above skills and collect up to 480 Mastery points Start quiz Discounts Learn Percent word problems: tax and discount (Opens a modal) Percent word problem: guavas (Opens a modal) Combining discounts (Opens a modal) Comparing discounts (Opens a modal) Practice Discount, markup, and commission word problemsGet 3 of 4 questions to level up! Practice Not started Compare discountsGet 3 of 4 questions to level up! Practice Not started Scaling a budget Learn Family budget calculator (Opens a modal) Practice Scale a budgetGet 3 of 4 questions to level up! Practice Not started Simple and compound interest Learn How to make your money grow (Opens a modal) Calculating simple & compound interest (Opens a modal) Compound growth (Opens a modal) Practice Variable expressions with exponentsGet 3 of 4 questions to level up! Practice Not started Simple and compound interestGet 3 of 4 questions to level up! Practice Not started Quiz 4 Level up on the above skills and collect up to 400 Mastery points Start quiz Unit test Level up on all the skills in this unit and collect up to 2,000 Mastery points!Start Unit test Texas 7th grade math is brought to you with support from the Exxon Mobile Foundation Texas 7th grade math is brought to you with support from the Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization. Donate or volunteer today! 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15401	https://dlmf.nist.gov/26.15	DLMF: §26.15 Permutations: Matrix Notation ‣ Properties ‣ Chapter 26 Combinatorial Analysis DLMF Index Notations Search Help? Citing Customize Annotate UnAnnotate About the Project 26 Combinatorial AnalysisProperties26.14 Permutations: Order Notation26.16 Multiset Permutations §26.15 Permutations: Matrix Notation ⓘ Keywords:inversion numbers, matrix notation, permutations, restricted position, rook polynomial, signNotes:See Stanley (1997, pp.71–76), Tucker (2006, pp.335–345).Referenced by:§26.18, Erratum (V1.0.19) for NotationPermalink: also:Annotations for Ch.26 The set 𝔖 n (§26.13) can be identified with the set of n×n matrices of 0’s and 1’s with exactly one 1 in each row and column. The permutation σ corresponds to the matrix in which there is a 1 at the intersection of row j with column σ⁡(j), and 0’s in all other positions. The permutation 35247816 corresponds to the matrix 26.15.1[0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0] ⓘ Permalink: pMML, pngSee also:Annotations for §26.15 and Ch.26 The sign of the permutation σ is the sign of the determinant of its matrix representation. The inversion number of σ is a sum of products of pairs of entries in the matrix representation of σ: 26.15.2 inv(σ)=∑a g⁢h⁢a k⁢ℓ, ⓘ Symbols:h: nonnegative integer, k: nonnegative integer, ℓ: nonnegative integer and σ: permutationPermalink: pMML, pngSee also:Annotations for §26.15 and Ch.26 where the sum is over 1≤gℓ≥1. The matrix represents the placement of n nonattacking rooks on an n×n chessboard, that is, rooks that share neither a row nor a column with any other rook. A permutation with restricted position specifies a subset B⊆{1,2,…,n}×{1,2,…,n}. If (j,k)∈B, then σ⁡(j)≠k. The number of derangements of n is the number of permutations with forbidden positions B={(1,1),(2,2),…,(n,n)}. Let r j⁡(B) be the number of ways of placing j nonattacking rooks on the squares of B. Define r 0⁡(B)=1. For the problem of derangements, r j⁡(B)=(n j). The rook polynomial is the generating function for r j⁡(B): 26.15.3 R⁡(x,B)=∑j=0 n r j⁡(B)⁢x j. ⓘ Symbols:x: real variable, j: nonnegative integer, n: nonnegative integer, B: subset, r j⁡(B): number of rook positions and R⁡(x,B): rook polynomialPermalink: pMML, pngSee also:Annotations for §26.15 and Ch.26 If B=B 1∪B 2, where no element of B 1 is in the same row or column as any element of B 2, then 26.15.4 R⁡(x,B)=R⁡(x,B 1)⁢R⁡(x,B 2). ⓘ Symbols:x: real variable, B: subset and R⁡(x,B): rook polynomialPermalink: pMML, pngSee also:Annotations for §26.15 and Ch.26 For (j,k)∈B, B∖[j,k] denotes B after removal of all elements of the form (j,t) or (t,k), t=1,2,…,n. B∖(j,k) denotes B with the element (j,k) removed. 26.15.5 R⁡(x,B)=x⁢R⁡(x,B∖[j,k])+R⁡(x,B∖(j,k)). ⓘ Symbols:[a,b]: closed interval, (a,b): open interval, ∖: set subtraction, x: real variable, j: nonnegative integer, k: nonnegative integer, B: subset and R⁡(x,B): rook polynomialPermalink: pMML, pngSee also:Annotations for §26.15 and Ch.26 N k⁡(B) is the number of permutations in 𝔖 n for which exactly k of the pairs (j,σ⁡(j)) are elements of B. N⁡(x,B) is the generating function: 26.15.6 N⁡(x,B)=∑k=0 n N k⁡(B)⁢x k, ⓘ Symbols:x: real variable, k: nonnegative integer, n: nonnegative integer, B: subset and N⁡(x,B): generating functionPermalink: pMML, pngSee also:Annotations for §26.15 and Ch.26 and 26.15.7 N⁡(x,B)=∑k=0 n r k⁡(B)⁢(n−k)!⁢(x−1)k. ⓘ Symbols:!: factorial (as in n!), x: real variable, k: nonnegative integer, n: nonnegative integer, B: subset, r j⁡(B): number of rook positions and N⁡(x,B): generating functionPermalink: pMML, pngSee also:Annotations for §26.15 and Ch.26 The number of permutations that avoid B is 26.15.8 N 0⁡(B)≡N⁡(0,B)=∑k=0 n(−1)k⁢r k⁡(B)⁢(n−k)!. ⓘ Symbols:≡: equals by definition, !: factorial (as in n!), k: nonnegative integer, n: nonnegative integer, B: subset, r j⁡(B): number of rook positions and N⁡(x,B): generating functionPermalink: pMML, pngSee also:Annotations for §26.15 and Ch.26 Example 1 ⓘ Keywords:problème des ménagesSee also:Annotations for §26.15 and Ch.26 The problème des ménages asks for the number of ways of seating n married couples around a circular table with labeled seats so that no men are adjacent, no women are adjacent, and no husband and wife are adjacent. There are 2⁢(n!) ways to place the wives. Let B={(j,j),(j,j+1)\| 1≤j<n}∪{(n,n),(n,1)}. Then 26.15.9 r k⁡(B)=2⁢n 2⁢n−k⁢(2⁢n−k k). ⓘ Symbols:(m n): binomial coefficient, k: nonnegative integer, n: nonnegative integer, B: subset and r j⁡(B): number of rook positionsPermalink: pMML, pngSee also:Annotations for §26.15, §26.15 and Ch.26 The solution is 26.15.10 2⁢(n!)⁢N 0⁡(B)=2⁢(n!)⁢∑k=0 n(−1)k⁢2⁢n 2⁢n−k⁢(2⁢n−k k)⁢(n−k)!. ⓘ Symbols:(m n): binomial coefficient, !: factorial (as in n!), k: nonnegative integer, n: nonnegative integer, B: subset and N⁡(x,B): generating functionPermalink: pMML, pngSee also:Annotations for §26.15, §26.15 and Ch.26 Example 2 ⓘ Keywords:Ferrers boardSee also:Annotations for §26.15 and Ch.26 The Ferrers board of shape (b 1,b 2,…,b n), 0≤b 1≤b 2≤⋯≤b n, is the set B={(j,k)\| 1≤j≤n,1≤k≤b j}. For this set, 26.15.11∑k=0 n r n−k⁡(B)⁢(x−k+1)k=∏j=1 n(x+b j−j+1). ⓘ Symbols:(a)n: Pochhammer’s symbol (or shifted factorial), x: real variable, j: nonnegative integer, k: nonnegative integer, n: nonnegative integer, B: subset and r j⁡(B): number of rook positionsPermalink: pMML, pngErrata (effective with 1.1.2): The Pochhammer symbol now links to its definition. See also:Annotations for §26.15, §26.15 and Ch.26 If B is the Ferrers board of shape (0,1,2,…,n−1), then 26.15.12∑k=0 n r n−k⁡(B)⁢(x−k+1)k=x n, ⓘ Symbols:(a)n: Pochhammer’s symbol (or shifted factorial), x: real variable, k: nonnegative integer, n: nonnegative integer, B: subset and r j⁡(B): number of rook positionsPermalink: pMML, pngErrata (effective with 1.1.2): The Pochhammer symbol now links to its definition. See also:Annotations for §26.15, §26.15 and Ch.26 and therefore by (26.8.10), 26.15.13 r n−k⁡(B)=S⁡(n,k). ⓘ Symbols:S⁡(n,k): Stirling number of the second kind, k: nonnegative integer, n: nonnegative integer, B: subset and r j⁡(B): number of rook positionsPermalink: pMML, pngSee also:Annotations for §26.15, §26.15 and Ch.26 26.14 Permutations: Order Notation26.16 Multiset Permutations © 2010–2025 NIST / Disclaimer / Feedback; Version 1.2.4; Release date 2025-03-15. Site PrivacyAccessibilityPrivacy ProgramCopyrightsVulnerability DisclosureNo Fear Act PolicyFOIAEnvironmental PolicyScientific IntegrityInformation Quality StandardsCommerce.govScience.govUSA.gov
15402	https://statproofbook.github.io/P/mean-tot.html	Loading [MathJax]/jax/output/CommonHTML/fonts/TeX/fontdata.js The Book of Statistical Proofs Proof: Law of total expectation Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Expected value ▷ Law of total expectation Theorem: (law of total expectation, also called “law of iterated expectations”) Let X be a random variable with expected value \mathrm{E}(X) and let Y be any random variable defined on the same probability space. Then, the expected value of the conditional expectation of X given Y is the same as the expected value of X: \label{eq:mean-tot} \mathrm{E}(X) = \mathrm{E}[\mathrm{E}(X \vert Y)] \; . Proof: Let X and Y be discrete random variables with sets of possible outcomes \mathcal{X} and \mathcal{Y}. Then, the expectation of the conditional expetectation can be rewritten as: \label{eq:mean-tot-s1} \begin{split} \mathrm{E}[\mathrm{E}(X \vert Y)] &= \mathrm{E}\left[ \sum_{x \in \mathcal{X}} x \cdot \mathrm{Pr}(X = x \vert Y) \right] \ &= \sum_{y \in \mathcal{Y}} \left[ \sum_{x \in \mathcal{X}} x \cdot \mathrm{Pr}(X = x \vert Y = y) \right] \cdot \mathrm{Pr}(Y = y) \ &= \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} x \cdot \mathrm{Pr}(X = x \vert Y = y) \cdot \mathrm{Pr}(Y = y) \; . \end{split} Using the law of conditional probability, this becomes: \label{eq:mean-tot-s2} \begin{split} \mathrm{E}[\mathrm{E}(X \vert Y)] &= \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} x \cdot \mathrm{Pr}(X = x, Y = y) \ &= \sum_{x \in \mathcal{X}} x \sum_{y \in \mathcal{Y}} \mathrm{Pr}(X = x, Y = y) \; . \end{split} Using the law of marginal probability, this becomes: \label{eq:mean-tot-s3} \begin{split} \mathrm{E}[\mathrm{E}(X \vert Y)] &= \sum_{x \in \mathcal{X}} x \cdot \mathrm{Pr}(X = x) \ &= \mathrm{E}(X) \; . \end{split} ∎ Sources: Wikipedia (2021): "Law of total expectation"; in: Wikipedia, the free encyclopedia, retrieved on 2021-11-26; URL: Metadata: ID: P291 \| shortcut: mean-tot \| author: JoramSoch \| date: 2021-11-26, 10:57. view/edit this proof cite this proof
15403	https://fchsmrsneal.files.wordpress.com/2013/06/calculus-book1.pdf	Calculus Single Variable Early Transcendentals james stewart 6E Visit Thomson Brooks/Cole at www.thomsonedu.com Calculus 6E stewart Single Variable Calculus Early Transcendentals AU S T R A L I A N B R A Z I L N C A N A DA N M E X I C O N S I N G A P O R E N S PA I N N U N I T E D K I N G D O M N U N I T E D S TAT E S S I N G L E VA R I A B L E CA L C U L U S E A R LY T R A N S C E N D E N TA L S SIXTH EDITION JAMES STEWART McMASTER UNIVERSITY Publisher N Bob Pirtle Assistant Editor N Stacy Green Editorial Assistant N Elizabeth Rodio Technology Project Manager N Sam Subity Marketing Manager N Mark Santee Marketing Assistant N Melissa Wong Marketing Communications Manager N Bryan Vann Project Manager, Editorial Production N Cheryll Linthicum Creative Director N Rob Hugel Art Director N Vernon T. 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K12T06 Single Variable Calculus: Early Transcendentals, Sixth Edition James Stewart FOR SALLY AND DON FOR ALAN AND SHARON FOR KELLY, KIM, AND CALLUM FOR JACKIE AND NINO Preface xi To the Student xxii Diagnostic Tests xxiv A PREVIEW OF CALCULUS 2 FUNCTIONS AND MODELS 10 1.1 Four Ways to Represent a Function 11 1.2 Mathematical Models: A Catalog of Essential Functions 24 1.3 New Functions from Old Functions 37 1.4 Graphing Calculators and Computers 46 1.5 Exponential Functions 52 1.6 Inverse Functions and Logarithms 59 Review 73 Principles of Problem Solving 76 LIMITS AND DERIVATIVES 82 2.1 The Tangent and Velocity Problems 83 2.2 The Limit of a Function 88 2.3 Calculating Limits Using the Limit Laws 99 2.4 The Precise Deﬁnition of a Limit 109 2.5 Continuity 119 2.6 Limits at Inﬁnity; Horizontal Asymptotes 130 2.7 Derivatives and Rates of Change 143 Writing Project N Early Methods for Finding Tangents 153 2.8 The Derivative as a Function 154 Review 165 Problems Plus 170 2 1 v CONTENTS vi \|\|\|\| CONTENTS DIFFERENTIATION RULES 172 3.1 Derivatives of Polynomials and Exponential Functions 173 Applied Project N Building a Better Roller Coaster 182 3.2 The Product and Quotient Rules 183 3.3 Derivatives of Trigonometric Functions 189 3.4 The Chain Rule 197 Applied Project N Where Should a Pilot Start Descent? 206 3.5 Implicit Differentiation 207 3.6 Derivatives of Logarithmic Functions 215 3.7 Rates of Change in the Natural and Social Sciences 221 3.8 Exponential Growth and Decay 233 3.9 Related Rates 241 3.10 Linear Approximations and Differentials 247 Laboratory Project N Taylor Polynomials 253 3.11 Hyperbolic Functions 254 Review 261 Problems Plus 265 APPLICATIONS OF DIFFERENTIATION 270 4.1 Maximum and Minimum Values 271 Applied Project N The Calculus of Rainbows 279 4.2 The Mean Value Theorem 280 4.3 How Derivatives Affect the Shape of a Graph 287 4.4 Indeterminate Forms and L’Hospital’s Rule 298 Writing Project N The Origins of L’Hospital’s Rule 307 4.5 Summary of Curve Sketching 307 4.6 Graphing with Calculus and Calculators 315 4.7 Optimization Problems 322 Applied Project N The Shape of a Can 333 4.8 Newton’s Method 334 4.9 Antiderivatives 340 Review 347 Problems Plus 351 4 3 0 y 0 π 2 m=1 m=_1 m=0 π 2 π π INTEGRALS 354 5.1 Areas and Distances 355 5.2 The Deﬁnite Integral 366 Discovery Project N Area Functions 379 5.3 The Fundamental Theorem of Calculus 379 5.4 Indeﬁnite Integrals and the Net Change Theorem 391 Writing Project N Newton, Leibniz, and the Invention of Calculus 399 5.5 The Substitution Rule 400 Review 408 Problems Plus 412 APPLICATIONS OF INTEGRATION 414 6.1 Areas between Curves 415 6.2 Volumes 422 6.3 Volumes by Cylindrical Shells 433 6.4 Work 438 6.5 Average Value of a Function 442 Applied Project N Where to Sit at the Movies 446 Review 446 Problems Plus 448 TECHNIQUES OF INTEGRATION 452 7.1 Integration by Parts 453 7.2 Trigonometric Integrals 460 7.3 Trigonometric Substitution 467 7.4 Integration of Rational Functions by Partial Fractions 473 7.5 Strategy for Integration 483 7.6 Integration Using Tables and Computer Algebra Systems 489 Discovery Project N Patterns in Integrals 494 7 6 5 CONTENTS \|\|\|\| VII viii \|\|\|\| CONTENTS 7.7 Approximate Integration 495 7.8 Improper Integrals 508 Review 518 Problems Plus 521 FURTHER APPLICATIONS OF INTEGRATION 524 8.1 Arc Length 525 Discovery Project N Arc Length Contest 532 8.2 Area of a Surface of Revolution 532 Discovery Project N Rotating on a Slant 538 8.3 Applications to Physics and Engineering 539 Discovery Project N Complementary Coffee Cups 550 8.4 Applications to Economics and Biology 550 8.5 Probability 555 Review 562 Problems Plus 564 DIFFERENTIAL EQUATIONS 566 9.1 Modeling with Differential Equations 567 9.2 Direction Fields and Euler’s Method 572 9.3 Separable Equations 580 Applied Project N How Fast Does a Tank Drain? 588 Applied Project N Which Is Faster, Going Up or Coming Down? 590 9.4 Models for Population Growth 591 Applied Project N Calculus and Baseball 601 9.5 Linear Equations 602 9.6 Predator-Prey Systems 608 Review 614 Problems Plus 618 9 8 PARAMETRIC EQUATIONS AND POLAR COORDINATES 620 10.1 Curves Deﬁned by Parametric Equations 621 Laboratory Project N Running Circles around Circles 629 10.2 Calculus with Parametric Curves 630 Laboratory Project N Bézier Curves 639 10.3 Polar Coordinates 639 10.4 Areas and Lengths in Polar Coordinates 650 10.5 Conic Sections 654 10.6 Conic Sections in Polar Coordinates 662 Review 669 Problems Plus 672 INFINITE SEQUENCES AND SERIES 674 11.1 Sequences 675 Laboratory Project N Logistic Sequences 687 11.2 Series 687 11.3 The Integral Test and Estimates of Sums 697 11.4 The Comparison Tests 705 11.5 Alternating Series 710 11.6 Absolute Convergence and the Ratio and Root Tests 714 11.7 Strategy for Testing Series 721 11.8 Power Series 723 11.9 Representations of Functions as Power Series 728 11.10 Taylor and Maclaurin Series 734 Laboratory Project N An Elusive Limit 748 Writing Project N How Newton Discovered the Binomial Series 748 11.11 Applications of Taylor Polynomials 749 Applied Project N Radiation from the Stars 757 Review 758 Problems Plus 761 11 10 CONTENTS \|\|\|\| ix APPENDIXES A1 A Numbers, Inequalities, and Absolute Values A2 B Coordinate Geometry and Lines A10 C Graphs of Second-Degree Equations A16 D Trigonometry A24 E Sigma Notation A34 F Proofs of Theorems A39 G The Logarithm Defined as an Integral A48 H Complex Numbers A55 I Answers to Odd-Numbered Exercises A63 INDEX A112 x \|\|\|\| CONTENTS A great discovery solves a great problem but there is a grain of discovery in the solution of any problem.Your problem may be modest; but if it challenges your curiosity and brings into play your inventive faculties, and if you solve it by your own means, you may experience the tension and enjoy the triumph of discovery. GEORGE POLYA PREFACE The art of teaching, Mark Van Doren said, is the art of assisting discovery. I have tried to write a book that assists students in discovering calculus—both for its practical power and its surprising beauty. In this edition, as in the ﬁrst ﬁve editions, I aim to convey to the stu-dent a sense of the utility of calculus and develop technical competence, but I also strive to give some appreciation for the intrinsic beauty of the subject. Newton undoubtedly experienced a sense of triumph when he made his great discoveries. I want students to share some of that excitement. The emphasis is on understanding concepts. I think that nearly everybody agrees that this should be the primary goal of calculus instruction. In fact, the impetus for the current calculus reform movement came from the Tulane Conference in 1986, which formulated as their ﬁrst recommendation: Focus on conceptual understanding. I have tried to implement this goal through the Rule of Three: “Topics should be pre-sented geometrically, numerically, and algebraically.” Visualization, numerical and graph-ical experimentation, and other approaches have changed how we teach conceptual reasoning in fundamental ways. More recently, the Rule of Three has been expanded to become the Rule of Four by emphasizing the verbal, or descriptive, point of view as well. In writing the sixth edition my premise has been that it is possible to achieve concep-tual understanding and still retain the best traditions of traditional calculus. The book con-tains elements of reform, but within the context of a traditional curriculum. ALTERNATIVE VERSIONS I have written several other calculus textbooks that might be preferable for some instruc-tors. Most of them also come in single variable and multivariable versions. N Calculus, Sixth Edition, is similar to the present textbook except that the exponential, logarithmic, and inverse trigonometric functions are covered in the second semester. N Essential Calculus is a much briefer book (800 pages), though it contains almost all of the topics in the present text. The relative brevity is achieved through briefer exposition of some topics and putting some features on the website. N Essential Calculus: Early Transcendentals resembles Essential Calculus, but the expo-nential, logarithmic, and inverse trigonometric functions are covered in Chapter 3. xi N Calculus: Concepts and Contexts, Third Edition, emphasizes conceptual understanding even more strongly than this book. The coverage of topics is not encyclopedic and the material on transcendental functions and on parametric equations is woven throughout the book instead of being treated in separate chapters. N Calculus: Early Vectors introduces vectors and vector functions in the ﬁrst semester and integrates them throughout the book. It is suitable for students taking Engineering and Physics courses concurrently with calculus. WHAT’S NEW IN THE SIXTH EDITION? Here are some of the changes for the sixth edition of Single Variable Calculus: Early Transcendentals: N At the beginning of the book there are four diagnostic tests, in Basic Algebra, Analytic Geometry, Functions, and Trigonometry. Answers are given and students who don’t do well are referred to where they should seek help (Appendixes, review sections of Chapter 1, and the website). N In response to requests of several users, the material motivating the derivative is briefer: Sections 2.7 and 2.8 are combined into a single section called Derivatives and Rates of Change. N The section on Higher Derivatives in Chapter 3 has disappeared and that material is integrated into various sections in Chapters 2 and 3. N Instructors who do not cover the chapter on differential equations have commented that the section on Exponential Growth and Decay was inconveniently located there. Accordingly, it is moved earlier in the book, to Chapter 3. This move precipitates a reorganization of Chapters 3 and 9. N Sections 4.7 and 4.8 are merged into a single section, with a briefer treatment of opti-mization problems in business and economics. N Sections 11.10 and 11.11 are merged into a single section. I had previously featured the binomial series in its own section to emphasize its importance. But I learned that some instructors were omitting that section, so I have decided to incorporate binomial series into 11.10. N New phrases and margin notes have been added to clarify the exposition. N A number of pieces of art have been redrawn. N The data in examples and exercises have been updated to be more timely. N Many examples have been added or changed. For instance, Example 2 on page 185 was changed because students are often bafﬂed when they see arbitrary constants in a problem and I wanted to give an example in which they occur. N Extra steps have been provided in some of the existing examples. N More than 25% of the exercises in each chapter are new. Here are a few of my favorites: 3.1.79, 3.1.80, 4.3.62, 4.3.83, 4.6.36 and 11.11.30. N There are also some good new problems in the Problems Plus sections. See, for instance, Problems 2 and 13 on page 413, Problem 13 on page 450, and Problem 24 on page 763. N The new project on page 550, Complementary Coffee Cups, comes from an article by Thomas Banchoff in which he wondered which of two coffee cups, whose convex and concave proﬁles ﬁt together snugly, would hold more coffee. xii \|\|\|\| PREFACE PREFACE \|\|\|\| xiii N Tools for Enriching Calculus (TEC) has been completely redesigned and is accessible on the Internet at www.stewartcalculus.com. It now includes what we call Visuals, brief animations of various ﬁgures in the text. See the description on page xiv. N The symbol has been placed beside examples (an average of three per section) for which there are videos of instructors explaining the example in more detail. This material is also available on DVD. See the description on page xx. FEATURES CONCEPTUAL EXERCISES The most important way to foster conceptual understanding is through the problems that we assign. To that end I have devised various types of problems. Some exercise sets begin with requests to explain the meanings of the basic concepts of the section. (See, for instance, the ﬁrst few exercises in Sections 2.2, 2.5, and 11.2.) Similarly, all the review sec-tions begin with a Concept Check and a True-False Quiz. Other exercises test conceptual understanding through graphs or tables (see Exercises 2.7.17, 2.8.33–38, 2.8.41–44, 9.1.11–12, 10.1.24–27, and 11.10.2). Another type of exercise uses verbal description to test conceptual understanding (see Exercises 2.5.8, 2.8.56, 4.3.63–64, and 7.8.67). I particularly value problems that combine and compare graphical, numerical, and algebraic approaches (see Exercises 2.6.37–38, 3.7.25, and 9.4.2). GRADED EXERCISE SETS Each exercise set is carefully graded, progressing from basic conceptual exercises and skill-development problems to more challenging problems involving applications and proofs. REAL-WORLD DATA My assistants and I spent a great deal of time looking in libraries, contacting companies and government agencies, and searching the Internet for interesting real-world data to intro-duce, motivate, and illustrate the concepts of calculus. As a result, many of the examples and exercises deal with functions deﬁned by such numerical data or graphs. See, for instance, Figure 1 in Section 1.1 (seismograms from the Northridge earthquake), Exercise 2.8.34 (percentage of the population under age 18), Exercise 5.1.14 (velocity of the space shuttle Endeavour), and Figure 4 in Section 5.4 (San Francisco power consumption). PROJECTS One way of involving students and making them active learners is to have them work (per-haps in groups) on extended projects that give a feeling of substantial accomplishment when completed. I have included four kinds of projects: Applied Projects involve applica-tions that are designed to appeal to the imagination of students. The project after Section 9.3 asks whether a ball thrown upward takes longer to reach its maximum height or to fall back to its original height. (The answer might surprise you.) Laboratory Projects involve technology; the one following Section 10.2 shows how to use Bézier curves to design shapes that represent letters for a laser printer. Writing Projects ask students to compare present-day methods with those of the founders of calculus—Fermat’s method for ﬁnding tangents, for instance. Suggested references are supplied. Discovery Projects anticipate results to be discussed later or encourage discovery through pattern recognition (see the one following Section 7.6). Additional projects can be found in the Instructor’s Guide (see, for instance, Group Exercise 5.1: Position from Samples). PROBLEM SOLVING Students usually have difﬁculties with problems for which there is no single well-deﬁned procedure for obtaining the answer. I think nobody has improved very much on George Polya’s four-stage problem-solving strategy and, accordingly, I have included a version of his problem-solving principles following Chapter 1. They are applied, both explicitly and implicitly, throughout the book. After the other chapters I have placed sections called Problems Plus, which feature examples of how to tackle challenging calculus problems. In V selecting the varied problems for these sections I kept in mind the following advice from David Hilbert: “A mathematical problem should be difﬁcult in order to entice us, yet not inaccessible lest it mock our efforts.” When I put these challenging problems on assign-ments and tests I grade them in a different way. Here I reward a student signiﬁcantly for ideas toward a solution and for recognizing which problem-solving principles are relevant. TECHNOLOGY The availability of technology makes it not less important but more important to clearly understand the concepts that underlie the images on the screen. But, when properly used, graphing calculators and computers are powerful tools for discovering and understanding those concepts. This textbook can be used either with or without technology and I use two special symbols to indicate clearly when a particular type of machine is required. The icon ; indicates an exercise that deﬁnitely requires the use of such technology, but that is not to say that it can’t be used on the other exercises as well. The symbol is reserved for problems in which the full resources of a computer algebra system (like Derive, Maple, Mathematica, or the TI-89/92) are required. But technology doesn’t make pencil and paper obsolete. Hand calculation and sketches are often preferable to technology for illustrating and reinforcing some concepts. Both instructors and students need to develop the ability to decide where the hand or the machine is appropriate. TEC is a companion to the text and is intended to enrich and complement its contents. (It is now accessible from the Internet at www.stewartcalculus.com.) Developed by Har-vey Keynes, Dan Clegg, Hubert Hohn, and myself, TEC uses a discovery and exploratory approach. In sections of the book where technology is particularly appropriate, marginal icons direct students to TEC modules that provide a laboratory environment in which they can explore the topic in different ways and at different levels. Visuals are animations of ﬁg-ures in text; Modules are more elaborate activities and include exercises. Instructors can choose to become involved at several different levels, ranging from simply encouraging students to use the Visuals and Modules for independent exploration, to assigning spe-ciﬁc exercises from those included with each Module, or to creating additional exercises, labs, and projects that make use of the Visuals and Modules. TEC also includes Homework Hints for representative exercises (usually odd-numbered) in every section of the text, indicated by printing the exercise number in red. These hints are usually presented in the form of questions and try to imitate an effective teaching assistant by functioning as a silent tutor. They are constructed so as not to reveal any more of the actual solution than is minimally necessary to make further progress. ENHANCED WEBASSIGN Technology is having an impact on the way homework is assigned to students, particu-larly in large classes. The use of online homework is growing and its appeal depends on ease of use, grading precision, and reliability. With the sixth edition we have been work-ing with the calculus community and WebAssign to develop an online homework system. Up to 70% of the exercises in each section are assignable as online homework, including free response, multiple choice, and multi-part formats. Some questions are multi-part problems based on simulations of the TEC Modules. The system also includes Active Examples, in which students are guided in step-by-step tutorials through text examples, with links to the textbook and to video solutions. WEBSITE: www.stewartcalculus.com This site has been renovated and now includes the following. N Algebra Review N Lies My Calculator and Computer Told Me N History of Mathematics, with links to the better historical websites N Additional Topics (complete with exercise sets): Fourier Series, Formulas for the Remainder Term in Taylor Series, Rotation of Axes TOOLS FOR ENRICHING™CALCULUS CAS xiv \|\|\|\| PREFACE PREFACE \|\|\|\| XV N Archived Problems (Drill exercises that appeared in previous editions, together with their solutions) N Challenge Problems (some from the Problems Plus sections from prior editions) N Links, for particular topics, to outside web resources N The complete Tools for Enriching Calculus (TEC) Modules, Visuals, and Homework Hints CONTENT Diagnostic Tests The book begins with four diagnostic tests, in Basic Algebra, Analytic Geometry, Func-tions, and Trigonometry. A Preview of Calculus This is an overview of the subject and includes a list of questions to motivate the study of calculus. 1 N Functions and Models From the beginning, multiple representations of functions are stressed: verbal, numerical, visual, and algebraic. A discussion of mathematical models leads to a review of the standard functions, including exponential and logarithmic functions, from these four points of view. 2 N Limits and Derivatives The material on limits is motivated by a prior discussion of the tangent and velocity prob-lems. Limits are treated from descriptive, graphical, numerical, and algebraic points of view. Section 2.4, on the precise ∑-∂deﬁnition of a limit, is an optional section. Sec-tions 2.7 and 2.8 deal with derivatives (especially with functions deﬁned graphically and numerically) before the differentiation rules are covered in Chapter 3. Here the examples and exercises explore the meanings of derivatives in various contexts. Higher derivatives are now introduced in Section 2.8. 3 N Differentiation Rules All the basic functions, including exponential, logarithmic, and inverse trigonometric func-tions, are differentiated here. When derivatives are computed in applied situations, students are asked to explain their meanings. Exponential growth and decay are now covered in this chapter. 4 N Applications of Differentiation The basic facts concerning extreme values and shapes of curves are deduced from the Mean Value Theorem. Graphing with technology emphasizes the interaction between cal-culus and calculators and the analysis of families of curves. Some substantial optimization problems are provided, including an explanation of why you need to raise your head 42° to see the top of a rainbow. 5 N Integrals The area problem and the distance problem serve to motivate the deﬁnite integral, with sigma notation introduced as needed. (Full coverage of sigma notation is provided in Appen-dix E.) Emphasis is placed on explaining the meanings of integrals in various contexts and on estimating their values from graphs and tables. 6 N Applications of Integration Here I present the applications of integration—area, volume, work, average value—that can reasonably be done without specialized techniques of integration. General methods are emphasized. The goal is for students to be able to divide a quantity into small pieces, esti-mate with Riemann sums, and recognize the limit as an integral. 7 N Techniques of Integration All the standard methods are covered but, of course, the real challenge is to be able to recog-nize which technique is best used in a given situation. Accordingly, in Section 7.5, I present a strategy for integration. The use of computer algebra systems is discussed in Section 7.6. Here are the applications of integration—arc length and surface area—for which it is use-ful to have available all the techniques of integration, as well as applications to biology, economics, and physics (hydrostatic force and centers of mass). I have also included a sec-tion on probability. There are more applications here than can realistically be covered in a given course. Instructors should select applications suitable for their students and for which they themselves have enthusiasm. 9 N Differential Equations Modeling is the theme that uniﬁes this introductory treatment of differential equations. Direction ﬁelds and Euler’s method are studied before separable and linear equations are solved explicitly, so that qualitative, numerical, and analytic approaches are given equal consideration. These methods are applied to the exponential, logistic, and other models for population growth. The ﬁrst four or ﬁve sections of this chapter serve as a good intro-duction to ﬁrst-order differential equations. An optional ﬁnal section uses predator-prey models to illustrate systems of differential equations. This chapter introduces parametric and polar curves and applies the methods of calculus to them. Parametric curves are well suited to laboratory projects; the two presented here involve families of curves and Bézier curves. A brief treatment of conic sections in polar coordinates prepares the way for Kepler’s Laws in Chapter 13. 11 N Infinite Sequences and Series The convergence tests have intuitive justiﬁcations (see page 697) as well as formal proofs. Numerical estimates of sums of series are based on which test was used to prove conver-gence. The emphasis is on Taylor series and polynomials and their applications to physics. Error estimates include those from graphing devices. ANCILLARIES Calculus, Early Transcendentals, Sixth Edition, is supported by a complete set of ancil-laries developed under my direction. Each piece has been designed to enhance student understanding and to facilitate creative instruction. The tables on pages xx–xxi describe each of these ancillaries. ACKNOWLEDGMENTS The preparation of this and previous editions has involved much time spent reading the reasoned (but sometimes contradictory) advice from a large number of astute reviewers. I greatly appreciate the time they spent to understand my motivation for the approach taken. I have learned something from each of them. SIXTH EDITION REVIEWERS Marilyn Belkin, Villanova University Philip L. Bowers, Florida State University Amy Elizabeth Bowman, University of Alabama in Huntsville M. Hilary Davies, University of Alaska Anchorage Frederick Gass, Miami University Nets Katz, Indiana University Bloomington James McKinney, California State Polytechnic University, Pomona Martin Nakashima, California State Polytechnic University, Pomona Lila Roberts, Georgia College and State University Paul Triantaﬁlos, Armstrong Atlantic State University 10 N Parametric Equations and Polar Coordinates 8 N Further Applications of Integration xvi \|\|\|\| PREFACE PREFACE \|\|\|\| xvii PREVIOUS EDITION REVIEWERS B. D. Aggarwala, University of Calgary John Alberghini, Manchester Community College Michael Albert, Carnegie-Mellon University Daniel Anderson, University of Iowa Donna J. Bailey, Northeast Missouri State University Wayne Barber, Chemeketa Community College Neil Berger, University of Illinois, Chicago David Berman, University of New Orleans Richard Biggs, University of Western Ontario Robert Blumenthal, Oglethorpe University Martina Bode, Northwestern University Barbara Bohannon, Hofstra University Philip L. Bowers, Florida State University Jay Bourland, Colorado State University Stephen W. Brady, Wichita State University Michael Breen, Tennessee Technological University Robert N. Bryan, University of Western Ontario David Buchthal, University of Akron Jorge Cassio, Miami-Dade Community College Jack Ceder, University of California, Santa Barbara Scott Chapman, Trinity University James Choike, Oklahoma State University Barbara Cortzen, DePaul University Carl Cowen, Purdue University Philip S. Crooke, Vanderbilt University Charles N. Curtis, Missouri Southern State College Daniel Cyphert, Armstrong State College Robert Dahlin Gregory J. Davis, University of Wisconsin–Green Bay Elias Deeba, University of Houston–Downtown Daniel DiMaria, Suffolk Community College Seymour Ditor, University of Western Ontario Greg Dresden, Washington and Lee University Daniel Drucker, Wayne State University Kenn Dunn, Dalhousie University Dennis Dunninger, Michigan State University Bruce Edwards, University of Florida David Ellis, San Francisco State University John Ellison, Grove City College Martin Erickson, Truman State University Garret Etgen, University of Houston Theodore G. Faticoni, Fordham University Laurene V. Fausett, Georgia Southern University Norman Feldman, Sonoma State University Newman Fisher, San Francisco State University José D. Flores, The University of South Dakota William Francis, Michigan Technological University James T. Franklin, Valencia Community College, East Stanley Friedlander, Bronx Community College Patrick Gallagher, Columbia University–New York Paul Garrett, University of Minnesota–Minneapolis Frederick Gass, Miami University of Ohio Bruce Gilligan, University of Regina Matthias K. Gobbert, University of Maryland, Baltimore County Gerald Goff, Oklahoma State University Stuart Goldenberg, California Polytechnic State University John A. Graham, Buckingham Browne & Nichols School Richard Grassl, University of New Mexico Michael Gregory, University of North Dakota Charles Groetsch, University of Cincinnati Salim M. Haïdar, Grand Valley State University D. W. Hall, Michigan State University Robert L. Hall, University of Wisconsin–Milwaukee Howard B. Hamilton, California State University, Sacramento Darel Hardy, Colorado State University Gary W. Harrison, College of Charleston Melvin Hausner, New York University/Courant Institute Curtis Herink, Mercer University Russell Herman, University of North Carolina at Wilmington Allen Hesse, Rochester Community College Randall R. Holmes, Auburn University James F. Hurley, University of Connecticut Matthew A. Isom, Arizona State University Gerald Janusz, University of Illinois at Urbana-Champaign John H. Jenkins, Embry-Riddle Aeronautical University, Prescott Campus Clement Jeske, University of Wisconsin, Platteville Carl Jockusch, University of Illinois at Urbana-Champaign Jan E. H. Johansson, University of Vermont Jerry Johnson, Oklahoma State University Zsuzsanna M. Kadas, St. Michael’s College Matt Kaufman Matthias Kawski, Arizona State University Frederick W. Keene, Pasadena City College Robert L. Kelley, University of Miami Virgil Kowalik, Texas A&I University Kevin Kreider, University of Akron Leonard Krop, DePaul University Mark Krusemeyer, Carleton College John C. Lawlor, University of Vermont Christopher C. Leary, State University of New York at Geneseo David Leeming, University of Victoria Sam Lesseig, Northeast Missouri State University Phil Locke, University of Maine Joan McCarter, Arizona State University Phil McCartney, Northern Kentucky University Igor Malyshev, San Jose State University Larry Mansﬁeld, Queens College xviii \|\|\|\| PREFACE Mary Martin, Colgate University Nathaniel F. G. Martin, University of Virginia Gerald Y. Matsumoto, American River College Tom Metzger, University of Pittsburgh Michael Montaño, Riverside Community College Teri Jo Murphy, University of Oklahoma Richard Nowakowski, Dalhousie University Hussain S. Nur, California State University, Fresno Wayne N. Palmer, Utica College Vincent Panico, University of the Paciﬁc F. J. Papp, University of Michigan–Dearborn Mike Penna, Indiana University–Purdue University Indianapolis Mark Pinsky, Northwestern University Lothar Redlin, The Pennsylvania State University Joel W. Robbin, University of Wisconsin–Madison E. Arthur Robinson, Jr., The George Washington University Richard Rockwell, Paciﬁc Union College Rob Root, Lafayette College Richard Ruedemann, Arizona State University David Ryeburn, Simon Fraser University Richard St. Andre, Central Michigan University Ricardo Salinas, San Antonio College Robert Schmidt, South Dakota State University Eric Schreiner, Western Michigan University Mihr J. Shah, Kent State University–Trumbull Theodore Shifrin, University of Georgia Wayne Skrapek, University of Saskatchewan Larry Small, Los Angeles Pierce College Teresa Morgan Smith, Blinn College William Smith, University of North Carolina Donald W. Solomon, University of Wisconsin–Milwaukee Edward Spitznagel, Washington University Joseph Stampﬂi, Indiana University Kristin Stoley, Blinn College M. B. Tavakoli, Chaffey College Paul Xavier Uhlig, St. Mary’s University, San Antonio Stan Ver Nooy, University of Oregon Andrei Verona, California State University–Los Angeles Russell C. Walker, Carnegie Mellon University William L. Walton, McCallie School Jack Weiner, University of Guelph Alan Weinstein, University of California, Berkeley Theodore W. Wilcox, Rochester Institute of Technology Steven Willard, University of Alberta Robert Wilson, University of Wisconsin–Madison Jerome Wolbert, University of Michigan–Ann Arbor Dennis H. Wortman, University of Massachusetts, Boston Mary Wright, Southern Illinois University–Carbondale Paul M. Wright, Austin Community College Xian Wu, University of South Carolina Virgil Kowalik, Texas A&I University Kevin Kreider, University of Akron Leonard Krop, DePaul University Mark Krusemeyer, Carleton College John C. Lawlor, University of Vermont Christopher C. Leary, State University of New York at Geneseo David Leeming, University of Victoria Sam Lesseig, Northeast Missouri State University Phil Locke, University of Maine Joan McCarter, Arizona State University Phil McCartney, Northern Kentucky University Igor Malyshev, San Jose State University Larry Mansﬁeld, Queens College Mary Martin, Colgate University Nathaniel F. G. Martin, University of Virginia Gerald Y. Matsumoto, American River College Tom Metzger, University of Pittsburgh Michael Montaño, Riverside Community College Teri Jo Murphy, University of Oklahoma Richard Nowakowski, Dalhousie University Hussain S. Nur, California State University, Fresno Wayne N. Palmer, Utica College Vincent Panico, University of the Paciﬁc F. J. Papp, University of Michigan–Dearborn Mike Penna, Indiana University–Purdue University Indianapolis Mark Pinsky, Northwestern University Lothar Redlin, The Pennsylvania State University John Ringland, State University of New York at Buffalo Tom Rishel, Cornell University Joel W. Robbin, University of Wisconsin–Madison E. Arthur Robinson, Jr., The George Washington University Richard Rockwell, Paciﬁc Union College Rob Root, Lafayette College Richard Ruedemann, Arizona State University David Ryeburn, Simon Fraser University Richard St. Andre, Central Michigan University Ricardo Salinas, San Antonio College Robert Schmidt, South Dakota State University Eric Schreiner, Western Michigan University Mihr J. Shah, Kent State University–Trumbull Theodore Shifrin, University of Georgia Wayne Skrapek, University of Saskatchewan Larry Small, Los Angeles Pierce College Teresa Morgan Smith, Blinn College William Smith, University of North Carolina Donald W. Solomon, University of Wisconsin–Milwaukee Edward Spitznagel, Washington University Joseph Stampﬂi, Indiana University Kristin Stoley, Blinn College PREFACE \|\|\|\| xix In addition, I would like to thank George Bergman, David Cusick, Stuart Goldenberg, Larry Peterson, Dan Silver, Norton Starr, Alan Weinstein, and Gail Wolkowicz for their suggestions; Dan Clegg for his research in libraries and on the Internet; Arnold Good for his treatment of optimization problems with implicit differentiation; Al Shenk and Dennis Zill for permission to use exercises from their calculus texts; COMAP for permission to use project material; George Bergman, David Bleecker, Dan Clegg, Victor Kaftal, Anthony Lam, Jamie Lawson, Ira Rosenholtz, Paul Sally, Lowell Smylie, and Larry Wallen for ideas for exercises; Dan Drucker for the roller derby project; Thomas Banchoff, Tom Farmer, Fred Gass, John Ramsay, Larry Riddle, and Philip Strafﬁn for ideas for projects; Dan Anderson, Jeff Cole, and Dan Drucker for solving the new exercises; and Marv Riedesel and Mary Johnson for accuracy in proofreading. I’m grateful to Jeff Cole for suggesting ways to improve the exercises. In addition, I thank those who have contributed to past editions: Ed Barbeau, Fred Brauer, Andy Bulman-Fleming, Bob Burton, Tom DiCiccio, Garret Etgen, Chris Fisher, Gene Hecht, Harvey Keynes, Kevin Kreider, E. L. Koh, Zdislav Kovarik, Emile LeBlanc, David Leep, Gerald Leibowitz, Lothar Redlin, Carl Riehm, Peter Rosenthal, Doug Shaw, and Saleem Watson. I also thank Kathi Townes, Stephanie Kuhns, and Brian Betsill of TECHarts for their production services and the following Brooks/Cole staff: Cheryll Linthicum, editorial pro-duction project manager; Mark Santee, Melissa Wong, and Bryan Vann, marketing team; Stacy Green, assistant editor, and Elizabeth Rodio, editorial assistant; Sam Subity, technol-ogy project manager; Rob Hugel, creative director, and Vernon Boes, art director; and Becky Cross, print buyer. They have all done an outstanding job. I have been very fortunate to have worked with some of the best mathematics editors in the business over the past two decades: Ron Munro, Harry Campbell, Craig Barth, Jeremy Hayhurst, Gary Ostedt, and now Bob Pirtle. Bob continues in that tradition of editors who, while offering sound advice and ample assistance, trust my instincts and allow me to write the books that I want to write. JAMES STEWART M. B. Tavakoli, Chaffey College Paul Xavier Uhlig, St. Mary’s University, San Antonio Stan Ver Nooy, University of Oregon Andrei Verona, California State University–Los Angeles Russell C. Walker, Carnegie Mellon University William L. Walton, McCallie School Jack Weiner, University of Guelph Alan Weinstein, University of California, Berkeley Theodore W. Wilcox, Rochester Institute of Technology Steven Willard, University of Alberta Robert Wilson, University of Wisconsin–Madison Jerome Wolbert, University of Michigan–Ann Arbor Dennis H. Wortman, University of Massachusetts, Boston Mary Wright, Southern Illinois University–Carbondale Paul M. Wright, Austin Community College Xian Wu, University of South Carolina ExamView ISBN 0-495-38240-X Create, deliver, and customize tests and study guides (both print and online) in minutes with this easy-to-use assessment and tutorial software on CD. Includes complete questions from the Printed Test Bank. JoinIn on TurningPoint ISBN 0-495-11894-X Enhance how your students interact with you, your lecture, and each other. Thomson Brooks/Cole is now pleased to offer you book-speciﬁc content for Response Systems tailored to Stewart’s Calculus, allowing you to transform your classroom and assess your students’ progress with instant in-class quizzes and polls. Contact your local Thomson representative to learn more about JoinIn on TurningPoint and our exclusive infrared and radio-frequency hardware solutions. Text-Specific DVDs ISBN 0-495-01243-2 Text-speciﬁc DVD set, available at no charge to adopters. Each disk features a 10- to 20-minute problem-solving lesson for each section of the chapter. Covers both single- and multi-variable calculus. Solution Builder www.thomsonedu.com/solutionbuilder The online Solution Builder lets instructors easily build and save personal solution sets either for printing or posting on password-protected class websites. Contact your local sales representative for more information on obtaining an account for this instructor-only resource. Stewart Specialty Website www.stewartcalculus.com Contents: Algebra Review N Additional Topics N Drill exercises N Challenge Problems N Web Links N History of Mathematics N Tools for Enriching Calculus (TEC) Enhanced WebAssign ISBN 0-495-10963-0 Instant feedback, grading precision, and ease of use are just three reasons why WebAssign is the most widely used home-work system in higher education. WebAssign’s homework deliv-ery system lets instructors deliver, collect, grade and record assignments via the web. And now, this proven system has been ANCILLARIES FOR INSTRUCTORS AND STUDENTS \|\|\|\| Electronic items \|\|\|\| Printed items xx Multimedia Manager Instructor’s Resource CD-ROM ISBN 0-495-01241-6 Contains all art from the text in both jpeg and PowerPoint formats, key equations and tables from the text, complete pre-built PowerPoint lectures, and an electronic version of the Instructor’s Guide. Tools for Enriching™ Calculus by James Stewart, Harvey Keynes, Dan Clegg, and developer Hu Hohn TEC provides a laboratory environment in which students can explore selected topics. TEC also includes homework hints for representative exercises. Available via the Enhanced WebAssign homework system and online at www.stewartcalculus.com. Instructor’s Guide by Douglas Shaw and James Stewart ISBN 0-495-01254-8 Each section of the main text is discussed from several view-points and contains suggested time to allot, points to stress, text discussion topics, core materials for lecture, workshop/discus-sion suggestions, group work exercises in a form suitable for handout, and suggested homework problems. An electronic version is available on the Multimedia Manager Instructor’s Resource CD-ROM. Instructor’s Guide for AP® Calculus by Douglas Shaw and Robert Gerver, contributing author ISBN 0-495-01223-8 Taking the perspective of optimizing preparation for the AP exam, each section of the main text is discussed from several viewpoints and contains suggested time to allot, points to stress, daily quizzes, core materials for lecture, workshop/ discussion suggestions, group work exercises in a form suitable for handout, tips for the AP exam, and suggested homework problems. Complete Solutions Manual Single Variable Early Transcendentals by Daniel Anderson, Jeffery A. Cole, and Daniel Drucker ISBN 0-495-01255-6 Includes worked-out solutions to all exercises in the text. Printed Test Bank ISBN 0-495-01242-4 Contains multiple-choice and short-answer test items that key directly to the text. TEC ANCILLARIES FOR INSTRUCTORS enhanced to include end-of-chapter problems from Stewart’s Calculus—incorporating exercises, examples, video skill-builders and quizzes to promote active learning and provide the immediate, relevant feedback students want. The Brooks/Cole Mathematics Resource Center Website www.thomsonedu.com/math When you adopt a Thomson–Brooks/Cole mathematics text, you and your students will have access to a variety of teaching and learning resources. This website features everything from book-speciﬁc resources to newsgroups. It’s a great way to make teaching and learning an interactive and intriguing experience. Maple CD-ROM ISBN 0-495-01492-3 Maple provides an advanced, high performance mathematical computation engine with fully integrated numerics & symbolics, all accessible from a WYSIWIG technical document environ-ment. Available for bundling with yout Stewart Calculus text at a special discount. Tools for Enriching™ Calculus by James Stewart, Harvey Keynes, Dan Clegg, and developer Hu Hohn TEC provides a laboratory environment in which students can explore selected topics. TEC also includes homework hints for representative exercises. Available online at www.stewartcalculus.com and via the Enhanced WebAssign homework system. Interactive Video SkillBuilder CD-ROM ISBN 0-495-01237-8 Think of it as portable ofﬁce hours! The Interactive Video Skillbuilder CD-ROM contains more than eight hours of video instruction. The problems worked during each video lesson are shown next to the viewing screen so that students can try work-ing them before watching the solution. To help students evalu-ate their progress, each section contains a ten-question web quiz (the results of which can be emailed to the instructor) and each chapter contains a chapter test, with answers to each problem. Study Guide Single Variable Early Transcendentals by Richard St. Andre ISBN 0-495-01239-4 Contains a short list of key concepts, a short list of skills to master, a brief introduction to the ideas of the section, an elaboration of the concepts and skills, including extra worked-out examples, and links in the margin to earlier and later material in the text and Study Guide. Student Solutions Manual Single Variable Early Transcendentals by Daniel Anderson, Jeffery A. Cole, and Daniel Drucker ISBN 0-495-01240-8 Provides completely worked-out solutions to all odd-numbered exercises within the text, giving students a way to check their answers and ensure that they took the correct steps to arrive at an answer. CalcLabs with Maple Single Variable by Philip Yasskin, Albert Boggess, David Barrow, Maurice Rahe, Jeffery Morgan, Michael Stecher, Art Belmonte, and Kirby Smith ISBN 0-495-01235-1 CalcLabs with Mathematica Single Variable by Selwyn Hollis ISBN 0-495-38245-0 Each of these comprehensive lab manuals will help students learn to effectively use the technology tools available to them. Each lab contains clearly explained exercises and a variety of labs and projects to accompany the text. A Companion to Calculus by Dennis Ebersole, Doris Schattschneider, Alicia Sevilla, and Kay Somers ISBN 0-495-01124-X Written to improve algebra and problem-solving skills of stu-dents taking a calculus course, every chapter in this companion is keyed to a calculus topic, providing conceptual background and speciﬁc algebra techniques needed to understand and solve calculus problems related to that topic. It is designed for calcu-lus courses that integrate the review of precalculus concepts or for individual use. Linear Algebra for Calculus by Konrad J. Heuvers, William P. Francis, John H. Kuisti, Deborah F. Lockhart, Daniel S. Moak, and Gene M. Ortner ISBN 0-534-25248-6 This comprehensive book, designed to supplement the calculus course, provides an introduction to and review of the basic ideas of linear algebra. TEC STUDENT RESOURCES \|\|\|\| Electronic items \|\|\|\| Printed items xxi TO THE STUDENT xxii Reading a calculus textbook is different from reading a newspaper or a novel, or even a physics book. Don’t be discouraged if you have to read a passage more than once in order to understand it. You should have pencil and paper and calculator at hand to sketch a dia-gram or make a calculation. Some students start by trying their homework problems and read the text only if they get stuck on an exercise. I suggest that a far better plan is to read and understand a section of the text before attempting the exercises. In particular, you should look at the deﬁnitions to see the exact meanings of the terms. And before you read each example, I suggest that you cover up the solution and try solving the problem yourself. You’ll get a lot more from looking at the solution if you do so. Part of the aim of this course is to train you to think logically. Learn to write the solu-tions of the exercises in a connected, step-by-step fashion with explanatory sentences— not just a string of disconnected equations or formulas. The answers to the odd-numbered exercises appear at the back of the book, in Appen-dix I. Some exercises ask for a verbal explanation or interpretation or description. In such cases there is no single correct way of expressing the answer, so don’t worry that you haven’t found the deﬁnitive answer. In addition, there are often several different forms in which to express a numerical or algebraic answer, so if your answer differs from mine, don’t immediately assume you’re wrong. For example, if the answer given in the back of the book is and you obtain , then you’re right and rationalizing the denominator will show that the answers are equivalent. The icon ; indicates an exercise that deﬁnitely requires the use of either a graphing calculator or a computer with graphing software. (Section 1.4 discusses the use of these graphing devices and some of the pitfalls that you may encounter.) But that doesn’t mean that graphing devices can’t be used to check your work on the other exercises as well. The symbol is reserved for problems in which the full resources of a computer algebra CAS 1(1 s2) s2 1 xxiii system (like Derive, Maple, Mathematica, or the TI-89/92) are required. You will also encounter the symbol \|, which warns you against committing an error. I have placed this symbol in the margin in situations where I have observed that a large proportion of my stu-dents tend to make the same mistake. Tools for Enriching Calculus, which is a companion to this text, is referred to by means of the symbol and can be accessed from www.stewartcalculus.com. It directs you to modules in which you can explore aspects of calculus for which the computer is particu-larly useful. TEC also provides Homework Hints for representative exercises that are indi-cated by printing the exercise number in red: These homework hints ask you questions that allow you to make progress toward a solution without actually giving you the answer. You need to pursue each hint in an active manner with pencil and paper to work out the details. If a particular hint doesn’t enable you to solve the problem, you can click to reveal the next hint. An optional CD-ROM that your instructor may have asked you to purchase is the Inter-active Video Skillbuilder, which contains videos of instructors explaining two or three of the examples in every section of the text. Also on the CD is a video in which I offer advice on how to succeed in your calculus course. I recommend that you keep this book for reference purposes after you ﬁnish the course. Because you will likely forget some of the speciﬁc details of calculus, the book will serve as a useful reminder when you need to use calculus in subsequent courses. And, because this book contains more material than can be covered in any one course, it can also serve as a valuable resource for a working scientist or engineer. Calculus is an exciting subject, justly considered to be one of the greatest achievements of the human intellect. I hope you will discover that it is not only useful but also intrinsi-cally beautiful. JAMES STEWART 15. TEC DIAGNOSTIC TESTS Success in calculus depends to a large extent on knowledge of the mathematics that precedes calculus: algebra, analytic geometry, functions, and trigonometry. The fol-lowing tests are intended to diagnose weaknesses that you might have in these areas. After taking each test you can check your answers against the given answers and, if necessary, refresh your skills by referring to the review materials that are provided. 1. Evaluate each expression without using a calculator. (a) (b) (c) (d) (e) (f) 2. Simplify each expression. Write your answer without negative exponents. (a) (b) (c) 3. Expand and simplfy. (a) (b) (c) (d) (e) 4. Factor each expression. (a) (b) (c) (d) (e) (f) 5. Simplify the rational expression. (a) (b) (c) (d) y x x y 1 y 1 x x 2 x 2 4 x 1 x 2 2x 2 x 1 x 2 9 x 3 2x 1 x 2 3x 2 x 2 x 2 x 3y 4xy 3x 32 9x 12 6x 12 x 4 27x x 3 3x 2 4x 12 2x 2 5x 12 4x 2 25 x 23 2x 32 (sa sb )(sa sb ) x 34x 5 3x 6 42x 5 3x 32y 3 x 2y12 2 3a3b34ab22 s200 s32 1634 2 3 2 523 521 34 34 34 DIAGNOSTIC TEST: ALGEBRA A xxiv DIAGNOSTIC TESTS \|\|\|\| xxv 6. Rationalize the expression and simplify. (a) (b) 7. Rewrite by completing the square. (a) (b) 8. Solve the equation. (Find only the real solutions.) (a) (b) (c) (d) (e) (f) (g) 9. Solve each inequality. Write your answer using interval notation. (a) (b) (c) (d) (e) 10. State whether each equation is true or false. (a) (b) (c) (d) (e) (f) 1x ax bx 1 a b 1 x y 1 x 1 y 1 TC C 1 T sa2 b2 a b sab sa sb p q2 p2 q 2 2x 3 x 1 1 x 4 3 xx 1x 2 0 x 2 2x 8 4 5 3x 17 2x4 x12 3s4 x 0 3 x 4 10 x 4 3x 2 2 0 2x 2 4x 1 0 x2 x 12 0 2x x 1 2x 1 x x 5 14 1 2x 2x 2 12x 11 x 2 x 1 s4 h 2 h s10 s5 2 6. (a) (b) 7. (a) (b) 8. (a) (b) (c) (d) (e) (f) (g) 9. (a) (b) (c) (d) (e) 10. (a) False (b) True (c) False (d) False (e) False (f) True 1, 4 1, 7 2, 0 1, 2, 4 4, 3 12 5 2 3, 22 3 1, s2 1 1 2s2 3, 4 1 6 2x 32 7 (x 1 2) 2 3 4 1 s4 h 2 5s2 2s10 1. (a) (b) (c) (d) (e) (f) 2. (a) (b) (c) 3. (a) (b) (c) (d) (e) 4. (a) (b) (c) (d) (e) (f) 5. (a) (b) (c) (d) x y 1 x 2 x 1 x 3 x 2 x 2 xyx 2x 2 3x 12x 1x 2 xx 3x 2 3x 9 x 3x 2x 2 2x 3x 4 2x 52x 5 x 3 6x 2 12x 8 4x 2 12x 9 a b 4x 2 7x 15 11x 2 x 9y7 48a5b7 6s2 1 8 9 4 25 1 81 81 81 If you have had difﬁculty with these problems, you may wish to consult the Review of Algebra on the website www.stewartcalculus.com. ANSWERS TO DIAGNOSTIC TEST A: ALGEBRA xxvi \|\|\|\| DIAGNOSTIC TESTS 1. Find an equation for the line that passes through the point and (a) has slope (b) is parallel to the -axis (c) is parallel to the -axis (d) is parallel to the line 2. Find an equation for the circle that has center and passes through the point . 3. Find the center and radius of the circle with equation . 4. Let and be points in the plane. (a) Find the slope of the line that contains and . (b) Find an equation of the line that passes through and . What are the intercepts? (c) Find the midpoint of the segment . (d) Find the length of the segment . (e) Find an equation of the perpendicular bisector of . (f) Find an equation of the circle for which is a diameter. 5. Sketch the region in the -plane deﬁned by the equation or inequalities. (a) (b) (c) (d) (e) (f) 9x 2 16y 2 144 x 2 y 2 4 y x 2 1 y 1 1 2x x 4 and y 2 1 y 3 xy AB AB AB AB B A B A B5, 12 A7, 4 x 2 y2 6x 10y 9 0 3, 2 1, 4 2x 4y 3 y x 3 2, 5 DIAGNOSTIC TEST: ANALYTIC GEOMETRY B 5. y x 1 2 0 y x 0 y x 0 4 3 _1 2 y x 0 y x 0 4 _4 y x 0 2 1 (a) (b) (c) (d) (e) (f) _1 3 2 _2 y=≈-1 ≈+¥=4 y=1- x 1 2 1. (a) (b) (c) (d) 2. 3. Center , radius 5 4. (a) (b) ; -intercept , -intercept (c) (d) (e) (f) x 12 y 42 100 3x 4y 13 20 1, 4 16 3 y 4 x 4x 3y 16 0 4 3 3, 5 x 12 y 42 52 y 1 2x 6 x 2 y 5 y 3x 1 ANSWERS TO DIAGNOSTIC TEST B: ANALYTIC GEOMETRY If you have had difﬁculty with these problems, you may wish to consult the Review of Analytic Geometry on the website www.stewartcalculus.com. DIAGNOSTIC TESTS \|\|\|\| xxvii 1. The graph of a function is given at the left. (a) State the value of . (b) Estimate the value of . (c) For what values of is ? (d) Estimate the values of such that . (e) State the domain and range of . 2. If , evaluate the difference quotient and simplify your answer. 3. Find the domain of the function. (a) (b) (c) 4. How are graphs of the functions obtained from the graph of ? (a) (b) (c) 5. Without using a calculator, make a rough sketch of the graph. (a) (b) (c) (d) (e) (f) (g) (h) 6. Let (a) Evaluate and . (b) Sketch the graph of . 7. If and , ﬁnd each of the following functions. (a) (b) (c) t t t t f f t tx 2x 3 f x x 2 2x 1 f f 1 f 2 f x 1 x 2 2x 1 if x 0 if x 0 y 1 x 1 y 2x y 2sx y sx y 4 x 2 y x 23 3 y x 13 y x 3 y f x 3 2 y 2 f x 1 y f x f hx s4 x sx 2 1 tx s 3 x x 2 1 f x 2x 1 x2 x 2 f 2 h f 2 h f x x 3 f f x 0 x f x 2 x f 2 f 1 f DIAGNOSTIC TEST: FUNCTIONS C 6. (a) 7. (a) (b) (b) (c) t t tx 8x 21 t f x 2x 2 4x 5 y x 0 _1 1 f tx 4x 2 8x 2 3, 3 y (h) x 0 1 1 (g) y x 0 1 _1 (f) y x 0 1 (e) y x 0 1 y (d) x 0 4 2 1. (a) (b) 2.8 (c) (d) (e) 2. 3. (a) (b) (c) 4. (a) Reﬂect about the -axis (b) Stretch vertically by a factor of 2, then shift 1 unit downward (c) Shift 3 units to the right and 2 units upward 5. (c) y x 0 (2, 3) y x 0 y (a) (b) 1 1 x 0 1 _1 x , 1 1, 4 , , 2 2, 1 1, 12 6h h2 3, 3, 2, 3 2.5, 0.3 3, 1 2 y 0 x 1 1 FIGURE FOR PROBLEM 1 ANSWERS TO DIAGNOSTIC TEST C: FUNCTIONS If you have had difﬁculty with these problems, you should look at Sections 1.1–1.3 of this book. 1. Convert from degrees to radians. (a) (b) 2. Convert from radians to degrees. (a) (b) 3. Find the length of an arc of a circle with radius 12 cm if the arc subtends a central angle of . 4. Find the exact values. (a) (b) (c) 5. Express the lengths and in the ﬁgure in terms of . 6. If and , where and lie between and , evaluate . 7. Prove the identities. (a) (b) 8. Find all values of such that and . 9. Sketch the graph of the function without using a calculator. y 1 sin 2x 0 x 2 sin 2x sin x x 2 tan x 1 tan2x sin 2x tan sin cos sec sinx y 2 0 y x sec y 5 4 sin x 1 3 b a sec5 3 sin7 6 tan 3 30 2 5 6 18 300 6. 8. 9. _π π x 0 2 y 0, 3, , 5 3, 2 1 15(4 6s2 ) 1. (a) (b) 2. (a) (b) 3. 4. (a) (b) (c) 5. (a) (b) 24 cos 24 sin 2 1 2 s3 2 cm 360 114.6 150 10 5 3 xxviii \|\|\|\| DIAGNOSTIC TESTS DIAGNOSTIC TEST: TRIGONOMETRY D a ¨ b 24 FIGURE FOR PROBLEM 5 ANSWERS TO DIAGNOSTIC TEST D: TRIGONOMETRY If you have had difﬁculty with these problems, you should look at Appendix D of this book. S I N G L E VA R I A B L E CA L C U L U S E A R LY T R A N S C E N D E N TA L S 2 Calculus is fundamentally different from the mathematics that you have studied pre-viously: calculus is less static and more dynamic. It is concerned with change and motion; it deals with quantities that approach other quantities. For that reason it may be useful to have an overview of the subject before beginning its intensive study. Here we give a glimpse of some of the main ideas of calculus by showing how the concept of a limit arises when we attempt to solve a variety of problems. A PREVIEW OF CALCULUS 3 THE AREA PROBLEM The origins of calculus go back at least 2500 years to the ancient Greeks, who found areas using the “method of exhaustion.” They knew how to ﬁnd the area of any polygon by dividing it into triangles as in Figure 1 and adding the areas of these triangles. It is a much more difﬁcult problem to ﬁnd the area of a curved ﬁgure. The Greek method of exhaustion was to inscribe polygons in the ﬁgure and circumscribe polygons about the ﬁgure and then let the number of sides of the polygons increase. Figure 2 illus-trates this process for the special case of a circle with inscribed regular polygons. Let be the area of the inscribed polygon with sides. As increases, it appears that becomes closer and closer to the area of the circle. We say that the area of the circle is the limit of the areas of the inscribed polygons, and we write The Greeks themselves did not use limits explicitly. However, by indirect reasoning, Eudoxus (ﬁfth century BC) used exhaustion to prove the familiar formula for the area of a circle: We will use a similar idea in Chapter 5 to ﬁnd areas of regions of the type shown in Fig-ure 3. We will approximate the desired area by areas of rectangles (as in Figure 4), let the width of the rectangles decrease, and then calculate as the limit of these sums of areas of rectangles. The area problem is the central problem in the branch of calculus called integral cal-culus. The techniques that we will develop in Chapter 5 for ﬁnding areas will also enable us to compute the volume of a solid, the length of a curve, the force of water against a dam, the mass and center of gravity of a rod, and the work done in pumping water out of a tank. FIGURE 3 1 n 1 0 x y (1, 1) 1 0 x y (1, 1) 1 4 1 2 3 4 0 x y 1 (1, 1) FIGURE 4 1 0 x y y=≈ A (1, 1) A A A r 2. A lim n l An An n n An A¡™ A¶ Aß A∞ A¢ A£ FIGURE 2 A FIGURE 1 A=A¡+A™+A£+A¢+A∞ A¡ A™ A£ A¢ A∞ In the Preview Visual, you can see how inscribed and circumscribed polygons approximate the area of a circle. TEC THE TANGENT PROBLEM Consider the problem of trying to ﬁnd an equation of the tangent line to a curve with equation at a given point . (We will give a precise deﬁnition of a tangent line in Chapter 2. For now you can think of it as a line that touches the curve at as in Figure 5.) Since we know that the point lies on the tangent line, we can ﬁnd the equation of if we know its slope . The problem is that we need two points to compute the slope and we know only one point, , on . To get around the problem we ﬁrst ﬁnd an approximation to by taking a nearby point on the curve and computing the slope of the secant line . From Figure 6 we see that Now imagine that moves along the curve toward as in Figure 7. You can see that the secant line rotates and approaches the tangent line as its limiting position. This means that the slope of the secant line becomes closer and closer to the slope of the tan-gent line. We write and we say that is the limit of as approaches along the curve. Since approaches as approaches , we could also use Equation 1 to write Speciﬁc examples of this procedure will be given in Chapter 2. The tangent problem has given rise to the branch of calculus called differential calcu-lus, which was not invented until more than 2000 years after integral calculus. The main ideas behind differential calculus are due to the French mathematician Pierre Fermat (1601–1665) and were developed by the English mathematicians John Wallis (1616–1703), Isaac Barrow (1630–1677), and Isaac Newton (1642–1727) and the German mathematician Gottfried Leibniz (1646–1716). The two branches of calculus and their chief problems, the area problem and the tan-gent problem, appear to be very different, but it turns out that there is a very close connec-tion between them. The tangent problem and the area problem are inverse problems in a sense that will be described in Chapter 5. VELOCITY When we look at the speedometer of a car and read that the car is traveling at 48 mih, what does that information indicate to us? We know that if the velocity remains constant, then after an hour we will have traveled 48 mi. But if the velocity of the car varies, what does it mean to say that the velocity at a given instant is 48 mih? In order to analyze this question, let’s examine the motion of a car that travels along a straight road and assume that we can measure the distance traveled by the car (in feet) at l-second intervals as in the following chart: m lim x l a f x f a x a 2 P Q a x P Q mPQ m m lim Q lP mPQ m mPQ P Q mPQ fx fa x a 1 PQ mPQ Q m t P m t P P P y fx t 4 \|\|\|\| A PREVIEW OF CALCULUS 0 y x P y=ƒ t P Q t 0 x y y 0 x a x ƒ-f(a) P{a, f(a)} x-a t Q{x, ƒ} FIGURE 5 The tangent line at P FIGURE 6 The secant line PQ FIGURE 7 Secant lines approaching the tangent line t Time elapsed (s) 0 1 2 3 4 5 d Distance (ft) 0 2 9 24 42 71 As a ﬁrst step toward ﬁnding the velocity after 2 seconds have elapsed, we ﬁnd the aver-age velocity during the time interval : Similarly, the average velocity in the time interval is We have the feeling that the velocity at the instant 2 can’t be much different from the average velocity during a short time interval starting at . So let’s imagine that the dis-tance traveled has been measured at 0.l-second time intervals as in the following chart: Then we can compute, for instance, the average velocity over the time interval : The results of such calculations are shown in the following chart: The average velocities over successively smaller intervals appear to be getting closer to a number near 10, and so we expect that the velocity at exactly is about 10 fts. In Chapter 2 we will deﬁne the instantaneous velocity of a moving object as the limiting value of the average velocities over smaller and smaller time intervals. In Figure 8 we show a graphical representation of the motion of the car by plotting the distance traveled as a function of time. If we write , then is the number of feet traveled after seconds. The average velocity in the time interval is which is the same as the slope of the secant line in Figure 8. The velocity when is the limiting value of this average velocity as approaches 2; that is, and we recognize from Equation 2 that this is the same as the slope of the tangent line to the curve at . P v lim t l 2 f t f 2 t 2 t t 2 v PQ average velocity change in position time elapsed ft f2 t 2 2, t t ft d ft t 2 average velocity 15.80 9.00 2.5 2 13.6 fts 2, 2.5 t 2 t average velocity 24 9 3 2 15 fts 2 t 3 16.5 fts 42 9 4 2 average velocity change in position time elapsed 2 t 4 A PREVIEW OF CALCULUS \|\|\|\| 5 t 2.0 2.1 2.2 2.3 2.4 2.5 d 9.00 10.02 11.16 12.45 13.96 15.80 Time interval Average velocity (fts) 15.0 13.6 12.4 11.5 10.8 10.2 2, 2.1 2, 2.2 2, 2.3 2, 2.4 2, 2.5 2, 3 FIGURE 8 t d 0 1 2 3 4 5 10 20 P{2, f(2)} Q{t, f(t)} Thus, when we solve the tangent problem in differential calculus, we are also solving problems concerning velocities. The same techniques also enable us to solve problems involving rates of change in all of the natural and social sciences. THE LIMIT OF A SEQUENCE In the ﬁfth century BC the Greek philosopher Zeno of Elea posed four problems, now known as Zeno’s paradoxes, that were intended to challenge some of the ideas concerning space and time that were held in his day. Zeno’s second paradox concerns a race between the Greek hero Achilles and a tortoise that has been given a head start. Zeno argued, as fol-lows, that Achilles could never pass the tortoise: Suppose that Achilles starts at position and the tortoise starts at position . (See Figure 9.) When Achilles reaches the point , the tortoise is farther ahead at position . When Achilles reaches , the tor-toise is at . This process continues indeﬁnitely and so it appears that the tortoise will always be ahead! But this deﬁes common sense. One way of explaining this paradox is with the idea of a sequence. The successive posi-tions of Achilles or the successive positions of the tortoise form what is known as a sequence. In general, a sequence is a set of numbers written in a deﬁnite order. For instance, the sequence can be described by giving the following formula for the th term: We can visualize this sequence by plotting its terms on a number line as in Fig-ure 10(a) or by drawing its graph as in Figure 10(b). Observe from either picture that the terms of the sequence are becoming closer and closer to 0 as increases. In fact, we can ﬁnd terms as small as we please by making large enough. We say that the limit of the sequence is 0, and we indicate this by writing In general, the notation is used if the terms approach the number as becomes large. This means that the num-bers can be made as close as we like to the number by taking sufﬁciently large. n L an n L an lim n l an L lim n l 1 n 0 n n an 1n an 1 n n {1, 1 2, 1 3, 1 4, 1 5, . . .} an t1, t2, t3, . . . a1, a2, a3, . . . FIGURE 9 Achilles tortoise a¡ a™ a£ a¢ a∞ t¡ t™ t£ t¢ . . . . . . t3 a3 t2 t2 a2 t1 t1 a1 6 \|\|\|\| A PREVIEW OF CALCULUS 1 n 1 2 3 4 5 6 7 8 FIGURE 10 1 0 a¡ a™ a£ a¢ (a) (b) The concept of the limit of a sequence occurs whenever we use the decimal represen-tation of a real number. For instance, if then The terms in this sequence are rational approximations to . Let’s return to Zeno’s paradox. The successive positions of Achilles and the tortoise form sequences and , where for all . It can be shown that both sequences have the same limit: It is precisely at this point that Achilles overtakes the tortoise. THE SUM OF A SERIES Another of Zeno’s paradoxes, as passed on to us by Aristotle, is the following: “A man standing in a room cannot walk to the wall. In order to do so, he would ﬁrst have to go half the distance, then half the remaining distance, and then again half of what still remains. This process can always be continued and can never be ended.” (See Figure 11.) Of course, we know that the man can actually reach the wall, so this suggests that per-haps the total distance can be expressed as the sum of inﬁnitely many smaller distances as follows: 1 1 2 1 4 1 8 1 16 1 2n 3 FIGURE 11 1 2 1 4 1 8 1 16 p lim n l an p lim n l tn n an tn tn an lim n l an a7 3.1415926 a6 3.141592 a5 3.14159 a4 3.1415 a3 3.141 a2 3.14 a1 3.1 A PREVIEW OF CALCULUS \|\|\|\| 7 Zeno was arguing that it doesn’t make sense to add inﬁnitely many numbers together. But there are other situations in which we implicitly use inﬁnite sums. For instance, in decimal notation, the symbol means and so, in some sense, it must be true that More generally, if denotes the nth digit in the decimal representation of a number, then Therefore some inﬁnite sums, or inﬁnite series as they are called, have a meaning. But we must deﬁne carefully what the sum of an inﬁnite series is. Returning to the series in Equation 3, we denote by the sum of the ﬁrst terms of the series. Thus Observe that as we add more and more terms, the partial sums become closer and closer to 1. In fact, it can be shown that by taking large enough (that is, by adding sufﬁciently many terms of the series), we can make the partial sum as close as we please to the num-ber 1. It therefore seems reasonable to say that the sum of the inﬁnite series is 1 and to write 1 2 1 4 1 8 1 2n 1 sn n s16 1 2 1 4 1 216 0.99998474 s10 1 2 1 4 1 1024 0.99902344 s7 1 2 1 4 1 8 1 16 1 32 1 64 1 128 0.9921875 s6 1 2 1 4 1 8 1 16 1 32 1 64 0.984375 s5 1 2 1 4 1 8 1 16 1 32 0.96875 s4 1 2 1 4 1 8 1 16 0.9375 s3 1 2 1 4 1 8 0.875 s2 1 2 1 4 0.75 s1 1 2 0.5 n sn 0.d1d2d3d4 . . . d1 10 d2 102 d3 103 dn 10n dn 3 10 3 100 3 1000 3 10,000 1 3 3 10 3 100 3 1000 3 10,000 0.3 0.3333 . . . 8 \|\|\|\| A PREVIEW OF CALCULUS In other words, the reason the sum of the series is 1 is that In Chapter 11 we will discuss these ideas further. We will then use Newton’s idea of combining inﬁnite series with differential and integral calculus. SUMMARY We have seen that the concept of a limit arises in trying to ﬁnd the area of a region, the slope of a tangent to a curve, the velocity of a car, or the sum of an inﬁnite series. In each case the common theme is the calculation of a quantity as the limit of other, easily calcu-lated quantities. It is this basic idea of a limit that sets calculus apart from other areas of mathematics. In fact, we could deﬁne calculus as the part of mathematics that deals with limits. After Sir Isaac Newton invented his version of calculus, he used it to explain the motion of the planets around the sun. Today calculus is used in calculating the orbits of satellites and spacecraft, in predicting population sizes, in estimating how fast coffee prices rise, in forecasting weather, in measuring the cardiac output of the heart, in calculating life insur-ance premiums, and in a great variety of other areas. We will explore some of these uses of calculus in this book. In order to convey a sense of the power of the subject, we end this preview with a list of some of the questions that you will be able to answer using calculus: 1. How can we explain the fact, illustrated in Figure 12, that the angle of elevation from an observer up to the highest point in a rainbow is 42°? (See page 279.) 2. How can we explain the shapes of cans on supermarket shelves? (See page 333.) 3. Where is the best place to sit in a movie theater? (See page 446.) 4. How far away from an airport should a pilot start descent? (See page 206.) 5. How can we ﬁt curves together to design shapes to represent letters on a laser printer? (See page 639.) 6. Where should an inﬁelder position himself to catch a baseball thrown by an out-ﬁelder and relay it to home plate? (See page 601.) 7. Does a ball thrown upward take longer to reach its maximum height or to fall back to its original height? (See page 590.) lim n l sn 1 A PREVIEW OF CALCULUS \|\|\|\| 9 rays from sun observer rays from sun 42° FIGURE 12 138° 10 The fundamental objects that we deal with in calculus are functions. This chapter prepares the way for calculus by discussing the basic ideas concerning functions, their graphs, and ways of transforming and combining them. We stress that a function can be represented in different ways: by an equation, in a table, by a graph, or in words. We look at the main types of functions that occur in calculus and describe the process of using these functions as mathematical models of real-world phenomena. We also discuss the use of graphing calculators and graphing software for computers. A graphical representation of a function––here the number of hours of daylight as a function of the time of year at various latitudes––is often the most natural and convenient way to represent the function. FUNCTIONS AND MODELS 1 0 2 4 6 8 10 12 14 16 18 20 Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. Hours 60° N 50° N 40° N 30° N 20° N FOUR WAYS TO REPRESENT A FUNCTION Functions arise whenever one quantity depends on another. Consider the following four situations. A. The area of a circle depends on the radius of the circle. The rule that connects and is given by the equation . With each positive number there is associ-ated one value of , and we say that is a function of . B. The human population of the world depends on the time . The table gives estimates of the world population at time for certain years. For instance, But for each value of the time there is a corresponding value of and we say that is a function of . C. The cost of mailing a ﬁrst-class letter depends on the weight of the letter. Although there is no simple formula that connects and , the post ofﬁce has a rule for determining when is known. D. The vertical acceleration of the ground as measured by a seismograph during an earthquake is a function of the elapsed time Figure 1 shows a graph generated by seismic activity during the Northridge earthquake that shook Los Angeles in 1994. For a given value of the graph provides a corresponding value of . Each of these examples describes a rule whereby, given a number ( , , , or ), another number ( , , , or ) is assigned. In each case we say that the second number is a func-tion of the ﬁrst number. A function is a rule that assigns to each element in a set exactly one ele-ment, called , in a set . We usually consider functions for which the sets and are sets of real numbers. The set is called the domain of the function. The number is the value of at and is read “ of .” The range of is the set of all possible values of as varies through-out the domain. A symbol that represents an arbitrary number in the domain of a function is called an independent variable. A symbol that represents a number in the range of is called a dependent variable. In Example A, for instance, r is the independent variable and A is the dependent variable. f f x f!x" f x f x f f!x" D E D E f!x" D x f a C P A t w t r FIGURE 1 Vertical ground acceleration during the Northridge earthquake {cm/s@} (seconds) Calif. Dept. of Mines and Geology 5 50 10 15 20 25 a t 100 30 _50 a t, t. a w C C w w C t P P, t P!1950" # 2,560,000,000 t, P!t" t P r A A r A ! !r 2 A r r A 1.1 11 Population Year (millions) 1900 1650 1910 1750 1920 1860 1930 2070 1940 2300 1950 2560 1960 3040 1970 3710 1980 4450 1990 5280 2000 6080 It’s helpful to think of a function as a machine (see Figure 2). If is in the domain of the function then when enters the machine, it’s accepted as an input and the machine produces an output according to the rule of the function. Thus we can think of the domain as the set of all possible inputs and the range as the set of all possible outputs. The preprogrammed functions in a calculator are good examples of a function as a machine. For example, the square root key on your calculator computes such a function. You press the key labeled (or ) and enter the input x. If , then is not in the domain of this function; that is, is not an acceptable input, and the calculator will indi-cate an error. If , then an approximation to will appear in the display. Thus the key on your calculator is not quite the same as the exact mathematical function deﬁned by . Another way to picture a function is by an arrow diagram as in Figure 3. Each arrow connects an element of to an element of . The arrow indicates that is associated with is associated with , and so on. The most common method for visualizing a function is its graph. If is a function with domain , then its graph is the set of ordered pairs (Notice that these are input-output pairs.) In other words, the graph of consists of all points in the coordinate plane such that and is in the domain of . The graph of a function gives us a useful picture of the behavior or “life history” of a function. Since the -coordinate of any point on the graph is , we can read the value of from the graph as being the height of the graph above the point (see Figure 4). The graph of also allows us to picture the domain of on the -axis and its range on the -axis as in Figure 5. EXAMPLE 1 The graph of a function is shown in Figure 6. (a) Find the values of and . (b) What are the domain and range of ? SOLUTION (a) We see from Figure 6 that the point lies on the graph of , so the value of at 1 is . (In other words, the point on the graph that lies above x ! 1 is 3 units above the x-axis.) When x ! 5, the graph lies about 0.7 unit below the x-axis, so we estimate that . (b) We see that is deﬁned when , so the domain of is the closed inter-val . Notice that takes on all values from "2 to 4, so the range of is M $y%"2 # y # 4& ! '"2, 4( f f '0, 7( f 0 # x # 7 f!x" f!5" # "0.7 f!1" ! 3 f f !1, 3" f f!5" f!1" f 0 y ! ƒ(x) domain range FIGURE 4 {x, ƒ} ƒ f(1) f(2) 0 1 2 x FIGURE 5 x y x y y x f f x f!x" y ! f!x" !x, y" y f f x y ! f!x" !x, y" f $!x, f!x""%x ! D& D f a f!a" x, f!x" E D f!x" ! sx f sx sx x $ 0 x x x % 0 sx s f!x" x f, x 12 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS FIGURE 2 Machine diagram for a function ƒ x (input) ƒ (output) f f D E ƒ f(a) a x FIGURE 3 Arrow diagram for ƒ FIGURE 6 x y 0 1 1 N The notation for intervals is given in Appendix A. EXAMPLE 2 Sketch the graph and ﬁnd the domain and range of each function. (a) (b) SOLUTION (a) The equation of the graph is , and we recognize this as being the equa-tion of a line with slope 2 and y-intercept "1. (Recall the slope-intercept form of the equation of a line: . See Appendix B.) This enables us to sketch a portion of the graph of in Figure 7. The expression is deﬁned for all real numbers, so the domain of is the set of all real numbers, which we denote by !. The graph shows that the range is also !. (b) Since and , we could plot the points and , together with a few other points on the graph, and join them to produce the graph (Figure 8). The equation of the graph is , which represents a parabola (see Appendix C). The domain of t is !. The range of t consists of all values of , that is, all numbers of the form . But for all numbers x and any positive number y is a square. So the range of t is . This can also be seen from Figure 8. M EXAMPLE 3 If and , evaluate . SOLUTION We ﬁrst evaluate by replacing by in the expression for : Then we substitute into the given expression and simplify: M REPRESENTATIONS OF FUNCTIONS There are four possible ways to represent a function: I verbally (by a description in words) I numerically (by a table of values) I visually (by a graph) I algebraically (by an explicit formula) If a single function can be represented in all four ways, it’s often useful to go from one representation to another to gain additional insight into the function. (In Example 2, for instance, we started with algebraic formulas and then obtained the graphs.) But certain ! 4ah & 2h2 " 5h h ! 4a & 2h " 5 ! 2a2 & 4ah & 2h2 " 5a " 5h & 1 " 2a2 & 5a " 1 h f!a & h" " f!a" h ! !2a2 & 4ah & 2h2 " 5a " 5h & 1" " !2a2 " 5a & 1" h ! 2a2 & 4ah & 2h2 " 5a " 5h & 1 ! 2!a2 & 2ah & h2" " 5!a & h" & 1 f!a & h" ! 2!a & h"2 " 5!a & h" & 1 f!x" a & h x f!a & h" f!a & h" " f!a" h h " 0 f!x" ! 2x 2 " 5x & 1 $y% y $ 0& ! '0, '" x 2 $ 0 x 2 t!x" y ! x 2 !"1, 1" !2, 4" t!"1" ! !"1"2 ! 1 t!2" ! 22 ! 4 f 2x " 1 f y ! mx & b y ! 2x " 1 t!x" ! x 2 f!x" ! 2x " 1 SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION \|\|\|\| 13 FIGURE 7 x y=2x-1 0 -1 y 1 2 (_1, 1) (2, 4) 0 y 1 x 1 y=≈ FIGURE 8 N The expression in Example 3 is called a difference quotient and occurs frequently in calculus. As we will see in Chapter 2, it represents the average rate of change of between and . x ! a & h x ! a f !x" f !a & h" " f !a" h functions are described more naturally by one method than by another. With this in mind, let’s reexamine the four situations that we considered at the beginning of this section. A. The most useful representation of the area of a circle as a function of its radius is probably the algebraic formula , though it is possible to compile a table of values or to sketch a graph (half a parabola). Because a circle has to have a positive radius, the domain is , and the range is also . B. We are given a description of the function in words: is the human population of the world at time t. The table of values of world population provides a convenient representation of this function. If we plot these values, we get the graph (called a scatter plot) in Figure 9. It too is a useful representation; the graph allows us to absorb all the data at once. What about a formula? Of course, it’s impossible to devise an explicit formula that gives the exact human population at any time t. But it is possible to ﬁnd an expression for a function that approximates . In fact, using methods explained in Section 1.2, we obtain the approximation and Figure 10 shows that it is a reasonably good “ﬁt.” The function is called a mathematical model for population growth. In other words, it is a function with an explicit formula that approximates the behavior of our given function. We will see, however, that the ideas of calculus can be applied to a table of values; an explicit formula is not necessary. The function is typical of the functions that arise whenever we attempt to apply calculus to the real world. We start with a verbal description of a function. Then we may be able to construct a table of values of the function, perhaps from instrument readings in a scientiﬁc experiment. Even though we don’t have complete knowledge of the values of the function, we will see throughout the book that it is still possible to perform the operations of calculus on such a function. C. Again the function is described in words: is the cost of mailing a ﬁrst-class letter with weight . The rule that the US Postal Service used as of 2007 is as follows: The cost is 39 cents for up to one ounce, plus 24 cents for each successive ounce up to 13 ounces. The table of values shown in the margin is the most convenient representation for this function, though it is possible to sketch a graph (see Example 10). D. The graph shown in Figure 1 is the most natural representation of the vertical acceler-ation function . It’s true that a table of values could be compiled, and it is even a!t" w C!w" P FIGURE 10 FIGURE 9 1900 6x10' P t 1920 1940 1960 1980 2000 1900 6x10' P t 1920 1940 1960 1980 2000 f P!t" # f!t" ! !0.008079266" ( !1.013731"t P!t" P!t" P!t" !0, '" $r%r ) 0& ! !0, '" A!r" ! !r 2 14 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS Population Year (millions) 1900 1650 1910 1750 1920 1860 1930 2070 1940 2300 1950 2560 1960 3040 1970 3710 1980 4450 1990 5280 2000 6080 (ounces) (dollars) 0.39 0.63 0.87 1.11 1.35 3.27 12 % w # 13 ( ( ( ( ( ( 4 % w # 5 3 % w # 4 2 % w # 3 1 % w # 2 0 % w # 1 C!w" w N A function deﬁned by a table of values is called a tabular function. possible to devise an approximate formula. But everything a geologist needs to know—amplitudes and patterns—can be seen easily from the graph. (The same is true for the patterns seen in electrocardiograms of heart patients and polygraphs for lie-detection.) In the next example we sketch the graph of a function that is deﬁned verbally. EXAMPLE 4 When you turn on a hot-water faucet, the temperature of the water depends on how long the water has been running. Draw a rough graph of as a function of the time that has elapsed since the faucet was turned on. SOLUTION The initial temperature of the running water is close to room temperature because the water has been sitting in the pipes. When the water from the hot-water tank starts ﬂowing from the faucet, increases quickly. In the next phase, is constant at the temperature of the heated water in the tank. When the tank is drained, decreases to the temperature of the water supply. This enables us to make the rough sketch of as a function of in Figure 11. M In the following example we start with a verbal description of a function in a physical situation and obtain an explicit algebraic formula. The ability to do this is a useful skill in solving calculus problems that ask for the maximum or minimum values of quantities. EXAMPLE 5 A rectangular storage container with an open top has a volume of 10 m . The length of its base is twice its width. Material for the base costs $10 per square meter; material for the sides costs $6 per square meter. Express the cost of materials as a function of the width of the base. SOLUTION We draw a diagram as in Figure 12 and introduce notation by letting and be the width and length of the base, respectively, and be the height. The area of the base is , so the cost, in dollars, of the material for the base is . Two of the sides have area and the other two have area , so the cost of the material for the sides is . The total cost is therefore To express as a function of alone, we need to eliminate and we do so by using the fact that the volume is 10 m . Thus which gives Substituting this into the expression for , we have Therefore, the equation expresses as a function of . M w C w ) 0 C!w" ! 20w2 & 180 w C ! 20w2 & 36w) 5 w2 ! 20w2 & 180 w C h ! 10 2w2 ! 5 w2 w!2w"h ! 10 3 h w C C ! 10!2w2" & 6'2!wh" & 2!2wh"( ! 20w2 & 36wh 6'2!wh" & 2!2wh"( 2wh wh 10!2w2" !2w"w ! 2w2 h 2w w 3 V t T T T T t T T SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION \|\|\|\| 15 t T 0 FIGURE 11 w 2w h FIGURE 12 N In setting up applied functions as in Example 5, it may be useful to review the principles of problem solving as discussed on page 76, particularly Step 1: Understand the Problem. EXAMPLE 6 Find the domain of each function. (a) (b) SOLUTION (a) Because the square root of a negative number is not deﬁned (as a real number), the domain of consists of all values of x such that . This is equivalent to , so the domain is the interval . (b) Since and division by is not allowed, we see that is not deﬁned when or . Thus the domain of is which could also be written in interval notation as M The graph of a function is a curve in the -plane. But the question arises: Which curves in the -plane are graphs of functions? This is answered by the following test. THE VERTICAL LINE TEST A curve in the -plane is the graph of a function of if and only if no vertical line intersects the curve more than once. The reason for the truth of the Vertical Line Test can be seen in Figure 13. If each ver-tical line intersects a curve only once, at , then exactly one functional value is deﬁned by . But if a line intersects the curve twice, at and , then the curve can’t represent a function because a function can’t assign two different val-ues to . For example, the parabola shown in Figure 14(a) on the next page is not the graph of a function of because, as you can see, there are vertical lines that intersect the parabola twice. The parabola, however, does contain the graphs of two functions of . Notice that the equation implies , so Thus the upper and lower halves of the parabola are the graphs of the functions [from Example 6(a)] and . [See Figures 14(b) and (c).] We observe that if we reverse the roles of and , then the equation does deﬁne as a function of (with as the independent variable and as the dependent variable) and the parabola now appears as the graph of the function . h x y y x x ! h!y" ! y 2 " 2 y x t!x" ! "sx & 2 f!x" ! sx & 2 y ! sx & 2 . y 2 ! x & 2 x ! y 2 " 2 x x x ! y 2 " 2 FIGURE 13 a x=a (a, b) 0 a (a, c) (a, b) x=a 0 x y x y a !a, c" !a, b" x ! a f!a" ! b !a, b" x ! a x xy xy xy !"', 0" " !0, 1" " !1, '" $x% x " 0, x " 1& t x ! 1 x ! 0 t!x" 0 t!x" ! 1 x 2 " x ! 1 x!x " 1" '"2, '" x $ "2 x & 2 $ 0 f t!x" ! 1 x 2 " x f!x" ! sx & 2 16 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS N If a function is given by a formula and the domain is not stated explicitly, the convention is that the domain is the set of all numbers for which the formula makes sense and deﬁnes a real number. PIECEWISE DEFINED FUNCTIONS The functions in the following four examples are deﬁned by different formulas in differ-ent parts of their domains. EXAMPLE 7 A function is deﬁned by Evaluate , , and and sketch the graph. SOLUTION Remember that a function is a rule. For this particular function the rule is the following: First look at the value of the input . If it happens that , then the value of is . On the other hand, if , then the value of is . How do we draw the graph of ? We observe that if , then , so the part of the graph of that lies to the left of the vertical line must coincide with the line , which has slope and -intercept 1. If , then , so the part of the graph of that lies to the right of the line must coincide with the graph of , which is a parabola. This enables us to sketch the graph in Figure 15. The solid dot indicates that the point is included on the graph; the open dot indi-cates that the point is excluded from the graph. M The next example of a piecewise deﬁned function is the absolute value function. Recall that the absolute value of a number , denoted by , is the distance from to on the real number line. Distances are always positive or , so we have for every number For example, In general, we have (Remember that if is negative, then is positive.) "a a if a % 0 %a% ! "a if a $ 0 %a% ! a %3 " !% ! ! " 3 %s2 " 1% ! s2 " 1 %0% ! 0 %"3% ! 3 %3% ! 3 a %a% $ 0 0 0 a %a% a !1, 1" !1, 0" y ! x 2 x ! 1 f f!x" ! x 2 x ) 1 y "1 y ! 1 " x x ! 1 f f!x" ! 1 " x x # 1 f Since 2 ) 1, we have f!2" ! 22 ! 4. Since 1 # 1, we have f!1" ! 1 " 1 ! 0. Since 0 # 1, we have f!0" ! 1 " 0 ! 1. x 2 f!x" x ) 1 1 " x f!x" # 1 x x f!2" f!1" f!0" f!x" !+ 1 " x x 2 if x # 1 if x ) 1 f V FIGURE 14 (b) y=œ„„„„ x+2 _2 0 x y (_2, 0) (a) x=¥-2 0 x y (c) y=_œ„„„„ x+2 _2 0 y x SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION \|\|\|\| 17 1 x y 1 FIGURE 15 N For a more extensive review of absolute values, see Appendix A. EXAMPLE 8 Sketch the graph of the absolute value function . SOLUTION From the preceding discussion we know that Using the same method as in Example 7, we see that the graph of coincides with the line to the right of the -axis and coincides with the line to the left of the -axis (see Figure 16). M EXAMPLE 9 Find a formula for the function graphed in Figure 17. SOLUTION The line through and has slope and -intercept , so its equation is . Thus, for the part of the graph of that joins to , we have The line through and has slope , so its point-slope form is So we have We also see that the graph of coincides with the -axis for . Putting this informa-tion together, we have the following three-piece formula for : M EXAMPLE 10 In Example C at the beginning of this section we considered the cost of mailing a ﬁrst-class letter with weight . In effect, this is a piecewise deﬁned function because, from the table of values, we have The graph is shown in Figure 18. You can see why functions similar to this one are called step functions—they jump from one value to the next. Such functions will be studied in Chapter 2. M ( ( ( 0.39 0.63 0.87 1.11 if 0 % w # 1 if 1 % w # 2 if 2 % w # 3 if 3 % w # 4 C!w" ! w C!w" f !x" !+ x 2 " x 0 if 0 # x # 1 if 1 % x # 2 if x ) 2 f x ) 2 x f if 1 % x # 2 f!x" ! 2 " x y ! 2 " x or y " 0 ! !"1"!x " 2" m ! "1 !2, 0" !1, 1" if 0 # x # 1 f!x" ! x !1, 1" !0, 0" f y ! x b ! 0 y m ! 1 !1, 1" !0, 0" FIGURE 17 x y 0 1 1 f y y ! "x y y ! x f %x% !+ x "x if x $ 0 if x % 0 f!x" ! %x% 18 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS x y=\|x \| 0 y FIGURE 16 N Point-slope form of the equation of a line: See Appendix B. y " y1 ! m!x " x1" FIGURE 18 C 1 1 0 2 3 4 5 w SYMMETRY If a function satisﬁes for every number in its domain, then is called an even function. For instance, the function is even because The geometric signiﬁcance of an even function is that its graph is symmetric with respect to the -axis (see Figure 19). This means that if we have plotted the graph of for , we obtain the entire graph simply by reﬂecting this portion about the -axis. If satisﬁes for every number in its domain, then is called an odd function. For example, the function is odd because The graph of an odd function is symmetric about the origin (see Figure 20). If we already have the graph of for , we can obtain the entire graph by rotating this portion through about the origin. EXAMPLE 11 Determine whether each of the following functions is even, odd, or neither even nor odd. (a) (b) (c) SOLUTION (a) Therefore is an odd function. (b) So is even. (c) Since and , we conclude that is neither even nor odd. M The graphs of the functions in Example 11 are shown in Figure 21. Notice that the graph of h is symmetric neither about the y-axis nor about the origin. FIGURE 21 1 1 x y h 1 1 y x g 1 _1 1 y x f _1 (a) (b) (c) h h!"x" " "h!x" h!"x" " h!x" h!"x" ! 2!"x" " !"x"2 ! "2x " x 2 t t!"x" ! 1 " !"x"4 ! 1 " x 4 ! t!x" f ! "f!x" ! "x 5 " x ! "!x 5 & x" f!"x" ! !"x"5 & !"x" ! !"1"5x 5 & !"x" h!x" ! 2x " x 2 t!x" ! 1 " x 4 f!x" ! x 5 & x V 180+ x $ 0 f f!"x" ! !"x"3 ! "x 3 ! "f!x" f!x" ! x 3 f x f!"x" ! "f!x" f y x $ 0 f y f!"x" ! !"x"2 ! x 2 ! f!x" f!x" ! x 2 f x f!"x" ! f!x" f SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION \|\|\|\| 19 0 x _x f(_x) ƒ FIGURE 19 An even function x y 0 x _x ƒ FIGURE 20 An odd function x y INCREASING AND DECREASING FUNCTIONS The graph shown in Figure 22 rises from to , falls from to , and rises again from to . The function is said to be increasing on the interval , decreasing on , and increasing again on . Notice that if and are any two numbers between and with , then . We use this as the deﬁning property of an increasing function. A function is called increasing on an interval if It is called decreasing on if In the deﬁnition of an increasing function it is important to realize that the inequality must be satisﬁed for every pair of numbers and in with . You can see from Figure 23 that the function is decreasing on the interval and increasing on the interval . !0, !" #"!, 0$ f#x" ! x 2 x1 # x2 I x2 x1 f#x1" # f#x2" whenever x1 # x2 in I f#x1" $ f#x2" I whenever x1 # x2 in I f#x1" # f#x2" I f A B C D y=ƒ f(x¡) f(x™) a y 0 x x¡ x™ b c d FIGURE 22 f#x1" # f#x2" x1 # x2 b a x2 x1 !c, d$ !b, c$ !a, b$ f D C C B B A 20 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS FIGURE 23 0 y x y=≈ y 0 x 1 1 1. The graph of a function is given. (a) State the value of . (b) Estimate the value of . (c) For what values of x is ? (d) Estimate the values of x such that . (e) State the domain and range of . (f) On what interval is increasing? f f f #x" ! 0 f #x" ! 2 f #2" f #"1" f EXERCISES 1.1 SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION \|\|\|\| 21 varies over time. What do you think happened when this person was 30 years old? 10. The graph shown gives a salesman’s distance from his home as a function of time on a certain day. Describe in words what the graph indicates about his travels on this day. You put some ice cubes in a glass, ﬁll the glass with cold water, and then let the glass sit on a table. Describe how the temperature of the water changes as time passes. Then sketch a rough graph of the temperature of the water as a function of the elapsed time. 12. Sketch a rough graph of the number of hours of daylight as a function of the time of year. Sketch a rough graph of the outdoor temperature as a function of time during a typical spring day. 14. Sketch a rough graph of the market value of a new car as a function of time for a period of 20 years. Assume the car is well maintained. 15. Sketch the graph of the amount of a particular brand of coffee sold by a store as a function of the price of the coffee. 16. You place a frozen pie in an oven and bake it for an hour. Then you take it out and let it cool before eating it. Describe how the temperature of the pie changes as time passes. Then sketch a rough graph of the temperature of the pie as a function of time. 17. A homeowner mows the lawn every Wednesday afternoon. Sketch a rough graph of the height of the grass as a function of time over the course of a four-week period. 18. An airplane takes off from an airport and lands an hour later at another airport, 400 miles away. If t represents the time in min-utes since the plane has left the terminal building, let be x#t" 13. 11. 8 AM 10 NOON 2 4 Time (hours) Distance from home (miles) 6 PM Age (years) Weight (pounds) 0 150 100 50 10 200 20 30 40 50 60 70 The graphs of and t are given. (a) State the values of and . (b) For what values of x is ? (c) Estimate the solution of the equation . (d) On what interval is decreasing? (e) State the domain and range of (f) State the domain and range of t. 3. Figure 1 was recorded by an instrument operated by the Cali-fornia Department of Mines and Geology at the University Hospital of the University of Southern California in Los Ange-les. Use it to estimate the range of the vertical ground accelera-tion function at USC during the Northridge earthquake. 4. In this section we discussed examples of ordinary, everyday functions: Population is a function of time, postage cost is a function of weight, water temperature is a function of time. Give three other examples of functions from everyday life that are described verbally. What can you say about the domain and range of each of your functions? If possible, sketch a rough graph of each function. 5–8 Determine whether the curve is the graph of a function of . If it is, state the domain and range of the function. 5. 6. 7. 8. The graph shown gives the weight of a certain person as a function of age. Describe in words how this person’s weight 9. y x 0 1 1 y x 0 1 1 y x 0 1 1 y x 0 1 1 x g x y 0 f 2 2 f. f f #x" ! "1 f #x" ! t#x" t#3" f #"4" f 2. 22 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS 32. Find the domain and range and sketch the graph of the function . 33–44 Find the domain and sketch the graph of the function. 33. 34. 35. 36. 37. 38. 40. 41. 42. 44. 45–50 Find an expression for the function whose graph is the given curve. 45. The line segment joining the points and 46. The line segment joining the points and The bottom half of the parabola 48. The top half of the circle 49. 50. 51–55 Find a formula for the described function and state its domain. 51. A rectangle has perimeter 20 m. Express the area of the rect-angle as a function of the length of one of its sides. y 0 x 1 1 y 0 x 1 1 x 2 % #y " 2"2 ! 4 x % #y " 1"2 ! 0 47. 7, "10" #"5, 10" #5, 7" #1, "3" f #x" !% x % 9 "2x "6 if x # "3 if & x& & 3 if x $ 3 f #x" !% x % 2 x 2 if x & "1 if x $ "1 43. f #x" !% 3 " 1 2 x 2x " 5 if x & 2 if x $ 2 f #x" !% x % 2 1 " x if x # 0 if x ' 0 t#x" ! & x& x 2 G#x" ! 3x % & x& x 39. F#x" ! & 2x % 1& t#x" ! sx " 5 H#t" ! 4 " t 2 2 " t f #t" ! t 2 " 6t F#x" ! 1 2#x % 3" f #x" ! 5 h#x" ! s4 " x 2 h#x" ! 1 s 4 x 2 " 5x 31. the horizontal distance traveled and be the altitude of the plane. (a) Sketch a possible graph of . (b) Sketch a possible graph of . (c) Sketch a possible graph of the ground speed. (d) Sketch a possible graph of the vertical velocity. 19. The number N (in millions) of cellular phone subscribers worldwide is shown in the table. (Midyear estimates are given.) (a) Use the data to sketch a rough graph of N as a function of (b) Use your graph to estimate the number of cell-phone sub-scribers at midyear in 1995 and 1999. 20. Temperature readings (in °F) were recorded every two hours from midnight to 2:00 PM in Dallas on June 2, 2001. The time was measured in hours from midnight. (a) Use the readings to sketch a rough graph of as a function of (b) Use your graph to estimate the temperature at 11:00 AM. 21. If , ﬁnd , , , , , , , , and . 22. A spherical balloon with radius r inches has volume . Find a function that represents the amount of air required to inﬂate the balloon from a radius of r inches to a radius of r % 1 inches. 23–26 Evaluate the difference quotient for the given function. Simplify your answer. , 24. , 25. , 26. , 27–31 Find the domain of the function. 27. 28. 29. 30. t#u" ! su % s4 " u f #t" ! st % s 3 t f #x" ! 5x % 4 x 2 % 3x % 2 f #x" ! x 3x " 1 f #x" " f #1" x " 1 f #x" ! x % 3 x % 1 f #x" " f #a" x " a f #x" ! 1 x f #a % h" " f #a" h f #x" ! x 3 f #3 % h" " f #3" h f #x" ! 4 % 3x " x 2 23. V#r" ! 4 3(r 3 f #a % h" [ f #a"]2, f #a2" f #2a" 2f #a" f #a % 1" f #"a" f #a" f #"2" f #2" f #x" ! 3x 2 " x % 2 t. T t T t. y#t" x#t" y#t" t 1990 1992 1994 1996 1998 2000 N 11 26 60 160 340 650 t 0 2 4 6 8 10 12 14 T 73 73 70 69 72 81 88 91 SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION \|\|\|\| 23 (b) How much tax is assessed on an income of $14,000? On $26,000? (c) Sketch the graph of the total assessed tax T as a function of the income I. 60. The functions in Example 10 and Exercises 58 and 59(a) are called step functions because their graphs look like stairs. Give two other examples of step functions that arise in everyday life. 61–62 Graphs of and are shown. Decide whether each function is even, odd, or neither. Explain your reasoning. 61. 62. 63. (a) If the point is on the graph of an even function, what other point must also be on the graph? (b) If the point is on the graph of an odd function, what other point must also be on the graph? 64. A function has domain and a portion of its graph is shown. (a) Complete the graph of if it is known that is even. (b) Complete the graph of if it is known that is odd. 65–70 Determine whether is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. 65. 66. 67. 68. 69. 70. f x 1 3x 3 x 5 f x 1 3x 2 x 4 f x x x f x x x 1 f x x 2 x 4 1 f x x x 2 1 f x 0 y 5 _5 f f f f 5, 5 f 5, 3 5, 3 y x f g y x f g t f 52. A rectangle has area 16 m . Express the perimeter of the rect-angle as a function of the length of one of its sides. 53. Express the area of an equilateral triangle as a function of the length of a side. 54. Express the surface area of a cube as a function of its volume. An open rectangular box with volume 2 m has a square base. Express the surface area of the box as a function of the length of a side of the base. 56. A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 ft, express the area of the window as a function of the width of the window. 57. A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 12 in. by 20 in. by cutting out equal squares of side at each corner and then folding up the sides as in the ﬁgure. Express the volume of the box as a function of . 58. A taxi company charges two dollars for the ﬁrst mile (or part of a mile) and 20 cents for each succeeding tenth of a mile (or part). Express the cost (in dollars) of a ride as a function of the distance traveled (in miles) for , and sketch the graph of this function. In a certain country, income tax is assessed as follows. There is no tax on income up to $10,000. Any income over $10,000 is taxed at a rate of 10%, up to an income of $20,000. Any income over $20,000 is taxed at 15%. (a) Sketch the graph of the tax rate R as a function of the income I. 59. 0 x 2 x C 20 12 x x x x x x x x x V x x © Catherine Karnow x A 3 55. 2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS A mathematical model is a mathematical description (often by means of a function or an equation) of a real-world phenomenon such as the size of a population, the demand for a product, the speed of a falling object, the concentration of a product in a chemical reac-tion, the life expectancy of a person at birth, or the cost of emission reductions. The pur-pose of the model is to understand the phenomenon and perhaps to make predictions about future behavior. Figure 1 illustrates the process of mathematical modeling. Given a real-world problem, our ﬁrst task is to formulate a mathematical model by identifying and naming the inde-pendent and dependent variables and making assumptions that simplify the phenomenon enough to make it mathematically tractable. We use our knowledge of the physical situa-tion and our mathematical skills to obtain equations that relate the variables. In situations where there is no physical law to guide us, we may need to collect data (either from a library or the Internet or by conducting our own experiments) and examine the data in the form of a table in order to discern patterns. From this numerical representation of a func-tion we may wish to obtain a graphical representation by plotting the data. The graph might even suggest a suitable algebraic formula in some cases. The second stage is to apply the mathematics that we know (such as the calculus that will be developed throughout this book) to the mathematical model that we have formu-lated in order to derive mathematical conclusions. Then, in the third stage, we take those mathematical conclusions and interpret them as information about the original real-world phenomenon by way of offering explanations or making predictions. The ﬁnal step is to test our predictions by checking against new real data. If the predictions don’t compare well with reality, we need to reﬁne our model or to formulate a new model and start the cycle again. A mathematical model is never a completely accurate representation of a physical situ-ation—it is an idealization. A good model simpliﬁes reality enough to permit mathemati-cal calculations but is accurate enough to provide valuable conclusions. It is important to realize the limitations of the model. In the end, Mother Nature has the ﬁnal say. There are many different types of functions that can be used to model relationships observed in the real world. In what follows, we discuss the behavior and graphs of these functions and give examples of situations appropriately modeled by such functions. LINEAR MODELS When we say that y is a linear function of x, we mean that the graph of the function is a line, so we can use the slope-intercept form of the equation of a line to write a formula for the function as where m is the slope of the line and b is the y-intercept. y ! f#x" ! mx % b FIGURE 1 The modeling process Real-world problem Mathematical model Real-world predictions Mathematical conclusions Test Formulate Solve Interpret 1.2 24 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS N The coordinate geometry of lines is reviewed in Appendix B. A characteristic feature of linear functions is that they grow at a constant rate. For instance, Figure 2 shows a graph of the linear function and a table of sam-ple values. Notice that whenever x increases by 0.1, the value of increases by 0.3. So increases three times as fast as x. Thus the slope of the graph , namely 3, can be interpreted as the rate of change of y with respect to x. EXAMPLE 1 (a) As dry air moves upward, it expands and cools. If the ground temperature is and the temperature at a height of 1 km is , express the temperature T (in °C) as a function of the height h (in kilometers), assuming that a linear model is appropriate. (b) Draw the graph of the function in part (a). What does the slope represent? (c) What is the temperature at a height of 2.5 km? SOLUTION (a) Because we are assuming that T is a linear function of h, we can write We are given that when , so In other words, the y-intercept is . We are also given that when , so The slope of the line is therefore and the required linear function is (b) The graph is sketched in Figure 3. The slope is , and this represents the rate of change of temperature with respect to height. (c) At a height of , the temperature is M If there is no physical law or principle to help us formulate a model, we construct an empirical model, which is based entirely on collected data. We seek a curve that “ﬁts” the data in the sense that it captures the basic trend of the data points. T ! "10#2.5" % 20 ! "5)C h ! 2.5 km m ! "10)C'km T ! "10h % 20 m ! 10 " 20 ! "10 10 ! m ! 1 % 20 h ! 1 T ! 10 b ! 20 20 ! m ! 0 % b ! b h ! 0 T ! 20 T ! mh % b 10)C 20)C V x y 0 y=3x-2 _2 FIGURE 2 y ! 3x " 2 f#x" f#x" f#x" ! 3x " 2 SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS \|\|\|\| 25 x 1.0 1.0 1.1 1.3 1.2 1.6 1.3 1.9 1.4 2.2 1.5 2.5 f #x" ! 3x " 2 FIGURE 3 T=_10h+20 T h 0 10 20 1 3 EXAMPLE 2 Table 1 lists the average carbon dioxide level in the atmosphere, measured in parts per million at Mauna Loa Observatory from 1980 to 2002. Use the data in Table 1 to ﬁnd a model for the carbon dioxide level. SOLUTION We use the data in Table 1 to make the scatter plot in Figure 4, where t repre-sents time (in years) and C represents the level (in parts per million, ppm). Notice that the data points appear to lie close to a straight line, so it’s natural to choose a linear model in this case. But there are many possible lines that approximate these data points, so which one should we use? From the graph, it appears that one possi-bility is the line that passes through the ﬁrst and last data points. The slope of this line is and its equation is or Equation 1 gives one possible linear model for the carbon dioxide level; it is graphed in Figure 5. Although our model ﬁts the data reasonably well, it gives values higher than most of the actual levels. A better linear model is obtained by a procedure from statistics CO2 Linear model through first and last data points FIGURE 5 340 350 360 1980 1985 1990 C t 1995 2000 370 C ! 1.5545t " 2739.21 1 C " 338.7 ! 1.5545#t " 1980" 372.9 " 338.7 2002 " 1980 ! 34.2 22 ( 1.5545 FIGURE 4 Scatter plot for the average CO™ level 340 350 360 1980 1985 1990 C t 1995 2000 370 CO2 V 26 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS TABLE 1 level level Year (in ppm) Year (in ppm) 1980 338.7 1992 356.4 1982 341.1 1994 358.9 1984 344.4 1996 362.6 1986 347.2 1998 366.6 1988 351.5 2000 369.4 1990 354.2 2002 372.9 CO2 CO2 called linear regression. If we use a graphing calculator, we enter the data from Table 1 into the data editor and choose the linear regression command. (With Maple we use the ﬁt[leastsquare] command in the stats package; with Mathematica we use the Fit com-mand.) The machine gives the slope and y-intercept of the regression line as So our least squares model for the level is In Figure 6 we graph the regression line as well as the data points. Comparing with Figure 5, we see that it gives a better ﬁt than our previous linear model. M EXAMPLE 3 Use the linear model given by Equation 2 to estimate the average level for 1987 and to predict the level for the year 2010. According to this model, when will the level exceed 400 parts per million? SOLUTION Using Equation 2 with t ! 1987, we estimate that the average level in 1987 was This is an example of interpolation because we have estimated a value between observed values. (In fact, the Mauna Loa Observatory reported that the average level in 1987 was 348.93 ppm, so our estimate is quite accurate.) With , we get So we predict that the average level in the year 2010 will be 384.8 ppm. This is an example of extrapolation because we have predicted a value outside the region of observations. Consequently, we are far less certain about the accuracy of our prediction. Using Equation 2, we see that the level exceeds 400 ppm when Solving this inequality, we get t $ 3134.55 1.55192 ( 2019.79 1.55192t " 2734.55 $ 400 CO2 CO2 C#2010" ! #1.55192"#2010" " 2734.55 ( 384.81 t ! 2010 CO2 C#1987" ! #1.55192"#1987" " 2734.55 ( 349.12 CO2 CO2 CO2 V FIGURE 6 The regression line 340 350 360 1980 1985 1990 C t 1995 2000 370 C ! 1.55192t " 2734.55 2 CO2 b ! "2734.55 m ! 1.55192 SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS \|\|\|\| 27 N A computer or graphing calculator ﬁnds the regression line by the method of least squares, which is to minimize the sum of the squares of the vertical distances between the data points and the line. The details are explained in Section 14.7. We therefore predict that the level will exceed 400 ppm by the year 2019. This prediction is somewhat risky because it involves a time quite remote from our observations. M POLYNOMIALS A function is called a polynomial if where is a nonnegative integer and the numbers are constants called the coefﬁcients of the polynomial. The domain of any polynomial is If the leading coefﬁcient , then the degree of the polynomial is . For example, the function is a polynomial of degree 6. A polynomial of degree 1 is of the form and so it is a linear function. A polynomial of degree 2 is of the form and is called a quadratic function. Its graph is always a parabola obtained by shifting the parabola , as we will see in the next section. The parabola opens upward if and downward if . (See Figure 7.) A polynomial of degree 3 is of the form and is called a cubic function. Figure 8 shows the graph of a cubic function in part (a) and graphs of polynomials of degrees 4 and 5 in parts (b) and (c). We will see later why the graphs have these shapes. FIGURE 8 (a) y=˛-x+1 x 1 y 1 0 (b) y=x$-3≈+x x 2 y 1 (c) y=3x%-25˛+60x x 20 y 1 #a " 0" P#x" ! ax 3 % bx 2 % cx % d The graphs of quadratic functions are parabolas. FIGURE 7 0 y 2 x 1 (a) y=≈+x+1 y 2 x 1 (b) y=_2≈+3x+1 a # 0 a $ 0 y ! ax 2 P#x" ! ax 2 % bx % c P#x" ! mx % b P#x" ! 2x 6 " x 4 % 2 5 x 3 % s2 n an " 0 ! ! #"!, !". a0, a1, a2, . . . , an n P#x" ! anx n % an"1x n"1 % % a2x 2 % a1x % a0 P CO2 28 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS Polynomials are commonly used to model various quantities that occur in the natural and social sciences. For instance, in Section 3.7 we will explain why economists often use a polynomial to represent the cost of producing units of a commodity. In the follow-ing example we use a quadratic function to model the fall of a ball. EXAMPLE 4 A ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground, and its height h above the ground is recorded at 1-second intervals in Table 2. Find a model to ﬁt the data and use the model to predict the time at which the ball hits the ground. SOLUTION We draw a scatter plot of the data in Figure 9 and observe that a linear model is inappropriate. But it looks as if the data points might lie on a parabola, so we try a quad-ratic model instead. Using a graphing calculator or computer algebra system (which uses the least squares method), we obtain the following quadratic model: In Figure 10 we plot the graph of Equation 3 together with the data points and see that the quadratic model gives a very good ﬁt. The ball hits the ground when , so we solve the quadratic equation The quadratic formula gives The positive root is , so we predict that the ball will hit the ground after about 9.7 seconds. M POWER FUNCTIONS A function of the form , where is a constant, is called a power function. We consider several cases. a f#x" ! x a t ( 9.67 t ! "0.96 + s#0.96"2 " 4#"4.90"#449.36" 2#"4.90" "4.90t 2 % 0.96t % 449.36 ! 0 h ! 0 FIGURE 10 Quadratic model for a falling ball 2 200 400 4 6 8 t 0 FIGURE 9 Scatter plot for a falling ball 200 400 t (seconds) 0 2 4 6 8 h h (meters) h ! 449.36 % 0.96t " 4.90t 2 3 x P#x" SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS \|\|\|\| 29 TABLE 2 Time Height (seconds) (meters) 0 450 1 445 2 431 3 408 4 375 5 332 6 279 7 216 8 143 9 61 (i) , where n is a positive integer The graphs of for , and are shown in Figure 11. (These are poly-nomials with only one term.) We already know the shape of the graphs of (a line through the origin with slope 1) and [a parabola, see Example 2(b) in Section 1.1]. The general shape of the graph of depends on whether is even or odd. If is even, then is an even function and its graph is similar to the parabola . If is odd, then is an odd function and its graph is similar to that of . Notice from Figure 12, however, that as increases, the graph of becomes ﬂatter near 0 and steeper when . (If is small, then is smaller, is even smaller, is smaller still, and so on.) (ii) , where n is a positive integer The function is a root function. For it is the square root func-tion , whose domain is and whose graph is the upper half of the parabola . [See Figure 13(a).] For other even values of n, the graph of is similar to that of . For we have the cube root function whose domain is (recall that every real number has a cube root) and whose graph is shown in Figure 13(b). The graph of for n odd is similar to that of . (b) ƒ=Œ„ x x y 0 (1, 1) (a) ƒ=œ„ x x y 0 (1, 1) FIGURE 13 Graphs of root functions y ! s 3 x !n ! 3" y ! s n x ! f !x" ! s 3 x n ! 3 y ! sx y ! s n x x ! y 2 #0, "" f !x" ! sx n ! 2 f !x" ! x 1$n ! s n x a ! 1$n FIGURE 12 Families of power functions y=x$ (1, 1) (_1, 1) y=x^ y=≈ y=x# y=x% (_1, _1) (1, 1) 0 y x x y 0 x 4 x 3 x 2 x %x% # 1 y ! x n n y ! x 3 f!x" ! x n n y ! x 2 f!x" ! x n n n f!x" ! x n Graphs of ƒ=x n for n=1, 2, 3, 4, 5 x 1 y 1 0 y=x% x 1 y 1 0 y=x# x 1 y 1 0 y=≈ x 1 y 1 0 y=x x 1 y 1 0 y=x$ FIGURE 11 y ! x 2 y ! x 5 2, 3, 4 n ! 1, f!x" ! x n a ! n 30 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS (iii) The graph of the reciprocal function is shown in Figure 14. Its graph has the equation , or , and is a hyperbola with the coordinate axes as its asymptotes. This function arises in physics and chemistry in connection with Boyle’s Law, which says that, when the temperature is constant, the volume of a gas is inversely proportional to the pressure : where C is a constant. Thus the graph of V as a function of P (see Figure 15) has the same general shape as the right half of Figure 14. Another instance in which a power function is used to model a physical phenomenon is discussed in Exercise 26. RATIONAL FUNCTIONS A rational function is a ratio of two polynomials: where and are polynomials. The domain consists of all values of such that . A simple example of a rational function is the function , whose domain is ; this is the reciprocal function graphed in Figure 14. The function is a rational function with domain . Its graph is shown in Figure 16. ALGEBRAIC FUNCTIONS A function is called an algebraic function if it can be constructed using algebraic oper-ations (such as addition, subtraction, multiplication, division, and taking roots) starting with polynomials. Any rational function is automatically an algebraic function. Here are two more examples: t!x" ! x 4 $ 16x 2 x % sx % !x $ 2"s 3 x % 1 f!x" ! sx 2 % 1 f &x% x " &2' f!x" ! 2x 4 $ x 2 % 1 x 2 $ 4 &x% x " 0' f !x" ! 1$x Q!x" " 0 x Q P f!x" ! P!x" Q!x" f P V 0 FIGURE 15 Volume as a function of pressure at constant temperature V ! C P P V xy ! 1 y ! 1$x f!x" ! x $1 ! 1$x a ! $1 SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS \|\|\|\| 31 FIGURE 14 The reciprocal function x 1 y 1 0 y=∆ FIGURE 16 ƒ=2x$-≈+1 ≈-4 x 20 y 2 0 When we sketch algebraic functions in Chapter 4, we will see that their graphs can assume a variety of shapes. Figure 17 illustrates some of the possibilities. An example of an algebraic function occurs in the theory of relativity. The mass of a particle with velocity is where is the rest mass of the particle and km$s is the speed of light in a vacuum. TRIGONOMETRIC FUNCTIONS Trigonometry and the trigonometric functions are reviewed on Reference Page 2 and also in Appendix D. In calculus the convention is that radian measure is always used (except when otherwise indicated). For example, when we use the function , it is under-stood that means the sine of the angle whose radian measure is . Thus the graphs of the sine and cosine functions are as shown in Figure 18. Notice that for both the sine and cosine functions the domain is and the range is the closed interval . Thus, for all values of , we have or, in terms of absolute values, Also, the zeros of the sine function occur at the integer multiples of ; that is, n an integer x ! n' when sin x ! 0 ' %cos x% ( 1 %sin x% ( 1 $1 ( cos x ( 1 $1 ( sin x ( 1 x #$1, 1( !$", "" (a) ƒ=sin x π 2 5π 2 3π 2 π 2 _ x y π 0 _π 1 _1 2π 3π (b) ©=cos x x y 0 1 _1 π _π 2π 3π π 2 5π 2 3π 2 π 2 _ x sin x f!x" ! sin x c ! 3.0 ) 105 m0 m ! f !v" ! m0 s1 $ v 2$c 2 v FIGURE 17 x 2 y 1 (a) ƒ=xœ„„„„ x+3 x 1 y 5 0 (b) ©=$ œ„„„„„„ ≈-25 x 1 y 1 0 (c) h(x)=x@?#(x-2)@ _3 32 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS FIGURE 18 N The Reference Pages are located at the front and back of the book. An important property of the sine and cosine functions is that they are periodic func-tions and have period . This means that, for all values of , The periodic nature of these functions makes them suitable for modeling repetitive phe-nomena such as tides, vibrating springs, and sound waves. For instance, in Example 4 in Section 1.3 we will see that a reasonable model for the number of hours of daylight in Philadelphia t days after January 1 is given by the function The tangent function is related to the sine and cosine functions by the equation and its graph is shown in Figure 19. It is undeﬁned whenever , that is, when , Its range is . Notice that the tangent function has period : The remaining three trigonometric functions (cosecant, secant, and cotangent) are the reciprocals of the sine, cosine, and tangent functions. Their graphs are shown in Appendix D. EXPONENTIAL FUNCTIONS The exponential functions are the functions of the form , where the base is a positive constant. The graphs of and are shown in Figure 20. In both cases the domain is and the range is . Exponential functions will be studied in detail in Section 1.5, and we will see that they are useful for modeling many natural phenomena, such as population growth (if ) and radioactive decay (if a 1". a ! 1 FIGURE 20 (a) y=2® (b) y=(0.5)® y x 1 1 0 y x 1 1 0 !0, "" !$", "" y ! !0.5"x y ! 2x a f!x" ! a x for all x tan!x % '" ! tan x ' !$", "" &3'$2, . . . . x ! &'$2 cos x ! 0 tan x ! sin x cos x L!t" ! 12 % 2.8 sin) 2' 365 !t $ 80" cos!x % 2'" ! cos x sin!x % 2'" ! sin x x 2' SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS \|\|\|\| 33 FIGURE 19 y=tan x x y π 0 _π 1 π 2 3π 2 π 2 _ 3π 2 _ LOGARITHMIC FUNCTIONS The logarithmic functions , where the base is a positive constant, are the inverse functions of the exponential functions. They will be studied in Section 1.6. Fig-ure 21 shows the graphs of four logarithmic functions with various bases. In each case the domain is , the range is , and the function increases slowly when . TRANSCENDENTAL FUNCTIONS These are functions that are not algebraic. The set of transcendental functions includes the trigonometric, inverse trigonometric, exponential, and logarithmic functions, but it also includes a vast number of other functions that have never been named. In Chapter 11 we will study transcendental functions that are deﬁned as sums of inﬁnite series. EXAMPLE 5 Classify the following functions as one of the types of functions that we have discussed. (a) (b) (c) (d) SOLUTION (a) is an exponential function. (The is the exponent.) (b) is a power function. (The is the base.) We could also consider it to be a polynomial of degree 5. (c) is an algebraic function. (d) is a polynomial of degree 4. M u!t" ! 1 $ t % 5t 4 h!x" ! 1 % x 1 $ sx x t!x" ! x 5 x f!x" ! 5x u!t" ! 1 $ t % 5t 4 h!x" ! 1 % x 1 $ sx t!x" ! x 5 f!x" ! 5x x ! 1 !$", "" !0, "" a f!x" ! loga x 34 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS FIGURE 21 0 y 1 x 1 y=log£ x y=log™ x y=log∞ x y=log¡¸ x 3–4 Match each equation with its graph. Explain your choices. (Don’t use a computer or graphing calculator.) (a) (b) (c) f 0 g h y x y ! x 8 y ! x 5 y ! x 2 3. 1–2 Classify each function as a power function, root function, polynomial (state its degree), rational function, algebraic function, trigonometric function, exponential function, or logarithmic function. 1. (a) (b) (c) (d) (e) (f) 2. (a) (b) (c) (d) (e) (f) y ! cos + % sin + y ! 2t 6 % t 4 $ ' y ! x 10 y ! 10 x y ! x % x 2 sx $ 1 y ! x $ 6 x % 6 t!x" ! log10 x s!x" ! tan 2x r!x" ! x 2 % 1 x 3 % x h!x" ! x 9 % x 4 t!x" ! s1 $ x 2 f !x" ! s 5 x EXERCISES 1.2 12. The manager of a weekend ﬂea market knows from past experience that if he charges dollars for a rental space at the market, then the number of spaces he can rent is given by the equation . (a) Sketch a graph of this linear function. (Remember that the rental charge per space and the number of spaces rented can’t be negative quantities.) (b) What do the slope, the y-intercept, and the x-intercept of the graph represent? 13. The relationship between the Fahrenheit and Celsius temperature scales is given by the linear function . (a) Sketch a graph of this function. (b) What is the slope of the graph and what does it represent? What is the F-intercept and what does it represent? 14. Jason leaves Detroit at 2:00 PM and drives at a constant speed west along I-96. He passes Ann Arbor, 40 mi from Detroit, at 2:50 PM. (a) Express the distance traveled in terms of the time elapsed. (b) Draw the graph of the equation in part (a). (c) What is the slope of this line? What does it represent? Biologists have noticed that the chirping rate of crickets of a certain species is related to temperature, and the relationship appears to be very nearly linear. A cricket produces 113 chirps per minute at and 173 chirps per minute at . (a) Find a linear equation that models the temperature T as a function of the number of chirps per minute N. (b) What is the slope of the graph? What does it represent? (c) If the crickets are chirping at 150 chirps per minute, esti-mate the temperature. 16. The manager of a furniture factory ﬁnds that it costs $2200 to manufacture 100 chairs in one day and $4800 to produce 300 chairs in one day. (a) Express the cost as a function of the number of chairs produced, assuming that it is linear. Then sketch the graph. (b) What is the slope of the graph and what does it represent? (c) What is the y-intercept of the graph and what does it represent? At the surface of the ocean, the water pressure is the same as the air pressure above the water, . Below the surface, the water pressure increases by for every 10 ft of descent. (a) Express the water pressure as a function of the depth below the ocean surface. (b) At what depth is the pressure ? 100 lb$in2 4.34 lb$in2 15 lb$in2 17. 80,F 70,F 15. F ! 9 5C % 32 !C" !F" y ! 200 $ 4x y x 4. (a) (b) (c) (d) (a) Find an equation for the family of linear functions with slope 2 and sketch several members of the family. (b) Find an equation for the family of linear functions such that and sketch several members of the family. (c) Which function belongs to both families? 6. What do all members of the family of linear functions have in common? Sketch several mem-bers of the family. 7. What do all members of the family of linear functions have in common? Sketch several members of the family. 8. Find expressions for the quadratic functions whose graphs are shown. 9. Find an expression for a cubic function if and . 10. Recent studies indicate that the average surface tempera-ture of the earth has been rising steadily. Some scientists have modeled the temperature by the linear function , where is temperature in and repre-sents years since 1900. (a) What do the slope and -intercept represent? (b) Use the equation to predict the average global surface temperature in 2100. 11. If the recommended adult dosage for a drug is (in mg), then to determine the appropriate dosage for a child of age , pharmacists use the equation . Suppose the dosage for an adult is 200 mg. (a) Find the slope of the graph of . What does it represent? (b) What is the dosage for a newborn? c c ! 0.0417D!a % 1" a c D T t ,C T T ! 0.02t % 8.50 f !$1" ! f !0" ! f !2" ! 0 f !1" ! 6 f y (0, 1) (1, _2.5) (_2, 2) y x 0 (4, 2) f g x 0 3 f !x" ! c $ x f !x" ! 1 % m!x % 3" f !2" ! 1 5. G f g F y x y ! s 3 x y ! x 3 y ! 3x y ! 3x SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS \|\|\|\| 35 (b) Find and graph a linear model using the ﬁrst and last data points. (c) Find and graph the least squares regression line. (d) Use the linear model in part (c) to estimate the ulcer rate for an income of $25,000. (e) According to the model, how likely is someone with an income of $80,000 to suffer from peptic ulcers? (f) Do you think it would be reasonable to apply the model to someone with an income of $200,000? ; 22. Biologists have observed that the chirping rate of crickets of a certain species appears to be related to temperature. The table shows the chirping rates for various temperatures. (a) Make a scatter plot of the data. (b) Find and graph the regression line. (c) Use the linear model in part (b) to estimate the chirping rate at . ; 23. The table gives the winning heights for the Olympic pole vault competitions in the 20th century. (a) Make a scatter plot and decide whether a linear model is appropriate. (b) Find and graph the regression line. (c) Use the linear model to predict the height of the winning pole vault at the 2000 Olympics and compare with the actual winning height of 19.36 feet. (d) Is it reasonable to use the model to predict the winning height at the 2100 Olympics? 100,F 18. The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her $380 to drive 480 mi and in June it cost her $460 to drive 800 mi. (a) Express the monthly cost as a function of the distance driven assuming that a linear relationship gives a suit-able model. (b) Use part (a) to predict the cost of driving 1500 miles per month. (c) Draw the graph of the linear function. What does the slope represent? (d) What does the y-intercept represent? (e) Why does a linear function give a suitable model in this situation? 19–20 For each scatter plot, decide what type of function you might choose as a model for the data. Explain your choices. 19. (a) (b) 20. (a) (b) ; 21. The table shows (lifetime) peptic ulcer rates (per 100 popula-tion) for various family incomes as reported by the National Health Interview Survey. (a) Make a scatter plot of these data and decide whether a linear model is appropriate. 0 x y 0 x y 0 x y 0 x y d, C 36 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS Ulcer rate Income (per 100 population) $4,000 14.1 $6,000 13.0 $8,000 13.4 $12,000 12.5 $16,000 12.0 $20,000 12.4 $30,000 10.5 $45,000 9.4 $60,000 8.2 Year Height (ft) Year Height (ft) 1900 10.83 1956 14.96 1904 11.48 1960 15.42 1908 12.17 1964 16.73 1912 12.96 1968 17.71 1920 13.42 1972 18.04 1924 12.96 1976 18.04 1928 13.77 1980 18.96 1932 14.15 1984 18.85 1936 14.27 1988 19.77 1948 14.10 1992 19.02 1952 14.92 1996 19.42 Temperature Chirping rate Temperature Chirping rate (°F) (chirps$min) (°F) (chirps$min) 50 20 75 140 55 46 80 173 60 79 85 198 65 91 90 211 70 113 ; 26. The table shows the mean (average) distances d of the planets from the sun (taking the unit of measurement to be the distance from the earth to the sun) and their periods T (time of revolution in years). (a) Fit a power model to the data. (b) Kepler’s Third Law of Planetary Motion states that “The square of the period of revolution of a planet is proportional to the cube of its mean distance from the sun.” Does your model corroborate Kepler’s Third Law? ; 24. A study by the US Ofﬁce of Science and Technology in 1972 estimated the cost (in 1972 dollars) to reduce auto-mobile emissions by certain percentages: Find a model that captures the “diminishing returns” trend of these data. ; 25. Use the data in the table to model the population of the world in the 20th century by a cubic function. Then use your model to estimate the population in the year 1925. SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS \|\|\|\| 37 Planet d T Mercury 0.387 0.241 Venus 0.723 0.615 Earth 1.000 1.000 Mars 1.523 1.881 Jupiter 5.203 11.861 Saturn 9.541 29.457 Uranus 19.190 84.008 Neptune 30.086 164.784 Reduction in Cost per Reduction in Cost per emissions (%) car (in $) emissions (%) car (in $) 50 45 75 90 55 55 80 100 60 62 85 200 65 70 90 375 70 80 95 600 Population Population Year (millions) Year (millions) 1900 1650 1960 3040 1910 1750 1970 3710 1920 1860 1980 4450 1930 2070 1990 5280 1940 2300 2000 6080 1950 2560 NEW FUNCTIONS FROM OLD FUNCTIONS In this section we start with the basic functions we discussed in Section 1.2 and obtain new functions by shifting, stretching, and reﬂecting their graphs. We also show how to combine pairs of functions by the standard arithmetic operations and by composition. TRANSFORMATIONS OF FUNCTIONS By applying certain transformations to the graph of a given function we can obtain the graphs of certain related functions. This will give us the ability to sketch the graphs of many functions quickly by hand. It will also enable us to write equations for given graphs. Let’s ﬁrst consider translations. If c is a positive number, then the graph of is just the graph of shifted upward a distance of c units (because each y-coordinate is increased by the same number c). Likewise, if , where , then the value of at x is the same as the value of at (c units to the left of x). Therefore, the graph of is just the graph of shifted units to the right (see Figure 1). VERTICAL AND HORIZONTAL SHIFTS Suppose . To obtain the graph of y ! f!x % c", shift the graph of y ! f!x" a distance c units to the left y ! f!x $ c", shift the graph of y ! f!x" a distance c units to the right y ! f!x" $ c, shift the graph of y ! f!x" a distance c units downward y ! f!x" % c, shift the graph of y ! f!x" a distance c units upward c ! 0 c y ! f!x" y ! f!x $ c" x $ c f t c ! 0 t!x" ! f!x $ c" y ! f!x" y ! f!x" % c 1.3 Now let’s consider the stretching and reﬂecting transformations. If , then the graph of is the graph of stretched by a factor of c in the vertical direction (because each y-coordinate is multiplied by the same number c). The graph of is the graph of reﬂected about the -axis because the point is replaced by the point . (See Figure 2 and the following chart, where the results of other stretching, compressing, and reﬂecting transformations are also given.) VERTICAL AND HORIZONTAL STRETCHING AND REFLECTING Suppose . To obtain the graph of Figure 3 illustrates these stretching transformations when applied to the cosine function with . For instance, in order to get the graph of we multiply the y-coor-dinate of each point on the graph of by 2. This means that the graph of gets stretched vertically by a factor of 2. FIGURE 3 x 1 2 y 0 y=cos x y=cos 2x y=cos x 1 2 x 1 2 y 0 y=2 cos x y=cos x y= cos x 1 2 1 y ! cos x y ! cos x y ! 2 cos x c ! 2 y ! f!$x", reflect the graph of y ! f!x" about the y-axis y ! $f!x", reflect the graph of y ! f!x" about the x-axis y ! f!x$c", stretch the graph of y ! f!x" horizontally by a factor of c y ! f!cx", compress the graph of y ! f!x" horizontally by a factor of c y ! !1$c"f!x", compress the graph of y ! f!x" vertically by a factor of c y ! cf!x", stretch the graph of y ! f!x" vertically by a factor of c c ! 1 !x, $y" !x, y" x y ! f!x" y ! $f!x" y ! f!x" y ! cf!x" c ! 1 FIGURE 2 Stretching and reﬂecting the graph of ƒ y= ƒ 1 c x y 0 y=f(_x) y=ƒ y=_ƒ y=cƒ (c>1) FIGURE 1 Translating the graph of ƒ x y 0 y=f(x-c) y=f(x+c) y =ƒ y=ƒ-c y=ƒ+c c c c c 38 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS EXAMPLE 1 Given the graph of , use transformations to graph , , , , and . SOLUTION The graph of the square root function , obtained from Figure 13(a) in Section 1.2, is shown in Figure 4(a). In the other parts of the ﬁgure we sketch by shifting 2 units downward, by shifting 2 units to the right, by reﬂecting about the -axis, by stretching vertically by a factor of 2, and by reﬂecting about the -axis. M EXAMPLE 2 Sketch the graph of the function . SOLUTION Completing the square, we write the equation of the graph as This means we obtain the desired graph by starting with the parabola and shifting 3 units to the left and then 1 unit upward (see Figure 5). M EXAMPLE 3 Sketch the graphs of the following functions. (a) (b) SOLUTION (a) We obtain the graph of from that of by compressing horizon-tally by a factor of 2 (see Figures 6 and 7). Thus, whereas the period of is , the period of is . FIGURE 6 x 0 y 1 π 2 π y=sin x FIGURE 7 x 0 y 1 π 2 π 4 π y=sin 2x 2'$2 ! ' y ! sin 2x 2' y ! sin x y ! sin x y ! sin 2x y ! 1 $ sin x y ! sin 2x FIGURE 5 (a) y=≈ (b) y=(x+3)@+1 x 0 _1 _3 1 y (_3, 1) x 0 y y ! x 2 y ! x 2 % 6x % 10 ! !x % 3"2 % 1 f(x) ! x 2 % 6x % 10 (a) y=œ„ x (b) y=œ„-2 x (c) y=œ„„„„ x-2 (d) y=_œ„ x (e) y=2œ„ x (f) y=œ„„ _x 0 x y 0 x y 0 x y 2 0 x y _2 0 x y 1 1 0 x y y y ! s$x y ! 2sx x y ! $sx y ! sx $ 2 y ! sx $ 2 y ! sx y ! s$x y ! 2sx y ! $sx y ! sx $ 2 y ! sx $ 2 y ! sx V SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS \|\|\|\| 39 FIGURE 4 (b) To obtain the graph of , we again start with . We reﬂect about the -axis to get the graph of and then we shift 1 unit upward to get (See Figure 8.) M EXAMPLE 4 Figure 9 shows graphs of the number of hours of daylight as functions of the time of the year at several latitudes. Given that Philadelphia is located at approximately latitude, ﬁnd a function that models the length of daylight at Philadelphia. SOLUTION Notice that each curve resembles a shifted and stretched sine function. By look-ing at the blue curve we see that, at the latitude of Philadelphia, daylight lasts about 14.8 hours on June 21 and 9.2 hours on December 21, so the amplitude of the curve (the factor by which we have to stretch the sine curve vertically) is . By what factor do we need to stretch the sine curve horizontally if we measure the time t in days? Because there are about 365 days in a year, the period of our model should be 365. But the period of is , so the horizontal stretching factor is . We also notice that the curve begins its cycle on March 21, the 80th day of the year, so we have to shift the curve 80 units to the right. In addition, we shift it 12 units upward. Therefore we model the length of daylight in Philadelphia on the tth day of the year by the function M Another transformation of some interest is taking the absolute value of a function. If , then according to the deﬁnition of absolute value, when and when . This tells us how to get the graph of from the graph of : The part of the graph that lies above the -axis remains the same; the part that lies below the -axis is reﬂected about the -axis. x x x y ! f!x" y ! # f!x"# f!x" ! 0 y ! "f!x" f!x" # 0 y ! f!x" y ! # f!x"# L!t" ! 12 $ 2.8 sin$ 2% 365 !t " 80"% c ! 2%&365 2% y ! sin t 1 2!14.8 " 9.2" ! 2.8 FIGURE 9 Graph of the length of daylight from March 21 through December 21 at various latitudes Lucia C. Harrison, Daylight, Twilight, Darkness and Time (New York: Silver, Burdett, 1935) page 40. 0 2 4 6 8 10 12 14 16 18 20 Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. Hours 60° N 50° N 40° N 30° N 20° N 40&N FIGURE 8 x 1 2 y π 0 2π y=1-sin x π 2 3π 2 y ! 1 " sin x. y ! "sin x x y ! sin x y ! 1 " sin x 40 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS EXAMPLE 5 Sketch the graph of the function . SOLUTION We ﬁrst graph the parabola in Figure 10(a) by shifting the parabola downward 1 unit. We see that the graph lies below the x-axis when , so we reﬂect that part of the graph about the x-axis to obtain the graph of in Figure 10(b). M COMBINATIONS OF FUNCTIONS Two functions and can be combined to form new functions , , , and in a manner similar to the way we add, subtract, multiply, and divide real numbers. The sum and difference functions are deﬁned by If the domain of is A and the domain of is B, then the domain of is the intersec-tion because both and have to be deﬁned. For example, the domain of is and the domain of is , so the domain of is . Similarly, the product and quotient functions are deﬁned by The domain of is , but we can’t divide by 0 and so the domain of is . For instance, if and , then the domain of the rational function is , or . There is another way of combining two functions to obtain a new function. For example, suppose that and . Since y is a function of u and u is, in turn, a function of x, it follows that is ultimately a function of x. We compute this by substitution: The procedure is called composition because the new function is composed of the two given functions and . In general, given any two functions and , we start with a number x in the domain of and ﬁnd its image . If this number is in the domain of , then we can calculate the value of . The result is a new function obtained by substituting into . It is called the composition (or composite) of and and is denoted by (“f circle t”). DEFINITION Given two functions and , the composite function (also called the composition of and ) is deﬁned by The domain of is the set of all in the domain of such that is in the domain of . In other words, is deﬁned whenever both and are deﬁned. Fig-ure 11 shows how to picture in terms of machines. f ! t f!t!x"" t!x" ! f ! t"!x" f t!x" t x f ! t ! f ! t"!x" ! f!t!x"" t f f ! t t f f ! t t f f t h!x" ! f!t!x"" f!t!x"" f t!x" t!x" t t f t f y ! f!u" ! f!t!x"" ! f!x 2 $ 1" ! sx 2 $ 1 y u ! t!x" ! x 2 $ 1 y ! f!u" ! su !"', 1" ! !1, '" 'x# x " 1( ! f&t"!x" ! x 2&!x " 1" t!x" ! x " 1 f!x" ! x 2 'x " A # B # t!x" " 0( f&t A # B ft ) f t!x" ! f!x" t!x" ! ft"!x" ! f!x"t!x" A # B ! +0, 2, ! f $ t"!x" ! sx $ s2 " x B ! !"', 2, t!x" ! s2 " x A ! +0, '" f!x" ! sx t!x" f!x" A # B f $ t t f ! f " t"!x" ! f!x" " t!x" ! f $ t"!x" ! f !x" $ t!x" f&t ft f " t f $ t t f y ! #x 2 " 1# "1 ! x ! 1 y ! x 2 y ! x 2 " 1 y ! #x 2 " 1# V SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS \|\|\|\| 41 FIGURE 10 0 x y _1 1 (a) y=≈-1 (b) y=\| ≈-1 \| 0 x y _1 1 f g FIGURE 11 f{©} f • g The f • g machine is composed of the g machine (first) and then the f machine. x © (input) (output) EXAMPLE 6 If and , ﬁnd the composite functions and . SOLUTION We have M \| You can see from Example 6 that, in general, . Remember, the notation means that the function is applied ﬁrst and then is applied second. In Example 6, is the function that ﬁrst subtracts 3 and then squares; is the function that ﬁrst squares and then subtracts 3. EXAMPLE 7 If and , ﬁnd each function and its domain. (a) (b) (c) (d) SOLUTION (a) The domain of is . (b) For to be deﬁned we must have . For to be deﬁned we must have If , then . , that is, , or . Thus we have , so the domain of is the closed interval . (c) The domain of is . (d) This expression is deﬁned when both and The ﬁrst inequality means , and the second is equivalent to , or , or . Thus , so the domain of is the closed interval . M It is possible to take the composition of three or more functions. For instance, the com-posite function is found by ﬁrst applying , then , and then as follows: EXAMPLE 8 Find if , and . SOLUTION M So far we have used composition to build complicated functions from simpler ones. But in calculus it is often useful to be able to decompose a complicated function into simpler ones, as in the following example. ! f!!x $ 3"10" ! !x $ 3"10 !x $ 3"10 $ 1 ! f ! t ! h"!x" ! f!t!h!x""" ! f!t!x $ 3"" h!x" ! x $ 3 f!x" ! x&!x $ 1", t!x" ! x 10 f ! t ! h ! f ! t ! h"!x" ! f!t!h!x""" f t h f ! t ! h +"2, 2, t ! t "2 ( x ( 2 x # "2 2 " x ( 4 s2 " x ( 2 x ( 2 2 " s2 " x # 0. 2 " x # 0 !t ! t"!x" ! t!t!x"" ! t(s2 " x ) ! s2 " s2 " x +0, '" f ! f ! f ! f "!x" ! f! f!x"" ! f (sx ) ! ssx ! s 4 x +0, 4, t ! f 0 ( x ( 4 x ( 4 sx ( 2 2 " sx # 0 a 2 ( b 2 0 ( a ( b s2 " sx x # 0 sx !t ! f "!x" ! t! f!x"" ! t(sx ) ! s2 " sx ! 'x# x ( 2( ! !"', 2, 'x#2 " x # 0( f ! t ! f ! t"!x" ! f!t!x"" ! f (s2 " x ) ! ss2 " x ! s 4 2 " x t ! t f ! f t ! f f ! t t!x" ! s2 " x f!x" ! sx V t ! f f ! t f t f ! t f ! t " t ! f NOTE !t ! f "!x" ! t! f!x"" ! t!x 2" ! x 2 " 3 ! f ! t"!x" ! f!t!x"" ! f!x " 3" ! !x " 3"2 t ! f f ! t t!x" ! x " 3 f!x" ! x 2 42 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS EXAMPLE 9 Given , ﬁnd functions , , and h such that . SOLUTION Since , the formula for F says: First add 9, then take the cosine of the result, and ﬁnally square. So we let Then M ! +cos!x $ 9",2 ! F!x" ! f ! t ! h"!x" ! f!t!h!x""" ! f!t!x $ 9"" ! f!cos!x $ 9"" f!x" ! x 2 t!x" ! cos x h!x" ! x $ 9 F!x" ! +cos!x $ 9",2 F ! f ! t ! h t f F!x" ! cos2!x $ 9" SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS \|\|\|\| 43 (c) (d) The graph of is given. Use it to graph the following functions. (a) (b) (c) (d) 6–7 The graph of is given. Use transformations to create a function whose graph is as shown. 6. _4 _1 _2.5 x y _1 0 7. 5 x y 2 0 3 1.5 y=œ„„„„„„ 3x-≈ x y 3 0 y ! s3x " x 2 x y 0 1 1 y ! "f !"x" y ! f !"x" y ! f ( 1 2x) y ! f !2x" f 5. x y 0 1 1 y ! " 1 2 f !x" $ 3 y ! 2f !x" Suppose the graph of is given. Write equations for the graphs that are obtained from the graph of as follows. (a) Shift 3 units upward. (b) Shift 3 units downward. (c) Shift 3 units to the right. (d) Shift 3 units to the left. (e) Reﬂect about the -axis. (f) Reﬂect about the -axis. (g) Stretch vertically by a factor of 3. (h) Shrink vertically by a factor of 3. 2. Explain how each graph is obtained from the graph of . (a) (b) (c) (d) (e) (f) 3. The graph of is given. Match each equation with its graph and give reasons for your choices. (a) (b) (c) (d) (e) 4. The graph of is given. Draw the graphs of the following functions. (a) (b) y ! f !x" $ 4 y ! f !x $ 4" f ! @ $ % # f y 3 _3 6 0 x 3 _3 _6 6 y ! 2f !x $ 6" y ! "f !x $ 4" y ! 1 3 f !x" y ! f !x" $ 3 y ! f !x " 4" y ! f !x" y ! 5f !x" " 3 y ! f !5x" y ! "5f !x" y ! "f !x" y ! f !x " 5" y ! 5f !x" y ! f !x" y x f f 1. EXERCISES 1.3 44 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS 29–30 Find , , , and and state their domains. , 30. , 31–36 Find the functions (a) , (b) , (c) , and (d) and their domains. 31. , 32. , 33. , 34. , , 36. , 37–40 Find 37. , , 38. , , 39. , , 40. , , 41–46 Express the function in the form 41. 42. 43. 44. 45. 47–49 Express the function in the form 47. 48. 49. 50. Use the table to evaluate each expression. (a) (b) (c) (d) (e) (f) ! f ! t"!6" !t ! f "!3" t!t!1"" f ! f !1"" t! f !1"" f !t!1"" H!x" ! sec4(sx ) H!x" ! s 8 2 $ # x# H!x" ! 1 " 3x2 f ! t ! h. u!t" ! tan t 1 $ tan t 46. u!t" ! scos t G!x" !-x 1 $ x 3 F!x" ! s 3 x 1 $ s 3 x F!x" ! sin(sx ) F!x" ! !x 2 $ 1"10 f ! t. h!x" ! s 3 x t!x" ! x x " 1 f !x" ! tan x h!x" ! x 3 $ 2 t!x" ! x 2 f !x" ! sx " 3 h!x" ! 1 " x t!x" ! x 2 f !x" ! 2x " 1 h!x" ! x " 1 t!x" ! 2x f !x" ! x $ 1 f ! t ! h. t!x" ! sin 2x f !x" ! x 1 $ x t!x" ! x $ 1 x $ 2 f !x" ! x $ 1 x 35. t!x" ! s 3 1 " x f !x" ! sx t!x" ! cos x f !x" ! 1 " 3x t!x" ! x 2 $ 3x $ 4 f !x" ! x " 2 t!x" ! 2x $ 1 f !x" ! x 2 " 1 t ! t f ! f t ! f f ! t t!x" ! sx 2 " 1 f !x" ! s3 " x t!x" ! 3x 2 " 1 f !x" ! x 3 $ 2x 2 29. f&t ft f " t f $ t 8. (a) How is the graph of related to the graph of ? Use your answer and Figure 6 to sketch the graph of . (b) How is the graph of related to the graph of ? Use your answer and Figure 4(a) to sketch the graph of . 9–24 Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions given in Sec-tion 1.2, and then applying the appropriate transformations. 9. 10. 11. 12. 13. 14. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. The city of New Orleans is located at latitude . Use Fig-ure 9 to ﬁnd a function that models the number of hours of daylight at New Orleans as a function of the time of year. To check the accuracy of your model, use the fact that on March 31 the sun rises at 5:51 AM and sets at 6:18 PM in New Orleans. 26. A variable star is one whose brightness alternately increases and decreases. For the most visible variable star, Delta Cephei, the time between periods of maximum brightness is 5.4 days, the average brightness (or magnitude) of the star is 4.0, and its brightness varies by magnitude. Find a function that models the brightness of Delta Cephei as a function of time. (a) How is the graph of related to the graph of ? (b) Sketch the graph of . (c) Sketch the graph of . 28. Use the given graph of to sketch the graph of . Which features of are the most important in sketching ? Explain how they are used. 1 1 0 x y y ! 1&f !x" f y ! 1&f !x" f y ! s# x# y ! sin # x# f y ! f (# x#) 27. )0.35 30&N y ! # x 2 " 2x# y ! #sin x# y ! 1 4 tan)x " % 4 y ! 2 x $ 1 y ! 1 $ s 3 x " 1 y ! 1 2 !x 2 $ 8x" y ! !x $ 2"4 $ 3 y ! sx $ 3 y ! 1 x " 4 y ! sin!x&2" 15. y ! 4 sin 3x y ! 1 $ 2 cos x y ! x 2 " 4x $ 3 y ! !x $ 1"2 y ! 1 " x 2 y ! "x 3 y ! 1 $ sx y ! sx y ! 1 $ sx y ! 2 sin x y ! sin x y ! 2 sin x x 1 2 3 4 5 6 3 1 4 2 2 5 6 3 2 1 2 3 t!x" f !x" SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS \|\|\|\| 45 57. The Heaviside function H is deﬁned by It is used in the study of electric circuits to represent the sudden surge of electric current, or voltage, when a switch is instantane-ously turned on. (a) Sketch the graph of the Heaviside function. (b) Sketch the graph of the voltage in a circuit if the switch is turned on at time and 120 volts are applied instantaneously to the circuit. Write a formula for in terms of . (c) Sketch the graph of the voltage in a circuit if the switch is turned on at time seconds and 240 volts are applied instantaneously to the circuit. Write a formula for in terms of . (Note that starting at corresponds to a translation.) 58. The Heaviside function deﬁned in Exercise 57 can also be used to deﬁne the ramp function , which represents a gradual increase in voltage or current in a circuit. (a) Sketch the graph of the ramp function . (b) Sketch the graph of the voltage in a circuit if the switch is turned on at time and the voltage is gradually increased to 120 volts over a 60-second time interval. Write a formula for in terms of for . (c) Sketch the graph of the voltage in a circuit if the switch is turned on at time seconds and the voltage is gradu-ally increased to 100 volts over a period of 25 seconds. Write a formula for in terms of for . 59. Let and be linear functions with equations and . Is also a linear function? If so, what is the slope of its graph? 60. If you invest dollars at 4% interest compounded annually, then the amount of the investment after one year is . Find , , and . What do these compo-sitions represent? Find a formula for the composition of copies of . 61. (a) If and , ﬁnd a function such that . (Think about what operations you would have to perform on the formula for to end up with the formula for .) (b) If and , ﬁnd a function such that . 62. If and , ﬁnd a function such that . 63. (a) Suppose and are even functions. What can you say about and ? (b) What if and are both odd? 64. Suppose is even and is odd. What can you say about ? Suppose t is an even function and let . Is h always an even function? 66. Suppose t is an odd function and let . Is h always an odd function? What if is odd? What if is even? f f h ! f ! t h ! f ! t 65. ft t f t f ft f $ t t f t ! f ! h t h!x" ! 4x " 1 f !x" ! x $ 4 f ! t ! h t h!x" ! 3x 2 $ 3x $ 2 f !x" ! 3x $ 5 h t f ! t ! h f h!x" ! 4x 2 $ 4x $ 7 t!x" ! 2x $ 1 A n A ! A ! A ! A A ! A ! A A ! A A!x" ! 1.04x A!x" x f ! t t!x" ! m2x $ b2 f !x" ! m1x $ b1 t f t ( 32 H!t" V!t" t ! 7 V!t" t ( 60 H!t" V!t" t ! 0 V!t" y ! tH!t" y ! ctH!t" t ! 5 H!t" V!t" t ! 5 V!t" H!t" V!t" t ! 0 V!t" H!t" !. 0 1 if t ! 0 if t # 0 51. Use the given graphs of and to evaluate each expression, or explain why it is undeﬁned. (a) (b) (c) (d) (e) (f) 52. Use the given graphs of and to estimate the value of for . Use these estimates to sketch a rough graph of . A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of . (a) Express the radius of this circle as a function of the time (in seconds). (b) If is the area of this circle as a function of the radius, ﬁnd and interpret it. 54. A spherical balloon is being inﬂated and the radius of the bal-loon is increasing at a rate of . (a) Express the radius of the balloon as a function of the time (in seconds). (b) If is the volume of the balloon as a function of the radius, ﬁnd and interpret it. 55. A ship is moving at a speed of parallel to a straight shoreline. The ship is 6 km from shore and it passes a light-house at noon. (a) Express the distance between the lighthouse and the ship as a function of , the distance the ship has traveled since noon; that is, ﬁnd so that . (b) Express as a function of , the time elapsed since noon; that is, ﬁnd so that . (c) Find . What does this function represent? 56. An airplane is ﬂying at a speed of at an altitude of one mile and passes directly over a radar station at time . (a) Express the horizontal distance (in miles) that the plane has ﬂown as a function of . (b) Express the distance between the plane and the radar station as a function of . (c) Use composition to express as a function of . t s d s t d t ! 0 350 mi&h f ! t d ! t!t" t t d s ! f !d" f d s 30 km&h V ! r V t r 2 cm&s A ! r A t r 60 cm&s 53. g f x y 0 1 1 f ! t x ! "5, "4, "3, . . . , 5 f !t!x"" t f x y 0 f g 2 2 ! f ! f "!4" !t ! t"!"2" !t ! f "!6" ! f ! t"!0" t! f !0"" f !t!2"" t f 46 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS GRAPHING CALCULATORS AND COMPUTERS In this section we assume that you have access to a graphing calculator or a computer with graphing software. We will see that the use of such a device enables us to graph more com-plicated functions and to solve more complex problems than would otherwise be possible. We also point out some of the pitfalls that can occur with these machines. Graphing calculators and computers can give very accurate graphs of functions. But we will see in Chapter 4 that only through the use of calculus can we be sure that we have uncovered all the interesting aspects of a graph. A graphing calculator or computer displays a rectangular portion of the graph of a func-tion in a display window or viewing screen, which we refer to as a viewing rectangle. The default screen often gives an incomplete or misleading picture, so it is important to choose the viewing rectangle with care. If we choose the -values to range from a mini-mum value of to a maximum value of and the -values to range from a minimum of to a maximum of , then the visible portion of the graph lies in the rectangle shown in Figure 1. We refer to this rectangle as the by viewing rectangle. The machine draws the graph of a function much as you would. It plots points of the form for a certain number of equally spaced values of between and . If an -value is not in the domain of , or if lies outside the viewing rectangle, it moves on to the next -value. The machine connects each point to the preceding plotted point to form a representation of the graph of . EXAMPLE 1 Draw the graph of the function in each of the following viewing rectangles. (a) by (b) by (c) by (d) by SOLUTION For part (a) we select the range by setting min , max , min and max . The resulting graph is shown in Figure 2(a). The display window is blank! A moment’s thought provides the explanation: Notice that for all , so for all . Thus the range of the function is . This means that the graph of lies entirely outside the viewing rectangle by . The graphs for the viewing rectangles in parts (b), (c), and (d) are also shown in Figure 2. Observe that we get a more complete picture in parts (c) and (d), but in part (d) it is not clear that the -intercept is 3. M y +"2, 2, +"2, 2, f +3, '" f!x" ! x2 $ 3 x x 2 $ 3 # 3 x x 2 # 0 ! 2 Y ! "2, Y ! 2 X ! "2 X +"100, 1000, +"50, 50, +"5, 30, +"10, 10, +"4, 4, +"4, 4, +"2, 2, +"2, 2, f!x" ! x 2 $ 3 f x f!x" f x b a x !x, f!x"" f +c, d, +a, b, +a, b, +c, d, ! '!x, y" # a ( x ( b, c ( y ( d( Ymax ! d Ymin ! c y Xmax ! b Xmin ! a x 1.4 FIGURE 1 The viewing rectangle +a, b, by +c, d, y=d x=a x=b y=c (a, d ) (b, d ) (a, c ) (b, c ) FIGURE 2 Graphs of ƒ=≈+3 (b) +4, 4, by +_4, 4, (a) +_2, 2, by +_2, 2, 2 _2 _2 2 4 _4 _4 4 (c) +_10, 10, by +_5, 30, 30 _5 _10 10 (d) +_50, 50, by +_100, 1000, 1000 _100 _50 50 We see from Example 1 that the choice of a viewing rectangle can make a big differ-ence in the appearance of a graph. Often it’s necessary to change to a larger viewing rectangle to obtain a more complete picture, a more global view, of the graph. In the next example we see that knowledge of the domain and range of a function sometimes provides us with enough information to select a good viewing rectangle. EXAMPLE 2 Determine an appropriate viewing rectangle for the function and use it to graph . SOLUTION The expression for is deﬁned when Therefore the domain of is the interval . Also, so the range of is the interval . We choose the viewing rectangle so that the -interval is somewhat larger than the domain and the -interval is larger than the range. Taking the viewing rectangle to be by , we get the graph shown in Figure 3. M EXAMPLE 3 Graph the function . SOLUTION Here the domain is , the set of all real numbers. That doesn’t help us choose a viewing rectangle. Let’s experiment. If we start with the viewing rectangle by , we get the graph in Figure 4. It appears blank, but actually the graph is so nearly vertical that it blends in with the -axis. If we change the viewing rectangle to by , we get the picture shown in Figure 5(a). The graph appears to consist of vertical lines, but we know that can’t be correct. If we look carefully while the graph is being drawn, we see that the graph leaves the screen and reappears during the graphing process. This indicates that we need to see more in the vertical direction, so we change the viewing rectangle to by . The resulting graph is shown in Figure 5(b). It still doesn’t quite reveal all the main features of the function, so we try by in Figure 5(c). Now we are more conﬁdent that we have arrived at an appropriate view-ing rectangle. In Chapter 4 we will be able to see that the graph shown in Figure 5(c) does indeed reveal all the main features of the function. M FIGURE 5 y=˛-150x (a) (c) (b) 1000 _1000 _20 20 500 _500 _20 20 20 _20 _20 20 +"1000, 1000, +"20, 20, +"500, 500, +"20, 20, +"20, 20, +"20, 20, y +"5, 5, +"5, 5, " y ! x 3 " 150x +"1, 4, +"3, 3, y x [0, 2s2 ] f 0 ( s8 " 2x 2 ( s8 ! 2s2 / 2.83 +"2, 2, f & ? #x# ( 2 & ? "2 ( x ( 2 8 " 2x 2 # 0 & ? 2x 2 ( 8 & ? x 2 ( 4 f!x" f f!x" ! s8 " 2x 2 SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS \|\|\|\| 47 FIGURE 3 4 _1 _3 3 5 _5 _5 5 FIGURE 4 EXAMPLE 4 Graph the function in an appropriate viewing rectangle. SOLUTION Figure 6(a) shows the graph of produced by a graphing calculator using the viewing rectangle by . At ﬁrst glance the graph appears to be rea-sonable. But if we change the viewing rectangle to the ones shown in the following parts of Figure 6, the graphs look very different. Something strange is happening. In order to explain the big differences in appearance of these graphs and to ﬁnd an appropriate viewing rectangle, we need to ﬁnd the period of the function We know that the function has period and the graph of is compressed horizontally by a factor of 50, so the period of is This suggests that we should deal only with small values of in order to show just a few oscillations of the graph. If we choose the viewing rectangle by , we get the graph shown in Figure 7. Now we see what went wrong in Figure 6. The oscillations of are so rapid that when the calculator plots points and joins them, it misses most of the maximum and minimum points and therefore gives a very misleading impression of the graph. M We have seen that the use of an inappropriate viewing rectangle can give a misleading impression of the graph of a function. In Examples 1 and 3 we solved the problem by changing to a larger viewing rectangle. In Example 4 we had to make the viewing rect-angle smaller. In the next example we look at a function for which there is no single view-ing rectangle that reveals the true shape of the graph. EXAMPLE 5 Graph the function . SOLUTION Figure 8 shows the graph of produced by a graphing calculator with viewing rectangle by . It looks much like the graph of , but per-haps with some bumps attached. If we zoom in to the viewing rectangle by , we can see much more clearly the shape of these bumps in Figure 9. The +"0.1, 0.1, +"0.1, 0.1, y ! sin x +"1.5, 1.5, +"6.5, 6.5, f f!x" ! sin x $ 1 100 cos 100x V y ! sin 50x +"1.5, 1.5, +"0.25, 0.25, x 2% 50 ! % 25 / 0.126 y ! sin 50x y ! sin 50x 2% y ! sin x y ! sin 50x. FIGURE 6 Graphs of ƒ=sin 50x in four viewing rectangles (a) (b) (c) (d) 1.5 _1.5 _10 10 1.5 _1.5 _12 12 1.5 _1.5 _9 9 1.5 _1.5 _6 6 +"1.5, 1.5, +"12, 12, f f!x" ! sin 50x V 48 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS N The appearance of the graphs in Figure 6 depends on the machine used. The graphs you get with your own graphing device might not look like these ﬁgures, but they will also be quite inaccurate. FIGURE 7 ƒ=sin 50x 1.5 _1.5 .25 .25 reason for this behavior is that the second term, , is very small in comparison with the ﬁrst term, . Thus we really need two graphs to see the true nature of this function. M EXAMPLE 6 Draw the graph of the function . SOLUTION Figure 10(a) shows the graph produced by a graphing calculator with view-ing rectangle by . In connecting successive points on the graph, the calculator produced a steep line segment from the top to the bottom of the screen. That line segment is not truly part of the graph. Notice that the domain of the function is . We can eliminate the extraneous near-vertical line by exper-imenting with a change of scale. When we change to the smaller viewing rectangle by on this particular calculator, we obtain the much better graph in Figure 10(b). M EXAMPLE 7 Graph the function . SOLUTION Some graphing devices display the graph shown in Figure 11, whereas others produce a graph like that in Figure 12. We know from Section 1.2 (Figure 13) that the graph in Figure 12 is correct, so what happened in Figure 11? The explanation is that some machines compute the cube root of using a logarithm, which is not deﬁned if is negative, so only the right half of the graph is produced. FIGURE 11 2 2 _3 3 FIGURE 12 2 _2 _3 3 x x y ! s 3 x (a) (b) 9 _9 _9 9 4.7 _4.7 _4.7 4.7 FIGURE 10 +"4.7, 4.7, +"4.7, 4.7, 'x# x " 1( y ! 1&!1 " x" +"9, 9, +"9, 9, y ! 1 1 " x FIGURE 9 0.1 _0.1 _0.1 0.1 FIGURE 8 1.5 _1.5 _6.5 6.5 sin x 1 100 cos 100x SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS \|\|\|\| 49 N Another way to avoid the extraneous line is to change the graphing mode on the calculator so that the dots are not connected. You should experiment with your own machine to see which of these two graphs is produced. If you get the graph in Figure 11, you can obtain the correct picture by graph-ing the function Notice that this function is equal to (except when ). M To understand how the expression for a function relates to its graph, it’s helpful to graph a family of functions, that is, a collection of functions whose equations are related. In the next example we graph members of a family of cubic polynomials. EXAMPLE 8 Graph the function for various values of the number . How does the graph change when is changed? SOLUTION Figure 13 shows the graphs of for , , , , and . We see that, for positive values of , the graph increases from left to right with no maximum or minimum points (peaks or valleys). When , the curve is ﬂat at the origin. When is negative, the curve has a maximum point and a minimum point. As decreases, the maximum point becomes higher and the minimum point lower. M EXAMPLE 9 Find the solution of the equation correct to two decimal places. SOLUTION The solutions of the equation are the -coordinates of the points of intersection of the curves and . From Figure 14(a) we see that there is only one solution and it lies between 0 and 1. Zooming in to the viewing rectangle by , we see from Figure 14(b) that the root lies between 0.7 and 0.8. So we zoom in further to the viewing rectangle by in Figure 14(c). By moving the cursor to the intersection point of the two curves, or by inspection and the fact that the -scale is 0.01, we see that the solution of the equation is about 0.74. (Many calculators have a built-in intersection feature.) M !0.7, 0.8" by !0.7, 0.8" x-scale=0.01 (c) !0, 1" by !0, 1" x-scale=0.1 (b) !_5, 5" by !_1.5, 1.5" x-scale=1 (a) 0.8 0.7 0.8 y=x 1 0 1 y=x 1.5 _1.5 _5 5 y=x y=cos x FIGURE 14 Locating the roots of cos x=x y=cos x y=cos x x !0.7, 0.8" !0.7, 0.8" !0, 1" !0, 1" y ! x y ! cos x x cos x ! x cos x ! x FIGURE 13 Several members of the family of functions y=˛+cx, all graphed in the viewing rectangle !_2, 2" by !_2.5, 2.5" (a) y=˛+2x (b) y=˛+x (c) y=˛ (d) y=˛-x (e) y=˛-2x c c c ! 0 c !2 !1 0 1 c ! 2 y ! x 3 " cx c c y ! x 3 " cx V x ! 0 s 3 x f#x$ ! x %x% ! %x% 1&3 50 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS In Visual 1.4 you can see an animation of Figure 13. TEC SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS \|\|\|\| 51 24. Use graphs to determine which of the functions and is eventually larger. 25. For what values of is it true that ? 26. Graph the polynomials and on the same screen, ﬁrst using the viewing rectangle by [ ] and then changing to by . What do you observe from these graphs? In this exercise we consider the family of root functions , where is a positive integer. (a) Graph the functions , , and on the same screen using the viewing rectangle by . (b) Graph the functions , , and on the same screen using the viewing rectangle by . (See Example 7.) (c) Graph the functions , , , and on the same screen using the viewing rectangle by . (d) What conclusions can you make from these graphs? 28. In this exercise we consider the family of functions , where is a positive integer. (a) Graph the functions and on the same screen using the viewing rectangle by . (b) Graph the functions and on the same screen using the same viewing rectangle as in part (a). (c) Graph all of the functions in parts (a) and (b) on the same screen using the viewing rectangle by . (d) What conclusions can you make from these graphs? Graph the function for several values of . How does the graph change when changes? 30. Graph the function for various values of . Describe how changing the value of affects the graph. 31. Graph the function , , for , and 6. How does the graph change as increases? 32. The curves with equations are called bullet-nose curves. Graph some of these curves to see why. What happens as increases? What happens to the graph of the equation as varies? 34. This exercise explores the effect of the inner function on a composite function . (a) Graph the function using the viewing rect-angle by . How does this graph differ from the graph of the sine function? !!1.5, 1.5" !0, 400" y ! sin(sx ) y ! f #t#x$$ t c y 2 ! cx 3 " x 2 33. c y ! % x% sc ! x 2 n n ! 1, 2, 3, 4, 5 x # 0 y ! x n2!x c c s1 " cx 2 f #x$ ! c c f #x$ ! x 4 " cx 2 " x 29. !!1, 3" !!1, 3" y ! 1&x 4 y ! 1&x 2 !!3, 3" !!3, 3" y ! 1&x 3 y ! 1&x n f #x$ ! 1&x n !!1, 2" !!1, 3" y ! s 5 x y ! s 4 x y ! s 3 x y ! sx !!2, 2" !!3, 3" y ! s 5 x y ! s 3 x y ! x !!1, 3" !!1, 4" y ! s 6 x y ! s 4 x y ! sx n f #x$ ! s n x 27. !!10,000, 10,000" !!10, 10" !2, 2 !!2, 2" Q#x$ ! 3x 5 P#x$ ! 3x 5 ! 5x 3 " 2x % sin x ! x% $ 0.1 x t#x$ ! x 3 f #x$ ! x 4 ! 100x 3 1. Use a graphing calculator or computer to determine which of the given viewing rectangles produces the most appropriate graph of the function . (a) by (b) by (c) by 2. Use a graphing calculator or computer to determine which of the given viewing rectangles produces the most appropriate graph of the function . (a) by (b) by (c) by (d) by 3–14 Determine an appropriate viewing rectangle for the given function and use it to draw the graph. 3. 4. 5. 6. 7. 10. 11. 12. 13. 14. 15. Graph the ellipse by graphing the functions whose graphs are the upper and lower halves of the ellipse. 16. Graph the hyperbola by graphing the functions whose graphs are the upper and lower branches of the hyperbola. 17–18 Do the graphs intersect in the given viewing rectangle? If they do, how many points of intersection are there? 17. , ; 18. , ; 19–21 Find all solutions of the equation correct to two decimal places. 19. 20. 21. 22. We saw in Example 9 that the equation has exactly one solution. (a) Use a graph to show that the equation has three solutions and ﬁnd their values correct to two decimal places. (b) Find an approximate value of such that the equation has exactly two solutions. Use graphs to determine which of the functions and is eventually larger (that is, larger when is very large). x t#x$ ! x 3&10 f #x$ ! 10x 2 23. cos x ! mx m cos x ! 0.3x cos x ! x x 2 ! sin x x 3 ! 4x ! 1 x 3 ! 9x 2 ! 4 ! 0 !!6, 2" by !!5, 20" y ! 3x " 18 y ! 6 ! 4x ! x 2 !!1, 3" by !!2.5, 1.5" y ! 0.23x ! 2.25 y ! 3x 2 ! 6x " 1 y 2 ! 9x 2 ! 1 4x 2 " 2y 2 ! 1 y ! x 2 " 0.02 sin 50x y ! 10 sin x " sin 100x f #x$ ! sec#20%x$ f #x$ ! sin sx f #x$ ! cos#0.001x$ f #x$ ! sin2#1000x$ 9. f #x$ ! x x 2 " 100 8. f #x$ ! x 3 ! 225x f #x$ ! s0.1x " 20 f #x$ ! s 4 81 ! x 4 f #x$ ! x 3 " 30x 2 " 200x f #x$ ! 5 " 20x ! x 2 !!50, 50" !!5, 5" !!50, 50" !!50, 50" !!10, 10" !!10, 10" !!3, 3" !!3, 3" f #x$ ! x 4 ! 16x 2 " 20 !0, 10" !0, 10" !0, 2" !0, 10" !!5, 5" !!5, 5" f #x$ ! sx 3 ! 5x 2 ; EXERCISES 1.4 36. The ﬁrst graph in the ﬁgure is that of as displayed by a TI-83 graphing calculator. It is inaccurate and so, to help explain its appearance, we replot the curve in dot mode in the second graph. What two sine curves does the calculator appear to be plotting? Show that each point on the graph of that the TI-83 chooses to plot is in fact on one of these two curves. (The TI-83’s graphing window is 95 pixels wide.) y ! sin 45x 0 2π 0 2π y ! sin 45x (b) Graph the function using the viewing rectangle by . How does this graph differ from the graph of the sine function? 35. The ﬁgure shows the graphs of and as displayed by a TI-83 graphing calculator. The ﬁrst graph is inaccurate. Explain why the two graphs appear identical. [Hint: The TI-83’s graphing window is 95 pixels wide. What speciﬁc points does the calculator plot?] y=sin 96x 0 2π y=sin 2x 0 2π y ! sin 2x y ! sin 96x !!1.5, 1.5" !!5, 5" y ! sin#x 2$ EXPONENTIAL FUNCTIONS The function is called an exponential function because the variable, x, is the exponent. It should not be confused with the power function , in which the vari-able is the base. In general, an exponential function is a function of the form where is a positive constant. Let’s recall what this means. If , a positive integer, then n factors If , and if , where is a positive integer, then If is a rational number, , where and are integers and , then But what is the meaning of if x is an irrational number? For instance, what is meant by or ? To help us answer this question we ﬁrst look at the graph of the function , where x is rational. A representation of this graph is shown in Figure 1. We want to enlarge the domain of to include both rational and irrational numbers. There are holes in the graph in Figure 1 corresponding to irrational values of x. We want to ﬁll in the holes by deﬁning , where , so that is an increasing function. In particular, since the irrational number satisﬁes 1.7 $ s3 $ 1.8 s3 f x " ! f#x$ ! 2x y ! 2x y ! 2x 5% 2s3 a x a x ! a p&q ! q sa p ! ( q sa ) p q & 0 q p x ! p&q x a !n ! 1 a n n x ! !n x ! 0, then a 0 ! 1 a n ! a ! a ! ' ' ' ! a x ! n a f#x$ ! a x t#x$ ! x 2 f#x$ ! 2x 1.5 52 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS FIGURE 1 Representation of y=2®, x rational x 0 y 1 1 N In Appendix G we present an alternative approach to the exponential and logarithmic functions using integral calculus. SECTION 1.5 EXPONENTIAL FUNCTIONS \|\|\|\| 53 we must have and we know what and mean because 1.7 and 1.8 are rational numbers. Similarly, if we use better approximations for , we obtain better approximations for : . . . . . . . . . . . . It can be shown that there is exactly one number that is greater than all of the numbers . . . and less than all of the numbers . . . We deﬁne to be this number. Using the preceding approximation process we can com-pute it correct to six decimal places: Similarly, we can deﬁne (or , if ) where x is any irrational number. Figure 2 shows how all the holes in Figure 1 have been ﬁlled to complete the graph of the function . The graphs of members of the family of functions are shown in Figure 3 for var-ious values of the base a. Notice that all of these graphs pass through the same point because for . Notice also that as the base a gets larger, the exponential func-tion grows more rapidly (for ). You can see from Figure 3 that there are basically three kinds of exponential functions . If , the exponential function decreases; if , it is a constant; and if , it increases. These three cases are illustrated in Figure 4. Observe that if , a " 1 a & 1 a ! 1 0 $ a $ 1 y ! a x FIGURE 3 0 1® 1.5® 2® 4® 10® ” ’® 1 4 ” ’® 1 2 x y 1 x & 0 a " 0 a 0 ! 1 #0, 1$ y ! a x f#x$ ! 2x, x " ! a & 0 a x 2x 2s3 ' 3.321997 2s3 21.73206, 21.7321, 21.733, 21.74, 21.8, 21.73205, 21.7320, 21.732, 21.73, 21.7, 1.73205 $ s3 $ 1.73206 ? 21.73205 $ 2s3 $ 21.73206 1.7320 $ s3 $ 1.7321 ? 21.7320 $ 2s3 $ 21.7321 1.732 $ s3 $ 1.733 ? 21.732 $ 2s3 $ 21.733 1.73 $ s3 $ 1.74 ? 21.73 $ 2s3 $ 21.74 2s3 s3 21.8 21.7 21.7 $ 2s3 $ 21.8 x 1 0 y 1 FIGURE 2 y=2®, x real N A proof of this fact is given in J. Marsden and A. Weinstein, Calculus Unlimited (Menlo Park, CA: Benjamin/Cummings, 1981). For an online version, see www.cds.caltech.edu/~marsden/ volume/cu/CU.pdf N If , then approaches as becomes large. If , then approaches as decreases through negative values. In both cases the -axis is a horizontal asymptote. These matters are discussed in Section 2.6. x x 0 a x a & 1 x 0 a x 0 $ a $ 1 then the exponential function has domain ! and range . Notice also that, since , the graph of is just the reﬂection of the graph of about the -axis. One reason for the importance of the exponential function lies in the following proper-ties. If x and y are rational numbers, then these laws are well known from elementary algebra. It can be proved that they remain true for arbitrary real numbers x and y. (See Appendix G.) LAWS OF EXPONENTS If a and b are positive numbers and x and y are any real numbers, then 1. 2. 3. 4. EXAMPLE 1 Sketch the graph of the function and determine its domain and range. SOLUTION First we reﬂect the graph of [shown in Figures 2 and 5(a)] about the x-axis to get the graph of in Figure 5(b). Then we shift the graph of upward 3 units to obtain the graph of in Figure 5(c). The domain is ! and the range is . M EXAMPLE 2 Use a graphing device to compare the exponential function and the power function . Which function grows more quickly when x is large? SOLUTION Figure 6 shows both functions graphed in the viewing rectangle by . We see that the graphs intersect three times, but for the graph of x & 4 !0, 40" !!2, 6" t#x$ ! x 2 f#x$ ! 2x V FIGURE 5 0 1 (a) y=2® x y 0 _1 (b) y=_2® x y y=3 0 2 (c) y=3-2® x y #!(, 3$ y ! 3 ! 2x y ! !2x y ! !2x y ! 2x y ! 3 ! 2x #ab$x ! a xb x #a x$y ! a xy a x!y ! a x a y a x"y ! a xa y 1 (0, 1) (a) y=a®, 0<a<1 (b) y=1® (c) y=a®, a>1 (0, 1) FIGURE 4 x 0 y x 0 y x 0 y y y ! a x y ! #1&a$x #1&a$x ! 1&a x ! a !x #0, ($ y ! a x 54 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS www.stewartcalculus.com For review and practice using the Laws of Exponents, click on Review of Algebra. N For a review of reﬂecting and shifting graphs, see Section 1.3. SECTION 1.5 EXPONENTIAL FUNCTIONS \|\|\|\| 55 stays above the graph of . Figure 7 gives a more global view and shows that for large values of x, the exponential function grows far more rapidly than the power function . M APPLICATIONS OF EXPONENTIAL FUNCTIONS The exponential function occurs very frequently in mathematical models of nature and society. Here we indicate brieﬂy how it arises in the description of population growth. In Chapter 3 we will pursue these and other applications in greater detail. First we consider a population of bacteria in a homogeneous nutrient medium. Suppose that by sampling the population at certain intervals it is determined that the population doubles every hour. If the number of bacteria at time t is , where t is measured in hours, and the initial population is , then we have It seems from this pattern that, in general, This population function is a constant multiple of the exponential function , so it exhibits the rapid growth that we observed in Figures 2 and 7. Under ideal conditions (unlimited space and nutrition and freedom from disease) this exponential growth is typi-cal of what actually occurs in nature. What about the human population? Table 1 shows data for the population of the world in the 20th century and Figure 8 shows the corresponding scatter plot. FIGURE 8 Scatter plot for world population growth 1900 6x10' P t 1920 1940 1960 1980 2000 y ! 2t p#t$ ! 2t ) 1000 ! #1000$2t p#3$ ! 2p#2$ ! 23 ) 1000 p#2$ ! 2p#1$ ! 22 ) 1000 p#1$ ! 2p#0$ ! 2 ) 1000 p#0$ ! 1000 p#t$ 250 0 8 y=2® y=≈ FIGURE 7 40 0 _2 6 y=2® y=≈ FIGURE 6 y ! x 2 y ! 2x t#x$ ! x 2 f#x$ ! 2x N Example 2 shows that increases more quickly than . To demonstrate just how quickly increases, let’s perform the following thought experiment. Suppose we start with a piece of paper a thousandth of an inch thick and we fold it in half 50 times. Each time we fold the paper in half, the thickness of the paper doubles, so the thickness of the resulting paper would be inches. How thick do you think that is? It works out to be more than 17 million miles! 250&1000 f #x$ ! 2x y ! x 2 y ! 2x TABLE 1 Population Year (millions) 1900 1650 1910 1750 1920 1860 1930 2070 1940 2300 1950 2560 1960 3040 1970 3710 1980 4450 1990 5280 2000 6080 The pattern of the data points in Figure 8 suggests exponential growth, so we use a graphing calculator with exponential regression capability to apply the method of least squares and obtain the exponential model Figure 9 shows the graph of this exponential function together with the original data points. We see that the exponential curve ﬁts the data reasonably well. The period of rela-tively slow population growth is explained by the two world wars and the Great Depres-sion of the 1930s. THE NUMBER e Of all possible bases for an exponential function, there is one that is most convenient for the purposes of calculus. The choice of a base a is inﬂuenced by the way the graph of crosses the y-axis. Figures 10 and 11 show the tangent lines to the graphs of and at the point . (Tangent lines will be deﬁned precisely in Section 2.7. For present purposes, you can think of the tangent line to an exponential graph at a point as the line that touches the graph only at that point.) If we measure the slopes of these tangent lines at , we ﬁnd that for and for . It turns out, as we will see in Chapter 3, that some of the formulas of calculus will be greatly simpliﬁed if we choose the base a so that the slope of the tangent line to at is exactly 1. (See Figure 12.) In fact, there is such a number and it is denoted by the letter e. (This notation was chosen by the Swiss mathematician Leonhard Euler in 1727, probably because it is the ﬁrst letter of the word exponential.) In view of Figures 10 and 11, it comes as no surprise that the number e lies between 2 and 3 and the graph of lies between the graphs of and . (See Figure 13.) In Chapter 3 we will see that the value of e, correct to ﬁve decimal places, is e ' 2.71828 y ! 3x y ! 2x y ! e x #0, 1$ y ! a x FIGURE 11 0 1 mÅ1.1 FIGURE 10 0 y=2® 1 mÅ0.7 x y y=3® x y y ! 3x m ' 1.1 y ! 2x m ' 0.7 #0, 1$ #0, 1$ y ! 3x y ! 2x y ! a x FIGURE 9 Exponential model for population growth 1900 6x10' P t 1920 1940 1960 1980 2000 P ! #0.008079266$ ! #1.013731$t 56 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS FIGURE 12 The natural exponential function crosses the y-axis with a slope of 1. 0 y=´ 1 m=1 x y SECTION 1.5 EXPONENTIAL FUNCTIONS \|\|\|\| 57 EXAMPLE 3 Graph the function and state the domain and range. SOLUTION We start with the graph of from Figures 12 and 14(a) and reﬂect about the y-axis to get the graph of in Figure 14(b). (Notice that the graph crosses the y-axis with a slope of !1). Then we compress the graph vertically by a factor of 2 to obtain the graph of in Figure 14(c). Finally, we shift the graph downward one unit to get the desired graph in Figure 14(d). The domain is ! and the range is . M How far to the right do you think we would have to go for the height of the graph of to exceed a million? The next example demonstrates the rapid growth of this func-tion by providing an answer that might surprise you. EXAMPLE 4 Use a graphing device to ﬁnd the values of x for which . SOLUTION In Figure 15 we graph both the function and the horizontal line . We see that these curves intersect when . Thus when . It is perhaps surprising that the values of the exponential function have already surpassed a million when x is only 14. M FIGURE 15 1.5x10^ 0 15 y=´ y=10^ x & 13.8 e x & 106 x ' 13.8 y ! 1,000,000 y ! e x e x & 1,000,000 y ! e x FIGURE 14 1 2 (d) y= e–®-1 y=_1 0 1 1 2 (c) y= e–® 0 1 0 (b) y=e–® 1 x 0 y (a) y=´ 1 y x y x y x #!1, ($ y ! 1 2e!x y ! e!x y ! e x y ! 1 2e!x ! 1 V FIGURE 13 0 1 y=2® y=e® y=3® y x Module 1.5 enables you to graph exponential functions with various bases and their tangent lines in order to estimate more closely the value of for which the tangent has slope . 1 a TEC 17–18 Find the exponential function whose graph is given. 18. 19. If , show that 20. Suppose you are offered a job that lasts one month. Which of the following methods of payment do you prefer? I. One million dollars at the end of the month. II. One cent on the ﬁrst day of the month, two cents on the second day, four cents on the third day, and, in general, cents on the th day. 21. Suppose the graphs of and are drawn on a coordinate grid where the unit of measurement is 1 inch. Show that, at a distance 2 ft to the right of the origin, the height of the graph of is 48 ft but the height of the graph of is about 265 mi. ; 22. Compare the functions and by graphing both functions in several viewing rectangles. Find all points of intersection of the graphs correct to one decimal place. Which function grows more rapidly when is large? ; Compare the functions and by graphing both and in several viewing rectangles. When does the graph of ﬁnally surpass the graph of ? ; 24. Use a graph to estimate the values of such that . e x & 1,000,000,000 x f t t f t#x$ ! e x f #x$ ! x 10 23. x t#x$ ! 5x f #x$ ! x 5 t f t#x$ ! 2x f #x$ ! x 2 n 2n!1 f(x " h) ! f(x) h ! 5x( 5h ! 1 h ) f #x$ ! 5x ”2, ’ 2 9 0 2 y x 0 (1, 6) (3, 24) y x 17. f #x$ ! Ca x 1. (a) Write an equation that deﬁnes the exponential function with base . (b) What is the domain of this function? (c) If , what is the range of this function? (d) Sketch the general shape of the graph of the exponential function for each of the following cases. (i) (ii) (iii) 2. (a) How is the number deﬁned? (b) What is an approximate value for ? (c) What is the natural exponential function? ; 3–6 Graph the given functions on a common screen. How are these graphs related? 3. , , , 4. , , , , , , 6. , , , 7–12 Make a rough sketch of the graph of the function. Do not use a calculator. Just use the graphs given in Figures 3 and 12 and, if necessary, the transformations of Section 1.3. 7. 8. 10. 12. Starting with the graph of , write the equation of the graph that results from (a) shifting 2 units downward (b) shifting 2 units to the right (c) reﬂecting about the x-axis (d) reﬂecting about the y-axis (e) reﬂecting about the x-axis and then about the y-axis 14. Starting with the graph of , ﬁnd the equation of the graph that results from (a) reﬂecting about the line (b) reﬂecting about the line 15–16 Find the domain of each function. 15. (a) (b) 16. (a) (b) t#t$ ! s1 ! 2t t#t$ ! sin#e!t$ f #x$ ! 1 1 ! e x f #x$ ! 1 1 " e x x ! 2 y ! 4 y ! e x y ! e x 13. y ! 2#1 ! e x$ y ! 1 ! 1 2e!x 11. y ! 1 " 2e x y ! !2!x 9. y ! 4x!3 y ! 4x ! 3 y ! 0.1x y ! 0.3x y ! 0.6x y ! 0.9 x y ! ( 1 10) x y ! ( 1 3) x y ! 10 x y ! 3x 5. y ! 8!x y ! 8x y ! e !x y ! e x y ! 20 x y ! 5x y ! e x y ! 2x e e 0 $ a $ 1 a ! 1 a & 1 a " 1 a & 0 EXERCISES 1.5 58 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS \|\|\|\| 59 population since 1900. Use the model to estimate the popula-tion in 1925 and to predict the population in the years 2010 and 2020. ; 29. If you graph the function you’ll see that appears to be an odd function. Prove it. ; 30. Graph several members of the family of functions where . How does the graph change when changes? How does it change when changes? a b a & 0 f #x$ ! 1 1 " ae bx f f #x$ ! 1 ! e 1&x 1 " e 1&x 25. Under ideal conditions a certain bacteria population is known to double every three hours. Suppose that there are initially 100 bacteria. (a) What is the size of the population after 15 hours? (b) What is the size of the population after hours? (c) Estimate the size of the population after 20 hours. ; (d) Graph the population function and estimate the time for the population to reach 50,000. 26. A bacterial culture starts with 500 bacteria and doubles in size every half hour. (a) How many bacteria are there after 3 hours? (b) How many bacteria are there after hours? (c) How many bacteria are there after 40 minutes? ; (d) Graph the population function and estimate the time for the population to reach 100,000. ; 27. Use a graphing calculator with exponential regression capa-bility to model the population of the world with the data from 1950 to 2000 in Table 1 on page 55. Use the model to esti-mate the population in 1993 and to predict the population in the year 2010. ; 28. The table gives the population of the United States, in mil-lions, for the years 1900–2000. Use a graphing calculator with exponential regression capability to model the US t t INVERSE FUNCTIONS AND LOGARITHMS Table 1 gives data from an experiment in which a bacteria culture started with 100 bacte-ria in a limited nutrient medium; the size of the bacteria population was recorded at hourly intervals. The number of bacteria N is a function of the time t: . Suppose, however, that the biologist changes her point of view and becomes interested in the time required for the population to reach various levels. In other words, she is think-ing of t as a function of N. This function is called the inverse function of f, denoted by , and read “f inverse.” Thus is the time required for the population level to reach N. The values of can be found by reading Table 1 from right to left or by consulting Table 2. For instance, because f#6$ ! 550. f !1#550$ ! 6 f !1 t ! f !1#N$ f !1 N ! f#t$ 1.6 Year Population Year Population 1900 76 1960 179 1910 92 1970 203 1920 106 1980 227 1930 123 1990 250 1940 131 2000 281 1950 150 TABLE 2 t as a function of N N ! time to reach N bacteria 100 0 168 1 259 2 358 3 445 4 509 5 550 6 573 7 586 8 t ! f !1#N$ TABLE 1 N as a function of t t (hours) ! population at time t 0 100 1 168 2 259 3 358 4 445 5 509 6 550 7 573 8 586 N ! f #t$ 60 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS Not all functions possess inverses. Let’s compare the functions and whose arrow diagrams are shown in Figure 1. Note that never takes on the same value twice (any two inputs in have different outputs), whereas does take on the same value twice (both 2 and 3 have the same output, 4). In symbols, but Functions that share this property with are called one-to-one functions. DEFINITION A function is called a one-to-one function if it never takes on the same value twice; that is, If a horizontal line intersects the graph of in more than one point, then we see from Figure 2 that there are numbers and such that . This means that is not one-to-one. Therefore we have the following geometric method for determining whether a function is one-to-one. HORIZONTAL LINE TEST A function is one-to-one if and only if no horizontal line intersects its graph more than once. EXAMPLE 1 Is the function one-to-one? SOLUTION 1 If , then (two different numbers can’t have the same cube). Therefore, by Deﬁnition 1, is one-to-one. SOLUTION 2 From Figure 3 we see that no horizontal line intersects the graph of more than once. Therefore, by the Horizontal Line Test, is one-to-one. M f f!x" ! x 3 f!x" ! x 3 x 3 1 " x 3 2 x1 " x 2 f!x" ! x 3 V FIGURE 2 This function is not one-to-one because f(⁄)=f(¤). 0 y x ⁄ ¤ ‡ ﬂ y=ƒ f f!x1" ! f!x2" x2 x1 f whenever x1 " x2 f!x1" " f!x2" f 1 FIGURE 1 4 3 2 1 10 4 2 A B g 4 3 2 1 10 7 4 2 A B f f is one-to-one; g is not f whenever x1 " x 2 f!x1" " f!x 2" t!2" ! t!3" t A f t f N In the language of inputs and outputs, this deﬁnition says that is one-to-one if each out-put corresponds to only one input. f FIGURE 3 ƒ=˛ is one-to-one. 0 y=˛ y x EXAMPLE 2 Is the function one-to-one? SOLUTION 1 This function is not one-to-one because, for instance, and so 1 and have the same output. SOLUTION 2 From Figure 4 we see that there are horizontal lines that intersect the graph of more than once. Therefore, by the Horizontal Line Test, is not one-to-one. M One-to-one functions are important because they are precisely the functions that pos-sess inverse functions according to the following deﬁnition. DEFINITION Let be a one-to-one function with domain and range . Then its inverse function has domain and range and is deﬁned by for any in . This deﬁnition says that if maps into , then maps back into . (If were not one-to-one, then would not be uniquely deﬁned.) The arrow diagram in Figure 5 indi-cates that reverses the effect of . Note that For example, the inverse function of is because if , then \| CAUTION Do not mistake the in for an exponent. Thus The reciprocal could, however, be written as . EXAMPLE 3 If , , and , ﬁnd and . SOLUTION From the deﬁnition of we have f!8" ! !10 because f !1!!10" ! 8 f!1" ! 5 because f !1!5" ! 1 f!3" ! 7 because f !1!7" ! 3 f !1 f !1!!10" f !1!5", f !1!7", f!8" ! !10 f!3" ! 7 f!1" ! 5 V # f!x"$!1 1%f!x" 1 f!x" does not mean f !1!x" f !1 !1 f !1!y" ! f !1!x 3" ! !x 3"1%3 ! x y ! x 3 f !1!x" ! x 1%3 f!x" ! x 3 range of f !1 ! domain of f domain of f !1 ! range of f f f !1 f !1 f x y f !1 y x f B y f!x" ! y & ? f !1!y" ! x A B f !1 B A f 2 t t !1 t!1" ! 1 ! t!!1" t!x" ! x 2 V SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS \|\|\|\| 61 x y A B f –! f FIGURE 5 FIGURE 4 ©=≈ is not one-to-one. 0 y=≈ x y The diagram in Figure 6 makes it clear how reverses the effect of in this case. M The letter is traditionally used as the independent variable, so when we concentrate on rather than on , we usually reverse the roles of and in Deﬁnition 2 and write By substituting for in Deﬁnition 2 and substituting for in (3), we get the following cancellation equations: The ﬁrst cancellation equation says that if we start with , apply , and then apply we arrive back at , where we started (see the machine diagram in Figure 7). Thus undoes what does. The second equation says that undoes what does. For example, if , then and so the cancellation equations become These equations simply say that the cube function and the cube root function cancel each other when applied in succession. Now let’s see how to compute inverse functions. If we have a function and are able to solve this equation for in terms of , then according to Deﬁnition 2 we must have . If we want to call the independent variable x, we then interchange and and arrive at the equation . HOW TO FIND THE INVERSE FUNCTION OF A ONE-TO-ONE FUNCTION f STEP 1 Write . STEP 2 Solve this equation for in terms of (if possible). STEP 3 To express as a function of x, interchange and . The resulting equation is . y ! f !1!x" y x f !1 y x y ! f!x" 5 y ! f !1!x" y x x ! f !1!y" y x y ! f!x" f! f !1!x"" ! !x 1%3"3 ! x f !1! f!x"" ! !x 3"1%3 ! x f !1!x" ! x 1%3 f!x" ! x 3 FIGURE 7 x x f ƒ f–! f !1 f f f !1 x f !1, f x f! f !1!x"" ! x for every x in B f !1! f!x"" ! x for every x in A 4 x y f!y" ! x & ? f !1!x" ! y 3 y x f f !1 x FIGURE 6 The inverse function reverses inputs and outputs. B 5 7 _10 f A 1 3 8 A 1 3 8 f –! B 5 7 _10 f f !1 62 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS EXAMPLE 4 Find the inverse function of . SOLUTION According to (5) we ﬁrst write Then we solve this equation for : Finally, we interchange and : Therefore the inverse function is . M The principle of interchanging and to ﬁnd the inverse function also gives us the method for obtaining the graph of from the graph of . Since if and only if , the point is on the graph of if and only if the point is on the graph of . But we get the point from by reﬂecting about the line . (See Figure 8.) Therefore, as illustrated by Figure 9: The graph of is obtained by reﬂecting the graph of about the line . EXAMPLE 5 Sketch the graphs of and its inverse function using the same coordinate axes. SOLUTION First we sketch the curve (the top half of the parabola , or ) and then we reﬂect about the line to get the graph of . (See Figure 10.) As a check on our graph, notice that the expression for is . So the graph of is the right half of the parabola and this seems reasonable from Figure 10. M LOGARITHMIC FUNCTIONS If and , the exponential function is either increasing or decreasing and so it is one-to-one by the Horizontal Line Test. It therefore has an inverse function , which is called the logarithmic function with base a and is denoted by . If we use the formulation of an inverse function given by (3), f!y" ! x & ? f !1!x" ! y loga f !1 f!x" ! a x a " 1 a " 0 y ! !x 2 ! 1 f !1 f !1!x" ! !x 2 ! 1, x # 0 f !1 f !1 y ! x x ! !y 2 ! 1 y 2 ! !1 ! x y ! s!1 ! x f!x" ! s!1 ! x y ! x f f !1 FIGURE 8 FIGURE 9 0 y x (b, a) (a, b) y=x 0 y x f –! y=x f y ! x !a, b" !b, a" f !1 !b, a" f !a, b" f !1!b" ! a f!a" ! b f f !1 y x f !1!x" ! s 3 x ! 2 y ! s 3 x ! 2 y x x ! s 3 y ! 2 x 3 ! y ! 2 x y ! x 3 $ 2 f!x" ! x 3 $ 2 V SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS \|\|\|\| 63 N In Example 4, notice how reverses the effect of . The function is the rule “Cube, then add 2”; is the rule “Subtract 2, then take the cube root.” f !1 f f f !1 0 y=x y=ƒ (0, _1) y=f –!(x) (_1, 0) FIGURE 10 y x then we have Thus, if , then is the exponent to which the base must be raised to give . For example, because . The cancellation equations (4), when applied to the functions and , become The logarithmic function has domain and range . Its graph is the reﬂection of the graph of about the line . Figure 11 shows the case where . (The most important logarithmic functions have base .) The fact that is a very rapidly increasing function for is reﬂected in the fact that is a very slowly increasing function for . Figure 12 shows the graphs of with various values of the base . Since , the graphs of all logarithmic functions pass through the point . The following properties of logarithmic functions follow from the corresponding prop-erties of exponential functions given in Section 1.5. LAWS OF LOGARITHMS If x and y are positive numbers, then 1. 2. 3. (where r is any real number) EXAMPLE 6 Use the laws of logarithms to evaluate . SOLUTION Using Law 2, we have because . M NATURAL LOGARITHMS Of all possible bases for logarithms, we will see in Chapter 3 that the most convenient choice of a base is the number , which was deﬁned in Section 1.5. The logarithm with base is called the natural logarithm and has a special notation: logex ! ln x e e a 24 ! 16 log2 80 ! log2 5 ! log2& 80 5' ! log2 16 ! 4 log2 80 ! log2 5 loga!x r" ! r logax loga& x y' ! loga x ! loga y loga!xy" ! loga x $ loga y !1, 0" loga 1 ! 0 a " 1 y ! loga x x " 1 y ! loga x x " 0 y ! a x a " 1 a " 1 y ! x y ! a x ! !0, %" loga aloga x ! x for every x " 0 loga!a x" ! x for every x ! ! 7 f !1!x" ! loga x f!x" ! a x 10!3 ! 0.001 log10 0.001 ! !3 x a loga x x " 0 a y ! x & ? loga x ! y 6 64 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS 0 y=x y=a®, a>1 y=loga x, a>1 FIGURE 11 y x FIGURE 12 0 y 1 x 1 y=log£ x y=log™ x y=log∞ x y=log¡¸ x N NOTATION FOR LOGARITHMS Most textbooks in calculus and the sciences, as well as calculators, use the notation for the natural logarithm and for the “common logarithm,” . In the more advanced mathe-matical and scientiﬁc literature and in computer languages, however, the notation usually denotes the natural logarithm. log x log10 x log x ln x If we put and replace with “ln” in (6) and (7), then the deﬁning properties of the natural logarithm function become In particular, if we set , we get EXAMPLE 7 Find if . SOLUTION 1 From (8) we see that Therefore . (If you have trouble working with the “ln” notation, just replace it by . Then the equation becomes ; so, by the deﬁnition of logarithm, .) SOLUTION 2 Start with the equation and apply the exponential function to both sides of the equation: But the second cancellation equation in (9) says that . Therefore, . M EXAMPLE 8 Solve the equation . SOLUTION We take natural logarithms of both sides of the equation and use (9): Since the natural logarithm is found on scientiﬁc calculators, we can approximate the solution: to four decimal places, . M x ( 0.8991 x ! 1 3!5 ! ln 10" 3x ! 5 ! ln 10 5 ! 3x ! ln 10 ln!e 5!3x" ! ln 10 e 5!3x ! 10 x ! e 5 e ln x ! x e ln x ! e 5 ln x ! 5 e 5 ! x logex ! 5 loge x ! e 5 e 5 ! x means ln x ! 5 ln x ! 5 x ln e ! 1 x ! 1 e ln x ! x x " 0 ln!e x" ! x x ! ! 9 e y ! x & ? ln x ! y 8 loge a ! e SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS \|\|\|\| 65 EXAMPLE 9 Express as a single logarithm. SOLUTION Using Laws 3 and 1 of logarithms, we have M The following formula shows that logarithms with any base can be expressed in terms of the natural logarithm. CHANGE OF BASE FORMULA For any positive number , we have PROOF Let . Then, from (6), we have . Taking natural logarithms of both sides of this equation, we get . Therefore M Scientiﬁc calculators have a key for natural logarithms, so Formula 10 enables us to use a calculator to compute a logarithm with any base (as shown in the following example). Similarly, Formula 10 allows us to graph any logarithmic function on a graphing calcula-tor or computer (see Exercises 41 and 42). EXAMPLE 10 Evaluate correct to six decimal places. SOLUTION Formula 10 gives M The graphs of the exponential function and its inverse function, the natural log-arithm function, are shown in Figure 13. Because the curve crosses the y-axis with a slope of 1, it follows that the reﬂected curve crosses the x-axis with a slope of 1. In common with all other logarithmic functions with base greater than 1, the natural logarithm is an increasing function deﬁned on and the y-axis is a vertical asymptote. (This means that the values of become very large negative as approaches 0.) EXAMPLE 11 Sketch the graph of the function . SOLUTION We start with the graph of as given in Figure 13. Using the transforma-tions of Section 1.3, we shift it 2 units to the right to get the graph of and then we shift it 1 unit downward to get the graph of . (See Figure 14.) y ! ln!x ! 2" ! 1 y ! ln!x ! 2" y ! ln x y ! ln!x ! 2" ! 1 x ln x !0, %" y ! ln x y ! e x y ! e x log8 5 ! ln 5 ln 8 ( 0.773976 log8 5 y ! ln x ln a y ln a ! ln x a y ! x y ! loga x loga x ! ln x ln a !a " 1" a 10 ! ln(asb ) ! ln a $ ln sb ln a $ 1 2 ln b ! ln a $ ln b 1%2 ln a $ 1 2 ln b V 66 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS y 1 0 x 1 y=x y=´ y=ln x FIGURE 13 M Although is an increasing function, it grows very slowly when . In fact, grows more slowly than any positive power of . To illustrate this fact, we compare approximate values of the functions and in the following table and we graph them in Figures 15 and 16. You can see that initially the graphs of and grow at comparable rates, but eventually the root function far surpasses the logarithm. INVERSE TRIGONOMETRIC FUNCTIONS When we try to ﬁnd the inverse trigonometric functions, we have a slight difﬁculty: Because the trigonometric functions are not one-to-one, they don’t have inverse functions. The difﬁculty is overcome by restricting the domains of these functions so that they become one-to-one. You can see from Figure 17 that the sine function is not one-to-one (use the Horizontal Line Test). But the function , is one-to-one (see Figure 18). The inverse function of this restricted sine function exists and is denot-ed by or . It is called the inverse sine function or the arcsine function. Since the deﬁnition of an inverse function says that f!y" ! x & ? f !1!x" ! y y 0 _π π x π 2 y=sin x FIGURE 17 0 y x _π 2 π 2 FIGURE 18 y=sin x, _ ¯x¯π 2 π 2 arcsin sin!1 f f!x" ! sin x, !&%2 ' x ' &%2 y ! sin x y ! ln x y ! sx y ! x 1%2 ! sx y ! ln x x ln x x " 1 ln x FIGURE 14 0 y 2 x (3, 0) x=2 y=ln(x-2) 0 y x y=ln x (1, 0) 0 y 2 x x=2 (3, _1) y=ln(x-2)-1 SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS \|\|\|\| 67 x 1 2 5 10 50 100 500 1000 10,000 100,000 0 0.69 1.61 2.30 3.91 4.6 6.2 6.9 9.2 11.5 1 1.41 2.24 3.16 7.07 10.0 22.4 31.6 100 316 0 0.49 0.72 0.73 0.55 0.46 0.28 0.22 0.09 0.04 ln x sx sx ln x x 0 y 1 1 y=œ„ x y=ln x FIGURE 15 x 0 y 1000 20 y=œ„ x y=ln x FIGURE 16 we have \| Thus, if , is the number between and whose sine is . EXAMPLE 12 Evaluate (a) and (b) . SOLUTION (a) We have because and lies between and . (b) Let , so . Then we can draw a right triangle with angle as in Figure 19 and deduce from the Pythagorean Theorem that the third side has length . This enables us to read from the triangle that M The cancellation equations for inverse functions become, in this case, The inverse sine function, , has domain and range , and its graph, shown in Figure 20, is obtained from that of the restricted sine function (Figure 18) by reﬂection about the line . The inverse cosine function is handled similarly. The restricted cosine function , , is one-to-one (see Figure 21) and so it has an inverse function denoted by or . 0 ' y ' & and cos y ! x & ? cos!1x ! y arccos cos!1 0 ' x ' & f!x" ! cos x 0 y x 1 π π 2 FIGURE 21 y=cos x, 0¯x¯π 0 y x 1 _1 π 2 _π 2 FIGURE 20 y=sin–! x=arcsin x y ! x #!&%2, &%2$ #!1, 1$ sin!1 for !1 ' x ' 1 sin!sin!1x" ! x for ! & 2 ' x ' & 2 sin!1!sin x" ! x tan(arcsin 1 3) ! tan ( ! 1 2s2 s9 ! 1 ! 2s2 ( sin ( ! 1 3 ( ! arcsin 1 3 &%2 !&%2 &%6 sin!&%6" ! 1 2 sin!1( 1 2) ! & 6 tan(arcsin 1 3) sin!1( 1 2) x &%2 !&%2 sin!1x !1 ' x ' 1 sin!1x " 1 sin x ! & 2 ' y ' & 2 and sin y ! x & ? sin!1x ! y 68 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS 2 œ„ 2 3 ¨ 1 FIGURE 19 The cancellation equations are The inverse cosine function, , has domain and range . Its graph is shown in Figure 22. The tangent function can be made one-to-one by restricting it to the interval . Thus the inverse tangent function is deﬁned as the inverse of the function . (See Figure 23.) It is denoted by or . EXAMPLE 13 Simplify the expression . SOLUTION 1 Let . Then and . We want to ﬁnd but, since is known, it is easier to ﬁnd ﬁrst: Thus SOLUTION 2 Instead of using trigonometric identities as in Solution 1, it is perhaps easier to use a diagram. If , then , and we can read from Figure 24 (which illustrates the case ) that M The inverse tangent function, , has domain and range . Its graph is shown in Figure 25. We know that the lines are vertical asymptotes of the graph of . Since the graph of is obtained by reﬂecting the graph of the restricted tangent function about the line , it follows that the lines and are horizontal asymptotes of the graph of . tan!1 y ! !&%2 y ! &%2 y ! x tan!1 tan x ! )&%2 FIGURE 25 y=tan–! x=arctan x π 2 _ π 2 y 0 x !!&%2, &%2" ! tan!1 ! arctan cos!tan!1x" ! cos y ! 1 s1 $ x 2 y " 0 tan y ! x y ! tan!1x cos!tan!1x" ! cos y ! 1 sec y ! 1 s1 $ x 2 !since sec y " 0 for !&%2 y &%2" sec y ! s1 $ x 2 sec2y ! 1 $ tan2y ! 1 $ x 2 sec y tan y cos y !&%2 y &%2 tan y ! x y ! tan!1x cos!tan!1x" ! & 2 y & 2 and tan y ! x & ? tan!1x ! y arctan tan!1 f!x" ! tan x, !&%2 x &%2 !!&%2, &%2" #0, &$ #!1, 1$ cos!1 for !1 ' x ' 1 cos!cos!1x" ! x for 0 ' x ' & cos!1!cos x" ! x SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS \|\|\|\| 69 œ„„„„„ 1+≈ 1 y x FIGURE 24 0 y x 1 π _1 π 2 FIGURE 22 y=cos–! x=arccos x π 2 π 2 _ y 0 x FIGURE 23 y=tan x, _ <x<π 2 π 2 The remaining inverse trigonometric functions are not used as frequently and are sum-marized here. The choice of intervals for in the deﬁnitions of and is not universally agreed upon. For instance, some authors use in the deﬁnition of . [You can see from the graph of the secant function in Figure 26 that both this choice and the one in (11) will work.] sec!1 y ! !0, ""2# " $""2, "% sec!1 csc!1 y y ! $0, "# and cot y ! x & ? y ! cot!1x $x ! !# y ! !0, ""2# " !", 3""2# and sec y ! x & ? y ! sec!1x $&x& # 1# y ! $0, ""2% " $", 3""2% and csc y ! x & ? y ! csc!1x $&x& # 1# 11 70 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS FIGURE 26 y=sec x 0 y x _1 2π π is the height of a football t seconds after kickoff. 14. is your height at age t. 15. If is a one-to-one function such that , what is ? 16. Let , where . (a) Find . (b) Find . If , ﬁnd . 18. The graph of is given. (a) Why is one-to-one? (b) What are the domain and range of ? (c) What is the value of ? (d) Estimate the value of . The formula , where , expresses the Celsius temperature C as a function of the Fahrenheit tem-perature F. Find a formula for the inverse function and interpret it. What is the domain of the inverse function? 20. In the theory of relativity, the mass of a particle with speed is where is the rest mass of the particle and is the speed of light in a vacuum. Find the inverse function of and explain its meaning. f c m 0 m ! f $v# ! m 0 s1 ! v 2"c 2 v F # !459.67 C ! 5 9$F ! 32# 19. y x 0 1 1 f !1$0# f !1$2# f !1 f f t!1$4# t$x# ! 3 $ x $ e x 17. f $ f !1$5## f !1$3# !1 % x % 1 f $x# ! 3 $ x 2 $ tan$"x"2# f !1$9# f $2# ! 9 f f $t# f $t# 13. 1. (a) What is a one-to-one function? (b) How can you tell from the graph of a function whether it is one-to-one? 2. (a) Suppose is a one-to-one function with domain and range . How is the inverse function deﬁned? What is the domain of ? What is the range of ? (b) If you are given a formula for , how do you ﬁnd a formula for ? (c) If you are given the graph of , how do you ﬁnd the graph of ? 3–14 A function is given by a table of values, a graph, a formula, or a verbal description. Determine whether it is one-to-one. 4. 5. 6. 7. 9. 10. 11. 12. t$x# ! cos x t$x# ! 1"x f $x# ! 10 ! 3x f $x# ! x 2 ! 2x y x 8. x y y x x y 3. f !1 f f !1 f f !1 f !1 f !1 B A f EXERCISES 1.6 x 1 2 3 4 5 6 1.5 2.0 3.6 5.3 2.8 2.0 f $x# x 1 2 3 4 5 6 1 2 4 8 16 32 f $x# SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS \|\|\|\| 71 ; 41–42 Use Formula 10 to graph the given functions on a common screen. How are these graphs related? 41. , , , 42. , , , Suppose that the graph of is drawn on a coordinate grid where the unit of measurement is an inch. How many miles to the right of the origin do we have to move before the height of the curve reaches ft? ; 44. Compare the functions and by graph-ing both and in several viewing rectangles. When does the graph of ﬁnally surpass the graph of ? 45–46 Make a rough sketch of the graph of each function. Do not use a calculator. Just use the graphs given in Figures 12 and 13 and, if necessary, the transformations of Section 1.3. (a) (b) 46. (a) (b) 47–50 Solve each equation for . 47. (a) (b) 48. (a) (b) (a) (b) 50. (a) (b) , where 51–52 Solve each inequality for . 51. (a) (b) 52. (a) (b) 53–54 Find (a) the domain of and (b) and its domain. 53. 54. 55. Graph the function and explain why it is one-to-one. Then use a computer algebra system to ﬁnd an explicit expression for . (Your CAS will produce three possible expressions. Explain why two of them are irrelevant in this context.) 56. (a) If , use a computer algebra system to ﬁnd an expression for . (b) Use the expression in part (a) to graph , and on the same screen. 57. If a bacteria population starts with 100 bacteria and doubles every three hours, then the number of bacteria after hours is . (See Exercise 25 in Section 1.5.) (a) Find the inverse of this function and explain its meaning. (b) When will the population reach 50,000? n f t 100 2t3 t y t 1x y tx, y x t 1x tx x 6 x 4, x 0 CAS f 1x f x sx 3 x 2 x 1 CAS fx ln2 ln x fx s3 e 2x f 1 f e23x 4 2 ln x 9 ln x 1 ex 10 x a b e ax Ce bx lnln x 1 ln x lnx 1 1 2x5 3 49. ln5 2x 3 e2x3 7 0 ex 5 2 ln x 1 x y ln x y lnx y ln x y log10x 5 45. t f t f tx ln x f x x 0.1 3 y log2 x 43. y 10 x y e x y log10 x y ln x y log50 x y log10 x y ln x y log1.5 x 21–26 Find a formula for the inverse of the function. 21. 24. 26. ; 27–28 Find an explicit formula for and use it to graph , and the line on the same screen. To check your work, see whether the graphs of and are reﬂections about the line. 27. , 28. 29–30 Use the given graph of to sketch the graph of . 29. 30. 31. (a) How is the logarithmic function deﬁned? (b) What is the domain of this function? (c) What is the range of this function? (d) Sketch the general shape of the graph of the function if . 32. (a) What is the natural logarithm? (b) What is the common logarithm? (c) Sketch the graphs of the natural logarithm function and the natural exponential function with a common set of axes. 33–36 Find the exact value of each expression. 33. (a) (b) 34. (a) (b) 35. (a) (b) 36. (a) (b) 37–39 Express the given quantity as a single logarithm. 37. 38. 39. 40. Use Formula 10 to evaluate each logarithm correct to six dec-imal places. (a) (b) log2 8.4 log12 10 ln1 x 2 1 2 ln x ln sin x lna b lna b 2 ln c ln 5 5 ln 3 ln(ln ee10) e2 ln 5 log3 100 log3 18 log3 50 log2 6 log2 15 log2 20 log10 s10 ln1e log3 1 27 log5 125 a 1 y loga x y loga x y x 0 2 1 y x 0 1 1 f 1 f f x 2 e x x 0 f x x 4 1 f 1 f y x f f 1, f 1 y e x 1 2e x y lnx 3 25. y 2x3 3 f x e x3 23. f x 4x 1 2x 3 22. f x s10 3x 72 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS 66–68 Simplify the expression. 66. 68. ; 69–70 Graph the given functions on the same screen. How are these graphs related? 69. , ; ; 70. , ; ; 71. Find the domain and range of the function ; 72. (a) Graph the function and explain the appearance of the graph. (b) Graph the function . How do you explain the appearance of this graph? 73. (a) If we shift a curve to the left, what happens to its reﬂec-tion about the line ? In view of this geometric principle, ﬁnd an expression for the inverse of , where is a one-to-one function. (b) Find an expression for the inverse of , where . c 0 hx f cx f tx f x c y x tx sin1sin x f x sinsin1x tx sin13x 1 y x y tan1x 2 x 2 y tan x y x y sin1x 2 x 2 y sin x cos2 tan1x sintan1x 67. tansin1x 58. When a camera ﬂash goes off, the batteries immediately begin to recharge the ﬂash’s capacitor, which stores electric charge given by (The maximum charge capacity is and is measured in seconds.) (a) Find the inverse of this function and explain its meaning. (b) How long does it take to recharge the capacitor to 90% of capacity if ? 59–64 Find the exact value of each expression. 59. (a) (b) 60. (a) (b) 61. (a) (b) 62. (a) (b) 63. (a) (b) 64. (a) (b) 65. Prove that . cossin1 x s1 x 2 sin(2 sin1 ( 3 5)) tansec1 4 sin1sin73 tanarctan 10 arccos( 1 2) cot1(s3 ) sin1(1s2 ) arctan 1 sec1 2 tan1(1s3 ) cos11 sin1(s3 2) a 2 t Q0 Qt Q01 e ta CHAPTER 1 REVIEW \|\|\|\| 73 REVIEW CONCEPT CHECK 1 (b) What is the domain of ? (c) What is the domain of ? 10. How is the composite function deﬁned? What is its domain? 11. Suppose the graph of is given. Write an equation for each of the graphs that are obtained from the graph of as follows. (a) Shift 2 units upward. (b) Shift 2 units downward. (c) Shift 2 units to the right. (d) Shift 2 units to the left. (e) Reﬂect about the x-axis. (f) Reﬂect about the y-axis. (g) Stretch vertically by a factor of 2. (h) Shrink vertically by a factor of 2. (i) Stretch horizontally by a factor of 2. (j) Shrink horizontally by a factor of 2. 12. (a) What is a one-to-one function? How can you tell if a func-tion is one-to-one by looking at its graph? (b) If is a one-to-one function, how is its inverse function deﬁned? How do you obtain the graph of from the graph of ? 13. (a) How is the inverse sine function deﬁned? What are its domain and range? (b) How is the inverse cosine function deﬁned? What are its domain and range? (c) How is the inverse tangent function deﬁned? What are its domain and range? f $x# ! tan!1x f $x# ! cos!1x f $x# ! sin!1x f f !1 f !1 f f f f " t f"t f t 1. (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How can you tell whether a given curve is the graph of a function? 2. Discuss four ways of representing a function. Illustrate your discussion with examples. 3. (a) What is an even function? How can you tell if a function is even by looking at its graph? (b) What is an odd function? How can you tell if a function is odd by looking at its graph? 4. What is an increasing function? 5. What is a mathematical model? 6. Give an example of each type of function. (a) Linear function (b) Power function (c) Exponential function (d) Quadratic function (e) Polynomial of degree 5 (f) Rational function 7. Sketch by hand, on the same axes, the graphs of the following functions. (a) (b) (c) (d) 8. Draw, by hand, a rough sketch of the graph of each function. (a) (b) (c) (d) (e) (f) (g) (h) 9. Suppose that has domain and has domain . (a) What is the domain of ? f $ t B t A f y ! tan!1x y ! sx y ! & x& y ! 1"x y ! ln x y ! e x y ! tan x y ! sin x j$x# ! x 4 h$x# ! x 3 t$x# ! x 2 f $x# ! x Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. If is a function, then . 2. If , then . 3. If is a function, then . 4. If and is a decreasing function, then . 5. A vertical line intersects the graph of a function at most once. 6. If and are functions, then . 7. If is one-to-one, then . f !1$x# ! 1 f $x# f f " t ! t " f t f f $x1# ' f $x2# f x1 % x2 f $3x# ! 3f $x# f s ! t f $s# ! f $t# f $s $ t# ! f $s# $ f $t# f 8. You can always divide by . 9. If , then . 10. If , then . 11. If and , then . 12. 13. tan!1x ! sin!1x cos!1x tan!1$!1# ! 3""4 ln x ln a !ln x a a ' 1 x ' 0 $ln x#6 ! 6 ln x x ' 0 ln a % ln b 0 % a % b e x TRUE-FALSE QUIZ 74 \|\|\|\| CHAPTER 1 FUNCTIONS AND MODELS 1. Let be the function whose graph is given. (a) Estimate the value of . (b) Estimate the values of such that . (c) State the domain of (d) State the range of (e) On what interval is increasing? (f) Is one-to-one? Explain. (g) Is even, odd, or neither even nor odd? Explain. 2. The graph of is given. (a) State the value of . (b) Why is one-to-one? (c) Estimate the value of . (d) Estimate the domain of . (e) Sketch the graph of . 3. If , evaluate the difference quotient 4. Sketch a rough graph of the yield of a crop as a function of the amount of fertilizer used. 5–8 Find the domain and range of the function. 5. 6. 7. 8. 9. Suppose that the graph of is given. Describe how the graphs of the following functions can be obtained from the graph of (a) (b) y ! f $x $ 8# y ! f $x# $ 8 f. f F$t# ! 3 $ cos 2t h$x# ! ln$x $ 6# t$x# ! s16 ! x 4 f $x# ! 2"$3x ! 1# f $a $ h# ! f $a# h f $x# ! x 2 ! 2x $ 3 g y x 0 1 1 t!1 t!1 t!1$2# t t$2# t y x 1 1 f f f f f. f. f $x# ! 3 x f $2# f (c) (d) (e) (f) 10. The graph of is given. Draw the graphs of the following functions. (a) (b) (c) (d) (e) (f) 11–16 Use transformations to sketch the graph of the function. 11. 12. 13. 14. 15. 16. 17. Determine whether is even, odd, or neither even nor odd. (a) (b) (c) (d) 18. Find an expression for the function whose graph consists of the line segment from the point to the point together with the top half of the circle with center the origin and radius 1. 19. If and , ﬁnd the functions (a) , (b) , (c) , (d) , and their domains. 20. Express the function as a composition of three functions. 21. Life expectancy improved dramatically in the 20th century. The table gives the life expectancy at birth (in years) of males born in the United States. F$x# ! 1"sx $sx t " t f " f t " f f " t t$x# ! x 2 ! 9 f $x# ! ln x $!1, 0# $!2, 2# f $x# ! 1 $ sin x f $x# ! e!x 2 f $x# ! x 3 ! x 7 f $x# ! 2x 5 ! 3x 2 $ 2 f f $x# !' !x e x ! 1 if x % 0 if x # 0 f $x# ! 1 x $ 2 y ! 2 ! sx y ! 1 2$1 $ e x# y ! 3 ln$x ! 2# y ! !sin 2x y x 0 1 1 y ! f !1$x $ 3# y ! f !1$x# y ! 1 2 f $x# ! 1 y ! 2 ! f $x# y ! !f $x# y ! f $x ! 8# f y ! f !1$x# y ! !f $x# y ! f $x ! 2# ! 2 y ! 1 $ 2f $x# EXERCISES Birth year Life expectancy Birth year Life expectancy 1900 48.3 1960 66.6 1910 51.1 1970 67.1 1920 55.2 1980 70.0 1930 57.4 1990 71.8 1940 62.5 2000 73.0 1950 65.6 CHAPTER 1 REVIEW \|\|\|\| 75 26. Solve each equation for x. (a) (b) (c) (d) 27. The population of a certain species in a limited environment with initial population 100 and carrying capacity 1000 is where is measured in years. ; (a) Graph this function and estimate how long it takes for the population to reach 900. (b) Find the inverse of this function and explain its meaning. (c) Use the inverse function to ﬁnd the time required for the population to reach 900. Compare with the result of part (a). ; 28. Graph the three functions , , and on the same screen for two or three values of . For large values of x, which of these functions has the largest values and which has the smallest values? a ' 1 y ! loga x y ! a x y ! x a t P$t# ! 100,000 100 $ 900e!t tan!1x ! 1 e e x ! 2 ln x ! 2 e x ! 5 Use a scatter plot to choose an appropriate type of model. Use your model to predict the life span of a male born in the year 2010. 22. A small-appliance manufacturer ﬁnds that it costs $9000 to produce 1000 toaster ovens a week and $12,000 to produce 1500 toaster ovens a week. (a) Express the cost as a function of the number of toaster ovens produced, assuming that it is linear. Then sketch the graph. (b) What is the slope of the graph and what does it represent? (c) What is the y-intercept of the graph and what does it represent? 23. If , ﬁnd . 24. Find the inverse function of . 25. Find the exact value of each expression. (a) (b) (c) (d) sin(cos!1 ( 4 5)) tan(arcsin 1 2) log10 25 $ log10 4 e 2 ln 3 f $x# ! x $ 1 2x $ 1 f !1$2# f $x# ! 2x $ ln x 76 There are no hard and fast rules that will ensure success in solving problems. However, it is possible to outline some general steps in the problem-solving process and to give some prin-ciples that may be useful in the solution of certain problems. These steps and principles are just common sense made explicit. They have been adapted from George Polya’s book How To Solve It. The ﬁrst step is to read the problem and make sure that you understand it clearly. Ask your-self the following questions: For many problems it is useful to draw a diagram and identify the given and required quantities on the diagram. Usually it is necessary to introduce suitable notation In choosing symbols for the unknown quantities we often use letters such as a, b, c, m, n, x, and y, but in some cases it helps to use initials as suggestive symbols; for instance, for volume or for time. Think of a Plan Find a connection between the given information and the unknown that will enable you to calculate the unknown. It often helps to ask yourself explicitly: “How can I relate the given to the unknown?” If you don’t see a connection immediately, the following ideas may be helpful in devising a plan. Try to Recognize Something Familiar Relate the given situation to previous knowledge. Look at the unknown and try to recall a more familiar problem that has a similar unknown. Try to Recognize Patterns Some problems are solved by recognizing that some kind of pat-tern is occurring. The pattern could be geometric, or numerical, or algebraic. If you can see regularity or repetition in a problem, you might be able to guess what the continuing pattern is and then prove it. Use Analogy Try to think of an analogous problem, that is, a similar problem, a related problem, but one that is easier than the original problem. If you can solve the similar, sim-pler problem, then it might give you the clues you need to solve the original, more difﬁcult problem. For instance, if a problem involves very large numbers, you could ﬁrst try a simi-lar problem with smaller numbers. Or if the problem involves three-dimensional geometry, you could look for a similar problem in two-dimensional geometry. Or if the problem you start with is a general one, you could ﬁrst try a special case. Introduce Something Extra It may sometimes be necessary to introduce something new, an auxiliary aid, to help make the connection between the given and the unknown. For instance, in a problem where a diagram is useful the auxiliary aid could be a new line drawn in a dia-gram. In a more algebraic problem it could be a new unknown that is related to the original unknown. 2 t V What is the unknown? What are the given quantities? What are the given conditions? Understand the Problem 1 P R I N C I P L E S O F P R O B L E M S O L V I N G P R I N C I P L E S O F P R O B L E M S O L V I N G 77 Take Cases We may sometimes have to split a problem into several cases and give a dif-ferent argument for each of the cases. For instance, we often have to use this strategy in deal-ing with absolute value. Work Backward Sometimes it is useful to imagine that your problem is solved and work backward, step by step, until you arrive at the given data. Then you may be able to reverse your steps and thereby construct a solution to the original problem. This procedure is com-monly used in solving equations. For instance, in solving the equation , we sup-pose that is a number that satisﬁes and work backward. We add 5 to each side of the equation and then divide each side by 3 to get . Since each of these steps can be reversed, we have solved the problem. Establish Subgoals In a complex problem it is often useful to set subgoals (in which the desired situation is only partially fulﬁlled). If we can ﬁrst reach these subgoals, then we may be able to build on them to reach our ﬁnal goal. Indirect Reasoning Sometimes it is appropriate to attack a problem indirectly. In using proof by contradiction to prove that implies , we assume that is true and is false and try to see why this can’t happen. Somehow we have to use this information and arrive at a contradiction to what we absolutely know is true. Mathematical Induction In proving statements that involve a positive integer , it is fre-quently helpful to use the following principle. PRINCIPLE OF MATHEMATICAL INDUCTION Let be a statement about the positive integer . Suppose that 1. is true. 2. is true whenever is true. Then is true for all positive integers . This is reasonable because, since is true, it follows from condition 2 (with ) that is true. Then, using condition 2 with , we see that is true. Again using condition 2, this time with , we have that is true. This procedure can be followed indeﬁnitely. Carry Out the Plan In Step 2 a plan was devised. In carrying out that plan we have to check each stage of the plan and write the details that prove that each stage is correct. Look Back Having completed our solution, it is wise to look back over it, partly to see if we have made errors in the solution and partly to see if we can think of an easier way to solve the problem. Another reason for looking back is that it will familiarize us with the method of solution and this may be useful for solving a future problem. Descartes said, “Every problem that I solved became a rule which served afterwards to solve other problems.” These principles of problem solving are illustrated in the following examples. Before you look at the solutions, try to solve these problems yourself, referring to these Principles of Problem Solving if you get stuck. You may ﬁnd it useful to refer to this section from time to time as you solve the exercises in the remaining chapters of this book. 4 3 S4 k ! 3 S3 k ! 2 S2 k ! 1 S1 n Sn Sk Sk$1 S1 n Sn n Q P Q P x ! 4 3x ! 5 ! 7 x 3x ! 5 ! 7 P R I N C I P L E S O F P R O B L E M S O L V I N G P R I N C I P L E S O F P R O B L E M S O L V I N G 78 EXAMPLE 1 Express the hypotenuse of a right triangle with area as a function of its perimeter P. SOLUTION Let’s ﬁrst sort out the information by identifying the unknown quantity and the data: It helps to draw a diagram and we do so in Figure 1. In order to connect the given quantities to the unknown, we introduce two extra vari-ables and , which are the lengths of the other two sides of the triangle. This enables us to express the given condition, which is that the triangle is right-angled, by the Pythago-rean Theorem: The other connections among the variables come by writing expressions for the area and perimeter: Since is given, notice that we now have three equations in the three unknowns , , and : Although we have the correct number of equations, they are not easy to solve in a straight-forward fashion. But if we use the problem-solving strategy of trying to recognize some-thing familiar, then we can solve these equations by an easier method. Look at the right sides of Equations 1, 2, and 3. Do these expressions remind you of anything familiar? Notice that they contain the ingredients of a familiar formula: Using this idea, we express in two ways. From Equations 1 and 2 we have From Equation 3 we have Thus This is the required expression for h as a function of P. M h ! P2 ! 100 2P 2Ph ! P2 ! 100 h 2 $ 100 ! P2 ! 2Ph $ h 2 $a $ b#2 ! $P ! h#2 ! P2 ! 2Ph $ h 2 $a $ b#2 ! $a 2 $ b 2# $ 2ab ! h 2 $ 4$25# $a $ b#2 $a $ b#2 ! a 2 $ 2ab $ b 2 P ! a $ b $ h 3 25 ! 1 2ab 2 h 2 ! a 2 $ b 2 1 h b a P P ! a $ b $ h 25 ! 1 2ab h 2 ! a 2 $ b 2 b a a h b FIGURE 1 Given quantities: perimeter P, area 25 m 2 Unknown: hypotenuse h 25 m2 h P R I N C I P L E S O F P R O B L E M S O L V I N G N Understand the problem N Relate to the familiar N Draw a diagram N Connect the given with the unknown N Introduce something extra 79 As the next example illustrates, it is often necessary to use the problem-solving prin-ciple of taking cases when dealing with absolute values. EXAMPLE 2 Solve the inequality . SOLUTION Recall the deﬁnition of absolute value: It follows that Similarly These expressions show that we must consider three cases: CASE I If , we have CASE II If the given inequality becomes (always true) CASE III If , the inequality becomes Combining cases I, II, and III, we see that the inequality is satisﬁed when . So the solution is the interval . M $!5, 6# !5 % x % 6 x % 6 2x % 12 x ! 3 $ x $ 2 % 11 x # 3 5 % 11 !x $ 3 $ x $ 2 % 11 !2 ( x % 3, x ' !5 !2x % 10 !x $ 3 ! x ! 2 % 11 &x ! 3& $ &x $ 2& % 11 x % !2 x # 3 !2 ( x % 3 x % !2 !' x $ 2 !x ! 2 if x # !2 if x % !2 &x $ 2& !' x $ 2 !$x $ 2# if x $ 2 # 0 if x $ 2 % 0 !' x ! 3 !x $ 3 if x # 3 if x % 3 &x ! 3& !' x ! 3 !$x ! 3# if x ! 3 # 0 if x ! 3 % 0 &x& !' x !x if x # 0 if x % 0 &x ! 3& $ &x $ 2& % 11 P R I N C I P L E S O F P R O B L E M S O L V I N G N Take cases 80 In the following example we ﬁrst guess the answer by looking at special cases and rec-ognizing a pattern. Then we prove it by mathematical induction. In using the Principle of Mathematical Induction, we follow three steps: STEP 1 Prove that is true when . STEP 2 Assume that is true when and deduce that is true when STEP 3 Conclude that is true for all n by the Principle of Mathematical Induction. EXAMPLE 3 If and for n ! 0, 1, 2, . . . , ﬁnd a formula for . SOLUTION We start by ﬁnding formulas for for the special cases n ! 1, 2, and 3. We notice a pattern: The coefﬁcient of x in the denominator of is n $ 1 in the three cases we have computed. So we make the guess that, in general, To prove this, we use the Principle of Mathematical Induction. We have already veriﬁed that (4) is true for n ! 1. Assume that it is true for , that is, fk$x# ! x $k $ 1#x $ 1 n ! k fn$x# ! x $n $ 1#x $ 1 4 fn$x# ! x 3x $ 1 x 3x $ 1 $ 1 ! x 3x $ 1 4x $ 1 3x $ 1 ! x 4x $ 1 f3$x# ! $ f0 " f2#$x# ! f0$ f2$x## ! f0( x 3x $ 1) ! x 2x $ 1 x 2x $ 1 $ 1 ! x 2x $ 1 3x $ 1 2x $ 1 ! x 3x $ 1 f2$x# ! $ f0 " f1#$x# ! f0$ f1$x## ! f0( x 2x $ 1) ! x x $ 1 x x $ 1 $ 1 ! x x $ 1 2x $ 1 x $ 1 ! x 2x $ 1 f1$x# ! $ f0 " f0#$x# ! f0$ f0$x## ! f0( x x $ 1) fn$x# fn$x# fn$1 ! f0 " fn f0$x# ! x"$x $ 1# Sn n ! k $ 1. Sn n ! k Sn n ! 1 Sn P R I N C I P L E S O F P R O B L E M S O L V I N G N Analogy: Try a similar, simpler problem N Look for a pattern 81 Then This expression shows that (4) is true for . Therefore, by mathematical induc-tion, it is true for all positive integers n. M 1. One of the legs of a right triangle has length 4 cm. Express the length of the altitude perpendi-cular to the hypotenuse as a function of the length of the hypotenuse. 2. The altitude perpendicular to the hypotenuse of a right triangle is 12 cm. Express the length of the hypotenuse as a function of the perimeter. 3. Solve the equation . 4. Solve the inequality . 5. Sketch the graph of the function . 6. Sketch the graph of the function . 7. Draw the graph of the equation 8. Draw the graph of the equation . 9. Sketch the region in the plane consisting of all points such that . 10. Sketch the region in the plane consisting of all points such that 11. Evaluate . 12. (a) Show that the function is an odd function. (b) Find the inverse function of 13. Solve the inequality . 14. Use indirect reasoning to prove that is an irrational number. 15. A driver sets out on a journey. For the ﬁrst half of the distance she drives at the leisurely pace of 30 mi"h; she drives the second half at 60 mi"h. What is her average speed on this trip? 16. Is it true that ? 17. Prove that if n is a positive integer, then is divisible by 6. 18. Prove that . 19. If and for ﬁnd a formula for . 20. (a) If and for ﬁnd an expression for and use mathematical induction to prove it. ; (b) Graph on the same screen and describe the effects of repeated composition. f0, f1, f2, f3 fn$x# n ! 0, 1, 2, . . . , fn$1 ! f0 " fn f0$x# ! 1 2 ! x fn$x# n ! 0, 1, 2, . . . , fn$1$x# ! f0$ fn$x## f0$x# ! x 2 1 $ 3 $ 5 $ &&& $ $2n ! 1# ! n2 7n ! 1 f " $t $ h# ! f " t $ f " h log2 5 ln$x 2 ! 2x ! 2# ( 0 f. f $x# ! ln(x $ sx 2 $ 1) $log2 3#$log3 4#$log4 5#&&&$log31 32# & x ! y& $ & x& ! & y& ( 2 $x, y# & x& $ & y& ( 1 $x, y# x 4 ! 4x 2 ! x 2y 2 $ 4y 2 ! 0 x $ & x& ! y $ & y&. t$x# ! & x 2 ! 1& ! & x 2 ! 4& & x 2 ! 4& x& $ 3& f $x# ! & x ! 1& ! & x ! 3& # 5 & x $ 5& ! 3 & 2x ! 1& ! PROBLEMS n ! k $ 1 ! x $k $ 1#x $ 1 x $k $ 1#x $ 1 $ 1 ! x $k $ 1#x $ 1 $k $ 2#x $ 1 $k $ 1#x $ 1 ! x $k $ 2#x $ 1 fk$1$x# ! $ f0 " fk#$x# ! f0$ fk$x## ! f0( x $k $ 1#x $ 1) P R I N C I P L E S O F P R O B L E M S O L V I N G 82 In A Preview of Calculus (page 2) we saw how the idea of a limit underlies the various branches of calculus. It is therefore appropriate to begin our study of calculus by investigating limits and their properties. The special type of limit that is used to ﬁnd tangents and velocities gives rise to the central idea in differential calculus, the derivative. The idea of a limit is illustrated by secant lines approaching a tangent line. LIMITS AND DERIVATIVES 2 THE TANGENT AND VELOCITY PROBLEMS In this section we see how limits arise when we attempt to ﬁnd the tangent to a curve or the velocity of an object. THE TANGENT PROBLEM The word tangent is derived from the Latin word tangens, which means “touching.” Thus a tangent to a curve is a line that touches the curve. In other words, a tangent line should have the same direction as the curve at the point of contact. How can this idea be made precise? For a circle we could simply follow Euclid and say that a tangent is a line that intersects the circle once and only once as in Figure 1(a). For more complicated curves this deﬁni-tion is inadequate. Figure l(b) shows two lines and passing through a point on a curve . The line intersects only once, but it certainly does not look like what we think of as a tangent. The line , on the other hand, looks like a tangent but it intersects twice. To be speciﬁc, let’s look at the problem of trying to ﬁnd a tangent line to the parabola in the following example. EXAMPLE 1 Find an equation of the tangent line to the parabola at the point . SOLUTION We will be able to ﬁnd an equation of the tangent line as soon as we know its slope . The difﬁculty is that we know only one point, , on , whereas we need two points to compute the slope. But observe that we can compute an approximation to by choosing a nearby point on the parabola (as in Figure 2) and computing the slope of the secant line . We choose so that . Then For instance, for the point we have The tables in the margin show the values of for several values of close to 1. The closer is to , the closer is to 1 and, it appears from the tables, the closer is to 2. This suggests that the slope of the tangent line should be . We say that the slope of the tangent line is the limit of the slopes of the secant lines, and we express this symbolically by writing and Assuming that the slope of the tangent line is indeed 2, we use the point-slope form of the equation of a line (see Appendix B) to write the equation of the tangent line through as y ! 2x ! 1 or y ! 1 ! 2!x ! 1" !1, 1" lim x l 1 x 2 ! 1 x ! 1 ! 2 lim Q lP mPQ ! m m ! 2 t mPQ x P Q x mPQ mPQ ! 2.25 ! 1 1.5 ! 1 ! 1.25 0.5 ! 2.5 Q!1.5, 2.25" mPQ ! x 2 ! 1 x ! 1 Q " P x " 1 PQ mPQ Q!x, x 2" m t P m t P!1, 1" y ! x 2 V y ! x 2 t C t C l C P t l 2.1 83 FIGURE 2 x y 0 y=≈ t Q{x, ≈} P(1, 1) (a) (b) t FIGURE 1 P C t l x 2 3 1.5 2.5 1.1 2.1 1.01 2.01 1.001 2.001 mPQ x 0 1 0.5 1.5 0.9 1.9 0.99 1.99 0.999 1.999 mPQ Figure 3 illustrates the limiting process that occurs in this example. As approaches along the parabola, the corresponding secant lines rotate about and approach the tangent line t. M Many functions that occur in science are not described by explicit equations; they are deﬁned by experimental data. The next example shows how to estimate the slope of the tangent line to the graph of such a function. EXAMPLE 2 The ﬂash unit on a camera operates by storing charge on a capacitor and releasing it suddenly when the ﬂash is set off. The data in the table describe the charge Q remaining on the capacitor (measured in microcoulombs) at time t (measured in seconds after the ﬂash goes off). Use the data to draw the graph of this function and estimate the slope of the tangent line at the point where t ! 0.04. [Note: The slope of the tangent line represents the electric current ﬂowing from the capacitor to the ﬂash bulb (measured in microamperes).] SOLUTION In Figure 4 we plot the given data and use them to sketch a curve that approxi-mates the graph of the function. FIGURE 4 t Q A B C P 0 0.02 0.04 0.06 0.08 0.1 90 100 60 70 80 50 (seconds) (microcoulombs) V Q approaches P from the right Q approaches P from the left P y x 0 Q t P y x 0 Q t P y x 0 Q t P y x 0 Q t P y x 0 Q t FIGURE 3 x 0 P y Q t P P Q 84 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES In Visual 2.1 you can see how the process in Figure 3 works for additional functions. TEC t Q 0.00 100.00 0.02 81.87 0.04 67.03 0.06 54.88 0.08 44.93 0.10 36.76 Given the points and on the graph, we ﬁnd that the slope of the secant line PR is The table at the left shows the results of similar calculations for the slopes of other secant lines. From this table we would expect the slope of the tangent line at to lie somewhere between 742 and 607.5. In fact, the average of the slopes of the two closest secant lines is So, by this method, we estimate the slope of the tangent line to be 675. Another method is to draw an approximation to the tangent line at P and measure the sides of the triangle ABC, as in Figure 4. This gives an estimate of the slope of the tan-gent line as M THE VELOCITY PROBLEM If you watch the speedometer of a car as you travel in city trafﬁc, you see that the needle doesn’t stay still for very long; that is, the velocity of the car is not constant. We assume from watching the speedometer that the car has a deﬁnite velocity at each moment, but how is the “instantaneous” velocity deﬁned? Let’s investigate the example of a falling ball. EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds. SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after seconds is denoted by and measured in meters, then Galileo’s law is expressed by the equation The difﬁculty in ﬁnding the velocity after 5 s is that we are dealing with a single instant of time , so no time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the brief time interval of a tenth of a second from to : 4.95.12 4.952 0.1 49.49 ms s5.1 s5 0.1 average velocity change in position time elapsed t 5.1 t 5 t 5 st 4.9t 2 st t V AB BC 80.4 53.6 0.06 0.02 670 1 2742 607.5 674.75 t 0.04 mPR 100.00 67.03 0.00 0.04 824.25 R0.00, 100.00 P0.04, 67.03 SECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS \|\|\|\| 85 R (0.00, 100.00) 824.25 (0.02, 81.87) 742.00 (0.06, 54.88) 607.50 (0.08, 44.93) 552.50 (0.10, 36.76) 504.50 mPR N The physical meaning of the answer in Example 2 is that the electric current ﬂowing from the capacitor to the ﬂash bulb after 0.04 second is about –670 microamperes. The CN T ower in T oronto is currently the tallest freestanding building in the world. © 2003 Brand X Pictures The following table shows the results of similar calculations of the average velocity over successively smaller time periods. It appears that as we shorten the time period, the average velocity is becoming closer to 49 m#s. The instantaneous velocity when is deﬁned to be the limiting value of these average velocities over shorter and shorter time periods that start at . Thus the (instantaneous) velocity after 5 s is M You may have the feeling that the calculations used in solving this problem are very sim-ilar to those used earlier in this section to ﬁnd tangents. In fact, there is a close connec-tion between the tangent problem and the problem of ﬁnding velocities. If we draw the graph of the distance function of the ball (as in Figure 5) and we consider the points and on the graph, then the slope of the secant line is which is the same as the average velocity over the time interval . Therefore, the velocity at time (the limit of these average velocities as approaches 0) must be equal to the slope of the tangent line at (the limit of the slopes of the secant lines). Examples 1 and 3 show that in order to solve tangent and velocity problems we must be able to ﬁnd limits. After studying methods for computing limits in the next ﬁve sections, we will return to the problems of ﬁnding tangents and velocities in Section 2.7. FIGURE 5 t s Q a a+h 0 slope of secant line ! average velocity P s=4.9t@ t s 0 a slope of tangent ! instantaneous velocity P s=4.9t@ P h t ! a &a, a " h' mPQ ! 4.9!a " h"2 ! 4.9a 2 !a " h" ! a PQ Q!a " h, 4.9!a " h"2" P!a, 4.9a 2" v ! 49 m#s t ! 5 t ! 5 86 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES Time interval Average velocity (m#s) 53.9 49.49 49.245 49.049 49.0049 5 # t # 5.001 5 # t # 5.01 5 # t # 5.05 5 # t # 5.1 5 # t # 6 SECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS \|\|\|\| 87 (c) Using the slope from part (b), ﬁnd an equation of the tangent line to the curve at . (d) Sketch the curve, two of the secant lines, and the tangent line. If a ball is thrown into the air with a velocity of 40 ft#s, its height in feet seconds later is given by . (a) Find the average velocity for the time period beginning when and lasting (i) 0.5 second (ii) 0.1 second (iii) 0.05 second (iv) 0.01 second (b) Estimate the instantaneous velocity when 6. If a rock is thrown upward on the planet Mars with a velocity of 10 m#s, its height in meters seconds later is given by (a) Find the average velocity over the given time intervals: (i) [1, 2] (ii) [1, 1.5] (iii) [1, 1.1] (iv) [1, 1.01] (v) [1, 1.001] (b) Estimate the instantaneous velocity when . 7. The table shows the position of a cyclist. (a) Find the average velocity for each time period: (i) (ii) (iii) (iv) (b) Use the graph of as a function of to estimate the instan-taneous velocity when . 8. The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion , where is measured in seconds. (a) Find the average velocity during each time period: (i) [1, 2] (ii) [1, 1.1] (iii) [1, 1.01] (iv) [1, 1.001] (b) Estimate the instantaneous velocity of the particle when . The point lies on the curve . (a) If is the point , ﬁnd the slope of the secant line (correct to four decimal places) for , 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9. Do the slopes appear to be approaching a limit? ; (b) Use a graph of the curve to explain why the slopes of the secant lines in part (a) are not close to the slope of the tan-gent line at . (c) By choosing appropriate secant lines, estimate the slope of the tangent line at . P P x ! 2 PQ !x, sin!10$#x"" Q y ! sin!10$#x" P!1, 0" 9. t ! 1 t s ! 2 sin $t " 3 cos $t t ! 3 t s &3, 4' &3, 5' &2, 3' &1, 3' t ! 1 y ! 10t ! 1.86t 2. t t ! 2. t ! 2 y ! 40t ! 16t 2 t 5. P!3, 1" 1. A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes. (a) If P is the point on the graph of V, ﬁnd the slopes of the secant lines PQ when Q is the point on the graph with , 10, 20, 25, and 30. (b) Estimate the slope of the tangent line at P by averaging the slopes of two secant lines. (c) Use a graph of the function to estimate the slope of the tangent line at P. (This slope represents the rate at which the water is ﬂowing from the tank after 15 minutes.) 2. A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after t min-utes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient’s heart rate after 42 minutes using the secant line between the points with the given values of t. (a) t ! 36 and t ! 42 (b) t ! 38 and t ! 42 (c) t ! 40 and t ! 42 (d) t ! 42 and t ! 44 What are your conclusions? The point lies on the curve . (a) If is the point , use your calculator to ﬁnd the slope of the secant line (correct to six decimal places) for the following values of : (i) 0.5 (ii) 0.9 (iii) 0.99 (iv) 0.999 (v) 1.5 (vi) 1.1 (vii) 1.01 (viii) 1.001 (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at . (c) Using the slope from part (b), ﬁnd an equation of the tangent line to the curve at . 4. The point lies on the curve . (a) If is the point , use your calculator to ﬁnd the slope of the secant line (correct to six decimal places) for the following values of : (i) 2.5 (ii) 2.9 (iii) 2.99 (iv) 2.999 (v) 3.5 (vi) 3.1 (vii) 3.01 (viii) 3.001 (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at . P!3, 1" x PQ (x, sx ! 2 ) Q y ! sx ! 2 P!3, 1" P(1, 1 2) P(1, 1 2) x PQ !x, x#!1 " x"" Q y ! x#!1 " x" P(1, 1 2) 3. t ! 5 !15, 250" EXERCISES 2.1 t (min) 5 10 15 20 25 30 V (gal) 694 444 250 111 28 0 t (min) 36 38 40 42 44 Heartbeats 2530 2661 2806 2948 3080 t (seconds) 0 1 2 3 4 5 s (meters) 0 1.4 5.1 10.7 17.7 25.8 88 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES THE LIMIT OF A FUNCTION Having seen in the preceding section how limits arise when we want to ﬁnd the tangent to a curve or the velocity of an object, we now turn our attention to limits in general and numerical and graphical methods for computing them. Let’s investigate the behavior of the function deﬁned by for val-ues of near 2. The following table gives values of for values of close to 2, but not equal to 2. From the table and the graph of (a parabola) shown in Figure 1 we see that when is close to 2 (on either side of 2), is close to 4. In fact, it appears that we can make the values of as close as we like to 4 by taking sufﬁciently close to 2. We express this by saying “the limit of the function as approaches 2 is equal to 4.” The notation for this is In general, we use the following notation. DEFINITION We write and say “the limit of , as approaches , equals ” if we can make the values of arbitrarily close to (as close to L as we like) by taking x to be sufﬁciently close to (on either side of ) but not equal to . Roughly speaking, this says that the values of tend to get closer and closer to the number as gets closer and closer to the number (from either side of ) but . (A more precise deﬁnition will be given in Section 2.4.) An alternative notation for is as which is usually read “ approaches as approaches .” a x L f!x" x l a f!x" l L lim x l a f!x" ! L x " a a a x L f!x" a a a L f!x" L a x f!x" lim x l a f!x" ! L 1 lim x l 2 !x 2 ! x " 2" ! 4 x f!x" ! x 2 ! x " 2 x f!x" f!x" x f x f!x" x f!x" ! x 2 ! x " 2 f 2.2 4 ƒ approaches 4. x y 2 As x approaches 2, y=≈-x+2 0 FIGURE 1 x 3.0 8.000000 2.5 5.750000 2.2 4.640000 2.1 4.310000 2.05 4.152500 2.01 4.030100 2.005 4.015025 2.001 4.003001 f !x" x 1.0 2.000000 1.5 2.750000 1.8 3.440000 1.9 3.710000 1.95 3.852500 1.99 3.970100 1.995 3.985025 1.999 3.997001 f !x" Notice the phrase “but ” in the deﬁnition of limit. This means that in ﬁnding the limit of as approaches , we never consider . In fact, need not even be deﬁned when . The only thing that matters is how is deﬁned near . Figure 2 shows the graphs of three functions. Note that in part (c), is not deﬁned and in part (b), . But in each case, regardless of what happens at , it is true that . EXAMPLE 1 Guess the value of . SOLUTION Notice that the function is not deﬁned when , but that doesn’t matter because the deﬁnition of says that we consider values of that are close to but not equal to . The tables at the left give values of (correct to six decimal places) for values of that approach 1 (but are not equal to 1). On the basis of the values in the tables, we make the guess that M Example 1 is illustrated by the graph of in Figure 3. Now let’s change slightly by giving it the value 2 when and calling the resulting function : This new function still has the same limit as approaches 1 (see Figure 4). 0 1 0.5 x-1 ≈-1 y= FIGURE 3 FIGURE 4 0 1 0.5 y=© 2 y x y x x t t(x) !( x ! 1 x 2 ! 1 if x " 1 2 if x ! 1 t x ! 1 f f lim x l 1 x ! 1 x 2 ! 1 ! 0.5 x f!x" a a x lim x l a f!x" x ! 1 f!x" ! !x ! 1"#!x 2 ! 1" lim x l1 x ! 1 x 2 ! 1 (c) x y 0 L a (b) x y 0 L a (a) x y 0 L a FIGURE 2 lim ƒ=L in all three cases x a lim x l a f!x" ! L a f!a" " L f!a" a f x ! a f!x" x ! a a x f!x" x " a SECTION 2.2 THE LIMIT OF A FUNCTION \|\|\|\| 89 0.5 0.666667 0.9 0.526316 0.99 0.502513 0.999 0.500250 0.9999 0.500025 f !x" x % 1 1.5 0.400000 1.1 0.476190 1.01 0.497512 1.001 0.499750 1.0001 0.499975 f !x" x & 1 EXAMPLE 2 Estimate the value of . SOLUTION The table lists values of the function for several values of near 0. As approaches 0, the values of the function seem to approach and so we guess that M In Example 2 what would have happened if we had taken even smaller values of The table in the margin shows the results from one calculator; you can see that something strange seems to be happening. If you try these calculations on your own calculator you might get different values, but eventually you will get the value 0 if you make sufﬁciently small. Does this mean that the answer is really 0 instead of ? No, the value of the limit is , as we will show in the \| next section. The problem is that the calculator gave false values because is very close to 3 when is small. (In fact, when t is sufﬁciently small, a calculator’s value for is to as many digits as the calculator is capable of carrying.) Something similar happens when we try to graph the function of Example 2 on a graphing calculator or computer. Parts (a) and (b) of Figure 5 show quite accurate graphs of , and when we use the trace mode (if available) we can estimate eas-ily that the limit is about . But if we zoom in too much, as in parts (c) and (d), then we get inaccurate graphs, again because of problems with subtraction. FIGURE 5 0.1 0.2 (a) &_5, 5' by &_0.1, 0.3' 0.1 0.2 (b) &_0.1, 0.1' by &_0.1, 0.3' (c) &_10–^, 10–^' by &_0.1, 0.3' (d) &_10–&, 10–&' by &_0.1, 0.3' 1 6 f f!t" ! st 2 " 9 ! 3 t 2 3.000. . . st 2 " 9 t st 2 " 9 1 6 1 6 t t? lim t l 0 st 2 " 9 ! 3 t 2 ! 1 6 0.1666666 . . . t t lim t l 0 st 2 " 9 ! 3 t 2 90 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES t '1.0 0.16228 '0.5 0.16553 '0.1 0.16662 '0.05 0.16666 '0.01 0.16667 st 2 " 9 ! 3 t 2 t '0.0005 0.16800 '0.0001 0.20000 '0.00005 0.00000 '0.00001 0.00000 st 2 " 9 ! 3 t 2 www.stewartcalculus.com For a further explanation of why calculators sometimes give false values, click on Lies My Calculator and Computer Told Me. In particular, see the section called The Perils of Subtraction. EXAMPLE 3 Guess the value of . SOLUTION The function is not deﬁned when . Using a calculator (and remembering that, if , means the sine of the angle whose radian mea-sure is ), we construct a table of values correct to eight decimal places. From the table at the left and the graph in Figure 6 we guess that This guess is in fact correct, as will be proved in Chapter 3 using a geometric argument. M EXAMPLE 4 Investigate . SOLUTION Again the function is undeﬁned at 0. Evaluating the function for some small values of , we get Similarly, On the basis of this information we might be tempted to guess that \| but this time our guess is wrong. Note that although for any integer , it is also true that for inﬁnitely many values of that approach 0. The graph of is given in Figure 7. FIGURE 7 y=sin(π/x) x y 1 1 _1 _1 f x f!x" ! 1 n f!1#n" ! sin n$ ! 0 lim x l 0 sin $ x ! 0 f!0.001" ! f!0.0001" ! 0. f!0.01" ! sin 100$ ! 0 f!0.1" ! sin 10$ ! 0 f ( 1 4) ! sin 4$ ! 0 f ( 1 3) ! sin 3$ ! 0 f ( 1 2) ! sin 2$ ! 0 f!1" ! sin $ ! 0 x f!x" ! sin!$#x" lim x l 0 sin $ x V 0 x _1 1 y sin x x y= 1 FIGURE 6 lim x l 0 sin x x ! 1 x sin x x ! ! x ! 0 f!x" ! !sin x"#x lim x l 0 sin x x V SECTION 2.2 THE LIMIT OF A FUNCTION \|\|\|\| 91 x '1.0 0.84147098 '0.5 0.95885108 '0.4 0.97354586 '0.3 0.98506736 '0.2 0.99334665 '0.1 0.99833417 '0.05 0.99958339 '0.01 0.99998333 '0.005 0.99999583 '0.001 0.99999983 sin x x N COMPUTER ALGEBRA SYSTEMS Computer algebra systems (CAS) have commands that compute limits. In order to avoid the types of pitfalls demonstrated in Examples 2, 4, and 5, they don’t ﬁnd limits by numerical experimentation. Instead, they use more sophisticated techniques such as com-puting inﬁnite series. If you have access to a CAS, use the limit command to compute the limits in the examples of this section and to check your answers in the exercises of this chapter. The dashed lines near the -axis indicate that the values of oscillate between 1 and inﬁnitely often as approaches 0. (See Exercise 39.) Since the values of do not approach a ﬁxed number as approaches 0, M EXAMPLE 5 Find . SOLUTION As before, we construct a table of values. From the ﬁrst table in the margin it appears that But if we persevere with smaller values of , the second table suggests that Later we will see that ; then it follows that the limit is 0.0001. M \| Examples 4 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is easy to guess the wrong value if we use inappropriate values of , but it is difﬁcult to know when to stop calculating values. And, as the discussion after Example 2 shows, sometimes calculators and computers give the wrong values. In the next section, however, we will develop foolproof methods for calculating limits. EXAMPLE 6 The Heaviside function is deﬁned by [This function is named after the electrical engineer Oliver Heaviside (1850–1925) and can be used to describe an electric current that is switched on at time .] Its graph is shown in Figure 8. As approaches 0 from the left, approaches 0. As approaches 0 from the right, approaches 1. There is no single number that approaches as approaches 0. Therefore, does not exist. M ONE-SIDED LIMITS We noticed in Example 6 that approaches 0 as approaches 0 from the left and approaches 1 as approaches 0 from the right. We indicate this situation symbolically by writing and The symbol “ ” indicates that we consider only values of that are less than 0. Like-wise, “ ” indicates that we consider only values of that are greater than 0. t t l 0! t t l 0" lim t l 0! H!t" ! 1 lim t l 0" H!t" ! 0 t H!t" t H!t" lim t l 0 H!t" t H!t" H!t" t H!t" t t ! 0 H!t" !# 0 1 if t # 0 if t $ 0 H V x lim x l 0 cos 5x ! 1 lim x l 0 $x 3 ! cos 5x 10,000% ! 0.000100 ! 1 10,000 x lim x l 0 $x 3 ! cos 5x 10,000% ! 0 lim x l 0 $x 3 ! cos 5x 10,000% lim x l 0 sin % x does not exist x f!x" x "1 sin!%&x" y 92 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES x 1 1.000028 0.5 0.124920 0.1 0.001088 0.05 0.000222 0.01 0.000101 x 3 ! cos 5x 10,000 x 0.005 0.00010009 0.001 0.00010000 x 3 ! cos 5x 10,000 t y 1 0 FIGURE 8 DEFINITION We write and say the left-hand limit of as approaches [or the limit of as approaches from the left] is equal to if we can make the values of arbi-trarily close to L by taking x to be sufﬁciently close to a and x less than a. Notice that Deﬁnition 2 differs from Deﬁnition 1 only in that we require to be less than . Similarly, if we require that be greater than , we get “the right-hand limit of as approaches is equal to ” and we write Thus the symbol “ ” means that we consider only . These deﬁnitions are illus-trated in Figure 9. By comparing Deﬁnition l with the deﬁnitions of one-sided limits, we see that the fol-lowing is true. if and only if and EXAMPLE 7 The graph of a function is shown in Figure 10. Use it to state the values (if they exist) of the following: (a) (b) (c) (d) (e) (f) SOLUTION From the graph we see that the values of approach 3 as x approaches 2 from the left, but they approach 1 as x approaches 2 from the right. Therefore (a) and (b) (c) Since the left and right limits are different, we conclude from (3) that does not exist. The graph also shows that (d) and (e) lim x l 5! t!x" ! 2 lim x l 5" t!x" ! 2 limx l 2 t!x" lim x l 2! t!x" ! 1 lim x l 2" t!x" ! 3 t!x" lim x l 5 t!x" lim x l 5! t!x" lim x l 5" t!x" lim x l 2 t!x" lim x l 2! t!x" lim x l 2" t!x" t V lim x l a! f!x" ! L lim x l a" f!x" ! L lim x l a f!x" ! L 3 0 x y L x a 0 x y ƒ L x a ƒ x a+ x a (a) lim ƒ=L (b) lim ƒ=L FIGURE 9 x & a x l a! lim x l a! f!x" ! L L a x f!x" a x a x f!x" L a x f!x" a x f!x" lim x l a" f!x" ! L 2 SECTION 2.2 THE LIMIT OF A FUNCTION \|\|\|\| 93 FIGURE 10 y 0 x y=© 1 2 3 4 5 1 3 4 (f) This time the left and right limits are the same and so, by (3), we have Despite this fact, notice that M INFINITE LIMITS EXAMPLE 8 Find if it exists. SOLUTION As becomes close to 0, also becomes close to 0, and becomes very large. (See the table in the margin.) In fact, it appears from the graph of the function shown in Figure 11 that the values of can be made arbitrarily large by taking close enough to 0. Thus the values of do not approach a number, so does not exist. M To indicate the kind of behavior exhibited in Example 8, we use the notation \| This does not mean that we are regarding as a number. Nor does it mean that the limit exists. It simply expresses the particular way in which the limit does not exist: can be made as large as we like by taking close enough to 0. In general, we write symbolically to indicate that the values of tend to become larger and larger (or “increase without bound”) as becomes closer and closer to . DEFINITION Let be a function deﬁned on both sides of , except possibly at itself. Then means that the values of can be made arbitrarily large (as large as we please) by taking sufﬁciently close to , but not equal to a. Another notation for is as Again the symbol is not a number, but the expression is often read as “the limit of , as approaches , is inﬁnity” or “ becomes inﬁnite as approaches ” or “ increases without bound as approaches ” This deﬁnition is illustrated graphically in Figure 12. a x f!x" a x f!x" a x f!x" lim x l a f!x" ! ' ' x l a f!x" l ' lim x l a f!x" ! ' a x f!x" lim x l a f!x" ! ' a a f 4 a x f!x" lim x l a f!x" ! ' x 1&x 2 ' lim x l 0 1 x 2 ! ' lim x l 0 !1&x 2" f!x" x f!x" f!x" ! 1&x 2 1&x 2 x 2 x lim x l 0 1 x 2 t!5" " 2. lim x l 5 t!x" ! 2 94 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES x (1 1 (0.5 4 (0.2 25 (0.1 100 (0.05 400 (0.01 10,000 (0.001 1,000,000 1 x2 FIGURE 11 y= 0 y x 1 ≈ x a FIGURE 12 lim ƒ=x y x=a y=ƒ a 0 A similar sort of limit, for functions that become large negative as gets close to , is deﬁned in Deﬁnition 5 and is illustrated in Figure 13. DEFINITION Let be deﬁned on both sides of , except possibly at itself. Then means that the values of can be made arbitrarily large negative by taking sufﬁciently close to , but not equal to a. The symbol can be read as “the limit of , as approaches , is negative inﬁnity” or “ decreases without bound as approaches .” As an example we have Similar deﬁnitions can be given for the one-sided inﬁnite limits remembering that “ ” means that we consider only values of that are less than , and similarly “ ” means that we consider only . Illustrations of these four cases are given in Figure 14. DEFINITION The line is called a vertical asymptote of the curve if at least one of the following statements is true: For instance, the -axis is a vertical asymptote of the curve because . In Figure 14 the line is a vertical asymptote in each of the four cases shown. In general, knowledge of vertical asymptotes is very useful in sketching graphs. x ! a lim x l 0 !1&x 2" ! ' y ! 1&x 2 y lim x l a! f!x" ! "' lim x l a" f!x" ! "' lim x l a f!x" ! "' lim x l a! f!x" ! ' lim x l a" f!x" ! ' lim x l a f!x" ! ' y ! f!x" x ! a 6 (d) lim ƒ=_ a y 0 x x a+ x a_ (c) lim ƒ=y 0 a x (a) lim ƒ= y 0 a x x a (b) lim ƒ=a y x x a+ 0 FIGURE 14 x & a x l a! a x x l a" lim x l a! f!x" ! "' lim x l a" f!x" ! "' lim x l a! f!x" ! ' lim x l a" f!x" ! ' lim x l 0 $" 1 x 2% ! "' a x f!x" a x f!x" lim x l a f!x" ! "' a x f!x" lim x l a f!x" ! "' a a f 5 a x SECTION 2.2 THE LIMIT OF A FUNCTION \|\|\|\| 95 0 x y x=a y=ƒ a FIGURE 13 lim ƒ=_ x a N When we say a number is “large negative,” we mean that it is negative but its magnitude (absolute value) is large. EXAMPLE 9 Find and . SOLUTION If is close to 3 but larger than 3, then the denominator is a small posi-tive number and is close to 6. So the quotient is a large positive number. Thus, intuitively, we see that Likewise, if is close to 3 but smaller than 3, then is a small negative number but is still a positive number (close to 6). So is a numerically large negative number. Thus The graph of the curve is given in Figure 15. The line is a verti-cal asymptote. M EXAMPLE 10 Find the vertical asymptotes of . SOLUTION Because there are potential vertical asymptotes where . In fact, since as and as , whereas is positive when x is near , we have and This shows that the line is a vertical asymptote. Similar reasoning shows that the lines , where n is an integer, are all vertical asymptotes of . The graph in Figure 16 conﬁrms this. M Another example of a function whose graph has a vertical asymptote is the natural log-arithmic function . From Figure 17 we see that and so the line (the y-axis) is a vertical asymptote. In fact, the same is true for provided that . (See Figures 11 and 12 in Section 1.6.) a & 1 y ! loga x x ! 0 lim x l 0! ln x ! "' y ! ln x f!x" ! tan x x ! !2n ! 1"%&2 x ! %&2 lim x l !%&2"! tan x ! "' lim x l !%&2"" tan x ! ' %&2 sin x x l !%&2"! cos x l 0" x l !%&2"" cos x l 0! cos x ! 0 tan x ! sin x cos x f!x" ! tan x x ! 3 y ! 2x&!x " 3" lim x l 3" 2x x " 3 ! "' 2x&!x " 3" 2x x " 3 x lim x l 3! 2x x " 3 ! ' 2x&!x " 3" 2x x " 3 x lim x l 3" 2x x " 3 lim x l 3! 2x x " 3 96 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES FIGURE 15 5 2x x-3 y= 0 x y x=3 _ _ x y π 0 π 1 π 2 3π 2 π 2 3π 2 FIGURE 16 y=tan x 2. Explain what it means to say that and In this situation is it possible that exists? Explain. limx l 1 f !x" lim x l 1! f !x" ! 7 lim x l 1" f !x" ! 3 1. Explain in your own words what is meant by the equation Is it possible for this statement to be true and yet ? Explain. f !2" ! 3 lim x l 2 f !x" ! 5 EXERCISES 2.2 FIGURE 17 The y-axis is a vertical asymptote of the natural logarithmic function. x 0 y 1 y=ln x SECTION 2.2 THE LIMIT OF A FUNCTION \|\|\|\| 97 7. For the function whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why. (a) (b) (c) (d) (e) (f) (g) (h) 8. For the function whose graph is shown, state the following. (a) (b) (c) (d) (e) The equations of the vertical asymptotes. 9. For the function whose graph is shown, state the following. (a) (b) (c) (d) (e) (f) The equations of the vertical asymptotes. 10. A patient receives a 150-mg injection of a drug every 4 hours. The graph shows the amount of the drug in the blood-f !t" x y 0 6 _3 _7 lim x l 6! f !x" lim x l 6" f !x" lim x l 0 f !x" lim x l"3 f !x" lim x l"7 f !x" f x y 0 2 5 _3 lim x l "3!R!x" lim x l "3" R!x" lim x l 5 R!x" lim x l2 R!x" R y t 2 4 4 2 lim t l 4 t!t" t!2" lim t l 2 t!t" lim t l 2! t!t" lim t l 2" t!t" lim t l 0 t!t" lim t l 0! t!t" lim t l 0" t!t" t 3. Explain the meaning of each of the following. (a) (b) For the function whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why. (a) (b) (c) (d) (e) 5. Use the given graph of to state the value of each quantity, if it exists. If it does not exist, explain why. (a) (b) (c) (d) (e) 6. For the function whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) y 0 x 2 _2 _4 4 6 lim x l 5" h!x" lim x l 5! h!x" h!2" lim x l 2 h!x" h!0" lim x l 0 h!x" lim x l 0! h!x" lim xl 0" h!x" h!"3" lim x l "3 h!x" lim x l "3! h!x" lim x l "3" h!x" h y 0 x 2 4 4 2 f !5" lim x l 5 f !x" lim x l 1 f !x" lim x l 1! f !x" lim x l 1" f !x" f y 0 x 2 4 4 2 f !3" lim x l 3 f !x" lim x l 3! f !x" lim x l 3" f !x" lim x l 0 f !x" f 4. lim x l 4! f !x" ! "' lim x l "3 f !x" ! ' 98 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES 18. , 19. , , , , , 20. , x ! 1, 0.5, 0.1, 0.05, 0.01, 0.005, 0.001 21–24 Use a table of values to estimate the value of the limit. If you have a graphing device, use it to conﬁrm your result graphically. 21. 22. 23. 24. 25–32 Determine the inﬁnite limit. 25. 26. 28. 29. 30. 31. 32. 33. Determine and (a) by evaluating for values of that approach 1 from the left and from the right, (b) by reasoning as in Example 9, and ; (c) from a graph of . 34. (a) Find the vertical asymptotes of the function ; (b) Conﬁrm your answer to part (a) by graphing the function. (a) Estimate the value of the limit to ﬁve decimal places. Does this number look familiar? ; (b) Illustrate part (a) by graphing the function . ; 36. (a) By graphing the function and zooming in toward the point where the graph crosses the y-axis, estimate the value of . (b) Check your answer in part (a) by evaluating for val-ues of x that approach 0. f !x" lim x l 0 f !x" f !x" ! !tan 4x"&x y ! !1 ! x"1&x lim x l 0 !1 ! x"1&x 35. y ! x 2 ! 1 3x " 2x 2 f x f !x" ! 1&!x 3 " 1" lim x l 1! 1 x 3 " 1 lim x l 1" 1 x 3 " 1 lim x l 2" x 2 " 2x x 2 " 4x ! 4 lim x l 2%" x csc x lim x l %" cot x lim x l 3! ln!x 2 " 9" lim x l 5" e x !x " 5"3 lim x l 1 2 " x !x " 1"2 27. lim x l "3" x ! 2 x ! 3 lim x l "3! x ! 2 x ! 3 lim x l 0 9 x " 5 x x lim x l 1 x6 " 1 x10 " 1 lim x l 0 tan 3x tan 5x lim x l 0 sx ! 4 " 2 x lim x l 0! x ln!x ! x2" (0.01 (0.05 (0.1 (0.5 x ! (1 lim x l 0 ex " 1 " x x 2 "2, "1.5, "1.1, "1.01, "1.001 x ! 0, "0.5, "0.9, "0.95, "0.99, "0.999, lim x l "1 x 2 " 2x x 2 " x " 2 stream after hours. Find and and explain the signiﬁcance of these one-sided limits. ; Use the graph of the function to state the value of each limit, if it exists. If it does not exist, explain why. (a) (b) (c) 12. Sketch the graph of the following function and use it to deter-mine the values of for which exists: 13–16 Sketch the graph of an example of a function that satisﬁes all of the given conditions. 13. , , 14. , , , , , is undeﬁned , , , , 16. , , , , 17–20 Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places). 17. , 1.9, 1.95, 1.99, 1.995, 1.999 x ! 2.5, 2.1, 2.05, 2.01, 2.005, 2.001, lim x l 2 x 2 " 2x x 2 " x " 2 f !4" ! "1 f !1" ! 1 lim x l 4! f !x" ! "3 lim x l 4" f !x" ! 3 lim x l 1 f !x" ! 3 f !"2" ! 1 f !3" ! 3 lim x l "2 f !x" ! 2 lim x l 3" f !x" ! 2 lim x l 3! f !x" ! 4 15. f !0" f !2" ! 1 lim x l 2! f !x" ! 1 lim x l 2" f !x" ! 0 lim x l 0! f !x" ! "1 lim x l 0" f !x" ! 1 f !1" ! 2 lim x l 1! f !x" ! "2 lim x l 1" f !x" ! 2 f f !x" !# 2 " x x !x " 1"2 if x # "1 if "1 ) x # 1 if x $ 1 limx l a f !x" a lim x l 0 f !x" lim x l 0! f !x" lim x l 0" f !x" f !x" ! 1&!1 ! e1&x" 11. 4 8 12 16 t f(t) 150 0 300 lim tl 12! f !t" lim tl 12" f !t" t SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS \|\|\|\| 99 CALCULATING LIMITS USING THE LIMIT LAWS In Section 2.2 we used calculators and graphs to guess the values of limits, but we saw that such methods don’t always lead to the correct answer. In this section we use the following properties of limits, called the Limit Laws, to calculate limits. LIMIT LAWS Suppose that is a constant and the limits exist. Then 1. 2. 3. 4. 5. lim x l a f!x" t!x" ! lim x l a f!x" lim x l a t!x" if lim x l a t!x" " 0 lim x l a ' f!x"t!x"( ! lim x l a f!x" ! lim x l a t!x" lim x l a 'cf!x"( ! c lim x l a f!x" lim x l a ' f!x" " t!x"( ! lim x l a f!x" " lim x l a t!x" lim x l a ' f!x" ! t!x"( ! lim x l a f!x" ! lim x l a t!x" lim x l a t!x" and lim x l a f!x" c 2.3 the origin several times. Comment on the behavior of this function. 40. In the theory of relativity, the mass of a particle with velocity is where is the mass of the particle at rest and is the speed of light. What happens as ? ; 41. Use a graph to estimate the equations of all the vertical asymptotes of the curve Then ﬁnd the exact equations of these asymptotes. ; (a) Use numerical and graphical evidence to guess the value of the limit (b) How close to 1 does have to be to ensure that the func-tion in part (a) is within a distance 0.5 of its limit? x lim x l 1 x3 " 1 sx " 1 42. y ! tan!2 sin x" "% ) x ) % v l c" c m0 m ! m0 s1 " v2&c2 v 37. (a) Evaluate the function for 1, 0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of (b) Evaluate for ! 0.04, 0.02, 0.01, 0.005, 0.003, and 0.001. Guess again. 38. (a) Evaluate for , 0.5, 0.1, 0.05, 0.01, and 0.005. (b) Guess the value of . (c) Evaluate for successively smaller values of until you ﬁnally reach a value of for . Are you still con-ﬁdent that your guess in part (b) is correct? Explain why you eventually obtained a value of . (In Section 4.4 a method for evaluating the limit will be explained.) ; (d) Graph the function h in the viewing rectangle by . Then zoom in toward the point where the graph crosses the y-axis to estimate the limit of as x approaches 0. Continue to zoom in until you observe distortions in the graph of h. Compare with the results of part (c). ; 39. Graph the function of Example 4 in the viewing rectangle by . Then zoom in toward '"1, 1( '"1, 1( f !x" ! sin!%&x" h!x" '0, 1( '"1, 1( 0 h!x" 0 x h!x" lim x l 0 tan x " x x 3 x ! 1 h!x" ! !tan x " x"&x 3 x f !x" lim x l 0 $x 2 " 2x 1000% x ! f !x" ! x 2 " !2x&1000" These ﬁve laws can be stated verbally as follows: SUM LAW 1. The limit of a sum is the sum of the limits. DIFFERENCE LAW 2. The limit of a difference is the difference of the limits. CONSTANT MULTIPLE LAW 3. The limit of a constant times a function is the constant times the limit of the function. PRODUCT LAW 4. The limit of a product is the product of the limits. QUOTIENT LAW 5. The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0). It is easy to believe that these properties are true. For instance, if is close to and is close to , it is reasonable to conclude that is close to . This gives us an intuitive basis for believing that Law 1 is true. In Section 2.4 we give a precise def-inition of a limit and use it to prove this law. The proofs of the remaining laws are given in Appendix F. EXAMPLE 1 Use the Limit Laws and the graphs of and t in Figure 1 to evaluate the following limits, if they exist. (a) (b) (c) SOLUTION (a) From the graphs of and t we see that Therefore, we have (by Law 1) (by Law 3) (b) We see that . But does not exist because the left and right limits are different: So we can’t use Law 4 for the desired limit. But we can use Law 4 for the one-sided limits: The left and right limits aren’t equal, so does not exist. (c) The graphs show that Because the limit of the denominator is 0, we can’t use Law 5. The given limit does not exist because the denominator approaches 0 while the numerator approaches a nonzero number. M lim x l 2 t!x" ! 0 and lim x l 2 f!x" ) 1.4 lim x l 1 ' f!x"t!x"( lim x l 1! ' f!x"t!x"( ! 2 ! !"1" ! "2 lim x l 1" ' f!x"t!x"( ! 2 ! !"2" ! "4 lim x l 1! t!x" ! "1 lim x l 1" t!x" ! "2 lim x l 1 t!x" lim x l 1 f!x" ! 2 ! 1 ! 5!"1" ! "4 ! lim x l "2 f!x" ! 5 lim x l "2 t!x" lim x l "2 ' f!x" ! 5t!x"( ! lim x l "2 f!x" ! lim x l "2 '5t!x"( lim x l "2 t!x" ! "1 and lim x l "2 f!x" ! 1 f lim x l 2 f!x" t!x" lim x l 1 ' f!x"t!x"( lim x l "2 ' f!x" ! 5t!x"( f L ! M f!x" ! t!x" M t!x" L f!x" 100 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES FIGURE 1 x y 0 f g 1 1 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS \|\|\|\| 101 If we use the Product Law repeatedly with , we obtain the following law. POWER LAW 6. where is a positive integer In applying these six limit laws, we need to use two special limits: 7. 8. These limits are obvious from an intuitive point of view (state them in words or draw graphs of and ), but proofs based on the precise deﬁnition are requested in the exercises for Section 2.4. If we now put in Law 6 and use Law 8, we get another useful special limit. 9. where is a positive integer A similar limit holds for roots as follows. (For square roots the proof is outlined in Exer-cise 37 in Section 2.4.) 10. where is a positive integer (If is even, we assume that .) More generally, we have the following law, which is proved as a consequence of Law 10 in Section 2.5. ROOT LAW 11. where is a positive integer [If is even, we assume that ] EXAMPLE 2 Evaluate the following limits and justify each step. (a) (b) SOLUTION (a) (by Laws 2 and 1) (by 3) (by 9, 8, and 7) ! 39 ! 2!52" " 3!5" ! 4 ! 2 lim x l 5 x 2 " 3 lim x l 5 x ! lim x l 5 4 lim x l 5 !2x 2 " 3x ! 4" ! lim x l 5 !2x 2" " lim x l 5 !3x" ! lim x l 5 4 lim x l "2 x 3 ! 2x 2 " 1 5 " 3x lim x l 5 !2x 2 " 3x ! 4" lim x l a f!x" & 0. n n lim x l as n f!x) ! s n lim x l a f!x) a & 0 n n lim x l a s n x ! s n a n lim x l a x n ! a n f!x" ! x y ! x y ! c lim x l a x ! a lim x l a c ! c n lim x l a ' f!x"(n ! [lim x l a f!x"] n t!x" ! f!x" (b) We start by using Law 5, but its use is fully justiﬁed only at the ﬁnal stage when we see that the limits of the numerator and denominator exist and the limit of the denomina-tor is not 0. (by Law 5) (by 1, 2, and 3) (by 9, 8, and 7) M If we let , then . In other words, we would have gotten the correct answer in Example 2(a) by substituting 5 for x. Similarly, direct substi-tution provides the correct answer in part (b). The functions in Example 2 are a polynomial and a rational function, respectively, and similar use of the Limit Laws proves that direct substitution always works for such functions (see Exercises 53 and 54). We state this fact as follows. DIRECT SUBSTITUTION PROPERTY If is a polynomial or a rational function and is in the domain of , then Functions with the Direct Substitution Property are called continuous at a and will be studied in Section 2.5. However, not all limits can be evaluated by direct substitution, as the following examples show. EXAMPLE 3 Find . SOLUTION Let . We can’t ﬁnd the limit by substituting because isn’t deﬁned. Nor can we apply the Quotient Law, because the limit of the denominator is 0. Instead, we need to do some preliminary algebra. We factor the numer-ator as a difference of squares: The numerator and denominator have a common factor of . When we take the limit as approaches 1, we have and so . Therefore we can cancel the com-mon factor and compute the limit as follows: The limit in this example arose in Section 2.1 when we were trying to ﬁnd the tangent to the parabola at the point . M In Example 3 we were able to compute the limit by replacing the given function by a simpler function, , with the same limit. This is t!x" ! x ! 1 f!x" ! !x 2 " 1"#!x " 1" NOTE !1, 1" y ! x 2 ! 1 ! 1 ! 2 ! lim x l 1 !x ! 1" lim x l 1 x 2 " 1 x " 1 ! lim x l 1 !x " 1"!x ! 1" x " 1 x " 1 " 0 x " 1 x x " 1 x 2 " 1 x " 1 ! !x " 1"!x ! 1" x " 1 f!1" x ! 1 f!x" ! !x 2 " 1"#!x " 1" lim x l 1 x 2 " 1 x " 1 lim x l a f!x" ! f!a" f a f f!5" ! 39 f!x" ! 2x 2 " 3x ! 4 NOTE ! " 1 11 ! !"2"3 ! 2!"2"2 " 1 5 " 3!"2" ! lim x l "2 x 3 ! 2 lim x l "2 x 2 " lim x l "2 1 lim x l "2 5 " 3 lim x l "2 x lim x l "2 x 3 ! 2x 2 " 1 5 " 3x ! lim x l "2 !x 3 ! 2x 2 " 1" lim x l "2 !5 " 3x" 102 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES Isaac Newton was born on Christmas Day in 1642, the year of Galileo’s death. When he entered Cambridge University in 1661 Newton didn’t know much mathematics, but he learned quickly by reading Euclid and Descartes and by attending the lectures of Isaac Barrow. Cambridge was closed because of the plague in 1665 and 1666, and Newton returned home to reﬂect on what he had learned. Those two years were amazingly productive for at that time he made four of his major discoveries: (1) his representation of functions as sums of inﬁnite series, including the binomial theorem; (2) his work on differential and integral calculus; (3) his laws of motion and law of universal gravitation; and (4) his prism experiments on the nature of light and color. Because of a fear of controversy and criticism, he was reluctant to publish his dis-coveries and it wasn’t until 1687, at the urging of the astronomer Halley, that Newton published Principia Mathematica. In this work, the greatest scientiﬁc treatise ever written, Newton set forth his version of calculus and used it to investigate mechanics, ﬂuid dynamics, and wave motion, and to explain the motion of planets and comets. The beginnings of calculus are found in the calculations of areas and volumes by ancient Greek scholars such as Eudoxus and Archimedes. Although aspects of the idea of a limit are implicit in their “method of exhaustion,” Eudoxus and Archimedes never explicitly formulated the concept of a limit. Likewise, mathematicians such as Cavalieri, Fermat, and Barrow, the imme-diate precursors of Newton in the development of calculus, did not actually use limits. It was Isaac Newton who was the ﬁrst to talk explicitly about limits. He explained that the main idea behind limits is that quantities “approach nearer than by any given difference.” Newton stated that the limit was the basic concept in calculus, but it was left to later mathematicians like Cauchy to clarify his ideas about limits. NEWTON AND LIMITS SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS \|\|\|\| 103 valid because except when , and in computing a limit as approaches 1 we don’t consider what happens when is actually equal to 1. In general, we have the fol-lowing useful fact. If when , then , provided the limits exist. EXAMPLE 4 Find where SOLUTION Here is deﬁned at and , but the value of a limit as approaches 1 does not depend on the value of the function at 1. Since for , we have M Note that the values of the functions in Examples 3 and 4 are identical except when (see Figure 2) and so they have the same limit as approaches 1. EXAMPLE 5 Evaluate . SOLUTION If we deﬁne then, as in Example 3, we can’t compute by letting since is undeﬁned. But if we simplify algebraically, we ﬁnd that (Recall that we consider only when letting approach 0.) Thus M EXAMPLE 6 Find . SOLUTION We can’t apply the Quotient Law immediately, since the limit of the denomi-nator is 0. Here the preliminary algebra consists of rationalizing the numerator: This calculation conﬁrms the guess that we made in Example 2 in Section 2.2. M ! lim t l 0 1 st 2 ! 9 ! 3 ! 1 slim t l 0 !t 2 ! 9" ! 3 ! 1 3 ! 3 ! 1 6 ! lim t l 0 !t 2 ! 9" " 9 t 2(st 2 ! 9 ! 3) ! lim t l 0 t 2 t 2(st 2 ! 9 ! 3) lim t l 0 st 2 ! 9 " 3 t 2 ! lim t l 0 st 2 ! 9 " 3 t 2 ! st 2 ! 9 ! 3 st 2 ! 9 ! 3 lim t l 0 st 2 ! 9 " 3 t 2 lim h l 0 !3 ! h"2 " 9 h ! lim h l 0 !6 ! h" ! 6 h h " 0 F!h" ! !9 ! 6h ! h 2" " 9 h ! 6h ! h 2 h ! 6 ! h F!h" F!0" h ! 0 lim h l 0 F!h" F!h" ! !3 ! h"2 " 9 h lim h l 0 !3 ! h"2 " 9 h V x x ! 1 lim x l 1 t!x" ! lim x l 1 !x ! 1" ! 2 x " 1 t!x" ! x ! 1 x t!1" ! # x ! 1 t t!x" !$ x ! 1 # if x " 1 if x ! 1 lim x l1 t!x" lim x l a f!x" ! lim x l a t!x" x " a f!x" ! t!x" x x x ! 1 f!x" ! t!x" y=© 1 2 3 1 x y 0 2 3 y=ƒ 1 2 3 1 x y 0 2 3 FIGURE 2 The graphs of the functions f (from Example 3) and g (from Example 4) Some limits are best calculated by ﬁrst ﬁnding the left- and right-hand limits. The fol-lowing theorem is a reminder of what we discovered in Section 2.2. It says that a two-sided limit exists if and only if both of the one-sided limits exist and are equal. THEOREM if and only if When computing one-sided limits, we use the fact that the Limit Laws also hold for one-sided limits. EXAMPLE 7 Show that . SOLUTION Recall that Since for , we have For we have and so Therefore, by Theorem 1, M EXAMPLE 8 Prove that does not exist. SOLUTION Since the right- and left-hand limits are different, it follows from Theorem 1 that does not exist. The graph of the function is shown in Figure 4 and supports the one-sided limits that we found. M EXAMPLE 9 If determine whether exists. SOLUTION Since for , we have lim x l 4! f!x" ! lim x l 4! sx " 4 ! s4 " 4 ! 0 x $ 4 f!x" ! sx " 4 lim x l 4 f!x" f!x" !$ sx " 4 8 " 2x if x $ 4 if x % 4 f!x" ! %x%#x lim x l 0 %x%#x lim x l 0" %x% x ! lim x l 0" "x x ! lim x l 0" !"1" ! "1 lim x l 0! %x% x ! lim x l 0! x x ! lim x l 0! 1 ! 1 lim x l 0 %x% x V lim x l 0 %x% ! 0 lim xl 0" %x% ! lim xl 0" !"x" ! 0 %x% ! "x x % 0 lim x l 0! %x% ! lim x l 0! x ! 0 x $ 0 %x% ! x %x% !$ x "x if x & 0 if x % 0 lim x l 0 %x% ! 0 lim x l a" f!x" ! L ! lim x l a! f!x" lim x l a f!x" ! L 1 104 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES N The result of Example 7 looks plausible from Figure 3. FIGURE 3 y x 0 y=\|x\| 1 _1 x y 0 y= \|x\| x FIGURE 4 N It is shown in Example 3 in Section 2.4 that . lim x l 0! sx ! 0 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS \|\|\|\| 105 Since for , we have The right- and left-hand limits are equal. Thus the limit exists and The graph of is shown in Figure 5. M EXAMPLE 10 The greatest integer function is deﬁned by the largest integer that is less than or equal to . (For instance, , , , , ) Show that does not exist. SOLUTION The graph of the greatest integer function is shown in Figure 6. Since for , we have Since for , we have Because these one-sided limits are not equal, does not exist by Theorem 1. M The next two theorems give two additional properties of limits. Their proofs can be found in Appendix F. THEOREM If when is near (except possibly at ) and the limits of and both exist as approaches , then THE SQUEEZE THEOREM If when is near (except possibly at ) and then The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the Pinch-ing Theorem, is illustrated by Figure 7. It says that if is squeezed between and near , and if and have the same limit at , then is forced to have the same limit at . a L t a L h f a h!x" f!x" t!x" lim x l a t!x" ! L lim x l a f!x" ! lim x l a h!x" ! L a a x f!x" ' t!x" ' h!x" 3 lim x l a f!x" ' lim x l a t!x" a x t f a a x f!x" ' t!x" 2 lim x l3 &x' lim x l 3" &x' ! lim x l 3" 2 ! 2 2 ' x % 3 &x' ! 2 lim x l 3! &x' ! lim x l 3! 3 ! 3 3 ' x % 4 &x' ! 3 lim x l3 &x' &" 1 2' ! "1. &s2 ' ! 1 &#' ! 3 &4.8' ! 4 &4' ! 4 x &x' ! f lim x l 4 f!x" ! 0 lim x l 4" f!x" ! lim x l 4" !8 " 2x" ! 8 " 2 ! 4 ! 0 x % 4 f!x" ! 8 " 2x FIGURE 5 4 x y 0 N Other notations for are and . The greatest integer function is sometimes called the ﬂoor function. x (x) &x' y=[x] 1 2 3 1 2 3 4 4 5 x y 0 FIGURE 6 Greatest integer function 0 x y a L f g h FIGURE 7 EXAMPLE 11 Show that . SOLUTION First note that we cannot use \| because does not exist (see Example 4 in Section 2.2). However, since we have, as illustrated by Figure 8, We know that Taking , , and in the Squeeze Theorem, we obtain M lim x l 0 x 2 sin 1 x ! 0 h!x" ! x 2 t!x" ! x 2 sin!1#x" f!x" ! "x 2 lim x l 0 !"x 2" ! 0 and lim x l 0 x 2 ! 0 "x 2 ' x 2 sin 1 x ' x 2 "1 ' sin 1 x ' 1 lim x l 0 sin!1#x" lim x l 0 x 2 sin 1 x ! lim x l 0 x 2 ! lim x l 0 sin 1 x lim x l 0 x 2 sin 1 x ! 0 V y=≈ y=≈ 0 x y FIGURE 8 y=≈ sin(1/x) (c) (d) (e) (f) 3–9 Evaluate the limit and justify each step by indicating the appropriate Limit Law(s). 3. 4. 5. 6. 7. 9. 10. (a) What is wrong with the following equation? x 2 ! x " 6 x " 2 ! x ! 3 lim x l 4" s16 " x 2 lim u l"2 su 4 ! 3u ! 6 8. lim x l 1 1 ! 3x 1 ! 4x2 ! 3x4+ 3 lim t l "1 !t2 ! 1"3!t ! 3"5 lim x l 8 (1 ! s 3 x )!2 " 6x 2 ! x 3" lim x l 2 2x 2 ! 1 x 2 ! 6x " 4 lim x l "2 !3x4 ! 2x 2 " x ! 1" lim x l 1 s3 ! f !x" lim x l 2 (x 3f !x") lim x l "1 f !x" t!x" lim x l 0 ( f !x"t!x") 1. Given that ﬁnd the limits that exist. If the limit does not exist, explain why. (a) (b) (c) (d) (e) (f) 2. The graphs of and t are given. Use them to evaluate each limit, if it exists. If the limit does not exist, explain why. (a) (b) lim x l 1 ( f !x" ! t!x") lim x l 2 ( f !x" ! t!x") x 1 y y=ƒ 1 0 x y 1 y=© 1 f lim x l 2 t!x"h!x" f !x" lim x l 2 t!x" h!x" lim x l 2 3f !x" t!x" lim x l 2 sf !x" lim x l 2 (t!x")3 lim x l 2 ( f !x" ! 5t!x") lim x l 2 h!x" ! 0 lim x l 2 t!x" ! "2 lim x l 2 f !x" ! 4 EXERCISES 2.3 106 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS \|\|\|\| 107 functions , and on the same screen. ; 34. Use the Squeeze Theorem to show that Illustrate by graphing the functions and (in the notation of the Squeeze Theorem) on the same screen. If for , ﬁnd . 36. If for all , evaluate . 37. Prove that 38. Prove that . 39–44 Find the limit, if it exists. If the limit does not exist, explain why. 40. 41. 42. 43. 44. 45. The signum (or sign) function, denoted by sgn, is deﬁned by (a) Sketch the graph of this function. (b) Find each of the following limits or explain why it does not exist. (i) (ii) (iii) (iv) 46. Let (a) Find and (b) Does exist? (c) Sketch the graph of . 47. Let . (a) Find (i) (ii) lim x l 1" F!x" lim x l 1! F!x" F!x" ! x 2 " 1 % x " 1% f lim x l2 f !x" lim x l2! f !x". lim x l2" f !x" f !x" !$ 4 " x 2 x " 1 if x ' 2 if x $ 2 lim x l 0 % sgn x% lim x l 0 sgn x lim x l 0" sgn x lim x l 0! sgn x sgn x !$ "1 "0 "1 if x % 0 if x ! 0 if x $ 0 lim x l 0! 1 x " 1 % x%+ lim x l 0" 1 x " 1 % x%+ lim x l "2 2 " % x% 2 ! x lim x l 0.5" 2x " 1 % 2x 3 " x 2% lim x l "6 2x ! 12 % x ! 6% lim x l 3 (2x ! % x " 3%) 39. lim x l 0! sx esin!##x" ! 0 lim x l 0 x 4 cos 2 x ! 0. limx l 1 t!x" x 2x ' t!x" ' x 4 " x 2 ! 2 limx l 4 f !x" x & 0 4x " 9 ' f !x" ' x 2 " 4x ! 7 35. h f, t, lim x l 0 sx 3 ! x 2 sin # x ! 0 h!x" ! x 2 f !x" ! "x 2, t!x" ! x 2 cos 20#x (b) In view of part (a), explain why the equation is correct. 11–30 Evaluate the limit, if it exists. 11. 12. 13. 14. 16. 17. 18. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. ; 31. (a) Estimate the value of by graphing the function . (b) Make a table of values of for x close to 0 and guess the value of the limit. (c) Use the Limit Laws to prove that your guess is correct. ; 32. (a) Use a graph of to estimate the value of to two decimal places. (b) Use a table of values of to estimate the limit to four decimal places. (c) Use the Limit Laws to ﬁnd the exact value of the limit. ; 33. Use the Squeeze Theorem to show that . Illustrate by graphing the limx l 0 !x 2 cos 20#x" ! 0 f !x" limx l 0 f !x" f !x" ! s3 ! x " s3 x f !x" f !x" ! x#(s1 ! 3x " 1) lim x l 0 x s1 ! 3x " 1 lim x l "4 sx 2 ! 9 " 5 x ! 4 lim t l 0 1 ts1 ! t " 1 t+ lim h l 0 !3 ! h""1 " 3"1 h lim x l 16 4 " sx 16x " x 2 lim t l 0 1 t " 1 t 2 ! t+ lim x l "4 1 4 ! 1 x 4 ! x lim x l "1 x 2 ! 2x ! 1 x 4 " 1 lim x l 7 sx ! 2 " 3 x " 7 lim h l 0 s1 ! h " 1 h lim t l 9 9 " t 3 " st lim h l 0 !2 ! h"3 " 8 h 20. lim x l "2 x ! 2 x 3 ! 8 19. lim x l 1 x 3 " 1 x 2 " 1 lim h l 0 !4 ! h"2 " 16 h lim x l "1 x 2 " 4x x 2 " 3x " 4 lim t l "3 t 2 " 9 2t 2 ! 7t ! 3 15. lim x l 4 x 2 " 4x x 2 " 3x " 4 lim x l 2 x 2 " x ! 6 x " 2 lim x l "4 x 2 ! 5x ! 4 x 2 ! 3x " 4 lim x l 2 x 2 ! x " 6 x " 2 lim x l 2 x 2 ! x " 6 x " 2 ! lim x l 2 !x ! 3" 55. If , ﬁnd . 56. If , ﬁnd the following limits. (a) (b) 57. If prove that . Show by means of an example that may exist even though neither nor exists. 59. Show by means of an example that may exist even though neither nor exists. 60. Evaluate . Is there a number a such that exists? If so, ﬁnd the value of a and the value of the limit. 62. The ﬁgure shows a ﬁxed circle with equation and a shrinking circle with radius and center the origin. P is the point , Q is the upper point of intersection of the two circles, and R is the point of intersection of the line PQ and the -axis. What happens to R as shrinks, that is, as ? x y 0 P Q C™ C¡ R r l 0! C2 x !0, r" r C2 !x " 1"2 ! y 2 ! 1 C1 lim x l"2 3x 2 ! ax ! a ! 3 x 2 ! x " 2 61. lim x l 2 s6 " x " 2 s3 " x " 1 limx l a t!x" limx l a f !x" limx l a ( f !x"t!x") limx l a t!x" limx l a f !x" limx l a ( f !x" ! t!x") 58. lim x l 0 f !x" ! 0 f !x" !$ x 2 0 if x is rational if x is irrational lim x l 0 f !x" x lim x l 0 f !x" lim x l 0 f !x" x 2 ! 5 lim x l 1 f !x" lim x l 1 f !x" " 8 x " 1 ! 10 (b) Does exist? (c) Sketch the graph of . 48. Let (a) Evaluate each of the following limits, if it exists. (i) (ii) (iii) (iv) (v) (vi) (b) Sketch the graph of . (a) If the symbol denotes the greatest integer function deﬁned in Example 10, evaluate (i) (ii) (iii) (b) If n is an integer, evaluate (i) (ii) (c) For what values of does exist? 50. Let , . (a) Sketch the graph of (b) Evaluate each limit, if it exists. (i) (ii) (iii) (iv) (c) For what values of does exist? 51. If , show that exists but is not equal to . 52. In the theory of relativity, the Lorentz contraction formula expresses the length L of an object as a function of its veloc-ity with respect to an observer, where is the length of the object at rest and c is the speed of light. Find and interpret the result. Why is a left-hand limit necessary? 53. If is a polynomial, show that . 54. If r is a rational function, use Exercise 53 to show that for every number a in the domain of r. limx l a r!x" ! r!a" lim xl a p!x" ! p!a" p lim v lc" L L0 v L ! L0s1 " v 2#c 2 f !2" limx l 2 f !x" f !x" ! &x' ! &"x' limx l a f !x" a lim x l ##2 f !x" lim x l !##2"! f !x" lim x l !##2"" f !x" lim x l 0 f !x" f. "# ' x ' # f !x" ! &cos x' limx l a &x' a lim x l n! &x' lim x l n" &x' lim x l "2.4 &x' lim x l "2 &x' lim x l "2! &x' & ' 49. t lim x l 2 t!x" lim x l 2! t!x" lim x l 2" t!x" t!1" lim x l 1 t!x" lim x l 1" t!x" x 3 2 " x 2 x " 3 if x % 1 if x ! 1 if 1 % x ' 2 if x $ 2 t!x" ! F limx l 1 F!x" 108 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT \|\|\|\| 109 THE PRECISE DEFINITION OF A LIMIT The intuitive deﬁnition of a limit given in Section 2.2 is inadequate for some purposes because such phrases as “ is close to 2” and “ gets closer and closer to L” are vague. In order to be able to prove conclusively that we must make the deﬁnition of a limit precise. To motivate the precise deﬁnition of a limit, let’s consider the function Intuitively, it is clear that when is close to 3 but , then is close to 5, and so . To obtain more detailed information about how varies when is close to 3, we ask the following question: How close to 3 does have to be so that differs from 5 by less than 0.l? The distance from to 3 is and the distance from to 5 is , so our problem is to ﬁnd a number such that If , then , so an equivalent formulation of our problem is to ﬁnd a num-ber such that Notice that if , then that is, Thus an answer to the problem is given by ; that is, if is within a distance of 0.05 from 3, then will be within a distance of 0.1 from 5. If we change the number 0.l in our problem to the smaller number 0.01, then by using the same method we ﬁnd that will differ from 5 by less than 0.01 provided that dif-fers from 3 by less than (0.01)#2 ! 0.005: Similarly, The numbers and that we have considered are error tolerances that we might allow. For 5 to be the precise limit of as approaches 3, we must not only be able to bring the difference between and 5 below each of these three numbers; we f!x" x f!x" 0.001 0.1, 0.01, 0 % %x " 3% % 0.0005 if % f!x" " 5% % 0.001 0 % %x " 3% % 0.005 if % f!x" " 5% % 0.01 x f!x" f!x" x ( ! 0.05 0 % %x " 3% % 0.05 if % f!x" " 5% % 0.1 % f!x" " 5% ! %!2x " 1" " 5% ! %2x " 6% ! 2%x " 3% % 0.1 0 % %x " 3% % !0.1"#2 ! 0.05 0 % %x " 3% % ( if % f!x" " 5% % 0.1 ( x " 3 %x " 3% $ 0 but x " 3 %x " 3% % ( if % f!x" " 5% % 0.1 ( % f!x" " 5% f!x" %x " 3% x f !x" x x f!x" lim x l3 f!x" ! 5 f!x" " 3 x x f!x" !$ 2x " 1 6 if x " 3 if x ! 3 lim x l 0 sin x x ! 1 or lim x l 0 x 3 ! cos 5x 10,000+ ! 0.0001 f!x" x 2.4 N It is traditional to use the Greek letter (delta) in this situation. ( must be able to bring it below any positive number. And, by the same reasoning, we can! If we write (the Greek letter epsilon) for an arbitrary positive number, then we ﬁnd as before that This is a precise way of saying that is close to 5 when is close to 3 because (1) says that we can make the values of within an arbitrary distance from 5 by taking the val-ues of within a distance from 3 (but ). Note that (1) can be rewritten as follows: then and this is illustrated in Figure 1. By taking the values of ( ) to lie in the interval we can make the values of lie in the interval . Using (1) as a model, we give a precise deﬁnition of a limit. DEFINITION Let be a function deﬁned on some open interval that contains the number , except possibly at itself. Then we say that the limit of as approaches is L, and we write if for every number there is a number such that then Since is the distance from to and is the distance from to , and since can be arbitrarily small, the deﬁnition of a limit can be expressed in words as follows: means that the distance between and can be made arbitrarily small by taking the distance from to sufﬁciently small (but not 0). Alternatively, means that the values of can be made as close as we please to by taking close enough to (but not equal to ). We can also reformulate Deﬁnition 2 in terms of intervals by observing that the in-equality is equivalent to , which in turn can be written as . Also is true if and only if , that is, . Similarly, the inequality is equivalent to the pair of inequalities . Therefore, in terms of intervals, Deﬁnition 2 can be stated as follows: means that for every (no matter how small is) we can ﬁnd such that if lies in the open interval and , then lies in the open interval We interpret this statement geometrically by representing a function by an arrow diagram as in Figure 2, where maps a subset of onto another subset of . ! ! f !L " ), L ! )". f !x" x " a !a " (, a ! (" x ( $ 0 ) ) $ 0 lim x l a f !x" ! L L ! ) % f!x" L " ) % % f!x" " L% % ) x " a x " a " 0 0 % %x " a% a " ( % x % a ! ( "( % x " a % ( %x " a% % ( a a x L f !x" lim x l a f !x" ! L a x L f !x" lim x l a f !x" ! L ) L f!x" % f!x" " L% a x %x " a% % f!x" " L% % ) 0 % %x " a% % ( if ( $ 0 ) $ 0 lim x l a f!x" ! L a x f!x" a a f 2 !5 " ), 5 ! )" f!x" !3 " (, 3 ! (" " 3 x 5 " ) % f!x" % 5 ! ) !x " 3" 3 " ( % x % 3 ! ( if x " 3 )#2 x ) f!x" x f!x" 0 % %x " 3% % ( ! ) 2 if % f!x" " 5% % ) 1 ) 110 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES FIGURE 1 0 x y 5+∑ 5 5-∑ 3 3+∂ 3-∂ ƒ is in here when x is in here (x≠3) SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT \|\|\|\| 111 The deﬁnition of limit says that if any small interval is given around , then we can ﬁnd an interval around such that maps all the points in (except possibly ) into the interval . (See Figure 3.) Another geometric interpretation of limits can be given in terms of the graph of a func-tion. If is given, then we draw the horizontal lines and and the graph of . (See Figure 4.) If , then we can ﬁnd a number such that if we restrict to lie in the interval and take , then the curve lies between the lines and . (See Figure 5.) You can see that if such a has been found, then any smaller will also work. It is important to realize that the process illustrated in Figures 4 and 5 must work for every positive number , no matter how small it is chosen. Figure 6 shows that if a smaller is chosen, then a smaller may be required. EXAMPLE 1 Use a graph to ﬁnd a number such that if then In other words, ﬁnd a number that corresponds to in the deﬁnition of a limit for the function with and . SOLUTION A graph of is shown in Figure 7; we are interested in the region near the point . Notice that we can rewrite the inequality as 1.8 % x 3 " 5x ! 6 % 2.2 %!x 3 " 5x ! 6" " 2% % 0.2 !1, 2" f L ! 2 a ! 1 f!x" ! x 3 " 5x ! 6 ) ! 0.2 ( %!x 3 " 5x ! 6" " 2% % 0.2 %x " 1% % ( ( FIGURE 4 a 0 x y y=ƒ y=L+∑ y=L-∑ ∑ ∑ L FIGURE 5 FIGURE 6 a 0 x y y=L+∑ y=L-∑ a-∂ a+∂ L+∑ L-∑ 0 x y a y=L+∑ y=L-∑ a-∂ a+∂ ∑ ∑ L when x is in here (x≠ a) ƒ is in here ( ) ) ( ( y ! L ! ) y ! L " ) y ! f!x" x " a !a " (, a ! (" x ( $ 0 lim x l a f!x" ! L f y ! L " ) y ! L ! ) ) $ 0 FIGURE 3 a-∂ a ƒ a+∂ x f L-∑ L L+∑ !L " ), L ! )" a !a " (, a ! (" f a !a " (, a ! (" L !L " ), L ! )" x a f(a) ƒ f FIGURE 2 FIGURE 7 15 _5 _3 3 So we need to determine the values of for which the curve lies between the horizontal lines and . Therefore we graph the curves , , and near the point in Figure 8. Then we use the cursor to estimate that the -coordinate of the point of intersection of the line and the curve is about . Similarly, intersects the line when . So, rounding to be safe, we can say that if This interval is not symmetric about . The distance from to the left endpoint is and the distance to the right endpoint is 0.12. We can choose to be the smaller of these numbers, that is, . Then we can rewrite our inequalities in terms of distances as follows: if This just says that by keeping within 0.08 of 1, we are able to keep within 0.2 of 2. Although we chose , any smaller positive value of would also have worked. M The graphical procedure in Example 1 gives an illustration of the deﬁnition for , but it does not prove that the limit is equal to 2. A proof has to provide a for every . In proving limit statements it may be helpful to think of the deﬁnition of limit as a chal-lenge. First it challenges you with a number . Then you must be able to produce a suit-able . You have to be able to do this for every , not just a particular . Imagine a contest between two people, A and B, and imagine yourself to be B. Person A stipulates that the ﬁxed number should be approximated by the values of to with-in a degree of accuracy (say, 0.01). Person B then responds by ﬁnding a number such that if , then . Then A may become more exacting and challenge B with a smaller value of (say, 0.0001). Again B has to respond by ﬁnding a corresponding . Usually the smaller the value of , the smaller the corresponding value of must be. If B always wins, no matter how small A makes , then EXAMPLE 2 Prove that . SOLUTION 1. Preliminary analysis of the problem (guessing a value for ). Let be a given positive number. We want to ﬁnd a number such that if But . Therefore, we want if that is, if then This suggests that we should choose . 2. Proof (showing that this works). Given , choose . If , then !"4x ! 5# ! 7! ! !4x ! 12! ! 4!x ! 3! " 4# ! 4$ $ 4% ! $ 0 " !x ! 3! " # # ! $&4 $ % 0 # # ! $&4 !x ! 3! " $ 4 0 " !x ! 3! " # 4!x ! 3! " $ then 0 " !x ! 3! " # !"4x ! 5# ! 7! ! !4x ! 12! ! !4"x ! 3#! ! 4!x ! 3! !"4x ! 5# ! 7! " $ then 0 " !x ! 3! " # # $ # lim x l3 "4x ! 5# ! 7 V lim x l a f"x# ! L. $ # $ # $ ! f"x# ! L! " $ 0 " !x ! a! " # # $ f"x# L $ $ % 0 # $ $ # $ ! 0.2 # # ! 0.08 f"x# x !"x 3 ! 5x & 6# ! 2! " 0.2 then !x ! 1! " 0.08 # ! 0.08 # 1 ! 0.92 ! 0.08 x ! 1 x ! 1 "0.92, 1.12# 1.8 " x 3 ! 5x & 6 " 2.2 then 0.92 " x " 1.12 x ' 1.124 y ! 1.8 y ! x 3 ! 5x & 6 0.911 y ! x 3 ! 5x & 6 y ! 2.2 x "1, 2# y ! 2.2 y ! 1.8 y ! x 3 ! 5x & 6 y ! 2.2 y ! 1.8 y ! x 3 ! 5x & 6 x 112 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES FIGURE 8 y=˛-5x+6 y=2.2 y=1.8 (1, 2) 0.8 1.2 2.3 1.7 Thus if Therefore, by the deﬁnition of a limit, This example is illustrated by Figure 9. M Note that in the solution of Example 2 there were two stages—guessing and proving. We made a preliminary analysis that enabled us to guess a value for . But then in the sec-ond stage we had to go back and prove in a careful, logical fashion that we had made a correct guess. This procedure is typical of much of mathematics. Sometimes it is neces-sary to ﬁrst make an intelligent guess about the answer to a problem and then prove that the guess is correct. The intuitive deﬁnitions of one-sided limits that were given in Section 2.2 can be pre-cisely reformulated as follows. DEFINITION OF LEFT-HAND LIMIT if for every number there is a number such that if DEFINITION OF RIGHT-HAND LIMIT if for every number there is a number such that if Notice that Deﬁnition 3 is the same as Deﬁnition 2 except that is restricted to lie in the left half of the interval . In Deﬁnition 4, is restricted to lie in the right half of the interval EXAMPLE 3 Use Deﬁnition 4 to prove that SOLUTION 1. Guessing a value for . Let be a given positive number. Here and , so we want to ﬁnd a number such that if that is, if sx " $ then 0 " x " # !sx ! 0! " $ then 0 " x " # # L ! 0 a ! 0 $ # lim x l 0& sx ! 0. V "a ! #, a & ##. "a, a & ## x "a ! #, a & ## "a ! #, a# x ! f"x# ! L! " $ then a " x " a & # # % 0 $ % 0 lim x l a& f"x# ! L 4 ! f"x# ! L! " $ then a ! # " x " a # % 0 $ % 0 lim x l a! f"x# ! L 3 # lim x l3 "4x ! 5# ! 7 !"4x ! 5# ! 7! " $ then 0 " !x ! 3! " # SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT \|\|\|\| 113 FIGURE 9 y 0 x 7+∑ 7 7-∑ 3-∂ 3+∂ 3 y=4x-5 After the invention of calculus in the 17th cen-tury, there followed a period of free development of the subject in the 18th century. Mathemati-cians like the Bernoulli brothers and Euler were eager to exploit the power of calculus and boldly explored the consequences of this new and wonderful mathematical theory without worrying too much about whether their proofs were com-pletely correct. The 19th century, by contrast, was the Age of Rigor in mathematics. There was a movement to go back to the foundations of the subject—to provide careful deﬁnitions and rigorous proofs. At the forefront of this movement was the French mathematician Augustin-Louis Cauchy (1789–1857), who started out as a military engi-neer before becoming a mathematics professor in Paris. Cauchy took Newton’s idea of a limit, which was kept alive in the 18th century by the French mathematician Jean d’Alembert, and made it more precise. His deﬁnition of a limit reads as follows: “When the successive values attributed to a variable approach indeﬁnitely a ﬁxed value so as to end by differing from it by as little as one wishes, this last is called the limit of all the others.” But when Cauchy used this deﬁnition in examples and proofs, he often employed delta-epsilon inequalities similar to the ones in this section. A typical Cauchy proof starts with: “Designate by and two very small numbers; . . .” He used because of the correspondence between epsilon and the French word erreur and because delta corresponds to diff´ erence. Later, the German mathematician Karl Weierstrass (1815–1897) stated the deﬁni-tion of a limit exactly as in our Deﬁnition 2. $ $ # CAUCHY AND LIMITS or, squaring both sides of the inequality , we get if This suggests that we should choose 2. Showing that this works. Given , let . If , then so According to Deﬁnition 4, this shows that . M EXAMPLE 4 Prove that . SOLUTION 1. Guessing a value for . Let be given. We have to ﬁnd a number such that if To connect with we write . Then we want if Notice that if we can ﬁnd a positive constant such that , then and we can make by taking . We can ﬁnd such a number if we restrict to lie in some interval centered at 3. In fact, since we are interested only in values of that are close to 3, it is reasonable to assume that is within a distance l from 3, that is, . Then , so . Thus we have , and so is a suitable choice for the constant. But now there are two restrictions on , namely To make sure that both of these inequalities are satisﬁed, we take to be the smaller of the two numbers 1 and . The notation for this is . 2. Showing that this works. Given , let . If , then (as in part l). We also have , so This shows that . M As Example 4 shows, it is not always easy to prove that limit statements are true using the deﬁnition. In fact, if we had been given a more complicated function such as , a proof would require a great deal of ingenuity. f"x# ! "6x 2 ! 8x & 9#&"2x 2 ! 1# $, # lim x l3 x 2 ! 9 !x 2 ! 9! ! !x & 3!!x ! 3! " 7 ! $ 7 ! $ !x ! 3! " $&7 !x ! 3! " 1 ? 2 " x " 4 ? !x & 3! " 7 0 " !x ! 3! " # # ! min(1, $&7) $ % 0 # # ! min(1, $&7) $&7 # !x ! 3! " $ C ! $ 7 and !x ! 3! " 1 !x ! 3! C ! 7 !x & 3! " 7 5 " x & 3 " 7 2 " x " 4 !x ! 3! " 1 x x x C ! # $&C " !x ! 3! C!x ! 3! " $ !x & 3!!x ! 3! " C!x ! 3! !x & 3! " C C !x & 3!!x ! 3! " $ then 0 " !x ! 3! " # !x 2 ! 9! ! !"x & 3#"x ! 3#! !x ! 3! !x 2 ! 9! !x 2 ! 9! " $ then 0 " !x ! 3! " # # % 0 $ % 0 # lim x l 3 x 2 ! 9 lim x l 0& sx ! 0 !sx ! 0! " $ sx " s# ! s$ 2 ! $ 0 " x " # # ! $2 $ % 0 # # ! $2. x " $2 then 0 " x " # sx " $ 114 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT \|\|\|\| 115 Fortunately this is unnecessary because the Limit Laws stated in Section 2.3 can be proved using Deﬁnition 2, and then the limits of complicated functions can be found rigorously from the Limit Laws without resorting to the deﬁnition directly. For instance, we prove the Sum Law: If and both exist, then The remaining laws are proved in the exercises and in Appendix F. PROOF OF THE SUM LAW Let be given. We must ﬁnd such that then Using the Triangle Inequality we can write We make less than by making each of the terms and less than . Since and , there exists a number such that if Similarly, since , there exists a number such that if Let . Notice that and so Therefore, by (5), To summarize, then Thus, by the deﬁnition of a limit, M lim x l a f"x# & t"x#+ ! L & M ! f"x# & t"x# ! "L & M#! " $ 0 " !x ! a! " # if " $ 2 & $ 2 ! $ ! f"x# & t"x# ! "L & M#! ' !f"x# ! L! & !t"x# ! M! !t"x# ! M! " $ 2 and ! f"x# ! L! " $ 2 0 " !x ! a! " # 2 and 0 " !x ! a! " #1 then 0 " !x ! a! " # if # ! min(#1, # 2) !t"x# ! M! " $ 2 then 0 " !x ! a! " # 2 # 2 % 0 lim x l a t"x# ! M ! f"x# ! L! " $ 2 then 0 " !x ! a! " #1 #1 % 0 lim x l a f"x# ! L % 0 $&2 $&2 !t"x# ! M! ! f"x# ! L! $ ! f"x# & t"x# ! "L & M#! ' ! f"x# ! L! & !t"x# ! M! ! f"x# & t"x# ! "L & M#! ! !" f"x# ! L# & "t"x# ! M#! 5 ! f"x# & t"x# ! "L & M#! " $ 0 " !x ! a! " # if # % 0 $ % 0 lim x l a f"x# & t"x#+ ! L & M lim x l a t"x# ! M lim x l a f"x# ! L N Triangle Inequality: (See Appendix A.) ! a & b! ' ! a! & ! b! INFINITE LIMITS Inﬁnite limits can also be deﬁned in a precise way. The following is a precise version of Deﬁnition 4 in Section 2.2. DEFINITION Let be a function deﬁned on some open interval that contains the number , except possibly at itself. Then means that for every positive number there is a positive number such that if This says that the values of can be made arbitrarily large (larger than any given number ) by taking close enough to (within a distance , where depends on , but with ). A geometric illustration is shown in Figure 10. Given any horizontal line , we can ﬁnd a number such that if we restrict to lie in the interval but , then the curve lies above the line . You can see that if a larger is chosen, then a smaller may be required. EXAMPLE 5 Use Deﬁnition 6 to prove that . SOLUTION Let be a given positive number. We want to ﬁnd a number such that if But So if we choose and , then . This shows that as . M Similarly, the following is a precise version of Deﬁnition 5 in Section 2.2. It is illus-trated by Figure 11. DEFINITION Let be a function deﬁned on some open interval that contains the number , except possibly at itself. Then means that for every negative number there is a positive number such that if f"x# " N then 0 " !x ! a! " # # N lim x l a f"x# ! !( a a f 7 x l 0 1&x 2 l ( 1&x 2 % M 0 " !x! " # ! 1&sM # ! 1&sM !x! " 1 sM & ? x 2 " 1 M & ? 1 x 2 % M 1&x 2 % M then 0 " !x! " # # M lim x l 0 1 x 2 ! ( V # M y ! M y ! f"x# x " a "a ! #, a & ## x # % 0 y ! M x " a M # # a x M f"x# f"x# % M then 0 " !x ! a! " # # M lim x l a f"x# ! ( a a f 6 116 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES FIGURE 10 0 x y y=M M a a+∂ a-∂ FIGURE 11 y y=N 0 x N a a+∂ a-∂ SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT \|\|\|\| 117 ; 5. Use a graph to ﬁnd a number such that if then ; 6. Use a graph to ﬁnd a number such that if then ; 7. For the limit illustrate Deﬁnition 2 by ﬁnding values of that correspond to and . ; 8. For the limit illustrate Deﬁnition 2 by ﬁnding values of that correspond to and ; 9. Given that , illustrate Deﬁnition 6 by ﬁnding values of that correspond to (a) and (b) . ; 10. Use a graph to ﬁnd a number such that if then 11. A machinist is required to manufacture a circular metal disk with area . (a) What radius produces such a disk? (b) If the machinist is allowed an error tolerance of in the area of the disk, how close to the ideal radius in part (a) must the machinist control the radius? (c) In terms of the deﬁnition of , what is ? What is ? What is ? What is ? What value of is given? What is the corresponding value of ? $ L a f "x# x limx l a f "x# ! L $, # )5 cm2 1000 cm2 x 2 sx ! 5 % 100 5 " x " 5 & # # M ! 10,000 M ! 1000 # limx l&2 tan2x ! ( $ ! 0.1. $ ! 0.5 # lim x l 0 e x ! 1 x ! 1 $ ! 0.1 $ ! 1 # lim x l 1 "4 & x ! 3x3# ! 2 , 2x x 2 & 4 ! 0.4 , " 0.1 ! x ! 1! " # # ! tan x ! 1! " 0.2 , x ! 4 , " # # x y ? 1 ? 0 1.5 1 0.5 y=≈ 1. Use the given graph of to ﬁnd a number such that if then 2. Use the given graph of to ﬁnd a number such that if then Use the given graph of to ﬁnd a number such that if then 4. Use the given graph of to ﬁnd a number such that if then ! x 2 ! 1! " 1 2 ! x ! 1! " # # f "x# ! x2 ? ? y=œ„ x x y 4 0 2 2.4 1.6 ! sx ! 2! " 0.4 ! x ! 4! " # # f "x# ! sx 3. 4 5.7 x y 5 0 3 3.6 2.4 ! f "x# ! 3! " 0.6 0 " ! x ! 5! " # # f 10 3 10 7 y= 1 x x y 2 0 1 0.5 0.7 0.3 , 1 x ! 0.5, " 0.2 ! x ! 2! " # # f "x# ! 1&x EXERCISES 2.4 33. Verify that another possible choice of for showing that in Example 4 is 34. Verify, by a geometric argument, that the largest possible choice of for showing that is . 35. (a) For the limit , use a graph to ﬁnd a value of that corresponds to (b) By using a computer algebra system to solve the cubic equation , ﬁnd the largest possible value of that works for any given . (c) Put in your answer to part (b) and compare with your answer to part (a). 36. Prove that . Prove that if Hint: 38. If is the Heaviside function deﬁned in Example 6 in Sec-tion 2.2, prove, using Deﬁnition 2, that does not exist. [Hint: Use an indirect proof as follows. Suppose that the limit is . Take in the deﬁnition of a limit and try to arrive at a contradiction.] 39. If the function is deﬁned by prove that does not exist. 40. By comparing Deﬁnitions 2, 3, and 4, prove Theorem 1 in Section 2.3. 41. How close to do we have to take so that 42. Prove, using Deﬁnition 6, that . Prove that . 44. Suppose that and , where is a real number. Prove each statement. (a) (b) if (c) if c " 0 lim x la f "x#t"x#+ ! !( c % 0 lim x la f "x#t"x#+ ! ( lim x la f "x# & t"x#+ ! ( c limx la t"x# ! c lim x l a f "x# ! ( lim x l 0& ln x ! !( 43. lim x l!3 1 "x & 3#4 ! ( 1 "x & 3#4 % 10,000 x !3 f "x# lim x l 0 f "x# !-0 1 if x is rational if x is irrational f $ ! 1 2 L lim t l 0 H"t# H Use \| sx ! sa \| ! ! x ! a! sx & sa .. / a % 0. lim x l a sx ! sa 37. lim x l2 1 x ! 1 2 $ ! 0.4 $ % 0 # x3 & x & 1 ! 3 & $ $ ! 0.4. limx l 1 "x3 & x & 1# ! 3 CAS # ! s9 & $ ! 3 limx l3 x2 ! 9 # # ! min(2, $&8). limx l3 x2 ! 9 # ; 12. A crystal growth furnace is used in research to determine how best to manufacture crystals used in electronic components for the space shuttle. For proper growth of the crystal, the temper-ature must be controlled accurately by adjusting the input power. Suppose the relationship is given by where is the temperature in degrees Celsius and is the power input in watts. (a) How much power is needed to maintain the temperature at ? (b) If the temperature is allowed to vary from by up to , what range of wattage is allowed for the input power? (c) In terms of the deﬁnition of , what is ? What is ? What is ? What is ? What value of is given? What is the corresponding value of ? 13. (a) Find a number such that if , then , where . (b) Repeat part (a) with . 14. Given that , illustrate Deﬁnition 2 by ﬁnding values of that correspond to , , and . 15–18 Prove the statement using the deﬁnition of limit and illustrate with a diagram like Figure 9. 15. 16. 18. 19–32 Prove the statement using the deﬁnition of limit. 19. 20. 21. 22. 23. 24. 26. 27. 28. 30. 32. lim x l2 x 3 ! 8 lim x l!2 "x 2 ! 1# ! 3 31. lim x l3 "x 2 & x ! 4# ! 8 lim x l2 "x 2 ! 4x & 5# ! 1 29. lim x l 9! s 4 9 ! x ! 0 lim x l 0 ! x! ! 0 lim x l 0 x 3 ! 0 lim x l 0 x 2 ! 0 25. lim x l a c ! c lim x l a x ! a lim x l!1.5 9 ! 4x 2 3 & 2x ! 6 lim x l2 x 2 & x ! 6 x ! 2 ! 5 lim x l 6 $ x 4 & 3% ! 9 2 lim x l 3 x 5 ! 3 5 $, # lim x l 4 "7 ! 3x# ! !5 lim x l !3 "1 ! 4x# ! 13 17. lim x l !2 ( 1 2 x & 3) ! 2 lim x l 1 "2x & 3# ! 5 $, # $ ! 0.01 $ ! 0.05 $ ! 0.1 # limx l 2 "5x ! 7# ! 3 $ ! 0.01 $ ! 0.1 ! 4x ! 8! " $ ! x ! 2! " # # # $ L a f "x# x limx l a f "x# ! L $, # )1+C 200+C 200+C w T T"w# ! 0.1w 2 & 2.155w & 20 118 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.5 CONTINUITY \|\|\|\| 119 CONTINUITY We noticed in Section 2.3 that the limit of a function as approaches can often be found simply by calculating the value of the function at . Functions with this property are called continuous at a. We will see that the mathematical deﬁnition of continuity corresponds closely with the meaning of the word continuity in everyday language. (A continuous process is one that takes place gradually, without interruption or abrupt change.) DEFINITION A function is continuous at a number a if Notice that Deﬁnition l implicitly requires three things if is continuous at a: 1. is deﬁned (that is, a is in the domain of ) 2. exists 3. The deﬁnition says that is continuous at if approaches as x approaches a. Thus a continuous function has the property that a small change in x produces only a small change in . In fact, the change in can be kept as small as we please by keep-ing the change in sufﬁciently small. If is deﬁned near (in other words, is deﬁned on an open interval containing , except perhaps at ), we say that is discontinuous at a (or has a discontinuity at ) if is not continuous at . Physical phenomena are usually continuous. For instance, the displacement or velocity of a vehicle varies continuously with time, as does a person’s height. But discontinuities do occur in such situations as electric currents. [See Example 6 in Section 2.2, where the Heaviside function is discontinuous at because does not exist.] Geometrically, you can think of a function that is continuous at every number in an interval as a function whose graph has no break in it. The graph can be drawn without removing your pen from the paper. EXAMPLE 1 Figure 2 shows the graph of a function f. At which numbers is f discontinu-ous? Why? SOLUTION It looks as if there is a discontinuity when a ! 1 because the graph has a break there. The ofﬁcial reason that f is discontinuous at 1 is that is not deﬁned. The graph also has a break when , but the reason for the discontinuity is differ-ent. Here, is deﬁned, but does not exist (because the left and right limits are different). So f is discontinuous at 3. What about ? Here, is deﬁned and exists (because the left and right limits are the same). But So is discontinuous at 5. M Now let’s see how to detect discontinuities when a function is deﬁned by a formula. f lim x l 5 f"x# " f"5# lim x l5 f"x# f"5# a ! 5 lim x l3 f"x# f"3# a ! 3 f"1# lim t l 0 H"t# 0 a f a f f a a f a f x f"x# f"x# f f"a# f"x# a f lim x l a f"x# ! f"a# lim x l a f"x# f f"a# f lim x l a f"x# ! f"a# f 1 a a x 2.5 N As illustrated in Figure 1, if is continuous, then the points on the graph of approach the point on the graph. So there is no gap in the curve. "a, f "a## f "x, f "x## f f(a) x 0 y a y=ƒ ƒ approaches f(a). As x approaches a, FIGURE 1 FIGURE 2 y 0 x 1 2 3 4 5 EXAMPLE 2 Where are each of the following functions discontinuous? (a) (b) (c) (d) SOLUTION (a) Notice that is not deﬁned, so f is discontinuous at 2. Later we’ll see why is continuous at all other numbers. (b) Here is deﬁned but does not exist. (See Example 8 in Section 2.2.) So f is discontinuous at 0. (c) Here is deﬁned and exists. But so is not continuous at 2. (d) The greatest integer function has discontinuities at all of the integers because does not exist if is an integer. (See Example 10 and Exercise 49 in Section 2.3.) M Figure 3 shows the graphs of the functions in Example 2. In each case the graph can’t be drawn without lifting the pen from the paper because a hole or break or jump occurs in the graph. The kind of discontinuity illustrated in parts (a) and (c) is called removable because we could remove the discontinuity by redeﬁning at just the single number 2. [The function is continuous.] The discontinuity in part (b) is called an inﬁ-nite discontinuity. The discontinuities in part (d) are called jump discontinuities because the function “jumps” from one value to another. 1 2 3 1 x y 0 (d) ƒ=[x] 1 2 1 x y 0 (c) ƒ= if x≠2 1 if x= 2 ≈-x-2 x-2 (b) ƒ= if x≠0 1 if 1 x=0 1 x y 0 1 2 x y 0 1 (a) ƒ=≈-x-2 x-2 FIGURE 3 Graphs of the functions in Example 2 ≈ t"x# ! x & 1 f n lim x ln 0x1 f"x# ! 0x1 f lim x l2 f"x# " f"2# lim x l2 f"x# ! lim x l2 x 2 ! x ! 2 x ! 2 ! lim x l2 "x ! 2#"x & 1# x ! 2 ! lim x l2 "x & 1# ! 3 f"2# ! 1 lim x l 0 f"x# ! lim x l 0 1 x 2 f"0# ! 1 f f"2# f"x# ! 0x1 f"x# !-x 2 ! x ! 2 x ! 2 if x " 2 1 if x ! 2 f"x# !-1 x 2 if x " 0 1 if x ! 0 f"x# ! x 2 ! x ! 2 x ! 2 V 120 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.5 CONTINUITY \|\|\|\| 121 DEFINITION A function is continuous from the right at a number a if and is continuous from the left at a if EXAMPLE 3 At each integer , the function [see Figure 3(d)] is continuous from the right but discontinuous from the left because but M DEFINITION A function is continuous on an interval if it is continuous at every number in the interval. (If f is deﬁned only on one side of an endpoint of the interval, we understand continuous at the endpoint to mean continuous from the right or continuous from the left.) EXAMPLE 4 Show that the function is continuous on the interval SOLUTION If , then using the Limit Laws, we have (by Laws 2 and 7) (by 11) (by 2, 7, and 9) Thus, by Deﬁnition l, is continuous at if . Similar calculations show that and so is continuous from the right at !1 and continuous from the left at 1. Therefore, according to Deﬁnition 3, is continuous on . The graph of is sketched in Figure 4. It is the lower half of the circle M Instead of always using Deﬁnitions 1, 2, and 3 to verify the continuity of a function as we did in Example 4, it is often convenient to use the next theorem, which shows how to build up complicated continuous functions from simple ones. x 2 & "y ! 1#2 ! 1 f !1, 1+ f f lim xl1! f"x# ! 1 ! f"1# lim x l !1& f"x# ! 1 ! f"!1# !1 " a " 1 a f ! f"a# ! 1 ! s1 ! a 2 ! 1 ! slim x l a "1 ! x 2# ! 1 ! lim x l a s1 ! x 2 lim x l a f"x# ! lim x l a (1 ! s1 ! x 2 ) !1 " a " 1 !1, 1+. f"x# ! 1 ! s1 ! x 2 f 3 lim x l n! f"x# ! lim x l n! 0x1 ! n ! 1 " f"n# lim x l n& f"x# ! lim x l n& 0x1 ! n ! f"n# f"x# ! 0x1 n lim x l a! f"x# ! f"a# f lim x l a& f"x# ! f"a# f 2 1 -1 1 x y 0 ƒ=1-œ„„„„„ 1-≈ FIGURE 4 THEOREM If and are continuous at and is a constant, then the follow-ing functions are also continuous at : 1. 2. 3. 4. 5. if PROOF Each of the ﬁve parts of this theorem follows from the corresponding Limit Law in Section 2.3. For instance, we give the proof of part 1. Since and are continuous at , we have Therefore (by Law 1) This shows that is continuous at . M It follows from Theorem 4 and Deﬁnition 3 that if and are continuous on an inter-val, then so are the functions , and (if is never 0) . The following theorem was stated in Section 2.3 as the Direct Substitution Property. THEOREM (a) Any polynomial is continuous everywhere; that is, it is continuous on . (b) Any rational function is continuous wherever it is deﬁned; that is, it is contin-uous on its domain. PROOF (a) A polynomial is a function of the form where are constants. We know that (by Law 7) and (by 9) This equation is precisely the statement that the function is a continuous function. Thus, by part 3 of Theorem 4, the function is continuous. Since is a sum of functions of this form and a constant function, it follows from part 1 of Theorem 4 that is continuous. P P t!x" ! cx m f!x" ! x m m ! 1, 2, . . . , n lim x l a x m ! a m lim x l a c0 ! c0 c0, c1, . . . , cn P!x" ! cnx n ! cn"1x n"1 ! # # # ! c1x ! c0 ! ! !"$, $" 5 f#t t f ! t, f " t, cf, ft t f a f ! t ! ! f ! t"!a" ! f!a" ! t!a" ! lim x l a f!x" ! lim x l a t!x" lim x l a ! f ! t"!x" ! lim x l a $ f!x" ! t!x"% lim x l a t!x" ! t!a" and lim x l a f!x" ! f!a" a t f t!a" " 0 f t ft cf f " t f ! t a c a t f 4 122 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.5 CONTINUITY \|\|\|\| 123 (b) A rational function is a function of the form where and are polynomials. The domain of is . We know from part (a) that and are continuous everywhere. Thus, by part 5 of Theorem 4, is continuous at every number in . M As an illustration of Theorem 5, observe that the volume of a sphere varies continuously with its radius because the formula shows that is a polynomial function of . Likewise, if a ball is thrown vertically into the air with a velocity of 50 ft#s, then the height of the ball in feet seconds later is given by the formula . Again this is a polynomial function, so the height is a continuous function of the elapsed time. Knowledge of which functions are continuous enables us to evaluate some limits very quickly, as the following example shows. Compare it with Example 2(b) in Section 2.3. EXAMPLE 5 Find . SOLUTION The function is rational, so by Theorem 5 it is continuous on its domain, which is . Therefore M It turns out that most of the familiar functions are continuous at every number in their domains. For instance, Limit Law 10 (page 101) is exactly the statement that root functions are continuous. From the appearance of the graphs of the sine and cosine functions (Figure 18 in Section 1.2), we would certainly guess that they are continuous. We know from the deﬁni-tions of and that the coordinates of the point P in Figure 5 are . As , we see that P approaches the point and so and . Thus Since and , the equations in (6) assert that the cosine and sine func-tions are continuous at 0. The addition formulas for cosine and sine can then be used to deduce that these functions are continuous everywhere (see Exercises 56 and 57). It follows from part 5 of Theorem 4 that tan x ! sin x cos x sin 0 ! 0 cos 0 ! 1 lim % l 0 cos % ! 1 lim % l 0 sin % ! 0 6 sin % l 0 cos % l 1 !1, 0" % l 0 !cos %, sin %" cos % sin % ! !"2"3 ! 2!"2"2 " 1 5 " 3!"2" ! " 1 11 lim x l"2 x 3 ! 2x 2 " 1 5 " 3x ! lim x l"2 f!x" ! f!"2" {x&x " 5 3} f!x" ! x 3 ! 2x 2 " 1 5 " 3x lim x l "2 x 3 ! 2x 2 " 1 5 " 3x h ! 50t " 16t 2 t r V V!r" ! 4 3&r 3 D f Q P D ! 'x ! !&Q!x" " 0( f Q P f!x" ! P!x" Q!x" ¨ 1 x 0 y (1, 0) P(cos ¨, sin ¨) FIGURE 5 N Another way to establish the limits in (6) is to use the Squeeze Theorem with the inequality (for ), which is proved in Section 3.3. % ' 0 sin % ( % is continuous except where . This happens when is an odd integer multiple of , so has inﬁnite discontinuities when and so on (see Figure 6). The inverse function of any continuous one-to-one function is also continuous. (This fact is proved in Appendix F, but our geometric intuition makes it seem plausible: The graph of is obtained by reﬂecting the graph of f about the line . So if the graph of f has no break in it, neither does the graph of .) Thus the inverse trigonometric func-tions are continuous. In Section 1.5 we deﬁned the exponential function so as to ﬁll in the holes in the graph of where x is rational. In other words, the very deﬁnition of makes it a continuous function on !. Therefore its inverse function is continuous on . THEOREM The following types of functions are continuous at every number in their domains: polynomials rational functions root functions trigonometric functions inverse trigonometric functions exponential functions logarithmic functions EXAMPLE 6 Where is the function continuous? SOLUTION We know from Theorem 7 that the function is continuous for and is continuous on !. Thus, by part 1 of Theorem 4, is continuous on . The denominator, , is a polynomial, so it is continuous everywhere. Therefore, by part 5 of Theorem 4, f is continuous at all positive numbers x except where . So f is continuous on the intervals and . M EXAMPLE 7 Evaluate . SOLUTION Theorem 7 tells us that is continuous. The function in the denomi-nator, , is the sum of two continuous functions and is therefore continu-ous. Notice that this function is never 0 because for all and so everywhere. Thus the ratio is continuous everywhere. Hence, by deﬁnition of a continuous function, M Another way of combining continuous functions and to get a new continuous func-tion is to form the composite function . This fact is a consequence of the following theorem. f " t t f lim x l & sin x 2 ! cos x ! lim x l & f!x" ! f !&" ! sin & 2 ! cos & ! 0 2 " 1 ! 0 f!x" ! sin x 2 ! cos x 2 ! cos x ' 0 x cos x ) "1 y ! 2 ! cos x y ! sin x lim x l & sin x 2 ! cos x !1, $" !0, 1" x 2 " 1 ! 0 y ! x 2 " 1 !0, $" y ! ln x ! tan"1x y ! tan"1x x ' 0 y ! ln x f!x" ! ln x ! tan"1x x 2 " 1 7 !0, $" y ! loga x y ! a x y ! a x y ! a x f "1 y ! x f "1 x ! , 3, 5, y ! tan x x cos x ! 0 124 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES N The inverse trigonometric functions are reviewed in Section 1.6. _ _ x y π 0 _π 1 π 2 3π 2 π 2 3π 2 FIGURE 6 y=tan x SECTION 2.5 CONTINUITY \|\|\|\| 125 THEOREM If is continuous at and then In other words, Intuitively, Theorem 8 is reasonable because if is close to , then is close to , and since is continuous at , if is close to , then is close to . A proof of Theorem 8 is given in Appendix F. EXAMPLE 8 Evaluate . SOLUTION Because is a continuous function, we can apply Theorem 8: M Let’s now apply Theorem 8 in the special case where , with being a posi-tive integer. Then and If we put these expressions into Theorem 8, we get and so Limit Law 11 has now been proved. (We assume that the roots exist.) THEOREM If is continuous at and is continuous at , then the compos-ite function given by is continuous at . This theorem is often expressed informally by saying “a continuous function of a con-tinuous function is a continuous function.” PROOF Since is continuous at , we have Since is continuous at , we can apply Theorem 8 to obtain lim x l a f!t!x"" ! f!t!a"" b ! t!a" f lim x l a t!x" ! t!a" a t a ! f " t"!x" ! f!t!x"" f " t t!a" f a t 9 lim x l a s n t!x" ! s n lim x l a t!x" f(lim x l a t!x") ! s n lim x l a t!x" f!t!x"" ! s n t!x" n f!x" ! s n x ! arcsin 1 2 ! & 6 ! arcsin)lim x l 1 1 1 ! sx ! arcsin)lim x l 1 1 " sx (1 " sx ) (1 ! sx ) lim x l 1 arcsin) 1 " sx 1 " x ! arcsin) lim xl1 1 " sx 1 " x arcsin lim x l 1 arcsin) 1 " sx 1 " x f!b" f!t!x"" b t!x" b f b t!x" a x lim x l a f!t!x"" ! f(lim x l a t!x") lim x l a f!t!x"" ! f!b". lim x l a t!x" ! b, b f 8 N This theorem says that a limit symbol can be moved through a function symbol if the function is continuous and the limit exists. In other words, the order of these two symbols can be reversed. which is precisely the statement that the function is continuous at ; that is, is continuous at . M EXAMPLE 9 Where are the following functions continuous? (a) (b) SOLUTION (a) We have , where Now is continuous on since it is a polynomial, and is also continuous everywhere. Thus is continuous on by Theorem 9. (b) We know from Theorem 7 that is continuous and is continuous (because both and are continuous). Therefore, by Theorem 9, is continuous wherever it is deﬁned. Now is deﬁned when . So it is undeﬁned when , and this happens when . Thus F has discontinuities when x is an odd multiple of and is continuous on the intervals between these values (see Figure 7). M An important property of continuous functions is expressed by the following theorem, whose proof is found in more advanced books on calculus. THE INTERMEDIATE VALUE THEOREM Suppose that is continuous on the closed interval and let be any number between and , where . Then there exists a number in such that . The Intermediate Value Theorem states that a continuous function takes on every inter-mediate value between the function values and . It is illustrated by Figure 8. Note that the value can be taken on once [as in part (a)] or more than once [as in part (b)]. If we think of a continuous function as a function whose graph has no hole or break, then it is easy to believe that the Intermediate Value Theorem is true. In geometric terms it says that if any horizontal line is given between and as in Fig-ure 9, then the graph of can’t jump over the line. It must intersect somewhere. It is important that the function in Theorem 10 be continuous. The Intermediate Value Theorem is not true in general for discontinuous functions (see Exercise 44). One use of the Intermediate Value Theorem is in locating roots of equations as in the following example. f y ! N f y ! f!b" y ! f!a" y ! N (b) 0 x y f(b) N f(a) a c£ b y=ƒ c™ c¡ (a) 0 x y f(b) N f(a) a c b y=ƒ FIGURE 8 N f!b" f!a" f!c" ! N !a, b" c f!a" " f!b" f!b" f!a" N $a, b% f 10 & x ! &, 3&, . . . cos x ! "1 1 ! cos x ' 0 ln!1 ! cos x" F!x" ! f!t!x"" y ! cos x y ! 1 t!x" ! 1 ! cos x f!x" ! ln x ! h ! f " t f ! t f!x" ! sin x and t!x" ! x 2 h!x" ! f!t!x"" F!x" ! ln!1 ! cos x" h!x" ! sin!x 2" V a f " t a h!x" ! f!t!x"" 126 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES FIGURE 7 y=ln(1+cos x) 2 6 _10 10 b 0 x y f(a) N f(b) a y=ƒ y=N FIGURE 9 SECTION 2.5 CONTINUITY \|\|\|\| 127 EXAMPLE 10 Show that there is a root of the equation between 1 and 2. SOLUTION Let . We are looking for a solution of the given equation, that is, a number between 1 and 2 such that . Therefore, we take , , and in Theorem 10. We have and Thus ; that is, is a number between and . Now is continuous since it is a polynomial, so the Intermediate Value Theorem says there is a number between 1 and 2 such that . In other words, the equation has at least one root in the interval . In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since a root must lie between 1.2 and 1.3. A calculator gives, by trial and error, so a root lies in the interval . M We can use a graphing calculator or computer to illustrate the use of the Intermediate Value Theorem in Example 10. Figure 10 shows the graph of in the viewing rectangle by and you can see that the graph crosses the -axis between 1 and 2. Fig-ure 11 shows the result of zooming in to the viewing rectangle by . In fact, the Intermediate Value Theorem plays a role in the very way these graphing devices work. A computer calculates a ﬁnite number of points on the graph and turns on the pixels that contain these calculated points. It assumes that the function is continuous and takes on all the intermediate values between two consecutive points. The computer therefore connects the pixels by turning on the intermediate pixels. 0.2 _0.2 1.2 1.3 FIGURE 11 FIGURE 10 3 _3 _1 3 $"0.2, 0.2% $1.2, 1.3% x $"3, 3% $"1, 3% f !1.22, 1.23" f!1.23" ! 0.056068 ' 0 and f!1.22" ! "0.007008 ( 0 f!1.3" ! 0.548 ' 0 and f!1.2" ! "0.128 ( 0 !1, 2" c 4x 3 " 6x 2 ! 3x " 2 ! 0 f!c" ! 0 c f f!2" f!1" N ! 0 f!1" ( 0 ( f!2" f!2" ! 32 " 24 ! 6 " 2 ! 12 ' 0 f!1" ! 4 " 6 ! 3 " 2 ! "1 ( 0 N ! 0 b ! 2 a ! 1 f!c" ! 0 c f!x" ! 4x 3 " 6x 2 ! 3x " 2 4x 3 " 6x 2 ! 3x " 2 ! 0 V (d) The cost of a taxi ride as a function of the distance traveled (e) The current in the circuit for the lights in a room as a function of time 9. If and are continuous functions with and , ﬁnd . 10–12 Use the deﬁnition of continuity and the properties of limits to show that the function is continuous at the given number . 10. , , 12. , 13–14 Use the deﬁnition of continuity and the properties of limits to show that the function is continuous on the given interval. 13. , 14. , 15–20 Explain why the function is discontinuous at the given number . Sketch the graph of the function. 15. 16. 17. 19. 20. 21–28 Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its domain. State the domain. 21. 22. G!x" ! s 3 x !1 ! x3" F!x" ! x x 2 ! 5x ! 6 a ! 3 f !x" !+ 2x 2 " 5x " 3 x " 3 6 if x " 3 if x ! 3 a ! 0 f !x" !+ cos x 0 1 " x 2 if x ( 0 if x ! 0 if x ' 0 a ! 1 f !x" !+ x 2 " x x 2 " 1 1 if x " 1 if x ! 1 18. a ! 0 f !x" !+ e x x2 if x ( 0 if x ) 0 a ! 1 f !x" !+ 1 x " 1 2 if x " 1 if x ! 1 a ! 2 f !x" ! ln & x " 2& a !"$, 3% t!x" ! 2 s3 " x !2, $" f !x" ! 2x ! 3 x " 2 a ! 1 h!t" ! 2t " 3t 2 1 ! t 3 a ! "1 f !x" ! !x ! 2x 3"4 11. a ! 4 f !x" ! x 2 ! s7 " x a t!3" lim x l 3 $2 f !x" " t!x"% ! 4 f !3" ! 5 t f 1. Write an equation that expresses the fact that a function is continuous at the number 4. 2. If is continuous on , what can you say about its graph? (a) From the graph of , state the numbers at which is dis-continuous and explain why. (b) For each of the numbers stated in part (a), determine whether is continuous from the right, or from the left, or neither. 4. From the graph of , state the intervals on which is continuous. 5. Sketch the graph of a function that is continuous everywhere except at x ! 3 and is continuous from the left at 3. 6. Sketch the graph of a function that has a jump discontinuity at and a removable discontinuity at , but is con-tinuous elsewhere. A parking lot charges $3 for the ﬁrst hour (or part of an hour) and $2 for each succeeding hour (or part), up to a daily maxi-mum of $10. (a) Sketch a graph of the cost of parking at this lot as a func-tion of the time parked there. (b) Discuss the discontinuities of this function and their signiﬁcance to someone who parks in the lot. 8. Explain why each function is continuous or discontinuous. (a) The temperature at a speciﬁc location as a function of time (b) The temperature at a speciﬁc time as a function of the dis-tance due west from New York City (c) The altitude above sea level as a function of the distance due west from New York City 7. x ! 4 x ! 2 y x _4 2 4 6 _2 8 t t y x _4 2 4 6 _2 0 f f f 3. !"$, $" f f EXERCISES 2.5 128 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.5 CONTINUITY \|\|\|\| 129 For what value of the constant is the function continuous on ? 42. Find the values of and that make continuous everywhere. 43. Which of the following functions has a removable disconti-nuity at ? If the discontinuity is removable, ﬁnd a function that agrees with for and is continuous at . (a) , (b) , (c) , 44. Suppose that a function is continuous on [0, 1] except at 0.25 and that and . Let N ! 2. Sketch two possible graphs of , one showing that might not satisfy the conclusion of the Intermediate Value Theorem and one showing that might still satisfy the conclusion of the Inter-mediate Value Theorem (even though it doesn’t satisfy the hypothesis). 45. If , show that there is a number such that . 46. Suppose is continuous on and the only solutions of the equation are and . If , explain why . 47–50 Use the Intermediate Value Theorem to show that there is a root of the given equation in the speciﬁed interval. , 48. , 49. , 50. , 51–52 (a) Prove that the equation has at least one real root. (b) Use your calculator to ﬁnd an interval of length 0.01 that con-tains a root. 51. 52. ; 53–54 (a) Prove that the equation has at least one real root. (b) Use your graphing device to ﬁnd the root correct to three deci-mal places. 54. arctan x ! 1 " x 100e"x#100 ! 0.01x2 53. ln x ! 3 " 2x cos x ! x 3 !1, 2" ln x ! e"x !0, 1" cos x ! x !0, 1" s 3 x ! 1 " x !1, 2" x 4 ! x " 3 ! 0 47. f !3" ' 6 f !2" ! 8 x ! 4 x ! 1 f !x" ! 6 $1, 5% f f !c" ! 1000 c f !x" ! x 2 ! 10 sin x f f f f !1" ! 3 f !0" ! 1 f a ! & f !x" ! ,sin x-a ! 2 f !x" ! x 3 " x 2 " 2x x " 2 a ! 1 f !x" ! x 4 " 1 x " 1 a x " a f t a f f !x" ! x 2 " 4 x " 2 ax 2 " bx ! 3 2x " a ! b if x ( 2 if 2 ( x ( 3 if x ) 3 f b a f !x" !+ cx 2 ! 2x x 3 " cx if x ( 2 if x ) 2 !"$, $" f c 41. 23. 24. 25. 26. 28. ; 29–30 Locate the discontinuities of the function and illustrate by graphing. 30. 31–34 Use continuity to evaluate the limit. 31. 33. 34. 35–36 Show that is continuous on . 35. 36. 37–39 Find the numbers at which is discontinuous. At which of these numbers is continuous from the right, from the left, or neither? Sketch the graph of . 37. 38. 40. The gravitational force exerted by the earth on a unit mass at a distance r from the center of the planet is where M is the mass of the earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r? if r ) R GM r 2 F!r" ! GMr R 3 if r ( R f !x" !+ x ! 2 ex 2 " x if x ( 0 if 0 + x + 1 if x ' 1 39. f !x" !+ x ! 1 1#x sx " 3 if x + 1 if 1 ( x ( 3 if x ) 3 f !x" !+ 1 ! x 2 2 " x !x " 2"2 if x + 0 if 0 ( x + 2 if x ' 2 f f f f !x" !+ sin x if x ( cos x if x ) f !x" !+ x 2 if x ( 1 sx if x ) 1 !"$, $" f lim x l 2 arctan) x 2 " 4 3x 2 " 6x lim x l 1 ex2"x lim x l & sin!x ! sin x" 32. lim x l 4 5 ! sx s5 ! x y ! ln!tan2x" y ! 1 1 ! e 1#x 29. H!x" ! cos(esx) G!t" ! ln!t 4 " 1" 27. F!x" ! sin"1!x 2 " 1" L!t" ! e"5t cos 2&t h!x" ! sin x x ! 1 R!x" ! x 2 ! s2x " 1 62. If and are positive numbers, prove that the equation has at least one solution in the interval . 63. Show that the function is continuous on . 64. (a) Show that the absolute value function is contin-uous everywhere. (b) Prove that if is a continuous function on an interval, then so is . (c) Is the converse of the statement in part (b) also true? In other words, if is continuous, does it follow that is continuous? If so, prove it. If not, ﬁnd a counterexample. 65. A Tibetan monk leaves the monastery at 7:00 AM and takes his usual path to the top of the mountain, arriving at 7:00 PM. The following morning, he starts at 7:00 AM at the top and takes the same path back, arriving at the monastery at 7:00 PM. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days. f & f & & f & f F!x" ! & x& !"$, $" f !x" !+ x 4 sin!1#x" 0 if x " 0 if x ! 0 !"1, 1" a x 3 ! 2x 2 " 1 ! b x 3 ! x " 2 ! 0 b a 55. Prove that is continuous at if and only if 56. To prove that sine is continuous, we need to show that for every real number . By Exercise 55 an equivalent statement is that Use (6) to show that this is true. 57. Prove that cosine is a continuous function. 58. (a) Prove Theorem 4, part 3. (b) Prove Theorem 4, part 5. 59. For what values of is continuous? 60. For what values of is continuous? Is there a number that is exactly 1 more than its cube? 61. t!x" !+ 0 x if x is rational if x is irrational t x f !x" !+ 0 1 if x is rational if x is irrational f x lim h l 0 sin!a ! h" ! sin a a limx la sin x ! sin a lim h l 0 f !a ! h" ! f !a" a f LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES In Sections 2.2 and 2.4 we investigated inﬁnite limits and vertical asymptotes. There we let approach a number and the result was that the values of became arbitrarily large (positive or negative). In this section we let become arbitrarily large (positive or nega-tive) and see what happens to . Let’s begin by investigating the behavior of the function deﬁned by as becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of has been drawn by a computer in Figure 1. As grows larger and larger you can see that the values of get closer and closer to 1. In fact, it seems that we can make the values of as close as we like to 1 by taking sufﬁciently large. This situation is expressed symbolically by writing lim x l $ x 2 " 1 x 2 ! 1 ! 1 x f!x" f!x" x x 1 0 y y=1 y=≈-1 ≈+1 FIGURE 1 f x f!x" ! x 2 " 1 x 2 ! 1 f y x y x 2.6 130 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES x 0 "1 0 0.600000 0.800000 0.882353 0.923077 0.980198 0.999200 0.999800 0.999998 1000 100 50 10 5 4 3 2 1 f !x" SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES \|\|\|\| 131 In general, we use the notation to indicate that the values of become closer and closer to as becomes larger and larger. DEFINITION Let be a function deﬁned on some interval . Then means that the values of can be made arbitrarily close to by taking suf-ﬁciently large. Another notation for is as The symbol does not represent a number. Nonetheless, the expression is often read as “the limit of , as approaches inﬁnity, is ” or “the limit of , as becomes inﬁnite, is ” or “the limit of , as increases without bound, is ” The meaning of such phrases is given by Deﬁnition 1. A more precise deﬁnition, similar to the deﬁnition of Section 2.4, is given at the end of this section. Geometric illustrations of Deﬁnition 1 are shown in Figure 2. Notice that there are many ways for the graph of to approach the line (which is called a horizontal asymptote) as we look to the far right of each graph. Referring back to Figure 1, we see that for numerically large negative values of , the values of are close to 1. By letting decrease through negative values without bound, we can make as close as we like to 1. This is expressed by writing The general deﬁnition is as follows. DEFINITION Let be a function deﬁned on some interval . Then means that the values of can be made arbitrarily close to by taking suf-ﬁciently large negative. x L f!x" lim x l"$ f!x" ! L !"$, a" f 2 lim x l"$ x 2 " 1 x 2 ! 1 ! 1 f!x" x f!x" x x y 0 y=ƒ y=L 0 x y y=ƒ y=L x y 0 y=ƒ y=L y ! L f ,, -L x f!x" L x f!x" L x f!x" lim x l$ f!x" ! L $ x l $ f!x" l L lim x l $ f!x" ! L x L f!x" lim x l $ f!x" ! L !a, $" f 1 x L f!x" lim x l $ f!x" ! L x FIGURE 2 Examples illustrating lim ƒ=L Again, the symbol does not represent a number, but the expression is often read as “the limit of , as x approaches negative inﬁnity, is L” Deﬁnition 2 is illustrated in Figure 3. Notice that the graph approaches the line as we look to the far left of each graph. DEFINITION The line is called a horizontal asymptote of the curve if either For instance, the curve illustrated in Figure 1 has the line as a horizontal asymp-tote because An example of a curve with two horizontal asymptotes is . (See Figure 4.) In fact, so both of the lines and are horizontal asymptotes. (This follows from the fact that the lines are vertical asymptotes of the graph of tan.) EXAMPLE 1 Find the inﬁnite limits, limits at inﬁnity, and asymptotes for the function f whose graph is shown in Figure 5. SOLUTION We see that the values of become large as from both sides, so Notice that becomes large negative as x approaches 2 from the left, but large posi-tive as x approaches 2 from the right. So Thus both of the lines and are vertical asymptotes. As x becomes large, it appears that approaches 4. But as x decreases through negative values, approaches 2. So This means that both y ! 4 and y ! 2 are horizontal asymptotes. M lim x l!" f!x" ! 2 and lim x l " f!x" ! 4 f!x" f!x" x ! 2 x ! !1 lim x l 2# f!x" ! " and lim x l 2! f!x" ! !" f!x" lim x l!1 f!x" ! " x l !1 f!x" x ! $%#2 y ! %#2 y ! !%#2 lim x l " tan!1x ! % 2 lim x l!" tan!1x ! ! % 2 4 y ! tan!1x lim x l " x 2 ! 1 x 2 # 1 ! 1 y ! 1 lim x l!" f!x" ! L or lim x l " f!x" ! L y ! f!x" y ! L 3 y ! L f!x" lim x l !" f!x" ! L !" 132 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES x _ FIGURE 3 Examples illustrating lim ƒ=L 0 y x y=ƒ y=L x 0 y y=ƒ y=L FIGURE 4 y=tan–!x y 0 x π 2 _ π 2 FIGURE 5 0 x y 2 2 SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES \|\|\|\| 133 EXAMPLE 2 Find and . SOLUTION Observe that when is large, is small. For instance, In fact, by taking large enough, we can make as close to 0 as we please. Therefore, according to Deﬁnition 1, we have Similar reasoning shows that when is large negative, is small negative, so we also have It follows that the line (the -axis) is a horizontal asymptote of the curve . (This is an equilateral hyperbola; see Figure 6.) M Most of the Limit Laws that were given in Section 2.3 also hold for limits at inﬁnity. It can be proved that the Limit Laws listed in Section 2.3 (with the exception of Laws 9 and 10) are also valid if “ ” is replaced by “ ” or “ .” In particular, if we combine Laws 6 and 11 with the results of Example 2, we obtain the following important rule for calculating limits. THEOREM If is a rational number, then If is a rational number such that is deﬁned for all x, then EXAMPLE 3 Evaluate and indicate which properties of limits are used at each stage. SOLUTION As becomes large, both numerator and denominator become large, so it isn’t obvious what happens to their ratio. We need to do some preliminary algebra. To evaluate the limit at inﬁnity of any rational function, we ﬁrst divide both the numerator and denominator by the highest power of that occurs in the denominator. x x lim x l " 3x 2 ! x ! 2 5x 2 # 4x # 1 V lim x l!" 1 x r ! 0 x r r & 0 lim x l " 1 x r ! 0 r & 0 5 x l !" x l " x l a y ! 1#x x y ! 0 lim x l!" 1 x ! 0 1#x x lim x l " 1 x ! 0 1#x x 1 1,000,000 ! 0.000001 1 10,000 ! 0.0001 1 100 ! 0.01 1#x x lim x l!" 1 x lim x l " 1 x x x _ 1 x 1 x 0 y x y=∆ FIGURE 6 lim =0, lim =0 (We may assume that , since we are interested only in large values of .) In this case the highest power of in the denominator is , so we have (by Limit Law 5) (by 1, 2, and 3) (by 7 and Theorem 5) A similar calculation shows that the limit as is also . Figure 7 illustrates the results of these calculations by showing how the graph of the given rational function approaches the horizontal asymptote . M EXAMPLE 4 Find the horizontal and vertical asymptotes of the graph of the function SOLUTION Dividing both numerator and denominator by and using the properties of lim-its, we have (since for ) Therefore the line is a horizontal asymptote of the graph of . In computing the limit as , we must remember that for , we have . So when we divide the numerator by , for we get 1 x s2x 2 # 1 ! ! 1 sx 2 s2x 2 # 1 ! !$2 # 1 x 2 x ' 0 x sx 2 ! %x% ! !x x ' 0 x l !" f y ! s2 #3 ! s2 # 0 3 ! 5 ! 0 ! s2 3 ! lim xl" $2 # 1 x 2 lim xl" &3 ! 5 x' !$ lim xl" 2 # lim xl" 1 x 2 lim xl" 3 ! 5 lim xl" 1 x x & 0 sx 2 ! x lim xl" s2x 2 # 1 3x ! 5 ! lim xl" $2 # 1 x 2 3 ! 5 x x f!x" ! s2x 2 # 1 3x ! 5 y ! 3 5 3 5 x l !" ! 3 5 ! 3 ! 0 ! 0 5 # 0 # 0 ! lim x l " 3 ! lim x l " 1 x ! 2 lim x l " 1 x 2 lim x l " 5 # 4 lim x l " 1 x # lim x l " 1 x 2 ! lim x l " &3 ! 1 x ! 2 x 2' lim x l " &5 # 4 x # 1 x 2' lim x l " 3x 2 ! x ! 2 5x 2 # 4x # 1 ! lim x l " 3x2 ! x ! 2 x2 5x2 # 4x # 1 x2 ! lim x l " 3 ! 1 x ! 2 x 2 5 # 4 x # 1 x 2 x 2 x x x " 0 134 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES 1 y=0.6 x y 0 y= 3≈-x-2 5≈+4x+1 SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES \|\|\|\| 135 Therefore Thus the line is also a horizontal asymptote. A vertical asymptote is likely to occur when the denominator, , is 0, that is, when . If is close to and , then the denominator is close to 0 and is positive. The numerator is always positive, so is positive. Therefore If is close to but , then and so is large negative. Thus The vertical asymptote is . All three asymptotes are shown in Figure 8. M EXAMPLE 5 Compute . SOLUTION Because both and x are large when x is large, it’s difﬁcult to see what happens to their difference, so we use algebra to rewrite the function. We ﬁrst multiply numerator and denominator by the conjugate radical: The Squeeze Theorem could be used to show that this limit is 0. But an easier method is to divide numerator and denominator by . Doing this and using the Limit Laws, we obtain Figure 9 illustrates this result. M The graph of the natural exponential function has the line (the x-axis) as a horizontal asymptote. (The same is true of any exponential function with base .) In a & 1 y ! 0 y ! e x ! lim x l " 1 x $1 # 1 x 2 # 1 ! 0 s1 # 0 # 1 ! 0 lim x l " (sx 2 # 1 ! x) ! lim x l " 1 sx 2 # 1 # x ! lim x l " 1 x sx 2 # 1 # x x x ! lim x l " !x 2 # 1" ! x 2 sx 2 # 1 # x ! lim x l " 1 sx 2 # 1 # x lim x l " (sx 2 # 1 ! x) ! lim x l " (sx 2 # 1 ! x) sx 2 # 1 # x sx 2 # 1 # x sx 2 # 1 lim x l " (sx 2 # 1 ! x) x ! 5 3 lim x l !5#3"! s2x 2 # 1 3x ! 5 ! !" f!x" 3x ! 5 ' 0 x ' 5 3 5 3 x lim x l !5#3"# s2x 2 # 1 3x ! 5 ! " f!x" s2x 2 # 1 3x ! 5 x & 5 3 5 3 x x ! 5 3 3x ! 5 y ! !s2#3 ! !$2 # lim x l!" 1 x 2 3 ! 5 lim x l!" 1 x ! ! s2 3 lim x l!" s2x 2 # 1 3x ! 5 ! lim x l!" !$2 # 1 x 2 3 ! 5 x FIGURE 8 y= œ„„„„„„ 3x-5 2≈+1 x y y= œ„ 2 3 y= œ„ 2 3 x=5 3 FIGURE 9 y= ≈+1 œ„„„„„-x x y 0 1 1 N We can think of the given function as having a denominator of 1. fact, from the graph in Figure 10 and the corresponding table of values, we see that Notice that the values of approach 0 very rapidly. EXAMPLE 6 Evaluate . SOLUTION If we let , we know that as . Therefore, by (6), (See Exercise 71.) M EXAMPLE 7 Evaluate . SOLUTION As x increases, the values of sin x oscillate between 1 and !1 inﬁnitely often and so they don’t approach any deﬁnite number. Thus does not exist. M INFINITE LIMITS AT INFINITY The notation is used to indicate that the values of become large as becomes large. Similar mean-ings are attached to the following symbols: EXAMPLE 8 Find and . SOLUTION When becomes large, also becomes large. For instance, In fact, we can make as big as we like by taking large enough. Therefore we can write Similarly, when is large negative, so is . Thus These limit statements can also be seen from the graph of in Figure 11. M y ! x 3 lim x l!" x 3 ! !" x 3 x lim x l " x 3 ! " x x 3 10003 ! 1,000,000,000 1003 ! 1,000,000 103 ! 1000 x 3 x lim x l!" x 3 lim x l " x 3 lim x l!" f!x" ! !" lim x l " f!x" ! !" lim x l!" f!x" ! " x f!x" lim x l " f!x" ! " lim x l" sin x lim x l " sin x lim x l 0! e 1#x ! lim t l !" e t ! 0 x l 0! t l !" t ! 1#x lim x l 0! e 1#x V y=´ x 0 1 y 1 FIGURE 10 e x lim x l!" e x ! 0 6 136 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES x 0 1.00000 !1 0.36788 !2 0.13534 !3 0.04979 !5 0.00674 !8 0.00034 !10 0.00005 e x N The problem-solving strategy for Example 6 is introducing something extra (see page 76). Here, the something extra, the auxiliary aid, is the new variable . t FIGURE 11 lim x#=, lim x#=_ x x _ x y 0 y=˛ SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES \|\|\|\| 137 Looking at Figure 10 we see that but, as Figure 12 demonstrates, becomes large as at a much faster rate than . EXAMPLE 9 Find . \| SOLUTION It would be wrong to write The Limit Laws can’t be applied to inﬁnite limits because is not a number ( can’t be deﬁned). However, we can write because both and become arbitrarily large and so their product does too. M EXAMPLE 10 Find . SOLUTION As in Example 3, we divide the numerator and denominator by the highest power of in the denominator, which is just x: because and as . M The next example shows that by using inﬁnite limits at inﬁnity, together with intercepts, we can get a rough idea of the graph of a polynomial without having to plot a large num-ber of points. EXAMPLE 11 Sketch the graph of by ﬁnding its inter-cepts and its limits as and as . SOLUTION The -intercept is and the -intercepts are found by setting : . Notice that since is positive, the function doesn’t change sign at ; thus the graph doesn’t cross the -axis at . The graph crosses the axis at and . When is large positive, all three factors are large, so When is large negative, the ﬁrst factor is large positive and the second and third factors are both large negative, so Combining this information, we give a rough sketch of the graph in Figure 13. M lim x l!" !x ! 2"4!x # 1"3!x ! 1" ! " x lim x l " !x ! 2"4!x # 1"3!x ! 1" ! " x 1 !1 2 x 2 !x ! 2"4 x ! 2, !1, 1 y ! 0 x f!0" ! !!2"4!1"3!!1" ! !16 y x l !" x l " y ! !x ! 2"4!x # 1"3!x ! 1" V x l " 3#x ! 1 l !1 x # 1 l " lim x l" x 2 # x 3 ! x ! lim x l" x # 1 3 x ! 1 ! !" x lim x l " x 2 # x 3 ! x x ! 1 x lim x l " !x 2 ! x" ! lim x l " x!x ! 1" ! " " ! " " ! " ! " lim x l " !x 2 ! x" ! lim x l " x 2 ! lim x l " x lim x l " !x 2 ! x" y ! x 3 x l " y ! e x lim x l" e x ! " y y=(x-2)$(x +1)#(x-1) FIGURE 13 0 x 1 _16 2 1 x 0 100 y 1 y=˛ y=´ FIGURE 12 ´ is much larger than ˛ when x is large. PRECISE DEFINITIONS Deﬁnition 1 can be stated precisely as follows. DEFINITION Let be a function deﬁned on some interval . Then means that for every there is a corresponding number such that then In words, this says that the values of can be made arbitrarily close to (within a distance , where is any positive number) by taking sufﬁciently large (larger than , where depends on ). Graphically it says that by choosing large enough (larger than some number ) we can make the graph of lie between the given horizontal lines and as in Figure 14. This must be true no matter how small we choose . Figure 15 shows that if a smaller value of is chosen, then a larger value of may be required. Similarly, a precise version of Deﬁnition 2 is given by Deﬁnition 8, which is illustrated in Figure 16. DEFINITION Let be a function deﬁned on some interval . Then means that for every there is a corresponding number such that then % f!x" ! L% ' ( x ' N if N ( & 0 lim x l!" f!x" ! L !!", a" f 8 FIGURE 14 lim ƒ=L x FIGURE 15 lim ƒ=L x y 0 x N L y=ƒ y=L-∑ y=L+∑ 0 y x N L when x is in here ƒ is in here y=L-∑ y=L+∑ ∑ ∑ y=ƒ N ( ( y ! L # ( y ! L ! ( f N x ( N N x ( ( L f!x" % f!x" ! L% ' ( x & N if N ( & 0 lim x l " f!x" ! L !a, "" f 7 138 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES \|\|\|\| 139 In Example 3 we calculated that In the next example we use a graphing device to relate this statement to Deﬁnition 7 with and . EXAMPLE 12 Use a graph to ﬁnd a number such that then SOLUTION We rewrite the given inequality as We need to determine the values of for which the given curve lies between the horizon-tal lines and . So we graph the curve and these lines in Figure 17. Then we use the cursor to estimate that the curve crosses the line when . To the right of this number the curve stays between the lines and . Round-ing to be safe, we can say that then In other words, for we can choose (or any larger number) in Deﬁnition 7. M EXAMPLE 13 Use Deﬁnition 7 to prove that . SOLUTION Given , we want to ﬁnd such that if then In computing the limit we may assume that . Then . Let’s choose . So if then ( 1 x ! 0( ! 1 x ' ( x & N ! 1 ( N ! 1#( 1#x ' ( & ? x & 1#( x & 0 ( 1 x ! 0( ' ( x & N N ( & 0 lim x l " 1 x ! 0 N ! 7 ( ! 0.1 ( 3x 2 ! x ! 2 5x 2 # 4x # 1 ! 0.6( ' 0.1 x & 7 if y ! 0.7 y ! 0.5 x ) 6.7 y ! 0.5 y ! 0.7 y ! 0.5 x 0.5 ' 3x 2 ! x ! 2 5x 2 # 4x # 1 ' 0.7 ( 3x 2 ! x ! 2 5x 2 # 4x # 1 ! 0.6( ' 0.1 x & N if N ( ! 0.1 L ! 3 5 lim x l " 3x 2 ! x ! 2 5x 2 # 4x # 1 ! 3 5 x FIGURE 16 lim ƒ=L x N y L y=L-∑ y=L+∑ y=ƒ 0 FIGURE 17 1 0 15 y=0.7 y=0.5 y= 3≈-x-2 5≈+4x+1 Therefore, by Deﬁnition 7, Figure 18 illustrates the proof by showing some values of and the corresponding values of . M Finally we note that an inﬁnite limit at inﬁnity can be deﬁned as follows. The geomet-ric illustration is given in Figure 19. DEFINITION Let be a function deﬁned on some interval . Then means that for every positive number there is a corresponding positive number N such that then Similar deﬁnitions apply when the symbol is replaced by . (See Exercise 70.) !" " f!x" & M x & N if M lim x l " f!x" ! " !a, "" f 9 N ( lim x l " 1 x ! 0 140 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES x y 0 N=5 ∑=0.2 FIGURE 18 x y 0 N=1 ∑=1 x y 0 N=10 ∑=0.1 FIGURE 19 lim ƒ= x ` 0 x y N M y=M (c) (d) (e) (f) The equations of the asymptotes x y 1 1 lim x l !" f !x" lim x l " f !x" lim x l !1# f !x" 1. Explain in your own words the meaning of each of the following. (a) (b) (a) Can the graph of intersect a vertical asymptote? Can it intersect a horizontal asymptote? Illustrate by sketching graphs. (b) How many horizontal asymptotes can the graph of have? Sketch graphs to illustrate the possibilities. 3. For the function whose graph is given, state the following. (a) (b) lim x l !1! f !x" lim x l 2 f !x" f y ! f !x" y ! f !x" 2. lim x l!" f !x" ! 3 lim x l" f !x" ! 5 EXERCISES 2.6 SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES \|\|\|\| 141 13–14 Evaluate the limit and justify each step by indicating the appropriate properties of limits. 13. 14. 15–36 Find the limit. 15. 16. 17. 18. 20. 21. 22. 23. 24. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. ; 37. (a) Estimate the value of by graphing the function . (b) Use a table of values of to guess the value of the limit. (c) Prove that your guess is correct. ; 38. (a) Use a graph of to estimate the value of to one decimal place. (b) Use a table of values of to estimate the limit to four decimal places. (c) Find the exact value of the limit. f !x" lim x l " f !x" f !x" ! s3x 2 # 8x # 6 ! s3x 2 # 3x # 1 f !x" f !x" ! sx 2 # x # 1 # x lim x l !" (sx 2 # x # 1 # x) lim x l !%#2"#etan x lim x l " !e!2x cos x" lim x l " tan!1!x 2 ! x 4" lim x l " 1 ! e x 1 # 2e x lim x l " x 3 ! 2x # 3 5 ! 2x2 lim x l !"!x 4 # x5" lim x l " sx 2 # 1 lim x l " x # x 3 # x 5 1 ! x 2 # x 4 lim x l " cos x lim x l " (sx 2 # ax ! sx2 # bx ) lim x l!" (x # sx 2 # 2x ) lim x l " (s9x 2 # x ! 3x) 25. lim x l !" s9x 6 ! x x 3 # 1 lim x l " s9x 6 ! x x 3 # 1 lim x l " x # 2 s9x 2 # 1 lim u l " 4u4 # 5 !u2 ! 2"!2u2 ! 1" lim t l !" t 2 # 2 t 3 # t 2 ! 1 lim x l " x 3 # 5x 2x 3 ! x 2 # 4 19. lim y l " 2 ! 3y2 5y2 # 4y lim x l !" 1 ! x ! x 2 2x 2 ! 7 lim x l " 3x # 5 x ! 4 lim x l " 1 2x # 3 lim x l "$ 12x 3 ! 5x # 2 1 # 4x 2 # 3x 3 lim x l " 3x2 ! x # 4 2x2 # 5x ! 8 4. For the function whose graph is given, state the following. (a) (b) (c) (d) (e) (f) The equations of the asymptotes 5–10 Sketch the graph of an example of a function that satisﬁes all of the given conditions. 5. is odd 6. 8. 9. 10. is even ; 11. Guess the value of the limit by evaluating the function for and . Then use a graph of to support your guess. ; 12. (a) Use a graph of to estimate the value of correct to two decimal places. (b) Use a table of values of to estimate the limit to four decimal places. f !x" lim x l " f !x" f !x" !&1 ! 2 x' x f 100 4, 5, 6, 7, 8, 9, 10, 20, 50, x ! 0, 1, 2, 3, f !x" ! x 2#2x lim x l " x 2 2x f f !0" ! 0, lim x l " f !x" ! 2, lim x l 3 f !x" ! !", lim x l " f !x" ! 3 lim x l 4# f !x" ! ", lim x l 4! f !x" ! !", lim x l !" f !x" ! !", lim x l 0# f !x" ! 2, lim x l 0! f !x" ! 4, f !0" ! 3, lim x l " f !x" ! !3 lim x l !" f !x" ! 3, lim x l !2 f !x" ! ", lim x l 0! f !x" ! !" lim x l 0# f !x" ! ", lim x l !" f !x" ! 0, lim x l " f !x" ! ", lim x l 2 f !x" ! !", 7. lim x l !" f !x" ! 1 lim x l " f !x" ! 1, lim x l 0! f !x" ! !", lim x l 0# f !x" ! ", f lim x l " f !x" ! 0, f !1" ! 1, f !0" ! 0, f 2 0 x y 1 lim x l !2# t!x" lim x l 0 t!x" lim x l 3 t!x" lim x l !" t!x" lim x l " t!x" t 142 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES 53. (a) Use the Squeeze Theorem to evaluate . ; (b) Graph . How many times does the graph cross the asymptote? ; 54. By the end behavior of a function we mean the behavior of its values as and as . (a) Describe and compare the end behavior of the functions by graphing both functions in the viewing rectangles by and by . (b) Two functions are said to have the same end behavior if their ratio approaches 1 as . Show that P and Q have the same end behavior. Let and be polynomials. Find if the degree of is (a) less than the degree of and (b) greater than the degree of . 56. Make a rough sketch of the curve ( an integer) for the following ﬁve cases: (i) (ii) , odd (iii) , even (iv) , odd (v) , even Then use these sketches to ﬁnd the following limits. (a) (b) (c) (d) Find if, for all , 58. (a) A tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water is pumped into the tank at a rate of 25 L!min. Show that the concentration of salt after minutes (in grams per liter) is (b) What happens to the concentration as ? 59. In Chapter 9 we will be able to show, under certain assump-tions, that the velocity of a falling raindrop at time t is where t is the acceleration due to gravity and is the terminal velocity of the raindrop. (a) Find . limt l ! v"t# v v"t# ! v"1 " e "tt!v# v"t# t l ! C"t# ! 30t 200 # t t 10e x " 21 2e x $ f "x# $ 5sx sx " 1 x % 1 limx l ! f "x# 57. lim x l "! x n lim x l ! x n lim x l 0" x n lim x l 0# x n n n $ 0 n n $ 0 n n % 0 n n % 0 n ! 0 n y ! x n Q Q P lim x l ! P"x# Q"x# Q P 55. x l ! $"10,000, 10,000% $"10, 10% $"2, 2% $"2, 2% Q"x# ! 3x 5 P"x# ! 3x 5 " 5x 3 # 2x x l "! x l ! f "x# ! "sin x#!x lim x l ! sin x x 39–44 Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes. 39. 40. 42. 43. 44. ; 45. Estimate the horizontal asymptote of the function by graphing for . Then calculate the equation of the asymptote by evaluating the limit. How do you explain the discrepancy? ; 46. (a) Graph the function How many horizontal and vertical asymptotes do you observe? Use the graph to estimate the values of the limits and (b) By calculating values of , give numerical estimates of the limits in part (a). (c) Calculate the exact values of the limits in part (a). Did you get the same value or different values for these two limits? [In view of your answer to part (a), you might have to check your calculation for the second limit.] 47. Find a formula for a function that satisﬁes the following conditions: , , , , 48. Find a formula for a function that has vertical asymptotes and and horizontal asymptote . 49–52 Find the limits as and as . Use this infor-mation, together with intercepts, to give a rough sketch of the graph as in Example 11. 49. 50. 51. 52. y ! x 2"x 2 " 1#2"x # 2# y ! "3 " x#"1 # x#2"1 " x#4 y ! x 3"x # 2#2"x " 1# y ! x 4 " x6 x l "! x l ! y ! 1 x ! 3 x ! 1 lim x l 3# f "x# ! "! lim x l 3" f "x# ! ! f "2# ! 0 lim x l 0 f "x# ! "! lim x l &! f "x# ! 0 f f "x# lim x l "! s2x 2 # 1 3x " 5 lim x l ! s2x 2 # 1 3x " 5 f "x# ! s2x 2 # 1 3x " 5 "10 ' x ' 10 f f "x# ! 3x 3 # 500x 2 x 3 # 500x 2 # 100x # 2000 y ! 2e x e x " 5 y ! x 3 " x x 2 " 6x # 5 y ! 1 # x 4 x 2 " x 4 y ! 2x 2 # x " 1 x 2 # x " 2 41. y ! x 2 # 1 2x 2 " 3x " 2 y ! 2x # 1 x " 2 SECTION 2.7 DERIVATIVES AND RATES OF CHANGE \|\|\|\| 143 DERIVATIVES AND RATES OF CHANGE The problem of ﬁnding the tangent line to a curve and the problem of ﬁnding the velocity of an object both involve ﬁnding the same type of limit, as we saw in Section 2.1. This spe-cial type of limit is called a derivative and we will see that it can be interpreted as a rate of change in any of the sciences or engineering. TANGENTS If a curve has equation and we want to ﬁnd the tangent line to at the point , then we consider a nearby point , where , and compute the slope of the secant line : Then we let approach along the curve by letting approach . If approaches a number , then we deﬁne the tangent t to be the line through with slope . (This m P m mPQ a x C P Q mPQ ! f"x# " f"a# x " a PQ x " a Q"x, f"x## P"a, f"a## C y ! f"x# C 2.7 (a) How large do we have to take so that ? (b) Taking in Theorem 5, we have the statement Prove this directly using Deﬁnition 7. 66. (a) How large do we have to take so that ? (b) Taking in Theorem 5, we have the statement Prove this directly using Deﬁnition 7. 67. Use Deﬁnition 8 to prove that . 68. Prove, using Deﬁnition 9, that . 69. Use Deﬁnition 9 to prove that . 70. Formulate a precise deﬁnition of Then use your deﬁnition to prove that 71. Prove that and if these limits exist. lim x l "! f "x# ! lim t l 0" f "1!t# lim x l ! f "x# ! lim t l 0# f "1!t# lim x l"! "1 # x 3# ! "! lim x l"! f "x# ! "! lim x l ! e x ! ! lim x l ! x 3 ! ! lim x l"! 1 x ! 0 lim x l ! 1 sx ! 0 r ! 1 2 1!sx $ 0.0001 x lim x l ! 1 x 2 ! 0 r ! 2 1!x 2 $ 0.0001 x 65. ; (b) Graph if and . How long does it take for the velocity of the raindrop to reach 99% of its terminal velocity? ; 60. (a) By graphing and y ! 0.1 on a common screen, discover how large you need to make x so that . (b) Can you solve part (a) without using a graphing device? ; 61. Use a graph to ﬁnd a number such that if ; 62. For the limit illustrate Deﬁnition 7 by ﬁnding values of that correspond to and . ; 63. For the limit illustrate Deﬁnition 8 by ﬁnding values of that correspond to and . ; 64. For the limit illustrate Deﬁnition 9 by ﬁnding a value of that corres-ponds to . M ! 100 N lim x l ! 2x # 1 sx # 1 ! ! ( ! 0.1 ( ! 0.5 N lim x l"! s4x 2 # 1 x # 1 ! "2 ( ! 0.1 ( ! 0.5 N lim x l ! s4x 2 # 1 x # 1 ! 2 & 3x 2 # 1 2x 2 # x # 1 " 1.5& $ 0.05 then x % N N e "x!10 $ 0.1 y ! e "x!10 t ! 9.8 m!s2 v ! 1 m!s v"t# amounts to saying that the tangent line is the limiting position of the secant line as approaches . See Figure 1.) DEFINITION The tangent line to the curve at the point is the line through with slope provided that this limit exists. In our ﬁrst example we conﬁrm the guess we made in Example 1 in Section 2.1. EXAMPLE 1 Find an equation of the tangent line to the parabola at the point . SOLUTION Here we have and , so the slope is Using the point-slope form of the equation of a line, we ﬁnd that an equation of the tangent line at is M We sometimes refer to the slope of the tangent line to a curve at a point as the slope of the curve at the point. The idea is that if we zoom in far enough toward the point, the curve looks almost like a straight line. Figure 2 illustrates this procedure for the curve in Example 1. The more we zoom in, the more the parabola looks like a line. In other words, the curve becomes almost indistinguishable from its tangent line. There is another expression for the slope of a tangent line that is sometimes easier to use. If , then and so the slope of the secant line is mPQ ! f"a # h# " f"a# h PQ x ! a # h h ! x " a FIGURE 2 Zooming in toward the point (1, 1) on the parabola y=≈ (1, 1) 2 0 2 (1, 1) 1.5 0.5 1.5 (1, 1) 1.1 0.9 1.1 y ! x 2 y ! 2x " 1 or y " 1 ! 2"x " 1# "1, 1# ! lim x l1 "x # 1# ! 1 # 1 ! 2 ! lim x l1 "x " 1#"x # 1# x " 1 m ! lim x l1 f"x# " f"1# x " 1 ! lim x l1 x 2 " 1 x " 1 f"x# ! x 2 a ! 1 P"1, 1# y ! x 2 V m ! lim x l a f"x# " f"a# x " a P P"a, f"a## y ! f"x# 1 P Q PQ 144 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES N Point-slope form for a line through the point with slope : y " y1 ! m"x " x1# m "x1, y1# FIGURE 1 0 x y P t Q Q Q 0 x y a x P{a, f(a)} ƒ-f(a) x-a Q{x, ƒ} Visual 2.7 shows an animation of Figure 2. TEC (See Figure 3 where the case is illustrated and is to the right of . If it happened that , however, would be to the left of .) Notice that as approaches , approaches (because ) and so the expres-sion for the slope of the tangent line in Deﬁnition 1 becomes EXAMPLE 2 Find an equation of the tangent line to the hyperbola at the point . SOLUTION Let . Then the slope of the tangent at is Therefore an equation of the tangent at the point is which simpliﬁes to The hyperbola and its tangent are shown in Figure 4. M VELOCITIES In Section 2.1 we investigated the motion of a ball dropped from the CN Tower and deﬁned its velocity to be the limiting value of average velocities over shorter and shorter time periods. In general, suppose an object moves along a straight line according to an equation of motion , where is the displacement (directed distance) of the object from the ori-gin at time . The function that describes the motion is called the position function of the object. In the time interval from to the change in position is . (See Figure 5.) The average velocity over this time interval is which is the same as the slope of the secant line in Figure 6. Now suppose we compute the average velocities over shorter and shorter time intervals . In other words, we let approach . As in the example of the falling ball, we deﬁne the velocity (or instantaneous velocity) at time to be the limit of these average velocities: v"a# ! lim h l 0 f"a # h# " f"a# h 3 t ! a v"a# 0 h $a, a # h% PQ average velocity ! displacement time ! f"a # h# " f"a# h f"a # h# " f"a# t ! a # h t ! a f t s s ! f"t# x # 3y " 6 ! 0 y " 1 ! " 1 3"x " 3# "3, 1# ! lim h l 0 "h h"3 # h# ! lim h l 0 " 1 3 # h ! " 1 3 ! lim h l 0 3 3 # h " 1 h ! lim h l 0 3 " "3 # h# 3 # h h m ! lim h l 0 f"3 # h# " f"3# h "3, 1# f"x# ! 3!x "3, 1# y ! 3!x m ! lim h l 0 f"a # h# " f"a# h 2 h ! x " a 0 h a x P Q h $ 0 P Q h % 0 SECTION 2.7 DERIVATIVES AND RATES OF CHANGE \|\|\|\| 145 FIGURE 3 0 x y a a+h P{a, f(a)} h Q{a+h, f(a+h)} t f(a+h)-f(a) FIGURE 4 y= (3, 1) x+3y-6=0 x y 0 3 x FIGURE 5 0 s f(a+h)-f(a) position at time t=a position at time t=a+h f(a) f(a+h) 0 P{a, f(a)} Q{a+h, f(a+h)} h a+h a s t mPQ= ! average velocity FIGURE 6 f(a+h)-f(a) h This means that the velocity at time is equal to the slope of the tangent line at (compare Equations 2 and 3). Now that we know how to compute limits, let’s reconsider the problem of the fall-ing ball. EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. (a) What is the velocity of the ball after 5 seconds? (b) How fast is the ball traveling when it hits the ground? SOLUTION We will need to ﬁnd the velocity both when and when the ball hits the ground, so it’s efﬁcient to start by ﬁnding the velocity at a general time . Using the equation of motion , we have (a) The velocity after 5 s is m!s. (b) Since the observation deck is 450 m above the ground, the ball will hit the ground at the time when , that is, This gives The velocity of the ball as it hits the ground is therefore M DERIVATIVES We have seen that the same type of limit arises in ﬁnding the slope of a tangent line (Equation 2) or the velocity of an object (Equation 3). In fact, limits of the form arise whenever we calculate a rate of change in any of the sciences or engineering, such as a rate of reaction in chemistry or a marginal cost in economics. Since this type of limit occurs so widely, it is given a special name and notation. DEFINITION The derivative of a function at a number , denoted by , is if this limit exists. f)"a# ! lim h l 0 f"a # h# " f"a# h f)"a# a f 4 lim h l 0 f"a # h# " f"a# h v"t1# ! 9.8t1 ! 9.8' 450 4.9 ( 94 m!s t1 !' 450 4.9 ( 9.6 s and t1 2 ! 450 4.9 4.9t1 2 ! 450 s"t1# ! 450 t1 v"5# ! "9.8#"5# ! 49 ! lim h l 0 4.9"2a # h# ! 9.8a ! lim h l 0 4.9"a 2 # 2ah # h 2 " a 2# h ! lim h l 0 4.9"2ah # h 2# h v"a# ! lim h l 0 f"a # h# " f"a# h ! lim h l 0 4.9"a # h#2 " 4.9a 2 h s ! f"t# ! 4.9t 2 t ! a t ! 5 V P t ! a 146 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES N Recall from Section 2.1: The distance (in meters) fallen after seconds is . 4.9t 2 t N is read “ prime of .” a f f)"a# SECTION 2.7 DERIVATIVES AND RATES OF CHANGE \|\|\|\| 147 If we write , then we have and approaches if and only if approaches . Therefore an equivalent way of stating the deﬁnition of the derivative, as we saw in ﬁnding tangent lines, is EXAMPLE 4 Find the derivative of the function at the number . SOLUTION From Deﬁnition 4 we have M We deﬁned the tangent line to the curve at the point to be the line that passes through and has slope given by Equation 1 or 2. Since, by Deﬁnition 4, this is the same as the derivative , we can now say the following. The tangent line to at is the line through whose slope is equal to , the derivative of at . If we use the point-slope form of the equation of a line, we can write an equation of the tangent line to the curve at the point : EXAMPLE 5 Find an equation of the tangent line to the parabola at the point . SOLUTION From Example 4 we know that the derivative of at the number is . Therefore the slope of the tangent line at is . Thus an equation of the tangent line, shown in Figure 7, is or M RATES OF CHANGE Suppose is a quantity that depends on another quantity . Thus is a function of and we write . If changes from to , then the change in (also called the incre-ment of ) is x ! x2 " x1 x x x2 x1 x y ! f"x# x y x y y ! "2x y " ""6# ! ""2#"x " 3# f)"3# ! 2"3# " 8 ! "2 "3, "6# f)"a# ! 2a " 8 a f"x# ! x 2 " 8x # 9 "3, "6# y ! x 2 " 8x # 9 V y " f"a# ! f)"a#"x " a# "a, f"a## y ! f"x# a f f)"a# "a, f"a## "a, f"a## y ! f"x# f)"a# m P P"a, f"a## y ! f"x# ! 2a " 8 ! lim h l 0 2ah # h 2 " 8h h ! lim h l 0 "2a # h " 8# ! lim h l 0 a 2 # 2ah # h 2 " 8a " 8h # 9 " a 2 # 8a " 9 h ! lim h l 0 $"a # h#2 " 8"a # h# # 9% " $a 2 " 8a # 9% h f)"a# ! lim h l 0 f"a # h# " f"a# h a f"x# ! x 2 " 8x # 9 V f)"a# ! lim x l a f"x# " f"a# x " a 5 a x 0 h h ! x " a x ! a # h y=≈-8x+9 (3, _6) y=_2x FIGURE 7 0 x y and the corresponding change in is The difference quotient is called the average rate of change of y with respect to x over the interval and can be interpreted as the slope of the secant line in Figure 8. By analogy with velocity, we consider the average rate of change over smaller and smaller intervals by letting approach and therefore letting approach . The limit of these average rates of change is called the (instantaneous) rate of change of y with respect to x at , which is interpreted as the slope of the tangent to the curve at : We recognize this limit as being the derivative . We know that one interpretation of the derivative is as the slope of the tangent line to the curve when . We now have a second interpretation: The derivative is the instantaneous rate of change of with respect to when . The connection with the ﬁrst interpretation is that if we sketch the curve , then the instantaneous rate of change is the slope of the tangent to this curve at the point where . This means that when the derivative is large (and therefore the curve is steep, as at the point in Figure 9), the -values change rapidly. When the derivative is small, the curve is relatively ﬂat and the -values change slowly. In particular, if is the position function of a particle that moves along a straight line, then is the rate of change of the displacement with respect to the time . In other words, is the velocity of the particle at time . The speed of the particle is the absolute value of the velocity, that is, In the next example we discuss the meaning of the derivative of a function that is deﬁned verbally. EXAMPLE 6 A manufacturer produces bolts of a fabric with a ﬁxed width. The cost of producing x yards of this fabric is dollars. (a) What is the meaning of the derivative ? What are its units? (b) In practical terms, what does it mean to say that ? (c) Which do you think is greater, or ? What about ? SOLUTION (a) The derivative is the instantaneous rate of change of C with respect to x; that is, means the rate of change of the production cost with respect to the number of yards produced. (Economists call this rate of change the marginal cost. This idea is dis-cussed in more detail in Sections 3.7 and 4.7.) f)"x# f)"x# f)"5000# f)"500# f)"50# f)"1000# ! 9 f)"x# C ! f"x# V ) f)"a#). t ! a f)"a# t s f)"a# s ! f"t# y y P x ! a y ! f"x# x ! a x y ! f"x# f)"a# x ! a y ! f"x# f)"a# f)"x1# ! lim x2 l x1 f"x2# " f"x1# x2 " x1 instantaneous rate of change ! lim x l 0 y x 6 P"x1, f"x1## y ! f"x# x ! x1 0 x x1 x2 PQ $x1, x2% y x ! f"x2# " f"x1# x2 " x1 y ! f"x2# " f"x1# y 148 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES average rate of change ! mPQ instantaneous rate of change ! slope of tangent at P FIGURE 8 0 x y ⁄ ¤ Q{¤, ‡} Îx Îy P{⁄, ﬂ} FIGURE 9 The y-values are changing rapidly at P and slowly at Q. P Q x y SECTION 2.7 DERIVATIVES AND RATES OF CHANGE \|\|\|\| 149 Because the units for are the same as the units for the difference quotient . Since is measured in dollars and in yards, it follows that the units for are dollars per yard. (b) The statement that means that, after 1000 yards of fabric have been manufactured, the rate at which the production cost is increasing is $9!yard. (When , C is increasing 9 times as fast as x.) Since is small compared with , we could use the approximation and say that the cost of manufacturing the 1000th yard (or the 1001st) is about $9. (c) The rate at which the production cost is increasing (per yard) is probably lower when x ! 500 than when x ! 50 (the cost of making the 500th yard is less than the cost of the 50th yard) because of economies of scale. (The manufacturer makes more efﬁcient use of the ﬁxed costs of production.) So But, as production expands, the resulting large-scale operation might become inefﬁcient and there might be overtime costs. Thus it is possible that the rate of increase of costs will eventually start to rise. So it may happen that M In the following example we estimate the rate of change of the national debt with respect to time. Here the function is deﬁned not by a formula but by a table of values. EXAMPLE 7 Let be the US national debt at time t. The table in the margin gives approximate values of this function by providing end of year estimates, in billions of dollars, from 1980 to 2000. Interpret and estimate the value of . SOLUTION The derivative means the rate of change of D with respect to t when , that is, the rate of increase of the national debt in 1990. According to Equation 5, So we compute and tabulate values of the difference quotient (the average rates of change) as follows. D)"1990# ! lim t l1990 D"t# " D"1990# t " 1990 t ! 1990 D)"1990# D)"1990# D"t# V f)"5000# % f)"500# f)"50# % f)"500# f)"1000# ( C x ! C 1 ! C x ! 1000 x ! 1 x ! 1000 f)"1000# ! 9 f)"x# x C C!x f)"x# f)"x# ! lim x l 0 C x N Here we are assuming that the cost function is well behaved; in other words, doesn’t oscillate rapidly near . x ! 1000 C"x# t 1980 930.2 1985 1945.9 1990 3233.3 1995 4974.0 2000 5674.2 D"t# t 1980 230.31 1985 257.48 1995 348.14 2000 244.09 D"t# " D"1990# t " 1990 From this table we see that lies somewhere between 257.48 and 348.14 billion dollars per year. [Here we are making the reasonable assumption that the debt didn’t ﬂuctuate wildly between 1980 and 2000.] We estimate that the rate of increase of the national debt of the United States in 1990 was the average of these two numbers, namely Another method would be to plot the debt function and estimate the slope of the tan-gent line when . M In Examples 3, 6, and 7 we saw three speciﬁc examples of rates of change: the veloci-ty of an object is the rate of change of displacement with respect to time; marginal cost is the rate of change of production cost with respect to the number of items produced; the rate of change of the debt with respect to time is of interest in economics. Here is a small sample of other rates of change: In physics, the rate of change of work with respect to time is called power. Chemists who study a chemical reaction are interested in the rate of change in the concentration of a reactant with respect to time (called the rate of reaction). A biologist is interested in the rate of change of the population of a colony of bacteria with respect to time. In fact, the computation of rates of change is important in all of the natu-ral sciences, in engineering, and even in the social sciences. Further examples will be given in Section 3.7. All these rates of change are derivatives and can therefore be interpreted as slopes of tangents. This gives added signiﬁcance to the solution of the tangent problem. Whenever we solve a problem involving tangent lines, we are not just solving a problem in geome-try. We are also implicitly solving a great variety of problems involving rates of change in science and engineering. t ! 1990 D)"1990# ( 303 billion dollars per year D)"1990# 150 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES 5–8 Find an equation of the tangent line to the curve at the given point. 6. 8. (a) Find the slope of the tangent to the curve at the point where . (b) Find equations of the tangent lines at the points and . ; (c) Graph the curve and both tangents on a common screen. 10. (a) Find the slope of the tangent to the curve at the point where . (b) Find equations of the tangent lines at the points and . ; (c) Graph the curve and both tangents on a common screen. 11. (a) A particle starts by moving to the right along a horizontal line; the graph of its position function is shown. When is the particle moving to the right? Moving to the left? Standing still? (4, 1 2) "1, 1# x ! a y ! 1!sx "2, 3# "1, 5# x ! a y ! 3 # 4x 2 " 2x 3 9. "0, 0# y ! 2x "x # 1#2 , (1, 1# y ! sx , 7. ""1, 3# y ! 2x 3 " 5x, "3, 2# y ! x " 1 x " 2 , 5. 1. A curve has equation . (a) Write an expression for the slope of the secant line through the points and . (b) Write an expression for the slope of the tangent line at P. ; 2. Graph the curve in the viewing rectangles by , by , and by . What do you notice about the curve as you zoom in toward the point ? 3. (a) Find the slope of the tangent line to the parabola at the point (i) using Deﬁnition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). ; (c) Graph the parabola and the tangent line. As a check on your work, zoom in toward the point until the parabola and the tangent line are indistinguishable. 4. (a) Find the slope of the tangent line to the curve at the point (i) using Deﬁnition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). ; (c) Graph the curve and the tangent line in successively smaller viewing rectangles centered at until the curve and the line appear to coincide. "1, 0# "1, 0# y ! x " x 3 "1, 3# "1, 3# y ! 4x " x 2 "0, 1# $0.9, 1.1% $"0.1, 0.1% $0.5, 1.5% $"0.5, 0.5% $0, 2% $"1, 1% y ! e x Q"x, f "x## P"3, f "3## y ! f "x# EXERCISES 2.7 N A NOTE ON UNITS The units for the average rate of change are the units for divided by the units for , namely, billions of dollars per year. The instan-taneous rate of change is the limit of the aver-age rates of change, so it is measured in the same units: billions of dollars per year. t D D!t SECTION 2.7 DERIVATIVES AND RATES OF CHANGE \|\|\|\| 151 (a) Find an equation of the tangent line to the graph of at if and . (b) If the tangent line to at (4, 3) passes through the point (0, 2), ﬁnd and . Sketch the graph of a function for which , , and . 20. Sketch the graph of a function for which , , and . 21. If , ﬁnd and use it to ﬁnd an equation of the tangent line to the parabola at the point . 22. If , ﬁnd and use it to ﬁnd an equation of the tangent line to the curve at the point . (a) If , ﬁnd and use it to ﬁnd an equation of the tangent line to the curve at the point . ; (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen. 24. (a) If , ﬁnd and use it to ﬁnd equa-tions of the tangent lines to the curve at the points and . ; (b) Illustrate part (a) by graphing the curve and the tangent lines on the same screen. 25–30 Find . 25. 26. 28. 29. 30. 31–36 Each limit represents the derivative of some function at some number . State such an and in each case. 31. 32. 33. 34. 36. lim t l 1 t 4 # t " 2 t " 1 lim h l 0 cos"+ # h# # 1 h 35. lim x l +!4 tan x " 1 x " +!4 lim x l 5 2x " 32 x " 5 lim h l 0 s 4 16 # h " 2 h lim h l 0 "1 # h#10 " 1 h a f a f f "x# ! s3x # 1 f "x# ! 1 sx # 2 f "x# ! x 2 # 1 x " 2 f "t# ! 2t # 1 t # 3 27. f "t# ! t 4 " 5t f "x# ! 3 " 2x # 4x 2 f )"a# "3, 9# "2, 8# y ! 4x 2 " x 3 G)"a# G"x# ! 4x 2 " x 3 "2, 2# y ! 5x!"1 # x 2# F)"2# F"x# ! 5x!"1 # x 2# 23. "0, 1# y ! 1 " x 3 t)"0# t"x# ! 1 " x 3 "2, 2# y ! 3x 2 " 5x f )"2# f "x# ! 3x 2 " 5x t)"2# ! 1 t)"1# ! 3 t)""1# ! "1 t"0# ! t)"0# ! 0, t f )"2# ! "1 f )"1# ! 0 f )"0# ! 3, f "0# ! 0 f 19. f )"4# f "4# y ! f "x# t)"5# ! 4 t"5# ! "3 x ! 5 y ! t"x# 18. y=© 1 3 4 _1 0 x 2 y (b) Draw a graph of the velocity function. 12. Shown are graphs of the position functions of two runners, and , who run a 100-m race and ﬁnish in a tie. (a) Describe and compare how the runners run the race. (b) At what time is the distance between the runners the greatest? (c) At what time do they have the same velocity? If a ball is thrown into the air with a velocity of 40 ft!s, its height (in feet) after seconds is given by . Find the velocity when . 14. If a rock is thrown upward on the planet Mars with a velocity of , its height (in meters) after seconds is given by . (a) Find the velocity of the rock after one second. (b) Find the velocity of the rock when . (c) When will the rock hit the surface? (d) With what velocity will the rock hit the surface? 15. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion , where is measured in seconds. Find the velocity of the particle at times , and . 16. The displacement (in meters) of a particle moving in a straight line is given by , where is mea-sured in seconds. (a) Find the average velocity over each time interval: (i) (ii) (iii) (iv) (b) Find the instantaneous velocity when . (c) Draw the graph of as a function of and draw the secant lines whose slopes are the average velocities in part (a) and the tangent line whose slope is the instantaneous velocity in part (b). For the function t whose graph is given, arrange the follow-ing numbers in increasing order and explain your reasoning: 0 t)""2# t)"0# t)"2# t)"4# 17. t s t ! 4 $4, 4.5% $4, 5% $3.5, 4% $3, 4% t s ! t 2 " 8t # 18 t ! 3 t ! a, t ! 1, t ! 2 t s ! 1!t 2 t ! a H ! 10t " 1.86t 2 t 10 m!s t ! 2 y ! 40t " 16t 2 t 13. s (meters) 0 4 8 12 80 40 t (seconds) A B B A s (meters) 0 2 4 6 4 2 t (seconds) (b) Estimate the instantaneous rate of growth in 2000 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 2000 by mea-suring the slope of a tangent. The cost (in dollars) of producing units of a certain com-modity is . (a) Find the average rate of change of with respect to when the production level is changed (i) from to (ii) from to (b) Find the instantaneous rate of change of with respect to when . (This is called the marginal cost. Its signiﬁcance will be explained in Section 3.7.) 44. If a cylindrical tank holds 100,000 gallons of water, which can be drained from the bottom of the tank in an hour, then Torri-celli’s Law gives the volume of water remaining in the tank after minutes as Find the rate at which the water is ﬂowing out of the tank (the instantaneous rate of change of with respect to ) as a func-tion of t. What are its units? For times t ! 0, 10, 20, 30, 40, 50, and 60 min, ﬁnd the ﬂow rate and the amount of water remain-ing in the tank. Summarize your ﬁndings in a sentence or two. At what time is the ﬂow rate the greatest? The least? The cost of producing x ounces of gold from a new gold mine is dollars. (a) What is the meaning of the derivative ? What are its units? (b) What does the statement mean? (c) Do you think the values of will increase or decrease in the short term? What about the long term? Explain. 46. The number of bacteria after t hours in a controlled laboratory experiment is . (a) What is the meaning of the derivative ? What are its units? (b) Suppose there is an unlimited amount of space and nutrients for the bacteria. Which do you think is larger, or ? If the supply of nutrients is limited, would that affect your conclusion? Explain. Let be the temperature (in ) in Dallas hours after mid-night on June 2, 2001. The table shows values of this function recorded every two hours. What is the meaning of ? Estimate its value. T!!10" t "F T!t" 47. f !!10" f !!5" f !!5" n ! f !t" f !!x" f !!800" ! 17 f !!x" C ! f !x" 45. t V 0 # t # 60 V!t" ! 100,000#1 $ t 60$ 2 t V x ! 100 x C x ! 101 x ! 100 x ! 105 x ! 100 x C C!x" ! 5000 % 10x % 0.05x 2 x 43. 37–38 A particle moves along a straight line with equation of motion , where is measured in meters and in seconds. Find the velocity and the speed when . 37. 38. A warm can of soda is placed in a cold refrigerator. Sketch the graph of the temperature of the soda as a function of time. Is the initial rate of change of temperature greater or less than the rate of change after an hour? 40. A roast turkey is taken from an oven when its temperature has reached 185°F and is placed on a table in a room where the temperature is 75°F. The graph shows how the temperature of the turkey decreases and eventually approaches room tempera-ture. By measuring the slope of the tangent, estimate the rate of change of the temperature after an hour. 41. The table shows the estimated percentage of the population of Europe that use cell phones. (Midyear estimates are given.) (a) Find the average rate of cell phone growth (i) from 2000 to 2002 (ii) from 2000 to 2001 (iii) from 1999 to 2000 In each case, include the units. (b) Estimate the instantaneous rate of growth in 2000 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 2000 by mea-suring the slope of a tangent. 42. The number of locations of a popular coffeehouse chain is given in the table. (The numbers of locations as of June 30 are given.) (a) Find the average rate of growth (i) from 2000 to 2002 (ii) from 2000 to 2001 (iii) from 1999 to 2000 In each case, include the units. N P P T (°F) 0 30 60 90 120 150 100 200 t (min) 39. f !t" ! t $1 $ t f !t" ! 100 % 50t $ 4.9t 2 t ! 5 t s s ! f !t" 152 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES Year 1998 1999 2000 2001 2002 2003 P 28 39 55 68 77 83 Year 1998 1999 2000 2001 2002 N 1886 2135 3501 4709 5886 t 0 2 4 6 8 10 12 14 T 73 73 70 69 72 81 88 91 WRITING PROJECT EARLY METHODS FOR FINDING TANGENTS \|\|\|\| 153 50. The graph shows the inﬂuence of the temperature on the maximum sustainable swimming speed of Coho salmon. (a) What is the meaning of the derivative ? What are its units? (b) Estimate the values of and and interpret them. 51–52 Determine whether exists. 52. f !x" !% x 2 sin 1 x if x " 0 0 if x ! 0 f !x" !% x sin 1 x if x " 0 0 if x ! 0 51. f !!0" 20 0 T (°C) 10 S (cm/s) 20 S!!25" S!!15" S!!T" S T 48. The quantity (in pounds) of a gourmet ground coffee that is sold by a coffee company at a price of p dollars per pound is . (a) What is the meaning of the derivative ? What are its units? (b) Is positive or negative? Explain. 49. The quantity of oxygen that can dissolve in water depends on the temperature of the water. (So thermal pollution inﬂuences the oxygen content of water.) The graph shows how oxygen solubility varies as a function of the water temperature . (a) What is the meaning of the derivative ? What are its units? (b) Estimate the value of and interpret it. (mg/L) 4 8 12 16 S 0 T (°C) Adapted from Environmental Science: Science: Living Within the System of Nature, 2d ed.; by Charles E. Kupchella, © 1989. Reprinted by permission of Prentice-Hall, Inc., Upper Saddle River, NJ. 8 16 24 32 40 S!!16" S!!T" T S f !!8" f !!8" Q ! f ! p" The ﬁrst person to formulate explicitly the ideas of limits and derivatives was Sir Isaac Newton in the 1660s. But Newton acknowledged that “If I have seen further than other men, it is because I have stood on the shoulders of giants.” Two of those giants were Pierre Fermat (1601–1665) and Newton’s teacher at Cambridge, Isaac Barrow (1630–1677). Newton was familiar with the meth-ods that these men used to ﬁnd tangent lines, and their methods played a role in Newton’s eventual formulation of calculus. The following references contain explanations of these methods. Read one or more of the references and write a report comparing the methods of either Fermat or Barrow to modern methods. In particular, use the method of Section 2.7 to ﬁnd an equation of the tangent line to the curve at the point (1, 3) and show how either Fermat or Barrow would have solved the same problem. Although you used derivatives and they did not, point out similarities between the methods. 1. Carl Boyer and Uta Merzbach, A History of Mathematics (New York: Wiley, 1989), pp. 389, 432. 2. C. H. Edwards, The Historical Development of the Calculus (New York: Springer-Verlag, 1979), pp. 124, 132. 3. Howard Eves, An Introduction to the History of Mathematics, 6th ed. (New York: Saunders, 1990), pp. 391, 395. 4. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford University Press, 1972), pp. 344, 346. y ! x 3 % 2x EARLY METHODS FOR FINDING TANGENTS W R I T I N G P R O J E C T THE DERIVATIVE AS A FUNCTION In the preceding section we considered the derivative of a function f at a ﬁxed number a: . Here we change our point of view and let the number a vary. If we replace a in Equation 1 by a variable x, we obtain Given any number x for which this limit exists, we assign to x the number . So we can regard as a new function, called the derivative of and deﬁned by Equation 2. We know that the value of at , , can be interpreted geometrically as the slope of the tangent line to the graph of at the point . The function is called the derivative of because it has been “derived” from by the limiting operation in Equation 2. The domain of is the set exists and may be smaller than the domain of . EXAMPLE 1 The graph of a function is given in Figure 1. Use it to sketch the graph of the derivative . SOLUTION We can estimate the value of the derivative at any value of by drawing the tangent at the point and estimating its slope. For instance, for x ! 5 we draw the tangent at in Figure 2(a) and estimate its slope to be about , so . This allows us to plot the point on the graph of directly beneath P. Repeating this procedure at several points, we get the graph shown in Figure 2(b). Notice that the tangents at , , and are horizontal, so the derivative is 0 there and the graph of crosses the -axis at the points , , and , directly beneath A, B, and C. Between and the tangents have positive slope, so is positive there. But between and the tangents have negative slope, so is negative there. f!!x" C B f!!x" B A C! B! A ! x f! C B A f! P!!5, 1.5" f!!5" & 1.5 3 2 P !x, f!x"" x FIGURE 1 x y 1 0 1 y=ƒ f! f V f ' (x) f!!x" f! f f f! !x, f!x"" f f!!x" x f! f f! f!!x" f!!x" ! lim h l 0 f!x % h" $ f!x" h 2 f!!a" ! lim h l 0 f!a % h" $ f!a" h 1 2.8 154 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES M EXAMPLE 2 (a) If , ﬁnd a formula for . (b) Illustrate by comparing the graphs of and . SOLUTION (a) When using Equation 2 to compute a derivative, we must remember that the variable is and that is temporarily regarded as a constant during the calculation of the limit. ! lim h l 0 !3x 2 % 3xh % h 2 $ 1" ! 3x 2 $ 1 ! lim h l 0 3x 2h % 3xh 2 % h 3 $ h h ! lim h l 0 x 3 % 3x 2h % 3xh 2 % h 3 $ x $ h $ x 3 % x h f!!x" ! lim h l 0 f!x % h" $ f!x" h ! lim h l 0 !x % h"3 $ !x % h"+ $ x 3 $ x+ h x h f! f f!!x" f!x" ! x 3 $ x V FIGURE 2 m=0 m=0 m=0 Pª(5, 1.5) y B A mÅ C P (a) x 1 1 0 5 y=ƒ y Aª Bª Cª (b) x 1 1 0 5 y=fª(x) 3 2 SECTION 2.8 THE DERIVATIVE AS A FUNCTION \|\|\|\| 155 Visual 2.8 shows an animation of Figure 2 for several functions. TEC (b) We use a graphing device to graph and in Figure 3. Notice that when has horizontal tangents and is positive when the tangents have positive slope. So these graphs serve as a check on our work in part (a). M EXAMPLE 3 If , ﬁnd the derivative of . State the domain of . SOLUTION We see that exists if , so the domain of is . This is smaller than the domain of , which is . M Let’s check to see that the result of Example 3 is reasonable by looking at the graphs of and in Figure 4. When is close to 0, is also close to , so is very large and this corresponds to the steep tangent lines near in Figure 4(a) and the large values of just to the right of 0 in Figure 4(b). When is large, is very small and this corresponds to the ﬂatter tangent lines at the far right of the graph of and the horizontal asymptote of the graph of . EXAMPLE 4 Find if . SOLUTION M ! lim h l 0 $3 !2 % x % h"!2 % x" ! $ 3 !2 % x"2 ! lim h l 0 $3h h!2 % x % h"!2 % x" ! lim h l 0 !2 $ x $ 2h $ x 2 $ xh" $ !2 $ x % h $ x 2 $ xh" h!2 % x % h"!2 % x" ! lim h l 0 !1 $ x $ h"!2 % x" $ !1 $ x"!2 % x % h" h!2 % x % h"!2 % x" ! lim h l 0 1 $ !x % h" 2 % !x % h" $ 1 $ x 2 % x h f!!x" ! lim h l 0 f!x % h" $ f!x" h f!x" ! 1 $ x 2 % x f! f! f f!!x" x f!!x" !0, 0" f!!x" ! 1,(2sx ) 0 sx x f! f 0, &" f !0, &" f! x ' 0 f!!x" ! 1 sx % sx ! 1 2sx ! lim h l 0 1 sx % h % sx ! lim h l 0 !x % h" $ x h(sx % h % sx ) ! lim h l 0 # sx % h $ sx h ! sx % h % sx sx % h % sx $ ! lim h l 0 sx % h $ sx h f!!x" ! lim h l 0 f!x % h" $ f!x" h f! f f!x" ! sx FIGURE 3 2 _2 _2 2 2 _2 _2 2 f fª f!!x" f f!!x" ! 0 f! f 156 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES Here we rationalize the numerator. FIGURE 4 (a) ƒ=œ„ x 1 2œ„ x (b) fª(x)= x 1 y 1 0 x 1 y 1 0 a b $ c d e ! ad $ bc bd ! 1 e OTHER NOTATIONS If we use the traditional notation to indicate that the independent variable is and the dependent variable is , then some common alternative notations for the derivative are as follows: The symbols and are called differentiation operators because they indicate the operation of differentiation, which is the process of calculating a derivative. The symbol , which was introduced by Leibniz, should not be regarded as a ratio (for the time being); it is simply a synonym for . Nonetheless, it is a very useful and suggestive notation, especially when used in conjunction with increment notation. Refer-ring to Equation 2.7.6, we can rewrite the deﬁnition of derivative in Leibniz notation in the form If we want to indicate the value of a derivative in Leibniz notation at a speciﬁc num-ber , we use the notation or which is a synonym for . DEFINITION A function is differentiable at a if exists. It is differen-tiable on an open interval [or or or ] if it is differ-entiable at every number in the interval. EXAMPLE 5 Where is the function differentiable? SOLUTION If , then and we can choose small enough that and hence . Therefore, for , we have and so is differentiable for any . Similarly, for we have and can be chosen small enough that . Therefore, for , and so is differentiable for any . x ( 0 f ! lim h l 0 $!x % h" $ !$x" h ! lim h l 0 $h h ! lim h l 0 !$1" ! $1 f!!x" ! lim h l 0 )x % h) $ )x) h x ( 0 x % h ( 0 and so )x % h) ! $!x % h" h )x) ! $x x ( 0 x ' 0 f ! lim h l 0 !x % h" $ x h ! lim h l 0 h h ! lim h l 0 1 ! 1 f!!x" ! lim h l 0 )x % h) $ )x) h x ' 0 )x % h) ! x % h x % h ' 0 h )x) ! x x ' 0 f!x" ! )x) V !$&, &" !$&, a" !a, &" !a, b" f!!a" f 3 f!!a" dy dx-x!a dy dx . x!a a dy,dx dy dx ! lim )x l 0 )y )x f!!x" dy,dx d,dx D f!!x" ! y! ! dy dx ! df dx ! d dx f!x" ! Df!x" ! Dx f!x" y x y ! f!x" SECTION 2.8 THE DERIVATIVE AS A FUNCTION \|\|\|\| 157 Gottfried Wilhelm Leibniz was born in Leipzig in 1646 and studied law, theology, philosophy, and mathematics at the university there, gradu-ating with a bachelor’s degree at age 17. After earning his doctorate in law at age 20, Leibniz entered the diplomatic service and spent most of his life traveling to the capitals of Europe on political missions. In particular, he worked to avert a French military threat against Germany and attempted to reconcile the Catholic and Protestant churches. His serious study of mathematics did not begin until 1672 while he was on a diplomatic mission in Paris. There he built a calculating machine and met scientists, like Huygens, who directed his attention to the latest developments in mathe-matics and science. Leibniz sought to develop a symbolic logic and system of notation that would simplify logical reasoning. In particular, the ver-sion of calculus that he published in 1684 estab-lished the notation and the rules for ﬁnding derivatives that we use today. Unfortunately, a dreadful priority dispute arose in the 1690s between the followers of Newton and those of Leibniz as to who had invented calculus ﬁrst. Leibniz was even accused of pla-giarism by members of the Royal Society in England. The truth is that each man invented calculus independently. Newton arrived at his version of calculus ﬁrst but, because of his fear of controversy, did not publish it immediately. So Leibniz’s 1684 account of calculus was the ﬁrst to be published. LEIBNIZ For we have to investigate Let’s compute the left and right limits separately: and Since these limits are different, does not exist. Thus is differentiable at all except 0. A formula for is given by and its graph is shown in Figure 5(b). The fact that does not exist is reﬂected geo-metrically in the fact that the curve does not have a tangent line at . [See Figure 5(a).] M Both continuity and differentiability are desirable properties for a function to have. The following theorem shows how these properties are related. THEOREM If is differentiable at , then is continuous at . PROOF To prove that is continuous at , we have to show that . We do this by showing that the difference approaches 0. The given information is that f is differentiable at a, that is, exists (see Equation 2.7.5). To connect the given and the unknown, we divide and multi-ply by (which we can do when ): Thus, using the Product Law and (2.7.5), we can write ! f!!a" ! 0 ! 0 ! lim x l a f!x" $ f!a" x $ a ! lim x l a !x $ a" lim x l a f!x" $ f!a"+ ! lim x l a f!x" $ f!a" x $ a !x $ a" f!x" $ f!a" ! f!x" $ f!a" x $ a !x $ a" x " a x $ a f!x" $ f!a" f!!a" ! lim x l a f!x" $ f!a" x $ a f!x" $ f!a" f!x" ! f!a" lim x l a a f a f a f 4 !0, 0" y ! )x) f!!0" f!!x" !% 1 $1 if x ' 0 if x ( 0 f! x f f!!0" lim h l 0$ )0 % h) $ )0) h ! lim h l 0$ )h) h ! lim h l 0$ $h h ! lim h l 0$ !$1" ! $1 lim h l 0% )0 % h) $ )0) h ! lim h l 0% )h) h ! lim h l 0% h h ! lim h l 0% 1 ! 1 ! lim h l 0 )0 % h) $ )0) h !if it exists" f!!0" ! lim h l 0 f!0 % h" $ f!0" h x ! 0 158 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES x 1 y _1 0 x y 0 FIGURE 5 (a) y=ƒ=\| x \| (b) y=fª(x) To use what we have just proved, we start with and add and subtract : Therefore is continuous at . M \| The converse of Theorem 4 is false; that is, there are functions that are continuous but not differentiable. For instance, the function is continuous at 0 because (See Example 7 in Section 2.3.) But in Example 5 we showed that is not differentiable at 0. HOW CAN A FUNCTION FAIL TO BE DIFFERENTIABLE? We saw that the function in Example 5 is not differentiable at 0 and Figure 5(a) shows that its graph changes direction abruptly when . In general, if the graph of a function has a “corner” or “kink” in it, then the graph of has no tangent at this point and is not differentiable there. [In trying to compute , we ﬁnd that the left and right limits are different.] Theorem 4 gives another way for a function not to have a derivative. It says that if is not continuous at , then is not differentiable at . So at any discontinuity (for instance, a jump discontinuity) fails to be differentiable. A third possibility is that the curve has a vertical tangent line when ; that is, is continuous at and This means that the tangent lines become steeper and steeper as . Figure 6 shows one way that this can happen; Figure 7(c) shows another. Figure 7 illustrates the three possi-bilities that we have discussed. A graphing calculator or computer provides another way of looking at differentiabil-ity. If is differentiable at , then when we zoom in toward the point the graph !a, f!a"" a f FIGURE 7 Three ways for ƒ not to be differentiable at a (a) A corner (c) A vertical tangent (b) A discontinuity x y a 0 x y a 0 x y a 0 x l a lim x l a ) f!!x") ! & a f x ! a f a f a f f!!a" f f f x ! 0 y ! )x) f lim x l 0 f!x" ! lim x l 0 )x) ! 0 ! f!0" f!x" ! )x) NOTE a f ! f!a" % 0 ! f!a" ! lim x l a f!a" % lim x l a f!x" $ f!a"+ lim x l a f!x" ! lim x l a f!a" % ! f!x" $ f!a""+ f!a" f!x" SECTION 2.8 THE DERIVATIVE AS A FUNCTION \|\|\|\| 159 FIGURE 6 vertical tangent line x y a 0 straightens out and appears more and more like a line. (See Figure 8. We saw a speciﬁc example of this in Figure 2 in Section 2.7.) But no matter how much we zoom in toward a point like the ones in Figures 6 and 7(a), we can’t eliminate the sharp point or corner (see Figure 9). HIGHER DERIVATIVES If is a differentiable function, then its derivative is also a function, so may have a derivative of its own, denoted by . This new function is called the second derivative of because it is the derivative of the derivative of . Using Leibniz notation, we write the second derivative of as EXAMPLE 6 If , ﬁnd and interpret . SOLUTION In Example 2 we found that the ﬁrst derivative is . So the sec-ond derivative is The graphs of , , are shown in Figure 10. We can interpret as the slope of the curve at the point . In other words, it is the rate of change of the slope of the original curve . Notice from Figure 10 that is negative when has negative slope and positive when has positive slope. So the graphs serve as a check on our calculations. M In general, we can interpret a second derivative as a rate of change of a rate of change. The most familiar example of this is acceleration, which we deﬁne as follows. If is the position function of an object that moves in a straight line, we know that its ﬁrst derivative represents the velocity of the object as a function of time: v!t" ! s!!t" ! ds dt v!t" s ! s!t" y ! f!!x" y ! f!!x" f "!x" y ! f!x" !x, f !!x"" y ! f!!x" f "!x" f " f! f ! lim h l 0 !6x # 3h" ! 6x ! lim h l 0 3x 2 # 6xh # 3h2 $ 1 $ 3x 2 # 1 h ! lim h l 0 #3!x # h"2 $ 1$ $ #3x 2 $ 1$ h f! !!x" ! ! f!"!!x" ! lim h l 0 f!!x # h" $ f!!x" h f!!x" ! 3x 2 $ 1 f "!x" f!x" ! x 3 $ x d dx % dy dx& ! d 2y dx 2 y ! f!x" f f f " ! f!"! ! f " f! f! f FIGURE 8 ƒ is differentiable at a. FIGURE 9 ƒ is not differentiable at a. x y a 0 x y a 0 160 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES FIGURE 10 f · fª f 1.5 _2 2 _1.5 In Module 2.8 you can see how changing the coefﬁcients of a polynomial affects the appearance of the graphs of , , and . f " f ! f f TEC The instantaneous rate of change of velocity with respect to time is called the acceleration of the object. Thus the acceleration function is the derivative of the velocity function and is therefore the second derivative of the position function: or, in Leibniz notation, The third derivative is the derivative of the second derivative: . So can be interpreted as the slope of the curve or as the rate of change of . If , then alternative notations for the third derivative are The process can be continued. The fourth derivative is usually denoted by . In gen-eral, the th derivative of is denoted by and is obtained from by differentiating times. If , we write EXAMPLE 7 If , ﬁnd and . SOLUTION In Example 6 we found that . The graph of the second derivative has equation and so it is a straight line with slope 6. Since the derivative is the slope of , we have for all values of . So is a constant function and its graph is a horizontal line. There-fore, for all values of , M We can interpret the third derivative physically in the case where the function is the position function of an object that moves along a straight line. Because , the third derivative of the position function is the derivative of the accel-eration function and is called the jerk: Thus the jerk j is the rate of change of acceleration. It is aptly named because a large jerk means a sudden change in acceleration, which causes an abrupt movement in a vehicle. We have seen that one application of second and third derivatives occurs in analyzing the motion of objects using acceleration and jerk. We will investigate another applica-tion of second derivatives in Section 4.3, where we show how knowledge of gives us information about the shape of the graph of . In Chapter 11 we will see how second and higher derivatives enable us to represent functions as sums of inﬁnite series. f f " j ! da dt ! d 3s dt 3 s% ! !s""! ! a! s ! s!t" f !4"!x" ! 0 x f % x f %!x" ! 6 f "!x" f %!x" y ! 6x f "!x" ! 6x f !4"!x" f %!x" f!x" ! x3 $ x y !n" ! f !n"!x" ! d ny dx n y ! f!x" n f f !n" f n f !4" f " y% ! f %!x" ! d dx % d 2y dx 2& ! d 3y dx 3 y ! f!x" f "!x" y ! f "!x" f %!x" f % ! ! f ""! f % a ! dv dt ! d 2s dt 2 a!t" ! v!!t" ! s"!t" a!t" SECTION 2.8 THE DERIVATIVE AS A FUNCTION \|\|\|\| 161 162 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES 4–11 Trace or copy the graph of the given function . (Assume that the axes have equal scales.) Then use the method of Example 1 to sketch the graph of below it. 4. 6. 7. 8. 9. 10. 12. Shown is the graph of the population function for yeast cells in a laboratory culture. Use the method of Example 1 to (yeast cells) t (hours) P 0 5 10 15 500 P!t" 0 x y 11. x y 0 0 x y 0 x y x y 0 0 x y x y 0 5. 0 x y f ! f 1–2 Use the given graph to estimate the value of each derivative. Then sketch the graph of . 1. (a) (b) (c) (d) (e) (f) (g) 2. (a) (b) (c) (d) (e) (f) Match the graph of each function in (a)–(d) with the graph of its derivative in I–IV. Give reasons for your choices. III IV y 0 x y 0 x II I y 0 y 0 x x y 0 y 0 y 0 y 0 x x x x (b) (a) (c) (d) 3. f !!5" f !!4" f !!3" f !!2" f !!1" y 0 x 1 1 y=f(x) f !!0" f !!3" f !!2" f !!1" f !!0" f !!$1" f !!$2" y 0 x 1 1 y=f(x) f !!$3" f ! EXERCISES 2.8 26. 28. 29. 30. (a) Sketch the graph of by starting with the graph of and using the transformations of Sec-tion 1.3. (b) Use the graph from part (a) to sketch the graph of . (c) Use the deﬁnition of a derivative to ﬁnd . What are the domains of f and ? ; (d) Use a graphing device to graph and compare with your sketch in part (b). 31. (a) If , ﬁnd . ; (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of and . 32. (a) If , ﬁnd . ; (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of and . The unemployment rate varies with time. The table (from the Bureau of Labor Statistics) gives the percentage of unemployed in the US labor force from 1993 to 2002. (a) What is the meaning of ? What are its units? (b) Construct a table of values for . 34. Let be the percentage of Americans under the age of 18 at time . The table gives values of this function in census years from 1950 to 2000. (a) What is the meaning of ? What are its units? (b) Construct a table of estimated values for . (c) Graph and . (d) How would it be possible to get more accurate values for ? P!!t" P! P P!!t" P!!t" t P!t" U!!t" U!!t" U!t" 33. f ! f f !!t" f !t" ! t 2 $ st f ! f f !!x" f !x" ! x 4 # 2x f ! f ! f !!x" f ! y ! sx f !x" ! s6 $ x f !x" ! x 4 t!t" ! 1 st G!t" ! 4t t # 1 27. f !x" ! 3 # x 1 $ 3x t!x" ! s1 # 2x 25. graph the derivative . What does the graph of tell us about the yeast population? 13. The graph shows how the average age of ﬁrst marriage of Japanese men has varied in the last half of the 20th century. Sketch the graph of the derivative function . During which years was the derivative negative? 14–16 Make a careful sketch of the graph of and below it sketch the graph of in the same manner as in Exercises 4–11. Can you guess a formula for from its graph? 14. 15. 16. ; Let . (a) Estimate the values of , , , and by using a graphing device to zoom in on the graph of f. (b) Use symmetry to deduce the values of , , and . (c) Use the results from parts (a) and (b) to guess a formula for . (d) Use the deﬁnition of a derivative to prove that your guess in part (c) is correct. ; 18. Let . (a) Estimate the values of , , , , and by using a graphing device to zoom in on the graph of f. (b) Use symmetry to deduce the values of , , , and . (c) Use the values from parts (a) and (b) to graph . (d) Guess a formula for . (e) Use the deﬁnition of a derivative to prove that your guess in part (d) is correct. 19–29 Find the derivative of the function using the deﬁnition of derivative. State the domain of the function and the domain of its derivative. 19. 20. 21. 22. 23. 24. f !x" ! x # sx f !x" ! x 3 $ 3x # 5 f !x" ! 1.5x 2 $ x # 3.7 f !t" ! 5t $ 9t 2 f !x" ! mx # b f !x" ! 1 2x $ 1 3 f !!x" f ! f !!$3" f !!$2" f !!$1" f !($ 1 2) f !!3" f !!2" f !!1" f !( 1 2) f !!0" f !x" ! x 3 f !!x" f !!$2" f !!$1" f !($ 1 2) f !!2" f !!1" f !( 1 2) f !!0" f !x" ! x 2 17. f !x" ! ln x f !x" ! e x f !x" ! sin x f !!x" f ! f 1990 2000 25 M 1960 1970 1980 27 t M!!t" P! P!!t" SECTION 2.8 THE DERIVATIVE AS A FUNCTION \|\|\|\| 163 t t 1993 6.9 1998 4.5 1994 6.1 1999 4.2 1995 5.6 2000 4.0 1996 5.4 2001 4.7 1997 4.9 2002 5.8 U!t" U!t" t t 1950 31.1 1980 28.0 1960 35.7 1990 25.7 1970 34.0 2000 25.7 P!t" P!t" 43. The ﬁgure shows the graphs of three functions. One is the position function of a car, one is the velocity of the car, and one is its acceleration. Identify each curve, and explain your choices. 44. The ﬁgure shows the graphs of four functions. One is the position function of a car, one is the velocity of the car, one is its acceleration, and one is its jerk. Identify each curve, and explain your choices. ; 45–46 Use the deﬁnition of a derivative to ﬁnd and . Then graph , , and on a common screen and check to see if your answers are reasonable. 45. 46. ; If , ﬁnd , , , and . Graph , , , and on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives? 48. (a) The graph of a position function of a car is shown, where s is measured in feet and t in seconds. Use it to graph the velocity and acceleration of the car. What is the accelera-tion at t ! 10 seconds? (b) Use the acceleration curve from part (a) to estimate the jerk at seconds. What are the units for jerk? t ! 10 10 0 t s 100 20 f % f " f ! f f !4"!x" f %!x" f "!x" f !!x" f !x" ! 2x 2 $ x3 47. f !x" ! 1'x f !x" ! 1 # 4x $ x 2 f " f ! f f "!x" f !!x" 0 t y a b c d t y a b c 0 35–38 The graph of is given. State, with reasons, the numbers at which is not differentiable. 36. 37. 38. ; 39. Graph the function . Zoom in repeatedly, ﬁrst toward the point ($1, 0) and then toward the origin. What is different about the behavior of in the vicinity of these two points? What do you conclude about the differentiability of f ? ; 40. Zoom in toward the points (1, 0), (0, 1), and ($1, 0) on the graph of the function . What do you notice? Account for what you see in terms of the differentiability of t. The ﬁgure shows the graphs of , , and . Identify each curve, and explain your choices. 42. The ﬁgure shows graphs of , , and . Identify each curve, and explain your choices. x y a b c d f % f " f ! f, x y a b c f " f ! f 41. t!x" ! !x 2 $ 1"2'3 f f !x" ! x # s( x( _2 2 x y 0 _2 4 x y 0 2 4 x y 0 _2 2 x y 0 35. f f 164 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES (a) Find and for the function (b) Sketch the graph of . (c) Where is discontinuous? (d) Where is not differentiable? 55. Recall that a function is called even if for all in its domain and odd if for all such . Prove each of the following. (a) The derivative of an even function is an odd function. (b) The derivative of an odd function is an even function. 56. When you turn on a hot-water faucet, the temperature of the water depends on how long the water has been running. (a) Sketch a possible graph of as a function of the time that has elapsed since the faucet was turned on. (b) Describe how the rate of change of with respect to varies as increases. (c) Sketch a graph of the derivative of . 57. Let be the tangent line to the parabola at the point . The angle of inclination of is the angle that makes with the positive direction of the -axis. Calculate correct to the nearest degree. & x ! & ! !1, 1" y ! x 2 ! T t t T t T T x f !$x" ! $f !x" x f !$x" ! f !x" f f f f 1 5 $ x if x ' 4 f !x" ! 0 5 $ x if x ( 0 if 0 ) x ) 4 f ! !4" f ! $!4" 49. Let . (a) If , use Equation 2.7.5 to ﬁnd . (b) Show that does not exist. (c) Show that has a vertical tangent line at . (Recall the shape of the graph of . See Figure 13 in Sec-tion 1.2.) 50. (a) If , show that does not exist. (b) If , ﬁnd . (c) Show that has a vertical tangent line at . ; (d) Illustrate part (c) by graphing . Show that the function is not differentiable at 6. Find a formula for and sketch its graph. 52. Where is the greatest integer function not differ-entiable? Find a formula for and sketch its graph. (a) Sketch the graph of the function . (b) For what values of is differentiable? (c) Find a formula for . 54. The left-hand and right-hand derivatives of at are deﬁned by and if these limits exist. Then exists if and only if these one-sided derivatives exist and are equal. f !!a" f ! !a" ! lim h l 0# f !a # h" $ f !a" h f ! $!a" ! lim h l 0$ f !a # h" $ f !a" h a f f ! f x f !x" ! x( x( 53 f ! f !x" ! ) x f ! f !x" ! ( x $ 6( 51. y ! x 2'3 !0, 0" y ! x 2'3 t!!a" a # 0 t!!0" t!x" ! x 2'3 f !0, 0" y ! s 3 x f !!0" f !!a" a # 0 f !x" ! s 3 x CHAPTER 2 REVIEW \|\|\|\| 165 REVIEW CONCEPT CHECK 2 (b) What does it mean to say that the line is a horizontal asymptote of the curve ? Draw curves to illustrate the various possibilities. 6. Which of the following curves have vertical asymptotes? Which have horizontal asymptotes? (a) (b) (c) (d) (e) (f) (g) (h) 7. (a) What does it mean for f to be continuous at a? (b) What does it mean for f to be continuous on the interval ? What can you say about the graph of such a function? 8. What does the Intermediate Value Theorem say? 9. Write an expression for the slope of the tangent line to the curve at the point . !a, f !a"" y ! f !x" !$, " y ! sx y ! 1'x y ! ln x y ! e x y ! tan$1x y ! tan x y ! sin x y ! x 4 y ! f !x" y ! L 1. Explain what each of the following means and illustrate with a sketch. (a) (b) (c) (d) (e) 2. Describe several ways in which a limit can fail to exist. Illustrate with sketches. 3. State the following Limit Laws. (a) Sum Law (b) Difference Law (c) Constant Multiple Law (d) Product Law (e) Quotient Law (f) Power Law (g) Root Law 4. What does the Squeeze Theorem say? 5. (a) What does it mean to say that the line is a vertical asymptote of the curve ? Draw curves to illustrate the various possibilities. y ! f !x" x ! a lim x l f !x" ! L lim x l a f !x" ! lim x l a$ f !x" ! L lim x l a# f !x" ! L lim x l a f !x" ! L 13. Deﬁne the second derivative of . If is the position function of a particle, how can you interpret the second derivative? 14. (a) What does it mean for to be differentiable at a? (b) What is the relation between the differentiability and conti-nuity of a function? (c) Sketch the graph of a function that is continuous but not differentiable at . 15. Describe several ways in which a function can fail to be differentiable. Illustrate with sketches. a ! 2 f f !t" f 10. Suppose an object moves along a straight line with position at time t. Write an expression for the instantaneous veloc-ity of the object at time . How can you interpret this velocity in terms of the graph of f ? 11. If and x changes from to , write expressions for the following. (a) The average rate of change of y with respect to x over the interval . (b) The instantaneous rate of change of y with respect to x at . 12. Deﬁne the derivative . Discuss two ways of interpreting this number. f !!a" x ! x1 #x1, x2$ x2 x1 y ! f !x" t ! a f !t" Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. 2. 3. 4. If and , then does not exist. 5. If and , then does not exist. 6. If exists, then the limit must be 7. If p is a polynomial, then 8. If and , then . 9. A function can have two different horizontal asymptotes. 10. If has domain and has no horizontal asymptote, then or . limx l f !x" ! $ limx l f !x" ! #0, " f limx l 0 # f !x" $ t!x"$ ! 0 limx l 0 t!x" ! limx l 0 f !x" ! limx l b p!x" ! p!b". f !6"t!6". limx l 6 # f !x"t!x"$ limx l 5 # f !x"'t!x"$ limx l 5 t!x" ! 0 lim x l5 f !x" ! 0 limx l 5 # f !x"'t!x"$ limx l 5 t!x" ! 0 limx l 5 f !x" ! 2 lim x l 1 x $ 3 x 2 # 2x $ 4 ! lim x l 1 !x $ 3" lim x l 1 !x 2 # 2x $ 4" lim x l 1 x 2 # 6x $ 7 x 2 # 5x $ 6 ! lim x l 1 !x 2 # 6x $ 7" lim x l 1 !x 2 # 5x $ 6" lim x l 4 % 2x x $ 4 $ 8 x $ 4& ! lim x l 4 2x x $ 4 $ lim x l 4 8 x $ 4 11. If the line is a vertical asymptote of , then is not deﬁned at 1. 12. If and , then there exists a number c between 1 and 3 such that . 13. If f is continuous at 5 and and , then 14. If f is continuous on and and then there exists a number r such that and . 15. Let be a function such that . Then there exists a number such that if , then . 16. If for all and exists, then . 17. If is continuous at a, then is differentiable at a. 18. If exists, then 19. 20. The equation has a root in the interval . !0, 2" x 10 $ 10x 2 # 5 ! 0 d 2y dx 2 !% dy dx& 2 limx l r f !x" ! f !r". f !!r" f f lim x l 0 f !x" + 1 lim x l 0 f !x" x f !x" + 1 ( f !x" $ 6( ) 1 0 ) ( x( ) , , lim x l 0 f !x" ! 6 f f !r" ! -( r( ) 1 f !1" ! 3, f !$1" ! 4 #$1, 1$ limx l 2 f !4x 2 $ 11" ! 2. f !4" ! 3 f !5" ! 2 f !c" ! 0 f !3" ) 0 f !1" + 0 f y ! f !x" x ! 1 TRUE-FALSE QUIZ 166 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES 1. The graph of is given. (a) Find each limit, or explain why it does not exist. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (b) State the equations of the horizontal asymptotes. (c) State the equations of the vertical asymptotes. (d) At what numbers is discontinuous? Explain. 2. Sketch the graph of an example of a function that satisﬁes all of the following conditions: , , , , , is continuous from the right at 3 3–20 Find the limit. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. lim x l e x$x2 lim x l (sx 2 # 4x # 1 $ x) lim x l $ 1 $ 2x 2 $ x 4 5 # x $ 3x 4 lim x l-$ ln!sin x" lim x l $ sx 2 $ 9 2x $ 6 lim x l sx 2 $ 9 2x $ 6 lim x l 3 sx # 6 $ x x 3 $ 3x 2 lim u l 1 u 4 $ 1 u3 # 5u 2 $ 6u lim v l 4# 4 $ v ( 4 $ v( lim r l 9 sr !r $ 9"4 lim t l 2 t 2 $ 4 t 3 $ 8 lim h l 0 !h $ 1"3 # 1 h lim x l 1# x 2 $ 9 x 2 # 2x $ 3 lim x l $3 x 2 $ 9 x 2 # 2x $ 3 lim x l 3 x 2 $ 9 x 2 # 2x $ 3 lim x l 1 e x3$x f lim x l 3# f !x" ! 2 lim x l 3$ f !x" ! $ lim x l $3 f !x" ! lim x l f !x" ! 0 lim x l $ f !x" ! $2 f 0 x y 1 1 f lim x l $ f !x" lim x l f !x" lim x l 2$ f !x" lim x l 0 f !x" lim x l 4 f !x" lim x l $3 f !x" lim x l $3# f !x" lim x l 2# f !x" f 19. 20. ; 21–22 Use graphs to discover the asymptotes of the curve. Then prove what you have discovered. 21. 22. 23. If for , ﬁnd . 24. Prove that . 25–28 Prove the statement using the precise deﬁnition of a limit. 25. 26. 27. 28. 29. Let (a) Evaluate each limit, if it exists. (i) (ii) (iii) (iv) (v) (vi) (b) Where is discontinuous? (c) Sketch the graph of . 30. Let (a) For each of the numbers 2, 3, and 4, discover whether is continuous from the left, continuous from the right, or con-tinuous at the number. (b) Sketch the graph of . 31–32 Show that each function is continuous on its domain. State the domain. 31. 32. t!x" ! sx 2 $ 9 x 2 $ 2 h!x" ! xesin x t t t!x" ! 2x $ x 2 2 $ x x $ 4 -if 0 ( x ( 2 if 2 ) x ( 3 if 3 ) x ) 4 if x ' 4 f f lim x l 3 f !x" lim x l 3# f !x" lim x l 3$ f !x" lim x l 0 f !x" lim x l 0$ f !x" lim x l 0# f !x" f !x" !+ s$x 3 $ x !x $ 3"2 if x ) 0 if 0 ( x ) 3 if x + 3 lim x l 4# 2 sx $ 4 ! lim x l 2 !x 2 $ 3x" ! $2 lim x l 0 s 3 x ! 0 lim x l 2 !14 $ 5x" ! 4 limx l 0 x 2 cos!1'x 2" ! 0 limx l1 f !x" 0 ) x ) 3 2x $ 1 ( f !x" ( x 2 y ! sx 2 # x # 1 $ sx 2 $ x y ! cos2x x 2 lim x l 1 % 1 x $ 1 # 1 x 2 $ 3x # 2& lim x l 0# tan$1!1'x" EXERCISES CHAPTER 2 REVIEW \|\|\|\| 167 42–44 Trace or copy the graph of the function. Then sketch a graph of its derivative directly beneath. 42. 43. 44. 45. (a) If , use the deﬁnition of a derivative to ﬁnd . (b) Find the domains of and . ; (c) Graph and on a common screen. Compare the graphs to see whether your answer to part (a) is reasonable. 46. (a) Find the asymptotes of the graph of and use them to sketch the graph. (b) Use your graph from part (a) to sketch the graph of . (c) Use the deﬁnition of a derivative to ﬁnd . ; (d) Use a graphing device to graph and compare with your sketch in part (b). 47. The graph of is shown. State, with reasons, the numbers at which is not differentiable. ; 48. The ﬁgure shows the graphs of , , and . Identify each curve, and explain your choices. x y a b c 0 f " f ! f x y 2 0 4 6 _1 f f f ! f !!x" f ! f !x" ! 4 $ x 3 # x f ! f f ! f f !!x" f !x" ! s3 $ 5x x y 0 x y 0 x y 33–34 Use the Intermediate Value Theorem to show that there is a root of the equation in the given interval. 33. 34. , 35. (a) Find the slope of the tangent line to the curve at the point . (b) Find an equation of this tangent line. 36. Find equations of the tangent lines to the curve at the points with -coordinates and 37. The displacement (in meters) of an object moving in a straight line is given by , where is mea-sured in seconds. (a) Find the average velocity over each time period. (i) (ii) (iii) (iv) (b) Find the instantaneous velocity when . 38. According to Boyle’s Law, if the temperature of a conﬁned gas is held ﬁxed, then the product of the pressure and the volume is a constant. Suppose that, for a certain gas, , where is measured in pounds per square inch and is measured in cubic inches. (a) Find the average rate of change of as increases from 200 in to 250 in . (b) Express as a function of and show that the instantan-eous rate of change of with respect to is inversely proportional to the square of . 39. (a) Use the deﬁnition of a derivative to ﬁnd , where . (b) Find an equation of the tangent line to the curve at the point (2, 4). ; (c) Illustrate part (b) by graphing the curve and the tangent line on the same screen. 40. Find a function and a number a such that 41. The total cost of repaying a student loan at an interest rate of r% per year is . (a) What is the meaning of the derivative ? What are its units? (b) What does the statement mean? (c) Is always positive or does it change sign? f !!r" f !!10" ! 1200 f !!r" C ! f !r" lim h l 0 !2 # h"6 $ 64 h ! f !!a" f y ! x 3 $ 2x f !x" ! x 3 $ 2x f !!2" P P V P V 3 3 V P V P PV ! 800 V P t ! 1 #1, 1.1$ #1, 1.5$ #1, 2$ #1, 3$ t s ! 1 # 2t # 1 4t 2 $1. 0 x y ! 2 1 $ 3x !2, 1" y ! 9 $ 2x 2 !0, 1" e$x2 ! x !$2, $1" 2x 3 # x 2 # 2 ! 0, 168 \|\|\|\| CHAPTER 2 LIMITS AND DERIVATIVES 51. Suppose that for all , where . Find . 52. Let . (a) For what values of does exist? (b) At what numbers is discontinuous? f lim x l a f !x" a f !x" ! )x # )$x lim x l a f !x" lim x l a t!x" ! 0 x ( f !x"( ( t!x" t y 1940 1960 1970 1980 1990 1950 1.5 2.0 2.5 3.0 3.5 y=F(t) baby boom baby bust baby boomlet 49. Let be the total value of US currency (coins and bank-notes) in circulation at time . The table gives values of this function from 1980 to 2000, as of September 30, in billions of dollars. Interpret and estimate the value of . 50. The total fertility rate at time t, denoted by , is an esti-mate of the average number of children born to each woman (assuming that current birth rates remain constant). The graph of the total fertility rate in the United States shows the ﬂuc-tuations from 1940 to 1990. (a) Estimate the values of , , and . (b) What are the meanings of these derivatives? (c) Can you suggest reasons for the values of these derivatives? F!!1987" F!!1965" F!!1950" F!t" C!!1990" t C!t" CHAPTER 2 REVIEW \|\|\|\| 169 t 1980 1985 1990 1995 2000 129.9 187.3 271.9 409.3 568.6 C!t" 170 In our discussion of the principles of problem solving we considered the problem-solving strategy of introducing something extra (see page 76). In the following example we show how this principle is sometimes useful when we evaluate limits. The idea is to change the variable—to introduce a new variable that is related to the original variable—in such a way as to make the problem simpler. Later, in Section 5.5, we will make more extensive use of this general idea. EXAMPLE 1 Evaluate , where c is a nonzero constant. SOLUTION As it stands, this limit looks challenging. In Section 2.3 we evaluated several lim-its in which both numerator and denominator approached 0. There our strategy was to per-form some sort of algebraic manipulation that led to a simplifying cancellation, but here it’s not clear what kind of algebra is necessary. So we introduce a new variable t by the equation We also need to express x in terms of t, so we solve this equation: Notice that is equivalent to . This allows us to convert the given limit into one involving the variable t: The change of variable allowed us to replace a relatively complicated limit by a simpler one of a type that we have seen before. Factoring the denominator as a difference of cubes, we get M The following problems are meant to test and challenge your problem-solving skills. Some of them require a considerable amount of time to think through, so don’t be discour-aged if you can’t solve them right away. If you get stuck, you might ﬁnd it helpful to refer to the discussion of the principles of problem solving on page 76. 1. Evaluate . 2. Find numbers a and b such that . lim x l 0 sax ! b " 2 x ! 1 lim x l 1 s 3 x " 1 sx " 1 PROBLEMS ! lim t l1 c t 2 ! t ! 1 ! c 3 lim t l1 c!t " 1" t 3 " 1 ! lim t l1 c!t " 1" !t " 1"!t 2 ! t ! 1" ! lim t l1 c!t " 1" t 3 " 1 lim x l 0 s 3 1 ! cx " 1 x ! lim t l1 t " 1 !t 3 " 1"#c t l 1 x l 0 x ! t 3 " 1 c t 3 ! 1 ! cx t ! s 3 1 ! cx lim x l 0 s 3 1 ! cx " 1 x P R O B L E M S P L U S 171 3. Evaluate . 4. The ﬁgure shows a point P on the parabola and the point Q where the perpendicular bisector of OP intersects the y-axis. As P approaches the origin along the parabola, what happens to Q? Does it have a limiting position? If so, ﬁnd it. 5. If denotes the greatest integer function, ﬁnd . 6. Sketch the region in the plane deﬁned by each of the following equations. (a) (b) (c) (d) 7. Find all values of a such that is continuous on !: 8. A ﬁxed point of a function is a number in its domain such that . (The function doesn’t move ; it stays ﬁxed.) (a) Sketch the graph of a continuous function with domain whose range also lies in . Locate a ﬁxed point of . (b) Try to draw the graph of a continuous function with domain and range in that does not have a ﬁxed point. What is the obstacle? (c) Use the Intermediate Value Theorem to prove that any continuous function with domain and range a subset of must have a ﬁxed point. 9. If and , ﬁnd . 10. (a) The ﬁgure shows an isosceles triangle with . The bisector of angle intersects the side at the point . Suppose that the base remains ﬁxed but the altitude of the triangle approaches 0, so approaches the midpoint of . What happens to during this process? Does it have a limiting position? If so, ﬁnd it. (b) Try to sketch the path traced out by during this process. Then ﬁnd an equation of this curve and use this equation to sketch the curve. 11. (a) If we start from latitude and proceed in a westerly direction, we can let denote the temperature at the point at any given time. Assuming that is a continuous function of , show that at any ﬁxed time there are at least two diametrically opposite points on the equator that have exactly the same temperature. (b) Does the result in part (a) hold for points lying on any circle on the earth’s surface? (c) Does the result in part (a) hold for barometric pressure and for altitude above sea level? 12. If is a differentiable function and , use the deﬁnition of a derivative to show that . 13. Suppose is a function that satisﬁes the equation for all real numbers x and y. Suppose also that (a) Find . (b) Find . (c) Find . 14. Suppose is a function with the property that for all x. Show that . Then show that . f #!0" ! 0 f !0" ! 0 $ f !x"$ $ x 2 f f #!x" f #!0" f !0" lim x l 0 f !x" x ! 1 f !x ! y" ! f !x" ! f !y" ! x 2y ! xy 2 f t#!x" ! xf #!x" ! f !x" t!x" ! xf !x" f x T x T!x" 0% P P BC M A $ AM$ BC P AC B "B ! "C ABC limx l a % f !x"t!x"& limx l a % f !x" " t!x"& ! 1 limx l a % f !x" ! t!x"& ! 2 %0, 1& %0, 1& %0, 1& %0, 1& f %0, 1& %0, 1& c f !c" ! c c f f !x" !' x ! 1 x 2 if x $ a if x & a f (x) ! ( y) ! 1 (x ! y) 2 ! 1 (x) 2 " ( y) 2 ! 3 (x) 2 ! ( y) 2 ! 1 lim x l ' x (x) (x) y ! x 2 lim x l 0 $ 2x " 1$ " $ 2x ! 1$ x P R O B L E M S P L U S FIGURE FOR PROBLEM 4 0 P Q y=≈ x y A C B M P FIGURE FOR PROBLEM 10 172 We have seen how to interpret derivatives as slopes and rates of change. We have seen how to estimate derivatives of functions given by tables of values. We have learned how to graph derivatives of functions that are deﬁned graphically. We have used the deﬁnition of a derivative to calculate the derivatives of functions deﬁned by formulas. But it would be tedious if we always had to use the deﬁnition, so in this chapter we develop rules for ﬁnding derivatives without having to use the deﬁnition directly. These differentiation rules enable us to calculate with relative ease the derivatives of poly-nomials, rational functions, algebraic functions, exponential and logarithmic functions, and trigonometric and inverse trigonometric functions. We then use these rules to solve problems involving rates of change and the approximation of functions. By measuring slopes at points on the sine curve, we get strong visual evidence that the derivative of the sine function is the cosine function. DIFFERENTIATION RULES 3 x ƒ= y= sin x 0 x y y fª(x y= ) 0 π 2 m=1 m=_1 m=0 π 2 π π DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS In this section we learn how to differentiate constant functions, power functions, polyno-mials, and exponential functions. Let’s start with the simplest of all functions, the constant function . The graph of this function is the horizontal line y ! c, which has slope 0, so we must have . (See Figure 1.) A formal proof, from the deﬁnition of a derivative, is also easy: In Leibniz notation, we write this rule as follows. DERIVATIVE OF A CONSTANT FUNCTION POWER FUNCTIONS We next look at the functions , where n is a positive integer. If , the graph of is the line y ! x, which has slope 1. (See Figure 2.) So (You can also verify Equation 1 from the deﬁnition of a derivative.) We have already investigated the cases and . In fact, in Section 2.8 (Exercises 17 and 18) we found that For we ﬁnd the derivative of as follows: Thus d dx !x 4" ! 4x 3 3 ! lim h l 0 !4x 3 ! 6x 2h ! 4xh 2 ! h 3" ! 4x 3 ! lim h l 0 4x 3h ! 6x 2h 2 ! 4xh 3 ! h 4 h ! lim h l 0 x 4 ! 4x 3h ! 6x 2h 2 ! 4xh 3 ! h 4 " x 4 h f#!x" ! lim h l 0 f!x ! h" " f!x" h ! lim h l 0 !x ! h"4 " x 4 h f!x" ! x 4 n ! 4 d dx !x 3" ! 3x 2 d dx !x 2" ! 2x 2 n ! 3 n ! 2 d dx !x" ! 1 1 f!x" ! x n ! 1 f!x" ! x n d dx !c" ! 0 ! lim h l 0 0 ! 0 f#!x" ! lim h l 0 f!x ! h" " f!x" h ! lim h l 0 c " c h f#!x" ! 0 f!x" ! c 3.1 173 FIGURE 1 The graph of ƒ=c is the line y=c, so fª(x)=0. y c 0 x y=c slope=0 6 030102 y 0 x y=x slope=1 FIGURE 2 The graph of ƒ=x is the line y=x, so fª(x)=1. Comparing the equations in (1), (2), and (3), we see a pattern emerging. It seems to be a reasonable guess that, when n is a positive integer, . This turns out to be true. THE POWER RULE If n is a positive integer, then FIRST PROOF The formula can be veriﬁed simply by multiplying out the right-hand side (or by summing the second factor as a geometric series). If , we can use Equation 2.7.5 for and the equation above to write SECOND PROOF In ﬁnding the derivative of we had to expand . Here we need to expand and we use the Binomial Theorem to do so: because every term except the ﬁrst has as a factor and therefore approaches 0. M We illustrate the Power Rule using various notations in Example 1. EXAMPLE 1 (a) If , then . (b) If , then ! . (c) If , then . (d) M d dr !r 3" ! 3r 2 dy dt ! 4t 3 y ! t 4 1000x 999 y# y ! x 1000 f#!x" ! 6x 5 f!x" ! x 6 h ! nx n"1 ! lim h l 0 #nx n"1 ! n!n " 1" 2 x n"2h ! $ $ $ ! nxh n"2 ! h n"1$ ! lim h l 0 nx n"1h ! n!n " 1" 2 x n"2h 2 ! $ $ $ ! nxh n"1 ! h n h f#!x" ! lim h l 0 #x n ! nx n"1h ! n!n " 1" 2 x n"2h 2 ! $ $ $ ! nxh n"1 ! h n$ " x n h !x ! h"n !x ! h"4 x 4 f#!x" ! lim h l 0 f!x ! h" " f!x" h ! lim h l 0 !x ! h"n " x n h ! na n"1 ! a n"1 ! a n"2a ! $ $ $ ! aa n"2 ! a n"1 ! lim x l a !x n"1 ! x n"2a ! $ $ $ ! xa n"2 ! a n"1" f#!a" ! lim x l a f!x" " f!a" x " a ! lim x l a x n " a n x " a f#!a" f!x" ! x n x n " a n ! !x " a"!x n"1 ! x n"2a ! $ $ $ ! xa n"2 ! a n"1" d dx !x n" ! nx n"1 !d%dx"!x n" ! nx n"1 174 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N The Binomial Theorem is given on Reference Page 1. What about power functions with negative integer exponents? In Exercise 61 we ask you to verify from the deﬁnition of a derivative that We can rewrite this equation as and so the Power Rule is true when . In fact, we will show in the next section [Exercise 58(c)] that it holds for all negative integers. What if the exponent is a fraction? In Example 3 in Section 2.8 we found that which can be written as This shows that the Power Rule is true even when . In fact, we will show in Sec-tion 3.6 that it is true for all real numbers n. THE POWER RULE (GENERAL VERSION) If n is any real number, then EXAMPLE 2 Differentiate: (a) (b) SOLUTION In each case we rewrite the function as a power of x. (a) Since , we use the Power Rule with : (b) M The Power Rule enables us to ﬁnd tangent lines without having to resort to the deﬁ-nition of a derivative. It also enables us to ﬁnd normal lines. The normal line to a curve at a point is the line through that is perpendicular to the tangent line at . (In the study of optics, one needs to consider the angle between a light ray and the normal line to a lens.) P P P C dy dx ! d dx (s 3 x 2 ) ! d dx !x 2%3" ! 2 3 x !2%3""1 ! 2 3 x"1%3 f#!x" ! d dx !x "2" ! "2x "2"1 ! "2x "3 ! " 2 x 3 n ! "2 f!x" ! x"2 y ! s 3 x 2 f!x" ! 1 x 2 d dx !x n" ! nx n"1 n ! 1 2 d dx !x1%2" ! 1 2 x"1%2 d dx sx ! 1 2sx n ! "1 d dx !x "1" ! !"1"x "2 d dx & 1 x' ! " 1 x 2 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS \|\|\|\| 175 2 _2 _3 3 y yª FIGURE 3 y=# œ„ ≈ N Figure 3 shows the function in Example 2(b) and its derivative . Notice that is not differ-entiable at ( is not deﬁned there). Observe that is positive when increases and is nega-tive when decreases. y y y# y# 0 y y# y EXAMPLE 3 Find equations of the tangent line and normal line to the curve at the point . Illustrate by graphing the curve and these lines. SOLUTION The derivative of is So the slope of the tangent line at (1, 1) is . Therefore an equation of the tan-gent line is The normal line is perpendicular to the tangent line, so its slope is the negative recipro-cal of , that is, . Thus an equation of the normal line is We graph the curve and its tangent line and normal line in Figure 4. M NEW DERIVATIVES FROM OLD When new functions are formed from old functions by addition, subtraction, or multipli-cation by a constant, their derivatives can be calculated in terms of derivatives of the old functions. In particular, the following formula says that the derivative of a constant times a function is the constant times the derivative of the function. THE CONSTANT MULTIPLE RULE If c is a constant and is a differentiable func-tion, then PROOF Let . Then (by Law 3 of limits) M EXAMPLE 4 (a) (b) M The next rule tells us that the derivative of a sum of functions is the sum of the derivatives. d dx !"x" ! d dx (!"1"x) ! !"1" d dx !x" ! "1!1" ! "1 d dx !3x 4" ! 3 d dx !x 4" ! 3!4x 3" ! 12x 3 ! cf#!x" ! c lim h l 0 f!x ! h" " f!x" h ! lim h l 0 c# f!x ! h" " f!x" h $ t#!x" ! lim h l 0 t!x ! h" " t!x" h ! lim h l 0 cf!x ! h" " cf!x" h t!x" ! cf!x" d dx (cf!x") ! c d dx f!x" f y ! " 2 3x ! 5 3 or y " 1 ! " 2 3!x " 1" " 2 3 3 2 y ! 3 2 x " 1 2 or y " 1 ! 3 2!x " 1" f#!1" ! 3 2 f#!x" ! 3 2 x !3%2""1 ! 3 2 x 1%2 ! 3 2 sx f!x" ! xsx ! xx 1%2 ! x 3%2 !1, 1" y ! xsx V 176 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES 3 _1 _1 3 tangent normal FIGURE 4 N GEOMETRIC INTERPRETATION OF THE CONSTANT MULTIPLE RULE x y 0 y=2ƒ y=ƒ Multiplying by stretches the graph verti-cally by a factor of 2. All the rises have been doubled but the runs stay the same. So the slopes are doubled, too. c ! 2 THE SUM RULE If f and t are both differentiable, then PROOF Let . Then (by Law 1) M The Sum Rule can be extended to the sum of any number of functions. For instance, using this theorem twice, we get By writing as and applying the Sum Rule and the Constant Multiple Rule, we get the following formula. THE DIFFERENCE RULE If f and t are both differentiable, then The Constant Multiple Rule, the Sum Rule, and the Difference Rule can be combined with the Power Rule to differentiate any polynomial, as the following examples demonstrate. EXAMPLE 5 M ! 8x 7 ! 60x 4 " 16x 3 ! 30x 2 " 6 ! 8x 7 ! 12!5x 4" " 4!4x 3" ! 10!3x 2" " 6!1" ! 0 ! d dx !x 8" ! 12 d dx !x 5" " 4 d dx !x 4" ! 10 d dx !x 3" " 6 d dx !x" ! d dx !5" d dx !x 8 ! 12x 5 " 4x 4 ! 10x 3 " 6x ! 5" d dx ( f!x" " t!x") ! d dx f!x" " d dx t!x" f ! !"1"t f " t ! f ! t ! h"# ! (! f ! t" ! h)# ! ! f ! t"# ! h# ! f# ! t# ! h# ! f#!x" ! t#!x" ! lim h l 0 f!x ! h" " f!x" h ! lim h l 0 t!x ! h" " t!x" h ! lim h l 0 # f!x ! h" " f!x" h ! t!x ! h" " t!x" h $ ! lim h l 0 ( f!x ! h" ! t!x ! h") " ( f!x" ! t!x") h F#!x" ! lim h l 0 F!x ! h" " F!x" h F!x" ! f!x" ! t!x" d dx ( f!x" ! t!x") ! d dx f!x" ! d dx t!x" SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS \|\|\|\| 177 N Using prime notation, we can write the Sum Rule as ! f ! t"# ! f # ! t# EXAMPLE 6 Find the points on the curve where the tangent line is horizontal. SOLUTION Horizontal tangents occur where the derivative is zero. We have Thus if x ! 0 or , that is, . So the given curve has horizontal tangents when x ! 0, , and . The corresponding points are , , and . (See Figure 5.) M EXAMPLE 7 The equation of motion of a particle is , where is measured in centimeters and in seconds. Find the acceleration as a function of time. What is the acceleration after 2 seconds? SOLUTION The velocity and acceleration are The acceleration after 2 s is . M EXPONENTIAL FUNCTIONS Let’s try to compute the derivative of the exponential function using the deﬁni-tion of a derivative: The factor doesn’t depend on h, so we can take it in front of the limit: Notice that the limit is the value of the derivative of at , that is, Therefore we have shown that if the exponential function is differentiable at 0, then it is differentiable everywhere and This equation says that the rate of change of any exponential function is proportional to the function itself. (The slope is proportional to the height.) f#!x" ! f#!0"a x 4 f!x" ! a x lim h l 0 a h " 1 h ! f#!0" 0 f f#!x" ! a x lim h l 0 a h " 1 h a x ! lim h l 0 a xa h " a x h ! lim h l 0 a x!a h " 1" h f#!x" ! lim h l 0 f!x ! h" " f!x" h ! lim h l 0 a x!h " a x h f!x" ! a x a!2" ! 14 cm%s2 a!t" ! dv dt ! 12t " 10 v!t" ! ds dt ! 6t 2 " 10t ! 3 t s s ! 2t 3 " 5t 2 ! 3t ! 4 ("s3, "5) (s3, "5) !0, 4" "s3 s3 x ! %s3 x 2 " 3 ! 0 dy%dx ! 0 ! 4x 3 " 12x ! 0 ! 4x!x 2 " 3" dy dx ! d dx !x 4" " 6 d dx !x 2" ! d dx !4" y ! x 4 " 6x 2 ! 4 V 178 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES FIGURE 5 The curve y=x$-6x@+4 and its horizontal tangents 0 x y (0, 4) {œ„ 3, _5} {_œ„ 3, _5} Numerical evidence for the existence of is given in the table at the left for the cases and . (Values are stated correct to four decimal places.) It appears that the limits exist and In fact, it can be proved that these limits exist and, correct to six decimal places, the val-ues are Thus, from Equation 4, we have Of all possible choices for the base in Equation 4, the simplest differentiation formula occurs when . In view of the estimates of for and , it seems rea-sonable that there is a number between 2 and 3 for which . It is traditional to denote this value by the letter . (In fact, that is how we introduced e in Section 1.5.) Thus we have the following deﬁnition. DEFINITION OF THE NUMBER e Geometrically, this means that of all the possible exponential functions , the function is the one whose tangent line at ( has a slope that is exactly 1. (See Figures 6 and 7.) If we put and, therefore, in Equation 4, it becomes the following impor-tant differentiation formula. f#!0" ! 1 a ! e FIGURE 7 0 y 1 x slope=1 slope=e® y=e® {x, e®} 0 y 1 x y=2® y=e® y=3® FIGURE 6 f#!0" 0, 1" f!x" ! e x y ! a x lim h l 0 e h " 1 h ! 1 e is the number such that e f#!0" ! 1 a a ! 3 a ! 2 f#!0" f#!0" ! 1 a d dx !3x" !1.10"3x d dx !2x" !0.69"2x 5 d dx !3x"+ x!0 1.098612 d dx !2x"+ x!0 0.693147 f#!0" ! lim h l 0 3h " 1 h 1.10 for a ! 3, f#!0" ! lim h l 0 2h " 1 h 0.69 for a ! 2, a ! 3 a ! 2 f#!0" SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS \|\|\|\| 179 h 0.1 0.7177 1.1612 0.01 0.6956 1.1047 0.001 0.6934 1.0992 0.0001 0.6932 1.0987 3h " 1 h 2h " 1 h N In Exercise 1 we will see that lies between and . Later we will be able to show that, correct to ﬁve decimal places, e 2.71828 2.8 2.7 e DERIVATIVE OF THE NATURAL EXPONENTIAL FUNCTION Thus the exponential function has the property that it is its own derivative. The geometrical signiﬁcance of this fact is that the slope of a tangent line to the curve is equal to the -coordinate of the point (see Figure 7). EXAMPLE 8 If , ﬁnd and . Compare the graphs of and . SOLUTION Using the Difference Rule, we have In Section 2.8 we deﬁned the second derivative as the derivative of , so The function f and its derivative are graphed in Figure 8. Notice that has a horizon-tal tangent when ; this corresponds to the fact that . Notice also that, for , is positive and is increasing. When , is negative and is decreasing. M EXAMPLE 9 At what point on the curve is the tangent line parallel to the line ? SOLUTION Since , we have . Let the x-coordinate of the point in question be a. Then the slope of the tangent line at that point is . This tangent line will be parallel to the line if it has the same slope, that is, 2. Equating slopes, we get Therefore the required point is . (See Figure 9.) M !a, e a" ! !ln 2, 2" a ! ln 2 e a ! 2 y ! 2x e a y# ! e x y ! e x y ! 2x y ! e x f f#!x" x & 0 f f#!x" x ' 0 f#!0" ! 0 x ! 0 f f# f (!x" ! d dx !e x " 1" ! d dx !e x" " d dx !1" ! e x f# f#!x" ! d dx !e x " x" ! d dx !e x" " d dx !x" ! e x " 1 f# f f ( f# f!x" ! e x " x V y y ! e x f!x" ! e x d dx !e x" ! e x FIGURE 8 3 _1 1.5 _1.5 f fª Visual 3.1 uses the slope-a-scope to illustrate this formula. TEC (b) What types of functions are and ? Compare the differentiation formulas for and t. (c) Which of the two functions in part (b) grows more rapidly when x is large? 3–32 Differentiate the function. 3. 4. 5. 6. 7. 8. f !t" ! 1 2t 6 " 3t 4 ! t f !x" ! x 3 " 4x ! 6 F!x" ! 3 4x 8 f !t" ! 2 " 2 3t f !x" ! s30 f !x" ! 186.5 f t!x" ! x e f !x" ! e x 1. (a) How is the number e deﬁned? (b) Use a calculator to estimate the values of the limits and correct to two decimal places. What can you conclude about the value of e? 2. (a) Sketch, by hand, the graph of the function , pay-ing particular attention to how the graph crosses the y-axis. What fact allows you to do this? f !x" ! e x lim h l 0 2.8h " 1 h lim h l 0 2.7h " 1 h EXERCISES 3.1 FIGURE 9 1 1 0 x 2 3 y y=´ y=2x (ln 2, 2) 180 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES (b) Using the graph in part (a) to estimate slopes, make a rough sketch, by hand, of the graph of . (See Example 1 in Section 2.8.) (c) Calculate and use this expression, with a graphing device, to graph . Compare with your sketch in part (b). ; 44. (a) Use a graphing calculator or computer to graph the func-tion in the viewing rectangle by . (b) Using the graph in part (a) to estimate slopes, make a rough sketch, by hand, of the graph of . (See Example 1 in Section 2.8.) (c) Calculate and use this expression, with a graphing device, to graph . Compare with your sketch in part (b). 45–46 Find the ﬁrst and second derivatives of the function. 45. ; 47–48 Find the ﬁrst and second derivatives of the function. Check to see that your answers are reasonable by comparing the graphs of , , and . 47. 48. The equation of motion of a particle is , where is in meters and is in seconds. Find (a) the velocity and acceleration as functions of , (b) the acceleration after 2 s, and (c) the acceleration when the velocity is 0. 50. The equation of motion of a particle is , where is in meters and is in seconds. (a) Find the velocity and acceleration as functions of . (b) Find the acceleration after 1 s. ; (c) Graph the position, velocity, and acceleration functions on the same screen. Find the points on the curve where the tangent is horizontal. 52. For what values of does the graph of have a horizontal tangent? 53. Show that the curve has no tangent line with slope 4. 54. Find an equation of the tangent line to the curve that is parallel to the line . 55. Find equations of both lines that are tangent to the curve and are parallel to the line . ; 56. At what point on the curve is the tangent line parallel to the line ? Illustrate by graphing the curve and both lines. 57. Find an equation of the normal line to the parabola that is parallel to the line . x " 3y ! 5 y ! x 2 " 5x ! 4 3x " y ! 5 y ! 1 ! 2e x " 3x 12x " y ! 1 y ! 1 ! x 3 y ! 1 ! 3x y ! xsx y ! 6x 3 ! 5x " 3 f !x" ! x 3 ! 3x 2 ! x ! 3 x y ! 2x 3 ! 3x 2 " 12x ! 1 51. t t s s ! 2t 3 " 7t 2 ! 4t ! 1 t t s s ! t 3 " 3t 49. f !x" ! e x " x 3 f !x" ! 2x " 5x 3%4 f ( f # f G!r" ! sr ! s 3 r 46. f !x" ! x 4 " 3x 3 ! 16x t# t#!x" t# ("8, 8) ("1, 4) t!x" ! e x " 3x 2 f # f #!x" f # 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 24. 25. 26. 27. 28. 29. 30. 32. 33–34 Find an equation of the tangent line to the curve at the given point. 33. , 34. , 35–36 Find equations of the tangent line and normal line to the curve at the given point. , 36. , ; 37–38 Find an equation of the tangent line to the curve at the given point. Illustrate by graphing the curve and the tangent line on the same screen. 37. , 38. , ; 39–42 Find . Compare the graphs of and and use them to explain why your answer is reasonable. 39. 40. 41. 42. ; 43. (a) Use a graphing calculator or computer to graph the func-tion in the viewing rectangle by . ("10, 50) ("3, 5) f !x" ! x 4 " 3x 3 " 6x 2 ! 7x ! 30 f !x" ! x ! 1 x f !x" ! 3x 15 " 5x 3 ! 3 f !x" ! 3x 5 " 20x 3 ! 50x f !x" ! e x " 5x f # f f #!x" !1, 0" y ! x " sx !1, 2" y ! 3x2 " x3 !1, 9" y ! !1 ! 2x"2 !0, 2" y ! x4 ! 2e x 35. !1, 2" y ! x 4 ! 2x 2 " x !1, 1" y ! s 4 x y ! e x!1 ! 1 z ! A y10 ! Be y 31. v !&sx ! 1 s 3 x ' 2 u ! s 5 t ! 4st 5 y ! aev ! b v ! c v 2 H!x" ! !x ! x"1"3 t!u" ! s2 u ! s3u y ! 4) 2 y ! x 2 " 2sx x y ! x 2 ! 4x ! 3 sx 23. y ! sx !x " 1" y ! ax 2 ! bx ! c f !t" ! st " 1 st F !x" ! ( 1 2 x) 5 y ! s 3 x G!x" ! sx " 2ex B!y" ! cy"6 A!s" ! " 12 s 5 R!t" ! 5t "3%5 V!r" ! 4 3)r 3 y ! 5e x ! 3 y ! x "2%5 h!x" ! !x " 2"!2x ! 3" f !t" ! 1 4!t 4 ! 8" SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS \|\|\|\| 181 69. (a) For what values of is the function dif-ferentiable? Find a formula for . (b) Sketch the graphs of and . 70. Where is the function differenti-able? Give a formula for and sketch the graphs of and . 71. Find the parabola with equation whose tangent line at (1, 1) has equation . 72. Suppose the curve has a tan-gent line when with equation and a tangent line when with equation . Find the values of , , , and . For what values of and is the line tangent to the parabola when ? 74. Find the value of such that the line is tangent to the curve . 75. Let Find the values of and that make differentiable every-where. 76. A tangent line is drawn to the hyperbola at a point . (a) Show that the midpoint of the line segment cut from this tangent line by the coordinate axes is . (b) Show that the triangle formed by the tangent line and the coordinate axes always has the same area, no matter where is located on the hyperbola. Evaluate . 78. Draw a diagram showing two perpendicular lines that intersect on the -axis and are both tangent to the parabola . Where do these lines intersect? 79. If , how many lines through the point are normal lines to the parabola ? What if ? 80. Sketch the parabolas and . Do you think there is a line that is tangent to both curves? If so, ﬁnd its equation. If not, why not? y ! x 2 ! 2x " 2 y ! x 2 c # 1 2 y ! x 2 !0, c" c $ 1 2 y ! x 2 y lim x l 1 x 1000 ! 1 x ! 1 77. P P P xy ! c f b m f !x" !# x 2 mx " b if x # 2 if x $ 2 y ! csx y ! 3 2x " 6 c x ! 2 y ! ax 2 2x " y ! b b a 73. d c b a y ! 2 ! 3x x ! 1 y ! 2x " 1 x ! 0 y ! x 4 " ax 3 " bx 2 " cx " d y ! 3x ! 2 y ! ax 2 " bx h% h h% h!x" ! $ x ! 1$ " $ x " 2$ f % f f % f !x" ! $ x 2 ! 9$ x 58. Where does the normal line to the parabola at the point (1, 0) intersect the parabola a second time? Illustrate with a sketch. Draw a diagram to show that there are two tangent lines to the parabola that pass through the point . Find the coordinates of the points where these tangent lines inter-sect the parabola. 60. (a) Find equations of both lines through the point that are tangent to the parabola . (b) Show that there is no line through the point that is tangent to the parabola. Then draw a diagram to see why. 61. Use the deﬁnition of a derivative to show that if , then . (This proves the Power Rule for the case .) 62. Find the derivative of each function by calculating the ﬁrst few derivatives and observing the pattern that occurs. (a) (b) 63. Find a second-degree polynomial such that , , and . 64. The equation is called a differential equation because it involves an unknown function and its derivatives and . Find constants such that the function satisﬁes this equation. (Differ-ential equations will be studied in detail in Chapter 9.) 65. Find a cubic function whose graph has horizontal tangents at the points and . 66. Find a parabola with equation that has slope 4 at , slope at , and passes through the point . 67. Let Is differentiable at 1? Sketch the graphs of and . 68. At what numbers is the following function differentiable? Give a formula for and sketch the graphs of and . t% t t% t!x" !# !1 ! 2x x 2 x if x & !1 if !1 # x # 1 if x $ 1 t f % f f f !x" !# 2 ! x x 2 ! 2x " 2 if x # 1 if x $ 1 !2, 15" x ! !1 !8 x ! 1 y ! ax 2 " bx " c !2, 0" !!2, 6" y ! ax 3 " bx 2 " cx " d y ! Ax 2 " Bx " C A, B, and C y' y% y y' " y% ! 2y ! x 2 P'!2" ! 2 P%!2" ! 3 P!2" ! 5 P f !x" ! 1%x f !x" ! x n nth n ! !1 f %!x" ! !1%x 2 f !x" ! 1%x !2, 7" y ! x 2 " x !2, !3" !0, !4" y ! x 2 59. y ! x ! x 2 182 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES Suppose you are asked to design the ﬁrst ascent and drop for a new roller coaster. By studying photographs of your favorite coasters, you decide to make the slope of the ascent 0.8 and the slope of the drop . You decide to connect these two straight stretches and with part of a parabola , where and are measured in feet. For the track to be smooth there can’t be abrupt changes in direction, so you want the linear f !x" x y ! f !x" ! ax 2 " bx " c y ! L 2!x" y ! L1!x" !1.6 BUILDING A BETTER ROLLER COASTER A P P L I E D P R O J E C T segments and to be tangent to the parabola at the transition points and . (See the ﬁg-ure.) To simplify the equations you decide to place the origin at . 1. (a) Suppose the horizontal distance between and is 100 ft. Write equations in , , and that will ensure that the track is smooth at the transition points. (b) Solve the equations in part (a) for to ﬁnd a formula for . ; (c) Plot , , and to verify graphically that the transitions are smooth. (d) Find the difference in elevation between and . 2. The solution in Problem 1 might look smooth, but it might not feel smooth because the piece-wise deﬁned function [consisting of for , for , and for ] doesn’t have a continuous second derivative. So you decide to improve the design by using a quadratic function only on the interval and con-necting it to the linear functions by means of two cubic functions: (a) Write a system of equations in 11 unknowns that ensure that the functions and their ﬁrst two derivatives agree at the transition points. (b) Solve the equations in part (a) with a computer algebra system to ﬁnd formulas for . (c) Plot , , , , and , and compare with the plot in Problem 1(c). L 2 h q t L1 q!x", t!x", and h!x" CAS 90 & x # 100 h!x" ! px 3 " qx 2 " rx " s t!x" ! kx 3 " lx 2 " mx " n 0 # x & 10 10 # x # 90 q!x" ! ax 2 " bx " c x $ 100 L 2!x" 0 # x # 100 f !x" x & 0 L1!x" Q P L 2 f L1 f !x" a, b, and c c b a Q P P Q P L 2 L1 SECTION 3.2 THE PRODUCT AND QUOTIENT RULES \|\|\|\| 183 L™ L¡ P f Q THE PRODUCT AND QUOTIENT RULES The formulas of this section enable us to differentiate new functions formed from old func-tions by multiplication or division. THE PRODUCT RULE \| By analogy with the Sum and Difference Rules, one might be tempted to guess, as Leibniz did three centuries ago, that the derivative of a product is the product of the derivatives. We can see, however, that this guess is wrong by looking at a particular example. Let and . Then the Power Rule gives and . But , so . Thus . The correct formula was discovered by Leibniz (soon after his false start) and is called the Product Rule. Before stating the Product Rule, let’s see how we might discover it. We start by assum-ing that and are both positive differentiable functions. Then we can interpret the product as an area of a rectangle (see Figure 1). If x changes by an amount , then the corresponding changes in u and are and the new value of the product, , can be interpreted as the area of the large rectangle in Figure 1 (provided that and happen to be positive). The change in the area of the rectangle is ! the sum of the three shaded areas (!uv" ! !u " (u"!v " (v" ! uv ! u (v " v (u " (u (v 1 (v (u !u " (u"!v " (v" (v ! t!x " (x" ! t!x" (u ! f!x " (x" ! f!x" v (x uv v ! t!x" u ! f!x" ! ft"% " f%t% ! ft"%!x" ! 3x 2 ! ft"!x" ! x 3 t%!x" ! 2x f%!x" ! 1 t!x" ! x 2 f!x" ! x 3.2 u Î√ Î√ √ u√ u Îu Î√ √ Îu Îu FIGURE 1 The geometry of the Product Rule If we divide by , we get If we now let , we get the derivative of : (Notice that as since is differentiable and therefore continuous.) Although we started by assuming (for the geometric interpretation) that all the quanti-ties are positive, we notice that Equation 1 is always true. (The algebra is valid whether u, , , and are positive or negative.) So we have proved Equation 2, known as the Product Rule, for all differentiable functions u and . THE PRODUCT RULE If and are both differentiable, then In words, the Product Rule says that the derivative of a product of two functions is the ﬁrst function times the derivative of the second function plus the second function times the derivative of the ﬁrst function. EXAMPLE 1 (a) If , ﬁnd . (b) Find the derivative, . SOLUTION (a) By the Product Rule, we have (b) Using the Product Rule a second time, we get ! !x " 1"e x " e x ! 1 ! !x " 2"e x f '!x" ! d dx &!x " 1"e x' ! !x " 1" d dx !e x" " e x d dx !x " 1" ! xe x " e x ) 1 ! !x " 1"e x f%!x" ! d dx !xe x" ! x d dx !e x" " e x d dx !x" f !n"!x" nth f %!x" f!x" ! xe x d dx & f!x"t!x"' ! f!x" d dx &t!x"' " t!x" d dx & f!x"' t f v (v (u v f (x l 0 (u l 0 d dx !uv" ! u dv dx " v du dx 2 ! u dv dx " v du dx " 0 ! dv dx ! u lim (x l 0 (v (x " v lim (x l 0 (u (x "( lim (x l 0 (u)( lim (x l 0 (v (x) d dx !uv" ! lim (x l 0 (!uv" (x ! lim (x l 0 (u (v (x " v (u (x " (u (v (x) uv (x l 0 (!uv" (x ! u (v (x " v (u (x " (u (v (x (x 184 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N Recall that in Leibniz notation the deﬁnition of a derivative can be written as dy dx ! lim (x l 0 (y (x N In prime notation: ! ft"% ! ft% " tf % 3 _1 _3 1.5 f fª FIGURE 2 N Figure 2 shows the graphs of the function of Example 1 and its derivative . Notice that is positive when is increasing and nega-tive when is decreasing. f f f %!x" f % f Further applications of the Product Rule give In fact, each successive differentiation adds another term , so M EXAMPLE 2 Differentiate the function . SOLUTION 1 Using the Product Rule, we have SOLUTION 2 If we ﬁrst use the laws of exponents to rewrite , then we can proceed directly without using the Product Rule. which is equivalent to the answer given in Solution 1. M Example 2 shows that it is sometimes easier to simplify a product of functions than to use the Product Rule. In Example 1, however, the Product Rule is the only possible method. EXAMPLE 3 If , where and , ﬁnd SOLUTION Applying the Product Rule, we get So M THE QUOTIENT RULE We ﬁnd a rule for differentiating the quotient of two differentiable functions and in much the same way that we found the Product Rule. If , , and change by amounts , , and , then the corresponding change in the quotient is ! v(u ! u(v v!v " (v" (( u v) ! u " (u v " (v ! u v ! !u " (u"v ! u!v " (v" v!v " (v" u%v (v (u (x v u x v ! t!x" u ! f!x" f%!4" ! s4 t%!4" " t!4" 2s4 ! 2 ) 3 " 2 2 ) 2 ! 6.5 ! sx t%!x" " t!x" 2sx ! sx t%!x" " t!x" ) 1 2 x !1%2 f%!x" ! d dx [sx t!x"] ! sx d dx &t!x"' " t!x" d dx [sx ] f%!4". t%!4" ! 3 t!4" ! 2 f!x" ! sx t!x" f%!t" ! 1 2at!1%2 " 3 2bt 1%2 f!t" ! ast " btst ! at 1%2 " bt 3%2 f!t" ! bst " a " bt 2st ! a " 3bt 2st ! st ! b " !a " bt" ! 1 2t !1%2 f%!t" ! st d dt !a " bt" " !a " bt" d dt (st ) f!t" ! st !a " bt" f !n"!x" ! !x " n"e x e x f !x" ! !x " 3"e x f !4"!x" ! !x " 4"e x SECTION 3.2 THE PRODUCT AND QUOTIENT RULES \|\|\|\| 185 N In Example 2, and are constants. It is customary in mathematics to use letters near the beginning of the alphabet to represent constants and letters near the end of the alphabet to repre-sent variables. b a so As , also, because is differentiable and therefore continuous. Thus, using the Limit Laws, we get THE QUOTIENT RULE If and are differentiable, then In words, the Quotient Rule says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. The Quotient Rule and the other differentiation formulas enable us to compute the derivative of any rational function, as the next example illustrates. EXAMPLE 4 Let . Then M EXAMPLE 5 Find an equation of the tangent line to the curve at the point . SOLUTION According to the Quotient Rule, we have ! !1 " x 2"e x ! e x!2x" !1 " x 2"2 ! e x!1 ! x"2 !1 " x 2"2 dy dx ! !1 " x 2" d dx !e x" ! e x d dx !1 " x 2" !1 " x 2"2 (1, 1 2e) y ! e x%!1 " x 2" V ! !x 4 ! 2x 3 " 6x 2 " 12x " 6 !x 3 " 6"2 ! !2x 4 " x 3 " 12x " 6" ! !3x 4 " 3x 3 ! 6x 2" !x 3 " 6"2 ! !x 3 " 6"!2x " 1" ! !x 2 " x ! 2"!3x 2" !x 3 " 6"2 y% ! !x 3 " 6" d dx !x 2 " x ! 2" ! !x 2 " x ! 2" d dx !x 3 " 6" !x 3 " 6"2 y ! x 2 " x ! 2 x 3 " 6 V d dx f!x" t!x"+ ! t!x" d dx & f!x"' ! f!x" d dx &t!x"' &t!x"' 2 t f d dx( u v) ! v lim (x l 0 (u (x ! u lim (x l 0 (v (x v lim (x l 0!v " (v" ! v du dx ! u dv dx v2 v ! t!x" (v l 0 (x l 0 d dx( u v) ! lim (x l 0 (!u%v" (x ! lim (x l 0 v (u (x ! u (v (x v!v " (v" 186 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N In prime notation: ( f t) % ! tf % ! ft% t2 N We can use a graphing device to check that the answer to Example 4 is plausible. Figure 3 shows the graphs of the function of Example 4 and its derivative. Notice that when grows rapidly (near ), is large. And when grows slowly, is near . 0 y% y y% !2 y 1.5 _1.5 _4 4 yª y FIGURE 3 So the slope of the tangent line at is This means that the tangent line at is horizontal and its equation is . [See Figure 4. Notice that the function is increasing and crosses its tangent line at .] M Don’t use the Quotient Rule every time you see a quotient. Sometimes it’s eas-ier to rewrite a quotient ﬁrst to put it in a form that is simpler for the purpose of differen-tiation. For instance, although it is possible to differentiate the function using the Quotient Rule, it is much easier to perform the division ﬁrst and write the func-tion as before differentiating. We summarize the differentiation formulas we have learned so far as follows. TABLE OF DIFFERENTIATION FORMULAS ( f t) % ! tf% ! ft% t2 ! ft"% ! ft% " tf% ! f ! t"% ! f% ! t% ! f " t"% ! f% " t% !cf "% ! cf% d dx !e x" ! e x d dx !x n" ! nx n!1 d dx !c" ! 0 F!x" ! 3x " 2x !1%2 F!x" ! 3x 2 " 2sx x NOTE (1, 1 2e) y ! 1 2e (1, 1 2e) dy dx , x!1 ! 0 (1, 1 2e) SECTION 3.2 THE PRODUCT AND QUOTIENT RULES \|\|\|\| 187 2.5 0 _2 3.5 y= ´ 1+≈ FIGURE 4 y= e 1 2 7. 8. 9. 10. 12. 13. 14. 15. 16. 17. 18. y ! 1 s " kes y ! !r 2 ! 2r"er y ! t !t ! 1"2 y ! t 2 " 2 t 4 ! 3t 2 " 1 y ! x " 1 x 3 " x ! 2 y ! x 3 1 ! x 2 R!t" ! !t " e t"(3 ! st ) F!y" !( 1 y2 ! 3 y4)!y " 5y3" 11. Y!u" ! !u!2 " u!3"!u5 ! 2u2" V!x" ! !2x3 " 3"!x4 ! 2x" f !t" ! 2t 4 " t 2 t!x" ! 3x ! 1 2x " 1 1. Find the derivative of in two ways: by using the Product Rule and by performing the multiplication ﬁrst. Do your answers agree? 2. Find the derivative of the function in two ways: by using the Quotient Rule and by simplifying ﬁrst. Show that your answers are equivalent. Which method do you prefer? 3–26 Differentiate. 3. 4. 5. 6. y ! e x 1 " x y ! e x x 2 t!x" ! sx e x f !x" ! !x 3 " 2x"e x F!x" ! x ! 3xsx sx y ! !x 2 " 1"!x 3 " 1" EXERCISES 3.2 39. (a) If , ﬁnd and . ; (b) Check to see that your answers to part (a) are reasonable by comparing the graphs of , , and . (a) If , ﬁnd and . ; (b) Check to see that your answers to part (a) are reasonable by comparing the graphs of , , and . 41. If , ﬁnd . 42. If , ﬁnd . Suppose that , , , and . Find the following values. (a) (b) (c) 44. Suppose that , , , and . Find . (a) (b) (c) (d) If , where and , ﬁnd . 46. If and , ﬁnd 47. If and are the functions whose graphs are shown, let and . (a) Find (b) Find 48. Let and , where and are the functions whose graphs are shown. (a) Find . (b) Find . F G x y 0 1 1 Q%!7" P%!2" G F Q!x" ! F!x"%G!x" P!x" ! F!x"G!x" f g x y 0 1 1 v%!5". u%!1". v!x" ! f !x"%t!x" u!x" ! f !x"t!x" t f d dx ( h!x" x ), x!2 h%!2" ! !3 h!2" ! 4 f %!0" t%!0" ! 5 t!0" ! 2 f !x" ! e xt!x" 45. h!x" ! t!x" 1 " f !x" h!x" ! f !x" t!x" h!x" ! f !x"t!x" h!x" ! 5f !x" ! 4t!x" h%!2" t%!2" ! 7 f %!2" ! !2 t!2" ! 4 f !2" ! !3 !t%f "%!5" ! f%t"%!5" ! ft"%!5" t%!5" ! 2 t!5" ! !3 f %!5" ! 6 f !5" ! 1 43. t !n"!x" t!x" ! x%e x f '!1" f !x" ! x 2%!1 " x" f ' f % f f '!x" f %!x" f !x" ! x%!x 2 " 1" 40. f ' f % f f '!x" f %!x" f !x" ! !x ! 1"e x 19. 20. 21. 22. 23. 24. 26. 27–30 Find and . 27. 28. 29. 30. 31–32 Find an equation of the tangent line to the given curve at the speciﬁed point. 31. , 32. , 33–34 Find equations of the tangent line and normal line to the given curve at the speciﬁed point. , 34. , 35. (a) The curve is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point . ; (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen. (a) The curve is called a serpentine. Find an equation of the tangent line to this curve at the point . ; (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen. 37. (a) If , ﬁnd . ; (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of and . 38. (a) If , ﬁnd . ; (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of and . f % f f %!x" f !x" ! x%!x 2 ! 1" f % f f %!x" f !x" ! e x%x 3 !3, 0.3" y ! x%!1 " x 2" 36. (!1, 1 2) y ! 1%!1 " x2" !4, 0.4" y ! sx x " 1 !0, 0" y ! 2xe x 33. !1, e" y ! e x x !1, 1" y ! 2x x " 1 f !x" ! x 3 " e x f !x" ! x 2 1 " 2x f !x" ! x 5%2e x f !x" ! x 4e x f '!x" f %!x" f !x" ! ax " b cx " d f !x" ! x x " c x 25. f !x" ! 1 ! xe x x " e x f !x" ! A B " Ce x t!t" ! t ! st t 1%3 f !t" ! 2t 2 " st z ! w 3%2!w " cew" y ! v3 ! 2vsv v 188 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS \|\|\|\| 189 write . Then the total revenue earned with selling price p is . (a) What does it mean to say that and ? (b) Assuming the values in part (a), ﬁnd and interpret your answer. (a) Use the Product Rule twice to prove that if , , and are differentiable, then . (b) Taking in part (a), show that (c) Use part (b) to differentiate . 56. (a) If , where and have derivatives of all orders, show that . (b) Find similar formulas for and . (c) Guess a formula for . 57. Find expressions for the ﬁrst ﬁve derivatives of . Do you see a pattern in these expressions? Guess a formula for and prove it using mathematical induction. 58. (a) If t is differentiable, the Reciprocal Rule says that Use the Quotient Rule to prove the Reciprocal Rule. (b) Use the Reciprocal Rule to differentiate the function in Exercise 18. (c) Use the Reciprocal Rule to verify that the Power Rule is valid for negative integers, that is, for all positive integers . n d dx !x!n" ! !nx!n!1 d dx 1 t!x"+ ! ! t%!x" &t!x"'2 f !n"!x" f !x" ! x 2e x F !n" F !4" F F ' ! f 't " 2f %t% " ft' t f F!x" ! f !x"t!x" y ! e 3x d dx & f!x"'3 ! 3& f!x"'2 f%!x" f ! t ! h ! fth"% ! f% th " ft%h " fth% h t f 55. R%!20" f %!20" ! !350 f !20" ! 10,000 R!p" ! pf !p" q ! f !p" 49. If is a differentiable function, ﬁnd an expression for the deriv-ative of each of the following functions. (a) (b) (c) If is a differentiable function, ﬁnd an expression for the derivative of each of the following functions. (a) (b) (c) (d) How many tangent lines to the curve ) pass through the point ? At which points do these tangent lines touch the curve? 52. Find equations of the tangent lines to the curve that are parallel to the line . 53. In this exercise we estimate the rate at which the total personal income is rising in the Richmond-Petersburg, Virginia, metro-politan area. In 1999, the population of this area was 961,400, and the population was increasing at roughly 9200 people per year. The average annual income was $30,593 per capita, and this average was increasing at about $1400 per year (a little above the national average of about $1225 yearly). Use the Product Rule and these ﬁgures to estimate the rate at which total personal income was rising in the Richmond-Petersburg area in 1999. Explain the meaning of each term in the Product Rule. 54. A manufacturer produces bolts of a fabric with a ﬁxed width. The quantity q of this fabric (measured in yards) that is sold is a function of the selling price p (in dollars per yard), so we can x ! 2y ! 2 y ! x ! 1 x " 1 !1, 2" y ! x%!x " 1 51. y ! 1 " xf !x" sx y ! x 2 f !x" y ! f !x" x 2 y ! x 2f !x" f 50. y ! t!x" x y ! x t!x" y ! xt!x" t DERIVATIVES OF TRIGONOMETRIC FUNCTIONS Before starting this section, you might need to review the trigonometric functions. In par-ticular, it is important to remember that when we talk about the function deﬁned for all real numbers by it is understood that means the sine of the angle whose radian measure is . A simi-lar convention holds for the other trigonometric functions cos, tan, csc, sec, and cot. Recall from Section 2.5 that all of the trigonometric functions are continuous at every number in their domains. If we sketch the graph of the function and use the interpretation of as the slope of the tangent to the sine curve in order to sketch the graph of (see Exer-f% f%!x" f!x" ! sin x x sin x f!x" ! sin x x f 3.3 N A review of the trigonometric functions is given in Appendix D. cise 14 in Section 2.8), then it looks as if the graph of may be the same as the cosine curve (see Figure 1). Let’s try to conﬁrm our guess that if , then . From the deﬁni-tion of a derivative, we have Two of these four limits are easy to evaluate. Since we regard x as a constant when com-puting a limit as , we have The limit of is not so obvious. In Example 3 in Section 2.2 we made the guess, on the basis of numerical and graphical evidence, that We now use a geometric argument to prove Equation 2. Assume ﬁrst that lies between 0 and . Figure 2(a) shows a sector of a circle with center O, central angle , and + ,%2 + lim + l 0 sin + + ! 1 2 !sin h"%h lim h l 0 cos x ! cos x and lim h l 0 sin x ! sin x h l 0 ! lim h l 0 sin x ! lim h l 0 cos h ! 1 h " lim h l 0 cos x ! lim h l 0 sin h h 1 ! lim h l 0 sin x ( cos h ! 1 h ) " cos x ( sin h h )+ ! lim h l 0 sin x cos h ! sin x h " cos x sin h h + ! lim h l 0 sin x cos h " cos x sin h ! sin x h ! lim h l 0 sin!x " h" ! sin x h f%!x" ! lim h l 0 f!x " h" ! f!x" h f%!x" ! cos x f!x" ! sin x x 0 2π x 0 π 2 FIGURE 1 π π 2 π ƒ= y= sin x y y fª(x y= ) f% 190 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES Visual 3.3 shows an animation of Figure 1. TEC N We have used the addition formula for sine. See Appendix D. radius 1. BC is drawn perpendicular to OA. By the deﬁnition of radian measure, we have arc . Also . From the diagram we see that Therefore so Let the tangent lines at and intersect at . You can see from Figure 2(b) that the circumference of a circle is smaller than the length of a circumscribed polygon, and so arc . Thus (In Appendix F the inequality is proved directly from the deﬁnition of the length of an arc without resorting to geometric intuition as we did here.) Therefore, we have so We know that and , so by the Squeeze Theorem, we have But the function is an even function, so its right and left limits must be equal. Hence, we have so we have proved Equation 2. We can deduce the value of the remaining limit in (1) as follows: (by Equation 2) ! !1 !( 0 1 " 1) ! 0 ! !lim + l 0 sin + + ! lim + l 0 sin + cos + " 1 ! lim + l 0 !sin2+ + !cos + " 1" ! !lim + l 0 ( sin + + ! sin + cos + " 1) lim + l 0 cos + ! 1 + ! lim + l 0 ( cos + ! 1 + ! cos + " 1 cos + " 1) ! lim + l 0 cos2+ ! 1 + !cos + " 1" lim + l 0 sin + + ! 1 !sin +"%+ lim +l 0" sin + + ! 1 lim + l 0 cos + ! 1 lim + l 0 1 ! 1 cos + & sin + + & 1 + & sin + cos + + # tan + ! tan + ! $AD$ ! $OA$ tan + & $AE$ " $ED$ + ! arc AB & $AE$ " $EB$ AB & $AE$ " $EB$ E B A sin + + & 1 sin + & + $BC$ & $AB$ & arc AB $BC$ ! $OB$sin + ! sin + AB ! + SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS \|\|\|\| 191 FIGURE 2 (b) (a) B A E O ¨ B A O 1 D E C N We multiply numerator and denominator by in order to put the function in a form in which we can use the limits we know. cos + " 1 If we now put the limits (2) and (3) in (1), we get So we have proved the formula for the derivative of the sine function: EXAMPLE 1 Differentiate . SOLUTION Using the Product Rule and Formula 4, we have M Using the same methods as in the proof of Formula 4, one can prove (see Exercise 20) that The tangent function can also be differentiated by using the deﬁnition of a derivative, but it is easier to use the Quotient Rule together with Formulas 4 and 5: ! 1 cos2x ! sec2x ! cos2x ! sin2x cos2x ! cos x ! cos x " sin x !"sin x" cos2x ! cos x d dx !sin x" " sin x d dx !cos x" cos2x d dx !tan x" ! d dx # sin x cos x$ d dx !cos x" ! "sin x 5 ! x 2 cos x ! 2x sin x dy dx ! x 2 d dx !sin x" ! sin x d dx !x 2" y ! x 2 sin x V d dx !sin x" ! cos x 4 ! !sin x" ! 0 ! !cos x" ! 1 ! cos x f#!x" ! lim h l 0 sin x ! lim h l 0 cos h " 1 h ! lim h l 0 cos x ! lim h l 0 sin h h lim $ l 0 cos $ " 1 $ ! 0 3 192 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N Figure 3 shows the graphs of the function of Example 1 and its derivative. Notice that whenever has a horizontal tangent. y y# ! 0 5 _5 _4 4 y yª FIGURE 3 The derivatives of the remaining trigonometric functions, , , and , can also be found easily using the Quotient Rule (see Exercises 17–19). We collect all the differentia-tion formulas for trigonometric functions in the following table. Remember that they are valid only when is measured in radians. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS EXAMPLE 2 Differentiate . For what values of x does the graph of have a horizontal tangent? SOLUTION The Quotient Rule gives In simplifying the answer we have used the identity . Since is never 0, we see that when , and this occurs when , where n is an integer (see Figure 4). M Trigonometric functions are often used in modeling real-world phenomena. In particu-lar, vibrations, waves, elastic motions, and other quantities that vary in a periodic manner can be described using trigonometric functions. In the following example we discuss an instance of simple harmonic motion. EXAMPLE 3 An object at the end of a vertical spring is stretched 4 cm beyond its rest position and released at time . (See Figure 5 and note that the downward direction is positive.) Its position at time t is s ! f!t" ! 4 cos t t ! 0 V x ! n% ! %%4 tan x ! 1 f#!x" ! 0 sec x tan2x ! 1 ! sec2x ! sec x !tan x " 1" !1 ! tan x"2 ! sec x !tan x ! tan2x " sec2x" !1 ! tan x"2 ! !1 ! tan x" sec x tan x " sec x ! sec2x !1 ! tan x"2 f#!x" ! !1 ! tan x" d dx !sec x" " sec x d dx !1 ! tan x" !1 ! tan x"2 f f!x" ! sec x 1 ! tan x d dx !cot x" ! "csc2x d dx !tan x" ! sec2x d dx !sec x" ! sec x tan x d dx !cos x" ! "sin x d dx !csc x" ! "csc x cot x d dx !sin x" ! cos x x cot sec csc d dx !tan x" ! sec2x 6 SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS \|\|\|\| 193 N When you memorize this table, it is helpful to notice that the minus signs go with the der-ivatives of the “cofunctions,” that is, cosine, cosecant, and cotangent. 3 _3 _3 5 FIGURE 4 The horizontal tangents in Example 2 s 0 4 FIGURE 5 Find the velocity and acceleration at time t and use them to analyze the motion of the object. SOLUTION The velocity and acceleration are The object oscillates from the lowest point to the highest point . The period of the oscillation is , the period of . The speed is , which is greatest when , that is, when . So the object moves fastest as it passes through its equilibrium position . Its speed is 0 when , that is, at the high and low points. The acceleration . It has greatest magnitude at the high and low points. See the graphs in Figure 6. M EXAMPLE 4 Find the 27th derivative of . SOLUTION The ﬁrst few derivatives of are as follows: We see that the successive derivatives occur in a cycle of length 4 and, in particular, whenever is a multiple of 4. Therefore and, differentiating three more times, we have M Our main use for the limit in Equation 2 has been to prove the differentiation formula for the sine function. But this limit is also useful in ﬁnding certain other trigonometric lim-its, as the following two examples show. EXAMPLE 5 Find . SOLUTION In order to apply Equation 2, we ﬁrst rewrite the function by multiplying and dividing by 7: sin 7x 4x ! 7 4# sin 7x 7x $ lim x l 0 sin 7x 4x f !27"!x" ! sin x f !24"!x" ! cos x n f !n"!x" ! cos x f !5"!x" ! "sin x f !4"!x" ! cos x f &!x" ! sin x f '!x" ! "cos x f #!x" ! "sin x f!x" ! cos x cos x a ! "4 cos t ! 0 when s ! 0 sin t ! 0 !s ! 0" cos t ! 0 &sin t& ! 1 & v& ! 4&sin t& cos t 2% !s ! "4 cm" !s ! 4 cm" a ! dv dt ! d dt !"4 sin t" ! "4 d dt !sin t" ! "4 cos t v ! ds dt ! d dt !4 cos t" ! 4 d dt !cos t" ! "4 sin t 194 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N Look for a pattern. FIGURE 6 2 _2 √ s a π 2π t 0 Note that . sin 7x " 7 sin x If we let , then as , so by Equation 2 we have M EXAMPLE 6 Calculate . SOLUTION Here we divide numerator and denominator by x: (by the continuity of cosine and Equation 2) M ! 1 ! cos 0 1 ! lim x l 0 cos x sin x x ! lim x l 0 cos x lim x l 0 sin x x lim x l 0 x cot x ! lim x l 0 x cos x sin x lim x l 0 x cot x V ! 7 4 lim $ l 0 sin $ $ ! 7 4 ! 1 ! 7 4 lim x l 0 sin 7x 4x ! 7 4 lim x l 0# sin 7x 7x $ x l 0 $ l 0 $ ! 7x 21–24 Find an equation of the tangent line to the curve at the given point. 21. 23. , 24. , 25. (a) Find an equation of the tangent line to the curve at the point . ; (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen. 26. (a) Find an equation of the tangent line to the curve at the point . ; (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen. 27. (a) If , ﬁnd . ; (b) Check to see that your answer to part (a) is reasonable by graphing both and for . 28. (a) If , ﬁnd and . ; (b) Check to see that your answers to part (a) are reasonable by graphing , , and . If , ﬁnd . 30. If , ﬁnd . f '!%%4" f !x" ! sec x H#!$" and H'!$" H!$" ! $ sin $ 29. f ' f # f f '!x" f #!x" f !x" ! e x cos x & x& ( %%2 f # f f #!x" f !x" ! sec x " x !%%3, 1" y ! sec x " 2 cos x !%%2, %" y ! 2x sin x !0, 1" y ! 1 sin x ! cos x !0, 1" y ! x ! cos x !0, 1" y ! e x cos x, 22. !%%3, 2" y ! sec x, 1–16 Differentiate. 1. 2. 3. 4. 5. 6. 7. 8. 10. 11. 12. 13. 14. 15. 16. 17. Prove that . 18. Prove that . 19. Prove that . 20. Prove, using the deﬁnition of derivative, that if , then . f #!x" ! "sin x f !x" ! cos x d dx !cot x" ! "csc2x d dx !sec x" ! sec x tan x d dx !csc x" ! "csc x cot x y ! x 2 sin x tan x f !x" ! xe x csc x y ! csc $ !$ ! cot $" y ! sin x x 2 y ! 1 " sec x tan x f !$" ! sec $ 1 ! sec $ y ! 1 ! sin x x ! cos x y ! x 2 " tan x 9. y ! e u !cos u ! cu" h!$" ! csc $ ! e$ cot $ t!t" ! 4 sec t ! tan t t!t" ! t 3 cos t y ! 2 csc x ! 5 cos x f !x" ! sin x ! 1 2 cot x f !x" ! sx sin x f !x" ! 3x 2 " 2 cos x EXERCISES 3.3 SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS \|\|\|\| 195 A ladder 10 ft long rests against a vertical wall. Let be the angle between the top of the ladder and the wall and let be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does change with respect to when ? 38. An object with weight is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle with the plane, then the magnitude of the force is where is a constant called the coefﬁcient of friction. (a) Find the rate of change of with respect to . (b) When is this rate of change equal to 0? ; (c) If lb and , draw the graph of as a func-tion of and use it to locate the value of for which . Is the value consistent with your answer to part (b)? 39–48 Find the limit. 39. 40. 42. 43. 44. 46. 47. 48. 49. Differentiate each trigonometric identity to obtain a new (or familiar) identity. (a) (b) (c) 50. A semicircle with diameter sits on an isosceles triangle to form a region shaped like a two-dimensional ice-PQR PQ sin x ! cos x ! 1 ! cot x csc x sec x ! 1 cos x tan x ! sin x cos x lim x l 1 sin!x " 1" x 2 ! x " 2 lim % l %%4 1 " tan x sin x " cos x lim x l 0 sin!x 2" x lim $ l 0 sin $ $ ! tan $ 45. lim t l 0 sin2 3t t 2 lim $ l 0 sin!cos $" sec $ lim $ l 0 cos $ " 1 sin $ lim t l 0 tan 6t sin 2t 41. lim x l 0 sin 4x sin 6x lim x l 0 sin 3x x dF%d$ ! 0 $ $ F ) ! 0.6 W ! 50 $ F ) F ! )W ) sin $ ! cos $ $ W $ ! %%3 $ x x $ 37. 31. (a) Use the Quotient Rule to differentiate the function (b) Simplify the expression for by writing it in terms of and , and then ﬁnd . (c) Show that your answers to parts (a) and (b) are equivalent. 32. Suppose and , and let and Find (a) and (b) . For what values of does the graph of have a horizontal tangent? 34. Find the points on the curve at which the tangent is horizontal. 35. A mass on a spring vibrates horizontally on a smooth level surface (see the ﬁgure). Its equation of motion is , where is in seconds and in centimeters. (a) Find the velocity and acceleration at time . (b) Find the position, velocity, and acceleration of the mass at time . In what direction is it moving at that time? ; 36. An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is , , where is measured in centimeters and in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time . (b) Graph the velocity and acceleration functions. (c) When does the mass pass through the equilibrium position for the ﬁrst time? (d) How far from its equilibrium position does the mass travel? (e) When is the speed the greatest? t t s t 0 s ! 2 cos t ! 3 sin t x x 0 equilibrium position t ! 2%%3 t x t x!t" ! 8 sin t y ! !cos x"%!2 ! sin x" f !x" ! x ! 2 sin x x 33. h#!%%3" t#!%%3" h!x" ! cos x f !x" t!x" ! f !x" sin x f #!%%3" ! "2 f !%%3" ! 4 f #!x" cos x sin x f !x" f !x" ! tan x " 1 sec x 196 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES SECTION 3.4 THE CHAIN RULE \|\|\|\| 197 The ﬁgure shows a circular arc of length and a chord of length , both subtended by a central angle . Find d ¨ s lim $l 0! s d $ d s 51. cream cone, as shown in the ﬁgure. If is the area of the semicircle and is the area of the triangle, ﬁnd P Q R B(¨) A(¨) ¨ 10 cm 10 cm lim $l 0! A!$" B!$" B!$" A!$" THE CHAIN RULE Suppose you are asked to differentiate the function The differentiation formulas you learned in the previous sections of this chapter do not enable you to calculate . Observe that is a composite function. In fact, if we let and let , then we can write , that is, . We know how to differentiate both and , so it would be useful to have a rule that tells us how to ﬁnd the derivative of in terms of the derivatives of and . It turns out that the derivative of the composite function is the product of the deriv-atives of and . This fact is one of the most important of the differentiation rules and is called the Chain Rule. It seems plausible if we interpret derivatives as rates of change. Regard as the rate of change of with respect to , as the rate of change of with respect to , and as the rate of change of with respect to . If changes twice as fast as and changes three times as fast as , then it seems reasonable that changes six times as fast as , and so we expect that THE CHAIN RULE If t is differentiable at and is differentiable at , then the composite function deﬁned by is differentiable at and is given by the product In Leibniz notation, if and are both differentiable functions, then dy dx ! dy du du dx u ! t!x" y ! f!u" F#!x" ! f#!t!x"" ! t#!x" F# x F!x" ! f!t!x"" F ! f ! t t!x" f x dy dx ! dy du du dx x y u y x u x y dy%dx u y dy%du x u du%dx t f f ! t t f F ! f ! t t f F ! f ! t y ! F!x" ! f!t!x"" u ! t!x" ! x 2 ! 1 y ! f!u" ! su F F#!x" F!x" ! sx 2 ! 1 3.4 N See Section 1.3 for a review of composite functions. COMMENTS ON THE PROOF OF THE CHAIN RULE Let be the change in corresponding to a change of in , that is, Then the corresponding change in is It is tempting to write The only ﬂaw in this reasoning is that in (1) it might happen that (even when ) and, of course, we can’t divide by 0. Nonetheless, this reasoning does at least suggest that the Chain Rule is true. A full proof of the Chain Rule is given at the end of this section. M The Chain Rule can be written either in the prime notation or, if and , in Leibniz notation: Equation 3 is easy to remember because if and were quotients, then we could cancel . Remember, however, that has not been deﬁned and should not be thought of as an actual quotient. EXAMPLE 1 Find if . SOLUTION 1 (using Equation 2): At the beginning of this section we expressed as where and . Since and we have ! 1 2sx 2 ! 1 ! 2x ! x sx 2 ! 1 F#!x" ! f#!t!x"" ! t#!x" t#!x" ! 2x f#!u" ! 1 2u"1%2 ! 1 2su t!x" ! x 2 ! 1 f!u" ! su F!x" ! ! f ! t"!x" ! f!t!x"" F F!x" ! sx 2 ! 1 F#!x" du%dx du du du%dx dy%du dy dx ! dy du du dx 3 u ! t!x" y ! f!u" ! f ! t"#!x" ! f#!t!x"" ! t#!x" 2 +x " 0 +u ! 0 ! dy du du dx ! lim +u l 0 +y +u ! lim +x l 0 +u +x ! lim +x l 0 +y +u ! lim +x l 0 +u +x ! lim +x l 0 +y +u ! +u +x 1 dy dx ! lim +x l 0 +y +x +y ! f!u ! +u" " f!u" y +u ! t!x ! +x" " t!x" x +x u +u 198 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES (Note that as since is continuous.) t +x l 0 +u l 0 SOLUTION 2 (using Equation 3): If we let and , then M When using Formula 3 we should bear in mind that refers to the derivative of when is considered as a function of (called the derivative of with respect to ), where-as refers to the derivative of when considered as a function of (the derivative of with respect to ). For instance, in Example 1, can be considered as a function of and also as a function of . Note that whereas In using the Chain Rule we work from the outside to the inside. Formula 2 says that we differentiate the outer function [at the inner function ] and then we multiply by the derivative of the inner function. EXAMPLE 2 Differentiate (a) and (b) . SOLUTION (a) If , then the outer function is the sine function and the inner function is the squaring function, so the Chain Rule gives (b) Note that . Here the outer function is the squaring function and the inner function is the sine function. So The answer can be left as or written as (by a trigonometric identity known as the double-angle formula). M In Example 2(a) we combined the Chain Rule with the rule for differentiating the sine function. In general, if , where is a differentiable function of , then, by the Chain Rule, dy dx ! dy du du dx ! cos u du dx x u y ! sin u sin 2x 2 sin x cos x dy dx ! d dx !sin x"2 ! 2 ! !sin x" ! cos x sin2x ! !sin x"2 ! 2x cos!x 2" dy dx ! d dx sin !x 2" ! cos !x 2" ! 2x y ! sin!x 2" y ! sin2x y ! sin!x 2" V d dx f !t!x"" ! f# !t!x"" ! t#!x" t!x" f NOTE dy du ! f#!u" ! 1 2su dy dx ! F#!x" ! x sx 2 ! 1 (y ! su ) u (y ! sx 2 ! 1) x y u y u y dy%du x y x y y dy%dx ! 1 2sx 2 ! 1 !2x" ! x sx 2 ! 1 F#!x" ! dy du du dx ! 1 2su !2x" y ! su u ! x 2 ! 1 SECTION 3.4 THE CHAIN RULE \|\|\|\| 199 outer function evaluated at inner function derivative of outer function evaluated at inner function derivative of inner function outer function evaluated at inner function derivative of outer function evaluated at inner function derivative of inner function inner function derivative of outer function evaluated at inner function derivative of inner function N See Reference Page 2 or Appendix D. Thus In a similar fashion, all of the formulas for differentiating trigonometric functions can be combined with the Chain Rule. Let’s make explicit the special case of the Chain Rule where the outer function is a power function. If , then we can write where . By using the Chain Rule and then the Power Rule, we get THE POWER RULE COMBINED WITH THE CHAIN RULE If is any real number and is differentiable, then Alternatively, Notice that the derivative in Example 1 could be calculated by taking in Rule 4. EXAMPLE 3 Differentiate . SOLUTION Taking and in (4), we have M EXAMPLE 4 Find if . SOLUTION First rewrite : Thus M EXAMPLE 5 Find the derivative of the function SOLUTION Combining the Power Rule, Chain Rule, and Quotient Rule, we get M ! 9# t " 2 2t ! 1$ 8 !2t ! 1" ! 1 " 2!t " 2" !2t ! 1"2 ! 45!t " 2"8 !2t ! 1"10 t#!t" ! 9# t " 2 2t ! 1$ 8 d dt # t " 2 2t ! 1$ t!t" !# t " 2 2t ! 1$ 9 ! " 1 3!x 2 ! x ! 1""4%3!2x ! 1" f#!x" ! " 1 3!x 2 ! x ! 1""4%3 d dx !x 2 ! x ! 1" f!x" ! !x 2 ! x ! 1""1%3 f f!x" ! 1 s 3 x 2 ! x ! 1 f#!x" V ! 100!x 3 " 1"99 ! 3x 2 ! 300x 2!x 3 " 1"99 dy dx ! d dx !x 3 " 1"100 ! 100!x 3 " 1"99 d dx !x 3 " 1" n ! 100 u ! t!x" ! x 3 " 1 y ! !x 3 " 1"100 n ! 1 2 d dx 't!x"(n ! n't!x"(n"1 ! t#!x" d dx !u n" ! nu n"1 du dx u ! t!x" n 4 dy dx ! dy du du dx ! nu n"1 du dx ! n't!x"(n"1t#!x" u ! t!x" y ! f!u" ! u n y ! 't!x"(n f d dx !sin u" ! cos u du dx 200 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES EXAMPLE 6 Differentiate . SOLUTION In this example we must use the Product Rule before using the Chain Rule: Noticing that each term has the common factor , we could factor it out and write the answer as M EXAMPLE 7 Differentiate . SOLUTION Here the inner function is and the outer function is the exponential function . So, by the Chain Rule, M We can use the Chain Rule to differentiate an exponential function with any base . Recall from Section 1.6 that . So and the Chain Rule gives because ln a is a constant. So we have the formula In particular, if , we get d dx !2x" ! 2x ln 2 6 a ! 2 d dx !a x" ! a x ln a 5 ! e !ln a"x , ln a ! a x ln a d dx !a x" ! d dx !e !ln a"x" ! e !ln a"x d dx !ln a"x a x ! !e ln a"x ! e !ln a"x a ! e ln a a - 0 dy dx ! d dx !e sin x" ! e sin x d dx !sin x" ! e sin x cos x f!x" ! e x t!x" ! sin x y ! e sin x dy dx ! 2!2x ! 1"4!x 3 " x ! 1"3!17x 3 ! 6x 2 " 9x ! 3" 2!2x ! 1"4!x 3 " x ! 1"3 ! 4!2x ! 1"5!x 3 " x ! 1"3!3x 2 " 1" ! 5!x 3 " x ! 1"4!2x ! 1"4 ! 2 ! !x 3 " x ! 1"4 ! 5!2x ! 1"4 d dx !2x ! 1" ! !2x ! 1"5 ! 4!x 3 " x ! 1"3 d dx !x 3 " x ! 1" dy dx ! !2x ! 1"5 d dx !x 3 " x ! 1"4 ! !x 3 " x ! 1"4 d dx !2x ! 1"5 y ! !2x ! 1"5!x 3 " x ! 1"4 SECTION 3.4 THE CHAIN RULE \|\|\|\| 201 10 _10 _2 1 y yª FIGURE 1 N The graphs of the functions and in Example 6 are shown in Figure 1. Notice that is large when increases rapidly and when has a horizontal tangent. So our answer appears to be reasonable. y y# ! 0 y y# y# y N More generally, the Chain Rule gives d dx !eu" ! eu du dx N Don’t confuse Formula 5 (where is the expo-nent) with the Power Rule (where is the base): d dx !x n" ! nx n"1 x x In Section 3.1 we gave the estimate This is consistent with the exact formula (6) because . The reason for the name “Chain Rule” becomes clear when we make a longer chain by adding another link. Suppose that , , and , where , , and are differentiable functions. Then, to compute the derivative of with respect to , we use the Chain Rule twice: EXAMPLE 8 If , then Notice that we used the Chain Rule twice. M EXAMPLE 9 Differentiate . SOLUTION The outer function is the exponential function, the middle function is the secant function and the inner function is the tripling function. So we have M HOW TO PROVE THE CHAIN RULE Recall that if and x changes from a to , we deﬁned the increment of y as According to the deﬁnition of a derivative, we have So if we denote by the difference between the difference quotient and the derivative, we obtain lim !x l 0 " ! lim !x l 0! !y !x # f$"a#$ ! f$"a# # f$"a# ! 0 " lim !x l 0 !y !x ! f$"a# !y ! f"a % !x# # f"a# a % !x y ! f"x# ! 3e sec 3& sec 3& tan 3& ! e sec 3& sec 3& tan 3& d d& "3&# dy d& ! e sec 3& d d& "sec 3&# y ! e sec 3& ! #cos"cos"tan x## sin"tan x# sec2x ! cos"cos"tan x## %#sin"tan x#& d dx "tan x# f$"x# ! cos"cos"tan x## d dx cos"tan x# f"x# ! sin"cos"tan x## V dy dt ! dy dx dx dt ! dy du du dx dx dt t y h t f x ! h"t# u ! t"x# y ! f"u# ln 2 ' 0.693147 d dx "2x# ' "0.69#2x 202 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES But If we deﬁne to be 0 when , then becomes a continuous function of . Thus, for a differentiable function f, we can write and is a continuous function of . This property of differentiable functions is what enables us to prove the Chain Rule. PROOF OF THE CHAIN RULE Suppose is differentiable at a and is differen-tiable at . If is an increment in x and and are the corresponding incre-ments in u and y, then we can use Equation 7 to write where as . Similarly where as . If we now substitute the expression for from Equation 8 into Equation 9, we get so As , Equation 8 shows that . So both and as . Therefore This proves the Chain Rule. M f bta f tata dy dx lim x l 0 y x lim x l 0 f b 2ta 1 x l 0 2 l 0 1 l 0 u l 0 x l 0 y x f b 2ta 1 y f b 2ta 1 x u u l 0 2 l 0 y f b u 2 u f b 2 u 9 x l 0 1 l 0 u ta x 1 x ta 1 x 8 y u x b ta y f u u tx x y f a x x where l 0 as x l 0 7 x x 0 y x f a ? y f a x x SECTION 3.4 THE CHAIN RULE \|\|\|\| 203 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 20. y x 2 1s 3 x 2 2 y 2x 548x 2 53 19. ht t 4 13t 3 14 tx 1 4x53 x x28 y 3 cotn y xekx y a3 cos3x y cosa3 x 3 f t s 3 1 tan t tt 1 t 4 13 f x 1 x 423 Fx s 4 1 2x x 3 1–6 Write the composite function in the form . [Identify the inner function and the outer function .] Then ﬁnd the derivative . 1. 2. 3. 4. 6. 7–46 Find the derivative of the function. 7. 8. Fx 4x x 2100 Fx x 4 3x 2 25 y sine x y esx 5. y tansin x y 1 x 210 y s4 3x y sin 4x dydx y f u u tx f tx EXERCISES 3.4 ; 58. The function , , arises in applications to frequency modulation (FM) synthesis. (a) Use a graph of produced by a graphing device to make a rough sketch of the graph of . (b) Calculate and use this expression, with a graphing device, to graph . Compare with your sketch in part (a). Find all points on the graph of the function at which the tangent line is horizontal. 60. Find the -coordinates of all points on the curve at which the tangent line is horizontal. If , where , , , , and , ﬁnd . 62. If , where and , ﬁnd . 63. A table of values for , , , and is given. (a) If , ﬁnd . (b) If , ﬁnd . 64. Let and be the functions in Exercise 63. (a) If , ﬁnd . (b) If , ﬁnd . If and are the functions whose graphs are shown, let , , and . Find each derivative, if it exists. If it does not exist, explain why. (a) (b) (c) 66. If is the function whose graph is shown, let and . Use the graph of to estimate the value of each derivative. (a) (b) x y 0 1 y=ƒ 1 t$"2# h$"2# f t"x# ! f "x 2# h"x# ! f " f "x## f x y 0 f g 1 1 w$"1# v$"1# u$"1# w"x# ! t"t"x## v"x# ! t" f "x## u"x# ! f "t"x## t f 65. G$"3# G"x# ! t"t"x## F$"2# F"x# ! f " f "x## t f H$"1# H"x# ! t" f "x## h$"1# h"x# ! f "t"x## t$ f $ t f h$"1# f $"1# ! 4 f "1# ! 7 h"x# ! s4 % 3f "x# F$"5# t$"5# ! 6 t"5# ! #2 f $"5# ! 3 f $"#2# ! 4 f "#2# ! 8 F"x# ! f "t"x## 61. y ! sin 2x # 2 sin x x f "x# ! 2 sin x % sin2x 59. f $ f $"x# f $ f 0 ' x ' ( f "x# ! sin"x % sin 2x# 21. 22. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47–50 Find the ﬁrst and second derivatives of the function. 47. 48. 49. 50. 51–54 Find an equation of the tangent line to the curve at the given point. 51. , 52. , 53. , 54. , (a) Find an equation of the tangent line to the curve at the point . ; (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen. 56. (a) The curve is called a bullet-nose curve. Find an equation of the tangent line to this curve at the point . ; (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen. 57. (a) If , ﬁnd . ; (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of and . f $ f f $"x# f "x# ! xs2 # x 2 "1, 1# y ! ) x)(s2 # x 2 "0, 1# y ! 2("1 % e#x# 55. "1, 1(e# y ! x 2e#x "(, 0# y ! sin"sin x# "0, 0# y ! sin x % sin2x "0, 1# y ! "1 % 2x#10 y ! e e x y ! e )x sin x y ! xe cx h"x# ! sx 2 % 1 y ! %x % "x % sin2x#3& 4 y ! cosssin"tan (x# y ! 23x2 t"x# ! "2ra rx % n#p y ! sx % sx % sx f "t# ! sin2"esin2t# y ! sin"sin"sin x## f "t# ! tan"e t# % etan t y ! ek tan sx y ! cot2"sin &# 37. f "t# ! t t 2 % 4 y ! cos! 1 # e2x 1 % e2x$ y ! x sin 1 x y ! sec2x % tan2x y ! tan2"3&# y ! 2sin (x G"y# !! y 2 y % 1$ 5 y ! sin"tan 2x# y ! e u # e#u e u % e#u y ! r sr 2 % 1 G"y# ! "y # 1#4 "y2 % 2y#5 F"z# ! z # 1 z % 1 y ! 101#x 2 y ! e x cos x 23. y ! e#5x cos 3x y !! x 2 % 1 x 2 # 1$ 3 204 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES x 1 3 2 4 6 2 1 8 5 7 3 7 2 7 9 t$"x# f $"x# t"x# f "x# Use this model to compare how the number of hours of day-light is increasing in Philadelphia on March 21 and May 21. ; 81. The motion of a spring that is subject to a frictional force or a damping force (such as a shock absorber in a car) is often modeled by the product of an exponential function and a sine or cosine function. Suppose the equation of motion of a point on such a spring is where is measured in centimeters and in seconds. Find the velocity after seconds and graph both the position and velocity functions for . 82. Under certain circumstances a rumor spreads according to the equation where is the proportion of the population that knows the rumor at time and and are positive constants. [In Sec-tion 9.4 we will see that this is a reasonable equation for .] (a) Find . (b) Find the rate of spread of the rumor. ; (c) Graph for the case , with measured in hours. Use the graph to estimate how long it will take for 80% of the population to hear the rumor. 83. A particle moves along a straight line with displacement velocity , and acceleration . Show that Explain the difference between the meanings of the deriv-atives . Air is being pumped into a spherical weather balloon. At any time , the volume of the balloon is and its radius is . (a) What do the derivatives and represent? (b) Express in terms of . ; 85. The ﬂash unit on a camera operates by storing charge on a capacitor and releasing it suddenly when the ﬂash is set off. The following data describe the charge remaining on the capacitor (measured in microcoulombs, +C) at time (mea-sured in seconds). (a) Use a graphing calculator or computer to ﬁnd an expo-nential model for the charge. (b) The derivative represents the electric current (mea-sured in microamperes, +A) ﬂowing from the capacitor to the ﬂash bulb. Use part (a) to estimate the current when s. Compare with the result of Example 2 in Section 2.1. t ! 0.04 Q$"t# t Q dr(dt dV(dt dV(dt dV(dr r"t# V"t# t 84. dv(dt and dv(ds a"t# ! v"t# dv ds a"t# v"t# s"t#, t k ! 0.5 a ! 10 p limt l , p"t# p"t# k a t p"t# p"t# ! 1 1 % ae #k t 0 ' t ' 2 t t s s"t# ! 2e#1.5t sin 2(t Suppose is differentiable on . Let and . Find expressions for (a) and (b) . 68. Suppose is differentiable on and is a real number. Let and . Find expressions for (a) and (b) . 69. Let , where , , , , and . Find . 70. If is a twice differentiable function and , ﬁnd in terms of , , and . 71. If , where and , ﬁnd . 72. If , where , , , , and , ﬁnd . 73. Show that the function satisﬁes the differ-ential equation . 74. For what values of does the function satisfy the equation ? Find the 50th derivative of . 76. Find the 1000th derivative of . The displacement of a particle on a vibrating string is given by the equation where is measured in centimeters and in seconds. Find the velocity of the particle after seconds. 78. If the equation of motion of a particle is given by , the particle is said to undergo simple harmonic motion. (a) Find the velocity of the particle at time . (b) When is the velocity 0? 79. A Cepheid variable star is a star whose brightness alternately increases and decreases. The most easily visible such star is Delta Cephei, for which the interval between times of maxi-mum brightness is 5.4 days. The average brightness of this star is 4.0 and its brightness changes by . In view of these data, the brightness of Delta Cephei at time , where is measured in days, has been modeled by the function (a) Find the rate of change of the brightness after days. (b) Find, correct to two decimal places, the rate of increase after one day. 80. In Example 4 in Section 1.3 we arrived at a model for the length of daylight (in hours) in Philadelphia on the th day of the year: L"t# ! 12 % 2.8 sin+ 2( 365 "t # 80#, t t B"t# ! 4.0 % 0.35 sin! 2(t 5.4$ t t -0.35 t s ! A cos".t % /# t t s s"t# ! 10 % 1 4 sin"10(t# 77. f"x# ! xe#x y ! cos 2x 75. y0 % 5y$ # 6y ! 0 y ! erx r y0 % 2y$ % y ! 0 y ! Ae#x % Bxe#x F$"1# f $"3# ! 6 f $"2# ! 5 f $"1# ! 4 f "2# ! 3 f "1# ! 2 F"x# ! f "x f "x f "x### F$"0# f $"0# ! 2 f "0# ! 0 F"x# ! f "3f "4 f "x### t0 t$ t f 0 f "x# ! xt"x 2# t r$"1# f $"3# ! 6 t$"2# ! 5 h$"1# ! 4 t"2# ! 3 h"1# ! 2 r"x# ! f "t"h"x### G$"x# F$"x# G"x# ! % f "x#&) F"x# ! f "x )# ) ! f G$"x# F$"x# G"x# ! e f "x# F"x# ! f "e x# ! f 67. SECTION 3.4 THE CHAIN RULE \|\|\|\| 205 t 0.00 0.02 0.04 0.06 0.08 0.10 Q 100.00 81.87 67.03 54.88 44.93 36.76 An approach path for an aircraft landing is shown in the ﬁgure on the next page and satisﬁes the following conditions: (i) The cruising altitude is when descent starts at a horizontal distance from touchdown at the origin. (ii) The pilot must maintain a constant horizontal speed throughout descent. v ! h WHERE SHOULD A PILOT START DESCENT? A P P L I E D P R O J E C T 90. Use the Chain Rule and the Product Rule to give an alternative proof of the Quotient Rule. [Hint: Write .] 91. (a) If is a positive integer, prove that (b) Find a formula for the derivative of that is similar to the one in part (a). 92. Suppose is a curve that always lies above the -axis and never has a horizontal tangent, where is differentiable everywhere. For what value of is the rate of change of with respect to eighty times the rate of change of with respect to ? Use the Chain Rule to show that if is measured in degrees, then (This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: The differentiation formulas would not be as simple if we used degree measure.) 94. (a) Write and use the Chain Rule to show that (b) If , ﬁnd and sketch the graphs of and . Where is not differentiable? (c) If , ﬁnd and sketch the graphs of and . Where is not differentiable? 95. If and , where and are twice differen-tiable functions, show that 96. If and , where and possess third deriva-tives, ﬁnd a formula for similar to the one given in Exercise 95. d 3y(dx 3 t f u ! t"x# y ! f "u# d 2y dx 2 ! d 2y du 2! du dx$ 2 % dy du d 2u dx 2 t f u ! t"x# y ! f "u# t t$ t t$"x# t"x# ! sin ) x) f f $ f f $"x# f "x# ! ) sin x) d dx ) x) ! x ) x) ) x) ! sx 2 d d& "sin &# ! ( 180 cos & & 93. x y x y 5 y f x y ! f "x# y ! cosnx cos nx d dx "sinnx cos nx# ! n sinn#1x cos"n % 1#x n f "x#(t"x# ! f "x#%t"x#&#1 ; 86. The table gives the US population from 1790 to 1860. (a) Use a graphing calculator or computer to ﬁt an exponen-tial function to the data. Graph the data points and the exponential model. How good is the ﬁt? (b) Estimate the rates of population growth in 1800 and 1850 by averaging slopes of secant lines. (c) Use the exponential model in part (a) to estimate the rates of growth in 1800 and 1850. Compare these estimates with the ones in part (b). (d) Use the exponential model to predict the population in 1870. Compare with the actual population of 38,558,000. Can you explain the discrepancy? 87. Computer algebra systems have commands that differentiate functions, but the form of the answer may not be convenient and so further commands may be necessary to simplify the answer. (a) Use a CAS to ﬁnd the derivative in Example 5 and com-pare with the answer in that example. Then use the sim-plify command and compare again. (b) Use a CAS to ﬁnd the derivative in Example 6. What hap-pens if you use the simplify command? What happens if you use the factor command? Which form of the answer would be best for locating horizontal tangents? 88. (a) Use a CAS to differentiate the function and to simplify the result. (b) Where does the graph of have horizontal tangents? (c) Graph and on the same screen. Are the graphs con-sistent with your answer to part (b)? 89. Use the Chain Rule to prove the following. (a) The derivative of an even function is an odd function. (b) The derivative of an odd function is an even function. f $ f f f "x# ! x 4 # x % 1 x 4 % x % 1 CAS CAS 206 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES Year Population Year Population 1790 3,929,000 1830 12,861,000 1800 5,308,000 1840 17,063,000 1810 7,240,000 1850 23,192,000 1820 9,639,000 1860 31,443,000 (iii) The absolute value of the vertical acceleration should not exceed a constant (which is much less than the acceleration due to gravity). 1. Find a cubic polynomial that satisﬁes condition (i) by imposing suitable conditions on and at the start of descent and at touchdown. 2. Use conditions (ii) and (iii) to show that 3. Suppose that an airline decides not to allow vertical acceleration of a plane to exceed mi(h . If the cruising altitude of a plane is 35,000 ft and the speed is 300 mi(h, how far away from the airport should the pilot start descent? ; 4. Graph the approach path if the conditions stated in Problem 3 are satisﬁed. 2 k ! 860 6hv 2 ! 2 ' k P$"x# P"x# P"x# ! ax 3 % bx 2 % cx % d k SECTION 3.5 IMPLICIT DIFFERENTIATION \|\|\|\| 207 IMPLICIT DIFFERENTIATION The functions that we have met so far can be described by expressing one variable explic-itly in terms of another variable—for example, or or, in general, . Some functions, however, are deﬁned implicitly by a relation between and such as or In some cases it is possible to solve such an equation for as an explicit function (or sev-eral functions) of . For instance, if we solve Equation 1 for , we get , so two of the functions determined by the implicit Equation l are and . The graphs of and are the upper and lower semicircles of the circle . (See Figure 1.) It’s not easy to solve Equation 2 for explicitly as a function of by hand. (A com-puter algebra system has no trouble, but the expressions it obtains are very complicated.) x y FIGURE 1 0 x y 0 x y 0 x y (c) ©=_œ„„„„„„ 25-≈ (b) ƒ=œ„„„„„„ 25-≈ (a) ≈+¥=25 x 2 % y 2 ! 25 t f t"x# ! #s25 # x 2 f"x# ! s25 # x 2 y ! -s25 # x 2 y x y x 3 % y 3 ! 6xy 2 x 2 % y 2 ! 25 1 y x y ! f"x# y ! x sin x y ! sx 3 % 1 3.5 y x 0 y=P(x) ! h Nonetheless, (2) is the equation of a curve called the folium of Descartes shown in Figure 2 and it implicitly deﬁnes as several functions of . The graphs of three such func-tions are shown in Figure 3. When we say that is a function deﬁned implicitly by Equa-tion 2, we mean that the equation is true for all values of in the domain of . Fortunately, we don’t need to solve an equation for in terms of in order to ﬁnd the derivative of . Instead we can use the method of implicit differentiation. This consists of differentiating both sides of the equation with respect to and then solving the resulting equation for . In the examples and exercises of this section it is always assumed that the given equation determines implicitly as a differentiable function of so that the method of implicit differentiation can be applied. EXAMPLE 1 (a) If , ﬁnd . (b) Find an equation of the tangent to the circle at the point . SOLUTION 1 (a) Differentiate both sides of the equation : Remembering that is a function of and using the Chain Rule, we have Thus Now we solve this equation for : dy dx ! # x y dy(dx 2x % 2y dy dx ! 0 d dx "y 2# ! d dy "y 2# dy dx ! 2y dy dx x y d dx "x 2# % d dx "y 2# ! 0 d dx "x 2 % y 2# ! d dx "25# x 2 % y 2 ! 25 "3, 4# x 2 % y 2 ! 25 dy dx x 2 % y 2 ! 25 V x y y$ x y x y x y 0 ˛+Á=6xy FIGURE 2 The folium of Descartes x y 0 FIGURE 3 Graphs of three functions defined by the folium of Descartes x y 0 x y 0 f x x 3 % % f"x#&3 ! 6x f"x# f x y 208 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES (b) At the point we have and , so An equation of the tangent to the circle at is therefore SOLUTION 2 (b) Solving the equation , we get . The point lies on the upper semicircle and so we consider the function . Differentiating using the Chain Rule, we have So and, as in Solution 1, an equation of the tangent is . M The expression in Solution 1 gives the derivative in terms of both and . It is correct no matter which function is determined by the given equation. For instance, for we have whereas for we have EXAMPLE 2 (a) Find if . (b) Find the tangent to the folium of Descartes at the point . (c) At what points in the ﬁrst quadrant is the tangent line horizontal? SOLUTION (a) Differentiating both sides of with respect to , regarding as a func-tion of , and using the Chain Rule on the term and the Product Rule on the term , we get or x 2 % y 2y$ ! 2xy$ % 2y 3x 2 % 3y 2y$ ! 6xy$ % 6y 6xy y 3 x y x x 3 % y 3 ! 6xy "3, 3# x 3 % y 3 ! 6xy x 3 % y 3 ! 6xy y$ V dy dx ! # x y ! # x #s25 # x 2 ! x s25 # x 2 y ! t"x# ! #s25 # x 2 dy dx ! # x y ! # x s25 # x 2 y ! f"x# ! s25 # x 2 y y x dy(dx ! #x(y NOTE 1 3x % 4y ! 25 f$"3# ! # 3 s25 # 32 ! # 3 4 ! 1 2"25 # x 2##1(2"#2x# ! # x s25 # x 2 f$"x# ! 1 2"25 # x 2##1(2 d dx "25 # x 2# f f"x# ! s25 # x 2 y ! s25 # x 2 "3, 4# y ! -s25 # x 2 x 2 % y 2 ! 25 3x % 4y ! 25 or y # 4 ! # 3 4"x # 3# "3, 4# dy dx ! # 3 4 y ! 4 x ! 3 "3, 4# SECTION 3.5 IMPLICIT DIFFERENTIATION \|\|\|\| 209 N Example 1 illustrates that even when it is possible to solve an equation explicitly for in terms of , it may be easier to use implicit differentiation. x y We now solve for : (b) When , and a glance at Figure 4 conﬁrms that this is a reasonable value for the slope at . So an equation of the tangent to the folium at is or (c) The tangent line is horizontal if . Using the expression for from part (a), we see that when (provided that . Substituting in the equation of the curve, we get which simpliﬁes to . Since in the ﬁrst quadrant, we have . If , then . Thus the tangent is horizontal at (0, 0) and at , which is approximately (2.5198, 3.1748). Looking at Figure 5, we see that our answer is reasonable. M There is a formula for the three roots of a cubic equation that is like the quad-ratic formula but much more complicated. If we use this formula (or a computer algebra system) to solve the equation for in terms of , we get three functions determined by the equation: and (These are the three functions whose graphs are shown in Figure 3.) You can see that the method of implicit differentiation saves an enormous amount of work in cases such as this. Moreover, implicit differentiation works just as easily for equations such as for which it is impossible to ﬁnd a similar expression for in terms of . EXAMPLE 3 Find if . SOLUTION Differentiating implicitly with respect to and remembering that is a function of , we get (Note that we have used the Chain Rule on the left side and the Product Rule and Chain cos"x % y# ! "1 % y$# ! y 2"#sin x# % "cos x#"2yy$# x y x sin"x % y# ! y 2 cos x y$ x y y 5 % 3x 2y 2 % 5x 4 ! 12 y ! 1 2[#f"x# - s#3(s 3 # 1 2 x 3 % s1 4 x 6 # 8x 3 # s 3 # 1 2 x 3 # s1 4 x 6 # 8x 3 )] y ! f"x# ! s 3 # 1 2 x 3 % s1 4 x 6 # 8x 3 % s 3 # 1 2 x 3 # s1 4 x 6 # 8x 3 x y x 3 % y 3 ! 6xy NOTE 2 "24(3, 25(3# y ! 1 2"28(3# ! 25(3 x ! 161(3 ! 24(3 x 3 ! 16 x " 0 x 6 ! 16x 3 x 3 % ( 1 2 x 2) 3 ! 6x( 1 2 x 2) y ! 1 2 x 2 y 2 # 2x " 0) 2y # x 2 ! 0 y$ ! 0 y$ y$ ! 0 x % y ! 6 y # 3 ! #1"x # 3# "3, 3# "3, 3# y$ ! 2 ! 3 # 32 32 # 2 ! 3 ! #1 x ! y ! 3 y$ ! 2y # x 2 y 2 # 2x "y 2 # 2x#y$ ! 2y # x 2 y 2y$ # 2xy$ ! 2y # x 2 y$ 210 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES FIGURE 4 0 y x (3, 3) 4 0 4 FIGURE 5 N The Norwegian mathematician Niels Abel proved in 1824 that no general formula can be given for the roots of a ﬁfth-degree equation in terms of radicals. Later the French mathemati-cian Evariste Galois proved that it is impossible to ﬁnd a general formula for the roots of an th-degree equation (in terms of algebraic operations on the coefﬁcients) if is any integer larger than 4. n n Rule on the right side.) If we collect the terms that involve , we get So Figure 6, drawn with the implicit-plotting command of a computer algebra system, shows part of the curve . As a check on our calculation, notice that when and it appears from the graph that the slope is approximately at the origin. M The following example shows how to ﬁnd the second derivative of a function that is deﬁned implicitly. EXAMPLE 4 Find if . SOLUTION Differentiating the equation implicitly with respect to , we get Solving for gives To ﬁnd we differentiate this expression for using the Quotient Rule and remember-ing that is a function of : If we now substitute Equation 3 into this expression, we get But the values of and must satisfy the original equation . So the answer simpliﬁes to M DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS The inverse trigonometric functions were reviewed in Section 1.6. We discussed their con-tinuity in Section 2.5 and their asymptotes in Section 2.6. Here we use implicit differenti-ation to ﬁnd the derivatives of the inverse trigonometric functions, assuming that these y0 ! # 3x 2"16# y 7 ! #48 x 2 y 7 x 4 % y 4 ! 16 y x ! # 3"x 2y 4 % x 6# y 7 ! # 3x 2"y 4 % x 4# y 7 y0 ! # 3x 2y 3 # 3x 3y 2!# x 3 y 3$ y 6 ! # y 3 ! 3x 2 # x 3"3y 2y$# y 6 y0 ! d dx !# x 3 y 3$ ! # y 3 "d(dx#"x 3# # x 3 "d(dx#"y 3# "y 3#2 x y y$ y0 y$ ! # x 3 y 3 3 y$ 4x 3 % 4y 3y$ ! 0 x x 4 % y 4 ! 16 y0 #1 x ! y ! 0 y$ ! #1 sin"x % y# ! y 2 cos x y$ ! y 2 sin x % cos"x % y# 2y cos x # cos"x % y# cos"x % y# % y 2 sin x ! "2y cos x#y$ # cos"x % y# ! y$ y$ SECTION 3.5 IMPLICIT DIFFERENTIATION \|\|\|\| 211 FIGURE 6 2 _2 _2 2 N Figure 7 shows the graph of the curve of Example 4. Notice that it’s a stretched and ﬂattened version of the circle . For this reason it’s sometimes called a fat circle. It starts out very steep on the left but quickly becomes very ﬂat. This can be seen from the expression y$ ! # x 3 y 3 ! #! x y$ 3 x 2 % y 2 ! 4 x 4 % y 4 ! 16 FIGURE 7 x 2 y 2 0 x$+y$=16 functions are differentiable. [In fact, if is any one-to-one differentiable function, it can be proved that its inverse function is also differentiable, except where its tangents are vertical. This is plausible because the graph of a differentiable function has no corner or kink and so if we reﬂect it about , the graph of its inverse function also has no cor-ner or kink.] Recall the deﬁnition of the arcsine function: Differentiating implicitly with respect to x, we obtain Now , since , so Therefore The formula for the derivative of the arctangent function is derived in a similar way. If , then . Differentiating this latter equation implicitly with respect to , we have EXAMPLE 5 Differentiate (a) and (b) . SOLUTION (a) (b) M ! sx 2!1 ! x" ! arctansx f"!x" ! x 1 1 ! (sx ) 2 ( 1 2 x#1#2) ! arctansx ! # 1 !sin#1x"2s1 # x 2 dy dx ! d dx !sin#1x"#1 ! #!sin#1x"#2 d dx !sin#1x" f!x" ! x arctansx y ! 1 sin#1x V d dx !tan#1x" ! 1 1 ! x 2 dy dx ! 1 sec2y ! 1 1 ! tan2y ! 1 1 ! x 2 sec2y dy dx ! 1 x tan y ! x y ! tan#1x d dx !sin#1x" ! 1 s1 # x 2 dy dx ! 1 cos y ! 1 s1 # x 2 cos y ! s1 # sin 2y ! s1 # x 2 #$#2 % y % $#2 cos y & 0 dy dx ! 1 cos y or cos y dy dx ! 1 sin y ! x # $ 2 % y % $ 2 and sin y ! x means y ! sin#1x y ! x f #1 f 212 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N The same method can be used to ﬁnd a formula for the derivative of any inverse function. See Exercise 67. N Recall that is an alternative notation for . tan#1x arctan x N Figure 8 shows the graph of and its derivative . Notice that is increasing and is always positive. The fact that as is reﬂected in the fact that as . x l '( f "!x" l 0 x l '( tan#1x l '$#2 f "!x" f f "!x" ! 1#!1 ! x 2" f !x" ! tan#1x 1.5 _1.5 _6 6 y=tan–! x y= 1 1+≈ FIGURE 8 The inverse trigonometric functions that occur most frequently are the ones that we have just discussed. The derivatives of the remaining four are given in the following table. The proofs of the formulas are left as exercises. DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS d dx !tan#1x" ! 1 1 ! x 2 d dx !cot#1x" ! # 1 1 ! x 2 d dx !cos#1x" ! # 1 s1 # x 2 d dx !sec#1x" ! 1 xsx 2 # 1 d dx !sin#1x" ! 1 s1 # x 2 d dx !csc#1x" ! # 1 xsx 2 # 1 SECTION 3.5 IMPLICIT DIFFERENTIATION \|\|\|\| 213 N The formulas for the derivatives of and depend on the deﬁnitions that are used for these functions. See Exercise 58. sec#1x csc#1x 25–30 Use implicit differentiation to ﬁnd an equation of the tangent line to the curve at the given point. 25. , (ellipse) 26. , (hyperbola) 28. (cardioid) (astroid) 29. 30. (3, 1) (0, #2) (lemniscate) (devil’s curve) 31. (a) The curve with equation is called a kampyle of Eudoxus. Find an equation of the tangent line to this curve at the point . ; (b) Illustrate part (a) by graphing the curve and the tangent line on a common screen. (If your graphing device will graph implicitly deﬁned curves, then use that capability. If !1, 2" y 2 ! 5x 4 # x 2 x y x y 0 y2!y2 # 4" ! x2!x2 # 5" 2!x 2 ! y 2"2 ! 25!x 2 # y 2" x y 0 8 x y (#3s3, 1) (0, 1 2) x 2#3 ! y 2#3 ! 4 x2 ! y2 ! !2x2 ! 2y2 # x"2 27. !1, 2" x2 ! 2xy # y2 ! x ! 2 !1, 1" x2 ! xy ! y2 ! 3 1–4 (a) Find by implicit differentiation. (b) Solve the equation explicitly for and differentiate to get in terms of . (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for into your solution for part (a). 1. 2. 3. 4. 5–20 Find by implicit differentiation. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 16. 17. 18. 19. 20. 21. If and , ﬁnd . 22. If , ﬁnd . 23–24 Regard as the independent variable and as the depen-dent variable and use implicit differentiation to ﬁnd . 23. 24. y sec x ! x tan y x 4y2 # x 3y ! 2xy3 ! 0 dx#dy x y t"!0" t!x" ! x sin t!x" ! x 2 f "!1" f !1" ! 2 f !x" ! x2 $ f !x"%3 ! 10 sin x ! cos y ! sin x cos y e y cos x ! 1 ! sin!xy" tan!x # y" ! y 1 ! x 2 sxy ! 1 ! x 2y sx ! y ! 1 ! x2y2 e x#y ! x # y 15. y sin!x 2" ! x sin!y 2" 4 cos x sin y ! 1 1 ! x ! sin!xy2" x2y2 ! x sin y ! 4 y 5 ! x 2y 3 ! 1 ! yex2 x 4!x ! y" ! y 2!3x # y" 2x 3 ! x 2y # xy3 ! 2 x 2 ! xy # y 2 ! 4 2sx ! sy ! 3 x 3 ! y3 ! 1 dy#dx cos x ! sy ! 5 1 x ! 1 y ! 1 4x 2 ! 9y 2 ! 36 xy ! 2x ! 3x 2 ! 4 y x y" y y" EXERCISES 3.5 44. The Power Rule can be proved using implicit differentiation for the case where is a rational number, , and is assumed beforehand to be a differentiable function. If , then . Use implicit differentia-tion to show that 45–54 Find the derivative of the function. Simplify where possible. 45. 46. 48. 49. 50. 51. 52. 53. 54. ; 55–56 Find . Check that your answer is reasonable by com-paring the graphs of and . 55. 56. 57. Prove the formula for by the same method as for . 58. (a) One way of deﬁning is to say that and or . Show that, with this deﬁnition, (b) Another way of deﬁning that is sometimes used is to say that and , . Show that, with this deﬁnition, 59–62 Two curves are orthogonal if their tangent lines are per-pendicular at each point of intersection. Show that the given fami-lies of curves are orthogonal trajectories of each other, that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes. 59. 60. 62. The equation represents a “rotated ellipse,” that is, an ellipse whose axes are not parallel to the coordinate axes. Find the points at which this ellipse crosses x 2 # xy ! y 2 ! 3 63. x 2 ! 3y 2 ! b y ! ax 3, x 2 ! 2y 2 ! k y ! cx 2, 61. x 2 ! y 2 ! by x 2 ! y 2 ! ax, ax ! by ! 0 x 2 ! y 2 ! r 2, d dx !sec#1x" ! 1 & x&sx 2 # 1 y " 0 0 % y % $ y ! sec#1x & ? sec y ! x sec#1x d dx !sec#1x" ! 1 xsx 2 # 1 $ % y ) 3$#2 0 % y ) $#2 sec y ! x y ! sec#1x & ? sec#1x !d#dx"!sin#1x" !d#dx"!cos#1x" f !x" ! arctan!x 2 # x" f !x" ! s1 # x 2 arcsin x f " f f "!x" y ! arctan' 1 # x 1 ! x y ! cos#1!e2x" F!" ! arcsin ssin h!t" ! cot#1!t" ! cot#1!1#t" y ! tan#1(x # s1 ! x 2 ) G!x" ! s1 # x 2 arccos x t!x" ! sx 2 # 1 sec#1x y ! sin#1!2x ! 1" 47. y ! stan#1x y ! tan#1sx y"! p q x ! p#q"#1 y q ! x p y ! x p#q y ! f !x" ! x n n ! p#q n not, you can still graph this curve by graphing its upper and lower halves separately.) 32. (a) The curve with equation is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point . (b) At what points does this curve have horizontal tangents? ; (c) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen. 33–36 Find by implicit differentiation. 33. 34. 35. 36. 37. Fanciful shapes can be created by using the implicit plotting capabilities of computer algebra systems. (a) Graph the curve with equation At how many points does this curve have horizontal tangents? Estimate the -coordinates of these points. (b) Find equations of the tangent lines at the points (0, 1) and (0, 2). (c) Find the exact -coordinates of the points in part (a). (d) Create even more fanciful curves by modifying the equa-tion in part (a). 38. (a) The curve with equation has been likened to a bouncing wagon. Use a computer algebra system to graph this curve and discover why. (b) At how many points does this curve have horizontal tangent lines? Find the -coordinates of these points. Find the points on the lemniscate in Exercise 29 where the tangent is horizontal. 40. Show by implicit differentiation that the tangent to the ellipse at the point is 41. Find an equation of the tangent line to the hyperbola at the point . 42. Show that the sum of the - and -intercepts of any tangent line to the curve is equal to . 43. Show, using implicit differentiation, that any tangent line at a point to a circle with center is perpendicular to the radius . OP O P c sx ! sy ! sc y x !x0, y0" x 2 a 2 # y 2 b2 ! 1 x0x a 2 ! y0y b2 ! 1 !x0, y0" x 2 a 2 ! y 2 b2 ! 1 39. x 2y 3 ! y 2 # y 5 ! x 4 # 2x 3 ! x 2 CAS x x y!y 2 # 1"!y # 2" ! x!x # 1"!x # 2" CAS x 4 ! y4 ! a4 x 3 ! y 3 ! 1 sx ! sy ! 1 9x2 ! y2 ! 9 y+ !1, #2" y 2 ! x 3 ! 3x 2 214 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES DERIVATIVES OF LOGARITHMIC FUNCTIONS In this section we use implicit differentiation to ﬁnd the derivatives of the logarithmic func-tions and, in particular, the natural logarithmic function . [It can be proved that logarithmic functions are differentiable; this is certainly plausible from their graphs (see Figure 12 in Section 1.6).] PROOF Let . Then Differentiating this equation implicitly with respect to x, using Formula 3.4.5, we get and so M If we put in Formula 1, then the factor on the right side becomes and we get the formula for the derivative of the natural logarithmic function : d dx !ln x" ! 1 x 2 loge x ! ln x ln e ! 1 ln a a ! e dy dx ! 1 a y ln a ! 1 x ln a a y!ln a" dy dx ! 1 a y ! x y ! loga x d dx !loga x" ! 1 x ln a 1 y ! ln x y ! loga x 3.6 SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS \|\|\|\| 215 (b) If and , ﬁnd . 68. (a) Show that is one-to-one. (b) What is the value of ? (c) Use the formula from Exercise 67(a) to ﬁnd . 69. The ﬁgure shows a lamp located three units to the right of the -axis and a shadow created by the elliptical region . If the point is on the edge of the shadow, how far above the -axis is the lamp located? ? x y 3 0 _5 ≈+4¥=5 x !#5, 0" x 2 ! 4y 2 % 5 y ! f #1""!1" f #1!1" f !x" ! 2x ! cos x ! f #1""!5" f "!4" ! 2 3 f !4" ! 5 the -axis and show that the tangent lines at these points are parallel. 64. (a) Where does the normal line to the ellipse at the point intersect the ellipse a second time? ; (b) Illustrate part (a) by graphing the ellipse and the normal line. 65. Find all points on the curve where the slope of the tangent line is . 66. Find equations of both the tangent lines to the ellipse that pass through the point . (a) Suppose is a one-to-one differentiable function and its inverse function is also differentiable. Use implicit differentiation to show that provided that the denominator is not 0. ! f #1""!x" ! 1 f "! f #1!x"" f #1 f 67. !12, 3" x 2 ! 4y 2 ! 36 #1 x 2y 2 ! xy ! 2 !#1, 1" x 2 # xy ! y 2 ! 3 x N Formula 3.4.5 says that d dx !a x" ! a x ln a By comparing Formulas 1 and 2, we see one of the main reasons that natural logarithms (logarithms with base e) are used in calculus: The differentiation formula is simplest when because . EXAMPLE 1 Differentiate . SOLUTION To use the Chain Rule, we let . Then , so M In general, if we combine Formula 2 with the Chain Rule as in Example 1, we get or EXAMPLE 2 Find . SOLUTION Using (3), we have M EXAMPLE 3 Differentiate . SOLUTION This time the logarithm is the inner function, so the Chain Rule gives M EXAMPLE 4 Differentiate . SOLUTION Using Formula 1 with , we have M EXAMPLE 5 Find . SOLUTION 1 ! x # 2 # 1 2!x ! 1" !x ! 1"!x # 2" ! x # 5 2!x ! 1"!x # 2" ! sx # 2 x ! 1 sx # 2 , 1 # !x ! 1"( 1 2)!x # 2"#1#2 x # 2 d dx ln x ! 1 sx # 2 ! 1 x ! 1 sx # 2 d dx x ! 1 sx # 2 d dx ln x ! 1 sx # 2 ! cos x !2 ! sin x" ln 10 f"!x" ! d dx log10!2 ! sin x" ! 1 !2 ! sin x" ln 10 d dx !2 ! sin x" a ! 10 f!x" ! log10!2 ! sin x" f"!x" ! 1 2!ln x"#1#2 d dx !ln x" ! 1 2sln x ! 1 x ! 1 2xsln x f!x" ! sln x d dx ln!sin x" ! 1 sin x d dx !sin x" ! 1 sin x cos x ! cot x d dx ln!sin x" d dx $ln t!x"% ! t"!x" t!x" d dx !ln u" ! 1 u du dx 3 dy dx ! dy du du dx ! 1 u du dx ! 1 x 3 ! 1 !3x 2" ! 3x 2 x 3 ! 1 y ! ln u u ! x 3 ! 1 y ! ln!x 3 ! 1" V ln e ! 1 a ! e 216 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N Figure 1 shows the graph of the function of Example 5 together with the graph of its deriv-ative. It gives a visual check on our calculation. Notice that is large negative when is rapidly decreasing. f f "!x" f x 0 y 1 f fª FIGURE 1 SOLUTION 2 If we ﬁrst simplify the given function using the laws of logarithms, then the differentiation becomes easier: (This answer can be left as written, but if we used a common denominator we would see that it gives the same answer as in Solution 1.) M EXAMPLE 6 Find if . SOLUTION Since it follows that Thus for all . M The result of Example 6 is worth remembering: LOGARITHMIC DIFFERENTIATION The calculation of derivatives of complicated functions involving products, quotients, or powers can often be simpliﬁed by taking logarithms. The method used in the following example is called logarithmic differentiation. EXAMPLE 7 Differentiate . SOLUTION We take logarithms of both sides of the equation and use the Laws of Loga-rithms to simplify: Differentiating implicitly with respect to gives 1 y dy dx ! 3 4 ! 1 x ! 1 2 ! 2x x 2 ! 1 # 5 ! 3 3x ! 2 x ln y ! 3 4 ln x ! 1 2 ln!x 2 ! 1" # 5 ln!3x ! 2" y ! x 3#4sx 2 ! 1 !3x ! 2"5 d dx ln &x& ! 1 x 4 x " 0 f"!x" ! 1#x 1 x if x - 0 1 #x !#1" ! 1 x if x ) 0 f"!x" ! f!x" !( ln x ln!#x" if x - 0 if x ) 0 f!x" ! ln &x& f"!x" V ! 1 x ! 1 # 1 2) 1 x # 2 d dx ln x ! 1 sx # 2 ! d dx [ln!x ! 1" # 1 2 ln!x # 2"] SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS \|\|\|\| 217 N Figure 2 shows the graph of the function in Example 6 and its derivative . Notice that when is small, the graph of is steep and so is large (positive or negative). f "!x" y ! ln & x& x f "!x" ! 1#x f !x" ! ln & x& 3 _3 _3 3 f fª FIGURE 2 Solving for , we get Because we have an explicit expression for , we can substitute and write M STEPS IN LOGARITHMIC DIFFERENTIATION 1. Take natural logarithms of both sides of an equation and use the Laws of Logarithms to simplify. 2. Differentiate implicitly with respect to . 3. Solve the resulting equation for . If for some values of , then is not deﬁned, but we can write and use Equation 4. We illustrate this procedure by proving the general ver-sion of the Power Rule, as promised in Section 3.1. THE POWER RULE If is any real number and , then PROOF Let and use logarithmic differentiation: Therefore Hence M \| You should distinguish carefully between the Power Rule , where the base is variable and the exponent is constant, and the rule for differentiating exponential functions , where the base is constant and the exponent is variable. In general there are four cases for exponents and bases: 1. ( and are constants) 2. 3. 4. To ﬁnd , logarithmic differentiation can be used, as in the next example. !d#dx"$ f!x"% t!x" d dx $a t!x"% ! a t!x"!ln a"t"!x" d dx $ f!x"%b ! b$ f!x"%b#1f"!x" b a d dx !a b" ! 0 $!a x"" ! a x ln a% $!x n"" ! nx n#1% y" ! n y x ! n x n x ! nx n#1 y" y ! n x x " 0 ln & y& ! ln &x& n ! n ln &x& y ! x n f"!x" ! nx n#1 f!x" ! x n n & y& ! & f!x"& ln f!x" x f!x" ) 0 y" x y ! f!x" dy dx ! x 3#4sx 2 ! 1 !3x ! 2"5 ) 3 4x ! x x 2 ! 1 # 15 3x ! 2 y dy dx ! y) 3 4x ! x x 2 ! 1 # 15 3x ! 2 dy#dx 218 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N If we hadn’t used logarithmic differentiation in Example 7, we would have had to use both the Quotient Rule and the Product Rule. The resulting calculation would have been horrendous. N If , we can show that for directly from the deﬁnition of a derivative. n - 1 f "!0" ! 0 x ! 0 EXAMPLE 8 Differentiate . SOLUTION 1 Using logarithmic differentiation, we have SOLUTION 2 Another method is to write : (as in Solution 1) M THE NUMBER e AS A LIMIT We have shown that if , then . Thus . We now use this fact to express the number as a limit. From the deﬁnition of a derivative as a limit, we have Because , we have Then, by Theorem 2.5.8 and the continuity of the exponential function, we have Formula 5 is illustrated by the graph of the function in Figure 4 and a table of values for small values of . This illustrates the fact that, correct to seven decimal places, e + 2.7182818 x y ! !1 ! x"1#x e ! lim x l 0 !1 ! x"1#x 5 e ! e1 ! elimx l 0 ln!1!x"1#x ! lim x l 0 eln!1!x"1#x ! lim x l 0 !1 ! x"1#x lim x l 0 ln!1 ! x"1#x ! 1 f"!1" ! 1 ! lim x l 0 ln!1 ! x"1#x ! lim x l 0 ln!1 ! x" # ln 1 x ! lim x l 0 1 x ln!1 ! x" f"!1" ! lim h l 0 f!1 ! h" # f!1" h ! lim x l 0 f!1 ! x" # f!1" x e f"!1" ! 1 f"!x" ! 1#x f!x" ! ln x ! xsx ) 2 ! ln x 2sx d dx (xsx ) ! d dx (esx ln x) ! esx ln x d dx (sx ln x) x sx ! !e ln x"sx y" ! y) 1 sx ! ln x 2sx ! x sx ) 2 ! ln x 2sx y" y ! sx ! 1 x ! !ln x" 1 2sx ln y ! ln x sx ! sx ln x y ! x sx V SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS \|\|\|\| 219 N Figure 3 illustrates Example 8 by showing the graphs of and its derivative. f !x" ! x sx FIGURE 3 1 1 f fª x 0 y FIGURE 4 2 3 y=(1+x)!?® 1 0 y x x 0.1 2.59374246 0.01 2.70481383 0.001 2.71692393 0.0001 2.71814593 0.00001 2.71826824 0.000001 2.71828047 0.0000001 2.71828169 0.00000001 2.71828181 (1 ! x)1/x If we put in Formula 5, then as and so an alternative expression for is e ! lim n l ( )1 ! 1 n n 6 e x l 0! n l ( n ! 1#x 220 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES 33–34 Find an equation of the tangent line to the curve at the given point. 33. , 34. ; 35. If , ﬁnd . Check that your answer is reasonable by comparing the graphs of and . ; 36. Find equations of the tangent lines to the curve at the points and . Illustrate by graphing the curve and its tangent lines. 37–48 Use logarithmic differentiation to ﬁnd the derivative of the function. 37. 38. 39. 40. 42. 44. 45. 46. 47. 48. 49. Find if . Find if . 51. Find a formula for if . 52. Find . Use the deﬁnition of derivative to prove that 54. Show that for any . x - 0 lim n l ( )1 ! x n n ! e x lim x l 0 ln!1 ! x" x ! 1 53. d 9 dx 9 !x 8 ln x" f !x" ! ln!x # 1" f !n"!x" x y ! y x y" 50. y ! ln!x 2 ! y 2" y" y ! !ln x"cos x y ! !tan x"1#x y ! !sin x"ln x y ! !cos x"x y ! sx x y ! x sin x 43. y ! x cos x y ! x x 41. y !' 4 x 2 ! 1 x 2 # 1 y ! sin2x tan4x !x 2 ! 1"2 y ! sx e x2!x 2 ! 1"10 y ! !2x ! 1"5!x 4 # 3"6 !e, 1#e" !1, 0" y ! !ln x"#x f " f f "!x" f !x" ! sin x ! ln x y ! ln!x 3 # 7", !2, 0" !1, 1" y ! ln(xe x2) 1. Explain why the natural logarithmic function is used much more frequently in calculus than the other logarithmic functions . 2–22 Differentiate the function. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 20. 21. 22. 23–26 Find and . 23. 24. 25. 26. 27–30 Differentiate and ﬁnd the domain of . 28. 29. 30. 31. If , ﬁnd . 32. If , ﬁnd . f "!0" f !x" ! ln!1 ! e 2x" f "!1" f !x" ! ln x x 2 f !x" ! ln ln ln x f !x" ! ln!x 2 # 2x" f !x" ! 1 1 ! ln x f !x" ! x 1 # ln!x # 1" 27. f f y ! ln!sec x ! tan x" y ! ln(x ! s1 ! x 2 ) y ! ln x x 2 y ! x 2 ln!2x" y+ y" y ! log2!e#x cos $x" y ! 2x log10 sx y ! $ln!1 ! e x"% 2 y ! ln!e#x ! xe#x" 19. H!z" ! ln' a 2 # z 2 a 2 ! z 2 y ! ln &2 # x # 5x 2& y ! 1 ln x f !u" ! ln u 1 ! ln!2u" F!y" ! y ln!1 ! e y" t!x" ! ln(xsx 2 # 1) h!x" ! ln(x ! sx 2 # 1) F!t" ! ln !2t ! 1"3 !3t # 1"4 f !t" ! 1 ! ln t 1 # ln t f !x" ! sin x ln!5x" f !x" ! ln s 5 x f !x" ! s 5 ln x f !x" ! log5!xe x" f !x" ! log2!1 # 3x" f !x" ! ln!sin2x" f !x" ! sin!ln x" f !x" ! ln!x 2 ! 10" y ! loga x y ! ln x EXERCISES 3.6 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES We know that if , then the derivative can be interpreted as the rate of change of with respect to . In this section we examine some of the applications of this idea to physics, chemistry, biology, economics, and other sciences. Let’s recall from Section 2.7 the basic idea behind rates of change. If changes from to , then the change in is and the corresponding change in is The difference quotient is the average rate of change of y with respect to x over the interval and can be interpreted as the slope of the secant line in Figure 1. Its limit as is the deriv-ative , which can therefore be interpreted as the instantaneous rate of change of y with respect to x or the slope of the tangent line at . Using Leibniz notation, we write the process in the form Whenever the function has a speciﬁc interpretation in one of the sciences, its derivative will have a speciﬁc interpretation as a rate of change. (As we discussed in Sec-tion 2.7, the units for are the units for y divided by the units for x.) We now look at some of these interpretations in the natural and social sciences. PHYSICS If is the position function of a particle that is moving in a straight line, then represents the average velocity over a time period , and represents the instan-taneous velocity (the rate of change of displacement with respect to time). The instanta-neous rate of change of velocity with respect to time is acceleration: . This was discussed in Sections 2.7 and 2.8, but now that we know the differentiation for-mulas, we are able to solve problems involving the motion of objects more easily. EXAMPLE 1 The position of a particle is given by the equation where is measured in seconds and in meters. (a) Find the velocity at time . (b) What is the velocity after 2 s? After 4 s? (c) When is the particle at rest? (d) When is the particle moving forward (that is, in the positive direction)? (e) Draw a diagram to represent the motion of the particle. (f) Find the total distance traveled by the particle during the ﬁrst ﬁve seconds. t s t s ! f!t" ! t 3 # 6t 2 ! 9t V a!t" ! v"!t" ! s+!t" v ! ds#dt .t .s#.t s ! f!t" dy#dx y ! f!x" dy dx ! lim .x l 0 .y .x P!x1, f!x1"" f"!x1" .x l 0 PQ $x1, x2% .y .x ! f!x2" # f!x1" x2 # x1 .y ! f!x2" # f!x1" y .x ! x2 # x1 x x2 x1 x x y dy#dx y ! f!x" 3.7 SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES \|\|\|\| 221 FIGURE 1 0 x y Îy ⁄ P{⁄, ﬂ} Q{¤, ‡} Îx ¤ mPQ ! average rate of change m=fª(⁄)=instantaneous rate of change (g) Find the acceleration at time and after 4 s. (h) Graph the position, velocity, and acceleration functions for . (i) When is the particle speeding up? When is it slowing down? SOLUTION (a) The velocity function is the derivative of the position function. (b) The velocity after 2 s means the instantaneous velocity when , that is, The velocity after 4 s is (c) The particle is at rest when , that is, and this is true when or . Thus the particle is at rest after 1 s and after 3 s. (d) The particle moves in the positive direction when , that is, This inequality is true when both factors are positive or when both factors are negative . Thus the particle moves in the positive direction in the time intervals and . It moves backward (in the negative direction) when . (e) Using the information from part (d) we make a schematic sketch in Figure 2 of the motion of the particle back and forth along a line (the -axis). (f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately. The distance traveled in the ﬁrst second is From to the distance traveled is From to the distance traveled is The total distance is . (g) The acceleration is the derivative of the velocity function: a!4" ! 6!4" ! 12 ! 12 m#s2 a!t" ! d 2s dt 2 ! dv dt ! 6t ! 12 4 " 4 " 20 ! 28 m $ f!5" ! f!3"$ ! $20 ! 0$ ! 20 m t ! 5 t ! 3 $ f!3" ! f!1"$ ! $0 ! 4$ ! 4 m t ! 3 t ! 1 $ f!1" ! f!0"$ ! $4 ! 0$ ! 4 m s 1 # t # 3 t $ 3 t # 1 !t # 1" !t $ 3" 3t 2 ! 12t " 9 ! 3!t ! 1"!t ! 3" $ 0 v!t" $ 0 t ! 3 t ! 1 3t 2 ! 12t " 9 ! 3!t 2 ! 4t " 3" ! 3!t ! 1"!t ! 3" ! 0 v!t" ! 0 v!4" ! 3!4"2 ! 12!4" " 9 ! 9 m#s v!2" ! ds dt % t!2 ! 3!2"2 ! 12!2" " 9 ! !3 m#s ! 2 t v!t" ! ds dt ! 3t 2 ! 12t " 9 s ! f!t" ! t 3 ! 6t 2 " 9t 0 % t % 5 t 222 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES t=0 s=0 t=1 s=4 s t=3 s=0 FIGURE 2 (h) Figure 3 shows the graphs of . (i) The particle speeds up when the velocity is positive and increasing ( are both positive) and also when the velocity is negative and decreasing ( are both nega-tive). In other words, the particle speeds up when the velocity and acceleration have the same sign. (The particle is pushed in the same direction it is moving.) From Figure 3 we see that this happens when and when . The particle slows down when have opposite signs, that is, when and when . Figure 4 sum-marizes the motion of the particle. M EXAMPLE 2 If a rod or piece of wire is homogeneous, then its linear density is uniform and is deﬁned as the mass per unit length and measured in kilograms per meter. Suppose, however, that the rod is not homogeneous but that its mass measured from its left end to a point is , as shown in Figure 5. The mass of the part of the rod that lies between and is given by , so the average density of that part of the rod is If we now let (that is, ), we are computing the average density over smaller and smaller intervals. The linear density at is the limit of these average densities as ; that is, the linear density is the rate of change of mass with respect to length. Symbolically, Thus the linear density of the rod is the derivative of mass with respect to length. & ! lim 'x l 0 'm 'x ! dm dx 'x l 0 x1 & x2 l x1 'x l 0 average density ! 'm 'x ! f!x2" ! f!x1" x2 ! x1 'm ! f!x2" ! f!x1" x ! x2 x ! x1 x¡ x™ This part of the rod has mass ƒ. x FIGURE 5 m ! f!x" x !& ! m#l" FIGURE 4 1 5 _5 √ s a forward slows down slows down backward speeds up speeds up forward t 0 2 # t # 3 0 % t # 1 v and a t $ 3 1 # t # 2 v and a v and a s, v, and a SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES \|\|\|\| 223 25 -12 0 5 √ s a FIGURE 3 In Module 3.7 you can see an anima-tion of Figure 4 with an expression for that you can choose yourself. s TEC For instance, if , where is measured in meters and in kilograms, then the average density of the part of the rod given by is while the density right at is M EXAMPLE 3 A current exists whenever electric charges move. Figure 6 shows part of a wire and electrons moving through a shaded plane surface. If is the net charge that passes through this surface during a time period , then the average current during this time interval is deﬁned as If we take the limit of this average current over smaller and smaller time intervals, we get what is called the current at a given time : Thus the current is the rate at which charge ﬂows through a surface. It is measured in units of charge per unit time (often coulombs per second, called amperes). M Velocity, density, and current are not the only rates of change that are important in physics. Others include power (the rate at which work is done), the rate of heat ﬂow, tem-perature gradient (the rate of change of temperature with respect to position), and the rate of decay of a radioactive substance in nuclear physics. CHEMISTRY EXAMPLE 4 A chemical reaction results in the formation of one or more substances (called products) from one or more starting materials (called reactants). For instance, the “equation” indicates that two molecules of hydrogen and one molecule of oxygen form two mole-cules of water. Let’s consider the reaction where A and B are the reactants and C is the product. The concentration of a reactant A is the number of moles ( 6.022 10 molecules) per liter and is denoted by . The concentration varies during a reaction, so , , and are all functions of &C' &B' &A' &A' 23 ( 1 mole ! A " B l C 2H2 " O2 l 2H2O I ! lim 't l 0 'Q 't ! dQ dt t1 I average current ! 'Q 't ! Q2 ! Q1 t2 ! t1 't 'Q V & ! dm dx % x!1 ! 1 2sx % x!1 ! 0.50 kg#m x ! 1 'm 'x ! f!1.2" ! f!1" 1.2 ! 1 ! s1.2 ! 1 0.2 ( 0.48 kg#m 1 % x % 1.2 m x m ! f!x" ! sx 224 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES ! ! ! ! ! ! ! FIGURE 6 time . The average rate of reaction of the product C over a time interval is But chemists are more interested in the instantaneous rate of reaction, which is obtained by taking the limit of the average rate of reaction as the time interval approaches 0: Since the concentration of the product increases as the reaction proceeds, the derivative will be positive, and so the rate of reaction of C is positive. The concentrations of the reactants, however, decrease during the reaction, so, to make the rates of reaction of A and B positive numbers, we put minus signs in front of the derivatives and . Since A and B each decrease at the same rate that C increases, we have More generally, it turns out that for a reaction of the form we have The rate of reaction can be determined from data and graphical methods. In some cases there are explicit formulas for the concentrations as functions of time, which enable us to compute the rate of reaction (see Exercise 22). M EXAMPLE 5 One of the quantities of interest in thermodynamics is compressibility. If a given substance is kept at a constant temperature, then its volume depends on its pres-sure . We can consider the rate of change of volume with respect to pressure—namely, the derivative . As increases, decreases, so . The compressibility is deﬁned by introducing a minus sign and dividing this derivative by the volume : Thus measures how fast, per unit volume, the volume of a substance decreases as the pressure on it increases at constant temperature. For instance, the volume (in cubic meters) of a sample of air at was found to be related to the pressure (in kilopascals) by the equation V ! 5.3 P P 25)C V isothermal compressibility ! ! ! 1 V dV dP V dV#dP # 0 V P dV#dP P V ! 1 a d&A' dt ! ! 1 b d&B' dt ! 1 c d&C' dt ! 1 d d&D' dt aA " bB l cC " dD rate of reaction ! d&C' dt ! ! d&A' dt ! ! d&B' dt ' & ' & ' & d&B'#dt d&A'#dt d&C'#dt rate of reaction ! lim 't l 0 '&C' 't ! d&C' dt 't '&C' 't ! &C'!t2" ! &C'!t1" t2 ! t1 t1 % t % t2 !t" SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES \|\|\|\| 225 The rate of change of with respect to when is The compressibility at that pressure is M BIOLOGY EXAMPLE 6 Let be the number of individuals in an animal or plant popula-tion at time . The change in the population size between the times and is , and so the average rate of growth during the time period is The instantaneous rate of growth is obtained from this average rate of growth by let-ting the time period approach 0: Strictly speaking, this is not quite accurate because the actual graph of a population function would be a step function that is discontinuous whenever a birth or death occurs and therefore not differentiable. However, for a large animal or plant population, we can replace the graph by a smooth approximating curve as in Figure 7. FIGURE 7 A smooth curve approximating a growth function t n 0 n ! f!t" growth rate ! lim 't l 0 'n 't ! dn dt 't average rate of growth ! 'n 't ! f!t2" ! f!t1" t2 ! t1 t1 % t % t2 'n ! f!t2" ! f!t1" t ! t2 t ! t1 t n ! f!t" ! ! 1 V dV dP % P!50 ! 0.00212 5.3 50 ! 0.02 !m3#kPa"#m3 ! ! 5.3 2500 ! !0.00212 m3#kPa dV dP % P!50 ! ! 5.3 P 2 % P!50 P ! 50 kPa P V 226 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES To be more speciﬁc, consider a population of bacteria in a homogeneous nutrient medium. Suppose that by sampling the population at certain intervals it is determined that the population doubles every hour. If the initial population is and the time is measured in hours, then and, in general, The population function is . In Section 3.4 we showed that So the rate of growth of the bacteria population at time t is For example, suppose that we start with an initial population of bacteria. Then the rate of growth after 4 hours is This means that, after 4 hours, the bacteria population is growing at a rate of about 1109 bacteria per hour. M EXAMPLE 7 When we consider the ﬂow of blood through a blood vessel, such as a vein or artery, we can model the shape of the blood vessel by a cylindrical tube with radius and length as illustrated in Figure 8. Because of friction at the walls of the tube, the velocity of the blood is greatest along the central axis of the tube and decreases as the distance from the axis increases until becomes 0 at the wall. The relationship between and is given by the law of laminar ﬂow discovered by the French physician Jean-Louis-Marie Poiseuille in 1840. This law states that where is the viscosity of the blood and is the pressure difference between the ends of the tube. If and are constant, then is a function of with domain . &0, R' r v l P P + v ! P 4+l !R2 ! r 2" 1 r v v r v FIGURE 8 Blood ﬂow in an artery R r l l R dn dt % t!4 ! 100 ! 24 ln 2 ! 1600 ln 2 ( 1109 n0 ! 100 dn dt ! d dt !n02t" ! n02t ln 2 d dx !ax" ! a x ln a n ! n02t f!t" ! 2tn0 f!3" ! 2f!2" ! 23n0 f!2" ! 2f!1" ! 22n0 f!1" ! 2f!0" ! 2n0 t n0 SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES \|\|\|\| 227 N For more detailed information, see W. Nichols and M. O’Rourke (eds.), McDonald’s Blood Flow in Arteries: Theoretic, Experimental, and Clinical Principles, 4th ed. (New York: Oxford University Press, 1998). The average rate of change of the velocity as we move from outward to is given by and if we let , we obtain the velocity gradient, that is, the instantaneous rate of change of velocity with respect to r: Using Equation 1, we obtain For one of the smaller human arteries we can take , cm, cm, and , which gives At cm the blood is ﬂowing at a speed of and the velocity gradient at that point is To get a feeling for what this statement means, let’s change our units from centi-meters to micrometers ( ,m). Then the radius of the artery is ,m. The velocity at the central axis is ,m#s, which decreases to ,m#s at a distance of ,m. The fact that (,m#s)#,m means that, when ,m, the velocity is decreasing at a rate of about ,m#s for each micrometer that we proceed away from the center. M ECONOMICS EXAMPLE 8 Suppose is the total cost that a company incurs in producing units of a certain commodity. The function is called a cost function. If the number of items produced is increased from to , then the additional cost is , and the average rate of change of the cost is The limit of this quantity as , that is, the instantaneous rate of change of cost l 0 'x 'C 'x ! C!x2" ! C!x1" x2 ! x1 ! C!x1 " 'x" ! C!x1" 'x 'C ! C!x2" ! C!x1" x2 x1 C x C!x" V 74 r ! 20 dv#dr ! !74 r ! 20 11,110 11,850 80 1 cm ! 10,000 dv dr % r!0.002 ! ! 4000!0.002" 2!0.027"2 ( !74 !cm#s"#cm ! 1.11 cm#s v!0.002" ( 1.85 ( 104!64 ( 10!6 ! 4 ( 10!6" r ! 0.002 ( 1.85 ( 104!6.4 ( 10!5 ! r 2" v ! 4000 4!0.027"2 !0.000064 ! r 2" P ! 4000 dynes#cm2 l ! 2 R ! 0.008 + ! 0.027 dv dr ! P 4+l !0 ! 2r" ! ! Pr 2+l velocity gradient ! lim 'r l 0 'v 'r ! dv dr 'r l 0 'v 'r ! v!r2" ! v!r1" r2 ! r1 r ! r2 r ! r1 228 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES with respect to the number of items produced, is called the marginal cost by economists: [Since often takes on only integer values, it may not make literal sense to let approach 0, but we can always replace by a smooth approximating function as in Example 6.] Taking and large (so that is small compared to ), we have Thus the marginal cost of producing units is approximately equal to the cost of pro-ducing one more unit [the st unit]. It is often appropriate to represent a total cost function by a polynomial where represents the overhead cost (rent, heat, maintenance) and the other terms represent the cost of raw materials, labor, and so on. (The cost of raw materials may be proportional to , but labor costs might depend partly on higher powers of because of overtime costs and inefﬁciencies involved in large-scale operations.) For instance, suppose a company has estimated that the cost (in dollars) of producing items is Then the marginal cost function is The marginal cost at the production level of 500 items is This gives the rate at which costs are increasing with respect to the production level when and predicts the cost of the 501st item. The actual cost of producing the 501st item is Notice that . M Economists also study marginal demand, marginal revenue, and marginal proﬁt, which are the derivatives of the demand, revenue, and proﬁt functions. These will be considered in Chapter 4 after we have developed techniques for ﬁnding the maximum and minimum values of functions. OTHER SCIENCES Rates of change occur in all the sciences. A geologist is interested in knowing the rate at which an intruded body of molten rock cools by conduction of heat into surrounding rocks. An engineer wants to know the rate at which water ﬂows into or out of a reservoir. An C-!500" ( C!501" ! C!500" ! $15.01 ! ! &10,000 " 5!500" " 0.01!500"2' C!501" ! C!500" ! &10,000 " 5!501" " 0.01!501"2' x ! 500 C-!500" ! 5 " 0.02!500" ! $15#item C-!x" ! 5 " 0.02x C!x" ! 10,000 " 5x " 0.01x 2 x x x a C!x" ! a " bx " cx 2 " dx 3 !n " 1" n C-!n" ( C!n " 1" ! C!n" n 'x n 'x ! 1 C!x" 'x x marginal cost ! lim 'x l 0 'C 'x ! dC dx SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES \|\|\|\| 229 urban geographer is interested in the rate of change of the population density in a city as the distance from the city center increases. A meteorologist is concerned with the rate of change of atmospheric pressure with respect to height (see Exercise 17 in Section 3.8). In psychology, those interested in learning theory study the so-called learning curve, which graphs the performance of someone learning a skill as a function of the train-ing time . Of particular interest is the rate at which performance improves as time passes, that is, . In sociology, differential calculus is used in analyzing the spread of rumors (or innova-tions or fads or fashions). If denotes the proportion of a population that knows a rumor by time , then the derivative represents the rate of spread of the rumor (see Exer-cise 82 in Section 3.4). A SINGLE IDEA, MANY INTERPRETATIONS Velocity, density, current, power, and temperature gradient in physics; rate of reaction and compressibility in chemistry; rate of growth and blood velocity gradient in biology; mar-ginal cost and marginal proﬁt in economics; rate of heat ﬂow in geology; rate of improve-ment of performance in psychology; rate of spread of a rumor in sociology—these are all special cases of a single mathematical concept, the derivative. This is an illustration of the fact that part of the power of mathematics lies in its abstractness. A single abstract mathematical concept (such as the derivative) can have dif-ferent interpretations in each of the sciences. When we develop the properties of the math-ematical concept once and for all, we can then turn around and apply these results to all of the sciences. This is much more efﬁcient than developing properties of special concepts in each separate science. The French mathematician Joseph Fourier (1768–1830) put it suc-cinctly: “Mathematics compares the most diverse phenomena and discovers the secret analogies that unite them.” dp#dt t p!t" dP#dt t P!t" 230 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES (a) (b) 6. Graphs of the position functions of two particles are shown, where is measured in seconds. When is each particle speed-ing up? When is it slowing down? Explain. (a) (b) 7. The position function of a particle is given by . (a) When does the particle reach a velocity of ? 5 m#s s ! t 3 ! 4.5t 2 ! 7t, t . 0 t s 0 1 t s 0 1 t t √ 0 1 t √ 0 1 1–4 A particle moves according to a law of motion , , where is measured in seconds and in feet. (a) Find the velocity at time . (b) What is the velocity after 3 s? (c) When is the particle at rest? (d) When is the particle moving in the positive direction? (e) Find the total distance traveled during the ﬁrst 8 s. (f) Draw a diagram like Figure 2 to illustrate the motion of the particle. (g) Find the acceleration at time and after 3 s. ; (h) Graph the position, velocity, and acceleration functions for . (i) When is the particle speeding up? When is it slowing down? 2. 3. , 4. 5. Graphs of the velocity functions of two particles are shown, where is measured in seconds. When is each particle speed-ing up? When is it slowing down? Explain. t f !t" ! te!t#2 t % 10 f !t" ! cos!/t#4" f !t" ! 0.01t 4 ! 0.04t 3 f !t" ! t 3 ! 12t 2 " 36t 1. 0 % t % 8 t t s t t . 0 s ! f !t" EXERCISES 3.7 A spherical balloon is being inﬂated. Find the rate of increase of the surface area with respect to the radius when is (a) 1 ft, (b) 2 ft, and (c) 3 ft. What conclusion can you make? 16. (a) The volume of a growing spherical cell is , where the radius is measured in micrometers (1 ,m ). Find the average rate of change of with respect to when changes from (i) 5 to 8 ,m (ii) 5 to 6 ,m (iii) 5 to 5.1 ,m (b) Find the instantaneous rate of change of with respect to when ,m. (c) Show that the rate of change of the volume of a sphere with respect to its radius is equal to its surface area. Explain geometrically why this result is true. Argue by analogy with Exercise 13(c). 17. The mass of the part of a metal rod that lies between its left end and a point meters to the right is kg. Find the linear density (see Example 2) when is (a) 1 m, (b) 2 m, and (c) 3 m. Where is the density the highest? The lowest? 18. If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli’s Law gives the volume of water remaining in the tank after minutes as Find the rate at which water is draining from the tank after (a) 5 min, (b) 10 min, (c) 20 min, and (d) 40 min. At what time is the water ﬂowing out the fastest? The slowest? Summarize your ﬁndings. The quantity of charge in coulombs (C) that has passed through a point in a wire up to time (measured in seconds) is given by . Find the current when (a) s and (b) s. [See Example 3. The unit of current is an ampere ( A C#s).] At what time is the cur-rent lowest? 20. Newton’s Law of Gravitation says that the magnitude of the force exerted by a body of mass on a body of mass is where is the gravitational constant and is the distance between the bodies. (a) Find and explain its meaning. What does the minus sign indicate? (b) Suppose it is known that the earth attracts an object with a force that decreases at the rate of 2 N#km when r ! 20,000 km. How fast does this force change when r ! 10,000 km? Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant: . (a) Find the rate of change of volume with respect to pressure. PV ! C 21. dF#dr r G F ! GmM r 2 M m F ! 1 1 t ! 1 t ! 0.5 Q!t" ! t 3 ! 2t 2 " 6t " 2 t Q 19. 0 % t % 40 V ! 5000)1 ! t 40 2 t V x 3x 2 x r ! 5 r V r r V ! 10!6 m r V ! 4 3/r 3 r r !S ! 4/r 2" 15. (b) When is the acceleration 0? What is the signiﬁcance of this value of ? 8. If a ball is given a push so that it has an initial velocity of down a certain inclined plane, then the distance it has rolled after seconds is . (a) Find the velocity after 2 s. (b) How long does it take for the velocity to reach ? 9. If a stone is thrown vertically upward from the surface of the moon with a velocity of , its height (in meters) after seconds is . (a) What is the velocity of the stone after 3 s? (b) What is the velocity of the stone after it has risen 25 m? 10. If a ball is thrown vertically upward with a velocity of 80 ft#s, then its height after seconds is . (a) What is the maximum height reached by the ball? (b) What is the velocity of the ball when it is 96 ft above the ground on its way up? On its way down? 11. (a) A company makes computer chips from square wafers of silicon. It wants to keep the side length of a wafer very close to 15 mm and it wants to know how the area of a wafer changes when the side length x changes. Find and explain its meaning in this situation. (b) Show that the rate of change of the area of a square with respect to its side length is half its perimeter. Try to explain geometrically why this is true by drawing a square whose side length x is increased by an amount . How can you approximate the resulting change in area if is small? 12. (a) Sodium chlorate crystals are easy to grow in the shape of cubes by allowing a solution of water and sodium chlorate to evaporate slowly. If V is the volume of such a cube with side length x, calculate when mm and explain its meaning. (b) Show that the rate of change of the volume of a cube with respect to its edge length is equal to half the surface area of the cube. Explain geometrically why this result is true by arguing by analogy with Exercise 11(b). 13. (a) Find the average rate of change of the area of a circle with respect to its radius as changes from (i) 2 to 3 (ii) 2 to 2.5 (iii) 2 to 2.1 (b) Find the instantaneous rate of change when . (c) Show that the rate of change of the area of a circle with respect to its radius (at any ) is equal to the circumfer-ence of the circle. Try to explain geometrically why this is true by drawing a circle whose radius is increased by an amount . How can you approximate the resulting change in area if is small? 14. A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm#s. Find the rate at which the area within the circle is increasing after (a) 1 s, (b) 3 s, and (c) 5 s. What can you conclude? 'r 'A 'r r r ! 2 r r x ! 3 dV#dx 'x 'A 'x A-!15" A!x" s ! 80t ! 16t 2 t h ! 10t ! 0.83t 2 t 10 m#s 35 m#s s ! 5t " 3t 2 t 5 m#s t SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES \|\|\|\| 231 (c) Use your model in part (b) to ﬁnd a model for the rate of population growth in the 20th century. (d) Use part (c) to estimate the rates of growth in 1920 and 1980. Compare with your estimates in part (a). (e) Estimate the rate of growth in 1985. ; 26. The table shows how the average age of ﬁrst marriage of Japanese women varied in the last half of the 20th century. (a) Use a graphing calculator or computer to model these data with a fourth-degree polynomial. (b) Use part (a) to ﬁnd a model for . (c) Estimate the rate of change of marriage age for women in 1990. (d) Graph the data points and the models for . 27. Refer to the law of laminar ﬂow given in Example 7. Consider a blood vessel with radius 0.01 cm, length 3 cm, pressure difference , and viscosity . (a) Find the velocity of the blood along the centerline , at radius cm, and at the wall . (b) Find the velocity gradient at , , and . (c) Where is the velocity the greatest? Where is the velocity changing most? The frequency of vibrations of a vibrating violin string is given by where is the length of the string, is its tension, and is its linear density. [See Chapter 11 in D. E. Hall, Musical Acoustics, 3d ed. (Paciﬁc Grove, CA: Brooks/Cole, 2002).] (a) Find the rate of change of the frequency with respect to (i) the length (when and are constant), (ii) the tension (when and are constant), and (iii) the linear density (when and are constant). (b) The pitch of a note (how high or low the note sounds) is determined by the frequency . (The higher the frequency, the higher the pitch.) Use the signs of the derivatives in part (a) to determine what happens to the pitch of a note (i) when the effective length of a string is decreased by placing a ﬁnger on the string so a shorter portion of the string vibrates, (ii) when the tension is increased by turning a tuning peg, (iii) when the linear density is increased by switching to another string. f T L ! L ! T ! T L f ! 1 2L ! T ! 28. r ! 0.01 r ! 0.005 r ! 0 r ! R ! 0.01 cm r ! 0.005 r ! 0 " ! 0.027 3000 dynes"cm2 A and A# A##t$ (b) A sample of gas is in a container at low pressure and is steadily compressed at constant temperature for 10 min-utes. Is the volume decreasing more rapidly at the begin-ning or the end of the 10 minutes? Explain. (c) Prove that the isothermal compressibility (see Example 5) is given by . 22. If, in Example 4, one molecule of the product C is formed from one molecule of the reactant A and one molecule of the reactant B, and the initial concentrations of A and B have a common value , then where is a constant. (a) Find the rate of reaction at time . (b) Show that if C , then (c) What happens to the concentration as ? (d) What happens to the rate of reaction as ? (e) What do the results of parts (c) and (d) mean in practical terms? 23. In Example 6 we considered a bacteria population that doubles every hour. Suppose that another population of bacteria triples every hour and starts with 400 bacteria. Find an expression for the number of bacteria after hours and use it to estimate the rate of growth of the bacteria population after 2.5 hours. 24. The number of yeast cells in a laboratory culture increases rapidly initially but levels off eventually. The population is modeled by the function where is measured in hours. At time the population is 20 cells and is increasing at a rate of . Find the values of and . According to this model, what happens to the yeast population in the long run? ; 25. The table gives the population of the world in the 20th century. (a) Estimate the rate of population growth in 1920 and in 1980 by averaging the slopes of two secant lines. (b) Use a graphing calculator or computer to ﬁnd a cubic function (a third-degree polynomial) that models the data. b a 12 cells"hour t ! 0 t n ! f #t$ ! a 1 $ be%0.7t t n t l & t l & dx dt ! k#a % x$2 % x ! & t k &C% ! a 2kt"#akt $ 1$ &A% ! &B% ! a moles"L ' ! 1"P 232 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES Population Population Year (in millions) Year (in millions) 1900 1650 1960 3040 1910 1750 1970 3710 1920 1860 1980 4450 1930 2070 1990 5280 1940 2300 2000 6080 1950 2560 t t 1950 23.0 1980 25.2 1955 23.8 1985 25.5 1960 24.4 1990 25.9 1965 24.5 1995 26.3 1970 24.2 2000 27.0 1975 24.7 A#t$ A#t$ SECTION 3.8 EXPONENTIAL GROWTH AND DECAY \|\|\|\| 233 33. The gas law for an ideal gas at absolute temperature (in kelvins), pressure (in atmospheres), and volume (in liters) is , where is the number of moles of the gas and is the gas constant. Suppose that, at a certain instant, atm and is increasing at a rate of 0.10 atm"min and and is decreasing at a rate of 0.15 L"min. Find the rate of change of with respect to time at that instant if mol. 34. In a ﬁsh farm, a population of ﬁsh is introduced into a pond and harvested regularly. A model for the rate of change of the ﬁsh population is given by the equation where is the birth rate of the ﬁsh, is the maximum population that the pond can sustain (called the carrying capacity), and is the percentage of the population that is harvested. (a) What value of corresponds to a stable population? (b) If the pond can sustain 10,000 ﬁsh, the birth rate is 5%, and the harvesting rate is 4%, ﬁnd the stable population level. (c) What happens if is raised to 5%? In the study of ecosystems, predator-prey models are often used to study the interaction between species. Consider populations of tundra wolves, given by , and caribou, given by , in northern Canada. The interaction has been modeled by the equations (a) What values of and correspond to stable populations? (b) How would the statement “The caribou go extinct” be represented mathematically? (c) Suppose that , , , and . Find all population pairs that lead to stable populations. According to this model, is it possible for the two species to live in balance or will one or both species become extinct? C, W$ d ! 0.0001 c ! 0.05 b ! 0.001 a ! 0.05 dW"dt dC"dt dW dt ! %cW $ dCW dC dt ! aC % bCW C#t$ W#t$ 35. ' dP"dt ' Pc r0 dP dt ! r0'1 % P#t$ Pc (P#t$ % 'P#t$ n ! 10 T V ! 10 L P ! 8.0 R ! 0.0821 n PV ! nRT V P T 29. The cost, in dollars, of producing yards of a certain fabric is (a) Find the marginal cost function. (b) Find and explain its meaning. What does it predict? (c) Compare with the cost of manufacturing the 201st yard of fabric. 30. The cost function for production of a commodity is (a) Find and interpret . (b) Compare with the cost of producing the 101st item. If is the total value of the production when there are workers in a plant, then the average productivity of the workforce at the plant is (a) Find . Why does the company want to hire more workers if ? (b) Show that if is greater than the average productivity. 32. If denotes the reaction of the body to some stimulus of strength , the sensitivity is deﬁned to be the rate of change of the reaction with respect to . A particular example is that when the brightness of a light source is increased, the eye reacts by decreasing the area of the pupil. The experimental formula has been used to model the dependence of on when is measured in square millimeters and is measured in appro-priate units of brightness. (a) Find the sensitivity. ; (b) Illustrate part (a) by graphing both and as functions of . Comment on the values of and at low levels of brightness. Is this what you would expect? S R x S R x R x R R ! 40 $ 24x 0.4 1 $ 4x 0.4 R x x S x R p##x$ A##x$ ( 0 A##x$ ( 0 A##x$ A#x$ ! p#x$ x x p#x$ 31. C##100$ C##100$ C#x$ ! 339 $ 25x % 0.09x 2 $ 0.0004x 3 C##200$ C##200$ C#x$ ! 1200 $ 12x % 0.1x 2 $ 0.0005x 3 x EXPONENTIAL GROWTH AND DECAY In many natural phenomena, quantities grow or decay at a rate proportional to their size. For instance, if is the number of individuals in a population of animals or bacte-ria at time , then it seems reasonable to expect that the rate of growth is proportion-al to the population ; that is, for some constant . Indeed, under ideal conditions (unlimited environment, adequate nutrition, immunity to disease) the mathe-matical model given by the equation predicts what actually happens fairly accurately. Another example occurs in nuclear physics where the mass of a radioactive substance decays at a rate proportional to the mass. In chemistry, the rate of a unimolecu-lar ﬁrst-order reaction is proportional to the concentration of the substance. In ﬁnance, the f##t$ ! kf#t$ k f##t$ ! kf#t$ f#t$ f##t$ t y ! f#t$ 3.8 value of a savings account with continuously compounded interest increases at a rate pro-portional to that value. In general, if is the value of a quantity at time and if the rate of change of with respect to is proportional to its size at any time, then where is a constant. Equation 1 is sometimes called the law of natural growth (if ) or the law of natural decay (if ). It is called a differential equation because it involves an unknown function and its derivative . It’s not hard to think of a solution of Equation 1. This equation asks us to ﬁnd a function whose derivative is a constant multiple of itself. We have met such functions in this chap-ter. Any exponential function of the form , where is a constant, satisﬁes We will see in Section 9.4 that any function that satsiﬁes must be of the form . To see the signiﬁcance of the constant , we observe that Therefore is the initial value of the function. THEOREM The only solutions of the differential equation are the exponential functions POPULATION GROWTH What is the signiﬁcance of the proportionality constant k? In the context of population growth, where is the size of a population at time , we can write The quantity is the growth rate divided by the population size; it is called the relative growth rate. According to (3), instead of saying “the growth rate is proportional to population size” we could say “the relative growth rate is constant.” Then (2) says that a population with con-stant relative growth rate must grow exponentially. Notice that the relative growth rate k appears as the coefﬁcient of t in the exponential function . For instance, if and t is measured in years, then the relative growth rate is and the population k ! 0.02 dP dt ! 0.02P Ce kt 1 P dP dt 1 P dP dt ! k or dP dt ! kP 3 t P#t$ y#t$ ! y#0$e kt dy"dt ! ky 2 C y#0$ ! Ce k!0 ! C C y ! Ce kt dy"dt ! ky y##t$ ! C#ke kt$ ! k#Ce kt$ ! ky#t$ C y#t$ ! Ce kt dy"dt y k ) 0 k ( 0 k dy dt ! ky 1 y#t$ t y t y y#t$ 234 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES grows at a relative rate of 2% per year. If the population at time 0 is , then the expres-sion for the population is EXAMPLE 1 Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 to model the population of the world in the second half of the 20th century. (Assume that the growth rate is proportional to the population size.) What is the relative growth rate? Use the model to estimate the world population in 1993 and to predict the population in the year 2020. SOLUTION We measure the time t in years and let t ! 0 in the year 1950. We measure the population in millions of people. Then and Since we are assuming that , Theorem 2 gives The relative growth rate is about 1.7% per year and the model is We estimate that the world population in 1993 was The model predicts that the population in 2020 will be The graph in Figure 1 shows that the model is fairly accurate to the end of the 20th cen-tury (the dots represent the actual population), so the estimate for 1993 is quite reliable. But the prediction for 2020 is riskier. M RADIOACTIVE DECAY Radioactive substances decay by spontaneously emitting radiation. If is the mass remaining from an initial mass of the substance after time t, then the relative decay rate % 1 m dm dt m0 m#t$ FIGURE 1 A model for world population growth in the second half of the 20th century 6000 P t 20 40 Years since 1950 Population (in millions) P=2560e0.017185t P#70$ ! 2560e 0.017185#70$ ) 8524 million P#43$ ! 2560e 0.017185#43$ ) 5360 million P#t$ ! 2560e 0.017185t k ! 1 10 ln 3040 2560 ) 0.017185 P#10$ ! 2560e 10k ! 3040 P#t$ ! P#0$e kt ! 2560e kt dP"dt ! kP P#10) ! 3040. P#0$ ! 2560 P#t$ V P#t$ ! P0e 0.02t P0 SECTION 3.8 EXPONENTIAL GROWTH AND DECAY \|\|\|\| 235 has been found experimentally to be constant. (Since is negative, the relative decay rate is positive.) It follows that where k is a negative constant. In other words, radioactive substances decay at a rate pro-portional to the remaining mass. This means that we can use (2) to show that the mass decays exponentially: Physicists express the rate of decay in terms of half-life, the time required for half of any given quantity to decay. EXAMPLE 2 The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of the sample that remains after years. (b) Find the mass after 1000 years correct to the nearest milligram. (c) When will the mass be reduced to 30 mg? SOLUTION (a) Let be the mass of radium-226 (in milligrams) that remains after years. Then and , so (2) gives In order to determine the value of , we use the fact that . Thus and Therefore We could use the fact that to write the expression for in the alternative form (b) The mass after 1000 years is (c) We want to ﬁnd the value of such that , that is, We solve this equation for by taking the natural logarithm of both sides: Thus M t ! %1590 ln 0.3 ln 2 ) 2762 years % ln 2 1590 t ! ln 0.3 t e%#ln 2$t"1590 ! 0.3 or 100e%#ln 2$t"1590 ! 30 m#t$ ! 30 t m#1000$ ! 100e%#ln 2$1000"1590 ) 65 mg m#t$ ! 100 2%t"1590 m#t$ e ln 2 ! 2 m#t$ ! 100e%#ln 2$t"1590 k ! % ln 2 1590 1590k ! ln 1 2 ! %ln 2 e 1590k ! 1 2 so 100e 1590k ! 50 y#1590$ ! 1 2#100$ k m#t$ ! m#0$e kt ! 100e kt y#0$ ! 100 dm"dt ! km t m#t$ t V m#t$ ! m0e kt dm dt ! km dm"dt 236 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES As a check on our work in Example 2, we use a graphing device to draw the graph of in Figure 2 together with the horizontal line . These curves intersect when , and this agrees with the answer to part (c). NEWTON’S LAW OF COOLING Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. (This law also applies to warming.) If we let be the tem-perature of the object at time and be the temperature of the surroundings, then we can formulate Newton’s Law of Cooling as a differential equation: where is a constant. This equation is not quite the same as Equation 1, so we make the change of variable . Because is constant, we have and so the equation becomes We can then use (2) to find an expression for , from which we can find . EXAMPLE 3 A bottle of soda pop at room temperature ( F) is placed in a refrigerator where the temperature is F. After half an hour the soda pop has cooled to F. (a) What is the temperature of the soda pop after another half hour? (b) How long does it take for the soda pop to cool to F? SOLUTION (a) Let be the temperature of the soda after minutes. The surrounding temperature is , so Newton’s Law of Cooling states that If we let , then , so satisﬁes and by (2) we have We are given that , so and Taking logarithms, we have k ! ln( 17 28) 30 ) %0.01663 e30k ! 17 28 28e30k ! 17 y#30$ ! 61 % 44 ! 17 T#30$ ! 61 y#t$ ! y#0$e kt ! 28e kt y#0$ ! 28 dy dt ! ky y y#0$ ! T#0$ % 44 ! 72 % 44 ! 28 y ! T % 44 dT dt ! k#T % 44) Ts ! 44+ F t T#t$ 50+ 61+ 44+ 72+ T y dy dt ! ky y##t$ ! T##t$ Ts y#t$ ! T#t$ % Ts k dT dt ! k#T % Ts$ Ts t T#t$ t ) 2800 m ! 30 m#t$ SECTION 3.8 EXPONENTIAL GROWTH AND DECAY \|\|\|\| 237 m=30 0 4000 150 m=100e_(ln 2)t/1590 FIGURE 2 Thus So after another half hour the pop has cooled to about F. (b) We have when The pop cools to F after about 1 hour 33 minutes. M Notice that in Example 3, we have which is to be expected. The graph of the temperature function is shown in Figure 3. CONTINUOUSLY COMPOUNDED INTEREST EXAMPLE 4 If $1000 is invested at 6% interest, compounded annually, then after 1 year the investment is worth , after 2 years it’s worth , and after years it’s worth . In general, if an amount is invested at an interest rate in this example), then after years it’s worth . Usually, however, interest is compounded more frequently, say, times a year. Then in each compounding period the interest rate is and there are compounding periods in years, so the value of the investment is For instance, after 3 years at 6% interest a $1000 investment will be worth with daily compounding $1000'1 $ 0.06 365( 365 ! 3 ! $1197.20 with monthly compounding $1000#1.005$36 ! $1196.68 with quarterly compounding $1000#1.015$12 ! $1195.62 with semiannual compounding $1000#1.03$6 ! $1194.05 with annual compounding $1000#1.06$3 ! $1191.02 A0'1 $ r n( nt t nt r"n n A0#1 $ r$t t #r ! 0.06 r A0 $1000#1.06$t t $&1000#1.06$%1.06 ! $1123.60 $1000#1.06$ ! $1060 lim t l & T#t$ ! lim t l & #44 $ 28e%0.01663t$ ! 44 $ 28 ! 0 ! 44 50+ t ! ln( 6 28) %0.01663 ) 92.6 e%0.01663t ! 6 28 44 $ 28e%0.01663t ! 50 T#t$ ! 50 54+ T#60$ ! 44 $ 28e%0.01663#60$ ) 54.3 T#t$ ! 44 $ 28e%0.01663t y#t$ ! 28e%0.01663t 238 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES FIGURE 3 72 T t 60 0 30 90 44 You can see that the interest paid increases as the number of compounding periods increases. If we let , then we will be compounding the interest continuously and the value of the investment will be (where ) But the limit in this expression is equal to the number e. (See Equation 3.6.6). So with continuous compounding of interest at interest rate r, the amount after t years is If we differentiate this function, we get which says that, with continuous compounding of interest, the rate of increase of an investment is proportional to its size. Returning to the example of $1000 invested for 3 years at 6% interest, we see that with continuous compounding of interest the value of the investment will be Notice how close this is to the amount we calculated for daily compounding, $1197.20. But the amount is easier to compute if we use continuous compounding. M A#3$ ! $1000e #0.06$3 ! $1197.22 dA dt ! rA0e rt ! rA#t$ A#t$ ! A0e rt m ! n"r ! A0lim m l & '1 $ 1 m( m+ rt ! A0lim n l & '1 $ r n( n"r+ rt A#t$ ! lim n l & A0'1 $ r n( nt ! lim n l & A0'1 $ r n( n"r+ rt n l & #n$ A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420. (a) Find an expression for the number of bacteria after hours. (b) Find the number of bacteria after 3 hours. (c) Find the rate of growth after 3 hours. (d) When will the population reach 10,000? 4. A bacteria culture grows with constant relative growth rate. After 2 hours there are 600 bacteria and after 8 hours the count is 75,000. (a) Find the initial population. (b) Find an expression for the population after hours. t t 3. 1. A population of protozoa develops with a constant relative growth rate of 0.7944 per member per day. On day zero the population consists of two members. Find the population size after six days. 2. A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 60 cells. (a) Find the relative growth rate. (b) Find an expression for the number of cells after hours. (c) Find the number of cells after 8 hours. (d) Find the rate of growth after 8 hours. (e) When will the population reach 20,000 cells? t EXERCISES 3.8 SECTION 3.8 EXPONENTIAL GROWTH AND DECAY \|\|\|\| 239 (b) How long will the reaction take to reduce the concentra-tion of N O to 90% of its original value? 8. Bismuth-210 has a half-life of 5.0 days. (a) A sample originally has a mass of 800 mg. Find a formula for the mass remaining after days. (b) Find the mass remaining after 30 days. (c) When is the mass reduced to 1 mg? (d) Sketch the graph of the mass function. The half-life of cesium-137 is 30 years. Suppose we have a 100-mg sample. (a) Find the mass that remains after years. (b) How much of the sample remains after 100 years? (c) After how long will only 1 mg remain? 10. A sample of tritium-3 decayed to 94.5% of its original amount after a year. (a) What is the half-life of tritium-3? (b) How long would it take the sample to decay to 20% of its original amount? 11. Scientists can determine the age of ancient objects by the method of radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, C, with a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates C through food chains. When a plant or animal dies, it stops replacing its carbon and the amount of C begins to decrease through radioactive decay. Therefore the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 74% as much C radioactivity as does plant material on the earth today. Estimate the age of the parchment. 12. A curve passes through the point and has the property that the slope of the curve at every point is twice the -coordinate of . What is the equation of the curve? A roast turkey is taken from an oven when its temperature has reached and is placed on a table in a room where the temperature is . (a) If the temperature of the turkey is after half an hour, what is the temperature after 45 minutes? (b) When will the turkey have cooled to ? 14. A thermometer is taken from a room where the temperature is C to the outdoors, where the temperature is . After one minute the thermometer reads C. (a) What will the reading on the thermometer be after one more minute? (b) When will the thermometer read C? 15. When a cold drink is taken from a refrigerator, its tempera-ture is C. After 25 minutes in a C room its temperature has increased to C. (a) What is the temperature of the drink after 50 minutes? (b) When will its temperature be C? 15+ 10+ 20+ 5+ 6+ 12+ 5+C 20+ 100+F 150+F 75+F 185+F 13. P y P #0, 5$ 14 14 14 14 t 9. t 5 2 (c) Find the number of cells after 5 hours. (d) Find the rate of growth after 5 hours. (e) When will the population reach 200,000? The table gives estimates of the world population, in millions, from 1750 to 2000: (a) Use the exponential model and the population ﬁgures for 1750 and 1800 to predict the world population in 1900 and 1950. Compare with the actual ﬁgures. (b) Use the exponential model and the population ﬁgures for 1850 and 1900 to predict the world population in 1950. Compare with the actual population. (c) Use the exponential model and the population ﬁgures for 1900 and 1950 to predict the world population in 2000. Compare with the actual population and try to explain the discrepancy. 6. The table gives the population of the United States, in millions, for the years 1900–2000. (a) Use the exponential model and the census ﬁgures for 1900 and 1910 to predict the population in 2000. Compare with the actual ﬁgure and try to explain the discrepancy. (b) Use the exponential model and the census ﬁgures for 1980 and 1990 to predict the population in 2000. Compare with the actual population. Then use this model to predict the population in the years 2010 and 2020. ; (c) Graph both of the exponential functions in parts (a) and (b) together with a plot of the actual population. Are these models reasonable ones? 7. Experiments show that if the chemical reaction takes place at , the rate of reaction of dinitrogen pent-oxide is proportional to its concentration as follows: (a) Find an expression for the concentration N O after seconds if the initial concentration is . C t 5% 2 & % d&N2O5% dt ! 0.0005&N2O5% 45+C N2O5 l 2NO2 $ 1 2O2 5. 240 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES Year Population Year Population 1750 790 1900 1650 1800 980 1950 2560 1850 1260 2000 6080 Year Population Year Population 1900 76 1960 179 1910 92 1970 203 1920 106 1980 227 1930 123 1990 250 1940 131 2000 275 1950 150 SECTION 3.9 RELATED RATES \|\|\|\| 241 ; (b) Suppose $1000 is borrowed and the interest is com-pounded continuously. If is the amount due after years, where , graph for each of the inter-est rates 6%, 8%, and 10% on a common screen. (a) If $3000 is invested at 5% interest, ﬁnd the value of the investment at the end of 5 years if the interest is com-pounded (i) annually, (ii) semiannually, (iii) monthly, (iv) weekly, (v) daily, and (vi) continuously. (b) If is the amount of the investment at time for the case of continuous compounding, write a differential equation and an initial condition satisﬁed by . 20. (a) How long will it take an investment to double in value if the interest rate is 6% compounded continuously? (b) What is the equivalent annual interest rate? A#t$ t A#t$ 19. A#t$ 0 , t , 3 t A#t$ 16. A freshly brewed cup of coffee has temperature C in a C room. When its temperature is C, it is cooling at a rate of C per minute. When does this occur? 17. The rate of change of atmospheric pressure with respect to altitude is proportional to , provided that the temperature is constant. At C the pressure is kPa at sea level and kPa at m. (a) What is the pressure at an altitude of 3000 m? (b) What is the pressure at the top of Mount McKinley, at an altitude of 6187 m? 18. (a) If $1000 is borrowed at 8% interest, ﬁnd the amounts due at the end of 3 years if the interest is compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) weekly, (v) daily, (vi) hourly, and (viii) continuously. h ! 1000 87.14 101.3 15+ P h P 1+ 70+ 20+ 95+ RELATED RATES If we are pumping air into a balloon, both the volume and the radius of the balloon are increasing and their rates of increase are related to each other. But it is much easier to measure directly the rate of increase of the volume than the rate of increase of the radius. In a related rates problem the idea is to compute the rate of change of one quantity in terms of the rate of change of another quantity (which may be more easily measured). The procedure is to ﬁnd an equation that relates the two quantities and then use the Chain Rule to differentiate both sides with respect to time. EXAMPLE 1 Air is being pumped into a spherical balloon so that its volume increases at a rate of . How fast is the radius of the balloon increasing when the diameter is 50 cm? SOLUTION We start by identifying two things: the given information: the rate of increase of the volume of air is and the unknown: the rate of increase of the radius when the diameter is 50 cm In order to express these quantities mathematically, we introduce some suggestive notation: Let V be the volume of the balloon and let r be its radius. The key thing to remember is that rates of change are derivatives. In this problem, the volume and the radius are both functions of the time . The rate of increase of the vol-ume with respect to time is the derivative , and the rate of increase of the radius is . We can therefore restate the given and the unknown as follows: Given: Unknown: dr dt when r ! 25 cm dV dt ! 100 cm3 "s dr"dt dV"dt t 100 cm3 "s 100 cm3 "s V 3.9 N According to the Principles of Problem Solving discussed on page 76, the ﬁrst step is to under-stand the problem. This includes reading the problem carefully, identifying the given and the unknown, and introducing suitable notation. In order to connect and , we ﬁrst relate and by the formula for the volume of a sphere: In order to use the given information, we differentiate each side of this equation with respect to . To differentiate the right side, we need to use the Chain Rule: Now we solve for the unknown quantity: If we put and in this equation, we obtain The radius of the balloon is increasing at the rate of cm!s. M EXAMPLE 2 A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft!s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall? SOLUTION We ﬁrst draw a diagram and label it as in Figure 1. Let feet be the distance from the bottom of the ladder to the wall and feet the distance from the top of the lad-der to the ground. Note that and are both functions of (time, measured in seconds). We are given that ft!s and we are asked to ﬁnd when ft (see Figure 2). In this problem, the relationship between and is given by the Pythagorean Theorem: Differentiating each side with respect to using the Chain Rule, we have and solving this equation for the desired rate, we obtain When , the Pythagorean Theorem gives and so, substituting these values and , we have The fact that is negative means that the distance from the top of the ladder to the ground is decreasing at a rate of . In other words, the top of the ladder is sliding down the wall at a rate of . M 3 4 ft!s 3 4 ft!s dy!dt dy dt ! ! 6 8 "1# ! ! 3 4 ft!s dx!dt ! 1 y ! 8 x ! 6 dy dt ! ! x y dx dt 2x dx dt " 2y dy dt ! 0 t x 2 " y 2 ! 100 y x x ! 6 dy!dt dx!dt ! 1 t y x y x 1!"25## $ 0.0127 dr dt ! 1 4#"25#2 100 ! 1 25# dV!dt ! 100 r ! 25 dr dt ! 1 4#r 2 dV dt dV dt ! dV dr dr dt ! 4#r 2 dr dt t V ! 4 3#r 3 r V dr!dt dV!dt N The second stage of problem solving is to think of a plan for connecting the given and the unknown. ground wall 10 y x FIGURE 1 y x dy dt =? dx dt =1 FIGURE 2 N Notice that, although is not constant, is constant. dV!dt dr!dt 242 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES EXAMPLE 3 A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m !min, ﬁnd the rate at which the water level is rising when the water is 3 m deep. SOLUTION We ﬁrst sketch the cone and label it as in Figure 3. Let , , and be the vol-ume of the water, the radius of the surface, and the height of the water at time , where is measured in minutes. We are given that m !min and we are asked to ﬁnd when is 3 m. The quantities and are related by the equation but it is very useful to express as a function of alone. In order to eliminate , we use the similar triangles in Figure 3 to write and the expression for becomes Now we can differentiate each side with respect to : so Substituting m and m !min, we have The water level is rising at a rate of . M STRATEGY It is useful to recall some of the problem-solving principles from page 76 and adapt them to related rates in light of our experience in Examples 1–3: 1. Read the problem carefully. 2. Draw a diagram if possible. 3. Introduce notation. Assign symbols to all quantities that are functions of time. 4. Express the given information and the required rate in terms of derivatives. 5. Write an equation that relates the various quantities of the problem. If necessary, use the geometry of the situation to eliminate one of the variables by substitution (as in Example 3). 6. Use the Chain Rule to differentiate both sides of the equation with respect to . 7. Substitute the given information into the resulting equation and solve for the unknown rate. The following examples are further illustrations of the strategy. t 8!"9## $ 0.28 m!min dh dt ! 4 #"3#2 ! 2 ! 8 9# 3 dV!dt ! 2 h ! 3 dh dt ! 4 #h 2 dV dt dV dt ! # 4 h 2 dh dt t V ! 1 3 #% h 2& 2 h ! # 12 h 3 V r ! h 2 r h ! 2 4 r h V V ! 1 3#r 2h h V h dh!dt 3 dV!dt ! 2 t t h r V 3 SECTION 3.9 RELATED RATES \|\|\|\| 243 FIGURE 3 2 r h 4 N Look back: What have we learned from Examples 1–3 that will help us solve future problems? \| WARNING: A common error is to sub-stitute the given numerical information (for quantities that vary with time) too early. This should be done only after the differentiation. (Step 7 follows Step 6.) For instance, in Example 3 we dealt with general values of until we ﬁnally substituted at the last stage. (If we had put earlier, we would have got-ten , which is clearly wrong.) dV!dt ! 0 h ! 3 h ! 3 h EXAMPLE 4 Car A is traveling west at and car B is traveling north at . Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection? SOLUTION We draw Figure 4, where is the intersection of the roads. At a given time let be the distance from car A to , let be the distance from car B to , and let be the distance between the cars, where , , and are measured in miles. We are given that mi!h and mi!h. (The derivatives are negative because and are decreasing.) We are asked to ﬁnd . The equation that relates , , and is given by the Pythagorean Theorem: Differentiating each side with respect to , we have When mi and mi, the Pythagorean Theorem gives mi, so The cars are approaching each other at a rate of 78 mi!h. M EXAMPLE 5 A man walks along a straight path at a speed of 4 ft!s. A searchlight is located on the ground 20 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight? SOLUTION We draw Figure 5 and let be the distance from the man to the point on the path closest to the searchlight. We let be the angle between the beam of the search-light and the perpendicular to the path. We are given that ft!s and are asked to ﬁnd when . The equa-tion that relates and can be written from Figure 5: Differentiating each side with respect to , we get so d$ dt ! 1 20 cos2$ dx dt ! 1 20 cos2$ "4# ! 1 5 cos2$ dx dt ! 20 sec2$ d$ dt t x ! 20 tan $ x 20 ! tan $ $ x x ! 15 d$!dt dx!dt ! 4 $ x V ! !78 mi!h dz dt ! 1 0.5 '0.3"!50# " 0.4"!60#( z ! 0.5 y ! 0.4 x ! 0.3 dz dt ! 1 z %x dx dt " y dy dt& 2z dz dt ! 2x dx dt " 2y dy dt t z2 ! x 2 " y 2 z y x dz!dt y x dy!dt ! !60 dx!dt ! !50 z y x z C y C x t, C 60 mi!h 50 mi!h V 244 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES FIGURE 4 C z y x B A FIGURE 5 x 20 ¨ When , the length of the beam is 25, so and The searchlight is rotating at a rate of 0.128 rad!s. M d$ dt ! 1 5 % 4 5& 2 ! 16 125 ! 0.128 cos $ ! 4 5 x ! 15 SECTION 3.9 RELATED RATES \|\|\|\| 245 which the distance from the plane to the station is increasing when it is 2 mi away from the station. If a snowball melts so that its surface area decreases at a rate of 1 cm !min, ﬁnd the rate at which the diameter decreases when the diameter is 10 cm. 13. A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft!s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole? 14. At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km!h and ship B is sailing north at 25 km!h. How fast is the distance between the ships changing at 4:00 PM? Two cars start moving from the same point. One travels south at 60 mi!h and the other travels west at 25 mi!h. At what rate is the distance between the cars increasing two hours later? 16. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m!s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building? 17. A man starts walking north at 4 ft!s from a point . Five min-utes later a woman starts walking south at 5 ft!s from a point 500 ft due east of . At what rate are the people moving apart 15 min after the woman starts walking? 18. A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward ﬁrst base with a speed of 24 ft!s. (a) At what rate is his distance from second base decreasing when he is halfway to ﬁrst base? (b) At what rate is his distance from third base increasing at the same moment? 90 ft P P 15. 2 12. 1. If is the volume of a cube with edge length and the cube expands as time passes, ﬁnd in terms of . 2. (a) If is the area of a circle with radius and the circle expands as time passes, ﬁnd in terms of . (b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of , how fast is the area of the spill increasing when the radius is 30 m? 3. Each side of a square is increasing at a rate of . At what rate is the area of the square increasing when the area of the square is ? 4. The length of a rectangle is increasing at a rate of and its width is increasing at a rate of . When the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing? 5. A cylindrical tank with radius 5 m is being ﬁlled with water at a rate of . How fast is the height of the water increasing? 6. The radius of a sphere is increasing at a rate of . How fast is the volume increasing when the diameter is 80 mm? 7. If and , ﬁnd when . 8. If and , ﬁnd when . 9. If , , and , ﬁnd when and . 10. A particle moves along the curve . As it reaches the point , the -coordinate is increasing at a rate of . How fast is the -coordinate of the point changing at that instant? 11–14 (a) What quantities are given in the problem? (b) What is the unknown? (c) Draw a picture of the situation for any time t. (d) Write an equation that relates the quantities. (e) Finish solving the problem. 11. A plane ﬂying horizontally at an altitude of 1 mi and a speed of passes directly over a radar station. Find the rate at 500 m!h x 4 cm!s y "2, 3# y ! s1 " x 3 y ! 12 x ! 5 dz!dt dy!dt ! 3 dx!dt ! 2 z2 ! x2 " y2 y ! 4 dx!dt dy!dt ! 6 x2 " y2 ! 25 x ! 2 dy!dt dx!dt ! 5 y ! x 3 " 2x 4 mm!s 3 m3!min 3 cm!s 8 cm!s 16 cm2 6 cm!s 1 m!s dr!dt dA!dt r A dx!dt dV!dt x V EXERCISES 3.9 246 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES equal. How fast is the height of the pile increasing when the pile is 10 ft high? 28. A kite 100 ft above the ground moves horizontally at a speed of 8 ft!s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out? 29. Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad!s. Find the rate at which the area of the triangle is increasing when the angle between the sides of ﬁxed length is . 30. How fast is the angle between the ladder and the ground chang-ing in Example 2 when the bottom of the ladder is 6 ft from the wall? Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure and volume satisfy the equation , where is a constant. Suppose that at a cer-tain instant the volume is 600 cm , the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa!min. At what rate is the volume decreasing at this instant? 32. When air expands adiabatically (without gaining or losing heat), its pressure and volume are related by the equation , where is a constant. Suppose that at a certain instant the volume is 400 cm and the pressure is 80 kPa and is decreasing at a rate of 10 kPa!min. At what rate is the volume increasing at this instant? 33. If two resistors with resistances and are connected in parallel, as in the ﬁgure, then the total resistance , measured in ohms ( ), is given by If and are increasing at rates of and , respectively, how fast is changing when and ? 34. Brain weight as a function of body weight in ﬁsh has been modeled by the power function , where and are measured in grams. A model for body weight W B B ! 0.007W 2!3 W B R¡ R™ R2 ! 100 % R1 ! 80 % R 0.2 %!s 0.3 %!s R2 R1 1 R ! 1 R1 " 1 R2 % R R2 R1 3 C PV 1.4 ! C V P 3 C PV ! C V P 31. !3 The altitude of a triangle is increasing at a rate of 1 cm!min while the area of the triangle is increasing at a rate of 2 cm !min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is ? A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m!s, how fast is the boat approaching the dock when it is 8 m from the dock? At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 km!h and ship B is sailing north at 25 km!h. How fast is the distance between the ships changing at 4:00 PM? A particle is moving along the curve . As the particle passes through the point , its -coordinate increases at a rate of . How fast is the distance from the particle to the origin changing at this instant? Water is leaking out of an inverted conical tank at a rate of 10,000 cm !min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm!min when the height of the water is 2 m, ﬁnd the rate at which water is being pumped into the tank. A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being ﬁlled with water at a rate of 12 ft !min, how fast is the water level rising when the water is 6 inches deep? A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. If the trough is being ﬁlled with water at the rate of 0.2 , how fast is the water level rising when the water is 30 cm deep? A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at its deepest point. A cross-section is shown in the ﬁgure. If the pool is being ﬁlled at a rate of 0.8 , how fast is the water level rising when the depth at the deepest point is 5 ft? Gravel is being dumped from a conveyor belt at a rate of 30 , and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always ft3 !min 27. 3 6 12 6 16 6 ft3 !min m3!min 25. 3 3 3 cm!s x "4, 2# y ! sx 100 cm2 2 19. SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS \|\|\|\| 247 (b) If the television camera is always kept aimed at the rocket, how fast is the camera’s angle of elevation changing at that same moment? A lighthouse is located on a small island 3 km away from the nearest point on a straight shoreline and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from ? A plane ﬂies horizontally at an altitude of and passes directly over a tracking telescope on the ground. When the angle of elevation is , this angle is decreasing at a rate of . How fast is the plane traveling at that time? A Ferris wheel with a radius of is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when his seat is 16 m above ground level? A plane ﬂying with a constant speed of 300 km!h passes over a ground radar station at an altitude of 1 km and climbs at an angle of 30 . At what rate is the distance from the plane to the radar station increasing a minute later? Two people start from the same point. One walks east at 3 mi!h and the other walks northeast at 2 mi!h. How fast is the distance between the people changing after 15 minutes? A runner sprints around a circular track of radius 100 m at a constant speed of 7 m!s. The runner’s friend is standing at a distance 200 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 m? The minute hand on a watch is 8 mm long and the hour hand is 4 mm long. How fast is the distance between the tips of the hands changing at one o’clock? 43. & 10 m #!6 rad!min #!3 5 km P P as a function of body length (measured in centimeters) is . If, over 10 million years, the average length of a certain species of ﬁsh evolved from 15 cm to 20 cm at a con-stant rate, how fast was this species’ brain growing when the average length was 18 cm? 35. Two sides of a triangle have lengths 12 m and 15 m. The angle between them is increasing at a rate of . How fast is the length of the third side increasing when the angle between the sides of ﬁxed length is 60 ? Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley (see the ﬁgure). The point is on the ﬂoor 12 ft directly beneath and between the carts. Cart A is being pulled away from at a speed of 2 ft!s. How fast is cart B moving toward at the instant when cart A is 5 ft from ? A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let’s assume the rocket rises vertically and its speed is 600 ft!s when it has risen 3000 ft. (a) How fast is the distance from the television camera to the rocket changing at that moment? 37. A B Q P 12 ft Q Q Q P Q P & 2& !min W ! 0.12L2.53 L LINEAR APPROXIMATIONS AND DIFFERENTIALS We have seen that a curve lies very close to its tangent line near the point of tangency. In fact, by zooming in toward a point on the graph of a differentiable function, we noticed that the graph looks more and more like its tangent line. (See Figure 2 in Section 2.7.) This observation is the basis for a method of ﬁnding approximate values of functions. The idea is that it might be easy to calculate a value of a function, but difﬁcult (or even impossible) to compute nearby values of f. So we settle for the easily computed val-ues of the linear function L whose graph is the tangent line of f at . (See Figure 1.) In other words, we use the tangent line at as an approximation to the curve when x is near a. An equation of this tangent line is and the approximation is called the linear approximation or tangent line approximation of f at a. The linear f"x# $ f"a# " f'"a#"x ! a# 1 y ! f"a# " f'"a#"x ! a# y ! f"x# "a, f"a## "a, f"a## f"a# 3.10 x 0 y {a, f(a)} y=ƒ y=L(x) FIGURE 1 function whose graph is this tangent line, that is, is called the linearization of f at a. EXAMPLE 1 Find the linearization of the function at and use it to approximate the numbers and . Are these approximations overestimates or underestimates? SOLUTION The derivative of is and so we have and . Putting these values into Equation 2, we see that the linearization is The corresponding linear approximation (1) is (when is near ) In particular, we have The linear approximation is illustrated in Figure 2. We see that, indeed, the tangent line approximation is a good approximation to the given function when is near l. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for and , but the linear approximation gives an approximation over an entire interval. M In the following table we compare the estimates from the linear approximation in Example 1 with the true values. Notice from this table, and also from Figure 2, that the tan-gent line approximation gives good estimates when x is close to 1 but the accuracy of the approximation deteriorates when x is farther away from 1. s4.05 s3.98 x s4.05 $ 7 4 " 1.05 4 ! 2.0125 and s3.98 $ 7 4 " 0.98 4 ! 1.995 1 x sx " 3 $ 7 4 " x 4 L"x# ! f"1# " f'"1#"x ! 1# ! 2 " 1 4"x ! 1# ! 7 4 " x 4 f'"1# ! 1 4 f"1# ! 2 f'"x# ! 1 2"x " 3#!1!2 ! 1 2sx " 3 f"x# ! "x " 3#1!2 s4.05 s3.98 a ! 1 f"x# ! sx " 3 V L"x# ! f"a# " f'"a#"x ! a# 2 248 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES y= x+3 _3 0 x y 1 (1, 2) y= + x 4 7 4 œ„„„„ FIGURE 2 x From Actual value 0.9 1.975 1.97484176 . . . 0.98 1.995 1.99499373 . . . 1 2 2.00000000 . . . 1.05 2.0125 2.01246117 . . . 1.1 2.025 2.02484567 . . . 2 2.25 2.23606797 . . . 3 2.5 2.44948974 . . . s6 s5 s4.1 s4.05 s4 s3.98 s3.9 L"x# How good is the approximation that we obtained in Example 1? The next example shows that by using a graphing calculator or computer we can determine an interval through-out which a linear approximation provides a speciﬁed accuracy. EXAMPLE 2 For what values of is the linear approximation accurate to within 0.5? What about accuracy to within 0.1? SOLUTION Accuracy to within 0.5 means that the functions should differ by less than 0.5: Equivalently, we could write This says that the linear approximation should lie between the curves obtained by shift-ing the curve upward and downward by an amount 0.5. Figure 3 shows the tangent line intersecting the upper curve at and . Zooming in and using the cursor, we estimate that the -coordinate of is about and the -coordinate of is about 8.66. Thus we see from the graph that the approximation is accurate to within 0.5 when . (We have rounded to be safe.) Similarly, from Figure 4 we see that the approximation is accurate to within 0.1 when . M APPLICATIONS TO PHYSICS Linear approximations are often used in physics. In analyzing the consequences of an equation, a physicist sometimes needs to simplify a function by replacing it with its linear approximation. For instance, in deriving a formula for the period of a pendulum, physics textbooks obtain the expression for tangential acceleration and then replace by with the remark that is very close to if is not too large. [See, for exam-ple, Physics: Calculus, 2d ed., by Eugene Hecht (Paciﬁc Grove, CA: Brooks/Cole, 2000), p. 431.] You can verify that the linearization of the function at a ! 0 is and so the linear approximation at 0 is (see Exercise 42). So, in effect, the derivation of the formula for the period of a pendulum uses the tangent line approximation for the sine function. Another example occurs in the theory of optics, where light rays that arrive at shallow angles relative to the optical axis are called paraxial rays. In paraxial (or Gaussian) optics, both and are replaced by their linearizations. In other words, the linear approx-imations and cos $ $ 1 sin $ $ $ cos $ sin $ sin x $ x L"x# ! x f"x# ! sin x $ $ sin $ $ sin $ aT ! !t sin $ !1.1 ( x ( 3.9 !2.6 ( x ( 8.6 sx " 3 $ 7 4 " x 4 Q x !2.66 P x Q P y ! sx " 3 " 0.5 y ! "7 " x#!4 y ! sx " 3 sx " 3 ! 0.5 ( 7 4 " x 4 ( sx " 3 " 0.5 ) sx " 3 !% 7 4 " x 4&) ( 0.5 sx " 3 $ 7 4 " x 4 x SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS \|\|\|\| 249 4.3 _1 _4 10 y= x+3-0.5 œ„„„„ Q P L(x) FIGURE 3 y= x+3+0.5 œ„„„„ 3 1 _2 y= x+3-0.1 œ„„„„ Q P 5 y= x+3+0.1 œ„„„„ FIGURE 4 are used because is close to 0. The results of calculations made with these approxi-mations became the basic theoretical tool used to design lenses. [See Optics, 4th ed., by Eugene Hecht (San Francisco: Addison-Wesley, 2002), p. 154.] In Section 11.11 we will present several other applications of the idea of linear approx-imations to physics. DIFFERENTIALS The ideas behind linear approximations are sometimes formulated in the terminology and notation of differentials. If , where is a differentiable function, then the differ-ential is an independent variable; that is, can be given the value of any real number. The differential is then deﬁned in terms of by the equation So is a dependent variable; it depends on the values of and . If is given a spe-ciﬁc value and is taken to be some speciﬁc number in the domain of , then the numer-ical value of is determined. The geometric meaning of differentials is shown in Figure 5. Let and be points on the graph of and let . The corresponding change in is The slope of the tangent line is the derivative . Thus the directed distance from S to R is . Therefore represents the amount that the tangent line rises or falls (the change in the linearization), whereas represents the amount that the curve rises or falls when changes by an amount . EXAMPLE 3 Compare the values of and if and changes (a) from 2 to 2.05 and (b) from 2 to 2.01. SOLUTION (a) We have In general, When and , this becomes (b) When , M dy ! '3"2#2 " 2"2# ! 2(0.01 ! 0.14 dx ! )x ! 0.01 )y ! f"2.01# ! f"2# ! 0.140701 f"2.01# ! "2.01#3 " "2.01#2 ! 2"2.01# " 1 ! 9.140701 dy ! '3"2#2 " 2"2# ! 2(0.05 ! 0.7 dx ! )x ! 0.05 x ! 2 dy ! f'"x# dx ! "3x 2 " 2x ! 2# dx )y ! f"2.05# ! f"2# ! 0.717625 f"2.05# ! "2.05#3 " "2.05#2 ! 2"2.05# " 1 ! 9.717625 f"2# ! 23 " 22 ! 2"2# " 1 ! 9 x y ! f"x# ! x 3 " x 2 ! 2x " 1 dy )y dx x y ! f"x# )y dy f'"x# dx ! dy f'"x# PR )y ! f"x " )x# ! f"x# y dx ! )x f Q"x " )x, f"x " )x## P"x, f"x## dy f x dx dx x dy dy ! f'"x# dx 3 dx dy dx dx f y ! f"x# $ 250 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N If , we can divide both sides of Equation 3 by to obtain We have seen similar equations before, but now the left side can genuinely be interpreted as a ratio of differentials. dy dx ! f '"x# dx dx " 0 R 0 x y Îy x P Q dx=Îx x+Îx y=ƒ S dy FIGURE 5 FIGURE 6 y=˛+≈-2x+1 (2, 9) dy Îy N Figure 6 shows the function in Example 3 and a comparison of and when . The viewing rectangle is by . '6, 18( '1.8, 2.5( a ! 2 )y dy Notice that the approximation becomes better as becomes smaller in Example 3. Notice also that was easier to compute than . For more complicated func-tions it may be impossible to compute exactly. In such cases the approximation by dif-ferentials is especially useful. In the notation of differentials, the linear approximation (1) can be written as For instance, for the function in Example 1, we have If a ! 1 and , then and just as we found in Example 1. Our ﬁnal example illustrates the use of differentials in estimating the errors that occur because of approximate measurements. EXAMPLE 4 The radius of a sphere was measured and found to be 21 cm with a pos-sible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere? SOLUTION If the radius of the sphere is , then its volume is . If the error in the measured value of is denoted by , then the corresponding error in the calcu-lated value of is , which can be approximated by the differential When and , this becomes The maximum error in the calculated volume is about 277 cm . M Although the possible error in Example 4 may appear to be rather large, a bet-ter picture of the error is given by the relative error, which is computed by dividing the error by the total volume: Thus the relative error in the volume is about three times the relative error in the radius. In Example 4 the relative error in the radius is approximately and it produces a relative error of about 0.007 in the volume. The errors could also be expressed as percentage errors of in the radius and in the volume. 0.7% 0.24% dr!r ! 0.05!21 $ 0.0024 )V V $ dV V ! 4#r 2 dr 4 3#r 3 ! 3 dr r NOTE 3 dV ! 4#"21#20.05 $ 277 dr ! 0.05 r ! 21 dV ! 4#r 2 dr )V V dr ! )r r V ! 4 3#r 3 r V s4.05 ! f"1.05# $ f"1# " dy ! 2.0125 dy ! 0.05 2s1 " 3 ! 0.0125 dx ! )x ! 0.05 dy ! f'"x# dx ! dx 2sx " 3 f"x# ! sx " 3 f"a " dx# $ f"a# " dy )y )y dy )x )y $ dy SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS \|\|\|\| 251 25. 26. 27. 28. 29–31 Explain, in terms of linear approximations or differentials, why the approximation is reasonable. 30. 31. Let and (a) Find the linearizations of , , and at . What do you notice? How do you explain what happened? ; (b) Graph , , and and their linear approximations. For which function is the linear approximation best? For which is it worst? Explain. The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing (a) the volume of the cube and (b) the sur-face area of the cube. The radius of a circular disk is given as 24 cm with a maxi-mum error in measurement of 0.2 cm. (a) Use differentials to estimate the maximum error in the calculated area of the disk. (b) What is the relative error? What is the percentage error? The circumference of a sphere was measured to be 84 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the maximum error in the calculated surface area. What is the relative error? (b) Use differentials to estimate the maximum error in the calculated volume. What is the relative error? Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05 cm thick to a hemispherical dome with diameter 50 m. (a) Use differentials to ﬁnd a formula for the approximate volume of a thin cylindrical shell with height , inner radius , and thickness . (b) What is the error involved in using the formula from part (a)? One side of a right triangle is known to be 20 cm long and the opposite angle is measured as , with a possible error of . (a) Use differentials to estimate the error in computing the length of the hypotenuse. (b) What is the percentage error? !1" 30" #r r h 33. h t f a ! 0 h t f h!x" ! 1 $ ln!1 % 2x" t!x" ! e%2x f !x" ! !x % 1"2 ln 1.05 # 0.05 !1.01"6 # 1.06 sec 0.08 # 1 29. s99.8 tan 44" 1$1002 !8.06"2$3 1–4 Find the linearization of the function at . 1. , 2. , , 4. , ; Find the linear approximation of the function at and use it to approximate the numbers and . Illustrate by graphing and the tangent line. ; 6. Find the linear approximation of the function at and use it to approximate the numbers and . Illustrate by graphing and the tangent line. ; 7–10 Verify the given linear approximation at . Then deter-mine the values of for which the linear approximation is accu-rate to within 0.1. 7. 8. 10. 11–14 Find the differential of each function. (a) (b) 12. (a) (b) (a) (b) 14. (a) (b) 15–18 (a) Find the differential and (b) evaluate for the given values of and . , , 16. , , 17. , , 18. , , 19–22 Compute and for the given values of and . Then sketch a diagram like Figure 5 showing the line segments with lengths , , and . 19. , , 20. , , 21. , , 22. , , 23–28 Use a linear approximation (or differentials) to estimate the given number. 23. e%0.015 !2.001"5 #x ! 0.5 x ! 0 y ! e x #x ! 1 x ! 4 y ! 2$x #x ! 1 x ! 1 y ! sx #x ! %0.4 x ! 2 y ! 2x % x 2 #y dy dx dx ! #x x dy #y dx ! 0.05 x ! &$3 y ! cos x dx ! %0.1 x ! &$4 y ! tan x dx ! %0.01 x ! 1 y ! 1$!x $ 1" dx ! 0.1 x ! 0 y ! e x $10 15. dx x dy dy y ! s1 $ ln z y ! etan &t y ! !1 $ r 3"%2 y ! u $ 1 u % 1 13. y ! e%u cos u y ! s$!1 $ 2s" y ! lns1 $ t 2 y ! x 2 sin 2x e x # 1 $ x 1$!1 $ 2x"4 # 1 % 8x 9. tan x # x s 3 1 % x # 1 % 1 3x x a ! 0 t s 3 1.1 s 3 0.95 a ! 0 t!x" ! s 3 1 $ x f s0.99 s0.9 a ! 0 f !x" ! s1 % x 5. a ! 16 f !x" ! x 3$4 a ! &$2 f !x" ! cos x 3. a ! 1 f !x" ! ln x a ! %1 f !x" ! x 4 $ 3x 2 a L!x" EXERCISES 3.10 252 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES LABORATORY PROJECT TAYLOR POLYNOMIALS \|\|\|\| 253 pendulum. He then says, “for small angles, the value of in radians is very nearly the value of ; they differ by less than 2% out to about 20°.” (a) Verify the linear approximation at 0 for the sine function: ; (b) Use a graphing device to determine the values of for which and differ by less than 2%. Then verify Hecht’s statement by converting from radians to degrees. Suppose that the only information we have about a function is that and the graph of its derivative is as shown. (a) Use a linear approximation to estimate and . (b) Are your estimates in part (a) too large or too small? Explain. Suppose that we don’t have a formula for but we know that and for all . (a) Use a linear approximation to estimate and . (b) Are your estimates in part (a) too large or too small? Explain. t!2.05" t!1.95" x t'!x" ! sx 2 $ 5 t!2" ! %4 t!x" y x 0 1 y=fª(x) 1 f !1.1" f !0.9" f !1" ! 5 f 43. x sin x x sin x # x sin ( ( 39. If a current passes through a resistor with resistance , Ohm’s Law states that the voltage drop is . If is constant and is measured with a certain error, use differen-tials to show that the relative error in calculating is approxi-mately the same (in magnitude) as the relative error in . When blood ﬂows along a blood vessel, the ﬂux (the volume of blood per unit time that ﬂows past a given point) is proportional to the fourth power of the radius of the blood vessel: (This is known as Poiseuille’s Law; we will show why it is true in Section 8.4.) A partially clogged artery can be expanded by an operation called angioplasty, in which a balloon-tipped catheter is inﬂated inside the artery in order to widen it and restore the normal blood ﬂow. Show that the relative change in is about four times the relative change in . How will a 5% increase in the radius affect the ﬂow of blood? Establish the following rules for working with differentials (where denotes a constant and and are functions of ). (a) (b) (c) (d) (e) (f) 42. On page 431 of Physics: Calculus, 2d ed., by Eugene Hecht (Paciﬁc Grove, CA: Brooks/Cole, 2000), in the course of deriving the formula for the period of a pendulum of length L, the author obtains the equation for the tangential acceleration of the bob of the aT ! %t sin ( T ! 2&sL$t d!x n" ! nx n%1 dx d% u v& ! v du % u dv v2 d!uv" ! u dv $ v du d!u $ v" ! du $ dv d!cu" ! c du dc ! 0 x v u c R F F ! kR 4 R F 40. R I R V V ! RI R I The tangent line approximation is the best ﬁrst-degree (linear) approximation to near because and have the same rate of change (derivative) at . For a better approxi-mation than a linear one, let’s try a second-degree (quadratic) approximation . In other words, we approximate a curve by a parabola instead of by a straight line. To make sure that the approximation is a good one, we stipulate the following: (i) ( and should have the same value at .) (ii) ( and should have the same rate of change at .) (iii) (The slopes of and should change at the same rate at .) 1. Find the quadratic approximation to the function that satisﬁes conditions (i), (ii), and (iii) with . Graph , , and the linear approximation on a common screen. Comment on how well the functions and approximate . Determine the values of for which the quadratic approximation in Problem 1 is accurate to within 0.1. [Hint: Graph , and on a common screen.] y ! cos x $ 0.1 y ! cos x % 0.1, y ! P!x" f !x" ! P!x" x f L P L!x" ! 1 f P a ! 0 f !x" ! cos x P!x" ! A $ Bx $ Cx 2 a f P P)!a" ! f )!a" a f P P'!a" ! f '!a" a f P P!a" ! f !a" P!x" a L!x" f !x" x ! a f !x" L!x" ; TAYLOR POLYNOMIALS L A B O R AT O R Y P R O J E C T 3. To approximate a function by a quadratic function near a number , it is best to write in the form Show that the quadratic function that satisﬁes conditions (i), (ii), and (iii) is 4. Find the quadratic approximation to near . Graph , the quadratic approximation, and the linear approximation from Example 2 in Section 3.10 on a common screen. What do you conclude? Instead of being satisﬁed with a linear or quadratic approximation to near , let’s try to ﬁnd better approximations with higher-degree polynomials. We look for an th-degree polynomial such that and its ﬁrst derivatives have the same values at as and its ﬁrst derivatives. By differentiating repeatedly and setting , show that these conditions are satisﬁed if , and in general where . The resulting polynomial is called the th-degree Taylor polynomial of centered at . Find the 8th-degree Taylor polynomial centered at for the function . Graph together with the Taylor polynomials in the viewing rectangle [%5, 5] by [%1.4, 1.4] and comment on how well they approximate . f T2, T4, T6, T8 f f !x" ! cos x a ! 0 a f n Tn!x" ! f !a" $ f '!a"!x % a" $ f )!a" 2! !x % a"2 $ $ f !n"!a" n! !x % a"n k! ! 1 ! 2 ! 3 ! 4 ! ! k ck ! f !k"!a" k! c0 ! f !a", c1 ! f '!a", c2 ! 1 2 f )!a" x ! a n f x ! a n Tn Tn!x" ! c0 $ c1!x % a" $ c2!x % a"2 $ c3!x % a"3 $ $ cn!x % a"n n x ! a f !x" f a ! 1 f !x" ! sx $ 3 P!x" ! f !a" $ f '!a"!x % a" $ 1 2 f )!a"!x % a"2 P!x" ! A $ B!x % a" $ C!x % a"2 P a P f 254 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES HYPERBOLIC FUNCTIONS Certain even and odd combinations of the exponential functions and arise so fre-quently in mathematics and its applications that they deserve to be given special names. In many ways they are analogous to the trigonometric functions, and they have the same relationship to the hyperbola that the trigonometric functions have to the circle. For this reason they are collectively called hyperbolic functions and individually called hyperbolic sine, hyperbolic cosine, and so on. DEFINITION OF THE HYPERBOLIC FUNCTIONS coth x ! cosh x sinh x tanh x ! sinh x cosh x sech x ! 1 cosh x cosh x ! e x $ e%x 2 csch x ! 1 sinh x sinh x ! e x % e%x 2 e%x e x 3.11 The graphs of hyperbolic sine and cosine can be sketched using graphical addition as in Figures 1 and 2. Note that has domain and range , while has domain and range . The graph of is shown in Figure 3. It has the horizontal asymptotes . (See Exercise 23.) Some of the mathematical uses of hyperbolic functions will be seen in Chapter 7. Applications to science and engineering occur whenever an entity such as light, velocity, electricity, or radioactivity is gradually absorbed or extinguished, for the decay can be rep-resented by hyperbolic functions. The most famous application is the use of hyperbolic cosine to describe the shape of a hanging wire. It can be proved that if a heavy ﬂexible cable (such as a telephone or power line) is suspended between two points at the same height, then it takes the shape of a curve with equation called a cate-nary (see Figure 4). (The Latin word catena means “chain.”) Another application of hyperbolic functions occurs in the description of ocean waves: The velocity of a water wave with length moving across a body of water with depth is modeled by the function where is the acceleration due to gravity. (See Figure 5 and Exercise 49.) The hyperbolic functions satisfy a number of identities that are similar to well-known trigonometric identities. We list some of them here and leave most of the proofs to the exercises. HYPERBOLIC IDENTITIES cosh!x $ y" ! cosh x cosh y $ sinh x sinh y sinh!x $ y" ! sinh x cosh y $ cosh x sinh y 1 % tanh2x ! sech2x cosh2x % sinh2x ! 1 cosh!%x" ! cosh x sinh!%x" ! %sinh x t v !' tL 2& tanh% 2&d L & d L y ! c $ a cosh!x$a" y ! !1 tanh (1, +" ! cosh ! ! sinh FIGURE 3 y=tanh x y 0 x y=1 y=1 FIGURE 1 y=sinh x= ´- e–® 1 2 1 2 1 2 y= ´ y= e–® 1 2 y=sinh x 0 y x FIGURE 2 y=cosh x= ´+ e–® 1 2 1 2 y= e–® 1 2 1 2 y= ´ y=cosh x 1 0 y x SECTION 3.11 HYPERBOLIC FUNCTIONS \|\|\|\| 255 FIGURE 4 A catenary y=c+a cosh(x/a) y 0 x L d FIGURE 5 Idealized ocean wave EXAMPLE 1 Prove (a) and (b) . SOLUTION (a) (b) We start with the identity proved in part (a): If we divide both sides by , we get or M The identity proved in Example 1(a) gives a clue to the reason for the name “hyper-bolic” functions: If is any real number, then the point lies on the unit circle because . In fact, can be interpreted as the radian measure of in Figure 6. For this reason the trigonometric functions are sometimes called circular functions. Likewise, if is any real number, then the point lies on the right branch of the hyperbola because and . This time, does not represent the measure of an angle. However, it turns out that represents twice the area of the shaded hyperbolic sector in Figure 7, just as in the trigonometric case rep-resents twice the area of the shaded circular sector in Figure 6. The derivatives of the hyperbolic functions are easily computed. For example, We list the differentiation formulas for the hyperbolic functions as Table 1. The remaining proofs are left as exercises. Note the analogy with the differentiation formulas for trigono-metric functions, but beware that the signs are different in some cases. DERIVATIVES OF HYPERBOLIC FUNCTIONS d dx tanh x sech2x d dx coth x csch2x d dx cosh x sinh x d dx sech x sech x tanh x d dx sinh x cosh x d dx csch x csch x coth x 1 d dx sinh x d dx e x ex 2 e x ex 2 cosh x t t t cosh t 1 cosh2t sinh2t 1 x 2 y 2 1 Pcosh t, sinh t t POQ t cos2t sin2t 1 x 2 y 2 1 Pcos t, sin t t 1 tanh2x sech2x 1 sinh2x cosh2x 1 cosh2x cosh2x cosh2x sinh2x 1 4 4 1 e 2x 2 e2x 4 e 2x 2 e2x 4 cosh2x sinh2x e x ex 2 2 e x ex 2 2 1 tanh2x sech2x cosh2x sinh2x 1 V 256 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES FIGURE 7 0 y x ≈-¥=1 P(cosh t, sinh t) FIGURE 6 O y x P(cos t, sin t) ≈+¥=1 Q The Gateway Arch in St. Louis was designed using a hyperbolic cosine function (Exercise 48). © 2006 Getty Images EXAMPLE 2 Any of these differentiation rules can be combined with the Chain Rule. For instance, M INVERSE HYPERBOLIC FUNCTIONS You can see from Figures 1 and 3 that and are one-to-one functions and so they have inverse functions denoted by and . Figure 2 shows that is not one-to-one, but when restricted to the domain it becomes one-to-one. The inverse hyper-bolic cosine function is deﬁned as the inverse of this restricted function. The remaining inverse hyperbolic functions are deﬁned similarly (see Exercise 28). We can sketch the graphs of , , and in Figures 8, 9, and 10 by using Figures 1, 2, and 3. Since the hyperbolic functions are deﬁned in terms of exponential functions, it’s not surprising to learn that the inverse hyperbolic functions can be expressed in terms of log-arithms. In particular, we have: EXAMPLE 3 Show that . SOLUTION Let . Then x ! sinh y ! e y % e%y 2 y ! sinh%1x sinh%1x ! ln(x $ sx 2 $ 1) tanh%1x ! 1 2 ln% 1 $ x 1 % x& %1 - x - 1 5 cosh%1x ! ln(x $ sx 2 % 1) x , 1 4 sinh%1x ! ln(x $ sx 2 $ 1) x " ! 3 FIGURE 8 y=sinh–! x domain=R range=R 0 y x FIGURE 9 y=cosh–! x domain=[1, } range=[0,} 0 y x 1 FIGURE 10 y=tanh–! x domain=(_1, 1) range=R 0 y x 1 _1 tanh%1 cosh%1 sinh%1 y ! tanh%1x & ? tanh y ! x y ! cosh%1x & ? cosh y ! x and y , 0 y ! sinh%1x & ? sinh y ! x 2 (0, +" cosh tanh%1 sinh%1 tanh sinh d dx (cosh sx ) ! sinh sx ! d dx sx ! sinh sx 2sx SECTION 3.11 HYPERBOLIC FUNCTIONS \|\|\|\| 257 N Formula 3 is proved in Example 3. The proofs of Formulas 4 and 5 are requested in Exercises 26 and 27. so or, multiplying by , This is really a quadratic equation in : Solving by the quadratic formula, we get Note that , but (because ). Thus the minus sign is inadmissible and we have Therefore (See Exercise 25 for another method.) M DERIVATIVES OF INVERSE HYPERBOLIC FUNCTIONS The inverse hyperbolic functions are all differentiable because the hyperbolic functions are differentiable. The formulas in Table 6 can be proved either by the method for inverse functions or by differentiating Formulas 3, 4, and 5. EXAMPLE 4 Prove that . SOLUTION 1 Let . Then . If we differentiate this equation implicitly with respect to , we get Since and , we have , so dy dx ! 1 cosh y ! 1 s1 $ sinh2y ! 1 s1 $ x 2 cosh y ! s1 $ sinh2y cosh y , 0 cosh2y % sinh2y ! 1 cosh y dy dx ! 1 x sinh y ! x y ! sinh%1x d dx !sinh%1x" ! 1 s1 $ x 2 V d dx !tanh%1x" ! 1 1 % x 2 d dx !coth%1x" ! 1 1 % x 2 d dx !cosh%1x" ! 1 sx 2 % 1 d dx !sech%1x" ! % 1 xs1 % x 2 d dx !sinh%1x" ! 1 s1 $ x 2 d dx !csch%1x" ! % 1 )x)sx 2 $ 1 6 y ! ln!e y" ! ln(x $ sx 2 $ 1) e y ! x $ sx 2 $ 1 x - sx 2 $ 1 x % sx 2 $ 1 - 0 e y . 0 e y ! 2x ! s4x 2 $ 4 2 ! x ! sx 2 $ 1 !e y"2 % 2x!e y" % 1 ! 0 e y e 2y % 2xe y % 1 ! 0 e y e y % 2x % e%y ! 0 258 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES N Notice that the formulas for the derivatives of and appear to be identical. But the domains of these functions have no numbers in common: is deﬁned for , whereas is deﬁned for ) x) . 1. coth%1x ) x) - 1 tanh%1x coth%1x tanh%1x SOLUTION 2 From Equation 3 (proved in Example 3), we have M EXAMPLE 5 Find . SOLUTION Using Table 6 and the Chain Rule, we have M 1 1 sin2x cos x cos x cos2x sec x d dx tanh1sin x 1 1 sin x2 d dx sin x d dx tanh1sin x V 1 sx 2 1 sx 2 1 x (x sx 2 1)sx 2 1 1 x sx 2 1 1 x sx 2 1 1 x sx 2 1 d dx (x sx 2 1) d dx sinh1x d dx ln(x sx 2 1) SECTION 3.11 HYPERBOLIC FUNCTIONS \|\|\|\| 259 13. 14. 16. 18. 19. ( any real number) 20. If , ﬁnd the values of the other hyperbolic functions at . If and , ﬁnd the values of the other hyperbolic functions at . (a) Use the graphs of , , and in Figures 1–3 to draw the graphs of , , and . coth sech csch tanh cosh sinh x x 0 cosh x 5 3 x tanh x 12 13 n cosh x sinh xn cosh nx sinh nx 1 tanh x 1 tanh x e 2x tanhln x x 2 1 x 2 1 17. cosh 2x cosh2x sinh2x sinh 2x 2 sinh x cosh x 15. tanhx y tanh x tanh y 1 tanh x tanh y coth2x 1 csch2x 1–6 Find the numerical value of each expression. (a) (b) 2. (a) (b) 3. (a) (b) 4. (a) (b) 5. (a) (b) 6. (a) (b) 7–19 Prove the identity. 7. (This shows that is an odd function.) 8. (This shows that is an even function.) 10. 11. coshx y cosh x cosh y sinh x sinh y sinhx y sinh x cosh y cosh x sinh y cosh x sinh x ex cosh x sinh x e x 9. cosh coshx cosh x sinh sinhx sinh x sinh1 1 sinh 1 cosh1 1 sech 0 coshln 3 cosh 3 sinh 2 sinhln 2 tanh 1 tanh 0 cosh 0 sinh 0 EXERCISES 3.11 for the central curve of the arch, where and are measured in meters and . ; (a) Graph the central curve. (b) What is the height of the arch at its center? (c) At what points is the height 100 m? (d) What is the slope of the arch at the points in part (c)? If a water wave with length moves with velocity in a body of water with depth , then where is the acceleration due to gravity. (See Figure 5.) Explain why the approximation is appropriate in deep water. ; 50. A ﬂexible cable always hangs in the shape of a catenary , where and are constants and (see Figure 4 and Exercise 52). Graph several members of the family of functions . How does the graph change as varies? A telephone line hangs between two poles 14 m apart in the shape of the catenary , where and are measured in meters. (a) Find the slope of this curve where it meets the right pole. (b) Find the angle between the line and the pole. Using principles from physics it can be shown that when a cable is hung between two poles, it takes the shape of a curve that satisﬁes the differential equation where is the linear density of the cable, is the acceleration due to gravity, and is the tension in the cable at its lowest point, and the coordinate system is chosen appropriately. Verify that the function is a solution of this differential equation. y ! f !x" ! T /t cosh% /tx T & T t / d 2y dx 2 ! /t T'1 $% dy dx& 2 y ! f !x" y 0 x _7 7 5 ¨ ( y x y ! 20 cosh!x$20" % 15 51. a y ! a cosh!x$a" a . 0 a c y ! c $ a cosh!x$a" v #' tL 2& t v !' tL 2& tanh% 2&d L & d v L ) x) 0 91.20 y x ; (b) Check the graphs that you sketched in part (a) by using a graphing device to produce them. Use the deﬁnitions of the hyperbolic functions to ﬁnd each of the following limits. (a) (b) (c) (d) (e) (f) (g) (h) (i) 24. Prove the formulas given in Table 1 for the derivatives of the functions (a) , (b) , (c) , (d) , and (e) . Give an alternative solution to Example 3 by letting and then using Exercise 9 and Example 1(a) with replaced by . Prove Equation 4. Prove Equation 5 using (a) the method of Example 3 and (b) Exercise 18 with replaced by . For each of the following functions (i) give a deﬁnition like those in (2), (ii) sketch the graph, and (iii) ﬁnd a formula sim-ilar to Equation 3. (a) (b) (c) 29. Prove the formulas given in Table 6 for the derivatives of the following functions. (a) (b) (c) (d) (e) 30–47 Find the derivative. Simplify where possible. 30. 31. 32. 33. 34. 36. 37. 38. 39. 40. 41. 42. 43. 44. 46. 47. The Gateway Arch in St. Louis was designed by Eero Saarinen and was constructed using the equation y ! 211.49 % 20.96 cosh 0.03291765x y ! coth%1sx 2 $ 1 y ! sech%1s1 % x 2 , x . 0 y ! x sinh%1!x$3" % s9 $ x 2 45. y ! x tanh%1x $ ln s1 % x 2 y ! tanh%1sx y ! x 2 sinh%1!2x" G!x" ! 1 % cosh x 1 $ cosh x y !' 1 $ tanh x 1 % tanh x 4 y ! arctan!tanh x" y ! sinh!cosh x" f !t" ! sech2!e t" f !t" ! csch t!1 % ln csch t" y ! e cosh 3x 35. y ! x coth!1 $ x 2" h!x" ! ln!cosh x" t!x" ! cosh!ln x" f !x" ! x sinh x % cosh x f !x" ! tanh!1 $ e 2x" coth%1 sech%1 csch%1 tanh%1 cosh%1 coth%1 sech%1 csch%1 y x y x y ! sinh%1x coth sech csch tanh cosh lim x l%+ csch x lim x l 0% coth x lim x l 0$ coth x lim x l + coth x lim x l + sech x lim x l%+ sinh x lim x l + sinh x lim x l%+ tanh x lim x l + tanh x 260 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES CHAPTER 3 REVIEW \|\|\|\| 261 55. At what point of the curve does the tangent have slope 1? If , show that . Show that if and , then there exist numbers and such that equals either or . In other words, almost every function of the form is a shifted and stretched hyperbolic sine or cosine function. f !x" ! ae x $ be%x 1 cosh!x $ 2" 1 sinh!x $ 2" ae x $ be%x 2 1 b " 0 a " 0 sec ( ! cosh x x ! ln!sec ( $ tan (" y ! cosh x (a) Show that any function of the form satisﬁes the differential equation . (b) Find such that , , and . Evaluate . lim x l + sinh x e x y'!0" ! 6 y!0" ! %4 y) ! 9y y ! y!x" y) ! m 2y y ! A sinh mx $ B cosh mx 53. REVIEW CONCEPT CHECK 3 3. (a) How is the number deﬁned? (b) Express as a limit. (c) Why is the natural exponential function used more often in calculus than the other exponential functions ? (d) Why is the natural logarithmic function used more often in calculus than the other logarithmic functions ? (a) Explain how implicit differentiation works. (b) Explain how logarithmic differentiation works. (a) Write an expression for the linearization of at . (b) If , write an expression for the differential . (c) If , draw a picture showing the geometric mean-ings of and . dy #y dx ! #x dy y ! f !x" a f y ! loga x y ! ln x y ! a x y ! e x e e 1. State each differentiation rule both in symbols and in words. (a) The Power Rule (b) The Constant Multiple Rule (c) The Sum Rule (d) The Difference Rule (e) The Product Rule (f) The Quotient Rule (g) The Chain Rule 2. State the derivative of each function. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r) (s) (t) y ! tanh%1x y ! cosh%1x y ! sinh%1x y ! tanh x y ! cosh x y ! sinh x y ! tan%1x y ! cos%1x y ! sin%1x y ! cot x y ! sec x y ! csc x y ! tan x y ! cos x y ! sin x y ! loga x y ! ln x y ! a x y ! e x y ! x n Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. If and are differentiable, then 2. If and are differentiable, then 3. If and are differentiable, then 4. If is differentiable, then . If is differentiable, then . d dx f(sx ) ! f '!x" 2sx f d dx sf !x" ! f '!x" 2sf !x" f d dx ( f !t!x"" ! f '!t!x""t'!x" t f d dx ( f !x"t!x" ! f '!x"t'!x" t f d dx ( f !x" $ t!x" ! f '!x" $ t'!x" t f 6. If , then . 7. 8. 9. 10. If , then . An equation of the tangent line to the parabola at is . y % 4 ! 2x!x $ 2" !%2, 4" y ! x 2 lim x l 2 t!x" % t!2" x % 2 ! 80 t!x" ! x 5 d dx ) x 2 $ x) ! ) 2x $ 1) d dx !tan2x" ! d dx !sec2x" d dx !ln 10" ! 1 10 d dx !10 x" ! x10 x%1 y' ! 2e y ! e 2 TRUE-FALSE QUIZ 262 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES 1–50 Calculate . 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. If , ﬁnd . f !!2" f !t" ! s4t " 1 y ! sin2(cosssin #x ) y ! cos(estan 3x ) y ! x tanh$1sx y ! cosh$1!sinh x" y ! ln # x 2 $ 4 2x " 5 # y ! ln!cosh 3x" y ! sin mx x y ! x sinh!x 2" y ! !x " %"4 x 4 " %4 y ! sx " 1 !2 $ x"5 !x " 3"7 xe y ! y $ 1 y ! tan2!sin &" y ! arctan(arcsin sx ) y ! sin(tan s1 " x 3 ) y ! st ln!t 4" y ! cot!3x 2 " 5" y ! 10tan #& y ! ln$ sec 5x " tan 5x$ y ! e cos x " cos!e x" y ! x tan$1!4x" y ! !x 2 " 1"4 !2x " 1"3!3x $ 1"5 y ! ln sin x $ 1 2 sin2x y ! !cos x"x y ! log 5!1 " 2x" y ! ssin sx sin!xy" ! x2 $ y y ! 1%s 3 x " sx y ! !1 $ x $1"$1 y ! sec!1 " x 2" y ! 3 x ln x y ! ln!x 2e x" y ! e cx !c sin x $ cos x" x 2 cos y " sin 2y ! xy y ! sec 2& 1 " tan 2& y ! ln!csc 5x" xy 4 " x 2y ! x " 3y y ! 1 sin!x $ sin x" y ! e1%x x 2 y ! !arcsin 2x"2 y ! sx cos sx y ! emx cos nx y ! t 1 $ t 2 y ! e$t!t 2 $ 2t " 2" y ! esin 2& y ! e x 1 " x 2 y ! 2xsx 2 " 1 y ! 3x $ 2 s2x " 1 y ! sx " 1 s 3 x 4 y ! cos!tan x" y ! !x 4 $ 3x 2 " 5"3 y' 52. If , ﬁnd . Find if . Find if . Use mathematical induction (page 77) to show that if , then . Evaluate . 57–59 Find an equation of the tangent to the curve at the given point. 57. , 58. , 59. , 60–61 Find equations of the tangent line and normal line to the curve at the given point. 60. , 61. , ; 62. If , ﬁnd . Graph and on the same screen and comment. (a) If , ﬁnd . (b) Find equations of the tangent lines to the curve at the points and . ; (c) Illustrate part (b) by graphing the curve and tangent lines on the same screen. ; (d) Check to see that your answer to part (a) is reasonable by comparing the graphs of and . (a) If , , ﬁnd and . ; (b) Check to see that your answers to part (a) are reasonable by comparing the graphs of , , and . At what points on the curve , , is the tangent line horizontal? Find the points on the ellipse where the tangent line has slope 1. If , show that 68. (a) By differentiating the double-angle formula obtain the double-angle formula for the sine function. (b) By differentiating the addition formula obtain the addition formula for the cosine function. sin!x " a" ! sin x cos a " cos x sin a cos 2x ! cos2x $ sin2x f '!x" f !x" ! 1 x $ a " 1 x $ b " 1 x $ c f !x" ! !x $ a"!x $ b"!x $ c" x 2 " 2y 2 ! 1 0 ( x ( 2# y ! sin x " cos x f ! f ' f f ! f ' $#%2 ) x ) #%2 f !x" ! 4x $ tan x f ' f !4, 4" !1, 2" y ! xs5 $ x f '!x" f !x" ! xs5 $ x f ' f f '!x" f !x" ! xesin x !0, 2" y ! !2 " x"e$x !2, 1" x2 " 4xy " y2 ! 13 !0, 1" y ! s1 " 4 sin x !0, $1" y ! x2 $ 1 x2 " 1 !#%6, 1" y ! 4 sin2x lim t l 0 t 3 tan3 !2t" f !n"!x" ! !x " n"e x f !x" ! xe x f !x" ! 1%!2 $ x" f !n"!x" x 6 " y 6 ! 1 y! t!!#%6" t!&" ! & sin & EXERCISES (b) Find , the rate at which the drug is cleared from circulation. (c) When is this rate equal to 0? An equation of motion of the form represents damped oscillation of an object. Find the velocity and acceleration of the object. A particle moves along a horizontal line so that its coordinate at time is , , where and are positive constants. (a) Find the velocity and acceleration functions. (b) Show that the particle always moves in the positive direction. A particle moves on a vertical line so that its coordinate at time is , . (a) Find the velocity and acceleration functions. (b) When is the particle moving upward and when is it moving downward? (c) Find the distance that the particle travels in the time interval . ; (d) Graph the position, velocity, and acceleration functions for . (e) When is the particle speeding up? When is it slowing down? The volume of a right circular cone is , where is the radius of the base and is the height. (a) Find the rate of change of the volume with respect to the height if the radius is constant. (b) Find the rate of change of the volume with respect to the radius if the height is constant. The mass of part of a wire is kilograms, where is measured in meters from one end of the wire. Find the linear density of the wire when m. The cost, in dollars, of producing units of a certain com-modity is (a) Find the marginal cost function. (b) Find and explain its meaning. (c) Compare with the cost of producing the 101st item. A bacteria culture contains 200 cells initially and grows at a rate proportional to its size. After half an hour the population has increased to 360 cells. (a) Find the number of bacteria after hours. (b) Find the number of bacteria after 4 hours. (c) Find the rate of growth after 4 hours. (d) When will the population reach 10,000? Cobalt-60 has a half-life of 5.24 years. (a) Find the mass that remains from a 100-mg sample after 20 years. (b) How long would it take for the mass to decay to 1 mg? t C'!100" C'!100" C!x" ! 920 " 2x $ 0.02x 2 " 0.00007x 3 x x ! 4 x x(1 " sx ) h r V ! #r 2h%3 0 ( t ( 3 0 ( t ( 3 t 0 y ! t 3 $ 12t " 3 t c b t 0 x ! sb 2 " c 2t 2 t s ! Ae$ct cos!+t " ," C'!t" 69. Suppose that and , where , , , , and . Find (a) and (b) . If and are the functions whose graphs are shown, let , , and . Find (a) , (b) , and (c) . 71–78 Find in terms of . 71. 72. 73. 74. 75. 76. 77. 78. 79–81 Find in terms of and . 79. 80. 81. ; 82. (a) Graph the function in the viewing rectangle by . (b) On which interval is the average rate of change larger: or ? (c) At which value of is the instantaneous rate of change larger: or ? (d) Check your visual estimates in part (c) by computing and comparing the numerical values of and . At what point on the curve is the tangent horizontal? (a) Find an equation of the tangent to the curve that is parallel to the line . (b) Find an equation of the tangent to the curve that passes through the origin. Find a parabola that passes through the point and whose tangent lines at and have slopes 6 and , respectively. The function , where a, b, and K are positive constants and , is used to model the concentra-tion at time t of a drug injected into the bloodstream. (a) Show that . limt l - C!t" ! 0 b . a C!t" ! K!e$at $ e$bt" $2 x ! 5 x ! $1 !1, 4" y ! ax 2 " bx " c y ! e x x $ 4y ! 1 y ! e x y ! &ln!x " 4"'2 f '!5" f '!2" f '!x" x ! 5 x ! 2 x &2, 3' &1, 2' &$2, 8' &0, 8' f !x" ! x $ 2 sin x h!x" ! f !t!sin 4x"" h!x" !( f !x" t!x" h!x" ! f !x"t!x" f !x" " t!x" t' f ' h' f !x" ! t!ln x" f !x" ! ln $ t!x"$ f !x" ! e t!x" f !x" ! t!e x" f !x" ! t!t!x"" f !x" ! &t!x"'2 f !x" ! t!x 2" f !x" ! x 2t!x" t' f ' 0 g f y x 1 1 C'!2" Q'!2" P'!2" C!x" ! f !t!x"" Q!x" ! f !x"%t!x" P!x" ! f !x"t!x" t f F'!2" h'!2" f '!5" ! 11 f '!2" ! $2 t'!2" ! 4 t!2" ! 5 f !2" ! 3 F!x" ! f !t!x"" h!x" ! f !x"t!x" CHAPTER 3 REVIEW \|\|\|\| 263 ;102. (a) Find the linear approximation to near 3. (b) Illustrate part (a) by graphing and the linear approximation. (c) For what values of is the linear approximation accurate to within 0.1? (a) Find the linearization of at . State the corresponding linear approximation and use it to give an approximate value for . ; (b) Determine the values of for which the linear approxima-tion given in part (a) is accurate to within 0.1. Evaluate if , , and . A window has the shape of a square surmounted by a semi-circle. The base of the window is measured as having width 60 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum error possible in com-puting the area of the window. 106–108 Express the limit as a derivative and evaluate. 106. 107. 108. Evaluate . Suppose is a differentiable function such that and . Show that . Find if it is known that 112. Show that the length of the portion of any tangent line to the astroid cut off by the coordinate axes is constant. x 2%3 " y 2%3 ! a 2%3 d dx & f !2x"' ! x 2 f '!x" t'!x" ! 1%!1 " x 2" f '!x" ! 1 " & f !x"'2 f !t!x"" ! x f lim x l 0 s1 " tan x $ s1 " sin x x 3 lim & l #%3 cos & $ 0.5 & $ #%3 lim h l 0 s 4 16 " h $ 2 h lim x l1 x 17 $ 1 x $ 1 dx ! 0.2 x ! 2 y ! x 3 $ 2x 2 " 1 dy x s 3 1.03 a ! 0 f !x" ! s 3 1 " 3x x f f !x" ! s25 $ x 2 95. Let be the concentration of a drug in the bloodstream. As the body eliminates the drug, decreases at a rate that is proportional to the amount of the drug that is present at the time. Thus , where is a positive number called the elimination constant of the drug. (a) If is the concentration at time , ﬁnd the concen-tration at time . (b) If the body eliminates half the drug in 30 hours, how long does it take to eliminate 90% of the drug? A cup of hot chocolate has temperature in a room kept at . After half an hour the hot chocolate cools to . (a) What is the temperature of the chocolate after another half hour? (b) When will the chocolate have cooled to ? The volume of a cube is increasing at a rate of 10 . How fast is the surface area increasing when the length of an edge is 30 cm? A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top). If water is poured into the cup at a rate of , how fast is the water level rising when the water is 5 cm deep? A balloon is rising at a constant speed of . A boy is cycling along a straight road at a speed of . When he passes under the balloon, it is 45 ft above him. How fast is the distance between the boy and the balloon increasing 3 s later? A waterskier skis over the ramp shown in the ﬁgure at a speed of . How fast is she rising as she leaves the ramp? The angle of elevation of the sun is decreasing at a rate of . How fast is the shadow cast by a 400-ft-tall building increasing when the angle of elevation of the sun is ? %6 0.25 rad%h 4 ft 15 ft 30 ft%s 15 ft%s 5 ft%s 2 cm3%s cm3%min 40/C 60/C 20/C 80/C t t ! 0 C0 k C'!t" ! $kC!t" C!t" C!t" 264 \|\|\|\| CHAPTER 3 DIFFERENTIATION RULES Before you look at the example, cover up the solution and try it yourself ﬁrst. EXAMPLE 1 How many lines are tangent to both of the parabolas and ? Find the coordinates of the points at which these tangents touch the parabolas. SOLUTION To gain insight into this problem, it is essential to draw a diagram. So we sketch the parabolas (which is the standard parabola shifted 1 unit upward) and (which is obtained by reﬂecting the ﬁrst parabola about the x-axis). If we try to draw a line tangent to both parabolas, we soon discover that there are only two possibilities, as illustrated in Figure 1. Let P be a point at which one of these tangents touches the upper parabola and let a be its x-coordinate. (The choice of notation for the unknown is important. Of course we could have used b or c or or instead of a. However, it’s not advisable to use x in place of a because that x could be confused with the variable x in the equation of the parabola.) Then, since P lies on the parabola , its y-coordinate must be Because of the symmetry shown in Figure 1, the coordinates of the point Q where the tangent touches the lower parabola must be . To use the given information that the line is a tangent, we equate the slope of the line PQ to the slope of the tangent line at P. We have If , then the slope of the tangent line at P is . Thus the condi-tion that we need to use is that Solving this equation, we get , so and . Therefore the points are (1, 2) and ($1, $2). By symmetry, the two remaining points are ($1, 2) and (1, $2). M EXAMPLE 2 For what values of does the equation have exactly one solution? SOLUTION One of the most important principles of problem solving is to draw a diagram, even if the problem as stated doesn’t explicitly mention a geometric situation. Our pres-ent problem can be reformulated geometrically as follows: For what values of does the curve intersect the curve in exactly one point? Let’s start by graphing and for various values of . We know that, for , is a parabola that opens upward if and downward if . Figure 2 shows the parabolas for several positive values of . Most of them don’t intersect at all and one intersects twice. We have the feeling that there must be a value of (somewhere between and ) for which the curves intersect exactly once, as in Figure 3. To ﬁnd that particular value of , we let be the -coordinate of the single point of intersection. In other words, , so is the unique solution of the given equa-tion. We see from Figure 3 that the curves just touch, so they have a common tangent a ln a ! ca 2 x a c 0.3 0.1 c y ! ln x c y ! cx 2 c ) 0 c . 0 y ! cx 2 c " 0 c y ! cx 2 y ! ln x y ! cx 2 y ! ln x c ln x ! cx 2 c a ! 01 a 2 ! 1 1 " a 2 ! 2a 2 1 " a 2 a ! 2a f'!a" ! 2a f!x" ! 1 " x 2 mPQ ! 1 " a 2 $ !$1 $ a 2" a $ !$a" ! 1 " a 2 a !$a, $!1 " a 2"" 1 " a 2. y ! 1 " x 2 x1 x0 y ! $1 $ x 2 y ! x 2 y ! 1 " x 2 y ! 1 " x 2 y ! $1 $ x 2 265 x y P Q 1 _1 FIGURE 1 0 3≈≈ 0.3≈ 0.1≈ y=ln x ≈ 1 2 x y FIGURE 2 y=ln x y=c≈ c=? y x 0 a FIGURE 3 P R O B L E M S P L U S P R O B L E M S P L U S 266 line when . That means the curves and have the same slope when . Therefore Solving the equations and , we get Thus and For negative values of we have the situation illustrated in Figure 4: All parabolas with negative values of intersect exactly once. And let’s not forget about : The curve is just the -axis, which intersects exactly once. To summarize, the required values of are and . M 1. Find points and on the parabola so that the triangle formed by the -axis and the tangent lines at and is an equilateral triangle. ; 2. Find the point where the curves and are tangent to each other, that is, have a common tangent line. Illustrate by sketching both curves and the common tangent. 3. Show that the tangent lines to the parabola at any two points with -coordinates and must intersect at a point whose -coordinate is halfway between and . 4. Show that d dx) sin2x 1 " cot x " cos2x 1 " tan x ! $cos 2x q p x q p x y ! ax 2 " bx " c y ! 3!x 2 $ x" y ! x 3 $ 3x " 4 x y P Q A 0 B C Q P x ABC y ! 1 $ x 2 Q P PROBLEMS c ( 0 c ! 1%!2e" c y ! ln x x y ! 0x 2 ! 0 c ! 0 y ! ln x c y ! cx 2 c c ! ln a a 2 ! ln e 1%2 e ! 1 2e a ! e 1%2 ln a ! ca 2 ! c ! 1 2c ! 1 2 1%a ! 2ca ln a ! ca 2 1 a ! 2ca x ! a y ! cx 2 y ! ln x x ! a y y=ln x x 0 FIGURE 4 P R O B L E M S P L U S 267 5. Show that . 6. A car is traveling at night along a highway shaped like a parabola with its vertex at the origin (see the ﬁgure). The car starts at a point 100 m west and 100 m north of the origin and travels in an easterly direction. There is a statue located 100 m east and 50 m north of the origin. At what point on the highway will the car’s headlights illuminate the statue? 7. Prove that . 8. Find the th derivative of the function . 9. The ﬁgure shows a circle with radius 1 inscribed in the parabola . Find the center of the circle. 10. If is differentiable at , where , evaluate the following limit in terms of : 11. The ﬁgure shows a rotating wheel with radius 40 cm and a connecting rod with length 1.2 m. The pin slides back and forth along the -axis as the wheel rotates counterclockwise at a rate of 360 revolutions per minute. (a) Find the angular velocity of the connecting rod, , in radians per second, when . (b) Express the distance in terms of . (c) Find an expression for the velocity of the pin in terms of . 12. Tangent lines and are drawn at two points and on the parabola and they intersect at a point . Another tangent line is drawn at a point between and ; it intersects at and at . Show that 13. Show that where and are positive numbers, , and . 14. Evaluate . lim x l # e sin x $ 1 x $ # & ! tan$1!b%a" r 2 ! a 2 " b 2 b a d n dx n !e ax sin bx" ! r ne ax sin!bx " n&" $ PQ1$ $ PP1$ " $ PQ2$ $ PP2$ ! 1 Q2 T2 Q1 T1 P2 P1 T P y ! x 2 P2 P1 T2 T1 & P & x ! $ OP$ & ! #%3 d1%dt x P AP lim x l a f !x" $ f !a" sx $ sa f '!a" a . 0 a f x 0 y 1 1 y=≈ y ! x 2 f !x" ! x n%!1 $ x" n d n dx n !sin4x " cos4x" ! 4n$1 cos!4x " n#%2" sin$1!tanh x" ! tan$1!sinh x" x y FIGURE FOR PROBLEM 6 A P(x, 0) ¨ å FIGURE FOR PROBLEM 11 x y O P R O B L E M S P L U S 268 15. Let and be the tangent and normal lines to the ellipse at any point on the ellipse in the ﬁrst quadrant. Let and be the - and -intercepts of and and be the intercepts of . As moves along the ellipse in the ﬁrst quadrant (but not on the axes), what values can , , , and take on? First try to guess the answers just by looking at the ﬁgure. Then use calculus to solve the problem and see how good your intuition is. 16. Evaluate . 17. (a) Use the identity for (see Equation 14b in Appendix D) to show that if two lines and intersect at an angle , then where and are the slopes of and , respectively. (b) The angle between the curves and at a point of intersection is deﬁned to be the angle between the tangent lines to and at (if these tangent lines exist). Use part (a) to ﬁnd, correct to the nearest degree, the angle between each pair of curves at each point of intersection. (i) and (ii) and 18. Let be a point on the parabola with focus . Let be the angle between the parabola and the line segment , and let be the angle between the horizontal line and the parabola as in the ﬁgure. Prove that . (Thus, by a principle of geo-metrical optics, light from a source placed at will be reﬂected along a line parallel to the -axis. This explains why paraboloids, the surfaces obtained by rotating parabolas about their axes, are used as the shape of some automobile headlights and mirrors for telescopes.) 0 x y F(p, 0) P(⁄, ›) ¥=4px y=› å ∫ x F 1 ! 2 y ! y1 2 FP 1 F!p, 0" y 2 ! 4px P!x1, y1" x 2 $ 4x " y 2 " 3 ! 0 x 2 $ y 2 ! 3 y ! !x $ 2"2 y ! x 2 P C2 C1 P C2 C1 L 2 L1 m2 m1 tan 1 ! m2 $ m1 1 " m1m2 1 L 2 L1 tan !x $ y" lim x l 0 sin!3 " x"2 $ sin 9 x xN xT yT yN 3 2 T N P x y 0 yN xN yT xT P N yN xN T y x yT xT P x 2%9 " y 2%4 ! 1 N T P R O B L E M S P L U S 269 19. Suppose that we replace the parabolic mirror of Problem 18 by a spherical mirror. Although the mirror has no focus, we can show the existence of an approximate focus. In the ﬁgure, is a semicircle with center . A ray of light coming in toward the mirror parallel to the axis along the line will be reﬂected to the point on the axis so that (the angle of incidence is equal to the angle of reﬂection). What happens to the point as is taken closer and closer to the axis? 20. If and are differentiable functions with and , show that 21. Evaluate . 22. (a) The cubic function has three distinct zeros: 0, 2, and 6. Graph and its tangent lines at the average of each pair of zeros. What do you notice? (b) Suppose the cubic function has three distinct zeros: , , and . Prove, with the help of a computer algebra system, that a tangent line drawn at the average of the zeros intersects the graph of at the third zero. 23. For what value of does the equation have exactly one solution? 24. For which positive numbers is it true that for all ? 25. If show that . 26. Given an ellipse , where , ﬁnd the equation of the set of all points from which there are two tangents to the curve whose slopes are (a) reciprocals and (b) nega-tive reciprocals. 27. Find the two points on the curve that have a common tangent line. 28. Suppose that three points on the parabola have the property that their normal lines intersect at a common point. Show that the sum of their -coordinates is 0. 29. A lattice point in the plane is a point with integer coordinates. Suppose that circles with radius are drawn using all lattice points as centers. Find the smallest value of such that any line with slope intersects some of these circles. 30. A cone of radius centimeters and height centimeters is lowered point ﬁrst at a rate of 1 cm%s into a tall cylinder of radius centimeters that is partially ﬁlled with water. How fast is the water level rising at the instant the cone is completely submerged? 31. A container in the shape of an inverted cone has height 16 cm and radius 5 cm at the top. It is partially ﬁlled with a liquid that oozes through the sides at a rate proportional to the area of the container that is in contact with the liquid. (The surface area of a cone is , where is the radius and is the slant height.) If we pour the liquid into the container at a rate of , then the height of the liquid decreases at a rate of 0.3 cm%min when the height is 10 cm. If our goal is to keep the liquid at a constant height of 10 cm, at what rate should we pour the liquid into the container? 2 cm3%min l r #rl R h r 2 5 r r x y ! x 2 y ! x 4 $ 2x 2 $ x a " b x 2%a 2 " y 2%b 2 ! 1 y' ! 1 a " cos x y ! x sa 2 $ 1 $ 2 sa 2 $ 1 arctan sin x a " sa 2 $ 1 " cos x x a x 1 " x a e2x ! ksx k f a and b c b a f !x" ! !x $ a"!x $ b"!x $ c" f f !x" ! x!x $ 2"!x $ 6" CAS lim x l 0 sin!a " 2x" $ 2 sin!a " x" " sin a x 2 lim x l 0 f !x" t!x" ! f '!0" t'!0" t'!0" " 0 f !0" ! t!0" ! 0 t f P R !PQO ! !OQR R PQ O C O R P Q ¨ ¨ C A FIGURE FOR PROBLEM 19 270 We have already investigated some of the applications of derivatives, but now that we know the differentiation rules we are in a better position to pursue the applications of differentiation in greater depth. Here we learn how derivatives affect the shape of a graph of a function and, in particular, how they help us locate maximum and minimum values of functions. Many practical problems require us to minimize a cost or maximize an area or somehow ﬁnd the best possible outcome of a situation. In particular, we will be able to investigate the optimal shape of a can and to explain the location of rainbows in the sky. Calculus reveals all the important aspects of graphs of functions. Members of the family of functions are illustrated. f !x" ! cx ! sin x APPLICATIONS OF DIFFERENTIATION 4 x y MAXIMUM AND MINIMUM VALUES Some of the most important applications of differential calculus are optimization prob-lems, in which we are required to ﬁnd the optimal (best) way of doing something. Here are examples of such problems that we will solve in this chapter: I What is the shape of a can that minimizes manufacturing costs? I What is the maximum acceleration of a space shuttle? (This is an important question to the astronauts who have to withstand the effects of acceleration.) I What is the radius of a contracted windpipe that expels air most rapidly during a cough? I At what angle should blood vessels branch so as to minimize the energy expended by the heart in pumping blood? These problems can be reduced to ﬁnding the maximum or minimum values of a function. Let’s ﬁrst explain exactly what we mean by maximum and minimum values. DEFINITION A function has an absolute maximum (or global maximum) at if for all in , where is the domain of . The number is called the maximum value of on . Similarly, has an absolute minimum at if for all in and the number is called the minimum value of on . The maximum and minimum values of are called the extreme values of . Figure 1 shows the graph of a function with absolute maximum at and absolute minimum at . Note that is the highest point on the graph and is the low-est point. If we consider only values of near [for instance, if we restrict our attention to the interval ], then is the largest of those values of and is called a local maximum value of . Likewise, is called a local minimum value of because for near [in the interval , for instance]. The function also has a local minimum at . In general, we have the following deﬁnition. DEFINITION A function has a local maximum (or relative maximum) at if when x is near c. [This means that for all in some open interval containing c.] Similarly, has a local minimum at if when is near c. EXAMPLE 1 The function takes on its (local and absolute) maximum value of 1 inﬁnitely many times, since for any integer and for all . Likewise, is its minimum value, where is any integer. M EXAMPLE 2 If , then because for all . Therefore is the absolute (and local) minimum value of . This corresponds to the fact that the origin is the lowest point on the parabola . (See Figure 2.) However, there is no highest point on the parabola and so this function has no maximum value. M EXAMPLE 3 From the graph of the function , shown in Figure 3, we see that this function has neither an absolute maximum value nor an absolute minimum value. In fact, it has no local extreme values either. M f!x" ! x 3 y ! x 2 f f!0" ! 0 x x 2 " 0 f!x" " f!0" f!x" ! x 2 n cos!2n ! 1"# ! $1 x $1 % cos x % 1 n cos 2n# ! 1 f!x" ! cos x x f!c" % f!x" c f x f!c" " f!x" f!c" " f!x" c f 2 e f !b, d" c x f!c" % f!x" f f!c" f f!x" f!b" !a, c" b x !a, f!a"" !d, f!d"" a d f f f D f f!c" D x f!c" % f!x" c f D f f!c" f D D x f!c" " f!x" c f 1 4.1 271 f(a) f(d) b x y 0 d e a c FIGURE 1 Minimum value f(a), maximum value f(d) FIGURE 2 Minimum value 0, no maximum x y 0 y=≈ FIGURE 3 No minimum, no maximum x y 0 y=˛ EXAMPLE 4 The graph of the function is shown in Figure 4. You can see that is a local maximum, whereas the absolute maximum is . (This absolute maximum is not a local maximum because it occurs at an endpoint.) Also, is a local minimum and is both a local and an absolute minimum. Note that has neither a local nor an absolute maximum at . M We have seen that some functions have extreme values, whereas others do not. The following theorem gives conditions under which a function is guaranteed to possess extreme values. THE EXTREME VALUE THEOREM If is continuous on a closed interval , then attains an absolute maximum value and an absolute minimum value at some numbers and in . The Extreme Value Theorem is illustrated in Figure 5. Note that an extreme value can be taken on more than once. Although the Extreme Value Theorem is intuitively very plau-sible, it is difﬁcult to prove and so we omit the proof. Figures 6 and 7 show that a function need not possess extreme values if either hypoth-esis (continuity or closed interval) is omitted from the Extreme Value Theorem. The function f whose graph is shown in Figure 6 is deﬁned on the closed interval [0, 2] but has no maximum value. (Notice that the range of f is [0, 3). The function takes on val-ues arbitrarily close to 3, but never actually attains the value 3.) This does not contradict the Extreme Value Theorem because f is not continuous. [Nonetheless, a discontinuous function could have maximum and minimum values. See Exercise 13(b).] 1 x y 0 FIGURE 7 This continuous function g has no maximum or minimum. 2 1 x y 0 FIGURE 6 This function has minimum value f(2)=0, but no maximum value. 2 3 FIGURE 5 x y 0 b a c d x y 0 b a c¡ d c™ x y 0 d=b a c #a, b$ d c f!d" f!c" f #a, b$ f 3 x ! 4 f f!3" ! $27 f!0" ! 0 f!$1" ! 37 f!1" ! 5 $1 % x % 4 f!x" ! 3x 4 $ 16x 3 ! 18x 2 V 272 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION FIGURE 4 (_1, 37) _1 1 2 3 4 5 (3, _27) (1, 5) y x y=3x$-16˛+18≈ The function t shown in Figure 7 is continuous on the open interval (0, 2) but has nei-ther a maximum nor a minimum value. [The range of t is . The function takes on arbitrarily large values.] This does not contradict the Extreme Value Theorem because the interval (0, 2) is not closed. The Extreme Value Theorem says that a continuous function on a closed interval has a maximum value and a minimum value, but it does not tell us how to ﬁnd these extreme values. We start by looking for local extreme values. Figure 8 shows the graph of a function with a local maximum at and a local minimum at . It appears that at the maximum and minimum points the tangent lines are horizontal and therefore each has slope 0. We know that the derivative is the slope of the tangent line, so it appears that and . The following theorem says that this is always true for differentiable functions. FERMAT’S THEOREM If has a local maximum or minimum at , and if exists, then . PROOF Suppose, for the sake of deﬁniteness, that has a local maximum at c. Then, according to Deﬁnition 2, if is sufﬁciently close to . This implies that if is sufﬁciently close to 0, with being positive or negative, then and therefore We can divide both sides of an inequality by a positive number. Thus, if and is sufﬁciently small, we have Taking the right-hand limit of both sides of this inequality (using Theorem 2.3.2), we get But since exists, we have and so we have shown that . If , then the direction of the inequality (5) is reversed when we divide by : So, taking the left-hand limit, we have f&!c" ! lim h l 0 f!c ! h" $ f!c" h ! lim h l 0$ f!c ! h" $ f!c" h " 0 h ' 0 f!c ! h" $ f!c" h " 0 h h ' 0 f&!c" % 0 f&!c" ! lim h l 0 f!c ! h" $ f!c" h ! lim h l 0! f!c ! h" $ f!c" h f&!c" lim h l 0! f!c ! h" $ f!c" h % lim h l 0! 0 ! 0 f!c ! h" $ f!c" h % 0 h h ( 0 f!c ! h" $ f!c" % 0 5 f!c" " f!c ! h" h h c x f!c" " f!x" f f&!c" ! 0 f&!c" c f 4 f&!d" ! 0 f&!c" ! 0 d c f !1, )" SECTION 4.1 MAXIMUM AND MINIMUM VALUES \|\|\|\| 273 N Fermat’s Theorem is named after Pierre Fermat (1601–1665), a French lawyer who took up mathematics as a hobby. Despite his amateur status, Fermat was one of the two inventors of analytic geometry (Descartes was the other). His methods for ﬁnding tangents to curves and maxi-mum and minimum values (before the invention of limits and derivatives) made him a forerunner of Newton in the creation of differential calculus. 0 x c d y {c, f(c)} {d, f(d)} FIGURE 8 We have shown that and also that . Since both of these inequalities must be true, the only possibility is that . We have proved Fermat’s Theorem for the case of a local maximum. The case of a local minimum can be proved in a similar manner, or we could use Exercise 76 to deduce it from the case we have just proved (see Exercise 77). M The following examples caution us against reading too much into Fermat’s Theorem. We can’t expect to locate extreme values simply by setting and solving for . EXAMPLE 5 If , then , so . But has no maximum or minimum at 0, as you can see from its graph in Figure 9. (Or observe that for but for .) The fact that simply means that the curve has a horizontal tangent at . Instead of having a maximum or minimum at , the curve crosses its horizontal tangent there. M EXAMPLE 6 The function has its (local and absolute) minimum value at 0, but that value can’t be found by setting because, as was shown in Example 5 in Section 2.8, does not exist. (See Figure 10.) M \| WARNING Examples 5 and 6 show that we must be careful when using Fermat’s Theorem. Example 5 demonstrates that even when there need not be a maximum or minimum at . (In other words, the converse of Fermat’s Theorem is false in general.) Furthermore, there may be an extreme value even when does not exist (as in Example 6). Fermat’s Theorem does suggest that we should at least start looking for extreme values of at the numbers where or where does not exist. Such numbers are given a special name. DEFINITION A critical number of a function is a number in the domain of such that either or does not exist. EXAMPLE 7 Find the critical numbers of . SOLUTION The Product Rule gives [The same result could be obtained by ﬁrst writing .] Therefore if , that is, , and does not exist when . Thus the critical numbers are and . M In terms of critical numbers, Fermat’s Theorem can be rephrased as follows (compare Deﬁnition 6 with Theorem 4): If f has a local maximum or minimum at c, then c is a critical number of f. 7 0 3 2 x ! 0 f&!x" x ! 3 2 12 $ 8x ! 0 f&!x" ! 0 f!x" ! 4x 3%5 $ x 8%5 ! $5x ! 3!4 $ x" 5x 2%5 ! 12 $ 8x 5x 2%5 f&!x" ! x 3%5!$1" ! !4 $ x"( 3 5 x$2%5) ! $x3%5 ! 3!4 $ x" 5x 2%5 f!x" ! x 3%5!4 $ x" V f&!c" f&!c" ! 0 f c f 6 f&!c" f&!c" ! 0 c f f&!c" c f&!c" ! 0 f&!0" f&!x" ! 0 f!x" ! &x& !0, 0" !0, 0" y ! x 3 f&!0" ! 0 x ' 0 x 3 ' 0 x ( 0 x 3 ( 0 f f&!0" ! 0 f&!x" ! 3x 2 f!x" ! x 3 x f&!x" ! 0 f&!c" ! 0 f&!c" % 0 f&!c" " 0 274 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION FIGURE 9 If ƒ=˛, then fª(0)=0 but ƒ has no maximum or minimum. y=˛ x y 0 FIGURE 10 If ƒ=\| x \|, then f(0)=0 is a minimum value, but fª(0) does not exist. x 0 y=\|x\| y FIGURE 11 3.5 _2 _0.5 5 N Figure 11 shows a graph of the function in Example 7. It supports our answer because there is a horizontal tangent when and a vertical tangent when . x ! 0 x ! 1.5 f To ﬁnd an absolute maximum or minimum of a continuous function on a closed interval, we note that either it is local [in which case it occurs at a critical number by (7)] or it occurs at an endpoint of the interval. Thus the following three-step procedure always works. THE CLOSED INTERVAL METHOD To ﬁnd the absolute maximum and minimum values of a continuous function on a closed interval : 1. Find the values of at the critical numbers of in . 2. Find the values of at the endpoints of the interval. 3. The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. EXAMPLE 8 Find the absolute maximum and minimum values of the function SOLUTION Since is continuous on , we can use the Closed Interval Method: Since exists for all , the only critical numbers of occur when , that is, or . Notice that each of these critical numbers lies in the interval . The values of at these critical numbers are The values of at the endpoints of the interval are Comparing these four numbers, we see that the absolute maximum value is and the absolute minimum value is . Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of is sketched in Figure 12. M If you have a graphing calculator or a computer with graphing software, it is possible to estimate maximum and minimum values very easily. But, as the next example shows, calculus is needed to ﬁnd the exact values. EXAMPLE 9 (a) Use a graphing device to estimate the absolute minimum and maximum values of the function . (b) Use calculus to ﬁnd the exact minimum and maximum values. SOLUTION (a) Figure 13 shows a graph of in the viewing rectangle by . By mov-ing the cursor close to the maximum point, we see that the -coordinates don’t change very much in the vicinity of the maximum. The absolute maximum value is about 6.97 and it occurs when . Similarly, by moving the cursor close to the minimum point, we see that the absolute minimum value is about and it occurs when . It is x ' 1.0 $0.68 x ' 5.2 y #$1, 8$ #0, 2#$ f f!x" ! x $ 2 sin x, 0 % x % 2# f f!2" ! $3 f!4" ! 17 f!4" ! 17 f ($ 1 2) ! 1 8 f f!2" ! $3 f!0" ! 1 f ($ 1 2, 4) x ! 2 x ! 0 f&!x" ! 0 f x f&!x" f&!x" ! 3x 2 $ 6x ! 3x!x $ 2" f!x" ! x 3 $ 3x 2 ! 1 [$ 1 2, 4] f $ 1 2 % x % 4 f!x" ! x 3 $ 3x 2 ! 1 V f !a, b" f f #a, b$ f SECTION 4.1 MAXIMUM AND MINIMUM VALUES \|\|\|\| 275 FIGURE 12 5 10 20 _5 15 1 2 3 4 (4, 17) (2, _3) _1 y=˛-3≈+1 x y 0 FIGURE 13 0 8 _1 2π possible to get more accurate estimates by zooming in toward the maximum and mini-mum points, but instead let’s use calculus. (b) The function is continuous on . Since , we have when and this occurs when or . The values of at these critical points are and The values of at the endpoints are Comparing these four numbers and using the Closed Interval Method, we see that the absolute minimum value is and the absolute maximum value is . The values from part (a) serve as a check on our work. M EXAMPLE 10 The Hubble Space Telescope was deployed on April 24, 1990, by the space shuttle Discovery. A model for the velocity of the shuttle during this mission, from liftoff at until the solid rocket boosters were jettisoned at , is given by (in feet per second). Using this model, estimate the absolute maximum and minimum values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters. SOLUTION We are asked for the extreme values not of the given velocity function, but rather of the acceleration function. So we ﬁrst need to differentiate to ﬁnd the acceleration: We now apply the Closed Interval Method to the continuous function a on the interval . Its derivative is The only critical number occurs when : Evaluating at the critical number and at the endpoints, we have So the maximum acceleration is about and the minimum acceleration is about . M 21.52 fts2 62.87 fts2 a126 62.87 at1 21.52 a0 23.61 at t1 0.18058 0.007812 23.12 at 0 at 0.007812t 0.18058 0 t 126 0.003906t 2 0.18058t 23.61 at vt d dt 0.001302t 3 0.09029t 2 23.61t 3.083 vt 0.001302t 3 0.09029t 2 23.61t 3.083 t 126 s t 0 f 53 53 s3 f 3 3 s3 f 2 2 6.28 and f 0 0 f f 53 5 3 2 sin 5 3 5 3 s3 6.968039 f 3 3 2 sin 3 3 s3 0.684853 f 53 x 3 cos x 1 2 f x 0 f x 1 2 cos x 0, 2 f x x 2 sin x 276 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION NASA SECTION 4.1 MAXIMUM AND MINIMUM VALUES \|\|\|\| 277 (c) Sketch the graph of a function that has a local maximum at 2 and is not continuous at 2. 12. (a) Sketch the graph of a function on [$1, 2] that has an absolute maximum but no local maximum. (b) Sketch the graph of a function on [$1, 2] that has a local maximum but no absolute maximum. (a) Sketch the graph of a function on [$1, 2] that has an absolute maximum but no absolute minimum. (b) Sketch the graph of a function on [$1, 2] that is discontin-uous but has both an absolute maximum and an absolute minimum. 14. (a) Sketch the graph of a function that has two local maxima, one local minimum, and no absolute minimum. (b) Sketch the graph of a function that has three local minima, two local maxima, and seven critical numbers. 15–28 Sketch the graph of by hand and use your sketch to ﬁnd the absolute and local maximum and minimum values of . (Use the graphs and transformations of Sections 1.2 and 1.3.) 15. , 16. , 17. , 18. , 19. , 20. , 21. , 22. , 23. , 24. , 26. 27. 28. 29–44 Find the critical numbers of the function. 29. 30. 31. 32. 33. 34. 35. 36. h!p" ! p $ 1 p2 ! 4 t!y" ! y $ 1 y 2 $ y ! 1 t!t" ! & 3t $ 4& s!t" ! 3t 4 ! 4t 3 $ 6t 2 f !x" ! x 3 ! x 2 ! x f !x" ! x 3 ! 3x 2 $ 24x f !x" ! x 3 ! x 2 $ x f !x" ! 5x 2 ! 4x f !x" !( 4 $ x2 2x $ 1 if $2 % x ' 0 if 0 % x % 2 f !x" !( 1 $ x 2x $ 4 if 0 % x ' 2 if 2 % x % 3 f !x" ! e x f !x" ! 1 $ sx 25. $3#%2 % t % 3#%2 f !t" ! cos t 0 ' x % 2 f !x" ! ln x $2 % x ' 5 f !x" ! 1 ! !x ! 1"2 $3 % x % 2 f !x" ! x 2 0 % x % 2 f !x" ! x 2 0 % x ' 2 f !x" ! x 2 0 ' x % 2 f !x" ! x 2 0 ' x ' 2 f !x" ! x 2 x % 5 f !x" ! 3 $ 2x x " 1 f !x" ! 8 $ 3x f f 13. 1. Explain the difference between an absolute minimum and a local minimum. 2. Suppose is a continuous function deﬁned on a closed interval . (a) What theorem guarantees the existence of an absolute max-imum value and an absolute minimum value for ? (b) What steps would you take to ﬁnd those maximum and minimum values? 3–4 For each of the numbers a, b, c, d, r, and s, state whether the function whose graph is shown has an absolute maximum or min-imum, a local maximum or minimum, or neither a maximum nor a minimum. 3. 4. 5–6 Use the graph to state the absolute and local maximum and minimum values of the function. 5. 6. 7–10 Sketch the graph of a function that is continuous on [1, 5] and has the given properties. 7. Absolute minimum at 2, absolute maximum at 3, local minimum at 4 8. Absolute minimum at 1, absolute maximum at 5, local maximum at 2, local minimum at 4 Absolute maximum at 5, absolute minimum at 2, local maximum at 3, local minima at 2 and 4 10. has no local maximum or minimum, but 2 and 4 are critical numbers (a) Sketch the graph of a function that has a local maximum at 2 and is differentiable at 2. (b) Sketch the graph of a function that has a local maximum at 2 and is continuous but not differentiable at 2. 11. f 9. f y 0 x y=© 1 1 y 0 x y=ƒ 1 1 x y 0 a b c d r s x y 0 a b c d r s f #a, b$ f EXERCISES 4.1 278 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 68. 69. Between and , the volume (in cubic centimeters) of 1 kg of water at a temperature is given approximately by the formula Find the temperature at which water has its maximum density. 70. An object with weight is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle with the plane, then the magnitude of the force is where is a positive constant called the coefﬁcient of friction and where . Show that is minimized when . 71. A model for the US average price of a pound of white sugar from 1993 to 2003 is given by the function where is measured in years since August of 1993. Estimate the times when sugar was cheapest and most expensive dur-ing the period 1993–2003. ; 72. On May 7, 1992, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. (a) Use a graphing calculator or computer to ﬁnd the cubic polynomial that best models the velocity of the shuttle for the time interval . Then graph this polynomial. (b) Find a model for the acceleration of the shuttle and use it to estimate the maximum and minimum values of the acceleration during the ﬁrst 125 seconds. t ! #0, 125$ t ! 0.03629t 2 $ 0.04458t ! 0.4074 S!t" ! $0.00003237t 5 ! 0.0009037t 4 $ 0.008956t 3 tan ! + F 0 % % #%2 + F ! +W + sin ! cos W V ! 999.87 $ 0.06426T ! 0.0085043T 2 $ 0.0000679T 3 T V 30,C 0,C f !x" ! x $ 2 cos x, $2 % x % 0 f !x" ! xsx $ x 2 67. 37. 38. 40. 42. 43. 44. ; 45–46 A formula for the derivative of a function is given. How many critical numbers does have? 45. 46. 47–62 Find the absolute maximum and absolute minimum values of on the given interval. 47. , 48. , , 50. , 51. , 52. , 53. , 54. , 55. , 56. , 57. , 58. , 59. , 60. , 61. 62. 63. If and are positive numbers, ﬁnd the maximum value of , . ; 64. Use a graph to estimate the critical numbers of correct to one decimal place. ; 65–68 (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. (b) Use calculus to ﬁnd the exact maximum and minimum values. 65. 66. f !x" ! ex3$x, $1 % x % 0 f !x" ! x5 $ x3 ! 2, $1 % x % 1 f !x" ! & x 3 $ 3x 2 ! 2& 0 % x % 1 f !x" ! x a!1 $ x"b b a f !x" ! e$x $ e$2x, #0, 1$ f !x" ! ln!x 2 ! x ! 1", #$1, 1$ [ 1 2, 2] f !x" ! x $ ln x #$1, 4$ f !x" ! xe$x2%8 ##%4, 7#%4$ f !t" ! t ! cot!t%2" #0,#%2$ f !t" ! 2cos t ! sin 2t #0, 8$ f !t" ! s 3 t !8 $ t" #$1, 2$ f !t" ! ts4 $ t 2 #$4, 4$ f !x" ! x2 $ 4 x2 ! 4 #0, 2$ f !x" ! x x 2 ! 1 #$1, 2$ f !x" ! !x2 $ 1"3 #$2, 3$ f !x" ! x 4 $ 2x 2 ! 3 #$1, 4$ f !x" ! x 3 $ 6x 2 ! 9x ! 2 #$2, 3$ f !x" ! 2x 3 $ 3x 2 $ 12x ! 1 49. 0, 3$ f !x" ! x 3 $ 3x ! 1 #0, 3$ f !x" ! 3x 2 $ 12x ! 5 f f &!x" ! 100 cos2x 10 ! x 2 $ 1 f &!x" ! 5e$0.1 & x& sinx $ 1 f f f!x" ! x $2 ln x f!x" ! x 2e$3x t!" ! 4 $ tan f !" ! 2 cos ! sin2 41. t!x" ! x 1%3 $ x$2%3 F!x" ! x 4%5!x $ 4"2 39. t!x" ! s1 $ x 2 h!t" ! t 3%4 $ 2t 1%4 Event Time (s) Velocity (ft%s) Launch 0 0 Begin roll maneuver 10 185 End roll maneuver 15 319 Throttle to 89% 20 447 Throttle to 67% 32 742 Throttle to 104% 59 1325 Maximum dynamic pressure 62 1445 Solid rocket booster separation 125 4151 APPLIED PROJECT THE CALCULUS OF RAINBOWS \|\|\|\| 279 (b) What is the absolute maximum value of on the interval? (c) Sketch the graph of on the interval . 74. Show that 5 is a critical number of the function but does not have a local extreme value at 5. 75. Prove that the function has neither a local maximum nor a local minimum. 76. If has a minimum value at , show that the function has a maximum value at . 77. Prove Fermat’s Theorem for the case in which has a local minimum at . A cubic function is a polynomial of degree 3; that is, it has the form , where . (a) Show that a cubic function can have two, one, or no critical number(s). Give examples and sketches to illustrate the three possibilities. (b) How many local extreme values can a cubic function have? a " 0 f !x" ! ax 3 ! bx 2 ! cx ! d 78. c f c t!x" ! $f !x" c f f !x" ! x 101 ! x 51 ! x ! 1 t t!x" ! 2 ! !x $ 5"3 #0, r0$ v v 73. When a foreign object lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. This is accompanied by a contraction of the trachea, making a narrower channel for the expelled air to ﬂow through. For a given amount of air to escape in a ﬁxed time, it must move faster through the narrower channel than the wider one. The greater the velocity of the airstream, the greater the force on the foreign object. X rays show that the radius of the circular tracheal tube contracts to about two-thirds of its normal radius during a cough. According to a mathematical model of coughing, the velocity of the airstream is related to the radius of the trachea by the equation where is a constant and is the normal radius of the trachea. The restriction on is due to the fact that the tracheal wall stiff-ens under pressure and a contraction greater than is prevented (otherwise the person would suffocate). (a) Determine the value of in the interval at which has an absolute maximum. How does this compare with experimental evidence? v [ 1 2r0, r0] r 1 2r0 r r0 k 1 2r0 % r % r0 v!r" ! k!r0 $ r"r 2 r v Rainbows are created when raindrops scatter sunlight. They have fascinated mankind since ancient times and have inspired attempts at scientiﬁc explanation since the time of Aristotle. In this project we use the ideas of Descartes and Newton to explain the shape, location, and colors of rainbows. 1. The ﬁgure shows a ray of sunlight entering a spherical raindrop at . Some of the light is reﬂected, but the line shows the path of the part that enters the drop. Notice that the light is refracted toward the normal line and in fact Snell’s Law says that , where is the angle of incidence, is the angle of refraction, and is the index of refraction for water. At some of the light passes through the drop and is refracted into the air, but the line shows the part that is reﬂected. (The angle of incidence equals the angle of reﬂection.) When the ray reaches , part of it is reﬂected, but for the time being we are more interested in the part that leaves the raindrop at . (Notice that it is refracted away from the normal line.) The angle of deviation is the amount of clockwise rotation that the ray has undergone during this three-stage process. Thus Show that the minimum value of the deviation is and occurs when . The signiﬁcance of the minimum deviation is that when we have , so . This means that many rays with become deviated by approximately the same amount. It is the concentration of rays coming from near the direction of minimum deviation that creates the brightness of the primary rainbow. The ﬁgure at the left shows that the angle of elevation from the observer up to the highest point on the rainbow is . (This angle is called the rainbow angle.) 2. Problem 1 explains the location of the primary rainbow, but how do we explain the colors? Sunlight comprises a range of wavelengths, from the red range through orange, yellow, 180, $ 138, ! 42, - ' 59.4, .D%.- ' 0 D&!-" ' 0 - ' 59.4, - ' 59.4, D!-" ' 138, D!-" ! !- $ /" ! !# $ 2/" ! !- $ /" ! # ! 2- $ 4/ D!-" C C BC B k ' 4 3 / -sin - ! k sin / AO AB A THE CALCULUS OF RAINBOWS A P P L I E D P R O J E C T å å D(å) ∫ A from sun Formation of the primary rainbow to observer C B O ∫ ∫ ∫ rays from sun rays from sun 42° 138° observer green, blue, indigo, and violet. As Newton discovered in his prism experiments of 1666, the index of refraction is different for each color. (The effect is called dispersion.) For red light the refractive index is whereas for violet light it is . By repeating the calculation of Problem 1 for these values of , show that the rainbow angle is about for the red bow and for the violet bow. So the rainbow really consists of seven individual bows corresponding to the seven colors. 3. Perhaps you have seen a fainter secondary rainbow above the primary bow. That results from the part of a ray that enters a raindrop and is refracted at , reﬂected twice (at and ), and refracted as it leaves the drop at (see the ﬁgure). This time the deviation angle is the total amount of counterclockwise rotation that the ray undergoes in this four-stage process. Show that and has a minimum value when Taking , show that the minimum deviation is about and so the rainbow angle for the secondary rainbow is about , as shown in the ﬁgure. 4. Show that the colors in the secondary rainbow appear in the opposite order from those in the primary rainbow. 42° 51° 51 129 k 4 3 cos k 2 1 8 D D 2 6 2 D D C B A 40.6 42.3 k k 1.3435 k 1.3318 Formation of the secondary rainbow D C B A å å ∫ ∫ ∫ ∫∫ ∫ to observer from sun THE MEAN VALUE THEOREM We will see that many of the results of this chapter depend on one central fact, which is called the Mean Value Theorem. But to arrive at the Mean Value Theorem we ﬁrst need the following result. ROLLE’S THEOREM Let be a function that satisﬁes the following three hypotheses: 1. is continuous on the closed interval . 2. is differentiable on the open interval . 3. Then there is a number in such that . f c 0 a, b c f a f b a, b f a, b f f 4.2 N Rolle’s Theorem was ﬁrst published in 1691 by the French mathematician Michel Rolle (1652–1719) in a book entitled Méthode pour résoudre les égalitéz. He was a vocal critic of the methods of his day and attacked calculus as being a “collection of ingenious fallacies.” Later, however, he became convinced of the essential correctness of the methods of calculus. © C. Donald Ahrens 280 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Before giving the proof let’s take a look at the graphs of some typical functions that sat-isfy the three hypotheses. Figure 1 shows the graphs of four such functions. In each case it appears that there is at least one point on the graph where the tangent is hori-zontal and therefore . Thus Rolle’s Theorem is plausible. PROOF There are three cases: CASE I N , a constant Then , so the number can be taken to be any number in . CASE II N for some x in [as in Figure 1(b) or (c)] By the Extreme Value Theorem (which we can apply by hypothesis 1), has a maxi-mum value somewhere in . Since , it must attain this maximum value at a number in the open interval . Then has a local maximum at and, by hypoth-esis 2, is differentiable at . Therefore by Fermat’s Theorem. CASE III N for some x in [as in Figure 1(c) or (d)] By the Extreme Value Theorem, has a minimum value in and, since , it attains this minimum value at a number in . Again by Fermat’s Theorem. M EXAMPLE 1 Let’s apply Rolle’s Theorem to the position function of a moving object. If the object is in the same place at two different instants and , then . Rolle’s Theorem says that there is some instant of time between and when ; that is, the velocity is 0. (In particular, you can see that this is true when a ball is thrown directly upward.) M EXAMPLE 2 Prove that the equation has exactly one real root. SOLUTION First we use the Intermediate Value Theorem (2.5.10) to show that a root exists. Let . Then and . Since is a polynomi-al, it is continuous, so the Intermediate Value Theorem states that there is a number between 0 and 1 such that . Thus the given equation has a root. To show that the equation has no other real root, we use Rolle’s Theorem and argue by contradiction. Suppose that it had two roots and . Then and, since is a polynomial, it is differentiable on and continuous on . Thus, by Rolle’s Theorem, there is a number between and such that . But (since ) so can never be 0. This gives a contradiction. Therefore the equation can’t have two real roots. M f'"x# x 2 ( 0 for all x f'"x# ! 3x 2 % 1 ( 1 f'"c# ! 0 b a c %a, b& "a, b# f f"a# ! 0 ! f"b# b a f"c# ! 0 c f f"1# ! 1 ) 0 f"0# ! #1 0 f"x# ! x 3 % x # 1 x 3 % x # 1 ! 0 f'"c# ! 0 b a t ! c f"a# ! f"b# t ! b t ! a s ! f"t# f '"c# ! 0 "a, b# c f "a# ! f "b# %a, b& f "a, b# f"x# f"a# f'"c# ! 0 c f c f "a, b# c f"a# ! f"b# %a, b& f "a, b# f "x# ) f "a# "a, b# c f '"x# ! 0 f "x# ! k FIGURE 1 (a) b a c¡ c™ x y 0 (b) a c b x y 0 (c) b a c¡ c™ x y 0 (d) b a c y x 0 f'"c# ! 0 "c, f"c## SECTION 4.2 THE MEAN VALUE THEOREM \|\|\|\| 281 N Take cases N Figure 2 shows a graph of the function discussed in Example 2. Rolle’s Theorem shows that, no matter how much we enlarge the viewing rectangle, we can never ﬁnd a second -intercept. x f "x# ! x 3 % x # 1 FIGURE 2 2 3 _3 2 Our main use of Rolle’s Theorem is in proving the following important theorem, which was ﬁrst stated by another French mathematician, Joseph-Louis Lagrange. THE MEAN VALUE THEOREM Let be a function that satisﬁes the following hypotheses: 1. is continuous on the closed interval . 2. is differentiable on the open interval . Then there is a number in such that or, equivalently, Before proving this theorem, we can see that it is reasonable by interpreting it geomet-rically. Figures 3 and 4 show the points and on the graphs of two dif-ferentiable functions. The slope of the secant line is which is the same expression as on the right side of Equation 1. Since is the slope of the tangent line at the point , the Mean Value Theorem, in the form given by Equa-tion 1, says that there is at least one point on the graph where the slope of the tangent line is the same as the slope of the secant line . In other words, there is a point where the tangent line is parallel to the secant line . PROOF We apply Rolle’s Theorem to a new function deﬁned as the difference between and the function whose graph is the secant line . Using Equation 3, we see that the equation of the line can be written as or as y ! f"a# % f"b# # f"a# b # a "x # a# y # f"a# ! f"b# # f"a# b # a "x # a# AB AB f h FIGURE 3 FIGURE 4 0 x y a c b B{b, f(b)} P{c, f(c)} A{a, f(a)} 0 x y c¡ c™ B P¡ A P™ b a AB P AB P"c, f"c## "c, f"c## f'"c# mAB ! f"b# # f"a# b # a 3 AB B"b, f"b## A"a, f"a## f"b# # f"a# ! f'"c#"b # a# 2 f'"c# ! f"b# # f"a# b # a 1 "a, b# c "a, b# f %a, b& f f 282 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION N The Mean Value Theorem is an example of what is called an existence theorem. Like the Intermediate Value Theorem, the Extreme Value Theorem, and Rolle’s Theorem, it guarantees that there exists a number with a certain property, but it doesn’t tell us how to ﬁnd the number. So, as shown in Figure 5, First we must verify that satisﬁes the three hypotheses of Rolle’s Theorem. 1. The function is continuous on because it is the sum of and a ﬁrst-degree polynomial, both of which are continuous. 2. The function is differentiable on because both and the ﬁrst-degree poly-nomial are differentiable. In fact, we can compute directly from Equation 4: (Note that and are constants.) 3. Therefore, . Since satisﬁes the hypotheses of Rolle’s Theorem, that theorem says there is a num-ber in such that . Therefore and so M EXAMPLE 3 To illustrate the Mean Value Theorem with a speciﬁc function, let’s con-sider . Since is a polynomial, it is continuous and differ-entiable for all , so it is certainly continuous on and differentiable on . Therefore, by the Mean Value Theorem, there is a number in such that Now , and , so this equation becomes which gives , that is, . But must lie in , so . Figure 6 illustrates this calculation: The tangent line at this value of is parallel to the secant line . M EXAMPLE 4 If an object moves in a straight line with position function , then the average velocity between and is f"b# # f"a# b # a t ! b t ! a s ! f"t# V OB c c ! 2's3 "0, 2# c c ! +2's3 c 2 ! 4 3 6 ! "3c 2 # 1#2 ! 6c 2 # 2 f'"x# ! 3x 2 # 1 f"2# ! 6, f"0# ! 0 f"2# # f"0# ! f'"c#"2 # 0# "0, 2# c "0, 2# %0, 2& x f f"x# ! x 3 # x, a ! 0, b ! 2 V f'"c# ! f"b# # f"a# b # a 0 ! h'"c# ! f'"c# # f"b# # f"a# b # a h'"c# ! 0 "a, b# c h h"a# ! h"b# ! f"b# # f"a# # % f"b# # f"a#& ! 0 h"b# ! f"b# # f"a# # f"b# # f"a# b # a "b # a# h"a# ! f"a# # f"a# # f"b# # f"a# b # a "a # a# ! 0 % f"b# # f"a#&'"b # a# f"a# h'"x# ! f'"x# # f"b# # f"a# b # a h' f "a, b# h f %a, b& h h h"x# ! f"x# # f"a# # f"b# # f"a# b # a "x # a# 4 SECTION 4.2 THE MEAN VALUE THEOREM \|\|\|\| 283 FIGURE 5 0 x y x h(x) y=ƒ ƒ A B f(a)+ (x-a) f(b)-f(a) b-a The Mean Value Theorem was ﬁrst formulated by Joseph-Louis Lagrange (1736–1813), born in Italy of a French father and an Italian mother. He was a child prodigy and became a professor in Turin at the tender age of 19. Lagrange made great con-tributions to number theory, theory of functions, theory of equations, and analytical and celestial mechanics. In particular, he applied calculus to the analysis of the stability of the solar system. At the invitation of Frederick the Great, he succeeded Euler at the Berlin Academy and, when Frederick died, Lagrange accepted King Louis XVI’s invitation to Paris, where he was given apartments in the Louvre and became a professor at the Ecole Poly-technique. Despite all the trappings of luxury and fame, he was a kind and quiet man, living only for science. LAGRANGE AND THE MEAN VALUE THEOREM FIGURE 6 y=˛- x B x y c 2 O and the velocity at is . Thus the Mean Value Theorem (in the form of Equa-tion 1) tells us that at some time between and the instantaneous velocity is equal to that average velocity. For instance, if a car traveled 180 km in 2 hours, then the speedometer must have read 90 km'h at least once. In general, the Mean Value Theorem can be interpreted as saying that there is a num-ber at which the instantaneous rate of change is equal to the average rate of change over an interval. M The main signiﬁcance of the Mean Value Theorem is that it enables us to obtain infor-mation about a function from information about its derivative. The next example provides an instance of this principle. EXAMPLE 5 Suppose that and for all values of . How large can possibly be? SOLUTION We are given that is differentiable (and therefore continuous) everywhere. In particular, we can apply the Mean Value Theorem on the interval . There exists a number such that so We are given that for all , so in particular we know that . Multiply-ing both sides of this inequality by 2, we have , so The largest possible value for is 7. M The Mean Value Theorem can be used to establish some of the basic facts of differen-tial calculus. One of these basic facts is the following theorem. Others will be found in the following sections. THEOREM If for all in an interval , then is constant on . PROOF Let and be any two numbers in with . Since is differen-tiable on , it must be differentiable on and continuous on . By apply-ing the Mean Value Theorem to on the interval , we get a number such that and Since for all , we have , and so Equation 6 becomes Therefore has the same value at any two numbers and in . This means that is constant on . M COROLLARY If for all in an interval , then is con-stant on ; that is, where is a constant. c f"x# ! t"x# % c "a, b# f # t "a, b# x f'"x# ! t'"x# 7 "a, b# f "a, b# x2 x1 f f"x2# ! f"x1# or f"x2# # f"x1# ! 0 f'"c# ! 0 x f'"x# ! 0 f"x2# # f"x1# ! f'"c#"x2 # x1# 6 x1 c x2 c %x1, x2& f %x1, x2& "x1, x2# "a, b# f x1 x2 "a, b# x2 x1 "a, b# f "a, b# x f'"x# ! 0 5 f"2# f"2# ! #3 % 2f'"c# , #3 % 10 ! 7 2f'"c# , 10 f'"c# , 5 x f'"x# , 5 f"2# ! f"0# % 2f'"c# ! #3 % 2f'"c# f"2# # f"0# ! f'"c#"2 # 0# c %0, 2& f f"2# x f'"x# , 5 f"0# ! #3 V f'"c# b a t ! c f'"c# t ! c 284 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION PROOF Let . Then for all in . Thus, by Theorem 5, is constant; that is, is constant. M Care must be taken in applying Theorem 5. Let The domain of is and for all in . But is obviously not a constant function. This does not contradict Theorem 5 because is not an interval. Notice that is constant on the interval and also on the interval . EXAMPLE 6 Prove the identity . SOLUTION Although calculus isn’t needed to prove this identity, the proof using calculus is quite simple. If , then for all values of . Therefore , a constant. To determine the value of , we put [because we can evaluate exactly]. Then Thus . M tan#1x % cot#1x ! &'2 C ! f"1# ! tan#1 1 % cot#1 1 ! & 4 % & 4 ! & 2 f"1# x ! 1 C f"x# ! C x f'"x# ! 1 1 % x 2 # 1 1 % x 2 ! 0 f"x# ! tan#1x % cot#1x tan#1x % cot#1x ! &'2 "#-, 0# "0, -# f D f D x f'"x# ! 0 D ! (x)x " 0 f f"x# ! x )x) !+ 1 #1 if x ) 0 if x 0 NOTE f # t F "a, b# x F'"x# ! f'"x# # t'"x# ! 0 F"x# ! f"x# # t"x# SECTION 4.2 THE MEAN VALUE THEOREM \|\|\|\| 285 7. Use the graph of to estimate the values of that satisfy the conclusion of the Mean Value Theorem for the interval . 8. Use the graph of given in Exercise 7 to estimate the values of that satisfy the conclusion of the Mean Value Theorem for the interval . %1, 7& c f y y =ƒ 1 x 0 1 %0, 8& c f 1–4 Verify that the function satisﬁes the three hypotheses of Rolle’s Theorem on the given interval. Then ﬁnd all numbers that satisfy the conclusion of Rolle’s Theorem. 1. 2. 3. 4. Let . Show that but there is no number in such that . Why does this not contradict Rolle’s Theorem? 6. Let . Show that but there is no number in such that . Why does this not contradict Rolle’s Theorem? f '"c# ! 0 "0, &# c f "0# ! f "&# f "x# ! tan x f '"c# ! 0 "#1, 1# c f "#1# ! f "1# f "x# ! 1 # x 2'3 5. %&'8, 7&'8& f "x# ! cos 2x, %0, 9& f "x# ! sx # 1 3 x, %0, 3& f "x# ! x 3 # x 2 # 6x % 2, %1, 3& f "x# ! 5 # 12x % 3x 2, c EXERCISES 4.2 (b) Suppose is twice differentiable on and has three roots. Show that has at least one real root. (c) Can you generalize parts (a) and (b)? If and for , how small can possibly be? 24. Suppose that for all values of . Show that . Does there exist a function such that , , and for all ? 26. Suppose that and are continuous on and differentiable on . Suppose also that and for . Prove that . [Hint: Apply the Mean Value Theorem to the function .] 27. Show that if . 28. Suppose is an odd function and is differentiable every-where. Prove that for every positive number , there exists a number in such that . 29. Use the Mean Value Theorem to prove the inequality 30. If (c a constant) for all , use Corollary 7 to show that for some constant . 31. Let and Show that for all in their domains. Can we conclude from Corollary 7 that is constant? 32. Use the method of Example 6 to prove the identity 33. Prove the identity 34. At 2:00 PM a car’s speedometer reads 30 mi'h. At 2:10 PM it reads 50 mi'h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mi'h . Two runners start a race at the same time and ﬁnish in a tie. Prove that at some time during the race they have the same speed. [Hint: Consider , where and are the position functions of the two runners.] 36. A number a is called a ﬁxed point of a function if . Prove that if for all real numbers x, then has at most one ﬁxed point. f f '"x# " 1 f "a# ! a f h t f "t# ! t"t# # h"t# 35. 2 arcsin x # 1 x % 1 ! 2 arctan sx # & 2 x ( 0 2 sin#1x ! cos#1"1 # 2x 2# f # t x f '"x# ! t'"x# t"x# ! 1 x 1 % 1 x if if x ) 0 x 0 f "x# ! 1'x d f "x# ! cx % d x f '"x# ! c for all a and b ) sin a # sin b) , ) a # b) f '"c# ! f "b#'b "#b, b# c b f x ) 0 s1 % x 1 % 1 2 x h ! f # t f "b# t"b# a x b f '"x# t'"x# f "a# ! t"a# "a, b# %a, b& t f x f '"x# , 2 f "2# ! 4 f "0# ! #1 f 25. 18 , f "8# # f "2# , 30 x 3 , f '"x# , 5 f "4# 1 , x , 4 f '"x# ( 2 f "1# ! 10 23. f . ! f ; 9. (a) Graph the function in the viewing rect-angle by . (b) Graph the secant line that passes through the points and on the same screen with . (c) Find the number that satisﬁes the conclusion of the Mean Value Theorem for this function and the interval . Then graph the tangent line at the point and notice that it is parallel to the secant line. ; 10. (a) In the viewing rectangle by , graph the function and its secant line through the points and . Use the graph to estimate the -coordinates of the points where the tangent line is parallel to the secant line. (b) Find the exact values of the numbers that satisfy the conclusion of the Mean Value Theorem for the interval and compare with your answers to part (a). 11–14 Verify that the function satisﬁes the hypotheses of the Mean Value Theorem on the given interval. Then ﬁnd all numbers that satisfy the conclusion of the Mean Value Theorem. , 12. , 13. 14. , 15. Let . Show that there is no value of in such that . Why does this not contradict the Mean Value Theorem? 16. Let . Show that there is no value of such that . Why does this not con-tradict the Mean Value Theorem? 17. Show that the equation has exactly one real root. 18. Show that the equation has exactly one real root. Show that the equation has at most one root in the interval . 20. Show that the equation has at most two real roots. 21. (a) Show that a polynomial of degree 3 has at most three real roots. (b) Show that a polynomial of degree has at most real roots. 22. (a) Suppose that is differentiable on and has two roots. Show that has at least one root. f ' ! f n n x 4 % 4x % c ! 0 %#2, 2& x 3 # 15x % c ! 0 19. 2x # 1 # sin x ! 0 1 % 2x % x 3 % 4x 5 ! 0 f "3# # f "0# ! f '"c#"3 # 0# c f "x# ! 2 # )2x # 1) f "4# # f "1# ! f '"c#"4 # 1# "1, 4# c f "x# ! "x # 3##2 %1, 4& f "x# ! x x % 2 f "x# ! e#2x, %0, 3& %0, 2& f "x# ! x 3 % x # 1 %#1, 1& f "x# ! 3x 2 % 2x % 5 11. c %#2, 2& c x "2, 4# "#2, #4# f "x# ! x 3 # 2x %#5, 5& %#3, 3& "c, f "c## %1, 8& f c f "8, 8.5# "1, 5# %0, 10& %0, 10& f "x# ! x % 4'x 286 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH Many of the applications of calculus depend on our ability to deduce facts about a func-tion f from information concerning its derivatives. Because represents the slope of the curve at the point , it tells us the direction in which the curve proceeds at each point. So it is reasonable to expect that information about will provide us with information about . WHAT DOES SAY ABOUT ? To see how the derivative of can tell us where a function is increasing or decreasing, look at Figure 1. (Increasing functions and decreasing functions were deﬁned in Section 1.1.) Between A and B and between C and D, the tangent lines have positive slope and so . Between B and C, the tangent lines have negative slope and so . Thus it appears that f increases when is positive and decreases when is negative. To prove that this is always the case, we use the Mean Value Theorem. INCREASING/DECREASING TEST (a) If on an interval, then is increasing on that interval. (b) If on an interval, then is decreasing on that interval. PROOF (a) Let and be any two numbers in the interval with . According to the deﬁ-nition of an increasing function (page 20) we have to show that . Because we are given that , we know that is differentiable on . So, by the Mean Value Theorem there is a number c between and such that Now by assumption and because . Thus the right side of Equation 1 is positive, and so This shows that f is increasing. Part (b) is proved similarly. M EXAMPLE 1 Find where the function is increasing and where it is decreasing. SOLUTION To use the I'D Test we have to know where and where . This depends on the signs of the three factors of , namely, , , and . We divide the real line into intervals whose endpoints are the critical numbers , and and arrange our work in a chart. A plus sign indicates that the given expression is posi-tive, and a minus sign indicates that it is negative. The last column of the chart gives the 2 #1, 0 x % 1 x # 2 12x f'"x# f'"x# 0 f'"x# ) 0 f'"x# ! 12x 3 # 12x 2 # 24x ! 12x"x # 2#"x % 1# f"x# ! 3x 4 # 4x 3 # 12x 2 % 5 V f"x1# f"x2# or f"x2# # f"x1# ) 0 x1 x2 x2 # x1 ) 0 f'"c# ) 0 f"x2# # f"x1# ! f'"c#"x2 # x1# 1 x2 x1 %x1, x2& f f'"x# ) 0 f"x1# f"x2# x1 x2 x2 x1 f f'"x# 0 f f'"x# ) 0 f'"x# f'"x# f'"x# 0 f'"x# ) 0 f f f ' f"x# f'"x# "x, f"x## y ! f"x# f'"x# 4.3 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH \|\|\|\| 287 D A B C y 0 x FIGURE 1 N Let’s abbreviate the name of this test to the I/D Test. conclusion based on the I'D Test. For instance, for , so is decreas-ing on (0, 2). (It would also be true to say that f is decreasing on the closed interval .) The graph of f shown in Figure 2 conﬁrms the information in the chart. M Recall from Section 4.1 that if has a local maximum or minimum at , then must be a critical number of (by Fermat’s Theorem), but not every critical number gives rise to a maximum or a minimum. We therefore need a test that will tell us whether or not has a local maximum or minimum at a critical number. You can see from Figure 2 that is a local maximum value of because increases on and decreases on . Or, in terms of derivatives, for and for . In other words, the sign of changes from positive to negative at . This observation is the basis of the following test. THE FIRST DERIVATIVE TEST Suppose that is a critical number of a continuous function . (a) If changes from positive to negative at , then has a local maximum at . (b) If changes from negative to positive at , then has a local minimum at . (c) If does not change sign at (for example, if is positive on both sides of c or negative on both sides), then has no local maximum or minimum at . The First Derivative Test is a consequence of the I'D Test. In part (a), for instance, since the sign of changes from positive to negative at c, is increasing to the left of c and decreasing to the right of c. It follows that has a local maximum at c. It is easy to remember the First Derivative Test by visualizing diagrams such as those in Figure 3. FIGURE 3 0 x y c fª(x)>0 fª(x)<0 (a) Local maximum c 0 x y fª(x)<0 fª(x)<0 (d) No maximum or minimum (c) No maximum or minimum c 0 x y fª(x)>0 fª(x)>0 c 0 x y fª(x)<0 fª(x)>0 (b) Local minimum f f f'"x# c f f' c f' c f c f' c f c f' f c 0 f'"x# 0 x 2 f'"x# 0 #1 x 0 f'"x# ) 0 "0, 2# "#1, 0# f f f"0# ! 5 f f c c f %0, 2& f 0 x 2 f'"x# 0 288 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Interval x # 2 x % 1 f # # # # decreasing on (#-, #1) # # % % increasing on (#1, 0) % # % # decreasing on (0, 2) % % % % increasing on (2, -) x ) 2 0 x 2 #1 x 0 x #1 f '"x# 12x 20 _30 _2 3 FIGURE 2 EXAMPLE 2 Find the local minimum and maximum values of the function f in Example 1. SOLUTION From the chart in the solution to Example 1 we see that changes from neg-ative to positive at #1, so is a local minimum value by the First Derivative Test. Similarly, changes from negative to positive at 2, so is also a local minimum value. As previously noted, is a local maximum value because changes from positive to negative at 0. M EXAMPLE 3 Find the local maximum and minimum values of the function SOLUTION To ﬁnd the critical numbers of , we differentiate: So when . The solutions of this equation are and . Because is differentiable everywhere, the only critical numbers are and and so we analyze in the following table. Because changes from positive to negative at , the First Derivative Test tells us that there is a local maximum at and the local maximum value is Likewise, changes from negative to positive at and so is a local minimum value. The graph of in Figure 4 supports our conclusion. M FIGURE 4 y=x+2 sin x 6 0 2π t t"4&'3# ! 4& 3 % 2 sin 4& 3 ! 4& 3 % 2,# s3 2 - ! 4& 3 # s3 $ 2.46 4&'3 t'"x# t"2&'3# ! 2& 3 % 2 sin 2& 3 ! 2& 3 % 2, s3 2 - ! 2& 3 % s3 $ 3.83 2&'3 2&'3 t'"x# t 4&'3 2&'3 t 4&'3 2&'3 cos x ! # 1 2 t'"x# ! 0 t'"x# ! 1 % 2 cos x t 0 , x , 2& t"x# ! x % 2 sin x f'"x# f"0# ! 5 f"2# ! #27 f' f"#1# ! 0 f'"x# V SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH \|\|\|\| 289 Interval % increasing on # decreasing on % increasing on "4&'3, 2&# 4&'3 x 2& "2'3, 4&'3# 2&'3 x 4&'3 "0, 2&'3# 0 x 2&'3 t t'"x# ! 1 % 2 cos x N The + signs in the table come from the fact that when . From the graph of , this is true in the indicated intervals. y ! cos x cos x ) # 1 2 t'"x# ) 0 WHAT DOES SAY ABOUT ? Figure 5 shows the graphs of two increasing functions on . Both graphs join point to point but they look different because they bend in different directions. How can we distinguish between these two types of behavior? In Figure 6 tangents to these curves have been drawn at several points. In (a) the curve lies above the tangents and is called con-cave upward on . In (b) the curve lies below the tangents and is called concave downward on . DEFINITION If the graph of lies above all of its tangents on an interval , then it is called concave upward on . If the graph of lies below all of its tangents on I, it is called concave downward on . Figure 7 shows the graph of a function that is concave upward (abbreviated CU) on the intervals , , and and concave downward (CD) on the intervals , , and . Let’s see how the second derivative helps determine the intervals of concavity. Looking at Figure 6(a), you can see that, going from left to right, the slope of the tangent increases. FIGURE 7 a b c d e p q B C D P x y 0 CD CU CD CU CD CU !p, q" !c, d" !a, b" !e, p" !d, e" !b, c" I f I I f FIGURE 5 FIGURE 6 a b f A B x y 0 a b g A B x y 0 g A B x y 0 f A B x y 0 (a) (b) (a) Concave upward (b) Concave downward !a, b" t !a, b" f B A !a, b" f f ! 290 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION This means that the derivative is an increasing function and therefore its derivative is positive. Likewise, in Figure 6(b) the slope of the tangent decreases from left to right, so decreases and therefore is negative. This reasoning can be reversed and suggests that the following theorem is true. A proof is given in Appendix F with the help of the Mean Value Theorem. CONCAVITY TEST (a) If for all in , then the graph of is concave upward on . (b) If for all in , then the graph of is concave downward on . EXAMPLE 4 Figure 8 shows a population graph for Cyprian honeybees raised in an apiary. How does the rate of population increase change over time? When is this rate highest? Over what intervals is P concave upward or concave downward? SOLUTION By looking at the slope of the curve as t increases, we see that the rate of increase of the population is initially very small, then gets larger until it reaches a maxi-mum at about t ! 12 weeks, and decreases as the population begins to level off. As the population approaches its maximum value of about 75,000 (called the carrying capac-ity), the rate of increase, , approaches 0. The curve appears to be concave upward on (0, 12) and concave downward on (12, 18). M In Example 4, the population curve changed from concave upward to concave down-ward at approximately the point (12, 38,000). This point is called an inﬂection point of the curve. The signiﬁcance of this point is that the rate of population increase has its maximum value there. In general, an inﬂection point is a point where a curve changes its direction of concavity. DEFINITION A point on a curve is called an inﬂection point if is continuous there and the curve changes from concave upward to concave down-ward or from concave downward to concave upward at . For instance, in Figure 7, , and are the points of inﬂection. Notice that if a curve has a tangent at a point of inﬂection, then the curve crosses its tangent there. In view of the Concavity Test, there is a point of inﬂection at any point where the sec-ond derivative changes sign. P B, C, D P f y ! f!x" P P"!t" t P 3 20 0 Time (in weeks) 6 9 12 15 40 60 80 Number of bees (in thousands) 18 FIGURE 8 I f I x f !!x" # 0 I f I x f !!x" $ 0 f ! f" f ! f" SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH \|\|\|\| 291 EXAMPLE 5 Sketch a possible graph of a function that satisﬁes the following conditions: SOLUTION Condition (i) tells us that is increasing on and decreasing on . Condition (ii) says that is concave upward on and , and concave down-ward on . From condition (iii) we know that the graph of has two horizontal asymptotes: and . We ﬁrst draw the horizontal asymptote as a dashed line (see Figure 9). We then draw the graph of approaching this asymptote at the far left, increasing to its maxi-mum point at and decreasing toward the x-axis at the far right. We also make sure that the graph has inﬂection points when and 2. Notice that we made the curve bend upward for and , and bend downward when x is between %2 and 2. M Another application of the second derivative is the following test for maximum and minimum values. It is a consequence of the Concavity Test. THE SECOND DERIVATIVE TEST Suppose is continuous near . (a) If and , then has a local minimum at . (b) If and , then has a local maximum at . For instance, part (a) is true because near c and so is concave upward near c. This means that the graph of lies above its horizontal tangent at c and so has a local minimum at c. (See Figure 10.) EXAMPLE 6 Discuss the curve with respect to concavity, points of inﬂection, and local maxima and minima. Use this information to sketch the curve. SOLUTION If , then To ﬁnd the critical numbers we set and obtain and . To use the Second Derivative Test we evaluate at these critical numbers: Since and , is a local minimum. Since , the Second Derivative Test gives no information about the critical number 0. But since for and also for , the First Derivative Test tells us that does not have a local maximum or minimum at 0. [In fact, the expression for shows that f decreases to the left of 3 and increases to the right of 3.] f"!x" f 0 # x # 3 x # 0 f"!x" # 0 f !!0" ! 0 f!3" ! %27 f !!3" $ 0 f"!3" ! 0 f !!3" ! 36 $ 0 f !!0" ! 0 f ! x ! 3 x ! 0 f"!x" ! 0 f !!x" ! 12x 2 % 24x ! 12x!x % 2" f"!x" ! 4x 3 % 12x 2 ! 4x 2!x % 3" f!x" ! x 4 % 4x 3 y ! x 4 % 4x 3 V f f f f !!x" $ 0 c f f !!c" # 0 f"!c" ! 0 c f f !!c" $ 0 f"!c" ! 0 c f ! x $ 2 x # %2 x ! %2 x ! 1 f y ! %2 y ! 0 y ! %2 f !%2, 2" !2, &" !%&, %2" f !1, &" !%&, 1" f !iii" lim x l%& f!x" ! %2, lim x l & f!x" ! 0 !ii" f !!x" $ 0 on !%&, %2" and !2, &", f !!x" # 0 on !%2, 2" !i" f"!x" $ 0 on !%&, 1", f"!x" # 0 on !1, &" f V 292 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION FIGURE 9 x y=_2 0 1 2 -2 y fª(c)=0 f(c) ƒ c P x x y 0 FIGURE 10 f·(c)>0, f is concave upward f Since when or , we divide the real line into intervals with these numbers as endpoints and complete the following chart. The point is an inﬂection point since the curve changes from concave upward to concave downward there. Also, is an inﬂection point since the curve changes from concave downward to concave upward there. Using the local minimum, the intervals of concavity, and the inﬂection points, we sketch the curve in Figure 11. M The Second Derivative Test is inconclusive when . In other words, at such a point there might be a maximum, there might be a minimum, or there might be nei-ther (as in Example 6). This test also fails when does not exist. In such cases the First Derivative Test must be used. In fact, even when both tests apply, the First Derivative Test is often the easier one to use. EXAMPLE 7 Sketch the graph of the function . SOLUTION You can use the differentiation rules to check that the ﬁrst two derivatives are Since when and does not exist when or , the critical numbers are , and . To ﬁnd the local extreme values we use the First Derivative Test. Since changes from negative to positive at 0, is a local minimum. Since changes from positive to negative at 4, is a local maximum. The sign of does not change at 6, so there is no minimum or maximum there. (The Second Derivative Test could be used at 4, but not at 0 or 6 since does not exist at either of these numbers.) Looking at the expression for and noting that for all , we have for and for and for . So is concave down-ward on and and concave upward on , and the only inﬂection point is . The graph is sketched in Figure 12. Note that the curve has vertical tangents at and because as and as . M EXAMPLE 8 Use the ﬁrst and second derivatives of , together with asymp-totes, to sketch its graph. SOLUTION Notice that the domain of is , so we check for vertical asymptotes by computing the left and right limits as . As , we know that , t ! 1#x l & x l 0' x l 0 $x%x " 0& f f!x" ! e 1#x x l 6 x l 0 % f"!x"% l & !6, 0" !0, 0" !6, 0" !6, &" !0, 6" !%&, 0" f x $ 6 f !!x" $ 0 0 # x # 6 x # 0 f !!x" # 0 x x 4#3 ( 0 f !!x" f ! f" f!4" ! 25#3 f" f!0" ! 0 f" 6 0, 4 x ! 6 x ! 0 f"!x" x ! 4 f"!x" ! 0 f !!x" ! %8 x 4#3!6 % x"5#3 f"!x" ! 4 % x x 1#3!6 % x"2#3 f!x" ! x 2#3!6 % x"1#3 f !!c" f !!c" ! 0 NOTE !2, %16" !0, 0" 2 x ! 0 f !!x" ! 0 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH \|\|\|\| 293 Interval Concavity (%&, 0) ' upward (0, 2) % downward (2, &) ' upward f !!x" ! 12x!x % 2" FIGURE 11 x y 2 3 (2, _16) (3, _27) y=x$-4˛ inﬂection points (0, 0) Interval 4 % x f ' % ' % decreasing on (%&, 0) ' ' ' ' increasing on (0, 4) % ' ' % decreasing on (4, 6) % ' ' % decreasing on (6, &) x $ 6 4 # x # 6 0 # x # 4 x # 0 f "!x" !6 % x"2#3 x 1#3 FIGURE 12 y x 0 2 3 4 1 2 3 4 5 7 (4, 2%?#) y=x@?#(6-x)!?# N Try reproducing the graph in Figure 12 with a graphing calculator or computer. Some machines produce the complete graph, some produce only the portion to the right of the -axis, and some produce only the portion between and . For an explanation and cure, see Example 7 in Section 1.4. An equivalent expression that gives the correct graph is y ! !x 2"1#3 ! 6 % x % 6 % x% % 6 % x% 1#3 x ! 6 x ! 0 y so and this shows that is a vertical asymptote. As , we have , so As , we have and so This shows that is a horizontal asymptote. Now let’s compute the derivative. The Chain Rule gives Since and for all , we have for all . Thus is decreasing on and on . There is no critical number, so the function has no maximum or minimum. The second derivative is Since and , we have when and when . So the curve is concave downward on and concave upward on and on . The inﬂection point is . To sketch the graph of we ﬁrst draw the horizontal asymptote (as a dashed line), together with the parts of the curve near the asymptotes in a preliminary sketch [Figure 13(a)]. These parts reﬂect the information concerning limits and the fact that is decreasing on both and . Notice that we have indicated that as even though does not exist. In Figure 13(b) we ﬁnish the sketch by incorpo-rating the information concerning concavity and the inﬂection point. In Figure 13(c) we check our work with a graphing device. M (a) Preliminary sketch (b) Finished sketch FIGURE 13 (c) Computer confirmation 4 0 _3 3 x 0 y y=1 y=‰ inﬂection point x 0 y y=1 f!0" x l 0% f!x" l 0 !0, &" !%&, 0" f y ! 1 f (% 1 2, e%2) !0, &" (% 1 2, 0) (%&, % 1 2) x # % 1 2 f !!x" # 0 !x " 0" x $ % 1 2 f !!x" $ 0 x 4 $ 0 e 1#x $ 0 f !!x" ! % x 2e 1#x!%1#x 2" % e 1#x!2x" x 4 ! e 1#x!2x ' 1" x 4 !0, &" !%&, 0" f x " 0 f"!x" # 0 x " 0 x 2 $ 0 e 1#x $ 0 f"!x" ! % e 1#x x 2 y ! 1 lim x l)& e 1#x ! e 0 ! 1 1#x l 0 x l )& lim x l 0% e 1#x ! lim t l %& e t ! 0 t ! 1#x l %& x l 0% x ! 0 lim x l 0' e 1#x ! lim t l & e t ! & 294 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION In Module 4.3 you can practice using graphical information about to determine the shape of the graph of . f f" TEC SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH \|\|\|\| 295 (c) On what intervals is concave upward or concave down-ward? Explain. (d) What are the -coordinates of the inﬂection points of ? Why? 9–18 (a) Find the intervals on which is increasing or decreasing. (b) Find the local maximum and minimum values of . (c) Find the intervals of concavity and the inﬂection points. 9. 10. 12. 13. 14. , 15. 16. 18. 19–21 Find the local maximum and minimum values of using both the First and Second Derivative Tests. Which method do you prefer? 19. 20. 21. 22. (a) Find the critical numbers of . (b) What does the Second Derivative Test tell you about the behavior of at these critical numbers? (c) What does the First Derivative Test tell you? 23. Suppose is continuous on . (a) If and , what can you say about ? (b) If and , what can you say about ? 24–29 Sketch the graph of a function that satisﬁes all of the given conditions. 24. for all , vertical asymptote , if or , if , if or , if or , if , if or x $ 3 x # 1 f !!x" # 0 1 # x # 3 f !!x" $ 0 x $ 4 0 # x # 2 f "!x" # 0 2 # x # 4 x # 0 f "!x" $ 0 f "!0" ! f "!2" ! f "!4" ! 0 25. 1 # x # 3 f !!x" # 0 x $ 3 x # 1 f !!x" $ 0 x ! 1 x " 1 f "!x" $ 0 f f !!6" ! 0 f "!6" ! 0 f f !!2" ! %5 f "!2" ! 0 !%&, &" f ! f f !x" ! x 4!x % 1"3 f !x" ! x ' s1 % x f !x" ! x x 2 ' 4 f !x" ! x 5 % 5x ' 3 f f !x" ! sx e%x f !x" ! !ln x"#sx 17. f !x" ! x 2 ln x f !x" ! e2x ' e%x 0 x 2+ f !x" ! cos2x % 2 sin x f !x" ! sin x ' cos x, 0 x 2+ f !x" ! x 2 x 2 ' 3 f !x" ! x4 % 2x2 ' 3 11. f !x" ! 4x 3 ' 3x 2 % 6x ' 1 f !x" ! 2x 3 ' 3x 2 % 36x f f 3 y 0 x 5 7 1 9 y=fª(x) f x f 1–2 Use the given graph of to ﬁnd the following. (a) The open intervals on which is increasing. (b) The open intervals on which is decreasing. (c) The open intervals on which is concave upward. (d) The open intervals on which is concave downward. (e) The coordinates of the points of inﬂection. 1. 2. 3. Suppose you are given a formula for a function . (a) How do you determine where is increasing or decreasing? (b) How do you determine where the graph of is concave upward or concave downward? (c) How do you locate inﬂection points? 4. (a) State the First Derivative Test. (b) State the Second Derivative Test. Under what circum-stances is it inconclusive? What do you do if it fails? 5–6 The graph of the derivative of a function is shown. (a) On what intervals is increasing or decreasing? (b) At what values of x does have a local maximum or minimum? 6. The graph of the second derivative of a function is shown. State the -coordinates of the inﬂection points of . Give reasons for your answers. 8. The graph of the ﬁrst derivative of a function is shown. (a) On what intervals is increasing? Explain. (b) At what values of does have a local maximum or minimum? Explain. f x f f f " y=f·(x) 2 y 0 x 4 6 8 f x f f ! 7. 2 4 6 x y 0 y=fª(x) 2 4 6 x y 0 y=fª(x) 5. f f f f " f f f y 0 x 1 1 y 0 x 1 1 f f f f f EXERCISES 4.3 (d) Use the information from parts (a)–(c) to sketch the graph. Check your work with a graphing device if you have one. 33. 34. 35. 36. 37. 38. 40. 42. 43. , 44. , 45–52 (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inﬂection points. (e) Use the information from parts (a)–(d) to sketch the graph of . 45. 46. 47. 48. , 49. 50. 52. 53. Suppose the derivative of a function is . On what interval is increasing? 54. Use the methods of this section to sketch the curve , where is a positive constant. What do the members of this family of curves have in common? How do they differ from each other? ; 55–56 (a) Use a graph of to estimate the maximum and minimum values. Then ﬁnd the exact values. (b) Estimate the value of at which increases most rapidly. Then ﬁnd the exact value. 56. ; 57–58 (a) Use a graph of to give a rough estimate of the intervals of concavity and the coordinates of the points of inﬂection. (b) Use a graph of to give better estimates. 57. , 0 x 2+ f !x" ! cos x ' 1 2 cos 2x f ! f f !x" ! x 2e%x f !x" ! x ' 1 sx 2 ' 1 55. f x f a y ! x3 % 3a2x ' 2a3 f f "!x" ! !x ' 1"2!x % 3"5!x % 6"4 f f !x" ! earctan x f !x" ! e %1#!x'1" 51. f !x" ! e x 1 ' e x f !x" ! ln!1 % ln x" %+#2 # x # +#2 f !x" ! x tan x f !x" ! sx 2 ' 1 % x f !x" ! x2 !x % 2"2 f !x" ! x 2 x 2 % 1 f %2+ t 2+ f!t" ! t ' cos t 0 , 2+ f !," ! 2 cos , ' cos2, f !x" ! ln!x 4 ' 27" C!x" ! x1#3!x ' 4" 41. B!x" ! 3x 2#3 % x A!x" ! xsx ' 3 39. h!x" ! x5 % 2x 3 ' x h!x" ! !x ' 1"5 % 5x % 2 t!x" ! 200 ' 8x 3 ' x 4 f !x" ! 2 ' 2x 2 % x 4 f !x" ! 2 ' 3x % x 3 f !x" ! 2x 3 % 3x 2 % 12x 26. , if , if , if , if , inﬂection point 27. if , if , , , if 28. if , if , , , , if , if 29. and for all 30. Suppose , and and for all . (a) Sketch a possible graph for . (b) How many solutions does the equation have? Why? (c) Is it possible that ? Why? 31–32 The graph of the derivative of a continuous function is shown. (a) On what intervals is increasing or decreasing? (b) At what values of x does have a local maximum or minimum? (c) On what intervals is concave upward or downward? (d) State the x-coordinate(s) of the point(s) of inﬂection. (e) Assuming that , sketch a graph of f. 32. 33–44 (a) Find the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inﬂection points. y 0 x 2 4 6 8 _2 y=fª(x) 2 2 4 6 8 y 0 x _2 y=fª(x) 2 31. f !0" ! 0 f f f f f " f "!2" ! 1 3 f !x" ! 0 f x f !!x" # 0 f "!x" $ 0 f!3" ! 2, f "!3" ! 1 2 x f !!x" # 0 f"!x" # 0 x $ 3 f !!x" $ 0 0 # x # 3 f !!x" # 0 f !%x" ! %f !x" lim x l & f !x" ! 1 f "!2" ! 0 % x% $ 2 f "!x" # 0 % x% # 2 f "!x" $ 0 x " 2 f !!x" $ 0 lim x l 2 % f "!x"% ! & f "!%2" ! 0 % x% $ 2 f "!x" # 0 % x% # 2 f "!x" $ 0 !0, 1" %2 # x # 0 f !!x" # 0 % x% $ 2 f "!x" ! %1 1 # % x% # 2 f "!x" $ 0 % x% # 1 f "!x" # 0 f "!1" ! f "!%1" ! 0 296 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION ; 65. A drug response curve describes the level of medication in the bloodstream after a drug is administered. A surge func-tion is often used to model the response curve, reﬂecting an initial surge in the drug level and then a more gradual decline. If, for a particular drug, , and is measured in minutes, estimate the times corresponding to the inﬂection points and explain their signiﬁcance. If you have a graphing device, use it to graph the drug response curve. 66. The family of bell-shaped curves occurs in probability and statistics, where it is called the nor-mal density function. The constant is called the mean and the positive constant is called the standard deviation. For simplicity, let’s scale the function so as to remove the factor and let’s analyze the special case where . So we study the function (a) Find the asymptote, maximum value, and inﬂection points of . (b) What role does play in the shape of the curve? ; (c) Illustrate by graphing four members of this family on the same screen. Find a cubic function that has a local maximum value of at and a local minimum value of 0 at 1. 68. For what values of the numbers and does the function have the maximum value ? 69. Show that the curve has three points of inﬂection and they all lie on one straight line. 70. Show that the curves and touch the curve at its inﬂection points. 71. Suppose is differentiable on an interval and for all numbers in except for a single number . Prove that is increasing on the entire interval . 72–74 Assume that all of the functions are twice differentiable and the second derivatives are never 0. 72. (a) If and are concave upward on , show that is concave upward on . (b) If is positive and concave upward on , show that the function is concave upward on . 73. (a) If and are positive, increasing, concave upward func-tions on , show that the product function is concave upward on . (b) Show that part (a) remains true if and are both decreasing. t f I ft I t f I t!x" ! ' f !x"( 2 I f I f ' t I t f I f c I x f "!x" $ 0 I f y ! e%x sin x y ! %e%x y ! e%x y ! !1 ' x"#!1 ' x 2" f !2" ! 1 f !x" ! axe bx2 b a %2 3 f !x" ! ax 3 ' bx 2 ' cx ' d 67. -f f !x" ! e%x 2#!2- 2" . ! 0 1#(-s2+ ) -. y ! 1 -s2+ e%!x%."2#!2- 2" t p ! 4, k ! 0.07 A ! 0.01, S!t" ! At pe%kt 58. 59–60 Estimate the intervals of concavity to one decimal place by using a computer algebra system to compute and graph . 59. 60. 61. A graph of a population of yeast cells in a new laboratory culture as a function of time is shown. (a) Describe how the rate of population increase varies. (b) When is this rate highest? (c) On what intervals is the population function concave upward or downward? (d) Estimate the coordinates of the inﬂection point. 62. Let be the temperature at time where you live and sup-pose that at time you feel uncomfortably hot. How do you feel about the given data in each case? (a) (b) (c) (d) Let be a measure of the knowledge you gain by studying for a test for t hours. Which do you think is larger, or ? Is the graph of K concave upward or concave downward? Why? Coffee is being poured into the mug shown in the ﬁgure at a constant rate (measured in volume per unit time). Sketch a rough graph of the depth of the coffee in the mug as a func-tion of time. Account for the shape of the graph in terms of concavity. What is the signiﬁcance of the inﬂection point? 64. K!3" % K!2" K!8" % K!7" K!t" 63. f "!3" ! %2, f !!3" ! %4 f "!3" ! %2, f !!3" ! 4 f "!3" ! 2, f !!3" ! %4 f "!3" ! 2, f !!3" ! 4 t ! 3 t f !t" 2 6 10 14 18 4 8 12 16 0 Time (in hours) Number of yeast cells 100 200 300 400 500 600 700 f !x" ! x 2 tan%1x 1 ' x 3 f !x" ! x 4 ' x 3 ' 1 sx 2 ' x ' 1 f ! CAS f !x" ! x 3!x % 2"4 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH \|\|\|\| 297 79. Prove that if is a point of inﬂection of the graph of and exists in an open interval that contains , then . [Hint: Apply the First Derivative Test and Fermat’s Theorem to the function .] 80. Show that if , then , but is not an inﬂection point of the graph of . 81. Show that the function has an inﬂection point at but does not exist. 82. Suppose that is continuous and , but . Does have a local maximum or minimum at ? Does have a point of inﬂection at ? 83. The three cases in the First Derivative Test cover the situations one commonly encounters but do not exhaust all possibilities. Consider the functions whose values at 0 are all 0 and, for (a) Show that 0 is a critical number of all three functions but their derivatives change sign inﬁnitely often on both sides of 0. (b) Show that has neither a local maximum nor a local mini-mum at 0, has a local minimum, and has a local maximum. h t f h!x" ! x 4 )%2 ' sin 1 x t!x" ! x 4 )2 ' sin 1 x f!x" ! x 4 sin 1 x x " 0, f, t, and h c f c f f /!c" $ 0 f "!c" ! f !!c" ! 0 f / t!!0" !0, 0" t!x" ! x% x% f !0, 0" f !!0" ! 0 f !x" ! x 4 t ! f " f !!c" ! 0 c f ! f !c, f !c"" (c) Suppose is increasing and is decreasing. Show, by giving three examples, that may be concave upward, concave downward, or linear. Why doesn’t the argument in parts (a) and (b) work in this case? 74. Suppose and are both concave upward on . Under what condition on will the composite function be concave upward? Show that for . [Hint: Show that is increasing on .] 76. (a) Show that for . (b) Deduce that for . (c) Use mathematical induction to prove that for and any positive integer , 77. Show that a cubic function (a third-degree polynomial) always has exactly one point of inﬂection. If its graph has three -intercepts , and , show that the -coordinate of the inﬂection point is . ; 78. For what values of does the polynomial have two inﬂection points? One inﬂec-tion point? None? Illustrate by graphing for several values of . How does the graph change as decreases? c c P P!x" ! x 4 ' cx 3 ' x 2 c !x1 ' x2 ' x3"#3 x x3 x1, x2 x e x ( 1 ' x ' x 2 2! ' 0 0 0 ' x n n! n x ( 0 x ( 0 e x ( 1 ' x ' 1 2 x 2 x ( 0 e x ( 1 ' x !0, +#2" f !x" ! tan x % x 0 # x # +#2 tan x $ x 75. h!x" ! f !t!x"" f !%&, &" t f ft t f 298 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION INDETERMINATE FORMS AND L’HOSPITAL’S RULE Suppose we are trying to analyze the behavior of the function Although is not deﬁned when , we need to know how behaves near 1. In partic-ular, we would like to know the value of the limit In computing this limit we can’t apply Law 5 of limits (the limit of a quotient is the quo-tient of the limits, see Section 2.3) because the limit of the denominator is 0. In fact, although the limit in (1) exists, its value is not obvious because both numerator and denom-inator approach and is not deﬁned. In general, if we have a limit of the form where both and as , then this limit may or may not exist and is called an indeterminate form of type . We met some limits of this type in Chapter 2. For 0 0 x l a t!x" l 0 f!x" l 0 lim x l a f!x" t!x" 0 0 0 lim x l1 ln x x % 1 1 F x ! 1 F F!x" ! ln x x % 1 4.4 rational functions, we can cancel common factors: We used a geometric argument to show that But these methods do not work for limits such as (1), so in this section we introduce a sys-tematic method, known as l’Hospital’s Rule, for the evaluation of indeterminate forms. Another situation in which a limit is not obvious occurs when we look for a horizontal asymptote of F and need to evaluate the limit It isn’t obvious how to evaluate this limit because both numerator and denominator become large as . There is a struggle between numerator and denominator. If the numerator wins, the limit will be ; if the denominator wins, the answer will be 0. Or there may be some compromise, in which case the answer may be some ﬁnite positive number. In general, if we have a limit of the form where both (or ) and (or ), then the limit may or may not exist and is called an indeterminate form of type . We saw in Section 2.6 that this type of limit can be evaluated for certain functions, including rational functions, by dividing numerator and denominator by the highest power of that occurs in the denominator. For instance, This method does not work for limits such as (2), but l’Hospital’s Rule also applies to this type of indeterminate form. L’HOSPITAL’S RULE Suppose and are differentiable and on an open interval that contains (except possibly at ). Suppose that and or that and (In other words, we have an indeterminate form of type or .) Then if the limit on the right side exists (or is or ). %& & lim x l a f!x" t!x" ! lim x l a f"!x" t"!x" &#& 0 0 lim x l a t!x" ! )& lim x l a f!x" ! )& lim x l a t!x" ! 0 lim x l a f!x" ! 0 a a I t"!x" " 0 t f lim x l & x 2 % 1 2x 2 ' 1 ! lim x l & 1 % 1 x 2 2 ' 1 x 2 ! 1 % 0 2 ' 0 ! 1 2 x "#" %& t!x" l & %& f!x" l & lim x l a f!x" t!x" & x l & lim x l & ln x x % 1 2 lim x l 0 sin x x ! 1 lim x l1 x 2 % x x 2 % 1 ! lim x l1 x!x % 1" !x ' 1"!x % 1" ! lim x l1 x x ' 1 ! 1 2 SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE \|\|\|\| 299 L’Hospital’s Rule is named after a French nobleman, the Marquis de l’Hospital (1661–1704), but was discovered by a Swiss mathematician, John Bernoulli (1667–1748). You might sometimes see l’Hospital spelled as l’Hôpital, but he spelled his own name l’Hospital, as was common in the 17th century. See Exer-cise 77 for the example that the Marquis used to illustrate his rule. See the project on page 307 for further historical details. L’HOSPITAL L’Hospital’s Rule says that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives, provided that the given conditions are satisﬁed. It is especially important to verify the conditions regarding the limits of and before using l’Hospital’s Rule. L’Hospital’s Rule is also valid for one-sided limits and for limits at inﬁnity or negative inﬁnity; that is, “ ” can be replaced by any of the symbols , , , or . For the special case in which , and are continuous, and , it is easy to see why l’Hospital’s Rule is true. In fact, using the alternative form of the deﬁnition of a derivative, we have It is more difﬁcult to prove the general version of l’Hospital’s Rule. See Appendix F. EXAMPLE 1 Find . SOLUTION Since we can apply l’Hospital’s Rule: M EXAMPLE 2 Calculate . SOLUTION We have and , so l’Hospital’s Rule gives Since and as , the limit on the right side is also indeterminate, but a second application of l’Hospital’s Rule gives M lim x l ! e x x 2 ! lim x l ! e x 2x ! lim x l ! e x 2 ! ! x l ! 2x l ! e x l ! lim x l ! e x x 2 ! lim x l ! d dx !e x" d dx !x 2" ! lim x l ! e x 2x lim x l ! x 2 ! ! lim x l ! e x ! ! lim x l ! e x x 2 ! lim x l1 1 x ! 1 lim x l1 ln x x " 1 ! lim x l1 d dx !ln x" d dx !x " 1" ! lim x l1 1#x 1 lim x l1 !x " 1" ! 0 and lim x l1 ln x ! ln 1 ! 0 lim x l1 ln x x " 1 V ! lim x l a f!x" t!x" ! lim x l a f!x" " f!a" t!x" " t!a" ! lim x l a f!x" " f!a" x " a t!x" " t!a" x " a lim x l a f#!x" t#!x" ! f#!a" t#!a" ! lim x l a f!x" " f!a" x " a lim x l a t!x" " t!a" x " a t#!a" " 0 t# f# f!a" ! t!a" ! 0 NOTE 3 x l "! x l ! x l a" x l a$ x l a NOTE 2 t f NOTE 1 300 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 0 y x a y=m¡(x-a) y=m™(x-a) 0 y x a f g FIGURE 1 N Figure 1 suggests visually why l’Hospital’s Rule might be true. The ﬁrst graph shows two differentiable functions and , each of which approaches as . If we were to zoom in toward the point , the graphs would start to look almost linear. But if the functions actually were linear, as in the second graph, then their ratio would be which is the ratio of their derivatives. This sug-gests that lim x l a f!x" t!x" ! lim x l a f#!x" t#!x" m1!x " a" m2!x " a" ! m1 m2 !a, 0" x l a 0 t f \| Notice that when using l’Hospital’s Rule we differentiate the numerator and denominator separately. We do not use the Quotient Rule. N The graph of the function of Example 2 is shown in Figure 2. We have noticed previously that exponential functions grow far more rapidly than power functions, so the result of Example 2 is not unexpected. See also Exercise 69. y=´ ≈ 10 20 0 FIGURE 2 EXAMPLE 3 Calculate . SOLUTION Since and as , l’Hospital’s Rule applies: Notice that the limit on the right side is now indeterminate of type . But instead of applying l’Hospital’s Rule a second time as we did in Example 2, we simplify the expression and see that a second application is unnecessary: M EXAMPLE 4 Find . (See Exercise 38 in Section 2.2.) SOLUTION Noting that both and as , we use l’Hospital’s Rule: Since the limit on the right side is still indeterminate of type , we apply l’Hospital’s Rule again: Because , we simplify the calculation by writing We can evaluate this last limit either by using l’Hospital’s Rule a third time or by writing as and making use of our knowledge of trigonometric limits. Putting together all the steps, we get M EXAMPLE 5 Find . SOLUTION If we blindly attempted to use l’Hospital’s Rule, we would get \| This is wrong! Although the numerator as , notice that the denomi-nator does not approach , so l’Hospital’s Rule can’t be applied here. 0 !1 " cos x" x l %" sin x l 0 lim x l % " sin x 1 " cos x ! lim x l % " cos x sin x ! "! lim x l % " sin x 1 " cos x ! 1 3 lim x l 0 tan x x ! 1 3 lim x l 0 sec2x 1 ! 1 3 lim x l 0 tan x " x x3 ! lim x l 0 sec2x " 1 3x2 ! lim x l 0 2 sec2x tan x 6x !sin x"#!cos x" tan x lim x l 0 2 sec2x tan x 6x ! 1 3 lim x l 0 sec2x lim x l 0 tan x x ! 1 3 lim x l 0 tan x x limx l 0 sec2x ! 1 lim x l 0 sec2x " 1 3x 2 ! lim x l 0 2 sec2x tan x 6x 0 0 lim x l 0 tan x " x x 3 ! lim x l 0 sec2x " 1 3x 2 x l 0 x 3 l 0 tan x " x l 0 lim x l 0 tan x " x x 3 lim x l ! ln x s 3 x ! lim x l ! 1#x 1 3 x"2#3 ! lim x l ! 3 s 3 x ! 0 0 0 lim x l ! ln x s 3 x ! lim x l ! 1#x 1 3 x"2#3 x l ! s 3 x l ! ln x l ! lim x l ! ln x s 3 x V SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE \|\|\|\| 301 N The graph of the function of Example 3 is shown in Figure 3. We have discussed previously the slow growth of logarithms, so it isn’t surpris-ing that this ratio approaches as . See also Exercise 70. x l ! 0 0 _1 2 10,000 y= ln x Œ„ x FIGURE 3 N The graph in Figure 4 gives visual conﬁrma-tion of the result of Example 4. If we were to zoom in too far, however, we would get an inaccurate graph because is close to when is small. See Exercise 38(d) in Section 2.2. x x tan x FIGURE 4 y=tan x-x ˛ 0 _1 1 1 The required limit is, in fact, easy to ﬁnd because the function is continuous at and the denominator is nonzero there: M Example 5 shows what can go wrong if you use l’Hospital’s Rule without thinking. Other limits can be found using l’Hospital’s Rule but are more easily found by other meth-ods. (See Examples 3 and 5 in Section 2.3, Example 3 in Section 2.6, and the discussion at the beginning of this section.) So when evaluating any limit, you should consider other methods before using l’Hospital’s Rule. INDETERMINATE PRODUCTS If and (or ), then it isn’t clear what the value of , if any, will be. There is a struggle between and . If wins, the answer will be ; if wins, the answer will be (or ). Or there may be a compromise where the answer is a ﬁnite nonzero number. This kind of limit is called an indeterminate form of type . We can deal with it by writing the product as a quotient: or This converts the given limit into an indeterminate form of type or so that we can use l’Hospital’s Rule. EXAMPLE 6 Evaluate . SOLUTION The given limit is indeterminate because, as , the ﬁrst factor approaches 0 while the second factor approaches . Writing , we have as , so l’Hospital’s Rule gives M In solving Example 6 another possible option would have been to write This gives an indeterminate form of the type , but if we apply l’Hospital’s Rule we get a more complicated expression than the one we started with. In general, when we rewrite an indeterminate product, we try to choose the option that leads to the simpler limit. INDETERMINATE DIFFERENCES If and , then the limit is called an indeterminate form of type . Again there is a contest between and . Will the answer be ( wins) or will it be ( wins) or will they compromise on a ﬁnite number? To ﬁnd out, we try to convert the difference into a quotient (for instance, by using t "! f ! t f ! " ! lim x l a $ f!x" " t!x"% lim x l a t!x" ! ! lim x l a f!x" ! ! 0#0 lim x l 0$ x ln x ! lim x l 0$ x 1#ln x NOTE ! lim x l 0$ !"x" ! 0 lim x l 0$ x ln x ! lim x l 0$ ln x 1#x ! lim x l 0$ 1#x "1#x 2 x l 0$ 1#x l ! x ! 1#!1#x" "! !ln x" !x" x l 0$ lim x l 0$ x ln x V !#! 0 0 ft ! t 1#f ft ! f 1#t ft 0 # ! "! ! t 0 f t f lim x l a f!x"t!x" "! lim x l a t!x" ! ! lim x l a f!x" ! 0 lim x l % " sin x 1 " cos x ! sin % 1 " cos % ! 0 1 " !"1" ! 0 % 302 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 0 y x 1 y=x ln x FIGURE 5 N Figure 5 shows the graph of the function in Example 6. Notice that the function is undeﬁned at ; the graph approaches the origin but never quite reaches it. x ! 0 a common denominator, or rationalization, or factoring out a common factor) so that we have an indeterminate form of type or . EXAMPLE 7 Compute . SOLUTION First notice that and as , so the limit is inde-terminate. Here we use a common denominator: Note that the use of l’Hospital’s Rule is justiﬁed because and as . M INDETERMINATE POWERS Several indeterminate forms arise from the limit 1. and type 2. and type 3. and type Each of these three cases can be treated either by taking the natural logarithm: , or by writing the function as an exponential: (Recall that both of these methods were used in differentiating such functions.) In either method we are led to the indeterminate product , which is of type . EXAMPLE 8 Calculate . SOLUTION First notice that as , we have and , so the given limit is indeterminate. Let Then so l’Hospital’s Rule gives So far we have computed the limit of , but what we want is the limit of . To ﬁnd this y ln y ! lim x l 0$ 4 cos 4x 1 $ sin 4x sec2x ! 4 lim x l 0$ ln y ! lim x l 0$ ln!1 $ sin 4x" tan x ln y ! ln$!1 $ sin 4x"cot x% ! cot x ln!1 $ sin 4x" y ! !1 $ sin 4x"cot x cot x l ! 1 $ sin 4x l 1 x l 0$ lim x l 0$ !1 $ sin 4x"cot x 0 # ! t!x" ln f!x" $ f!x"%t!x" ! e t!x" ln f !x" ln y ! t!x" ln f!x" then y ! $ f!x"%t!x" let 1! lim x l a t!x" ! &! lim x l a f!x" ! 1 ! 0 lim x l a t!x" ! 0 lim x l a f!x" ! ! 00 lim x l a t!x" ! 0 lim x l a f!x" ! 0 lim x l a $ f!x"%t!x" x l !%#2"" cos x l 0 1 " sin x l 0 ! lim x l !%#2"" 1 " sin x cos x ! lim x l !%#2"" "cos x "sin x ! 0 lim x l !%#2"" !sec x " tan x" ! lim x l !%#2"" & 1 cos x " sin x cos x' x l !%#2"" tan x l ! sec x l ! lim x l !%#2"" !sec x " tan x" !#! 0 0 SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE \|\|\|\| 303 we use the fact that : M EXAMPLE 9 Find . SOLUTION Notice that this limit is indeterminate since for any but for any . We could proceed as in Example 8 or by writing the function as an exponential: In Example 6 we used l’Hospital’s Rule to show that Therefore M lim x l 0$ x x ! lim x l 0$ e x ln x ! e 0 ! 1 lim x l 0$ x ln x ! 0 x x ! !e ln x"x ! e x ln x x " 0 x 0 ! 1 x ' 0 0 x ! 0 lim x l 0$ x x lim x l 0$ !1 $ sin 4x"cot x ! lim x l 0$ y ! lim x l 0$ e ln y ! e 4 y ! e ln y 304 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION N The graph of the function , , is shown in Figure 6. Notice that although is not deﬁned, the values of the function approach as . This conﬁrms the result of Example 9. x l 0$ 1 00 x ' 0 y ! x x 2 0 2 _1 FIGURE 6 5–64 Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 22. lim x l 0 e x " 1 " x " 1 2x 2 x 3 lim x l 0 e x " 1 " x x 2 21. lim x l 1 ln x sin %x lim x l ! e x x 3 lim x l ! ln ln x x lim x l 0$ ln x x lim x l ! x $ x 2 1 " 2x 2 lim x l ! ln x sx lim ( l %#2 1 " sin ( csc ( lim x l 0 tan px tan qx lim t l 0 e3t " 1 t lim t l 0 et " 1 t 3 lim xl 0 sin 4x tan 5x lim x l !%#2"$ cos x 1 " sin x lim x l 1 x a " 1 x b " 1 lim x l1 x 9 " 1 x 5 " 1 lim x l 2 x 2 $ x " 6 x " 2 lim x l 1 x 2 " 1 x 2 " x 1–4 Given that which of the following limits are indeterminate forms? For those that are not an indeterminate form, evaluate the limit where possible. (a) (b) (c) (d) (e) 2. (a) (b) (c) 3. (a) (b) (c) 4. (a) (b) (c) (d) (e) (f) lim x l a q!x" sp!x" lim x l a $ p!x"%q!x" lim x l a $ p!x"% f !x" lim x l a $h!x"% p!x" lim x l a $ f !x"% p!x" lim x l a $ f !x"% t!x" lim x l a $ p!x" $ q!x"% lim x l a $ p!x" " q!x"% lim x l a $ f !x" " p!x"% lim x l a $ p!x"q!x"% lim x l a $h!x"p!x"% lim x l a $ f !x"p!x"% lim x l a p!x" q!x" lim x l a p!x" f !x" lim x l a h!x" p!x" lim x l a f !x" p!x" lim x l a f !x" t!x" 1. lim x l a p!x" ! ! lim x l a q!x" ! ! lim x l a f !x" ! 0 lim x l a t!x" ! 0 lim x l a h!x" ! 1 EXERCISES 4.4 63. 64. ; 65–66 Use a graph to estimate the value of the limit. Then use l’Hospital’s Rule to ﬁnd the exact value. 65. 66. ; 67–68 Illustrate l’Hospital’s Rule by graphing both and near to see that these ratios have the same limit as . Also calculate the exact value of the limit. 67. , 68. , Prove that for any positive integer . This shows that the exponential function approaches inﬁnity faster than any power of . 70. Prove that for any number . This shows that the logarithmic func-tion approaches more slowly than any power of . 71. What happens if you try to use l’Hospital’s Rule to evaluate Evaluate the limit using another method. 72. If an object with mass is dropped from rest, one model for its speed after seconds, taking air resistance into account, is where is the acceleration due to gravity and is a positive constant. (In Chapter 9 we will be able to deduce this equa-tion from the assumption that the air resistance is propor-tional to the speed of the object; is the proportionality constant.) (a) Calculate . What is the meaning of this limit? (b) For ﬁxed , use l’Hospital’s Rule to calculate . What can you conclude about the velocity of a falling object in a vacuum? limc l 0$ v t limt l ! v c c t v ! mt c !1 " e "ct#m" t v m lim x l ! x sx 2 $ 1 x ! p ' 0 lim x l ! ln x x p ! 0 x n lim x l ! e x x n ! ! 69. t!x" ! sec x " 1 f !x" ! 2x sin x t!x" ! x 3 $ 4x f !x" ! e x " 1 x l 0 x ! 0 f #!x"#t#!x" f !x"#t!x" lim x l 0 5x " 4x 3x " 2x lim x l ! &1 $ 2 x' x lim x l ! & 2x " 3 2x $ 5' 2x$1 lim x l 0$ !cos x"1#x2 23. 24. 25. 26. 27. 28. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 44. 45. 46. 47. 49. 50. 51. 52. 53. 54. 56. 57. 58. 59. 60. 61. 62. lim x l 1 !2 " x"tan!% x#2" lim x l 0$ !4x $ 1"cot x lim x l ! !e x $ x"1#x lim x l ! x 1#x lim x l ! x !ln 2"#!1 $ ln x" lim x l ! &1 $ 3 x $ 5 x 2' x lim x l ! &1 $ a x' bx lim x l 0 !1 " 2x"1#x 55. lim x l 0$ !tan 2x"x lim x l 0$ x x2 lim x l ! !xe 1#x " x" lim x l ! !x " ln x" lim x l 0 &cot x " 1 x' lim x l ! (sx2 $ x " x) lim x l 0 !csc x " cot x" 48. lim x l 1 & x x " 1 " 1 ln x' lim x l ! x tan!1#x" lim x l 1$ ln x tan!%x#2" lim x l %#4 !1 " tan x"sec x lim x l ! x 3e "x 2 43. lim x l 0$ sin x ln x lim x l 0 cot 2x sin 6x lim x l "! x 2e x lim x l ! x sin!%#x" lim x l a$ cos x ln!x " a" ln!e x " ea" lim x l 0 cos x " 1 $ 1 2x 2 x 4 lim x l 0 e x " e"x " 2x x " sin x lim x l 1 x a " ax $ a " 1 !x " 1"2 lim x l ! sx 2 $ 2 s2x 2 $ 1 lim x l 1 1 " x $ ln x 1 $ cos %x lim x l 0 x tan"1!4x" lim x l 0 x $ sin x x $ cos x lim x l 0 cos mx " cos nx x 2 lim x l 0 1 " cos x x 2 29. lim x l ! !ln x"2 x lim x l 0 sin"1x x lim x l 0 sin x " x x 3 lim t l 0 5t " 3t t lim x l 0 x " sin x x " tan x lim x l 0 tanh x tan x SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE \|\|\|\| 305 the arc . Let be the area of the triangle . Find . 79. If is continuous, , and , evaluate 80. For what values of and is the following equation true? If is continuous, use l’Hospital’s Rule to show that Explain the meaning of this equation with the aid of a diagram. 82. If is continuous, show that 83. Let (a) Use the deﬁnition of derivative to compute . (b) Show that has derivatives of all orders that are deﬁned on . [Hint: First show by induction that there is a polynomial and a nonnegative integer such that for .] ; 84. Let (a) Show that is continuous at . (b) Investigate graphically whether is differentiable at by zooming in several times toward the point on the graph of . (c) Show that is not differentiable at . How can you reconcile this fact with the appearance of the graphs in part (b)? 0 f f !0, 1" 0 f 0 f f !x" !() x) x 1 if x " 0 if x ! 0 x " 0 f !n"!x" ! pn!x"f !x"#x kn kn pn!x" ! f f #!0" f !x" !( e"1#x 2 0 if x " 0 if x ! 0 lim h l 0 f !x $ h" " 2f !x" $ f !x " h" h 2 ! f )!x" f ) lim h l 0 f !x $ h" " f !x " h" 2h ! f #!x" f # 81. lim x l 0 & sin 2x x3 $ a $ b x2' ! 0 b a lim x l 0 f !2 $ 3x" $ f !2 $ 5x" x f #!2" ! 7 f !2" ! 0 f # 6 0404 78 P Q R A(¨) B(¨) O ¨ lim( l 0$ !("#+!(" PQR B!(" PR 73. If an initial amount of money is invested at an interest rate compounded times a year, the value of the investment after years is If we let , we refer to the continuous compounding of interest. Use l’Hospital’s Rule to show that if interest is compounded continuously, then the amount after years is 74. If a metal ball with mass is projected in water and the force of resistance is proportional to the square of the velocity, then the distance the ball travels in time is where is a positive constant. Find . 75. If an electrostatic ﬁeld acts on a liquid or a gaseous polar dielectric, the net dipole moment per unit volume is Show that . 76. A metal cable has radius and is covered by insulation, so that the distance from the center of the cable to the exterior of the insulation is . The velocity of an electrical impulse in the cable is where is a positive constant. Find the following limits and interpret your answers. (a) (b) 77. The ﬁrst appearance in print of l’Hospital’s Rule was in the book Analyse des Inﬁniment Petits published by the Marquis de l’Hospital in 1696. This was the ﬁrst calculus textbook ever published and the example that the Marquis used in that book to illustrate his rule was to ﬁnd the limit of the function as approaches , where . (At that time it was common to write instead of .) Solve this problem. 78. The ﬁgure shows a sector of a circle with central angle . Let be the area of the segment between the chord and PR A!(" ( a 2 aa a ' 0 a x y ! s2a 3x " x 4 " as 3 aax a " s 4 ax 3 lim r l 0$ v lim R l r$ v c v ! "c& r R' 2 ln& r R' v R r lim E l 0$ P!E" ! 0 P!E" ! e E $ e"E e E " e"E " 1 E P E lim c l 0$ s!t" c s!t" ! m c ln cosh tc mt t m A ! A0ert t n l ! A ! A0&1 $ r n' nt t n r A0 306 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION L’Hospital’s Rule was ﬁrst published in 1696 in the Marquis de l’Hospital’s calculus textbook Analyse des Inﬁniment Petits, but the rule was discovered in 1694 by the Swiss mathematician John (Johann) Bernoulli. The explanation is that these two mathematicians had entered into a curious business arrangement whereby the Marquis de l’Hospital bought the rights to Bernoulli’s mathematical discoveries. The details, including a translation of l’Hospital’s letter to Bernoulli proposing the arrangement, can be found in the book by Eves . Write a report on the historical and mathematical origins of l’Hospital’s Rule. Start by pro-viding brief biographical details of both men (the dictionary edited by Gillispie is a good source) and outline the business deal between them. Then give l’Hospital’s statement of his rule, which is found in Struik’s sourcebook and more brieﬂy in the book of Katz . Notice that l’Hospital and Bernoulli formulated the rule geometrically and gave the answer in terms of dif-ferentials. Compare their statement with the version of l’Hospital’s Rule given in Section 4.4 and show that the two statements are essentially the same. 1. Howard Eves, In Mathematical Circles (Volume 2: Quadrants III and IV) (Boston: Prindle, Weber and Schmidt, 1969), pp. 20–22. 2. C. C. Gillispie, ed., Dictionary of Scientiﬁc Biography (New York: Scribner’s, 1974). See the article on Johann Bernoulli by E. A. Fellmann and J. O. Fleckenstein in Volume II and the article on the Marquis de l’Hospital by Abraham Robinson in Volume VIII. 3. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993), p. 484. 4. D. J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, NJ: Princeton Uni-versity Press, 1969), pp. 315–316. THE ORIGINS OF L’HOSPITAL’S RULE W R I T I N G P R O J E C T www.stewartcalculus.com The Internet is another source of infor-mation for this project. Click on History of Mathematics for a list of reliable websites. SECTION 4.5 SUMMARY OF CURVE SKETCHING \|\|\|\| 307 SUMMARY OF CURVE SKETCHING So far we have been concerned with some particular aspects of curve sketching: domain, range, and symmetry in Chapter 1; limits, continuity, and asymptotes in Chapter 2; deriva-tives and tangents in Chapters 2 and 3; and extreme values, intervals of increase and decrease, concavity, points of inﬂection, and l’Hospital’s Rule in this chapter. It is now time to put all of this information together to sketch graphs that reveal the important fea-tures of functions. You might ask: Why don’t we just use a graphing calculator or computer to graph a curve? Why do we need to use calculus? It’s true that modern technology is capable of producing very accurate graphs. But even the best graphing devices have to be used intelligently. We saw in Section 1.4 that it is extremely important to choose an appropriate viewing rectangle to avoid getting a mis-leading graph. (See especially Examples 1, 3, 4, and 5 in that section.) The use of calculus enables us to discover the most interesting aspects of graphs and in many cases to calcu-late maximum and minimum points and inﬂection points exactly instead of approximately. For instance, Figure 1 shows the graph of . At ﬁrst glance it seems reasonable: It has the same shape as cubic curves like , and it appears to have no maximum or minimum point. But if you compute the derivative, you will see that there is a maximum when and a minimum when . Indeed, if we zoom in to this portion of the graph, we see that behavior exhibited in Figure 2. Without calculus, we could easily have overlooked it. In the next section we will graph functions by using the interaction between calculus and graphing devices. In this section we draw graphs by ﬁrst considering the following x 1 x 0.75 y x 3 f x 8x 3 21x 2 18x 2 4.5 FIGURE 1 30 _10 _2 4 y=8˛-21≈+18x+2 Thomas Fisher Rare Book Library FIGURE 2 y=8˛-21≈+18x+2 8 6 0 2 information. We don’t assume that you have a graphing device, but if you do have one you should use it as a check on your work. GUIDELINES FOR SKETCHING A CURVE The following checklist is intended as a guide to sketching a curve by hand. Not every item is relevant to every function. (For instance, a given curve might not have an asymptote or possess symmetry.) But the guidelines provide all the information you need to make a sketch that displays the most important aspects of the function. A. Domain It’s often useful to start by determining the domain of , that is, the set of values of for which is deﬁned. B. Intercepts The -intercept is and this tells us where the curve intersects the -axis. To ﬁnd the -intercepts, we set and solve for . (You can omit this step if the equation is difﬁcult to solve.) C. Symmetry (i) If for all in , that is, the equation of the curve is unchanged when is replaced by , then is an even function and the curve is symmetric about the -axis. This means that our work is cut in half. If we know what the curve looks like for , then we need only reﬂect about the -axis to obtain the complete curve [see Figure 3(a)]. Here are some examples: , and . (ii) If for all in , then is an odd function and the curve is symmetric about the origin. Again we can obtain the complete curve if we know what it looks like for . [Rotate 180° about the origin; see Figure 3(b).] Some simple examples of odd functions are , and . (iii) If for all in , where is a positive constant, then is called a periodic function and the smallest such number is called the period. For instance, has period and has period . If we know what the graph looks like in an interval of length , then we can use translation to sketch the entire graph (see Figure 4). D. Asymptotes (i) Horizontal Asymptotes. Recall from Section 2.6 that if either or , then the line is a horizontal asymptote of the curve . If it turns out that (or ), then we do not have an asymptote to the right, but that is still useful information for sketching the curve. (ii) Vertical Asymptotes. Recall from Section 2.2 that the line is a vertical asymptote if at least one of the following statements is true: lim x l a$ f!x" ! "! lim x l a" f!x" ! "! lim x l a$ f!x" ! ! lim x l a" f!x" ! ! 1 x ! a "! lim x l ! f!x" ! ! y ! f!x" y ! L lim x l" ! f!x" ! L lim x l ! f!x" ! L FIGURE 4 Periodic function: translational symmetry a-p a a+p a+2p x y 0 p % y ! tan x 2% y ! sin x p f p D x f!x $ p" ! f!x" y ! sin x y ! x, y ! x 3, y ! x 5 x , 0 f D x f!"x" ! "f!x" y ! cos x y ! x 2, y ! x 4, y ! )x) y x , 0 y f "x x D x f!"x" ! f!x" x y ! 0 x y f!0" y f!x" x f D y ! f!x" 308 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION FIGURE 3 (a) Even function: reﬂectional symmetry (b) Odd function: rotational symmetry x y 0 x y 0 (For rational functions you can locate the vertical asymptotes by equating the denomi-nator to 0 after canceling any common factors. But for other functions this method does not apply.) Furthermore, in sketching the curve it is very useful to know exactly which of the statements in (1) is true. If is not deﬁned but is an endpoint of the domain of , then you should compute or , whether or not this limit is inﬁnite. (iii) Slant Asymptotes. These are discussed at the end of this section. E. Intervals of Increase or Decrease Use the I/D Test. Compute and ﬁnd the intervals on which is positive ( is increasing) and the intervals on which is negative ( is decreasing). F. Local Maximum and Minimum Values Find the critical numbers of [the numbers where or does not exist]. Then use the First Derivative Test. If changes from positive to negative at a critical number , then is a local maximum. If changes from negative to positive at , then is a local minimum. Although it is usually preferable to use the First Derivative Test, you can use the Second Derivative Test if and . Then implies that is a local minimum, whereas implies that is a local maximum. G. Concavity and Points of Inﬂection Compute and use the Concavity Test. The curve is concave upward where and concave downward where . Inﬂec-tion points occur where the direction of concavity changes. H. Sketch the Curve Using the information in items A–G, draw the graph. Sketch the asymptotes as dashed lines. Plot the intercepts, maximum and minimum points, and inﬂection points. Then make the curve pass through these points, rising and falling according to E, with concavity according to G, and approaching the asymptotes. If addi-tional accuracy is desired near any point, you can compute the value of the derivative there. The tangent indicates the direction in which the curve proceeds. EXAMPLE 1 Use the guidelines to sketch the curve . A. The domain is B. The - and -intercepts are both 0. C. Since , the function is even. The curve is symmetric about the -axis. D. Therefore the line is a horizontal asymptote. Since the denominator is 0 when , we compute the following limits: Therefore the lines and are vertical asymptotes. This information about limits and asymptotes enables us to draw the preliminary sketch in Figure 5, showing the parts of the curve near the asymptotes. x ! "1 x ! 1 lim x l "1$ 2x 2 x 2 " 1 ! "! lim x l "1" 2x 2 x 2 " 1 ! ! lim x l 1$ 2x 2 x 2 " 1 ! ! lim x l 1" 2x 2 x 2 " 1 ! "! x ! &1 y ! 2 lim x l&! 2x 2 x 2 " 1 ! lim x l&! 2 1 " 1#x 2 ! 2 y f f!"x" ! f!x" y x +x)x 2 " 1 " 0, ! +x)x " &1, ! !"!, "1" $ !"1, 1" $ !1, !" y ! 2x 2 x 2 " 1 V f )!x" - 0 f )!x" ' 0 f )!x" f!c" f )!c" - 0 f!c" f )!c" ' 0 f )!c" " 0 f#!c" ! 0 f!c" c f# f!c" c f# f#!c" f#!c" ! 0 c f f f#!x" f f#!x" f#!x" lim x l a$ f!x" lim x l a" f!x" f a f!a" SECTION 4.5 SUMMARY OF CURVE SKETCHING \|\|\|\| 309 FIGURE 5 Preliminary sketch x=1 x=_1 y=2 x y 0 N We have shown the curve approaching its horizontal asymptote from above in Figure 5. This is conﬁrmed by the intervals of increase and decrease. 310 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION E. Since when and when , is increasing on and and decreasing on and . F. The only critical number is . Since changes from positive to negative at 0, is a local maximum by the First Derivative Test. G. Since for all , we have and . Thus the curve is concave upward on the intervals and and concave downward on . It has no point of inﬂection since 1 and are not in the domain of . H. Using the information in E–G, we ﬁnish the sketch in Figure 6. M EXAMPLE 2 Sketch the graph of . A. Domain B. The - and -intercepts are both 0. C. Symmetry: None D. Since there is no horizontal asymptote. Since as and is always positive, we have and so the line is a vertical asymptote. E. We see that when (notice that is not in the domain of ), so the only critical number is 0. Since when and when , is decreasing on and increasing on . F. Since and changes from negative to positive at 0, is a local (and absolute) minimum by the First Derivative Test. G. Note that the denominator is always positive. The numerator is the quadratic , which is always positive because its discriminant is , which is negative, and the coefﬁcient of is positive. Thus for all in the domain of , which means that is concave upward on and there is no point of inﬂection. H. The curve is sketched in Figure 7. M !!1, "" f f x f #!x" $ 0 x 2 b 2 ! 4ac ! !32 3x 2 % 8x % 8 f #!x" ! 2!x % 1"3#2!6x % 4" ! !3x 2 % 4x"3!x % 1"1#2 4!x % 1"3 ! 3x 2 % 8x % 8 4!x % 1"5#2 f!0" ! 0 f& f&!0" ! 0 !0, "" !!1, 0" f x $ 0 f&!x" $ 0 !1 ' x ' 0 f&!x" ' 0 f ! 4 3 x ! 0 f&!x" ! 0 f&!x" ! 2xsx % 1 ! x 2 ! 1#(2sx % 1) x % 1 ! x!3x % 4" 2!x % 1"3#2 x ! !1 lim x l !1% x 2 sx % 1 ! " f!x" x l !1% sx % 1 l 0 lim x l " x 2 sx % 1 ! " y x ! $x%x % 1 $ 0& ! $x%x $ !1& ! !!1, "" f!x" ! x 2 sx % 1 f !1 !!1, 1" !1, "" !!", !1" f #!x" ' 0 & ? %x% ' 1 %x% $ 1 & ? x 2 ! 1 $ 0 & ? f #!x" $ 0 x 12x 2 % 4 $ 0 f #!x" ! !4!x 2 ! 1"2 % 4x ! 2!x 2 ! 1"2x !x 2 ! 1"4 ! 12x 2 % 4 !x 2 ! 1"3 f!0" ! 0 f& x ! 0 !1, "" !0, 1" !!1, 0" !!", !1" f !x " 1" x $ 0 f&!x" ' 0 !x " !1" x ' 0 f&!x" $ 0 f&!x" ! 4x!x 2 ! 1" ! 2x 2 ! 2x !x 2 ! 1"2 ! !4x !x 2 ! 1"2 FIGURE 6 Finished sketch of y= x=1 x=_1 y=2 2≈ ≈-1 x y 0 FIGURE 7 x=_1 x y 0 œ„„„„ y= ≈ x+1 EXAMPLE 3 Sketch the graph of . A. The domain is . B. The x- and -intercepts are both 0. C. Symmetry: None D. Because both x and become large as , we have . As , however, and so we have an indeterminate product that requires the use of l’Hospital’s Rule: Thus the x-axis is a horizontal asymptote. E. Since is always positive, we see that when , and when . So f is increasing on and decreasing on . F. Because and changes from negative to positive at , is a local (and absolute) minimum. G. Since if and if , is concave upward on and concave downward on . The inﬂection point is . H. We use this information to sketch the curve in Figure 8. M EXAMPLE 4 Sketch the graph of . A. The domain is . B. The -intercept is . The -intercepts occur when , that is, , where is an integer. C. is neither even nor odd, but for all and so is periodic and has period . Thus, in what follows, we need to consider only and then extend the curve by translation in part H. D. Asymptotes: None E. Thus when . So is increasing on and decreasing on and . F. From part E and the First Derivative Test, we see that the local minimum value is and the local maximum value is . G. If we use the Quotient Rule again and simplify, we get Because and for all , we know that when , that is, . So is concave upward on and concave downward on and . The inﬂection points are . and !3(#2, 0" !(#2, 0" !3(#2, 2(" !0, (#2" !(#2, 3(#2" f (#2 ' x ' 3(#2 cos x ' 0 f #!x" $ 0 x 1 ! sin x ) 0 !2 % sin x"3 $ 0 f #!x" ! ! 2 cos x !1 ! sin x" !2 % sin x"3 f!11(#6" ! 1#s3 f!7(#6" ! !1#s3 !11(#6, 2(" !0, 7(#6" !7(#6, 11(#6" f 7(#6 ' x ' 11(#6 2 sin x % 1 ' 0 & ? sin x ' ! 1 2 & ? f&!x" $ 0 f&!x" ! !2 % sin x"!!sin x" ! cos x !cos x" !2 % sin x"2 ! ! 2 sin x % 1 !2 % sin x"2 0 x 2( 2( f x f!x % 2(" ! f!x" f n x ! !2n % 1"(#2 cos x ! 0 x f!0" ! 1 2 y ! f!x" ! cos x 2 % sin x !!2, !2e!2" !!", !2" !!2, "" f x ' !2 f #!x" ' 0 x $ !2 f #!x" $ 0 f #!x" ! !x % 1"e x % e x ! !x % 2"e x f!!1" ! !e!1 x ! !1 f& f&!!1" ! 0 !!", !1" !!1, "" x % 1 ' 0 f&!x" ' 0 x % 1 $ 0 f&!x" $ 0 e x f&!x" ! xe x % e x ! !x % 1"e x lim x l!" xe x ! lim x l!" x e!x ! lim x l!" 1 !e!x ! lim x l!" !!e x" ! 0 e x l 0 x l !" lim x l " xe x ! " x l " e x y ! f!x" ! xe x V SECTION 4.5 SUMMARY OF CURVE SKETCHING \|\|\|\| 311 FIGURE 8 x y 1 _1 _2 y=x´ (_1, _1/e) H. The graph of the function restricted to is shown in Figure 9. Then we extend it, using periodicity, to the complete graph in Figure 10. M EXAMPLE 5 Sketch the graph of . A. The domain is B. The -intercept is . To ﬁnd the -intercept we set We know that , so we have and therefore the -intercepts are . C. Since , is even and the curve is symmetric about the -axis. D. We look for vertical asymptotes at the endpoints of the domain. Since as and also as , we have Thus the lines and are vertical asymptotes. E. Since when and when , is increasing on and decreasing on . F. The only critical number is . Since changes from positive to negative at , is a local maximum by the First Derivative Test. G. Since for all , the curve is concave downward on and has no inﬂection point. H. Using this information, we sketch the curve in Figure 11. M SLANT ASYMPTOTES Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical. If then the line is called a slant asymptote because the vertical distance y ! mx % b lim x l " ' f!x" ! !mx % b"( ! 0 !!2, 2" x f #!x" ' 0 f #!x" ! !4 ! x 2"!!2" % 2x!!2x" !4 ! x 2"2 ! !8 ! 2x 2 !4 ! x 2"2 f!0" ! ln 4 0 f& x ! 0 !0, 2" !!2, 0" f 0 ' x ' 2 f&!x" ' 0 !2 ' x ' 0 f&!x" $ 0 f&!x" ! !2x 4 ! x 2 x ! !2 x ! 2 lim x l !2% ln!4 ! x 2" ! !" lim x l 2! ln!4 ! x 2" ! !" x l !2% x l 2! 4 ! x 2 l 0% y f f!!x" ! f!x" +s3 x 4 ! x 2 ! 1 ? x 2 ! 3 ln 1 ! 0 y ! ln!4 ! x 2" ! 0 x f!0" ! ln 4 y $x%4 ! x 2 $ 0& ! $x%x 2 ' 4& ! $x%%x% ' 2& ! !!2, 2" y ! ln!4 ! x 2" FIGURE 9 y x π π 2 1 2 2π 3π 2 ” , ’ 11π 6 1 œ„ 3 -” 7π 6 1 œ„ 3 , ’ FIGURE 10 y x π _π 1 2 2π 3π 0 x 2( 312 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 0 y x {œ„ 3, 0} {_œ„ 3, 0} x=2 x=_2 (0, ln 4) y=ln(4 -≈) FIGURE 11 Interval x f ! ! % % CU on ! % % ! CD on % % % % CU on % ! % ! CD on (s3 , ") x $ s3 (0, s3 ) 0 ' x ' s3 (!s3 , 0) !s3 ' x ' 0 (!", !s3 ) x ' !s3 f #!x" !x 2 % 1"3 3 ! x 2 between the curve and the line approaches 0, as in Figure 12. (A similar situation exists if we let .) For rational functions, slant asymptotes occur when the degree of the numerator is one more than the degree of the denominator. In such a case the equation of the slant asymptote can be found by long division as in the follow-ing example. EXAMPLE 6 Sketch the graph of . A. The domain is . B. The - and -intercepts are both 0. C. Since , is odd and its graph is symmetric about the origin. D. Since is never 0, there is no vertical asymptote. Since as and as , there is no horizontal asymptote. But long division gives as So the line is a slant asymptote. E. Since for all (except 0), is increasing on . F. Although , does not change sign at 0, so there is no local maximum or minimum. G. Since when or , we set up the following chart: The points of inﬂection are , and . H. The graph of is sketched in Figure 13. M f (s3 , 3 4s3 ) (!s3 , ! 3 4s3 ), !0, 0" x ! +s3 x ! 0 f #!x" ! 0 f #!x" ! !4x 3 % 6x"!x 2 % 1"2 ! !x 4 % 3x 2" ! 2!x 2 % 1"2x !x 2 % 1"4 ! 2x!3 ! x 2" !x 2 % 1"3 f& f&!0" ! 0 !!", "" f x f&!x" $ 0 f&!x" ! 3x 2!x 2 % 1" ! x 3 ! 2x !x 2 % 1"2 ! x 2!x 2 % 3" !x 2 % 1"2 y ! x x l +" f!x" ! x ! ! x x 2 % 1 ! ! 1 x 1 % 1 x 2 l 0 f!x" ! x 3 x 2 % 1 ! x ! x x 2 % 1 x l !" f!x" l !" x l " f!x" l " x 2 % 1 f f!!x" ! !f!x" y x ! ! !!", "" f!x" ! x 3 x 2 % 1 V x l !" y ! mx % b y ! f!x" SECTION 4.5 SUMMARY OF CURVE SKETCHING \|\|\|\| 313 FIGURE 12 y=ƒ x y 0 y=mx+b ƒ-(mx+b) FIGURE 13 y=x ”_œ„ 3, _ ’ 3œ„ 3 4 inflection points y= ˛ ≈+1 x y 0 ”œ„ 3, ’ 3œ„ 3 4 314 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 51. 52. 53. In the theory of relativity, the mass of a particle is where is the rest mass of the particle, is the mass when the particle moves with speed relative to the observer, and is the speed of light. Sketch the graph of as a function of . 54. In the theory of relativity, the energy of a particle is where is the rest mass of the particle, is its wave length, and is Planck’s constant. Sketch the graph of as a function of . What does the graph say about the energy? 55. The ﬁgure shows a beam of length embedded in concrete walls. If a constant load is distributed evenly along its length, the beam takes the shape of the deﬂection curve where and are positive constants. ( is Young’s modulus of elasticity and is the moment of inertia of a cross-section of the beam.) Sketch the graph of the deﬂection curve. 56. Coulomb’s Law states that the force of attraction between two charged particles is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The ﬁgure shows particles with charge 1 located at positions 0 and 2 on a coordinate line and a particle with charge at a position between them. It follows from Cou-lomb’s Law that the net force acting on the middle particle is where is a positive constant. Sketch the graph of the net force function. What does the graph say about the force? _1 x x +1 2 +1 0 k Fx k x 2 k x 22 0 x 2 x 1 W y 0 L I E I E y W 24EI x 4 WL 12EI x 3 WL2 24EI x 2 W L E h m0 E sm0 2c4 h2c 22 v m c v m m0 m m0 s1 v2c2 y tan1 x 1 x 1 y e3x e2x 1–52 Use the guidelines of this section to sketch the curve. 1. 2. 3. 4. 6. 7. 8. 10. 11. 12. 13. 14. 15. 16. 18. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. , 34. , 35. , 36. , 37. 38. 39. 40. , 42. 43. 44. 45. 46. 47. 48. 49. 50. y x 2 3ex y xex 2 y ln x x 2 y lnsin x y lnx 2 3x 2 y 1 e x2 y e xx y x ln x y e2 x e x y 11 e x 41. 0 x 2 y ex sin x y esin x y sin x 2 cos x y sin x 1 cos x 0 x 2 y sec x tan x 0 x 3 y 1 2 x sin x 2 x 2 y 2x tan x 2 x 2 y x tan x 33. y x cos x y 3 sin x sin3x y s 3 x 3 1 y s 3 x 2 1 y x 53 5x 23 y x 3x13 y x sx 2 1 y s1 x 2 x y xs2 x 2 y x sx 2 1 y sx 2 x x y sx 2 x 2 y 2sx x y xs5 x 19. y x x 3 1 y x 2 x 2 3 17. y 1 1 x 1 x 2 y x 1 x 2 y x 2 x 2 9 y x x 2 9 y x x 2 9 y 1 x 2 9 y x 2 4 x 2 2x y x x 1 9. y 4 x 25 y 2x 5 5x 2 1 y xx 23 y x 4 4x 3 5. y 8x 2 x 4 y 2 15x 9x 2 x 3 y x 3 6x 2 9x y x 3 x EXERCISES 4.5 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS \|\|\|\| 315 68. Show that the curve has two slant asymptotes: and . Use this fact to help sketch the curve. 69. Show that the lines and are slant asymptotes of the hyperbola . 70. Let . Show that This shows that the graph of approaches the graph of , and we say that the curve is asymptotic to the parabola . Use this fact to help sketch the graph of . 71. Discuss the asymptotic behavior of in the same manner as in Exercise 70. Then use your results to help sketch the graph of . 72. Use the asymptotic behavior of to sketch its graph without going through the curve-sketching procedure of this section. f !x" ! cos x % 1#x 2 f f !x" ! !x 4 % 1"#x f y ! x 2 y ! f !x" y ! x 2 f lim x l+" ' f !x" ! x 2( ! 0 f !x" ! !x 3 % 1"#x !x 2#a 2" ! !y 2#b 2" ! 1 y ! !!b#a"x y ! !b#a"x y ! !x ! 2 y ! x % 2 y ! sx2 % 4x 57–60 Find an equation of the slant asymptote. Do not sketch the curve. 58. 59. 60. 61–66 Use the guidelines of this section to sketch the curve. In guideline D ﬁnd an equation of the slant asymptote. 61. 62. 63. 64. 65. 66. 67. Show that the curve has two slant asymptotes: and . Use this fact to help sketch the curve. y ! x ! (#2 y ! x % (#2 y ! x ! tan!1x y ! !x % 1"3 !x ! 1"2 y ! 2x 3 % x 2 % 1 x 2 % 1 y ! e x ! x xy ! x 2 % 4 y ! x 2 % 12 x ! 2 y ! !2x 2 % 5x ! 1 2x ! 1 y ! 5x 4 % x 2 % x x 3 ! x 2 % 2 y ! 4x 3 ! 2x 2 % 5 2x 2 % x ! 3 y ! 2x 3 % x 2 % x % 3 x 2 % 2x y ! x2 % 1 x % 1 57. GRAPHING WITH CALCULUS AND CALCULATORS The method we used to sketch curves in the preceding section was a culmination of much of our study of differential calculus. The graph was the ﬁnal object that we produced. In this section our point of view is completely different. Here we start with a graph produced by a graphing calculator or computer and then we reﬁne it. We use calculus to make sure that we reveal all the important aspects of the curve. And with the use of graphing devices we can tackle curves that would be far too complicated to consider without technology. The theme is the interaction between calculus and calculators. EXAMPLE 1 Graph the polynomial . Use the graphs of and to estimate all maximum and minimum points and intervals of concavity. SOLUTION If we specify a domain but not a range, many graphing devices will deduce a suitable range from the values computed. Figure 1 shows the plot from one such device if we specify that . Although this viewing rectangle is useful for showing that the asymptotic behavior (or end behavior) is the same as for , it is obviously hiding some ﬁner detail. So we change to the viewing rectangle by shown in Figure 2. From this graph it appears that there is an absolute minimum value of about 3 when (by using the cursor) and is decreasing on and increas-ing on . Also, there appears to be a horizontal tangent at the origin and inﬂec-tion points when and when is somewhere between and . Now let’s try to conﬁrm these impressions using calculus. We differentiate and get f #!x" ! 60x 4 % 60x 3 % 18x ! 4 f&!x" ! 12x 5 % 15x 4 % 9x 2 ! 4x !1 !2 x x ! 0 !!1.62, "" !!", !1.62" f x + !1.62 !15.3 '!50, 100( '!3, 2( y ! 2x 6 !5 x 5 f # f& f!x" ! 2x 6 % 3x 5 % 3x 3 ! 2x 2 4.6 41,000 _1000 _5 5 y=ƒ FIGURE 1 100 _50 _3 2 y=ƒ FIGURE 2 N If you have not already read Section 1.4, you should do so now. In particular, it explains how to avoid some of the pitfalls of graphing devices by choosing appropriate viewing rectangles. When we graph in Figure 3 we see that changes from negative to positive when ; this conﬁrms (by the First Derivative Test) the minimum value that we found earlier. But, perhaps to our surprise, we also notice that changes from posi-tive to negative when and from negative to positive when . This means that has a local maximum at 0 and a local minimum when , but these were hidden in Figure 2. Indeed, if we now zoom in toward the origin in Figure 4, we see what we missed before: a local maximum value of 0 when and a local minimum value of about when . What about concavity and inﬂection points? From Figures 2 and 4 there appear to be inﬂection points when is a little to the left of and when is a little to the right of 0. But it’s difﬁcult to determine inﬂection points from the graph of , so we graph the sec-ond derivative in Figure 5. We see that changes from positive to negative when and from negative to positive when . So, correct to two decimal places, is concave upward on and and concave downward on . The inﬂection points are and . We have discovered that no single graph reveals all the important features of this polynomial. But Figures 2 and 4, when taken together, do provide an accurate picture. M EXAMPLE 2 Draw the graph of the function in a viewing rectangle that contains all the important features of the function. Estimate the maximum and minimum values and the intervals of concavity. Then use calculus to ﬁnd these quantities exactly. SOLUTION Figure 6, produced by a computer with automatic scaling, is a disaster. Some graphing calculators use by as the default viewing rectangle, so let’s try it. We get the graph shown in Figure 7; it’s a major improvement. The -axis appears to be a vertical asymptote and indeed it is because Figure 7 also allows us to estimate the -intercepts: about and . The exact val-ues are obtained by using the quadratic formula to solve the equation ; we get . To get a better look at horizontal asymptotes, we change to the viewing rectangle by in Figure 8. It appears that is the horizontal asymptote and this is easily conﬁrmed: lim x l+" x 2 % 7x % 3 x 2 ! lim x l+" )1 % 7 x % 3 x 2 ! 1 y ! 1 '!5, 10( '!20, 20( 3 - 10! _5 5 y=ƒ FIGURE 6 10 _10 _10 10 y=ƒ FIGURE 7 10 _5 _20 20 y=ƒ y=1 FIGURE 8 x ! (!7 + s37 )#2 x 2 % 7x % 3 ! 0 !6.5 !0.5 x lim x l 0 x 2 % 7x % 3 x 2 ! " y '!10, 10( '!10, 10( f!x" ! x 2 % 7x % 3 x 2 V !0.19, !0.05" !!1.23, !10.18" !!1.23, 0.19" !0.19, "" !!", !1.23" f x + 0.19 x + !1.23 f # f # f x !1 x x + 0.35 !0.1 x ! 0 x + 0.35 f x + 0.35 x ! 0 f&!x" x + !1.62 f&!x" f& 316 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 10 _30 _3 2 y=f·(x) FIGURE 5 20 _5 _3 2 y=fª(x) FIGURE 3 1 _1 _1 1 y=ƒ FIGURE 4 To estimate the minimum value we zoom in to the viewing rectangle by in Figure 9. The cursor indicates that the absolute minimum value is about when , and we see that the function decreases on and and increases on . The exact values are obtained by differentiating: This shows that when and when and when . The exact minimum value is . Figure 9 also shows that an inﬂection point occurs somewhere between and . We could estimate it much more accurately using the graph of the second deriv-ative, but in this case it’s just as easy to ﬁnd exact values. Since we see that when . So is concave upward on and and concave downward on . The inﬂection point is . The analysis using the ﬁrst two derivatives shows that Figures 7 and 8 display all the major aspects of the curve. M EXAMPLE 3 Graph the function . SOLUTION Drawing on our experience with a rational function in Example 2, let’s start by graphing in the viewing rectangle by . From Figure 10 we have the feeling that we are going to have to zoom in to see some ﬁner detail and also zoom out to see the larger picture. But, as a guide to intelligent zooming, let’s ﬁrst take a close look at the expression for . Because of the factors and in the denominator, we expect and to be the vertical asymptotes. Indeed To ﬁnd the horizontal asymptotes we divide numerator and denominator by : This shows that , so the -axis is a horizontal asymptote. It is also very useful to consider the behavior of the graph near the -intercepts using an analysis like that in Example 11 in Section 2.6. Since is positive, does not change sign at 0 and so its graph doesn’t cross the -axis at 0. But, because of the factor , the graph does cross the -axis at and has a horizontal tangent there. Putting all this information together, but without using derivatives, we see that the curve has to look something like the one in Figure 11. !1 x !x % 1"3 x f!x" x 2 x x f!x"l 0 as xl +" x 2!x % 1"3 !x ! 2"2!x ! 4"4 ! x 2 x 3 ! !x % 1"3 x 3 !x ! 2"2 x 2 ! !x ! 4"4 x 4 ! 1 x)1 % 1 x 3 )1 ! 2 x 2)1 ! 4 x 4 x 6 lim x l 4 x 2!x % 1"3 !x ! 2"2!x ! 4"4 ! " and lim x l2 x 2!x % 1"3 !x ! 2"2!x ! 4"4 ! " x ! 4 x ! 2 !x ! 4"4 !x ! 2"2 f!x" '!10, 10( '!10, 10( f f!x" ! x 2!x % 1"3 !x ! 2"2!x ! 4"4 V (! 9 7, ! 71 27) (!", ! 9 7) !0, "" (! 9 7, 0) f !x " 0" x $ ! 9 7 f #!x" $ 0 f #!x" ! 14 x 3 % 18 x 4 ! 2(7x % 9" x 4 x ! !2 x ! !1 f (! 6 7) ! ! 37 12 + !3.08 x $ 0 x ' ! 6 7 f&!x" ' 0 ! 6 7 ' x ' 0 f&!x" $ 0 f&!x" ! ! 7 x 2 ! 6 x 3 ! ! 7x % 6 x 3 !!0.9, 0" !0, "" !!", !0.9" x + !0.9 !3.1 '!4, 2( '!3, 0( SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS \|\|\|\| 317 2 _4 _3 0 y=ƒ FIGURE 9 10 _10 _10 10 y=ƒ FIGURE 10 FIGURE 11 x y 1 2 3 _1 4 Now that we know what to look for, we zoom in (several times) to produce the graphs in Figures 12 and 13 and zoom out (several times) to get Figure 14. We can read from these graphs that the absolute minimum is about and occurs when . There is also a local maximum when and a local minimum when . These graphs also show three inﬂection points near and and two between and . To estimate the inﬂection points closely we would need to graph , but to compute by hand is an unreasonable chore. If you have a computer algebra system, then it’s easy to do (see Exercise 15). We have seen that, for this particular function, three graphs (Figures 12, 13, and 14) are necessary to convey all the useful information. The only way to display all these features of the function on a single graph is to draw it by hand. Despite the exaggera-tions and distortions, Figure 11 does manage to summarize the essential nature of the function. M EXAMPLE 4 Graph the function . For , estimate all maximum and minimum values, intervals of increase and decrease, and inﬂection points correct to one decimal place. SOLUTION We ﬁrst note that is periodic with period . Also, is odd and for all . So the choice of a viewing rectangle is not a problem for this function: We start with by . (See Figure 15.) It appears that there are three local maximum values and two local minimum values in that window. To conﬁrm this and locate them more accurately, we calculate that and graph both and in Figure 16. f& f f&!x" ! cos!x % sin 2x" ! !1 % 2 cos 2x" 1.1 _1.1 0 FIGURE 15 1.2 _1.2 0 π π y=ƒ y=fª(x) FIGURE 16 '!1.1, 1.1( '0, (( x % f!x"% 1 f 2( f 0 x ( f!x" ! sin!x % sin 2x" f # f # 0 !1 !1 !5, !35, x + 2.5 +211 x + !0.3 +0.00002 x + !20 !0.02 0.05 _0.05 _100 1 y=ƒ FIGURE 12 0.0001 _0.0001 _1.5 0.5 y=ƒ FIGURE 13 500 _10 _1 10 y=ƒ FIGURE 14 318 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION N The family of functions where is a constant, occurs in applications to frequency modulation (FM) synthesis. A sine wave is modulated by a wave with a different frequency . The case where is studied in Example 4. Exercise 25 explores another special case. c ! 2 !sin cx" c f !x" ! sin!x % sin cx" Using zoom-in and the First Derivative Test, we ﬁnd the following values to one deci-mal place. The second derivative is Graphing both and in Figure 17, we obtain the following approximate values: Having checked that Figure 15 does indeed represent accurately for , we can state that the extended graph in Figure 18 represents accurately for . M Our ﬁnal example is concerned with families of functions. As discussed in Section 1.4, this means that the functions in the family are related to each other by a formula that con-tains one or more arbitrary constants. Each value of the constant gives rise to a member of the family and the idea is to see how the graph of the function changes as the constant changes. EXAMPLE 5 How does the graph of vary as varies? SOLUTION The graphs in Figures 19 and 20 (the special cases and ) show two very different-looking curves. Before drawing any more graphs, let’s see what mem-bers of this family have in common. Since for any value of , they all have the -axis as a horizontal asymptote. A vertical asymp-tote will occur when . Solving this quadratic equation, we get . When , there is no vertical asymptote (as in Figure 19). When , the graph has a single vertical asymptote because When , there are two vertical asymptotes: (as in Figure 20). Now we compute the derivative: This shows that when (if ), when , and x ' !1 f&!x" $ 0 c " 1 x ! !1 f&!x" ! 0 f&!x" ! ! 2x % 2 !x 2 % 2x % c"2 x ! !1 + s1 ! c c ' 1 lim x l!1 1 x 2 % 2x % 1 ! lim x l!1 1 !x % 1"2 ! " x ! !1 c ! 1 c $ 1 x ! !1 + s1 ! c x 2 % 2x % c ! 0 x c lim x l+" 1 x 2 % 2x % c ! 0 c ! !2 c ! 2 c f!x" ! 1#!x 2 % 2x % c" V !2( x 2( f 0 x ( f Inflection points: !0, 0", !0.8, 0.97", !1.3, 0.97", !1.8, 0.97", !2.3, 0.97" Concave downward on: !0, 0.8", !1.3, 1.8", !2.3, (" Concave upward on: !0.8, 1.3", !1.8, 2.3" f # f f #!x" ! !!1 % 2 cos 2x"2 sin!x % sin 2x" ! 4 sin 2x cos!x % sin 2x" Local minimum values: f!1.0" + 0.94, f!2.1" + 0.94 Local maximum values: f!0.6" + 1, f!1.6" + 1, f!2.5" + 1 Intervals of decrease: !0.6, 1.0", !1.6, 2.1", !2.5, (" Intervals of increase: !0, 0.6", !1.0, 1.6", !2.1, 2.5" SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS \|\|\|\| 319 FIGURE 19 c=2 y= 1 ≈+2x+2 2 _2 _5 4 1.2 _1.2 0 π f f · FIGURE 17 1.2 _1.2 _2π 2π FIGURE 18 FIGURE 20 c=_2 2 _2 _5 4 y= 1 ≈+2x-2 when . For , this means that increases on and decreases on . For , there is an absolute maximum value . For , is a local maximum value and the intervals of increase and decrease are interrupted at the vertical asymptotes. Figure 21 is a “slide show” displaying ﬁve members of the family, all graphed in the viewing rectangle by . As predicted, is the value at which a transi-tion takes place from two vertical asymptotes to one, and then to none. As increases from , we see that the maximum point becomes lower; this is explained by the fact that as . As decreases from , the vertical asymptotes become more widely separated because the distance between them is , which becomes large as . Again, the maximum point approaches the -axis because as . There is clearly no inﬂection point when . For we calculate that and deduce that inﬂection points occur when . So the inﬂection points become more spread out as increases and this seems plausible from the last two parts of Figure 21. M c x ! !1 " s3!c ! 1"#3 f #!x" ! 2!3x 2 $ 6x $ 4 ! c" !x 2 $ 2x $ c"3 c % 1 c & 1 c=3 c=2 c=1 c=0 c=_1 FIGURE 21 The family of functions ƒ=1/(≈+2x+c) c l !' 1#!c ! 1" l 0 x c l !' 2s1 ! c 1 c c l ' 1#!c ! 1" l 0 1 c c ! 1 $!2, 2% $!5, 4% f!!1" ! 1#!c ! 1" c ( 1 f!!1" ! 1#!c ! 1" c % 1 !!1, '" !!', !1" f c ) 1 x % !1 f!x" ( 0 320 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION See an animation of Figure 21 in Visual 4.6. TEC 9–10 Produce graphs of that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavity, and use calculus to ﬁnd these intervals exactly. 9. 10. 11–12 (a) Graph the function. (b) Use l’Hospital’s Rule to explain the behavior as . (c) Estimate the minimum value and intervals of concavity. Then use calculus to ﬁnd the exact values. 12. f !x" ! xe1#x f !x" ! x 2 ln x 11. x l 0 f !x" ! 1 x 8 ! 2 + 108 x 4 f !x" ! 1 $ 1 x $ 8 x 2 $ 1 x 3 f 1–8 Produce graphs of that reveal all the important aspects of the curve. In particular, you should use graphs of and to esti-mate the intervals of increase and decrease, extreme values, inter-vals of concavity, and inﬂection points. 1. 2. 3. 4. 5. 6. 7. , 8. f !x" ! e x x 2 ! 9 !4 & x & 4 f !x" ! x 2 ! 4x $ 7 cos x f !x" ! tan x $ 5 cos x f !x" ! x x 3 ! x 2 ! 4x $ 1 f !x" ! x 2 ! 1 40x 3 $ x $ 1 f !x" ! x 6 ! 10x 5 ! 400x 4 $ 2500x 3 f !x" ! x6 ! 15x 5 $ 75x 4 ! 125x 3 ! x f !x" ! 4x 4 ! 32x 3 $ 89x 2 ! 95x $ 29 f # f f ; EXERCISES 4.6 the same time. Find all the maximum and minimum values and inﬂection points. Then graph in the viewing rectangle by and comment on symmetry. 26–33 Describe how the graph of varies as varies. Graph several members of the family to illustrate the trends that you dis-cover. In particular, you should investigate how maximum and minimum points and inﬂection points move when changes. You should also identify any transitional values of at which the basic shape of the curve changes. 26. 27. 28. 29. 32. 33. The family of functions , where , , and are positive numbers and , has been used to model the concentration of a drug injected into the blood-stream at time . Graph several members of this family. What do they have in common? For ﬁxed values of and , discover graphically what happens as increases. Then use calculus to prove what you have discovered. 35. Investigate the family of curves given by , where is a real number. Start by computing the limits as . Identify any transitional values of where the basic shape changes. What happens to the maximum or minimum points and inﬂection points as changes? Illustrate by graphing sev-eral members of the family. 36. Investigate the family of curves given by the equation . Start by determining the transitional value of at which the number of inﬂection points changes. Then graph several members of the family to see what shapes are possible. There is another transitional value of at which the number of critical numbers changes. Try to discover it graphically. Then prove what you have discovered. 37. (a) Investigate the family of polynomials given by the equa-tion . For what values of does the curve have minimum points? (b) Show that the minimum and maximum points of every curve in the family lie on the parabola . Illus-trate by graphing this parabola and several members of the family. 38. (a) Investigate the family of polynomials given by the equa-tion . For what values of does the curve have maximum and minimum points? (b) Show that the minimum and maximum points of every curve in the family lie on the curve . Illustrate by graphing this curve and several members of the family. y ! x ! x 3 c f !x" ! 2x 3 $ cx 2 $ 2x y ! 1 ! x 2 c f !x" ! cx 4 ! 2x 2 $ 1 c c f !x" ! x 4 $ cx 2 $ x c c x l "' c f !x" ! xe!cx b a C t ! 0 b % a C b a f !t" ! C!e!at ! e!bt" 34. f !x" ! cx $ sin x f !x" ! 1 !1 ! x 2"2 $ cx 2 f !x" ! cx 1 $ c 2x 2 31. f !x" ! ln!x 2 $ c" 30. f !x" ! e!c#x 2 f !x" ! xs c 2 ! x 2 f !x" ! x 4 $ cx 2 f !x" ! x 3 $ cx c c c f $!1.2, 1.2% $!2,, 2,% f 13–14 Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to producing graphs (with a graphing device) that display the major features of the curve. Use these graphs to estimate the maximum and mini-mum values. 14. 15. If is the function considered in Example 3, use a computer algebra system to calculate and then graph it to conﬁrm that all the maximum and minimum values are as given in the example. Calculate and use it to estimate the intervals of concavity and inﬂection points. 16. If is the function of Exercise 14, ﬁnd and and use their graphs to estimate the intervals of increase and decrease and concavity of . 17–22 Use a computer algebra system to graph and to ﬁnd and . Use graphs of these derivatives to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inﬂection points of . 17. 18. 19. , 20. 21. 22. 23–24 (a) Graph the function. (b) Explain the shape of the graph by computing the limit as or as . (c) Estimate the maximum and minimum values and then use calculus to ﬁnd the exact values. (d) Use a graph of to estimate the x-coordinates of the inﬂec-tion points. 24. 25. In Example 4 we considered a member of the family of func-tions that occur in FM synthesis. Here we investigate the function with . Start by graphing in the viewing rectangle by . How many local maximum points do you see? The graph has more than are visible to the naked eye. To discover the hidden maximum and minimum points you will need to examine the graph of very carefully. In fact, it helps to look at the graph of at f # f $!1.2, 1.2% $0, ,% f c ! 3 f !x" ! sin!x $ sin cx" f !x" ! !sin x"sin x f !x" ! x 1#x 23. f # x l ' x l 0$ CAS f !x" ! 1 1 $ etan x f !x" ! 1 ! e1#x 1 $ e1#x f !x" ! !x 2 ! 1"e arctan x x & 20 f !x" ! sx $ 5 sin x f !x" ! x 2#3 1 $ x $ x 4 f !x" ! sx x 2 $ x $ 1 f f # f f CAS f f # f f CAS f # f f CAS f !x" ! !2x $ 3"2!x ! 2"5 x 3!x ! 5"2 f !x" ! !x $ 4"!x ! 3"2 x 4!x ! 1" 13. SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS \|\|\|\| 321 322 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION OPTIMIZATION PROBLEMS The methods we have learned in this chapter for ﬁnding extreme values have practical applications in many areas of life. A businessperson wants to minimize costs and maxi-mize proﬁts. A traveler wants to minimize transportation time. Fermat’s Principle in optics states that light follows the path that takes the least time. In this section and the next we solve such problems as maximizing areas, volumes, and proﬁts and minimizing distances, times, and costs. In solving such practical problems the greatest challenge is often to convert the word problem into a mathematical optimization problem by setting up the function that is to be maximized or minimized. Let’s recall the problem-solving principles discussed on page 76 and adapt them to this situation: STEPS IN SOLVING OPTIMIZATION PROBLEMS 1. Understand the Problem The ﬁrst step is to read the problem carefully until it is clearly understood. Ask yourself: What is the unknown? What are the given quanti-ties? What are the given conditions? 2. Draw a Diagram In most problems it is useful to draw a diagram and identify the given and required quantities on the diagram. 3. Introduce Notation Assign a symbol to the quantity that is to be maximized or minimized (let’s call it for now). Also select symbols for other unknown quantities and label the diagram with these symbols. It may help to use initials as suggestive symbols—for example, for area, for height, for time. 4. Express in terms of some of the other symbols from Step 3. 5. If has been expressed as a function of more than one variable in Step 4, use the given information to ﬁnd relationships (in the form of equations) among these variables. Then use these equations to eliminate all but one of the variables in the expression for . Thus will be expressed as a function of one variable , say, . Write the domain of this function. 6. Use the methods of Sections 4.1 and 4.3 to ﬁnd the absolute maximum or mini-mum value of . In particular, if the domain of is a closed interval, then the Closed Interval Method in Section 4.1 can be used. EXAMPLE 1 A farmer has 2400 ft of fencing and wants to fence off a rectangular ﬁeld that borders a straight river. He needs no fence along the river. What are the dimensions of the ﬁeld that has the largest area? SOLUTION In order to get a feeling for what is happening in this problem, let’s experiment with some special cases. Figure 1 (not to scale) shows three possible ways of laying out the 2400 ft of fencing. FIGURE 1 100 2200 100 Area=100 · 2200=220,000 ft@ 700 1000 700 Area=700 · 1000=700,000 ft@ 1000 400 1000 Area=1000 · 400=400,000 ft@ f f Q ! f!x" x Q Q Q Q t h A !a, b, c, . . . , x, y" Q 4.7 N Understand the problem N Analogy: Try special cases N Draw diagrams SECTION 4.7 OPTIMIZATION PROBLEMS \|\|\|\| 323 We see that when we try shallow, wide ﬁelds or deep, narrow ﬁelds, we get relatively small areas. It seems plausible that there is some intermediate conﬁguration that pro-duces the largest area. Figure 2 illustrates the general case. We wish to maximize the area of the rectangle. Let and be the depth and width of the rectangle (in feet). Then we express in terms of and : We want to express as a function of just one variable, so we eliminate by expressing it in terms of . To do this we use the given information that the total length of the fenc-ing is 2400 ft. Thus From this equation we have , which gives Note that 0 and (otherwise ). So the function that we wish to maxi-mize is The derivative is , so to ﬁnd the critical numbers we solve the equation which gives . The maximum value of must occur either at this critical number or at an endpoint of the interval. Since , and , the Closed Interval Method gives the maximum value as . [Alternatively, we could have observed that for all , so is always concave downward and the local maximum at must be an absolute maximum.] Thus the rectangular ﬁeld should be 600 ft deep and 1200 ft wide. M EXAMPLE 2 A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. SOLUTION Draw the diagram as in Figure 3, where is the radius and the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). From Figure 4 we see that the sides are made from a rectangular sheet with dimensions and h. So the surface area is To eliminate we use the fact that the volume is given as 1 L, which we take to be 1000 cm . Thus which gives . Substitution of this into the expression for gives A ! 2,r 2 $ 2,r& 1000 ,r 2 ' ! 2,r 2 $ 2000 r A h ! 1000#!,r 2" ,r 2h ! 1000 3 h A ! 2,r 2 $ 2,rh 2,r h r V x ! 600 A x A#!x" ! !4 ( 0 A!600" ! 720,000 A!1200" ! 0 A!0" ! 0, A!600" ! 720,000 A x ! 600 2400 ! 4x ! 0 A !x" ! 2400 ! 4x 0 & x & 1200 A!x" ! 2400x ! 2x 2 A ( 0 x & 1200 x ) A ! x!2400 ! 2x" ! 2400x ! 2x 2 y ! 2400 ! 2x 2x $ y ! 2400 x y A A ! xy y x A y x A x y A x FIGURE 2 N Introduce notation FIGURE 3 r h r Area 2{πr@} FIGURE 4 Area (2πr)h 2πr h Therefore the function that we want to minimize is To ﬁnd the critical numbers, we differentiate: Then when , so the only critical number is . Since the domain of is , we can’t use the argument of Example 1 concerning endpoints. But we can observe that for and for , so is decreasing for all to the left of the critical number and increas-ing for all to the right. Thus must give rise to an absolute minimum. [Alternatively, we could argue that as and as , so there must be a minimum value of , which must occur at the critical number. See Figure 5.] The value of corresponding to is Thus, to minimize the cost of the can, the radius should be cm and the height should be equal to twice the radius, namely, the diameter. M The argument used in Example 2 to justify the absolute minimum is a variant of the First Derivative Test (which applies only to local maximum or minimum values) and is stated here for future reference. FIRST DERIVATIVE TEST FOR ABSOLUTE EXTREME VALUES Suppose that is a criti-cal number of a continuous function deﬁned on an interval. (a) If for all and for all , then is the absolute maximum value of . (b) If for all and for all , then is the absolute minimum value of . An alternative method for solving optimization problems is to use implicit dif-ferentiation. Let’s look at Example 2 again to illustrate the method. We work with the same equations but instead of eliminating h, we differentiate both equations implicitly with respect to r: The minimum occurs at a critical number, so we set , simplify, and arrive at the equations and subtraction gives , or . h ! 2r 2r ! h ! 0 2h $ rh ! 0 2r $ h $ rh ! 0 A ! 0 2,rh $ ,r 2h ! 0 A ! 4,r $ 2,h $ 2,rh ,r 2h ! 100 A ! 2,r 2 $ 2,rh NOTE 2 f f!c" x % c f!x" % 0 x ( c f!x" ( 0 f f!c" x % c f!x" ( 0 x ( c f!x" % 0 f c NOTE 1 s 3 500#, h ! 1000 ,r 2 ! 1000 ,!500#,"2#3 ! 2( 3 500 , ! 2r r ! s 3 500#, h A!r" r l ' A!r" l ' r l 0$ A!r" l ' r ! s 3 500#, r r A r % s 3 500#, A!r" % 0 r ( s 3 500#, A!r" ( 0 !0, '" A r ! s 3 500#, ,r 3 ! 500 A!r" ! 0 A !r" ! 4,r ! 2000 r 2 ! 4!,r 3 ! 500" r 2 r % 0 A!r" ! 2,r 2 $ 2000 r 324 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION r y 0 10 1000 y=A(r) FIGURE 5 N In the Applied Project on page 333 we investi-gate the most economical shape for a can by taking into account other manufacturing costs. Module 4.7 takes you through six additional optimization problems, including animations of the physical situations. TEC EXAMPLE 3 Find the point on the parabola that is closest to the point . SOLUTION The distance between the point and the point is (See Figure 6.) But if lies on the parabola, then , so the expression for becomes (Alternatively, we could have substituted to get in terms of alone.) Instead of minimizing , we minimize its square: (You should convince yourself that the minimum of occurs at the same point as the minimum of , but is easier to work with.) Differentiating, we obtain so when . Observe that when and when , so by the First Derivative Test for Absolute Extreme Values, the absolute mini-mum occurs when . (Or we could simply say that because of the geometric nature of the problem, it’s obvious that there is a closest point but not a farthest point.) The corresponding value of is . Thus the point on closest to is . M EXAMPLE 4 A man launches his boat from point on a bank of a straight river, 3 km wide, and wants to reach point , 8 km downstream on the opposite bank, as quickly as possible (see Figure 7). He could row his boat directly across the river to point and then run to , or he could row directly to , or he could row to some point between and and then run to . If he can row 6 km#h and run 8 km#h, where should he land to reach as soon as possible? (We assume that the speed of the water is negligible com-pared with the speed at which the man rows.) SOLUTION If we let be the distance from to , then the running distance is and the Pythagorean Theorem gives the rowing distance as . We use the equation Then the rowing time is and the running time is , so the total time as a function of is The domain of this function is . Notice that if , he rows to and if , he rows directly to . The derivative of is T!x" ! x 6sx 2 $ 9 ! 1 8 T B x ! 8 C x ! 0 $0, 8% T T!x" ! sx 2 $ 9 6 $ 8 ! x 8 x T !8 ! x"#8 sx 2 $ 9 #6 time ! distance rate )AD) ! sx 2 $ 9 )DB) ! 8 ! x D C x B B B C D B B C B A !2, 2" !1, 4" y 2 ! 2x x ! 1 2 y 2 ! 2 x y ! 2 y % 2 f!y" % 0 y ( 2 f!y" ( 0 y ! 2 f!y" ! 0 f!y" ! 2( 1 2 y 2 ! 1)y $ 2!y ! 4" ! y 3 ! 8 d 2 d 2 d d 2 ! f!y" ! ( 1 2 y 2 ! 1) 2 $ !y ! 4"2 d x d y ! s2x d ! s( 1 2 y 2 ! 1)2 $ !y ! 4"2 d x ! 1 2 y 2 !x, y" d ! s!x ! 1"2 $ !y ! 4"2 !x, y" !1, 4" !1, 4" y 2 ! 2x V SECTION 4.7 OPTIMIZATION PROBLEMS \|\|\|\| 325 x y 0 1 1 2 3 4 ¥=2x (1, 4) (x, y) FIGURE 6 8 km C D B A 3 km FIGURE 7 Thus, using the fact that , we have The only critical number is . To see whether the minimum occurs at this criti-cal number or at an endpoint of the domain , we evaluate at all three points: Since the smallest of these values of occurs when , the absolute minimum value of must occur there. Figure 8 illustrates this calculation by showing the graph of . Thus the man should land the boat at a point km ( km) downstream from his starting point. M EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of radius . SOLUTION 1 Let’s take the semicircle to be the upper half of the circle with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the -axis as shown in Figure 9. Let be the vertex that lies in the ﬁrst quadrant. Then the rectangle has sides of lengths and , so its area is To eliminate we use the fact that lies on the circle and so . Thus The domain of this function is . Its derivative is which is 0 when , that is, (since ). This value of gives a maximum value of since and . Therefore the area of the largest inscribed rectangle is SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let be the angle shown in Figure 10. Then the area of the rectangle is A!-" ! !2r cos -"!r sin -" ! r 2!2 sin - cos -" ! r 2 sin 2--A& r s2 ' ! 2 r s2 (r 2 ! r 2 2 ! r 2 A!r" ! 0 A!0" ! 0 A x x ) 0 x ! r#s2 2x 2 ! r 2 A ! 2sr 2 ! x 2 ! 2x 2 sr 2 ! x 2 ! 2!r 2 ! 2x 2" sr 2 ! x 2 0 & x & r A ! 2xsr 2 ! x 2 y ! sr 2 ! x 2 x 2 $ y 2 ! r 2 !x, y" y A ! 2xy y 2x !x, y" x x 2 $ y 2 ! r 2 r V 3.4 9#s7 T T x ! 9#s7 T T!8" ! s73 6 1.42 T& 9 s7 ' ! 1 $ s7 8 1.33 T!0" ! 1.5 T $0, 8% x ! 9#s7 x ! 9 s7 & ? 7x 2 ! 81 & ? 16x 2 ! 9!x 2 $ 9" & ? 4x ! 3sx 2 $ 9 & ? x 6sx 2 $ 9 ! 1 8 & ? T!x" ! 0 x ) 0 326 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION FIGURE 8 x T 0 1 2 4 6 y=T(x) x y 0 2x (x, y) y _r r FIGURE 9 r ¨ r cos ¨ r sin ¨ FIGURE 10 We know that has a maximum value of 1 and it occurs when . So has a maximum value of and it occurs when . Notice that this trigonometric solution doesn’t involve differentiation. In fact, we didn’t need to use calculus at all. M APPLICATIONS TO BUSINESS AND ECONOMICS In Section 3.7 we introduced the idea of marginal cost. Recall that if , the cost func-tion, is the cost of producing units of a certain product, then the marginal cost is the rate of change of with respect to . In other words, the marginal cost function is the deriva-tive, , of the cost function. Now let’s consider marketing. Let be the price per unit that the company can charge if it sells units. Then is called the demand function (or price function) and we would expect it to be a decreasing function of . If units are sold and the price per unit is , then the total revenue is and is called the revenue function. The derivative of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold. If units are sold, then the total proﬁt is and is called the proﬁt function. The marginal proﬁt function is , the derivative of the proﬁt function. In Exercises 53–58 you are asked to use the marginal cost, revenue, and proﬁt functions to minimize costs and maximize revenues and proﬁts. EXAMPLE 6 A store has been selling 200 DVD burners a week at each. A mar-ket survey indicates that for each rebate offered to buyers, the number of units sold will increase by 20 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize its revenue? SOLUTION If is the number of DVD burners sold per week, then the weekly increase in sales is . For each increase of 20 units sold, the price is decreased by . So for each additional unit sold, the decrease in price will be and the demand function is The revenue function is Since , we see that when . This value of gives an absolute maximum by the First Derivative Test (or simply by observing that the graph of is a parabola that opens downward). The corresponding price is and the rebate is . Therefore, to maximize revenue, the store should offer a rebate of . M $125 350 ! 225 ! 125 p!450" ! 450 ! 1 2!450" ! 225 R x x ! 450 R!x" ! 0 R!x" ! 450 ! x R!x" ! xp!x" ! 450x ! 1 2 x 2 p!x" ! 350 ! 10 20!x ! 200" ! 450 ! 1 2 x 1 20 + 10 $10 x ! 200 x $10 $350 V P P P!x" ! R!x" ! C!x" x R R R!x" ! xp!x" p!x" x x p x p!x" C!x" x C x C!x" - ! ,#4 r 2 A!-" 2- ! ,#2 sin 2-SECTION 4.7 OPTIMIZATION PROBLEMS \|\|\|\| 327 328 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the total area as a function of one variable. (f) Finish solving the problem and compare the answer with your estimate in part (a). 10. Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have. (a) Draw several diagrams to illustrate the situation, some short boxes with large bases and some tall boxes with small bases. Find the volumes of several such boxes. Does it appear that there is a maximum volume? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols. (c) Write an expression for the volume. (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the volume as a function of one variable. (f) Finish solving the problem and compare the answer with your estimate in part (a). 11. A farmer wants to fence an area of 1.5 million square feet in a rectangular ﬁeld and then divide it in half with a fence parallel to one of the sides of the rectangle. How can he do this so as to minimize the cost of the fence? 12. A box with a square base and open top must have a volume of 32,000 cm . Find the dimensions of the box that minimize the amount of material used. If 1200 cm of material is available to make a box with a square base and an open top, ﬁnd the largest possible volume of the box. 14. A rectangular storage container with an open top is to have a volume of 10 m . The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container. 15. Do Exercise 14 assuming the container has a lid that is made from the same material as the sides. (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square. (b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square. Find the point on the line that is closest to the origin. 18. Find the point on the line that is closest to the point . Find the points on the ellipse that are farthest away from the point . 1, 0 4x 2 y 2 4 19. 3, 1 6x y 9 y 4x 7 17. 16. 3 2 13. 3 1. Consider the following problem: Find two numbers whose sum is 23 and whose product is a maximum. (a) Make a table of values, like the following one, so that the sum of the numbers in the ﬁrst two columns is always 23. On the basis of the evidence in your table, estimate the answer to the problem. (b) Use calculus to solve the problem and compare with your answer to part (a). 2. Find two numbers whose difference is 100 and whose product is a minimum. 3. Find two positive numbers whose product is 100 and whose sum is a minimum. 4. Find a positive number such that the sum of the number and its reciprocal is as small as possible. 5. Find the dimensions of a rectangle with perimeter 100 m whose area is as large as possible. 6. Find the dimensions of a rectangle with area whose perimeter is as small as possible. 7. A model used for the yield of an agricultural crop as a func-tion of the nitrogen level in the soil (measured in appropriate units) is where is a positive constant. What nitrogen level gives the best yield? 8. The rate at which photosynthesis takes place for a species of phytoplankton is modeled by the function where is the light intensity (measured in thousands of foot-candles). For what light intensity is a maximum? 9. Consider the following problem: A farmer with 750 ft of fenc-ing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? (a) Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these conﬁgurations. Does it appear that there is a maximum area? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols. (c) Write an expression for the total area. P I P 100I I 2 I 4 in mg carbonm3h k Y kN 1 N 2 N Y 1000 m2 EXERCISES 4.7 First number Second number Product 1 22 22 2 21 42 3 20 60 . . . . . . . . . der that will reach from the ground over the fence to the wall of the building? 37. A cone-shaped drinking cup is made from a circular piece of paper of radius by cutting out a sector and joining the edges and . Find the maximum capacity of such a cup. 38. A cone-shaped paper drinking cup is to be made to hold of water. Find the height and radius of the cup that will use the smallest amount of paper. 39. A cone with height is inscribed in a larger cone with height so that its vertex is at the center of the base of the larger cone. Show that the inner cone has maximum volume when . 40. An object with weight is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle with a plane, then the magnitude of the force is where is a constant called the coefﬁcient of friction. For what value of is smallest? 41. If a resistor of ohms is connected across a battery of volts with internal resistance ohms, then the power (in watts) in the external resistor is If and are ﬁxed but varies, what is the maximum value of the power? 42. For a ﬁsh swimming at a speed relative to the water, the energy expenditure per unit time is proportional to . It is believed that migrating ﬁsh try to minimize the total energy required to swim a ﬁxed distance. If the ﬁsh are swimming against a current , then the time required to swim a distance is and the total energy required to swim the distance is given by where is the proportionality constant. (a) Determine the value of that minimizes . (b) Sketch the graph of . Note: This result has been veriﬁed experimentally; migrating ﬁsh swim against a current at a speed greater than the current speed. 50% E E v a E!v" ! av 3 ! L v ! u E L#!v ! u" L !u ( v" u v 3 v R r E P ! E 2R !R $ r"2 r E R F -. F ! .W . sin - $ cos --W h ! 1 3 H H h 27 cm3 A B R C CB CA R ; 20. Find, correct to two decimal places, the coordinates of the point on the curve that is closest to the point . 21. Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius . Find the area of the largest rectangle that can be inscribed in the ellipse . 23. Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side if one side of the rectangle lies on the base of the triangle. 24. Find the dimensions of the rectangle of largest area that has its base on the -axis and its other two vertices above the -axis and lying on the parabola . 25. Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius . 26. Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 4 cm if two sides of the rectangle lie along the legs. 27. A right circular cylinder is inscribed in a sphere of radius . Find the largest possible volume of such a cylinder. 28. A right circular cylinder is inscribed in a cone with height and base radius . Find the largest possible volume of such a cylinder. 29. A right circular cylinder is inscribed in a sphere of radius . Find the largest possible surface area of such a cylinder. A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See Exercise 56 on page 23.) If the perimeter of the window is 30 ft, ﬁnd the dimensions of the window so that the greatest possible amount of light is admitted. 31. The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is ﬁxed at 384 cm , ﬁnd the dimensions of the poster with the smallest area. 32. A poster is to have an area of 180 in with 1-inch margins at the bottom and sides and a 2-inch margin at the top. What dimensions will give the largest printed area? A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) A minimum? 34. Answer Exercise 33 if one piece is bent into a square and the other into a circle. 35. A cylindrical can without a top is made to contain of liquid. Find the dimensions that will minimize the cost of the metal to make the can. 36. A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest lad-V cm3 33. 2 2 30. r r h r r y ! 8 ! x 2 x x L x 2#a2 $ y 2#b 2 ! 1 22. r !1, 1" y ! tan x SECTION 4.7 OPTIMIZATION PROBLEMS \|\|\|\| 329 330 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 47. An oil reﬁnery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the reﬁnery to storage tanks located on the south bank of the river 6 km east of the reﬁnery. The cost of laying pipe is over land to a point on the north bank and under the river to the tanks. To minimize the cost of the pipeline, where should be located? ; 48. Suppose the reﬁnery in Exercise 47 is located 1 km north of the river. Where should be located? The illumination of an object by a light source is directly propor-tional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 ft apart, where should an object be placed on the line between the sources so as to receive the least illumination? Find an equation of the line through the point that cuts off the least area from the ﬁrst quadrant. 51. Let and be positive numbers. Find the length of the shortest line segment that is cut off by the ﬁrst quadrant and passes through the point . 52. At which points on the curve does the tangent line have the largest slope? (a) If is the cost of producing units of a commodity, then the average cost per unit is . Show that if the average cost is a minimum, then the marginal cost equals the average cost. (b) If , in dollars, ﬁnd (i) the cost, average cost, and marginal cost at a production level of 1000 units; (ii) the production level that will minimize the average cost; and (iii) the minimum average cost. 54. (a) Show that if the proﬁt is a maximum, then the marginal revenue equals the marginal cost. (b) If is the cost function and is the demand function, ﬁnd the production level that will maximize proﬁt. A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at , the average attendance had been 27,000. When ticket prices were lowered to , the average attendance rose to 33,000. (a) Find the demand function, assuming that it is linear. (b) How should ticket prices be set to maximize revenue? 56. During the summer months Terry makes and sells necklaces on the beach. Last summer he sold the necklaces for each and his sales averaged 20 per day. When he increased the price by , he found that the average decreased by two sales per day. (a) Find the demand function, assuming that it is linear. (b) If the material for each necklace costs Terry , what should the selling price be to maximize his proﬁt? $6 $1 $10 $8 $10 55. px 1700 7x Cx 16,000 500x 1.6x 2 0.004x 3 Px Cx 16,000 200x 4x 32 cx Cxx x Cx 53. y 1 40x 3 3x 5 a, b b a 3, 5 50. 49. P P $800,000km P $400,000km 43. In a beehive, each cell is a regular hexagonal prism, open at one end with a trihedral angle at the other end as in the ﬁg-ure. It is believed that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of wax in cell construction. Examination of these cells has shown that the measure of the apex angle is amaz-ingly consistent. Based on the geometry of the cell, it can be shown that the surface area is given by where , the length of the sides of the hexagon, and , the height, are constants. (a) Calculate . (b) What angle should the bees prefer? (c) Determine the minimum surface area of the cell (in terms of and ). Note: Actual measurements of the angle in beehives have been made, and the measures of these angles seldom differ from the calculated value by more than . 44. A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 kmh. Another boat has been heading due east at 15 kmh and reaches the same dock at 3:00 PM. At what time were the two boats closest together? 45. Solve the problem in Example 4 if the river is 5 km wide and point is only 5 km downstream from . 46. A woman at a point on the shore of a circular lake with radius 2 mi wants to arrive at the point diametrically opposite on the other side of the lake in the shortest possible time. She can walk at the rate of 4 mih and row a boat at 2 mih. How should she proceed? ¨ B A C 2 2 A C A A B s trihedral angle ¨ rear of cell front of cell h b 2 h s dSd h s S 6sh 3 2s2 cot (3s 2s3 2) csc S this consumption . Using the graph, estimate the speed at which has its minimum value. Let be the velocity of light in air and the velocity of light in water. According to Fermat’s Principle, a ray of light will travel from a point in the air to a point in the water by a path that minimizes the time taken. Show that where (the angle of incidence) and (the angle of refrac-tion) are as shown. This equation is known as Snell’s Law. 64. Two vertical poles and are secured by a rope going from the top of the ﬁrst pole to a point on the ground between the poles and then to the top of the second pole as in the ﬁgure. Show that the shortest length of such a rope occurs when . 65. The upper right-hand corner of a piece of paper, 12 in. by 8 in., as in the ﬁgure, is folded over to the bottom edge. How would you fold it so as to minimize the length of the fold? In other words, how would you choose to minimize ? x y 8 12 y x Q R T P S ¨¡ ¨™ $1 ! $ 2 R PRS ST PQ C A B ¨¡ ¨™ $2 $1 sin $1 sin $2 ! v1 v2 ACB B A v2 v1 63. √ c 0 20 40 60 G G 57. A manufacturer has been selling 1000 television sets a week at each. A market survey indicates that for each rebate offered to the buyer, the number of sets sold will increase by 100 per week. (a) Find the demand function. (b) How large a rebate should the company offer the buyer in order to maximize its revenue? (c) If its weekly cost function is , how should the manufacturer set the size of the rebate in order to maximize its proﬁt? 58. The manager of a 100-unit apartment complex knows from experience that all units will be occupied if the rent is per month. A market survey suggests that, on average, one additional unit will remain vacant for each increase in rent. What rent should the manager charge to maximize revenue? 59. Show that of all the isosceles triangles with a given perimeter, the one with the greatest area is equilateral. 60. The frame for a kite is to be made from six pieces of wood. The four exterior pieces have been cut with the lengths indicated in the ﬁgure. To maximize the area of the kite, how long should the diagonal pieces be? ; 61. A point needs to be located somewhere on the line so that the total length of cables linking to the points , , and is minimized (see the ﬁgure). Express as a function of and use the graphs of and to estimate the minimum value. 62. The graph shows the fuel consumption of a car (measured in gallons per hour) as a function of the speed of the car. At very low speeds the engine runs inefﬁciently, so initially decreases as the speed increases. But at high speeds the fuel consumption increases. You can see that is minimized for this car when mi#h. However, for fuel efﬁciency, what must be minimized is not the consumption in gallons per hour but rather the fuel consumption in gallons per mile. Let’s call v $ 30 c!v" c v c B C P A 2 m 3 m D 5 m dL#dx L x ! % AP% L C B A P L AD P a a b b CAS $10 $800 C!x" ! 68,000 " 150x $10 $450 SECTION 4.7 OPTIMIZATION PROBLEMS \|\|\|\| 331 observer stand so as to maximize the angle subtended at his eye by the painting?) 71. Find the maximum area of a rectangle that can be circum-scribed about a given rectangle with length and width . [Hint: Express the area as a function of an angle .] 72. The blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart. This system should work so as to minimize the energy expended by the heart in pumping the blood. In particular, this energy is reduced when the resistance of the blood is lowered. One of Poiseuille’s Laws gives the resistance of the blood as where is the length of the blood vessel, is the radius, and is a positive constant determined by the viscosity of the blood. (Poiseuille established this law experimentally, but it also follows from Equation 8.4.2.) The ﬁgure shows a main blood vessel with radius branching at an angle into a smaller vessel with radius © Manfred Cage / Peter Arnold b A B r¡ r™ ¨ C a vascular branching r2 r1 C r L R C L r 4 R W L ¨ h d 66. A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? 67. An observer stands at a point , one unit away from a track. Two runners start at the point in the ﬁgure and run along the track. One runner runs three times as fast as the other. Find the maximum value of the observer’s angle of sight between the runners. [Hint: Maximize .] 68. A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle . How should be chosen so that the gut-ter will carry the maximum amount of water? Where should the point be chosen on the line segment so as to maximize the angle ? 70. A painting in an art gallery has height and is hung so that its lower edge is a distance above the eye of an observer (as in the ﬁgure). How far from the wall should the observer stand to get the best view? (In other words, where should the d h 5 2 A B P ¨ 3 AB P 69. 10 cm 10 cm 10 cm ¨ ¨ S 1 P ¨ tan S P 6 ¨ 9 332 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (d) If the ornithologists observe that birds of a certain species reach the shore at a point 4 km from B, how many times more energy does it take a bird to ﬂy over water than land? ; 74. Two light sources of identical strength are placed 10 m apart. An object is to be placed at a point on a line ! parallel to the line joining the light sources and at a distance meters from it (see the ﬁgure). We want to locate on ! so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. (a) Find an expression for the intensity at the point . (b) If m, use graphs of and to show that the intensity is minimized when m, that is, when is at the midpoint of !. (c) If m, show that the intensity (perhaps surpris-ingly) is not minimized at the midpoint. (d) Somewhere between m and m there is a transitional value of at which the point of minimal illu-mination abruptly changes. Estimate this value of by graphical methods. Then ﬁnd the exact value of . ! P d 10 m x d d d d ! 10 d ! 5 d ! 10 P x ! 5 I%!x" I!x" d ! 5 P I!x" P d P 13 km B C D island 5 km nest (a) Use Poiseuille’s Law to show that the total resistance of the blood along the path is where and are the distances shown in the ﬁgure. (b) Prove that this resistance is minimized when (c) Find the optimal branching angle (correct to the nearest degree) when the radius of the smaller blood vessel is two-thirds the radius of the larger vessel. 73. Ornithologists have determined that some species of birds tend to avoid ﬂights over large bodies of water during daylight hours. It is believed that more energy is required to ﬂy over water than land because air generally rises over land and falls over water during the day. A bird with these tenden-cies is released from an island that is 5 km from the nearest point on a straight shoreline, ﬂies to a point on the shore-line, and then ﬂies along the shoreline to its nesting area . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points and are 13 km apart. (a) In general, if it takes 1.4 times as much energy to ﬂy over water as land, to what point should the bird ﬂy in order to minimize the total energy expended in returning to its nesting area? (b) Let and L denote the energy (in joules) per kilometer ﬂown over water and land, respectively. What would a large value of the ratio W#L mean in terms of the bird’s ﬂight? What would a small value mean? Determine the ratio corresponding to the minimum expenditure of energy. (c) What should the value of be in order for the bird to ﬂy directly to its nesting area ? What should the value of be for the bird to ﬂy to and then along the shore to ? D B W#L D W#L W#L W C D B D C B cos $ ! r 4 2 r 4 1 b a R ! C& a ! b cot $ r1 4 " b csc $ r2 4 ' ABC APPLIED PROJECT THE SHAPE OF A CAN \|\|\|\| 333 In this project we investigate the most economical shape for a can. We ﬁrst interpret this to mean that the volume of a cylindrical can is given and we need to ﬁnd the height and radius that minimize the cost of the metal to make the can (see the ﬁgure). If we disregard any waste metal in the manufacturing process, then the problem is to minimize the surface area of the cylinder. We solved this problem in Example 2 in Section 4.7 and we found that ; that is, the height should be the same as the diameter. But if you go to your cupboard or your supermarket with a ruler, you will discover that the height is usually greater than the diameter and the ratio varies from 2 up to about 3.8. Let’s see if we can explain this phenomenon. 1. The material for the cans is cut from sheets of metal. The cylindrical sides are formed by bending rectangles; these rectangles are cut from the sheet with little or no waste. But if the h#r h ! 2r r h V THE SHAPE OF A CAN A P P L I E D P R O J E C T r h top and bottom discs are cut from squares of side (as in the ﬁgure), this leaves considerable waste metal, which may be recycled but has little or no value to the can makers. If this is the case, show that the amount of metal used is minimized when 2. A more efﬁcient packing of the discs is obtained by dividing the metal sheet into hexagons and cutting the circular lids and bases from the hexagons (see the ﬁgure). Show that if this strategy is adopted, then 3. The values of that we found in Problems 1 and 2 are a little closer to the ones that actually occur on supermarket shelves, but they still don’t account for everything. If we look more closely at some real cans, we see that the lid and the base are formed from discs with radius larger than that are bent over the ends of the can. If we allow for this we would increase . More signiﬁcantly, in addition to the cost of the metal we need to incorporate the manufacturing of the can into the cost. Let’s assume that most of the expense is incurred in joining the sides to the rims of the cans. If we cut the discs from hexagons as in Problem 2, then the total cost is proportional to where is the reciprocal of the length that can be joined for the cost of one unit area of metal. Show that this expression is minimized when ; 4. Plot as a function of and use your graph to argue that when a can is large or joining is cheap, we should make approximately 2.21 (as in Problem 2). But when the can is small or joining is costly, should be substantially larger. 5. Our analysis shows that large cans should be almost square but small cans should be tall and thin. Take a look at the relative shapes of the cans in a supermarket. Is our conclusion usually true in practice? Are there exceptions? Can you suggest reasons why small cans are not always tall and thin? h#r h#r x ! h#r s 3 V #k s 3 V k !( &h r ! 2& ! h#r &h#r ! 4s3 k 4s3 r 2 " 2&rh " k!4&r " h" h#r r h#r h r ! 4s3 & $ 2.21 h r ! 8 & $ 2.55 2r 334 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Discs cut from hexagons Discs cut from squares NEWTON’S METHOD Suppose that a car dealer offers to sell you a car for $18,000 or for payments of $375 per month for ﬁve years. You would like to know what monthly interest rate the dealer is, in effect, charging you. To ﬁnd the answer, you have to solve the equation (The details are explained in Exercise 41.) How would you solve such an equation? For a quadratic equation there is a well-known formula for the roots. For third- and fourth-degree equations there are also formulas for the roots, but they are ax 2 " bx " c ! 0 48x!1 " x"60 ! !1 " x"60 " 1 ! 0 1 4.8 3 extremely complicated. If f is a polynomial of degree 5 or higher, there is no such formula (see the note on page 210). Likewise, there is no formula that will enable us to ﬁnd the exact roots of a transcendental equation such as . We can ﬁnd an approximate solution to Equation 1 by plotting the left side of the equa-tion. Using a graphing device, and after experimenting with viewing rectangles, we pro-duce the graph in Figure 1. We see that in addition to the solution x ! 0, which doesn’t interest us, there is a solu-tion between 0.007 and 0.008. Zooming in shows that the root is approximately 0.0076. If we need more accuracy we could zoom in repeatedly, but that becomes tiresome. A faster alternative is to use a numerical rootﬁnder on a calculator or computer algebra system. If we do so, we ﬁnd that the root, correct to nine decimal places, is 0.007628603. How do those numerical rootﬁnders work? They use a variety of methods, but most of them make some use of Newton’s method, also called the Newton-Raphson method. We will explain how this method works, partly to show what happens inside a calculator or computer, and partly as an application of the idea of linear approximation. The geometry behind Newton’s method is shown in Figure 2, where the root that we are trying to ﬁnd is labeled . We start with a ﬁrst approximation , which is obtained by guessing, or from a rough sketch of the graph of , or from a computer-generated graph of f. Consider the tangent line to the curve at the point and look at the -intercept of , labeled . The idea behind Newton’s method is that the tangent line is close to the curve and so its x-intercept, , is close to the x-intercept of the curve (namely, the root r that we are seeking). Because the tangent is a line, we can easily ﬁnd its x-intercept. To ﬁnd a formula for in terms of we use the fact that the slope of L is , so its equation is Since the -intercept of is , we set and obtain If , we can solve this equation for : We use as a second approximation to r. Next we repeat this procedure with replaced by , using the tangent line at . This gives a third approximation: If we keep repeating this process, we obtain a sequence of approximations as shown in Figure 3. In general, if the th approximation is and , then the next approximation is given by xn"1 ! xn ! f!xn" f%!xn" 2 f%!xn" " 0 xn n x1, x2, x3, x4, . . . x3 ! x2 ! f!x2" f%!x2" !x2, f!x2"" x2 x1 x2 x2 ! x1 ! f!x1" f%!x1" x2 f%!x1" " 0 0 ! f!x1" ! f%!x1"!x2 ! x1" y ! 0 x2 L x y ! f!x1" ! f%!x1"!x ! x1" f%!x1" x1 x2 x2 x2 L x !x1, f!x1"" y ! f!x" L f x1 r cos x ! x SECTION 4.8 NEWTON’S METHOD \|\|\|\| 335 0.15 _0.05 0 0.012 FIGURE 1 N Try to solve Equation 1 using the numerical rootﬁnder on your calculator or computer. Some machines are not able to solve it. Others are suc-cessful but require you to specify a starting point for the search. FIGURE 2 y 0 x {x¡, f(x¡)} x™ x¡ L r y=ƒ y 0 x x™ x¡ x£ x¢ r FIGURE 3 {x™, f(x™)} {x¡, f(x¡)} If the numbers become closer and closer to as becomes large, then we say that the sequence converges to and we write \| Although the sequence of successive approximations converges to the desired root for functions of the type illustrated in Figure 3, in certain circumstances the sequence may not converge. For example, consider the situation shown in Figure 4. You can see that is a worse approximation than . This is likely to be the case when is close to 0. It might even happen that an approximation (such as in Figure 4) falls outside the domain of . Then Newton’s method fails and a better initial approximation should be chosen. See Exercises 31–34 for speciﬁc examples in which Newton’s method works very slowly or does not work at all. EXAMPLE 1 Starting with , ﬁnd the third approximation to the root of the equation . SOLUTION We apply Newton’s method with and Newton himself used this equation to illustrate his method and he chose after some experimentation because , , and . Equation 2 becomes With we have Then with we obtain It turns out that this third approximation is accurate to four decimal places. M Suppose that we want to achieve a given accuracy, say to eight decimal places, using Newton’s method. How do we know when to stop? The rule of thumb that is generally used is that we can stop when successive approximations and agree to eight decimal places. (A precise statement concerning accuracy in Newton’s method will be given in Exercise 37 in Section 11.11.) Notice that the procedure in going from to is the same for all values of . (It is called an iterative process.) This means that Newton’s method is particularly convenient for use with a programmable calculator or a computer. n n " 1 n xn"1 xn x3 $ 2.0946 ! 2.1 ! !2.1"3 ! 2!2.1" ! 5 3!2.1"2 ! 2 $ 2.0946 x3 ! x2 ! x2 3 ! 2x2 ! 5 3x2 2 ! 2 n ! 2 ! 2 ! 23 ! 2!2" ! 5 3!2"2 ! 2 ! 2.1 x2 ! x1 ! x1 3 ! 2x1 ! 5 3x1 2 ! 2 n ! 1 xn"1 ! xn ! x n 3 ! 2xn ! 5 3x n 2 ! 2 f!3" ! 16 f!2" ! !1 f!1" ! !6 x1 ! 2 f%!x" ! 3x 2 ! 2 f!x" ! x 3 ! 2x ! 5 x 3 ! 2x ! 5 ! 0 x3 x1 ! 2 V x1 f x3 f%!x1" x1 x2 lim n l ' xn ! r r n r xn 336 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION x y 0 r x™ x£ x¡ FIGURE 4 In Module 4.8 you can investigate how Newton’s Method works for several functions and what happens when you change . x1 TEC N Sequences were brieﬂy introduced in A Preview of Calculus on page 6. A more thorough discussion starts in Section 11.1. FIGURE 5 1 1.8 2.2 _2 y=10x-21 x™ N Figure 5 shows the geometry behind the ﬁrst step in Newton’s method in Example 1. Since , the tangent line to at has equation so its -intercept is . x 2 ! 2.1 x y ! 10x ! 21 !2, !1" y ! x3 ! 2x ! 5 f %!2" ! 10 EXAMPLE 2 Use Newton’s method to ﬁnd correct to eight decimal places. SOLUTION First we observe that ﬁnding is equivalent to ﬁnding the positive root of the equation so we take . Then and Formula 2 (Newton’s method) becomes If we choose as the initial approximation, then we obtain Since and agree to eight decimal places, we conclude that to eight decimal places. M EXAMPLE 3 Find, correct to six decimal places, the root of the equation . SOLUTION We ﬁrst rewrite the equation in standard form: Therefore we let . Then , so Formula 2 becomes In order to guess a suitable value for we sketch the graphs of and in Figure 6. It appears that they intersect at a point whose -coordinate is somewhat less than 1, so let’s take as a convenient ﬁrst approximation. Then, remembering to put our calculator in radian mode, we get Since and agree to six decimal places (eight, in fact), we conclude that the root of the equation, correct to six decimal places, is . M Instead of using the rough sketch in Figure 6 to get a starting approximation for Newton’s method in Example 3, we could have used the more accurate graph that a calcu-0.739085 x5 x4 x5 $ 0.73908513 x4 $ 0.73908513 x3 $ 0.73911289 x2 $ 0.75036387 x1 ! 1 x y ! x y ! cos x x1 xn"1 ! xn ! cos xn ! xn !sin xn ! 1 ! xn " cos xn ! xn sin xn " 1 f%!x" ! !sin x ! 1 f!x" ! cos x ! x cos x ! x ! 0 cos x ! x V s 6 2 $ 1.12246205 x6 x5 x6 $ 1.12246205 x5 $ 1.12246205 x4 $ 1.12249707 x3 $ 1.12644368 x2 $ 1.16666667 x1 ! 1 xn"1 ! xn ! xn 6 ! 2 6xn 5 f%!x" ! 6x 5 f!x" ! x 6 ! 2 x 6 ! 2 ! 0 s 6 2 s 6 2 V SECTION 4.8 NEWTON’S METHOD \|\|\|\| 337 FIGURE 6 1 y x π y=cos x y=x π 2 lator or computer provides. Figure 7 suggests that we use as the initial approx-imation. Then Newton’s method gives and so we obtain the same answer as before, but with one fewer step. You might wonder why we bother at all with Newton’s method if a graphing device is available. Isn’t it easier to zoom in repeatedly and ﬁnd the roots as we did in Section 1.4? If only one or two decimal places of accuracy are required, then indeed Newton’s method is inappropriate and a graphing device sufﬁces. But if six or eight decimal places are required, then repeated zooming becomes tiresome. It is usually faster and more efﬁcient to use a computer and Newton’s method in tandem—the graphing device to get started and Newton’s method to ﬁnish. x4 $ 0.73908513 x3 $ 0.73908513 x2 $ 0.73911114 x1 ! 0.75 338 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 5–8 Use Newton’s method with the speciﬁed initial approxima-tion to ﬁnd , the third approximation to the root of the given equation. (Give your answer to four decimal places.) 5. , 6. , 7. , 8. , ; 9. Use Newton’s method with initial approximation to ﬁnd , the second approximation to the root of the equation . Explain how the method works by ﬁrst graphing the function and its tangent line at . ; 10. Use Newton’s method with initial approximation to ﬁnd , the second approximation to the root of the equa-tion . Explain how the method works by ﬁrst graphing the function and its tangent line at . 11–12 Use Newton’s method to approximate the given number correct to eight decimal places. 11. 12. 13–16 Use Newton’s method to approximate the indicated root of the equation correct to six decimal places. 13. The root of in the interval 14. The root of in the interval 15. The positive root of 16. The positive root of 17–22 Use Newton’s method to ﬁnd all roots of the equation cor-rect to six decimal places. 17. 18. e x ! 3 ! 2x x4 ! 1 " x 2 cos x ! x 4 sin x ! x 2 )!2, !1 2.2x 5 ! 4.4x 3 " 1.3x 2 ! 0.9x ! 4.0 ! 0 )1, 2 x 4 ! 2x 3 " 5x 2 ! 6 ! 0 100 s100 s 5 20 !1, !1" x4 ! x ! 1 ! 0 x2 x1 ! 1 !!1, 1" x 3 " x " 3 ! 0 x2 x1 ! !1 x1 ! !1 x 5 " 2 ! 0 x1 ! 1 x 5 ! x ! 1 ! 0 x1 ! !3 1 3x 3 " 1 2x 2 " 3 ! 0 x1 ! 1 x 3 " 2x ! 4 ! 0 x3 x1 1. The ﬁgure shows the graph of a function . Suppose that Newton’s method is used to approximate the root of the equation with initial approximation . (a) Draw the tangent lines that are used to ﬁnd and , and estimate the numerical values of and . (b) Would be a better ﬁrst approximation? Explain. 2. Follow the instructions for Exercise 1(a) but use as the starting approximation for ﬁnding the root . 3. Suppose the line is tangent to the curve when . If Newton’s method is used to locate a root of the equation and the initial approximation is , ﬁnd the second approximation . For each initial approximation, determine graphically what happens if Newton’s method is used for the function whose graph is shown. (a) (b) (c) (d) (e) 3 y 0 5 1 x x1 ! 5 x1 ! 4 x1 ! 3 x1 ! 1 x1 ! 0 4. x2 x1 ! 3 f !x" ! 0 x ! 3 y ! f !x" y ! 5x ! 4 s x1 ! 9 x y 0 r 1 1 s x1 ! 5 x3 x2 x3 x2 x1 ! 1 f !x" ! 0 r f EXERCISES 4.8 FIGURE 7 1 0 1 y=x y=cos x approximation is used. Illustrate your explanation with a sketch. 35. (a) Use Newton’s method to ﬁnd the critical numbers of the function correct to six deci-mal places. (b) Find the absolute minimum value of correct to four decimal places. 36. Use Newton’s method to ﬁnd the absolute maximum value of the function , correct to six decimal places. Use Newton’s method to ﬁnd the coordinates of the inﬂection point of the curve , , correct to six deci-mal places. 38. Of the inﬁnitely many lines that are tangent to the curve and pass through the origin, there is one that has the largest slope. Use Newton’s method to ﬁnd the slope of that line correct to six decimal places. 39. Use Newton’s method to ﬁnd the coordinates, correct to six decimal places, of the point on the parabola that is closest to the origin. 40. In the ﬁgure, the length of the chord is 4 cm and the length of the arc is 5 cm. Find the central angle , in radi-ans, correct to four decimal places. Then give the answer to the nearest degree. A car dealer sells a new car for . He also offers to sell the same car for payments of per month for ﬁve years. What monthly interest rate is this dealer charging? To solve this problem you will need to use the formula for the present value of an annuity consisting of equal pay-ments of size with interest rate per time period: Replacing by , show that Use Newton’s method to solve this equation. 42. The ﬁgure shows the sun located at the origin and the earth at the point . (The unit here is the distance between the centers of the earth and the sun, called an astronomical unit: 1 AU km.) There are ﬁve locations , , , , and in this plane of rotation of the earth about the sun where a satellite remains motionless with respect to the earth because the forces acting on the satellite (including the gravi-L 5 L 4 L 3 L 2 L1 $ 1.496 ( 108 !1, 0" 48x!1 " x"60 ! !1 " x"60 " 1 ! 0 x i A ! R i )1 ! !1 " i"!n i R n A $375 $18,000 41. 5 cm 4 cm ¨ B A $ AB AB y ! !x ! 1"2 y ! !sin x 0 ) x ) & y ! ecos x 37. f !x" ! x cos x, 0 ) x ) & f f !x" ! x 6 ! x 4 " 3x 3 ! 2x x1 " 0 19. 20. 21. 22. ; 23–28 Use Newton’s method to ﬁnd all the roots of the equation correct to eight decimal places. Start by drawing a graph to ﬁnd initial approximations. 23. 25. 26. 27. 28. 29. (a) Apply Newton’s method to the equation to derive the following square-root algorithm used by the ancient Babylonians to compute : (b) Use part (a) to compute correct to six decimal places. 30. (a) Apply Newton’s method to the equation to derive the following reciprocal algorithm: (This algorithm enables a computer to ﬁnd reciprocals without actually dividing.) (b) Use part (a) to compute correct to six decimal places. Explain why Newton’s method doesn’t work for ﬁnding the root of the equation if the initial approxi-mation is chosen to be . 32. (a) Use Newton’s method with to ﬁnd the root of the equation correct to six decimal places. (b) Solve the equation in part (a) using as the initial approximation. (c) Solve the equation in part (a) using . (You deﬁ-nitely need a programmable calculator for this part.) ; (d) Graph and its tangent lines at , 0.6, and 0.57 to explain why Newton’s method is so sen-sitive to the value of the initial approximation. 33. Explain why Newton’s method fails when applied to the equation with any initial approximation . Illustrate your explanation with a sketch. 34. If then the root of the equation is . Explain why Newton’s method fails to ﬁnd the root no matter which initial x ! 0 f !x" ! 0 f !x" !+ sx !s!x if x 0 if x + 0 x1 " 0 s 3 x ! 0 x1 ! 1 f !x" ! x 3 ! x ! 1 x1 ! 0.57 x1 ! 0.6 x 3 ! x ! 1 x1 ! 1 x1 ! 1 x 3 ! 3x " 6 ! 0 31. 1#1.6984 xn"1 ! 2xn ! axn 2 1#x ! a ! 0 s1000 xn"1 ! 1 2&xn " a xn' sa ) ( x 2 ! a ! 0 e arctan x ! sx 3 " 1 4e!x 2 sin x ! x 2 ! x " 1 3 sin!x 2" ! 2x x2s2 ! x ! x 2 ! 1 x 2!4 ! x 2" ! 4 x 2 " 1 24. x 6 ! x 5 ! 6x 4 ! x 2 " x " 10 ! 0 tan x ! s1 ! x 2 cos x ! sx 1 x ! 1 " x 3 !x ! 2"2 ! ln x SECTION 4.8 NEWTON’S METHOD \|\|\|\| 339 ANTIDERIVATIVES A physicist who knows the velocity of a particle might wish to know its position at a given time. An engineer who can measure the variable rate at which water is leaking from a tank wants to know the amount leaked over a certain time period. A biologist who knows the rate at which a bacteria population is increasing might want to deduce what the size of the population will be at some future time. In each case, the problem is to ﬁnd a function F whose derivative is a known function f. If such a function F exists, it is called an anti-derivative of f. DEFINITION A function is called an antiderivative of on an interval if for all in . For instance, let . It isn’t difﬁcult to discover an antiderivative of if we keep the Power Rule in mind. In fact, if , then . But the function also satisﬁes . Therefore both and are antiderivatives of . Indeed, any function of the form , where is a constant, is an anti-derivative of . The question arises: Are there any others? To answer this question, recall that in Section 4.2 we used the Mean Value Theorem to prove that if two functions have identical derivatives on an interval, then they must differ by a constant (Corollary 4.2.7). Thus if and are any two antiderivatives of , then so , where is a constant. We can write this as , so we have the following result. THEOREM If is an antiderivative of on an interval , then the most general antiderivative of on is where is an arbitrary constant. Going back to the function , we see that the general antiderivative of is . By assigning speciﬁc values to the constant , we obtain a family of functions whose graphs are vertical translates of one another (see Figure 1). This makes sense because each curve must have the same slope at any given value of . x C 1 3x 3 ! C f f!x" ! x 2 C F!x" ! C I f I f F 1 G!x" ! F!x" ! C C G!x" " F!x" ! C F#!x" ! f!x" ! G#!x" f G F f C H!x" ! 1 3 x 3 ! C f G F G#!x" ! x 2 G!x" ! 1 3 x 3 ! 100 F#!x" ! x 2 ! f!x" F!x" ! 1 3 x 3 f f!x" ! x 2 I x F#!x" ! f!x" I f F 4.9 x y 0 y= ˛ 3 y= -2 ˛ 3 y= -1 ˛ 3 y= +1 ˛ 3 y= +2 ˛ 3 y= +3 ˛ 3 FIGURE 1 Members of the family of antiderivatives of ƒ=≈ 340 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Using the value , ﬁnd the locations of the libration points (a) and (b) . L¡ L™ L∞ L¢ L£ sun earth x y L 2 L 1 r # 3.04042 $ 10"6 tational attractions of the earth and the sun) balance each other. These locations are called libration points. (A solar research satellite has been placed at one of these libration points.) If is the mass of the sun, is the mass of the earth, and , it turns out that the -coordi-nate of is the unique root of the ﬁfth-degree equation and the -coordinate of is the root of the equation p!x" " 2rx 2 ! 0 L 2 x ! ! 2!1 " r"x ! r " 1 ! 0 p!x" ! x 5 " !2 ! r"x 4 ! !1 ! 2r"x 3 " !1 " r"x 2 L 1 x r ! m2$!m1 ! m2" m2 m1 SECTION 4.9 ANTIDERIVATIVES \|\|\|\| 341 EXAMPLE 1 Find the most general antiderivative of each of the following functions. (a) (b) (c) , SOLUTION (a) If , then , so an antiderivative of is . By Theorem 1, the most general antiderivative is . (b) Recall from Section 3.6 that So on the interval the general antiderivative of is . We also learned that for all . Theorem 1 then tells us that the general antiderivative of is on any interval that doesn’t contain 0. In particular, this is true on each of the intervals and . So the general antiderivative of is (c) We use the Power Rule to discover an antiderivative of . In fact, if , then Thus the general antiderivative of is This is valid for since then is deﬁned on an interval. If n is negative (but ), it is valid on any interval that doesn’t contain 0. M As in Example 1, every differentiation formula, when read from right to left, gives rise to an antidifferentiation formula. In Table 2 we list some particular antiderivatives. Each formula in the table is true because the derivative of the function in the right column appears in the left column. In particular, the ﬁrst formula says that the antiderivative of a constant times a function is the constant times the antiderivative of the function. The sec-ond formula says that the antiderivative of a sum is the sum of the antiderivatives. (We use the notation , .) G# ! t F# ! f n " "1 f!x" ! x n n % 0 F!x" ! x n!1 n ! 1 ! C f!x" ! x n d dx % x n!1 n ! 1& ! !n ! 1"x n n ! 1 ! x n n " "1 x n F!x" !' ln x ! C1 ln!"x" ! C2 if x & 0 if x ' 0 f !0, (" !"(, 0" ln (x( ! C f!x" ! 1$x x " 0 d dx !ln (x(" ! 1 x ln x ! C 1$x !0, (" d dx !ln x" ! 1 x G!x" ! "cos x ! C "cos x sin x F#!x" ! sin x F!x" ! "cos x n " "1 f!x" ! x n f!x" ! 1$x f!x" ! sin x Function Particular antiderivative Function Particular antiderivative cos x sin x e x e x ln ( x( 1$x x n!1 n ! 1 x n !n " "1" F!x" ! G!x" f !x" ! t!x" cF!x" cf !x" sin x "cos x tan x sec x tan x sec x tan"1x 1 1 ! x 2 sin"1x 1 s1 " x 2 sec2x TABLE OF ANTIDIFFERENTIATION FORMULAS 2 N To obtain the most general antiderivative from the particular ones in Table 2, we have to add a constant (or constants), as in Example 1. EXAMPLE 2 Find all functions such that SOLUTION We ﬁrst rewrite the given function as follows: Thus we want to ﬁnd an antiderivative of Using the formulas in Table 2 together with Theorem 1, we obtain M In applications of calculus it is very common to have a situation as in Example 2, where it is required to ﬁnd a function, given knowledge about its derivatives. An equation that involves the derivatives of a function is called a differential equation. These will be studied in some detail in Chapter 9, but for the present we can solve some elementary dif-ferential equations. The general solution of a differential equation involves an arbitrary constant (or constants) as in Example 2. However, there may be some extra conditions given that will determine the constants and therefore uniquely specify the solution. EXAMPLE 3 Find if . SOLUTION The general antiderivative of is To determine we use the fact that : Thus we have , so the particular solution is M EXAMPLE 4 Find if , , and . SOLUTION The general antiderivative of is Using the antidifferentiation rules once more, we ﬁnd that f!x" ! 4 x 4 4 ! 3 x 3 3 " 4 x 2 2 ! Cx ! D ! x 4 ! x 3 " 2x 2 ! Cx ! D f#!x" ! 12 x 3 3 ! 6 x 2 2 " 4x ! C ! 4x 3 ! 3x 2 " 4x ! C f )!x" ! 12x 2 ! 6x " 4 f!1" ! 1 f!0" ! 4 f )!x" ! 12x 2 ! 6x " 4 f V f!x" ! e x ! 20 tan"1x " 3 C ! "2 " 1 ! "3 f!0" ! e 0 ! 20 tan"1 0 ! C ! "2 f!0" ! "2 C f!x" ! e x ! 20 tan"1x ! C f#!x" ! e x ! 20 1 ! x 2 f#!x" ! e x ! 20!1 ! x 2""1 and f!0" ! "2 f ! "4 cos x ! 2 5 x 5 " 2sx ! C t!x" ! 4!"cos x" ! 2 x 5 5 " x1$2 1 2 ! C t#!x" ! 4 sin x ! 2x 4 " x"1$2 t#!x" ! 4 sin x ! 2x 5 x " sx x ! 4 sin x ! 2x 4 " 1 sx t#!x" ! 4 sin x ! 2x 5 " sx x t 342 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 40 _2 3 f fª _25 FIGURE 2 N Figure 2 shows the graphs of the function in Example 3 and its antiderivative . Notice that , so is always increasing. Also notice that when has a maximum or minimum, appears to have an inﬂection point. So the graph serves as a check on our calculation. f f # f f #!x" & 0 f f # To determine and we use the given conditions that and . Since , we have . Since we have . Therefore the required function is M If we are given the graph of a function , it seems reasonable that we should be able to sketch the graph of an antiderivative . Suppose, for instance, that we are given that . Then we have a place to start, the point , and the direction in which we move our pencil is given at each stage by the derivative . In the next example we use the principles of this chapter to show how to graph even when we don’t have a formula for . This would be the case, for instance, when is determined by experi-mental data. EXAMPLE 5 The graph of a function is given in Figure 3. Make a rough sketch of an antiderivative , given that . SOLUTION We are guided by the fact that the slope of is . We start at the point and draw as an initially decreasing function since is negative when . Notice that , so has horizontal tangents when and . For , is positive and so is increasing. We see that has a local minimum when and a local maximum when . For , is negative and so is decreasing on . Since as , the graph of becomes ﬂat-ter as . Also notice that changes from positive to negative at and from negative to positive at , so has inﬂection points when and . We use this information to sketch the graph of the antiderivative in Figure 4. M RECTILINEAR MOTION Antidifferentiation is particularly useful in analyzing the motion of an object moving in a straight line. Recall that if the object has position function , then the velocity func-tion is . This means that the position function is an antiderivative of the veloc-ity function. Likewise, the acceleration function is , so the velocity function is an antiderivative of the acceleration. If the acceleration and the initial values and are known, then the position function can be found by antidifferentiating twice. EXAMPLE 6 A particle moves in a straight line and has acceleration given by . Its initial velocity is cm$s and its initial displacement is cm. Find its position function . SOLUTION Since , antidifferentiation gives Note that . But we are given that , so and v!t" ! 3t 2 ! 4t " 6 C ! "6 v!0" ! "6 v!0" ! C v!t" ! 6 t 2 2 ! 4t ! C ! 3t 2 ! 4t ! C v#!t" ! a!t" ! 6t ! 4 s!t" s!0" ! 9 v!0" ! "6 a!t" ! 6t ! 4 V v!0" s!0" a!t" ! v#!t" v!t" ! s#!t" s ! f!t" x ! 4 x ! 2 F x ! 4 x ! 2 F)!x" ! f#!x" x l ( F x l ( f!x" l 0 !3, (" F f!x" x & 3 x ! 3 x ! 1 F F f!x" 1 ' x ' 3 x ! 3 x ! 1 F f!1" ! f!3" ! 0 0 ' x ' 1 f!x" F !0, 2" f!x" y ! F!x" F!0" ! 2 F f V f!x" f F F#!x" ! f!x" !0, 1" F!0" ! 1 F f f!x" ! x 4 ! x 3 " 2x 2 " 3x ! 4 C ! "3 f!1" ! 1 ! 1 " 2 ! C ! 4 ! 1 D ! 4 f!0" ! 0 ! D ! 4 f!1" ! 1 f!0" ! 4 D C SECTION 4.9 ANTIDERIVATIVES \|\|\|\| 343 1 2 3 0 4 x y y=ƒ FIGURE 3 FIGURE 4 x y 1 2 0 y=F(x) 1 Since , is the antiderivative of : This gives . We are given that , so and the required position function is M An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by . For motion close to the ground we may assume that is constant, its value being about (or ft$s ). EXAMPLE 7 A ball is thrown upward with a speed of ft$s from the edge of a cliff ft above the ground. Find its height above the ground seconds later. When does it reach its maximum height? When does it hit the ground? SOLUTION The motion is vertical and we choose the positive direction to be upward. At time the distance above the ground is and the velocity is decreasing. Therefore, the acceleration must be negative and we have Taking antiderivatives, we have To determine we use the given information that . This gives , so The maximum height is reached when , that is, after s. Since , we antidifferentiate again and obtain Using the fact that , we have and so The expression for is valid until the ball hits the ground. This happens when , that is, when or, equivalently, Using the quadratic formula to solve this equation, we get We reject the solution with the minus sign since it gives a negative value for . Therefore the ball hits the ground after s. M 3(1 ! s13 )$2 # 6.9 t t ! 3 3s13 2 t 2 " 3t " 27 ! 0 "16t 2 ! 48t ! 432 ! 0 s!t" ! 0 s!t" s!t" ! "16t 2 ! 48t ! 432 432 ! 0 ! D s!0" ! 432 s!t" ! "16t 2 ! 48t ! D s#!t" ! v!t" 1.5 v!t" ! 0 v!t" ! "32t ! 48 48 ! 0 ! C v!0" ! 48 C v!t" ! "32t ! C a!t" ! dv dt ! "32 v!t" s!t" t t 432 48 2 32 9.8 m$s2 t t s!t" ! t 3 ! 2t 2 " 6t ! 9 D ! 9 s!0" ! 9 s!0" ! D s!t" ! 3 t 3 3 ! 4 t 2 2 " 6t ! D ! t 3 ! 2t 2 " 6t ! D v s v!t" ! s#!t" 344 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 500 0 8 FIGURE 5 N Figure 5 shows the position function of the ball in Example 7. The graph corroborates the conclusions we reached: The ball reaches its maximum height after and hits the ground after . 6.9 s 1.5 s 43. , , 44. , , 45. , , , 46. , , , 47. Given that the graph of passes through the point and that the slope of its tangent line at is , ﬁnd . 48. Find a function such that and the line is tangent to the graph of . 49–50 The graph of a function is shown. Which graph is an antiderivative of and why? 50. 51. The graph of a function is shown in the ﬁgure. Make a rough sketch of an antiderivative , given that . 52. The graph of the velocity function of a particle is shown in the ﬁgure. Sketch the graph of the position function. The graph of is shown in the ﬁgure. Sketch the graph of if is continuous and . _1 x y 0 1 2 1 2 y=fª(x) f !0" ! "1 f f f # 53. √ 0 t y y=ƒ 0 x 1 F!0" ! 1 F x y f b c a y x f b c a 49. f f f x ! y ! 0 f #!x" ! x 3 f f !2" 2x ! 1 !x, f !x"" !1, 6" f f )!0" ! 3 f#!0" ! 2 f!0" ! 1 f +!x" ! cos x f !2" ! 0 f !1" ! 0 x & 0 f )!x" ! x "2 f !," ! 0 f !0" ! 0 f )!t" ! 2e t ! 3 sin t f !,$2" ! 0 f !0" ! "1 f )!x" ! 2 ! cos x 1–20 Find the most general antiderivative of the function. (Check your answer by differentiation.) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 14. 15. 16. 17. 18. 19. 20. ; 21–22 Find the antiderivative of that satisﬁes the given con-dition. Check your answer by comparing the graphs of and . 22. 23–46 Find . 24. 25. 26. 27. 28. 29. 30. 31. , 32. , , 33. , , 34. , , 35. , , 36. 37. , , 38. , , , , 40. , , 41. , , 42. , , f !1" ! 5 f !0" ! 8 f )!x" ! 20x 3 ! 12x 2 ! 4 f !2" ! 15 f !0" ! 9 f )!x" ! 2 " 12x f #!4" ! 7 f !4" ! 20 f )!t" ! 3$st f #!0" ! 4 f !0" ! 3 f )!-" ! sin - ! cos -39. f #!0" ! 1 f !0" ! 2 f )!x" ! 4 " 6x " 40x 3 f #!1" ! "3 f !1" ! 5 f )!x" ! 24x 2 ! 2x ! 10 f #!x" ! 4$s1 " x 2 , f( 1 2) ! 1 f !"1" ! "1 f!1" ! 1 f#!x" ! x"1$3 f!"1" ! 0 f!1" ! 1 2 f#!x" ! !x 2 " 1"$x f !,$3" ! 4 ",$2 ' t ' ,$2 f #!t" ! 2 cos t ! sec2t f !1" ! 3 x & 0 f #!x" ! 2x " 3$x 4 f !1" ! 10 f #!x" ! sx !6 ! 5x" f #!x" ! 8x 3 ! 12x ! 3, f !1" ! 6 f #!x" ! 1 " 6x, f !0" ! 8 f +!t" ! t " st f +!t" ! e t f )!x" ! 6x ! sin x f )!x" ! 2 3x 2$3 f )!x" ! 2 ! x 3 ! x 6 f )!x" ! 6x ! 12x 2 23. f f !x" ! 4 " 3!1 ! x 2""1, F!1" ! 0 f !x" ! 5x 4 " 2x 5, F!0" ! 4 21. F f f F f !x" ! 2 ! x 2 1 ! x 2 f !x" ! x 5 " x 3 ! 2x x 4 f !x" ! 2sx ! 6 cos x f !x" ! 5e x " 3 cosh x f !t" ! sin t ! 2 sinh t t!-" ! cos - " 5 sin -f !x" ! 3e x ! 7 sec2x f !u" ! u4 ! 3su u2 13. t!x" ! 5 " 4x 3 ! 2x 6 x 6 f !x" ! 10 x 9 f !x" ! s 4 x3 ! s 3 x 4 f !x" ! 6sx " s 6 x f !x" ! 2x ! 3x 1.7 f !x" ! 5x 1$4 " 7x 3$4 f !x" ! x!2 " x"2 f !x" ! !x ! 1"!2x " 1" f !x" ! 8x 9 " 3x 6 ! 12x 3 f !x" ! 1 2 ! 3 4x 2 " 4 5x 3 f !x" ! 1 2x 2 " 2x ! 6 f !x" ! x " 3 SECTION 4.9 ANTIDERIVATIVES \|\|\|\| 345 EXERCISES 4.9 and are positive constants that depend on the material of the board and is the acceleration due to gravity. (a) Find an expression for the shape of the curve. (b) Use to estimate the distance below the horizontal at the end of the board. 69. A company estimates that the marginal cost (in dollars per item) of producing items is . If the cost of producing one item is , ﬁnd the cost of producing items. 70. The linear density of a rod of length m is given by , in grams per centimeter, where is measured in centimeters from one end of the rod. Find the mass of the rod. 71. Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A rain-drop has an initial downward velocity of 10 m$s and its downward acceleration is If the raindrop is initially m above the ground, how long does it take to fall? 72. A car is traveling at 50 mi$h when the brakes are fully applied, producing a constant deceleration of 22 ft$s . What is the distance traveled before the car comes to a stop? What constant acceleration is required to increase the speed of a car from 30 mi$h to 50 mi$h in 5 s? 74. A car braked with a constant deceleration of 16 ft$s , pro-ducing skid marks measuring 200 ft before coming to a stop. How fast was the car traveling when the brakes were ﬁrst applied? 75. A car is traveling at when the driver sees an acci-dent 80 m ahead and slams on the brakes. What constant deceleration is required to stop the car in time to avoid a pileup? 76. A model rocket is ﬁred vertically upward from rest. Its accel-eration for the ﬁrst three seconds is , at which time the fuel is exhausted and it becomes a freely “falling” body. Fourteen seconds later, the rocket’s parachute opens, and the (downward) velocity slows linearly to ft$s in 5 s. The rocket then “ﬂoats” to the ground at that rate. (a) Determine the position function and the velocity func-tion (for all times ). Sketch the graphs of and . v s t v s "18 a!t" ! 60t 100 km$h 2 73. 2 500 a !' 9 " 0.9t 0 if 0 . t . 10 if t & 10 x /!x" ! 1$sx 1 100 $562 1.92 " 0.002x x y x 0 f !L" t !' 0" I E ; 54. (a) Use a graphing device to graph . (b) Starting with the graph in part (a), sketch a rough graph of the antiderivative that satisﬁes . (c) Use the rules of this section to ﬁnd an expression for . (d) Graph using the expression in part (c). Compare with your sketch in part (b). ; 55–56 Draw a graph of and use it to make a rough sketch of the antiderivative that passes through the origin. 55. , 56. , 57–62 A particle is moving with the given data. Find the posi-tion of the particle. 58. 59. 60. , , 61. , , 62. , , 63. A stone is dropped from the upper observation deck (the Space Deck) of the CN Tower, m above the ground. (a) Find the distance of the stone above ground level at time . (b) How long does it take the stone to reach the ground? (c) With what velocity does it strike the ground? (d) If the stone is thrown downward with a speed of 5 m$s, how long does it take to reach the ground? 64. Show that for motion in a straight line with constant accelera-tion , initial velocity , and initial displacement , the dis-placement after time is An object is projected upward with initial velocity meters per second from a point meters above the ground. Show that 66. Two balls are thrown upward from the edge of the cliff in Example 7. The ﬁrst is thrown with a speed of ft$s and the other is thrown a second later with a speed of ft$s. Do the balls ever pass each other? 67. A stone was dropped off a cliff and hit the ground with a speed of 120 ft$s. What is the height of the cliff? 68. If a diver of mass stands at the end of a diving board with length and linear density , then the board takes on the shape of a curve , where EIy) ! mt!L " x" ! 1 2 /t!L " x"2 y ! f !x" / L m 24 48 )v!t"2 ! v0 2 " 19.6)s!t" " s0 s0 v0 65. s ! 1 2at 2 ! v0t ! s0 t s0 v0 a t 450 s!1" ! 20 s!0" ! 0 a!t" ! t 2 " 4t ! 6 s!2," ! 12 s!0" ! 0 a!t" ! 10 sin t ! 3 cos t v!0" ! 5 s!0" ! 0 a!t" ! cos t ! sin t a!t" ! t " 2, s!0" ! 1, v!0" ! 3 v!t" ! 1.5st , s!4" ! 10 v!t" ! sin t " cos t, s!0" ! 0 57. "1.5 . x . 1.5 f !x" ! sx 4 " 2x 2 ! 2 " 1 "2, . x . 2, f !x" ! sin x 1 ! x 2 f F F!x" F!0" ! 1 F f !x" ! 2x " 3sx 346 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (b) Suppose that the train starts from rest and must come to a complete stop in 15 minutes. What is the maximum dis-tance it can travel under these conditions? (c) Find the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart. (d) The trip from one station to the next takes 37.5 minutes. How far apart are the stations? (b) At what time does the rocket reach its maximum height, and what is that height? (c) At what time does the rocket land? 77. A high-speed bullet train accelerates and decelerates at the rate of . Its maximum cruising speed is 90 mi$h. (a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes? 4 ft$s2 CHAPTER 4 REVIEW \|\|\|\| 347 REVIEW CONCEPT CHECK 4 where and as ? (c) How can you use l’Hospital’s Rule if you have a difference where and as ? (d) How can you use l’Hospital’s Rule if you have a power where and as ? 8. If you have a graphing calculator or computer, why do you need calculus to graph a function? 9. (a) Given an initial approximation to a root of the equation , explain geometrically, with a diagram, how the second approximation in Newton’s method is obtained. (b) Write an expression for in terms of , , and . (c) Write an expression for in terms of , and . (d) Under what circumstances is Newton’s method likely to fail or to work very slowly? 10. (a) What is an antiderivative of a function ? (b) Suppose and are both antiderivatives of on an inter-val . How are and related? F2 F1 I f F2 F1 f f #!xn" xn, f !xn" xn!1 f #!x1" f !x1" x1 x2 x2 f !x" ! 0 x1 x l a t!x" l 0 f !x" l 0 ) f !x"t!x" x l a t!x" l ( f !x" l ( f !x" " t!x" x l a t!x" l ( f !x" l 0 f !x"t!x" 1. Explain the difference between an absolute maximum and a local maximum. Illustrate with a sketch. 2. (a) What does the Extreme Value Theorem say? (b) Explain how the Closed Interval Method works. 3. (a) State Fermat’s Theorem. (b) Deﬁne a critical number of . 4. (a) State Rolle’s Theorem. (b) State the Mean Value Theorem and give a geometric interpretation. 5. (a) State the Increasing/Decreasing Test. (b) What does it mean to say that is concave upward on an interval ? (c) State the Concavity Test. (d) What are inﬂection points? How do you ﬁnd them? 6. (a) State the First Derivative Test. (b) State the Second Derivative Test. (c) What are the relative advantages and disadvantages of these tests? 7. (a) What does l’Hospital’s Rule say? (b) How can you use l’Hospital’s Rule if you have a product I f f Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. If , then has a local maximum or minimum at . 2. If has an absolute minimum value at , then . 3. If is continuous on , then attains an absolute maxi-mum value and an absolute minimum value at some numbers and in . 4. If is differentiable and , then there is a number such that and . 5. If for , then is decreasing on (1, 6). 6. If , then is an inﬂection point of the curve . y ! f !x" !2, f !2"" f )!2" ! 0 f 1 ' x ' 6 f #!x" ' 0 f #!c" ! 0 ( c( ' 1 c f !"1" ! f !1" f !a, b" d c f !d" f !c" f !a, b" f f #!c" ! 0 c f c f f #!c" ! 0 7. If for , then for 8. There exists a function such that , and for all . 9. There exists a function such that , , and for all . 10. There exists a function such that , , and for all . 11. If and are increasing on an interval , then is increasing on . 12. If and are increasing on an interval , then is increasing on . I f " t I t f I f ! t I t f x f )!x" & 0 f #!x" ' 0 f !x" ' 0 f x f )!x" & 0 f #!x" ' 0 f !x" & 0 f x f #!x" & 1 f !1" ! "2, f !3" ! 0 f 0 ' x ' 1. f !x" ! t!x" 0 ' x ' 1 f #!x" ! t#!x" TRUE-FALSE QUIZ 18. The ﬁgure shows the graph of the derivative of a function . (a) On what intervals is increasing or decreasing? (b) For what values of does have a local maximum or minimum? (c) Sketch the graph of . (d) Sketch a possible graph of . 19–34 Use the guidelines of Section 4.5 to sketch the curve. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. ; 35–38 Produce graphs of that reveal all the important aspects of the curve. Use graphs of and to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inﬂection points. In Exercise 35 use calculus to ﬁnd these quantities exactly. 35. 36. 37. f !x" ! 3x 6 " 5x 5 ! x 4 " 5x 3 " 2x 2 ! 2 f !x" ! x 3 " x x 2 ! x ! 3 f !x" ! x 2 " 1 x 3 f ) f # f y ! x ! ln!x 2 ! 1" y ! xe"2x y ! e2x"x 2 y ! sin"1!1$x" y ! 4x " tan x, ",$2 ' x ' ,$2 y ! sin2x " 2 cos x y ! s 3 x 2 ! 1 y ! xs2 ! x y ! s1 " x ! s1 ! x y ! x 2$!x ! 8" y ! 1 x 2 " 1 !x " 2"2 y ! 1 x!x " 3"2 y ! 1 1 " x 2 y ! x 4 " 3x 3 ! 3x 2 " x y ! x 3 " 6x 2 " 15x ! 4 y ! 2 " 2x " x 3 0 x y 1 2 3 4 5 6 7 _1 _2 y=fª(x) f f ) f x f f f # 1–6 Find the local and absolute extreme values of the function on the given interval. 1. , 2. , 3. , 4. , 5. , 6. , 7–14 Evaluate the limit. 7. 8. 9. 10. 11. 12. 13. 14. 15–17 Sketch the graph of a function that satisﬁes the given conditions: 15. , on , and on and on and on and 16. , is continuous and even, if if , if 17. is odd, for , for , for , for , lim x l ( f !x" ! "2 x & 3 f )!x" ' 0 0 ' x ' 3 f )!x" & 0 x & 2 f #!x" & 0 0 ' x ' 2 f #!x" ' 0 f x & 3 f #!x" ! 1 1 ' x ' 3 0 ' x ' 1, f #!x" ! "1 f #!x" ! 2x f f !0" ! 0 !6, 12" !0, 6" f )!x" ' 0 !12, (", !"(, 0" f )!x" & 0 !6, 9", !"2, 1" f #!x" & 0 !9, (", !"(, "2", !1, 6" f #!x" ' 0 limx l ( f !x" ! 0, limx l 6 f !x" ! "(, f !0" ! 0, f #!"2" ! f #!1" ! f #!9" ! 0 lim x l !,$2" "!tan x"cos x lim x l 1! % x x " 1 " 1 ln x& lim x l 0! x2 ln x lim x l ( x3e"x lim x l ( e4x " 1 " 4x x2 lim x l 0 e4x " 1 " 4x x2 lim x l 0 1 " cos x x 2 ! x lim x l 0 tan ,x ln!1 ! x" )1, 3 f !x" ! !ln x"$x 2 )0, , f !x" ! x ! sin 2x )"2, 1 f !x" ! !x 2 ! 2x"3 )"2, 2 f !x" ! 3x " 4 x 2 ! 1 )"1, 1 f !x" ! xs1 " x )2, 4 f !x" ! x 3 " 6x 2 ! 9x ! 1 EXERCISES 348 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 17. If is periodic, then is periodic. 18. The most general antiderivative of is 19. If exists and is nonzero for all , then . 20. lim x l 0 x e x ! 1 f !1" " f !0" x f #!x" F!x" ! " 1 x ! C f !x" ! x "2 f # f 13. If and are increasing on an interval , then is increasing on . 14. If and are positive increasing functions on an interval , then is increasing on . 15. If is increasing and on , then is decreasing on . 16. If is even, then is even. f # f I t!x" ! 1$f !x" I f !x" & 0 f I ft I t f I ft I t f CHAPTER 4 REVIEW \|\|\|\| 349 51. Show that the shortest distance from the point to the straight line is 52. Find the point on the hyperbola that is closest to the point . 53. Find the smallest possible area of an isosceles triangle that is circumscribed about a circle of radius . 54. Find the volume of the largest circular cone that can be inscribed in a sphere of radius . 55. In , lies on , , cm, and cm. Where should a point be chosen on so that the sum is a minimum? 56. Solve Exercise 55 when cm. 57. The velocity of a wave of length in deep water is where and are known positive constants. What is the length of the wave that gives the minimum velocity? 58. A metal storage tank with volume is to be constructed in the shape of a right circular cylinder surmounted by a hemi-sphere. What dimensions will require the least amount of metal? 59. A hockey team plays in an arena with a seating capacity of 15,000 spectators. With the ticket price set at , average attendance at a game has been 11,000. A market survey indi-cates that for each dollar the ticket price is lowered, average attendance will increase by 1000. How should the owners of the team set the ticket price to maximize their revenue from ticket sales? ; 60. A manufacturer determines that the cost of making units of a commodity is and the demand function is . (a) Graph the cost and revenue functions and use the graphs to estimate the production level for maximum proﬁt. (b) Use calculus to ﬁnd the production level for maximum proﬁt. (c) Estimate the production level that minimizes the average cost. 61. Use Newton’s method to ﬁnd the root of the equation in the interval correct to six decimal places. 62. Use Newton’s method to ﬁnd all roots of the equation correct to six decimal places. 63. Use Newton’s method to ﬁnd the absolute maximum value of the function correct to eight decimal places. f !t" ! cos t ! t " t 2 sin x ! x 2 " 3x ! 1 )1, 2 x5 " x4 ! 3x2 " 3x " 2 ! 0 p!x" ! 48.2 " 0.03x C!x" ! 1800 ! 25x " 0.2x 2 ! 0.001x 3 x $12 V C K v ! K+ L C ! C L L ( CD( ! 2 ( PA( ! ( PB( ! ( PC( CD P ( CD( ! 5 ( AD( ! ( BD( ! 4 CD ! AB AB D 0ABC r r !3, 0" xy ! 8 ( Ax1 ! By1 ! C( sA2 ! B2 Ax ! By ! C ! 0 !x1, y1" 38. ; 39. Graph in a viewing rectangle that shows all the main aspects of this function. Estimate the inﬂection points. Then use calculus to ﬁnd them exactly. 40. (a) Graph the function . (b) Explain the shape of the graph by computing the limits of as approaches , , , and . (c) Use the graph of to estimate the coordinates of the inﬂection points. (d) Use your CAS to compute and graph . (e) Use the graph in part (d) to estimate the inﬂection points more accurately. 41–42 Use the graphs of to estimate the -coordinates of the maximum and minimum points and inﬂection points of . 41. , 42. ; 43. Investigate the family of functions . What features do the members of this family have in common? How do they differ? For which values of is continuous on ? For which values of does have no graph at all? What happens as ? ; 44. Investigate the family of functions . What hap-pens to the maximum and minimum points and the inﬂection points as changes? Illustrate your conclusions by graphing several members of the family. 45. Show that the equation has exactly one real root. 46. Suppose that is continuous on , and for all in . Show that . 47. By applying the Mean Value Theorem to the function on the interval , show that 48. For what values of the constants and is a point of inﬂection of the curve ? 49. Let , where is twice differentiable for all , for all , and is concave downward on and concave upward on . (a) At what numbers does have an extreme value? (b) Discuss the concavity of . 50. Find two positive integers such that the sum of the ﬁrst num-ber and four times the second number is 1000 and the product of the numbers is as large as possible. t t !0, (" !"(, 0" f x " 0 f #!x" & 0 x f t!x" ! f !x 2" y ! x 3 ! ax 2 ! bx ! 1 !1, 6" b a 2 ' s 5 33 ' 2.0125 )32, 33 f !x" ! x 1$5 9 . f !4" . 21 !0, 4" x 2 . f #!x" . 5 )0, 4, f !0" ! 1 f 3x ! 2 cos x ! 5 ! 0 c f !x" ! cxe "cx 2 C l ( f C !"(, (" f C f !x" ! ln!sin x ! C" f!x" ! e"0.1x ln!x 2 " 1" ", . x . , f !x" ! cos2 x sx 2 ! x ! 1 f x f, f #, and f ) CAS f ) f 0" 0! "( ( x f !x" f !x" ! 1$!1 ! e 1$x" CAS f !x" ! e "1$x 2 f !x" ! x 2 ! 6.5 sin x, "5 . x . 5 350 \|\|\|\| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 64. Use the guidelines in Section 4.5 to sketch the curve , . Use Newton’s method when necessary. 65–72 Find . 65. 66. 67. 68. , 69. , 70. , 71. , , 72. , , 73–74 A particle is moving with the given data. Find the position of the particle. 73. , 74. , , ; 75. (a) If , use a graph of to sketch a rough graph of the antiderivative of that satisﬁes . (b) Find an expression for . (c) Graph using the expression in part (b). Compare with your sketch in part (a). ; 76. Investigate the family of curves given by In particular you should determine the transitional value of at which the number of critical numbers changes and the transitional value at which the number of inﬂection points changes. Illustrate the various possible shapes with graphs. 77. A canister is dropped from a helicopter m above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of m!s. Will it burst? 78. In an automobile race along a straight road, car A passed car B twice. Prove that at some time during the race their accelerations were equal. State the assumptions that you make. 79. A rectangular beam will be cut from a cylindrical log of radius 10 inches. (a) Show that the beam of maximal cross-sectional area is a square. 100 500 c f "x# ! x 4 ! x 3 ! cx 2 F F"x# F"0# ! 0 f F f f "x# ! 0.1e x ! sin x, "4 # x # 4 v"0# ! 2 s"0# ! 0 a"t# ! sin t ! 3 cos t s"0# ! 1 v"t# ! 2t " 1!"1 ! t 2# f "1# ! 0 f "0# ! 2 f $"x# ! 2x 3 ! 3x 2 " 4x ! 5 f %"0# ! 2 f "0# ! 1 f $"x# ! 1 " 6x ! 48x 2 f "1# ! 3 f %"u# ! u2 ! su u f "0# ! 5 f %"t# ! 2t " 3 sin t f"0# ! 2 f %"x# ! sinh x ! 2 cosh x f %"x# ! sx 3 ! s 3 x 2 f %"x# ! 2e x ! sec x tan x f %"x# ! cos x " "1 " x 2#"1!2 f 0 # x # 2& y ! x sin x (b) Four rectangular planks will be cut from the four sections of the log that remain after cutting the square beam. Deter-mine the dimensions of the planks that will have maximal cross-sectional area. (c) Suppose that the strength of a rectangular beam is propor-tional to the product of its width and the square of its depth. Find the dimensions of the strongest beam that can be cut from the cylindrical log. 80. If a projectile is ﬁred with an initial velocity at an angle of inclination from the horizontal, then its trajectory, neglect-ing air resistance, is the parabola (a) Suppose the projectile is ﬁred from the base of a plane that is inclined at an angle , , from the horizontal, as shown in the ﬁgure. Show that the range of the projec-tile, measured up the slope, is given by (b) Determine so that is a maximum. (c) Suppose the plane is at an angle below the horizontal. Determine the range in this case, and determine the angle at which the projectile should be ﬁred to maximize . 81. Show that, for , 82. Sketch the graph of a function such that for all for for , and . lim x l'( $ f "x# ! x% ! 0 & x& ) 1 f $"x# ) 0 & x& 1, x, f $"x# 0 f %"x# ) 0 f x 1 ! x 2 ) tan"1x ) x x 0 ¨ å x y 0 R R R + R , R",# ! 2v 2 cos , sin", " +# t cos2+ + 0 + 0 # , # & 2 y ! "tan ,#x " t 2v2 cos2, x 2 , v depth width 10 351 One of the most important principles of problem solving is analogy (see page 76). If you are having trouble getting started on a problem, it is sometimes helpful to start by solving a similar, but simpler, problem. The following example illustrates the principle. Cover up the solution and try solving it yourself ﬁrst. EXAMPLE 1 If x, y, and are positive numbers, prove that SOLUTION It may be difﬁcult to get started on this problem. (Some students have tackled it by multiplying out the numerator, but that just creates a mess.) Let’s try to think of a similar, simpler problem. When several variables are involved, it’s often helpful to think of an analogous problem with fewer variables. In the present case we can reduce the number of variables from three to one and prove the analogous inequality In fact, if we are able to prove (1), then the desired inequality follows because The key to proving (1) is to recognize that it is a disguised version of a minimum prob-lem. If we let then , so when x ! 1. Also, for and for . Therefore the absolute minimum value of is . This means that for all positive values of x and, as previously mentioned, the given inequality follows by multiplication. The inequality in (1) could also be proved without calculus. In fact, if , we have Because the last inequality is obviously true, the ﬁrst one is true too. M & ? "x " 1#2 - 0 x 2 ! 1 x - 2 & ? x 2 ! 1 - 2x & ? x 2 " 2x ! 1 - 0 x 0 x 2 ! 1 x - 2 f"1# ! 2 f x 1 f%"x# 0 0 ) x ) 1 f%"x# ) 0 f%"x# ! 0 f%"x# ! 1 " "1!x 2# x 0 f"x# ! x 2 ! 1 x ! x ! 1 x "x 2 ! 1#"y 2 ! 1#"z2 ! 1# xyz !' x 2 ! 1 x (' y 2 ! 1 y (' z 2 ! 1 z ( - 2 ! 2 ! 2 ! 8 x 2 ! 1 x - 2 for x 0 1 "x 2 ! 1#"y 2 ! 1#"z 2 ! 1# xyz - 8 z P R O B L E M S P L U S Look Back What have we learned from the solution to this example? N To solve a problem involving several variables, it might help to solve a similar problem with just one variable. N When trying to prove an inequality, it might help to think of it as a maximum or minimum problem. 1. If a rectangle has its base on the -axis and two vertices on the curve , show that the rectangle has the largest possible area when the two vertices are at the points of inﬂection of the curve. 2. Show that for all . 3. Show that, for all positive values of and , 4. Show that for all numbers and such that and . 5. If , , , and are constants such that ﬁnd the value of the sum . 6. Find the point on the parabola at which the tangent line cuts from the ﬁrst quad-rant the triangle with the smallest area. 7. Find the highest and lowest points on the curve . 8. Sketch the set of all points such that . 9. If is any point on the parabola , except for the origin, let be the point where the normal line intersects the parabola again. Show that the line segment has the shortest possible length when . 10. For what values of does the curve have inﬂection points? 11. Determine the values of the number for which the function has no critical number: 12. Sketch the region in the plane consisting of all points such that 13. The line intersects the parabola in points and (see the ﬁgure). Find the point on the arc of the parabola that maximizes the area of the triangle . 14. is a square piece of paper with sides of length 1 m. A quarter-circle is drawn from to with center . The piece of paper is folded along , with on and on , so that falls on the quarter-circle. Determine the maximum and minimum areas that the triangle can have. 15. For which positive numbers does the curve intersect the line ? 16. For what value of is the following equation true? 17. Let , where , , . . . , are real numbers and is a positive integer. If it is given that for all , show that & a1 ! 2a2 ! . . . ! nan& # 1 x & f "x#& # & sin x& n an a2 a1 f "x# ! a1 sin x ! a2 sin 2x ! . . . ! an sin nx lim x l ( ' x ! a x " a( x ! e a y ! x y ! a x a AEF A AD F AB E EF A D B ABCD PAB AOB P B A y ! x 2 y ! mx ! b 2xy # & x " y& # x 2 ! y 2 "x, y# f "x# ! "a 2 ! a " 6# cos 2x ! "a " 2#x ! cos 1 f a y ! cx 3 ! e x c a ! 1!s2 PQ Q y ! x 2 P"a, a 2# & x ! y& # e x "x, y# x 2 ! xy ! y 2 ! 12 y ! 1 " x 2 a ! b ! c ! d lim x l 0 ax 2 ! sin bx ! sin cx ! sin dx 3x 2 ! 5x 4 ! 7x 6 ! 8 d c b a & y& # 2 & x& # 2 y x x 2y 2"4 " x 2#"4 " y 2# # 16 e x!y xy - e 2 y x x & sin x " cos x& # s2 y ! e "x 2 x PROBLEMS 352 P R O B L E M S P L U S FIGURE FOR PROBLEM 9 0 x y P Q FIGURE FOR PROBLEM 13 O y x y=≈ y=mx+b P B A 18. An arc of a circle subtends a central angle as in the ﬁgure. Let be the area between the chord and the arc . Let be the area between the tangent lines , and the arc. Find 19. The speeds of sound in an upper layer and in a lower layer of rock and the thickness of the upper layer can be determined by seismic exploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamite charge is detonated at a point and the transmitted signals are recorded at a point , which is a distance from . The ﬁrst signal to arrive at travels along the surface and takes seconds. The next signal travels from to a point , from to in the lower layer, and then to taking seconds. The third signal is reﬂected off the lower layer at the midpoint of and takes seconds to reach . (a) Express in terms of . (b) Show that is a minimum when . (c) Suppose that , , , and . Find . Note: Geophysicists use this technique when studying the structure of the earth’s crust, whether searching for oil or examining fault lines. 20. For what values of is there a straight line that intersects the curve in four distinct points? 21. One of the problems posed by the Marquis de l’Hospital in his calculus textbook Analyse des Inﬁniment Petits concerns a pulley that is attached to the ceiling of a room at a point by a rope of length . At another point on the ceiling, at a distance from (where ), a rope of length ! is attached and passed through the pulley at and connected to a weight . The weight is released and comes to rest at its equilibrium position . As l’Hospital argued, this happens when the distance is maximized. Show that when the system reaches equi-librium, the value of is Notice that this expression is independent of both and !. 22. Given a sphere with radius , ﬁnd the height of a pyramid of minimum volume whose base is a square and whose base and triangular faces are all tangent to the sphere. What if the base of the pyramid is a regular -gon? (A regular -gon is a polygon with equal sides and angles.) (Use the fact that the volume of a pyramid is , where is the area of the base.) 23. Assume that a snowball melts so that its volume decreases at a rate proportional to its surface area. If it takes three hours for the snowball to decrease to half its original volume, how much longer will it take for the snowball to melt completely? 24. A hemispherical bubble is placed on a spherical bubble of radius 1. A smaller hemispherical bubble is then placed on the ﬁrst one. This process is continued until chambers, including the sphere, are formed. (The ﬁgure shows the case .) Use mathematical induction to prove that the maximum height of any bubble tower with chambers is . 1 ! sn n n ! 4 n A 1 3 Ah n n n r W r 4d (r ! sr 2 ! 8d 2 ) x & ED& D W F d r C d B r C y ! x 4 ! cx 3 ! 12x 2 " 5x ! 2 c D h R ¨ Speed of sound=c™ Q O S ¨ Speed of sound=c¡ P c1, c2, and h T3 ! 0.34 s T2 ! 0.32 s T1 ! 0.26 s D ! 1 km sin , ! c1!c2 T2 D, h, c1, c2, and , T1, T2, and T3 Q T3 RS O T2 Q, S R R P T1 Q P D Q P h c2 c1 lim , l 0! A",# B",# PR, QR B",# PQ PQ A",# , PQ 353 P R O B L E M S P L U S P Q B(¨) A(¨) FIGURE FOR PROBLEM 18 ¨ R FIGURE FOR PROBLEM 24 r C F D d x FIGURE FOR PROBLEM 21 B E 354 In Chapter 2 we used the tangent and velocity problems to introduce the derivative, which is the central idea in differential calculus. In much the same way, this chapter starts with the area and distance problems and uses them to formulate the idea of a deﬁnite integral, which is the basic concept of integral calculus. We will see in Chapters 6 and 8 how to use the integral to solve problems concerning volumes, lengths of curves, population predictions, cardiac output, forces on a dam, work, consumer surplus, and baseball, among many others. There is a connection between integral calculus and differential calculus. The Funda-mental Theorem of Calculus relates the integral to the derivative, and we will see in this chapter that it greatly simpliﬁes the solution of many problems. T o compute an area we approximate a region by rectangles and let the number of rectangles become large.The precise area is the limit of these sums of areas of rectangles. INTEGRALS 5 AREAS AND DISTANCES In this section we discover that in trying to ﬁnd the area under a curve or the distance traveled by a car, we end up with the same special type of limit. THE AREA PROBLEM We begin by attempting to solve the area problem: Find the area of the region that lies under the curve from to . This means that , illustrated in Figure 1, is bounded by the graph of a continuous function [where ], the vertical lines and , and the -axis. In trying to solve the area problem we have to ask ourselves: What is the meaning of the word area? This question is easy to answer for regions with straight sides. For a rect-angle, the area is deﬁned as the product of the length and the width. The area of a triangle is half the base times the height. The area of a polygon is found by dividing it into tri-angles (as in Figure 2) and adding the areas of the triangles. However, it isn’t so easy to ﬁnd the area of a region with curved sides. We all have an intuitive idea of what the area of a region is. But part of the area problem is to make this intuitive idea precise by giving an exact deﬁnition of area. Recall that in deﬁning a tangent we ﬁrst approximated the slope of the tangent line by slopes of secant lines and then we took the limit of these approximations. We pursue a sim-ilar idea for areas. We ﬁrst approximate the region by rectangles and then we take the limit of the areas of these rectangles as we increase the number of rectangles. The follow-ing example illustrates the procedure. EXAMPLE 1 Use rectangles to estimate the area under the parabola from 0 to 1 (the parabolic region S illustrated in Figure 3). SOLUTION We ﬁrst notice that the area of S must be somewhere between 0 and 1 because is contained in a square with side length 1, but we can certainly do better than that. S y ! x 2 V S FIGURE 2 h b A= bh A¡ A™ A£ A¢ A=A¡+A™+A£+A¢ A=lw l w 1 2 0 y a b x y=ƒ S x=a x=b FIGURE 1 S=s(x, y) \| a¯x¯b, 0¯y¯ƒd x x ! b x ! a f!x" ! 0 f S b a y ! f!x" S 5.1 355 N Now is a good time to read (or reread) A Preview of Calculus (see page 2). It discusses the unifying ideas of calculus and helps put in perspective where we have been and where we are going. FIGURE 3 0 y x 1 (1, 1) y=≈ S Suppose we divide S into four strips , , , and by drawing the vertical lines , , and as in Figure 4(a). We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the strip [see Figure 4(b)]. In other words, the heights of these rectangles are the values of the function at the right end-points of the subintervals , , , and . Each rectangle has width and the heights are , , , and . If we let be the sum of the areas of these approximating rectangles, we get From Figure 4(b) we see that the area A of S is less than , so Instead of using the rectangles in Figure 4(b) we could use the smaller rectangles in Figure 5 whose heights are the values of at the left endpoints of the subintervals. (The leftmost rectangle has collapsed because its height is 0.) The sum of the areas of these approximating rectangles is We see that the area of S is larger than , so we have lower and upper estimates for A: We can repeat this procedure with a larger number of strips. Figure 6 shows what happens when we divide the region S into eight strips of equal width. FIGURE 6 Approximating S with eight rectangles (a) Using left endpoints (b) Using right endpoints 0 1 (1, 1) 1 8 0 1 1 8 y=≈ (1, 1) y x y x 0.21875 " A " 0.46875 L4 L4 ! 1 4 ! 02 # 1 4 ! ( 1 4) 2 # 1 4 ! ( 1 2) 2 # 1 4 ! ( 3 4) 2 ! 7 32 ! 0.21875 f A " 0.46875 R4 R4 ! 1 4 ! ( 1 4) 2 # 1 4 ! ( 1 2) 2 # 1 4 ! ( 3 4) 2 # 1 4 ! 12 ! 15 32 ! 0.46875 R4 12 ( 3 4) 2 ( 1 2) 2 ( 1 4) 2 1 4 [ 3 4, 1] [ 1 2, 3 4] [ 1 4, 1 2] [0, 1 4] f!x" ! x 2 FIGURE 4 (b) 0 1 (1, 1) 3 4 1 2 1 4 (a) 0 y x 1 (1, 1) y=≈ 3 4 1 2 1 4 S¢ S£ S™ S¡ y x x ! 3 4 x ! 1 2 x ! 1 4 S4 S3 S2 S1 356 \|\|\|\| CHAPTER 5 INTEGRALS 0 y x 1 (1, 1) 3 4 1 2 1 4 y=≈ FIGURE 5 By computing the sum of the areas of the smaller rectangles and the sum of the areas of the larger rectangles , we obtain better lower and upper estimates for A: So one possible answer to the question is to say that the true area of S lies somewhere between 0.2734375 and 0.3984375. We could obtain better estimates by increasing the number of strips. The table at the left shows the results of similar calculations (with a computer) using n rectangles whose heights are found with left endpoints or right endpoints . In particular, we see by using 50 strips that the area lies between 0.3234 and 0.3434. With 1000 strips we narrow it down even more: A lies between 0.3328335 and 0.3338335. A good estimate is obtained by averaging these numbers: . M From the values in the table in Example 1, it looks as if is approaching as n increases. We conﬁrm this in the next example. EXAMPLE 2 For the region S in Example 1, show that the sum of the areas of the upper approximating rectangles approaches , that is, SOLUTION is the sum of the areas of the rectangles in Figure 7. Each rectangle has width and the heights are the values of the function at the points ; that is, the heights are . Thus Here we need the formula for the sum of the squares of the ﬁrst n positive integers: Perhaps you have seen this formula before. It is proved in Example 5 in Appendix E. Putting Formula 1 into our expression for , we get Thus we have M ! 1 6 ! 1 ! 2 ! 1 3 ! lim n l $ 1 6 #1 # 1 n$#2 # 1 n$ ! lim n l $ 1 6 # n # 1 n $# 2n # 1 n $ lim n l $ Rn ! lim n l $ !n # 1"!2n # 1" 6n 2 Rn ! 1 n 3 % n!n # 1"!2n # 1" 6 ! !n # 1"!2n # 1" 6n 2 Rn 12 # 22 # 32 # % % % # n 2 ! n!n # 1"!2n # 1" 6 1 ! 1 n 3 !12 # 22 # 32 # % % % # n 2" ! 1 n % 1 n 2 !12 # 22 # 32 # % % % # n 2" Rn ! 1 n # 1 n$ 2 # 1 n # 2 n$ 2 # 1 n # 3 n$ 2 # % % % # 1 n # n n$ 2 !1%n"2, !2%n"2, !3%n"2, . . . , !n%n"2 1%n, 2%n, 3%n, . . . , n%n f!x" ! x 2 1%n n Rn lim n l $ Rn ! 1 3 1 3 V 1 3 Rn A & 0.3333335 !Rn" !Ln" 0.2734375 " A " 0.3984375 !R8" !L8" SECTION 5.1 AREAS AND DISTANCES \|\|\|\| 357 n 10 0.2850000 0.3850000 20 0.3087500 0.3587500 30 0.3168519 0.3501852 50 0.3234000 0.3434000 100 0.3283500 0.3383500 1000 0.3328335 0.3338335 Rn Ln FIGURE 7 1 n 0 y x 1 (1, 1) y=≈ N Here we are computing the limit of the sequence . Sequences were discussed in A Preview of Calculus and will be studied in detail in Chapter 11. Their limits are calculated in the same way as limits at inﬁnity (Section 2.6). In particular, we know that lim n l $ 1 n ! 0 'Rn( It can be shown that the lower approximating sums also approach , that is, From Figures 8 and 9 it appears that, as n increases, both and become better and bet-ter approximations to the area of S. Therefore, we deﬁne the area A to be the limit of the sums of the areas of the approximating rectangles, that is, Let’s apply the idea of Examples 1 and 2 to the more general region S of Figure 1. We start by subdividing into strips of equal width as in Figure 10. FIGURE 10 b a 0 y x . . . . . . y=ƒ S¡ S™ S£ Si Sn xi xi-1 xn-1 ¤ ⁄ ‹ S1, S2, . . . , Sn n S 1 0 y n=10 L¡¸=0.285 x 1 0 x y n=30 L£¸Å0.3169 1 0 x y n=50 L∞¸=0.3234 FIGURE 9 FIGURE 8 1 0 y n=30 R£¸Å0.3502 1 0 x x y n=10 R¡¸=0.385 1 0 y n=50 R∞¸=0.3434 x A ! lim n l $ Rn ! lim n l $ Ln ! 1 3 Rn Ln lim n l $ Ln ! 1 3 1 3 358 \|\|\|\| CHAPTER 5 INTEGRALS In Visual 5.1 you can create pic-tures like those in Figures 8 and 9 for other values of . n TEC The area is the number that is smaller than all upper sums and larger than all lower sums The width of the interval is , so the width of each of the n strips is These strips divide the interval [a, b] into n subintervals where and . The right endpoints of the subintervals are Let’s approximate the th strip by a rectangle with width and height , which is the value of at the right endpoint (see Figure 11). Then the area of the rectangle is . What we think of intuitively as the area of is approximated by the sum of the areas of these rectangles, which is Figure 12 shows this approximation for and . Notice that this approxi-mation appears to become better and better as the number of strips increases, that is, as . Therefore we deﬁne the area of the region in the following way. FIGURE 12 0 y x a ⁄ (a) n=2 b 0 y x a ⁄ ¤ ‹ (b) n=4 b 0 y x a (c) n=8 b 0 y x a b (d) n=12 S A n l $ 12 n ! 2, 4, 8, FIGURE 11 0 y x Îx f(xi) xi xi-1 a b ¤ ⁄ ‹ Rn ! f!x1" &x # f!x2" &x # % % % # f!xn" &x S f!xi" &x ith f f!xi" &x Si i % % % x3 ! a # 3 &x, x2 ! a # 2 &x, x1 ! a # &x, xn ! b x0 ! a )x0, x1, )x1, x2, )x2, x3, . . . , )xn'1, xn &x ! b ' a n b ' a )a, b SECTION 5.1 AREAS AND DISTANCES \|\|\|\| 359 DEFINITION The area A of the region S that lies under the graph of the contin-uous function is the limit of the sum of the areas of approximating rectangles: It can be proved that the limit in Deﬁnition 2 always exists, since we are assuming that is continuous. It can also be shown that we get the same value if we use left endpoints: In fact, instead of using left endpoints or right endpoints, we could take the height of the ith rectangle to be the value of f at any number in the ith subinterval . We call the numbers , , . . . , the sample points. Figure 13 shows approximating rectangles when the sample points are not chosen to be endpoints. So a more general expression for the area of S is We often use sigma notation to write sums with many terms more compactly. For instance, So the expressions for area in Equations 2, 3, and 4 can be written as follows: A ! lim n l $ + n i!1 f!xi " &x A ! lim n l $ + n i!1 f!xi'1" &x A ! lim n l $ + n i!1 f!xi" &x + n i!1 f!xi" &x ! f!x1" &x # f!x2" &x # % % % # f!xn" &x FIGURE 13 xi xi-1 0 y x a b ¤ ⁄ ‹ xn-1 x¡ x™ x£ xn xi Îx f(xi ) A ! lim n l $ ) f!x1 " &x # f!x2 " &x # % % % # f!xn " &x 4 xn x2 x1 )xi'1, xi xi A ! lim n l $ Ln ! lim n l $ ) f!x0" &x # f!x1" &x # % % % # f!xn'1" &x 3 f A ! lim n l $ Rn ! lim n l $ ) f!x1" &x # f!x2" &x # % % % # f!xn" &x f 2 360 \|\|\|\| CHAPTER 5 INTEGRALS N If you need practice with sigma notation, look at the examples and try some of the exercises in Appendix E. This tells us to end with i=n. This tells us to add. This tells us to start with i=m. µ f(xi) Îx n i=m We can also rewrite Formula 1 in the following way: EXAMPLE 3 Let A be the area of the region that lies under the graph of between and . (a) Using right endpoints, ﬁnd an expression for A as a limit. Do not evaluate the limit. (b) Estimate the area by taking the sample points to be midpoints and using four sub-intervals and then ten subintervals. SOLUTION (a) Since and , the width of a subinterval is So , and . The sum of the areas of the approximating rectangles is According to Deﬁnition 2, the area is Using sigma notation we could write It is difﬁcult to evaluate this limit directly by hand, but with the aid of a computer alge-bra system it isn’t hard (see Exercise 24). In Section 5.3 we will be able to ﬁnd A more easily using a different method. (b) With the subintervals of equal width are , , , and . The midpoints of these subintervals are , , , and , and the sum of the areas of the four approximating rectangles (see Fig-ure 14) is So an estimate for the area is A & 0.8557 ! 1 2!e'0.25 # e'0.75 # e'1.25 # e'1.75" & 0.8557 ! e'0.25!0.5" # e'0.75!0.5" # e'1.25!0.5" # e'1.75!0.5" ! f!0.25" &x # f!0.75" &x # f!1.25" &x # f!1.75" &x M4 ! + 4 i!1 f!xi " &x x4 ! 1.75 x3 ! 1.25 x2 ! 0.75 x1 ! 0.25 )1.5, 2 )1, 1.5 )0.5, 1 )0, 0.5 &x ! 0.5 n ! 4 A ! lim n l $ 2 n + n i!1 e'2i%n A ! lim n l $ Rn ! lim n l $ 2 n !e'2%n # e'4%n # e'6%n # % % % # e'2n%n" ! e'2%n# 2 n$ # e'4%n# 2 n$ # % % % # e'2n%n# 2 n$ ! e'x1 &x # e'x2 &x # % % % # e'xn &x Rn ! f!x1" &x # f!x2" &x # % % % # f!xn" &x xn ! 2n%n x1 ! 2%n, x2 ! 4%n, x3 ! 6%n, xi ! 2i%n &x ! 2 ' 0 n ! 2 n b ! 2 a ! 0 x ! 2 x ! 0 f!x" ! e'x + n i!1 i 2 ! n!n # 1"!2n # 1" 6 SECTION 5.1 AREAS AND DISTANCES \|\|\|\| 361 FIGURE 14 1 2 1 y=e–® 0 y x With the subintervals are , , . . . , and the midpoints are . Thus From Figure 15 it appears that this estimate is better than the estimate with . M THE DISTANCE PROBLEM Now let’s consider the distance problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times. (In a sense this is the inverse problem of the velocity problem that we discussed in Section 2.1.) If the velocity remains constant, then the distance problem is easy to solve by means of the formula But if the velocity varies, it’s not so easy to ﬁnd the distance traveled. We investigate the problem in the following example. EXAMPLE 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30-second time interval. We take speedometer readings every ﬁve seconds and record them in the following table: In order to have the time and the velocity in consistent units, let’s convert the velocity readings to feet per second (1 mi%h ! 5280%3600 ft%s): During the ﬁrst ﬁve seconds the velocity doesn’t change very much, so we can estimate the distance traveled during that time by assuming that the velocity is constant. If we take the velocity during that time interval to be the initial velocity (25 ft%s), then we obtain the approximate distance traveled during the ﬁrst ﬁve seconds: Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t ! 5 s. So our estimate for the distance traveled from to is If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: !25 ( 5" # !31 ( 5" # !35 ( 5" # !43 ( 5" # !47 ( 5" # !46 ( 5" ! 1135 ft 31 ft%s ( 5 s ! 155 ft t ! 10 s t ! 5 s 25 ft%s ( 5 s ! 125 ft V distance ! velocity ( time n ! 4 ! 0.2!e'0.1 # e'0.3 # e'0.5 # % % % # e'1.9" & 0.8632 A & M10 ! f!0.1" &x # f!0.3" &x # f!0.5" &x # % % % # f!1.9" &x x1 ! 0.1, x2 ! 0.3, x3 ! 0.5, . . . , x10 ! 1.9 )1.8, 2 )0.2, 0.4 )0, 0.2 n ! 10 362 \|\|\|\| CHAPTER 5 INTEGRALS y=e–® 1 1 0 y x FIGURE 15 2 Time (s) 0 5 10 15 20 25 30 Velocity (mi%h) 17 21 24 29 32 31 28 Time (s) 0 5 10 15 20 25 30 Velocity (ft%s) 25 31 35 43 47 46 41 We could just as well have used the velocity at the end of each time period instead of the velocity at the beginning as our assumed constant velocity. Then our estimate becomes If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second. M Perhaps the calculations in Example 4 remind you of the sums we used earlier to esti-mate areas. The similarity is explained when we sketch a graph of the velocity function of the car in Figure 16 and draw rectangles whose heights are the initial velocities for each time interval. The area of the ﬁrst rectangle is , which is also our estimate for the distance traveled in the ﬁrst ﬁve seconds. In fact, the area of each rectangle can be interpreted as a distance because the height represents velocity and the width represents time. The sum of the areas of the rectangles in Figure 16 is , which is our ini-tial estimate for the total distance traveled. In general, suppose an object moves with velocity , where and (so the object always moves in the positive direction). We take velocity readings at times so that the velocity is approximately constant on each subinterval. If these times are equally spaced, then the time between consecutive readings is . During the ﬁrst time interval the velocity is approximately and so the distance traveled is approximately . Similarly, the distance traveled during the second time interval is about and the total distance traveled during the time inter-val is approximately If we use the velocity at right endpoints instead of left endpoints, our estimate for the total distance becomes The more frequently we measure the velocity, the more accurate our estimates become, so it seems plausible that the exact distance d traveled is the limit of such expressions: We will see in Section 5.4 that this is indeed true. Because Equation 5 has the same form as our expressions for area in Equations 2 and 3, it follows that the distance traveled is equal to the area under the graph of the velocity function. In Chapters 6 and 8 we will see that other quantities of interest in the natural and social sciences—such as the work done by a variable force or the cardiac output of the heart—can also be interpreted as the area under a curve. So when we compute areas in this chapter, bear in mind that they can be interpreted in a variety of practical ways. d ! lim n l $ + n i!1 f!ti'1" &t ! lim n l $ + n i!1 f!ti" &t 5 f!t1" &t # f!t2" &t # % % % # f!tn" &t ! + n i!1 f!ti" &t f!t0" &t # f!t1" &t # % % % # f!tn'1" &t ! + n i!1 f!ti'1" &t )a, b f!t1" &t f!t0" &t f!t0" &t ! !b ' a"%n t0 !! a", t1, t2, . . . , tn !! b" f!t" ! 0 a ) t ) b v ! f!t" L6 ! 1135 25 ( 5 ! 125 !31 ( 5" # !35 ( 5" # !43 ( 5" # !47 ( 5" # !46 ( 5" # !41 ( 5" ! 1215 ft SECTION 5.1 AREAS AND DISTANCES \|\|\|\| 363 FIGURE 16 10 20 20 40 30 0 √ t points. Then improve your estimate by using six rectangles. Sketch the curve and the approximating rectangles. (b) Repeat part (a) using left endpoints. (c) Repeat part (a) using midpoints. (d) From your sketches in parts (a)–(c), which appears to be the best estimate? ; 6. (a) Graph the function . (b) Estimate the area under the graph of using four approx-imating rectangles and taking the sample points to be (i) right endpoints and (ii) midpoints. In each case sketch the curve and the rectangles. (c) Improve your estimates in part (b) by using 8 rectangles. 7–8 With a programmable calculator (or a computer), it is pos-sible to evaluate the expressions for the sums of areas of approx-imating rectangles, even for large values of , using looping. (On a TI use the Is command or a For-EndFor loop, on a Casio use Isz, on an HP or in BASIC use a FOR-NEXT loop.) Compute the sum of the areas of approximating rectangles using equal subintervals and right endpoints for , 30, 50, and 100. Then guess the value of the exact area. 7. The region under from to 8. The region under from to 9. Some computer algebra systems have commands that will draw approximating rectangles and evaluate the sums of their areas, at least if is a left or right endpoint. (For instance, in Maple use leftbox, rightbox, leftsum, and right-sum.) (a) If , ﬁnd the left and right sums for and . (b) Illustrate by graphing the rectangles in part (a). (c) Show that the exact area under lies between 0.780 and 0.791. 10. (a) If , use the commands discussed in Exercise 9 to ﬁnd the left and right sums for 30, and . (b) Illustrate by graphing the rectangles in part (a). (c) Show that the exact area under lies between 2.50 and 2.59. The speed of a runner increased steadily during the ﬁrst three seconds of a race. Her speed at half-second intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds. 11. f 50 n ! 10, f !x" ! ln x, 1 ! x ! 4 CAS f 50 n ! 10, 30, f !x" ! 1#!x 2 " 1", 0 ! x ! 1 xi CAS ##2 0 y ! cos x 1 0 y ! x 4 n ! 10 $ n f f !x" ! e%x 2, %2 ! x ! 2 1. (a) By reading values from the given graph of , use ﬁve rectangles to ﬁnd a lower estimate and an upper estimate for the area under the given graph of from to . In each case sketch the rectangles that you use. (b) Find new estimates using ten rectangles in each case. (a) Use six rectangles to ﬁnd estimates of each type for the area under the given graph of from to . (i) (sample points are left endpoints) (ii) (sample points are right endpoints) (iii) (sample points are midpoints) (b) Is an underestimate or overestimate of the true area? (c) Is an underestimate or overestimate of the true area? (d) Which of the numbers , , or gives the best estimate? Explain. 3. (a) Estimate the area under the graph of from to using four approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? (b) Repeat part (a) using left endpoints. 4. (a) Estimate the area under the graph of from to using four approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? (b) Repeat part (a) using left endpoints. (a) Estimate the area under the graph of from to using three rectangles and right end-x ! 2 x ! %1 f !x" ! 1 " x 2 5. x ! 4 x ! 0 f !x" ! sx x ! ##2 x ! 0 f !x" ! cos x y x 0 4 4 8 y=ƒ 8 12 M6 R6 L6 R6 L6 M6 R6 L6 x ! 12 x ! 0 f 2. y x 0 5 5 y=ƒ 10 x ! 10 x ! 0 f f EXERCISES 5.1 t (s) 0 0.5 1.0 1.5 2.0 2.5 3.0 (ft#s) 0 6.2 10.8 14.9 18.1 19.4 20.2 v 364 \|\|\|\| CHAPTER 5 INTEGRALS 16. The velocity graph of a car accelerating from rest to a speed of over a period of 30 seconds is shown. Estimate the distance traveled during this period. 17–19 Use Deﬁnition 2 to ﬁnd an expression for the area under the graph of as a limit. Do not evaluate the limit. 17. , 18. , 19. , 20–21 Determine a region whose area is equal to the given limit. Do not evaluate the limit. 20. 22. (a) Use Deﬁnition 2 to ﬁnd an expression for the area under the curve from 0 to 1 as a limit. (b) The following formula for the sum of the cubes of the ﬁrst integers is proved in Appendix E. Use it to evaluate the limit in part (a). 23. (a) Express the area under the curve from 0 to 2 as a limit. (b) Use a computer algebra system to ﬁnd the sum in your expression from part (a). (c) Evaluate the limit in part (a). 24. Find the exact area of the region under the graph of from 0 to 2 by using a computer algebra system to evaluate the sum and then the limit in Example 3(a). Compare your answer with the estimate obtained in Example 3(b). y ! e%x CAS y ! x 5 CAS 13 " 23 " 33 " & & & " n 3 !$ n!n " 1" 2 % 2 n y ! x 3 lim n l ' & n i!1 # 4n tan i# 4n 21. lim n l ' & n i!1 2 n '5 " 2i n( 10 0 ! x ! ##2 f !x" ! x cos x 3 ! x ! 10 f !x" ! ln x x 1 ! x ! 16 f !x" ! s 4 x f 40 80 √ (km/h) t (seconds) 0 10 20 30 120 km#h 12. Speedometer readings for a motorcycle at 12-second intervals are given in the table. (a) Estimate the distance traveled by the motorcycle during this time period using the velocities at the beginning of the time intervals. (b) Give another estimate using the velocities at the end of the time periods. (c) Are your estimates in parts (a) and (b) upper and lower estimates? Explain. 13. Oil leaked from a tank at a rate of liters per hour. The rate decreased as time passed and values of the rate at two-hour time intervals are shown in the table. Find lower and upper estimates for the total amount of oil that leaked out. 14. When we estimate distances from velocity data, it is some-times necessary to use times that are not equally spaced. We can still estimate distances using the time periods . For example, on May 7, 1992, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table, provided by NASA, gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Use these data to estimate the height above the earth’s surface of the Endeavour, 62 seconds after liftoff. The velocity graph of a braking car is shown. Use it to esti-mate the distance traveled by the car while the brakes are applied. √ t (seconds) 0 2 20 40 60 4 6 (ft/s) 15. (ti ! ti % ti%1 t0, t1, t2, t3, . . . r!t" t (s) 0 12 24 36 48 60 (ft#s) 30 28 25 22 24 27 v 0 2 4 6 8 10 (L#h) 8.7 7.6 6.8 6.2 5.7 5.3 r!t" t !h" Event Time (s) Velocity (ft#s) Launch 0 0 Begin roll maneuver 10 185 End roll maneuver 15 319 Throttle to 89% 20 447 Throttle to 67% 32 742 Throttle to 104% 59 1325 Maximum dynamic pressure 62 1445 Solid rocket booster separation 125 4151 SECTION 5.1 AREAS AND DISTANCES \|\|\|\| 365 THE DEFINITE INTEGRAL We saw in Section 5.1 that a limit of the form arises when we compute an area. We also saw that it arises when we try to ﬁnd the dis-tance traveled by an object. It turns out that this same type of limit occurs in a wide vari-ety of situations even when is not necessarily a positive function. In Chapters 6 and 8 we will see that limits of the form (1) also arise in ﬁnding lengths of curves, volumes of solids, centers of mass, force due to water pressure, and work, as well as other quantities. We therefore give this type of limit a special name and notation. DEFINITION OF A DEFINITE INTEGRAL If is a function deﬁned for , we divide the interval into n subintervals of equal width . We let (! b) be the endpoints of these subintervals and we let be any sample points in these subintervals, so lies in the ith subinterval . Then the deﬁnite integral of f from a to b is provided that this limit exists. If it does exist, we say that is integrable on . The precise meaning of the limit that deﬁnes the integral is as follows: For every number there is an integer such that for every integer and for every choice of in . The symbol was introduced by Leibniz and is called an integral sign. It is an elongated and was chosen because an integral is a limit of sums. In the notation is called the integrand and and are called the limits of integration; is the lower limit and is the upper limit. For now, the symbol has no meaning by itself; is all one symbol. The simply indicates that the independent vari-able is . The procedure of calculating an integral is called integration. x dx xb a f!x" dx dx b a b a f!x" xb a f!x" dx, S x NOTE 1 )xi%1, xi xi n $ N + y b a f!x" dx % & n i!1 f!xi " (x+ ) N $ 0 )a, b f y b a f!x" dx ! lim n l ' & n i!1 f!xi " (x )xi%1, xi x i x1 , x2 , . . . , xn x0 !! a", x1, x2, . . . , xn (x ! !b % a"#n )a, b a ! x ! b f 2 f lim n l ' & n i!1 f!xi " (x ! lim n l ' ) f!x1 " (x " f!x2 " (x " & & & " f!xn " (x 1 5.2 into congruent triangles with central angle , show that (b) Show that . [Hint: Use Equation 3.3.2.] limn l ' An ! #r 2 An ! 1 2nr 2 sin' 2# n( 2##n n 25. Find the exact area under the cosine curve from to , where . (Use a computer alge-bra system both to evaluate the sum and compute the limit.) In particular, what is the area if ? 26. (a) Let be the area of a polygon with equal sides inscribed in a circle with radius . By dividing the polygon r n An b ! ##2 0 ! b ! ##2 x ! b x ! 0 y ! cos x CAS 366 \|\|\|\| CHAPTER 5 INTEGRALS The deﬁnite integral is a number; it does not depend on . In fact, we could use any letter in place of without changing the value of the integral: The sum that occurs in Deﬁnition 2 is called a Riemann sum after the German mathematician Bernhard Riemann (1826–1866). So Deﬁnition 2 says that the deﬁnite integral of an inte-grable function can be approximated to within any desired degree of accuracy by a Riemann sum. We know that if happens to be positive, then the Riemann sum can be interpreted as a sum of areas of approximating rectangles (see Figure 1). By comparing Deﬁnition 2 with the deﬁnition of area in Section 5.1, we see that the deﬁnite integral can be inter-preted as the area under the curve from a to b. (See Figure 2.) If takes on both positive and negative values, as in Figure 3, then the Riemann sum is the sum of the areas of the rectangles that lie above the -axis and the negatives of the areas of the rectangles that lie below the -axis (the areas of the gold rectangles minus the areas of the blue rectangles). When we take the limit of such Riemann sums, we get the situa-tion illustrated in Figure 4. A deﬁnite integral can be interpreted as a net area, that is, a difference of areas: where is the area of the region above the -axis and below the graph of , and is the area of the region below the -axis and above the graph of . Although we have deﬁned by dividing into subintervals of equal width, there are situations in which it is advantageous to work with subintervals of unequal width. For instance, in Exercise 14 in Section 5.1 NASA provided velocity data at times that were not equally spaced, but we were still able to estimate the distance trav-eled. And there are methods for numerical integration that take advantage of unequal subintervals. )a, b xb a f!x" dx NOTE 4 f x A2 f x A1 y b a f!x" dx ! A1 % A2 x x f xi 0 y x a Îx FIGURE 1 If ƒ˘0, the Riemann sum µ f(xi ) Îx is the sum of areas of rectangles. y=ƒ 0 y x a b b FIGURE 2 If ƒ˘0, the integral j ƒ dx is the area under the curve y=ƒ from a to b. a b y ! f!x" xb a f!x" dx f & n i!1 f!xi " (x NOTE 3 y b a f!x" dx ! y b a f!t" dt ! y b a f!r" dr x x xb a f!x" dx NOTE 2 Bernhard Riemann received his Ph.D. under the direction of the legendary Gauss at the University of Göttingen and remained there to teach. Gauss, who was not in the habit of praising other mathe-maticians, spoke of Riemann’s “creative, active, truly mathematical mind and gloriously fertile originality.” The deﬁnition (2) of an integral that we use is due to Riemann. He also made major contributions to the theory of functions of a complex variable, mathematical physics, num-ber theory, and the foundations of geometry. Riemann’s broad concept of space and geometry turned out to be the right setting, 50 years later, for Einstein’s general relativity theory. Riemann’s health was poor throughout his life, and he died of tuberculosis at the age of 39. RIEMANN FIGURE 3 µ f(xi ) Îx is an approximation to the net area 0 y=ƒ y a b x y=ƒ y x a b 0 FIGURE 4 j ƒ dx is the net area a b SECTION 5.2 THE DEFINITE INTEGRAL \|\|\|\| 367 If the subinterval widths are , we have to ensure that all these widths approach 0 in the limiting process. This happens if the largest width, , approaches 0. So in this case the deﬁnition of a deﬁnite integral becomes We have deﬁned the deﬁnite integral for an inegrable function, but not all func-tions are integrable (see Exercises 67–68). The following theorem shows that the most commonly occurring functions are in fact integrable. It is proved in more advanced courses. THEOREM If is continuous on , or if has only a ﬁnite number of jump discontinuities, then is integrable on ; that is, the deﬁnite integral exists. If is integrable on , then the limit in Deﬁnition 2 exists and gives the same value no matter how we choose the sample points . To simplify the calculation of the integral we often take the sample points to be right endpoints. Then and the deﬁnition of an integral simpliﬁes as follows. THEOREM If is integrable on , then where and EXAMPLE 1 Express as an integral on the interval . SOLUTION Comparing the given limit with the limit in Theorem 4, we see that they will be identical if we choose . We are given that and . Therefore, by Theorem 4, we have M Later, when we apply the deﬁnite integral to physical situations, it will be important to recognize limits of sums as integrals, as we did in Example 1. When Leibniz chose the notation for an integral, he chose the ingredients as reminders of the limiting process. In general, when we write we replace by , by x, and by dx. (x xi x lim , lim n l ' & n i!1 f!x i " (x ! y b a f!x" dx lim n l ' & n i!1 !xi 3 " xi sin xi" (x ! y # 0 !x 3 " x sin x" dx b ! # a ! 0 f!x" ! x 3 " x sin x )0, # lim n l ' & n i!1 !xi 3 " xi sin xi" (x xi ! a " i (x (x ! b % a n y b a f!x" dx ! lim n l ' & n i!1 f!xi" (x )a, b f 4 xi ! xi xi )a, b f xb a f!x" dx )a, b f f )a, b f 3 NOTE 5 y b a f!x" dx ! lim max (xi l 0 & n i!1 f!xi " (xi max (xi (x1, (x2, . . . , (xn 368 \|\|\|\| CHAPTER 5 INTEGRALS EVALUATING INTEGRALS When we use a limit to evaluate a deﬁnite integral, we need to know how to work with sums. The following three equations give formulas for sums of powers of positive integers. Equation 5 may be familiar to you from a course in algebra. Equations 6 and 7 were dis-cussed in Section 5.1 and are proved in Appendix E. The remaining formulas are simple rules for working with sigma notation: EXAMPLE 2 (a) Evaluate the Riemann sum for taking the sample points to be right endpoints and a ! 0, b ! 3, and n ! 6. (b) Evaluate . SOLUTION (a) With n ! 6 the interval width is and the right endpoints are , , , , , and . So the Riemann sum is ! %3.9375 ! 1 2!%2.875 % 5 % 5.625 % 4 " 0.625 " 9" ! f!0.5" (x " f!1.0" (x " f!1.5" (x " f!2.0" (x " f!2.5" (x " f!3.0" (x R6 ! & 6 i!1 f!xi" (x x6 ! 3.0 x5 ! 2.5 x4 ! 2.0 x3 ! 1.5 x2 ! 1.0 x1 ! 0.5 (x ! b % a n ! 3 % 0 6 ! 1 2 y 3 0 !x 3 % 6x" dx f!x" ! x 3 % 6x & n i!1 !ai % bi" ! & n i!1 ai % & n i!1 bi 11 & n i!1 !ai " bi" ! & n i!1 ai " & n i!1 bi 10 & n i!1 cai ! c & n i!1 ai 9 & n i!1 c ! nc 8 & n i!1 i 3 !$ n!n " 1" 2 % 2 7 & n i!1 i 2 ! n!n " 1"!2n " 1" 6 6 & n i!1 i ! n!n " 1" 2 5 N Formulas 8–11 are proved by writing out each side in expanded form. The left side of Equation 9 is The right side is These are equal by the distributive property. The other formulas are discussed in Appendix E. c!a1 " a2 " & & & " an" ca1 " ca2 " & & & " can SECTION 5.2 THE DEFINITE INTEGRAL \|\|\|\| 369 Notice that f is not a positive function and so the Riemann sum does not represent a sum of areas of rectangles. But it does represent the sum of the areas of the gold rectangles (above the x-axis) minus the sum of the areas of the blue rectangles (below the x-axis) in Figure 5. (b) With n subintervals we have Thus , , , , and, in general, . Since we are using right endpoints, we can use Theorem 4: (Equation 9 with ) (Equations 11 and 9) (Equations 7 and 5) This integral can’t be interpreted as an area because takes on both positive and nega-tive values. But it can be interpreted as the difference of areas , where and are shown in Figure 6. Figure 7 illustrates the calculation by showing the positive and negative terms in the right Riemann sum for . The values in the table show the Riemann sums approaching the exact value of the integral, , as . M 0 y 3 x 5 y=˛-6x FIGURE 7 R¢¸Å_6.3998 n l ' %6.75 n ! 40 Rn A2 A1 A1 % A2 f ! 81 4 % 27 ! % 27 4 ! %6.75 ! lim n l ' $ 81 4 '1 " 1 n( 2 % 27'1 " 1 n(% ! lim n l ' -81 n 4 $ n!n " 1" 2 % 2 % 54 n 2 n!n " 1" 2 . ! lim n l ' $ 81 n 4 & n i!1 i 3 % 54 n 2 & n i!1 i% ! lim n l ' 3 n & n i!1 $ 27 n 3 i 3 % 18 n i% c ! 3#n ! lim n l ' 3 n & n i!1 $' 3i n( 3 % 6' 3i n(% y 3 0 !x 3 % 6x" dx ! lim n l ' & n i!1 f!xi" (x ! lim n l ' & n i!1 f' 3i n( 3 n xi ! 3i#n x3 ! 9#n x2 ! 6#n x1 ! 3#n x0 ! 0 (x ! b % a n ! 3 n 0 y 3 x 5 y=˛-6x FIGURE 5 FIGURE 6 j (˛-6x) dx=A¡-A™=_6.75 0 3 A™ A¡ 0 y 3 x 5 y=˛-6x n 40 %6.3998 100 %6.6130 500 %6.7229 1000 %6.7365 5000 %6.7473 Rn 370 \|\|\|\| CHAPTER 5 INTEGRALS N In the sum, is a constant (unlike ), so we can move in front of the sign. , 3#n i n A much simpler method for evaluating the integral in Example 2 will be given in Section 5.3. EXAMPLE 3 (a) Set up an expression for as a limit of sums. (b) Use a computer algebra system to evaluate the expression. SOLUTION (a) Here we have , , , and So , , , , and From Theorem 4, we get (b) If we ask a computer algebra system to evaluate the sum and simplify, we obtain Now we ask the computer algebra system to evaluate the limit: We will learn a much easier method for the evaluation of integrals in the next section. M EXAMPLE 4 Evaluate the following integrals by interpreting each in terms of areas. (a) (b) SOLUTION (a) Since , we can interpret this integral as the area under the curve from 0 to 1. But, since , we get , which shows that the graph of is the quarter-circle with radius 1 in Figure 9. Therefore (In Section 7.3 we will be able to prove that the area of a circle of radius r is .) #r 2 y 1 0 s1 % x 2 dx ! 1 4#!1"2 ! # 4 f x 2 " y 2 ! 1 y 2 ! 1 % x 2 y ! s1 % x 2 f!x" ! s1 % x 2 + 0 y 3 0 !x % 1" dx y 1 0 s1 % x 2 dx V y 3 1 e x dx ! lim n l ' 2 n ! e !3n"2"#n % e !n"2"#n e 2#n % 1 ! e 3 % e & n i!1 e 1"2i#n ! e !3n"2"#n % e !n"2"#n e 2#n % 1 ! lim n l ' 2 n & n i!1 e1"2i#n ! lim n l ' & n i!1 f'1 " 2i n( 2 n y 3 1 e x dx ! lim n l ' & n i!1 f!xi" (x xi ! 1 " 2i n x3 ! 1 " 6#n x2 ! 1 " 4#n x1 ! 1 " 2#n x0 ! 1 (x ! b % a n ! 2 n b ! 3 a ! 1 f!x" ! e x x3 1 e x dx N Because is positive, the integral in Example 3 represents the area shown in Figure 8. f !x" ! e x x y 0 1 3 10 y=´ FIGURE 8 N A computer algebra system is able to ﬁnd an explicit expression for this sum because it is a geometric series. The limit could be found using l’Hospital’s Rule. y 1 0 1 x y= 1-≈ or ≈+¥=1 œ„„„„„ FIGURE 9 SECTION 5.2 THE DEFINITE INTEGRAL \|\|\|\| 371 (b) The graph of is the line with slope 1 shown in Figure 10. We compute the integral as the difference of the areas of the two triangles: M THE MIDPOINT RULE We often choose the sample point to be the right endpoint of the th subinterval because it is convenient for computing the limit. But if the purpose is to ﬁnd an approximation to an integral, it is usually better to choose to be the midpoint of the interval, which we denote by . Any Riemann sum is an approximation to an integral, but if we use midpoints we get the following approximation. MIDPOINT RULE where and EXAMPLE 5 Use the Midpoint Rule with to approximate . SOLUTION The endpoints of the ﬁve subintervals are , , , , , and , so the midpoints are , , , , and . The width of the subintervals is , so the Midpoint Rule gives Since for , the integral represents an area, and the approxi-mation given by the Midpoint Rule is the sum of the areas of the rectangles shown in Figure 11. M 1 ! x ! 2 f!x" ! 1#x $ 0 / 0.691908 ! 1 5 ' 1 1.1 " 1 1.3 " 1 1.5 " 1 1.7 " 1 1.9( y 2 1 1 x dx / (x ) f!1.1" " f!1.3" " f!1.5" " f!1.7" " f!1.9" (x ! !2 % 1"#5 ! 1 5 1.9 1.7 1.5 1.3 1.1 2.0 1.8 1.6 1.4 1.2 1 y 2 1 1 x dx n ! 5 V xi ! 1 2!xi%1 " xi" ! midpoint of )xi%1, xi (x ! b % a n y b a f!x" dx / & n i!1 f!xi" (x ! (x ) f!x1" " & & & " f!xn" xi xi i xi x y 1 0 _1 3 y=x-1 A¡ (3, 2) A™ FIGURE 10 y 3 0 !x % 1" dx ! A1 % A2 ! 1 2!2 & 2" % 1 2!1 & 1" ! 1.5 y ! x % 1 FIGURE 11 0 x y 1 2 y= 1 x Module 5.2/7.7 shows how the Midpoint Rule estimates improve as increases. n TEC 372 \|\|\|\| CHAPTER 5 INTEGRALS At the moment we don’t know how accurate the approximation in Example 5 is, but in Section 7.7 we will learn a method for estimating the error involved in using the Midpoint Rule. At that time we will discuss other methods for approximating deﬁnite integrals. If we apply the Midpoint Rule to the integral in Example 2, we get the picture in Fig-ure 12. The approximation is much closer to the true value than the right endpoint approximation, , shown in Figure 7. PROPERTIES OF THE DEFINITE INTEGRAL When we deﬁned the deﬁnite integral , we implicitly assumed that . But the deﬁnition as a limit of Riemann sums makes sense even if . Notice that if we reverse a and b, then changes from to . Therefore If , then and so We now develop some basic properties of integrals that will help us to evaluate integrals in a simple manner. We assume that f and t are continuous functions. PROPERTIES OF THE INTEGRAL 1. , where c is any constant 2. 3. , where c is any constant 4. Property 1 says that the integral of a constant function is the constant times the length of the interval. If and , this is to be expected because is the area of the shaded rectangle in Figure 13. c!b % a" a ) b c $ 0 f!x" ! c y b a ) f!x" % t!x" dx ! y b a f!x" dx % y b a t!x" dx y b a cf!x" dx ! c y b a f!x" dx y b a ) f!x" " t!x" dx ! y b a f!x" dx " y b a t!x" dx y b a c dx ! c!b % a" y a a f!x" dx ! 0 (x ! 0 a ! b y a b f!x" dx ! %y b a f!x" dx !a % b"#n !b % a"#n (x a $ b a ) b xb a f!x" dx FIGURE 12 M¢¸Å_6.7563 0 y 3 x 5 y=˛-6x R40 / %6.3998 %6.75 M40 / %6.7563 In Visual 5.2 you can compare left, right, and midpoint approximations to the integral in Example 2 for different values of . n TEC FIGURE 13 j c dx=c(b-a) a b 0 y x a b c y=c area=c(b-a) SECTION 5.2 THE DEFINITE INTEGRAL \|\|\|\| 373 Property 2 says that the integral of a sum is the sum of the integrals. For positive func-tions it says that the area under is the area under plus the area under t. Figure 14 helps us understand why this is true: In view of how graphical addition works, the corre-sponding vertical line segments have equal height. In general, Property 2 follows from Theorem 4 and the fact that the limit of a sum is the sum of the limits: Property 3 can be proved in a similar manner and says that the integral of a constant times a function is the constant times the integral of the function. In other words, a con-stant (but only a constant) can be taken in front of an integral sign. Property 4 is proved by writing and using Properties 2 and 3 with . EXAMPLE 6 Use the properties of integrals to evaluate . SOLUTION Using Properties 2 and 3 of integrals, we have We know from Property 1 that and we found in Example 2 in Section 5.1 that . So M The next property tells us how to combine integrals of the same function over adjacent intervals: 5. This is not easy to prove in general, but for the case where and Property 5 can be seen from the geometric interpretation in Figure 15: The area under from a to c plus the area from c to b is equal to the total area from a to b. y ! f!x" a ! c ! b f!x" " 0 y c a f!x" dx # y b c f!x" dx ! y b a f!x" dx ! 4 # 3 $ 1 3 ! 5 y 1 0 !4 # 3x 2" dx ! y 1 0 4 dx # 3 y 1 0 x 2 dx y 1 0 x 2 dx ! 1 3 y 1 0 4 dx ! 4!1 % 0" ! 4 y 1 0 !4 # 3x 2" dx ! y 1 0 4 dx # y 1 0 3x 2 dx ! y 1 0 4 dx # 3 y 1 0 x 2 dx y 1 0 !4 # 3x 2" dx c ! %1 f % t ! f # !%t" ! y b a f!x" dx # y b a t!x" dx ! lim n l & # n i!1 f!xi" 'x # lim n l & # n i!1 t!xi" 'x ! lim n l & $# n i!1 f!xi" 'x # # n i!1 t!xi" 'x% y b a & f!x" # t!x"' dx ! lim n l & # n i!1 & f!xi" # t!xi"' 'x f f # t 374 \|\|\|\| CHAPTER 5 INTEGRALS y 0 x a b f g f+g FIGURE 14 j [ƒ+©] dx= j ƒ dx+j © dx a b a b a b N Property 3 seems intuitively reasonable because we know that multiplying a function by a positive number stretches or shrinks its graph vertically by a factor of . So it stretches or shrinks each approximating rectangle by a factor and therefore it has the effect of multiplying the area by . c c c c FIGURE 15 0 y x a b c y=ƒ EXAMPLE 7 If it is known that and , ﬁnd . SOLUTION By Property 5, we have so M Properties 1–5 are true whether , , or . The following properties, in which we compare sizes of functions and sizes of integrals, are true only if COMPARISON PROPERTIES OF THE INTEGRAL 6. If for , then . 7. If for , then . 8. If for , then If , then represents the area under the graph of , so the geometric interpretation of Property 6 is simply that areas are positive. But the property can be proved from the deﬁnition of an integral (Exercise 64). Property 7 says that a bigger func-tion has a bigger integral. It follows from Properties 6 and 4 because Property 8 is illustrated by Figure 16 for the case where . If is continuous we could take and to be the absolute minimum and maximum values of on the inter-val . In this case Property 8 says that the area under the graph of is greater than the area of the rectangle with height and less than the area of the rectangle with height . PROOF OF PROPERTY 8 Since , Property 7 gives Using Property 1 to evaluate the integrals on the left and right sides, we obtain M Property 8 is useful when all we want is a rough estimate of the size of an integral with-out going to the bother of using the Midpoint Rule. EXAMPLE 8 Use Property 8 to estimate . SOLUTION Because is a decreasing function on , its absolute maximum value is and its absolute minimum value is . Thus, by m ! f!1" ! e%1 M ! f!0" ! 1 &0, 1' f!x" ! e%x 2 y 1 0 e%x 2 dx m!b % a" ( y b a f!x" dx ( M!b % a" y b a m dx ( y b a f!x" dx ( y b a M dx m ( f!x" ( M M m f &a, b' f M m f f!x" " 0 f % t " 0. f xb a f!x" dx f!x" " 0 m!b % a" ( y b a f!x" dx ( M!b % a" a ( x ( b m ( f!x" ( M y b a f!x" dx " y b a t!x" dx a ( x ( b f!x" " t!x" y b a f!x" dx " 0 a ( x ( b f!x" " 0 a ( b. a ) b a ! b a ! b y 10 8 f!x" dx ! y 10 0 f!x" dx % y 8 0 f!x" dx ! 17 % 12 ! 5 y 8 0 f!x" dx # y 10 8 f!x" dx ! y 10 0 f!x" dx x10 8 f!x" dx x8 0 f!x" dx ! 12 x10 0 f!x" dx ! 17 V SECTION 5.2 THE DEFINITE INTEGRAL \|\|\|\| 375 0 y m M x a b y=ƒ FIGURE 16 Property 8, or Since , we can write M The result of Example 8 is illustrated in Figure 17. The integral is greater than the area of the lower rectangle and less than the area of the square. 0.367 ( y 1 0 e%x 2 dx ( 1 e%1 ( 0.3679 e%1 ( y 1 0 e%x 2dx ( 1 e%1!1 % 0" ( y 1 0 e%x 2 dx ( 1!1 % 0" 376 \|\|\|\| CHAPTER 5 INTEGRALS 6. The graph of is shown. Estimate with six sub-intervals using (a) right endpoints, (b) left endpoints, and (c) midpoints. 7. A table of values of an increasing function is shown. Use the table to ﬁnd lower and upper estimates for . 8. The table gives the values of a function obtained from an experiment. Use them to estimate using three equal subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints. If the function is known to be an increasing function, can you say whether your estimates are less than or greater than the exact value of the integral? x9 3 f !x" dx x25 0 f !x" dx f x y 0 g 1 1 x3 %3 t!x" dx t 1. Evaluate the Riemann sum for , , with six subintervals, taking the sample points to be left end-points. Explain, with the aid of a diagram, what the Riemann sum represents. 2. If , , evaluate the Riemann sum with , taking the sample points to be right endpoints. What does the Riemann sum represent? Illustrate with a diagram. 3. If , , ﬁnd the Riemann sum with correct to six decimal places, taking the sample points to be midpoints. What does the Riemann sum represent? Illustrate with a diagram. 4. (a) Find the Riemann sum for , , with six terms, taking the sample points to be right endpoints. (Give your answer correct to six decimal places.) Explain what the Riemann sum represents with the aid of a sketch. (b) Repeat part (a) with midpoints as sample points. The graph of a function is given. Estimate using four subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints. x y 0 f 1 1 x8 0 f !x" dx f 5. 0 ( x ( 3)2 f !x" ! sin x n ! 4 0 ( x ( 2 f !x" ! e x % 2 n ! 6 0 ( x ( 3 f !x" ! x 2 % 2x 2 ( x ( 14 f !x" ! 3 % 1 2x EXERCISES 5.2 FIGURE 17 y x 1 0 1 y=1 y=1/e y=e–x2 x 0 5 10 15 20 25 %42 %37 %25 %6 15 36 f !x" x 3 4 5 6 7 8 9 0.3 0.9 1.4 1.8 %0.6 %2.1 %3.4 f !x" 26. (a) Find an approximation to the integral using a Riemann sum with right endpoints and . (b) Draw a diagram like Figure 3 to illustrate the approxi-mation in part (a). (c) Use Theorem 4 to evaluate . (d) Interpret the integral in part (c) as a difference of areas and illustrate with a diagram like Figure 4. 27. Prove that . 28. Prove that . 29–30 Express the integral as a limit of Riemann sums. Do not evaluate the limit. 29. 30. 31–32 Express the integral as a limit of sums. Then evaluate, using a computer algebra system to ﬁnd both the sum and the limit. 31. 32. The graph of is shown. Evaluate each integral by inter-preting it in terms of areas. (a) (b) (c) (d) 34. The graph of t consists of two straight lines and a semicircle. Use it to evaluate each integral. (a) (b) (c) x y 0 2 4 7 4 y=© y 7 0 t!x" dx y 6 2 t!x" dx y 2 0 t!x" dx x y 0 2 4 6 8 2 y=ƒ y 9 0 f !x" dx y 7 5 f !x" dx y 5 0 f !x" dx y 2 0 f !x" dx f 33. y 10 2 x 6 dx y 0 sin 5x dx CAS y 10 1 !x % 4 ln x" dx y 6 2 x 1 # x 5 dx y b a x 2 dx ! b 3 % a 3 3 y b a x dx ! b 2 % a 2 2 x4 0 !x 2 % 3x" dx n ! 8 x4 0 !x 2 % 3x" dx 9–12 Use the Midpoint Rule with the given value of to approx-imate the integral. Round the answer to four decimal places. 10. , 11. 12. 13. If you have a CAS that evaluates midpoint approximations and graphs the corresponding rectangles (use middlesum and middlebox commands in Maple), check the answer to Exercise 11 and illustrate with a graph. Then repeat with and . 14. With a programmable calculator or computer (see the instruc-tions for Exercise 7 in Section 5.1), compute the left and right Riemann sums for the function on the interval with . Explain why these estimates show that Deduce that the approximation using the Midpoint Rule with in Exercise 11 is accurate to two decimal places. 15. Use a calculator or computer to make a table of values of right Riemann sums for the integral with , 10, 50, and 100. What value do these numbers appear to be approaching? 16. Use a calculator or computer to make a table of values of left and right Riemann sums and for the integral with , 10, 50, and 100. Between what two numbers must the value of the integral lie? Can you make a similar statement for the integral ? Explain. 17–20 Express the limit as a deﬁnite integral on the given interval. 17. 18. , 20. , 21–25 Use the form of the deﬁnition of the integral given in Theorem 4 to evaluate the integral. 21. 22. 24. 25. y 2 1 x 3 dx y 5 0 !1 # 2x 3" dx y 2 0 !2 % x 2" dx 23. y 4 1 !x 2 # 2x % 5" dx y 5 %1 !1 # 3x" dx &0, 2' lim n l & # n i!1 &4 % 3!xi "2 # 6!xi "5' 'x &1, 8] lim n l & # n i!1 s2xi # !xi "2 'x 19. lim n l & # n i!1 cos xi xi 'x, &, 2' lim n l & # n i!1 xi ln!1 # xi 2" 'x, &2, 6' x2 %1 e%x 2 dx n ! 5 x2 0 e%x 2 dx Rn Ln n ! 5 x 0 sin x dx Rn n ! 5 0.306 ! y 1 0 sin!x 2" dx ! 0.315 n ! 100 &0, 1' f !x" ! sin!x 2" n ! 20 n ! 10 CAS y 5 1 x 2e%x dx, n ! 4 y 1 0 sin!x 2" dx, n ! 5 n ! 4 y )2 0 cos4x dx y 10 2 sx 3 # 1 dx, n ! 4 9. n SECTION 5.2 THE DEFINITE INTEGRAL \|\|\|\| 377 378 \|\|\|\| CHAPTER 5 INTEGRALS 54. 55– 60 Use Property 8 to estimate the value of the integral. 55. 56. 57. 58. 59. 60. 61– 62 Use properties of integrals, together with Exercises 27 and 28, to prove the inequality. 61. 62. 63. Prove Property 3 of integrals. 64. Prove Property 6 of integrals. 65. If is continuous on , show that [Hint: .] 66. Use the result of Exercise 65 to show that 67. Let if is any rational number and if is any irrational number. Show that is not integrable on . 68. Let and if . Show that is not integrable on . [Hint: Show that the ﬁrst term in the Rie-mann sum, , can be made arbitrarily large.] 69– 70 Express the limit as a deﬁnite integral. 69. [Hint: Consider .] 70. 71. Find . Hint: Choose to be the geometric mean of and (that is, ) and use the identity 1 m!m # 1" ! 1 m % 1 m # 1 xi ! sxi%1xi xi xi%1 xi x2 1 x %2 dx lim n l & 1 n # n i!1 1 1 # !i)n"2 f !x" ! x 4 lim n l & # n i!1 i 4 n 5 f !xi " 'x &0, 1' f 0 ! x ( 1 f !x" ! 1)x f !0" ! 0 &0, 1' f x f !x" ! 1 x f !x" ! 0 y 2 0 f !x" sin 2x dx ( y 2 0 f !x" dx % f !x" ( f !x" ( f !x" y b a f !x" dx ( y b a f !x" dx &a, b' f y )2 0 x sin x dx ( 2 8 y 3 1 sx 4 # 1 dx " 26 3 y 2 !x % 2 sin x" dx y 2 0 xe%x dx y 2 0 !x 3 % 3x # 3" dx y )3 )4 tan x dx y 2 0 1 1 # x 2 dx y 4 1 sx dx s2 24 ( y )4 )6 cos x dx ( s3 24 35–40 Evaluate the integral by interpreting it in terms of areas. 35. 36. 38. 39. 40. 41. Evaluate . 42. Given that , what is ? 43. In Example 2 in Section 5.1 we showed that . Use this fact and the properties of integrals to evaluate . 44. Use the properties of integrals and the result of Example 3 to evaluate . 45. Use the result of Example 3 to evaluate . 46. Use the result of Exercise 27 and the fact that (from Exercise 25 in Section 5.1), together with the properties of integrals, to evaluate . Write as a single integral in the form : 48. If and , ﬁnd . If and , ﬁnd . 50. Find if 51. Suppose has absolute minimum value and absolute max-imum value . Between what two values must lie? Which property of integrals allows you to make your conclusion? 52–54 Use the properties of integrals to verify the inequality with-out evaluating the integrals. 52. 2 ( y 1 %1 s1 # x 2 dx ( 2s2 53. y 1 0 s1 # x 2 dx ( y 1 0 s1 # x dx x2 0 f !x" dx M m f f !x" !+ 3 for x ! 3 x for x " 3 x5 0 f !x" dx x9 0 &2 f !x" # 3t!x"' dx x9 0 t!x" dx ! 16 x9 0 f !x" dx ! 37 49. x4 1 f !x" dx x5 4 f !x" dx ! 3.6 x5 1 f !x" dx ! 12 y 2 %2 f !x" dx # y 5 2 f !x" dx % y %1 %2 f !x" dx xb a f !x" dx 47. x)2 0 !2 cos x % 5x" dx x)2 0 cos x dx ! 1 x3 1 e x#2 dx x3 1 !2e x % 1" dx x1 0 !5 % 6x 2" dx x1 0 x 2 dx ! 1 3 y 0 1 3usu 2 # 4 du y 1 0 3xsx 2 # 4 dx ! 5s5 % 8 y sin2x cos4x dx y 10 0 x % 5 dx y 2 %1 x dx y 3 %1 !3 % 2x" dx y 0 %3 (1 # s9 % x 2 ) dx 37. y 2 %2 s4 % x 2 dx y 3 0 ( 1 2 x % 1" dx 1. (a) Draw the line and use geometry to ﬁnd the area under this line, above the -axis, and between the vertical lines and . (b) If , let be the area of the region that lies under the line between and . Sketch this region and use geometry to ﬁnd an expression for . (c) Differentiate the area function . What do you notice? 2. (a) If , let represents the area of a region. Sketch that region. (b) Use the result of Exercise 28 in Section 5.2 to ﬁnd an expression for . (c) Find . What do you notice? (d) If and h is a small positive number, then represents the area of a region. Describe and sketch the region. (e) Draw a rectangle that approximates the region in part (d). By comparing the areas of these two regions, show that (f) Use part (e) to give an intuitive explanation for the result of part (c). ; 3. (a) Draw the graph of the function in the viewing rectangle by . (b) If we deﬁne a new function by then is the area under the graph of from 0 to [until becomes negative, at which point becomes a difference of areas]. Use part (a) to determine the value of at which starts to decrease. [Unlike the integral in Problem 2, it is impossible to evaluate the integral deﬁning to obtain an explicit expression for .] (c) Use the integration command on your calculator or computer to estimate , , , . . . , , . Then use these values to sketch a graph of . (d) Use your graph of from part (c) to sketch the graph of using the interpretation of as the slope of a tangent line. How does the graph of compare with the graph of ? 4. Suppose is a continuous function on the interval and we deﬁne a new function by the equation Based on your results in Problems 1–3, conjecture an expression for . t+!x" t!x" ! y x a f !t" dt t &a, b' f f t+ t+!x" t+ t t t!2" t!1.8" t!0.6" t!0.4" t!0.2" t!x" t t!x" x t!x" f !x" x f t!x" t!x" ! y x 0 cos!t 2" dt t &%1.25, 1.25' &0, 2' f !x" ! cos!x 2" A!x # h" % A!x" h ( 1 # x 2 A!x # h" % A!x" x " %1 A+!x" A!x" A!x" A!x" ! y x %1 !1 # t 2" dt x " %1 A!x" A!x" t ! x t ! 1 y ! 2t # 1 A!x" x ) 1 t ! 3 t ! 1 t y ! 2t # 1 AREA FUNCTIONS D I S C O V E R Y P R O J E C T SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS \|\|\|\| 379 THE FUNDAMENTAL THEOREM OF CALCULUS The Fundamental Theorem of Calculus is appropriately named because it establishes a connection between the two branches of calculus: differential calculus and integral calcu-lus. Differential calculus arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, the area problem. Newton’s mentor at Cambridge, 5.3 Isaac Barrow (1630–1677), discovered that these two problems are actually closely related. In fact, he realized that differentiation and integration are inverse processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the derivative and the integral. It was Newton and Leibniz who exploited this relationship and used it to develop calculus into a systematic mathematical method. In particular, they saw that the Fundamental Theorem enabled them to compute areas and integrals very easily without having to compute them as limits of sums as we did in Sections 5.1 and 5.2. The ﬁrst part of the Fundamental Theorem deals with functions deﬁned by an equation of the form where is a continuous function on and varies between and . Observe that depends only on , which appears as the variable upper limit in the integral. If is a ﬁxed number, then the integral is a deﬁnite number. If we then let vary, the number also varies and deﬁnes a function of denoted by . If happens to be a positive function, then can be interpreted as the area under the graph of from to , where can vary from to . (Think of as the “area so far” func-tion; see Figure 1.) EXAMPLE 1 If is the function whose graph is shown in Figure 2 and , ﬁnd the values of , , , , , and . Then sketch a rough graph of . SOLUTION First we notice that . From Figure 3 we see that is the area of a triangle: To ﬁnd we add to the area of a rectangle: We estimate that the area under from 2 to 3 is about 1.3, so FIGURE 3 g(1)=1 t 0 1 1 2 2 y t 0 1 1 2 2 2 y g(2)=3 t 0 1 1 2 2 2 y 3 g(3)Å4.3 t 0 1 1 2 2 4 2 y g(4)Å3 t 0 1 1 2 2 4 2 y g(5)Å1.7 t!3" ! t!2" # y 3 2 f!t" dt ( 3 # 1.3 ! 4.3 f t!2" ! y 2 0 f!t" dt ! y 1 0 f!t" dt # y 2 1 f!t" dt ! 1 # !1 ! 2" ! 3 t!1" t!2" t!1" ! y 1 0 f!t" dt ! 1 2 !1 ! 2" ! 1 t!1" t!0" ! x0 0 f!t" dt ! 0 t t!5" t!4" t!3" t!2" t!1" t!0" t!x" ! xx 0 f!t" dt f V t b a x x a f t!x" f t!x" x xx a f!t" dt x xx a f!t" dt x x t b a x &a, b' f t!x" ! y x a f!t" dt 1 380 \|\|\|\| CHAPTER 5 INTEGRALS 0 y t a b x area=© y=f(t) FIGURE 1 t 0 1 1 2 2 4 2 y y=f(t) FIGURE 2 For , is negative and so we start subtracting areas: We use these values to sketch the graph of in Figure 4. Notice that, because is positive for , we keep adding area for and so is increasing up to , where it attains a maximum value. For , decreases because is negative. M If we take and , then, using Exercise 27 in Section 5.2, we have Notice that , that is, . In other words, if is deﬁned as the integral of by Equation 1, then turns out to be an antiderivative of , at least in this case. And if we sketch the derivative of the function shown in Figure 4 by estimating slopes of tangents, we get a graph like that of in Figure 2. So we suspect that in Example 1 too. To see why this might be generally true we consider any continuous function with . Then can be interpreted as the area under the graph of from to , as in Figure 1. In order to compute from the deﬁnition of derivative we ﬁrst observe that, for is obtained by subtracting areas, so it is the area under the graph of from to (the gold area in Figure 5). For small you can see from the ﬁgure that this area is approximately equal to the area of the rectangle with height and width : so Intuitively, we therefore expect that The fact that this is true, even when is not necessarily positive, is the ﬁrst part of the Fun-damental Theorem of Calculus. THE FUNDAMENTAL THEOREM OF CALCULUS, PART 1 If is continuous on , then the function deﬁned by is continuous on and differentiable on , and . t+!x" ! f!x" !a, b" &a, b' a ( x ( b t!x" ! y x a f!t" dt t &a, b' f f t+!x" ! lim h l 0 t!x # h" % t!x" h ! f!x" t!x # h" % t!x" h ( f!x" t!x # h" % t!x" ( hf!x" h f!x" h x # h x f h ) 0, t!x # h" % t!x" t+!x" x a f t!x" ! xx a f!t" dt f!x" " 0 f t+! f f t f t f t t+ ! f t+!x" ! x t!x" ! y x 0 t dt ! x 2 2 a ! 0 f!t" ! t f!t" t x ) 3 x ! 3 t t ! 3 t ! 3 f!t" t t!5" ! t!4" # y 5 4 f!t" dt ( 3 # !%1.3" ! 1.7 t!4" ! t!3" # y 4 3 f!t" dt ( 4.3 # !%1.3" ! 3.0 f!t" t ) 3 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS \|\|\|\| 381 FIGURE 4 ©=j f(t) dt a x x 0 1 1 2 4 2 y 3 4 5 3 g y 0 t a b x x+h h ƒ FIGURE 5 N We abbreviate the name of this theorem as FTC1. In words, it says that the derivative of a deﬁnite integral with respect to its upper limit is the integrand evaluated at the upper limit. PROOF If and are in , then (by Property 5) and so, for , For now let us assume that . Since is continuous on , the Extreme Value Theorem says that there are numbers and in such that and , where and are the absolute minimum and maximum values of on . (See Figure 6.) By Property 8 of integrals, we have that is, Since , we can divide this inequality by : Now we use Equation 2 to replace the middle part of this inequality: Inequality 3 can be proved in a similar manner for the case . (See Exercise 67.) Now we let . Then and , since and lie between and . Therefore and because is continuous at . We conclude, from (3) and the Squeeze Theorem, that x f lim h l 0 f!v" ! lim v lx f!v" ! f!x" lim h l 0 f!u" ! lim u lx f!u" ! f!x" x # h x v u v l x u l x h l 0 h ! 0 f!u" ( t!x # h" % t!x" h ( f!v" 3 f!u" ( 1 h y x#h x f!t" dt ( f!v" h h ) 0 f!u"h ( y x#h x f!t" dt ( f!v"h mh ( y x#h x f!t" dt ( Mh &x, x # h' f M m f!v" ! M f!u" ! m &x, x # h' v u &x, x # h' f h ) 0 t!x # h" % t!x" h ! 1 h y x#h x f!t" dt 2 h " 0 ! y x#h x f!t" dt !,y x a f!t" dt # y x#h x f!t" dt- % y x a f!t" dt t!x # h" % t!x" ! y x#h a f!t" dt % y x a f!t" dt !a, b" x # h x 382 \|\|\|\| CHAPTER 5 INTEGRALS FIGURE 6 0 y x x u √=x+h y=ƒ m M Module 5.3 provides visual evidence for FTC1. TEC If or , then Equation 4 can be interpreted as a one-sided limit. Then Theo-rem 2.8.4 (modiﬁed for one-sided limits) shows that is continuous on . M Using Leibniz notation for derivatives, we can write FTC1 as when is continuous. Roughly speaking, Equation 5 says that if we ﬁrst integrate and then differentiate the result, we get back to the original function . EXAMPLE 2 Find the derivative of the function . SOLUTION Since is continuous, Part 1 of the Fundamental Theorem of Calculus gives M EXAMPLE 3 Although a formula of the form may seem like a strange way of deﬁning a function, books on physics, chemistry, and statistics are full of such functions. For instance, the Fresnel function is named after the French physicist Augustin Fresnel (1788–1827), who is famous for his works in optics. This function ﬁrst appeared in Fresnel’s theory of the diffraction of light waves, but more recently it has been applied to the design of highways. Part 1 of the Fundamental Theorem tells us how to differentiate the Fresnel function: This means that we can apply all the methods of differential calculus to analyze (see Exercise 61). Figure 7 shows the graphs of and the Fresnel function . A computer was used to graph by computing the value of this integral for many values of . It does indeed look as if is the area under the graph of from to [until , when becomes a difference of areas]. Figure 8 shows a larger part of the graph of . If we now start with the graph of in Figure 7 and think about what its derivative should look like, it seems reasonable that . [For instance, is increasing when and decreasing when .] So this gives a visual conﬁrmation of Part 1 of the Fundamental Theorem of Calculus. M f!x" ! 0 f!x" ) 0 S S+!x" ! f!x" S S S!x" x ( 1.4 x 0 f S!x" x S S!x" ! xx 0 f!t" dt f!x" ! sin!x 2)2" S S+!x" ! sin!x 2)2" S!x" ! y x 0 sin!t 2)2" dt t!x" ! xx a f!t" dt t+!x" ! s1 # x 2 f!t" ! s1 # t 2 t!x" ! y x 0 s1 # t 2 dt V f f f d dx y x a f!t" dt ! f!x" 5 &a, b' t b x ! a t+!x" ! lim h l 0 t!x # h" % t!x" h ! f!x" 4 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS \|\|\|\| 383 FIGURE 7 ƒ=sin(π≈/2) S(x)=j sin(πt@/2) dt 0 x 1 0 x 1 y f S FIGURE 8 The Fresnel function S(x)=j sin(πt@/2) dt 0 x y 1 0.5 x EXAMPLE 4 Find . SOLUTION Here we have to be careful to use the Chain Rule in conjunction with FTC1. Let . Then (by the Chain Rule) (by FTC1) M In Section 5.2 we computed integrals from the deﬁnition as a limit of Riemann sums and we saw that this procedure is sometimes long and difﬁcult. The second part of the Fundamental Theorem of Calculus, which follows easily from the ﬁrst part, provides us with a much simpler method for the evaluation of integrals. THE FUNDAMENTAL THEOREM OF CALCULUS, PART 2 If is continuous on , then where is any antiderivative of , that is, a function such that . PROOF Let . We know from Part 1 that ; that is, is an anti-derivative of . If is any other antiderivative of on , then we know from Corol-lary 4.2.7 that and differ by a constant: for . But both and are continuous on and so, by taking limits of both sides of Equation 6 (as and ), we see that it also holds when and . If we put in the formula for , we get So, using Equation 6 with and , we have M ! y b a f!t" dt ! t!b" ! t!a" ! t!b" F!b" ! F!a" ! #t!b" " C$ ! #t!a" " C$ x ! a x ! b t!a" ! y a a f!t" dt ! 0 t!x" x ! a x ! b x ! a x l b! x l a" #a, b$ t F a # x # b F!x" ! t!x" " C 6 t F #a, b$ f F f t t$!x" ! f!x" t!x" ! xx a f!t" dt F$ ! f f F y b a f!x" dx ! F!b" ! F!a" #a, b$ f ! sec!x 4" ! 4x 3 ! sec u du dx ! d du %y u 1 sec t dt& du dx d dx y x 4 1 sec t dt ! d dx y u 1 sec t dt u ! x 4 d dx y x 4 1 sec t dt 384 \|\|\|\| CHAPTER 5 INTEGRALS N We abbreviate this theorem as FTC2. Part 2 of the Fundamental Theorem states that if we know an antiderivative of , then we can evaluate simply by subtracting the values of at the endpoints of the interval . It’s very surprising that , which was deﬁned by a complicated pro-cedure involving all of the values of for , can be found by knowing the val-ues of at only two points, and . Although the theorem may be surprising at ﬁrst glance, it becomes plausible if we inter-pret it in physical terms. If is the velocity of an object and is its position at time t, then , so s is an antiderivative of . In Section 5.1 we considered an object that always moves in the positive direction and made the guess that the area under the velocity curve is equal to the distance traveled. In symbols: That is exactly what FTC2 says in this context. EXAMPLE 5 Evaluate the integral . SOLUTION The function is continuous everywhere and we know that an anti-derivative is , so Part 2 of the Fundamental Theorem gives Notice that FTC2 says we can use any antiderivative F of f. So we may as well use the simplest one, namely , instead of or . M We often use the notation So the equation of FTC2 can be written as Other common notations are and . EXAMPLE 6 Find the area under the parabola from 0 to 1. SOLUTION An antiderivative of is . The required area is found using Part 2 of the Fundamental Theorem: M If you compare the calculation in Example 6 with the one in Example 2 in Section 5.1, you will see that the Fundamental Theorem gives a much shorter method. A ! y 1 0 x 2 dx ! x 3 3' 0 1 ! 13 3 ! 03 3 ! 1 3 A F!x" ! 1 3 x 3 f!x" ! x 2 y ! x 2 #F!x"$ a b F!x"( a b F$! f where y b a f!x" dx ! F!x"] a b F!x"] a b ! F!b" ! F!a" e x " C e x " 7 F!x" ! e x y 3 1 e x dx ! F!3" ! F!1" ! e 3 ! e F!x" ! e x f!x" ! e x y 3 1 e x dx V y b a v!t" dt ! s!b" ! s!a" v v!t" ! s$!t" s!t" v!t" b a F!x" a % x % b f!x" xb a f!x" dx #a, b$ F xb a f!x" dx f F SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS \|\|\|\| 385 N Compare the calculation in Example 5 with the much harder one in Example 3 in Section 5.2. N In applying the Fundamental Theorem we use a particular antiderivative of . It is not neces-sary to use the most general antiderivative. f F EXAMPLE 7 Evaluate . SOLUTION The given integral is an abbreviation for An antiderivative of is and, because , we can write . So M EXAMPLE 8 Find the area under the cosine curve from to , where . SOLUTION Since an antiderivative of is , we have In particular, taking , we have proved that the area under the cosine curve from 0 to is . (See Figure 9.) M When the French mathematician Gilles de Roberval ﬁrst found the area under the sine and cosine curves in 1635, this was a very challenging problem that required a great deal of ingenuity. If we didn’t have the beneﬁt of the Fundamental Theorem, we would have to compute a difﬁcult limit of sums using obscure trigonometric identities (or a computer algebra system as in Exercise 25 in Section 5.1). It was even more difﬁcult for Roberval because the apparatus of limits had not been invented in 1635. But in the 1660s and 1670s, when the Fundamental Theorem was discovered by Barrow and exploited by Newton and Leibniz, such problems became very easy, as you can see from Example 8. EXAMPLE 9 What is wrong with the following calculation? \| SOLUTION To start, we notice that this calculation must be wrong because the answer is negative but and Property 6 of integrals says that when . The Fundamental Theorem of Calculus applies to continuous functions. It can’t be applied here because is not continuous on . In fact, f has an inﬁ-nite discontinuity at , so M does not exist y 3 !1 1 x 2 dx x ! 0 #!1, 3$ f!x" ! 1)x 2 f & 0 xb a f!x" dx & 0 f!x" ! 1)x 2 & 0 y 3 !1 1 x 2 dx ! x!1 !1' 3 !1 ! ! 1 3 ! 1 ! ! 4 3 sin!')2" ! 1 ')2 b ! ')2 A ! y b 0 cos x dx ! sin x] 0 b ! sin b ! sin 0 ! sin b F!x" ! sin x f!x" ! cos x 0 % b % ')2 b 0 ! ln 6 3 ! ln 2 y 6 3 1 x dx ! ln x]3 6 ! ln 6 ! ln 3 F!x" ! ln x 3 % x % 6 F!x" ! ln (x( f!x" ! 1)x y 6 3 1 x dx y 6 3 dx x 386 \|\|\|\| CHAPTER 5 INTEGRALS FIGURE 9 y 0 1 x y=cos x area=1 π 2 DIFFERENTIATION AND INTEGRATION AS INVERSE PROCESSES We end this section by bringing together the two parts of the Fundamental Theorem. THE FUNDAMENTAL THEOREM OF CALCULUS Suppose is continuous on . 1. If , then . 2. , where is any antiderivative of , that is, . We noted that Part 1 can be rewritten as which says that if is integrated and then the result is differentiated, we arrive back at the original function . Since , Part 2 can be rewritten as This version says that if we take a function , ﬁrst differentiate it, and then integrate the result, we arrive back at the original function , but in the form . Taken together, the two parts of the Fundamental Theorem of Calculus say that differentiation and integration are inverse processes. Each undoes what the other does. The Fundamental Theorem of Calculus is unquestionably the most important theorem in calculus and, indeed, it ranks as one of the great accomplishments of the human mind. Before it was discovered, from the time of Eudoxus and Archimedes to the time of Galileo and Fermat, problems of ﬁnding areas, volumes, and lengths of curves were so difﬁcult that only a genius could meet the challenge. But now, armed with the systematic method that Newton and Leibniz fashioned out of the Fundamental Theorem, we will see in the chapters to come that these challenging problems are accessible to all of us. F!b" ! F!a" F F y b a F$!x" dx ! F!b" ! F!a" F$!x" ! f!x" f f d dx y x a f!t" dt ! f!x" F$! f f F y b a f!x" dx ! F!b" ! F!a" t$!x" ! f!x" t!x" ! y x a f!t" dt #a, b$ f SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS \|\|\|\| 387 (a) Evaluate for and 6. (b) Estimate . (c) Where does have a maximum value? Where does it have a minimum value? (d) Sketch a rough graph of . Let , where is the function whose graph is shown. (a) Evaluate , , , , and . (b) On what interval is increasing? t t!6" t!3" t!2" t!1" t!0" f t!x" ! xx 0 f !t" dt 3. t t t!7" x ! 0, 1, 2, 3, 4, 5, t!x" 1. Explain exactly what is meant by the statement that “differenti-ation and integration are inverse processes.” 2. Let , where is the function whose graph is shown. t y 0 1 1 4 6 f t!x" ! xx 0 f!t" dt EXERCISES 5.3 388 \|\|\|\| CHAPTER 5 INTEGRALS 15. 16. 18. 19–42 Evaluate the integral. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. where 42. where ; 43–46 What is wrong with the equation? 43. 44. 45. 46. y 0 sec2x dx tan x]0 0 y 3 sec tan d sec ] 3 3 y 2 1 4 x 3 dx 2 x 2 2 1 3 2 y 1 2 x4 dx x3 3 1 2 3 8 f x 2 4 x 2 if 2 x 0 if 0 x 2 y 2 2 f x dx f x sin x cos x if 0 x 2 if 2 x y 0 f x dx y 2 1 4 u2 u3 du y 1 1 eu1 du y 1 0 4 t 2 1 dt y s32 12 6 s1 t 2 dt y 1 0 10 x dx y 9 1 1 2x dx y 1 0 cosh t dt y 2 1 1 2y2 dy y 4 0 sec tan d y 4 0 sec2t dt y 2 0 y 12y 1 dy y 9 1 x 1 sx dx y 1 0 (3 xsx ) dx y 2 0 x2 x5 dx y 2 cos d y 2 1 3 t 4 dt y 8 1 s 3 x dx y 1 0 x45 dx y 1 0 (1 1 2u 4 2 5u9) du y 4 1 5 2t 3t 2 dt y 5 2 6 dx y 2 1 x 3 2x dx y y 0 ex sin3t dt y y 1 13x u 3 1 u 2 du 17. y y cos x 1 1 v 210 dv y y tan x 0 st st dt (c) Where does have a maximum value? (d) Sketch a rough graph of . 4. Let , where is the function whose graph is shown. (a) Evaluate and . (b) Estimate , and . (c) On what interval is increasing? (d) Where does have a maximum value? (e) Sketch a rough graph of . (f) Use the graph in part (e) to sketch the graph of . Compare with the graph of . 5–6 Sketch the area represented by . Then ﬁnd in two ways: (a) by using Part 1 of the Fundamental Theorem and (b) by evaluating the integral using Part 2 and then differentiating. 5. 6. 7–18 Use Part 1 of the Fundamental Theorem of Calculus to ﬁnd the derivative of the function. 7. 8. 10. 11. 12. 14. hx y x 2 0 s1 r 3 dr hx y 1x 2 arctan t dt 13. Gx y 1 x cos st dt Hint: y x s1 sec t dt y x s1 sec t dt Fx y x s1 sec t dt tr y r 0 sx 2 4 dx ty y y 2 t 2 sin t dt 9. tx y x 3 et 2t dt tx y x 1 1 t 3 1 dt tx y x 0 (1 st ) dt tx y x 1 t 2 dt tx tx 1 t y 1 0 f f tx t t t t0 t2, t1 t3 t3 f tx xx 3 f t dt 1 5 t y 1 0 f t t The sine integral function is important in electrical engineering. [The integrand is not deﬁned when , but we know that its limit is 1 when . So we deﬁne and this makes a continuous function everywhere.] (a) Draw the graph of . (b) At what values of does this function have local maxi-mum values? (c) Find the coordinates of the ﬁrst inﬂection point to the right of the origin. (d) Does this function have horizontal asymptotes? (e) Solve the following equation correct to one decimal place: 63–64 Let , where is the function whose graph is shown. (a) At what values of do the local maximum and minimum values of occur? (b) Where does attain its absolute maximum value? (c) On what intervals is concave downward? (d) Sketch the graph of . 64. 65– 66 Evaluate the limit by ﬁrst recognizing the sum as a Rie-mann sum for a function deﬁned on . 65. 66. lim n l ) 1 n %, 1 n ", 2 n ", 3 n " ", n n & lim n l ) -n i!1 i 3 n 4 #0, 1$ y 1 t 0 7 3 5 9 f _0.2 0.2 0.4 y 2 t 0 _1 _2 1 2 4 6 8 3 f 63. t t t t x f t!x" ! xx 0 f !t" dt y x 0 sin t t dt ! 1 x Si f f !0" ! 1 t l 0 t ! 0 f !t" ! !sin t")t Si!x" ! y x 0 sin t t dt 62. CAS ; 47– 50 Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then ﬁnd the exact area. 47. , 48. , 49. , 50. , 51–52 Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch. 52. 53–56 Find the derivative of the function. 54. 55. 56. 57. If , where , ﬁnd . 58. Find the interval on which the curve is concave upward. 59. If , is continuous, and , what is the value of ? 60. The error function is used in probability, statistics, and engineering. (a) Show that . (b) Show that the function satisﬁes the differ-ential equation . 61. The Fresnel function was deﬁned in Example 3 and graphed in Figures 7 and 8. (a) At what values of does this function have local maxi-mum values? (b) On what intervals is the function concave upward? (c) Use a graph to solve the following equation correct to two decimal places: y x 0 sin!'t 2)2" dt ! 0.2 CAS x S y$ ! 2xy " 2)s' y ! e x2erf!x" xb a e!t2 dt ! 1 2 s' #erf!b" ! erf!a"$ erf!x" ! 2 s' y x 0 e!t 2 dt f !4" x4 1 f $!x" dx ! 17 f $ f !1" ! 12 y ! y x 0 1 1 " t " t 2 dt F+!2" f !t" ! y t 2 1 s1 " u 4 u du F!x" ! y x 1 f !t" dt y ! y 5x cos x cos!u 2" du y ! y x 3 sx st sin t dt t!x" ! y x 2 tan x 1 s2 " t 4 dt + Hint: y 3x 2x f !u" du ! y 0 2x f !u" du " y 3x 0 f !u" du' t!x" ! y 3x 2x u 2 ! 1 u 2 " 1 du 53. y 5')2 ')4 sin x dx y 2 !1 x 3 dx 51. 0 % x % ')3 y ! sec2x 0 % x % ' y ! sin x 1 % x % 6 y ! x !4 0 % x % 27 y ! s 3 x SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS \|\|\|\| 389 75. A manufacturing company owns a major piece of equipment that depreciates at the (continuous) rate , where is the time measured in months since its last overhaul. Because a ﬁxed cost is incurred each time the machine is overhauled, the company wants to determine the optimal time (in months) between overhauls. (a) Explain why represents the loss in value of the machine over the period of time since the last overhaul. (b) Let be given by What does represent and why would the company want to minimize ? (c) Show that has a minimum value at the numbers where . 76. A high-tech company purchases a new computing system whose initial value is . The system will depreciate at the rate and will accumulate maintenance costs at the rate , where is the time measured in months. The company wants to determine the optimal time to replace the system. (a) Let Show that the critical numbers of occur at the numbers where . (b) Suppose that and Determine the length of time for the total depreciation to equal the initial value . (c) Determine the absolute minimum of on . (d) Sketch the graphs of and in the same coordinate system, and verify the result in part (a) in this case. f " t C !0, T $ C V D!t" ! xt 0 f !s" ds T t , 0 t!t" ! Vt 2 12,900 f !t" ! 0 V 15 ! V 450 t if if 0 # t % 30 t , 30 C!t" ! f !t" " t!t" t C C!t" ! 1 t y t 0 # f !s" " t!s"$ ds t t ! t!t" f ! f !t" V C!T" ! f !T" t ! T C C C C!t" ! 1 t +A " y t 0 f !s" ds' C ! C!t" t xt 0 f !s" ds T A t f ! f !t" 67. Justify (3) for the case . 68. If is continuous and and are differentiable functions, ﬁnd a formula for 69. (a) Show that for . (b) Show that . 70. (a) Show that for . (b) Deduce that . 71. Show that by comparing the integrand to a simpler function. Let and (a) Find an expression for similar to the one for . (b) Sketch the graphs of and . (c) Where is differentiable? Where is differentiable? Find a function and a number such that for all The area labeled is three times the area labeled . Express in terms of . 0 a A y=´ 0 b B y=´ y x y x a b A B 74. x , 0 6 " y x a f !t" t 2 dt ! 2sx a f 73. t f t f f !x" t!x" t!x" ! y x 0 f !t" dt 0 x 2 ! x 0 if x # 0 if 0 % x % 1 if 1 # x % 2 if x , 2 f !x" ! 72. 0 % y 10 5 x 2 x 4 " x 2 " 1 dx % 0.1 x ')6 0 cos!x 2" dx & 1 2 0 % x % 1 cos!x 2" & cos x 1 % x1 0 s1 " x 3 dx % 1.25 x & 0 1 % s1 " x 3 % 1 " x 3 d dx y h!x" t!x" f !t" dt h t f h # 0 390 \|\|\|\| CHAPTER 5 INTEGRALS INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM We saw in Section 5.3 that the second part of the Fundamental Theorem of Calculus pro-vides a very powerful method for evaluating the deﬁnite integral of a function, assuming that we can ﬁnd an antiderivative of the function. In this section we introduce a notation for antiderivatives, review the formulas for antiderivatives, and use them to evaluate deﬁ-nite integrals. We also reformulate FTC2 in a way that makes it easier to apply to science and engineering problems. INDEFINITE INTEGRALS Both parts of the Fundamental Theorem establish connections between antiderivatives and deﬁnite integrals. Part 1 says that if is continuous, then is an antiderivative of . Part 2 says that can be found by evaluating , where F is an antideriv-ative of f. We need a convenient notation for antiderivatives that makes them easy to work with. Because of the relation given by the Fundamental Theorem between antiderivatives and integrals, the notation is traditionally used for an antiderivative of and is called an indeﬁnite integral. Thus means For example, we can write So we can regard an indeﬁnite integral as representing an entire family of functions (one antiderivative for each value of the constant C). \| You should distinguish carefully between deﬁnite and indeﬁnite integrals. A deﬁnite integral is a number, whereas an indeﬁnite integral is a function (or family of functions). The connection between them is given by Part 2 of the Fundamental Theorem. If is continuous on , then The effectiveness of the Fundamental Theorem depends on having a supply of anti-derivatives of functions. We therefore restate the Table of Antidifferentiation Formulas from Section 4.9, together with a few others, in the notation of indeﬁnite integrals. Any formula can be veriﬁed by differentiating the function on the right side and obtaining the integrand. For instance because d dx !tan x " C" ! sec2x y sec2x dx ! tan x " C y b a f!x" dx ! y f!x" dx]a b #a, b$ f x f!x" dx xb a f!x" dx y x 2 dx ! x 3 3 " C because d dx % x 3 3 " C& ! x 2 F$!x" ! f!x" y f!x" dx ! F!x" f x f!x" dx F!b" ! F!a" xb a f!x" dx f xx a f!t" dt f 5.4 SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM \|\|\|\| 391 TABLE OF INDEFINITE INTEGRALS Recall from Theorem 4.9.1 that the most general antiderivative on a given interval is obtained by adding a constant to a particular antiderivative. We adopt the convention that when a formula for a general indeﬁnite integral is given, it is valid only on an inter-val. Thus we write with the understanding that it is valid on the interval or on the interval . This is true despite the fact that the general antiderivative of the function , , is EXAMPLE 1 Find the general indeﬁnite integral SOLUTION Using our convention and Table 1, we have You should check this answer by differentiating it. M ! 2x 5 ! 2 tan x " C ! 10 x 5 5 ! 2 tan x " C y !10x 4 ! 2 sec2x" dx ! 10 y x 4 dx ! 2 y sec2x dx y !10x 4 ! 2 sec2x" dx ! 1 x " C2 if x , 0 F!x" ! ! 1 x " C1 if x # 0 x " 0 f!x" ! 1)x 2 !!), 0" !0, )" y 1 x 2 dx ! ! 1 x " C y cosh x dx ! sinh x " C y sinh x dx ! cosh x " C y 1 s1 ! x 2 dx ! sin!1x " C y 1 x 2 " 1 dx ! tan!1x " C y csc x cot x dx ! !csc x " C y sec x tan x dx ! sec x " C y csc2x dx ! !cot x " C y sec2x dx ! tan x " C y cos x dx ! sin x " C y sin x dx ! !cos x " C y a x dx ! a x ln a " C y e x dx ! e x " C y 1 x dx ! ln(x( " C !n " !1" y x n dx ! x n"1 n " 1 " C y k dx ! kx " C y # f!x" " t!x"$ dx ! y f!x" dx " y t!x" dx y cf!x" dx ! c y f!x" dx 1 392 \|\|\|\| CHAPTER 5 INTEGRALS N The indeﬁnite integral in Example 1 is graphed in Figure 1 for several values of . The value of is the -intercept. y C C 4 _4 _1.5 1.5 FIGURE 1 EXAMPLE 2 Evaluate . SOLUTION This indeﬁnite integral isn’t immediately apparent in Table 1, so we use trigo-nometric identities to rewrite the function before integrating: M EXAMPLE 3 Evaluate . SOLUTION Using FTC2 and Table 1, we have Compare this calculation with Example 2(b) in Section 5.2. M EXAMPLE 4 Find and interpret the result in terms of areas. SOLUTION The Fundamental Theorem gives This is the exact value of the integral. If a decimal approximation is desired, we can use a calculator to approximate . Doing so, we get M EXAMPLE 5 Evaluate . SOLUTION First we need to write the integrand in a simpler form by carrying out the division: M ! 18 " 18 " 1 9 ! 2 ! 2 3 ! 1 ! 32 4 9 ! (2 ! 9 " 2 3 ! 9 3)2 " 1 9) ! (2 ! 1 " 2 3 ! 13)2 " 1 1) ! 2t " 2 3t 3)2 " 1 t' 1 9 ! 2t " t 3)2 3 2 ! t!1 !1' 1 9 y 9 1 2t 2 " t 2st ! 1 t 2 dt ! y 9 1 !2 " t 1)2 ! t!2" dt y 9 1 2t 2 " t 2st ! 1 t 2 dt y 2 0 %2x 3 ! 6x " 3 x 2 " 1& dx . !0.67855 tan!1 2 ! !4 " 3 tan!1 2 ! 1 2!24" ! 3!22" " 3 tan!1 2 ! 0 ! 1 2 x 4 ! 3x 2 " 3 tan!1x] 2 0 y 2 0 %2x 3 ! 6x " 3 x 2 " 1& dx ! 2 x 4 4 ! 6 x 2 2 " 3 tan!1x' 0 2 y 2 0 %2x 3 ! 6x " 3 x 2 " 1& dx V ! 81 4 ! 27 ! 0 " 0 ! !6.75 ! ( 1 4 ! 34 ! 3 ! 32) ! ( 1 4 ! 04 ! 3 ! 02) y 3 0 !x 3 ! 6x" dx ! x 4 4 ! 6 x 2 2' 0 3 y 3 0 !x 3 ! 6x" dx ! y csc ( cot ( d( ! !csc ( " C y cos ( sin2( d( ! y % 1 sin (&% cos ( sin (& d( y cos ( sin2( d( V SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM \|\|\|\| 393 N Figure 2 shows the graph of the integrand in Example 4. We know from Section 5.2 that the value of the integral can be interpreted as the sum of the areas labeled with a plus sign minus the area labeled with a minus sign. 0 y 2 x 3 FIGURE 2 APPLICATIONS Part 2 of the Fundamental Theorem says that if f is continuous on , then where F is any antiderivative of f. This means that , so the equation can be rewrit-ten as We know that represents the rate of change of with respect to x and is the change in y when x changes from a to b. [Note that could, for instance, increase, then decrease, then increase again. Although might change in both directions, represents the net change in .] So we can reformulate FTC2 in words as follows. THE NET CHANGE THEOREM The integral of a rate of change is the net change: This principle can be applied to all of the rates of change in the natural and social sci-ences that we discussed in Section 3.7. Here are a few instances of this idea: N If is the volume of water in a reservoir at time t, then its derivative is the rate at which water ﬂows into the reservoir at time t. So is the change in the amount of water in the reservoir between time and time . N If is the concentration of the product of a chemical reaction at time t, then the rate of reaction is the derivative . So is the change in the concentration of C from time to time . N If the mass of a rod measured from the left end to a point x is , then the linear density is . So is the mass of the segment of the rod that lies between and . N If the rate of growth of a population is , then is the net change in population during the time period from to . (The popu-lation increases when births happen and decreases when deaths occur. The net change takes into account both births and deaths.) t2 t1 y t2 t1 dn dt dt ! n!t2" ! n!t1" dn#dt x ! b x ! a y b a "!x" dx ! m!b" ! m!a" "!x" ! m#!x" m!x" t2 t1 y t2 t1 d$C% dt dt ! $C%!t2" ! $C%!t1" d$C%#dt $C%!t" t2 t1 y t2 t1 V#!t" dt ! V!t2" ! V!t1" V#!t" V!t" y b a F#!x" dx ! F!b" ! F!a" y F!b" ! F!a" y y F!b" ! F!a" y ! F!x" F#!x" y b a F#!x" dx ! F!b" ! F!a" F#! f y b a f!x" dx ! F!b" ! F!a" $a, b% 394 \|\|\|\| CHAPTER 5 INTEGRALS N If is the cost of producing x units of a commodity, then the marginal cost is the derivative . So is the increase in cost when production is increased from units to units. N If an object moves along a straight line with position function , then its velocity is , so is the net change of position, or displacement, of the particle during the time period from to . In Section 5.1 we guessed that this was true for the case where the object moves in the positive direction, but now we have proved that it is always true. N If we want to calculate the distance the object travels during that time interval, we have to consider the intervals when (the particle moves to the right) and also the intervals when (the particle moves to the left). In both cases the distance is computed by integrating , the speed. Therefore Figure 3 shows how both displacement and distance traveled can be interpreted in terms of areas under a velocity curve. N The acceleration of the object is , so is the change in velocity from time to time . EXAMPLE 6 A particle moves along a line so that its velocity at time is (measured in meters per second). (a) Find the displacement of the particle during the time period . (b) Find the distance traveled during this time period. SOLUTION (a) By Equation 2, the displacement is This means that the particle moved 4.5 m toward the left. !& t 3 3 ! t 2 2 ! 6t' 1 4 ! ! 9 2 s!4" ! s!1" ! y 4 1 v!t" dt ! y 4 1 !t 2 ! t ! 6" dt 1 $ t $ 4 v!t" ! t 2 ! t ! 6 t V t2 t1 y t2 t1 a!t" dt ! v!t2" ! v!t1" a!t" ! v#!t" FIGURE 3 t¡ t™ distance=j \|√(t)\| dt=A¡+A™+A£ t¡ t™ displacement=j √(t) dt=A¡-A™+A£ √ 0 t A¡ A™ A£ t¡ t™ √(t) y t2 t1 ( v!t"(dt ! total distance traveled 3 ( v!t"( v!t" $ 0 v!t" % 0 t2 t1 y t2 t1 v!t" dt ! s!t2" ! s!t1" 2 v!t" ! s#!t" s!t" x2 x1 y x2 x1 C#!x" dx ! C!x2" ! C!x1" C#!x" C!x" SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM \|\|\|\| 395 (b) Note that and so on the interval and on . Thus, from Equation 3, the distance traveled is M EXAMPLE 7 Figure 4 shows the power consumption in the city of San Francisco for a day in September (P is measured in megawatts; t is measured in hours starting at mid-night). Estimate the energy used on that day. SOLUTION Power is the rate of change of energy: . So, by the Net Change Theorem, is the total amount of energy used that day. We approximate the value of the integral using the Midpoint Rule with 12 subintervals and : The energy used was approximately 15,840 megawatt-hours. M How did we know what units to use for energy in Example 7? The integral is deﬁned as the limit of sums of terms of the form . Now is measured in megawatts and is measured in hours, so their product is measured in megawatt-hours. The same is true of the limit. In general, the unit of measurement for is the prod-uct of the unit for and the unit for x. f!x" xb a f!x" dx &t P!ti " P!ti " &t x24 0 P!t" dt ! 15,840 ' 840 ' 810 ' 690 ' 670 ' 550"!2" ) !440 ' 400 ' 420 ' 620 ' 790 ' 840 ' 850 y 24 0 P!t" dt ) $P!1" ' P!3" ' P!5" ' ( ( ( ' P!21" ' P!23"% &t &t ! 2 y 24 0 P!t" dt ! y 24 0 E#!t" dt ! E!24" ! E!0" P!t" ! E#!t" FIGURE 4 P 0 18 15 12 9 6 3 t 21 400 600 800 200 Pacific Gas & Electric ! 61 6 ) 10.17 m !&! t 3 3 ' t 2 2 ' 6t' 3 1 '& t 3 3 ! t 2 2 ! 6t' 3 4 ! y 3 1 !!t 2 ' t ' 6" dt ' y 4 3 !t 2 ! t ! 6" dt y 4 1 ( v!t"(dt ! y 3 1 $!v!t"% dt ' y 4 3 v!t" dt $3, 4% v!t" % 0 $1, 3% v!t" $ 0 v!t" ! t 2 ! t ! 6 ! !t ! 3"!t ' 2" 396 \|\|\|\| CHAPTER 5 INTEGRALS N A note on units N To integrate the absolute value of , we use Property 5 of integrals from Section 5.2 to split the integral into two parts, one where and one where . v!t" % 0 v!t" $ 0 v!t" 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 44. ; 45. Use a graph to estimate the -intercepts of the curve . Then use this information to estimate the area of the region that lies under the curve and above the -axis. ; 46. Repeat Exercise 45 for the curve . 47. The area of the region that lies to the right of the -axis and to the left of the parabola (the shaded region in the ﬁgure) is given by the integral . (Turn your head clockwise and think of the region as lying below the curve from to .) Find the area of the region. The boundaries of the shaded region are the y-axis, the line , and the curve . Find the area of this region by writing x as a function of y and integrating with respect to y (as in Exercise 47). y=$ œ„ x y=1 0 y x 1 1 y ! s 4 x y ! 1 48. 0 y x 1 x=2y-¥ 2 y ! 2 y ! 0 x ! 2y ! y 2 x2 0 !2y ! y 2" dy x ! 2y ! y 2 y y ! 2x ' 3x 4 ! 2x 6 x y ! x ' x 2 ! x 4 x y 3)#2 0 ( sin x( dx y 2 !1 (x ! 2( x() dx 43. y 2 1 !x ! 1"3 x 2 dx y 1#s3 0 t 2 ! 1 t 4 ! 1 dt y 10 !10 2e x sinh x ' cosh x dx y 64 1 1 ' s 3 x sx dx y )#3 0 sin ' sin tan2 sec2 d y )#4 0 1 ' cos2 cos2 d y )#3 )#4 sec tan d y ) 0 !4 sin ! 3 cos " d y 9 1 3x ! 2 sx dx y 4 1 5 x dx y 5 0 !2e x ' 4 cos x" dx y 1 0 x(s 3 x ' s 4 x ) dx 31. 1–4 Verify by differentiation that the formula is correct. 1. 3. 4. 5–18 Find the general indeﬁnite integral. 5. 6. 7. 8. 10. 11. 12. 13. 14. 15. 16. 17. 18. ; 19–20 Find the general indeﬁnite integral. Illustrate by graphing several members of the family on the same screen. 19. 20. 21–44 Evaluate the integral. 21. 22. 24. 25. 26. 27. 28. 29. 30. y 2 1 y ' 5y7 y 3 dy y !1 !2 +4y 3 ' 2 y 3, dy y 9 0 s2t dt y 4 1 st !1 ' t" dt y 4 0 !2v ' 5"!3v ! 1" dv y 2 !2 !3u ' 1"2 du y 0 !2 !u5 ! u3 ' u2" du y 0 !1 !2x ! e x" dx 23. y 3 1 !1 ' 2x ! 4x 3" dx y 2 0 !6x 2 ! 4x ' 5" dx y !e x ! 2x2" dx y (cos x ' 1 2x) dx y sin 2x sin x dx y !1 ' tan2+" d+ y sec t !sec t ' tan t" dt y ! ! csc cot " d y !csc2t ! 2et" dt y !sin x ' sinh x" dx y +x 2 ' 1 ' 1 x 2 ' 1, dx y x 3 ! 2sx x dx y v!v 2 ' 2"2 dv y !1 ! t"!2 ' t 2" dt 9. y !y 3 ' 1.8y 2 ! 2.4y" dy y (x 4 ! 1 2x 3 ' 1 4x ! 2) dx y (sx 3 ' s 3 x 2 ) dx y !x 2 ' x!2" dx y x sa ' bx dx ! 2 3b2 !bx ! 2a"sa ' bx ' C y cos3 x dx ! sin x ! 1 3 sin3 x ' C y x cos x dx ! x sin x ' cos x ' C 2. y x sx 2 ' 1 dx ! sx 2 ' 1 ' C EXERCISES 5.4 SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM \|\|\|\| 397 63. The velocity of a car was read from its speedometer at 10-second intervals and recorded in the table. Use the Midpoint Rule to estimate the distance traveled by the car. 64. Suppose that a volcano is erupting and readings of the rate at which solid materials are spewed into the atmosphere are given in the table. The time is measured in seconds and the units for are tonnes (metric tons) per second. (a) Give upper and lower estimates for the total quantity of erupted materials after 6 seconds. (b) Use the Midpoint Rule to estimate . 65. The marginal cost of manufacturing yards of a certain fabric is (in dollars per yard). Find the increase in cost if the production level is raised from 2000 yards to 4000 yards. 66. Water ﬂows into and out of a storage tank. A graph of the rate of change of the volume of water in the tank, in liters per day, is shown. If the amount of water in the tank at time is 25,000 L, use the Midpoint Rule to estimate the amount of water four days later. 67. Economists use a cumulative distribution called a Lorenz curve to describe the distribution of income between house-holds in a given country. Typically, a Lorenz curve is deﬁned on with endpoints and , and is continuous, increasing, and concave upward. The points on this curve are determined by ranking all households by income and then computing the percentage of households whose income is less than or equal to a given percentage of the total income of the country. For example, the point is on the Lorenz curve if the bottom of the households receive less than or equal to of the total income. Absolute equality of income distribution would occur if the bottom of the a% b% a% !a#100, b#100" !1, 1" !0, 0" $0, 1% 3 2000 _1000 r t 0 1 2 4 1000 t ! 0 r!t" C#!x" ! 3 ! 0.01x ' 0.000006x 2 x Q!6" Q!6" r!t" t r!t" 49. If is the rate of growth of a child in pounds per year, what does represent? 50. The current in a wire is deﬁned as the derivative of the charge: . (See Example 3 in Section 3.7.) What does represent? If oil leaks from a tank at a rate of gallons per minute at time , what does represent? 52. A honeybee population starts with 100 bees and increases at a rate of bees per week. What does represent? 53. In Section 4.7 we deﬁned the marginal revenue function as the derivative of the revenue function , where is the number of units sold. What does represent? 54. If is the slope of a trail at a distance of miles from the start of the trail, what does represent? 55. If is measured in meters and is measured in newtons, what are the units for ? 56. If the units for are feet and the units for are pounds per foot, what are the units for ? What units does have? 57–58 The velocity function (in meters per second) is given for a particle moving along a line. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval. , 58. , 59–60 The acceleration function (in m#s ) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time and (b) the distance traveled during the given time interval. , , 60. , , 61. The linear density of a rod of length 4 m is given by measured in kilograms per meter, where is measured in meters from one end of the rod. Find the total mass of the rod. 62. Water ﬂows from the bottom of a storage tank at a rate of liters per minute, where . Find the amount of water that ﬂows from the tank during the ﬁrst 10 minutes. 0 $ t $ 50 r!t" ! 200 ! 4t x "!x" ! 9 ' 2sx 0 $ t $ 3 v!0" ! !4 a!t" ! 2t ' 3 0 $ t $ 10 v!0" ! 5 a!t" ! t ' 4 59. t 2 1 $ t $ 6 v!t" ! t 2 ! 2t ! 8 0 $ t $ 3 v!t" ! 3t ! 5 57. x8 2 a!x" dx da#dx a!x" x x100 0 f !x" dx f !x" x x5 3 f !x" dx x f !x" x5000 1000 R#!x" dx x R!x" R#!x" 100 ' x15 0 n#!t" dt n#!t" x120 0 r!t" dt t r!t" 51. xb a I!t" dt I!t" ! Q#!t" x10 5 w#!t" dt w#!t" 398 \|\|\|\| CHAPTER 5 INTEGRALS t (s) (mi#h) t (s) (mi#h) 0 0 60 56 10 38 70 53 20 52 80 50 30 58 90 47 40 55 100 45 50 51 v v t 0 1 2 3 4 5 6 2 10 24 36 46 54 60 r!t" We sometimes read that the inventors of calculus were Sir Isaac Newton (1642–1727) and Gottfried Wilhelm Leibniz (1646–1716). But we know that the basic ideas behind integration were investigated 2500 years ago by ancient Greeks such as Eudoxus and Archimedes, and methods for ﬁnding tangents were pioneered by Pierre Fermat (1601–1665), Isaac Barrow (1630–1677), and others. Barrow––who taught at Cambridge and was a major inﬂuence on Newton––was the ﬁrst to understand the inverse relationship between differentiation and integra-tion. What Newton and Leibniz did was to use this relationship, in the form of the Fundamental Theorem of Calculus, in order to develop calculus into a systematic mathematical discipline. It is in this sense that Newton and Leibniz are credited with the invention of calculus. Read about the contributions of these men in one or more of the given references and write a report on one of the following three topics. You can include biographical details, but the main thrust of your report should be a description, in some detail, of their methods and notations. In particular, you should consult one of the sourcebooks, which give excerpts from the original publications of Newton and Leibniz, translated from Latin to English. N The Role of Newton in the Development of Calculus N The Role of Leibniz in the Development of Calculus N The Controversy between the Followers of Newton and Leibniz over Priority in the Invention of Calculus References 1. Carl Boyer and Uta Merzbach, A History of Mathematics (New York: Wiley, 1987), Chapter 19. NEWTON, LEIBNIZ, AND THE INVENTION OF CALCULUS W R I T I N G P R O J E C T What is the percentage of total income received by the bottom of the households? Find the coefﬁcient of inequality. ; 68. On May 7, 1992, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. (a) Use a graphing calculator or computer to model these data by a third-degree polynomial. (b) Use the model in part (a) to estimate the height reached by the Endeavour, 125 seconds after liftoff. 50% households receive of the income, in which case the Lorenz curve would be the line . The area between the Lorenz curve and the line measures how much the income distribution differs from absolute equality. The coefﬁcient of inequality is the ratio of the area between the Lorenz curve and the line to the area under . (a) Show that the coefﬁcient of inequality is twice the area between the Lorenz curve and the line , that is, show that (b) The income distribution for a certain country is repre-sented by the Lorenz curve deﬁned by the equation L!x" ! 5 12 x 2 ' 7 12 x coefficient of inequality ! 2 y 1 0 $x ! L!x"% dx y ! x x 1 y 0 1 y=x y=L(x) (1, 1) y ! x y ! x y ! x y ! x a% WRITING PROJECT NEWTON, LEIBNIZ, AND THE INVENTION OF CALCULUS \|\|\|\| 399 Event Time (s) Velocity (ft#s) Launch 0 0 Begin roll maneuver 10 185 End roll maneuver 15 319 Throttle to 89% 20 447 Throttle to 67% 32 742 Throttle to 104% 59 1325 Maximum dynamic pressure 62 1445 Solid rocket booster separation 125 4151 2. Carl Boyer, The History of the Calculus and Its Conceptual Development (New York: Dover, 1959), Chapter V. 3. C. H. Edwards, The Historical Development of the Calculus (New York: Springer-Verlag, 1979), Chapters 8 and 9. 4. Howard Eves, An Introduction to the History of Mathematics, 6th ed. (New York: Saunders, 1990), Chapter 11. 5. C. C. Gillispie, ed., Dictionary of Scientiﬁc Biography (New York: Scribner’s, 1974). See the article on Leibniz by Joseph Hofmann in Volume VIII and the article on Newton by I. B. Cohen in Volume X. 6. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993), Chapter 12. 7. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford University Press, 1972), Chapter 17. Sourcebooks 1. John Fauvel and Jeremy Gray, eds., The History of Mathematics: A Reader (London: MacMillan Press, 1987), Chapters 12 and 13. 2. D. E. Smith, ed., A Sourcebook in Mathematics (New York: Dover, 1959), Chapter V. 3. D. J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, N.J.: Princeton University Press, 1969), Chapter V. 400 \|\|\|\| CHAPTER 5 INTEGRALS THE SUBSTITUTION RULE Because of the Fundamental Theorem, it’s important to be able to ﬁnd antiderivatives. But our antidifferentiation formulas don’t tell us how to evaluate integrals such as To ﬁnd this integral we use the problem-solving strategy of introducing something extra. Here the “something extra” is a new variable; we change from the variable x to a new vari-able u. Suppose that we let be the quantity under the root sign in (1), . Then the differential of is . Notice that if the in the notation for an integral were to be interpreted as a differential, then the differential would occur in (1) and so, formally, without justifying our calculation, we could write But now we can check that we have the correct answer by using the Chain Rule to differ-entiate the ﬁnal function of Equation 2: In general, this method works whenever we have an integral that we can write in the form . Observe that if , then y F#!t!x""t#!x" dx ! F!t!x"" ' C 3 F# ! f x f!t!x""t#!x" dx d dx [ 2 3!x 2 ' 1"3#2 ' C] ! 2 3 ! 3 2!x 2 ' 1"1#2 ! 2x ! 2xsx 2 ' 1 ! 2 3u3#2 ' C ! 2 3!x 2 ' 1"3#2 ' C y 2xs1 ' x 2 dx ! y s1 ' x 2 2x dx ! y su du 2 2x dx dx du ! 2x dx u u ! 1 ' x 2 u y 2xs1 ' x 2 dx 1 5.5 N Differentials were deﬁned in Section 3.10. If , then du ! f #!x" dx u ! f !x" because, by the Chain Rule, If we make the “change of variable” or “substitution” , then from Equation 3 we have or, writing , we get Thus we have proved the following rule. THE SUBSTITUTION RULE If is a differentiable function whose range is an interval and is continuous on , then Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation. Notice also that if , then , so a way to remember the Substitution Rule is to think of and in (4) as differentials. Thus the Substitution Rule says: It is permissible to operate with dx and du after integral signs as if they were differentials. EXAMPLE 1 Find . SOLUTION We make the substitution because its differential is , which, apart from the constant factor 4, occurs in the integral. Thus, using and the Substitution Rule, we have Notice that at the ﬁnal stage we had to return to the original variable . M The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral. This is accomplished by changing from the original variable to a new variable that is a function of . Thus, in Example 1, we replaced the integral by the simpler integral . The main challenge in using the Substitution Rule is to think of an appropriate substi-tution. You should try to choose to be some function in the integrand whose differential also occurs (except for a constant factor). This was the case in Example 1. If that is not u 1 4 x cos u du x x 3 cos!x 4 ' 2" dx x u x x ! 1 4 sin!x 4 ' 2" ' C ! 1 4 sin u ' C y x 3 cos!x 4 ' 2" dx ! y cos u ! 1 4 du ! 1 4 y cos u du x 3 dx ! du#4 du ! 4x 3 dx u ! x 4 ' 2 y x 3 cos!x 4 ' 2" dx du dx du ! t#!x" dx u ! t!x" y f!t!x""t#!x" dx ! y f!u" du I f I u ! t!x" 4 y f!t!x""t#!x" dx ! y f!u" du F# ! f y F#!t!x""t#!x" dx ! F!t!x"" ' C ! F!u" ' C ! y F#!u" du u ! t!x" d dx $F!t!x""% ! F#!t!x""t#!x" SECTION 5.5 THE SUBSTITUTION RULE \|\|\|\| 401 N Check the answer by differentiating it. possible, try choosing to be some complicated part of the integrand (perhaps the inner function in a composite function). Finding the right substitution is a bit of an art. It’s not unusual to guess wrong; if your ﬁrst guess doesn’t work, try another substitution. EXAMPLE 2 Evaluate . SOLUTION 1 Let . Then , so . Thus the Substitution Rule gives SOLUTION 2 Another possible substitution is . Then so (Or observe that so .) Therefore M EXAMPLE 3 Find . SOLUTION Let . Then , so and M The answer to Example 3 could be checked by differentiation, but instead let’s check it with a graph. In Figure 1 we have used a computer to graph both the integrand and its indeﬁnite integral (we take the case ). Notice that decreases when is negative, increases when is positive, and has its minimum value when . So it seems reasonable, from the graphical evi-dence, that is an antiderivative of . EXAMPLE 4 Calculate . SOLUTION If we let , then , so . Therefore M y e 5x dx ! 1 5 y e u du ! 1 5 e u ' C ! 1 5 e 5x ' C dx ! 1 5 du du ! 5 dx u ! 5x y e 5x dx f t f!x" ! 0 f!x" f!x" t!x" C ! 0 t!x" ! ! 1 4s1 ! 4x 2 f!x" ! x#s1 ! 4x 2 ! ! 1 8(2su ) ' C ! ! 1 4s1 ! 4x 2 ' C y x s1 ! 4x 2 dx ! ! 1 8 y 1 su du ! ! 1 8 y u !1#2 du x dx ! ! 1 8 du du ! !8x dx u ! 1 ! 4x 2 y x s1 ! 4x 2 dx V ! u 3 3 ' C ! 1 3!2x ' 1"3#2 ' C y s2x ' 1 dx ! y u ! u du ! y u 2 du 2u du ! 2 dx u 2 ! 2x ' 1, dx ! s2x ' 1 du ! u du du ! dx s2x ' 1 u ! s2x ' 1 ! 1 3!2x ' 1"3#2 ' C ! 1 2 ! u 3#2 3#2 ' C ! 1 3u 3#2 ' C y s2x ' 1 dx ! y su du 2 ! 1 2 y u 1#2 du dx ! du#2 du ! 2 dx u ! 2x ' 1 y s2x ' 1 dx u 402 \|\|\|\| CHAPTER 5 INTEGRALS 1 _1 _1 1 ©=- ƒ dx f FIGURE 1 ©=j ƒ dx= œ„„„„„„ x œ„„„„„„ 1-4≈ 1-4≈ 1 4 ƒ= EXAMPLE 5 Find . SOLUTION An appropriate substitution becomes more obvious if we factor as . Let . Then , so . Also , so : M EXAMPLE 6 Calculate . SOLUTION First we write tangent in terms of sine and cosine: This suggests that we should substitute , since then and so : M Since , the result of Example 6 can also be written as DEFINITE INTEGRALS When evaluating a deﬁnite integral by substitution, two methods are possible. One method is to evaluate the indeﬁnite integral ﬁrst and then use the Fundamental Theorem. For instance, using the result of Example 2, we have Another method, which is usually preferable, is to change the limits of integration when the variable is changed. ! 1 3!9"3#2 ! 1 3!1"3#2 ! 1 3!27 ! 1" ! 26 3 y 4 0 s2x ' 1 dx ! y s2x ' 1 dx]0 4 ! 1 3!2x ' 1"3#2]0 4 y tan x dx ! ln(sec x( ' C 5 !ln(cos x( ! ln!(cos x( !1" ! ln!1#(cos x(" ! ln(sec x( ! !ln (u( ' C ! !ln (cos x( ' C y tan x dx ! y sin x cos x dx ! !y du u sin x dx ! !du du ! !sin x dx u ! cos x y tan x dx ! y sin x cos x dx y tan x dx V ! 1 7!1 ' x 2"7#2 ! 2 5!1 ' x 2"5#2 ' 1 3!1 ' x 2"3#2 ' C ! 1 2( 2 7u 7#2 ! 2 ( 2 5u 5#2 ' 2 3u 3#2) ' C ! 1 2 y !u 5#2 ! 2u 3#2 ' u 1#2" du ! y su !u ! 1"2 du 2 ! 1 2 y su !u 2 ! 2u ' 1" du y s1 ' x 2 x 5 dx ! y s1 ' x 2 x 4 ( x dx x 4 ! !u ! 1"2 x 2 ! u ! 1 x dx ! du#2 du ! 2x dx u ! 1 ' x 2 x 4 ! x x 5 y s1 ' x 2 x 5 dx SECTION 5.5 THE SUBSTITUTION RULE \|\|\|\| 403 THE SUBSTITUTION RULE FOR DEFINITE INTEGRALS If is continuous on and is continuous on the range of , then PROOF Let be an antiderivative of . Then, by (3), is an antiderivative of , so by Part 2 of the Fundamental Theorem, we have But, applying FTC2 a second time, we also have M EXAMPLE 7 Evaluate using (6). SOLUTION Using the substitution from Solution 1 of Example 2, we have and . To ﬁnd the new limits of integration we note that and Therefore Observe that when using (6) we do not return to the variable x after integrating. We simply evaluate the expression in u between the appropriate values of u. M EXAMPLE 8 Evaluate . SOLUTION Let . Then , so . When , and u ! !2 x ! 1 dx ! !du!5 du ! !5 dx u ! 3 ! 5x y 2 1 dx "3 ! 5x#2 FIGURE 2 y x 0 1 2 3 4 y=œ„„„„„ 2x+1 y u 0 1 2 3 9 1 œ„ u 2 y= ! 1 3"93!2 ! 13!2# ! 26 3 ! 1 2 ! 2 3u 3!2]1 9 y 4 0 s2x " 1 dx ! y 9 1 1 2 su du when x ! 4, u ! 2"4# " 1 ! 9 when x ! 0, u ! 2"0# " 1 ! 1 dx ! du!2 u ! 2x " 1 y 4 0 s2x " 1 dx y t"b# t"a# f"u# du ! F"u#]t"a# t"b# ! F"t"b## ! F"t"a## y b a f"t"x##t#"x# dx ! F"t"x##] b a ! F"t"b## ! F"t"a## f"t"x##t#"x# F"t"x## f F y b a f"t"x##t#"x# dx ! y t"b# t"a# f"u# du u ! t"x# f $a, b% t# 6 404 \|\|\|\| CHAPTER 5 INTEGRALS N The geometric interpretation of Example 7 is shown in Figure 2. The substitution stretches the interval by a factor of and translates it to the right by unit. The Substitu-tion Rule shows that the two areas are equal. 1 2 $0, 4% u ! 2x " 1 N This rule says that when using a substitution in a deﬁnite integral, we must put everything in terms of the new variable , not only and but also the limits of integration. The new limits of integration are the values of that correspond to and . x ! b x ! a u dx x u N The integral given in Example 8 is an abbreviation for y 2 1 1 "3 ! 5x#2 dx when , . Thus M EXAMPLE 9 Calculate . SOLUTION We let because its differential occurs in the integral. When , ; when , . Thus M SYMMETRY The next theorem uses the Substitution Rule for Deﬁnite Integrals (6) to simplify the cal-culation of integrals of functions that possess symmetry properties. INTEGRALS OF SYMMETRIC FUNCTIONS Suppose is continuous on . (a) If is even , then . (b) If is odd , then . PROOF We split the integral in two: In the ﬁrst integral on the far right side we make the substitution . Then and when , . Therefore !y !a 0 f"x# dx ! !y a 0 f"!u#"!du# ! y a 0 f"!u# du u ! a x ! !a du ! !dx u ! !x y a !a f"x# dx ! y 0 !a f"x# dx " y a 0 f"x# dx ! !y !a 0 f"x# dx " y a 0 f"x# dx 8 xa !a f"x# dx ! 0 $ f"!x# ! !f"x#% f xa !a f"x# dx ! 2 xa 0 f"x# dx $ f"!x# ! f"x#% f $!a, a% f 7 FIGURE 3 x 0 y 0.5 1 e y=ln x x y e 1 ln x x dx ! y 1 0 u du ! u 2 2& 0 1 ! 1 2 u ! ln e ! 1 x ! e u ! ln 1 ! 0 x ! 1 du ! dx!x u ! ln x y e 1 ln x x dx V ! 1 5 '! 1 7 " 1 2( ! 1 14 ! ! 1 5 )! 1 u& !2 !7 ! 1 5u& !2 !7 y 2 1 dx "3 ! 5x#2 ! ! 1 5 y !7 !2 du u 2 u ! !7 x ! 2 SECTION 5.5 THE SUBSTITUTION RULE \|\|\|\| 405 N Since the function in Example 9 is positive for , the integral represents the area of the shaded region in Figure 3. x $ 1 f "x# ! "ln x#!x and so Equation 8 becomes (a) If is even, then so Equation 9 gives (b) If is odd, then and so Equation 9 gives M Theorem 7 is illustrated by Figure 4. For the case where is positive and even, part (a) says that the area under from to is twice the area from to because of symmetry. Recall that an integral can be expressed as the area above the -axis and below minus the area below the axis and above the curve. Thus part (b) says the integral is because the areas cancel. EXAMPLE 10 Since satisﬁes , it is even and so M EXAMPLE 11 Since satisﬁes , it is odd and so M y 1 !1 tan x 1 " x 2 " x 4 dx ! 0 f"!x# ! !f"x# f"x# ! "tan x#!"1 " x 2 " x 4# ! 2[ 1 7 x 7 " x]0 2 ! 2( 128 7 " 2) ! 284 7 y 2 !2 "x 6 " 1# dx ! 2 y 2 0 "x 6 " 1# dx f"!x# ! f"x# f"x# ! x 6 " 1 0 y ! f"x# x xb a f"x# dx a 0 a !a y ! f"x# f y a !a f"x# dx ! !y a 0 f"u# du " y a 0 f"x# dx ! 0 f"!u# ! !f"u# f y a !a f"x# dx ! y a 0 f"u# du " y a 0 f"x# dx ! 2 y a 0 f"x# dx f"!u# ! f"u# f y a !a f"x# dx ! y a 0 f"!u# du " y a 0 f"x# dx 9 406 \|\|\|\| CHAPTER 5 INTEGRALS 7–46 Evaluate the indeﬁnite integral. 7. 8. 9. 10. 11. 12. 14. 15. 16. 17. 18. y sec 2% tan 2% d% y a " bx 2 s3ax " bx 3 dx y x x 2 " 1 dx y sin &t dt y e x sin"e x# dx y dx 5 ! 3x 13. y x "x 2 " 1#2 dx y "x " 1#s2x " x 2 dx y "3t " 2#2.4 dt y "3x ! 2#20 dx y x 2"x 3 " 5#9 dx y x sin"x 2# dx 1–6 Evaluate the integral by making the given substitution. 1. 2. 4. 5. 6. y sec2"1!x# x 2 dx, u ! 1!x y cos3% sin % d%, u ! cos % y dt "1 ! 6t#4 , u ! 1 ! 6t y x 2sx 3 " 1 dx, u ! x 3 " 1 3. y x 3"2 " x 4#5 dx, u ! 2 " x 4 y e!x dx, u ! !x EXERCISES 5.5 0 y x a a FIGURE 4 (a) ƒ even, j ƒ dx=2 j ƒ dx 0 a _a a 0 x _a a y (b) ƒ odd, j ƒ dx=0 _a a 55. 56. 57. 58. 60. 61. 62. 63. 65. 66. 68. 69. 70. ; 71–72 Use a graph to give a rough estimate of the area of the region that lies under the given curve. Then ﬁnd the exact area. 71. , 72. , 73. Evaluate by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area. 74. Evaluate by making a substitution and inter-preting the resulting integral in terms of an area. Which of the following areas are equal? Why? 76. A model for the basal metabolism rate, in , of a young man is , where is the time in hours measured from 5:00 AM. What is the total basal metab-olism of this man, , over a 24-hour time period? x24 0 Rt dt t Rt 85 0.18 cost12 kcalh y=2x´ 0 x y 1 y=esin x sin 2x 0 x y π 2 1 y=eœ„ x 0 x y 1 75. x1 0 xs1 x 4 dx x2 2 x 3s4 x 2 dx 0 x y 2 sin x sin 2x 0 x 1 y s2x 1 y T2 0 sin2tT dt y 1 0 e z 1 e z z dz y 12 0 sin1x s1 x 2 dx y e4 e dx xsln x 67. y 4 0 x s1 2x dx y 2 1 xsx 1 dx y a 0 xsa 2 x 2 dx 64. y a 0 xsx 2 a 2 dx a 0 y 2 0 cos x sinsin x dx y 13 0 dx s 3 1 2x2 y 2 2 x 2 sin x 1 x 6 dx y 2 1 e1x x 2 dx 59. y 1 0 xex2 dx y 6 6 tan3 d y 12 16 csc t cot t dt y 0 sec2t4 dt 20. 22. 23. 24. 26. 27. 28. 29. 30. 31. 32. 34. 35. 36. 37. 38. 39. 40. 41. 42. 44. 45. 46. ; 47–50 Evaluate the indeﬁnite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take ). 47. 48. 49. 50. 51–70 Evaluate the deﬁnite integral. 51. 52. 53. 54. y s 0 x cosx 2 dx y 1 0 x 21 2x 35 dx y 7 0 s4 3x dx y 2 0 x 125 dx y tan2 sec2 d y sin3x cos x dx y sin sx sx dx y xx 2 13 dx C 0 y x 3sx 2 1 dx y x s 4 x 2 dx y x 2 s1 x dx y 1 x 1 x 2 dx 43. y x 1 x 4 dx y dx s1 x 2 sin1x y sin t sec2cos t dt y sec3x tan x dx y dt cos2 ts1 tan t y cot x dx y sin x 1 cos2x dx y sin 2x 1 cos2x dx y cosx x 2 dx y scot x csc2x dx 33. y e x e x 1 dx y cos x sin2x dx y sinln x x dx y etan x sec2 x dx y tan1x 1 x 2 dx y z2 s 3 1 z3 dz y ecos t sin t dt y e xs1 e x dx 25. y 1 tan 5 sec2 d y cos sin6 d y sx sin1 x 32 dx y cos st st dt 21. y dx ax b a 0 y ln x2 x dx 19. SECTION 5.5 THE SUBSTITUTION RULE \|\|\|\| 407 83. If is continuous on , prove that For the case where and , draw a diagram to interpret this equation geometrically as an equality of areas. 84. If is continuous on , prove that For the case where , draw a diagram to interpret this equation geometrically as an equality of areas. If and are positive numbers, show that 86. If is continuous on , use the substitution to show that 87. Use Exercise 86 to evaluate the integral 88. (a) If is continuous, prove that (b) Use part (a) to evaluate and . x&!2 0 sin2x dx x&!2 0 cos2x dx y &!2 0 f "cos x# dx ! y &!2 0 f "sin x# dx f y & 0 x sin x 1 " cos2x dx y & 0 xf "sin x# dx ! & 2 y & 0 f "sin x# dx u ! & ! x $0, &% f y 1 0 x a"1 ! x#b dx ! y 1 0 x b"1 ! x#a dx b a 85. f "x# ) 0 y b a f "x " c# dx ! y b"c a"c f "x# dx ! f 0 a b f "x# ) 0 y b a f "!x# dx ! y !a !b f "x# dx ! f 77. An oil storage tank ruptures at time and oil leaks from the tank at a rate of liters per minute. How much oil leaks out during the ﬁrst hour? 78. A bacteria population starts with 400 bacteria and grows at a rate of bacteria per hour. How many bacteria will there be after three hours? 79. Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 5 s. The maximum rate of air ﬂow into the lungs is about 0.5 L!s. This explains, in part, why the function has often been used to model the rate of air ﬂow into the lungs. Use this model to ﬁnd the volume of inhaled air in the lungs at time . 80. Alabama Instruments Company has set up a production line to manufacture a new calculator. The rate of production of these calculators after weeks is (Notice that production approaches 5000 per week as time goes on, but the initial production is lower because of the workers’ unfamiliarity with the new techniques.) Find the number of cal-culators produced from the beginning of the third week to the end of the fourth week. If is continuous and , ﬁnd . 82. If is continuous and , ﬁnd . y 3 0 xf "x 2# dx y 9 0 f "x# dx ! 4 f y 2 0 f "2x# dx y 4 0 f "x# dx ! 10 f 81. dx dt ! 5000'1 ! 100 "t " 10#2( calculators!week t t f "t# ! 1 2 sin"2&t!5# r"t# ! "450.268#e1.12567t r"t# ! 100e!0.01t t ! 0 408 \|\|\|\| CHAPTER 5 INTEGRALS REVIEW CONCEPT CHECK 5 (b) If is the rate at which water ﬂows into a reservoir, what does represent? 5. Suppose a particle moves back and forth along a straight line with velocity , measured in feet per second, and accelera-tion . (a) What is the meaning of ? (b) What is the meaning of ? (c) What is the meaning of ? 6. (a) Explain the meaning of the indeﬁnite integral . (b) What is the connection between the deﬁnite integral and the indeﬁnite integral ? 7. Explain exactly what is meant by the statement that “differen-tiation and integration are inverse processes.” 8. State the Substitution Rule. In practice, how do you use it? x f "x# dx xb a f "x# dx x f "x# dx x120 60 a"t# dt x120 60 v"t# dt x120 60 v"t# dt a"t# v"t# xt2 t1 r"t# dt r"t# 1. (a) Write an expression for a Riemann sum of a function . Explain the meaning of the notation that you use. (b) If , what is the geometric interpretation of a Riemann sum? Illustrate with a diagram. (c) If takes on both positive and negative values, what is the geometric interpretation of a Riemann sum? Illustrate with a diagram. 2. (a) Write the deﬁnition of the deﬁnite integral of a function from to . (b) What is the geometric interpretation of if ? (c) What is the geometric interpretation of if takes on both positive and negative values? Illustrate with a diagram. 3. State both parts of the Fundamental Theorem of Calculus. 4. (a) State the Net Change Theorem. f "x# xb a f "x# dx f "x# ) 0 xb a f "x# dx b a f "x# f "x# ) 0 f Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. If and are continuous on , then 2. If and are continuous on , then 3. If is continuous on , then 4. If is continuous on , then 5. If is continuous on and , then 6. If is continuous on , then . y 3 1 f #"v# dv ! f "3# ! f "1# $1, 3% f # y b a sf "x# dx !+y b a f "x# dx f "x# ) 0 $a, b% f y b a xf "x# dx ! x y b a f "x# dx $a, b% f y b a 5f "x# dx ! 5 y b a f "x# dx $a, b% f y b a $ f "x#t"x#% dx !'y b a f "x# dx('y b a t"x# dx( $a, b% t f y b a $ f "x# " t"x#% dx ! y b a f "x# dx " y b a t"x# dx $a, b% t f 7. If and are continuous and for , then 8. If and are differentiable and for , then for . 9. 10. 11. 12. represents the area under the curve from 0 to 2. 13. All continuous functions have derivatives. 14. All continuous functions have antiderivatives. 15. If is continuous on , then d dx'y b a f "x# dx( ! f "x# $a, b% f y ! x ! x 3 x2 0 "x ! x 3# dx y 1 !2 1 x 4 dx ! ! 3 8 y 5 !5 "ax 2 " bx " c# dx ! 2 y 5 0 "ax 2 " c# dx y 1 !1 'x 5 ! 6x 9 " sin x "1 " x 4#2( dx ! 0 a x b f #"x# ) t#"x# a x b f "x# ) t"x# t f y b a f "x# dx ) y b a t"x# dx a ' x ' b f "x# ) t"x# t f TRUE-FALSE QUIZ 1. Use the given graph of to ﬁnd the Riemann sum with six subintervals. Take the sample points to be (a) left endpoints and (b) midpoints. In each case draw a diagram and explain what the Riemann sum represents. 2. (a) Evaluate the Riemann sum for with four subintervals, taking the sample points to be right endpoints. Explain, with the aid of a diagram, what the Riemann sum represents. 0 ' x ' 2 f "x# ! x 2 ! x 2 x y 2 0 6 y=ƒ f (b) Use the deﬁnition of a deﬁnite integral (with right end-points) to calculate the value of the integral (c) Use the Fundamental Theorem to check your answer to part (b). (d) Draw a diagram to explain the geometric meaning of the integral in part (b). 3. Evaluate by interpreting it in terms of areas. 4. Express as a deﬁnite integral on the interval and then evaluate the integral. $0, &% lim n l + , n i!1 sin xi ,x y 1 0 (x " s1 ! x 2 ) dx y 2 0 "x 2 ! x# dx EXERCISES CHAPTER 5 REVIEW \|\|\|\| 409 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. ; 39–40 Evaluate the indeﬁnite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take ). 39. 40. ; 41. Use a graph to give a rough estimate of the area of the region that lies under the curve . Then ﬁnd the exact area. ; 42. Graph the function and use the graph to guess the value of the integral . Then evaluate the integral to conﬁrm your guess. 43–48 Find the derivative of the function. 43. 44. 45. 46. 47. 48. 49–50 Use Property 8 of integrals to estimate the value of the integral. 49. 50. 51–54 Use the properties of integrals to verify the inequality. 51. 52. 53. 54. 55. Use the Midpoint Rule with to approximate . x3 0 sin"x 3# dx n ! 6 y 1 0 x sin!1x dx ' &!4 y 1 0 e x cos x dx ' e ! 1 y &!2 &!4 sin x x dx ' s2 2 y 1 0 x 2 cos x dx ' 1 3 y 5 3 1 x " 1 dx y 3 1 sx 2 " 3 dx y ! y 3x"1 2x sin"t 4# dt y ! y x sx e t t dt t"x# ! y sin x 1 1 ! t 2 1 " t 4 dt t"x# ! y x4 0 cos"t 2# dt F"x# ! y 1 x st " sin t dt F"x# ! y x 0 t 2 1 " t 3 dt x2& 0 f "x# dx f "x# ! cos2x sin3x y ! xsx , 0 ' x ' 4 y x 3 sx 2 " 1 dx y cos x s1 " sin x dx C ! 0 y 4 0 sx ! 1 dx y 3 0 x 2 ! 4 dx y &!4 0 "1 " tan t#3 sec2t dt y sec % tan % 1 " sec % d% y sinh"1 " 4x# dx y x 3 1 " x 4 dx y x s1 ! x 4 dx y tan x ln"cos x# dx y cos"ln x# x dx y e sx sx dx 5. If and , ﬁnd . 6. (a) Write as a limit of Riemann sums, taking the sample points to be right endpoints. Use a computer algebra system to evaluate the sum and to com-pute the limit. (b) Use the Fundamental Theorem to check your answer to part (a). 7. The following ﬁgure shows the graphs of , and . Identify each graph, and explain your choices. 8. Evaluate: (a) (b) (c) 9–38 Evaluate the integral, if it exists. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. y sin x cos"cos x# dx y sin &t cos &t dt y csc2 x 1 " cot x dx y x " 2 sx 2 " 4x dx y 10 1 x x 2 ! 4 dx y ' 1 ! x x ( 2 dx y 1 0 e x 1 " e 2x dx y &!4 !&!4 t 4 tan t 2 " cos t dt y 1 !1 sin x 1 " x 2 dx y 1 0 v2 cos"v3# dv y 1 0 sin"3&t# dt y 5 1 dt "t ! 4#2 y 2 0 y 2s1 " y 3 dy y 1 0 y"y2 " 1#5 dy y 1 0 (s 4 u " 1#2 du y 9 1 su ! 2u2 u du y 1 0 "1 ! x#9 dx y 1 0 "1 ! x 9# dx y T 0 "x 4 ! 8x " 7# dx y 2 1 "8x 3 " 3x 2# dx d dx y x 0 earctan t dt d dx y 1 0 earctan x dx y 1 0 d dx "earctan x# dx y x a b c xx 0 f "t# dt f, f # x5 1 "x " 2x 5# dx CAS x6 4 f "x# dx x4 0 f "x# dx ! 7 x6 0 f "x# dx ! 10 410 \|\|\|\| CHAPTER 5 INTEGRALS (b) On what intervals is concave upward? (c) Use a graph to solve the following equation correct to two decimal places: (d) Plot the graphs of and on the same screen. How are these graphs related? ; 63. Estimate the value of the number c such that the area under the curve between and is equal to 1. 64. Suppose that the temperature in a long, thin rod placed along the -axis is initially if and 0 if . It can be shown that if the heat diffusivity of the rod is , then the temperature of the rod at the point at time is To ﬁnd the temperature distribution that results from an initial hot spot concentrated at the origin, we need to compute Use l’Hospital’s Rule to ﬁnd this limit. 65. If is a continuous function such that for all , ﬁnd an explicit formula for . 66. Suppose h is a function such that , , , , , , and is contin-uous everywhere. Evaluate . 67. If is continuous on , show that 68. Find . 69. If is continuous on , prove that 70. Evaluate 71. Suppose is continuous, , , , and . Find the value of the integral . x1 0 f !1"y# dy x1 0 f "x# dx ! 1 3 f #"x# $ 0 f "1# ! 1 f "0# ! 0 f lim n l + 1 n )' 1 n( 9 "' 2 n( 9 "' 3 n( 9 " - - - "' n n( 9& y 1 0 f "x# dx ! y 1 0 f "1 ! x# dx $0, 1% f lim h l 0 1 h y 2"h 2 s1 " t 3 dt 2 y b a f "x#f #"x# dx ! $ f "b#%2 ! $ f "a#%2 $a, b% f # x2 1 h."u# du h. h."2# ! 13 h#"2# ! 5 h"2# ! 6 h."1# ! 3 h#"1# ! 2 h"1# ! !2 f "x# x y x 0 f "t# dt ! xe 2x " y x 0 e !tf "t# dt f lim a l 0 T"x, t# T"x, t# ! C as4&kt y a 0 e !"x!u#2!"4kt# du t x k x $ a x ' a C!"2a# x x ! 1 x ! 0 y ! sinh cx S C CAS y x 0 cos( 1 2&t 2) dt ! 0.7 CAS C 56. A particle moves along a line with velocity function , where is measured in meters per second. Find (a) the displacement and (b) the distance traveled by the particle during the time interval . 57. Let be the rate at which the world’s oil is consumed, where is measured in years starting at on January 1, 2000, and is measured in barrels per year. What does represent? 58. A radar gun was used to record the speed of a runner at the times given in the table. Use the Midpoint Rule to estimate the distance the runner covered during those 5 seconds. 59. A population of honeybees increased at a rate of bees per week, where the graph of r is as shown. Use the Midpoint Rule with six subintervals to estimate the increase in the bee population during the ﬁrst 24 weeks. 60. Let Evaluate by interpreting the integral as a difference of areas. 61. If is continuous and , evaluate . 62. The Fresnel function was introduced in Section 5.3. Fresnel also used the function in his theory of the diffraction of light waves. (a) On what intervals is increasing? C C"x# ! y x 0 cos( 1 2 &t 2) dt S"x# ! xx 0 sin( 1 2&t 2) dt x&!2 0 f "2 sin %# cos % d% x2 0 f "x# dx ! 6 f x1 !3 f "x# dx f "x# !-!x ! 1 !s1 ! x 2 if !3 ' x ' 0 if 0 ' x ' 1 r 0 24 20 16 12 8 4 (weeks) t 4000 8000 12000 r"t# x8 0 r"t# dt r"t# t ! 0 t r"t# $0, 5% v v"t# ! t 2 ! t CHAPTER 5 REVIEW \|\|\|\| 411 t (s) (m!s) t (s) (m!s) 0 0 3.0 10.51 0.5 4.67 3.5 10.67 1.0 7.34 4.0 10.76 1.5 8.86 4.5 10.81 2.0 9.73 5.0 10.81 2.5 10.22 v v 412 Before you look at the solution of the following example, cover it up and ﬁrst try to solve the problem yourself. EXAMPLE 1 Evaluate . SOLUTION Let’s start by having a preliminary look at the ingredients of the function. What happens to the ﬁrst factor, , when approaches ? The numerator approaches and the denominator approaches , so we have and The second factor approaches , which is . It’s not clear what happens to the function as a whole. (One factor is becoming large while the other is becoming small.) So how do we proceed? One of the principles of problem solving is recognizing something familiar. Is there a part of the function that reminds us of something we’ve seen before? Well, the integral has as its upper limit of integration and that type of integral occurs in Part 1 of the Fundamental Theorem of Calculus: This suggests that differentiation might be involved. Once we start thinking about differentiation, the denominator reminds us of something else that should be familiar: One of the forms of the deﬁnition of the deriva-tive in Chapter 2 is and with this becomes So what is the function in our situation? Notice that if we deﬁne then . What about the factor in the numerator? That’s just a red herring, so let’s factor it out and put together the calculation: (FTC1) M ! sin 3 ! 3 sin 3 3 ! 3F#"3# ! 3 lim x l3 F"x# ! F"3# x ! 3 lim x l3 ' x x ! 3 y x 3 sin t t dt( ! lim x l3 x ! lim x l3 y x 3 sin t t dt x ! 3 x F"3# ! 0 F"x# ! y x 3 sin t t dt F F#"3# ! lim x l3 F"x# ! F"3# x ! 3 a ! 3 F#"a# ! lim x l a F"x# ! F"a# x ! a "x ! 3# d dx y x a f"t# dt ! f"x# x y x 3 sin t t dt 0 x3 3 "sin t#!t dt x l 3! as x x ! 3 l !+ x l 3" as x x ! 3 l + 0 3 3 x x!"x ! 3# lim x l3 ' x x ! 3 y x 3 sin t t dt( P R O B L E M S P L U S N The principles of problem solving are discussed on page 76. N Another approach is to use l’Hospital’s Rule. 413 1. If , where is a continuous function, ﬁnd . 2. Find the minimum value of the area of the region under the curve from to , for all . 3. If is a differentiable function such that is never and for all , ﬁnd . ; 4. (a) Graph several members of the family of functions for and look at the regions enclosed by these curves and the -axis. Make a conjecture about how the areas of these regions are related. (b) Prove your conjecture in part (a). (c) Take another look at the graphs in part (a) and use them to sketch the curve traced out by the vertices (highest points) of the family of functions. Can you guess what kind of curve this is? (d) Find an equation of the curve you sketched in part (c). 5. If , where , ﬁnd . 6. If , ﬁnd . 7. Evaluate . 8. The ﬁgure shows two regions in the ﬁrst quadrant: is the area under the curve from to , and is the area of the triangle with vertices , , and . Find . 9. Find the interval for which the value of the integral is a maximum. 10. Use an integral to estimate the sum . 11. (a) Evaluate , where is a positive integer. (b) Evaluate , where and are real numbers with . 12. Find . 13. Suppose the coefﬁcients of the cubic polynomial satisfy the equation Show that the equation has a root between 0 and 1. Can you generalize this result for an -degree polynomial? 14. A circular disk of radius is used in an evaporator and is rotated in a vertical plane. If it is to be partially submerged in the liquid so as to maximize the exposed wetted area of the disk, show that the center of the disk should be positioned at a height above the surface of the liquid. 15. Prove that if is continuous, then . 16. The ﬁgure shows a region consisting of all points inside a square that are closer to the center than to the sides of the square. Find the area of the region. 17. Evaluate . 18. For any number , we let be the smaller of the two numbers and . Then we deﬁne . Find the maximum and minimum values of if . !2 ' c ' 2 t"c# t"c# ! x1 0 fc"x# dx "x ! c ! 2#2 "x ! c#2 fc"x# c lim n l + ' 1 sn sn " 1 " 1 sn sn " 2 " - - - " 1 sn sn " n ( y x 0 f "u#"x ! u# du ! y x 0 'y u 0 f "t# dt( du f r!s1 " & 2 r nth P"x# ! 0 a " b 2 " c 3 " d 4 ! 0 P"x# ! a " bx " cx 2 " dx 3 d 2 dx 2 y x 0 'y sin t 1 s1 " u4 du(dt 0 ' a b b a xb a .x/ dx n xn 0 .x/ dx , 10000 i!1 si xb a "2 " x ! x 2# dx $a, b% lim t l 0" A"t#!B"t# "t, 0# P O B"t# t 0 y ! sin"x 2# A"t# lim x l 0 1 x y x 0 "1 ! tan 2t#1!t dt f #"x# f "x# ! xx 0 x 2 sin"t 2# dt f #"&!2# t"x# ! y cos x 0 $1 " sin"t 2#% dt f "x# ! y t"x# 0 1 s1 " t 3 dt x c $ 0 f "x# ! "2cx ! x 2#!c 3 f x xx 0 f "t# dt ! $ f "x#%2 0 f "x# f a $ 0 x ! a " 1.5 x ! a y ! x " 1!x f "4# f x sin &x ! y x2 0 f "t# dt PROBLEMS P R O B L E M S P L U S O y x t y=sin{≈} A(t) P{t, sin(t@)} O y x t B(t) P{t, sin(t@)} FIGURE FOR PROBLEM 8 FIGURE FOR PROBLEM 16 2 2 2 2 414 In this chapter we explore some of the applications of the deﬁnite integral by using it to compute areas between curves, volumes of solids, and the work done by a varying force. The common theme is the following general method, which is similar to the one we used to ﬁnd areas under curves: We break up a quantity into a large number of small parts. We next approximate each small part by a quantity of the form and thus approximate by a Riemann sum. Then we take the limit and express as an integral. Finally we evaluate the integral using the Fundamental Theorem of Calculus or the Midpoint Rule. Q Q f xi x Q The volume of a sphere is the limit of sums of volumes of approximating cylinders. APPLICATIONS OF INTEGRATION 6 0 y=© y=ƒ S FIGURE 1 S=s(x, y) \| a¯x¯b, ©¯y¯ƒd x y b a AREAS BETWEEN CURVES In Chapter 5 we deﬁned and calculated areas of regions that lie under the graphs of functions. Here we use integrals to ﬁnd areas of regions that lie between the graphs of two functions. Consider the region that lies between two curves and and be-tween the vertical lines and , where and are continuous functions and for all in . (See Figure 1.) Just as we did for areas under curves in Section 5.1, we divide S into n strips of equal width and then we approximate the ith strip by a rectangle with base and height . (See Figure 2. If we like, we could take all of the sample points to be right endpoints, in which case .) The Riemann sum is therefore an approximation to what we intuitively think of as the area of S. This approximation appears to become better and better as . Therefore we deﬁne the area of the region as the limiting value of the sum of the areas of these approxi-mating rectangles. We recognize the limit in (1) as the deﬁnite integral of . Therefore we have the fol-lowing formula for area. The area A of the region bounded by the curves , and the lines , , where and are continuous and for all in , is Notice that in the special case where , is the region under the graph of and our general deﬁnition of area (1) reduces to our previous deﬁnition (Deﬁnition 2 in Section 5.1). f S t x 0 A y b a f x tx dx a, b x f x t x t f x b x a y f x, y tx 2 f t A lim n l n i1 f xi t xi x 1 S A n l (a) Typical rectangle x y b 0 a f(xi ) f(xi )-g(xi ) _g(xi ) xi Îx (b) Approximating rectangles x y b 0 a FIGURE 2 n i1 f xi t xi x xi xi f xi t xi x a, b x f x t x t f x b x a y t x y f x S 6.1 415 In the case where both and are positive, you can see from Figure 3 why (2) is true: EXAMPLE 1 Find the area of the region bounded above by , bounded below by , and bounded on the sides by x 0 and x 1. SOLUTION The region is shown in Figure 4. The upper boundary curve is and the lower boundary curve is . So we use the area formula (2) with , , and : M In Figure 4 we drew a typical approximating rectangle with width as a reminder of the procedure by which the area is deﬁned in (1). In general, when we set up an integral for an area, it’s helpful to sketch the region to identify the top curve , the bottom curve , and a typical approximating rectangle as in Figure 5. Then the area of a typical rect-angle is and the equation summarizes the procedure of adding (in a limiting sense) the areas of all the typical rectangles. Notice that in Figure 5 the left-hand boundary reduces to a point, whereas in Figure 3 the right-hand boundary reduces to a point. In the next example both of the side bound-aries reduce to a point, so the ﬁrst step is to ﬁnd a and b. EXAMPLE 2 Find the area of the region enclosed by the parabolas and . SOLUTION We ﬁrst ﬁnd the points of intersection of the parabolas by solving their equa-tions simultaneously. This gives , or . Thus , so or 1. The points of intersection are and . We see from Figure 6 that the top and bottom boundaries are and The area of a typical rectangle is and the region lies between and . So the total area is M 2 x 2 2 x 3 3 0 1 A y 1 0 2x 2x 2 dx 2 y 1 0 x x 2 dx x 1 x 0 yT yB x 2x x 2 x 2 x yB x 2 yT 2x x 2 1, 1 0, 0 x 0 2xx 1 0 2x 2 2x 0 x 2 2x x 2 y 2x x 2 y x 2 V A lim n l n i1 yT yB x y b a yT yB dx yT yB x yB yT x e 1 2 1 e 1.5 A y 1 0 e x x dx e x 1 2 x 2] 1 0 b 1 a 0, t x x f x e x y x y e x y x y e x y b a f x dx y b a tx dx y b a f x tx dx A area under y f x area under y tx t f 416 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION 0 x y a b yT yB yT-yB Îx FIGURE 5 FIGURE 3 A=j ƒ dx-j © dx a j b a j b 0 x y a b y=ƒ y=© S 0 x y 1 y=´ y=x Îx x=1 1 FIGURE 4 Îx (0, 0) (1, 1) FIGURE 6 yT=2x-≈ yB=≈ x y Sometimes it’s difﬁcult, or even impossible, to ﬁnd the points of intersection of two curves exactly. As shown in the following example, we can use a graphing calculator or computer to ﬁnd approximate values for the intersection points and then proceed as before. EXAMPLE 3 Find the approximate area of the region bounded by the curves and SOLUTION If we were to try to ﬁnd the exact intersection points, we would have to solve the equation This looks like a very difﬁcult equation to solve exactly (in fact, it’s impossible), so instead we use a graphing device to draw the graphs of the two curves in Figure 7. One intersection point is the origin. We zoom in toward the other point of intersection and ﬁnd that . (If greater accuracy is required, we could use Newton’s method or a rootﬁnder, if available on our graphing device.) Thus an approximation to the area between the curves is To integrate the ﬁrst term we use the subsitution . Then , and when . So M EXAMPLE 4 Figure 8 shows velocity curves for two cars, A and B, that start side by side and move along the same road. What does the area between the curves represent? Use the Midpoint Rule to estimate it. SOLUTION We know from Section 5.4 that the area under the velocity curve A represents the distance traveled by car A during the ﬁrst 16 seconds. Similarly, the area under curve B is the distance traveled by car B during that time period. So the area between these curves, which is the difference of the areas under the curves, is the distance between the cars after 16 seconds. We read the velocities from the graph and convert them to feet per second . 1 mi h 5280 3600 ft s 0.785 s2.39 1 1.185 5 1.182 2 su ]1 2.39 x 5 5 x 2 2 0 1.18 A 1 2 y 2.39 1 du su y 1.18 0 x 4 x dx x 1.18, we have u 2.39 du 2x dx u x 2 1 A y 1.18 0 x sx 2 1 x 4 x dx x 1.18 x sx 2 1 x 4 x y x 4 x. y x sx 2 1 SECTION 6.1 AREAS BETWEEN CURVES \|\|\|\| 417 1.5 _1 _1 2 y=x$-x x œ„„„„„ ≈+1 FIGURE 7 y= t 0 2 4 6 8 10 12 14 16 0 34 54 67 76 84 89 92 95 0 21 34 44 51 56 60 63 65 0 13 20 23 25 28 29 29 30 vA vB vB vA FIGURE 8 0 10 20 30 40 50 60 A B 2 4 6 8 10 12 14 16 t (seconds) √ (mi/h) We use the Midpoint Rule with intervals, so that . The midpoints of the intervals are , , , and . We estimate the distance between the cars after 16 seconds as follows: M If we are asked to ﬁnd the area between the curves and where for some values of but for other values of , then we split the given region into several regions , , . . . with areas , , . . . as shown in Figure 9. We then deﬁne the area of the region to be the sum of the areas of the smaller regions , , . . . , that is, . Since we have the following expression for A. The area between the curves and and between and is When evaluating the integral in (3), however, we must still split it into integrals corre-sponding to , , . . . . EXAMPLE 5 Find the area of the region bounded by the curves , , , and . SOLUTION The points of intersection occur when , that is, when (since ). The region is sketched in Figure 10. Observe that when but when . Therefore the required area is In this particular example we could have saved some work by noticing that the region is symmetric about and so M A 2A1 2 y 4 0 cos x sin x dx x 4 2s2 2 1 s2 1 s2 0 1 0 1 1 s2 1 s2 [sin x cos x]0 4 [cos x sin x] 4 2 y 4 0 cos x sin x dx y 2 4 sin x cos x dx A y 2 0 cos x sin x dx A1 A2 4 x 2 sin x cos x 0 x 4 cos x sin x 0 x 2 x 4 sin x cos x x 2 x 0 y cos x y sin x V A2 A1 A y b a f x tx dx x b x a y tx y f x 3 f x tx f x tx tx f x when f x tx when tx f x A A1 A2 S2 S1 S A2 A1 S2 S1 S x t x f x x f x t x y t x y f x 493 372 ft y 16 0 vA vB dt t 13 23 28 29 t4 14 t3 10 t2 6 t1 2 t 4 n 4 418 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION 0 x y a b y=ƒ y=© S¡ S™ S£ FIGURE 9 FIGURE 10 0 x y x=0 A¡ y =cos x y=sin x A™ π 4 π 2 x=π 2 Some regions are best treated by regarding x as a function of y. If a region is bounded by curves with equations , , , and , where and are contin-uous and for (see Figure 11), then its area is If we write for the right boundary and for the left boundary, then, as Figure 12 illustrates, we have Here a typical approximating rectangle has dimensions and . EXAMPLE 6 Find the area enclosed by the line and the parabola . SOLUTION By solving the two equations we ﬁnd that the points of intersection are and . We solve the equation of the parabola for x and notice from Figure 13 that the left and right boundary curves are We must integrate between the appropriate -values, and . Thus M We could have found the area in Example 6 by integrating with respect to x instead of y, but the calculation is much more involved. It would have meant splitting the region in two and computing the areas labeled and in Figure 14. The method we used in Example 6 is much easier. A2 A1 1 664 8 16 ( 4 3 2 8) 18 1 2 y 3 3 y 2 2 4y 2 4 y 4 2 ( 1 2 y 2 y 4) dy y 4 2 [y 1 ( 1 2 y 2 3)] dy A y 4 2 xR xL dy y 4 y 2 y xR y 1 xL 1 2 y 2 3 5, 4 1, 2 y 2 2x 6 y x 1 V y xR xL A y d c xR xL dy xL xR x c d y 0 y=d x=g(y) x=f(y) y=c Îy FIGURE 11 0 x y c d xR xL xR-xL Îy FIGURE 12 A y d c f y ty dy c y d f y ty t f y d y c x ty x f y SECTION 6.1 AREAS BETWEEN CURVES \|\|\|\| 419 x y _2 4 0 (_1, _2) (5, 4) xR=y+1 1 2 xL= ¥-3 FIGURE 13 3 (5, 4) (_1, _2) y=x-1 A¡ y= 2x+6 A™ y= 2x+6 œ„„„„„ œ„„„„„ FIGURE 14 0 x y 420 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION , 22. , 23. , , , 24. , , 25. 26. 27. , , , 28. , , , 29–30 Use calculus to ﬁnd the area of the triangle with the given vertices. , , 30. , , 31–32 Evaluate the integral and interpret it as the area of a region. Sketch the region. 31. 32. 33–34 Use the Midpoint Rule with to approximate the area of the region bounded by the given curves. 33. , , 34. , , ; 35–38 Use a graph to ﬁnd approximate -coordinates of the points of intersection of the given curves. Then ﬁnd (approximately) the area of the region bounded by the curves. 35. , 36. 37. , 38. , y x 10 y x cos x y x 3 3x 4 y 3x 2 2x y e x, y 2 x 2 y x 4 y x sinx 2 x x 0 y x y s 3 16 x 3 0 x 1 y cos2x 4 y sin2x 4 n 4 y 4 0 sx 2 x dx y 2 0 sin x cos 2x dx 5, 1 2, 2 0, 5 1, 6 2, 1 0, 0 29. x 0 4x y 4 y 8x 2 y 3x 2 x 0 y 1 4x y x y 1 x y x , y x2 2 y x 2, y 2 x 2 1 0 x y 1 cos x y cos x x 2 x 0 y sin 2x y cos x y x y sinx 2 x y 2 1 x 1 y 2 21. 1–4 Find the area of the shaded region. 1. 2. 4. 5–28 Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approx-imating rectangle and label its height and width. Then ﬁnd the area of the region. 5. 6. 7. , 8. 10. 11. , 12. , , 14. , , 15. , , 16. 17. , , 18. , , , 19. , 20. , x y 4x y2 12 x 4 y 2 x 2y 2 x 3 x 3 y x 2 y 8 x 2 x 9 y 1 2x y sx y x 3 x, y 3x 3 x 3 y 2 sin x y tan x 0 x 2 y 2 cos x y cos x y x 2 6 y 12 x 2 13. y 4x x 2 y x 2 y 2 x y x 2 y 1 sx , y 3 x 3 y 1 x, y 1 x 2, x 2 9. y x 2 2x, y x 4 y x 2 y x y sin x, y e x, x 0, x 2 y x 1, y 9 x 2, x 1, x 2 x y (3, 3) x=2y-¥ x=¥-4y x x=¥-2 x=ey y=1 y=_1 y 3. x=2 y=œ„„„„ x+2 y= 1 x+1 x y y=x y=5x-≈ x y (4, 4) EXERCISES 6.1 SECTION 6.1 AREAS BETWEEN CURVES \|\|\|\| 421 (c) Which car is ahead after two minutes? Explain. (d) Estimate the time at which the cars are again side by side. 46. The ﬁgure shows graphs of the marginal revenue function and the marginal cost function for a manufacturer. [Recall from Section 4.7 that and represent the revenue and cost when units are manufactured. Assume that and are measured in thousands of dollars.] What is the meaning of the area of the shaded region? Use the Midpoint Rule to estimate the value of this quantity. ; 47. The curve with equation is called Tschirn-hausen’s cubic. If you graph this curve you will see that part of the curve forms a loop. Find the area enclosed by the loop. 48. Find the area of the region bounded by the parabola , the tangent line to this parabola at , and the -axis. 49. Find the number such that the line divides the region bounded by the curves and into two regions with equal area. 50. (a) Find the number such that the line bisects the area under the curve , (b) Find the number such that the line bisects the area in part (a). Find the values of such that the area of the region bounded by the parabolas and is 576. 52. Suppose that . For what value of is the area of the region enclosed by the curves , , and equal to the area of the region enclosed by the curves , , and ? For what values of do the line and the curve enclose a region? Find the area of the region. y x x 2 1 y mx m 53. y 0 x y cosx c x 0 y cosx c y cos x c 0 c 2 y c 2 x 2 y x 2 c 2 c 51. y b b 1 x 4. y 1 x 2 x a a y 4 y x 2 y b b x 1, 1 y x 2 y 2 x 2x 3 Cª(x) y x 0 100 50 1 2 3 Rª(x) C R x Cx Rx C R 0 A B 2 1 √ t (min) 39. Use a computer algebra system to ﬁnd the exact area enclosed by the curves and . 40. Sketch the region in the -plane deﬁned by the inequalities , and ﬁnd its area. 41. Racing cars driven by Chris and Kelly are side by side at the start of a race. The table shows the velocities of each car (in miles per hour) during the ﬁrst ten seconds of the race. Use the Midpoint Rule to estimate how much farther Kelly travels than Chris does during the ﬁrst ten seconds. 42. The widths (in meters) of a kidney-shaped swimming pool were measured at 2-meter intervals as indicated in the ﬁgure. Use the Midpoint Rule to estimate the area of the pool. 43. A cross-section of an airplane wing is shown. Measurements of the height of the wing, in centimeters, at 20-centimeter intervals are , , , , , , , , , , and . Use the Midpoint Rule to estimate the area of the wing’s cross-section. 44. If the birth rate of a population is people per year and the death rate is people per year, ﬁnd the area between these curves for . What does this area represent? Two cars, A and B, start side by side and accelerate from rest. The ﬁgure shows the graphs of their velocity functions. (a) Which car is ahead after one minute? Explain. (b) What is the meaning of the area of the shaded region? 45. 0 t 10 dt 1460e0.018t bt 2200e0.024t 200 cm 2.8 8.7 15.1 20.5 23.8 27.3 27.6 29.0 26.7 20.3 5.8 6.2 5.0 7 .2 6.8 5.6 4.8 4.8 1 x y 0 x 2y 2 0 xy y x y x5 6x3 4x CAS t t 0 0 0 6 69 80 1 20 22 7 75 86 2 32 37 8 81 93 3 46 52 9 86 98 4 54 61 10 90 102 5 62 71 vK vC vK vC VOLUMES In trying to ﬁnd the volume of a solid we face the same type of problem as in ﬁnding areas. We have an intuitive idea of what volume means, but we must make this idea precise by using calculus to give an exact deﬁnition of volume. We start with a simple type of solid called a cylinder (or, more precisely, a right cylin-der). As illustrated in Figure 1(a), a cylinder is bounded by a plane region , called the base, and a congruent region in a parallel plane. The cylinder consists of all points on line segments that are perpendicular to the base and join to . If the area of the base is and the height of the cylinder (the distance from to ) is , then the volume of the cylinder is deﬁned as In particular, if the base is a circle with radius , then the cylinder is a circular cylinder with volume [see Figure 1(b)], and if the base is a rectangle with length and width , then the cylinder is a rectangular box (also called a rectangular parallelepiped) with volume [see Figure 1(c)]. For a solid S that isn’t a cylinder we ﬁrst “cut” S into pieces and approximate each piece by a cylinder. We estimate the volume of S by adding the volumes of the cylinders. We arrive at the exact volume of S through a limiting process in which the number of pieces becomes large. We start by intersecting S with a plane and obtaining a plane region that is called a cross-section of Let be the area of the cross-section of in a plane perpen-dicular to the -axis and passing through the point , where . (See Figure 2. Think of slicing with a knife through and computing the area of this slice.) The cross-sectional area will vary as increases from to . FIGURE 2 y x 0 a b x A(b) A Px P b a x Ax x S a x b x x P x S Ax S. FIGURE 1 h B¡ B™ h r h l (a) Cylinder V=Ah (b) Circular cylinder V=πr@h (c) Rectangular box V=lwh w V lwh w l V r 2h r V Ah V h B2 B1 A B2 B1 B2 B1 6.2 422 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION Let’s divide S into n “slabs” of equal width by using the planes , , . . . to slice the solid. (Think of slicing a loaf of bread.) If we choose sample points in , we can approximate the th slab (the part of that lies between the planes and ) by a cylinder with base area and “height” . (See Figure 3.) FIGURE 3 The volume of this cylinder is , so an approximation to our intuitive concep-tion of the volume of the th slab is Adding the volumes of these slabs, we get an approximation to the total volume (that is, what we think of intuitively as the volume): This approximation appears to become better and better as . (Think of the slices as becoming thinner and thinner.) Therefore, we deﬁne the volume as the limit of these sums as . But we recognize the limit of Riemann sums as a deﬁnite integral and so we have the following deﬁnition. DEFINITION OF VOLUME Let be a solid that lies between and . If the cross-sectional area of in the plane , through x and perpendicular to the x-axis, is , where is a continuous function, then the volume of is When we use the volume formula , it is important to remember that is the area of a moving cross-section obtained by slicing through perpendicular to the -axis. Notice that, for a cylinder, the cross-sectional area is constant: for all . So our deﬁnition of volume gives ; this agrees with the formula EXAMPLE 1 Show that the volume of a sphere of radius is . SOLUTION If we place the sphere so that its center is at the origin (see Figure 4), then the plane intersects the sphere in a circle whose radius (from the Pythagorean Theorem) P x V 4 3r 3 r V Ah. V xb a A dx Ab a x Ax A x x Ax V xb a Ax dx V lim n l n i1 Axi x y b a Ax dx S A Ax P x S x b x a S n l n l V n i1 Axi x VSi Axi x Si i Axi x y 0 x x¶=b a=x¸ ⁄ ¤ ‹ x¢ x x∞ xß xi-1 xi y 0 x x i Îx S a b x Axi P xi P xi1 S Si i xi1, xi xi P x2 P x1 x SECTION 6.2 VOLUMES \|\|\|\| 423 FIGURE 4 y _r r x N It can be proved that this deﬁnition is inde-pendent of how is situated with respect to the -axis. In other words, no matter how we slice with parallel planes, we always get the same answer for . V S x S is . So the cross-sectional area is Using the deﬁnition of volume with and , we have (The integrand is even.) M Figure 5 illustrates the deﬁnition of volume when the solid is a sphere with radius . From the result of Example 1, we know that the volume of the sphere is . Here the slabs are circular cylinders, or disks, and the three parts of Fig-ure 5 show the geometric interpretations of the Riemann sums when n 5, 10, and 20 if we choose the sample points to be the midpoints . Notice that as we increase the number of approximating cylinders, the corresponding Riemann sums become closer to the true volume. EXAMPLE 2 Find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to 1. Illustrate the deﬁnition of volume by sketch-ing a typical approximating cylinder. SOLUTION The region is shown in Figure 6(a). If we rotate about the x-axis, we get the solid shown in Figure 6(b). When we slice through the point x, we get a disk with radius . The area of this cross-section is and the volume of the approximating cylinder (a disk with thickness ) is Ax x x x x Ax (sx ) 2 x sx y sx V (a) Using 5 disks, VÅ4.2726 (b) Using 10 disks, VÅ4.2097 (c) Using 20 disks, VÅ4.1940 FIGURE 5 Approximating the volume of a sphere with radius 1 xi xi n i1 Axi x n i1 12 xi 2 x 4 3 4.18879 r 1 4 3r 3 2r 2x x 3 3 0 r 2r 3 r 3 3 2 y r 0 r 2 x 2 dx V y r r Ax dx y r r r 2 x 2 dx b r a r Ax y 2 r 2 x 2 y sr 2 x 2 424 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION Visual 6.2A shows an animation of Figure 5. TEC The solid lies between and , so its volume is M EXAMPLE 3 Find the volume of the solid obtained by rotating the region bounded by , , and about the -axis. SOLUTION The region is shown in Figure 7(a) and the resulting solid is shown in Figure 7(b). Because the region is rotated about the y-axis, it makes sense to slice the solid perpendicular to the y-axis and therefore to integrate with respect to y. If we slice at height y, we get a circular disk with radius x, where . So the area of a cross-section through y is and the volume of the approximating cylinder pictured in Figure 7(b) is Since the solid lies between y 0 and y 8, its volume is M FIGURE 7 y=8 x=0 y=˛ or (a) 0 (x, y) Îy (b) x y 0 x y 8 x=œ„ y 3 [ 3 5 y 5 3]0 8 96 5 V y 8 0 Ay dy y 8 0 y 2 3 dy Ay y y 2 3 y Ay x 2 (s 3 y )2 y 2 3 x s 3 y y x 0 y 8 y x 3 V FIGURE 6 (a) x 0 x y y=œ„ œ 1 œ„ œ (b) Îx 0 x y 1 V y 1 0 Ax dx y 1 0 x dx x 2 2 0 1 2 x 1 x 0 SECTION 6.2 VOLUMES \|\|\|\| 425 N Did we get a reasonable answer in Example 2? As a check on our work, let’s replace the given region by a square with base and height . If we rotate this square, we get a cylinder with radius , height , and volume . We computed that the given solid has half this volume. That seems about right. 12 1 1 1 1 0, 1 EXAMPLE 4 The region enclosed by the curves and is rotated about the -axis. Find the volume of the resulting solid. SOLUTION The curves and intersect at the points and . The region between them, the solid of rotation, and a cross-section perpendicular to the -axis are shown in Figure 8. A cross-section in the plane has the shape of a washer (an annular ring) with inner radius and outer radius , so we ﬁnd the cross-sectional area by sub-tracting the area of the inner circle from the area of the outer circle: Therefore we have M EXAMPLE 5 Find the volume of the solid obtained by rotating the region in Example 4 about the line . SOLUTION The solid and a cross-section are shown in Figure 9. Again the cross-section is a washer, but this time the inner radius is and the outer radius is . FIGURE 9 0 y=2 y=2 4 2 x 1 x y=≈ y=x y x x 2-≈ ≈ 2-x x 2 x 2 2 x y 2 FIGURE 8 (1, 1) y=≈ y=x (b) x y 0 (a) (c) x ≈ A(x) x y (0, 0) x 3 3 x 5 5 0 1 2 15 V y 1 0 Ax dx y 1 0 x 2 x 4 dx Ax x 2 x 22 x 2 x 4 x x 2 P x x 1, 1 0, 0 y x 2 y x x y x 2 y x 426 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION Visual 6.2B shows how solids of revolution are formed. TEC The cross-sectional area is and so the volume of is M The solids in Examples 1–5 are all called solids of revolution because they are obtained by revolving a region about a line. In general, we calculate the volume of a solid of revo-lution by using the basic deﬁning formula and we ﬁnd the cross-sectional area or in one of the following ways: N If the cross-section is a disk (as in Examples 1–3), we ﬁnd the radius of the disk (in terms of x or y) and use N If the cross-section is a washer (as in Examples 4 and 5), we ﬁnd the inner radius and outer radius from a sketch (as in Figures 8, 9, and 10) and compute the area of the washer by subtracting the area of the inner disk from the area of the outer disk: The next example gives a further illustration of the procedure. FIGURE 10 rin rout A outer radius2 inner radius2 rout rin A radius2 Ay Ax V y b a Ax dx or V y d c Ay dy 8 15 x 5 5 5 x 3 3 4 x 2 2 0 1 y 1 0 x 4 5x 2 4x dx y 1 0 2 x 22 2 x2 dx V y 1 0 Ax dx S Ax 2 x 22 2 x2 SECTION 6.2 VOLUMES \|\|\|\| 427 EXAMPLE 6 Find the volume of the solid obtained by rotating the region in Example 4 about the line . SOLUTION Figure 11 shows a horizontal cross-section. It is a washer with inner radius and outer radius , so the cross-sectional area is The volume is M We now ﬁnd the volumes of three solids that are not solids of revolution. EXAMPLE 7 Figure 12 shows a solid with a circular base of radius 1. Parallel cross-sections perpendicular to the base are equilateral triangles. Find the volume of the solid. SOLUTION Let’s take the circle to be . The solid, its base, and a typical cross-section at a distance from the origin are shown in Figure 13. FIGURE 13 y y 60° 60° B A C œ„ œ3y (c) A cross-section A B(x, y) y=œ„„„„„„ ≈ (b) Its base x y 0 y x (a) The solid 0 A B 1 _1 x y C FIGURE 12 Computer-generated picture of the solid in Example 7 y x x x 2 y 2 1 FIGURE 11 x=_1 y y x 0 x=œ„ y y x=y y 1 1+y 1+œ„ 4y 3 2 3 y 2 2 y 3 3 0 1 2 y 1 0 (2sy y y 2) dy y 1 0 [(1 sy )2 1 y2] dy V y 1 0 Ay dy (1 sy )2 1 y2 Ay outer radius2 inner radius2 1 sy 1 y x 1 428 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION Visual 6.2C shows how the solid in Figure 12 is generated. TEC Since lies on the circle, we have and so the base of the triangle is . Since the triangle is equilateral, we see from Figure 13(c) that its height is . The cross-sectional area is therefore and the volume of the solid is M EXAMPLE 8 Find the volume of a pyramid whose base is a square with side and whose height is . SOLUTION We place the origin at the vertex of the pyramid and the -axis along its cen-tral axis as in Figure 14. Any plane that passes through and is perpendicular to the -axis intersects the pyramid in a square with side of length , say. We can express in terms of by observing from the similar triangles in Figure 15 that and so . [Another method is to observe that the line has slope and so its equation is .] Thus the cross-sectional area is The pyramid lies between and , so its volume is M We didn’t need to place the vertex of the pyramid at the origin in Example 8. We did so merely to make the equations simple. If, instead, we had placed the center of the base at the origin and the vertex on the positive -axis, as in Figure 16, you can verify that y NOTE L2 h 2 x 3 3 0 h L2h 3 V y h 0 Ax dx y h 0 L2 h 2 x 2 dx x h x 0 O x h FIGURE 14 s L O P FIGURE 15 x y x y x h Ax s 2 L2 h 2 x 2 y Lx 2h L 2h OP s Lx h x h s 2 L 2 s L x s s x x Px x O h L V 2 y 1 0 s3 1 x 2 dx 2s3 x x 3 3 0 1 4s3 3 V y 1 1 Ax dx y 1 1 s3 1 x 2 dx Ax 1 2 2s1 x 2 s3 s1 x 2 s3 1 x 2 s3 y s3 s1 x 2 AB 2s1 x 2 ABC y s1 x 2 B SECTION 6.2 VOLUMES \|\|\|\| 429 h 0 y FIGURE 16 x y we would have obtained the integral EXAMPLE 9 A wedge is cut out of a circular cylinder of radius 4 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the ﬁrst at an angle of 30 along a diameter of the cylinder. Find the volume of the wedge. SOLUTION If we place the -axis along the diameter where the planes meet, then the base of the solid is a semicircle with equation , . A cross-section perpendicular to the -axis at a distance from the origin is a triangle , as shown in Figure 17, whose base is and whose height is . Thus the cross-sectional area is and the volume is For another method see Exercise 64. M 128 3s3 1 s3 y 4 0 16 x 2 dx 1 s3 16x x 3 3 0 4 V y 4 4 Ax dx y 4 4 16 x 2 2s3 dx 16 x 2 2s3 Ax 1 2s16 x 2 1 s3 s16 x 2 BC y tan 30 s16 x 2 s3 y s16 x 2 ABC x x 4 x 4 y s16 x 2 x V y h 0 L2 h 2 h y2 dy L2h 3 430 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION y=œ„„„„„„ 16-≈ x y 0 A B C 4 FIGURE 17 A B C y 30° 10. , , ; about the -axis , ; about 12. , , ; about 13. , ; about 14. , , , ; about 15. , ; about 16. , ; about 17. , ; about 18. , , , ; about x 1 x 4 x 2 y 0 y x x 1 x y 2 y x 2 x 2 y sx y x x 1 x 1 x y 2 y 1 x 3 x 1 y 0 y 1 x y 1 y 3 y 1 sec x y 2 x 2 y 1 y ex y 1 y sx y x 11. y y 0 x 2 y 1 4x 2 1–18 Find the volume of the solid obtained by rotating the region bounded by the given curves about the speciﬁed line. Sketch the region, the solid, and a typical disk or washer. 1. , , , ; about the -axis 2. , ; about the -axis 3. , , , ; about the -axis 4. , , , ; about the -axis 5. , , ; about the -axis 6. , , , ; about the -axis , , ; about the -axis 8. , ; about the -axis , ; about the -axis y x 2y y 2 x 9. x y 5 x 2 y 1 4x 2 x x 0 y x y x 3 7. y x 0 y 2 y 1 y ln x y y 9 x 0 x 2sy x x 4 x 2 y 0 y s25 x 2 x y 0 x 2 x 1 y 1 x x y 0 y 1 x 2 x x 2 x 1 y 0 y 2 1 2x EXERCISES 6.2 44. 45. A CAT scan produces equally spaced cross-sectional views of a human organ that provide information about the organ other-wise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas, in square centimeters, are 0, 18, 58, 79, 94, 106, 117, 128, 63, 39, and 0. Use the Midpoint Rule to estimate the volume of the liver. 46. A log 10 m long is cut at 1-meter intervals and its cross-sectional areas (at a distance from the end of the log) are listed in the table. Use the Midpoint Rule with to esti-mate the volume of the log. 47. (a) If the region shown in the ﬁgure is rotated about the -axis to form a solid, use the Midpoint Rule with to estimate the volume of the solid. (b) Estimate the volume if the region is rotated about the -axis. Again use the Midpoint Rule with . 48. (a) A model for the shape of a bird’s egg is obtained by rotating about the -axis the region under the graph of Use a CAS to ﬁnd the volume of such an egg. (b) For a Red-throated Loon, , , , and . Graph and ﬁnd the volume of an egg of this species. 49–61 Find the volume of the described solid . A right circular cone with height and base radius 50. A frustum of a right circular cone with height , lower base radius , and top radius R h r r R h r h 49. S f d 0.54 c 0.1 b 0.04 a 0.06 f x ax 3 bx 2 cx ds1 x 2 x CAS n 4 y 0 4 4 10 2 8 6 2 y x n 4 x n 5 x A y 2 0 1 cos x2 12 dx y 1 0 y 4 y 8 dy 43. 19–30 Refer to the ﬁgure and ﬁnd the volume generated by rotating the given region about the speciﬁed line. 19. about 20. about 21. about 22. about 23. about 24. about 25. about 26. about 27. about 28. about 29. about 30. about 31–36 Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the speciﬁed line. 31. 32. , ; about 33. , , ; about 34. , , ; about 35. , ; about 36. , , ; about ; 37–38 Use a graph to ﬁnd approximate -coordinates of the points of intersection of the given curves. Then use your calcula-tor to ﬁnd (approximately) the volume of the solid obtained by rotating about the -axis the region bounded by these curves. 37. , 38. 39–40 Use a computer algebra system to ﬁnd the exact volume of the solid obtained by rotating the region bounded by the given curves about the speciﬁed line. 39. , , ; 40. , ; 41–44 Each integral represents the volume of a solid. Describe the solid. 41. 42. y 5 2 y dy y 2 0 cos2x dx about y 3 y xe1x 2 y x about y 1 0 x y 0 y sin2x CAS y 3 sinx 2, y e x 2 e2x y x 4 x 1 y 2 x 2 cos x x x y 4 0 x 2 y 2 cos x y cos x x 2 x 3 x 2 y 2 1 y 2 0 x y sin x y 0 y 1 0 x y sin x y 0 x 10 8x y 16 y x 24 y tan3x, y 1, x 0; about y 1 BC 3 AB 3 OC 3 OA 3 BC 2 AB 2 OC 2 OA 2 BC 1 AB 1 OC 1 OA 1 O x y T™ y=˛ T£ T¡ B(1, 1) A(1, 0) y=œ„ x C(0, 1) SECTION 6.2 VOLUMES \|\|\|\| 431 x (m) A ( ) x (m) A ( ) 0 0.68 6 0.53 1 0.65 7 0.55 2 0.64 8 0.52 3 0.61 9 0.50 4 0.58 10 0.48 5 0.59 m2 m2 432 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION 62. The base of is a circular disk with radius . Parallel cross-sections perpendicular to the base are isosceles triangles with height and unequal side in the base. (a) Set up an integral for the volume of . (b) By interpreting the integral as an area, ﬁnd the volume of . (a) Set up an integral for the volume of a solid torus (the donut-shaped solid shown in the ﬁgure) with radii and . (b) By interpreting the integral as an area, ﬁnd the volume of the torus. 64. Solve Example 9 taking cross-sections to be parallel to the line of intersection of the two planes. 65. (a) Cavalieri’s Principle states that if a family of parallel planes gives equal cross-sectional areas for two solids and , then the volumes of and are equal. Prove this principle. (b) Use Cavalieri’s Principle to ﬁnd the volume of the oblique cylinder shown in the ﬁgure. 66. Find the volume common to two circular cylinders, each with radius , if the axes of the cylinders intersect at right angles. Find the volume common to two spheres, each with radius , if the center of each sphere lies on the surface of the other sphere. 68. A bowl is shaped like a hemisphere with diameter 30 cm. A ball with diameter 10 cm is placed in the bowl and water is poured into the bowl to a depth of centimeters. Find the vol-ume of water in the bowl. 69. A hole of radius is bored through a cylinder of radius at right angles to the axis of the cylinder. Set up, but do not evaluate, an integral for the volume cut out. R r r h r 67. r h r S2 S1 S2 S1 r R R r 63. S S h r S A cap of a sphere with radius and height 52. A frustum of a pyramid with square base of side , square top of side , and height What happens if ? What happens if ? 53. A pyramid with height and rectangular base with dimensions and 54. A pyramid with height and base an equilateral triangle with side (a tetrahedron) 55. A tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5 cm 56. The base of is a circular disk with radius . Parallel cross-sections perpendicular to the base are squares. The base of is an elliptical region with boundary curve . Cross-sections perpendicular to the -axis are isosceles right triangles with hypotenuse in the base. 58. The base of is the triangular region with vertices , , and . Cross-sections perpendicular to the -axis are equilateral triangles. 59. The base of is the same base as in Exercise 58, but cross-sections perpendicular to the -axis are squares. 60. The base of is the region enclosed by the parabola and the -axis. Cross-sections perpendicular to the -axis are squares. 61. The base of is the same base as in Exercise 60, but cross-sections perpendicular to the -axis are isosceles triangles with height equal to the base. x S y x y 1 x 2 S x S y 0, 1 1, 0 0, 0 S x 9x 2 4y 2 36 S 57. r S a a a a h 2b b h a 0 a b a b h a b r h h r 51. SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS \|\|\|\| 433 constant. Show that the radius of each end of the barrel is , where . (b) Show that the volume enclosed by the barrel is 72. Suppose that a region has area and lies above the -axis. When is rotated about the -axis, it sweeps out a solid with volume . When is rotated about the line (where is a positive number), it sweeps out a solid with volume . Express in terms of , , and . A k V1 V2 V2 k y k V1 x x A V 1 3h(2R2 r 2 2 5d 2) d ch 2 4 r R d 70. A hole of radius is bored through the center of a sphere of radius . Find the volume of the remaining portion of the sphere. 71. Some of the pioneers of calculus, such as Kepler and Newton, were inspired by the problem of ﬁnding the volumes of wine barrels. (In fact Kepler published a book Stereometria doliorum in 1715 devoted to methods for ﬁnding the volumes of barrels.) They often approximated the shape of the sides by parabolas. (a) A barrel with height and maximum radius is con-structed by rotating about the -axis the parabola , , where is a positive c h 2 x h 2 y R cx 2 x R h R r r VOLUMES BY CYLINDRICAL SHELLS Some volume problems are very difﬁcult to handle by the methods of the preceding sec-tion. For instance, let’s consider the problem of ﬁnding the volume of the solid obtained by rotating about the -axis the region bounded by and . (See Figure 1.) If we slice perpendicular to the y-axis, we get a washer. But to compute the inner radius and the outer radius of the washer, we would have to solve the cubic equation for x in terms of y; that’s not easy. Fortunately, there is a method, called the method of cylindrical shells, that is easier to use in such a case. Figure 2 shows a cylindrical shell with inner radius , outer radius , and height . Its volume is calculated by subtracting the volume of the inner cylinder from the volume of the outer cylinder: If we let (the thickness of the shell) and (the average radius of the shell), then this formula for the volume of a cylindrical shell becomes and it can be remembered as Now let be the solid obtained by rotating about the -axis the region bounded by [where ], and , where . (See Figure 3.) FIGURE 3 x y a b 0 y=ƒ a b x y 0 y=ƒ b a 0 x b y 0, x a, fx 0 y fx y S V [circumference][height][thickness] V 2rh r 1 r 1 2r2 r1 r r2 r1 2 r2 r1 2 hr2 r1 r2 r1r2 r1h r2 2 h r2 1 h r 2 2 r 2 1h V V2 V1 V2 V1 V h r2 r1 y 2x 2 x 3 y 0 y 2x 2 x 3 y 6.3 FIGURE 1 y x 0 2 1 y=2≈-˛ xL x =? xR x =? FIGURE 2 Îr h We divide the interval into n subintervals of equal width and let be the midpoint of the ith subinterval. If the rectangle with base and height is rotated about the y-axis, then the result is a cylindrical shell with average radius , height , and thickness (see Figure 4), so by Formula 1 its volume is Therefore an approximation to the volume of is given by the sum of the volumes of these shells: This approximation appears to become better as . But, from the deﬁnition of an inte-gral, we know that Thus the following appears plausible: The volume of the solid in Figure 3, obtained by rotating about the y-axis the region under the curve from a to b, is The argument using cylindrical shells makes Formula 2 seem reasonable, but later we will be able to prove it (see Exercise 67 in Section 7.1). The best way to remember Formula 2 is to think of a typical shell, cut and ﬂattened as in Figure 5, with radius x, circumference , height , and thickness or : This type of reasoning will be helpful in other situations, such as when we rotate about lines other than the y-axis. EXAMPLE 1 Find the volume of the solid obtained by rotating about the -axis the region bounded by and . SOLUTION From the sketch in Figure 6 we see that a typical shell has radius x, circumfer-ence , and height . So, by the shell method, the volume is fx 2x 2 x 3 2x y 0 y 2x 2 x 3 y FIGURE 5 2πx Îx ƒ y x x ƒ thickness height circumference dx fx 2x y b a dx x fx 2x where 0 a b V y b a 2xfx dx y fx 2 lim n l n i1 2 xi fxi x y b a 2xfx dx n l V n i1 Vi n i1 2 xi fxi x S V Vi 2 xi fxi x x fxi xi fxi xi1, xi xi x xi1, xi a, b 434 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION FIGURE 4 x y b y=ƒ xi – b a 0 x y xi-1 xi y=ƒ It can be veriﬁed that the shell method gives the same answer as slicing. M Comparing the solution of Example 1 with the remarks at the beginning of this section, we see that the method of cylindrical shells is much easier than the washer method for this problem. We did not have to ﬁnd the coordinates of the local maximum and we did not have to solve the equation of the curve for in terms of . However, in other examples the methods of the preceding section may be easier. EXAMPLE 2 Find the volume of the solid obtained by rotating about the -axis the region between and . SOLUTION The region and a typical shell are shown in Figure 8. We see that the shell has radius x, circumference , and height . So the volume is M As the following example shows, the shell method works just as well if we rotate about the x-axis. We simply have to draw a diagram to identify the radius and height of a shell. EXAMPLE 3 Use cylindrical shells to ﬁnd the volume of the solid obtained by rotating about the -axis the region under the curve from 0 to 1. SOLUTION This problem was solved using disks in Example 2 in Section 6.2. To use shells we relabel the curve (in the ﬁgure in that example) as in Figure 9. For rotation about the x-axis we see that a typical shell has radius y, circumference , and height . So the volume is In this problem the disk method was simpler. M 2 y 2 2 y 4 4 0 1 2 V y 1 0 2y1 y 2 dy 2 y 1 0 y y 3 dy 1 y 2 2y x y 2 y sx y sx x V 2 x 3 3 x 4 4 0 1 6 V y 1 0 2xx x 2 dx 2 y 1 0 x 2 x 3 dx x x 2 2x y x 2 y x y V y x NOTE FIGURE 7 y x 2[ 1 2 x 4 1 5 x 5]0 2 2(8 32 5 ) 16 5 V y 2 0 2x2x 2 x 3 dx 2 y 2 0 2x 3 x 4 dx SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS \|\|\|\| 435 FIGURE 6 y x 2≈-˛ x 2 N Figure 7 shows a computer-generated picture of the solid whose volume we computed in Example 1. FIGURE 9 1 y y shell radius=y shell height=1-¥ 0 x x=1 1 x= =¥ FIGURE 8 0 x y y=x y=≈ x shell height=x-≈ EXAMPLE 4 Find the volume of the solid obtained by rotating the region bounded by and about the line . SOLUTION Figure 10 shows the region and a cylindrical shell formed by rotation about the line . It has radius , circumference , and height . The volume of the given solid is M 2 x 4 4 x 3 x 2 0 1 2 V y 1 0 22 xx x 2 dx 2 y 1 0 x 3 3x 2 2x dx FIGURE 10 0 y x y=x-≈ 0 y x x 1 2 3 4 2-x x=2 x x 2 22 x 2 x x 2 x 2 y 0 y x x 2 V 436 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION 4. , , 6. , 7. , 8. Let be the volume of the solid obtained by rotating about the -axis the region bounded by and . Find both by slicing and by cylindrical shells. In both cases draw a dia-gram to explain your method. 9–14 Use the method of cylindrical shells to ﬁnd the volume of the solid obtained by rotating the region bounded by the given curves about the -axis. Sketch the region and a typical shell. 9. 10. 11. , , 12. , , 14. 15–20 Use the method of cylindrical shells to ﬁnd the volume gen-erated by rotating the region bounded by the given curves about the speciﬁed axis. Sketch the region and a typical shell. 15. , ; about x 2 y 0, x 1 y x 4 x y 3, x 4 y 12 x 2 x 1 y 22 13. x 0 x 4y 2 y 3 x 0 y 8 y x 3 x sy , x 0, y 1 x 1 y 2, x 0, y 1, y 2 x V y x 2 y sx y V y x 2 4x 7 y 4x 22 x y 3 y 3 2x x 2 y ex2, y 0, x 0, x 1 5. x 1 y 0 y x 2 1. Let be the solid obtained by rotating the region shown in the ﬁgure about the -axis. Explain why it is awkward to use slicing to ﬁnd the volume of . Sketch a typical approxi-mating shell. What are its circumference and height? Use shells to ﬁnd . 2. Let be the solid obtained by rotating the region shown in the ﬁgure about the -axis. Sketch a typical cylindrical shell and ﬁnd its circumference and height. Use shells to ﬁnd the volume of . Do you think this method is preferable to slicing? Explain. 3–7 Use the method of cylindrical shells to ﬁnd the volume gener-ated by rotating the region bounded by the given curves about the -axis. Sketch the region and a typical shell. 3. , , , x 2 x 1 y 0 y 1 x y 0 x y œ„ π y=sin{≈} S y S 0 x y 1 y=x(x-1)@ V S V y S EXERCISES 6.3 ; 33–34 Use a graph to estimate the -coordinates of the points of intersection of the given curves. Then use this information and your calculator to estimate the volume of the solid obtained by rotating about the -axis the region enclosed by these curves. 33. , 34. , 35–36 Use a computer algebra system to ﬁnd the exact volume of the solid obtained by rotating the region bounded by the given curves about the speciﬁed line. 35. , , ; about 36. , , ; about 37– 42 The region bounded by the given curves is rotated about the speciﬁed axis. Find the volume of the resulting solid by any method. 37. , ; about the -axis 38. , ; about the -axis 39. , ; about 40. , ; about ; about the -axis 42. , ; about 43–45 Use cylindrical shells to ﬁnd the volume of the solid. 43. A sphere of radius 44. The solid torus of Exercise 63 in Section 6.2 A right circular cone with height and base radius 46. Suppose you make napkin rings by drilling holes with differ-ent diameters through two wooden balls (which also have dif-ferent diameters). You discover that both napkin rings have the same height , as shown in the ﬁgure. (a) Guess which ring has more wood in it. (b) Check your guess: Use cylindrical shells to compute the volume of a napkin ring created by drilling a hole with radius through the center of a sphere of radius and express the answer in terms of . h h R r h r h 45. r y 1 x 4 x y 32 y x 2 y 12 1 41. x 2 x 0 x 1 y 4 x 1 y x 4 x y 5 x y 0 y x 2 6x 8 y y 0 y x 2 6x 8 x 1 0 x y 0 y x 3sin x x 2 0 x y sin4x y sin2x CAS y x 4 4x 1 y x 3 x 1 y sx 1 y e x y x 16. , ; about , ; about 18. , ; about 19. , , ; about 20. 21–26 Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the speciﬁed axis. 21. 22. , ; about 23. , ; about 24. about about 26. about 27. Use the Midpoint Rule with to estimate the volume obtained by rotating about the -axis the region under the curve , . 28. If the region shown in the ﬁgure is rotated about the -axis to form a solid, use the Midpoint Rule with to estimate the volume of the solid. 29–32 Each integral represents the volume of a solid. Describe the solid. 30. 31. 32. y 4 0 2 xcos x sin x dx y 1 0 23 y1 y2 dy 2 y 2 0 y 1 y 2 dy y 3 0 2x 5 dx 29. 0 x y 1 1 2 3 4 5 2 3 4 5 6 7 8 9 10 11 12 n 5 y 0 x 1 y s1 x 3 y n 5 y 5 x 2 y 2 7, x 4; y 4 x ssin y , 0 y , x 0; 25. x 2 y 1 1 x 2, y 0, x 0, x 2; x 1 y sinx 2 y x 4 x 7 y 4x x 2 y x y ln x, y 0, x 2; about the y-axis y x 2, x y 2; about y 1 y 1 x 1 y 0 y x 3 x 1 y 2 x 2 y x 2 x 1 y 3 y 4x x 2 17. x 1 y 0, x 1 y sx SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS \|\|\|\| 437 WORK The term work is used in everyday language to mean the total amount of effort required to perform a task. In physics it has a technical meaning that depends on the idea of a force. Intuitively, you can think of a force as describing a push or pull on an object—for example, a horizontal push of a book across a table or the downward pull of the earth’s gravity on a ball. In general, if an object moves along a straight line with position function , then the force F on the object (in the same direction) is deﬁned by Newton’s Second Law of Motion as the product of its mass and its acceleration: In the SI metric system, the mass is measured in kilograms (kg), the displacement in meters (m), the time in seconds (s), and the force in newtons ( ). Thus a force of 1 N acting on a mass of 1 kg produces an acceleration of 1 m s . In the US Customary system, the fundamental unit is chosen to be the unit of force, which is the pound. In the case of constant acceleration, the force is also constant and the work done is deﬁned to be the product of the force and the distance that the object moves: If is measured in newtons and in meters, then the unit for is a newton-meter, which is called a joule (J). If is measured in pounds and in feet, then the unit for is a foot-pound (ft-lb), which is about 1.36 J. EXAMPLE 1 (a) How much work is done in lifting a 1.2-kg book off the ﬂoor to put it on a desk that is 0.7 m high? Use the fact that the acceleration due to gravity is m s . (b) How much work is done in lifting a 20-lb weight 6 ft off the ground? SOLUTION (a) The force exerted is equal and opposite to that exerted by gravity, so Equation 1 gives and then Equation 2 gives the work done as (b) Here the force is given as lb, so the work done is Notice that in part (b), unlike part (a), we did not have to multiply by because we were given the weight (which is a force) and not the mass of the object. M Equation 2 deﬁnes work as long as the force is constant, but what happens if the force is variable? Let’s suppose that the object moves along the -axis in the positive direction, from to , and at each point between and a force acts on the object, where is a continuous function. We divide the interval into n subintervals with end-points and equal width . We choose a sample point in the th sub-interval . Then the force at that point is . If is large, then is small, and x n fxi xi1, xi i xi x x0, x1, . . . , xn a, b f fx b a x x b x a x t W Fd 20 6 120 ft-lb F 20 W Fd 11.760.7 8.2 J F mt 1.29.8 11.76 N 2 t 9.8 V W d F W d F work force distance W Fd 2 d F F 2 N kg m s2 F m d 2s dt 2 1 m st 6.4 438 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION since is continuous, the values of don’t change very much over the interval . In other words, is almost constant on the interval and so the work that is done in mov-ing the particle from to is approximately given by Equation 2: Thus we can approximate the total work by It seems that this approximation becomes better as we make larger. Therefore we deﬁne the work done in moving the object from a to b as the limit of this quantity as . Since the right side of (3) is a Riemann sum, we recognize its limit as being a deﬁnite inte-gral and so EXAMPLE 2 When a particle is located a distance feet from the origin, a force of pounds acts on it. How much work is done in moving it from to ? SOLUTION The work done is ft-lb. M In the next example we use a law from physics: Hooke’s Law states that the force required to maintain a spring stretched units beyond its natural length is proportional to : where is a positive constant (called the spring constant). Hooke’s Law holds provided that is not too large (see Figure 1). EXAMPLE 3 A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How much work is done in stretching the spring from 15 cm to 18 cm? SOLUTION According to Hooke’s Law, the force required to hold the spring stretched meters beyond its natural length is . When the spring is stretched from 10 cm to 15 cm, the amount stretched is cm m. This means that , so Thus and the work done in stretching the spring from 15 cm to 18 cm is M 4000.082 0.052 1.56 J W y 0.08 0.05 800x dx 800 x 2 2 0.05 0.08 fx 800x k 40 0.05 800 0.05k 40 f0.05 40 0.05 5 fx kx x V x k fx kx x x 16 2 3 W y 3 1 x 2 2x dx x 3 3 x 2 1 3 50 3 x 3 x 1 x 2 2x x W lim n l n i1 fxi x y b a fx dx 4 n l n W n i1 fxi x 3 Wi fxi x xi xi1 Wi f xi1, xi f f SECTION 6.4 WORK \|\|\|\| 439 FIGURE 1 Hooke’s Law x 0 frictionless surface x 0 x ƒ=kx (a) Natural position of spring (b) Stretched position of spring EXAMPLE 4 A 200-lb cable is 100 ft long and hangs vertically from the top of a tall building. How much work is required to lift the cable to the top of the building? SOLUTION Here we don’t have a formula for the force function, but we can use an argu-ment similar to the one that led to Deﬁnition 4. Let’s place the origin at the top of the building and the -axis pointing downward as in Figure 2. We divide the cable into small parts with length . If is a point in the such interval, then all points in the interval are lifted by approximately the same amount, namely . The cable weighs 2 pounds per foot, so the weight of the part is . Thus the work done on the part, in foot-pounds, is We get the total work done by adding all these approximations and letting the number of parts become large (so ): M EXAMPLE 5 A tank has the shape of an inverted circular cone with height 10 m and base radius 4 m. It is ﬁlled with water to a height of 8 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg m .) SOLUTION Let’s measure depths from the top of the tank by introducing a vertical coordi-nate line as in Figure 3. The water extends from a depth of 2 m to a depth of 10 m and so we divide the interval into n subintervals with endpoints and choose in the th subinterval. This divides the water into layers. The th layer is approximated by a circular cylinder with radius and height . We can compute from similar triangles, using Figure 4, as follows: Thus an approximation to the volume of the th layer of water is and so its mass is The force required to raise this layer must overcome the force of gravity and so Each particle in the layer must travel a distance of approximately . The work done to raise this layer to the top is approximately the product of the force and the distance : Wi Fixi 1570xi 10 xi 2 x xi Fi Wi xi 157010 xi 2 x Fi mit 9.816010 xi 2 x 1000 4 25 10 xi 2 x 16010 xi 2 x mi density volume Vi ri 2 x 4 25 10 xi 2 x i ri 2 510 xi ri 10 xi 4 10 ri x ri i n i xi x0, x1, . . . , xn 2, 10 3 x 2] 100 0 10,000 ft-lb W lim nl n i1 2xi x y 100 0 2x dx x l 0 distance force 2x xi 2xi x ith 2x ith xi ith xi x x V 440 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION N If we had placed the origin at the bottom of the cable and the -axis upward, we would have gotten which gives the same answer. W y 100 0 2100 x dx x 0 100 x i x Îx FIGURE 2 FIGURE 3 4 10 10-xi FIGURE 4 0 x 2 m 4 m 10 m xi ri Îx ri To ﬁnd the total work done in emptying the entire tank, we add the contributions of each of the layers and then take the limit as : M 1570( 2048 3 ) 3.4 106 J 1570 y 10 2 100x 20x 2 x 3 dx 157050x 2 20x 3 3 x 4 4 2 10 W lim n l n i1 1570xi 10 xi 2 x y 10 2 1570x10 x2 dx n l n SECTION 6.4 WORK \|\|\|\| 441 Suppose that 2 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. (a) How much work is needed to stretch the spring from 35 cm to 40 cm? (b) How far beyond its natural length will a force of 30 N keep the spring stretched? 10. If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch it 9 in. beyond its natural length? 11. A spring has natural length 20 cm. Compare the work done in stretching the spring from 20 cm to 30 cm with the work done in stretching it from 30 cm to 40 cm. How are and related? 12. If 6 J of work is needed to stretch a spring from 10 cm to 12 cm and another 10 J is needed to stretch it from 12 cm to 14 cm, what is the natural length of the spring? 13–20 Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it. A heavy rope, 50 ft long, weighs and hangs over the edge of a building 120 ft high. (a) How much work is done in pulling the rope to the top of the building? (b) How much work is done in pulling half the rope to the top of the building? 14. A chain lying on the ground is 10 m long and its mass is 80 kg. How much work is required to raise one end of the chain to a height of 6 m? 15. A cable that weighs is used to lift 800 lb of coal up a mine shaft 500 ft deep. Find the work done. 16. A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is ﬁlled with 40 lb of water and is pulled up at a rate of , but water leaks out of a hole in the bucket at a rate of . Find the work done in pulling the bucket to the top of the well. A leaky 10-kg bucket is lifted from the ground to a height of 12 m at a constant speed with a rope that weighs . Initially the bucket contains 36 kg of water, but the water 0.8 kg m 17. 0.2 lb s 2 ft s 2 lb ft 0.5 lb ft 13. W1 W2 W2 W1 9. 1. How much work is done in lifting a 40-kg sandbag to a height of 1.5 m? 2. Find the work done if a constant force of 100 lb is used to pull a cart a distance of 200 ft. 3. A particle is moved along the -axis by a force that measures pounds at a point feet from the origin. Find the work done in moving the particle from the origin to a distance of 9 ft. 4. When a particle is located a distance meters from the origin, a force of newtons acts on it. How much work is done in moving the particle from to ? Interpret your answer by considering the work done from to and from to . 5. Shown is the graph of a force function (in newtons) that increases to its maximum value and then remains constant. How much work is done by the force in moving an object a distance of 8 m? 6. The table shows values of a force function , where is measured in meters and in newtons. Use the Midpoint Rule to estimate the work done by the force in moving an object from to . A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length? 8. A spring has a natural length of 20 cm. If a 25-N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm? 7. x 20 x 4 f x x f x 0 x (m) F 10 1 20 30 2 3 4 5 6 7 8 (N) x 2 x 1.5 x 1.5 x 1 x 2 x 1 cosx 3 x x 10 1 x2 x EXERCISES 6.4 x 4 6 8 10 12 14 16 18 20 5 5.8 7.0 8.8 9.6 8.2 6.7 5.2 4.1 f x 26. Solve Exercise 22 if the tank is half full of oil that has a den-sity of . When gas expands in a cylinder with radius , the pressure at any given time is a function of the volume: . The force exerted by the gas on the piston (see the ﬁgure) is the product of the pressure and the area: . Show that the work done by the gas when the volume expands from volume to volume is 28. In a steam engine the pressure and volume of steam satisfy the equation , where is a constant. (This is true for adiabatic expansion, that is, expansion in which there is no heat transfer between the cylinder and its surroundings.) Use Exer-cise 27 to calculate the work done by the engine during a cycle when the steam starts at a pressure of 160 lb in and a volume of 100 in and expands to a volume of 800 in . 29. Newton’s Law of Gravitation states that two bodies with masses and attract each other with a force where is the distance between the bodies and is the gravi-tational constant. If one of the bodies is ﬁxed, ﬁnd the work needed to move the other from to . 30. Use Newton’s Law of Gravitation to compute the work required to launch a 1000-kg satellite vertically to an orbit 1000 km high. You may assume that the earth’s mass is kg and is concentrated at its center. Take the radius of the earth to be m and . G 6.67 1011 N m2 kg2 6.37 106 5.98 1024 r b r a G r F G m1m2 r 2 m2 m1 3 3 2 k PV 1.4 k V P x piston head W y V2 V1 P dV V2 V1 F r 2P P PV r 27. 900 kg m3 leaks at a constant rate and ﬁnishes draining just as the bucket reaches the 12 m level. How much work is done? 18. A 10-ft chain weighs 25 lb and hangs from a ceiling. Find the work done in lifting the lower end of the chain to the ceiling so that it’s level with the upper end. An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is .) 20. A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? (Use the fact that water weighs .) 21–24 A tank is full of water. Find the work required to pump the water out of the spout. In Exercises 23 and 24 use the fact that water weighs 62.5 lb ft . 21. 22. 23. 24. ; 25. Suppose that for the tank in Exercise 21 the pump breaks down after J of work has been done. What is the depth of the water remaining in the tank? 4.7 105 10 ft 12 ft 6 ft 6 ft frustum of a cone 3 ft 8 ft 3 m 1 m 2 m 3 m 8 m 3 m 3 62.5 lb ft3 1000 kg m3 19. 442 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION AVERAGE VALUE OF A FUNCTION It is easy to calculate the average value of ﬁnitely many numbers , , . . . , : But how do we compute the average temperature during a day if inﬁnitely many tempera-ture readings are possible? Figure 1 shows the graph of a temperature function , where is measured in hours and in C, and a guess at the average temperature, . In general, let’s try to compute the average value of a function , . We start by dividing the interval into equal subintervals, each with length . Then we choose points , . . . , in successive subintervals and cal-x n x 1 x b a n n a, b a x b y fx Tave T t Tt yave y1 y2 yn n yn y2 y1 6.5 0 t T Tave 5 10 15 12 6 18 24 FIGURE 1 culate the average of the numbers , . . . , : (For example, if represents a temperature function and , this means that we take temperature readings every hour and then average them.) Since , we can write and the average value becomes If we let increase, we would be computing the average value of a large number of closely spaced values. (For example, we would be averaging temperature readings taken every minute or even every second.) The limiting value is by the deﬁnition of a deﬁnite integral. Therefore we deﬁne the average value of f on the interval as EXAMPLE 1 Find the average value of the function on the interval . SOLUTION With and we have M If is the temperature at time , we might wonder if there is a speciﬁc time when the temperature is the same as the average temperature. For the temperature function graphed in Figure 1, we see that there are two such times––just before noon and just before mid-night. In general, is there a number at which the value of a function is exactly equal to the average value of the function, that is, ? The following theorem says that this is true for continuous functions. THE MEAN VALUE THEOREM FOR INTEGRALS If is continuous on , then there exists a number in such that that is, y b a fx dx fcb a fc fave 1 b a y b a fx dx a, b c a, b f fc fave f c t Tt 1 3 x x 3 3 2 1 2 fave 1 b a y b a fx dx 1 2 1 y 2 1 1 x 2 dx b 2 a 1 1, 2 fx 1 x 2 V fave 1 b a y b a fx dx a, b lim n l 1 b a n i1 fx i x 1 b a y b a fx dx n 1 b a n i1 fxi x fx1 fxn b a x 1 b a fx1 x fxn x n b a x x b a n n 24 f fx1 fxn n fx n fx1 SECTION 6.5 AVERAGE VALUE OF A FUNCTION \|\|\|\| 443 N For a positive function, we can think of this deﬁnition as saying area width average height The Mean Value Theorem for Integrals is a consequence of the Mean Value Theorem for derivatives and the Fundamental Theorem of Calculus. The proof is outlined in Exer-cise 23. The geometric interpretation of the Mean Value Theorem for Integrals is that, for posi-tive functions , there is a number such that the rectangle with base and height has the same area as the region under the graph of from to . (See Figure 2 and the more picturesque interpretation in the margin note.) EXAMPLE 2 Since is continuous on the interval , the Mean Value Theorem for Integrals says there is a number in such that In this particular case we can ﬁnd explicitly. From Example 1 we know that , so the value of c satisﬁes Therefore So in this case there happen to be two numbers in the interval that work in the Mean Value Theorem for Integrals. M Examples 1 and 2 are illustrated by Figure 3. EXAMPLE 3 Show that the average velocity of a car over a time interval is the same as the average of its velocities during the trip. SOLUTION If is the displacement of the car at time , then, by deﬁnition, the average velocity of the car over the interval is On the other hand, the average value of the velocity function on the interval is (by the Net Change Theorem) M st2 st1 t2 t1 average velocity 1 t2 t1 st2 st1 vave 1 t2 t1 y t2 t1 vt dt 1 t2 t1 y t2 t1 st dt s t st2 st1 t2 t1 t st t1, t2 V 1, 2 c 1 c 2 1 so 1 c 2 2 fc fave 2 fave 2 c y 2 1 1 x 2 dx fc2 1 1, 2 c 1, 2 fx 1 x 2 V FIGURE 2 0 x y a c b y=ƒ f(c)=fave b a f fc a, b c f 444 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION N You can always chop off the top of a (two-dimensional) mountain at a certain height and use it to ﬁll in the valleys so that the mountaintop becomes completely ﬂat. 0 1 2 _1 (_1, 2) (2, 5) y=1+≈ fave=2 FIGURE 3 x y SECTION 6.5 AVERAGE VALUE OF A FUNCTION \|\|\|\| 445 In a certain city the temperature (in F) hours after 9 AM was modeled by the function Find the average temperature during the period from 9 AM to 9 PM. 18. (a) A cup of coffee has temperature 95 C and takes 30 min-utes to cool to 61 C in a room with temperature 20 C. Use Newton’s Law of Cooling (Section 3.8) to show that the temperature of the coffee after minutes is where . (b) What is the average temperature of the coffee during the ﬁrst half hour? 19. The linear density in a rod 8 m long is , where is measured in meters from one end of the rod. Find the average density of the rod. 20. If a freely falling body starts from rest, then its displacement is given by . Let the velocity after a time be . Show that if we compute the average of the velocities with respect to we get , but if we compute the average of the velocities with respect to we get . 21. Use the result of Exercise 79 in Section 5.5 to compute the average volume of inhaled air in the lungs in one respiratory cycle. 22. The velocity of blood that ﬂows in a blood vessel with radius and length at a distance from the central axis is where is the pressure difference between the ends of the vessel and is the viscosity of the blood (see Example 7 in Section 3.7). Find the average velocity (with respect to ) over the interval . Compare the average velocity with the maximum velocity. Prove the Mean Value Theorem for Integrals by applying the Mean Value Theorem for derivatives (see Section 4.2) to the function . 24. If denotes the average value of on the interval and , show that favea, b c a b a favea, c b c b a favec, b a c b a, b f favea, b Fx xx a f t dt 23. 0 r R r P vr P 4 l R2 r 2 r l R v vave 2 3 vT s vave 1 2 vT t vT T s 1 2tt 2 x kgm 12sx 1 k 0.02 Tt 20 75ekt t Tt 50 14 sin t 12 t 17. 1–8 Find the average value of the function on the given interval. 1. 2. 3. 4. 5. 6. 8. 9–12 (a) Find the average value of on the given interval. (b) Find such that . (c) Sketch the graph of and a rectangle whose area is the same as the area under the graph of . , 10. , ; 11. , ; 12. , If is continuous and , show that takes on the value 4 at least once on the interval . 14. Find the numbers such that the average value of on the interval is equal to 3. 15. The table gives values of a continuous function. Use the Mid-point Rule to estimate the average value of on . 16. The velocity graph of an accelerating car is shown. (a) Estimate the average velocity of the car during the ﬁrst 12 seconds. (b) At what time was the instantaneous velocity equal to the average velocity? 4 t (seconds) 20 0 8 12 40 60 √ (km/h) 20, 50 f 0, b f x 2 6x 3x 2 b 1, 3 f x3 1 f x dx 8 f 13. 0, 2 f x 2x1 x 22 0, f x 2 sin x sin 2x 0, 4 f x sx 2, 5 f x x 32 9. f f fave f c c f hu 3 2u1, 1, 1 hx cos4x sin x, 0, 7. f sec22, 0, 2 f t tet 2, 0, 5 tx x 2s1 x 3 , 0, 2 tx s 3 x , 1, 8 f x sin 4x, , f x 4x x 2, 0, 4 EXERCISES 6.5 x 20 25 30 35 40 45 50 42 38 31 29 35 48 60 f x A movie theater has a screen that is positioned 10 ft off the ﬂoor and is 25 ft high. The ﬁrst row of seats is placed 9 ft from the screen and the rows are set 3 ft apart. The ﬂoor of the seating area is inclined at an angle of above the horizontal and the distance up the incline that you sit is . The theater has 21 rows of seats, so . Suppose you decide that the best place to sit is in the row where the angle subtended by the screen at your eyes is a maximum. Let’s also suppose that your eyes are 4 ft above the ﬂoor, as shown in the ﬁgure. (In Exercise 70 in Sec-tion 4.7 we looked at a simpler version of this problem, where the ﬂoor is horizontal, but this project involves a more complicated situation and requires technology.) 1. Show that where and 2. Use a graph of as a function of to estimate the value of that maximizes . In which row should you sit? What is the viewing angle in this row? 3. Use your computer algebra system to differentiate and ﬁnd a numerical value for the root of the equation . Does this value conﬁrm your result in Problem 2? 4. Use the graph of to estimate the average value of on the interval . Then use your CAS to compute the average value. Compare with the maximum and minimum values of . 0 x 60 ddx 0 x x b 2 9 x cos 2 x sin 62 a 2 9 x cos 2 31 x sin 2 arccos a 2 b 2 625 2ab 0 x 60 x 20 WHERE TO SIT AT THE MOVIES CAS A P P L I E D P R O J E C T 446 \|\|\|\| CHAPTER 6 APPLICATIONS OF INTEGRATION 25 ft 10 ft 9 ft ¨ x å 4 ft REVIEW CONCEPT CHECK 6 (b) If is a solid of revolution, how do you ﬁnd the cross-sectional areas? 4. (a) What is the volume of a cylindrical shell? (b) Explain how to use cylindrical shells to ﬁnd the volume of a solid of revolution. (c) Why might you want to use the shell method instead of slicing? 5. Suppose that you push a book across a 6-meter-long table by exerting a force at each point from to . What does represent? If is measured in newtons, what are the units for the integral? 6. (a) What is the average value of a function on an interval ? (b) What does the Mean Value Theorem for Integrals say? What is its geometric interpretation? a, b f f x x6 0 f x dx x 6 x 0 f x S 1. (a) Draw two typical curves and , where for . Show how to approximate the area between these curves by a Riemann sum and sketch the corresponding approximating rectangles. Then write an expression for the exact area. (b) Explain how the situation changes if the curves have equations and , where for . 2. Suppose that Sue runs faster than Kathy throughout a 1500-meter race. What is the physical meaning of the area between their velocity curves for the ﬁrst minute of the race? 3. (a) Suppose is a solid with known cross-sectional areas. Explain how to approximate the volume of by a Riemann sum. Then write an expression for the exact volume. S S c y d f y ty x ty x f y a x b f x tx y tx y f x 1–6 Find the area of the region bounded by the given curves. 1. 2. y 1x, y x 2, y 0, x e y x 2, y 4x x 2 3. 4. , 5. , y x 2 2x y sinx2 x y 2 3y x y 0 y 1 2x 2, y x EXERCISES CHAPTER 6 REVIEW \|\|\|\| 447 the base are isosceles right triangles with hypotenuse lying along the base. 24. The base of a solid is the region bounded by the parabolas and . Find the volume of the solid if the cross-sections perpendicular to the -axis are squares with one side lying along the base. 25. The height of a monument is 20 m. A horizontal cross-section at a distance meters from the top is an equilateral triangle with side meters. Find the volume of the monument. 26. (a) The base of a solid is a square with vertices located at , and . Each cross-section per-pendicular to the -axis is a semicircle. Find the volume of the solid. (b) Show that by cutting the solid of part (a), we can rearrange it to form a cone. Thus compute its volume more simply. 27. A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm? 28. A 1600-lb elevator is suspended by a 200-ft cable that weighs 10 lbft. How much work is required to raise the elevator from the basement to the third ﬂoor, a distance of 30 ft? 29. A tank full of water has the shape of a paraboloid of revolu-tion as shown in the ﬁgure; that is, its shape is obtained by rotating a parabola about a vertical axis. (a) If its height is 4 ft and the radius at the top is 4 ft, ﬁnd the work required to pump the water out of the tank. ; (b) After 4000 ft-lb of work has been done, what is the depth of the water remaining in the tank? 30. Find the average value of the function on the interval . 31. If is a continuous function, what is the limit as of the average value of on the interval ? 32. Let be the region bounded by , , and , where . Let be the region bounded by , , and . (a) Is there a value of such that and have the same area? (b) Is there a value of such that sweeps out the same volume when rotated about the -axis and the -axis? (c) Is there a value of such that and sweep out the same volume when rotated about the -axis? (d) Is there a value of such that and sweep out the same volume when rotated about the -axis? y 2 1 b x 2 1 b y x 1 b 2 1 b y b 2 x 0 y x 2 2 b 0 x b y 0 y x 2 1 x, x h f h l 0 f 0, 10 f t t sint2 4 ft 4 ft x 0, 1 1, 0, 0, 1, 1, 0 1 4x x x y 2 x 2 y x 2 6. , , 7–11 Find the volume of the solid obtained by rotating the region bounded by the given curves about the speciﬁed axis. 7. , ; 8. , ; 9. , ; 10. , ; 11. , (where , ); about the -axis 12–14 Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the speciﬁed axis. 12. , , ; about the -axis 13. , , ; about 14. , ; about 15. Find the volumes of the solids obtained by rotating the region bounded by the curves and about the following lines. (a) The -axis (b) The -axis (c) 16. Let be the region in the ﬁrst quadrant bounded by the curves and . Calculate the following quantities. (a) The area of (b) The volume obtained by rotating about the -axis (c) The volume obtained by rotating about the -axis 17. Let be the region bounded by the curves , and . Use the Midpoint Rule with to esti-mate the following quantities. (a) The area of (b) The volume obtained by rotating about the -axis ; 18. Let be the region bounded by the curves and . Estimate the following quantities. (a) The -coordinates of the points of intersection of the curves (b) The area of (c) The volume generated when is rotated about the -axis (d) The volume generated when is rotated about the -axis 19–22 Each integral represents the volume of a solid. Describe the solid. 19. 20. 21. 22. 23. The base of a solid is a circular disk with radius 3. Find the volume of the solid if parallel cross-sections perpendicular to y 4 0 26 y4y y 2 dy y 0 2 sin x2 dx y 2 0 2 cos2x dx y 2 0 2x cos x dx y x x y x 6 x 1 y 1 x 2 x n 4 y 0 x 1 y tanx 2, y x y 2x x 2 y x 3 y 2 y x y x 2 y x y 2 y x 2 y sx x 2 y 1 4 x 2 y cos2 x y x 3 y x y tan x y h 0 a 0 x a h x 2 y 2 a2 about y 1 y 9 x 2 y x 2 1 about x 1 x 9 y 2 x 0 about the y-axis y x 3 x 1 y 2 about the x-axis y x 2 y 2x x 2 y x 2 y sx 448 1. (a) Find a positive continuous function such that the area under the graph of from 0 to is for all . (b) A solid is generated by rotating about the -axis the region under the curve , where is a positive function and . The volume generated by the part of the curve from to is for all . Find the function . 2. There is a line through the origin that divides the region bounded by the parabola and the -axis into two regions with equal area. What is the slope of that line? 3. The ﬁgure shows a horizontal line intersecting the curve . Find the num-ber c such that the areas of the shaded regions are equal. 4. A cylindrical glass of radius and height is ﬁlled with water and then tilted until the water remaining in the glass exactly covers its base. (a) Determine a way to “slice” the water into parallel rectangular cross-sections and then set up a deﬁnite integral for the volume of the water in the glass. (b) Determine a way to “slice” the water into parallel cross-sections that are trapezoids and then set up a deﬁnite integral for the volume of the water. (c) Find the volume of water in the glass by evaluating one of the integrals in part (a) or part (b). (d) Find the volume of the water in the glass from purely geometric considerations. (e) Suppose the glass is tilted until the water exactly covers half the base. In what direction can you “slice” the water into triangular cross-sections? Rectangular cross-sections? Cross-sections that are segments of circles? Find the volume of water in the glass. 5. (a) Show that the volume of a segment of height of a sphere of radius is (b) Show that if a sphere of radius 1 is sliced by a plane at a distance from the center in such a way that the volume of one segment is twice the volume of the other, then is a solution of the equation where . Use Newton’s method to ﬁnd accurate to four decimal places. (c) Using the formula for the volume of a segment of a sphere, it can be shown that the depth to which a ﬂoating sphere of radius sinks in water is a root of the equation where is the speciﬁc gravity of the sphere. Suppose a wooden sphere of radius 0.5 m has speciﬁc gravity 0.75. Calculate, to four-decimal-place accuracy, the depth to which the sphere will sink. s x 3 3rx 2 4r 3s 0 r x x 0 x 1 3x 3 9x 2 0 x x V 1 3h 23r h r h r L L r L r y 8x 27x 3 y c x y x x 2 f b 0 b 2 x b x 0 x 0 f y f x x t 0 At t 3 t f f P R O B L E M S P L U S 0 x y y=8x-27˛ y=c FIGURE FOR PROBLEM 3 FIGURE FOR PROBLEM 5 r h 449 (d) A hemispherical bowl has radius 5 inches and water is running into the bowl at the rate of . (i) How fast is the water level in the bowl rising at the instant the water is 3 inches deep? (ii) At a certain instant, the water is 4 inches deep. How long will it take to ﬁll the bowl? 6. Archimedes’ Principle states that the buoyant force on an object partially or fully submerged in a ﬂuid is equal to the weight of the ﬂuid that the object displaces. Thus, for an object of density ﬂoating partly submerged in a ﬂuid of density , the buoyant force is given by , where is the acceleration due to gravity and is the area of a typi-cal cross-section of the object. The weight of the object is given by (a) Show that the percentage of the volume of the object above the surface of the liquid is (b) The density of ice is and the density of seawater is . What percent-age of the volume of an iceberg is above water? (c) An ice cube ﬂoats in a glass ﬁlled to the brim with water. Does the water overﬂow when the ice melts? (d) A sphere of radius 0.4 m and having negligible weight is ﬂoating in a large freshwater lake. How much work is required to completely submerge the sphere? The density of the water is . 7. Water in an open bowl evaporates at a rate proportional to the area of the surface of the water. (This means that the rate of decrease of the volume is proportional to the area of the surface.) Show that the depth of the water decreases at a constant rate, regardless of the shape of the bowl. 8. A sphere of radius 1 overlaps a smaller sphere of radius in such a way that their intersection is a circle of radius . (In other words, they intersect in a great circle of the small sphere.) Find so that the volume inside the small sphere and outside the large sphere is as large as possible. 9. The ﬁgure shows a curve with the property that, for every point on the middle curve , the areas and are equal. Find an equation for . 10. A paper drinking cup ﬁlled with water has the shape of a cone with height and semivertical angle (see the ﬁgure). A ball is placed carefully in the cup, thereby displacing some of the water and making it overﬂow. What is the radius of the ball that causes the greatest volume of water to spill out of the cup? h C B A y 2x 2 P C r r r 1000 kgm3 1030 kgm3 917 kgm3 100 f 0 f W 0t y Lh h Ay dy Ay t F f t x0 h Ay dy f 0 0.2 in3s P R O B L E M S P L U S FIGURE FOR PROBLEM 6 y=0 h y=_h y=L-h L 0 x y P y=≈ A B C y=2≈ FIGURE FOR PROBLEM 9 450 11. A clepsydra, or water clock, is a glass container with a small hole in the bottom through which water can ﬂow. The “clock” is calibrated for measuring time by placing markings on the container corresponding to water levels at equally spaced times. Let be continu-ous on the interval and assume that the container is formed by rotating the graph of about the -axis. Let denote the volume of water and the height of the water level at time . (a) Determine as a function of . (b) Show that (c) Suppose that is the area of the hole in the bottom of the container. It follows from Torricelli’s Law that the rate of change of the volume of the water is given by where is a negative constant. Determine a formula for the function such that is a constant . What is the advantage in having ? 12. A cylindrical container of radius and height is partially ﬁlled with a liquid whose volume is . If the container is rotated about its axis of symmetry with constant angular speed , then the container will induce a rotational motion in the liquid around the same axis. Eventually, the liquid will be rotating at the same angular speed as the container. The surface of the liquid will be convex, as indicated in the ﬁgure, because the centrifugal force on the liquid particles increases with the distance from the axis of the container. It can be shown that the surface of the liquid is a paraboloid of revolution generated by rotating the parabola about the -axis, where is the acceleration due to gravity. (a) Determine as a function of . (b) At what angular speed will the surface of the liquid touch the bottom? At what speed will it spill over the top? (c) Suppose the radius of the container is 2 ft, the height is 7 ft, and the container and liquid are rotating at the same constant angular speed. The surface of the liquid is 5 ft below the top of the tank at the central axis and 4 ft below the top of the tank 1 ft out from the cen-tral axis. (i) Determine the angular speed of the container and the volume of the ﬂuid. (ii) How far below the top of the tank is the liquid at the wall of the container? 13. Suppose the graph of a cubic polynomial intersects the parabola when , , and , where . If the two regions between the curves have the same area, how is related to ? a b 0 a b x b x a x 0 y x 2 h t y y h 2x 2 2t V L r x y x=f(y) h b dhdt C C dhdt f k dV dt kAsh A dV dt f h2 dh dt h V t h V y f 0, b x f y P R O B L E M S P L U S FIGURE FOR PROBLEM 12 x y h L r v 451 14. Suppose we are planning to make a taco from a round tortilla with diameter 8 inches by bend-ing the tortilla so that it is shaped as if it is partially wrapped around a circular cylinder. We will ﬁll the tortilla to the edge (but no more) with meat, cheese, and other ingredients. Our problem is to decide how to curve the tortilla in order to maximize the volume of food it can hold. (a) We start by placing a circular cylinder of radius along a diameter of the tortilla and folding the tortilla around the cylinder. Let represent the distance from the center of the tortilla to a point on the diameter (see the ﬁgure). Show that the cross-sectional area of the ﬁlled taco in the plane through perpendicular to the axis of the cylinder is and write an expression for the volume of the ﬁlled taco. (b) Determine (approximately) the value of that maximizes the volume of the taco. (Use a graphical approach with your CAS.) 15. If the tangent at a point on the curve intersects the curve again at , let be the area of the region bounded by the curve and the line segment . Let be the area of the region deﬁned in the same way starting with instead of . What is the relationship between and ? B A P Q B PQ A Q y x 3 P P x r Ax rs16 x 2 1 2 r 2 sin 2 r s16 x 2 P P x r CAS P R O B L E M S P L U S 452 Because of the Fundamental Theorem of Calculus, we can integrate a function if we know an antiderivative, that is, an indeﬁnite integral. We summarize here the most important integrals that we have learned so far. In this chapter we develop techniques for using these basic integration formulas to obtain indeﬁnite integrals of more complicated functions. We learned the most important method of integration, the Substitution Rule, in Section 5.5. The other general technique, integration by parts, is presented in Section 7.1. Then we learn methods that are special to particular classes of functions, such as trigonometric functions and rational functions. Integration is not as straightforward as differentiation; there are no rules that absolutely guarantee obtaining an indeﬁnite integral of a function. Therefore we discuss a strategy for integration in Section 7.5. y 1 sa 2 x 2 dx sin1 x a C y 1 x 2 a 2 dx 1 a tan1 x a C y cot x dx lnsin x C y tan x dx lnsec x C y cosh x dx sinh x C y sinh x dx cosh x C y csc x cot x dx csc x C y sec x tan x dx sec x C y csc2x dx cot x C y sec2x dx tan x C y cos x dx sin x C y sin x dx cos x C y a x dx a x ln a C y e x dx e x C y 1 x dx lnx C n 1 y x n dx x n1 n 1 C Simpson’s Rule estimates integrals by approximating graphs with parabolas. TECHNIQUES OF INTEGRATION 7 INTEGRATION BY PARTS Every differentiation rule has a corresponding integration rule. For instance, the Substi-tution Rule for integration corresponds to the Chain Rule for differentiation. The rule that corresponds to the Product Rule for differentiation is called the rule for integration by parts. The Product Rule states that if and are differentiable functions, then In the notation for indeﬁnite integrals this equation becomes or We can rearrange this equation as Formula 1 is called the formula for integration by parts. It is perhaps easier to remem-ber in the following notation. Let and . Then the differentials are and , so, by the Substitution Rule, the formula for integration by parts becomes EXAMPLE 1 Find . SOLUTION USING FORMULA 1 Suppose we choose and . Then and . (For we can choose any antiderivative of .) Thus, using Formula 1, we have It’s wise to check the answer by differentiating it. If we do so, we get , as expected. x sin x x cos x sin x C x cos x y cos x dx xcos x y cos x dx y x sin x dx fxtx y txfx dx t t tx cos x fx 1 tx sin x fx x y x sin x dx y u dv uv y v du 2 dv tx dx du fx dx v tx u fx y fxtx dx fxtx y txfx dx 1 y fxtx dx y txfx dx fxtx y fxtx txfx dx fxtx d dx fxtx fxtx txfx t f 7.1 453 SOLUTION USING FORMULA 2 Let Then and so M Our aim in using integration by parts is to obtain a simpler integral than the one we started with. Thus in Example 1 we started with and expressed it in terms of the simpler integral . If we had instead chosen and , then and , so integration by parts gives Although this is true, is a more difﬁcult integral than the one we started with. In general, when deciding on a choice for and , we usually try to choose to be a function that becomes simpler when differentiated (or at least not more complicated) as long as can be readily integrated to give . EXAMPLE 2 Evaluate . SOLUTION Here we don’t have much choice for and . Let Then Integrating by parts, we get Integration by parts is effective in this example because the derivative of the function is simpler than . M f fx ln x x ln x x C x ln x y dx y ln x dx x ln x y x dx x du 1 x dx v x u ln x dv dx dv u y ln x dx V v dv tx dx u fx dv u x x 2 cos x dx y x sin x dx sin x x 2 2 1 2 y x 2 cos x dx v x 2 2 du cos x dx dv x dx u sin x x cos x dx x x sin x dx NOTE x cos x sin x C x cos x y cos x dx y x sin x dx y x sin x dx x cos x y cos x dx v cos x du dx dv sin x dx u x 454 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION u d√ u √ √ du N It is helpful to use the pattern: v du dv u N It’s customary to write as . x dx x 1 dx N Check the answer by differentiating it. EXAMPLE 3 Find . SOLUTION Notice that becomes simpler when differentiated (whereas is unchanged when differentiated or integrated), so we choose Then Integration by parts gives The integral that we obtained, , is simpler than the original integral but is still not obvious. Therefore, we use integration by parts a second time, this time with and . Then , , and Putting this in Equation 3, we get M EXAMPLE 4 Evaluate . SOLUTION Neither nor becomes simpler when differentiated, but we try choosing and anyway. Then and , so integration by parts gives The integral that we have obtained, , is no simpler than the original one, but at least it’s no more difﬁcult. Having had success in the preceding example integrating by parts twice, we persevere and integrate by parts again. This time we use and . Then , , and At ﬁrst glance, it appears as if we have accomplished nothing because we have arrived at , which is where we started. However, if we put the expression for from Equation 5 into Equation 4 we get y e x sin x dx e x cos x e x sin x y e x sin x dx x e x cos x dx x e x sin x dx y e x cos x dx e x sin x y e x sin x dx 5 v sin x du e x dx dv cos x dx u e x x e x cos x dx y e x sin x dx e x cos x y e x cos x dx 4 v cos x du e x dx dv sin x dx u e x sin x e x y e x sin x dx V where C1 2C t 2et 2tet 2et C1 t 2et 2tet et C y t 2et dt t 2et 2 y tet dt tet et C y tet dt tet y et dt v et du dt dv et dt u t x tet dt y t 2et dt t 2et 2 y tet dt 3 du 2t dt v et u t 2 dv et dt et t 2 y t 2et dt V SECTION 7.1 INTEGRATION BY PARTS \|\|\|\| 455 N An easier method, using complex numbers, is given in Exercise 50 in Appendix H. This can be regarded as an equation to be solved for the unknown integral. Adding to both sides, we obtain Dividing by 2 and adding the constant of integration, we get M If we combine the formula for integration by parts with Part 2 of the Fundamental Theorem of Calculus, we can evaluate deﬁnite integrals by parts. Evaluating both sides of Formula 1 between and , assuming and are continuous, and using the Fundamental Theorem, we obtain EXAMPLE 5 Calculate . SOLUTION Let Then So Formula 6 gives To evaluate this integral we use the substitution (since has another meaning in this example). Then , so . When , ; when , ; so Therefore M y 1 0 tan1x dx 4 y 1 0 x 1 x 2 dx 4 ln 2 2 1 2ln 2 ln 1 1 2 ln 2 y 1 0 x 1 x 2 dx 1 2 y 2 1 dt t 1 2 ln t]1 2 t 2 x 1 t 1 x 0 x dx 1 2 dt dt 2x dx u t 1 x 2 4 y 1 0 x 1 x 2 dx 1 tan1 1 0 tan1 0 y 1 0 x 1 x 2 dx y 1 0 tan1x dx x tan1x]0 1 y 1 0 x 1 x 2 dx du dx 1 x 2 v x u tan1x dv dx y 1 0 tan1x dx y b a fxtx dx fxtx]a b y b a txfx dx 6 t f b a y e x sin x dx 1 2e xsin x cos x C 2 y e x sin x dx e x cos x e x sin x x ex sin x dx 456 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION N Since for , the integral in Example 5 can be interpreted as the area of the region shown in Figure 2. x 0 tan1x 0 y 0 x 1 y=tan–!x FIGURE 2 N Figure 1 illustrates Example 4 by show-ing the graphs of and . As a visual check on our work, notice that when has a maximum or minimum. F f x 0 Fx 1 2 e xsin x cos x f x e x sin x _3 _4 12 6 F f FIGURE 1 EXAMPLE 6 Prove the reduction formula where is an integer. SOLUTION Let Then so integration by parts gives Since , we have As in Example 4, we solve this equation for the desired integral by taking the last term on the right side to the left side. Thus we have or M The reduction formula (7) is useful because by using it repeatedly we could eventually express in terms of (if is odd) or (if is even). n x sin x0 dx x dx n x sin x dx x sinnx dx y sinnx dx 1 n cos x sinn1x n 1 n y sinn2x dx n y sinnx dx cos x sinn1x n 1 y sinn2x dx y sinnx dx cos x sinn1x n 1 y sinn2x dx n 1 y sinnx dx cos2x 1 sin2x y sinnx dx cos x sinn1x n 1 y sinn2x cos2x dx v cos x du n 1 sinn2x cos x dx dv sin x dx u sinn1x n 2 y sinnx dx 1 n cos x sinn1x n 1 n y sinn2x dx 7 SECTION 7.1 INTEGRATION BY PARTS \|\|\|\| 457 N Equation 7 is called a reduction formula because the exponent has been reduced to and . n 2 n 1 n 11. 12. 13. 14. 16. 18. 19. 21. 22. 23. 24. y 0 x 3 cos x dx y 2 1 ln x x 2 dx y 9 4 ln y sy dy y 1 0 t cosh t dt y 1 0 x 2 1ex dx 20. y 0 t sin 3t dt y e cos 2 d y e 2 sin 3 d 17. y t sinh mt dt y ln x2 dx 15. y s 2s ds y t sec2 2t dt y p5 ln p dp y arctan 4t dt 1–2 Evaluate the integral using integration by parts with the indicated choices of and . 1. ; , 2. ; , 3–32 Evaluate the integral. 4. 5. 6. 7. 8. 9. 10. y sin1x dx y ln2x 1 dx y x 2 cos mx dx y x 2 sin x dx y t sin 2t dt y rer2 dr y xex dx y x cos 5x dx 3. dv cos d u y cos d dv x 2 dx u ln x y x 2 ln x dx dv u EXERCISES 7.1 (b) Use part (a) to evaluate and . (c) Use part (a) to show that, for odd powers of sine, 46. Prove that, for even powers of sine, 47–50 Use integration by parts to prove the reduction formula. 48. 49. 50. 51. Use Exercise 47 to ﬁnd . 52. Use Exercise 48 to ﬁnd . 53–54 Find the area of the region bounded by the given curves. 53. , , 54. ; 55–56 Use a graph to ﬁnd approximate -coordinates of the points of intersection of the given curves. Then ﬁnd (approxi-mately) the area of the region bounded by the curves. 55. , 56. , 57–60 Use the method of cylindrical shells to ﬁnd the volume generated by rotating the region bounded by the given curves about the speciﬁed axis. , , ; about the -axis 58. , , ; about the -axis 59. , , , ; about 60. , , ; about the -axis x y x 0 y e x x 1 x 0 x 1 y 0 y ex y x 1 y ex y e x y 0 x 1 y 0 y cosx 2 57. y 1 2x y arctan 3x y x 22 y x sin x x y x ln x y 5 ln x, x 5 y 0 y xe0.4x x x 4e x dx x ln x3 dx n 1 y secnx dx tan x secn2x n 1 n 2 n 1 y secn2x dx n 1 tann x dx tann1 x n 1 y tann2 x dx y x ne x dx x ne x n y x n1e x dx y ln xn dx xln xn n y ln xn1 dx 47. y 2 0 sin2nx dx 1 3 5 2n 1 2 4 6 2n 2 y 2 0 sin2n1x dx 2 4 6 2n 3 5 7 2n 1 x 2 0 sin5x dx x 2 0 sin3x dx 25. 26. 27. 28. 29. 30. 31. 32. 33–38 First make a substitution and then use integration by parts to evaluate the integral. 33. 34. 36. 37. 38. ; 39–42 Evaluate the indeﬁnite integral. Illustrate, and check that your answer is reasonable, by graphing both the function and its antiderivative (take ). 39. 40. 41. 42. 43. (a) Use the reduction formula in Example 6 to show that (b) Use part (a) and the reduction formula to evaluate . 44. (a) Prove the reduction formula (b) Use part (a) to evaluate . (c) Use parts (a) and (b) to evaluate . 45. (a) Use the reduction formula in Example 6 to show that where is an integer. n 2 y 2 0 sinnx dx n 1 n y 2 0 sinn2x dx x cos4x dx x cos2x dx y cosnx dx 1 n cosn1x sin x n 1 n y cosn2x dx x sin4x dx y sin2x dx x 2 sin 2x 4 C y x 2 sin 2x dx y x 3s1 x 2 dx y x 3 2 ln x dx y 2x 3e x dx C 0 y sinln x dx y x ln1 x dx y 0 e cos t sin 2t dt y s s 2 3 cos 2 d 35. y t 3et2 dt y cos sx dx y t 0 e s sint s ds y 2 1 x 4ln x2 dx y 1 0 r 3 s4 r 2 dr y cos x lnsin x dx y 2 1 ln x2 x 3 dx y 1 2 0 cos1x dx y s3 1 arctan1 x dx y 1 0 y e2y dy 458 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION SECTION 7.1 INTEGRATION BY PARTS \|\|\|\| 459 parts on the resulting integral to prove that 68. Let . (a) Show that . (b) Use Exercise 46 to show that (c) Use parts (a) and (b) to show that and deduce that . (d) Use part (c) and Exercises 45 and 46 to show that This formula is usually written as an inﬁnite product: and is called the Wallis product. (e) We construct rectangles as follows. Start with a square of area 1 and attach rectangles of area 1 alternately beside or on top of the previous rectangle (see the ﬁgure). Find the limit of the ratios of width to height of these rectangles. 2 2 1 2 3 4 3 4 5 6 5 6 7 lim n l 2 1 2 3 4 3 4 5 6 5 6 7 2n 2n 1 2n 2n 1 2 limn l I2n1 I2n 1 2n 1 2n 2 I2n1 I2n 1 I2n2 I2n 2n 1 2n 2 I2n2 I2n1 I2n In x 2 0 sinnx dx y 0 x a b c d x=a x=b y=ƒ x=g(y) V y b a 2x f x dx 61. Find the average value of on the interval . 62. A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is , the fuel is consumed at rate , and the exhaust gases are ejected with constant velocity (relative to the rocket). A model for the velocity of the rocket at time is given by the equation where is the acceleration due to gravity and is not too large. If , kg, kg s, and , ﬁnd the height of the rocket one minute after liftoff. A particle that moves along a straight line has velocity meters per second after seconds. How far will it travel during the ﬁrst seconds? 64. If and and are continuous, show that 65. Suppose that , , , , and is continuous. Find the value of . (a) Use integration by parts to show that (b) If and are inverse functions and is continuous, prove that [Hint: Use part (a) and make the substitution .] (c) In the case where and are positive functions and , draw a diagram to give a geometric interpre-tation of part (b). (d) Use part (b) to evaluate . 67. We arrived at Formula 6.3.2, , by using cylindrical shells, but now we can use integration by parts to prove it using the slicing method of Section 6.2, at least for the case where is one-to-one and therefore has an inverse function . Use the ﬁgure to show that Make the substitution and then use integration by y f x V b 2d a 2c y d c ty2 dy t f V xb a 2x f x dx xe 1 ln x dx b a 0 t f y f x y b a f x dx bf b af a y f b f a ty dy f t f y f x dx xf x y xf x dx 66. x4 1 xf x dx f f 4 3 f 1 5 f 4 7 f 1 2 y a 0 f xtx dx f ata f ata y a 0 f xtx dx t f f 0 t0 0 t t vt t 2et 63. ve 3000 m s r 160 m 30,000 t 9.8 m s2 t t vt tt ve ln m rt m t ve r m 1, 3 f x x 2 ln x TRIGONOMETRIC INTEGRALS In this section we use trigonometric identities to integrate certain combinations of trigo-nometric functions. We start with powers of sine and cosine. EXAMPLE 1 Evaluate . SOLUTION Simply substituting isn’t helpful, since then . In order to integrate powers of cosine, we would need an extra factor. Similarly, a power of sine would require an extra factor. Thus here we can separate one cosine factor and convert the remaining factor to an expression involving sine using the identity : We can then evaluate the integral by substituting , so and M In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor (and the remainder of the expression in terms of cosine) or only one cosine factor (and the remainder of the expression in terms of sine). The identity enables us to convert back and forth between even powers of sine and cosine. EXAMPLE 2 Find . SOLUTION We could convert to , but we would be left with an expression in terms of with no extra factor. Instead, we separate a single sine factor and rewrite the remaining factor in terms of : Substituting , we have and so M 1 3 cos3x 2 5 cos5x 1 7 cos7x C u 3 3 2 u 5 5 u 7 7 C y 1 u 22u 2du y u 2 2u 4 u 6 du y 1 cos2x2 cos2x sin x dx y sin5x cos2x dx y sin2x2 cos2x sin x dx du sin x dx u cos x sin5x cos2x sin2x2 cos2x sin x 1 cos2x2 cos2x sin x cos x sin4x cos x sin x 1 sin2x cos2x y sin5x cos2x dx V sin2x cos2x 1 sin x 1 3 sin3x C y 1 u 2 du u 1 3u 3 C y cos3x dx y cos2x cos x dx y 1 sin2x cos x dx du cos x dx u sin x cos3x cos2x cos x 1 sin2x cos x sin2x cos2x 1 cos2x cos x sin x du sin x dx u cos x y cos3x dx 7.2 460 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION N Figure 1 shows the graphs of the integrand in Example 2 and its indeﬁnite inte-gral (with ). Which is which? C 0 sin5x cos2x FIGURE 1 _π _0.2 0.2 π In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. If the integrand contains even powers of both sine and cosine, this strategy fails. In this case, we can take advantage of the fol-lowing half-angle identities (see Equations 17b and 17a in Appendix D): and EXAMPLE 3 Evaluate . SOLUTION If we write , the integral is no simpler to evaluate. Using the half-angle formula for , however, we have Notice that we mentally made the substitution when integrating . Another method for evaluating this integral was given in Exercise 43 in Section 7.1. M EXAMPLE 4 Find . SOLUTION We could evaluate this integral using the reduction formula for (Equation 7.1.7) together with Example 3 (as in Exercise 43 in Section 7.1), but a better method is to write and use a half-angle formula: Since occurs, we must use another half-angle formula This gives M To summarize, we list guidelines to follow when evaluating integrals of the form , where and are integers. n 0 m 0 x sinmx cosnx dx 1 4( 3 2 x sin 2x 1 8 sin 4x) C 1 4 y ( 3 2 2 cos 2x 1 2 cos 4x) dx y sin4x dx 1 4 y 1 2 cos 2x 1 21 cos 4x dx cos2 2x 1 21 cos 4x cos2 2x 1 4 y 1 2 cos 2x cos2 2x dx y 1 cos 2x 2 2 dx y sin4x dx y sin2x2 dx sin4x sin2x2 x sinnx dx y sin4x dx cos 2x u 2x 1 2( 1 2 sin 2) 1 2(0 1 2 sin 0) 1 2 y 0 sin2x dx 1 2 y 0 1 cos 2x dx [ 1 2(x 1 2 sin 2x)]0 sin2x sin2x 1 cos2x y 0 sin2x dx V cos2x 1 21 cos 2x sin2x 1 21 cos 2x SECTION 7.2 TRIGONOMETRIC INTEGRALS \|\|\|\| 461 N Example 3 shows that the area of the region shown in Figure 2 is . 2 FIGURE 2 0 _0.5 1.5 π y=sin@ x STRATEGY FOR EVALUATING (a) If the power of cosine is odd , save one cosine factor and use to express the remaining factors in terms of sine: Then substitute . (b) If the power of sine is odd , save one sine factor and use to express the remaining factors in terms of cosine: Then substitute . [Note that if the powers of both sine and cosine are odd, either (a) or (b) can be used.] (c) If the powers of both sine and cosine are even, use the half-angle identities It is sometimes helpful to use the identity We can use a similar strategy to evaluate integrals of the form . Since , we can separate a factor and convert the remaining (even) power of secant to an expression involving tangent using the identity . Or, since , we can separate a factor and convert the remaining (even) power of tangent to secant. EXAMPLE 5 Evaluate . SOLUTION If we separate one factor, we can express the remaining factor in terms of tangent using the identity . We can then evaluate the integral by substituting so that : M 1 7 tan7x 1 9 tan9x C u 7 7 u 9 9 C y u 61 u 2 du y u 6 u 8 du y tan6x 1 tan2x sec2x dx y tan6x sec4x dx y tan6x sec2x sec2x dx du sec2x dx u tan x sec2x 1 tan2x sec2x sec2x y tan6x sec4x dx V sec x tan x ddx sec x sec x tan x sec2x 1 tan2x sec2x ddx tan x sec2x x tanmx secnx dx sin x cos x 1 2 sin 2x cos2x 1 21 cos 2x sin2x 1 21 cos 2x u cos x y 1 cos2xk cosnx sin x dx y sin2k1x cosnx dx y sin2xk cosnx sin x dx sin2x 1 cos2x m 2k 1 u sin x y sinmx 1 sin2xk cos x dx y sinmx cos2k1x dx y sinmx cos2xk cos x dx cos2x 1 sin2x n 2k 1 y sin mx cos nx dx 462 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION EXAMPLE 6 Find . SOLUTION If we separate a factor, as in the preceding example, we are left with a factor, which isn’t easily converted to tangent. However, if we separate a factor, we can convert the remaining power of tangent to an expression involving only secant using the identity . We can then evaluate the integral by substituting , so : M The preceding examples demonstrate strategies for evaluating integrals of the form for two cases, which we summarize here. STRATEGY FOR EVALUATING (a) If the power of secant is even , save a factor of and use to express the remaining factors in terms of : Then substitute . (b) If the power of tangent is odd , save a factor of and use to express the remaining factors in terms of : Then substitute . For other cases, the guidelines are not as clear-cut. We may need to use identities, inte-gration by parts, and occasionally a little ingenuity. We will sometimes need to be able to u sec x y sec2x 1k secn1x sec x tan x dx y tan2k1x secnx dx y tan2xk secn1x sec x tan x dx sec x tan2x sec2x 1 sec x tan x m 2k 1 u tan x y tanmx 1 tan2xk1 sec2x dx y tanmx sec2kx dx y tanmx sec2xk1 sec2x dx tan x sec2x 1 tan2x sec2x n 2k, k 2 y tanmx secnx dx x tanmx secnx dx 1 11 sec11 2 9 sec9 1 7 sec7 C u 11 11 2 u 9 9 u 7 7 C y u 10 2u 8 u 6 du y u 2 12u 6 du y sec2 12 sec6 sec tan d y tan5 sec7 d y tan4 sec6 sec tan d du sec tan d u sec tan2 sec2 1 sec tan sec5 sec2 y tan5 sec7 d SECTION 7.2 TRIGONOMETRIC INTEGRALS \|\|\|\| 463 integrate by using the formula established in (5.5.5): We will also need the indeﬁnite integral of secant: We could verify Formula 1 by differentiating the right side, or as follows. First we multi-ply numerator and denominator by : If we substitute , then , so the integral becomes . Thus we have EXAMPLE 7 Find . SOLUTION Here only occurs, so we use to rewrite a factor in terms of : In the ﬁrst integral we mentally substituted so that . M If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of . Powers of may require integration by parts, as shown in the following example. EXAMPLE 8 Find . SOLUTION Here we integrate by parts with du sec x tan x dx v tan x u sec x dv sec2x dx y sec3x dx sec x sec x du sec2x dx u tan x tan2x 2 ln sec x C y tan x sec2x dx y tan x dx y tan x sec2x 1 dx y tan3x dx y tan x tan2x dx sec2x tan2x tan2x sec2x 1 tan x y tan3x dx y sec x dx ln sec x tan x C x 1u du ln u C du sec x tan x sec2x dx u sec x tan x y sec2x sec x tan x sec x tan x dx y sec x dx y sec x sec x tan x sec x tan x dx sec x tan x y sec x dx ln sec x tan x C 1 y tan x dx ln sec x C tan x 464 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION Then Using Formula 1 and solving for the required integral, we get M Integrals such as the one in the preceding example may seem very special but they occur frequently in applications of integration, as we will see in Chapter 8. Integrals of the form can be found by similar methods because of the identity . Finally, we can make use of another set of trigonometric identities: To evaluate the integrals (a) , (b) , or (c) , use the corresponding identity: (a) (b) (c) EXAMPLE 9 Evaluate . SOLUTION This integral could be evaluated using integration by parts, but it’s easier to use the identity in Equation 2(a) as follows: M 1 2(cos x 1 9 cos 9x C 1 2 y sin x sin 9x dx y sin 4x cos 5x dx y 1 2sinx sin 9x dx y sin 4x cos 5x dx cos A cos B 1 2cosA B cosA B sin A sin B 1 2cosA B cosA B sin A cos B 1 2sinA B sinA B x cos mx cos nx dx x sin mx sin nx dx x sin mx cos nx dx 2 1 cot2x csc2x x cotmx cscnx dx y sec3x dx 1 2(sec x tan x ln sec x tan x) C sec x tan x y sec3x dx y sec x dx sec x tan x y sec x sec2x 1 dx y sec3x dx sec x tan x y sec x tan2x dx SECTION 7.2 TRIGONOMETRIC INTEGRALS \|\|\|\| 465 N These product identities are discussed in Appendix D. 9. 10. 11. 12. 14. 15. 16. y cos cos5sin d y cos5 ssin d y 0 sin2 t cos4 t dt y 2 0 sin2x cos2x dx 13. y x cos2x dx y 1 cos 2 d y 0 cos6 d y 0 sin43t dt 1–49 Evaluate the integral. 1. 2. 4. 5. 6. 8. y 2 0 sin22 d y 2 0 cos2 d 7. y sin3(sx ) sx dx y sin2x cos5x dx y 2 0 cos5x dx y 34 2 sin5x cos3x dx 3. y sin6x cos3x dx y sin3x cos2x dx EXERCISES 7.2 53. 54. Find the average value of the function on the interval . 56. Evaluate by four methods: (a) the substitution (b) the substitution (c) the identity (d) integration by parts Explain the different appearances of the answers. 57–58 Find the area of the region bounded by the given curves. 57. 58. , , ; 59–60 Use a graph of the integrand to guess the value of the integral. Then use the methods of this section to prove that your guess is correct. 59. 60. 61–64 Find the volume obtained by rotating the region bounded by the given curves about the speciﬁed axis. , , ; about the -axis 62. , , ; about the -axis 63. , , ; about 64. , , ; about 65. A particle moves on a straight line with velocity function . Find its position function if 66. Household electricity is supplied in the form of alternating current that varies from V to V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation where is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of over one cycle. (a) Calculate the RMS voltage of household current. (b) Many electric stoves require an RMS voltage of 220 V. Find the corresponding amplitude needed for the volt-age . Et A sin120t A Et2 t Et 155 sin120t 155 155 f 0 0. s f t vt sin t cos2t y 1 0 x 3 y cos x y sec x y 1 0 x 4 y cos x y sin x x 0 x y 0 y sin2 x x 2 x y 0 y sin x 61. y 2 0 sin 2x cos 5x dx y 2 0 cos3x dx 4 x 54 y cos3 x y sin3 x 4 x 4 y cos2 x, y sin2 x, sin 2x 2 sin x cos x u sin x u cos x x sin x cos x dx , f x sin2x cos3x 55. y sec4 x 2 dx y sin 3x sin 6x dx 17. 18. 19. 20. 21. 22. 24. 25. 26. 27. 28. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 44. 45. 46. 47. 48. 49. 50. If , express the value of in terms of . ; 51–54 Evaluate the indeﬁnite integral. Illustrate, and check that your answer is reasonable, by graphing both the integrand and its antiderivative (taking . 51. 52. y sin3 x cos4 x dx y x sin2x 2 dx C 0 I x4 0 tan8x sec x dx x4 0 tan6x sec x dx I y t sec2t 2 tan4t 2 dt y dx cos x 1 y 1 tan2x sec2x dx y cos x sin x sin 2x dx y sin 5 sin d y cos x cos 4x dx y sin 8x cos 5x dx 43. y 3 6 csc3x dx y csc x dx y csc 4x cot 6x dx y cot 3 csc3 d y 2 4 cot3x dx y 2 6 cot2x dx y sin cos3 d y x sec x tan x dx y tan2x sec x dx y tan3 cos4 d y tan6ay dy y tan5x dx y 3 0 tan5x sec6x dx y tan3x sec x dx 29. y tan32x sec52x dx y 3 0 tan5x sec4x dx y 4 0 sec4 tan4 d y sec6t dt y tan2 x tan4 x dx y tan2x dx 23. y 2 0 sec4t2 dt y sec2x tan x dx y cos2x sin 2x dx y cos x sin 2x sin x dx y cot5 sin4 d y cos2x tan3x dx 466 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION SECTION 7.3 TRIGONOMETRIC SUBSTITUTION \|\|\|\| 467 70. A ﬁnite Fourier series is given by the sum Show that the th coefﬁcient is given by the formula am 1 y f x sin mx dx am m a1 sin x a2 sin 2x aN sin Nx f x N n1 an sin nx 67–69 Prove the formula, where and are positive integers. 67. 68. 69. y cos mx cos nx dx 0 if m n if m n y sin mx sin nx dx 0 if m n if m n y sin mx cos nx dx 0 n m TRIGONOMETRIC SUBSTITUTION In ﬁnding the area of a circle or an ellipse, an integral of the form arises, where . If it were , the substitution would be effective but, as it stands, is more difﬁcult. If we change the variable from to by the substitution , then the identity allows us to get rid of the root sign because Notice the difference between the substitution (in which the new variable is a function of the old one) and the substitution (the old variable is a function of the new one). In general we can make a substitution of the form by using the Substitution Rule in reverse. To make our calculations simpler, we assume that has an inverse func-tion; that is, is one-to-one. In this case, if we replace by and by in the Substitution Rule (Equation 5.5.4), we obtain This kind of substitution is called inverse substitution. We can make the inverse substitution provided that it deﬁnes a one-to-one function. This can be accomplished by restricting to lie in the interval . In the following table we list trigonometric substitutions that are effective for the given radical expressions because of the speciﬁed trigonometric identities. In each case the restric-tion on is imposed to ensure that the function that deﬁnes the substitution is one-to-one. (These are the same intervals used in Section 1.6 in deﬁning the inverse functions.) TABLE OF TRIGONOMETRIC SUBSTITUTIONS 2, 2 x a sin y fx dx y ftttt dt t x x u t t x tt x a sin u a 2 x 2 sa 2 x 2 sa 2 a 2 sin2 sa 21 sin 2 sa 2 cos2 acos 1 sin2 cos2 x a sin x x sa 2 x 2 dx u a 2 x 2 x xsa 2 x 2 dx a 0 x sa 2 x 2 dx 7.3 Expression Substitution Identity sec2 1 tan2 x a sec , 0 2 or 3 2 sx 2 a 2 1 tan2 sec2 x a tan , 2 2 sa 2 x 2 1 sin2 cos2 x a sin , 2 2 sa 2 x 2 EXAMPLE 1 Evaluate . SOLUTION Let , where . Then and (Note that because .) Thus the Inverse Substitution Rule gives Since this is an indeﬁnite integral, we must return to the original variable . This can be done either by using trigonometric identities to express in terms of or by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right tri-angle. Since , we label the opposite side and the hypotenuse as having lengths and . Then the Pythagorean Theorem gives the length of the adjacent side as , so we can simply read the value of from the ﬁgure: (Although in the diagram, this expression for is valid even when .) Since , we have and so M EXAMPLE 2 Find the area enclosed by the ellipse SOLUTION Solving the equation of the ellipse for , we get Because the ellipse is symmetric with respect to both axes, the total area is four times the area in the ﬁrst quadrant (see Figure 2). The part of the ellipse in the ﬁrst quadrant is given by the function and so 1 4 A y a 0 b a sa 2 x 2 dx 0 x a y b a sa 2 x 2 A y b a sa 2 x 2 or y 2 b 2 1 x 2 a 2 a 2 x 2 a 2 y x 2 a 2 y 2 b 2 1 V y s9 x 2 x 2 dx s9 x 2 x sin1 x 3 C sin1x3 sin x3 0 cot 0 cot s9 x 2 x cot s9 x 2 3 x sin x3 sin x3 cot x cot C y csc2 1 d y cos2 sin2 d y cot2 d y s9 x 2 x 2 dx y 3 cos 9 sin2 3 cos d 2 2 cos 0 s9 x 2 s9 9 sin2 s9 cos2 3cos 3 cos dx 3 cos d 2 2 x 3 sin y s9 x 2 x 2 dx V 468 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION 3 ¨ x œ„„„„„ 9-≈ FIGURE 1 sin ¨= x 3 FIGURE 2 ≈ a@ ¥ b@ + =1 y 0 x (0, b) (a, 0) To evaluate this integral we substitute . Then . To change the limits of integration we note that when , , so ; when , , so . Also since . Therefore We have shown that the area of an ellipse with semiaxes and is . In particular, taking , we have proved the famous formula that the area of a circle with radius is . M Since the integral in Example 2 was a deﬁnite integral, we changed the limits of integration and did not have to convert back to the original variable . EXAMPLE 3 Find . SOLUTION Let . Then and Thus we have To evaluate this trigonometric integral we put everything in terms of and : Therefore, making the substitution , we have We use Figure 3 to determine that and so M y dx x 2sx 2 4 sx 2 4 4x C csc sx 2 4 x csc 4 C 1 4 1 u C 1 4 sin C y dx x 2sx 2 4 1 4 y cos sin2 d 1 4 y du u 2 u sin sec tan2 1 cos cos2 sin2 cos sin2 cos sin y dx x 2sx 2 4 y 2 sec2 d 4 tan2 2 sec 1 4 y sec tan2 d sx 2 4 s4tan 2 1 s4 sec 2 2sec 2 sec dx 2 sec2 d x 2 tan , 2 2 y 1 x 2sx 2 4 dx V x NOTE r 2 r a b r ab b a 2ab[ 1 2 sin 2]0 2 2ab 2 0 0 ab 4ab y 2 0 cos2 d 4ab y 2 0 1 21 cos 2 d A 4 b a y a 0 sa 2 x 2 dx 4 b a y 2 0 a cos a cos d 0 2 sa 2 x 2 sa 2 a 2 sin2 sa 2 cos2 acos a cos 2 sin 1 x a 0 sin 0 x 0 dx a cos d x a sin SECTION 7.3 TRIGONOMETRIC SUBSTITUTION \|\|\|\| 469 œ„„„„„ ≈+4 2 ¨ x FIGURE 3 tan ¨= x 2 EXAMPLE 4 Find . SOLUTION It would be possible to use the trigonometric substitution here (as in Example 3). But the direct substitution is simpler, because then and M Example 4 illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution. You should look for a simpler method ﬁrst. EXAMPLE 5 Evaluate , where . SOLUTION 1 We let , where or . Then and Therefore The triangle in Figure 4 gives , so we have Writing , we have SOLUTION 2 For the hyperbolic substitution can also be used. Using the identity , we have Since , we obtain Since , we have and y dx sx 2 a 2 cosh1 x a C 2 t cosh1xa cosh t xa y dx sx 2 a 2 y a sinh t dt a sinh t y dt t C dx a sinh t dt sx 2 a 2 sa 2cosh2t 1 sa 2 sinh2t a sinh t cosh2y sinh2y 1 x a cosh t x 0 y dx sx 2 a 2 ln x sx 2 a 2 C1 1 C1 C ln a lnx sx 2 a 2 ln a C y dx sx 2 a 2 ln x a sx 2 a 2 a C tan sx 2 a 2 a y sec d lnsec tan C y dx sx 2 a 2 y a sec tan a tan d sx 2 a 2 sa 2sec 2 1 sa 2 tan2 atan a tan dx a sec tan d 32 0 2 x a sec a 0 y dx sx 2 a 2 NOTE y x sx 2 4 dx 1 2 y du su su C sx 2 4 C du 2x dx u x 2 4 x 2 tan y x sx 2 4 dx 470 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION FIGURE 4 sec ¨= x a œ„„„„„ a ¨ x ≈-a@ Although Formulas 1 and 2 look quite different, they are actually equivalent by Formula 3.11.4. M As Example 5 illustrates, hyperbolic substitutions can be used in place of trigo-nometric substitutions and sometimes they lead to simpler answers. But we usually use trigonometric substitutions because trigonometric identities are more familiar than hyper-bolic identities. EXAMPLE 6 Find . SOLUTION First we note that so trigonometric substitution is appropriate. Although is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitu-tion . When we combine this with the tangent substitution, we have , which gives and When , , so ; when , , so . Now we substitute so that . When , ; when . Therefore M EXAMPLE 7 Evaluate . SOLUTION We can transform the integrand into a function for which trigonometric substitu-tion is appropriate by ﬁrst completing the square under the root sign: This suggests that we make the substitution . Then and , so y x s3 2x x 2 dx y u 1 s4 u 2 du x u 1 du dx u x 1 4 x 12 3 2x x 2 3 x 2 2x 3 1 x 2 2x 1 y x s3 2x x 2 dx 3 16 u 1 u 1 12 3 16 [( 1 2 2) 1 1] 3 32 y 3 s32 0 x 3 4x 2 932 dx 3 16 y 12 1 1 u 2 u 2 du 3 16 y 12 1 1 u 2 du 3, u 1 2 u 1 0 du sin d u cos 3 16 y 3 0 1 cos2 cos2 sin d 3 16 y 3 0 tan3 sec d 3 16 y 3 0 sin3 cos2 d y 3 s32 0 x 3 4x 2 932 dx y 3 0 27 8 tan3 27 sec3 3 2 sec2 d 3 tan s3 x 3s3 2 0 tan 0 x 0 s4x 2 9 s9 tan2 9 3 sec dx 3 2 sec2 d x 3 2 tan u 2x s4x 2 9 4x 2 932 s4x 2 9 )3 y 3 s32 0 x 3 4x 2 932 dx NOTE SECTION 7.3 TRIGONOMETRIC SUBSTITUTION \|\|\|\| 471 We now substitute , giving and , so M s3 2x x 2 sin1 x 1 2 C s4 u 2 sin1 u 2 C 2 cos C y 2 sin 1 d y x s3 2x x 2 dx y 2 sin 1 2 cos 2 cos d s4 u 2 2 cos du 2 cos d u 2 sin 472 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION N Figure 5 shows the graphs of the integrand in Example 7 and its indeﬁnite integral (with ). Which is which? C 0 _4 _5 3 2 FIGURE 5 21. 23. 24. 25. 26. 27. 28. 29. 30. (a) Use trigonometric substitution to show that (b) Use the hyperbolic substitution to show that These formulas are connected by Formula 3.11.3. 32. Evaluate (a) by trigonometric substitution. (b) by the hyperbolic substitution . 33. Find the average value of , . 34. Find the area of the region bounded by the hyperbola and the line . x 3 9x 2 4y 2 36 1 x 7 f x sx 2 1x x a sinh t y x 2 x 2 a 232 dx y dx sx 2 a 2 sinh1 x a C x a sinh t y dx sx 2 a 2 ln(x sx 2 a 2 ) C 31. y 2 0 cos t s1 sin2t dt y xs1 x 4 dx y x 2 1 x 2 2x 22 dx y sx 2 2x dx y x 2 3 4x 4x 232 dx y x sx 2 x 1 dx y dt st 2 6t 13 y s5 4x x 2 dx y 1 0 sx 2 1 dx 22. y 0.6 0 x 2 s9 25x 2 dx 1–3 Evaluate the integral using the indicated trigonometric sub-stitution. Sketch and label the associated right triangle. 1. ; 2. ; ; 4–30 Evaluate the integral. 4. 5. 6. 8. 9. 10. 11. 12. 14. 15. 16. 18. 19. 20. y t s25 t 2 dt y s1 x 2 x dx y dx ax2 b 232 y x sx 2 7 dx 17. y 23 s23 dx x 5s9x 2 1 y a 0 x 2sa 2 x 2 dx y du us5 u 2 y sx 2 9 x 3 dx 13. y 1 0 xsx 2 4 dx y s1 4x 2 dx y t 5 st 2 2 dt y dx sx 2 16 y x 3 sx 2 100 dx y 1 x 2s25 x 2 dx 7. y 2 1 sx 2 1 x dx y 2 s2 1 t 3st 2 1 dt y 2 s3 0 x 3 s16 x 2 dx x 3 tan y x 3 sx 2 9 dx 3. x 3 sin y x 3s9 x 2 dx x 3 sec y 1 x 2sx 2 9 dx EXERCISES 7.3 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS \|\|\|\| 473 39. (a) Use trigonometric substitution to verify that (b) Use the ﬁgure to give trigonometric interpretations of both terms on the right side of the equation in part (a). 40. The parabola divides the disk into two parts. Find the areas of both parts. 41. Find the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii and . (See the ﬁgure.) 42. A water storage tank has the shape of a cylinder with diam-eter 10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage of the total capacity is being used? 43. A torus is generated by rotating the circle about the -axis. Find the volume enclosed by the torus. x x 2 y R2 r 2 R r R r x 2 y 2 8 y 1 2x 2 ¨ ¨ y=œ„„„„„ a@-t@ t 0 y a x y x 0 sa 2 t 2 dt 1 2a 2 sin1xa 1 2x sa 2 x 2 35. Prove the formula for the area of a sector of a circle with radius and central angle . [Hint: Assume and place the center of the circle at the origin so it has the equation . Then is the sum of the area of the triangle and the area of the region in the ﬁgure.] ; 36. Evaluate the integral Graph the integrand and its indeﬁnite integral on the same screen and check that your answer is reasonable. ; 37. Use a graph to approximate the roots of the equation . Then approximate the area bounded by the curve and the line . 38. A charged rod of length produces an electric ﬁeld at point given by where is the charge density per unit length on the rod and is the free space permittivity (see the ﬁgure). Evaluate the integral to determine an expression for the electric ﬁeld . 0 x y L P (a, b) EP 0 EP y La a b 40x 2 b 232 dx Pa, b L y 2 x y x 2s4 x 2 x 2s4 x 2 2 x y dx x 4sx 2 2 O x y R Q ¨ P PQR POQ A x 2 y 2 r 2 0 2 r A 1 2r 2 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions and to a common denominator we obtain If we now reverse the procedure, we see how to integrate the function on the right side of 2 x 1 1 x 2 2x 2 x 1 x 1x 2 x 5 x 2 x 2 1x 2 2x 1 7.4 this equation: To see how the method of partial fractions works in general, let’s consider a rational function where and are polynomials. It’s possible to express as a sum of simpler fractions provided that the degree of is less than the degree of . Such a rational function is called proper. Recall that if where , then the degree of is and we write . If is improper, that is, , then we must take the preliminary step of dividing into (by long division) until a remainder is obtained such that . The division statement is where and are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required. EXAMPLE 1 Find . SOLUTION Since the degree of the numerator is greater than the degree of the denominator, we ﬁrst perform the long division. This enables us to write M The next step is to factor the denominator as far as possible. It can be shown that any polynomial can be factored as a product of linear factors (of the form ) and irreducible quadratic factors (of the form , where ). For instance, if , we could factor it as The third step is to express the proper rational function (from Equation 1) as a sum of partial fractions of the form Ax B ax 2 bx c j or A ax bi RxQx Qx x 2 4x 2 4 x 2x 2x 2 4 Qx x 4 16 b 2 4ac 0 ax 2 bx c ax b Q Qx x 3 3 x 2 2 2x 2 ln x 1 C y x 3 x x 1 dx y x 2 x 2 2 x 1 dx y x 3 x x 1 dx V R S fx Px Qx Sx Rx Qx 1 degR degQ Rx P Q degP degQ f degP n n P an 0 Px anx n an1x n1 a1x a0 Q P f Q P fx Px Qx 2 ln x 1 ln x 2 C y x 5 x 2 x 2 dx y 2 x 1 1 x 2 dx 474 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION x-1 ≈+x +2 ˛-≈ ≈+x ≈-x 2x 2x-2 2 ˛ +x ) A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur. CASE I N The denominator Q(x) is a product of distinct linear factors. This means that we can write where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants such that These constants can be determined as in the following example. EXAMPLE 2 Evaluate . SOLUTION Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide. We factor the denominator as Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form To determine the values of , , and , we multiply both sides of this equation by the product of the denominators, , obtaining Expanding the right side of Equation 4 and writing it in the standard form for polyno-mials, we get The polynomials in Equation 5 are identical, so their coefﬁcients must be equal. The coefﬁcient of on the right side, , must equal the coefﬁcient of on the left side—namely, 1. Likewise, the coefﬁcients of are equal and the constant terms are equal. This gives the following system of equations for , , and : 2A 2B 2C 1 3A 2B C 2 2A B 2C 1 C B A x x 2 2A B 2C x 2 x 2 2x 1 2A B 2Cx 2 3A 2B Cx 2A 5 x 2 2x 1 A2x 1x 2 Bxx 2 Cx2x 1 4 x2x 1x 2 C B A x 2 2x 1 x2x 1x 2 A x B 2x 1 C x 2 3 2x 3 3x 2 2x x2x 2 3x 2 x2x 1x 2 y x 2 2x 1 2x 3 3x 2 2x dx V Rx Qx A1 a1x b1 A2 a2x b2 Ak akx bk 2 A1, A2, . . . , Ak Qx a1x b1a2x b2 akx bk SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS \|\|\|\| 475 N Another method for ﬁnding , , and is given in the note after this example. C B A Solving, we get , , and , and so In integrating the middle term we have made the mental substitution , which gives and . M We can use an alternative method to ﬁnd the coefﬁcients , , and in Example 2. Equation 4 is an identity; it is true for every value of . Let’s choose values of that simplify the equation. If we put in Equation 4, then the second and third terms on the right side vanish and the equation then becomes , or . Likewise, gives and gives , so and . (You may object that Equation 3 is not valid for , , or , so why should Equation 4 be valid for those values? In fact, Equation 4 is true for all values of , even , , and . See Exercise 69 for the reason.) EXAMPLE 3 Find , where . SOLUTION The method of partial fractions gives and therefore Using the method of the preceding note, we put in this equation and get , so . If we put , we get , so . Thus Since , we can write the integral as See Exercises 55–56 for ways of using Formula 6. M CASE 11 N Q(x) is a product of linear factors, some of which are repeated. Suppose the ﬁrst linear factor is repeated times; that is, occurs in the factorization of . Then instead of the single term in Equation 2, we A1a1x b1 Qx a1x b1r r a1x b1 y dx x 2 a 2 1 2a ln x a x a C 6 ln x ln y lnxy 1 2a (ln x a ln x a ) C y dx x 2 a 2 1 2a y 1 x a 1 x a dx B 12a B2a 1 x a A 12a A2a 1 x a Ax a Bx a 1 1 x 2 a 2 1 x ax a A x a B x a a 0 y dx x 2 a 2 2 1 2 x 0 x 2 1 2 x 0 C 1 10 B 1 5 10C 1 x 2 5B4 1 4 x 1 2 A 1 2 2A 1 x 0 x x C B A NOTE dx du2 du 2 dx u 2x 1 1 2 ln x 1 10 ln 2x 1 1 10 ln x 2 K y x 2 2x 1 2x 3 3x 2 2x dx y 1 2 1 x 1 5 1 2x 1 1 10 1 x 2 dx C 1 10 B 1 5 A 1 2 476 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION N We could check our work by taking the terms to a common denominator and adding them. N Figure 1 shows the graphs of the integrand in Example 2 and its indeﬁnite integral (with ). Which is which? K 0 FIGURE 1 _3 _2 2 3 would use By way of illustration, we could write but we prefer to work out in detail a simpler example. EXAMPLE 4 Find . SOLUTION The ﬁrst step is to divide. The result of long division is The second step is to factor the denominator . Since , we know that is a factor and we obtain Since the linear factor occurs twice, the partial fraction decomposition is Multiplying by the least common denominator, , we get Now we equate coefﬁcients: Solving, we obtain , , and , so M x 2 2 x 2 x 1 ln x 1 x 1 K x 2 2 x ln x 1 2 x 1 ln x 1 K y x 4 2x 2 4x 1 x 3 x 2 x 1 dx y x 1 1 x 1 2 x 12 1 x 1 dx C 1 B 2 A 1 A B C 0 A B 2C 4 A B C 0 A Cx 2 B 2Cx A B C 4x Ax 1x 1 Bx 1 Cx 12 8 x 12x 1 4x x 12x 1 A x 1 B x 12 C x 1 x 1 x 12x 1 x 3 x 2 x 1 x 1x 2 1 x 1x 1x 1 x 1 Q1 0 Qx x 3 x 2 x 1 x 4 2x 2 4x 1 x 3 x 2 x 1 x 1 4x x 3 x 2 x 1 y x 4 2x 2 4x 1 x 3 x 2 x 1 dx x 3 x 1 x 2x 13 A x B x 2 C x 1 D x 12 E x 13 A1 a1x b1 A2 a1x b12 Ar a1x b1r 7 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS \|\|\|\| 477 N Another method for ﬁnding the coefﬁcients: Put in (8): . Put : . Put : . A B C 1 x 0 C 1 x 1 B 2 x 1 CASE III N Q(x) contains irreducible quadratic factors, none of which is repeated. If has the factor , where , then, in addition to the partial fractions in Equations 2 and 7, the expression for will have a term of the form where and are constants to be determined. For instance, the function given by has a partial fraction decomposition of the form The term given in (9) can be integrated by completing the square and using the formula EXAMPLE 5 Evaluate . SOLUTION Since can’t be factored further, we write Multiplying by , we have Equating coefﬁcients, we obtain Thus , , and and so In order to integrate the second term we split it into two parts: We make the substitution in the ﬁrst of these integrals so that . We evaluate the second integral by means of Formula 10 with : M ln x 1 2 lnx 2 4 1 2 tan1x2 K y 2x 2 x 4 xx 2 4 dx y 1 x dx y x x 2 4 dx y 1 x 2 4 dx a 2 du 2x dx u x 2 4 y x 1 x 2 4 dx y x x 2 4 dx y 1 x 2 4 dx y 2x 2 x 4 x 3 4x dx y 1 x x 1 x 2 4 dx C 1 B 1 A 1 4A 4 C 1 A B 2 A Bx 2 Cx 4A 2x 2 x 4 Ax 2 4 Bx Cx xx 2 4 2x 2 x 4 xx 2 4 A x Bx C x 2 4 x 3 4x xx 2 4 y 2x 2 x 4 x 3 4x dx V y dx x 2 a 2 1 a tan1 x a C 10 x x 2x 2 1x 2 4 A x 2 Bx C x 2 1 Dx E x 2 4 fx xx 2x 2 1x 2 4 B A Ax B ax 2 bx c 9 RxQx b 2 4ac 0 ax 2 bx c Qx 478 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION EXAMPLE 6 Evaluate . SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we ﬁrst divide and obtain Notice that the quadratic is irreducible because its discriminant is . This means it can’t be factored, so we don’t need to use the partial fraction technique. To integrate the given function we complete the square in the denominator: This suggests that we make the substitution . Then, and , so M Example 6 illustrates the general procedure for integrating a partial fraction of the form We complete the square in the denominator and then make a substitution that brings the integral into the form Then the ﬁrst integral is a logarithm and the second is expressed in terms of . CASE IV N Q(x) contains a repeated irreducible quadratic factor. If has the factor , where , then instead of the single partial fraction (9), the sum A1x B1 ax 2 bx c A2x B2 ax 2 bx c2 Ar x Br ax 2 bx cr 11 b 2 4ac 0 ax 2 bx cr Qx tan1 y Cu D u2 a2 du C y u u2 a 2 du D y 1 u2 a2 du where b 2 4ac 0 Ax B ax 2 bx c NOTE x 1 8 ln4x 2 4x 3 1 4s2 tan1 2x 1 s2 C x 1 8 lnu2 2 1 4 1 s2 tan1 u s2 C x 1 4 y u u2 2 du 1 4 y 1 u2 2 du x 1 2 y 1 2u 1 1 u2 2 du x 1 4 y u 1 u2 2 du y 4x 2 3x 2 4x 2 4x 3 dx y 1 x 1 4x 2 4x 3 dx x 1 2u 1 du 2 dx u 2x 1 4x 2 4x 3 2x 12 2 b 2 4ac 32 0 4x 2 4x 3 4x 2 3x 2 4x 2 4x 3 1 x 1 4x 2 4x 3 y 4x 2 3x 2 4x 2 4x 3 dx SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS \|\|\|\| 479 occurs in the partial fraction decomposition of . Each of the terms in (11) can be integrated by ﬁrst completing the square. EXAMPLE 7 Write out the form of the partial fraction decomposition of the function SOLUTION M EXAMPLE 8 Evaluate . SOLUTION The form of the partial fraction decomposition is Multiplying by , we have If we equate coefﬁcients, we get the system which has the solution , , , , and . Thus M We note that sometimes partial fractions can be avoided when integrating a rational func-tion. For instance, although the integral y x 2 1 xx 2 3 dx ln x 1 2 lnx 2 1 tan1x 1 2x 2 1 K y dx x y x x 2 1 dx y dx x 2 1 y x dx x 2 12 y 1 x 2x 2 x 3 xx 2 12 dx y 1 x x 1 x 2 1 x x 2 12 dx E 0 D 1 C 1 B 1 A 1 A 1 C E 1 2A B D 2 C 1 A B 0 A Bx 4 Cx 3 2A B Dx 2 C Ex A Ax 4 2x 2 1 Bx 4 x 2 Cx 3 x Dx 2 Ex x 3 2x 2 x 1 Ax 2 12 Bx Cxx 2 1 Dx Ex xx 2 12 1 x 2x 2 x 3 xx 2 12 A x Bx C x 2 1 Dx E x 2 12 y 1 x 2x 2 x 3 xx 2 12 dx A x B x 1 Cx D x 2 x 1 Ex F x 2 1 Gx H x 2 12 Ix J x 2 13 x 3 x 2 1 xx 1x 2 x 1x 2 13 x 3 x 2 1 xx 1x 2 x 1x 2 13 RxQx 480 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION N It would be extremely tedious to work out by hand the numerical values of the coefﬁcients in Example 7. Most computer algebra systems, however, can ﬁnd the numerical values very quickly. For instance, the Maple command or the Mathematica command gives the following values: I 1 2, J 1 2 E 15 8 , F 1 8, G H 3 4, A 1, B 1 8, C D 1, Apart[f] convertf, parfrac, x N In the second and fourth terms we made the mental substitution . u x 2 1 could be evaluated by the method of Case III, it’s much easier to observe that if , then and so RATIONALIZING SUBSTITUTIONS Some nonrational functions can be changed into rational functions by means of appropri-ate substitutions. In particular, when an integrand contains an expression of the form , then the substitution may be effective. Other instances appear in the exercises. EXAMPLE 9 Evaluate . SOLUTION Let . Then , so and . Therefore We can evaluate this integral either by factoring as and using partial fractions or by using Formula 6 with : M 2sx 4 2 ln sx 4 2 sx 4 2 C 2u 8 1 2 2 ln u 2 u 2 C y sx 4 x dx 2 y du 8 y du u 2 4 a 2 u 2u 2 u 2 4 2 y 1 4 u 2 4 du y sx 4 x dx y u u 2 4 2u du 2 y u 2 u 2 4 du dx 2u du x u 2 4 u 2 x 4 u sx 4 y sx 4 x dx u s n tx s n tx y x 2 1 xx 2 3 dx 1 3 ln x 3 3x C du 3x 2 3 dx u xx 2 3 x 3 3x SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS \|\|\|\| 481 (a) (b) 6. (a) (b) 7–38 Evaluate the integral. 7. 8. 9. 10. y 1 t 4t 1 dt y x 9 x 5x 2 dx y r 2 r 4 dr y x x 6 dx 1 x 6 x 3 x 4 x 3 xx 2 x 3 t 4 t 2 1 t 2 1t 2 42 x 4 x 4 1 5. 1–6 Write out the form of the partial fraction decomposition of the function (as in Example 7). Do not determine the numerical values of the coefﬁcients. 1. (a) (b) 2. (a) (b) 3. (a) (b) 4. (a) (b) 2x 1 x 13x 2 42 x 3 x 2 4x 3 1 x 2 92 x 4 1 x 5 4x 3 x 2 x 2 x 2 x x 2 x 2 1 x 3 2x 2 x 2x x 33x 1 EXERCISES 7.4 482 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION 48. 49. 50. 51–52 Use integration by parts, together with the techniques of this section, to evaluate the integral. 51. 52. ; 53. Use a graph of to decide whether is positive or negative. Use the graph to give a rough estimate of the value of the integral and then use partial fractions to ﬁnd the exact value. ; 54. Graph both and an antiderivative on the same screen. 55–56 Evaluate the integral by completing the square and using Formula 6. 56. 57. The German mathematician Karl Weierstrass (1815–1897) noticed that the substitution will convert any rational function of and into an ordinary rational function of . (a) If , , sketch a right triangle or use trigonometric identities to show that (b) Show that (c) Show that 58–61 Use the substitution in Exercise 57 to transform the inte-grand into a rational function of and then evaluate the integral. 58. 59. 60. y 2 3 1 1 sin x cos x dx y 1 3 sin x 4 cos x dx y dx 3 5 sin x t dx 2 1 t 2 dt cos x 1 t 2 1 t 2 and sin x 2t 1 t 2 cos x 2 1 s1 t 2 and sin x 2 t s1 t 2 x t tanx2 t cos x sin x t tanx2 y 2x 1 4x 2 12x 7 dx y dx x 2 2x 55. y 1x 3 2x 2 x2 0 f x dx f x 1x 2 2x 3 y x tan1x dx y lnx 2 x 2 dx y e x e x 2e 2x 1 dx y sec 2 t tan2 t 3 tan t 2 dt y cos x sin2x sin x dx y e 2x e 2x 3e x 2 dx 47. 12. 13. 14. 15. 16. 18. 19. 20. 21. 22. 23. 24. 26. 27. 28. 30. 32. 33. 34. 35. 36. 37. 38. 39–50 Make a substitution to express the integrand as a rational function and then evaluate the integral. 39. 40. 41. 42. 44. 45. [Hint: Substitute .] 46. y s1 sx x dx u 6 sx y 1 sx s 3 x dx y 3 13 sx x 2 x dx y x 3 s 3 x 2 1 dx 43. y 1 0 1 1 s 3 x dx y 16 9 sx x 4 dx y dx 2sx 3 x y 1 xsx 1 dx y x 3 2x 2 3x 2 x 2 2x 22 dx y x 2 3x 7 x 2 4x 62 dx y x 4 3x 2 1 x 5 5x 3 5x dx y dx xx 2 42 y x 3 x 3 1 dx y 1 0 x 3 2x x 4 4x 2 3 dx y 1 0 x x 2 4x 13 dx y 1 x 3 1 dx 31. y 3x 2 x 4 x 4 3x 2 2 dx y x 4 x 2 2x 5 dx 29. y x 2 2x 1 x 12x 2 1 dx y x 3 x 2 2x 1 x 2 1x 2 2 dx y x 2 x 1 x 2 12 dx y 10 x 1x 2 9 dx 25. y x 2 x 6 x 3 3x dx y 5x 2 3x 2 x 3 2x 2 dx y ds s 2s 12 y x 3 4 x 2 4 dx y x 2 5x 16 2x 1x 22 dx y 1 x 52x 1 dx y x 2 2x 1 x 3 x dx y 2 1 4y 2 7y 12 yy 2y 3 dy 17. y 1 0 x 3 4x 10 x 2 x 6 dx y 4 3 x 3 2x 2 4 x 3 2x 2 dx y 1 x ax b dx y ax x 2 bx dx y 1 0 x 1 x 2 3x 2 dx y 3 2 1 x 2 1 dx 11. SECTION 7.5 STRATEGY FOR INTEGRATION \|\|\|\| 483 67. (a) Use a computer algebra system to ﬁnd the partial fraction decomposition of the function (b) Use part (a) to ﬁnd (by hand) and compare with the result of using the CAS to integrate directly. Com-ment on any discrepancy. 68. (a) Find the partial fraction decomposition of the function (b) Use part (a) to ﬁnd and graph and its indeﬁnite integral on the same screen. (c) Use the graph of to discover the main features of the graph of . 69. Suppose that , and are polynomials and for all except when . Prove that for all . [Hint: Use continuity.] 70. If is a quadratic function such that and is a rational function, ﬁnd the value of . f 0 y f x x 2x 13 dx f 0 1 f x Fx Gx Qx 0 x Fx Qx Gx Qx Q F, G x f x dx f f x f x dx f x 12x 5 7x 3 13x 2 8 100x 6 80x 5 116x 4 80x 3 41x 2 20x 4 CAS f x f x dx f x 4x 3 27x 2 5x 32 30x 5 13x 4 50x 3 286x 2 299x 70 CAS 61. 62–63 Find the area of the region under the given curve from 1 to 2. 62. 63. 64. Find the volume of the resulting solid if the region under the curve from to is rotated about (a) the -axis and (b) the -axis. 65. One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. If represents the number of female insects in a population, the number of sterile males introduced each generation, and the population’s natural growth rate, then the female population is related to time by Suppose an insect population with 10,000 females grows at a rate of and 900 sterile males are added. Evaluate the integral to give an equation relating the female population to time. (Note that the resulting equation can’t be solved explic-itly for .) 66. Factor as a difference of squares by ﬁrst adding and subtracting the same quantity. Use this factorization to evalu-ate . x 1x 4 1 dx x 4 1 P r 0.10 t y P S Pr 1P S dP t r S P y x x 1 x 0 y 1x 2 3x 2 y x 2 1 3x x 2 y 1 x 3 x y 2 0 sin 2x 2 cos x dx STRATEGY FOR INTEGRATION As we have seen, integration is more challenging than differentiation. In ﬁnding the deriv-ative of a function it is obvious which differentiation formula we should apply. But it may not be obvious which technique we should use to integrate a given function. Until now individual techniques have been applied in each section. For instance, we usually used substitution in Exercises 5.5, integration by parts in Exercises 7.1, and partial fractions in Exercises 7.4. But in this section we present a collection of miscellaneous inte-grals in random order and the main challenge is to recognize which technique or formula to use. No hard and fast rules can be given as to which method applies in a given situation, but we give some advice on strategy that you may ﬁnd useful. A prerequisite for strategy selection is a knowledge of the basic integration formulas. In the following table we have collected the integrals from our previous list together with several additional formulas that we have learned in this chapter. Most of them should be memorized. It is useful to know them all, but the ones marked with an asterisk need not be 7.5 memorized since they are easily derived. Formula 19 can be avoided by using partial frac-tions, and trigonometric substitutions can be used in place of Formula 20. TABLE OF INTEGRATION FORMULAS Constants of integration have been omitted. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. Once you are armed with these basic integration formulas, if you don’t immediately see how to attack a given integral, you might try the following four-step strategy. 1. Simplify the Integrand if Possible Sometimes the use of algebraic manipula-tion or trigonometric identities will simplify the integrand and make the method of integration obvious. Here are some examples: y 1 2 sin x cos x dx y sin x cos x2 dx y sin2x 2 sin x cos x cos2x dx y sin cos d 1 2 y sin 2 d y tan sec2 d y sin cos cos2 d y sx (1 sx ) dx y (sx x) dx y dx sx 2 a2 ln x sx 2 a2 y dx x 2 a2 1 2a ln x a x a y dx sa 2 x 2 sin1 x a y dx x 2 a 2 1 a tan1 x a y cosh x dx sinh x y sinh x dx cosh x y cot x dx ln sin x y tan x dx ln sec x y csc x dx ln csc x cot x y sec x dx ln sec x tan x y csc x cot x dx csc x y sec x tan x dx sec x y csc2x dx cot x y sec2x dx tan x y cos x dx sin x y sin x dx cos x y a x dx a x ln a y e x dx e x y 1 x dx ln x n 1 y x n dx x n1 n 1 484 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION 2. Look for an Obvious Substitution Try to ﬁnd some function in the integrand whose differential also occurs, apart from a constant fac-tor. For instance, in the integral we notice that if , then . Therefore we use the substitu-tion instead of the method of partial fractions. 3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the solution, then we take a look at the form of the integrand . (a) Trigonometric functions. If is a product of powers of and , of and , or of and , then we use the substitutions recom-mended in Section 7.2. (b) Rational functions. If is a rational function, we use the procedure of Sec-tion 7.4 involving partial fractions. (c) Integration by parts. If is a product of a power of (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing and according to the advice given in Section 7.1. If you look at the functions in Exercises 7.1, you will see that most of them are the type just described. (d) Radicals. Particular kinds of substitutions are recommended when certain radicals appear. (i) If occurs, we use a trigonometric substitution according to the table in Section 7.3. (ii) If occurs, we use the rationalizing substitution . More generally, this sometimes works for . 4. Try Again If the ﬁrst three steps have not produced the answer, remember that there are basically only two methods of integration: substitution and parts. (a) Try substitution. Even if no substitution is obvious (Step 2), some inspiration or ingenuity (or even desperation) may suggest an appropriate substitution. (b) Try parts. Although integration by parts is used most of the time on products of the form described in Step 3(c), it is sometimes effective on single func-tions. Looking at Section 7.1, we see that it works on , , and , and these are all inverse functions. (c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the denominator or using trigonometric identities) may be useful in transforming the integral into an easier form. These manipulations may be more substantial than in Step 1 and may involve some ingenuity. Here is an example: (d) Relate the problem to previous problems. When you have built up some expe-rience in integration, you may be able to use a method on a given integral that is similar to a method you have already used on a previous integral. Or you may even be able to express the given integral in terms of a previous one. For y 1 cos x sin2x dx y csc2x cos x sin2x dx y dx 1 cos x y 1 1 cos x 1 cos x 1 cos x dx y 1 cos x 1 cos2x dx ln x sin1x tan1x s n tx u s n ax b s n ax b sx 2 a 2 dv u x fx f csc x cot x sec x tan x cos x sin x fx fx u x 2 1 du 2x dx u x 2 1 y x x 2 1 dx du tx dx u tx SECTION 7.5 STRATEGY FOR INTEGRATION \|\|\|\| 485 instance, is a challenging integral, but if we make use of the iden-tity , we can write and if has previously been evaluated (see Example 8 in Section 7.2), then that calculation can be used in the present problem. (e) Use several methods. Sometimes two or three methods are required to evalu-ate an integral. The evaluation could involve several successive substitutions of different types, or it might combine integration by parts with one or more substitutions. In the following examples we indicate a method of attack but do not fully work out the integral. EXAMPLE 1 In Step 1 we rewrite the integral: The integral is now of the form with odd, so we can use the advice in Section 7.2. Alternatively, if in Step 1 we had written then we could have continued as follows with the substitution : M EXAMPLE 2 According to (ii) in Step 3(d), we substitute . Then , so and The integrand is now a product of and the transcendental function so it can be inte-grated by parts. M e u u y esx dx 2 y ue u du dx 2u du x u 2 u sx y esx dx V y u 2 1 u 6 du y u 4 u 6 du y sin3x cos6x dx y 1 cos2x cos6x sin x dx y 1 u 2 u 6 du u cos x y tan3x cos3x dx y sin3x cos3x 1 cos3x dx y sin3x cos6x dx m x tanmx secnx dx y tan3x cos3x dx y tan3x sec3x dx y tan3x cos3x dx x sec3x dx y tan2x sec x dx y sec3x dx y sec x dx tan2x sec2x 1 x tan2x sec x dx 486 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION EXAMPLE 3 No algebraic simpliﬁcation or substitution is obvious, so Steps 1 and 2 don’t apply here. The integrand is a rational function so we apply the procedure of Section 7.4, remember-ing that the ﬁrst step is to divide. M EXAMPLE 4 Here Step 2 is all that is needed. We substitute because its differential is , which occurs in the integral. M EXAMPLE 5 Although the rationalizing substitution works here [(ii) in Step 3(d)], it leads to a very complicated rational function. An easier method is to do some algebraic manipulation [either as Step 1 or as Step 4(c)]. Multiply-ing numerator and denominator by , we have M CAN WE INTEGRATE ALL CONTINUOUS FUNCTIONS? The question arises: Will our strategy for integration enable us to ﬁnd the integral of every continuous function? For example, can we use it to evaluate ? The answer is No, at least not in terms of the functions that we are familiar with. The functions that we have been dealing with in this book are called elementary func-tions. These are the polynomials, rational functions, power functions , exponential functions , logarithmic functions, trigonometric and inverse trigonometric functions, hyperbolic and inverse hyperbolic functions, and all functions that can be obtained from these by the ﬁve operations of addition, subtraction, multiplication, division, and compo-sition. For instance, the function is an elementary function. If is an elementary function, then is an elementary function but need not be an elementary function. Consider . Since is continuous, its integral exists, and if we deﬁne the function by Fx y x 0 et 2 dt F f fx ex2 x fx dx f f fx x 2 1 x 3 2x 1 lncosh x xe sin 2x a x x a x ex2 dx sin1x s1 x 2 C y 1 s1 x 2 dx y x s1 x 2 dx y 1 x 1 x dx y 1 x s1 x 2 dx s1 x u 1 x 1 x y 1 x 1 x dx V du dxx u ln x y dx xsln x V y x 5 1 x 3 3x 2 10x dx SECTION 7.5 STRATEGY FOR INTEGRATION \|\|\|\| 487 then we know from Part 1 of the Fundamental Theorem of Calculus that Thus, has an antiderivative , but it has been proved that is not an elemen-tary function. This means that no matter how hard we try, we will never succeed in evalu-ating in terms of the functions we know. (In Chapter 11, however, we will see how to express as an inﬁnite series.) The same can be said of the following integrals: In fact, the majority of elementary functions don’t have elementary antiderivatives. You may be assured, though, that the integrals in the following exercises are all elementary functions. y sx 3 1 dx y 1 ln x dx y sin x x dx y e x x dx y sinx 2 dx y cose x dx x ex2 dx x ex2 dx F F fx ex2 Fx ex2 488 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION 25. 26. 27. 28. 29. 30. 32. 33. 34. 35. 36. 37. 38. 39. 40. 42. 43. 44. 46. 47. 48. y x x 4 a 4 dx y x 3x 14 dx y 1 sin x 1 sin x dx y x 5e x3dx 45. y s1 e x dx y e xs1 e x dx y tan1 x x 2 dx y tan2 d 41. y 1 s4y 2 4y 3 dy y sec tan sec2 sec d y 4 0 tan5 sec3 d y 4 0 cos2 tan2 d y sin 4x cos 3x dx y 1 1 x 8 sin x dx y 2 4 1 4 cot x 4 cot x dx y s3 2x x 2 dx y s2x 1 2x 3 dx y 1 x 1 x dx 31. y 2 2 x 2 4x dx y 5 0 3w 1 w 2 dw y sin sat dt y dx 1 e x y 3x 2 2 x 3 2x 8 dx y 3x 2 2 x 2 2x 8 dx 1–80 Evaluate the integral. 1. 2. 3. 4. 5. 6. 8. 9. 10. 11. 12. 13. 14. 15. 16. 18. 19. 20. 21. 22. 24. y lnx 2 1 dx y 1 0 (1 sx )8 dx 23. y ln x xs1 ln x2 dx y arctan sx dx y e 2 dx y e xe xdx y e2t 1 e4t dt y x sin2x dx 17. y s22 0 x 2 s1 x 2 dx y dx 1 x 232 y x 3 s1 x 2 dx y sin3 cos5 d y x x 4 x 2 1 dx y x 1 x 2 4x 5 dx y 4 0 x 1 x 2 4x 5 dx y 3 1 r 4 ln r dr y x csc x cot x dx y 1 1 e arctan y 1 y 2 dy 7. y x s3 x 4 dx y 2 0 2t t 32 dt y tan3 d y sin x sec x tan x dx y sin3x cos x dx y cos x 1 sin2x dx EXERCISES 7.5 SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS \|\|\|\| 489 67. 68. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. The functions and don’t have elementary antiderivatives, but does. Evaluate . x 2x 2 1e x2 dx y 2x 2 1e x2 y x 2e x2 y e x2 y sin x cos x sin4x cos4x dx y x sin2x cos x dx y sec x cos 2x sin x sec x dx y sx 1 x 3 dx y x 2 bx sin 2x dx y xe x s1 e x dx y dx sx (2 sx )4 y 1 x 2x 2 4 dx y 4 x 10 x 2 x dx y x arcsin x s1 x 2 dx y lnx 1 x 2 dx y e 2x 1 e x dx 69. y 1 1 2e x ex dx y s3 1 s1 x 2 x 2 dx 50. 51. 52. 53. 54. 55. 56. 58. 59. 60. 62. 63. 64. 65. 66. y 3 2 u 3 1 u 3 u 2 du y 1 sx 1 sx dx y 3 4 lntan x sin x cos x dx y sin 2x 1 cos4x dx y 1 x s 3 x dx y sx esx dx 61. y dx x 2s4x 2 1 y cos x cos3sin x dx y x ln x sx 2 1 dx y xs 3 x c dx 57. y dx sx xsx y dx x xsx y x sin x2 dx y x 2 sinh mx dx y dx xx 4 1 y 1 xs4x 2 1 dx y 1 x 2s4x 1 dx y 1 xs4x 1 dx 49. INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS In this section we describe how to use tables and computer algebra systems to integrate functions that have elementary antiderivatives. You should bear in mind, though, that even the most powerful computer algebra systems can’t ﬁnd explicit formulas for the antideriv-atives of functions like or the other functions described at the end of Section 7.5. TABLES OF INTEGRALS Tables of indeﬁnite integrals are very useful when we are confronted by an integral that is difﬁcult to evaluate by hand and we don’t have access to a computer algebra system. A rel-atively brief table of 120 integrals, categorized by form, is provided on the Reference Pages at the back of the book. More extensive tables are available in CRC Standard Mathe-matical Tables and Formulae, 31st ed. by Daniel Zwillinger (Boca Raton, FL: CRC Press, 2002) (709 entries) or in Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products, 6e (San Diego: Academic Press, 2000), which contains hundreds of pages of integrals. It should be remembered, however, that integrals do not often occur in exactly the form listed in a table. Usually we need to use substitution or algebraic manipulation to transform a given integral into one of the forms in the table. EXAMPLE 1 The region bounded by the curves , and is rotated about the -axis. Find the volume of the resulting solid. SOLUTION Using the method of cylindrical shells, we see that the volume is V y 1 0 2x arctan x dx y x 1 y arctan x, y 0 e x2 7.6 In the section of the Table of Integrals titled Inverse Trigonometric Forms we locate Formula 92: Thus the volume is M EXAMPLE 2 Use the Table of Integrals to ﬁnd . SOLUTION If we look at the section of the table titled Forms involving , we see that the closest entry is number 34: This is not exactly what we have, but we will be able to use it if we ﬁrst make the substi-tution : Then we use Formula 34 with (so ): M EXAMPLE 3 Use the Table of Integrals to ﬁnd . SOLUTION If we look in the section called Trigonometric Forms, we see that none of the entries explicitly includes a factor. However, we can use the reduction formula in entry 84 with : We now need to evaluate . We can use the reduction formula in entry 85 with , followed by entry 82: x 2 sin x 2sin x x cos x K y x 2 cos x dx x 2 sin x 2 y x sin x dx n 2 x x 2 cos x dx y x 3 sin x dx x 3 cos x 3 y x 2 cos x dx n 3 u 3 y x 3 sin x dx x 8 s5 4x 2 5 16 sin1 2x s5 C y x 2 s5 4x 2 dx 1 8 y u 2 s5 u 2 du 1 8 u 2 s5 u 2 5 2 sin1 u s5 C a s5 a 2 5 y x 2 s5 4x 2 dx y u22 s5 u 2 du 2 1 8 y u 2 s5 u 2 du u 2x y u 2 sa 2 u 2 du u 2 sa 2 u 2 a 2 2 sin1 u a C sa 2 u 2 y x 2 s5 4x 2 dx V 24 1 1 2 2 [x 2 1 tan1x x]0 1 2 tan1 1 1 V 2 y 1 0 x tan1x dx 2 x 2 1 2 tan1x x 2 0 1 y u tan1u du u 2 1 2 tan1u u 2 C 490 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION N The Table of Integrals appears on Reference Pages 6–10 at the back of the book. 85. u n sin u n y u n1 sin u du y u n cos u du Combining these calculations, we get where . M EXAMPLE 4 Use the Table of Integrals to ﬁnd . SOLUTION Since the table gives forms involving , , and , but not , we ﬁrst complete the square: If we make the substitution (so ), the integrand will involve the pattern : The ﬁrst integral is evaluated using the substitution : For the second integral we use Formula 21 with : Thus M COMPUTER ALGEBRA SYSTEMS We have seen that the use of tables involves matching the form of the given integrand with the forms of the integrands in the tables. Computers are particularly good at matching pat-terns. And just as we used substitutions in conjunction with tables, a CAS can perform sub-stitutions that transform a given integral into one that occurs in its stored formulas. So it isn’t surprising that computer algebra systems excel at integration. That doesn’t mean that integration by hand is an obsolete skill. We will see that a hand computation sometimes produces an indeﬁnite integral in a form that is more convenient than a machine answer. To begin, let’s see what happens when we ask a machine to integrate the relatively simple function . Using the substitution , an easy calculation by hand gives y 1 3x 2 dx 1 3 ln 3x 2 C u 3x 2 y 13x 2 1 3x 2 2x 432 x 1 2 sx 2 2x 4 3 2 ln(x 1 sx 2 2x 4 ) C y xsx 2 2x 4 dx y su 2 3 du u 2 su 2 3 3 2 ln(u su 2 3 ) a s3 y usu 2 3 du 1 2 y st dt 1 2 2 3t 32 1 3u 2 332 t u 2 3 y usu 2 3 du y su 2 3 du y xsx 2 2x 4 dx y u 1 su 2 3 du sa 2 u 2 x u 1 u x 1 x 2 2x 4 x 12 3 sax 2 bx c sx 2 a 2 sa 2 x 2 sa 2 x 2 y xsx 2 2x 4 dx V C 3K y x 3 sin x dx x 3 cos x 3x 2 sin x 6x cos x 6 sin x C SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS \|\|\|\| 491 21. a 2 2 ln(u sa 2 u 2 ) C y sa 2 u 2 du u 2 sa 2 u 2 whereas Derive, Mathematica, and Maple all return the answer The ﬁrst thing to notice is that computer algebra systems omit the constant of integra-tion. In other words, they produce a particular antiderivative, not the most general one. Therefore, when making use of a machine integration, we might have to add a constant. Second, the absolute value signs are omitted in the machine answer. That is ﬁne if our problem is concerned only with values of greater than . But if we are interested in other values of , then we need to insert the absolute value symbol. In the next example we reconsider the integral of Example 4, but this time we ask a machine for the answer. EXAMPLE 5 Use a computer algebra system to ﬁnd . SOLUTION Maple responds with the answer This looks different from the answer we found in Example 4, but it is equivalent because the third term can be rewritten using the identity Thus The resulting extra term can be absorbed into the constant of integration. Mathematica gives the answer Mathematica combined the ﬁrst two terms of Example 4 (and the Maple result) into a single term by factoring. Derive gives the answer The ﬁrst term is like the ﬁrst term in the Mathematica answer, and the second term is identical to the last term in Example 4. M EXAMPLE 6 Use a CAS to evaluate . SOLUTION Maple and Mathematica give the same answer: 1 18 x 18 5 2 x 16 50x 14 1750 3 x 12 4375x 10 21875x 8 218750 3 x 6 156250x 4 390625 2 x 2 y xx 2 58 dx 1 6sx 2 2x 4 2x 2 x 5 3 2 ln(sx 2 2x 4 x 1) 5 6 x 6 x 2 3 sx 2 2x 4 3 2 arcsinh 1 x s3 3 2 ln(1s3 ) ln 1 s3 ln(x 1 sx 2 2x 4 ) ln 1 s3 [1 x s1 x2 3 ] arcsinh s3 3 1 x ln s3 3 1 x s\|1 31 x2 1 arcsinh x ln(x sx 2 1) 1 3x 2 2x 432 1 42x 2sx 2 2x 4 3 2 arcsinh s3 3 1 x y xsx 2 2x 4 dx x 2 3 x 1 3 ln3x 2 492 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION N This is Equation 3.11.3. It’s clear that both systems must have expanded by the Binomial Theorem and then integrated each term. If we integrate by hand instead, using the substitution , we get For most purposes, this is a more convenient form of the answer. M EXAMPLE 7 Use a CAS to ﬁnd . SOLUTION In Example 2 in Section 7.2 we found that Derive and Maple report the answer whereas Mathematica produces We suspect that there are trigonometric identities which show these three answers are equivalent. Indeed, if we ask Derive, Maple, and Mathematica to simplify their expres-sions using trigonometric identities, they ultimately produce the same form of the answer as in Equation 1. M 5 64 cos x 1 192 cos 3x 3 320 cos 5x 1 448 cos 7x 1 7 sin4x cos3x 4 35 sin2x cos3x 8 105 cos3x y sin5x cos2x dx 1 3 cos3x 2 5 cos5x 1 7 cos7x C 1 y sin5x cos2x dx y xx 2 58 dx 1 18x 2 59 C u x 2 5 x 2 58 SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS \|\|\|\| 493 N Derive and the TI-89/92 also give this answer. 11. 12. 13. 14. 15. 16. 18. 20. 21. 22. 23. 24. 25. y 1 0 x 4ex dx 26. y s4 ln x2 x dx y sin6 2x dx y sec5x dx y 2 0 x 3s4x 2 x 4 dx y e x 3 e2x dx y sin 2 s5 sin d y sin2x cos x lnsin x dx 19. y dx 2x 3 3x 2 y ys6 4y 4y2 dy 17. y x sinx 2 cos3x 2 dx y e 2x arctane x dx y sin1sx dx y tan31z z 2 dz y x 2 cschx 3 1 dx y 0 1 t 2et dt 1–4 Use the indicated entry in the Table of Integrals on the Reference Pages to evaluate the integral. 1. ; entry 33 2. ; entry 55 3. ; entry 71 4. ; entry 98 5–30 Use the Table of Integrals on Reference Pages 6–10 to evalu-ate the integral. 5. 6. 7. 8. 9. y s2y 2 3 y 2 dy 10. y dx x 2s4x 2 9 y ln(1 sx ) sx dx y tan3x dx y 3 2 1 x 2s4x 2 7 dx y 1 0 2x cos1x dx y e 2 sin 3 d y sec3x dx y 3x s3 2x dx y s7 2x 2 x 2 dx EXERCISES 7.6 43. (a) Use the table of integrals to evaluate , where What is the domain of and ? (b) Use a CAS to evaluate . What is the domain of the function that the CAS produces? Is there a discrepancy between this domain and the domain of the function that you found in part (a)? 44. Computer algebra systems sometimes need a helping hand from human beings. Try to evaluate with a computer algebra system. If it doesn’t return an answer, make a substitution that changes the integral into one that the CAS can evaluate. 45–48 Use a CAS to ﬁnd an antiderivative of such that . Graph and and locate approximately the -coordinates of the extreme points and inﬂection points of . 45. 46. 47. , 48. f x x 3 x x 6 1 0 x f x sin4x cos6x f x xex sin x, 5 x 5 f x x 2 1 x 4 x 2 1 F x F f F0 0 f F CAS y 1 ln x s1 x ln x2 dx CAS F F Fx F f f x 1 xs1 x 2 Fx x f x dx CAS 28. 30. 31. Find the volume of the solid obtained when the region under the curve , , is rotated about the -axis. 32. The region under the curve from 0 to is rotated about the -axis. Find the volume of the resulting solid. Verify Formula 53 in the Table of Integrals (a) by differentia-tion and (b) by using the substitution . 34. Verify Formula 31 (a) by differentiation and (b) by substi-tuting . 35–42 Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent. 35. 36. 37. 38. 39. 40. 41. 42. y 1 s1 s 3 x dx y tan5x dx y sin4x dx y xs1 2x dx y dx e x3e x 2 y x 2sx 2 4 dx y csc5x dx y sec4x dx CAS u a sin t a bu 33. x 4 y tan2x y 0 x 2 y xs4 x 2 y sec2 tan2 s9 tan2 d y x 4 dx sx 10 2 29. y e t sint 3 dt y se 2x 1 dx 27. 494 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION In this project a computer algebra system is used to investigate indeﬁnite integrals of families of functions. By observing the patterns that occur in the integrals of several members of the family, you will ﬁrst guess, and then prove, a general formula for the integral of any member of the family. 1. (a) Use a computer algebra system to evaluate the following integrals. (i) (ii) (iii) (iv) (b) Based on the pattern of your responses in part (a), guess the value of the integral if . What if ? (c) Check your guess by asking your CAS to evaluate the integral in part (b). Then prove it using partial fractions. a b a b y 1 x ax b dx y 1 x 22 dx y 1 x 2x 5 dx y 1 x 1x 5 dx y 1 x 2x 3 dx PATTERNS IN INTEGRALS CAS D I S C O V E R Y P R O J E C T 2. (a) Use a computer algebra system to evaluate the following integrals. (i) (ii) (iii) (b) Based on the pattern of your responses in part (a), guess the value of the integral (c) Check your guess with a CAS. Then prove it using the techniques of Section 7.2. For what values of and is it valid? 3. (a) Use a computer algebra system to evaluate the following integrals. (i) (ii) (iii) (iv) (v) (b) Based on the pattern of your responses in part (a), guess the value of (c) Use integration by parts to prove the conjecture that you made in part (b). For what values of is it valid? 4. (a) Use a computer algebra system to evaluate the following integrals. (i) (ii) (iii) (iv) (v) (b) Based on the pattern of your responses in part (a), guess the value of . Then use your CAS to check your guess. (c) Based on the patterns in parts (a) and (b), make a conjecture as to the value of the integral when is a positive integer. (d) Use mathematical induction to prove the conjecture you made in part (c). n y x ne x dx x x 6e x dx y x 5e x dx y x 4e x dx y x 3e x dx y x 2e x dx y xe x dx n y x n ln x dx y x7 ln x dx y x 3 ln x dx y x 2 ln x dx y x ln x dx y ln x dx b a y sin ax cos bx dx y sin 8x cos 3x dx y sin 3x cos 7x dx y sin x cos 2x dx SECTION 7.7 APPROXIMATE INTEGRATION \|\|\|\| 495 APPROXIMATE INTEGRATION There are two situations in which it is impossible to ﬁnd the exact value of a deﬁnite integral. The ﬁrst situation arises from the fact that in order to evaluate using the Fundamental Theorem of Calculus we need to know an antiderivative of . Sometimes, however, it is difﬁcult, or even impossible, to ﬁnd an antiderivative (see Section 7.5). For example, it is impossible to evaluate the following integrals exactly: y 1 1 s1 x 3 dx y 1 0 ex2 dx f xb a fx dx 7.7 The second situation arises when the function is determined from a scientiﬁc experi-ment through instrument readings or collected data. There may be no formula for the func-tion (see Example 5). In both cases we need to ﬁnd approximate values of deﬁnite integrals. We already know one such method. Recall that the deﬁnite integral is deﬁned as a limit of Riemann sums, so any Riemann sum could be used as an approximation to the integral: If we divide into subintervals of equal length , then we have where is any point in the th subinterval . If is chosen to be the left endpoint of the interval, then and we have If , then the integral represents an area and (1) represents an approximation of this area by the rectangles shown in Figure 1(a). If we choose to be the right endpoint, then and we have [See Figure 1(b).] The approximations and deﬁned by Equations 1 and 2 are called the left endpoint approximation and right endpoint approximation, respectively. In Section 5.2 we also considered the case where is chosen to be the midpoint of the subinterval . Figure 1(c) shows the midpoint approximation , which appears to be better than either or . MIDPOINT RULE and Another approximation, called the Trapezoidal Rule, results from averaging the approx-imations in Equations 1 and 2: x 2 fx0 2 fx1 2 fx2 2 fxn1 fxn x 2 [( fx0 fx1) ( fx1 fx2) ( fxn1 fxn)] y b a fx dx 1 2 n i1 fxi1 x n i1 fxi x x 2 n i1 ( fxi1 fxi) xi 1 2xi1 xi midpoint of xi1, xi x b a n where y b a fx dx Mn x fx1 fx2 fxn Rn Ln Mn xi1, xi xi x i Rn Ln y b a fx dx Rn n i1 fxi x 2 x i xi x i fx 0 y b a fx dx Ln n i1 fxi1 x 1 x i xi1 x i xi1, xi i x i y b a fx dx n i1 fxi x x b an n a, b 496 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION ⁄ ¤ – – – – (a) Left endpoint approximation y x¸ ⁄ ¤ ‹ x¢ x¸ ⁄ ¤ ‹ x¢ ‹ x¢ x 0 (b) Right endpoint approximation y x 0 x (c) Midpoint approximation y 0 FIGURE 1 TRAPEZOIDAL RULE where and . The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates the case . The area of the trapezoid that lies above the th subinterval is and if we add the areas of all these trapezoids, we get the right side of the Trapezoidal Rule. EXAMPLE 1 Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with to approximate the integral . SOLUTION (a) With , and , we have , and so the Trape-zoidal Rule gives This approximation is illustrated in Figure 3. (b) The midpoints of the ﬁve subintervals are , , , , and , so the Midpoint Rule gives This approximation is illustrated in Figure 4. M In Example 1 we deliberately chose an integral whose value can be computed explicitly so that we can see how accurate the Trapezoidal and Midpoint Rules are. By the Funda-mental Theorem of Calculus, The error in using an approximation is deﬁned to be the amount that needs to be added to the approximation to make it exact. From the values in Example 1 we see that the errors in the Trapezoidal and Midpoint Rule approximations for are EM 0.001239 and ET 0.002488 n 5 y 2 1 1 x dx ln x]1 2 ln 2 0.693147 . . . 0.691908 1 5 1 1.1 1 1.3 1 1.5 1 1.7 1 1.9 y 2 1 1 x dx x f1.1 f1.3 f1.5 f1.7 f1.9 1.9 1.7 1.5 1.3 1.1 0.695635 0.1 1 1 2 1.2 2 1.4 2 1.6 2 1.8 1 2 y 2 1 1 x dx T5 0.2 2 f1 2 f1.2 2 f1.4 2 f1.6 2 f1.8 f2 x 2 15 0.2 b 2 n 5, a 1 x2 1 1x dx n 5 x fxi1 fxi 2 x 2 fxi1 fxi i fx 0 xi a i x x b an y b a fx dx Tn x 2 fx0 2 fx1 2 fx2 2 fxn1 fxn SECTION 7.7 APPROXIMATE INTEGRATION \|\|\|\| 497 FIGURE 3 0 y x x¸ ⁄ ¤ ‹ x¢ FIGURE 2 Trapezoidal approximation FIGURE 4 1 2 1 2 1 x y= 1 x y= y b a f x dx approximation error In general, we have The following tables show the results of calculations similar to those in Example 1, but for , and and for the left and right endpoint approximations as well as the Trapezoidal and Midpoint Rules. We can make several observations from these tables: 1. In all of the methods we get more accurate approximations when we increase the value of . (But very large values of result in so many arithmetic operations that we have to beware of accumulated round-off error.) 2. The errors in the left and right endpoint approximations are opposite in sign and appear to decrease by a factor of about 2 when we double the value of . 3. The Trapezoidal and Midpoint Rules are much more accurate than the endpoint approximations. 4. The errors in the Trapezoidal and Midpoint Rules are opposite in sign and appear to decrease by a factor of about 4 when we double the value of . 5. The size of the error in the Midpoint Rule is about half the size of the error in the Trapezoidal Rule. Figure 5 shows why we can usually expect the Midpoint Rule to be more accurate than the Trapezoidal Rule. The area of a typical rectangle in the Midpoint Rule is the same as the area of the trapezoid whose upper side is tangent to the graph at . The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid used in the Trapezoidal Rule. [The midpoint error (shaded red) is smaller than the trape-zoidal error (shaded blue).] FIGURE 5 C P D A B R Q C P D A B xi-1 xi i-1 x –i AQRD P ABCD n n n n 20 n 5, 10 EM y b a fx dx Mn and ET y b a fx dx Tn 498 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION n 5 0.745635 0.645635 0.695635 0.691908 10 0.718771 0.668771 0.693771 0.692835 20 0.705803 0.680803 0.693303 0.693069 Mn Tn Rn Ln n 5 0.052488 0.047512 0.002488 0.001239 10 0.025624 0.024376 0.000624 0.000312 20 0.012656 0.012344 0.000156 0.000078 EM ET ER EL Approximations to y 2 1 1 x dx Corresponding errors N It turns out that these observations are true in most cases. Module 5.2/7.7 allows you to compare approximation methods. TEC These observations are corroborated in the following error estimates, which are proved in books on numerical analysis. Notice that Observation 4 corresponds to the in each denominator because . The fact that the estimates depend on the size of the second derivative is not surprising if you look at Figure 5, because measures how much the graph is curved. [Recall that measures how fast the slope of changes.] ERROR BOUNDS Suppose for . If and are the errors in the Trapezoidal and Midpoint Rules, then Let’s apply this error estimate to the Trapezoidal Rule approximation in Example 1. If , then and . Since , we have , so Therefore, taking , and in the error estimate (3), we see that Comparing this error estimate of with the actual error of about , we see that it can happen that the actual error is substantially less than the upper bound for the error given by (3). EXAMPLE 2 How large should we take in order to guarantee that the Trapezoidal and Midpoint Rule approximations for are accurate to within ? SOLUTION We saw in the preceding calculation that for , so we can take , , and in (3). Accuracy to within means that the size of the error should be less than . Therefore we choose so that Solving the inequality for , we get or Thus will ensure the desired accuracy. n 41 n 1 s0.0006 40.8 n2 2 120.0001 n 213 12n2 0.0001 n 0.0001 0.0001 b 2 a 1 K 2 1 x 2 f x 2 0.0001 x2 1 1x dx n V 0.002488 0.006667 ET 22 13 1252 1 150 0.006667 n 5 K 2, a 1, b 2 f x 2 x 3 2 13 2 1x 1 1 x 2 f x 2x 3 fx 1x 2 fx 1x EM Kb a3 24n2 and ET Kb a3 12n2 EM ET a x b f x K 3 y fx f x f x 2n2 4n2 n2 SECTION 7.7 APPROXIMATE INTEGRATION \|\|\|\| 499 N It’s quite possible that a lower value for would sufﬁce, but is the smallest value for which the error bound formula can guarantee us accuracy to within . 0.0001 41 n N can be any number larger than all the values of , but smaller values of give better error bounds. K f x K For the same accuracy with the Midpoint Rule we choose so that which gives M EXAMPLE 3 (a) Use the Midpoint Rule with to approximate the integral . (b) Give an upper bound for the error involved in this approximation. SOLUTION (a) Since , and , the Midpoint Rule gives Figure 6 illustrates this approximation. (b) Since , we have and . Also, since , we have and so Taking , , , and in the error estimate (3), we see that an upper bound for the error is M SIMPSON’S RULE Another rule for approximate integration results from using parabolas instead of straight line segments to approximate a curve. As before, we divide into subintervals of equal length , but this time we assume that is an even number. Then on each consecutive pair of intervals we approximate the curve by a parabola as shown in Figure 7. If , then is the point on the curve lying above . A typical parabola passes through three consecutive points , and . 0 y x a=x¸ ⁄ x™ x¢ x£ xß=b x∞ P¸ P¡ P™ P¢ P£ Pß P∞ FIGURE 7 0 y x h _h P¸(_h, y¸) P¡(0, ›) P™(h, ﬁ) FIGURE 8 Pi2 Pi, Pi1 xi Pixi, yi yi fxi y fx 0 n h x b an n a, b 6e13 24102 e 400 0.007 n 10 b 1 a 0 K 6e 0 f x 2 4x 2ex2 6e x 2 1 0 x 1 f x 2 4x 2ex2 fx 2xex2 fx ex2 1.460393 e 0.4225 e 0.5625 e 0.7225 e 0.9025 0.1 e 0.0025 e 0.0225 e 0.0625 e 0.1225 e 0.2025 e 0.3025 y 1 0 ex2dx x f0.05 f0.15 f0.85 f0.95 n 10 a 0, b 1 x1 0 ex2dx n 10 V n 1 s0.0012 29 213 24n2 0.0001 n 500 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION 0 y x 1 FIGURE 6 y=ex2 N Error estimates give upper bounds for the error. They are theoretical, worst-case scenarios. The actual error in this case turns out to be about . 0.0023 To simplify our calculations, we ﬁrst consider the case where , and . (See Figure 8.) We know that the equation of the parabola through , and is of the form and so the area under the parabola from to is But, since the parabola passes through , , and , we have and therefore Thus we can rewrite the area under the parabola as Now, by shifting this parabola horizontally we do not change the area under it. This means that the area under the parabola through , and from to in Figure 7 is still Similarly, the area under the parabola through from to is If we compute the areas under all the parabolas in this manner and add the results, we get Although we have derived this approximation for the case in which , it is a rea-sonable approximation for any continuous function and is called Simpson’s Rule after the English mathematician Thomas Simpson (1710–1761). Note the pattern of coefﬁ-cients: . 1, 4, 2, 4, 2, 4, 2, . . . , 4, 2, 4, 1 f fx 0 h 3 y0 4y1 2y2 4y3 2y4 2yn2 4yn1 yn h 3 yn2 4yn1 yn y b a fx dx h 3 y0 4y1 y2 h 3 y2 4y3 y4 h 3 y2 4y3 y4 x x4 x x2 P2, P3, and P4 h 3 y0 4y1 y2 x x2 x x0 P2 P0, P1 h 3 y0 4y1 y2 y0 4y1 y2 2Ah 2 6C y2 Ah 2 Bh C y1 C y0 Ah2 Bh C Ah 2 Bh C P2h, y2 P10, y1 P0h, y0 2A h 3 3 Ch h 3 2Ah 2 6C 2A x 3 3 Cx 0 h y h h Ax 2 Bx C dx 2 y h 0 Ax 2 C dx x h x h y Ax 2 Bx C P2 P0, P1 x2 h x0 h, x1 0 SECTION 7.7 APPROXIMATE INTEGRATION \|\|\|\| 501 N Here we have used Theorem 5.5.7. Notice that is even and is odd. Bx Ax 2 C SIMPSON’S RULE where is even and . EXAMPLE 4 Use Simpson’s Rule with to approximate . SOLUTION Putting , and in Simpson’s Rule, we obtain M Notice that, in Example 4, Simpson’s Rule gives us a much better approximation to the true value of the integral than does the Trapezoidal Rule or the Midpoint Rule . It turns out (see Exercise 48) that the approximations in Simpson’s Rule are weighted averages of those in the Trapezoidal and Midpoint Rules: (Recall that and usually have opposite signs and is about half the size of .) In many applications of calculus we need to evaluate an integral even if no explicit for-mula is known for y as a function of x. A function may be given graphically or as a table of values of collected data. If there is evidence that the values are not changing rapidly, then the Trapezoidal Rule or Simpson’s Rule can still be used to ﬁnd an approximate value for , the integral of y with respect to x. EXAMPLE 5 Figure 9 shows data trafﬁc on the link from the United States to SWITCH, the Swiss academic and research network, on February 10, 1998. is the data through-put, measured in megabits per second . Use Simpson’s Rule to estimate the total amount of data transmitted on the link up to noon on that day. FIGURE 9 0 2 4 6 D 8 3 6 9 12 15 18 21 24 t (hours) Mbs Dt V xb a y dx ET EM EM ET S2n 1 3Tn 2 3Mn M10 0.692835 T10 0.693771 ln 2 0.693147. . . S10 0.693150 0.693150 0.1 3 1 1 4 1.1 2 1.2 4 1.3 2 1.4 4 1.5 2 1.6 4 1.7 2 1.8 4 1.9 1 2 x 3 f1 4 f1.1 2 f1.2 4 f1.3 2 f1.8 4 f1.9 f2 y 2 1 1 x dx S10 x 0.1 fx 1x, n 10 x2 1 1x dx n 10 x b an n 2 fxn2 4 fxn1 fxn y b a fx dx Sn x 3 fx0 4 fx1 2 fx2 4 fx3 502 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION Thomas Simpson was a weaver who taught himself mathematics and went on to become one of the best English mathematicians of the 18th century. What we call Simpson’s Rule was actually known to Cavalieri and Gregory in the 17th century, but Simpson popularized it in his best-selling calculus textbook, A New Treatise of Fluxions. SIMPSON SOLUTION Because we want the units to be consistent and is measured in megabits per second, we convert the units for from hours to seconds. If we let be the amount of data (in megabits) transmitted by time , where is measured in seconds, then . So, by the Net Change Theorem (see Section 5.4), the total amount of data transmitted by noon (when ) is We estimate the values of at hourly intervals from the graph and compile them in the table. Then we use Simpson’s Rule with and to estimate the integral: Thus the total amount of data transmitted up to noon is about 144,000 megabits, or 144 gigabits. M The table in the margin shows how Simpson’s Rule compares with the Midpoint Rule for the integral , whose true value is about 0.69314718. The second table shows how the error in Simpson’s Rule decreases by a factor of about 16 when is doubled. (In Exercises 27 and 28 you are asked to verify this for two additional integrals.) That is consistent with the appearance of in the denominator of the following error estimate for Simpson’s Rule. It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules, but it uses the fourth derivative of . ERROR BOUND FOR SIMPSON’S RULE Suppose that for . If is the error involved in using Simpson’s Rule, then ES Kb a5 180n 4 ES a x b f 4x K 4 f n 4 n Es x2 1 1x dx 143,880 21.1 41.3 22.8 45.7 27.1 47.7 7.9 3600 3 3.2 42.7 21.9 41.7 21.3 41.0 y 43,200 0 At dt t 3 D0 4D3600 2D7200 4D39,600 D43,200 t 3600 n 12 Dt A43,200 y 43,200 0 Dt dt t 12 602 43,200 At Dt t t At t Dt SECTION 7.7 APPROXIMATE INTEGRATION \|\|\|\| 503 0 0 3.2 7 25,200 1.3 1 3,600 2.7 8 28,800 2.8 2 7,200 1.9 9 32,400 5.7 3 10,800 1.7 10 36,000 7.1 4 14,400 1.3 11 39,600 7.7 5 18,000 1.0 12 43,200 7.9 6 21,600 1.1 Dt t seconds t hours Dt t seconds t hours 4 0.69121989 0.69315453 8 0.69266055 0.69314765 16 0.69302521 0.69314721 Sn Mn n 4 0.00192729 8 0.00048663 16 0.00012197 0.00000003 0.00000047 0.00000735 ES EM n EXAMPLE 6 How large should we take in order to guarantee that the Simpson’s Rule approximation for is accurate to within ? SOLUTION If , then . Since , we have and so Therefore we can take in (4). Thus, for an error less than , we should choose so that This gives or Therefore ( must be even) gives the desired accuracy. (Compare this with Example 2, where we obtained for the Trapezoidal Rule and for the Midpoint Rule.) M EXAMPLE 7 (a) Use Simpson’s Rule with to approximate the integral . (b) Estimate the error involved in this approximation. SOLUTION (a) If , then and Simpson’s Rule gives (b) The fourth derivative of is and so, since , we have Therefore, putting , and in (4), we see that the error is at most (Compare this with Example 3.) Thus, correct to three decimal places, we have M y 1 0 ex2 dx 1.463 76e15 180104 0.000115 n 10 K 76e, a 0, b 1 0 f 4x 12 48 16e 1 76e 0 x 1 f 4x 12 48x 2 16x 4ex2 fx ex2 1.462681 4e 0.49 2e 0.64 4e 0.81 e 1 0.1 3 e 0 4e 0.01 2e 0.04 4e 0.09 2e 0.16 4e 0.25 2e 0.36 y 1 0 ex2dx x 3 f0 4 f0.1 2 f0.2 2 f0.8 4 f0.9 f1 x 0.1 n 10 x1 0 ex2 dx n 10 n 29 n 41 n n 8 n 1 s 4 0.00075 6.04 n 4 24 1800.0001 2415 180n 4 0.0001 n 0.0001 K 24 f 4x 24 x 5 24 1x 1 x 1 f 4x 24x 5 fx 1x 0.0001 x2 1 1x dx n 504 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION N Many calculators and computer algebra sys-tems have a built-in algorithm that computes an approximation of a deﬁnite integral. Some of these machines use Simpson’s Rule; others use more sophisticated techniques such as adaptive numerical integration. This means that if a func-tion ﬂuctuates much more on a certain part of the interval than it does elsewhere, then that part gets divided into more subintervals. This strategy reduces the number of calculations required to achieve a prescribed accuracy. N Figure 10 illustrates the calculation in Example 7. Notice that the parabolic arcs are so close to the graph of that they are practically indistinguishable from it. y ex2 0 y x 1 y=ex2 FIGURE 10 SECTION 7.7 APPROXIMATE INTEGRATION \|\|\|\| 505 (Round your answers to six decimal places.) Compare your results to the actual value to determine the error in each approximation. 5. , 6. , 7–18 Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule to approximate the given integral with the speciﬁed value of . (Round your answers to six decimal places.) 7. , 8. , 9. , 10. , 11. , 12. , 13. , 14. , 15. , 16. , 17. , 18. , 19. (a) Find the approximations and for the integral . (b) Estimate the errors in the approximations of part (a). (c) How large do we have to choose so that the approxima-tions and to the integral in part (a) are accurate to within ? 20. (a) Find the approximations and for . (b) Estimate the errors in the approximations of part (a). (c) How large do we have to choose so that the approxima-tions and to the integral in part (a) are accurate to within ? 21. (a) Find the approximations , , and for and the corresponding errors , , and . (b) Compare the actual errors in part (a) with the error esti-mates given by (3) and (4). (c) How large do we have to choose so that the approxima-tions , , and to the integral in part (a) are accurate to within ? 22. How large should be to guarantee that the Simpson’s Rule approximation to is accurate to within ? 23. The trouble with the error estimates is that it is often very difﬁcult to compute four derivatives and obtain a good upper bound for by hand. But computer algebra systems f 4x K CAS 0.00001 x1 0 e x 2dx n 0.00001 Sn Mn Tn n ES EM ET x 0 sin x dx S10 M10 T10 0.0001 Mn Tn n x2 1 e 1x dx M10 T10 0.0001 Mn Tn n x1 0 cos x 2 dx M8 T8 n 10 y 4 0 cos sx dx n 6 y 3 0 1 1 y 5 dy n 10 y 6 4 lnx 3 2 dx n 8 y 5 1 cos x x dx n 10 y 1 0 sz ezdz n 8 y 4 0 e st sin t dt n 8 y 4 0 s1 sx dx n 8 y 12 0 sine t2 dt n 6 y 3 0 dt 1 t 2 t 4 n 10 y 2 1 ln x 1 x dx n 4 y 12 0 sinx 2 dx n 8 y 2 0 s 4 1 x 2 dx n n 6 y 1 0 esx dx n 8 y 0 x 2 sin x dx Let , where is the function whose graph is shown. (a) Use the graph to ﬁnd . (b) Are these underestimates or overestimates of ? (c) Use the graph to ﬁnd . How does it compare with ? (d) For any value of , list the numbers and in increasing order. 2. The left, right, Trapezoidal, and Midpoint Rule approxi-mations were used to estimate , where is the function whose graph is shown. The estimates were 0.7811, 0.8675, 0.8632, and 0.9540, and the same number of sub-intervals were used in each case. (a) Which rule produced which estimate? (b) Between which two approximations does the true value of lie? ; Estimate using (a) the Trapezoidal Rule and (b) the Midpoint Rule, each with . From a graph of the integrand, decide whether your answers are underestimates or overestimates. What can you conclude about the true value of the integral? ; Draw the graph of in the viewing rectangle by and let . (a) Use the graph to decide whether , and under-estimate or overestimate . (b) For any value of , list the numbers and in increasing order. (c) Compute . From the graph, which do you think gives the best estimate of ? 5–6 Use (a) the Midpoint Rule and (b) Simpson’s Rule to approximate the given integral with the speciﬁed value of . n I L5, R5, M5, and T5 I Ln, Rn, Mn, Tn, n I T2 L2, R2, M2 I x1 0 f x dx 0, 0.5 0, 1 f x sin( 1 2x 2) 4. n 4 x1 0 cosx 2 dx 3. y x 0 1 2 y=ƒ x2 0 f x dx f x2 0 f x dx f x 1 y 2 3 1 0 2 3 4 I Ln, Rn, Mn, Tn, n I T2 I L2, R2, and M2 f I x4 0 f x dx 1. EXERCISES 7.7 ﬁgure. Use Simpson’s Rule to estimate the area of the pool. 31. (a) Use the Midpoint Rule and the given data to estimate the value of the integral . (b) If it is known that for all , estimate the error involved in the approximation in part (a). 32. A radar gun was used to record the speed of a runner during the ﬁrst 5 seconds of a race (see the table). Use Simpson’s Rule to estimate the distance the runner covered during those 5 seconds. The graph of the acceleration of a car measured in is shown. Use Simpson’s Rule to estimate the increase in the velocity of the car during the 6-second time interval. 34. Water leaked from a tank at a rate of liters per hour, where the graph of is as shown. Use Simpson’s Rule to estimate the total amount of water that leaked out during the ﬁrst 6 hours. r 0 6 4 2 2 4 t (seconds) r rt a 0 6 4 2 4 8 12 t (seconds) fts2 at 33. x 4 f x 1 x3.2 0 f x dx 6.2 5.0 7 .2 6.8 5.6 4.8 4.8 have no problem computing and graphing it, so we can easily ﬁnd a value for from a machine graph. This exercise deals with approximations to the integral , where . (a) Use a graph to get a good upper bound for . (b) Use to approximate . (c) Use part (a) to estimate the error in part (b). (d) Use the built-in numerical integration capability of your CAS to approximate . (e) How does the actual error compare with the error esti-mate in part (c)? (f) Use a graph to get a good upper bound for . (g) Use to approximate . (h) Use part (f) to estimate the error in part (g). (i) How does the actual error compare with the error esti-mate in part (h)? (j) How large should be to guarantee that the size of the error in using is less than ? 24. Repeat Exercise 23 for the integral . 25–26 Find the approximations , and for , and . Then compute the corresponding errors , and . (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when is doubled? 25. 26. 27–28 Find the approximations , , and for and . Then compute the corresponding errors , and . (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when is doubled? 27. 28. 29. Estimate the area under the graph in the ﬁgure by using (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule, each with . 30. The widths (in meters) of a kidney-shaped swimming pool were measured at 2-meter intervals as indicated in the 1 x y 0 4 3 6 5 2 1 n 6 y 4 1 1 sx dx y 2 0 x 4 dx n ES ET, EM 12 n 6 Sn Mn Tn y 2 1 1 x 2 dx y 1 0 xe x dx n EM EL, ER, ET 20 n 5, 10 Mn Ln, Rn, Tn y 1 1 s4 x 3 dx CAS 0.0001 Sn n I S10 f 4x I I M10 f x f x e cos x I x2 0 f x dx K f 4 506 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION x x 0.0 6.8 2.0 7.6 0.4 6.5 2.4 8.4 0.8 6.3 2.8 8.8 1.2 6.4 3.2 9.0 1.6 6.9 f x f x t (s) (ms) t (s) (ms) 0 0 3.0 10.51 0.5 4.67 3.5 10.67 1.0 7.34 4.0 10.76 1.5 8.86 4.5 10.81 2.0 9.73 5.0 10.81 2.5 10.22 v v 39. The region bounded by the curves , , , and is rotated about the -axis. Use Simpson’s Rule with to estimate the volume of the resulting solid. 40. The ﬁgure shows a pendulum with length that makes a maximum angle with the vertical. Using Newton’s Second Law, it can be shown that the period (the time for one complete swing) is given by where and is the acceleration due to gravity. If m and , use Simpson’s Rule with to ﬁnd the period. 41. The intensity of light with wavelength traveling through a diffraction grating with slits at an angle is given by , where and is the distance between adjacent slits. A helium-neon laser with wavelength is emitting a narrow band of light, given by , through a grating with 10,000 slits spaced apart. Use the Midpoint Rule with to estimate the total light intensity emerging from the grating. 42. Use the Trapezoidal Rule with to approximate . Compare your result to the actual value. Can you explain the discrepancy? 43. Sketch the graph of a continuous function on for which the Trapezoidal Rule with is more accurate than the Midpoint Rule. 44. Sketch the graph of a continuous function on for which the right endpoint approximation with is more accurate than Simpson’s Rule. If is a positive function and for , show that 46. Show that if is a polynomial of degree 3 or lower, then Simpson’s Rule gives the exact value of . 47. Show that . 48. Show that . 1 3Tn 2 3Mn S2n 1 2Tn Mn T2n xb a f x dx f Tn y b a f x dx Mn a x b f x 0 f 45. n 2 0, 2 n 2 0, 2 x20 0 cosx dx n 10 x106 106 I d n 10 104 m 106 106 632.8 109 m d k Nd sin I N 2 sin2kk 2 N ¨¸ n 10 0 42 L 1 t k sin( 1 20) T 4 L t y 2 0 dx s1 k 2 sin2x T 0 L CAS n 8 x x 5 x 1 y 0 y e1x The table (supplied by San Diego Gas and Electric) gives the power consumption in megawatts in San Diego County from midnight to 6:00 AM on December 8, 1999. Use Simp-son’s Rule to estimate the energy used during that time period. (Use the fact that power is the derivative of energy.) 36. Shown is the graph of trafﬁc on an Internet service pro-vider’s T1 data line from midnight to 8:00 AM. is the data throughput, measured in megabits per second. Use Simpson’s Rule to estimate the total amount of data transmitted during that time period. 37. If the region shown in the ﬁgure is rotated about the -axis to form a solid, use Simpson’s Rule with to estimate the volume of the solid. 38. The table shows values of a force function , where is measured in meters and in newtons. Use Simpson’s Rule to estimate the work done by the force in moving an object a distance of 18 m. f x x f x 0 4 4 10 2 8 6 2 y x n 8 y 0 0.4 4 6 0.8 2 8 D t (hours) D P 35. SECTION 7.7 APPROXIMATE INTEGRATION \|\|\|\| 507 t P t P 0:00 1814 3:30 1611 0:30 1735 4:00 1621 1:00 1686 4:30 1666 1:30 1646 5:00 1745 2:00 1637 5:30 1886 2:30 1609 6:00 2052 3:00 1604 x 0 3 6 9 12 15 18 9.8 9.1 8.5 8.0 7.7 7.5 7.4 f x IMPROPER INTEGRALS In deﬁning a deﬁnite integral we dealt with a function deﬁned on a ﬁnite inter-val and we assumed that does not have an inﬁnite discontinuity (see Section 5.2). In this section we extend the concept of a deﬁnite integral to the case where the interval is inﬁnite and also to the case where has an inﬁnite discontinuity in . In either case the integral is called an improper integral. One of the most important applications of this idea, probability distributions, will be studied in Section 8.5. TYPE 1: INFINITE INTERVALS Consider the inﬁnite region that lies under the curve , above the -axis, and to the right of the line . You might think that, since is inﬁnite in extent, its area must be inﬁnite, but let’s take a closer look. The area of the part of that lies to the left of the line (shaded in Figure 1) is Notice that no matter how large is chosen. We also observe that The area of the shaded region approaches as (see Figure 2), so we say that the area of the inﬁnite region is equal to and we write Using this example as a guide, we deﬁne the integral of (not necessarily a positive function) over an inﬁnite interval as the limit of integrals over ﬁnite intervals. f 0 y x 1 2 area= 1 2 0 y x 1 3 area= 2 3 0 y x 1 area=1 0 y x 1 5 4 5 area= FIGURE 2 y 1 1 x 2 dx lim t l y t 1 1 x 2 dx 1 1 S t l 1 lim t l At lim t l 1 1 t 1 FIGURE 1 0 y x 1 t y= x=1 area=1-=1 1 t 1 ≈ t At 1 At y t 1 1 x 2 dx 1 x 1 t 1 1 t x t S S x 1 x y 1x 2 S a, b f f a, b f xb a fx dx 7.8 508 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1 (a) If exists for every number , then provided this limit exists (as a ﬁnite number). (b) If exists for every number , then provided this limit exists (as a ﬁnite number). The improper integrals and are called convergent if the corresponding limit exists and divergent if the limit does not exist. (c) If both and are convergent, then we deﬁne In part (c) any real number can be used (see Exercise 74). Any of the improper integrals in Deﬁnition 1 can be interpreted as an area provided that is a positive function. For instance, in case (a) if and the integral is convergent, then we deﬁne the area of the region in Figure 3 to be This is appropriate because is the limit as of the area under the graph of from to . EXAMPLE 1 Determine whether the integral is convergent or divergent. SOLUTION According to part (a) of Deﬁnition 1, we have The limit does not exist as a ﬁnite number and so the improper integral is divergent. M x 1 1x dx lim t l ln t ln 1 lim t l ln t y 1 1 x dx lim t l y t 1 1 x dx lim t l lnx]1 t x 1 1x dx V FIGURE 3 0 y x a S y=ƒ t a f t l x a fx dx AS y a fx dx S x, yx a, 0 y fx x a fx dx fx 0 f a y fx dx y a fx dx y a fx dx xa fx dx x a fx dx xb fx dx x a fx dx y b fx dx lim t l y b t fx dx t b xb t fx dx y a fx dx lim t l y t a fx dx t a xt a fx dx 1 SECTION 7.8 IMPROPER INTEGRALS \|\|\|\| 509 Let’s compare the result of Example 1 with the example given at the beginning of this section: Geometrically, this says that although the curves and look very similar for , the region under to the right of (the shaded region in Figure 4) has ﬁnite area whereas the corresponding region under (in Figure 5) has inﬁnite area. Note that both and approach as but approaches faster than . The values of 1x don’t decrease fast enough for its integral to have a ﬁnite value. EXAMPLE 2 Evaluate . SOLUTION Using part (b) of Deﬁnition 1, we have We integrate by parts with , so that , : We know that as , and by l’Hospital’s Rule we have Therefore M EXAMPLE 3 Evaluate . SOLUTION It’s convenient to choose in Deﬁnition 1(c): We must now evaluate the integrals on the right side separately: lim t l tan1t tan1 0 lim t l tan1t 2 y 0 1 1 x 2 dx lim tl y t 0 dx 1 x 2 lim tl tan1x]0 t y 1 1 x 2 dx y 0 1 1 x 2 dx y 0 1 1 x 2 dx a 0 y 1 1 x 2 dx 0 1 0 1 y 0 xe x dx lim t l te t 1 e t lim t l e t 0 lim t l te t lim t l t et lim t l 1 et t l e t l 0 te t 1 e t y 0 t xe x dx xe x]t 0 y 0 t e x dx v e x du dx dv e x dx u x y 0 xe x dx lim t l y 0 t xe x dx y 0 xe x dx 1x 0 1x 2 x l 0 1x 1x 2 y 1x x 1 y 1x 2 x 0 y 1x y 1x 2 y 1 1 x dx diverges y 1 1 x 2 dx converges 510 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION 1 x FIGURE 4 FIGURE 5 inﬁnite area 0 y x 1 y= 0 y x 1 ﬁnite area y= 1 ≈ In Module 7.8 you can investigate visually and numerically whether several improper integrals are convergent or divergent. TEC Since both of these integrals are convergent, the given integral is convergent and Since , the given improper integral can be interpreted as the area of the inﬁnite region that lies under the curve and above the -axis (see Figure 6). M EXAMPLE 4 For what values of is the integral convergent? SOLUTION We know from Example 1 that if , then the integral is divergent, so let’s assume that . Then If , then , so as , and . Therefore and so the integral converges. But if , then and so and the integral diverges. M We summarize the result of Example 4 for future reference: TYPE 2: DISCONTINUOUS INTEGRANDS Suppose that is a positive continuous function deﬁned on a ﬁnite interval but has a vertical asymptote at . Let be the unbounded region under the graph of and above the -axis between and . (For Type 1 integrals, the regions extended indeﬁnitely in a b a x f S b a, b f y 1 1 x p dx is convergent if p 1 and divergent if p 1. 2 as t l 1 t p1 t 1p l p 1 0 p 1 if p 1 y 1 1 x p dx 1 p 1 1t p1 l 0 t p1 l t l p 1 0 p 1 lim t l 1 1 p 1 t p1 1 lim t l xp1 p 1 x1 xt y 1 1 x p dx lim t l y t 1 x p dx p 1 p 1 y 1 1 x p dx p x y 11 x 2 11 x 2 0 y 1 1 x 2 dx 2 2 0 2 2 lim t l tan1 0 tan1t y 0 1 1 x 2 dx lim t l y 0 t dx 1 x 2 lim t l tan1x] t 0 SECTION 7.8 IMPROPER INTEGRALS \|\|\|\| 511 0 y x y= area=π FIGURE 6 1 1+≈ horizontal direction. Here the region is inﬁnite in a vertical direction.) The area of the part of between and (the shaded region in Figure 7) is If it happens that approaches a deﬁnite number as , then we say that the area of the region is and we write We use this equation to deﬁne an improper integral of Type 2 even when is not a posi-tive function, no matter what type of discontinuity has at . DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2 (a) If is continuous on and is discontinuous at , then if this limit exists (as a ﬁnite number). (b) If is continuous on and is discontinuous at , then if this limit exists (as a ﬁnite number). The improper integral is called convergent if the corresponding limit exists and divergent if the limit does not exist. (c) If has a discontinuity at , where , and both and are convergent, then we deﬁne EXAMPLE 5 Find . SOLUTION We note ﬁrst that the given integral is improper because has the vertical asymptote . Since the inﬁnite discontinuity occurs at the left end-point of , we use part (b) of Deﬁnition 3: Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure 10. M 2s3 lim t l 2 2(s3 st 2 ) lim t l 2 2sx 2 ]t 5 y 5 2 dx sx 2 lim t l 2 y 5 t dx sx 2 2, 5 x 2 fx 1sx 2 y 5 2 1 sx 2 dx y b a fx dx y c a fx dx y b c fx dx xb c fx dx xc a fx dx a c b c f xb a fx dx y b a fx dx lim t l a y b t fx dx a a, b f y b a fx dx lim t l b y t a fx dx b a, b f 3 b f f y b a fx dx lim t l b y t a fx dx A S t l b A At At y t a fx dx t a S 512 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION FIGURE 7 0 y x b t a x=b y=ƒ N Parts (b) and (c) of Deﬁnition 3 are illustrated in Figures 8 and 9 for the case where and has vertical asymptotes at and , respectively. c a f f x 0 0 y x a t b FIGURE 8 0 y x a c b FIGURE 9 0 y x 1 2 4 5 3 y= 1 œ„„„„ x-2 area=2œ„ 3 FIGURE 10 EXAMPLE 6 Determine whether converges or diverges. SOLUTION Note that the given integral is improper because . Using part (a) of Deﬁnition 3 and Formula 14 from the Table of Integrals, we have because and as . Thus the given improper integral is divergent. M EXAMPLE 7 Evaluate if possible. SOLUTION Observe that the line is a vertical asymptote of the integrand. Since it occurs in the middle of the interval , we must use part (c) of Deﬁnition 3 with : where because as . Thus is divergent. This implies that is divergent. [We do not need to evaluate .] M \| WARNING If we had not noticed the asymptote in Example 7 and had instead confused the integral with an ordinary integral, then we might have made the following erroneous calculation: This is wrong because the integral is improper and must be calculated in terms of limits. From now on, whenever you meet the symbol you must decide, by looking at the function on , whether it is an ordinary deﬁnite integral or an improper integral. EXAMPLE 8 Evaluate . SOLUTION We know that the function has a vertical asymptote at 0 since . Thus the given integral is improper and we have y 1 0 ln x dx lim t l 0 y 1 t ln x dx lim x l 0 ln x fx ln x y 1 0 ln x dx a, b f xb a fx dx y 3 0 dx x 1 ln x 1] 3 0 ln 2 ln 1 ln 2 x 1 x3 1 dxx 1 x3 0 dxx 1 x1 0 dxx 1 t l 1 1 t l 0 lim t l 1 ln1 t lim t l 1 (ln t 1 ln 1) y 1 0 dx x 1 lim t l 1 y t 0 dx x 1 lim t l 1 ln x 1]0 t y 3 0 dx x 1 y 1 0 dx x 1 y 3 1 dx x 1 c 1 0, 3 x 1 y 3 0 dx x 1 t l 2 tan t l sec t l lim t l 2 lnsec t tan t ln 1 lim t l 2 ln sec x tan x]0 t y 2 0 sec x dx lim t l 2 y t 0 sec x dx lim x l2 sec x y 2 0 sec x dx V SECTION 7.8 IMPROPER INTEGRALS \|\|\|\| 513 Now we integrate by parts with , , , and : To ﬁnd the limit of the ﬁrst term we use l’Hospital’s Rule: Therefore Figure 11 shows the geometric interpretation of this result. The area of the shaded region above and below the -axis is . M A COMPARISON TEST FOR IMPROPER INTEGRALS Sometimes it is impossible to ﬁnd the exact value of an improper integral and yet it is important to know whether it is convergent or divergent. In such cases the following the-orem is useful. Although we state it for Type 1 integrals, a similar theorem is true for Type 2 integrals. COMPARISON THEOREM Suppose that and are continuous functions with for . (a) If is convergent, then is convergent. (b) If is divergent, then is divergent. We omit the proof of the Comparison Theorem, but Figure 12 makes it seem plausible. If the area under the top curve is ﬁnite, then so is the area under the bottom curve . And if the area under is inﬁnite, then so is the area under . [Note that the reverse is not necessarily true: If is convergent, may or may not be convergent, and if is divergent, may or may not be divergent.] EXAMPLE 9 Show that is convergent. SOLUTION We can’t evaluate the integral directly because the antiderivative of is not an elementary function (as explained in Section 7.5). We write and observe that the ﬁrst integral on the right-hand side is just an ordinary deﬁnite inte-gral. In the second integral we use the fact that for we have , so and therefore . (See Figure 13.) The integral of is easy to evaluate: lim t l e1 et e1 y 1 ex dx lim t l y t 1 ex dx ex ex2 ex x 2 x x 2 x x 1 y 0 ex2 dx y 1 0 ex2 dx y 1 ex2 dx ex2 y 0 ex2dx V x a tx dx x a fx dx x a fx dx x a tx dx y fx y tx y tx y fx x a fx dx x a tx dx x a tx dx x a fx dx x a fx tx 0 t f 1 x y ln x 0 1 0 1 y 1 0 ln x dx lim t l 0 t ln t 1 t lim t l 0 t 0 lim t l 0 1t 1t 2 lim t l 0 t ln t lim t l 0 ln t 1t t ln t 1 t 1 ln 1 t ln t 1 t y 1 t ln x dx x ln x]t 1 y 1 t dx v x du dxx dv dx u ln x 514 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION FIGURE 11 0 y x 1 area=1 y=ln x 0 y x a g f FIGURE 12 0 y x 1 y=e_x FIGURE 13 y=e_x2 Thus, taking and in the Comparison Theorem, we see that is convergent. It follows that is convergent. M In Example 9 we showed that is convergent without computing its value. In Exercise 70 we indicate how to show that its value is approximately 0.8862. In probabil-ity theory it is important to know the exact value of this improper integral, as we will see in Section 8.5; using the methods of multivariable calculus it can be shown that the exact value is . Table 1 illustrates the deﬁnition of an improper integral by showing how the (computer-generated) values of approach as t becomes large. In fact, these values converge quite quickly because very rapidly as . EXAMPLE 10 The integral is divergent by the Comparison Theorem because and is divergent by Example 1 [or by (2) with ]. M Table 2 illustrates the divergence of the integral in Example 10. It appears that the values are not approaching any ﬁxed number. p 1 x 1 1x dx 1 ex x 1 x y 1 1 ex x dx x l ex2 l 0 s 2 xt 0 ex2dx s 2 x 0 ex2 dx x 0 ex2 dx x 1 ex2 dx tx ex2 fx ex SECTION 7.8 IMPROPER INTEGRALS \|\|\|\| 515 TABLE 1 t 1 0.7468241328 2 0.8820813908 3 0.8862073483 4 0.8862269118 5 0.8862269255 6 0.8862269255 xt 0 ex2 dx TABLE 2 t 2 0.8636306042 5 1.8276735512 10 2.5219648704 100 4.8245541204 1000 7.1271392134 10000 9.4297243064 xt 1 1 exx dx 8. 9. 10. 11. 12. 14. 15. 16. 17. 18. 19. 20. 22. 23. 24. 25. 26. 27. 28. y 3 2 1 s3 x dx y 1 0 3 x 5 dx y 0 x arctan x 1 x 22 dx y e 1 xln x3 dx y 0 e x e 2x 3 dx y x 2 9 x 6 dx y x 3ex4 dx y 1 ln x x dx 21. y 6 re r3 dr y 0 se5s ds y 0 dz z 2 3z 2 y 1 x 1 x 2 2x dx y cos t dt y 2 sin d y 1 esx sx dx y xex2 dx 13. y 2 v 4 dv y x 1 x 2 dx y 1 e2t dt y 4 ey2 dy y 0 x x 2 22 dx y 1 1 s2 w dw 7. Explain why each of the following integrals is improper. (a) (b) (c) (d) 2. Which of the following integrals are improper? Why? (a) (b) (c) (d) 3. Find the area under the curve from to and evaluate it for , , and . Then ﬁnd the total area under this curve for . ; 4. (a) Graph the functions and in the viewing rectangles by and by . (b) Find the areas under the graphs of and from to and evaluate for , , , , , and . (c) Find the total area under each curve for , if it exists. 5–40 Determine whether each integral is convergent or divergent. Evaluate those that are convergent. 5. 6. y 0 1 2x 5 dx y 1 1 3x 12 dx x 1 1020 1010 106 104 100 t 10 x t x 1 t f 0, 1 0, 100 0, 1 0, 10 tx 1x 0.9 f x 1x 1.1 x 1 1000 100 t 10 x t x 1 y 1x 3 y 2 1 lnx 1 dx y sin x 1 x 2 dx y 1 0 1 2x 1 dx y 2 1 1 2x 1 dx y 0 1 x 2 5 dx y 2 0 x x 2 5x 6 dx y 2 0 sec x dx y 1 x 4ex4 dx 1. EXERCISES 7.8 51. 52. 53. 54. 55. The integral is improper for two reasons: The interval is inﬁnite and the integrand has an inﬁnite discontinuity at 0. Evaluate it by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows: 56. Evaluate by the same method as in Exercise 55. 57–59 Find the values of for which the integral converges and evaluate the integral for those values of . 58. 59. 60. (a) Evaluate the integral for , , , and . (b) Guess the value of when is an arbitrary posi-tive integer. (c) Prove your guess using mathematical induction. (a) Show that is divergent. (b) Show that This shows that we can’t deﬁne 62. The average speed of molecules in an ideal gas is where is the molecular weight of the gas, is the gas con-stant, is the gas temperature, and is the molecular speed. Show that v 8RT M v T R M v 4 s M 2RT 32 y 0 v 3eMv22RT dv y f x dx lim t l y t t f x dx lim t l y t t x dx 0 x x dx 61. n x 0 x nex dx 3 2 1 n 0 x 0 x nex dx y 1 0 x p ln x dx y e 1 xln x p dx y 1 0 1 x p dx 57. p p y 2 1 xsx 2 4 dx y 1 0 1 sx 1 x dx y 1 1 sx 1 x dx y 0 1 sx 1 x dx 0, y 0 1 sx 1 x dx y 0 sin2 x sx dx y 1 0 sec 2 x xsx dx y 0 arctan x 2 e x dx y 1 x 1 sx 4 x dx 30. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41–46 Sketch the region and ﬁnd its area (if the area is ﬁnite). 41. 42. ; ; 44. ; 45. ; 46. ; 47. (a) If , use your calculator or computer to make a table of approximate values of for , 5, 10, 100, 1000, and 10,000. Does it appear that is convergent? (b) Use the Comparison Theorem with to show that is convergent. (c) Illustrate part (b) by graphing and on the same screen for . Use your graph to explain intuitively why is convergent. ; 48. (a) If , use your calculator or computer to make a table of approximate values of for , 10, 100, 1000, and 10,000. Does it appear that is convergent or divergent? (b) Use the Comparison Theorem with to show that is divergent. (c) Illustrate part (b) by graphing and on the same screen for . Use your graph to explain intuitively why is divergent. 49–54 Use the Comparison Theorem to determine whether the integral is convergent or divergent. 50. y 1 2 ex x dx y 0 x x 3 1 dx 49. x 2 tx dx 2 x 20 t f x 2 tx dx f x 1sx x 2 tx dx t 5 xt 2 tx dx tx 1(sx 1) x 1 tx dx 1 x 10 t f x 1 tx dx f x 1x 2 x 1 tx dx t 2 xt 1 tx dx tx sin2xx 2 S {x, y 2 x 0, 0 y 1sx 2 } S x, y 0 x 2, 0 y sec2x S x, y x 0, 0 y xx 2 9 S x, y 0 y 2x 2 9 43. S x, y x 2, 0 y ex2 S x, y x 1, 0 y e x y 1 0 ln x sx dx y 2 0 z 2 ln z dz y 1 0 e1x x 3 dx y 0 1 e1x x 3 dx y 2 csc x dx y 3 0 dx x 2 6x 5 y 1 0 1 4y 1 dy y 33 0 x 115 dx y 1 0 dx s1 x 2 y 3 2 1 x 4 dx 31. y 8 6 4 x 63 dx y 14 2 dx s 4 x 2 29. 516 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION 70. Estimate the numerical value of by writing it as the sum of and . Approximate the ﬁrst inte-gral by using Simpson’s Rule with and show that the second integral is smaller than , which is less than 0.0000001. 71. If is continuous for , the Laplace transform of is the function deﬁned by and the domain of is the set consisting of all numbers for which the integral converges. Find the Laplace transforms of the following functions. (a) (b) (c) 72. Show that if for , where and are constants, then the Laplace transform exists for . 73. Suppose that and for , where is continuous. If the Laplace transform of is and the Laplace transform of is , show that 74. If is convergent and and are real numbers, show that 75. Show that . 76. Show that by interpreting the integrals as areas. 77. Find the value of the constant for which the integral converges. Evaluate the integral for this value of . 78. Find the value of the constant for which the integral converges. Evaluate the integral for this value of . 79. Suppose is continuous on and . Is it possible that is convergent? 80. Show that if and , then the following inte-gral is convergent. y 0 x a 1 x b dx b a 1 a 1 x 0 f x dx limx l f x 1 0, f C y 0 x x 2 1 C 3x 1 dx C C y 0 1 sx 2 4 C x 2 dx C x 0 ex 2 dx x1 0 sln y dy x 0 x 2ex 2 dx 1 2 x 0 ex 2 dx y a f x dx y a f x dx y b f x dx y b f x dx b a x f x dx s a Gs sFs f 0 Gs f t Fs f t f t 0 0 f t Ke at 0 f t Me at s a Fs a M t 0 0 f t Me at f t t f t e t f t 1 s F Fs y 0 f test dt F f t 0 f t x 4 e4x dx n 8 x 4 ex2 dx x4 0 ex2 dx x 0 ex2 dx 63. We know from Example 1 that the region has inﬁnite area. Show that by rotating about the -axis we obtain a solid with ﬁnite volume. 64. Use the information and data in Exercises 29 and 30 of Sec-tion 6.4 to ﬁnd the work required to propel a 1000-kg satellite out of the earth’s gravitational ﬁeld. 65. Find the escape velocity that is needed to propel a rocket of mass out of the gravitational ﬁeld of a planet with mass and radius . Use Newton’s Law of Gravitation (see Exer-cise 29 in Section 6.4) and the fact that the initial kinetic energy of supplies the needed work. 66. Astronomers use a technique called stellar stereography to determine the density of stars in a star cluster from the observed (two-dimensional) density that can be analyzed from a photograph. Suppose that in a spherical cluster of radius the density of stars depends only on the distance from the center of the cluster. If the perceived star density is given by , where is the observed planar distance from the center of the cluster, and is the actual density, it can be shown that If the actual density of stars in a cluster is , ﬁnd the perceived density . 67. A manufacturer of lightbulbs wants to produce bulbs that last about 700 hours but, of course, some bulbs burn out faster than others. Let be the fraction of the company’s bulbs that burn out before hours, so always lies between 0 and 1. (a) Make a rough sketch of what you think the graph of might look like. (b) What is the meaning of the derivative ? (c) What is the value of ? Why? 68. As we saw in Section 3.8, a radioactive substance decays exponentially: The mass at time is , where is the initial mass and is a negative constant. The mean life of an atom in the substance is For the radioactive carbon isotope, , used in radiocarbon dating, the value of is . Find the mean life of a atom. Determine how large the number has to be so that y a 1 x 2 1 dx 0.001 a 69. 14C 0.000121 k 14C M k y 0 te kt dt M k m0 mt m0e kt t x 0 rt dt rt Ft F Ft t Ft ys xr 1 2R r2 ys y R s 2r sr 2 s 2 xr dr xr s ys r R 1 2mv 2 0 R M m v0 x x, y x 1, 0 y 1x SECTION 7.8 IMPROPER INTEGRALS \|\|\|\| 517 Note: Additional practice in techniques of integration is provided in Exercises 7.5. 1–40 Evaluate the integral. 1. 2. 3. 4. y 4 1 dt 2t 13 y 2 0 cos 1 sin d y 5 0 ye0.6y dy y 5 0 x x 10 dx 5. 6. 7. 8. 9. 10. y 1 0 sarctan x 1 x 2 dx y 4 1 x 32 ln x dx y dx se x 1 y sinln t t dt y 1 y 2 4y 12 dy y 2 0 sin3 cos2 d EXERCISES 518 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION REVIEW CONCEPT CHECK 7 5. State the rules for approximating the deﬁnite integral with the Midpoint Rule, the Trapezoidal Rule, and Simpson’s Rule. Which would you expect to give the best estimate? How do you approximate the error for each rule? 6. Deﬁne the following improper integrals. (a) (b) (c) 7. Deﬁne the improper integral for each of the follow-ing cases. (a) has an inﬁnite discontinuity at . (b) has an inﬁnite discontinuity at . (c) has an inﬁnite discontinuity at , where . 8. State the Comparison Theorem for improper integrals. a c b c f b f a f xb a f x dx y f x dx y b f x dx y a f x dx xb a f x dx 1. State the rule for integration by parts. In practice, how do you use it? 2. How do you evaluate if is odd? What if is odd? What if and are both even? 3. If the expression occurs in an integral, what sub-stitution might you try? What if occurs? What if occurs? 4. What is the form of the partial fraction expansion of a rational function if the degree of is less than the degree of and has only distinct linear factors? What if a linear factor is repeated? What if has an irreducible quadratic factor (not repeated)? What if the quadratic factor is repeated? Qx Qx Q P PxQx sx 2 a 2 sa 2 x 2 sa 2 x 2 n m n m x sinmx cosnx dx Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. can be put in the form . 2. can be put in the form . 3. can be put in the form . 4. can be put in the form . 5. 6. is convergent. 7. If is continuous, then . x f x dx limt l xt t f x dx f y 1 1 xs2 dx y 4 0 x x 2 1 dx 1 2 ln 15 A x B x 2 4 x 2 4 xx 2 4 A x 2 B x 4 x 2 4 x 2x 4 A x B x 2 C x 2 x 2 4 xx 2 4 A x 2 B x 2 xx 2 4 x 2 4 8. The Midpoint Rule is always more accurate than the Trapezoidal Rule. 9. (a) Every elementary function has an elementary derivative. (b) Every elementary function has an elementary anti-derivative. 10. If is continuous on and is convergent, then is convergent. 11. If is a continuous, decreasing function on and , then is convergent. 12. If and are both convergent, then is convergent. 13. If and are both divergent, then is divergent. 14. If and diverges, then also diverges. x 0 f x dx x 0 tx dx f x tx x a f x tx dx x a tx dx x a f x dx x a f x tx dx x a tx dx x a f x dx x 1 f x dx limx l f x 0 1, f x 0 f x dx x 1 f x dx 0, f TRUE-FALSE QUIZ 49. 50. ; 51–52 Evaluate the indeﬁnite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take ). 51. 52. ; 53. Graph the function and use the graph to guess the value of the integral . Then evaluate the integral to conﬁrm your guess. 54. (a) How would you evaluate by hand? (Don’t actually carry out the integration.) (b) How would you evaluate using tables? (Don’t actually do it.) (c) Use a CAS to evaluate . (d) Graph the integrand and the indeﬁnite integral on the same screen. 55–58 Use the Table of Integrals on the Reference Pages to evaluate the integral. 55. 56. 57. 58. 59. Verify Formula 33 in the Table of Integrals (a) by differentia-tion and (b) by using a trigonometric substitution. 60. Verify Formula 62 in the Table of Integrals. 61. Is it possible to ﬁnd a number such that is convergent? 62. For what values of is convergent? Evaluate the integral for those values of . 63–64 Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule with to approximate the given integral. Round your answers to six decimal places. 63. 64. 65. Estimate the errors involved in Exercise 63, parts (a) and (b). How large should be in each case to guarantee an error of less than 0.00001? 66. Use Simpson’s Rule with to estimate the area under the curve from to . x 4 x 1 y e xx n 6 n y 4 1 sx cos x dx y 4 2 1 ln x dx n 10 a x 0 e ax cos x dx a x 0 x n dx n y cot x s1 2 sin x dx y cos x s4 sin2 x dx y csc5t dt y s4x 2 4x 3 dx x x 5e2x dx x x 5e2x dx x x 5e2x dx CAS x2 0 f x dx f x cos2x sin3x y x 3 sx 2 1 dx y lnx 2 2x 2 dx C 0 y 1 tan1x x 2 dx y dx 4x 2 4x 5 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41–50 Evaluate the integral or show that it is divergent. 41. 42. 43. 44. 45. 46. 47. 48. y 1 1 dx x 2 2x y 1 0 x 1 sx dx y 1 0 1 2 3x dx y 4 0 ln x sx dx y 6 2 y sy 2 dy y 2 dx x ln x y 1 ln x x 4 dx y 1 1 2x 13 dx y 3 4 stan sin 2 d y 12 0 xe 2x 1 2x2 dx y x 2 x 23 dx y cos x sin x2 cos 2x dx y 1 tan 1 tan d y 1 sx x 32 dx y arcsin x2 dx y x 2 4 x 232 dx y 4 0 x sin x cos3x dx y ln 10 0 e xse x 1 e x 8 dx y dx e xs1 e2x y 1 1 x 5 sec x dx y s 3 x 1 s 3 x 1 dx y 2 0 cos3x sin 2x dx y x sin x cos x dx y 3x 3 x 2 6x 4 x 2 1x 2 2 dx y e x cos x dx y dx xsx 2 1 y test dt y dx sx 2 4x y tan5 sec3 d y x 1 9x 2 6x 5 dx y x 2 8x 3 x 3 3x 2 dx y x sec x tan x dx y sec6 tan2 d y x 1 x 2 2x dx y x 2 2 x 2 dx y es 3 x dx y 1 1 sin x 1 x 2 dx y 2 1 sx 2 1 x dx CHAPTER 7 REVIEW \|\|\|\| 519 71. Use the Comparison Theorem to determine whether the integral is convergent or divergent. 72. Find the area of the region bounded by the hyperbola and the line . 73. Find the area bounded by the curves and between and . 74. Find the area of the region bounded by the curves , , and . 75. The region under the curve , is rotated about the -axis. Find the volume of the resulting solid. 76. The region in Exercise 75 is rotated about the -axis. Find the volume of the resulting solid. 77. If is continuous on and , show that 78. We can extend our deﬁnition of average value of a continuous function to an inﬁnite interval by deﬁning the average value of on the interval to be (a) Find the average value of on the interval . (b) If and is divergent, show that the average value of on the interval is , if this limit exists. (c) If is convergent, what is the average value of on the interval ? (d) Find the average value of on the interval . 79. Use the substitution to show that 80. The magnitude of the repulsive force between two point charges with the same sign, one of size 1 and the other of size , is where is the distance between the charges and is a con-stant. The potential at a point due to the charge is deﬁned to be the work expended in bringing a unit charge to from inﬁnity along the straight line that joins and . Find a formula for . V P q P q P V 0 r F q 40r 2 q y 0 ln x 1 x 2 dx 0 u 1x 0, y sin x a, f x a f x dx lim x l f x a, f x a f x dx f x 0 0, y tan1x lim t l 1 t a y t a f x dx a, f y 0 f x dx f 0 lim x l f x 0 0, f y x y cos2x, 0 x 2 x 1 y 1(2 sx ) y 1(2 sx ) x x 0 y cos2x y cos x y 3 y 2 x 2 1 y 1 x 3 x 5 2 dx 67. The speedometer reading ( ) on a car was observed at 1-minute intervals and recorded in the chart. Use Simpson’s Rule to estimate the distance traveled by the car. 68. A population of honeybees increased at a rate of bees per week, where the graph of is as shown. Use Simpson’s Rule with six subintervals to estimate the increase in the bee popu-lation during the ﬁrst 24 weeks. 69. (a) If , use a graph to ﬁnd an upper bound for . (b) Use Simpson’s Rule with to approximate and use part (a) to estimate the error. (c) How large should be to guarantee that the size of the error in using is less than ? 70. Suppose you are asked to estimate the volume of a football. You measure and ﬁnd that a football is 28 cm long. You use a piece of string and measure the circumference at its widest point to be 53 cm. The circumference 7 cm from each end is 45 cm. Use Simpson’s Rule to make your estimate. 28 cm 0.00001 Sn n x 0 f x dx n 10 f 4x f x sinsin x CAS r 0 24 20 16 12 8 4 (weeks) t 4000 8000 12000 r rt v 520 \|\|\|\| CHAPTER 7 TECHNIQUES OF INTEGRATION t (min) (mih) t (min) (mih) 0 40 6 56 1 42 7 57 2 45 8 57 3 49 9 55 4 52 10 56 5 54 v v 521 EXAMPLE 1 (a) Prove that if is a continuous function, then (b) Use part (a) to show that for all positive numbers . SOLUTION (a) At ﬁrst sight, the given equation may appear somewhat bafﬂing. How is it possible to connect the left side to the right side? Connections can often be made through one of the principles of problem solving: introduce something extra. Here the extra ingredient is a new variable. We often think of introducing a new variable when we use the Substitu-tion Rule to integrate a speciﬁc function. But that technique is still useful in the present circumstance in which we have a general function . Once we think of making a substitution, the form of the right side suggests that it should be . Then . When , ; when , . So But this integral on the right side is just another way of writing . So the given equation is proved. (b) If we let the given integral be and apply part (a) with , we get A well-known trigonometric identity tells us that and , so we get Notice that the two expressions for are very similar. In fact, the integrands have the same denominator. This suggests that we should add the two expressions. If we do so, we get Therefore, . M I 4 2I y 2 0 sinnx cosnx sinnx cosnx dx y 2 0 1 dx 2 I I y 2 0 cosnx cosnx sinnx dx cos2 x sin x sin2 x cos x I y 2 0 sinnx sinnx cosnx dx y 2 0 sinn2 x sinn2 x cosn2 x dx a 2 I xa 0 fx dx y a 0 fa x dx y 0 a fu du y a 0 fu du u 0 x a u a x 0 du dx u a x f n y 2 0 sinnx sinnx cosnx dx 4 y a 0 fx dx y a 0 fa x dx f P R O B L E M S P L U S N The principles of problem solving are discussed on page 76. N Cover up the solution to the example and try it yourself ﬁrst. N The computer graphs in Figure 1 make it seem plausible that all of the integrals in the example have the same value. The graph of each integrand is labeled with the corresponding value of . n 1 0 1 2 4 3 π 2 FIGURE 1 522 ; 1. Three mathematics students have ordered a 14-inch pizza. Instead of slicing it in the tradi-tional way, they decide to slice it by parallel cuts, as shown in the ﬁgure. Being mathematics majors, they are able to determine where to slice so that each gets the same amount of pizza. Where are the cuts made? 2. Evaluate . The straightforward approach would be to start with partial fractions, but that would be brutal. Try a substitution. 3. Evaluate . 4. The centers of two disks with radius 1 are one unit apart. Find the area of the union of the two disks. 5. An ellipse is cut out of a circle with radius . The major axis of the ellipse coincides with a diameter of the circle and the minor axis has length . Prove that the area of the remaining part of the circle is the same as the area of an ellipse with semiaxes and . 6. A man initially standing at the point walks along a pier pulling a rowboat by a rope of length . The man keeps the rope straight and taut. The path followed by the boat is a curve called a tractrix and it has the property that the rope is always tangent to the curve (see the ﬁgure). (a) Show that if the path followed by the boat is the graph of the function , then (b) Determine the function . 7. A function is deﬁned by Find the minimum value of . 8. If is a positive integer, prove that 9. Show that Hint: Start by showing that if denotes the integral, then Ik1 2k 2 2k 3 Ik In y 1 0 1 x 2n dx 22nn!2 2n 1! y 1 0 ln xn dx 1nn! n f 0 x 2 f x y 0 cos t cosx t dt f y f x f x dy dx sL 2 x 2 x y f x L O a b a 2b a y 1 0 (s 3 1 x7 s 7 1 x 3 ) dx y 1 x 7 x dx PROBLEMS P R O B L E M S P L U S FIGURE FOR PROBLEM 1 14 in y x O (L, 0) (x, y) L pier FIGURE FOR PROBLEM 6 523 ; 10. Suppose that is a positive function such that is continuous. (a) How is the graph of related to the graph of ? What happens as ? (b) Make a guess as to the value of the limit based on graphs of the integrand. (c) Using integration by parts, conﬁrm the guess that you made in part (b). [Use the fact that, since is continuous, there is a constant such that for .] 11. If , ﬁnd . ; 12. Graph and use the graph to estimate the value of such that is a maximum. Then ﬁnd the exact value of that maximizes this integral. 13. The circle with radius 1 shown in the ﬁgure touches the curve twice. Find the area of the region that lies between the two curves. 14. A rocket is ﬁred straight up, burning fuel at the constant rate of kilograms per second. Let be the velocity of the rocket at time and suppose that the velocity of the exhaust gas is constant. Let be the mass of the rocket at time and note that decreases as the fuel burns. If we neglect air resistance, it follows from Newton’s Second Law that where the force . Thus Let be the mass of the rocket without fuel, the initial mass of the fuel, and . Then, until the fuel runs out at time , the mass is . (a) Substitute into Equation 1 and solve the resulting equation for . Use the initial condition to evaluate the constant. (b) Determine the velocity of the rocket at time . This is called the burnout velocity. (c) Determine the height of the rocket at the burnout time. (d) Find the height of the rocket at any time . 15. Use integration by parts to show that, for all , 16. Suppose , is continuous on and for all . Show that y 1 0 f x dx 1 2 x f x 3 0, 1 f f 1 f 1 0 0 y 0 sin t ln1 x t dt 2 ln1 x x 0 t y yt t M2b v0 0 v M M0 bt M M0 bt t M2b M0 M1 M2 M2 M1 M dv dt ub Mt 1 F Mt F M dv dt ub M t M Mt u t v vt b y 2x t xt1 t f x dx t f x sine x lim t l 0 y 1 0 bx a1 xt dx 1t 0 a b 0 x 1 f x M M f lim n l y 1 0 f x sin nx dx n l y f x y f x sin nx f f P R O B L E M S P L U S FIGURE FOR PROBLEM 13 y=\| 2x \| y 0 x 524 We looked at some applications of integrals in Chapter 6: areas, volumes, work, and average values. Here we explore some of the many other geometric applications of integration—the length of a curve, the area of a surface—as well as quantities of interest in physics, engineering, biology, economics, and statistics. For instance, we will investigate the center of gravity of a plate, the force exerted by water pressure on a dam, the ﬂow of blood from the human heart, and the average time spent on hold during a customer support telephone call. The length of a curve is the limit of lengths of inscribed polygons. FURTHER APPLICATIONS OF INTEGRATION 8 ARC LENGTH What do we mean by the length of a curve? We might think of ﬁtting a piece of string to the curve in Figure 1 and then measuring the string against a ruler. But that might be difﬁcult to do with much accuracy if we have a complicated curve. We need a precise deﬁnition for the length of an arc of a curve, in the same spirit as the deﬁnitions we devel-oped for the concepts of area and volume. If the curve is a polygon, we can easily ﬁnd its length; we just add the lengths of the line segments that form the polygon. (We can use the distance formula to ﬁnd the distance between the endpoints of each segment.) We are going to deﬁne the length of a general curve by ﬁrst approximating it by a polygon and then taking a limit as the number of seg-ments of the polygon is increased. This process is familiar for the case of a circle, where the circumference is the limit of lengths of inscribed polygons (see Figure 2). Now suppose that a curve is deﬁned by the equation , where f is continuous and . We obtain a polygonal approximation to by dividing the interval into n subintervals with endpoints and equal width . If , then the point lies on and the polygon with vertices , , . . . , , illustrated in Figure 3, is an approximation to . The length L of is approximately the length of this polygon and the approximation gets better as we let n increase. (See Figure 4, where the arc of the curve between and has been magniﬁed and approximations with successively smaller values of are shown.) Therefore we deﬁne the length of the curve with equation , , as the limit of the lengths of these inscribed polygons (if the limit exists): Notice that the procedure for deﬁning arc length is very similar to the procedure we used for deﬁning area and volume: We divided the curve into a large number of small parts. We then found the approximate lengths of the small parts and added them. Finally, we took the limit as . The deﬁnition of arc length given by Equation 1 is not very convenient for compu-tational purposes, but we can derive an integral formula for in the case where has a continuous derivative. [Such a function is called smooth because a small change in produces a small change in .] If we let , then Pi1Pi sxi xi12 yi yi12 sx2 yi2 yi yi yi1 fx x f f L n l L lim n l n i1 Pi1Pi 1 a x b y fx C L x Pi Pi1 C FIGURE 3 y P¸ P¡ P™ Pi-1 Pi Pn y=ƒ 0 x i ¤ i-1 b x¡ a x x C Pn P1 P0 C Pixi, yi yi fxi x x0, x1, . . . , xn a, b C a x b y fx C 8.1 525 FIGURE 1 Pi-1 Pi Pi-1 Pi Pi-1 Pi Pi-1 Pi FIGURE 4 FIGURE 2 Visual 8.1 shows an animation of Figure 2. TEC By applying the Mean Value Theorem to on the interval , we ﬁnd that there is a number between and such that that is, Thus we have (since ) Therefore, by Deﬁnition 1, We recognize this expression as being equal to by the deﬁnition of a deﬁnite integral. This integral exists because the function is continuous. Thus we have proved the following theorem: THE ARC LENGTH FORMULA If is continuous on , then the length of the curve , , is If we use Leibniz notation for derivatives, we can write the arc length formula as follows: EXAMPLE 1 Find the length of the arc of the semicubical parabola between the points and . (See Figure 5.) SOLUTION For the top half of the curve we have and so the arc length formula gives If we substitute , then . When , ; when , . u 10 x 4 u 13 4 x 1 du 9 4 dx u 1 9 4x L y 4 1 1 dy dx 2 dx y 4 1 s1 9 4 x dx dy dx 3 2 x 12 y x 32 4, 8 1, 1 y 2 x 3 L y b a 1 dy dx 2 dx 3 L y b a s1 fx2 dx a x b y fx a, b f 2 tx s1 fx2 y b a s1 fx2 dx lim n l n i1 s1 fxi 2 x L lim n l n i1 Pi1Pi x 0 s1 fxi 2 x s1 [ fxi 2 sx2 sx2 fxi x2 Pi1Pi sx2 yi2 yi fxi x fxi fxi1 fxi xi xi1 xi xi1 xi xi1, xi f 526 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION (4, 8) FIGURE 5 0 x y (1, 1) ¥=˛ Therefore M If a curve has the equation , , and is continuous, then by inter-changing the roles of and in Formula 2 or Equation 3, we obtain the following formula for its length: EXAMPLE 2 Find the length of the arc of the parabola from to . SOLUTION Since , we have , and Formula 4 gives We make the trigonometric substitution , which gives and . When , , so ; when , , so , say. Thus (from Example 8 in Section 7.2) (We could have used Formula 21 in the Table of Integrals.) Since , we have , so and M 0 x y 1 1 x=¥ FIGURE 6 L s5 2 ln(s5 2) 4 sec s5 sec2 1 tan2 5 tan 2 1 4(sec tan ln sec tan ) 1 2 1 2[sec tan ln sec tan ]0 L y 0 sec 1 2 sec2 d 1 2 y 0 sec3 d tan1 2 tan 2 y 1 0 tan 0 y 0 s1 4y 2 s1 tan2 sec dy 1 2 sec2 d y 1 2 tan L y 1 0 1 dx dy 2 dy y 1 0 s1 4y 2 dy dxdy 2y x y2 1, 1 0, 0 y 2 x V L y d c s1 ty2 dy y d c 1 dx dy 2 dy 4 y x ty c y d x ty 1 27(80s10 13s13 ) 8 27[1032 ( 13 4 ) 32] L 4 9 y 10 134 su du 4 9 2 3u 32]134 10 SECTION 8.1 ARC LENGTH \|\|\|\| 527 N As a check on our answer to Example 1, notice from Figure 5 that the arc length ought to be slightly larger than the distance from to , which is According to our calculation in Example 1, we have Sure enough, this is a bit greater than the length of the line segment. L 1 27(80s10 13s13 ) 7.633705 s58 7.615773 4, 8 1, 1 N Figure 6 shows the arc of the parabola whose length is computed in Example 2, together with polygonal approximations having and line segments, respectively. For the approximate length is , the diago-nal of a square. The table shows the approxima-tions that we get by dividing into equal subintervals. Notice that each time we double the number of sides of the polygon, we get closer to the exact length, which is L s5 2 ln(s5 2) 4 1.478943 n 0, 1 Ln L1 s2 n 1 n 2 n 1 n 1 1.414 2 1.445 4 1.464 8 1.472 16 1.476 32 1.478 64 1.479 Ln Because of the presence of the square root sign in Formulas 2 and 4, the calculation of an arc length often leads to an integral that is very difﬁcult or even impossible to evaluate explicitly. Thus we sometimes have to be content with ﬁnding an approximation to the length of a curve, as in the following example. EXAMPLE 3 (a) Set up an integral for the length of the arc of the hyperbola from the point to the point . (b) Use Simpson’s Rule with to estimate the arc length. SOLUTION (a) We have and so the arc length is (b) Using Simpson’s Rule (see Section 7.7) with , , , , and , we have M THE ARC LENGTH FUNCTION We will ﬁnd it useful to have a function that measures the arc length of a curve from a par-ticular starting point to any other point on the curve. Thus if a smooth curve has the equation , , let be the distance along from the initial point to the point . Then is a function, called the arc length function, and, by Formula 2, (We have replaced the variable of integration by so that does not have two meanings.) We can use Part 1 of the Fundamental Theorem of Calculus to differentiate Equation 5 (since the integrand is continuous): ds dx s1 fx2 1 dy dx 2 6 x t sx y x a s1 ft2 dt 5 s Qx, fx P0a, fa C sx a x b y fx C 1.1321 x 3 f1 4 f1.1 2 f1.2 4 f1.3 2 f1.8 4 f1.9 f2 L y 2 1 1 1 x 4 dx fx s1 1x 4 x 0.1 n 10 b 2 a 1 L y 2 1 1 dy dx 2 dx y 2 1 1 1 x 4 dx y 2 1 sx 4 1 x 2 dx dy dx 1 x 2 y 1 x n 10 (2, 1 2) 1, 1 xy 1 V 528 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION N Checking the value of the deﬁnite integral with a more accurate approximation produced by a computer algebra system, we see that the approximation using Simpson’s Rule is accurate to four decimal places. Equation 6 shows that the rate of change of with respect to is always at least 1 and is equal to 1 when , the slope of the curve, is 0. The differential of arc length is and this equation is sometimes written in the symmetric form The geometric interpretation of Equation 8 is shown in Figure 7. It can be used as a mnemonic device for remembering both of the Formulas 3 and 4. If we write , then from Equation 8 either we can solve to get (7), which gives (3), or we can solve to get which gives (4). EXAMPLE 4 Find the arc length function for the curve taking as the starting point. SOLUTION If , then Thus the arc length function is given by For instance, the arc length along the curve from to is M s3 32 1 8 ln 3 1 8 ln 3 8 8.1373 3, f3 1, 1 x 2 1 8 ln x 1 y x 1 2t 1 8t dt t 2 1 8 ln t]1 x sx y x 1 s1 ft2 dt s1 fx2 2x 1 8x 4x 2 1 2 1 64x 2 2x 1 8x 2 1 fx2 1 2x 1 8x 2 1 4x 2 1 2 1 64x 2 fx 2x 1 8x fx x 2 1 8 ln x P01, 1 y x 2 1 8 ln x V ds 1 dx dy 2 dy L x ds ds2 dx2 dy2 8 ds 1 dy dx 2 dx 7 fx x s SECTION 8.1 ARC LENGTH \|\|\|\| 529 FIGURE 7 0 x y dx ds dy Îs Îy FIGURE 9 FIGURE 8 P¸ 1 8 y=≈- ln x 1 8 0 x y 1 1 0 x y 1 x 1 s(x)=≈+ ln x-1 s(x) 530 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION 15. , 16. 17. , 18. , , ; 19–20 Find the length of the arc of the curve from point to point . 19. , , 20. , , ; 21–22 Graph the curve and visually estimate its length. Then ﬁnd its exact length. 21. , 22. , 23–26 Use Simpson’s Rule with to estimate the arc length of the curve. Compare your answer with the value of the integral produced by your calculator. 23. , 24. , 25. , 26. , 1 x 3 y x ln x 0 x 3 y sec x 1 y 2 x y sy 0 x 5 y xex n 10 1 2 x 1 y x 3 6 1 2x 1 x 3 y 2 3x 2 132 Q8, 8 P1, 5 x 2 y 43 Q(1, 1 2) P(1, 1 2) y 1 2x 2 Q P a 0 a x b y ln e x 1 e x 1 0 x 1 y e x y sx x 2 sin1(sx ) 0 x 1 2 y ln1 x 2 1. Use the arc length formula (3) to ﬁnd the length of the curve , . Check your answer by noting that the curve is a line segment and calculating its length by the distance formula. 2. Use the arc length formula to ﬁnd the length of the curve , . Check your answer by noting that the curve is part of a circle. 3–6 Set up, but do not evaluate, an integral for the length of the curve. 3. , 4. , 5. , 6. 7–18 Find the length of the curve. , 8. , , 9. , 10. , , 12. , , 14. , 0 x 1 y 3 1 2 cosh 2x 0 x 4 y lnsec x 13. 0 x 3 y lncos x 1 y 9 x 1 3 sy y 3 11. 1 y 2 x y 4 8 1 4y 2 1 x 2 y x 5 6 1 10x 3 y 0 0 x 2 y 2 4x 43 0 x 1 y 1 6x 32 7. x 2 a 2 y 2 b 2 1 1 y 4 x y y 3 0 x 1 y xex2 0 x 2 y cos x 0 x 1 y s2 x 2 1 x 3 y 2x 5 EXERCISES 8.1 N Figure 8 shows the interpretation of the arc length function in Example 4. Figure 9 shows the graph of this arc length function. Why is negative when is less than ? 1 x sx SECTION 8.1 ARC LENGTH \|\|\|\| 531 the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter. 38. The Gateway Arch in St. Louis (see the photo on page 256) was constructed using the equation for the central curve of the arch, where and are measured in meters and . Set up an integral for the length of the arch and use your calculator to estimate the length correct to the nearest meter. A manufacturer of corrugated metal rooﬁng wants to produce panels that are 28 in. wide and 2 in. thick by processing ﬂat sheets of metal as shown in the ﬁgure. The proﬁle of the roof-ing takes the shape of a sine wave. Verify that the sine curve has equation and ﬁnd the width of a ﬂat metal sheet that is needed to make a 28-inch panel. (Use your calculator to evaluate the integral correct to four signiﬁcant digits.) 40. (a) The ﬁgure shows a telephone wire hanging between two poles at and . It takes the shape of a catenary with equation . Find the length of the wire. ; (b) Suppose two telephone poles are 50 ft apart and the length of the wire between the poles is 51 ft. If the lowest point of the wire must be 20 ft above the ground, how high up on each pole should the wire be attached? 41. Find the length of the curve ; The curves with equations , , , , . . . , are called fat circles. Graph the curves with , , , , and to see why. Set up an integral for the length of the fat circle with . Without attempting to evaluate this inte-gral, state the value of . limk l L 2k n 2k L2k 10 8 6 4 n 2 8 6 n 4 x n y n 1 42. 1 x 4 y xx 1 st 3 1 dt y 0 x _b b y c a coshxa x b x b 28 in 2 in w w y sinx7 39. x 91.20 y x y 211.49 20.96 cosh 0.03291765x ; 27. (a) Graph the curve , . (b) Compute the lengths of inscribed polygons with , , and sides. (Divide the interval into equal subintervals.) Illustrate by sketching these polygons (as in Figure 6). (c) Set up an integral for the length of the curve. (d) Use your calculator to ﬁnd the length of the curve to four decimal places. Compare with the approximations in part (b). ; 28. Repeat Exercise 27 for the curve 29. Use either a computer algebra system or a table of integrals to ﬁnd the exact length of the arc of the curve that lies between the points and . 30. Use either a computer algebra system or a table of integrals to ﬁnd the exact length of the arc of the curve that lies between the points and . If your CAS has trouble evaluating the integral, make a substitution that changes the integral into one that the CAS can evaluate. Sketch the curve with equation and use sym-metry to ﬁnd its length. 32. (a) Sketch the curve . (b) Use Formulas 3 and 4 to set up two integrals for the arc length from to . Observe that one of these is an improper integral and evaluate both of them. (c) Find the length of the arc of this curve from to . Find the arc length function for the curve with starting point . ; 34. (a) Graph the curve , . (b) Find the arc length function for this curve with starting point . (c) Graph the arc length function. 35. Find the arc length function for the curve with starting point . 36. A steady wind blows a kite due west. The kite’s height above ground from horizontal position to is given by . Find the distance traveled by the kite. 37. A hawk ﬂying at at an altitude of 180 m accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation until it hits the ground, where is its height above the ground and is the horizontal distance traveled in meters. Calculate x y y 180 x 2 45 15 ms y 150 1 40x 502 x 80 ft x 0 0, 1 y sin1 x s1 x 2 P0(1, 7 12) x 0 y 1 3 x 3 14x P01, 2 y 2x 32 33. 8, 4 1, 1 1, 1 0, 0 y 3 x 2 x 23 y 23 1 31. 1, 1 0, 0 y x 43 CAS 2, ln 2 1, 0 y ln x CAS 0 x 2 y x sin x 4 2 n 1 0 x 4 y x s 3 4 x The curves shown are all examples of graphs of continuous functions that have the following properties. 1. 2. 3. The area under the graph of from 0 to 1 is equal to 1. The lengths of these curves, however, are different. Try to discover formulas for two functions that satisfy the given conditions 1, 2, and 3. (Your graphs might be similar to the ones shown or could look quite different.) Then calculate the arc length of each graph. The winning entry will be the one with the smallest arc length. LÅ3.249 x y 0 1 1 LÅ2.919 x y 0 1 1 LÅ3.152 x y 0 1 1 LÅ3.213 x y 0 1 1 L f f x 0 for 0 x 1 f 0 0 and f 1 0 f ARC LENGTH CONTEST D I S C O V E R Y P R O J E C T 532 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION AREA OF A SURFACE OF REVOLUTION A surface of revolution is formed when a curve is rotated about a line. Such a surface is the lateral boundary of a solid of revolution of the type discussed in Sections 6.2 and 6.3. We want to deﬁne the area of a surface of revolution in such a way that it corresponds to our intuition. If the surface area is , we can imagine that painting the surface would require the same amount of paint as does a ﬂat region with area . Let’s start with some simple surfaces. The lateral surface area of a circular cylinder with radius and height is taken to be because we can imagine cutting the cylin-der and unrolling it (as in Figure 1) to obtain a rectangle with dimensions and . Likewise, we can take a circular cone with base radius and slant height , cut it along the dashed line in Figure 2, and ﬂatten it to form a sector of a circle with radius and cen-tral angle . We know that, in general, the area of a sector of a circle with radius and angle is (see Exercise 35 in Section 7.3) and so in this case the area is Therefore we deﬁne the lateral surface area of a cone to be . A rl A 1 2l 2 1 2l 2 2r l rl 1 2l 2 l 2rl l l r h 2r A 2rh h r A A 8.2 h 2πr FIGURE 1 h r cut What about more complicated surfaces of revolution? If we follow the strategy we used with arc length, we can approximate the original curve by a polygon. When this polygon is rotated about an axis, it creates a simpler surface whose surface area approximates the actual surface area. By taking a limit, we can determine the exact surface area. The approximating surface, then, consists of a number of bands, each formed by rotat-ing a line segment about an axis. To ﬁnd the surface area, each of these bands can be considered a portion of a circular cone, as shown in Figure 3. The area of the band (or frus-tum of a cone) with slant height and upper and lower radii and is found by sub-tracting the areas of two cones: From similar triangles we have which gives or Putting this in Equation 1, we get or where is the average radius of the band. Now we apply this formula to our strategy. Consider the surface shown in Figure 4, which is obtained by rotating the curve , , about the -axis, where is positive and has a continuous derivative. In order to deﬁne its surface area, we divide the interval into n subintervals with endpoints and equal width , as we did in determining arc length. If , then the point lies on the curve. The part of the surface between and is approximated by taking the line segment and rotating it about the -axis. The result is a band with slant height and aver-age radius so, by Formula 2, its surface area is 2 yi1 yi 2 Pi1Pi r 1 2yi1 yi l Pi1Pi x Pi1Pi xi xi1 Pixi, yi yi fxi x x0, x1, . . . , xn a, b f x a x b y fx r 1 2r1 r2 A 2rl 2 A r1l r2l r2 r1l1 r1l r2l1 r1l1 r1l l1 r1 l1 l r2 A r2l1 l r1l1 r2 r1l1 r2l 1 r2 r1 l l ¨ 2πr FIGURE 2 l r cut SECTION 8.2 AREA OF A SURFACE OF REVOLUTION \|\|\|\| 533 r¡ r™ l¡ l FIGURE 3 FIGURE 4 0 x y y=ƒ (a) Surface of revolution P¸ Pi-1 Pi Pn yi 0 x y (b) Approximating band As in the proof of Theorem 8.1.2, we have where is some number in . When is small, we have and also , since is continuous. Therefore and so an approximation to what we think of as the area of the complete surface of revo-lution is This approximation appears to become better as and, recognizing (3) as a Riemann sum for the function , we have Therefore, in the case where is positive and has a continuous derivative, we deﬁne the surface area of the surface obtained by rotating the curve , , about the -axis as With the Leibniz notation for derivatives, this formula becomes If the curve is described as , , then the formula for surface area becomes and both Formulas 5 and 6 can be summarized symbolically, using the notation for arc length given in Section 8.1, as S y 2y ds 7 S y d c 2y1 dx dy 2 dy 6 c y d x ty S y b a 2y1 dy dx 2 dx 5 S y b a 2 fx s1 fx2 dx 4 x a x b y fx f lim n l n i1 2 fxi s1 fxi 2 x y b a 2 fx s1 fx2 dx tx 2 fx s1 fx2 n l n i1 2 fxi s1 fxi 2 x 3 2 yi1 yi 2 Pi1Pi 2 fxi s1 fxi 2 x f yi1 fxi1 fxi yi fxi fxi x xi1, xi xi Pi1Pi s1 fxi 2 x 534 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION For rotation about the -axis, the surface area formula becomes where, as before, we can use either or These formulas can be remembered by thinking of or as the circumference of a circle traced out by the point on the curve as it is rotated about the -axis or -axis, respectively (see Figure 5). EXAMPLE 1 The curve , , is an arc of the circle . Find the area of the surface obtained by rotating this arc about the -axis. (The surface is a portion of a sphere of radius 2. See Figure 6.) SOLUTION We have and so, by Formula 5, the surface area is M 4 y 1 1 1 dx 42 8 2 y 1 1 s4 x 2 2 s4 x 2 dx 2 y 1 1 s4 x 2 1 x 2 4 x 2 dx S y 1 1 2y 1 dy dx 2 dx dy dx 1 24 x 2122x x s4 x 2 x x 2 y 2 4 1 x 1 y s4 x 2 V FIGURE 5 (a) Rotation about x-axis: S=j 2πy ds (x, y) y circumference=2πy x 0 y (b) Rotation about y-axis: S=j 2πx ds (x, y) x circumference=2πx x 0 y y x x, y 2x 2y ds 1 dx dy 2 dy ds 1 dy dx 2 dx S y 2x ds 8 y SECTION 8.2 AREA OF A SURFACE OF REVOLUTION \|\|\|\| 535 N Figure 6 shows the portion of the sphere whose surface area is computed in Example 1. 1 x y FIGURE 6 EXAMPLE 2 The arc of the parabola from to is rotated about the -axis. Find the area of the resulting surface. SOLUTION 1 Using and we have, from Formula 8, Substituting , we have . Remembering to change the limits of integration, we have SOLUTION 2 Using and we have (where ) (as in Solution 1) M EXAMPLE 3 Find the area of the surface generated by rotating the curve , , about the -axis. SOLUTION Using Formula 5 with and dy dx e x y e x x 0 x 1 y e x V 6 (17s17 5s5 ) u 1 4y 4 y 17 5 su du 2 y 4 1 sy 1 1 4y dy y 4 1 s4y 1 dy S y 2x ds y 4 1 2x 1 dx dy 2 dy dx dy 1 2sy x sy 6 (17s17 5s5 ) S 4 y 17 5 su du 4 [ 2 3u 32]5 17 du 8x dx u 1 4x 2 2 y 2 1 x s1 4x 2 dx y 2 1 2x 1 dy dx 2 dx S y 2x ds dy dx 2x y x 2 y 2, 4 1, 1 y x 2 V 536 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION (2, 4) y=≈ x 0 y 1 2 FIGURE 7 N Figure 7 shows the surface of revolution whose area is computed in Example 2. N As a check on our answer to Example 2, notice from Figure 7 that the surface area should be close to that of a circular cylinder with the same height and radius halfway between the upper and lower radius of the surface: . We computed that the surface area was which seems reasonable. Alternatively, the sur-face area should be slightly larger than the area of a frustum of a cone with the same top and bottom edges. From Equation 2, this is . 21.5(s10 ) 29.80 6 (17s17 5s5 ) 30.85 21.53 28.27 N Another method: Use Formula 6 with . x ln y we have (where ) (where and ) (by Example 8 in Section 7.2) Since , we have and M S [es1 e 2 ln(e s1 e 2 ) s2 ln(s2 1)] sec2 1 tan2 1 e 2 tan e [sec tan lnsec tan s2 ln(s2 1)] 2 1 2[sec tan ln sec tan ]4 tan1e u tan 2 y 4 sec3 d u e x 2 y e 1 s1 u2 du S y 1 0 2y 1 dy dx 2 dx 2 y 1 0 e xs1 e 2x dx SECTION 8.2 AREA OF A SURFACE OF REVOLUTION \|\|\|\| 537 17–20 Use Simpson’s Rule with to approximate the area of the surface obtained by rotating the curve about the -axis. Compare your answer with the value of the integral produced by your calculator. 17. , 18. , 19. , 20. , 21–22 Use either a CAS or a table of integrals to ﬁnd the exact area of the surface obtained by rotating the given curve about the -axis. 21. , 22. , 23–24 Use a CAS to ﬁnd the exact area of the surface obtained by rotating the curve about the -axis. If your CAS has trouble evaluating the integral, express the surface area as an integral in the other variable. 23. , 24. , If the region is rotated about the -axis, the volume of the resulting solid is ﬁnite (see Exercise 63 in Section 7.8). Show that the surface area is inﬁnite. (The surface is shown in the ﬁgure and is known as Gabriel’s horn.) 0 1 1 x y= y x x x, y x 1, 0 y 1x 25. 0 x 1 y lnx 1 0 y 1 y x 3 y CAS 0 x 3 y sx 2 1 1 x 2 y 1x x CAS 0 x 1 y ex2 0 x 3 y sec x 1 x 2 y x sx 1 x 3 y ln x x n 10 1–4 Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about (a) the -axis and (b) the -axis. , 2. , 3. , 4. 5–12 Find the area of the surface obtained by rotating the curve about the -axis. , 6. , 7. , 8. , 9. , 10. , , 12. , 13–16 The given curve is rotated about the -axis. Find the area of the resulting surface. 13. , 14. , , 16. , 1 x 2 y 1 4x 2 1 2 ln x 0 y a2 x sa 2 y 2 15. 0 x 1 y 1 x 2 1 y 2 y s 3 x y 1 y 2 x 1 2y 2 1 y 2 x 1 3y 2 232 11. 1 2 x 1 y x 3 6 1 2x 0 x 1 y sin x 0 x a y c a coshxa 1 x 5 y s1 4x 2 x 6 9x y 2 18 0 x 2 y x 3 5. x x sy y 2 0 x 1 y tan1 x 1 x 3 y xex 0 x 1 y x 4 1. y x EXERCISES 8.2 N Or use Formula 21 in the Table of Integrals. 32. Use the result of Exercise 31 to set up an integral to ﬁnd the area of the surface generated by rotating the curve , , about the line . Then use a CAS to evaluate the integral. 33. Find the area of the surface obtained by rotating the circle about the line . 34. Show that the surface area of a zone of a sphere that lies between two parallel planes is , where is the diam-eter of the sphere and is the distance between the planes. (Notice that depends only on the distance between the planes and not on their location, provided that both planes intersect the sphere.) 35. Formula 4 is valid only when . Show that when is not necessarily positive, the formula for surface area becomes 36. Let be the length of the curve , , where is positive and has a continuous derivative. Let be the surface area generated by rotating the curve about the -axis. If is a positive constant, deﬁne and let be the corresponding surface area generated by the curve , . Express in terms of and . L Sf St a x b y tx St tx f x c c x Sf f a x b y f x L S y b a 2 f x s1 f x2 dx f x f x 0 S h d S dh y r x 2 y 2 r 2 y 4 0 x 4 y sx CAS 26. If the inﬁnite curve , , is rotated about the -axis, ﬁnd the area of the resulting surface. 27. (a) If , ﬁnd the area of the surface generated by rotating the loop of the curve about the -axis. (b) Find the surface area if the loop is rotated about the -axis. 28. A group of engineers is building a parabolic satellite dish whose shape will be formed by rotating the curve about the -axis. If the dish is to have a 10-ft diameter and a maximum depth of 2 ft, ﬁnd the value of and the surface area of the dish. 29. (a) The ellipse is rotated about the -axis to form a surface called an ellipsoid, or prolate spheroid. Find the surface area of this ellipsoid. (b) If the ellipse in part (a) is rotated about its minor axis (the -axis), the resulting ellipsoid is called an oblate spheroid. Find the surface area of this ellipsoid. 30. Find the surface area of the torus in Exercise 63 in Section 6.2. If the curve , , is rotated about the horizontal line , where , ﬁnd a formula for the area of the resulting surface. f x c y c a x b y f x 31. y x a b x 2 a 2 y 2 b 2 1 a y y ax 2 y x 3ay 2 xa x2 a 0 x x 0 y ex 538 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION We know how to ﬁnd the volume of a solid of revolution obtained by rotating a region about a horizontal or vertical line (see Section 6.2). We also know how to ﬁnd the surface area of a sur-face of revolution if we rotate a curve about a horizontal or vertical line (see Section 8.2). But what if we rotate about a slanted line, that is, a line that is neither horizontal nor vertical? In this project you are asked to discover formulas for the volume of a solid of revolution and for the area of a surface of revolution when the axis of rotation is a slanted line. Let be the arc of the curve between the points and and let be the region bounded by , by the line (which lies entirely below ), and by the perpendiculars to the line from and . P 0 x y q p C Q y=ƒ y=mx+b Îu Q P C y mx b C Qq, f q Pp, f p y f x C ROTATING ON A SLANT D I S C O V E R Y P R O J E C T 1. Show that the area of is [Hint: This formula can be veriﬁed by subtracting areas, but it will be helpful throughout the project to derive it by ﬁrst approximating the area using rectangles perpendicular to the line, as shown in the ﬁgure. Use the ﬁgure to help express in terms of .] 2. Find the area of the region shown in the ﬁgure at the left. 3. Find a formula similar to the one in Problem 1 for the volume of the solid obtained by rotating about the line . 4. Find the volume of the solid obtained by rotating the region of Problem 2 about the line . 5. Find a formula for the area of the surface obtained by rotating about the line . 6. Use a computer algebra system to ﬁnd the exact area of the surface obtained by rotating the curve , , about the line . Then approximate your result to three decimal places. y 1 2 x 0 x 4 y sx CAS y mx b C y x 2 y mx b y=mx+b Îu å tangent to C at {xi, f(xi)} xi ∫ ? Îx ? x u 1 1 m 2 y q p f x mx b1 mf x dx SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING \|\|\|\| 539 y x 0 (2π, 2π) y=x+sin x y=x-2 APPLICATIONS TO PHYSICS AND ENGINEERING Among the many applications of integral calculus to physics and engineering, we consider two here: force due to water pressure and centers of mass. As with our previous applica-tions to geometry (areas, volumes, and lengths) and to work, our strategy is to break up the physical quantity into a large number of small parts, approximate each small part, add the results, take the limit, and then evaluate the resulting integral. HYDROSTATIC FORCE AND PRESSURE Deep-sea divers realize that water pressure increases as they dive deeper. This is because the weight of the water above them increases. In general, suppose that a thin horizontal plate with area square meters is submerged in a ﬂuid of density kilograms per cubic meter at a depth meters below the surface of the ﬂuid as in Figure 1. The ﬂuid directly above the plate has volume , so its mass is . The force exerted by the ﬂuid on the plate is therefore F mt tAd m V Ad V Ad d A 8.3 surface of ﬂuid FIGURE 1 where is the acceleration due to gravity. The pressure on the plate is deﬁned to be the force per unit area: The SI unit for measuring pressure is newtons per square meter, which is called a pascal (abbreviation: 1 Nm Pa). Since this is a small unit, the kilopascal (kPa) is often used. For instance, because the density of water is , the pressure at the bottom of a swimming pool 2 m deep is An important principle of ﬂuid pressure is the experimentally veriﬁed fact that at any point in a liquid the pressure is the same in all directions. (A diver feels the same pressure on nose and both ears.) Thus the pressure in any direction at a depth in a ﬂuid with mass density is given by This helps us determine the hydrostatic force against a vertical plate or wall or dam in a ﬂuid. This is not a straightforward problem because the pressure is not constant but increases as the depth increases. EXAMPLE 1 A dam has the shape of the trapezoid shown in Figure 2. The height is 20 m, and the width is 50 m at the top and 30 m at the bottom. Find the force on the dam due to hydrostatic pressure if the water level is 4 m from the top of the dam. SOLUTION We choose a vertical -axis with origin at the surface of the water as in Figure 3(a). The depth of the water is 16 m, so we divide the interval into sub-intervals of equal length with endpoints and we choose . The hori-zontal strip of the dam is approximated by a rectangle with height and width , where, from similar triangles in Figure 3(b), or and so If is the area of the strip, then If is small, then the pressure on the strip is almost constant and we can use Equation 1 to write The hydrostatic force acting on the strip is the product of the pressure and the area: Fi PiAi 1000txi 46 xi x ith Fi Pi 1000txi ith Pi x Ai wi x 46 xi x ith Ai wi 215 a 2(15 8 1 2 xi ) 46 xi a 16 xi 2 8 xi 2 a 16 xi 10 20 wi x ith xi xi1, xi xi 0, 16 x V P td d 1 d 19,600 Pa 19.6 kPa P td 1000 kgm3 9.8 ms2 2 m 1000 kgm3 2 1 P F A td P t 540 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION 50 m 20 m 30 m FIGURE 2 FIGURE 3 (b) a 10 16-xi 20 (a) x 0 _4 15 15 10 Îx N When using US Customary units, we write , where is the weight density (as opposed to , which is the mass density). For instance, the weight density of water is . 62.5 lbft3 t P td d Adding these forces and taking the limit as , we obtain the total hydrostatic force on the dam: M EXAMPLE 2 Find the hydrostatic force on one end of a cylindrical drum with radius 3 ft if the drum is submerged in water 10 ft deep. SOLUTION In this example it is convenient to choose the axes as in Figure 4 so that the origin is placed at the center of the drum. Then the circle has a simple equation, . As in Example 1 we divide the circular region into horizontal strips of equal width. From the equation of the circle, we see that the length of the strip is and so its area is The pressure on this strip is approximately and so the force on the strip is approximately The total force is obtained by adding the forces on all the strips and taking the limit: The second integral is 0 because the integrand is an odd function (see Theorem 5.5.7). The ﬁrst integral can be evaluated using the trigonometric substitution , but it’s simpler to observe that it is the area of a semicircular disk with radius 3. Thus M 7875 2 12,370 lb F 875 y 3 3 s9 y 2 dy 875 1 232 y 3 sin 125 7 y 3 3 s9 y 2 dy 125 y 3 3 ys9 y 2 dy 125 y 3 3 7 y s9 y 2 dy F lim n l n i1 62.57 yi 2s9 yi 2 y diAi 62.57 yi 2s9 yi 2 y di 62.57 yi Ai 2s9 yi 2 y 2s9 yi 2 ith x 2 y 2 9 4.43 107 N 980023x 2 x 3 3 0 16 10009.8 y 16 0 46x x 2 dx y 16 0 1000tx46 x dx F lim n l n i1 1000txi 46 xi x n l SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING \|\|\|\| 541 FIGURE 4 œ„„„„„„„ œ œ œ„ œ„ œ„ œ„ œ ( œ yi ) x 0 y 10 10 10 10 7 di ≈+¥=9 9 =9 =9 9 yi Îy MOMENTS AND CENTERS OF MASS Our main objective here is to ﬁnd the point on which a thin plate of any given shape bal-ances horizontally as in Figure 5. This point is called the center of mass (or center of grav-ity) of the plate. We ﬁrst consider the simpler situation illustrated in Figure 6, where two masses and are attached to a rod of negligible mass on opposite sides of a fulcrum and at distances and from the fulcrum. The rod will balance if This is an experimental fact discovered by Archimedes and called the Law of the Lever. (Think of a lighter person balancing a heavier one on a seesaw by sitting farther away from the center.) Now suppose that the rod lies along the -axis with at and at and the center of mass at . If we compare Figures 6 and 7, we see that and and so Equation 2 gives The numbers and are called the moments of the masses and (with respect to the origin), and Equation 3 says that the center of mass is obtained by adding the moments of the masses and dividing by the total mass . In general, if we have a system of particles with masses , . . . , located at the points , . . . , on the -axis, it can be shown similarly that the center of mass of the system is located at where is the total mass of the system, and the sum of the individual moments is called the moment of the system about the origin. Then Equation 4 could be rewritten as , which says that if the total mass were considered as being concentrated at the center of mass , then its moment would be the same as the moment of the system. x mx M M n i1 mixi m mi x n i1 mixi n i1 mi n i1 mixi m 4 x xn x2, x1 mn m2, m1 n 0 ⁄ – x ¤ ¤-x – m¡ m™ x – x-⁄ FIGURE 7 m m1 m2 x m2 m1 m2x2 m1x1 x m1x1 m2x2 m1 m2 3 m1x m2x m1x1 m2x2 m1x x1 m2x2 x d2 x2 x d1 x x1 x x2 m2 x1 m1 x m1d1 m2d2 2 d2 d1 m2 m1 P 542 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION FIGURE 5 P FIGURE 6 m¡ m™ d¡ fulcrum d™ Now we consider a system of particles with masses , . . . , located at the points , , . . . , in the -plane as shown in Figure 8. By analogy with the one-dimensional case, we deﬁne the moment of the system about the y-axis to be and the moment of the system about the x-axis as Then measures the tendency of the system to rotate about the -axis and measures the tendency to rotate about the -axis. As in the one-dimensional case, the coordinates of the center of mass are given in terms of the moments by the formulas where is the total mass. Since and , the center of mass is the point where a single particle of mass would have the same moments as the system. EXAMPLE 3 Find the moments and center of mass of the system of objects that have masses 3, 4, and 8 at the points , , and , respectively. SOLUTION We use Equations 5 and 6 to compute the moments: Since , we use Equations 7 to obtain Thus the center of mass is . (See Figure 9.) M Next we consider a ﬂat plate (called a lamina) with uniform density that occupies a region of the plane. We wish to locate the center of mass of the plate, which is called the centroid of . In doing so we use the following physical principles: The symmetry principle says that if is symmetric about a line , then the centroid of lies on . (If is reﬂected about , then remains the same so its centroid remains ﬁxed. But the only ﬁxed points lie on .) Thus the centroid of a rectangle is its center. Moments should be deﬁned so that if the entire mass of a region is concentrated at the center of mass, then its moments remain unchanged. Also, the moment of the union of two nonoverlapping regions should be the sum of the moments of the individual regions. l l l l (1 14 15, 1) y Mx m 15 15 1 x My m 29 15 m 3 4 8 15 Mx 31 41 82 15 My 31 42 83 29 3, 2 2, 1 1, 1 V m x, y my Mx mx My m mi y Mx m x My m 7 x, y x Mx y My Mx n i1 miyi 6 My n i1 mixi 5 xy xn, yn x2, y2 x1, y1 mn m2, m1 n SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING \|\|\|\| 543 m£ m¡ m™ y 0 x ‹ y£ ⁄ › ¤ ﬁ FIGURE 8 FIGURE 9 y 0 x 8 4 3 center of mass Suppose that the region is of the type shown in Figure 10(a); that is, lies between the lines and , above the -axis, and beneath the graph of , where is a continuous function. We divide the interval into n subintervals with endpoints and equal width . We choose the sample point to be the midpoint of the th subinterval, that is, . This determines the polygonal approxima-tion to shown in Figure 10(b). The centroid of the th approximating rectangle is its center . Its area is , so its mass is The moment of about the -axis is the product of its mass and the distance from to the -axis, which is Adding these moments, we obtain the moment of the polygonal approximation to , and then by taking the limit as we obtain the moment of itself about the -axis: In a similar fashion we compute the moment of about the -axis as the product of its mass and the distance from to the -axis: Again we add these moments and take the limit to obtain the moment of about the -axis: Just as for systems of particles, the center of mass of the plate is deﬁned so that and . But the mass of the plate is the product of its density and its area: and so Notice the cancellation of the ’s. The location of the center of mass is independent of the density. y Mx m y b a 1 2 fx2 dx y b a fx dx y b a 1 2 fx2 dx y b a fx dx x My m y b a xfx dx y b a fx dx y b a xfx dx y b a fx dx m A y b a fx dx my Mx mx My Mx lim n l n i1 1 2 fxi2 x y b a 1 2 fx2 dx x MxRi fxi x 1 2 fxi 1 2 fxi2 x x Ci x Ri My lim n l n i1 xi fxi x y b a x fx dx y n l MyRi fxi x xi xi fxi x xi. Thus y Ci y Ri fxi x fxi x Ci(xi, 1 2 fxi) Ri i xi xi1 xi2 i xi xi x x0, x1, . . . , xn a, b f f x x b x a 544 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION Ci”xi, f(xi)’ xi FIGURE 10 y 0 x a b R¡ R™R£ xi_1 xi {xi, f(xi)} 1 2 (b) y 0 x a b y=ƒ (a) In summary, the center of mass of the plate (or the centroid of ) is located at the point , where EXAMPLE 4 Find the center of mass of a semicircular plate of radius . SOLUTION In order to use (8) we place the semicircle as in Figure 11 so that and , . Here there is no need to use the formula to calcu-late because, by the symmetry principle, the center of mass must lie on the -axis, so . The area of the semicircle is , so The center of mass is located at the point . M EXAMPLE 5 Find the centroid of the region bounded by the curves , , , and . SOLUTION The area of the region is so Formulas 8 give (by integration by parts) The centroid is and is shown in Figure 12. M ( 1 2 1, 1 8) 8 1 4 y 2 0 1 cos 2x dx 1 4[x 1 2 sin 2x]0 2 y 1 A y 2 0 1 2 fx2 dx 1 2 y 2 0 cos2x dx 2 1 x sin x]0 2 y 2 0 sin x dx x 1 A y 2 0 xfx dx y 2 0 x cos x dx A y 2 0 cos x dx sin x]0 2 1 x 2 x 0 y 0 y cos x 0, 4r3 2 r 2 2r 3 3 4r 3 2 r 2 y r 0 r 2 x 2 dx 2 r 2r 2x x 3 3 0 r 1 1 2r 2 1 2 y r r (sr 2 x 2 ) 2 dx y 1 A y r r 1 2 fx2 dx A 1 2r 2 x 0 y x b r a r fx sr 2 x 2 r y 1 A y b a 1 2 fx2 dx x 1 A y b a xfx dx 8 x, y SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING \|\|\|\| 545 x y r 0 r ”0, ’ 4r 3π y= r@-≈ œ„„„„„ FIGURE 11 x y 0 ” -1, ’ π 8 π 2 π 2 FIGURE 12 y=cos x If the region lies between two curves and , where , as illustrated in Figure 13, then the same sort of argument that led to Formulas 8 can be used to show that the centroid of is , where (See Exercise 47.) EXAMPLE 6 Find the centroid of the region bounded by the line and the parabola . SOLUTION The region is sketched in Figure 14. We take , , , and in Formulas 9. First we note that the area of the region is Therefore The centroid is . M We end this section by showing a surprising connection between centroids and volumes of revolution. THEOREM OF PAPPUS Let be a plane region that lies entirely on one side of a line in the plane. If is rotated about , then the volume of the resulting solid is the product of the area of and the distance traveled by the centroid of . PROOF We give the proof for the special case in which the region lies between and as in Figure 13 and the line is the -axis. Using the method of cylindrical shells y l y tx y fx d A l l ( 1 2, 2 5) 3 x 3 3 x 5 5 0 1 2 5 y 1 A y 1 0 1 2 fx2 tx2 dx 1 1 6 y 1 0 1 2x 2 x 4 dx 6 y 1 0 x 2 x 3 dx 6 x 3 3 x 4 4 0 1 1 2 x 1 A y 1 0 x fx tx dx 1 1 6 y 1 0 xx x 2 dx A y 1 0 x x 2 dx x 2 2 x 3 3 0 1 1 6 b 1 a 0 tx x 2 fx x y x 2 y x y 1 A y b a 1 2 fx2 tx2 dx x 1 A y b a x fx tx dx 9 x, y fx tx y tx y fx 546 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION FIGURE 13 xi x y 0 a b xi y=ƒ y=© 1 2 Ci”xi, 2 f(xi)+g(xi)’ x y 0 ” , ’ 2 5 1 2 y=x y=≈ (1, 1) FIGURE 14 N This theorem is named after the Greek mathematician Pappus of Alexandria, who lived in the fourth century AD. (see Section 6.3), we have (by Formulas 9) where is the distance traveled by the centroid during one rotation about the -axis. M EXAMPLE 7 A torus is formed by rotating a circle of radius about a line in the plane of the circle that is a distance from the center of the circle. Find the volume of the torus. SOLUTION The circle has area . By the symmetry principle, its centroid is its cen-ter and so the distance traveled by the centroid during a rotation is . Therefore, by the Theorem of Pappus, the volume of the torus is M The method of Example 7 should be compared with the method of Exercise 63 in Section 6.2. V Ad 2Rr 2 2 2r 2R d 2R A r 2 r R r V y d 2x 2 xA Ad 2xA 2 y b a x fx tx dx V y b a 2x fx tx dx SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING \|\|\|\| 547 5. 6. 8. 9. 10. 4 m 1 m 2 m 7. 6 m 6 m 1 m 1. An aquarium 5 ft long, 2 ft wide, and 3 ft deep is full of water. Find (a) the hydrostatic pressure on the bottom of the aquarium, (b) the hydrostatic force on the bottom, and (c) the hydrostatic force on one end of the aquarium. 2. A tank is 8 m long, 4 m wide, 2 m high, and contains kerosene with density to a depth of 1.5 m. Find (a) the hydro-static pressure on the bottom of the tank, (b) the hydrostatic force on the bottom, and (c) the hydrostatic force on one end of the tank. 3–11 A vertical plate is submerged (or partially submerged) in water and has the indicated shape. Explain how to approximate the hydrostatic force against one side of the plate by a Riemann sum. Then express the force as an integral and evaluate it. 3. 4. 820 kgm3 EXERCISES 8.3 19. A vertical, irregularly shaped plate is submerged in water. The table shows measurements of its width, taken at the indi-cated depths. Use Simpson’s Rule to estimate the force of the water against the plate. 20. (a) Use the formula of Exercise 18 to show that where is the -coordinate of the centroid of the plate and is its area. This equation shows that the hydrostatic force against a vertical plane region is the same as if the region were horizontal at the depth of the centroid of the region. (b) Use the result of part (a) to give another solution to Exercise 10. 21–22 Point-masses are located on the -axis as shown. Find the moment of the system about the origin and the center of mass . 21. 22. 23–24 The masses are located at the points . Find the moments and and the center of mass of the system. 23. , , ; , , 24. , , , ; , , , 25–28 Sketch the region bounded by the curves, and visually esti-mate the location of the centroid. Then ﬁnd the exact coordinates of the centroid. 25. , 26. , , , , , 28. , , , x 2 x 1 y 0 y 1x x 1 x 0 y 0 y e x 27. x 0 y 0 3x 2y 6 y 0 y 4 x 2 P46, 1 P33, 7 P23, 4 P11, 2 m4 4 m3 1 m2 5 m1 6 P32, 1 P23, 2 P11, 5 m3 10 m2 5 m1 6 My Mx Pi mi x 0 3 7 m™=20 m£=10 _2 m¡=25 x 0 2 5 m¡=40 m™=30 x M x mi A x x F txA 11. 12. A large tank is designed with ends in the shape of the region between the curves and , measured in feet. Find the hydrostatic force on one end of the tank if it is ﬁlled to a depth of 8 ft with gasoline. (Assume the gasoline’s den-sity is .) A trough is ﬁlled with a liquid of density 840 kgm . The ends of the trough are equilateral triangles with sides 8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. 14. A vertical dam has a semicircular gate as shown in the ﬁgure. Find the hydrostatic force against the gate. 15. A cube with 20-cm-long sides is sitting on the bottom of an aquarium in which the water is one meter deep. Estimate the hydrostatic force on (a) the top of the cube and (b) one of the sides of the cube. 16. A dam is inclined at an angle of from the vertical and has the shape of an isosceles trapezoid 100 ft wide at the top and 50 ft wide at the bottom and with a slant height of 70 ft. Find the hydrostatic force on the dam when it is full of water. 17. A swimming pool is 20 ft wide and 40 ft long and its bottom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9 ft. If the pool is full of water, estimate the hydrostatic force on (a) the shallow end, (b) the deep end, (c) one of the sides, and (d) the bottom of the pool. 18. Suppose that a plate is immersed vertically in a ﬂuid with density and the width of the plate is at a depth of meters beneath the surface of the ﬂuid. If the top of the plate is at depth and the bottom is at depth , show that the hydrostatic force on one side of the plate is F y b a txwx dx b a x wx 30 12 m 2 m 4 m water level 3 13. 42.0 lbft3 y 12 y 1 2x 2 2a 548 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION Depth (m) 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Plate width (m) 0 0.8 1.7 2.4 2.9 3.3 3.6 40–41 Find the centroid of the region shown, not by integration, but by locating the centroids of the rectangles and triangles (from Exercise 39) and using additivity of moments. 40. 42. A rectangle with sides and is divided into two parts and by an arc of a parabola that has its vertex at one corner of and passes through the opposite corner. Find the centroids of both and . 43. If is the -coordinate of the centroid of the region that lies under the graph of a continuous function , where , show that 44–46 Use the Theorem of Pappus to ﬁnd the volume of the given solid. 44. A sphere of radius (Use Example 4.) 45. A cone with height and base radius 46. The solid obtained by rotating the triangle with vertices , , and about the -axis 47. Prove Formulas 9. 48. Let be the region that lies between the curves and , , where and are integers with . (a) Sketch the region . (b) Find the coordinates of the centroid of . (c) Try to ﬁnd values of and such that the centroid lies outside . n m 0 n m n m 0 x 1 y x n y x m x 5, 4 2, 5 2, 3 r h r y b a cx d f x dx cx dy b a f x dx a x b f x x x y 0 a R™ R¡ b R2 R1 R R2 R1 b a R x y 0 1 _1 _1 2 1 2 _2 41. x y 0 1 1 3 2 3 _2 29–33 Find the centroid of the region bounded by the given curves. 29. , 30. , , , , 32. , , 33. , 34–35 Calculate the moments and and the center of mass of a lamina with the given density and shape. 34. 35. 36. Use Simpson’s Rule to estimate the centroid of the region shown. ; 37. Find the centroid of the region bounded by the curves and , , to three decimal places. Sketch the region and plot the centroid to see if your answer is reasonable. ; 38. Use a graph to ﬁnd approximate -coordinates of the points of intersection of the curves and . Then ﬁnd (approximately) the centroid of the region bounded by these curves. 39. Prove that the centroid of any triangle is located at the point of intersection of the medians. [Hints: Place the axes so that the vertices are , , and . Recall that a median is a line segment from a vertex to the midpoint of the oppo-site side. Recall also that the medians intersect at a point two-thirds of the way from each vertex (along the median) to the opposite side.] c, 0 0, b a, 0 y x 3 x y x ln x x 0 x 2 y x 2 y 2x 2 4 x y 0 8 6 4 2 x (4, 3) y 0 x y 0 1 _1 1 10 3 My Mx x 0 x 5 y 2 y 0 x y 2 y x 3 x 4 x 0 y cos x y sin x 31. y x 2 y x 2 x y 2 y x 2 SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING \|\|\|\| 549 550 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION APPLICATIONS TO ECONOMICS AND BIOLOGY In this section we consider some applications of integration to economics (consumer sur-plus) and biology (blood ﬂow, cardiac output). Others are described in the exercises. CONSUMER SURPLUS Recall from Section 4.7 that the demand function is the price that a company has to charge in order to sell units of a commodity. Usually, selling larger quantities requires lowering prices, so the demand function is a decreasing function. The graph of a typical demand function, called a demand curve, is shown in Figure 1. If is the amount of the commodity that is currently available, then is the current selling price. We divide the interval into subintervals, each of length , and let be the right endpoint of the th subinterval, as in Figure 2. If, after the ﬁrst units were sold, a total of only units had been available and the price per unit had been set at dollars, then the additional units could have been sold (but no more). The consumers who would have paid dollars placed a high value on the product; they would have paid what it was worth to them. So, in paying only dollars they have saved an amount of savings per unitnumber of units pxi P x P pxi x pxi xi xi1 i xi xi x Xn n 0, X P pX X x px 8.4 0 x p P X (X, P) p=p(x) FIGURE 1 A typical demand curve Suppose you have a choice of two coffee cups of the type shown, one that bends outward and one inward, and you notice that they have the same height and their shapes ﬁt together snugly. You wonder which cup holds more coffee. Of course you could ﬁll one cup with water and pour it into the other one but, being a calculus student, you decide on a more mathematical approach. Ignoring the handles, you observe that both cups are surfaces of revolution, so you can think of the coffee as a volume of revolution. 1. Suppose the cups have height , cup A is formed by rotating the curve about the -axis, and cup B is formed by rotating the same curve about the line . Find the value of such that the two cups hold the same amount of coffee. 2. What does your result from Problem 1 say about the areas and shown in the ﬁgure? 3. Use Pappus’s Theorem to explain your result in Problems 1 and 2. 4. Based on your own measurements and observations, suggest a value for and an equation for and calculate the amount of coffee that each cup holds. x f y h A2 A1 k x k y x f y h x y 0 h k x=k A¡ x=f(y) A™ Cup A Cup B COMPLEMENTARY COFFEE CUPS D I S C O V E R Y P R O J E C T Considering similar groups of willing consumers for each of the subintervals and adding the savings, we get the total savings: (This sum corresponds to the area enclosed by the rectangles in Figure 2.) If we let , this Riemann sum approaches the integral which economists call the consumer surplus for the commodity. The consumer surplus represents the amount of money saved by consumers in pur-chasing the commodity at price , corresponding to an amount demanded of . Figure 3 shows the interpretation of the consumer surplus as the area under the demand curve and above the line . EXAMPLE 1 The demand for a product, in dollars, is Find the consumer surplus when the sales level is 500. SOLUTION Since the number of products sold is , the corresponding price is Therefore, from Deﬁnition 1, the consumer surplus is M BLOOD FLOW In Example 7 in Section 3.7 we discussed the law of laminar ﬂow: which gives the velocity of blood that ﬂows along a blood vessel with radius and length at a distance from the central axis, where is the pressure difference between the ends of the vessel and is the viscosity of the blood. Now, in order to compute the rate of blood ﬂow, or ﬂux (volume per unit time), we consider smaller, equally spaced radii . . . . r1, r2, P r l R v vr P 4 l R2 r 2 $33,333.33 125500 0.15002 0.00015003 3 125x 0.1x 2 0.0001 x 3 3 0 500 y 500 0 125 0.2x 0.0001x 2 dx y 500 0 px P dx y 500 0 1200 0.2x 0.0001x 2 1075 dx P 1200 0.2500 0.00015002 1075 X 500 p 1200 0.2x 0.0001x 2 V p P X P y X 0 px P dx 1 n l n i1 pxi P x SECTION 8.4 APPLICATIONS TO ECONOMICS AND BIOLOGY \|\|\|\| 551 0 x p P ⁄ xi X (X, P) FIGURE 2 0 x p (X, P) P X p=p(x) p=P consumer surplus FIGURE 3 The approximate area of the ring (or washer) with inner radius and outer radius is where (See Figure 4.) If is small, then the velocity is almost constant throughout this ring and can be approximated by . Thus the volume of blood per unit time that ﬂows across the ring is approximately and the total volume of blood that ﬂows across a cross-section per unit time is approxi-mately This approximation is illustrated in Figure 5. Notice that the velocity (and hence the vol-ume per unit time) increases toward the center of the blood vessel. The approximation gets better as n increases. When we take the limit we get the exact value of the ﬂux (or dis-charge), which is the volume of blood that passes a cross-section per unit time: The resulting equation is called Poiseuille’s Law; it shows that the ﬂux is proportional to the fourth power of the radius of the blood vessel. CARDIAC OUTPUT Figure 6 shows the human cardiovascular system. Blood returns from the body through the veins, enters the right atrium of the heart, and is pumped to the lungs through the pul-monary arteries for oxygenation. It then ﬂows back into the left atrium through the pulmo-nary veins and then out to the rest of the body through the aorta. The cardiac output of the heart is the volume of blood pumped by the heart per unit time, that is, the rate of ﬂow into the aorta. The dye dilution method is used to measure the cardiac output. Dye is injected into the right atrium and ﬂows through the heart into the aorta. A probe inserted into the aorta measures the concentration of the dye leaving the heart at equally spaced times over a time interval until the dye has cleared. Let be the concentration of the dye at time If we divide into subintervals of equal length , then the amount of dye that ﬂows past the measuring point during the subinterval from to is approximately concentrationvolume ctiF t t ti t ti1 t 0, T t. ct 0, T F PR4 8 l 2 P 2 l R4 2 R4 4 PR4 8 l P 2 l y R 0 R2r r 3 dr P 2 l R2 r 2 2 r 4 4 r0 rR y R 0 2r P 4 l R2 r 2 dr F lim n l n i1 2ri vri r y R 0 2r vr dr n i1 2ri vri r 2ri r vri 2ri vri r vri r r ri ri1 2ri r ri ri1 552 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION Îr ri FIGURE 4 FIGURE 5 FIGURE 6 aorta vein right atrium pulmonary arteries left atrium pulmonary veins pulmonary veins vein pulmonary arteries where is the rate of ﬂow that we are trying to determine. Thus the total amount of dye is approximately and, letting , we ﬁnd that the amount of dye is Thus the cardiac output is given by where the amount of dye is known and the integral can be approximated from the con-centration readings. EXAMPLE 2 A 5-mg bolus of dye is injected into a right atrium. The concentration of the dye (in milligrams per liter) is measured in the aorta at one-second intervals as shown in the chart. Estimate the cardiac output. SOLUTION Here , , and . We use Simpson’s Rule to approximate the integral of the concentration: Thus Formula 3 gives the cardiac output to be M 0.12 Ls 7.2 Lmin F A y 10 0 ct dt 5 41.87 41.87 26.1 44.0 22.3 41.1 0 y 10 0 ct dt 1 30 40.4 22.8 46.5 29.8 48.9 T 10 t 1 A 5 V A F A y T 0 ct dt 3 A F y T 0 ct dt n l n i1 ctiF t F n i1 cti t F SECTION 8.4 APPLICATIONS TO ECONOMICS AND BIOLOGY \|\|\|\| 553 t t 0 0 6 6.1 1 0.4 7 4.0 2 2.8 8 2.3 3 6.5 9 1.1 4 9.8 10 0 5 8.9 ct ct 4. The demand function for a certain commodity is . Find the consumer surplus when the sales level is 300. Illustrate by drawing the demand curve and identifying the consumer surplus as an area. A demand curve is given by . Find the con-sumer surplus when the selling price is . 6. The supply function for a commodity gives the rela-tion between the selling price and the number of units that manufacturers will produce at that price. For a higher price, manufacturers will produce more units, so is an increasing function of . Let be the amount of the commodity currently produced and let be the current price. Some pro-ducers would be willing to make and sell the commodity for a lower selling price and are therefore receiving more than their minimal price. The excess is called the producer surplus. An P pSX X x pS pSx $10 p 450x 8 5. p 20 0.05x 1. The marginal cost function was deﬁned to be the derivative of the cost function. (See Sections 3.7 and 4.7.) If the marginal cost of maufacturing meters of a fabric is (measured in dollars per meter) and the ﬁxed start-up cost is , use the Net Change Theorem to ﬁnd the cost of producing the ﬁrst 2000 units. 2. The marginal revenue from the sale of units of a product is . If the revenue from the sale of the ﬁrst 1000 units is $12,400, ﬁnd the revenue from the sale of the ﬁrst 5000 units. The marginal cost of producing units of a certain product is (in dollars per unit). Find the increase in cost if the production level is raised from 1200 units to 1600 units. 74 1.1x 0.002x 2 0.00004x 3 x 3. 12 0.0004x x C0 $20,000 Cx 5 0.008x 0.000009x 2 x Cx EXERCISES 8.4 14. A hot, wet summer is causing a mosquito population explo-sion in a lake resort area. The number of mosquitos is increasing at an estimated rate of per week (where is measured in weeks). By how much does the mos-quito population increase between the ﬁfth and ninth weeks of summer? 15. Use Poiseuille’s Law to calculate the rate of ﬂow in a small human artery where we can take , cm, cm, and dynescm . 16. High blood pressure results from constriction of the arteries. To maintain a normal ﬂow rate (ﬂux), the heart has to pump harder, thus increasing the blood pressure. Use Poiseuille’s Law to show that if and are normal values of the radius and pressure in an artery and the constricted values are and , then for the ﬂux to remain constant, and are related by the equation Deduce that if the radius of an artery is reduced to three-fourths of its former value, then the pressure is more than tripled. The dye dilution method is used to measure cardiac output with 6 mg of dye. The dye concentrations, in , are mod-eled by , , where is measured in seconds. Find the cardiac output. 18. After an 8-mg injection of dye, the readings of dye concentra-tion, in , at two-second intervals are as shown in the table. Use Simpson’s Rule to estimate the cardiac output. 19. The graph of the concentration function is shown after a 7-mg injection of dye into a heart. Use Simpson’s Rule to estimate the cardiac output. 0 y (mg/L) t (seconds) 4 6 2 4 10 2 8 14 12 6 ct mgL t 0 t 10 ct 20te0.6t mgL 17. P P0 R0 R 4 R P P R P0 R0 2 P 4000 l 2 R 0.008 0.027 t 2200 10e0.8t argument similar to that for consumer surplus shows that the surplus is given by the integral Calculate the producer surplus for the supply function at the sales level . Illustrate by drawing the supply curve and identifying the producer surplus as an area. 7. If a supply curve is modeled by the equation , ﬁnd the producer surplus when the selling price is $400. 8. For a given commodity and pure competition, the number of units produced and the price per unit are determined as the coordinates of the point of intersection of the supply and demand curves. Given the demand curve and the supply curve , ﬁnd the consumer surplus and the producer surplus. Illustrate by sketching the supply and demand curves and identifying the surpluses as areas. ; 9. A company modeled the demand curve for its product (in dollars) by the equation Use a graph to estimate the sales level when the selling price is $16. Then ﬁnd (approximately) the consumer surplus for this sales level. A movie theater has been charging $7.50 per person and sell-ing about 400 tickets on a typical weeknight. After surveying their customers, the theater estimates that for every 50 cents that they lower the price, the number of moviegoers will increase by 35 per night. Find the demand function and calcu-late the consumer surplus when the tickets are priced at $6.00. 11. If the amount of capital that a company has at time is , then the derivative, , is called the net investment ﬂow. Suppose that the net investment ﬂow is million dollars per year (where is measured in years). Find the increase in capi-tal (the capital formation) from the fourth year to the eighth year. 12. If revenue ﬂows into a company at a rate of , where is measured in years and is measured in dollars per year, ﬁnd the total revenue obtained in the ﬁrst four years. 13. Pareto’s Law of Income states that the number of people with incomes between and is , where and are constants with and . The average income of these people is Calculate . x x 1 N y b a Ax 1k dx k 1 A 0 k A N xb a Axk dx x b x a f t t f t 9000s1 2t t st f t f t t 10. p 800,000ex5000 x 20,000 p 20 1 10x p 50 1 20x p 200 0.2x 3 / 2 X 10 pSx 3 0.01x 2 y X 0 P pSx dx 554 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION t t 0 0 12 3.9 2 2.4 14 2.3 4 5.1 16 1.6 6 7.8 18 0.7 8 7.6 20 0 10 5.4 ct ct SECTION 8.5 PROBABILITY \|\|\|\| 555 PROBABILITY Calculus plays a role in the analysis of random behavior. Suppose we consider the choles-terol level of a person chosen at random from a certain age group, or the height of an adult female chosen at random, or the lifetime of a randomly chosen battery of a certain type. Such quantities are called continuous random variables because their values actually range over an interval of real numbers, although they might be measured or recorded only to the nearest integer. We might want to know the probability that a blood cholesterol level is greater than 250, or the probability that the height of an adult female is between 60 and 70 inches, or the probability that the battery we are buying lasts between 100 and 200 hours. If X represents the lifetime of that type of battery, we denote this last probability as follows: According to the frequency interpretation of probability, this number is the long-run pro-portion of all batteries of the speciﬁed type whose lifetimes are between 100 and 200 hours. Since it represents a proportion, the probability naturally falls between 0 and 1. Every continuous random variable X has a probability density function . This means that the probability that X lies between a and b is found by integrating from a to b: For example, Figure 1 shows the graph of a model for the probability density function for a random variable X deﬁned to be the height in inches of an adult female in the United States (according to data from the National Health Survey). The probability that the height of a woman chosen at random from this population is between 60 and 70 inches is equal to the area under the graph of from 60 to 70. In general, the probability density function of a random variable X satisﬁes the con-dition for all x. Because probabilities are measured on a scale from 0 to 1, it fol-lows that EXAMPLE 1 Let for and for all other values of . (a) Verify that is a probability density function. (b) Find . P4 X 8 f x fx 0 0 x 10 fx 0.006x10 x y fx dx 1 2 fx 0 f x y 0 65 60 70 y=ƒ area=probability that the height of a woman is between 60 and 70 inches Probability density function for the height of an adult female FIGURE 1 f f Pa X b y b a fx dx 1 f f P100 X 200 8.5 SOLUTION (a) For we have , so for all . We also need to check that Equation 2 is satisﬁed: Therefore is a probability density function. (b) The probability that lies between 4 and 8 is M EXAMPLE 2 Phenomena such as waiting times and equipment failure times are com-monly modeled by exponentially decreasing probability density functions. Find the exact form of such a function. SOLUTION Think of the random variable as being the time you wait on hold before an agent of a company you’re telephoning answers your call. So instead of x, let’s use t to represent time, in minutes. If is the probability density function and you call at time , then, from Deﬁnition 1, represents the probability that an agent answers within the ﬁrst two minutes and is the probability that your call is answered during the ﬁfth minute. It’s clear that for (the agent can’t answer before you place the call). For we are told to use an exponentially decreasing function, that is, a function of the form , where A and c are positive constants. Thus We use Equation 2 to determine the value of A: Therefore and so . Thus every exponential density function has the form A typical graph is shown in Figure 2. M ft 0 cect if t 0 if t 0 A c Ac 1 A c lim x l A c ect 0 x lim x l A c 1 ecx y 0 Aect dt lim x l y x 0 Aect dt 1 y ft dt y 0 ft dt y 0 ft dt ft 0 Aect if t 0 if t 0 ft Aect t 0 t 0 ft 0 x5 4 ft dt x2 0 ft dt t 0 f V 0.006[5x 2 1 3 x 3]4 8 0.544 P4 X 8 y 8 4 fx dx 0.006 y 8 4 10x x 2 dx X f 0.006[5x 2 1 3 x 3]0 10 0.006(500 1000 3 ) 1 y fx dx y 10 0 0.006x10 x dx 0.006 y 10 0 10x x 2 dx x fx 0 0.006x10 x 0 0 x 10 556 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION FIGURE 2 An exponential density function 0 f(t)= 0 ce_ct if t<0 if t˘0 t y c AVERAGE VALUES Suppose you’re waiting for a company to answer your phone call and you wonder how long, on average, you can expect to wait. Let be the corresponding density function, where t is measured in minutes, and think of a sample of N people who have called this company. Most likely, none of them had to wait more than an hour, so let’s restrict our attention to the interval . Let’s divide that interval into n intervals of length and endpoints . . . . (Think of as lasting a minute, or half a minute, or 10 sec-onds, or even a second.) The probability that somebody’s call gets answered during the time period from to is the area under the curve from to , which is approximately equal to . (This is the area of the approximating rectangle in Fig-ure 3, where is the midpoint of the interval.) Since the long-run proportion of calls that get answered in the time period from to is , we expect that, out of our sample of N callers, the number whose call was answered in that time period is approximately and the time that each waited is about . Therefore the total time they waited is the product of these numbers: approxi-mately . Adding over all such intervals, we get the approximate total of every-body’s waiting times: If we now divide by the number of callers N, we get the approximate average waiting time: We recognize this as a Riemann sum for the function . As the time interval shrinks (that is, and ), this Riemann sum approaches the integral This integral is called the mean waiting time. In general, the mean of any probability density function is deﬁned to be The mean can be interpreted as the long-run average value of the random variable X. It can also be interpreted as a measure of centrality of the probability density function. The expression for the mean resembles an integral we have seen before. If is the region that lies under the graph of , we know from Formula 8.3.8 that the x-coordinate of the centroid of is because of Equation 2. So a thin plate in the shape of balances at a point on the vertical line . (See Figure 4.) x x y x fx dx y fx dx y x fx dx f y x fx dx f y 60 0 t ft dt n l t l 0 t ft n i1 ti fti t n i1 N ti fti t tiN fti t ti N fti t fti t ti ti1 ti fti t ti ti1 y ft ti ti1 t t60 0, t1, t2, t 0 t 60 ft SECTION 8.5 PROBABILITY \|\|\|\| 557 0 t ti ti y=f(t) FIGURE 3 t y Ît ti-1 N It is traditional to denote the mean by the Greek letter (mu). FIGURE 4 T balances at a point on the line x=m 0 m y=ƒ x=m T t y EXAMPLE 3 Find the mean of the exponential distribution of Example 2: SOLUTION According to the deﬁnition of a mean, we have To evaluate this integral we use integration by parts, with and : The mean is , so we can rewrite the probability density function as M EXAMPLE 4 Suppose the average waiting time for a customer’s call to be answered by a company representative is ﬁve minutes. (a) Find the probability that a call is answered during the ﬁrst minute. (b) Find the probability that a customer waits more than ﬁve minutes to be answered. SOLUTION (a) We are given that the mean of the exponential distribution is min and so, from the result of Example 3, we know that the probability density function is Thus the probability that a call is answered during the ﬁrst minute is So about 18% of customers’ calls are answered during the ﬁrst minute. (b) The probability that a customer waits more than ﬁve minutes is About 37% of customers wait more than ﬁve minutes before their calls are answered. M 1 e 0.368 lim x l y x 5 0.2et5 dt lim x l e1 ex5 PT 5 y 5 ft dt y 5 0.2et5 dt 1 e15 0.1813 0.25et5]0 1 y 1 0 0.2et5 dt P0 T 1 y 1 0 ft dt ft 0 0.2et5 if t 0 if t 0 5 V ft 0 1et if t 0 if t 0 1c lim x l xecx 1 c ecx c 1 c y 0 tcect dt lim x l y x 0 tcect dt lim x l tect] x 0 y x 0 ect dt dv cect dt u t y t ft dt y 0 tcect dt ft 0 cect if t 0 if t 0 558 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION N The limit of the ﬁrst term is by l’Hospital’s Rule. 0 Notice the result of Example 4(b): Even though the mean waiting time is 5 minutes, only 37% of callers wait more than 5 minutes. The reason is that some callers have to wait much longer (maybe 10 or 15 minutes), and this brings up the average. Another measure of centrality of a probability density function is the median. That is a number m such that half the callers have a waiting time less than m and the other callers have a waiting time longer than m. In general, the median of a probability density func-tion is the number m such that This means that half the area under the graph of lies to the right of m. In Exercise 9 you are asked to show that the median waiting time for the company described in Example 4 is approximately 3.5 minutes. NORMAL DISTRIBUTIONS Many important random phenomena—such as test scores on aptitude tests, heights and weights of individuals from a homogeneous population, annual rainfall in a given loca-tion—are modeled by a normal distribution. This means that the probability density function of the random variable X is a member of the family of functions You can verify that the mean for this function is . The positive constant is called the standard deviation; it measures how spread out the values of X are. From the bell-shaped graphs of members of the family in Figure 5, we see that for small values of the values of X are clustered about the mean, whereas for larger values of the values of X are more spread out. Statisticians have methods for using sets of data to estimate and . The factor is needed to make a probability density function. In fact, it can be veriﬁed using the methods of multivariable calculus that EXAMPLE 5 Intelligence Quotient (IQ) scores are distributed normally with mean 100 and standard deviation 15. (Figure 6 shows the corresponding probability density function.) (a) What percentage of the population has an IQ score between 85 and 115? (b) What percentage of the population has an IQ above 140? V y 1 s2 ex22 2 dx 1 f 1( s2 ) FIGURE 5 Normal distributions x y 0 m 1 2 s=2 s=1 s= fx 1 s2 ex22 2 3 f y m fx dx 1 2 SECTION 8.5 PROBABILITY \|\|\|\| 559 N The standard deviation is denoted by the lowercase Greek letter (sigma). FIGURE 6 Distribution of IQ scores x y 0 60 0.01 80 100 120 140 0.02 SOLUTION (a) Since IQ scores are normally distributed, we use the probability density function given by Equation 3 with and : Recall from Section 7.5 that the function doesn’t have an elementary anti-derivative, so we can’t evaluate the integral exactly. But we can use the numerical integration capability of a calculator or computer (or the Midpoint Rule or Simpson’s Rule) to estimate the integral. Doing so, we ﬁnd that So about 68% of the population has an IQ between 85 and 115, that is, within one stan-dard deviation of the mean. (b) The probability that the IQ score of a person chosen at random is more than 140 is To avoid the improper integral we could approximate it by the integral from 140 to 200. (It’s quite safe to say that people with an IQ over 200 are extremely rare.) Then Therefore about 0.4% of the population has an IQ over 140. M PX 140 y 200 140 1 15s2 ex1002 450 dx 0.0038 PX 140 y 140 1 15s2 ex1002 450 dx P85 X 115 0.68 y ex2 P85 X 115 y 115 85 1 15s2 ex10022152 dx 15 100 560 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION 5. Let . (a) For what value of is a probability density function? (b) For that value of , ﬁnd . 6. Let if and if or . (a) For what value of is a probability density function? (b) For that value of , ﬁnd . (c) Find the mean. A spinner from a board game randomly indicates a real number between 0 and 10. The spinner is fair in the sense that it indi-cates a number in a given interval with the same probability as it indicates a number in any other interval of the same length. (a) Explain why the function is a probability density function for the spinner’s values. (b) What does your intuition tell you about the value of the mean? Check your guess by evaluating an integral. f x 0.1 0 if 0 x 10 if x 0 or x 10 7. P(X 1 2) k f k x 1 x 0 f x 0 0 x 1 f x kx 21 x P1 X 1 c f c f x c1 x 2 Let be the probability density function for the lifetime of a manufacturer’s highest quality car tire, where is measured in miles. Explain the meaning of each integral. (a) (b) 2. Let be the probability density function for the time it takes you to drive to school in the morning, where is measured in minutes. Express the following probabilities as integrals. (a) The probability that you drive to school in less than 15 minutes (b) The probability that it takes you more than half an hour to get to school 3. Let for and for all other values of . (a) Verify that is a probability density function. (b) Find . 4. Let if and if . (a) Verify that is a probability density function. (b) Find . P1 X 2 f x 0 f x 0 x 0 f x xex P(X 2 f x f x 0 0 x 4 f x 3 64 xs16 x 2 t f t y 25,000 f x dx y 40,000 30,000 f x dx x f x 1. EXERCISES 8.5 of 500 g. At what target weight should the manufacturer set its ﬁlling machine? 15. The speeds of vehicles on a highway with speed limit are normally distributed with mean and standard deviation . (a) What is the probability that a randomly chosen vehicle is traveling at a legal speed? (b) If police are instructed to ticket motorists driving or more, what percentage of motorists are targeted? 16. Show that the probability density function for a normally dis-tributed random variable has inﬂection points at . 17. For any normal distribution, ﬁnd the probability that the random variable lies within two standard deviations of the mean. 18. The standard deviation for a random variable with probability density function and mean is deﬁned by Find the standard deviation for an exponential density function with mean . 19. The hydrogen atom is composed of one proton in the nucleus and one electron, which moves about the nucleus. In the quan-tum theory of atomic structure, it is assumed that the electron does not move in a well-deﬁned orbit. Instead, it occupies a state known as an orbital, which may be thought of as a “cloud” of negative charge surrounding the nucleus. At the state of lowest energy, called the ground state, or 1s-orbital, the shape of this cloud is assumed to be a sphere centered at the nucleus. This sphere is described in terms of the probability density function where is the Bohr radius . The integral gives the probability that the electron will be found within the sphere of radius meters centered at the nucleus. (a) Verify that is a probability density function. (b) Find . For what value of does have its maximum value? ; (c) Graph the density function. (d) Find the probability that the electron will be within the sphere of radius centered at the nucleus. (e) Calculate the mean distance of the electron from the nucleus in the ground state of the hydrogen atom. 4a0 pr r limr l pr pr r Pr y r 0 4 a 3 0 s 2e2sa0 ds a0 5.59 1011 m a0 r 0 pr 4 a 3 0 r 2e2ra0 y x 2f x dx 12 f x 125 kmh 8 kmh 112 kmh 100 kmh (a) Explain why the function whose graph is shown is a proba-bility density function. (b) Use the graph to ﬁnd the following probabilities: (i) (ii) (c) Calculate the mean. 9. Show that the median waiting time for a phone call to the com-pany described in Example 4 is about 3.5 minutes. 10. (a) A type of lightbulb is labeled as having an average lifetime of 1000 hours. It’s reasonable to model the probability of failure of these bulbs by an exponential density function with mean . Use this model to ﬁnd the probability that a bulb (i) fails within the ﬁrst 200 hours, (ii) burns for more than 800 hours. (b) What is the median lifetime of these lightbulbs? 11. The manager of a fast-food restaurant determines that the average time that her customers wait for service is 2.5 min-utes. (a) Find the probability that a customer has to wait more than 4 minutes. (b) Find the probability that a customer is served within the ﬁrst 2 minutes. (c) The manager wants to advertise that anybody who isn’t served within a certain number of minutes gets a free ham-burger. But she doesn’t want to give away free hamburgers to more than 2% of her customers. What should the adver-tisement say? 12. According to the National Health Survey, the heights of adult males in the United States are normally distributed with mean 69.0 inches and standard deviation 2.8 inches. (a) What is the probability that an adult male chosen at random is between 65 inches and 73 inches tall? (b) What percentage of the adult male population is more than 6 feet tall? The “Garbage Project” at the University of Arizona reports that the amount of paper discarded by households per week is normally distributed with mean 9.4 lb and standard deviation 4.2 lb. What percentage of households throw out at least 10 lb of paper a week? 14. Boxes are labeled as containing 500 g of cereal. The machine ﬁlling the boxes produces weights that are normally distributed with standard deviation 12 g. (a) If the target weight is 500 g, what is the probability that the machine produces a box with less than 480 g of cereal? (b) Suppose a law states that no more than 5% of a manufac-turer’s cereal boxes can contain less than the stated weight 13. 1000 y=ƒ 4 6 8 10 x y 0 2 0.1 0.2 P3 X 8 PX 3 8. SECTION 8.5 PROBABILITY \|\|\|\| 561 1–2 Find the length of the curve. 1. , 2. , 3. (a) Find the length of the curve (b) Find the area of the surface obtained by rotating the curve in part (a) about the -axis. 4. (a) The curve , , is rotated about the -axis. Find the area of the resulting surface. (b) Find the area of the surface obtained by rotating the curve in part (a) about the -axis. 5. Use Simpson’s Rule with to estimate the length of the curve , . 6. Use Simpson’s Rule with to estimate the area of the surface obtained by rotating the curve in Exercise 5 about the -axis. x n 6 0 x 3 y ex2 n 6 x y 0 x 1 y x 2 y 1 x 2 y x 4 16 1 2x 2 3 x y 2 ln(sin 1 2x) 0 x 3 y 1 6x 2 432 7. Find the length of the curve 8. Find the area of the surface obtained by rotating the curve in Exercise 7 about the -axis. 9. A gate in an irrigation canal is constructed in the form of a trapezoid 3 ft wide at the bottom, 5 ft wide at the top, and 2 ft high. It is placed vertically in the canal so that the water just covers the gate. Find the hydrostatic force on one side of the gate. 10. A trough is ﬁlled with water and its vertical ends have the shape of the parabolic region in the ﬁgure. Find the hydrostatic force on one end of the trough. 4 ft 8 ft y 1 x 16 y y x 1 sst 1 dt EXERCISES 562 \|\|\|\| CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION REVIEW CONCEPT CHECK 8 6. Given a demand function , explain what is meant by the consumer surplus when the amount of a commodity currently available is and the current selling price is . Illustrate with a sketch. 7. (a) What is the cardiac output of the heart? (b) Explain how the cardiac output can be measured by the dye dilution method. 8. What is a probability density function? What properties does such a function have? 9. Suppose is the probability density function for the weight of a female college student, where is measured in pounds. (a) What is the meaning of the integral ? (b) Write an expression for the mean of this density function. (c) How can we ﬁnd the median of this density function? 10. What is a normal distribution? What is the signiﬁcance of the standard deviation? x130 0 f x dx x f x P X px 1. (a) How is the length of a curve deﬁned? (b) Write an expression for the length of a smooth curve given by , . (c) What if is given as a function of ? 2. (a) Write an expression for the surface area of the surface obtained by rotating the curve , , about the -axis. (b) What if is given as a function of ? (c) What if the curve is rotated about the -axis? 3. Describe how we can ﬁnd the hydrostatic force against a verti-cal wall submersed in a ﬂuid. 4. (a) What is the physical signiﬁcance of the center of mass of a thin plate? (b) If the plate lies between and , where , write expressions for the coordinates of the center of mass. 5. What does the Theorem of Pappus say? a x b y 0 y f x y y x x a x b y f x y x a x b y f x 19. (a) Explain why the function is a probability density function. (b) Find . (c) Calculate the mean. Is the value what you would expect? 20. Lengths of human pregnancies are normally distributed with mean 268 days and standard deviation 15 days. What per-centage of pregnancies last between 250 days and 280 days? 21. The length of time spent waiting in line at a certain bank is modeled by an exponential density function with mean 8 minutes. (a) What is the probability that a customer is served in the ﬁrst 3 minutes? (b) What is the probability that a customer has to wait more than 10 minutes? (c) What is the median waiting time? PX 4 f x 0 20 sin x 10 if if 0 x 10 x 0 or x 10 11–12 Find the centroid of the region bounded by the given curves. 11. , 12. , , , 13–14 Find the centroid of the region shown 13. 14. 15. Find the volume obtained when the circle of radius 1 with center is rotated about the -axis. 16. Use the Theorem of Pappus and the fact that the volume of a sphere of radius is to ﬁnd the centroid of the semi-circular region bounded by the curve and the -axis. 17. The demand function for a commodity is given by Find the consumer surplus when the sales level is 100. 18. After a 6-mg injection of dye into a heart, the readings of dye concentration at two-second intervals are as shown in the table. Use Simpson’s Rule to estimate the cardiac output. p 2000 0.1x 0.01x 2 x y sr 2 x 2 4 3r 3 r y 1, 0 x y 0 1 2 3 3 _2 (3, 2) x y 0 x 34 x 4 y 0 y sin x y sx y 1 2x CHAPTER 8 REVIEW \|\|\|\| 563 t t 0 0 14 4.7 2 1.9 16 3.3 4 3.3 18 2.1 6 5.1 20 1.1 8 7.6 22 0.5 10 7.1 24 0 12 5.8 ct ct 564 1. Find the area of the region . 2. Find the centroid of the region enclosed by the loop of the curve . 3. If a sphere of radius is sliced by a plane whose distance from the center of the sphere is , then the sphere is divided into two pieces called segments of one base. The corresponding surfaces are called spherical zones of one base. (a) Determine the surface areas of the two spherical zones indicated in the ﬁgure. (b) Determine the approximate area of the Arctic Ocean by assuming that it is approximately circular in shape, with center at the North Pole and “circumference” at north latitude. Use mi for the radius of the earth. (c) A sphere of radius is inscribed in a right circular cylinder of radius . Two planes perpen-dicular to the central axis of the cylinder and a distance apart cut off a spherical zone of two bases on the sphere. Show that the surface area of the spherical zone equals the surface area of the region that the two planes cut off on the cylinder. (d) The Torrid Zone is the region on the surface of the earth that is between the Tropic of Cancer ( north latitude) and the Tropic of Capricorn ( south latitude). What is the area of the Torrid Zone? 4. (a) Show that an observer at height above the north pole of a sphere of radius can see a part of the sphere that has area (b) Two spheres with radii and are placed so that the distance between their centers is , where . Where should a light be placed on the line joining the centers of the spheres in order to illuminate the largest total surface? 5. Suppose that the density of seawater, , varies with the depth below the surface. (a) Show that the hydrostatic pressure is governed by the differential equation where is the acceleration due to gravity. Let and be the pressure and density at . Express the pressure at depth as an integral. (b) Suppose the density of seawater at depth is given by , where is a positive constant. Find the total force, expressed as an integral, exerted on a vertical circular port-hole of radius whose center is located at a distance below the surface. L r r H 0e zH z z z 0 0 P0 t dP dz zt z z d r R d R r 2r 2H r H r H h d 23.45 23.45 h r r r 3960 75 d r y 2 x 3 x 4 S x, y x 0, y 1, x 2 y 2 4y P R O B L E M S P L U S 565 6. The ﬁgure shows a semicircle with radius 1, horizontal diameter , and tangent lines at and . At what height above the diameter should the horizontal line be placed so as to mini-mize the shaded area? 7. Let be a pyramid with a square base of side and suppose that is a sphere with its center on the base of and is tangent to all eight edges of . Find the height of . Then ﬁnd the volume of the intersection of and . 8. Consider a ﬂat metal plate to be placed vertically under water with its top 2 m below the surface of the water. Determine a shape for the plate so that if the plate is divided into any number of horizontal strips of equal height, the hydrostatic force on each strip is the same. 9. A uniform disk with radius 1 m is to be cut by a line so that the center of mass of the smaller piece lies halfway along a radius. How close to the center of the disk should the cut be made? (Express your answer correct to two decimal places.) 10. A triangle with area is cut from a corner of a square with side 10 cm, as shown in the ﬁgure. If the centroid of the remaining region is 4 cm from the right side of the square, how far is it from the bottom of the square? 11. In a famous 18th-century problem, known as Buffon’s needle problem, a needle of length is dropped onto a ﬂat surface (for example, a table) on which parallel lines units apart, , have been drawn. The problem is to determine the probability that the needle will come to rest intersecting one of the lines. Assume that the lines run east-west, parallel to the -axis in a rectangular coordinate system (as in the ﬁgure). Let be the distance from the “southern” end of the needle to the nearest line to the north. (If the needle’s southern end lies on a line, let . If the needle happens to lie east-west, let the “western” end be the “southern” end.) Let be the angle that the needle makes with a ray extending eastward from the “southern” end. Then and . Note that the needle intersects one of the lines only when . The total set of possibilities for the needle can be identiﬁed with the rectangular region , , and the proportion of times that the needle intersects a line is the ratio This ratio is the probability that the needle intersects a line. Find the probability that the needle will intersect a line if . What if ? 12. If the needle in Problem 11 has length , it’s possible for the needle to intersect more than one line. (a) If , ﬁnd the probability that a needle of length 7 will intersect at least one line. [Hint: Proceed as in Problem 11. Deﬁne as before; then the total set of possibilities for the needle can be identiﬁed with the same rectangular region , . What portion of the rectangle corresponds to the needle intersecting a line?] (b) If , ﬁnd the probability that a needle of length 7 will intersect two lines. (c) If , ﬁnd a general formula for the probability that the needle intersects three lines. 2L h 3L L 4 0 0 y L y L 4 h L h 1 2L h L area under y h sin area of rectangle 0 0 y L y h sin 0 0 y L y 0 y x L h L h 30 cm 2 P S P P S P S 2b P Q P PQ P R O B L E M S P L U S P Q FIGURE FOR PROBLEM 6 10 cm FIGURE FOR PROBLEM 10 y h sin ¨ ¨ h L FIGURE FOR PROBLEM 11 π 2 π y ¨ L h 566 Perhaps the most important of all the applications of calculus is to differential equa-tions. When physical scientists or social scientists use calculus, more often than not it is to analyze a differential equation that has arisen in the process of modeling some phenomenon that they are studying. Although it is often impossible to ﬁnd an explicit formula for the solution of a differential equation, we will see that graphical and numer-ical approaches provide the needed information. Direction ﬁelds enable us to sketch solutions of differential equations with-out an explicit formula. DIFFERENTIAL EQUATIONS 9 MODELING WITH DIFFERENTIAL EQUATIONS In describing the process of modeling in Section 1.2, we talked about formulating a math-ematical model of a real-world problem either through intuitive reasoning about the phe-nomenon or from a physical law based on evidence from experiments. The mathematical model often takes the form of a differential equation, that is, an equation that contains an unknown function and some of its derivatives. This is not surprising because in a real-world problem we often notice that changes occur and we want to predict future behavior on the basis of how current values change. Let’s begin by examining several examples of how differential equations arise when we model physical phenomena. MODELS OF POPULATION GROWTH One model for the growth of a population is based on the assumption that the population grows at a rate proportional to the size of the population. That is a reasonable assumption for a population of bacteria or animals under ideal conditions (unlimited environment, ade-quate nutrition, absence of predators, immunity from disease). Let’s identify and name the variables in this model: The rate of growth of the population is the derivative . So our assumption that the rate of growth of the population is proportional to the population size is written as the equation where is the proportionality constant. Equation 1 is our ﬁrst model for population growth; it is a differential equation because it contains an unknown function P and its derivative . Having formulated a model, let’s look at its consequences. If we rule out a population of 0, then for all t. So, if , then Equation 1 shows that for all t. This means that the population is always increasing. In fact, as increases, Equation 1 shows that becomes larger. In other words, the growth rate increases as the popula-tion increases. Equation 1 asks us to ﬁnd a function whose derivative is a constant multiple of itself. We know from Chapter 3 that exponential functions have that property. In fact, if we let , then Thus any exponential function of the form is a solution of Equation 1. In Sec-tion 9.4 we will see that there is no other solution. Allowing C to vary through all the real numbers, we get the family of solutions whose graphs are shown in Figure 1. But populations have only positive values and so we are interested only in the solutions with . And we are probably con-C 0 Pt Ce kt Pt Ce kt Pt Cke kt kCe kt kPt Pt Ce kt dPdt Pt Pt 0 k 0 Pt 0 dPdt k dP dt kP 1 dPdt P the number of individuals in the population the dependent variable t time the independent variable 9.1 567 N Now is a good time to read (or reread) the dis-cussion of mathematical modeling on page 24. t P FIGURE 1 The family of solutions of dP/dt=kP cerned only with values of t greater than the initial time . Figure 2 shows the physi-cally meaningful solutions. Putting , we get , so the constant C turns out to be the initial population, . Equation 1 is appropriate for modeling population growth under ideal conditions, but we have to recognize that a more realistic model must reﬂect the fact that a given environ-ment has limited resources. Many populations start by increasing in an exponential man-ner, but the population levels off when it approaches its carrying capacity K (or decreases toward K if it ever exceeds K). For a model to take into account both trends, we make two assumptions: N if P is small (Initially, the growth rate is proportional to P.) N if (P decreases if it ever exceeds K.) A simple expression that incorporates both assumptions is given by the equation Notice that if P is small compared with K, then is close to 0 and so . If , then is negative and so . Equation 2 is called the logistic differential equation and was proposed by the Dutch mathematical biologist Pierre-François Verhulst in the 1840s as a model for world popula-tion growth. We will develop techniques that enable us to ﬁnd explicit solutions of the logistic equation in Section 9.4, but for now we can deduce qualitative characteristics of the solutions directly from Equation 2. We ﬁrst observe that the constant functions and are solutions because, in either case, one of the factors on the right side of Equation 2 is zero. (This certainly makes physical sense: If the population is ever either 0 or at the carrying capacity, it stays that way.) These two constant solutions are called equilibrium solutions. If the initial population lies between 0 and K, then the right side of Equation 2 is positive, so and the population increases. But if the population exceeds the car-rying capacity , then is negative, so and the population decreases. Notice that, in either case, if the population approaches the carrying capacity , then , which means the population levels off. So we expect that the solutions of the logistic differential equation have graphs that look something like the ones in Figure 3. Notice that the graphs move away from the equilibrium solution and move toward the equilibrium solution . A MODEL FOR THE MOTION OF A SPRING Let’s now look at an example of a model from the physical sciences. We consider the motion of an object with mass m at the end of a vertical spring (as in Figure 4). In Sec-tion 6.4 we discussed Hooke’s Law, which says that if the spring is stretched (or com-pressed) x units from its natural length, then it exerts a force that is proportional to x: where k is a positive constant (called the spring constant). If we ignore any external resist-ing forces (due to air resistance or friction) then, by Newton’s Second Law (force equals restoring force kx P K P 0 dPdt l 0 P l K dPdt 0 1 PK P K dPdt 0 P0 Pt K Pt 0 dPdt 0 1 PK P K dPdt kP PK dP dt kP1 P K 2 P K dP dt 0 dP dt kP P0 P0 Ce k0 C t 0 t 0 568 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS 0 t P FIGURE 2 The family of solutions P(t)=Cekt with C>0 and t˘0 FIGURE 3 Solutions of the logistic equation t P 0 P=K P=0 equilibrium solutions mass times acceleration), we have This is an example of what is called a second-order differential equation because it involves second derivatives. Let’s see what we can guess about the form of the solution directly from the equation. We can rewrite Equation 3 in the form which says that the second derivative of x is proportional to x but has the opposite sign. We know two functions with this property, the sine and cosine functions. In fact, it turns out that all solutions of Equation 3 can be written as combinations of certain sine and cosine functions (see Exercise 4). This is not surprising; we expect the spring to oscil-late about its equilibrium position and so it is natural to think that trigonometric functions are involved. GENERAL DIFFERENTIAL EQUATIONS In general, a differential equation is an equation that contains an unknown function and one or more of its derivatives. The order of a differential equation is the order of the high-est derivative that occurs in the equation. Thus, Equations 1 and 2 are ﬁrst-order equations and Equation 3 is a second-order equation. In all three of those equations the independent variable is called t and represents time, but in general the independent variable doesn’t have to represent time. For example, when we consider the differential equation it is understood that y is an unknown function of x. A function is called a solution of a differential equation if the equation is satisﬁed when and its derivatives are substituted into the equation. Thus is a solution of Equation 4 if for all values of x in some interval. When we are asked to solve a differential equation we are expected to ﬁnd all possible solutions of the equation. We have already solved some particularly simple differential equations, namely, those of the form For instance, we know that the general solution of the differential equation is given by where C is an arbitrary constant. But, in general, solving a differential equation is not an easy matter. There is no sys-tematic technique that enables us to solve all differential equations. In Section 9.2, how-ever, we will see how to draw rough graphs of solutions even when we have no explicit formula. We will also learn how to ﬁnd numerical approximations to solutions. y x 4 4 C y x 3 y fx fx x fx f y fx f y xy 4 d 2x dt 2 k m x m d 2x dt 2 kx 3 SECTION 9.1 MODELING WITH DIFFERENTIAL EQUATIONS \|\|\|\| 569 FIGURE 4 m x 0 x m equilibrium position EXAMPLE 1 Show that every member of the family of functions is a solution of the differential equation . SOLUTION We use the Quotient Rule to differentiate the expression for y: The right side of the differential equation becomes Therefore, for every value of c, the given function is a solution of the differential equation. M When applying differential equations, we are usually not as interested in ﬁnding a fam-ily of solutions (the general solution) as we are in ﬁnding a solution that satisﬁes some additional requirement. In many physical problems we need to ﬁnd the particular solution that satisﬁes a condition of the form . This is called an initial condition, and the problem of ﬁnding a solution of the differential equation that satisﬁes the initial condition is called an initial-value problem. Geometrically, when we impose an initial condition, we look at the family of solution curves and pick the one that passes through the point . Physically, this corresponds to measuring the state of a system at time and using the solution of the initial-value prob-lem to predict the future behavior of the system. EXAMPLE 2 Find a solution of the differential equation that satisﬁes the initial condition . SOLUTION Substituting the values and into the formula from Example 1, we get Solving this equation for c, we get , which gives . So the solution of the initial-value problem is M y 1 1 3e t 1 1 3e t 3 e t 3 e t c 1 3 2 2c 1 c 2 1 ce 0 1 ce 0 1 c 1 c y 1 ce t 1 ce t y 2 t 0 y0 2 y 1 2y 2 1 V t0 t0, y0 yt0 y0 1 2 4cet 1 cet2 2cet 1 cet2 1 2y 2 1 1 2 1 ce t 1 ce t 2 1 1 2 1 ce t2 1 ce t2 1 ce t2 ce t c 2e 2t ce t c 2e 2t 1 ce t2 2ce t 1 ce t2 y 1 ce tce t 1 ce tce t 1 ce t2 y 1 2y 2 1 y 1 ce t 1 ce t V 570 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS N Figure 5 shows graphs of seven members of the family in Example 1. The differential equation shows that if , then . That is borne out by the ﬂatness of the graphs near and . y 1 y 1 y 0 y 1 5 _5 _5 5 FIGURE 5 SECTION 9.1 MODELING WITH DIFFERENTIAL EQUATIONS \|\|\|\| 571 A population is modeled by the differential equation (a) For what values of is the population increasing? (b) For what values of is the population decreasing? (c) What are the equilibrium solutions? 10. A function satisﬁes the differential equation (a) What are the constant solutions of the equation? (b) For what values of is increasing? (c) For what values of is decreasing? Explain why the functions with the given graphs can’t be solutions of the differential equation 12. The function with the given graph is a solution of one of the following differential equations. Decide which is the correct equation and justify your answer. A. B. C. Psychologists interested in learning theory study learning curves. A learning curve is the graph of a function , the performance of someone learning a skill as a function of the training time . The derivative represents the rate at which performance improves. (a) When do you think increases most rapidly? What hap-pens to as increases? Explain. (b) If is the maximum level of performance of which the learner is capable, explain why the differential equation is a reasonable model for learning. k a positive constant dP dt kM P M t dPdt P dPdt t Pt 13. y 1 2xy y 2xy y 1 xy 0 x y y t 1 1 y t 1 1 (a) (b) dy dt e ty 12 11. y y y y dy dt y 4 6y 3 5y 2 yt P P dP dt 1.2P1 P 4200 9. 1. Show that is a solution of the differential equa-tion . 2. Verify that is a solution of the initial-value problem on the interval . (a) For what values of does the function satisfy the differential equation ? (b) If and are the values of that you found in part (a), show that every member of the family of functions is also a solution. 4. (a) For what values of does the function satisfy the differential equation ? (b) For those values of , verify that every member of the family of functions is also a solution. 5. Which of the following functions are solutions of the differ-ential equation ? (a) (b) (c) (d) 6. (a) Show that every member of the family of functions is a solution of the differential equation . ; (b) Illustrate part (a) by graphing several members of the family of solutions on a common screen. (c) Find a solution of the differential equation that satisﬁes the initial condition . (d) Find a solution of the differential equation that satisﬁes the initial condition . (a) What can you say about a solution of the equation just by looking at the differential equation? (b) Verify that all members of the family are solutions of the equation in part (a). (c) Can you think of a solution of the differential equation that is not a member of the family in part (b)? (d) Find a solution of the initial-value problem 8. (a) What can you say about the graph of a solution of the equation when is close to 0? What if is large? (b) Verify that all members of the family are solutions of the differential equation . ; (c) Graph several members of the family of solutions on a common screen. Do the graphs conﬁrm what you pre-dicted in part (a)? (d) Find a solution of the initial-value problem y0 2 y xy 3 y xy 3 y c x 212 x x y xy 3 y0 0.5 y y 2 y y 2 y 1x C y y 2 7. y2 1 y1 2 x 2y xy 1 y ln x Cx y 1 2 x cos x y 1 2 x sin x y cos x y sin x y y sin x y A sin kt B cos kt k 4y 25y y cos kt k y ae r1x be r2x r r2 r1 2y y y 0 y e rx r 3. 2 x 2 y0 1 y tan xy cos2x y sin x cos x cos x xy y 2x y x x1 EXERCISES 9.1 an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton’s Law of Cooling for this particular situ-ation. What is the initial condition? In view of your answer to part (a), do you think this differential equation is an appropriate model for cooling? (c) Make a rough sketch of the graph of the solution of the initial-value problem in part (b). (c) Make a rough sketch of a possible solution of this differen-tial equation. 14. Suppose you have just poured a cup of freshly brewed coffee with temperature in a room where the temperature is . (a) When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain. (b) Newton’s Law of Cooling states that the rate of cooling of 20 C 95 C 572 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS DIRECTION FIELDS AND EULER’S METHOD Unfortunately, it’s impossible to solve most differential equations in the sense of obtain-ing an explicit formula for the solution. In this section we show that, despite the absence of an explicit solution, we can still learn a lot about the solution through a graphical approach (direction ﬁelds) or a numerical approach (Euler’s method). DIRECTION FIELDS Suppose we are asked to sketch the graph of the solution of the initial-value problem We don’t know a formula for the solution, so how can we possibly sketch its graph? Let’s think about what the differential equation means. The equation tells us that the slope at any point on the graph (called the solution curve) is equal to the sum of the x- and y-coordinates of the point (see Figure 1). In particular, because the curve passes through the point , its slope there must be . So a small portion of the solu-tion curve near the point looks like a short line segment through with slope 1. (See Figure 2.) As a guide to sketching the rest of the curve, let’s draw short line segments at a num-ber of points with slope . The result is called a direction ﬁeld and is shown in Figure 3. For instance, the line segment at the point has slope . The direc-tion ﬁeld allows us to visualize the general shape of the solution curves by indicating the direction in which the curves proceed at each point. 0 x 2 1 y FIGURE 3 Direction field for yª=x+y 0 x 2 1 y FIGURE 4 The solution curve through (0, 1) (0, 1) 1 2 3 1, 2 x y x, y 0, 1 0, 1 0 1 1 0, 1 x, y y x y y0 1 y x y 9.2 Slope at (¤, ﬁ) is ¤+ﬁ. Slope at (⁄, ›) is ⁄+›. 0 x y FIGURE 1 A solution of yª=x+y 0 x y (0, 1) Slope at (0, 1) is 0+1=1. FIGURE 2 Beginning of the solution curve through (0, 1) Now we can sketch the solution curve through the point by following the direc-tion ﬁeld as in Figure 4. Notice that we have drawn the curve so that it is parallel to near-by line segments. In general, suppose we have a ﬁrst-order differential equation of the form where is some expression in and . The differential equation says that the slope of a solution curve at a point on the curve is . If we draw short line segments with slope at several points , the result is called a direction ﬁeld (or slope ﬁeld). These line segments indicate the direction in which a solution curve is heading, so the direction ﬁeld helps us visualize the general shape of these curves. EXAMPLE 1 (a) Sketch the direction ﬁeld for the differential equation . (b) Use part (a) to sketch the solution curve that passes through the origin. SOLUTION (a) We start by computing the slope at several points in the following chart: Now we draw short line segments with these slopes at these points. The result is the direction ﬁeld shown in Figure 5. (b) We start at the origin and move to the right in the direction of the line segment (which has slope ). We continue to draw the solution curve so that it moves parallel to the nearby line segments. The resulting solution curve is shown in Figure 6. Returning to the origin, we draw the solution curve to the left as well. M The more line segments we draw in a direction ﬁeld, the clearer the picture becomes. Of course, it’s tedious to compute slopes and draw line segments for a huge number of points by hand, but computers are well suited for this task. Figure 7 shows a more detailed, computer-drawn direction ﬁeld for the differential equation in Example 1. It enables us to draw, with reasonable accuracy, the solution curves shown in Figure 8 with -intercepts , , , , and . FIGURE 7 3 _3 _3 3 FIGURE 8 3 _3 _3 3 2 1 0 1 2 y 1 y x 2 y 2 1 V x, y Fx, y Fx, y x, y y x Fx, y y Fx, y 0, 1 SECTION 9.2 DIRECTION FIELDS AND EULER’S METHOD \|\|\|\| 573 x 2 1 0 1 2 2 1 0 1 2 . . . y 0 0 0 0 0 1 1 1 1 1 . . . 3 0 1 0 3 4 1 0 1 4 . . . y x 2 y 2 1 0 x y 1 _1 _2 1 2 -1 _2 FIGURE 5 2 Module 9.2A shows direction ﬁelds and solution curves for a variety of differential equations. TEC 0 x y 1 2 _1 _2 1 2 -1 _2 FIGURE 6 Now let’s see how direction ﬁelds give insight into physical situations. The simple elec-tric circuit shown in Figure 9 contains an electromotive force (usually a battery or gener-ator) that produces a voltage of volts (V) and a current of amperes (A) at time t. The circuit also contains a resistor with a resistance of R ohms ( ) and an inductor with an inductance of L henries (H). Ohm’s Law gives the drop in voltage due to the resistor as RI. The voltage drop due to the inductor is . One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage . Thus we have which is a ﬁrst-order differential equation that models the current at time . EXAMPLE 2 Suppose that in the simple circuit of Figure 9 the resistance is , the inductance is 4 H, and a battery gives a constant voltage of 60 V. (a) Draw a direction ﬁeld for Equation 1 with these values. (b) What can you say about the limiting value of the current? (c) Identify any equilibrium solutions. (d) If the switch is closed when so the current starts with , use the direc-tion ﬁeld to sketch the solution curve. SOLUTION (a) If we put , , and in Equation 1, we get The direction ﬁeld for this differential equation is shown in Figure 10. (b) It appears from the direction ﬁeld that all solutions approach the value 5 A, that is, (c) It appears that the constant function is an equilibrium solution. Indeed, we can verify this directly from the differential equation . If , then the left side is and the right side is . 15 35 0 dIdt 0 It 5 dIdt 15 3I It 5 lim t l It 5 FIGURE 10 0 t 1 I 2 3 2 4 6 dI dt 15 3I or 4 dI dt 12I 60 Et 60 R 12 L 4 I0 0 t 0 12 V t I L dI dt RI Et 1 Et LdIdt It Et 574 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS R E switch L FIGURE 9 (d) We use the direction ﬁeld to sketch the solution curve that passes through , as shown in red in Figure 11. M Notice from Figure 10 that the line segments along any horizontal line are parallel. That is because the independent variable t does not occur on the right side of the equation . In general, a differential equation of the form in which the independent variable is missing from the right side, is called autonomous. For such an equation, the slopes corresponding to two different points with the same y-coordinate must be equal. This means that if we know one solution to an autonomous differential equation, then we can obtain inﬁnitely many others just by shifting the graph of the known solution to the right or left. In Figure 11 we have shown the solutions that result from shifting the solution curve of Example 2 one and two time units (namely, seconds) to the right. They correspond to closing the switch when or . EULER’S METHOD The basic idea behind direction ﬁelds can be used to ﬁnd numerical approximations to solutions of differential equations. We illustrate the method on the initial-value problem that we used to introduce direction ﬁelds: The differential equation tells us that , so the solution curve has slope 1 at the point . As a ﬁrst approximation to the solution we could use the linear approx-imation . In other words, we could use the tangent line at as a rough approximation to the solution curve (see Figure 12). Euler’s idea was to improve on this approximation by proceeding only a short distance along this tangent line and then making a midcourse correction by changing direction as indicated by the direction ﬁeld. Figure 13 shows what happens if we start out along the tangent line but stop when . (This horizontal distance traveled is called the step size.) Since , we have and we take as the starting point for a new line segment. The differential equation tells us that , so we use the linear function y 1.5 2x 0.5 2x 0.5 y0.5 0.5 1.5 2 0.5, 1.5 y0.5 1.5 L0.5 1.5 x 0.5 0, 1 Lx x 1 0, 1 y0 0 1 1 y0 1 y x y t 2 t 1 y fy I 15 3I FIGURE 11 0 t 1 I 2 3 2 4 6 0, 0 SECTION 9.2 DIRECTION FIELDS AND EULER’S METHOD \|\|\|\| 575 y x 0 1 1 y=L(x) solution curve FIGURE 12 First Euler approximation y x 0 1 1 0.5 1.5 FIGURE 13 Euler approximation with step size 0.5 as an approximation to the solution for (the orange segment in Figure 13). If we decrease the step size from to , we get the better Euler approximation shown in Figure 14. In general, Euler’s method says to start at the point given by the initial value and pro-ceed in the direction indicated by the direction ﬁeld. Stop after a short time, look at the slope at the new location, and proceed in that direction. Keep stopping and changing direc-tion according to the direction ﬁeld. Euler’s method does not produce the exact solution to an initial-value problem—it gives approximations. But by decreasing the step size (and therefore increasing the number of midcourse corrections), we obtain successively better approximations to the exact solution. (Compare Figures 12, 13, and 14.) For the general ﬁrst-order initial-value problem , , our aim is to ﬁnd approximate values for the solution at equally spaced numbers , , , . . . , where is the step size. The differential equation tells us that the slope at is , so Figure 15 shows that the approximate value of the solution when is Similarly, In general, EXAMPLE 3 Use Euler’s method with step size to construct a table of approximate values for the solution of the initial-value problem SOLUTION We are given that , , , and . So we have This means that if is the exact solution, then . Proceeding with similar calculations, we get the values in the table: M For a more accurate table of values in Example 3 we could decrease the step size. But for a large number of small steps the amount of computation is considerable and so we need to program a calculator or computer to carry out these calculations. The following table shows the results of applying Euler’s method with decreasing step size to the initial-value problem of Example 3. y0.3 1.362 yx y3 y2 hFx2, y2 1.22 0.10.2 1.22 1.362 y2 y1 hFx1, y1 1.1 0.10.1 1.1 1.22 y1 y0 hFx0, y0 1 0.10 1 1.1 Fx, y x y y0 1 x0 0 h 0.1 y0 1 y x y 0.1 yn yn1 hFxn1, yn1 y2 y1 hFx1, y1 y1 y0 hFx0, y0 x x1 y Fx0, y0 x0, y0 h x2 x1 h x1 x0 h x0 yx0 y0 y Fx, y 0.25 0.5 x 0.5 576 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS n n 1 0.1 1.100000 6 0.6 1.943122 2 0.2 1.220000 7 0.7 2.197434 3 0.3 1.362000 8 0.8 2.487178 4 0.4 1.528200 9 0.9 2.815895 5 0.5 1.721020 10 1.0 3.187485 yn xn yn xn y x ⁄ x¸ 0 y¸ h h F(x¸, y¸) (⁄, ›) slope=F(x¸, y¸) FIGURE 15 y x 0 1 1 0.25 FIGURE 14 Euler approximation with step size 0.25 Module 9.2B shows how Euler’s method works numerically and visually for a variety of differential equations and step sizes. TEC Notice that the Euler estimates in the table seem to be approaching limits, namely, the true values of and . Figure 16 shows graphs of the Euler approximations with step sizes 0.5, 0.25, 0.1, 0.05, 0.02, 0.01, and 0.005. They are approaching the exact solu-tion curve as the step size h approaches 0. EXAMPLE 4 In Example 2 we discussed a simple electric circuit with resistance , inductance 4 H, and a battery with voltage 60 V. If the switch is closed when , we modeled the current I at time t by the initial-value problem Estimate the current in the circuit half a second after the switch is closed. SOLUTION We use Euler’s method with , and step size second: So the current after 0.5 seconds is M I0.5 4.16 A I5 3.7995 0.115 3 3.7995 4.15965 I4 3.285 0.115 3 3.285 3.7995 I3 2.55 0.115 3 2.55 3.285 I2 1.5 0.115 3 1.5 2.55 I1 0 0.115 3 0 1.5 h 0.1 Ft, I 15 3I, t0 0, I0 0 I0 0 dI dt 15 3I t 0 12 V 0 x y 0.5 1 1 FIGURE 16 Euler approximations approaching the exact solution y1 y0.5 SECTION 9.2 DIRECTION FIELDS AND EULER’S METHOD \|\|\|\| 577 Step size Euler estimate of Euler estimate of 0.500 1.500000 2.500000 0.250 1.625000 2.882813 0.100 1.721020 3.187485 0.050 1.757789 3.306595 0.020 1.781212 3.383176 0.010 1.789264 3.409628 0.005 1.793337 3.423034 0.001 1.796619 3.433848 y1 y0.5 N Computer software packages that produce numerical approximations to solutions of differential equations use methods that are reﬁnements of Euler’s method. Although Euler’s method is simple and not as accurate, it is the basic idea on which the more accurate methods are based. 578 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS 5. 6. 7. Use the direction ﬁeld labeled II (above) to sketch the graphs of the solutions that satisfy the given initial conditions. (a) (b) (c) 8. Use the direction ﬁeld labeled IV (above) to sketch the graphs of the solutions that satisfy the given initial conditions. (a) (b) (c) 9–10 Sketch a direction ﬁeld for the differential equation. Then use it to sketch three solution curves. 9. 10. 11–14 Sketch the direction ﬁeld of the differential equation. Then use it to sketch a solution curve that passes through the given point. , 12. , , 14. , 15–16 Use a computer algebra system to draw a direction ﬁeld for the given differential equation. Get a printout and sketch on it the solution curve that passes through . Then use the CAS to draw the solution curve and compare it with your sketch. 15. 16. 17. Use a computer algebra system to draw a direction ﬁeld for the differential equation . Get a printout and y y 3 4y CAS y xy 2 4 y x 2 sin y 0, 1 CAS 1, 0 y x xy 0, 1 y y xy 13. 0, 0 y 1 xy 1, 0 y y 2x 11. y x 2 y 2 y 1 y y0 1 y0 0 y0 1 y0 1 y0 2 y0 1 y 0 x 4 2 _2 2 y 0 x 2 _2 2 _2 y 0 x 4 2 _2 2 y 0 x 2 _2 2 _2 I II III IV y sin x sin y y x y 1 1. A direction ﬁeld for the differential equation is shown. (a) Sketch the graphs of the solutions that satisfy the given initial conditions. (i) (ii) (iii) (iv) (b) Find all the equilibrium solutions. 2. A direction ﬁeld for the differential equation is shown. (a) Sketch the graphs of the solutions that satisfy the given initial conditions. (i) (ii) (iii) (iv) (v) (b) Find all the equilibrium solutions. 3–6 Match the differential equation with its direction ﬁeld (labeled I–IV). Give reasons for your answer. 4. y x2 y y 2 y 3. y 0 x 3 3 _3 5 4 1 2 _1 _2 1 2 y0 5 y0 4 y0 y0 2 y0 1 y x sin y y 0 x 3 3 _3 _3 1 2 _1 _2 1 2 _1 _2 y0 3 y0 3 y0 1 y0 1 y y(1 1 4 y 2) EXERCISES 9.2 22. Use Euler’s method with step size to estimate , where is the solution of the initial-value problem , . Use Euler’s method with step size to estimate , where is the solution of the initial-value problem , . 24. (a) Use Euler’s method with step size to estimate , where is the solution of the initial-value problem , . (b) Repeat part (a) with step size . ; 25. (a) Program a calculator or computer to use Euler’s method to compute , where is the solution of the initial-value problem (i) (ii) (iii) (iv) (b) Verify that is the exact solution of the differential equation. (c) Find the errors in using Euler’s method to compute with the step sizes in part (a). What happens to the error when the step size is divided by 10? 26. (a) Program your computer algebra system, using Euler’s method with step size 0.01, to calculate , where is the solution of the initial-value problem (b) Check your work by using the CAS to draw the solution curve. 27. The ﬁgure shows a circuit containing an electromotive force, a capacitor with a capacitance of farads (F), and a resistor with a resistance of ohms ( ). The voltage drop across the capacitor is , where is the charge (in coulombs), so in this case Kirchhoff’s Law gives But , so we have Suppose the resistance is , the capacitance is F, and a battery gives a constant voltage of 60 V. (a) Draw a direction ﬁeld for this differential equation. (b) What is the limiting value of the charge? C E R 0.05 5 R dQ dt 1 C Q Et I dQdt RI Q C Et Q QC R C y0 1 y x 3 y 3 y y2 CAS y1 y 2 ex3 h 0.001 h 0.01 h 0.1 h 1 y0 3 dy dx 3x 2y 6x 2 yx y1 0.1 y1 0 y x xy yx y1.4 0.2 y0 1 y y xy yx y0.5 0.1 23. y0 0 y 1 xy yx y1 0.2 sketch on it solutions that satisfy the initial condition for various values of . For what values of does exist? What are the possible values for this limit? Make a rough sketch of a direction ﬁeld for the autonomous differential equation , where the graph of is as shown. How does the limiting behavior of solutions depend on the value of ? (a) Use Euler’s method with each of the following step sizes to estimate the value of , where is the solution of the initial-value problem . (i) (ii) (iii) (b) We know that the exact solution of the initial-value problem in part (a) is . Draw, as accurately as you can, the graph of , together with the Euler approximations using the step sizes in part (a). (Your sketches should resemble Figures 12, 13, and 14.) Use your sketches to decide whether your estimates in part (a) are underestimates or overestimates. (c) The error in Euler’s method is the difference between the exact value and the approximate value. Find the errors made in part (a) in using Euler’s method to estimate the true value of , namely . What happens to the error each time the step size is halved? 20. A direction ﬁeld for a differential equation is shown. Draw, with a ruler, the graphs of the Euler approximations to the solution curve that passes through the origin. Use step sizes and . Will the Euler estimates be under-estimates or overestimates? Explain. Use Euler’s method with step size to compute the approx-imate -values of the solution of the initial-value problem , . y1 0 y y 2x y1, y2, y3, and y4 y 0.5 21. y 2 1 1 2 x 0 h 0.5 h 1 e 0.4 y0.4 y e x, 0 x 0.4 y e x h 0.1 h 0.2 h 0.4 y y, y0 1 y y0.4 19. 0 y 2 1 _1 _2 f(y) y0 f y f y 18. limt l yt c c y0 c SECTION 9.2 DIRECTION FILEDS AND EULER’S METHOD \|\|\|\| 579 at a rate of per minute when its temperature is . (a) What does the differential equation become in this case? (b) Sketch a direction ﬁeld and use it to sketch the solution curve for the initial-value problem. What is the limiting value of the temperature? (c) Use Euler’s method with step size minutes to estimate the temperature of the coffee after 10 minutes. h 2 70 C 1 C (c) Is there an equilibrium solution? (d) If the initial charge is , use the direction ﬁeld to sketch the solution curve. (e) If the initial charge is , use Euler’s method with step size 0.1 to estimate the charge after half a second. 28. In Exercise 14 in Section 9.1 we considered a cup of cof-fee in a room. Suppose it is known that the coffee cools 20 C 95 C Q0 0 C Q0 0 C 580 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS SEPARABLE EQUATIONS We have looked at ﬁrst-order differential equations from a geometric point of view (direc-tion ﬁelds) and from a numerical point of view (Euler’s method). What about the symbolic point of view? It would be nice to have an explicit formula for a solution of a differential equation. Unfortunately, that is not always possible. But in this section we examine a cer-tain type of differential equation that can be solved explicitly. A separable equation is a ﬁrst-order differential equation in which the expression for can be factored as a function of x times a function of y. In other words, it can be written in the form The name separable comes from the fact that the expression on the right side can be “sep-arated” into a function of and a function of . Equivalently, if , we could write where . To solve this equation we rewrite it in the differential form so that all ’s are on one side of the equation and all ’s are on the other side. Then we inte-grate both sides of the equation: Equation 2 deﬁnes implicitly as a function of . In some cases we may be able to solve for in terms of . We use the Chain Rule to justify this procedure: If and satisfy (2), then so and Thus Equation 1 is satisﬁed. hy dy dx tx d dy y hy dy dy dx tx d dx y hy dy d dx y tx dx t h x y x y y hy dy y tx dx 2 x y hy dy tx dx hy 1fy dy dx tx hy 1 fy 0 y x dy dx txfy dydx 9.3 N The technique for solving separable differen-tial equations was ﬁrst used by James Bernoulli (in 1690) in solving a problem about pendulums and by Leibniz (in a letter to Huygens in 1691). John Bernoulli explained the general method in a paper published in 1694. EXAMPLE 1 (a) Solve the differential equation . (b) Find the solution of this equation that satisﬁes the initial condition . SOLUTION (a) We write the equation in terms of differentials and integrate both sides: where is an arbitrary constant. (We could have used a constant on the left side and another constant on the right side. But then we could combine these constants by writing .) Solving for , we get We could leave the solution like this or we could write it in the form where . (Since is an arbitrary constant, so is .) (b) If we put in the general solution in part (a), we get . To satisfy the initial condition , we must have and so . Thus the solution of the initial-value problem is M EXAMPLE 2 Solve the differential equation . SOLUTION Writing the equation in differential form and integrating both sides, we have where is a constant. Equation 3 gives the general solution implicitly. In this case it’s impossible to solve the equation to express explicitly as a function of . M EXAMPLE 3 Solve the equation . SOLUTION First we rewrite the equation using Leibniz notation: dy dx x 2y y x 2y x y C y 2 sin y 2x 3 C 3 y 2y cos ydy y 6x 2 dx 2y cos ydy 6x 2 dx dy dx 6x 2 2y cos y V y s 3 x 3 8 K 8 s 3 K 2 y0 2 y0 s 3 K x 0 K C K 3C y s 3 x 3 K y s 3 x 3 3C y C C2 C1 C2 C1 C 1 3 y 3 1 3 x 3 C y y 2dy y x 2dx y 2dy x 2dx y0 2 dy dx x 2 y 2 SECTION 9.3 SEPARABLE EQUATIONS \|\|\|\| 581 N Figure 1 shows graphs of several members of the family of solutions of the differential equation in Example 1. The solution of the initial-value problem in part (b) is shown in red. 3 _3 _3 3 FIGURE 1 N Some computer algebra systems can plot curves deﬁned by implicit equations. Figure 2 shows the graphs of several members of the family of solutions of the differential equation in Example 2. As we look at the curves from left to right, the values of are , , , , , , and . 3 2 1 0 1 2 3 C 4 _4 _2 2 FIGURE 2 If , we can rewrite it in differential notation and integrate: This equation deﬁnes implicitly as a function of . But in this case we can solve explicitly for as follows: so We can easily verify that the function is also a solution of the given differential equation. So we can write the general solution in the form where is an arbitrary constant ( , or , or ). M EXAMPLE 4 In Section 9.2 we modeled the current in the electric circuit shown in Figure 5 by the differential equation Find an expression for the current in a circuit where the resistance is , the induc-tance is 4 H, a battery gives a constant voltage of 60 V, and the switch is turned on when . What is the limiting value of the current? SOLUTION With L 4, R 12, and , the equation becomes or dI dt 15 3I 4 dI dt 12I 60 Et 60 t 0 12 L dI dt RI Et It V 6 _6 _2 2 FIGURE 4 FIGURE 3 2 _4 0 x y 1 2 _1 _2 4 6 _2 _6 A 0 A e C A e C A y Ae x 33 y 0 y e Ce x33 y e ln y e x33C e Ce x33 y x y ln y x 3 3 C y dy y y x 2 dx dy y x 2 dx y 0 y 0 582 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS N If a solution is a function that satisﬁes for some , it follows from a uniqueness theorem for solutions of differential equations that for all . x yx 0 x yx 0 y N Figure 3 shows a direction ﬁeld for the differ-ential equation in Example 3. Compare it with Figure 4, in which we use the equation to graph solutions for several values of . If you use the direction ﬁeld to sketch solution curves with -intercepts , , , , and , they will resemble the curves in Figure 4. 2 1 1 2 5 y A y Ae x 33 R E switch L FIGURE 5 and the initial-value problem is We recognize this equation as being separable, and we solve it as follows: Since , we have , so A 15 and the solution is The limiting current, in amperes, is M ORTHOGONAL TRAJECTORIES An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family orthogonally, that is, at right angles (see Figure 7). For instance, each member of the family of straight lines through the origin is an orthogonal trajectory of the family of concentric circles with center the origin (see Figure 8). We say that the two families are orthogonal trajectories of each other. EXAMPLE 5 Find the orthogonal trajectories of the family of curves , where is an arbitrary constant. SOLUTION The curves form a family of parabolas whose axis of symmetry is the -axis. The ﬁrst step is to ﬁnd a single differential equation that is satisﬁed by all x x ky 2 k x ky 2 V x y FIGURE 8 orthogonal trajectory FIGURE 7 x 2 y 2 r 2 y mx 5 0 5 lim t l It lim t l 5 5e3t 5 5 lim t l e3t It 5 5e3t 5 1 3A 0 I0 0 I 5 1 3Ae3t 15 3I e3Ce3t Ae3t 15 3I e3tC 1 3 ln 15 3I t C 15 3I 0 y dI 15 3I y dt I0 0 dI dt 15 3I SECTION 9.3 SEPARABLE EQUATIONS \|\|\|\| 583 N Figure 6 shows how the solution in Example 4 (the current) approaches its limiting value. Com-parison with Figure 11 in Section 9.2 shows that we were able to draw a fairly accurate solution curve from the direction ﬁeld. 6 0 2.5 y=5 FIGURE 6 members of the family. If we differentiate , we get This differential equation depends on , but we need an equation that is valid for all values of simultaneously. To eliminate we note that, from the equation of the given general parabola , we have and so the differential equation can be written as or This means that the slope of the tangent line at any point on one of the parabolas is . On an orthogonal trajectory the slope of the tangent line must be the nega-tive reciprocal of this slope. Therefore the orthogonal trajectories must satisfy the differ-ential equation This differential equation is separable, and we solve it as follows: where is an arbitrary positive constant. Thus the orthogonal trajectories are the family of ellipses given by Equation 4 and sketched in Figure 9. M Orthogonal trajectories occur in various branches of physics. For example, in an elec-trostatic ﬁeld the lines of force are orthogonal to the lines of constant potential. Also, the streamlines in aerodynamics are orthogonal trajectories of the velocity-equipotential curves. MIXING PROBLEMS A typical mixing problem involves a tank of ﬁxed capacity ﬁlled with a thoroughly mixed solution of some substance, such as salt. A solution of a given concentration enters the tank at a ﬁxed rate and the mixture, thoroughly stirred, leaves at a ﬁxed rate, which may differ from the entering rate. If denotes the amount of substance in the tank at time t, then is the rate at which the substance is being added minus the rate at which it is being removed. The mathematical description of this situation often leads to a ﬁrst-order sepa-rable differential equation. We can use the same type of reasoning to model a variety of phenomena: chemical reactions, discharge of pollutants into a lake, injection of a drug into the bloodstream. yt yt C x 2 y 2 2 C 4 y 2 2 x 2 C y y dy y 2x dx dy dx 2x y y y2x x, y dy dx y 2x dy dx 1 2ky 1 2 x y 2 y k xy 2 x ky 2 k k k dy dx 1 2ky or 1 2ky dy dx x ky 2 584 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS x y FIGURE 9 EXAMPLE 6 A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that con-tains 0.03 kg of salt per liter of water enters the tank at a rate of 25 Lmin. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt remains in the tank after half an hour? SOLUTION Let be the amount of salt (in kilograms) after minutes. We are given that and we want to ﬁnd . We do this by ﬁnding a differential equation satis-ﬁed by . Note that is the rate of change of the amount of salt, so where (rate in) is the rate at which salt enters the tank and (rate out) is the rate at which salt leaves the tank. We have The tank always contains 5000 L of liquid, so the concentration at time is (measured in kilograms per liter). Since the brine ﬂows out at a rate of 25 Lmin, we have Thus, from Equation 5, we get Solving this separable differential equation, we obtain Since , we have , so Therefore Since is continuous and and the right side is never 0, we deduce that is always positive. Thus and so The amount of salt after 30 min is M y30 150 130e30200 38.1 kg yt 150 130et200 150 y 150 y 150 yt y0 20 yt 150 y 130et200 ln 150 y t 200 ln 130 ln 130 C y0 20 ln 150 y t 200 C y dy 150 y y dt 200 dy dt 0.75 yt 200 150 yt 200 rate out yt 5000 kg L25 L min yt 200 kg min yt5000 t rate in 0.03 kg L25 L min 0.75 kg min dy dt rate in rate out 5 dydt yt y30 y0 20 t yt SECTION 9.3 SEPARABLE EQUATIONS \|\|\|\| 585 N Figure 10 shows the graph of the function of Example 6. Notice that, as time goes by, the amount of salt approaches 150 kg. yt t y 0 200 400 50 100 150 FIGURE 10 586 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS ; 24. Solve the equation and graph several members of the family of solutions. How does the solution curve change as the constant varies? Solve the initial-value problem , , and graph the solution (if your CAS does implicit plots). 26. Solve the equation and graph several members of the family of solutions (if your CAS does implicit plots). How does the solution curve change as the constant varies? 27–28 (a) Use a computer algebra system to draw a direction ﬁeld for the differential equation. Get a printout and use it to sketch some solution curves without solving the differential equation. (b) Solve the differential equation. (c) Use the CAS to draw several members of the family of solu-tions obtained in part (b). Compare with the curves from part (a). 27. 28. ; 29–32 Find the orthogonal trajectories of the family of curves. Use a graphing device to draw several members of each family on a common screen. 29. 30. 32. 33. Solve the initial-value problem in Exercise 27 in Section 9.2 to ﬁnd an expression for the charge at time . Find the limit-ing value of the charge. 34. In Exercise 28 in Section 9.2 we discussed a differential equation that models the temperature of a cup of coffee in a room. Solve the differential equation to ﬁnd an expression for the temperature of the coffee at time . In Exercise 13 in Section 9.1 we formulated a model for learning in the form of the differential equation where measures the performance of someone learning a skill after a training time , is the maximum level of per-formance, and is a positive constant. Solve this differential equation to ﬁnd an expression for . What is the limit of this expression? Pt k M t Pt dP dt kM P 35. t 20C 95C t y x 1 kx y k x 31. y 2 kx 3 x 2 2y 2 k 2 y x 2y y 1y CAS C y xsx 2 1ye y CAS y0 2 y sin xsin y 25. CAS C eyy cos x 0 1–10 Solve the differential equation. 1. 2. 3. 4. 5. 6. 7. 8. 9. 11–18 Find the solution of the differential equation that satisﬁes the given initial condition. 11. , 12. , 13. , 14. , , 16. , 17. , , 18. , 19. Find an equation of the curve that passes through the point and whose slope at is . 20. Find the function such that and . 21. Solve the differential equation by making the change of variable . 22. Solve the differential equation by making the change of variable . 23. (a) Solve the differential equation . ; (b) Solve the initial-value problem , , and graph the solution. (c) Does the initial-value problem , , have a solution? Explain. y0 2 y 2xs1 y 2 y0 0 y 2xs1 y 2 y 2xs1 y 2 v yx xy y xe yx u x y y x y f 0 1 2 f x f x1 f x f xy x, y 0, 1 L1 1 dL dt kL2 ln t 0 x 2 y3 a ytan x a y y1 1 xy y y 2 u0 5 du dt 2t sec2t 2u 15. P1 2 dP dt sPt y0 0 x cos x 2y e 3yy y0 1 dy dx y cos x 1 y 2 y0 3 dy dx x y dz dt e tz 0 10. du dt 2 2u t tu dy d e y sin2 y sec dy dt te t ys1 y 2 du dr 1 sr 1 su 1 tan yy x 2 1 y y 2 sin x x 2 1y xy dy dx sx e y dy dx y x EXERCISES 9.3 (a) Suppose that the concentration at time is . Deter-mine the concentration at any time by solving the differ-ential equation. (b) Assuming that , ﬁnd and interpret your answer. 40. A certain small country has $10 billion in paper currency in circulation, and each day $50 million comes into the country’s banks. The government decides to introduce new currency by having the banks replace old bills with new ones whenever old currency comes into the banks. Let denote the amount of new currency in circulation at time , with . (a) Formulate a mathematical model in the form of an initial-value problem that represents the “ﬂow” of the new currency into circulation. (b) Solve the initial-value problem found in part (a). (c) How long will it take for the new bills to account for of the currency in circulation? 41. A tank contains 1000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 Lmin. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank (a) after minutes and (b) after 20 minutes? 42. The air in a room with volume contains carbon dioxide initially. Fresher air with only 0.05% carbon dioxide ﬂows into the room at a rate of and the mixed air ﬂows out at the same rate. Find the percentage of carbon dioxide in the room as a function of time. What happens in the long run? 43. A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour? 44. A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 Lmin. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 Lmin. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 Lmin. How much salt is in the tank (a) after minutes and (b) after one hour? When a raindrop falls, it increases in size and so its mass at time is a function of , . The rate of growth of the mass is for some positive constant . When we apply New-ton’s Law of Motion to the raindrop, we get , where is the velocity of the raindrop (directed downward) and is the acceleration due to gravity. The terminal velocity of the raindrop is . Find an expression for the ter-minal velocity in terms of and . 46. An object of mass is moving horizontally through a medium which resists the motion with a force that is a func-tion of the velocity; that is, m d 2s dt 2 m dv dt f v m k t lim t l vt t v mv tm k kmt mt t t 45. t 5 galmin 2 m3min 0.15% 180 m3 t 90% x0 0 t x xt lim t l Ct C0 rk t C0 t 0 36. In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: . The law of mass action states that the rate of reaction is proportional to the product of the concen-trations of A and B: (See Example 4 in Section 3.7.) Thus, if the initial concentra-tions are A molesL and B molesL and we write C , then we have (a) Assuming that , ﬁnd as a function of . Use the fact that the initial concentration of C is 0. (b) Find assuming that . How does this expres-sion for simplify if it is known that after 20 seconds? 37. In contrast to the situation of Exercise 36, experiments show that the reaction satisﬁes the rate law and so for this reaction the differential equation becomes where and and are the initial concentrations of hydrogen and bromine. (a) Find as a function of in the case where . Use the fact that . (b) If , ﬁnd as a function of . Hint: In performing the integration, make the substitution 38. A sphere with radius 1 m has temperature . It lies inside a concentric sphere with radius 2 m and temperature . The temperature at a distance from the common center of the spheres satisﬁes the differential equation If we let , then satisﬁes a ﬁrst-order differential equation. Solve it to ﬁnd an expression for the temperature between the spheres. A glucose solution is administered intravenously into the bloodstream at a constant rate . As the glucose is added, it is converted into other substances and removed from the blood-stream at a rate that is proportional to the concentration at that time. Thus a model for the concentration of the glucose solution in the bloodstream is where is a positive constant. k dC dt r kC C Ct r 39. Tr S S dTdr d 2T dr 2 2 r dT dr 0 r Tr 25 C 15 C u sb x .] [ x t a b x0 0 a b t x b a x HBr dx dt ka xb x12 dHBr dt k H 2Br212 H2 Br2 l 2HBr C 1 2a xt a b xt t x a b CAS dx dt ka xb x x b a dC dt k AB A B l C SECTION 9.3 SEPARABLE EQUATIONS \|\|\|\| 587 If water (or other liquid) drains from a tank, we expect that the ﬂow will be greatest at ﬁrst (when the water depth is greatest) and will gradually decrease as the water level decreases. But we need a more precise mathematical description of how the ﬂow decreases in order to answer the kinds of questions that engineers ask: How long does it take for a tank to drain completely? How much water should a tank hold in order to guarantee a certain minimum water pressure for a sprinkler system? Let and be the height and volume of water in a tank at time . If water drains through a hole with area at the bottom of the tank, then Torricelli’s Law says that where is the acceleration due to gravity. So the rate at which water ﬂows from the tank is propor-tional to the square root of the water height. 1. (a) Suppose the tank is cylindrical with height 6 ft and radius 2 ft and the hole is circular with radius 1 inch. If we take fts , show that satisﬁes the differential equation (b) Solve this equation to ﬁnd the height of the water at time , assuming the tank is full at time . (c) How long will it take for the water to drain completely? t 0 t dh dt 1 72 sh h 2 t 32 t dV dt as2th 1 a t Vt ht HOW FAST DOES A TANK DRAIN? A P P L I E D P R O J E C T 48. According to Newton’s Law of Universal Gravitation, the gravitational force on an object of mass that has been pro-jected vertically upward from the earth’s surface is where is the object’s distance above the surface at time , is the earth’s radius, and is the acceleration due to gravity. Also, by Newton’s Second Law, and so (a) Suppose a rocket is ﬁred vertically upward with an initial velocity . Let be the maximum height above the sur-face reached by the object. Show that [Hint: By the Chain Rule, .] (b) Calculate . This limit is called the escape velocity for the earth. (c) Use mi and fts to calculate in feet per second and in miles per second. ve 2 t 32 R 3960 ve lim h l v0 m dvdt mv dvdx v0 2tRh R h h v0 m dv dt mtR 2 x R2 F ma mdvdt t R t x xt F mtR 2 x R2 m where and represent the velocity and position of the object at time , respectively. For example, think of a boat moving through the water. (a) Suppose that the resisting force is proportional to the velocity, that is, , a positive constant. (This model is appropriate for small values of .) Let and be the initial values of and . Determine and at any time . What is the total distance that the object travels from time ? (b) For larger values of a better model is obtained by sup-posing that the resisting force is proportional to the square of the velocity, that is, , . (This model was ﬁrst proposed by Newton.) Let and be the initial values of and . Determine and at any time . What is the total distance that the object travels in this case? 47. Let be the area of a tissue culture at time and let be the ﬁnal area of the tissue when growth is complete. Most cell divisions occur on the periphery of the tissue and the number of cells on the periphery is proportional to . So a reasonable model for the growth of tissue is obtained by assuming that the rate of growth of the area is jointly propor-tional to and . (a) Formulate a differential equation and use it to show that the tissue grows fastest when . (b) Solve the differential equation to ﬁnd an expression for . Use a computer algebra system to perform the integration. At CAS At 1 3M M At sAt sAt M t At t s v s v s0 v0 k 0 f v kv2 v t 0 t s v s v s0 s0 v0 v0 v k f v kv t s st v vt 588 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS 2. Because of the rotation and viscosity of the liquid, the theoretical model given by Equation 1 isn’t quite accurate. Instead, the model is often used and the constant (which depends on the physical properties of the liquid) is determined from data concerning the draining of the tank. (a) Suppose that a hole is drilled in the side of a cylindrical bottle and the height of the water (above the hole) decreases from 10 cm to 3 cm in 68 seconds. Use Equation 2 to ﬁnd an expression for . Evaluate for . (b) Drill a 4-mm hole near the bottom of the cylindrical part of a two-liter plastic soft-drink bottle. Attach a strip of masking tape marked in centimeters from 0 to 10, with 0 corre-sponding to the top of the hole. With one ﬁnger over the hole, ﬁll the bottle with water to the 10-cm mark. Then take your ﬁnger off the hole and record the values of for seconds. (You will probably ﬁnd that it takes 68 seconds for the level to decrease to .) Compare your data with the values of from part (a). How well did the model predict the actual values? 3. In many parts of the world, the water for sprinkler systems in large hotels and hospitals is supplied by gravity from cylindrical tanks on or near the roofs of the buildings. Suppose such a tank has radius 10 ft and the diameter of the outlet is 2.5 inches. An engineer has to guarantee that the water pressure will be at least 2160 for a period of 10 minutes. (When a ﬁre happens, the electrical system might fail and it could take up to 10 minutes for the emer-gency generator and ﬁre pump to be activated.) What height should the engineer specify for the tank in order to make such a guarantee? (Use the fact that the water pressure at a depth of feet is . See Section 8.3.) 4. Not all water tanks are shaped like cylinders. Suppose a tank has cross-sectional area at height . Then the volume of water up to height is and so the Fundamental Theorem of Calculus gives . It follows that and so Torricelli’s Law becomes (a) Suppose the tank has the shape of a sphere with radius 2 m and is initially half full of water. If the radius of the circular hole is 1 cm and we take ms , show that satisﬁes the differential equation (b) How long will it take for the water to drain completely? 4h h2 dh dt 0.0001s20h h 2 t 10 Ah dh dt as2th dV dt dV dh dh dt Ah dh dt dVdh Ah V xh 0 Au du h h Ah P 62.5d d lbft 2 ht h 3 cm t 10, 20, 30, 40, 50, 60 ht t 10, 20, 30, 40, 50, 60 ht ht h k dh dt ksh 2 APPLIED PROJECT HOW FAST DOES A TANK DRAIN? \|\|\|\| 589 N This part of the project is best done as a classroom demonstration or as a group project with three students in each group: a time-keeper to call out seconds, a bottle keeper to estimate the height every 10 seconds, and a record keeper to record these values. Suppose you throw a ball into the air. Do you think it takes longer to reach its maximum height or to fall back to earth from its maximum height? We will solve the problem in this proj-ect but, before getting started, think about that situation and make a guess based on your physical intuition. 1. A ball with mass is projected vertically upward from the earth’s surface with a positive initial velocity . We assume the forces acting on the ball are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude , where is a positive constant and is the velocity of the ball at time . In both the ascent and the descent, the total force acting on the ball is . [During ascent, is positive and the resistance acts downward; during descent, is negative and the resistance acts upward.] So, by Newton’s Second Law, the equation of motion is Solve this differential equation to show that the velocity is 2. Show that the height of the ball, until it hits the ground, is 3. Let be the time that the ball takes to reach its maximum height. Show that Find this time for a ball with mass 1 kg and initial velocity 20 ms. Assume the air resistance is of the speed. ; 4. Let be the time at which the ball falls back to earth. For the particular ball in Problem 3, estimate by using a graph of the height function . Which is faster, going up or coming down? 5. In general, it’s not easy to ﬁnd because it’s impossible to solve the equation explicitly. We can, however, use an indirect method to determine whether ascent or descent is faster; we determine whether is positive or negative. Show that where . Then show that and the function is increasing for . Use this result to decide whether is positive or negative. What can you conclude? Is ascent or descent faster? y2t1 x 1 f x x 1 x 2 ln x x 1 x e pt1m y2t1 m 2t p 2 x 1 x 2 ln x y2t1 yt 0 t2 yt t2 t2 1 10 t1 m p ln mt pv0 mt t1 yt v0 mt p m p 1 eptm mtt p vt v0 mt p eptm mt p mv pv mt vt vt pv mt t vt p p vt v0 m WHICH IS FASTER, GOING UP OR COMING DOWN? A P P L I E D P R O J E C T 590 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS N In modeling force due to air resistance, various functions have been used, depending on the physical characteristics and speed of the ball. Here we use a linear model, , but a quadratic model ( on the way up and on the way down) is another possibility for higher speeds (see Exercise 46 in Section 9.3). For a golf ball, experiments have shown that a good model is going up and coming down. But no matter which force func-tion is used [where for and for ], the answer to the question remains the same. See F. Brauer, “What Goes Up Must Come Down, Eventually,” Amer. Math. Monthly 108 (2001), pp. 437–440. v 0 f v 0 v 0 f v 0 f v p v 1.3 pv 1.3 pv2 pv 2 pv MODELS FOR POPULATION GROWTH In this section we investigate differential equations that are used to model population growth: the law of natural growth, the logistic equation, and several others. THE LAW OF NATURAL GROWTH One of the models for population growth that we considered in Section 9.1 was based on the assumption that the population grows at a rate proportional to the size of the population: Is that a reasonable assumption? Suppose we have a population (of bacteria, for instance) with size and at a certain time it is growing at a rate of bacteria per hour. Now let’s take another 1000 bacteria of the same type and put them with the ﬁrst pop-ulation. Each half of the new population was growing at a rate of 300 bacteria per hour. We would expect the total population of 2000 to increase at a rate of 600 bacteria per hour initially (provided there’s enough room and nutrition). So if we double the size, we double the growth rate. In general, it seems reasonable that the growth rate should be pro-portional to the size. In general, if is the value of a quantity at time and if the rate of change of with respect to is proportional to its size at any time, then where is a constant. Equation 1 is sometimes called the law of natural growth. If is pos-itive, then the population increases; if is negative, it decreases. Because Equation 1 is a separable differential equation, we can solve it by the methods of Section 9.3: where A ( or 0) is an arbitrary constant. To see the signiﬁcance of the constant A, we observe that Therefore A is the initial value of the function. The solution of the initial-value problem is Pt P0e kt P0 P0 dP dt kP 2 P0 Ae k 0 A e C P Ae kt P e ktC e Ce kt ln P kt C y dP P y k dt k k k dP dt kP 1 Pt t P t y Pt P 300 P 1000 dP dt kP 9.4 SECTION 9.4 MODELS FOR POPULATION GROWTH \|\|\|\| 591 N Examples and exercises on the use of (2) are given in Section 3.8. Another way of writing Equation 1 is which says that the relative growth rate (the growth rate divided by the population size) is constant. Then (2) says that a population with constant relative growth rate must grow exponentially. We can account for emigration (or “harvesting”) from a population by modifying Equa-tion 1: If the rate of emigration is a constant , then the rate of change of the population is modeled by the differential equation See Exercise 13 for the solution and consequences of Equation 3. THE LOGISTIC MODEL As we discussed in Section 9.1, a population often increases exponentially in its early stages but levels off eventually and approaches its carrying capacity because of limited resources. If is the size of the population at time t, we assume that This says that the growth rate is initially close to being proportional to size. In other words, the relative growth rate is almost constant when the population is small. But we also want to reﬂect the fact that the relative growth rate decreases as the population P increases and becomes negative if P ever exceeds its carrying capacity K, the maximum population that the environment is capable of sustaining in the long run. The simplest expression for the relative growth rate that incorporates these assumptions is Multiplying by P, we obtain the model for population growth known as the logistic differ-ential equation: Notice from Equation 4 that if P is small compared with K, then is close to 0 and so . However, if (the population approaches its carrying capacity), then , so . We can deduce information about whether solutions increase or decrease directly from Equation 4. If the population P lies between 0 and K, then the right side of the equation is positive, so and the population increases. But if the pop-ulation exceeds the carrying capacity , then is negative, so and the population decreases. Let’s start our more detailed analysis of the logistic differential equation by looking at a direction ﬁeld. dPdt 0 1 PK P K dPdt 0 dPdt l 0 PK l 1 P l K dPdt kP PK dP dt kP1 P K 4 1 P dP dt k1 P K if P is small dP dt kP Pt dP dt kP m 3 m 1 P dP dt k 592 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS EXAMPLE 1 Draw a direction ﬁeld for the logistic equation with and carry-ing capacity . What can you deduce about the solutions? SOLUTION In this case the logistic differential equation is A direction ﬁeld for this equation is shown in Figure 1. We show only the ﬁrst quadrant because negative populations aren’t meaningful and we are interested only in what hap-pens after . The logistic equation is autonomous ( depends only on P, not on t), so the slopes are the same along any horizontal line. As expected, the slopes are positive for and negative for . The slopes are small when P is close to 0 or 1000 (the carrying capacity). Notice that the solutions move away from the equilibrium solution and move toward the equilibrium solution . In Figure 2 we use the direction ﬁeld to sketch solution curves with initial populations , , and . Notice that solution curves that start below are increasing and those that start above are decreasing. The slopes are greatest when and therefore the solution curves that start below have inﬂection points when . In fact we can prove that all solution curves that start below have an inﬂection point when P is exactly 500. (See Exercise 9.) M 0 t P 80 1400 60 40 20 1200 1000 800 600 400 200 FIGURE 2 Solution curves for the logistic equation in Example 1 P 500 P 500 P 1000 P 500 P 1000 P 1000 P0 1300 P0 400 P0 100 P 1000 P 0 P 1000 0 P 1000 dPdt 0 t P 80 1400 60 40 20 1200 1000 800 600 400 200 FIGURE 1 Direction field for the logistic equation in Example 1 t 0 dP dt 0.08P1 P 1000 K 1000 k 0.08 V SECTION 9.4 MODELS FOR POPULATION GROWTH \|\|\|\| 593 The logistic equation (4) is separable and so we can solve it explicitly using the method of Section 9.3. Since we have To evaluate the integral on the left side, we write Using partial fractions (see Section 7.4), we get This enables us to rewrite Equation 5: where . Solving Equation 6 for P, we get so We ﬁnd the value of A by putting in Equation 6. If , then (the initial population), so K P0 P0 Ae 0 A P P0 t 0 t 0 P K 1 Aekt P K 1 1 Aekt ? K P 1 Aekt A eC K P P Aekt 6 K P P ektC eCekt ln K P P kt C ln P ln K P kt C y 1 P 1 K P dP y k dt K PK P 1 P 1 K P 1 P1 PK K PK P y dP P1 PK y k dt 5 dP dt kP1 P K 594 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS Thus the solution to the logistic equation is Using the expression for in Equation 7, we see that which is to be expected. EXAMPLE 2 Write the solution of the initial-value problem and use it to ﬁnd the population sizes and . At what time does the population reach 900? SOLUTION The differential equation is a logistic equation with , carrying capac-ity , and initial population . So Equation 7 gives the population at time t as Thus So the population sizes when and 80 are The population reaches 900 when Solving this equation for t, we get So the population reaches 900 when t is approximately 55. As a check on our work, we graph the population curve in Figure 3 and observe where it intersects the line . The cursor indicates that . M t 55 P 900 t ln 81 0.08 54.9 0.08t ln 1 81 ln 81 e0.08t 1 81 1 9e0.08t 10 9 1000 1 9e0.08t 900 P80 1000 1 9e6.4 985.3 P40 1000 1 9e3.2 731.6 t 40 Pt 1000 1 9e0.08t where A 1000 100 100 9 Pt 1000 1 Ae0.08t P0 100 K 1000 k 0.08 P80 P40 P0 100 dP dt 0.08P1 P 1000 lim t l Pt K Pt where A K P0 P0 Pt K 1 Aekt 7 SECTION 9.4 MODELS FOR POPULATION GROWTH \|\|\|\| 595 N Compare the solution curve in Figure 3 with the lowest solution curve we drew from the direction ﬁeld in Figure 2. 1000 0 80 P= 1000 1+9e_0.08t P=900 FIGURE 3 COMPARISON OF THE NATURAL GROWTH AND LOGISTIC MODELS In the 1930s the biologist G. F. Gause conducted an experiment with the protozoan Para-mecium and used a logistic equation to model his data. The table gives his daily count of the population of protozoa. He estimated the initial relative growth rate to be 0.7944 and the carrying capacity to be 64. EXAMPLE 3 Find the exponential and logistic models for Gause’s data. Compare the predicted values with the observed values and comment on the ﬁt. SOLUTION Given the relative growth rate and the initial population , the exponential model is Gause used the same value of k for his logistic model. [This is reasonable because is small compared with the carrying capacity ( ). The equation shows that the value of k for the logistic model is very close to the value for the expo-nential model.] Then the solution of the logistic equation in Equation 7 gives where So We use these equations to calculate the predicted values (rounded to the nearest integer) and compare them in the following table. We notice from the table and from the graph in Figure 4 that for the ﬁrst three or four days the exponential model gives results comparable to those of the more sophisticated logistic model. For , however, the exponential model is hopelessly inaccurate, but the logistic model ﬁts the observations reasonably well. t 5 Pt 64 1 31e 0.7944t A K P0 P0 64 2 2 31 Pt K 1 Aekt 64 1 Ae0.7944t 1 P0 dP dt t0 k1 2 64 k K 64 P0 2 Pt P0e kt 2e 0.7944t P0 2 k 0.7944 V 596 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS t (days) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P (observed) 2 3 22 16 39 52 54 47 50 76 69 51 57 70 53 59 57 P (logistic model) 2 4 9 17 28 40 51 57 61 62 63 64 64 64 64 64 64 P (exponential model) 2 4 10 22 48 106 . . . t (days) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P (observed) 2 3 22 16 39 52 54 47 50 76 69 51 57 70 53 59 57 M Many countries that formerly experienced exponential growth are now ﬁnding that their rates of population growth are declining and the logistic model provides a better model. The table in the margin shows midyear values of , the population of Belgium, in thou-sands, at time , from 1980 to 2000. Figure 5 shows these data points together with a shifted logistic function obtained from a calculator with the ability to ﬁt a logistic function to these points by regression. We see that the logistic model provides a very good ﬁt. OTHER MODELS FOR POPULATION GROWTH The Law of Natural Growth and the logistic differential equation are not the only equa-tions that have been proposed to model population growth. In Exercise 18 we look at the Gompertz growth function and in Exercises 19 and 20 we investigate seasonal-growth models. Two of the other models are modiﬁcations of the logistic model. The differential equation has been used to model populations that are subject to harvesting of one sort or another. (Think of a population of ﬁsh being caught at a constant rate.) This equation is explored in Exercises 15 and 16. For some species there is a minimum population level m below which the species tends to become extinct. (Adults may not be able to ﬁnd suitable mates.) Such populations have been modeled by the differential equation where the extra factor, , takes into account the consequences of a sparse popula-tion (see Exercise 17). 1 mP dP dt kP1 P K1 m P dP dt kP1 P K c FIGURE 5 Logistic model for the population of Belgium 0 t P 1980 1984 1988 1992 1996 2000 P=9840+ 350 1+2.05e_0.48(t-1990) 9,800 9,900 10,000 10,100 t Bt FIGURE 4 The exponential and logistic models for the Paramecium data 0 t P 16 12 8 4 60 40 20 P= 64 1+31e_0.7944t P=2e0.7944t SECTION 9.4 MODELS FOR POPULATION GROWTH \|\|\|\| 597 t t 1980 9,847 1992 10,036 1982 9,856 1994 10,109 1984 9,855 1996 10,152 1986 9,862 1998 10,175 1988 9,884 2000 10,186 1990 9,962 Bt Bt 598 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS where is the biomass (the total mass of the members of the population) in kilograms at time (measured in years), the carrying capacity is estimated to be , and per year. (a) If , ﬁnd the biomass a year later. (b) How long will it take for the biomass to reach ? 4. The table gives the number of yeast cells in a new laboratory culture. (a) Plot the data and use the plot to estimate the carrying capacity for the yeast population. (b) Use the data to estimate the initial relative growth rate. (c) Find both an exponential model and a logistic model for these data. (d) Compare the predicted values with the observed values, both in a table and with graphs. Comment on how well your models ﬁt the data. (e) Use your logistic model to estimate the number of yeast cells after 7 hours. 5. The population of the world was about 5.3 billion in 1990. Birth rates in the 1990s ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Let’s assume that the carrying capacity for world population is 100 billion. (a) Write the logistic differential equation for these data. (Because the initial population is small compared to the carrying capacity, you can take to be an estimate of the initial relative growth rate.) (b) Use the logistic model to estimate the world population in the year 2000 and compare with the actual population of 6.1 billion. (c) Use the logistic model to predict the world population in the years 2100 and 2500. (d) What are your predictions if the carrying capacity is 50 billion? 6. (a) Make a guess as to the carrying capacity for the US population. Use it and the fact that the population was 250 million in 1990 to formulate a logistic model for the US population. (b) Determine the value of in your model by using the fact that the population in 2000 was 275 million. (c) Use your model to predict the US population in the years 2100 and 2200. k k 4 107 kg y0 2 107 kg k 0.71 K 8 107 kg t yt Suppose that a population develops according to the logistic equation where is measured in weeks. (a) What is the carrying capacity? What is the value of ? (b) A direction ﬁeld for this equation is shown. Where are the slopes close to 0? Where are they largest? Which solutions are increasing? Which solutions are decreasing? (c) Use the direction ﬁeld to sketch solutions for initial popu-lations of 20, 40, 60, 80, 120, and 140. What do these solutions have in common? How do they differ? Which solutions have inﬂection points? At what population levels do they occur? (d) What are the equilibrium solutions? How are the other solutions related to these solutions? ; 2. Suppose that a population grows according to a logistic model with carrying capacity 6000 and per year. (a) Write the logistic differential equation for these data. (b) Draw a direction ﬁeld (either by hand or with a computer algebra system). What does it tell you about the solution curves? (c) Use the direction ﬁeld to sketch the solution curves for initial populations of 1000, 2000, 4000, and 8000. What can you say about the concavity of these curves? What is the signiﬁcance of the inﬂection points? (d) Program a calculator or computer to use Euler’s method with step size to estimate the population after 50 years if the initial population is 1000. (e) If the initial population is 1000, write a formula for the population after years. Use it to ﬁnd the population after 50 years and compare with your estimate in part (d). (f) Graph the solution in part (e) and compare with the solu-tion curve you sketched in part (c). The Paciﬁc halibut ﬁshery has been modeled by the differen-tial equation dy dt ky1 y K 3. t h 1 k 0.0015 0 t P 60 40 20 150 100 50 k t dP dt 0.05P 0.0005P 2 1. EXERCISES 9.4 Time (hours) Yeast cells Time (hours) Yeast cells 0 18 10 509 2 39 12 597 4 80 14 640 6 171 16 664 8 336 18 672 ;12.The table gives the midyear population of Spain, in thousands, from 1955 to 2000. Use a graphing calculator to ﬁt both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models. [Hint: Subtract 29,000 from each of the population ﬁgures. Then, after obtaining a model from your calculator, add 29,000 to get your ﬁnal model. It might be helpful to choose to correspond to 1955 or 1975.] 13. Consider a population with constant relative birth and death rates and , respectively, and a constant emigra-tion rate , where , , and are positive constants. Assume that . Then the rate of change of the population at time is modeled by the differential equation where (a) Find the solution of this equation that satisﬁes the initial condition (b) What condition on will lead to an exponential expan-sion of the population? (c) What condition on will result in a constant population? A population decline? (d) In 1847, the population of Ireland was about 8 million and the difference between the relative birth and death rates was 1.6% of the population. Because of the potato famine in the 1840s and 1850s, about 210,000 inhabitants per year emigrated from Ireland. Was the population expanding or declining at that time? Let be a positive number. A differential equation of the form where is a positive constant, is called a doomsday equation because the exponent in the expression is larger than the exponent 1 for natural growth. (a) Determine the solution that satisﬁes the initial condition (b) Show that there is a ﬁnite time (doomsday) such that . (c) An especially proliﬁc breed of rabbits has the growth term . If 2 such rabbits breed initially and the warren has 16 rabbits after three months, then when is doomsday? ky 1.01 lim t l T yt t T y0 y0. ky 1c k dy dt ky 1c c 14. m m P0 P0. k dP dt kP m t m m P Pt t 0 (d) Use your model to predict the year in which the US popu-lation will exceed 350 million. One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction of the popula-tion who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisﬁed by . (b) Solve the differential equation. (c) A small town has 1000 inhabitants. At 8 AM, 80 people have heard a rumor. By noon half the town has heard it. At what time will of the population have heard the rumor? 8. Biologists stocked a lake with 400 ﬁsh and estimated the carrying capacity (the maximal population for the ﬁsh of that species in that lake) to be 10,000. The number of ﬁsh tripled in the ﬁrst year. (a) Assuming that the size of the ﬁsh population satisﬁes the logistic equation, ﬁnd an expression for the size of the population after years. (b) How long will it take for the population to increase to 5000? (a) Show that if satisﬁes the logistic equation (4), then (b) Deduce that a population grows fastest when it reaches half its carrying capacity. ; 10. For a ﬁxed value of (say ), the family of logistic functions given by Equation 7 depends on the initial value and the proportionality constant . Graph several members of this family. How does the graph change when varies? How does it change when varies? ; 11. The table gives the midyear population of Japan, in thousands, from 1960 to 2005. Use a graphing calculator to ﬁt both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models. [Hint: Subtract 94,000 from each of the population ﬁgures. Then, after obtaining a model from your calculator, add 94,000 to get your ﬁnal model. It might be helpful to choose to correspond to 1960 or 1980.] t 0 k P0 k P0 K 10 K d 2P dt 2 k 2P1 P K1 2P K P 9. t 90% y y 7. SECTION 9.4 MODELS FOR POPULATION GROWTH \|\|\|\| 599 Year Population Year Population 1960 94,092 1985 120,754 1965 98,883 1990 123,537 1970 104,345 1995 125,341 1975 111,573 2000 126,700 1980 116,807 2005 127,417 Year Population Year Population 1955 29,319 1980 37,488 1960 30,641 1985 38,535 1965 32,085 1990 39,351 1970 33,876 1995 39,750 1975 35,564 2000 40,016 (d) Use the solution in part (c) to show that if , then the species will become extinct. [Hint: Show that the numerator in your expression for is 0 for some value of .] 18. Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of the differential equation where is a constant and is the carrying capacity. (a) Solve this differential equation. (b) Compute . (c) Graph the Gompertz growth function for , , and , and compare it with the logistic function in Example 2. What are the similarities? What are the differences? (d) We know from Exercise 9 that the logistic function grows fastest when . Use the Gompertz differential equation to show that the Gompertz function grows fastest when . 19. In a seasonal-growth model, a periodic function of time is introduced to account for seasonal variations in the rate of growth. Such variations could, for example, be caused by seasonal changes in the availability of food. (a) Find the solution of the seasonal-growth model where , , and are positive constants. ; (b) By graphing the solution for several values of , , and , explain how the values of , , and affect the solution. What can you say about ? 20. Suppose we alter the differential equation in Exercise 19 as follows: (a) Solve this differential equation with the help of a table of integrals or a CAS. ; (b) Graph the solution for several values of , , and . How do the values of , , and affect the solution? What can you say about in this case? 21. Graphs of logistic functions (Figures 2 and 3) look suspi-ciously similar to the graph of the hyperbolic tangent function (Figure 3 in Section 3.11). Explain the similarity by showing that the logistic function given by Equation 7 can be written as where . Thus the logistic function is really just a shifted hyperbolic tangent. c ln Ak Pt 1 2K [1 tanh( 1 2kt c)] lim t l Pt r k r k P0 P0 dP dt kP cos2rt lim t l Pt r k r k r k P0 P0 dP dt kP cosrt P Ke P K2 c 0.05 P0 100 K 1000 lim t l Pt K c dP dt c ln K PP t Pt P0 m Let’s modify the logistic differential equation of Example 1 as follows: (a) Suppose represents a ﬁsh population at time , where is measured in weeks. Explain the meaning of the term . (b) Draw a direction ﬁeld for this differential equation. (c) What are the equilibrium solutions? (d) Use the direction ﬁeld to sketch several solution curves. Describe what happens to the ﬁsh population for various initial populations. (e) Solve this differential equation explicitly, either by using partial fractions or with a computer algebra system. Use the initial populations 200 and 300. Graph the solutions and compare with your sketches in part (d). 16. Consider the differential equation as a model for a ﬁsh population, where is measured in weeks and is a constant. (a) Use a CAS to draw direction ﬁelds for various values of . (b) From your direction ﬁelds in part (a), determine the values of for which there is at least one equilibrium solution. For what values of does the ﬁsh population always die out? (c) Use the differential equation to prove what you dis-covered graphically in part (b). (d) What would you recommend for a limit to the weekly catch of this ﬁsh population? There is considerable evidence to support the theory that for some species there is a minimum population such that the species will become extinct if the size of the population falls below . This condition can be incorporated into the logistic equation by introducing the factor . Thus the mod-iﬁed logistic model is given by the differential equation (a) Use the differential equation to show that any solution is increasing if and decreasing if . (b) For the case where , , and , draw a direction ﬁeld and use it to sketch several solu-tion curves. Describe what happens to the population for various initial populations. What are the equilibrium solutions? (c) Solve the differential equation explicitly, either by using partial fractions or with a computer algebra system. Use the initial population . P0 m 200 K 1000 k 0.08 0 P m m P K dP dt kP1 P K1 m P 1 mP m m 17. c c c c t dP dt 0.08P1 P 1000 c CAS CAS 15 t t Pt dP dt 0.08P1 P 1000 15 15. 600 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS In this project we explore three of the many applications of calculus to baseball. The physical interactions of the game, especially the collision of ball and bat, are quite complex and their models are discussed in detail in a book by Robert Adair, The Physics of Baseball, 3d ed. (New York: HarperPerennial, 2002). 1. It may surprise you to learn that the collision of baseball and bat lasts only about a thou-sandth of a second. Here we calculate the average force on the bat during this collision by ﬁrst computing the change in the ball’s momentum. The of an object is the product of its mass and its velocity , that is, . Suppose an object, moving along a straight line, is acted on by a force that is a continuous function of time. (a) Show that the change in momentum over a time interval is equal to the integral of from to ; that is, show that This integral is called the impulse of the force over the time interval. (b) A pitcher throws a 90-mih fastball to a batter, who hits a line drive directly back to the pitcher. The ball is in contact with the bat for 0.001 s and leaves the bat with velocity 110 mih. A baseball weighs 5 oz and, in US Customary units, its mass is measured in slugs: where . (i) Find the change in the ball’s momentum. (ii) Find the average force on the bat. 2. In this problem we calculate the work required for a pitcher to throw a 90-mih fastball by ﬁrst considering kinetic energy. The kinetic energy of an object of mass and velocity is given by . Sup-pose an object of mass , moving in a straight line, is acted on by a force that depends on its position . According to Newton’s Second Law where and denote the acceleration and velocity of the object. (a) Show that the work done in moving the object from a position to a position is equal to the change in the object’s kinetic energy; that is, show that where and are the velocities of the object at the positions and . Hint: By the Chain Rule, (b) How many foot-pounds of work does it take to throw a baseball at a speed of 90 mih? 3. (a) An outﬁelder ﬁelds a baseball 280 ft away from home plate and throws it directly to the catcher with an initial velocity of 100 fts. Assume that the velocity of the ball after seconds satisﬁes the differential equation because of air resistance. How long does it take for the ball to reach home plate? (Ignore any vertical motion of the ball.) (b) The manager of the team wonders whether the ball will reach home plate sooner if it is relayed by an inﬁelder. The shortstop can position himself directly between the out-ﬁelder and home plate, catch the ball thrown by the outﬁelder, turn, and throw the ball to dvdt 1 10 v t vt m dv dt m dv ds ds dt mv dv ds s1 s0 v1 vs1 v0 vs0 W y s1 s 0 Fs ds 1 2mv1 2 1 2mv0 2 s1 s0 v a Fs ma m dv dt s F Fs m K 1 2mv2 v m K t 32 fts2 m wt pt1 pt0 y t 1 t 0 Ft dt t1 t0 F t0, t1 F Ft p mv v m momentum p CALCULUS AND BASEBALL A P P L I E D P R O J E C T APPLIED PROJECT CALCULUS AND BASEBALL \|\|\|\| 601 An overhead view of the position of a baseball bat, shown every ﬁftieth of a second during a typical swing. (Adapted from The Physics of Baseball) Batter’s box the catcher with an initial velocity of 105 fts. The manager clocks the relay time of the shortstop (catching, turning, throwing) at half a second. How far from home plate should the shortstop position himself to minimize the total time for the ball to reach home plate? Should the manager encourage a direct throw or a relayed throw? What if the shortstop can throw at 115 fts? ; (c) For what throwing velocity of the shortstop does a relayed throw take the same time as a direct throw? 602 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS LINEAR EQUATIONS A ﬁrst-order linear differential equation is one that can be put into the form where and are continuous functions on a given interval. This type of equation occurs frequently in various sciences, as we will see. An example of a linear equation is because, for , it can be written in the form Notice that this differential equation is not separable because it’s impossible to factor the expression for as a function of x times a function of y. But we can still solve the equa-tion by noticing, by the Product Rule, that and so we can rewrite the equation as If we now integrate both sides of this equation, we get or If we had been given the differential equation in the form of Equation 2, we would have had to take the preliminary step of multiplying each side of the equation by x. It turns out that every ﬁrst-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function called an integrating factor. We try to ﬁnd so that the left side of Equation 1, when multiplied by , becomes the derivative of the product : If we can ﬁnd such a function , then Equation 1 becomes Ixy IxQx I Ixy Pxy Ixy 3 Ixy Ix I Ix y x C x xy x 2 C xy 2x xy y xy y y 1 x y 2 2 x 0 xy y 2x Q P dy dx Pxy Qx 1 9.5 Integrating both sides, we would have so the solution would be To ﬁnd such an , we expand Equation 3 and cancel terms: This is a separable differential equation for , which we solve as follows: where . We are looking for a particular integrating factor, not the most general one, so we take and use Thus a formula for the general solution to Equation 1 is provided by Equation 4, where is given by Equation 5. Instead of memorizing this formula, however, we just remember the form of the integrating factor. To solve the linear differential equation , multiply both sides by the integrating factor and integrate both sides. EXAMPLE 1 Solve the differential equation . SOLUTION The given equation is linear since it has the form of Equation 1 with and . An integrating factor is Multiplying both sides of the differential equation by , we get or d dx e x3y 6x 2e x3 e x3 dy dx 3x 2e x3y 6x 2e x3 e x3 Ix ex 3x 2 dx e x3 Qx 6x 2 Px 3x 2 dy dx 3x 2y 6x 2 V Ix ex Px dx y Pxy Qx I Ix ex Px dx 5 A 1 A e C I Aex Px dx ln I y Px dx y dI I y Px dx I IxPx Ix Ixy IxPxy Ixy Ixy Ixy I yx 1 Ix y IxQx dx C 4 Ixy y IxQx dx C SECTION 9.5 LINEAR EQUATIONS 603 Integrating both sides, we have M EXAMPLE 2 Find the solution of the initial-value problem SOLUTION We must ﬁrst divide both sides by the coefﬁcient of to put the differential equation into standard form: The integrating factor is Multiplication of Equation 6 by gives Then and so Since , we have Therefore the solution to the initial-value problem is M EXAMPLE 3 Solve . SOLUTION The given equation is in the standard form for a linear equation. Multiplying by the integrating factor we get or Therefore e x2y y e x2 dx C (e x2y) e x2 e x2y 2xe x2y e x2 ex 2x dx e x2 y 2xy 1 y ln x 2 x 2 ln 1 C 1 C y1 2 y ln x C x xy y 1 x dx ln x C xy 1 x or xy y 1 x x Ix ex 1x dx e ln x x x 0 y 1 x y 1 x 2 6 y y1 2 x 0 x 2y xy 1 V y 2 Cex3 e x3y y 6x 2e x3 dx 2e x3 C 604 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS N The solution of the initial-value problem in Example 2 is shown in Figure 2. FIGURE 2 (1, 2) 5 _5 0 4 FIGURE 1 6 _3 _1.5 1.8 C=2 C=1 C=_2 C=_1 C=0 N Figure 1 shows the graphs of several members of the family of solutions in Example 1. Notice that they all approach as . x l 2 Recall from Section 7.5 that can’t be expressed in terms of elementary functions. Nonetheless, it’s a perfectly good function and we can leave the answer as Another way of writing the solution is (Any number can be chosen for the lower limit of integration.) M APPLICATION TO ELECTRIC CIRCUITS In Section 9.2 we considered the simple electric circuit shown in Figure 4: An electro-motive force (usually a battery or generator) produces a voltage of volts (V) and a cur-rent of amperes (A) at time . The circuit also contains a resistor with a resistance of ohms ( ) and an inductor with an inductance of henries (H). Ohm’s Law gives the drop in voltage due to the resistor as . The voltage drop due to the inductor is . One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage . Thus we have which is a ﬁrst-order linear differential equation. The solution gives the current at time . EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is and the inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed when so the current starts with , ﬁnd (a) , (b) the current after 1 s, and (c) the limiting value of the current. SOLUTION (a) If we put , , and in Equation 7, we obtain the initial-value problem or Multiplying by the integrating factor , we get It 5 Ce3t e 3tI y 15e 3t dt 5e 3t C d dt e 3tI 15e 3t e 3t dI dt 3e 3tI 15e 3t ex 3 dt e 3t I0 0 dI dt 3I 15 I0 0 4 dI dt 12I 60 Et 60 R 12 L 4 It I0 0 t 0 12 V t I L dI dt RI Et 7 Et LdIdt RI L R t It Et y ex2 y x 0 e t 2 dt Cex2 y ex2 y e x2 dx Cex2 x e x2 dx SECTION 9.5 LINEAR EQUATIONS \|\|\|\| 605 N Even though the solutions of the differential equation in Example 3 are expressed in terms of an integral, they can still be graphed by a com-puter algebra system (Figure 3). FIGURE 3 C=2 C=_2 2.5 _2.5 _2.5 2.5 FIGURE 4 R E switch L N The differential equation in Example 4 is both linear and separable, so an alternative method is to solve it as a separable equation (Example 4 in Section 9.3). If we replace the battery by a gen-erator, however, we get an equation that is linear but not separable (Example 5). Since , we have , so and (b) After 1 second the current is (c) The limiting value of the current is given by M EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4 but, instead of the battery, we use a generator that produces a variable voltage of volts. Find . SOLUTION This time the differential equation becomes The same integrating factor gives Using Formula 98 in the Table of Integrals, we have Since , we get so M It 5 101sin 30t 10 cos 30t 50 101e3t 50 101 C 0 I0 0 I 5 101 sin 30t 10 cos 30t Ce3t e 3tI y 15e 3t sin 30t dt 15 e 3t 909 3 sin 30t 30 cos 30t C d dt e 3tI e 3t dI dt 3e 3tI 15e 3t sin 30t e 3t dI dt 3I 15 sin 30t or 4 dI dt 12I 60 sin 30t It Et 60 sin 30t 5 0 5 5 5 lim t l e3t lim t l It lim t l 51 e3t I1 51 e3 4.75 A It 51 e3t C 5 5 C 0 I0 0 606 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS FIGURE 5 6 0 2.5 y=5 N Figure 5 shows how the current in Example 4 approaches its limiting value. N Figure 6 shows the graph of the current when the battery is replaced by a generator. FIGURE 6 2 _2 2.5 0 11. 12. 13. , 14. 15–20 Solve the initial-value problem. 15. , y0 2 y x y t ln t dr dt r te t t 0 1 t du dt u 1 t x dy dx 4y x 4e x sin x dy dx cos xy sinx 2 1–4 Determine whether the differential equation is linear. 1. 2. 3. 4. 5–14 Solve the differential equation. 6. 7. 8. 10. y y sine x xy y sx 9. x 2y 2xy cos 2x xy 2y x 2 y x 5y y 2y 2e x 5. xy sx e xy yy xy x 2 y cos y tan x y cos x y EXERCISES 9.5 this case Kirchhoff’s Law gives But (see Example 3 in Section 3.7), so we have Suppose the resistance is , the capacitance is F, a battery gives a constant voltage of 60 V, and the initial charge is C. Find the charge and the current at time . 30. In the circuit of Exercise 29, , , , and . Find the charge and the current at time . Let be the performance level of someone learning a skill as a function of the training time . The graph of is called a learning curve. In Exercise 13 in Section 9.1 we proposed the differential equation as a reasonable model for learning, where is a positive con-stant. Solve it as a linear differential equation and use your solution to graph the learning curve. 32. Two new workers were hired for an assembly line. Jim pro-cessed 25 units during the ﬁrst hour and 45 units during the second hour. Mark processed 35 units during the ﬁrst hour and 50 units the second hour. Using the model of Exercise 31 and assuming that , estimate the maximum number of units per hour that each worker is capable of processing. In Section 9.3 we looked at mixing problems in which the volume of ﬂuid remained constant and saw that such prob-lems give rise to separable equations. (See Example 6 in that section.) If the rates of ﬂow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable. A tank contains 100 L of water. A solution with a salt con-centration of is added at a rate of . The solution is kept mixed and is drained from the tank at a rate of . If is the amount of salt (in kilograms) after minutes, show that satisﬁes the differential equation Solve this equation and ﬁnd the concentration after 20 minutes. 34. A tank with a capacity of 400 L is full of a mixture of water and chlorine with a concentration of 0.05 g of chlorine per dy dt 2 3y 100 2t y t yt 3 Lmin 5 Lmin 0.4 kgL 33. P0 0 k dP dt kM Pt P t Pt 31. t Et 10 sin 60t Q0 0 C 0.01 F R 2 C E R t Q0 0 0.05 5 R dQ dt 1 C Q Et I dQdt RI Q C Et 16. , , 17. , 18. , , , 20. , ; 21–22 Solve the differential equation and use a graphing cal-culator or computer to graph several members of the family of solutions. How does the solution curve change as varies? 21. 22. 23. A Bernoulli differential equation (named after James Bernoulli) is of the form Observe that, if or , the Bernoulli equation is linear. For other values of , show that the substitution transforms the Bernoulli equation into the linear equation 24–25 Use the method of Exercise 23 to solve the differential equation. 24. 26. Solve the second-order equation by making the substitution . 27. In the circuit shown in Figure 4, a battery supplies a constant voltage of 40 V, the inductance is 2 H, the resistance is , and . (a) Find . (b) Find the current after s. 28. In the circuit shown in Figure 4, a generator supplies a volt-age of volts, the inductance is H, the resistance is , and A. (a) Find . (b) Find the current after s. ; (c) Use a graphing device to draw the graph of the current function. 29. The ﬁgure shows a circuit containing an electromotive force, a capacitor with a capacitance of farads (F), and a resistor with a resistance of ohms ( ). The voltage drop across the capacitor is , where is the charge (in coulombs), so in Q QC R C 0.1 It I0 1 20 1 Et 40 sin 60t 0.1 It I0 0 10 u y xy 2y 12x 2 y 2 x y y 3 x 2 25. xy y xy 2 du dx 1 nPxu 1 nQx u y 1n n 1 n 0 dy dx Pxy Qxy n y cos xy cos x xy 2y e x C y0 2 x 2 1 dy dx 3xy 1 0 y 0 xy y x 2 sin x 19. y4 20 x 0 2xy y 6x v0 5 dv dt 2tv 3t 2e t 2 y1 0 t 0 t dy dt 2y t 3 SECTION 9.5 LINEAR EQUATIONS \|\|\|\| 607 (a) Solve this as a linear equation to show that (b) What is the limiting velocity? (c) Find the distance the object has fallen after seconds. 36. If we ignore air resistance, we can conclude that heavier objects fall no faster than lighter objects. But if we take air resistance into account, our conclusion changes. Use the expression for the velocity of a falling object in Exercise 35(a) to ﬁnd and show that heavier objects do fall faster than lighter ones. dvdm t v mt c 1 ectm liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of . The mixture is kept stirred and is pumped out at a rate of . Find the amount of chlorine in the tank as a function of time. 35. An object with mass is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If is the distance dropped after seconds, then the speed is and the acceleration is . If is the accelera-tion due to gravity, then the downward force on the object is , where is a positive constant, and Newton’s Second Law gives m dv dt mt cv c mt cv t a vt v st t st m 10 Ls 4 Ls 608 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS PREDATOR-PREY SYSTEMS We have looked at a variety of models for the growth of a single species that lives alone in an environment. In this section we consider more realistic models that take into account the interaction of two species in the same habitat. We will see that these models take the form of a pair of linked differential equations. We ﬁrst consider the situation in which one species, called the prey, has an ample food supply and the second species, called the predator, feeds on the prey. Examples of prey and predators include rabbits and wolves in an isolated forest, food ﬁsh and sharks, aphids and ladybugs, and bacteria and amoebas. Our model will have two dependent variables and both are functions of time. We let be the number of prey (using R for rabbits) and be the number of predators (with W for wolves) at time t. In the absence of predators, the ample food supply would support exponential growth of the prey, that is, In the absence of prey, we assume that the predator population would decline at a rate pro-portional to itself, that is, With both species present, however, we assume that the principal cause of death among the prey is being eaten by a predator, and the birth and survival rates of the predators depend on their available food supply, namely, the prey. We also assume that the two species encounter each other at a rate that is proportional to both populations and is therefore pro-portional to the product RW. (The more there are of either population, the more encoun-ters there are likely to be.) A system of two differential equations that incorporates these assumptions is as follows: where k, r, a, and b are positive constants. Notice that the term aRW decreases the nat-ural growth rate of the prey and the term bRW increases the natural growth rate of the predators. dW dt rW bRW dR dt kR aRW 1 where r is a positive constant dW dt rW where k is a positive constant dR dt kR Wt Rt 9.6 W represents the predator. R represents the prey. The equations in (1) are known as the predator-prey equations, or the Lotka-Volterra equations. A solution of this system of equations is a pair of functions and that describe the populations of prey and predator as functions of time. Because the system is coupled (R and W occur in both equations), we can’t solve one equation and then the other; we have to solve them simultaneously. Unfortunately, it is usually impossible to ﬁnd explicit formulas for R and W as functions of t. We can, however, use graphical methods to analyze the equations. EXAMPLE 1 Suppose that populations of rabbits and wolves are described by the Lotka-Volterra equations (1) with , , , and . The time is measured in months. (a) Find the constant solutions (called the equilibrium solutions) and interpret the answer. (b) Use the system of differential equations to ﬁnd an expression for . (c) Draw a direction ﬁeld for the resulting differential equation in the RW-plane. Then use that direction ﬁeld to sketch some solution curves. (d) Suppose that, at some point in time, there are 1000 rabbits and 40 wolves. Draw the corresponding solution curve and use it to describe the changes in both population levels. (e) Use part (d) to make sketches of R and W as functions of t. SOLUTION (a) With the given values of k, a, r, and b, the Lotka-Volterra equations become Both R and W will be constant if both derivatives are 0, that is, One solution is given by R 0 and W 0. (This makes sense: If there are no rabbits or wolves, the populations are certainly not going to increase.) The other constant solution is So the equilibrium populations consist of 80 wolves and 1000 rabbits. This means that 1000 rabbits are just enough to support a constant wolf population of 80. There are neither too many wolves (which would result in fewer rabbits) nor too few wolves (which would result in more rabbits). (b) We use the Chain Rule to eliminate t: so dW dR dW dt dR dt 0.02W 0.00002RW 0.08R 0.001RW dW dt dW dR dR dt R 0.02 0.00002 1000 W 0.08 0.001 80 W W0.02 0.00002R 0 R R0.08 0.001W 0 dW dt 0.02W 0.00002RW dR dt 0.08R 0.001RW dWdR t b 0.00002 r 0.02 a 0.001 k 0.08 V Wt Rt SECTION 9.6 PREDATOR-PREY SYSTEMS \|\|\|\| 609 N The Lotka-Volterra equations were proposed as a model to explain the variations in the shark and food-ﬁsh populations in the Adriatic Sea by the Italian mathematician Vito Volterra (1860–1940). (c) If we think of W as a function of R, we have the differential equation We draw the direction ﬁeld for this differential equation in Figure 1 and we use it to sketch several solution curves in Figure 2. If we move along a solution curve, we observe how the relationship between R and W changes as time passes. Notice that the curves appear to be closed in the sense that if we travel along a curve, we always return to the same point. Notice also that the point (1000, 80) is inside all the solution curves. That point is called an equilibrium point because it corresponds to the equilibrium solu-tion R 1000, W 80. When we represent solutions of a system of differential equations as in Figure 2, we refer to the RW-plane as the phase plane, and we call the solution curves phase trajec-tories. So a phase trajectory is a path traced out by solutions as time goes by. A phase portrait consists of equilibrium points and typical phase trajectories, as shown in Figure 2. (d) Starting with 1000 rabbits and 40 wolves corresponds to drawing the solution curve through the point . Figure 3 shows this phase trajectory with the direction ﬁeld removed. Starting at the point at time and letting increase, do we move clockwise or counterclockwise around the phase trajectory? If we put and FIGURE 3 Phase trajectory through (1000, 40) 0 R W 1000 140 2000 3000 120 100 80 60 40 20 500 1500 2500 P¸ (1000, 40) P¡ P™ P£ R 1000 t t 0 P0 P01000, 40 R, W 0 R W 1000 150 100 50 2000 3000 FIGURE 1 Direction ﬁeld for the predator-prey system FIGURE 2 Phase portrait of the system 0 R W 1000 150 100 50 2000 3000 dW dR 0.02W 0.00002RW 0.08R 0.001RW 610 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS in the ﬁrst differential equation, we get Since , we conclude that R is increasing at and so we move counter-clockwise around the phase trajectory. We see that at there aren’t enough wolves to maintain a balance between the popu-lations, so the rabbit population increases. That results in more wolves and eventually there are so many wolves that the rabbits have a hard time avoiding them. So the number of rabbits begins to decline (at , where we estimate that R reaches its maximum popu-lation of about 2800). This means that at some later time the wolf population starts to fall (at , where and ). But this beneﬁts the rabbits, so their popula-tion later starts to increase (at , where and ). As a consequence, the wolf population eventually starts to increase as well. This happens when the populations return to their initial values of and , and the entire cycle begins again. (e) From the description in part (d) of how the rabbit and wolf populations rise and fall, we can sketch the graphs of and . Suppose the points , , and in Figure 3 are reached at times , , and . Then we can sketch graphs of R and W as in Figure 4. To make the graphs easier to compare, we draw the graphs on the same axes but with different scales for R and W, as in Figure 5. Notice that the rabbits reach their maximum populations about a quarter of a cycle before the wolves. M FIGURE 5 Comparison of the rabbi and wolf populations 0 t R t¡ t£ W 120 80 40 t™ 2000 1000 R W Number of wolves Number of rabbits 3000 t 0 t W 140 t¡ t£ 120 100 80 60 40 20 t™ 0 t R 2500 t¡ t£ t™ 2000 1500 1000 500 t3 t2 t1 P3 P2 P1 Wt Rt W 40 R 1000 R 210 W 80 P3 W 140 R 1000 P2 P1 P0 P0 dRdt 0 dR dt 0.081000 0.001100040 80 40 40 W 40 SECTION 9.6 PREDATOR-PREY SYSTEMS \|\|\|\| 611 FIGURE 4 Graphs of the rabbit and wolf populations as functions of time In Module 9.6 you can change the coefﬁcients in the Lotka-Volterra equations and observe the resulting changes in the phase trajectory and graphs of the rabbit and wolf populations. TEC An important part of the modeling process, as we discussed in Section 1.2, is to inter-pret our mathematical conclusions as real-world predictions and to test the predictions against real data. The Hudson’s Bay Company, which started trading in animal furs in Canada in 1670, has kept records that date back to the 1840s. Figure 6 shows graphs of the number of pelts of the snowshoe hare and its predator, the Canada lynx, traded by the com-pany over a 90-year period. You can see that the coupled oscillations in the hare and lynx populations predicted by the Lotka-Volterra model do actually occur and the period of these cycles is roughly 10 years. Although the relatively simple Lotka-Volterra model has had some success in explain-ing and predicting coupled populations, more sophisticated models have also been pro-posed. One way to modify the Lotka-Volterra equations is to assume that, in the absence of predators, the prey grow according to a logistic model with carrying capacity K. Then the Lotka-Volterra equations (1) are replaced by the system of differential equations This model is investigated in Exercises 9 and 10. Models have also been proposed to describe and predict population levels of two species that compete for the same resources or cooperate for mutual beneﬁt. Such models are explored in Exercise 2. dW dt rW bRW dR dt kR1 R K aRW 0 1850 9 6 3 160 120 80 40 hare lynx Thousands of lynx Thousands of hares 1875 1900 1925 FIGURE 6 Relative abundance of hare and lynx from Hudson’s Bay Company records 612 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS 2. Each system of differential equations is a model for two species that either compete for the same resources or cooperate for mutual beneﬁt (ﬂowering plants and insect pollinators, for instance). Decide whether each system describes competition or cooperation and explain why it is a reasonable model. (Ask yourself what effect an increase in one species has on the growth rate of the other.) (a) (b) dy dt 0.2y 0.00008y 2 0.0002xy dx dt 0.15x 0.0002x 2 0.0006xy dy dt 0.08x 0.00004xy dx dt 0.12x 0.0006x 2 0.00001xy For each predator-prey system, determine which of the vari-ables, or , represents the prey population and which repre-sents the predator population. Is the growth of the prey restricted just by the predators or by other factors as well? Do the predators feed only on the prey or do they have additional food sources? Explain. (a) (b) dy dt 0.015y 0.00008xy dx dt 0.2x 0.0002x 2 0.006xy dy dt 0.1y 0.005xy dx dt 0.05x 0.0001xy y x 1. EXERCISES 9.6 6. In Example 1(b) we showed that the rabbit and wolf popula-tions satisfy the differential equation By solving this separable differential equation, show that where is a constant. It is impossible to solve this equation for as an explicit function of (or vice versa). If you have a computer algebra system that graphs implicitly deﬁned curves, use this equation and your CAS to draw the solution curve that passes through the point and compare with Figure 3. 8. Populations of aphids and ladybugs are modeled by the equations (a) Find the equilibrium solutions and explain their signiﬁcance. (b) Find an expression for . (c) The direction ﬁeld for the differential equation in part (b) is shown. Use it to sketch a phase portrait. What do the phase trajectories have in common? 0 A L 200 5000 10000 15000 400 100 300 dLdA dL dt 0.5L 0.0001AL dA dt 2A 0.01AL 1000, 40 R W C R0.02W 0.08 e 0.00002Re 0.001W C dW dR 0.02W 0.00002RW 0.08R 0.001RW 7. 0 t y 800 5 400 species 1 species 2 10 15 1200 600 200 1000 3–4 A phase trajectory is shown for populations of rabbits and foxes . (a) Describe how each population changes as time goes by. (b) Use your description to make a rough sketch of the graphs of R and F as functions of time. 4. 5–6 Graphs of populations of two species are shown. Use them to sketch the corresponding phase trajectory. species 1 species 2 0 t y 200 150 1 100 50 5. t=0 0 R F 400 160 120 80 800 1200 1600 40 t=0 0 R F 400 300 200 100 800 1200 1600 2000 3. F R SECTION 9.6 PREDATOR-PREY SYSTEMS \|\|\|\| 613 (b) Find all the equilibrium solutions and explain their signiﬁcance. (c) The ﬁgure shows the phase trajectory that starts at the point . Describe what eventually happens to the rabbit and wolf populations. (d) Sketch graphs of the rabbit and wolf populations as func-tions of time. 10. In Exercise 8 we modeled populations of aphids and ladybugs with a Lotka-Volterra system. Suppose we modify those equa-tions as follows: (a) In the absence of ladybugs, what does the model predict about the aphids? (b) Find the equilibrium solutions. (c) Find an expression for . (d) Use a computer algebra system to draw a direction ﬁeld for the differential equation in part (c). Then use the direction ﬁeld to sketch a phase portrait. What do the phase trajectories have in common? (e) Suppose that at time there are 1000 aphids and 200 ladybugs. Draw the corresponding phase trajectory and use it to describe how both populations change. (f) Use part (e) to make rough sketches of the aphid and ladybug populations as functions of . How are the graphs related to each other? t t 0 dLdA dL dt 0.5L 0.0001AL dA dt 2A1 0.0001A 0.01AL CAS 1000, 40 (d) Suppose that at time there are 1000 aphids and 200 ladybugs. Draw the corresponding phase trajectory and use it to describe how both populations change. (e) Use part (d) to make rough sketches of the aphid and ladybug populations as functions of . How are the graphs related to each other? 9. In Example 1 we used Lotka-Volterra equations to model populations of rabbits and wolves. Let’s modify those equations as follows: (a) According to these equations, what happens to the rabbit population in the absence of wolves? R W 800 70 60 50 1000 1200 1400 40 1600 dW dt 0.02W 0.00002RW dR dt 0.08R1 0.0002R 0.001RW t t 0 614 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS REVIEW CONCEPT CHECK 9 7. (a) Write a differential equation that expresses the law of natural growth. What does it say in terms of relative growth rate? (b) Under what circumstances is this an appropriate model for population growth? (c) What are the solutions of this equation? 8. (a) Write the logistic equation. (b) Under what circumstances is this an appropriate model for population growth? 9. (a) Write Lotka-Volterra equations to model populations of food ﬁsh and sharks . (b) What do these equations say about each population in the absence of the other? S F 1. (a) What is a differential equation? (b) What is the order of a differential equation? (c) What is an initial condition? 2. What can you say about the solutions of the equation just by looking at the differential equation? 3. What is a direction ﬁeld for the differential equation ? 4. Explain how Euler’s method works. 5. What is a separable differential equation? How do you solve it? 6. What is a ﬁrst-order linear differential equation? How do you solve it? y Fx, y y x 2 y 2 Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. All solutions of the differential equation are decreasing functions. 2. The function is a solution of the differential equation . 3. The equation is separable. 4. The equation is separable. y 3y 2x 6xy 1 y x y x 2y xy 1 f x ln xx y 1 y 4 5. The equation is linear. 6. The equation is linear. 7. If is the solution of the initial-value problem then . lim t l y 5 y0 1 dy dt 2y1 y 5 y y xy e y e xy y TRUE-FALSE QUIZ 1. (a) A direction ﬁeld for the differential equation is shown. Sketch the graphs of the solutions that satisfy the given initial conditions. (i) (ii) (iii) (iv) (b) If the initial condition is , for what values of is ﬁnite? What are the equilibrium solutions? 2. (a) Sketch a direction ﬁeld for the differential equation . Then use it to sketch the four solutions that satisfy the initial conditions , , , and . (b) Check your work in part (a) by solving the differential equation explicitly. What type of curve is each solution curve? 3. (a) A direction ﬁeld for the differential equation is shown. Sketch the solution of the initial-value problem Use your graph to estimate the value of . y0.3 y0 1 y x 2 y 2 y x 2 y 2 y2 1 y2 1 y0 1 y0 1 y xy 0 x y 1 2 2 4 6 limt l yt c y0 c y0 4.3 y0 3 y0 1 y0 0.3 y yy 2y 4 (b) Use Euler’s method with step size 0.1 to estimate where is the solution of the initial-value problem in part (a). Compare with your estimate from part (a). (c) On what lines are the centers of the horizontal line segments of the direction ﬁeld in part (a) located? What happens when a solution curve crosses these lines? 4. (a) Use Euler’s method with step size 0.2 to estimate , where is the solution of the initial-value problem (b) Repeat part (a) with step size 0.1. (c) Find the exact solution of the differential equation and compare the value at 0.4 with the approximations in parts (a) and (b). 5–8 Solve the differential equation. 5. 6. 7. 8. x 2y y 2x 3e1x 2ye y2y 2x 3sx dx dt 1 t x tx y xesin x y cos x y0 1 y 2xy 2 yx y0.4 yx y0.3 0 x y 1 2 _1 _2 1 2 _1 _2 3 _3 3 _3 EXERCISES CHAPTER 9 REVIEW \|\|\|\| 615 people and the number of uninfected people. In an isolated town of 5000 inhabitants, 160 people have a disease at the beginning of the week and 1200 have it at the end of the week. How long does it take for of the population to become infected? 20. The Brentano-Stevens Law in psychology models the way that a subject reacts to a stimulus. It states that if represents the reaction to an amount of stimulus, then the relative rates of increase are proportional: where is a positive constant. Find as a function of . 21. The transport of a substance across a capillary wall in lung physiology has been modeled by the differential equation where is the hormone concentration in the bloodstream, is time, is the maximum transport rate, is the volume of the capillary, and is a positive constant that measures the afﬁn-ity between the hormones and the enzymes that assist the process. Solve this differential equation to ﬁnd a relationship between and . 22. Populations of birds and insects are modeled by the equations (a) Which of the variables, or , represents the bird pop-ulation and which represents the insect population? Explain. (b) Find the equilibrium solutions and explain their signiﬁcance. (c) Find an expression for . (d) The direction ﬁeld for the differential equation in part (c) is shown. Use it to sketch the phase trajectory corre-0 x y 20000 40000 100 200 300 60000 400 dydx y x dy dt 0.2y 0.000008xy dx dt 0.4x 0.002xy t h k V R t h dh dt R V h k h S R k 1 R dR dt k S dS dt S R 80% 9–11 Solve the initial-value problem. 9. , 10. , 11. , ; 12. Solve the initial-value problem , , and graph the solution. 13–14 Find the orthogonal trajectories of the family of curves. 13. 14. 15. (a) Write the solution of the initial-value problem and use it to ﬁnd the population when . (b) When does the population reach 1200? 16. (a) The population of the world was 5.28 billion in 1990 and 6.07 billion in 2000. Find an exponential model for these data and use the model to predict the world population in the year 2020. (b) According to the model in part (a), when will the world population exceed 10 billion? (c) Use the data in part (a) to ﬁnd a logistic model for the population. Assume a carrying capacity of 100 billion. Then use the logistic model to predict the population in 2020. Compare with your prediction from the exponential model. (d) According to the logistic model, when will the world pop-ulation exceed 10 billion? Compare with your prediction in part (b). 17. The von Bertalanffy growth model is used to predict the length of a ﬁsh over a period of time. If is the largest length for a species, then the hypothesis is that the rate of growth in length is proportional to , the length yet to be achieved. (a) Formulate and solve a differential equation to ﬁnd an expression for . (b) For the North Sea haddock it has been determined that , cm, and the constant of propor-tionality is . What does the expression for become with these data? 18. A tank contains 100 L of pure water. Brine that contains 0.1 kg of salt per liter enters the tank at a rate of 10 Lmin. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank after 6 minutes? 19. One model for the spread of an epidemic is that the rate of spread is jointly proportional to the number of infected Lt 0.2 L0 10 L 53 cm Lt L L L Lt t 20 P0 100 dP dt 0.1P 1 P 2000 y e kx y ke x y0 1 y 3x 2e y y1 2 xy y x ln x y0 0 1 cos xy 1 eysin x r0 5 dr dt 2tr r 616 \|\|\|\| CHAPTER 9 DIFFERENTIAL EQUATIONS (d) Sketch graphs of the bird and insect populations as func-tions of time. 24. Barbara weighs 60 kg and is on a diet of 1600 calories per day, of which 850 are used automatically by basal metabolism. She spends about 15 calkgday times her weight doing exer-cise. If 1 kg of fat contains 10,000 cal and we assume that the storage of calories in the form of fat is efﬁcient, formulate a differential equation and solve it to ﬁnd her weight as a function of time. Does her weight ultimately approach an equilibrium weight? 25. When a ﬂexible cable of uniform density is suspended between two ﬁxed points and hangs of its own weight, the shape of the cable must satisfy a differential equa-tion of the form where is a positive constant. Consider the cable shown in the ﬁgure. (a) Let in the differential equation. Solve the resulting ﬁrst-order differential equation (in ), and then integrate to ﬁnd . (b) Determine the length of the cable. x b 0 y _b (0, a) (b, h) (_b, h) y z z dydx k d 2y dx 2 k1 dy dx 2 y f x 100% sponding to initial populations of 100 birds and 40,000 insects. Then use the phase trajectory to describe how both populations change. (e) Use part (d) to make rough sketches of the bird and insect populations as functions of time. How are these graphs related to each other? 23. Suppose the model of Exercise 22 is replaced by the equations (a) According to these equations, what happens to the insect population in the absence of birds? (b) Find the equilibrium solutions and explain their signiﬁcance. (c) The ﬁgure shows the phase trajectory that starts with 100 birds and 40,000 insects. Describe what eventually happens to the bird and insect populations. x y 15000 100 45000 25000 35000 120 140 160 180 200 220 240 260 dy dt 0.2y 0.000008xy dx dt 0.4x1 0.000005x 0.002xy CHAPTER 9 REVIEW \|\|\|\| 617 P R O B L E M S P L U S 618 1. Find all functions such that is continuous and 2. A student forgot the Product Rule for differentiation and made the mistake of thinking that . However, he was lucky and got the correct answer. The function that he used was and the domain of his problem was the interval . What was the function ? 3. Let be a function with the property that , , and for all real numbers and . Show that for all and deduce that . 4. Find all functions that satisfy the equation 5. Find the curve such that , , , and the area under the graph of from to is proportional to the power of . 6. A subtangent is a portion of the -axis that lies directly beneath the segment of a tangent line from the point of contact to the -axis. Find the curves that pass through the point and whose subtangents all have length . 7. A peach pie is removed from the oven at 5:00 PM. At that time it is piping hot, . At 5:10 PM its temperature is ; at 5:20 PM it is . What is the temperature of the room? 8. Snow began to fall during the morning of February 2 and continued steadily into the after-noon. At noon a snowplow began removing snow from a road at a constant rate. The plow traveled 6 km from noon to 1 PM but only 3 km from 1 PM to 2 PM. When did the snow begin to fall? [Hints: To get started, let be the time measured in hours after noon; let be the distance traveled by the plow at time ; then the speed of the plow is . Let be the num-ber of hours before noon that it began to snow. Find an expression for the height of the snow at time . Then use the given information that the rate of removal (in ) is constant.] 9. A dog sees a rabbit running in a straight line across an open ﬁeld and gives chase. In a rectan-gular coordinate system (as shown in the ﬁgure), assume: (i) The rabbit is at the origin and the dog is at the point at the instant the dog ﬁrst sees the rabbit. (ii) The rabbit runs up the -axis and the dog always runs straight for the rabbit. (iii) The dog runs at the same speed as the rabbit. (a) Show that the dog’s path is the graph of the function , where satisﬁes the differ-ential equation (b) Determine the solution of the equation in part (a) that satisﬁes the initial conditions when . [Hint: Let in the differential equation and solve the resulting ﬁrst-order equation to ﬁnd ; then integrate to ﬁnd .] (c) Does the dog ever catch the rabbit? y z z z dydx x L y y 0 x d 2y dx 2 1 dy dx 2 y y f x y L, 0 m3h R t b dxdt t xt t 65 C 80 C 100 C c c, 1 x x f x n 1st x 0 f f 1 1 f 0 0 f x 0 y f x y f x dxy 1 f x dx 1 f f x e x x f x f x b a f a b f af b f 0 1 f 0 1 f t ( 1 2, ) f x e x2 f ft f t for all real x [ f x]2 100 y x 0 [ f t]2 [ f t]2 dt f f FIGURE FOR PROBLEM 9 (L, 0) (x, y) x 0 y P R O B L E M S P L U S 619 10. (a) Suppose that the dog in Problem 9 runs twice as fast as the rabbit. Find a differential equation for the path of the dog. Then solve it to ﬁnd the point where the dog catches the rabbit. (b) Suppose the dog runs half as fast as the rabbit. How close does the dog get to the rabbit? What are their positions when they are closest? 11. A planning engineer for a new alum plant must present some estimates to his company regard-ing the capacity of a silo designed to contain bauxite ore until it is processed into alum. The ore resembles pink talcum powder and is poured from a conveyor at the top of the silo. The silo is a cylinder 100 ft high with a radius of 200 ft. The conveyor carries and the ore maintains a conical shape whose radius is 1.5 times its height. (a) If, at a certain time , the pile is 60 ft high, how long will it take for the pile to reach the top of the silo? (b) Management wants to know how much room will be left in the ﬂoor area of the silo when the pile is 60 ft high. How fast is the ﬂoor area of the pile growing at that height? (c) Suppose a loader starts removing the ore at the rate of when the height of the pile reaches 90 ft. Suppose, also, that the pile continues to maintain its shape. How long will it take for the pile to reach the top of the silo under these conditions? 12. Find the curve that passes through the point and has the property that if the tangent line is drawn at any point on the curve, then the part of the tangent line that lies in the ﬁrst quadrant is bisected at . 13. Recall that the normal line to a curve at a point on the curve is the line that passes through and is perpendicular to the tangent line at . Find the curve that passes through the point and has the property that if the normal line is drawn at any point on the curve, then the -intercept of the normal line is always 6. 14. Find all curves with the property that if the normal line is drawn at any point on the curve, then the part of the normal line between and the -axis is bisected by the -axis. y x P P y 3, 2 P P P P P 3, 2 20,000 ft3 h t 60,000 ft3 h 620 So far we have described plane curves by giving as a function of or as a function of or by giving a relation between and that deﬁnes implicitly as a function of . In this chapter we discuss two new methods for describing curves. Some curves, such as the cycloid, are best handled when both and are given in terms of a third variable called a parameter . Other curves, such as the cardioid, have their most convenient description when we use a new coordinate system, called the polar coordinate system. x ft, y tt t y x fx, y 0 x y y x x ty y x y fx x y Parametric equations and polar coordinates enable us to describe a great variety of new curves—some practical, some beautiful, some fanciful, some strange. PARAMETRIC EQUATIONS AND POLAR COORDINATES 10 CURVES DEFINED BY PARAMETRIC EQUATIONS Imagine that a particle moves along the curve C shown in Figure 1. It is impossible to describe C by an equation of the form because C fails the Vertical Line Test. But the x- and y-coordinates of the particle are functions of time and so we can write and . Such a pair of equations is often a convenient way of describing a curve and gives rise to the following deﬁnition. Suppose that and are both given as functions of a third variable (called a param-eter) by the equations (called parametric equations). Each value of determines a point , which we can plot in a coordinate plane. As varies, the point varies and traces out a curve , which we call a parametric curve. The parameter t does not necessarily repre-sent time and, in fact, we could use a letter other than t for the parameter. But in many applications of parametric curves, t does denote time and therefore we can interpret as the position of a particle at time t. EXAMPLE 1 Sketch and identify the curve deﬁned by the parametric equations SOLUTION Each value of gives a point on the curve, as shown in the table. For instance, if , then , and so the corresponding point is . In Figure 2 we plot the points determined by several values of the parameter and we join them to produce a curve. A particle whose position is given by the parametric equations moves along the curve in the direction of the arrows as increases. Notice that the consecutive points marked on the curve appear at equal time intervals but not at equal distances. That is because the particle slows down and then speeds up as increases. It appears from Figure 2 that the curve traced out by the particle may be a parabola. This can be conﬁrmed by eliminating the parameter as follows. We obtain from the second equation and substitute into the ﬁrst equation. This gives and so the curve represented by the given parametric equations is the parabola . M x y 2 4y 3 x t 2 2t y 12 2y 1 y 2 4y 3 t y 1 t t t FIGURE 2 0 t=0 t=1 t=2 t=3 t=4 t=_1 t=_2 (0, 1) y x 8 x, y 0, 1 y 1 x 0 t 0 t y t 1 x t 2 2t x, y ft, tt C x, y ft, tt t x, y t y tt x ft t y x y tt x ft y fx 10.1 t x y 2 8 1 1 3 0 0 0 1 1 1 2 2 0 3 3 3 4 4 8 5 N This equation in and describes where the particle has been, but it doesn’t tell us when the particle was at a particular point. The parametric equations have an advantage––they tell us when the particle was at a point. They also indi-cate the direction of the motion. y x C 0 (x, y)={f(t), g(t)} FIGURE 1 y x 621 No restriction was placed on the parameter in Example 1, so we assumed that t could be any real number. But sometimes we restrict t to lie in a ﬁnite interval. For instance, the parametric curve shown in Figure 3 is the part of the parabola in Example 1 that starts at the point and ends at the point . The arrowhead indicates the direction in which the curve is traced as increases from 0 to 4. In general, the curve with parametric equations has initial point and terminal point . EXAMPLE 2 What curve is represented by the following parametric equations? SOLUTION If we plot points, it appears that the curve is a circle. We can conﬁrm this impression by eliminating Observe that Thus the point moves on the unit circle . Notice that in this example the parameter can be interpreted as the angle (in radians) shown in Figure 4. As increases from 0 to , the point moves once around the circle in the counterclockwise direction starting from the point . M EXAMPLE 3 What curve is represented by the given parametric equations? SOLUTION Again we have so the parametric equations again represent the unit circle . But as increases from 0 to , the point starts at and moves twice around the circle in the clockwise direction as indicated in Figure 5. M Examples 2 and 3 show that different sets of parametric equations can represent the same curve. Thus we distinguish between a curve, which is a set of points, and a parametric curve, in which the points are traced in a particular way. 0, 1 x, y sin 2t, cos 2t 2 t x 2 y 2 1 x 2 y 2 sin2 2t cos2 2t 1 0 t 2 y cos 2t x sin 2t FIGURE 4 3π 2 t= π 2 t= 0 t t=0 (1, 0) (cos t, sin t) t=2π t=π x y 1, 0 x, y cos t, sin t 2 t t x 2 y 2 1 x, y x 2 y 2 cos2t sin2t 1 t. 0 t 2 y sin t x cos t V fb, tb fa, ta a t b y tt x ft t 8, 5 0, 1 0 t 4 y t 1 x t 2 2t t 622 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES FIGURE 3 0 (8, 5) (0, 1) y x 0 t=0, π, 2π FIGURE 5 x y (0, 1) EXAMPLE 4 Find parametric equations for the circle with center and radius . SOLUTION If we take the equations of the unit circle in Example 2 and multiply the expres-sions for and by , we get , . You can verify that these equations represent a circle with radius and center the origin traced counterclockwise. We now shift units in the -direction and units in the -direction and obtain parametric equa-tions of the circle (Figure 6) with center and radius : M EXAMPLE 5 Sketch the curve with parametric equations , . SOLUTION Observe that and so the point moves on the parabola . But note also that, since , we have , so the para-metric equations represent only the part of the parabola for which . Since is periodic, the point moves back and forth inﬁnitely often along the parabola from to . (See Figure 7.) M y=sin 2t x=cos t y=sin 2t x=cos t FIGURE 8 t x y t y x 1, 1 1, 1 x, y sin t, sin2t sin t 1 x 1 1 x 1 1 sin t 1 y x 2 x, y y sin t2 x 2 y sin2t x sin t V FIGURE 6 x=h+r cos t, y=k+r sin t 0 (h, k) r x y 0 t 2 y k r sin t x h r cos t r h, k y k x h r y r sin t x r cos t r y x r h, k SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS \|\|\|\| 623 Module 10.1A gives an animation of the relationship between motion along a parametric curve , and motion along the graphs of and as functions of . Clicking on TRIG gives you the family of parametric curves If you choose and click on animate, you will see how the graphs of and relate to the circle in Example 2. If you choose , , you will see graphs as in Figure 8. By clicking on animate or moving the -slider to the right, you can see from the color coding how motion along the graphs of and corresponds to motion along the para-metric curve, which is called a Lissajous ﬁgure. y sin 2t x cos t t d 2 a b c 1 y sin t x cos t a b c d 1 y c sin dt x a cos bt t t f y tt x ft TEC FIGURE 7 0 (1, 1) (_1, 1) x y GRAPHING DEVICES Most graphing calculators and computer graphing programs can be used to graph curves deﬁned by parametric equations. In fact, it’s instructive to watch a parametric curve being drawn by a graphing calculator because the points are plotted in order as the correspon-ding parameter values increase. EXAMPLE 6 Use a graphing device to graph the curve . SOLUTION If we let the parameter be , then we have the equations Using these parametric equations to graph the curve, we obtain Figure 9. It would be possible to solve the given equation for y as four functions of x and graph them individually, but the parametric equations provide a much easier method. M In general, if we need to graph an equation of the form , we can use the para-metric equations Notice also that curves with equations (the ones we are most familiar with— graphs of functions) can also be regarded as curves with parametric equations Graphing devices are particularly useful when sketching complicated curves. For instance, the curves shown in Figures 10, 11, and 12 would be virtually impossible to pro-duce by hand. One of the most important uses of parametric curves is in computer-aided design (CAD). In the Laboratory Project after Section 10.2 we will investigate special parametric curves, called Bézier curves, that are used extensively in manufacturing, especially in the automotive industry. These curves are also employed in specifying the shapes of letters and other symbols in laser printers. THE CYCLOID EXAMPLE 7 The curve traced out by a point on the circumference of a circle as the circle rolls along a straight line is called a cycloid (see Figure 13). If the circle has radius and rolls along the -axis and if one position of is the origin, ﬁnd parametric equations for the cycloid. P x r P 8 _8 _6.5 6.5 FIGURE 10 x=t+2 sin 2t y=t+2 cos 5t 2.5 _2.5 2.5 FIGURE 11 x=1.5 cos t-cos 30t y=1.5 sin t-sin 30t _2.5 1 _1 1 FIGURE 12 x=sin(t+cos 100t) y=cos(t+sin 100t) _1 y ft x t y fx y t x tt x ty x y 4 3y 2 y t x t 4 3t 2 t y x y 4 3y 2 624 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES An animation in Module 10.1B shows how the cycloid is formed as the circle moves. TEC 3 _3 _3 3 FIGURE 9 SOLUTION We choose as parameter the angle of rotation of the circle when is at the origin). Suppose the circle has rotated through radians. Because the circle has been in contact with the line, we see from Figure 14 that the distance it has rolled from the origin is Therefore the center of the circle is . Let the coordinates of be . Then from Figure 14 we see that Therefore parametric equations of the cycloid are One arch of the cycloid comes from one rotation of the circle and so is described by . Although Equations 1 were derived from Figure 14, which illustrates the case where , it can be seen that these equations are still valid for other values of (see Exercise 39). Although it is possible to eliminate the parameter from Equations 1, the resulting Cartesian equation in and is very complicated and not as convenient to work with as the parametric equations. M One of the ﬁrst people to study the cycloid was Galileo, who proposed that bridges be built in the shape of cycloids and who tried to ﬁnd the area under one arch of a cycloid. Later this curve arose in connection with the brachistochrone problem: Find the curve along which a particle will slide in the shortest time (under the inﬂuence of gravity) from a point to a lower point not directly beneath . The Swiss mathematician John Bernoulli, who posed this problem in 1696, showed that among all possible curves that join to , as in Figure 15, the particle will take the least time sliding from to if the curve is part of an inverted arch of a cycloid. The Dutch physicist Huygens had already shown that the cycloid is also the solution to the tautochrone problem; that is, no matter where a particle is placed on an inverted cycloid, it takes the same time to slide to the bottom (see Figure 16). Huygens proposed that pendulum clocks (which he invented) swing in cycloidal arcs because then the pendu-lum takes the same time to make a complete oscillation whether it swings through a wide or a small arc. FAMILIES OF PARAMETRIC CURVES EXAMPLE 8 Investigate the family of curves with parametric equations What do these curves have in common? How does the shape change as increases? a y a tan t sin t x a cos t V P B A B A A B A y x 0 2 0 2 y r1 cos x r sin 1 y TC QC r r cos r1 cos x OT PQ r r sin r sin x, y P Cr, r OT arc PT r P 0 FIGURE 13 P P P SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS \|\|\|\| 625 FIGURE 15 A B cycloid FIGURE 14 x O y T C(r¨, r) r ¨ x y r¨ P Q P P P P P FIGURE 16 SOLUTION We use a graphing device to produce the graphs for the cases , , , , , , , and shown in Figure 17. Notice that all of these curves (except the case ) have two branches, and both branches approach the vertical asymptote as approaches from the left or right. When , both branches are smooth; but when reaches , the right branch acquires a sharp point, called a cusp. For between and 0 the cusp turns into a loop, which becomes larger as approaches 0. When , both branches come together and form a circle (see Example 2). For between 0 and 1, the left branch has a loop, which shrinks to become a cusp when . For , the branches become smooth again, and as increases further, they become less curved. Notice that the curves with posi-tive are reﬂections about the -axis of the corresponding curves with negative. These curves are called conchoids of Nicomedes after the ancient Greek scholar Nicomedes. He called them conchoids because the shape of their outer branches resembles that of a conch shell or mussel shell. M a y a a a 1 a 1 a a 0 a 1 a 1 a a 1 a=_2 a=_1 a=_0.5 a=_0.2 a=2 a=1 a=0.5 a=0 a x x a a 0 2 1 0.5 0 0.2 0.5 1 a 2 8. , , 10. , 11–18 (a) Eliminate the parameter to ﬁnd a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. 11. , , 12. , , , , 14. , 15. , 16. , , 17. , y cosh t x sinh t t 1 y st x ln t y t 1 x e 2t y e 2t x et 1 0 t 2 y csc t x sin t 13. 2 2 y 5 sin x 4 cos 0 y cos x sin y t 3 x t 2 y 1 t x st 9. y 2 t 2 x 1 3t 1–4 Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as increases. 1. , , 2. , , 3. , , , , 5–10 (a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases. (b) Eliminate the parameter to ﬁnd a Cartesian equation of the curve. 5. , 6. , , 7. , , 3 t 4 y 5 2t x t 2 2 2 t 3 y 5 2t x 1 t y 2t 1 x 3t 5 2 t 2 y e t t x et t 4. t y t 2 x 5 sin t 0 t 2 y t cos t x 2 cos t 0 t 5 y t 2 4t x 1 st t EXERCISES 10.1 626 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES FIGURE 17 Members of the family x=a+cos t, y=a tan t+sin t, all graphed in the viewing rectangle _4, 4 by _4, 4 25–27 Use the graphs of and to sketch the parametric curve , . Indicate with arrows the direction in which the curve is traced as increases. 25. 26. 27. 28. Match the parametric equations with the graphs labeled I-VI. Give reasons for your choices. (Do not use a graphing device.) (a) , (b) , (c) , (d) , (e) , (f) , ; 29. Graph the curve . ; 30. Graph the curves and and ﬁnd their points of intersection correct to one decimal place. x yy 12 y x 5 x y 3y 3 y 5 x y x y x y x y x y x y I II III IV V VI y cos 2t 4 t 2 x sin 2t 4 t 2 y t 2 cos 3t x t sin 4t y sin 2t x cos 5t y sint sin 2t x sin 2t y st x t 2 2t y t 2 x t 4 t 1 t y 1 1 t x 1 1 t x 1 1 t y 1 1 t x _1 1 t y 1 1 t y tt x ft y tt x ft 18. , 19–22 Describe the motion of a particle with position as varies in the given interval. 19. , , 20. , , , , 22. , , 23. Suppose a curve is given by the parametric equations , , where the range of is and the range of is . What can you say about the curve? 24. Match the graphs of the parametric equations and in (a)–(d) with the parametric curves labeled I–IV. Give reasons for your choices. (c) III t 2 2 y x t 2 2 (d) IV t 2 2 y x t 2 2 y x 2 2 1 y x 1 2 t x 2 1 1 t y 1 1 y x 2 2 (a) I (b) II x t 2 1 t 2 1 y y x 2 2 y tt x ft 2, 3 t 1, 4 f y tt x ft 2 t 2 y cos2t x sin t t 5 y 2 cos t x 5 sin t 21. 0 t 32 y 4 cos t x 2 sin t 2 t 32 y 1 2 sin t x 3 2 cos t t x, y y 5 sinh t x 2 cosh t SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS \|\|\|\| 627 If and are ﬁxed numbers, ﬁnd parametric equations for the curve that consists of all possible positions of the point in the ﬁgure, using the angle as the parameter. Then elimi-nate the parameter and identify the curve. 42. If and are ﬁxed numbers, ﬁnd parametric equations for the curve that consists of all possible positions of the point in the ﬁgure, using the angle as the parameter. The line seg-ment is tangent to the larger circle. 43. A curve, called a witch of Maria Agnesi, consists of all pos-sible positions of the point in the ﬁgure. Show that para-metric equations for this curve can be written as Sketch the curve. 44. (a) Find parametric equations for the set of all points as shown in the ﬁgure such that . (This curve is called the cissoid of Diocles after the Greek scholar Diocles, who introduced the cissoid as a graphical method for constructing the edge of a cube whose volume is twice that of a given cube.) OP AB P O x a A P y=2a ¨ y C y 2a sin2 x 2a cot P O x y ¨ a b A B P AB P b a O y x ¨ a b P P b a 41. (a) Show that the parametric equations where , describe the line segment that joins the points and . (b) Find parametric equations to represent the line segment from to . ; 32. Use a graphing device and the result of Exercise 31(a) to draw the triangle with vertices , , and . Find parametric equations for the path of a particle that moves along the circle in the manner described. (a) Once around clockwise, starting at (b) Three times around counterclockwise, starting at (c) Halfway around counterclockwise, starting at ; (a) Find parametric equations for the ellipse . [Hint: Modify the equations of the circle in Example 2.] (b) Use these parametric equations to graph the ellipse when and b 1, 2, 4, and 8. (c) How does the shape of the ellipse change as b varies? ; 35–36 Use a graphing calculator or computer to reproduce the picture. 35. 36. 37–38 Compare the curves represented by the parametric equa-tions. How do they differ? 37. (a) , (b) , (c) , 38. (a) , (b) , (c) , 39. Derive Equations 1 for the case . 40. Let be a point at a distance from the center of a circle of radius . The curve traced out by as the circle rolls along a straight line is called a trochoid. (Think of the motion of a point on a spoke of a bicycle wheel.) The cycloid is the spe-cial case of a trochoid with . Using the same parameter as for the cycloid and, assuming the line is the -axis and when is at one of its lowest points, show that parametric equations of the trochoid are Sketch the trochoid for the cases and . d r d r y r d cos x r d sin P 0 x d r P r d P 2 y e2t x e t y sec2t x cos t y t 2 x t y e2t x e3t y t 4 x t 6 y t 2 x t 3 0 y x 2 3 8 4 0 2 y x 2 a 3 x 2a 2 y 2b 2 1 34. 0, 3 2, 1 2, 1 x 2 y 12 4 33. C1, 5 B4, 2 A1, 1 3, 1 2, 7 P2x 2, y2 P1x1, y1 0 t 1 y y1 y2 y1t x x1 x 2 x1t 31. 628 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES LABORATORY PROJECT RUNNING CIRCLES AROUND CIRCLES \|\|\|\| 629 given by the parametric equations where is the acceleration due to gravity ( ms ). (a) If a gun is ﬁred with and ms, when will the bullet hit the ground? How far from the gun will it hit the ground? What is the maximum height reached by the bullet? ; (b) Use a graphing device to check your answers to part (a). Then graph the path of the projectile for several other values of the angle to see where it hits the ground. Summarize your ﬁndings. (c) Show that the path is parabolic by eliminating the parameter. ; Investigate the family of curves deﬁned by the parametric equations , . How does the shape change as increases? Illustrate by graphing several members of the family. ; 48. The swallowtail catastrophe curves are deﬁned by the para-metric equations , . Graph several of these curves. What features do the curves have in common? How do they change when increases? ; The curves with equations , are called Lissajous ﬁgures. Investigate how these curves vary when , , and vary. (Take to be a positive integer.) ; 50. Investigate the family of curves deﬁned by the parametric equations , , where . Start by letting be a positive integer and see what happens to the shape as increases. Then explore some of the possibilities that occur when is a fraction. c c c c 0 y sin t sin ct x cos t n n b a y b cos t x a sin nt 49. c y ct 2 3t 4 x 2ct 4t 3 c y t 3 ct x t 2 47. v0 500 30 2 9.8 t y v0 sin t 1 2tt 2 x v0 cos t (b) Use the geometric description of the curve to draw a rough sketch of the curve by hand. Check your work by using the parametric equations to graph the curve. ; 45. Suppose that the position of one particle at time is given by and the position of a second particle is given by (a) Graph the paths of both particles. How many points of intersection are there? (b) Are any of these points of intersection collision points? In other words, are the particles ever at the same place at the same time? If so, ﬁnd the collision points. (c) Describe what happens if the path of the second particle is given by 46. If a projectile is ﬁred with an initial velocity of meters per second at an angle above the horizontal and air resistance is assumed to be negligible, then its position after seconds is t v0 x 2 3 cos t y2 1 sin t 0 t 2 0 t 2 y2 1 sin t x 2 3 cos t 0 t 2 y1 2 cos t x1 3 sin t t x O y A P x=2a B a In this project we investigate families of curves, called hypocycloids and epicycloids, that are generated by the motion of a point on a circle that rolls inside or outside another circle. 1. A hypocycloid is a curve traced out by a ﬁxed point P on a circle C of radius b as C rolls on the inside of a circle with center O and radius a. Show that if the initial position of P is and the parameter is chosen as in the ﬁgure, then parametric equations of the hypocycloid are 2. Use a graphing device (or the interactive graphic in TEC Module 10.1B) to draw the graphs of hypocycloids with a a positive integer and b 1. How does the value of a affect the graph? Show that if we take a 4, then the parametric equations of the hypocycloid reduce to This curve is called a hypocycloid of four cusps, or an astroid. y 4 sin3 x 4 cos3 y a b sin b sin a b b x a b cos b cos a b b a, 0 ; RUNNING CIRCLES AROUND CIRCLES L A B O R AT O R Y P R O J E C T x O y a C P b (a, 0) ¨ A Look at Module 10.1B to see how hypocycloids and epicycloids are formed by the motion of rolling circles. TEC 630 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 3. Now try b 1 and , a fraction where n and d have no common factor. First let n 1 and try to determine graphically the effect of the denominator d on the shape of the graph. Then let n vary while keeping d constant. What happens when ? 4. What happens if and is irrational? Experiment with an irrational number like or . Take larger and larger values for and speculate on what would happen if we were to graph the hypocycloid for all real values of . 5. If the circle rolls on the outside of the ﬁxed circle, the curve traced out by is called an epicycloid. Find parametric equations for the epicycloid. 6. Investigate the possible shapes for epicycloids. Use methods similar to Problems 2–4. P C e 2 s2 a b 1 n d 1 a nd CALCULUS WITH PARAMETRIC CURVES Having seen how to represent curves by parametric equations, we now apply the methods of calculus to these parametric curves. In particular, we solve problems involving tangents, area, arc length, and surface area. TANGENTS In the preceding section we saw that some curves deﬁned by parametric equations and can also be expressed, by eliminating the parameter, in the form . (See Exercise 67 for general conditions under which this is possible.) If we substitute and in the equation , we get and so, if , , and are differentiable, the Chain Rule gives If , we can solve for : Since the slope of the tangent to the curve at is , Equation 1 enables us to ﬁnd tangents to parametric curves without having to eliminate the parameter. Using Leibniz notation, we can rewrite Equation 1 in an easily remembered form: It can be seen from Equation 2 that the curve has a horizontal tangent when (provided that ) and it has a vertical tangent when (provided that ). This information is useful for sketching parametric curves. dydt 0 dxdt 0 dxdt 0 dydt 0 dx dt 0 if dy dx dy dt dx dt 2 Fx x, Fx y Fx Fx tt ft 1 Fx ft 0 tt F ftft Fxft f F t tt F ft y Fx y tt x ft y Fx y tt x ft 10.2 N If we think of a parametric curve as being traced out by a moving particle, then and are the vertical and horizontal velocities of the particle and Formula 2 says that the slope of the tangent is the ratio of these velocities. dxdt dydt As we know from Chapter 4, it is also useful to consider . This can be found by replacing y by dydx in Equation 2: EXAMPLE 1 A curve is deﬁned by the parametric equations , (a) Show that has two tangents at the point (3, 0) and ﬁnd their equations. (b) Find the points on where the tangent is horizontal or vertical. (c) Determine where the curve is concave upward or downward. (d) Sketch the curve. SOLUTION (a) Notice that when or . Therefore the point on arises from two values of the parameter, and . This indicates that crosses itself at . Since the slope of the tangent when is , so the equa-tions of the tangents at are (b) has a horizontal tangent when , that is, when and . Since , this happens when , that is, . The corresponding points on are and (1, 2). has a vertical tangent when , that is, . (Note that there.) The corresponding point on is (0, 0). (c) To determine concavity we calculate the second derivative: Thus the curve is concave upward when and concave downward when . (d) Using the information from parts (b) and (c), we sketch in Figure 1. M EXAMPLE 2 (a) Find the tangent to the cycloid , at the point where . (See Example 7 in Section 10.1.) (b) At what points is the tangent horizontal? When is it vertical? SOLUTION (a) The slope of the tangent line is dy dx dyd dxd r sin r1 cos sin 1 cos 3 y r1 cos x r sin V C t 0 t 0 d 2y dx 2 d dt dy dx dx dt 3 2 1 1 t 2 2t 3t 2 1 4t 3 C dydt 0 t 0 dxdt 2t 0 C 1, 2 C t 1 t 2 1 dydt 3t 2 3 dxdt 0 dydt 0 dydx 0 C y s3 x 3 and y s3 x 3 3, 0 dydx 6(2s3 ) s3 t s3 dy dx dydt dxdt 3t 2 3 2t 3 2 t 1 t 3, 0 C t s3 t s3 C 3, 0 t s3 t 0 y t 3 3t tt 2 3 0 C C y t 3 3t. x t 2 C d 2y dx 2 d dx dy dx d dt dy dx dx dt d 2ydx 2 \| Note that d 2y dx2 d 2y dt 2 d 2x dt 2 SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES \|\|\|\| 631 0 y x (3, 0) (1, _2) (1, 2) t=1 t=_1 y=œ„ 3(x-3) y= œ„ 3(x-3) FIGURE 1 When , we have and Therefore the slope of the tangent is and its equation is The tangent is sketched in Figure 2. (b) The tangent is horizontal when , which occurs when and , that is, , an integer. The corresponding point on the cycloid is . When , both and are 0. It appears from the graph that there are vertical tangents at those points. We can verify this by using l’Hospital’s Rule as follows: A similar computation shows that as , so indeed there are verti-cal tangents when , that is, when . M AREAS We know that the area under a curve from to is , where . If the curve is traced out once by the parametric equations and , , then we can calculate an area formula by using the Substitution Rule for Deﬁnite Integrals as follows: EXAMPLE 3 Find the area under one arch of the cycloid (See Figure 3.) y r1 cos x r sin V or y tt ft dt A y b a y dx y tt ft dt t y tt x ft Fx 0 A xb a Fx dx b a y Fx x 2nr 2n l 2n dydx l lim l 2n dy dx lim l 2n sin 1 cos lim l 2n cos sin dyd dxd 2n 2n 1r, 2r n 2n 1 1 cos 0 sin 0 dydx 0 FIGURE 2 0 y x 2πr 4πr (πr, 2r) (_πr, 2r) (3πr, 2r) (5πr, 2r) π 3 ¨= s3 x y r s3 2 or y r 2 s3 x r 3 rs3 2 s3 dy dx sin3 1 cos3 s3 2 1 1 2 s3 y r1 cos 3 r 2 x r 3 sin 3 r 3 s3 2 3 632 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES N The limits of integration for are found as usual with the Substitution Rule. When is either . When is the remaining value. x b, t or t x a, t SOLUTION One arch of the cycloid is given by . Using the Substitution Rule with and , we have M ARC LENGTH We already know how to ﬁnd the length of a curve given in the form , . Formula 8.1.3 says that if is continuous, then Suppose that can also be described by the parametric equations and , , where . This means that is traversed once, from left to right, as increases from to and , . Putting Formula 2 into Formula 3 and using the Substitution Rule, we obtain Since , we have Even if can’t be expressed in the form , Formula 4 is still valid but we obtain it by polygonal approximations. We divide the parameter interval into n subintervals of equal width . If , , , . . . , are the endpoints of these subintervals, then and are the coordinates of points that lie on and the polygon with ver-tices , , . . . , approximates . (See Figure 4.) As in Section 8.1, we deﬁne the length of to be the limit of the lengths of these approximating polygons as : The Mean Value Theorem, when applied to on the interval , gives a number in such that If we let and , this equation becomes xi fti t yi yi yi1 xi xi xi1 fti fti1 ftiti ti1 ti1, ti ti ti1, ti f L lim nl n i1 Pi1Pi n l C L C Pn P1 P0 C Pixi, yi yi tti xi fti tn t2 t1 t0 t , y Fx C L y dx dt 2 dy dt 2 dt 4 dxdt 0 L y b a 1 dy dx 2 dx y 1 dydt dxdt 2 dx dt dt f b f a t C dxdt ft 0 t y tt x ft C L y b a 1 dy dx 2 dx 3 F a x b y Fx C L r 2( 3 2 2) 3r 2 r 2[ 3 2 2 sin 1 4 sin 2]0 2 r 2 y 2 0 [1 2 cos 1 21 cos 2] d r 2 y 2 0 1 cos 2 d r 2 y 2 0 1 2 cos cos2 d A y 2r 0 y dx y 2 0 r1 cos r1 cos d dx r1 cos d y r1 cos 0 2 SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES \|\|\|\| 633 N The result of Example 3 says that the area under one arch of the cycloid is three times the area of the rolling circle that generates the cycloid (see Example 7 in Section 10.1). Galileo guessed this result but it was ﬁrst proved by the French mathematician Roberval and the Italian mathematician Torricelli. FIGURE 3 0 y x 2πr 0 y x P¸ P¡ P™ Pi _1 Pi Pn C FIGURE 4 Similarly, when applied to , the Mean Value Theorem gives a number in such that Therefore and so The sum in (5) resembles a Riemann sum for the function but it is not exactly a Riemann sum because in general. Nevertheless, if and are contin-uous, it can be shown that the limit in (5) is the same as if and were equal, namely, Thus, using Leibniz notation, we have the following result, which has the same form as Formula (4). THEOREM If a curve is described by the parametric equations , , , where and are continuous on and is traversed exactly once as increases from to , then the length of is Notice that the formula in Theorem 6 is consistent with the general formulas and of Section 8.1. EXAMPLE 4 If we use the representation of the unit circle given in Example 2 in Sec-tion 10.1, then and , so Theorem 6 gives as expected. If, on the other hand, we use the representation given in Example 3 in Sec-tion 10.1, then , , and the integral in Theorem 6 gives y 2 0 dx dt 2 dy dt 2 dt y 2 0 s4 cos2 2t 4 sin2 2t dt y 2 0 2 dt 4 dydt 2 sin 2t dxdt 2 cos 2t 0 t 2 y cos 2t x sin 2t y 2 0 dt 2 L y 2 0 dx dt 2 dy dt 2 dt y 2 0 ssin2t cos2t dt dydt cos t dxdt sin t 0 t 2 y sin t x cos t ds2 dx2 dy2 L x ds L y dx dt 2 dy dt 2 dt C t C , t f t y tt x ft C 6 L y s ft 2 tt 2 dt ti ti t f ti ti s ft 2 tt 2 L lim n l n i1 s fti 2 tti 2 t 5 s fti 2 tti 2 t s fti t 2 tti t 2 Pi1Pi sxi2 yi2 yi tti t ti1, ti ti t 634 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES \| Notice that the integral gives twice the arc length of the circle because as increases from 0 to , the point traverses the circle twice. In general, when ﬁnding the length of a curve from a parametric representation, we have to be careful to ensure that is traversed only once as increases from to . M EXAMPLE 5 Find the length of one arch of the cycloid , SOLUTION From Example 3 we see that one arch is described by the parameter interval . Since we have To evaluate this integral we use the identity with , which gives . Since , we have and so . Therefore and so M SURFACE AREA In the same way as for arc length, we can adapt Formula 8.2.5 to obtain a formula for surface area. If the curve given by the parametric equations , , , is rotated about the -axis, where , are continuous and , then the area of the resulting surface is given by The general symbolic formulas and (Formulas 8.2.7 and 8.2.8) are still valid, but for parametric curves we use EXAMPLE 6 Show that the surface area of a sphere of radius is . SOLUTION The sphere is obtained by rotating the semicircle 0 t y r sin t x r cos t 4r 2 r ds dx dt 2 dy dt 2 dt S x 2x ds S x 2y ds S y 2y dx dt 2 dy dt 2 dt 7 tt 0 t f x t y tt x ft 2r 2 2 8r L 2r y 2 0 sin2 d 2r 2 cos2]0 2 s21 cos s4 sin22 2sin2 2 sin2 sin2 0 0 2 0 2 1 cos 2 sin22 2x sin2x 1 21 cos 2x y 2 0 sr 21 2 cos cos2 sin2 d r y 2 0 s21 cos d L y 2 0 dx d 2 dy d 2 d y 2 0 sr 21 cos 2 r 2 sin2 d dy d r sin and dx d r1 cos 0 2 y r1 cos . x r sin V t C C sin 2t, cos 2t 2 t SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES \|\|\|\| 635 N The result of Example 5 says that the length of one arch of a cycloid is eight times the radius of the generating circle (see Figure 5). This was ﬁrst proved in 1658 by Sir Christopher Wren, who later became the architect of St. Paul’s Cathedral in London. FIGURE 5 0 y x 2πr r L=8r about the -axis. Therefore, from Formula 7, we get M 2r 2cos t]0 4r 2 2r 2 y 0 sin t dt 2 y 0 r sin t r dt 2 y 0 r sin t sr 2sin2t cos2t dt S y 0 2r sin t sr sin t2 r cos t2 dt x 636 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 19. , 20. , ; 21. Use a graph to estimate the coordinates of the rightmost point on the curve , . Then use calculus to ﬁnd the exact coordinates. ; 22. Use a graph to estimate the coordinates of the lowest point and the leftmost point on the curve , . Then ﬁnd the exact coordinates. ; 23–24 Graph the curve in a viewing rectangle that displays all the important aspects of the curve. , 24. , Show that the curve , has two tangents at and ﬁnd their equations. Sketch the curve. ; 26. Graph the curve , to discover where it crosses itself. Then ﬁnd equations of both tangents at that point. 27. (a) Find the slope of the tangent line to the trochoid , in terms of . (See Exer-cise 40 in Section 10.1.) (b) Show that if , then the trochoid does not have a vertical tangent. 28. (a) Find the slope of the tangent to the astroid , in terms of . (Astroids are explored in the Laboratory Project on page 629.) (b) At what points is the tangent horizontal or vertical? (c) At what points does the tangent have slope 1 or ? 29. At what points on the curve , does the tangent line have slope ? 30. Find equations of the tangents to the curve , that pass through the point . Use the parametric equations of an ellipse, , , , to ﬁnd the area that it encloses. 0 2 y b sin x a cos 31. 4, 3 y 2t 3 1 x 3t 2 1 1 y 1 4t t 2 x 2t 3 1 y a sin3 x a cos3 d r y r d cos x r d sin y sin t 2 sin 2t x cos t 2 cos 2t 0, 0 y sin t cos t x cos t 25. y 2t 2 t x t 4 4t 3 8t 2 y t 3 t x t 4 2t 3 2t 2 23. y t t 4 x t 4 2t y e t x t t 6 y 2 sin x cos 3 y sin 2 x 2 cos 1–2 Find . 1. , 2. , 3–6 Find an equation of the tangent to the curve at the point corre-sponding to the given value of the parameter. 3. , ; 4. , ; , ; 6. , ; 7–8 Find an equation of the tangent to the curve at the given point by two methods: (a) without eliminating the parameter and (b) by ﬁrst eliminating the parameter. 7. , ; 8. , ; ; 9–10 Find an equation of the tangent(s) to the curve at the given point. Then graph the curve and the tangent(s). 9. , ; 10. , ; 11–16 Find and . For which values of is the curve concave upward? , 12. , 13. , 14. , 15. , , 16. , , 17–20 Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. 17. , 18. , y 2t 3 3t 2 1 x 2t 3 3t 2 12t y t 3 12t x 10 t 2 0 t y cos t x cos 2t 0 t 2 y 3 cos t x 2 sin t y t ln t x t ln t y t et x t e t y t 2 1 x t 3 12t y t 2 t 3 x 4 t 2 11. t d 2ydx 2 dydx 1, 1 y sin t sin 2t x cos t cos 2t 0, 0 y t 2 t x 6 sin t (1, s2) y sec x tan 1, 3 y t 2 2 x 1 ln t 0 y sin cos 2 x cos sin 2 t 1 y t ln t 2 x est 5. t 1 y 1 t 2 x t t1 t 1 y t 3 t x t 4 1 y st et x 1t y t 2 t x t sin t dydx EXERCISES 10.2 49. Use Simpson’s Rule with to estimate the length of the curve , , . 50. In Exercise 43 in Section 10.1 you were asked to derive the parametric equations , for the curve called the witch of Maria Agnesi. Use Simpson’s Rule with to estimate the length of the arc of this curve given by . 51–52 Find the distance traveled by a particle with position as varies in the given time interval. Compare with the length of the curve. 51. , , 52. , , 53. Show that the total length of the ellipse , , , is where is the eccentricity of the ellipse , where . 54. Find the total length of the astroid , , where 55. (a) Graph the epitrochoid with equations What parameter interval gives the complete curve? (b) Use your CAS to ﬁnd the approximate length of this curve. 56. A curve called Cornu’s spiral is deﬁned by the parametric equations where and are the Fresnel functions that were introduced in Chapter 5. (a) Graph this curve. What happens as and as ? (b) Find the length of Cornu’s spiral from the origin to the point with parameter value . 57–58 Set up an integral that represents the area of the surface obtained by rotating the given curve about the -axis. Then use your calculator to ﬁnd the surface area correct to four decimal places. 57. , , 58. , , 0 t 3 y sin 3t x sin2t 0 t 1 y t 2 1e t x 1 te t x t t l t l S C y St y t 0 sinu 22 du x Ct y t 0 cosu 22 du CAS y 11 sin t 4 sin11t2 x 11 cos t 4 cos11t2 CAS a 0. y a sin3 x a cos3 c sa 2 b 2 ) (e ca e L 4a y 2 0 s1 e 2 sin2 d a b 0 y b cos x a sin 0 t 4 y cos t x cos2t 0 t 3 y cos2t x sin2t t x, y 4 2 n 4 y 2a sin2 x 2a cot 6 t 6 y t e t x t e t n 6 32. Find the area enclosed by the curve , and the . 33. Find the area enclosed by the and the curve , . 34. Find the area of the region enclosed by the astroid , . (Astroids are explored in the Labo-ratory Project on page 629.) 35. Find the area under one arch of the trochoid of Exercise 40 in Section 10.1 for the case . 36. Let be the region enclosed by the loop of the curve in Example 1. (a) Find the area of . (b) If is rotated about the -axis, ﬁnd the volume of the resulting solid. (c) Find the centroid of . 37–40 Set up an integral that represents the length of the curve. Then use your calculator to ﬁnd the length correct to four decimal places. 37. , , 38. , , 39. , , 40. , , 41–44 Find the exact length of the curve. , , 42. , , 43. , , 44. , , ; 45–47 Graph the curve and ﬁnd its length. , , 46. , , 47. , , 48. Find the length of the loop of the curve , . y 3t 2 x 3t t 3 8 t 3 y 4e t2 x e t t 4 t 34 y sin t x cos t ln(tan 1 2t) 0 t y e t sin t x e t cos t 45. 0 t y 3 sin t sin 3t x 3 cos t cos 3t 0 t 2 y ln1 t x t 1 t 0 t 3 y 5 2t x et et 0 t 1 y 4 2t 3 x 1 3t 2 41. 1 t 5 y st 1 x ln t 0 t 2 y t sin t x t cos t 3 t 3 y t 2 x 1 e t 1 t 2 y 4 3t 32 x t t 2 x d r y x 0 a a _a a y a sin3 x a cos3 y t t 2 x 1 e t x-axis y-axis y st x t 2 2t SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES \|\|\|\| 637 (b) By regarding a curve as the parametric curve , , with parameter , show that the formula in part (a) becomes 70. (a) Use the formula in Exercise 69(b) to ﬁnd the curvature of the parabola at the point . (b) At what point does this parabola have maximum curvature? 71. Use the formula in Exercise 69(a) to ﬁnd the curvature of the cycloid , at the top of one of its arches. 72. (a) Show that the curvature at each point of a straight line is . (b) Show that the curvature at each point of a circle of radius is . 73. A string is wound around a circle and then unwound while being held taut. The curve traced by the point at the end of the string is called the involute of the circle. If the circle has radius and center and the initial position of is , and if the parameter is chosen as in the ﬁgure, show that parametric equations of the involute are 74. A cow is tied to a silo with radius by a rope just long enough to reach the opposite side of the silo. Find the area available for grazing by the cow. r x O y r ¨ P T y rsin cos x rcos sin r, 0 P O r P 1r r 0 y 1 cos x sin 1, 1 y x 2 0 x y P ˙ d 2ydx 2 1 dydx2 32 x y f x x x y f x 59–61 Find the exact area of the surface obtained by rotating the given curve about the -axis. 59. , , 60. , , , , ; 62. Graph the curve If this curve is rotated about the -axis, ﬁnd the area of the resulting surface. (Use your graph to help ﬁnd the correct parameter interval.) 63. If the curve is rotated about the -axis, use your calculator to estimate the area of the resulting surface to three decimal places. 64. If the arc of the curve in Exercise 50 is rotated about the -axis, estimate the area of the resulting surface using Simp-son’s Rule with . 65–66 Find the surface area generated by rotating the given curve about the -axis. , , 66. , , 67. If is continuous and for , show that the parametric curve , , , can be put in the form . [Hint: Show that exists.] 68. Use Formula 2 to derive Formula 7 from Formula 8.2.5 for the case in which the curve can be represented in the form , . 69. The curvature at a point of a curve is deﬁned as where is the angle of inclination of the tangent line at , as shown in the ﬁgure. Thus the curvature is the absolute value of the rate of change of with respect to arc length. It can be regarded as a measure of the rate of change of direc-tion of the curve at and will be studied in greater detail in Chapter 13. (a) For a parametric curve , , derive the formula where the dots indicate derivatives with respect to , so . [Hint: Use and Formula 2 to ﬁnd . Then use the Chain Rule to ﬁnd .] dds ddt tan1dydx x dxdt t x y x y x 2 y 2 32 y yt x xt P P d ds P a x b y Fx f 1 y Fx a t b y tt x f t a t b f t 0 f 0 t 1 y 4e t2 x e t t 0 t 5 y 2t 3 x 3t 2 65. y n 4 x x 1 t 2 y t 1 t 2 x t t 3 x y 2 sin sin 2 x 2 cos cos 2 0 2 y a sin3 x a cos3 61. 0 t 1 y 3t 2 x 3t t 3 0 t 1 y t 2 x t 3 x 638 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES The Bézier curves are used in computer-aided design and are named after the French mathema-tician Pierre Bézier (1910–1999), who worked in the automotive industry. A cubic Bézier curve is determined by four control points, and , and is deﬁned by the parametric equations where . Notice that when we have and when we have , so the curve starts at and ends at . 1. Graph the Bézier curve with control points , , , and Then, on the same screen, graph the line segments , , and . (Exercise 31 in Section 10.1 shows how to do this.) Notice that the middle control points and don’t lie on the curve; the curve starts at , heads toward and without reaching them, and ends at 2. From the graph in Problem 1, it appears that the tangent at passes through and the tangent at passes through . Prove it. 3. Try to produce a Bézier curve with a loop by changing the second control point in Problem 1. 4. Some laser printers use Bézier curves to represent letters and other symbols. Experiment with control points until you ﬁnd a Bézier curve that gives a reasonable representation of the letter C. 5. More complicated shapes can be represented by piecing together two or more Bézier curves. Suppose the ﬁrst Bézier curve has control points and the second one has con-trol points . If we want these two pieces to join together smoothly, then the tangents at should match and so the points , , and all have to lie on this common tangent line. Using this principle, ﬁnd control points for a pair of Bézier curves that repre-sent the letter S. P4 P3 P2 P3 P3, P4, P5, P6 P0, P1, P2, P3 P2 P3 P1 P0 P3. P2 P1 P0 P2 P1 P2P3 P1P2 P0P1 P340, 5. P250, 42 P128, 48 P04, 1 P3 P0 x, y x3, y3 t 1 x, y x0, y0 t 0 0 t 1 y y01 t3 3y1t1 t2 3y2t 21 t y3t 3 x x01 t3 3x1t1 t2 3x2t 21 t x3t 3 P3x3, y3 P0x0, y0, P1x1, y1, P2x2, y2, SECTION 10.3 POLAR COORDINATES \|\|\|\| 639 POLAR COORDINATES A coordinate system represents a point in the plane by an ordered pair of numbers called coordinates. Usually we use Cartesian coordinates, which are directed distances from two perpendicular axes. Here we describe a coordinate system introduced by Newton, called the polar coordinate system, which is more convenient for many purposes. We choose a point in the plane that is called the pole (or origin) and is labeled . Then we draw a ray (half-line) starting at called the polar axis. This axis is usually drawn hor-izontally to the right and corresponds to the positive -axis in Cartesian coordinates. If is any other point in the plane, let be the distance from to and let be the angle (usually measured in radians) between the polar axis and the line as in Figure 1. Then the point is represented by the ordered pair and , are called polar coordi-nates of . We use the convention that an angle is positive if measured in the counter-clockwise direction from the polar axis and negative in the clockwise direction. If , then and we agree that represents the pole for any value of . 0, r 0 P O P r r, P OP P O r P x O O 10.3 x O ¨ r polar axis P(r, ¨) FIGURE 1 ; BE ´ZIER CURVES L A B O R AT O R Y P R O J E C T We extend the meaning of polar coordinates to the case in which is negative by agreeing that, as in Figure 2, the points and lie on the same line through and at the same distance from , but on opposite sides of . If , the point lies in the same quadrant as ; if , it lies in the quadrant on the opposite side of the pole. Notice that represents the same point as . EXAMPLE 1 Plot the points whose polar coordinates are given. (a) (b) (c) (d) SOLUTION The points are plotted in Figure 3. In part (d) the point is located three units from the pole in the fourth quadrant because the angle is in the second quadrant and is negative. M In the Cartesian coordinate system every point has only one representation, but in the polar coordinate system each point has many representations. For instance, the point in Example 1(a) could be written as or or . (See Figure 4.) In fact, since a complete counterclockwise rotation is given by an angle 2 , the point represented by polar coordinates is also represented by where is any integer. The connection between polar and Cartesian coordinates can be seen from Figure 5, in which the pole corresponds to the origin and the polar axis coincides with the positive -axis. If the point has Cartesian coordinates and polar coordinates , then, from the ﬁgure, we have and so Although Equations 1 were deduced from Figure 5, which illustrates the case where and , these equations are valid for all values of and (See the gen-eral deﬁnition of and in Appendix D.) cos sin . r 0 2 r 0 y r sin x r cos 1 sin y r cos x r r, x, y P x n r, 2n 1 and r, 2n r, O 13π 4 ”1, ’ 13π 4 O _ 3π 4 ”1, _ ’ 3π 4 O ”1, ’ 5π 4 5π 4 FIGURE 4 O ”_1, ’ π 4 π 4 1, 4 1, 134 1, 34 1, 54 O ”_3, ’ 3π 4 3π 4 (2, 3π) O 3π ”1, ’ 5π 4 5π 4 O FIGURE 3 O ”2, _ ’ 2π 3 2π 3 _ r 3 34 3, 34 3, 34 2, 23 2, 3 1, 54 r, r, r 0 r, r 0 O O r O r, r, r r, 640 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES (_r, ¨) O ¨ (r, ¨) ¨+π FIGURE 2 O y x ¨ x y r P(r, ¨ )=P(x, y) FIGURE 5 Equations 1 allow us to ﬁnd the Cartesian coordinates of a point when the polar coor-dinates are known. To ﬁnd and when and are known, we use the equations which can be deduced from Equations 1 or simply read from Figure 5. EXAMPLE 2 Convert the point from polar to Cartesian coordinates. SOLUTION Since and , Equations 1 give Therefore the point is in Cartesian coordinates. M EXAMPLE 3 Represent the point with Cartesian coordinates in terms of polar coordinates. SOLUTION If we choose to be positive, then Equations 2 give Since the point lies in the fourth quadrant, we can choose or . Thus one possible answer is ; another is . M Equations 2 do not uniquely determine when and are given because, as increases through the interval , each value of occurs twice. Therefore, in converting from Cartesian to polar coordinates, it’s not good enough just to ﬁnd and that satisfy Equations 2. As in Example 3, we must choose so that the point lies in the correct quadrant. POLAR CURVES The graph of a polar equation , or more generally , consists of all points that have at least one polar representation whose coordinates satisfy the equation. EXAMPLE 4 What curve is represented by the polar equation ? SOLUTION The curve consists of all points with . Since represents the distance from the point to the pole, the curve represents the circle with center and radius . In general, the equation represents a circle with center and radius . (See Figure 6.) M a O r a 2 O r 2 r r 2 r, r 2 V r, P Fr, 0 r f r, r tan 0 2 y x NOTE s2 , 74 (s2 , 4) 74 4 1, 1 tan y x 1 r sx 2 y 2 s1 2 12 s2 r 1, 1 (1, s3 ) y r sin 2 sin 3 2 s3 2 s3 x r cos 2 cos 3 2 1 2 1 3 r 2 2, 3 tan y x r 2 x 2 y 2 2 y x r SECTION 10.3 POLAR COORDINATES \|\|\|\| 641 FIGURE 6 x r= 1 2 r=1 r=2 r=4 EXAMPLE 5 Sketch the polar curve . SOLUTION This curve consists of all points such that the polar angle is 1 radian. It is the straight line that passes through and makes an angle of 1 radian with the polar axis (see Figure 7). Notice that the points on the line with are in the ﬁrst quadrant, whereas those with are in the third quadrant. M EXAMPLE 6 (a) Sketch the curve with polar equation . (b) Find a Cartesian equation for this curve. SOLUTION (a) In Figure 8 we ﬁnd the values of for some convenient values of and plot the corresponding points . Then we join these points to sketch the curve, which appears to be a circle. We have used only values of between 0 and , since if we let increase beyond , we obtain the same points again. (b) To convert the given equation to a Cartesian equation we use Equations 1 and 2. From we have , so the equation becomes , which gives or Completing the square, we obtain which is an equation of a circle with center and radius 1. M FIGURE 9 O y x 2 ¨ r P Q 1, 0 x 12 y 2 1 x 2 y 2 2x 0 2x r 2 x 2 y 2 r 2xr r 2 cos cos xr x r cos FIGURE 8 Table of values and graph of r=2 cos ¨ (2, 0) 2 ”_1, ’ 2π 3 ”0, ’ π 2 ”1, ’ π 3 ”œ„, ’ π 4 ”œ„, ’ π 6 3 ” œ„, ’ 5π 6 3 ” œ„, ’ 3π 4 2 r, r r 2 cos r 0 r 0 r, 1 O r, 1 642 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES O x 1 (_1, 1) (_2, 1) (1, 1) (2, 1) (3, 1) ¨=1 FIGURE 7 0 2 1 0 1 2 s3 56 s2 34 23 2 3 s2 4 s3 6 r 2 cos N Figure 9 shows a geometrical illustration that the circle in Example 6 has the equation . The angle is a right angle (Why?) and so . r2 cos OPQ r 2 cos EXAMPLE 7 Sketch the curve . SOLUTION Instead of plotting points as in Example 6, we ﬁrst sketch the graph of in Cartesian coordinates in Figure 10 by shifting the sine curve up one unit. This enables us to read at a glance the values of that correspond to increasing values of . For instance, we see that as increases from 0 to , (the distance from ) increases from 1 to 2, so we sketch the corresponding part of the polar curve in Figure 11(a). As increases from to , Figure 10 shows that decreases from 2 to 1, so we sketch the next part of the curve as in Figure 11(b). As increases from to , decreases from 1 to 0 as shown in part (c). Finally, as increases from to , increases from 0 to 1 as shown in part (d). If we let increase beyond or decrease beyond 0, we would simply retrace our path. Putting together the parts of the curve from Figure 11(a)–(d), we sketch the complete curve in part (e). It is called a cardioid, because it’s shaped like a heart. M EXAMPLE 8 Sketch the curve . SOLUTION As in Example 7, we ﬁrst sketch , , in Cartesian coordi-nates in Figure 12. As increases from 0 to , Figure 12 shows that decreases from 1 to 0 and so we draw the corresponding portion of the polar curve in Figure 13 (indi-cated by !). As increases from to , goes from 0 to . This means that the distance from increases from 0 to 1, but instead of being in the ﬁrst quadrant this por-tion of the polar curve (indicated by @) lies on the opposite side of the pole in the third quadrant. The remainder of the curve is drawn in a similar fashion, with the arrows and numbers indicating the order in which the portions are traced out. The resulting curve has four loops and is called a four-leaved rose. M ¨=0 ¨=π ⑧ ¨=3π 4 ¨=π 2 ¨=π 4 FIGURE 12 r=cos 2¨ in Cartesian coordinates FIGURE 13 Four-leaved rose r=cos 2¨ r 1 ¨ 2π π 5π 4 π 2 π 4 3π 4 3π 2 7π 4 ! @ # ^ & % $ ! @ # $ % & ^ O 1 r 2 4 r 4 0 2 r cos 2 r cos 2 (a) (b) (c) (d) (e) FIGURE 11 Stages in sketching the cardioid r=1+sin ¨ O ¨=π ¨=π 2 O ¨=π ¨=3π 2 O ¨=2π ¨= 3π 2 O O ¨=0 ¨=π 2 1 2 2 r 2 32 r 32 r 2 O r 2 r r 1 sin r 1 sin V SECTION 10.3 POLAR COORDINATES \|\|\|\| 643 FIGURE 10 r=1+sin ¨ in Cartesian coordinates, 0¯¨¯2π 0 r 1 2 ¨ π 2π 3π 2 π 2 Module 10.3 helps you see how polar curves are traced out by showing animations similar to Figures 10–13. TEC SYMMETRY When we sketch polar curves, it is sometimes helpful to take advantage of symmetry. The following three rules are explained by Figure 14. (a) If a polar equation is unchanged when is replaced by , the curve is symmetric about the polar axis. (b) If the equation is unchanged when is replaced by , or when is replaced by , the curve is symmetric about the pole. (This means that the curve remains unchanged if we rotate it through 180° about the origin.) (c) If the equation is unchanged when is replaced by , the curve is symmetric about the vertical line . The curves sketched in Examples 6 and 8 are symmetric about the polar axis, since . The curves in Examples 7 and 8 are symmetric about because and . The four-leaved rose is also symmetric about the pole. These symmetry properties could have been used in sketching the curves. For instance, in Example 6 we need only have plotted points for and then reﬂected about the polar axis to obtain the complete circle. TANGENTS TO POLAR CURVES To ﬁnd a tangent line to a polar curve , we regard as a parameter and write its parametric equations as Then, using the method for ﬁnding slopes of parametric curves (Equation 10.2.2) and the Product Rule, we have We locate horizontal tangents by ﬁnding the points where (provided that ). Likewise, we locate vertical tangents at the points where (pro-vided that ). Notice that if we are looking for tangent lines at the pole, then and Equation 3 sim-pliﬁes to dr d 0 if dy dx tan r 0 dyd 0 dxd 0 dxd 0 dyd 0 dy dx dy d dx d dr d sin r cos dr d cos r sin 3 y r sin f sin x r cos f cos r f 0 2 cos 2 cos 2 sin sin 2 cos cos O (r, ¨) (_r, ¨) O (r, ¨) (r, ¨) _¨ ¨ (a) (b) (c) FIGURE 14 O (r, ¨) (r, π-¨) π-¨ ¨ 2 r r 644 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES For instance, in Example 8 we found that when or . This means that the lines and (or and ) are tangent lines to at the origin. EXAMPLE 9 (a) For the cardioid of Example 7, ﬁnd the slope of the tangent line when . (b) Find the points on the cardioid where the tangent line is horizontal or vertical. SOLUTION Using Equation 3 with , we have (a) The slope of the tangent at the point where is (b) Observe that Therefore there are horizontal tangents at the points , , and vertical tangents at and . When , both and are 0, so we must be careful. Using l’Hospital’s Rule, we have By symmetry, Thus there is a vertical tangent line at the pole (see Figure 15). M lim l 32 dy dx 1 3 lim l 32 sin cos 1 3 lim l 32 cos 1 sin lim l 32 dy dx lim l 32 1 2 sin 1 2 sin lim l 32 cos 1 sin dxd dyd 32 ( 3 2, 56) ( 3 2, 6) ( 1 2, 116) ( 1 2, 76) 2, 2 when 3 2 , 6 , 5 6 dx d 1 sin 1 2 sin 0 when 2 , 3 2 , 7 6 , 11 6 dy d cos 1 2 sin 0 1 s3 1 s3 1 1 s3 (2 s3 )(1 s3 ) 1 2(1 s3 ) (1 s3 2)(1 s3 ) dy dx 3 cos31 2 sin3 1 sin31 2 sin3 3 cos 1 2 sin 1 2 sin2 sin cos 1 2 sin 1 sin 1 2 sin dy dx dr d sin r cos dr d cos r sin cos sin 1 sin cos cos cos 1 sin sin r 1 sin 3 r 1 sin r cos 2 y x y x 34 4 34 4 r cos 2 0 SECTION 10.3 POLAR COORDINATES \|\|\|\| 645 5π 6 ” , ’ 3 2 7π 6 ” , ’ 1 2 ” , ’ 11π 6 1 2 3 2 π 6 ” , ’ (0, 0) m=_1 ”1+ , ’ π 3 œ„ 3 2 ”2, ’ π 2 FIGURE 15 Tangent lines for r=1+sin ¨ Instead of having to remember Equation 3, we could employ the method used to derive it. For instance, in Example 9 we could have written Then we would have which is equivalent to our previous expression. GRAPHING POLAR CURVES WITH GRAPHING DEVICES Although it’s useful to be able to sketch simple polar curves by hand, we need to use a graphing calculator or computer when we are faced with a curve as complicated as the ones shown in Figures 16 and 17. Some graphing devices have commands that enable us to graph polar curves directly. With other machines we need to convert to parametric equations ﬁrst. In this case we take the polar equation and write its parametric equations as Some machines require that the parameter be called rather than . EXAMPLE 10 Graph the curve . SOLUTION Let’s assume that our graphing device doesn’t have a built-in polar graphing command. In this case we need to work with the corresponding parametric equations, which are In any case, we need to determine the domain for . So we ask ourselves: How many complete rotations are required until the curve starts to repeat itself? If the answer is , then sin 8 2n 5 sin 8 5 16n 5 sin 8 5 n y r sin sin85 sin x r cos sin85 cos r sin85 t y r sin f sin x r cos f cos r f FIGURE 17 r=sin@(1.2¨)+cos#(6¨) 1.7 _1.7 _1.9 1.9 FIGURE 16 r=sin@(2.4¨)+cos$(2.4¨) 1 _1 _1 1 dy dx dyd dxd cos 2 sin cos sin cos 2 cos sin 2 sin cos 2 y r sin 1 sin sin sin sin2 x r cos 1 sin cos cos 1 2 sin 2 NOTE 646 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES and so we require that be an even multiple of . This will ﬁrst occur when . Therefore we will graph the entire curve if we specify that . Switching from to , we have the equations and Figure 18 shows the resulting curve. Notice that this rose has 16 loops. M EXAMPLE 11 Investigate the family of polar curves given by . How does the shape change as changes? (These curves are called limaçons, after a French word for snail, because of the shape of the curves for certain values of .) SOLUTION Figure 19 shows computer-drawn graphs for various values of . For there is a loop that decreases in size as decreases. When the loop disappears and the curve becomes the cardioid that we sketched in Example 7. For between 1 and the cardioid’s cusp is smoothed out and becomes a “dimple.” When decreases from to , the limaçon is shaped like an oval. This oval becomes more circular as , and when the curve is just the circle . The remaining parts of Figure 19 show that as becomes negative, the shapes change in reverse order. In fact, these curves are reﬂections about the horizontal axis of the corre-sponding curves with positive . M c c FIGURE 19 Members of the family of limaçons r=1+c sin ¨ c=2.5 c=0 c=_0.2 c=_0.5 c=_0.8 c=_1 c=_2 c=1.7 c=1 c=0.7 c=0.5 c=0.2 r 1 c 0 c l 0 0 1 2 c 1 2 c c 1 c c 1 c c c r 1 c sin V 0 t 10 y sin8t5 sin t x sin8t5 cos t t 0 10 n 5 16n5 SECTION 10.3 POLAR COORDINATES \|\|\|\| 647 1 _1 _1 1 FIGURE 18 r=sin(8¨/5) N In Exercise 55 you are asked to prove analyti-cally what we have discovered from the graphs in Figure 19. 4. (a) (b) (c) 5–6 The Cartesian coordinates of a point are given. (i) Find polar coordinates of the point, where and . (ii) Find polar coordinates of the point, where and . 5. (a) (b) 6. (a) (b) 1, 2 (3s3 , 3) (1, s3 ) 2, 2 0 2 r 0 r, 0 2 r 0 r, 2, 76 1, 52 (s2 , 54) 1–2 Plot the point whose polar coordinates are given. Then ﬁnd two other pairs of polar coordinates of this point, one with and one with . 1. (a) (b) (c) 2. (a) (b) (c) 3–4 Plot the point whose polar coordinates are given. Then ﬁnd the Cartesian coordinates of the point. 3. (a) (b) (c) 2, 34 (2, 23) 1, 1, 1 3, 6 1, 74 1, 2 1, 34 2, 3 r 0 r 0 EXERCISES 10.3 7–12 Sketch the region in the plane consisting of points whose polar coordinates satisfy the given conditions. 7. 8. , 9. , 10. , , 12. , 13. Find the distance between the points with polar coordinates and . 14. Find a formula for the distance between the points with polar coordinates and . 15–20 Identify the curve by ﬁnding a Cartesian equation for the curve. 15. 16. 18. 19. 20. 21–26 Find a polar equation for the curve represented by the given Cartesian equation. 21. 22. 23. 24. 26. 27–28 For each of the described curves, decide if the curve would be more easily given by a polar equation or a Cartesian equation. Then write an equation for the curve. 27. (a) A line through the origin that makes an angle of with the positive -axis (b) A vertical line through the point 28. (a) A circle with radius 5 and center (b) A circle centered at the origin with radius 4 29–48 Sketch the curve with the given polar equation. 29. 30. 31. 32. 33. , 34. , 36. , 37. 38. 40. 41. 42. r 2 sin r 1 2 sin r 3 cos 6 r 2 cos 4 39. r cos 5 r 4 sin 3 1 r ln 0 r 35. r 1 3 cos 0 r 21 sin r 3 cos r sin r 2 3r 2 0 6 2, 3 3, 3 x 6 xy 4 x 2 y 2 2cx 25. x y 9 x y 2 x 2 y 2 9 x 3 r tan sec r csc r 2 sin 2 cos r 3 sin 17. r cos 1 r 2 r2, 2 r1, 1 4, 23 2, 3 2 r 1 53 73 2 r 3 11. 34 54 2 r 5 2 6 0 r 4 3 23 r 0 1 r 2 648 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 43. 44. 45. 46. 47. 48. 49–50 The ﬁgure shows the graph of as a function of in Carte-sian coordinates. Use it to sketch the corresponding polar curve. 50. 51. Show that the polar curve (called a conchoid) has the line as a vertical asymptote by showing that . Use this fact to help sketch the conchoid. 52. Show that the curve (also a conchoid) has the line as a horizontal asymptote by showing that . Use this fact to help sketch the conchoid. 53. Show that the curve (called a cissoid of Diocles) has the line as a vertical asymptote. Show also that the curve lies entirely within the vertical strip . Use these facts to help sketch the cissoid. 54. Sketch the curve . (a) In Example 11 the graphs suggest that the limaçon has an inner loop when . Prove that this is true, and ﬁnd the values of that correspond to the inner loop. (b) From Figure 19 it appears that the limaçon loses its dimple when . Prove this. 56. Match the polar equations with the graphs labeled I–VI. Give reasons for your choices. (Don’t use a graphing device.) (a) (b) (c) (d) (e) (f) I II III IV V VI r 1 2 sin 3 r 2 sin 3 r 1 2 cos r cos3 r 2, 0 16 r s , 0 16 c 1 2 c 1 r 1 c sin 55. x 2 y 23 4x 2y 2 0 x 1 x 1 r sin tan lim r l y 1 y 1 r 2 csc lim r l x 2 x 2 r 4 2 sec ¨ r 0 π 2π 1 2 ¨ r 0 π 2π 2 _2 49. r r 1 2 cos2 r 1 2 cos 2 r 2 1 r 2 cos32 r 2 cos 4 r 2 9 sin 2 SECTION 10.3 POLAR COORDINATES \|\|\|\| 649 ; 80. A family of curves is given by the equations , where is a real number and is a positive integer. How does the graph change as increases? How does it change as changes? Illustrate by graphing enough members of the fam-ily to support your conclusions. ; 81. A family of curves has polar equations Investigate how the graph changes as the number changes. In particular, you should identify the transitional values of for which the basic shape of the curve changes. ; 82. The astronomer Giovanni Cassini (1625–1712) studied the family of curves with polar equations where and are positive real numbers. These curves are called the ovals of Cassini even though they are oval shaped only for certain values of and . (Cassini thought that these curves might represent planetary orbits better than Kepler’s ellipses.) Investigate the variety of shapes that these curves may have. In particular, how are and related to each other when the curve splits into two parts? 83. Let be any point (except the origin) on the curve . If is the angle between the tangent line at and the radial line , show that [Hint: Observe that in the ﬁgure.] 84. (a) Use Exercise 83 to show that the angle between the tan-gent line and the radial line is at every point on the curve . ; (b) Illustrate part (a) by graphing the curve and the tangent lines at the points where and . (c) Prove that any polar curve with the property that the angle between the radial line and the tangent line is a constant must be of the form , where and are constants. k C r Ce k r f 2 0 r e 4 O P ÿ ¨ ˙ r=f(¨) tan r drd OP P r f P c a c a c a r 4 2c2r 2 cos 2 c 4 a 4 0 a a r 1 a cos 1 a cos c n n c r 1 c sin n 57–62 Find the slope of the tangent line to the given polar curve at the point speciﬁed by the value of . 57. , 58. , , 60. , 61. , 62. , 63–68 Find the points on the given curve where the tangent line is horizontal or vertical. 64. 65. 66. 67. 68. Show that the polar equation , where , represents a circle, and ﬁnd its center and radius. 70. Show that the curves and intersect at right angles. ; 71–76 Use a graphing device to graph the polar curve. Choose the parameter interval to make sure that you produce the entire curve. 71. (nephroid of Freeth) 72. (hippopede) 73. (butterﬂy curve) 74. 75. 76. ; 77. How are the graphs of and related to the graph of ? In general, how is the graph of related to the graph of ? ; 78. Use a graph to estimate the -coordinate of the highest points on the curve . Then use calculus to ﬁnd the exact value. ; 79. (a) Investigate the family of curves deﬁned by the polar equa-tions , where is a positive integer. How is the number of loops related to ? (b) What happens if the equation in part (a) is replaced by ? r sin n n n r sin n r sin 2 y r f r f r 1 sin r 1 sin 3 r 1 sin 6 r cos2 cos3 r 2 5 sin6 r sin24 cos4 r e sin 2 cos4 r s1 0.8 sin 2 r 1 2 sin2 r a cos r a sin ab 0 r a sin b cos 69. r 2 sin 2 r 2 sin r e r 1 cos r 1 sin r 3 cos 63. 3 r 1 2 cos 4 r cos 2 r cos3 r 1 59. 3 r 2 sin 6 r 2 sin AREAS AND LENGTHS IN POLAR COORDINATES In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the area of a sector of a circle where, as in Figure 1, is the radius and is the radian measure of the central angle. Formula 1 follows from the fact that the area of a sector is proportional to its central angle: . (See also Exercise 35 in Section 7.3.) Let be the region, illustrated in Figure 2, bounded by the polar curve and by the rays and , where is a positive continuous function and where . We divide the interval into subintervals with endpoints , , , . . . , and equal width . The rays then divide into smaller regions with central angle . If we choose in the subinterval , then the area of the th region is approximated by the area of the sector of a circle with central angle and radius . (See Figure 3.) Thus from Formula 1 we have and so an approximation to the total area of is It appears from Figure 3 that the approximation in (2) improves as . But the sums in (2) are Riemann sums for the function , so It therefore appears plausible (and can in fact be proved) that the formula for the area of the polar region is Formula 3 is often written as with the understanding that . Note the similarity between Formulas 1 and 4. When we apply Formula 3 or 4, it is helpful to think of the area as being swept out by a rotating ray through that starts with angle and ends with angle . EXAMPLE 1 Find the area enclosed by one loop of the four-leaved rose . SOLUTION The curve was sketched in Example 8 in Section 10.3. Notice from Figure 4 that the region enclosed by the right loop is swept out by a ray that rotates from r cos 2 r cos 2 V b a O r f A y b a 1 2r 2 d 4 A y b a 1 2 f2 d 3 A lim n l n i1 1 2 fi 2 y b a 1 2 f2 d t 1 2 f2 n l A n i1 1 2 fi 2 2 A Ai 1 2 fi 2 fi i Ai i1, i ith i i i1 n i n 2 1 0 a, b 0 b a 2 f b a r f A 2r 2 1 2r 2 r A 1 2r 2 1 10.4 650 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES ¨ r FIGURE 1 FIGURE 2 O ¨=b b ¨=a r=f(¨) a O ¨=b ¨=a ¨=¨i-1 ¨=¨i Î¨ f(¨i ) FIGURE 3 to . Therefore Formula 4 gives M EXAMPLE 2 Find the area of the region that lies inside the circle and outside the cardioid . SOLUTION The cardioid (see Example 7 in Section 10.3) and the circle are sketched in Figure 5 and the desired region is shaded. The values of and in Formula 4 are deter-mined by ﬁnding the points of intersection of the two curves. They intersect when , which gives , so , . The desired area can be found by subtracting the area inside the cardioid between and from the area inside the circle from to . Thus Since the region is symmetric about the vertical axis , we can write [because ] M Example 2 illustrates the procedure for ﬁnding the area of the region bounded by two polar curves. In general, let be a region, as illustrated in Figure 6, that is bounded by curves with polar equations , , , and , where and . The area of is found by subtracting the area inside from the area inside , so using Formula 3 we have \| CAUTION The fact that a single point has many representations in polar coordinates sometimes makes it difﬁcult to ﬁnd all the points of intersection of two polar curves. For instance, it is obvious from Figure 5 that the circle and the cardioid have three points of intersection; however, in Example 2 we solved the equations and and found only two such points, and . The origin is also a point of intersection, but we can’t ﬁnd it by solving the equations of the curves because the origin has no single representation in polar coordinates that satisﬁes both equations. Notice that, when represented as or , the origin satisﬁes and so it lies on the circle; when represented as , it satisﬁes and so it lies on the cardioid. Think of two points moving along the curves as the parameter value increases from 0 to . On one curve the origin is reached at and ; on the 0 2 r 1 sin 0, 32 r 3 sin 0, 0, 0 ( 3 2, 56) ( 3 2, 6) r 1 sin r 3 sin 1 2 y b a f2 t2 d A y b a 1 2 f2 d y b a 1 2t2 d r f r t A 0 b a 2 f t 0 b a r t r f 3 2 sin 2 2 cos ]6 2 sin2 1 2 1 cos 2 y 2 6 3 4 cos 2 2 sin d y 2 6 8 sin2 1 2 sin d A 2 1 2 y 2 6 9 sin2 d 1 2 y 2 6 1 2 sin sin2 d 2 A 1 2 y 56 6 3 sin 2 d 1 2 y 56 6 1 sin 2 d 56 6 56 6 56 6 sin 1 2 3 sin 1 sin b a r 1 sin r 3 sin V 1 2[ 1 4 sin 4]0 4 8 y 4 0 1 21 cos 4 d y 4 0 cos2 2 d A y 4 4 1 2r 2 d 1 2 y 4 4 cos2 2 d 4 4 SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES \|\|\|\| 651 r=cos 2¨ ¨=π 4 ¨=π 4 FIGURE 4 FIGURE 5 O ¨=5π 6 ¨=π 6 r=3 sin ¨ r=1+sin ¨ O ¨=b ¨=a r=f(¨) r=g(¨) FIGURE 6 other curve it is reached at . The points don’t collide at the origin because they reach the origin at different times, but the curves intersect there nonetheless. Thus, to ﬁnd all points of intersection of two polar curves, it is recommended that you draw the graphs of both curves. It is especially convenient to use a graphing calculator or computer to help with this task. EXAMPLE 3 Find all points of intersection of the curves and . SOLUTION If we solve the equations and , we get and, there-fore, , , , . Thus the values of between 0 and that sat-isfy both equations are , , , . We have found four points of intersection: , , and . However, you can see from Figure 7 that the curves have four other points of inter-section—namely, , , , and . These can be found using symmetry or by noticing that another equation of the circle is and then solving the equations and . M ARC LENGTH To ﬁnd the length of a polar curve , , we regard as a parameter and write the parametric equations of the curve as Using the Product Rule and differentiating with respect to , we obtain so, using , we have Assuming that is continuous, we can use Theorem 10.2.6 to write the arc length as Therefore the length of a curve with polar equation , , is EXAMPLE 4 Find the length of the cardioid . SOLUTION The cardioid is shown in Figure 8. (We sketched it in Example 7 in Section 10.3.) Its full length is given by the parameter interval , so 0 2 r 1 sin V L y b a r 2 dr d 2 d 5 a b r f L y b a dx d 2 dy d 2 d f dr d 2 r 2 dr d 2 sin2 2r dr d sin cos r 2 cos2 dx d 2 dy d 2 dr d 2 cos2 2r dr d cos sin r 2 sin2 cos2 sin2 1 dy d dr d sin r cos dx d dr d cos r sin y r sin f sin x r cos f cos a b r f r 1 2 r cos 2 r 1 2 ( 1 2, 53) ( 1 2, 43) ( 1 2, 23) ( 1 2, 3) ( 1 2, 116) ( 1 2, 56), ( 1 2, 76) ( 1 2, 6) 116 76 56 6 2 113 73 53 3 2 cos 2 1 2 r 1 2 r cos 2 r 1 2 r cos 2 32 652 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES FIGURE 7 r=cos 2¨ 1 2 r= ” , ’ 1 2 π 3 ” , ’ 1 2 π 6 Formula 5 gives We could evaluate this integral by multiplying and dividing the integrand by , or we could use a computer algebra system. In any event, we ﬁnd that the length of the cardioid is . M L 8 s2 2 sin y 2 0 s2 2 sin d L y 2 0 r 2 dr d 2 d y 2 0 s1 sin 2 cos 2 d SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES \|\|\|\| 653 O FIGURE 8 r=1+sin ¨ 19. 20. (inner loop) 22. Find the area enclosed by the loop of the strophoid . 23–28 Find the area of the region that lies inside the ﬁrst curve and outside the second curve. 23. , 24. , 25. , 26. , , 28. , 29–34 Find the area of the region that lies inside both curves. 29. , 30. , , 32. , 33. , 34. , , , 35. Find the area inside the larger loop and outside the smaller loop of the limaçon . 36. Find the area between a large loop and the enclosed small loop of the curve . 37–42 Find all points of intersection of the given curves. 37. , 38. , 39. , 40. , , 42. , r 2 cos 2 r 2 sin 2 r sin 2 r sin 41. r sin 3 r cos 3 r 1 r 2 sin 2 r 1 sin r 1 cos r 3 sin r 1 sin r 1 2 cos 3 r 1 2 cos b 0 a 0 r b cos r a sin r 2 cos 2 r 2 sin 2 r 3 2 sin r 3 2 cos r cos 2 r sin 2 31. r 1 cos r 1 cos r sin r s3 cos r 2 sin r 3 sin r 1 cos r 3 cos 27. r 3 sin r 2 sin r 2 r 2 8 cos 2 r 1 r 1 sin r 1 r 2 cos r 2 cos sec r 1 2 sin 21. r 2 sin 6 r 3 cos 5 1–4 Find the area of the region that is bounded by the given curve and lies in the speciﬁed sector. 1. , 2. , 3. , 4. , 5–8 Find the area of the shaded region. 5. 6. 8. 9–14 Sketch the curve and ﬁnd the area that it encloses. 9. 10. 12. 13. 14. ; 15–16 Graph the curve and ﬁnd the area that it encloses. 15. 16. 17–21 Find the area of the region enclosed by one loop of the curve. 17. 18. r 4 sin 3 r sin 2 r 2 sin 3 sin 9 r 1 2 sin 6 r 2 cos 2 r 2 cos 3 r 2 sin r 2 4 cos 2 11. r 31 cos r 3 cos r=sin 2¨ r=4+3 sin ¨ 7. r=1+cos ¨ r=œ„ ¨ 0 r ssin 3 23 r sin 2 r e 2 0 4 r 2 EXERCISES 10.4 49–52 Use a calculator to ﬁnd the length of the curve correct to four decimal places. 49. 50. 51. 52. ; 53–54 Graph the curve and ﬁnd its length. 53. 54. 55. (a) Use Formula 10.2.7 to show that the area of the surface generated by rotating the polar curve (where is continuous and ) about the polar axis is (b) Use the formula in part (a) to ﬁnd the surface area gener-ated by rotating the lemniscate about the polar axis. 56. (a) Find a formula for the area of the surface generated by rotating the polar curve , (where is continuous and ), about the line . (b) Find the surface area generated by rotating the lemniscate about the line . 2 r 2 cos 2 2 0 a b f a b r f r 2 cos 2 S y b a 2r sin r 2 dr d 2 d 0 a b f a b r f r cos22 r cos44 r 1 cos3 r sin2 r 4 sin 3 r 3 sin 2 ; 43. The points of intersection of the cardioid and the spiral loop , , can’t be found exactly. Use a graphing device to ﬁnd the approximate values of at which they intersect. Then use these values to estimate the area that lies inside both curves. 44. When recording live performances, sound engineers often use a microphone with a cardioid pickup pattern because it sup-presses noise from the audience. Suppose the microphone is placed 4 m from the front of the stage (as in the ﬁgure) and the boundary of the optimal pickup region is given by the car-dioid , where is measured in meters and the microphone is at the pole. The musicians want to know the area they will have on stage within the optimal pickup range of the microphone. Answer their question. 45–48 Find the exact length of the polar curve. 45. , 46. , , 48. , 0 2 r 0 2 r 2 47. 0 2 r e 2 0 3 r 3 sin stage audience microphone 12 m 4 m r r 8 8 sin 2 2 r 2 r 1 sin CONIC SECTIONS In this section we give geometric deﬁnitions of parabolas, ellipses, and hyperbolas and derive their standard equations. They are called conic sections, or conics, because they result from intersecting a cone with a plane as shown in Figure 1. FIGURE 1 Conics ellipse parabola hyperbola 10.5 654 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES PARABOLAS A parabola is the set of points in a plane that are equidistant from a ﬁxed point (called the focus) and a ﬁxed line (called the directrix). This deﬁnition is illustrated by Figure 2. Notice that the point halfway between the focus and the directrix lies on the parabola; it is called the vertex. The line through the focus perpendicular to the directrix is called the axis of the parabola. In the 16th century Galileo showed that the path of a projectile that is shot into the air at an angle to the ground is a parabola. Since then, parabolic shapes have been used in designing automobile headlights, reﬂecting telescopes, and suspension bridges. (See Problem 18 on page 268 for the reﬂection property of parabolas that makes them so useful.) We obtain a particularly simple equation for a parabola if we place its vertex at the ori-gin and its directrix parallel to the -axis as in Figure 3. If the focus is the point , then the directrix has the equation . If is any point on the parabola, then the distance from to the focus is and the distance from to the directrix is . (Figure 3 illustrates the case where .) The deﬁning property of a parabola is that these distances are equal: We get an equivalent equation by squaring and simplifying: An equation of the parabola with focus and directrix is If we write , then the standard equation of a parabola (1) becomes . It opens upward if and downward if [see Figure 4, parts (a) and (b)]. The graph is symmetric with respect to the -axis because (1) is unchanged when is replaced by . FIGURE 4 0 x y (p, 0) x=_p (d) ¥=4px, p<0 0 x y (p, 0) x=_p (c) ¥=4px, p>0 0 x y (0, p) y=_p (b) ≈=4py, p<0 0 x y (0, p) y=_p (a) ≈=4py, p>0 x x y p 0 p 0 y ax 2 a 14p x 2 4py y p 0, p 1 x 2 4py x 2 y 2 2py p 2 y 2 2py p 2 x 2 y p2 y p 2 y p2 sx 2 y p2 y p p 0 y p P PF sx 2 y p2 P Px, y y p 0, p x O F SECTION 10.5 CONIC SECTIONS \|\|\|\| 655 axis F focus parabola vertex directrix FIGURE 2 FIGURE 3 x y O F(0, p) y=_p P(x, y) y p If we interchange and in (1), we obtain which is an equation of the parabola with focus and directrix . (Inter-changing and amounts to reﬂecting about the diagonal line .) The parabola opens to the right if and to the left if [see Figure 4, parts (c) and (d)]. In both cases the graph is symmetric with respect to the -axis, which is the axis of the parabola. EXAMPLE 1 Find the focus and directrix of the parabola and sketch the graph. SOLUTION If we write the equation as and compare it with Equation 2, we see that , so . Thus the focus is and the directrix is . The sketch is shown in Figure 5. M ELLIPSES An ellipse is the set of points in a plane the sum of whose distances from two ﬁxed points and is a constant (see Figure 6). These two ﬁxed points are called the foci (plural of focus). One of Kepler’s laws is that the orbits of the planets in the solar system are ellipses with the sun at one focus. In order to obtain the simplest equation for an ellipse, we place the foci on the -axis at the points and as in Figure 7 so that the origin is halfway between the foci. Let the sum of the distances from a point on the ellipse to the foci be . Then is a point on the ellipse when that is, or Squaring both sides, we have which simpliﬁes to We square again: which becomes a 2 c 2x 2 a 2y 2 a 2a 2 c 2 a 2x 2 2cx c 2 y 2 a 4 2a 2cx c 2x 2 asx c2 y 2 a 2 cx x 2 2cx c 2 y 2 4a 2 4asx c2 y 2 x 2 2cx c 2 y 2 sx c2 y 2 2a sx c2 y 2 sx c2 y 2 sx c2 y 2 2a PF1 PF2 2a Px, y 2a 0 c, 0 c, 0 x FIGURE 7 F¡(_c, 0) F™(c, 0) 0 x y P(x, y) FIGURE 6 F¡ F™ P F2 F1 x 5 2 p, 0 ( 5 2, 0) p 5 2 4p 10 y 2 10x y 2 10x 0 x p 0 p 0 y x y x x p p, 0 y 2 4px 2 y x 656 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES FIGURE 5 0 x y x=5 2 ¥+10x=0 ” , 0’ 5 2 From triangle in Figure 7 we see that , so and therefore . For convenience, let . Then the equation of the ellipse becomes or, if both sides are divided by , Since , it follows that . The -intercepts are found by setting . Then , or , so . The corresponding points and are called the vertices of the ellipse and the line segment joining the vertices is called the major axis. To ﬁnd the -intercepts we set and obtain , so . Equation 3 is unchanged if is replaced by or is replaced by , so the ellipse is symmetric about both axes. Notice that if the foci coincide, then , so and the ellipse becomes a circle with radius . We summarize this discussion as follows (see also Figure 8). The ellipse has foci , where , and vertices . If the foci of an ellipse are located on the -axis at , then we can ﬁnd its equa-tion by interchanging and in (4). (See Figure 9.) The ellipse has foci , where , and vertices . EXAMPLE 2 Sketch the graph of and locate the foci. SOLUTION Divide both sides of the equation by 144: The equation is now in the standard form for an ellipse, so we have , , , and . The -intercepts are and the -intercepts are . Also, , so and the foci are . The graph is sketched in Figure 10. M EXAMPLE 3 Find an equation of the ellipse with foci and vertices . SOLUTION Using the notation of (5), we have and . Then we obtain , so an equation of the ellipse is Another way of writing the equation is . M 9x 2 5y 2 45 x 2 5 y 2 9 1 b 2 a 2 c 2 9 4 5 a 3 c 2 0, 3 0, 2 V ( s7 , 0) c s7 c 2 a 2 b 2 7 3 y 4 x b 3 a 4 b 2 9 a 2 16 x 2 16 y 2 9 1 9x 2 16y 2 144 V 0, a c 2 a 2 b 2 0, c a b 0 x 2 b 2 y 2 a 2 1 5 y x 0, c y a, 0 c 2 a 2 b 2 c, 0 a b 0 x 2 a 2 y 2 b 2 1 4 r a b a b c 0 y y x x y b y 2 b 2 x 0 y a, 0 a, 0 x a x 2 a 2 x 2a 2 1 y 0 x b a b 2 a 2 c 2 a 2 x 2 a 2 y 2 b 2 1 3 a 2b 2 b 2x 2 a 2y 2 a 2b 2 b 2 a 2 c 2 a 2 c 2 0 c a 2c 2a F1F2P SECTION 10.5 CONIC SECTIONS \|\|\|\| 657 + =1, FIGURE 8 ≈ a@ ¥ b@ (c, 0) 0 x y a b c (0, b) (c, 0) (0, _b) (a, 0) (_a, 0) a˘b 0 x y (0, a) (0, c) (b, 0) (0, _c) (_b, 0) (0, _a) ≈ b@ ¥ a@ + =1, a˘b FIGURE 9 0 x y (0, 3) {œ„ 7, 0} (4, 0) (_4, 0) (0, _3) {_œ„ 7, 0} FIGURE 10 9≈+16¥=144 Like parabolas, ellipses have an interesting reﬂection property that has practical conse-quences. If a source of light or sound is placed at one focus of a surface with elliptical cross-sections, then all the light or sound is reﬂected off the surface to the other focus (see Exercise 63). This principle is used in lithotripsy, a treatment for kidney stones. A reﬂec-tor with elliptical cross-section is placed in such a way that the kidney stone is at one focus. High-intensity sound waves generated at the other focus are reﬂected to the stone and destroy it without damaging surrounding tissue. The patient is spared the trauma of sur-gery and recovers within a few days. HYPERBOLAS A hyperbola is the set of all points in a plane the difference of whose distances from two ﬁxed points and (the foci) is a constant. This deﬁnition is illustrated in Figure 11. Hyperbolas occur frequently as graphs of equations in chemistry, physics, biology, and economics (Boyle’s Law, Ohm’s Law, supply and demand curves). A particularly signiﬁ-cant application of hyperbolas is found in the navigation systems developed in World Wars I and II (see Exercise 51). Notice that the deﬁnition of a hyperbola is similar to that of an ellipse; the only change is that the sum of distances has become a difference of distances. In fact, the derivation of the equation of a hyperbola is also similar to the one given earlier for an ellipse. It is left as Exercise 52 to show that when the foci are on the -axis at and the difference of distances is , then the equation of the hyperbola is where . Notice that the -intercepts are again and the points and are the vertices of the hyperbola. But if we put in Equation 6 we get , which is impossible, so there is no -intercept. The hyperbola is symmetric with respect to both axes. To analyze the hyperbola further, we look at Equation 6 and obtain This shows that , so . Therefore we have or . This means that the hyperbola consists of two parts, called its branches. When we draw a hyperbola it is useful to ﬁrst draw its asymptotes, which are the dashed lines and shown in Figure 12. Both branches of the hyperbola approach the asymptotes; that is, they come arbitrarily close to the asymptotes. [See Exercise 69 in Section 4.5, where these lines are shown to be slant asymptotes.] The hyperbola has foci , where , vertices , and asymptotes . y bax a, 0 c 2 a 2 b 2 c, 0 x 2 a 2 y 2 b 2 1 7 y bax y bax x a x a x sx 2 a x 2 a 2 x 2 a 2 1 y 2 b 2 1 y y 2 b 2 x 0 a, 0 a, 0 a x c 2 a 2 b 2 x 2 a 2 y 2 b 2 1 6 PF1 PF2 2a c, 0 x F2 F1 658 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES FIGURE 11 P is on the hyperbola when \|PF¡PF™\|= 2a. F™(c, 0) F¡(_c, 0) 0 x y P(x, y) (a, 0) FIGURE 12 ¥ b@ - =1 ≈ a@ (c, 0) 0 x y (_c, 0) (_a, 0) y= x b a y= x b a If the foci of a hyperbola are on the -axis, then by reversing the roles of and we obtain the following information, which is illustrated in Figure 13. The hyperbola has foci , where , vertices , and asymptotes . EXAMPLE 4 Find the foci and asymptotes of the hyperbola and sketch its graph. SOLUTION If we divide both sides of the equation by 144, it becomes which is of the form given in (7) with and . Since , the foci are . The asymptotes are the lines and . The graph is shown in Figure 14. M EXAMPLE 5 Find the foci and equation of the hyperbola with vertices and asymptote . SOLUTION From (8) and the given information, we see that and . Thus and . The foci are and the equation of the hyperbola is M SHIFTED CONICS As discussed in Appendix C, we shift conics by taking the standard equations (1), (2), (4), (5), (7), and (8) and replacing and by and . EXAMPLE 6 Find an equation of the ellipse with foci , and vertices , . SOLUTION The major axis is the line segment that joins the vertices , and has length , so . The distance between the foci is , so . Thus . Since the center of the ellipse is , we replace and in (4) by and to obtain as the equation of the ellipse. M x 32 4 y 22 3 1 y 2 x 3 y x 3, 2 b 2 a 2 c 2 3 c 1 2 a 2 4 5, 2 1, 2 5, 2 1, 2 4, 2 2, 2 y k x h y x y 2 4x 2 1 (0, s52) c 2 a 2 b 2 5 4 b a2 1 2 ab 2 a 1 y 2x 0, 1 y 3 4 x y 3 4 x 5, 0 c 2 16 9 25 b 3 a 4 x 2 16 y 2 9 1 9x 2 16y 2 144 y abx 0, a c 2 a 2 b 2 0, c y 2 a 2 x 2 b 2 1 8 y x y SECTION 10.5 CONIC SECTIONS \|\|\|\| 659 0 x y (0, c) (0, c) (0, a) (0, _a) y= x a b a b y= x FIGURE 13 ≈ b@ - =1 ¥ a@ FIGURE 14 9≈-16¥=144 0 x y (5, 0) (5, 0) (4, 0) (_4, 0) y= x 3 4 y= x 3 4 17–18 Find an equation of the ellipse. Then ﬁnd its foci. 17. 18. 19–24 Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph. 20. 21. 22. 23. 24. 25–30 Identify the type of conic section whose equation is given and ﬁnd the vertices and foci. 25. 26. 28. 29. 30. 4x 2 4x y 2 0 y 2 2y 4x 2 3 y 2 8y 6x 16 x 2 4y 2y 2 27. x 2 y 2 1 x 2 y 1 y2 4x 2 2y 16x 31 4x 2 y2 24x 4y 28 0 9x 2 4y 2 36 y 2 x 2 4 y 2 16 x 2 36 1 x 2 144 y 2 25 1 19. y x 1 2 y x 1 1 0 1–8 Find the vertex, focus, and directrix of the parabola and sketch its graph. 1. 2. 3. 4. 6. 7. 8. 9–10 Find an equation of the parabola. Then ﬁnd the focus and directrix. 9. 10. 11–16 Find the vertices and foci of the ellipse and sketch its graph. 11. 12. 13. 14. 16. x 2 3y2 2x 12y 10 0 9x 2 18x 4y 2 27 15. 4x 2 25y 2 25 4x 2 y 2 16 x 2 64 y 2 100 1 x 2 9 y 2 5 1 y x 1 2 0 y x 1 2 y 12x 2x 2 16 y 2 2y 12x 25 0 x 1 y 52 x 22 8y 3 5. y 2 12x 4x 2 y 4y x 2 0 x 2y 2 EXERCISES 10.5 660 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES EXAMPLE 7 Sketch the conic and ﬁnd its foci. SOLUTION We complete the squares as follows: This is in the form (8) except that and are replaced by and . Thus , , and . The hyperbola is shifted four units to the right and one unit upward. The foci are and and the vertices are and . The asymptotes are . The hyperbola is sketched in Figure 15. M y 1 3 2x 4 4, 2 4, 4 (4, 1 s13) (4, 1 s13) c 2 13 b 2 4 a 2 9 y 1 x 4 y x y 12 9 x 42 4 1 4y 12 9x 42 36 4y 2 2y 1 9x 2 8x 16 176 4 144 4y 2 2y 9x 2 8x 176 9x 2 4y 2 72x 8y 176 0 V FIGURE 15 9≈-4¥-72x+8y+176=0 0 x y y-1= (x-4) 3 2 y-1= (x-4) 3 2 (4, 4) (4, 2) (4, 1) SECTION 10.5 CONIC SECTIONS \|\|\|\| 661 51. In the LORAN (LOng RAnge Navigation) radio navigation system, two radio stations located at and transmit simul-taneous signals to a ship or an aircraft located at . The onboard computer converts the time difference in receiving these signals into a distance difference , and this, according to the deﬁnition of a hyperbola, locates the ship or aircraft on one branch of a hyperbola (see the ﬁgure). Suppose that station B is located 400 mi due east of station A on a coastline. A ship received the signal from B 1200 micro-seconds (s) before it received the signal from A. (a) Assuming that radio signals travel at a speed of 980 ft s, ﬁnd an equation of the hyperbola on which the ship lies. (b) If the ship is due north of , how far off the coastline is the ship? 52. Use the deﬁnition of a hyperbola to derive Equation 6 for a hyperbola with foci and vertices . 53. Show that the function deﬁned by the upper branch of the hyperbola is concave upward. 54. Find an equation for the ellipse with foci and and major axis of length 4. 55. Determine the type of curve represented by the equation in each of the following cases: (a) , (b) , and (c) . (d) Show that all the curves in parts (a) and (b) have the same foci, no matter what the value of is. 56. (a) Show that the equation of the tangent line to the parabola at the point can be written as (b) What is the -intercept of this tangent line? Use this fact to draw the tangent line. 57. Show that the tangent lines to the parabola drawn from any point on the directrix are perpendicular. 58. Show that if an ellipse and a hyperbola have the same foci, then their tangent lines at each point of intersection are perpendicular. 59. Use Simpson’s Rule with to estimate the length of the ellipse . 60. The planet Pluto travels in an elliptical orbit around the sun (at one focus). The length of the major axis is km 1.18 1010 x 2 4y 2 4 n 10 x 2 4py x y0y 2px x 0 x0, y0 y 2 4px k k 0 0 k 16 k 16 x 2 k y 2 k 16 1 1, 1 1, 1 y 2a 2 x 2b 2 1 a, 0 c, 0 400 mi transmitting stations coastline A B P B PA PB P B A 31–48 Find an equation for the conic that satisﬁes the given conditions. 31. Parabola, vertex , focus 32. Parabola, vertex , directrix Parabola, focus , directrix 34. Parabola, focus , vertex 35. Parabola, vertex , vertical axis, passing through 36. Parabola, horizontal axis, passing through , , and Ellipse, foci , vertices 38. Ellipse, foci , vertices 39. Ellipse, foci , , vertices , 40. Ellipse, foci , , vertex 41. Ellipse, center , vertex , focus 42. Ellipse, foci , passing through 43. Hyperbola, vertices , foci 44. Hyperbola, vertices , foci 45. Hyperbola, vertices , , foci , 46. Hyperbola, vertices , , foci , Hyperbola, vertices , asymptotes 48. Hyperbola, foci , , asymptotes and 49. The point in a lunar orbit nearest the surface of the moon is called perilune and the point farthest from the surface is called apolune. The Apollo 11 spacecraft was placed in an elliptical lunar orbit with perilune altitude 110 km and apolune altitude 314 km (above the moon). Find an equation of this ellipse if the radius of the moon is 1728 km and the center of the moon is at one focus. 50. A cross-section of a parabolic reﬂector is shown in the ﬁgure. The bulb is located at the focus and the opening at the focus is 10 cm. (a) Find an equation of the parabola. (b) Find the diameter of the opening , 11 cm from the vertex. 5 cm 5 cm A B C D V F 11 cm CD y 5 1 2 x y 3 1 2 x 2, 8 2, 0 y 2x 3, 0 47. 8, 2 2, 2 7, 2 1, 2 3, 9 3, 7 3, 6 3, 4 0, 5 0, 2 5, 0 3, 0 4, 1.8 4, 0 1, 6 1, 0 1, 4 9, 1 8, 1 0, 1 0, 8 0, 0 0, 6 0, 2 0, 13 0, 5 5, 0 2, 0 37. 3, 1 1, 1 1, 0 1, 5 2, 3 3, 2 3, 6 x 2 4, 0 33. x 5 1, 0 0, 2 0, 0 CONIC SECTIONS IN POLAR COORDINATES In the preceding section we deﬁned the parabola in terms of a focus and directrix, but we deﬁned the ellipse and hyperbola in terms of two foci. In this section we give a more uni-ﬁed treatment of all three types of conic sections in terms of a focus and directrix. Further-more, if we place the focus at the origin, then a conic section has a simple polar equation, which provides a convenient description of the motion of planets, satellites, and comets. THEOREM Let be a ﬁxed point (called the focus) and be a ﬁxed line (called the directrix) in a plane. Let be a ﬁxed positive number (called the eccentricity). The set of all points in the plane such that (that is, the ratio of the distance from to the distance from is the constant ) is a conic section. The conic is (a) (b) (c) a hyperbola if e 1 a parabola if e 1 an ellipse if e 1 e l F PF Pl e P e l F 1 10.6 and the length of the minor axis is km. Use Simp-son’s Rule with to estimate the distance traveled by the planet during one complete orbit around the sun. 61. Find the area of the region enclosed by the hyperbola and the vertical line through a focus. 62. (a) If an ellipse is rotated about its major axis, ﬁnd the volume of the resulting solid. (b) If it is rotated about its minor axis, ﬁnd the resulting volume. 63. Let be a point on the ellipse with foci and and let and be the angles between the lines , and the ellipse as shown in the ﬁgure. Prove that . This explains how whispering galleries and lithotripsy work. Sound coming from one focus is reﬂected and passes through the other focus. [Hint: Use the formula in Problem 17 on page 268 to show that .] F¡ F™ 0 x y ∫ å + =1 ≈ a@ ¥ b@ P(⁄, ›) tan tan PF2 PF1 F2 F1 x 2a 2 y 2b 2 1 P1x1, y1 x 2a 2 y 2b 2 1 n 10 1.14 1010 662 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 64. Let be a point on the hyperbola with foci and and let and be the angles between the lines , and the hyperbola as shown in the ﬁgure. Prove that . (This is the reﬂection property of the hyperbola. It shows that light aimed at a focus of a hyperbolic mirror is reﬂected toward the other focus .) 0 x y å ∫ F™ F¡ P F™ F¡ P F1 F2 PF2 PF1 F2 F1 x 2a 2 y 2b 2 1 Px1, y1 SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES \|\|\|\| 663 PROOF Notice that if the eccentricity is , then and so the given condi-tion simply becomes the deﬁnition of a parabola as given in Section 10.5. Let us place the focus at the origin and the directrix parallel to the -axis and units to the right. Thus the directrix has equation and is perpendicular to the polar axis. If the point has polar coordinates , we see from Figure 1 that Thus the condition , or , becomes If we square both sides of this polar equation and convert to rectangular coordinates, we get or After completing the square, we have If , we recognize Equation 3 as the equation of an ellipse. In fact, it is of the form where In Section 10.5 we found that the foci of an ellipse are at a distance from the center, where This shows that and conﬁrms that the focus as deﬁned in Theorem 1 means the same as the focus deﬁned in Section 10.5. It also follows from Equations 4 and 5 that the eccentricity is given by If , then and we see that Equation 3 represents a hyperbola. Just as we did before, we could rewrite Equation 3 in the form and see that M where c 2 a 2 b 2 e c a x h2 a 2 y 2 b 2 1 1 e 2 0 e 1 e c a c e 2d 1 e 2 h c 2 a 2 b 2 e 4d 2 1 e 22 5 c b 2 e 2d 2 1 e 2 a 2 e 2d 2 1 e 22 h e 2d 1 e 2 4 x h2 a 2 y 2 b 2 1 e 1 x e 2d 1 e 2 2 y 2 1 e 2 e 2d 2 1 e 22 3 1 e 2x 2 2de 2x y 2 e 2d 2 x 2 y 2 e 2d x2 e 2d 2 2dx x 2 r ed r cos 2 PF ePl PFPl e Pl d r cos PF r r, P x d d y F PF Pl e 1 FIGURE 1 y x F l (directrix) x=d r cos ¨ P ¨ r d C By solving Equation 2 for , we see that the polar equation of the conic shown in Fig-ure 1 can be written as If the directrix is chosen to be to the left of the focus as , or if the directrix is cho-sen to be parallel to the polar axis as , then the polar equation of the conic is given by the following theorem, which is illustrated by Figure 2. (See Exercises 21–23.) THEOREM A polar equation of the form represents a conic section with eccentricity . The conic is an ellipse if , a parabola if , or a hyperbola if . EXAMPLE 1 Find a polar equation for a parabola that has its focus at the origin and whose directrix is the line . SOLUTION Using Theorem 6 with and , and using part (d) of Figure 2, we see that the equation of the parabola is M EXAMPLE 2 A conic is given by the polar equation Find the eccentricity, identify the conic, locate the directrix, and sketch the conic. SOLUTION Dividing numerator and denominator by 3, we write the equation as r 10 3 1 2 3 cos r 10 3 2 cos V r 6 1 sin d 6 e 1 y 6 V e 1 e 1 e 1 e r ed 1 e sin or r ed 1 e cos 6 FIGURE 2 Polar equations of conics (a) r= ed 1+e cos ¨ y x F x=d directrix (b) r= ed 1-e cos ¨ x F y x=_d directrix (c) r= ed 1+e sin ¨ y F x y=d directrix (d) r= ed 1-e sin ¨ x y y=_d directrix F y d x d r ed 1 e cos r 664 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES \|\|\|\| 665 From Theorem 6 we see that this represents an ellipse with . Since , we have so the directrix has Cartesian equation . When , ; when , . So the vertices have polar coordinates and . The ellipse is sketched in Figure 3. M EXAMPLE 3 Sketch the conic . SOLUTION Writing the equation in the form we see that the eccentricity is and the equation therefore represents a hyperbola. Since , and the directrix has equation . The vertices occur when and , so they are and . It is also useful to plot the -intercepts. These occur when , ; in both cases . For additional accuracy we could draw the asymptotes. Note that when or and when . Thus the asymptotes are parallel to the rays and . The hyperbola is sketched in Figure 4. M When rotating conic sections, we ﬁnd it much more convenient to use polar equations than Cartesian equations. We just use the fact (see Exercise 77 in Section 10.3) that the graph of is the graph of rotated counterclockwise about the origin through an angle . EXAMPLE 4 If the ellipse of Example 2 is rotated through an angle about the origin, ﬁnd a polar equation and graph the resulting ellipse. SOLUTION We get the equation of the rotated ellipse by replacing with in the equation given in Example 2. So the new equation is We use this equation to graph the rotated ellipse in Figure 5. Notice that the ellipse has been rotated about its left focus. M r 10 3 2 cos 4 4 4 V r f r f FIGURE 4 r= 12 2+4 sin ¨ x 0 y (6, π) (6, 0) y=3 (directrix) focus ”2, ’ π 2 ”6, ’ π 2 116 76 sin 1 2 1 2 sin 0 0 1 2 sin l 0 r l r 6 0 x 6, 32 6, 2 2, 2 32 2 y 3 d 3 ed 6 e 2 r 6 1 2 sin r 12 2 4 sin 2, 10, 0 r 2 r 10 0 x 5 d 10 3 e 10 3 2 3 5 ed 10 3 e 2 3 FIGURE 5 11 _6 _5 15 r= 10 3-2 cos(¨-π/4) r= 10 3-2 cos ¨ FIGURE 3 y 0 x r= 10 3-2 cos ¨ x=_5 (directrix) (10, 0) (2, π) focus In Figure 6 we use a computer to sketch a number of conics to demonstrate the effect of varying the eccentricity . Notice that when is close to 0 the ellipse is nearly circular, whereas it becomes more elongated as . When , of course, the conic is a parabola. KEPLER’S LAWS In 1609 the German mathematician and astronomer Johannes Kepler, on the basis of huge amounts of astronomical data, published the following three laws of planetary motion. KEPLER’S LAWS 1. A planet revolves around the sun in an elliptical orbit with the sun at one focus. 2. The line joining the sun to a planet sweeps out equal areas in equal times. 3. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit. Although Kepler formulated his laws in terms of the motion of planets around the sun, they apply equally well to the motion of moons, comets, satellites, and other bodies that orbit subject to a single gravitational force. In Section 13.4 we will show how to deduce Kepler’s Laws from Newton’s Laws. Here we use Kepler’s First Law, together with the polar equation of an ellipse, to calculate quantities of interest in astronomy. For purposes of astronomical calculations, it’s useful to express the equation of an ellipse in terms of its eccentricity and its semimajor axis . We can write the distance from the focus to the directrix in terms of if we use (4): So . If the directrix is , then the polar equation is r ed 1 e cos a1 e2 1 e cos x d ed a1 e2 a2 e2d 2 1 e 22 ? d 2 a 21 e22 e2 ? d a1 e2 e a d a e FIGURE 6 e=1 e=1.1 e=1.4 e=4 e=0.96 e=0.86 e=0.68 e=0.1 e=0.5 e 1 e l 1 e e 666 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES \|\|\|\| 667 The polar equation of an ellipse with focus at the origin, semimajor axis , eccentricity , and directrix can be written in the form The positions of a planet that are closest to and farthest from the sun are called its peri-helion and aphelion, respectively, and correspond to the vertices of the ellipse. (See Figure 7.) The distances from the sun to the perihelion and aphelion are called the peri-helion distance and aphelion distance, respectively. In Figure 1 the sun is at the focus , so at perihelion we have and, from Equation 7, Similarly, at aphelion and . The perihelion distance from a planet to the sun is and the aphelion distance is . EXAMPLE 5 (a) Find an approximate polar equation for the elliptical orbit of the earth around the sun (at one focus) given that the eccentricity is about and the length of the major axis is about . (b) Find the distance from the earth to the sun at perihelion and at aphelion. SOLUTION (a) The length of the major axis is , so . We are given that and so, from Equation 7, an equation of the earth’s orbit around the sun is or, approximately, (b) From (8), the perihelion distance from the earth to the sun is and the aphelion distance is M a1 e 1.495 1081 0.017 1.52 108 km a1 e 1.495 1081 0.017 1.47 108 km r 1.49 108 1 0.017 cos r a1 e2 1 e cos 1.495 108 1 0.0172 1 0.017 cos e 0.017 a 1.495 108 2a 2.99 108 2.99 108 km 0.017 a1 e a1 e 8 r a1 e r a1 e2 1 e cos 0 a1 e1 e 1 e a1 e 0 F r a1 e2 1 e cos x d e a 7 perihelion aphelion sun planet ¨ r FIGURE 7 Show that a conic with focus at the origin, eccentricity , and directrix has polar equation 22. Show that a conic with focus at the origin, eccentricity , and directrix has polar equation 23. Show that a conic with focus at the origin, eccentricity , and directrix has polar equation 24. Show that the parabolas and intersect at right angles. 25. The orbit of Mars around the sun is an ellipse with eccen-tricity and semimajor axis . Find a polar equation for the orbit. 26. Jupiter’s orbit has eccentricity and the length of the major axis is . Find a polar equation for the orbit. The orbit of Halley’s comet, last seen in 1986 and due to return in 2062, is an ellipse with eccentricity 0.97 and one focus at the sun. The length of its major axis is 36.18 AU. [An astronomical unit (AU) is the mean distance between the earth and the sun, about 93 million miles.] Find a polar equa-tion for the orbit of Halley’s comet. What is the maximum distance from the comet to the sun? 28. The Hale-Bopp comet, discovered in 1995, has an elliptical orbit with eccentricity 0.9951 and the length of the major axis is 356.5 AU. Find a polar equation for the orbit of this comet. How close to the sun does it come? 29. The planet Mercury travels in an elliptical orbit with eccen-tricity . Its minimum distance from the sun is km. Find its maximum distance from the sun. 30. The distance from the planet Pluto to the sun is km at perihelion and km at aphelion. Find the eccentricity of Pluto’s orbit. 31. Using the data from Exercise 29, ﬁnd the distance traveled by the planet Mercury during one complete orbit around the sun. (If your calculator or computer algebra system evaluates deﬁ-nite integrals, use it. Otherwise, use Simpson’s Rule.) 7.37 109 4.43 109 4.6 107 0.206 27. 1.56 109 km 0.048 2.28 108 km 0.093 r d1 cos r c1 cos r ed 1 e sin y d e r ed 1 e sin y d e r ed 1 e cos x d e 21. 1–8 Write a polar equation of a conic with the focus at the origin and the given data. 1. Hyperbola, eccentricity , directrix 2. Parabola, directrix Ellipse, eccentricity , directrix 4. Hyperbola, eccentricity 2, directrix 5. Parabola, vertex 6. Ellipse, eccentricity , vertex 7. Ellipse, eccentricity , directrix 8. Hyperbola, eccentricity 3, directrix 9–16 (a) Find the eccentricity, (b) identify the conic, (c) give an equation of the directrix, and (d) sketch the conic. 9. 10. 11. 12. 14. 15. 16. ; 17. (a) Find the eccentricity and directrix of the conic and graph the conic and its directrix. (b) If this conic is rotated counterclockwise about the origin through an angle , write the resulting equation and graph its curve. ; 18. Graph the conic and its directrix. Also graph the conic obtained by rotating this curve about the ori-gin through an angle . ; 19. Graph the conics with , , , and on a common screen. How does the value of affect the shape of the curve? ; 20. (a) Graph the conics for and var-ious values of . How does the value of affect the shape of the conic? (b) Graph these conics for and various values of . How does the value of affect the shape of the conic? e e d 1 d d e 1 r ed1 e sin e 1.0 0.8 0.6 e 0.4 r e1 e cos 3 r 45 6 cos 34 r 11 2 sin r 10 5 6 sin r 3 4 8 cos r 8 4 5 sin r 9 6 2 cos 13. r 3 2 2 cos r 12 4 sin r 12 3 10 cos r 1 1 sin r 6 csc r 4 sec 1 2 1, 2 0.8 4, 32 y 2 x 5 3 4 3. x 4 y 6 7 4 668 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES EXERCISES 10.6 CHAPTER 10 REVIEW \|\|\|\| 669 REVIEW CONCEPT CHECK 10 6. (a) Give a geometric deﬁnition of a parabola. (b) Write an equation of a parabola with focus and direc-trix . What if the focus is and the directrix is ? 7. (a) Give a deﬁnition of an ellipse in terms of foci. (b) Write an equation for the ellipse with foci and vertices . 8. (a) Give a deﬁnition of a hyperbola in terms of foci. (b) Write an equation for the hyperbola with foci and vertices . (c) Write equations for the asymptotes of the hyperbola in part (b). 9. (a) What is the eccentricity of a conic section? (b) What can you say about the eccentricity if the conic section is an ellipse? A hyperbola? A parabola? (c) Write a polar equation for a conic section with eccentricity and directrix . What if the directrix is ? ? ? y d y d x d x d e a, 0 c, 0 a, 0 c, 0 x p p, 0 y p 0, p 1. (a) What is a parametric curve? (b) How do you sketch a parametric curve? 2. (a) How do you ﬁnd the slope of a tangent to a parametric curve? (b) How do you ﬁnd the area under a parametric curve? 3. Write an expression for each of the following: (a) The length of a parametric curve (b) The area of the surface obtained by rotating a parametric curve about the 4. (a) Use a diagram to explain the meaning of the polar coordi-nates of a point. (b) Write equations that express the Cartesian coordinates of a point in terms of the polar coordinates. (c) What equations would you use to ﬁnd the polar coordinates of a point if you knew the Cartesian coordinates? 5. (a) How do you ﬁnd the slope of a tangent line to a polar curve? (b) How do you ﬁnd the area of a region bounded by a polar curve? (c) How do you ﬁnd the length of a polar curve? x, y r, x-axis Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. If the parametric curve , satisﬁes , then it has a horizontal tangent when . 2. If and are twice differentiable, then 3. The length of the curve , , , is . 4. If a point is represented by in Cartesian coordinates (where ) and in polar coordinates, then . tan 1 yx r, x 0 x, y xb a s f t 2 tt 2 dt a t b y tt x f t d 2y dx 2 d 2ydt 2 d 2xdt 2 y tt x f t t 1 t1 0 y tt x f t 5. The polar curves and have the same graph. 6. The equations , , and , all have the same graph. 7. The parametric equations , have the same graph as , . 8. The graph of is a parabola. 9. A tangent line to a parabola intersects the parabola only once. 10. A hyperbola never intersects its directrix. y 2 2y 3x y t 6 x t 3 y t 4 x t 2 0 t 2 y 2 cos 3t x 2 sin 3t x 2 y 2 4 r 2 r sin 2 1 r 1 sin 2 TRUE-FALSE QUIZ TRUE-FALSE QUIZ 670 \|\|\|\| CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 1–4 Sketch the parametric curve and eliminate the parameter to ﬁnd the Cartesian equation of the curve. 1. , , 2. , 3. , , 4. , 5. Write three different sets of parametric equations for the curve . 6. Use the graphs of and to sketch the para-metric curve , . Indicate with arrows the direction in which the curve is traced as increases. 7. (a) Plot the point with polar coordinates . Then ﬁnd its Cartesian coordinates. (b) The Cartesian coordinates of a point are . Find two sets of polar coordinates for the point. 8. Sketch the region consisting of points whose polar coor-dinates satisfy . 9–16 Sketch the polar curve. 9. 10. 11. 12. 13. 14. 15. 16. 17–18 Find a polar equation for the curve represented by the given Cartesian equation. 17. 18. ; 19. The curve with polar equation is called a cochleoid. Use a graph of as a function of in Cartesian coordinates to sketch the cochleoid by hand. Then graph it with a machine to check your sketch. ; 20. Graph the ellipse and its directrix. Also graph the ellipse obtained by rotation about the origin through an angle . 23 r 24 3 cos r r sin x 2 y 2 2 x y 2 r 3 2 2 cos r 3 1 2 sin r 2 cos2 r 1 cos 2 r 3 cos 3 r cos 3 r sin 4 r 1 cos 1 r 2 and 6 56 3, 3 4, 23 t x _1 1 t y 1 1 t y tt x f t y tt x f t y sx y 1 sin x 2 cos 0 2 y sec x cos y e t x 1 e 2t 4 t 1 y 2 t x t 2 4t 21–24 Find the slope of the tangent line to the given curve at the point corresponding to the speciﬁed value of the parameter. 21. , ; 22. , ; 23. ; 24. ; 25–26 Find and . 25. , 26. , ; 27. Use a graph to estimate the coordinates of the lowest point on the curve , . Then use calculus to ﬁnd the exact coordinates. 28. Find the area enclosed by the loop of the curve in Exercise 27. 29. At what points does the curve have vertical or horizontal tangents? Use this information to help sketch the curve. 30. Find the area enclosed by the curve in Exercise 29. 31. Find the area enclosed by the curve . 32. Find the area enclosed by the inner loop of the curve . 33. Find the points of intersection of the curves and . 34. Find the points of intersection of the curves and . 35. Find the area of the region that lies inside both of the circles and . 36. Find the area of the region that lies inside the curve but outside the curve . 37–40 Find the length of the curve. 37. , , 38. , , 39. , 40. , 0 r sin33 2 r 1 0 t 1 y cosh 3t x 2 3t 0 t 2 y 2t 3 x 3t 2 r 2 sin r 2 cos 2 r sin cos r 2 sin r 2 cos r cot r 4 cos r 2 r 1 3 sin r 2 9 cos 5 y 2a sin t a sin 2t x 2a cos t a cos 2t y t 2 t 1 x t 3 3t y t t 3 x 1 t 2 y t cos t x t sin t d 2ydx 2 dydx 2 r 3 cos 3 r e t 1 y 2t t 2 x t 3 6t 1 t 1 y 1 t 2 x ln t EXERCISES 52. Find an equation of the ellipse with foci and major axis with length 8. 53. Find an equation for the ellipse that shares a vertex and a focus with the parabola and that has its other focus at the origin. 54. Show that if is any real number, then there are exactly two lines of slope that are tangent to the ellipse and their equations are . 55. Find a polar equation for the ellipse with focus at the origin, eccentricity . 56. Show that the angles between the polar axis and the asymptotes of the hyperbola , , are given by . 57. In the ﬁgure the circle of radius is stationary, and for every , the point is the midpoint of the segment . The curve traced out by for is called the longbow curve. Find parametric equations for this curve. 0 y 2a a x y=2a ¨ R P Q 0 P QR P a cos11e e 1 r ed1 e cos 1 3, and directrix with equation r 4 sec y mx sa 2m 2 b 2 x 2a 2 y 2b 2 1 m m x 2 y 100 3, 2 41–42 Find the area of the surface obtained by rotating the given curve about the -axis. 41. , , 42. , , ; 43. The curves deﬁned by the parametric equations are called strophoids (from a Greek word meaning “to turn or twist”). Investigate how these curves vary as varies. ; 44. A family of curves has polar equations where is a positive number. Investigate how the curves change as changes. 45–48 Find the foci and vertices and sketch the graph. 45. 46. 47. 48. 49. Find an equation of the ellipse with foci and vertices . 50. Find an equation of the parabola with focus and direc-trix . 51. Find an equation of the hyperbola with foci and asymptotes . y 3x 0, 4 x 4 2, 1 5, 0 4, 0 25x 2 4y 2 50x 16y 59 6y 2 x 36y 55 0 4x 2 y 2 16 x 2 9 y 2 8 1 a a r a sin 2 c y tt 2 c t 2 1 x t 2 c t 2 1 0 t 1 y cosh 3t x 2 3t 1 t 4 y t 3 3 1 2t 2 x 4st x CHAPTER 10 REVIEW \|\|\|\| 671 672 1. A curve is deﬁned by the parametric equations Find the length of the arc of the curve from the origin to the nearest point where there is a vertical tangent line. 2. (a) Find the highest and lowest points on the curve . (b) Sketch the curve. (Notice that it is symmetric with respect to both axes and both of the lines , so it sufﬁces to consider initially.) (c) Use polar coordinates and a computer algebra system to ﬁnd the area enclosed by the curve. ; 3. What is the smallest viewing rectangle that contains every member of the family of polar curves , where ? Illustrate your answer by graphing several mem-bers of the family in this viewing rectangle. 4. Four bugs are placed at the four corners of a square with side length . The bugs crawl counterclockwise at the same speed and each bug crawls directly toward the next bug at all times. They approach the center of the square along spiral paths. (a) Find the polar equation of a bug’s path assuming the pole is at the center of the square. (Use the fact that the line joining one bug to the next is tangent to the bug’s path.) (b) Find the distance traveled by a bug by the time it meets the other bugs at the center. 5. A curve called the folium of Descartes is deﬁned by the parametric equations (a) Show that if lies on the curve, then so does ; that is, the curve is symmetric with respect to the line . Where does the curve intersect this line? (b) Find the points on the curve where the tangent lines are horizontal or vertical. (c) Show that the line is a slant asymptote. (d) Sketch the curve. (e) Show that a Cartesian equation of this curve is . (f) Show that the polar equation can be written in the form (g) Find the area enclosed by the loop of this curve. (h) Show that the area of the loop is the same as the area that lies between the asymptote and the inﬁnite branches of the curve. (Use a computer algebra system to evaluate the integral.) CAS r 3 sec tan 1 tan3 x 3 y 3 3xy y x 1 y x b, a a, b x 3t 1 t 3 y 3t 2 1 t 3 a a a a a 0 c 1 r 1 c sin CAS y x 0 y x x 4 y 4 x 2 y 2 x y t 1 cos u u du y y t 1 sin u u du P R O B L E M S P L U S 673 6. A circle of radius has its center at the origin. A circle of radius rolls without slipping in the counterclockwise direction around . A point is located on a ﬁxed radius of the rolling circle at a distance from its center, . [See parts (i) and (ii) of the ﬁgure.] Let be the line from the center of to the center of the rolling circle and let be the angle that makes with the positive -axis. (a) Using as a parameter, show that parametric equations of the path traced out by are Note: If , the path is a circle of radius ; if , the path is an epicycloid. The path traced out by for is called an epitrochoid. ; (b) Graph the curve for various values of between and . (c) Show that an equilateral triangle can be inscribed in the epitrochoid and that its centroid is on the circle of radius centered at the origin. Note: This is the principle of the Wankel rotary engine. When the equilateral triangle rotates with its vertices on the epitrochoid, its centroid sweeps out a circle whose center is at the center of the curve. (d) In most rotary engines the sides of the equilateral triangles are replaced by arcs of circles centered at the opposite vertices as in part (iii) of the ﬁgure. (Then the diameter of the rotor is constant.) Show that the rotor will ﬁt in the epitrochoid if . (ii) y x P¸ ¨ P y x r b P=P¸ 2r (i) (iii) b 3 2(2 s3 )r b r 0 b 0 b r P b r 3r b 0 x b cos 3 3r cos y b sin 3 3r sin P x L C L 0 b r b P C r 2r C P R O B L E M S P L U S 674 INFINITE SEQUENCES AND SERIES 11 Inﬁnite sequences and series were introduced brieﬂy in A Preview of Calculus in connection with Zeno’s paradoxes and the decimal representation of numbers. Their importance in calculus stems from Newton’s idea of representing functions as sums of inﬁnite series. For instance, in ﬁnding areas he often integrated a function by ﬁrst expressing it as a series and then integrating each term of the series. We will pursue his idea in Section 11.10 in order to integrate such functions as . (Recall that we have previously been unable to do this.) Many of the functions that arise in mathematical physics and chemistry, such as Bessel functions, are deﬁned as sums of series, so it is important to be familiar with the basic concepts of convergence of inﬁnite sequences and series. Physicists also use series in another way, as we will see in Section 11.11. In studying ﬁelds as diverse as optics, special relativity, and electromagnetism, they analyze phe-nomena by replacing a function with the ﬁrst few terms in the series that represents it. ex 2 x y T∞ y=sin x T¡ T¶ T£ The partial sums of a Taylor series provide better and better approximations to a function as increases. n Tn SEQUENCES A sequence can be thought of as a list of numbers written in a deﬁnite order: The number is called the ﬁrst term, is the second term, and in general is the nth term. We will deal exclusively with inﬁnite sequences and so each term will have a successor . Notice that for every positive integer there is a corresponding number and so a sequence can be deﬁned as a function whose domain is the set of positive integers. But we usually write instead of the function notation for the value of the function at the number . The sequence { , , , . . .} is also denoted by EXAMPLE 1 Some sequences can be deﬁned by giving a formula for the nth term. In the following examples we give three descriptions of the sequence: one by using the preced-ing notation, another by using the deﬁning formula, and a third by writing out the terms of the sequence. Notice that doesn’t have to start at 1. (a) (b) (c) (d) M EXAMPLE 2 Find a formula for the general term of the sequence assuming that the pattern of the ﬁrst few terms continues. SOLUTION We are given that Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the th term will have numerator . The denominators are the powers of 5, so has denominator . The signs of the terms are alternately positive and negative, so 5 n an n 2 n a 5 7 3125 a 4 6 625 a 3 5 125 a 2 4 25 a 1 3 5 3 5 , 4 25 , 5 125 , 6 625 , 7 3125 , . . . an V 1, s3 2 , 1 2 , 0, . . . , cos n 6 , . . . an cos n 6 , n 0 cos n 6 n0 {0, 1, s2 , s3 , . . . , sn 3 , . . .} an sn 3 , n 3 {sn 3 }n3 2 3 , 3 9 , 4 27 , 5 81 , . . . , 1nn 1 3n , . . . an 1nn 1 3n 1nn 1 3n 1 2 , 2 3 , 3 4 , 4 5 , . . . , n n 1 , . . . an n n 1 n n 1 n1 n an n1 or an a3 a2 a1 NOTATION n fn an an n an1 an an a2 a1 a1, a2, a3, a4, . . . , an, . . . 11.1 675 we need to multiply by a power of . In Example 1(b) the factor meant we started with a negative term. Here we want to start with a positive term and so we use or . Therefore M EXAMPLE 3 Here are some sequences that don’t have a simple deﬁning equation. (a) The sequence , where is the population of the world as of January 1 in the year . (b) If we let be the digit in the th decimal place of the number , then is a well-deﬁned sequence whose ﬁrst few terms are (c) The Fibonacci sequence is deﬁned recursively by the conditions Each term is the sum of the two preceding terms. The ﬁrst few terms are This sequence arose when the 13th-century Italian mathematician known as Fibonacci solved a problem concerning the breeding of rabbits (see Exercise 71). M A sequence such as the one in Example 1(a), , can be pictured either by plotting its terms on a number line as in Figure 1 or by plotting its graph as in Figure 2. Note that, since a sequence is a function whose domain is the set of positive integers, its graph consists of isolated points with coordinates . . . . . . From Figure 1 or 2 it appears that the terms of the sequence are approaching 1 as becomes large. In fact, the difference can be made as small as we like by taking sufﬁciently large. We indicate this by writing In general, the notation means that the terms of the sequence approach as becomes large. Notice that the following deﬁnition of the limit of a sequence is very similar to the deﬁnition of a limit of a function at inﬁnity given in Section 2.6. n L an lim n l an L lim nl n n 1 1 n 1 n n 1 1 n 1 n an nn 1 n, an 3, a3 2, a2 1, a1 an nn 1 1, 1, 2, 3, 5, 8, 13, 21, . . . n 3 fn fn1 fn2 f2 1 f1 1 fn 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . . an e n an n pn pn an 1 n1 n 2 5 n 1 n1 1 n1 1 n 1 676 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES 0 1 1 2 a¡ a™a£ a¢ FIGURE 1 FIGURE 2 0 n an 1 1 2 3 4 5 6 7 7 8 a¶= DEFINITION A sequence has the limit and we write if we can make the terms as close to as we like by taking sufﬁciently large. If exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent). Figure 3 illustrates Deﬁnition 1 by showing the graphs of two sequences that have the limit . A more precise version of Deﬁnition 1 is as follows. DEFINITION A sequence has the limit and we write if for every there is a corresponding integer such that if then Deﬁnition 2 is illustrated by Figure 4, in which the terms , , , . . . are plotted on a number line. No matter how small an interval is chosen, there exists an such that all terms of the sequence from onward must lie in that interval. Another illustration of Deﬁnition 2 is given in Figure 5. The points on the graph of must lie between the horizontal lines and if . This picture must be valid no matter how small is chosen, but usually a smaller requires a larger . FIGURE 5 2 0 n y 1 3 4 L y=L+∑ N y=L-∑ N n N y L y L an FIGURE 4 0 L-∑ L L+∑ a¡ a£ a¢ a™ a∞ aß a¶ aˆ a˜ aN+1 aN+2 aN1 N L , L a3 a2 a1 an L n N N 0 an l L as n l or lim nl an L L an 2 0 n an L 0 n an L FIGURE 3 Graphs of two sequences with lim an= L n L limn l an n L an an l L as n l or lim n l an L L an 1 SECTION 11.1 SEQUENCES \|\|\|\| 677 N Compare this deﬁnition with Deﬁnition 2.6.7. If you compare Deﬁnition 2 with Deﬁnition 2.6.7, you will see that the only difference between and is that is required to be an integer. Thus we have the following theorem, which is illustrated by Figure 6. THEOREM If and when is an integer, then . In particular, since we know that when (Theorem 2.6.5), we have if If becomes large as n becomes large, we use the notation . The fol-lowing precise deﬁnition is similar to Deﬁnition 2.6.9. DEFINITION means that for every positive number there is an integer such that if then If , then the sequence is divergent but in a special way. We say that diverges to . The Limit Laws given in Section 2.3 also hold for the limits of sequences and their proofs are similar. If and are convergent sequences and is a constant, then lim n l an p [lim n l an] p if p 0 and an 0 lim n l an bn lim n l an lim n l bn if lim n l bn 0 lim n l anbn lim n l an lim n l bn lim n l c c lim n l can c lim n l an lim n l an bn lim n l an lim n l bn lim n l an bn lim n l an lim n l bn c bn an an an lim n l an an M n N N M limnl an 5 lim n l an an r 0 lim nl 1 nr 0 4 r 0 limx l 1x r 0 FIGURE 6 2 0 x y 1 3 4 L y=ƒ limnl an L n fn an lim x l fx L 3 n limx l fx L limn l an L 678 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES LIMIT LAWS FOR SEQUENCES The Squeeze Theorem can also be adapted for sequences as follows (see Figure 7). If for and , then . Another useful fact about limits of sequences is given by the following theorem, whose proof is left as Exercise 75. THEOREM If , then . EXAMPLE 4 Find . SOLUTION The method is similar to the one we used in Section 2.6: Divide numerator and denominator by the highest power of and then use the Limit Laws. Here we used Equation 4 with . M EXAMPLE 5 Calculate . SOLUTION Notice that both numerator and denominator approach inﬁnity as . We can’t apply l’Hospital’s Rule directly because it applies not to sequences but to func-tions of a real variable. However, we can apply l’Hospital’s Rule to the related function and obtain Therefore, by Theorem 3, we have M EXAMPLE 6 Determine whether the sequence is convergent or divergent. SOLUTION If we write out the terms of the sequence, we obtain The graph of this sequence is shown in Figure 8. Since the terms oscillate between 1 and inﬁnitely often, does not approach any number. Thus does not exist; that is, the sequence is divergent. M 1n limnl 1n an 1 1, 1, 1, 1, 1, 1, 1, . . . an 1n lim nl ln n n 0 lim xl ln x x lim xl 1x 1 0 fx ln xx n l lim nl ln n n r 1 1 1 0 1 lim n l n n 1 lim n l 1 1 1 n lim n l 1 lim n l 1 lim n l 1 n n lim nl n n 1 lim n l an 0 lim n l an 0 6 lim nl bn L lim n l an lim n l cn L n n0 an bn cn SECTION 11.1 SEQUENCES \|\|\|\| 679 N This shows that the guess we made earlier from Figures 1 and 2 was correct. FIGURE 7 The sequence bn is squeezed between the sequences an and cn. 0 n cn an bn n n n SQUEEZE THEOREM FOR SEQUENCES 0 n an 1 1 2 3 4 _1 FIGURE 8 EXAMPLE 7 Evaluate if it exists. SOLUTION Therefore, by Theorem 6, M The following theorem says that if we apply a continuous function to the terms of a con-vergent sequence, the result is also convergent. The proof is left as Exercise 76. THEOREM If and the function is continuous at , then EXAMPLE 8 Find . SOLUTION Because the sine function is continuous at , Theorem 7 enables us to write M EXAMPLE 9 Discuss the convergence of the sequence , where . SOLUTION Both numerator and denominator approach inﬁnity as , but here we have no corresponding function for use with l’Hospital’s Rule ( is not deﬁned when is not an integer). Let’s write out a few terms to get a feeling for what happens to as gets large: It appears from these expressions and the graph in Figure 10 that the terms are decreas-ing and perhaps approach 0. To conﬁrm this, observe from Equation 8 that Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator. So We know that as . Therefore as by the Squeeze Theorem. M n l an l 0 n l 1n l 0 0 an 1 n an 1 n 2 3 n n n n an 1 2 3 n n n n n 8 a3 1 2 3 3 3 3 a2 1 2 2 2 a1 1 n an x x! n l n! 1 2 3 n an n!n n V lim n l sinn sin lim n l n sin 0 0 0 lim n l sinn lim n l fan fL L f lim n l an L 7 lim n l 1n n 0 lim n l 1n n lim n l 1 n 0 lim n l 1n n 680 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N CREATING GRAPHS OF SEQUENCES Some computer algebra systems have special commands that enable us to create sequences and graph them directly. With most graphing calculators, however, sequences can be graphed by using parametric equations. For instance, the sequence in Example 9 can be graphed by enter-ing the parametric equations and graphing in dot mode, starting with and setting the -step equal to . The result is shown in Figure 10. 1 t t 1 x t y t!t t N The graph of the sequence in Example 7 is shown in Figure 9 and supports our answer. FIGURE 10 1 0 10 FIGURE 9 0 n an 1 1 _1 EXAMPLE 10 For what values of is the sequence convergent? SOLUTION We know from Section 2.6 and the graphs of the exponential functions in Section 1.5 that for and for . Therefore, putting and using Theorem 3, we have It is obvious that and If , then , so and therefore by Theorem 6. If , then diverges as in Example 6. Figure 11 shows the graphs for various values of . (The case is shown in Figure 8.) M The results of Example 10 are summarized for future use as follows. The sequence is convergent if and divergent for all other values of . DEFINITION A sequence is called increasing if for all , that is, It is called decreasing if for all . It is called monotonic if it is either increasing or decreasing. EXAMPLE 11 The sequence is decreasing because and so for all . M n 1 an an1 3 n 5 3 n 1 5 3 n 6 3 n 5 n 1 an an1 a1 a2 a3 . n 1 an an1 an 10 lim n l r n 0 1 if 1 r 1 if r 1 r 1 r 1 r n 9 r>1 r=1 0<r<1 0 r<_1 _1<r<0 0 n an 1 1 n an 1 1 FIGURE 11 The sequence an=rn r 1 r r n r 1 lim n l r n 0 lim n l r n lim n l r n 0 0 r 1 1 r 0 lim n l 0 n 0 lim nl 1n 1 lim n l r n 0 if r 1 if 0 r 1 a r 0 a 1 limxl ax 0 a 1 limxl ax r n r V SECTION 11.1 SEQUENCES \|\|\|\| 681 N The right side is smaller because it has a larger denominator. EXAMPLE 12 Show that the sequence is decreasing. SOLUTION 1 We must show that , that is, This inequality is equivalent to the one we get by cross-multiplication: Since , we know that the inequality is true. Therefore and so is decreasing. SOLUTION 2 Consider the function : Thus is decreasing on and so . Therefore is decreasing. M DEFINITION A sequence is bounded above if there is a number such that It is bounded below if there is a number such that If it is bounded above and below, then is a bounded sequence. For instance, the sequence is bounded below but not above. The sequence is bounded because for all . We know that not every bounded sequence is convergent [for instance, the sequence satisﬁes but is divergent from Example 6] and not every mono-tonic sequence is convergent . But if a sequence is both bounded and monotonic, then it must be convergent. This fact is proved as Theorem 12, but intuitively you can understand why it is true by looking at Figure 12. If is increasing and for all , then the terms are forced to crowd together and approach some number . The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers, which says that if is a nonempty set of real numbers that has an upper bound ( for all in ), then has a least upper bound . (This means that is an upper bound for , but if is any other upper bound, then .) The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line. b M M S b b S S x x M M S L n an M an an n l 1 an 1 an 1n n 0 an 1 an nn 1 an 0 an n an for all n 1 m an m for all n 1 an M M an 11 an fn fn 1 1, f whenever x2 1 fx x 2 1 2x 2 x 2 12 1 x 2 x 2 12 0 fx x x 2 1 an an1 an n2 n 1 n 1 1 n2 n & ? n3 n2 n 1 n3 2n2 2n & ? n 1n2 1 n n 12 1 & ? n 1 n 12 1 n n2 1 n 1 n 12 1 n n 2 1 an1 an an n n 2 1 682 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES 2 0 n an 1 3 L M FIGURE 12 MONOTONIC SEQUENCE THEOREM Every bounded, monotonic sequence is convergent. PROOF Suppose is an increasing sequence. Since is bounded, the set has an upper bound. By the Completeness Axiom it has a least upper bound . Given , is not an upper bound for (since is the least upper bound). Therefore But the sequence is increasing so for every . Thus if , we have so since . Thus so . A similar proof (using the greatest lower bound) works if is decreasing. M The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent. (Likewise, a decreasing sequence that is bounded below is convergent.) This fact is used many times in dealing with inﬁnite series. EXAMPLE 13 Investigate the sequence deﬁned by the recurrence relation SOLUTION We begin by computing the ﬁrst several terms: These initial terms suggest that the sequence is increasing and the terms are approaching 6. To conﬁrm that the sequence is increasing, we use mathematical induction to show that for all . This is true for because . If we assume that it is true for , then we have so and Thus ak2 ak1 1 2ak1 6 1 2ak 6 ak1 6 ak 6 ak1 ak n k a2 4 a1 n 1 n 1 an1 an a9 5.984375 a8 5.96875 a7 5.9375 a6 5.875 a5 5.75 a4 1 25 6 5.5 a3 1 24 6 5 a2 1 22 6 4 a1 2 for n 1, 2, 3, . . . an1 1 2an 6 a1 2 an an lim n l an L n N whenever L an an L 0 L an an L n N n N an aN for some integer N aN L L S L 0 L S an n 1 an an 12 SECTION 11.1 SEQUENCES \|\|\|\| 683 N Mathematical induction is often used in dealing with recursive sequences. See page 77 for a discussion of the Principle of Mathematical Induction. We have deduced that is true for . Therefore the inequality is true for all by induction. Next we verify that is bounded by showing that for all . (Since the sequence is increasing, we already know that it has a lower bound: for all .) We know that , so the assertion is true for . Suppose it is true for . Then so Thus This shows, by mathematical induction, that for all . Since the sequence is increasing and bounded, Theorem 12 guarantees that it has a limit. The theorem doesn’t tell us what the value of the limit is. But now that we know exists, we can use the recurrence relation to write Since , it follows that , too (as , too). So we have Solving this equation for , we get , as predicted. M L 6 L L 1 2L 6 n 1 l n l an1 l L an l L lim n l an1 lim n l 1 2an 6 1 2(lim n l an 6) 1 2L 6 L limnl an an n an 6 ak1 6 1 2ak 6 1 212 6 ak 6 12 ak 6 n k n 1 a1 6 n an a1 2 n an 6 an n n k 1 an1 an 684 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES EXERCISES 11.1 1. (a) What is a sequence? (b) What does it mean to say that ? (c) What does it mean to say that ? 2. (a) What is a convergent sequence? Give two examples. (b) What is a divergent sequence? Give two examples. 3–8 List the ﬁrst ﬁve terms of the sequence. 3. 4. 5. 6. 7. , 8. , 9–14 Find a formula for the general term of the sequence, assuming that the pattern of the ﬁrst few terms continues. 9. 10. {1, 1 3, 1 9, 1 27, 1 81, . . .} {1, 1 3, 1 5, 1 7, 1 9, . . .} an an1 an an 1 a1 4 an1 2an 1 a1 3 2 4 6 2n an 31n n! an n 1 3n 1 an 1 0.2n limn l an limn l an 8 11. 12. 14. 15. List the ﬁrst six terms of the sequence deﬁned by Does the sequence appear to have a limit? If so, ﬁnd it. 16. List the ﬁrst nine terms of the sequence . Does this sequence appear to have a limit? If so, ﬁnd it. If not, explain why. 17–46 Determine whether the sequence converges or diverges. If it converges, ﬁnd the limit. 17. 18. an n3 n3 1 an 1 0.2n cosn 3 an n 2n 1 5, 1, 5, 1, 5, 1, . . . {1, 2 3, 4 9, 8 27, . . .} 13. { 1 4, 2 9, 3 16, 4 25, . . .} 2, 7, 12, 17, . . . N A proof of this fact is requested in Exercise 58. SECTION 11.1 SEQUENCES \|\|\|\| 685 (a) Determine whether the sequence deﬁned as follows is convergent or divergent: (b) What happens if the ﬁrst term is ? 55. If $1000 is invested at 6% interest, compounded annually, then after years the investment is worth dollars. (a) Find the ﬁrst ﬁve terms of the sequence . (b) Is the sequence convergent or divergent? Explain. 56. Find the ﬁrst 40 terms of the sequence deﬁned by and . Do the same if . Make a conjecture about this type of sequence. 57. For what values of is the sequence convergent? 58. (a) If is convergent, show that (b) A sequence is deﬁned by and for . Assuming that is convergent, ﬁnd its limit. Suppose you know that is a decreasing sequence and all its terms lie between the numbers 5 and 8. Explain why the sequence has a limit. What can you say about the value of the limit? 60–66 Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? 60. 62. 63. 64. 65. 66. 67. Find the limit of the sequence 68. A sequence is given by , . (a) By induction or otherwise, show that is increasing and bounded above by 3. Apply the Monotonic Sequence Theorem to show that exists. (b) Find . limn l an limn l an an an1 s2 an a1 s2 an {s2 , s2s2 , s2s2s2 , . . .} an n 1 n an n n 2 1 an ne n an n1n an 2n 3 3n 4 an 1 2n 3 61. an 2n1 an 59. an n 1 an1 11 an a1 1 an lim n l an1 lim n l an an nr n r a1 25 a1 11 an1 1 2an 3an 1 if an is an even number if an is an odd number an an 10001.06n n a1 2 a1 1 an1 4 an for n 1 54. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 37. 38. 39. 40. 41. 42. 44. 45. 46. ; 47–53 Use a graph of the sequence to decide whether the sequence is convergent or divergent. If the sequence is conver-gent, guess the value of the limit from the graph and then prove your guess. (See the margin note on page 680 for advice on graphing sequences.) 47. 48. 49. 50. 51. 52. 53. an 1 3 5 2n 1 2nn an 1 3 5 2n 1 n! an n2 cos n 1 n2 an s n 3n 5n an 3 2n2 8n2 n an sn sin( sn ) an 1 2en an 3n n! an n! 2n { 1 1, 1 3, 1 2, 1 4, 1 3, 1 5, 1 4, 1 6, . . .} 0, 1, 0, 0, 1, 0, 0, 0, 1, . . . 43. an ln n2 n an ln2n2 1 lnn2 1 an sin 2n 1 sn an 1 2 n n an s n 213n an n sin1n an lnn 1 ln n 36. an cos2n 2n 35. n cos n n2e n ln n ln 2n e n e n e 2n 1 arctan 2n 2n 1! 2n 1! an cos2n an cosn2 an 1nn3 n3 2n2 1 an 1n1n n2 1 an n 1 9n 1 an tan 2n 1 8n an 3n2 5n an e1n an n3 n 1 an 3 5n2 n n2 19. 686 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES (e) Use parts (c) and (d) to show that for all . (f) Use Theorem 12 to show that exists. (The limit is . See Equation 3.6.6.) 79. Let and be positive numbers with . Let be their arithmetic mean and their geometric mean: Repeat this process so that, in general, (a) Use mathematical induction to show that (b) Deduce that both and are convergent. (c) Show that . Gauss called the common value of these limits the arithmetic-geometric mean of the numbers and . 80. (a) Show that if and , then is convergent and . (b) If and ﬁnd the ﬁrst eight terms of the sequence . Then use part (a) to show that . This gives the continued fraction expansion 81. The size of an undisturbed ﬁsh population has been modeled by the formula where is the ﬁsh population after years and and are positive constants that depend on the species and its environ-ment. Suppose that the population in year 0 is . (a) Show that if is convergent, then the only possible values for its limit are 0 and . (b) Show that . (c) Use part (b) to show that if , then ; in other words, the population dies out. (d) Now assume that . Show that if , then is increasing and . Show also that if , then is decreasing and . Deduce that if , then . limn l pn b a a b pn b a pn p 0 b a 0 pn b a pn p 0 b a a b limn l pn 0 a b pn1 bapn b a pn p0 0 b a n pn pn1 bpn a pn s2 1 1 2 1 2 lim n l an s2 an an1 1 1 1 an a1 1 lim n l an L an lim n l a2n1 L lim n l a2n L b a limn l an limn l bn bn an an an1 bn1 bn bn1 sanbn an1 an bn 2 b1 sab a1 a b 2 b1 a1 a b b a e lim n l 1 1nn n an 4 Show that the sequence deﬁned by is increasing and for all . Deduce that is conver-gent and ﬁnd its limit. 70. Show that the sequence deﬁned by satisﬁes and is decreasing. Deduce that the sequence is convergent and ﬁnd its limit. 71. (a) Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair which becomes productive at age 2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the month? Show that the answer is , where is the Fibonacci sequence deﬁned in Example 3(c). (b) Let and show that . Assuming that is convergent, ﬁnd its limit. 72. (a) Let , , , . . . , , where is a continuous function. If , show that . (b) Illustrate part (a) by taking , , and estimating the value of to ﬁve decimal places. ; 73. (a) Use a graph to guess the value of the limit (b) Use a graph of the sequence in part (a) to ﬁnd the smallest values of that correspond to and in Deﬁnition 2. 74. Use Deﬁnition 2 directly to prove that when . 75. Prove Theorem 6. [Hint: Use either Deﬁnition 2 or the Squeeze Theorem.] 76. Prove Theorem 7. 77. Prove that if and is bounded, then . 78. Let . (a) Show that if , then (b) Deduce that . (c) Use and in part (b) to show that is increasing. (d) Use and in part (b) to show that . a2n 4 b 1 12n a 1 an b 1 1n a 1 1n 1 b nn 1a nb a n1 b n1 a n1 b a n 1b n 0 a b an 1 1 n n limn l anbn 0 bn limn l an 0 r 1 lim n l r n 0 0.001 0.1 N lim n l n5 n! L a 1 f x cos x f L L limn l an L f an1 f an a3 f a2 f f a a2 f a a1 a an an1 1 1an2 an fn1fn fn fn nth 0 an 2 an1 1 3 an a1 2 an n an 3 an1 3 1 an a1 1 69. A sequence that arises in ecology as a model for population growth is deﬁned by the logistic difference equation where measures the size of the population of the generation of a single species. To keep the numbers manageable, is a fraction of the maximal size of the population, so . Notice that the form of this equation is similar to the logistic differential equation in Section 9.4. The discrete model—with sequences instead of continuous functions—is preferable for modeling insect populations, where mating and death occur in a periodic fashion. An ecologist is interested in predicting the size of the population as time goes on, and asks these questions: Will it stabilize at a limiting value? Will it change in a cyclical fashion? Or will it exhibit random behavior? Write a program to compute the ﬁrst terms of this sequence starting with an initial population . Use this program to do the following. 1. Calculate 20 or 30 terms of the sequence for and for two values of such that . Graph the sequences. Do they appear to converge? Repeat for a different value of between 0 and 1. Does the limit depend on the choice of ? Does it depend on the choice of ? 2. Calculate terms of the sequence for a value of between 3 and 3.4 and plot them. What do you notice about the behavior of the terms? 3. Experiment with values of between 3.4 and 3.5. What happens to the terms? 4. For values of between 3.6 and 4, compute and plot at least 100 terms and comment on the behavior of the sequence. What happens if you change by 0.001? This type of behavior is called chaotic and is exhibited by insect populations under certain conditions. p0 k k k k p0 p0 1 k 3 k p0 1 2 p0, where 0 p0 1 n 0 pn 1 pn nth pn pn1 kpn1 pn LOGISTIC SEQUENCES CAS L A B O R AT O R Y P R O J E C T SERIES If we try to add the terms of an inﬁnite sequence we get an expression of the form which is called an inﬁnite series (or just a series) and is denoted, for short, by the symbol But does it make sense to talk about the sum of inﬁnitely many terms? It would be impossible to ﬁnd a ﬁnite sum for the series because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15, 21, . . . and, after the term, we get , which becomes very large as increases. However, if we start to add the terms of the series 1 2 1 4 1 8 1 16 1 32 1 64 1 2n n nn 12 nth 1 2 3 4 5 n an or n1 an a1 a2 a3 an 1 ann1 11.2 SECTION 11.2 SERIES \|\|\|\| 687 we get , , , , , , . . . , , . . . . The table shows that as we add more and more terms, these partial sums become closer and closer to 1. (See also Figure 11 in A Preview of Calculus, page 7.) In fact, by adding sufﬁciently many terms of the series we can make the partial sums as close as we like to 1. So it seems reasonable to say that the sum of this inﬁnite series is 1 and to write We use a similar idea to determine whether or not a general series (1) has a sum. We consider the partial sums and, in general, These partial sums form a new sequence , which may or may not have a limit. If exists (as a ﬁnite number), then, as in the preceding example, we call it the sum of the inﬁnite series . DEFINITION Given a series , let denote its partial sum: If the sequence is convergent and exists as a real number, then the series is called convergent and we write The number is called the sum of the series. Otherwise, the series is called divergent. Thus the sum of a series is the limit of the sequence of partial sums. So when we write , we mean that by adding sufﬁciently many terms of the series we can get as close as we like to the number . Notice that EXAMPLE 1 An important example of an inﬁnite series is the geometric series a 0 a ar ar 2 ar 3 ar n1 n1 ar n1 n1 an lim n l n i1 ai s n1 an s s n1 an s or a1 a2 an s an lim n l sn s sn sn n i1 ai a1 a2 an nth sn n1 an a1 a2 a3 2 an lim n l sn s sn sn a1 a2 a3 an n i1 ai s4 a1 a2 a3 a4 s3 a1 a2 a3 s2 a1 a2 s1 a1 n1 1 2n 1 2 1 4 1 8 1 16 1 2n 1 1 12n 63 64 31 32 15 16 7 8 3 4 1 2 688 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES n Sum of ﬁrst n terms 1 0.50000000 2 0.75000000 3 0.87500000 4 0.93750000 5 0.96875000 6 0.98437500 7 0.99218750 10 0.99902344 15 0.99996948 20 0.99999905 25 0.99999997 N Compare with the improper integral To ﬁnd this integral, we integrate from 1 to and then let . For a series, we sum from 1 to and then let . nl n tl t y 1 f x dx lim t l y t 1 fx dx Each term is obtained from the preceding one by multiplying it by the common ratio . (We have already considered the special case where and on page 687.) If , then . Since doesn’t exist, the geometric series diverges in this case. If , we have and Subtracting these equations, we get If , we know from (11.1.9) that as , so Thus when the geometric series is convergent and its sum is . If or , the sequence is divergent by (11.1.9) and so, by Equation 3, does not exist. Therefore the geometric series diverges in those cases. M We summarize the results of Example 1 as follows. The geometric series is convergent if and its sum is If , the geometric series is divergent. EXAMPLE 2 Find the sum of the geometric series SOLUTION The ﬁrst term is and the common ratio is . Since , the series is convergent by (4) and its sum is M 5 10 3 20 9 40 27 5 1 ( 2 3) 5 5 3 3 r 2 3 1 r 2 3 a 5 5 10 3 20 9 40 27 V r 1 r 1 n1 ar n1 a 1 r r 1 n1 ar n1 a ar ar 2 4 lim n l sn r n r 1 r 1 a1 r r 1 lim n l sn lim n l a1 r n 1 r a 1 r a 1 r lim n l r n a 1 r n l r n l 0 1 r 1 sn a1 r n 1 r 3 sn rsn a ar n rsn ar ar 2 ar n1 ar n sn a ar ar 2 ar n1 r 1 lim n l sn sn a a a na l r 1 r 1 2 a 1 2 r SECTION 11.2 SERIES \|\|\|\| 689 N Figure 1 provides a geometric demonstration of the result in Example 1. If the triangles are constructed as shown and is the sum of the series, then, by similar triangles, s a a a ar so s a 1 r s FIGURE 1 a a a ar a-ar ar ar@ ar# ar@ s N In words: The sum of a convergent geometric series is first term 1 common ratio EXAMPLE 3 Is the series convergent or divergent? SOLUTION Let’s rewrite the term of the series in the form : We recognize this series as a geometric series with and . Since , the series diverges by (4). M EXAMPLE 4 Write the number . . . as a ratio of integers. SOLUTION After the ﬁrst term we have a geometric series with and . Therefore M EXAMPLE 5 Find the sum of the series , where . SOLUTION Notice that this series starts with and so the ﬁrst term is . (With series, we adopt the convention that even when .) Thus This is a geometric series with and . Since , it converges and (4) gives M n0 x n 1 1 x 5 r x 1 r x a 1 n0 x n 1 x x 2 x 3 x 4 x 0 x 0 1 x 0 1 n 0 x 1 n0 x n 23 10 17 990 1147 495 2.317 2.3 17 103 1 1 102 2.3 17 1000 99 100 r 1102 a 17103 2.3171717. . . 2.3 17 103 17 105 17 107 2.317 2.3171717 V r 1 r 4 3 a 4 n1 22n31n n1 2 2n 3n1 n1 4n 3n1 n1 4( 4 3) n1 ar n1 nth n1 22n31n FIGURE 2 0 n sn 20 3 690 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N What do we really mean when we say that the sum of the series in Example 2 is 3? Of course, we can’t literally add an inﬁnite number of terms, one by one. But, according to Deﬁnition 2, the total sum is the limit of the sequence of partial sums. So, by taking the sum of sufﬁciently many terms, we can get as close as we like to the number 3. The table shows the ﬁrst ten par-tial sums and the graph in Figure 2 shows how the sequence of partial sums approaches 3. sn N Another way to identify and is to write out the ﬁrst few terms: 4 16 3 64 9 r a Module 11.2 explores a series that depends on an angle in a triangle and enables you to see how rapidly the series converges when varies. TEC n 1 5.000000 2 1.666667 3 3.888889 4 2.407407 5 3.395062 6 2.736626 7 3.175583 8 2.882945 9 3.078037 10 2.947975 sn EXAMPLE 6 Show that the series is convergent, and ﬁnd its sum. SOLUTION This is not a geometric series, so we go back to the deﬁnition of a convergent series and compute the partial sums. We can simplify this expression if we use the partial fraction decomposition (see Section 7.4). Thus we have and so Therefore the given series is convergent and M EXAMPLE 7 Show that the harmonic series is divergent. SOLUTION For this particular series it’s convenient to consider the partial sums , , , and show that they become large. 1 1 2 1 2 1 2 1 2 1 4 2 1 1 2 ( 1 4 1 4) ( 1 8 1 8) ( 1 16 1 16) s16 1 1 2 ( 1 3 1 4) ( 1 5 1 8) ( 1 9 1 16) 1 1 2 1 2 1 2 1 3 2 1 1 2 ( 1 4 1 4) ( 1 8 1 8 1 8 1 8) s8 1 1 2 ( 1 3 1 4) ( 1 5 1 6 1 7 1 8) s4 1 1 2 ( 1 3 1 4) 1 1 2 ( 1 4 1 4) 1 2 2 s2 1 1 2 s1 1 s32, . . . s16 s8, s4 s2 n1 1 n 1 1 2 1 3 1 4 V n1 1 nn 1 1 lim n l sn lim n l 1 1 n 1 1 0 1 1 1 n 1 1 1 2 1 2 1 3 1 3 1 4 1 n 1 n 1 sn n i1 1 ii 1 n i1 1 i 1 i 1 1 ii 1 1 i 1 i 1 sn n i1 1 ii 1 1 1 2 1 2 3 1 3 4 1 nn 1 n1 1 nn 1 SECTION 11.2 SERIES \|\|\|\| 691 N Notice that the terms cancel in pairs. This is an example of a telescoping sum: Because of all the cancellations, the sum collapses (like a pirate’s collapsing telescope) into just two terms. N Figure 3 illustrates Example 6 by show-ing the graphs of the sequence of terms and the sequence of partial sums. Notice that and . See Exercises 62 and 63 for two geometric inter-pretations of Example 6. sn l 1 an l 0 sn an 1[nn 1] FIGURE 3 0 1 an n sn Similarly, , , and in general This shows that as and so is divergent. Therefore the harmonic series diverges. M THEOREM If the series is convergent, then . PROOF Let . Then . Since is convergent, the sequence is convergent. Let . Since as , we also have . Therefore M With any series we associate two sequences: the sequence of its par-tial sums and the sequence of its terms. If is convergent, then the limit of the sequence is (the sum of the series) and, as Theorem 6 asserts, the limit of the sequence is 0. \| The converse of Theorem 6 is not true in general. If , we cannot conclude that is convergent. Observe that for the harmonic series we have as , but we showed in Example 7 that is divergent. THE TEST FOR DIVERGENCE If does not exist or if , then the series is divergent. The Test for Divergence follows from Theorem 6 because, if the series is not divergent, then it is convergent, and so . EXAMPLE 8 Show that the series diverges. SOLUTION So the series diverges by the Test for Divergence. M If we ﬁnd that , we know that is divergent. If we ﬁnd that , we know nothing about the convergence or divergence of . Remem-ber the warning in Note 2: If , the series might converge or it might diverge. an lim n l an 0 an lim n l an 0 an lim n l an 0 NOTE 3 lim n l an lim n l n2 5n2 4 lim n l 1 5 4n2 1 5 0 n1 n2 5n2 4 lim n l an 0 n1 an lim n l an 0 lim n l an 7 1n n l an 1n l 0 1n an lim n l an 0 NOTE 2 an s sn an an sn an NOTE 1 s s 0 lim n l an lim n l sn sn1 lim n l sn lim n l sn1 lim n l sn1 s n l n 1 l lim n l sn s sn an an sn sn1 sn a1 a2 an lim n l an 0 n1 an 6 sn n l s2n l s2n 1 n 2 s64 1 6 2 s32 1 5 2 692 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N The method used in Example 7 for showing that the harmonic series diverges is due to the French scholar Nicole Oresme (1323–1382). THEOREM If and are convergent series, then so are the series (where is a constant), , and , and (i) (ii) (iii) These properties of convergent series follow from the corresponding Limit Laws for Sequences in Section 11.1. For instance, here is how part (ii) of Theorem 8 is proved: Let The nth partial sum for the series is and, using Equation 5.2.10, we have Therefore is convergent and its sum is M EXAMPLE 9 Find the sum of the series . SOLUTION The series is a geometric series with and , so In Example 6 we found that So, by Theorem 8, the given series is convergent and M A ﬁnite number of terms doesn’t affect the convergence or divergence of a series. For instance, suppose that we were able to show that the series n4 n n 3 1 NOTE 4 3 1 1 4 n1 3 nn 1 1 2n 3 n1 1 nn 1 n1 1 2n n1 1 nn 1 1 n1 1 2n 1 2 1 1 2 1 r 1 2 a 1 2 12n n1 3 nn 1 1 2n n1 an bn s t n1 an n1 bn an bn lim n l sn lim n l tn s t lim n l n i1 ai lim n l n i1 bi lim n l un lim n l n i1 ai bi lim n l n i1 ai n i1 bi un n i1 ai bi an bn t n1 bn tn n i1 bi s n1 an sn n i1 ai n1 an bn n1 an n1 bn n1 an bn n1 an n1 bn n1 can c n1 an an bn an bn c can bn an 8 SECTION 11.2 SERIES \|\|\|\| 693 is convergent. Since it follows that the entire series is convergent. Similarly, if it is known that the series converges, then the full series is also convergent. n1 an N n1 an nN1 an nN1 an n1 nn 3 1 n1 n n 3 1 1 2 2 9 3 28 n4 n n 3 1 694 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES 18. 19. 20. 21–34 Determine whether the series is convergent or divergent. If it is convergent, ﬁnd its sum. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 32. 33. 34. 35–40 Determine whether the series is convergent or divergent by expressing as a telescoping sum (as in Example 6). If it is convergent, ﬁnd its sum. 36. 37. 38. n1 ln n n 1 n1 3 nn 3 n1 2 n 2 4n 3 n2 2 n 2 1 35. sn n1 en n2 n1 1 en 1 nn 1 n1 3 5 n 2 n n1 arctan n 31. k1 cos 1k n1 ln n2 1 2n2 1 n1 0.8n1 0.3n n1 s n 2 n1 1 3 n 2 n n1 1 2 n 3 n k1 kk 2 k 32 k2 k 2 k 2 1 n1 n 1 2n 3 n1 1 2n n1 e n 3 n1 n0 n 3 n1 n0 1 (s2 ) n n1 3n1 4 n 17. 1. (a) What is the difference between a sequence and a series? (b) What is a convergent series? What is a divergent series? 2. Explain what it means to say that . ; 3–8 Find at least 10 partial sums of the series. Graph both the sequence of terms and the sequence of partial sums on the same screen. Does it appear that the series is convergent or divergent? If it is convergent, ﬁnd the sum. If it is divergent, explain why. 4. 5. 6. 7. 8. Let . (a) Determine whether is convergent. (b) Determine whether is convergent. 10. (a) Explain the difference between (b) Explain the difference between 11–20 Determine whether the geometric series is convergent or divergent. If it is convergent, ﬁnd its sum. 11. 12. 13. 14. 15. 16. n1 10 n 9n1 n1 60.9n1 1 0.4 0.16 0.064 3 4 16 3 64 9 1 8 1 4 1 2 1 3 2 4 3 8 9 n i1 aj and n i1 ai n j1 aj and n i1 ai n1 an an an 2n 3n 1 9. n2 1 n(n 2) n1 1 sn 1 sn 1 n1 0.6n1 n1 tan n n1 2n2 1 n2 1 n1 12 5n 3. n1 an 5 EXERCISES 11.2 SECTION 11.2 SERIES \|\|\|\| 695 56. If the partial sum of a series is , ﬁnd and . 57. When money is spent on goods and services, those who receive the money also spend some of it. The people receiv-ing some of the twice-spent money will spend some of that, and so on. Economists call this chain reaction the multiplier effect. In a hypothetical isolated community, the local govern-ment begins the process by spending dollars. Suppose that each recipient of spent money spends and saves of the money that he or she receives. The values and s are called the marginal propensity to consume and the mar-ginal propensity to save and, of course, . (a) Let be the total spending that has been generated after transactions. Find an equation for . (b) Show that , where . The number is called the multiplier. What is the multiplier if the marginal propensity to consume is ? Note: The federal government uses this principle to justify deﬁcit spending. Banks use this principle to justify lending a large percentage of the money that they receive in deposits. 58. A certain ball has the property that each time it falls from a height onto a hard, level surface, it rebounds to a height , where . Suppose that the ball is dropped from an initial height of meters. (a) Assuming that the ball continues to bounce indeﬁnitely, ﬁnd the total distance that it travels. (Use the fact that the ball falls in .) (b) Calculate the total time that the ball travels. (c) Suppose that each time the ball strikes the surface with velocity it rebounds with velocity , where . How long will it take for the ball to come to rest? Find the value of if 60. Find the value of such that 61. In Example 7 we showed that the harmonic series is diver-gent. Here we outline another method, making use of the fact that for any . (See Exercise 4.3.76.) If is the partial sum of the harmonic series, show that . Why does this imply that the harmonic series is divergent? ; 62. Graph the curves , , for on a common screen. By ﬁnding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that n1 1 nn 1 1 4, . . . n 0, 1, 2, 3, 0 x 1 y x n e sn n 1 nth sn x 0 e x 1 x n0 e nc 10 c n2 1 cn 2 c 59. 0 k 1 kv v t seconds 1 2tt 2 meters H 0 r 1 rh h 80% k k 1s limn l Sn kD Sn n Sn c s 1 c 100s% 100c% D n1 an an sn 3 n2n n1 an nth 39. 40. 41–46 Express the number as a ratio of integers. 42. 43. 44. 45. 46. 47–51 Find the values of for which the series converges. Find the sum of the series for those values of . 48. 49. 50. 51. 52. We have seen that the harmonic series is a divergent series whose terms approach 0. Show that is another series with this property. 53–54 Use the partial fraction command on your CAS to ﬁnd a convenient expression for the partial sum, and then use this expression to ﬁnd the sum of the series. Check your answer by using the CAS to sum the series directly. 53. 54. If the partial sum of a series is ﬁnd and . n1 an an sn n 1 n 1 n1 an nth 55. n2 1 n3 n n1 3n2 3n 1 n2 n3 CAS n1 ln1 1 n n0 cos nx 2n n0 x 3n 2n n0 4nx n n1 x 4n n1 x n 3n 47. x x 7.12345 1.5342 6.254 6.2545454 . . . 3.417 3.417417417 . . . 0.73 0.73737373 . . . 0.2 0.2222 . . . 41. n1 cos 1 n2 cos 1 n 12 n1 (e 1n e1n1) 69. If is convergent and is divergent, show that the series is divergent. [Hint: Argue by contradiction.] 70. If and are both divergent, is neces-sarily divergent? Suppose that a series has positive terms and its partial sums satisfy the inequality for all . Explain why must be convergent. 72. The Fibonacci sequence was deﬁned in Section 11.1 by the equations Show that each of the following statements is true. (a) (b) (c) The Cantor set, named after the German mathematician Georg Cantor (1845–1918), is constructed as follows. We start with the closed interval and remove the open inter-val . That leaves the two intervals and and we remove the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indeﬁnitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the num-bers that remain in after all those intervals have been removed. (a) Show that the total length of all the intervals that are removed is 1. Despite that, the Cantor set contains inﬁ-nitely many numbers. Give examples of some numbers in the Cantor set. (b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It 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. (The ﬁgure shows the ﬁrst three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0. [0, 1] [ 2 3, 1] [0, 1 3] ( 1 3, 2 3) [0, 1] 73. n2 fn fn1 fn1 2 n2 1 fn1 fn1 1 1 fn1 fn1 1 fn1 fn 1 fn fn1 n 3 fn fn1 fn2 f2 1, f1 1, an n sn 1000 sn an 71. an bn bn an an bn bn an 63. The ﬁgure shows two circles and of radius 1 that touch at . is a common tangent line; is the circle that touches , , and ; is the circle that touches , , and ; is the circle that touches , , and . This procedure can be continued indeﬁnitely and produces an inﬁnite sequence of circles . Find an expression for the diameter of and thus provide another geometric demonstration of Example 6. 64. A right triangle is given with and . is drawn perpendicular to , is drawn perpen-dicular to , , and this process is continued indeﬁnitely, as shown in the ﬁgure. Find the total length of all the perpendiculars in terms of and . What is wrong with the following calculation? (Guido Ubaldus thought that this proved the existence of God because “something has been created out of nothing.”) 66. Suppose that is known to be a convergent series. Prove that is a divergent series. 67. Prove part (i) of Theorem 8. 68. If is divergent and , show that is divergent. can c 0 an n1 1an n1 an an 0 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 65. A C E G B F H D ¨ b b CD DE EF FG EF AB BC DE AB CD AC b A ABC 1 1 P C£ C™ C¡ D T C Cn Cn C2 D C C3 C1 D C C2 T D C C1 T P D C 696 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES THE INTEGRAL TEST AND ESTIMATES OF SUMS In general, it is difﬁcult to ﬁnd the exact sum of a series. We were able to accomplish this for geometric series and the series because in each of those cases we could ﬁnd a simple formula for the partial sum . But usually it isn’t easy to compute . Therefore, in the next few sections, we develop several tests that enable us to determine whether a series is convergent or divergent without explicitly ﬁnding its sum. (In some cases, however, our methods will enable us to ﬁnd good estimates of the sum.) Our ﬁrst test involves improper integrals. We begin by investigating the series whose terms are the reciprocals of the squares of the positive integers: There’s no simple formula for the sum of the ﬁrst n terms, but the computer-generated table of values given in the margin suggests that the partial sums are approaching a num-ber near 1.64 as and so it looks as if the series is convergent. We can conﬁrm this impression with a geometric argument. Figure 1 shows the curve and rectangles that lie below the curve. The base of each rectangle is an interval of length 1; the height is equal to the value of the function at the right endpoint of the interval. So the sum of the areas of the rectangles is FIGURE 1 x y 0 2 1 3 4 5 y= 1 ≈ area= 1 1@ area= 1 2@ area= 1 3@ area= 1 4@ area= 1 5@ 1 12 1 22 1 32 1 42 1 52 n1 1 n 2 y 1x 2 y 1x 2 n l sn n1 1 n 2 1 12 1 22 1 32 1 42 1 52 lim n l sn sn nth 1 nn 1 11.3 SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS \|\|\|\| 697 circles and sides of the triangle. If the triangle has sides of length 1, ﬁnd the total area occupied by the circles. 74. (a) A sequence is deﬁned recursively by the equation for , where and can be any real numbers. Experiment with various values of and and use your calculator to guess the limit of the sequence. (b) Find in terms of and by expressing in terms of and summing a series. 75. Consider the series (a) Find the partial sums and . Do you recognize the denominators? Use the pattern to guess a formula for . (b) Use mathematical induction to prove your guess. (c) Show that the given inﬁnite series is convergent, and ﬁnd its sum. 76. In the ﬁgure there are inﬁnitely many circles approaching the vertices of an equilateral triangle, each circle touching other sn s4 s1, s2, s3, n1 n n 1! a2 a1 an1 an a2 a1 limn l an a2 a1 a2 a1 n 3 an 1 2an1 an2 an n 5 1.4636 10 1.5498 50 1.6251 100 1.6350 500 1.6429 1000 1.6439 5000 1.6447 sn n i1 1 i 2 If we exclude the ﬁrst rectangle, the total area of the remaining rectangles is smaller than the area under the curve for , which is the value of the integral . In Section 7.8 we discovered that this improper integral is convergent and has value 1. So the picture shows that all the partial sums are less than Thus the partial sums are bounded. We also know that the partial sums are increasing (because all the terms are positive). Therefore the partial sums converge (by the Monotonic Sequence Theorem) and so the series is convergent. The sum of the series (the limit of the partial sums) is also less than 2: [The exact sum of this series was found by the Swiss mathematician Leonhard Euler (1707–1783) to be , but the proof of this fact is quite difﬁcult. (See Problem 6 in the Problems Plus following Chapter 15.)] Now let’s look at the series The table of values of suggests that the partial sums aren’t approaching a ﬁnite number, so we suspect that the given series may be divergent. Again we use a picture for conﬁrma-tion. Figure 2 shows the curve , but this time we use rectangles whose tops lie above the curve. The base of each rectangle is an interval of length 1. The height is equal to the value of the function at the left endpoint of the interval. So the sum of the areas of all the rectangles is This total area is greater than the area under the curve for , which is equal to the integral . But we know from Section 7.8 that this improper integral is divergent. In other words, the area under the curve is inﬁnite. So the sum of the series must be inﬁnite; that is, the series is divergent. The same sort of geometric reasoning that we used for these two series can be used to prove the following test. (The proof is given at the end of this section.) x 1 (1sx ) dx x 1 y 1sx 1 s1 1 s2 1 s3 1 s4 1 s5 n1 1 sn y 1sx FIGURE 2 x y 0 2 1 3 4 5 area= 1 œ„ 1 œ„ 1 œ„ 1 œ„ 1 y= 1 œ„ x area= 2 area= 3 area= 4 y 1sx sn n1 1 sn 1 s1 1 s2 1 s3 1 s4 1 s5 26 n1 1 n 2 1 12 1 22 1 32 1 42 2 1 12 y 1 1 x 2 dx 2 x 1 1x 2 dx x 1 y 1x 2 698 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES n 5 3.2317 10 5.0210 50 12.7524 100 18.5896 500 43.2834 1000 61.8010 5000 139.9681 sn n i1 1 si THE INTEGRAL TEST Suppose is a continuous, positive, decreasing function on and let . Then the series is convergent if and only if the improper integral is convergent. In other words: (i) If is convergent, then is convergent. (ii) If is divergent, then is divergent. When we use the Integral Test, it is not necessary to start the series or the inte-gral at . For instance, in testing the series Also, it is not necessary that be always decreasing. What is important is that be ulti-mately decreasing, that is, decreasing for larger than some number . Then is convergent, so is convergent by Note 4 of Section 11.2. EXAMPLE 1 Test the series for convergence or divergence. SOLUTION The function is continuous, positive, and decreasing on so we use the Integral Test: Thus is a convergent integral and so, by the Integral Test, the series is convergent. M EXAMPLE 2 For what values of is the series convergent? SOLUTION If , then . If , then . In either case, , so the given series diverges by the Test for Divergence (11.2.7). If , then the function is clearly continuous, positive, and decreasing on . We found in Chapter 7 [see (7.8.2)] that It follows from the Integral Test that the series converges if and diverges if . (For , this series is the harmonic series discussed in Example 7 in Section 11.2.) M The series in Example 2 is called the p-series. It is important in the rest of this chapter, so we summarize the results of Example 2 for future reference as follows. p 1 0 p 1 p 1 1n p y 1 1 x p dx converges if p 1 and diverges if p 1 1, fx 1x p p 0 lim n l 1n p 0 lim n l 1n p 1 p 0 lim n l 1n p p 0 n1 1 n p p V 1n2 1 x 1 1x 2 1 dx lim t l tan1t 4 2 4 4 y 1 1 x 2 1 dx lim t l y t 1 1 x 2 1 dx lim t l tan1x]1 t 1, fx 1x 2 1 n1 1 n2 1 n1 an nN an N x f f y 4 1 x 32 dx we use n4 1 n 32 n 1 NOTE n1 an y 1 fx dx n1 an y 1 fx dx x 1 fx dx n1 an an fn 1, f SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS \|\|\|\| 699 N In order to use the Integral Test we need to be able to evaluate and therefore we have to be able to ﬁnd an antiderivative of . Frequently this is difﬁcult or impossible, so we need other tests for convergence too. f x 1 f x dx The -series is convergent if and divergent if . EXAMPLE 3 (a) The series is convergent because it is a p-series with . (b) The series is divergent because it is a p-series with . M We should not infer from the Integral Test that the sum of the series is equal to the value of the integral. In fact, Therefore, in general, EXAMPLE 4 Determine whether the series converges or diverges. SOLUTION The function is positive and continuous for because the logarithm function is continuous. But it is not obvious whether or not is decreasing, so we compute its derivative: Thus when , that is, . It follows that is decreasing when and so we can apply the Integral Test: Since this improper integral is divergent, the series is also divergent by the Integral Test. M ESTIMATING THE SUM OF A SERIES Suppose we have been able to use the Integral Test to show that a series is conver-gent and we now want to ﬁnd an approximation to the sum of the series. Of course, any partial sum is an approximation to because . But how good is such an approximation? To ﬁnd out, we need to estimate the size of the remainder Rn s sn an1 an2 an3 lim n l sn s s sn s an ln nn lim t l ln t2 2 y 1 ln x x dx lim t l y t 1 ln x x dx lim t l ln x2 2 1 t x e f x e ln x 1 fx 0 fx 1xx ln x x 2 1 ln x x 2 f x 1 fx ln xx n1 ln n n V n1 an y 1 fx dx y 1 1 x 2 dx 1 whereas n1 1 n2 2 6 NOTE p 1 3 1 n1 1 n 13 n1 1 s 3 n 1 1 s 3 2 1 s 3 3 1 s 3 4 p 3 1 n1 1 n 3 1 13 1 23 1 33 1 43 p 1 p 1 n1 1 n p p 1 700 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES The remainder is the error made when , the sum of the ﬁrst terms, is used as an approximation to the total sum. We use the same notation and ideas as in the Integral Test, assuming that is decreas-ing on . Comparing the areas of the rectangles with the area under for in Figure 3, we see that Similarly, we see from Figure 4 that So we have proved the following error estimate. REMAINDER ESTIMATE FOR THE INTEGRAL TEST Suppose , where is a continuous, positive, decreasing function for and is convergent. If , then EXAMPLE 5 (a) Approximate the sum of the series by using the sum of the ﬁrst 10 terms. Estimate the error involved in this approximation. (b) How many terms are required to ensure that the sum is accurate to within ? SOLUTION In both parts (a) and (b) we need to know . With , which satisﬁes the conditions of the Integral Test, we have (a) According to the remainder estimate in (2), we have So the size of the error is at most . (b) Accuracy to within means that we have to ﬁnd a value of such that . Since we want 1 2n2 0.0005 Rn y n 1 x 3 dx 1 2n2 Rn 0.0005 n 0.0005 0.005 R10 y 10 1 x 3 dx 1 2102 1 200 n1 1 n3 s10 1 13 1 23 1 33 1 103 1.1975 y n 1 x 3 dx lim t l 1 2x 2 n t lim t l 1 2t 2 1 2n2 1 2n2 fx 1x 3 x n fx dx 0.0005 1n 3 V y n1 fx dx Rn y n fx dx Rn s sn an x n f fk ak 2 Rn an1 an2 y n1 fx dx Rn an1 an2 y n fx dx x n y fx n, f n sn Rn SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS \|\|\|\| 701 FIGURE 3 0 x y n . . . y=ƒ an+1 an+2 FIGURE 4 0 x y an+1 an+2 n+1 . . . y=ƒ Solving this inequality, we get We need 32 terms to ensure accuracy to within . M If we add to each side of the inequalities in (2), we get because . The inequalities in (3) give a lower bound and an upper bound for . They provide a more accurate approximation to the sum of the series than the partial sum does. EXAMPLE 6 Use (3) with to estimate the sum of the series . SOLUTION The inequalities in (3) become From Example 5 we know that so Using , we get If we approximate by the midpoint of this interval, then the error is at most half the length of the interval. So M If we compare Example 6 with Example 5, we see that the improved estimate in (3) can be much better than the estimate . To make the error smaller than we had to use 32 terms in Example 5 but only 10 terms in Example 6. PROOF OF THE INTEGRAL TEST We have already seen the basic idea behind the proof of the Integral Test in Figures 1 and 2 for the series and . For the general series , look at Figures 5 and 6. The area of the ﬁrst shaded rectangle in Figure 5 is the value of at the right endpoint of , 1, 2 f an 1sn 1n2 0.0005 s sn with error 0.0005 n1 1 n3 1.2021 s 1.201664 s 1.202532 s10 1.197532 s10 1 2112 s s10 1 2102 y n 1 x 3 dx 1 2n2 s10 y 11 1 x 3 dx s s10 y 10 1 x 3 dx n1 1 n3 n 10 sn s sn Rn s sn y n1 fx dx s sn y n fx dx 3 sn 0.0005 n s1000 31.6 or n2 1 0.001 1000 702 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES that is, . So, comparing the areas of the shaded rectangles with the area under from 1 to , we see that (Notice that this inequality depends on the fact that is decreasing.) Likewise, Figure 6 shows that (i) If is convergent, then (4) gives since . Therefore Since for all , the sequence is bounded above. Also since . Thus is an increasing bounded sequence and so it is convergent by the Monotonic Sequence Theorem (11.1.12). This means that is convergent. (ii) If is divergent, then as because . But (5) gives and so . This implies that and so diverges. M an sn l sn1 l y n 1 fx dx n1 i1 ai sn1 fx 0 n l xn 1 fx dx l x 1 fx dx an sn an1 fn 1 0 sn1 sn an1 sn sn n sn M sn a1 n i2 ai a1 y 1 fx dx M, say fx 0 n i2 ai y n 1 fx dx y 1 fx dx y 1 fx dx y n 1 fx dx a1 a2 an1 5 f a2 a3 an y n 1 fx dx 4 n y fx f2 a2 SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS \|\|\|\| 703 0 x y 1 2 3 4 5 . . . n y=ƒ an a™ a£ a¢ a∞ FIGURE 5 FIGURE 6 0 x y 1 2 3 4 5 . . . n y=ƒ a™ a£ a¢ a¡ an-1 3–8 Use the Integral Test to determine whether the series is convergent or divergent. 3. 4. 5. 6. 8. n1 n 2 n 1 n1 nen 7. n1 1 sn 4 n1 1 2n 13 n1 1 n5 n1 1 s 5 n 1. Draw a picture to show that What can you conclude about the series? 2. Suppose is a continuous positive decreasing function for and . By drawing a picture, rank the following three quantities in increasing order: 6 i2 ai 5 i1 ai y 6 1 f x dx an f n x 1 f n2 1 n 1.3 y 1 1 x 1.3 dx EXERCISES 11.3 (a) Use the sum of the ﬁrst 10 terms to estimate the sum of the series . How good is this estimate? (b) Improve this estimate using (3) with . (c) Find a value of that will ensure that the error in the approximation is less than . 34. Find the sum of the series correct to three decimal places. 35. Estimate correct to ﬁve decimal places. 36. How many terms of the series would you need to add to ﬁnd its sum to within ? 37. Show that if we want to approximate the sum of the series so that the error is less than 5 in the ninth decimal place, then we need to add more than terms! 38. (a) Show that the series is convergent. (b) Find an upper bound for the error in the approximation . (c) What is the smallest value of such that this upper bound is less than ? (d) Find for this value of . (a) Use (4) to show that if is the partial sum of the harmonic series, then (b) The harmonic series diverges, but very slowly. Use part (a) to show that the sum of the ﬁrst million terms is less than 15 and the sum of the ﬁrst billion terms is less than 22. 40. Use the following steps to show that the sequence has a limit. (The value of the limit is denoted by and is called Euler’s constant.) (a) Draw a picture like Figure 6 with and interpret as an area [or use (5)] to show that for all . (b) Interpret as a difference of areas to show that . There-fore, is a decreasing sequence. (c) Use the Monotonic Sequence Theorem to show that is convergent. 41. Find all positive values of for which the series converges. 42. Find all values of for which the following series converges. n1 c n 1 n 1 c n1 b ln n b tn tn tn tn1 0 tn tn1 ln n 1 ln n 1 n 1 n tn 0 tn f x 1x tn 1 1 2 1 3 1 n ln n sn 1 ln n nth sn 39. n sn 0.05 n s sn n1 ln n 2n 2 CAS 1011,301 n1 n1.001 0.01 n2 1n ln n 2 n1 2n 1 6 n1 1n5 0.001 s sn n n 10 n1 1n2 33. 9–26 Determine whether the series is convergent or divergent. 9. 10. 12. 13. 14. 15. 16. 18. 19. 20. 22. 23. 24. 25. 26. 27–30 Find the values of for which the series is convergent. 27. 28. 29. 30. 31. The Riemann zeta-function is deﬁned by and is used in number theory to study the distribution of prime numbers. What is the domain of ? 32. (a) Find the partial sum of the series . Estimate the error in using as an approximation to the sum of the series. (b) Use (3) with to give an improved estimate of the sum. (c) Find a value of so that is within of the sum. 0.00001 sn n n 10 s10 n1 1n4 s10 x n1 1 n x n1 ln n n p n1 n 1 n2 p n3 1 n ln n ln ln n p n2 1 n ln n p p n1 n n4 1 n1 1 n3 n n3 n2 e n n1 e1n n2 n2 1 n ln n 2 n2 1 n ln n 21. n1 1 n2 4n 5 n1 ln n n3 n1 3n 2 n n 1 n1 1 n2 4 17. n1 n2 n3 1 n1 5 2sn n3 1 5 1 8 1 11 1 14 1 17 1 1 3 1 5 1 7 1 9 1 1 2s2 1 3s3 1 4s4 1 5s5 1 1 8 1 27 1 64 1 125 11. n1 n 1.4 3n 1.2 n1 2 n0.85 704 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES THE COMPARISON TESTS In the comparison tests the idea is to compare a given series with a series that is known to be convergent or divergent. For instance, the series reminds us of the series , which is a geometric series with and and is therefore convergent. Because the series (1) is so similar to a convergent series, we have the feeling that it too must be convergent. Indeed, it is. The inequality shows that our given series (1) has smaller terms than those of the geometric series and therefore all its partial sums are also smaller than 1 (the sum of the geometric series). This means that its partial sums form a bounded increasing sequence, which is convergent. It also follows that the sum of the series is less than the sum of the geometric series: Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive. The ﬁrst part says that if we have a series whose terms are smaller than those of a known convergent series, then our series is also convergent. The second part says that if we start with a series whose terms are larger than those of a known divergent series, then it too is divergent. THE COMPARISON TEST Suppose that and are series with positive terms. (i) If is convergent and for all , then is also convergent. (ii) If is divergent and for all , then is also divergent. PROOF (i) Let Since both series have positive terms, the sequences and are increasing . Also , so for all . Since . Thus for all . This means that is increasing and bounded above and therefore converges by the Monotonic Sequence Theorem. Thus converges. (ii) If is divergent, then (since is increasing). But so . Thus . Therefore diverges. M In using the Comparison Test we must, of course, have some known series for the purpose of comparison. Most of the time we use one of these series: N A -series [ converges if and diverges if ; see (11.3.1)] N A geometric series [ converges if and diverges if ; see (11.2.4)] r 1 r 1 ar n1 p 1 p 1 1n p p bn an sn l sn tn ai bi tn tn l bn an sn n sn t ai bi, we have sn tn n tn t tn l t sn1 sn an1 sn tn sn t n1 bn tn n i1 bi sn n i1 ai an n an bn bn an n an bn bn bn an n1 1 2n 1 1 1 2n 1 1 2n r 1 2 a 1 2 n1 12n n1 1 2n 1 1 11.4 SECTION 11.4 THE COMPARISON TESTS \|\|\|\| 705 N It is important to keep in mind the distinction between a sequence and a series. A sequence is a list of numbers, whereas a series is a sum. With every series there are associated two sequences: the sequence of terms and the sequence of partial sums. sn an an Standard Series for Use with the Comparison Test EXAMPLE 1 Determine whether the series converges or diverges. SOLUTION For large the dominant term in the denominator is so we compare the given series with the series . Observe that because the left side has a bigger denominator. (In the notation of the Comparison Test, is the left side and is the right side.) We know that is convergent because it’s a constant times a -series with . Therefore is convergent by part (i) of the Comparison Test. M Although the condition or in the Comparison Test is given for all , we need verify only that it holds for , where is some ﬁxed integer, because the convergence of a series is not affected by a ﬁnite number of terms. This is illustrated in the next example. EXAMPLE 2 Test the series for convergence or divergence. SOLUTION This series was tested (using the Integral Test) in Example 4 in Section 11.3, but it is also possible to test it by comparing it with the harmonic series. Observe that for and so We know that is divergent ( -series with ). Thus the given series is divergent by the Comparison Test. M The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesn’t apply. Consider, for instance, the series The inequality is useless as far as the Comparison Test is concerned because is convergent and . Nonetheless, we have the feeling that ought to be convergent because it is very similar to the convergent geometric series . In such cases the fol-lowing test can be used. ( 1 2) n 1 2n 1 an bn bn ( 1 2) n 1 2n 1 1 2n n1 1 2n 1 NOTE 2 p 1 p 1n n 3 ln n n 1 n n 3 ln n 1 n1 ln n n V N n N n an bn an bn NOTE 1 n1 5 2n2 4n 3 p 2 1 p n1 5 2n2 5 2 n1 1 n2 bn an 5 2n2 4n 3 5 2n2 5 2n2 2n2 n n1 5 2n2 4n 3 V 706 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES THE LIMIT COMPARISON TEST Suppose that and are series with positive terms. If where c is a ﬁnite number and , then either both series converge or both diverge. PROOF Let m and M be positive numbers such that . Because is close to c for large n, there is an integer N such that and so If converges, so does . Thus converges by part (i) of the Comparison Test. If diverges, so does and part (ii) of the Comparison Test shows that diverges. M EXAMPLE 3 Test the series for convergence or divergence. SOLUTION We use the Limit Comparison Test with and obtain Since this limit exists and is a convergent geometric series, the given series con-verges by the Limit Comparison Test. M EXAMPLE 4 Determine whether the series converges or diverges. SOLUTION The dominant part of the numerator is and the dominant part of the denomi-nator is . This suggests taking lim n l 2 3 n 2 5 n5 1 2 0 2s0 1 1 lim n l an bn lim n l 2n2 3n s5 n 5 n 12 2 lim n l 2n52 3n32 2s5 n 5 bn 2n2 n 52 2 n 12 an 2n2 3n s5 n 5 sn5 n 52 2n2 n1 2n2 3n s5 n 5 12n lim n l an bn lim n l 1 2n 1 12n lim n l 2n 2n 1 lim n l 1 1 12n 1 0 bn 1 2n an 1 2n 1 n1 1 2n 1 an mbn bn an Mbn bn mbn an Mbn when n N m an bn M when n N anbn m c M c 0 lim n l an bn c bn an SECTION 11.4 THE COMPARISON TESTS \|\|\|\| 707 N Exercises 40 and 41 deal with the cases and . c c 0 Since is divergent ( -series with ), the given series diverges by the Limit Comparison Test. M Notice that in testing many series we ﬁnd a suitable comparison series by keeping only the highest powers in the numerator and denominator. ESTIMATING SUMS If we have used the Comparison Test to show that a series converges by comparison with a series , then we may be able to estimate the sum by comparing remainders. As in Section 11.3, we consider the remainder For the comparison series we consider the corresponding remainder Since for all , we have . If is a -series, we can estimate its remain-der as in Section 11.3. If is a geometric series, then is the sum of a geometric series and we can sum it exactly (see Exercises 35 and 36). In either case we know that is smaller than . EXAMPLE 5 Use the sum of the ﬁrst 100 terms to approximate the sum of the series . Estimate the error involved in this approximation. SOLUTION Since the given series is convergent by the Comparison Test. The remainder for the compari-son series was estimated in Example 5 in Section 11.3 using the Remainder Esti-mate for the Integral Test. There we found that Therefore the remainder for the given series satisﬁes With we have Using a programmable calculator or a computer, we ﬁnd that with error less than . M 0.00005 n1 1 n3 1 100 n1 1 n3 1 0.6864538 R100 1 2 100 2 0.00005 n 100 Rn Tn 1 2n2 Rn Tn y n 1 x 3 dx 1 2n2 1n3 Tn 1 n3 1 1 n3 1 n3 1 V Tn Rn Tn bn Tn p bn Rn Tn n an bn Tn t tn bn1 bn2 bn Rn s sn an1 an2 an bn an bn p 1 2 1 p bn 2 1n 12 708 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES 33–36 Use the sum of the ﬁrst 10 terms to approximate the sum of the series. Estimate the error. 33. 34. 35. 36. The meaning of the decimal representation of a number (where the digit is one of the numbers 0, 1, 2, . . . , 9) is that Show that this series always converges. 38. For what values of does the series converge? 39. Prove that if and converges, then also converges. 40. (a) Suppose that and are series with positive terms and is convergent. Prove that if then is also convergent. (b) Use part (a) to show that the series converges. (i) (ii) (a) Suppose that and are series with positive terms and is divergent. Prove that if then is also divergent. (b) Use part (a) to show that the series diverges. (i) (ii) 42. Give an example of a pair of series and with positive terms where and diverges, but converges. (Compare with Exercise 40.) 43. Show that if and then is divergent. 44. Show that if and is convergent, then is convergent. 45. If is a convergent series with positive terms, is it true that is also convergent? 46. If and are both convergent series with positive terms, is it true that is also convergent? anbn bn an sin an an ln 1 an an an 0 an lim n l nan 0, an 0 an bn lim n l anbn 0 bn an n1 ln n n n2 1 ln n an lim n l an bn bn bn an 41. n1 ln n sn en n1 ln n n3 an lim n l an bn 0 bn bn an an 2 an an 0 n2 1 n p ln n p 0.d1d2d3d4 . . . d1 10 d2 102 d3 103 d4 104 di 0.d1d2d3 . . . 37. n1 n n 1 3n n1 1 1 2n n1 sin 2 n n3 n1 1 sn4 1 Suppose and are series with positive terms and is known to be convergent. (a) If for all , what can you say about ? Why? (b) If for all , what can you say about ? Why? 2. Suppose and are series with positive terms and is known to be divergent. (a) If for all n, what can you say about ? Why? (b) If for all n, what can you say about ? Why? 3–32 Determine whether the series converges or diverges. 3. 4. 5. 6. 7. 9. 11. 12. 13. 14. 15. 16. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 32. n1 1 n 11n n1 sin 1 n 31. n1 n! n n n1 1 n! n1 e 1n n n1 1 1 n 2 en n1 n 5 s 3 n7 n2 n1 1 n n2 s1 n2 n6 n1 n2 5n n3 n 1 n1 5 2n 1 n 2 2 n3 n 2 n 1 3 n1 sn 2 2n2 n 1 n1 n 4 n n 6 n n1 1 4n 1 3n n1 1 2n 3 n1 1 sn 2 1 17. n1 1 sn 3 1 n1 2 1 n nsn n2 sn n 1 n1 arctan n n1.2 n0 1 sin n 10 n n1 n 1 n4 n n1 n2 1 3n4 1 10. n1 cos2n n2 1 n1 4 3n 2n 8. n1 9n 3 10 n n1 n 1 n2sn n1 n 1 nsn n2 n3 n4 1 n1 n 2n3 1 an an bn an an bn bn bn an an n an bn an n an bn bn bn an 1. EXERCISES 11.4 SECTION 11.4 THE COMPARISON TESTS \|\|\|\| 709 ALTERNATING SERIES The convergence tests that we have looked at so far apply only to series with positive terms. In this section and the next we learn how to deal with series whose terms are not necessarily positive. Of particular importance are alternating series, whose terms alternate in sign. An alternating series is a series whose terms are alternately positive and negative. Here are two examples: We see from these examples that the th term of an alternating series is of the form where is a positive number. (In fact, .) The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. THE ALTERNATING SERIES TEST If the alternating series satisﬁes (i) (ii) then the series is convergent. Before giving the proof let’s look at Figure 1, which gives a picture of the idea behind the proof. We ﬁrst plot on a number line. To ﬁnd we subtract , so is to the left of . Then to ﬁnd we add , so is to the right of . But, since , is to the left of . Continuing in this manner, we see that the partial sums oscillate back and forth. Since , the successive steps are becoming smaller and smaller. The even par-tial sums , , , . . . are increasing and the odd partial sums , , , . . . are decreasing. Thus it seems plausible that both are converging to some number , which is the sum of the series. Therefore we consider the even and odd partial sums separately in the follow-ing proof. FIGURE 1 0 s¡ s™ s£ s¢ s∞ sß s b¡ -b™ +b£ -b¢ +b∞ -bß s s5 s3 s1 s6 s4 s2 bn l 0 s1 s3 b3 b2 s2 s3 b3 s3 s1 s2 b2 s2 s1 b1 lim n l bn 0 for all n bn1 bn bn 0 n1 1 n1bn b1 b2 b3 b4 b5 b6 bn an bn an 1 nbn or an 1 n1bn n 1 2 2 3 3 4 4 5 5 6 6 7 n1 1 n n n 1 1 1 2 1 3 1 4 1 5 1 6 n1 1 n1 n 11.5 710 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES PROOF OF THE ALTERNATING SERIES TEST We ﬁrst consider the even partial sums: In general Thus But we can also write Every term in brackets is positive, so for all . Therefore the sequence of even partial sums is increasing and bounded above. It is therefore convergent by the Monotonic Sequence Theorem. Let’s call its limit , that is, Now we compute the limit of the odd partial sums: [by condition (ii)] Since both the even and odd partial sums converge to , we have [see Exercise 80(a) in Section 11.1] and so the series is convergent. M EXAMPLE 1 The alternating harmonic series satisﬁes (i) because (ii) so the series is convergent by the Alternating Series Test. M EXAMPLE 2 The series is alternating but lim n l bn lim n l 3n 4n 1 lim n l 3 4 1 n 3 4 n1 1 n3n 4n 1 V lim n l bn lim n l 1 n 0 1 n 1 1 n bn1 bn 1 1 2 1 3 1 4 n1 1 n1 n V lim n l sn s s s s 0 lim n l s2n lim n l b2n1 lim n l s2n1 lim n l s2n b2n1 lim n l s2n s s s2n n s2n b1 s2n b1 b2 b3 b4 b5 b2n2 b2n1 b2n 0 s2 s4 s6 s2n since b2n b2n1 s2n s2n2 b2n1 b2n s2n2 since b4 b3 s4 s2 b3 b4 s2 since b2 b1 s2 b1 b2 0 SECTION 11.5 ALTERNATING SERIES \|\|\|\| 711 FIGURE 2 0 n 1 an sn N Figure 2 illustrates Example 1 by showing the graphs of the terms and the partial sums . Notice how the values of zigzag across the limiting value, which appears to be about . In fact, it can be proved that the exact sum of the series is (see Exercise 36). ln 2 0.693 0.7 sn sn an 1 n1n so condition (ii) is not satisﬁed. Instead, we look at the limit of the nth term of the series: This limit does not exist, so the series diverges by the Test for Divergence. M EXAMPLE 3 Test the series for convergence or divergence. SOLUTION The given series is alternating so we try to verify conditions (i) and (ii) of the Alternating Series Test. Unlike the situation in Example 1, it is not obvious that the sequence given by is decreasing. However, if we consider the related function , we ﬁnd that Since we are considering only positive , we see that if , that is, . Thus is decreasing on the interval . This means that and therefore when . (The inequality can be veriﬁed directly but all that really matters is that the sequence is eventually decreasing.) Condition (ii) is readily veriﬁed: Thus the given series is convergent by the Alternating Series Test. M ESTIMATING SUMS A partial sum of any convergent series can be used as an approximation to the total sum , but this is not of much use unless we can estimate the accuracy of the approximation. The error involved in using is the remainder . The next theorem says that for series that satisfy the conditions of the Alternating Series Test, the size of the error is smaller than , which is the absolute value of the ﬁrst neglected term. ALTERNATING SERIES ESTIMATION THEOREM If is the sum of an alternating series that satisﬁes (i) and (ii) then PROOF We know from the proof of the Alternating Series Test that s lies between any two consecutive partial sums and . It follows that M s sn sn1 sn bn1 sn1 sn Rn s sn bn1 lim n l bn 0 0 bn1 bn s 1 n1bn bn1 Rn s sn s sn s sn lim n l bn lim n l n2 n3 1 lim n l 1 n 1 1 n3 0 bn b2 b1 n 2 bn1 bn f n 1 f n (s 3 2 , ) f x s 3 2 2 x 3 0 f x 0 x f x x 2 x 3 x 3 1 2 f x x 2 x 3 1 bn n2 n3 1 n1 1 n1 n2 n3 1 lim n l an lim n l 1 n3n 4n 1 712 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N Instead of verifying condition (i) of the Alter-nating Series Test by computing a derivative, we could verify that directly by using the technique of Solution 1 of Example 12 in Section 11.1. bn1 bn N You can see geometrically why the Alternating Series Estimation Theorem is true by looking at Figure 1 (on page 710). Notice that and so on. Notice also that lies between any two consecutive partial sums. s s s5 b6, s s4 b5, EXAMPLE 4 Find the sum of the series correct to three decimal places. (By deﬁnition, .) SOLUTION We ﬁrst observe that the series is convergent by the Alternating Series Test because (i) (ii) so as To get a feel for how many terms we need to use in our approximation, let’s write out the ﬁrst few terms of the series: Notice that and By the Alternating Series Estimation Theorem we know that This error of less than does not affect the third decimal place, so we have correct to three decimal places. M \| The rule that the error (in using to approximate ) is smaller than the ﬁrst neglected term is, in general, valid only for alternating series that satisfy the conditions of the Alternating Series Estimation Theorem. The rule does not apply to other types of series. s sn NOTE s 0.368 0.0002 s s6 b7 0.0002 s6 1 1 1 2 1 6 1 24 1 120 1 720 0.368056 b7 1 5040 1 5000 0.0002 1 1 1 2 1 6 1 24 1 120 1 720 1 5040 s 1 0! 1 1! 1 2! 1 3! 1 4! 1 5! 1 6! 1 7! n l 1 n! l 0 0 1 n! 1 n l 0 1 n 1 ! 1 n! n 1 1 n! 0! 1 n0 1 n n! V SECTION 11.5 ALTERNATING SERIES \|\|\|\| 713 N In Section 11.10 we will prove that for all , so what we have obtained in Example 4 is actually an approxi-mation to the number . e 1 x e x n0 x nn! 9. 10. 12. 14. 15. 16. 18. 19. 20. n1 n 5 n n1 1 n n n n! n1 1 n cos n n1 1 n sin n 17. n1 sin n 2 n! n1 cos n n 34 n1 1 n1 ln n n n2 1 n n ln n 13. n1 1 n1 e 1n n n1 1 n1 n 2 n 3 4 11. n1 1 n sn 1 2sn n1 1 n n 10 n 1. (a) What is an alternating series? (b) Under what conditions does an alternating series converge? (c) If these conditions are satisﬁed, what can you say about the remainder after terms? 2–20 Test the series for convergence or divergence. 2. 4. 5. 6. 8. n1 1 n n sn3 2 n1 1 n 3n 1 2n 1 7. n1 1 n1 ln n 4 n1 1 n1 2n 1 1 s2 1 s3 1 s4 1 s5 1 s6 4 7 4 8 4 9 4 10 4 11 3. 1 3 2 4 3 5 4 6 5 7 n EXERCISES 11.5 31. Is the 50th partial sum of the alternating series an overestimate or an underestimate of the total sum? Explain. 32–34 For what values of is each series convergent? 33. 34. 35. Show that the series , where if is odd and if is even, is divergent. Why does the Alter-nating Series Test not apply? 36. Use the following steps to show that Let and be the partial sums of the harmonic and alter-nating harmonic series. (a) Show that . (b) From Exercise 40 in Section 11.3 we have as and therefore as Use these facts together with part (a) to show that as . n l s2n l ln 2 n l h2n ln2n l n l hn ln n l s2n h2n hn sn hn n1 1n1 n ln 2 n bn 1n2 n bn 1n 1n1bn n2 1n1 ln n p n n1 1n n p n1 1n1 n p 32. p n1 1n1n s50 ; 21–22 Calculate the ﬁrst 10 partial sums of the series and graph both the sequence of terms and the sequence of partial sums on the same screen. Estimate the error in using the 10th partial sum to approximate the total sum. 21. 22. 23–26 Show that the series is convergent. How many terms of the series do we need to add in order to ﬁnd the sum to the indi-cated accuracy? 24. 25. 26. 27–30 Approximate the sum of the series correct to four decimal places. 27. 28. 29. 30. n1 1n 3 nn! n1 1n1n2 10 n n1 1nn 8n n1 1n1 n 5 ( error 0.01) n1 1n1nen ( error 0.000005) n0 1n 10n n! ( error 0.0001) n1 1n n 5n ( error 0.00005) n1 1n1 n6 23. n1 1n1 n 3 n1 1n1 n 32 714 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS Given any series , we can consider the corresponding series whose terms are the absolute values of the terms of the original series. DEFINITION A series is called absolutely convergent if the series of absolute values is convergent. Notice that if is a series with positive terms, then and so absolute con-vergence is the same as convergence in this case. EXAMPLE 1 The series n1 1n1 n2 1 1 22 1 32 1 42 an an an an an 1 n1 an a1 a2 a3 an 11.6 N We have convergence tests for series with positive terms and for alternating series. But what if the signs of the terms switch back and forth irregularly? We will see in Example 3 that the idea of absolute convergence sometimes helps in such cases. is absolutely convergent because is a convergent -series ( ). M EXAMPLE 2 We know that the alternating harmonic series is convergent (see Example 1 in Section 11.5), but it is not absolutely convergent because the corresponding series of absolute values is which is the harmonic series ( -series with ) and is therefore divergent. M DEFINITION A series is called conditionally convergent if it is conver-gent but not absolutely convergent. Example 2 shows that the alternating harmonic series is conditionally convergent. Thus it is possible for a series to be convergent but not absolutely convergent. However, the next theorem shows that absolute convergence implies convergence. THEOREM If a series is absolutely convergent, then it is convergent. PROOF Observe that the inequality is true because is either or . If is absolutely convergent, then is convergent, so is convergent. Therefore, by the Comparison Test, is convergent. Then is the difference of two convergent series and is therefore convergent. M EXAMPLE 3 Determine whether the series is convergent or divergent. SOLUTION This series has both positive and negative terms, but it is not alternating. (The ﬁrst term is positive, the next three are negative, and the following three are posi-tive: The signs change irregularly.) We can apply the Comparison Test to the series of absolute values n1 cos n n2 n1 cos n n2 n1 cos n n2 cos 1 12 cos 2 22 cos 3 32 V an (an an) an (an an) 2an an an an an an 0 an an 2an an 3 an 2 p 1 p n1 1n1 n n1 1 n 1 1 2 1 3 1 4 n1 1n1 n 1 1 2 1 3 1 4 p 2 p n1 1n1 n2 n1 1 n2 1 1 22 1 32 1 42 SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS \|\|\|\| 715 FIGURE 1 0 n 0.5 an sn N Figure 1 shows the graphs of the terms and partial sums of the series in Example 3. Notice that the series is not alternating but has positive and negative terms. sn an Since for all , we have We know that is convergent ( -series with ) and therefore is convergent by the Comparison Test. Thus the given series is absolutely convergent and therefore convergent by Theorem 3. M The following test is very useful in determining whether a given series is absolutely convergent. THE RATIO TEST (i) If , then the series is absolutely convergent (and therefore convergent). (ii) If or , then the series is divergent. (iii) If , the Ratio Test is inconclusive; that is, no conclusion can be drawn about the convergence or divergence of . PROOF (i) The idea is to compare the given series with a convergent geometric series. Since , we can choose a number such that . Since the ratio will eventually be less than ; that is, there exists an integer such that or, equivalently, Putting successively equal to , , , . . . in (4), we obtain and, in general, for all k 1 aNk aNr k 5 aN3 aN2r aNr 3 aN2 aN1r aNr 2 aN1 aNr N 2 N 1 N n an1 anr whenever n N 4 an1 an r whenever n N N r an1an L r and lim n l an1 an L L r 1 r L 1 an lim n l an1 an 1 n1 an lim n l an1 an lim n l an1 an L 1 n1 an lim n l an1 an L 1 cos nn2 cos nn2 p 2 p 1n2 cos n n2 1 n2 n cos n 1 716 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES Now the series is convergent because it is a geometric series with . So the inequality (5), together with the Comparison Test, shows that the series is also convergent. It follows that the series is convergent. (Recall that a ﬁnite number of terms doesn’t affect convergence.) Therefore is absolutely convergent. (ii) If or , then the ratio will eventually be greater than 1; that is, there exists an integer such that This means that whenever and so Therefore diverges by the Test for Divergence. M Part (iii) of the Ratio Test says that if , the test gives no information. For instance, for the convergent series we have whereas for the divergent series we have Therefore, if , the series might converge or it might diverge. In this case the Ratio Test fails and we must use some other test. EXAMPLE 4 Test the series for absolute convergence. SOLUTION We use the Ratio Test with : 1 3 n 1 n 3 1 3 1 1 n 3 l 1 3 1 an1 an \| 1n1n 13 3n1 1n n3 3n \| n 13 3n1 3n n3 an 1n n 33n n1 1n n3 3n an lim n l an1an 1 as n l an1 an 1 n 1 1 n n n 1 1 1 1 n l 1 1n as n l an1 an 1 n 12 1 n2 n2 n 12 1 1 1 n 2 l 1 1n2 lim n l an1an 1 NOTE an lim n l an 0 n N an1 an n N whenever an1 an 1 N an1an an1anl an1anl L 1 an n1 an nN1 an k1 aNk aN1 aN2 aN3 0 r 1 k1 aNr k aNr aNr 2 aNr 3 SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS \|\|\|\| 717 N ESTIMATING SUMS In the last three sections we used various meth-ods for estimating the sum of a series—the method depended on which test was used to prove convergence. What about series for which the Ratio Test works? There are two possibilities: If the series happens to be an alternating series, as in Example 4, then it is best to use the meth-ods of Section 11.5. If the terms are all positive, then use the special methods explained in Exercise 34. Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent. M EXAMPLE 5 Test the convergence of the series . SOLUTION Since the terms are positive, we don’t need the absolute value signs. (See Equation 3.6.6.) Since , the given series is divergent by the Ratio Test. M Although the Ratio Test works in Example 5, an easier method is to use the Test for Divergence. Since it follows that does not approach 0 as . Therefore the given series is divergent by the Test for Divergence. The following test is convenient to apply when th powers occur. Its proof is similar to the proof of the Ratio Test and is left as Exercise 37. THE ROOT TEST (i) If , then the series is absolutely convergent (and therefore convergent). (ii) If or , then the series is divergent. (iii) If , the Root Test is inconclusive. If , then part (iii) of the Root Test says that the test gives no infor-mation. The series could converge or diverge. (If in the Ratio Test, don’t try the Root Test because will again be 1. And if in the Root Test, don’t try the Ratio Test because it will fail too.) EXAMPLE 6 Test the convergence of the series . SOLUTION Thus the given series converges by the Root Test. M s n an 2n 3 3n 2 2 3 n 3 2 n l 2 3 1 an 2n 3 3n 2 n n1 2n 3 3n 2 n V L 1 L L 1 an lim n l s n an 1 lim n l s n an 1 n1 an lim n l s n an lim n l s n an L 1 n1 an lim n l s n an L 1 n n l an an nn n! n n n n 1 2 3 n n NOTE e 1 n 1 n n 1 1 n n l e as n l an1 an n 1n1 n 1! n! n n n 1n 1n n 1n! n! n n an n nn! n1 n n n! V 718 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES REARRANGEMENTS The question of whether a given convergent series is absolutely convergent or condi-tionally convergent has a bearing on the question of whether inﬁnite sums behave like ﬁnite sums. If we rearrange the order of the terms in a ﬁnite sum, then of course the value of the sum remains unchanged. But this is not always the case for an inﬁnite series. By a rearrangement of an inﬁnite series we mean a series obtained by simply changing the order of the terms. For instance, a rearrangement of could start as follows: It turns out that if is an absolutely convergent series with sum s, then any rearrangement of has the same sum s. However, any conditionally convergent series can be rearranged to give a different sum. To illustrate this fact let’s consider the alternating harmonic series (See Exercise 36 in Section 11.5.) If we multiply this series by , we get Inserting zeros between the terms of this series, we have Now we add the series in Equations 6 and 7 using Theorem 11.2.8: Notice that the series in (8) contains the same terms as in (6), but rearranged so that one negative term occurs after each pair of positive terms. The sums of these series, however, are different. In fact, Riemann proved that if is a conditionally convergent series and r is any real number what-soever, then there is a rearrangement of that has a sum equal to r. A proof of this fact is outlined in Exercise 40. an an 1 1 3 1 2 1 5 1 7 1 4 3 2 ln 2 8 0 1 2 0 1 4 0 1 6 0 1 8 1 2 ln 2 7 1 2 1 4 1 6 1 8 1 2 ln 2 1 2 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 ln 2 6 an an a1 a2 a5 a3 a4 a15 a6 a7 a20 an an SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS \|\|\|\| 719 N Adding these zeros does not affect the sum of the series; each term in the sequence of partial sums is repeated, but the limit is the same. 5. 6. 7. 8. 9. 10. 11. 12. 14. n1 1n1 n 2 2n n! n1 10n n 142n1 13. n1 sin 4n 4n n1 1n e 1n n3 n1 1 n n sn3 2 n1 1 n 1.1n n4 n1 n! 100 n k1 k( 2 3) k n1 1n n 4 n1 1n1 s 4 n 1. What can you say about the series in each of the following cases? (a) (b) (c) 2–28 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. 2. 3. n1 1 n1 2 n n4 4. n0 10n n! n1 n2 2n lim n l an1 an 1 lim n l an1 an 0.8 lim n l an1 an 8 an EXERCISES 11.6 720 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES Ratio Test. As usual, we let be the remainder after terms, that is, (a) If is a decreasing sequence and , show, by summing a geometric series, that (b) If is an increasing sequence, show that 35. (a) Find the partial sum of the series . Use Exer-cise 34 to estimate the error in using as an approximation to the sum of the series. (b) Find a value of so that is within of the sum. Use this value of to approximate the sum of the series. 36. Use the sum of the ﬁrst 10 terms to approximate the sum of the series Use Exercise 34 to estimate the error. 37. Prove the Root Test. [Hint for part (i): Take any number such that and use the fact that there is an integer such that whenever .] 38. Around 1910, the Indian mathematician Srinivasa Ramanujan discovered the formula William Gosper used this series in 1985 to compute the ﬁrst 17 million digits of . (a) Verify that the series is convergent. (b) How many correct decimal places of do you get if you use just the ﬁrst term of the series? What if you use two terms? 39. Given any series , we deﬁne a series whose terms are all the positive terms of and a series whose terms are all the negative terms of . To be speciﬁc, we let Notice that if , then and , whereas if , then and . (a) If is absolutely convergent, show that both of the series and are convergent. (b) If is conditionally convergent, show that both of the series and are divergent. 40. Prove that if is a conditionally convergent series and is any real number, then there is a rearrangement of whose sum is . [Hints: Use the notation of Exercise 39. Take just enough positive terms so that their sum is greater than . Then add just enough negative terms so that the cumulative sum is less than . Continue in this manner and use Theorem 11.2.6.] r an r an r an r an an an an an an an an 0 an an an 0 an 0 an an an 0 an an an 2 an an an 2 an an an an an 1 2s2 9801 n0 4n!1103 26390n n!43964n n N s n an r N L r 1 r n1 n 2n n 0.00005 sn n s5 n1 1n2n s5 Rn an1 1 L rn Rn an1 1 rn1 rn1 1 rn Rn an1 an2 an3 n Rn 15. 16. 17. 18. 20. 22. 23. 24. 25. 26. 27. 28. The terms of a series are deﬁned recursively by the equations Determine whether converges or diverges. 30. A series is deﬁned by the equations Determine whether converges or diverges. For which of the following series is the Ratio Test inconclusive (that is, it fails to give a deﬁnite answer)? (a) (b) (c) (d) 32. For which positive integers is the following series convergent? (a) Show that converges for all . (b) Deduce that for all . 34. Let be a series with positive terms and let . Suppose that , so converges by the an limn l rn L 1 rn an1an an x limn l x nn! 0 x n0 x nn! 33. n1 n!2 kn! k n1 sn 1 n2 n1 3n1 sn n1 n 2n n1 1 n3 31. an an1 2 cos n sn an a1 1 an an an1 5n 1 4n 3 an a1 2 29. n1 1n 2nn! 5 8 11 3n 2 n1 2 4 6 2n n! 2 5 2 6 5 8 2 6 10 5 8 11 2 6 10 14 5 8 11 14 1n1 1 3 5 2n 1 2n 1! 1 1 3 3! 1 3 5 5! 1 3 5 7 7! n2 n ln nn n1 1 1 n n2 n2 2n n 1 5n n1 n2 1 2n2 1 n 21. n1 2n n n n1 cosn3 n! 19. n1 n! n n n2 1n ln n n1 3 cos n n 23 2 n1 1n arctan n n2 SECTION 11.7 STRATEGY FOR TESTING SERIES \|\|\|\| 721 STRATEGY FOR TESTING SERIES We now have several ways of testing a series for convergence or divergence; the problem is to decide which test to use on which series. In this respect, testing series is similar to integrating functions. Again there are no hard and fast rules about which test to apply to a given series, but you may ﬁnd the following advice of some use. It is not wise to apply a list of the tests in a speciﬁc order until one ﬁnally works. That would be a waste of time and effort. Instead, as with integration, the main strategy is to classify the series according to its form. 1. If the series is of the form , it is a -series, which we know to be convergent if and divergent if . 2. If the series has the form or , it is a geometric series, which converges if and diverges if . Some preliminary algebraic manipulation may be required to bring the series into this form. 3. If the series has a form that is similar to a -series or a geometric series, then one of the comparison tests should be considered. In particular, if is a rational function or an algebraic function of (involving roots of polynomials), then the series should be compared with a -series. Notice that most of the series in Exer-cises 11.4 have this form. (The value of should be chosen as in Section 11.4 by keeping only the highest powers of in the numerator and denominator.) The com-parison tests apply only to series with positive terms, but if has some negative terms, then we can apply the Comparison Test to and test for absolute convergence. 4. If you can see at a glance that , then the Test for Divergence should be used. 5. If the series is of the form or , then the Alternating Series Test is an obvious possibility. 6. Series that involve factorials or other products (including a constant raised to the power) are often conveniently tested using the Ratio Test. Bear in mind that as for all -series and therefore all rational or algebraic functions of . Thus the Ratio Test should not be used for such series. 7. If is of the form , then the Root Test may be useful. 8. If , where is easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisﬁed). In the following examples we don’t work out all the details but simply indicate which tests should be used. EXAMPLE 1 Since as , we should use the Test for Divergence. M EXAMPLE 2 Since is an algebraic function of , we compare the given series with a -series. The p n an n1 sn3 1 3n3 4n2 2 n l an l 1 2 0 n1 n 1 2n 1 V x 1 fx dx an fn bnn an n p n l an1anl 1 nth 1nbn 1n1bn lim n l an 0 an an n p p n an p r 1 r 1 ar n ar n1 p 1 p 1 p 1n p 11.7 comparison series for the Limit Comparison Test is , where M EXAMPLE 3 Since the integral is easily evaluated, we use the Integral Test. The Ratio Test also works. M EXAMPLE 4 Since the series is alternating, we use the Alternating Series Test. M EXAMPLE 5 Since the series involves , we use the Ratio Test. M EXAMPLE 6 Since the series is closely related to the geometric series , we use the Comparison Test. M 13n n1 1 2 3n k! k1 2k k! V n1 1n n3 n4 1 x 1 xex2 dx n1 nen2 V bn sn3 3n3 n32 3n3 1 3n32 bn 722 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. n1 (s n 2 1) n1 (s n 2 1) n n2 1 ln nln n n1 n n 1 n2 n1 1 n n cos2n n1 sin1n sn n1 n!n n4n k1 5 k 3 k 4 k j1 1 j sj j 5 n1 1n cosh n n1 e 1n n2 k1 k ln k k 13 n1 n2 1 5n n1 n! e n2 n1 n sin1n n1 tan1n n1 sn 2 1 n3 2n2 5 n1 22n n n 1–38 Test the series for convergence or divergence. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. k1 k 5 5k n1 1n ln n sn n2 1n1 sn 1 n1 1n 21n n1 n2 1 n3 1 n0 n! 2 5 8 3n 2 n1 sin 2n 1 2n n1 3n n2 n! n1 sin n n2 1n1 n ln n n1 n2en3 k1 k 2ek k1 2 kk! k 2! n2 1 nsln n n1 1 2n 1 n1 n2 2n1 5n n1 1n n n2 2 n1 1n n n 2 n1 2n 1n n 2n n1 1 n 3n EXERCISES 11.7 POWER SERIES A power series is a series of the form where is a variable and the ’s are constants called the coefﬁcients of the series. For each ﬁxed , the series (1) is a series of constants that we can test for convergence or divergence. A power series may converge for some values of and diverge for other values of . The sum of the series is a function whose domain is the set of all for which the series converges. Notice that resembles a polynomial. The only difference is that has inﬁnitely many terms. For instance, if we take for all , the power series becomes the geometric series which converges when and diverges when (see Equation 11.2.5). More generally, a series of the form is called a power series in or a power series centered at a or a power series about a. Notice that in writing out the term corresponding to in Equations 1 and 2 we have adopted the convention that even when . Notice also that when , all of the terms are 0 for and so the power series (2) always converges when . EXAMPLE 1 For what values of is the series convergent? SOLUTION We use the Ratio Test. If we let , as usual, denote the nth term of the series, then . If , we have By the Ratio Test, the series diverges when . Thus the given series converges only when . M EXAMPLE 2 For what values of does the series converge? SOLUTION Let . Then 1 1 1 n x 3 l x 3 as n l an1 an x 3n1 n 1 n x 3n an x 3nn n1 x 3n n x V x 0 x 0 lim n l n 1x lim n l an1 an lim n l n 1!x n1 n!x n x 0 an n!x n an n0 n!x n x V x a n 1 x a x a x a0 1 n 0 x a n0 cnx an c0 c1x a c2x a2 2 x 1 1 x 1 n0 x n 1 x x 2 x n n cn 1 f f x fx c0 c1x c2x 2 cnx n x x x cn x n0 cnx n c0 c1x c2x 2 c3x 3 1 11.8 SECTION 11.8 POWER SERIES \|\|\|\| 723 N TRIGONOMETRIC SERIES A power series is a series in which each term is a power function. A trigonometric series is a series whose terms are trigonometric func-tions. This type of series is discussed on the website www.stewartcalculus.com Click on Additional Topics and then on Fourier Series. n0 an cos nx bn sin nx N Notice that n 1n! n 1! n 1nn 1 . . . 3 2 1 By the Ratio Test, the given series is absolutely convergent, and therefore convergent, when and divergent when . Now so the series converges when and diverges when or . The Ratio Test gives no information when so we must consider and separately. If we put in the series, it becomes , the harmonic series, which is divergent. If , the series is , which converges by the Alternating Series Test. Thus the given power series converges for . M We will see that the main use of a power series is that it provides a way to represent some of the most important functions that arise in mathematics, physics, and chemistry. In particular, the sum of the power series in the next example is called a Bessel function, after the German astronomer Friedrich Bessel (1784–1846), and the function given in Exer-cise 35 is another example of a Bessel function. In fact, these functions ﬁrst arose when Bessel solved Kepler’s equation for describing planetary motion. Since that time, these functions have been applied in many different physical situations, including the tempera-ture distribution in a circular plate and the shape of a vibrating drumhead. EXAMPLE 3 Find the domain of the Bessel function of order 0 deﬁned by SOLUTION Let . Then Thus, by the Ratio Test, the given series converges for all values of . In other words, the domain of the Bessel function is . M Recall that the sum of a series is equal to the limit of the sequence of partial sums. So when we deﬁne the Bessel function in Example 3 as the sum of a series we mean that, for every real number , where The ﬁrst few partial sums are s4x 1 x 2 4 x 4 64 x 6 2304 x 8 147,456 s3x 1 x 2 4 x 4 64 x 6 2304 s2x 1 x 2 4 x 4 64 s1x 1 x 2 4 s0x 1 snx n i0 1i x 2i 22i i!2 J0x lim n l snx x , J0 x x 2 4n 12 l 0 1 for all x x 2n2 22n2n 12n!2 22nn!2 x 2n an1 an 1n1x 2n1 22n1n 1!2 22nn!2 1nx 2n an 1nx 2n22nn!2 J0x n0 1nx 2n 22nn!2 2 x 4 1nn x 2 1n x 4 x 4 x 2 x 3 1 x 4 x 2 2 x 4 2 x 4 & ? 1 x 3 1 & ? x 3 1 x 3 1 x 3 1 724 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N Notice how closely the computer-generated model (which involves Bessel functions and cosine functions) matches the photograph of a vibrating rubber membrane. National Film Board of Canada Figure 1 shows the graphs of these partial sums, which are polynomials. They are all approximations to the function , but notice that the approximations become better when more terms are included. Figure 2 shows a more complete graph of the Bessel function. For the power series that we have looked at so far, the set of values of for which the series is convergent has always turned out to be an interval [a ﬁnite interval for the geometric series and the series in Example 2, the inﬁnite interval in Example 3, and a collapsed interval in Example 1]. The following theorem, proved in Appendix F, says that this is true in general. THEOREM For a given power series , there are only three possibilities: (i) The series converges only when . (ii) The series converges for all . (iii) There is a positive number such that the series converges if and diverges if . The number in case (iii) is called the radius of convergence of the power series. By convention, the radius of convergence is in case (i) and in case (ii). The interval of convergence of a power series is the interval that consists of all values of for which the series converges. In case (i) the interval consists of just a single point . In case (ii) the interval is . In case (iii) note that the inequality can be rewrit-ten as . When is an endpoint of the interval, that is, , anything can happen—the series might converge at one or both endpoints or it might diverge at both endpoints. Thus in case (iii) there are four possibilities for the interval of convergence: The situation is illustrated in Figure 3. We summarize here the radius and interval of convergence for each of the examples already considered in this section. FIGURE 3 a-R a a+R convergence for \|x-a\|<R divergence for \|x-a\|>R a R, a R a R, a R a R, a R a R, a R x a R x a R x a R x a R , a x R R 0 R x a R x a R R x x a n0 cnx an 3 0, 0 0 , x J0 SECTION 11.8 POWER SERIES \|\|\|\| 725 Series Radius of convergence Interval of convergence Geometric series Example 1 Example 2 Example 3 , R n0 1n x 2n 22nn!2 2, 4 R 1 n1 x 3n n 0 R 0 n0 n! x n 1, 1 R 1 n0 x n s¢ 0 x 1 y 1 s¡ s™ s£ s¸ J¸ FIGURE 1 Partial sums of the Bessel function J¸ FIGURE 2 0 x 1 y 10 _10 y=J¸(x) In general, the Ratio Test (or sometimes the Root Test) should be used to determine the radius of convergence . The Ratio and Root Tests always fail when is an endpoint of the interval of convergence, so the endpoints must be checked with some other test. EXAMPLE 4 Find the radius of convergence and interval of convergence of the series SOLUTION Let . Then By the Ratio Test, the given series converges if and diverges if . Thus it converges if and diverges if . This means that the radius of con-vergence is . We know the series converges in the interval , but we must now test for con-vergence at the endpoints of this interval. If , the series becomes which diverges. (Use the Integral Test or simply observe that it is a -series with .) If , the series is which converges by the Alternating Series Test. Therefore the given power series con-verges when , so the interval of convergence is . M EXAMPLE 5 Find the radius of convergence and interval of convergence of the series SOLUTION If , then Using the Ratio Test, we see that the series converges if and it diverges if . So it converges if and diverges if . Thus the radius of convergence is . R 3 x 2 3 x 2 3 x 2 3 1 x 2 3 1 1 1 n x 2 3 l x 2 3 as n l an1 an n 1x 2n1 3n2 3n1 nx 2n an nx 2n3n1 n0 nx 2n 3n1 V ( 1 3, 1 3] 1 3 x 1 3 n0 3n( 1 3) n sn 1 n0 1n sn 1 x 1 3 p 1 2 1 p n0 3n( 1 3) n sn 1 n0 1 sn 1 1 s1 1 s2 1 s3 1 s4 x 1 3 ( 1 3, 1 3) R 1 3 x 1 3 x 1 3 3 x 1 3 x 1 3 1 1n 1 2n x l 3 x as n l an1 an 3n1x n1 sn 2 sn 1 3nx n 3x n 1 n 2 an 3nx nsn 1 n0 3nx n sn 1 x R 726 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES The inequality can be written as , so we test the series at the endpoints and 1. When , the series is which diverges by the Test for Divergence [ doesn’t converge to 0]. When , the series is which also diverges by the Test for Divergence. Thus the series converges only when , so the interval of convergence is . M 5, 1 5 x 1 n0 n3n 3n1 1 3 n0 n x 1 1nn n0 n3n 3n1 1 3 n0 1nn x 5 5 5 x 1 x 2 3 SECTION 11.8 POWER SERIES \|\|\|\| 727 25. 26. 27. 28. If is convergent, does it follow that the following series are convergent? (a) (b) 30. Suppose that converges when and diverges when . What can be said about the convergence or diver-gence of the following series? (a) (b) (c) (d) 31. If is a positive integer, ﬁnd the radius of convergence of the series 32. Let and be real numbers with . Find a power series whose interval of convergence is (a) (b) (c) (d) 33. Is it possible to ﬁnd a power series whose interval of convergence is ? Explain. 0, p, q p, q p, q p, q p q q p n0 n!k kn! x n k n0 1ncn 9n n0 cn3n n0 cn8n n0 cn x 6 x 4 n0 cnx n n0 cn4n n0 cn2n cn4n n0 29. n1 n!x n 1 3 5 2n 1 n1 x n 1 3 5 2n 1 n2 x 2n nln n2 n1 4x 1n n2 n1 n 2x n 2 4 6 2n 24. n1 n!2x 1n 23. 1. What is a power series? 2. (a) What is the radius of convergence of a power series? How do you ﬁnd it? (b) What is the interval of convergence of a power series? How do you ﬁnd it? 3–28 Find the radius of convergence and interval of convergence of the series. 4. 5. 6. 8. 9. 10. 11. 12. 13. 14. 16. 17. 18. 19. 20. 21. , 22. n1 nx 4n n3 1 b 0 n1 n bn x an n1 3x 2n n 3n n1 x 2n n n n1 n 4n x 1n n1 3nx 4n sn n0 1n x 3 n 2n 1 n0 x 2n n2 1 15. n0 1n x 2n 2n! n2 1n x n 4n ln n n1 x n 5 nn 5 n1 2nx n s 4 n n1 10 nx n n3 n1 1n n 2 x n 2 n n1 n nx n n0 x n n! 7. n1 sn x n n1 1n1x n n 3 n0 1nx n n 1 n1 x n sn 3. EXERCISES 11.8 (c) If your CAS has built-in Airy functions, graph on the same screen as the partial sums in part (b) and observe how the partial sums approximate . A function is deﬁned by that is, its coefﬁcients are and for all . Find the interval of convergence of the series and ﬁnd an explicit formula for . 38. If , where for all , ﬁnd the interval of convergence of the series and a formula for . 39. Show that if , where , then the radius of convergence of the power series is . 40. Suppose that the power series satisﬁes for all . Show that if exists, then it is equal to the radius of convergence of the power series. 41. Suppose the series has radius of convergence 2 and the series has radius of convergence 3. What is the radius of convergence of the series ? 42. Suppose that the radius of convergence of the power series is . What is the radius of convergence of the power series ? cnx 2n R cnx n cn dnx n dnx n cnx n lim n l cncn1 n cn 0 cnx a n R 1c cnx n c 0 lim n l s n cn c f x n 0 cn4 cn f x n0 cnx n f x n 0 c2n1 2 c2n 1 f x 1 2x x 2 2x 3 x 4 f 37. A A CAS ; 34. Graph the ﬁrst several partial sums of the series , together with the sum function , on a com-mon screen. On what interval do these partial sums appear to be converging to ? 35. The function deﬁned by is called the Bessel function of order 1. (a) Find its domain. ; (b) Graph the ﬁrst several partial sums on a common screen. (c) If your CAS has built-in Bessel functions, graph on the same screen as the partial sums in part (b) and observe how the partial sums approximate . 36. The function deﬁned by is called the Airy function after the English mathematician and astronomer Sir George Airy (1801–1892). (a) Find the domain of the Airy function. ; (b) Graph the ﬁrst several partial sums on a common screen. Ax 1 x 3 2 3 x 6 2 3 5 6 x 9 2 3 5 6 8 9 A J1 J1 CAS J1x n0 1nx 2n1 n!n 1!22n1 J1 f x f x 11 x n0 x n snx 728 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES REPRESENTATIONS OF FUNCTIONS AS POWER SERIES In this section we learn how to represent certain types of functions as sums of power series by manipulating geometric series or by differentiating or integrating such a series. You might wonder why we would ever want to express a known function as a sum of inﬁnitely many terms. We will see later that this strategy is useful for integrating functions that don’t have elementary antiderivatives, for solving differential equations, and for approximating func-tions by polynomials. (Scientists do this to simplify the expressions they deal with; com-puter scientists do this to represent functions on calculators and computers.) We start with an equation that we have seen before: We ﬁrst encountered this equation in Example 5 in Section 11.2, where we obtained it by observing that it is a geometric series with and . But here our point of view is different. We now regard Equation 1 as expressing the function as a sum of a power series. FIGURE 1 ƒ= 1 1-x and some partial sums 0 x y 1 _1 f s™ s∞ sˆ s¡¡ fx 11 x r x a 1 x 1 1 1 x 1 x x 2 x 3 n0 x n 1 11.9 N A geometric illustration of Equation 1 is shown in Figure 1. Because the sum of a series is the limit of the sequence of partial sums, we have where is the th partial sum. Notice that as increases, becomes a better approxima-tion to for . 1 x 1 f x snx n n snx 1 x x2 x n 1 1 x lim n l snx EXAMPLE 1 Express as the sum of a power series and ﬁnd the interval of convergence. SOLUTION Replacing by in Equation 1, we have Because this is a geometric series, it converges when , that is, , or . Therefore the interval of convergence is . (Of course, we could have determined the radius of convergence by applying the Ratio Test, but that much work is unnecessary here.) M EXAMPLE 2 Find a power series representation for . SOLUTION In order to put this function in the form of the left side of Equation 1 we ﬁrst factor a 2 from the denominator: This series converges when , that is, . So the interval of convergence is . M EXAMPLE 3 Find a power series representation of . SOLUTION Since this function is just times the function in Example 2, all we have to do is to multiply that series by : Another way of writing this series is as follows: As in Example 2, the interval of convergence is . M DIFFERENTIATION AND INTEGRATION OF POWER SERIES The sum of a power series is a function whose domain is the inter-val of convergence of the series. We would like to be able to differentiate and integrate such functions, and the following theorem (which we won’t prove) says that we can do so by differentiating or integrating each individual term in the series, just as we would for a polynomial. This is called term-by-term differentiation and integration. fx n0 cnx an 2, 2 x 3 x 2 n3 1n1 2n2 x n 1 2 x 3 1 4 x 4 1 8 x 5 1 16 x 6 x 3 x 2 x 3 1 x 2 x 3 n0 1n 2n1 x n n0 1n 2n1 x n3 x 3 x 3 x 3x 2 2, 2 x 2 x2 1 1 2 n0 x 2 n n0 1n 2n1 x n 1 2 x 1 2 1 x 2 1 21 x 2 1x 2 1, 1 x 1 x 2 1 x 2 1 n0 1nx 2n 1 x 2 x 4 x 6 x 8 1 1 x 2 1 1 x 2 n0 x 2n x 2 x 11 x 2 V SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES \|\|\|\| 729 N It’s legitimate to move across the sigma sign because it doesn’t depend on . [Use Theorem 11.2.8(i) with .] c x 3 n x 3 THEOREM If the power series has radius of convergence , then the function deﬁned by is differentiable (and therefore continuous) on the interval and (i) (ii) The radii of convergence of the power series in Equations (i) and (ii) are both . Equations (i) and (ii) in Theorem 2 can be rewritten in the form (iii) (iv) We know that, for ﬁnite sums, the derivative of a sum is the sum of the derivatives and the integral of a sum is the sum of the integrals. Equations (iii) and (iv) assert that the same is true for inﬁnite sums, provided we are dealing with power series. (For other types of series of functions the situation is not as simple; see Exercise 36.) Although Theorem 2 says that the radius of convergence remains the same when a power series is differentiated or integrated, this does not mean that the interval of convergence remains the same. It may happen that the original series converges at an end-point, whereas the differentiated series diverges there. (See Exercise 37.) The idea of differentiating a power series term by term is the basis for a power-ful method for solving differential equations. We will discuss this method in Chapter 17. EXAMPLE 4 In Example 3 in Section 11.8 we saw that the Bessel function is deﬁned for all . Thus, by Theorem 2, is differentiable for all and its derivative is found by term-by-term differentiation as follows: M J0 x n0 d dx 1nx 2n 22nn!2 n1 1n 2nx 2n1 22nn!2 x J0 x J0x n0 1nx 2n 22nn!2 NOTE 3 NOTE 2 y n0 cnx andx n0 y cnx an dx d dx n0 cnx an n0 d dx cnx an NOTE 1 R C n0 cn x an1 n 1 y fx dx C c0x a c1 x a2 2 c2 x a3 3 fx c1 2c2x a 3c3x a2 n1 ncnx an1 a R, a R fx c0 c1x a c2x a2 n0 cnx an f R 0 cnx an 2 730 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N In part (ii), is written as , where , so all the terms of the series have the same form. C C1 ac0 c0x a C x c0 dx c0x C1 EXAMPLE 5 Express as a power series by differentiating Equation 1. What is the radius of convergence? SOLUTION Differentiating each side of the equation we get If we wish, we can replace by and write the answer as According to Theorem 2, the radius of convergence of the differentiated series is the same as the radius of convergence of the original series, namely, . M EXAMPLE 6 Find a power series representation for and its radius of convergence. SOLUTION We notice that, except for a factor of , the derivative of this function is . So we integrate both sides of Equation 1: To determine the value of we put in this equation and obtain . Thus and The radius of convergence is the same as for the original series: . M Notice what happens if we put in the result of Example 6. Since , we see that EXAMPLE 7 Find a power series representation for . SOLUTION We observe that and ﬁnd the required series by integrating the power series for found in Example 1. C x x 3 3 x 5 5 x 7 7 tan1x y 1 1 x 2 dx y 1 x 2 x 4 x 6 dx 11 x 2 fx 11 x 2 fx tan1x V ln 2 1 2 1 8 1 24 1 64 n1 1 n 2n ln 1 2 ln 2 x 1 2 R 1 x 1 ln1 x x x 2 2 x 3 3 n1 x n n C 0 ln1 0 C x 0 C n1 x n n C x 1 x x 2 2 x 3 3 C n0 x n1 n 1 C ln1 x y 1 1 x dx y 1 x x 2 dx 11 x 1 ln1 x R 1 1 1 x2 n0 n 1x n n 1 n 1 1 x2 1 2x 3x 2 n1 nx n1 1 1 x 1 x x 2 x 3 n0 x n 11 x2 V SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES \|\|\|\| 731 To ﬁnd we put and obtain . Therefore Since the radius of convergence of the series for is 1, the radius of conver-gence of this series for is also 1. M EXAMPLE 8 (a) Evaluate as a power series. (b) Use part (a) to approximate correct to within . SOLUTION (a) The ﬁrst step is to express the integrand, , as the sum of a power series. As in Example 1, we start with Equation 1 and replace by : Now we integrate term by term: This series converges for , that is, for . (b) In applying the Fundamental Theorem of Calculus, it doesn’t matter which anti-derivative we use, so let’s use the antiderivative from part (a) with C 0: This inﬁnite series is the exact value of the deﬁnite integral, but since it is an alternating series, we can approximate the sum using the Alternating Series Estimation Theorem. If we stop adding after the term with , the error is smaller than the term with n 4: So we have M y 0.5 0 1 1 x 7 dx 1 2 1 8 28 1 15 215 1 22 222 0.49951374 1 29 229 6.4 1011 n 3 1 2 1 8 28 1 15 215 1 22 222 1n 7n 127n1 y 0.5 0 1 1 x 7 dx x x 8 8 x 15 15 x 22 22 0 12 x 1 x 7 1 C x x 8 8 x 15 15 x 22 22 y 1 1 x 7 dx y n0 1nx 7n dx C n0 1n x 7n1 7n 1 n0 1nx 7n 1 x 7 x 14 1 1 x 7 1 1 x 7 n0 x 7n x 7 x 11 x 7 107 x0.5 0 11 x 7 dx x 11 x 7 dx tan1x 11 x 2 tan1x x x 3 3 x 5 5 x 7 7 n0 1n x 2n1 2n 1 C tan1 0 0 x 0 C 732 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N The power series for obtained in Exam-ple 7 is called Gregory’s series after the Scottish mathematician James Gregory (1638–1675), who had anticipated some of Newton’s discoveries. We have shown that Gregory’s series is valid when , but it turns out (although it isn’t easy to prove) that it is also valid when . Notice that when the series becomes This beautiful result is known as the Leibniz formula for . 4 1 1 3 1 5 1 7 x 1 x 1 1 x 1 tan1x N This example demonstrates one way in which power series representations are useful. Integrating by hand is incredibly dif-ﬁcult. Different computer algebra systems return different forms of the answer, but they are all extremely complicated. (If you have a CAS, try it yourself.) The inﬁnite series answer that we obtain in Example 8(a) is actually much easier to deal with than the ﬁnite answer provided by a CAS. 11 x 7 SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES \|\|\|\| 733 15–18 Find a power series representation for the function and determine the radius of convergence. 16. 17. 18. ; 19–22 Find a power series representation for , and graph and several partial sums on the same screen. What happens as increases? 19. 20. 22. 23–26 Evaluate the indeﬁnite integral as a power series. What is the radius of convergence? 24. 25. 26. 27–30 Use a power series to approximate the deﬁnite integral to six decimal places. 27. 28. 29. 30. 31. Use the result of Example 6 to compute correct to ﬁve decimal places. 32. Show that the function is a solution of the differential equation 33. (a) Show that (the Bessel function of order 0 given in Example 4) satisﬁes the differential equation (b) Evaluate correct to three decimal places. x1 0 J0x dx x 2J0x xJ0 x x 2J0x 0 J0 f x f x 0 f x n0 1nx 2n 2n! ln 1.1 y 0.3 0 x 2 1 x 4 dx y 0.1 0 x arctan3x dx y 0.4 0 ln1 x 4 dx y 0.2 0 1 1 x 5 dx y tan1x 2 dx y x tan 1x x 3 dx y ln1 t t dt y t 1 t 8 dt 23. f x tan12x f x ln 1 x 1 x 21. f x lnx 2 4 f x x x 2 16 n snx f f f x arctanx3 f x x 3 x 22 f x x 2 1 2x2 f x ln5 x 15. 1. If the radius of convergence of the power series is 10, what is the radius of convergence of the series ? Why? 2. Suppose you know that the series converges for . What can you say about the following series? Why? 3–10 Find a power series representation for the function and determine the interval of convergence. 3. 4. 6. 7. 9. 10. 11–12 Express the function as the sum of a power series by ﬁrst using partial fractions. Find the interval of convergence. 11. 12. (a) Use differentiation to ﬁnd a power series representation for What is the radius of convergence? (b) Use part (a) to ﬁnd a power series for (c) Use part (b) to ﬁnd a power series for 14. (a) Find a power series representation for . What is the radius of convergence? (b) Use part (a) to ﬁnd a power series for . (c) Use part (a) to ﬁnd a power series for . f x lnx 2 1 f x x ln1 x f x ln1 x f x x 2 1 x3 f x 1 1 x3 f x 1 1 x2 13. f x x 2 2x 2 x 1 f x 3 x 2 x 2 f x x 2 a 3 x 3 f x 1 x 1 x f x x 2x 2 1 8. f x x 9 x 2 f x 1 x 10 f x 2 3 x 5. f x 3 1 x 4 f x 1 1 x n0 bn n 1 x n1 x 2 n0 bnx n n1 ncnx n1 n0 cnx n EXERCISES 11.9 38. (a) Starting with the geometric series , ﬁnd the sum of the series (b) Find the sum of each of the following series. (i) , (ii) (c) Find the sum of each of the following series. (i) , (ii) (iii) 39. Use the power series for to prove the following expres-sion for as the sum of an inﬁnite series: 40. (a) By completing the square, show that (b) By factoring as a sum of cubes, rewrite the integral in part (a). Then express as the sum of a power series and use it to prove the following formula for : 3s3 4 n0 1n 8 n 2 3n 1 1 3n 2 1x 3 1 x 3 1 y 12 0 dx x 2 x 1 3s3 2s3 n0 1n 2n 13n tan 1x n1 n2 2n n2 n2 n 2n x 1 n2 nn 1x n n1 n 2n x 1 n1 nx n x 1 n1 nx n1 n0 x n 34. The Bessel function of order 1 is deﬁned by (a) Show that satisﬁes the differential equation (b) Show that . (a) Show that the function is a solution of the differential equation (b) Show that . 36. Let . Show that the series converges for all values of but the series of derivatives diverges when , an integer. For what values of does the series converge? Let Find the intervals of convergence for , , and . f f f f x n1 x n n2 37. fnx x n x 2n fnx x fnx fnx sin nxn2 f x e x f x f x f x n0 x n n! 35. J0 x J1x x 2J1x xJ1x x 2 1J1x 0 J1 J1x n0 1nx 2n1 n!n 1!22n1 734 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES TAYLOR AND MACLAURIN SERIES In the preceding section we were able to ﬁnd power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions have power series representations? How can we ﬁnd such representations? We start by supposing that is any function that can be represented by a power series Let’s try to determine what the coefﬁcients must be in terms of . To begin, notice that if we put in Equation 1, then all terms after the ﬁrst one are 0 and we get By Theorem 11.9.2, we can differentiate the series in Equation 1 term by term: and substitution of in Equation 2 gives fa c1 x a x a R fx c1 2c2x a 3c3x a2 4c4x a3 2 fa c0 x a f cn x a R fx c0 c1x a c2x a2 c3x a3 c4x a4 1 f 11.10 SECTION 11.10 TAYLOR AND MACLAURIN SERIES \|\|\|\| 735 Now we differentiate both sides of Equation 2 and obtain Again we put in Equation 3. The result is Let’s apply the procedure one more time. Differentiation of the series in Equation 3 gives and substitution of in Equation 4 gives By now you can see the pattern. If we continue to differentiate and substitute , we obtain Solving this equation for the coefﬁcient This formula remains valid even for if we adopt the conventions that and . Thus we have proved the following theorem. THEOREM If has a power series representation (expansion) at , that is, if then its coefﬁcients are given by the formula Substituting this formula for back into the series, we see that if has a power series expansion at , then it must be of the following form. fa fa 1! x a f a 2! x a2 f a 3! x a3 fx n0 f na n! x an 6 a f cn cn f na n! x a R fx n0 cnx an a f 5 f 0 f 0! 1 n 0 cn f na n! cn, we get nth f na 2 3 4 ncn n!cn x a f a 2 3c3 3!c3 x a x a R f x 2 3c3 2 3 4c4x a 3 4 5c5x a2 4 f a 2c2 x a x a R f x 2c2 2 3c3x a 3 4c4x a2 3 The series in Equation 6 is called the Taylor series of the function f at a (or about a or centered at a). For the special case the Taylor series becomes This case arises frequently enough that it is given the special name Maclaurin series. We have shown that if can be represented as a power series about , then is equal to the sum of its Taylor series. But there exist functions that are not equal to the sum of their Taylor series. An example of such a function is given in Exercise 70. EXAMPLE 1 Find the Maclaurin series of the function and its radius of convergence. SOLUTION If , then , so for all . Therefore the Taylor series for at 0 (that is, the Maclaurin series) is To ﬁnd the radius of convergence we let . Then so, by the Ratio Test, the series converges for all and the radius of convergence is . M The conclusion we can draw from Theorem 5 and Example 1 is that if has a power series expansion at 0, then So how can we determine whether does have a power series representation? Let’s investigate the more general question: Under what circumstances is a function equal to the sum of its Taylor series? In other words, if has derivatives of all orders, when is it true that As with any convergent series, this means that is the limit of the sequence of partial sums. In the case of the Taylor series, the partial sums are fa fa 1! x a f a 2! x a2 f na n! x an Tnx n i0 f ia i! x ai fx fx n0 f na n! x an f e x e x n0 x n n! e x R x an1 an x n1 n 1! n! x n x n 1 l 0 1 an x nn! n0 f n0 n! x n n0 x n n! 1 x 1! x 2 2! x 3 3! f n f n0 e 0 1 f nx e x fx e x fx e x V f a f NOTE fx n0 f n0 n! x n f0 f0 1! x f 0 2! x 2 7 a 0 736 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES The Taylor series is named after the English mathematician Brook Taylor (1685–1731) and the Maclaurin series is named in honor of the Scot-tish mathematician Colin Maclaurin (1698–1746) despite the fact that the Maclaurin series is really just a special case of the Taylor series. But the idea of representing particular functions as sums of power series goes back to Newton, and the general Taylor series was known to the Scot-tish mathematician James Gregory in 1668 and to the Swiss mathematician John Bernoulli in the 1690s. Taylor was apparently unaware of the work of Gregory and Bernoulli when he published his discoveries on series in 1715 in his book Methodus incrementorum directa et inversa. Maclaurin series are named after Colin Maclau-rin because he popularized them in his calculus textbook Treatise of Fluxions published in 1742. TAYLOR AND MACLAURIN Notice that is a polynomial of degree called the nth-degree Taylor polynomial of f at a. For instance, for the exponential function , the result of Example 1 shows that the Taylor polynomials at 0 (or Maclaurin polynomials) with , 2, and 3 are The graphs of the exponential function and these three Taylor polynomials are drawn in Figure 1. In general, is the sum of its Taylor series if If we let so that then is called the remainder of the Taylor series. If we can somehow show that , then it follows that We have therefore proved the following. THEOREM If , where is the nth-degree Taylor polyno-mial of at and for , then is equal to the sum of its Taylor series on the interval . In trying to show that for a speciﬁc function , we usually use the following fact. TAYLOR’S INEQUALITY If for , then the remainder of the Taylor series satisﬁes the inequality To see why this is true for n 1, we assume that . In particular, we have , so for we have An antiderivative of is , so by Part 2 of the Fundamental Theorem of Calculus, we have fx fa Mx a or fx fa Mx a f f y x a f t dt y x a M dt a x a d f x M f x M for x a d Rnx M n 1! x a n1 Rnx x a d f n1x M 9 f lim n l Rnx 0 x a R f x a R lim n l Rnx 0 a f Tn fx Tnx Rnx 8 lim n l Tnx lim n l fx Rnx fx lim n l Rnx fx lim n l Rnx 0 Rnx fx Tnx Rnx Rnx fx Tnx fx lim n l Tnx fx T3x 1 x x 2 2! x 3 3! T2x 1 x x 2 2! T1x 1 x n 1 fx e x n Tn SECTION 11.10 TAYLOR AND MACLAURIN SERIES \|\|\|\| 737 0 x y y=´ y=T£(x) (0, 1) y=T™(x) y=T¡(x) y=T™(x) y=T£(x) FIGURE 1 N As increases, appears to approach in Figure 1. This suggests that is equal to the sum of its Taylor series. e x e x Tnx n Thus But . So A similar argument, using , shows that So Although we have assumed that , similar calculations show that this inequality is also true for . This proves Taylor’s Inequality for the case where . The result for any n is proved in a similar way by integrating times. (See Exercise 69 for the case .) In Section 11.11 we will explore the use of Taylor’s Inequality in approxi-mating functions. Our immediate use of it is in conjunction with Theorem 8. In applying Theorems 8 and 9 it is often helpful to make use of the following fact. for every real number x This is true because we know from Example 1 that the series converges for all and so its term approaches 0. EXAMPLE 2 Prove that is equal to the sum of its Maclaurin series. SOLUTION If , then for all n. If d is any positive number and , then . So Taylor’s Inequality, with and , says that for Notice that the same constant works for every value of n. But, from Equa-tion 10, we have lim n l e d n 1! x n1 e d lim n l x n1 n 1! 0 M e d x d Rnx e d n 1! x n1 M e d a 0 f n1x e x e d x d f n1x e x fx e x e x V nth x x nn! lim n l x n n! 0 10 NOTE n 2 n 1 n 1 x a x a R1x M 2 x a 2 R1x M 2 x a2 f x M R1x M 2 x a2 R1x fx T1x fx fa fax a fx fa fax a M 2 x a2 fx fa fax a M x a2 2 y x a ft dt y x a fa Mt a dt 738 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N As alternatives to Taylor’s Inequality, we have the following formulas for the remainder term. If is continuous on an interval and , then This is called the integral form of the remainder term. Another formula, called Lagrange’s form of the remainder term, states that there is a number between and such that This version is an extension of the Mean Value Theorem (which is the case ). Proofs of these formulas, together with dis-cussions of how to use them to solve the exam-ples of Sections 11.10 and 11.11, are given on the website www.stewartcalculus.com Click on Additional Topics and then on Formulas for the Remainder Term in Taylor series. n 0 Rnx f n1z n 1! x an1 a x z Rnx 1 n! y x a x tn f n1t dt x I I f n1 It follows from the Squeeze Theorem that and therefore for all values of x. By Theorem 8, is equal to the sum of its Maclaurin series, that is, M In particular, if we put in Equation 11, we obtain the following expression for the number as a sum of an inﬁnite series: EXAMPLE 3 Find the Taylor series for at . SOLUTION We have and so, putting in the deﬁnition of a Taylor series (6), we get Again it can be veriﬁed, as in Example 1, that the radius of convergence is . As in Example 2 we can verify that , so M We have two power series expansions for , the Maclaurin series in Equation 11 and the Taylor series in Equation 13. The ﬁrst is better if we are interested in values of near 0 and the second is better if is near 2. EXAMPLE 4 Find the Maclaurin series for and prove that it represents for all . SOLUTION We arrange our computation in two columns as follows: Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows: x x 3 3! x 5 5! x 7 7! n0 1n x 2n1 2n 1! f0 f0 1! x f 0 2! x 2 f 0 3! x 3 f 4x sin x f 40 0 f x cos x f 0 1 f x sin x f 0 0 fx cos x f0 1 fx sin x f0 0 x sin x sin x x x e x for all x e x n0 e 2 n! x 2n 13 lim n l Rnx 0 R n0 f n2 n! x 2n n0 e 2 n! x 2n a 2 f n2 e 2 a 2 fx e x e n0 1 n! 1 1 1! 1 2! 1 3! 12 e x 1 for all x e x n0 x n n! 11 e x lim n l Rnx 0 lim n l Rnx 0 SECTION 11.10 TAYLOR AND MACLAURIN SERIES \|\|\|\| 739 N In 1748 Leonard Euler used Equation 12 to ﬁnd the value of correct to digits. In 2003 Shigeru Kondo, again using the series in (12), computed to more than 50 billion decimal places. The special techniques employed to speed up the computation are explained on the web page numbers.computation.free.fr e 23 e Since is or , we know that for all x. So we can take in Taylor’s Inequality: By Equation 10 the right side of this inequality approaches 0 as , so by the Squeeze Theorem. It follows that as , so is equal to the sum of its Maclaurin series by Theorem 8. M We state the result of Example 4 for future reference. EXAMPLE 5 Find the Maclaurin series for . SOLUTION We could proceed directly as in Example 4 but it’s easier to differentiate the Maclaurin series for given by Equation 15: Since the Maclaurin series for converges for all , Theorem 2 in Section 11.9 tells us that the differentiated series for also converges for all . Thus M EXAMPLE 6 Find the Maclaurin series for the function . SOLUTION Instead of computing derivatives and substituting in Equation 7, it’s easier to multiply the series for (Equation 16) by : M EXAMPLE 7 Represent as the sum of its Taylor series centered at . 3 fx sin x x cos x x n0 1n x 2n 2n! n0 1n x 2n1 2n! x cos x fx x cos x for all x n0 1n x 2n 2n! cos x 1 x 2 2! x 4 4! x 6 6! 16 x cos x x sin x 1 3x 2 3! 5x 4 5! 7x 6 7! 1 x 2 2! x 4 4! x 6 6! cos x d dx sin x d dx x x 3 3! x 5 5! x 7 7! sin x cos x for all x n0 1n x 2n1 2n 1! sin x x x 3 3! x 5 5! x 7 7! 15 sin x n l Rnx l 0 Rnxl 0 n l Rnx M n 1! x n1 x n1 n 1! 14 M 1 f n1x 1 cos x sin x f n1x 740 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES FIGURE 2 0 x y 1 1 y=sin x T∞ T£ T¡ N Figure 2 shows the graph of together with its Taylor (or Maclaurin) polynomials Notice that, as increases, becomes a better approximation to . sin x Tnx n T5x x x 3 3! x 5 5! T3x x x 3 3! T1x x sin x N The Maclaurin series for , , and that we found in Examples 2, 4, and 5 were dis-covered, using different methods, by Newton. These equations are remarkable because they say we know everything about each of these functions if we know all its derivatives at the single number 0. cos x sin x e x SOLUTION Arranging our work in columns, we have and this pattern repeats indeﬁnitely. Therefore the Taylor series at is The proof that this series represents for all is very similar to that in Example 4. [Just replace by in (14).] We can write the series in sigma notation if we separate the terms that contain : M The power series that we obtained by indirect methods in Examples 5 and 6 and in Section 11.9 are indeed the Taylor or Maclaurin series of the given functions because Theorem 5 asserts that, no matter how a power series representation is obtained, it is always true that . In other words, the coefﬁcients are uniquely determined. EXAMPLE 8 Find the Maclaurin series for , where is any real number. SOLUTION Arranging our work in columns, we have . . . . . . Therefore the Maclaurin series of is n0 f n0 n! x n n0 kk 1 k n 1 n! x n fx 1 xk f nx kk 1 k n 11 xkn f n0 kk 1 k n 1 f x kk 1k 21 xk3 f 0 kk 1k 2 f x kk 11 xk2 f 0 kk 1 fx k1 xk1 f0 k fx 1 xk f0 1 k fx 1 xk cn f nan! fx cnx an sin x n0 1ns3 22n! x 3 2n n0 1n 22n 1!x 3 2n1 s3 x 3 x x sin x s3 2 1 2 1! x 3 s3 2 2! x 3 2 1 2 3! x 3 3 f 3 f 3 1! x 3 f 3 2! x 3 2 f 3 3! x 3 3 3 f 3 1 2 f x cos x f 3 s3 2 f x sin x f 3 1 2 fx cos x f 3 s3 2 fx sin x SECTION 11.10 TAYLOR AND MACLAURIN SERIES \|\|\|\| 741 0 x y π 3 y=sin x T£ FIGURE 3 N We have obtained two different series repre-sentations for , the Maclaurin series in Example 4 and the Taylor series in Example 7. It is best to use the Maclaurin series for values of near 0 and the Taylor series for near . Notice that the third Taylor polynomial in Fig-ure 3 is a good approximation to near but not as good near 0. Compare it with the third Maclaurin polynomial in Figure 2, where the opposite is true. T3 3 sin x T3 3 x x sin x This series is called the binomial series. If its th term is Thus, by the Ratio Test, the binomial series converges if and diverges if . M The traditional notation for the coefﬁcients in the binomial series is and these numbers are called the binomial coefﬁcients. The following theorem states that is equal to the sum of its Maclaurin series. It is possible to prove this by showing that the remainder term approaches 0, but that turns out to be quite difﬁcult. The proof outlined in Exercise 71 is much easier. THE BINOMIAL SERIES If is any real number and , then Although the binomial series always converges when , the question of whether or not it converges at the endpoints, , depends on the value of . It turns out that the series converges at 1 if and at both endpoints if . Notice that if is a positive integer and , then the expression for contains a factor , so for . This means that the series terminates and reduces to the ordinary Binomial Theorem when is a positive integer. (See Reference Page 1.) EXAMPLE 9 Find the Maclaurin series for the function and its radius of convergence. SOLUTION We write in a form where we can use the binomial series: 1 s4 x 1 41 x 4 1 2 1 x 4 1 2 1 x 4 12 fx fx 1 s4 x V k n k 0 ( k n) k k ( k n) n k k k 0 1 k 0 k 1 x 1 1 xk n0 k nx n 1 kx kk 1 2! x 2 kk 1k 2 3! x 3 x 1 k 17 Rnx 1 xk k n kk 1k 2 k n 1 n! x 1 x 1 k n n 1 x 1 k n 1 1 n x l x as n l an1 an kk 1 k n 1k nx n1 n 1! n! kk 1 k n 1x n an, then n 742 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES Using the binomial series with and with replaced by , we have We know from (17) that this series converges when , that is, , so the radius of convergence is . M We collect in the following table, for future reference, some important Maclaurin series that we have derived in this section and the preceding one. One reason that Taylor series are important is that they enable us to integrate functions that we couldn’t previously handle. In fact, in the introduction to this chapter we men-tioned that Newton often integrated functions by ﬁrst expressing them as power series and then integrating the series term by term. The function can’t be integrated by techniques discussed so far because its antiderivative is not an elementary function (see Section 7.5). In the following example we use Newton’s idea to integrate this function. fx ex2 R 1 1 xk n0 k nx n 1 kx kk 1 2! x 2 kk 1k 2 3! x 3 R 1 tan1x n0 1n x 2n1 2n 1 x x 3 3 x 5 5 x 7 7 R cos x n0 1n x 2n 2n! 1 x 2 2! x 4 4! x 6 6! R sin x n0 1n x 2n1 2n 1! x x 3 3! x 5 5! x 7 7! R e x n0 x n n! 1 x 1! x 2 2! x 3 3! R 1 1 1 x n0 x n 1 x x 2 x 3 R 4 x 4 x4 1 1 2 1 1 8 x 1 3 2!82 x 2 1 3 5 3!83 x 3 1 3 5 2n 1 n!8n x n ( 1 2)( 3 2)( 5 2) ( 1 2 n 1) n! x 4 n 1 2 1 1 2 x 4 ( 1 2)( 3 2) 2! x 4 2 ( 1 2)( 3 2)( 5 2) 3! x 4 3 1 s4 x 1 2 1 x 4 12 1 2 n0 1 2 n x 4 n x4 x k 1 2 SECTION 11.10 TAYLOR AND MACLAURIN SERIES \|\|\|\| 743 TABLE 1 Important Maclaurin Series and Their Radii of Convergence Module 11.10/11.11 enables you to see how successive Taylor polynomials approach the original function. TEC EXAMPLE 10 (a) Evaluate as an inﬁnite series. (b) Evaluate correct to within an error of . SOLUTION (a) First we ﬁnd the Maclaurin series for . Although it’s possible to use the direct method, let’s ﬁnd it simply by replacing with in the series for given in Table 1. Thus, for all values of x, Now we integrate term by term: This series converges for all because the original series for converges for all . (b) The Fundamental Theorem of Calculus gives The Alternating Series Estimation Theorem shows that the error involved in this approxi-mation is less than M Another use of Taylor series is illustrated in the next example. The limit could be found with l’Hospital’s Rule, but instead we use a series. EXAMPLE 11 Evaluate . SOLUTION Using the Maclaurin series for , we have because power series are continuous functions. M lim x l 0 1 2 x 3! x 2 4! x 3 5! 1 2 lim x l 0 x 2 2! x 3 3! x4 4! x 2 lim x l 0 e x 1 x x 2 lim x l 0 1 x 1! x 2 2! x 3 3! 1 x x 2 e x lim x l 0 e x 1 x x 2 1 11 5! 1 1320 0.001 1 1 3 1 10 1 42 1 216 0.7475 1 1 3 1 10 1 42 1 216 y 1 0 ex2 dx x x 3 3 1! x 5 5 2! x 7 7 3! x 9 9 4! 0 1 x ex2 x C x x 3 3 1! x 5 5 2! x 7 7 3! 1n x 2n1 2n 1n! y ex2 dx y 1 x 2 1! x 4 2! x 6 3! 1n x 2n n! dx ex2 n0 x 2n n! n0 1n x 2n n! 1 x 2 1! x 4 2! x 6 3! e x x 2 x fx ex2 0.001 x1 0 ex2 dx x ex2 dx V 744 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N We can take in the antiderivative in part (a). C 0 N Some computer algebra systems compute limits in this way. MULTIPLICATION AND DIVISION OF POWER SERIES If power series are added or subtracted, they behave like polynomials (Theorem 11.2.8 shows this). In fact, as the following example illustrates, they can also be multiplied and divided like polynomials. We ﬁnd only the ﬁrst few terms because the calculations for the later terms become tedious and the initial terms are the most important ones. EXAMPLE 12 Find the ﬁrst three nonzero terms in the Maclaurin series for (a) and (b) . SOLUTION (a) Using the Maclaurin series for and in Table 1, we have We multiply these expressions, collecting like terms just as for polynomials: Thus (b) Using the Maclaurin series in Table 1, we have We use a procedure like long division: Thus M Although we have not attempted to justify the formal manipulations used in Exam-ple 12, they are legitimate. There is a theorem which states that if both and converge for and the series are multiplied as if they were polyno-mials, then the resulting series also converges for and represents . For division we require ; the resulting series converges for sufﬁciently small . x b0 0 fxtx x R x R tx bnx n fx cnx n tan x x 1 3 x 3 2 15 x 5 tan x sin x cos x x x 3 3! x 5 5! 1 x 2 2! x 4 4! e x sin x x x 2 1 3 x 3 1 3 x 3 x 2 x 1 6 x 4 1 6 x 3 1 6 x 4 1 2 x 3 x 2 x 1 6 x 3 x 1 6 x 3 1 2 x 2 x 1 e x sin x 1 x 1! x 2 2! x 3 3! x x 3 3! sin x e x tan x e x sin x SECTION 11.10 TAYLOR AND MACLAURIN POLYNOMIALS \|\|\|\| 745 2 15 x 5 1 3 x 3 1 6 x 5 1 3 x 3 1 30 x 5 x 1 2 x 3 1 24 x 5 1 1 2 x 2 1 24 x 4 )x 1 6 x 3 1 120x 5 x 1 3 x 3 2 15 x 5 746 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES 17. , 18. , 19. , 20. , 21. Prove that the series obtained in Exercise 7 represents for all . 22. Prove that the series obtained in Exercise 18 represents for all . 23. Prove that the series obtained in Exercise 11 represents for all . 24. Prove that the series obtained in Exercise 12 represents for all . 25–28 Use the binomial series to expand the function as a power series. State the radius of convergence. 25. 26. 28. 29–38 Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function. 29. 30. 31. 32. 34. 36. 37. Hint: Use 38. ; 39–42 Find the Maclaurin series of (by any method) and its radius of convergence. Graph and its ﬁrst few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and ? 40. 41. 42. 43. Use the Maclaurin series for to calculate correct to ﬁve decimal places. e0.2 e x f x ln1 x 2 f x xex f x ex2 cos x f x cosx 2 39. f f f f x 1 6 x sin x x 3 if x 0 if x 0 sin2x 1 21 cos 2x.] [ f x sin2x f x x 2 s2 x f x x s4 x 2 35. f x x 2 tan1x 3 f x x cos( 1 2 x 2) 33. f x e x 2ex f x e x e 2x f x cosx 2 f x sin x 1 x2 3 1 2 x3 27. 1 1 x4 s1 x x cosh x x sinh x x sin x x sin x a 1 f x x 2 a 9 f x 1 sx a 2 f x sin x a f x cos x 1. If for all , write a formula for . 2. The graph of is shown. (a) Explain why the series is not the Taylor series of centered at 1. (b) Explain why the series is not the Taylor series of centered at 2. 3. If for ﬁnd the Maclaurin series for and its radius of convergence. 4. Find the Taylor series for centered at 4 if What is the radius of convergence of the Taylor series? 5–12 Find the Maclaurin series for using the deﬁnition of a Maclaurin series. [Assume that has a power series expan-sion. Do not show that .] Also ﬁnd the associated radius of convergence. 6. 7. 8. 9. 10. 11. 12. 13–20 Find the Taylor series for centered at the given value of . [Assume that has a power series expansion. Do not show that .] 13. , 14. , , 16. , a 3 f x 1 x a 3 f x e x 15. a 2 f x x x 3 a 1 f x x 4 3x 2 1 Rnx l 0 f a f x f x cosh x f x sinh x f x xe x f x e5x f x cos 3x f x sin x f x ln1 x f x 1 x2 5. Rnx l 0 f f x f n4 1n n! 3nn 1 f f n 0, 1, 2, . . . , f n0 n 1! f 2.8 0.5x 2 1.5x 22 0.1x 23 f 1.6 0.8x 1 0.4x 12 0.1x 13 y 0 x f 1 1 f b8 x f x n0 bnx 5n EXERCISES 11.10 SECTION 11.10 TAYLOR AND MACLAURIN SERIES \|\|\|\| 747 61. 62. 63–68 Find the sum of the series. 64. 65. 66. 67. 68. 69. Prove Taylor’s Inequality for , that is, prove that if for , then 70. (a) Show that the function deﬁned by is not equal to its Maclaurin series. ; (b) Graph the function in part (a) and comment on its behavior near the origin. 71. Use the following steps to prove (17). (a) Let . Differentiate this series to show that (b) Let and show that . (c) Deduce that . 72. In Exercise 53 in Section 10.2 it was shown that the length of the ellipse , , where , is where is the eccentricity of the ellipse. Expand the integrand as a binomial series and use the result of Exercise 46 in Section 7.1 to express as a series in powers of the eccentricity up to the term in . e 6 L e sa 2 b 2 a L 4a y 2 0 s1 e 2 sin2 d a b 0 y b cos x a sin tx 1 xk hx 0 hx 1 xktx 1 x 1 tx ktx 1 x x n ( k n) tx n0 f x e1 x2 0 if x 0 if x 0 R2x M 6 x a 3 for x a d x a d f x M n 2 1 ln 2 ln 22 2! ln 23 3! 3 9 2! 27 3! 81 4! n0 3n 5nn! n0 1n 2n1 42n12n 1! n0 1n 2n 62n2n! n0 1n x 4n n! 63. y e x ln1 x y x sin x 44. Use the Maclaurin series for to compute correct to ﬁve decimal places. (a) Use the binomial series to expand . (b) Use part (a) to ﬁnd the Maclaurin series for . 46. (a) Expand as a power series. (b) Use part (a) to estimate correct to three decimal places. 47–50 Evaluate the indeﬁnite integral as an inﬁnite series. 47. 48. 49. 50. 51–54 Use series to approximate the deﬁnite integral to within the indicated accuracy. 51. (three decimal places) 52. (ﬁve decimal places) 53. 54. 55–57 Use series to evaluate the limit. 55. 56. 58. Use the series in Example 12(b) to evaluate We found this limit in Example 4 in Section 4.4 using l’Hospi-tal’s Rule three times. Which method do you prefer? 59–62 Use multiplication or division of power series to ﬁnd the ﬁrst three nonzero terms in the Maclaurin series for the function. 60. y sec x y ex2 cos x 59. lim x l 0 tan x x x 3 lim x l 0 sin x x 1 6 x 3 x 5 57. lim x l 0 1 cos x 1 x e x lim x l 0 x tan1x x 3 ( error 0.001) y 0.5 0 x 2ex2 dx ( error 5 106) y 0.4 0 s1 x 4 dx y 0.2 0 tan 1x 3 sinx 3 dx y 1 0 x cosx 3 dx y arctanx 2 dx y cos x 1 x dx y e x 1 x dx y x cosx 3 dx 1 s 4 1.1 1 s 4 1 x sin1x 1 s1 x 2 45. sin 3 sin x The Binomial Theorem, which gives the expansion of , was known to Chinese mathe-maticians many centuries before the time of Newton for the case where the exponent k is a positive integer. In 1665, when he was 22, Newton was the ﬁrst to discover the inﬁnite series expansion of when k is a fractional exponent (positive or negative). He didn’t publish his discovery, but he stated it and gave examples of how to use it in a letter (now called the epistola prior) dated June 13, 1676, that he sent to Henry Oldenburg, secretary of the Royal Society of London, to transmit to Leibniz. When Leibniz replied, he asked how Newton had discovered the binomial series. Newton wrote a second letter, the epistola posterior of Octo-ber 24, 1676, in which he explained in great detail how he arrived at his discovery by a very indirect route. He was investigating the areas under the curves from 0 to x for , 1, 2, 3, 4, . . . . These are easy to calculate if n is even. By observing patterns and inter-polating, Newton was able to guess the answers for odd values of n. Then he realized he could get the same answers by expressing as an inﬁnite series. Write a report on Newton’s discovery of the binomial series. Start by giving the statement of the binomial series in Newton’s notation (see the epistola prior on page 285 of or page 402 of ). Explain why Newton’s version is equivalent to Theorem 17 on page 742. Then read Newton’s epistola posterior (page 287 in or page 404 in ) and explain the patterns that Newton discovered in the areas under the curves . Show how he was able to guess the areas under the remaining curves and how he veriﬁed his answers. Finally, explain how these discoveries led to the binomial series. The books by Edwards and Katz contain commentaries on Newton’s letters. 1. C. H. Edwards, The Historical Development of the Calculus (New York: Springer-Verlag, 1979), pp. 178–187. 2. John Fauvel and Jeremy Gray, eds., The History of Mathematics: A Reader (London: MacMillan Press, 1987). 3. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993), pp. 463–466. 4. D. J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, NJ: Princeton University Press, 1969). y 1 x 2n 2 1 x 2n 2 n 0 y 1 x 2n 2 a bk a bk HOW NEWTON DISCOVERED THE BINOMIAL SERIES W R I T I N G P R O J E C T 748 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES This project deals with the function 1. Use your computer algebra system to evaluate for and . Does it appear that has a limit as ? 2. Use the CAS to graph near . Does it appear that has a limit as ? 3. Try to evaluate with l’Hospital’s Rule, using the CAS to ﬁnd derivatives of the numerator and denominator. What do you discover? How many applications of l’Hospital’s Rule are required? 4. Evaluate by using the CAS to ﬁnd sufﬁciently many terms in the Taylor series of the numerator and denominator. (Use the command taylor in Maple or Series in Mathematica.) 5. Use the limit command on your CAS to ﬁnd directly. (Most computer algebra systems use the method of Problem 4 to compute limits.) 6. In view of the answers to Problems 4 and 5, how do you explain the results of Problems 1 and 2? limx l 0 fx limx l 0 fx limx l 0 fx x l 0 f x 0 f x l 0 f 0.0001 x 1, 0.1, 0.01, 0.001, fx f x sintan x tansin x arcsinarctan x arctanarcsin x AN ELUSIVE LIMIT CAS L A B O R AT O R Y P R O J E C T APPLICATIONS OF TAYLOR POLYNOMIALS In this section we explore two types of applications of Taylor polynomials. First we look at how they are used to approximate functions––computer scientists like them because polynomials are the simplest of functions. Then we investigate how physicists and engi-neers use them in such ﬁelds as relativity, optics, blackbody radiation, electric dipoles, the velocity of water waves, and building highways across a desert. APPROXIMATING FUNCTIONS BY POLYNOMIALS Suppose that is equal to the sum of its Taylor series at a: In Section 11.10 we introduced the notation for the th partial sum of this series and called it the th-degree Taylor polynomial of at . Thus Since is the sum of its Taylor series, we know that as and so can be used as an approximation to : . Notice that the ﬁrst-degree Taylor polynomial is the same as the linearization of f at a that we discussed in Section 3.10. Notice also that and its derivative have the same values at a that and have. In general, it can be shown that the derivatives of at agree with those of up to and including derivatives of order (see Exercise 38). To illustrate these ideas let’s take another look at the graphs of and its ﬁrst few Taylor polynomials, as shown in Figure 1. The graph of is the tangent line to at ; this tangent line is the best linear approximation to near . The graph of is the parabola , and the graph of is the cubic curve , which is a closer ﬁt to the exponential curve than . The next Taylor polynomial would be an even better approximation, and so on. The values in the table give a numerical demonstration of the convergence of the Taylor polynomials to the function . We see that when the convergence is very rapid, but when it is somewhat slower. In fact, the farther is from 0, the more slowly converges to . When using a Taylor polynomial to approximate a function , we have to ask the questions: How good an approximation is it? How large should we take to be in order to achieve a desired accuracy? To answer these questions we need to look at the absolute value of the remainder: Rnx fx Tnx n f Tn e x Tnx x x 3 x 0.2 y e x Tnx T4 T2 y e x y 1 x x 2 2 x 3 6 T3 y 1 x x 2 2 T2 0, 1 e x 0, 1 y e x T1 y e x n f a Tn f f T1 T1x fa fax a fx Tnx f Tn n l Tnx l fx f fa fa 1! x a f a 2! x a2 f na n! x an Tnx n i0 f ia i! x ai a f n n Tnx fx n0 f na n! x an fx 11.11 SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS \|\|\|\| 749 1.220000 8.500000 1.221400 16.375000 1.221403 19.412500 1.221403 20.009152 1.221403 20.079665 1.221403 20.085537 e x T10x T8x T6x T4x T2x x 3.0 x 0.2 0 x y y=´ y=T£(x) (0, 1) y=T™(x) y=T¡(x) FIGURE 1 There are three possible methods for estimating the size of the error: 1. If a graphing device is available, we can use it to graph and thereby esti-mate the error. 2. If the series happens to be an alternating series, we can use the Alternating Series Estimation Theorem. 3. In all cases we can use Taylor’s Inequality (Theorem 11.10.9), which says that if , then EXAMPLE 1 (a) Approximate the function by a Taylor polynomial of degree 2 at . (b) How accurate is this approximation when ? SOLUTION (a) Thus the second-degree Taylor polynomial is The desired approximation is (b) The Taylor series is not alternating when , so we can’t use the Alternating Series Estimation Theorem in this example. But we can use Taylor’s Inequality with and : where . Because , we have and so Therefore we can take . Also , so and . Then Taylor’s Inequality gives Thus, if , the approximation in part (a) is accurate to within . M 0.0004 7 x 9 R2x 0.0021 3! 13 0.0021 6 0.0004 x 8 1 1 x 8 1 7 x 9 M 0.0021 f x 10 27 1 x 8 3 10 27 1 78 3 0.0021 x 8 3 78 3 x 7 f x M R2x M 3! x 8 3 a 8 n 2 x 8 s 3 x T2x 2 1 12x 8 1 288x 82 2 1 12x 8 1 288x 82 T2x f8 f8 1! x 8 f 8 2! x 82 f x 10 27 x8 3 f x 2 9 x5 3 f 8 1 144 fx 1 3 x2 3 f8 1 12 fx s 3 x x 1 3 f8 2 7 x 9 a 8 fx s 3 x V Rnx M n 1! x a n1 f n1x M Rnx 750 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES Let’s use a graphing device to check the calculation in Example 1. Figure 2 shows that the graphs of and are very close to each other when is near 8. Fig-ure 3 shows the graph of computed from the expression We see from the graph that when . Thus the error estimate from graphical methods is slightly better than the error estimate from Taylor’s Inequality in this case. EXAMPLE 2 (a) What is the maximum error possible in using the approximation when ? Use this approximation to ﬁnd correct to six decimal places. (b) For what values of is this approximation accurate to within ? SOLUTION (a) Notice that the Maclaurin series is alternating for all nonzero values of , and the successive terms decrease in size because , so we can use the Alternating Series Estimation Theorem. The error in approximating by the ﬁrst three terms of its Maclaurin series is at most If , then , so the error is smaller than To ﬁnd we ﬁrst convert to radian measure. Thus, correct to six decimal places, . (b) The error will be smaller than if x 7 5040 0.00005 0.00005 sin 12 0.207912 15 15 3 1 3! 15 5 1 5! 0.20791169 sin 12 sin 12 180 sin 15 sin 12 0.37 5040 4.3 108 x 0.3 0.3 x 0.3 x 7 7! x 7 5040 sin x x 1 x sin x x x 3 3! x 5 5! x 7 7! 0.00005 x sin 12 0.3 x 0.3 sin x x x 3 3! x 5 5! V 7 x 9 R2x 0.0003 R2x s 3 x T2x R2x x y T2x y s 3 x SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS \|\|\|\| 751 2.5 0 15 T™ y=# œ„ x FIGURE 2 0.0003 7 9 y=\|R™(x)\| 0 FIGURE 3 Solving this inequality for , we get So the given approximation is accurate to within when . M What if we use Taylor’s Inequality to solve Example 2? Since , we have and so So we get the same estimates as with the Alternating Series Estimation Theorem. What about graphical methods? Figure 4 shows the graph of and we see from it that when . This is the same estimate that we obtained in Example 2. For part (b) we want , so we graph both and in Figure 5. By placing the cursor on the right intersection point we ﬁnd that the inequality is satisﬁed when . Again this is the same esti-mate that we obtained in the solution to Example 2. If we had been asked to approximate instead of in Example 2, it would have been wise to use the Taylor polynomials at (instead of ) because they are better approximations to for values of close to . Notice that is close to (or radians) and the derivatives of are easy to compute at . Figure 6 shows the graphs of the Maclaurin polynomial approximations to the sine curve. You can see that as increases, is a good approximation to on a larger and larger interval. One use of the type of calculation done in Examples 1 and 2 occurs in calculators and computers. For instance, when you press the or key on your calculator, or when a computer programmer uses a subroutine for a trigonometric or exponential or Bessel func-tion, in many machines a polynomial approximation is calculated. The polynomial is often a Taylor polynomial that has been modiﬁed so that the error is spread more evenly through-out an interval. APPLICATIONS TO PHYSICS Taylor polynomials are also used frequently in physics. In order to gain insight into an equation, a physicist often simpliﬁes a function by considering only the ﬁrst two or three terms in its Taylor series. In other words, the physicist uses a Taylor polynomial as an e x sin FIGURE 6 0 x y T¶ T∞ T£ y=sin x T¡ sin x Tnx n T5x x x 3 3! x 5 5! T7x x x 3 3! x 5 5! x 7 7! T1x x T3x x x 3 3! 3 sin x 3 60 72 3 x sin x a 0 a 3 sin 12 sin 72 x 0.82 y 0.00005 y R6x R6x 0.00005 x 0.3 R6x 4.3 108 R6x sin x (x 1 6 x 3 1 120 x 5) R6x 1 7! x 7 f 7x 1 f 7x cos x x 0.82 0.00005 x 0.2521 7 0.821 or x 7 0.252 x 752 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES Module 11.10/11.11 graphically shows the remainders in Taylor polynomial approximations. TEC 4.3 10– _0.3 0.3 0 y=\|Rß(x)\| FIGURE 4 0.00006 _1 1 y=\|Rß(x)\| 0 y=0.00005 FIGURE 5 approximation to the function. Taylor’s Inequality can then be used to gauge the accuracy of the approximation. The following example shows one way in which this idea is used in special relativity. EXAMPLE 3 In Einstein’s theory of special relativity the mass of an object moving with velocity is where is the mass of the object when at rest and is the speed of light. The kinetic energy of the object is the difference between its total energy and its energy at rest: (a) Show that when is very small compared with , this expression for agrees with classical Newtonian physics: . (b) Use Taylor’s Inequality to estimate the difference in these expressions for when m s. SOLUTION (a) Using the expressions given for and , we get With , the Maclaurin series for is most easily computed as a binomial series with . (Notice that because .) Therefore we have and If is much smaller than , then all terms after the ﬁrst are very small when compared with the ﬁrst term. If we omit them, we get (b) If , , and M is a number such that , then we can use Taylor’s Inequality to write We have and we are given that m s, so f x 3m0c2 41 v 2 c25 2 3m0c2 41 1002 c25 2 M v 100 f x 3 4m0c21 x5 2 R1x M 2! x 2 f x M fx m0c21 x1 2 1 x v 2 c2 K m0c2 1 2 v2 c2 1 2m0v2 c v m0c2 1 2 v2 c2 3 8 v4 c 4 5 16 v6 c 6 K m0c21 1 2 v2 c2 3 8 v4 c 4 5 16 v6 c 6 1 1 1 2 x 3 8 x 2 5 16 x 3 1 x1 2 1 1 2 x ( 1 2)( 3 2) 2! x 2 ( 1 2)( 3 2)( 5 2) 3! x 3 v c x 1 k 1 2 1 x1 2 x v2 c2 m0c21 v2 c2 1 2 1 K mc2 m0c2 m0c2 s1 v 2 c2 m0c2 m K v 100 K K 1 2m0v2 K c v K mc2 m0c2 c m0 m m0 s1 v 2 c 2 v V SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS \|\|\|\| 753 N The upper curve in Figure 7 is the graph of the expression for the kinetic energy of an object with velocity in special relativity. The lower curve shows the function used for in classical Newtonian physics. When is much smaller than the speed of light, the curves are practically identical. v K v K FIGURE 7 √ K 0 K=mc@-m¸c@ K= m¸√@ 1 2 c Thus, with , So when ms, the magnitude of the error in using the Newtonian expression for kinetic energy is at most . M Another application to physics occurs in optics. Figure 8 is adapted from Optics, 4th ed., by Eugene Hecht (San Francisco: Addison-Wesley, 2002), page 153. It depicts a wave from the point source S meeting a spherical interface of radius R centered at C. The ray SA is refracted toward P. Using Fermat’s principle that light travels so as to minimize the time taken, Hecht derives the equation where and are indexes of refraction and , , , and are the distances indicated in Figure 8. By the Law of Cosines, applied to triangles ACS and ACP, we have Because Equation 1 is cumbersome to work with, Gauss, in 1841, simpliﬁed it by using the linear approximation for small values of . (This amounts to using the Taylor polynomial of degree 1.) Then Equation 1 becomes the following simpler equation [as you are asked to show in Exercise 34(a)]: The resulting optical theory is known as Gaussian optics, or ﬁrst-order optics, and has become the basic theoretical tool used to design lenses. A more accurate theory is obtained by approximating by its Taylor polynomial of degree 3 (which is the same as the Taylor polynomial of degree 2). This takes into account rays for which is not so small, that is, rays that strike the surface at greater distances h above the axis. In Exercise 34(b) you are asked to use this approximation to derive the cos n1 so n2 si n2 n1 R 3 cos 1 i sR2 si R2 2Rsi R cos o sR2 so R2 2Rso R cos 2 si so i o n2 n1 n1 o n2 i 1 R n2si i n1so o 1 A V h C P R S ¨t ¨r ¨i ˙ Lo so si Li n¡ n™ Courtesy of Eugene Hecht FIGURE 8 Refraction at a spherical interface 4.2 1010m0 v 100 R1x 1 2 3m0c2 41 1002c252 1004 c 4 4.17 1010m0 c 3 108 ms 754 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES N Here we use the identity cos cos more accurate equation The resulting optical theory is known as third-order optics. Other applications of Taylor polynomials to physics and engineering are explored in Exercises 32, 33, 35, 36, and 37 and in the Applied Project on page 757. n1 so n2 si n2 n1 R h 2 n1 2so 1 so 1 R 2 n2 2si 1 R 1 si 2 4 SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS \|\|\|\| 755 ; (c) Check your result in part (b) by graphing . 13. , , , 14. , , , 15. , , , 16. , , , 17. , , , , , , , , , 20. , , , 21. , , , 22. , , , 23. Use the information from Exercise 5 to estimate cor-rect to ﬁve decimal places. 24. Use the information from Exercise 16 to estimate correct to ﬁve decimal places. Use Taylor’s Inequality to determine the number of terms of the Maclaurin series for that should be used to estimate to within . 26. How many terms of the Maclaurin series for do you need to use to estimate to within ? ; 27–29 Use the Alternating Series Estimation Theorem or Taylor’s Inequality to estimate the range of values of for which the given approximation is accurate to within the stated error. Check your answer graphically. 27. 28. 29. ( error 0.05) arctan x x x 3 3 x 5 5 ( error 0.005) cos x 1 x 2 2 x 4 24 ( error 0.01) sin x x x 3 6 x 0.001 ln 1.4 ln1 x 0.00001 e 0.1 e x 25. sin 38 cos 80 1 x 1 n 5 a 0 f x sinh 2x 1 x 1 n 4 a 0 f x x sin x 0.5 x 1.5 n 3 a 1 f x x ln x 0 x 0.1 n 3 a 0 f x ex2 19. 0.5 x 1.5 n 3 a 1 f x ln1 2x 18. 0.2 x 0.2 n 2 a 0 f x sec x 0 x 3 n 4 a 6 f x sin x 0.8 x 1.2 n 3 a 1 f x x 23 0.9 x 1.1 n 2 a 1 f x x2 4 x 4.2 n 2 a 4 f x sx Rnx ; 1. (a) Find the Taylor polynomials up to degree 6 for centered at . Graph and these polynomials on a common screen. (b) Evaluate and these polynomials at , , and . (c) Comment on how the Taylor polynomials converge to . ; 2. (a) Find the Taylor polynomials up to degree 3 for centered at . Graph and these polynomials on a common screen. (b) Evaluate and these polynomials at and 1.3. (c) Comment on how the Taylor polynomials converge to . ; 3–10 Find the Taylor polynomial for the function at the number . Graph and on the same screen. 3. , 4. , , 6. , 7. , 8. , , 10. , 11–12 Use a computer algebra system to ﬁnd the Taylor poly-nomials centered at for . Then graph these polynomials and on the same screen. 11. , 12. , 13–22 (a) Approximate by a Taylor polynomial with degree at the number . (b) Use Taylor’s Inequality to estimate the accuracy of the approximation when x lies in the given interval. f x Tnx a n f a 0 f x s 3 1 x 2 a 4 f x cot x f n 2, 3, 4, 5 a Tn CAS a 1 f x tan 1x a 0 f x xe 2x 9. a 1 f x ln x x a 0 f x arcsin x a 0 f x ex sin x a 2 f x cos x 5. a 0 f x x ex a 2 f x 1x T3 f a f Tnx f x x 0.9 f f a 1 f x 1x f x 2 x 4 f f a 0 f x cos x EXERCISES 11.11 756 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES Equation 4 for third-order optics. [Hint: Use the ﬁrst two terms in the binomial series for and . Also, use .] 35. If a water wave with length moves with velocity across a body of water with depth , as in the ﬁgure, then (a) If the water is deep, show that . (b) If the water is shallow, use the Maclaurin series for to show that . (Thus in shallow water the veloc-ity of a wave tends to be independent of the length of the wave.) (c) Use the Alternating Series Estimation Theorem to show that if , then the estimate is accurate to within . 36. The period of a pendulum with length that makes a maxi-mum angle with the vertical is where and is the acceleration due to gravity. (In Exercise 40 in Section 7.7 we approximated this integral using Simpson’s Rule.) (a) Expand the integrand as a binomial series and use the result of Exercise 46 in Section 7.1 to show that If is not too large, the approximation , obtained by using only the ﬁrst term in the series, is often used. A better approximation is obtained by using two terms: (b) Notice that all the terms in the series after the ﬁrst one have coefﬁcients that are at most . Use this fact to com-pare this series with a geometric series and show that (c) Use the inequalities in part (b) to estimate the period of a pendulum with meter and . How does it compare with the estimate ? What if ? 0 42 T 2sLt 0 10 L 1 2 L t (1 1 4k 2) T 2 L t 4 3k 2 4 4k 2 1 4 T 2 L t (1 1 4k 2) T 2sLt 0 T 2 L t 1 12 22 k 2 1232 2242 k 4 123252 224262 k 6 t k sin( 1 2 0) T 4 L t y 2 0 dx s1 k 2 sin2x 0 L L d 0.014tL v 2 td L 10d v std tanh v stL2 v 2 tL 2 tanh 2d L d v L sin i 1 o 1 30. Suppose you know that and the Taylor series of centered at 4 converges to for all in the interval of convergence. Show that the ﬁfth-degree Taylor polynomial approximates with error less than 0.0002. A car is moving with speed 20 ms and acceleration 2 ms at a given instant. Using a second-degree Taylor polynomial, estimate how far the car moves in the next second. Would it be reasonable to use this polynomial to estimate the distance traveled during the next minute? 32. The resistivity of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters ( -m). The resistivity of a given metal depends on the temperature according to the equation where is the temperature in . There are tables that list the values of (called the temperature coefﬁcient) and (the resistivity at C) for various metals. Except at very low temperatures, the resistivity varies almost linearly with tem-perature and so it is common to approximate the expression for by its ﬁrst- or second-degree Taylor polynomial at . (a) Find expressions for these linear and quadratic approximations. ; (b) For copper, the tables give C and -m. Graph the resistivity of copper and the linear and quadratic approximations for C C. ; (c) For what values of does the linear approximation agree with the exponential expression to within one percent? An electric dipole consists of two electric charges of equal magnitude and opposite sign. If the charges are and and are located at a distance from each other, then the electric ﬁeld at the point in the ﬁgure is By expanding this expression for as a series in powers of , show that is approximately proportional to when is far away from the dipole. 34. (a) Derive Equation 3 for Gaussian optics from Equation 1 by approximating in Equation 2 by its ﬁrst-degree Taylor polynomial. (b) Show that if is replaced by its third-degree Taylor polynomial in Equation 2, then Equation 1 becomes cos cos P D d q _q P 1D 3 E dD E E q D2 q D d2 P E d q q 33. t t 1000 250 20 1.7 108 0.0039 t 20 t 20 20 C t t 20e t20 2 31. f 5 x f x f f n4 1nn! 3nn 1 Any object emits radiation when heated. A blackbody is a system that absorbs all the radiation that falls on it. For instance, a matte black surface or a large cavity with a small hole in its wall (like a blastfurnace) is a blackbody and emits blackbody radiation. Even the radiation from the sun is close to being blackbody radiation. Proposed in the late 19th century, the Rayleigh-Jeans Law expresses the energy density of blackbody radiation of wavelength as where is measured in meters, is the temperature in kelvins (K), and is Boltzmann’s con-stant. The Rayleigh-Jeans Law agrees with experimental measurements for long wavelengths but disagrees drastically for short wavelengths. [The law predicts that as but experiments have shown that .] This fact is known as the ultraviolet catastrophe. In 1900 Max Planck found a better model (known now as Planck’s Law) for blackbody radiation: where is measured in meters, is the temperature (in kelvins), and 1. Use l’Hospital’s Rule to show that for Planck’s Law. So this law models blackbody radiation better than the Rayleigh-Jeans Law for short wavelengths. lim l 0 f 0 and lim l f 0 k Boltzmann’s constant 1.3807 1023 JK c speed of light 2.997925 108 ms h Planck’s constant 6.6262 1034 Js T f 8hc5 e hckT 1 f l 0 l 0 f l k T f 8kT 4 RADIATION FROM THE STARS A P P L I E D P R O J E C T 38. Show that and have the same derivatives at up to order . 39. In Section 4.8 we considered Newton’s method for approxi-mating a root of the equation , and from an initial approximation we obtained successive approxi-mations , , . . . , where Use Taylor’s Inequality with , , and to show that if exists on an interval containing , , and , and , for all , then [This means that if is accurate to decimal places, then is accurate to about decimal places. More precisely, if the error at stage is at most , then the error at stage is at most .] M2K102m n 1 10m n 2d xn1 d xn xn1 r M 2K xn r 2 x I f x K f x M xn1 xn r I f x x r a xn n 1 xn1 xn f xn f xn x3 x2 x1 f x 0 r n a f Tn 37. If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the earth. (a) If is the radius of the earth and is the length of the highway, show that the correction is (b) Use a Taylor polynomial to show that (c) Compare the corrections given by the formulas in parts (a) and (b) for a highway that is 100 km long. (Take the radius of the earth to be 6370 km.) R L C R C L 2 2R 5L 4 24R 3 C R secLR R L R APPLIED PROJECT RADIATION FROM THE STARS \|\|\|\| 757 © Luke Dodd, Photo Researchers, Inc. REVIEW CONCEPT CHECK 11 (b) If a series is convergent by the Comparison Test, how do you estimate its sum? (c) If a series is convergent by the Alternating Series Test, how do you estimate its sum? 8. (a) Write the general form of a power series. (b) What is the radius of convergence of a power series? (c) What is the interval of convergence of a power series? 9. Suppose is the sum of a power series with radius of con-vergence . (a) How do you differentiate ? What is the radius of conver-gence of the series for ? (b) How do you integrate ? What is the radius of convergence of the series for ? 10. (a) Write an expression for the -degree Taylor polynomial of centered at . (b) Write an expression for the Taylor series of centered at . (c) Write an expression for the Maclaurin series of . (d) How do you show that is equal to the sum of its Taylor series? (e) State Taylor’s Inequality. 11. Write the Maclaurin series and the interval of convergence for each of the following functions. (a) (b) (c) (d) (e) 12. Write the binomial series expansion of . What is the radius of convergence of this series? 1 xk tan1x cos x sin x e x 11 x f x f a f a f nth x f x dx f f f R f x 1. (a) What is a convergent sequence? (b) What is a convergent series? (c) What does mean? (d) What does mean? 2. (a) What is a bounded sequence? (b) What is a monotonic sequence? (c) What can you say about a bounded monotonic sequence? 3. (a) What is a geometric series? Under what circumstances is it convergent? What is its sum? (b) What is a -series? Under what circumstances is it convergent? 4. Suppose and is the partial sum of the series. What is ? What is ? 5. State the following. (a) The Test for Divergence (b) The Integral Test (c) The Comparison Test (d) The Limit Comparison Test (e) The Alternating Series Test (f) The Ratio Test (g) The Root Test 6. (a) What is an absolutely convergent series? (b) What can you say about such a series? (c) What is a conditionally convergent series? 7. (a) If a series is convergent by the Integral Test, how do you estimate its sum? limn l sn limn l an nth sn an 3 p n1 an 3 limn l an 3 758 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES 2. Use a Taylor polynomial to show that, for large wavelengths, Planck’s Law gives approxi-mately the same values as the Rayleigh-Jeans Law. ; 3. Graph as given by both laws on the same screen and comment on the similarities and differences. Use K (the temperature of the sun). (You may want to change from meters to the more convenient unit of micrometers: m m.) 4. Use your graph in Problem 3 to estimate the value of for which is a maximum under Planck’s Law. ; 5. Investigate how the graph of changes as varies. (Use Planck’s Law.) In particular, graph for the stars Betelgeuse ( ), Procyon ( ), and Sirius ( ) as well as the sun. How does the total radiation emitted (the area under the curve) vary with ? Use the graph to comment on why Sirius is known as a blue star and Betelgeuse as a red star. T T 9200 K T 6400 K T 3400 K f T f f 106 1 T 5700 f Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. If , then is convergent. 2. The series is convergent. 3. If , then . 4. If is convergent, then is convergent. 5. If is convergent, then is convergent. 6. If diverges when , then it diverges when . 7. The Ratio Test can be used to determine whether converges. 8. The Ratio Test can be used to determine whether converges. 9. If and diverges, then diverges. 10. n0 1n n! 1 e an bn 0 an bn 1n! 1n 3 x 10 x 6 cnx n cn6n cn6n cn2n cn6n limn l a2n1 L limn l an L n1 n sin 1 an limn l an 0 11. If , then . 12. If is divergent, then is divergent. 13. If converges for all , then . 14. If and are divergent, then is divergent. 15. If and are divergent, then is divergent. 16. If is decreasing and for all , then is convergent. 17. If and converges, then converges. 18. If and , then . 19. 20. If and , then . n1 anbn AB n1 bn B n1 an A 0.99999 . . . 1 limn l an 0 limn l an1an 1 an 0 1nan an an 0 an n an 0 an anbn bn an an bn bn an f 0 2 x f x 2x x 2 1 3 x 3 an an limn l n 0 1 1 TRUE-FALSE QUIZ 1–8 Determine whether the sequence is convergent or divergent. If it is convergent, ﬁnd its limit. 1. 2. 3. 4. 5. 6. 7. 8. 9. A sequence is deﬁned recursively by the equations , . Show that is increasing and for all . Deduce that is convergent and ﬁnd its limit. ; 10. Show that and use a graph to ﬁnd the smallest value of that corresponds to in the pre-cise deﬁnition of a limit. 11–22 Determine whether the series is convergent or divergent. 11. 12. 13. 14. 15. 16. n1 ln n 3n 1 n2 1 nsln n n1 1n sn 1 n1 n3 5n n1 n2 1 n3 1 n1 n n3 1 0.1 N lim n l n 4e n 0 an n an 2 an an1 1 3an 4 a1 1 10nn! 1 3n4n an ln n sn an n sin n n2 1 an cosn2 an n3 1 n2 an 9n1 10n an 2 n3 1 2n3 17. 18. 19. 20. 21. 22. 23–26 Determine whether the series is conditionally conver-gent, absolutely convergent, or divergent. 23. 24. 25. 26. 27–31 Find the sum of the series. 27. 28. 29. 30. n0 1n n 32n2n! n1 tan1n 1 tan1n n1 1 nn 3 n1 3n1 23n n2 1nsn ln n n1 1nn 13n 22n1 n1 1n1n 3 n1 1n1n 13 n1 sn 1 sn 1 n n1 1n1 sn n 1 n1 52n n 29n n1 1 3 5 2n 1 5nn! n1 n2n 1 2n2n n1 cos 3n 1 1.2n EXERCISES CHAPTER 11 REVIEW \|\|\|\| 759 760 \|\|\|\| CHAPTER 11 INFINITE SEQUENCES AND SERIES 49. 50. 51. 52. 53. 54. 55. Evaluate as an inﬁnite series. 56. Use series to approximate correct to two deci-mal places. 57–58 (a) Approximate by a Taylor polynomial with degree at the number . ; (b) Graph and on a common screen. (c) Use Taylor’s Inequality to estimate the accuracy of the approxi-mation when lies in the given interval. ; (d) Check your result in part (c) by graphing . 57. , , , 58. , , , 59. Use series to evaluate the following limit. 60. The force due to gravity on an object with mass at a height above the surface of the earth is where is the radius of the earth and is the acceleration due to gravity. (a) Express as a series in powers of . ; (b) Observe that if we approximate by the ﬁrst term in the series, we get the expression that is usually used when is much smaller than . Use the Alternating Series Estimation Theorem to estimate the range of values of for which the approximation is accurate to within one percent. (Use km.) 61. Suppose that for all . (a) If is an odd function, show that (b) If is an even function, show that 62. If , show that . f 2n0 2n! n! f x ex2 c1 c3 c5 0 f c0 c2 c4 0 f x f x n0 cnx n R 6400 F mt h R h F mt F hR F t R F mtR2 R h2 h m lim x l 0 sin x x x 3 0 x 6 n 2 a 0 f x sec x 0.9 x 1.1 n 3 a 1 f x sx Rnx x f x Tnx Tn f a n f x1 0 s1 x 4 dx y e x x dx f x 1 3x5 f x 1s 4 16 x f x 10 x f x sinx 4 f x xe 2x f x ln1 x 31. 32. Express the repeating decimal as a fraction. 33. Show that for all . 34. For what values of does the series converge? 35. Find the sum of the series correct to four decimal places. 36. (a) Find the partial sum of the series and estimate the error in using it as an approximation to the sum of the series. (b) Find the sum of this series correct to ﬁve decimal places. 37. Use the sum of the ﬁrst eight terms to approximate the sum of the series . Estimate the error involved in this approximation. 38. (a) Show that the series is convergent. (b) Deduce that . 39. Prove that if the series is absolutely convergent, then the series is also absolutely convergent. 40–43 Find the radius of convergence and interval of convergence of the series. 40. 41. 42. 43. 44. Find the radius of convergence of the series 45. Find the Taylor series of at . 46. Find the Taylor series of at . 47–54 Find the Maclaurin series for and its radius of conver-gence. You may use either the direct method (deﬁnition of a Maclaurin series) or known series such as geometric series, binomial series, or the Maclaurin series for , , and . 47. 48. f x tan1x 2 f x x 2 1 x tan1x sin x e x f a 3 f x cos x a 6 f x sin x n1 2n! n!2 x n n0 2nx 3n sn 3 n1 2nx 2n n 2! n1 x 2n n 4n n1 1n x n n25n n1 n 1 n an n1 an lim n l n n 2n! 0 n1 n n 2n! n1 2 5n1 n1 1n6 s5 n1 1n1 n 5 n1 ln xn x x cosh x 1 1 2x 2 4.17326326326 . . . 1 e e 2 2! e 3 3! e 4 4! 761 1. If , ﬁnd . 2. A function is deﬁned by Where is continuous? 3. (a) Show that . (b) Find the sum of the series 4. Let be a sequence of points determined as in the ﬁgure. Thus , , and angle is a right angle. Find . 5. To construct the snowﬂake curve, start with an equilateral triangle with sides of length . Step 1 in the construction is to divide each side into three equal parts, construct an equilateral triangle on the middle part, and then delete the middle part (see the ﬁgure). Step 2 is to repeat step 1 for each side of the resulting polygon. This process is repeated at each succeeding step. The snowﬂake curve is the curve that results from repeating this process indeﬁnitely. (a) Let , , and represent the number of sides, the length of a side, and the total length of the th approximating curve (the curve obtained after step of the construction), respec-tively. Find formulas for , , and . (b) Show that as . (c) Sum an inﬁnite series to ﬁnd the area enclosed by the snowﬂake curve. Note: Parts (b) and (c) show that the snowﬂake curve is inﬁnitely long but encloses only a ﬁnite area. 6. Find the sum of the series where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s. 7. (a) Show that for , if the left side lies between and . (b) Show that (c) Deduce the following formula of John Machin (1680–1751): (d) Use the Maclaurin series for to show that (e) Show that 0.004184075 arctan 1 239 0.004184077 0.197395560 arctan 1 5 0.197395562 arctan 4 arctan 1 5 arctan 1 239 4 arctan 120 119 arctan 1 239 4 2 2 arctan x arctan y arctan x y 1 xy xy 1 1 1 2 1 3 1 4 1 6 1 8 1 9 1 12 n l pn l pn ln sn n n pn ln sn 1 limn l Pn APn1 APnPn1 PnPn1 2n1 AP1 1 Pn n1 1 2n tan x 2n tan 1 2 x cot 1 2 x 2 cot x f f x lim n l x 2n 1 x 2n 1 f f 150 f x sinx 3 P R O B L E M S P L U S P∞ 8 P¢ P£ P™ P¡ A 4 2 1 1 FIGURE FOR PROBLEM 4 FIGURE FOR PROBLEM 5 2 1 3 762 (f) Deduce that, correct to seven decimal places, Machin used this method in 1706 to ﬁnd correct to 100 decimal places. Recently, with the aid of computers, the value of has been computed to increasingly greater accuracy. Yasumada Kanada of the University of Tokyo recently computed the value of to a trillion decimal places! 8. (a) Prove a formula similar to the one in Problem 7(a) but involving instead of . (b) Find the sum of the series 9. Find the interval of convergence of and ﬁnd its sum. 10. If , show that If you don’t see how to prove this, try the problem-solving strategy of using analogy (see page 76). Try the special cases and ﬁrst. If you can see how to prove the asser-tion for these cases, then you will probably see how to prove it in general. 11. Find the sum of the series . 12. Suppose you have a large supply of books, all the same size, and you stack them at the edge of a table, with each book extending farther beyond the edge of the table than the one beneath it. Show that it is possible to do this so that the top book extends entirely beyond the table. In fact, show that the top book can extend any distance at all beyond the edge of the table if the stack is high enough. Use the following method of stacking: The top book extends half its length beyond the second book. The second book extends a quarter of its length beyond the third. The third extends one-sixth of its length beyond the fourth, and so on. (Try it yourself with a deck of cards.) Consider centers of mass. 13. If the curve , is rotated about the , the resulting solid looks like an inﬁnite decreasing string of beads. (a) Find the exact volume of the bead. (Use either a table of integrals or a computer algebra system.) (b) Find the total volume of the beads. 14. If , evaluate the expression 15. Suppose that circles of equal diameter are packed tightly in rows inside an equilateral tri-angle. (The ﬁgure illustrates the case .) If is the area of the triangle and is the total area occupied by the rows of circles, show that lim n l An A 2s3 n An A n 4 n 1 1 2 p 1 3p 1 4 p 1 1 2 p 1 3p 1 4 p p 1 nth x-axis y e x10 sin x, x 0 n2 ln1 1 n 2 k 2 k 1 lim n l (a0sn a1sn 1 a2sn 2 aksn k ) 0 a0 a1 a2 ak 0 n1 n3x n n0 arccotn 2 n 1 arctan arccot 3.1415927 P R O B L E M S P L U S FIGURE FOR PROBLEM 12 1 2 1 4 1 6 1 8 FIGURE FOR PROBLEM 15 763 16. A sequence is deﬁned recursively by the equations Find the sum of the series . 17. Taking the value of at 0 to be 1 and integrating a series term by term, show that 18. Starting with the vertices , , , of a square, we construct further points as shown in the ﬁgure: is the midpoint of is the midpoint of is the midpoint of , and so on. The polygonal spiral path approaches a point inside the square. (a) If the coordinates of are , show that and ﬁnd a similar equation for the -coordinates. (b) Find the coordinates of . 19. If has positive radius of convergence and , show that 20. Right-angled triangles are constructed as in the ﬁgure. Each triangle has height 1 and its base is the hypotenuse of the preceding triangle. Show that this sequence of triangles makes indeﬁ-nitely many turns around by showing that is a divergent series. 21. Consider the series whose terms are the reciprocals of the positive integers that can be written in base 10 notation without using the digit 0. Show that this series is convergent and the sum is less than 90. 22. (a) Show that the Maclaurin series of the function is where is the Fibonacci number, that is, , , and for . [Hint: Write and multiply both sides of this equation by .] (b) By writing as a sum of partial fractions and thereby obtaining the Maclaurin series in a different way, ﬁnd an explicit formula for the Fibonacci number. 23. Let Show that . 24. Prove that if , the partial sum of the harmonic series is not an integer. Hint: Let be the largest power of 2 that is less than or equal to and let be the product of all odd integers that are less than or equal to . Suppose that , an integer. Then . The right side of this equation is even. Prove that the left side is odd by showing that each of its terms is an even integer, except for the last one. M2ksn M2km sn m n M n 2k nth n 1 u 3 v3 w3 3uvw 1 w x 2 2! x 5 5! x 8 8! v x x 4 4! x 7 7! x 10 10! u 1 x 3 3! x 6 6! x 9 9! nth f x 1 x x 2 x1 x x 2 c0 c1x c2x 2 . . . n 3 fn fn1 fn2 f2 1 f1 1 nth fn n1 fnx n f x x 1 x x 2 n P n 1 ndn n i1 ici dni e f x n0 dnx n f x m0 cmx m P y 1 2 xn xn1 xn2 xn3 2 xn, yn Pn P P1P2P3P4P5P6P7 . . . P3P4 P2P3, P7 P1P2, P6 P5 P40, 0 P31, 0 P21, 1 P10, 1 y 1 0 x x dx n1 1n1 n n x x n0 an nn 1an n 1n 2an1 n 3an2 a0 a1 1 an P R O B L E M S P L U S P¡ P™ P¢ P£ P∞ Pß P¶ Pˆ P˜ P¡¸ FIGURE FOR PROBLEM 18 ¨¡ ¨™ ¨£ P 1 1 1 1 1 1 FIGURE FOR PROBLEM 20 A1 APPENDIXES A Numbers, Inequalities, and Absolute Values B Coordinate Geometry and Lines C Graphs of Second-Degree Equations D Trigonometry E Sigma Notation F Proofs of Theorems G The Logarithm Deﬁned as an Integral H Complex Numbers I Answers to Odd-Numbered Exercises NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES Calculus is based on the real number system. We start with the integers: Then we construct the rational numbers, which are ratios of integers. Thus any rational number can be expressed as Examples are (Recall that division by is always ruled out, so expressions like and are undeﬁned.) Some real numbers, such as , can’t be expressed as a ratio of integers and are therefore called irrational numbers. It can be shown, with varying degrees of difﬁculty, that the fol-lowing are also irrational numbers: The set of all real numbers is usually denoted by the symbol . When we use the word number without qualiﬁcation, we mean “real number.” Every number has a decimal representation. If the number is rational, then the corre-sponding decimal is repeating. For example, (The bar indicates that the sequence of digits repeats forever.) On the other hand, if the number is irrational, the decimal is nonrepeating: If we stop the decimal expansion of any number at a certain place, we get an approxima-tion to the number. For instance, we can write where the symbol is read “is approximately equal to.” The more decimal places we retain, the better the approximation we get. The real numbers can be represented by points on a line as in Figure 1. The positive direction (to the right) is indicated by an arrow. We choose an arbitrary reference point , called the origin, which corresponds to the real number . Given any convenient unit of measurement, each positive number is represented by the point on the line a distance of units to the right of the origin, and each negative number is represented by the point units to the left of the origin. Thus every real number is represented by a point on the line, and every point on the line corresponds to exactly one real number. The number associated with the point is called the coordinate of and the line is then called a coor-P P P x x x x 0 O 3.14159265 3.141592653589793 . . . s2 1.414213562373095 . . . 157 495 0.317171717 . . . 0.317 9 7 1.285714285714 . . . 1.285714 1 2 0.5000 . . . 0.50 2 3 0.66666 . . . 0.6 log10 2 sin 1 s 3 2 s5 s3 s2 0 0 3 0 0 0.17 17 100 46 46 1 3 7 1 2 where m and n are integers and n 0 r m n r . . . , 3, 2, 1, 0, 1, 2, 3, 4, . . . A A2 \|\|\|\| APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES dinate line, or a real number line, or simply a real line. Often we identify the point with its coordinate and think of a number as being a point on the real line. The real numbers are ordered. We say is less than and write if is a pos-itive number. Geometrically this means that lies to the left of on the number line. (Equivalently, we say is greater than and write .) The symbol (or ) means that either or and is read “ is less than or equal to .” For instance, the following are true inequalities: In what follows we need to use set notation. A set is a collection of objects, and these objects are called the elements of the set. If is a set, the notation means that is an element of , and means that is not an element of . For example, if repre-sents the set of integers, then but . If and are sets, then their union is the set consisting of all elements that are in or (or in both and ). The inter-section of and is the set consisting of all elements that are in both and . In other words, is the common part of and . The empty set, denoted by ∅, is the set that contains no element. Some sets can be described by listing their elements between braces. For instance, the set consisting of all positive integers less than 7 can be written as We could also write in set-builder notation as which is read “ is the set of such that is an integer and .” INTERVALS Certain sets of real numbers, called intervals, occur frequently in calculus and correspond geometrically to line segments. For example, if , the open interval from to con-sists of all numbers between and and is denoted by the symbol . Using set-builder notation, we can write Notice that the endpoints of the interval—namely, and —are excluded. This is indicated by the round brackets and by the open dots in Figure 2. The closed interval from to is the set Here the endpoints of the interval are included. This is indicated by the square brackets and by the solid dots in Figure 3. It is also possible to include only one endpoint in an inter-val, as shown in Table 1. a, b xa x b b a b a a, b xa x b a, b b a b a a b 0 x 7 x x A A xx is an integer and 0 x 7 A A 1, 2, 3, 4, 5, 6 A T S S T T S S T T S T S T S S T T S Z 3 Z Z S a a S S a a S S 2 2 s2 2 s2 2 3 7 7.4 7.5 b a a b a b b a a b b a a b b a b a a b b a FIGURE 1 0 1 2 3 4 _1 _2 _3 _2.63 2 π _ œ„ 1 2 3 7 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES \|\|\|\| A3 FIGURE 2 Open interval (a, b) a b FIGURE 3 Closed interval [a, b] a b TABLE OF INTERVALS We also need to consider inﬁnite intervals such as This does not mean that (“inﬁnity”) is a number. The notation stands for the set of all numbers that are greater than , so the symbol simply indicates that the interval extends indeﬁnitely far in the positive direction. INEQUALITIES When working with inequalities, note the following rules. RULES FOR INEQUALITIES 1. If , then . 2. If and , then . 3. If and , then . 4. If and , then . 5. If , then . Rule 1 says that we can add any number to both sides of an inequality, and Rule 2 says that two inequalities can be added. However, we have to be careful with multiplication. Rule 3 says that we can multiply both sides of an inequality by a positive number, but \| Rule 4 says that if we multiply both sides of an inequality by a negative number, then we reverse the direction of the inequality. For example, if we take the inequality and multiply by , we get , but if we multiply by , we get . Finally, Rule 5 says that if we take reciprocals, then we reverse the direction of an inequality (provided the numbers are positive). EXAMPLE 1 Solve the inequality . SOLUTION The given inequality is satisﬁed by some values of but not by others. To solve an inequality means to determine the set of numbers for which the inequality is true. This is called the solution set. x x 1 x 7x 5 6 10 2 6 10 2 3 5 1 a 1 b 0 a b ac bc c 0 a b ac bc c 0 a b a c b d c d a b a c b c a b 2 a a, a, xx a 1 A4 \|\|\|\| APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES Notation Set description Picture (set of all real numbers) , x x b , b x x b , b x x a a, x x a a, x a x b a, b x a x b a, b x a x b a, b x a x b a, b N Table 1 lists the nine possible types of inter-vals. When these intervals are discussed, it is always assumed that . a b a b a b a b a b a a b b First we subtract 1 from each side of the inequality (using Rule 1 with ): Then we subtract from both sides (Rule 1 with ): Now we divide both sides by (Rule 4 with ): These steps can all be reversed, so the solution set consists of all numbers greater than . In other words, the solution of the inequality is the interval . M EXAMPLE 2 Solve the inequalities . SOLUTION Here the solution set consists of all values of that satisfy both inequalities. Using the rules given in (2), we see that the following inequalities are equivalent: (add 2) (divide by 3) Therefore the solution set is . M EXAMPLE 3 Solve the inequality . SOLUTION First we factor the left side: We know that the corresponding equation has the solutions 2 and 3. The numbers 2 and 3 divide the real line into three intervals: On each of these intervals we determine the signs of the factors. For instance, Then we record these signs in the following chart: Another method for obtaining the information in the chart is to use test values. For instance, if we use the test value for the interval , then substitution in gives 12 51 6 2 x 2 5x 6 , 2 x 1 x 2 0 ? x 2 ? x , 2 3, 2, 3 , 2 x 2 x 3 0 x 2 x 3 0 x 2 5x 6 0 2, 5 2 x 5 6 3x 15 4 3x 2 13 x 4 3x 2 13 ( 2 3, ) 2 3 x 4 6 2 3 c 1 6 6 6x 4 c 7x 7x x 7x 4 c 1 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES \|\|\|\| A5 FIGURE 4 x 0 y y=≈-5x+6 1 2 3 4 N A visual method for solving Example 3 is to use a graphing device to graph the parabola (as in Figure 4) and observe that the curve lies on or below the -axis when . 2 x 3 x y x 2 5x 6 Interval x 3 2 x 3 x 2 x 2 x 3 x 3 x 2 The polynomial doesn’t change sign inside any of the three intervals, so we conclude that it is positive on . Then we read from the chart that is negative when . Thus the solution of the inequality is Notice that we have included the endpoints 2 and 3 because we are looking for values of such that the product is either negative or zero. The solution is illustrated in Figure 5. M EXAMPLE 4 Solve . SOLUTION First we take all nonzero terms to one side of the inequality sign and factor the resulting expression: As in Example 3 we solve the corresponding equation and use the solutions , , and to divide the real line into four intervals , , , and . On each interval the product keeps a constant sign as shown in the following chart: Then we read from the chart that the solution set is The solution is illustrated in Figure 6. M ABSOLUTE VALUE The absolute value of a number , denoted by , is the distance from to on the real number line. Distances are always positive or , so we have For example, In general, we have a a if a 0 a a if a 0 3 3 3 s2 1 s2 1 0 0 3 3 3 3 for every number a a 0 0 0 a a a x4 x 0 or x 1 4, 0 1, 1, 0, 1 4, 0 , 4 x 1 x 0 x 4 xx 1 x 4 0 xx 1 x 4 0 or x 3 3x 2 4x 0 x 3 3x 2 4x x x2 x 3 2, 3 x 2 x 3 0 2 x 3 x 2 x 3 , 2 x 2 5x 6 A6 \|\|\|\| APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES 0 2 3 + -+ FIGURE 5 x Interval x x 1 0 x 1 4 x 0 x 4 xx 1 x 4 x 4 x 1 0 1 _4 FIGURE 6 N Remember that if is negative, then is positive. a a APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES \|\|\|\| A7 0 a _a x a a \| x \| FIGURE 7 \|a-b \| a b \|a-b \| b a FIGURE 8 Length of a line segment=\|a-b \| EXAMPLE 5 Express without using the absolute-value symbol. SOLUTION M Recall that the symbol means “the positive square root of.” Thus means \| and . Therefore, the equation is not always true. It is true only when . If , then , so we have . In view of (3), we then have the equation which is true for all values of . Hints for the proofs of the following properties are given in the exercises. PROPERTIES OF ABSOLUTE VALUES Suppose and are any real numbers and is an integer. Then 1. 2. 3. For solving equations or inequalities involving absolute values, it’s often very helpful to use the following statements. Suppose . Then 4. if and only if 5. if and only if 6. if and only if or For instance, the inequality says that the distance from to the origin is less than , and you can see from Figure 7 that this is true if and only if lies between and . If and are any real numbers, then the distance between and is the absolute value of the difference, namely, , which is also equal to . (See Figure 8.) EXAMPLE 6 Solve . SOLUTION By Property 4 of (6), is equivalent to So or . Thus or . M x 1 x 4 2x 2 2x 8 2x 5 3 or 2x 5 3 2x 5 3 2x 5 3 b a a b b a b a a a x a x x a x a x a x a a x a x a x a x a a 0 6 a n a n b 0 a b a b ab ab n b a 5 a sa 2 a 4 sa 2 a a 0 a 0 a 0 sa 2 a s 0 s 2 r sr s s1 3x 2 2 3x if x 2 3 if x 2 3 3x 2 3x 2 3x 2 if 3x 2 0 if 3x 2 0 3x 2 EXAMPLE 7 Solve . SOLUTION 1 By Property 5 of (6), is equivalent to Therefore, adding 5 to each side, we have and the solution set is the open interval . SOLUTION 2 Geometrically the solution set consists of all numbers whose distance from 5 is less than 2. From Figure 9 we see that this is the interval . M EXAMPLE 8 Solve . SOLUTION By Properties 4 and 6 of (6), is equivalent to In the ﬁrst case , which gives . In the second case , which gives . So the solution set is M Another important property of absolute value, called the Triangle Inequality, is used fre-quently not only in calculus but throughout mathematics in general. THE TRIANGLE INEQUALITY If and are any real numbers, then Observe that if the numbers and are both positive or both negative, then the two sides in the Triangle Inequality are actually equal. But if and have opposite signs, the left side involves a subtraction and the right side does not. This makes the Triangle Inequality seem reasonable, but we can prove it as follows. Notice that is always true because equals either or . The corresponding statement for is Adding these inequalities, we get If we now apply Properties 4 and 5 (with replaced by and by ), we obtain which is what we wanted to show. a b a b a b a a b x (a b) a b a b b b b b a a a a a a b a b a a b a b b a 7 {xx 2 or x 2 3} , 2 [ 2 3, ) x 2 3x 6 x 2 3 3x 2 3x 2 4 or 3x 2 4 3x 2 4 3x 2 4 3, 7 x 3, 7 3 x 7 2 x 5 2 x 5 2 x 5 2 A8 \|\|\|\| APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES 3 5 7 2 2 FIGURE 9 EXAMPLE 9 If and , use the Triangle Inequality to estimate . SOLUTION In order to use the given information, we use the Triangle Inequality with and : Thus M x y 11 0.3 0.1 0.2 0.3 x 4 y 7 x y 11 x 4 y 7 b y 7 a x 4 x y 11 y 7 0.2 x 4 0.1 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES \|\|\|\| A9 ature in degrees Celsius and is the temperature in degrees Fahrenheit. What interval on the Celsius scale corresponds to the temperature range ? 40. Use the relationship between and given in Exercise 39 to ﬁnd the interval on the Fahrenheit scale corresponding to the temperature range . 41. As dry air moves upward, it expands and in so doing cools at a rate of about C for each 100-m rise, up to about 12 km. (a) If the ground temperature is C, write a formula for the temperature at height . (b) What range of temperature can be expected if a plane takes off and reaches a maximum height of 5 km? 42. If a ball is thrown upward from the top of a building 128 ft high with an initial velocity of 16 ft s, then the height above the ground seconds later will be During what time interval will the ball be at least 32 ft above the ground? 43–46 Solve the equation for . 43. 44. 45. 46. 47–56 Solve the inequality. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 0 x 5 1 2 1 x 4 5x 2 6 2x 3 0.4 x 1 3 x 5 2 x 6 0.1 x 4 1 x 3 x 3 2x 1 x 1 3 x 3 2x 1 3x 5 1 2x 3 x h 128 16t 16t 2 t h h 20 1 20 C 30 F C 50 F 95 F 1–12 Rewrite the expression without using the absolute value symbol. 1. 2. 3. 4. 5. 6. 7. if 8. if 9. 10. 11. 12. 13–38 Solve the inequality in terms of intervals and illustrate the solution set on the real number line. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. The relationship between the Celsius and Fahrenheit tempera-ture scales is given by , where is the temper-C C 5 9F 32 3 1 x 1 1 x 4 x 3 3x 4x 2 x 3 x x 1 x 2 x 3 0 x 3 x 2 0 x 2 5 x 2 3 x 2 x 1 x 2 x 1 0 2x 3 x 1 0 2x 2 x 1 2x 3 x 1 0 x 1 x 2 0 2x 3 x 4 3x 2 4x 2x 1 3x 2 5 3 2x 9 0 1 x 1 1 3x 4 16 1 2x 5 7 1 5x 5 3x 2x 1 5x 8 4 3x 6 1 x 2 3x 11 4 2x 7 3 1 2x 2 x 2 1 2x 1 x 1 x 2 x 2 x 2 x 2 2 3 s5 5 2 5 23 5 23 EXERCISES A 65. Prove that . [Hint: Use Equation 4.] 66. Prove that . 67. Show that if , then . 68. Prove that . [Hint: Use the Triangle Inequality with and .] 69. Show that the sum, difference, and product of rational numbers are rational numbers. 70. (a) Is the sum of two irrational numbers always an irrational number? (b) Is the product of two irrational numbers always an irrational number? b y a x y x y x y a 2 b 2 0 a b a b a b ab a b 57–58 Solve for , assuming , , and are positive constants. 57. 58. 59–60 Solve for , assuming , , and are negative constants. 59. 60. 61. Suppose that and . Use the Triangle Inequality to show that . 62. Show that if , then . 63. Show that if , then . 64. Use Rule 3 to prove Rule 5 of (2). a a b 2 b a b 4x 13 3 x 3 1 2 x y 5 0.05 y 3 0.04 x 2 0.01 ax b c b ax b c c b a x a bx c 2a abx c bc c b a x A10 \|\|\|\| APPENDIX B COORDINATE GEOMETRY AND LINES COORDINATE GEOMETRY AND LINES Just as the points on a line can be identiﬁed with real numbers by assigning them coordi-nates, as described in Appendix A, so the points in a plane can be identiﬁed with ordered pairs of real numbers. We start by drawing two perpendicular coordinate lines that inter-sect at the origin on each line. Usually one line is horizontal with positive direction to the right and is called the -axis; the other line is vertical with positive direction upward and is called the -axis. Any point in the plane can be located by a unique ordered pair of numbers as follows. Draw lines through perpendicular to the - and -axes. These lines intersect the axes in points with coordinates and as shown in Figure 1. Then the point is assigned the ordered pair . The ﬁrst number is called the x-coordinate of ; the second number is called the y-coordinate of . We say that is the point with coordinates , and we denote the point by the symbol . Several points are labeled with their coordi-nates in Figure 2. By reversing the preceding process we can start with an ordered pair and arrive at the corresponding point . Often we identify the point with the ordered pair and refer to “the point .” [Although the notation used for an open interval is the a, b a, b a, b P P a, b 0 x 1 2 3 4 5 _1 _2 _3 1 2 3 4 _2 _3 _1 y _4 (5, 0) (1, 3) (_2, 2) (_3, _2) ) (2, _4) FIGURE 2 x 1 2 3 4 5 _1 _2 _3 a O 2 4 _2 _1 b y 1 3 P(a, b) I II IV III _3 FIGURE 1 _4 Pa, b a, b P P b P a a, b P b a y x P P y x O B same as the notation used for a point , you will be able to tell from the context which meaning is intended.] This coordinate system is called the rectangular coordinate system or the Cartesian coordinate system in honor of the French mathematician René Descartes (1596–1650), even though another Frenchman, Pierre Fermat (1601–1665), invented the principles of analytic geometry at about the same time as Descartes. The plane supplied with this coor-dinate system is called the coordinate plane or the Cartesian plane and is denoted by . The - and -axes are called the coordinate axes and divide the Cartesian plane into four quadrants, which are labeled I, II, III, and IV in Figure 1. Notice that the ﬁrst quad-rant consists of those points whose - and -coordinates are both positive. EXAMPLE 1 Describe and sketch the regions given by the following sets. (a) (b) (c ) SOLUTION (a) The points whose -coordinates are 0 or positive lie on the -axis or to the right of it as indicated by the shaded region in Figure 3(a). (b) The set of all points with -coordinate 1 is a horizontal line one unit above the -axis [see Figure 3(b)]. (c) Recall from Appendix A that The given region consists of those points in the plane whose -coordinates lie between and . Thus the region consists of all points that lie between (but not on) the hori-zontal lines and . [These lines are shown as dashed lines in Figure 3(c) to indicate that the points on these lines don’t lie in the set.] M Recall from Appendix A that the distance between points and on a number line is . Thus the distance between points and on a hor-izontal line must be and the distance between and on a ver-tical line must be . (See Figure 4.) To ﬁnd the distance between any two points and , we note that triangle in Figure 4 is a right triangle, and so by the Pythagorean Theorem we have sx2 x1 2 y2 y1 2 P1P2 sP1P3 2 P2P3 2 sx2 x1 2 y2 y1 2 P1P2P3 P2x2, y2 P1x1, y1 P1P2 y2 y1 P3x2, y1 P2x2, y2 x2 x1 P3x2, y1 P1x1, y1 a b b a b a y 1 y 1 1 1 y 1 y 1 if and only if y 1 x y FIGURE 3 x 0 y x 0 y y=1 x 0 y y=1 y=_1 (a) x 0 (b) y=1 (c) \| y \|<1 y x {x, y y 1} x, y y 1 x, y x 0 y x y x 2 a, b APPENDIX B COORDINATE GEOMETRY AND LINES \|\|\|\| A11 FIGURE 4 P¡(⁄, ›) x ⁄ ¤ 0 › ﬁ y P™(¤, ﬁ) P£(¤, ›) \|¤-⁄\| \|ﬁ-›\| A12 \|\|\|\| APPENDIX B COORDINATE GEOMETRY AND LINES DISTANCE FORMULA The distance between the points and is EXAMPLE 2 The distance between and is M LINES We want to ﬁnd an equation of a given line ; such an equation is satisﬁed by the coordi-nates of the points on and by no other point. To ﬁnd the equation of we use its slope, which is a measure of the steepness of the line. DEFINITION The slope of a nonvertical line that passes through the points and is The slope of a vertical line is not deﬁned. Thus the slope of a line is the ratio of the change in , , to the change in , . (See Figure 5.) The slope is therefore the rate of change of y with respect to x. The fact that the line is straight means that the rate of change is constant. Figure 6 shows several lines labeled with their slopes. Notice that lines with positive slope slant upward to the right, whereas lines with negative slope slant downward to the right. Notice also that the steepest lines are the ones for which the absolute value of the slope is largest, and a horizontal line has slope 0. Now let’s ﬁnd an equation of the line that passes through a given point and has slope . A point with lies on this line if and only if the slope of the line through and is equal to ; that is, This equation can be rewritten in the form and we observe that this equation is also satisﬁed when and . Therefore it is an equation of the given line. POINT-SLOPE FORM OF THE EQUATION OF A LINE An equation of the line passing through the point and having slope is y y1 mx x1 m P1x1, y1 3 y y1 x x1 y y1 mx x1 y y1 x x1 m m P P1 x x1 Px, y m P1x1, y1 x x y y m y x y2 y1 x2 x1 P2x2, y2 P1x1, y1 2 L L L s5 1 2 3 2 2 s4 2 5 2 s41 5, 3 1, 2 P1P2 sx2 x12 y2 y12 P2x2, y2 P1x1, y1 1 FIGURE 5 P™(x™, y™) P¡(x¡, y¡) L Îy=ﬁ-› =rise Îx=¤-⁄ =run x 0 y x 0 y m=1 m=0 m=_1 m=_2 m=_5 m=2 m=5 m=1 2 m=_ 1 2 FIGURE 6 EXAMPLE 3 Find an equation of the line through with slope . SOLUTION Using with , , and , we obtain an equation of the line as which we can rewrite as M EXAMPLE 4 Find an equation of the line through the points and . SOLUTION By Deﬁnition 2 the slope of the line is Using the point-slope form with and , we obtain which simpliﬁes to M Suppose a nonvertical line has slope and -intercept . (See Figure 7.) This means it intersects the -axis at the point , so the point-slope form of the equation of the line, with and , becomes This simpliﬁes as follows. SLOPE-INTERCEPT FORM OF THE EQUATION OF A LINE An equation of the line with slope and -intercept is In particular, if a line is horizontal, its slope is , so its equation is , where is the -intercept (see Figure 8). A vertical line does not have a slope, but we can write its equation as , where is the -intercept, because the -coordinate of every point on the line is . Observe that the equation of every line can be written in the form because a vertical line has the equation or ( , , ) and a nonvertical line has the equation or ( , , ). Conversely, if we start with a general ﬁrst-degree equation, that is, an equation of the form (5), where , , and are constants and and are not both 0, then we can show that it is the equation of a line. If , the equation becomes or , which represents a vertical line with -intercept . If , the equation B 0 CA x x CA Ax C 0 B 0 B A C B A C b B 1 A m mx y b 0 y mx b C a B 0 A 1 x a 0 x a Ax By C 0 5 a x x a x a y b y b m 0 y mx b b y m 4 y b mx 0 y1 b x1 0 0, b y b y m 3x 2y 1 y 2 3 2x 1 y1 2 x1 1 m 4 2 3 1 3 2 3, 4 1, 2 x 2y 13 0 or 2y 14 x 1 y 7 1 2x 1 y1 7 x1 1 m 1 2 3 1 2 1, 7 APPENDIX B COORDINATE GEOMETRY AND LINES \|\|\|\| A13 x 0 y b y=mx+b FIGURE 7 0 y b x a x=a y=b FIGURE 8 can be rewritten by solving for : and we recognize this as being the slope-intercept form of the equation of a line ( , ). Therefore an equation of the form (5) is called a linear equation or the general equation of a line. For brevity, we often refer to “the line ” instead of “the line whose equation is .” EXAMPLE 5 Sketch the graph of the equation . SOLUTION Since the equation is linear, its graph is a line. To draw the graph, we can sim-ply ﬁnd two points on the line. It’s easiest to ﬁnd the intercepts. Substituting (the equation of the -axis) in the given equation, we get , so is the -intercept. Substituting in the equation, we see that the -intercept is . This allows us to sketch the graph as in Figure 9. M EXAMPLE 6 Graph the inequality . SOLUTION We are asked to sketch the graph of the set and we do so by solving the inequality for : Compare this inequality with the equation , which represents a line with slope and -intercept . We see that the given graph consists of points whose -coordinates are larger than those on the line . Thus the graph is the region that lies above the line, as illustrated in Figure 10. M PARALLEL AND PERPENDICULAR LINES Slopes can be used to show that lines are parallel or perpendicular. The following facts are proved, for instance, in Precalculus: Mathematics for Calculus, Fifth Edition by Stewart, Redlin, and Watson (Thomson BrooksCole, Belmont, CA, 2006). PARALLEL AND PERPENDICULAR LINES 1. Two nonvertical lines are parallel if and only if they have the same slope. 2. Two lines with slopes and are perpendicular if and only if ; that is, their slopes are negative reciprocals: EXAMPLE 7 Find an equation of the line through the point that is parallel to the line . SOLUTION The given line can be written in the form y 2 3 x 5 6 4x 6y 5 0 5, 2 m2 1 m1 m1m2 1 m2 m1 6 y 1 2 x 5 2 y 5 2 y 1 2 y 1 2 x 5 2 y 1 2 x 5 2 2y x 5 x 2y 5 y x, yx 2y 5 x 2y 5 3 y x 0 x x 5 3x 15 x y 0 3x 5y 15 Ax By C 0 Ax By C 0 b CB m AB y A B x C B y A14 \|\|\|\| APPENDIX B COORDINATE GEOMETRY AND LINES FIGURE 9 y 0 x (5, 0) (0, _3) 3x-5y=15 FIGURE 10 0 y 2.5 x 5 y=_ x+ 1 2 5 2 which is in slope-intercept form with . Parallel lines have the same slope, so the required line has slope and its equation in point-slope form is We can write this equation as . M EXAMPLE 8 Show that the lines and are perpendicular. SOLUTION The equations can be written as from which we see that the slopes are Since , the lines are perpendicular. M m1m2 1 m2 3 2 and m1 2 3 y 3 2 x 1 4 and y 2 3 x 1 3 6x 4y 1 0 2x 3y 1 2x 3y 16 y 2 2 3x 5 2 3 m 2 3 APPENDIX B COORDINATE GEOMETRY AND LINES \|\|\|\| A15 19. 20. 21–36 Find an equation of the line that satisﬁes the given conditions. 21. Through , slope 22. Through , slope 23. Through , slope 24. Through , slope 25. Through and 26. Through and 27. Slope , -intercept 28. Slope , -intercept 29. -intercept , -intercept 30. -intercept , -intercept 31. Through , parallel to the -axis 32. Through , parallel to the -axis 33. Through , parallel to the line 34. -intercept , parallel to the line 35. Through , perpendicular to the line 36. Through , perpendicular to the line 37–42 Find the slope and -intercept of the line and draw its graph. 37. 38. 2x 5y 0 x 3y 0 y 4x 8y 1 ( 1 2, 2 3) 2x 5y 8 0 1, 2 2x 3y 4 0 6 y x 2y 6 1, 6 y 4, 5 x 4, 5 6 y 8 x 3 y 1 x 4 y 2 5 2 y 3 4, 3 1, 2 1, 6 2, 1 7 2 3, 5 2 3 1, 7 3 1, 4 6 2, 3 y 1 xy 0 1–6 Find the distance between the points. 1. , 2. , 3. , 4. , 5. , 6. , 7–10 Find the slope of the line through and . 7. 8. , 9. , 10. , 11. Show that the triangle with vertices , , and is isosceles. 12. (a) Show that the triangle with vertices , , and is a right triangle using the converse of the Pythagorean Theorem. (b) Use slopes to show that is a right triangle. (c) Find the area of the triangle. 13. Show that the points , , , and are the vertices of a square. 14. (a) Show that the points , , and are collinear (lie on the same line) by showing that . (b) Use slopes to show that , , and are collinear. 15. Show that , , , and are vertices of a parallelogram. 16. Show that , , , and are vertices of a rectangle. 17–20 Sketch the graph of the equation. 17. 18. y 2 x 3 D0, 6 C10, 8 B11, 3 A1, 1 D1, 7 C5, 10 B7, 4 A1, 1 C B A AB BC AC C5, 15 B3, 11 A1, 3 5, 3 1, 0 4, 6 2, 9 ABC C2, 2 B11, 3 A6, 7 C4, 3 B3, 1 A0, 2 Q6, 0 P1, 4 Q1, 6 P3, 3 Q4, 3 P1, 6 Q4, 11 P1, 5, Q P b, a a, b 4, 7 2, 5 1, 3 1, 6 1, 3 6, 2 5, 7 1, 3 4, 5 1, 1 EXERCISES B 57. Show that the lines and are not parallel and ﬁnd their point of intersection. 58. Show that the lines and are perpendicular and ﬁnd their point of intersection. 59. Find an equation of the perpendicular bisector of the line seg-ment joining the points and . 60. (a) Find equations for the sides of the triangle with vertices , , and . (b) Find equations for the medians of this triangle. Where do they intersect? 61. (a) Show that if the - and -intercepts of a line are nonzero numbers and , then the equation of the line can be put in the form This equation is called the two-intercept form of an equa-tion of a line. (b) Use part (a) to ﬁnd an equation of the line whose -intercept is 6 and whose -intercept is . 62. A car leaves Detroit at 2:00 PM, traveling at a constant speed west along I-96. It passes Ann Arbor, 40 mi from Detroit, at 2:50 PM. (a) Express the distance traveled in terms of the time elapsed. (b) Draw the graph of the equation in part (a). (c) What is the slope of this line? What does it represent? 8 y x x a y b 1 b a y x R1, 6 Q3, 4 P1, 0 B7, 2 A1, 4 10x 6y 50 0 3x 5y 19 0 6x 2y 10 2x y 4 39. 40. 41. 42. 43–52 Sketch the region in the -plane. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. Find a point on the -axis that is equidistant from and . 54. Show that the midpoint of the line segment from to is 55. Find the midpoint of the line segment joining the given points. (a) and (b) and 56. Find the lengths of the medians of the triangle with vertices , , and . (A median is a line segment from a vertex to the midpoint of the opposite side.) C8, 2 B3, 6 A1, 0 8, 12 1, 6 7, 15 1, 3 x1 x2 2 , y1 y2 2 P2x2, y2 P1x1, y1 1, 1 5, 5 y {x, y x y 1 2x 3} x, y 1 x y 1 2x x, y y 2x 1 x, y 0 y 4 and x 2 {x, y x 3 and y 2} {x, y x 2} x, y x 1 and y 3 x, y xy 0 x, y y 0 x, y x 0 xy 4x 5y 10 3x 4y 12 2x 3y 6 0 y 2 A16 \|\|\|\| APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS GRAPHS OF SECOND-DEGREE EQUATIONS In Appendix B we saw that a ﬁrst-degree, or linear, equation represents a line. In this section we discuss second-degree equations such as which represent a circle, a parabola, an ellipse, and a hyperbola, respectively. The graph of such an equation in and is the set of all points that satisfy the equation; it gives a visual representation of the equation. Conversely, given a curve in the -plane, we may have to ﬁnd an equation that represents it, that is, an equation satisﬁed by the coordinates of the points on the curve and by no other point. This is the other half of the basic principle of analytic geometry as formulated by Descartes and Fermat. The idea is that if a geometric curve can be represented by an algebraic equation, then the rules of algebra can be used to analyze the geometric problem. CIRCLES As an example of this type of problem, let’s ﬁnd an equation of the circle with radius and center . By deﬁnition, the circle is the set of all points whose distance from Px, y h, k r xy x, y y x x 2 y 2 1 x 2 9 y 2 4 1 y x 2 1 x 2 y 2 1 Ax By C 0 C the center is . (See Figure 1.) Thus is on the circle if and only if . From the distance formula, we have or equivalently, squaring both sides, we get This is the desired equation. EQUATION OF A CIRCLE An equation of the circle with center and radius is In particular, if the center is the origin , the equation is EXAMPLE 1 Find an equation of the circle with radius 3 and center . SOLUTION From Equation 1 with , , and , we obtain M EXAMPLE 2 Sketch the graph of the equation by ﬁrst show-ing that it represents a circle and then ﬁnding its center and radius. SOLUTION We ﬁrst group the -terms and -terms as follows: Then we complete the square within each grouping, adding the appropriate constants to both sides of the equation: or Comparing this equation with the standard equation of a circle (1), we see that , , and , so the given equation represents a circle with center and radius . It is sketched in Figure 2. M x 0 y 1 (_1, 3) FIGURE 2 ≈+¥+2x-6y+7=0 s3 1, 3 r s3 k 3 h 1 x 12 y 32 3 x 2 2x 1 y 2 6y 9 7 1 9 x 2 2x y 2 6y 7 y x x 2 y 2 2x 6y 7 0 x 22 y 52 9 k 5 h 2 r 3 2, 5 x 2 y 2 r 2 0, 0 x h2 y k2 r 2 r h, k 1 x h2 y k2 r 2 sx h2 y k2 r PC r P r Ch, k APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS \|\|\|\| A17 C(h, k) x 0 y r P(x, y) FIGURE 1 PARABOLAS The geometric properties of parabolas are reviewed in Section 10.5. Here we regard a parabola as a graph of an equation of the form . EXAMPLE 3 Draw the graph of the parabola . SOLUTION We set up a table of values, plot points, and join them by a smooth curve to obtain the graph in Figure 3. . M Figure 4 shows the graphs of several parabolas with equations of the form for various values of the number . In each case the vertex, the point where the parabola changes direction, is the origin. We see that the parabola opens upward if and downward if (as in Figure 5). Notice that if satisﬁes , then so does . This corresponds to the geo-metric fact that if the right half of the graph is reﬂected about the -axis, then the left half of the graph is obtained. We say that the graph is symmetric with respect to the y-axis. The graph of an equation is symmetric with respect to the -axis if the equation is unchanged when is replaced by . If we interchange and in the equation , the result is , which also represents a parabola. (Interchanging and amounts to reﬂecting about the diagonal line .) The parabola opens to the right if and to the left if . (See a 0 a 0 x ay 2 y x y x x ay 2 y ax 2 y x x x y y x, y y ax 2 x, y FIGURE 5 x 0 y (_x, y) (x, y) x 0 y (a) y=a≈, a>0 (b) y=a≈, a<0 y x y=2≈ y=≈ y=_≈ y=_2≈ y= ≈ 1 2 y=_ ≈ 1 2 FIGURE 4 a 0 a 0 y ax 2 a y ax 2 FIGURE 3 0 y 1 x 1 y=≈ y x 2 y ax 2 bx c A18 \|\|\|\| APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS x 0 0 1 4 9 3 2 1 1 4 1 2 y x 2 Figure 6.) This time the parabola is symmetric with respect to the -axis because if satisﬁes , then so does . The graph of an equation is symmetric with respect to the -axis if the equation is unchanged when is replaced by . EXAMPLE 4 Sketch the region bounded by the parabola and the line . SOLUTION First we ﬁnd the points of intersection by solving the two equations. Substituting into the equation , we get , which gives so or . Thus the points of intersection are and , and we draw the line passing through these points. We then sketch the parabola by referring to Figure 6(a) and having the parabola pass through and . The region bounded by and means the ﬁnite region whose boundaries are these curves. It is sketched in Figure 7. M ELLIPSES The curve with equation where and are positive numbers, is called an ellipse in standard position. (Geometric properties of ellipses are discussed in Section 10.5.) Observe that Equation 2 is unchanged if is replaced by or is replaced by , so the ellipse is symmetric with respect to both axes. As a further aid to sketching the ellipse, we ﬁnd its intercepts. The x-intercepts of a graph are the -coordinates of the points where the graph intersects the -axis. They are found by setting in the equation of the graph. The y-intercepts are the -coordinates of the points where the graph intersects the -axis. They are found by setting in its equation. If we set in Equation 2, we get and so the -intercepts are . Setting , we get , so the -intercepts are . Using this information, together with symmetry, we sketch the ellipse in Figure 8. If , the ellipse is a circle with radius . a a b b y y 2 b 2 x 0 a x x 2 a 2 y 0 x 0 y y y 0 x x y y x x b a x 2 a 2 y 2 b 2 1 2 y x 2 x y 2 1, 1 4, 2 x y 2 y x 2 1, 1 4, 2 1 y 2 0 y 2 y 2 y 2y 1 y 2 y 2 x y 2 x y 2 y x 2 x y 2 y y x FIGURE 6 x 0 y x 0 y (a) x=a¥, a>0 (b) x=a¥, a<0 x, y x ay 2 x, y x APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS \|\|\|\| A19 FIGURE 7 x 0 y 1 2 4 y=x-2 x=¥ (1, _1) (4, 2) 0 x y (0, b) (0, _b) (a, 0) (_a, 0) FIGURE 8 ≈ a@ ¥ b@ + =1 EXAMPLE 5 Sketch the graph of . SOLUTION We divide both sides of the equation by 144: The equation is now in the standard form for an ellipse (2), so we have , , , and . The -intercepts are ; the -intercepts are . The graph is sketched in Figure 9. M HYPERBOLAS The curve with equation is called a hyperbola in standard position. Again, Equation 3 is unchanged when is replaced by or is replaced by , so the hyperbola is symmetric with respect to both axes. To ﬁnd the -intercepts we set and obtain and . However, if we put in Equation 3, we get , which is impossible, so there is no -inter-cept. In fact, from Equation 3 we obtain which shows that and so . Therefore we have or . This means that the hyperbola consists of two parts, called its branches. It is sketched in Figure 10. In drawing a hyperbola it is useful to draw ﬁrst its asymptotes, which are the lines and shown in Figure 10. Both branches of the hyperbola approach the asymptotes; that is, they come arbitrarily close to the asymptotes. This involves the idea of a limit, which is discussed in Chapter 2. (See also Exercise 69 in Section 4.5.) By interchanging the roles of and we get an equation of the form which also represents a hyperbola and is sketched in Figure 11. y 2 a 2 x 2 b 2 1 y x y bax y bax x a x a x sx 2 a x 2 a 2 x 2 a 2 1 y 2 b 2 1 y y 2 b 2 x 0 x a x 2 a 2 y 0 x y y x x x 2 a 2 y 2 b 2 1 3 0 x y (0, 3) (4, 0) (_4, 0) (0, _3) FIGURE 9 9≈+16¥=144 3 y 4 x b 3 a 4 b 2 9 a 2 16 x 2 16 y 2 9 1 9x 2 16y 2 144 A20 \|\|\|\| APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS 0 y x (_a, 0) (a, 0) y=_ x b a y= x b a FIGURE 10 The hyperbola - =1 ≈ a@ ¥ b@ FIGURE 11 The hyperbola - =1 ¥ a@ ≈ b@ y 0 x (0, a) (0, _a) y=_ x a b y= x a b EXAMPLE 6 Sketch the curve . SOLUTION Dividing both sides by 36, we obtain which is the standard form of the equation of a hyperbola (Equation 3). Since , the -intercepts are . Since , we have and the asymptotes are . The hyperbola is sketched in Figure 12. M If , a hyperbola has the equation (or ) and is called an equilateral hyperbola [see Figure 13(a)]. Its asymptotes are , which are perpendi-cular. If an equilateral hyperbola is rotated by , the asymptotes become the - and -axes, and it can be shown that the new equation of the hyperbola is , where is a constant [see Figure 13(b)]. SHIFTED CONICS Recall that an equation of the circle with center the origin and radius is , but if the center is the point , then the equation of the circle becomes Similarly, if we take the ellipse with equation x 2 a 2 y 2 b 2 1 4 x h2 y k2 r 2 h, k x 2 y 2 r 2 r FIGURE 13 Equilateral hyperbolas (a) ≈-¥=a@ (b) xy=k (k>0) 0 y x 0 y x y=x y=_x k xy k y x 45 y x y 2 x 2 a 2 x 2 y 2 a 2 b a FIGURE 12 The hyperbola 9≈-4¥=36 0 y (_2, 0) (2, 0) x y= x 3 2 y=_ x 3 2 y ( 3 2)x b 3 b 2 9 2 x a 2 4 x 2 4 y 2 9 1 9x 2 4y 2 36 APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS \|\|\|\| A21 A22 \|\|\|\| APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS and translate it (shift it) so that its center is the point , then its equation becomes (See Figure 14.) Notice that in shifting the ellipse, we replaced by and by in Equation 4 to obtain Equation 5. We use the same procedure to shift the parabola so that its vertex (the origin) becomes the point as in Figure 15. Replacing by and by , we see that the new equation is EXAMPLE 7 Sketch the graph of the equation . SOLUTION First we complete the square: In this form we see that the equation represents the parabola obtained by shifting so that its vertex is at the point . The graph is sketched in Figure 16. M FIGURE 16 y=2≈-4x+1 x 0 y 1 3 2 1 (1, _1) 1, 1 y 2x 2 y 2x 2 2x 1 2x 12 1 y 2x 2 4x 1 FIGURE 15 y 0 x (h, k) y=a(x-h)@+k y=a≈ y ax h2 k or y k ax h2 y k y x h x h, k y ax 2 y k y x h x FIGURE 14 (0, 0) y (x-h, y-k) (h, k) (x, y) x ¥ b@ ≈ a@ + =1 (x-h)@ a@ (y-k)@ b@ + =1 b a h b k a x h2 a 2 y k2 b 2 1 5 h, k EXAMPLE 8 Sketch the curve . SOLUTION This time we start with the parabola (as in Figure 6 with ) and shift one unit to the right to get the graph of . (See Figure 17.) M FIGURE 17 (a) x=_¥ 0 y x (b) x=1-¥ x 1 0 y x 1 y 2 a 1 x y 2 x 1 y 2 APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS \|\|\|\| A23 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33–34 Sketch the region bounded by the curves. 33. , 34. , 35. Find an equation of the parabola with vertex that passes through the points and . 36. Find an equation of the ellipse with center at the origin that passes through the points and . 37– 40 Sketch the graph of the set. 37. 38. 39. 40. x, y x 2 4y 2 4 x, y y x 2 1 x, y x 2 y 2 4 x, y x 2 y 2 1 (2, 5s5 3) (1, 10s2 3) 3, 3 1, 3 1, 1 x 2y 2 y 4 x 2 y x 2 y 3x 4x 2 9y 2 16x 54y 61 0 x 2 4y 2 6x 5 0 y 2 2x 6y 5 0 x 4 y 2 x 2 y 2 4x 3 0 y x 2 6x 13 16x 2 9y 2 36y 108 9x 12 4y 22 36 y x 2 2x xy 4 2x 2 5y 2 10 9y 2 x 2 9 9x 2 25y 2 225 x y 2 1 y x 2 2 4x 2 y 2 1 25x 2 4y 2 100 16x 2 25y 2 400 1–4 Find an equation of a circle that satisﬁes the given conditions. 1. Center , radius 2. Center , radius 3. Center at the origin, passes through 4. Center , passes through 5–9 Show that the equation represents a circle and ﬁnd the center and radius. 5. 6. 7. 8. 9. 10. Under what condition on the coefﬁcients , , and does the equation represent a circle? When that condition is satisﬁed, ﬁnd the center and radius of the circle. 11–32 Identify the type of curve and sketch the graph. Do not plot points. Just use the standard graphs given in Figures 5, 6, 8, 10, and 11 and shift if necessary. 11. 12. 13. 14. x 2y 2 x 2 4y 2 16 y 2 x 2 1 y x 2 x 2 y 2 ax by c 0 c b a 2x 2 2y 2 x y 1 16x 2 16y 2 8x 32y 1 0 x 2 y 2 x 0 x 2 y 2 6y 2 0 x 2 y 2 4x 10y 13 0 4, 6 1, 5 4, 7 10 2, 8 5 3, 1 EXERCISES C TRIGONOMETRY ANGLES Angles can be measured in degrees or in radians (abbreviated as rad). The angle given by a complete revolution contains , which is the same as rad. Therefore and EXAMPLE 1 (a) Find the radian measure of . (b) Express rad in degrees. SOLUTION (a) From Equation 1 or 2 we see that to convert from degrees to radians we multiply by . Therefore (b) To convert from radians to degrees we multiply by . Thus M In calculus we use radians to measure angles except when otherwise indicated. The fol-lowing table gives the correspondence between degree and radian measures of some com-mon angles. Figure 1 shows a sector of a circle with central angle and radius subtending an arc with length . Since the length of the arc is proportional to the size of the angle, and since the entire circle has circumference and central angle , we have Solving this equation for and for , we obtain Remember that Equations 3 are valid only when is measured in radians. a r a r 3 a 2 a 2r 2 2r a r 5 4 rad 5 4 180 225 180 60 60 180 3 rad 180 54 60 1 180 rad 0.017 rad 1 rad 180 57.3 2 rad 180 1 2 360 D A24 \|\|\|\| APPENDIX D TRIGONOMETRY Degrees 0° 30° 45° 60° 90° 120° 135° 150° 180° 270° 360° Radians 0 2 3 2 5 6 3 4 2 3 2 3 4 6 r r a ¨ FIGURE 1 In particular, putting in Equation 3, we see that an angle of 1 rad is the angle sub-tended at the center of a circle by an arc equal in length to the radius of the circle (see Figure 2). EXAMPLE 2 (a) If the radius of a circle is 5 cm, what angle is subtended by an arc of 6 cm? (b) If a circle has radius 3 cm, what is the length of an arc subtended by a central angle of rad? SOLUTION (a) Using Equation 3 with and , we see that the angle is (b) With cm and rad, the arc length is M The standard position of an angle occurs when we place its vertex at the origin of a coordinate system and its initial side on the positive -axis as in Figure 3. A positive angle is obtained by rotating the initial side counterclockwise until it coincides with the termi-nal side. Likewise, negative angles are obtained by clockwise rotation as in Figure 4. Figure 5 shows several examples of angles in standard position. Notice that different angles can have the same terminal side. For instance, the angles , , and have the same initial and terminal sides because and rad represents a complete revolution. FIGURE 5 Angles in standard position y x 0 ¨=_ 5π 4 0 y x ¨=11π 4 0 y x ¨=3π 4 0 y x ¨=_ π 2 0 y x ¨=1 2 3 4 2 11 4 3 4 2 5 4 114 54 34 0 y x ¨ initial side terminal side FIGURE 3 ¨˘0 0 y x ¨ initial side terminal side FIGURE 4 ¨<0 x a r 3 3 8 9 8 cm 38 r 3 6 5 1.2 rad r 5 a 6 38 a r APPENDIX D TRIGONOMETRY \|\|\|\| A25 r r r 1 rad FIGURE 2 THE TRIGONOMETRIC FUNCTIONS For an acute angle the six trigonometric functions are deﬁned as ratios of lengths of sides of a right triangle as follows (see Figure 6). This deﬁnition doesn’t apply to obtuse or negative angles, so for a general angle in standard position we let be any point on the terminal side of and we let be the distance as in Figure 7. Then we deﬁne Since division by 0 is not deﬁned, and are undeﬁned when and and are undeﬁned when . Notice that the deﬁnitions in (4) and (5) are consis-tent when is an acute angle. If is a number, the convention is that means the sine of the angle whose radian measure is . For example, the expression implies that we are dealing with an angle of 3 rad. When ﬁnding a calculator approximation to this number, we must remember to set our calculator in radian mode, and then we obtain If we want to know the sine of the angle we would write and, with our calculator in degree mode, we ﬁnd that The exact trigonometric ratios for certain angles can be read from the triangles in Fig-ure 8. For instance, tan 4 1 tan 6 1 s3 tan 3 s3 cos 4 1 s2 cos 6 s3 2 cos 3 1 2 sin 4 1 s2 sin 6 1 2 sin 3 s3 2 sin 3 0.05234 sin 3 3 sin 3 0.14112 sin 3 sin y 0 cot csc x 0 sec tan cot x y tan y x sec r x cos x r csc r y sin y r 5 OP r Px, y cot adj opp tan opp adj sec hyp adj cos adj hyp csc hyp opp sin opp hyp 4 A26 \|\|\|\| APPENDIX D TRIGONOMETRY opposite hypotenuse adjacent ¨ FIGURE 6 P(x, y) O y x r ¨ FIGURE 7 1 1 2 œ„ π 4 π 4 1 2 π 3 œ„ 3 π 6 FIGURE 8 The signs of the trigonometric functions for angles in each of the four quadrants can be remembered by means of the rule “All Students Take Calculus” shown in Figure 9. EXAMPLE 3 Find the exact trigonometric ratios for . SOLUTION From Figure 10 we see that a point on the terminal line for is . Therefore, taking in the deﬁnitions of the trigonometric ratios, we have M The following table gives some values of and found by the method of Example 3. EXAMPLE 4 If and , ﬁnd the other ﬁve trigonometric functions of . SOLUTION Since , we can label the hypotenuse as having length 5 and the adjacent side as having length 2 in Figure 11. If the opposite side has length , then the Pythagorean Theorem gives and so , . We can now use the diagram to write the other ﬁve trigonometric functions: M EXAMPLE 5 Use a calculator to approximate the value of in Figure 12. SOLUTION From the diagram we see that Therefore M x 16 tan 40 19.07 tan 40 16 x x cot 2 s21 sec 5 2 csc 5 s21 tan s21 2 sin s21 5 x s21 x 2 21 x 2 4 25 x cos 2 5 0 2 cos 2 5 cos sin csc 2 3 2 s3 sec 2 3 2 cot 2 3 1 s3 sin 2 3 s3 2 cos 2 3 1 2 tan 2 3 s3 r 2 y s3 x 1 P(1, s3 ) 23 23 APPENDIX D TRIGONOMETRY \|\|\|\| A27 0 0 1 0 0 1 0 0 1 1 s3 2 1 s2 1 2 1 2 1 s2 s3 2 cos 1 1 2 1 s2 s3 2 s3 2 1 s2 1 2 sin 2 3 2 5 6 3 4 2 3 2 3 4 6 0 y x sin ¨>0 tan ¨>0 all ratios>0 cos ¨>0 FIGURE 9 y 0 x 2π 3 π 3 2 œ„ 3 1 P{_1, œ„ 3} FIGURE 10 16 40° x FIGURE 12 5 2 ¨ x=œ„„ 21 FIGURE 11 TRIGONOMETRIC IDENTITIES A trigonometric identity is a relationship among the trigonometric functions. The most ele-mentary are the following, which are immediate consequences of the deﬁnitions of the trig-onometric functions. For the next identity we refer back to Figure 7. The distance formula (or, equivalently, the Pythagorean Theorem) tells us that . Therefore We have therefore proved one of the most useful of all trigonometric identities: If we now divide both sides of Equation 7 by and use Equations 6, we get Similarly, if we divide both sides of Equation 7 by , we get The identities show that is an odd function and is an even function. They are easily proved by drawing a diagram showing and in standard position (see Exercise 39). Since the angles and have the same terminal side, we have These identities show that the sine and cosine functions are periodic with period . The remaining trigonometric identities are all consequences of two basic identities called the addition formulas: 2 cos 2 cos sin 2 sin 11 2 cos sin cos cos 10b sin sin 10a 1 cot2 csc2 9 sin2 tan2 1 sec2 8 cos2 sin2 cos2 1 7 sin2 cos2 y 2 r 2 x 2 r 2 x 2 y 2 r 2 r 2 r 2 1 x 2 y 2 r 2 cot cos sin tan sin cos cot 1 tan sec 1 cos csc 1 sin 6 A28 \|\|\|\| APPENDIX D TRIGONOMETRY N Odd functions and even functions are discussed in Section 1.1. The proofs of these addition formulas are outlined in Exercises 85, 86, and 87. By substituting for in Equations 12a and 12b and using Equations 10a and 10b, we obtain the following subtraction formulas: Then, by dividing the formulas in Equations 12 or Equations 13, we obtain the corre-sponding formulas for : If we put in the addition formulas (12), we get the double-angle formulas: Then, by using the identity , we obtain the following alternate forms of the double-angle formulas for : If we now solve these equations for and , we get the following half-angle for-mulas, which are useful in integral calculus: Finally, we state the product formulas, which can be deduced from Equations 12 and 13: sin2x 1 cos 2x 2 17b cos2x 1 cos 2x 2 17a sin2x cos2x cos 2x 1 2 sin2x 16b cos 2x 2 cos2x 1 16a cos 2x sin2x cos2x 1 cos 2x cos2x sin2x 15b sin 2x 2 sin x cos x 15a y x tanx y tan x tan y 1 tan x tan y 14b tanx y tan x tan y 1 tan x tan y 14a tanx y cosx y cos x cos y sin x sin y 13b sinx y sin x cos y cos x sin y 13a y y cosx y cos x cos y sin x sin y 12b sinx y sin x cos y cos x sin y 12a APPENDIX D TRIGONOMETRY \|\|\|\| A29 There are many other trigonometric identities, but those we have stated are the ones used most often in calculus. If you forget any of them, remember that they can all be deduced from Equations 12a and 12b. EXAMPLE 6 Find all values of in the interval such that . SOLUTION Using the double-angle formula (15a), we rewrite the given equation as Therefore, there are two possibilities: The given equation has ﬁve solutions: , , , , and . M GRAPHS OF THE TRIGONOMETRIC FUNCTIONS The graph of the function , shown in Figure 13(a), is obtained by plotting points for and then using the periodic nature of the function (from Equa-tion 11) to complete the graph. Notice that the zeros of the sine function occur at the FIGURE 13 y 1 _1 x π _π 2π 3π 0 _π 2 π 2 3π 2 5π 2 (b) ©=cos x y 1 _1 0 x π _π 2π 3π _π 2 π 2 3π 2 5π 2 (a) ƒ=sin x 0 x 2 fx sin x 2 53 3 0 x or x 3 , 5 3 x 0, , 2 or cos x 1 2 sin x 0 or 1 2 cos x 0 sin x1 2 cos x 0 or sin x 2 sin x cos x sin x sin 2x 0, 2 x sin x sin y 1 2cosx y cosx y 18c cos x cos y 1 2cosx y cosx y 18b sin x cos y 1 2sinx y sinx y 18a A30 \|\|\|\| APPENDIX D TRIGONOMETRY integer multiples of , that is, Because of the identity (which can be veriﬁed using Equation 12a), the graph of cosine is obtained by shifting the graph of sine by an amount to the left [see Figure 13(b)]. Note that for both the sine and cosine functions the domain is and the range is the closed interval . Thus, for all values of , we have The graphs of the remaining four trigonometric functions are shown in Figure 14 and their domains are indicated there. Notice that tangent and cotangent have range , whereas cosecant and secant have range . All four functions are peri-odic: tangent and cotangent have period , whereas cosecant and secant have period . FIGURE 14 (c) y=csc x y 1 _1 0 x π y=sin x _π 2 π 2 3π 2 (d) y=sec x y 0 x π _π _1 1 y=cos x _π 2 π 2 3π 2 (a) y=tan x (b) y=cot x y 0 x π _π _π 2 π 2 3π 2 y 1 _1 0 x π _π _π 2 π 2 3π 2 2 , 1 1, , 1 cos x 1 1 sin x 1 x 1, 1 , 2 cos x sinx 2 whenever x n, n an integer sin x 0 APPENDIX D TRIGONOMETRY \|\|\|\| A31 A32 \|\|\|\| APPENDIX D TRIGONOMETRY 34. , 35–38 Find, correct to ﬁve decimal places, the length of the side labeled . 35. 36. 37. 38. 39–41 Prove each equation. 39. (a) Equation 10a (b) Equation 10b 40. (a) Equation 14a (b) Equation 14b 41. (a) Equation 18a (b) Equation 18b (c) Equation 18c 42–58 Prove the identity. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. sin 1 cos csc cot sin2x sin2y sinx y sinx y sin x sin 2x cos x cos 2x cos x 1 1 sin 1 1 sin 2 sec2 tan 2 2 tan 1 tan2 2 csc 2t sec t csc t cot2 sec2 tan2 csc2 tan2 sin2 tan2 sin2 sec y cos y tan y sin y sin x cos x2 1 sin 2x sin cot cos sin x sin x sin 2 x cos x cos 2 x sin x 22 cm x 3π 8 8 cm x 2π 5 25 cm x 40° 10 cm x 35° x 3 2 2 csc 4 3 1–6 Convert from degrees to radians. 1. 2. 3. 4. 5. 6. 7–12 Convert from radians to degrees. 7. 8. 9. 10. 11. 12. 13. Find the length of a circular arc subtended by an angle of rad if the radius of the circle is 36 cm. 14. If a circle has radius 10 cm, ﬁnd the length of the arc subtended by a central angle of . 15. A circle has radius m. What angle is subtended at the center of the circle by an arc 1 m long? 16. Find the radius of a circular sector with angle and arc length 6 cm. 17–22 Draw, in standard position, the angle whose measure is given. 17. 18. 19. rad 20. rad 21. rad 22. rad 23–28 Find the exact trigonometric ratios for the angle whose radian measure is given. 23. 24. 25. 26. 27. 28. 29–34 Find the remaining trigonometric ratios. 29. , 30. , 31. , 32. , 33. , 2 cot 3 x 3 2 cos x 1 3 2 sec 1.5 0 2 tan 2 0 2 sin 3 5 11 4 5 6 5 9 2 4 3 3 4 3 2 7 3 3 4 150 315 34 1.5 72 12 5 3 8 8 3 5 12 7 2 4 36 900 315 9 300 210 EXERCISES D APPENDIX D TRIGONOMETRY \|\|\|\| A33 position as in the ﬁgure. Express and in terms of and then use the distance formula to compute .] 84. In order to ﬁnd the distance across a small inlet, a point is located as in the ﬁgure and the following measurements were recorded: m m Use the Law of Cosines from Exercise 83 to ﬁnd the required distance. 85. Use the ﬁgure to prove the subtraction formula [Hint: Compute in two ways (using the Law of Cosines from Exercise 83 and also using the distance formula) and compare the two expressions.] 86. Use the formula in Exercise 85 to prove the addition formula for cosine (12b). 87. Use the addition formula for cosine and the identities to prove the subtraction formula for the sine function. 88. Show that the area of a triangle with sides of lengths and and with included angle is 89. Find the area of triangle , correct to ﬁve decimal places, if cm cm ABC 107 BC 3 AB 10 ABC A 1 2ab sin b a sin 2 cos cos 2 sin 0 y B(cos ∫, sin ∫) ∫ 1 A(cos å, sin å) 1 å c x c2 cos cos cos sin sin A C B BC 910 AC 820 C 103 C AB c y x 56. 57. 58. 59–64 If and , where and lie between and , evaluate the expression. 59. 60. 61. 62. 63. 64. 65–72 Find all values of in the interval that satisfy the equation. 65. 66. 67. 68. 69. 70. 71. 72. 73–76 Find all values of in the interval that satisfy the inequality. 73. 74. 75. 76. 77–82 Graph the function by starting with the graphs in Figures 13 and 14 and applying the transformations of Section 1.3 where appropriate. 77. 78. 79. 80. 81. 82. 83. Prove the Law of Cosines: If a triangle has sides with lengths , , and , and is the angle between the sides with lengths and , then [Hint: Introduce a coordinate system so that is in standard 0 y P(x, y) ¨ c b (a, 0) x c 2 a 2 b 2 2ab cos b a c b a y 2 sinx 4 y sin x y 1 sec x y 1 3 tanx 2 y tan 2x y cosx 3 sin x cos x 1 tan x 1 2 cos x 1 0 sin x 1 2 0, 2 x 2 cos 2x 3 cos x sin x tan x 2 cos x sin 2x 0 sin 2x cos x tan x 1 2 sin2x 1 3 cot2x 1 2 cos x 1 0 0, 2 x cos 2y sin 2y sinx y cosx y cosx y sinx y 2 0 y x sec y 5 4 sin x 1 3 cos 3 4 cos3 3 cos sin 3 sin 2 sin 2 cos tan x tan y sinx y cos x cos y SIGMA NOTATION A convenient way of writing sums uses the Greek letter (capital sigma, corresponding to our letter S) and is called sigma notation. DEFINITION If are real numbers and and are integers such that then With function notation, Deﬁnition 1 can be written as Thus the symbol indicates a summation in which the letter (called the index of summation) takes on consecutive integer values beginning with m and ending with n, that is, . Other letters can also be used as the index of summation. EXAMPLE 1 (a) (b) (c) (d) (e) (f) M EXAMPLE 2 Write the sum in sigma notation. SOLUTION There is no unique way of writing a sum in sigma notation. We could write or or M The following theorem gives three simple rules for working with sigma notation. 23 33 n 3 n2 k0 k 23 23 33 n 3 n1 j1 j 13 23 33 n 3 n i2 i 3 23 33 n 3 4 i1 2 2 2 2 2 8 3 i1 i 1 i 2 3 1 1 12 3 2 1 22 3 3 1 32 3 0 1 7 1 6 13 42 n k1 1 k 1 1 2 1 3 1 n 5 j0 2 j 20 21 22 23 24 25 63 n i3 i 3 4 5 n 1 n 4 i1 i 2 12 22 32 42 30 m, m 1, . . . , n i n im n im fi fm fm 1 fm 2 fn 1 fn n im ai am am1 am2 an1 an m n, n m am, am1, . . . , an 1 E A34 \|\|\|\| APPENDIX E SIGMA NOTATION This tells us to end with i=n. This tells us to add. This tells us to start with i=m. μ ai n im THEOREM If is any constant (that is, it does not depend on ), then (a) (b) (c) PROOF To see why these rules are true, all we have to do is write both sides in expanded form. Rule (a) is just the distributive property of real numbers: Rule (b) follows from the associative and commutative properties: Rule (c) is proved similarly. M EXAMPLE 3 Find SOLUTION M EXAMPLE 4 Prove the formula for the sum of the ﬁrst positive integers: SOLUTION This formula can be proved by mathematical induction (see page 77) or by the following method used by the German mathematician Karl Friedrich Gauss (1777–1855) when he was ten years old. Write the sum twice, once in the usual order and once in reverse order: Adding all columns vertically, we get On the right side there are terms, each of which is , so M EXAMPLE 5 Prove the formula for the sum of the squares of the ﬁrst positive integers: n i1 i 2 12 22 32 n 2 nn 12n 1 6 n S nn 1 2 or 2S nn 1 n 1 n 2S n 1 n 1 n 1 n 1 n 1 S n n 1 n 2 2 1 S 1 2 3 n 1 n S n i1 i 1 2 3 n nn 1 2 n n i1 1 1 1 1 n n i1 1. am am1 an bm bm1 bn am bm am1 bm1 an bn cam cam1 can cam am1 an n im ai bi n im ai n im bi n im ai bi n im ai n im bi n im cai c n im ai i c 2 APPENDIX E SIGMA NOTATION \|\|\|\| A35 n terms SOLUTION 1 Let be the desired sum. We start with the telescoping sum (or collapsing sum): On the other hand, using Theorem 2 and Examples 3 and 4, we have Thus we have Solving this equation for , we obtain or SOLUTION 2 Let be the given formula. 1. is true because 2. Assume that is true; that is, Then So is true. By the Principle of Mathematical Induction, is true for all . M n Sn Sk1 k 1k 1 1 2k 1 1 6 k 1k 22k 3 6 k 1 2k 2 7k 6 6 k 1 k2k 1 6k 1 6 kk 12k 1 6 k 12 12 22 32 k 12 12 22 32 k 2 k 12 12 22 32 k 2 kk 12k 1 6 Sk 12 11 12 1 1 6 S1 Sn S 2n 3 3n 2 n 6 nn 12n 1 6 3S n 3 3 2n 2 1 2n S n 3 3n 2 3n 3S 3 2n 2 5 2n 3S 3 nn 1 2 n 3S 3 2n 2 5 2n n i1 1 i3 i 3 n i1 3i 2 3i 1 3 n i1 i 2 3 n i1 i n i1 1 n 13 13 n 3 3n 2 3n n i1 1 i3 i 3 23 13 33 23 43 33 n 13 n 3 S A36 \|\|\|\| APPENDIX E SIGMA NOTATION Most terms cancel in pairs. N See pages 77 and 80 for a more thorough discussion of mathematical induction. N PRINCIPLE OF MATHEMATICAL INDUCTION Let be a statement involving the positive inte-ger . Suppose that 1. is true. 2. If is true, then is true. Then is true for all positive integers . n Sn Sk1 Sk S1 n Sn We list the results of Examples 3, 4, and 5 together with a similar result for cubes (see Exercises 37–40) as Theorem 3. These formulas are needed for ﬁnding areas and evalu-ating integrals in Chapter 5. THEOREM Let be a constant and a positive integer. Then (a) (b) (c) (d) (e) EXAMPLE 6 Evaluate . SOLUTION Using Theorems 2 and 3, we have M EXAMPLE 7 Find . SOLUTION M 1 2 1 1 2 3 4 lim nl 1 2 11 1 n2 1 n 3 lim nl 1 2 n n n 1 n 2n 1 n 3 lim nl 3 n 3 nn 12n 1 6 3 n n lim n l 3 n 3 n i1 i 2 3 n n i1 1 lim nl n i1 3 n i n 2 1 lim nl n i1 3 n 3 i 2 3 n lim n l n i1 3 n i n 2 1 nn 12n 2 2n 3 2 nn 12nn 1 3 2 4 nn 1 2 2 3 nn 1 2 n i1 i4i 2 3 n i1 4i 3 3i 4 n i1 i 3 3 n i1 i n i1 i4i 2 3 n i1 i 3 nn 1 2 2 n i1 i 2 nn 12n 1 6 n i1 i nn 1 2 n i1 c nc n i1 1 n n c 3 APPENDIX E SIGMA NOTATION \|\|\|\| A37 N The type of calculation in Example 7 arises in Chapter 5 when we compute areas. A38 \|\|\|\| APPENDIX E SIGMA NOTATION 35. 36. Find the number such that . 37. Prove formula (b) of Theorem 3. 38. Prove formula (e) of Theorem 3 using mathematical induction. 39. Prove formula (e) of Theorem 3 using a method similar to that of Example 5, Solution 1 [start with . 40. Prove formula (e) of Theorem 3 using the following method published by Abu Bekr Mohammed ibn Alhusain Alkarchi in about AD 1010. The ﬁgure shows a square in which sides and have been divided into segments of lengths , , , . . . , Thus the side of the square has length so the area is . But the area is also the sum of the areas of the n “gnomons” , , . . . , shown in the ﬁgure. Show that the area of is and conclude that formula (e) is true. 41. Evaluate each telescoping sum. (a) (b) (c) (d) 42. Prove the generalized triangle inequality: 43–46 Find the limit. 43. 44. 45. lim n l n i1 2 n 2i n 3 5 2i n lim n l n i1 1 n i n 3 1 lim n l n i1 1 n i n 2 n i1 ai n i1 ai n i1 ai ai1 99 i3 1 i 1 i 1 100 i1 5i 5i1 n i1 i 4 i 14 1 2 3 4 5 . . . n B A 1 2 3 4 5 n D . . . C Gn G™ G£ G¢ G∞ . . . i 3 Gi Gn G2 G1 nn 12 2 nn 12 n. 3 2 1 AD AB ABCD 1 i4 i 4 n i1 i 78 n n i1 i 3 i 2 1–10 Write the sum in expanded form. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11–20 Write the sum in sigma notation. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21–35 Find the value of the sum. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. n i1 ii 1i 2 n i1 i 1i 2 n i1 3 2i2 n i1 i 2 3i 4 n i1 2 5i n i1 2i 4 i2 23i 4 i0 2i i 2 100 i1 4 20 n1 1n 8 k0 cos k 6 j1 3 j1 6 i3 ii 2 8 i4 3i 2 1 x x 2 x 3 1nx n x x 2 x 3 x n 1 1 1 4 1 9 1 16 1 25 1 36 1 2 4 8 16 32 1 3 5 7 2n 1 2 4 6 8 2n 3 7 4 8 5 9 6 10 23 27 1 2 2 3 3 4 4 5 19 20 s3 s4 s5 s6 s7 1 2 3 4 10 n i1 f xi xi n1 j0 1 j n3 jn j 2 n i1 i 10 8 k5 x k 4 k0 2k 1 2k 1 6 i4 i 3 6 i4 3i 6 i1 1 i 1 5 i1 si EXERCISES E APPENDIX F PROOFS OF THEOREMS \|\|\|\| A39 48. Evaluate . 49. Evaluate . 50. Evaluate . m i1 n j1 i j n i1 2i 2i n i1 3 2i1 46. 47. Prove the formula for the sum of a ﬁnite geometric series with ﬁrst term and common ratio : n i1 ar i1 a ar ar 2 ar n1 ar n 1 r 1 r 1 a lim n l n i1 3 n 1 3i n 3 21 3i n PROOFS OF THEOREMS In this appendix we present proofs of several theorems that are stated in the main body of the text. The sections in which they occur are indicated in the margin. SECTION 2.3 LIMIT LAWS Suppose that is a constant and the limits and exist. Then 1. 2. 3. 4. 5. if PROOF OF LAW 4 Let be given. We want to ﬁnd such that In order to get terms that contain and , we add and subtract as follows: (Triangle Inequality) We want to make each of these terms less than . Since , there is a number such that Also, there is a number such that if , then and therefore tx tx M M tx M M 1 M tx M 1 0 x a 2 2 0 tx M 2(1 L) then 0 x a 1 if 1 0 lim x l a tx M 2 fx Ltx Ltx M fx L tx Ltx M fx L tx Ltx M fxtx LM fxtx Ltx Ltx LM Ltx tx M fx L fxtx LM then 0 x a if 0 0 M 0 lim x l a fx tx L M lim xla fxtx LM lim x l a cfx cL lim x l a fx tx L M lim x l a fx tx L M lim x l a tx M lim x l a fx L c F Since , there is a number such that Let min . If , then we have , , and , so we can combine the inequalities to obtain This shows that . M PROOF OF LAW 3 If we take in Law 4, we get (by Law 7) M PROOF OF LAW 2 Using Law 1 and Law 3 with , we have M PROOF OF LAW 5 First let us show that To do this we must show that, given , there exists such that Observe that We know that we can make the numerator small. But we also need to know that the denominator is not small when is near . Since , there is a number such that, whenever , we have and therefore M 2 tx M M tx tx M tx tx tx M M 2 0 x a 1 1 0 lim x l a tx M a x 1 tx 1 M M tx Mtx 1 tx 1 M then 0 x a if 0 0 lim x l a 1 tx 1 M lim x l a fx 1 lim x l a tx lim x l a fx lim x l a tx lim x l a fx tx lim x l a fx 1tx lim x l a fx lim x l a 1tx c 1 c lim x l a fx lim x l a c lim x l a fx lim x l a cfx lim x l a txfx lim x l a tx lim x l a fx tx c lim x l a fxtx LM 2 2 2(1 M) (1 M) L 2(1 L) fxtx LM fx Ltx Ltx M 0 x a 3 0 x a 2 0 x a 1 0 x a 1, 2, 3 fx L 2(1 M) then 0 x a 3 if 3 0 lim x l a fx L A40 \|\|\|\| APPENDIX F PROOFS OF THEOREMS This shows that and so, for these values of , Also, there exists such that Let min . Then, for , we have It follows that . Finally, using Law 4, we obtain M THEOREM If for all in an open interval that contains (except possibly at ) and and then . PROOF We use the method of proof by contradiction. Suppose, if possible, that . Law 2 of limits says that Therefore, for any , there exists such that In particular, taking (noting that by hypothesis), we have a number such that then Since for any number , we have then which simpliﬁes to then But this contradicts . Thus the inequality must be false. Therefore . M L M L M fx tx tx fx 0 x a if tx fx M L L M 0 x a if a a a tx fx M L L M 0 x a if 0 L M 0 L M tx fx M L then 0 x a if 0 0 lim x l a tx fx M L L M L M lim x l a tx M lim x l a fx L a a x fx tx 2 lim x l a fx lim x l a 1 tx L 1 M L M lim x l a fx tx lim x l a fx 1 tx lim x l a 1tx 1M 1 tx 1 M M tx Mtx 2 M 2 M 2 2 0 x a 1, 2 tx M M 2 2 then 0 x a 2 if 2 0 1 Mtx 1 Mtx 1 M 2 M 2 M 2 x tx M 2 then 0 x a 1 if APPENDIX F PROOFS OF THEOREMS \|\|\|\| A41 THE SQUEEZE THEOREM If for all in an open interval that contains (except possibly at ) and then PROOF Let be given. Since , there is a number such that if then that is, if then Since , there is a number such that then that is, then Let min . If , then and , so In particular, and so . Therefore . M SECTION 2.5 THEOREM If is a one-to-one continuous function deﬁned on an interval , then its inverse function is also continuous. PROOF First we show that if is both one-to-one and continuous on , then it must be either increasing or decreasing on . If it were neither increasing nor decreasing, then there would exist numbers , , and in with such that does not lie between and . There are two possibilities: either (1) lies between and or (2) lies between and . (Draw a picture.) In case (1) we apply the Intermediate Value Theorem to the continuous function to get a number between and such that . In case (2) the Intermediate Value Theorem gives a number between and such that . In either case we have contradicted the fact that is one-to-one. Let us assume, for the sake of deﬁniteness, that is increasing on . We take any number in the domain of and we let ; that is, is the number in such that . To show that is continuous at we take any such that the interval is contained in the interval . Since is increasing, it maps the numbers in the interval onto the numbers in the interval and reverses the correspondence. If we let denote the smaller of the numbers and , then the interval is contained in the interval and so is mapped into the interval by . (See the arrow diagram in Figure 1.) We have f 1 x0 , x0 fx0 , fx0 y0 , y0 2 fx0 y0 1 y0 fx0 f 1 fx0 , fx0 x0 , x0 f a, b x0 , x0 0 y0 f 1 fx0 y0 a, b x0 f 1y0 x0 f 1 y0 a, b f f fc fx1 x3 x2 c fc fx3 x2 x1 c f fx3 fx2 fx1 fx2 fx1 fx3 fx3 fx1 fx2 x1 x2 x3 a, b x3 x2 x1 a, b a, b f f 1 a, b f lim x l a tx L tx L L tx L L fx tx hx L 0 x a 2 0 x a 1 0 x a 1, 2 L hx L 0 x a 2 if hx L 0 x a 2 if 2 0 lim x l a hx L L fx L 0 x a 1 fx L 0 x a 1 1 0 lim x l a fx L 0 lim x l a tx L lim x l a fx lim x l a hx L a a x fx tx hx 3 A42 \|\|\|\| APPENDIX F PROOFS OF THEOREMS therefore found a number such that This shows that and so is continuous at any number in its domain. M THEOREM If is continuous at and , then PROOF Let be given. We want to ﬁnd a number such that then Since is continuous at , we have and so there exists such that then Since , there exists such that then Combining these two statements, we see that whenever we have , which implies that . Therefore we have proved that . M SECTION 3.3 The proof of the following result was promised when we proved that . THEOREM If , then . PROOF Figure 2 shows a sector of a circle with center , central angle , and radius 1. Then We approximate the arc by an inscribed polygon consisting of equal line segments n AB AD OA tan tan O tan 0 2 lim l 0 sin 1 limx l a ftx fb ftx fb tx b 1 0 x a tx b 1 0 x a if 0 lim x l a tx b fy fb 0 y b 1 if 1 0 lim y l b fy fb b f ftx fb 0 x a if 0 0 lim x l a ftx fb lim x l a tx b b f 8 y0 f 1 lim y l y0 f 1y f 1y0 x y x¸ y¸ f f –! b a f(x¸-∑) f(x¸+∑) x¸-∑ x¸+∑ ∂¡ ∂™ } { } { } { FIGURE 1 f f 1y f 1y0 then y y0 if 0 APPENDIX F PROOFS OF THEOREMS \|\|\|\| A43 and we look at a typical segment . We extend the lines and to meet in the points and . Then we draw as in Figure 2. Observe that and so . Therefore we have If we add such inequalities, we get where is the length of the inscribed polygon. Thus, by Theorem 2.3.2, we have But the arc length is deﬁned in Equation 8.1.1 as the limit of the lengths of inscribed polygons, so M SECTION 4.3 CONCAVITY TEST (a) If for all in , then the graph of is concave upward on . (b) If for all in , then the graph of is concave downward on . PROOF OF (a) Let be any number in . We must show that the curve lies above the tangent line at the point . The equation of this tangent is So we must show that whenever . (See Figure 3.) First let us take the case where . Applying the Mean Value Theorem to on the interval , we get a number , with , such that Since on , we know from the Increasing/Decreasing Test that is increasing on . Thus, since , we have and so, multiplying this inequality by the positive number , we get fax a fcx a 2 x a fa fc a c I f I f 0 fx fa fcx a 1 a c x c a, x f x a x a x I fx fa fax a y fa fax a a, fa y fx I a I f I x f x 0 I f I x f x 0 lim n l Ln tan lim n l Ln tan Ln Ln AD tan n PQ RT RS RTS 90 RTO PQO 90 RT PQ S R AD OQ OP PQ A44 \|\|\|\| APPENDIX F PROOFS OF THEOREMS Q T S ° ° B D ° ° P R A O 1 FIGURE 2 ¨ FIGURE 3 a x f(a)+fª(a)(x-a) ƒ y=ƒ x y 0 Now we add to both sides of this inequality: But from Equation 1 we have . So this inequality becomes which is what we wanted to prove. For the case where we have , but multiplication by the negative number reverses the inequality, so we get (2) and (3) as before. M SECTION 4.4 In order to give the promised proof of l’Hospital’s Rule, we ﬁrst need a generalization of the Mean Value Theorem. The following theorem is named after another French mathema-tician, Augustin-Louis Cauchy (1789–1857). CAUCHY’S MEAN VALUE THEOREM Suppose that the functions and are con-tinuous on and differentiable on , and for all in . Then there is a number in such that Notice that if we take the special case in which , then and Theorem 1 is just the ordinary Mean Value Theorem. Furthermore, Theorem 1 can be proved in a sim-ilar manner. You can verify that all we have to do is change the function given by Equa-tion 4.2.4 to the function and apply Rolle’s Theorem as before. L’HOSPITAL’S RULE Suppose and are differentiable and on an open interval that contains (except possibly at ). Suppose that and or that and (In other words, we have an indeterminate form of type or .) Then if the limit on the right side exists (or is or ). lim x l a fx tx lim x l a fx tx 0 0 lim x l a tx lim x l a fx lim x l a tx 0 lim x l a fx 0 a a I tx 0 t f hx fx fa fb fa tb ta tx ta h tc 1 tx x fc tc fb fa tb ta a, b c a, b x tx 0 a, b a, b t f 1 x a fc fa x a fx fa fax a 3 fx fa fcx a fa fax a fa fcx a fa APPENDIX F PROOFS OF THEOREMS \|\|\|\| A45 N See the biographical sketch of Cauchy on page 113. PROOF OF L’HOSPITAL’S RULE We are assuming that and . Let We must show that . Deﬁne Then is continuous on since is continuous on and Likewise, is continuous on . Let and . Then and are continuous on and differentiable on and there (since and ). There-fore, by Cauchy’s Mean Value Theorem, there is a number such that and Here we have used the fact that, by deﬁnition, and . Now, if we let , then (since ), so A similar argument shows that the left-hand limit is also . Therefore This proves l’Hospital’s Rule for the case where is ﬁnite. If is inﬁnite, we let . Then as , so we have (by l’Hospital’s Rule for ﬁnite a) M SECTION 11.8 In order to prove Theorem 11.8.3, we ﬁrst need the following results. THEOREM 1. If a power series converges when (where ), then it converges whenever . 2. If a power series diverges when (where ), then it diverges whenever . x d d 0 x d cnx n x b b 0 x b cnx n lim t l 0 f1t t1t lim x l fx tx lim t l 0 f1t1t 2 t1t1t 2 lim x l fx tx lim t l 0 f1t t1t x l t l 0 t 1x a a lim x l a fx tx L L lim x l a fx tx lim x l a Fx Gx lim yl a Fy Gy lim yl a fy ty L a y x y l a x l a Ga 0 Fa 0 Fy Gy Fx Fa Gx Ga Fx Gx a y x y G t F f G 0 a, x a, x G F x a x I I G lim x l a Fx lim x l a fx 0 Fa x Ix a f I F Fx fx 0 if x a if x a Gx tx 0 if x a if x a lim x l a fxtx L L lim x l a fx tx lim x l a tx 0 lim x l a fx 0 A46 \|\|\|\| APPENDIX F PROOFS OF THEOREMS PROOF OF 1 Suppose that converges. Then, by Theorem 11.2.6, we have . According to Deﬁnition 11.1.2 with , there is a positive integer such that whenever . Thus, for , we have If , then , so is a convergent geometric series. Therefore, by the Comparison Test, the series is convergent. Thus the series is absolutely convergent and therefore convergent. M PROOF OF 2 Suppose that diverges. If is any number such that , then cannot converge because, by part 1, the convergence of would imply the convergence of . Therefore diverges whenever . M THEOREM For a power series there are only three possibilities: 1. The series converges only when . 2. The series converges for all . 3. There is a positive number such that the series converges if and diverges if . PROOF Suppose that neither case 1 nor case 2 is true. Then there are nonzero numbers and such that converges for and diverges for . Therefore the set is not empty. By the preceding theorem, the series diverges if , so for all . This says that is an upper bound for the set . Thus, by the Completeness Axiom (see Section 11.1), has a least upper bound . If , then , so diverges. If , then is not an upper bound for and so there exists such that . Since , converges, so by the preceding theorem converges. M THEOREM For a power series there are only three possibilities: 1. The series converges only when . 2. The series converges for all . 3. There is a positive number such that the series converges if and diverges if . PROOF If we make the change of variable , then the power series becomes and we can apply the preceding theorem to this series. In case 3 we have con-vergence for and divergence for . Thus we have convergence for and divergence for . M x a R x a R u R u R cnu n u x a x a R x a R R x x a cnx an 3 cnx n cnb n b S b x b S S x x R cnx n x S x R R S S d x S x d x d S x cnx n converges x d x b cnx n d b x R x R R x x 0 cnx n x d cnx n cnd n cnx n cnx n x d x cnd n cnx n nN cnx n xb n xb 1 x b cnx n cnb nx n b n cnb n x b n x b n n N n N cnb n 1 N 1 lim n l cnb n 0 cnb n APPENDIX F PROOFS OF THEOREMS \|\|\|\| A47 THE LOGARITHM DEFINED AS AN INTEGRAL Our treatment of exponential and logarithmic functions until now has relied on our intuition, which is based on numerical and visual evidence. (See Sections 1.5, 1.6, and 3.1.) Here we use the Fundamental Theorem of Calculus to give an alternative treatment that provides a surer footing for these functions. Instead of starting with and deﬁning as its inverse, this time we start by deﬁn-ing as an integral and then deﬁne the exponential function as its inverse. You should bear in mind that we do not use any of our previous deﬁnitions and results concerning exponential and logarithmic functions. THE NATURAL LOGARITHM We ﬁrst deﬁne as an integral. DEFINITION The natural logarithmic function is the function deﬁned by The existence of this function depends on the fact that the integral of a continuous func-tion always exists. If , then can be interpreted geometrically as the area under the hyperbola from to . (See Figure 1.) For , we have For , and so is the negative of the area shown in Figure 2. EXAMPLE 1 (a) By comparing areas, show that . (b) Use the Midpoint Rule with to estimate the value of . SOLUTION (a) We can interpret as the area under the curve from 1 to 2. From Figure 3 we see that this area is larger than the area of rectangle and smaller than the area of trapezoid . Thus we have (b) If we use the Midpoint Rule with , and , we get M 0.1 1 1.05 1 1.15 1 1.95 0.693 ln 2 y 2 1 1 t dt 0.1 f1.05 f1.15 f1.95 t 0.1 ft 1t, n 10 1 2 ln 2 3 4 1 2 1 ln 2 1 1 2(1 1 2) ABCD BCDE y 1t ln 2 ln 2 n 10 1 2 ln 2 3 4 V ln x ln x y x 1 1 t dt y 1 x 1 t dt 0 0 x 1 ln 1 y 1 1 1 t dt 0 x 1 t x t 1 y 1t ln x x 1 x 0 ln x y x 1 1 t dt 1 ln x ln x loga x a x G A48 \|\|\|\| APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL FIGURE 2 FIGURE 1 FIGURE 3 y=1 t 0 y 1 x t area=ln x y=1 t 0 y 1 x t area=_ln x y=1 t 0 y 1 2 t A B C D E Notice that the integral that deﬁnes is exactly the type of integral discussed in Part 1 of the Fundamental Theorem of Calculus (see Section 5.3). In fact, using that theorem, we have and so We now use this differentiation rule to prove the following properties of the logarithm function. LAWS OF LOGARITHMS If and are positive numbers and is a rational number, then 1. 2. 3. PROOF 1. Let , where is a positive constant. Then, using Equation 2 and the Chain Rule, we have Therefore and have the same derivative and so they must differ by a constant: Putting in this equation, we get . Thus If we now replace the constant by any number , we have 2. Using Law 1 with , we have and so Using Law 1 again, we have The proof of Law 3 is left as an exercise. M ln x y lnx 1 y ln x ln 1 y ln x ln y ln 1 y ln y ln 1 y ln y ln 1 y y ln 1 0 x 1y lnxy ln x ln y y a lnax ln x ln a ln a ln 1 C 0 C C x 1 lnax ln x C ln x fx fx 1 ax d dx ax 1 ax a 1 x a fx lnax lnx r r ln x ln x y ln x ln y lnxy ln x ln y r y x 3 d dx ln x 1 x 2 d dx y x 1 1 t dt 1 x ln x APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL \|\|\|\| A49 In order to graph , we ﬁrst determine its limits: (a) (b) PROOF (a) Using Law 3 with and (where n is any positive integer), we have . Now , so this shows that as . But is an increasing function since its derivative . Therefore as . (b) If we let , then as . Thus, using (a), we have M If , then and which shows that is increasing and concave downward on . Putting this informa-tion together with (4), we draw the graph of in Figure 4. Since and is an increasing continuous function that takes on arbitrarily large values, the Intermediate Value Theorem shows that there is a number where takes on the value 1. (See Figure 5.) This important number is denoted by . DEFINITION is the number such that . We will show (in Theorem 19) that this deﬁnition is consistent with our previous deﬁ-nition of e. THE NATURAL EXPONENTIAL FUNCTION Since ln is an increasing function, it is one-to-one and therefore has an inverse function, which we denote by exp. Thus, according to the deﬁnition of an inverse function, and the cancellation equations are In particular, we have We obtain the graph of by reﬂecting the graph of about the line y x. y ln x y exp x exp1 e since ln e 1 exp0 1 since ln 1 0 expln x x and lnexp x x 7 expx y & ? ln y x 6 ln e 1 e 5 e ln x ln x ln 1 0 y ln x 0, ln x d 2y dx 2 1 x 2 0 dy dx 1 x 0 y ln x, x 0 lim x l 0 ln x lim t l ln 1 t lim t l ln t x l 0 t l t 1x x l ln x l 1x 0 ln x n l ln2n l ln 2 0 ln2n n ln 2 r n x 2 lim x l 0 ln x lim x l ln x 4 y ln x A50 \|\|\|\| APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL FIGURE 4 0 y x 1 y=ln x FIGURE 5 0 y 1 x 1 e y=ln x f 1x y & ? f y x f f 1x x f 1 f x x (See Figure 6.) The domain of is the range of ln, that is, ; the range of exp is the domain of ln, that is, . If is any rational number, then the third law of logarithms gives Therefore, by (6), Thus whenever is a rational number. This leads us to deﬁne , even for irra-tional values of , by the equation In other words, for the reasons given, we deﬁne to be the inverse of the function . In this notation (6) becomes and the cancellation equations (7) become The natural exponential function is one of the most frequently occurring functions in calculus and its applications, so it is important to be familiar with its graph (Figure 7) and its properties (which follow from the fact that it is the inverse of the natural logarithmic function). PROPERTIES OF THE EXPONENTIAL FUNCTION The exponential function is an increasing continuous function with domain and range . Thus for all . Also So the -axis is a horizontal asymptote of . We now verify that has the other properties expected of an exponential function. LAWS OF EXPONENTS If and are real numbers and is rational, then 1. 2. 3. e xr e rx e xy e x e y e xy e xe y r y x 11 f fx e x x lim x l e x lim x l e x 0 x e x 0 0, fx e x fx e x lne x x for all x 10 e ln x x x 0 9 e x y & ? ln y x 8 ln x e x e x expx x e x x expx e x expr e r lne r r ln e r r 0, , exp APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL \|\|\|\| A51 FIGURE 6 y 1 0 x y=x y=ln x y=exp x 1 y=´ x 0 1 y 1 FIGURE 7 The natural exponential function A52 \|\|\|\| APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL PROOF OF LAW 1 Using the ﬁrst law of logarithms and Equation 10, we have Since ln is a one-to-one function, it follows that Laws 2 and 3 are proved similarly (see Exercises 6 and 7). As we will soon see, Law 3 actually holds when is any real number. M We now prove the differentiation formula for . PROOF The function is differentiable because it is the inverse function of , which we know is differentiable with nonzero derivative. To ﬁnd its derivative, we use the inverse function method. Let . Then and, differentiating this latter equation implicitly with respect to , we get M GENERAL EXPONENTIAL FUNCTIONS If and is any rational number, then by (9) and (11), Therefore, even for irrational numbers , we deﬁne Thus, for instance, The function is called the exponential function with base a. Notice that is positive for all because is positive for all . Deﬁnition 13 allows us to extend one of the laws of logarithms. We already know that when is rational. But if we now let be any real number we have, from Deﬁnition 13, Thus ln a r r ln a for any real number r 14 ln a r lne r ln a r ln a r r lna r r ln a x e x x a x fx a x 2 s3 e s3 ln 2 e1.20 3.32 a x e x ln a 13 x a r e ln ar e r ln a r a 0 dy dx y e x 1 y dy dx 1 x ln y x y e x y ln x y e x d dx e x e x 12 e x r e xe y e xy. lne xe y lne x lne y x y lne xy The general laws of exponents follow from Deﬁnition 13 together with the laws of expo-nents for . LAWS OF EXPONENTS If and are real numbers and , , then 1. 2. 3. 4. PROOF 1. Using Deﬁnition 13 and the laws of exponents for , we have 3. Using Equation 14 we obtain The remaining proofs are left as exercises. M The differentiation formula for exponential functions is also a consequence of Deﬁni-tion 13: PROOF M If , then , so , which shows that is increasing (see Figure 8). If , then and so is decreasing (see Figure 9). GENERAL LOGARITHMIC FUNCTIONS If and , then is a one-to-one function. Its inverse function is called the logarithmic function with base a and is denoted by . Thus In particular, we see that loge x ln x loga x y & ? a y x 17 loga fx a x a 1 a 0 y a x ln a 0 0 a 1 y a x ddx a x a x ln a 0 ln a 0 a 1 a x ln a d dx a x d dx e x ln a e x ln a d dx x ln a d dx a x a x ln a 16 e xy ln a a xy a xy e y lnax e yx ln a e x ln ae y ln a a xa y a xy e xy ln a e x ln a y ln a e x abx a xb x a xy a xy a xy a xa y a xy a xa y b 0 a y x 15 e x APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL \|\|\|\| A53 FIGURE 8 y=a®, a>1 x lim a®=0, lim a®= x x 0 y 1 FIGURE 9 y=a®, 0<a<1 x lim a®=, lim a®=0 x _ x ` 0 y 1 (b) Use part (a) to show that . By comparing areas, show that 4. (a) By comparing areas, show that . (b) Deduce that . 2 e 3 ln 2 1 ln 3 1 2 1 3 1 n ln n 1 1 2 1 3 1 n 1 3. ln 2 0.66 (a) By comparing areas, show that (b) Use the Midpoint Rule with to estimate . 2. Refer to Example 1. (a) Find the equation of the tangent line to the curve that is parallel to the secant line . AD y 1t ln 1.5 n 10 1 3 ln 1.5 5 12 1. EXERCISES G The laws of logarithms are similar to those for the natural logarithm and can be deduced from the laws of exponents (see Exercise 10). To differentiate , we write the equation as . From Equation 14 we have , so Since is a constant, we can differentiate as follows: THE NUMBER e EXPRESSED AS A LIMIT In this section we deﬁned e as the number such that ln e 1. The next theorem shows that this is the same as the number e deﬁned in Section 3.1 (see Equation 3.6.5). PROOF Let . Then , so . But, by the deﬁnition of derivative, Because , we have Then, by Theorem 2.5.8 and the continuity of the exponential function, we have M e e1 elimx l 0 ln1x1x lim x l 0 eln1x1x lim x l 0 1 x1x lim x l 0 ln1 x1x 1 f1 1 lim x l 0 ln1 x ln 1 x lim x l 0 1 x ln1 x lim x l 0 ln1 x1x f1 lim hl0 f1 h f1 h lim xl0 f1 x f1 x f1 1 fx 1x fx ln x e lim x l 0 1 x1x 19 d dx loga x 1 x ln a 18 d dx loga x d dx ln x ln a 1 ln a d dx ln x 1 x ln a ln a loga x y ln x ln a y ln a ln x a y x y loga x A54 \|\|\|\| APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL APPENDIX H COMPLEX NUMBERS \|\|\|\| A55 COMPLEX NUMBERS A complex number can be represented by an expression of the form , where and are real numbers and is a symbol with the property that . The complex num-ber can also be represented by the ordered pair and plotted as a point in a plane (called the Argand plane) as in Figure 1. Thus the complex number is identiﬁed with the point . The real part of the complex number is the real number and the imaginary part is the real number . Thus the real part of is and the imaginary part is . Two complex numbers and are equal if and ; that is, their real parts are equal and their imaginary parts are equal. In the Argand plane the horizontal axis is called the real axis and the vertical axis is called the imaginary axis. The sum and difference of two complex numbers are deﬁned by adding or subtracting their real parts and their imaginary parts: For instance, The product of complex numbers is deﬁned so that the usual commutative and distributive laws hold: Since , this becomes EXAMPLE 1 M Division of complex numbers is much like rationalizing the denominator of a rational expression. For the complex number , we deﬁne its complex conjugate to be . To ﬁnd the quotient of two complex numbers we multiply numerator and denominator by the complex conjugate of the denominator. EXAMPLE 2 Express the number in the form . a bi 1 3i 2 5i z a bi z a bi 2 5i 6i 151 13 11i 1 3i2 5i 12 5i 3i2 5i a bic di ac bd ad bci i 2 1 ac adi bci bdi 2 a bic di ac di bic di 1 i 4 7i 1 4 1 7i 5 6i a bi c di a c b di a bi c di a c b di b d a c c di a bi 3 4 4 3i b a a bi 0, 1 i 0 1 i a, b a bi i 2 1 i b a a bi H FIGURE 1 Complex numbers as points in the Argand plane Re Im 0 i 2-2i _i 3-2i 2+3i _4+2i 1 9. Prove the fourth law of exponents [see (15)]. 10. Deduce the following laws of logarithms from (15): (a) (b) (c) logax y y loga x logaxy loga x loga y logaxy loga x loga y Prove the third law of logarithms. [Hint: Start by showing that both sides of the equation have the same derivative.] 6. Prove the second law of exponents for [see (11)]. 7. Prove the third law of exponents for [see (11)]. 8. Prove the second law of exponents [see (15)]. e x e x 5. SOLUTION We multiply numerator and denominator by the complex conjugate of , namely , and we take advantage of the result of Example 1: M The geometric interpretation of the complex conjugate is shown in Figure 2: is the reﬂection of in the real axis. We list some of the properties of the complex conjugate in the following box. The proofs follow from the deﬁnition and are requested in Exercise 18. PROPERTIES OF CONJUGATES The modulus, or absolute value, of a complex number is its distance from the origin. From Figure 3 we see that if , then Notice that and so This explains why the division procedure in Example 2 works in general: Since , we can think of as a square root of . But notice that we also have and so is also a square root of . We say that is the principal square root of and write . In general, if is any positive number, we write With this convention, the usual derivation and formula for the roots of the quadratic equa-tion are valid even when : EXAMPLE 3 Find the roots of the equation . SOLUTION Using the quadratic formula, we have M x 1 s1 2 4 1 2 1 s3 2 1 s3 i 2 x 2 x 1 0 x b sb 2 4ac 2a b 2 4ac 0 ax 2 bx c 0 sc sc i c s1 i 1 i 1 i i2 i 2 1 1 i i 2 1 z w zw ww zw w 2 zz z 2 zz a bia bi a 2 abi abi b 2i 2 a 2 b 2 z sa 2 b 2 z a bi z a bi z z n z n zw z w z w z w z z 1 3i 2 5i 1 3i 2 5i 2 5i 2 5i 13 11i 22 52 13 29 11 29 i 2 5i 2 5i A56 \|\|\|\| APPENDIX H COMPLEX NUMBERS FIGURE 3 Re Im 0 bi a b z=a+bi \|z\|= a@+b@ œ„„„„„ Re Im 0 i _i z=a-bi – z=a+bi FIGURE 2 We observe that the solutions of the equation in Example 3 are complex conjugates of each other. In general, the solutions of any quadratic equation with real coefﬁcients , , and are always complex conjugates. (If is real, , so is its own conjugate.) We have seen that if we allow complex numbers as solutions, then every quadratic equa-tion has a solution. More generally, it is true that every polynomial equation of degree at least one has a solution among the complex numbers. This fact is known as the Fundamental Theorem of Algebra and was proved by Gauss. POLAR FORM We know that any complex number can be considered as a point and that any such point can be represented by polar coordinates with . In fact, as in Figure 4. Therefore we have Thus we can write any complex number in the form where The angle is called the argument of and we write . Note that is not unique; any two arguments of differ by an integer multiple of . EXAMPLE 4 Write the following numbers in polar form. (a) (b) SOLUTION (a) We have and , so we can take . Therefore the polar form is (b) Here we have and . Since lies in the fourth quadrant, we take and The numbers and are shown in Figure 5. M w z w 2cos 6 i sin 6 6 w tan 1s3 r w s3 1 2 z s2 cos 4 i sin 4 4 tan 1 r z s12 12 s2 w s3 i z 1 i 2 z argz argz z tan b a and r z sa 2 b 2 z rcos i sin z z a bi r cos r sin i b r sin a r cos r 0 r, a, b z a bi anx n an1x n1 a1x a0 0 z z z z c b a ax 2 bx c 0 APPENDIX H COMPLEX NUMBERS \|\|\|\| A57 Re Im 0 a+bi b ¨ r a FIGURE 4 Re Im 0 œ„ 3-i 2 1+i œ„ 2 π 4 _π 6 FIGURE 5 The polar form of complex numbers gives insight into multiplication and division. Let be two complex numbers written in polar form. Then Therefore, using the addition formulas for cosine and sine, we have This formula says that to multiply two complex numbers we multiply the moduli and add the arguments. (See Figure 6.) A similar argument using the subtraction formulas for sine and cosine shows that to divide two complex numbers we divide the moduli and subtract the arguments. In particular, taking and (and therefore and ), we have the fol-lowing, which is illustrated in Figure 7. EXAMPLE 5 Find the product of the complex numbers and in polar form. SOLUTION From Example 4 we have and So, by Equation 1, This is illustrated in Figure 8. M 2s2 cos 12 i sin 12 1 i(s3 i) 2s2 cos 4 6 i sin 4 6 s3 i 2cos 6 i sin 6 1 i s2 cos 4 i sin 4 s3 i 1 i 1 z 1 r cos i sin . then z rcos i sin , If 2 1 0 z2 z z1 1 z2 0 z1 z2 r1 r2 cos 1 2 i sin 1 2 z1z2 r1r2cos 1 2 i sin 1 2 1 r1r2cos 1 cos 2 sin 1 sin 2 isin 1 cos 2 cos 1 sin 2 z1z2 r1r2cos 1 i sin 1cos 2 i sin 2 z2 r2cos 2 i sin 2 z1 r1cos 1 i sin 1 A58 \|\|\|\| APPENDIX H COMPLEX NUMBERS Re Im 0 r z ¨ ¨ 1 r 1 z FIGURE 7 z¡ FIGURE 6 Re Im z¡z™ ¨¡+¨™ z™ ¨¡ ¨™ 0 2 z=1+i w=œ„ 3-i zw 2œ„ 2 œ„ 2 FIGURE 8 Re Im π 12 Repeated use of Formula 1 shows how to compute powers of a complex number. If then and In general, we obtain the following result, which is named after the French mathematician Abraham De Moivre (1667–1754). DE MOIVRE’S THEOREM If and is a positive integer, then This says that to take the nth power of a complex number we take the nth power of the modulus and multiply the argument by n. EXAMPLE 6 Find . SOLUTION Since , it follows from Example 4(a) that has the polar form So by De Moivre’s Theorem, M De Moivre’s Theorem can also be used to ﬁnd the th roots of complex numbers. An th root of the complex number is a complex number such that Writing these two numbers in trigonometric form as and using De Moivre’s Theorem, we get The equality of these two complex numbers shows that and sin n sin and cos n cos s r 1n or s n r s ncos n i sin n rcos i sin z rcos i sin and w scos i sin wn z w z n n 25 210 cos 5 2 i sin 5 2 1 32 i 1 2 1 2 i 10 s2 2 10cos 10 4 i sin 10 4 1 2 1 2 i s2 2 cos 4 i sin 4 1 2 1 2i 1 2 1 2i 1 21 i ( 1 2 1 2i) 10 z n rcos i sin n r ncos n i sin n n z rcos i sin 2 z 3 zz 2 r 3cos 3 i sin 3 z 2 r 2cos 2 i sin 2 z rcos i sin APPENDIX H COMPLEX NUMBERS \|\|\|\| A59 From the fact that sine and cosine have period it follows that Thus Since this expression gives a different value of for , 1, 2, . . . , , we have the following. ROOTS OF A COMPLEX NUMBER Let and let be a posi-tive integer. Then has the distinct th roots where , 1, 2, . . . , . Notice that each of the th roots of has modulus . Thus all the th roots of lie on the circle of radius in the complex plane. Also, since the argument of each successive th root exceeds the argument of the previous root by , we see that the th roots of are equally spaced on this circle. EXAMPLE 7 Find the six sixth roots of and graph these roots in the complex plane. SOLUTION In trigonometric form, . Applying Equation 3 with , we get We get the six sixth roots of by taking in this formula: All these points lie on the circle of radius as shown in Figure 9. M s2 w5 816cos 11 6 i sin 11 6 s2 s3 2 1 2 i w4 816cos 3 2 i sin 3 2 s2 i w3 816cos 7 6 i sin 7 6 s2 s3 2 1 2 i w2 816cos 5 6 i sin 5 6 s2 s3 2 1 2 i w1 816cos 2 i sin 2 s2 i w0 816cos 6 i sin 6 s2 s3 2 1 2 i k 0, 1, 2, 3, 4, 5 8 wk 816cos 2k 6 i sin 2k 6 n 6 z 8cos i sin z 8 z n 2 n n r 1n z n wk r 1n z n n 1 k 0 wk r 1ncos 2k n i sin 2k n n n z n z rcos i sin 3 n 1 k 0 w w r 1ncos 2k n i sin 2k n 2k n or n 2k 2 A60 \|\|\|\| APPENDIX H COMPLEX NUMBERS FIGURE 9 The six sixth roots of z=8 0 w¡ w¢ w∞ w¸ w™ w£ _œ„ 2 œ„ 2 _œ„ 2i œ„ 2i Re Im COMPLEX EXPONENTIALS We also need to give a meaning to the expression when is a complex num-ber. The theory of inﬁnite series as developed in Chapter 11 can be extended to the case where the terms are complex numbers. Using the Taylor series for (11.10.11) as our guide, we deﬁne and it turns out that this complex exponential function has the same properties as the real exponential function. In particular, it is true that If we put , where is a real number, in Equation 4, and use the facts that . . . we get Here we have used the Taylor series for and (Equations 11.10.16 and 11.10.15). The result is a famous formula called Euler’s formula: Combining Euler’s formula with Equation 5, we get EXAMPLE 8 Evaluate: (a) (b) SOLUTION (a) From Euler’s equation (6) we have (b) Using Equation 7 we get M Finally, we note that Euler’s equation provides us with an easier method of proving De Moivre’s Theorem: rcos i sin n re i n r ne in r ncos n i sin n e1i 2 e1cos 2 i sin 2 1 e 0 i1 i e e i cos i sin 1 i0 1 e1i 2 e i e xiy e xe iy e xcos y i sin y 7 e iy cos y i sin y 6 sin y cos y cos y i sin y 1 y 2 2! y 4 4! y 6 6! iy y 3 3! y 5 5! 1 iy y 2 2! i y 3 3! y 4 4! i y 5 5! e iy 1 iy iy2 2! iy3 3! iy4 4! iy5 5! i 5 i, i 4 1, i 3 i 2i i, i 2 1, y z iy e z1z2 e z1e z2 5 e z n0 zn n! 1 z z2 2! z3 3! 4 e x z x iy e z APPENDIX H COMPLEX NUMBERS \|\|\|\| A61 N We could write the result of Example 8(a) as This equation relates the ﬁve most famous num-bers in all of mathematics: and . 0, 1, e, i, e i 1 0 A62 \|\|\|\| APPENDIX H COMPLEX NUMBERS 33–36 Find the indicated power using De Moivre’s Theorem. 33. 34. 35. 36. 37–40 Find the indicated roots. Sketch the roots in the complex plane. 37. The eighth roots of 1 38. The ﬁfth roots of 32 39. The cube roots of 40. The cube roots of 41–46 Write the number in the form . 41. 42. 43. 44. 45. 46. 47. Use De Moivre’s Theorem with to express and in terms of and . 48. Use Euler’s formula to prove the following formulas for and : 49. If is a complex-valued function of a real variable and the real and imaginary parts and are differentiable functions of , then the derivative of is deﬁned to be . Use this together with Equation 7 to prove that if , then when is a complex number. 50. (a) If is a complex-valued function of a real variable, its indeﬁnite integral is an antiderivative of . Evaluate (b) By considering the real and imaginary parts of the integral in part (a), evaluate the real integrals and (c) Compare with the method used in Example 4 in Sec-tion 7.1. y e x sin x dx y e x cos x dx y e 1i x dx u x ux dx u r a bi Fx re rx Fx e rx ux f x itx u x tx f x x ux f x itx sin x eix eix 2i cos x eix eix 2 sin x cos x sin cos sin 3 cos 3 n 3 e i e 2i e i e i 3 e 2 i e i 2 a bi 1 i i 1 i8 (2s3 2i) 5 (1 s3i) 5 1 i20 1–14 Evaluate the expression and write your answer in the form . 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15–17 Find the complex conjugate and the modulus of the number. 15. 16. 17. 18. Prove the following properties of complex numbers. (a) (b) (c) , where is a positive integer [Hint: Write , .] 19–24 Find all solutions of the equation. 19. 20. 21. 22. 23. 24. 25–28 Write the number in polar form with argument between and . 25. 26. 27. 28. 29–32 Find polar forms for , , and by ﬁrst putting and into polar form. 29. , 30. , 31. , 32. , w 3 3i z 4(s3 i) w 1 i z 2s3 2i w 8i z 4s3 4i w 1 s3 i z s3 i w z 1z zw zw 8i 3 4i 1 s3 i 3 3i 2 0 z2 1 2z 1 4 0 z2 z 2 0 2x 2 2x 1 0 x 2 2x 5 0 x 4 1 4x 2 9 0 w c di z a bi n z n z n zw z w z w z w 4i 1 2s2 i 12 5i s3 s12 s25 i 100 i 3 3 4 3i 1 1 i 3 2i 1 4i 1 4i 3 2i 2i( 1 2 i) 12 7i 1 2i8 3i 2 5i4 i (4 1 2i) (9 5 2i) 5 6i 3 2i a bi EXERCISES H ANSWERS TO ODD-NUMBERED EXERCISES H CHAPTER 1 EXERCISES 1.1 N PAGE 20 1. (a) 2 (b) 2.8 (c) 3, 1 (d) 2.5, 0.3 (e) (f ) 3. 5. No 7. Yes, 9. Diet, exercise, or illness 11. 13. 15. 17. 19. (a) (b) In millions: 92; 485 21. 12, 16, , , , , , , , 23. 25. 27. 29. 31. , 0 5, 0, {x x 1 3} (, 1 3) ( 1 3 , ) 1ax 3 h 3a2 6ah 3h2 a h 2 9a4 6a3 13a2 4a 4 3a4 a2 2 12a2 2a 2 6a2 2a 4 3a2 5a 4 3a2 a 2 3a2 a 2 N 0 t 1992 1990 1996 1994 1998 2000 100 200 300 400 500 600 Height of grass t Wed. Wed. Wed. Wed. Wed. 0 amount price T t midnight noon T 0 t 3, 2, 3, 2 1, 3 85, 115 1, 3 3, 3, 2, 3 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. (a) (b) $400, $1900 (c) 61. is odd, is even 63. (a) (b) 65. Odd 67. Neither 69. Even EXERCISES 1.2 N PAGE 34 1. (a) Root (b) Algebraic (c) Polynomial (degree 9) (d) Rational (e) Trigonometric (f) Logarithmic 3. (a) h (b) f (c) t 5, 3 5, 3 t f T (in dollars) 0 I (in dollars) 10,000 20,000 1000 2500 30,000 R (%) 0 I (in dollars) 10,000 20,000 10 15 Vx 4x 3 64x 2 240x, 0 x 6 Sx x 2 8x, x 0 Ax s3x 24, x 0 AL 10L L2, 0 L 10 f x x 3 2x 6 if 0 x 3 if 3 x 5 f x 1 sx f x 5 2 x 11 2 , 1 x 5 x y 1 1 0 x (0, 2) (0, 1) _2 1 y 0 , , x 2 y 0 4 y x 0 5 , 0 0, 5, y t 0 6 _9 y x 0 5 , , APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A63 A64 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 5. (a) , where b is the y-intercept. (b) , where m is the slope. See graph at right. (c) 7. Their graphs have slope . 9. 11. (a) 8.34, change in mg for every 1 year change (b) 8.34 mg 13. (a) (b) , change in for every change; 32, Fahrenheit temperature corresponding to 15. (a) (b) , change in for every chirp per minute change (c) 17. (a) (b) 196 ft 19. (a) Cosine (b) Linear 21. (a) Linear model is appropriate (b) 15 0 61,000 (b) (c) y 0.000105x 14.521 15 0 61,000 P 0.434d 15 76F F 1 6 T 1 6N 307 6 0C 1C F 9 5 F C (100, 212) F= C+32 9 5 (_40, _40) 32 f x 3xx 1x 2 y x c=_2 c=_1 0 c=2 c=1 c=0 1 y 2x 3 y x m=_1 m=1 m=0 y-1=m(x-2) (2, 1) y mx 1 2m y x b=3 b=0 b=_1 y=2x+b y 2x b (c) [See graph in (b).] (d) About 11.5 per 100 population (e) About 6% (f) No 23. (a) Linear model is appropriate (b) (c) 20 ft (d) No 25. 1914 million EXERCISES 1.3 N PAGE 43 1. (a) (b) (c) (d) (e) (f) (g) (h) 3. (a) 3 (b) 1 (c) 4 (d) 5 (e) 2 5. (a) (b) (c) (d) 7. 9. 11. 13. y x 0 3 π y=1+2 cos x y x 0 _1 1 y=(x+1)@ y x 0 y=_x# y sx 2 5x 4 1 y 0 x y 0 x y 0 x y 0 x y 1 3 f x y 3f x y f x y f x y f x 3 y f x 3 y f x 3 y f x 3 y 0.0012937x 3 7.06142x 2 12,823x 7,743,770; 20 (ft) 10 2000 (year) 1896 y 0.08912x 158.24 20 (ft) 10 2000 (year) 1896 y 0.00009979x 13.951 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A65 15. 17. 19. 21. 23. 25. 27. (a) The portion of the graph of to the right of the y-axis is reﬂected about the y-axis. (b) (c) 29. , , , , 31. (a) , (b) , (c) , (d) , 33. (a) , (b) , (c) , (d) , 35. (a) , (b) , (c) , (d) , 37. 39. 41. 43. tx s 3 x , fx x1 x tx x 2 1, fx x 10 f t hx sx 6 4x 3 1 f t hx 2x 1 {x x 2, 5 3} t tx 2x 33x 5 {x x 0 f f x x 4 3x 2 1xx 2 1 {x x 1, 0 t f x x 2 x 1x 12 x x 2, 1 2x 2 6x 5x 2x 1 f tx , t tx coscos x , f f x 9x 2 , t f x cos1 3x , f tx 1 3 cos x , t tx 4x 3 , f f x x 4 2x 2 , t f x 2x 2 1 , f tx 4x 2 4x {x x 1s3} ftx x 3 2x 23x 2 1 , ftx 3x 5 6x 4 x 3 2x 2 , f tx x 3 x 2 1 , f tx x 3 5x 2 1 y x 0 y=œ„„ \|x\| x y y= \|x\| 0 sin y f x Lt 12 2 sin 2 365 t 80 y x 0 y=\|sin x\| π 1 y x 0 y= 2 x+1 2 x=_1 y=(≈+8x) 1 2 y x 0 _4 _8 y x 0 _3 y=œ„„„„ x+3 y x 0 1 2π y=sin(x/2) 45. 47. , , 49. , , 51. (a) 4 (b) 3 (c) 0 (d) Does not exist; is not in the domain of . (e) 4 (f) 53. (a) (b) ; the area of the circle as a function of time 55. (a) (b) (c) ; the distance between the lighthouse and the ship as a function of the time elapsed since noon 57. (a) (b) (c) 59. Yes; 61. (a) (b) 63. (a) Even; even (b) Odd; even 65. Yes EXERCISES 1.4 N PAGE 51 1. (c) 3. 5. 7. 9. 11. 1.5 _1.5 100 0 _0.01 0.01 0 1.1 3500 _3500 20 _20 4 1 4 4 150 _50 30 _10 tx x 2 x 1 f(x x 2 6 m1m2 Vt 240Ht 5 V t 240 0 5 Vt 120Ht V t 120 0 H t 1 0 s s900t 2 36 d 30t s sd 2 36 A rt 3600 t 2 rt 60t 2 t f 6 6 f x x 4 tx sec x hx sx f x 1 x tx 3x hx x 2 tt cos t, f t st A66 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 13. 15. 17. No 19. 9.05 21. 0, 0.88 23. 25. 27. (a) (b) (c) (d) Graphs of even roots are similar to , graphs of odd roots are similar to . As n increases, the graph of becomes steeper near 0 and ﬂatter for . 29. If , the graph has three humps: two minimum points and a maximum point. These humps get ﬂatter as c increases until at two of the humps disappear and there is only one mini-mum point. This single hump then moves to the right and approaches the origin as c increases. 31. The hump gets larger and moves to the right. 33. If , the loop is to the right of the origin; if , the loop is to the left. The closer c is to 0, the larger the loop. EXERCISES 1.5 N PAGE 58 1. (a) (b) (c) (d) See Figures 4(c), 4(b), and 4(a), respectively. 0, f x a x, a 0 c 0 c 0 c 1.5 c 1.5 2 _4 _2.5 2.5 _1.5 -1 -2 -3 1 x 1 y s n x s 3 x sx 2 _1 _1 3 $ œ„ x œ„ x Œ„ x % œ„ x 2 _2 _3 3 x % œ„ x Œ„ x 3 _1 _1 4 ^ œ„ x $ œ„ x œ„ x 0.85 x 0.85 t 1 _1 _1 1 2 _2 _ π 25 π 25 11 _11 2π _2π 3. All approach 0 as , all pass through , and all are increasing. The larger the base, the faster the rate of increase. 5. The functions with base greater than 1 are increasing and those with base less than 1 are decreasing. The latter are reﬂections of the former about the y-axis. 7. 9. 11. 13. (a) (b) (c) (d) (e) 15. (a) (b) 17. 23. At 25. (a) 3200 (b) (c) 10,159 (d) 27. , where and ; 5498 million; 7417 million EXERCISES 1.6 N PAGE 70 1. (a) See Deﬁnition 1. (b) It must pass the Horizontal Line Test. 3. No 5. Yes 7. No 9. No 11. Yes 13. No 15. 2 17. 0 19. ; the Fahrenheit temperature as a function of the Celsius temperature; 21. 23. f 1x s 3 ln x f 1x 1 3 x 2 10 3 , x 0 273.15, F 9 5 C 32 b 1.017764706 a 3.154832569 1012 y ab t t 26.9 h 60,000 0 40 100 2t3 x 35.8 f x 3 2x , 0 0, , y ex y ex y e x y e x2 y e x 2 x y 0 y=1- e–® y=1 1 2 ”0, ’ 1 2 x _1 y 0 y=_2–® x _2 y 0 y=4®-3 y=_3 5 _2 2 y=3® y=10® 0 y=” ’® 1 3 y=” ’® 1 10 0, 1 x l y=20® y=5® y=´ y=2® 5 _1 2 0 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A67 25. 27. 29. 31. (a) It’s deﬁned as the inverse of the exponential function with base a, that is, . (b) (c) (d) See Figure 11. 33. (a) 3 (b) 35. (a) 3 (b) 37. 39. 41. All graphs approach as , all pass through , and all are increasing. The larger the base, the slower the rate of increase. 43. About 1,084,588 mi 45. (a) (b) 47. (a) (b) 49. (a) (b) 51. (a) (b) 53. (a) (b) , 55. The graph passes the Horizontal Line Test. , where ; two of the expressions are complex. 57. (a) ; the time elapsed when there are n bacteria (b) After about 26.9 hours 59. (a) (b) 61. (a) (b) 63. (a) 10 (b) 67. xs1 x 2 3 4 4 3 f 1n 3ln 2 lnn100 D 3s3s27x 4 40x 2 16 f 1(x) (s 3 46)(s 3 D 27x 2 20 s 3 D 27x 2 20 s 3 2) 5 _1 4 _2 [0, s3) f 1x 1 2 ln3 x2 (, 1 2 ln 3] x 1e x ln 10 1 2(1 s1 4e) 5 log2 3 or 5 ln 3ln 2 ln 5 se 1 x y 0 y= -ln x _5 _4 x y 0 y=log10 (x+5) 1, 0 x l 0 3 5 4 y=log1.5 x y=log10 x 0 y=ln x y=log50 x ln (1 x2)sx sin x ln 1215 2 3 0, loga x y & ? a y x x y f f–! 0 6 6 0 f–! f f 1x s 4 x 1 y e x 3 69. The second graph is the reﬂection of the ﬁrst graph about the line . 71. (a) (b) 73. (a) (b) CHAPTER 1 REVIEW N PAGE 73 True-False Quiz 1. False 3. False 5. True 7. False 9. True 11. False 13. False Exercises 1. (a) 2.7 (b) 2.3, 5.6 (c) (d) (e) (f) No; it fails the Horizontal Line Test. (g) Odd; its graph is symmetric about the origin. 3. 5. , 7. 9. (a) Shift the graph 8 units upward. (b) Shift the graph 8 units to the left. (c) Stretch the graph vertically by a factor of 2, then shift it 1 unit upward. (d) Shift the graph 2 units to the right and 2 units downward. (e) Reﬂect the graph about the x-axis. (f) Reﬂect the graph about the line (assuming f is one-to-one). 11. 13. 15. 17. (a) Neither (b) Odd (c) Even (d) Neither 19. (a) , (b) , (c) , (d) , 21. ; about 77.6 years y 0.2493x 423.4818 , t tx x 2 92 9 1, f f x ln ln x 0, t f x ln x2 9 , 3 3, f tx lnx 2 9 y x 0 y= 1 x+2 1 2 x=_2 x y 0 (1+´) 1 y= 1 2 y= 1 2 y x 0 y=_sin 2x π y x 6, , , 0 0, (, 1 3) ( 1 3, ) 2a h 2 4, 4 4, 4 6, 6 h1x 1c f 1x t1x f 1x c 2, 2 2 3, 0 y x y=sin–! x π 2 π 2 _ y=sin x π 2 _ π 2 A68 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 23. 1 25. (a) 9 (b) 2 (c) (d) 27. (a) years (b) the time required for the population to reach a given number . (c) years PRINCIPLES OF PROBLEM SOLVING N PAGE 81 1. , where a is the length of the altitude and h is the length of the hypotenuse 3. 5. 7. 9. 11. 5 13. 15. 19. CHAPTER 2 EXERCISES 2.1 N PAGE 87 1. (a) , , , , (b) (c) 3. (a) (i) 0.333333 (ii) 0.263158 (iii) 0.251256 (iv) 0.250125 (v) 0.2 (vi) 0.238095 (vii) 0.248756 (viii) 0.249875 (b) (c) 5. (a) (i) (ii) (iii) (iv) (b) 7. (a) (i) (ii) (iii) (iv) (b) 9. (a) 0, 1.7321, 1.0847, 2.7433, 4.3301, 2.8173, 0, 2.1651, 2.6061, 5, 3.4202; no (c) 31.4 EXERCISES 2.2 N PAGE 96 1. Yes 3. (a) means that the values of can be made arbitrarily large (as large as we please) by taking x sufﬁcient-ly close to 3 (but not equal to 3). f x lim x l3 f x 6.3 ms 7 ms 7.55 ms 5.6 ms 4.65 ms 24 fts 24.16 fts 24.8 fts 25.6 fts 32 fts y 1 4x 1 4 1 4 33 1 3 33.3 16.6 22.2 27.8 38.8 44.4 fnx x 2 n1 40 mih x [1, 1 s3) (1 s3, 3] y x 1 0 1 y x 0 1 x y 7 3, 9 a 4sh 2 16h ln 81 4.4 P t ln 1000 P 9P ; 4.4 1000 10 0 3 5 1s3 (b) means that the values of can be made arbitrarily large negative by taking x sufﬁciently close to 4 through values larger than 4. 5. (a) 2 (b) (c) Does not exist (d) 4 (e) Does not exist 7. (a) (b) (c) Does not exist (d) 2 (e) 0 (f) Does not exist (g) 1 (h) 3 9. (a) (b) (c) (d) (e) (f) 11. (a) (b) (c) Does not exist 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. (a) 2.71828 (b) 37. (a) 0.998000, 0.638259, 0.358484, 0.158680, 0.038851, 0.008928, 0.001465; 0 (b) 0.000572, 0.000614, 0.000907, 0.000978, 0.000993, 0.001000; 0.001 39. No matter how many times we zoom in toward the origin, the graph appears to consist of almost-vertical lines. This indicates more and more frequent oscillations as . 41. , ; , EXERCISES 2.3 N PAGE 106 1. (a) (b) (c) 2 (d) (e) Does not exist (f) 0 3. 5. 390 7. 9. 0 11. 5 13. Does not exist 15. 17. 19. 21. 23. 25. 27. 29. 31. (a), (b) 35. 39. 41. 43. Does not exist 45. (a) (b) (i) (ii) (iii) Does not exist (iv) 47. (a) (i) (ii) (b) No (c) 1 x 2 y 0 2 2 1 1 1 x 1 y 0 4 6 7 2 3 1 2 1 128 1 16 1 6 6 1 12 8 6 5 1 8 59 6 8 6 sin1 4 x sin1 4 2.24 x 0.90 x l 0 6 4 _4 _2 ; 3 5 1 4 1 2 2 3 y 0 x 1 y x 0 1 x 7, x 3, x 0, x 6 2 1 3 f x lim x l 4 f x APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A69 49. (a) (i) (ii) Does not exist (iii) (b) (i) (ii) (c) is not an integer. 55. 61. EXERCISES 2.4 N PAGE 117 1. (or any smaller positive number) 3. 1.44 (or any smaller positive number) 5. (or any smaller positive number) 7. 0.11, 0.012 (or smaller positive numbers) 9. (a) (b) 11. (a) (b) Within approximately 0.0445 cm (c) Radius; area; ; 1000; 5; 13. (a) (b) 35. (a) (b) , where 41. Within 0.1 EXERCISES 2.5 N PAGE 128 1. 3. (a) (removable), ( jump), 2 ( jump), 4 (inﬁnite) (b) , neither; , left; 2, right; 4, right 5. 7. (a) (b) Discontinuous at , 2, 3, 4 9. 15. is not deﬁned. 17. does not exist. 19. 21. 23. 25. 27. , 1 1, , [ 1 2, ) y 0 x 1 _π x x 3, 2 lim x l 0 f x f 0 x y 0 y=´ y=≈ 1 x y 0 x=2 lim x l 0 f x f2 6 t 1 Cost (in dollars) 0 Time (in hours) 1 1 y 0 x 2 1 2 4 2 4 lim x l 4 f x f 4 B 216 108 12s336 324 81 2 B 23 126B 13 1 0.093 0.0025 0.025 0.0445 s1000 s1000 cm 0.010 0.031 0.0906 4 7 15; 1 8 a n n 1 3 2 29. 31. 33. 37. 0, left 39. 0, right; 1, left 41. 43. (a) (b) 51. (b) 53. (b) 59. None 61. Yes EXERCISES 2.6 N PAGE 140 1. (a) As becomes large, approaches . (b) As becomes large negative, approaches 3. 3. (a) (b) (c) (d) 1 (e) 2 (f) 5. 7. 9. 11. 13. 15. 0 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. (a), (b) 39. 41. 43. 45. 47. 49. 51. 0 1 3 y x 0 1 y x , , f x 2 x x 2x 3 y 3 x 5 y 2; x 2, x 1 y 2; x 2 1 2 0 1 2 1 2a b 1 6 3 2 1 2 1 2 3 2 0 x y 0 y=3 x=4 x y 0 x=2 x y 0 1 1 x 1, x 2, y 1, y 2 f x x 5 f x x 70.347 0.86, 0.87 tx x 2 x tx x 3 x 2 x 1 2 3 y x 0 (1, e) (1, 1) (0, 1) (0, 2) x y 0 (0, 1) (0, 2) (2, 0) 1 7 3 3 4 _4 _1 x 0 A70 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 53. (a) 0 (b) An inﬁnite number of times 55. (a) (b) 57. 59. (a) (b) s 61. 63. 65. (a) EXERCISES 2.7 N PAGE 150 1. (a) (b) 3. (a) (b) (c) 5. 7. 9. (a) (b) (c) 11. (a) Right: and ; left: ; standing still: and (b) 13. 15. ; ; ; 17. t0, 0, t4, t2, t2 2 27 ms 1 4 ms 2 ms 2a3 ms 24 fts t (seconds) v (m/s) 0 1 1 3 t 4 1 t 2 2 t 3 4 t 6 0 t 1 10 _3 4 _2 y 2x 3, y 8x 19 8a 6a 2 y 1 2x 1 2 y x 5 6 0 5 _1 y 2x 1 2 lim x l 3 f x f 3 x 3 f x f 3 x 3 x 100 N 6, N 22 N 15 0.47 1.2 0 1 v 5 0 _0.5 1 -25 25 19. 21. 7; 23. (a) (b) 25. 27. 29. 31. or , 33. 35. or , 37. ; 39. Greater (in magnitude) 41. (a) (i) (ii) (iii) (b) (c) 43. (a) (i) (ii) (b) 45. (a) The rate at which the cost is changing per ounce of gold produced; dollars per ounce (b) When the 800th ounce of gold is produced, the cost of production is (c) Decrease in the short term; increase in the long term 47. The rate at which the temperature is changing at 10:00 AM; 49. (a) The rate at which the oxygen solubility changes with respect to the water temperature; (b) as the temperature increases past , the oxygen solubility is decreasing at a rate of . 51. Does not exist EXERCISES 2.8 N PAGE 162 1. (a) 1.5 (b) 1 (c) 0 (d) 4 (e) 0 (f) 1 (g) 1.5 x 0 y fª 0.25 mgLC 16C S16 0.25; mgLC 4Fh $17oz. $20unit $20.05unit $20.25unit 15 percentyear 14.5 percentyear 16 percentyear 13 percentyear 11 percentyear Temperature (in °F) 0 Time (in hours) 1 38 2 72 1 ms 1 ms a 0 f x cos x f x cos x, a f x 2x, a 5 a 0 f x 1 x10 f x x 10, a 1 1 2a 232 5 a 32 2 8a 4 _2 6 _1 3 5; y 3 5x 16 5 y 7x 12 y 0 x 1 1 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A71 3. (a) II (b) IV (c) I (d) III 5. 7. 9. 11. 13. 1963 to 1971 15. 17. (a) 0, 1, 2, 4 (b) 1, 2, 4 (c) 19. , 21. 23. 25. 27. , , 29. 31. (a) 33. (a) The rate at which the unemployment rate is changing, in percent unemployed per year (b) 35. 37. 1 vertical tangent; 4 corner 4 corner; 0 discontinuity f x 4x 3 2 f x 4x 3, , , 1 1, , 1 1, Gt 4 t 12 tx 1s1 2x, [ 1 2, ), ( 1 2, ) f x 3x 2 3, , f t 5 18t, , , f x 1 2 f x 2x f x e x y x 1 1 0 f, fª 0.05 1990 1980 1970 1960 1950 _0.03 t y=Mª(t) 0.1 y x 0 y fª y 0 x fª x y 0 fª fª x y 0 39. Differentiable at 1; not differentiable at 0 41. 43. 45. , 47. , , , 49. (a) 51. or 53. (a) (b) All x (c) 57. CHAPTER 2 REVIEW N PAGE 166 True-False Quiz 1. False 3. True 5. False 7. True 9. True 11. False 13. True 15. True 17. False 19. False Exercises 1. (a) (i) 3 (ii) 0 (iii) Does not exist (iv) 2 (v) (vi) (vii) 4 (viii) 1 (b) , (c) , (d) 3, 0, 2, 4 3. 1 5. 7. 3 9. 11. 13. 15. 17. 19. 21. 23. 1 29. (a) (i) 3 (ii) 0 (iii) Does not exist (iv) 0 (v) 0 (vi) 0 x 0, y 0 2 2 1 2 4 7 3 2 x 2 x 0 y 1 y 4 63 f x 2 x x 0 y f x x 6 x 6 x y 0 6 1 _1 fª f x 1 1 if x 6 if x 6 1 3a23 f 4x 0 f x 6 f x 4 6x f x 4x 3x 2 3 6 4 7 f fª f· fªªª f x 2 f x 4 2x 6 10 6 10 f fª f· a acceleration, b velocity, c position a f, b f , c f 2 _1 _2 1 t t 1993 1998 1994 1999 1995 2000 1996 2001 1997 2002 1.10 0.45 0.90 0.35 0.25 0.35 0.25 0.65 0.35 0.80 Ut Ut A72 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES (b) At 0 and 3 (c) 31. 35. (a) (b) 37. (a) (i) (ii) (iii) (iv) (b) 39. (a) (b) (c) 41. (a) The rate at which the cost changes with respect to the interest rate; dollars(percent per year) (b) As the interest rate increases past 10%, the cost is increasing at a rate of $1200(percent per year). (c) Always positive 43. 45. (a) (b) (c) 47. 4 (discontinuity), 1 (corner), 2 (discontinuity), 5 (vertical tangent) 49. The rate at which the total value of US currency in circulation is changing in billions of dollars per year; 51. 0 PROBLEMS PLUS N PAGE 170 1. 3. 4 5. 1 7. 9. 11. (b) Yes (c) Yes; no 13. (a) 0 (b) 1 (c) CHAPTER 3 EXERCISES 3.1 N PAGE 180 1. (a) See Deﬁnition of the Number e (page 179). (b) 0.99, 1.03; 3. 5. 7. f x 3x 2 4 ft 2 3 f x 0 2.7 e 2.8 f x x 2 1 3 4 a 1 2 1 2s5 2 3 $22.2 billionyear 6 1 _3 _6 f fª (, 3 5], (, 3 5) f x 5 23 5x12 x y 0 fª 4 –4 –12 12 y 10x 16 10 2.5 ms 2.525 ms 2.625 ms 2.75 ms 3 ms y 8x 17 8 x 0 y 3 3 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. Tangent: ; normal: 37. 39. 41. 43. (a) (c) 45. 47. 49. (a) (b) (c) 51. 55. , 57. 59. 63. 65. 67. No 69. (a) Not differentiable at 3 or 3 (b) 71. 73. 75. 77. 79. EXERCISES 3.2 N PAGE 187 1. 3. 5. 7. 9. 11. 13. 15. y 2tt 4 4t 2 7 t 4 3t 2 12 y x 23 x 2 1 x 22 Fy 5 14y 2 9y 4 Vx 14x 6 4x 3 6 tx 52x 12 y x 2e xx 3 fx e xx 3 3x 2 2x 2 y 5x 4 3x 2 2x 3; 1 1000 m 4, b 4 a 1 2, b 2 y 2x 2 x 3 3 x 0 y ƒ 3 3 x 0 9 y ƒ f x 2x 2x if x 3 if x 3 x y 0 (1, 1) ƒ x y 0 1 ƒ y 3 16 x 3 9 4 x 3 Px x 2 x 3 2, 4 y 1 3x 1 3 y 12x 17 y 12x 15 2, 21, 1, 6 a1 6 ms2 12 ms2 vt 3t 2 3, at 6t f x 2 15 4 x14, f x 15 16 x54 f x 4x 3 9x 2 16, f x 12x 2 18x 100 40 3 5 50 10 3 5 4x 3 9x 2 12x 7 45x 14 15x 2 e x 5 y 3x 1 y 1 2x 2 y 2x 2 y 1 4x 3 4 z 10Ay11 Be y u 1 5t 45 10t 32 Hx 3x 2 3 3x 2 3x 4 y 0 y 3 2sx (2sx) 3(2xsx) y 2ax b Fx 5 32 x 4 Gx 1(2sx) 2e x As 60s 6 Vr 4 r 2 y 2 5 x75 f t t 3 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A73 17. 19. 21. 23. 25. 27. 29. 31. 33. ; 35. (a) (b) 37. (a) 39. , 41. 43. (a) 16 (b) (c) 20 45. 7 47. (a) 0 (b) 49. (a) (b) (c) 51. Two, 53. 55. (c) 57. , , EXERCISES 3.3 N PAGE 195 1. 3. 5. 7. 9. 11. 13. 15. 21. 23. 25. (a) (b) 27. (a) 29. ; 31. (a) (b) 33. , n an integer 35. (a) , (b) ; to the left 37. 39. 41. 43. 45. 47. s2 1 2 sin 1 3 3 5 ftrad 4s3, 4, 4s3 at 8 sin t vt 8 cos t 2n 1 1 3 f x cos x sin x f x 1 tan xsec x 2 cos sin cos sin sec x tan x 1 π 0 3π 2 ” , π’ π 2 y 2x y x 1 y 2s3x 2 3s3 2 f x e x csc x x cot x x 1 y x cos x 2 sin xx 3 f sec tan 1 sec 2 y 2 tan x x sec2x 2 tan x2 h csc cot e cot csc2 tt 3t 2 cos t t 3 sin t f x cos x 1 2 csc2x f x 6x 2 sin x f (n)x x 2 2nx nn 1e x f 5x x 2 10x 20e x; f 4x x 2 8x 12e x, f x x 2 6x 6e x f x x 2 2xe x, f x x 2 4x 2e x 3e 3x $1.627 billionyear (2 s3, (1 s3)2) y xtx txx 2 y tx xtxtx 2 y xtx tx 2 3 20 9 1 4 x 1e x xe x e xx 3x 4 (_1, 0.5) 1.5 0.5 4 4 y 1 2 x 1 y 1 2x y 2x y 1 2 x 1 2 2x 2 2x 1 2x2 ; 2 1 2x3 x 4 4x 3e x; x4 8x3 12x 2e x f x 2cxx 2 c2 f x ACe xB Ce x2 ft 4 t 12 (2 st)2 y 2v 1sv y r 2 2e r 49. (a) (b) (c) 51. 1 EXERCISES 3.4 N PAGE 203 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. ; 51. 53. 55. (a) (b) 57. (a) 59. , n an integer 61. 24 63. (a) 30 (b) 36 65. (a) (b) Does not exist (c) 2 67. (a) (b) 69. 120 71. 96 75. 77. 79. (a) (b) 0.16 81. 15 7 0 2 √ 2 1 0 2 s vt 2e1.5t2 cos 2 t 1.5 sin 2 t dB dt 7 54 cos 2 t 5.4 vt 5 2 cos10 t cms 250 cos 2x Gx e f x f x Fx e x f e x 3 4 2 2n , 3, 3 2 2n , 1 f x 2 2x 2s2 x 2 3 _1.5 _3 3 (0, 1) y 1 2 x 1 y x y 20x 1 ex 2 2 sin x 2 cos x ex cos x sin x hx xsx 2 1, hx 1x 2 132 y costan x sec2 x sinssintan x 2ssintan x tx 2r 2 pln a 2ra rx n p1a rx f t 4 sinesin 2t cosesin 2t esin 2t sin t cos t ft sec2etet etan t sec2t y 2 cos cotsin csc2sin y 4e2x 1 e2x2 sin 1 e2x 1 e2x y 4 sec 2x tan x y 2sin x ln 2 cos x y 2 costan 2x sec22x y r 2 132 Fz 1z 112z 132 y cos x x sin xe x cos x y 12xx 2 12 x 2 14 y 82x 538x2 54 4x2 30x 5 tx 41 4x43 x x 2717 9x 21x 2 y ekxkx 1 y 3x 2 sina 3 x 3 tt 12t 3 t 4 14 Fx 2 3x 2 41 2x x334 Fx 10xx4 3x 2 242x 2 3 esx(2sx) 20x1 x 29 4 cos 4x cos x sin x cot x 1csc x sec x tan x sin xcos2x sec2x 1cos2x A74 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 83. is the rate of change of velocity with respect to time; is the rate of change of velocity with respect to displacement 85. (a) where and (b) A 87. (b) The factored form 91. (b) EXERCISES 3.5 N PAGE 213 1. (a) (b) , 3. (a) (b) , 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. (a) (b) 33. 35. 37. (a) Eight; (b) , (c) 39. 41. 45. 47. 49. 51. 53. 55. 59. 61. x y x y 1 x arcsin x s1 x 2 y 2e2xs1 e 4x ht 0 Gx 1 x arccos x s1 x 2 y 1 sx 2 x y 1 2sx1 x x0xa 2 y0yb 2 1 ( 5 4s3, 5 4) 1 1 3s3 y 1 3 x 2 y x 1 x 0.42, 1.58 4 5 _2 _3 2xy 5 81y 3 5 2 _2 _2 (1, 2) y 9 2 x 5 2 y 9 13 x 40 13 y x 1 2 y x 2 x 2x4y x3 6xy2 4x3y2 3x 2y 2y3 16 13 y e y sin x y cosxy e y cos x x cosxy y 4xysxy y x 2x 2sxy y yy e xy y 2 xe xy y tan x tan y y 2xy 2 sin y 2x 2y x cos y y 3y2 5x4 4x 3y x 4 3y 2 6xy y 2x y 2y x y x 2y 2 y 1x 12 y xx 1 y y 2x 2 y 4x 2 3 y 4x 2 3x y y 2 6xx n cos n1x sinn 1x 670.63 b 0.000045146 a 100.01244 y abt dvds dvdt 63. 65. 67. (b) 69. EXERCISES 3.6 N PAGE 220 1. The differentiation formula is simplest. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. ; 29. 31. 1 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. EXERCISES 3.7 N PAGE 230 1. (a) (b) (c) (d) (e) (f) (g) (h) (i) Speeding up when ; slowing down when 0 t 2 or 4 t 6 2 t 4 or t 6 s 40 8 0 25 √ a 6t 24; 6 ms2 t 2, s 32 t 0, s 0 t 6, s 0 t 8, s 32 s 0 20 96 ft 0 t 2, t 6 t 2, 6 9 fts 3t 2 24t 36 f nx 1n1n 1! x 1n y 2x x 2 y 2 2y y tan x1x sec2x x tan x ln tan x x 2 y cos xxx tan x ln cos x y x sin x sin x x cos x ln x y x x1 ln x y sin2x tan4x x 2 12 2 cot x 4 sec2x tan x 4x x 2 1 y 2x 15x 4 36 10 2x 1 24x 3 x 4 3 cos x 1x y 3x 2 f x 2x 1 xx 2 ; , 0 2, 1, 1 e 1 e, f x 2x 1 x 1 lnx 1 x 11 lnx 1 2 y 1 s1 x 2 ; y x 1 x 232 y x 2x ln2x; y 3 2 ln2x y 1 ln 10 log10 x y x 1 x y 10x 1 5x 2 x 2 f u 1 ln 2 u1 ln2u2 tx 2x 2 1 xx 2 1 Ft 6 2t 1 12 3t 1 f x sin x x cos x ln5x f x 1 5xs 5 ln x4 f x 3 3x 1 ln 2 fx cosln x x 2 3 2 1, 1, 1, 1 (s3, 0) APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A75 3. (a) (b) (c) (d) (e) 4 ft (f) (g) (h) (i) Speeding up when , ; slowing down when , 5. (a) Speeding up when or ; slowing down when (b) Speeding up when or ; slowing down when or 7. (a) (b) ; the velocity has an absolute minimum. 9. (a) (b) 11. (a) ; the rate at which the area is increasing with respect to side length as x reaches 15 mm (b) 13. (a) (i) (ii) (iii) (b) (c) 15. (a) (b) (c) The rate increases as the radius increases. 17. (a) (b) (c) At the right end; at the left end 19. (a) 4.75 A (b) 5 A; 21. (a) (b) At the beginning 23. 25. (a) (b) , where , , , (c) (d) ; (smaller) (e) 27. (a) ; ; 0 (b) 0; ; (c) At the center; at the edge 29. (a) (b) ; the cost of producing the 201st yard (c) $32.20 31. (a) ; the average productivity increases as new workers are added. 33. 35. (a) 0 and 0 (b) (c) it is possible for the species to coexist. 0, 0, 500, 50; C 0 0.2436 Kmin xpx pxx 2 $32yard Cx 12 0.2x 0.0015x 2 185.2 cmscm 92.6 cmscm 0.694 cms 0.926 cms 81.62 millionyear 75.29 millionyear 14.48 millionyear Pt 3at 2 2bt c d 7,743,770 c 12,822.979 b 7.061422 a 0.00129371 Pt at 3 bt 2 ct d 78.5 millionyear 16 millionyear; 4003t ln 3; 6850 bacteriah dVdP CP 2 t 2 3 s 18 kgm 12 kgm 6 kgm 24 ft2ft 16 ft2ft 8 ft2ft A 2r r 4 4.1 4.5 5 A 2x x 30 mm2mm s17 ms 5.02 ms t 1.5 s t 4 s 2 t 3 0 t 1 3 t 4 1 t 2 1 t 2 2 t 3 0 t 1 6 t 8 2 t 4 4 t 6 0 t 2 _1 1 8 0 a √ s 1 32 2s2 fts2 1 16 2 cost4; t 0, s 1 t 4, s _1 t 8, s 1 s 0 t =10, s=0 4 t 8 t 0, 4, 8 1 8s2 fts 4 sin t 4 EXERCISES 3.8 N PAGE 239 1. About 235 3. (a) (b) (c) (d) 5. (a) 1508 million, 1871 million (b) 2161 million (c) 3972 million; wars in the ﬁrst half of century, increased life expectancy in second half 7. (a) (b) 9. (a) (b) (c) 11. 13. (a) (b) 15. (a) (b) 17. (a) (b) 19. (a) (i) $3828.84 (ii) $3840.25 (iii) $3850.08 (iv) $3851.61 (v) $3852.01 (vi) $3852.08 (b) , EXERCISES 3.9 N PAGE 245 1. 3. 5. 7. 9. 11. (a) The plane’s altitude is 1 mi and its speed is . (b) The rate at which the distance from the plane to the station is increasing when the plane is 2 mi from the station (c) (d) (e) 13. (a) The height of the pole (15 ft), the height of the man (6 ft), and the speed of the man (5 fts) (b) The rate at which the tip of the man’s shadow is moving when he is 40 ft from the pole (c) (d) (e) 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. (a) (b) 39. 41. 43. EXERCISES 3.10 N PAGE 252 1. 3. 5. ; , 7. 9. 11. (a) (b) dy t 1 t 2 dt dy 2xx cos 2x sin 2x dx 0.045 x 0.055 1.204 x 0.706 s0.99 0.995 s0.9 0.95 3 3 _3 _1 (0, 1) (1, 0) y=œ„„„„ 1-x y=1- x 1 2 s1 x 1 1 2 x Lx x 2 Lx 10x 6 7 4s15 6.78 ms 1650s31 296 kmh 10 9 kmmin 0.096 rads 360 fts 0.396 mmin 107 810 0.132 s 80 cm3min 0.3 m2s 65 0.38 ftmin 10 3 cmmin 10,000 800,0009 2.89 105 cm3min 720 13 55.4 kmh 1.6 cmmin 837s8674 8.99 fts 65 mih 25 3 fts 15 6 x y y y x 15 6 250s3 mih y 2 x 2 1 y x 1 500 mih 46 13 70 325 mmin 48 cm2s dVdt 3x 2 dxdt A0 3000 dAdt 0.05A 39.9 kPa 64.5 kPa 67.74 min 13.3C 116 min 137F 2500 years 199.3 years 9.92 mg 100 2t30 mg 2000 ln 0.9 211 s Ce0.0005t ln 100ln 4.2 3.2 h 10,632 bacteriah 7409 1004.2t A76 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 13. (a) (b) 15. (a) (b) ; 17. (a) (b) 19. 21. 23. 25. 27. 33. (a) , 0.01, 1% (b) , , 35. (a) ; (b) ; 37. (a) (b) 43. (a) 4.8, 5.2 (b) Too large EXERCISES 3.11 N PAGE 259 1. (a) 0 (b) 1 3. (a) (b) 5. (a) 1 (b) 0 21. 23. (a) 1 (b) (c) (d) (e) 0 (f ) 1 (g) (h) (i) 0 31. 33. 35. 37. 39. 41. 43. 45. 47. 51. (a) 0.3572 (b) 70.34° 53. (b) 55. CHAPTER 3 REVIEW N PAGE 261 True-False Quiz 1. True 3. True 5. False 7. False 9. True 11. True (ln 1 s2), s2) y 2 sinh 3x 4 cosh 3x y 1 xsx 2 1 y sinh1x3 y 1 2sx1 x Gx 2 sinh x 1 cosh x2 y sech2x 1 tanh2x f t 2et sech2et tanhe t y 3e cosh 3x sinh 3x hx tanh x f x x cosh x 1 csch x 3 4, tanh x 4 5, coth x 5 4 sech x 3 5, sinh x 4 3, 1 2e 2 e2 3.62686 3 4 r2h 2rh r 1 56 0.018 1764 2 179 cm3 1 84 0.012 84 27 cm2 0.6% 0.006 36 cm2 270 cm3 1 90 0.965 4.02 32.08 y x 1 0 1 2 y= 2 x Îy dy dx=Îx y 0.1, dy 0.125 y 1 0 1 y=2x-≈ Îy dy dx=Îx y 0.64, dy 0.8 0.2 dy sec2x dx 0.0101 0.01 dy 1 10 e x10 dx dy 6r 2 1 r 33 dr dy 2 u 12 du Exercises 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 57. 59. 61. ; 63. (a) (b) (c) 65. 69. (a) 2 (b) 44 71. 73. 75. 77. 79. 81. 83. 85. 87. , 89. (a) (b) (c) (d) (e) ; 0 t 2 t 2 20 0 3 15 a v position t y 23 t 2; 0 t 2 vt 3t 2 12; at 6t at Aectc 2 2 cost 2c sint vt Aectc cost sint y 2 3 x 2 14 3 x 3, 0 ftsin 4xtsin 4xcos 4x4 fxtx 2 tx fx 2 f x tx 2 txtx te xe x 2txtx 2xtx x 2tx (4, s2), (54, s2) (4, 4) 10 _10 _10 10 (1, 2) ƒ y 7 4 x 1 4, y x 8 10 3x 2s5 x y x 2 y x 2 y 2x 1 y 2s3x 1 s33 5x 4y 11 4 27 3 sin(estan 3x)estan 3x sec23x 2stan 3x cosh x ssinh2x 1 3 tanh 3x 2x 2 coshx 2 sinhx 2 x 243x 2 55x 52 2sx 1x 38 2 cos tansin sec2sin cos(tan s1 x 3)(sec2s1 x 3) 3x 2 2s1 x 3 6x csc2 3x2 5 5 sec 5x 4x 1 16x 2 tan14x cot x sin x cos x 2 1 2x ln 5 2x y cosxy x cosxy 1 x 12 3x ln xln 31 ln x 1 c 2e cx sin x 2 sec 2 tan 2 1 1 tan 22 1 y 4 2xy 4xy 3 x 2 3 e1x1 2x x 4 cossx sx sin sx 2sx t 2 1 1 t 22 2 cos 2 esin 2 22x 2 1 sx 2 1 1 2sx 4 3s 3 x7 6xx 4 3x 2 522x 2 3 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A77 91. 93. (a) (b) (c) (d) 95. (a) (b) 97. 99. 101. 103. (a) ; (b) 105. 107. 109. 111. PROBLEMS PLUS N PAGE 266 1. 9. 11. (a) (b) (c) 15. 17. (b) (i) (or ) (ii) (or ) 19. R approaches the midpoint of the radius AO. 21. 23. 27. 29. 31. CHAPTER 4 EXERCISES 4.1 N PAGE 277 Abbreviations: abs., absolute; loc., local; max., maximum; min., minimum 1. Absolute minimum: smallest function value on the entire domain of the function; local minimum at c: smallest function value when x is near c 3. Abs. max. at , abs. min. at , loc. max. at , loc. min. at and 5. Abs. max. , loc. max. and , loc. min. and 7. 9. 11. (a) (b) (c) y 0 x 1 _1 2 1 2 3 y 0 x 1 _1 2 1 3 y 0 x 1 _1 2 1 3 y x 0 5 4 3 2 1 3 2 1 y x 0 5 1 2 3 4 1 2 3 f5 3 f2 2 f6 4 f4 5 f4 5 r b c r s 11.204 cm3min s2958 1, 2, 1, 0 2se sin a 117 63 127 53 xT 3, , yT 2, , xN (0, 5 3), yN ( 5 2, 0) 480 sin (1 cos s8 cos2) cms 40(cos s8 cos2) cm 4s3s11 rads (0, 5 4) ( 1 2s3, 1 4) 1 8x 2 1 4 1 32 12 3 2 16.7 cm2 0.23 x 0.40 s 3 1.03 1.01 Lx 1 x; s 3 1 3x 1 x 400 fth 13 fts 4 3 cm2min 100 h C0ekt ln 50ln 3.24 3.33 h 25,910 bacteriah 22,040 2003.24t 4 kgm 13. (a) (b) 15. Abs. max. 17. None 19. Abs. min. 21. Abs. max. , abs. and loc. min. 23. Abs. max. 25. Abs. max. 27. Abs. max. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. , 49. , 51. , 53. , 55. , 57. 59. , 61. 63. 65. (a) (b) , 67. (a) 0.32, 0.00 (b) 69. 71. Cheapest, (June 1994); most expensive, (March 1998) 73. (a) (b) (c) EXERCISES 4.2 N PAGE 285 1. 3. 5. is not differentiable on 7. 0.8, 3.2, 4.4, 6.1 9. (a), (b) (c) 11. 0 13. 15. is not continous at 23. 16 25. No 31. No 3 f 1 2 ln[ 1 6(1 e6)] 10 0 10 10 0 10 2s2 (1, 1) f 9 4 2 √ 0 r kr# ¸ 4 27 r¸ 2 3 r¸ v 4 27kr 0 3 r 2 3r0 t 4.618 t 0.855 3.9665C 3 16s3, 0 6 25s3 5 2 6 25s3 5 2 2.19, 1.81 f a a b a abb a bab f 1 ln 3, f ( 1 2) ln 3 4 f 1 1s 8 e f 2 2se f6 3 2s3, f2 0 f 1 s3 f (s2) 2 f 0 0 f 1 1 2 f 1 2 f 3 66 f 2 19 f 1 8 f 2 7 f 0 5 10 0, 2 3 n n an integer 0, 8 7, 4 0, 4 9 0, 2 0, 1 2(1 s5) 4, 2 2 5 f 3 2 f 0 1 f 2 ln 2 f 0 0 f 3 9 f 0 0 f 1 5 y 0 x y 0 x 2 _1 A78 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES EXERCISES 4.3 N PAGE 295 Abbreviations: inc., increasing; dec., decreasing; CD, concave downward; CU, concave upward; HA, horizontal asymptote; VA, vertical asymptote; IP, inﬂection point(s) 1. (a) (b) (c) (d) (e) 3. (a) I/D Test (b) Concavity Test (c) Find points at which the concavity changes. 5. (a) Inc. on ; dec. on (b) Loc. max. at , loc. min. at 7. 9. (a) Inc. on ; (b) Loc. max. ; loc. min. (c) CU on ; CD on ; IP 11. (a) Inc. on , ; dec. on , (b) Loc. max. ; loc. min. (c) CU on , ; CD on ; IP 13. (a) Inc. on , ; dec. on (b) Loc. max. ; loc. min. (c) CU on ; CD on , ; IP 15. (a) Inc. on ; dec. on (b) Loc. min. (c) CU on 17. (a) Inc. on ; dec. on (b) Loc. max. (c) CU on ; CD on ; IP 19. Loc. max. , loc. min. 21. Loc. max. 23. (a) has a local maximum at 2. (b) has a horizontal tangent at 6. 25. 27. 29. y 0 x x y 0 _2 x=2 x y 0 _2 2 x y 0 1 2 3 4 f f f( 3 4) 5 4 f 1 1 f 1 7 (e 83, 8 3e43) 0, e 83 e 83, f e 2 2e e 2, 0, e 2 , f ( 1 3 ln 2) 223 213 (, 1 3 ln 2) ( 1 3 ln 2, ) 34, 0, 74, 0 74, 2 0, 34 34, 74 f54 s2 f4 s2 4, 54 54, 2 0, 4 (s33, 22 9 ) (s33, s33) (s33, ) (, s33) f 1 2 f 0 3 0, 1 , 1 1, 1, 0 ( 1 2, 37 2 ) (, 1 2) 1 2, f2 44 f3 81 dec. on 3, 2 , 3, 2, x 1, 7 x 1 x 5 0, 1 and 5, 6 1, 5 2, 3 2, 4, 4, 6 0, 2 0, 1, 3, 4 1, 3, 4, 6 31. (a) Inc. on (0, 2), (4, 6), ; dec. on (2, 4), (6, 8) (b) Loc. max. at ; loc. min. at (c) CU on (3, 6), ; CD on (0, 3) (d) 3 (e) See graph at right. 33. (a) Inc. on ; dec. on (b) Loc. max. ; loc. min. (c) CU on ; CD on ; IP (d) See graph at right. 35. (a) Inc. on ; dec. on (b) Loc. max. ; loc. min. (c) CU on ; CD on ; IP (d) See graph at right. 37. (a) Inc. on ; dec. on (b) Loc. max. ; loc. min. (c) CU on ; CD on ; IP (d) See graph at right. 39. (a) Inc. on ; dec. on (b) Loc. min. (c) CU on (d) See graph at right. 41. (a) Inc. on ; dec. on (b) Loc. min. (c) CU on , ; CD on ; IPs , (d) See graph at right. 43. (a) Inc. on ; dec. on (b) Loc. min. (c) CU on ; CD on , ; IP , (d) See graph at right. (53, 5 4) (3, 5 4) 53, 2 0, 3 3, 53 f 1 0, ¨ (π, _1) ” , ’ y π 3 5 4 ” , ’ 5π 3 5 4 1 _1 0 π 2π , 2 (2, 6s 3 2) 0, 0 0, 2 2, , 0 C1 3 , 1 x y _4 0 { 2, 6 Œ„ 2 } (_1, _3) 1, 3, A2 2 3, 2 x y _3 _2 _2 2 2, 1, 3 , 1 1, h0 1 h2 7 2, 0 x _1 (_1, 3) (0, _1) (_2, 7) y 7 , 2, 0, (1s3, 23 9 ) (, 1s3), 1s3, ) (1s3, 1s3) f 0 2 f 1 3, f 1 3 1, 0, 1, x 1 0 (1, 3) (_1, 3) ” , ’ y 1 23 9 1 œ„ 3 ” , ’ 23 9 1 œ„ 3 , 1, 0, 1 ( 1 2, 13 2 ) (, 1 2) ( 1 2, ) f 2 20 f 1 7 1, 2 y 0 x (1, 7) (2, _20) ” , _ ’ 1 2 13 2 , 1, 2, 6, x 4, 8 x 2, 6 x y 0 2 4 6 8 8, APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A79 45. (a) HA , VA , (b) Inc. on , ; dec. on (c) Loc. max. (d) CU on , ; CD on (e) See graph at right. 47. (a) HA (b) Dec. on (c) None (d) CU on (e) See graph at right. 49. (a) VA (b) Dec. on (c) None (d) CU on (0, 1); CD on ; IP (1, 0) (e) See graph at right. 51. (a) HA , VA (b) Inc. on , (c) None (d) CU on , ; CD on ; IP (e) See graph at right. 53. 55. (a) Loc. and abs. max. , no min. (b) 57. (b) CU on , ; CD on , , ; IP , , , 59. CU on ; CD on 61. (a) The rate of increase is initially very small, increases to a maximum at , then decreases toward 0. (b) When (c) CU on ; CD on (d) 63. ; CD 65. 28.57 min, when the rate of increase of drug level in the blood-stream is greatest; 85.71 min, when rate of decrease is greatest 67. EXERCISES 4.4 N PAGE 304 1. (a) Indeterminate (b) 0 (c) 0 (d) , , or does not exist (e) Indeterminate 3. (a) (b) Indeterminate (c) 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 1 25. 27. 1 29. 31. 0 33. 35. 37. 39. 41. 3 43. 0 45. 47. 49. 51. 53. 1 1 2 1 2 2 1 24 1 2aa 1 1 2 1 2 ln 5 3 1 2 0 pq 9 5 2 f x 1 92x 3 3x 2 12x 7 K3 K2 8, 350 8, 18 0, 8 t 8 t 8 h 0.6, 0.0 , 0.6, 0.0, 5.35, 0.44 3.71, 0.63 2.57, 0.63 0.94, 0.44 5.35, 2 2.57, 3.71 0, 0.94 3.71, 5.35 0.94, 2.57 1 4(3 s17) f 1 s2 3, ( 1 2, 1e 2) ( 1 2, ) (1, 1 2) , 1 1, , 1 x y 0 x=_1 y=1 x 1 y 1 1, e 0, e y 0 x (1, 0) 1 x=e x=0 x 0, x e , , x y 0 1 y 0 1, 1 1, , 1 f 0 0 0, 1, 1, 1, 0 , 1 x y 0 x=1 x=_1 y=1 x 1 x 1 y 1 55. 57. 59. 61. 63. 65. 67. 71. 77. 79. 83. (a) EXERCISES 4.5 N PAGE 314 1. A. B. y-int. 0; x-int. 0 C. About D. None E. Inc. on F. None G. CU on ; CD on ; IP (0, 0) H. See graph at right. 3. A. B. y-int. 2; x-int. 2, C. None D. None E. Inc. on (1, 5); dec. on F. Loc. min. ; loc. max. G. CU on ; CD on ; IP H. See graph at right. 5. A. B. y-int. 0; x-int. 4, 0 C. None D. None E. Inc. on ; dec. on F. Loc. min. G. CU on , ; CD on ; IP (0, 0), H. See graph at right. 7. A. B. y-int. 1 C. None D. None E. Inc. on , ; dec. on F. Loc. max. ; loc. min. G. CU on ; CD on IP H. See graph at right. 9. A. B. y-int. 0; x-int. 0 C. None D. VA , HA E. Dec. on F. None G. CU on ; CD on H. See graph at right. , 1 1, , 1, 1, y 1 x 1 x y 0 x 1 y1 x x 1 (1s 3 4, 1 9(2 s 3 16)) (, 1s 3 4); (1s 3 4, ) f 1 2 f 0 1 0, 1 1, , 0 y x 0 1, _2 0, 1 2, 16 2, 0 0, , 2 f 3 27 , 3 3, y 0 x (_3, _27) 3, 11 3, , 3 f 5 27 f 1 5 , 1, 5, y 0 x (1, _5) (5, 27) 1 2(7 3s5) , 0 0, , 0, 0 y x 1 1 0 56 16 9 a 1 1 4 e2 1se e 4 1 e 3 e2 A80 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 11. A. B. y-int. C. About y-axis D. VA , HA E. Inc. on , ; dec. on (0, 3), F. Loc. max. G. CU on ; CD on H. See graph at right. 13. A. B. y-int. 0; x-int. 0 C. About (0, 0) D. HA E. Inc. on ; dec. on F. Loc. min. ; loc. max. ; G. CU on , ; CD on , ; IP (0, 0), H. See graph at right. 15. A. B. x-int. 1 C. None D. HA ; VA E. Inc. on ; dec. on F. Loc. max. G. CU on ; CD on , ; IP H. See graph at right 17. A. B. y-int. 0, x-int. 0 C. About y-axis D. HA E. Inc. on ; dec. on F. Loc. min. G. CU on ; CD on , ; IP H. See graph at right 19. A. B. y-int. 0; x-int. 0, 5 C. None D. None E. Inc. on ; dec. on F. Loc. max. G. CD on H. See graph at right. 21. A. B. x-int. C. None D. None E. Inc. on ; dec. on F. None G. CD on H. See graph at right. , 2, 1, (, 2) (1, ) 2, 1 y 0 x 1 _2 , 2 1, , 5 f ( 10 3 ) 10 9 s15 ( 10 3 , 5) (, 10 3 ) y x ” , ’ 10 3 10œ„„ 9 15 , 5 (1, 1 4) 1, , 1 1, 1 f 0 0 , 0 0, y 1 x y (0, 0) y=1 1 4 ”1, ’ 1 4 ”_1, ’ (3, 2 9) 0, 3 , 0 3, f 2 1 4 , 0, 2, 0, 2 x 0 y 0 ”3, ’ 2 9 x y 0 1 ”2, ’ 1 4 , 0 0, (3s3, s312) (0, 3s3) (, 3s3) (3s3, ) (3s3, 0) f 3 1 6 f 3 1 6 , 3, 3, 3, 3 y 0 y x ”3, ’ 1 6 ”_3, _ ’ 1 6 3, 3 , 3, 3, f 0 1 9 3, x y x 3 x 3 3, 0 , 3 y 0 x 3 1 9 x x 3 23. A. B. y-int. 0; x-int. 0 C. About the origin D. HA E. Inc. on F. None G. CU on ; CD on ; IP H. See graph at right. 25. A. B. x-int. C. About (0, 0) D. VA E. Dec. on , F. None G. CU on , ; CD on , ; IP H. See graph at right. 27. A. B. y-int. 0; x-int. C. About the origin D. None E. Inc. on , ; dec. on F. Loc. max. ; loc. min. G. CU on ; CD on ; IP H. See graph at right. 29. A. B. y-int. ; x-int. C. About -axis D. None E. Inc. on ; dec. on F. Loc. min. G. CU on ; CD on ; IP H. See graph at right. 31. A. B. y-int. 0; x-int. ( an integer) C. About the origin, period D. None E. Inc. on ; dec. on F. Loc. max. ; loc. min. G. CU on ; CD on ; IP H. See graph at right. 33. A. B. y-int. 0; x-int. 0 C. About y-axis D. VA E. Inc. on ; dec. on F. Loc. min. G. CU on H. See graph at right. 2, 2 f 0 0 2, 0 0, 2 x y 0 x π 2 x π 2 x 2 2, 2 n, 0 2n, 2n 1 2n 1, 2n f 2n 32 2 f 2n 2 2 2n 2, 2n 32 x y 0 2π _2π 1 2 _2 _1 ” , 2’ π 2 ” , 2’ π 2 2n 2, 2n 2 2 n n 1, 0 , 1, 1, 1, 1 f 0 1 , 0 0, y y 0 x (_1, 0) (1, 0) (0, _1) 1 1 0, 0 , 0 0, f 1 2 x y 0 _3œ„ 3, 0 3œ„ 3, 0 1, _2 _1, 2 0, 0 f 1 2 1, 1 1, , 1 0, 3s3 (s23, 1s2) (s23, 1) (s23, 0) (0, s23) (1, s23) 0, 1 1, 0 x 0 1 1 x y 0 1 {x x 1, x 0} 1, 0 0, 1 0, 0 0, , 0 , y 1 x y (0, 0) y=_1 y=1 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A81 35. A. C. None D. None E. Inc. on , ; dec. on , F. Loc. min. , ; loc. max. G. CU on , ; CD on ; IP , H. See graph at right. 37. A. All reals except ( an integer) B. y-int. 0; x-int. C. About the origin, period D. VA E. Inc. on F. None G. CU on ; CD on ; IP H. 39. A. B. y-int. C. Period D. None Answers for E–G are for the interval . E. Inc. on , ; dec. on F. Loc. max. ; loc. min. G. CU on , where , ; CD on ; IP when H. 41. A. B. y-int. C. None D. HA E. Inc. on F. None G. CU on ; CD on ; IP H. See graph at right. 43. A. B. None C. None D. VA E. Inc. on ; dec. on F. Loc. min. G. CU on H. See graph at right. 0, f1 1 0, 1 1, x 0 y 0 x (1, 1) 0, (0, 1 2) 0, , 0 y 0, y 1 x y 0 y 1 1 2 _2π 1 e 2π 4π y x x , , sin1( 1 2(1 s5)) , 2 0, f32 e1 f2 e 2, 32 32, 2 0, 2 0, 2 2 1 x y _2π 0 2π x=_3π x=_π x=π x=3π 2n, 0 2n 1, 2n 2n, 2n 1 2n 1, 2n 1 x 2n 1 2 2n n 2n 1 2, , 2 , 2 2, 3 0, 3π x y 0 π 2π 7π 3 5π 3 π 3 f 53 56 1 2s3 f 73 76 1 2s3 f 3 6 1 2s3 53, 73 0, 3 73, 3 3, 53 0, 3 45. A. B. y-int. C. None D. HA E. Dec. on F. None G. CU on ; CD on ; IP H. See graph at right. 47. A. All in ( an integer) B. x-int. C. Period D. VA E. Inc. on ; dec. on F. Loc. max. G. CD on H. 49. A. B. y-int. 0; x-int. 0 C. About (0, 0) D. HA E. Inc. on ; dec. on , F. Loc. min. ; loc. max. G. CU on , ; CD on , ; IP , H. 51. A. B. y-int. 2 C. None D. None E. Inc. on ; dec. on F. Loc. min. G. CU on H. See graph at right. 53. 55. 57. 59. y 2x 2 y x 1 y x L L/2 0 m 0 √ (0, m¸) √=c , f ( 1 5 ln 2 3) ( 2 3) 35 ( 2 3) 25 (, 1 5 ln 2 3) ( 1 5 ln 2 3, ) x y 0 local minimum (0, 2) 1 œ„ 2 1 œ„„ 2e ” , ’ x y 0 0, 0 (s32, s32e32) (0, s32) (, s32) (s32, ) (s32, 0) f(1s2) 1s2e f(1s2) 1s2e (1s2, ) (, 1s2) (1s2, 1s2) y 0 x y _4π _3π _2π _π π 2π 3π 4π 0 2n, 2n 1 f 2 2n 0 2 2n, 2n 1 2n, 2 2n x n 2 2 2n n 2n, 2n 1 x (ln 1 2, 4 9) , ln 1 2) (ln 1 2, ) y 0, y 1 y 0 x y=1 ”ln , ’ 4 9 1 2 1 4 A82 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 61. A. B. y-int. 1; x-int. C. None D. VA ; SA E. Dec. on , F. None G. CU on ; CD on H. See graph at right 63. A. B. None C. About (0, 0) D. VA ; SA E. Inc. on , ; dec. on , F. Loc. max. ; loc. min. G. CU on ; CD on H. See graph at right. 65. A. B. y-int. 1; x-int. C. None D. SA E. Inc. on F. None G. CU on , CD on , ; IP , H. See graph at right. 67. 71. VA , asymptotic to EXERCISES 4.6 N PAGE 320 1. Inc. on , ; dec. on , ; loc. max. ; loc. min. , ; CU on , ; CD on ; IP , 3. Inc. on , ; dec. on , ; loc. max. ; loc. min. , ; CU on , , ; CD on , ; 2.92, 15.08 11.34, 0 15.08, 0, 2.92 , 11.34 f 18.93 12,700,000 f 15 9,700,000 f 4.40 53,800 4.40, 18.93 , 15 18.93, 15, 4.40 _6 10 0 4 ƒ 2.7 3.96 4.04 2.4 ƒ 2.54, 3.999 1.46, 1.40 1.46, 2.54 2.54, , 1.46 f 2.58 3.998 f 0.92 5.12 f 2.5 4 2.5, 2.58 , 0.92 2.58, 0.92, 2.5 2 2 x 10 10 y 0 ƒ y=˛ y x 3 x 0 x y 0 y x π 2 y x π 2 0, 1 (s3, 1 3 2s3) (s3, ) (s3, 0) (0, s3); (, s3) , y x 0 _1 (0, 1) y=2x+1 {œ„ 3, œ„ 3+1} 3 2 {_œ„ 3, _ œ„ 3+1} 3 2 y 2x 1 1 , 0 0, f 2 4 f 2 4 0, 2 2, 0 2, , 2 y x x 0 0 x y (2, 4) y=x (_2, _4) x x 0 (, 1 2) ( 1 2, ) ( 1 2, ) (, 1 2) y x 2 x 1 2 1 4(5 s17) x y 0 y=_x+2 x= 1 2 (, 1 2) ( 1 2, ) IP , , , 5. Inc. on , , ; dec. on , ; loc. max. ; CU on , , ; CD on , ; IP 7. Inc. on , ; dec. on , ; loc. max. ; loc. min. , ; CU on , CD on ; IP , 9. Inc. on ; dec. on , , ; CU on , CD on , 11. (a) (b) (c) Loc. min. ; CD on ; CU on e32, 0, e32 f(1se) 12e limx l 0 f x 0 1 0.25 _0.25 1.75 , 1 2e 1 œ„ e 75 _10 f _1 1 1 0.95 f _100 _1 (12 s138, 0) (, 12 s138) 0, ; (12 s138, 12 s138) 0, 8 s61, 0 (, 8 s61) (8 s61, 8 s61) _4 4 _10 30 ƒ _2.5 0 10 6 ƒ 1.28, 1.48 1.28, 8.77 1.28, 1.28 1.28, 4; 4, 1.28 f 2.89 9.99 f 1.49 8.75 f 1.07 8.79 1.07, 2.89 4, 1.49 2.89, 4 1.49, 1.07 3 _3 _5 5 0.506, 0.192 0.24, 2.46 1.7, 0.506 2.46, 0.506, 0.24 , 1.7 f 1 1 3 2.46, 1, 2.46 0.24, 1 1.7, 0.24 , 1.7 60,000 f _30,000 10 _10 10,000,000 f _13,000,000 25 _25 15.08, 8,150,000 2.92, 31,800 11.34, 6,250,000 0, 0 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A83 13. Loc. max. , , ; loc. min. 15. CU on , , , , ; CD on , , ; IP , , , , 17. Inc. on ; dec. on ; loc. max. ; CU on ; CD on ; IP 19. Inc. on , , , , ; dec. on , , , ; loc. max. , , , ; loc. min. , ; CU on , ; CD on , , ; IPs at , 5 f 0 20 _5 15.81, 3.91, 18.65, 4.20 9.60, 2.95, 12.25, 3.27 18.65, 20 12.25, 15.81 4.91, 4.10, 0, 4.10, 4.91, 9.60 15.81, 18.65 9.60, 12.25 f 17.08 3.49 f 10.79 2.43 f 14.34 4.39 f 8.06 3.60 f 1.77 2.58 f4.51 0.62 14.34, 17.08 8.06, 10.79 1.77, 4.10 4.51, 4.10 17.08, 20 10.79, 14.34 4.91, 8.06 0, 1.77 4.91, 4.51 0.5 f 5 0 0.94, 0.34 0, 0.94 0.94, f 0.43 0.41 0.43, 0, 0.43 0.1, 0.0000066 0.5, 0.00001 1, 0 5.0, 0.005 35.3, 0.015 0.5, 0.1 5.0, 1 , 35.3 4, 2, 4 0.1, 2 1, 0.5 35.3, 5.0 f x 2 x 1x 6 36x 5 6x 4 628x 3 684x 2 672x 64 x 24x 46 f x xx 12x 3 18x 2 44x 16 x 23x 45 0.03 8 2.5 0 500 2 1 1500 0.02 3.5 8 0.04 y x 1 f 3 0 f 5.2 0.0145 f 0.82 281.5 f 5.6 0.018 21. Inc. on , ; CU on , ; CD on , ; IP 23. (a) (b) , (c) Loc. max. (d) IP at 25. Max. , , ; min. , , ; IP , , , , 27. For , there is no IP and only one extreme point, the origin. For , there is a maximum point at the origin, two minimum points, and two IPs, which move downward and away from the origin as . 4 _2.3 _2.1 2.1 _3 _2 _1 1 4 c l c 0 c 0 1 0.9997 0.55 0.73 1.2 _1.2 _2π 2π 1.2 1.2 0 π f 2.28, 0.34 1.75, 0.77 1.17, 0.72 0.66, 0.99998 0.61, 0.99998 f 2.73 0.51 f 1.46 0.49 f 0.64 0.99996 f 1.96 1 f 0.68 1 f 0.59 1 x 0.58, 4.37 f e e 1e lim x l x 1x 1 lim x l0 x 1x 0 2 _1 0 8 0.4, 0.8 0.4, 0.4, 0 0, 0.4 , 0.4 _3 3 _1 1 ƒ ƒ 0, , 0 A84 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 29. There is no maximum or minimum, regardless of the value of c. For , there is a vertical asymptote at , , and . is a transitional value at which for . For , , , and there are two IPs, which move away from the y-axis as . 31. For , the maximum and minimum values are always , but the extreme points and IPs move closer to the y-axis as c increases. is a transitional value: when c is replaced by , the curve is reﬂected in the x-axis. 33. For , the graph has local maximum and minimum values; for it does not. The function increases for and decreases for . As c changes, the IPs move vertically but not horizontally. 35. For , and . For , and . As increases, the maximum and minimum points and the IPs get closer to the origin. c lim x l f x 0 lim x l f x c 0 lim x l f x lim x l f x 0 c 0 3 3 _3 _3 _2 1 0 1 2 10 _10 _15 1 5 c=3 c=1 c=0.5 c=_3 c=_1 c=_0.5 c=0 c 1 c 1 c 1 c 1 0.6 0.6 5 5 0.2 0.5 1 2 1 4 c c 0 1 2 c 0 c=_0.5 c=_2 c=_1 4 _1 _4 4 c=0.5 c=2 c=1 2 _1 _4 4 c l lim x l f x 1 lim x l 0 fx 0 c 0 x 0 f x 1 c 0 lim x l f x 1 lim x l 0 fx x 0 c 0 37. (a) Positive (b) EXERCISES 4.7 N PAGE 328 1. (a) 11, 12 (b) 11.5, 11.5 3. 10, 10 5. 25 m by 25 m 7. 9. (a) (b) (c) (d) (e) (f) 11. 1000 ft by 1500 ft 13. 15. $191.28 17. 19. 21. Square, side 23. 25. Base , height 27. 29. 31. 24 cm, 36 cm 33. (a) Use all of the wire for the square (b) m for the square 35. 37. 41. 43. (a) (b) (c) 45. Row directly to B 47. km east of the reﬁnery 49. ft from the stronger source 51. 53. (b) (i) $342,491; $342unit; $390unit (ii) 400 (iii) $320unit 55. (a) (b) $9.50 57. (a) (b) $175 (c) $100 61. 9.35 m 65. 67. 69. At a distance from A 71. 73. (a) About 5.1 km from B (b) C is close to B; C is close to D; , where (c) ; no such value (d) s414 1.6 1.07 x BC WL s25 x 2x 1 2L W2 5 2s5 6 x 6 in. px 550 1 10 x px 19 1 3000 x a 23 b 2332 10s 3 3(1 s 3 3) 4.85 6s[h s(2s2)] cos1(1s3) 55 3 2 S 2 csc csc s3 cot E 24r V 2 R3(9s3) Height radius s 3 V cm 40s3(9 4s3) r 2(1 s5) 4 r 3(3s3) 3r2 s3r L2, s3 L4 s2r ( 1 3, 4 3s2) ( 28 17, 7 17) 4000 cm3 14,062.5 ft 2 Ax 375x 5 2 x 2 5x 2y 750 A xy y x 75 120 9000 ft@ 250 50 12,500 ft@ 125 100 12,500 ft@ N 1 12 _12 _6 6 c=4 c=1 c=0.5 c=_1 c=0.1 c=0.2 c=0 c=_4 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A85 EXERCISES 4.8 N PAGE 338 1. (a) (b) No 3. 5. 1.1797 7. 9. 11. 13. 1.217562 15. 0.876726 17. 19. 21. 0.641714 23. , 25. 27. 29. (b) 31.622777 35. (a) (b) 37. 39. 41. 0.76286% EXERCISES 4.9 N PAGE 345 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. (a) (b) (c) (d) About 9.09 s 67. 225 ft 69. $742.08 71. 130 11 11.8 s 9.8s4504.9 93.9 ms s4504.9 9.58 s st 450 4.9t 2 st 10 sin t 3 cos t 6 t 3 st 1 6t 3 t 2 3t 1 st 1 cos t sin t x 0 F 2π _2π y y 0 x 1 _1 2 1 2 3 (1, 1) (2, 2) (3, 1) y 0 x 1 F 1 b 10 ln x ln 2x ln 2 x 2 cos x 1 2 x x 2 2x 3 9x 9 sin cos 5 4 2x 4 1 3x 3 5x 2 22x 59 3 3 2x 23 1 2 if x 0; 3 2x 23 5 2 if x 0 2 sin t tan t 4 2s3 4x 32 2x 52 4 x 3x 2 8 e t 1 2Ct 2 Dt E 3 20 x 83 Cx D x 3 x 4 Cx D Fx x 5 1 3 x 6 4 Fx 1 2x 2 ln x 1x 2 C Fx 5e x 3 sinh x C G sin 5 cos C Fu 1 3u 3 6u12 C Fx 54x 8 C1 if x 0 54x 8 C2 if x 0 Fx 4x 32 6 7x 76 C Fx 4x 54 4x 74 C Fx 2 3x 3 1 2x 2 x C Fx 1 2x 1 4x3 1 5x4 C Fx 1 2x 2 3x C 0.410245, 0.347810 0.904557, 1.855277 2.0212 1.293227, 0.441731, 0.507854 0.21916368, 1.08422462 1.97806681, 0.82646233 1.13929375, 2.98984102 1.93822883, 1.21997997 1.412391, 3.057104 0.724492, 1.220744 1.82056420 1.25 1.1785 4 5 x2 2.3, x3 3 73. 75. 77. (a) 22.9125 mi (b) 21.675 mi (c) 30 min 33 s (d) 55.425 mi CHAPTER 4 REVIEW N PAGE 347 True-False Quiz 1. False 3. False 5. True 7. False 9. True 11. True 13. False 15. True 17. True 19. True Exercises 1. Abs. max. , abs. and loc. min. ; loc. min. 3. Abs. max. , abs. and loc. min. 5. Abs. max. ; abs. min. ; loc. max. ; loc. min. 7. 9. 8 11. 0 13. 15. 17. 19. A. B. y-int. 2 C. None D. None E. Dec. on F. None G. CU on ; CD on ; IP H. See graph at right. 21. A. B. y-int. 0; x-int. 0, 1 C. None D. None E. Inc. on , dec. on F. Loc. min. G. CU on , ; CD on ; IP , H. See graph at right. 23. A. B. None C. None D. HA ; VA , E. Inc. on ; dec. on , , F. Loc. min. G. CU on , ; CD on H. See graph at right. , 0 3, 0, 3 f 1 1 4 3, 0, 1 , 0 1, 3 x 3 x 0 y 0 y 0 x x 3 x x 0, 3 1, 0 ( 1 2, 1 16) ( 1 2, 1) 1, (, 1 2) f( 1 4) 27 256 (, 1 4) ( 1 4, ) y 0 x 1 1 2 2 0, 2 0, , 0 , y x 2 y x y=_2 y=2 y 0 x 1 2 9 12 x 6 1 2 f 2 3 2 3 1 2s3 f 3 3 1 2s3 f 0 0 f f( 1 3) 9 2 f2 2 5 f3 1 f3 1 f4 5 62,500 kmh2 4.82 ms2 88 15 5.87 fts2 A86 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 25. A. B. y-int. 0, x-int. 0 C. None D. VA ; SA E. Inc. on , ; dec. on , F. Loc. max. ; loc. min. G. CU on ; CD on H. See graph at right. 27. A. B. y-int. 0; x-int. C. None D. None E. Inc. on , dec. on F. Loc. min. G. CU on H. See graph at right. 29. A. B. y-int. C. About -axis, period D. None E. Inc. on , n an integer; dec. on F. Loc. max. ; loc. min. G. CU on ; CD on ; IP H. 31. A. B. None C. About (0, 0) D. HA E. Dec. on , F. None G. CU on ; CD on H. See graph at right. 33. A. B. y-int. , C. None D. HA E. Inc. on , dec. on F. Loc. max. G. CU on ; CD on ; IP H. 35. Inc. on , ; dec. on , ; loc. max. , loc. min. ; CU on , ; CD on , ; IP , (s6, 5 36 s6) (s6, 5 36 s6) (0, s6) (, s6) (s6, ) (s6, 0) f(s3) 2 9s3 f(s3) 2 9 s3 (s3, ) (, s3) ƒ 1.5 _1.5 _5 5 (0, s3) (s3, 0) 0 ” , e–!’ 1 2 1 2 {1, e–@} y x 1 1, e2 , 1 1, f( 1 2) 12e ( 1 2, ) (, 1 2) y 0 x-int. 0 0 , 1 1, 1, , 1 y 0 x y 0 1 _1 π 2 π 2 _ {x x 1} y x 2 π _2 _π 2π _2π (2n 3, 1 4) 2n 3, 2n 5 3 2n 3, 2n 3 f 2n 2 f 2n 1 2 2n 1 , 2n 2n , 2n 1 2 y 2 2, f( 4 3) 4 9s6 (2, 4 3) ( 4 3, ) 2, 0 y x ” , _ ’ 4 3 4œ„ 6 9 2, , 8 8, f 0 0 f 16 32 8, 0 16, 8 0, , 16 y x 8 x 8 0 x y x 8 y x 8 16, 32 x x 8 37. Inc. on , ; dec. on , loc. max. ; loc. min. , ; CU on , ; CD on ; IP , 39. ; 41. 43. For , f is periodic with period and has local maxima at , n an integer. For , f has no graph. For , f has vertical asymptotes. For , f is con-tinuous on . As C increases, f moves upward and its oscillations become less pronounced. 49. (a) 0 (b) 53. 55. cm from D 57. 59. $11.50 61. 1.297383 63. 1.16718557 65. 67. 69. 71. 73. 75. (b) (c) 77. No 79. (b) About 8.5 in. by 2 in. (c) , PROBLEMS PLUS N PAGE 352 5. 7. 11. 13. 15. 19. (a) , , (c) 23. 3(s 3 2 1) 11 1 2 h c1 3.85 kms, c2 7.66 kms, h 0.42 km T3 s4h2 D 2c1 T2 2h sec c1 D 2h tan c2 T1 Dc1 a e 1e m2, m 24 3.5 a 2.5 2, 4, 2, 4 24 20s23 in. 20s3 in. 5 4 1 _4 F 0.1e x cos x 0.9 st t 2 tan1t 1 f (x 1 2x 2 x 3 4x 4 2x 1 f t t 2 3 cos t 2 f x 2 5x 52 3 5x 53 C f x sin x sin1x C L C 4s3 3s3r 2 CU on C 1 1 C 1 C 1 2n 2 2 C 1 2.16, 0.75, 0.46, 2.21 2.96, 0.18, 3.01; 1.57, 1.57; (s23, e32) 0.82, 0.22 5 0 _5 1 2.5 0.4 _0.5 1.5 f f 15 2.1 _1 _20 1.24, 12.1 0.12, 1.98 0.12, 1.24 1.24, , 0.12 f 1.62 19.2 f 0.23 1.96 f 0 2 0, 1.62; , 0.23 1.62, 0.23, 0 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A87 CHAPTER 5 EXERCISES 5.1 N PAGE 364 1. (a) 40, 52 (b) 43.2, 49.2 3. (a) 0.7908, underestimate (b) 1.1835, overestimate 5. (a) 8, 6.875 (b) 5, 5.375 (c) 5.75, 5.9375 (d) 7. 0.2533, 0.2170, 0.2101, 0.2050; 0.2 9. (a) Left: 0.8100, 0.7937, 0.7904; right: 0.7600, 0.7770, 0.7804 11. 34.7 ft, 44.8 ft 13. 63.2 L, 70 L 15. 155 ft 17. 19. 21. The region under the graph of from 0 to 23. (a) (b) (c) 25. sin b, 1 32 3 n 2n 122n 2 2n 1 12 lim n l 64 n 6 n i1 i 5 4 y tan x lim n l n i1 i 2n cos i 2n 2n lim n l n i1 s 4 1 15in 15n M6 y x 0 1 2 y x 0 1 2 y x 0 1 2 y x 0 1 2 y x 0 1 2 y x 0 1 2 y 0 x 1 π 2 3π 8 π 4 π 8 ƒ=cos x y 0 x 1 π 2 3π 8 π 4 π 8 ƒ=cos x y x 0 5 5 10 y x 0 5 5 10 y=ƒ y=ƒ EXERCISES 5.2 N PAGE 376 1. The Riemann sum represents the sum of the areas of the two rectangles above the -axis minus the sum of the areas of the three rectangles below the -axis; that is, the net area of the rectangles with respect to the -axis. 3. 2.322986 The Riemann sum represents the sum of the areas of the three rectangles above the -axis minus the area of the rectangle below the -axis. 5. (a) 4 (b) 6 (c) 10 7. , 9. 124.1644 11. 0.3084 13. 0.30843908, 0.30981629, 0.31015563 15. The values of appear to be approaching 2. 17. 19. 21. 42 23. 25. 3.75 29. 31. 33. (a) 4 (b) 10 (c) 3 (d) 2 35. 37. 39. 41. 0 43. 3 45. 47. 49. 51. by Comparison Property 8 55. 57. 59. 69. 71. EXERCISES 5.3 N PAGE 387 1. One process undoes what the other one does. See the Fundamental Theorem of Calculus, page 387. 3. (a) 0, 2, 5, 7, 3 (d) (b) (0, 3) (c) x 3 y 0 x 1 1 g 1 2 x1 0 x 4 dx 0 y 2 0 xex dx 2e 12 y 3 4 tan x dx 12 s3 3 x4 1 sx dx 6 2m x2 0 f x dx 2M 122 x5 1 f x dx e 5 e 3 2.5 3 9 4 3 4 lim n l n i1 sin 5 i n n 2 5 lim n l n i1 2 4in 1 2 4in5 4 n 4 3 x8 1 s2x x2 dx x6 2 x ln1 x 2 dx Rn 85 475 x x y 0 x 2 3 4 5 6 1 _1 1 2 ƒ=´-2 x x x y 0 x 2 3 1 2 4 6 ƒ=3- x 1 2 8 10 12 14 6 n 5 1.933766 10 1.983524 50 1.999342 100 1.999836 Rn A88 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 5. (a), (b) 7. 9. 11. 13. 15. 17. 19. 21. 63 23. 25. 27. 29. 31. 1 33. 35. 37. 39. 41. 0 43. The function is not continuous on the interval , so FTC2 cannot be applied. 45. The function is not continuous on the interval , so FTC2 cannot be applied. 47. 49. 2 51. 3.75 53. 55. 57. 59. 29 61. (a) , n an integer (b) , , and , an integer (c) 0.74 63. (a) Loc. max. at 1 and 5; loc. min. at 3 and 7 (b) (c) (d) See graph at right. 65. 73. 75. (b) Average expenditure over ; minimize average expenditure EXERCISES 5.4 N PAGE 397 5. 7. 9. 11. 13. 15. 17. tan C 1 2 2 csc C cos x cosh x C 1 3x 3 4sx C 2t t 2 1 3t 3 1 4t 4 C 1 5x 5 1 8x 4 1 8x 2 2x C 1 3x 3 1x C 0, t f x x 32, a 9 1 4 ( 1 2, 2), 4, 6, 8, 9 x 9 x 8 6 4 2 1 0 _1 y _2 0 n (s4n 1, s4n 1) (s4n 1, s4n 3) 0, 1 0 2sn, s4n 2 s257 y 3x 72 sinx 3 sin sx 2 s 4 x tx 24x 2 1 4x 2 1 39x 2 1 9x 2 1 x y 0 2 1 y=˛ 243 4 3, f sec tan 2, 1 f x x 4 e 2 1 ln 3 49 3 40 3 156 7 7 8 5 9 3 4 y 31 3x3 1 1 3x2 y stan x stan x sec2x hx arctan1x x 2 Fx s1 sec x ty y 2 sin y tx 1x 3 1 x 2 0 1 y t x y=t@ 19. 21. 18 23. 25. 52 27. 29. 31. 33. 35. 8 37. 39. 41. 43. 45. 0, 1.32; 0.84 47. 49. The increase in the child’s weight (in pounds) between the ages of 5 and 10 51. Number of gallons of oil leaked in the ﬁrst 2 hours 53. Increase in revenue when production is increased from 1000 to 5000 units 55. Newton-meters (or joules) 57. (a) (b) 59. (a) (b) 61. 63. 1.4 mi 65. $58,000 67. (b) At most 40%; EXERCISES 5.5 N PAGE 406 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 0 53. 55. 57. 59. 61. 3 63. 65. 67. 69. 71. 73. 75. All three areas are equal. 77. 79. 81. 5 87. CHAPTER 5 REVIEW N PAGE 409 True-False Quiz 1. True 3. True 5. False 7. True 9. True 11. False 13. False 15. False 24 5 4 1 cos 2 t 5 L 4512 L 6 s3 1 3 lne 1 2 16 15 1 3(2s2 1)a 3 e se 0 4 182 9 0.35 _0.35 π 0 F ƒ 1 _1 2 _2 F f 1 4 sin4x C 1 8x 2 14 C 4 7x 274 8 3x 234 C tan1x 1 2 ln1 x 2 C ln sin1x C 1 3 sec3x C lnsin x C ln1 cos2x C 2 3cot x32 C 1sin x C e tan x C 1 21 z323 C 2 31 e x32 C 1 7 sin7 C 2 sin st C 1 3ln x3 C 2 3s3ax bx 3 C (1 cos t C 1 3 ln 5 3x C 1 32x x 232 C 1 63 3x 221 C 1 2 cosx 2 C 1 4 cos4 C 2 9x 3 132 C ex C 5 36 46 2 3 kg 416 2 3 m vt 1 2t 2 4t 5 ms 41 6 m 3 2 m 4 3 3.5 6 256 5 1 4 2s5 55 63 63 4 256 15 2 1e 20 10 _5 0 5 _6 10 _10 sin x 1 4x 2 C APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A89 Exercises 1. (a) 8 (b) 5.7 3. 5. 3 7. 9. 37 11. 13. 15. 17. Does not exist 19. 21. 0 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 55. 0.280981 57. Number of barrels of oil consumed from Jan. 1, 2000, through Jan. 1, 2008 59. 72,400 61. 3 63. 65. 71. PROBLEMS PLUS N PAGE 413 1. 3. 5. 7. 9. 11. (a) (b) 17. CHAPTER 6 EXERCISES 6.1 N PAGE 420 1. 3. 5. 19.5 7. 9. 11. 13. 72 15. 17. 19. 21. 23. 25. 27. 29. 6.5 31. 33. 0.6407 35. 0, 0.90; 0.04 37. 8.38 39. 41. 43. 45. (a) Car A (b) The distance by which A is ahead of B after 1 minute (c) Car A (d) 47. 49. 51. 53. ; EXERCISES 6.2 N PAGE 430 1. y 1 0 x 2 1 y=0 x=1 x=2 y 0 x y=2- 1 2 x 19 12 m ln m 1 0 m 1 6 423 24 5 s3 t 2.2 min 4232 cm2 117 1 3 ft 12s6 9 3 2s3 1 ln 2 2 3 1 2 8 3 32 3 59 12 2 2 ln 2 1 3 ln 2 1 2 1 6 e 1e 10 3 32 3 2(s2 1) 1 2b2b b 1 1 2a2a a 1 1 2n 1n 1, 2 e2 1 f x 1 2x 2 2 3 f x e2x1 2x1 ex c 1.62 4 x3 1 sx 2 3 dx 4s3 y (2e x e sx)2x tx 4x 3 cosx 8 Fx x 21 x 3 64 5 2s1 sin x C 23 3 ln 1 sec C 1 4 ln1 x 4 C 1 2 lncos x2 C 2e sx C 1(2 sin2 t C sx2 4x C 1x 2 ln x x C 1 3 sin 1 21 4 76 9 10 f is c, f is b, xx 0 f t dt is a 1 2 4 6 2 x 2 0 y=ƒ y 2 x 2 0 y=ƒ 6 y 3. 5. 7. 9. 11. 13. y 0 x y=1 y=1 y=3 y=1+sec x y 0 x ” , 3’ π 3 ” , 3’ π 3 2 ( 4 3 s3) y 0 y=x y=1 x y 0 x (1, 1) y=œ„ x 6 x 0 y (4, 2) x 0 y x=2y ¥=x 64 15 y 0 x (1, 1) y=˛ y=x y 0 x 4 21 y 0 x (6, 9) x=2œ„ y y=9 x=0 y 0 x 162 2 A90 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. Solid obtained by rotating the region , about the x-axis 43. Solid obtained by rotating the region above the bounded by about the y-axis 45. 47. (a) (b) 838 49. 51. 53. 55. 57. 24 59. 61. 63. (a) (b) 65. (b) 67. 69. EXERCISES 6.3 N PAGE 436 1. Circumference , height ; 3. 2 15 xx 12 2 x 8 xr 0 sR 2 y 2sr 2 y 2 dy 5 12 r 3 r 2h 2 2r 2R 8 R xr 0 sr 2 y 2 dy 8 15 1 3 10 cm3 2 3b 2h h2(r 1 3h) 1 3 r 2h 196 1110 cm3 x y 2 and x y4 x-axis 0 x 2 0 y cos x 11 8 2 1.288, 0.884; 23.780 x2s2 2s2 [5 2 (s1 y 2 2) 2] dy x 0 12 1 sin x2 dx x 4 0 1 tan3x2 dx 13 30 5 14 7 15 2 10 7 (1, 1) x 0 y x=¥ y=≈ 1 x 0 y x=_1 29 30 x 0 y x=1 (1, 1) x 0 y x=¥ x=1 (1, _1) 16 15 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 3.68 29. Solid obtained by rotating the region , about the -axis 31. Solid obtained by rotating the region bounded by (i) , , and , or (ii) , , and about the line 33. 0.13 35. 37. 39. 41. 43. 45. EXERCISES 6.4 N PAGE 441 1. 588 J 3. 5. 180 J 7. 9. (a) (b) 10.8 cm 11. 13. (a) 625 ft-lb (b) 15. 17. 19. 21. 23. 25. 2.0 m 29. Gm1m2 1 a 1 b 1.04 105 ft-lb 1.06 106 J 2450 J 3857 J 650,000 ft-lb 1875 4 ft-lb W2 3W1 25 24 1.04 J 15 4 ft-lb 9 ft-lb 1 3 r 2h 4 3 r 3 4 3 2 12 4 ln 4 8 1 32 3 y 3 y 0 x 1 x y 2 y 0 x 0 x 1 y 2 y 0 x 3 0 y x 4 x 0 2 4 yssin y dy x1 0 2 x 1 sin x2 x 4 dx x 2 1 2 x ln x dx 5 14 8 3 7 15 16 3 768 7 21 2 0 x y (1, 4) (3, 4) y=4(x-2)@ y=≈-4x+7 2 7 x y x 2 16 0 x y y=e≈ 1 1 x y 0 x 1 1e APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A91 EXERCISES 6.5 N PAGE 445 1. 3. 5. 7. 9. (a) 1 (b) 2, 4 (c) 11. (a) (b) (c) 15. 17. 19. 21. CHAPTER 6 REVIEW N PAGE 446 Exercises 1. 3. 5. 7. 9. 11. 13. 15. (a) (b) (c) 17. (a) 0.38 (b) 0.87 19. Solid obtained by rotating the region , about the y-axis 21. Solid obtained by rotating the region , about the -axis 23. 36 25. 27. 29. (a) (b) 2.1 ft 31. PROBLEMS PLUS N PAGE 448 1. (a) (b) 3. 5. (b) 0.2261 (c) 0.6736 m (d) (i) (ii) 9. 11. (a) (c) Advantage: the markings on the container are equally spaced. 13. 15. CHAPTER 7 EXERCISES 7.1 N PAGE 457 1. 3. 5. 2r 2er2 C 1 5x sin 5x 1 25 cos 5x C 1 3 x 3 ln x 1 9 x 3 C B 16A b 2a f y skA C y 14 V xh 0 f y2 dy y 32 9 x 2 370 3 s 6.5 min 1105 0.003 ins 32 27 f x s2x f t 3t 2 f x 8000 3 8378 ft-lb 3.2 J 125 3 s3 m3 x 0 y 2 sin x 0 x 0 x 2 0 y cos x 8 15 6 2 15 x 3 3 2 ( 2 x)(cos2x 1 4) dx 4 3 2ah h 232 1656 5 64 15 4 3 4 7 12 8 3 54 0.4 L 6 kgm 50 28 F 59F 38 1 3 1.24, 2.81 4 25 1 101 e25 45 28 8 3 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. (b) 45. (b) 51. 53. 55. 57. 59. 61. 63. 65. 2 EXERCISES 7.2 N PAGE 465 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 1 10 tan5t 2 C 1 2 sin 2x C 1 8 sin 4 1 12 sin 6 C 1 6 cos 3x 1 26 cos 13x C ln csc x cot x C 1 3 csc3 1 5 csc5 C s3 1 3 x sec x ln sec x tan x C 1 6 tan6 1 4 tan4 C 1 4 sec4x tan2x ln sec x C 1 3 sec3x sec x C 117 8 1 5 tan5t 2 3 tan3t tan t C tan x x C 1 2 tan2x C ln sin x 2 sin x C 1 2 cos2x ln cos x C 2 45ssin 45 18 sin2 15 sin4 C 16 3 2 2 sin 1 4 sin 2 C 3 8 4 1 3 sin3 x 2 5 sin5 x 1 7 sin7 x C 11 384 1 5 cos5x 1 3 cos3x C 2 ett 2 2t 2 m 9 2 ln 3 13 9 2 e 4 8 1.0475, 2.8731; 2.1828 25 4 75 4 e2 xln x3 3xln x2 6x ln x 6x C 2 3, 8 15 1 4 cos x sin3x 3 8 x 3 16 sin 2x C 4 4 2 _2 F f 1 3x 21 x 232 2 151 x 252 C 7 1 3.5 1.5 ƒ F 2x 1e x C 1 2x 2 1 ln1 x 1 4x 2 1 2x 3 4 C 1 2 4 2sx sin sx 2 cos sx C 32 5 ln 22 64 25 ln 2 62 125 sin x ln sin x 1 C 1 6( 6 3s3) 1 4 3 4e2 1 2 1 2 ln 2 1 1e 3 1 13e 22 sin 3 3 cos 3 C xln x2 2x ln x 2x C 1 2t tan 2t 1 4 lnsec 2t C t arctan 4t 1 8 ln1 16t 2 C 1 22x 1 ln2x 1 x C 1 x 2 cos x 2 2 x sin x 2 3 cos x C A92 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 51. 53. 55. 0 57. 1 59. 0 61. 63. 65. EXERCISES 7.3 N PAGE 472 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 33. 37. 0.81, 2; 2.10 41. 43. EXERCISES 7.4 N PAGE 481 1. (a) (b) 3. (a) (b) 5. (a) (b) 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 1 16 ln x 1 32 lnx 2 4 1 8x 2 4 C 1 4 ln 8 3 1 3 ln x 1 1 6 lnx 2 x 1 1 s3 tan1 2x 1 s3 C 1 2 lnx 2 2x 5 3 2 tan1 x 1 2 C 1 2 lnx 2 1 (1s2 ) tan1(xs2 ) C ln x 1 1 2 lnx 2 9 1 3 tan1x3 C 2 ln x 1x 3 ln x 2 C 1 2x 2 2 lnx 2 4 2 tan1x2 C 1 36 ln x 5 1 6 1 x 5 1 36 ln x 1 C 27 5 ln 2 9 5 ln 3 (or 9 5 ln 8 3) 7 6 ln 2 3 a ln x b C 1 2 ln 3 2 2 ln x 5 ln x 2 C x 6 ln x 6 C At B t 2 1 Ct D t 2 4 Et F t 2 42 1 A x 1 B x 1 Cx D x 2 1 A x 3 B x 32 C x 3 D x 32 A x B x 2 C x 3 Dx E x 2 4 A x B x 1 C x 12 A x 3 B 3x 1 2 2Rr 2 rsR 2 r 2 r 22 R 2 arcsinrR 1 6(s48 sec1 7) 1 4 sin1x 2 1 4 x 2s1 x 4 C 1 2x 1sx 2 2x 1 2 ln x 1 sx 2 2x C sx 2 x 1 1 2 ln(sx 2 x 1 x 1 2) C 9 2 sin1x 23 1 2x 2s5 4x x 2 C 9 500 ln (s1 x 2 1)x s1 x 2 C sx 2 7 C 1 16 a4 1 6 sec1x3 sx 2 92x 2 C 1 4 sin12x 1 2 xs1 4x 2 C ln(sx 2 16 x) C s25 x 225x C 24 s38 1 4 1 3x 2 18sx 2 9 C sx 2 99x C s 1 cos3t3 (2s2 5 2) 24 ƒ 1 1 _2 2 F π _π π _π F f 1 6 sin 3x 1 18 sin 9x C 1 4x 2 1 4 sinx 2 cosx 2 C 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 59. 61. 63. 65. , where 67. (a) (b) The CAS omits the absolute value signs and the constant of integration. EXERCISES 7.5 N PAGE 488 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 0 37. 39. 41. 43. 45. 47. 49. 51. 53. 1 m x 2 coshmx 2 m2 x sinhmx 2 m3 coshmx C ln s4x 2 1 1 2x C ln s4x 1 1 s4x 1 1 C ln x 1 3x 11 3 2x 12 1 3x 13 C 1 3x 3 1ex 3 C 2 31 e x32 C tan 1 2 2 ln sec C ln sec 1 ln sec C 8 1 4 2 sin1 x 1 2 x 1 2 s3 2x x 2 C sin1x s1 x 2 C 15 7 ln 2 7 x ln 1 e x C 3x 23 3 ln x 4 5 3 ln x 2 C 4097 45 x 1 arctan sx sx C e e x C (or 1 4x 2 1 4 x sin 2x 1 8 cos 2x C) 1 4x 2 1 2x sin x cos x 1 4 sin2x C xs1 x 2 C 1 8 cos8 1 6 cos6 C (or 1 4 sin4 1 3 sin6 1 8 sin8 C) 1 2 lnx 2 4x 5 tan1x 2 C 243 5 ln 3 242 25 e 4 e 4 4 ln 9 sin x ln csc x cot x C sin x 1 3 sin3x C 75,772 260,015s19 tan1 2x 1 s19 C 11,049 260,015 lnx 2 x 5 3146 80,155 ln 3x 7 4822 4879 ln 5x 2 334 323 ln 2x 1 1 260,015 22,098x 48,935 x 2 x 5 24,110 4879 1 5x 2 668 323 1 2x 1 9438 80,155 1 3x 7 C 10.23 t ln P 1 9 ln0.9P 900 C 1 11 3 ln 2 4 ln 2 3 2 1 5 ln 2 tanx2 1 tanx2 2 C 1 2 ln x 2 x C 1 2 ln 3 0.55 (x 1 2) lnx 2 x 2 2x s7 tan1 2x 1 s7 C ln tan t 1 ln tan t 2 C ln e x 22 e x 1 C 2sx 3s 3 x 6s 6 x 6 ln s 6 x 1 C 3 10x 2 153 3 4x 2 123 C 2 ln 25 9 ln sx 1 1 sx 1 1 C 7 8s2 tan1 x 2 s2 3x 8 4x 2 4x 6 C APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A93 55. 57. 59. 61. 63. 65. 67. 69. 71. 73. 75. 77. 79. 81. EXERCISES 7.6 N PAGE 493 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 35. 37. 39. 41. 43. (a) ; both have domain 45. ; max. at , min. at 1; IP at , 0, and 1.7 4 f F 4 1.1 0.6 1.7 1 Fx 1 2 lnx 2 x 1 1 2 lnx 2 x 1 1, 0 0, 1 ln 1 s1 x 2 x C ln cos x 1 2 tan2x 1 4 tan4x C 1 101 2x52 1 61 2x32 C 1 4xx 2 2sx 2 4 2 ln(sx 2 4 x) C 1 3 tan x sec2x 2 3 tan x C 2 2 1 5 ln x 5 sx 10 2 C se 2x 1 cos1ex C 1 2ln xs4 ln x2 2 ln[ln x s4 ln x2] C 1 4 tan x sec3x 3 8 tan x sec x 3 8 ln sec x tan x C 1 2s3 ln e x s3 e x s3 C 1 9 sin3x 3 lnsin x 1 C 1 126 4y 4y 232 C 2y 1 8 s6 4y 4y 2 7 8 sin1 2y 1 s7 1 2e2x 1 arctane x 1 2e x C 1 2 tan21z lncos1z C e 2 s4x 2 99x C 1 2 tan2 x 1 ln cos x C 4 1 2 sec x tan x 1 2 ln sec x tan x C 1xs7 2x 2 s2 sin1(s2xs7) C xe x2 C 1 3x sin3x 1 3 cos x 1 9 cos3x C 2 3 tan1x 32 C 2x 2s1 e x 2 ln s1 e x 1 s1 e x 1 C 1 8 ln x 2 1 16 lnx 2 4 1 8 tan1x2 C s1 x 2 1 2arcsin x2 C e x ln1 e x C s2 2s3 ln(2 s3) ln(1 s2) 2 3 x 132 x 32 C tan1cos2x C 2(x 2sx 2)esx C sinsin x 1 3 sin3sin x C 3 7x c73 3 4cx c43 C 2 ln sx 2 ln(1 sx ) C 47. ; max. at , min. at 0; IP at , , and 2.5 EXERCISES 7.7 N PAGE 505 1. (a) (b) is an underestimate, and are overestimates. (c) (d) 3. (a) (underestimate) (b) (overestimate) 5. (a) (b) 7. (a) 2.413790 (b) 2.411453 (c) 2.412232 9. (a) 0.146879 (b) 0.147391 (c) 0.147219 11. (a) 0.451948 (b) 0.451991 (c) 0.451976 13. (a) 4.513618 (b) 4.748256 (c) 4.675111 15. (a) (b) (c) 17. (a) 1.064275 (b) 1.067416 (c) 1.074915 19. (a) (b) , (c) for , for 21. (a) , ; , ; , (b) , (c) for , for , for 23. (a) 2.8 (b) 7.954926518 (c) 0.2894 (d) 7.954926521 (e) The actual error is much smaller. (f) 10.9 (g) 7.953789422 (h) 0.0593 (i) The actual error is smaller. (j) 25. Observations are the same as after Example 1. n 50 Sn n 22 Mn n 360 Tn n 509 ES 0.000170 ET 0.025839, EM 0.012919 ES 0.000110 S10 2.000110 EM 0.008248 M10 2.008248 ET 0.016476 T10 1.983524 Mn n 50 Tn n 71 EM 0.0039 ET 0.0078 T8 0.902333, M8 0.905620 0.526123 0.543321 0.495333 5.869247, E S 0.000357 5.932957, EM 0.063353 T4 I M4 M4 0.908907 T4 0.895759 Ln Tn I Mn Rn T2 9 I M2 R2 L2 L2 6, R2 12, M2 9.6 0.04 π 0 F ƒ 2 0.7 1 128 sin x cos3x 3 256 sin x cos x 3 256 x Fx 1 10 sin3x cos7x 3 80 sin x cos7x 1 160 sin x cos5x n 5 0.742943 1.286599 1.014771 0.992621 10 0.867782 1.139610 1.003696 0.998152 20 0.932967 1.068881 1.000924 0.999538 Mn Tn Rn Ln n 5 0.257057 0.286599 0.014771 0.007379 10 0.132218 0.139610 0.003696 0.001848 20 0.067033 0.068881 0.000924 0.000462 EM ET ER EL A94 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 27. Observations are the same as after Example 1. 29. (a) 19.8 (b) 20.6 (c) 31. (a) 23.44 (b) 33. 35. 37. 828 39. 6.0 41. 59.4 43. EXERCISES 7.8 N PAGE 515 Abbreviations: C, convergent; D, divergent 1. (a) Inﬁnite interval (b) Inﬁnite discontinuity (c) Inﬁnite discontinuity (d) Inﬁnite interval 3. ; 0.495, 0.49995, 0.4999995; 0.5 5. 7. D 9. 11. D 13. 0 15. D 17. D 19. 21. D 23. 25. 27. D 29. 31. D 33. 35. D 37. 39. 41. e 43. 45. Inﬁnite area 47. (a) It appears that the integral is convergent. 20 0 π 2 y=sec@ x 0.5 _7 7 2 9 y= 2 ≈+9 0 x y 0 x 1 y e x 1 23 8 3 ln 2 8 9 2e 75 4 32 3 1 2 9 1 25 2e2 1 12 1 2 12t 2 0 x y 1 1 2 0.5 1.5 10,177 megawatt-hours 37.73 fts 0.3413 20.53 (c) 49. C 51. D 53. D 55. 57. 59. 65. 67. (a) (b) The rate at which the fraction increases as t increases (c) 1; all bulbs burn out eventually 69. 1000 71. (a) (b) (c) 77. 79. No CHAPTER 7 REVIEW N PAGE 518 True-False Quiz 1. False 3. False 5. False 7. False 9. (a) True (b) False 11. False 13. False Exercises 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 0 31. 33. 35. 37. 39. 41. 43. D 45. 47. 49. 51. 53. 0 55. ln 2x 1 s4x 2 4x 3 C 1 42x 1s4x 2 4x 3 x 1 lnx 2 2x 2 2 arctanx 1 2x C 4 4 3 4 ln 4 8 1 36 1 8e 1 4 1 2 sin 2x 1 8 cos 4x C 4s1 sx C x s4 x 2 sin1 x 2 C 6 3 2 2 5 3 2 lnx 2 1 3 tan1x s2 tan1(xs2) C ln sx 2 1 1 x C ln x 2 sx 2 4x C 1 18 ln9x 2 6x 5 1 9 tan1[ 1 2(3x 1)] C x sec x ln sec x tan x C 1 2 ln x 3 2 ln x 2 C 3es 3 x(s 3 x 2 2s 3 x 2) C s3 1 3 64 5 ln 4 124 25 cosln t C 2 15 ln 2 5 10 ln 2 3 C 1; ln 2 Fs 1s 2, s 0 Fs 1s 1, s 1 Fs 1s, s 0 Ft 1 700 t 0 (in hours) y y=F(t) s2GMR p 1, 1p 12 p 1, 11 p 1 0.1 1 10 ©=sin@ x ≈ ƒ= 1 ≈ t 2 0.447453 5 0.577101 10 0.621306 100 0.668479 1,000 0.672957 10,000 0.673407 y t 1 sin2xx 2 dx n 6 6.695473 6.252572 6.403292 12 6.474023 6.363008 6.400206 Sn Mn Tn n 6 0.295473 0.147428 0.003292 12 0.074023 0.036992 0.000206 ES EM ET APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A95 57. 61. No 63. (a) 1.925444 (b) 1.920915 (c) 1.922470 65. (a) 0.01348, (b) , 67. 8.6 mi 69. (a) 3.8 (b) 1.7867, 0.000646 (c) 71. C 73. 2 75. PROBLEMS PLUS N PAGE 521 1. About 1.85 inches from the center 3. 0 7. 11. 13. CHAPTER 8 EXERCISES 8.1 N PAGE 530 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 5.115840 25. 1.569619 27. (a), (b) , , (c) (d) 7.7988 29. 31. 6 33. 35. 37. 209.1 m 39. 29.36 in. 41. 12.4 EXERCISES 8.2 N PAGE 537 1. (a) (b) 3. (a) (b) 5. 7. 98 3 1 27(145s145 1) y 1 0 2x1 1 1 x 22 dx y 1 0 2 tan1x1 1 1 x 22 dx x1 0 2xs1 16x 6 dx x1 0 2x 4s1 16x 6 dx 2s2(s1 x 1) sx 2 27[1 9x32 10s10] s5 ln( 1 2(1 s5)) s2 ln(1 s2) x4 0 s1 43 x34 x23 2 dx L4 7.50 L2 6.43 L1 4 46 3 s2 ln(1 s2) s1 e 2 s2 ln(s1 e 2 1) 1 ln(s2 1) ln 3 1 2 ln(s2 1) 32 3 1261 240 2 243(82s82 1) y 4 1 s9y 4 6y 2 2 dy x2 0 s1 sin2x dx 4s5 2 sin1(2s5) b baa1bae1 f 2 3 16 2 n 30 n 260 0.00674 n 368 1 2 sin xs4 sin2x 2 ln(sin x s4 sin2x ) C 9. 11. 13. 15. 17. 19. 21. 23. 27. (a) (b) 29. (a) (b) 31. 33. EXERCISES 8.3 N PAGE 547 1. (a) (b) 1875 lb (c) 562.5 lb 3. 6000 lb 5. 7. 9. 11. 13. 15. (a) 314 N (b) 353 N 17. (a) (b) (c) (d) 19. 21. 23. 25. 27. 29. 31. 33. (2, 0) 35. 37. 41. 45. EXERCISES 8.4 N PAGE 553 1. $38,000 3. $43,866,933.33 5. $407.25 7. $12,000 9. 3727; $37,753 11. 13. 15. 17. 19. EXERCISES 8.5 N PAGE 560 1. (a) The probability that a randomly chosen tire will have a lifetime between 30,000 and 40,000 miles (b) The probability that a randomly chosen tire will have a lifetime of at least 25,000 miles 3. (a) for all x and (b) 5. (a) (b) 7. (a) for all x and (b) 5 11. (a) (b) (c) If you aren’t served within 10 minutes, you get a free hamburger. 13. 15. (a) (b) 17. 0.9545 5.21% 0.0668 44% 1 e22.5 0.55 e42.5 0.20 x f x dx 1 f x 0 1 2 1 1 3 8s3 0.35 x f x dx 1 f x 0 5.77 Lmin 6.60 Lmin 1.19 104 cm3s 1 kb 2k a 2k 2 kb1k a1k 2 3(16s2 8) $9.75 million 1 3r 2h (0, 1 12) 0.781, 1.330 60; 160; ( 8 3, 1) s2 4 4(s2 1) , 1 4(s2 1) ( 9 20, 9 20) 1 e 1 , e 1 4 0, 1.6 10; 1; ( 1 21, 10 21) 230; 23 7 2.5 10 5 N 3.03 105 lb 4.88 104 lb 5.06 104 lb 5.63 103 lb 5.27 105 N 2 3ah 1.2 10 4 lb 9.8 103 N 6.7 10 4 N 187.5 lbft2 4 2r 2 xb a 2 c f x s1 f x 2 dx 2 a2 ab 2 sin1(sb 2 a 2b) sb 2 a 2 2 b 2 a 2b sin1(sa 2 b 2a) sa 2 b 2 56 45s3a 2 1 3a 2 1 6[ln(s10 3) 3s10] 1 4[4 ln(s17 4) 4 ln(s2 1) s17 4s2] 13.527296 9.023754 a 2 1 27(145s145 10s10) 21 2 2s1 2 2 ln( s1 2) A96 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 19. (b) 0; (c) (d) (e) CHAPTER 8 REVIEW N PAGE 562 Exercises 1. 3. (a) (b) 5. 7. 9. 11. 13. 15. 17. $7166.67 19. (a) for all x and (b) (c) 5, yes 21. (a) (b) (c) PROBLEMS PLUS N PAGE 564 1. 3. (a) (b) (d) 5. (a) (b) 7. Height , volume 9. 11. CHAPTER 9 EXERCISES 9.1 N PAGE 571 3. (a) 5. (d) 7. (a) It must be either 0 or decreasing (c) (d) 9. (a) (b) (c) 13. (a) At the beginning; stays positive, but decreases (c) 0 M P(t) t P(0) P 0, P 4200 P 4200 0 P 4200 y 1x 2 y 0 1 2, 1 2, 1 0.14 m ( 28 27 s6 2)b 3 s2 b P0 0tHr 2 0tHe LH xr r e xH 2sr 2 x 2 dx Pz P0 t xz 0 x dx 7.84 107 mi2 3.36 106 mi2 2rr d 2 3 1 2s3 8 ln 2 5.55 min e54 0.29 1 e38 0.31 0.3455 x f x dx 1 f x 0 2 2 (2, 2 3) ( 8 5, 1) 458 lb 124 5 3.292287 41 10 21 16 15 2 3 2a0 1 41e8 0.986 1x1010 0 4x10–10 a0 EXERCISES 9.2 N PAGE 578 1. (a) (b) 3. III 5. IV 7. 9. 11. 13. 15. 17. ; 2, 0, 2 2 c 2 y 0 _2 _1 t 1 2 c=3 c=_3 c=_1 c=1 4 _2 3 _3 y x 3 _3 3 _3 y x 3 _3 3 _3 y x 3 _3 3 _3 y x 3 _3 _3 (c) (a) (b) y 2 y 2, y 0, y x 3 3 _3 _3 (i) (ii) (iv) (iii) APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A97 19. (a) (i) 1.4 (ii) 1.44 (iii) 1.4641 (b) Underestimates (c) (i) 0.0918 (ii) 0.0518 (iii) 0.0277 It appears that the error is also halved (approximately). 21. 23. 25. (a) (i) 3 (ii) 2.3928 (iii) 2.3701 (iv) 2.3681 (c) (i) 0.6321 (ii) 0.0249 (iii) 0.0022 (iv) 0.0002 It appears that the error is also divided by 10 (approximately). 27. (a), (d) (b) 3 (c) Yes; (e) 2.77 C EXERCISES 9.3 N PAGE 586 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. (a) (b) , (c) No 25. 5 2.5 0 2.5 cos y cos x 1 1 0 y=sin ≈ _œ„„„ π/2 _œ „„„ π/2 œ s2 x s2 y sinx 2 sin1y x 2 C y Ke x x 1 y e x22 y 4a s3 sin x a u st 2 tan t 25 cos x x sin x y 2 1 3e 3y 2 3 y sx 2 9 u Ae 2tt 22 1 y s 3te t e t C 23 1 y ln sec y 1 3x 3 x C y Ksx 2 1 y Kx Q 3 Q 0 2 2 t 4 4 6 1.7616 1, 3, 6.5, 12.25 y 0 0.2 x 0.4 0.1 0.3 y=´ h=0.1 h=0.2 h=0.4 1.0 1.1 1.2 1.3 1.4 1.5 27. (a), (c) (b) 29. 31. 33. ; 3 35. ; M 37. (a) (b) 39. (a) (b) ; the concentration approaches regardless of the value of 41. (a) (b) 43. About 4.9% 45. 47. (a) (b) , where and EXERCISES 9.4 N PAGE 598 1. (a) 100; 0.05 (b) Where is close to or ; on the line ; ; (c) Solutions approach 100; some increase and some decrease, some have an inﬂection point but others don’t; solutions with and have inﬂection points at (d) , ; other solutions move away from and toward 3. (a) (b) 1.55 years 3.23 107 kg P 100 P 0 P 100 P 0 P 50 P0 40 P0 20 P¸=140 P¸=120 P¸=80 P¸=40 P¸=20 P¸=60 0 t P 60 40 20 150 100 50 P0 100 0 P0 100 P 50 100 0 P A0 A0 C sM sA0 sM sA0 At M CesM kt 1 Ce sM kt 1 2 dAdt ksA M A tk 15e0.2 12.3 kg 15et100 kg C0 rk rk Ct C0 rkekt rk t 2 ksa btan1 b a b tan1 b x a b x a 4 (kt 2sa) 2 Pt M Mekt Qt 3 3e4t ≈-¥=C xy=k 4 4 _4 _4 4 4 _4 _4 x 2 y2 C y Cx 2 y s2x C 5 5 _5 _5 A98 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 5. (a) , in billions (b) 5.49 billion (c) In billions: 7.81, 27.72 (d) In billions: 5.48, 7.61, 22.41 7. (a) (b) (c) 3:36 PM 11. ; 13. (a) (b) (c) , (d) Declining 15. (a) Fish are caught at a rate of 15 per week. (b) See part (d) (c) (d) ; ; (e) where 17. (b) ; ; (c) 19. (a) (b) Does not exist Pt P0e kr sinrt sin Pt mK P0 KP0 me KmkK t K P0 P0 me KmkK t P0 200: P l 1000 P0 200: P l 200 0 P0 200: P l 0 0 t P 100 80 60 1400 800 400 40 20 600 200 1200 1000 k 1 11, 1 9 0 120 1200 Pt 250 750ke t25 1 ke t25 P0 250: P l 750 P0 250: P l 250 0 P0 250: P l 0 0 t P 120 80 40 1200 800 400 P 250, P 750 m kP0 m kP0 m kP0 Pt m k P0 m ke kt t (year) 1960 1980 2000 90,000 0 45 P (in thousands) 130,000 P L PE PLt 32,658.5 1 12.75e0.1706t 94,000 PEt 1578.31.0933t 94,000 y y0 y0 1 y0ekt dydt ky1 y P dPdt 1 265P1 P100 EXERCISES 9.5 N PAGE 606 1. Yes 3. No 5. 7. 9. 11. 13. 15. 17. 19. 21. 25. 27. (a) (b) 29. 31. 33. ; 0.2275 kgL 35. (b) (c) EXERCISES 9.6 N PAGE 612 1. (a) , ; growth is restricted only by predators, which feed only on prey. (b) , ; growth is restricted by carrying capacity and by predators, which feed only on prey. 3. (a) The rabbit population starts at about 300, increases to 2400, then decreases back to 300. The fox population starts at 100, decreases to about 20, increases to about 315, decreases to 100, and the cycle starts again. (b) 0 t R 2000 t¡ 1000 F 200 t™t£ 1500 500 2500 300 100 R F y predators x prey y prey x predators mtc t mcectm m 2tc 2 mtc y 2 5100 2t 40,000100 2t32 0 M P(t) t P(0) Pt M Cekt Qt 31 e4t, It 12e4t 4 4e12 1.57 A It 4 4e5t y Cx 4 2 5x 12 5 _5 3 _3 c=_1 c=_1 c=_3 c=_3 c=1 c=3 c=3 c=5 c=5 c=7 c=7 c=_5 c=_5 y x 1e x C x 2 y x cos x x v t 3e t 2 5e t 2 y x 1 3e x u t 2 2t 2C 2t 1 y x sinx 2 dx C sin x y 2 3sx Cx y x 2 ln x Cx 2 y 2 3e x Ce2x APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A99 5. 9. (a) Population stabilizes at 5000. (b) (i) , : Zero populations (ii) , : In the absence of wolves, the rabbit population is always 5000. (iii) , : Both populations are stable. (c) The populations stabilize at 1000 rabbits and 64 wolves. (d) CHAPTER 9 REVIEW N PAGE 615 True-False Quiz 1. True 3. False 5. True 7. True Exercises 1. (a) (b) ; , , 3. (a) (b) 0.75676 (c) and ; there is a local maximum or minimum 5. 7. 9. 11. 13. x C 1 2y2 y 1 2xln x2 2x rt 5ett 2 y slnx 2 2x 32 C y ( 1 2x 2 C)esin x y x y x y0.3 0.8 0 x y 1 2 _1 _2 1 2 3 _3 3 y 4 y 2 y 0 0 c 4 6 1 0 t y 2 4 (i) (ii) (iv) (iii) 0 t R 1000 W 40 1500 500 60 20 80 W R R 1000 W 64 R 5000 W 0 R 0 W 0 0 Species 1 Species 2 50 200 100 50 100 150 200 250 t=3 t=0, 5 150 t=1 t=2 t=4 15. (a) ; (b) 17. (a) (b) 19. 15 days 21. 23. (a) Stabilizes at 200,000 (b) (i) , : Zero populations (ii) , : In the absence of birds, the insect population is always 200,000. (iii) , : Both populations are stable. (c) The populations stabilize at 25,000 insects and 175 birds. (d) 25. (a) or (b) PROBLEMS PLUS N PAGE 618 1. 5. 7. 9. (b) (c) No 11. (a) 9.8 h (b) ; (c) 5.1 h 13. CHAPTER 10 EXERCISES 10.1 N PAGE 626 1. 3. 5. (a) (b) y 2 3x 13 3 (_8, _1) t=_1 x y 0 (_5, 1) t=0 (_2, 3) t=1 (1, 5) t=2 x y t=0 (0, 0) t=π {0, π@} 5 5 x y t=0 (1, 0) t=5 {1+œ„ 5, 5} t=4 (3, 0) x 2 y 62 25 6283 ft2h 31,900 100,000 ft2 f x x 2 L2 4L 1 2L ln x L 20 C y x 1n f x 10e x 2k sinh kb y 1k cosh kx 1k cosh kb h y 1k cosh kx a 1k 0 t x 35,000 15,000 y 150 25,000 5,000 45,000 200 100 250 (insects) (birds) 50 birds insects y 175 x 25,000 y 0 x 200,000 y 0 x 0 k ln h h RVt C Lt 53 43e0.2t Lt L L L0 ekt t 10 ln 2 57 33.5 560 Pt 2000 1 19e0.1t A100 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 7. (a) (b) , 9. (a) (b) 11. (a) 13. (a) , (b) (b) 15. (a) 17. (a) (b) (b) 19. Moves counterclockwise along the circle from to 21. Moves 3 times clockwise around the ellipse , starting and ending at 23. It is contained in the rectangle described by and . 25. 27. 29. 3 3 3 3 x y t=0 t=1 2 1 1 y x (0, _1) t=_1 (0, 1) t=1 (_1, 0) t=0 2 y 3 1 x 4 0, 2 x 225 y 24 1 3, 1 3, 3 x 32 y 12 4 x 0 1 y y x 0 1 1 y 2 x 2 1, y 1 y 1 2ln x 1 y x 0 (1, 1) x (0, 1) y (0, _1) 0 y 1 y 1x x 2 y 2 1, x 0 y 1 x 2, x 0 y 0 x (0, 1) t=0 (1, 0) t=1 (2, _3) t=4 3 y 11 x 1 4y 52 2 y x (7 , 11) t=_3 (14, _3) t=4 (_2, 5) t=0 ”4 , 0’ t= 1 4 5 2 31. (b) , 33. (a) (b) (c) 37. The curve is generated in (a). In (b), only the portion with is generated, and in (c) we get only the portion with . 41. , ellipse 43. 45. (a) Two points of intersection (b) One collision point at when (c) There are still two intersection points, but no collision point. 47. For , there is a cusp; for , there is a loop whose size increases as c increases. 49. As n increases, the number of oscillations increases; a and b determine the width and height. EXERCISES 10.2 N PAGE 636 1. 3. 5. 7. 9. 11. 13. 15. 17. Horizontal at , vertical at 19. Horizontal at (four points), vertical at 21. ; (5 665, e615) 0.6, 2 2, 0 (s2, 1) 10, 0 6, 16 3 2 tan t, 3 4 sec3t, 2 t 32 et, et1 et, t 0 1 3 2t, 34t, t 0 20 _2 10 _10 y 1 6x y 2x 1 y 2ex 3 y x 2t 1 t cos t sin t 3 0 1.5 _3 _1 0 0 1.5 1 _1 1 1 2 c 0 c 0 t 32 3, 0 4 4 6 6 y O x 2a x a cos , y b sin ; x 2a 2 y 2b 2 1 x 0 x 0 y x 23 x 2 cos t, y 1 2 sin t, 2 t 32 x 2 cos t, y 1 2 sin t, 0 t 6 x 2 cos t, y 1 2 sin t, 0 t 2 0 t 1 y 7 8t x 2 5t, APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A101 23. 25. 27. (a) 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 612.3053 51. 55. (a) (b) 57. 59. 61. 63. 59.101 65. 71. EXERCISES 10.3 N PAGE 647 1. (a) (b) 1, 54, 1, 4 2, 73, 2, 43 O _3π 4 ”1, _ ’ 3π 4 O π 3 π 3 ”2, ’ 1 4 24 5 (949s26 1) 6 5a 2 2 1215(247s13 64) x1 0 2t 2 1e tse 2tt 12t 2 2t 2 dt 103.5999 294 t 0, 4 15 15 15 15 6s2, s2 21 1 1 21 e 3 11 e8 8 0 25 2.5 s2 e 1 s103 ln(3 s10) s2 ln(1 s2) 4s2 2 x2 0 s3 2 sin t 2 cos t dt 10.0367 x2 1 s1 4t 2 dt 3.1678 2r 2 d 2 3 e ab ( 16 27, 29 9 ), 2, 4 d sin r d cos 0 y x y x, y x 7.5 1 8.5 3 (c) 3. (a) (b) (c) 5. (a) (i) (ii) (b) (i) (ii) 7. 9. 11. 13. 15. Circle, center , radius 2 17. Circle, center , radius 19. Horizontal line, 1 unit above the -axis 21. 23. 25. 27. (a) (b) 29. 31. O ”1, ’ π 2 O ¨=_π 6 x 3 6 r 2c cos r cot csc r 3 sec x 3 2 (0, 3 2) O 2s3 O r=2 r=3 ¨=7π 3 ¨=5π 3 O r=4 ¨=π 6 ¨=_π 2 O r=1 r=2 2, 53 2, 23 (2s2, 34) (2s2, 74) (s2, s2) O 3π 4 ”_2, ’ 3π 4 (1, s3) 1, 0 O _2π 3 ”2, _ ’ 2π 3 π O (1, π) 1, 32, 1, 52 O π 2 ”_1, ’ π 2 A102 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. (a) For , the inner loop begins at and ends at ; for , it begins at and ends at . 57. 59. 61. 1 63. Horizontal at , ; vertical at 65. Horizontal at , [the pole], and ; vertical at (2, 0), , 67. Horizontal at ; vertical at , where sin1( 1 2 1 2s3) ( 3 2 1 2s3, ) ( 3 2 1 2s3, ) 3, 2, 1, 32 ( 1 2, 43) ( 1 2, 23) ( 3 2, 53) 0, ( 3 2, 3) 3, 0, 0, 2 (3s2, 34) (3s2, 4) s3 2 sin11c sin11c c 1 sin1 1/c sin1 1c c 1 O (2, 0) (6, 0) O 1 1 2 ¨=π 3 ¨=2π 3 (3, 0) (3, π) O O ¨= π 6 ¨=5π 6 ¨=π 8 1 3 4 5 6 2 ¨=π 3 O O 69. Center , radius 71. 73. 75. 77. By counterclockwise rotation through angle , , or about the origin 79. (a) A rose with n loops if n is odd and 2n loops if n is even (b) Number of loops is always 2n 81. For , the curve is an oval, which develops a dimple as . When , the curve splits into two parts, one of which has a loop. EXERCISES 10.4 N PAGE 653 1. 3. 5. 7. 9. 11. 4 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. , and the pole 39. where , , , and where , , , 2312 1912 1112 712 1, 1712 1312 512 12 1, ( 3 2, 6), ( 3 2, 56) 1 4( 3s3) 1 8 1 4 1 2 1 5 24 1 4s3 4s3 4 3 1 3 1 2s3 3 2s3 9 20 1 8 3 3 3 3 ¨=π 6 3 O O 9 4 41 4 2 12 1 8s3 510,240 a 1 a l1 0 a 1 3 6 7 7 7 7 _3 3 _2.5 3.5 _3.4 1.8 _2.6 2.6 sa 2 b 22 b2, a2 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A103 41. , and the pole 43. Intersection at 45. 47. 49. 51. 53. 55. (b) EXERCISES 10.5 N PAGE 660 1. , , 3. , , 5. , , 7. , , 9. , focus , directrix 11. , 13. , 15. 17. , foci x 0 y 1 3 (1,_3) (1, 3) (0, s5) x 2 4 y 2 9 1 1, 3, (1, s5) 0 x y _2 2 4 _4 0 x y œ„ 5 _œ„ 5 _3 3 (0, 2s3) 0, 4 2, 0 3, 0 x 1 4 ( 1 4, 0) x y 2 0 x y x=1 (_2, _1) (_5, _1) y x y=1 (_2, 5) x 1 5, 1 2, 1 y 1 2, 5 2, 3 y x y= 1 16 ”0, _ ’ 1 16 y x x= 1 8 ” , 0’ 1 8 y 1 16 (0, 1 16) 0, 0 x 1 8 ( 1 8, 0) 0, 0 2(2 s2) 1 1 _0.75 1.25 16 3 9.6884 29.0653 8 3 2 132 1 0.89, 2.25; area 3.46 ( 1 2s3, 3), ( 1 2s3, 23) 19. , 21. , 23. 25. Parabola, 27. Ellipse, , 29. Hyperbola, 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. (a) (b) 55. (a) Ellipse (b) Hyperbola (c) No curve 59. 9.69 61. where EXERCISES 10.6 N PAGE 668 1. 3. 5. 7. 9. (a) 1 (b) Parabola (c) (d) O y=1 ” , ’ 1 2 π 2 y 1 r 4 2 cos r 8 1 sin r 15 4 3 cos r 42 4 7 sin c 2 a 2 b2 b2c a ab ln a b c 248 mi 121x 2 1,500,625 121y 2 3,339,375 1 x 2 3,763,600 y 2 3,753,196 1 x 2 9 y 2 36 1 y 12 25 x 32 39 1 x 2 9 y 2 16 1 x 12 12 y 42 16 1 x 2 12 y 42 16 1 x 2 25 y 2 21 1 y 3 2x 22 y 2 12x 1 x 2 8y 0, 1, 0, 3; (0, 1 s5) 1, 1 (s2, 1) 0, 1, (0, 3 4) x 0 y (4, _2) (3+œ„ 5, _2) (3-œ„ 5, _2) (2, _2) y 2 2x 3 (3s5, 2); 4, 2, 2, 2; 0 x y 2 _2 y=x x 12 y 0 y= x 5 12 y x y 5 12 x 0, 2, (0, 2s2) 12, 0, 13, 0 A104 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 11. (a) (b) Ellipse (c) (d) 13. (a) (b) Ellipse (c) (d) 15. (a) 2 (b) Hyperbola (c) (d) 17. (a) (b) 19. The ellipse is nearly circular when e is close to 0 and becomes more elongated as . At , the curve becomes a parabola. e 1 e l 1 e=0.4 e=1.0 e=0.8 e=0.6 2 _2 _2 2 r 1 1 2 sin 34 1 _3 _2 2 -y= 1 2 2, y 1 2 O x=_3 8 ”- , 0’ 3 4 ” , π’ 1 4 x 3 8 O x=9 2 ” , ’ π 2 3 2 ” , 0’ 9 8 ” , π’ 9 4 ” , ’ 3π 2 3 2 x 9 2 1 3 O (3, 0) (3, π) ”4, ’ π 2 ” , ’ 3π 2 12 5 y=_12 y 12 1 4 25. 27. 35.64 AU 29. 31. CHAPTER 10 REVIEW N PAGE 669 True-False Quiz 1. False 3. False 5. True 7. False 9. True Exercises 1. 3. 5. , ; , ; , , 7. (a) (b) , 9. 11. 13. 15. 17. 19. 0.75 -0.3 1.2 -0.75 r= sin ¨ ¨ r 2 cos sin ”_3, ’ 3π 2 ”1, ’ π 2 3 2 y= O O 1 _1 (2, π) (2, 0) ¨=π 6 (1, 0) O (2, π) ”1, ’ π 2 ”1, ’ 3π 2 (2, 2s3) (3s2, 74) (3s2, 34) O 2π 3 ”4, ’ 2π 3 0 t 2 y tan t x tan2t y t 2 x t 4 y st x t x y (1, 1), ¨=0 y x (0, 6), t=_4 (5, 1), t=1 y 1x x y 2 8y 12 3.6 108 km 7.0 107 km r 2.26 108 1 0.093 cos APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A105 21. 2 23. 25. 27. 29. Vertical tangent at ; horizontal tangent at 31. 18 33. 35. 37. 39. 41. 43. All curves have the vertical asymptote . For , the curve bulges to the right. At , the curve is the line . For , it bulges to the left. At there is a cusp at (0, 0). For , there is a loop. 45. 47. 49. 51. 53. 55. 57. , PROBLEMS PLUS N PAGE 672 1. 3. 5. (a) At (0, 0) and (b) Horizontal tangents at (0, 0) and ; vertical tangents at (0, 0) and (d) (g) 3 2 y x y x 1 (s 3 4, s 3 2) (s 3 2, s 3 4) ( 3 2, 3 2) [ 3 4s3, 3 4s3] 1, 2 ln2 y a1 sin2 x acot sin cos r 4 3 cos x 2 25 8y 3992 160,801 1 y 2 725 x 2 85 1 x 2 25 y 2 9 1 x (_1, 3) y 0 x y 0 (1, 0) 2œ„ 2 2œ„ 2 3 3 ( 25 24, 3), 1, 3 1, 0, 3, 0 c 0 c 0 1 c 0 x 1 c 1 c 1 x 1 471,2951024 2s 2 1 s4 2 1 2 ln 2 s4 2 1 s 2 1 2(5s5 1) 1 2 1 2, 3 a, 0, ( 1 2a, 3 2s3a) ( 3 2a, 1 2s3 a), 3a, 0 x y 0 (3a, 0) (a, 0) ( 11 8 , 3 4) 1 sin t 1 cos t, 1 cos t sin t 1 cos t3 1 CHAPTER 11 EXERCISES 11.1 N PAGE 684 Abbreviations: C, convergent; D, divergent 1. (a) A sequence is an ordered list of numbers. It can also be deﬁned as a function whose domain is the set of positive integers. (b) The terms approach 8 as becomes large. (c) The terms become large as becomes large. 3. 0.8, 0.96, 0.992, 0.9984, 0.99968 5. 7. 3, 5, 9, 17, 33 9. 11. 13. 15. ; yes; 17. 1 19. 5 21. 1 23. 1 25. 0 27. D 29. 0 31. 0 33. 0 35. 0 37. 1 39. 41. 43. D 45. D 47. 1 49. 51. D 53. 0 55. (a) 1060, 1123.60, 1191.02, 1262.48, 1338.23 (b) D 57. 59. Convergent by the Monotonic Sequence Theorem; 61. Decreasing; yes 63. Not monotonic; no 65. Decreasing; yes 67. 2 69. 71. (b) 73. (a) 0 (b) 9, 11 EXERCISES 11.2 N PAGE 694 1. (a) A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers. (b) A series is convergent if the sequence of partial sums is a con-vergent sequence. A series is divergent if it is not convergent. 3. , , , , , , , , , ; convergent, 5. , , , , , , , , , ; divergent 7. , , , , , , , , , ; convergent, sum 1 0.69849 0.68377 0.66667 0.64645 0.62204 0.59175 0.55279 0.50000 0 1 11 {an} {sn} 0.42265 0.29289 9.01610 9.66446 9.21214 2.41243 3.28388 2.99287 0.38764 0.77018 2 0 10 _10 ssnd sand 0.62763 1.55741 sum 2 2.00000 2.00000 1.99999 2.00003 1.99987 2.00064 1.99680 2.01600 ssnd 1 0 10 _3 sand 1.92000 2.40000 1 2(1 s5) 1 2(3 s5) 5 L 8 1 r 1 1 2 ln 2 e 2 1 2 1 3, 2 5, 3 7, 4 9, 5 11, 6 13 an ( 2 3) n1 an 5n 3 an 12n 1 3, 3 2, 1 2, 1 8, 1 40 n an n an A106 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 9. (a) C (b) D 11. 9 13. D 15. 60 17. 19. D 21. D 23. D 25. 27. D 29. D 31. D 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. All ; 53. 1 55. for , 57. (a) (b) 5 59. 63. 65. The series is divergent. 71. is bounded and increasing. 73. (a) 75. (a) (c) 1 EXERCISES 11.3 N PAGE 703 1. C 3. D 5. C 7. C 9. D 11. C 13. D 15. C 17. C 19. C 21. D 23. C 25. C 27. 29. 31. 33. (a) 1.54977, (b) 1.64522, (c) 35. 0.00145 41. EXERCISES 11.4 N PAGE 709 1. (a) Nothing (b) C 3. C 5. D 7. C 9. C 11. C 13. C 15. C 17. D 19. D 21. C 23. C 25. D 27. C 29. C 31. D 33. 1.249, error 35. 0.76352, 45. Yes EXERCISES 11.5 N PAGE 713 1. (a) A series whose terms are alternately positive and negative (b) and , where (c) 3. C 5. C 7. D 9. C 11. C 13. D 15. C 17. C 19. D Rn bn1 bn an limn l bn 0 0 bn1 bn error 0.001 0.1 b 1e n 1000 error 0.005 error 0.1 1, p 1 p 1 0 x y 1 . . . a™ a£ a¢ a∞ 2 3 4 y= 1 x1.3 1 2, 5 6, 23 24, 119 120; n 1! 1 n 1! 0, 1 9, 2 9, 1 3, 2 3, 7 9, 8 9, 1 sn 1 nn 1 1 2(s3 1) Sn D1 c n 1 c sum 1 n 1 a1 0, an 2 nn 1 2 2 cos x x 1 4 x 1 4; 1 1 4x 3 x 3; x 3 x 50633300 1138333 2 9 e 1 11 6 3 2 ee 1 5 2 1 7 21. 1.0000, 0.6464, 0.8389, 0.7139, 0.8033, 0.7353, 0.7893, 0.7451, 0.7821, 0.7505; 23. 5 25. 4 27. 29. 31. An underestimate 33. is not a negative integer 35. is not decreasing EXERCISES 11.6 N PAGE 719 Abbreviations: AC, absolutely convergent; CC, conditionally convergent 1. (a) D (b) C (c) May converge or diverge 3. AC 5. CC 7. AC 9. D 11. AC 13. AC 15. AC 17. CC 19. AC 21. AC 23. D 25. AC 27. D 29. D 31. (a) and (d) 35. (a) , (b) , 0.693109 EXERCISES 11.7 N PAGE 722 1. C 3. D 5. C 7. D 9. C 11. C 13. C 15. C 17. D 19. C 21. C 23. D 25. C 27. C 29. C 31. D 33. C 35. C 37. C EXERCISES 11.8 N PAGE 727 1. A series of the form , where is a variable and and the ’s are constants 3. 1, 5. 1, 7. 9. 11. 13. 15. 17. 19. 21. 23. 0, 25. 27. 29. (a) Yes (b) No 31. 33. No 35. (a) (b), (c) 37. , 41. 2 EXERCISES 11.9 N PAGE 733 1. 10 3. 5. 7. 9. 1 2 n1 x n, 1, 1 n0 1n 1 9 n1 x 2n1, 3, 3 2 n0 1 3 n1 x n, 3, 3 n0 1nx n, 1, 1 f x 1 2x1 x 2 1, 1 2 8 _2 _8 s¸ J¡ s£ s∞ s¡ s™s¢ , k k , , 1 4, [ 1 2, 0] { 1 2} b, a b, a b , , 1 3, [ 13 3 , 11 3 ) 1, 1, 3 4, 4, 4 1 2, ( 1 2, 1 2] 2, 2, 2 , , 1, 1 1, 1 cn a x n0 cnx an n 11 error 0.00521 661 960 0.68854 bn p 0.0676 0.9721 error 0.0275 1 _1 0 10 ssnd sand APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A107 11. 13. (a) (b) (c) 15. 17. 19. 21. 23. 25. 27. 0.199989 29. 0.000983 31. 0.09531 33. (b) 0.920 37. EXERCISES 11.10 N PAGE 746 1. 3. 5. 7. 9. 11. 13. , R 1 2x 1 3x 12 4x 13 x 14 n0 x 2n1 2n 1! , R n0 5 n n! x n, R n0 1n 2n1 2n 1! x 2n1, R n0 n 1x n, R 1 n0 n 1x n, R 1 b8 f 858! 1, 1, 1, 1, 1, 1 C n1 1n1 x 2n1 4n2 1 , R 1 C n0 t 8n2 8n 2 , R 1 3 2 3 2 s¡ f s£ s™ n0 2x 2n1 2n 1 , R 1 0.25 _0.25 4 _4 s¡ s¡ s™ s™ s£ s£ s¢ s¢ s∞ s∞ f f n0 1n 1 16n1 x 2n1, R 4 n3 n 2 2n1 x n, R 2 ln 5 n1 x n n5n , R 5 1 2 n2 1nnn 1x n, R 1 1 2 n0 1nn 2n 1x n, R 1 n0 1nn 1x n, R 1 n0 1n1 1 2n1x n, 1, 1 15. 17. 19. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 0.81873 45. (a) (b) x n1 1 3 5 2n 1 2n 12nn! x 2n1 1 n1 1 3 5 2n 1 2nn! x 2n 6 _6 4 _3 T¡ T¡ T£ T£ T™ T™ T¢ T¢ Tß Tß T∞ T∞ f f n1 1n1 n 1! x n, R 1.5 1.5 _1.5 _1.5 Tˆ=T˜=T¡¸=T¡¡ T¢=T∞=Tß=T¶ T¸=T¡=T™=T£ f n0 1n 1 2n! x 4n, R n1 1n1 22n1 2n! x 2n, R 1 2 x n1 1n 1 3 5 2n 1 n!23n1 x 2n1, R 2 n0 1n 1 2 2n2n! x 4n1, R n0 2n 1 n! x n, R n0 1n 2n1 2n 1! x 2n1, R n0 1n n 1n 2 2n4 x n, R 2 1 x 2 n2 1n1 1 3 5 2n 3 2nn! x n, R 1 n0 1n 1 3 5 2n 1 2 n 32n1 n! x 9n, R 9 n0 1n1 1 2n! x 2n, R n0 e 3 n! x 3n, R A108 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 47. 49. 51. 0.440 53. 0.40102 55. 57. 59. 61. 63. 65. 67. EXERCISES 11.11 N PAGE 755 1. (a) , , (b) (c) As increases, is a good approximation to on a larger and larger interval. 3. 5. 1.1 _1.1 T£ T£ f f π 0 π 2 x 2 1 6x 2 3 2 4 0 T£ f 1 2 1 4x 2 1 8x 22 1 16x 23 f x Tnx n 2 2π _2 _2π T¢=T∞ T™=T£ T¸=T¡ Tß f T6x 1 1 2 x 2 1 24 x 4 1 720 x 6 T4x 1 1 2 x 2 1 24 x 4 T5x T0x 1 T1x, T2x 1 1 2 x 2 T3x e 3 1 1s2 ex 4 1 1 6x 2 7 360x 4 1 3 2 x 2 25 24 x 4 1 120 1 3 C n1 1n 1 2n 2n! x 2n, R C n0 1n x 6n2 6n 22n!, R 7. 9. 11. 13. (a) (b) 15. (a) (b) 17. (a) (b) 19. (a) (b) 0.00006 21. (a) (b) 0.042 23. 25. Four 27. 29. 31. 21 m, no 37. (c) They differ by about CHAPTER 11 REVIEW N PAGE 759 True-False Quiz 1. False 3. True 5. False 7. False 9. False 11. True 13. True 15. False 17. True 19. True Exercises 1. 3. D 5. 0 7. 9. 2 11. C 13. C 15. D 17. C 19. C 21. C 23. CC 25. AC 27. 29. 31. 35. 0.9721 ee 4 1 11 e 12 1 2 8 109 km. 0.86 x 0.86 1.037 x 1.037 0.17365 x 2 1 6x 4 1 x 2 0.0015 1 1 2x 2 0.000097 1 2 3x 1 1 9x 12 4 81x 13 1.5625 105 2 1 4x 4 1 64x 42 5 _2 T™ T¢ T™ T£ T£ T¢ T∞ T∞ f f 2 0 π 4 π 2 10 3 x 4 4 64 15 x 4 5 T5x 1 2 x 4 2 x 4 2 8 3 x 4 3 _4 f 3 _1 1.5 T£ x 2x 2 2x 3 _1.6 T£ f 1.6 _1 1 x 1 6x 3 x f 0.7071 1 0.6916 0.7074 0.7071 0 1 0.2337 0.0200 0.0009 1 1 3.9348 0.1239 1.2114 2 4 T6 T4 T5 T2 T3 T0 T1 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A109 37. , error 41. 43. 0.5, [2.5, 3.5) 45. 47. 49. 51. 53. 55. 57. (a) (b) (c) 0.000006 59. PROBLEMS PLUS N PAGE 762 1. 3. (b) 0 if , if , k an integer 5. (a) (c) 9. 11. 13. (a) (b) APPENDIXES EXERCISES A N PAGE A9 1. 18 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. _1 1 2 1 2 [1, 1 2] , 1 2, _1 1 2 0 1 [1, 1 2) 0, 1 2 6 3 2, 6 3, 0 _1 0 _2 1, 2, x 2 1 x 1 x 1 x 1 for x 1 for x 1 2 x 5 s5 250 101 250 101en15 en5 ln 1 2 1, 1, x 3 4x 2 x 1 x4 2 5s3 sn 3 4n, ln 13n, pn 4n3n1 x k 1x cot x x 0 15!5! 10,897,286,400 1 6 1.5 2 0 T£ f 1 1 2x 1 1 8x 12 1 16x 13 C ln x n1 x n n n! 1 2 n1 1 5 9 4n 3 n!26n1 x n, R 16 n0 1n x 8n4 2n 1! , R n1 x n n , R 1 n0 1nx n2, R 1 1 2 n0 1n 1 2n! x 6 2n s3 2n 1! x 6 2n1 4, 6, 2 6.4 107 0.18976224 29. 31. 33. 35. 37. 39. 41. (a) (b) 43. 45. 47. 49. 51. 53. 55. 57. 59. EXERCISES B N PAGE A15 1. 5 3. 5. 7. 2 9. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. , 39. , 41. , 43. 45. 47. 49. 0 y x x 2 y 4 0 y x _2 2 0 y x 0 y x 0 x y _3 0 _2 x y y=_2 0 x y b 3 b 2 b 0 m 3 4 m 0 m 1 3 5x 2y 1 0 x 2y 11 0 y 5 y 3x 3 y 3x 2 5x y 11 2x 3y 19 0 y 6x 15 0 x y xy=0 0 3 x y x=3 9 2 2s37 s74 x c ba x a bcab 4, 1 1, 4 1.3, 1.7 , 7 3, 3, 5 3, 3 2, 4 3 3 2 30C T 20C T 20 10h, 0 h 12 10 C 35 0 1 4 , 0 ( 1 4, ) _1 1 0 0 1 1, 0 1, , 1 _œ„ 3 0 œ„ 3 (s3, s3) , A110 \|\|\|\| APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES 51. 53. 55. (a) (b) 57. 59. 61. (b) EXERCISES C N PAGE A23 1. 3. 5. 7. 9. 11. Parabola 13. Ellipse 15. Hyperbola 17. Ellipse 19. Parabola 21. Hyperbola 23. Hyperbola 25. Ellipse y 0 x (1, 2) y 0 x y 0 x y=x 3 y= x 3 1 1 0 x y _1 1 _1 0 x y 1 _1 1 2 1 2 _ 0 x y _5 5 y= x 4 5 y= x 4 5 0 x y _4 4 2 _2 0 x y ( 1 4, 1 4), s104 ( 1 2, 0), 1 2 2, 5, 4 x 2 y 2 65 x 32 y 12 25 4x 3y 24 0 y x 3 1, 2 3.5, 3 4, 9 0, 4 0 y x y=1-2x y=1+x 0, 1 27. Parabola 29. Parabola 31. Ellipse 33. 35. 37. 39. EXERCISES D N PAGE A32 1. 3. 5. 7. 720° 9. 75° 11. 13. 15. 17. 19. 21. 23. , , , , , 25. , , , , and undeﬁned 27. 29. , , , , cot 4 3 sec 5 4 csc 5 3 tan 3 4 cos 4 5 cot56 s3 sec56 2s3, csc56 2, tan56 1s3, cos56 s32, sin56 1 2, sec92 tan92 cot92 0 csc92 1 cos92 0 sin92 1 cot34 1 sec34 s2 csc34 s2 tan34 1 cos34 1s2 sin34 1s2 0 x y 2 rad 0 x y _ 3π 4 0 x y 315° 2 3 rad 120 3 cm 67.5 5 20 76 y 0 x 1 1 y 0 x 1 1 y x 2 2x 0 x y (3, 9) 0 x y 1 3 5 y 0 x 2 2 4 (3, 4) y 0 x APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES \|\|\|\| A111 31. , , , , 33. , , , , 35. 5.73576 cm 37. 24.62147 cm 59. 61. 63. 65. 67. 69. 71. 73. and 75. 77. 79. 81. 89. EXERCISES E N PAGE A38 1. 3. 5. 7. 9. 11. 10 i1 i 1 1 1 1 1n1 110 210 310 n10 1 1 3 3 5 5 7 7 9 34 35 36 s1 s2 s3 s4 s5 14.34457 cm2 y 0 x 1 π 2π _π y 0 x 3π 2 2π π π 2 5π 2 3π y 0 x 1 1 2 π 3 5π 6 0 x 4, 34 x 54, 74 x 2 56 x 2 0 x 6 0, , 2 6, 2, 56, 32 4, 34, 54, 74 3, 53 24 25 1 15(3 8s2) 1 15(4 6s2) sec s103 csc s10 tan 1 3 cos 3s10 sin 1s10 cot 2s5 csc 3s5 tan s52 cos 2 3 sin s53 13. 15. 17. 19. 21. 80 23. 3276 25. 0 27. 61 29. 31. 33. 35. 41. (a) (b) (c) (d) 43. 45. 14 49. EXERCISES G N PAGE A54 1. (b) 0.405 EXERCISES H N PAGE A62 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. , 31. , , 33. 35. 37. 39. 41. 43. 45. 47. , sin 3 3 cos2 sin sin3 cos 3 cos3 3 cos sin2 e 2 1 2 (s32)i i 0 Im Re _i 0 Im Re i 1 (s32) 1 2i, i 1, i, (1s2) 1 i 512s3 512i 1024 1 4cos6 i sin6 (2s2)cos1312 i sin1312 4s2 cos712 i sin712 1 2cos6 i sin6 4cos2 i sin2, cos6 i sin6 5{cos[tan1( 4 3)] i sin[tan1( 4 3)]} 3s2 cos34 i sin34 1 2 (s72)i 1 2i 3 2i 4i, 4 12 5i, 13 5i i 1 2 1 2i 11 13 10 13i 12 7i 13 18i 8 4i 2n1 n 2 n 2 1 3 an a0 97 300 5100 1 n 4 nn 3 2n 2 n 104 nn 2 6n 113 nn 2 6n 173 nn 1 n i1 x i 5 i0 2i n i1 2i 19 i1 i i 1 INDEX A113 Abel, Niels, 210 absolutely convergent series, 740 absolute maximum and minimum, 271 absolute value, 17, A6, A56 absolute value function, 17 acceleration, 160, 221 Achilles and the tortoise, 6 adaptive numerical integration, 504 addition formulas for sine and cosine, A28, A29 algebraic function, 31 alternating harmonic series, 711, 714 alternating series, 710 Alternating Series Estimation Theorem, 712 Alternating Series Test, 711 analytic geometry, A10 angle, A24 between curves, 268 of deviation, 279 negative, A25 positive, A25 standard position, A25 antiderivative, 340 antidifferentiation formulas, 351 ? aphelion, 667 apolune, 661 approach path of an aircraft, 206 approximate integration, 495 approximating cylinder, 424 approximating surface, 532 approximation by differentials, 250 to e, 179 linear, 247 by the Midpoint Rule, 496 by Newton’s method, 335 by an nth-degree Taylor polynomial, 253 quadratic, 253 by Riemann sums, 367 by Simpson’s Rule, 500, 501 tangent line, 247 by Taylor’s Inequality, 737 by the Trapezoidal Rule, 497 Archimedes’ Principle, 449 arc length, 525, 633, 634, 652 arc length contest, 532 arc length formula, 526 arc length function, 528 area, 3, 355 of a circle, 469 under a curve, 355, 360, 365 between curves, 415, 418 of an ellipse, 469 by exhaustion, 3 under a parametric curve, 632 in polar coordinates, 639 of a sector of a circle, 650 surface, 635 of a surface of a revolution, 532, 538 area function, 379 Area Problem, 3, 355 argument of a complex number, A57 arithmetic-geometric mean, 686 arrow diagram, 12 astroid, 213, 629 asymptote, 308 horizontal, 132, 308 of a hyperbola, 658, A20 slant, 312 vertical, 95, 308 asymptotic curve, 315 autonomous differential equation, 575 average cost function, 330 average rate of change, 148, 221 average speed of molecules, 516 average value of a function, 443, 557 average velocity, 5, 85, 145, 221 axes, coordinate, A11 axes of ellipse, A19 bacterial growth, 591, 598 Barrow, Isaac, 4, 153, 380 baseball and calculus, 601 base of a cylinder, 422 base of logarithm, 63, 428, A53 change of, 66 Bernoulli, James, 580, 607 Bernoulli, John, 307, 580 Bernoulli differential equation, 607 Bessel, Friedrich, 724 Bessel function, 674, 724 Bézier, Pierre, 639 Bézier curves, 624, 639 binomial series, 742, 748 Binomial Theorem, 772 discovery by Newton, 748 blackbody radiation, 757 blood ﬂow, 227, 332, 551 bounded sequence, 682 Boyle’s Law, 221, 246 brachistochrone problem, 625 branches of hyperbola, 629, A20 Buffon’s needle problem, 565 bullet-nose curve, 51, 204 RP denotes Reference Page numbers. cable (hanging), 255 calculator, graphing, 46, 315, 624, 646 calculus, 9 invention of, 399 cancellation equations, 62 for inverse trigonometric functions, 67 for logarithms, 64 cans, manufacturing, 333 Cantor, Georg, 696 Cantor set, 696 capital formation, 554 cardiac output, 552 cardioid, 213, 643 carrying capacity, 233, 568 Cartesian coordinate system, A11 Cartesian plane, A11 Cassini, Giovanni, 649 catenary, 255 Cauchy, Augustin-Louis, 113, A45 Cauchy’s Mean Value Theorem, A45 Cavalieri’s Principle, 432 center of gravity, 542 center of mass, 542 centroid of a plane region, 543 Chain Rule, 197 change of base, 66 change of variables in integration, 401 charge, 224 chemical reaction, 224 circle, A16 area of, 469 circular cylinder, 422 cissoid, 629, 648 closed interval, A3 Closed Interval Method, 275 cochleoid, 670 coefﬁcient of friction, 196, 278 of inequality, 399 of a polynomial, 28 of a power series, 723 combinations of functions, 41 comets, orbits of, 668 comparison properties of the integral, 375 Comparison Test, 514, 706 comparison tests for series, 705 Comparison Theorem for integrals, 514 Completeness Axiom, 682 complex conjugate, A55 complex exponentials, A57 complex number(s), A55 argument of, A57 division of, A55, A58 equality of, A55 imaginary part of, A55 modulus of, A56 multiplication of, A55, A58 polar form, A57 powers of, A59 principal square root of, A56 real part of, A46 roots of, A60 composition of functions, 41, 197 continuity of, 125 derivative of, 199 compound interest, 238, 306, 618 compressibility, 225 computer algebra system, 91, 624 for graphing sequences, 680 for integration, 491 computer, graphing with, 46, 315, 646 concavity, 291 Concavity Test, 291, A44 concentration, 224 conchoid, 626, 648 conditionally convergent series, 715 cone, 431 conic section, 654, 662 directrix, 662 eccentricity, 662 focus, 662 polar equation, 663 shifted, 659, A21 conjugates, properties of, A56 constant function, 173 Constant Multiple Law of limits, 100 Constant Multiple Rule, 176 consumer surplus, 550 continued fraction expansion, 686 continuity of a function, 119 on an interval, 121 from the left, 121 from the right, 121 continuous compounding of interest, 306, 618 continuous random variable, 555 convergence absolute, 714 conditional, 715 of an improper integral, 509, 512 interval of, 725 radius of, 725 of a sequence, 677 of a series, 688 convergent improper integral, 509, 512 convergent sequence, 677 convergent series, 688 properties of, 693 coordinate(s), A2 Cartesian, A11 polar, 639 rectangular, A11 coordinate axes, A11 Cornu’s spiral, 637 cosine function, A26 derivative, 192 graph, 33, A31 power series, 740 cost function, 228, 327 critical number, 274 cross-section, 422 cubic function, 28 current, 201 curvature, 638 curve(s) asymptotic, 315 Bézier, 624, 639 bullet-nose, 51, 204 Devil’s, 213 length of, 525 orthogonal, 214 parametric, 621 polar, 641 smooth, 525 swallowtail catastrophe, 629 curve ﬁtting, 25 curve-sketching procedure, 308 cusp, 626 cycloid, 624 cylinder, 422 cylindrical shell, 433 decay, law of natural, 236 decay, radioactive, 236 decreasing function, 20 decreasing sequence, 681 deﬁnite integral, 366 properties of, 373 Substitution Rule for, 404 deﬁnite integration by parts, 456 by substitution, 404 degree of a polynomial, 28 delta () notation, 148 demand curve, 327, 550 demand function, 327, 550 De Moivre, Abraham, A59 De Moivre’s Theorem, A59 density linear, 223 liquid, 540 A114 \|\|\|\| INDEX dependent variable, 12 derivative(s), 143,154 of a composite function, 199 of a constant function, 173 domain of, 154 of exponential functions, 180, 201, A51, A53 as a function, 154 higher, 160 of hyperbolic functions, 254 of an integral, 380 of an inverse function, 212 of inverse trigonometric func-tions, 211, 213 left-hand, 165 of logarithmic functions, 215, A49, A54 notation, 157 of a power function, 174 of a power series, 729 of a product, 183, 184 of a quotient, 185 as a rate of change, 148 right-hand, 165 second, 160 as the slope of a tangent, 146 third, 161 of trigonometric functions, 189, 193 Descartes, René, A11 descent of aircraft, determining start of, 206 Devil’s curve, 213 Difference Law of limits, 100 Difference Rule, 177 differentiable function, 157 differential, 250 differential equation, 234, 342, 566, 569 autonomous, 575 Bernoulli, 607 ﬁrst-order, 580 general solution of, 569 linear, 602 logistic, 592, 687 order of, 569 second-order, 569 separable, 580 solution of, 569 differentiation, 157 formulas for, 187, RP5 implicit, 207, 208 logarithmic, 217 of a power series, 729 term by term, 729 differentiation operator, 157 Direct Substitution Property, 102 direction ﬁeld, 572, 573, 600 directrix, 655, 662 discontinuity, 119, 120 discontinuous function, 119 discontinuous integrand, 511 disk method, 424 dispersion, 280 displacement, 145, 395 distance between points in a plane, A11 between real numbers, A7 distance formula, A12 distance problem, 362 divergence of an improper integral, 509, 512 of an inﬁnite series, 688 of a sequence, 677 Test for, 692 divergent improper integral, 509, 512 divergent sequence, 677 divergent series, 688 division of power series, 745 domain of a function, 11 double-angle formulas, A29 drumhead, vibration of, 724 dye dilution method, 552 e (the number), 56, 179, A50 as a limit, 219 eccentricity, 662 electric circuit, 605 to a ﬂash bulb, 84, 205 elementary function, 487 ellipse, 213, 656, 662, A19 area, 491 directrix, 662 eccentricity, 662 foci, 656, 662 major axis, 667, 657 polar equation, 663 reﬂection property, 658 rotated, 214 vertices, 666, 657 empirical model, 25 end behavior of a function, 142 endpoint extreme values, 272 epicycloid, 630 equation(s) of a circle, A17 differential (see Differential equation) of an ellipse, A19 of a graph, A10, A16 of a hyperbola, A20 of a line, A12, A13, A14, A16 linear, A14 logistic, 568 logistic differential, 568, 600 Lotka-Volterra, 609 nth-degree, 210 of a parabola, A18 parametric, 631 point-slope, 19, A12 polar, 641 predator-prey, 609 second-degree, A16 slope-intercept, A13 two-intercept form, A16 equilateral hyperbola, A21 equilibrium point, 610 equilibrium position, 196 equilibrium solution, 568, 609 error in approximate integration, 521 percentage, 251 relative, 251 in Taylor approximation, 749 error bounds, 499, 503 error estimate for alternating series, 712 for the Midpoint Rule, 521 for Simpson’s Rule, 525 for the Trapezoidal Rule, 521 escape velocity, 539 estimate of the sum of a series, 700, 708, 712, 717 Eudoxus, 3 Euler’s constant, 704 Euler’s formula, A61 Euler’s Method, 575 even function, 19, 308 exponential decay, 591 exponential function(s), 33, 92, 180 with base a, A53 derivatives of, 180, 201, A51, A53 graphs of, 53, 180 integration of, 370, 385, 400 limits of, 137, A51 power series for, 736 properties of, A51 exponential graph, 53 exponential growth, 591 exponents, laws of, 54, A51, A53 extrapolation, 27 extreme value, 271 Extreme Value Theorem, 272 family of functions, 50, 320 family of hypocycloids, 629 family of solutions, 568 fat circles, 211, 531 Fermat, Pierre, 4, 153, 273 INDEX \|\|\|\| A115 Fermat’s Principle, 331 Fermat’s Theorem, 273 Fibonacci, 686 Fibonacci sequence, 676 First Derivative Test, 288 for Absolute Extreme Values, 324 ﬁrst-order linear differential equation, 602 ﬁxed point of a function, 171, 286 ﬂash bulb, current to, 84 ﬂux, 551, 552 FM synthesis, 318 focus of a conic section, 662 of an ellipse, 656, 662 of a hyperbola, 658 of a parabola, 655 folium of Descartes, 208, 672 force, 438 exerted by ﬂuid, 540 Fourier, Joseph, 230 Fourier sequence, 467 four-leaved rose, 643 fractions (partial), 473, 474 Fresnel, Augustin, 383 Fresnel function, 383 frustum of a cone, 431 of a pyramid, 432 function(s), 12 absolute value, 17 algebraic, 31 arc length, 528 arrow diagram of, 12 average cost, 330 average value of, 433, 557 Bessel, 674, 724 combinations of, 41 composite, 41, 197 constant, 173 continuous, 119 cost, 228, 327 cubic, 28 decreasing, 20 demand, 327 derivative of, 146 differentiable, 157 discontinuous, 119 domain of, 11 elementary, 487 even, 19, 308 exponential, 33, 52, 180 extreme values of, 271 family of, 50, 320 ﬁxed point of, 171, 286 Fresnel, 397 Gompertz, 600 graph of, 12 greatest integer, 105 Heaviside, 45, 92 hyperbolic, 254 implicit, 207 increasing, 20 inverse, 61 inverse hyperbolic, 254 inverse trigonometric, 67, 68 limit of, 88, 110 linear, 24 logarithmic, 34, 63, A48, A53 machine diagram of, 12 marginal cost, 229, 327 marginal proﬁt, 327 marginal revenue, 327 natural logarithmic, 64 nondifferentiable, 60 odd, 19, 308 one-to-one, 60 periodic, 308 piecewise deﬁned, 17 polynomial, 28 position, 145 power, 29, 173 proﬁt, 327 quadratic, 28 ramp, 45 range of, 11 rational, 31, 473 reciprocal, 31 reﬂected, 38 representation as a power series, 728 representations of, 11, 13 revenue, 327 root, 30 shifted, 37 sine integral, 389 step, 18 stretched, 38 transcendental, 34 transformation of, 37, 38 translation of, 38 trigonometric, 32, A26 value of, 12 Fundamental Theorem of Calculus, 381, 384, 387 G (gravitational constant), 231, 442 Gabriel’s horn, 537 Galileo, 625, 633 Galois, Evariste, 210 Gause, G. F., 596 Gauss, Karl Friedrich, A35 geometric series, 688 Gompertz function, 600 Gourdon, Xavier, 739 graph(s) of an equation, A10, A16 of exponential functions, 53 of a function, 12 of logarithmic functions, 66 of a parametric curve, 632 polar, 641, 646 of power functions, 30, RP3 of trignometric functions, A30, RP2 graphing calculator, 46, 315, 624, 646 gravitational acceleration, 438 gravitation law, 442 greatest integer function, 105 Gregory, James, 732 Gregory’s series, 732 growth, law of natural, 234, 591 growth rate, 226 relative, 234 half-angle formulas, A29 half-life, 236 hare-lynx system, 612 harmonic series, 791 Heaviside, Oliver, 92 Heaviside function, 45, 92 Hecht, Eugene, 250, 754 height of a rocket 459 higher derivatives, 160 Hooke’s Law, 439 horizontal asymptote, 132 horizontal line, A13 Horizontal Line Test, 60 Hubble Space Telescope, 276 Huygens, 625 hydrostatic pressure and force, 539 hyperbola, 213, 658, 662, A20 asymptotes, 629, A20 branches, 629, A20 directrix, 662 eccentricity, 662 equation, 658, A20 equilateral, A21 foci, 658, 662 polar equation, 663 reﬂection property, 662 vertices, 658 hyperbolic function(s), 254 derivatives, 254 inverse, 257, 258 hyperbolic identities, 255 hyperbolic substitution, 471 hypocycloid, 629 A116 \|\|\|\| INDEX i (imaginary number), A46 I/D Test, 287 ideal gas law, 233 implicit differentiation, 207, 208 implicit function, 207 improper integral, 508 impulse of a force, 601 Increasing/Decreasing Test, 287 increasing function, 20 increasing sequence, 681 increment, 147 indeﬁnite integrals, 392 table of, 392 independent variable, 11 indeterminate difference, 298 indeterminate forms of limits, 298 indeterminate product, 302 indeterminate power, 303 index of summation, A34 inequalities, rules for, A4 inﬁnite discontinuity, 120 inﬁnite interval, 530, 531 inﬁnite limit, 94, 116, 136 inﬁnite sequence (see Sequence) inﬁnite series (see Series) inﬂection point, 219 initial condition, 570 initial point of a parametric curve, 662 initial-value problem, 570 instantaneous rate of change, 85, 148, 148, 221 instantaneous rate of growth, 221 instantaneous rate of reaction, 225 instantaneous velocity, 85, 145, 221 integer, A2 integral(s) approximations to, 372 change of variables in, 400 comparison properties of, 375 deﬁnite, 366 derivative of, 401 evaluating, 369 improper, 508 indeﬁnite, 391 patterns in, 494 properties of, 373 of symmetric functions, 405 table of, 484, RP6–10 units for, 396 Integral Test, 697 integrand, 366 discontinuous, 511 integration, 366 approximate, 495 by computer algebra system, 491 of exponential functions, 370, 385, 400 formulas, 452, 484, RP6–10 indeﬁnite, 391 limits of, 366 numerical, 496 by partial fractions, 473 left-hand derivative, 165 left-hand limit, 192, 113 Leibniz, Gottfried Wilhelm, 4, 157, 399, 580, 748 Leibniz notation, 157 lemniscate, 213 length of a curve, 325 of a line segment, A7, A12 of a parametric curve, 635 of a polar curve, 652 l’Hospital, Marquis de, 299, 307 l’Hospital’s Rule, 299, 307 origins of, 307 libration point, 340 limaçon, 647 Limit Comparison Test, 707 Limit Laws, 99, A39 Limit Laws for sequences, 778 limit(s), 3 calculating, 99 of exponential functions, 136, 137 of a function, 88, 110 inﬁnite, 94, 116, 136 at inﬁnity, 130, 131, 137 of integration, 366 left-hand, 93, 113 one-sided, 93, 113 precise deﬁnitions, 109, 113, 116, 138, 140 properties of, 99 right-hand, 93, 163 of a sequence, 6, 359, 677 of a trigonometric function, 190 linear approximation, 246 linear density, 223 linear differential equation, 602 linear equation, A14 linear function, 24 linearization, 48 linear model, 24 linear regression, 27 line(s) in the plane, A12 equations of, A12, A13, A14 horizontal, A13 normal, 181 parallel, A14 perpendicular, A14 secant, 4, 83 slope of, A12 tangent, 4, 83, 144 liquid force, 539, 540 Lissajous ﬁgure, 673 lithotripsy, 658 local maximum and minimum, 271 logarithm(s), 34, 63 laws of, 64, A49 natural, 64, A48 notation for, 64 logarithmic differentiation, 217 logarithmic function(s), 34, 63 with base a, A53 derivatives of, 215, A49, A53 graphs of, 64, 66 integration of, 218 limits of, 101, A50 properties of, 68, A49 logistic difference equation, 687 logistic differential equation, 568, 592 logistic model, 600 logistic sequence, 687 LORAN system, 661 Lotka-Volterra equations, 609 machine diagram of a function, 12 Maclaurin, Colin, 736 Maclaurin series, 734, 736 table of, 741 major axis of ellipse, 657 marginal cost function, 228, 327 marginal proﬁt function, 327 marginal propensity to consume or save, 695 marginal revenue function, 327 mass, center of, 542 mathematical induction, 695 principle of, 77 mathematical model, 14, 24 maximum and minimum values, 271 mean life of an atom, 517 mean of a probability density function, 557 Mean Value Theorem, 282 for integrals, 443 mean waiting time, 557 median of a probability density function, 558 method of cylindrical shells, 433 method of exhaustion, 3, 102 method of least squares, 27 midpoint formula, A16 Midpoint Rule, 372, 496 error in using, 497 mixing problems, 584 INDEX \|\|\|\| A117 modeling with differential equations, 567 motion of a spring, 568 population growth, 55, 567, 591, 600, 597, 616 vibration of membrane, 724 model(s), mathematical, 24 comparison of natural growth vs. logistic, 596 empirical, 25 exponential, 33 Gompertz function, 600 linear, 24 logarithmic, 34 polynomial, 28 power function, 29 predator-prey, 233, 609 rational function, 31 seasonal growth, 600 trigonometric, 32, 33 von Bertalanffy, 616 modulus, A56 moment about an axis, 543 of a lamina, 543 of a mass, 543 of a system of particles, 543 momentum of an object, 601 monotonic sequence, 681 Monotonic Sequence Theorem, 683 movie theater seating, 446 multiplication of power series, 745 multiplier effect, 695 natural exponential function, 56, A50 derivatives of, 180, A51 graph of, 180 properties of, A51 natural growth law, 234, 591 natural logarithm function, 64, A48 derivatives of, 215, A49 limits of, 424 properties of, A49 negative angle, A25 net area, 367 Net Change Theorem, 394 net investment ﬂow, 554 Newton, Sir Isaac, 4, 9, 102, 153, 157, 380, 399, 748 newton (unit of force), 438 Newton’s Law of Cooling, 237 Newton’s Law of Gravitation, 231, 442 Newton’s method, 334, 335 Newton’s Second Law, 438 Nicomedes, 626 nondifferentiable function, 160 normal distribution, 559 normal line, 181 nth-degree equation, ﬁnding roots of, 210 nth-degree Taylor polynomial, 254, 737 number complex, A55 integer, A2 irrational, A2 rational, A2 real, A2 numerical integration, 496 odd function, 19, 308 one-sided limits, 92, 113 one-to-one function, 60 open interval, A3 optics ﬁrst-order, 754 Gaussian, 7554 third-order, 755 optimization problems, 271, 322 order of a differential equation, 569 ordered pair, A10 Oresme, Nicole, 692 origin, A2, A10 orthogonal curves, 214 orthogonal trajectory, 214, 583 ovals of Cassini, 649 Pappus, Theorem of, 546 Pappus of Alexandria, 546 parabola, 655, 663, A18 axis, 655 directrix, 655 equation, 656 focus, 655, 662 polar equation, 663 reﬂection property, 268, 269 vertex, 655 paradoxes of Zeno, 6, 7 parallelepiped, 422 parallel lines, A14 parameter, 621 parametric curve, 621 parametric equations, 621 paraxial rays, 249 partial fractions, 473, 474 partial integration, 453, 454 partial sum of a series, 688 parts, integration by, 453, 454 patterns in integrals, 494 pendulum, approximating the period of, 249, 253 percentage error, 251 perihelion, 667 perilune, 661 period, 308 periodic function, 308 perpendicular lines, A14 phase plane, 610 phase portrait, 610 phase trajectory, 610 piecewise deﬁned function, 17 Planck’s Law, 757 point of inﬂection, 291 point-slope equation of a line, 18, A12 Poiseuille, Jean-Louis-Marie, 227 Poiseuille’s Laws, 253, 332, 552 polar axis, 639 polar coordinates, 639 polar equation(s), 639 of conics, 663 graph of, 641 polar form of a complex number, A57 pole, 639 polynomial, 28 population growth, 591 of bacteria, 227, 591, 598 of ﬁsh, 599 of insects, 483 models, 567 world, 55, 235 position function, 145 positive angle, A25 potential, 520 pound, 438 power consumption, approximation of, 396 power function, 29, 173 Power Law of limits, 101 Power Rule, 174, 218 power series, 723 coefﬁcient of, 723 differentiation of, 729 division of, 745 integration of, 729 interval of convergence, 725 multiplication of, 745 radius of convergence, 725 representations of functions as, 728 predator, 608 predator-prey model, 233, 609 pressure exerted by a ﬂuid, 540 prey, 609 prime notation, 146, 177 principal square root of a complex number, A56 principle of mathematical induction, 77, 80, A36 A118 \|\|\|\| INDEX probability density function, 555 for customer waiting time, 561 problem-solving principles, 76 producer surplus, 553 product formulas, A29 Product Law of limits, 100 Product Rule, 183 proﬁt function, 327 projectile, 629 p-series, 699 quadrant, A11 quadratic approximation, 254 quadratic function, 28 Quotient Law of limits, 100 Quotient Rule, 185 radian measure, 189, A24 radiation from stars, 757 radioactive decay, 235 radiocarbon dating, 240 radius of convergence, 725 rainbow, formation and location of, 279 rainbow angle, 279 ramp function, 45 range of a function, 11 rate of change average, 148, 221 derivative as, 148 instantaneous, 85, 148, 221 rate of growth, 226 rate of reaction, 225 rates, related, 241 rational function, 31 integration of, 473 rationalizing substitution, 481 rational number, A2 Ratio Test, 716 Rayleigh-Jeans Law, 757 real line, A3 real number, A2 rearrangement of a series, 719 reciprocal function, 31 Reciprocal Rule, 189 rectangular coordinate system, A11 rectilinear motion, 343 reduction formula, 457 reﬂecting a function, 38 reﬂection property of an ellipse, 658 of a hyperbola, 662 of a parabola, 268, 269 region under a graph, 355, 360 between two graphs, 415 related rates, 241 relative error, 251 relative growth rate, 234, 592 relative maximum and minimum, 271 remainder estimates for the Alternating Series, 712 for the Comparison Test, 707 for the Integral Test, 701 for the Ratio Test, 716 remainder of the Taylor series, 737 removable discontinuity, 120 representations of functions, 11, 13 revenue function, 327 revolution, solid of, 427 revolution, surface of, 532 Riemann, Georg Bernhard, 367 Riemann sum(s), 367 right circular cylinder, 422 right-hand derivative, 165 right-hand limit, 92, 113 RMS voltage, 466 Roberval, Gilles de, 386, 633 Rolle, Michel, 280 roller coaster, design of, 182 Rolle’s Theorem, 280 root function, 30 roots of a complex number, A60 roots of an nth-degree equation, 210 Root Test, 714 rumors, rate of spread of, 230 sample point, 360 scatter plot, 14 seasonal growth model, 600 secant function, A26 derivative, 193 graph, A31 secant line, 4, 83 second derivative, 160 Second Derivative Test, 292 second-order differential equation, 569 sector of a circle, 650 sensitivity, 233 separable differential equation, 580 sequence, 6, 675 bounded, 682 convergent, 677 decreasing, 681 divergent, 677 Fibonacci, 676 graph of, 681 increasing, 681 limit of, 6, 357, 677 monotonic, 681 of partial sums, 688 term of, 675 series, 7, 687 absolutely convergent, 714 alternating, 710 alternating harmonic, 711, 715 binomial, 742, 748 coefﬁcient of, 723 conditionally convergent, 715 convergent, 688 divergent, 688 geometric, 688 Gregory’s, 732 harmonic, 691 inﬁnite, 687 Maclaurin, 734, 736 p-, 699 partial sum of, 688 power, 723 rearrangement of, 719 strategy for testing, 721 sum of, 7, 688 Taylor, 734, 736 term of, 687 trigonometric, 723 set, A3 serpentine, 188 shell method, 433 shifted conics, 659, A21 shift of a function, 37 Sierpinski carpet, 696 sigma notation, 360, A34 simple harmonic motion, 205 Simpson, Thomas, 501, 502 Simpson’s Rule, 500, 502 error bounds for, 503 sine function, A26 derivative, 193 graph, 32, A31 power series, 740 sine integral function, 389 slant asymptote, 312 slope, A12 slope-intercept equation of a line, A13 slope ﬁeld, 573 smooth curve, 525 smooth function, 525 Snell’s Law, 331 snowﬂake curve, 761 solid, 422 volume of, 423 solid of revolution, 427 rotated on a slant, 538 volume of, 430, 434, 538 INDEX \|\|\|\| A119 solution of predator-prey equations, 609 solution curve, 572 speed, 148 speedometer readings, interpre-tation of, 85 spherical zones, 564 spring constant, 439, 568 Squeeze Theorem, 105, A42 for sequences, 679 standard deviation, 559 stellar stereography, 539 step function, 18 strategy for integration, 483, 484 for optimization problems, 322 for problem solving, 76 for related rates, 243 for testing series, 721 for trigonometric integrals, 462, 463 stretching a function, 38 strophoid, 653, 671 Substitution Rule, 400, 401, 404 subtraction formulas for sine and cosine, A29 sum of a geometric series, 689 of an inﬁnite series, 688 of partial fractions, 474 Riemann, 367 telescoping, 791 Sum Law of limits, 100 summation notation, A34 Sum Rule, 177 supply function, 553 surface area, 534, 538 of a parametric surface, 635 surface of revolution, 532 area of, 534, 538 swallowtail catastrophe curve, 629 symmetric functions, integrals of, 405 symmetry, 19, 308, 405 in polar graphs, 644 symmetry principle, 543 table of differentiation formulas, 187, RP5 tables of integrals, 484, RP6–10 use of, 489 tangent function, A26 derivative, 192 graph, 33, A31 tangent line(s), 143 to a curve, 4, 88, 144 early methods of ﬁnding, 153 to a parametric curve, 630, 631 to a polar curve, 644 tangent line approximation, 247 tangent problem, 4, 83, 144 tautochrone problem, 625 Taylor, Brook, 736 Taylor polynomial, 254, 737, 749 Taylor series, 734, 736 Taylor’s Inequality, 737 techniques of integration, summary, 484 telescoping sum, 691 term of a sequence, 675 term of a series, 687 terminal point of a curve, 622 terminal velocity, 587 term-by-term differentiation and integration, 729 tests for convergence and divergence of series Alternating Series Test, 710 Comparison Test, 705 Integral Test, 697 Limit Comparison Test, 707 Ratio Test, 716 Root Test, 714 summary of tests, 721 Test for Divergence, 692 third derivative, 161 Torricelli, Evangelista, 633 Torricelli’s Law, 231 torus, 432, 473 total fertility rate, 169 transcendental function, 34 transformation of a function, 37 translation of a function, 38 Trapezoidal Rule, 497 error in, 497 Triangle Inequality, A8 trigonometric functions, 32, A26 derivatives of, 189, 193 graphs of, A30 integrals of, 460 inverse, 67 limits of, 190 trigonometric identities, A28 trigonometric integrals, 460 strategy for evaluating, 462, 463 trigonometric series, 723 trigonometric substitutions, 467 table of, 467 trochoid, 628 Tschirnhausen cubic, 214, 421 ultraviolet catastrophe, 757 union of sets, A3 value of a function, 11 variable dependent, 11 independent, 11 variables, change of, 401 vascular branching, 332 velocity, 4, 83, 145, 221 average, 5, 86, 145, 221 instantaneous, 86, 145, 221 velocity gradient, 228 velocity problem, 85, 145 Verhulst, 568 vertex of a parabola, 655 vertical asymptote, 95, 308 Vertical Line Test, 16 vertical tangent line, 159 vertical translation of a graph, 37 vertices of an ellipse, 656, 657 of a hyperbola, 658 vibration of a rubber membrane, model of, 724 viewing rectangle, 46 volume, 423 by cross-sections, 422 by cylindrical shells, 433 by disks, 424, 425 of a solid, 422 of a solid of revolution, 427, 538 of a solid on a slant, 538 by washers, 426, 427 Volterra, Vito, 609 Von Bertalanffy model, 616 Wallis, John, 4 Wallis product, 459 washer method, 426 Weierstrass, Karl, 482 weight, 438 witch of Maria Agnesi, 188 work, 438 Wren, Sir Christopher, 635 x-axis, A10 x-coordinate, A10 x-intercept, A19 y-axis, A10 y-coordinate, A10 y-intercept, A19 Zeno, 6 Zeno’s paradoxes, 6, 7 zone of a sphere, 538 A120 \|\|\|\| INDEX
15404	https://www.amazon.com/Matrix-Analysis-Roger-Horn/dp/0521386322	Matrix Analysis: Horn, Roger A., Johnson, Charles R.: 9780521386326: Amazon.com: Books Skip to Main content About this item About this item About this item Buying options Compare with similar items Videos Reviews Keyboard shortcuts Search opt+/ Cart shift+opt+C Home shift+opt+H Orders shift+opt+O Add to cart shift+opt+K Show/Hide shortcuts shift+opt+Z To move between items, use your keyboard's up or down arrows. .us Delivering to Brentwood 94513 Update location Books Select the department you want to search in Search Amazon EN Hello, sign in Account & Lists Returns& Orders0 Cart Sign in New customer? Start here. 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Download the free Kindle app and start reading Kindle books instantly on your smartphone, tablet, or computer - no Kindle device required. Read instantly on your browser with Kindle for Web. Using your mobile phone camera - scan the code below and download the Kindle app. Image Unavailable Image not available for Color: To view this video download Flash Player VIDEOS 360° VIEW IMAGES Read sample Follow the authors Charles R. Johnson Charles R. John… Follow Roger A. Horn Roger A. Horn Follow Something went wrong. Please try your request again later. OK Matrix Analysis Reprint Edition by Roger A. Horn(Author), Charles R. Johnson(Author) 4.3 4.3 out of 5 stars31 ratings Sorry, there was a problem loading this page. Try again. See all formats and editions {"desktop_buybox_group_1":[{"displayPrice":"$53.00","priceAmount":53.00,"currencySymbol":"$","integerValue":"53","decimalSeparator":".","fractionalValue":"00","symbolPosition":"left","hasSpace":false,"showFractionalPartIfEmpty":true,"offerListingId":"H948viJ5Q%2FXlo7etM8b0P6sADSp5y%2FNp6J6qJ1SHTnwnKZvJio%2FVGjj6mT7StaDmQQCs0Le1tl8WZEpGeSTJJtmVfBceTJS9wfThdrImd3mdCts%2FYi4xOwzypF01MXIoekbINQj2UDX%2FnTc%2FTT%2F3gHUddzaLMy3LvMxj6Gzaix5DjlzzLwzCwDPMGEkfxZ33","locale":"en-US","buyingOptionType":"NEW","aapiBuyingOptionIndex":0}, {"displayPrice":"$18.19","priceAmount":18.19,"currencySymbol":"$","integerValue":"18","decimalSeparator":".","fractionalValue":"19","symbolPosition":"left","hasSpace":false,"showFractionalPartIfEmpty":true,"offerListingId":"H948viJ5Q%2FXlo7etM8b0P6sADSp5y%2FNpiffmQ43KswPVa74odx1xjpQR0x0FrSswy9IapxLk%2B1fjQERfIbp1VBuKVPUCCKJ7v1kC5UJY9Hy0VgdPp3gODN0ECh5RnEWm8ZcsnmN7hdmtq%2FbC2U5UM5HPOHEClHqUuQ8VOwmkRBVC8mvwdjNlh4nZzL13EAcN","locale":"en-US","buyingOptionType":"USED","aapiBuyingOptionIndex":1}]} Purchase options and add-ons [x] Linear algebra and matrix theory have long been fundamental tools in mathematical disciplines as well as fertile fields for research. In this book the authors present classical and recent results of matrix analysis that have proved to be important to applied mathematics. Facts about matrices, beyond those found in an elementary linear algebra course, are needed to understand virtually any area of mathematical science, but the necessary material has appeared only sporadically in the literature and in university curricula. As interest in applied mathematics has grown, the need for a text and reference offering a broad selection of topics in matrix theory has become apparent, and this book meets that need. This volume reflects two concurrent views of matrix analysis. First, it encompasses topics in linear algebra that have arisen out of the needs of mathematical analysis. Second, it is an approach to real and complex linear algebraic problems that does not hesitate to use notions from analysis. Both views are reflected in its choice and treatment of topics. Read more Report an issue with this product or seller Previous slide of product details ISBN-10 0521386322 ISBN-13 978-0521386326 Edition Reprint Publisher Cambridge University Press Publication date February 23, 1990 Language English Dimensions 6.25 x 1 x 9.5 inches Print length 575 pages Next slide of product details See all details There is a newer edition of this item: Matrix Analysis $50.80 (101) In Stock Frequently purchased items with fast delivery Page 1 of 22Start over Previous set of slides Matrix AnalysisRoger A. Horn 4.6 out of 5 stars 101 Paperback $50.80$50.80 Get it as soon as Friday, Oct 3 FREE Shipping by Amazon Complex Analysis (AMS Chelsea Publishing)Lars Ahlfors 4.4 out of 5 stars 17 Paperback $60.00$60.00 Get it as soon as Friday, Oct 3 FREE Shipping by Amazon Only 13 left in stock (more on the way). Real AnalysisN. L. Carothers 4.5 out of 5 stars 94 Paperback $69.71$69.71 Get it as soon as Friday, Oct 3 FREE Shipping by Amazon Next set of slides Customers who viewed this item also viewed Page 1 of 1Start over Previous set of slides Matrix AnalysisRoger A. Horn 4.6 out of 5 stars 101 Paperback $50.80$50.80 Get it as soon as Friday, Oct 3 FREE Shipping by Amazon Next set of slides Editorial Reviews Review "There seems little doubt that the book will become a standard reference for research workers in numerical mathematics." Computing Reviews "The reviewer strongly recommends that those working in either pure or applied linear algebra have this book on their desks." SIAM Review "The presentation is straightforward and extremely readable. The authors' enthusiasm pervades the book, and the printing is what we expect from this publisher. This will doubtless be the standard text for years to come." American Scientist Book Description Matrix Analysis presents the classical and recent results for matrix analysis that have proved to be important to applied mathematics. Book Description A presentation of classical as well as recent results of matrix analysis that have proven important to applied mathematics and reflect two concurrent views of the subject. Read more Product details Publisher ‏ : ‎ Cambridge University Press Publication date ‏ : ‎ February 23, 1990 Edition ‏ : ‎ Reprint Language ‏ : ‎ English Print length ‏ : ‎ 575 pages ISBN-10 ‏ : ‎ 0521386322 ISBN-13 ‏ : ‎ 978-0521386326 Item Weight ‏ : ‎ 1.7 pounds Dimensions ‏ : ‎ 6.25 x 1 x 9.5 inches Best Sellers Rank: #699,259 in Books (See Top 100 in Books) 17 in Mathematical Matrices 94 in Algebra & Trigonometry 199 in Algebra Customer Reviews: 4.3 4.3 out of 5 stars31 ratings Brief content visible, double tap to read full content. Full content visible, double tap to read brief content. Videos Help others learn more about this product by uploading a video! Upload your video About the authors Follow authors to get new release updates, plus improved recommendations. Previous page Follow Charles R. Johnson ------------------ Brief content visible, double tap to read full content. Full content visible, double tap to read brief content. Discover more of the author’s books, see similar authors, read book recommendations and more. See more on the author's page Follow Roger A. Horn ------------- Brief content visible, double tap to read full content. Full content visible, double tap to read brief content. Discover more of the author’s books, see similar authors, read book recommendations and more. 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Instead, our system considers things like how recent a review is and if the reviewer bought the item on Amazon. It also analyzed reviews to verify trustworthiness. Learn more how customers reviews work on Amazon Sponsored View Image Gallery Amazon Customer 5.0 out of 5 stars Images in this review Top reviews from the United States There was a problem filtering reviews. Please reload the page. Zaid Al-Shayeb ##### 5.0 out of 5 stars Good Reviewed in the United States on December 8, 2023 Format: PaperbackVerified Purchase Good Read more Helpful Report CalvinNHobbes ##### 5.0 out of 5 stars Excellent text! Reviewed in the United States on July 18, 2022 Format: PaperbackVerified Purchase The is one of the best books on matrix analysis. Extremely well written! Read more Helpful Report Amcust ##### 5.0 out of 5 stars Good Introductory text Reviewed in the United States on January 1, 2011 Format: PaperbackVerified Purchase This book has a very good introductory material on matrices and related topics like eigenvalues, canonical forms and location of eigenvalues etc.. It starts from basics like definitions, types of matrices etc.. and goes on to touch some advanced topics as well. A very good text. Read more Helpful Report Patrick Thompson ##### 3.0 out of 5 stars Matrix Analysis Reviewed in the United States on July 22, 2009 Format: PaperbackVerified Purchase What an uninteresting book! The content of the book may be good, but the style is very dry and boring. It's really abstract and I don't think the problems in the book are very interesting. I would recommend this book as a reference rather than a textbook. Read more 5 people found this helpful Helpful Report teclado ##### 5.0 out of 5 stars A brilliant guided tour Reviewed in the United States on October 20, 2012 Format: PaperbackVerified Purchase Just excellent. A lot of material, and a lot of problems that allow you verify that you are really understanding. Difficult to find another book dealing with many of the subjects present in this book. Read more One person found this helpful Helpful Report Ricardo Alberto Marques Pereira ##### 5.0 out of 5 stars Excellent service. Reviewed in the United States on July 19, 2019 Format: HardcoverVerified Purchase The product has met my expectations. Read more Helpful Report Srinivasa ##### 3.0 out of 5 stars Good collection of results in the subject Reviewed in the United States on October 18, 2007 Format: PaperbackVerified Purchase Definitely, not a lot of attention has been paid to pedagogy by the authors of this book. However it forms an excellent summary of most of the theory and is very good for one who understands it and one who wants a ready reference in the subject. Read more 4 people found this helpful Helpful Report Nancy ) Africh ##### 5.0 out of 5 stars Five Stars Reviewed in the United States on August 13, 2015 Format: PaperbackVerified Purchase Good book - a hidden gem Read more Helpful Report See more reviews Top reviews from other countries Translate all reviews to English Amazon Customer ##### 5.0 out of 5 stars Useful book for users of matrices Reviewed in the United Kingdom on January 13, 2018 Format: HardcoverVerified Purchase Very useful for professional mathematicians who use matrix analysis; full of useful results. Read more Report bachpacker ##### 4.0 out of 5 stars 痒いところに手が届く Reviewed in Japan on May 16, 2007 Format: PaperbackVerified Purchase 著者らは、発展的な線型代数についての本が世の中にあまり存在しないことを理由に本書を執筆している。また、「線型代数」という用語（線型性のある代数構造を分析する学問）では、行列の織り成す世界を表すのには不十分であると考え、敢えて「行列解析」という題名にしている。さて本書は、標準的な線型代数は既知の読者を対象に、0章で簡単にこれらの概念を復習し、それ以降の章で発展的な内容を扱っている。具体的には、1章は固有値・固有ベクトルと相似性、2章はユニタリ同値性と正規行列、3章は標準形、4章はエルミート行列と対称行列、5章はノルム、6章は固有値の局在と摂動、7章は正定行列、8章は非負行列である。証明はテクニカルなことがある（例えば3章のジョルダン標準形は、広義固有空間を用いる方法とも単因子論を用いる方法とも異なり、成分計算で証明している）が、扱っている話題が非常に豊富かつ詳細で、まさに痒いところに手が届くという印象がある。通読するにはあまり適さないかもしれないが、応用家にとっては有用な参考書ではなかろうか。 Read more ReportTranslate review to English Andreas GrÃ¼ndinger ##### 5.0 out of 5 stars Gutes wenn auch nicht selbsterklärendes Buch Reviewed in Germany on May 18, 2015 Format: PaperbackVerified Purchase Das Buch enthält zuwar einige kleinere Fehler, aber im großen und ganzen kann man sehr viel beim durcharbeiten lernen. Speziell bei der Eigenwertu und Singulärwertzerlegung wird man definitiv immer wieder neue Herleitungen und Ungleichungen finden. Read more ReportTranslate review to English Jane Ye ##### 4.0 out of 5 stars Four Stars Reviewed in Canada on December 9, 2014 Format: PaperbackVerified Purchase Good but slow. Read more Report See more reviews TopAbout this itemSimilarReviews Your recently viewed items and featured recommendations › View or edit your browsing history After viewing product detail pages, look here to find an easy way to navigate back to pages you are interested in. 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15405	https://www.mathworksheets4kids.com/grams-kilograms.php	Child Login Main Menu Math Language Arts Science Social Studies Interactive Worksheets Browse By Grade Become a Member Become a Member Math Interactive Worksheets Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Worksheets by Grade Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Number Sense and Operations Number Recognition Counting Skip Counting Place Value Number Lines Addition Subtraction Multiplication Division Word Problems Comparing Numbers Ordering Numbers Odd and Even Prime and Composite Roman Numerals Ordinal Numbers Properties Patterns Rounding Estimation In and Out Boxes Number System Conversions More Number Sense Worksheets Measurement Size Comparison Time Calendar Money Measuring Length Weight Capacity Metric Unit Conversion Customary Unit Conversion Temperature More Measurement Worksheets Financial Literacy Writing Checks Profit and Loss Discount Sales Tax Simple Interest Compound Interest Statistics and Data Analysis Tally Marks Pictograph Line Plot Bar Graph Line Graph Pie Graph Scatter Plot Mean, Median, Mode, Range Mean Absolute Deviation Stem-and-leaf Plot Box-and-whisker Plot Factorial Permutation and Combination Probability Venn Diagram More Statistics Worksheets Geometry Positions Shapes - 2D Shapes - 3D Lines, Rays and Line Segments Points, Lines and Planes Angles Symmetry Transformation Area Perimeter Rectangle Triangle Circle Quadrilateral Polygon Ordered Pairs Midpoint Formula Distance Formula Slope Parallel, Perpendicular and Intersecting Lines Scale Factor Surface Area Volume Pythagorean Theorem More Geometry Worksheets Pre-Algebra Factoring GCF LCM Fractions Decimals Converting between Fractions and Decimals Integers Significant Figures Percents Convert between Fractions, Decimals, and Percents Ratio Proportions Direct and Inverse Variation Order of Operations Exponents Radicals Squaring Numbers Square Roots Scientific Notations Logarithms Speed, Distance, and Time Absolute Value More Pre-Algebra Worksheets Algebra Translating Algebraic Phrases Evaluating Algebraic Expressions Simplifying Algebraic Expressions Algebraic Identities Equations Quadratic Equations Systems of Equations Functions Polynomials Inequalities Sequence and Series Matrices Complex Numbers Vectors More Algebra Worksheets Trigonometry Calculus Math Workbooks English Language Arts Summer Review Packets Science Social Studies Holidays and Events Support Worksheets> Math> Measurement> Metric Unit Conversion> Gram and Kilogram Gram and Kilogram Worksheets Explore this array of metric unit conversion worksheets and expedite your practice of converting between grams and kilograms! Including a plethora of printable exercises such as comparing two masses in grams and kilograms, converting measures in grams to kilograms and vice versa, and filling the missing measure on a balance scale, this set of exercises guarantees students of grade 3, grade 4, and grade 5 an instantaneous-yet-exponential progress in their scores. Access some of these worksheets for free, and subscribe for complete access! Conversion from Gram to Kilogram Assuring key upgrades in converting from grams to kilograms and comparing two measures and offering a few one-liner questions on conversions, this pdf practice set is a must-have for 4th grade children. Conversion from Kilogram to Gram How many grams make 3 kilograms? Respond to such questions like an ace conversion scholar, with these pdf worksheets. Multiply the given measures by 1000. That's all it takes to successfully convert into grams! Conversion between Gram and Kilogram \| Balance Scales Help 3rd grade children ace the concept of equivalence between grams and kilograms with the balance scales here! The task is to fill in the right pan of the scale with a mass equivalent to the one on the left pan. Conversion between Kilogram and Gram Refine your skills in conversion from grams to kilograms and kilograms to grams, and boost your grasp of comparison of the measures in the two units, with the wealth of exercises included in each printable here. Conversion between Gram and Kilogram \| Type 1 If you are lagging behind your peers in quickly converting between grams and kilograms, look no further! This 5th grade resource, which contain both examples and 45+ practice problems, is a surefire cure! Conversion between Gram and Kilogram \| Type 2 Progress like never before and convert to and from kilograms and grams with utmost confidence, using these printable worksheets! 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15406	https://math.stackexchange.com/questions/2388898/probabilities-traffic-lights	probability - Probabilities - Traffic Lights - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Probabilities - Traffic Lights Ask Question Asked 8 years, 1 month ago Modified8 years, 1 month ago Viewed 6k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. A person passes through three intersections monitored by traffic lights. The location and operation of these traffic lights is such that, to all intents and purposes, they appear to operate independently to a person travelling from one to another. The probability of a red light is 0.4 0.4, 0.8 0.8, and 0.5 0.5 respectively for each of the traffic lights. (a) Find the probability function of X X, the number of red lights the person encounters in a single trip. (b) Compute the mean of X X. (c) Assume that the waiting time for each red light is two minutes. What is the mean waiting time in one trip? What I got so far: P(R 1)=0.4 P(R 1)=0.4 P(R 2)=0.8 P(R 2)=0.8 P(R 3)=0.5 P(R 3)=0.5 P(X=3)=0.4∗0.8∗0.5=0.16 P(X=3)=0.4∗0.8∗0.5=0.16 P(X=0)=1−P(X=3)=1−0.16=0.84 P(X=0)=1−P(X=3)=1−0.16=0.84 probability combinatorics Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 10, 2017 at 10:19 maliiaButterflymaliiaButterfly asked Aug 10, 2017 at 9:46 maliiaButterflymaliiaButterfly 413 6 6 silver badges 14 14 bronze badges 7 Let X X be the number of red lights encountered. Then X X is a discrete random variable with possible values 0,1,2,3 0,1,2,3. You need to find the following probabilities: P(X=0),P(X=1),P(X=2),P(X=3)P(X=0),P(X=1),P(X=2),P(X=3) quasi –quasi 2017-08-10 09:52:23 +00:00 Commented Aug 10, 2017 at 9:52 Also, there's no information about yellow lights, so you should assume that the only possibilities are red and non-red. So for example, when encountering the first light, it will be red with probability 0.4, and non-red with probability 0.6.quasi –quasi 2017-08-10 09:54:55 +00:00 Commented Aug 10, 2017 at 9:54 So on a single three lane intersection, isn't the maximum number of red lights only 1 or lets say 2? I mean, I cannot encounter 3 red lights on a single intersection. Maybe if it was a straight long lane with many traffic lights then it would be more logical to me? The intersection confuses me in this problem maliiaButterfly –maliiaButterfly 2017-08-10 09:57:34 +00:00 Commented Aug 10, 2017 at 9:57 There are no "lanes", just three separate intersections. In other words, the vehicle will encounter 3 3 separate lights. The reason for the term :"intersection" is that lights usually occur at intersections. So yes, just one long lane with 3 3 separate lights, each with their own probability of being red.quasi –quasi 2017-08-10 09:58:28 +00:00 Commented Aug 10, 2017 at 9:58 Oh, makes more sense now! Great. Thanks!maliiaButterfly –maliiaButterfly 2017-08-10 10:02:25 +00:00 Commented Aug 10, 2017 at 10:02 \|Show 2 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Part (a): For convenience of notation, let u=0.4,v=0.8,w=0.5 u=0.4,v=0.8,w=0.5. P(X=0)=P(X=1)=P(X=2)=P(X=3)=(1−u)(1−v)(1−w)=.06 u(1−v)(1−w)+(1−u)v(1−w)+(1−u)(1−v)w=.34(1−u)v w+u(1−v)w+u v(1−w)=.44 u v w=.16 P(X=0)=(1−u)(1−v)(1−w)=.06 P(X=1)=u(1−v)(1−w)+(1−u)v(1−w)+(1−u)(1−v)w=.34 P(X=2)=(1−u)v w+u(1−v)w+u v(1−w)=.44 P(X=3)=u v w=.16 Part (b): The mean of X X is 0⋅P(X=0)+1⋅P(X=1)+2⋅P(X=2)+3⋅P(X=3)=1.7 minutes 0⋅P(X=0)+1⋅P(X=1)+2⋅P(X=2)+3⋅P(X=3)=1.7 minutes Part (c): Assume each red light last for 2 2 minutes . Then on average, if a given light is red, the waiting time for that light is 1 1 minute (half of 2 2). Hence, the mean waiting time will be u⋅1+v⋅1+w⋅1=1.7 minutes u⋅1+v⋅1+w⋅1=1.7 minutes Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 10, 2017 at 11:02 answered Aug 10, 2017 at 10:40 quasiquasi 61.3k 3 3 gold badges 46 46 silver badges 105 105 bronze badges 0 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. As we've got only 3 lights with probabilities of red light p 1,p 2 p 1,p 2 and p 3 p 3, the solution seems to be easy P(X=0)=∏i=1 3(1−p i)P(X=1)=∑i=1 3 p i∏k≠i(1−p k)P(X=2)=∑i=1 3(1−p i)∏k≠i p k P(X=3)=∏i=1 3 p i P(X=0)=∏i=1 3(1−p i)P(X=1)=∑i=1 3 p i∏k≠i(1−p k)P(X=2)=∑i=1 3(1−p i)∏k≠i p k P(X=3)=∏i=1 3 p i Notice, that it is not true, that P(X=0)=1−P(X=3)P(X=0)=1−P(X=3) As You've got these probabilities, you can easily compute the mean value of X X Computing the mean waiting time could be calculated using yhis formula: T=∑i=1 3(t i p i+0(1−p i))T=∑i=1 3(t i p i+0(1−p i)) where t i t i is waiting time on i i'th light. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 10, 2017 at 10:40 Jaroslaw MatlakJaroslaw Matlak 4,965 16 16 silver badges 34 34 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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15407	https://openstax.org/books/anatomy-and-physiology-2e/pages/18-5-hemostasis	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in Anatomy and Physiology 2e 18.5 Hemostasis Anatomy and Physiology 2e 18.5 Hemostasis Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Describe the three mechanisms involved in hemostasis Explain how the extrinsic and intrinsic coagulation pathways lead to the common pathway, and the coagulation factors involved in each Discuss disorders affecting hemostasis Platelets are key players in hemostasis, the process by which the body seals a ruptured blood vessel and prevents further loss of blood. Although rupture of larger vessels usually requires medical intervention, hemostasis is quite effective in dealing with small, simple wounds. There are three steps to the process: vascular spasm, the formation of a platelet plug, and coagulation (blood clotting). Failure of any of these steps will result in hemorrhage—excessive bleeding. Vascular Spasm When a vessel is severed or punctured, or when the wall of a vessel is damaged, vascular spasm occurs. In vascular spasm, the smooth muscle in the walls of the vessel contracts dramatically. This smooth muscle has both circular layers; larger vessels also have longitudinal layers. The circular layers tend to constrict the flow of blood, whereas the longitudinal layers, when present, draw the vessel back into the surrounding tissue, often making it more difficult for a surgeon to locate, clamp, and tie off a severed vessel. The vascular spasm response is believed to be triggered by several chemicals called endothelins that are released by vessel-lining cells and by pain receptors in response to vessel injury. This phenomenon typically lasts for up to 30 minutes, although it can last for hours. Formation of the Platelet Plug In the second step, platelets, which normally float free in the plasma, encounter the area of vessel rupture with the exposed underlying connective tissue and collagenous fibers. The platelets begin to clump together, become spiked and sticky, and bind to the exposed collagen and endothelial lining. This process is assisted by a glycoprotein in the blood plasma called von Willebrand factor, which helps stabilize the growing platelet plug. As platelets collect, they simultaneously release chemicals from their granules into the plasma that further contribute to hemostasis. Among the substances released by the platelets are: adenosine diphosphate (ADP), which helps additional platelets to adhere to the injury site, reinforcing and expanding the platelet plug serotonin, which maintains vasoconstriction prostaglandins and phospholipids, which also maintain vasoconstriction and help to activate further clotting chemicals, as discussed next A platelet plug can temporarily seal a small opening in a blood vessel. Plug formation, in essence, buys the body time while more sophisticated and durable repairs are being made. In a similar manner, even modern naval warships still carry an assortment of wooden plugs to temporarily repair small breaches in their hulls until permanent repairs can be made. Coagulation Those more sophisticated and more durable repairs are collectively called coagulation, the formation of a blood clot. The process is sometimes characterized as a cascade, because one event prompts the next as in a multi-level waterfall. The result is the production of a gelatinous but robust clot made up of a mesh of fibrin—an insoluble filamentous protein derived from fibrinogen, the plasma protein introduced earlier—in which platelets and blood cells are trapped. Figure 18.14 summarizes the three steps of hemostasis. Figure 18.14 Hemostasis (a) An injury to a blood vessel initiates the process of hemostasis. Blood clotting involves three steps. First, vascular spasm constricts the flow of blood. Next, a platelet plug forms to temporarily seal small openings in the vessel. Coagulation then enables the repair of the vessel wall once the leakage of blood has stopped. (b) The synthesis of fibrin in blood clots involves either an intrinsic pathway or an extrinsic pathway, both of which lead to a common pathway. (credit a: Kevin MacKenzie) Clotting Factors Involved in Coagulation In the coagulation cascade, chemicals called clotting factors (or coagulation factors) prompt reactions that activate still more coagulation factors. The process is complex, but is initiated along two basic pathways: The extrinsic pathway, which normally is triggered by trauma. The intrinsic pathway, which begins in the bloodstream and is triggered by internal damage to the wall of the vessel. Both of these merge into a third pathway, referred to as the common pathway (see Figure 18.14b). All three pathways are dependent upon the 12 known clotting factors, including Ca2+ and vitamin K (Table 18.1). Clotting factors are secreted primarily by the liver and the platelets. The liver requires the fat-soluble vitamin K to produce many of them. Vitamin K (along with biotin and folate) is somewhat unusual among vitamins in that it is not only consumed in the diet but is also synthesized by bacteria residing in the large intestine. The calcium ion, considered factor IV, is derived from the diet and from the breakdown of bone. Some recent evidence indicates that activation of various clotting factors occurs on specific receptor sites on the surfaces of platelets. The 12 clotting factors are numbered I through XIII according to the order of their discovery. Factor VI was once believed to be a distinct clotting factor, but is now thought to be identical to factor V. Rather than renumber the other factors, factor VI was allowed to remain as a placeholder and also a reminder that knowledge changes over time. Clotting Factors \| Factor number \| Name \| Type of molecule \| Source \| Pathway(s) \| --- --- \| I \| Fibrinogen \| Plasma protein \| Liver \| Common; converted into fibrin \| \| II \| Prothrombin \| Plasma protein \| Liver \| Common; converted into thrombin \| \| III \| Tissue thromboplastin or tissue factor \| Lipoprotein mixture \| Damaged cells and platelets \| Extrinsic \| \| IV \| Calcium ions \| Inorganic ions in plasma \| Diet, platelets, bone matrix \| Entire process \| \| V \| Proaccelerin \| Plasma protein \| Liver, platelets \| Extrinsic and intrinsic \| \| VI \| Not used \| Not used \| Not used \| Not used \| \| VII \| Proconvertin \| Plasma protein \| Liver \| Extrinsic \| \| VIII \| Antihemolytic factor A \| Plasma protein factor \| Platelets and endothelial cells \| Intrinsic; deficiency results in hemophilia A \| \| IX \| Antihemolytic factor B (plasma thromboplastin component) \| Plasma protein \| Liver \| Intrinsic; deficiency results in hemophilia B \| \| X \| Stuart–Prower factor (thrombokinase) \| Protein \| Liver \| Extrinsic and intrinsic \| \| XI \| Antihemolytic factor C (plasma thromboplastin antecedent) \| Plasma protein \| Liver \| Intrinsic; deficiency results in hemophilia C \| \| XII \| Hageman factor \| Plasma protein \| Liver \| Intrinsic; initiates clotting in vitro also activates plasmin \| \| XIII \| Fibrin-stabilizing factor \| Plasma protein \| Liver, platelets \| Stabilizes fibrin; slows fibrinolysis \| Table 18.1 Vitamin K required. Extrinsic Pathway The quicker responding and more direct extrinsic pathway (also known as the tissue factor pathway) begins when damage occurs to the surrounding tissues, such as in a traumatic injury. Upon contact with blood plasma, the damaged extravascular cells, which are extrinsic to the bloodstream, release factor III (thromboplastin). Sequentially, Ca2+ then factor VII (proconvertin), which is activated by factor III, are added, forming an enzyme complex. This enzyme complex leads to activation of factor X (Stuart–Prower factor), which activates the common pathway discussed below. The events in the extrinsic pathway are completed in a matter of seconds. Intrinsic Pathway The intrinsic pathway (also known as the contact activation pathway) is longer and more complex. In this case, the factors involved are intrinsic to (present within) the bloodstream. The pathway can be prompted by damage to the tissues, resulting from internal factors such as arterial disease; however, it is most often initiated when factor XII (Hageman factor) comes into contact with foreign materials, such as when a blood sample is put into a glass test tube. Within the body, factor XII is typically activated when it encounters negatively charged molecules, such as inorganic polymers and phosphate produced earlier in the series of intrinsic pathway reactions. Factor XII sets off a series of reactions that in turn activates factor XI (antihemolytic factor C or plasma thromboplastin antecedent) then factor IX (antihemolytic factor B or plasma thromboplasmin). In the meantime, chemicals released by the platelets increase the rate of these activation reactions. Finally, factor VIII (antihemolytic factor A) from the platelets and endothelial cells combines with factor IX (antihemolytic factor B or plasma thromboplasmin) to form an enzyme complex that activates factor X (Stuart–Prower factor or thrombokinase), leading to the common pathway. The events in the intrinsic pathway are completed in a few minutes. Common Pathway Both the intrinsic and extrinsic pathways lead to the common pathway, in which fibrin is produced to seal off the vessel. Once factor X has been activated by either the intrinsic or extrinsic pathway, the enzyme prothrombinase converts factor II, the inactive enzyme prothrombin, into the active enzyme thrombin. (Note that if the enzyme thrombin were not normally in an inactive form, clots would form spontaneously, a condition not consistent with life.) Then, thrombin converts factor I, the soluble fibrinogen, into the insoluble fibrin protein strands. Factor XIII then stabilizes the fibrin clot. Fibrinolysis The stabilized clot is acted upon by contractile proteins within the platelets. As these proteins contract, they pull on the fibrin threads, bringing the edges of the clot more tightly together, somewhat as we do when tightening loose shoelaces (see Figure 18.14a). This process also wrings out of the clot a small amount of fluid called serum, which is blood plasma without its clotting factors. To restore normal blood flow as the vessel heals, the clot must eventually be removed. Fibrinolysis is the gradual degradation of the clot. Again, there is a fairly complicated series of reactions that involves factor XII and protein-catabolizing enzymes. During this process, the inactive protein plasminogen is converted into the active plasmin, which gradually breaks down the fibrin of the clot. Additionally, bradykinin, a vasodilator, is released, reversing the effects of the serotonin and prostaglandins from the platelets. This allows the smooth muscle in the walls of the vessels to relax and helps to restore the circulation. Plasma Anticoagulants An anticoagulant is any substance that opposes coagulation. Several circulating plasma anticoagulants play a role in limiting the coagulation process to the region of injury and restoring a normal, clot-free condition of blood. For instance, a cluster of proteins collectively referred to as the protein C system inactivates clotting factors involved in the intrinsic pathway. TFPI (tissue factor pathway inhibitor) inhibits the conversion of the inactive factor VII to the active form in the extrinsic pathway. Antithrombin inactivates factor X and opposes the conversion of prothrombin (factor II) to thrombin in the common pathway. And as noted earlier, basophils release heparin, a short-acting anticoagulant that also opposes prothrombin. Heparin is also found on the surfaces of cells lining the blood vessels. A pharmaceutical form of heparin is often administered therapeutically, for example, in surgical patients at risk for blood clots. Interactive Link View these animations to explore the intrinsic, extrinsic, and common pathways that are involved the process of coagulation. The coagulation cascade restores hemostasis by activating coagulation factors in the presence of an injury. Disorders of Clotting Either an insufficient or an excessive production of platelets can lead to severe disease or death. As discussed earlier, an insufficient number of platelets, called thrombocytopenia, typically results in the inability of blood to form clots. This can lead to excessive bleeding, even from minor wounds. Another reason for failure of the blood to clot is the inadequate production of functional amounts of one or more clotting factors. This is the case in the genetic disorder hemophilia, which is actually a group of related disorders, the most common of which is hemophilia A, accounting for approximately 80 percent of cases. This disorder results in the inability to synthesize sufficient quantities of factor VIII. Hemophilia B is the second most common form, accounting for approximately 20 percent of cases. In this case, there is a deficiency of factor IX. Both of these defects are linked to the X chromosome and are typically passed from a healthy (carrier) female to their male offspring, since males are XY. Females would need to inherit a defective gene from each parent to manifest the disease, since they are XX. Patients with hemophilia bleed from even minor internal and external wounds, and leak blood into joint spaces after exercise and into urine and stool. Hemophilia C is a rare condition that is triggered by an autosomal (not sex) chromosome that renders factor XI nonfunctional. It is not a true recessive condition, since even individuals with a single copy of the mutant gene show a tendency to bleed. Regular infusions of clotting factors isolated from healthy donors can help prevent bleeding in hemophiliac patients. At some point, genetic therapy will become a viable option. In contrast to the disorders characterized by coagulation failure is thrombocytosis, also mentioned earlier, a condition characterized by excessive numbers of platelets that increases the risk for excessive clot formation, a condition known as thrombosis. A thrombus (plural = thrombi) is an aggregation of platelets, erythrocytes, and even WBCs typically trapped within a mass of fibrin strands. While the formation of a clot is normal following the hemostatic mechanism just described, thrombi can form within an intact or only slightly damaged blood vessel. In a large vessel, a thrombus will adhere to the vessel wall and decrease the flow of blood, and is referred to as a mural thrombus. In a small vessel, it may actually totally block the flow of blood and is termed an occlusive thrombus. Thrombi are most commonly caused by vessel damage to the endothelial lining, which activates the clotting mechanism. These may include venous stasis, when blood in the veins, particularly in the legs, remains stationary for long periods. This is one of the dangers of long airplane flights in crowded conditions and may lead to deep vein thrombosis. Thrombophilia, also called hypercoagulation, is a condition in which there is a tendency to form thrombosis. This may be familial (genetic) or acquired. Acquired forms include the autoimmune disease lupus, immune reactions to heparin, polycythemia vera, thrombocytosis, sickle cell disease, pregnancy, and even obesity. A thrombus can seriously impede blood flow to or from a region and will cause a local increase in blood pressure. If flow is to be maintained, the heart will need to generate a greater pressure to overcome the resistance. When a portion of a thrombus breaks free from the vessel wall and enters the circulation, it is referred to as an embolus. An embolus that is carried through the bloodstream can be large enough to block a vessel critical to a major organ. When it becomes trapped, an embolus is called an embolism. In the heart, brain, or lungs, an embolism may accordingly cause a heart attack, a stroke, or a pulmonary embolism. These are medical emergencies. Among the many known biochemical activities of aspirin is its role as an anticoagulant. Aspirin (acetylsalicylic acid) is very effective at inhibiting the aggregation of platelets. It is routinely administered during a heart attack or stroke to reduce the adverse effects. Physicians sometimes recommend that patients at risk for cardiovascular disease take a low dose of aspirin on a daily basis as a preventive measure. However, aspirin can also lead to serious side effects, including increasing the risk of ulcers. A patient is well advised to consult a physician before beginning any aspirin regimen. A class of drugs collectively known as thrombolytic agents can help speed up the degradation of an abnormal clot. If a thrombolytic agent is administered to a patient within 3 hours following a thrombotic stroke, the patient’s prognosis improves significantly. However, some strokes are not caused by thrombi, but by hemorrhage. Thus, the cause must be determined before treatment begins. Tissue plasminogen activator is an enzyme that catalyzes the conversion of plasminogen to plasmin, the primary enzyme that breaks down clots. It is released naturally by endothelial cells but is also used in clinical medicine. New research is progressing using compounds isolated from the venom of some species of snakes, particularly vipers and cobras, which may eventually have therapeutic value as thrombolytic agents. 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15408	https://www.prestwickhouse.com/blog/post/2024/01/get-to-the-root-of-the-reading-problem-growing-your-vocabulary?srsltid=AfmBOopSSX4Hv4vmLxKlLa1A9kJFUqd2ObuFRO5fE93OFeoYygzb1X7s	Get to the Root of the Reading Problem \| Prestwick House 9/29/2025 Search 0 Loading... Menu Prestwick House Blog Free Library Teaching Guides Grammar & Writing Vocabulary Reading Literature Vocabulary \| Reading Get to the Root of the Reading Problem: Growing Your Vocabulary by Prestwick House In the quest to improve reading proficiency among American students, schools and districts across the country are adopting new teaching strategies based on the science of reading. Some states, like New York, Georgia, Delaware, and Virginia, have even passed legislation that ensures research-backed methods of reading instruction are part of public school curricula. Which brings us to today’s topic: using direct vocabulary instruction to improve reading skills. Vocabulary has always been an essential part of English language arts, but with the science of reading gaining traction, it’s become even more valuable. And it makes sense, too. When a student knows a lot of words and has the ability to decode the meanings of unfamiliar words, they’re more likely to be better readers. But the type of vocabulary instruction matters, especially for students at the elementary and middle-school levels who are old enough to have a grasp on language but not quite ready to tackle harder texts. One of the most effective types of vocabulary instruction is based on etymology—the study of word origins and meanings—and specifically, the study of Latin and Greek roots. With more than 60% of all words in the English language derived from these roots, it’s essential for students to understand how they work. Plus, if students are introduced to word roots when they’re young, they'll have a huge advantage when it comes to reading complex texts later in life. That’s at the core of Growing Your Vocabulary, a complete program that’s helped nearly 700,000 students cultivate their vocabulary skills. Growing Your Vocabulary is intended for students in fourth through sixth grade, though it can be adapted for older students who may need remedial instruction. This series presents dozens of new words alongside related Latin and Greek roots, helping students understand the conventions of word formation. Practice exercises and humorous illustrations throughout encourage students to think critically while having fun! There are three books in the series. The twenty chapters in each book are organized by themes. For instance, the first chapter of Level 5 is all about plants and focuses on the roots flor, dendr, foli, and foil. Each chapter covers up to five roots and up to twelve grade-appropriate vocabulary words. All chapters begin with an introduction to the roots and words students will learn, and when applicable, any affixes they’ll encounter. Descriptive text explains the root’s historical origins and meanings. Written in a conversational tone, the accompanying paragraphs define each vocabulary word and give real-world examples of usage. Cartoon illustrations in each chapter add a visual element, offering students another way to commit new vocabulary words and roots to memory. After the introduction, there are ten exercises designed to ensure students retain the words and roots they learn. The types of exercises may vary among chapters and grade levels. Basic matching drills help students review definitions, while fill-in-the-blank activities reinforce associations between words and roots. Growing Your Vocabulary gives plenty of opportunities to improve writing skills, too. Students are tasked with writing open-ended responses to different styles of prompts, all of which require them to use newly acquired vocabulary words in context. This is their chance to demonstrate word mastery. Each chapter wraps up with lower-stakes brain teaser activities, including word searches, riddles, and crossword puzzles. There is a cumulative review after every five chapters. The reviews consist of multiple-choice questions that ask students to recall word and root definitions, identify related antonyms, and use vocabulary words correctly in context. The Teacher’s Editions lay out a five-day lesson plan for each chapter and give additional advice on helping English language learners and students currently working below the 5th-grade level. Exercise answers are also included. To purchase a Teacher’s Edition, please contact Customer Service directly. Take a Closer Look Interested in previewing Growing Your Vocabulary for yourself or sharing information about it with a colleague? Free sample pages, including the ones shown in this post, are available for download. Simply fill out the form to access pages for all three levels of the program! 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15409	https://www.stata.com/links/stata-basics/simple-linear-regression/	Home / Resources & Support / Introduction to Stata basics / Simple linear regression Simple linear regression Linear regression is a popular tool used to quantify the relationship between a continuous outcome variable and one or more predictor variables. The outcome variable is often called the “dependent” variable, and the predictor variables are often called “independent” variables. The predictor variables can be binary, categorical, or continuous. Here we will learn how to use Stata's regress command to fit simple linear regression models, and we will explore more sophisticated features later. Let's begin by opening the nhanes2l dataset. ``` . webuse nhanes2l (Second National Health and Nutrition Examination Survey) ``` Simple linear regression is often used to explore the linear relationship between two continuous variables. It can be useful to create a scatterplot to examine the relationship between the variables before fitting a linear regression model. Let's create a scatterplot for body mass index (bmi) and age. ``` . twoway scatter bmi age ``` We will use linear regression to identify a straight line that best describes the cloud of points in the scatterplot. “Best describes” is defined to be the line that has the smallest sum of squared vertical differences between each observation and the line. The line is defined by the equation below, and we will use our data to estimate the intercept and slope. bmi = intercept + slopeage Let's estimate the intercept and slope by typing regress followed by the outcome variable bmi followed by the predictor variable age. ``` . regress bmi age ``` \| \| \| \| \| --- --- \| \| Source \| \| SS df MS \| Number of obs = 10,351 \| \| \| \| \| F(1, 10349) = 312.45 \| \| Model \| \| 7327.24496 1 7327.24496 \| Prob > F = 0.0000 \| \| Residual \| \| 242696.917 10,349 23.4512433 \| R-squared = 0.0293 \| \| \| \| \| Adj R-squared = 0.0292 \| \| Total \| \| 250024.162 10,350 24.1569239 \| Root MSE = 4.8426 \| \| \| \| \| --- \| \| \| \| \| bmi \| \| Coefficient Std. err. t P>\|t\| [95% conf. interval] \| \| \| \| \| \| age \| \| .0488762 .0027651 17.68 0.000 .0434561 .0542963 \| \| _cons \| \| 23.21209 .1399079 165.91 0.000 22.93784 23.48633 \| \| \| \| \| The intercept is the coefficient labeled _cons, and the slope is the coefficient labeled age. Let's use generate to create a new variable named bmi_predicted, which defines the regression line using the intercept and slope from our output. ``` . generate double bmi_predicted = 23.21209 + 0.0488762 age ``` Next let's use twoway line to plot our regression line with our scatterplot. ``` . twoway (scatter bmi age) (line bmi_predicted age) ``` Our regression line has a slight upward slope, which also suggests that larger values of age tend to be associated with larger values of bmi. The slope tells us that each additional year of age is associated with 0.049 more bmi. In the regression table, the p-value for the estimated age coefficient is labeled P>\|t\| and tests the null hypothesis that the coefficient equals 0 (no relationship). The p-value equals 0.000, which suggests that our result is inconsistent with the null hypothesis that there is no relationship between age and bmi. That is the basic idea of linear regression. But there are many other numbers in the regression output table, and I've also used the term “null hypothesis” without defining it. Let's dig a little deeper to learn what it all means. Digging deeper First, let's define our null and alternative hypotheses. Our null hypothesis is that there is no relationship between age and bmi and our regression line is flat. In other words, our best prediction of bmi is simply the mean of bmi for any value of age. We use egen to create variable bmi_mean containing the mean of bmi, and then we use twoway line to draw a flat line for the mean of bmi. ``` . egen double bmi_mean = mean(bmi) . twoway (scatter bmi age) (line bmi_predicted age) (line bmi_mean age) ``` Our alternative hypothesis is that there is a relationship between age and bmi and our regression line is not flat, which is what the red line in our graph shows. We might ask ourselves, “is the red line really different from the green line?” or “is the slope of our regression line large enough to describe a meaningful relationship between age and bmi, or is our regression line basically flat?” We used a p-value to answer that question above, but where did the p-value come from? We can answer this question using the output from our regress command. The output contains a lot of numbers bundled into three groups. The group at the top left is often called the ANOVA table, where ANOVA stands for “ANalysis Of VAriance”. The group at the top right displays information about the entire model. And the group at the bottom displays information about the relationship of each predictor variable with the outcome variable. The ANOVA table The ANOVA table includes columns labeled SS, df, and MS, which stand for “sum of squares”, “degrees of freedom”, and “mean squares”, respectively. The first column, labeled Source, includes rows labeled Model, Residual, and Total. Let's calculate the numbers in the table manually to develop our intuition. The SS in the Total row equals 250,024.162, and it is called the “total sum of squares”. It is calculated by adding up the squared vertical distances between each blue dot and the green line in our graph. It quantifies how far the observed values of bmi deviate from the predicted bmi assuming the null hypothesis is true. ``` . generate double ss_total = (bmi - bmi_mean)^2 . table, statistic(total ss_total) nformat(%16.3f total) ``` \| \| \| \| --- \| \| \| \| \| Total \| \| 250024.162 \| \| \| \| \| Our calculation is off slightly because of a rounding error. Stata's regress command uses much more accurate algorithms than our simple calculation. But the idea is the same. Next let's calculate the value of SS in the Residual row, which equals 242,696.917. It is calculated by adding up the squared vertical distances between each blue dot and the red line in our graph. It quantifies how far the observed values of bmi deviate from the predicted values of bmi assuming the alternative hypothesis is true. ``` . generate double ss_residual = (bmi - bmi_predicted)^2 . table, statistic(total ss_residual) nformat(%16.3f total) ``` \| \| \| \| --- \| \| \| \| \| Total \| \| 242696.917 \| \| \| \| \| Finally, let's calculate the value of SS in the Model row, which equals 7,327.24. It is calculated by adding up the squared vertical distances between the red line and the green line for each blue dot in our graph. It quantifies how far the predicted values of bmi deviate from each other assuming the alternative hypothesis is true and assuming the null hypothesis is true, respectively. ``` . generate double ss_model = (bmi_predicted - bmi_mean)^2 . table, statistic(total ss_model) nformat(%16.3f total) ``` \| \| \| \| --- \| \| \| \| \| Total \| \| 7327.244 \| \| \| \| \| Note that the sum of ss_model and ss_residual always equals ss_total. ``` . table, statistic(total ss_model ss_residual ss_total) nformat(%16.3f total) ``` \| \| \| \| --- \| \| \| \| \| ss_model \| \| 7327.244 \| \| ss_residual \| \| 242696.917 \| \| ss_total \| \| 250024.162 \| \| \| \| \| The df column in the ANOVA table displays the degrees of freedom for each source. The model df equals the number of estimated parameters, k, minus 1. We estimated an intercept and a slope for age, so k - 1 = 2 - 1 = 1. The residual df equals the sample size, n, minus the number of estimated parameters, k. Our sample size is 10,351, and we estimated two parameters, so n - k = 10,351 - 2 = 10,349. The total df equals the sample size, n, minus 1. In our example, n - 1 = 10,351 - 1 = 10,350. The MS column displays the mean sum of squares for each source. MS equals SS divided by df. We can verify this using display. ``` . display "Model MS = " 7327.24496 / 1 Model MS = 7327.245 . display "Residual MS = " 242696.917 / 10349 Residual MS = 23.451243 . display "Total MS = " 250024.162 / 10350 Total MS = 24.156924 ``` There is some rounding error when we do the calculations by hand, but the numbers are basically the same as the output from regress. The model statistics The group of numbers at the top right of the regression output displays information about the model. The Number of obs tells us that Stata used 10,351 observations to fit the model. The next row, labeled F(1, 10349), is the F statistic. The F statistic is the ratio of the model MS and the residual MS, which, in our example, is 7327.24 / 23.45 = 312.45. The numbers in parentheses are the numerator and denominator degrees of freedom, respectively, also shown in the ANOVA table. The value of the F statistic does not have a direct interpretation, but we can use it to calculate the p-value labeled Prob > F. The F statistic has a central F distribution, and we can calculate the p-value manually using Stata's Ftail() function. ``` . display "Prob > F = " Ftail(1, 10349, 312.45) Prob > F = 6.547e-69 ``` The result, 6.547e-69, is scientific notation for a very small number that is essentially 0. The p-value is used to test the null hypothesis that all slope coefficients (but not the intercept) in the model are simultaneously equal to 0. Our small p-value suggests that our result is inconsistent with the null hypothesis that the age coefficient is 0. Another interpretation is that our model fits the data better than a model that includes only the intercept. We often hear that the residuals in a regression model must be normally distributed. Statistical theory tells us that the square of a normally distributed variable with a mean of 0 has a central χ2 distribution. Statistical theory also tells us that the ratio of two independent χ2 divided by their respective degrees of freedom has an F distribution. We can square normally distributed residuals to get sums of squares, divide the sums of squares by their degrees of freedom to get mean squares, and take the ratio of the mean squares to get an F statistic. The “normal residuals” assumption allows us to use an F distribution to calculate the p-value for our F statistic. The next number in the table is labeled R-squared. The R2 is the ratio of the model SS and the total SS. We can use display to calculate the R2 manually. ``` . display "R-squared = " 7327.24496 / 250024.162 R-squared = .02930615 ``` The R2 is often interpreted as the proportion of variability in the outcome variable that is explained by the model. Using this interpretation, our model explains 2.9% of the variability of bmi. The next number, labeled Adj R-squared, is similar to the R2 but is penalized, or reduced, to account for variables in the model that may be unnecessary. The next number, labeled Root MSE, is the square root of the residual MS or “Mean Squared Error”. Let's verify this manually. ``` . display "Root MSE = " sqrt(23.4512433) Root MSE = 4.8426484 ``` The root MSE tells us that the observed values of bmi deviate from our regression line by an average of 4.84 points. The regression statistics The group of numbers at the bottom of the regression output displays information about the relationship between each predictor variable and the outcome variable. The row labeled _cons tells us about the intercept, but we are usually focused on the slope coefficients. The column labeled Coefficient displays the estimated slope for each variable in the model. The slope coefficient for age equals 0.0488762 and, as noted above, we interpret this to mean that each additional year of age is associated with 0.049 more bmi. Stata also estimates the standard error for the coefficient, and it is displayed in the column labeled Std. err. Different methods can be used to estimate standard errors, and we will discuss them later. By default, Stata uses ordinary least squares to estimate standard errors. The column labeled t is the t statistic, which is used to test the null hypothesis that the coefficient equals 0. The t statistic is the difference between the estimated coefficient and the value of the coefficient assuming the null hypothesis (0) divided by the standard error. We usually just divide the estimated coefficient by the standard error because the null value is always 0. ``` . display "t = " (0.0488762 - 0) / .0027651 t = 17.676106 . display "t = " 0.0488762 / .0027651 t = 17.676106 ``` The next column, labeled P>\|t\|, is the p-value for the t statistic. We can use Stata's ttail() function to calculate the p-value. The first argument in the ttail() function is the degrees of freedom, which is n - 1 = 10,351 - 1 = 10,350. The second argument is the absolute value of the t statistic. We can use the abs() function to calculate absolute value. We must multiply the result by two because we are calculating a two-sided p-value. ``` . display "P>\|t\| = " 2ttail(10350, abs(17.68)) P>\|t\| = 6.136e-69 ``` The result is, again, scientific notation for a very small number that is effectively 0. This suggests that we should reject the null hypothesis that the coefficient equals 0. The last two columns, jointly labeled [95% conf. interval], are the lower and upper bounds of the 95% confidence interval. Let's calculate them manually to develop our intuition. We will need the “critical value” for a t distribution that corresponds to a two-sided hypothesis test with an alpha level of 0.05. We can use Stata's invt() function to calculate the critical value. The first argument is the degrees of freedom, which is n - 1 = 10,351 - 1 = 10,350. The second argument is 1 - alpha/2 = 1 - 0.05/2 = 0.975. We multiply this number by the standard error and subtract it from the coefficient to estimate the lower bound. ``` . display "Lower 95% CI bound = " .0488762 - invt(10350, 0.975) .0027651 Lower 95% CI bound = .04345607 ``` We multiply the critical value by the standard error and add it to the coefficient to estimate the upper bound. ``` . display "Upper 95% CI bound = " .0488762 + invt(10350, 0.975) .0027651 Upper 95% CI bound = .05429633 ``` It is tempting to interpret the 95% confidence interval using the word “confidence” but that is not correct. The interval simply tells us that intervals estimated from many samples would include the true population parameter 95% of the time. Let's interpret the results of our regression model now that we understand the numbers in our output and how to calculate them. Interpreting the model Our model examined the relationship between the outcome variable bmi and the predictor variable age. The model included 10,351 observations, explained roughly 3% of the variability of bmi, and fit the data better than a model that included only the intercept (F(1, 10349) = 312.45, p < 0.000). The results suggest that each additional year of age is associated with 0.049 more bmi (t = 17.68, df = 10,350, p < 0.000). You can watch a demonstration of these commands by clicking on the link to the YouTube video below. You can read more about regress and the many other options available with regress by clicking on the links to the Stata manual entries below. See it in action Watch Simple linear regression in Stata. Tell me more Read more in the Stata Base Reference Manual; see [R] regress. And in the Stata Graphics Reference Manual, see [G-2] graph twoway scatter. 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15410	https://dmtcs.episciences.org/2411	2411 - Descents of $\lambda$-unimodal cyclic permutations This site uses cookies and collects personal data Learn more Close Toggle navigation Create account Sign in Sign in Forgotten password ? Forgotten username ? Discrete Mathematics & Theoretical Computer Science Home Volumes Articles Browse by author Browse by section Search an article Browse latest articles Accepted articles to be published About the journal Contact Editorial Staff members Editorial policies Indexing Open Access & Archiving Submissions Other Help Support Documentation My Account Sign in Create account Forgotten username ? Forgotten password ? Permissions Kassie Archer - Descents of λ λ-unimodal cyclic permutations dmtcs:2411 - Discrete Mathematics & Theoretical Computer Science, January 1, 2014, DMTCS Proceedings vol. AT, 26th International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2014) - Descents of λ λ-unimodal cyclic permutationsConference paper Authors: Kassie Archer 1 NULL Kassie Archer 1 Department of Mathematics [Dartmouth] [en] We prove an identity conjectured by Adin and Roichman involving the descent set of λ λ-unimodal cyclic permutations. These permutations appear in the character formulas for certain representations of the symmetric group and these formulas are usually proven algebraically. Here, we give a combinatorial proof for one such formula and discuss the consequences for the distribution of the descent set on cyclic permutations. [fr] Nous prouvons une identité conjecturée par Adin et Roichman impliquant les ensembles des descentes des permutations cycliques λ λ-unimodales. Ces permutations apparaissent dans les formules des caractères pour certaines représentations du groupe symétrique, et ces formules sont généralement prouvées dans une manière algébrique. Ici, nous donnons une preuve combinatoire pour une telle formule et discutons les conséquences pour la distribution de l’ensemble des descentes sur des permutations cycliques. Download articleSee the document's page on HAL Source: HAL:hal-01207601v1 Volume: DMTCS Proceedings vol. AT, 26th International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2014) Section: Proceedings Published on: January 1, 2014 Imported on: November 21, 2016 Keywords: [INFO.INFO-DM]Computer Science [cs]/Discrete Mathematics [cs.DM], [MATH.MATH-CO]Mathematics [math]/Combinatorics [math.CO], [en]cyclic permutation, descent, necklaces, characters of representations of the symmetric group. Bibliographic References Share and export Export BibTeX TEI Dublin Core OpenAIRE Crossref DOAJ ZbMATH Open JSON JSON-V2 Consultation statistics This page has been seen 423 times. This article's PDF has been downloaded 563 times. × Close Submit About RSS E-mail Contact eISSN 1365-8050 Support Documentation Technical support Legal mentions Privacy Service status About Episciences Episciences Hosted journals Acknowledgements Episciences v1.0.52-633666e6c Terms Of Use
15411	https://www.quora.com/What-is-the-difference-among-molarity-mole-fraction-weight-percent-volume-percent-molality-parts-per-million-normality-strength-basicity-and-acidity	Something went wrong. Wait a moment and try again. Volume Percent Solutions & Mixtures Strength of Acids Mole Fraction Parts Per Million Acidity and Basicity Normality (Chemistry) 5 What is the difference among molarity, mole fraction, weight percent, volume percent, molality, parts per million, normality, strength, basicity, and acidity? · Sep 6 Molarity, mole fraction, weight percent, volume percent, molality, parts per million, normality, strength, basicity, and acidity are different ways to express concentration or reactive capacity of chemical solutions. Below are precise definitions, typical units, how they’re calculated, when to use each, and short notes on limitations. Molarity (M) Definition: moles of solute per liter of solution. Formula: M = n(solute) / V(solution, L). Units: mol·L⁻¹. Use when: reactions in solution, stoichiometry with volumes at a given temperature. Limitations: depends on temperature (volume chang Molarity, mole fraction, weight percent, volume percent, molality, parts per million, normality, strength, basicity, and acidity are different ways to express concentration or reactive capacity of chemical solutions. Below are precise definitions, typical units, how they’re calculated, when to use each, and short notes on limitations. Molarity (M) Definition: moles of solute per liter of solution. Formula: M = n(solute) / V(solution, L). Units: mol·L⁻¹. Use when: reactions in solution, stoichiometry with volumes at a given temperature. Limitations: depends on temperature (volume changes with T). Molality (m) Definition: moles of solute per kilogram of solvent. Formula: m = n(solute) / mass(solvent, kg). Units: mol·kg⁻¹. Use when: temperature-independent concentration, colligative properties (boiling point elevation, freezing point depression). Advantages: not affected by temperature because based on mass. Mole fraction (χ) Definition: moles of component divided by total moles of all components. Formula: χi = n_i / Σ n_j. Units: dimensionless. Use when: thermodynamic calculations (Raoult’s law, partial pressures, activity coefficients). Properties: sum of all mole fractions = 1. Weight percent (mass percent, wt%) Definition: mass of solute divided by total mass of solution, multiplied by 100. Formula: wt% = (mass solute / mass solution) × 100%. Units: % (w/w). Use when: formulations specified by mass (paints, alloys, some lab protocols). Note: independent of volume, temperature. Volume percent (vol%) Definition: volume of solute divided by total volume of solution, multiplied by 100. Formula: vol% = (volume solute / volume solution) × 100%. Units: % (v/v). Use when: mixing miscible liquids (ethanol in water), where volumes are additive or specified by volume. Limitation: volumes are not strictly additive for all mixtures; temperature dependent. Parts per million (ppm) and parts per billion (ppb) Definition: very dilute concentration measures. ppm = mass (or volume) of solute per 10^6 parts of solution; ppb per 10^9. Approximate formulas: ppm ≈ (mass solute / mass solution) × 10^6 (for dilute aqueous solutions 1 mg/L ≈ 1 ppm). ppb ≈ (mass solute / mass solution) × 10^9. Units: ppm or ppb (dimensionless ratio expressed per million/billion). Use when: trace contaminants in water, air, soils. Normality (N) Definition: equivalents of reactive species per liter of solution. Formula: N = equivalents / V(solution, L). Equivalents = moles × (n factor). Units: eq·L⁻¹. Use when: acid–base titrations, redox reactions, where reaction depends on transferable protons or electrons. Caveat: depends on the definition of equivalent (acid–base: number of H+ or OH− transferable; redox: electrons transferred). Normality changes with reaction considered; less favored in modern practice. Strength (in volumetric chemical context) Definition: often used in industrial or pharmaceutical contexts to mean mass or concentration per volume (usually g per 100 mL, mg per mL, or % w/v). Example: “strength of a solution = 5% w/v” means 5 g solute per 100 mL solution. Use when: prescriptions, reagent labels. Note: “strength” is ambiguous—check units on the label. Basicity (in acid–base / titration context) Two senses: a) For acids: number of ionizable protons per molecule (e.g., H2SO4 has acidity = 2 in protic count). b) For bases: number of OH− groups per base formula unit that can react (e.g., Ca(OH)2 has basicity = 2). Role: used to convert between molarity and normality: Normality = Molarity × (basicity or acidity, i.e., equivalents per mole). Use when: calculating titration equivalents. Acidity (in this list’s context) Two senses: a) As proticity: number of acidic protons per molecule (same concept as “acidity” used for acids to get equivalents). b) As pH-related property: tendency to donate protons (quantified by pH, Ka). Role in concentration units: used to compute equivalents for normality. Quick conversion examples Molarity to normality (acid with n H+ transferable): N = M × n. Mass percent to molality: convert mass percent to masses per 100 g solution, compute moles solute and kg solvent, then molality. For dilute aqueous solutions: 1 ppm (mass/mass) ≈ 1 mg·L⁻¹ if density ≈ 1 g·mL⁻¹. Practical selection guide Use molarity for everyday lab stoichiometry where volumes are measured. Use molality for colligative properties and when temperature varies. Use mole fraction for vapor–liquid equilibrium and thermodynamics. Use weight percent/volume percent or strength for formulations and labeling. Use ppm/ppb for trace analysis. Use normality/basicity/acidity when dealing with titrations and equivalents (or replace normality with molarity plus stoichiometric factors for clarity). Pitfalls to remember Molarity and volume percent depend on temperature (because volume changes). Normality depends on the chemical reaction considered; it is not an intrinsic property of the solution alone. Volume percent assumes additive volumes; some mixtures are non‑ideal. “Strength,” “basicity,” and “acidity” are context-dependent terms—always confirm their intended definition in the specific application. Related questions If a bottle of wine contains 13% alcohol, what will be the alcohol percentage for a glass of wine? Why is molarity better than molality? What are the main differences between using mole fraction per molality and molarity to describe the concentration of a solution? What is the difference between molality, molarity, normality, and valence? What is molality, molarity and normality? Koti Thimappiah Author has 60 answers and 60K answer views · 6y Molarity refers to the number of moles of solute present in one litre of the solution. Normality refers to the number of gram equivalents present in one litre of solution. Molality refers to number of moles of solute present in one kg of solvent. Weight percent is ratio of weight of solute to the weight of solution multiplied by 100 Volume percent is the ratio of weight of solute to the volume of solution multiplied by 100 Mole fraction is the ratio of number of moles of solute or solvent to the total moles of solute and solvent present in solution Parts per million is the weight of solute divided b Molarity refers to the number of moles of solute present in one litre of the solution. Normality refers to the number of gram equivalents present in one litre of solution. Molality refers to number of moles of solute present in one kg of solvent. Weight percent is ratio of weight of solute to the weight of solution multiplied by 100 Volume percent is the ratio of weight of solute to the volume of solution multiplied by 100 Mole fraction is the ratio of number of moles of solute or solvent to the total moles of solute and solvent present in solution Parts per million is the weight of solute divided by the weight of solution multiplied by 1000000 Parts per billion is the weight of solute divided by the weight of solution multiplied by 1000000000 Basicity refers to the number of replaceable hydrogen atoms present in one mole of an acid. Acidity refers to the number of replaceable hydroxyl radicals present in one mole of an alkali or base. Sponsored by MEDvi Prescription Weight Loss Tirzepatide can drop weight 2x faster than Ozempic For half the cost of popular weight loss drugs on the market, Americans use this site for GLP-1 Meds. Sri Sandhya 7y Related How do you relate mole fraction and molality? The mole fraction is the ratio of moles of solute to moles of solvent. Molality is the number of moles of solute per mass of solvent. The mole fraction is the ratio of moles of solute to moles of solvent. Molality is the number of moles of solute per mass of solvent. Harsh Jain M.Sc from Dr. Bhim Rao Ambedkar University, Agra (Graduated 1991) · Author has 86 answers and 130.3K answer views · 6y Related How do I convert mole fraction into molarity and molality? Mole fraction is the ratio of moles of solute to total moles. So it can be assumed that if mole fraction of solute is X, than X mole of solute are dissolved in (1-X) moles of solvent. Hence Mole fraction is the ratio of moles of solute to total moles. So it can be assumed that if mole fraction of solute is X, than X mole of solute are dissolved in (1-X) moles of solvent. Hence Related questions Can you explain the difference between molarity, molality, and mole fraction? Can you explain the differences between molarity, normality and mole fraction? What is the difference between molality, mole fraction, and percent composition in a solution? What is the relationship between molarity, molality, and mole fraction in an ideal solution? What is the advantage of molarity over molality? Alexander Mathey Former Chemical Engineer, retired, lives in Athens, GR · Author has 5.6K answers and 10.9M answer views · 8y Related What is the relation between molarity, molality and mole fraction? What is the formula(s)? Molarity (M), is the number of moles of solute (soluted substance) per unit volume of solution. M(A) = N(A) / V(solution) Example: Molecular weight of salt (NaCl) is 23 + 35,5 = 58,5, therefore 1 Mole of NaCl = 58,5 g. Put that amount of salt in a Volumetric Cup of 1 Liter, add some water, stir until it is totally solved, then add water until the surface of the solution reaches the Volumetric Mark. You have a Sodium Chloride solution with a Molarity of 1. Molarity is very practical for doing chemical reactions in solutions by just measuring volumes. Molality (m), is the number of moles of solute per Molarity (M), is the number of moles of solute (soluted substance) per unit volume of solution. M(A) = N(A) / V(solution) Example: Molecular weight of salt (NaCl) is 23 + 35,5 = 58,5, therefore 1 Mole of NaCl = 58,5 g. Put that amount of salt in a Volumetric Cup of 1 Liter, add some water, stir until it is totally solved, then add water until the surface of the solution reaches the Volumetric Mark. You have a Sodium Chloride solution with a Molarity of 1. Molarity is very practical for doing chemical reactions in solutions by just measuring volumes. Molality (m), is the number of moles of solute per unit mass of solvent. m(A) = N(A) / Mass of solvent Example: A solution of 58,5 g of salt (1 Mole of NaCl) in 1 kg of water has a Molality of 1. Mole fraction (X) If a solution consists of n substances A(1), A(2), . . A(n), and N(1), N(2), . . N(n) are the number of moles of each substance present in the solution, then Mole Fraction (or Molar Fraction) of a substance is the ratio of the number of moles of that substance to the total number of moles present. It is evident that the sum of all the molar fractions of a solution is equal to 1. Hope this clears the confusion. Amit Kundu Former Talent Acquisition Specalist at Tata Consultancy Services (company) (2020–2022) · Author has 60 answers and 308.3K answer views · 7y Related How do you relate mole fraction and molality? Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. 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Rahul Dev B.Tech in Electronics and Communication Engineering, BNM Institute of Technology · Author has 137 answers and 982.9K answer views · 7y Related How can you describe the difference between molality and molarity? By definition, MOLARITY is the number of moles of solute dissolved per liter of solution. We use capital letter “M” to represent molarity and its formula is M= (# mol SOLUTE)/ (Liters of SOLUTION). MOLALITY is then the number of moles of solute per kilogram of the SOLVENT, NOT solution! We use lower case letter “m” to represent molality and its formula can be represented as: m= (# mole SOLUTE) / (Kilograms of SOLVENT). MOLALITY is preferred when the temperature of the solution varies. That is because MOLALITY does not depend on temperature, (Neither number of moles of solute nor mass of solvent By definition, MOLARITY is the number of moles of solute dissolved per liter of solution. We use capital letter “M” to represent molarity and its formula is M= (# mol SOLUTE)/ (Liters of SOLUTION). MOLALITY is then the number of moles of solute per kilogram of the SOLVENT, NOT solution! We use lower case letter “m” to represent molality and its formula can be represented as: m= (# mole SOLUTE) / (Kilograms of SOLVENT). MOLALITY is preferred when the temperature of the solution varies. That is because MOLALITY does not depend on temperature, (Neither number of moles of solute nor mass of solvent will be affected by changes of temperature.) while MOLARITY changes as temperature changes. (Volume of solution in the formula changes as temperature changes this is the basic difference b/w molality and molarity. Ayush Addhyayan Former Competitive Coder \| Budding Engineer \| Explorer · Author has 140 answers and 682.3K answer views · 7y Related What is the difference between molarity molality and normality? Molarity, Molality and Normality are the different terms that are used for representation of concentration of any solution there is slight difference between them. Let us define each term separately :- Molarity = It is defined as moles of solute / Volume of solution in litre. Ex:- 5M soln of NaCl (aq) means that 5 moles of NaCl is dissolved in 1 litre of water Molality = It is defined as moles of solute / Mass of solvent in kg. Ex-5 Molal solution of NaCl(aq) means that 5 moles of NaCl is dissolved in 1 kg of water. Normality :- It is defined as equivalents of solute/volume of soln in litre . You Molarity, Molality and Normality are the different terms that are used for representation of concentration of any solution there is slight difference between them. Let us define each term separately :- Molarity = It is defined as moles of solute / Volume of solution in litre. Ex:- 5M soln of NaCl (aq) means that 5 moles of NaCl is dissolved in 1 litre of water Molality = It is defined as moles of solute / Mass of solvent in kg. Ex-5 Molal solution of NaCl(aq) means that 5 moles of NaCl is dissolved in 1 kg of water. Normality :- It is defined as equivalents of solute/volume of soln in litre . You can also denote it by formula molarity n factor. Hope it will help. Promoted by Quest Quest www.questhealth.com · Sep 10 Can I buy STD tests online? Yes, you can buy various STD tests online at Quest without a doctor's visit. Purchase your test online, schedule an appointment at a Quest near you. See your results online as soon as they're available. Get tested on your own terms, when you need it. Sexual Health \| Quest® Get 10% off your first purchase Thank you! × Looking for health insights? Sign up to receive your discount. We're excited to share the latest updates from Quest with you. As a thank you, use your special code at checkout to receive 10% off. 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(more) Dyuti Prasad Mondal B.Sc in Chemistry, Midnapore College · Author has 327 answers and 532.9K answer views · 3y Related What is the relation between molality, molar mass, and mole fraction? Let, a solution is made where - Solvent weight [math]= W_{1} [/math]g Solute weight [math]= W_{2}[/math] g Solvent molecular weight [math]= M_{1} [/math]g/mole Solute molecular weight [math]= M_{2}[/math] g/mole Mole number of the Solvent, [math]n_{1}=\frac{W_{1}}{M_{1}}[/math] Mole number of the Solute, [math]n_{2}=\frac{W_{2}}{M_{2}}[/math] Mole fraction of the Solvent, [math]x_{1}=\frac{n_{1}}{n_{1}+n_{2}}[/math] Mole fraction of the Solute, [math]x_{2}=\frac{n_{2}}{n_{1}+n_{2}}[/math] Now, as per the definition, molality (m) is the number of moles of Solute present in 1000g of the Solvent. [math]W_{1}[/math] g solvent contains, [math]n_{2} [/math]moles of solute. So, 1000g of solvent contains, [math]\frac{1000×n_{2}}{W_{1}}[/math] moles of Let, a solution is made where - Solvent weight [math]= W_{1} [/math]g Solute weight [math]= W_{2}[/math] g Solvent molecular weight [math]= M_{1} [/math]g/mole Solute molecular weight [math]= M_{2}[/math] g/mole Mole number of the Solvent, [math]n_{1}=\frac{W_{1}}{M_{1}}[/math] Mole number of the Solute, [math]n_{2}=\frac{W_{2}}{M_{2}}[/math] Mole fraction of the Solvent, [math]x_{1}=\frac{n_{1}}{n_{1}+n_{2}}[/math] Mole fraction of the Solute, [math]x_{2}=\frac{n_{2}}{n_{1}+n_{2}}[/math] Now, as per the definition, molality (m) is the number of moles of Solute present in 1000g of the Solvent. [math]W_{1}[/math] g solvent contains, [math]n_{2} [/math]moles of solute. So, 1000g of solvent contains, [math]\frac{1000×n_{2}}{W_{1}}[/math] moles of solute. Hence, molality, [math]m=\frac{1000×n_{2}}{W_{1}}[/math] [math]m=\frac{1000×W_{2}}{W_{1}×M_{2}}[/math] [math]m=\frac{1000×\frac{W_{2}}{M_{2}}}{\frac{W_{1}}{M_{1}}×M_{1}}[/math] [math]m=\frac{1000×n_{2}}{n_{1}×M_{1}}[/math] [math]m=\frac{1000×x_{2}}{x_{1}×M_{1}}[/math] This is the relation of molality, molar mass and mole fraction. Surya Pratap Singh Studied Chemistry at Kirori Mal College, Delhi University (Graduated 2016) · 7y Related What is the relation between molarity, molality and mole fraction? What is the formula(s)? There is no direct formula which relates the three concentration terms, but you can formulate one. But I think there is no such need. I, hereby, give you the formulae for the three concentration terms. Molarity: M = (w/M) x (1000/V) where; M = Molarity, w= Weight of solute(g), M = Molecular mass of the solute(g/mol), V = Volume of solution(mL). Molality: m = (w/M) x (1000/A) where; m = Molality of solution and A = Mass of solvent(g). Mole Fraction: x = n/(n+N) and X = N/(n+N) where; x= Mole Fraction of solute, X = Mole Fraction of Solvent, n = number of moles of solute, and, N = number of moles of solve There is no direct formula which relates the three concentration terms, but you can formulate one. But I think there is no such need. I, hereby, give you the formulae for the three concentration terms. Molarity: M = (w/M) x (1000/V) where; M = Molarity, w= Weight of solute(g), M = Molecular mass of the solute(g/mol), V = Volume of solution(mL). Molality: m = (w/M) x (1000/A) where; m = Molality of solution and A = Mass of solvent(g). Mole Fraction: x = n/(n+N) and X = N/(n+N) where; x= Mole Fraction of solute, X = Mole Fraction of Solvent, n = number of moles of solute, and, N = number of moles of solvent. Using these three expressions, you can calculate any of the three concentration terms, i.e., Molarity, Molarity and Mole-fraction provided enough data is given. Hope this helps. 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Rel.b/w. molality(m) and mole fraction (X): m=1000×X of solute /X of solvent × molar mass of solvent Rel. b/w M and X of solute: M= X of solute ×1000×D/X of solvent ×molar mass of solvent +X of solute ×molar mass of solute M ,when density is given:- M=% of solution /molar mass of solute ×10×D I hope these relationship are helpful for u Thanks a lot for this question . Diana Black Studies chemistry · 6y Related Is solubility = molarity = molality? Sadly, no. Solubility in simple terms is the property of substances (called the solvent) to dissolve the solutes. For example, salt is soluble in water. Here water is the solvent because it dissolves the salt (solute). Molarity and molality both are used to express the concentration of the solute. Again, despite being notoriously similar in their spelling, Molarity (aka Molar concentration) is the number of moles of the solute that is dissolved in 1 Litre of the solution, hence its expressed as (number of moles) / volume of solution (in Litres) Its unit is mol/L or M (molar) Molality is the concentra Sadly, no. Solubility in simple terms is the property of substances (called the solvent) to dissolve the solutes. For example, salt is soluble in water. Here water is the solvent because it dissolves the salt (solute). Molarity and molality both are used to express the concentration of the solute. Again, despite being notoriously similar in their spelling, Molarity (aka Molar concentration) is the number of moles of the solute that is dissolved in 1 Litre of the solution, hence its expressed as (number of moles) / volume of solution (in Litres) Its unit is mol/L or M (molar) Molality is the concentration of the solution expressed as the number of moles per kilogram of solvent. (number of moles) / kilogram of solvent Its unit is mol/kg or m [Notice how the denominator in molality is about the solvent, not the solution as in the case of molarity. ] Confused yet? ;) In most cases, we prefer to use molality over molarity because it uses the weight (in the denominator) which doesn’t change with an increase in temperature and pressure, whereas volume (molarity) may change. Hope it helped!! Jason Slaton Author has 220 answers and 562K answer views · 8y Related How do molarity and molality formulas differ? molarity=molality=concentration In chemistry, concentration is the measure of how much of a given substance there is mixed with another substance. This can apply to any sort of chemical mixture, but most frequently is used in relation to solutions, where it refers to the amount of solute dissolved in a solvent. To concentrate a solution, one must add more solute, or reduce the amount of solvent (for instance, by selective evaporation). By contrast, to dilute a solution, one must add more solvent, or reduce the amount of solute. Unless two substances are fully miscible there exists a molarity=molality=concentration In chemistry, concentration is the measure of how much of a given substance there is mixed with another substance. This can apply to any sort of chemical mixture, but most frequently is used in relation to solutions, where it refers to the amount of solute dissolved in a solvent. To concentrate a solution, one must add more solute, or reduce the amount of solvent (for instance, by selective evaporation). By contrast, to dilute a solution, one must add more solvent, or reduce the amount of solute. Unless two substances are fully miscible there exists a concentration at which no further solute will dissolve in a solution. At this point, the solution is said to be saturated. If additional solute is added to a saturated solution, it will not dissolve. Instead, phase separation will occur, leading to either coexisting phases or a suspension. The point of saturation depends on many variables such as ambient temperature and the precise chemical nature of the solvent and solute. Related questions If a bottle of wine contains 13% alcohol, what will be the alcohol percentage for a glass of wine? Why is molarity better than molality? What are the main differences between using mole fraction per molality and molarity to describe the concentration of a solution? What is the difference between molality, molarity, normality, and valence? What is molality, molarity and normality? Can you explain the difference between molarity, molality, and mole fraction? Can you explain the differences between molarity, normality and mole fraction? What is the difference between molality, mole fraction, and percent composition in a solution? What is the relationship between molarity, molality, and mole fraction in an ideal solution? What is the advantage of molarity over molality? How do you convert molarity to percent by weight or percent by volume? Can you give some examples of calculations involving molarity, molality, normality and mole fractions? Why do we use molarity more than molality? Which molality and molarity of the same solution is large values? What is the SI unit of molality, molarity and normality? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15412	https://en.wikipedia.org/wiki/Fredholm_determinant	Published Time: Tue, 01 Jul 2025 05:48:15 GMT Fredholm determinant - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 DefinitionToggle Definition subsection 1.1 Setup 1.2 Definition by exponential trace 1.3 Definition by exterior powers 2 Properties 3 Integral operatorsToggle Integral operators subsection 3.1 Integral equation 4 Commutators 5 Szegő limit formula 6 History 7 References [x] Toggle the table of contents Fredholm determinant [x] 4 languages Deutsch Français Italiano 日本語 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia In mathematics, the Fredholm determinant is a complex-valued function which generalizes the determinant of a finite dimensional linear operator. It is defined for bounded operators on a Hilbert space which differ from the identity operator by a trace-class operator (i.e. an operator whose singular values sum up to a finite number). The function is named after the mathematicianErik Ivar Fredholm. Fredholm determinants have had many applications in mathematical physics, the most celebrated example being Gábor Szegő's limit formula, proved in response to a question raised by Lars Onsager and C. N. Yang on the spontaneous magnetization of the Ising model. Definition [edit] Setup [edit] Let H{\displaystyle H} be a Hilbert space and G{\displaystyle G} the set of bounded invertible operators on H{\displaystyle H} of the form I+T{\displaystyle I+T}, where T{\displaystyle T} is a trace-class operator. G{\displaystyle G} is a group because The set of trace-class operators is an ideal in the algebra of bounded linear operators, so (I+T)(I+T′)−I=T+T′+T T′{\displaystyle (I+T)(I+T')-I=T+T'+TT'} is trace-class. (I+T)−1−I=−T(I+T)−1,{\textstyle (I+T)^{-1}-I=-T(I+T)^{-1},} so (I+T)−1−I{\displaystyle (I+T)^{-1}-I} is trace class if T{\displaystyle T} is. G{\displaystyle G} has a natural metric given by d(X,Y)=‖X−Y‖1{\displaystyle d(X,Y)=\\|X-Y\\|{1}}, where ‖X‖1=∑i\|λ i(X)\|{\displaystyle \\|X\\|{1}=\sum {i}\|\lambda {i}(X)\|} is the trace-class norm. Definition by exponential trace [edit] One definition uses the exponential trace formula. For finite-dimensional matrices, we have det(I+A)=e Tr⁡(ln⁡(I+A)){\textstyle \det(I+A)=e^{\operatorname {Tr} (\ln(I+A))}}, which expands in Taylor series to det⁡(I+A)=exp⁡(∑n=1∞(−1)n+1 n Tr⁡(A n)){\displaystyle \operatorname {det} (I+A)=\exp \left(\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}\operatorname {Tr} \left(A^{n}\right)\right)}This then generalizes directly to trace-class operators. Definition by exterior powers [edit] The exterior product of up to 3 vectors. In the finite-dimensional case, the determinant of an operator can be interpreted as the factor by which it scales the (oriented) volume of a parallelepiped. This can be generalized to infinite dimensions. In finite dimensions, by expanding the definition of determinant as a sum over permutations,det(I+A)=∑S det(A S S){\displaystyle \det(I+A)=\sum {S}\det(A{SS})}where S{\displaystyle S} ranges over all subsets of the index set of A{\displaystyle A}. For example, when the index set is {1,2}{\displaystyle {1,2}} then S={},{1},{2},{1,2}{\displaystyle S={},{1},{2},{1,2}}. If H{\displaystyle H} is an n{\displaystyle n}-dimensional Hilbert space with inner product(⋅,⋅){\displaystyle (\cdot ,\cdot )}, then the k{\displaystyle k}-th exterior powerΛ k H{\displaystyle \Lambda ^{k}H} is also a (n k){\displaystyle {\binom {n}{k}}}-dimensional Hilbert space, with inner product (v 1∧v 2∧⋯∧v k,w 1∧w 2∧⋯∧w k)=det(v i,w j).{\displaystyle (v_{1}\wedge v_{2}\wedge \cdots \wedge v_{k},w_{1}\wedge w_{2}\wedge \cdots \wedge w_{k})=\det(v_{i},w_{j}).}In particular e i 1∧e i 2∧⋯∧e i k,(i 1<i 2<⋯<i k){\displaystyle e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}},\qquad (i_{1}<i_{2}<\cdots <i_{k})}gives an orthonormal basis of Λ k H{\displaystyle \Lambda ^{k}H} if (e i){\displaystyle (e_{i})} is an orthonormal basis of H{\displaystyle H}. If A{\displaystyle A} is an operator on H{\displaystyle H}, then A{\displaystyle A}functorially defines a bounded operator Λ k(A){\displaystyle \Lambda ^{k}(A)} on Λ k H{\displaystyle \Lambda ^{k}H} by Λ k(A)v 1∧v 2∧⋯∧v k=A v 1∧A v 2∧⋯∧A v k.{\displaystyle \Lambda ^{k}(A)v_{1}\wedge v_{2}\wedge \cdots \wedge v_{k}=Av_{1}\wedge Av_{2}\wedge \cdots \wedge Av_{k}.}By definition of trace, we have Tr⁡(Λ k A)=∑1≤i 1<⋯<i k≤n(e i 1∧e i 2∧⋯∧e i k,A e i 1∧A e i 2∧⋯∧A e i k){\displaystyle \operatorname {Tr} \left(\Lambda ^{k}A\right)=\sum {1\leq i{1}<\cdots <i_{k}\leq n}(e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}},Ae_{i_{1}}\wedge Ae_{i_{2}}\wedge \cdots \wedge Ae_{i_{k}})}The summand simplifies to det[(e i j,A e i j′)]=det(A S S){\displaystyle \det[(e_{i_{j}},Ae_{i_{j'}})]=\det(A_{SS})} where S={i 1,…,i k}{\displaystyle S={i_{1},\dots ,i_{k}}}. Thus Tr⁡Λ k(A)=∑\|S\|=k det⁡(A S S).det⁡(I+A)=∑k=0 n Tr⁡Λ k(A).{\displaystyle {\begin{aligned}&\operatorname {Tr} \Lambda ^{k}(A)=\sum {\|S\|=k}\operatorname {det} \left(A{SS}\right).\&\operatorname {det} (I+A)=\sum {k=0}^{n}\operatorname {Tr} \Lambda ^{k}(A).\end{aligned}}}This generalizes to infinite-dimensional Hilbert spaces, and bounded trace-class operators, allowing us to define the Fredholm determinant by det(I+A)=∑k=0∞Tr⁡Λ k(A){\displaystyle \det(I+A)=\sum {k=0}^{\infty }\operatorname {Tr} \Lambda ^{k}(A)}To show that the definition makes sense, note that if A{\displaystyle A} is trace-class, then Λ k(A){\displaystyle \Lambda ^{k}(A)} is also trace-class with ‖Λ k(A)‖1≤‖A‖1 k/k!{\textstyle \\|\Lambda ^{k}(A)\\|{1}\leq \\|A\\|{1}^{k}/k!}, thus ∑k=0∞\|Tr⁡Λ k(A)\|≤e‖A‖1{\displaystyle \sum {k=0}^{\infty }\|\operatorname {Tr} \Lambda ^{k}(A)\|\leq e^{\\|A\\|{1}}}. Proof Proof We have ‖A‖1=∑i σ i{\textstyle \\|A\\|{1}=\sum {i}\sigma {i}} where σ i{\textstyle \sigma {i}} are the singular values of A{\textstyle A}. The singular values of Λ k(A){\textstyle \Lambda ^{k}(A)} are exactly the products of k{\textstyle k} distinct singular values of A{\textstyle A}. In other words, if you list all k{\textstyle k}-tuples with 1≤i 1<i 2<⋯<i k{\textstyle 1\leq i_{1}<i_{2}<\cdots <i_{k}}, then the corresponding singular value of Λ k(A){\textstyle \Lambda ^{k}(A)} is σ i 1 σ i 2⋯σ i k{\textstyle \sigma {i{1}}\sigma {i{2}}\cdots \sigma {i{k}}} Thus, ‖Λ k(A)‖1=∑1≤i 1<i 2<⋯<i k σ i 1 σ i 2⋯σ i k{\displaystyle \left\\|\Lambda ^{k}(A)\right\\|{1}=\sum {1\leq i_{1}<i_{2}<\cdots <i_{k}}\sigma {i{1}}\sigma {i{2}}\cdots \sigma {i{k}}} This is the k{\textstyle k} th elementary symmetric function of the singular values of A{\textstyle A}. Let a i≥0{\textstyle a_{i}\geq 0} (in our case a i=σ i{\textstyle a_{i}=\sigma _{i}} ) then by expanding the right side, we have ∑1≤i 1<⋯<i k a i 1⋯a i k≤1 k!(∑i a i)k{\displaystyle \sum {1\leq i{1}<\cdots <i_{k}}a_{i_{1}}\cdots a_{i_{k}}\leq {\frac {1}{k!}}\left(\sum {i}a{i}\right)^{k}} Properties [edit] By default, all operators are assumed trace-class. det(I+A)⋅det(I+B)=det(I+A)(I+B).{\textstyle \det(I+A)\cdot \det(I+B)=\det(I+A)(I+B).} z↦det(I+z A)=∑k=0∞z k Tr⁡Λ k(A){\textstyle z\mapsto \det(I+zA)=\sum {k=0}^{\infty }z^{k}\operatorname {Tr} \Lambda ^{k}(A)} defines an entire function, with \|det(I+z A)\|≤exp⁡(\|z\|⋅‖A‖1).{\textstyle \left\|\det(I+zA)\right\|\leq \exp(\|z\|\cdot \\|A\\|{1}).} The function A↦det(I+A){\displaystyle A\mapsto \det(I+A)} is continuous on trace-class operators, with \|det(I+A)−det(I+B)\|≤‖A−B‖1 exp⁡(‖A‖1+‖B‖1+1).{\displaystyle \left\|\det(I+A)-\det(I+B)\right\|\leq \\|A-B\\|{1}\exp(\\|A\\|{1}+\\|B\\|{1}+1).} One can improve this inequality slightly to the following, as noted in (Simon 2005, Chapter 5): \|det(I+A)−det(I+B)\|≤‖A−B‖1 exp⁡(max(‖A‖1,‖B‖1)+1).{\displaystyle \left\|\det(I+A)-\det(I+B)\right\|\leq \\|A-B\\|{1}\exp(\max(\\|A\\|{1},\\|B\\|{1})+1).} The function det{\displaystyle \det } defines a homomorphism of type G→C×{\displaystyle G\to \mathbb {C} ^{\times }} where C×{\displaystyle \mathbb {C} ^{\times }} the multiplicative group of nonzero complex numbers (since elements of G{\displaystyle G} are invertible). If T{\displaystyle T} is in G{\displaystyle G} and X{\displaystyle X} is invertible, det X T X−1=det T.{\textstyle \det XTX^{-1}=\det T.} det e A=exp Tr⁡(A).{\textstyle \det e^{A}=\exp \,\operatorname {Tr} (A).} log⁡det(I+z A)=Tr⁡(log⁡(I+z A))=∑k=1∞(−1)k+1 Tr⁡A k k z k{\textstyle \log \det(I+zA)=\operatorname {Tr} (\log {(I+zA)})=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {\operatorname {Tr} A^{k}}{k}}z^{k}} Integral operators [edit] The Fredholm determinant is often applied to integral operators. Let the trace-class operator T{\displaystyle T} be an integral operator given by a kernel K(x,y){\displaystyle K(x,y)}, then the Fredholm determinant is defined, like before, by det(I−λ T)=∑n=0∞(−λ)n Tr⁡Λ n(T)=exp⁡(−∑n=1∞Tr⁡(T n)n λ n){\displaystyle \det(I-\lambda T)=\sum {n=0}^{\infty }(-\lambda )^{n}\operatorname {Tr} \Lambda ^{n}(T)=\exp {\left(-\sum {n=1}^{\infty }{\frac {\operatorname {Tr} (T^{n})}{n}}\lambda ^{n}\right)}}where T{\displaystyle T} is an integral operator. The trace of the operator T{\displaystyle T} and its alternating powers is given in terms of the kernel K{\displaystyle K} by Tr⁡T=∫K(x,x)d x{\displaystyle \operatorname {Tr} T=\int K(x,x)\,dx} and Tr⁡Λ 2(T)=1 2!∬(K(x,x)K(y,y)−K(x,y)K(y,x))d x d y{\displaystyle \operatorname {Tr} \Lambda ^{2}(T)={\frac {1}{2!}}\iint \left(K(x,x)K(y,y)-K(x,y)K(y,x)\right)dx\,dy} and in general Tr⁡Λ n(T)=1 n!∫⋯∫det[K(x i,x j)]i,j∈1:n d x 1:n{\displaystyle \operatorname {Tr} \Lambda ^{n}(T)={\frac {1}{n!}}\int \cdots \int \det[K(x_{i},x_{j})]{i,j\in 1:n}\,dx{1:n}}The trace is well-defined for these kernels, since these are trace-class or nuclear operators. To see that this is a special case of the previous section's general definition, note that,Tr⁡(Λ k A)=∑1≤i 1<⋯<i k≤n(e i 1∧e i 2∧⋯∧e i k,A e i 1∧A e i 2∧⋯∧A e i k){\displaystyle \operatorname {Tr} \left(\Lambda ^{k}A\right)=\sum {1\leq i{1}<\cdots <i_{k}\leq n}(e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}},Ae_{i_{1}}\wedge Ae_{i_{2}}\wedge \cdots \wedge Ae_{i_{k}})}is equivalent to 1 k!∑i 1,⋯,i k∈1:n,all different det(A S S){\displaystyle {\frac {1}{k!}}\sum {i{1},\cdots ,i_{k}\in 1:n,{\text{ all different}}}\det(A_{SS})}where S{\displaystyle S} is the ordered sequence i 1,…,i k{\displaystyle i_{1},\dots ,i_{k}}. Now, to convert this to integral equations, a matrix becomes a kernel, and a summation over indices becomes an integral over coordinates. The above argument is intuitive. A proper definition requires a presentation showing that each of the manipulations are well-defined, convergent, and so on, for the given situation for which the Fredholm determinant is contemplated. Since the kernel K{\displaystyle K} may be defined for a large variety of Hilbert spaces and Banach spaces, this is a non-trivial exercise. Integral equation [edit] The original (Fredholm 1903) considered the integral equation u(x)+z∫a b K(x,y)u(y)d y=f(x)(x∈(a,b)){\displaystyle u(x)+z\int _{a}^{b}K(x,y)u(y)dy=f(x)\quad (x\in (a,b))}which can be written as (I+z A)u=f{\displaystyle (I+zA)u=f}. Fredholm proved that this equation has a unique solution iff det(I+z A)≠0{\displaystyle \det(I+zA)\neq 0}. Commutators [edit] A function F(t){\displaystyle F(t)} from (a,b){\displaystyle (a,b)} into G{\displaystyle G} is said to be differentiable if F(t)−I{\displaystyle F(t)-I} is differentiable as a map into the trace-class operators, i.e. if the limit F˙(t)=lim h→0 F(t+h)−F(t)h{\displaystyle {\dot {F}}(t)=\lim _{h\to 0}{F(t+h)-F(t) \over h}} exists in trace-class norm. If g(t){\displaystyle g(t)} is a differentiable function with values in trace-class operators, then so too is exp⁡g(t){\displaystyle \exp g(t)} and F−1 F˙=id−exp−ad⁡g(t)ad⁡g(t)⋅g˙(t),{\displaystyle F^{-1}{\dot {F}}={\operatorname {id} -\exp -\operatorname {ad} g(t) \over \operatorname {ad} g(t)}\cdot {\dot {g}}(t),} where ad⁡(X)⋅Y=X Y−Y X.{\displaystyle \operatorname {ad} (X)\cdot Y=XY-YX.} Israel Gohberg and Mark Krein proved that if F{\displaystyle F} is a differentiable function into G{\displaystyle G}, then f=det F{\displaystyle f=\det F} is a differentiable map into C∗{\displaystyle \mathbb {C} ^{}} with f−1 f˙=Tr⁡F−1 F˙.{\displaystyle f^{-1}{\dot {f}}=\operatorname {Tr} F^{-1}{\dot {F}}.} This result was used by Joel Pincus, William Helton and Roger Howe to prove that if A{\displaystyle A} and B{\displaystyle B} are bounded operators with trace-class commutator A B−B A{\displaystyle AB-BA}, then det e A e B e−A e−B=exp⁡Tr⁡(A B−B A).{\displaystyle \det e^{A}e^{B}e^{-A}e^{-B}=\exp \operatorname {Tr} (AB-BA).} Szegő limit formula [edit] See also: Szegő limit theorems Let H=L 2(S 1){\displaystyle H=L^{2}(S^{1})} and let P{\displaystyle P} be the orthogonal projection onto the Hardy spaceH 2(S 1){\displaystyle H^{2}(S^{1})}. If f{\displaystyle f} is a smooth function on the circle, let m(f){\displaystyle m(f)} denote the corresponding multiplication operator on H{\displaystyle H}. The commutator P m(f)−m(f)P{\displaystyle Pm(f)-m(f)P} is trace-class. Let T(f){\displaystyle T(f)} be the Toeplitz operator on H 2(S 1){\displaystyle H^{2}(S^{1})} defined by T(f)=P m(f)P,{\displaystyle T(f)=Pm(f)P,} then the additive commutator T(f)T(g)−T(g)T(f){\displaystyle T(f)T(g)-T(g)T(f)} is trace-class if f{\displaystyle f} and g{\displaystyle g} are smooth. Berger and Shaw proved that tr⁡(T(f)T(g)−T(g)T(f))=1 2 π i∫0 2 π f d g.{\displaystyle \operatorname {tr} (T(f)T(g)-T(g)T(f))={1 \over 2\pi i}\int _{0}^{2\pi }f\,dg.} If f{\displaystyle f} and g{\displaystyle g} are smooth, then T(e f+g)T(e−f)T(e−g){\displaystyle T(e^{f+g})T(e^{-f})T(e^{-g})} is in G{\displaystyle G}. Harold Widom used the result of Pincus-Helton-Howe to prove that det T(e f)T(e−f)=exp⁡∑n>0 n a n a−n,{\displaystyle \det T(e^{f})T(e^{-f})=\exp \sum {n>0}na{n}a_{-n},} where f(z)=∑a n z n.{\displaystyle f(z)=\sum a_{n}z^{n}.} He used this to give a new proof of Gábor Szegő's celebrated limit formula: lim N→∞det P N m(e f)P N=exp⁡∑n>0 n a n a−n,{\displaystyle \lim {N\to \infty }\det P{N}m(e^{f})P_{N}=\exp \sum {n>0}na{n}a_{-n},} where P N{\displaystyle P_{N}} is the projection onto the subspace of H{\displaystyle H} spanned by 1,z,…,z N{\displaystyle 1,z,\ldots ,z^{N}} and a 0=0{\displaystyle a_{0}=0}. Szegő's limit formula was proved in 1951 in response to a question raised by the work Lars Onsager and C. N. Yang on the calculation of the spontaneous magnetization for the Ising model. The formula of Widom, which leads quite quickly to Szegő's limit formula, is also equivalent to the duality between bosons and fermions in conformal field theory. A singular version of Szegő's limit formula for functions supported on an arc of the circle was proved by Widom; it has been applied to establish probabilistic results on the eigenvalue distribution of random unitary matrices. History [edit] The Fredholm determinant was first used in (Fredholm 1903) to solve an integral equation. Realizing the potential, Hilbert wrote 6 papers during 1904 to 1910 (collected in (Hilbert 1924)), beginning the theory of compact operators on Hilbert spaces. See (Bornemann 2010) and references therein. The Fredholm determinant was used by physicist John A. Wheeler (1937, Phys. Rev. 52:1107) to help provide mathematical description of the wavefunction for a composite nucleus composed of antisymmetrized combination of partial wavefunctions by the method of Resonating Group Structure. This method corresponds to the various possible ways of distributing the energy of neutrons and protons into fundamental boson and fermion nucleon cluster groups or building blocks such as the alpha-particle, helium-3, deuterium, triton, di-neutron, etc. When applied to the method of Resonating Group Structure for beta and alpha stable isotopes, use of the Fredholm determinant: (1) determines the energy values of the composite system, and (2) determines scattering and disintegration cross sections. The method of Resonating Group Structure of Wheeler provides the theoretical bases for all subsequent Nucleon Cluster Models and associated cluster energy dynamics for all light and heavy mass isotopes (see review of Cluster Models in physics in N.D. Cook, 2006). References [edit] Fredholm, Ivar (1903). "Sur une classe d'équations fonctionnelles"(PDF). Acta Mathematica. 27 (0): 365–390. doi:10.1007/BF02421317. ISSN0001-5962. Retrieved February 7, 2025. Hilbert, D. (1924). Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen. Fortschritte der mathematischen Wissenschaften in Monographien (in German). B. G. Teubner. Gohberg, Israel; Goldberg, Seymour; Krupnik, Nahum (2000). Traces and Determinants of Linear Operators. Basel: Birkhäuser Basel. doi:10.1007/978-3-0348-8401-3. ISBN978-3-0348-9551-4. Simon, Barry (2005), Trace Ideals and Their Applications, Mathematical Surveys and Monographs, vol.120, American Mathematical Society, ISBN0-8218-3581-5 Wheeler, John A. (1937-12-01). "On the Mathematical Description of Light Nuclei by the Method of Resonating Group Structure". Physical Review. 52 (11). American Physical Society (APS): 1107–1122. Bibcode:1937PhRv...52.1107W. doi:10.1103/physrev.52.1107. ISSN0031-899X. Bornemann, Folkmar (2010), "On the numerical evaluation of Fredholm determinants", Math. Comp., 79 (270), Springer: 871–915, arXiv:0804.2543, doi:10.1090/s0025-5718-09-02280-7 \| v t e Hilbert spaces \| \| Basic concepts \| Adjoint Inner product and L-semi-inner product Hilbert space and Prehilbert space Orthogonal complement Orthonormal basis \| \| Main results \| Bessel's inequality Cauchy–Schwarz inequality Riesz representation \| \| Other results \| Hilbert projection theorem Parseval's identity Polarization identity (Parallelogram law) \| \| Maps \| Compact operator on Hilbert space Densely defined Hermitian form Hilbert–Schmidt Normal Self-adjoint Sesquilinear form Trace class Unitary \| \| Examples \| C n(K) with K compact &n<∞ Segal–Bargmann F \| \| v t e Topological tensor products and nuclear spaces \| \| Basic concepts \| Auxiliary normed spaces Nuclear space Tensor product Topological tensor product of Hilbert spaces \| \| Topologies \| Inductive tensor product Injective tensor product Projective tensor product \| \| Operators/Maps \| Fredholm determinant Fredholm kernel Hilbert–Schmidt operator Hypocontinuity Integral Nuclear between Banach spaces Trace class \| \| Theorems \| Grothendieck trace theorem Schwartz kernel theorem \| \| v t e Functional analysis(topics – glossary) \| \| Spaces \| Banach Besov Fréchet Hilbert Hölder Nuclear Orlicz Schwartz Sobolev Topological vector Properties Barrelled Complete Dual (Algebraic / Topological) Locally convex Reflexive Separable \| \| Theorems \| Hahn–Banach Riesz representation Closed graph Uniform boundedness principle Kakutani fixed-point Krein–Milman Min–max Gelfand–Naimark Banach–Alaoglu \| \| Operators \| Adjoint Bounded Compact Hilbert–Schmidt Normal Nuclear Trace class Transpose Unbounded Unitary \| \| Algebras \| Banach algebra C-algebra Spectrum of a C-algebra Operator algebra Group algebra of a locally compact group Von Neumann algebra \| \| Open problems \| Invariant subspace problem Mahler's conjecture \| \| Applications \| Hardy space Spectral theory of ordinary differential equations Heat kernel Index theorem Calculus of variations Functional calculus Integral linear operator Jones polynomial Topological quantum field theory Noncommutative geometry Riemann hypothesis Distribution (or Generalized functions) \| \| Advanced topics \| Approximation property Balanced set Choquet theory Weak topology Banach–Mazur distance Tomita–Takesaki theory \| \| Category \| Retrieved from " Categories: Determinants Fredholm theory Hilbert spaces Topological tensor products Hidden category: CS1 German-language sources (de) This page was last edited on 7 February 2025, at 02:31(UTC). 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15413	https://alfred.stlouisfed.org/series?seid=IORB	Skip to main content Home > Releases > Interest Rate on Reserve Balances > Interest Rate on Reserve Balances (IORB Rate) Interest Rate on Reserve Balances (IORB Rate) (IORB) Chart Bar chart with 2 data series. The chart has 1 X axis displaying xAxis. Data ranges from 2021-07-29 2:00:00 to 2025-08-22 2:00:00. The chart has 2 Y axes displaying Percent and yAxisRight. Interest Rate on Reserve Balances (IORB Rate) Vintage: 2025-08-20 Interest Rate on Reserve Balances (IORB Rate) Vintage: 2025-08-21 Percent End of interactive chart. 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Custom Graph Link Embed in Website Graph Image Link Save Graph Add to Dashboard Add to Data List Get Email Notification Save Map Notes \| \| \| \| --- \| Title \| Release Dates \| \| \| \| \| Interest Rate on Reserve Balances \| 2021-07-28 \| 2024-09-23 \| \| Interest Rate on Reserve Balances (IORB Rate) \| 2024-09-24 \| 2025-08-21 \| \| \| \| \| \| Source \| \| \| \| \| \| Board of Governors of the Federal Reserve System (US) \| 2021-07-28 \| 2025-08-21 \| \| \| \| \| \| Release \| \| \| \| \| \| Interest Rate on Reserve Balances \| 2021-07-28 \| 2025-08-21 \| \| \| \| \| \| Units \| \| \| \| \| \| Percent \| 2021-07-28 \| 2025-08-21 \| \| \| \| \| \| Frequency \| \| \| \| \| \| Daily, 7-Day \| 2021-07-28 \| 2025-08-21 \| \| \| \| \| \| Seasonal Adjustment \| \| \| \| \| \| Not Seasonally Adjusted \| 2021-07-28 \| 2025-08-21 \| \| \| \| \| \| Notes \| \| \| \| \| \| Starting July 29, 2021, the interest rate on excess reserves (IOER ( and the interest rate on required reserves (IORR ( were replaced with a single rate, the interest rate on reserve balances (IORB ( See the source's announcement ( for more details. The interest rate on reserve balances (IORB rate) is the rate of interest that the Federal Reserve pays on balances maintained by or on behalf of eligible institutions in master accounts at Federal Reserve Banks. The interest rate is set by the Board of Governors, and it is an important tool of monetary policy. See Policy Tools ( and the IORB FAQs ( for more information. \| 2021-07-28 \| 2024-09-23 \| \| IOER ( and the interest rate on required reserves (IORR ( were replaced with a single rate, the interest rate on reserve balances (IORB ( See the source's announcement ( for more details. The interest rate on reserve balances (IORB rate) is the rate of interest that the Federal Reserve pays on balances maintained by or on behalf of eligible institutions in master accounts at Federal Reserve Banks. The interest rate is set by the Board of Governors, and it is an important tool of monetary policy. See Policy Tools ( and the IORB FAQs ( for more information. 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15414	https://mathoverflow.net/questions/287011/a-conjecture-regarding-prime-numbers	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange A conjecture regarding prime numbers Ask Question Asked Modified 7 years, 10 months ago Viewed 4k times 62 $\begingroup$ For $n,m \geq 3$, define $ P_n = { p : p$ is a prime such that $ p\leq n$ and $ p \nmid n }$ . For example : $P_3= { 2 }$ $P_4= { 3 }$ $P_5= { 2, 3 }$, $P_6= { 5 }$ and so on. Claim: $P_n \neq P_m$ for $m\neq n$. While working on prime numbers I formulated this problem and it has eluded me for a while so I decided to post it here. I am not sure if this is an open problem or solved one. I couldn't find anything that looks like it. My attempts haven't come to fruition though I have been trying to prove it for a while. If $m$ and $n$ are different primes then it's clear. If $m \geq 2n$, I think we can find a prime in between so that case is also taken care of. My opinion is that it eventually boils down to proving this statement for integers that share the same prime factors. My coding is kind of rusty so would appreciate anybody checking if there is a counterexample to this claim. Any ideas if this might be true or false? Thanks. PS: I asked this question on mathstackexchage and somebody recommended I post it here as well. Here is the link to the original post nt.number-theory prime-numbers conjectures Share Improve this question asked Nov 26, 2017 at 15:32 Basanta PahariBasanta Pahari 65355 silver badges77 bronze badges $\endgroup$ 11 15 $\begingroup$ This is equivalent to say that among two numbers having the same radical, there is at least one prime number. $\endgroup$ Konstantinos Gaitanas – Konstantinos Gaitanas 2017-11-26 15:45:08 +00:00 Commented Nov 26, 2017 at 15:45 4 $\begingroup$ This does not seem terribly easy... in fact, I don't have a good intuition as to whether it should be true or not. Let $S = {p_1, \cdots, p_k}$. The question can be thought of whether there exist two $S$-units $m, n$, say $N < m < n \leq 2N$, such that $\|n - m\| \ll \log N$ say. If the answer is yes, then one would expect that there are no primes between $m$ and $n$ typically, so the conjecture would be false. $\endgroup$ Stanley Yao Xiao – Stanley Yao Xiao ♦ 2017-11-26 15:51:23 +00:00 Commented Nov 26, 2017 at 15:51 15 $\begingroup$ If you just consider $n^2(n+1)$ and $n(n+1)^2$, then this would imply that there always is a prime between $n^3$ and $(n+1)^3$. That is known, but highly non-trivial. However, those two are just the simplest products with guaranteed same radical that jump to one's mind. Whether a more clever construction would allow one to deduce something that we certainly do not know today remains to be seen. $\endgroup$ fedja – fedja 2017-11-26 16:28:50 +00:00 Commented Nov 26, 2017 at 16:28 2 $\begingroup$ Maybe then an easier and more interesting question would be to prove some bound function $b(m)$ on $n-m$ if $m $\endgroup$ Yaakov Baruch – Yaakov Baruch 2017-11-26 16:52:20 +00:00 Commented Nov 26, 2017 at 16:52 1 $\begingroup$ Say, based on the highest quality $abc$ triple known to date, you have that $48407720661$ and $48407720661+15042$ have the same radical. Now even such a small number as $15042$ is so much larger than $\log(48407720661)$ that the chance of not there being primes in that interval is practically nil. So my expectation is that the OP conjecture is true, but this reasoning clearly needs work to be made into a clean euristic argument. $\endgroup$ Yaakov Baruch – Yaakov Baruch 2017-11-26 17:46:21 +00:00 Commented Nov 26, 2017 at 17:46 \| Show 6 more comments 2 Answers 2 Reset to default 44 $\begingroup$ This is true (for large $m$ and $n$) under ABC plus the assumption that there is a prime in $[x,x+x^{1/2-\delta}]$ for some positive $\delta$ (which is widely believed, but beyond RH). To see this, suppose $m The problem is likely very hard, as fedja's observation in the comments already shows. There is a conjecture of Hall that $\|x^3-y^2\| \gg x^{1/2-\epsilon}$ which is wide open. The best results that are known here (going back to Baker's method) are of the flavor $\|x^3-y^2\| \gg (\log x)^C$ for some $C$. If $x^3-y^2$ does get as small as in the Baker result, then take $n=x^4y$ and $m=xy^3$, which clearly have the same radical and then $\|n-m\|$ is of size essentially $n^{5/11}$. In other words, either you have to improve work towards Hall's conjecture, or work towards gaps between primes! Added Thanks to Pasten's comment, I learned that this problem is already in the literature and is known as Dressler's conjecture. The conditional proof above is recorded in work of Cochrane and Dressler who give more information on the conjecture. Share Improve this answer edited Nov 27, 2017 at 18:14 answered Nov 26, 2017 at 17:38 LuciaLucia 44.2k66 gold badges198198 silver badges222222 bronze badges $\endgroup$ 10 2 $\begingroup$ If $n$ is divisible by a prime larger then $m$, then won't $n$ have to be that prime? (or one would have $n\ge 2m$ and then it should be obvious) $\endgroup$ Lucia – Lucia 2017-11-26 17:58:20 +00:00 Commented Nov 26, 2017 at 17:58 5 $\begingroup$ Nice! I am surprised that it is beyond the reach of RH. But that explains why my elementary approach (since I don't have any analytic number theory background ) weren't getting me anywhere. Regardless, it's really satisfying to see that the claim is actually true assuming the conjectures you mentioned. Great work! $\endgroup$ Basanta Pahari – Basanta Pahari 2017-11-26 23:24:01 +00:00 Commented Nov 26, 2017 at 23:24 1 $\begingroup$ There is a simple sequence where $nm<(\frac{3}{4}n)^{1/2}$: $n=\frac{3}{4}9^k(9^k1)$ and $m=\frac{3}{4}(9^k1)^2$. $\endgroup$ Yaakov Baruch – Yaakov Baruch 2017-11-27 13:18:49 +00:00 Commented Nov 27, 2017 at 13:18 7 $\begingroup$ cf. Dressler's conjecture. Often misquoted as a consequence of abc, while in fact it is a well-known consequence of abc and existence of primes in short intervals. $\endgroup$ Pasten – Pasten 2017-11-27 18:00:25 +00:00 Commented Nov 27, 2017 at 18:00 1 $\begingroup$ @BasantaR.Pahari I think your formulation is very elegant and it makes the problem look more natural. $\endgroup$ Pasten – Pasten 2017-11-28 00:12:45 +00:00 Commented Nov 28, 2017 at 0:12 \| Show 5 more comments 9 $\begingroup$ I am working on a related project involving Grimm's conjecture. The hope is to show that every interval of consecutive composite numbers below $10^{12}$ contains an injective divisor map, see On comparing two almost injective divisor maps for more detail. The upshot is that there are about 700 opportunities for your event to happen (because the map $L(m)$ being largest prime factor of m is often injective, and in your case it won't be) below 2.5 times $10^{10}$, and that your event won't happen because the numbers involved are too close. (Specifically, $L(m)=L(n)=p$, and $m-n =kp$ where $L(k)$ is less than $p$ and usually less than 3, and in those cases $m/p$ and $n/p$ have sufficiently different sets of prime factors.). If I can achieve my aims while offloading data regarding your claim (e.g. a data file of the estimated 3000 $L$ pairs below $10^{12}$), I will do so and report back. If you have several months of computer cycles to spare, I can provide a program so that you can join in the fun, AND get some data on your problem. Gerhard "Another Opportunity For Communal Computing" Paseman, 2017.11.26. Share Improve this answer edited Nov 27, 2017 at 13:21 Glorfindel 2,83366 gold badges3030 silver badges3939 bronze badges answered Nov 26, 2017 at 22:49 Gerhard PasemanGerhard Paseman 13.1k33 gold badges3434 silver badges6464 bronze badges $\endgroup$ 1 $\begingroup$ I think it should be the case (that there is an elementary proof) that for fixed j, there are only finitely many cases where your conjecture fails with all primes involved less than j. You might even be able to give an explicit bound. If I get any more along this line I will also report it. Gerhard "Maybe Only Finitely Many J?" Paseman, 2017.11.26. $\endgroup$ Gerhard Paseman – Gerhard Paseman 2017-11-26 23:54:24 +00:00 Commented Nov 26, 2017 at 23:54 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions nt.number-theory prime-numbers conjectures See similar questions with these tags. 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15415	https://www.ism.ac.jp/editsec/aism/pdf/055_4_0865.pdf	Ann. Inst. Statist. Math. Vol. 55, No. 4, 865-884 (2003) (~)2003 The Institute of Statistical Mathematics ON THE DISTRIBUTION OF THE TOTAL NUMBER OF RUN LENGTHS D. L. ANTZOULAKOS, S. BERSIMIS AND M. V. KOUTRAS Department of Statistics and I'nsurance Science, University of Piraeus, Karaoli ~4 Dimitriou 80, Piraeus 18534, Greece (Received September 9, 2002; revised December 16, 2002) Abstract. In the present paper, we study the distribution of a statistic utilizing the runs length of "reasonably long" series of alike elements (success runs) in a sequence of binary trials. More specifically, we are looking at the sum of exact lengths of subsequences (strings) consisting of k or more consecutive successes (k is a given positive integer). The investigation of the statistic of interest is accomplished by exploiting an appropriate generalization of the Markov chain embedding technique introduced by Fu and Koutras (1994, Y. Amer. Statist. Assoc., 89, 1050-1058) and Koutras and Alexandrou (1995, Ann. Inst. Statist. Math., 47, 743-766). In addition, we explore the conditional distribution of the same statistic, given the number of successes and establish statistical tests for the detection of the null hypothesis of randomness versus the alternative hypothesis of systematic clustering of successes in a sequence of binary outcomes. Key words and phrases: Success runs, run lengths, Markov chains, Markov chain embeddable variable of polynomial type, randomness tests. 1. Introduction In the analysis of experimental trials whose outcomes can be classified into two exclusive categories, a question that comes in naturally is whether reasonable criteria providing evidence of clustering of any of the two categories could be established. These criteria could then be used to detect changes in the underlying process which gener- ates the series of outcomes. Many commonly used criteria for the statistical analysis of such phenomena involve the concept of runs i.e. uninterrupted sequences of alike ele- ments bordered at each end by other types of elements or by the beginning or by the end of the complete sequence. For example, many quality control plans base the accep- tance/rejection of the sample lot on the occurrence of prolonged sequences of successive working/failed components, Wolfowitz (1943), Balakrishnan et al. (1993). For a mechan- ical engineer performing a start-up test for a new machine, it is reasonable to couch his decision (accepting the machine or rejecting it) on the number of consecutive successful or unsuccessful attempted start-ups, Hahn and Gage (1983), Viveros and Balakrishnan (1993), Balakrishnan et al. (1995, 1997). The same model, in the context of reliability, leads to the well known consecutive-k-out-of-n: F system and its variations (for a review refer to Chao et al. (1995)). Finally, an additional interesting application of the concept Research supported by General Secretary of Research and Technology of Greece under grand PENED 2001. 865 866 D. L. ANTZOULAKOS ET AL. of runs comes from the area of non-parametric runs tests, Gibbons and Chakraborti (1992), Agin and Godbole (1992). In this case, the interest is focused on the condi- tional distribution of runs or equivalently, on runs defined in a sequence of outcomes of prespecified composition. In the traditional runs/patterns literature, the criteria used take into account the number of runs/patterns observed in the experimental sequence or the number of runs of specified length or the waiting time for the occurrence of a prespecified number of runs. The distributions of the number of fixed size success runs and the associated waiting time distributions have been termed in the statistical bibliography as distributions of order k and have been extensively studied by Philippou and Makri (1986), Aki and Hirano (1988), Godbole (1990, 1992), Hirano et al. (1991), Hirano and Aki (1993) etc. For a detailed and systematic exposition of the distribution theory of runs the interested reader may wish to consult the recent monograph by Balakrishnan and Koutras (2002). None of the aforementioned criteria makes use of the exact length of runs appearing in the outcome sequence. Agin and Godbole (1992), aiming at the development of non-parametric randomness tests, suggested using run lengths of variable sizes (see also Koutras and Alexandrou (1997)). Of a similar flavour can be considered the approach set in O'Brien and Dyck (1985), where a test based on longest runs is investigated. Motivated b~} those works, we proceed here to the investigation of a statistic utilizing the runs length of "reasonably long" series of alike elements (successes) in a sequence of binary trials. More specifically, we are looking at the sum of the lengths of subsequences (strings) consisting of k or more consecutive successes (k is a given positive integer). It is clear that theoretical results on the distribution of this statistic are of major practical importance for establishing and investigating appropriate statistical tests which would detect the null hypothesis of randomness in the sequence of outcomes versus the alternative hypothesis of systematic clustering of successes. The investigation of the statistic of interest here is derived by exploiting a proper Markov chain embedding technique. In Section 2 we present in brief the general family of Markov chain embeddable variables (c.f. Fu and Koutras (1994)) and introduce the wide subclass of Markov chain embeddable variables of polynomial type (MVP) which may be efficiently used for the evaluation of the exact distribution of enumerating random variables. This class gen- eralizes the family of Markov chain embeddable variables of binomial type introduced in Koutras and Alexandrou (1995) and offers a very functional "working environment". Next we establish compact and computationally tractable formulae for obtaining the exact distribution (probability mass function and generating functions) of MVP's. In Section 3 we apply the MVP methodology to the problem of establishing the exact dis- tribution of the sum of the lengths of success runs whose length exceed a prespecified level. Finally in Section 4 we assume that the composition of the observed sequence is known, that is to say, the number of successes and failures are fixed quantities, and pro- ceed to the investigation of the conditional distribution of the aforementioned statistic. The results are then used (in Section 5) to investigate the performance of a new test of randomness. 2. General results Recently, Fu and Koutras (1994) developed a unified method for capturing the exact distribution of the number of runs of specified length by employing a Markov chain embedding technique. Koutras and Alexandrou (1995) refined the method and ON THE DISTRIBUTION OF RUN LENGTH 867 expressed these distributions in terms of multidimensional binomial type probability vectors. Fu (1996) extended the original method to cover the case of arbitrary patterns (instead of runs) whereas Koutras (1997) treated several waiting time problems within this framework. Finally Doi and Yamamoto (1998) and Han and Aki (1999) considered the case of multivariate run related distributions and offered simple solutions to the problem by exploiting proper extensions of the Markov chain embedding technique. We shall first introduce the notion of a Markov chain embeddable variable, in a way similar to the one used by Fu and Koutras (1994). Let Xn (n a non-negative integer) be a non-negative finite integer-valued random variable and denote by l, = sup{x : Pr(X, = x) > 0} its upper end point. DEFINITION 2.1. The random variable Xn will be called a Markov Chain embed- dable variable if (a) there exists a Markov chain {Yt : t > 0} defined on a state space D = {at, a2,... } which can be partitioned as ft = (-J,>0 Cx, (b) the probability mass function of Xn can be captured by considering the pro- jection of the probability space of Ks onto C,, i.e. Pr(Xn = x) = Pr(Yn C Cx), x = O, 1,...,In. If At is the one-step transition probability matrix of the Markov chain ({Yt : t > 0}, ft), then the exact distribution of Xn can be derived by the aid of the formula (2.1) Pr(X~ = x) = lr0 At e~, x = 0, 1,...,l~ t=l i:o~i cCx where ei are unit (row) vectors and rr0 = (Pr(Yo = c~,), Pr(Y0 = ct2), Pr(Y0 = c~3),... ) is the vector of initial probabilities. For typographical convenience we impose the convention Pr(X0 = 0) = 1 and set b I]t=a At = I for a > b. Should it be possible to partition At in the form of a bidiagonal blocked matrix with non-zero blocks appearing only on the main diagonal and the diagonal next to it, the investigation of Xn's distribution could be easily carried out by considering appropriate probability vectors describing the overall state formulation of the Markovian process at time t. Motivated by this observation, Koutras and Alexandrou (1995) proceeded to the introduction of a significant subclass of the family of Markov chain embeddable variables which offered a computationally efficient framework for tackling problems of this nature. Unfortunately, this class is not wide enough to accommodate the distributional problem we are aiming at in this article. For this reason we proceed to the introduction of a more general family of variables which will be called Markov chain embeddable variables of polynomial type. To start with, let us first observe that without loss of generality we may assume that the state subspaces Cx, x = 0, 1,... have the same finite cardinality s = IC, I. DEFINITION 2.2. The random variable Xn will be called Markov chain embeddable variable of polynomial type (MVP) if 868 D.L. ANTZOULAKOS ET AL. (a) there exists a Markov chain {Yt, t > 0} defined on a discrete state space fl which can be partitioned as ~'~ = U Cx' Cx : {Cx,o, Cx,l,...,Cx,s-l} x>O (b) there exists a positive integer m such that for t > 1 Pr(Yt E Cy [ Yt_I E Cx) = O for all y C x, x + l,... ,x + m (c) the probability mass function of X~, can be captured by considering the pro- jection of the probability space of Yn onto Cx i.e. Pr(X~ = x) = Pr(Yn E C~), n > 0, x > 0. For m = 1, Definition 2.2 reduces to the definition of the Markov chain embeddable variables of binomial type introduced by Koutras and Alexandrou (1995). Roughly speaking, a MVP is characterized by the following property: the state subclasses Cx, x > > O, can be ordered in such a way that once the chain enters Cx, the feasible one step transitions lead either to the same subclass Cx or to one of the next m (main state) subclasses C~+1,..., Cx+m. For the m + 1 transition probability matrices At,i(x) = (Pr(Yt = Cz+i,j, I Y t-1 = cx,j))~• Ol, x>O it is clear (c.f. condition (b) of Definition 2.2) that the matrix y~m_0 At,~(x) is stochastic. Moreover, on introducing the probability (row) vectors ft(x) = (Pr(Yt = cx,0),Pr(Yt = Cx,1),... ,Pr(Yt = Cx,s-1)), t ~ 0, x ~ 0 it follows directly from condition (c) of Definition 2.2 that Pr(Xn = x) = fn(X)(1,1,..., 1)' = f n(X)l', n > O, X > O. Finally, convention Pr(Xo = 0) = 1 implies that rrol' = f0(0)l' = (Pr(Y0 = c0,0), Pr(}5 = CO,l),..., Pr(Y0 = c0,s-1))l' = 1 7rxl' = fo(x)l' = 0, x > 1. Before proceeding to the development of general results facilitating the investigation of the exact distribution of a MVP, let us discuss in brief some potential applications where the approach taken here can be fruitfully used. Let g be an event (single or composite) associated with a sequence of binary (or multistate) trials and introduce a score function fi ($) denoting the points earned if event E occurs at the i-th triM. Then a MVP offers an appropriate methodological tool for investigating the exact distribution of the total score Xn achieved in a series of n outcomes. This setup is wide enough to accommodate the number of fixed length runs model (fi(g) = 1 if a run of prespecified length has been registered at the i-th trial), the sum of run lengths model introduced in Section 1 (fi(C) equals the exact length of a run completed at the i-th trial provided that the length exceeds a prespecified level), or even more complex models pertaining ON THE DISTRIBUTION OF RUN LENGTH 869 to the occurrence of specific patterns. By way of example assume that we are trying to investigate the efficacy of n questions. Each time we observe a cluster of correct responses (e.g. a certain number of correct answers in a row, or segments containing a prespecified percentage of correct answers) c points are added to the subject's score, i.e. fi (C) = c. For each subsequent correct response d extra points are earned. Apparently, the total number of points collected upon the completion of the test may be considered as an indication of method's efficacy. The same statistic, under a slightly different description may be used for deciding whether a quality control process is out of control, whether a disease can be declared as contagious based on patterns of infected among non-infected plants in a transect through a field, etc. It is clear that in all these situations the knowledge of the distribution of the test statistic will help the practitioner to setup reasonable statistical procedures guarantying prespecified levels of type I error. The general results presented in this section are quite useful for the investigation of the aforementioned models and models of similar nature encountered in numerous areas of applied sciences. Let us start our study with the next theorem which provides a method for the evaluation of the probability mass function of a MVP. THEOREM 2.1. The sequence of vectors f t(x) satisfies the recurrence relation min(x,m) ft(X) = E ft-l(X -- i)At#(x - i), t > 1, x > O. i=0 PROOF. Let t > 1, x > 0 and 0 < j < s - 1. The total probability theorem yields min(x,m) s-1 Pr(Yt = cx,j) = E E Pr(Yt = cx,j ] Yt-1 = Cx-i,r) Pr(Yt-1 = cx-i,r) i=0 r=O which can be equivalently written as min(x,m) s-1 Pr(Yt = c~,j) = ~ E er+lAt,i(x - i)e}+ 1 Pr(Yt-1 = cx-i,r) i=O r=0 min(x,rn) = E ,ft_l(X- i)At,i(x- i)e}+ 1 i=0 (ei denote the unit row vectors of II~S), and the proof is complete. [] Next, let ~t(z) and O(z, w) be the single and double generating functions oo (x) Ft(z) = E Pr(Xt = x)z x, O(z,w) = E~at(zlwt x=0 t=O and denote by ~t(z) and ~(z,w) the single (row) and double (row) vector generating functions of ft (x), respectively, that is o~ oo ~at(z) = E f t(x)zX , t>0, 'I~(z, w) = E ~ot(z)wt. x:0 t:0 870 D. L. ANTZOULAKOS ET AL. It is clear that r = 7r0, ~t(z) = ~t(z)l', t > 1 and (I)(z,w) = (I)(z,w)l'. We mention that, it is the rule rather than the exception that matrices At,i(x) do not depend on x, that is At,i(x) = At,i for all t __ 1 and x > 0. In this case the vector generating function ~t(z) can be expressed in the form of a product as stated in the following theorem. THEOREM 2.2. If At,i(x) = At# for all t > 1 and x > 0 then the (single) vector generating function of Xt is given by t(z) = l-I r=l \i=0 / t~l. PROOF. For t > 1 and upon using Theorem 2.1 we may write Or ?n X CX:) TYt r : E ft(x)zx ----- EE ft-l(x-i)nt'izx -}- E E ft-l(x-i)nt'izx x=0 x=0 i=0 x=m+l i=0 ) ) =Ez i ft_l(x-i)z x-i At,i+ z' ft_l(x-i)z x-i At# i=0 =" i=0 x=m+l ~ _ _ EZ i t-l(y)z y At,i = ~t_l(Z) At,iz i i=0 \y=0 / \ i=0 / The proof may easily be completed by repeated application of the last formula. [] It is well known that the probability generating function ~(z) of the (generalized) binomial distribution is the product of the binomial terms Pro + zPrl, where Pri denotes the probability of occurrence of outcome "i", i = 0, 1 at the r-th trial (two possible outcomes). Replacing the binomial terms by the polynomial terms ~-~m=op,.izi , where p~i denotes the probability of occurrence of outcome "i", i = 0, 1,..., m at the r-th trial (m+ 1 possible outcomes), we obtain the probability generating function of the univariate multinomial distribution introduced by Steyn (1956) (see also Panaretos and Xekalaki rn i (1986) and Philippou et al. (1990)). In Theorem 2.2, ~-~i=oPriZ has been replaced by the polynomial term ~-~,~n_0 Ar,iz i, a fact justifying the nomenclature polynomial type used for the random variables studied here. In the case of an homogeneous MVP, we have the next theorem. THEOREM 2.3. If At,i(x) -- Ai for all t > 1 and x > 0 then the double vector generating function of Xt can be expressed as (I)(z, w) = 7to I - w Aiz i i=0 / where I is the identity s • s matrix. --1 PROOF. Follows readily from Theorem 2.2 on observing that t=o t=o \ i=O -- w Aiz i i=0 -1 ON THE DISTRIBUTION OF RUN LENGTH 871 the last equality being valid in an appropriate neighborhood of zero for w. [] For an homogeneous MVP Xt let Pt = E(Xt), t > 1, denote the mean of Xt and M(w) ~- Eta~ ~t wt, its generating function. The next theorem provides two compact formulae for the evaluation of #t and M(w) through the transition probability matrices Ai, i = 0, 1,...,m. THEOREM 2.4. If At,i(x) = Ai for all t > 1 and x > 0 then Pt = E(Xt) = ~ro Ai iA~ 1' r=l i=1 1 - i=o \~=1 / 1'. PROOF. Exploiting the formula (m d ZAiz i =E Aiz~ iAiz~-I Aizi dz \i=0 / r=l i=0 i=1 \i=0 / we deduce, by virtue of Theorem 2.2 #t = ~-:~[~t(z)l'] = 7to A~ iAi Ai 1'. r=l i=0 \i=1 / m A The first result follows readily by recalling that matrix )-~i=o i is stochastic. The generating function M(w) may be written as ) t=l r=l \i=i /J = 7row Ai w~-i E wt-r iA~ r=l i=0 t=r 11 and the desired formula is effortlessly established by virtue of (2.2) E Ai w r-1 = I- w Ai r=l i=O i=0 [] It goes without saying that for m --- 1 the outcomes of Theorems 2.1-2.4 produce the respective results which have been developed by Koutras and Alexandrou (1997) for the Markov chain embeddable variables of binomial type. 872 D. L. ANTZOULAKOS ET AL. 3. The distribution of the sum of the exact lengths of runs of length at least k Consider a sequence of Bernoulli trials Z1, Z2,... with success probabilities pt = Pr(Zt = 1), and failure probabilities qt = Pr(Zt = O) = 1 -Pt, t > 1 and let n, k be any positive integers with n > k. For k < t < n we define k + ~, if Zt-k-e+l = Zt-k-e+2 ..... Zt = 1 and Zt-k-i = Zt+l = 0 Ut = O, otherwise (convention: Zo = Zn+l = 0). Then the sum of the exact lengths of substrings of the sequence Z1, Z2,. 9 Zn containing k or more consecutive successes, can be expressed as n U . -- Xn = }-:~t=k t, n > 1. It is clear that the support of X~ is {0, k,k+l,.. ,n}. For n < k we set Xn = 0 and the support of X~ reduces to {0}. In order to view the random variable Xn as a MVP, we set Cx = {cx,0, Cxa,..., cx,k} where cx,i = (x,i), 0 < i < k, x > 0 and define a Markov chain {Yt,t ~ 0} on ~ = Ux>o Cx as follows: Yt = Cx# (or equivalently Yt = (x, i)) if in the first t outcomes, say 1001 9 9 9 011~, the observed sum of the exact lengths of runs of k or more consecutive T successes is x and i={ r'k, if if r>k.r=O'l''"'k-1 It is apparent that, once the chain enters C~, the one step transitions may lead only to the subclasses C~, Cx+l or Cx+k. Hence the random variable Xn belongs to the class of MVP. The transition probability matrices At,i, i = O, 1,..., k, can be easily identified by observing that if Yt = c~,k the feasible one step transitions of the chain lead either to substate cx+l,k (if Zt+l = 1) or given by qt At, 0 ~- qt to substate pt O"'O Opt"" 0 qtO 0...0 qtO0"'O qtO0'''O Cx,o (if Zt+l = 0). Therefore, At,o will be 00" 00 Pt 0 O0 O0 (k+l) x (k+l) while At,2,... ,At,k-1 will be (k + 1) x (k + 1) matrices with all their entries 0. Matrix At,1 will have all its entries 0 except for the entry (k + 1, k + 1) which equals Pt. Finally, At,k will have all its entries 0 except for the entry (k, k + 1) which equals Pt. The appropriate initial probability vector of the Markov chain established here is given by 7r0 = (1,0,0,...,0). Recalling now Theorem 2.1 we may readily evaluate the probability mass function of X~. Moreover exploiting Theorem 2.2 we may derive its probability generating function as oo n ~n(z) = E P(Xn = x)z x = 1to I-I (A,~,o + zA~,l + zkAr,k)l '. x=0 r=l ON THE DISTRIBUTION OF RUN LENGTH 873 In the case of iid trials with success probabilities p (Pt = P, qt = q), Theorem 2.3 yields, after some routine calculations, (3.1) where O(z,w) = ~ ~n(Z)W n = fro(I- w(Ao + zA1 + zkAk))-ll ' -- Pl(z,w) P2(z, ~) rt=O Pl (Z, W) ~- 1 - wpz - (wp)k(1 -- z k) -- (wp)k+l (zk -- Z) P2(z,w) = 1 - w(1 + pz) + w2pz + wk+lqpk(1 -- z k) + Wk+2qpk+l(Z k -- Z). (I denotes the identity (k + 1) x (k + 1) matrix). It is not difficult to verify that (I)(z, w) may be written in the form e(z,~) = ~ ~(z)w ~ = n=0 1 -- wal(Z ) -- wka2(z) - wk+la3(z) 1 - [wbl(z) + w2b2(z) + wk+lb3(z) q- wk+2b4(z)] where ai(z), i = 1, 2, 3 and bi(z), i = 1, 2, 3, 4, are appropriate functions of z. Following the methodology employed by Antzoulakos and Chadjiconstantinidis (2001), we may express ~n(Z) as where ~n(Z) =~o(Z) -- al(Z)~l(Z) -- a2(z)~k(Z) - a3(z)~k+l(z) nl+2n2+(k+l)na+(k+2)n4=n-i j=l j=l nj! --, i = 0, 1, k,k + 1. Since the generating function of pn(Z), n > 0, is a rational function of the form (3.1) a recursive scheme may be readily established by the aid of standard combinatorial techniques (see e.g. Chapter 4.1 in Stanley (1997)). More specifically we have the next result. THEOREM 3.1. If Z1, Z2,..., Zn is a sequence of lid Bernoulli trials the probability generating function ~n(Z) of the random variable Xn satisfies the recursive scheme ~n(z) = (1 + pz)ifln_l(Z ) -- pZ)gn_2(Z ) - qpk(1 -- Zk)~On_k_l(Z) -- qpk+l(zk -- Z)(~n_k_2(Z), n>k+2 with initial conditions 1, if OK nKk ~n(z) = 1 - pk + (pz)k, if n = k 1--pk(l+q)+2qpkz k+(pz) k+l, /f n=k+l. PROOF. The desired result follows by writing (3.1) in the form oo P~(z, w) ~ ~n(Z)W n = Pl(z, w), n=0 874 D. L. ANTZOULAKOS ET AL. performing the multiplication in the LHS and considering the coefficients of w ~, n = 0, 1,2,... in the resulting power series equality. [] As far as the probability mass function g~(x) = Pr(X~ = x), x > 0 is concerned, its numerical computation can be easily achieved by launching the vector recursive scheme given in Theorem 2.1 (with matrices At,i, i = 0, 1,..., k being replaced by the special forms described earlier in this paragraph) and using the expression Pr(Xn=x)=fn(x)l' , n>O, x = 0,1,2,...,n; note that, for x = 1, 2,... ,k - 1 no calculations are necessary since in this range we always have f~ (x) = 0. In the special case of iid Bernoulli trials one could avoid working with vector re- currences. Instead he may exploit the following effective recursive scheme which ensues easily from the result established in Theorem 3.1. THEOREM 3.2. If Z1, Z2,... , Z n is a sequence of iid Bernoulli trials, the probabil- ity mass function gn(X) = Pr(Xn = x) of the random va~iable Xn satisfies the recursive scheme gn(X) = gn-l(X) -~- pgn-l(X -- 1) - pgn-2(x - 1) - qpk(gn-k-l(X) --gn-k-l(X -- k)) - qpk+l(g~_k_2(x -- k) - g~-k-2(x - 1)), with initial conditions n>k+2, x>0 g~(x) = O, ff x<0 or x > n 1, if x=0 00 /or 1-p/, if x=O gk(x) = pk if x = k 0, if l~x~k-1 1--pk(l +q), i/ x=0 2qp k, if x = k gk+l(X) = pk+l, if X = k + 1 O, if l\ 0 in the recursive formula given in Theo- rem 3.1 by the power series oo ~.(z)=~g~(x)z x x=O and then consider the coefficients of z x on both sides of the resulting identity. [] Although one can always resort to Theorem 2.2 to evaluate the exact distribution of Xn, when n and k become large the calculations might get time consuming. In this case ON THE DISTRIBUTION OF RUN LENGTH 875 n=15, k=2,1~0.5 0.16 ............................................................................................................................................................ Jl 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 n-lS, k,,2, p-O.S 0.35 ............................................................................................................................................. o++ r 0.25 i 0.20 ! 0.15 i 0.10 0,05 i o+1 . . . . . . . . - o 0 1 2 3 4 5 6 7 8 0 10 11 12 13 14 15 n=15, k'--3, p=0.5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 X n=lS, ~ 1~0.9 0 1 2 3 4 5 5 7 8 9 10 11 12 13 14 15 x n:15, k=S, 1~-0.5 0.90 ................................................................................................................................................. 0+00 0,70 ~ 0.60 ~' o,so 0.30 ~ 0.20 0.10 0.00 . . . . . \|~ . ~ + ~ . -- . . . . . . . 0 1 2 3 4 5 6 7 5 9 10 11 12 13 14 15 x 0.25 O.2O 0.15 o. 0.10 0.05 0.00 n=lS, k---S, i~,Q.9 0 1 2 3 4 5 6 7 0 9 10 11 12 13 14 15 Fig. 1. Probability mass function of Xn for various n, k. the investigation of the asymptotic distribution of Xn, will be quite valuable. Results of this flavour can be developed by appealing to the celebrated Chen-Stein method to settle an adequate Poisson convergence. Barbour et al. (1992) have provided a total variation distance bound for the joint distribution of runs of several lengths (see p. 244). Since applying a functional on the multivariate random variables involved does not increase the total variation distance, the upper bound offered there can be exploited for deriving an (asymptotic) estimate of the distribution of X~. However we are not going to pursue this issue in the present article. The interested reader can urged to consult the monograph of Barbour et al. (1992) and work out the details of the aforementioned approach. In Fig. 1 the probability mass function of Xn has been pictured for several values of n, k. Theorem 3.1 can also be used for the derivation of a recursive formula for the raw moments of Xn. To this end, we observe first that the moment generating function E[exp(zXn)] of Xn can be expressed as E[exp(zXn)] = ~n(eZ). Accordingly, replacing z by e z in the recursive formula provided by Theorem 3.1 we may easily derive a recursive 876 D. L. ANTZOULAKOS ET AL. scheme for it. Recalling next that dr z=0 -d-zTzE[exp(zXn)] , if r _> 1 E(X~) Pn,r 1, if r=O and making use of the well known formula dz r (ekZE[exp(zXn)]) = kr-i Pn,i i=0 we may readily verify the following theorem. THEOaEM 3.3. The raw moments Pn,~, r > 1, of the random variable Xn satisfy the recursive scheme #,~,~ = #n-l,~ + P Z (#n-l,i - #n-2,i) - qpk#n-k-l,r i=0 § (:)(kr-i(.n_k_l,i--P]~n_k_2,i)-~p]An_k_2,i) , n ~ k§ i=0 with initial conditions {o, ]~n,r ~--- kr p k, 2krqp k + (k + 1)rp k+l, if O k + 2. Applying this formula for s = k + 2, k + 3,... , n and summing up all the resulting equations we get ~tn -- ~tk-I-1 z P(Pn-1 -- Pk) + (n-- k -- 1)qpk(kq + p), n > k+2. If we replace next #k, Pk+l (see Theorem 3.3) we may easily obtain the following first order difference equation for the means Pn = E(Xn): t-tn : P#n--1 + kqp k + (kq + p)pk (p + (n - kq)), n > k + 1. For numerical calculation of Pn, n > 0 by the aid of the last formula it suffices to recall the initial conditions #n = 0, 0 < n < k, #k = kP k. Moreover, one could easily derive the solution of the above difference equation as #~ = E(Xn) = pk(k + (n - k)(kq + p)), n > k. ON THE DISTRIBUTION OF RUN LENGTH 877 Needless to say, the same expression can be established by expanding the means generating function oo = E(Xn)w n : n=O 1)) (1 - w) 2 which is effortlessly deduced by a direct application of Theorem 2.4. It is noteworthy that Theorem 3.1 can also be used for the derivation of a recursive scheme for the factorial moments dr z=l E(Xn(Xn - 1)..-(X. - r + 1)) = ~z~n(Z) . The details are left to the reader. Closing this section we mention that the approach used here for the study of the random variable Xn can be easily modified to cover the more general case where the sequence of trials exhibits a first order Markov dependence. To this goal, let us assume that Z1, Z2, 9 9 9 Zn is a sequence of Markov dependent trials, with transition probabilities defined by Pij = Pr(Zt+l = j I Zt = i), t > 1, 0 < i, j < 1 and initial probabilities Pr(Z1 = j) = pj, j = 0, 1. Using exactly the same state definition as in the lid case it can be readily verified that the transition probability matrix At,o = Ao takes on the form P00 Pol Pl0 0 A0 = PlO 0 PlO 0 PlO 0 0 ...0 0 0- pn--. 0 0 0 0 ... Opn 0 0-..0 0 0 0 ---0 0 0 (k+l) • while At# = At, i = 1, 2,..., k - 1, will be (k + 1) x (k + 1) matrices with all their entries 0. Matrix At,1 = A1 will have all its entries 0 except for the entry (k + 1, k + 1) which equals Pn. Finally, At,k = Ak will have all its entries 0 except for the entry (k, k + 1) which equals Pl 1. Now making use of the formula ( L )1 9 (z,w)=l+W~l(Z) I-w Aiz i 1' i=0 / with t~l (Z) = (P0, Pl, 0,... , 0)1• we get 9 (z, w) - Q1 (z, w) Q2(z, ) 878 D.L. ANTZOULAKOS ET AL. where and Q1 (z; w) = 1 - w(c~ + pllZ) Jr- w2~P11 z -- Wkplpk111 (1 - z k) -- wk+lpkll[Pl(Zk -- Z) + ~(1 - zk)]- wk+2pk?l'~(zk -- Z), Q2(z;w) = 1 - w(1 + c~ +puz) + w2[c~ +plxz(1 + c~)] - 3 pllz + k+iz(1 - z k) + k+2ZPii(zk -- z) a ---- Pll -- POl, /3 ---- PlOPO1Pk111, O ~ = POl -- Pl. These expressions can be exploited to establish recurrence relations for the probabil- ity generating functions, probability mass functions and means of the random variable X~ under the Maxkovian set up. The interested reader may carry out the respective calculations in exactly the same way as in the iid case. 4. Conditional distribution In this section it is assumed that the composition of the observed sequence is known, that is to say, the number of successes and failures are fixed quantities. The probabilities of our interest, therefore, become conditional ones. As elucidated in the next section outcomes of this nature are of primary interest in the development of tests of randomness in sequences of independent binary trials. Let us assume again that we have a fixed number of Bernoulli trims Z1, Z2,..., Zn with success probability p = P(Xi = 1) and failure probability q = P(Xi = O) = 1 - p, i = 1, 2,..., n. Our intention is to investigate the conditional distribution of the run statistic Xn given the number Sn -- n - y (0 < y < n) of successes in the n lid trials. Since Sn is a sufficient statistic for p, the conditional distribution we axe looking at does not depend on p. In this section we shall use the notation O(z, w;p) and ~o~(z; p) instead of ~(z, w) and 9~n(Z), respectively, that is o~ oo (4.1) (Pn(Z;p)= EPr(Xn=x)zX, r E(fln(Z;p)wn. x~O n=O Let also oo (4.2) r = EPr(X~ = x I Sn = n- y)z x x=O denote the probability generating function of the conditional distribution of Xn given that Sn = n- y. The next theorem provides a formula for the double generating function of the quantity y THEOREM 4.1. The double generating function of an(z;y), y = 0, 1,..., n = y, y + 1,..., is given by as(z; y) n = 9 z, (i + V;- y=O n=y ON THE DISTRIBUTION OF RUN LENGTH 879 PROOF. Replacing Pr(X~ = x) in (4.1) by the sum Pr(Xn = X) ~t = EPr(Xn=xlSn=n-y) Pr(Sn=n-y) y=O = pn Pr(X~ = x I S~ = n- y) y----O and making use of the expression (4.2) we deduce ~n(z;P) : ~-~ (y)pn (q)Y~)n(Z;B) y=O or equivalently 9 (z,w;p)=~ a~(z;y) n=0 y=0 Setting t = q/p in the last expression we obtain (~)n. z, w; = a,~(z; y)t y n=0 y=0 and the required result follows immediately on replacing w by (1 + t)w. [] The outcome of Theorem 4.1 can be exploited to derive an explicit formula for the conditional distribution of X~ given the number of successes S,~. Specifically we have the following interesting result. THEOREM 4.2. The conditional probability Pr(Xn = x I S,~ = n - y) is given by (~)--1 y+l r r--i i Pr(Xn . . . . x[Sn n y) E E E E(-1)~+i+Y'-'= r=0 i=0 jl=0 j2=0 i +a- x (y+l)r (~)(r jl i)(j2)(r a 1)(y +b)b where a = x - i + j2 - k(jl + j2) and b = n - y- kr- i - a. PROOF. Making use of (3.1) we deduce ~ (Z' (I + t)w;1--~ ) = ~ (1-- WZ -- Wk(I -- zk) -- Wk+I(zk -- Z) ) ~ W)--O---W--~) (wt)Y which can be written in the form p 1 - w~- (1 - z ~) - w~+l(z ~ - z) "+~ ---- wY \ ~ -w ] ( 1 ~Y+' (l_w k(1-zk)+w(zkl_wz -z)) y+I 880 D. L. ANTZOULAKOS ET AL. Expanding the RHS in a power series with respect to w and employing the conven- tions (•) = 0 if m < 0 and (01) = 1, we may easily arrive at the expression an(z;y) -- E E E (-1)r y§ y+ l m=0 r=0 i=0 S r • m n>y wheres=n y kr - i - m. The desired result follows immediately by a further expansion, by the aid of the binomial formula, of the powers appearing in the summand; c.f. (4.1), (4.2). [] 5. Non-parametric tests of randomness One of the widely known, oldest and easiest method of testing for random versus non-random ordering in a sequence of two types of symbols, is the classical runs test which has become a necessary addition in all contemporary non-parametric statistics textbooks, Bradley (1968), Gibbons and Chakraborti (1992). This test is based on the total number of runs, a run being any string of identical symbols which are followed and preceded by a different symbol or no symbol at all. An alternative test can be established by working with the length of the longest run in the observed sequence. Since an unusually long run indicates a tendency for like objects to cluster and, hence, the presence of a trend, Mosteller (1941) suggested a test for randomness based on the length of the longest run. The computation of the critical values of this test calls for the evaluation of the conditional distribution of the length of the longest run in n trials, given the number S~ = n - y of the successes. We are now going to investigate a new test of randomness based on the statistic X~ introduced in the previous section (with k being a fixed pre-determined integer). Using an upper tailed test, the null distribution will be directly related to the conditional event Xn>clS~=n-y, where c is specified in terms of the significance level of the test. It is therefore apparent why an outcome like the one established in Theorem 4.2 is of special importance. As mentioned earlier, in the 1940s, when the interest in the theory of runs was quite high, two different randomness tests were proposed: The classical runs test which was based on the total number Rn of runs of either type and the longest-run test which utilizes the length Ln of the longest success run. Recently, Agin and Godbole (1992) using the classical runs test as a model, developed a new exact test based on (a conditional version of) the total number Nn,k of non-overlapping success runs of length k. This new test was found to be significantly more powerful in detecting certain types of clustering (non-randomness), than the classical runs test. Motivated by this result, Koutras and Alexandrou (1997), explored the performance of tests based on the total number Gn,k of success runs of length at least k, and on the total number Mn,k of overlapping success runs of length k. The test based on M~,k was found to be more powerful than the tests based on Gn,k and Nn,k. In the sequel, we conduct a systematic numerical experimentation in order to assess the performance of the randomness test based on the conditional distribution of the ON THE DISTRIBUTION OF RUN LENGTH Table 1. Empirical power/first-order Markov dependence model. Parameters a = 0.10 a = 0.05 a = 0.01 p n Xn Mn,k Xn Mn,k Xn Mn,k 0.99 50 0.98 0.99 0.99 1.00 0.96 0.88 0.99 100 0.98 0.98 0.98 0.97 0.98 0.85 0.99 150 0.94 1.00 0.97 1.00 0.93 0.81 0.95 50 0.92 0.92 0.91 0.92 0.90 0.76 0.95 100 0.86 0.99 0.80 0.95 0.87 0.87 0.95 150 0.82 0.98 0.74 0.97 0.69 0.91 0.65 50 0.74 0.59 0.46 0.25 0.39 0.14 0.65 100 0.62 0.65 0.48 0.54 0.39 0.14 0.65 150 0.63 0.66 0.49 0.54 0.41 0.15 881 Table 2. Empirical power/cyclical clustering model (with cycle length equal to 10). Parameters a = 0.10 a = 0.05 a = 0.01 p n Xn Mn,k Xn Mn,k Xn Mn,k 0.99 50 0.65 0.32 0.68 0.35 0.66 0.04 0.99 100 0.78 0.38 0.51 0.26 0.50 O.11 0.90 50 0.68 0.27 0.69 0.14 0.65 0.02 0.90 100 0.49 0.29 0.47 0.18 0.48 0.29 0.80 50 0.62 0.16 0.55 0.10 0.46 0.09 0.80 100 0.51 0.17 0.45 0.11 0.38 0.02 0.65 50 0.77 0.16 0.59 0.08 0.56 0.02 0.65 100 0.51 0.13 0.43 0.07 0.42 0.01 Xn (sum of the exact lengths of runs of length at least k). As already indicated, our tests are upper tailed and the critical values for rejection are determined by the aid of Theorem 4.2. The empirical power of the new randomness test was compared to the empirical power of the randomness test based on Mn,k. The evaluation of the operational char- acteristics curves of the test, was achieved by the aid of Monte Carlo techniques. Thus, using specific alternatives 100 non random sequences were generated, and the probability (proportion) of rejecting the null hypothesis was computed. The parametric configurations upon which the comparisons were performed are the following: A. First-order Markov dependence: Pl = 0.5 and p, if the (i - 1)-th trial is a success Pi =- Pl, if the (i - 1)-th trial is a failure for i = 2, 3,.... B. Cyclical clustering (with cycle length equal to 10): The success probabilities 882 D. L. ANTZOULAKOS ET AL. 0.90 i 0.80 0.70 0.60 Empirical Power Against k 3 4 5 6 7 8 9 k 9 W(H1(0.99)/alpha=0.01/n=50) - 1- -W(Hl(0.80)/alpha=0.10/n=100) I I -- -,-- - O(Hl(0.90)/atpha=0.01/n=50) -- e- -O(Hl(0.90)/alpha=0.10/n=50) \| J Fig. 2. Empirical power for various n, k. Ernplrlcal Power Against k 0.80 ~...~- ..................................................................................... -~ 0.. 9 . . . . . "tl-" . . . . . 9 ! 0.70 "'""A ..... i 3 4 5 7 8 9 k I 9 W(H 1(0.65)/a~pha=0.10/n= 100) - -a- - W(H l(O.gO)/alpha--O. 10/n=50) - -A- -- O(H 1(0.65)/alpha=0.10/n= 100) -- o- - O(H 1 (0.90)/alpha=0.10/n=50) Fig. 3. Empirical power for various n, k. Pi, i = 1, 2,... are given by p, if lOr+l<i<lOr+c, r=0,1,2,3,... P/= 0.5, otherwise where c _< 10 is a fixed integer. It is worth mentioning that these sequences are indicative of real situations and have appeared before in certain practical applications (c.f. discussion in Koutras and Alexandrou (1997)). The results of the simulation study, in the case where the sequence of outcomes exhibits a first-order Markov dependence, are displayed in Table 1. The empirical power recorded there was obtained by applying each test for all k = 2, 3,..., 9 and choosing the largest power attained. Apparently the randomness tests based on Xn, are proved to be significantly more powerful than the ones based on Mn,k when the type I error must be kept low (a = 0.01). For higher values of the significance level (a = 0.05 or a = 0.10) the performance of the two tests is comparable. Shifting to the cyclical clustering model (Table 2) we observe that the Xn-based ON THE DISTRIBUTION OF RUN LENGTH Table 3. Empirical power against k/first-order Markov dependence model. 883 Parameters k p a n Statistic 2 3 5 6 7 9 max - rain 0.90 .10 50 xn 0.79 0.77 0.73 0.67 0.67 0.70 0.12 0.90 .10 50 Mn,k 0.82 0.82 0.74 0.78 0.61 0.62 0.21 0.90 .05 50 Xn 0.71 0.69 0.68 0.58 0.60 0.66 0.13 0.90 .05 50 Mn,k 0.86 0.82 0.67 0.63 0.48 0.54 0.38 0.95 .05 100 X~ 0.64 0.67 0.63 0.67 0.60 0.70 0.07 0.95 .05 100 Mn, k 0.88 0.82 0.71 0.65 0.59 0.50 0.38 0.65 .10 100 X,~ 0.52 0.52 0.46 0.37 0.56 0.62 0.16 0.65 .10 100 Mn,k 0.65 0.60 0.41 0.26 0.32 0.28 0.39 test is always superior. In this case, there are instances where the Mn,k-based test leads to extremely low empirical values, while the new one attains significantly higher levels. Closing we mention, that another interesting feature of the Xn-based test is that it is not very sensitive in changes on k, a property that is not present in the Mn,k-based test. 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Nonparamet~ic Statistical Inference, 3rd ed., Marcel Deker, New York. Godbole, A. P. (1990). On hypergeometric and related distributions of order k, Comm. Statist. Theory Methods, 19, 1291-1301. Godbole, A. P. (1992). The exact and asymptotic distribution of overlapping success runs, Comm. Statist. Theory Methods, 21,953-967. Hahn, G. J. and Gage, J. B. (1983). Evaluation of a start-up demonstration test, Journal of Quality Technology, 15, 103-105. Han, Q. and Aki, S. (1999). Joint distributions of runs in a sequence of multistate trials, Ann. Inst. Statist. Math., 51, 419-447. Hirano, K. and Aki, S. (1993). On number of occurrences of success runs of specified length in a two-state Markov chain, Statist. Sinica, 3, 313-320. Hirano, K., Aki, S., Kashiwagi, N. and Kuboki, H. (1991). On Ling's binomial and negative binomial distributions of order k, Statist. Probab. Lett., 11,503 509. Koutras, M. V. (1997). Waiting time distributions associated with runs of fixed length in two-state Markov chains, Ann. Inst. Statist. Math., 49, 123-139. Koutras, M. V. and Alexandrou, V. A. (1995). Runs, scans and urn model distributions: A unified Markov chain approach, Ann. Inst. Statist. Math., 47, 743-766. Koutras, M. V. and Alexandrou, V. A. (1997). Nonparametric statistical randomness tests based on success runs of fixed length, Statist. Probab. Lett., 32, 393-404. Mosteller, F. (1941). Note on an application of runs to quality control charts, Ann. Math. Statist., 12, 228-232. O'Brien, P. C. and Dyck, P. J. (1985). A runs test based on run lengths, Biometrics, 41, 237-244. Panaretos, J. and Xekalaki, E. (1986). On generalized binomial and multinomial distributions and their relation to generalized Poisson distributions, Ann. Inst. Statist. Math., 38, 223-231. Philippou, A. N. and Makri, F. S. (1986). Success runs and longest runs, Statist. Probab. Left., 4, 211-215. Philippou, A. N., Antzoulakos, D. L. and Tripsiannis, G. A. (1990). Multivariate distributions of order k, Part II, Statist. Probab. Lett., 10, 29-35. Stanley, R. P. (1997). Enumerative Combinatorics, Vol. I, Cambridge University Press. Steyn, H. S. (1956). On the univariate series F(t) = F(a; bl, b2,..., bk; t, t2,..., t k) and its applications in probability theory, Proc. Konink. Nederl. Akad. Wetensch. Ser. A, 59, 190-197. Viveros, R. and Balakrishnan, N. (1993). Statistical inference from start-up demonstration test data, Journal of Quality Technology, 25, 119-130. Wolfowitz~ J. (1943). On the theory of runs with some applications to quality control, Ann. Math. Statist., 14, 280-288.
15416	https://artofproblemsolving.com/community/c2307h1026427_inequalities_1?srsltid=AfmBOooI6_xaWIfh-SG797BsaWoQMhFUE1tOVDKHirMDG6bJf9DIB5GA	Olympiad topics : Inequalities #1 Community » Blogs » Olympiad topics » Inequalities #1 Sign In • Join AoPS • Blog Info Olympiad topics =============== Inequalities #1 by proglote, Jan 24, 2012, 11:24 PM (IMO 2001, Problem 2) Prove that for all positive real numbers , Solution. Since the inequality is homogeneous, we may assume From AM-GM, we obtain : But , hence it suffices to prove : I claim that the following inequality holds : Indeed, after dividing both sides by and cross-multiplying, it becomes : But and , so it is convenient to make the substitution The inequality is rewritten as : For this, lest squaring both sides, we apply AM-GM. The equality case clearly occurs at Here, we have and Observe also that is always positive. So it is convenient to use AM-GM in the following form : This may seem more complicated than what we previously had, but fear not, for it is actually much simpler. It suffices, thus, to prove : Multiplying both sides by and taking their square root, one obtains : A bit of awareness reveals that the above follows directly from AM-GM : Equality holds for (Remember wbhy is zero included) Recall, reader, that we have just proven . Summing up the similar ones, we get : We are left to prove the following : But and also , so it's equivalent to : Note also that For convenience, denote and The inequality becomes : But this is obvious because we have (This is easily proven by AM-GM) Remark. The "magic" cubic in is obtained by taking the expression on the left-hand side of that inequality and evaluating it at , and , namely the equality points. The cubic should have the same value in all such points. Furthermore, it should also lie always "beneath" the graph of the left function, so that the derivative of both at should also be the same. This is how it was found. See here for an illustration : (IMO 2008, Problem 2) Let be three real numbers different than such that Prove the inequality : And prove that equality is achieved for infinitely many triples of rational numbers. Solution. Perform the substitution and similarly define Note that we also have Then the condition becomes This is equivalent to Denote and . The inequality rewrites as : Equality holds for Note that if is rational then so is , so it suffices to prove that there are infinitely many triples of rational numbers with and Let Then and So are the roots of From the quadratic formula, those are: Pick rational. Then it suffices to prove that the expression inside the radical (namely the discriminant) is a square of an rational. Pick , with not necessarily coprime. The discriminant is rewritten as : [ \begin{align}(a-1)(a-1 - 4a) = (a-1)(-3a-1) & = (1-a)(3a+1) \ & = \left(1 - \frac{p}{q}\right)\left(\frac{3p}{q} + 1\right) \ & = \frac{(q-p)(3p+q)}{q^2}.] Since the denominator can be factored out of the radical, it suffices to prove that the numerator is a square. We will prove a stronger result, that both and are squares for infinitely many pairs Let and Then multiplying the first equation by and summing it with the second, we have : , and multiplying the first by and summing with the second gives Let and be even, with and We have and Hence, for any choice of two positive integers and with , we obtain a possible value for . Also, all of these satisfy , which is easily seen. For example, choosing and we get , and as we have seen, are also rational from the quadratic formula. Hence, there are infinitely many triples for which equality holds, and thus infinitely many triples I think the second problem is easy for IMO #2 This post has been edited 2 times. Last edited by proglote, Jan 24, 2012, 11:26 PM inequalities Many comments Comment Comment 3 Comments The post below has been deleted. Click to close. This post has been deleted. Click here to see post. The first problem has like a 5 line solution using Jensen's Inequality. by dinoboy, Jan 25, 2012, 12:54 AM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Also, the first is destroyed by Holder's inequality. by GlassBead, Jan 25, 2012, 1:03 AM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Yes, I am aware of other methods, in fact there is one with assumption which is more simple. But, I like the one with AM-GM, and I am not very prone to Jensen or Holder. I also hadn't yet seen a solution with the method I used, so rather than retyping well-known ones I decided be original Thanks for the comments. by proglote, Jan 25, 2012, 2:22 AM Report I'll add some problems from various topics, i.e. number theory, functional equations, algebra, inequalities, combinatorics, and maybe even geometry. proglote Archives December 2012 Geometry #7: A nice triangle center August 2012 Geometry #6: Tangencies July 2012 Geometry #5: Altitudes and medians June 2012 Algebra #3 April 2012 Nontrivial diophantine equations March 2012 Geometry #4 February 2012 Algebra #2 Geometry #3 Number Theory #3 January 2012 Geometry #2 Inequalities #1 Algebra #1 Combinatorics #1 December 2011 Number Theory #2 Geometry #1 Number Theory #1 Shouts Submit nice i arrived only 6 days later by Kanep, Sep 5, 2020, 1:34 PM 6 years! by justin1228, Aug 30, 2020, 5:58 PM oh hi..... by NewAlbionAcademy, Jul 18, 2014, 2:52 AM great blog!! by spikerboy, Mar 2, 2014, 8:01 AM Hope everything is going well at Romanian Masters! Good Luck! by IDMasterz, Mar 1, 2013, 9:16 AM Nice blog and good problems I;m suprised I'm the 2nd shout by Dynamite127, Jan 9, 2013, 10:36 PM nice blog! by tc1729, Jan 30, 2012, 1:30 AM 7 shouts Tags geometrynumber theoryalgebraprojective geometrycombinatoricsFunctional EquationsinequalitiesInvariants Monovariantsprobabilityrandomroots of unity About Owner Posts: 959 Joined: Jun 9, 2011 Blog Stats Blog created: Jun 9, 2011 Total entries: 16 Total visits: 18949 Total comments: 27 Search Blog Something appears to not have loaded correctly. Click to refresh. a
15417	https://en.wikipedia.org/wiki/R2	Jump to content R2 Català Čeština Deutsch Español Français 한국어 Italiano Nederlands 日本語 Polski Slovenčina Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska 中文 Edit links From Wikipedia, the free encyclopedia For the Wikipedia R2 criterion for speedy deletion, see Wikipedia:Criteria for speedy deletion § R2. R2, R02, R.II, R.2 or R-2 may refer to: Entertainment [edit] R2 (Rock'n'Reel), a British music magazine R2-D2, a character from Star Wars films and books, nickname R2 Code Geass: Lelouch of the Rebellion R2, a 2008 anime series Resistance 2, a video game Relapse 2, a 2009 album by rapper Eminem Patricio Rey y sus Redonditos de Ricota or R2, an Argentine rock band Region 2, a DVD region code Science and technology [edit] R2 or r2 (pronounced R-squared), the coefficient of determination of a linear regression in statistics R2, the two-dimensional real coordinate space in mathematics R2: Risk of explosion by shock, friction, fire or other sources of ignition, a risk phrase in chemistry R2 RNA element, a cis-acting element identified in R2 retrotransposons which is involved in priming the reverse transcription process ATC code R02 Throat preparations, a section of the Anatomical Therapeutic Chemical Classification System Haplogroup R2 (Y-DNA), a human Y-chromosome haplogroup receptor 2, the second in line of a series of cellular receptors, generally at the end of an acronym Gangrene (ICD-10 code) Robonaut 2, a NASA humanoid robotic development project R2 signalling, a family of telephony protocols for line and register signalling Computing [edit] .r02, a RAR file extension Radare2, an open source reverse engineering framework also known as r2 Transportation [edit] R2 (Rodalies de Catalunya), a commuter rail line in Barcelona, Catalonia, Spain R2 (RER Vaud), an S-Bahn line in the canton of Vaud Wilmington/Newark Line (R2 Wilmington/Newark line) Warminster Line (R2 Warminster line) Radial Road 2 or R-2, an arterial road of Manila, Philippines Autopista Radial R-2, a Spanish radial motorway connecting Madrid to Guadalajara and passing through Alcobendas and connecting to N-320 R2 expressway (Slovakia) R2 Marine Dr, an express bus route in Metro Vancouver, British Columbia, Canada R2 road (Zimbabwe), a road connecting Plumtree and Harare Orenburg Airlines (IATA code) Vehicles [edit] R2, a sub-class of Group R rally cars DFW R.II, a 1918 German bomber aircraft Fiat R.2, a 1919 Italian reconnaissance aircraft HMS Zest (R02), a World War II British Royal Navy Z-class destroyer Jaguar R2, Jaguar Racing's car for the 2001 Formula One season Linke-Hofmann R.II, a World War I German bomber aircraft ORA R2, a Chinese electric city car Panzer 35(t), a Czech tank whose Romanian model was known as the R-2. Polikarpov R-2, a Soviet Union copy of the 1931 British Airco DH.9A light bomber aircraft Ross R-2 Ibis, a glider Subaru R2, a 2003 Japanese car USS R-2 (SS-79), a 1918 R-class coastal and harbor defense submarine of the US Navy Other uses [edit] BBC Radio 2, a British radio station R-2 (missile), an improved version of German V-2 rocket manufactured by the Soviet Union R2 is a rank of United States research university in the Carnegie Classification of Institutions of Higher Education See also [edit] Windows Server 2003 R2 Windows Server 2008 R2 SQL Server 2008 R2 2R (disambiguation) R–2R ladder, a resistor ladder Retrieved from " Category: Letter–number combination disambiguation pages Hidden categories: Short description is different from Wikidata All article disambiguation pages All disambiguation pages
15418	http://principalsguide.org/importance-of-designated-open-forum-status/	Importance of designated open forum status \| Principal's Guide to Scholastic Journalism Navigation Menu Home About Principal & Adviser Law & Ethics Philosophy Media Content Resources Home Home About About Principal & Adviser Principal & Adviser Law & Ethics Law & Ethics Philosophy Philosophy Media Content Media Content Resources Resources Home»Law & Ethics» Importance of designated open forum status Sun 28 Importance of designated open forum status Posted by John Bowen in Law & Ethics \| 0 comments Why choose open public form status for student media? Consider these options. There is no requirement that any government agency establish a forum of any kind. But once a government does establish a forum, it cannot dictate the content of that forum. Jurisprudence sees three types of forums: open, limited, closed. The closed forum is a place that traditionally has not been open to public expression. Examples, in schools, could be newsletters or other means of communication not open to public use. So long as restrictions are reasonable and not based on a desire to suppress certain viewpoints, the government may close public access to them. The open or traditional public forum is a place with a long history of expression, such as a public park or street corner. The government can only impose content-neutral time, place and manner restrictions on speech in this forum. To override the open, public forum status, the government would have to show a compelling interest. The limited forum has the most problematic history. It is a space with a limited history of expression activity, usually only for certain topics or groups. A meeting hall or public-owned theater are examples. The government may limit access when setting up a forum, but may still not restrict expression unless there is a compelling interest. Schools, as government institutions, may, by policy or practice open student media for indiscriminate use by the public or some segment of the public. A designated public forum enables students to make decisions of content, thus empowering them to practice critical thinking and civic engagement roles. Educational value of the designated open forum is mirrored by the fact most schools have mission statements identifying these as essential life skills for students to learn while in school. Prior review and a lack of trust in the product (students) schools are expected to produce undermines the very missions school officials say are among their most important. Studies have clearly shown that students, and communities in general, do not understand the importance of the First Amendment. One reason may be that students are not allowed to practice what they are taught while in schools and thus do not believe the theories of the democratic system. Search for: Blogroll • Quill and Scroll Society • Journalism Education Association • JEA Scholastic Press Rights Commission Ordering print copies of The Principal’s Guide to Scholastic Journalism To order print copies of The Principal's Guide, go to quillandscroll.org. Print copies should be available around Nov. 1. Tags 4C'sAdviser Code of EthicsASNEbudgetingChildren's Internet Protection Actcoaching writersCommon CoreCommon Core State Standards. Social Media ToolboxCuring HeazelwoodcyberbullyingDeanElements of Journalismethical fitnessethicsFirst AmendmentFreedom ForumFuture of the First Amendmentgreen light ethicsHazelwoodInternet filteringJEA Certification programjournalismJournalism Education AssociationKnight Commissionlegal definitionsmaestroModel editorial policyMorseNASP Student Code of EthicsNational School Boards AssociationPartnership for 21st Century Skillsprior restraintprior reviewprior viewpublishing names and photos in student mediaRed lightresourcesRushworth Kidderstate free expression lawsstate legislationstudent decision makingstudent empowermentStudent Press Law CenterTechology needsTinker Recent Comments TRENDWATCH 20/20: OUTLINE FONTS - Journalism leadership education consultant on About The [Censored] History of Student Journalism in America - Watch Us Rise on The First Amendment and student media Teaching ISTE standards and 21st Century Skills through scholastic journalism – Digital Education Leadership on The First Amendment and student media 1. Law and Ethics – SoozKnowzNewz on Six principles behind news literacy Prior review imposes ineffective educational limits on learning, citizenship \| jeasprc.org on Why avoiding prior review is educationally sound Recent Posts A message from the American Society of News Editors Introduction Yearbook ethical guidelines for student media Journalism organization resources Importance of designated open forum status Archives October 2013 July 2013 June 2013 Designed by Elegant Themes \| Powered by WordPress
15419	https://math.stackexchange.com/questions/851728/what-does-extension-mean-in-the-axiom-of-extension	elementary set theory - What does "extension" mean in the Axiom of extension - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more What does "extension" mean in the Axiom of extension Ask Question Asked 11 years, 3 months ago Modified8 years ago Viewed 5k times This question shows research effort; it is useful and clear 21 Save this question. Show activity on this post. I am learning Set Theory from the book Naive Set Theory by Halmos as part of my course. The first chapter is on the Axiom of Extension. I understand what it is but what I don't understand is why it has the word "extension" in the title for this axiom. I can't understand how equality has the same meaning as extension. Please can someone clarify this. elementary-set-theory terminology axioms Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 31, 2017 at 17:10 user124485 user124485 asked Jun 29, 2014 at 23:26 user124485 user124485 5 2 The name is axiom of extensionality.Asaf Karagila –Asaf Karagila♦ 2014-06-29 23:28:48 +00:00 Commented Jun 29, 2014 at 23:28 2 See Axiom of extensionality : the role in the theory is simple; its historical background is more complex : see this post. In "ancient times" a property or predicate has an extension, formed by all the objects which satisfy the predicate. Thus, in modern set-theory, the meaning of extensionality in the corrisponding axiom is " what is relevant for the 'identity' of a set are only its elements, i.e. its extension".Mauro ALLEGRANZA –Mauro ALLEGRANZA 2014-06-30 06:22:16 +00:00 Commented Jun 30, 2014 at 6:22 Related article: en.wikipedia.org/wiki/Extensionalityuser9464 –user9464 2017-01-31 17:19:17 +00:00 Commented Jan 31, 2017 at 17:19 @AsafKaragila It’s called “axiom of extension” in the book. books.google.it/…egreg –egreg 2017-09-25 07:58:58 +00:00 Commented Sep 25, 2017 at 7:58 @egreg: I used to hear how great is this book. But as time progresses forward, I grow to dislike it more and more.Asaf Karagila –Asaf Karagila♦ 2017-09-25 09:08:36 +00:00 Commented Sep 25, 2017 at 9:08 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 33 Save this answer. Show activity on this post. The axiom states that two sets are equal if they have the same elements, i.e. they are equal in "extension" (scope, content), as opposed to equality in "intension" (meaning, concept). For example, the set of black US presidents is currently equal in extension to the set containing Barack Obama as a single element, but they are different in intension. The axiom of extension means that the set theory only deals with the content of sets, not with the concepts used to form them. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 29, 2014 at 23:52 ConifoldConifold 12.1k 3 3 gold badges 35 35 silver badges 61 61 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. I like to think of it as: Two sets A A and B B are equal if and only if B B is an extension of A A, and A A is an extension of B B. That is, two sets are equal iff A⊂B A⊂B and B⊂A B⊂A. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 25, 2017 at 7:47 user5826user5826 12.6k 17 17 gold badges 72 72 silver badges 156 156 bronze badges 2 2 The term extension is a technical term in logic.Asaf Karagila –Asaf Karagila♦ 2017-09-25 08:06:27 +00:00 Commented Sep 25, 2017 at 8:06 1 @AsafKaragila I was just trying to provide some context and another way of using the word extension to give intuition to the OP.user5826 –user5826 2017-09-25 16:34:23 +00:00 Commented Sep 25, 2017 at 16:34 Add a comment\| You must log in to answer this question. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 10Precise meaning of "extension"? 1Abstract Objects in Logic 1I need some assistance in certain sections of Naive set theory by Paul R Halmos Related 5Need help understanding Axiom of Extensionality 4What does Halmos mean when he says "to get something for nothing" (in the context of the axiom of specification) 5Axiom of Powers 6What are all underlying assumptions in the definition of Cartesian Product in Naive Set Theory? 0Axiom of Specification as in Halmos' Naive set theory 0Can we take the intersection of ALL successor(inductive) sets? 1How does the axiom of extension relate belonging to equality? 1Axiom of Specification 2Why doesn't Halmos include the Axiom of Regularity? 7On The Role of The Axiom of Specification Hot Network Questions Origin of Australian slang exclamation "struth" meaning greatly surprised How different is Roman Latin? 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15420	https://projecteuclid.org/journals/analysis-and-pde/volume-13/issue-6/On-uniqueness-results-for-Dirichlet-problems-of-elliptic-systems-without/10.2140/apde.2020.13.1605.full	On uniqueness results for Dirichlet problems of elliptic systems without de Giorgi–Nash–Moser regularity Sign InView CartHelp Email Password Forgot your password? [x] Show [x] Remember Email on this computer [x] Remember Password Please wait... No Project Euclid account? Create an account or Sign in with your institutional credentials We can help you reset your password using the email address linked to your Project Euclid account. Email Registered users receive a variety of benefits including the ability to customize email alerts, create favorite journals list, and save searches. Please note that a Project Euclid web account does not automatically grant access to full-text content. An institutional or society member subscription is required to view non-Open Access content. 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DOI: 10.2140/apde.2020.13.1605 ABOUT FIRST PAGE CITED BY REFERENCES DOWNLOAD PAPERSAVE TO MY LIBRARY PERSONAL SIGN IN Full access may be available with your subscription Email Password Forgot your password? [x] Show [x] Remember Email on this computer [x] Remember Password No Project Euclid account? Create an account or Sign in with your institutional credentials PURCHASE THIS CONTENT PURCHASE SINGLE ARTICLE Price: ADD TO CART Includes PDF & HTML, when available PURCHASE SINGLE ARTICLE This article is only available to subscribers. It is not available for individual sale. This will count as one of your downloads. You will have access to both the presentation and article (if available). DOWNLOAD NOW This content is available for download via your institution's subscription. To access this item, please sign in to your personal account. Email Password Forgot your password? [x] Show [x] Remember Email on this computer [x] Remember Password No Project Euclid account? Create an account My Library [x] You currently do not have any folders to save your paper to! Create a new folder below. Create New Folder SAVE > Abstract We study uniqueness of Dirichlet problems of second-order divergence-form elliptic systems with transversally independent coefficients on the upper half-space in the absence of regularity of solutions. To this end, we develop a substitute for the fundamental solution used to invert elliptic operators on the whole space by means of a representation via abstract single-layer potentials. We also show that such layer potentials are uniquely determined. Citation Download Citation Pascal Auscher.Moritz Egert."On uniqueness results for Dirichlet problems of elliptic systems without de Giorgi–Nash–Moser regularity."Anal. PDE 13(6)1605 - 1632,2020. Information Received: 28 March 2017; Accepted: 13 August 2019; Published: 2020 First available in Project Euclid: 22 September 2020 zbMATH: 07271841 MathSciNet: MR4150257 Digital Object Identifier: 10.2140/apde.2020.13.1605 Subjects: Primary: 35A02 , 35J57 Secondary: 35C15 , 35J50 , 42B25 Keywords: Dirichlet problems , elliptic systems , single-layer operators , uniqueness of solutions Rights: Copyright © 2020 Mathematical Sciences Publishers JOURNAL ARTICLE 28 PAGES DOWNLOAD PDF+ SAVE TO MY LIBRARY GET CITATION My Library [x] You currently do not have any folders to save your paper to! Create a new folder below. Create New Folder SAVE > Folder Name Folder Description SAVE Next Article> Anal. PDE Vol.13 • No. 6 • 2020 MSP Subscribe to Project Euclid Receive erratum alerts for this article Pascal Auscher, Moritz Egert "On uniqueness results for Dirichlet problems of elliptic systems without de Giorgi–Nash–Moser regularity," Analysis & PDE, Anal. 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15421	https://schools.utah.gov/curr/mathematics/_mathematics_/_core/_math_core_guides_tab_/CoreGuidesGrade3OperationsAlgebraic.pdf	Operations and Algebraic Thinking Core Guide Grade 3 Represent and solve problems involving multiplication and division within 100 (Standards 3.OA.1–4 and Standard 3.OA.7). Standard 3.OA.1 Interpret products of whole numbers, such as interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7. Concepts and Skills to Master • Understand multiplication as combining equal groups of objects • Model skip counting on a number line • Understand that in a multiplication equation, the first factor equals the number of groups and the second factor equals the number in each group • Find the total number of objects within equal groups (5 × 7 = 35; 5 groups of 7 is 35) • Write multiplication expressions and equations to represent pictures • Draw pictures to represent multiplication expressions and equations Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.2 Interpret whole-number quotients of whole numbers 4.OA.1, 4.OA.2 Interpret and solve a 3.OA.3 Use multiplication and division to solve word problems involving equal groups, arrays, and measurement multiplication equation as a comparison quantities 4.NBT.5 Multiply multi-digit whole 3.OA.4 Determine the unknown whole number in a multiplication or division equation relating three whole numbers numbers. 5.NBT.5 Fluently multiply multi-digit 3.OA.5 Apply properties of operations as strategies to multiply and divide whole numbers 3.OA.6 Understand both division as an unknown-factor problem and the relationship between multiplication and 4.NF.4, 5.NF.4 Apply and extend division previous understandings of 3.OA.7 Fluently multiply and divide within 100 multiplication to fractions Critical Background Knowledge from Previous Grade Levels • Use addition to find the total number of objects in an array (2.OA.4) • Skip count by fives and tens (2.NBT.2) Academic Vocabulary equal groups, array, multiplication, factor, product, equation Suggested Models Suggested Strategies Write an equation that can help you find the total number of points on the stars. 3 × 5 = 15 Frank bought six boxes of crayons. Each box of crayons has 8 crayons in it. How many crayons does he have? • Model equal groups with various counters • Discuss real-life situations where objects are in groups • Use and compare number lines, bar models, and area models 3.OA.1 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Represent and solve problems involving multiplication and division within 100 (Standards 3.OA.1–4 and Standard 3.OA.7). Standard 3.OA.2 Interpret whole-number quotients of whole numbers. For example, interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into eight shares (partitive), or as a number of shares when 56 objects are partitioned into equal shares of eight objects each (quotative). Concepts and Skills to Master ● Understand that division may represent two different situations: partitive (fair sharing) and quotative (measurement) ● Understand division as repeated subtraction to find the number of equal groups ● Find how many equal groups can be made from a certain number of objects ● Find how many objects can be shared equally among a certain number of groups ● Solve and interpret division problems ● Model a division equations using pictures, objects, or numbers ● Use objects and drawings to represent equal groups ● Use objects, drawings, expressions, and equations to represent division situations Teacher Note: This standard focuses on two distinct models of division: partitive and quotative. Partitive or fair share models provide students with the total number of objects and the number of groups. Students must solve for the number in each group. Quotative or measurement models provide students with the total number of objects and the number of objects in each group. Students must solve for the number of groups. Students are not expected to know or produce the terms partitive and quotative but should be exposed to them. Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.1 Interpret the products of whole numbers 4.OA.2 Multiply or divide to solve word problems involving multiplicative 3.OA.3 Use multiplication and division to solve word problems involving comparison equal groups, arrays, and measurement quantities 4.OA.3 Solve multi step word problems with all operations 3.OA.4 Determine the unknown number in a multiplication or division 4.NBT.6 Find whole-number quotients with up to 4-digit dividends and 1-digit equation relating three whole numbers divisors 3.OA.5 Apply properties of operations as strategies to multiply and divide 5.NBT.6 Find whole-number quotients with up to 4-digit dividends and 2-digit 3.OA.6 Understand both division as an unknown-factor problem and the divisors relationship between multiplication and division 5.NBT.7 Solve equations involving decimals with all operations 3.OA.7 Fluently multiply and divide 5.NF.7 Apply and extend previous understandings of division to fractions Critical Background Knowledge from Previous Grade Levels ● Add and subtract within 20 (2.OA.2) ● Use addition to find the total number of objects arranged in an array (2.OA.4) Academic Vocabulary quotient, dividend, divisor, divide, equal groups, whole numbers 3.OA.2 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Suggested Models Suggested Strategies Partitive Division: There are 12 cookies. If you put them in three bags, how many cookies will be in each bag? Quotative Division: There are 12 cookies. If you give put 3 cookies in each bag, how many bags will you fill? Partitive Quotative ● Use manipulatives/objects or other models ● Use repeated subtraction ● Drawing pictures ● Model equal groups ● Model equal groups with various counters ● Discuss real-life situations where objects are in groups ● Use and compare number lines, bar models, and area models 3.OA.2 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Represent and solve problems involving multiplication and division within 100 (Standards 3.OA.1–4 and Standard 3.OA.7). Standard 3.OA.3 Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities. For example, use drawings and equations with a symbol for the unknown number to represent the problem. Concepts and Skills to Master • Determine the operation based on the situation in the context of a word problem (avoid relying on keyword strategies) • Use numbers and symbols to represent word problems (×, ÷, =, and a variety of symbols for unknowns) • Solve the following multiplication and division situations. (See: TABLE 2. Common multiplication and division situations): o Equal Groups of Objects/Product Unknown word problems (There are 3 bags with 4 plums in each bag. How many plums are there in all?) o Equal Groups of Objects/Group Size Unknown word problems (24 plums are shared equally into 3 bags. How many plums will be in each bag?) o Equal Groups of Objects/Number of Groups Unknown word problems (24 plums are packed equally into some bags. 8 plums are packed into each bag. How many bags are needed?) o Arrays of Objects/Product Unknown word problems (The apples in the grocery window are in 3 rows and 4 columns. How many apples are there?) o Arrays of Objects/Group Size Unknown word problems (If 12 apples are arranged into an array with 3 rows, how many columns of apples are there?) o Arrays of Objects/Number of Groups Unknown word problems (If 12 apples are arranged into an array with 4 columns, how many rows are there?) Teacher Note: In this standard emphasis should be placed in solving for products of two one-digit numbers. Students may also be expected to solve problems in which a two-digit number is multiplied by a one-digit with a product less than or equal to 100. Emphasis should be placed on one-digit numbers multiplied by one-digit numbers; however, students should be exposed to a variety of problems with products less than or equal to 100. Examples may include problems such as: 12 × 5 = 60, 25 × 4 = 100, 33 × 3 = 99, etc. Multiplicative comparison situations (35 is 5 times as many as 7 and 7 times as many as 5) should not be introduced in third grade. This concept will be introduced in fourth grade in Standards 4.OA.1 and 4.OA.2. Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.1, 3.OA.2 Interpret products of whole numbers and whole-number 4.OA.2 Multiply and divide to solve word problems involving multiplicative quotients comparisons 3.OA.4 Determine the unknown whole number in a multiplication or 4.OA.3 Solve multi-step word problems using whole numbers and having whole-division equation number answers using the four operations 3.OA.5 Apply properties of operations as strategies to multiply and divide 4.NBT.5, 4.NBT.6 Multiply and divide with multi-digit numbers 3.OA.6 Understand the relationship between multiplication and division 4.NF.4 Extend understandings of multiplication to multiply a fraction by a whole 3.OA.7 Fluently multiply and divide within 100 number 3.OA.8 Solve two-step word problems 5.NF.4, 5.NF.6, 5.NF.7 Extend understandings of multiplication and division to 3.MD.2 Multiply and divide to solve measurement word problems multiply and divide with fractions 3.MD.7 Relate area to multiplication 5.NBT.5 Fluently multiply multi-digit whole numbers 5.NBT.6 Find whole-number quotients 3.OA.3 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Critical Background Knowledge from Previous Grade Levels • Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends (2.OA.4) • Partition a rectangle into rows and column of same-sized squares and count to find the total number of squares (2.G.2) • Use addition and subtraction to solve word problems (1.OA.1, 2.OA.1) Academic Vocabulary equal groups, array, row, column, area model, multiply, product, factor, divide, quotient, divisor, dividend Suggested Models Suggested Strategies 3 × 4 = 12 • Use objects and drawings to represent equal groups and arrays; Describe factors, 3 groups of 4 is 12 products, etc. in these models • Use bar models • Use counting all, skip counting, repeated addition to multiply 4 + 4 + 4 + 4 = 12 • Write equations to represent drawings and objects; Explain connections between 4 × 3 = 12 physical/visual models and equations • Use the relationship between multiplication and division to solve problems • Use a multiplication strategy (compensation, distributive property) to solve word 4 problems • Apply the commutative or associative properties of multiplication 4 × 3 = 12 • Students may create their own word problems 3 • Use equal groups, arrays, area models, bar models to solve problems • Use repeated subtraction to divide 4 × 3 =? 4 × 3 = 12 3.OA.3 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Represent and solve problems involving multiplication and division within 100 (Standards 3.OA.1–4 and Standard 3.OA.7). Standard 3.OA.4 Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number— product, factor, quotient, dividend, or divisor—that makes the equation true in each of the equations 8 × ? = 48, 5 = ? ÷ 3, 6 × 6 = ?. Concepts and Skills to Master • Solve the following multiplication and division situations (See: TABLE 2. Common multiplication and division situations): o Equal groups / unknown product word problems (There are 3 bags with 6 plums in each bag. How many plums are there in all?) o Equal groups / group size unknown word problems (If 18 plums are shared equally into 3 bags, then how many plums will be in each bag) o Equal groups / number of groups unknown word problems (If 18 plums are to be packed 6 to a bag, then how many bags are needed?) o Array or area / unknown product word problems (There are 3 rows of apples with 6 apples in each row. How many apples are there?) o Array or area / group size unknown word problems (If 18 apples are arranged into 3 equal rows, how many apples will be in each row?) o Array or area / number of groups unknown word problems (If 18 apples are arranged into equal rows of 6 apples, how many rows will there be?) • Understand that equations involving multiplication and division relate three whole numbers in related facts (3 × __ = 15; 15 ÷ __ = 3; 15 ÷ 3 = __) • Use a symbol to represent an unknown number • Apply multiplication or division to solve for an unknown in an equation Teacher Note: Comparison problem types are not introduced until 4th grade. Equations in the form of a × b = c and c = a × b should be used interchangeably, with the unknown in different positions. Examples: 24 = ? × 6, 72 ÷ __ = 9, or the following problem: Rachel has 3 bags. There are 4 marbles in each bag. How many marbles does Rachel have altogether? 3 × 4 = m Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.3 Use multiplication and division within 100 with symbols for the unknown number 3.OA.7 Fluently multiply and divide using the relationship between multiplication and division 3.MD.8 Solve real-world and mathematical problems involving perimeters 4.NBT.5 Multiply a whole number of up to four digits 4.NBT.6 Find whole number quotients 4.OA.3 Solve multi-step word problems posed with whole numbers 4.OA.2 Multiply or divide to solve word problems 4.MD.3 Apply the area and perimeter formulas for rectangles; view the area formula as a multiplication equation with an unknown factor 5.NBT.5 Fluently multiply multi-digit whole numbers 5.NBT.6 Find whole digit quotients using the relationship between multiplication and division Critical Background Knowledge from Previous Grade Levels • Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends (2.OA.4) • Partition a rectangle into rows and columns of same-sized squares and count to find the total number of squares (2.G.2) • Use addition and subtraction within 20 to solve word problems involving situations with unknowns in all positions. (1.OA.1) • Understand the meaning of the equal sign (1.OA.7) • Determine the unknown whole number in an addition and subtraction equation relating three whole numbers (1.OA.8) 3.OA.4 ADA Compliant 11/18/2019 Academic Vocabulary symbol, equal, multiplication, product, factor, quotient, dividend, divisor, division Suggested Models Suggested Strategies Part Part Whole/Multiplication and Division one part × number of parts = whole whole ÷ number of parts = one part ● Use a bar model to solve for the unknown whole number in an equation ● Use counters to model the relationship between multiplication and division ● Use base ten blocks to represent array and area models ● When given an equation such as 4 × ? = 40, students explain their thinking, for example: -4 groups of some number is the same as 40 -4 times some number is the same as 40 -I know 4 groups of 10 is 40 so the unknown number is 10 -• The missing factor is 10 because 4 times 10 equals 40 3.OA.4 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Demonstrate understanding of the properties of multiplication and the relationship between multiplication and division (Standards 3.OA.5–6). Standard 3.OA.5 Apply properties of operations as strategies to multiply and divide. For example: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known (commutative property of multiplication). 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30 (associative property of multiplication). Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56 (distributive property). (Third grade students may, but need not, use formal terms for these properties.) Concepts and Skills to Master • Understand that multiplication is commutative and division is not commutative (the order of the factors does not change the product of an equation) • Understand and apply the associative property of multiplication (factors can be grouped differently without changing the product) • Understand and apply the distributive property of multiplication over addition (to support students in solving for products by breaking apart the numbers) • Understand and apply the multiplicative identity property of one (8 × 1 = 8) • Understand and apply the zero property of multiplication (8 × 0 = 0) • Apply properties to simplify an expression into smaller problems (3 × 7 = (3 × 2) + (3 × 5); 3 × 8 = 3 × 2 × 4) Teacher Note: Emphasis should be placed on understanding of the properties and why each property applies to a particular operation rather than memorizing names and definitions. Convention defines arrays as rows by columns, however students should be allowed flexibility in describing arrays as either rows by columns or columns by rows and should understand how rotating an array demonstrates the commutative property. Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.1 Interpret the products of whole numbers 4.NBT.5 Multiply whole numbers using 3.OA.2 Interpret whole-number quotients of whole numbers strategies based on the properties of operations 3.OA.3 Use multiplication and division to solve word problems 4.NBT.6 Find whole-number quotients and 3.OA.4 Determine the unknown whole number in a multiplication or division equation remainders based on the properties of 3.OA.6 Understand division as an unknown-factor problem operations 3.OA.7 Fluently multiply and divide 4.OA.3 Solve multi-step word problems 3.MD.7 Relate area to the operations of multiplication and addition 5.OA.1 Use parenthesis, brackets, and braces in 3.OA.8 Solve two-step word problem numerical expressions 3.OA.9 Identify arithmetic patterns and explain them using properties of operations 5.MD.5 Relate volume to the operations of 3.NBT.3 Multiply one-digit whole numbers by multiples of 10 using strategies based on place value and multiplication and addition properties of operations Critical Background Knowledge from Previous Grade Levels • Explain why addition and subtraction strategies work, using place value and the properties of operations (2.NBT.9) • Use addition to find the total number of objects in a rectangular array (2.OA.4) • Apply properties of operations as strategies to add and subtract (2.NBT.5, 1.OA.3, 1.NBT.4) 3.OA.5 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Academic Vocabulary multiplication, division, product, factor, dividend, divisor, quotient, commutative property of multiplication, associative property of multiplication, parentheses, distributive property of multiplication over addition, zero property of multiplication, multiplicative identity property of one, array Suggested Models Suggested Strategies Commutative Property of Multiplication Model Distributive Property of Multiplication Model 4 × 7 = 4 × (5 + 2) = (4 × 5) + (4 × 2) = 20 + 8 = 28 Associative Property of Multiplication Model • Use an array or grouping to model the commutative property • Model the distributive property of multiplication over addition (see Suggested Models) • Use base-ten blocks, multiplication charts, grid/graph paper, and area models 3.OA.5 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Demonstrate understanding of the properties of multiplication and the relationship between multiplication and division (Standards 3.OA.5–6). Standard 3.OA.6 Understand division as an unknown-factor problem. Understand the relationship between multiplication and division (multiplication and division are inverse operations). For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8. Concepts and Skills to Master • Understand the relationship between multiplication and division as inverse operations, one operation can help solve the other • Understand and solve unknown-factor problems • Solve a division equation by using related multiplication facts Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.1 Interpret the products of whole numbers 3.OA.2 Interpret whole-number quotients of whole numbers 3.OA.3 Use multiplication and division within 100 to solve word problems 3.OA.4 Determine the unknown whole number in a multiplication or division equation relating three whole numbers 3.OA.5 Apply properties of operations as strategies to multiply and divide 3.OA.7 Fluently multiply and divide 3.OA.8 Solve two-step word problem 3.OA.9 Identify arithmetic patterns and explain them using properties of operations 3.MD.7 Relate area to the operations of multiplication and addition 4.NBT.6 Find whole-number quotients and remainders based on place value, the properties of operations, and the relationship between multiplication and division 5.NF.3 Interpret a fraction as division, solving real-world problems involving division of whole numbers 5.NF.6 Solve real-world problems involving multiplication of fractions and mixed numbers 5.NF.7 Apply and extend previous understandings of division to unit fractions and whole numbers Critical Background Knowledge from Previous Grade Levels • Use addition and subtraction with unknowns in all positions (2.OA.1) • Use addition and subtraction within 20 to solve word problems involving situations with unknowns in all positions. (1.OA.1) • Determine the unknown whole number in an addition and subtraction equation relating three whole numbers (1.OA.8) Academic Vocabulary related facts, multiplication, division, inverse operation, factor Suggested Models Suggested Strategies ● Use fact families and/or number bonds ● Model arrays to show related multiplication and division equations (e.g., 3 × 2 = 6; 2 × 3 = 6; 6 ÷ 2 = 3; 6 ÷ 3 = 2). ● Use equal groups, number lines, and area models 3.OA.6 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Represent and solve problems involving multiplication and division within 100 (Standards 3.OA.1–4 and Standard 3.OA.7). Standard 3.OA.7 Fluently multiply and divide. a. Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division or properties of operations. (For example, knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8). b. By the end of Grade 3, know from memory all products of two one-digit numbers. Concepts and Skills to Master  Apply multiplication and division strategies flexibly, accurately and efficiently  Understand the inverse relationship of multiplication and division  Understand and apply commutative and distributive properties  Know from memory all products of two one-digit numbers Teacher Note: Students should have exposure to multiplication and division problems presented in both vertical and horizontal forms. Students develop fluency over time as they have repeated experiences that build conceptual understanding of multiplication (concrete and pictorial representations, patterns, context, etc.). Learning is enhanced when practice is organized to focus most heavily on understood but not yet fluent facts. Fluency may be reached by becoming fluent for each number (2s, 5s, etc. by noticing patterns, not through memorization) and then extending the fluency to several, then all numbers mixed together. To achieve fluency by the end of third grade, students must begin working toward fluency as early as possible. This is not a matter of instilling facts divorced from their meanings, but rather the outcome of a carefully designed learning process that heavily involves the interplay of practice and reasoning. (Adapted from: p. 27) Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.1 Interpret the products of whole numbers 4.OA.4 Find all factor pairs for a whole number between 1-100 3.OA.2 Interpret whole-number quotients 4.NBT.5 Multiply up to four-digit numbers by one-digit numbers and two-digit 3.OA.3 Use multiplication and division within 100 to solve word numbers by two-digit numbers problems 4.NBT.6 Find whole-number quotients and remainders with up to four-digit 3.OA.4 Determine the unknown whole number in a multiplication or dividends and one-digit divisors division equation relating three whole numbers 5.NBT.5 Fluently multiply multi-digit whole numbers 3.OA.5 Apply properties of operations as strategies to multiply and 5.NBT.6 Find whole-number quotients divide 4.OA.1–3, 4.NF.1–2 and 4, 5.NF.4 and 6–7 Fluency with multiplication is a 3.OA.6 Understand division as an unknown-factor problem foundation for extending strategies when multiplying and dividing multi-digit whole numbers, fractions, and decimals Critical Background Knowledge from Previous Grade Levels  See Related Standards: Current Grade Level  Fluently add and subtract within 20 (2.OA.2)  Work with equal groups (2.OA.4)  Partition rectangles into squares (2.G.2)  Apply properties of operations as strategies to add and subtract (1.OA.3) Academic Vocabulary product, factor, dividend, divisor, quotient, multiplication, multiply, division, divide, commutative property of multiplication, distributive property 3.OA.7 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Suggested Models Suggested Strategies See models listed in the Core Guide for 3.OA.3 as students work to build fluency. Area model for 3 × 4 Base ten blocks used to represent 4 × 13  Model and/or count  Apply the Commutative Property  Find missing factors  Engage in number talk or math discourse  Play games for practice  Analyze multiplication by zeros and ones  Skip count (counting groups of __ and knowing how many groups have been counted)  Use doubles (2s), doubling twice (4s), doubling three times (8s)  Use tens facts (relating to place value, 5 × 10 is 5 tens or 50)  Use five facts (half of tens)  Recognize square numbers (e.g., 3 × 3)  Identify patterns in multiples of nines (10 groups less one group, e.g., 9 × 3 is 10 groups of 3 minus one group of 3)  Decompose into known facts (6 × 7 is 6 × 6 plus one more group of 6)  Use related facts (e.g., 6 × 4 = 24; 24 ÷ 6 = 4; 24 ÷ 4 = 6; 4 × 6 = 24)  Recognize and use patterns in multiplication table Fluently multiply and divide within 100 Knowing from memory all products of two one-Fluency involves a mixture of just knowing some answers, knowing some answers from patterns (for digit numbers includes the following facts: example, multiplying by one yields the same number), and knowing some answers from the use of strategies. It is important to push sensitively and encouragingly toward fluency of the designated numbers, recognizing that fluency will be a mixture of these kinds of thinking which may differ across students. Emphasis should be placed on one-digit numbers multiplied by one-digit numbers; however, students should be exposed to a variety of problems with products less than or equal to 100. Students are expected to use concrete models and reasoning strategies to solve problems in which a two-digit number is multiplied by a one-digit with a product less than or equal to 100. Examples may include problems such as: 15 × 5 = 75, 25 × 4 = 100, 33 × 3 = 99, etc. The standard algorithm for multiplication is introduced in fifth grade in standard 5.NBT.5 and should not be taught in third grade. Text Source: 3.OA.7 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Students use the four operations to identify and explain patterns in arithmetic (Standards 3.OA.8–9). Standard 3.OA.8 Solve two-step word problems. a. Solve two-step word problems using the four operations. Know how to perform operations in the conventional order when there are no parentheses to specify a particular order (Order of Operations). (Limit to problems posed with whole numbers and having whole number answers.) b. Represent two-step problems using equations with a letter standing for the unknown quantity. Create accurate equations to match word problems. c. Assess the reasonableness of answers using mental computation and estimation strategies, including rounding. Concepts and Skills to Master • Differentiate between one-step and two-step word problems (Two-step word problems may include any combination of two operations in the same problem) • Determine the operation(s) based on the actions in the context of two-step word problems (avoid relying on keyword strategies) • Use numbers and symbols to represent word problems (+, -, ×, ÷, =, and a letter for unknowns) • Know that multiplication and division are performed (in the order they appear in the problem; from left to right) prior to addition and subtraction (in the order they appear in the problem; from left to right) • Solve and apply the addition, subtraction, multiplication, and division situations listed in Standards K.OA.2, 1.OA.1, and 2.OA.1, and 3.OA.3 Related Standards: Current Grade Level Related Standards: Future Grade Level 3.OA.1, 3.OA.2 Interpret products of whole numbers and whole-number quotients 4.OA.2 Multiply and divide to solve word problems involving 3.OA.4 Determine the unknown whole number in a multiplication or division equations multiplicative comparisons 3.OA.5 Apply properties of operations as strategies to multiply and divide 4.OA.3 Solve multi-step word problems using whole numbers and 3.OA.6 Understand the relationship between multiplication and division having whole-number answers using the four operations 3.OA.7 Fluently multiply and divide 5.NF.4 Apply and extend previous understandings of 3.OA.8 Solve two-step word problems multiplication and division to multiply and fraction or a whole 3.MD.2 Multiply and divide to solve measurement word problems number by a fraction 3.MD.7 Relate area to multiplication Critical Background Knowledge from Previous Grade Levels • Interpret products of whole numbers and whole-number quotients (3.OA.1, 3.OA.2) • Understand and use the associative and commutative properties • Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends (2.OA.4) • Partition a rectangle into rows and column of same-sized squares and count to find the total number of squares (2.G.2) • Use addition and subtraction to solve word problems (1.OA.1, 2.OA.1) Academic Vocabulary Addends, sum, difference, round, estimate, equation, difference, multiplication, factors, product, array, multiples, division, divisor, dividend, quotient, reasonableness, symbol, 3.OA.8 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Suggested Models Suggested Strategies • Use drawings, objects, and equations • Use a bar model • Apply Part/Part/Whole • Create student-generated word problems • Skip count • Use the relationship between multiplication and division Image Source: 3.OA.8 ADA Compliant 11/18/2019 Operations and Algebraic Thinking Core Guide Grade 3 Students use the four operations to identify and explain patterns in arithmetic (Standards 3.OA.8–9). Standard 3.OA.9 Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that four times a number is always even, and explain why four times a number can be decomposed into two equal addends. Concepts and Skills to Master • Recognize arithmetic patterns that can be found on a hundreds chart, a number line, an addition and a multiplication table • Recognize multiplication patterns that can be found on a hundreds chart and a multiplication table • Know that multiplication by an even number results in an even number • Know that multiplication of an odd number by another odd number results in an odd number • Know that multiplication of an odd number by an even number results in an even number • Explain arithmetic patterns using properties of operations • Find the products of the commutative property on the multiplication chart • Model addition and multiplication patterns with a number line, hundreds chart, multiplication chart Related Standards: Current Grade Level Related Standards: Future Grade Levels 3.OA.5 Apply properties of operations as strategies to multiply and divide 4.OA.5 Generate number or shape patterns that follow a given rule 5.OA.3 Generate two numerical patterns using two given rules Critical Background Knowledge from Previous Grade Levels • Determine whether a group of objects is odd or even (2.OA.3) • Recognize patterns of skip counting with fives, tens, and hundreds (2.NBT.2) Academic Vocabulary sum, multiplication, multiples, factors, product, sequence, pattern , row, column Suggested Models Suggested Strategies Highlight a given factor and discuss patterns noticed ● Use number lines ● Use hundreds charts ● Highlight and discuss patterns on multiplication and addition charts ● Analyze patterns in basic facts 3.OA.9 ADA Compliant 11/18/2019 UTAH CORE S TABLE 2. Common multiplication and division situations. 1 TATESTANDARDS forMATHEMATICS Unknown Product Group Size Unknown (“How many in each group?” Division) Number of Groups Unknown (“How many groups?” Division) 3 × 6 = ? 3 × ? = 18 and 18 ÷ 3 = ? ? × 6 = 18 and 18 ÷ 6 = ? EQUAL GROUPS There are 3 bags with 6 plums in each bag. How many plums are there in all? Measurement example. You need 3 lengths of string, each 6 inches long. How much string will you need altogether? If 18 plums are shared equally into 3 bags, then how many plums will be in each bag? Measurement example. You have 18 inches of string, which you will cut into 3 equal pieces. How long will each piece of string be? If 18 plums are to be packed 6 to a bag, then how many bags are needed? Measurement example. You have 18 inches of string, which you will cut into pieces that are 6 inches long. How many pieces of string will you have? ARRAYS2 There are 3 rows of apples with 6 apples in each row. How many apples are there? If 18 apples are arranged into 3 equal rows, how many apples will be in each row? If 18 apples are arranged into equal rows of 6 apples, how many rows will there be? What is the area of a 3 cm by A rectangle has area 18 A rectangle has area 18 AREA3 6 cm rectangle? square centimeters. If one side is 3 cm long, how long is a side next to it? square centimeters. If one side is 6 cm long, how long is a side next to it? COMPARE4 A blue hat costs $6. A red hat costs 3 times as much as the blue hat. How much does the red hat cost? Measurement example. A rubber band is 6 cm long. How long will the rubber band be when it is stretched to be 3 times as long? A red hat costs $18 and that is 3 times as much as a blue hat costs. How much does a blue hat cost? Measurement example. A rubber band is stretched to be 18 cm long and that is 3 times as long as it was at first. How long was the rub ber band at first? A red hat costs $18 and a blue hat costs $6. How many times as much does the red hat cost as the blue hat? Measurement example. A rubber band was 6 cm long at first. Now it is stretched to be 18 cm long. How many times as long is the rubber band now as it was at first? GENERAL a × b = ? a × ? = p and p ÷ a = ? ? × b = p and p ÷ b = ? 1 The first examples in each cell are examples of discrete things. These are easier for students and should be given before the measurement examples. 2 The language in the array examples shows the easiest form of array problems. A harder form is to use the terms rows and columns: The apples in the grocery window are in 3 rows and 6 columns. How many apples are in there? Both forms are valuable. 3 Area involves arrays of squares that have been pushed together so that there are no gaps or overlaps, so array problems include these especially important measurement situations. 4 Multiplicative Compare problems appear first in Grade 4, with whole-number values in all places, and with the “times as much” language in the table. In Grade 5, unit fractions language such as “one third as much” may be used. Multiplying and unit fraction language change the subject of the comparing sentence, e.g., “A red hat costs A times as much as the blue hat” results in the same comparison as “A blue hat costs 1/A times as much as the red hat,” but has a different subject. ADA Compliant 11/18/2019
15422	https://www.thoughtco.com/protista-kingdom-of-life-4120782	The Protista Kingdom: Characteristics and Examples Skip to content Menu Home Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Search Close Search the site GO Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Contact Us Editorial Guidelines Privacy Policy Science, Tech, Math› Science› Biology› Organisms› All About the Protista Kingdom Print Diatoms (Kingdom Protista) may be extremely abundant in both freshwater and marine ecosystems; it is estimated that 20% to 25% of all organic carbon fixation on the planet is carried out by diatoms.STEVE GSCHMEISSNER/SCIENCE PHOTO LIBRARY/Getty Images Science Biology Organisms Basics Cell Biology Genetics Anatomy Physiology Botany Ecology Chemistry Physics Geology Astronomy Weather & Climate By Regina Bailey Regina Bailey Biology Expert B.A., Biology, Emory University A.S., Nursing, Chattahoochee Technical College Regina Bailey is a board-certified registered nurse, science writer and educator. Her work has been featured in "Kaplan AP Biology" and "The Internet for Cellular and Molecular Biologists." Learn about ourEditorial Process Updated on May 01, 2025 Close Key Takeaways Protists are eukaryotic organisms that can be unicellular or multicellular, and they live in moist environments. Protists can acquire nutrition through photosynthesis, absorbing nutrients, or feeding on other organisms. Some protists move using flagella, cilia, or pseudopodia, while others do not move at all. The Kingdom Protista consists of eukaryotic protists. Members of this very diverse kingdom are typically unicelluar and less complex in structure than other eukaryotes. In a superficial sense, these organisms are often described based on their similarities to the other groups of eukaryotes: animals, plants, and fungi. Protists do not share many similarities, but are grouped together because they do not fit into any of the other kingdoms. Some protists are capable of photosynthesis; some live in mutualistic relationships with other protists; some are single-celled; some are multicellular or form colonies; some are microscopic; some are enormous (giant kelp); some are bioluminescent; and some are responsible for a number of diseases that occur in plants and animals. Protists live in aquatic environments, moist land habitats, and even inside other eukaryotes. Protista Characteristics This is a photomicrograph of a paramecium.NNehring/E+/Getty Images Protists reside under the Eukarya Domain and are thus classified as eukaryotes. Eukaryotic organisms are distinguished from prokaryotes in that they have a nucleus that is surrounded by a membrane. In addition to a nucleus, protists have additional organelles in their cytoplasm. The endoplasmic reticulum and Golgi complexes are important for the synthesis of proteins and exocytosis of cellular molecules. Many protists also have lysosomes, which aid in the digestion of ingested organic material. Certain organelles may be found in some protist cells and not in others. Protists that have characteristics in common with animal cells also have mitochondria, which provide energy for the cell. Protists that are similar to plant cells have a cell wall and chloroplasts. Chloroplasts make photosynthesis possible in these cells. Protista Nutrition Acquisition Protists exhibit different methods of acquiring nutrition. Some are photosynthetic autotrophs, meaning that they are self-feeders and capable of using sunlight to generate carbohydrates for nutrition. Other protists are heterotrophs, which acquire nutrition through feeding on other organisms. This is accomplished by phagocytosis, the process in which particles are engulfed and digested internally. Still, other protists acquire nutrition predominately by absorbing nutrients from their environment. Some protists may exhibit both photosynthetic and heterotrophic forms of nutrient acquisition. Protista Locomotion While some protists are non-motile, others exhibit locomotion through different methods. Some protists have flagella or cilia. These organelles are protrusions formed from specialized groupings of microtubules that move to propel protists through their moist environment. Other protists move by using temporary extensions of their cytoplasm known as pseudopodia. These extensions are also valuable in allowing the protist to capture other organisms that they feed on. Protista Reproduction The most common method of reproduction displayed in protists is asexual reproduction. Sexual reproduction is possible, but typically only occurs during times of stress. Some protists reproduce asexually by binary fission or multiple fission. Others reproduce asexually by budding or through spore formation. In sexual reproduction, gametes are produced by meiosis and unite at fertilization to produce new individuals. Other protists, such as algae, exhibit a type of alternation of generations in which they alternate between haploid and diploid stages in their life cycles. Photosynthetic Protista Diatom and Dinoflagellate Protists.Oxford Scientific/Photodisc/Getty Images Protists can be grouped according to similarities in many different categories including nutrition acquisition, mobility, and reproduction. Examples of protists include algae, amoebas, euglena, plasmodium, and slime molds. Protists that are capable of photosynthesis include various types of algae, diatoms, dinoflagellates, and euglena. These organisms are often unicellular but can form colonies. They also contain chlorophyll, a pigment that absorbs light energy for photosynthesis. Photosynthetic protists are considered plant-like protists. Protists known as dinoflagellates or fire algae, are plankton that live in marine and freshwater environments. At times they can reproduce rapidly producing harmful algae blooms. Some dinoflagellates are also bioluminescent. Diatoms are among the most abundant types of unicellular algae known as phytoplankton. They are encased within a silicon shell and are abundant in marine and freshwater aquatic habitats. Photosynthetic euglena are similar to plant cells in that they contain chloroplasts. It is thought that the chloroplasts were acquired as a result of endosymbiotic relationships with green algae. Heterotrophic Protista This is an amoeba with finger-like pseudopodia (dactylopodia). These freshwater single-celled organisms feed on bacteria and smaller protozoa. They use their pseudopodia to engulf their food and for locomotion. Although the cell shape is extremely flexible, and most amoeba look 'naked' in the light microscope, SEM reveals many are covered by a coat of scales.Science Photo Library - STEVE GSCHMEISSNER/ Brand X Pictures/Getty Images Heterotrophic protists must obtain nutrition by taking in organic compounds. These protists feed on bacteria, decaying organic matter, and other protists. Heterotrophic protists can be categorized based on their type of movement or lack of locomotion. Examples of heterotrophic protists include amoebas, paramecia, sporozoans, water molds, and slime molds. Protista Movement With Pseudopodia Amoebas are examples of protists that move using pseudopodia. These temporary extensions of the cytoplasm allow the organism to move as well as to capture and engulf organic material through a type of endocytosis known as phagocytosis, or cell eating. Amoebas are amorphous and move by changing their shape. They reside in aquatic and moist environments, and some species are parasitic. Heterotrophic Protista With Flagella or Cilia Trypanosoma Parasite (Kingdom Protista), illustration.ROYALTYSTOCKPHOTO/Science Photo Library/Getty Images Trypanosomes are examples of heterptrophic protists that move with flagella. These long, whip-like appendages move back and forth enabling movement. Trypanosomes are parasites that can infect animals and humans. Some species cause African sleeping sickness which is transmitted to humans by biting flies. Paramecia are examples of protists that move with cilia. Cilia are short, thread-like protrusions that extend from the body and move in a sweeping motion. This motion allows the organism to move and also pulls food (bacteria, algae. ect.) toward the paramecium's mouth. Some paramecia live in mutualistic symbiotic relationships with green algae or with certain bacteria. Heterotrophic Protista With Limited Movement This is a magnified image of slime mold fruiting bodies.Joao Paulo Burini/Moment Open/Getty Images Slime molds and water molds are examples of protists that exhibit limited motion. These protists are similar to fungi in that they decompose organic matter and recycle nutrients back into the environment. They live in moist soils among decaying leaves or wood. There are two types of slime molds: plasmodial and cellular slime molds. A plasmodial slime mold exists as an enormous cell formed by the fusion of several individual cells. This huge blob of cytoplasm with many nuclei resembles slime that moves slowly in an amoeba-like fashion. Under harsh conditions, plasmodial slime molds produce reproductive stalks called sporangia that contain spores. When released into the environment, these spores may germinate producing more plasmodial slime molds. Cellular slime molds spend most of their life cycle as single-celled organisms. They too are capable of amoeba-like movement. When under stressful conditions, these cells unite forming a large group of individual cells that resemble a slug. The cells form a reproductive stalk or fruiting body that produces spores. Water molds live in aquatic and moist terrestrial environments. They feed on decaying matter, and some are parasites that live off of plants, animals, algae, and fungi. Species of the Oomycota phylum exhibit filamentous or thread-like growth, similar to fungi. However, unlike fungi, oomycetes have a cell wall that is composed of cellulose and not chitin. They can also reproduce both sexually and asexually. Non-motile Heterotrophic Protista This is a scanning electron microscopic image of parasitic protozoans (Plasmodium sp.) that cause malaria being released from a red blood cell.MedicalRF.com/Getty Images Sporozoans are examples of protists that do not have structures that are used for locomotion. These protists are parasites that feed off of their host and reproduce by the formation of spores. Sporozoans exhibit a type of alternation of generations in their life cycle, in which they alternate between sexual and asexual phases. Sporozoans are transmitted to humans by insect or other animal vectors. Toxoplasmosis is a disease caused by the sporozoan Toxoplasma gondii that can be transmitted to humans by animals or can be contracted by ingesting contaminated food or water. In severe toxoplasmosis, T. gondii damage eyes or other organs, such as the brain. Toxoplasmosis does not typically develop in people with healthy immune systems. Another sporozoan, known as plasmodium, causes malaria in humans. These protists are transmitted to mammals by insect bites, commonly by mosquitoes, and infect red blood cells. Plasmodium, in the merozoites stage of their life cycle, multiply within infected blood cells causing them to rupture. Once released, the merozoites can infect other red blood cells. Cite this Article Format mlaapachicago Your Citation Bailey, Regina. "All About the Protista Kingdom." ThoughtCo, May. 1, 2025, thoughtco.com/protista-kingdom-of-life-4120782.Bailey, Regina. (2025, May 1). All About the Protista Kingdom. Retrieved from Bailey, Regina. "All About the Protista Kingdom." ThoughtCo. 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15423	https://www.expii.com/t/base-ternary-numbers-9194	Expii Base 3: Ternary Numbers - Expii Base 3 uses powers of 3. The possible digits are 0, 1, and 2. Explanations (6) Ternary Ternary is a numeral system with base 3. This essentially means it only uses 3 digits to make up its numbers: 0,1,2 Our normal number system is called the decimal or base-10. As you may have guessed it is because we use 10 digits. 0,1,2,3,4,5,6,7,8,9 From Decimal to Ternary To convert from a regular decimal number to a ternary number can be a little tedious, but it only requires division. Let's work through an example to see how we do it. Suppose we want to convert 42 from base-10 to base-3. One way we show a number is base-10 is to put a subscript after it with 10. 4210 To begin the conversion to base-3, we divide our number by 3. We don't need to worry about decimal points, just what the remainder is. Let's divide 42 by 3. We get 14 with remainder 0. The remainder will be the digit in the "one's place" of our ternary number. Next, take the number we get as our solution (minus the remainder) and divide that by 3. Let's divide 14 by 3. We get 4 remainder 2. This remainder will be the digit in the "ten's place" of our ternary number. We again divide our result by 3. This will go till we get a result of 0. So let's divide 4 by 3. We get 1 remainder 1. This will be the digit in the "hundred's place" of our ternary number. We're almost done! One more division. Divide 1 by 3. We get another remainder of 1. This will be the digit in the "thousand's place" of our ternary number. We know to stop here because we finally get 0 up top. So, we got the remainders 0,2,1,1. The ternary version of 42 will be these numbers but written in reverse order. So: 4210=11203 From Ternary to Decimal Don't worry, this way is much simpler in my opinion. Let's use the number we just found, 1120, and convert it back to 42. Starting with the last digit, 0, we want to multiply each digit by 3 raised to a power starting from 0 and then add them all together. In other words, 0 is multiplied by 30, 2 is multiplied by 31, 1 is multiplied by 32, and 1 is multiplied by 33. 1120=1⋅33+1⋅32+2⋅31+0⋅30=1⋅27+1⋅9+2⋅3+0⋅1=27+9+6+0=42 Practice Try converting each way on your own. Report Share 13 Like Related Lessons Base 16: Hexadecimal Numbers Combinatorics: What Are Graphs? Base 2: Binary Numbers Eulerian Paths View All Related Lessons Joshua Siktar Text 9 All your life you've been counting in what is called the base 10 number system, where we build numbers using the digits 0,1,2,3,4,5,6,7,8, and 9. But it turns out we don't have to use all these digits. In particular, if we only use the first three, then we are writing numbers using a base 3 number system. It is important to know that we don't use the number 3 in this system, but only 0,1, and 2. Also, "ternary" is just a fancy word for base-3. So, let's try counting in base 3 for a little while. 0,1,2... The first three numbers are easy, but what's next? We can't use 3. To get an answer, remember that in base 10, we can write every number with an invisible 0 in front of it. That is, counting in base 10 gives 00,01,02,03,04,05,06,07,08,09... We can actually do the same thing in base 3, so in base 3 we start counting as 00,01,02... In base 3, the 1 still comes after 0. This makes us think we can keep counting in base 3 as 00,01,02,10,11,12,... and in fact that's exactly what we do. And we can do this as many times as we want, adding more digits as we go along: 00,01,02,10,11,12,20,21,22,100... Report Share 9 Like Joshua Siktar Text 9 When we count we usually use the digits 0,1,2,3,4,5,6,7,8, and 9 to make numbers. Well, as it might surprise you, that's not the only way to do it. The way we normally do it is called base 10 representation because we use 10 different digits, but we can use more or less digits to count. One common example, sometimes used in computer operations, is the base 3 representation, which only uses the digits 0, 1, and 2. We can count any number in base 3 that we could in base 10, and we usually use subscripts to indicate which base we're in. A base-10 number will have a 10 stuck to the end, while a base-3 number will have a 3 stuck to the end. The reason for this is clear upon answering the next question. Report Share 9 Like Katie Jergens Text 4 Ternary (aka base-3) numbers have 2 properties: 1. Each place numeral is in the set {0,1,2}. 2. Each place value is (number x 3place). To get the base-10 value add up the place values. E.g. What is base-10 equivalent to the base-3 number 2210? =(2×33)+(2×32)+(1×31)+(0×30) =(2×27)+(2×9)+(1×3)+(0×1) =54+18+3+0 =75 Report Share 4 Like Ivan Hu Text 2 Base 3 The base 3, or ternary, system, uses only the digits 0,1, and 2. For each place, instead of multiplying by the power of 10, you multiply by the power of 3. For example, 120123→1×34+2×33+0×32+1×31+2. You've reached the end How can we improve? General Bug Feature Send Feedback
15424	https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_10?srsltid=AfmBOor4q-2cflb50KEAvrhAXnQNHgiEFDic4CouhZx0mefz2TsQJ0t6	Art of Problem Solving 2024 AIME II Problems/Problem 10 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2024 AIME II Problems/Problem 10 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2024 AIME II Problems/Problem 10 Contents 1 Problem 2 Solution 1 (Similar Triangles and PoP) 3 Solution 1.1 4 Solution 1.2 5 Solution 1 (Continued) 6 Solution 2 (Excenters) 7 Solution 3 8 Solution 4 (Trig) 9 Solution 5 (Trig) 10 Solution 6 (Close to Solution 3) 11 Solution 7 12 Solution 8 13 Solution 9 14 Solution 10 15 Video Solution 16 Video Solution 17 See also Problem Let have incenter , circumcenter , inradius , and circumradius . Suppose that . Find . Solution 1 (Similar Triangles and PoP) Start off by (of course) drawing a diagram! Let and be the incenter and circumcenters of triangle , respectively. Furthermore, extend to meet at and the circumcircle of triangle at . We'll tackle the initial steps of the problem in two different manners, both leading us to the same final calculations. Solution 1.1 Since is the incenter, . Furthermore, and are both subtended by the same arc , so Therefore by AA similarity, . From this we can say that Since is a chord of the circle and is a perpendicular from the center to that chord, must bisect . This can be seen by drawing and recognizing that this creates two congruent right triangles. Therefore, We have successfully represented in terms of and . Solution 1.2 will explain an alternate method to get a similar relationship, and then we'll rejoin and finish off the solution. Solution 1.2 by vertical angles and because both are subtended by arc . Thus . Thus Symmetrically, we get , so Substituting, we get Proof: We commence angle chasing: we know . Therefore . Looking at triangle , we see that , and . Therefore because the sum of the angles must be , . Now is a straight line, so . Since , triangle is isosceles and thus . A similar argument should suffice to show by symmetry, so thus . Now we regroup and get Now note that and are part of the same chord in the circle, so we can use Power of a point to express their product differently. Solution 1 (Continued) Now we have some sort of expression for in terms of and . Let's try to find first. Drop an altitude from to , to , and to : Since and , . Furthermore, we know and , so . Since we have two right similar triangles and the corresponding sides are equal, these two triangles are actually congruent: this implies that since is the inradius. Now notice that because of equal vertical angles and right angles. Furthermore, is the inradius so it's length is , which equals the length of . Therefore these two triangles are congruent, so . Since , . Furthermore, . We can now plug back into our initial equations for : From , Alternatively, from , Now all we need to do is find . The problem now becomes very simple if one knows Euler's Formula for the distance between the incenter and the circumcenter of a triangle. This formula states that , where is the circumradius and is the inradius. We will prove this formula first, but if you already know the proof, skip this part. Theorem: in any triangle, let be the distance from the circumcenter to the incenter of the triangle. Then , where is the circumradius of the triangle and is the inradius of the triangle. Proof: Construct the following diagram: Let , , . By the Power of a Point, . and , so Now consider . Since all three points lie on the circumcircle of , the two triangles have the same circumcircle. Thus we can apply law of sines and we get . This implies Also, , and is right. Therefore Plugging in, we have Thus Now we can finish up our solution. We know that . Since , . Since is right, we can apply the pythagorean theorem: . Plugging in from Euler's formula, . Thus . Finally . ~KingRavi Solution 2 (Excenters) By Euler's formula , we have . Thus, by the Pythagorean theorem, . Let ; notice is isosceles and which is enough to imply that is the midpoint of , and itself is the midpoint of where is the -excenter of . Therefore, and Note that this problem is extremely similar to 2019 CIME I/14. Solution 3 Denote . By the given condition, , where is the area of . Moreover, since , the second intersection of the line and is the reflection of about , denote that as . By the incenter-excenter lemma with Ptolemy's Theorem, . Thus, we have . Now, we have ~Bluesoul Solution 4 (Trig) Denote by and the circumradius and inradius, respectively. First, we have Second, because , A I=A O cos⁡∠I A O=A O cos⁡(90∘−C−A 2)=A O sin⁡(C+A 2)=R sin⁡(C+180∘−B−C 2)=R cos⁡B−C 2. Thus, r=A I sin⁡A 2=R sin⁡A 2 cos⁡B−C 2(2) Taking , we get We have 2 sin⁡B 2 sin⁡C 2=−cos⁡B+C 2+cos⁡B−C 2. Plugging this into the above equation, we get Now, we analyze Equation (2). We have r R=sin⁡A 2 cos⁡B−C 2=sin⁡180∘−B−C 2 cos⁡B−C 2=cos⁡B+C 2 cos⁡B−C 2(4) Solving Equations (3) and (4), we get Now, we compute . We have A B⋅A C=2 R sin⁡C⋅2 R sin⁡B=2 R 2(−cos⁡(B+C)+cos⁡(B−C))=2 R 2(−(2(cos⁡B+C 2)2−1)+(2(cos⁡B−C 2)2−1))=6 R r=(468) where the first equality follows from the law of sines, the fourth equality follows from (5). ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) Solution 5 (Trig) Firstly, we can construct the triangle by drawing the circumcirlce (centered at with radius ) and incircle (centered at with radius ). Next, from , construct tangent lines to the incircle meeting the circumcirlce at point and , say, as shown in the diagram. By Euler's theorem (relating the distance between and to the circumradius and inradius), we have This leads to Let be the point of tangency where the incircle meets the side . Now we denote Notice that . Finally, the crux move is to recognize since is the circumcenter. Then multiply these two expressions and apply the compound-angle formula to get A B⋅A C=4 R 2 cos⁡(θ−ϕ)cos⁡(θ+ϕ)=4 R 2(cos 2⁡θ cos 2⁡ϕ−sin 2⁡θ sin 2⁡ϕ)=4 cos 2⁡θ(R cos⁡ϕ⏟A I=156)2−4 sin 2⁡θ(R sin⁡ϕ⏟O I=13)2=52(12 cos 2⁡θ−sin 2⁡θ)A B⋅A C=52(12−13 sin 2⁡θ), where in the last equality, we make use of the substitution . Looking at , we learn that which means . Hence we have This completes the solution -- VensL. Solution 6 (Close to Solution 3) Denote is inradius. It is known that Points on bisectors vladimir.shelomovskii@gmail.com, vvsss Solution 7 Call side , and similarly label the other sides. Note that . Also note that , so by the right angle, . However, we can double Angle Bisector theorem. The length of the angle bisector from A is . As a direct result, the length AI simplifies down to . Draw the incircle and call the tangent to side AB F. Then, . But this length, by Pythagorean, is , so . Also note that the area of the triangle is , by . By the incircle, we know that , and similarly, . By double-angle, . But the area of the triangle is simply , which is also . But we know this is from above, so . As a direct result, . Apply this to the formula listed above to get , so . We're done. - sepehr2010 Solution 8 Let the intersection of the -angle bisector and the circumcircle be , and denote the -excenter as . Denote the tangent to the incircle from as and the tangent to the excircle from as . Notice that our perpendicular condition implies , and Incenter-Excenter gives . Thus we have . From similar triangles we get . This implies . Using areas we have that . Substituting gives and we're done. - thoom Solution 9 We know that the area of is equal to , but is also equal to , where R is the circumcircle and r is the incircle. So, . Let's extend so it intersects the circumcircle of at . Something that we see is that is congruent to . Something else that we notice that since is the angle bisector of , is the midpoint of arc . Now, let's try calculating . By Euler's Theorem, where R is the circumcircle and r is the incircle, so . Using Pythagorean Theorem on gives us as we know that is 13. However, since is congruent to , . Since we know that is the midpoint of arc , we can apply the Incenter-Excenter Lemma to get that and . Now, we can use Ptolemy's Theorem on quadrilateral ABPC: However, we know that , so we can solve for a! So, . Dividing gives us . Substituting and cancelling into our equation, . Multiplying, So, = 312. Our answer is 312 + 156 = . ~aleyang Solution 10 We know by Euler's theorem Since we have Now, extend to meet at and the circumcircle of at By the Incenter-Excenter lemma, (Note that ) Using Ptolemy in the cyclic quadrilateral we have Also using the angle-bisector theorem we get, so call Since Thus, (as ), and In this problem, we want to find yielding an answer of ~anduran Video Solution ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) Video Solution ~MathProblemSolvingSkills.com See also 2024 AIME II (Problems • Answer Key • Resources) Preceded by Problem 9Followed by Problem 11 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 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15425	https://blog.csdn.net/weixin_46498268/article/details/126447093	求y=sinx在[0,π]上的反函数_求函数y=sinx在[0,π]上的反函数-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作求y=sinx在[0,π]上的反函数最新推荐文章于 2025-08-20 20:23:08 发布原创于 2022-08-21 09:17:53 发布·6k 阅读 · 5 · 10· CC 4.0 BY-SA版权版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。文章标签： #几何学社区：2048 AI社区加入数学分享专栏收录该内容 2 篇文章订阅专栏本文详细介绍了正弦函数sinx和反正弦函数arcsinx的基本性质，包括定义域、值域和周期，并展示了如何求y=sinx在特定区间上的反函数。此外，通过实例解析了如何计算由y=sinx定义的曲边梯形绕y轴旋转的立体体积，涉及两种不同的方法。 1.先介绍sinx和arcsinx相关知识 sinx：正弦函数 arcsinx：反正弦函数 y=sinx图像 y=arcsinx图像性质对比 \| 函数类型 \| 定义域 \| 值域 \| 周期 \| 对乘轴 \| --- --- \| y=sinx \| R \| [-1,1] \| 2π \| kπ+(π/2) \| \| y=arcsinx \| [-1,1] \| [-π/2,π/2] \| \| 中心对称 \| 2.应用（1）求y=sinx[0，π]的反函数解：当x∈[0,π/2]，y=sinx的反函数x=arcsiny 当x∈[π/2,π]，y=sinx的定义域不是其反函数的值域，所以我们首先要利用诱导公式把sinx在[π/2,π]的定义域变成[0,π/2]。易得π-x∈[0,π/2]且sin(π-x)=sinx=y，所以其反函数x=π-arcsiny (2)求0≤y≤sinx，0≤x≤π所示平面图形绕y轴旋转所得立体的体积解析：把图中阴影部分绕y轴旋转所得立体，y∈[-1,1]，我们可以把V看做两部分组成，[π/2,π]绕y轴旋转的体积V1减去[0,π/2]/绕y轴旋转的体积【方法一】解：当x∈[0,π/2]，y=sinx的反函数x=arcsiny；当x∈[π/2,π]，y=sinx反函数x=π-arcsiny。 (\begin{align} V &=\pi\int_{0}^{1}\left ( \pi-\arcsin y \right )^{2}-\left ( \arcsin y \right )^{2}dy\ &=\pi\int_{0}^{1}\left ( \pi^{2}-2\pi\arcsin y\right )dy\ &=2\pi^{2} \end{align}) 【方法二】解析：曲边梯形0≤y≤f(x),a≤x≤b绕y轴旋转所得立体体积的公式为 V=2 π∫b a x f(x)d x 直接带公式： (\begin{align} V&=2\pi\int_{0}^{\pi}xsinxdx\ &=2\pi(xcosx-sinx)\big\|_{0}^{\pi}\ &=2\pi^{2} \end{}) 确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用被数学狂虐的小金关注关注 5点赞踩 10 收藏觉得还不错? 一键收藏 0评论分享复制链接分享到 QQ 分享到新浪微博扫一扫打赏打赏打赏举报举报专栏目录 x ,1/x的关系；反函数的实质：xy可以互换，就是进行坐标轴变化，就是关于y = x对称反函数性质：f-1(f(x))= x；反三角函数‘ ZJQ的博客 10-14 1万+ 强烈的纠正x ,1/x之间不是反函数的关系，这是由于定义给的公式造成的。反函数一般来说，你认准：关于y = x对称。也就反函数之间x，y值互换成立：下面举例： sin（π/2）= 0;arcsin(1)= π/2； tan(π/4)= 1;arctan(1)= π/4; 设函数y = f(x)的定义域是D，值域是f(D)。如果对于值域f(D)中的每一个y，在D中有且只有一个x使得f(y)= x，则按... y = sinx 在[0,2 π]上的反函数？y = sinx 在[π/2,π]上的反函数是x = π-arcsiny?通过此文弄清楚三角函数反函数中的关系不负时光可怜人 09-23 2万+ 三角函数中反函数相关性质的应用；诱导公式的应用。在考研数学中，交换积分次序，遇到三角函数反函数如何求解问题参与评论您还未登录，请先登录后发表或查看评论 y = sinx 在[0,2 π]上的反函数?y = sinx 在[π/2,π]上的反函数是x = π-arcsin... 9-27 举例:求 sinx[π/2,π]上的反函数 [0,π/2]时,x = arcsin y,结合arcsin y的定义域,可知无法直接应用在[π/2,π]上; 于是利用诱导公式,当x在[π/2,π]时,π-x属于[0,π/2],y = sinx = sin(π-x); 此时,对应的反函数 arcsin y = π-x,易得x = π-arcsiny. 总结关键在于利用好诱导公式与理解三角函数... 关于 sinX 与y的大小比较取值范围计算_ sinx<y 9-22 通过 sinX 的图像推导出正确范围,并强调了反函数在这一过程中的限制。关于 sinX 与y的大小比较取值范围计算 @(概率论) 在求分布函数的时候,常常有已知XX的分布函数求 Y = g(X)Y = g(X)的分布函数类型。往往不小心就会计算出错误的范围,从而导致分布函数求错。比如:X∈[0,π]X∈[0,π] 则 sinX≤y sinX≤y的范围... 如何用Python 求函数 y = sinx 在区间[0, pi/2]上的弧长 Alex_mercer_boy的博客 07-25 4133 [今天在高数课上老师讲到求光滑曲线弧长问题，老师自己想了一个例子：求函数 y = sinx 在区间[0, pi/2]上的弧长但是经过微分和积分发现没法求出确定值，百度后发现不可积，是椭圆积分，只能求近似值。于是我就心血来潮用Python 进行求近似基本原理：（弧长公式）基本步骤可以如下：将区间分割为n份（可以等分）取每一小区间 Δx （）以及在x1处求得导函数于是我们可以写出以下代码： import math times_n = 1, 1 0, 1 0 0, 1 0 0 0. 高等数学：三角函数。反三角函数 Boki 11-22 4万+ 三角函数，反三角函数怎么用科学计算机算反三角函数值域,反三角函数值域怎么求 9-28 [正弦函数y = sinx 在[-π/2,π/2]上的反函数,叫做反正弦函数。记作arc sinx,表示一个正弦值为x的角,该角的范围在[-π/2,π/2]区间内。定义域[-1,1] ,值域[-π/2,π/2]。 2.反余弦函数余弦函数y = cosx在[0,π]上的反函数,叫做反余弦函数。记作arccosx,表示一个余弦值为x的角,该角的范围在0... 用计算机怎么求反三角函数图像及性质,反三角函数图像及性质 9-21 arc sinx =-arcsin(-x) arcsin(sinx)= x ,x属于[0,π/2] arccosx 反三角函数中的反余弦。含意为:余弦的反函数,函数为y = arccosx,函数图象如右下图。便是已知余弦数值,反求角度,例如cos(a) = b,则arccos(b) = a; 它的值是以弧度表现的角度。定义域:[-1,1]。微积分-反函数 6.6（反三角函数） weixin_45911156的博客 10-09 2665 在本节中，我们应用 6.1 节的思想来找到所谓反三角函数的导数。在这个任务中，我们遇到了一些困难：由于三角函数不是一对一的，它们没有反函数。这个困难通过限制这些函数的定义域，使其成为一对一的函数，从而得到解决。从图 1 可以看到，正弦函数 y = sin⁡xy = \sin xy = sinx 不是一对一的函数（使用水平线测试）。但是函数 f(x)= sin⁡xf(x) = \sin xf(x)= sinx 在 −π 2≤x≤π 2-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}−2 π≤ arccotx图像在matlab,反三角函数图像与性质是什么？ weixin_32349561的博客 03-29 5万+ 0 1反三角函数是反正弦arc sinx，反余弦arccosx，反正切arctanx，反余切arccotx，反正割arcsecx，反余割arccscx这些函数的统称，各自表示其反正弦、反余弦、反正切、反余切，反正割，反余割为x的角。三角函数的反函数是个多值函数，因为它并不满足一个自变量对应一个函数值的要求，其图像与其原函数关于函数 y = x 对称。欧拉提出反三角函数的概念，并且首先使用了“arc+函数名... 蔡高厅高等数学19-初等函数的连续性、反函数的连续性、复合函数的连续性... 9-17 例如: y = sinx 在Ix [-π/2,π/2] , 上是单调增且连续,因此其反函数 y = arc sinx 在[-1,1]上单调增, 且连续。 y = cosx在Ix[0,π]上是单调减, 因此其反函数在[-1,+1]这个区间是单调减且连续 y = tanx 在(-π/2,π/2) 内单调增且连续,其反函数在y = arctanx在(-∞,+∞)去区间内单调增且连续 ... 反三角函数解析 9-26 acrcot(反余切函数) y = cot x在(0 ,π)内单减,—>y = arccot x在(-∞ , +∞)也单减具体如下图例题设f(x+1) = x2-x,求 f(x). 解: 令x + 1 = t,则 x = t - 1带入上式中得 f(t) = (t - 1)2- (t - 1) = t2- 3t + 2 ... 考研数学二复习笔记-高等数学-第一章函数极限连续 qq_61249949的博客 03-02 1391 反函数: 1.y = sinx -->反函数为 y = arc sinx x∈[-π 2\frac{π}{2}2 π,π 2\frac{π}{2}2 π] 推广: y = π-arcsin x,x∈[π 2\frac{π}{2}2 π,32\frac{3}{2}23π] 推算过程: y = sinx = sin(π-x) 因此反函数 1. x = arcsiny 2.π-x = arcsiny 2.y = cosx -->反函数 y = arccos x 推广:y = arccos x,x∈[0,π] & y = sinx 求反函数 06-07 函数 y = sinx 在区间 [-π/2, π/2] 内是单调增加的，因此它存在反函数。反函数记为 x = sin⁻¹y，意思是求出一个角度 x，使得 sinx = y。则有 sinx = y，两边同时对 x 求导： cosx dx/dy = 1 dx/dy = 1/... 函数解析与极限理论 9-13 当题目给出反函数,让我们求原函数的时候,可以根据书上所说反函数记作x = f-1(y),而通常我们习惯上写y = f-1(x),这样我们就知道其实给出的反函数只要将它的x变成y,y变成x就可以了。反函数引申 :反三角函数是不是三角函数的反函数呢? 答案是否定的,因为三角函数没有反函数,或者说在一定的定义域内是有反三... 已知三角函数值用计算机如何求角度,已知三角函数值,如何来求角度? 9-26 sinx 的局部区间内,例如在[- , ]内,考虑函数y = sinx, xÎ[- , ] (2) 因为它在定义域上单调增加,反函数是存在的(图6-19).把值域是[-1, 1]的函数(2)(注意它不是正弦函数)的反函数称为反正弦函数. 我们已经知道,“sin”本来就是一个函数记号,你一看见函数 sinx,尽管没有具体的x的数学式,但立即... 用MATLAB绘制显函数方程 y = sin(tan x) -tan(sinx)在x∈[-π，π ] 区间上的曲线图形. 10-24 在MATLAB中，你可以按照以下步骤来绘制函数y = sin(tan(x)) - tan(sin(x))在区间x ∈ [-π, π] 上的曲线： 1. 首先，你需要打开MATLAB并创建一个新的工作空间。 2. 使用syms x命令定义变量x为符号变量，因为... 反函数与周期函数.doc 12-25 A选项y =\|sinx\|在(0,π)上是增函数，但不是偶函数，因为\|sin(-x)\|=\|sinx\|。B选项y =\|cosx\|是偶函数，且周期为π，但在(0,π)上先减后增。C选项y =\|sin2x\|的周期是π/2，不满足条件。D选项y = cos2x是偶函数，周期为π，... 【概率论与数理统计】第二章随机变量及其分布(4) 夏明亮的博客 01-06 1604 第二章随机变量及其分布 4 随机变量函数的概率分布 4.1 离散型随机变量函数的概率分布有时候我们所关心的随机变量不能直接测量得到，而他确是某个能直接测量的随机变量的函数。（可以某个简单函数的符合函数，大概就是这么个意思）例如：我们能测量圆的直径XXX，而关心的却是其面接Y = π(X2)2 = π 4X2Y = \pi (\frac{X}{2})^2 = \frac{\pi}{4} X^2Y = π(2X)2 = 4 πX2；这里随机变量YYY就是随机变量XXX的函数。设g(x)g(x)g(x)是一个给定的连续函反三角函数 qq_51519554的博客 07-11 1万+ / / / y = sinx 与 y = arc sinx 的图像（忽略定义域） / / / y = cosx 与 y = arccosx 的图像（忽略定义域） / / / y = tanx 与 y = arctanx 的图像（忽略定义域） / / / 反三角函数公式 arc sinx 的图_arctanx的图像(arc sinx 的图像) weixin_39915210的博客 01-26 3万+ 简介：反三角函数中的反正切。意思为：tan(a) = b；等价于 Arctan(b) = a。计算性. + arctan (1/x) = π/2 反三角函数在无穷小替换公式中的应用：当x→0 时，arctanx~x谁能画出arctanx的图像？麻烦发到我邮箱：cycy-0 17@163.com 谢谢！！！.y = arctan(x)的定义域：{x∣x∈r} 值域：(-π/2，π/2)把cotx在定义域(-... 反函数的求导法则热门推荐白水的博客 01-13 15万+ 如果函数x = f(y)x = f(y)在区间IyI_y内单调、可导且f′(y)≠0 f'(y) \neq 0，那么它的反函数 y = f−1(x)y = f^{-1}(x)在区间Ix ={x\|x = f(y)，y∈Iy}I_x = {x \| x = f(y)，y \in I_y}内也可导，且[f−1(x)]′= 1f′(y)或dydx = 1dxdy[f^{-1}(x)]' = \frac{1}{f'(y)} 或《算法导论》第 33 章 - 计算几何学 2302_80961196的博客 08-20 2147 本文摘要：《算法导论》第33章深入讲解计算几何学核心算法，包括线段相交检测（叉积判断方向）、凸包寻找（Graham扫描法）和最近点对查找（分治法）。每部分均提供完整C++实现，包含关键函数如叉积计算、凸包排序和分治合并策略。特别强调浮点数精度处理（EPS常量）和算法优化技巧（如提前排序减少开销）。文章还给出实际应用场景（图形学、机器人导航）和两道思考题（共线点处理、排序优化），适合通过代码实践加深理解。所有功能整合为可交互式菜单程序，便于测试不同算法模块。 PDF Gallery_2 0 25 0 927_143 0 51.pdf 09-27 PDF Gallery_2 0 25 0 927_143 0 51.pdf Audio categories sample 09-27 Shows how to create apps that use various audio categories. 基于Matlab平台开发的六轴工业机器人轨迹规划与运动学仿真系统_六轴机器人正逆解直线圆弧轨迹规划姿态插补S型速度曲线规划三次样条插值算法_为工业机器人控制算法开发提供完整的仿真验.zip 09-27 基于Matlab平台开发的六轴工业机器人轨迹规划与运动学仿真系统_六轴机器人正逆解直线圆弧轨迹规划姿态插补S型速度曲线规划三次样条插值算法_为工业机器人控制算法开发提供完整的仿真验.zip LED16X16点阵显示最新发布 09-27 LED16X16 关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司被数学狂虐的小金博客等级码龄6年 7 原创28 点赞 101 收藏 13 粉丝关注私信猜你想问 sinx的反函数存在条件是什么？如何求sinx在指定区间的反函数？计算sinx曲边梯形绕y轴旋转体体积的方法有哪些？ 🔥码云GVP开源项目 16k starUniapp + ElementUI 功能强大支持多语言、二开方便广告 TA的精选新 Frullani(傅汝兰尼积分) 2077 阅读新今天说说rand和randi的区别 8254 阅读热 matlab中变量的命名规则 14511 阅读热 matlab中行向量和列向量相乘的两种情况 13652 阅读热 MATLAB绘制散点图——plot 12194 阅读查看更多 2022年 2篇 2021年 5篇大家在看如何为便携大功率音箱设计高效升压供电电路？！ k8s-pod的资源限制基于SpringBoot的校园共享电动车短租平台系统(毕业设计项目源码+文档) 两款高效实用的在线工具推荐基于Spring Boot的中医养生系统设计 840 分类专栏数学分享2篇笔记5篇上一篇：今天说说rand和randi的区别下一篇： Frullani(傅汝兰尼积分) 🔥码云GVP开源项目 16k starUniapp + ElementUI 功能强大支持多语言、二开方便广告上一篇：今天说说rand和randi的区别下一篇： Frullani(傅汝兰尼积分) 分类专栏数学分享2篇笔记5篇展开全部收起登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包打赏作者被数学狂虐的小金你的鼓励将是我创作的最大动力 ¥1¥2¥4¥6¥10¥20 扫码支付：¥1 获取中扫码支付您的余额不足，请更换扫码支付或充值打赏作者实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
15426	https://www.dynamedex.com/condition/osteoarthritis-oa-of-the-carpometacarpal-cmc-joint-of-the-thumb	Osteoarthritis (OA) of the Carpometacarpal (CMC) Joint of the Thumb - DynaMedex Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Cookie Policy Specialties Alerts Drug Resources General Drug Monographs NeoFax®& Pediatrics Drug Monographs Drug Interaction Checker IV Compatibility Tool Chemo Regimens Patient Info Calculators EBM Čeština Español Deutsch English Français Italiano Nederlands Norsk Português Русский Suomi Svenska 中文日本語 한국어 CME Čeština Español Deutsch English Français Italiano Nederlands Norsk Português Русский Suomi Svenska 中文日本語 한국어 Top of Page Condition Updated 28 Oct 2024 About CME Osteoarthritis (OA) of the Carpometacarpal (CMC) Joint of the Thumb Editors: Brent R. DeGeorge MD; Alan Ehrlich MD, FAAFP; Amir Qaseem MD, PhD, MHA, MRCP (London), FACP All Editors & Disclosures Topic Updates Follow Cite Share Overview and Recommendations Background Information Description Also Called Definitions Types Classification Systems Normal Anatomy and Function Epidemiology Etiology and Pathogenesis History and Physical Diagnosis Management Complications Prognosis Prevention Outcome Measures Quality Improvement Guidelines and Resources Patient Information References Topic Menu < Previous Section Next Section > Background Information Description Osteoarthritis (OA) of the carpometacarpal (CMC) joint of the thumb is a degenerative disorder of the cartilage between the trapezium and base of the first metacarpal, which can result in subluxation and functional limitations.1 Also Called Thumb OA Thumb basal joint arthritis or thumb basilar joint arthritis CMC joint arthritis or CMC arthritis First CMC joint osteoarthritis Trapeziometacarpal joint osteoarthritis Degenerative arthritis of the carpometacarpal joint of the thumb Osteoarthrosis of the carpometacarpal joint of the thumb Definitions Background for terminology used when referring to digits/fingers: The term digit is a more general term commonly used when referring to the thumb, any finger, or any toe. Digits are numbered 1-5; on the hand, numbering starts with the thumb and moves towards the ulnar side, while on the foot it starts with the great toe moving towards the lateral side. The term finger is a more specific term. It is typically used when referring to digits 2-5 of the hand. Finger is often distinct from the term thumb, which refers to the first digit of the hand. Reference - Am J Med Genet A 2009 Jan;149A(1):93 Common terminology used for each digit/finger of the hand: Thumb: first digit Index finger: second digit Middle finger (also commonly referred to as long finger): third digit Ring finger: fourth digit Little finger (also commonly referred to as small finger or pinky finger): fifth digit PubMed 31210604 The Journal of hand surgery, European volume J Hand Surg Eur Vol 20190701 44 6 657-658 657 Reference - J Hand Surg Eur Vol 2019 Jul;44(6):657 Types OA can be classified into primary and secondary types. Primary (idiopathic) OA: Primary OA is the most common type of OA. It is defined as OA that is associated with no known anatomic abnormality, trauma, or disease process. Primary OA typically affects multiple joints. It is considered a diagnosis of exclusion. Secondary OA: It is defined as OA that is associated with an anatomic abnormality, preceding trauma, or disease that predisposes the hand to mechanical factors, leading to joint degeneration. Secondary OA typically affects a single joint. Reference - Perm J 2018;22:17 Classification Systems Eaton and Littler classification of arthritis of the first CMC joint: Stage 1: Normal or subtle joint space widening, subluxation ≤ 1/3 of the joint surface, and normal articular contours Stage 2: Narrowing of the joint space, subluxation ≤ 1/3 of the joint surface, and the presence of osteophytes or loose bodies < 2 mm in size Stage 3: Advanced narrowing of the joint space, subluxation > 1/3 of the joint surface, presence of osteophytes or loose bodies ≥ 2 mm in size, and subchondral cysts or sclerosis Stage 4: Findings associated with Stage 3 (above) plus arthritis of the scaphotrapeziotrapezoidal (STT) joint, trapeziotrapezoid joint, or trapeziometacarpal joint of the index finger References - 2 ,4 , J Hand Surg Am 1985 Sep;10(5):645 Noyes and Stabler classification of arthritis: Stage 1a: Softening of cartilage with residual resilience Stage 1b: Softening of cartilage with no resilience Stage 2a: < 50% loss of vertical substance cartilage Stage 2b: > 50% loss of vertical substance cartilage Stage 3a: Total cartilage loss with an intact subchondral zone Stage 3b: Total cartilage loss with a damaged subchondral zone References - 2 , Am J Sports Med 1989 Jul-Aug;17(4):505 Normal Anatomy and Function Osseous Anatomy Hand metacarpals: The 5 metacarpals (1 thumb metacarpal and 4 finger metacarpals) are tubular bones which connect the carpus to the phalanges and form the volar concavity of the palm. The thumb (first) metacarpal is shorter and thicker compared to the other metacarpals, articulating exclusively with the trapezium. PubMed 28654615 Plastic and reconstructive surgery Plast Reconstr Surg 20170701 140 1 140e-151e 140e Reference - Plast Reconstr Surg 2017 Jul;140(1):140e The thumb attaches to the wrist via the trapezium bone, and it is composed of successive joints and phalangeal bones. The trapezium has 4 basal joints, articulating with the scaphoid, trapezoid, and first and second metacarpal bases. The first carpometacarpal joint connects the trapezium to the metacarpal bone. The metacarpophalangeal joint connects the metacarpal bone to the proximal phalanx. The interphalangeal joint connects the proximal phalanx to the distal phalanx. Reference - Pol J Radiol. 2020;85:e461-e488 The scaphoid, trapezium, and trapezoid combine to form the scaphotrapeziotrapezoidal (STT) joint, which may be affected by OA of the CMC joint of the thumb.4 Tendinous Anatomy Flexor and extensor tendons stabilize the thumb and contribute to its motion. Flexor pollicis longus tendon: The flexor pollicis longus tendon originates in the forearm along the palmar aspect of the radius and extends through the radial portion of the carpal tunnel to the thumb's palmar surface between the metacarpophalangeal sesamoid bones that overlie the palmar plate. It inserts on the palmar distal phalangeal base and flexes the interphalangeal and metacarpophalangeal joints. The thumb pulleys help stabilize the tendon and attach it to the palmar surfaces of the metacarpophalangeal joint, proximal phalanx, and interphalangeal joint. The thumb has only A pulleys. A1 is at the metacarpophalangeal joint, and A2 is at the interphalangeal joint. The oblique annular pulley (A0) is at the mid proximal phalanx level. A variable pulley, located between the A1 and A0 pulleys, is reported in 93% of cadavers. Extensor tendons: 2 wrist extensor compartments abduct and extend the thumb. The first extensor compartment is comprised of 2 tendons. The abductor pollicis longus tendon has a dominant insertion on the radial margin of the first metacarpal bone. The extensor pollicis brevis tendon inserts on the proximal phalangeal base or the dorsal first metacarpophalangeal joint capsule. The third extensor compartment is comprised of the extensor pollicis longus tendon. It inserts on the dorsal distal phalangeal base and allows interphalangeal joint extension. It is stabilized by the extensor hood at the metacarpophalangeal joint, which has radial and ulnar sagittal bands and slightly more distal triangular expansions that extend distally to insert on the distal phalangeal base. Reference - Pol J Radiol. 2020;85:e461-e488 Ligamentous Anatomy The carpometacarpal joint is a biconcave, saddle-shaped joint between the trapezium and first metacarpal base. It allows for range of motion including flexion with medial rotation, extension with lateral rotation, abduction and adduction, and circumduction. The carpometacarpal joint is stabilized by multiple ligaments. Anterior oblique ligament (also called the beak ligament): The anterior oblique ligament is a palmar structure that goes between the palmar tubercle of the trapezium and the ulnar margin of the first metacarpal base. It contains superficial and deep bands, and it is the chief restraint to dorsal radial subluxation. Dorsal ligaments: These ligaments play a role in resisting dorsal metacarpal base subluxation. The dorsal ligaments include: Posterior oblique ligament Dorsoradial ligament (thickest and strongest of the dorsal ligaments) Intermetacarpal ligament Ulnar collateral ligament Radial and ulnar collateral ligaments: The radial collateral ligament resists varus angular force, and the ulnar collateral ligament resists valgus angular force. The radial and ulnar collateral ligaments are composed of a proper ligament and an accessory ligament. The proper ligament runs from the dorsal metacarpal head to the palmar proximal phalangeal base and is taut in flexion and lax in extension. The accessory ligament is more palmar and runs from the metacarpal head to the volar plate and sesamoid bones. The adductor aponeurosis arises from the adductor pollicis and runs superficially to the ulnar collateral ligament and attaches onto the proximal shaft of the proximal phalanx. Reference - Pol J Radiol. 2020;85:e461-e488 Hand/Finger Function The metacarpophalangeal (MCP) joints are condylar joints that allow for flexion (most stable position), extension, radial, and ulnar motions (Hand Clin 2013 Nov;29(4):507, Clin Sports Med 2015 Jan;34(1):69). The thumb plays a role in both power grip and precise fine motor movements (for example, pinching and opposition) (Clin Sports Med 2015 Jan;34(1):69, Proc (Bayl Univ Med Cent) 2000 Oct;13(4):343). Finger-specific functions: The index finger plays a very important role in hand function due to its: Proximity to the thumb Ability to abduct and adduct Role in precision pinch and directional grip The middle finger produces the most individual flexion force, and it is involved in both power grip and precision movements. The ring and little fingers are the fingers primarily involved in power grip. They are less commonly involved in precision grip or pinch maneuvers. The little finger plays an important role in holding objects in the palm, as well as assisting with grasping while spanning an object (due to its ability to abduct). PubMed 24209947 Hand clinics Hand Clin 20131101 29 4 483-92 483 References - Hand Clin 2013 Nov;29(4):483, Proc (Bayl Univ Med Cent) 2000 Oct;13(4):343 Most hand functions are made up of 7 primary movements. Precision pinch (terminal pinch): A flexed thumb interphalangeal (IP) joint and second distal interphalangeal (DIP) joint come together. Precision pinch allows the tips of the thumb and index finger to pick up small objects such as a pen. Oppositional pinch (subterminal pinch): The distal aspects (pulp) of the thumb and index finger come together with the IP and DIP joints extended. Oppositional pinch allows for greater forces to be generated through thumb opposition. Key pinch: The distal aspect of the thumb and radial aspect of the middle phalanx of the index finger come together. Key pinch allows objects, such as a key, to be held. Chuck grip (directional grip): The first, second, and third phalanges come together by applying a rotational and axial force. Chuck grip allows cylindrical objects, such as a screwdriver, to be held. Hook grip: The second through fifth DIP and proximal interphalangeal (PIP) joints are flexed, while the MCP joints are extended (no function is required by the thumb). Hook grip allows for objects, such as a briefcase or suitcase, to be picked up. Power grasp: The fingers are flexed and the thumb is flexed and opposed (relative to other fingers). Power grasp allows for gripping of objects such as a bat or club. Span grasp: The DIP and PIP joints are flexed to about 30 degrees and the thumb is abducted, creating forces between the thumb and fingers. It requires stability at the thumb, MCP, and IP joints. Span grasp allows for the gripping of objects such as a ball. PubMed 24209947 Hand clinics Hand Clin 20131101 29 4 483-92 483 Reference - Hand Clin 2013 Nov;29(4):483 Epidemiology Incidence/Prevalence In patients with OA of the hand, the CMC joint of the thumb is commonly affected due to repetitive loading during activities of daily living (J Hand Surg Glob Online 2023 Jan;5(1):108, Nat Rev Rheumatol 2018 Nov;14(11):641). Female adults aged 50-70 years are most commonly affected.4 In adults > 50 years old, prevalence is reported to be 25% in female patients (up to 36% in postmenopausal female patients) and 8% in male patients.1 Prevalence varies by age, with a reported prevalence of about 15% in patients aged > 30 years and 85% in patients aged 71-80 years (J Hand Surg Glob Online 2023 Jan;5(1):108). Risk Factors Risk factors for OA of the CMC joint of the thumb include:2 Female sex Age > 40 years Menopause Family history of arthritis Overweight or obesity Ligamentous laxity Occupational exposures (for example, repetitive thumb use or heavy manual labor) Previous joint injury Knuckle cracking may not be associated with OA of the hand (J Am Board Fam Med 2011 Mar-Apr;24(2):169). Associated Conditions Carpal tunnel syndrome may be present in up to 43% of patients.4 De Quervain tenosynovitis may be present in up to 15% of patients.4 Etiology and Pathogenesis Causes The cause of OA of the CMC joint of the thumb is not fully understood but is likely due to ligament deterioration of the CMC joint, leading to metacarpal subluxation, joint instability, and pain.1 Causes of secondary OA of the hand may include prior joint changes, metabolic disorders, and endocrine disorders. Prior joint changes may be caused by: Previous trauma Previous septic arthritis Inflammatory arthropathies, such as rheumatoid arthritis (RA) Metabolic disorders that may cause OA of the hand include: Calcium pyrophosphate deposition (CPPD) disease (pseudogout) Hemochromatosis Wilson disease Gaucher disease Endocrine disorders that may cause OA of the hand include: Acromegaly Diabetes mellitus Hypothyroidism Hyperparathyroidism PubMed 17043444 Journal of clinical rheumatology : practical reports on rheumatic & musculoskeletal diseases J Clin Rheumatol 20031201 9 6 359-61 359 References - J Pharm Pharmacol 2014 Mar;66(3):339, J Clin Rheumatol 2003 Dec;9(6):359 Pathogenesis Proposed pathogenesis of OA: The disease process may start with localized, nonuniform degradation of articular cartilage followed by: Sclerosis of subchondral bone Development of osteophytes Synovial inflammation Precipitating events are unclear, but may be due to age-related morphological and metabolic changes affecting chondrocytes and the extracellular matrix, resulting in fragmentation of cartilage matrix proteins and release of inflammatory mediators. Inflammatory mediators (such as cytokines) appear to play a significant role in both the development and progression of OA. Soluble inflammatory mediators are elevated in synovial fluid in patients with OA, leading to synovitis. The source of inflammatory mediators may be from joint cells (local) or from other tissues (systemic). It is controversial whether inflammatory mediators play a primary or secondary role in cartilage damage and defective repair mechanisms in the OA disease process. The mechanism of pain production is not fully elucidated, but may be due to: Synovitis and effusion Bone marrow edema Changes in subchondral bone structure Instability Reference - World J Orthop 2014 Jul 18;5(3):351 Advanced stages of OA of the CMC joint of the thumb can lead to hand deformities.4 The thumb metacarpal migrates dorsally and proximally, forcing the distal portion of the metacarpal into adduction. This may cause compensatory hyperextension of the first metacarpophalangeal (MCP) joint and flexion of the thumb interphalangeal joint, resulting in the zigzag deformity. The scaphotrapeziotrapezoidal (STT) joint also becomes involved with progressive arthritis and compensatory extension of the scaphoid. < Previous Section Next Section > Related Topics Operative Management of Osteoarthritis (OA) of the Carpometacarpal (CMC) Joint of the ThumbThumb Pain - Approach to the PatientWrist Pain - Approach to the PatientOsteoarthritis (OA) of the Hand Related Topics Operative Management of Osteoarthritis (OA) of the Carpometacarpal (CMC) Joint of the ThumbThumb Pain - Approach to the PatientWrist Pain - Approach to the PatientOsteoarthritis (OA) of the Hand Published by EBSCO Information Services. Copyright © 2025, EBSCO Information Services. All rights reserved. No part of this may be reproduced or utilized in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without permission. EBSCO Information Services accepts no liability for advice or information given herein or errors/omissions in the text. It is merely intended as a general informational overview of the subject for the healthcare professional. Editorial Leadership About Mobile System Check Help Privacy Policy Terms of Use Copyright Manage Cookies © 2025 EBSCO Industries, Inc. All rights reserved. Unlock full access to evidence-based medical guidance. Sign inSubscribe Or, sign up for a FREE Trial Subscribe for unlimited access to DynaMedex content. Subscribe Already subscribed? Sign in
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Using transpose versus ctranspose in MATLAB Ask Question Asked 11 years, 1 month ago Modified9 years ago Viewed 3k times This question shows research effort; it is useful and clear 26 Save this question. Show activity on this post. When transposing vectors/matrices in MATLAB, I've seen and used just the ' (apostropohe) operator for a long time. For example: ```matlab v = [ 1 2 3 ]' v = 1 2 3 ``` However this is the conjugate transpose as I've recently found out, or ctranspose. This seems to only matter when there are complex numbers involved, where if you want to transpose a matrix without getting the conjugate, you need to use the .' opertator. Is it good practice to also use the .' for real matrices and vectors then? What should we be teaching MATLAB beginners? arrays matlab matrix Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Sep 23, 2015 at 14:09 Luis Mendo 113k 13 13 gold badges 80 80 silver badges 154 154 bronze badges asked Aug 5, 2014 at 23:38 legaslegas 406 4 4 silver badges 10 10 bronze badges 8 1 Hi @rayryeng, thank you! I saw your replies to a few Matlab posts recently! I'm in awe. I've been lurking around, finally have a bit of courage to start contributing. Great dialogue between you, Luis Mendo, and Yvon.legas –legas 2014-08-06 00:42:12 +00:00 Commented Aug 6, 2014 at 0:42 5 @legas It's a good idea to contribute. You'll find you learn a lot by answering and seeing other people's answers. My Matlab skills have improved a lot that way. But watch out, it can be very addictive :-)Luis Mendo –Luis Mendo 2014-08-06 01:16:11 +00:00 Commented Aug 6, 2014 at 1:16 3 @LuisMendo - I definitely agree. I've had a SO account for 6 months, but I didn't start actively answering questions until about 3 months ago. Over 3 months, I gained about 6500 reputation.... answering questions is very addictive!rayryeng –rayryeng 2014-08-06 01:20:19 +00:00 Commented Aug 6, 2014 at 1:20 2 +1 nice question. It's a good practice distinguishing between ' and .' even when you have nothing to do with complex numbers. Much like avoiding the use of i and j as variable names.Shai –Shai 2014-08-06 05:57:50 +00:00 Commented Aug 6, 2014 at 5:57 2 One thing just forgot to mention. I'm really confused by the . Try to follow the pattern - "The . in A.B means to perform on each element in A and B; the . in A./B means to perform / on each element in A and B; the .' in A.' means to perform ' on ??? in A ?? How much sense does it make that "element-wise transpose" a matrix? If I were the student how could I establish an analogy between . and .'?Yvon –Yvon 2014-08-06 08:20:06 +00:00 Commented Aug 6, 2014 at 8:20 \|Show 3 more comments 4 Answers 4 Sorted by: Reset to default This answer is useful 28 Save this answer. Show activity on this post. Interesting question! I would definitely say it's good practice to use .' when you just want to transpose, even if the numbers are real and thus ' would have the same effect. The mains reasons for this are: Conceptual clarity: if you need to transpose, just transpose. Don't throw in an unnecessary conjugation. It's bad practice. You'll get used to writing ' to transpose and will fail to notice the difference. One day you will write ' when .' should be used. As probable illustrations of this, see this question or this one. Future-proofness. If one day in the future you apply your function to complex inputs the behaviour will suddenly change, and you will have a hard time finding the cause. Believe me, I know what I say 1. Of course, if you are using real inputs but a conjugation would make sense for complex inputs, do use '. For example, if you are defining a dot product for real vectors, it may be appropriate to use ', because should you want to use complex inputs in the future, the conjugate transpose would make more sense. 1 In my early Matlab days, it took me quite a while to trace back a certain problem in my code, which turned out to be caused by using ' when I should have used .'. What really got me upset is, it was my professor who had actually said that ' meant transpose! He forgot to mention the conjugate, and hence my error. Lessons I learned: ' is not .'; and professors can tell you things that are plain wrong :-) Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited May 23, 2017 at 12:34 CommunityBot 1 1 1 silver badge answered Aug 6, 2014 at 0:09 Luis MendoLuis Mendo 113k 13 13 gold badges 80 80 silver badges 154 154 bronze badges 10 Comments Add a comment rayryeng rayryengOver a year ago Hahahah! I made a reference to you in my post... and look who decides to write one :) +1 from me 2014-08-06T00:12:34.433Z+00:00 1 Reply Copy link Luis Mendo Luis MendoOver a year ago @rayryeng I couldn't resist answering. You know what strong feelings I have on this :-) 2014-08-06T00:17:15.387Z+00:00 2 Reply Copy link legas legasOver a year ago Thanks you, yes I was also taught ' and the exact same problem happened to me, took me SO LONG to find it. :) 2014-08-06T00:26:30.23Z+00:00 3 Reply Copy link rayryeng rayryengOver a year ago @legas - I remember the first time I encountered this slight (but disastrous) difference. I was implementing the DFT on my own without using any built-in functions. To do this vectorized, this requires you to compute a matrix of complex coefficients. To do the inverse, you simply have to transpose the matrix. It took me several hours to figure out why it wasn't working... and that's because I used ' instead of .'... d'oh! Like Luis, it's an experience like this that ingrains the concept of ' and .' to be forever stuck in your head. 2014-08-06T00:41:22.15Z+00:00 2 Reply Copy link legas legasOver a year ago I'm still fairly new to Stack Overflow etiquette, can I accept everyone's answer? Otherwise I choose this one because @LuisMendo has "Matlab's ' is not transpose; .' is." in his profile blurb haha! 2014-08-06T00:48:16.45Z+00:00 1 Reply Copy link Add a comment\|Show 5 more comments This answer is useful 12 Save this answer. Show activity on this post. My very biased view: Most cases I use ' are purely "formal", aka not related to mathematical calculations. Most likely I want to rotate a vector like the index sequence 1:10 by 90 degrees. I seldomly use ' to matrices since it's ambiguous - the first question you've to answer is why you want to make a transpose? If the matrix is originally defined in a wrong direction, I would rather define the matrix in the correct one it should be, but not turning it afterwards. To transpose a matrix for a mathematical calculation, I explicitly use transpose and ctranspose. Because by doing so the code is easier to read (don't have to focus on those tiny dots) and to debug (don't have to care about missing dots). Do the following jobs such as dot product as usual. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Aug 6, 2014 at 0:21 answered Aug 6, 2014 at 0:15 YvonYvon 3,003 2 2 gold badges 19 19 silver badges 41 41 bronze badges 6 Comments Add a comment rayryeng rayryengOver a year ago Very nicely put. I agree with the non-ambiguity. +1 2014-08-06T00:15:50.297Z+00:00 0 Reply Copy link craigim craigimOver a year ago I mostly agree with you here except that I have found that sometimes you will define a matrix correctly, use a built-in function on it, and the built-in function will have mysteriously transposed it. I have started peppering my code with statements such as vectorOut = reshape(vectorOut,size(vectorIn)); to enforce a consistent matrix shape. 2014-08-06T00:21:36.92Z+00:00 3 Reply Copy link Yvon YvonOver a year ago @rayryeng I also learned ambiguity and avoiding it in a hard way. But luckily it took me not so long to learn to be prudent when naming variables and choosing functions (before actually keyboarding the code!). 2014-08-06T00:25:42.493Z+00:00 0 Reply Copy link Yvon YvonOver a year ago @legas Will it help if you add an example with complex numbers? 2014-08-06T00:35:59.227Z+00:00 1 Reply Copy link rayryeng rayryengOver a year ago @Yvon - Whenever I give MATLAB seminars, this is one of the points I always stress, so yes an example is always good. A bonus is to show them that using ' to compute the magnitude squared of a complex vector. 2014-08-06T00:38:41.937Z+00:00 2 Reply Copy link Add a comment\|Show 1 more comment This answer is useful 10 Save this answer. Show activity on this post. This is actually a subject of debate among many MATLAB programmers. Some say that if you know what you're doing, then you can go ahead and use ' if you know that your data is purely real, and to use .' if your data is complex. However, some people (such as Luis Mendo) advocate the fact that you should definitely use .' all the time so that you don't get confused. This allows for the proper handling of input into functions in case the data that are expected for your inputs into these functions do turn out to be complex. There is a time when complex transposition is required, such as compute the magnitude squared of a complex vector. In fact, Loren Shure in one of her MATLAB digests (I can't remember which one...) stated that this is one of the reasons why the complex transpose was introduced. My suggestion is that you should use .' always if your goal is to transpose data. If you want to do some complex arithmetic, then use ' and .' depending on what operation / computation you're doing. Obviously, Luis Mendo's good practices have rubbed off on me. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited May 23, 2017 at 12:26 CommunityBot 1 1 1 silver badge answered Aug 6, 2014 at 0:12 rayryengrayryeng 105k 22 22 gold badges 200 200 silver badges 204 204 bronze badges 1 Comment Add a comment Yvon YvonOver a year ago It might be too detailed for a beginner to distinguish ' and .' for his/her first day with Matlab. But once introduced with a real problem, it should be the teacher's duty to stress the student on the differences (poor Luis :( 2014-08-06T00:32:12.923Z+00:00 3 Reply Copy link This answer is useful -1 Save this answer. Show activity on this post. There are two cases to distinguish here: Taking transpose for non-mathematical reasons, like you have a function that is treating the data as arrays, rather than mathematical vectors, and you need your error correcting input to get it in the format that you expect. Taking transpose as a mathematical operation. In the latter case, the situation has to dictate which is correct, and probably only one of the two choice is correct in that situation. Most often that will be to take the conjugate transpose, which corresponds to ', but there are cases where you must take the straight transpose and then, of course, you need to use .'. In the former case, I suggest not using either transpose operator. Instead you should either use reshape or just insist that the input be make correctly and throw an error if it is not. This clearly distinguishes these "computer science" instance from true mathematical instances. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 8, 2016 at 13:58 BrickBrick 4,299 8 8 gold badges 31 31 silver badges 55 55 bronze badges 1 Comment Add a comment Luis Mendo Luis MendoOver a year ago For an arbitrary matrix, I don't think you can achieve transpose behaviour with reshape (you can for vectors) 2017-03-31T00:18:27.067Z+00:00 2 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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15428	https://arxiv.org/pdf/1606.08042	On Families of Numerical Semigroups with two Coprime Generators and Dimension three Antoine Mhanna Kfardebian, Lebanon, tmhanat@yahoo.com Abstract This paper examines in a new way some known facts about numerical semigroups especially when the number of minimal generators (that is the embedding dimension) is at most three and at least two minimal gener-ators are coprime. For such semigroups, an algorithm is re-investigated to find the pseudo-Frobenius numbers. A certain family of n dimensional numerical semigroups of type at most n − 1 is also given. MSC 2020: 11D07; 11A05. Keywords: Numerical Semigroup, pseudo-Frobenius Number. 1 Introduction and preliminaries Let a1, . . . , a n be n positive integers with gcd( a1, . . . , a n) = 1. The set S =( nX i=1 λiai n ∈ N, λ i ≥ 0, for all i ) is called the numerical semigroup S and a1, . . . , a n are called the generators of S. They are minimal generators if we cannot take out a generator ai without changing the set S; in general we denote S by ⟨a1, . . . , a n⟩. For a numerical semigroup S̸ = N the number F (S) := max {n ∈ N\|n / ∈ S} exists (see [7, Theorem 1.0.1] for proof) and is called the Frobenius number of S. We will suppose hereafter that a1 < a 2. The interested reader can see and the references therein (in particular ) for an extensive reading on numerical semigroups. There are different algorithms computing F (S) with S a three minimally generated semigroup; among them is , also [4, 9], and R¨ odseth’s formula . We refer to for a further discussion on these and other generalized algorithms. In this section we revisit the classical semigroup ⟨a1, a 2⟩. We will represent a proof from [8, p. 134] for two main purposes: one, this makes the article more self-contained and two, reformulates in details an often cited result of Sylvester on the Frobenius problem. 1 arXiv:1606.08042v4 [math.NT] 2 Sep 2023 Definition 1. Let a1 and a2 be positive integers with gcd( a1, a 2) = 1 . The set of positive integers x of the form x = αa i − βa 3−i, i = 1 or 2 with 0 < α < a 3−i and 0 < β < a i is called the non-compound set and is denoted by N C . The set of integers x of the form x = αa 1 + βa 2 with α > 0, β > 0, and x < a 1a2 is called the compound set and is denoted by C. Lemma 1. Let S = ⟨a1, a 2⟩ and let x be a positive integer; we have x ∈ N C if and only if 0 < x < a 1a2 and x / ∈ S.Proof. Suppose x ∈ S and in N C with x < a 1a2. Consequently, x = aa 1 +ba 2 = αa i − βa 3−i for some non-negative numbers a, b ; but this contradicts the fact that α < a 3−i and β < a i (for example if i = 1, we have ( a − α)a1 = −(β + b)a2 and \|a − α\| < a 2). Take the positive integers less than a1a2 having neither a1 nor a2 as a divisor, there are ( a1 − 1)( a2 − 1) such integers. Arrange these numbers as pairs ( x, y ) summing a1a2; if x ∈ N C , then a1a2 − x = y ∈ S and if y ∈ C, then ( a1a2 − y) = x / ∈ S, which means that x ∈ N C .We have proven that \|N C \| = \|C\| = (a1 − 1)( a2 − 1) 2 . The number F(S) = (a1 − 1)( a2 − 1) − 1 = a1a2 − a1 − a2 does not belong to S since a1a2 − F (S) is compound. Any number y > a 1a2 can be written as αa 1a2 + r and it is easily seen that since r < a 1a2, the remainder r is either in S or in N C . Therefore all n > F(S) do belong to S.Now we can characterize all the numbers that are not in S = ⟨a1, a 2⟩. Proposition 1. The set N C is the set {F (S) − y\|y ∈ S, y < F(S)}.Proof. A direct application of previous result. Remark 1. Notice at last that any integer x < a 1a2 in S has a unique repre-sentation as x = αa 1 + βa 2 with (α, β ) ∈ N2, α < a 2, and β < a 1; assuming the contrary αa 1 + βa 2 = νa 1 + γa 2, that is (α − ν)a1 = ( γ − β)a2 and we can not have a2 divides (α − ν) or a1 divides (γ − β). A number x in N C can be written as a1a2 − wa 1 − ra 2 = ( a2 − w)a1 − ra 2, where 1 ≤ w < a 2 and 1 ≤ r < a 1.Hence, the uniqueness of (w, r ) ∈ N2 follows similarly by contradiction. We end this expository section by the following definition. Definition 2. For a numerical semigroup S, we have T (S) := {x ∈ N\|x̸ ∈ S, x + s ∈ S, for all s ∈ S, s > 0}. The cardinality of T (S), denoted t(S), is called the type of S and a number in T (S) is called a pseudo-Frobenius number. The Ap´ ery set of S with respect to n ∈ S is the set Ap( S, n ) = {s ∈ S\|s−n / ∈ S} and the genus of S, denoted g(S), is the cardinality of {N\S}. 2 Adding minimal generator Here we consider numerical semigroups S minimally generated by a1, a 2, a 3; the integers a1 and a2 are coprime with a1 < a 2. The next remark is already well known and is very useful, see [6, Chapter 1]. 2Remark 2. [14, 13] Let a1, a 2, . . . , a n be positive integers. If gcd( a2, . . . , a n) = d and aj = da ′ j for each j > 1, then the following holds: • F (⟨a1, a 2, . . . , a n⟩) = dF (⟨a1, a ′ 2 , . . . , a ′ n ⟩) + a1(d − 1) (Johnson’s formula). • The type of ⟨a1, a 2, . . . , a n⟩ equals type of ⟨a1, a ′ 2 , . . . , a ′ n ⟩. • The type of S = ⟨a1, a 2, a 3⟩ is at most two. • We have the equivalence: w ∈ Ap( ⟨a1, a ′ 2 , . . . , a ′ n ⟩, a 1) ⇐⇒ dw ∈ Ap( ⟨a1, a 2, . . . , a n⟩, a 1). Remark 3. A number x in N C is written as a1a2 − ka 2 − ja 1 and we have 0 < k < a 1 if and only if 0 < j < a 2. This is useful when ix ∈ N C (i ∈ N∗) and 0 < ik < a 1 so 0 < ij − (i − 1) a2 < a 2 or 0 < ij < a 2 so 0 < ik − (i − 1) a1 < a 1. From the unique representation of N C numbers we have Lemma 2. Lemma 2. Let S = ⟨a1, a 2, a 3⟩, Q = ⟨a1, a 2⟩, and a3 = a1a2 − ka 2 − ja 1 for some (k, j ) with 1 ≤ k < a 1 and 1 ≤ j < a 2. If 2a3 ∈ Q, then the possible pseudo-Frobenius numbers are {a1a2 − (k + 1) a2 − a1, a 1a2 − a2 − (j + 1) a1}. Corollary 1. Let S = ⟨a, ka + d, ha + 2 d⟩ with d ≥ 1 and h a non-negative integer. For a = 2 i + 1 odd, if gcd( a, d ) = 1 and ⌊ h 2 ⌋ < k < h ( a+1 2 ) + d, then T (S) = {F1 := a(ha + 2 d) − a − ( a−12 + 1)( ha + 2 d), F 0 := a(ha + 2 d) − (h a−12 + d + 1 − k + h)a − (ha + 2 d)}.Proof. Write first ka + d = a(ha + 2 d) − h a − 12 + d − k + h a − a − 12 (ha + 2 d)and ha + 2 d = 2( ka + d) − (2 k − h)a. From these expressions and the conditions stated, we know that the generators of S are minimal. Take Q = ⟨a, ha + 2 d⟩, where a1 = a and a2 = ( ha + 2 d). Apply Lemma 2; since F0 + a3 = (2 k − h − 1)( a1) + i(a2) ∈ Q and F1 + a3 = a1(d − 1 + hi + k) ∈ Q we get the result. The case k = h is well known . See [7, Theorem 5.4.3] and for a general family of numerical semigroups with an arithmetic progression of generators. Definition 3. Let S = ⟨a1, a 2, a 3⟩ and Q = ⟨a1, a 2⟩. If x := a1a2 − ka 2 − ja 1, 1 ≤ k < a 1, and 1 ≤ j < a 2, then the integers k and j are called the level and the under-level of x, respectively. Let m be the least integer such that m · a3 ∈ Q with m ≥ 3, x1 := a3 and let xi := ia 3 /∈ Q for every i, 2 ≤ i ≤ m − 1. By Remark 3 we have either ik < a 1 (so ij > a 2) with ij − (i − 1) a2 < a 2, which we refer to as an upgrade start or ij < a 2 (so ik > a 1) with ik − (i − 1) a1 < a 1 referred to as a downgrade start, where i verifies 2 ≤ i ≤ s1 ≤ m − 1 for a certain s1. 3Next we represent an algorithm to find T (S). Definition 4. Keeping the previous notation, if h denotes the highest level and l the highest under-level among xi, then our pseudo-Frobenius numbers candidates are of the form F0 := a1a2 − a2 − (l + 1) a1 + wa 3 for some w ≥ 0 and F1 := a1a2 − (h + 1) a2 − a1 + ra 3 for some r ≥ 0, where the integers w and r are the least integers such that F1 + a3 ∈ Q and F0 + a3 ∈ Q, respectively. It is worth mentioning that the two previous pseudo-Frobenius candidates are the precise formulas for pseudo-Frobenius numbers of S. This is proved by Ralf Fr¨ oberg et al. [13, Theorem 11, second proof] where similar characteriza-tions of pseudo-Frobenius numbers for any S = ⟨a1, a 2, a 3⟩ are given, yielding that the type of ⟨a1, a 2, a 3⟩ is at most two. Example 1. We illustrate the above definitions here; take ⟨11 , 13 , 10 ⟩, a3 =10 = 11 · 13 − 5 · 11 − 6 · 13 , and it can be verified that m = 5 , (5 · a3 ∈ Q = ⟨11 , 13 ⟩). The semigroup has a downgrade start with s1 = 2 , h = 7 , and l = 10 . The pseudo-Frobenius numbers are 11 · 13 − 11 − 8 · 13 + 10 = 38 and 11 · 13 − 11 · 11 − 13 + 20 = 29 . The previous lemma has the following extension: Theorem 1. Let S = ⟨a1, . . . , a n⟩ and Q = ⟨a1, a 2⟩, where a1 and a2 are two coprime generators. If ai + aj ∈ Q for all i and j ≥ 3, then the type of S is at most n − 1.Proof. The idea is simple in the sense that we get at most n−1 possible pseudo-Frobenius numbers Fi /∈ S verifying both Fi + a1 ∈ S and Fi + a2 ∈ S. By Lemma 1 we can write ai := a1a2 − αia2 − βia1 with i ≥ 3 for some 1 ≤ αi < a 1 and 1 ≤ βi < a 2. Arrange these generators so sequence ( αi)i≥3 is decreasing (strictly) and thus, ( βi)i≥3 is strictly increasing (since otherwise the difference between two consecutive minimal generators is in Q). Set F1 = a1a2 − a2 − max 3≤i≤n (βi + 1) a1,F2 = a1a2 − max 3≤i≤n (αi + 1) a2 − a1, and for 3 ≤ i ≤ n − 1, set Fi = a1a2 − (min( αi, α i+1 ) + 1) a2 − (min( βi, β i+1 ) + 1) a1. The proof is direct for possible pseudo-Frobenius numbers having their level or under-level equals 1 (the numbers F1 and F2 verify F1 /∈ S, F2 /∈ S, and Fi + aj ∈ S for any i = 1 , 2 and j = 1 , 2). Otherwise we have the result from: {ai + s\|s ∈ Q, i ≥ 3} ∪ Q = S and Remark 1. For example, one can prove that if Fi = a1a2 − ka 2 − ja 1 is a pseudo-Frobenius number, k > 1, and j > 1, then for some i we have αi+1 < k < α i and βi < j < β i+1 , to get the necessity of the given expressions. Remark 4. It is well known that numerical semigroups S of embedding dimen-sion n with t(S) ≤ n − 1 verify Wilf ’s conjecture; see [6, Chapter 1]. 43 Algorithm’s Applications Before giving some examples and applications, we point out that we do not discuss or compare other algorithms with the one in Definition 4. For example, the algorithm formulas [4, Corollary 12], proved by giving Ap´ ery sets , require additional parameters assuming pairwise coprime minimal generators. Theorem 2. Let S = ⟨a, b, c ⟩, a < b < c , d = gcd( a, b ), n = bv d − 1, p = n( cdv a − 1) , v ∈ N such that bv ≡ d (mod a) with vd < a , then F (⟨a, b, c ⟩) = ab (F (⟨n, n + 1 , n + p⟩) + 2 n + 1) dn (n + 1) + ( d − 1) c − (a + b). As stated by Igor Kan et al. [3, Theorem 1], Theorem 2 follows by applying Johnson’s formula (Remark 2) three times; it could be better seen if one starts considering b = db ′, a = da ′. Replacing the different variables n, p by the given values, the computation should be applied two times to the semigroup ⟨n, n + 1 , n + p⟩. Lastly the formula is applied for ⟨a, b, c ⟩ with gcd( a, b ) = d. Theorem 3. Let S = ⟨a, a + 1 , a + p⟩. If a ≥ p(p − 4) + 2 , then F (S) = ap (a + 1) + ( p − 1) a + 1 p ( p − 3) a − 1. 3.1 A downgrade semigroup Next we present particular semigroups that were introduced by A. M. Robles-P´ erez and J. C. Rosales . Lemma 3. Let S = ⟨a, a + 1 , ia + d⟩, a = dk + j, 0 < i < d < a , and 1 ≤ j < d .If (k + 1) i + j + 1 = d, then the type of S is one. Proof. The proof makes use of Remark 2: by finding a common factor f of a + 1, ia + d and showing that a ∈ ⟨ a+1 f , ia +df ⟩. Notice that a = j(k + 1) + k(i(1 + k) + 1) from the definition of a and d with a + 1 = ( d − i)( k + 1) and ia + d = ( d − i)(1 + i(1 + k)). The result follows taking f = d − i. Theorem 4. Let S = ⟨a, b, c ⟩ with pairwise coprime minimal generators and so c = bh − ar = ab − (a − h)b − ra = ( b − r)a − (a − h)b. If br ≥ l ah m , then T (S) = a − 1 h c + ra − b − a, (a − 1) b − a − 1 h ra − a . Using Remark 2 and [6, Proposition 2.20], it is not hard to prove that the formulas in the previous theorem are invariant with gcd( a, c ) = d or gcd( b, c ) = d and d > 1 (gcd( a, b ) = 1 necessarily). Consequently, by Theorem 2, we see that Theorem 4 and Theorem 5 are equivalent for those numerical semigroups of type 5Theorem 5. Let S = ⟨a, a + 1 , ia + d⟩, Q = ⟨a, a + 1 ⟩, where a = dk + j, a3 = ia + d, 0 < i < d < a , 1 ≤ j < d , and d ≥ 3. If (k + 1) i + j ≥ d − 1, then T (S) = {F1, F 2} for: F1 = a(a + 1) − (k(d − i) + 1) a − (a + 1) + wa 3,F2 = a(a + 1) − a − (a − d + 1)( a + 1) + ( k − 1) a3, where w = 0 if (k + 1) i + j ≥ d and w = k otherwise. Proof. Write ia + d = a(a + 1) − (d − i)a − (a − d)( a + 1), we are in a downgrade start and na 3 = a(a + 1) − n(d − i)a − (a − dn )( a + 1) for all n with 1 ≤ n ≤ k. The number ( k + 1) a3 ∈ Q since (k + 1) a3 = a(a + 1) − (a − (i(k + 1) + j − d)) a + ( d − j)( a + 1) . Set by definition wd := 3 + x, if F = a(a + 1) − (k(d − i) + 1) a − (a + 1) , then F / ∈ S and F + wa 3 = ( −2 + j − x + wi − 1 + ki )( dk + j) + (2 + x)( dk + j + 1) is in Q if, in particular: ki + j ≥ w(d − i) and 2 + x = wd − 1 ≥ 0 (1) , or − (a + 1) < ki + j − w(d − i) < 0 and wd − 1 ≥ a (2) . We call w0 the smallest such integer ≥ 1; but it can be verified that w0 exists as w0 = 1 if ( k + 1) i + j ≥ d from (1) and w0 = k + 1 otherwise (2). Thus, first F1 = a(a + 1) − (k(d − i) + 1) a − (a + 1) + ( w0 − 1)( ia + d) ∈ T (S). Take G = a(a + 1) − a − (a − d + 1)( a + 1) + k(ia + d) and set y = d − 1 − j, G = a(ki + j) + y(a + 1) ∈ Q. Now take F = a(d − 2) + d − 1 + ( k − t)( ia + d)with 1 ≤ t ≤ k, clearly F − G = −(td + ita ) thus, F = a(ki + j + td + ita ) − (−d + 1 + j + td + ita )( a + 1) = a(ki + j + td − it ) − (1 + j + ( t − 1) d)( a + 1) and clearly F is not in Q if and only if ki + j + t(d − i) < a + 1 and 1 + j + ( t − 1) d > 0. We call t0 the smallest such integer ≥ 1; but it can be verified that (k − 1) i + j + d < a + 1, that is ( k − 1) i < (k − 1) d + 1 and so if r = k − t0 = k − 1, the second pseudo-Frobenius is F2 = a(d − 2) + d − 1 + ( k − 1)( ia + d). Rearranging terms we get F1 = ( j + ki − w(d − i)) a + ( wd − 1)( a + 1) and F2 = a(( k − 1) i + j + d) − (1 + j)( a + 1); if w = 0, then F1̸ = F2 since 1 < 1 + j and for w = k we have ( k + 1) i + j + 1 = d with F1 = F2.6Remark 5. By Remark 2, the case d divides a in Theorem 5 is almost trivial. Note also that the condition (k + 1) i + j ≥ d − 1 of Theorem 5 is equivalent to a − (d − i)( k + 1) ≥ − 1 ⇐⇒ a + 1 d − i ≥ ad (T heorem 4) . See also [15, Corollary 1]. Example 2. For S = ⟨34 , 35 , 46 ⟩, we cannot apply Theorem 3, applying The-orem 5 we have d = 12 , a = 34 = 2 · 12 + 10 , i = 1 , k = 2 , and j = 10 . Here w = 0 and T (S) = {373 , 397 }. Notice that Theorem 4 gives the same result without the pairwise coprime condition. 3.2 Semigroup of three consecutive squares The Frobenius number formulas of numerical semigroups generated by three consecutive squares were given by M. Lepilov et al. and also by Fel . Here we compute their two pseudo-Frobenius numbers. One can verify that (n + 1)( n + 2) 2 = ( n + 1) n2 + 4( n + 1) 2 and n(2 n + 1) 2 = n(n + 1) 2 + (3 n + 2) n2 for any n. Lemma 4. Let S = ⟨(4 s − 1) 2, (4 s)2, (4 s + 1) 2⟩ and Q = ⟨n2, (n + 1) 2⟩ with n = 4 s−1 and s > 1; we have T = {256 s3 −144 s2 +8 s−1, 272 s3 −168 s2 +s−2}.Proof. First let us write (n + 2) 2 = n2(n + 1) 2 − (n2 − 2n − 4) n2 − (4 n − 4)( n + 1) 2;if n = 4 s − 1 and s > 1, then we verify that s(n + 2) 2 ∈ Q and (s − 1)( n + 2) 2 = (16 s2 − 15 s − 1) n2 − (n2 − 4n + 3)( n + 1) 2 /∈ Q. From Definition 3 and Definition 4 notation, we have that S has an upgrade start with s1 = s − 1, l = n2 − 2n − 4, and h = n2 − 4n + 3. Finally we have: n2(n + 1) 2 − (n2 − 2n − 3) n2 − (n + 1) 2 + ( α)( n + 2) 2 = (4 n + 4 + α(4 n + 5)) n2 − (1 + α(4 n − 4))( n + 1) 2, we verify n2(n+1) 2 −(n2 −2n−3) n2 −(n+1) 2 +( s−1)( n+2) 2 = ( s−1) n2 +(16 s−8)( n+1) 2 and n2(n + 1) 2 − n2 − (n2 − 4n + 4)( n + 1) 2 + ( n + 2) 2 = 16 sn 2. Lemma 5. Let S = ⟨(4 s − 3) 2, (4 s − 2) 2, (4 s − 1) 2⟩ and Q = ⟨n2, (n + 1) 2⟩ with n = 4 s − 3 and s > 1; we have T (S) = {144 s3 − 296 s2 + 193 s − 42 , 160 s3 − 352 s2 + 234 s − 51 }. 7Proof. If n = 4 s − 3 for some s > 1, then (2 s − 1)( n + 2) 2 ∈ Q, the highest under-level is for s(n + 2) 2 = n2(n + 1) 2 − (16 s2 − 25 s + 8) n2 − (8 s − 9)( n + 1) 2, thus, we get l = 16 s2 − 25 s + 8 and the highest level is for h = 16 s2 − 32 s + 16 with (s − 1)( n + 2) 2 = n2(n + 1) 2 − (7 s − 3) n2 − (16 s2 − 32 s + 16)( n + 1) 2 and (2 s − 2)( n + 2) 2 = n2(n + 1) 2 − (14 s − 6) n2 − (16 s2 − 40 s + 23)( n + 1) 2. Here s1 = s − 1 and we verify that n2(n + 1) 2 − n2 − (16 s2 − 32 s + 17)( n + 1) 2 + α(n + 2) 2 = ( α(4 n + 5) − 1) n2 − (8 − 8s + α(4 n − 4))( n + 1) 2,n2(n + 1) 2 − (16 s2 − 25 s + 9) n2 − (n + 1) 2 + α(n + 2) 2 = (9 s − 5 + α(4 n + 5)) n2 − (1 + α(4 n − 4))( n + 1) 2,n2(n+1) 2 −n2 −(16 s2 −32 s+17)( n+1) 2 +s(n+2) 2 = (9( s−1)+4) n2 +( n+1) 2, and n2(n + 1) 2 − (16 s2 − 25 s + 9) n2 − (n + 1) 2 + ( s − 1)( n + 2) 2 = (2( s − 1)) n2 +(8( s − 1))( n + 1) 2. Lemma 6. Let S = ⟨(4 s)2, (4 s + 1) 2, (4 s + 2) 2⟩ and Q = ⟨n2, (n + 1) 2⟩ with n = 4 s and s ≥ 2; we have T (S) = {112 s3 + 48 s2 + 8 s − 1, 128 s3 − 20 s − 5}.Proof. Write ( n + 2) 2 = n2(n + 1) 2 − (n2 − 2n − 4) n2 − (4 n − 4)( n + 1) 2; here we need the following expressions:  [l]s(n + 2) 2 = n2(n + 1) 2 − (3 s + 1) n2 − (16 s2 − 4s)( n + 1) 2, (s + 1)( n + 2) 2 = n2(n + 1) 2 − (16 s2 − 5s − 3) n2 − (12 s − 4)( n + 1) 2, 2s(n + 2) 2 = n2(n + 1) 2 − (2 + 6 s)n2 − (16 s2 − 8s)( n + 1) 2, (2 s + 1)( n + 2) 2 = n2(n + 1) 2 − (16 s2 − 2s − 2) n2 − (8 s − 4)( n + 1) 2, 3s(n + 2) 2 = n2(n + 1) 2 − (9 s + 3) n2 − (16 s2 − 12 s)( n + 1) 2, (3 s + 1)( n + 2) 2 = n2(n + 1) 2 − (16 s2 + s − 1) n2 − (4 s − 4)( n + 1) 2,n(n + 2) 2 = n2(n + 1) 2 − (12 s + 4) n2 − (16 s2 − 16 s)( n + 1) 2. With ( n + 1)( n + 2) 2 ∈ Q, we have s1 = s, h = 16 s2 − 4s, and l = 16 s2 + s − 1. Finally n2(n + 1) 2 − n2 − (16 s2 − 4s + 1)( n + 1) 2 + α(n + 2) 2 = ( α(4 n + 5) − 1) n2 − (1 − 4s + 16 αs − 4α)( n + 1) 2, 8n2(n + 1) 2 − (16 s2 + s)n2 − (n + 1) 2 + α(n + 2) 2 = (2 n + 1 − s + α(4 n + 5)) n2 − (1 + α(4 n − 4))( n + 1) 2,n2(n+1) 2 −n2 −(16 s2 −4s+1)( n+1) 2 +(3 s+1)( n+2) 2 = 3( n+1) 2 +(7 s+1) n2, and n2(n + 1) 2 − (16 s2 + s)n2 − (n + 1) 2 + s(n + 2) 2 = (4 s)n2 + (4 s − 1)( n + 1) 2. Lemma 7. Let S = ⟨(4 s − 2) 2, (4 s − 1) 2, (4 s)2⟩ and Q = ⟨n2, (n + 1) 2⟩ with n =4s−2 for s > 1; we have T (S) = {80 s3 −80 s2 +28 s−5, 128 s3 −160 s2 +60 s−9}.Proof. We check that ( n + 2) 2 = n2(n + 1) 2 − (n2 − 2n − 4) n2 − (4 n − 4)( n + 1) 2.The potential highest level and under-level are for the following multiples:  (s − 1)( n + 2) 2 = n2(n + 1) 2 − (11 s − 2) n2 − (16 s2 − 28 s + 12)( n + 1) 2,s(n + 2) 2 = n2(n + 1) 2 − (16 s2 − 13 s + 2) n2 − (4 s − 4)( n + 1) 2, (2 s − 1)( n + 2) 2 = n2(n + 1) 2 − (6 s − 1) n2 − (16 s2 − 24 s + 8)( n + 1) 2, (2 s)( n + 2) 2 = n2(n + 1) 2 − (16 s2 − 18 s + 3) n2 − (8 s − 8)( n + 1) 2, (3 s − 1)( n + 2) 2 = n2(n + 1) 2 − (s)n2 − (16 s2 − 20 s + 4)( n + 1) 2, (3 s)( n + 2) 2 = n2(n + 1) 2 − (16 s2 − 23 s + 4) n2 − (12 s − 12)( n + 1) 2,n(n + 2) 2 = n2(n + 1) 2 − (12 s − 2) n2 − (16 s2 − 32 s + 12)( n + 1) 2, with ( n + 1)( n + 2) 2 ∈ Q. Consequently, s1 = s − 1, h = 16 s2 − 20 s + 4, and l = 16 s2 − 13 s + 2. Finally we have n2(n + 1) 2 − n2 − (16 s2 − 20 s + 5)( n + 1) 2 + α(n + 2) 2 = ( α(4 n + 5) − 1) n2 − (1 − 4s + 16 αs − 12 α)( n + 1) 2,n2(n + 1) 2 − (16 s2 − 13 s + 3) n2 − (n + 1) 2 + α(n + 2) 2 = (5 s − 2 + α(4 n + 5)) n2 − (1 + α(4 n − 4))( n + 1) 2. Thus, n2(n + 1) 2 − n2 − (16 s2 − 20 s + 5)( n + 1) 2 + s(n + 2) 2 = 3( n + 1) 2 + (5 s − 2) n2 and n2(n+1) 2 −(16 s2 −13 s+3) n2 −(n+1) 2 +(3 s−1)( n+2) 2 = ( n)n2 +( n+1)( n+1) 2. Lemma 8 and Lemma 9 can be considered the reduction by 4 ( d = 4 see Remark 2) of Lemma 6 and Lemma 7, respectively . 9Lemma 8. Let S = ⟨(2 s)2, (4 s + 1) 2, (2 s + 1) 2⟩ and Q = ⟨n2, (n + 1) 2⟩ with n = 2 s > 2; we have T (S) = {32 s3 − 12 s2 − 11 s − 2, 28 s3 − 4s − 1}.Proof. First (2 n + 1) 2 = n2(n + 1) 2 − (2 n + 1) n2 − (n2 − 2n − 1)( n + 1) 2. For n = 2 s, the semigroup S has a downgrade start with s1 = s − 1, l = n2 − 3s − 1, and h = n2 − 2n − 1. By computation we have n2(n + 1) 2 − n2 − (n2 − 2n)( n + 1) 2 + α(2 n + 1) 2 = (2 n + α(2 n + 1))( n + 1) 2 − (1 + α(2 n + 1)) n2,n2(n+1) 2 −n2 −(n2 −2n)( n+1) 2 +( s−1)(2 n+1) 2 = ( s−1)( n+1) 2 +(7 s+1) n2, and n2(n + 1) 2 − (n2 − 3s)n2 − (n + 1) 2 + (2 n + 1) 2 = 3 sn 2 + 4 s(n + 1) 2. Lemma 9. Let S = ⟨(2 s + 1) 2, (4 s + 3) 2, (2 s + 2) 2⟩ and Q = ⟨n2, (n + 1) 2⟩ with n = 2 s + 1 > 1; we have T (S) = {20 s3 + 28 s2 + 9 s − 1, 32 s3 + 44 s2 + 13 s − 2}.Proof. First we got (2 n + 1) 2 = n2(n + 1) 2 − (2 n + 1) n2 − (n2 − 2n − 1)( n + 1) 2. Also if n = 2 s + 1, then s(2 n + 1) 2 /∈ Q and (s + 1)(2 n + 1) 2 = ( s + 1) n2 + (3 s + 2)( n + 1) 2, so s1 = s, l = n2 − s − 1, and h = n2 − 2n − 1. From computation: n2(n + 1) 2 − n2 − (n2 − 2n)( n + 1) 2 + α(2 n + 1) 2 = (2 n + α(2 n + 1))( n + 1) 2 − (1 + α(2 n + 1)) n2. By a direct verification n2(n + 1) 2 − n2 − (n2 − 2n)( n + 1) 2 + s(2 n + 1) 2 = (5 s + 3) n2 + (3 s + 1)( n + 1) 2 and n2(n + 1) 2 − (n2 − s)n2 − (n + 1) 2 + (2 n + 1) 2 = sn 2 + (4 s + 2)( n + 1) 2. 3.3 A cyclic semigroup The next lemma is the particular three dimensional case of Prof. Dao’s question at: . Lemma 10. [6, Proposition 10.16] Let a, b, c be three positive integers such that ab + a + 1 and ca + c + 1 are coprime. For S = ⟨ab + a + 1 , bc + b + 1 , ca + c + 1 ⟩,we have T (S) = {abc − 1, 2( abc − 1) }. 10 Proof. First it is clear from the cyclic expression that, if any two of the gener-ators share a common factor d, the third one will be divisible by d. Assuming the hypothesis, we write bc + b + 1 = ( ab + a + 1)( ca + c + 1) − c(ab + a + 1) − (( b + 1) a − b)( ca + c + 1) . The generators are a3 = bc + b + 1 and Q = ⟨ab + a + 1 , ca + c + 1 ⟩.We get m = a + 1 (see Definition 3) and (ab + a + 1)( ca + c + 1) −(ab + a + 1) − (( b + 1) a − b + 1)( ca + c + 1) + αa 3 = ( b + α(b + 1))( ca + c + 1) − (1 + αc )( ab + a + 1) . Finally we verify F1 = ( ab + a + 1)( ca + c + 1) − (ab + a + 1) − (( b + 1) a − b + 1)( ca + c + 1) + ( a − 1)( bc + b + 1) = 2( abc − 1) and F2 = ( ab + a + 1)( ca + c + 1) − (ca + 1)( ab + a + 1) − (ca + c + 1) = abc − 1. 4 Acknowledgments I thank the site and the reviewers for the corrections and advice. Computations were assisted through Maple software. References A. M. Robles-P´ erez and J. C. Rosales, The Frobenius problem for some numerical semigroups with embedding dimension equal to three, Hacettepe Journal of Mathematics and Statistics 44 (2015), 901–908. A. Tripathi, The Frobenius problem for modified arithmetic progressions, J. Integer Seq. 16(7) (2013), Article 13.7.4. I. D. Kan, B. S. Stechkin, and I. V. Sharkov, On the Frobenius problem for three arguments, Math. Notes 62 (1997), 521–523. J. C. Rosales and P. A. Garc´ ıa-S´ anchez, Numerical semigroups with embed-ding dimension three, Archiv der Mathematik 83 (2004), 488–496. J. C. Rosales and M. B. Branco, The Frobenius problem for numerical semi-groups, Journal of Number Theory 131 (2011), 2310–2319. J. C. Rosales and P. A. Garc´ ıa-S´ anchez, Numerical Semigroups , Develop-ments in Mathematics 20 , Springer, New York, 2009. J. L. Ram´ ırez Alfons´ ın, The Diophantine Frobenius Problem , Oxford Univ. Press, Oxford, 2005. 11 J. J. Sylvester. Excursus on rational fractions and partitions, Amer J. Math. 5 (1882), 119–136. L. G. Fel, Frobenius Problem for Semigroups S(d1, d 2, d 3), Funct. Anal. Other Math. 1(2006), 119–157. L. G. Fel, Numerical semigroups generated by squares, cubes and quartics of three consecutive integers, Conference paper, International meeting on numerical semigroups with applications , Levico-Terme Italy, 2016. M. Lepilov, J. O’Rourke, and I. Swanson, Frobenius numbers of numeri-cal semigroups generated by three consecutive squares or cubes, Semigroup Forum , 91 (2015), 238–259. ¨O. J. R¨ odseth, On a linear Diophantine problem of Frobenius, J. Reine Angew. Math. 301 , (1978) 171–178. R. Fr¨ oberg, C. Gottlieb, and R. H¨ aggkvist, On numerical semigroups, Semi-group Forum 35 (1987), 63–83. S. M. Johnson, A Linear Diophantine Problem, Can. J. Math. 12 (1960), 390–398. S. S. Batra, N. Kumar, and A. Tripathi, Some problems concerning the Frobenius number for extensions of an arithmetic progression, The Ramanu-jan Journal 48 (2019), 545–565. 12
15429	https://arxiv.org/html/2403.17110v1	Fixed Points and Cycles of Parking Functions 1 Introduction 2 Classical parking functions 3 Prime parking functions 4 Further discussions Fixed Points and Cycles of Parking Functions Martin Rubey Fakultät für Mathematik und Geoinformation, TU Wien, Vienna, Austria martin.rubey@tuwien.ac.atand Mei Yin Department of Mathematics, University of Denver, Denver, CO, USA mei.yin@du.edu Abstract. A parking function of length n 𝑛 n italic_n is a sequence π=(π 1,…,π n)𝜋 subscript 𝜋 1…subscript 𝜋 𝑛\pi=(\pi_{1},\dots,\pi_{n})italic_π = ( italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) of positive integers such that if λ 1≤⋯≤λ n subscript 𝜆 1⋯subscript 𝜆 𝑛\lambda_{1}\leq\cdots\leq\lambda_{n}italic_λ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ ⋯ ≤ italic_λ start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT is the increasing rearrangement of π 1,…,π n subscript 𝜋 1…subscript 𝜋 𝑛\pi_{1},\dots,\pi_{n}italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT, then λ i≤i subscript 𝜆 𝑖 𝑖\lambda_{i}\leq i italic_λ start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ≤ italic_i for 1≤i≤n 1 𝑖 𝑛 1\leq i\leq n 1 ≤ italic_i ≤ italic_n. In this paper we obtain some exact results on the number of fixed points and cycles of parking functions. Our proofs will be based on generalizations of Pollak’s argument. Extensions of our techniques are discussed. Key words and phrases: Parking function; Pollak’s circle argument 2010 Mathematics Subject Classification: 05A15; 05A19, 60C05 M.Yin was supported by the University of Denver’s Professional Research Opportunities for Faculty Fund 80369-145601. 1. Introduction Parking functions were introduced by Konheim and Weiss in the study of the linear probes of random hashing functions. In the classical parking function scenario, we have n 𝑛 n italic_n parking spaces on a one-way street, labelled 1,2,…,n 1 2…𝑛 1,2,\dots,n 1 , 2 , … , italic_n in consecutive order as we drive down the street. There are n 𝑛 n italic_n cars C 1,…,C n subscript 𝐶 1…subscript 𝐶 𝑛 C_{1},\dots,C_{n}italic_C start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_C start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT. Each car C i subscript 𝐶 𝑖 C_{i}italic_C start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT has a preferred space 1≤π i≤n 1 subscript 𝜋 𝑖 𝑛 1\leq\pi_{i}\leq n 1 ≤ italic_π start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ≤ italic_n. The cars drive down the street one at a time in the order C 1,…,C n subscript 𝐶 1…subscript 𝐶 𝑛 C_{1},\dots,C_{n}italic_C start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_C start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT. The car C i subscript 𝐶 𝑖 C_{i}italic_C start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT drives immediately to space i 𝑖 i italic_i and then parks in the first available space. Thus if i 𝑖 i italic_i is empty, then C i subscript 𝐶 𝑖 C_{i}italic_C start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT parks there; otherwise C i subscript 𝐶 𝑖 C_{i}italic_C start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT next goes to space i+1 𝑖 1 i+1 italic_i + 1, etc. If all cars are able to park, then the sequence π=(π 1,…,π n)𝜋 subscript 𝜋 1…subscript 𝜋 𝑛\pi=(\pi_{1},\dots,\pi_{n})italic_π = ( italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) is called a parking function of length n 𝑛 n italic_n. It is well-known and easy to see that if λ 1≤⋯≤λ n subscript 𝜆 1⋯subscript 𝜆 𝑛\lambda_{1}\leq\cdots\leq\lambda_{n}italic_λ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ ⋯ ≤ italic_λ start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT is the (weakly) increasing rearrangement of π 1,…,π n subscript 𝜋 1…subscript 𝜋 𝑛\pi_{1},\dots,\pi_{n}italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT, then π 𝜋\pi italic_π is a parking function if and only if λ i≤i subscript 𝜆 𝑖 𝑖\lambda_{i}\leq i italic_λ start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ≤ italic_i for 1≤i≤n 1 𝑖 𝑛 1\leq i\leq n 1 ≤ italic_i ≤ italic_n. In particular, any permutation of a parking function is a parking function. Write PF n subscript PF 𝑛\operatorname{PF}_{n}roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT for the set of parking functions of length n 𝑛 n italic_n. The first significant result on parking functions, due to Pyke in another context and then to Konheim and Weiss , is that the number of parking functions of length n 𝑛 n italic_n is equal to (n+1)n−1 superscript 𝑛 1 𝑛 1(n+1)^{n-1}( italic_n + 1 ) start_POSTSUPERSCRIPT italic_n - 1 end_POSTSUPERSCRIPT. A famous combinatorial proof was given by Pollak (unpublished but recounted in and ). It boils down to the following easily verified statement: Let G 𝐺 G italic_G denote the group of all n 𝑛 n italic_n-tuples (a 1,…,a n)∈[n+1]n subscript 𝑎 1…subscript 𝑎 𝑛 superscript delimited-[]𝑛 1 𝑛(a_{1},\dots,a_{n})\in[n+1]^{n}( italic_a start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_a start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) ∈ [ italic_n + 1 ] start_POSTSUPERSCRIPT italic_n end_POSTSUPERSCRIPT with componentwise addition modulo n+1 𝑛 1 n+1 italic_n + 1. Let H 𝐻 H italic_H be the subgroup generated by (1,1,…,1)1 1…1(1,1,\dots,1)( 1 , 1 , … , 1 ). Then every coset of H 𝐻 H italic_H contains exactly one parking function. We will be interested in deriving some exact formulas for the number of fixed points and cycles of parking functions in this paper. For a parking function π=(π 1,…,π n)𝜋 subscript 𝜋 1…subscript 𝜋 𝑛\pi=(\pi_{1},\dots,\pi_{n})italic_π = ( italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ), we say i 𝑖 i italic_i is a fixed point of π 𝜋\pi italic_π if π i=i subscript 𝜋 𝑖 𝑖\pi_{i}=i italic_π start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT = italic_i. More generally, for m≥1 𝑚 1 m\geq 1 italic_m ≥ 1, we say (i 1,…,i m)subscript 𝑖 1…subscript 𝑖 𝑚(i_{1},\dots,i_{m})( italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ) where the i j subscript 𝑖 𝑗 i_{j}italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT’s are all distinct is an m 𝑚 m italic_m-cycle of π 𝜋\pi italic_π if π i 1=i 2,π i 2=i 3,…,π i m−1=i m,π i m=i 1 formulae-sequence subscript 𝜋 subscript 𝑖 1 subscript 𝑖 2 formulae-sequence subscript 𝜋 subscript 𝑖 2 subscript 𝑖 3…formulae-sequence subscript 𝜋 subscript 𝑖 𝑚 1 subscript 𝑖 𝑚 subscript 𝜋 subscript 𝑖 𝑚 subscript 𝑖 1\pi_{i_{1}}=i_{2},\pi_{i_{2}}=i_{3},\dots,\pi_{i_{m-1}}=i_{m},\pi_{i_{m}}=i_{1}italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_m - 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT, so fixed points are simply 1 1 1 1-cycles. Our proofs will be based on generalizations of Pollak’s argument. In Section 2 we study classical parking functions, in Section 3 we study prime parking functions, and in Section 4 we discuss extensions of our techniques to (r,k)𝑟 𝑘(r,k)( italic_r , italic_k )-parking functions. 2. Classical parking functions Our first result identifies the number of parking functions with a prescribed number of fixed points. Theorem 2.1. Let 0≤k≤n 0 𝑘 𝑛 0\leq k\leq n 0 ≤ italic_k ≤ italic_n. The number of parking functions of length n 𝑛 n italic_n with k 𝑘 k italic_k fixed points is given by 1(n+1)2⁢(n+1 k)⁢(n n−k+1+(−1)n+k).1 superscript 𝑛 1 2 binomial 𝑛 1 𝑘 superscript 𝑛 𝑛 𝑘 1 superscript 1 𝑛 𝑘\frac{1}{(n+1)^{2}}\binom{n+1}{k}\left(n^{n-k+1}+(-1)^{n+k}\right).divide start_ARG 1 end_ARG start_ARG ( italic_n + 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG ( FRACOP start_ARG italic_n + 1 end_ARG start_ARG italic_k end_ARG ) ( italic_n start_POSTSUPERSCRIPT italic_n - italic_k + 1 end_POSTSUPERSCRIPT + ( - 1 ) start_POSTSUPERSCRIPT italic_n + italic_k end_POSTSUPERSCRIPT ) . Proof. For 1≤i≤n 1 𝑖 𝑛 1\leq i\leq n 1 ≤ italic_i ≤ italic_n, let A i=\|{π∈PF n:π i=i}\|subscript 𝐴 𝑖 conditional-set 𝜋 subscript PF 𝑛 subscript 𝜋 𝑖 𝑖 A_{i}=\|{\pi\in\operatorname{PF}{n}:\pi{i}=i}\|italic_A start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT = \| { italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : italic_π start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT = italic_i } \|. By De Morgan’s law and the inclusion-exclusion principle, the number of parking functions of length n 𝑛 n italic_n with k 𝑘 k italic_k fixed points is ∑1≤i 1<⋯<i k≤n\|(⋂j=1 k A i j)⁢⋂i≠i j⁢∀1≤j≤k A i c\|subscript 1 subscript 𝑖 1⋯subscript 𝑖 𝑘 𝑛 superscript subscript 𝑗 1 𝑘 subscript 𝐴 subscript 𝑖 𝑗 subscript 𝑖 subscript 𝑖 𝑗 for-all 1 𝑗 𝑘 superscript subscript 𝐴 𝑖 𝑐\displaystyle\hskip 56.9055pt\sum_{1\leq i_{1}<\cdots<i_{k}\leq n}\left\|\left(% \bigcap_{j=1}^{k}A_{i_{j}}\right)\bigcap_{i\neq i_{j}\,\forall 1\leq j\leq k}A% {i}^{c}\right\|∑ start_POSTSUBSCRIPT 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_i start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT ≤ italic_n end_POSTSUBSCRIPT \| ( ⋂ start_POSTSUBSCRIPT italic_j = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_k end_POSTSUPERSCRIPT italic_A start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT ) ⋂ start_POSTSUBSCRIPT italic_i ≠ italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∀ 1 ≤ italic_j ≤ italic_k end_POSTSUBSCRIPT italic_A start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_c end_POSTSUPERSCRIPT \| =\displaystyle==∑1≤i 1<⋯<i k≤n∑ℓ=0 n−k(−1)ℓ⁢∑1≤s 1<⋯<s ℓ≤n s t≠i j⁢∀1≤t≤ℓ⁢∀1≤j≤k\|(⋂j=1 k A i j)⁢⋂(⋂t=1 ℓ A s t)\|subscript 1 subscript 𝑖 1⋯subscript 𝑖 𝑘 𝑛 superscript subscript ℓ 0 𝑛 𝑘 superscript 1 ℓ subscript 1 subscript 𝑠 1⋯subscript 𝑠 ℓ 𝑛 subscript 𝑠 𝑡 subscript 𝑖 𝑗 for-all 1 𝑡 ℓ for-all 1 𝑗 𝑘 superscript subscript 𝑗 1 𝑘 subscript 𝐴 subscript 𝑖 𝑗 superscript subscript 𝑡 1 ℓ subscript 𝐴 subscript 𝑠 𝑡\displaystyle\sum{1\leq i_{1}<\cdots<i_{k}\leq n}\sum_{\ell=0}^{n-k}(-1)^{% \ell}\sum_{\begin{subarray}{c}1\leq s_{1}<\cdots<s_{\ell}\leq n\ s_{t}\neq i_{j}\,\forall 1\leq t\leq\ell\,\forall 1\leq j\leq k\end{subarray}}% \left\|\left(\bigcap_{j=1}^{k}A_{i_{j}}\right)\bigcap\left(\bigcap_{t=1}^{\ell}% A_{s_{t}}\right)\right\|∑ start_POSTSUBSCRIPT 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_i start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT ≤ italic_n end_POSTSUBSCRIPT ∑ start_POSTSUBSCRIPT roman_ℓ = 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_n - italic_k end_POSTSUPERSCRIPT ( - 1 ) start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT ∑ start_POSTSUBSCRIPT start_ARG start_ROW start_CELL 1 ≤ italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_s start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT ≤ italic_n end_CELL end_ROW start_ROW start_CELL italic_s start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ≠ italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∀ 1 ≤ italic_t ≤ roman_ℓ ∀ 1 ≤ italic_j ≤ italic_k end_CELL end_ROW end_ARG end_POSTSUBSCRIPT \| ( ⋂ start_POSTSUBSCRIPT italic_j = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_k end_POSTSUPERSCRIPT italic_A start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT ) ⋂ ( ⋂ start_POSTSUBSCRIPT italic_t = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT italic_A start_POSTSUBSCRIPT italic_s start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT end_POSTSUBSCRIPT ) \| =\displaystyle==∑ℓ=0 n−k(−1)ℓ⁢∑1≤i 1<⋯<i k≤n∑1≤s 1<⋯<s ℓ≤n s t≠i j⁢∀1≤t≤ℓ⁢∀1≤j≤k\|(⋂j=1 k A i j)⁢⋂(⋂t=1 ℓ A s t)\|superscript subscript ℓ 0 𝑛 𝑘 superscript 1 ℓ subscript 1 subscript 𝑖 1⋯subscript 𝑖 𝑘 𝑛 subscript 1 subscript 𝑠 1⋯subscript 𝑠 ℓ 𝑛 subscript 𝑠 𝑡 subscript 𝑖 𝑗 for-all 1 𝑡 ℓ for-all 1 𝑗 𝑘 superscript subscript 𝑗 1 𝑘 subscript 𝐴 subscript 𝑖 𝑗 superscript subscript 𝑡 1 ℓ subscript 𝐴 subscript 𝑠 𝑡\displaystyle\sum_{\ell=0}^{n-k}(-1)^{\ell}\sum_{1\leq i_{1}<\cdots<i_{k}\leq n% }\sum_{\begin{subarray}{c}1\leq s_{1}<\cdots<s_{\ell}\leq n\ s_{t}\neq i_{j}\,\forall 1\leq t\leq\ell\,\forall 1\leq j\leq k\end{subarray}}% \left\|\left(\bigcap_{j=1}^{k}A_{i_{j}}\right)\bigcap\left(\bigcap_{t=1}^{\ell}% A_{s_{t}}\right)\right\|∑ start_POSTSUBSCRIPT roman_ℓ = 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_n - italic_k end_POSTSUPERSCRIPT ( - 1 ) start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT ∑ start_POSTSUBSCRIPT 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_i start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT ≤ italic_n end_POSTSUBSCRIPT ∑ start_POSTSUBSCRIPT start_ARG start_ROW start_CELL 1 ≤ italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_s start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT ≤ italic_n end_CELL end_ROW start_ROW start_CELL italic_s start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ≠ italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∀ 1 ≤ italic_t ≤ roman_ℓ ∀ 1 ≤ italic_j ≤ italic_k end_CELL end_ROW end_ARG end_POSTSUBSCRIPT \| ( ⋂ start_POSTSUBSCRIPT italic_j = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_k end_POSTSUPERSCRIPT italic_A start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT ) ⋂ ( ⋂ start_POSTSUBSCRIPT italic_t = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT italic_A start_POSTSUBSCRIPT italic_s start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT end_POSTSUBSCRIPT ) \| =\displaystyle==∑ℓ=0 n−k(−1)ℓ⁢∑1≤i 1<⋯<i k≤n∑1≤s 1<⋯<s ℓ≤n s t≠i j⁢∀1≤t≤ℓ⁢∀1≤j≤k\|{π∈PF n:π 1<⋯<π k+ℓ{π 1,…,π k+ℓ}={i 1,…,i k,s 1,…,s ℓ}}\|,superscript subscript ℓ 0 𝑛 𝑘 superscript 1 ℓ subscript 1 subscript 𝑖 1⋯subscript 𝑖 𝑘 𝑛 subscript 1 subscript 𝑠 1⋯subscript 𝑠 ℓ 𝑛 subscript 𝑠 𝑡 subscript 𝑖 𝑗 for-all 1 𝑡 ℓ for-all 1 𝑗 𝑘 conditional-set 𝜋 subscript PF 𝑛 subscript 𝜋 1⋯subscript 𝜋 𝑘 ℓ subscript 𝜋 1…subscript 𝜋 𝑘 ℓ subscript 𝑖 1…subscript 𝑖 𝑘 subscript 𝑠 1…subscript 𝑠 ℓ\displaystyle\sum_{\ell=0}^{n-k}(-1)^{\ell}\sum_{1\leq i_{1}<\cdots<i_{k}\leq n% }\sum_{\begin{subarray}{c}1\leq s_{1}<\cdots<s_{\ell}\leq n\ s_{t}\neq i_{j}\,\forall 1\leq t\leq\ell\,\forall 1\leq j\leq k\end{subarray}}% \left\|\left{\pi\in\operatorname{PF}{n}:\begin{subarray}{c}\pi{1}<\cdots<\pi% {k+\ell}\ \left{\pi{1},\ldots,\pi_{k+\ell}\right}=\left{i_{1},\ldots,i_{k},s_{1},% \ldots,s_{\ell}\right}\end{subarray}\right}\right\|,∑ start_POSTSUBSCRIPT roman_ℓ = 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_n - italic_k end_POSTSUPERSCRIPT ( - 1 ) start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT ∑ start_POSTSUBSCRIPT 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_i start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT ≤ italic_n end_POSTSUBSCRIPT ∑ start_POSTSUBSCRIPT start_ARG start_ROW start_CELL 1 ≤ italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_s start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT ≤ italic_n end_CELL end_ROW start_ROW start_CELL italic_s start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ≠ italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∀ 1 ≤ italic_t ≤ roman_ℓ ∀ 1 ≤ italic_j ≤ italic_k end_CELL end_ROW end_ARG end_POSTSUBSCRIPT \| { italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : start_ARG start_ROW start_CELL italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_π start_POSTSUBSCRIPT italic_k + roman_ℓ end_POSTSUBSCRIPT end_CELL end_ROW start_ROW start_CELL { italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_k + roman_ℓ end_POSTSUBSCRIPT } = { italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT , italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_s start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT } end_CELL end_ROW end_ARG } \| , where the last equality follows by symmetry of parking coordinates, and by {π 1,…,π k+ℓ}={i 1,…,i k,s 1,…,s ℓ},subscript 𝜋 1…subscript 𝜋 𝑘 ℓ subscript 𝑖 1…subscript 𝑖 𝑘 subscript 𝑠 1…subscript 𝑠 ℓ\left{\pi_{1},\ldots,\pi_{k+\ell}\right}=\left{i_{1},\ldots,i_{k},s_{1},% \ldots,s_{\ell}\right},{ italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_k + roman_ℓ end_POSTSUBSCRIPT } = { italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT , italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_s start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT } , we mean equality as a set. To obtain a simplified expression for this last equation, we use an extension of Pollak’s circle argument. Add an additional parking spot n+1 𝑛 1 n+1 italic_n + 1, and arrange the spots in a circle. Allow n+1 𝑛 1 n+1 italic_n + 1 also as a preferred spot. We first select k+ℓ 𝑘 ℓ k+\ell italic_k + roman_ℓ spots for the first k+ℓ 𝑘 ℓ k+\ell italic_k + roman_ℓ cars. Note that it matters whether these spots correspond to i j subscript 𝑖 𝑗 i_{j}italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT for 1≤j≤k 1 𝑗 𝑘 1\leq j\leq k 1 ≤ italic_j ≤ italic_k or to s t subscript 𝑠 𝑡 s_{t}italic_s start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT for 1≤t≤ℓ 1 𝑡 ℓ 1\leq t\leq\ell 1 ≤ italic_t ≤ roman_ℓ, and altogether this can be done in (n+1 k,ℓ,n−k−ℓ+1)binomial 𝑛 1 𝑘 ℓ 𝑛 𝑘 ℓ 1\binom{n+1}{k,\ell,n-k-\ell+1}( FRACOP start_ARG italic_n + 1 end_ARG start_ARG italic_k , roman_ℓ , italic_n - italic_k - roman_ℓ + 1 end_ARG ) ways. Then for the remaining n−k−ℓ 𝑛 𝑘 ℓ n-k-\ell italic_n - italic_k - roman_ℓ cars, there are (n+1)n−k−ℓ superscript 𝑛 1 𝑛 𝑘 ℓ(n+1)^{n-k-\ell}( italic_n + 1 ) start_POSTSUPERSCRIPT italic_n - italic_k - roman_ℓ end_POSTSUPERSCRIPT possible preference sequences. Out of the n+1 𝑛 1 n+1 italic_n + 1 rotations for any preference sequence, only one rotation becomes a valid parking function. Standard circular symmetry argument yields ∑1≤i 1<⋯<i k≤n∑1≤s 1<⋯<s ℓ≤n s t≠i j⁢∀1≤t≤ℓ⁢∀1≤j≤k\|{π∈PF n:π 1<⋯<π k+ℓ{π 1,…,π k+ℓ}={i 1,…,i k,s 1,…,s ℓ}}\|subscript 1 subscript 𝑖 1⋯subscript 𝑖 𝑘 𝑛 subscript 1 subscript 𝑠 1⋯subscript 𝑠 ℓ 𝑛 subscript 𝑠 𝑡 subscript 𝑖 𝑗 for-all 1 𝑡 ℓ for-all 1 𝑗 𝑘 conditional-set 𝜋 subscript PF 𝑛 subscript 𝜋 1⋯subscript 𝜋 𝑘 ℓ subscript 𝜋 1…subscript 𝜋 𝑘 ℓ subscript 𝑖 1…subscript 𝑖 𝑘 subscript 𝑠 1…subscript 𝑠 ℓ\displaystyle\sum_{1\leq i_{1}<\cdots<i_{k}\leq n}\sum_{\begin{subarray}{c}1% \leq s_{1}<\cdots<s_{\ell}\leq n\ s_{t}\neq i_{j}\,\forall 1\leq t\leq\ell\,\forall 1\leq j\leq k\end{subarray}}% \left\|\left{\pi\in\operatorname{PF}{n}:\begin{subarray}{c}\pi{1}<\cdots<\pi% {k+\ell}\ \left{\pi{1},\ldots,\pi_{k+\ell}\right}=\left{i_{1},\ldots,i_{k},s_{1},% \ldots,s_{\ell}\right}\end{subarray}\right}\right\|∑ start_POSTSUBSCRIPT 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_i start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT ≤ italic_n end_POSTSUBSCRIPT ∑ start_POSTSUBSCRIPT start_ARG start_ROW start_CELL 1 ≤ italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_s start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT ≤ italic_n end_CELL end_ROW start_ROW start_CELL italic_s start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ≠ italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∀ 1 ≤ italic_t ≤ roman_ℓ ∀ 1 ≤ italic_j ≤ italic_k end_CELL end_ROW end_ARG end_POSTSUBSCRIPT \| { italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : start_ARG start_ROW start_CELL italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_π start_POSTSUBSCRIPT italic_k + roman_ℓ end_POSTSUBSCRIPT end_CELL end_ROW start_ROW start_CELL { italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_k + roman_ℓ end_POSTSUBSCRIPT } = { italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT , italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_s start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT } end_CELL end_ROW end_ARG } \| =(n+1 k,ℓ,n−k−ℓ+1)⁢(n+1)n−k−ℓ−1.absent binomial 𝑛 1 𝑘 ℓ 𝑛 𝑘 ℓ 1 superscript 𝑛 1 𝑛 𝑘 ℓ 1\displaystyle\hskip 56.9055pt=\binom{n+1}{k,\ell,n-k-\ell+1}(n+1)^{n-k-\ell-1}.= ( FRACOP start_ARG italic_n + 1 end_ARG start_ARG italic_k , roman_ℓ , italic_n - italic_k - roman_ℓ + 1 end_ARG ) ( italic_n + 1 ) start_POSTSUPERSCRIPT italic_n - italic_k - roman_ℓ - 1 end_POSTSUPERSCRIPT . So what we need to compute is ∑ℓ=0 n−k(−1)ℓ⁢(n+1 k,ℓ,n−k−ℓ+1)⁢(n+1)n−k−ℓ−1=(n+1 k)⁢((n+1)−1)n−k+1+(−1)n+k(n+1)2.superscript subscript ℓ 0 𝑛 𝑘 superscript 1 ℓ binomial 𝑛 1 𝑘 ℓ 𝑛 𝑘 ℓ 1 superscript 𝑛 1 𝑛 𝑘 ℓ 1 binomial 𝑛 1 𝑘 superscript 𝑛 1 1 𝑛 𝑘 1 superscript 1 𝑛 𝑘 superscript 𝑛 1 2\sum_{\ell=0}^{n-k}(-1)^{\ell}\binom{n+1}{k,\ell,n-k-\ell+1}(n+1)^{n-k-\ell-1}% =\binom{n+1}{k}\frac{((n+1)-1)^{n-k+1}+(-1)^{n+k}}{(n+1)^{2}}.∑ start_POSTSUBSCRIPT roman_ℓ = 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_n - italic_k end_POSTSUPERSCRIPT ( - 1 ) start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT ( FRACOP start_ARG italic_n + 1 end_ARG start_ARG italic_k , roman_ℓ , italic_n - italic_k - roman_ℓ + 1 end_ARG ) ( italic_n + 1 ) start_POSTSUPERSCRIPT italic_n - italic_k - roman_ℓ - 1 end_POSTSUPERSCRIPT = ( FRACOP start_ARG italic_n + 1 end_ARG start_ARG italic_k end_ARG ) divide start_ARG ( ( italic_n + 1 ) - 1 ) start_POSTSUPERSCRIPT italic_n - italic_k + 1 end_POSTSUPERSCRIPT + ( - 1 ) start_POSTSUPERSCRIPT italic_n + italic_k end_POSTSUPERSCRIPT end_ARG start_ARG ( italic_n + 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG . ∎ Extending the notion of fixed points to cycles, we have a more general result. Theorem 2.2. Let m≥1 𝑚 1 m\geq 1 italic_m ≥ 1 and k≥0 𝑘 0 k\geq 0 italic_k ≥ 0 with k⁢m≤n 𝑘 𝑚 𝑛 km\leq n italic_k italic_m ≤ italic_n. The number of parking functions of length n 𝑛 n italic_n with k 𝑘 k italic_k m 𝑚 m italic_m-cycles is given by ∑ℓ:(k+ℓ)⁢m≤n(−1)ℓ⁢((m−1)!)k+ℓ k!⁢ℓ!⁢(n+1 m,⋯,m﹈k+ℓ⁢m’s,n−(k+ℓ)⁢m+1)⁢(n+1)n−(k+ℓ)⁢m−1.subscript:ℓ 𝑘 ℓ 𝑚 𝑛 superscript 1 ℓ superscript 𝑚 1 𝑘 ℓ 𝑘 ℓ binomial 𝑛 1 subscript﹈𝑚⋯𝑚 𝑘 ℓ m’s 𝑛 𝑘 ℓ 𝑚 1 superscript 𝑛 1 𝑛 𝑘 ℓ 𝑚 1\sum_{\ell:(k+\ell)m\leq n}(-1)^{\ell}\frac{\left((m-1)!\right)^{k+\ell}}{k!% \ell!}\binom{n+1}{\underbracket{m,\cdots,m}_{k+\ell\hskip 2.84544pt\text{$m$'s% }},n-(k+\ell)m+1}(n+1)^{n-(k+\ell)m-1}.∑ start_POSTSUBSCRIPT roman_ℓ : ( italic_k + roman_ℓ ) italic_m ≤ italic_n end_POSTSUBSCRIPT ( - 1 ) start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT divide start_ARG ( ( italic_m - 1 ) ! ) start_POSTSUPERSCRIPT italic_k + roman_ℓ end_POSTSUPERSCRIPT end_ARG start_ARG italic_k ! roman_ℓ ! end_ARG ( FRACOP start_ARG italic_n + 1 end_ARG start_ARG under﹈ start_ARG italic_m , ⋯ , italic_m end_ARG start_POSTSUBSCRIPT italic_k + roman_ℓ italic_m ’s end_POSTSUBSCRIPT , italic_n - ( italic_k + roman_ℓ ) italic_m + 1 end_ARG ) ( italic_n + 1 ) start_POSTSUPERSCRIPT italic_n - ( italic_k + roman_ℓ ) italic_m - 1 end_POSTSUPERSCRIPT . Proof. The proof follows similarly as in Theorem 2.1. We will not include all the technical details, but instead walk through the key ideas. For distinct 1≤i 1,…,i m≤n formulae-sequence 1 subscript 𝑖 1…subscript 𝑖 𝑚 𝑛 1\leq i_{1},\dots,i_{m}\leq n 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ≤ italic_n, let A(i 1,…,i m)=\|{π∈PF n:π i 1=i 2,π i 2=i 3,…,π i m−1=i m,π i m=i 1}\|.subscript 𝐴 subscript 𝑖 1…subscript 𝑖 𝑚 conditional-set 𝜋 subscript PF 𝑛 formulae-sequence subscript 𝜋 subscript 𝑖 1 subscript 𝑖 2 formulae-sequence subscript 𝜋 subscript 𝑖 2 subscript 𝑖 3…formulae-sequence subscript 𝜋 subscript 𝑖 𝑚 1 subscript 𝑖 𝑚 subscript 𝜋 subscript 𝑖 𝑚 subscript 𝑖 1 A_{(i_{1},\dots,i_{m})}=\|{\pi\in\operatorname{PF}{n}:\pi{i_{1}}=i_{2},\pi_{% i_{2}}=i_{3},\dots,\pi_{i_{m-1}}=i_{m},\pi_{i_{m}}=i_{1}}\|.italic_A start_POSTSUBSCRIPT ( italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ) end_POSTSUBSCRIPT = \| { italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_m - 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT } \| . We still apply De Morgan’s law, the inclusion-exclusion principle, and an extension of Pollak’s circle argument. The major difference is that instead of selecting k+ℓ 𝑘 ℓ k+\ell italic_k + roman_ℓ spots on a circle with n+1 𝑛 1 n+1 italic_n + 1 spots, we select k+ℓ 𝑘 ℓ k+\ell italic_k + roman_ℓ m 𝑚 m italic_m-tuples on the circle. We also include an additional scalar factor (m−1)!𝑚 1(m-1)!( italic_m - 1 ) ! for each such tuple, which accounts for the number of ways of arranging an m 𝑚 m italic_m-cycle with distinct entries. ∎ As is standard, we denote the probability distribution and expectation with respect to the uniform measure on the set of classical parking functions of length n 𝑛 n italic_n by ℙ n subscript ℙ 𝑛\mathbb{P}{n}blackboard_P start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT and 𝔼 n subscript 𝔼 𝑛\mathbb{E}{n}blackboard_E start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT, respectively. The following theorem provides an exact result on the expected number of m 𝑚 m italic_m-cycles in a classical parking function drawn uniformly at random. Theorem 2.3. Take 1≤m≤n 1 𝑚 𝑛 1\leq m\leq n 1 ≤ italic_m ≤ italic_n. Let π∈PF n 𝜋 subscript PF 𝑛\pi\in\operatorname{PF}{n}italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT be a parking function chosen uniformly at random and C m⁢(π)subscript 𝐶 𝑚 𝜋 C{m}(\pi)italic_C start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ( italic_π ) be the number of m 𝑚 m italic_m-cycles of π 𝜋\pi italic_π. The expected number of m 𝑚 m italic_m-cycles is given by 𝔼 n⁢(C m⁢(π))=(m−1)!⁢(n+1 m)/(n+1)m.subscript 𝔼 𝑛 subscript 𝐶 𝑚 𝜋 𝑚 1 binomial 𝑛 1 𝑚 superscript 𝑛 1 𝑚\mathbb{E}{n}(C{m}(\pi))=(m-1)!\binom{n+1}{m}/(n+1)^{m}.blackboard_E start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ( italic_C start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ( italic_π ) ) = ( italic_m - 1 ) ! ( FRACOP start_ARG italic_n + 1 end_ARG start_ARG italic_m end_ARG ) / ( italic_n + 1 ) start_POSTSUPERSCRIPT italic_m end_POSTSUPERSCRIPT . Remark 2.4. From Theorem 2.3, we may readily derive that for fixed m 𝑚 m italic_m, as n 𝑛 n italic_n gets large, the expected number of m 𝑚 m italic_m-cycles in a random parking function is asymptotically 1/m 1 𝑚 1/m 1 / italic_m. This asymptotic result was obtained earlier by Paguyo using other means. Proof. For distinct 1≤i 1,…,i m≤n formulae-sequence 1 subscript 𝑖 1…subscript 𝑖 𝑚 𝑛 1\leq i_{1},\dots,i_{m}\leq n 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ≤ italic_n, let A(i 1,…,i m)=\|{π∈PF n:π i 1=i 2,π i 2=i 3,…,π i m−1=i m,π i m=i 1}\|.subscript 𝐴 subscript 𝑖 1…subscript 𝑖 𝑚 conditional-set 𝜋 subscript PF 𝑛 formulae-sequence subscript 𝜋 subscript 𝑖 1 subscript 𝑖 2 formulae-sequence subscript 𝜋 subscript 𝑖 2 subscript 𝑖 3…formulae-sequence subscript 𝜋 subscript 𝑖 𝑚 1 subscript 𝑖 𝑚 subscript 𝜋 subscript 𝑖 𝑚 subscript 𝑖 1 A_{(i_{1},\dots,i_{m})}=\|{\pi\in\operatorname{PF}{n}:\pi{i_{1}}=i_{2},\pi_{% i_{2}}=i_{3},\dots,\pi_{i_{m-1}}=i_{m},\pi_{i_{m}}=i_{1}}\|.italic_A start_POSTSUBSCRIPT ( italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ) end_POSTSUBSCRIPT = \| { italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_m - 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT , italic_π start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT } \| . By linearity of expectation and symmetry of parking coordinates, 𝔼 n⁢(C m⁢(π))subscript 𝔼 𝑛 subscript 𝐶 𝑚 𝜋\displaystyle\mathbb{E}{n}(C{m}(\pi))blackboard_E start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ( italic_C start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ( italic_π ) )=∑1≤i 1<⋯<i m≤n(m−1)!⁢ℙ n⁢(A(i 1,…,i m))absent subscript 1 subscript 𝑖 1⋯subscript 𝑖 𝑚 𝑛 𝑚 1 subscript ℙ 𝑛 subscript 𝐴 subscript 𝑖 1…subscript 𝑖 𝑚\displaystyle=\sum_{1\leq i_{1}<\cdots<i_{m}\leq n}(m-1)!\,\mathbb{P}{n}(A{(% i_{1},\dots,i_{m})})= ∑ start_POSTSUBSCRIPT 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ≤ italic_n end_POSTSUBSCRIPT ( italic_m - 1 ) ! blackboard_P start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ( italic_A start_POSTSUBSCRIPT ( italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ) end_POSTSUBSCRIPT ) =∑1≤i 1<⋯<i m≤n(m−1)!⁢ℙ n⁢(π 1=i 1,…,π m=i m)absent subscript 1 subscript 𝑖 1⋯subscript 𝑖 𝑚 𝑛 𝑚 1 subscript ℙ 𝑛 formulae-sequence subscript 𝜋 1 subscript 𝑖 1…subscript 𝜋 𝑚 subscript 𝑖 𝑚\displaystyle=\sum_{1\leq i_{1}<\cdots<i_{m}\leq n}(m-1)!\,\mathbb{P}{n}(\pi% {1}=i_{1},\dots,\pi_{m}=i_{m})= ∑ start_POSTSUBSCRIPT 1 ≤ italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < ⋯ < italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ≤ italic_n end_POSTSUBSCRIPT ( italic_m - 1 ) ! blackboard_P start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ( italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ) =(m−1)!(n+1)n−1⁢\|{π∈PF n:π 1<π 2<⋯<π m}\|,absent 𝑚 1 superscript 𝑛 1 𝑛 1 conditional-set 𝜋 subscript PF 𝑛 subscript 𝜋 1 subscript 𝜋 2⋯subscript 𝜋 𝑚\displaystyle=\frac{(m-1)!}{(n+1)^{n-1}}\,\left\|\left{\pi\in\operatorname{PF}% {n}:\pi{1}<\pi_{2}<\cdots<\pi_{m}\right}\right\|,= divide start_ARG ( italic_m - 1 ) ! end_ARG start_ARG ( italic_n + 1 ) start_POSTSUPERSCRIPT italic_n - 1 end_POSTSUPERSCRIPT end_ARG \| { italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < italic_π start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT < ⋯ < italic_π start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT } \| , where the additional scalar factor (m−1)!𝑚 1(m-1)!( italic_m - 1 ) ! in the first equation accounts for the number of ways of arranging an m 𝑚 m italic_m-cycle with distinct entries i 1,…,i m subscript 𝑖 1…subscript 𝑖 𝑚 i_{1},\dots,i_{m}italic_i start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_i start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT. To obtain a simplified expression for this last equation, we use an extension of Pollak’s circle argument as in the proof of Theorem 2.1. Add an additional parking spot n+1 𝑛 1 n+1 italic_n + 1, and arrange the spots in a circle. Allow n+1 𝑛 1 n+1 italic_n + 1 also as a preferred spot. We first select m 𝑚 m italic_m spots for the first m 𝑚 m italic_m cars, which can be done in (n+1 m)binomial 𝑛 1 𝑚\binom{n+1}{m}( FRACOP start_ARG italic_n + 1 end_ARG start_ARG italic_m end_ARG ) ways. Then for the remaining n−m 𝑛 𝑚 n-m italic_n - italic_m cars, there are (n+1)n−m superscript 𝑛 1 𝑛 𝑚(n+1)^{n-m}( italic_n + 1 ) start_POSTSUPERSCRIPT italic_n - italic_m end_POSTSUPERSCRIPT possible preference sequences. Out of the n+1 𝑛 1 n+1 italic_n + 1 rotations for any preference sequence, only one rotation becomes a valid parking function. Standard circular symmetry argument yields \|{π∈PF n:π 1<π 2<⋯<π m}\|=(n+1 m)⁢(n+1)n−m−1.conditional-set 𝜋 subscript PF 𝑛 subscript 𝜋 1 subscript 𝜋 2⋯subscript 𝜋 𝑚 binomial 𝑛 1 𝑚 superscript 𝑛 1 𝑛 𝑚 1\left\|\left{\pi\in\operatorname{PF}{n}:\pi{1}<\pi_{2}<\cdots<\pi_{m}\right% }\right\|=\binom{n+1}{m}(n+1)^{n-m-1}.\| { italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < italic_π start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT < ⋯ < italic_π start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT } \| = ( FRACOP start_ARG italic_n + 1 end_ARG start_ARG italic_m end_ARG ) ( italic_n + 1 ) start_POSTSUPERSCRIPT italic_n - italic_m - 1 end_POSTSUPERSCRIPT . Our conclusion follows. ∎ 3. Prime parking functions A classical parking function π=(π 1,…,π n)𝜋 subscript 𝜋 1…subscript 𝜋 𝑛\pi=(\pi_{1},\dots,\pi_{n})italic_π = ( italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) is said to be prime if for all 1≤j≤n−1 1 𝑗 𝑛 1 1\leq j\leq n-1 1 ≤ italic_j ≤ italic_n - 1, at least j+1 𝑗 1 j+1 italic_j + 1 cars want to park in the first j 𝑗 j italic_j places. (Equivalently, if we remove some term of π 𝜋\pi italic_π equal to 1, then we still have a parking function.) Denote the set of prime parking functions of length n 𝑛 n italic_n by PPF n subscript PPF 𝑛\operatorname{PPF}_{n}roman_PPF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT. Note that for prime parking functions it is impossible to have n 𝑛 n italic_n fixed points or to have an n 𝑛 n italic_n-cycle. As with classical parking functions, we can also study prime parking functions via circular rotation. The modified circular symmetry argument was first due to Kalikow [7, pp.141-142], who provided the following observation: Let G 𝐺 G italic_G denote the group of all n 𝑛 n italic_n-tuples (a 1,…,a n)∈[n−1]n subscript 𝑎 1…subscript 𝑎 𝑛 superscript delimited-[]𝑛 1 𝑛(a_{1},\dots,a_{n})\in[n-1]^{n}( italic_a start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_a start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) ∈ [ italic_n - 1 ] start_POSTSUPERSCRIPT italic_n end_POSTSUPERSCRIPT with componentwise addition modulo n−1 𝑛 1 n-1 italic_n - 1. Let H 𝐻 H italic_H be the subgroup generated by (1,1,…,1)1 1…1(1,1,\dots,1)( 1 , 1 , … , 1 ). Then every coset of H 𝐻 H italic_H contains exactly one prime parking function. The results we present in this section are thus largely parallel to the results for classical parking functions in our previous section. Theorem 3.1. Let 0≤k≤n−1 0 𝑘 𝑛 1 0\leq k\leq n-1 0 ≤ italic_k ≤ italic_n - 1. The number of prime parking functions of length n 𝑛 n italic_n with k 𝑘 k italic_k fixed points is given by (n−1 k)⁢(n−2)n−k−1.binomial 𝑛 1 𝑘 superscript 𝑛 2 𝑛 𝑘 1\binom{n-1}{k}(n-2)^{n-k-1}.( FRACOP start_ARG italic_n - 1 end_ARG start_ARG italic_k end_ARG ) ( italic_n - 2 ) start_POSTSUPERSCRIPT italic_n - italic_k - 1 end_POSTSUPERSCRIPT . Proof. We proceed as in the proof of Theorem 2.1. The only difference is that the circular symmetry argument is now for a circle with n−1 𝑛 1 n-1 italic_n - 1 spots. So we have ∑ℓ=0 n−k−1(−1)ℓ⁢(n−1 k,ℓ,n−k−ℓ−1)⁢(n−1)n−k−ℓ−1=(n−1 k)⁢(n−2)n−k−1.superscript subscript ℓ 0 𝑛 𝑘 1 superscript 1 ℓ binomial 𝑛 1 𝑘 ℓ 𝑛 𝑘 ℓ 1 superscript 𝑛 1 𝑛 𝑘 ℓ 1 binomial 𝑛 1 𝑘 superscript 𝑛 2 𝑛 𝑘 1\sum_{\ell=0}^{n-k-1}(-1)^{\ell}\binom{n-1}{k,\ell,n-k-\ell-1}(n-1)^{n-k-\ell-% 1}=\binom{n-1}{k}(n-2)^{n-k-1}.∑ start_POSTSUBSCRIPT roman_ℓ = 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_n - italic_k - 1 end_POSTSUPERSCRIPT ( - 1 ) start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT ( FRACOP start_ARG italic_n - 1 end_ARG start_ARG italic_k , roman_ℓ , italic_n - italic_k - roman_ℓ - 1 end_ARG ) ( italic_n - 1 ) start_POSTSUPERSCRIPT italic_n - italic_k - roman_ℓ - 1 end_POSTSUPERSCRIPT = ( FRACOP start_ARG italic_n - 1 end_ARG start_ARG italic_k end_ARG ) ( italic_n - 2 ) start_POSTSUPERSCRIPT italic_n - italic_k - 1 end_POSTSUPERSCRIPT . ∎ As in the case of classical parking functions, a more general theorem exists for prime parking functions when we extend the notion of fixed points to cycles. Theorem 3.2. Let m≥1 𝑚 1 m\geq 1 italic_m ≥ 1 and k≥0 𝑘 0 k\geq 0 italic_k ≥ 0 with k⁢m≤n−1 𝑘 𝑚 𝑛 1 km\leq n-1 italic_k italic_m ≤ italic_n - 1. The number of prime parking functions of length n 𝑛 n italic_n with k 𝑘 k italic_k m 𝑚 m italic_m-cycles is given by ∑ℓ:(k+ℓ)⁢m≤n−1(−1)ℓ⁢((m−1)!)k+ℓ k!⁢ℓ!⁢(n−1 m,⋯,m﹈k+ℓ⁢m’s,n−(k+ℓ)⁢m−1)⁢(n−1)n−(k+ℓ)⁢m−1.subscript:ℓ 𝑘 ℓ 𝑚 𝑛 1 superscript 1 ℓ superscript 𝑚 1 𝑘 ℓ 𝑘 ℓ binomial 𝑛 1 subscript﹈𝑚⋯𝑚 𝑘 ℓ m’s 𝑛 𝑘 ℓ 𝑚 1 superscript 𝑛 1 𝑛 𝑘 ℓ 𝑚 1\sum_{\ell:(k+\ell)m\leq n-1}(-1)^{\ell}\frac{\left((m-1)!\right)^{k+\ell}}{k!% \ell!}\binom{n-1}{\underbracket{m,\cdots,m}_{k+\ell\hskip 2.84544pt\text{$m$'s% }},n-(k+\ell)m-1}(n-1)^{n-(k+\ell)m-1}.∑ start_POSTSUBSCRIPT roman_ℓ : ( italic_k + roman_ℓ ) italic_m ≤ italic_n - 1 end_POSTSUBSCRIPT ( - 1 ) start_POSTSUPERSCRIPT roman_ℓ end_POSTSUPERSCRIPT divide start_ARG ( ( italic_m - 1 ) ! ) start_POSTSUPERSCRIPT italic_k + roman_ℓ end_POSTSUPERSCRIPT end_ARG start_ARG italic_k ! roman_ℓ ! end_ARG ( FRACOP start_ARG italic_n - 1 end_ARG start_ARG under﹈ start_ARG italic_m , ⋯ , italic_m end_ARG start_POSTSUBSCRIPT italic_k + roman_ℓ italic_m ’s end_POSTSUBSCRIPT , italic_n - ( italic_k + roman_ℓ ) italic_m - 1 end_ARG ) ( italic_n - 1 ) start_POSTSUPERSCRIPT italic_n - ( italic_k + roman_ℓ ) italic_m - 1 end_POSTSUPERSCRIPT . Proof. We proceed as in the proof of Theorem 2.2 and note that the circular symmetry argument is now for a circle with n−1 𝑛 1 n-1 italic_n - 1 spots. ∎ As is standard, we denote the probability distribution and expectation with respect to the uniform measure on the set of prime parking functions of length n 𝑛 n italic_n by ℙ n subscript ℙ 𝑛\mathbb{P}{n}blackboard_P start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT and 𝔼 n subscript 𝔼 𝑛\mathbb{E}{n}blackboard_E start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT, respectively. The following theorem provides an exact result on the expected number of m 𝑚 m italic_m-cycles in a prime parking function drawn uniformly at random. Theorem 3.3. Take 1≤m≤n−1 1 𝑚 𝑛 1 1\leq m\leq n-1 1 ≤ italic_m ≤ italic_n - 1. Let π∈PPF n 𝜋 subscript PPF 𝑛\pi\in\operatorname{PPF}{n}italic_π ∈ roman_PPF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT be a prime parking function chosen uniformly at random and C m⁢(π)subscript 𝐶 𝑚 𝜋 C{m}(\pi)italic_C start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ( italic_π ) be the number of m 𝑚 m italic_m-cycles of π 𝜋\pi italic_π. The expected number of m 𝑚 m italic_m-cycles is given by 𝔼 n⁢(C m⁢(π))=(m−1)!⁢(n−1 m)/(n−1)m.subscript 𝔼 𝑛 subscript 𝐶 𝑚 𝜋 𝑚 1 binomial 𝑛 1 𝑚 superscript 𝑛 1 𝑚\mathbb{E}{n}(C{m}(\pi))=(m-1)!\binom{n-1}{m}/(n-1)^{m}.blackboard_E start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ( italic_C start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ( italic_π ) ) = ( italic_m - 1 ) ! ( FRACOP start_ARG italic_n - 1 end_ARG start_ARG italic_m end_ARG ) / ( italic_n - 1 ) start_POSTSUPERSCRIPT italic_m end_POSTSUPERSCRIPT . Remark 3.4. From Theorem 3.3, we may readily derive that for fixed m 𝑚 m italic_m, as n 𝑛 n italic_n gets large, the expected number of m 𝑚 m italic_m-cycles in a random prime parking function is asymptotically 1/m 1 𝑚 1/m 1 / italic_m. Proof. We proceed as in the proof of Theorem 2.3. The only difference is that the circular symmetry argument is now for a circle with n−1 𝑛 1 n-1 italic_n - 1 spots. So we have 𝔼 n⁢(C m⁢(π))=(m−1)!(n−1)n−1⁢(n−1 m)⁢(n−1)n−m−1=(m−1)!⁢(n−1 m)/(n−1)m.subscript 𝔼 𝑛 subscript 𝐶 𝑚 𝜋 𝑚 1 superscript 𝑛 1 𝑛 1 binomial 𝑛 1 𝑚 superscript 𝑛 1 𝑛 𝑚 1 𝑚 1 binomial 𝑛 1 𝑚 superscript 𝑛 1 𝑚\mathbb{E}{n}(C{m}(\pi))=\frac{(m-1)!}{(n-1)^{n-1}}\binom{n-1}{m}(n-1)^{n-m-% 1}=(m-1)!\binom{n-1}{m}/(n-1)^{m}.blackboard_E start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ( italic_C start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT ( italic_π ) ) = divide start_ARG ( italic_m - 1 ) ! end_ARG start_ARG ( italic_n - 1 ) start_POSTSUPERSCRIPT italic_n - 1 end_POSTSUPERSCRIPT end_ARG ( FRACOP start_ARG italic_n - 1 end_ARG start_ARG italic_m end_ARG ) ( italic_n - 1 ) start_POSTSUPERSCRIPT italic_n - italic_m - 1 end_POSTSUPERSCRIPT = ( italic_m - 1 ) ! ( FRACOP start_ARG italic_n - 1 end_ARG start_ARG italic_m end_ARG ) / ( italic_n - 1 ) start_POSTSUPERSCRIPT italic_m end_POSTSUPERSCRIPT . ∎ 4. Further discussions Fix some positive integers k 𝑘 k italic_k and r 𝑟 r italic_r. Define an (r,k)𝑟 𝑘(r,k)( italic_r , italic_k )-parking function of length n 𝑛 n italic_n to be a sequence (π 1,…,π n)subscript 𝜋 1…subscript 𝜋 𝑛(\pi_{1},\dots,\pi_{n})( italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) of positive integers such that if λ 1≤⋯≤λ n subscript 𝜆 1⋯subscript 𝜆 𝑛\lambda_{1}\leq\cdots\leq\lambda_{n}italic_λ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ ⋯ ≤ italic_λ start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT is the (weakly) increasing rearrangement of π 1,…,π n subscript 𝜋 1…subscript 𝜋 𝑛\pi_{1},\dots,\pi_{n}italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_π start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT, then λ i≤k+(i−1)⁢r subscript 𝜆 𝑖 𝑘 𝑖 1 𝑟\lambda_{i}\leq k+(i-1)r italic_λ start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ≤ italic_k + ( italic_i - 1 ) italic_r for 1≤i≤n 1 𝑖 𝑛 1\leq i\leq n 1 ≤ italic_i ≤ italic_n. Denote the set of (r,k)𝑟 𝑘(r,k)( italic_r , italic_k )-parking functions of length n 𝑛 n italic_n by PF n⁡(r,k)subscript PF 𝑛 𝑟 𝑘\operatorname{PF}_{n}(r,k)roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ( italic_r , italic_k ). There is a similar interpretation for such (r,k)𝑟 𝑘(r,k)( italic_r , italic_k )-parking functions in terms of the classical parking function scenario: One wishes to park n 𝑛 n italic_n cars on a street with k+(n−1)⁢r 𝑘 𝑛 1 𝑟 k+(n-1)r italic_k + ( italic_n - 1 ) italic_r spots, but only n 𝑛 n italic_n spots are still empty, which are at positions no later than k,…,k+(n−1)⁢r 𝑘…𝑘 𝑛 1 𝑟 k,\dots,k+(n-1)r italic_k , … , italic_k + ( italic_n - 1 ) italic_r. Pollak’s original circle argument may be further extended to (r,k)𝑟 𝑘(r,k)( italic_r , italic_k )-parking functions. Let G 𝐺 G italic_G denote the group of all n 𝑛 n italic_n-tuples (a 1,…,a n)∈[k+n⁢r]n subscript 𝑎 1…subscript 𝑎 𝑛 superscript delimited-[]𝑘 𝑛 𝑟 𝑛(a_{1},\dots,a_{n})\in[k+nr]^{n}( italic_a start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_a start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) ∈ [ italic_k + italic_n italic_r ] start_POSTSUPERSCRIPT italic_n end_POSTSUPERSCRIPT with componentwise addition modulo k+n⁢r 𝑘 𝑛 𝑟 k+nr italic_k + italic_n italic_r. Let H 𝐻 H italic_H be the subgroup generated by (1,1,…,1)1 1…1(1,1,\dots,1)( 1 , 1 , … , 1 ). Then every coset of H 𝐻 H italic_H contains exactly k 𝑘 k italic_k(r,k)𝑟 𝑘(r,k)( italic_r , italic_k )-parking functions. See for more details. The following proposition is then immediate. Proposition 4.1. Let 0≤s≤n 0 𝑠 𝑛 0\leq s\leq n 0 ≤ italic_s ≤ italic_n. We have \|{π∈PF n⁡(r,k):π 1<π 2<⋯<π s}\|=k⁢(k+n⁢r s)⁢(k+n⁢r)n−s−1.conditional-set 𝜋 subscript PF 𝑛 𝑟 𝑘 subscript 𝜋 1 subscript 𝜋 2⋯subscript 𝜋 𝑠 𝑘 binomial 𝑘 𝑛 𝑟 𝑠 superscript 𝑘 𝑛 𝑟 𝑛 𝑠 1\|{\pi\in\operatorname{PF}{n}(r,k):\pi{1}<\pi_{2}<\dots<\pi_{s}}\|=k\binom{k% +nr}{s}(k+nr)^{n-s-1}.\| { italic_π ∈ roman_PF start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ( italic_r , italic_k ) : italic_π start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < italic_π start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT < ⋯ < italic_π start_POSTSUBSCRIPT italic_s end_POSTSUBSCRIPT } \| = italic_k ( FRACOP start_ARG italic_k + italic_n italic_r end_ARG start_ARG italic_s end_ARG ) ( italic_k + italic_n italic_r ) start_POSTSUPERSCRIPT italic_n - italic_s - 1 end_POSTSUPERSCRIPT . Proof. We apply the generalized circle argument as described above. Arrange k+n⁢r 𝑘 𝑛 𝑟 k+nr italic_k + italic_n italic_r spots in a circle. We first select s 𝑠 s italic_s spots for the first s 𝑠 s italic_s cars, which can be done in (k+n⁢r s)binomial 𝑘 𝑛 𝑟 𝑠\binom{k+nr}{s}( FRACOP start_ARG italic_k + italic_n italic_r end_ARG start_ARG italic_s end_ARG ) ways. Then for the remaining n−s 𝑛 𝑠 n-s italic_n - italic_s cars, there are (k+n⁢r)n−s superscript 𝑘 𝑛 𝑟 𝑛 𝑠(k+nr)^{n-s}( italic_k + italic_n italic_r ) start_POSTSUPERSCRIPT italic_n - italic_s end_POSTSUPERSCRIPT possible preference sequences. Out of the k+n⁢r 𝑘 𝑛 𝑟 k+nr italic_k + italic_n italic_r rotations for any preference sequence, exactly k 𝑘 k italic_k rotations become valid (r,k)𝑟 𝑘(r,k)( italic_r , italic_k )-parking functions. The rest is standard circular symmetry argument. ∎ acknowledgements We thank Richard Stanley for helpful discussions. Some of the results in this paper were independently conjectured by Ari Cruz. References Foata, D., Riordan, J.: Mappings of acyclic and parking functions. Aequationes Math.10: 10-22 (1974). Konheim, A.G., Weiss, B.: An occupancy discipline and applications. SIAM J. Appl.Math.14: 1266-1274 (1966). Paguyo, J.E.: Cycle structure of random parking functions. Adv.Appl. Math.144: 102458 (2023). Pyke, R.: The supremum and infimum of the Poisson process. Ann.Math.Statist. 30: 568–576 (1959). Riordan, J.: Ballots and trees. J. Combin.Theory 6: 408-411 (1969). Stanley, R.P.: Parking functions and noncrossing partitions. Electron.J. Combin.4: Research Paper 20, 14 pp. (1997). Stanley, R.P.: Enumerative Combinatorics Volume 2. Cambridge University Press, Cambridge. (1999). Stanley, R.P., Wang, Y.: Some aspects of (r,k)𝑟 𝑘(r,k)( italic_r , italic_k )-parking functions. J. Combin.Theory Ser.A 159: 54-78 (2018). Generated on Fri May 3 00:09:16 2024 by L a T e XML
15430	https://www.quora.com/What-is-the-difference-between-atomic-mass-mass-number-atomic-weight-and-relative-atomic-mass	Something went wrong. Wait a moment and try again. Mass Number Atomic Structure Relative Atomic Mass Atomic Weight Physical Chemistry The Sciences Science Chemistry 5 What is the difference between atomic mass, mass number, atomic weight, and relative atomic mass? Upvoted by Bob Spillman , PhD Chemistry, University of Illinois at Urbana-Champaign (1975) · 7y Here are some important points you should know that will clear all your doubts: Atomic Number: It is the number of protons (i.e. positive charger) which is also equal to the number of electrons (i.e. negative charges) in a neutral atom. It is just a number and has no unit. Also note that a neutron has no charge. Mass Number: Mass number is nothing but the number of protons added to the number of neutrons. We can also say that it is the number of electrons plus number of neutrons. Again, it is just a number and has no unit. Now, if you sit in a physics class you'll see that weight is a product of Here are some important points you should know that will clear all your doubts: Atomic Number: It is the number of protons (i.e. positive charger) which is also equal to the number of electrons (i.e. negative charges) in a neutral atom. It is just a number and has no unit. Also note that a neutron has no charge. Mass Number: Mass number is nothing but the number of protons added to the number of neutrons. We can also say that it is the number of electrons plus number of neutrons. Again, it is just a number and has no unit. Now, if you sit in a physics class you'll see that weight is a product of the mass and gravity and being a force, it is expressed in newtons (N). Mass however gives us an idea about the amout of substance/matter there is in a body so its S.I. Unit is Kg. Here in chemistry however these are slightly different. Atomic Mass: It is simply the mass of a particular atom expressed in a.m.u. It does not take into consideration the various isotopes. Mass of 1 proton = Mass of 1 neutrom = 1 a.m.u. Also, the mass of an electron is negligible so it may/ may not be considered when calculating atomic mass. So we have, Atomic Mass = (No. of protons + neutrons) a.m.u. = (No. of neucleons) a.m.u. = (Mass Number) a.m.u. Thus, atomic mass is approximately equal to the mass number (except it has a unit) and so is generally a whole number. We also know, 1.66 × 10^-24 g = 1 a.m.u. 1 g = 1/(1.66 × 10^-24) a.m.u. But 1 a.m.u = Actual mass of C-12 ÷ 12 Let us say that the actual mass in grams of an atom is x g. 1 g = 1/(1.66×10^-24) a.m.u = 1/(Actual mass of C-12 ÷ 12) a.m.u. x g = x/(Actual mass of C-12 ÷12) In other words, Atomic mass = Actual mass of an atom/ 1 a.m.u. That is atomic mass done. Average Atomic Mass: It is basically a weighted average of all the atomic masses of various isotopes of an element. It is also measured in a.m.u. Many a times atomic mass and average atomic mass are used interchangeably. Relative isotopic mass: This is unitless/dimensionless as it is a ratio. It is exactly the same as the atomic mass expressed without a unit. This loss of unit occurs as atomic mass divided by 1/12th the mass of C-12 (or simply 1 a.m.u.). It is therefore a ratio between the two. That is why its called 'relative' as it is in relation with 1/12th the mass of C-12 which is taken as standard. Note that we are using the atomic mass here and not the average atomic mass. That is why this quantity too does not take account of the various isotopes. Relative Atomic Mass/Atomic weight (They are the same)- It is the same as relative isotopic mass except that instead of using the atomic mass, we use the average atomic mass. So again it is a ratio, it is unitless/dimensionless and is in relation to something. Hope this helped Sponsored by MEDvi Prescription Weight Loss Tirzepatide can drop weight 2x faster than Ozempic For half the cost of popular weight loss drugs on the market, Americans use this site for GLP-1 Meds. Related questions What is the difference between atomic mass and relative atomic mass? What is the difference between relative atomic mass and average atomic mass? What is atomic mass and relative atomic mass? What is the difference between atomic mass and average atomic mass? What is the difference between atomic mass and atomic weight? Bala V CEO - Tech Startup, Enthusiast in Personal Finance Advise · Author has 381 answers and 349.6K answer views · 6y ATOMIC NUMBER - denoted by “Z” [OTHER NAMES - PROTON NUMBER] No of protons in an atom This decides what the element is MASS NUMBER - denoted by “A” [OTHER NAMES - ATOMIC MASS NUMBER, NUCLEON NUMBER] No of protons + No of Neutrons in the nucleus Ignores the electrons in the atom The above two - ATOMIC NUMBER & MASS NUMBER do not have any units as they are just the “count” of certain sub-atomic particles. Relation among them: A = Z + N where N is the no of Neutrons ATOMIC NUMBER - denoted by “Z” [OTHER NAMES - PROTON NUMBER] No of protons in an atom This decides what the element is MASS NUMBER - denoted by “A” [OTHER NAMES - ATOMIC MASS NUMBER, NUCLEON NUMBER] No of protons + No of Neutrons in the nucleus Ignores the electrons in the atom The above two - ATOMIC NUMBER & MASS NUMBER do not have any units as they are just the “count” of certain sub-atomic particles. Relation among them: A = Z + N where N is the no of Neutrons Before going to atomic mass, let us define atomic mass unit. ATOMIC MASS UNIT - denoted by by mu (actually m subscript u) Abbreviated as amu [OTHER NAMES _ ATOMIC MASS CONSTANT] amu is defined as 1/12th the mass of single atom of carbon C-12 at rest Numerical value equals 1.66 X 10 exp(-27) Kg or 1.66 X 10(-24) grams and is defined as one “u” or one Dalton “da” ATOMIC MASS - denoted by ma (actually m subscript a) Mass of an atom expressed in amu Unit is “u” or “da” Includes the mass of atom including electrons Atomic mass of C-12 carbon is 12.01 u NOTE: Atomic Number and Mass Number - always integers with no units Atomic Mass is always a non integer real number with “u” as its unit Atomic Weight and Relative Atomic Mass are one and the same. RELATIVE ATOMIC MASS - denoted by Ar (actually A subscript r) [OTHER NAMES - ATOMIC WEIGHT] There are two definitions depending upon the case I. FOR ELEMENT WITH NO ISOTOPES: Ratio of Average mass per atom of an element to 1/12th the mass of C-12 carbon atom or in other words, mass of an atom divided by 1 amu Gives a measure of how heavy the atom of an element is Has no units as it is just the ratio II. FOR A SAMPLE OF ELEMENT CONTAINING MIX OF ITS ISOTOPES: Ratio of Average mass of atoms of an element in a given sample to that of 1 amu Weighted arithmetic mean of all isotopes in a given sample Julie Bodette Studied Chemistry at Ohio University · 7y Originally Answered: What is the difference between atomic number, atomic mass, atomic weight, mass number, and relative atomic mass? I’ve read the answers to similar questions posted here, but I can’t find anything that clears it up nicely for me. · I’ll use Sodium (Na) as an example. Atomic Number- the number of protons in an atom. Unique for each element. 11 Atomic mass- The total mass of the protons+neutrons+electrons in the atom. Note that the number of neutrons can vary, so this is an average of all the isotopes. 22.989 Atomic weight- same thing as atomic mass. 22.989 Mass number- the total number of protons and neutrons in a nucleus Sodium has 11 protons and 12 neutrons, so mass number is 23. This can vary because the number of neutrons is not always the same! Relative atomic mass- same thing as atomic mass 22.989 I’ve never heard the terms at I’ll use Sodium (Na) as an example. Atomic Number- the number of protons in an atom. Unique for each element. 11 Atomic mass- The total mass of the protons+neutrons+electrons in the atom. Note that the number of neutrons can vary, so this is an average of all the isotopes. 22.989 Atomic weight- same thing as atomic mass. 22.989 Mass number- the total number of protons and neutrons in a nucleus Sodium has 11 protons and 12 neutrons, so mass number is 23. This can vary because the number of neutrons is not always the same! Relative atomic mass- same thing as atomic mass 22.989 I’ve never heard the terms atomic weight and relative atomic mass before, but Google told me they were the same thing as atomic mass. I don’t think they are very commonly used. Ganesh Subramaniam Amateur Radio Operator, Amateur Astronomer · Upvoted by Proactive Intellectual , M.sc (Chemistry) Science & India, Mahatma Jyotiba Phule Rohilkhand University (2016) · Author has 11.3K answers and 22.7M answer views · 9y As you are aware, "mass" is a physical property of matter. The mass of an atom is known as it's Atomic Mass. Atomic mass is calculated by adding the number of protons and neutrons in an atom. The electrons are not counted because of their small size. The unified atomic mass unit (symbol: u) or Dalton (symbol: Da) is the standard unit that is used for indicating mass on an atomic or molecular scale. The atomic mass of an atom will change depending on its amount of protons and neutrons and will be different for each individual element. Atomic mass is also known as the Atomic Weight, with the abbr As you are aware, "mass" is a physical property of matter. The mass of an atom is known as it's Atomic Mass. Atomic mass is calculated by adding the number of protons and neutrons in an atom. The electrons are not counted because of their small size. The unified atomic mass unit (symbol: u) or Dalton (symbol: Da) is the standard unit that is used for indicating mass on an atomic or molecular scale. The atomic mass of an atom will change depending on its amount of protons and neutrons and will be different for each individual element. Atomic mass is also known as the Atomic Weight, with the abbreviation at.wt. Actually the two terms are not exact equivalent. "Weight" implies a force exerted in a gravitational field, which would be measured in units of force, like Newtons. The term "atomic weight" is in use for two centuries or more, so it continues to be used; however, to avoid any confusion, atomic weight is more commonly known now, as relative atomic mass. As the name indicates, it is a ratio of two masses. The mass of an atom is compared with that of an atom of Carbon-12. The relative atomic mass of Carbon-12 is taken to be 12. Relative masses have no units. Mass number is simply a count of the total number of protons and neutrons in an atom's nucleus. The Mass Number of an atom is represented by the symbol A and is also known as the Nucleon Number. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. 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They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions What is the difference between atomic mass, atomic weight, and atomic number? What is the relation between mass number and atomic number? How are atomic mass, average atomic mass, atomic weight, relative atomic mass, and relative isotopic mass different from or similar to each other? What is the difference between mass number and atomic number? What is the atomic weight of mass? Assistant Bot · 1y The terms atomic mass, mass number, atomic weight, and relative atomic mass are often used in chemistry and physics, but they refer to different concepts. Here’s a breakdown of each term: Atomic Mass Definition: The atomic mass of an element is the mass of a single atom of that element, usually expressed in atomic mass units (amu). Measurement: It takes into account the mass of protons, neutrons, and electrons, but since electrons have negligible mass, it primarily reflects the combined mass of protons and neutrons. Mass Number Definition: The mass number is the total number of protons and ne The terms atomic mass, mass number, atomic weight, and relative atomic mass are often used in chemistry and physics, but they refer to different concepts. Here’s a breakdown of each term: Atomic Mass Definition: The atomic mass of an element is the mass of a single atom of that element, usually expressed in atomic mass units (amu). Measurement: It takes into account the mass of protons, neutrons, and electrons, but since electrons have negligible mass, it primarily reflects the combined mass of protons and neutrons. Mass Number Definition: The mass number is the total number of protons and neutrons in the nucleus of an atom. It is a whole number and is specific to a particular isotope of an element. Notation: It is often denoted as A and is used to identify isotopes (e.g., Carbon-12 has a mass number of 12). Atomic Weight Definition: Atomic weight is the weighted average of the atomic masses of an element's isotopes, taking into account their natural abundance. It is a dimensionless quantity and typically expressed in atomic mass units. Example: Chlorine has two stable isotopes, 35Cl and 37Cl. The atomic weight of chlorine reflects the relative abundances of these isotopes. Relative Atomic Mass Definition: Relative atomic mass (also called atomic weight in some contexts) is a dimensionless quantity that compares the mass of an atom to one-twelfth of the mass of a carbon-12 atom. Usage: Like atomic weight, it reflects the weighted average of the isotopes of an element. Summary Atomic Mass: Mass of a single atom (in amu). Mass Number: Total number of protons and neutrons (whole number). Atomic Weight: Weighted average of atomic masses of isotopes (in amu). Relative Atomic Mass: Dimensionless ratio of the mass of an atom to a standard (carbon-12). Understanding these differences is crucial for interpreting chemical data and performing calculations in chemistry. John Richardson Freelance Writer at Upwork (2012–present) · Author has 909 answers and 2.2M answer views · 6y The atomic number is equal to the number of protons in the nucleus or the number of orbiting electrons. The atomic weight is equal to the number of protons plus the number of neutrons present in the nucleus. The neutron is a nuclear particle with a mass approximately equal to that of the proton but with no electrical charge. The various particles found within the nucleus are known as nucleons, but for biologists it is the neutrons and protons that are of principal interest. Physicists believe that many of the nucleons, once thought to be fundamental particles, are'themselves composed of much sm The atomic number is equal to the number of protons in the nucleus or the number of orbiting electrons. The atomic weight is equal to the number of protons plus the number of neutrons present in the nucleus. The neutron is a nuclear particle with a mass approximately equal to that of the proton but with no electrical charge. The various particles found within the nucleus are known as nucleons, but for biologists it is the neutrons and protons that are of principal interest. Physicists believe that many of the nucleons, once thought to be fundamental particles, are'themselves composed of much smaller units called quarks. Adesh Tamrakar Always a student. · Author has 68 answers and 182.8K answer views · 9y Relative atomic mass/Atomic weight (both are exactly the same) - It is the mass of atom calculated as taking the sum of all the masses of the subatomic particles and dividing it with mass of (the same technique calculated) 1/12th of C-12 atom. Atomic mass- It is the mass of the atom calculated as taking the mean abundances of atomic weight of the different isotopes present in that sample. Mass Number- It is the mass calculated as the sum of the nucleons , as there is only a slight difference in the mass of the proton and neutron, and the mass of electron is negligible as compared with the mass o Relative atomic mass/Atomic weight (both are exactly the same) - It is the mass of atom calculated as taking the sum of all the masses of the subatomic particles and dividing it with mass of (the same technique calculated) 1/12th of C-12 atom. Atomic mass- It is the mass of the atom calculated as taking the mean abundances of atomic weight of the different isotopes present in that sample. Mass Number- It is the mass calculated as the sum of the nucleons , as there is only a slight difference in the mass of the proton and neutron, and the mass of electron is negligible as compared with the mass of the proton and neutron. EXAMPLE : Li-7, 3 protons, 4 neutrons, 3 electrons. R.A.M. - 7.016 u (calculated as- {mass of proton+neutron+electron of Li atom}/{mass of proton+neutron+electron of 1/12th of C-12 atom}) Mass number- 7 Li-6, 3 protons, 3 neutrons, 3 electrons. R.A.M. - 6.015 u (calculated as- {mass of proton+neutron+electron of Li atom}/{mass of proton+neutron+electron of 1/12th of C-12 atom}) Mass number- 6 Now the Atomic Mass will be - mean of the masses according to the abundances of the two isotopes in nature. Atomic mass- 6.941 Promoted by The Hartford The Hartford We help protect over 1 million small businesses · Updated Sep 19 What is small business insurance? Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickl Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickly and maintaining operations. Choosing the right insurance for your small business involves assessing your unique needs and consulting with an advisor to pick from comprehensive policy options. With over 200 years of experience and more than 1 million small business owners served, The Hartford is dedicated to providing personalized solutions that help you focus on growth and success. Learn about our coverage options! Prithwish Biswas Studied at Himalaya International School , Ratlam · Upvoted by Proactive Intellectual , M.sc (Chemistry) Science & India, Mahatma Jyotiba Phule Rohilkhand University (2016) · Updated 7y Atomic mass:- sum of the mass of all particles of atom. Generally mass are taken as Proton 1.6727 x 10^-24 g ~1 amu Neutron 1.6750 x 10^-24 g ~1 amu 1 amu (atomic mass unit) is defined so that both protons and neutrons have a mass of approximately 1 amu Atomic mass number:- sum of number of proton and neutron in nuclues of atom. Atomic weight and relative atomic mass are different terms for same thing. It is defined as the ratio of the average mass of one atom of an element to one twelfth of the mass of an atom of carbon-12. Ex. Atomic weight of Oxygen Atomic mass:- sum of the mass of all particles of atom. Generally mass are taken as Proton 1.6727 x 10^-24 g ~1 amu Neutron 1.6750 x 10^-24 g ~1 amu 1 amu (atomic mass unit) is defined so that both protons and neutrons have a mass of approximately 1 amu Atomic mass number:- sum of number of proton and neutron in nuclues of atom. Atomic weight and relative atomic mass are different terms for same thing. It is defined as the ratio of the average mass of one atom of an element to one twelfth of the mass of an atom of carbon-12. Ex. Atomic weight of Oxygen Eckart Mildenstein Author has 232 answers and 555.7K answer views · Updated 10y 1) atomic mass: mass of 1 atom of an element's isotope. e.g. m(Mg-24) =41.490010^-27 [kg] 2) mass number A = Z+N = # protons + # neutrons Mg-24: mass number A = 24, Z=12 N=12 Mg-25 A = 25 Z =12 N 13 Mg-26 A = 25 Z =12 N 14 3) atomic weight (or see 4) = weighted average atomic mass of the isotopes of an element m_wAv(Mg)= m(Mg-24)w(Mg-24)+m(Mg-25)w(Mg-25)+ m(Mg-26)w(Mg-26) = 39.828110^-27 [kg] 78.99/100 + 41.490010^-27 [kg] 10.00/100 + 43.145110^-27 [kg] 11.0 1) atomic mass: mass of 1 atom of an element's isotope. e.g. m(Mg-24) =41.490010^-27 [kg] 2) mass number A = Z+N = # protons + # neutrons Mg-24: mass number A = 24, Z=12 N=12 Mg-25 A = 25 Z =12 N 13 Mg-26 A = 25 Z =12 N 14 3) atomic weight (or see 4) = weighted average atomic mass of the isotopes of an element m_wAv(Mg)= m(Mg-24)w(Mg-24)+m(Mg-25)w(Mg-25)+ m(Mg-26)w(Mg-26) = 39.828110^-27 [kg] 78.99/100 + 41.490010^-27 [kg] 10.00/100 + 43.145110^-27 [kg] 11.01/100 = 31.460210^-27 + 4.149010^-27+4.750310^-27 [kg] = 40.359510^-27 [kg] = 40.359510^-24 [g] w(): natural abundance 4) relative atomic mass (also sometimes called atomic weight) A_R weighted average atomic mass[kg] /1.660510^-27 [kg] 1.660510^-27 [kg] = 1 [u] = m(C-12)/12 (atomic mass unit) Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? 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Arjun Raju Marine Consultant at Kartech Maritime Services · Author has 213 answers and 240.7K answer views · 6y Atoms are the basic units of matter. They are the smallest components of a chemical element, which is a substance that cannot be broken down into simpler material. Atoms have specific properties that will determine their chemical and physical nature. One of these properties is their atomic mass. Atomic mass is equal to the sum of the individual particle masses of an atom. Atoms have three basic components: protons (positively charged particles), neutrons (non-charged particles), and electrons (negatively charged particles). Protons and neutrons are the larger particles, and are found in the nucl Atoms are the basic units of matter. They are the smallest components of a chemical element, which is a substance that cannot be broken down into simpler material. Atoms have specific properties that will determine their chemical and physical nature. One of these properties is their atomic mass. Atomic mass is equal to the sum of the individual particle masses of an atom. Atoms have three basic components: protons (positively charged particles), neutrons (non-charged particles), and electrons (negatively charged particles). Protons and neutrons are the larger particles, and are found in the nucleus, which is the core of the atom. Atomic mass is typically calculated by adding the number of protons and neutrons together, ignoring the electrons because of their small size. Daltons are the standard units used for measuring atomic mass. Atomic mass units, or amu, are also used to measure atomic masses, and they are equal to Daltons. 1 unified atomic mass unit (amu) is defined as ¹⁄₁₂ of the mass of a single carbon-12 atom, at rest. Relative Atomic Mass: In chemistry, it is important to quantify particles. For the fact that the size of an atom is too small to be weighed practically, relative atomic mass is therefore used to represent the mass or weight of an atom of an element. The relative atomic mass (symbol: Ar) of an element is the mass of one atom of the element compared with the mass of an atom of the carbon-12 isotope whose mass is exactly 12. For naturally occurring elements with isotopes, the above definition is adjusted to accommodate the fact that each isotope of the element has a different mass. The relative atomic mass of a naturally occurring element with isotopes can be defined as the weighted average of the masses of its isotopes compared with the mass of an atom of the carbon-12 isotope whose mass is taken to be exactly 12. Relative atomic mass is a dimensionless quantity and so does not have a unit. Relative atomic mass is also called atomic weight. The mass number also called atomic mass number or nucleon number is the total number of protons and neutrons (together known as nucleons) in an atomic nucleus. This is always an integer. The mass number should also not be confused with the Relative Atomic Mass or Atomic Weight of an element. The atomic weight is an actual mass, while the mass number is a counted number. Robert Kern BA in Chemistry & Mathematics, College of the Holy Cross (Graduated 1974) · Author has 2.5K answers and 2M answer views · 5y What is the difference between a relative atomic mass and an atomic mass unit? There is no difference numerically, but there is a slight difference in meaning. Relative atomic mass, as its name suggests, is the mass of an atom relative to another and has no unit. Just for simplicity, let’s assume the relative atomic mass of carbon is 12 and that of helium is 4. All this means is that a carbon atom is 3 times heavier than a helium atom. It is a comparison of their masses, not an actual measure of mass. An atomic mass unit (1 amu) is an actual mass that is equal to about 1.66 x 10^-24 grams. So the mass of a carbon atom is 12 amu and the mass of a helium atom is 4 amu. It There is no difference numerically, but there is a slight difference in meaning. Relative atomic mass, as its name suggests, is the mass of an atom relative to another and has no unit. Just for simplicity, let’s assume the relative atomic mass of carbon is 12 and that of helium is 4. All this means is that a carbon atom is 3 times heavier than a helium atom. It is a comparison of their masses, not an actual measure of mass. An atomic mass unit (1 amu) is an actual mass that is equal to about 1.66 x 10^-24 grams. So the mass of a carbon atom is 12 amu and the mass of a helium atom is 4 amu. It makes sense that the amu masses are in the same proportion as the relative masses, but using amu’s, you can also calculate the actual masses of the atoms. This same idea works for molar masses as well. Continuing with our example, the molar mass of carbon is 12 g and the molar mass of helium is 4 grams. These quantities are in the same proportion as the relative masses because there are 6.02 x 10^23 atoms in 1 mole of carbon and helium. If you take the inverse of 6.02 x 10^23, guess what you obtain? That’s right, it’s 1.66 x 10^-24. Co-Founder and Director of Operations at Baba Herons (2017–present) · Author has 53 answers and 153.4K answer views · 7y If you are facing any doubts regarding atomic mass, gram atomic mass, relative atomic mass , average atomic mass, and same with the molecules then you can go through this video taught by me and still if you are facing any doubt then feel free to contact me however I have attempted to my full to solve all doubts… Initially there is introduction .. ur topic starts at 20:33 and ends at 59:18 Ali Abdulla professor of physics · Author has 22.6K answers and 6.2M answer views · 8y The atomic mass is in general is the total mass of its contituents,nucleons and electrons,for example an atom with 7 protons,8 neutrons and 7 electrons its atomic mass is almost the sum of the masses of these subatomic particles which is highly dominated by the masses of the nucleons(p&n).The mass number is A= Z+N,where Z is the atomic number(protons number=electrons number) and N is the neutrons number. The relatie atomic mass is equivalent to the atomic weight,which is the ratio of the average mass of atoms to 1/12 of the mass of an atom of Carbon 12, it can be considered as a statistical te The atomic mass is in general is the total mass of its contituents,nucleons and electrons,for example an atom with 7 protons,8 neutrons and 7 electrons its atomic mass is almost the sum of the masses of these subatomic particles which is highly dominated by the masses of the nucleons(p&n).The mass number is A= Z+N,where Z is the atomic number(protons number=electrons number) and N is the neutrons number. The relatie atomic mass is equivalent to the atomic weight,which is the ratio of the average mass of atoms to 1/12 of the mass of an atom of Carbon 12, it can be considered as a statistical term. Related questions What is the difference between atomic mass and relative atomic mass? What is the difference between relative atomic mass and average atomic mass? What is atomic mass and relative atomic mass? What is the difference between atomic mass and average atomic mass? What is the difference between atomic mass and atomic weight? What is the difference between atomic mass, atomic weight, and atomic number? What is the relation between mass number and atomic number? How are atomic mass, average atomic mass, atomic weight, relative atomic mass, and relative isotopic mass different from or similar to each other? What is the difference between mass number and atomic number? What is the atomic weight of mass? What is the difference between atomic mass, relative atomic mass and standard atomic weight? What does the atomic mass represent? Why is atomic mass called a relative atomic mass? What is the difference between atomic mass and formula mass? What is the difference between atomic weight and relative atomic mass of an element/atom? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15431	https://www.youtube.com/watch?v=Q_klLmTSyyw	Scientific Notation - Conversion Math Meeting 512000 subscribers 7905 likes Description 827396 views Posted: 28 Oct 2013 Learn how to convert in and out of scientific notation in these step by step examples. Check out my channel to see all of my videos. 758 comments Transcript: hey everybody Welcome to my video on scientific notation and what if I asked you the question what is the weight of the Earth without using scientific notation and assuming that you actually knew the answer to that question here's how you would have to write your answer the weight of the Earth is the number 13 followed by a whole bunch of zeros and I wrote Our answer in pounds and I think many of you just by looking at this number realize why we need to use scientific notation imagine if we had to multiply this large number by another number or or imagine if we had to type this number into a computer or a calculator most most calculators don't even have the capacity to to enter this many digits and even if it did um You probably would would enter the wrong number of zeros because there's so many um and the Earth isn't even that big you know the sun is 340,000 times larger than the Earth so this is why we need to use scientific notation um if we rewrote the weight of the Earth using scientific notation our answer would be 1.3 multiplied by 10 with a 25 exponent on top of the 10 and once again our answer is in pounds so scientific notation allows us to take this large number and to rewrite it in a shorter much simpler way and this also works for really small numbers as well um scientific notation allows us to take really small numbers and rewrite them in a shorter much simpler way as well so let's get started on some examples and I think this will make a lot more sense so first let's take a number that is in scientific notation and convert it into a regular number so let's say that we have 5.7 multiplied by 10 with a exponent of 8 and whenever you multiply by 10 you're actually moving the decimal point one time to the right um since we're multiplying by 10 eight times we're actually moving the decimal eight times to the right so so we have our number 5.7 and we're moving the decimal point eight times to the right so let's take our decimal point and move it eight times to the right 1 2 3 4 5 6 7 8 and our new decimal point is going to be right here and after you move the decimal point you need to fill in zeros in all of the empty spaces so our final answer after we move the decimal point is 570 million and our decimal point is going to be all the way on the right which is unnecessary so I'm not going to write it so 5.7 10 to the 8 is equal to 500 and 70 million so now let's go over an example with a really small number let's say we have 5.7 10 with ative 8 exponent many people think when they see this negative exponent that the answer is going to be negative but that is completely wrong uh this negative exponent uh just means that we have a really small number and we're going to move our decimal point to the left so so we have we have the number 5.7 and we need to move the decimal point eight times to the left all right so if we we take our decimal and we move it eight times to the left it's going to be one 2 3 4 5 6 7 8 it's going to be in this new position right here and once again we need to fill in zeros in all of the empty spaces so our final answer is 0.0000 000000 0 57 and I want you to notice how with a negative exponent we had a really small final answer and with a positive exponent we had a really large Final Answer um so that is a good way to double check if you actually move the decimal point in the right direction now let's go over some examples of converting regular numbers into scientific notation let's say we had the number 18, 500,000 and we needed to convert this number into scientific notation and right now since you don't see a decimal point you can assume that it's on the right now we need to move this decimal point to the nonzero digit that is farthest to the left so the non-zero digit that is farthest to the left is this number one and if we move the decimal point in front of that number we have to move it one 2 3 4 5 6 seven times and now we can place our decimal in front of the leading digit so now let's take a look at all of our nonzero digits notice how inside of our box now we have the number 1.85 so we have the number 1.85 and we moved the decimal point seven times so we need to put a seven exponent on top of the 10 and notice how I put a positive seven because we started with a really large number so a large number has a positive exponent and if you wanted to round to one decimal point you could round up and you can also say 1.9 10 to the 7th now let's go over over another example with a really small number let's say we had 0.085 so we need to convert this into scientific notation so we need to move the decimal points in front of the leading digit or in front of the nonzero digit that is farthest to the left so once again this is the number one so we need to move the decimal point in front of the number one so how many spaces did the decimal point just move well it started over here and it moved 1 2 3 4 five 6 seven spaces and once again if we look at all of our non-zero digits we have the number 1.85 so we have the number 1.85 and we moved the decimal point seven units so we know that on top of the 10 we're going to have an exponent of -7 since we started with a small number we know that our our exponent is going to be negative um so this is our final answer 1.85 10 to the -7 so I hope this gave you a good idea on converting in and out of scientific notation in my next video I'm going to talk about multiplying and dividing in scientific notation so stay tuned for that I really hope that you're enjoying these and I will see you in my next one
15432	https://www.mathworks.com/matlabcentral/answers/493783-how-to-find-the-equation-of-a-tangent-line-to-a-circle-known-radius-from-a-known-point-and-how-t	How to find the equation of a tangent line to a circle ( known radius) from a known point? And How to find the intersection p... - MATLAB Answers - MATLAB Central Skip to content MATLAB Answers MATLAB Help Center Community Learning Get MATLABMATLAB Sign In My Account My Community Profile Link License Sign Out Contact MathWorks Support Visit mathworks.com Search Answers Answers Help Center Answers MathWorks MATLAB Help Center MathWorks MATLAB Answers File Exchange Videos Online Training Blogs Cody MATLAB Drive ThingSpeak Bug Reports Community MATLAB Answers File Exchange Cody AI Chat Playground Discussions Contests Blogs More Communities Treasure Hunt People Community Advisors Virtual Badges About Home Ask Answer Browse MATLAB FAQs More Contributors Recent Activity Flagged Content Manage Spam Help You are now following this question You will see updates in your followed content feed. You may receive emails, depending on your communication preferences. How to find the equation of a tangent line to a circle ( known radius) from a known point? And How to find the intersection point on the circle? Follow 1 view (last 30 days) Show older comments Zarak khon 28 Nov 2019 Vote 0 Link × Direct link to this question Cancel Copy to Clipboard ⋮ Vote 0 Link × Direct link to this question Cancel Copy to Clipboard Commented:Image Analyst on 29 Nov 2019 Accepted Answer:Image Analyst How to find the equation of a tangent line to a circle ( known radius) from a known point? And How to find the intersection point on the circle? 0 Comments Show -2 older comments Hide -2 older comments Sign in to comment. Sign in to answer this question. Accepted Answer Image Analyston 28 Nov 2019 Vote 2 Link × Direct link to this answer Cancel Copy to Clipboard ⋮ Vote 2 Link × Direct link to this answer Cancel Copy to Clipboard Edited: Image Analyston 29 Nov 2019 This is not a MATLAB question - it's just analytical geometry formulas and algebra. Just draw it out and make sure you see the right angle between the center, the point off the circle, and the point on the circle. Then use the Pythagorean theorem. [EDIT] OK some more hints: Let's say R is the radius, and (xr, yr) is on the circle, and (xc, yc) is the center of the circle, and (xp, yp) is the point away from and off the circle. So by the Pythagorean theorem length1 = sqrt((xp-xr)^2 + (yp-yr)^2) % Length of side #1 length2 = sqrt((xp-xc)^2 + (yp-yc)^2) % Length of side #2 R^2 + length1^2 = length2^2 % The right triangle. Plug in for length1 and length 2 : R^2 + (xp-xr)^2 + (yp-yr)^2 = (xp-xc)^2 + (yp-yc)^2 % The right triangle. Multiply out and cancel and combine terms and you'll be further along. 2 Comments Show None Hide None Zarak khon 29 Nov 2019 × Direct link to this comment Cancel Copy to Clipboard ⋮ Link× Direct link to this comment Cancel Copy to Clipboard Hello and thanks for your answer, As you said I used mathematical and geometry formulation for my question as below: I calculate the slope of the tangent line according to the fact that the slop of the tangent line is equal to derivation of the circle at that point. Then I have a point off the circle and the slope and I need to find the point on the circle. I also have the equation of the circle. so I have 2 equations and two unknown variables which are (xr, yr) and by solving them I get (xr, yr). Regarding your answer as you wrote in the last line, you provide one equation with two unkown parameters which are (xr, yr). Should I add circle equation to your last equation? Image Analyston 29 Nov 2019 × Direct link to this comment Cancel Copy to Clipboard ⋮ Link× Direct link to this comment Cancel Copy to Clipboard Try this web site for the formulas Sign in to comment. More Answers (0) Sign in to answer this question. FEATURED DISCUSSION ###### MATLAB EXPO 2025 Registration is Now Open! November 12 – 13, 2025 Registration is now open for MathWorks annual virtual event MATLAB EXPO 2025... 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15433	https://www.dummies.com/article/academics-the-arts/math/pre-calculus/understanding-the-rules-of-exponential-functions-167736/	Understanding the Rules of Exponential Functions \| dummies Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Fantastic Four For Dummies is here, and we're hosting a sweepstakes to celebrate! Enter now for a chance to win! 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Share Pre-Calculus All-in-One For Dummies Explore Book Buy Now Buy on AmazonBuy on Wiley Subscribe on Perlego Pre-Calculus All-in-One For Dummies Explore Book Buy Now Buy on AmazonBuy on Wiley Subscribe on Perlego Exponential functions follow all the rules of functions. However, because they also make up their own unique family, they have their own subset of rules. The following list outlines some basic rules that apply to exponential functions: The parent exponential function _f_(_x_) = _b__x_always has a horizontal asymptote at _y_ = 0, except when _b_= 1. You can’t raise a positive number to any power and get 0 or a negative number. The domain of any exponential function is This rule is true because you can raise a positive number to any power. However, the range of exponential functions reflects that all exponential functions have horizontal asymptotes. All parent exponential functions (except when _b_= 1) have ranges greater than 0, or The order of operations still governs how you act on the function. When the idea of a vertical transformation applies to an exponential function, most people take the order of operations and throw it out the window. Avoid this mistake. For example, You can’t multiply before you deal with the exponent. You can’t have a base that’s negative. For example, _y_ = (–2)_x_ isn’t an equation you have to worry about graphing in pre-calculus. If you’re asked to graph _y_ = –2 _x_ _,_ don’t fret. You read this as “the opposite of 2 to the _x,_” which means that (remember the order of operations) you raise 2 to the power first and then multiply by –1 _._ This simple change flips the graph upside down and changes its range to A number with a negative exponent is the reciprocal of the number to the corresponding positive exponent. For instance, _y_ = 2–3 doesn’t equal (–2)3 or –2 3. Raising any number to a negative power takes the reciprocal of the number to the positive power: When you multiply monomials with exponents, you add the exponents.For instance, If you break down the problem, the function is easier to see: When you have multiple factors inside parentheses raised to a power, you raise every single term to that power. For instance, (4 _x_ 3 _y_ 5)2 isn’t 4 _x_ 3 _y_ 10; it’s 16 _x_ 6 _y_ 10. When graphing an exponential function, remember that the graph of an exponential function whose base number is greater than 1 always increases (or _rise_ _s_) as it moves to the right; as the graph moves to the left, it always approaches 0 but never actually get there. For example, _f_(_x_) = 2 _x_ is an exponential function, as is The table shows the _x_ and _y_ values of these exponential functions. These parent functions illustrate that, as long as the exponent is positive, the graph of an exponential function whose base is greater than 1 increases as _x_ increases — an example of exponential growth — whereas the graph of an exponential function whose base is between 0 and 1 decreases towards the _x_-axis as _x_ increases — an example of exponential decay. The graph of an exponential function who base numbers is fractions between 0 and 1 always rise to the left and approach 0 to the right. This rule holds true until you start to transform the parent graphs. About This Article This article is from the book: No items found. About the book author: This article can be found in the category: Pre-Calculus No items found. Quick Links About For DummiesContact UsActivate Online ContentSite Map Connect About Dummies Dummies has always stood for taking on complex concepts and making them easy to understand. 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15434	https://www.scribd.com/doc/256002084/Physics-Olympiad-Error-and-Data-Analysis-Note	Physics Olympiad Error and Data Analysis Note \| PDF \| Accuracy And Precision \| Observational Error Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 100%(1)100% found this document useful (1 vote) 2K views 7 pages Physics Olympiad Error and Data Analysis Note This document provides an overview of data analysis and error analysis in scientific experiments. It discusses key aspects of analyzing experimental results including understanding physical … Full description Uploaded by Science Olympiad Blog AI-enhanced description Go to previous items Go to next items Download Save Save Physics Olympiad Error and Data Analysis Note For Later Share 100%100% found this document useful, undefined 0%, undefined Print Embed Translate Ask AI Report Download Save Physics Olympiad Error and Data Analysis Note For Later You are on page 1/ 7 Search Fullscreen D a t a&E r r o r A n a l y s i s 1 DATA and ERROR ANALYSIS Performing the experiment and collecting data is only the beginning of the process of completing an experiment in science. Understanding t he results of any given experiment is always the central goal of the experiment. Presenting those results in a clear concise manner completes the ex periment. This overview of the complete process is as valid in an instructional laboratory course as in a research environm ent. You will not have learned any physics if you did not understand the experiment. Presenting the results of your experimental work is fundamentally saying, "This is what I did and this is what I learned." Putting together your presentation of the results should help you clarify the results to yourself. (If your instructor can clearly see what you did and what you learned, you might get a better grade.) Data analysis should NOT be delayed until all of the data is recorded. Take a low point, a high point and maybe a middle point, and do a quick analysis and plot. This will help one avo id the problem of spending an entire class collecting bad data because of a mistake in experimental procedure or an equipment failure. First and foremost, data analysis means understanding what your results mean. When analyzing your data, try to think through the physical processes which are occurring. Write your train of thought down. Ultimately, the g oal is for you to understand physics and the world a bit bette r. Your understanding of your results probably occurs in stages, with each stage being a re finement and possibly more mathematical than the previous stage. For example, one might first note that as time increases so does dist ance. Next a quick graph of distance vs time might verify this understanding but the relationship is NOT linear, i.e. the data does not form a straight line. By further work, one mig ht discover that distance increase linearly with the square of the tim e. Or sometimes the mathem atical relationship may remain hidden. Relate each successive stage of your understanding and interpretation of your results to the physical principles that are involved . In the above example, one might note that the change in position with time is caused by velocity that is in turn caused by an acceleration from the gravitational force. Finally, develop the related mathem atics. Equations are nearly meaningly unless they are related to the physical laws. (Remember to identify all the variables and constants in you equations.) Sometimes, your results will not support and may even contradict the physical explanation suggested by the manual or your instructor. Say so! But of course then a few sug gestions as to the reason for this apparent failure of the physical laws,would be in order. Do NO T just say " The equipment was a piece of sh_t!" Try to explain what went w rong or what competing effects have come into play.One of the reasons that you are encouraged to record everything that is going on as it is going on, is that this information may help exp lain bad results. For example, partly for fun, you note ea ch time your lab partner sneezes. Later while looking at the data, you discover that each data point that was being collected during a sneeze deviates from the pattern of the rest of the data. This m ay g ive yo u good reason for dropping"bad" data. The quality of the data, determines to a great extent, what conclusions can be reached from them . If you are look ing for a small effect, say a total change of 1 mm, and the uncerta inties in your data is 2 mm then you really can not make any solid conclusions. (See the section on error analy sis below.)When one considers the quality of a measurement there are two aspects to consider. The first is if one were to repeat the measurement, how close would new results be to the old, i.e., how reproducible is th e measurement? Scientists refer to this as the precision of the measurement. Secondly, a measurement is considered “good”if it agrees with the true value. This is known as the accuracy of the m easurement. But there is a potential problem in that one needs to know the “true value” to determine the accuracy. A good measurement must be close to the“true value” and be reproducible. In this experiment, if someone made one measurement of g 2 and got 9.79 m/s, it would be an accur ate measurement. But if next time they tried they got 4.1 2 m/s, no one w ould believe that they were any thing but lucky in the first measurement. Similarly, if one group 2 got values of 7.31, 7.30, 7.33, and 7.29 m/s their results are reproducible but not really very good. adDownload to read ad-free D a t a&E r r o r A n a l y s i s 2 Accuracy vs. Precision: These two words do not mean the sam e thing. " Accuracy " deals with how close is a measured value to an accep ted or "true"value. " Precision " deals with how reproducible is a given measureme nt. Because, precision is a measure of how reproducible a measurement is, one ca n gain some knowledge of the precision simply by taking a number of measurements and comparing them. If the true value is not known, the accuracy of a measurement is more difficult to know. Here are a few things to consider when trying to estimate (or explain) the quality of a measurement:1.  How well does your equipment make the needed measurements? Two examples of problems: a meter stick with 2 mm worn off one e nd and trying to measure 0.01 mm with a meter s tick.2.  How does the lack of precision, or uncertainty in any one measured variable effect the final calculated value? For example, if the data is expected to fall on a straight line, some uncertainties may only shift the intercept and leave the slope unchanged. 3.  Making a plot that shows each of your measurements, can help you “see” your uncertainty. Also certain uncertainties are sidestepped by extracting a slope from a plot and calculating the final value from this slope. 4.  One can estimate the uncertainty a fter making multiple measurem ents. First, note that when plotted, about a of the data points will be outside of the error bars as they are normally drawn at one standard deviation . For example if there are six measurements,one expects that 2 of the points ( probably one too high and one too low) will be outside the normal error bars. One can draw a reasonable set of error bars based on this assumption, by drawing the upper part of the error bar between the highest two points and the lower part of the error bar between the lowest two points. Data analysis is seldom a straight forward process because of the presence of uncertainties. Data can not be fully understood until the associated uncertainties are understood.  ERROR ANALYSIS The words “error” and “uncertainty” are used to describe the same concept in measurement. It is unfortunate that the term, “error’ is the standard scientific word because usually there is no mistake or error in mak ing a measurement. Frequently, the uncertainties are dominated by natural irregularities or differences in what is being measured. Types of Error: All measurements have errors. Errors may arise from three sources:a) Careless errors : These are due to mistak es in reading scales or careless setting of markers,etc. They can be eliminated by repetition of readings by one or two observers.b) Systematic errors: These are due to built-in errors in the instrument either in design or calibration. Repetition of observation with the same instrument will not show a spread in the measurements. They are t he hardest source of errors to detect.c) Random errors: These always lead to a spread or distribution of results on repetition of the p articular measurement. They may arise from fluctuations in either the physical parameters due to the statistical nature of the particular phenomenon or the judgement of the experimenter, such as variation in re sponse time or estimation in scale reading. Taking multiple measurements helps reduce uncertainties.  DETERMINING ERRORS: Although it is interesting and reassuring to compare your results against “accepted values,” it is suggested that error analysis be done, before this comparison. One reason is that the accuracy of a set of measurements depends on how well the experiment was done, not how close the measurement was to the accepted value. One could get close to the accepted value, by sloppiness and luck. Do not base your experimental uncertainties on the accepted values. adDownload to read ad-free D a t a&E r r o r A n a l y s i s 3 Determining the source of uncertainty and the magnitude of this uncertainty is often difficult. Some errors or uncertainties are caused by natural fluctuations or irregularities These can not be eliminated. To estim ate these uncertainties one frequently uses mathematical methods similar to those discussed in the section below titled, Averages and Deviation. Another method of estimating uncertainties is to assign an uncertainty to the measurement equal to the finest scale r eading on the measuring instrument. For example, if a ruler may be marked in millimeters. then the uncertainty in any measurement with this ruler can be given as 1 mm. But with practice, one might be able to interpolate the scale and reduce the error to 0.25 mm.  EXPRESSING ERRORS: For each measured value, A , there is an estimated error,  A . The complete result is given by A ±  A . This means that the “true value” prob ably lies between a maximum value of A +  A and a minimum value of A -  A. Sometimes the terms relative error and percent error are used, where:Errors are expressed in graphs by using error bars. Consider a data point ( A,B ) and the associated uncertainties are  A and  B respectively. The vertical error bar is drawn from B-  B to B+  B. Similarly, the horizontal error bar is drawn from A-  A to A+  A . For example if A = 4.2±0.8 and B =3.2±0.5 then this point would be plotted like:  PROPAGATION OF ERRORS When measured values are used to calculate other values, the uncertainties in these measured values causes uncertainties in the calculated values. Calculating the uncertainties in the calculated values is called error propagation. For the few sim ple cases that are discussed below, let C be a function of A and B and the associated uncertainties are  C ,  A , and  B , respectively. 1) Product with a Constant Here C = k A where k is a con stant. Then This rule can be applied if k is a measured quantity with a relatively negligible uncertainty, for example, if k were the gravitational constant, g . 2) Addition and Subtraction Here C = A + B or C = A - B. In either case:Note that  C is less than  A  B . This is an expression of the fact that the uncertainties in A and B are independent of each o ther. ( In m ath lingo, one could say that  A , and  B are orthogonal or perpendicular to each other. Note how the calculation of  C is identical to the Pythagorean theorem for the sides of a right triangle.) Why should the same formula work for additon and subtraction? Notice that the original uncertainties are squared. 3)M u l t i p l i c a t i o n o r D i v i s i o n If C = A B or C = A/B then:Again the assumption is that the uncertainties in B and B are independant. An example of when this is not true is C = A A. This bring s up another rule. adDownload to read ad-free D a t a&E r r o r A n a l y s i s 4 4)R a i s e d t o a P o w e r This is the case were C = A , where n is a n constant. In this case: 5) Graphical Analysis of Uncertainties in ,Slopes and Intercepts. If the slope or intercept of a line on a plot is the required calculated value (or the required value is calculated from these values) then the uncertainty of the slope and intercept will also be required. Graphically one can estimate th ese uncertainties. First draw the best line possible, and then draw the two lines that just barely pas s through the data. The differences of these slopes and intercepts from those of the best fit line provide an estimate of the uncertainties in these quantities. In the graph below the solid line is a good fit to the data and the while the dotted line and the dashed line represent the extremes for lines that just fit the data.. The dashed line has the steepest slope and will be referred to as the max. line, while the dotted line will be referred to as the m in. line. The slopes and y- intercepts for these three lines are:S l o p e(d i f.)y-i n t e r.(d i f.)B e s t f i t 1.0 2.0 M a x. l i n e 1.1 6( 0.1 6) -1.5(-3.5)M i n. L i n e 0.8 1(-0.1 9) 5.2( 3.2)Thus if the data was expected to fit the equation:then one would estimate the constants as:a = 2.0 ± 3.4 b = 1.0 ± 0.2.The uncertainties are based on averaging the absolute value of the differences (labeled (dif.)). adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Read this document in other languages English Português Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Analytical Mechanics Seventh Edition Grant R. Fowles Download No ratings yet Analytical Mechanics Seventh Edition Grant R. 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15435	https://geo.libretexts.org/Bookshelves/Sedimentology/Introduction_to_Fluid_Motions_and_Sediment_Transport_(Southard)/03%3A_Flow_Past_a_Sphere_II_-_Stokes'_Law_The_Bernoulli_Equation_Turbulence_Boundary_Layers_Flow_Separation/3.04%3A_The_Bernoulli_Equation	Skip to main content 3.4: The Bernoulli Equation Last updated : Mar 5, 2021 Save as PDF 3.3: Inviscid Flow 3.5: Turbulence Page ID : 4165 John Southard Massachusetts Institute of Technology via MIT OpenCourseware ( \newcommand{\kernel}{\mathrm{null}\,}) In the example of inviscid flow past a sphere described in the preceding section, the pressure is high at points where the velocity is low, and vice versa. It is not difficult to derive an equation, called the Bernoulli equation, that accounts for this relationship. Because this will be useful later on, I will show you here how it comes about. Flowlines First I have to be more specific about what I have casually been calling flow lines. Fluid velocity is a vector quantity, and, because the fluid behaves as a continuum, a velocity vector can be associated with every point in the flow. (Mathematically, this is described as a vector field.) Continuous and smooth curves that can be drawn to be everywhere tangent to the velocity vectors throughout the vector field are called streamlines (Figure ). One and only one streamline passes through each point in the flow, and at any given time there is only one such set of curves in the flow. There obviously is an infinity of streamlines passing through any region of flow, no matter how small; usually only a few representative streamlines are shown in sketches and diagrams. An important property of streamlines follows directly from their definition: the flow can never cross streamlines. If the flow is steady, the streamline pattern does not change with time; if the streamline pattern changes with time, the flow is unsteady. But note that the converse of each of these statements is not necessarily true, because an unsteady flow can exhibit an unchanging pattern of streamlines as velocities everywhere increase or decrease with time. There are two other kinds of flow lines, with which you should not confuse streamlines (Figure ): pathlines, which are the trajectories traced out by individual tiny marker particles emitted from some point within the flow that is fixed relative to the stationary boundaries of the flow, and streaklines, which are the streaks formed by a whole stream of tiny marker particles being emitted continuously from some point within the flow that is fixed relative to the stationary boundaries of the flow. In steady flow, streamlines and pathlines and streaklines are all the same; in unsteady flow, they are generally all different. Streamtubes and the Bernoulli Equation You also can imagine a tube-like surface formed by streamlines, called a stream tube, passing through some region (Figure ). This surface or set of streamlines can be viewed as functioning as if it were a real tube or conduit, in that there is flow through the tube but there is no flow either inward or outward across its surface. Consider a short segment of one such tiny stream tube in a flow of incompressible fluid (Figure ). Write the equation of motion (Newton’s second law) for the fluid contained at some instant in this stream-tube segment. The cross-sectional area of the tube is , and the length of the segment is . If the pressure at cross section 1, at the left-hand end of the segment, is , then the force exerted on this end of the segment is . It is not important that the area of the cross section might be slightly different at the two ends (if the flow is expanding or contracting), or that might vary slightly over the cross section, because you can make the cross-sectional area of the stream tube as small as you please. What is the force on the other end of the tube? The pressure at cross section 2 is different from that at cross section 1 by , the rate of change of pressure in the flow direction times the distance between the two cross sections, so the force on the right-hand end of the tube is The net force on the stream tube in the flow direction is then The pressure on the lateral surface of the tube is of no concern, because the pressure force on it acts normal to the flow direction. Newton’s second law, , for the fluid in the segment of the stream tube, where is the velocity of the fluid at any point (in this section is used not as the component of velocity in the direction but as the component of velocity tangent to the streamline at a given point), is then Simplifying Equation and making use of the fact that is constant and so can be moved outside the derivative, The derivative on the right side of Equation can be put into more convenient form by use of the chain rule and a simple “undifferentiation” of one of the resulting terms: Equation is strictly true only for the single streamline to which the stream tube collapses as we let go to , because only then need we not worry about possible variation of either or over the cross sections. Assuming further that the flow is steady, , and Equation becomes It is easy to integrate Equation between two points and on the streamline (remember that this equation holds for any streamline in the flow): or, viewed in another way, You can see from Equation that if the flow is steady and incompressible there is an inverse relationship between fluid pressure and fluid velocity along any streamline. Equation or Equation is called the Bernoulli equation. Remember that it holds only along individual streamlines, not through the entire flow. In other words, the constant in the Equation is generally different for each streamline in the flow. And it holds only for inviscid flow, because if the fluid is viscous there are shearing forces across the lateral surfaces of stream tubes, and Newton’s second law cannot be written and manipulated so simply. But often in flow of a real fluid the viscous forces are small enough outside the boundary layer that the Bernoulli equation is a good approximation. Note Note that the right-hand side of Equation is the negative of the increase in kinetic energy per unit volume of fluid between point 1 and point 2. The Bernoulli equation is just a statement of the work–energy theorem, whereby the work done by a force acting on a body is equal to the change in kinetic energy of the body. In this case, fluid pressure is the only force acting on the fluid. In discussing inviscid flow around a sphere I called the front and rear points of the sphere the stagnation points, because velocities relative to the sphere are zero there. Using the Bernoulli equation it is easy to find the corresponding stagnation pressures. Taking the free-stream values of pressure and velocity to be po and vo, writing Equation in the form and substituting at the stagnation points, the stagnation pressures (the same for front and rear points) are 3.3: Inviscid Flow 3.5: Turbulence
15436	https://math.fandom.com/wiki/Quotient_rule	Quotient rule The quotient rule is used to determine the derivative of a function expressed as the quotient of 2 differentiable functions. It is defined as shown: Also written as: This can also be done as a Product rule (with an inlaid Chain rule): d d x ( u v − 1 ) = u ( − 1 v − 2 v ′ ) + ( v − 1 ) u ′ {\displaystyle {d\over dx} (u v^{-1}) = u(-1v^{-2} v') + (v^{-1})u'} d d x ( u v − 1 ) = − u v ′ v 2 + u ′ v {\displaystyle {d\over dx} (u v^{-1}) = {-uv'\over v^{2}} + {u'\over v}} d d x ( u v − 1 ) = u ′ v − u v ′ v 2 {\displaystyle {d\over dx} (u v^{-1}) = {u'v-uv'\over v^{2}}} You may do this whichever way you prefer. Proof[] We know that the two following limits exist as f ( x ) , g ( x ) {\displaystyle f(x), g(x)} are differentiable. We also have the condition that g ( x ) ≠ 0 {\displaystyle g(x) \ne 0} . Applying the first principles definition of differentiation we get We can combine the two fractions into one fraction by cross-multiplying. If we add and subtract a f ( x ) g ( x ) {\displaystyle f(x) g(x)} , it will change nothing. If we bring the g ( x + h ) g ( x ) {\displaystyle g(x+h)g(x)} term in the denominator to the front, and separate the numerator with algebra of limits we will have We now break this into a product of two limits, and a sum of two limits. Evaluating all limits yields Putting this in a more familiar form, we have This completes the proof. Fandom logo Explore properties Follow Us Overview Community Advertise Fandom Apps
15437	https://tutorial.math.lamar.edu/classes/calci/areaproblem.aspx	Calculus I - Area Problem Pauls NotesClose menu Pauls NotesPauls Notes Home Classes Open submenu (Algebra)Algebra Open submenu (Calculus I)Calculus I Open submenu (Calculus II)Calculus II Open submenu (Calculus III)Calculus III Open submenu (Differential Equations)Differential Equations Extras Open submenu (Algebra & Trig Review)Algebra & Trig Review Open submenu (Common Math Errors)Common Math Errors Open submenu (Complex Number Primer)Complex Number Primer Open submenu (How To Study Math)How To Study Math Cheat Sheets & Tables Misc Links Contact Me Links MathJax Help and Configuration Privacy Statement Site Help & FAQ Terms of Use No results found. Close submenu (Algebra)AlgebraPauls Notes/Algebra Open submenu (1. Preliminaries)1. Preliminaries Open submenu (2. Solving Equations and Inequalities)2. Solving Equations and Inequalities Open submenu (3. Graphing and Functions)3. Graphing and Functions Open submenu (4. 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Learn From Your Errors Paul's Online Notes Notes Quick Nav Download × Custom Search Go To Notes Practice Problems Assignment Problems Show/Hide Show all Solutions/Steps/etc. Hide all Solutions/Steps/etc. Sections More Substitution Rule Definition of the Definite Integral Chapters Applications of Derivatives Applications of Integrals Classes Algebra Calculus I Calculus II Calculus III Differential Equations Extras Algebra & Trig Review Common Math Errors Complex Number Primer How To Study Math Cheat Sheets & Tables Misc Contact Me MathJax Help and Configuration Notes Downloads Complete Book Practice Problems Downloads Complete Book - Problems Only Complete Book - Solutions Assignment Problems Downloads Complete Book Other Items Get URL's for Download Items Print Page in Current Form (Default) Show all Solutions/Steps and Print Page Hide all Solutions/Steps and Print Page Paul's Online Notes Home/ Calculus I/ Integrals / Area Problem Prev. SectionNotesPractice ProblemsAssignment ProblemsNext Section Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width. Section 5.5 : Area Problem As noted in the first section of this section there are two kinds of integrals and to this point we’ve looked at indefinite integrals. It is now time to start thinking about the second kind of integral : Definite Integrals. However, before we do that we’re going to take a look at the Area Problem. The area problem is to definite integrals what the tangent and rate of change problems are to derivatives. The area problem will give us one of the interpretations of a definite integral and it will lead us to the definition of the definite integral. To start off we are going to assume that we’ve got a function f(x)f(x) that is positive on some interval [a,b][a,b]. What we want to do is determine the area of the region between the function and the x x-axis. It’s probably easiest to see how we do this with an example. So, let’s determine the area between f(x)=x 2+1 f(x)=x 2+1 on [0,2][0,2]. In other words, we want to determine the area of the shaded region below. Now, at this point, we can’t do this exactly. However, we can estimate the area. We will estimate the area by dividing up the interval into n n subintervals each of width, Δ x=b−a n Δ x=b−a n Then in each interval we can form a rectangle whose height is given by the function value at a specific point in the interval. We can then find the area of each of these rectangles, add them up and this will be an estimate of the area. It’s probably easier to see this with a sketch of the situation. So, let’s divide up the interval into 4 subintervals and use the function value at the right endpoint of each interval to define the height of the rectangle. This gives, Note that by choosing the height as we did each of the rectangles will over estimate the area since each rectangle takes in more area than the graph each time. Now let’s estimate the area. First, the width of each of the rectangles is 1 2 1 2. The height of each rectangle is determined by the function value at the right endpoint and so the height of each rectangle is nothing more that the function value at the right endpoint. Here is the estimated area. A r=1 2 f(1 2)+1 2 f(1)+1 2 f(3 2)+1 2 f(2)=1 2(5 4)+1 2(2)+1 2(13 4)+1 2(5)=5.75 A r=1 2 f(1 2)+1 2 f(1)+1 2 f(3 2)+1 2 f(2)=1 2(5 4)+1 2(2)+1 2(13 4)+1 2(5)=5.75 Of course, taking the rectangle heights to be the function value at the right endpoint is not our only option. We could have taken the rectangle heights to be the function value at the left endpoint. Using the left endpoints as the heights of the rectangles will give the following graph and estimated area. A l=1 2 f(0)+1 2 f(1 2)+1 2 f(1)+1 2 f(3 2)=1 2(1)+1 2(5 4)+1 2(2)+1 2(13 4)=3.75 A l=1 2 f(0)+1 2 f(1 2)+1 2 f(1)+1 2 f(3 2)=1 2(1)+1 2(5 4)+1 2(2)+1 2(13 4)=3.75 In this case we can see that the estimation will be an underestimation since each rectangle misses some of the area each time. There is one more common point for getting the heights of the rectangles that is often more accurate. Instead of using the right or left endpoints of each sub interval we could take the midpoint of each subinterval as the height of each rectangle. Here is the graph for this case. So, it looks like each rectangle will over and under estimate the area. This means that the approximation this time should be much better than the previous two choices of points. Here is the estimation for this case. A m=1 2 f(1 4)+1 2 f(3 4)+1 2 f(5 4)+1 2 f(7 4)=1 2(17 16)+1 2(25 16)+1 2(41 16)+1 2(65 16)=4.625 A m=1 2 f(1 4)+1 2 f(3 4)+1 2 f(5 4)+1 2 f(7 4)=1 2(17 16)+1 2(25 16)+1 2(41 16)+1 2(65 16)=4.625 We’ve now got three estimates. For comparison’s sake the exact area is A=14 3=4.66¯¯¯6 A=14 3=4.66 6¯ So, both the right and left endpoint estimation did not do all that great of a job at the estimation. The midpoint estimation however did quite well. Be careful to not draw any conclusion about how choosing each of the points will affect our estimation. In this case, because we are working with an increasing function choosing the right endpoints will overestimate and choosing left endpoint will underestimate. If we were to work with a decreasing function we would get the opposite results. For decreasing functions the right endpoints will underestimate and the left endpoints will overestimate. Also, if we had a function that both increased and decreased in the interval we would, in all likelihood, not even be able to determine if we would get an overestimation or underestimation. Now, let’s suppose that we want a better estimation, because none of the estimations above really did all that great of a job at estimating the area. We could try to find a different point to use for the height of each rectangle but that would be cumbersome and there wouldn’t be any guarantee that the estimation would in fact be better. Also, we would like a method for getting better approximations that would work for any function we would chose to work with and if we just pick new points that may not work for other functions. The easiest way to get a better approximation is to take more rectangles (i.e. increase n n). Let’s double the number of rectangles that we used and see what happens. Here are the graphs showing the eight rectangles and the estimations for each of the three choices for rectangle heights that we used above. Here are the area estimations for each of these cases. A r=5.1875 A l=4.1875 A m=4.65625 A r=5.1875 A l=4.1875 A m=4.65625 So, increasing the number of rectangles did improve the accuracy of the estimation as we’d guessed that it would. Let’s work a slightly more complicated example. Example 1 Estimate the area between f(x)=x 3−5 x 2+6 x+5 f(x)=x 3−5 x 2+6 x+5 and the x x-axis on [0,4][0,4] using n=5 n=5 subintervals and all three cases above for the heights of each rectangle. Show Solution First, let’s get the graph to make sure that the function is positive. So, the graph is positive and the width of each subinterval will be, Δ x=4 5=0.8 Δ x=4 5=0.8 This means that the endpoints of the subintervals are, 0,0.8,1.6,2.4,3.2,4 0,0.8,1.6,2.4,3.2,4 Let’s first look at using the right endpoints for the function height. Here is the graph for this case. Notice, that unlike the first area we looked at, the choosing the right endpoints here will both over and underestimate the area depending on where we are on the curve. This will often be the case with a more general curve that the one we initially looked at. The area estimation using the right endpoints of each interval for the rectangle height is, A r=0.8 f(0.8)+0.8 f(1.6)+0.8 f(2.4)+0.8 f(3.2)+0.8 f(4)=28.96 A r=0.8 f(0.8)+0.8 f(1.6)+0.8 f(2.4)+0.8 f(3.2)+0.8 f(4)=28.96 Now let’s take a look at left endpoints for the function height. Here is the graph. The area estimation using the left endpoints of each interval for the rectangle height is, A l=0.8 f(0)+0.8 f(0.8)+0.8 f(1.6)+0.8 f(2.4)+0.8 f(3.2)=22.56 A l=0.8 f(0)+0.8 f(0.8)+0.8 f(1.6)+0.8 f(2.4)+0.8 f(3.2)=22.56 Finally, let’s take a look at the midpoints for the heights of each rectangle. Here is the graph, The area estimation using the midpoint is then, A m=0.8 f(0.4)+0.8 f(1.2)+0.8 f(2)+0.8 f(2.8)+0.8 f(3.6)=25.12 A m=0.8 f(0.4)+0.8 f(1.2)+0.8 f(2)+0.8 f(2.8)+0.8 f(3.6)=25.12 For comparison purposes the exact area is, A=76 3=25.33¯¯¯3 A=76 3=25.33 3¯ So, again the midpoint did a better job than the other two. While this will be the case more often than not, it won’t always be the case and so don’t expect this to always happen. Now, let’s move on to the general case. Let’s start out with f(x)≥0 f(x)≥0 on [a,b][a,b] and we’ll divide the interval into n n subintervals each of length, Δ x=b−a n Δ x=b−a n Note that the subintervals don’t have to be equal length, but it will make our work significantly easier. The endpoints of each subinterval are, x 0=a x 1=a+Δ x x 2=a+2 Δ x⋮x i=a+i Δ x⋮x n−1=a+(n−1)Δ x x n=a+n Δ x=b x 0=a x 1=a+Δ x x 2=a+2 Δ x⋮x i=a+i Δ x⋮x n−1=a+(n−1)Δ x x n=a+n Δ x=b Next in each interval, [x 0,x 1],[x 1,x 2],…,[x i−1,x i],…,[x n−1,x n][x 0,x 1],[x 1,x 2],…,[x i−1,x i],…,[x n−1,x n] we choose a point x∗1,x∗2,…,x∗i,…x∗n x 1∗,x 2∗,…,x i∗,…x n∗. These points will define the height of the rectangle in each subinterval. Note as well that these points do not have to occur at the same point in each subinterval. However, they are usually the left end point of the interval, right end point of the interval or the midpoint of the interval. Here is a sketch of this situation. Image 12: This is the graph of some unknown function on the domain a<x<b. It is completely in the 1st quadrant. It initially decreases until it hits a valley, then increases to a peak, then decreases to a new valley and finally increases for a short distance before ending. Along the x-axis are points labeled $x_{0}$, $x_{1}$, $x_{2}$, $x_{3}$ on the left end and $x_{n-1}$, $x_{n}$ on the right end. Between each pair of points is another set of points labeled $x_{1}^{}$, $x_{2}^{}$, $x_{3}^{}$ on the left and , $x_{n}^{}$ on the right. Each of the “” points are used to get the height of a rectangle above it. So, there are three rectangles on the left and one on the right that are used to represent the area. There is a blank space between the three rectangles on the left and the one on the right to indicate it would be filled in with other rectangles if we actually a specific value for n. The area under the curve on the given interval is then approximately, A≈f(x∗1)Δ x+f(x∗2)Δ x+⋯+f(x∗i)Δ x+⋯+f(x∗n)Δ x A≈f(x 1∗)Δ x+f(x 2∗)Δ x+⋯+f(x i∗)Δ x+⋯+f(x n∗)Δ x We will use summation notation or sigma notation at this point to simplify up our notation a little. If you need a refresher on summation notation check out the section devoted to this in the Extras chapter. Using summation notation the area estimation is, A≈n∑i=1 f(x∗i)Δ x A≈∑i=1 n f(x i∗)Δ x The summation in the above equation is called a Riemann Sum. To get a better estimation we will take n n larger and larger. In fact, if we let n n go out to infinity we will get the exact area. In other words, A=lim n→∞n∑i=1 f(x∗i)Δ x A=lim n→∞⁡∑i=1 n f(x i∗)Δ x Before leaving this section let’s address one more issue. To this point we’ve required the function to be positive in our work. Many functions are not positive however. Consider the case of f(x)=x 2−4 f(x)=x 2−4 on [0,2]. If we use n=8 n=8 and the midpoints for the rectangle height we get the following graph, In this case let’s notice that the function lies completely below the x x-axis and hence is always negative. If we ignore the fact that the function is always negative and use the same ideas above to estimate the area between the graph and the x x-axis we get, A m=1 4 f(1 8)+1 4 f(3 8)+1 4 f(5 8)+1 4 f(7 8)+1 4 f(9 8)+1 4 f(11 8)+1 4 f(13 8)+1 4 f(15 8)=−5.34375 A m=1 4 f(1 8)+1 4 f(3 8)+1 4 f(5 8)+1 4 f(7 8)+1 4 f(9 8)+1 4 f(11 8)+1 4 f(13 8)+1 4 f(15 8)=−5.34375 Our answer is negative as we might have expected given that all the function evaluations are negative. So, using the technique in this section it looks like if the function is above the x x-axis we will get a positive area and if the function is below the x x-axis we will get a negative area. Now, what about a function that is both positive and negative in the interval? For example, f(x)=x 2−2 f(x)=x 2−2 on [0,2]. Using n=8 n=8 and midpoints the graph is, Some of the rectangles are below the x x-axis and so will give negative areas while some are above the x x-axis and will give positive areas. Since more rectangles are below the x x-axis than above it looks like we should probably get a negative area estimation for this case. In fact that is correct. Here the area estimation for this case. A m=1 4 f(1 8)+1 4 f(3 8)+1 4 f(5 8)+1 4 f(7 8)+1 4 f(9 8)+1 4 f(11 8)+1 4 f(13 8)+1 4 f(15 8)=−1.34375 A m=1 4 f(1 8)+1 4 f(3 8)+1 4 f(5 8)+1 4 f(7 8)+1 4 f(9 8)+1 4 f(11 8)+1 4 f(13 8)+1 4 f(15 8)=−1.34375 In cases where the function is both above and below the x x-axis the technique given in the section will give the net area between the function and the x x-axis with areas below the x x-axis negative and areas above the x x-axis positive. So, if the net area is negative then there is more area under the x x-axis than above while a positive net area will mean that more of the area is above the x x-axis. [Contact Me][Privacy Statement][Site Help & FAQ][Terms of Use] © 2003 - 2025 Paul Dawkins Page Last Modified : 11/2/2023
15438	https://www.youtube.com/watch?v=VH1lAfZL_fU	Worked examples: Punnett squares \| Inheritance and variation \| Middle school biology \| Khan Academy Khan Academy 8970000 subscribers 128 likes Description 12548 views Posted: 22 Jul 2022 Keep going! Check out the next lesson and practice what you’re learning: Solve Punnett squares that include heterozygous and homozygous genotypes. Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help! Donate or volunteer today! Donate here: Volunteer here: 5 comments Transcript: Noxud bitkilərinin populyasiyasında bəzilərinin yumru, digərlərinin isə qırışıq toxumlara sahib olduğu bizə məlumdur. Bu populyasiyada toxum formasının iki mümkün alleli mövcuddur. Allellərin genin versiyaları olduğunu xatırlayaq. Bir allel və ya bir versiya yumru toxumlar üçündür, onu böyük R hərfi ilə işarə edirik, digər allel isə qırışıq toxumlar üçündür, onu isə kiçik r hərfi ilə qeyd edirik. Böyük R hərfi ilə işarə etdiyimiz yumru toxum allelinin dominant olduğunu qeyd edək. İki heteroziqot noxud bitkisinin çarpazlaşdırılması üçün, yumru toxumların qırışıq toxumlara nisbətini müəyyənləşdirmək lazımdır. İndi videonu dayandırın, baxın, birlikdə etməzdən əvvəl təkbaşına edə bilərsinizmi? Davam edək. İki heteroziqot noxud bitkisinin genotipi haqqında düşünək. Heteroziqot, bir genin iki müxtəlif alleli və ya versiyası deməkdir. Bunlar böyük R və kiçik r ilə işarə olunan həmin noxud bitkiləridir. Hər birindən bir ədəd var. Çarpazlaşma zamanı nələr baş verdiyini görmək üçün Pennet cədvəli çəkəcəyəm. Bu formada çəkirəm, indi isə sizə burada nə baş verdiyini izah edəcəyəm. Bu formada çəkirəm. İki fərdin olduğunu düşünərək onların hər birini böyük R və kiçik r ilə qeyd edək. Bu o deməkdir ki, fərdlərdən biri həm böyük R və həm də kiçik r alleli verə bilər, həmçinin eyni proses digər fərd üçün də keçərlidir. Yəni böyük R alleli də verə bilir, kiçik r alleli də. Bəs bütün bunların nəsilvermə üçün yaratdığı imkanlar eynidir mi? Yeni nəsil hər iki fərddən böyük R, yəni dominant allel olan yumru toxum allelini alır. İndi gəlin sütun və cərgədəki hərfləri birləşdirək. Bu fərddən böyük R hərfini götürürəm, digər fərddən isə kiçik r hərfini. Buradan isə kiçik r hərfini götürürəm, böyük R hərfini də buradan, burada isə hər ikisində iki ədəd kiçik r alınır. Beləliklə, bu dördü bərabər genotiplərdir. Hazırda bizdən genotiplər soruşulmur. Biz yumru toxumlarla qırışıq toxumların nisbətini müəyyən etməliyik. Yumru toxumlara sahib olmaq üçün genotip necə olmalıdır? Yumru toxum allellərinin dominant olduğunu bilirsiniz. Burada böyük R yaranır , bu, dominant gendir. Həmçinin böyük R və kiçik r genləri də yumru toxumlarda mövcuddur. Qırışıq toxumlara sahib olmağın yeganə yolu homoziqot qırışıq toxum allellərinin olmasıdır. Bu yalnız qırışıq toxumlara aiddir. Buraya baxsanız görərsiniz ki, mövcud 4 fenotipdən 3ü yumru toxumların fenotipidir, yalnız biri digərlərindən fərqlənir. Fenotip görünüşdə ifadə olunan əlamətlərdir. Bu 4 gendən yalnız biri qırışıq toxum fenotipinə sahibdir. Beləliklə nəsil üçün gözlənilən nisbət alınır, yumru toxumlar üç, qırışıq toxumlar isə bir nisbətindədir. İndi isə başqa bir sual meydana gəlir, noxud bitkisinin homoziqot və heteroziqot toxumları arasında çarpazlaşma necə olur? Videonu yenidən dayandırın, bunu tək edə bilərsiniz. Burada vacib məsələ hər bir fərdin genotipində nə baş verdiyidir. Qırışıq toxumların homoziqot formalarından danışırıqsa, xatırlayaq ki, biz qırışıq toxum allellərini kiçik r ilə qeyd edirik. Onun genotipi kiçik rr kimi qeyd olunur ki və heteroziqot olduğu deyilir. Əvvəl də dediyimiz kimi, Böyük R və kiçik r ilə qeyd olunur. Bunu aydın görə bilmək üçün Pennet cədvəlini yenidən çəkək. Cədvələ iki sətir və iki sütün çəkirəm, homoziqot fərdləri bura yazıram. Burada o, kiçik r hərfi ilə qeyd olunur, həmçinin buraya da kiçik r hərfi yazılır. Bu hərflərdən biri ilə də yazıla bilər. indi isə heteroziqot fərd haqqında düşünək. Onlar dominant allel olan böyük R hərfi, və kiçik r hərfi ilə qeyd olunur. Digər oxşar xanaların genotipini yazaq. Bütün xanalar buradakı heteroziqot fərddən böyük R hərfini götürür. Böyük R hərfini yazaq. Hər iki xanada homoziqot fərddən kiçik r hərfini alırıq. Bu xanalarda isə, hər iki valideyndən qırışıq toxum allelini alırıq. Heteroziqot fərddən kiçik r genini alırıq, digər xanalarda isə kiçik r genini homoziqot fərdlərdən alırıq. Bunu burada qeyd edirəm. Beləliklə, bu genotiplərinin hansının yumru toxum fenotipi olacaq? Ən az bir yumru toxum allellərinin olduğu xanalar yumru toxumlardır. Bu xanadakılar yumru toxumlarındır. Yumru toxum geninin allelləri dominant olduğu üçün onlar həmin fenotipə malik olacaqlar. Onlardan yalnız birinin olmasına baxmayaraq, gördüyünüz dominant allelin fenotipi buradadır. Bu ikisi isə qırışıq toxumlar olacaq. Yumru toxumların qırışıqlara nisbəti nə qədər olacaq? 1:1 nisbətində olacaq. Hər iki yumru toxuma iki qırışıq toxum, hər bir yumru toxuma isə 1 qırışıq toxum. 2:2 və 1:1 isə nisbət baxımından eynidir. Beləliklə, işimiz bitdi.
15439	https://pmc.ncbi.nlm.nih.gov/articles/PMC2929977/	Central respiratory chemoreception - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Published in final edited form as: J Comp Neurol. 2010 Oct 1;518(19):3883–3906. doi: 10.1002/cne.22435 Search in PMC Search in PubMed View in NLM Catalog Add to search Central respiratory chemoreception Patrice G Guyenet Patrice G Guyenet 1 Department of Pharmacology, University of Virginia, Charlottesville, VA 22908 Find articles by Patrice G Guyenet 1, Ruth L Stornetta Ruth L Stornetta 1 Department of Pharmacology, University of Virginia, Charlottesville, VA 22908 Find articles by Ruth L Stornetta 1, Douglas A Bayliss Douglas A Bayliss 1 Department of Pharmacology, University of Virginia, Charlottesville, VA 22908 Find articles by Douglas A Bayliss 1 Author information Copyright and License information 1 Department of Pharmacology, University of Virginia, Charlottesville, VA 22908 ✉ Address correspondence to: Patrice G. Guyenet, PhD, University of Virginia Health System, P.O. Box 800735, 1300 Jefferson Park Avenue, Charlottesville, VA 22908-0735. Tel: (434) 924-7433; Fax: (434) 982-3878; pgg@virginia.edu PMC Copyright notice PMCID: PMC2929977 NIHMSID: NIHMS221311 PMID: 20737591 The publisher's version of this article is available at J Comp Neurol Summary By definition central respiratory chemoreceptors (CRCs) are cells that are sensitive to changes in brain PCO 2 or pH and contribute to the stimulation of breathing elicited by hypercapnia or metabolic acidosis. CO 2 most likely works by lowering pH. The pertinent proton receptors have not been identified and may be ion channels. CRCs are probably neurons but may also include acid-sensitive glia and vascular cells that communicate with neurons via paracrine mechanisms. Retrotrapezoid nucleus (RTN) neurons are the most completely characterized CRCs. Their high sensitivity to CO 2 in vivo presumably relies on their intrinsic acid-sensitivity, excitatory inputs from the carotid bodies and brain regions such as raphe and hypothalamus, and facilitating influences from neighboring astrocytes. RTN neurons are necessary for the respiratory network to respond to CO 2 during the perinatal period and under anesthesia. In conscious adults, RTN neurons contribute to an unknown degree to the pH-dependent regulation of breathing rate, inspiratory and expiratory activity. The abnormal prenatal development of RTN neurons probably contributes to the congenital central hypoventilation syndrome. Other CRCs presumably exist but the supportive evidence is less complete. The proposed locations of these CRCs are the medullary raphe, the nucleus tractus solitarius, the ventrolateral medulla, the fastigial nucleus and the hypothalamus. Several wake-promoting systems (serotonergic and catecholaminergic neurons, orexinergic neurons) are also putative CRCs. Their contribution to central respiratory chemoreception may be behavior-dependent or vary according to the state of vigilance. Introduction Central respiratory chemoreception is the mechanism by which an increase in brain PCO 2 stimulates breathing. The term also refers to the respiratory stimulation caused by metabolic acidosis (blood acidification at normal levels of CO 2). Under normal circumstances (absence of metabolic acidosis), central respiratory chemoreception operates as a sensitive feedback that helps to maintain arterial PCO 2 within a few mmHg of the steady-state (∼40 mmHg) regardless of the metabolic production of this gas and the level of vigilance (Nattie, 1999; Feldman et al., 2003; Nattie and Li, 2009). Central respiratory chemoreception normally operates in concert with peripheral chemoreceptors (Smith et al., 2006). Central respiratory chemoreception has a very slow time constant (around 50s) attributed to the time needed for brain extracellular pH to equilibrate with a change in arterial PCO 2 (Ahmad and Loeschcke, 1982; Eldridge et al., 1984; Smith et al., 2006). Central respiratory chemoreception also has a very high gain. For example, in a conscious goat, a rise in brain PCO 2 of approximately 2 mmHg (0.5% change from normal values) increases resting ventilation by around 50% (Pappenheimer et al., 1965) and presumably causes a reduction of no more than 0.01 pH unit in the vicinity of the central chemoreceptors (Nattie, 1999). In man at rest, ventilation approximately doubles for a 1.5 mmHg rise in alveolar (presumed arterial) PCO 2 (Haldane and Priestley, 1905). Central respiratory chemoreception also refers to the effects produced by abnormally high levels of CO 2 to which mammals and man are exposed only by accident (airway blockade of some sort, including sleep apnea in man) or because of intentional administration of high levels of CO 2 as is commonly done in experiments designed to study the central respiratory chemoreflex. Under such conditions, arterial PCO 2 may rise by tens of mmHg and, in intact unanesthetized mammals, this rise typically produces arousal and some form of interoceptive awareness in addition to respiratory stimulation (Phillipson et al., 1977; Berthon-Jones and Sullivan, 1984; Moosavi et al., 2003). These behavioral effects and or sensations indicate that high levels of CO 2 recruit neural pathways that are not normally influenced by the small variations of PCO 2 that regulate breathing under physiological conditions. This fact should be taken into consideration when interpreting breathing data from animals that have been exposed to high levels of CO 2. The phenomenon is not unique to the central chemoreflex. Incremental levels of stimulation of the peripheral chemoreceptors also produce a hierarchy of responses that range from simple cardiorespiratory adjustments to arousal and, finally, to behaviors denoting obvious discomfort (defense reaction, vocalizations, escape behavior) (Marshall, 1994). At this time, the dominant theory of central respiratory chemoreception is that CO 2 works via the proxy of pH, breathing stimulation derives from the simultaneous recruitment of numerous types of acid-sensitive CNS neurons (the central respiratory chemoreceptors, CRCs) and CRCs detect pH via a cell-specific combination of several acid-sensitive channels (Jiang et al., 2005; Chernov et al., 2008; Nattie and Li, 2009). As this review will indicate, this apparently straightforward summary masks an inordinate number of uncertainties. 1. Theories of central respiratory chemoreception 1.1. What is a central respiratory chemoreceptor, CRC? Central respiratory chemoreception is a reflex initiated by sensors located within the CNS. Like all reflexes, central respiratory chemoreception has three defining aspects: molecular (the receptors), cellular (the cells that express the receptors, a.k.a. the respiratory chemoreceptors) and integrative (the brain circuit engaged by the respiratory chemoreceptors). The first of many problems that this field of research faces is that the molecules that are presumably being sensed are protons. Protons, unlike most other intercellular signaling molecules (odorant molecules, hormones, transmitters, NO etc.) have the potential to modify the activity of countless regulatory proteins that are expressed not only by neurons but by glial cells and by blood vessels. At this time, the proton sensors that are germane to respiratory chemoreception are unidentified. Even the type of cell that expresses the pertinent proton sensors is not settled definitively (Figure 1). Many brain neurons are intrinsically sensitive to acid in vitro (Putnam et al., 2004; Nichols et al., 2008; Corcoran et al., 2009) but there is no hard evidence that this property persists in vivo or that the intrinsic effect of acid on neurons in vivo is sufficiently powerful and sensitive to account for respiratory chemoreception. In fact, serious consideration is still given to the possibility that CRCs are glial cells or possibly vascular cells that regulate the activity of surrounding neurons via paracrine mechanisms (Gourine et al., 2005). Given these uncertainties, it is premature to define a unique set of ideal criteria that a CNS cell must possess in order to qualify as a CRC. Figure 1 represents a few possibilities and an attempt will be made in this review to fit the various proposed chemoreceptors into one of these categories. Option 1 depicts the simplest concept, called here specialized CRCs, (Loeschcke, 1982). The specialized CRCs are viewed as separate from the respiratory rhythm and pattern network (the respiratory controller in Figure 1 which includes the motor neurons). These CRCs would selectively regulate the activity of the breathing network according to the level of acidity of the parenchyma that surrounds them. These specialized neurons could be intrinsically sensitive to acid (option 1a) and or could respond to the acidification of satellite non-neuronal cells (glia, vascular cells) via a paracrine mechanism (option 1b; factor X being one or several mediators). In such a case, the satellite cells would also qualify as CRCs. There could be one type or many types of specialized CRCs, an issue that is hotly debated at present. They need not all be excitatory neurons since some neurons are inhibited by acid in vitro and could increase breathing by disinhibition. Broad-spectrum CRCs (Figure 1, option 2) could be defined as pH-sensitive neurons that target multiple downstream networks in addition to the respiratory controller. As depicted in Figure 1 option 3, respiratory chemoreception could also reflect the widespread, possibly ubiquitous, acid-sensitivity of the neurons that make up the respiratory network or the ubiquitous location of the paracrine mechanism depicted in option 1b. In this view, all neurons are to some extent CRCs although the contribution of the various types of neurons to the overall pH-sensitivity of the respiratory motor outflow could be unequal and the degree to which each class of neurons contributes to the overall chemosensitivity of the respiratory outflow could vary according to the behavior or the state of consciousness of the organism. Option 4 (regulators of breathing) illustrates a scenario that is useful to consider when interpreting the literature on central respiratory chemoreception. The two hypothetical neurons that are represented regulate the activity of the breathing network but are unresponsive to acidification of the surrounding brain parenchyma. The activation of these neurons would be expected to influence the activity of the respiratory controller and its response to CO 2. The destruction of such neurons would also be expected to modify the overall reflex. However, because these hypothetical neurons do not respond to changes in the surrounding pH, neither of them qualifies as a CRC. Figure 1. what is a central respiratory chemoreceptor (CRC)? Open in a new tab The concept of central respiratory chemoreceptors is still being refined as more pertinent experimental evidence accumulates. Schemes 1-3 represent various CRC concepts. Scheme 4 represents neurons that should not be considered as CRCs. Specialized CRCs (option 1) could be defined as neurons that detect the concentration of protons in the surrounding brain parenchyma and drive the respiratory rhythm and pattern generating network selectively and in a graded manner according to the level of acidity. Option 1a assumes that these specialized neurons express the proton receptors. Option 1b assumes that the proton receptors reside both on these specialized neurons and on satellite cells (e.g. glia) that communicate with the neurons via a messenger X (possibly ATP). In this second case, the satellite cells would also qualify as CRCs. Broad-spectrum CRCs (option 2) could be defined as proton-sensitive neurons that modulate the activity of numerous targets besides the respiratory controller. Subsets of serotonergic neurons, the locus coeruleus and orexin neurons may be in this category. Ubiquitous chemoreception (option 3) refers to the possibility that central respiratory chemoreception could be an emergent property caused by the summation of small effects of pH almost everywhere in the respiratory network. Option 4 illustrates hypothetical examples of neurons that regulate the respiratory controller and its reactivity to protons. These neurons do not qualify as CRCs because they are not directly affected by the extracellular proton concentration. 1.2. CO 2 stimulates respiration via the proxy of pH No fundamentally new supportive evidence for this highly plausible yet unproven hypothesis has accrued in recent years (for prior discussions of this issue see (Loeschcke, 1982; Nattie, 1999; Feldman et al., 2003)). The hypothesis rests on the following evidence. Hypercapnia reduces the pH of brain extracellular fluid (ECF) with possible regional variations (Eldridge et al., 1984; Arita et al., 1989; Nattie, 2007). Metabolic acidosis also activates breathing and topical brain acidification, in particular at the level of the ventrolateral medulla, increases breathing (Loeschcke, 1982; Nattie, 1999; Feldman et al., 2003). Finally, in vitro, metabolic acidosis generally activates / inhibits neurons and depolarizes certain glial cells to an extent roughly comparable to a rise in CO 2 that produces the same change in extracellular pH (Wang et al., 2002; Putnam et al., 2004; Kawai et al., 2006). It is generally assumed that the proton-sensitive molecules responsible for respiratory chemosensitivity are ion channels and that the relevant channels are expressed by neurons. The reason is quite simply that many brainstem neurons express such pH-sensitive channels and respond to acidification in vitro under conditions of complete or semi synaptic isolation (Dean et al., 1990; Scheid et al., 2001; Mulkey et al., 2004; Corcoran et al., 2009). The interpretation of this evidence is subject to some degree of circular reasoning since putative CRCs are identified by their acid-sensitivity in vitro and the fact that neurons respond to acid in vitro is taken as evidence that central respiratory chemosensitivity operates via the effect of acid on these neurons. There are alternative theories regarding how CO 2 works. One is that the proton sensors are indeed channels but that these channels are expressed by glial cells which influence neurons by releasing gliotransmitters such as ATP (Gourine et al., 2005; Spyer and Gourine, 2009) (option 1b in Figure 1). Other proposed mechanisms include pH-sensing G-protein coupled receptors (GPCRs, (Ludwig et al., 2003)), and receptors to inorganic carbon (inorganic carbon is defined as CO 2 or bicarbonate) such as adenylate cyclases (Chen et al., 2000; Townsend et al., 2009). In Drosophila, CO 2 detection operates within a range of CO 2 concentration not much different from that present in the arterial blood of mammals and this sensory process requires the presence of two seven-transmembrane gustatory receptors (Gr21a and Gr63a) (Jones et al., 2007). Both cognate genes are required to confer CO 2 response to olfactory neurons from which the original odorant receptors have been removed (Kwon et al., 2007). This evidence suggests that Gr21a and Gr63a form a heterodimeric receptor for the detection of CO 2. Whether this dimer is a receptor for CO 2, protons or bicarbonate is not known. No mammalian homolog of these insect receptors has been identified yet but proton-sensing GPCRs are expressed by the central nervous system of mammals (Huang et al., 2007) and could conceivably serve an equivalent function (Ludwig et al., 2003). In vitro studies designed to determine whether internal or external acidification plays the dominant role in mediating the effects of CO 2 or metabolic acidosis have produced mixed results. Raphe neurons grown in culture on a bed of glia are activated to the same degree by hypercapnic acidosis (increase in CO 2 at constant bicarbonate level), isocapnic acidosis (reduced bicarbonate at constant CO 2), isohydric hypercapnia (increased CO 2 with lowered bicarbonate to maintain extracellular pH, pHo, constant) and by changes in pH produced in a CO 2-bicarbonate-free HEPES buffer (Wang et al., 2002). Thus, the chemosensitivity of raphe serotonergic neurons in culture can occur independently of changes in pHo, PCO2 or bicarbonate. Because intracellular acidification is a common denominator of all these manipulations, pHi would seem to be the primary stimulus for chemosensitivity in serotonergic neurons in culture. Similar findings and interpretations have been reported for retrotrapezoid neurons in slices (Ritucci et al., 2005b). However, the primacy of the change in pHi for neuronal responses to acidification is not clear because the latter authors also found that the degree to which RTN neurons were activated did not correlate with the magnitude of the global intracellular acidification (Ritucci et al., 2005b). Furthermore, in other cells such as the locus coeruleus, hypercapnic acidosis is also able to produce neuronal activation when pHi is clamped (Hartzler et al., 2008). This effect may rely on the closure of a TEA-resistant potassium conductance, presumably due to TASK-1 channels (Sirois et al., 2000; Filosa and Putnam, 2003). From these studies it appears that a change in pHi can be sufficient but is not absolutely required for cells to respond to CO 2. In isolated rabbit glomus cells, an increase in PCO 2 activates a Ca channel current by a mechanism that seems independent of the proton concentration inside or outside but appears to rely on the production of cAMP and protein kinase A activation (Summers et al., 2002). One interpretation is that some form of adenylyl cyclase functions as a sensor for inorganic carbon (CO 2 or bicarbonate) (Chen et al., 2000; Townsend et al., 2009). Bicarbonate is very unlikely to mediate the effects of CO 2 in brain because CO 2-activated neurons are inhibited by increasing extracellular bicarbonate at constant CO 2 level, a procedure that should also increase the intracellular concentration of this anion (Mulkey et al., 2004; Bouyer et al., 2004; Kawai et al., 2006). However, the possibility that molecular CO 2 could contribute to the response to hypercapnia has not been eliminated. Evidence that chemosensitivity could involve several proton receptors on a given neuron comes from studies in which the permeability changes elicited by metabolic acidosis (constant CO 2, reduced bicarbonate) vs. hypercapnic acidosis (constant bicarbonate, elevated CO 2) have been compared in the same neurons. In the neonate retrotrapezoid nucleus for example, both stimuli depolarize the cells equally after action potential blockade with tetrodotoxin but the depolarization caused by hypercapnic acidosis is cadmium-insensitive, blocked by high barium (1mM) and associated with significant increase in membrane resistance suggesting that a decrease in potassium conductance is dominant (Kawai et al., 2006). In contrast, the depolarization produced by metabolic acidosis is less barium sensitive and is associated with a smaller decrease in input resistance suggesting that it could be mediated by a mix of potassium and other channels (Kawai et al., 2006). A concern with all these studies is that they typically rely on small numbers of neurons that are often only assumed to contribute to respiratory chemoreception in vivo. The range of CO 2 or pH to which the cells are exposed is large (up to 15 % CO 2; 0.6 pH units) compared to normal physiological variations (<0.1 pH unit). Neurons are typically studied in the presence of astrocytes that are also pH-responsive (Ritucci et al., 2005b) and may contribute to central respiratory chemosensitivity (Gourine et al., 2005; Spyer and Gourine, 2009). Finally, different recording conditions have been used (age, temperature, presence or absence of glia) and, more often than not, heterogeneous populations of neurons were recorded within a single study. The relevant sites of intracellular acidification are unknown. They could conceivably be membrane-associated microdomains whose pH changes are not adequately reflected by whole cell pH measurements. This possibility could account for the above mentioned discrepancies between apparent intracellular acidification and neuronal responses and for the slight differences between the effects of acidification by CO 2 vs. fixed acids in vitro due to a different mode of penetration into cells. In any event, given all these uncertainties and the possibility that different cells have different mechanisms of chemosensitivity, channels whose proton sensitivity resides on either side of the neuronal membrane could potentially contribute to central chemosensitivity. 1.3. How many respiratory chemoreceptors are there? Most investigators in the field are convinced that CO 2 stimulates breathing via simultaneous effects of pH on multiple types of acid-sensitive neurons (Chernov et al., 2008; Nattie and Li, 2009). In our estimation, most putative CRCs are not characterized well enough to pass judgment on this issue. The notion that there are many CRCs is supported by the following evidence. Lesions of several regions of the pontomedullary respiratory network attenuate the chemoreflex (e.g. (St John, 1972; Berger and Cooney, 1982)). This approach is largely of historical interest because it does not discriminate between neurons that are intrinsically pH-sensitive and those that are activated synaptically during exposure to hypercapnia. At present, the main evidence supporting a wide anatomical distribution of respiratory chemoreceptors is that topical acidification of many brainstem regions increases breathing (Nattie, 1999; Hodges et al., 2004a). Historically, topical acidification was first used at the ventral medullary surface because of its accessibility (Mitchell et al., 1963b; Mitchell, 2004). Strong breathing stimulation was caused by the application of acid-soaked material to two large regions of the ventral medullary surface, one located roughly under the facial motor nucleus and the other towards the caudal medulla (Mitchell et al., 1963b; Loeschcke, 1982; Mitchell, 2004). These responses were interpreted as evidence that CRCs were very superficially located. More recently, acidification of many brainstem or cerebellar regions with dialysis probes (NTS, RTN, ventral respiratory column, midline medulla, fastigial nucleus) was found to activate breathing to some degree and simultaneous acidification of two regions usually produced additive effects (e.g.(Solomon et al., 2000; Hodges et al., 2004a; Nattie and Li, 2008)). Of note, respiration is activated by acidifying the ventrolateral medulla anywhere from its rostral end, the RTN (Li and Nattie, 2002), to the caudal region presumed to overlay the “Loeschke area” (da Silva et al., 2010) and in-between for example within the pre-Bötzinger complex (Solomon et al., 2000). The degree of respiratory stimulation caused by acidifying any given site is typically not very large in conscious animals (∼20% increase in minute-volume) and does not vary much between regions. The small magnitude of the responses may be due to the fact that direct activation of a subset of CRCs exposes the other CRCs to a more alkaline environment due to the resulting hyperventilation and deactivates them (Dias et al., 2008). The interpretation of the evidence rests on the sensible but unverified assumption that focal tissue acidification, for example using dialysis probes, correctly reproduces the effects of respiratory or metabolic acidosis on the surrounding neurons. It has been demonstrated that the tissue surrounding a dialysis probe is indeed acidified to an appropriate degree. However, evidence that neurons located close to the dialysis probe respond identically to focal acidification and to a commensurate rise in arterial PCO 2 has not been produced nor has it been shown that the neurons that are unresponsive to hypercapnia are also unaffected by local acidification. The second argument supporting a wide distribution of central respiratory chemoreceptors is that hypercapnia increases the level of expression of the protooncogene product c-Fos in many brain regions including regions located close to the ventral medullary surface (e.g.(Sato et al., 1992)). Although, putative respiratory chemoreceptors such as retrotrapezoid neurons are among the cells that express c-Fos in animals exposed to hypercapnia (Sato et al., 1992; Fortuna et al., 2009), the c-Fos method does not distinguish between neurons that are directly pH-sensitive in vivo, neurons that are activated synaptically during exposure to hypercapnia and neurons that are directly pH-activated but play no role in breathing. The third argument supporting a wide distribution of central respiratory chemoreceptors is that, by all measures (electrical recordings or voltage-sensitive dyes), every region of the brainstem contains a substantial proportion of neurons that respond to acid in vitro, often by a depolarization but sometimes with the opposite response (Scheid et al., 2001; Ritucci et al., 2005b; Nichols et al., 2008; Erlichman et al., 2009). This characteristic is not unique to the brainstem, however. Other brain regions, including cortex and cerebellum, contain neurons that respond to acid and CO 2 either by an excitation or an inhibition, the proportion depending on the recording conditions and the magnitude of the pH change (Wang and Richerson, 2000). In the neonate brainstem preparation, the vast majority of respiratory neurons located in the ventrolateral medulla were found to respond to acidification either by a depolarization or by a hyperpolarization, even after action potential blockade with TTX (Kawai et al., 1996). The CO 2-sensitivity of many respiratory neurons located in the ventral respiratory column of adult cats is attenuated by point source delivery (iontophoresis) of the purinergic receptor antagonist suramin (Thomas and Spyer, 2000). Because iontophoretic delivery of a drug can only exert spatially restricted effects, this evidence has been interpreted to suggest that pH sensitivity is local, is due to the release of ATP and is widespread in the respiratory network (Thomas and Spyer, 2000). The dubious selectivity of suramin weakens the credibility of this evidence. Finally, experiments done on cultures of brainstem neurons indicate that synaptically connected neurons are generally more strongly activated by acidification than isolated ones, suggesting that networks amplify the small effects of acid observed in a high proportion of isolated neurons (Su et al., 2007). The last pieces of evidence discussed here can be interpreted as suggesting that central respiratory chemoreception is virtually ubiquitous (scheme 3 of Figure 1). Other evidence contradicts the notion that respiratory neurons cooperate widely and on an equal footing to the stimulation of the respiratory network by acid. This evidence is not as extensive as that which supports the opposite view but it is highly significant. In the neonate brainstem-spinal cord preparation, a.k.a. Suzue preparation (Suzue, 1984), selective genetic deletion of the retrotrapezoid nucleus eliminates the activation of the phrenic nerve outflow by CO 2 but does not eliminate the ability of the brainstem to generate a regular inspiratory activity (Dubreuil et al., 2008; 2009b). Thus, the apparently ubiquitous pH sensitivity of the respiratory neurons located in the ventrolateral medulla (Kawai et al., 1996) does not produce a pH-sensitive network unless the retrotrapezoid nucleus is present. These results were obtained using late embryonic or early postnatal tissue and their interpretation may not apply later on in life. In anesthetized adult rats, the integrity of the retrotrapezoid nucleus is necessary for the respiratory network to be activated by CO 2 (Takakura et al., 2008). The respiratory pathology observed in the congenital central hypoventilation syndrome (CCHS) also supports the notion that central respiratory chemosensitivity may rely on relatively few specialized neurons. In this genetic disease, central chemosensitivity is essentially absent, severe hypoventilation occurs during sleep but relatively adequate breathing persists during waking and some degree of exercise-induced hyperventilation remains, at least in the milder cases (Paton et al., 1989; 1993; Shea et al., 1993; Gozal, 1998; Amiel et al., 2003; 2009; Carroll et al., 2010). The severity of the symptoms increases according to the number of extra alanine residues in the mutated polyalanine track of Phox2b (Carroll et al., 2010). The respiratory symptoms of the disease suggest that the central respiratory controller of CCHS patients is largely intact but that the brain of these individuals lacks neurons that are either uniquely specialized in detecting CO 2 or are specialized in funneling the excitatory influence of CO 2 to the respiratory controller. The mouse model of the disease (Phox2b 27ala/+) suggests that these missing neurons could be the ccRTN neurons (Dubreuil et al., 2008), a chemically defined subset of retrotrapezoid nucleus neurons whose properties will be extensively discussed later on. In short, current evidence is insufficient to assert whether central respiratory chemoreception relies on just a few cells or is a property widely distributed throughout the brain. Recent animal data, reinforced by the natural experiment provided by the congenital central hypoventilation syndrome, suggest that the high gain central chemosensitivity responsible for CO 2 stability under physiological conditions (pH changes of considerably less than 0.1 unit) could rely on relatively few cells, the retrotrapezoid nucleus being most critical (Guyenet, 2008; Goridis and Brunet, 2010), whereas a much larger number of cells could be implicated in the effects produced by high levels of CO 2 (behavioral arousal, dyspnea or panic attacks). However, the high gain of the central chemoreflex, especially in the adult, could also be due to a network summation of individually small effects of pH on vast numbers of neurons, a concept that is generally dominant at this time (see for example (Nattie and Li, 2009)). 2. Molecular basis of central respiratory chemoreception Acid-sensitive ion channels located on neurons or satellite cells are currently believed to be the proton sensors responsible for respiratory chemosensitivity (e.g.(Chernov et al., 2008)). However, to interpret the literature optimally, one should keep in mind that ion channels could theoretically contribute to the pH response of central chemoreceptor neurons in three distinct ways. The proton receptors could be ion channels as is currently hypothesized. A subset of pH-insensitive ion channels could be intracellular effectors of a membrane-associated proton receptor, e.g. a GPCR. Finally, the vast majority of the ion channels expressed by chemoreceptor neurons are likely to be pH-insensitive channels that shape the intrinsic properties of these cells. The presence of these channels will influence the response of the chemoreceptor neurons to acid and by way of consequence the chemoreflex. A few of these channels may be somewhat selectively expressed by central chemoreceptor neurons (TASK-2 may be an example) but most are probably not. The channels most frequently mentioned as proton receptor candidates are considered in this section of the review. Although the hypothesis “makes sense” no channel has been shown convincingly to contribute to respiratory chemosensitivity by virtue of its proton binding ability and alternative hypotheses should still be pursued at this time. 2.1. Acid-sensitive channels Potassium channels: 2 pore channels and others Two-pore potassium channels (K2P channels) regulate the excitability of neurons by setting their resting membrane potential and changing their input resistance (Lesage and Lazdunski, 2000; Duprat et al., 2007; Bayliss and Barrett, 2008). The family includes around 15 subunits that have been grouped into six structurally and functionally different subclasses (Holzer, 2009). Many of these channels are activated or inhibited, in vitro at least, by relatively small deviations of extra- or intracellular pH from physiological levels. The TASK subfamily of K2P channels (TASK1, KCNK3; TASK3, KCNK9) form homo- or heterodimers and are widely expressed in brain (Lesage and Lazdunski, 2000; Talley et al., 2001; Duprat et al., 2007). They are constitutively active, voltage-insensitive, non-inactivating and inhibited by the actions of many GPCRs (serotonin type 2 receptors, TRH, alpha-1 adrenergic etc.) (Talley and Bayliss, 2002). TASK1/3 homo- or hetero-dimer current is reduced by acidification with slight variations in pH 50 according to the sub-unit composition. TASK1-/-, TASK3-/- and double knock-out mice have normal ventilatory responses to hyperoxic hypercapnia (Mulkey et al., 2007b). This result suggests that neither isoform of the channel makes a critical contribution to central respiratory chemosensitivity. The fact is surprising given the very wide distribution of these channels in the ventrolateral medulla oblongata and elsewhere in the brainstem, including in motor neurons and the prediction, based on modeling, that TASK channels should make the greatest single contribution to neuronal pH sensitivity (Washburn et al., 2003; Chernov et al., 2008). Yet, in agreement with the whole animal observation, the pH-induced potassium current recorded in retrotrapezoid neurons is the same in TASK1-/-, TASK3-/- and double knock-out mice as in control mice (Mulkey et al., 2007b). It seems unlikely that the absence of TASK1/3 channels since conception could be fully compensated by the up-regulation of other acid-sensitive potassium channels in RTN with exactly the same properties (Mulkey et al., 2007b). TASK1 deletion in mice reduces the sensitivity of the carotid body to both oxygen and CO 2 (Trapp et al., 2008). This observation does not demonstrate that TASK1 is the pH sensor of the carotid body but it highlights an important molecular difference between central and peripheral pH-detection. Despite their name, TASK2 (KCNK5) channels are only distantly related to TASK 1 or 3 from an evolutionary viewpoint. They belong to the TALK subgroup of alkaline-activated K-2P channels because their pKa is 8.0 or above (Morton et al., 2005; Niemeyer et al., 2007; Bayliss and Barrett, 2008). TASK-2 mRNA levels are very low in brain (Reyes et al., 1998; Talley et al., 2001; Aller and Wisden, 2008) and, according to recent genetic evidence, the channel is expressed in only a few neuronal clusters including the retrotrapezoid neurons (Dubreuil et al., 2009a; Gestreau et al., 2010). Oddly, TASK-2-like immunoreactivity has been reported as quite extensive in brain (Gabriel et al., 2002). The selectivity of the antibody used in these studies should be verified in a TASK-2 knock-out mouse. TASK-2 channel deletion does not change the pH sensitivity of the neonate breathing network in vitro, contrary to the deletion of the RTN neurons (Dubreuil et al., 2009b; Gestreau et al., 2010) therefore TASK-2 does not seem responsible for the pH sensitivity of RTN neurons under these conditions. This result is consistent with the fact that the pKa of TASK-2 channels is far outside the physiological range on the alkaline side. Because of this characteristic, TASK-2 channel conductance is unlikely to be modulated by acidification below the normal brain pH of 7.3. The Gestreau study suggests that TASK-2 activation by hypoxia contributes to post-hypoxic respiratory depression. Given that its expression is potentially down-regulated by chronic hypoxia (Brazier et al., 2005), TASK-2 may serve to increase the excitability of RTN neurons under chronic hypoxic conditions where lowering of the CO 2 set-point might be advantageous to increase oxygen intake. TASK-2 is abundantly expressed in the kidney where it plays a major role in acid-base homeostasis (Morton et al., 2005; Bayliss and Barrett, 2008). Due to their inability to retain bicarbonate, TASK-2 KO mice are in metabolic acidosis (Warth et al., 2004). This peculiarity may account for the left shift of their breathing response to CO 2 and the reduced effect of high levels of hypercapnia (Gestreau et al., 2010). Inwardly rectifying potassium channels (Kir) In heterologous expression systems, inwardly rectifying potassium channels (e.g. Kir1.1, Kir4.1–Kir5.1) can generate acid-sensitive currents with a pKa appropriately close to physiological pH (7.4) (Xu et al., 2000; Su et al., 2007). Kir4.1 is a primary molecular substrate of astroglial K+ transport (Neusch et al., 2006). This channel underlies most of the inwardly rectifying current of astrocytes and is essential to maintain their very hyperpolarized membrane potential (Neusch et al., 2006). Inwardly rectifying potassium channels are closed by intracellular acidification and appear to contribute to the acid sensitivity of some brainstem neurons in culture because acid sensitivity is suppressed by a concentration of barium that blocks Kir channels with some selectivity in heterologous expression systems (Su et al., 2007). The activation of mature locus coeruleus neurons by hypercapnic acidosis in slices has also been attributed to the closure of an inwardly rectifying potassium current although, in this case, intracellular protonation of a polyamine has been invoked as the cause of the reduced current (Pineda and Aghajanian, 1997). However, in native RTN neurons, the acid-sensitive potassium current is neither inwardly rectifying nor sensitive to 40 mM tetraethylammonium which excludes the contribution of Kir (Mulkey et al., 2007b). The ASIC family of acid-sensitive channels Acid-sensing ion channel (ASIC) subunits associate as trimers to form acid-activated channels permeable to Na+ and Ca 2+ (Yermolaieva et al., 2004; Jasti et al., 2007). ASICs and transient receptor potential vanilloid-1 (TRPV1) underlie the detection of peripheral tissue acidosis by primary sensory afferents, ASICs being responsive to moderate decreases in extracellular pH whereas TRPV1 requires pH to drop below 6.0 (Holzer, 2009). ASIC1a, -2a, and −2b are the most commonly expressed subunits in the CNS, where they assemble into homo- and heterotrimeric complexes. Disrupting the ASIC1a gene (ACCN2) in mice eliminates currents evoked in vitro by pH as low as 5.0 in brain regions such as the hippocampus or the amygdala (Wemmie et al., 2002; Askwith et al., 2004). ASIC1a-/- mice have attenuated fear behavior but normal ventilatory responses to moderate levels of normoxic hypercapnia suggesting that ASIC1a is not critical for central respiratory chemosensitivity (Ziemann et al., 2009). This interpretation is also in line with evidence that acidification most often operates selectively by closing a potassium conductance in putative respiratory chemoreceptors (Dean et al., 1989; Mulkey et al., 2004; 2007b; Nichols et al., 2008). Other acid-sensitive channels Other acid-sensitive channels include members of the TRP family, ionotropic purinoceptors (P2X), voltage-activated potassium channels, L-type Ca 2+ channels, hyperpolarization-activated cyclic nucleotide gated channels, gap junction channels, and Cl- channels (Shi et al., 1998; Vacher et al., 2008; Holzer, 2009). A large fraction of these channels are expressed in the brainstem. Their contribution to respiratory chemosensitivity is typically untested. Increased calcium current may contribute to the response of locus coeruleus neurons and the carotid body to acid (Summers et al., 2002; Filosa and Putnam, 2003). 2.2. GPCRs as pH sensors Four structurally related G protein-coupled receptors (OGR1, GPR4, TDAG8, and G2A) originally believed to respond to lipid messengers have recently been promoted to proton detectors (Ludwig et al., 2003; Holzer, 2009). Their pH-sensitivity, like that of many ion channels, is attributed to a strategic subset of histidine residues. In vitro, the receptors are inactive at pH 7.8 and fully activated at pH 6.8. OGR1 family members (OGR1, GPR4, TDAG8, and G2A) are widely expressed in the nervous system (CNS and sensory afferents) and in many other organs or tissues (bone, vascular system). At this time, there is still no evidence that any member of the OGR1 receptor family is implicated in central respiratory chemosensitivity. 3. Contribution of glial cells to central respiratory chemosensitivity Central respiratory chemoreception could be a multicellular process along the lines depicted in Figure 1, option 1b. In this section, we discuss the possibility that the satellite cells might be glia. There is also anatomical and functional evidence for interaction between brain blood flow and CRCs (Xie et al., 2006; Ainslie and Duffin, 2009; Lazarenko et al., 2009). This aspect of central respiratory chemoreception will not be covered here. 3.1. pH-sensitivity of glial cells The first evidence that some brainstem glial cells are depolarized by acidification is usually traced back to the experiments of Fukuda and Honda (1975). These authors recorded intracellularly from ventral medullary surface cells located in the “intermediate chemosensitive zone” just lateral to the rootlets of the hypoglossal nerve in thin horizontal slices. Many cells located at the ventral surface i.e. less than 200 microns from the pia mater were depolarized by isocapnic acidosis (acidification at constant CO 2 via bicarbonate reduction). The recorded cells were unexcitable, with notably negative membrane potential at alkaline pH and could therefore have been glia. Isohydric hypercapnia was ineffective in these cells suggesting that CO 2 operated strictly via the proxy of pH and that the cells were selectively sensitive to changes in extracellular pH (Fukuda and Honda, 1975). This result is noteworthy because, in neurons, acid sensitivity is generally attributed to a change in intracellular pH (Wang et al., 2002; Ritucci et al., 2005b). If the Fukuda and Honda results are indeed representative of all glial cells, Kir4.1, the major barium-sensitive inwardly rectifying channel of astrocytes (Neusch et al., 2006) could not be responsible for astrocyte depolarization by acid since this channel is responsive to intracellular pH. More work is obviously required because of the presumed heterogeneity of the potassium channels expressed by astrocytes (Fukuda et al., 1978; Grass et al., 2004). 3.2. Amplification of CO 2-induced extracellular fluid pH changes by glia The idea that CO 2-dependent changes in pH could be amplified by glial cells is based on the phenomenon called depolarization-induced alkalosis (DIA), whereby an increase in extracellular potassium and or acidification causes glial cell depolarization and the extrusion of protons from these cells (Fukuda et al., 1978; Erlichman et al., 2008). This process may be limited to a subset of astrocytes and matures a couple of weeks after birth in rodents (Erlichman et al., 2008). This mechanism, which should boost the response of central chemoreceptor neurons to a given level of PCO 2, seems to be at work in the retrotrapezoid nucleus but not in the nucleus of the solitary tract (NTS) (Erlichman et al., 2008). Because it matures later in development, this phenomenon could also perhaps explain the higher acid sensitivity of RTN neurons in the adult in vivo (∼0.4 Hz per 0.01 pH unit (Guyenet et al., 2005a)) compared to the neonate in vitro at comparable temperature (∼0.17 Hz per 0.01 pH unit (Guyenet et al., 2005b)). 3.3. Acid-induced release of ATP by glial cells Large subsets of glial cells (presumably astrocytes) are depolarized by acidification, in vitro at least (Fukuda and Honda, 1975; Fukuda et al., 1978). ATP is released by glial cells and this release is required for the intercellular propagation of calcium waves in these cells (Guthrie et al., 1999). ATP is released by hypercapnia at the ventral medullary surface in vivo (Spyer et al., 2004; Gourine et al., 2005). Topical application of P2 receptor antagonists at the ventral medullary surface reduces somewhat the respiratory response to CO 2 in anesthetized rats (Gourine et al., 2005). Finally, the purinergic receptor antagonist suramin delivered by iontophoresis attenuates the activation by CO 2 of various types of respiratory neurons in vivo (Thomas and Spyer, 2000). The ATP hypothesis of central chemosensitivity faces a number of unresolved issues. As mentioned previously, suramin is not a selective purinergic antagonist. There is uncertainty concerning the cellular origin of the ATP released at the central medullary surface. For example, pia-arachnoid cells could be the source of this ATP since these cells communicate via this nucleotide very much like glial cells (Grafstein et al., 2000). More importantly, purinergic P2-receptor antagonists do not change the pH-sensitivity of RTN neurons in slices (Mulkey et al., 2004; Mulkey et al., 2006) and these drugs do not change the pH/CO 2 response of the Suzue preparation, a neonate brainstem-spinal cord preparation that generates a respiratory like outflow (Lorier et al., 2004). Finally PPADS, the most commonly used non-selective P2 receptor blocker does not attenuate the acid sensitivity of brainstem cultures (Su et al., 2007). The lack of effect of P2 antagonists in these in vitro models could mean that acidification causes glial cells to release ATP only in vivo. Possible reasons include the immaturity of the brain tissue used in vitro, especially the glia (see previous section), the use of a low recording temperature and/or the absence of blood perfusion. Two findings support the possibility that ATP release by glial cells might be occurring in vivo only. Although the activation of the ccRTN neurons by acid does not depend on ATP release in vitro, these cells do express purinergic 2Y receptors (Mulkey et al., 2006). Secondly, the pH-sensitivity of these cells is roughly 2-3 times higher in vivo, a difference that could conceivably represent the contribution of ATP release from surrounding mature glial cells. 4. Central respiratory chemoreception and level of vigilance Under normal conditions, the regulation of breathing operates within a very narrow range of PCO 2 and minimal CO 2 fluctuations have no effect on the level of consciousness or vigilance or on the depth of sleep. In contrast, severe hypercapnia produces arousal from sleep in animals and man (Phillipson et al., 1977; Berthon-Jones and Sullivan, 1984). In awake individuals, hypercapnia also causes an aversive form of interoceptive awareness called air hunger (Moosavi et al., 2003) and in susceptible individuals, CO 2 can even trigger outright panic. In addition, rodents are capable of detecting very low levels of atmospheric CO 2 and they express fear behavior when exposed to hypercapnia (Hu et al., 2007; Ziemann et al., 2009). Studies in man demonstrated long ago that the gain of the chemoreflex is greater in awake than in sleeping subjects (reviewed in (Ainslie and Duffin, 2009)). In the awake state, breathing is also activated by CO 2-independent drives collectively grouped under the name of the “waking drive to breathe” (Ainslie and Duffin, 2009). Thus, the state of vigilance influences the gain of the chemoreflex at all levels of CO 2 including around the normal operating range of 40 mmHg but the normal homeostatic regulation of breathing by CO 2 is evidently not caused by continual fluctuations of the state of vigilance (Kuwaki, 2008). By contrast, general CNS arousal and its associated somatic and autonomic responses presumably make a significant contribution to the breathing stimulation caused by high levels of CO 2 under experimental or pathological conditions (airway obstruction or exposure to abnormally elevated levels of CO 2) (Berthon-Jones and Sullivan, 1984). The chemoreflex studied in awake experimental animals is always elicited by high levels of CO 2 (5% minimum), therefore this reflex presumably depends on general arousal mechanisms engaged by this stimulus, fear responses etc., in addition to the chemosensory mechanisms that intervene around physiological levels of CO 2. 5. Putative respiratory chemoreceptor neurons This section focuses on the properties of neurons that are currently believed to make a particularly important contribution to central respiratory chemoreception. 5.1. The retrotrapezoid nucleus (RTN) The discovery of the RTN, the earliest evidence that this brain region controls breathing and may contain chemoreceptors, and the potential relationship between the RTN and the parafacial respiratory group of the neonate are well documented and outside the scope of this review (Smith et al., 1989; Nattie et al., 2001; Nattie, 2001; Li and Nattie, 2002; Feldman et al., 2003; Onimaru and Homma, 2003; Guyenet, 2008; Guyenet and Mulkey, 2010). The rest of this section focuses on a group of 2000 chemically defined neurons (in rats; ∼ 800 in mice) located within the region originally defined as the RTN (Connelly et al., 1989). These cells have a distinctive biochemical phenotype that includes the presence of Phox2b, neurokinin-1 receptors, vesicular glutamate transporter2 mRNA and TASK-2, and the lack of tyrosine hydroxylase, choline acetyl transferase, glutamic acid decarboxylase and tryptophan-hydroxylase (Mulkey et al., 2004; Stornetta et al., 2006; Takakura et al., 2008; Gestreau et al., 2010; Guyenet and Mulkey, 2010). We refer to these specific neurons as the ccRTN (chemically coded RTN) neurons (Lazarenko et al., 2009) to distinguish them from other types of neurons that may also reside in this heterogeneous region of the reticular formation. ccRTN neurons are a subset of a heterogeneous collection of neurons called the parafacial respiratory group in the neonate (Onimaru et al., 2008; Guyenet and Mulkey, 2010). The developmental lineage of the ccRTN neurons has also been worked out recently in considerable detail (Thoby-Brisson et al., 2009; Dubreuil et al., 2009b). The key evidence that the ccRTN neurons are CRCs is illustrated in Figure 2. First, these neurons respond vigorously and sensitively to hypercapnia in vivo (Figure 2C) (Mulkey et al., 2004; Guyenet et al., 2005a). Under anesthesia, their discharge rate varies by an average of 2.8 Hz for a 1% change in arterial PCO 2 or an estimated ∼0.5 Hz (5% of full range) per 0.01 unit change in arterial pH (Guyenet et al., 2005a). These cells are silent at low levels of CO 2 and many of them have a CO 2 recruitment threshold (CO 2 level above which the cells are active) that is below that of the phrenic nerve and below that of the respiratory neurons of the ventral respiratory column, regardless of the anesthetic used (Figure 2C2) (Guyenet et al., 2005a; Fortuna et al., 2009). Such a characteristic is expected from neurons that drive the respiratory controller, at least the portion of this network that generates the motor outflow to the diaphragm. Figure 2. the retrotrapezoid nucleus (RTN). Open in a new tab A. The RTN contains a neurochemically well-defined group of glutamatergic neurons, the ccRTN neurons (Takakura et al., 2008; Lazarenko et al., 2009) so named to distinguish them from the overlapping but phenotypically heterogeneous collection of cells called the parafacial respiratory group (Guyenet and Mulkey, 2010). The axonal projections of the ccRTN neurons in the rat are represented schematically based on the anterograde transport of the membrane bound fusion protein channelrhodopsin2-mCherry (for primary data see (Abbott et al., 2009). The scale (1 mm) is an approximation based on the rostrocaudal extent of the facial motor nucleus in the adult rat. B. transverse histological section along the plane identified by the dotted line in panel A (after (Stornetta et al., 2006)). The ccRTN neurons express the transcription factor Phox2b (nuclei in green) and lack tyrosine hydroxylase (TH, in red). The TH-ir neurons located at this level are the C1 neurons, which regulate the heart and vasomotor tone. The location of the facial motor neurons (7) is revealed by the presence of fluorogold (in blue). The scale bar is 100 microns. The ventral medullary surface is recognizable by a green edge artifact. C. CO 2-sensitivity of the ccRTN neurons in an anesthetized rat in vivo. C1: experimental design for unit recording in anesthetized rats. The ccRTN neurons are located under the facial motor nucleus the lower boundary of which is identified by antidromic field potentials elicited by stimulation of the facial nerve. C2: single ccRTN neuron recorded alongside end-expiratory CO 2 (top trace), arterial blood pressure (AP) and the mass activity of the phrenic nerve (iPND; a reporter of the level of activity of the respiratory controller). ccRTN neurons are robustly activated by adding CO 2 to the breathing mixture. The expanded time-scale insert shows that the neuron has only a mild propensity to discharge in synchrony with the central respiratory controller. C3 is an example of one such neuron labeled in vivo with biotinamide following its physiological characterization. Note the profusion of dendrites at the ventral medullary surface (VMS) (after (Mulkey et al., 2004). D: selective stimulation of the ccRTN neurons activates breathing in vivo. D1, experimental design. A thin optical fiber is introduced in the vicinity of a group of ccRTN neurons that were selectively transfected with the channelrhodopsin2-mCherry (ChR2) fusion protein. ccRTN neurons are recorded while blue laser light is applied to activate ChR2-expressing neurons. The rat is anesthetized with isoflurane. D2, transverse section showing several superficial Phox2b-expressing ccRTN neurons that were transfected with ChR2 (m-Cherry exhibits a natural red fluorescence). The transgene was expressed almost exclusively by Phox2b-containing neurons (Phox2b immunoreactivity appears green in untransfected neurons, green arrow, and yellow in transfected cells, white arrows). The transfected neurons line the ventral medullary surface. Note that the overlying facial motor neurons (7) did not express the transgene (scale bar: 0.1 mm). D3, activation of ChR2-transfected ccRTN neurons (one of them is being simultaneously recorded) activates breathing. In this excerpt, the end-expiratory CO 2 was below the apneic threshold therefore both the RTN unit and the phrenic nerve were silent prior to shining the laser light (after (Abbott et al., 2009)). During light application, the RTN unit was entrained 1 to 1 to the laser pulses (bottom two traces). E: pH sensitivity of ccRTN neurons in brain slices. E1, ccRTN neurons in a Phox2b-eGFP transgenic mouse (the B/G mouse). In this strain, GFP identifies selectively the ccRTN neurons (scale bar: 50 microns). E2, thick section (∼300 μm) through the RTN of the B/G mouse. Two ccRTN neurons were recorded in vitro and were filled with biotinamide (scale bar: 100 microns). E3, typical response of a ccRTN neuron to acidification in vitro (B/G mouse, temperature: 22°C). The cell was silent at pH 7.5. It was activated by acidification and by substance P (SP; 100 nM), the latter due to the presence of neurokinin1 receptors on the cells. Panels E1-3 after (Lazarenko et al., 2009). F1, ccRTN neurons fit the concept of specialized chemoreceptors. The molecular and cellular mechanism by which they detect surrounding protons (1a or 1b) is uncertain however. F2, a slightly more elaborate concept of how ccRTN neurons might regulate breathing. ccRTN neurons drive the breathing rate, the amplitude of inspiration and, probably active expiration. ccRTN neurons regulate breathing according to the level of acidity that surrounds the neurons but also according to the strength of the input that they receive from the carotid bodies, the raphe and the hypothalamus, including from orexinergic neurons. Distinct subsets of ccRTN neurons may drive the various respiratory motor outflows (inspiratory vs. expiratory muscles). By definition, the activation of CRCs should stimulate breathing. The ccRTN neurons pass this essential test since their selective activation in vivo using genetically targeted channelrhodopsin2 produces a vigorous increase in breathing rate and amplitude (Figure 2D) (Abbott et al., 2009). The ccRTN neurons are therefore CO 2-sensitive, histologically glutamatergic (VGLUT2 mRNA-expressing) and functionally excitatory with respect to the respiratory network. Also, the excitatory drive contributed by the ccRTN neurons is notably powerful because a large respiratory response can be produced by activating at most 15% of the cell population (Abbott et al., 2009). Whether or not the ccRTN neurons are part of the respiratory rhythm and pattern generator is important from a conceptual standpoint. Three pieces of evidence support our contention that these neurons are not part of the rhythm and pattern generating network but neurons that simply regulate the level of activity of this circuitry according to the level of CO 2/pH as depicted in Figure 2F1,2. First, the ccRTN neurons have a modest and phase-spanning respiratory entrainment at all but extreme levels of CO 2 in vivo (Figure 2C2, insert) (Guyenet et al., 2005a). Second, when a subset of ccRTN neurons are activated at a constant rate of around 20 Hz using the optogenetic approach, the resulting effect on respiration is a normal rhythmic activation of the respiratory controller similar to what is observed during central chemoreceptor stimulation (Figure 2 D3) (Abbott et al., 2009). Finally, the time constant of the excitatory effects of ccRTN neurons on the rodent respiratory network is long (T 1/2 ∼11 s) relative to the breathing period (1 s or less) (Abbott et al., 2009). In the aggregate, this evidence suggests that the exact pattern of discharge of the ccRTN neurons is less important than the mean rate of discharge of these cells. This characteristic is in contrast with the precisely timed phasic discharges of the neurons that are assumed to directly contribute to rhythm and pattern generation (Rybak et al., 2007). The third essential attribute of CRCs (Figure 2F1) is pH-sensitivity which could theoretically be either intrinsic (option 1a) or due to a local autocrine process (option 1b). Proving that a neuron that responds to hypercapnia in vivo does so exclusively via such mechanisms would require deleting the proton sensor in a cell specific manner and showing that the targeted neurons no longer respond to hypercapnia in vivo. This type of evidence has not yet been produced for any candidate CRC. However, in the case of the ccRTN neurons and only in this case, it has at least been possible to show that hypercapnia in vivo remains capable of activating these neurons normally under conditions of partial synaptic blockade which silence the respiratory controller (Mulkey et al., 2004). This evidence makes the essential point that the activation of the ccRTN neurons by CO 2 is not secondary to the stimulation of the respiratory controller. In brain slices, ccRTN neurons are activated by hypercapnic acidosis and metabolic acidosis even under conditions of partial synaptic blockade (blockade of glutamate GABA, glycine and purinergic transmission) and acidification produces a reduction in potassium conductance in the presence of tetrodotoxin (Mulkey et al., 2004; 2006). This evidence partially illustrated in Figure 2E indicates that the ccRTN neurons are responsive to local changes in pH. It does not prove that their acid-sensitivity is (entirely) an intrinsic property because the participation of satellite cells (glia, blood vessels) has not been ruled out (Gourine et al., 2005; Erlichman et al., 2008). In addition, there is still an unexplained mismatch between the dynamic range of the response of ccRTN neurons to acidification in vitro vs. in vivo (∼0.16 Hz - at 33 °C in slices vs. ∼ 0.5 Hz per hundredth pH unit in vivo). This discrepancy may have to do with the limitations inherent to experimentation with slices (dendritic pruning, lower temperature). It could also simply be related to the age of the animals (6-13 day-old rodents for in vitro recordings vs. young adults for in vivo experiments). Indeed rodents have notably less powerful chemoreflexes at and soon after birth than later on (Conrad et al., 2009). Finally, in vivo, the ccRTN neurons may also integrate information from additional central chemoreceptors (Figure 2F2) and their response to acidification could be enhanced by unknown factors that modify their intrinsic properties and increase the gain of their action potential response to acid-induced current. The projection pattern of candidate CRCs is also an important consideration. Neurons that innervate selectively the respiratory controller are more likely to be specifically dedicated to central respiratory chemoreception than neurons that have projections throughout the brain, even if these projections include elements of the respiratory controller. The ccRTN neurons innervate exclusively the pontomedullary regions known to contain the respiratory controller, i.e. ventrolateral medulla, dorsal pons and ventrolateral portions of the NTS (Figure 2A) (Mulkey et al., 2004; Rosin et al., 2006; Abbott et al., 2009). Finally, as alluded to previously, the ccRTN neurons are of special relevance to CCHS, a human disease caused by polyalanine expansion of transcription factor Phox2b (Amiel et al., 2009; Carroll et al., 2010). CCHS is a disease in which the newborn child presents with extremely depressed chemoreflexes and a severe hypoventilation that would be rapidly lethal in the absence of mechanical ventilation in all but the milder forms of the disease (Amiel et al., 2009). The mouse model of the disease (Phox2b 27ala/+ strain) presents similar breathing problems at birth (Dubreuil et al., 2008; Dubreuil et al., 2009b). The only brainstem defect that has been identified so far in the Phox2b 27ala/+ mouse is a massive and selective loss of the ccRTN neurons (Dubreuil et al., 2008; 2009b). This evidence suggests that the ccRTN neurons are fundamentally important for central respiratory chemoreception during the perinatal period. Because CCHS patients with 7 extra alanines in the PHOX2B sequence experience severe if not complete sleep apnea, it is possible that the ccRTN neurons are indispensable to maintain breathing automaticity during sleep. This working hypothesis has not been tested adequately in laboratory animals. The RTN region contributes to ventilation and to the central chemoreflex in unanesthetized rats regardless of the level of vigilance (Nattie and Li, 2002a). Sleep apnea was not observed in these experiments but the RTN lesions performed in these experiments were probably very partial and compensatory mechanisms are likely to have developed. In short, the ccRTN neurons contribute a powerful excitatory drive to the respiratory controller and their firing characteristics suggest that they are not directly involved in the generation of the respiratory rhythm or the classic 3-phase respiratory pattern. Their activity is up-regulated by the level of acidity of the neuropil that surrounds them. These properties qualify the ccRTN neurons as CRCs. However, it is also clear that the level of activity of the ccRTN neurons is not solely determined by the surrounding pH (Figure 2F2). Of particular significance, ccRTN neurons are also activated by peripheral chemoreceptor stimulation (Takakura et al., 2006), they are up-regulated by inputs originating from the hypothalamus (Fortuna et al., 2009; Dias et al., 2009b), they are subject to a negative feedback from the respiratory controller (Guyenet et al., 2005a) and their activity is probably regulated by the raphe (Mulkey et al., 2007a; Dias et al., 2008). Finally, a growing amount of evidence suggests that several subsets of ccRTN CRCs may exist, which differentially regulate the breathing rate vs. the inspiratory amplitude and/or active expiration (Abdala et al., 2009; Guyenet and Mulkey, 2010). Other targets, notably cardiovascular, are also not excluded. 5.2. Serotonergic neurons and central respiratory chemosensitivity Richerson and colleagues have proposed the following three-part theory (for a recent review see (Corcoran et al., 2009)). 1: All serotonergic neurons are CO 2/pH detectors (Richerson, 2004). 2: Midbrain raphe serotonergic neurons contribute to the arousal-promoting effects of hypercapnia. 3: The lower brainstem serotonergic neurons mediate the respiratory chemoreflex or some large portion of it. The main supportive arguments, briefly summarized, are as follows (Corcoran et al., 2009). Collectively, serotonergic neurons innervate every region of the brain including the respiratory controller and they also target respiratory motor neurons. Some serotonergic neurons are acid-activated in slices. In neuronal culture, serotonergic neurons generally respond vigorously to CO 2 and acid (Figure 3B). Hypercapnia increases the overflow of serotonin in the hypoglossal nucleus in vivo and experimental acidification of the midline raphe can increase breathing (Nattie and Li, 2001; Hodges et al., 2004b; 2009; Corcoran et al., 2009). Serotonergic neurons reside in close proximity to blood vessels which can be viewed as potentially advantageous for the detection of arterial CO 2 (Figure 3A). Finally, genetic deletion of serotonergic neurons markedly attenuates the chemoreflex (Hodges et al., 2008). Figure 3. serotonergic neurons and central respiratory chemosensitivity. Open in a new tab A, ventral view of the rat brainstem showing the location of the serotonergic neurons that reside close to the ventral medullary surface (tryptophan-hydroxylase, TpOHase, immunoreactivity in green; calibration bar: 1.5 mm). The blood vessels are red because they have been filled with a resin. The superficial serotonergic neurons are close to blood vessels and reside in regions of the ventral medullary surface assumed to contain CRCs because experimental acidification of these regions increases breathing. We added the white ovals that outline the retrotrapezoid nucleus. B, typical serotonin neuron in culture that responds briskly to extracellular acidification. A and B reprinted with slight modification from Respiration Physiology and Neurobiology, Volume 168, Corcoran et al., Medullary serotonin neurons and central CO 2 chemoreception, pp. 49-58, Copyright (2009), with permission from Elsevier. C. the lateral B3 group of serotonergic neurons is insensitive to hypercapnia in anesthetized rats. C1, example of a serotonergic neuron (tryptophan-hydroxylase-immunoreactive) filled with biotinamide after electrophysiological study in vivo (calibration bar: 20 microns). C2, location of 24 biochemically identified serotonergic neurons found unresponsive to hypercapnia in vivo (red dots). These neurons were located in the lateral band of serotonergic neurons shown in panel A at the level of the RTN (pyr, pyramidal tract; RPa, raphe pallidus; TpOHase, tryptophan hydroxylase). C3 example of a single identified lateral B3 group serotonergic neuron. The neuron was unaffected by increasing end-expiratory CO 2 by 5% from just below the apneic threshold. The activity of the respiratory controller, monitored at the level of the phrenic nerve (iPND) was robustly activated by CO 2 as expected. C1-3 from (Mulkey et al., 2004). D, raphe obscurus serotonergic neurons driven by the respiratory controller. These recordings were obtained in a coronal slice of neonate medulla oblongata that generates a respiratory-like activity (the “breathing slice”) (Smith et al., 1991). In this slice, the activity of the residual respiratory network was monitored by the mass discharge of the hypoglossal motor neurons (integral XII). In the lower two traces the membrane potential of the serotonergic cell was deliberately hyperpolarized to reveal the excitatory drive potential synchronized with the inspiratory phase. Reproduced with permission from the Journal of Neuroscience from (Ptak et al., 2009). E, putative raphe obscurus serotonergic neurons recorded in a conscious cat. Most cells (21/27) did not respond to hypercapnia. Small subsets of putative serotonergic neurons, one of which is illustrated here, were activated by CO 2 but only while the cats were awake. Reproduced with permission from the Journal of Neuroscience from (Veasey et al., 1995). F, the release of serotonin contributes to the activity of the respiratory controller in vitro. The figure shows the mass activity of the preBötzinger region of the ventral respiratory column in the “breathing slice” (ipreBot, an indication of the inspiratory phase of the breathing cycle; top trace) and an inspiratory neuron that was recorded intracellularly (lower trace). Superfusion of the slice with a serotonin receptor 2A antagonist slowed the fictive breathing rate and the amplitude of the respiratory bursts indicating that the ongoing release of serotonin was contributing to the activity of the network under these in vitro conditions. Reproduced with permission from the Journal of Neuroscience from (Pena and Ramirez, 2002). G, putative role of serotonergic neurons in central chemoreception. Left: a small subset of serotonergic neurons, the location of which needs to be clarified, may be central respiratory chemoreceptors, i.e. may be acid-sensitive in vivo and may activate the respiratory controller among other targets. Right, most serotonergic neurons recorded so far in adult mammals in vivo were not CO 2-responsive. A few serotonergic neurons were CO 2–activated, either because of an intrinsic acid-sensitivity or because they receive excitatory inputs from the respiratory controller. The serotonergic system at large activates the breathing network and regulates its response to CO 2. RTN neurons receive an excitatory input from serotonergic cells (Mulkey et al., 2007a). In theory, this input could originate from pH-sensitive (left) or pH-insensitive serotonergic cells (right). Serotonergic neurons are at least one order of magnitude more numerous than the ccRTN neurons (∼20,000 vs. ∼2,000, in rats) and they are extremely diverse, embryologically, anatomically and functionally (Hökfelt et al., 2000; Kocsis et al., 2006; Jensen et al., 2008; Wylie et al., 2010). Recent evidence that only a small fraction of these neurons may have properties consistent with putative CRCs is therefore not surprising. In brainstem slices from neonate rodents, serotonergic neurons are virtually insensitive to acid at birth (Corcoran et al., 2009). This insensitivity may be related to the fact that the chemoreflex is weak at birth in rodents. However, only about 18% of brainstem serotonergic neurons recorded in slices are acid-sensitive two weeks after birth when the chemoreflex is reasonably mature (Corcoran et al., 2009). In the adult in vivo, the proportion of CO 2-responsive serotonergic neurons seems to be equally low and is dependent on which portion of the raphe is being considered. For example, an extremely small fraction of medullary midline raphe neurons respond consistently to activation of peripheral and central chemoreceptors in anesthetized cats (5 neurons excited and 4 inhibited by both stimuli out of a sample of 125) (Nuding et al., 2009). The interpretation of this data should be qualified because no attempt was made to determine whether these neurons were serotonergic in these studies. However, in a study on unanesthetized cats where the characterization of the serotonergic cells was convincing, if not definitive, only twenty-two percent of the putative serotonergic neurons located in the midline medullary raphe (raphe obscurus or magnus) were activated by hypercapnia (Veasey et al., 1995). One of the few putatively serotonergic neurons (6/27) that responded to hypercapnia in this study is depicted in Figure 3E. The general conclusion of these studies in conscious cats was that the activity of midline medulla serotonergic neurons correlates best with the overall motor activity of the animals (Jacobs et al., 2002). In anesthetized rats, the serotonergic neurons of the parapyramidal region, including those that are located at the ventral medullary surface in an area presumed to be “chemosensitive”, are remarkably unresponsive to even severe hypercapnia (sample: 24 chemically identified serotonergic neurons; Figure 3C) (Mulkey et al., 2004). Nonetheless, the possibility that a small subset of serotonergic neurons are acid-sensitive is supported by evidence that in conscious rats, acidification of the raphe causes small and region-specific increases in breathing with best results produced in the rostral midline raphe (presumably r. obscurus and magnus) (Dias et al., 2008). However, also in rats, acidification of more caudal regions of the raphe obscurus that are believed to contain serotonergic neurons which innervate the respiratory region of the ventrolateral medulla (Ptak et al., 2009) produced no effect on breathing (Dias et al., 2008). The results are also apparently species specific. Midline raphe acidification increases breathing during waking but not during non-REM sleep in goats whereas, in rats, breathing stimulation is seen only during sleep (Nattie and Li, 2001; Hodges et al., 2004a). Regions of the raphe where serotonergic cells are apparently CO 2 insensitive in vivo (parapyramidal region, nucleus interfascicularis hypoglossi, a.k.a. lateral B3 group) innervate sympathetic efferents (Strack et al., 1989) and primarily regulate thermogenesis and skin blood flow (Morrison, 2004). The response of raphe obscurus serotonergic neurons to CO 2 in vivo could also be due to synaptic inputs. Indeed, in unanesthetized cats, this response was state-dependent therefore unlikely to be a cell autonomous pH response (4/4 cells tested; Figure 3E) (Veasey et al., 1995). In support of this interpretation, many raphe obscurus serotonergic neurons receive respiratory phasic excitatory inputs in brain slices (Figure 3D) (Ptak et al., 2009). This observation means that the activation of a subset of serotonergic neurons by CO 2 in vivo could be a consequence of the activation of the respiratory controller rather than an intrinsic sensitivity to acid. In slices, the ongoing activity of the residual respiratory network is maintained by the release of serotonin (Figure 3F) (Pena and Ramirez, 2002) which suggests that some of the serotonergic cells of the medullary raphe may participate in some feed-forward loop involving the pre-Bötzinger complex (Figure 3G, right). Finally, serotonergic neurons receive an orexinergic input which could also contribute to their activation by hypercapnia in vivo since the emotional control of breathing and the chemoreflex depends on the integrity of the orexin system (Kuwaki, 2008; Dias et al., 2009a). To our knowledge, the hypothesis that the intrinsic pH sensitivity of serotonergic neurons (and other wake-promoting systems) is responsible for CO 2-induced arousal is untested. The notion seems at odds with the fact that severe hypercapnia fails to wake humans with central congenital hypoventilation syndrome (CCHS) from sleep (Gozal, 1998; Spengler et al., 2001; Amiel et al., 2009). This genetic disease is not known to interfere with the sleep cycle per se and the PHOX2B mutation that underlies it produces no change in the number of serotonergic neurons in the mouse model of the disease (Pattyn et al., 2003; Dubreuil et al., 2008; Amiel et al., 2009). In summary, subsets of serotonergic neurons make an important contribution to the activity and plasticity of the respiratory network (Pena and Ramirez, 2002; Richter et al., 2003; McCrimmon et al., 2008; Ptak et al., 2009). Furthermore, in the absence of serotonergic neurons or when brainstem serotonergic cells are acutely inhibited, the chemoreflex is markedly attenuated (Taylor et al., 2005; Taylor et al., 2006; Hodges et al., 2008). Yet, evidence that CO 2 increases the activity of more than a small minority of the serotonergic neurons in vivo is yet to be produced and there is reason to believe that this activation could be due, in part at least, to synaptic inputs (Veasey et al., 1995; Nattie and Li, 2001; Hodges et al., 2004b; Ptak et al., 2009). Figure 3G summarizes a few of the many possibilities that could account for the role of brainstem serotonergic neurons in breathing and CO 2 regulation based on current experimental evidence. One of these possibilities is that a small subset of serotonergic neurons, possibly located in the rostral medullary midline raphe, encodes pH changes in vivo and functions as CRCs. 5.3. Orexinergic neurons These neurons, like the pontine noradrenergic neurons and the serotonergic neurons have extremely widespread projections and are wake promoting (Lu et al., 2006). Orexin-KO mice have lowered BP and breathing during the waking state (Deng et al., 2007; Kuwaki, 2008; Kuwaki et al., 2008). The orexin system targets the ventrolateral medulla and the RTN among many other regions (Kuwaki, 2008; Dias et al., 2009b). Orexin neurons are activated by acidification in vitro (Williams et al., 2007). This effect, initially attributed to TASK channel inhibition, persists in TASK1/3 KO mice and is therefore still unexplained (Williams et al., 2007; Gonzalez et al., 2009). There is no information regarding the sensitivity of orexin neurons to CO 2 in vivo except for the fact that severe hypercapnia increase Fos expression in a relatively small fraction of these neurons (Sunanaga et al., 2009). The latter result could be also explained by the general arousal effect of CO 2. Because of the widespread projection pattern, the orexin neurons are either broad-spectrum central respiratory chemoreceptors (Figure 1, option 2) or simply neurons that regulate breathing along the lines depicted in Figure 1, option 4. 5.4. Locus coeruleus (LC) This major collection of pontine noradrenergic neurons has an activity level that is state-dependent and is activated by a large number of intero- and exteroceptive inputs (Aston-Jones et al., 1996; 1999). The LC is one of several groups of wake-promoting neurons that include subsets of serotonergic neurons, orexinergic neurons and histaminergic neurons. The LC innervates vast tracts of the brain from cortex to spinal cord (Aston-Jones et al., 1996; 1999). The LC is mildly activated by hypercapnia in anesthetized animals and by hypercapnic acidosis in slices (Elam et al., 1981; Ritucci et al., 2005a). The activation of LC neurons by CO 2 observed in vivo could be partially or wholly due to the activation of synaptic inputs since these cells receive a massive innervation from the C1 neurons and from the orexin neurons which are activated by hypercapnia (Horvath et al., 1999; Aston-Jones et al., 2001; Card et al., 2006; Kuwaki, 2008; Sunanaga et al., 2009). Like the serotonergic neurons, LC neurons are more responsive to CO 2 in culture than in slices (Johnson et al., 2008). This characteristic, as in the case of serotonergic neurons, could be due to a higher input resistance in vitro because of low ongoing synaptic activity or it could be due to the expression of different channels in culture. Lesions of LC and other NE neurons reduce the respiratory chemoreflex induced by hypercapnia (Li et al., 2008; Biancardi et al., 2008). Relatively selective lesions of the locus coeruleus (with 6-hydroxydopamine) do not modify the stimulation of breathing induced by hypoxia and a similar selectivity towards CO 2-induced breathing stimulation has been consistently observed after inhibition or lesions of the serotonergic system (Taylor et al., 2005; Hodges et al., 2008; Biancardi et al., 2010). In the aggregate, these results suggest that both noradrenergic neurons and serotonergic cells contribute to the respiratory stimulation elicited by hypercapnia but not hypoxia. Given that both stimuli activate the retrotrapezoid nucleus (Takakura et al., 2006), the difference could be due to the fact that hypoxia is sedative in rodents whereas CO 2 or asphyxia cause arousal. CO 2 may also directly activate noradrenergic (Sirois et al., 2000; Filosa and Putnam, 2003) and a subset of serotonergic cells (see above discussion). Evidence that selective activation of the locus coeruleus neurons activates breathing in vivo has not been produced but indirect evidence suggests that the locus coeruleus has a tonic facilitatory influence on the neonate breathing network (Hilaire et al., 2004). Because of their widespread brain projections and the lack of evidence that subsets of locus coeruleus neurons target the respiratory network with any degree of selectivity, locus coeruleus neurons should provisionally be viewed as either broad-spectrum central respiratory chemoreceptors (Figure 1, option 2) or neurons that regulate breathing along the lines depicted in Figure 1, option 4, possibly in a state-dependent context. 5.5 Other putative chemoreceptors CRCs may be located at the ventral medullary surface caudal to the retrotrapezoid nucleus, within the respiratory column, the NTS and the fastigial nucleus of the cerebellum (Nattie and Li, 2009). The raphe has been mentioned above. The presence of non-serotonergic CRCs within the region of the raphe could conceivably account for the mild breathing stimulation caused by raphe acidification (Nattie and Li, 2001; Hodges et al., 2004a). The regions of the ventral medullary surface that were originally described as chemosensitive based on the respiratory effect produced by topical acidification in cats extend from the caudal-most hypoglossal rootlets to the trapezoid body (Mitchell et al., 1963a; Schlaefke et al., 1979; Loeschcke, 1982). Only the rostral chemosensitive region also called area M can be viewed as roughly in register with the retrotrapezoid nucleus. The intermediate or caudal chemosensitive regions which overlap considerably with the serotonin rich areas of the ventral medullary surface (Figure 3A) contain tonically active neurons that are excited by injection of 100% CO 2-saturated saline into the vertebral artery of anesthetized cats (Arita et al., 1988) or iontophoretic applications of acid (Ribas-Salgueiro et al., 2003). These observations are difficult to interpret because the identity, function and connectivity of these acid-sensitive cells are unknown and, in the case of the Ribas-Salgueiro paper, there is no evidence that these cells respond to hypercapnia. This general area of the brain also expresses c-Fos in animals exposed to hypercapnia (Teppema et al., 1994; 1997; Berquin et al., 2000). The NTS contains large numbers of neurons that are activated by acidification in slices and acidification of the NTS, particularly its caudal part, increases breathing in unanesthetized rats (Dean et al., 1989; 1990; Nattie and Li, 2002b). This evidence suggests that the caudal aspect of the NTS could harbor CRCs although, paradoxically, lesions of this region increase the ventilatory response to CO 2 (Sinclair et al., 1985). The caudal NTS mediates the effect of peripheral chemoreceptors on breathing and on the circulation (Boscan et al., 2002). Transmission of peripheral chemoreceptor information through the NTS could conceivably be also influenced by ECF pH. 6. Contribution of peripheral chemoreceptors to respiratory chemosensitivity The long-standing and still unanswered question is how the motor outflow of the respiratory network could achieve such a high sensitivity to CO 2. Glial mechanisms that amplify the response of certain neurons to acid have been already been evoked as well as the possibility that multiple types of acid-sensitive neurons cooperate to stimulate breathing. Synaptic gain between the putative CRCs and their targets within the respiratory network has to be another critical variable but this aspect of the biology of CRCs remains essentially undocumented. Another very important contributing factor is the input from the carotid bodies. In the absence of anesthesia, peripheral chemoreceptors provide a tonic stimulus to the central respiratory controller, even at rest and under normocapnic conditions (Smith et al., 2006; Forster et al., 2008). Surgical excision of the carotid bodies causes chronic hypoventilation, reduction of the gain of the chemoreflex gain and respiratory acidosis in awake rats, goats and man (Olson, Jr. et al., 1988; Pan et al., 1998; Dahan et al., 2007). These deficits are attenuated with the passage of time presumably because of some form of CNS plasticity. In addition, virtually all studies indicate that the ventilatory effects produced by simultaneous activation of peripheral and central chemoreceptors are at least additive (Nattie, 1999). The recent study by Day and Wilson (2009) is a very rare exception that may be explained by a ceiling effect of each stimulus applied individually in an extremely responsive preparation. In the aggregate, this evidence indicates that the central respiratory drive depends both on brain parenchymal pH and the ongoing resting activity of the carotid bodies at all times. The ccRTN neurons are a site of convergence between peripheral and central chemoreceptor inputs (Takakura et al., 2006). The carotid body input to the RTN is via a presumably direct glutamatergic projection from the commissural portion of the NTS (Takakura et al., 2006). In theory, the loss of this glutamatergic input should elevate the CO 2 recruitment threshold of the ccRTN neurons and reduce the gain of their activation by CO 2. Thus, the loss of a tonic carotid body input to the RTN neurons should contribute to produce the type of respiratory deficits that have been observed after carotid body excision in man and animals (hypoventilation, reduced central chemoreflex and chronic elevation in arterial PCO 2). Other sites of convergence between carotid body input and putative central chemoreceptors are suspected but are less thoroughly documented. Locus coeruleus neurons and certain serotonergic neurons for example are activated by stimulation of peripheral chemoreceptors (Elam et al., 1981; Erickson and Millhorn, 1994) but the pathway is not worked out and the question remains as to whether these neurons would respond to very small deviations in PCO 2 around physiological levels. Summary and conclusions CO 2 most likely works by changing brain pH but the existence of receptors for molecular CO 2 remains a possibility. The number of potential proton receptors is large. In peripheral tissues, acidity is detected by many cell types (blood vessels, glial cells, bone, sensory afferents) through a variety of proton-sensitive molecules (mostly channels plus a few GPCRs). Most of these proteins are expressed in brain and could, in theory, mediate the pH sensitivity of CNS neurons and, more specifically, central respiratory chemosensitivity. Despite the virtually ubiquitous presence of pH-sensitive channels in neurons, in the brainstem and elsewhere, a substantial proportion of neurons do not respond in vitro to even severe (>0.2 pH units) acidification. The fact is noteworthy because it suggests that some sort of neuronal homeostasis (Bucher, 2009) may render most neurons insensitive or very weakly sensitive to the small pH fluctuations (<0.1 pH unit) that are relevant to the homeostatic regulation of CO 2 via breathing. There is uncertainty regarding which cells (neurons, glia, others) are the primary proton detectors involved in central respiratory chemosensitivity. Neurons are likely contributors but the pH-response of the few putative CRC neurons that have been examined in some detail in vitro is fairly modest and varies relatively little between various chemoreceptor candidates. A significant problem with this evidence is that the properties of neurons in slices and especially in culture may be very different from their properties in vivo. This quandary is illustrated by the specific case of the serotonergic neurons which are robustly activated by acid in culture and generally insensitive to CO 2 in vivo (Figure 3). In vitro, the quasi absence of active synaptic inputs likely increases neuronal input resistance which should greatly enhance the effects of minor changes in membrane current relative to what would be expected in vivo (Destexhe et al., 2003). This phenomenon could blur the difference between the real CRCs and other cells in unpredictable ways when the cells are studied in vitro. In addition, the channel expression profile of neurons is likely to be different in culture. Finally, the pH-induced changes in conductance present in individual neurons may have no impact at the level of the respiratory network. Proof of concept was provided by the recent experiments of Dubreuil et al. (2009b) who showed that the apparently uniform acid-sensitivity of the ventral respiratory column neurons (Kawai et al., 1996) does not add up to an acid-sensitive respiratory network in the absence of the ccRTN neurons. Evidence that the pH sensitivity of the CRCs relies on acid-sensitive channels is also weak although the hypothesis is eminently plausible. The same can be said of the theory that multiple pH-sensitive channels are involved simultaneously and that a different mix of channels contributes to the acid response of different CRCs. However, if correct, the latter possibility adds an additional level of complexity to an already challenging issue because many of the candidate proton-sensitive channels (e.g. TASK channels, calcium channels) are modulated by extracellular signals, which means that their contribution to the overall pH sensitivity of the cells could be synaptically regulated. On the other hand, the complexity of proton detection in the context of central respiratory chemoreception may also be greatly exaggerated. One or very few pH-sensitive channels could conceivably be essential for central respiratory chemoreception if they were expressed by neurons that play a pivotal role in this function. This hypothesis does not presuppose the existence of an esoteric proton detector in such cells although this possibility is not ruled out. The right anatomical connections combined with intrinsic properties that are appropriate to give these cells a robust response range to pH-induced current would be enough. If this view is correct, cell-specific deletion of an already well described and widely expressed pH-sensitive channel or receptor may provide a simultaneous answer to two of the most critical questions in this field, namely the real importance of a particular cell group to central respiratory chemoreception and the importance of a particular channel for this function. This type of approach has already been carried out to a limited extent using the retrotrapezoid nucleus as a model (Mulkey et al., 2007b; Gestreau et al., 2010) but it has not yet borne fruit. Accordingly, the issue of whether central respiratory chemosensitivity relies on one or a few specialized cell groups or is a diffuse (“emergent”) property of the respiratory controller and the many brain pathways that regulate its activity (Nattie and Li, 2009) remains open. These options are supported by very different types of evidence, none of which is incontrovertible, singly or in the aggregate. The highly sensitive response of the respiratory network to very small deviations of CO 2 around the normal level could be the purview of a limited number of specialized cells with a low CO 2 threshold and high response gain to acid whereas higher levels of CO 2, which produce an alerting response and conscious aversive sensations, could be recruiting additional groups of cells that are responsive only to much larger deviations of brain pH. An alternate view is that the sensitive response of the respiratory network to very small deviations of CO 2 is a network effect caused by the summation of very small effects of pH on a very large number of neurons (Figure 1, option c). The retrotrapezoid nucleus contains the most thoroughly documented group of CRCs although many important aspects of the biology of these neurons are still missing. The most notable missing piece is the molecular mechanism responsible for their activation by CO 2 and one can argue that the definitive demonstration that these neurons are CRCs will require deleting the proton receptor selectively from these cells. The ccRTN neurons contribute a powerful pH-modulated excitatory drive to the breathing network and their absence at birth is lethal (Dubreuil et al., 2008; Guyenet, 2008; Amiel et al., 2009; 2009b). The function of the ccRTN neurons is clearly far more complex than that of a mere pH detector. The notion that RTN is a major convergence point for the various pathways that automatically regulate breathing to maintain arterial PCO 2 constant is, at present, the most sensible explanation of the respiratory deficits observed in the CCHS (Guyenet, 2008; Weese-Mayer et al., 2010; Goridis and Brunet, 2010). The other putative CRCs are in various stages of characterization at this time. In all cases, one or more critical pieces of evidence are missing, for example evidence that these neurons respond to CO 2 in vivo or that their selective activation in vivo increases breathing. The case of the serotonergic neurons is the most intriguing because of the unresolved discrepancy between their obvious importance to breathing and the chemoreflex and their general insensitivity to CO 2 in vivo. One possible interpretation of the literature is that the other postulated chemosensitive regions of the brainstem regulate breathing primarily by modulating the activity of the ccRTN neurons. This notion could explain why, despite the existence of multiple other CRCs, the chemoreflex is absent when these few RTN neurons are destroyed, as is presumably the case in CCHS patients. This possibility could also explain why the sensitivity of RTN neurons to acid is lower in vitro than in vivo. The chemoreflex is enhanced during waking compared to slow-wave sleep. This reflex is depressed by genetic deletion and other forms of insult to several major wake-promoting systems (orexinergic, serotoninergic, noradrenergic). Large increases in CO 2 above the normal range produce arousal from sleep and the arousal presumably contributes to the stimulation of breathing in part by recruiting the orexinergic, serotonergic and noradrenergic neurons many of which are known to innervate and activate several components of the breathing network down to the motor neurons. How hypercapnia causes cortical / behavioral arousal is not known. One possibility is that the orexinergic, pontine serotoninergic and pontine noradrenergic neurons are directly activated by high level of CO 2 in vivo. This suggestion is supported by physiological evidence obtained in vitro but there is also contrary evidence. If CO 2-induced arousal and the interoceptive awareness of hypercapnia depended on an effect of CO 2 on these systems, these effects of CO 2 should be preserved in CCHS which is a disease that impacts the brainstem selectively and does not or at least is not known to affect the sleep cycle markedly. In fact, a remarkable feature of the disease is the loss of the interoceptive awareness of asphyxia and the inability of hypercapnia to produce arousal from sleep (Carroll et al., 2010). CO 2-mediated arousal may therefore depend primarily on ascending signals from peripheral and brainstem chemoreceptors that reach the vigilance-regulating hypothalamic centers either without further processing or on efferent copies of the activity of the central respiratory controller or as a result of sensory inputs from the chest and lungs, a combination of mechanisms that may also underlie the sensation of dyspnea (Chen et al., 1992; Flume et al., 1996; Moosavi et al., 2004). 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Calculate change from paid amount Ask Question Asked 5 years, 10 months ago Modified5 years, 10 months ago Viewed 435 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Why is this code creating an infinite loop? I would think this should be an appropriate solution for this type of problem. For example, if the price was $5 and you paid $5.47, the program would print: python Quarters: 1 Dimes: 2 Nickels: 0 Pennies: 2 However, an infinite loop occurs and I'm not sure why. Anyone know the reason? python price = round(float(input("Enter the price: ")), 2) print price paid = round(float(input("Enter the amount paid: ")), 2) print paid change = round(float(paid - price), 2) print change quarters = 0 dimes = 0 nickels = 0 pennies = 0 while change > 0.00: print change if change >= .25: change = change - .25 quarters += 1 continue elif change >= .1: change = change - .1 dimes += 1 continue elif change >= .05: change = change - .05 nickels += 1 elif change >= .01: change = change - .01 pennies += 1 print "Quarters: " + str(quarters) print "Dimes: " + str(dimes) print "Nickels: " + str(nickels) print "Pennies: " + str(pennies) python python-2.7 Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Dec 4, 2019 at 14:03 robert lewisonrobert lewison asked Dec 3, 2019 at 17:13 robert lewisonrobert lewison 127 8 8 bronze badges 12 3 Shouldn't it be change = paid - price?OneCricketeer –OneCricketeer 2019-12-03 17:17:32 +00:00 Commented Dec 3, 2019 at 17:17 Note: Python2 is end of life by the end of the year. You may want to learn Python3 OneCricketeer –OneCricketeer 2019-12-03 17:18:32 +00:00 Commented Dec 3, 2019 at 17:18 2 I recommend reading the following article: ericlippert.com/2014/03/05/how-to-debug-small-programs.AMC –AMC 2019-12-03 17:20:59 +00:00 Commented Dec 3, 2019 at 17:20 2 Also, just ran the code, and entered 5, then 5.47 and there is no loop.OneCricketeer –OneCricketeer 2019-12-03 17:21:43 +00:00 Commented Dec 3, 2019 at 17:21 1 As the previous commented stated you should have "change = paid - price" but in addition your code runs into a common issue subtracting floats where you get more digits than expected. From your example float(5.47) - float(5) = 0.46999999999999975. There are a few libraries that will make working with money a little less problematic.KILLtheWEEZEL –KILLtheWEEZEL 2019-12-03 17:24:21 +00:00 Commented Dec 3, 2019 at 17:24 \|Show 7 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Rather than dealing with loops, I would suggest just subtracing off the change that you already gathered, prioritizing larger coins. ```python price = float(input("Enter the price: ")) paid = float(input("Enter the amount paid: ")) change = paid - price if change < 0: raise ValueError('Not enough paid') quarters = change // 0.25 dimes = (change - (0.25 quarters)) // 0.10 nickels = (change - (0.25 quarters) - (0.10 dimes)) // 0.05 pennies = 100 (change - (0.25 quarters) - (0.10 dimes) - (0.05 nickels)) print("Quarters: {:.0f}".format(quarters)) print("Dimes: {:.0f}".format(dimes)) print("Nickels: {:.0f}".format(nickels)) print("Pennies: {:.0f}".format(pennies)) ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Dec 3, 2019 at 17:43 OneCricketeerOneCricketeer 193k 20 20 gold badges 145 145 silver badges 276 276 bronze badges 1 Comment Add a comment robert lewison robert lewisonOver a year ago thanks! works great, just got set on the idea that i would need to do it with loops 2019-12-04T14:10:31.983Z+00:00 0 Reply Copy link This answer is useful 0 Save this answer. Show activity on this post. There's one minor bug in the code which causes the program to only work correctly if price and amount paid are interchanged (e.g. price = 2, paid = 1). But that is not the issue causing the infinite loop. Your code creates an infinite loop for e.g. the following arguments: price: 5.6 paid: 5.4 The reason for the infinite loop can be seen from your own print output: python 0.009999999999999275 0.009999999999999275 0.009999999999999275 0.009999999999999275 0.009999999999999275 0.009999999999999275 0.009999999999999275 ... Since change < 0.01, no if clause applies and thus the loop is never left. How could you solve the problem more robustly? Here's a sketch ```python from math import floor change = paid - price quarters = int(floor(change / 0.25)) change -= quarters 0.25 dimes = int(floor(change / 0.1)) change -= dimes 0.1 nickels = int(floor(change / 0.05)) change -= nickels 0.05 pennies = int(floor(change / 0.01)) change -= pennies 0.01 remaining = change print("Quarters:", quarters) print("Dimes:", dimes) print("Nickels:", nickels) print("Pennies:", pennies) ``` Personally I would also condense this into a loop over the coin type: ```python increments = {"quarter":0.25, "dimes": 0.1, "nickels": 0.05, "pennies": 0.01} change_parts = {} for inc_name, inc in increments.items(): amount = int(floor(change / inc)) print(inc_name, inc, amount) change -= amount inc change_parts[inc_name] = amount for inc_name, amount in change_parts.items(): print(inc_name + ":", amount) ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Dec 3, 2019 at 17:45 OmniOmni 1,022 6 6 silver badges 12 12 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! 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15441	https://byjus.com/maths/quadratic-formula/	Thequadratic formula is used to find the roots of a quadratic equation. This formula helps to evaluate the solution of quadratic equations replacing the factorization method. If a quadratic equation does not contain real roots, then the quadratic formula helps to find the imaginary roots of that equation. The quadratic formula is also known as Shreedhara Acharya’s formula. In this article, you will learn the quadratic formula, derivation and proof of the quadratic formula, along with a video lesson and solved examples. Let’s learn what a quadratic equation is and how to solve the quadratic equation using the quadratic formula. Learn: Quadratic equation What is Quadratic Formula? An algebraic expression of degree 2 is called the quadratic equation. The general form of a quadratic equation is ax 2+ bx + c = 0, where a, b and c are real numbers, also called “numeric coefficients”and a ≠ 0. Here, x is an unknown variable for which we need to find the solution.We know that thequadratic formula used to find the solutions (or roots) of the quadratic equation ax 2 + bx + c = 0 is given by: Here, a, b, c = Constants (real numbers) a ≠ 0 x = Unknown, i.e. variable The above formula can also be written as: x=−b 2 a±b 2−4 a c 4 a 2 Or x=−b 2 a±(b 2 a)2−c a What is the Quadratic Formula used for? The quadratic formula is used to find the roots of a quadratic equation and these roots are called the solutions of the quadratic equation. However, there are several methods of solving quadratic equations such as factoring, completing the square, graphing, etc. Also, check:Quadratic Equation Calculator Roots of Quadratic Equation by Quadratic Formula We know that a second-degree polynomial will have at most two zeros, and therefore a quadratic equation will have at most two roots. In general, if α is a root of the quadratic equation ax 2 + bx + c = 0, a ≠ 0; then, aα 2 + bα + c = 0. We can also say that x = α is a solution of the quadratic equation or α satisfies the equation, ax 2 + bx + c = 0. Note:Roots of the quadratic equation ax 2 + bx + c = 0 are the same as zeros of the polynomial ax 2 + bx + c. One of the easiest ways to find the roots of a quadratic equation is to apply the quadratic formula. Quadratic formula: x=−b±b 2−4 a c 2 a Here, b 2 – 4ac is called the discriminant and is denoted by D. The sign of plus (+) and minus (-) in the quadratic formula represents that there are two solutions for quadratic equations and are called the roots of the quadratic equation. Root 1: x 1=−b+b 2−4 a c 2 a And Root 2: x 2=−b−b 2−4 a c 2 a These are the formulas for quadratic equations to find its solution. Also, read: Nature Of Roots Quadratic Quadratic Equation Solver Quadratic Equation For Class 10 Quadratic Equation Worksheet Quadratic Equations Class 11 Derivation of Quadratic Formula We can derive the quadratic formula in different ways using various techniques. Derivation Using Completing the Square Technique Let us write the standard form of a quadratic equation. ax 2 + bx + c = 0 Divide the equation by the coefficient of x 2, i.e., a. x 2 + (b/a)x + (c/a) = 0 Subtract c/a from both sides of this equation. x 2 + (b/a)x = -c/a Now, apply the method of completing the square. Add a constant to both sides of the equation to make the LHS of the equation as complete square. Adding (b/2a)2 on both sides, x 2 + (b/a)x + (b/2a)2 = (-c/a) + (b/2a)2 Using the identity a 2 + 2ab + b 2 = (a + b)2, [x + (b/2a)]2 = (-c/a) + (b 2/4a 2) [x + (b/2a)]2 = (b 2 – 4ac)/4a 2 Take the square root on both sides, x+b 2 a=±b 2−4 a c 2 a Therefore, x=−b±b 2−4 a c 2 a This is the most commonly used method to derive the quadratic formula in maths. Shortcut Method of Derivation Write the standard form of a quadratic equation. ax 2 + bx + c = 0 Multiply both sides of the equation by 4a. 4a(ax 2 + bx + c) = 4a(0) 4a 2 x 2 + 4abx + 4ac = 0 4a 2 x 2 +4abx = -4ac Add a constant on sides such that LHS will become a complete square. Adding b 2 on both sides, 4a 2 x 2 + 4abx + b 2 = b 2 – 4ac (2ax)2 + 2(2ax)(b) + b 2 = b 2 – 4ac Using algebraic identity a 2 + 2ab + b 2 = (a + b)2, (2ax + b)2 = b 2 – 4ac Taking square root on both sides, 2ax + b = ±√(b 2 – 4ac) 2ax = -b ±√(b 2 – 4ac) x = [-b ±√(b 2 – 4ac)]/2a How to Solve Using Quadratic Formula – Steps Let us understand how to use the quadratic formula with the help of the steps given below. These steps will help you to understand the method of s olving quadratic equations using the quadratic formula Let us consider a quadratic equation: Step1:Consider a quadratic equation x 2 + 4x – 13 = 0 Step 2:Compare with the standard form and write the coefficients. Here, a = 1, b = 4, c = -13 Step 3: Now. find the value of b 2 – 4ac. b 2 – 4ac = (4)2 – 4(1)(-13) = 16 + 52 = 68 Step 4: Substitute the values in the quadratic formula to get the roots of the given quadratic equation. x = [-b ± √(b 2 – 4ac)]/ 2a = [-4 ± √68]/ 2(1) Step 5:Simplify the expression to get the values for x. = [-4 ± 2√17]/2 = -2 ± √17 Therefore, x = -2 – √7 and x = -2 + √17 are the roots of the given quadratic equation. S ome of the important points about quadratic formula and the nature of roots of a quadratic equation are listed below: The expression under the radical in the quadratic formula is called the discriminant, i.e., D = b 2 – 4ac The nature of the roots of a quadratic equation can be determined based on the value of D. If D = 0, the two roots are real and equal If D > 0, the roots are real and unequal If D < 0, the roots are not real, i.e. imaginary If the value of discriminant is 0, then the roots of the quadratic equation ax 2 + bx + c = 0 are -b/2a and -b/2a. Solved Examples on Quadratic Formula Example 1: Find the roots of the equation x 2– 5x + 6= 0 using the quadratic formula. Solution: Given quadratic equation is: x 2– 5x + 6= 0 Comparing the equation with ax 2+bx+c= 0 gives, a = 1, b = -5 and c = 6 b 2– 4ac = (-5)2– 4 × 1 × 6= 25 – 24 = 1 > 0 The roots of the given equation are real. Using quadratic formula, x = [-b ± √(b 2– 4ac)]/ 2a = [-(-5) ± √1]/ 2(1) = [5 ± 1]/ 2 i.e. x = (5 + 1)/2 and x = (5 – 1)/2 x = 6/2, x = 4/2 x = 3, 2 Hence, the roots of the given quadratic equation are 3 and 2. Example 2: Find the roots of 4x 2+ 3x + 5 = 0 using quadratic formula. Solution: Given quadratic equation is: 4x 2+ 3x + 5 = 0 Comparing with the standard form ax 2+ bx + c = 0, a = 4, b = 3, c = 5 Determinant (D) = b 2– 4ac = (3)2– 4(4)(5) = 9 – 80 = -71 < 0 That means, the roots are complex (not real). Using quadratic formula, x = [-b ± √(b 2– 4ac)]/ 2a = [-3 ± √(-71)]/ 2(4) = [-3 ± √(i 2 71)]/ 8 = (-3 ± i√71)/8 Therefore, the complex roots of the given equation are x = (-3 + i√71)/8 and x (-3 – i√71)/8. Quadratic Formula Questions Solve with a quadratic formula and write the roots of equations. 2x 2 – 3x + 4 = 0 x 2 + 6x – 5 = 0 7x 2 + 6x – 4 = 0 How to Solve Quadratic Equations? – Video Lesson 2,98,622 To know more about quadratic equations, download BYJU’S – The Learning A pp from Google play store. Also Access NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations NCERT Exemplar for Class 10 Maths Chapter 4 Quadratic Equations CBSE Notes for Class 10 Maths Chapter 4 Quadratic Equations Comments Leave a Comment Cancel reply Your Mobile number and Email id will not be published.Required fields are marked Send OTP Did not receive OTP? Request OTP on Voice Call Website Post My Comment Register with BYJU'S & Download Free PDFs Send OTP Download Now Register with BYJU'S & Watch Live Videos Send OTP Watch Now × To continue watching the video please share the details. Send OTP Did not receive OTP? Request OTP on Voice Call Play Now
15442	https://www.youtube.com/watch?v=49cnkJBrS9k	Subtracting 614 - 257 Using Partial Differences National Center on Intensive Intervention 4810 subscribers 49 likes Description 25893 views Posted: 13 Jan 2017 This video describes how to use the partial differences strategy to solve multi-digit subtraction. The partial differences strategy eliminates the need to regroup numbers. The partial differences strategy is typically performed left to right and focuses on subtracting only part of each multi-digit number at a time (e.g., only subtracting digits in the hundreds column to determine the partial difference of hundreds). It may be important for students to know and understand the partial differences strategies if they have not yet developed the understanding for regrouping that is necessary to use the standard algorithm; however, this strategy does introduce some complexities. Students may encounter situations where they need to use negative integers because they may be moving from the positive side of the number line to the negative side of the number line when determining differences. Transcript: I can use the partial differences algorithm to subtract 614 minus 257. With the partial differences algorithm I start in the hundreds column. 600 minus 200 is 400. I’m going to mark that as it's positive 400. Now I move to the tens column. One ten minus five tens actually moves over into negative integers. So one ten minus five tens is actually negative forty. And four ones minus seven ones also moves across the number line into negative territory. Four ones minus seven ones is negative three. Now I am going to do the calculation here. 400 minus 40 is 360, minus three is 357. So when I have 614 and I subtract 257 with the partial differences algorithm my answer is 357. Now one comment about this algorithm: Many times students at this point have not learned about negative integers so they do learn a trick where if the bottom number is greater than the top number, they switch the order of the numbers and they have to write a minus sign. That works, but mathematically the partial differences algorithm only works because we are crossing from positive territory into negative territory.
15443	https://emedicine.medscape.com/article/321273-overview	close Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel processing.... Tools & Reference>Physical Medicine and Rehabilitation Bladder Dysfunction Updated: Dec 23, 2024 Author: Gregory T Carter, MD, MS; Chief Editor: Elizabeth A Moberg-Wolff, MD more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Bladder Dysfunction Sections Bladder Dysfunction Overview Practice Essentials Pathophysiology Etiology Epidemiology Prognosis Patient Education Show All Presentation History Physical Examination Complications Show All DDx Workup Laboratory Studies Radiography and Ultrasonography Cystometry Electromyography Other Tests Show All Treatment Rehabilitation Facilitative Techniques and Maneuvers Catheterization Pharmacologic Therapy Surgery on Bladder Outlet Surgery on Bladder Other Treatments Long-Term Monitoring Show All Guidelines Medication Medication Summary Cholinergics, Genitourinary BPH, Alpha Blockers Anticholinergic Drugs Tricyclic Antidepressants Alpha/Beta Adrenergic Agonists Show All Media Gallery;) References;) Overview Practice Essentials In the practice of physical medicine and rehabilitation, voiding disorders are usually a result of neurologic conditions such as spinal cord injury (SCI) or disease, cerebrovascular accident (CVA), traumatic brain injury (TBI), multiple sclerosis (MS), or dementia. Incontinence and urinary retention can cause social embarrassment and added morbidity, such as infections, stones, or renal injury. Evaluation of bladder dysfunction can involve laboratory studies; imaging studies, including with radiography and ultrasonography; and other procedures, such as cystometry and electromyography. Management can include rehabilitation, bladder training, facilitative techniques and maneuvers, catheterization, pharmacologic therapy, and surgery. [2, 3, 4, 5, 6, 7] Signs of bladder dysfunction In general, patients may be symptomatic or asymptomatic and can present with urinary retention, urinary incontinence, or a mixed picture of incomplete emptying and incontinence. Workup in bladder dysfunction Laboratory studies indicated in the workup of a patient with neurogenic bladder dysfunction include the following: Urinalysis and urine culture with sensitivity to rule out infection 24-hour creatinine clearance Residual urine volume Ultrasonography is used for the routine assessment of the upper urinary tract (which consists of the kidneys and ureters) and may be employed to identify the presence of ureteral obstruction, scarring, masses, and either renal or bladder calculi. Computed tomography (CT) scanning or plain radiography used with ultrasonography can determine the presence of radiopaque calculi in the urinary tract, bladder, and kidneys. Excretory urography or intravenous pyelography (IVP) may be used for visualization of the collecting system. Isotope studies (eg, technetium-99m dimercaptosuccinic acid [DMSA]) are used for the evaluation of renal cortex function. Cystometry is the measurement of bladder volume and intravesical pressure during filling and storage phases for the purpose of evaluating detrusor function. Electromyography (EMG) is used to diagnose the mechanistic cause of urinary retention and incontinence through the measurement of electrical potentials generated by depolarization of the detrusor muscle and the urethral sphincter. Management Early mobilization and transfer training are recommended to minimize urinary incontinence and other complications such as pressure sores. Various techniques are used to maintain continence or empty the bladder, including the Credé maneuver (involving manual compression of the bladder) and pinching or stimulation of the lumbar and sacral dermatomal levels to provoke reflex bladder contraction (which may be used in spinal cord injuries [SCIs] if there is no outlet obstruction or detrusor-sphincter dyssynergia). The practice of clean intermittent catheterization (CIC) is used primarily in patients with neurogenic bladder disease, such as is seen in cases of SCI. Cholinergic agonists are used in patients with detrusor areflexia. The alpha-adrenergic blocking agent phenoxybenzamine is useful for reducing bladder outlet resistance in patients with SCIs, as long as detrusor bladder contractions are present, while anticholinergic agents may help to alleviate symptoms in patients with urinary incontinence resulting from uninhibited bladder contractions secondary to suprasacral lesions. Tricyclic antidepressants (TCAs) may (1) have peripheral alpha-adrenergic and central anticholinergic effects, (2) suppress bladder contractions, and (3) enhance bladder neck resistance. Transurethral resection of the bladder neck is indicated in patients who have obstruction at the bladder neck, when medical therapy has failed to produce satisfactory results. External sphincterotomy is indicated in patients who have suprasacral lesions that cause failure to empty, when other therapeutic modalities have not been successful. Candidates for this procedure should have adequate detrusor contractions. Bladder augmentation is performed primarily in patients with refractory hyperreflexic bladders, when medical treatment has failed to alleviate symptoms. In this procedure, the bladder is opened and patched with a reconfigured segment of bowel. Augmentation also is used to achieve a normal bladder capacity in children and adolescents, often in conjunction with an artificial sphincter. Next: Pathophysiology Pathophysiology Control of typical bladder function (ie, filling and emptying) involves action of the somatic and autonomic nervous systems on the neuromuscular junctions of the muscles of the bladder, bladder neck, urethra, diaphragm, abdomen, and pelvic floor. Somatic efferent, autonomic sympathetic, and parasympathetic systems are selectively activated by a filling (ie, guarding) reflex or an emptying (ie, voiding) reflex. These reflexes are controlled by the central nervous system (CNS), with involvement of the spinal cord, the pontine micturition center, and the sacral micturition center; further oversight is provided by the medial prefrontal cortex. The sympathetic nervous system regulates the process of urine storage in the bladder by activating the muscles of the bladder neck and urethra where they form an internal sphincter known as the trigone sphincter, and by relaxing the muscles of the bladder wall. In contrast, the parasympathetic nervous system controls bladder contractions and the passage of urine through a relaxed bladder neck and urethra. The somatic efferent system permits voluntary control over the external periurethral sphincter, also known as the urogenital sphincter. For bladder contraction to occur, parasympathetic nerve impulses must travel from the S2-S4 ventral gray matter of the CNS via the preganglionic pelvic nerves, to the ganglia near the bladder wall, where acetylcholine is released to stimulate postganglionic neurons that then travel to the smooth muscle cholinergic receptors at the bladder wall and cause muscle contraction. Sympathetic efferent nerve fibers originate from the lateral gray column of the spinal cord between T11 and L2. The sympathetic system has a long postganglionic chain that runs with the hypogastric nerve to synapse with alpha and beta receptors in the bladder wall and the bladder neck or internal sphincter. Beta receptors are responsible for mediating relaxation of the bladder for filling. Alpha receptors are responsible for tonically contracting the internal sphincter during bladder filling. The somatic efferent nerve fibers originate from the pudendal nucleus of S2-S4 and supply the external periurethral sphincter. The external sphincter is under voluntary control and normally contracts in response to coughing or the Valsalva maneuver or when a person actively tries to prevent or halt urine flow. When the bladder retains urine, sympathetic activation of alpha1 receptors on the trigone, bladder neck, and urethra causes contraction of the internal sphincter, while beta-adrenergic receptors in the bladder body induce relaxation of the detrusor muscle to permit filling. Somatic efferent fibers from the cerebral cortex permit voluntary contraction of the external sphincter to provide extra support. Alpha-adrenergic receptors in the trigone, bladder neck, and urethra stimulate sphincter relaxation, while the muscarinic receptors in the detrusor smooth muscle body stimulate contraction to facilitate bladder emptying. Central control of micturition is performed by three areas: the sacral micturition center, the pontine micturition center, and the cerebral cortex. The sacral micturition center is at the S2-S4 levels and is responsible for bladder contraction. The pontine micturition center is found in the brain stem, and while its primary role is to stimulate urination when activated, it also acts as a central relay and may play a role in the coordination of external sphincter relaxation with bladder contraction. The cerebral cortex plays an inhibitory role in relation to the sacral micturition center. See the image below. The pons is a major relay center between the brain and bladder. The mechanical process of urination is coordinated by the pons in an area known as the pontine micturition center (PMC). View Media Gallery) Previous Next: Pathophysiology Etiology Classification of a neurogenic bladder depends on the location of the lesion along the central nerve pathway and can be categorized as having sacral, suprasacral, or suprapontine lesions, based on level of injury. [3, 9] Lesions of the peripheral nerves or at the level of the sacral micturition center cause detrusor areflexia (ie, flaccid bladder). These lesions may affect the conus medullaris, the cauda equina, and the S2-S4 peripheral nerves. Common causes of sacral lesions are trauma, stenosis, tumors, peripheral neuropathy, and infection. In general, these lead to variable loss of parasympathetic and somatic nerve function. Detrusor areflexia, bladder neck incompetence, or loss of external sphincter function may occur and present as bladder distention with overflow incontinence. Lesions of the spinal cord or brain stem below the pontine micturition center but above the sacral micturition center lead to uninhibited bladder contractions with uncoordinated sphincter activity. These suprasacral lesions are associated with the group of neurogenic bladder problems caused by spinal cord lesions from trauma, tumors, or spina bifida, as well as by multiple sclerosis or transverse myelitis. Such lesions interrupt the spinobulbospinal pathway required for the micturition reflex. Two pathologic mechanisms are in play here. The first is acute areflexia and then detrusor hyperreflexia. The second is detrusor-sphincter dyssynergia, where the external sphincter contracts at the same time that the detrusor contracts. Individuals with lesions below the pontine micturition center have detrusor hyperreflexia and detrusor-sphincter dyssynergia. Suprapontine lesions lead to uninhibited bladder contractions, which may be secondary to loss of cerebral cortex inhibition at the sacral micturition center. These lesions are often due to cardiovascular events, multiple sclerosis, dementia, brain tumors, or trauma. Suprapontine lesions are located at or between the pontine micturition center and the cerebral cortex and involve the spinobulbospinal reflex. They may or may not coincide with sacral lesions. These lesions present as uninhibited bladder contractions with retained voluntary urethral sphincter relaxation during micturition. Previous Next: Pathophysiology Epidemiology The incidence of neurogenic bladder dysfunction depends on the primary cause. The etiology and the level of CNS or peripheral nervous system injury correlate with different causes and classifications of bladder dysfunction. For instance, estimates of the incidence of urologic symptoms in patients who have sustained a CVA vary, ranging from 33-60% in the acute setting; symptoms persist in 15% at 6 months to 1 year. Bladder disorders are reported in 40-90% of patients with multiple sclerosis. The reported rate of urologic dysfunction in patients with Parkinson disease ranges from 37-72%, while the rate of urinary incontinence is higher in patients with dementia and other types of cognitive impairment (eg, TBI, CVA, and Parkinson disease) than in the general population. Bladder disorders are nearly universal in children with myelomeningocele and in patients with SCI. [11, 10] Age-, sex-, and race-related demographics The age at which bladder disorders occur is related to the age at the time of neurologic disorder onset. The male-to-female ratio varies greatly between disease entities causing neurogenic bladder dysfunction. In one study, the reported incidence of urinary incontinence, regardless of etiology, was 8.5% of females aged 15-64 years; in the same age group, only 1.6% of the male population reported being affected by urinary incontinence. No racial predilection is reported. A retrospective cohort study by Mckellar et al reported that within a racially diverse population of women aged 18 years or older, the prevalence of overactive bladder (excluding in women with neurogenic bladder or pelvic cancer) was 4.41%, with the highest prevalence found in Hispanic women. A study by Kinlaw et al found that between 2000 and 2013, antimuscarinic prescriptions were filled for about 1 in 500 children in the United States, most frequently for oxybutynin (78%) and tolterodine (17%), both of which are used for overactive bladders. Previous Next: Pathophysiology Prognosis The prognosis for recovery depends on the type, severity, and location of the lesion causing the bladder problem and can be organized by dysfunctions that are permanent, dysfunctions that resolve with treatment of the condition, and dysfunctions that require medical and/or surgical intervention. Patients with permanent bladder dysfunction include those with complete transection of the spinal cord. These individuals remain on intermittent or indwelling catheterization for the rest of their lives. Causes of temporary, syndromic urinary incontinence include polyneuropathy secondary to vitamin B-12 deficiency, which typically improves with nutritional correction. Dysfunctions that require medical and/or surgical interventions include upper motor neuron lesions (eg, stroke, multiple sclerosis, spinal cord pathology), for which patients may have to depend on medications for the rest of their lives (although some degree of recovery may be expected). Individuals with anatomic derangements like pelvic floor weakness may find symptomatic improvement with surgical correction. Previous Next: Pathophysiology Patient Education Patients with bladder dysfunction should receive education on the following aspects of care: Proper techniques of intermittent self-catheterization (including timing, frequency, and proper hygiene) Effects of oral medications on bladder function, as well as possible systemic side effects Management of some emergencies (eg, absence of urine output secondary to a kinked catheter) Prevention and awareness of potential complications and long-term comorbidities (eg, urinary tract infection, bladder cancer, urolithiasis) Use of facilitative bladder emptying techniques, such as the Credé and Valsalva maneuvers, as a conservative measure, although these techniques are generally not recommended due to adverse effects to other abdominal structures, such as increased bladder pressure and the development of pressure-related hernias and hemorrhoids Previous Clinical Presentation References Gleicher S, Sebesta EM, Dmochowski RR. 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Hoffman BB, Lefkowitz RT. Adrenergic receptor antagonists: The Pharmacologic Basis of Therapeutics. New York:. Pergamon Press. 1990. Welk B, Lenherr S, Santiago-Lastra Y, Norman HS, Keiser MG, Elliott CS. Differences in the incidence of urinary tract infections between neurogenic and non-neurogenic bladder dysfunction individuals performing intermittent catheterization. Neurourol Urodyn. 2022 Apr. 41 (4):1002-11. [QxMD MEDLINE Link]. Sung BM, Oh DJ, Choi MH, Choi HM. Chronic Kidney Disease in Neurogenic Bladder. Nephrology (Carlton). 2016 Dec 30. [QxMD MEDLINE Link]. Cardenas DD, Hooton TM. Urinary tract infection in persons with spinal cord injury. Arch Phys Med Rehabil. 1995 Mar. 76(3):272-80. [QxMD MEDLINE Link]. Wein AJ. Lower urinary tract function and pharmacologic management of lower urinary tract dysfunction. Urol Clin North Am. 1987 May. 14(2):273-96. [QxMD MEDLINE Link]. Chen Y, DeVivo MJ, Roseman JM. Current trend and risk factors for kidney stones in persons with spinal cord injury: a longitudinal study. Spinal Cord. 2000 Jun. 38 (6):346-53. [QxMD MEDLINE Link]. [Full Text]. Kalisvaart JF, Katsumi HK, Ronningen LD, Hovey RM. Bladder cancer in spinal cord injury patients. Spinal Cord. 2010 Mar. 48 (3):257-61. [QxMD MEDLINE Link]. [Full Text]. Wittig L, Carlson KV, Andrews JM, Crump RT, Baverstock RJ. Diabetic Bladder Dysfunction: A Review. Urology. 2019 Jan. 123:1-6. [QxMD MEDLINE Link]. [Full Text]. Duncan PW, Zorowitz R, Bates B, et al. Management of Adult Stroke Rehabilitation Care: a clinical practice guideline. Stroke. 2005 Sep. 36(9):e100-43. Neville CE, Beneciuk J, Bishop M, Alappattu M. Analysis of Physical Therapy Intervention Outcomes for Urinary Incontinence in Women Older Than 65 Years in Outpatient Clinical Settings. Top Geriatr Rehabil. 2016 Oct-Dec. 32 (4):251-7. [QxMD MEDLINE Link]. [Full Text]. Chaudhry R, Balsara ZR, Madden-Fuentes RJ, et al. Risk Factors Associated with Recurrent Urinary Tract Infection in Neurogenic Bladders Managed by Clean Intermittent Catheterization. Urology. 2017 Jan 5. [QxMD MEDLINE Link]. Giannantoni A, Di Stasi SM, Stephen RL, et al. Intravesical capsaicin versus resiniferatoxin in patients with detrusor hyperreflexia: a prospective randomized study. J Urol. 2002 Apr. 167(4):1710-4. [QxMD MEDLINE Link]. de Sèze M, Wiart L, Joseph PA, et al. Capsaicin and neurogenic detrusor hyperreflexia: a double-blind placebo-controlled study in 20 patients with spinal cord lesions. Neurourol Urodyn. 1998. 17(5):513-23. [QxMD MEDLINE Link]. de Seze M, Wiart L, de Seze MP, Soyeur L, Dosque JP, Blajezewski S, et al. Intravesical capsaicin versus resiniferatoxin for the treatment of detrusor hyperreflexia in spinal cord injured patients: a double-blind, randomized, controlled study. J Urol. 2004 Jan. 171 (1):251-5. [QxMD MEDLINE Link]. Gutierrez-Martin P, Virseda-Chamorro M, Salinas Casado J, et al. Factors that influence the urodynamic results of botulinum toxin in the treatment of neurogenic hyperactivity. Actas Urol Esp. 2015 Jan 9. [QxMD MEDLINE Link]. Komesu YM, Amundsen CL, Richter HE, et al. Refractory urgency urinary incontinence treatment in women: impact of age on outcomes and complications. Am J Obstet Gynecol. 2018 Jan. 218 (1):111.e1-e9. [QxMD MEDLINE Link]. Faure Walker NA, Syed O, Malde S, Taylor C, Sahai A. Onabotulinum toxin A Injections in Men With Refractory Idiopathic Detrusor Overactivity. Urology. 2019 Jan. 123:242-6. [QxMD MEDLINE Link]. Basu M, Khullar V, Duckett J. Urethral dilatation: Is there any benefit over cystoscopy and distension? A randomized trial in women with overactive bladder symptoms. Neurourol Urodyn. 2014 Mar. 33 (3):283-8. [QxMD MEDLINE Link]. Abedi A, Ojeda LM, Montero S, et al. Long-Term Decision Regret and Associated Factors Following Urinary Reconstruction in Underserved Patients with Spinal Cord Injury. 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The possible role of percutaneous tibial nerve stimulation using adhesive skin surface electrodes in patients with neurogenic detrusor overactivity secondary to spinal cord injury. Int Urol Nephrol. 2015 Jan 22. [QxMD MEDLINE Link]. Vecchio M, Chiaramonte R, DI Benedetto P. Management of bladder dysfunction in multiple sclerosis: a systematic review and meta-analysis of studies regarding bladder rehabilitation. Eur J Phys Rehabil Med. 2022 Jun. 58 (3):387-96. [QxMD MEDLINE Link]. [Full Text]. Parittotokkaporn S, Varghese C, O'Grady G, Lawrence A, Svirskis D, O'Carroll SJ. Transcutaneous Electrical Stimulation for Neurogenic Bladder Dysfunction Following Spinal Cord Injury: Meta-Analysis of Randomized Controlled Trials. Neuromodulation. 2021 May 19. [QxMD MEDLINE Link]. Maltagliati M, Sampogna G, Citeri M, et al. A consecutive series of patients undergoing trans-urethral cystolithotripsy with ballistic lithotripsy by a tertiary referral center for neurogenic bladder. Acta Biomed. 2020 Jul 15. 91 (4):e2020112. [QxMD MEDLINE Link]. [Full Text]. Kuhlemeier KV, Lloyd LK, Stover SL. Long-term followup of renal function after spinal cord injury. J Urol. 1985 Sep. 134(3):510-3. [QxMD MEDLINE Link]. Stein R, Bogaert G, Dogan HS, et al. EAU/ESPU guidelines on the management of neurogenic bladder in children and adolescent part I diagnostics and conservative treatment. Neurourol Urodyn. 2020 Jan. 39 (1):45-57. [QxMD MEDLINE Link]. Stein R, Bogaert G, Dogan HS, et al. EAU/ESPU guidelines on the management of neurogenic bladder in children and adolescent part II operative management. Neurourol Urodyn. 2020 Feb. 39 (2):498-506. [QxMD MEDLINE Link]. Gotoh M, Kobayashi T, Sogabe K. Impact of symptom improvement on patients' bother and quality of life in female patients with overactive bladder treated by solifenacin (SET-Q). Int J Urol. 2014 May. 21 (5):505-11. [QxMD MEDLINE Link]. [Full Text]. Media Gallery The pons is a major relay center between the brain and bladder. The mechanical process of urination is coordinated by the pons in an area known as the pontine micturition center (PMC). Large stellate urinary bladder stone. Image courtesy of Wikimedia Commons. of 2 Tables Back to List Contributor Information and Disclosures Author Gregory T Carter, MD, MS Chief Medical Officer and PM&R Residency Program Director, Providence St Luke's Rehabilitation Medical Center; Clinical Professor, Department of Medical Education and Clinical Sciences, Fellow, Gleason Institute for Neuroscience Gregory T Carter, MD, MS is a member of the following medical societies: American Academy of Physical Medicine and Rehabilitation, American Association of Neuromuscular and Electrodiagnostic Medicine, Association of Academic Physiatrists Disclosure: Nothing to disclose. Coauthor(s) Kate E Whiteneck, DO Chief Resident Physician, Department of Physical Medicine and Rehabilitation, Providence St Luke’s Rehabilitation Medical Center Disclosure: Nothing to disclose. Chief Editor Elizabeth A Moberg-Wolff, MD Medical Director, Pediatric Rehabilitation Medicine Associates Elizabeth A Moberg-Wolff, MD is a member of the following medical societies: American Academy for Cerebral Palsy and Developmental Medicine, American Academy of Physical Medicine and Rehabilitation Disclosure: Nothing to disclose. Acknowledgements Andrew C Krouskop, MD Assistant Professor and Chair, Department of Orthopedics, Division of Physical Medicine and Rehabilitation, University of Tennessee College of Medicine at Chattanooga Disclosure: Nothing to disclose. Ramon S Lansang Jr, MD Consulting Staff, Department of Orthopedics, Charleston Area Medical Center Ramon S Lansang Jr, MD is a member of the following medical societies: American Academy of Pediatrics, American Academy of Physical Medicine and Rehabilitation, and American Medical Association Disclosure: Nothing to disclose. Teresa L Massagli, MD Professor of Rehabilitation Medicine and Pediatrics, University of Washington School of Medicine Teresa L Massagli, MD is a member of the following medical societies: American Academy of Pediatrics, American Academy of Physical Medicine and Rehabilitation, and Association of Academic Physiatrists Disclosure: Nothing to disclose. Richard Salcido, MD Chairman, Erdman Professor of Rehabilitation, Department of Physical Medicine and Rehabilitation, University of Pennsylvania School of Medicine Richard Salcido, MD is a member of the following medical societies: American Academy of Pain Medicine, American Academy of Physical Medicine and Rehabilitation, American College of Physician Executives, American Medical Association, and American Paraplegia Society Disclosure: Nothing to disclose. Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Medscape Reference Salary Employment Close;) What would you like to print? What would you like to print? 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15444	https://askfilo.com/user-question-answers-smart-solutions/calculate-the-percentage-shift-in-the-1s-to-2p-transition-3335313931333736	Question asked by Filo student Calculate the percentage shift in the 1s→2p transition energy for a (μ−π+) exotic atom due to the finite size of the π+'s spherical shell charge distribution (radius r0=1fm), compared to the point-charge case. Use first-order perturbation theory with the given parameters: mμ=206.77me, mπ=273.2me, ℏ=c=1, and α≈1371. Express your answer as a numerical percentage. Also mention the reference from where you chose the solution. Views: 5,267 students Updated on: Jun 3, 2025 Text SolutionText solutionverified iconVerified Concepts Hydrogen-like atoms, reduced mass, Bohr radius, first-order perturbation theory, finite-size correction, dimensionless scaling Explanation We estimate the fractional shift in the transition energy due to the non–pointlike charge distribution in the nucleus (here, the π+) by comparing the finite size correction in the 1s state with the unperturbed 1s→2p transition energy. Since the 2p state has negligible probability density at the origin, only the 1s energy gets a correction. In leading order perturbation theory the correction, expressed as a fraction of the unperturbed splitting, scales as (r0/a0)2 (with a numerical coefficient). For a hydrogenic system the Bohr radius is a0=μα1, so that (r0/a0)2=(μαr0)2. The finite size correction to the energy is known to give a fractional shift ΔE0δE1s≈916(μαr0)2. The percentage shift is then Percentage shift≈916(μαr0)2×100%. Step 1 Calculate the reduced mass. Given mμ=206.77me,mπ=273.2me, we have μ=mμ+mπmμmπ≈117.8me. Step 2 Determine the Bohr radius for a hydrogen-like system: a0=μα1. Using me=0.511MeV (so that μ≈117.8me≈60.2MeV) and α≈1371, we get (in natural units with ℏ=c=1): a0=60.2/1371=60.2137≈2.275MeV−1. To convert to femtometers (fm), recall that 1MeV−1≈197.3fm. Thus a0≈2.275×197.3≈448.8fm. Step 3 The given π+ charge radius is r0=1fm. Then the ratio is a0r0≈448.8fm1fm≈0.00223. Step 4 Plug into the fractional energy shift expression: ΔE0δE1s≈916(μαr0)2=916(a0r0)2. Computing the square: (a0r0)2≈(0.00223)2≈4.97×10−6. Then ΔE0δE1s≈916×4.97×10−6≈1.78×4.97×10−6≈8.84×10−6. Expressed as a percentage, the shift is Percentage shift≈8.84×10−6×100%≈8.84×10−4%. Final Answer The percentage shift in the 1s→2p transition energy due to the finite size of the π+ (with r0=1fm) is approximately 8.84×10−4%. Reference: Griffiths, Introduction to Quantum Mechanics, Example on finite-size corrections using perturbation theory; similar treatments appear in advanced quantum mechanics texts on exotic atoms. Students who ask this question also asked Views: 5,530 Topic: Smart Solutions View solution Views: 5,075 Topic: Smart Solutions View solution Views: 5,836 Topic: Smart Solutions View solution Views: 5,277 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| Calculate the percentage shift in the 1s→2p transition energy for a (μ−π+) exotic atom due to the finite size of the π+'s spherical shell charge distribution (radius r0=1fm), compared to the point-charge case. Use first-order perturbation theory with the given parameters: mμ=206.77me, mπ=273.2me, ℏ=c=1, and α≈1371. Express your answer as a numerical percentage. Also mention the reference from where you chose the solution. \| \| Updated On \| Jun 3, 2025 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 12 \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
15445	https://utppublishing.com/doi/book/10.3138/9781487582586	Lattice Path Combinatorics with Statistical Applications; Mathematical Expositions 23 - University of Toronto Press Search Anywhere Anywhere This Book Quick Search anywhere Enter words / phrases / DOI / ISBN / keywords / authors / etc Search Quick Search in Books Enter words / phrases / DOI / ISBN / keywords / authors / etc Search Advanced Search 0 Skip main navigationOpen Drawer Menu Close Drawer Menu Home About Books Browse Imprints Series Subjects Collections Journals Publish with Us Book Authors Journals and Societies Call for Papers Resources Librarian Resources Bookseller Resources Orders and Subscriptions Rights and Permissions Editorial Policies Open Access Events Contact Lattice Path Combinatorics with Statistical Applications; Mathematical Expositions 23 T.V. Narayana Share Share on Facebook X LinkedIn Copy Link Email HomeHeritageLattice Path Combinatorics with Statistical Applications; Mathematical Expositions 23BUY THIS BOOK Description Details Author Lattice path combinatorics has developed greatly as a branch of probability studies recently, and the need for new books on the subject is obvious. It treats several recent results and it offers a powerful new tool for studying many problems in mathematical statistics. Lattice path combinatorics has developed greatly as a branch of probability studies recently, and the need for new books on the subject is obvious. The present monograph, by one who has made significant contributions to combinatorics and its applications to probability and statistics, will be useful to research workers, teachers, professional statisticians, and advanced students alike. It treats several recent results and it offers a powerful new tool for studying many problems in mathematical statistics. The emphasis in the five chapters is on ‘dominance.’ From a consideration of exceedances in the lattice path problem, the text goes on to provide solutions to tests of hypotheses and simple sampling plans, displaying the usefulness of Young chains in the enumeration of the latter. The fourth chapter, on knock-out tournaments, represents one approach to paired comparisons quite close in spirit to dominance and lattice path combinatorics, and the final chapter considers the advantages of using combinatorial methods in statistical problems (including the Frame-Robinson-Thrall theorem to derive properties of non-parametric tests) and mentions current trends of research. Numerous examples, exercises, and references round out the text. Imprint: University of Toronto Press Published: December 1979 Pages: 120 pdf: 9781487582586 paperback: 9781487587284 T.V. NARAYANA is a member of the Department of Mathematics at the University of Alberta. Chapters PDF PREFACE I LATTICE PATH PROBLEMS AND VECTORS OF INTEGERS II THE DOMINANCE THEOREM AND SMIRNOV TEST-STATISTICS III SOME APPLICATIONS OF DOMINANCE TO STATISTICAL PROBLEMS IV THE COMBINATORICS OF KNOCK-OUT TOURNAMENTS V A MISCELLANY OF FURTHER RESEARCH PROBLEMS APPENDIX On some convolution identities from lattice path combinatorics NOTES AND SOLUTIONS SUPPLEMENTARY BIBLIOGRAPHY INDEX Recommend to Library Request a Desk Copy Favourite Purchase Options PurchaseItem saved, go to cart eBook $24.95 Add to cart eBook Checkout Related Titles ##### Mathematical preliminaries ###### Weighting Evidence in Language and Literature October 2024 ##### The Form of Mathematical Inference 1 ###### Shorter Papers October 2024 ##### The Development and Limits of Mathematical Logic 1 ###### Phenomenology and Logic October 2024 ##### The General Character of Mathematical Logic 1 ###### Phenomenology and Logic October 2024 ##### Second Fundamental Form. 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15446	https://pmc.ncbi.nlm.nih.gov/articles/PMC275000/	Oxygen tolerance of fresh clinical anaerobic bacteria - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice J Clin Microbiol . 1975 Feb;1(2):161–164. doi: 10.1128/jcm.1.2.161-164.1975 Search in PMC Search in PubMed View in NLM Catalog Add to search Oxygen tolerance of fresh clinical anaerobic bacteria. F P Tally F P Tally Find articles by F P Tally , P R Stewart P R Stewart Find articles by P R Stewart , V L Sutter V L Sutter Find articles by V L Sutter , J E Rosenblatt J E Rosenblatt Find articles by J E Rosenblatt Copyright and License information PMC Copyright notice PMCID: PMC275000 PMID: 1176601 Abstract The oxygen tolerance and sensitivity of 57 freshly isolated anaerobic bacteria from clinical specimens was studied. All the organisms tolerated 8 h or more of exposure to oxygen in room air. Growth of the isolates in increasing oxygen concentrations demonstrated that the 57 isolates varied in oxygen sensitivity from strict to aerotolerant anaerobes. Comparison of the oxygen tolerance and sensitivity showed that the most tolerant organisms (best survival after prolonged exposure) included anaerobes capable of growth at only 0.4% or less O2 (strict) as well as those able to grow in as much as 10% O2. The least tolerant were predominately strict anaerobes. Decrease in the inoculum size from a concentration of 10(8) to 10(6) colony-forming units per ml had only a minor effect. The data indicate that the brief oxygen exposure with bench techniques in clinical laboratories would not be deleterious to the anaerobic bacteria present in clinical specimens. Full text PDF Previous 161 162 163 164 Next Selected References These references are in PubMed. This may not be the complete list of references from this article. Arank A., Syed S. A., Kenney E. B., Freter R. Isolation of anaerobic bacteria from human gingiva and mouse cecum by means of a simplified glove box procedure. Appl Microbiol. 1969 Apr;17(4):568–576. doi: 10.1128/am.17.4.568-576.1969. [DOI] [PMC free article] [PubMed] [Google Scholar] Attebery H. R., Finegold S. M. Combined screw-cap and rubber-stopper closure for Hungate tubes (pre-reduced anaerobically sterilized roll tubes and liquid media). Appl Microbiol. 1969 Oct;18(4):558–561. doi: 10.1128/am.18.4.558-561.1969. [DOI] [PMC free article] [PubMed] [Google Scholar] Gorbach S. L., Bartlett J. G. Anaerobic infections. 1. N Engl J Med. 1974 May 23;290(21):1177–1184. doi: 10.1056/NEJM197405232902106. [DOI] [PubMed] [Google Scholar] Hill G. B., Osterhout S. Experimental effects of hyperbaric oxgen on selected clostridial species. I. In-vitro studies. J Infect Dis. 1972 Jan;125(1):17–25. doi: 10.1093/infdis/125.1.17. [DOI] [PubMed] [Google Scholar] Loesche W. J. Oxygen sensitivity of various anaerobic bacteria. Appl Microbiol. 1969 Nov;18(5):723–727. doi: 10.1128/am.18.5.723-727.1969. [DOI] [PMC free article] [PubMed] [Google Scholar] Martin W. J. Isolation and indentification of anaerobic bacteria in the clinical laboratory. A 2-year experience. Mayo Clin Proc. 1974 May;49(5):300–308. [PubMed] [Google Scholar] McCord J. M., Keele B. B., Jr, Fridovich I. An enzyme-based theory of obligate anaerobiosis: the physiological function of superoxide dismutase. Proc Natl Acad Sci U S A. 1971 May;68(5):1024–1027. doi: 10.1073/pnas.68.5.1024. [DOI] [PMC free article] [PubMed] [Google Scholar] Rosenblatt J. E., Fallon A., Finegold S. M. Comparison of methods for isolation of anaerobic bacteria from clinical specimens. Appl Microbiol. 1973 Jan;25(1):77–85. doi: 10.1128/am.25.1.77-85.1973. [DOI] [PMC free article] [PubMed] [Google Scholar] Starr S. E., Thompson F. S., Dowell V. R., Jr, Balows A. Micromethod system for identification of anaerobic bacteria. Appl Microbiol. 1973 May;25(5):713–717. doi: 10.1128/am.25.5.713-717.1973. [DOI] [PMC free article] [PubMed] [Google Scholar] Sutter V. L., Kwok Y. Y., Finegold S. M. Standardized antimicrobial disc susceptibility testing of anaerobic bacteria. I. Susceptibility of Bacteroides fragilis to tetracycline. Appl Microbiol. 1972 Feb;23(2):268–275. doi: 10.1128/am.23.2.268-275.1972. [DOI] [PMC free article] [PubMed] [Google Scholar] Articles from Journal of Clinical Microbiology are provided here courtesy of American Society for Microbiology (ASM) ACTIONS View on publisher site PDF (586.8 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Full text Selected References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15447	https://geekymedics.com/extrapyramidal-side-effects-epse-examination-osce-guide/	Save 15% on our Everything Bundle with discount code EVERY15 🎓 Login Register Extrapyramidal Side Effects (EPSE) Examination – OSCE Guide Access all our resources with a subscription ⚡️ Supercharge your learning with access to OSCE stations, virtual patients, question banks, flashcards and more. Learn more ✨ This guide discusses how to approach performing a focused examination of a patient to identify extrapyramidal side effects (EPSE) from antipsychotic medications in an OSCE setting. Background The extrapyramidal system plays a critical role in fine-tuning motor control, influencing speech, facial expressions, posture, and spontaneous motor activity.1 Extrapyramidal side effectsare associated with antipsychotic medications and are a type of drug-induced movement disorder. They are most commonly associated with typical antipsychotics (e.g. haloperidol, chlorpromazine). These medications often cause EPSEs due to their antagonistic effect on dopamine D2 receptors, disrupting the normal balance of neurotransmission in the basal ganglia, which is integral to controlling movement and coordination.2 However, it is crucial to recognise that other pharmaceutical agents, including certain antidepressants, lithium, anticonvulsants, and antiemetics, can also precipitate these effects.2,3 EPSEs may be divided into two broad categories: hypokinetic (may resemble Parkinson’s disease) and hyperkinetic (may resemble Huntington’s disease).2 There are four common extrapyramidal side effects. These can be remembered through the ADAPT mnemonic:4 Acute Dystonia Akathisia Parkinsonism Tardive dyskinesia Parkinson’s disease vs Huntington’s disease Parkinson’s disease and Huntington’s disease are two motor pathologies that both affect the basal ganglia. Parkinson’s disease results from degenerations of neurons in the substantia nigra, resulting in a dopamine deficiency in the basal ganglia. It is characterised by the mnemonic TRAP:5 Tremor (rest) Rigidity (cogwheel) Akinesia (or bradykinesia) Postural instability Huntington’s disease results from neuronal degeneration in the basal ganglia and cerebral cortex. It is characterised by choreiform movements and cognitive decline. Hypokinetic EPSEs Acute dystonia Acute dystonia is characterised by involuntary muscle contractions leading to abnormal postures or repetitive movements, affecting various body parts. This can be incredibly painful and is thus important to treat immediately. It typically presents within 1 – 5 days after drug exposure. It may affect any skeletal muscle, including the back and extremities (opisthotonos), neck (torticollis), jaw (trismus), eyes (oculogyric crisis), abdominal wall, pelvic muscles (tortipelvic crisis), facial, and tongue muscles (buccolingual crisis).2,4 Intramuscular administration of antimuscarinic (e.g. benztropine) or diphenhydramine may relieve dystonia in minutes.4 Parkinsonism Parkinsonism is a clinical syndrome of resting tremor, cogwheel rigidity and general slowing of motor functions. Notable characteristics include masked facies, a stooped posture, and a slow, shuffling gait.2,4 Patients often experience gait imbalance and difficulty rising from a seated position. Although the most common cause of parkinsonism is Parkinson’s disease, antipsychotics (drug-induced parkinsonism) and other conditions can also cause parkinsonism due to interfering with the dopaminergic pathway. Management of drug-induced parkinsonism involves discontinuation or dose reduction of the causative medication. Patients can be switched to an atypical antipsychotic. In some cases, medications used for Parkinson’s disease can be used, including amantadine, antimuscarinic agents (e.g. benztropine), dopamine agonists and levodopa.4 Hyperkinetic EPSEs Akathisia Akathisia is characterised by a subjective feeling of internal restlessness and a compelling urge to move, leading to the objective observation of repetitive movements comprising leg crossing, swinging, or shifting from one foot to another.2,4 It typically presents within four weeks of starting or increasing the dosage of the offending medication. Misdiagnosis is common due to its non-specific symptoms that are often difficult to describe by patients and can often be confused with anxiety or agitation, leading to inadvertent exacerbation of the condition when antidepressant or antipsychotic medications are increased. Withdrawal akathisia can also occur with the discontinuation or dose reduction of antipsychotic medications and is typically self-limited, often resolving within six weeks.2,4 Administration of antimuscarinic (e.g. benztropine) or diphenhydramine may relieve akathisia. If caused by a first-generation antipsychotic, patients can be switched to an atypical antipsychotic. Additional therapeutic strategies specific to akathisia include beta-blocker (ex: propranolol), amantadine, clonidine, benzodiazepines and mirtazapine.4 Differentiating akathisia from anxiety and agitation Akathisia, anxiety and agitation all cause restlessness. However, it is important to differentiate these as they have different treatments. Confusing akathisia for agitation could lead to the inappropriate administration of antipsychotics, which would further exacerbate the symptoms. Anxiety is a feeling of worry or unease. It may present with physical signs such as nervousness, restlessness, tension, tachycardia, sweating and gastrointestinal discomfort. However, anxiety primarily affects thoughts and feelings rather than directing a particular movement. The restlessness associated with anxiety is generally less intense than akathisia and is less driven by an urge to move. Agitation is a state of increased arousal or restlessness, which can manifest as both psychological and physical symptoms. Individuals may present with irritability, anger or aggression that may include verbal outbursts. It does not typically arise as a side effect of medications and is common in conditions such as dementia, manic episodes of bipolar disorder, depression, and psychosis. It is also associated with substance use. In akathisia, the restlessness is largely driven by an internal sensation that is predominantly physical rather than emotional. Tardive dyskinesia Tardive dyskinesia is characterised by involuntary, choreoathetoid movements that primarily affect the orofacial and tongue muscles, extending to the trunk and extremities.2,4 Symptoms are typically painless but may impede social interactions and impact daily functions such as chewing, swallowing, and speaking. It is crucial to detect and treat early as there are cases where symptoms can become irreversible. Tardive dyskinesia may improve temporarily with an increase in antipsychotic medication and temporarily worsen with a decrease. Tardive dyskinesia is much less common with modern antipsychotics, and the incidence is less common today. It is also important to note that it may be seen in other populations, such as patients with dementia. Treatment options include reducing the dose or discontinuing the causative medication and possibly switching to an atypical antipsychotic. Other interventions include dopamine-depleting medications such as tetrabenazine.4 Neuroleptic malignant syndrome (NMS) NMS is a life-threatening reaction to medications with dopamine receptor-antagonist properties characterised by altered mental status, fever, muscle rigidity and autonomic dysfunction.6 Patients typically develop NMS within hours or days after exposure to causative drugs, with most cases falling between 2 weeks and 30 days. The clinical course typically begins with muscle rigidity followed by fever within several hours of onset and mental status changes that can range from mild drowsiness, agitation or confusion to delirium or coma. NMS is accompanied by autonomic instability, including labile blood pressure, tachypnoea, tachycardia, excessive salivation (sialorrhea), sweating (diaphoresis), flushing, skin pallor and incontinence.6 Immediate management is critical. The causative medication should be discontinued. General management includes aggressive hydration, cooling blankets to manage hyperthermia, and correcting metabolic abnormalities. Bromocriptine (dopamine agonist) and dantrolene (muscle relaxant) can be used in severe cases.6 Explore our premium collection of 1,300+ OSCE stations, including clinical examination stations ✨ Introduction Wash your hands and don PPE if appropriate. Introduce yourself to the patient, including your name and role. Confirm the patient’s name and date of birth. Briefly explain what the examination will involve using patient-friendly language. Gain consent to proceed with the examination. Ask the patient if they have any pain before proceeding with clinical examination. Painful muscle spasms may occur in acute dystonia General inspection Most EPSEs can be observed through observation alone. Perform a brief general inspection of the patient, looking for signs of EPSEs: Pain: acute dystonia can be incredibly painful Difficulty breathing, swallowing, or speaking could indicate signs of acute dystonia. Posture: abnormal posture, such as abnormal muscular contractions, twisted neck (torticollis), back (opisthotonus) or jaw (trismus) that may be seen in acute dystonia Facial expression: look for blunted affect or grimacing. Blunted affect may present as reduced emotional expression, while grimacing could manifest as involuntary contractions of facial muscles Blink rate: observe the patient’s blink rate, noting that less than ten blinks per minute could indicate a hypokinetic state, while more than 30 could suggest a hyperkinetic condition. Vigorous blinking could be a sign of a Tic. Motor restlessness (akathisia): akathisia presents as a subjective urge to move, coupled with observable excessive movement. Patients may engage in excessive pacing or vocalise the need to move. Abnormal involuntary movements: these can be seen in tardive dyskinesia. Look for chorea (sudden, quick involuntary movements in any skeletal muscle), athetosis (slow, writhing movements), and tics (sudden, quick, patterned movements). While chorea could affect any skeletal muscle, athetosis is noticeable mainly in the shoulders, fingers, and lower face. Common tics include throat clearing, vigorous blinking, sniffing, and shaking head. Differentiating a blunted affect from mask-like facies Mask-like facies and affective blunting both cause changes in facial expressions that may be picked up in a physical examination. Affective blunting is a reduction in the expression and experience of emotions. This may also be associated with reduced vocal inflection (prosody), decreased expressive gestures, and may appear emotionally indifferent. Their reactions to both positive and negative events may be severely limited or absent. Affective blunting is generally more common in psychiatric disorders such as schizophrenia and depression. Mask-like facies (hypomimia) is a reduction in facial expressivity and is a motor symptom. Mask-like facies is more common in patients with parkinsonism and is generally associated with other signs such as rigidity and bradykinesia. Unlike affective blunting, a patient with mask-like facies may experience emotions and display appropriate emotional responses in conversation. Tremor Observe for a resting tremor associated with parkinsonism, characterised by involuntary movements in a limb at rest, typically slowing down at around 4 – 6 cycles per second. A prominent tremor in the second digit that causes it to rub against the palmar surface of the thumb may be observed (“pill-rolling tremor”). A resting tremor can be confused with a postural tremor and intention tremor: Postural tremors appear promptly with unsupported postures (such as in extended hands) and are more rapid than resting tremors. It is the most common tremor seen in psychiatry and can be attributed to a range of causes, such as essential tremor, anxiety, caffeine, theophylline, lithium, valproate, beta-adrenergic agonists, and/or withdrawal from alcohol, benzodiazepines, or barbiturates. Intention tremor is a broad, coarse, low-frequency tremor seen when a body part approaches a target. It consists of rhythmic shifting from one side to the other of the target (dysmetria). Gait Sitting to standing Ask the patient tostand from their seated position with their arms across their chest to screen for postural instability. Make sure to stand close to the patient so that you can intervene if they lose their balance. Imbalance and difficulty rising from a seated position may also be present in parkinsonism. Observe gait Ask the patient to walk to the end of the examination room and then turn and walk back whilst you observe their gait: Parkinsonism gait will have short steps, diminished arm swinging, and rushed (festinating). When turning, several steps are required. Patients may appear unable to rotate at the waist (en bloc turning”). The patient may get stuck (freeze). Tone Assess tone in the muscle groups of the neck, upper limbs, and lower limbs, comparing each side. Looking for rigidity(parkinsonism) and dystonia. Before beginning, it is important to explain to the patient what you will be doing and why. Rigidity is a velocity-independent increase in muscle tone associated with parkinsonism. During the assessment, it is crucial to move the limbs both quickly and slowly to observe any differences in resistance. Cogwheel rigidity: characterised by a ratcheting sensation, often due to an underlying tremor superimposed on hypertonia Lead pipe rigidity: a uniform increase in muscle tone throughout the movement, often associated with neuroleptic malignant syndrome Dystonia may manifest as sustained muscle contractions or intermittent spasms. Pay attention to any unintended contractions or abnormal postures. Dystonia more often affects the axial muscles, causing torticollis or rotation of the lower spine. Assessment Neck 1. Support the patient’s head gently. 2. Ask the patient to relax and allow you to control the movement. 3. Passively rotate the head towards the shoulder, observing for rigidity or dystonia. Upper limb 1. Support the patient’s arm. 2. Guide through circumduction of the shoulder, flexion/extension of the elbow, and wrist circumduction. 3. Compare the tone on both sides, noting any asymmetry or unilateral abnormalities. Lower limb 1. Roll each leg on the examination bed, observing for any resistance or abnormal postures. 2. Lift each knee briskly off the bed while ensuring the heel remains in contact with the bed. A rise in the knee with the heel lifting off could indicate increased tone. 3. Compare the tone on both sides, noting any asymmetry or unilateral abnormalities. To complete the examination Explain to the patient that the examination is now finished. Thank the patient for their time. Dispose of PPE appropriately and wash your hands. Summarise your findings. Example summary “Today I performed a physical examination on a 29-year-old man who is taking antipsychotic medication for schizophrenia to assess clinical features of EPSE. On general inspection, the patient demonstrated a blunted affect and marked restlessness, getting up and pacing before and after the examination. A tremor was noted in his right hand that was present at rest and resolved with movement. Observation of gait was notable for bradykinesia, reduced arm swinging and hesitation when turning. Assessment of Tone revealed rigidity in the neck. The vitals were stable. These findings are consistent with the clinical features of EPSE. For completeness, I would like to perform the following further assessments and investigations.” Further assessments and investigations Consider completing a movement disorder assessment scale. Several scales can be used to evaluate different types of movement disorders:2,8 For assessing hypokinetic or parkinsonian signs, commonly used scales include the Unified Parkinson’s Disease Rating Scale motor exam section and the Simpson-Angus scale The Unified Huntington’s Disease Rating Scale and Abnormal Involuntary Movement Scale (AIMS) evaluate hyperkinetic or tardive-like signs The Barnes Akathisia Rating Scale (BARS) may be used to assess the presence and severity of akathisia The Extrapyramidal Symptoms Rating Scale addresses both hyperkinetic and hypokinetic signs, providing a comprehensive assessment tool for various movement disorders Perform a neurological and cerebellar examination if concerned about neurological pathology. Assess the drug chart for neuroleptics, dopamine-blocking antiemetics and sodium valproate. Reviewer Dr Aisling Campbell Consultant Psychiatrist University College Cork References Lee, J., & Muzio, M. R. (2022). Neuroanatomy, Extrapyramidal System. In StatPearls. StatPearls Publishing. Available from: [LINK] Sanders, R. D., & Gillig, P. M. (2012). Extrapyramidal examinations in psychiatry. Innovations in clinical neuroscience, 9(7-8), 10–16. Available from Blair, D. T., & Dauner, A. (1992). Extrapyramidal symptoms are serious side-effects of antipsychotic and other drugs. The Nurse practitioner, 17(11), 56–67. D’Souza, R. S., & Hooten, W. M. (2023). Extrapyramidal Symptoms. In StatPearls. StatPearls Publishing. Available from: [LINK] Jankovic J. (2008). Parkinson’s disease: clinical features and diagnosis. Journal of neurology, neurosurgery, and psychiatry, 79(4), 368–376. Berman B. D. (2011). Neuroleptic malignant syndrome: a review for neurohospitalists. The Neurohospitalist, 1(1), 41–47. Oregon Health & Science University. Abnormal Involuntary Movement Scale (AIMS). Available from: [LINK] Barnes T. R. (2003). The Barnes Akathisia Rating Scale–revisited. Journal of psychopharmacology (Oxford, England), 17(4), 365–370. 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15448	https://study.com/learn/lesson/average-speed-formula.html	Average Speed \| Definition, Formula & Calculation - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Math Courses ASSET Numerical Skills Test: Practice & Study Guide Average Speed \| Definition, Formula & Calculation Contributors: Claire Drigotas, Ria Yambao, Matthew Bergstresser Author Author: Claire Drigotas Show more Instructor Instructor: Ria Yambao Show more Expert Contributor Expert Contributor: Matthew Bergstresser Show more What is average speed and how is average speed calculated? Give the average speed equation, explain its terms and its application with some examples. Updated: 11/21/2023 Table of Contents Definition of Average Speed Average Speed Formula How to calculate average speed Using the Speed formula to calculate distance and time traveled. Average Speed Vs. Instantaneous Speed Case Studies for Calculating Average Speed Example Problems Lesson Summary Show FAQActivities How do u calculate average speed? You can calculate average speed by dividing the total distance traveled by the total time spent traveling. If there are intervals of travel, you must add up the individual distances and times to find total distance traveled and total time traveled. How do you find the average speed with distance and time? You can find the average speed of a traveling object based on its traveled distance and time spent traveling. First, add up the total distance traveled. Then divide this by the total time spent traveling. This will give you the average speed of the traveling object. When would you use the formula for average speed? You can use the average speed formula when you want to know how fast something traveled during a given period of time. You need the distance and time spent traveling in order to use the formula. Average Speed Modern cars have computers in them that complete all sorts of functions. One of them is to calculate the average speed of the vehicle. This might help in determining the gasoline consumption rate, or the quickest route to get to the grocery store. Let's get some extra practice by working some problems solving for the average speed of a vehicle. Problems A person driving to the store travels 200 meters in 300 seconds. Then they go to another store traveling an extra 400 meters in 500 seconds. What is their average speed? A truck delivering supplies to a construction site is late with the delivery. The foreman at the construction site will check the average speed of the truck when it gets there. If it isn't higher than 20 m/s they get a reduction in the cost of the delivery. The truck covered the firs 10000 meters of the trip in 600 seconds. How much time does the truck have to cover the remaining 4000 meters so its average speed is 20 m/s? Is this possible? A vehicle makes a trip to the store in 30 minutes. Its average speed was 30 m/s. How far did the car travel? Answers 1. 2. 3. Create an account LessonTranscript VideoQuizCourse Click for sound 7:41 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. 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Your next lesson will play in 10 seconds 0:02 What is Average Speed? 1:22 Calculating Average Speed 3:16 Examples 7:16 Lesson Summary QuizCourseView Video Only Save Timeline 1.3M views Recommended lessons and courses for you Related LessonsRelated Courses ##### Calculating Total Cost: Relationship with Units & Rate 5:23 ##### How to Solve a One-Step Problem Involving Rates 5:06 ##### Math Concept \| List, Facts & Examples 4:31 ##### Math Mnemonics Devices & Examples 5:21 ##### How to Solve Problems with Money 8:29 ##### Money Facts: Lesson for Kids 3:09 ##### Adding & Subtracting Money Amounts 4:29 ##### Taxes & Discounts: Calculations & Examples 8:07 ##### How to Calculate a Tip 7:08 ##### Solving Money Word Problems \| Strategies & Examples 4:31 ##### How to Write Amounts of Money: Lesson for Kids 3:14 ##### Dividing Money Amounts 5:14 ##### How to Solve Problems with Time 6:18 ##### Converting 9 pm GMT to EST 3:56 ##### Strategies for Teaching Time Problems 4:36 ##### Military Time \| Definition, Format & Use 4:53 ##### Time Measurement Units \| Types, Conversion & Examples 4:16 ##### Average Velocity \| Definition, Formula & Calculation 4:56 ##### Instantaneous Speed \| Definition, Formula & Examples 3:29 ##### Speed in Physics \| Overview, Formula & Calculation 3:53 ##### ELM: CSU Math Study Guide ##### THEA Study Guide and Test Prep ##### CUNY Assessment Test in Math: Practice & Study Guide ##### DSST Math for Liberal Arts Study Guide and Test Prep ##### GED Math: Quantitative, Arithmetic & Algebraic Problem Solving ##### LSAT Study Guide and Test Prep ##### SAT Subject Test Mathematics Level 1: Practice and Study Guide ##### NY Regents Exam - Integrated Algebra: Test Prep & Practice ##### DSST Business Mathematics Study Guide and Test Prep ##### CBEST Mathematics Study Guide - CBEST 096 Exam Prep ##### DSST Principles of Statistics Study Guide and Test Prep ##### ASVAB Mathematics Knowledge & Arithmetic Reasoning Study Guide and Test Prep ##### ORELA Middle Grades Mathematics Study Guide and Test Prep ##### WEST Middle Grades Mathematics (203) Study Guide and Test Prep ##### NMTA Middle Grades Mathematics (203): Practice & Study Guide ##### GACE Middle Grades Mathematics (013) Study Guide and Test Prep ##### OAE Middle Grades Mathematics (030) Study Guide and Test Prep ##### Smarter Balanced Assessments - Math Grade 6 Study Guide and Test Prep ##### Accuplacer Arithmetic Placement Study Guide and Test Prep ##### Smarter Balanced Assessments - Math Grade 7 Study Guide and Test Prep Definition of Average Speed --------------------------- To determine the average speed at which someone is traveling, it is important to understand both the concepts of average and speed alone. Average is a statistical term that is also often called mean. It is one way to calculate the approximate middle of a series of data points. To calculate the average of a series of points, one must add all the points together then divide by the number of points. Averages are used to describe many types of data. For example, the average test score is a common way to say how a class did on a test as a whole. The average of a group of numbers uses the same unit as the numbers themselves. For example, if one were to find the average between 7 different heights, the average would be represented in feet, inches just like the heights themselves. A v e r a g e S p e e d=(T o t a l D i s t a n c e T r a v e l e d)/(T i m e S p e n t T r a v e l i n g) Speed is a measure of how fast something is moving over a moment or a longer amount of time. When taken together, Average Speed is how fast something is moving overall during a given amount of time. It is measured by finding the total distance traveled by the moving object and the amount of total time the object spent moving. The SI unit for average speed is meters per second (or m/s). In the United States, using miles per hour (mph) is also quite common. The average speed formula can be used to calculate how fast a person must walk to finish a hike by sunset. Scalar And Vector Quantities Average speed is known as a scalar quantity. This means that it does not consider direction. The speed of a car does not consider the direction in which it is traveling (whether that be North, South, East, or West). Thus, speed always has a positive value. The speed of a car traveling 35mph North is 35mph just as it would be for the same car traveling South. It is important not to confuse average speed with average velocity. Velocity is a vector quantity, meaning that it does consider the direction of travel. For example, if a car traveling East has a velocity of 35mph, a car traveling in the opposite direction would have a velocity of -35mph. The average velocity of such a car would be lower than 35mph because of this travel in opposite directions. Thus, if a car changes directions, its speed does not change but its velocity does. When calculating average speed, it is important to make sure that one is not looking at velocity. Speed is a scalar quantity and should always be positive. Instantaneous Speed Speed can be considered in several different ways. One such way is through the idea of instantaneous speed. Instantaneous speed is how fast someone or something is moving at an exact moment. For example, a car could have an instantaneous speed of 35 miles per hour if at an instantaneous moment it is going 35mph. Calculating instantaneous speed can be fairly complicated because it must consider an exact moment. It involves using calculus. In contrast to instantaneous speed, average speed calculates the speed someone or something is going over a longer timeframe. It is calculated by dividing the total distance something travels by the total amount of time it spends traveling. For example, one can consider the car from earlier. If the car travels 70 miles in 2 hours, it would have an average speed of 70 / 2 = 35 miles per hour. It would thus have a speed of 35mph. This calculation does not tell any information about how fast the car was going at any moment during those 2 hours. Theoretically, the car could have been going 90mph for a while then 10mph for the rest of the trip. The description of 35mph only describes the average speed of the car over the course of the entire trip. To unlock this lesson you must be a Study.com memberCreate an account Average Speed Formula --------------------- The average speed equation is one simple way to determine the average speed of a traveling object. The formula for average speed is shown here and describes how to find average speed. A v e r a g e S p e e d=(D i s t a n c e T r a v e l e d)/(T i m e T r a v e l e d) To unlock this lesson you must be a Study.com memberCreate an account How to calculate average speed ------------------------------ Average speed is calculated by dividing the total distance something travels over the total amount of time it spends traveling. Direction is not considered in the total distance so all distances traveled should be positive numbers even if they are in different directions. For example, if a car travels 10 miles east in 1 hour, its average speed would be 10 miles / 1 hour or 10mph. This calculation does not say how fast the car was going at any given moment during the hour, just that its average speed was 10mph. Here is another example. If a car travels 50 miles East then 50 miles West over the course of 4 hours, one would have to add the distance traveled before finding the average speed. Both distances would be positive because speed is scalar and does not include a direction. The total distance the car traveled would be 100 miles, and the amount of time it spent traveling would be 4 hours. Thus, the car would have an average speed of 100 miles / 4 hours = 25 mph. To unlock this lesson you must be a Study.com memberCreate an account Using the Speed formula to calculate distance and time traveled. ---------------------------------------------------------------- The speed formula can also be rearranged to calculate the distance formula traveled and time traveled. Average speed is the distance traveled divided by the time traveled. The distance traveled is thus equal to the time traveled multiplied by the speed of travel. Time traveled is equal to the distance traveled divided by the speed of travel. D i s t a n c e T r a v e l e d=(T i m e T r a v e l e d)x(S p e e d o f T r a v e l) T i m e T r a v e l e d=(D i s t a n c e T r a v e l e d)/(S p e e d o f T r a v e l) The amount of time a car spends traveling can be found from its distance and speed. If a car travels 40 miles at an average speed of 10 mph, its time-traveled would be 40 miles / 10 mph = 4 hours. The distance a car travels can be found from its time and average speed of travel. For example, a car that travels 30 mph for 3 hours would travel 30 mph 3 h = 90 total miles. To unlock this lesson you must be a Study.com memberCreate an account Average Speed Vs. Instantaneous Speed ------------------------------------- In the article so far, the method to find instantaneous speed has not been presented. This has been done for two major reasons. Instantaneous speed is difficult to calculate. It involves the calculus concept of finding the derivative (or instantaneous slope) of a curve. Average speed is a more useful measurement of speed than instantaneous speed. Instantaneous speed gives a tiny snapshot of exactly how fast something is moving in a single instant. Average speed paints a broader picture of how fast an object is moving overall. This makes average speed much more useful for determining the length of a road trip or the average speed someone traveled during a race. To know who will win the race, the average speed is much more useful than instantaneous speed. To unlock this lesson you must be a Study.com memberCreate an account Case Studies for Calculating Average Speed ------------------------------------------ Case-1: For a single distance traveled for a single time duration In the case of a single distance traveled for a single time duration, one must simply divide the listed distance by the time listed. For example, if a child walked 10 miles in 5 hours, their average speed would be 10 miles / 5 hours = 2mph. Case-2: For various distances traveled in respectively given timings In the case of an object or person traveling several distances in several different timings, the calculation of average speed is slightly more complex. One must add all the individual distances traveled to find the total distance traveled and all the individual timings traveled to find the total travel time. Then, one divides total distance / total travel time. For example, if a car travels 100 miles in 1.5 hours and 20 miles in .5 hours, their total distance traveled would be 100+20 = 120 miles and their total time traveled would be 1.5 + .5 = 2 hours. Their average speed would thus be 120 miles / 2 hours = 60mph. Case-3: If an object travels with various given speeds The case of an object traveling at different speeds for different amounts of time is a more complicated average speed case. First, one must use the speed and time traveled in each interval to find the distance in said interval. Then, one must add up the distances traveled and divide by the total time traveled. For example, if a car goes for 4 hours at 40mph and 2 hours at 60 mph, the average speed can be calculated as such: Interval 1: 4 hours x 40 mph = 160 miles Interval 2: 2 hours x 60 mph = 120 miles Total distance = 160 miles + 120 miles = 280 miles Total time = 4 hours + 2 hours = 6 hours Average Speed = 280 miles / 6 hours = 46.7 mph To unlock this lesson you must be a Study.com memberCreate an account Example Problems ---------------- A runner races 27 miles in 3 hours, what was their average speed? A train leaves a station at 6:00 am and arrives at its destination at 10:00 am. What was its average speed if the trip was 400 miles long? A car goes 50mph for 30 minutes and 75 mph for one hour. What was its average speed? A child walks 2 miles at a speed of 4mph. How much time did they spend walking? A train leaves a station at 2:00 pm and arrives at its destination at 5:00 pm. It traveled at an average speed of 50mph. How far away are the station and the destination? Answers to Example Problems 27 miles / 3 hours = 9 mph Total time traveled = 10-6 = 4 hours. Total distance = 400 miles. Average speed = 400 miles / 4 hours = 100mph. Interval 1 distance = 50mph x .5 hours = 25 miles. Interval 2 distance = 75mph x 1 hour = 75 miles. Total distance = 100 miles. Total time = 1.5 hours. 100 miles / 1.5 hours = 66.7 mph. 2 miles / 4 mph= .5 hours (30 minutes) Time traveled = 5-2 = 3 hours. 3 hours x 50 mph = 150 miles. To unlock this lesson you must be a Study.com memberCreate an account Lesson Summary -------------- Key Ideas average speed can be used to describe how far an object travels during a given period of time average speed is measured as a scalar quantity and is always positive average speed is calculated as distance traveled / time traveled the average speed equation can be rearranged to solve for distance and time traveled average speed is a more useful speed measurement than instantaneous speed Vocabulary average - a statistical measurement of multiple data points in which the data points are added together and divided by the number of data points speed - a scalar representation of how far something travels in a given amount of time velocity - a vector representation of how far something travels in a given amount of time scalar- a quantity that is always positive and is only measured in one dimension velocity - a quantity that is measured in two dimensions and can be either positive or negative average speed equation - a formula used to determine the average speed of a traveling object instantaneous speed - the speed of an object in a specific moment Understanding the difference between instantaneous speed, average speed, and velocity is important in being able to use these calculations for their true purpose. Learning how to calculate average speed will allow one to determine the speed needed to arrive on time, or the time remaining on a long trip. To unlock this lesson you must be a Study.com memberCreate an account Video Transcript What is Average Speed? You and your friend decided to take your brand new red sports car out for a spin. Your car is capable of speeds up to 220 mph. You drove 45 miles in 1.25 hours. At the end of your trip, your friend tells you that your average speed during the trip is 36 miles per hour. You were appalled. You asked yourself, 'How can a sports car have such a pitiful average speed?' and tried to recall how much you paid for it. What is average speed anyway? The average speed of an object is the total distance traveled by the object divided by the elapsed time to cover that distance. It's a scalar quantity which means it is defined only by magnitude. A related concept, average velocity, is a vector quantity. A vector quantity is defined by magnitude and direction. For example, we might say that a car has an average speed of 25 miles per hour. Its average velocity might be 25 miles per hour due east. Average speed can be viewed as the rate of change in distance with respect to time. A car traveling at an average speed of 25 miles per hour covers an average distance of 25 miles every hour. Calculating Average Speed If an object travels with constant speed, then the formula for the speed of the object is given by, Total distance is the distance traveled by the object at the constant speed. Elapsed time is the time the object took to cover the total distance. In most instances, an object will travel with varying speeds over a certain distance. For example, a car traveling from one city to another will rarely move at a constant speed. It is more likely that the car's speed will fluctuate during the trip. The car might travel at 65 mph for some time and then slow down to 25 mph. It's possible that at certain times, the car is even at full stop (such as, at a red light). To calculate the car's average speed, we don't really care about the fluctuations in its speed. We only care about the total distance traveled by the car and the elapsed time to cover that distance. The formula for average speed is It's important to note that this formula is identical to that of constant speed. Average speed is measured in units of distance per time. Common units include miles per hour (mph), kilometers per hour (kph), meters per second (m/s), or feet per second (ft/s). As for your brand-new red sports car, your friend was exactly right in his calculation of the average speed. He used the distance traveled by the car (45 miles) divided by elapsed time (1.25 hours). The construction on the highway and the series of red lights on the local roads really slowed you down. The high elapsed time resulted in a low average speed. Examples Let's look at some other examples of average speed: Suppose a freight train travels a distance of 120 miles in 3 hours. What is the average speed of the train? Answer: Its average speed is Suppose a truck travels in segments that are described in the following table: \| Segment \| Distance (miles) \| Time (hours) \| --- \| 1 \| 30 \| 1 \| \| 2 \| 45 \| 2 \| \| 3 \| 50 \| 1 \| \| 4 \| 65 \| 2 \| What is the average speed of the truck? Answer: Based on the information given, its average speed over the four segments can be calculated as A car travels 50 mph on a trip from Chicago, IL, to Minneapolis, MN, and 65 mph on the return trip. What is the average speed of the car for the entire round trip? Answer: In this example, we are given two speeds. Suppose the distance traveled during the trip from Chicago to Minneapolis is D, then the distance traveled in the return trip is also D. The total distance for the entire trip is 2 D. The diagram below illustrates the situation. Using the formula for average speed, the elapsed time from Chicago to Minneapolis (t1) and back (t2), can be calculated separately. The calculations are shown below. Using the average speed formula, total distance and elapsed time, the average speed for the entire trip can be calculated as, Note that the average speed is not (50 + 65) / 2 = 57.5 mph! Remember that the average speed is defined as total distance traveled divided by elapsed time. The two elements we need to calculate average speed are: 1) Total distance traveled 2) Elapsed time to cover that distance The round-trip distance can be represented by 2 D (if we let one-way distance be D), and the round-trip elapsed time is (D / s1) + (D / s2) (since one-way elapsed time is one-way distance divided by one-way speed). A runner completes a race according to the distance-time graph shown below. What is the average speed of the runner for the first 10 seconds? What is the average speed of the runner for the entire race? Answer: The distance-time graph shows the distance traveled by the runner over elapsed time. Using the graph, we can see that the runner ran 0 m at the beginning of the race. After 10 seconds, he ran 75 meters. The runner's average speed for the first 10 seconds is calculated as: After 30 seconds, he completes the race and ran a total distance of 200 meters. The runner's average speed for the entire race is calculated as: Lesson Summary Let's review. The average speed of an object is the total distance traveled by the object divided by the elapsed time to cover that distance. To calculate the average speed of an object, we do not care about fluctuations in its speed. The key elements that we need are: 1) Total distance traveled 2) Elapsed time to cover that distance Points to Remember \| Average Speed \| \| The total distance traveled by an object divided by the elapsed time to cover that distance \| \| A scalar quantity defined only by magnitude \| \| Speed fluctuations don't matter \| Learning Outcomes When you are done, you should be able to: State the difference between average speed and average velocity Write the equation for calculating average speed Calculate the average speed of an object Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Create an account to start this course today Start today. Try it now ASSET Numerical Skills Test: Practice & Study Guide 16 chapters 116 lessons Chapter 1 ASSET Numerical Skills Test: Solving Word Problems in Math What is the Correct Setup to Solve Math Problems?: Writing Arithmetic Expressions 5:50 min Solving Word Problems: Steps & Examples 5:54 min Solving Word Problems with Multiple Steps 5:30 min Restating Word Problems Using Words or Images 4:32 min Personalizing a Word Problem to Increase Understanding 4:50 min How to Simplify Word Problems with Fractions Using Whole Numbers 3:38 min Chapter 2 ASSET Numerical Skills Test: Basic Math Operations Whole Number \| Definition, Properties & Examples 3:11 min Arithmetic with Whole Numbers 9:43 min What is Addition? \| Definition, Operation & Examples 4:07 min How to Perform Subtraction: Steps & Examples 3:46 min How to Perform Multiplication: Steps & Examples 5:22 min Multiplying Large Numbers \| Overview, Steps & Examples 7:43 min How to Perform Division: Steps & Examples 3:56 min Performing Long Division with Large Numbers: Steps and Examples 9:12 min Arithmetic Calculations with Signed Numbers 5:21 min Commutative Property \| Definition, Examples & Applications 3:53 min Associative Property \| Definition & Examples 4:28 min The Zero Property of Multiplication \| Definition & Examples 2:40 min What is the Greatest Common Factor? \| GCF Examples 4:56 min Least Common Multiple \| Definition, Formula & Examples 5:37 min Chapter 3 ASSET Numerical Skills Test: Using Decimals Decimal Place Value \| Definition & Examples 6:19 min Comparing & Ordering Decimals \| Definition & Examples 8:56 min Adding & Subtracting Decimals \| Steps, Examples & Word Problems 6:53 min Multiplying & Dividing Decimals \| Overview & Word Problems 5:29 min Solving Problems Using Decimal Numbers 6:57 min How to Estimate with Decimals to Solve Math Problems 8:51 min Decimal & Mixed Number Conversion \| Overview & Examples 5:44 min Chapter 4 ASSET Numerical Skills Test: Understanding Fractions Fraction \| Definition, Concept & Examples 6:20 min How to Raise and Reduce Fractions 6:17 min Relating Fractions and Decimals 6:32 min Finding the Least Common Denominator \| Overview & Examples 4:30 min Comparing & Ordering Fractions \| Method & Examples 7:33 min Using the Number Line to Compare Decimals, Fractions, and Whole Numbers 6:46 min Improper Fractions & Mixed Numbers \| Conversion & Examples 4:55 min How to Change Mixed Numbers to Improper Fractions 3:31 min Chapter 5 ASSET Numerical Skills Test: Operations with Fractions Adding & Subtracting Fractions \| Rules & Examples 4:14 min How to Add and Subtract Unlike Fractions and Mixed Numbers 6:46 min Multiplying Mixed Numbers \| Steps & Examples 7:23 min Dividing Fractions & Mixed Numbers \| Overview & Examples 7:12 min Addition, Subtraction, Multiplication, and Division with Decimal Notation 4:50 min Practice with Fraction and Mixed Number Arithmetic 7:50 min Estimation Problems using Fractions 7:37 min Solving Problems using Fractions and Mixed Numbers 7:08 min How to Solve Complex Fractions 5:20 min Chapter 6 ASSET Numerical Skills Test: Using Percents Percentage \| Definition & Calculation 4:20 min Converting Percent Notation to Fraction Notation 3:16 min Converting Fractions to Percents 3:43 min Changing Between Decimals and Percents 4:53 min Solve Problems Using Percents 7:50 min How to Solve Word Problems That Use Percents 6:30 min Chapter 7 ASSET Numerical Skills Test: Calculating Taxes, Discounts & Interest Taxes & Discounts: Calculations & Examples 8:07 min Simple Interest Problems \| Definition, Formula & Examples 6:05 min Compounding Interest \| Formula, Types & Examples 7:45 min Chapter 8 ASSET Numerical Skills Test: Ratios & Proportions Ratios & Rates \| Differences & Examples 6:37 min How to Find the Unit Rate 3:57 min How to Solve a One-Step Problem Involving Ratios 4:32 min How to Solve a One-Step Problem Involving Rates 5:06 min Proportion \| Definition, Formula & Types 6:05 min Calculations with Ratios and Proportions 5:35 min How to Find an Unknown in a Proportion 6:06 min Proportional Relationship \| Definition, Equation & Examples Graphing Proportional Relationships Proportional Relationships in Multistep Ratio & Percent Problems Constructing Proportions to Solve Problems \| Steps & Examples 5:20 min Chapter 9 ASSET Numerical Skills Test: Understanding Math Formulas Real Number \| Definition, Types & Examples 4:50 min Parentheses in Math \| Definition & Examples 4:01 min Order of Operations in Math \| Steps & Examples 5:50 min Common Math Formulas \| Overview, Uses & Importance 7:08 min Viewing now Average Speed \| Definition, Formula & Calculation 7:40 min Chapter 10 ASSET Numerical Skills Test: Factoring Prime & Composite Numbers Up next Factoring in Algebra \| Definition, Equations & Examples 5:32 min Watch next lesson Prime Number Definition & Examples 5:49 min Composite Number \| Definition, Types & Examples 4:38 min Prime & Composite Numbers \| Differences & Identification 5:07 min Finding the Prime Factorization of a Number \| Meaning & Examples 5:36 min Using Prime Factorizations to Find the Least Common Multiples 7:28 min Equivalent Expressions and Fraction Notation 5:46 min Using Fraction Notation: Addition, Subtraction, Multiplication & Division 6:12 min Factoring Out Variables: Instructions & Examples 6:46 min Combining Numbers and Variables When Factoring 6:35 min Transforming Factoring Into A Division Problem 5:11 min Factoring by Grouping \| Definition, Steps & Examples 7:46 min Chapter 11 ASSET Numerical Skills Test: Positive & Negative Numbers Signed Number \| Definition, Uses & Examples 5:04 min Positive Numbers \| Definition, Properties & Examples 3:33 min Introduction to Negative Numbers 6:27 min Adding & Subtracting Negative Numbers 4:35 min Multiplying & Dividing Negative Numbers 4:50 min Chapter 12 ASSET Numerical Skills Test: Absolute Value Absolute Value \| Explanation & Examples 4:42 min Solving Absolute Value Functions & Equations \| Rules & Examples 5:26 min Absolute Value \| Overview & Practice Problems 7:09 min Absolute Value \| Graph & Transformations 8:14 min Graphing Absolute Value Functions \| Definition & Translation 6:08 min Chapter 13 ASSET Numerical Skills Test: Inequalities in Math Inequality Signs in Math \| Symbols, Examples & Variation 7:09 min Inequality Notation \| Overview & Examples 8:16 min Graphing Inequalities \| Definition, Rules & Examples 7:59 min Graphing Inequalities \| Overview & Examples 12:06 min Solve & Graph an Absolute Value Inequality \| Formula & Examples 8:02 min Absolute Value Inequalities \| Definition, Calculation & Examples 9:06 min Chapter 14 ASSET Numerical Skills Test: Scientific Notation Scientific Notation \| Definition, Conversion & Examples 6:49 min Converting Numbers to Scientific Notation 5:19 min Comparing Numbers Written in Scientific Notation 5:15 min Scientific Notation: Practice Problems 6:31 min Chapter 15 ASSET Numerical Skills Test: Square Roots Square Root \| Definition, Formula & Examples 7:05 min Estimating Square Roots \| Overview & Examples 5:10 min Simplifying Square Roots When not a Perfect Square 4:45 min Simplifying Square Root Expressions \| Steps & Examples 7:03 min Radicands and Radical Expressions 4:29 min Evaluating Square Roots of Perfect Squares 5:12 min Simplifying Square Roots of Powers in Radical Expressions 3:51 min Simplifying Square Roots \| Overview & Examples 4:49 min Chapter 16 ASSET Numerical Skills Test Flashcards Related Study Materials Average Speed \| Definition, Formula & Calculation LessonsCoursesTopics ##### Calculating Total Cost: Relationship with Units & Rate 5:23 ##### How to Solve a One-Step Problem Involving Rates 5:06 ##### Math Concept \| List, Facts & Examples 4:31 ##### Math Mnemonics Devices & Examples 5:21 ##### How to Solve Problems with Money 8:29 ##### Money Facts: Lesson for Kids 3:09 ##### Adding & Subtracting Money Amounts 4:29 ##### Taxes & Discounts: Calculations & 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15449	https://files.vipulnaik.com/math-152/concaveinflectioncusptangentasymptote.pdf	CONCAVITY, INFLECTIONS, CUSPS, TANGENTS, AND ASYMPTOTES MATH 152, SECTION 55 (VIPUL NAIK) Corresponding material in the book: Section 4.6, 4.7. Diﬃculty level: Moderate to hard. If you have seen these topics in AP Calculus, then moderate diﬃculty; if you haven’t, then hard. What students should deﬁnitely get: The deﬁnitions of concave up, concave down, and point of inﬂection. The strategies to determine limits at inﬁnity, limits valued at inﬁnity, vertical tangents, cusps, vertical asymptotes, and horizontal asymptotes. What students should hopefully get: The intuitive meanings of these concepts, important examples and boundary cases, the signiﬁcance of concavity in determining local extrema, the use of higher derivative tests. Important tricks for calculating limits at inﬁnity. Executive summary 0.1. Concavity and points of inﬂection. Words ... (1) A function is called concave up on an interval if it is continuous and its ﬁrst derivative is continuous and increasing on the interval. If the function is twice diﬀerentiable, this is equivalent to requiring that the second derivative be positive except possibly at isolated points, where it can be zero. (Think x4, whose ﬁrst derivative, 4x3, is increasing, and the second derivative is positive everywhere except at 0, where it is zero). (2) A function is called concave down on an interval if it is continuous and its ﬁrst derivative is continuous and decreasing on the interval. If the function is twice diﬀerentiable, this is equivalent to requiring that the second derivative be negative except possibly at isolated points, where it can be zero. (3) A point of inﬂection is a point where the sense of concavity of the function changes. A point of inﬂection for a function is a point of local extremum for the ﬁrst derivative. (4) Geometrically, at a point of inﬂection, the tangent line to the graph of the function cuts through the graph. Actions ... (1) To determine points of inﬂection, we ﬁrst ﬁnd critical points for the ﬁrst derivative (which are points where this derivative is zero or undeﬁned) and then use the ﬁrst or second derivative test at these points. Note that these derivative tests are applied to the ﬁrst derivative, so the ﬁrst derivative here is the second derivative and the second derivative here is the third derivative. (2) In particular, if the second derivative is zero and the third derivative exists and is nonzero, we have a point of inﬂection. (3) A point where the ﬁrst two derivatives are zero could be a point of local extremum or a point of inﬂection. To ﬁnd out which one it is, we either use sign changes of the derivatives, or we use higher derivatives. (4) Most importantly, the second derivative being zero does not automatically imply that we have a point of inﬂection. 0.2. Tangents, cusps, and asymptotes. Words... (1) We say that f has a horizontal asymptote with value L if limx→∞f(x) = L or limx→−∞f(x) = L. Sometimes, both might occur. (In fact, in almost all the examples you have seen, the limits at ±∞, if ﬁnite, are both equal). (2) We say that f has a vertical asymptote at c if limx→c−f(x) = ±∞and/or limx→c+ f(x) = ±∞. Note that in this case, it usually also happens that f ′(x) →±∞on the relevant side, with the sign 1 the same as that of f(x)’s approach if the approach is from the left and opposite to that of f(x)’s approach if the approach is from the right. However, this is not a foregone conclusion. (3) We say that f has a vertical tangent at the point c if f is continuous (and ﬁnite) at c and limx→c f ′(x) = ±∞, with the same sign of inﬁnity from both sides. If f is increasing, this sign is +∞, and if f is decreasing, this sign is −∞. Geometrically, points of vertical tangent behave a lot like points of inﬂection (in the sense that the tangent line cuts through the graph). Think x1/3. (4) We say that f has a vertical cusp at the point c if f is continuous (and ﬁnite) at c and limx→c−f ′(x) and limx→c+ f ′(x) are inﬁnities of opposite sign. In other words, f takes a sharp about-turn at the x-value of c. Think x2/3. (5) We say that f is asymptotic to g if limx→∞f(x)−g(x) = limx→−∞f(x)−g(x) = 0. In other words, the graphs of f and g come progressively closer as \|x\| becomes larger. (We can also talk of one-sided asymptoticity, i.e., asymptotic only in the positive direction or only in the negative direction). When g is a nonconstant linear function, we say that f has an oblique asymptote. Horizontal asymptotes are a special case, where one of the functions is a constant function. Actions... (1) Finding the horizontal asymptotes involves computing limits as the domain value goes to inﬁnity. Finding the vertical asymptotes involves locating points in the domain, or the boundary of the domain, where the function limits oﬀto inﬁnity. For both of these, it is useful to remember the various rules for limits related to inﬁnities. (2) Remember that for a vertical tangent or vertical cusp at a point, it is necessary that the function be continuous (and take a ﬁnite value). So, we not only need to ﬁnd the points where the derivative goes oﬀto inﬁnity, we also need to make sure those are points where the function is continuous. Thus, for the function f(x) = 1/x, f ′(x) →−∞on both sides as x →0, but we do not obtain a vertical tangent – rather, we obtain a vertical asymptote. 1. Concavity and points of inflection 1.1. Concavity. Concave up means that the derivative of the function (which measures its rate of change) is itself increasing. Formally, a function f diﬀerentiable on an open interval I is termed concave up on I if f ′ is increasing on I. I hope you remember the deﬁnition of an increasing function: it means that for two points x1, x2 ∈I, with x1 < x2, we have f ′(x1) < f ′(x2). Here’s three points: (1) If f itself is increasing (so that f ′ is positive), then being concave up means that f is increasing at an increasing rate. In other words, the slope of the tangent line to the graph of f becomes steeper and steeper (up) as we go from left to right. Here’s a typical picture: 2 (2) If f itself is decreasing (so that f ′ is negative), then being concave up means that f is decreasing at a decreasing rate. In other words, the slope of the tangent line to the graph is negative, but it is becoming less and less steep as we go from left to right. Here’s a typical picture: (3) If f is twice diﬀerentiable, i.e., f ′ is diﬀerentiable, then we can deduce whether f ′ is increasing by looking at f ′′. Speciﬁcally, if f ′ is continuous on I, and f ′′ > 0 everywhere on I except at a few isolated points, then f is concave up throughout. Similarly, if f is diﬀerentiable on an open interval I, we say that f is concave down on I if f ′ is decreasing on the interval I. I hope you remember the deﬁnition of a decreasing function: it means that for two points x1, x2 ∈I, with x1 < x2, we have f ′(x1) > f ′(x2). Here’s three points: (1) If f itself is decreasing (so that f ′ is negative), then being concave down means that f is decreasing at an increasing rate. In other words, the slope of the tangent line to the graph of f becomes steeper and steeper (downward) as we go from left to right. (2) If f itself is increasing (so that f ′ is positive), then being concave down means that f is increasing at a decreasing rate. In other words, the slope of the tangent line to the graph is positive, but it is becoming less and less steep as we go from left to right. 3 (3) If f is twice diﬀerentiable, i.e., f ′ is diﬀerentiable, then we can deduce whether f ′ is increasing by looking at f ′′. Speciﬁcally, if f ′ is continuous on I, and f ′′ < 0 everywhere on I except at a few isolated points, then f is concave down throughout. 1.2. Points of inﬂection. A point of inﬂection is a point c in the interior of the domain of a diﬀerentiable function (i.e., the function is deﬁned and diﬀerentiable on an open interval containing that point) such that the function is concave in one sense to the immediate left of c and concave in the other sense to the immediate right of c. Another way of thinking of this is that points of inﬂection of a function are points where the derivative is increasing to the immediate left and decreasing to the immediate right, or decreasing to the immediate left and increasing to the immediate right. In other words, it is a point of local maximum or a point of local minimum for the derivative of the function. Recall that earlier, we noted that for a point of local maximum or a point of local minimum, either the derivative is zero or the derivative does not exist. Since everything we’re talking about now is related to f ′, we have that for a point of inﬂection, either f ′′ = 0 or f ′′ does not exist. So the upshot: concave up means the derivative is increasing, concave down means the derivative is decreasing, point of inﬂection means the sense in which the derivative is changing changes at the point. 1.3. A point of inﬂection where the ﬁrst two derivatives are zero. We now consider one kind of point of inﬂection: where the ﬁrst derivative and the second derivative are both zero. Let’s begin with the example. Consider the function f(x) := x3. Recall ﬁrst that since f is a cubic function, it has odd degree, so as x →−∞, f(x) →−∞, and as x →∞, we also have f(x) →∞. Further, if we compute f ′(x), we get 3x2. Note that the function 3x2 is positive for x ̸= 0, and is 0 at x = 0. So, from our prior discussion of increasing and decreasing functions, we see that f is increasing on (−∞, 0] and then again on [0, ∞). And since the point 0 is common to the two intervals, f is in fact increasing everywhere on (−∞, ∞). If you remember, this was an important and somewhat weird example because, although f ′(0) = 0, f does not attain a local extreme value at 0. This is because the derivative of f is positive on both sides of 0. 4 This was the picture we had from our earlier analysis. But now, with the concepts of concave up, concave down, and points of inﬂection, we can get a better understanding of what’s going on. Speciﬁcally, we see that the second derivative f ′′ is 6x, which is negative for x < 0, zero for x = 0, and positive for x > 0. Thus, f is concave down for x < 0, x = 0 is a point of inﬂection, and f is concave up for x > 0. So, here’s the picture: for x < 0, f is negative, f ′ is positive, and f ′′ is again negative. Thus, the graph of f is below the x-axis (approaching 0), it is going upward, and it is going up at a decreasing rate. So, as x →0, the graph becomes ﬂatter and ﬂatter. For x > 0, f is positive, f ′ is positive, and f ′′ is again positive. Thus, the graph of f is above the x-axis (starting from the origin), it is going upward, and it is going up at an increasing rate. The graph starts out from ﬂat and becomes steeper and steeper. So, x = 0 is a no sign change point for f ′, which is why it is not a point of local maximum or local minimum. This is because on both sides of 0, f ′ is positive. What happens is that it is going down from positive to zero and then up again from zero to positive. But on a related note, because f ′ itself dips down to zero and then goes back up, the point 0 is a point of local minimum for f ′, so it is a point of inﬂection for f. The main thing you should remember is that when we have a critical point for a function, where the derivative is zero, but it is neither a point of local maximum nor a point of local minimum, then it is likely to be a point of inﬂection. In other words, this idea of something that is increasing (or decreasing) and momentarily stops in its tracks, is the picture of neither a local maximum nor a local minimum but a point of inﬂection. 1.4. Points of inﬂection where the derivative is not zero. Let’s review the graph of the sine function. The sine function starts with sin(0) = 0, goes up from 0 to π/2, where it reaches the value 1, then drops down to 0, drops down further to −1 at 3π/2, and then turns back up to reach 0 at 2π. And this pattern repeats periodically. 5 So far, you’ve taken me on faith about the way the graph curves. But we can now start looking at things in terms of concave up and concave down. The derivative of the sin function is the cos function. Let’s graph the cos function. This starts with the value 1 at x = 0, goes down to zero at x = π/2, dips down to −1 at x = π, goes back up to 0 at x = 3π/2, and then goes up to 1 at x = 2π. We see that cos is positive from 0 to π/2, and sin is increasing on that interval. cos is negative from π/2 to 3π/2, and sin is decreasing on that interval. cos is again positive from 3π/2 to 2π, and sin is again increasing on that interval. Next, we want to know where sin is concave up and where it is concave down. And for this, we look at the second derivative of sin, which is the function −sin. As you know, the graph of this function is the same as the graph of sin, but ﬂipped about the x-axis. This means that where sin is positive, its second derivative is negative, and where sin is negative, its second derivative is positive. So, from the interval between 0 and π, sin is concave down and on the interval between π and 2π, sin is concave up. Breaking the interval down further, sin is increasing and concave down on (0, π/2), decreasing and concave down on (π/2, π), decreasing and concave up on (π, 3π/2), and increasing and concave up on (3π/2, 2π). The behavior repeats periodically. Now, let’s concentrate on the points of inﬂection. Note that the sense of concavity changes at multiples of π – at the point 0, the function changes from concave up to concave down. At the point π, the function changes from concave down to concave up. Another way of thinking about this is that just before π, the function is decreasing at an increasing rate – it is becoming progressively steeper. But from π onwards, it starts decreasing at a decreasing rate, in the sense that it starts becoming less steep. So π is the point where the way the tangent line is turning starts changing. 1.5. A graphical characterization of inﬂection points. Inﬂection points are graphically special because they are points where the way the tangent line is turning changes sense. There’s a related characterization. If you draw the tangent line through an inﬂection point, the tangent line cuts through the curve. Equivalently, the curve crosses the tangent line. This is opposed to any other point, where the curve locally lies to one side of the tangent line. For instance, for the cube function f(x) := x3, the tangent line is the x-axis, and the curve crosses the x-axis at x = 0. We see something similar for the tangent lines at the points of inﬂection 0 and π for the sin function. 1.6. Third and higher derivatives: exploration. (I may not get time to cover this in class). A while ago, we had developed criteria to determine whether a critical point is a point where a local extreme value is attained. We discussed two tests that could be used: the ﬁrst derivative test and the second derivative test. The ﬁrst derivative test said that if f ′ changes sign across the critical point, it is a point 6 where a local extreme value occurs: a local maximum if the sign change is from positive on the left to negative on the right, and a local minimum if the sign change is from negative on the left to positive on the right. The second derivative test was a test specially suited for functions that are twice diﬀerentiable at the critical point. This test states that if the second derivative at a critical point is negative, the function attains a local maximum, and if the second derivative is positive, the function attains a local minimum. This leaves one case open: what happens if the second derivative at the critical point is zero? In this case, things are inconclusive. We might have a point of local maximum, a point of local minimum, an inﬂection point, or none of the above. How do we ﬁgure this out? I will give two general principles of alternation, and then we will look at some examples: (1) If c is a point of inﬂection for f ′ and and f ′(c) = 0, then c is a point of local extremum for f. If the point of inﬂection is a change from concave up to concave down, we get a local maximum and if the change is from concave down to concave up, we get a local minimum. (2) If c is a point of local maximum or minimum for f ′, then c is a point of inﬂection for f. Local maximum implies a change from concave up to concave down and local minimum implies a change from concave down to concave up. Let’s illustrate this with the function f(x) := x5 and the point c = 0. Let’s also assume you knew nothing except diﬀerentiation and applying the derivative tests. We have f ′(x) = 5x4, f ′′(x) = 20x3, f ′′′(x) = 60x2, f (4)(x) = 120x, and f (5)(x) = 120. At c = 0, f (5) is the ﬁrst nonzero derivative. Now, 0 is a point at which f (4) = 0 and f (5) > 0. Thus, by the second derivative test, 0 is a point of local minimum for f (3). So, 0 is a point of inﬂection for f (2), by point (2) above. Thus, 0 is a point of local minimum for f ′, by point (1) above. Thus, 0 is a point of inﬂection for f, by point (2) above. So, the upshot of this is the alternating behavior between derivatives. 1.7. Higher derivative tests. The discussion above gives a practical criterion to simply use evaluation of derivatives to determine whether a critical point, where a function is inﬁnitely diﬀerentiable, is a point of local maximum, point of local minimum, point of inﬂection, or none of these. Suppose f is an inﬁnitely diﬀerentiable function around a critical point c for f.. Let k be the smallest integer for which f (k)(c) ̸= 0 and let L be the nonzero value of the kth derivative. Then: (1) If k is odd, then c is a point of inﬂection for f and hence neither a point of local maximum nor a point of local minimum. (2) If k is even and L > 0, then c is a point of local minimum for f. (3) If k is even and L < 0, then c is a point of local maximum for f. For instance, for the power function f(x) := xn, n ≥2. 0 is a critical point and f is inﬁnitely diﬀerentiable. In this case, c = 0, k = n, and L = n! > 0. Thus, if n is even, then f does attain a local minimum at 0. if n is odd, 0 is a point of inﬂection. In this simple situation, we could have deduced this directly from the ﬁrst derivative test – for n even, the ﬁrst derivative changes sign fron negative to positive at 0, because n −1 is odd. For n odd, n −1 is even, so the ﬁrst derivative has positive sign on both sides of 0. However, the good news is that this general method is applicable for other situations where the ﬁrst derivative test is harder to apply. 1.8. Notion of concave up and concave down for one-sided diﬀerentiable. So far, we have deﬁned the notion of concave up and concave down on an interval assmuing the function is diﬀerentiable everywhere on the interval. In higher mathematics, a somewhat more general deﬁnition is used, and this makes sense for functions that have one-sided derivatives everywhere. Note: Please, please, please, please make sure you understand this clearly: we can calculate the left-hand derivative and right-hand derivative using the formal expressions only after we have checked that the function is continuous from that side! If there is a piecewise description of the function and it is not continuous from one side where the deﬁnition is changing, then the corresponding one-sided derivative is not, repeat not deﬁned. Suppose f is a function on an interval I = (a, b) such that both the left-hand derivative and the right-hand derivative of f are deﬁned everywhere on I. Note that both one-sided derivatives being deﬁned at every point in particular means that the function is continuous at each point, and hence on I. However, f need not be 7 diﬀerentiable at every point, because it is possible that the left-hand derivative and right-hand derivative diﬀer at diﬀerent points. We then say that: (1) f is concave up if, at every point, the right-hand derivative is greater than or equal to the left-hand derivative, and both one-sided derivative are increasing functions on R. (2) f is concave down if, at every point, the right-hand derivative is less than or equal to the left-hand derivative, and both one-sided derivatives are decreasing functions on R. For instance, ﬁrst consider the function f on (0, ∞): f(x) := { x2, 0 < x ≤1 x3, 1 < x Before we proceed further, we check/note that the function is continuous at 1. Indeed it is. Hence, to calculate the left-hand derivative and right-hand derivative at 1, we can formally diﬀerentiate the expressions at 1. We obtain that the left-hand derivative at 1 is 2 · 1 = 2 and the right-hand derivative at 1 is 3 · 12 = 3. We thus obtain the following piecewise deﬁnitions for the left-hand derivative and right-hand derivative: LHD of f at x = { 2x, 0 < x ≤1 3x2, 1 < x and: RHD of f at x = { 2x, 0 < x < 1 3x2, 1 ≤x The derivative is undeﬁned at 1. Note that both one-sided derivatives are increasing everywhere, and at the point 1, where the function is not diﬀerentiable, the right-hand derivative is bigger. Thus, the function is concave up on (0, ∞). Here’s the graph, with dashed lines indicating the one-sided derivatives: On the other hand, consider the function g on (0, ∞): g(x) := { x3, 0 < x ≤1 x2, 1 < x 8 The function g is continuous and has one-sided derivatives everywhere. Also note that on the intervals (0, 1) and (1, ∞), g is concave up. However, at the critical point 1 where g′ is undeﬁned, the right-hand derivative is smaller than the left-hand derivative. Thus, the function is not concave up overall on (0, ∞), because at the critical point, its rate of increase takes a plunge for the worse. Here’s the picture of g: 1.9. Graphical properties of concave functions. Here are some properties of the graphs of functions that are concave up, which are particularly important in the context of optimization. You should be able to do suitable role changes and obtain corresponding properties for concave down functions. In all the points below, f is a continuous function on an interval [a, b] and is concave up on the interior (a, b). (1) The only possibilities for the increase-decrease behavior of f are: increasing throughout, decreasing throughout, or decreasing ﬁrst and then increasing. (2) In particular, f either has exactly one local minimum or exactly one endpoint minimum, and this local or endpoint minimum is also the absolute minimum. (3) Also, f cannot have a local maximum in its interior. It has exactly one endpoint maximum, and this is also the absolute maximum. (4) For any two points x1, x2 in the domain of f, the part of the graph of f between (x1, f(x1)) and (x2, f(x2)) lies below the chord joining the points (x1, f(x1)) and (x2, f(x2)). (5) If we assume that f is diﬀerentiable on (a, b), the tangent line through any point (x, f(x)) for a < x < b does not intersect the curve at any other point. In the more general notion where f has one-sided derivatives, both the left and right tangent line satisfy this property. For concave down functions, the role of minimum and maximum gets interchanged, and in point (3) above, the graph is now above the chord rather than below. 1.10. Addendum: concave and convex. The book uses the terminology concave up and concave down, but it’s worth knowing that in much of mathematics as well as applications of mathematics, the term convex is used for concave up and the term concave is used for concave down. However, there is some confusion about this since some people use concave for concave up and convex for concave down. 9 2. Infinity and asymptotes 2.1. Limits to inﬁnity and vertical asymptotes. We have already discussed what it means to say limx→c f(x) = +∞, but here’s a friendly reminder. It means that as x comes closer and closer to c (from either side), f(x) goes above every ﬁnite value and does not then come back down. You can similarly understand what it means to say that limx→c f(x) = −∞. You should also be able to understand the one-sided versions of these concepts: limx→c−f(x) = ∞, limx→c+ f(x) = ∞, limx→c−f(x) = −∞, and limx→c+ f(x) = −∞. Now, here’s a little guesswork question: if limx→c−f(x) = ∞, what can you say about limx→c−f ′(x)? As a general rule, nothing, but in most of the situations that we see, it turns out that f ′(x) also approaches +∞1. The reason is easy to see graphically: for f(x) to head to +∞as x approaches a ﬁnite value, f needs to climb faster and faster and faster. Similarly, it is usually the case that if limx→c−f(x) = −∞, then limx→c−f ′(x) = −∞. Also, if limx→c+ f(x) = ∞, then it is likely that limx→c+ f ′(x) = −∞(because it has to drop very quickly down from inﬁnity) and if limx→c+ f(x) = −∞, then limx→c+ f ′(x) = ∞(because it has to rise very quickly back up from −∞). Again, these things are not always true, but for most of the typical examples you’ll see, they will be. Now here’s the meaning of vertical asymptote. If limx→c−f(x) = ±∞and/or limx→c+ f(x) = ±∞, then the line x = c is termed a vertical asymptote for f. This is because the graph of f is approaching the vertical line x = c. In some sense, if we think of f(c) = +∞or −∞as the case may be, the vertical line becomes the tangent line to the curve at that inﬁnite point. Some of the typical situations worth noting are: (1) limx→c f(x) = +∞from both sides. An example of this is the function f(x) = 1/x2 with c = 0. The vertical asymptote is the y-axis, i.e., the line x = 0. In this case, and in most other representative examples, limx→c−f ′(x) = +∞, and limx→c+ f ′(x) = −∞. (2) limx→c f(x) = −∞from both sides. An example of this is the function f(x) = −1/x2 with c = 0. The vertical asymptote is the y-axis, i.e., the line x = 0. In this case, and in most other representative examples, limx→c−f(x) = −∞, and limx→c+ f(x) = +∞. 1More precisely, it turns out that if f′ is continuous and does approach something, that something must be +∞. However, there are weird examples where it oscillates 10 (3) limx→c−f(x) = ∞and limx→c+ f(x) = −∞. Examples include f = tan at c = π/2 (vertical asymptote x = π/2) and f(x) = −1/x at c = 0 (vertical asymptote x = 0). In both these cases, as in most others, limx→c f ′(x) = +∞. 11 (4) limx→c−f(x) = −∞and limx→c+ f(x) = ∞. Examples include f = cot at c = 0 and f(x) := 1/x at c = 0. In both these cases, as in most others, limx→c f ′(x) = −∞. 2.2. Horizontal asymptotes. Horizontal asymptotes are horizontal lines that the graph comes closer and closer to, just as vertical asymptotes are vertical lines that the graph comes closer and closer to. We saw that vertical asymptotes arose when the range value was approaching ±∞for a ﬁnite limiting value of the domain. Horizontal asymptotes arise where the domain value approaches ±∞for a ﬁnite limiting value of the range. Explicitly, if limx→∞f(x) = L (with L a ﬁnite number), then the line y = L is a horizontal asymptote for the graph of f, because as x →∞, the graph comes closer and closer to this horizontal line. Similarly, if limx→−∞f(x) = M, then the line y = M is a horizontal asymptote for the graph of f. Thus, a function whose domain extends to inﬁnity in both directions could have zero, one, or two horizontal asymptotes. 12 We will discuss some of the computational aspects of vertical and horizontal asymptotes in the problem sessions. Later in the lecture, and in the addendum, we look at some computational tips and guidelines over and above what is there in the book. 2.3. Vertical tangents. A vertical tangent to the graph of a function f occurs at a point (c, f(c)) if f is continuous but not diﬀerentiable at c, and limx→c f ′(x) = +∞or limx→c f ′(x) = −∞. It is important that the sign of inﬁnity in the limit is the same from both the left and the right side. An example is the function f(x) := x1/3 at the point c = 0. The function is continuous at 0. The derivative functions is (1/3)x−2/3, and the limit of this as x →0 (from either side) is +∞. Graphically, what this means is that the tangent is vertical. In this case, the vertical tangent coincides with the y-axis, because it is attained at the point 0. 13 Points of vertical tangent are points of inﬂection, as we can see from the x1/3 example. Recall that the horizontal tangent case of the point of inﬂection was typiﬁed by x3, and the general slogan was that the function slows down for an instant to speed zero. For vertical tangents, we can think of it as the function speeding up instantaneously to speed inﬁnity before returning to the realm of ﬁnite speed. It is important to note that the situation of a vertical tangent requires that the function itself be deﬁned and continuous, and hence ﬁnite-valued, at the point. Thus, for instance, the function f(x) := 1/x satisﬁes limx→0 f ′(x) = −∞but does not have a vertical tangent at zero because the function is undeﬁned at zero. 2.4. Vertical cusps. A vertical cusp in the graph of f occurs at a point c if f is continuous at c, and both one-sided limits of f ′ at c are inﬁnities of opposite sign. There are two possibilities: (1) The left-hand limit of the derivative is +∞and the right-hand limit of the derivative is −∞. Then, (c, f(c)) is a point of local maximum. An example is f(x) := −x2/3 and c = 0 What happens in this situation is the the graph has a sharp peak (picking up to speed inﬁnity) at the point c, after which it rapidly starts dropping. (2) The left-hand limit of the derivative is −∞and the right-hand limit of the derivative is +∞. In this case we get a local minimum. An example is f(x) := x2/3 and c = 0. There is a special curve called the astroid curve (I had planned to put this on the homework, but it went on the chopping block when I needed to trim the homeworks to size), given by the equation x2/3 + y2/3 = a2/3. This curve is not the graph of a function, since every value of x in (−a, a) has two corresponding values of y. Nonetheless, the curve is a good illustration of the concept of cusps: there are two vertical cusps at the points (0, a) and (0, −a) respectively, and two horizontal cusps at the points (a, 0) and (−a, 0) respectively. Shown below is the astroid curve for a = 1: Note that for the graph of a function, the only kind of cusp that can occur is a vertical cusp, because a horizontal or oblique cusp would result in the curve intersecting a vertical line at multiple points, which would contradict the meaning of a function. It is important to note that the situation of a vertical cusp requires that the function itself be deﬁned and continuous, and hence ﬁnite-valued, at the point. Thus, for instance, the function f(x) := 1/x2 satisﬁes limx→0−f ′(x) = +∞and limx→0+ f ′(x) = −∞but does not have a vertical tangent at zero because the function is undeﬁned at zero. 3. Computational aspects 3.1. Computing limits at inﬁnity: a review. We review the main results that you have probably seen and add some more: (1) (→∞)(→∞) =→∞. In other words, if limx→c f(x) = ∞and limx→c g(x) = ∞, then limx→c f(x)g(x) = ∞. c could be ﬁnite or ±∞here, and we could take one-sided limits instead. 14 (2) (→∞)(→−∞) =→−∞, and (→−∞)(→−∞) =→∞. (3) (→a)(→∞) =→∞if a > 0 and (→a)(→∞) =→−∞if a < 0. Similarly, (→a)(→−∞) =→−∞ if a > 0 and (→a)(→−∞) =→∞if a < 0. (4) The previous point can be generalized somewhat: (→∞), times a function that eventually has a positive lower bound (even if it keeps oscillating), is also →∞. Analogous results hold for negative upper bounds. (5) (→0)(→∞) is an indeterminate form: it is not clear what it tends to without doing more work. (6) (→∞) + (→∞) =→∞. (7) (→∞) + (→a) =→∞where a is ﬁnite. More generally, →∞plus anything that is bounded from below is also →∞. (8) (→∞) −(→∞) and (→∞) + (→−∞) are indeterminate forms. Apart from this, the main facts you need to remember are that if a > 0, then limx→∞xa = ∞and limx→∞x−a = limx→−∞x−a = 0. Note that this holds regardless of whether a is an integer. When a is an odd integer or a rational number with odd numerator and odd denominator, limx→−∞xa = −∞. When a is an even integer or a rational number with even numerator and odd denominator, limx→−∞xa = ∞. Also worth noting: limx→0+ x−a = ∞for a > 0 and limx→0−x−a = limx→−∞xa, which is computed by the rule above. We can use these facts to explain most of the limits involving polynomial and rational functions. Earlier, we had noted that when calculating the limit of a polynomial, it is enough to calculate the limit of its leading monomial. Let’s now see why. Consider the function f(x) := x7 −5x5 +3x+2. Then, we can write f(x) = x7 1 −5x−2 + 3x−6 + 2x−7 . The expression on the inside is 1 plus various negative powers of x. Each of those negative powers of x goes to 0 as x →∞. So, we obtain: lim x→∞[1 −5x−2 + 3x−6 + 2x−7] = 1 We also have limx→∞x7 = ∞. Thus, the limit of the product is ∞. Let’s now consider an example of a rational function: 9x3 −3x + 2 103x2 −17x −99 Earlier, we had discussed that when computing such limits at ±∞, we can simply calculate thel imits of the leading terms and ignore the rest. We now have a better understanding of the rationale behind this. Formally: lim x→∞ 9x3 −3x + 2 103x2 −17x −99 = lim x→∞ x3(9 −3x−2 + 2x−3) x2(103 −17x−1 −99x−2) = lim x →∞x lim x→∞ 9 −3x−2 + 2x−3 103 −17x−1 −99x−2 = lim x→∞x · 9 103 = + ∞ More generally, we see that if the degree of the numerator is greater than the degree of the denominator, the fraction approaches ±∞as x →±∞, with the sign depending on the signs of the leading coeﬃcients and the parity (even versus odd) of the exponents. If the numerator and denominator have equal degree, the limit is a ﬁnite number. For x →±∞, it is the ratio of the leading coeﬃcients (notice that it is the same on both sides). This is the case where we get horizontal asymptotes. In this case, the horizontal asymptotes on both ends coincide. For instance: 15 lim x→−∞ 2x2 −3x + 5 23x2 −x −1 = lim x→−∞ x2(2 −3x−1 + 5x−2) x2(23 −x−1 −x−2) = lim x→−∞ x2 x2 lim x→−∞ 2 −3x−1 + 5x−2 23 −x−1 −x−2 =1 · 2 23 = 2 23 Finally, when the degree of the numerator is less than the degree of the denominator, then the fraction tends to 0 as x →∞and also tends to 0 as x →−∞. Thus, in this case, we get the x-axis as the horizontal asymptote on both sides. 3.2. The 1/x substitution trick. Consider the limit: lim x→∞x sin(1/x) This limit cannot be computed by plugging in values, because x →∞and 1/x →0, so sin(1/x) →0, and we get the indeterminate form (→∞)(→0). The approach we use here is to set t = 1/x. As x →∞, t →0+. Since t = 1/x, we get x = 1/t. Plugging in, we get: lim t→0+ sin t t This limit is 1, as we know well. Note that with this general substitution, limits to inﬁnity correspond to right-hand limits at 0 for the reciprocal and limits at −∞correspond to left-hand limits at 0 for the reciprocal. If there is a two-sided limit at 0 for the reciprocal, the limits at ±∞are the same. In fact, in the x sin(1/x) example, the limits at ∞and −∞are both 1 since limt→0 sin t/t = 1. 3.3. Diﬀerence of square roots. Consider the limit: lim x→∞( √ x + 1 −√x) There are many ways to compute this limit, but the easiest is to use the general 1/x substitution trick. Let t = 1/x. Then the above limit becomes: lim t→0+ √t + 1 −1 √ t This is an indeterminate form (speciﬁcally, a 0/0 form). However, we can do the rationalization trick and rewrite this as: 16 lim t→0+ t √ t(√t + 1 + 1) The t and √ t cancel to give a √ t in the numerator, and we can evaluate and ﬁnd the limit to be 0. A similar approach can be used to handle, for instance, a diﬀerence of cube roots. 3.4. Combinations of polynomial and trigonometric functions. We illustrate using some examples: (1) Consider the function f(x) := x + 25 sin x. As x →∞, this is the sum of a function that tends to inﬁnity and a function that oscillates. The oscillating component, however, has a ﬁnite lower bound, and hence, limx→∞f(x) = ∞. Similarly, limx→−∞f(x) = −∞. (2) Consider the function f(x) := x sin x. As x →∞, this is the product of a function that goes to ∞and a function that oscillates between −1 and 1. The oscillating part causes the sign of the whole expression to shift, and so as x →∞, f(x) is oscillating with an ever-increasing magnitude of oscillation. A similar observation holds for x →−∞. (3) Consider the function f(x) := x(3 + sin x) As x →∞, this is the product of a function that tends to ∞and a function that oscillates between 2 and 4. The important point here is that the latter oscillation has a positive lower bound, so the product still tends to ∞. (4) Consider the function f(x) := x sin(1/x). As x →∞, this is the product of a function that tends to ∞and a function that tends to 0, so it is an indeterminate form. We already discussed above how this particular indeterminate form can be handled. 17
15450	https://artofproblemsolving.com/wiki/index.php/Proofs_of_AM-GM?srsltid=AfmBOoqW2Sr1YkiVHIksrnzKvSCjhpk3ahc3r7ArBTu-q-Y708fdXmAI	Art of Problem Solving Proofs of AM-GM - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. 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We first note that we may disregard any for which , as they contribute to neither side of the desired inequality. We also note that if and , for some , then the right-hand side of the inequality is zero and the left hand of the inequality is greater or equal to zero, with equality if and only if whenever . Thus we may henceforth assume that all and are strictly positive. Contents 1 Proofs of Unweighted AM-GM 1.1 Proof by Cauchy Induction 1.2 Proof by Rearrangement 1.3 Proof by Calculus 1.4 Proof by An Easier Induction 2 Proof of Weighted AM-GM 2.1 Proof by Convexity 2.2 Alternate Proof by Convexity Proofs of Unweighted AM-GM These proofs use the assumption that , for all integers . Proof by Cauchy Induction We use Cauchy Induction, a variant of induction in which one proves a result for , all powers of , and then that implies . Base Case: The smallest nontrivial case of AM-GM is in two variables. By the properties of perfect squares (or by the Trivial Inequality), with equality if and only if , or . Then because and are nonnegative, we can perform the following manipulations: with equality if and only if , just as before. This completes the proof of the base case. Powers of Two: We use induction. Suppose that AM-GM is true for variables; we will then prove that the inequality is true for . Let be any list of nonnegative reals. Then, because the two lists and , each have variables, Adding these two inequalities together and dividing by yields From here, we perform AM-GM in two variables on and to get Combining this inequality with the previous one yields AM-GM in variables, with one exception — equality. For equality, note that every AM-GM application mentioned must have equality as well; thus, inequality holds if and only if all the numbers in are the same, all the numbers in are the same, and . From here, it is trivial to show that this implies , which is the equality condition for AM-GM in variables. This completes the induction and proves that the inequality holds for all powers of . Backward Step: Assume that AM-GM holds for variables. We will then use a substitution to derive AM-GM for variables. Letting , we have that Because we assumed AM-GM in variables, equality holds if and only if . However, note that the last equality is implied if all the numbers of are the same; thus, equality holds if and only if . We first simplify the lefthand side. Multiplying both sides of the fraction by and combining like terms, we get that Plugging this into the earlier inequality yields Raising both sides to the th power yields From here, we divide by and take the root to get that This is the inequality in variables. Note that every step taken also preserves equality, which completes the backward step. Then by Cauchy Induction, the AM-GM inequality holds. Proof by Rearrangement Define the sequence as , for all integers . Evidently, these sequences are similarly sorted. Then by the Rearrangement Inequality, where we take our indices modulo , with equality exactly when all the , and therefore all the , are equal. Dividing both sides by gives the desired inequality. Proof by Calculus We will start the proof by considering the function . We will now find the maximum of this function. We can do this simply using calculus. We need to find the critical points of , we can do that by finding and setting it equal to . Using the linearity of the derivative . We need Note that this is the only critical point of . We can confirm it is the maximum by finding its second derivative and making sure it is negative. letting we get . Since the second derivative , is a maximum. . Now that we have that is a maximum of , we can safely say that or in other words . We will now define a few more things and do some manipulations with them. Letting , notice that . This fact will come into play later. Now we can do the following. Letting and plugging this into , we get Adding all these results together we get Now exponentiating both sides we get This proves the AM-GM inequality. Proof by An Easier Induction We can always rearrange the order of without changing its geometric mean and arithmetic mean. Hence, WLOG, let . This follows by For the case of , , so its true. For , since , Hence GM-AM Inequality is true for . If we assume it is true for , then for : By our inductive hypothesis, Now we need to prove a lemma to finish our induction process. Recall that , this implies that Hence, Thus, substitute this part in, By mathematical induction, the statement is true for as it's true for , and since it's true for and , we can conclude that the statement is true for all . Proof of Weighted AM-GM Proof by Convexity We note that the function is strictly concave. Then by Jensen's Inequality, with equality if and only if all the are equal. Since is a strictly increasing function, it then follows that with equality if and only if all the are equal, as desired. Alternate Proof by Convexity This proof is due to G. Pólya. Note that the function is strictly convex. Let be the line tangent to at ; then . Since is also a continuous, differentiable function, it follows that for all , with equality exactly when , i.e., with equality exactly when . Now, set for all integers . Our earlier bound tells us that so Multiplying such inequalities gives us Evaluating the left hand side: for Evaluating the right hand side: Substituting the results for the left and right sides: as desired. 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15451	https://www.ncbi.nlm.nih.gov/books/NBK562188/	Postmenopausal Bleeding - StatPearls - NCBI Bookshelf Warning: The NCBI web site requires JavaScript to function. more... An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Postmenopausal Bleeding Sharon Sung; Karen Carlson; Aaron Abramovitz. Author Information and Affiliations Authors Sharon Sung 1; Karen Carlson 2; Aaron Abramovitz 3. Affiliations 1 Sparrow Hospital 2 University of Nebraska Medical Center 3 Sparrow Hospital Last Update: January 22, 2025. Go to: Continuing Education Activity Menopause is characterized by the complete absence of menstrual cycles, which occurs when a female has no ovarian follicles left in reserve and is clinically diagnosed when a woman has had amenorrhea for 1 year.In the US, the average age of menopause is 51. Bleeding after menopause has been established, termed postmenopausal bleeding, is considered abnormal and is the reason for approximately two-thirds of all gynecologic office visits in postmenopausal women. The differential diagnoses associated with postmenopausal bleeding (PMB) include several conditions. Though the most common cause of PMB is atrophy of the lower reproductive tract, 90% of postmenopausal women diagnosed with endometrial cancer presented with vaginal bleeding. Early diagnosis and management lead to a significantly more favorable prognosis, as with most malignancies.Therefore, any postmenopausal woman with vaginal bleeding should be promptly and appropriately evaluated through a comprehensive clinical examination and diagnostic studies, including endometrial biopsy and imaging. Typically, management depends on the etiology identified.This activity for healthcare professionals aims to enhance learners' competence in selecting appropriate diagnostic tests, managing the identified etiology, and fostering effective interprofessional teamwork to improve outcomes in patients with postmenopausal bleeding. Objectives: Identify the various etiologies of postmenopausal bleeding. Differentiate the differential diagnoses of postmenopausal bleeding. Implement the recommended management of postmenopausal bleeding. Apply interprofessional team strategies to improve care coordination and outcomes for patients with postmenopausal bleeding. Access free multiple choice questions on this topic. Go to: Introduction Menopause is characterized by the complete absence of menstrual cycles, which occurs when a female has no ovarian follicles left in reserve and is clinically diagnosed when a woman has had amenorrhea for 1 year. In the United States, the average age of menopause is 51.Bleeding after menopause has been established, termed postmenopausal bleeding, is considered abnormal and is the reason for approximately two-thirds of all gynecologic office visits in postmenopausal women. The differential diagnoses associated with postmenopausal bleeding (PMB) include several conditions. Though the most common cause of PMB is atrophy of the lower reproductive tract, 90% of postmenopausal women diagnosed with endometrial cancer presented with vaginal bleeding.As with most malignancies, early diagnosis and management lead to a significantly more favorable prognosis. Therefore, any postmenopausal woman with vaginal bleeding should be promptly and appropriately evaluated through a comprehensive clinical examination and diagnostic studies, including endometrial biopsy and imaging. Typically, management depends on the etiology identified. Due to the frequency with which clinicians encounter postmenopausal bleeding,healthcare professionals should have enhanced knowledge in selecting appropriate diagnostic tests, managing the etiology, and fostering effective interprofessional teamwork to improve outcomes in patients with this common condition. Go to: Etiology PMB is often attributed to the uterus, with women mistakenly attributing bleeding to menstrual bleeding despite not having had menses for over 1 year. However, the bleeding may also arise from the urethra, vulva, vagina, cervix, or rectum. The most common cause of postmenopausal bleeding is genitourinary atrophy, accounting for 60%.The etiology of PMB may also be nongynecologic (eg, the urethra, bladder, or GI tract) and mistaken for vaginal bleeding.Some common underlying causes of postmenopausal bleeding include: Vaginal or endometrial atrophy Urogenital infections (eg, endometrial tuberculosis, vaginitis, cystitis, or cervicitis) Medications (eg, estrogen, tamoxifen, and anticoagulants) Uterine leiomyomas Genital tract malignancies Vaginal foreign bodies Endometrial polyps Genitourinary atrophy Endometrial hyperplasia with or without atypia Go to: Epidemiology Vaginal bleeding is reported in up to 10% of postmenopausal women and is the presenting symptom for approximately two-thirds of gynecologic office visits in this population.However, the incidence of PMB may decrease with age. With the onset of menopause, bleeding is reported in approximately 40% of women per year, but 3 years after menopause, PMB decreases to 4% per year. Endometrial cancer is the fifth most common cause of death due to malignancy in the US and the fourth most common overall cancer in females. Furthermore, endometrial cancer is the most commonly diagnosed cancer in women and the most common site of uterine cancer, accounting for 92% of cases.Over 90% of postmenopausal women with endometrial cancer present with PMB.In women younger than 50, <1% of PMB is secondary to endometrial cancer. However, the incidence of PMB due to endometrial cancer increases to 24% in women older than 80.The global incidence of endometrial cancer is increasing, primarily due to the increased prevalence of endometrial cancer risk factors (eg, obesity and late menopause). The number of patients diagnosed with endometrial cancer is anticipated to double by the year 2030. Go to: Pathophysiology The most common etiology for postmenopausal bleeding is an atrophic endometrium. The hypoestrogenic environment following menopause leads to genitourinary atrophy. The collapsed and atrophic endometrial lining within the uterus contains scant or no fluid to prevent friction inside the cavity, causing epithelial microerosions and subsequent chronic inflammation. Chronic endometritis secondary to atrophy can present with vaginal spotting or light bleeding. Pelvic ultrasounds performed in these patients typically reveal a thin endometrial stripe with an otherwise normal appearance, a small uterus, and small ovaries. Conversely, unopposed estrogen exposure often develops premalignant or malignant endometrial conditions. Systemic estrogen-only therapy, obesity, and estrogen-secreting tumors can lead to abnormal endometrial changes. Some women have genetic predispositions to endometrial cancer, eg, Lynch syndrome and Cowden disease.More recent studies demonstrate that BRCA gene mutation carriers may have a slightly increased risk of endometrial cancer, especially those with a BRCA1 mutation. Go to: Histopathology Postmenopausal Endometrial Histologic Findings The hypoestrogenic environment has several effects on the postmenopausal endometrium that are apparent on microscopic examination. Normal premenopausal proliferative endometrium contains regularly spaced, ordered glands lined by simple epithelial cells within the stroma with an approximate 1 to 1 gland-to-stroma ratio. Furthermore, histologically, mitotic activity is a characteristic feature of proliferative endometrium.Conversely, normal histologic findings in the postmenopausal endometrium show widely spaced glands that vary from small to cystically dilated. Moreover, the mitotic proliferative activity characteristic of a premenopausal endometrium is not apparent in postmenopausal endometrial cells.Pathologic endometrial findings have varying histologic findings depending on the underlying etiology. For instance, the histopathology of endometrial polyps frequently shows cellular immaturity with cystic hyperplasia, while smooth muscle fibers are a characteristic finding of uterine fibroids. Both conditions have a higher risk of malignant transformation in postmenopausal women. In women with PMB, the most concerning abnormal histologic findings are endometrial hyperplasia and carcinoma. Endometrial Hyperplasia Histologic Findings Characteristic histologic findings of endometrial hyperplasia include widespread crowding of endometrial glands, a disordered proliferation of glands, and an increased gland-to-stroma ratio. The precise ratio diagnostic of endometrial hyperplasia is debated; however, many pathologists use a gland-to-stroma ratio of 2 to 1.For benign endometrial hyperplasia, gland cytology is normal, with only occasional mitotic figures found. However, on histological examination, gland crowding and abnormal gland nuclei are typically seen in endometrial intraepithelial neoplasia (EIN), considered a premalignant condition. Endometrial CancerHistologic Findings One of the most concerning etiologies of postmenopausal bleeding is endometrial cancer, which is typically subclassified into type I or II based on histologic morphology, grade, and hormone receptors. Endometrial carcinoma type 1:The most common histologic type of endometrial carcinomas, accounting for up to 90% of uterine cancers, is primarily made up of grade I or II endometrioid adenocarcinomas which involves the endometrial glands. Histologically, solid areas, maze-like glands, or appreciable cribriforming are observed. Most of these endometrioid adenocarcinomas are low-grade and confined to the uterus. Endometrial carcinoma type 2: These are rare, high-grade, poorly differentiated, and more aggressive types of uterine cancers that include clear cell carcinoma, sarcoma, carcinosarcoma, and papillary serous histologies. Type 2 endometrial cancers have a higher risk of extrauterine disease at diagnosis and a worse prognosis than type I tumors. For instance, though only 10% of uterine cancers are papillary serous, they cause approximately 40% of deaths. Go to: History and Physical A comprehensive history is essential when evaluating PMB to assess both underlying etiologies and establish the patient's menopausal status. Excluding a malignant etiology, primarily endometrial cancer, is the most critical aspect of PMB evaluation. Clinicians should elicit a thorough history, including the following standard elements: History of Present Illness To evaluate the differential diagnoses for PMB and clinically confirm the patient's menopausal state, clinicians should obtain a history regarding the nature of a patient's current bleeding and any associated symptoms. Heavy menstrual bleeding or other abnormal uterine bleeding patterns may indicate structural abnormalities (eg, leiomyomas, polyps), hyperplasia, or malignancy.PMB onset, duration, heaviness, and precipitating factors (eg, bleeding after intercourse or after wiping) are all essential to characterize vaginal bleeding symptoms. Associated symptoms, including fever, pelvic pain, dysuria, vasomotor symptoms, dyspareunia, and vaginal dryness, should also be investigated. Past medical history: Potential etiologies for abnormal bleeding include obesity, polycystic ovarian syndrome, diabetes, thyroid disease, pelvic infection, or coagulopathies. Several of these conditions are also risk factors for endometrial cancer or hyperplasia (eg, obesity and diabetes). Radiation exposure may also cause PMB. Gynecologic history: Primarily, a thorough review of the menstrual history, including the last menstrual period, is critical when assessing an individual's postmenopausal status. Menarche younger than 12 years (ie, early menarche) or menopause beginning in patients older than 55 (ie, late menopause) increase the risk for endometrial cancer.Additionally, recent Pap and HPV testing may indicate cervical or endometrial etiologies. Surgical history: Past surgical procedures can provide clinicians with potential etiologies for vaginal bleeding and menopausal amenorrhea. Recent pelvic procedures can result in postoperative bleeding, and surgically induced menopause (eg, hysterectomy, oophorectomy) can assist in definitively establishing a patient's menopausal status. Social history: Genitourinary infections, such as sexually transmitted diseases, are also causes of PMB. Additionally, individuals who smoke have an increased risk of bladder cancer, mesh erosion, and hematuria, which may cause bleeding. Family history: Due to the increased malignancy risk, a family history of breast, gynecologic, urologic, or gastrointestinal cancers should be assessed, as well as inherited mutations (eg, Lynch or Cowden syndrome) that can predispose individuals to endometrial cancer. Medications: Clinicians should also elicit medication history from patients as hormone replacement therapy, tamoxifen, anticoagulants, and certain herbal supplements can all affect the endometrial lining, leading to postmenopausal bleeding. Physical Exam On physical exam, it is crucial to thoroughly evaluate the internal and external anatomy of the genital tract. A speculum exam should be performed to visualize bleeding sites, genital lesions, lacerations, urethral prolapse, and signs of genitourinary atrophy, which typically include pale, dry vaginal epithelium with loss of rugae. Erythema, petechiae, friability, and discharge may indicate inflammation. Furthermore, clinicians can palpate for pelvic masses, abdominal distention, and enlarged lymph nodes through a bimanual exam. Go to: Evaluation Diagnostic Studies In addition to the clinical assessment, diagnostic evaluation of PMB is primarily directed toward excluding endometrial hyperplasia or malignancy. The American College of Obstetricians and Gynecologists (ACOG) recommends either transvaginal ultrasound or endometrial biopsy initially to evaluate low-risk women presenting for PMB; performing both studies simultaneously is unnecessary. However, endometrial biopsy is recommended as the first-line test in individuals with endometrial cancer risk factors or recurrent PMB.Laboratory studies may be considered to assess for complications secondary to heavy vaginal bleeding and to help exclude differential diagnoses. Endometrial Biopsy Endometrial sampling is a first-line test for evaluating PMB in any patient because this study provides tissue for histologic diagnosis, which is critical in identifying malignancies. Other indications for endometrial sampling in patients with PMB include: Persistent or recurrent bleeding PMB, despite a thin endometrial stripe on ultrasound Risk factors for endometrial cancer (eg, obesity, smoking, exposure to unopposed estrogen) Endometrial lining thickness by transvaginal ultrasound of >4 mm in a woman with postmenopausal bleeding Inadequate visualization of the endometrium in imaging studies ACOG also recommends performing endometrial sampling on women with abnormal uterine bleeding older than 45.The primary modalities utilized for endometrial sampling include dilation and curettage (D&C) and outpatient disposable thin plastic devices inserted through the endocervical opening into the endometrial cavity without anesthesia.Clinicians have performed D&Cs with or without hysteroscopy for years for histologic assessment as the diagnostic sensitivity for endometrial cancer exceeds 90%.In studies, office endometrial biopsy using flexible plastic samplers has been found to have similar diagnostic accuracy.However, endometrial biopsy can result in findings that are insufficient for diagnosis, with rates of sampling failure up to 54%.The diagnostic accuracy of endometrial sampling correlates positively with the amount of tissue that is collected.For inadequate sampling that was performed first, a follow-up ultrasound may be performed; if subsequent transvaginal ultrasound reveals a thin endometrial stripe and vaginal bleeding does not persist, further evaluation is unnecessary. Imaging Studies Transvaginal imaging to measure endometrial thickness may also be used as a first-line modality to evaluate PMB in patients without indications for histologic assessment. The technique ACOG recommends to measure endometrial thickness consists of using the thickest portion of the endometrial stripe seen on a long-axis uterine view, measuring anterior to posterior.An endometrial stripe thickness of≤4 mm has a negative predictive value >99% for endometrial carcinoma.The probability of having endometrial cancer with an endometrial stripe thickness of <4 mm is approximately 0.3%. Using this 4 mm threshold, only 1 in 339 endometrial cancers are missed, according to studies.Ultrasound imaging may also identify other pathologic pelvic etiologies (eg, leiomyomas or adnexal masses). However, further evaluation with endometrial sampling is indicated if the endometrial stripe is inadequately visualized. Because endometrial imaging is not diagnostic and endometrial thickness does not always exclude malignancy, clinicians should perform an endometrial biopsy for continued or recurrent PMB, even for patients with an endometrial lining <4 mm. A thickened endometrial stripe may be caused by hyperplasia, malignancy, and intracavitary lesions (eg, leiomyomas and endometrial polyps). For ultrasound findings suggestive of intracavitary lesions or history-indicated suspicion (eg, previous polyps), additional imaging may be useful, including saline-infused ultrasonography or hysterosalpingogram. Pelvic computed tomography (CT) and magnetic resonance imaging (MRI) are sometimes used to characterize better urogenital pathology noted on ultrasound. However, hysteroscopy with dilation and curettage is the gold standard as diagnostic sampling and therapeutic excision for some etiologies of PMB (eg, endometrial polyps) can be performed simultaneously. Furthermore, blind endometrial sampling without hysteroscopy may miss focal lesions or intrauterine pathology, such as polyps, and mass lesions may deflect flexible disposable devices. Consequently, further imaging evaluation should be considered for patients with insufficient sampling or persistent vaginal bleeding in whom focal lesions may have been missed; hysteroscopy with dilation and curettage or directed biopsy may be warranted. Laboratory Studies A CBC, TSH, and coagulation studies (eg, prothrombin time and partial thromboplastin time) should be considered for individuals with PMB to help exclude some differential diagnoses and assess for secondary anemia due to abnormal bleeding.Vaginal cultures for sexually transmitted diseases and other genital infections should be considered if clinically indicated. In women who have undergone early menopause (ie, older than 40), a pregnancy test is typically performed also.Papanicolaou smears are not diagnostic but can suggest PMB etiologies (eg, cervicitis, sexually transmitted diseases, cervical and endometrial cancers). Go to: Treatment / Management The underlying cause primarily directs the treatment of PMB. However, other clinical factors, including patient comorbidities and preferences, as well as PMB characteristics (eg, heaviness and duration), are considered when determining management. Genitourinary atrophy: Bleeding is usually self-limited and requires no treatment. Vaginal dryness may be treated with nonhormonal vaginal moisturizers and lubricants to maintain sexual activity.Vulvar and vaginal atrophy symptoms can be effectively reversed by topical estrogen and are the preferred pharmacologic therapy for genitourinary atrophy. Oral hormone replacement and hormonal receptor modulators (eg, ospemifene) also may be considered if there is no improvement with other treatments. Endometrial polyps: Thirty percent of PMB cases are caused by endometrial polyps; however, polyps can be asymptomatic. Furthermore, the severity of PMB is not affected by the number or size of polyps. One percent of all endometrial polyps are malignant, most commonly occurring in postmenopausal women.Therefore, endometrial polyps in symptomatic postmenopausal women should be removed and histologically assessed. Surgical excision should be considered in asymptomatic women at higher risk of malignancy (eg, large polyps, tamoxifen use, obesity, or diabetes).Hysteroscopic polypectomy is the preferred treatment because the clinician can obtain directed biopsies and excise polyps at once. Uterine leiomyoma: Typically, leiomyomas (ie, fibroids) are benign and regress during menopause and may not need treatment if patients are asymptomatic. A small percentage of fibroids are malignant, primarily in postmenopausal women. Occasionally, benign leiomyomas may grow or become symptomatic even in postmenopausal patients, particularly in obese women, due to peripheral conversion of estrogen from adipose stores.In women with PMB found to have uterine fibroids with an otherwise normal evaluation, pharmacologic (eg, aromatase inhibitors and selective estrogen receptor modulators) or surgical (eg, myomectomy and hysterectomy) therapy may be considered. However, leiomyosarcomas can not be definitively excluded through laboratory or imaging studies, and symptomatic postmenopausal women are at higher risk; therefore, clinicians should thoroughly counsel patients and determine management through shared decision-making. Genitourinary infection: For sexually transmitted diseases and other genital infections, treatment is guided by vaginal culture results.Clinicians may consider oral doxycycline to treat endometritis. Cervical, vaginal, and vulvar carcinomas:Most treatment modalities for these PMB etiologies include surgery and chemoradiotherapy based on the stage. Endometrial hyperplasia or malignancy: Endometrial hyperplasia is classified as benign endometrial hyperplasia or intraepithelial neoplasia. Management of hyperplasia consists of surgical and nonsurgical therapies; the treatment approach is individualized based on clinical factors (eg, comorbid conditions). However, endometrial carcinoma requires surgical treatment and staging by gynecologic oncology specialists. Additionally, chemotherapy and radiation may be indicated for some patients. Benign endometrial hyperplasia: Also referred to as nonatypical endometrial hyperplasia, this type of hyperplasia is typically managed with hormonal therapy and/or curettage. Endometrial intraepithelial neoplasia: The approach to treatment depends on various clinical factors, including patient preferences. Generally, in postmenopausal women, fertility conservation is not desired; therefore, minimally invasive hysterectomy with bilateral salpingectomy is the preferred treatment. Shared decision-making should be utilized when determining if bilateral oophorectomy is also performed.Medical management is an option for patients who are declining surgery or are poor surgical candidates. Patients should be cautioned that many women with EIN have concurrent endometrial malignancy. If medical treatment is chosen, repeat endometrial sampling to assess therapeutic response should be done in 3 to 6 months for a minimum of 1 year. Nonsurgical therapies include: Megestrol 80 mg orally twice a day Depot medroxyprogesterone acetate 150 mg intramuscularly every 3 months Micronized vaginal progesterone 100 to 200 mg daily Levonorgestrel intrauterine device, 52 mg Medroxyprogesterone acetate orally 10 to 20 mg daily Endometrial adenocarcinoma: Definitive treatment with hysterectomy and comprehensive staging is the standard of care. Prognosis and appropriate adjuvant therapy are determined by staging. Hematuria:Genitourinary atrophy can lead to asymptomatic microscopic hematuria. Because the risk of urinary malignancy is lower in women without risk factors (eg, smokers, no gross hematuria) than in men, ACOG only recommends evaluation in asymptomatic low-risk women if urine microscopy shows >25 red blood cells per high-power field. Clinical findings and diagnostic studies consistent with acute cystitis should be treated with appropriate antibiotics. Gastrointestinal Bleeding: The differential diagnoses for PMB include gastrointestinal etiologies (eg, hemorrhoids and diverticulitis), which can be mistaken for vaginal bleeding. Management may consist of surgery, anti-inflammatories, or antibiotics. Recurrent or severe bleeding may necessitate referral to a gastrointestinal specialist for additional evaluation and treatment. Medications:Postmenopausal hormone replacement therapy frequently causes PMB for the first 2 to 3 months after initiation, which spontaneously resolves in most women. However, clinicians should evaluate for endometrial pathology in women with persistent or recurrent PMB following the first few months of therapy.For anticoagulants leading to vaginal bleeding, progestin therapy may improve bleeding until anticoagulants can be discontinued. Longer-term solutions may need to be discussed in patients on lifelong anticoagulation. Go to: Differential Diagnosis Clinicians evaluating patients with PMB must consider the various differential diagnoses that have a similar presentation to PMB which may arise from nongynecologic (eg, the urethra, bladder, or GI tract) or gynecologic areas.Common conditions that should also be considered when evaluating PMB include: Urogenital infections (eg, endometritis, vaginitis, cystitis, or cervicitis) Uterine leiomyomas Genital tract malignancies Vaginal foreign bodies Gastrointestinal conditions (eg, diverticulitis, colitis, hemorrhoids, malignancy) Genitourinary atrophy Radiation effects on adjacent organs (eg, hemorrhagic cystitis, proctitis, necrosis) Go to: Prognosis PMB has a favorable prognosis, as the most common etiologies are benign and treatable. Furthermore, the prognosis for the most common malignant PMB etiology, endometrial cancer, is significantly better than other cancers, with a 5-year survival rate of 90%.In patients diagnosed with endometrial hyperplasia who underwent hysterectomy, approximately 43% had undiagnosed concurrent endometrial carcinoma. Go to: Complications The primary complication of PMB is secondary anemia, which occurs in approximately 10% of postmenopausal women.Other complications that may occur are typically associated with the underlying etiology. For instance, genitourinary atrophy can decrease quality of life due to reduced sexual intimacy and self-esteem, while uterine fibroids may cause pelvic discomfort.Management of PMB etiologies, such as pharmacologic therapy of endometrial hyperplasia with megestrol, can also be associated with adverse effects, including weight gain, nausea, venous thromboembolism, and continued vaginal bleeding. Go to: Consultations Gynecologists are frequently consulted to evaluate and manage patients with PMB. Urologists or urogynecologists may also be helpful if a urologic etiology is suspected. Consultation with a gynecologic oncologist should be obtained for patients with a gynecologic malignancy as the prognosis is better for patients cared for by gynecologic oncologists. Go to: Deterrence and Patient Education Clinicians should provide perimenopausal patients with anticipatory education regarding menopausal transition and typical changes that may occur during menstrual cycles. Patients should be educated that any bleeding after menopause is established is considered abnormal. Patients should be instructed when to seek medical attention and asked about abnormal vaginal bleeding during routine office visits. Because primary care clinicians are often the first to evaluate patients with PMP, they should educate patients on addressing endometrial hyperplasia risk factors (eg, obesity, unopposed estrogen exposure). Furthermore, preventive therapies, including intrauterine devices, have been shown to decrease the rate of uterine cancer by 50%. Go to: Enhancing Healthcare Team Outcomes Postmenopausal bleeding presents a diagnostic challenge due to its diverse etiologies, requiring a comprehensive and collaborative approach by an interprofessional team to ensure patient-centered care. Physicians, advanced practitioners, nurses, pharmacists, radiologists, and pathologists each play a critical role in evaluating and managing this condition. Radiology and pathology specialists collaborate closely with primary care clinicians, gynecologists, and other ancillary team members to accurately identify the underlying cause and develop a tailored treatment plan. Effective interprofessional communication and shared decision-making are pivotal in aligning the team’s efforts, improving outcomes, and enhancing patient satisfaction. By fostering a holistic and integrated care approach, the team ensures not only the identification and treatment of the condition but also the provision of support and safety throughout the patient journey. For premalignant gynecologic conditions, timely consultation with gynecologic oncologists is crucial to optimizing care and improving clinical outcomes. This level of coordination highlights the importance of mutual respect, clear communication, and shared responsibilities among healthcare professionals to deliver exceptional care that prioritizes patient safety, enhances outcomes, and strengthens team performance. Go to: Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Go to: References 1. Carugno J. 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[PubMed: 29470333] Disclosure:Sharon Sung declares no relevant financial relationships with ineligible companies. Disclosure:Karen Carlson declares no relevant financial relationships with ineligible companies. Disclosure:Aaron Abramovitz declares no relevant financial relationships with ineligible companies. Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology Histopathology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Consultations Deterrence and Patient Education Enhancing Healthcare Team Outcomes Review Questions References Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. 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15452	https://en.wikipedia.org/wiki/Indian_numbering_system	Jump to content Indian numbering system বাংলা Català Deutsch Ελληνικά Español Français ગુજરાતી 한국어 हिन्दी Íslenska Italiano Latina Bahasa Melayu नेपाली 日本語 Norsk bokmål ଓଡ଼ିଆ پنجابی Português Romani čhib தமிழ் Türkçe اردو Edit links From Wikipedia, the free encyclopedia Indian convention of naming large numbers \| \| \| --- \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Indian numbering system" – news · newspapers · books · scholar · JSTOR (January 2018) (Learn how and when to remove this message) \| This article contains Indic text. Without proper rendering support, you may see question marks or boxes, misplaced vowels or missing conjuncts instead of Indic text. The Indian numbering system is used in India, Pakistan, Nepal, Sri Lanka, and Bangladesh to express large numbers, which differs from the International System of Units. Commonly used quantities include lakh (one hundred thousand, 105) and crore (ten million, 107) – written as 1,00,000 and 1,00,00,000 respectively in some locales. For example: 150,000 rupees is "1.5 lakh rupees" which can be written as "1,50,000 rupees", and 30,000,000 (thirty million) rupees is referred to as "3 crore rupees" which can be written as "3,00,00,000 rupees". There are names for numbers larger than crore, but they are less commonly used. These include arab (100 crore, 109), kharab (100 arab, 1011), nil or sometimes transliterated as neel (100 kharab, 1013), padma (100 nil, 1015), shankh (100 padma, 1017), and mahashankh (100 shankh, 1019). In common parlance (though inconsistent), the lakh and crore terminology repeats for larger numbers. Thus lakh crore is 1012. In the ancient Indian system, still in use in regional languages of India, there are words for[clarification needed] (1062). These names respectively starting at 1000 are sahasra, ayuta, laksha, niyuta, koti, arbhudha, abhja, karva, nikarva, mahapadma, shanmkhu, jaladhi, amtya, madhya, paraardha. In the Indian system, now prevalent in the northern parts,[clarification needed] the next powers of ten are one lakh, ten lakh, one crore, ten crore, one arab (or one hundred crore), and so on. Multiples [edit] The Indian system is decimal (base-10), same as in the International System of Units, and the first five orders of magnitude are named in a similar way: one (100), ten (101), one hundred (102), one thousand (103), and ten thousand (104). For higher powers of ten, naming diverges. The Indian system uses names for every second power of ten: lakh (105), crore (107), arab (109), kharab (1011), etc. In the rest of the world, long and short scales, there are names for every third power of ten. The short scale uses million (106), billion (109), trillion (1012), etc. Decimal formatting [edit] The Indian system groups digits of a large decimal is represented differently than the International System of Units. The Indian system does group the first three digits to the left of the decimal point, but thereafter, groups by two digits to align with the naming of quantities at multiples of 100. \| Indian \| English \| --- \| \| 5,00,000 \| 500,000 \| \| 12,34,56,789 \| 123,456,789 \| \| 17,00,00,00,000 \| 17,000,000,000 \| \| 6,78,90,00,00,00,00,000 \| 6,789,000,000,000,000 \| Like English and other locales, the Indian system uses a period as the decimal separator and the comma for grouping, while others use a comma for decimal separator and a thin space or point to group digits. Pronunciation in English [edit] When speakers of indigenous Indian languages are speaking English, the pronunciations may be closer to their mother tongue; e.g. "lakh" and "crore" might be pronounced /lɑkʰ/, /kɑrɔːr/, respectively. lakh /lɑːkʰ/ crore /kɹɔːɹ/ (or /kɹoʊɹ/ in American English) arab /ʌˈɾʌb/ kharab /kʰʌˈɾʌb/ Names of numbers [edit] The table below includes the spelling and pronunciation of numbers in various Indian languages along with corresponding short scale names. \| Value \| Short scale \| Indian English \| Hindustani Hindi / Urdu \| Marathi \| Bengali \| \| Odia \| \| Nepali \| Tamil \| Telugu \| Kannada \| Malayalam \| --- --- --- --- --- --- --- \| \| 1 \| one \| one \| एक / ایک (ēk) \| एक (ēk) \| এক (êk) \| \| ଏକ (ekå) \| \| एक (ēk) \| ஒன்று (oṉṟu) \| ఒకటి (okaṭi) \| ಒಂದು (ಬಿಡಿ) (ondu) \| ഒന്ന് (onn) \| \| 10 \| ten \| ten \| दस / دس (das) \| दहा (dahā) \| দশ (dôś) \| \| ଦଶ (dåśå) \| \| दश (daś) \| பத்து (pattu) \| పది (padi) \| ಹತ್ತು (hattu) \| പത്ത് (patt) \| \| 102 \| hundred \| hundred \| सौ / سو (sau) \| शंभर (śambhar) \| শত (śato) \| \| ଶହେ (śåhe) \| \| सय (saya) \| நூறு (nūṟu) \| వంద/నూరు (vanda/nūru) \| ನೂರು (nūru) \| നൂറ് (nuṟ) \| \| 103 \| thousand \| thousand \| हज़ार / ہزار (hazār) \| एक हजार (ēk hajār) \| হাজার (hāzār) \| \| ହଜାର (håjārå) \| ସହସ୍ର (såhåsrå) \| एक हजार (ēk hajār) \| ஆயிரம் (āyiram) \| వెయ్యి (veyyi) \| ಸಾವಿರ (sāvira) \| ആയിരം (āyiraṁ) \| \| 104 \| ten thousand \| ten thousand \| दस हज़ार / دس ہزار (das hazār) \| दहा हजार (dahā hajār) \| অযুত (ōjut) \| দশ হাজার (dôś hāzār) \| ଦଶ ହଜାର (dåśå håjārå) \| ଅୟୁତ (åyutå) \| दश हजार (daś hajār) \| பத்தாயிரம் (pattāyiram) ஆயுதம் (āyutam) \| పది వేలు (padi vēlu) \| ಹತ್ತು ಸಾವಿರ (ಅಯುತ) (hattu sāvira) \| പതിനായിരം (patināyiraṁ) \| \| 105 \| hundred thousand \| lakh \| लाख / لاکھ (lākh) \| एक लाख (ēk lākh) \| লক্ষ (lôkkhō) \| লাখ (lākh) \| ଲକ୍ଷ (låkṣå) \| \| एक लाख (ēk lākh) \| இலட்சம் (ilaṭcam) நியுதம் (niyutam) \| లక్ష (lakṣa) \| ಲಕ್ಷ (lakṣa) \| ലക്ഷം (lakṣaṁ) \| \| 106 \| million \| ten lakh \| दस लाख / دس لاکھ (das lākh) \| दहा लाख (dahā lākh) \| নিযুত (nijut) \| দশ লাখ (dôś lākh) \| ଦଶ ଲକ୍ଷ (dåśå låkṣå) \| ନିୟୁତ (niyutå) \| दश लाख (daś lākh) \| பத்து இலட்சம் (pattu ilaṭcam) \| పది లక్షలు (padi lakṣalu) \| ಹತ್ತು ಲಕ್ಷ (ನಿಯುತ) (hattu lakṣa) \| പത്തുലക്ഷം (pattulakṣaṁ) \| \| 107 \| ten million \| crore \| करोड़ / کروڑ (karōṛ) \| एक कोटी (ēk kōṭī) \| কোটি (kōṭi) \| \| କୋଟି (koṭi) \| \| एक करोड (ēk karoḍ) \| கோடி (kōṭi) \| కోటి (kōṭi) \| ಕೋಟಿ (kōṭi) \| കോടി (kōṭi) \| \| 108 \| hundred million \| ten crore \| दस करोड़ / دس کروڑ (das karōṛ) \| दहा कोटी (dahā kōṭī) \| অর্বুদ (ōrbud) \| দশ কোটি (dôś kōṭi) \| ଦଶ କୋଟି (dåśå koṭi) \| ଅର୍ବୁଦ (årbudå) \| दश करोड (daś karoḍ) \| அற்புதம் (aṟputam) \| పది కోట్లు (padi kōṭlu) \| ಹತ್ತು ಕೋಟಿ (ಅರ್ಭುಧ) (hattu kōṭi) \| പത്തുകോടി (pattukōṭi) \| \| 109 \| billion \| arab / hundred crore \| अरब / ارب (arab) सौ करोड़ / سو کروڑ (sau karōṛ) \| एक अब्ज (ēk abja) \| মহার্বুদ (môhārbud) \| একশ কোটি (êkśō kōṭi) \| ଶହେ କୋଟି (śåhe koṭi) \| ବୃନ୍ଦ (brundå) \| एक अर्ब (ēk arba) \| நிகற்புதம் (nikaṟputam) \| వంద కోట్లు (vanda kōṭlu) \| ನೂರು ಕೋಟಿ (ಅಭ್ಜ) (nūru kōṭi) \| നൂറുകോടി (nūṟukōṭi) \| \| 1010 \| ten billion \| ten arab / thousand crore \| दस अरब / دس ارب (das arab) एक हज़ार करोड़ / ایک ہزار کروڑ (ēk hazār karōṛ) \| एक खर्व (ek kharva) \| খর্ব (khôrbō) \| হাজার কোটি (hāzār kōṭi) \| ହଜାର କୋଟି (håjārå koṭi) \| ଖର୍ବ (khårbå) \| दश अर्ब (daś arba) \| கும்பம் (kumpam) \| వెయ్యి కోట్లు (veyyi kōṭlu) \| ಒಂದು ಸಾವಿರ ಕೋಟಿ (ಕರ್ವ) (ondu sāvira kōṭi) \| ആയിരം കോടി (āyiraṁ kōṭi) \| \| 1011 \| hundred billion \| kharab / hundred arab / ten thousand crore \| खरब / کھرب (kharab) \| एक निखर्व (ek nikharva) \| মহাখর্ব (môhākhôrbō) দশ হাজার কোটি (dôś hājār kōṭi) \| \| ଦଶ ହଜାର କୋଟି (dåśå håjārå koṭi) \| ନିଖର୍ବ (nikhårbå) \| एक खर्ब (ēk kharba) \| கணம் (kaṇam) \| పది వేల కోట్లు (padi vēla kōṭlu) \| ಹತ್ತು ಸಾವಿರ ಕೋಟಿ (ನಿಕರ್ವ) (hattu sāvira kōṭi) \| പതിനായിരം കോടി (patināyiraṁ kōṭi) \| \| 1012 \| trillion \| ten kharab / one thousand arab / one lakh crore \| दस खरब / دس کھرب (das kharab) एक लाख करोड़ / ایک لاکھ کروڑ (ēk lākh karōṛ) \| एक पद्म (ēk padma) \| শঙ্খ (śôṅkhō) লাখ কোটি (lākh kōṭi) \| \| ଲକ୍ଷ କୋଟି (låkṣå koṭi) \| ଶଙ୍ଖ (śåṅkhå) \| दश खर्ब (daś kharba) \| கற்பம் (kaṟpam) \| లక్ష కోట్లు (lakṣa kōṭlu) \| ಒಂದು ಲಕ್ಷ ಕೋಟಿ (ಮಹಾಪದ್ಮ) (ondu lakṣa kōṭi) \| ഒരു ലക്ഷം കോടി (oru lakṣaṁ kōṭi) \| \| 1013 \| ten trillion \| nil / hundred kharab / ten thousand arab / ten lakh crore \| नील / نیل (nīl) \| एक महापद्म (ek mahāpadma) \| মহাশঙ্খ (môhāśôṅkhō) দশ লাখ কোটি (dôś lākh kōṭi) \| \| ଦଶ ଲକ୍ଷ କୋଟି (dåśå låkṣå koṭi) \| ପଦ୍ମ (pådmå) \| नील (nīl) \| நிகற்பம் (nikaṟpam) \| పది లక్షల కోట్లు (padi lakṣala kōṭlu) \| ಹತ್ತು ಲಕ್ಷ ಕೋಟಿ (ಶಂಖು) (hattu lakṣa kōṭi) \| പത്തുലക്ഷം കോടി (pattulakṣaṁ kōṭi) \| \| 1014 \| hundred trillion \| ten nil / crore crore \| दस नील / دس نیل (das nīl) एक करोड़ करोड़ / ایک کروڑ کروڑ (ēk karōṛ karōṛ) \| एक शंखू (ēk śaṅkhū) \| পদ্ম (pôddō) একশ লাখ কোটি (êkśō lākh kōṭi) শতলক্ষ কোটি (śôtôkōṭi lôkkō) \| \| ଶହେ ଲକ୍ଷ କୋଟି (śåhe låkṣå koṭi) \| ସାଗର (sāgårå) \| दश नील (daś nīl) \| பதுமம் (patumam) \| కోటి కోట్లు (kōṭi kōṭlu) \| ಒಂದು ಕೋಟಿ ಕೋಟಿ (ಜಲಧಿ) (ondu kōṭi kōṭi) \| നൂറ് ലക്ഷം കോടി (nuṟ lakṣaṁ kōṭi) \| \| 1015 \| quadrillion \| padma / hundred nil / ten crore crore \| पद्म / پدم (padma) \| एक जलधि शंखू (eka jaladhi śaṅkhū) \| মহাপদ্ম (môhāpôddō) হাজার লাখ কোটি (hāzār lākh kōṭi) \| \| ହଜାର ଲକ୍ଷ କୋଟି (håjārå låkṣå koṭi) \| ଅନ୍ତ୍ୟ (åntyå) \| पद्म (padma) \| சங்கம் (caṅkam) \| పది కోట్ల కోట్లు (padi kōṭla kōṭlu) \| ಹತ್ತು ಕೋಟಿ ಕೋಟಿ (ಅಂತ್ಯ) (hattu kōṭi kōṭi) \| ആയിരം ലക്ഷം കോടി (āyiraṁ lakṣaṁ kōṭi) \| \| 1016 \| ten quadrillion \| ten padma / hundred crore crore \| दस पद्म / دس پدم (das padma) \| एक अंत्य (eka antya) \| বঙ্গ (bongo) দশ হাজার লাখ কোটি (dôś hāzār lākh kōṭi) \| \| ଦଶ ହଜାର ଲକ୍ଷ କୋଟି (dåśå håjārå låkṣå koṭi) ମଧ୍ୟ (mådhyå) \| \| दश पद्म (daś padma) \| வெள்ளம் (veḷḷam) சமுத்திரம் (camuttiram) \| వంద కోట్ల కోట్లు (vanda kōṭla kōṭlu) \| ನೂರು ಕೋಟಿ ಕೋಟಿ (ಮಧ್ಯ) (nūru kōṭi kōṭi) \| പതിനായിരം ലക്ഷം കോടി (patināyiraṁ lakṣaṁ kōṭi) \| \| 1017 \| hundred quadrillion \| shankh / hundred padma / thousand crore crore / lakh lakh crore \| शंख / شنکھ (śaṅkh) \| एक परार्ध (eka parārdha) \| মহাবঙ্গ (mohabongo) শত হাজার লাখ কোটি (śoto hāzār lākh kōṭi) \| \| ଶହେ ହଜାର ଲକ୍ଷ କୋଟି (śåhe håjārå låkṣå koṭi) ପରାର୍ଦ୍ଧ (pårārddhå) \| \| शंख (śaṅkha) \| அந்நியம் (anniyam) \| వెయ్యి కోట్ల కోట్లు (veyyi kōṭla kōṭlu) \| ಒಂದು ಸಾವಿರ ಕೋಟಿ ಕೋಟಿ (ಪರಾರ್ಧ) (ondu sāvira kōṭi kōṭi) \| ലക്ഷം ലക്ഷം കോടി (lakṣaṁ lakṣaṁ kōṭi) \| \| 1018 \| quintillion \| ten shankh / ten thousand crore crore \| दस शंख / دس شنکھ (das śaṅkh) गुलशन / گلشن (gulśan) \| एक महापरार्ध (eka mahāparārdha) \| গুলশান (gulśān) দশ শত হাজার লাখ কোটি (doś śoto hāzār lākh kōṭi) \| \| ଦଶ ଶହେ ହଜାର ଲକ୍ଷ କୋଟି (dåśå śåhe håjārå låkṣå koṭi) ଦଶ ପରାର୍ଦ୍ଧ (dåśå pårārddhå) \| \| दश शंख (daś śaṅkha) \| அர்த்தம் (arttam) \| పది వేల కోట్ల కోట్లు (padi vēla kōṭla kōṭlu) \| ಹತ್ತು ಸಾವಿರ ಕೋಟಿ ಕೋಟಿ (hattu sāvira kōṭi kōṭi) \| പത്തുലക്ഷം ലക്ഷം കോടി (pattulakṣaṁ lakṣaṁ kōṭi) \| Historic numbering systems [edit] Numbering systems in Hindu epics [edit] There are various systems of numeration found in various ancient epic literature of India (itihasas). The following table gives one such system used in the Valmiki Ramayana. \| Name \| Indian decimal \| Value \| Short scale \| --- --- \| \| एक (ēka) \| 0,00,001 \| 1 \| one \| \| दश (daśa) \| 0,00,010 \| 10 \| ten \| \| शत (śata) \| 0,00,100 \| 102 \| hundred \| \| सहस्र (sahasra) \| 0,01,000 \| 103 \| thousand \| \| लक्ष (lakṣa) \| 1,00,000 \| 105 \| hundred thousand \| \| कोटि (kōṭi) \| 1,00,00,000 \| 107 \| ten million \| \| शङ्कु (śaṅku) \| 1,00,000 koṭi \| 1012 \| trillion \| \| महाशङ्कु (mahāśaṅku) \| 1,00,000 śaṅku \| 1017 \| hundred quadrillion \| \| वृन्द (vr̥nda) \| 1,00,000 mahāśaṅku \| 1022 \| ten sextillion \| \| महावृन्द (mahāvr̥nda) \| 1,00,000 vr̥nda \| 1027 \| octillion \| \| पद्म (padma) \| 1,00,000 mahāvr̥nda \| 1032 \| hundred nonillion \| \| महापद्म (mahāpadma) \| 1,00,000 padma \| 1037 \| ten undecillion \| \| खर्व (kharva) \| 1,00,000 mahāpadma \| 1042 \| tredecillion \| \| महाखर्व (mahākharva) \| 1,00,000 kharva \| 1047 \| hundred quattuordecillion \| \| समुद्र (samudra) \| 1,00,000 mahākharva \| 1052 \| ten sexdecillion \| \| ओघ (ogha) \| 1,00,000 samudra \| 1057 \| octodecillion \| \| महौघ (mahaugha) \| 1,00,000 ogha \| 1062 \| hundred novemdecillion \| Other numbering systems [edit] The denominations by which land was measured in the Kumaon Kingdom were based on arable lands and thus followed an approximate system with local variations. The most common of these was a vigesimal (base-20) numbering system with the main denomination called a bisi (see Hindustani number bīs), which corresponded to the land required to sow 20 nalis of seed. Consequently, its actual land measure varied based on the quality of the soil. This system became the established norm in Kumaon by 1891. Usage in different languages [edit] Main articles: lakh and crore Below is a list of translations for the words lakh and crore in other languages spoken in the Indian subcontinent: Usage of lakh and crore in other Indic languages \| Language \| Lakh \| Crore \| 100 crores (a billion) \| 1000 crores (10 billion) \| 10,000 crores (100 billion) \| \| Assamese \| লক্ষ lokhyo, or লাখ lakh \| কৌটি kouti, or কোটি koti \| — \| \| \| \| Bengali \| লাখ lākh[a], or লক্ষ lokkho[b] \| কোটি kōṭi \| \| Burmese \| lakh[c] \| ကုဋေ [ɡədè] \| \| Dhivehi \| ލައްކަ la'kha \| ކްރޯރް kroaru \| \| Gujarati \| લાખ lākh \| કરોડ karoḍ \| અબજ abaj \| — \| \| \| Hindi \| लाख lākh \| करोड karoḍ \| अरब arab \| \| Kannada \| ಲಕ್ಷ lakṣha \| ಕೋಟಿ kōṭi \| — \| \| \| \| Khasi \| lak \| klur or krur \| arab \| — \| kharab \| \| Malayalam \| ലക്ഷം laksham \| കോടി kodi \| — \| \| \| \| Marathi \| लाख/लक्ष lākh \| कोटी koṭi, or करोड karoḍ \| अब्ज abja \| — \| \| \| Nepali \| लाख lākh \| करोड karoḍ \| — \| \| \| \| Odia \| ଲକ୍ଷ lôkhyô \| କୋଟି koṭi \| \| Punjabi \| lakkh (Gurmukhi: ਲੱਖ, Shahmukhi: لکھ) \| karoṛ (Gurmukhi: ਕਰੋੜ, Shahmukhi: کروڑ) \| \| Rohingya \| lák \| kurul \| — \| kuthí \| — \| \| Sinhala \| ලක්ෂ lakṣa \| කෝටි kōṭi \| — \| \| \| \| Tamil \| இலட்சம் ilaṭcam \| கோடி kōṭi \| \| Telugu \| లక్ష lakṣha \| కోటి kōṭi \| \| Urdu \| لاکھ lākh \| کروڑ karoṛ \| ارب arab \| — \| کھرب kharab \| \| Swahili \| laki \| — \| \| \| \| Formal written publications in English in India tend to use lakh/crore for Indian currency and International numbering for foreign currencies. Current usage [edit] The official usage of this system is limited to the nations of India, Pakistan and Bangladesh. It is universally employed within these countries, and is preferred to the International numbering system. Sri Lanka and Nepal used this system in the past but has switched to the International numbering system in recent years. In the Maldives, the term lakh is widely used in official documents and local speech. However, the International System of Units is preferred for higher denominations (such as millions).[citation needed] Most institutions and citizens in India use the Indian number system. The Reserve Bank of India was noted as a rare exception in 2015, whereas by 2024 the Indian system was used for amounts in rupees and the International system for foreign currencies throughout the Reserve Bank's website. See also [edit] Scientific notation Japanese notation References [edit] Notes ^ Tadbhava ^ Ardha-tatsama ^ in Burmese English References ^ "Knowing our Numbers". Department Of School Education And Literacy. National Repository of Open Educational Resources. Archived from the original on 16 February 2016. Retrieved 13 February 2016. ^ Emmons, John (25 March 2018). "UNICODE LOCALE DATA MARKUP LANGUAGE (LDML) PART 3: NUMBERS". Unicode.org. Archived from the original on 25 July 2018. Retrieved 25 March 2018. ^ "Decimal and Thousands Separators (International Language Environments Guide)". docs.oracle.com. Retrieved 29 April 2021. ^ "Valmiki Ramayana - Yuddha Kanda, Chapter (Sarga) 28,(Verses 33 – 38)". Retrieved 15 July 2021. ^ Traill, G.W. (1828). Batten, J.H. (ed.). Statistical Sketch of Kamaon. John Murray. p. 34. ^ "North Indian Notes and Queries". Google Books. 1–3. Pioneer Press: 216. 1891. ^ Shapiro, Richard (16 August 2012). "The most distinctive counting system in English? Indian cardinal numbers". Oxford English Dictionary. Archived from the original on 30 June 2020. Retrieved 24 May 2020. Shapiro is/was an OED employee. The article states: "The opinions and other information contained in the OED blog posts and comments do not necessarily reflect the opinions or positions of Oxford University Press." ^ Krulwich, Robert; Block, Ezra (21 October 2010). "Hey! Who Can Explain What India Does With Its Commas? (Not Commies. Commas.)". NPR. Retrieved 3 August 2020. ^ Gurpur (10 August 2015). "Can we follow Indian numbering system for simplicity and good order?". Moneylife News & Views. Retrieved 3 September 2020. ^ Reserve Bank of India, old and new website (known on archive.org since 5 Apr 2024). Retrieved 16 Apr 2024. 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15453	https://www.cliffsnotes.com/study-notes/16924722	Lit NotesStudy GuidesDocumentsQ&A Chat PDF Log In Biochem chapter 19 .pdf Boston CollegeWe aren't endorsed by this school BIOCHEMIST 2 Aug 1, 2024 Uploaded by EarlNightingalePerson230 Home/ Biology Lehninger principles of biochemistry test bank ch 19pdf Biochemistry 2 (The University of British Columbia) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Lehninger principles of biochemistry test bank ch 19pdf Biochemistry 2 (The University of British Columbia) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Stephany Arevalo (satiffy1@gmail.com) lOMoARcPSD\|36807583 Chapter 19 Oxidative Phosphorylation Multiple Choice Questions 1. Electron-transfer reactions in mitochondria Almost all of the oxygen (O 2 ) one consumes in breathing is converted to: A) acetyl- CoA. B) carbon dioxide (CO 2 ). C) carbon monoxide and then to carbon dioxide. D) none of the above. E) water. 2. Electron-transfer reactions in mitochondria A new compound isolated from mitochondria is claimed to represent a previously unrecognized carrier in the electron transfer chain. It is given the name coenzyme Z. Which line of evidence do you feel is the least conclusive in assigning this compound a position in the electron transfer chain? A) Alternate oxidation and reduction of the mitochondrion-bound coenzyme Z can be readily demonstrated. B) Removal of coenzyme Z from the mitochondria results in a decreased rate of oxygen consumption. C) The rate of oxidation and reduction of mitochondrion-bound coenzyme is of the same order of magnitude as the overall rate of electron transfer in mitochondria as measured by oxygen consumption. D) The reduction potential of Z is between that of two compounds known to participate in the electron transport chain E) When added to a mitochondrial suspension, coenzyme Z is taken up very rapidly and specifically by the mitochondria. 3. Electron-transfer reactions in mitochondria Antimycin A blocks electron transfer between cytochromes b and c 1 . If intact mitochondria were incubated with antimycin A, excess NADH, and an adequate supply of O 2 , which of the following would be found in the oxidized state? A) Coenzyme Q B) Cytochrome a 3 C) Cytochrome b D) Cytochrome e E) Cytochrome f Downloaded by Stephany Arevalo (satiffy1@gmail.com) lOMoARcPSD\|36807583 Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. Chapter 19 Oxidative Phosphorylation and Photophosphorylation 221 4. Electron-transfer reactions in mitochondria Cyanide, oligomycin, and 2,4-dinitrophenol (DNP) are inhibitors of mitochondrial aerobic phosphorylation. Which of the following statements correctly describes the mode of action of the three inhibitors? A) Cyanide and 2,4-dinitrophenol inhibit the respiratory chain, and oligomycin inhibits the synthesis of ATP. B) Cyanide inhibits the respiratory chain, whereas oligomycin and 2,4-dinitrophenol inhibit the synthesis of ATP. C) Cyanide, oligomycin, and 2,4-dinitrophenol compete with O 2 for cytochrome oxidase (Complex IV). D) Oligomycin and cyanide inhibit synthesis of ATP; 2,4-dinitrophenol inhibits the respiratory chain. E) Oligomycin inhibits the respiratory chain, whereas cyanide and 2,4-dinitrophenol prevent the synthesis of ATP. 5. Electron-transfer reactions in mitochondria In the reoxidation of QH 2 by purified ubiquinone-cytochrome c reductase (Complex III) from heart muscle, the overall stoichiometry of the reaction requires 2 mol of cytochrome c per mole of QH 2 because: A) cytochrome c is a one-electron acceptor, whereas QH 2 is a two-electron donor. B) cytochrome c is a two-electron acceptor, whereas QH 2 is a one-electron donor. C) cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes D) heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH 2 for electron transfer to proceed normally. E) two molecules of cytochrome c must first combine physically before they are catalytically active. 6. ATP synthesis If electron transfer in tightly coupled mitochondria is blocked (with antimycin A) between cytochrome b and cytochrome c 1 , then: A) all ATP synthesis will stop. B) ATP synthesis will continue, but the P/O ratio will drop to one. C) electron transfer from NADH will cease, but O 2 uptake will continue. D) electron transfer from succinate to O 2 will continue unabated. E) energy diverted from the cytochromes will be used to make ATP, and the P/O ratio will rise. Downloaded by Stephany Arevalo (satiffy1@gmail.com) lOMoARcPSD\|36807583 Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. Page1of 14 Students also studied Honors Biology Midterm Exam Review Guide.docx Honors Biology Midterm Exam Review Guide Unit 1 - Introduction to Science 1. Know the parts of the scientific method and their definitions Observation: using your 5 senses. Hypothesis: an educated guess based on the facts that you believe will happen in t Our Lady of Fatima University, Antipolo City BIO 101 Assignment 5 Ch 11&12.xlsx Problem 1 5 points Demand for an item is 1,000 units per year. Each order placed costs $10; the annual cost to carry items in invento Note: If needed, please check the related example (one formular with three variables) in our textbook as we did no SQRT(2 Houston Community College ALL 10 Chapter22.pdf Chapter 22 Biosynthesis of Amino Acids, Nucleotides, and Related Molecules Multiple Choice Questions 1. Overview of nitrogen metabolism Pages: 854-855 Difficulty: 2 Ans: D Which of the following statements about the fixation of atmospheric nitrogen (N2) i Korea University CS, BIOLOG 101. 238 Photosynthesis Leaf Disk Practical Report FINAL SINA MARDANI .pdf Photosynthesis Leaf Disk Practical Report SACE ID: 251116J Sina Mardani \| Biology \| 22/03/2023 Introduction Photosynthesis is the process that occurs in plants, algae, and certain bacteria. Photosynthesis is the process in which these organisms convert l The University of Adelaide BIOLOGY 3664 (2) Mini Virtual Lab Calculating GPP and NPP.docx.pdf Mini Lab: Calculating GPP and NPP Name _ Open this Video and answer the questions while watching the video: https:/youtu.be/voSQZcqgYNY 1. Where and what is the sample of water used in this lab? The sample water is from an algae covered fish tank. 2. What Juan N. Corpas University Foundation DA 1 Kami Export - Germani Reed - Case Study - Killing Chloroplast.pdf NATIONAL CENTER FOR CASE STUDY TEACHING IN SCIENCE Killing Chloroplasts: Herbicides Targeting Photosynthesis by Angela K. Hartsock Department of Biology University of Akron Wayne College, Orrville, OH Part I - Illuminating Photosynthesis Welcome to our pl John Tyler H S BIOLOGY 12 Study your doc or PDF Upload your materials to get instant AI help Students also studied Honors Biology Midterm Exam Review Guide.docx Honors Biology Midterm Exam Review Guide Unit 1 - Introduction to Science 1. Know the parts of the scientific method and their definitions Observation: using your 5 senses. Hypothesis: an educated guess based on the facts that you believe will happen in t Our Lady of Fatima University, Antipolo City BIO 101Assignment 5 Ch 11&12.xlsx Problem 1 5 points Demand for an item is 1,000 units per year. Each order placed costs $10; the annual cost to carry items in invento Note: If needed, please check the related example (one formular with three variables) in our textbook as we did no SQRT(2 Houston Community College ALL 10Chapter22.pdf Chapter 22 Biosynthesis of Amino Acids, Nucleotides, and Related Molecules Multiple Choice Questions 1. Overview of nitrogen metabolism Pages: 854-855 Difficulty: 2 Ans: D Which of the following statements about the fixation of atmospheric nitrogen (N2) i Korea University CS, BIOLOG 101. 238Photosynthesis Leaf Disk Practical Report FINAL SINA MARDANI .pdf Photosynthesis Leaf Disk Practical Report SACE ID: 251116J Sina Mardani \| Biology \| 22/03/2023 Introduction Photosynthesis is the process that occurs in plants, algae, and certain bacteria. Photosynthesis is the process in which these organisms convert l The University of Adelaide BIOLOGY 3664(2) Mini Virtual Lab Calculating GPP and NPP.docx.pdf Mini Lab: Calculating GPP and NPP Name _ Open this Video and answer the questions while watching the video: https:/youtu.be/voSQZcqgYNY 1. Where and what is the sample of water used in this lab? The sample water is from an algae covered fish tank. 2. What Juan N. Corpas University Foundation DA 1Kami Export - Germani Reed - Case Study - Killing Chloroplast.pdf NATIONAL CENTER FOR CASE STUDY TEACHING IN SCIENCE Killing Chloroplasts: Herbicides Targeting Photosynthesis by Angela K. 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15454	https://dokumen.pub/additive-combinatorics-9780521853866-9780511245305-0511245300-0521853869.html	Additive Combinatorics 9780521853866, 9780511245305, 0511245300, 0521853869 - DOKUMEN.PUB Anmelden Registrierung Deutsch English Español Português Français Dom Najlepsze kategorie CAREER & MONEY PERSONAL GROWTH POLITICS & CURRENT AFFAIRS SCIENCE & TECH HEALTH & FITNESS LIFESTYLE ENTERTAINMENT BIOGRAPHIES & HISTORY FICTION Najlepsze historie Najlepsze historie Dodaj historię Moje historie Home Additive Combinatorics 9780521853866, 9780511245305, 0511245300, 0521853869 Additive Combinatorics 9780521853866, 9780511245305, 0511245300, 0521853869 Additive combinatorics is the theory of counting additive structures in sets. This theory has seen exciting developments 499 63 3MB English Pages 532 Year 2006 Report DMCA / Copyright DOWNLOAD FILE Polecaj historie ###### Graph theory and additive combinatorics: exploring structure and randomness 9781009310949 1,230 185 37MB Read more ###### Combinatorics 475 143 393KB Read more ###### Additive Manufacturing Technologies 991 67 8MB Read more ###### Additive Manufacturing Technologies 9781441911193 877 120 6MB Read more ###### Machining and Additive Manufacturing 125 70 31MB Read more ###### Olympiad Combinatorics 2,546 305 6MB Read more ###### Additive Manufacturing Technologies 1,646 95 6MB Read more ###### Non Additive Geometry 114 42 13MB Read more ###### Industrializing Additive Manufacturing 118 44 95MB Read more ###### Design for Additive Manufacturing 125 77 81MB Read more Author / Uploaded Terence Tao _Table of contents : Additive Combinatorics Half-title Series-title Title Copyright Contents Prolouge Acknowledgements General notation Sets and functions Number systems Landau asymptotic notation Progressions Other notation 1 The probabilistic method 1.1 The first moment method 1.1.1 Sum-free sets 1.2 The second moment method 1.2.1 The number of prime divisors 1.3 The exponential moment method 1.3.1 Sidon’s problem on thin bases 1.3.2 Complementary bases 1.4 Correlation inequalities 1.4.1 Asymptotic complementary bases 1.5 The Lovász local lemma 1.5.1 Colorings of the real line 1.6 Janson’s inequality 1.7 Concentration of polynomials 1.7.1 Bh[g] sets 1.8 Thin bases of higher order 1.9 Thin Waring bases 1.10 Appendix: the distribution of the primes 2 Sum set estimates 2.1 Sum sets 2.2 Doubling constants 2.3 Ruzsa distance and additive energy 2.4 Covering lemmas 2.5 The Balog–Szemerédi–Gowers theorem 2.6 Symmetry sets and imbalanced partial sum sets 2.7 Non-commutative analogues 2.8 Elementary sum-product estimates Chapter 3 Additive geometry 3.1 Additive groups 3.1.1 Lattices 3.1.2 Quotients of lattices 3.2 Progressions 3.3 Convex bodies 3.4 The Brunn–Minkowski inequality 3.5 Intersecting a convex set with a lattice 3.6 Progressions and proper progressions 4 Fourier-analytic methods 4.1 Basic theory 4.2 L theory 4.3 Linear bias 4.4 Bohr sets 4.5 Lambda( p) constants, Bh[g] sets, and dissociated sets 4.6 The spectrum of an additive set 4.7 Progressions in sum sets Chapter 5 Inverse sum set theorems 5.1 Minimal size of sum sets and the e-transform 5.2 Sum sets in vector spaces 5.3 Freiman homomorphisms 5.4 Torsion and torsion-free inverse theorems 5.5 Universal ambient groups 5.6 Freiman’s theorem in an arbitrary group 6 Graph-theoretic methods 6.1 Basic Notions 6.2 Independent sets, sum-free subsets, and Sidon sets 6.2.1 Sum-free subsets 6.2.2 Turán’s theorem and triangle-free graphs 6.2.3 Sidon sets 6.3 Ramsey theory 6.4 Proof of the Balog–Szemerédi–Gowers theorem 6.5 Plünnecke’s theorem 6.5.1 Main ideas of the proof 6.5.2 The first step 6.5.3 The second step Chapter 7 The Littlewood–Offord problem 7.1 The combinatorial approach 7.2 The Fourier-analytic approach 7.3 The Esséen concentration inequality 7.4 Inverse Littlewood–Offord results 7.5 Random Bernoulli matrices 7.6 The quadratic Littlewood–Offord problem Chapter 8 Incidence geometry 8.1 The crossing number of a graph 8.2 The Szemerédi–Trotter theorem 8.3 The sum-product problem in R 8.4 Cell decompositions and the distinct distances problem 8.5 The sum-product problem in other fields Chapter 9 Algebraic methods 9.1 The combinatorial Nullstellensatz 9.2 Restricted sum sets 9.3 Snevily’s conjecture 9.4 Finite fields 9.5 Davenport’s problem 9.6 Kemnitz’s conjecture 9.7 Stepanov’s method 9.8 Cyclotomic fields, and the uncertainty principle Chapter 10 Szemerédi’s theorem for k = 3 10.1 General strategy 10.2 The small torsion case 10.3 The integer case 10.4 Quantitative bounds 10.5 An ergodic argument 10.6 The Szemerédi regularity lemma 10.7 Szemerédi’s argument Chapter 11 Szemerédi’s theorem for k > 3 11.1 Gowers uniformity norms 11.2 Hard obstructions to uniformity 11.3 Proof of Theorem 11.6 11.3.1 Locating a somewhat linear phase derivative 11.3.2 Obtaining a perfectly linear phase derivative 11.3.3 Symmetrizing the derivative 11.3.4 Eliminating the quadratic phase 11.4 Soft obstructions to uniformity 11.5 The infinitary ergodic approach 11.6 The hypergraph approach 11.7 Arithmetic progressions in the primes Chapter 12 Long arithmetic progressions in sum sets 12.1 Introduction 12.2 Proof of Theorem 12.4 12.3 Generalizations and variants 12.4 Complete and subcomplete sequences 12.5 Proof of Theorem 12.17 12.6 Further applications 12.6.1 Olson’s problem 12.6.2 Monochromatic sum sets Bibliography Index Пустая страница Пустая страница_ Citation preview This page intentionally left blank CAMBRIDGE STUDIES IN ADVANCED MATHEMATICS 105 EDITORIAL BOARD B. BOLLOBAS, W. FULTON, A. KATOK, F. KIRWAN, P. SARNAK, B. SIMON, B. TOTARO ADDITIVE COMBINATORICS Additive combinatorics is the theory of counting additive structures in sets. This theory has seen exciting developments and dramatic changes in direction in recent years, thanks to its connections with areas such as number theory, ergodic theory and graph theory. This graduate level textbook will allow students and researchers easy entry into this fascinating field. Here, for the first time, the authors bring together, in a self-contained and systematic manner, the many different tools and ideas that are used in the modern theory, presenting them in an accessible, coherent, and intuitively clear manner, and providing immediate applications to problems in additive combinatorics. The power of these tools is well demonstrated in the presentation of recent advances such as the Green-Tao theorem on arithmetic progressions and Erd˝os distance problems, and the developing field of sum-product estimates. The text is supplemented by a large number of exercises and new material. Te rence Tao is a professor in the Department of Mathematics at the University of California, Los Angeles. Van Vu is a professor in the Department of Mathematics at Rutgers University, New Jersey. CAMBRIDGE STUDIES IN ADVANCED MATHEMATICS Editorial Board: B. Bollob´as, W. Fulton, A. Katok, F. Kirwan, P. Sarnak, B. Simon, B. Totaro Au the title, listed below can be obtained from good booksellers or from Cambridge University Press for a complete listing visit www.cambridge.org/uk/series/&Series.asp?code=CSAM. 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 81 82 83 84 85 86 87 89 90 91 92 93 95 R. Stanley Enumerative combinatorics I I. Porteous Clifford algebras and the classical groups M. Audin Spinning tops V. Jurdjevic Geometric control theory H. Volklein Groups as Galois groups J. Le Potier Lectures on vector bundles D. Bump Automorphic forms and representations G. Laumon Cohomology of Drinfeld modular varieties II D.M. Clark & B.A. Davey Natural dualities for the working algebraist J. McCleary A user’s guide to spectral sequences II P. Taylor Practical foundations of mathematics M.P. Brodmann & R.Y. Sharp Local cohomology J.D. Dixon et al. Analytic pro-P groups R. Stanley Enumerative combinatorics II R.M. Dudley Uniform central limit theorems J. Jost & X. Li-Jost Calculus of variations A.J. Berrick & M.E. Keating An introduction to rings and modules S. Morosawa Holomorphic dynamics A.J. Berrick & M.E. Keating Categories and modules with K-theory in view K. Sato Levy processes and infinitely divisible distributions H. Hida Modular forms and Galois cohomology R. Iorio & V. Iorio Fourier analysis and partial differential equations R. Blei Analysis in integer and fractional dimensions F. Borceaux & G. Janelidze Galois theories B. Bollobas Random graphs R.M. Dudley Real analysis and probability T. Sheil-Small Complex polynomials C. Voisin Hodge theory and complex algebraic geometry I C. Voisin Hodge theory and complex algebraic geometry II V. Paulsen Completely bounded maps and operator algebras F. Gesztesy & H. Holden Soliton Equations and their Algebro-Geometric Solutions Volume 1 Shigeru Mukai An Introduction to Invariants and Moduli G. Tourlakis Lectures in logic and set theory I G. Tourlakis Lectures in logic and set theory II R.A. Bailey Association Schemes James Carlson, Stefan M¨uller-Stach, & Chris Peters Period Mappings and Period Domains J.J. Duistermaat & J.A.C. Kolk Multidimensional Real Analysis I J.J. Duistermaat & J.A.C. Kolk Multidimensional Real Analysis II M. Golumbic & A.N. Trenk Tolerance Graphs L.H. Harper Global Methods for Combinatorial Isoperimetric Problems I. Moerdijk & J. Mrcun Introduction to Foliations and Lie Groupoids J´anos Koll´ar, Karen E. Smith, & Alessio Corti Rational and Nearly Rational Varieties David Applebaum L´evy Processes and Stochastic Calculus Martin Schechter An Introduction to Nonlinear Analysis Additive Combinatorics TERENCE TAO, VAN VU cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge cb2 2ru, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521853866 © Cambridge University Press 2006 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2006 isbn-13 isbn-10 978-0-511-24530-5 eBook (EBL) 0-511-24530-0 eBook (EBL) isbn-13 isbn-10 978-0-521-85386-6 hardback 0-521-85386-9 hardback Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. To our families Contents Prologue page xi 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 The probabilistic method The first moment method The second moment method The exponential moment method Correlation inequalities The Lov´asz local lemma Janson’s inequality Concentration of polynomials Thin bases of higher order Thin Waring bases Appendix: the distribution of the primes 1 2 6 9 19 23 27 33 37 42 45 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 Sum set estimates Sum sets Doubling constants Ruzsa distance and additive energy Covering lemmas The Balog–Szemer´edi–Gowers theorem Symmetry sets and imbalanced partial sum sets Non-commutative analogs Elementary sum-product estimates 51 54 57 59 69 78 83 92 99 3 3.1 3.2 3.3 Additive geometry Additive groups Progressions Convex bodies 112 113 119 122 vii viii Contents 3.4 3.5 3.6 The Brunn–Minkowski inequality Intersecting a convex set with a lattice Progressions and proper progressions 127 130 143 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 Fourier-analytic methods Basic theory L p theory Linear bias Bohr sets ( p) constants, Bh [g] sets, and dissociated sets The spectrum of an additive set Progressions in sum sets 149 150 156 160 165 172 181 189 5 5.1 5.2 5.3 5.4 5.5 5.6 Inverse sum set theorems Minimal size of sum sets and the e-transform Sum sets in vector spaces Freiman homomorphisms Torsion and torsion-free inverse theorems Universal ambient groups Freiman’s theorem in an arbitrary group 198 198 211 220 227 233 239 6 6.1 6.2 6.3 6.4 6.5 Graph-theoretic methods Basic Notions Independent sets, sum-free subsets, and Sidon sets Ramsey theory Proof of the Balog–Szemer´edi–Gowers theorem Pl¨unnecke’s theorem 246 247 248 254 261 267 7 7.1 7.2 7.3 7.4 7.5 7.6 The Littlewood–Offord problem The combinatorial approach The Fourier-analytic approach The Ess´een concentration inequality Inverse Littlewood–Offord results Random Bernoulli matrices The quadratic Littlewood–Offord problem 276 277 281 290 292 297 304 8 8.1 8.2 8.3 8.4 8.5 Incidence geometry The crossing number of a graph The Szemer´edi–Trotter theorem The sum-product problem in R Cell decompositions and the distinct distances problem The sum-product problem in other fields 308 308 311 315 319 325 Contents ix 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 Algebraic methods The combinatorial Nullstellensatz Restricted sum sets Snevily’s conjecture Finite fields Davenport’s problem Kemnitz’s conjecture Stepanov’s method Cyclotomic fields, and the uncertainty principle 329 330 333 342 345 350 354 356 362 10 10.1 10.2 10.3 10.4 10.5 10.6 10.7 Szemer´edi’s theorem for k = 3 General strategy The small torsion case The integer case Quantitative bounds An ergodic argument The Szemer´edi regularity lemma Szemer´edi’s argument 369 372 378 386 389 398 406 411 11 11.1 11.2 11.3 11.4 11.5 11.6 11.7 Szemer´edi’s theorem for k > 3 Gowers uniformity norms Hard obstructions to uniformity Proof of Theorem 11.6 Soft obstructions to uniformity The infinitary ergodic approach The hypergraph approach Arithmetic progressions in the primes 414 417 424 432 440 448 454 463 12 12.1 12.2 12.3 12.4 12.5 12.6 Long arithmetic progressions in sum sets Introduction Proof of Theorem 12.4 Generalizations and variants Complete and subcomplete sequences Proof of Theorem 12.17 Further applications 470 470 473 477 480 482 484 Bibliography Index 488 505 Prologue This book arose out of lecture notes developed by us while teaching courses on additive combinatorics at the University of California, Los Angeles and the University of California, San Diego. Additive combinatorics is currently a highly active area of research for several reasons, for example its many applications to additive number theory. One remarkable feature of the field is the use of tools from many diverse fields of mathematics, including elementary combinatorics, harmonic analysis, convex geometry, incidence geometry, graph theory, probability, algebraic geometry, and ergodic theory; this wealth of perspectives makes additive combinatorics a rich, fascinating, and multi-faceted subject. There are still many major problems left in the field, and it seems likely that many of these will require a combination of tools from several of the areas mentioned above in order to solve them. The main purpose of this book is to gather all these diverse tools in one location, present them in a self-contained and introductory manner, and illustrate their application to problems in additive combinatorics. Many aspects of this material have already been covered in other papers and texts (and in particular several earlier books , , have focused on some of the aspects of additive combinatorics), but this book attempts to present as many perspectives and techniques as possible in a unified setting. Additive combinatorics is largely concerned with the additive structure1 of sets. To clarify what we mean by “additive structure”, let us introduce the following definitions. Definition 0.1 An additive group is any abelian group Z with group operation +. Note that we can define a multiplication operation nx ∈ Z whenever n ∈ Z and 1 We will also occasionally consider the multiplicative structure of sets as well; we will refer to the combined study of such structures as arithmetic combinatorics. xi Prologue xii x ∈ Z in the usual manner: thus 3x = x + x + x, −2x = −x − x, etc. An additive set is a pair (A, Z ), where Z is an additive group, and A is a finite non-empty subset of Z . We often abbreviate an additive set (A, Z ) simply as A, and refer to Z as the ambient group of the additive set. If A, B are additive sets in Z , we define the sum set A + B := {a + b : a ∈ A, b ∈ B} and difference set A − B := {a − b : a ∈ A, b ∈ B}. Also, we define the iterated sumset k A for k ∈ Z+ by k A := {a1 + · · · + ak : a1 , . . . , ak ∈ A}. We caution that the sumset k A is usually distinct from the dilation k · A of A, defined by k · A := {ka : a ∈ A}. For us, typical examples of additive groups Z will be the integers Z, a cyclic group Z N , a Euclidean space Rn , or a finite field geometry F pn . As the notation suggests, we will eventually be viewing additive sets as “intrinsic” objects, which can be embedded inside any number of different ambient groups; this is somewhat similar to how a manifold can be thought of intrinsically, or alternatively can be embedded into an ambient space. To make these ideas rigorous we will need to develop the theory of Freiman homomorphisms, but we will defer this to Section 5.3. Additive sets may have a large or small amount of additive structure. A good example of a set with little additive structure would be a randomly chosen subset A of a finite additive group Z with some fixed cardinality. At the other extreme, examples of sets with very strong additive structure would include arithmetic progressions a + [0, N ) · r := {a, a + r, . . . , a + (N − 1)r } where a, r ∈ Z and N ∈ Z+ ; or d-dimensional generalized arithmetic progressions a + [0, N ) · v := {a + n 1 v1 + · · · + n d vd : 0 ≤ n j < N j for all 1 ≤ j ≤ d} where a ∈ Z , v = (v1 , . . . , vd ) ∈ Z d , and N = (N1 , . . . , Nd ) ∈ (Z+ )d ; or ddimensional cubes a + {0, 1}d · v = {a + 1 v1 + · · · + d vd : 1 , . . . , d ∈ {0, 1}}; or the subset sums F S(A) := { a∈B a : B ⊆ A} of a finite set A. Prologue xiii A fundamental task in this subject is to give some quantitative measures of additive structure in a set, and then investigate to what extent these measures are equivalent to each other. For example, one could try to quantify each of the following informal statements as being some version of the assertion “A has additive structure”: r r r r r r r r r r r r r A + A is small; A − A is small; A − A can be covered by a small number of translates of A; k A is small for any fixed k; there are many quadruples (a1 , a2 , a3 , a4 ) ∈ A × A × A × A such that a1 + a2 = a3 + a4 ; there are many quadruples (a1 , a2 , a3 , a4 ) ∈ A × A × A × A such that a1 − a2 = a3 − a4 ; the convolution 1 A ∗ 1 A is highly concentrated; the subset sums F S(A) := { a∈B a : B ⊆ A} have high multiplicity; the Fourier transform 1A is highly concentrated; the Fourier transform 1A is highly concentrated in a cube; A has a large intersection with a generalized arithmetic progression, of size comparable to A; A is contained in a generalized arithmetic progression, of size comparable to A; A (or perhaps A − A, or 2A − 2A) contains a large generalized arithmetic progression. The reader is invited to investigate to what extent these informal statements are true for sets such as progressions and cubes, and false for sets such as random sets. As it turns out, once one makes the above assertions more quantitative, there are a number of deep and important equivalences between them; indeed, to oversimplify tremendously, all of the above criteria for additive structure are “essentially” equivalent. There is also a similar heuristic to quantify what it would mean for two additive sets A, B of comparable size to have a large amount of “shared additive structure” (e.g. A and B are progressions with the same step size v); we invite the reader to devise analogs of the above criteria to capture this concept. Making the above heuristics precise and rigorous will require some work, and in fact will occupy large parts of Chapters 2, 3, 4, 5, 6. In deriving these basic tools of the field, we shall need to develop and combine techniques from elementary combinatorics, additive geometry, harmonic analysis, and graph theory; many of these methods are of independent interest in their own right, and so we have devoted some space to treating them in detail. Of course, a “typical” additive set will most likely behave like a random additive set, which one expects to have very little additive structure. Nevertheless, it is a xiv Prologue deep and surprising fact that as long as an additive set is dense enough in its ambient group, it will always have some level of additive structure. The most famous example of this principle is Szemer´edi’s theorem, which asserts that every subset of the integers of positive upper density will contain arbitrarily long arithmetic progressions; we shall devote all of Chapter 11 to this beautiful and important theorem. A variant of this fact is the very recent Green–Tao theorem, which asserts that every subset of the prime numbers of positive upper relative density also contains arbitrarily long arithmetic progressions; in particular, the primes themselves have this property. If one starts with an even sparser set A than the primes, then it is not yet known whether A will necessarily contain long progressions; however, if one forms sum sets such as A + A, A + A + A, 2A − 2A, F S(A) then these sets contain extraordinarily long arithmetic progressions (see in particular Section 4.7 and Chapter 12). This basic principle – that sumsets have much more additive structure than general sets – is closely connected to the equivalences between the various types of additive structure mentioned previously; indeed results of the former type can be used to deduce results of the latter type, and conversely. We now describe some other topics covered in this text. In Chapter 1 we recall the simple yet powerful probabilistic method, which is very useful in additive combinatorics for constructing sets with certain desirable properties (e.g. thin additive bases of the integers), and provides an important conceptual framework that complements more classical deterministic approaches to such constructions. In Chapter 6 we present some ways in which graph theory interacts with additive combinatorics, for instance in the theory of sum-free sets, or via Ramsey theory. Graph theory is also decisive in establishing two important results in the theory of sum sets, the Balog–Szemer´edi–Gowers theorem and the Pl¨unnecke inequalities. Two other important tools from graph theory, namely the crossing number inequality and the Szemer´edi regularity lemma, will also be covered in Chapter 8 and Sections 10.6, 11.6 respectively. In Chapter 7 we view sum sets from the perspective of random walks, and give some classical and recent results concerning the distribution of these sum sets, and in particular recent applications to random matrices. Last, but not least, in Chapter 9 we describe some algebraic methods, notably the combinatorial Nullstellensatz and Chevalley–Waring type methods, which have led to several deep arithmetical results (often with very sharp bounds) not obtainable by other means. Acknowledgements The authors would like to thank Shimon Brooks, Robin Chapman, Michael Cowling, Andrew Granville, Ben Green, Timothy Gowers, Harald Helfgott, Martin Klazar, Mariah Hamel, Vsevolod Lev, Roy Meshulam, Melvyn Nathanson, Imre Ruzsa, Roman Sasyk, and Benny Sudakov for helpful comments and corrections, Prologue xv and to the Australian National University and the University of Edinburgh for their hospitality while portions of this book were being written. Parts of this work were inspired by the lecture notes of Ben Green , the expository article of Imre Ruzsa , and the book by Melvyn Nathanson . TT is also particularly indebted to Roman Sasyk and Hillel Furstenberg for explaining the ergodic theory proof of Szemer´edi’s theorem. VV would like to thank Endre Szemer´edi for many useful discussions on mathematics and other aspects of life. Last, and most importantly, the authors thank their wives, Laura and Huong, without whom this book would not be finished. General notation The following general notational conventions will be used throughout the book. Sets and functions For any set A, we use Ad := A × · · · × A = {(a1 , . . . , ad ) : a1 , . . . , ad ∈ A} to denote the Cartesian product of d copies of A: thus for instance Zd is the ddimensional integer lattice. We shall occasionally denote Ad by A⊕d , in order to distinguish this Cartesian product from the d-fold product set A·d = A · . . . · A of A, or the d-fold powers A∧ d := {a d : a ∈ A} of A. If A, B are sets, we use A\B := {a ∈ A : a ∈ B} to denote the set-theoretic difference of A and B; and B A to denote the space of functions f : A → B from A to B. We also use 2 A := {B : B ⊂ A} to denote the power set of A. We use \|A\| to denote the cardinality of A. (We shall also use \|x\| to denote the magnitude of a real or complex number x, and \|v\| = v12 + · · · + vd2 to denote the magnitude of a vector v = (v1 , . . . , vd ) in a Euclidean space Rd . The meaning of the absolute value signs should be clear from context in all cases.) If A ⊂ Z , we use 1 A : Z → {0, 1} to denote the indicator function of A: thus 1 A (x) = 1 when x ∈ A and 1 A (x) = 0 otherwise. Similarly if P is a property, we let I(P) denote the quantity 1 if P holds and 0 otherwise; thus for instance 1 A (x) = I(x n ∈ A). n! We use k = k!(n−k)! to denote the number of k-element subsets of an n-element set. In particular we have the natural convention that nk = 0 if k > n or k < 0. Number systems We shall rely frequently on the integers Z, the positive integers Z+ := {1, 2, . . .}, the natural numbers N := Z≥0 = {0, 1, . . .}, the reals R, the positive reals xvi Prologue R+ := {x ∈ R : x > 0}, the non-negative reals R≥0 := {x ∈ R : x ≥ 0}, and the complex numbers C, as well as the circle group R/Z := {x + Z : x ∈ R}. For any natural number N ∈ N, we use Z N := Z/N Z to denote the cyclic group of order N , and use n → n mod N to denote the canonical projection from Z to Z N . If q is a prime power, we use Fq to denote the finite field of order q (see Section 9.4). In particular if p is a prime then F p is identifiable with Z p . If x is a real number, we use x to denote the greatest integer less than or equal to x. Landau asymptotic notation Let n be a positive variable (usually taking values on N, Z+ , R≥0 , or R+ , and often assumed to be large) and let f (n) and g(n) be real-valued functions of n. r g(n) = O( f (n)) means that f is non-negative, and there is a positive constant C such that \|g(n)\| ≤ C f (n) for all n. r g(n) = ( f (n)) means that f, g are non-negative, and there is a positive constant c such that g(n) ≥ c f (n) for all sufficiently large n. r g(n) = ( f (n)) means that f, g are non-negative and both g(n) = O( f (n)) and g(n) = ( f (n)) hold; that is, there are positive constants c and C such that c f (n) ≥ g(n) ≥ C f (n) for all n. r g(n) = o n→∞ ( f (n)) means that f is non-negative and g(n) = O(a(n) f (n)) for some a(n) which tends to zero as n → ∞; if f is strictly positive, this is equivalent to limn→∞ g(n)/ f (n) = 0. r g(n) = ω n→∞ ( f (n)) means that f, g are non-negative and f (n) = on→∞ (g(n)). In most cases the asymptotic variable n will be clear from context, and we shall simply write on→∞ ( f (n)) as o( f (n)), and similarly write ωn→∞ ( f (n)) as ω( f (n)). In some cases the constants c,C and the decaying function a(n) will depend on some other parameters, in which case we indicate this by subscripts. Thus for instance g(n) = Ok ( f (n)) would mean that g(n) ≤ Ck f (n) for all n, where Ck depends on the parameter k; similarly, g(n) = on→∞;k ( f (n)) would mean that g(n) = O(ak (n) f (n)) for some ak (n) which tends to zero as n → ∞ for each fixed k. ˜ f (n)) has been used widely in the combinatorics and The notation g(n) = O( ˜ f (n)) means theoretical computer science community in recent years; g(n) = O( c that there is a constant c such that g(n) ≤ f (n) log n for all sufficiently large n. ˜ and , ˜ though this notation will only be We can define, in a similar manner, used occasionally here. Here and throughout the rest of the book, log shall denote y the natural logarithm unless specified by subscripts, thus logx y = log . log x Prologue xvii Progressions We have already encountered the concept of a generalized arithmetic progression. We now make this concept more precise. Definition 0.2 (Progressions) For any integers a ≤ b, we let [a, b] denote the discrete closed interval [a, b] := {n ∈ Z : a ≤ n ≤ b}; similarly define the halfopen discrete interval [a, b), etc. More generally, if a = (a1 , . . . , ad ) and b = (b1 , . . . , bd ) are elements of Zd such that a j ≤ b j , we define the discrete box [a, b] := {(n 1 , . . . , n d ) ∈ Zd : a j ≤ n j ≤ b j for all 1 ≤ j ≤ d}, and similarly [a, b) := {(n 1 , . . . , n d ) ∈ Zd : a j ≤ n j < b j for all 1 ≤ j ≤ d}, etc. If Z is an additive group, we define a generalized arithmetic progression (or just progression for short) in Z to be any set1 of the form P = a + [0, N ] · v, where a ∈ Z , N = (N1 , . . . , Nd ) is a tuple, [0, N ] ⊂ Zd is a discrete box, v = (v1 , . . . , vd ) ∈ Z d , the map · : Zd × Z d → Z is the dot product (n 1 , . . . , n d ) · (v1 , . . . , vd ) := n 1 v1 + · · · + n d vd , and [0, N ] · v := {n · v : n ∈ [0, N ]}. In other words, P = {a + n 1 v1 + · · · + n d vd : 0 ≤ n j ≤ N j for all 1 ≤ j ≤ d}. We call a the base point of P, v = (v1 , . . . , vd ) the basis vectors of P, N the dimen sion of P, d the dimension or rank of P, and vol(P) := \|[0, N ]\| = dj=1 (N j + 1) the volume of P. We say that the progression P is proper if the map n → n · v is injective on [0, N ], or equivalently if the cardinality of P is equal to its volume (as opposed to being strictly smaller than the volume, which can occur if the basis vectors are linearly dependent over Z). We say that P is symmetric if −P = P; for instance [−N , N ] · v = −N · v + [0, 2N ] · v is a symmetric progression. Other notation There are a number of other definitions that we shall introduce at appropriate junctures and which will be used in more than one chapter of the book. These include the probabilistic notation (such as E(), P(), I(), Var(), Cov()) that we introduce 1 Strictly speaking, this is an abuse of notation; the arithmetic progression should really be the sextuple (P, d, N , a, v, Z ), because the set P alone does not always uniquely determine the base point, step, ambient space or even length (if the progression is improper) of the progression P. However, as it would be cumbersome continually to use this sextuple, we shall usually just P to denote the progression. Prologue xviii at the start of Chapter 1, and measures of additive structure such as the doubling constant σ [A] (Definition 2.4), the Ruzsa distance d(A, B) (Definition 2.5), and the additive energy E(A, B) (Definition 2.8). We also introduce the concept of a G partial sum set A + B in Definition 2.28. The Fourier transform and the averaging notation Ex∈Z f (x), P Z A is defined in Section 4.1, Fourier bias Au is defined in Definition 4.12, Bohr sets Bohr(S, ρ) are defined in Definition 4.17, and ( p) constants are defined in Definition 4.26. The important notion of a Freiman homomorphism is defined in Definition 5.21. The notation for group theory (e.g. ord(x) and x) is summarized in Section 3.1, while the notation for finite fields is summarized in Section 9.4. 1 The probabilistic method In additive number theory, one frequently faces the problem of showing that a set A contains a subset B with a certain property P. A very powerful tool for such a problem is Erd˝os’ probabilistic method. In order to show that such a subset B exists, it suffices to prove that a properly defined random subset of A satisfies P with positive probability. The power of the probabilistic method has been justified by the fact that in most problems solved using this approach, it seems impossible to come up with a deterministically constructive proof of comparable simplicity. In this chapter we are going to present several basic probabilistic tools together with some representative applications of the probabilistic method, particularly with regard to additive bases and the primes. We shall require several standard facts about the distribution of primes P = {2, 3, 5, . . .}; so as not to disrupt the flow of the chapter we have placed these facts in an appendix (Section 1.10). Notation. We assume the existence of some sample space (usually this will be finite). If E is an event in this sample space, we use P(E) to denote the probability of E, and I(E) to denote the indicator function (thus I(E) = 1 if E occurs and 0 otherwise). If E, F are events, we use E ∧ F to denote the event that E, F both hold, E ∨ F to denote the event that at least one of E, F hold, and E¯ to denote the event that E does not hold. In this chapter all random variables will be assumed to be real-valued (and usually denoted by X or Y ) or set-valued (and usually denoted by B). If X is a real-valued random variable with discrete support, we use E(X ) := xP(X = x) x to denote the expectation of X , and Var(X ) := E(\|X − E(X )\|2 ) = E(\|X \|2 ) − E(\|X \|)2 1 2 1 The probabilistic method to denote the variance. Thus for instance E(I(E)) = P(E); Var(I(E)) = P(E) − P(E)2 . (1.1) If F is an event of non-zero probability, we define the conditional probability of another event E with respect to F by: P(E\|F) := P(E ∧ F) P(F) and similarly the conditional expectation of a random variable X by E(X I(F)) E(X \|F) := xP(X = x\|F). = E(I(F)) x A random variable is boolean if it takes values in {0, 1}, or equivalently if it is an indicator function I(E) for some event E. 1.1 The first moment method The simplest instance of the probabilistic method is the first moment method, which seeks to control the distribution of a random variable X in terms of its expectation (or first moment) E(X ). Firstly, we make the trivial observation (essentially the pigeonhole principle) that X ≤ E(X ) with positive probability, and X ≥ E(X ) with positive probability. A more quantitative variant of this is Theorem 1.1 (Markov’s inequality) Let X be a non-negative random variable. Then for any positive real λ > 0 E(X ) . (1.2) λ Proof Start with the trivial inequality X ≥ λI(X ≥ λ) and take expectations of both sides. P(X ≥ λ) ≤ Informally, this inequality asserts that X = O(E(X )) with high probability; for instance, X ≤ 10E(X ) with probability at least 0.9. Note that this is only an upper tail estimate; it gives an upper bound for how likely X is to be much larger than E(X ), but does not control how likely X is to be much smaller than E(X ). Indeed, if all one knows is the expectation E(X ), it is easy to see that X could be as small as zero with probability arbitrarily close to 1, so the first moment method cannot give any non-trivial lower tail estimate. Later on we shall introduce more refined methods, such as the second moment method, that give further upper and lower tail estimates. 1.1 The first moment method 3 To apply the first moment method, we of course need to compute the expectations of random variables. A fundamental tool in doing so is linearity of expectation, which asserts that E(c1 X 1 + · · · + cn X n ) = c1 E(X 1 ) + · · · + cn E(X n ) (1.3) whenever X 1 , . . . , X n are random variables and c1 , . . . , cn are real numbers. The power of this principle comes from there being no restriction on the independence or dependence between the X i s. A very typical application of (1.3) is in estimating the size \|B\| of a subset B of a given set A, where B is generated in some random manner. From the obvious identity \|B\| = I(a ∈ B) a∈A and (1.3), (1.1) we see that E(\|B\|) = P(a ∈ B). (1.4) a∈A Again, we emphasize that the events a ∈ B do not need to be independent in order for (1.4) to apply. A weaker version of the linearity of expectation principle is the union bound P(E 1 ∨ · · · ∨ E n ) ≤ P(E 1 ) + · · · + P(E n ) (1.5) for arbitrary events E 1 , . . . , E n (compare this with (1.3) with X i := I(E i ) and ci := 1). This trivial bound is still useful, especially in the case when the events E 1 , . . . , E n are rare and not too strongly correlated (see Exercise 1.1.3). A related estimate is as follows. Lemma 1.2 (Borel–Cantelli lemma) Let E 1 , E 2 , . . . be a sequence of events (possibly infinite or dependent), such that n P(E n ) < ∞. Then for any integer M, we have P(E n ) P(Fewer than M of the events E 1 , E 2 , . . . hold) ≥ 1 − n . M In particular, with probability 1 at most finitely many of the events E 1 , E 2 , . . . hold. Another useful way of phrasing the Borel–Cantelli lemma is that if F1 , F2 , . . . are events such that n (1 − P(Fn )) < ∞, then, with probability n, all but finitely many of the events Fn hold. Proof By monotone convergence it suffices to prove the claim when there are only finitely many events. From (1.3) we have E( n I(E n )) = n P(E n ). If one now applies Markov’s inequality with λ = M, the claim follows. 1 The probabilistic method 4 1.1.1 Sum-free sets We now apply the first moment method to the theory of sum-free sets. An additive set A is called sum-free iff it does not contain three elements x, y, z such that x + y = z; equivalently, A is sum-free iff A ∩ 2A = ∅. Theorem 1.3 Let A be an additive set of non-zero integers. Then A contains a sum-free subset B of size \|B\| > \|A\|/3. Proof Choose a prime number p = 3k + 2, where k is sufficiently large so that A ⊂ [− p/3, p/3]{0}. We can thus view A as a subset of the cyclic group Z p rather than the integers Z, and observe that a subset B of A will be sum-free in Z p if and only if 1 it is sum-free in Z. Now choose a random number x ∈ Z p {0} uniformly, and form the random set B := A ∩ (x · [k + 1, 2k + 1]) = {a ∈ A : x −1 a ∈ {k + 1, . . . , 2k + 1}}. Since [k + 1, 2k + 1] is sum-free in Z p , we see that x · [k + 1, 2k + 1] is too, and thus B is a sum-free subset of A. We would like to show that \|B\| > \|A\|/3 with positive probability; by the first moment method it suffices to show that E(\|B\|) > \|A\|/3. From (1.4) we have E(\|B\|) = P(a ∈ B) = P(x −1 a ∈ [k + 1, 2k + 1]). a∈A a∈A If a ∈ A, then a is an invertible element of Z p , and thus x −1 a is uniformly distributed in Z p {0}. Since \|[k + 1, 2k + 1]\| > p−1 , we conclude that P(x −1 a ∈ 3 1 [k + 1, 2k + 1]) > 3 for all a ∈ A. Thus we have E(\|B\|) > \|A\| as desired. 3 Theorem 1.3 was proved by Erd˝os in 1965 . Several years later, Bourgain used harmonic analysis arguments to improve the bound slightly. It is surprising that the following question is open. Question 1.4 Can one replace n/3 by (n/3) + 10? Alon and Kleiman considered the case of more general additive sets (not necessarily in Z). They showed that in this case A always contains a sum-free subset of 2\|A\|/7 elements and the constant 2/7 is best possible. Another classical problem concerning sum-free sets is the Erd˝os–Moser problem. Consider a finite additive set A. A subset B of A is sum-free with respect to A if 2∗ B ∩ A = ∅, where 2∗ B = {b1 + b2 \|b1 , b2 ∈ B, b1 = b2 }. Erd˝os and Moser asked for an estimate of the size of the largest sum-free subset of any given set A of cardinality n. We will discuss this problem in Section 6.2.1. 1 This trick can be placed in a more systematic context using the theory of Freiman homomorphisms: see Section 5.3. 1.1 The first moment method 5 Exercises 1.1.1 If X is a non-negative random variable, establish the identity ∞ E(X ) = P(X > λ) dλ (1.6) 0 and more generally for any 0 < p < ∞ ∞ E(X p ) = p λ p−1 P(X > λ) dλ. (1.7) 0 1.1.2 1.1.3 Thus the probability distribution function P(X > λ) controls all the moments E(X p ) of X . When does equality hold in Markov’s inequality? If E 1 , . . . , E n are arbitrary probabilistic events, establish the lower bound P(E 1 ∨ · · · ∨ E n ) ≥ n P(E i ) − i=1 P(E i ∧ E j ); 1≤i< j≤n this bound should be compared with (1.5), and can be thought of as a variant of the second moment method which we discuss in the next section. n (Hint: consider the random variable i=1 I(E i ) − 1≤i< j≤n I(E i )I(E j ).) More generally, establish the Bonferroni inequalities P(E 1 ∨ · · · ∨ E n ) ≥ (−1)k P Ei i∈A A⊂[1,n]:1≤\|A\|≤k when k is even, and P(E 1 ∨ · · · ∨ E n ) ≤ A⊂[1,n]:1≤\|A\|≤k 1.1.4 1.1.5 k (−1) P Ei i∈A when k is odd. Let X be a non-negative random variable. Establish the popularity principle E(X I(X > 12 E(X ))) ≥ 12 E(X ). In particular, if X is bounded by some 1 constant M, then P(X > 12 E(X )) ≥ 2M E(X ). Thus while there is in general no lower tail estimate on the event X ≤ 12 E(X ), we can say that the majority of the expectation of X is generated outside of this tail event, which does lead to a lower tail estimate if X is bounded. Let A, B be non-empty subsets of a finite additive group Z . Show that there exists an x ∈ Z such that \|A ∩ (B + x)\| \|A\| \|B\| 1− ≤ 1− 1− , \|Z \| \|Z \| \|Z \| 1 The probabilistic method 6 and a y ∈ Z such that 1− 1.1.6 \|A ∩ (B + y)\| \|A\| \|B\| ≥ 1− 1− . \|Z \| \|Z \| \|Z \| Consider a set A as above. Show that there exists a subset {v1 , . . . , vd } of \| Z with d = O(log \|Z ) such that \|A\| \|A + [0, 1]d · (v1 , . . . , vd )\| ≥ \|Z \|/2. 1.1.7 Consider a set A as above. Show that there exists a subset {v1 , . . . , vd } of \| Z with d := O(log \|Z + log log(10 + \|Z \|)) such that \|A\| A + [0, 1]d · (v1 , . . . , vd ) = Z . 1.2 The second moment method The first moment method allows one to control the order of magnitude of a random variable X by its expectation E(X ). In many cases, this control is insufficient, and one also needs to establish that X usually does not deviate too greatly from its expected value. These types of estimates are known as large deviation inequalities, and are a fundamental set of tools in the subject. They can be significantly more powerful than the first moment method, but often require some assumptions concerning independence or approximate independence. The simplest such large deviation inequality is Chebyshev’s inequality, which controls the deviation in terms of the variance Var(X ): Theorem 1.5 (Chebyshev’s inequality) Let X be a random variable. Then for any positive λ 1 P \|X − E(X )\| > λVar(X )1/2 ≤ 2 . (1.8) λ Proof We may assume Var(X ) > 0 as the case Var(X ) = 0 is trivial. From Markov’s inequality we have P(\|X − E(X )\|2 > λ2 Var(X )) ≤ and the claim follows. 1 E(\|X − E(X )\|2 ) = 2 λ2 Var(X ) λ Thus Chebyshev’s inequality asserts that X = E(X ) + O(Var(X )1/2 ) with high probability, while in the converse direction it is clear that \|X − E(X )\| ≥ Var(X )1/2 with positive probability. The application of these facts is referred to as the second moment method. Note that Chebyshev’s inequality provides both upper tail and lower tail bounds on X , with the tail decaying like 1/λ2 rather than 1/λ. Thus 1.2 The second moment method 7 the second moment method tends to give better distributional control than the first moment method. The downside is that the second moment method requires computing the variance, which is often trickier than computing the expectation. Assume that X = X 1 + · · · + X n , where X i s are random variables. In view of (1.3), one might wonder whether Var(X ) = Var(X 1 ) + · · · + Var(X n ). (1.9) This equality holds in the special case when the X i s are pairwise independent (and in particular when they are jointly independent), but does not hold in general. For arbitrary X i s, we instead have n Var(X ) = Var(X i ) + Cov(X i , X j ), (1.10) i, j∈[1,n]:i = j i=1 where the covariance Cov(X i , X j ) is defined as Cov(X i , X j ) := E((X i − E(X i ))(X j − E(X j )) = E(X i X j ) − E(X i )E(X j ). Applying (1.9) to the special case when X = \|B\|, where B is some randomly generated subset of a set A, we see from (1.1) that if the events a ∈ B are pairwise independent for all a ∈ A, then Var(\|B\|) = P(a ∈ B) − P(a ∈ B)2 (1.11) a∈A and in particular we see from (1.4) that Var(\|B\|) ≤ E(\|B\|). (1.12) In the case when the events a ∈ B are not pairwise independent, we must replace (1.11) by the more complicated identity Var(\|B\|) = P(a ∈ B) − P(a ∈ B)2 + Cov(I(a ∈ B), I(a ∈ B)). a,a ∈A:a =a a∈A (1.13) 1.2.1 The number of prime divisors Now we present a nice application of the second moment method to classical number theory. To this end, let1 ν(n) := I( p\|n) p≤n 1 We shall adopt the convention that whenever a summation is over the index p, then p is understood to be prime. 1 The probabilistic method 8 denote the number of prime divisors of n. This function is among the most studied objects in classical number theory. Hardy and Ramanujan in the 1920s showed that “almost” all n have about log log n prime divisors. We give a very simple proof of this result, found by Tur´an in 1934 . Theorem 1.6 Let ω(n) tend to infinity arbitrarily slowly. Then \|{x ∈ [1, n] : \|ν(x) − log log x\| > ω(n) log log n}\| = o(n). (1.14) Informally speaking, this result asserts that for a “generic” integer x, we have √ ν(x) = log log x + O( log log x) with high probability. Proof Let x be chosen uniformly at random from the interval {1, 2, . . . , n}. Our task is now to show that P(\|ν(x) − log log x\| > ω(n) log log n) = o(1). Due to a technical reason, instead of ν(x) we shall consider the related quantity \|B\|, where B := p prime : p ≤ n 1/10 , p\|x . Since x cannot have 10 different prime divisors larger than n 1/10 , it follows that \|B\| − 10 ≤ ν(x) ≤ \|B\|. Thus, to prove (1.14), it suffices to show P(\|\|B\| − log log n\| ≥ ω(n) ln log n) = o(1). Note that log log x = log log n + O(1) with probability 1 − o(1). In light of Chebyshev’s inequality, this will follow from the following expectation and variance estimates: E(\|B\|), Var(\|B\|) = log log n + O(1). It remains to verify the expectation and variance estimate. From linearity of expectation (1.4) we have E(\|B\|) = P( p\|x) p≤n 1/10 while from the variance identity (1.13) we have Var(\|B\|) = (P( p\|x) − P( p\|x)2 ) + p≤n 1/10 Cov(I( p\|x), I(q\|x)). p,q≤n 1/10 : p =q Observe that I( p\|x)I(q\|x) = I( pq\|x). Since P(d\|x) = we conclude that 1 1 P( p\|x) = + O p n 1 d O( n1 ) for any d ≥ 1, 1.3 The exponential moment method and 1 +O Cov(I( p\|x), I(q\|x)) = pq 1 1 1 1 1 1 − +O +O =O . n p n q n n We thus conclude that E(\|B\|) = and 9 1 O n −9/10 p p≤n 1/10 1 1 Var(\|B\|) = − 2 + O n −8/10 . p p p≤n 1/10 The expectation and variance estimates now follow from Mertens’ theorem (see Proposition 1.51) and the convergence of the sum k k12 . Exercises 1.2.1 1.2.2 1.2.3 1.2.4 1.2.5 1.2.6 When does equality hold in Chebyshev’s inequality? If X and Y are two random variables, verify the Cauchy–Schwarz inequality \|Cov(X, Y )\| ≤ Var(X )1/2 Var(Y )1/2 and the triangle inequality Var(X + Y )1/2 ≤ Var(X )1/2 + Var(Y )1/2 . When does equality occur? Prove (1.10). If φ : R → R is a convex function and X is a random variable, verify Jensen’s inequality E(φ(X )) ≤ φ(E(X )). If φ is strictly convex, when does equality occur? Generalize Chebyshev’s inequality using higher moments E(\|X − E(X )\| p ) instead of the variance. By obtaining an upper bound on the fourth moment, improve Theorem 1.6 to 1 \|{x ∈ [1, N ] : \|ν(x) − log log N \| > K log log N }\| = O(K −4 ). N Can you generalize this to obtain a bound of Om (K −m ) for any even integer m ≥ 2, where the constant in the O() notation is allowed to depend on m? 1.3 The exponential moment method Chebyshev’s inequality shows that if one has control of the second moment Var(X ) = E(\|X − E(X )\|2 ), then a random variable X takes the value E(X ) + O(λVar(X )1/2 ) with probability 1 − O(λ−2 ). If one uses higher moments, one 10 1 The probabilistic method can obtain better decay of the tail probability than O(λ−2 ). In particular, if one can control exponential moments1 such as E(et X ) for some real parameter t, then one can obtain exponential decay in upper and lower tail probabilities, since Markov’s inequality yields P(X ≥ λ) = P(et X ≥ etλ ) ≤ E(et X ) etλ (1.15) for t > 0 and λ ∈ R, and similarly P(X ≤ −λ) = P(e−t X ≥ etλ ) ≤ E(e−t X ) etλ (1.16) for the same range of t, λ. The quantity E(et X ) is known as an exponential moment of X , and the function t → E(et X ) is known as the moment generating function, thanks to the Taylor expansion t2 t3 E(X 2 ) + E(X 3 ) + · · · . 2! 3! The application of (1.15) or (1.16) is known as the exponential moment method. Of course, to use it effectively one needs to be able to compute the exponential moments E(et X ). A preliminary tool for doing so is E(et X ) = 1 + tE(X ) + Lemma 1.7 Let X be a random variable with \|X \| ≤ 1 and E(X ) = 0. Then for any −1 ≤ t ≤ 1 we have E(et X ) ≤ exp(t 2 Var(X )). Proof Since \|t X \| ≤ 1, a simple comparison of Taylor series gives the inequality et X ≤ 1 + t X + t 2 X 2 . Taking expectations of both sides and using linearity of expectation and the hypothesis E(X ) = 0 we obtain E(et X ) ≤ 1 + t 2 Var(X ) ≤ exp(t 2 Var(X )) as desired. This lemma by itself is not terribly effective as it requires both X and t to be bounded. However the power of this lemma can be amplified considerably when applied to random variables X which are sums of bounded random variables, X = X 1 + · · · + X n , provided that we have the very strong assumption of joint independence between the X 1 , . . . , X n . More precisely, we have 1 To avoid questions of integrability or measurability, let us assume for sake of discussion that the random variable X here only takes finitely many values; this is the case of importance in combinatorial applications. 1.3 The exponential moment method 11 Theorem 1.8 (Chernoff’s inequality) Assume that X 1 , . . . , X n are jointly independent random variables where \|X i − E(X i )\| ≤ 1 for all i. Set X := X 1 + · · · + √ X n and let σ := Var(X ) be the standard deviation of X . Then for any λ > 0 2 P(\|X − E(X )\| ≥ λσ ) ≤ 2 max e−λ /4 , e−λσ/2 . (1.17) Informally speaking, (1.17) asserts that X = E(X ) + O(Var(X )1/2 ) with high probability, and X = E(X ) + O(ln1/2 nVar(X )1/2 ) with extremely high probability (1 − O(n −C ) for some large C). The bound in Chernoff’s theorem provides a huge improvement over Chebyshev’s inequality when λ is large. However the joint independence of the X i is essential (Exercise 1.3.8). Later on we shall develop several variants of Chernoff’s inequality in which there is some limited interaction between the X i . Proof By subtracting a constant from each of the X i we may normalize E(X i ) = 0 for each i. Observe that P(\|X \| ≥ λσ ) = P(X ≥ λσ ) + P(X ≤ −λσ ). By symmetry, it thus suffices to prove that P(X ≥ λσ ) ≤ e−tλσ/2 (1.18) where t := min(λ/2σ, 1). Applying (1.15) we have P(X ≥ λσ ) ≤ e−tλσ E et X 1 · · · et X n . Since the X i are jointly independent, so are the et X i . Using this and Lemma 1.7 we obtain E et X 1 · · · et X n = E et X 1 · · · E et X n ≤ exp(t 2 Var(X 1 )) · · · exp(t 2 Var(X n )). On the other hand, from (1.9) we have Var(X 1 ) + · · · + Var(X n ) = σ 2 . Putting all this together, we obtain P(X ≥ λσ ) ≤ e−tλσ et 2 σ2 . Since t ≤ λ/2σ , the claim follows. Now let us consider a special, but important case when X i s are independent boolean (or Bernoulli) variables. Corollary 1.9 Let X = t1 + · · · + tn where the ti are independent boolean random variables. Then for any > 0 P(\|X − E(X )\| ≥ E(X )) ≤ 2e− min( 2 /4,/2)E(X ) . (1.19) 12 1 The probabilistic method Applying this with = 1/2 (for instance), we conclude in particular that P(X = (E(X ))) ≥ 1 − 2e−E(X )/16 . (1.20) Proof From (1.1) we have that \|ti − E(ti )\| ≤ 1 and Var(ti ) ≤ E(ti ). Summing this using (1.3), (1.9), we conclude that Var(X ) ≤ E(X ) (cf. (1.12)). The claim now follows from Theorem 1.8 with λ := E(X )/σ . As an immediate consequence of Corollary 1.9 and (1.4) we obtain the following concentration of measure property for the distribution of certain types of random sets. Corollary 1.10 Let A be a set (possibly infinite), and let B ⊂ A be a random subset of A with the property that the events a ∈ B are independent for every a ∈ A. Then for any > 0 and any finite A ⊆ A we have 2 P \|\|B ∩ A \| − pa \| ≥ pa ≤ 2e− min( /4,/2) a∈A pa a∈A a∈A where pa := P(a ∈ B). In particular 1 3 P pa ≤ \|B ∩ A \| ≤ pa ≥ 1 − 2e− a∈A pa /16 . 2 a∈A 2 a∈A 1.3.1 Sidon’s problem on thin bases We now apply Chernoff’s inequality to the study of thin bases in additive combinatorics. Definition 1.11 (Bases) Let B ⊂ N be an (infinite) set of natural numbers, and let k ∈ Z+ . We define the counting function rk,B (n) for any n ∈ N as rk,B (n) := \|{(b1 , . . . , bk ) ∈ B k : b1 + · · · + bk = n}\|. We say that B is a basis of order k if every sufficiently large positive integer can be represented as sum of k (not necessarily distinct) elements of B, or equivalently if rk,B (n) ≥ 1 for all sufficiently large n. Alternatively, B is a basis of order k if and only if N\k B is finite. Examples 1.12 The squares N∧ 2 = {0, 1, 4, 9, . . .} are known to be a basis of order 4 (Legendre’s theorem), while the primes P = {2, 3, 5, 7, . . .} are conjectured to be a basis of order 3 (Goldbach’s conjecture) and are known to be a basis of order 4 (Vinogradov’s theorem). Furthermore, for any k ≥ 1, the kth powers N∧ k = {0k , 1k , 2k , . . .} are known to be a basis of order C(k) for some finite C(k) (Waring’s conjecture, first proven by Hilbert). Indeed in this case, the powerful Hardy–Littlewood circle method yields the stronger result that 1.3 The exponential moment method 13 rm,N∧ k (n) = m,k (n k −1 ) for all large n, if m is sufficiently large depending on k (see for instance for a discussion). On the other hand, the powers of k k ∧ N = {k 0 , k 1 , k 2 , . . .} and the infinite progression k · N = {0, k, 2k, . . .} are not bases of any order when k > 1. m The function rk,B is closely related to the density of the set B. Indeed, we have the easy inequalities rk,B (n) ≤ \|B ∩ [0, N ]\|k ≤ rk,B (n) (1.21) n≤N n≤k N for any N ≥ 1; this reflects the obvious fact that if n = b1 + · · · + bk is a decomposition of a natural number n into k natural numbers b1 , . . . , bk , then n ≤ N implies that b1 , . . . , bk ∈ [0, N ], and conversely b1 , . . . , bk ∈ [0, N ] implies n ≤ k N . In particular if B is a basis of order k then \|B ∩ [0, N ]\| = (N 1/k ). (1.22) Let us say that a basis B of order k is thin if rk,B (n) = O(log n) for all large n. This would mean that \|B ∩ [0, N ]\| = N 1/k+ok (1) , thus the basis B would be nearly as “thin” as possible given (1.22). In the 1930s, Sidon asked the question of whether thin bases actually exist (or more generally, any basis which is “high quality” in the sense that rk,B (n) = n o(1) for all n). As Erd˝os recalled in one of his memoirs, he thought he could provide an answer within a few days. It took a little bit longer. In 1956, Erd˝os positively answered Sidon’s question. Theorem 1.13 There exists a basis B ⊂ Z+ of order 2 so that r2,B (n) = (log n) for every sufficiently large n. In particular, there exists a thin basis of order 2. Remark 1.14 A very old, but still unsolved conjecture of Erd˝os and Tur´an states that if B ⊂ N is a basis of order 2, then lim supn→∞ r2,B (n) = ∞. In fact, Erd˝os later conjectured that lim supn→∞ r2,B (n)/ log n > 0 (so that the thin basis constructed above is essentially as thin as possible). Nothing is known concerning these conjectures (though see Exercise 1.3.10 for a much weaker result). Proof Define1 a set B ⊂ Z+ randomly by requiring the events n ∈ B (for n ∈ Z+ ) to be jointly independent with probability log n P(n ∈ B) = min C ,1 n 1 Strictly speaking, to make this argument rigorous one needs an infinite probability space such as Wiener space, which in turn requires a certain amount of measure theory to construct. One can avoid this by proving a “finitary” version of Theorem 1.13 to provide a thin basis for an interval [1, N ] for all sufficiently large N , and then gluing those bases together; we leave the details to the interested reader. A similar remark applies to other random subsets of Z+ which we shall construct later in this chapter. 1 The probabilistic method 14 where C > 0 is a large constant to be chosen later. We now show that r2,B (n) = (log n) for all sufficiently large n with positive probability (indeed, it is true with probability 1). Writing r2,B (n) = I(i ∈ B)I( j ∈ B) = I(i ∈ B)I(n − i ∈ B) + O(1) i+ j=n 1≤i 0, prove the reflection principle j n P max εi ≥ λ = 2P εi ≥ λ . 1≤ j≤n i=1 i=1 Hint: Let A ⊂ {−1, 1} be the set of n-tuples (ε1 , . . . , εn ) such that n λ, and let B ⊂ {−1, 1}n be the set of n-tuples (ε1 , . . . , εn ) i=1 εi ≥ j n such that i=1 εi < λ but i=1 εi ≥ λ for some 1 ≤ j < n. Create a “reflection map” which exhibits a bijection between A and B. With the same notation as the previous exercise, show that j n P max ai εi ≥ λ ≤ 2P ai εi ≥ λ n 1.3.2 1≤ j≤n 1.3.3 1.3.4 i=1 i=1 for all non-negative real numbers a1 , . . . , an . By considering the case when X 1 , . . . , X n ∈ {−1, 1} are independent variables taking values +1 and −1 with equal probability 1/2, show that Theorem 1.8 cannot be improved except for the constant in the exponent. Let the hypotheses be as in Theorem 1.8, but with the X i complex-valued instead of real-valued. Show that √ 2 E(\|X − E(X )\| ≥ λσ ) ≤ 4 max e−λ /8 , eλσ/2 2 for all λ > 0. (Hint: if \|z\| ≥ λσ , then either \|Re(z)\| ≥ 1.3.5 √1 λσ .) 2 √1 λσ 2 or \|Im(z)\| ≥ The constants here can be improved slightly. (Hoeffding’s inequality) Let X 1 , . . . , X n be jointly independent random variables, taking finitely many values, with ai ≤ X i ≤ bi for all i and some real numbers ai < bi . Let X := X 1 + · · · + X n . Using the exponential moment method, show that ⎛ 1/2 ⎞ n ⎠ ≤ 2e−2λ2 . P ⎝\|X − E(X )\| ≥ λ \|bi − ai \|2 i=1 1.3.6 (Azuma’s inequality) Let X 1 , . . . , X n be random variables taking finitely many values with \|X i \| ≤ 1 for all i. We do not assume that the X i are jointly independent, however we do require that the X i form a martingale difference sequence, by which we mean that E(X i \|X 1 = x1 , . . . , X i−1 = 1 The probabilistic method 18 xi−1 ) = 0 for all 1 ≤ i ≤ n and all x1 , . . . , xi−1 . Using the exponential moment method, establish the large deviation inequality √ 2 P(\|X 1 + · · · + X n \| ≥ λ n) ≤ 2e−λ /4 . (1.24) Let n be a sufficiently large integer, and color each of the elements in [1, n] red or blue, uniformly and independently at random (so each element is red with probability 1/2 and blue with probability 1/2). Show that the following statements hold with probability at least 0.9: n (a) there is a red arithmetic progression of length at least log ; 10 (b) there is no monochromatic arithmetic progression of length exceeding 10 log n; (c) the number of red elements and the number of blue elements in [1, n] differ by O(n 1/2 ); (d) in every arithmetic progression in [1, n], the numbers of red and blue elements differ by O(n 1/2 log1/2 n). 1.3.8 Let us color the elements of [1, n] red or blue as in the preceding exercise. For each A ⊂ [1, n], let t A denote the parity of the red elements in A; thus t A = 1 if there are an odd number of red elements in A, and t A = 0 otherwise. Let X = A⊆[1,n] t A . Show that the t A are pairwise (but not necessarily jointly) independent, that E(X ) = 2n−1 , and that Var(X ) = 2n−2 . Furthermore, show that P(X = 0) = 2−n . This shows that Chernoff’s inequality can fail dramatically if one only assumes pairwise independence instead of joint independence (though Chebyshev’s inequality is of course still valid in this case). 1.3.9 For any k ≥ 1, find a basis B ⊂ N of order k such that \|B ∩ [0, n]\| = k (n 1/k ) for all large n. (This can be done constructively, without recourse to the probabilistic method, for instance by taking advantage of the base k representation of the integers.) 1.3.10 Prove that there do not exist positive integers k, m ≥ 1, and a set B ⊂ N such that rk,B (n) = m for all sufficiently large n; thus a base of order k cannot be perfectly regular. (Hint: consider the complex-analytic func tion n∈B z n , defined for \|z\| < 1, and compute the kth power of this function. It is rather challenging to find an elementary proof of this fact that does not use complex analysis, or the closely related tools of Fourier analysis.) 1.3.11 With the hypotheses of Theorem 1.8, establish the moment estimates √ E(\|X \| p )1/ p = O( pσ + p) 1.3.7 for all p ≥ 1. 1.4 Correlation inequalities 1.3.12 19 With the hypotheses of Corollary 1.9, establish the inequality X 1 E ≤ E(X )n n n! X for all n ∈ N. (Hint: expand n as i1 0 and X ⊂ N, we say that X is a (1 − ε)-complementary base of A if σ (A + k X ) ≥ 1 − ε, and that X is an asymptotic complementary base of order k of A if σ (A + k X ) = 1. Theorem 1.22 Let P = {2, 3, 5, . . .} be the primes. For any 0 < < 1, there is an (1 − )-complementary base X ⊂ Z+ of order 1 of P with \|X ∩ [1, n]\| = O (log n) for all large n. It follows that (the proof is left as an exercise) Corollary 1.23 For any function ω(n) tending to infinity with n, there is an asymptotic complementary base X ∈ Z+ of order 1 of P with \|X ∩ [1, n]\| ≤ ω(n) log n for all large n. Corollary 1.23 improves an earlier result of Kolountzakis , and should also be compared with Theorem 1.16 (note that every complementary basis is automatically an asymptotic complementary basis). Since P has density (n/ log n), it is clear that an asymptotic complementary base of P should have density (log n). Thus, Corollary 1.23 is nearly best possible. Proof of Theorem 1.22 The theorem follows from the following finite statement. Lemma 1.24 For every ε > 0, and all natural numbers n which are sufficiently large depending on ε, there exists a set B ⊂ [n 2/3 , 2n 2/3 ] with \|B\| = Oε (log n) such that \|1, x\| ≤ εx, for all n 3/4 ≤ x ≤ n. (1.27) 22 1 The probabilistic method The deduction of Theorem 1.22 from Lemma 1.24 is straightforward and is left as an exercise. To prove Lemma 1.24, we use the probabilistic method. We choose B ⊂ [n 2/3 , 2n 2/3 ] randomly, by letting the events l ∈ B with l ∈ [n 2/3 , 2n 2/3 ] be jointly independent with probability K log n n 2/3 where K = K ε is a large constant to be chosen later. From Corollary 1.10 we have P(l ∈ B) = P (\|B\| ≥ 100K log n) This is particularly useful when the Ai are bad events that we would like to avoid. If the Ai are mutually independent, then the problem is trivial, as we have ¯ P Ai = P( A¯ i ) = (1 − P(Ai )), (1.30) i∈V i∈V v∈V which is positive if P(Ai ) are all strictly less than one. On the other hand, mutual independence is a very strong assumption which rarely holds. One may expect that something similar to (1.30) is still true if we allow a sufficiently “local” dependence among the Ai s, so that we still have good control ¯ j . This is indeed possible, on P(Ai ) even after conditioning on most of the events A 1 The probabilistic method 24 as shown by Lov´asz in 1975 in a joint paper with Erd˝os . We present a modern version of this lemma as follows. Lemma 1.25 (Lov´asz local lemma) Let V be a finite set, and for each i ∈ V let Ai be a probabilistic event. Assume that there is a directed graph G(V, E) (without loops) on the vertex set V (which is known as the dependency graph of the Ai ); and a sequence of numbers 0 ≤ xi < 1 for each i ∈ V such that the estimate P Ai \| A¯ j ≤ xi (1 − x j ) (1.31) j∈S (i, j)∈E holds whenever i ∈ V ; and S ⊆ V {i} is such that j∈S A¯ j has non-zero probability and (i, j) ∈ / E for all j ∈ S. Then for any disjoint S, S ⊆ V we have P (1 − xi ) > 0. (1.32) A¯ i \| A¯ i ≥ i∈S i∈S In particular we have P i∈V i∈S A¯ i ≥ (1 − xi ) > 0. i∈V The graph G is usually referred to as the dependency graph of the Ai . Note that (1.31) will hold if we have P(Ai ) ≤ xi (1 − x j ) (i, j)∈E and each Ai is mutually independent to all of the A j with (i, j) ∈ / E and j = i. This was in fact the hypothesis stated in the original formulation of the lemma. However, there are situations where these rather strong mutual independence hypotheses are not available and one needs the full strength of Lemma 1.25. Alon and Spencer’s book Chapter 5 contains many interesting applications. Proof of Lemma 1.25 We shall induce on the total cardinality \|S\| + \|S \|. If \|S\| + \|S \| = 0 then S, S are empty, and the claim (1.32) is trivial. Now assume inductively that \|S\| + \|S \| ≥ 1, and the claim has already been proven for smaller values of \|S\| + \|S \|. Note that the case \|S\| = 0 is trivial. To establish the claim for \|S\| ≥ 1, it suffices to do so for the case \|S\| = 1. Indeed, if \|S\| ≥ 1, then we can split S = { j} ∪ (S{ j}) for some j ∈ S. From the definition of conditional probability we have P A¯ i \| A¯ i = P A¯ i \| A¯ i P A¯ i \| A¯ i i∈S i∈S i∈S ∪S{ j} i∈S{ j} i∈S and the claim (1.32) then follows by applying the induction hypothesis to estimate the second factor. 1.5 The Lov´asz local lemma 25 Thus it remains to verify the \|S\| = 1 case of (1.32). Writing S = {i}, we reduce to showing that P Ai \| A¯ j ≥ xi . j∈S We split S = S1 ∪ S2 where S1 := { j ∈ S\|(i, j) ∈ E} are those indices j which are adjacent to i in the dependency graph, and S2 := S \S1 . From the definition of conditional probability again we have P Ai , j∈S1 A¯ j \| j∈S2 A¯ j . P Ai \| A¯ j = ¯ j\| ¯j A A P j∈S j∈S1 j∈S2 Note that by induction hypothesis, j∈S2 A¯ j occurs with positive probability. From (1.31) we have P Ai , A¯ j \| A¯ j ≤ P Ai \| A¯ j ≤ xi (1 − x j ). j∈S1 j∈S2 j∈S2 j∈V :(i, j)∈E On the other hand, from the induction hypothesis (since \|S1 \| + \|S2 \| < 1 + \|S \|) we have P A¯ j \| A¯ j ≥ (1 − x j ) ≥ (1 − x j ). j∈S1 j∈S2 j∈S1 j∈V :(i, j)∈E Combining the two, we obtain the claim. In practice, the following corollary of Lemma 1.25 is sometimes easier to apply. Corollary 1.26 Let d ≥ 1 and 0 < p < 1 be numbers such that p≤ 1 , e(d + 1) where e = 2.718 . . . is the base of the natural logarithm. Let V be a finite set, and for each i ∈ V let Ai be a probabilistic event with P(Ai ) ≤ p. Assume also that each Ai is mutually independent of all but at most d of the other events A j . Then \|V \| 1 ¯ P Ai ≥ 1 − > 0. d +1 i∈V If d = 0, then Corollary 1.26 follows from (1.30). For d ≥ 1, the corollary 1 1 d follows from Lemma 1.25 by setting xi = d+1 and using the fact that (1 − d+1 ) > 1 . The constant e is best possible as shown by Shearer. e 1.5.1 Colorings of the real line We now give an application of Corollary 1.26. This is the original result from the paper of Erd˝os and Lov´asz, which motivated the development of the local lemma. 26 1 The probabilistic method Let us use k colors [1, k] to color the real numbers. (Thus, a coloring is a map from R to [1, k].) A subset T of R is called colorful if it contains all k colors. Theorem 1.27 Let m and k be two positive integers satisfying 1 m e(m(m − 1) + 1)k 1 − ≤ 1. k (1.33) Then for any set S of real numbers with \|S\| = m, and any set X ⊂ R (possibly infinite), there is a k-coloring of R such that the translates x + S of S are colorful for every x ∈ X . Proof We first prove this theorem in the special case when X is finite, and then use a compactness argument to handle the general case (of course, the theorem is strongest when X = R). The point is that the bound (1.33) does not depend on the cardinality of X . Fix X to be finite; thus X + S is also finite. Note that we only need to color the real numbers in X + S, since the real numbers outside of X + S are irrelevant. For each element y in X + S, we color it randomly and independently: y receives each of the colors in [1, k] with the same probability 1/k. Let A x be the event that the translate x + S is not colorful. We need to show that P A¯ x > 0. x∈X In order to apply Corollary 1.26, we first estimate P(A x ). If x S is not colorful, then at least one color is missing. The probability that a particular color (say 1) is missing is (1 − k1 )\|x+S\| = (1 − k1 )m . As there are k colors, we conclude 1 m P(A x ) ≤ k 1 − . k (In fact we have a strict inequality as there is a positive chance that more than one color is missing.) Next, observe that if two translates x + S and x + S are disjoint, then the events A x and A x are independent. On the other hand, x + S and x ∈ S intersect if and only if there are two elements s1 , s2 ∈ S such that x + s1 = x + s2 . It follows that x = x + (s1 − s2 ). Since that number of (ordered) pairs (s1 , s2 ) with s1 = s2 and s1 , s2 ∈ S is m(m − 1), we conclude that each A x is independent from all but at most m(m − 1) events A x . Set p = k(1 − k1 )m and d = m(m − 1). The condition (1.33) guarantees that the condition of Corollary 1.26 is met and this corollary implies that P( x∈X A¯ x ) > 0, as desired. A routine way of passing from a finite statement to an infinite one is to use a compactness argument and that is what we do next. The space of colorings of R can be identified with the product space [1, k]R , which is compact in the product topology by Tychonoff’s theorem. In this product space, for each x ∈ R we set 1.6 Janson’s inequality 27 K x to be the set of all k-colorings such that x + S is colorful. It is easy to see that each K x is closed. The finite statement proved above asserts that any finite collection of the K x has a non-empty intersection. It follows, by compactness, that all K x , x ∈ R, have a non-empty intersection. Any element in this intersection is a coloring desired by the theorem. Exercise 1.5.1 Show that there exists a positive constant c such that the following holds. For every sufficiently large n, there is a graph on n points which does not contain the following two objects: a triangle and an independent set of √ size c n log n. (An independent set is a set of vertices, no two of which are connected by an edge.) 1.6 Janson’s inequality Let t1 , . . . , tn be jointly independent boolean random variables. In Corollary 1.9 we established a large deviation inequality for the polynomial t1 + · · · + tn . In many applications, it is also of interest to obtain large deviation inequalities for more general polynomials P(t1 , . . . , tn ) of the boolean variables t1 , . . . , tn . One particularly important case is that of a boolean polynomial X := tj, A∈A j∈A where A is some collection of non-empty subsets of [1, n]. Observe that boolean polynomials are automatically positive and monotone increasing, and hence any two boolean polynomials are positively correlated via the FKG inequality (Theorem 1.19). More generally, if X and Y are boolean polynomials, then f (X ) and f (Y ) will be positively correlated whenever f is a monotone increasing or decreasing function. In particular, we see that E e−s(X +Y ) ≥ E e−s X E e−sY (1.34) for any real number s. Using this fact, the exponential moment method, and some additional convexity arguments, Janson derived a powerful bound for the lower tail probability P(X ≤ E(X ) − T ): Theorem 1.28 (Janson’s inequality) Let t1 , . . . , tn , A, X be as above. Then for any 0 ≤ T ≤ E(X ) we have the lower tail estimate T2 P(X ≤ E(X ) − T ) ≤ exp − 2 1 The probabilistic method 28 where = E A,B∈A:A∩B =∅ In particular, we have tj . j∈A∪B E(X )2 P(X = 0) ≤ exp − 2 . Remark 1.29 Informally, Janson’s inequality asserts that if = O(E(X )2 ), then X = (E(X )) with large probability. In the case where A is just the collection of singletons {1}, . . . , {n}, then X = t1 + · · · + tn , = E(X ), and the above claim is then essentially (the lower half of) Corollary 1.9. The quantity is somewhat inconvenient to work with directly. Using the independence of the t j , one can rewrite it as = E tj E tj . Since E(X ) = A∈A E( B∈A:A∩B =∅ j∈A A∈A j∈A t j ), j∈B\A we thus have ≤ E(X ) sup A∈A B∈A:A∩B =∅ E tj . (1.35) j∈B\A We record a particular consequence of this estimate concerning quadratic boolean polynomials that we shall use shortly. Corollary 1.30 Let t1 , . . . , tn be as above, and let X = 1≤i≤ j≤n:i∼ j ti t j , where i ∼ j is some symmetric relation on [1, n]. Then we have E(X ) P(X = 0) ≤ exp − . 2 + 4 supi j:i∼ j E(t j ) Proof We take A := {{i, j} : i ∼ j}. For any A ∈ A, it is easy to verify that E t j ≤ 1 + 2 sup E(t j ) B∈A:A∩B =∅ j∈B\A i j:i∼ j and so the claim follows from (1.35) and Theorem 1.28. Before presenting the proof of Theorem 1.28, let us give an application. This application again concerns complementary bases of primes, but this time of order 2 rather than 1. The following result (which should be compared with Theorems 1.16 and 1.22) in the case k = 2 was recently proved by Vu . 1.6 Janson’s inequality 29 Theorem 1.31 For any k ≥ 2, P has a complementary base B ∈ Z+ of order k with \|B ∩ [1, n]\| = O(log n) for all large n. Proof It suffices to establish the claim when k = 2. To construct B we shall again use the probabilistic method. More precisely, we let B ⊂ Z+ be a random set with the events n ∈ B being independent with probability c P(n ∈ B) = min , 1 n for all n ∈ Z+ , where c is a positive constant to be determined. As before, we will not discuss the measure-theoretic issues associated with requiring infinitely-many independent random variables, as they can be dealt with by a suitable finitization of this argument. Let tn be the boolean random variable tn := I(n ∈ B). By Corollary 1.10 we have 1 P(\|B ∩ [1, m]\| ≤ 10c log m) = 1 − O m2 for all large m, and hence by the Borel–Cantelli lemma (Lemma 1.2) we have with probability 1 that \|B ∩ [1, m]\| = Oc (log m) for all sufficiently large m > 1. (1.36) Now for each n ∈ Z+ , consider the counting function r P+B+B (n) = \|{( p, i, j) ∈ P × B × B : n = p + i + j}\| = ti t j . p j≥n 2/3 :i+ j∈n−P Clearly we have Yn ≤ r P+B+B (n), and so it suffices to show that 1 P(Yn = 0) = O . n2 30 1 The probabilistic method We now apply Corollary 1.30 (using the relation i ∼ j if i = j and i + j ∈ n − P) to give E(Yn ) P(Yn = 0) ≤ exp − . 2 + 4 supi≥n 2/3 j≥n 2/3 :i+ j∈n−P E(t j ) By construction of the t j , and Proposition 1.54 from the Appendix, we have for any i ≥ n 2/3 c E(t j ) = min ,1 n−i − p j≥n 2/3 :i+ j∈n−P p≤n−i−n 2/3 = O(c). On the other hand, from linearity of expectation (1.3) and independence, we have E(Yn ) = E(ti t j ) i> j≥n 2/3 :i+ j∈n−P = i> j≥n 2/3 :i+ j∈n−P c2 ij 1 ij p≤n−2n 2/3 i> j≥n 2/3 :i+ j=n− p log(n − p) = c2 n−p p≤n−2n 2/3 = c2 = (c2 log n), where in the last line we again used Proposition 1.54 from the Appendix. Putting all of these estimates together we obtain P(Yn = 0) ≤ exp(− (c log n)) and the claim follows by choosing c to be suitably large. Now we are going to prove Theorem 1.28. Proof of Theorem 1.28 We shall use the exponential moment method. By a limiting argument we may assume that P(t j = 0), P(t j = 1) > 0 for all j. We introduce the moment generating function F(t) := E(e−t X ) for any t > 0. By (1.16) we have P(X ≤ E(X ) − T ) ≤ F(t) e−t(E(X )−T ) . 1.6 Janson’s inequality 31 Taking logarithms, we see that we only need to establish the inequality T2 2 for some t > 0. Unlike the situation in Theorem 1.8, the summands in X are not necessarily independent, so we cannot factorize F(t) = E(e−t X ) easily. Janson found a beautiful argument to get around this difficulty. Since F(0) = 1, we see from the fundamental theorem of calculus that t F (s) log F(t) = ds. 0 F(s) log F(t) + t(E(X ) − T ) ≤ − Direct calculation shows that F (s) = −E(X e−s X ) −s X =− E e tj j∈A A∈A =− E(e −s X \|E A )P(E A ), A∈A where E A is the event that t j = 1 for all j ∈ A. Thus it suffices to show that t E(e−s X \|E A ) T2 P(E A ) ds − t(E(X ) − T ) ≥ F(s) 2 0 A∈A for some t > 0. We now exploit the fact that some of the factors of e−s X are independent of E A . For each A ∈ A, we split X as Y A + Z A , which are the boolean polynomials Y A := tj; ZA = tj. B∈A:A∩B =∅ j∈B B∈A:A∩B=∅ j∈B By (1.34) (conditioning on the variables in E A ), we conclude E(e−s X \|E A ) ≥ E(e−sY A \|E A )E(e−s Z A \|E A ). On the other hand, Z A is independent from E A and is bounded from above by X ; thus E(e−s Z A \|E A ) = E(e−s Z A ) ≥ E(e−s X ) = F(s). Combining all these estimates, we have reduced to showing that t T2 P(E A ) E(e−sY A \|E A ) ds − t(E(X ) − T ) ≥ 2 0 A∈A for some t > 0. 1 The probabilistic method 32 Next, we exploit the convexity of the function x → e−sx via Jensen’s inequality (Exercise 1.2.4), concluding that E(e−sY A \|E A ) ≥ e−sE(Y A \|E A ) . From linearity of expectation we have A∈A P(E A ) = E(X ), and so another application of Jensen’s inequality gives P(E A ) P(E A )e−sE(Y A \|E A ) ≥ E(X )e−s A∈A E(X ) E(Y A \|E A ) . A∈A On the other hand, from the definition of conditional probability we have P(E A )E(Y A \|E A ) = E I(E A ) t j = . A∈A B∈A:A∩B =∅ A∈A We thus have t P(E A ) E(e−sY A \|E A ) ds − t(E(X ) − T ) 0 A∈A j∈B t ≥ E(X ) (1.37) e−s /E(X ) ds − t(E(X ) − T ) 0 E(X )2 1 − e−t /E(X ) − t(E(X ) − T ). If we set t := T / , then t /E(X ) = T /E(X ) ≤ 1, and we have = (1.38) 1 − e−t /E(X ) = 1 − e−T /E(X ) ≥ T /E(X ) − T 2 /2E(X )2 and hence A∈A t P(E A ) E(e−sY A \|E A ) ds − t(E(X ) − T ) 0 T E(X ) T2 T − − (E(X ) − T ) 2 T2 = 2 ≥ as desired. Remark 1.32 Choosing t = T / might be convenient, but may not be optimal. One can have a slightly better bound by optimizing the right hand side of (1.38) over t. Remark 1.33 The proof of Janson’s inequality is not symmetric. In other words, it cannot be extended to give a bound for the upper tail probability P(X ≥ μ + T ). This probability will be addressed in the next section. 1.7 Concentration of polynomials 33 Exercises 1.6.1 1.6.2 1.6.3 By refining the argument, show that the complementary base B constructed in the proof of Theorem 1.31 has (with high probability) the property that r P+B+B (n) = (log n) for all sufficiently large n. Define a random graph G(n, p) on the vertex set [1, n] as follows. For each pair i, j (1 ≤ i < j ≤ n) draw an edge between i and j with probability p, independently. (a) Prove that if p = o(n −1 ), then with probability 1 − o(1), G(n, p) does not contain a triangle. (b) Assume that p = n −1+ for some small positive constant . Bound the probability that G does not contain a triangle. Prove that for any k ≥ 2 there is a basis B of order k with with \|B ∩ [1, n]\| = O(n 1/2 log1/k n) for all large n. 1.7 Concentration of polynomials In previous sections, we often considered a polynomial Y = Y (t1 , . . . , tn ) of n independent random variables t1 , . . . , tn , and wished to control the tail distribution of Y . For instance Chernoff’s inequality shows that the polynomial t1 + · · · + tn is concentrated around its mean, while Janson’s inequality shows that the values of certain polynomials (especially those of low degree) could very rarely be significantly less than the mean. In this section, we present some further results of this type, that assert that certain polynomials with small degrees are strongly concentrated. These results can be seen as generalizing Chernoff’s bound, and also provide (in certain cases) the missing half (upper tail bound) of Janson’s inequality. To motivate the results, let us first give a classical result which works for any function Y (not just a polynomial) provided that the Lipschitz constant of Y is small. Lemma 1.34 (Lipschitz concentration inequality) Let Y : {0, 1}n → R be a function such that \|Y (t) − Y (t )\| ≤ K whenever t, t ∈ {0, 1}n differ in only one coordinate. Then if t1 , . . . , tn are independent boolean variables, we have √ 2 P(\|Y (t1 , . . . , tn ) − E(Y (t1 , . . . , tn ))\| ≥ λK n) ≤ 2e−λ /2 for all λ > 0. Remark 1.35 This inequality asserts that if each ti can only influence the random variable Y (t1 , . . . , tn ) by at most O(K ), then Y (t1 , . . . , tn ) itself is concentrated √ in an interval of length O(K n) around its mean. It should be compared with Hoeffding’s inequality, which deals with the case Y (t1 , . . . , tn ) := t1 + · · · + tn , and also with Corollary 1.30. 34 1 The probabilistic method Proof By dividing Y by K we may renormalize K = 1. Introduce the partiallyconditioned random variables Y0 , Y1 (t1 ), . . . , Yn (t1 , . . . , tn ) = Y (t1 , . . . , tn ) by Y j (t1 , . . . , t j ) := E(Y \|t1 , . . . , t j ); thus Y j is the conditional expectation of Y with the first j boolean variables t j fixed. In particular Y0 = E(Y ) and Yn = Y (t1 , . . . , tn ). We can thus write Y (t1 , . . . , tn ) − E(Y (t1 , . . . , tn )) = X 1 + · · · + X n where X j := Y j − Y j−1 . One then easily verifies (using the Lipschitz property) that \|X j \| ≤ 1 and X 1 , . . . , X n form a martingale difference sequence in the sense of Exercise 1.3.6. The claim then follows from Azuma’s inequality (1.24). The above lemma is very useful when one has uniform Lipschitz control on Y , for instance if Y = Y (t1 , . . . , tn ) is a polynomial for which the partial derivatives ∂Y are small for all t1 , . . . , tn in the unit cube. However in many applications ∂ti (especially to thin bases), these partial derivatives will only be small on the average. Fortunately there are analogs of the above lemma which apply in this case, though they also require some average control on higher derivatives of Y . To state the results we need some notation. Let Y = Y (t1 , . . . , tn ) be a polynomial of n real variables. We say that Y is totally positive if all of its coefficients are non-negative, and furthermore that Y is regular if all the coefficients are between zero and one. We also say that Y is simplified if all of its monomials are square-free (i.e. do not contain any factor of ti2 ), and homogeneous if all the monomials have the same degree. Thus for instance a boolean polynomial is automatically regular and simplified, though not necessarily homogeneous. Given any multi-index α = (α1 , . . . , αn ) ∈ Zn+ , we define the partial derivative ∂ α Y as α1 αn ∂ ∂ ∂ α Y := ··· Y (t1 , . . . , tn ), ∂t1 ∂tn and denote the order of α as \|α\| := α1 + · · · + αn . For any order d ≥ 0, we denote Ed (Y ) := maxα:\|α\|=d E(∂ α Y ); thus for instance E0 (Y ) = E(Y ), and Ed (Y ) = 0 if d exceeds the degree of Y . These quantities are vaguely reminiscent of Sobolev norms for the random variable Y . We also define E≥d (Y ) := maxd ≥d Ed (Y ). The following result is due to Kim and Vu . Theorem 1.36 Let k ≥ 1, and let Y = Y (t1 , . . . , tn ) be a totally positive polynomial of n independent boolean variables t1 , . . . , tn . Then there exists a constant Ck > 0 depending only on k such that P \|Y − E(Y )\| ≥ Ck λk−1/2 E≥0 (Y )E≥1 (Y ) = Ok e−λ/4+(k−1) log n for all λ > 0. Informally Theorem 1.36 asserts that when the derivatives of Y are smaller on average than Y itself, and the degree of Y is small, then Y is concentrated around 1.7 Concentration of polynomials 35 (Y ) k−1/2 its mean, and in fact we have Y = (1 + Ok EE≥1 log n E(Y ) with high (Y ) ≥0 probability. In applications in additive number theory, we frequently deal with the case when Y is roughly of size log n. In this case, the error term e(k−1) log n renders Theorem 1.36 ineffective. We, however, have a variant which is designed to handle this case: Theorem 1.37 Let k, n ≥ 1 and β, γ , > 0. If Y = Y (t1 , . . . , tn ) is a regular polynomial (not necessarily simplified) of n independent boolean variables t1 , . . . , tn , which is homogeneous of degree k and obeys the expectation bounds Q log n ≤ E(Y ) ≤ n/Q; E1 (Y ), . . . , Ek−1 (Y ) ≤ n −γ for some sufficiently large Q = Q(k, , β, γ ) (independent of n), then P(\|Y − E(Y )\| ≥ E(Y )) ≤ n −β . In the next section, we will use this theorem to prove Theorem 1.15. The next theorem deals with the case when the expectation of Y is less than one. In this case it is convenient to remove the constant term from any derivative of Y which appears. More precisely, introduce the renormalized derivative ∂∗α Y (t) := ∂ α Y (t) − ∂ α Y (0). Theorem 1.38 Let Y = Y (t1 , . . . , tn ) be a simplified regular polynomial of n independent boolean variables (not necessarily homogeneous) such that E(∂∗α Y ) ≤ n −γ for some γ > 0 and all α. Then, for any β > 0, we have the bound P(Y ≥ K β,γ ) < n −β for some K β,γ which is independent of n and Y . Notice that the assumption implies that Y has small expectation. Taking α to be all zero, we have E(Y ) ≤ n −γ . The proof of Theorem 1.36 relies on the so-called “divide and conquer martingale” technique, together with the exponential moment method. It is not too technical but requires lots of introduction. We thus skip it and refer the reader to . The proof of Theorem 1.37 is more complicated. Besides the abovementioned martingale technique, it also requires some non-trivial combinatorial considerations. Theorem 1.38 is a by-product of this proof (for details see ). These theorems have a wide range of applications in several areas and we refer the reader to for a survey. 1.7.1 Bh [g] sets Let us conclude this section by an application of Theorem 1.38. A set A ⊂ N is called a Bh [g] set or a Bh [g] sequence if for any positive integer m, the equation m = x1 + · · · + x h , x1 ≤ x2 ≤ · · · ≤ x h , xi ∈ A, has at most g solutions; up to a 1 The probabilistic method 36 factor of h!, this is equivalent to requiring that rh,A (m) be bounded by g for all m. Bh [g] sets were studied by Erd˝os and Tur´an in . From (1.21) we see that if A is a Bh [g] set, then \|A ∩ [0, n]\| = Oh,g (n 1/ h ) for all n. In the converse direction, Erd˝os and Tur´an proved Theorem 1.39 For any h ≥ 1 and > 0, there exists a set A ⊂ Z+ with \|A ∩ [0, n]\| = h (n 1/ h− ) for all large n, which is a Bh [g] set for some g = gh, (or in other words, rh,A (n) is uniformly bounded in n). Proof By using Theorem 1.38 we can give a short proof of this theorem. As before, we construct A randomly, letting the events n ∈ A be independent with probability P(n ∈ A) = n 1/ h−1− . A simple application of Corollary 1.9 and the Borel–Cantelli lemma also gives \|A ∩ [0, n]\| = h, (n 1/ h− ) for all but finitely many n with probability 1. Thus it will suffice to show that A is a Bh [g] set with probability 1 (perhaps after removing finitely many elements), for some suitably large g = gh depending only on h. Let tn denote the indicator variables tn := I(n ∈ A). For each m, we observe that the random variable Ym = Ym (t1 , . . . , tm ) = tn 1 · · · tn h n 1 ≤···≤n h :n 1 +···+n h =m will become a regular polynomial of degree h in the t1 , . . . , tm once we use the identity tia = ti for a = 2, 3, . . . to make the monomials square-free. To show that A is a Bh [g] set after removing finitely many elements, it will suffice to show that Ym ≤ g for all but finitely many m; by the Borel–Cantelli lemma, it is enough to establish the upper tail estimate P(Ym > g) ≤ m −2 for all large m. From linearity of expectation and independence we have 1/ h−1− 1/ h−1− E(Ym ) = n1 · · · nh n 1 ≤···≤n h :n 1 +···+n h =m ≤ Oh m 1/ h−1− n 1 ,...,n h−1 ≤m ⎛ ≤ m 1/ h−1− Oh ⎝ 1/ h−1− n1 1/ h−1− · · · n h−1 h−1 ⎞ ⎠ n 1/ h−1− n≤m ≤ Oh (n −h ). This already gives some non-trivial bound on P(Ym > g) from Markov’s inequality, but does not give the required decay in m. However, a similar computation to the 1.8 Thin bases of higher order 37 above (which we leave as an exercise) establishes that E(∂∗α Ym ) = Oh (m −1/ h ) for all non-zero α. The claim now follows from Theorem 1.38. The study of Bh [g] sets is a popular topic in additive combinatorics. A detailed discussion of this topic is beyond the scope of our book. Let us, however, mention one new result of Cilleruelo, Ruzsa and Trujillo from . Many other recent results can be found in [62, 191, 213, 61, 145, 272]. Let A ⊂ [1, N ] be a Bh [g] set. A simple counting argument (related to (1.21)) gives \|A\|+h−1 ≤ gh N , which in turn yields the trivial bound \|A\| ≤ (ghh!N )1/ h . h Cilleruelo, Ruzsa and Trujillo gave the first non-trivial bounds for the case g ≥ 2. They prove that \|A\| ≤ 1.864(g N )1/2 + 1 when h = 2, and that Fh (g, N ) ≤ (1 + cosh (π/ h))−1/ h (hh!g N )1/ h when h > 2. The proofs made use of harmonic analysis methods via the con sideration of the trigonometric polynomials f (t) = a∈X eiat . The authors also constructed sets to establish for any g, the existence of a B2 [g] set A ⊂ [1, N ] with g + [g/2] \|A\| ≥ √ + og (1) N 1/2 . g + 2[g/2] Exercises 1.7.1 Consider the random graph G(n, p) defined in Exercise 1.6.2, and set p := n −1+ . Let Y be the number of triangles in G(n, p). Give an upper bound and a lower bound for 3 P Y ≥ E(Y ) . 2 1.7.2 Verify the bound E(∂∗α Ym ) = Oh (n −1/ h ) claimed in the Proof of Theorem 1.39. 1.8 Thin bases of higher order We now return to the study of thin bases B and their associated counting functions rk,B (n), initiated in Section 1.3. However, in this section we can use Theorem 1.37 to present a proof of Theorem 1.15, which asserted for each k ≥ 1 the existence of a base B of order k with rk,B (n) = Ok (log n) for all large n. This was proven in the k = 2 case (see Theorem 1.13) using Chernoff’s inequality, but that method does not directly apply for higher k because rk,B (n) cannot be easily expressed as the sum of independent random variables. 38 1 The probabilistic method We begin with a simple lemma on boolean polynomials that shows that if E(X ) is not too large, then at most points (t1 , . . . , tn ) of the sample space, the polynomial X does not contain too many independent terms (cf. Exercise 1.3.12). Lemma 1.40 Let X = A∈A j∈A t j be a boolean polynomial of n independent boolean variables t1 , . . . , tn , let B ⊆ [1, n] be the random set B := { j ∈ [1, n] : t j = 1}, and let D ∈ N be the random variable, defined as the largest number of disjoint sets in A which are contained in B. Then for any integer K ≥ 1 we have E(X ) K . K! Observe that for A1 , . . . , Ak disjoint, 1 I(D ≥ K ) ≤ tj . . . tj. K ! A ,...,A ∈A,disjoint j∈A1 j∈Ak P(D ≥ K ) ≤ Proof K 1 Taking expectations of both sides and using linearity of expectation (1.3) followed by independence, we conclude 1 P(D ≥ K ) ≤ E tj . . . E tj . K ! A ,...,A ∈A j∈A1 j∈Ak 1 K But by linearity of expectation again, the left-hand side is just E(X ) K /K !, and the claim follows. This lemma is particularly useful when combined with the sunflower lemma of Erd˝os and Rado . A collection of sets A1 , . . . , Al forms a sunflower if the pairwise intersections Ai ∩ A j for i = j are all the same (the Ai are called the petals of the flower). We allow this common pairwise intersection to be empty. Lemma 1.41 (Sunflower lemma) If A is a collection of sets, each of size at most k, and \|A\| > (l − 1)k k!, then A contains l sets forming a sunflower. This lemma can be proven by elementary combinatorics and is left as an exercise. It has the following consequence for the counting function rk,B (n). Corollary 1.42 Let B ⊂ Z+ and k ≥ 2, and for each n ∈ Z+ let Dk,n be the largest number of disjoint multisets2 {x1 , . . . , xk } of elements of B which sum to n. Then k . rk,B (n) ≤ k!k k max Dk,n , sup rk−1,B (m) − 1 m 1. If C is sufficiently large depending on k, then with probability 1, we have rk,B (n) = C,k (log n) for all but finitely many n. In particular, B is a thin basis of order k with probability 1. Proof We shall estimate rk,B (n) in terms of two related expressions: R(n) := {(x1 , . . . , xk ) ∈ B : x1 + · · · + xk = n; n 0.1 < x1 < x2 < · · · < xk } (1.40) E(n) := {(x1 , . . . , xk ) ∈ B : x1 + · · · + xk = n; x1 = x2 or x1 ≤ n 0.1 }. (1.41) It is clear (using the symmetry of x1 + · · · + xk under permutations) that k!R(n) ≤ rk,B (n) ≤ k!R(n) + k 2 E(n). We view R(n) as the main term and E(n) as the error term; this reflects the intuitive fact that for most representations n = x1 + · · · + xk , the xi will be distinct and comparable in magnitude to n. It will suffice to show that with probability 1 we have E(n) = OC,k,B (1); R(n) = C,k,B (log n) for all but finitely many n. Let us deal first with the error term E(n). We argue as in the proof of Proposition 1.43. Let An denote those sets which arise from the multisets {x1 , · · · , xk } with x1 + · · · + xk = n and either x1 = x2 or x1 ≤ n 0.1 . By arguing as in Corollary 1.42, we have k k E(n) ≤ k!k max Dn , sup rk−1,B (m) − 1 m E(Y ) = OC,k 2 n2 for all large n. Applying Theorem 1.37 (and choosing C sufficiently large), we see that it suffices to show the derivative estimates E1 (Y ), . . . , Ek−1 (Y ) ≤ n −γ for all large n and some γ > 0. In other words, we need to establish α1 αn ∂ ∂ E ... Y (t1 , . . . , tn ) ≤ n −γ ∂t1 ∂tn 1 The probabilistic method 42 whenever n is large and 1 ≤ α1 + · · · + αn ≤ k − 1. From the definition of A n we see that we may take α j = 0 for all j ≤ n 0.1 , and all the other α j equal to 0 or 1, since the above partial derivative vanishes otherwise. One can then compute the partial derivative and reduce our problem to showing that E t j ≤ n −γ A∈A n :A⊃A0 j∈A\A0 whenever A0 is any subset of [n 0.1 , n] of cardinality 1 ≤ \|A0 \| ≤ k − 1 (this is the set of indices where α j = 1). Applying linearity of expectation and independence, and noting that j ∈ [n 0.1 , n] for all j ∈ A\A0 , we conclude that k−\|A0 \| E tj ≤ OC,k n 1/k−1 log1/k n A∈A n :A⊃A0 j∈A\A0 A∈A n :A⊃A0 k−\|A0 \| ≤ Ok n k−\|A0 \|−1 OC,k n 1/k−1 log1/k n ≤ OC,k n −1/k log n and the claim follows for large n. Remark 1.45 The proof above is from and is based on the proof of Theorem 1.48 in . The original proof in was different and did not use Theorem 1.37. Exercises 1.8.1 1.8.2 Let A ∈ Z+ be a set of n different integers. Prove that A contains a subset B of cardinality (log n) with the following property. No two elements of B add up to an element of A (thus r2,B (m) vanishes for all m ∈ A, or equivalently A ∩ 2B = ∅). Prove Lemma 1.41. (Hint: first use the pigeonhole principle to show that if \|A\| > (l − 1)k, then either A contains l disjoint sets, or that there exist at least \|A\|/(l − 1)k sets in A which all have a common element x0 . Then use induction on k.) 1.9 Thin Waring bases Recall that a thin basis of order k is a set B ⊂ N such that rk,B (n) = O(log n) for all large n. Theorem 1.15, proved above, asserts that N contains a thin basis of any order. Given the abundance of classical bases such as the squares and primes, it is then natural to pose the following question: 1.9 Thin Waring bases 43 Question 1.46 Let A be any fixed basis of order k. Does A contain a thin subbasis B? Note that Sidon’s original question can be viewed as the k = 2, A = N case of this question. From (1.21) we know that a thin basis B enjoys the bounds \|B ∩ [0, N ]\| = k N 1/k ; \|B ∩ [0, N ]\| = Ok N 1/k log1/k N for all large N . Thus we can consider the following weaker version of Question 1.46: Question 1.47 Let A be any fixed basis of order k. Does A contain a subbasis B with \|B ∩ [0, N ]\| = Ok (N 1/k log1/k N ) for all large N ? Question 1.47 has been investigated intensively for the Waring bases N∧r = {0 , 1r , 2r , . . .}, especially when r = 2 [90, 56, 387, 388, 384, 331]. For these bases it is known that if k is sufficiently large depending on r , then N∧r is a basis of order k, and furthermore that k rk,N∧ r (n) = k,r n r −1 ; (1.42) r note that this is consistent with (1.21). Choi, Erd˝os and Nathanson proved in that N∧ 2, the set of squares, contains a subbasis B of order 4, with \|B ∩ [0, N ]\| = Oε (N 1/3 + ε) for all N > 1 and all ε > 0. This was generalized by Z¨ollner [387, 388], who showed that for any k ≥ 4 there was a subbasis B ⊂ N∧ 2 of order k with \|B ∩ [0, N ]\| = Ok,ε (N 1/k+ε ) for any ε > 0 and N > 1. This bound was then sharpened further to \|B ∩ [0, N ]\| = Ok (N 1/k log1/k N ); from (1.21) we know that this is sharp except for the logarithmic factor. A short proof of Wirsing’s result for the case k = 4 was given by Spencer in . For r ≥ 3, much less was known. In 1980, Nathanson proved that N∧r contains a subbasis of some order with density o(N 1/r ). In the same paper, he posed a special case of Question 1.47, when A = N∧r . In , Vu positively answered Question 1.46 (and hence Question 1.47) for the case A = N∧r for any r ≥ 1: Theorem 1.48 For any fixed r there is an integer k0 such that the following holds. For any k ≥ k0 , the set N∧r of all r th powers contains a thin basis B of order k. In particular, from (1.21) we have \|B ∩ [0, n)\| = Ok (N 1/k log1/k N ) for all large N . Remark 1.49 The sharp concentration result in Theorem 1.37 was first developed in order to prove Theorem 1.48. Just as Theorem 1.15 followed from Proposition 1.44, Theorem 1.48 is an immediate consequence of 44 1 The probabilistic method Proposition 1.50 Let k, r ≥ 2, and let B be a random subset of (Z+ )∧r , defined by letting x r ∈ B be independent with probability r P(x r ∈ B) = min C x k −1 log1/k x, 1 for some positive constant C > 1. If k is sufficiently large depending on r , and C is sufficiently large depending on k, r , then with probability 1 we have rk,B (n) = C,k,r,B (log n) for all but finitely many n. In particular, B is a thin basis of order k with probability 1. Proof (Sketch) As in the proof of Proposition 1.44, it suffices to show that with probability 1 we have E(n) = OC,k,r,B (1); R(n) = C,k,r,B (log n) for all but finitely many n, where R(n) and E(n) were defined in (1.40), (1.41). The contribution of E(n) can be dealt with by similar arguments to the previous section and is left as an exercise, so we focus on R(n). As before we can write R(n) as a boolean polynomial Yn = Yn (t1 , . . . , tm ), where m = n 1/k , tx = I(x r ∈ B), and Yn = tx A∈An x∈A where An is the collection of sets {x1 , . . . , xk } of positive integers with x1r + · · · + xkr = n and n 0.1 < x1r < · · · < xkr . Given the framework presented in the last section, the substantial difficulty remaining is to estimate the expectations of Yn and its partial derivatives. In the following, we shall focus on the expectation of Yn , establishing in particular that E(Yn ) = k,r (C k log n). This is the main estimate, and the remainder of the argument proceeds as in Proposition 1.44. Notice that E(Yn ) = C k k x1 p) . p t log2 t 1 p≤n p p≤n Swapping the sum and integral, we obtain ∞ 1 dt log p = . p t log2 t 1 p≤n p p≤t Applying (1.47), we obtain 1 p≤n p ∞ = 1 (log t + O(1)) dt . t log2 t Since log t logt 2 t is the antiderivative of log log t, and t log1 2 t is absolutely convergent, the claim follows. We now turn to a deeper fact concerning the distribution of primes in intervals. Theorem 1.53 For all sufficiently large n, we have \|P ∩ [n − x, n)\| = ( logx n ) for all n 2/3 < x < n. Results of this type first appeared by Hoheisel ; the result as claimed is due to Ingham . Note that this theorem follows immediately from the Riemann hypothesis (1.45). However, this theorem can be proven without using the Riemann hypothesis, rather some weaker (but still very non-trivial) facts on the distribution of zeroes of the Riemann zeta function: see . We remark that if one only seeks the upper bound on \|P ∩ [n − x, n)\| then one can use relatively elementary sieve theory methods to establish the claim. The constant 2/3 has been lowered 1 The probabilistic method 48 (the current record is 7/12, see , ). However, for the applications here, any exponent less than 1 will suffice. We now combine this theorem with the Abel summation method to establish some further estimates on sums involving primes. Proposition 1.54 Let n be a large integer. Then we have the estimates 1 = (1) n − p p∈P∩[1,n−n 2/3 ) log(n − p) = (log n). n−p p∈P∩[1,n−n 2/3 ) (1.50) (1.51) Proof We begin by proving (1.50). From the fundamental theorem of calculus we have ∞ 1 1 1 p∈[n−x,n−n 2/3 ) 2 d x = n−p x 1 for all p ∈ P ∩ [1, n − n 2/3 ), and hence ∞ 1 P ∩ n − x, n − n 2/3 d x . = n−p x2 1 p∈P∩[1,n−n 2/3 ) The integrand vanishes when x ≤ n 2/3 . When n 2/3 < x ≤ 2n 2/3 , Theorem 1.53 shows that the integrand is O( n 2/3 1log n ), while for x ≥ n 2/3 another application of 1 n Theorem 1.53 shows that the integrand is ( x log ) when x ≤ n and ( x 2 log ) n n when x > n. Putting all these estimates together we obtain (1.50). The estimate (1.51) then follows immediately from (1.50) since log(n − p) = (log n) when p ∈ [1, n − n 2/3 ]. Exercises 1.10.1 By approximating the sum Stirling’s formula n m=1 log m by the integral log n! = n log n − n + O(log n) 1.10.2 1.10.3 n 1 log x d x, prove (1.52) for all n > 1. Using Proposition 1.51, show that there is a constant c so that there is always a prime between n and cn for every positive integer n. By being more careful in the proof of (1.46), show that log p ≤ 2n log 2 + O n 1/2 p 1. Also, use (1.53) to give an alternative proof of (1.49). Using the preceding exercise, show that ∞ log p n=1 ps = 1 + O(1) s−1 for all s > 1; integrate this to conclude ∞ 1 1 + O(1) = log s p s − 1 n=1 1.10.8 (1.54) for all s > 1. Show that these estimates can also be deduced from Proposition 1.51 via Abel’s method. Conversely, use (1.54) and (1.46) to give an alternative proof of (1.48). Using Abel’s summation method, show that the prime number theo rem π (x) = (1 + o(1)) logx x is equivalent to the estimate n≤x (n) = (1 + o(1))x. 50 1.10.9 1 The probabilistic method By being more careful in the proof of (1.48), show that 1 1 = log log n + C + O log n p m, then n A − m A will contain (n − m)A but will generally be larger. A very fundamental question in this topic is the following: under what conditions is A + B “small”, and under what conditions is it “large”? More precisely, we will be interested in the cardinality \|A + B\| of the sum set A + B. We have the following trivial estimates: Lemma 2.1 (Trivial sum set estimates) Let A, B be additive sets with common ambient group Z , and let x ∈ Z . Then we have the identities \|A + x\| = \| − A\| = \|A\|, the inequalities max(\|A\|, \|B\|) ≤ \|A + B\|, \|A − B\| ≤ \|A\|\|B\| (2.1) and the inequalities \|A\|(\|A\| + 1) . 2 More generally, for any integer n ≥ 1, we have \|(n + 1)A\| ≥ \|n A\| and \|A\| + n − 1 \|A\|(\|A\| + 1) · · · (\|A\| + n − 1) \|n A\| ≤ = . n n! \|A\| ≤ \|A + A\| ≤ (2.2) (2.3) We remark that the lower bound in (2.1) can be improved for specific groups Z , or when A and B have large “dimension”; see Theorem 3.16, Lemma 5.3, Theorem 5.17, Corollary 5.13, Theorem 5.4. Proof We shall just prove (2.3), as all the other inequalities either follow from this inequality or are trivial. We argue by induction on \|A\|. If \|A\| = 1 then both sides of (2.3) are equal to 1. If \|A\| > 1, then we can write A = B ∪ {x} where B is a non-empty set with \|B\| = \|A\| − 1. Then nA = n j=0 ( j B + (n − j) · x) 2.1 Sum sets 55 and hence by the induction hypothesis and Pascal’s triangle identity n \|A\| − 1 + j − 1 \|A\| + n − 1 \|n A\| ≤ = \| j B\| ≤ j n j=0 j=0 n as claimed. (We adopt the convention that 0B = {0}.) Observe from the above facts that the magnitude of sum sets such as A + B, A − B, k A are unaffected if one translates A or B by an arbitrary amount. This gives much of the theory of sum sets a “translation-invariant” or “affine” flavor. We will sometimes take advantage of this translation invariance to normalize one of the sets, for instance to contain the origin 0. For “generic” additive sets A and B, the cardinalities of the sum sets considered in Lemma 2.1 are much more likely to be closer to the upper bounds listed above than the lower bounds; see for instance Exercise 2.1.1. This suggests that the lower bounds are only attainable, or close to being attainable, when the sets A and B have a considerable amount of structure; we shall develop this theme in the remainder of this chapter, by introducing tools such as doubling and difference constants, Ruzsa distance, additive energy, and K -approximate groups to quantify some of these notions of “structure”. For now, we at least settle the question of when the lower bound in (2.1) is attained. Proposition 2.2 (Exact inverse sum set theorem) Suppose that A, B are additive sets with common ambient group Z . Then the following are equivalent: r r r r r \|A + B\| = \|A\|; \|A − B\| = \|A\|; \|A + n B − m B\| = \|A\| for at least one pair of integers (n, m) = (0, 0); \|A + n B − m B\| = \|A\| for all integers n, m; there exists a finite subgroup G of Z such that B is contained in a coset of G, and A is a union of cosets of G. Proof We shall just show that the first claim implies the fifth; the remaining claims are either similar or easy and are left to the exercises. By translating B if necessary we may assume that B contains 0. Then A + B ⊃ {0} + A = A, but since \|A + B\| = \|A\| we have A + B = A. In particular A + b = A for all b ∈ B. Thus if we define the symmetry group Sym1 (A) (also known as the period of A) to be the set Sym1 (A) := {h ∈ Z : A + h = A}, then we have B ⊆ Sym1 (A). We leave as an exercise for the reader the verification that Sym1 (A) is a finite group, and A is the union of cosets of Sym1 (A); the claim then follows by setting G := Sym1 (A). 2 Sum set estimates 56 We shall study the symmetry group Sym1 (A), as well as the more general symmetry sets Symα (A), more systematically in Section 2.6. As to when the upper bound is attained, we do not have as explicit a description, but we can give a number of equivalent formulations of the condition. Proposition 2.3 Suppose that A, B are additive sets with common ambient group Z . Then the following are equivalent: r r r r r r r \|A + B\| = \|A\|\|B\|; \|A − B\| = \|A\|\|B\|; \|{(a, a , b, b ) ∈ A × A × B × B : a + b = a + b }\| = \|A\|\|B\|; \|{(a, a , b, b ) ∈ A × A × B × B : a − b = a − b }\| = \|A\|\|B\|; \|A ∩ (x − B)\| = 1 for all x ∈ A + B; \|A ∩ (B + y)\| = 1 for all y ∈ A − B; (A − A) ∩ (B − B) = {0}. We leave the easy proof of this proposition to the exercises. For a partial generalization of it, see Corollary 2.10 below. In Proposition 2.2 and Proposition 2.3, the sets A + B and A − B have the same size (see also Exercise 2.1.6). However, this is not true in general. A basic example is the set A = {0, 1, 3} ⊂ Z; then A + A = {0, 1, 2, 3, 4, 6} has six elements and A − A = {−3, −2, −1, 0, 1, 2, 3} has seven elements. More generally, if A = {0, 1, 3}d ⊂ Zd , then A + A has 6d elements and A − A has 7d . Thus A − A can be larger than A + A by an arbitrarily large amount. In the converse direction, the set A := {(0, 0), (1, 0), (2, 0), (3, 1), (4, 0), (5, 1), (6, 1), (7, 0), (8, 1), (9, 1)} ∈ Z10 × Z2 is such that A + A = Z10 × Z2 has 20 elements, but A − A = Z10 × Z2 {(0, 1)} has only 19 elements; one can amplify this example as before by raising to the power d. Despite these examples, however, there are still several relationships between the size of \|A + A\| and \|A − A\|; see in particular (2.11) below. Exercises 2.1.1 2.1.2 2.1.3 2.1.4 2.1.5 Let N , M ≥ 1 be integers, and let A and B be sets of cardinality N and M respectively chosen uniformly at random from the real interval {x ∈ R : 0 ≤ x ≤1}. Show that with probability 1 we have \|A + B\| = \|A\|\|B\| and \|n A\| = \|A\|+n−1 for all n ≥ 1. n Prove the remaining claims in Proposition 2.2. Let A be an additive set. Show that A is a group if and only if 2A = A. Prove Proposition 2.3. Find an additive set A of integers such that \|A − A\| < \|A + A\|. (Hint: there are several ways to proceed. One way is to tile the lattice Z2 with the Z10 × Z2 example given above, and somehow truncate and then project this back to Z.) 2.2 Doubling constants 2.1.6 2.1.7 2.1.8 57 Let A, B be additive sets in a finite additive group Z , such that \|A\| + \|B\| > \|Z \|. Prove that A + B = A − B = Z . Give an example to show that the condition \|A\| + \|B\| > \|Z \| cannot be improved. Show that for any additive set A, the symmetry group Sym1 (A) of A as defined in the proof of Proposition 2.2 is a finite group contained in A − A, obeys the identity A = A + Sym1 (A), and that A is a union of cosets of Sym1 (A). (We shall define a more general notion of symmetry sets Symα (A) of an additive set in Section 2.6.) Let d ≥ 1. Give an example of an additive set A of integers such that \|A + A\| = 6d and \|A − A\| = 7d . (see also Lemma 5.25.) 2.2 Doubling constants The traditional way to measure the additive structure inside an additive set A is via doubling constants σ [A], which we now define. We will shortly develop two other measures of additive structure, namely the additive energy E(A, A), and the concept of a K -approximate group, which are also useful, and are closely related to the doubling constant. Definition 2.4 (Doubling constant) For an additive set A, the doubling constant σ [A] is defined to be the quantity σ [A] := \|2A\| \|A + A\| = . \|A\| \|A\| Similarly we define the difference constant δ[A] as δ[A] := \|A − A\| . \|A\| From (2.2) we thus have the bounds 1 ≤ σ [A] ≤ \|A\| + 1 \|A\| − 1 1 and 1 ≤ δ[A] ≤ + . 2 2 \|A\| The upper bound here is quite easy to attain; for instance if A = 2∧ [0, N ) = {1, 2, 22 , . . . , 2 N −1 } ⊂ Z, then \|A\| = N , \|A + A\| = N (N2+1) , and \|A − A\| = N (N −1) + 1, hence σ [A] = N 2+1 and δ[A] = N 2−1 + N1 . In the converse direction, 2 Proposition 2.2 shows that σ [A] = 1 (or δ[A] = 1) if and only if A is a coset of a group; we shall elaborate upon this in Proposition 2.7 below. An additive set A with the maximal value of doubling constant σ [A] = (\|A\| + 1)/2 (or equivalently, with maximal difference constant δ[A] = \|A\|−1 + 2 1 ) is known as a Sidon set or a B set. Informally, this means that all the pairwise 2 \|A\| sums of A are distinct, excluding the trivial equalities coming from the identity a + b = b + a; see Exercise 2.2.1. We will revisit Sidon sets in Section 4.5. 2 Sum set estimates 58 There are various senses in which this behavior is “generic”; for instance, if A is a set of N real numbers chosen uniformly at random from the unit interval {x ∈ R : 0 ≤ x ≤ 1}, then we see from Exercise 2.1.1 that A is a Sidon set with probability 1, and so \|A + A\| = N (N2+1) ; the point is that if {a, b} = {c, d} then a + b and c + d will “generically” be distinct. A more interesting question is to understand the conditions under which the doubling constant σ [A] (or difference constant δ[A]) can be small. As mentioned earlier, σ [A] = 1 if and only if A is the coset of a finite subgroup G of Z . We thus expect that if A has a doubling constant which is small, but not actually equal to 1, then it should behave “approximately” like a group (up to translations); we shall see several manifestations of this heuristic throughout this book, when we develop more tools with which to analyze the doubling constant. Indeed, the study of sets of small doubling constant can be thought of as a kind of “approximate group theory”, with the inverse sum set theorems of Chapter 5 then being analogous to a classification theorem for groups. The study of sets with close to maximal doubling appears to be hopeless at present. A probabilistic construction of Ruzsa shows that there exist large additive sets A with \|A − A\| very close to the maximal value of \|A\|2 , but \|A + A\| < \|A\|2−c for some explicit absolute constant c > 0; and similarly with the roles of A − A and A + A reversed. Exercises 2.2.1 2.2.2 2.2.3 2.2.4 2.2.5 2.2.6 Let A be an additive set. Show that A is a Sidon set if and only if, for any a, b, c, d ∈ A, we have a + b = c + d unless {a, b} = {c, d}. Let Z be an additive group, let a, r ∈ Z , and let N ≥ 1 be an integer. Let P = {a, a + r, . . . , a + (N − 1)r } be an arithmetic progression in Z . Show that σ [P] ≤ 2 − N1 , with equality if and only if ord(r ) ≥ 2N − 1, where ord(r ) is the order of the group element r in Z . If φ : Z → Z is a surjective group homomorphism whose kernel ker(φ) := φ −1 ({0}) is finite, and A is an additive set in Z , show that σ [φ −1 (A)] = σ [A]. If A, A are additive sets in Z , Z respectively, show that σ [A × A ] = σ [A]σ [A ]. In particular σ [A⊕d ] = σ [A]d for all d ≥ 1. Let A be any additive set. Show that a non-empty subset of A can have √ doubling constant at most σ [A]\|A\|/2. Give examples that show that this bound cannot be improved except by an absolute constant. What is the analogous statement for the difference constant? Let A be any additive set. Show that a Sidon set contained in A √ can have cardinality at most 2σ [A]\|A\|. (Thus sets with small doubling 2.3 Ruzsa distance and additive energy 2.2.7 2.2.8 2.2.9 2.2.10 2.2.11 59 constant cannot contain very large Sidon sets.) What is the analogous statement for the difference constant? Let p be a prime, let θ ∈ Z p \0 be a multiplicative generator of Z p , and let Z := Z p−1 × Z p . Let A ⊂ Z be the set A := {(t, θ t ) : t = 1, . . . , p − 1}. Show that A is a Sidon set, and compare this to Exercise 2.2.6. Modify this construction to give an example of a Sidon set A ⊂ [0, N ] for a large integer N such that \|A\| is comparable to N 1/2 . A similar example can be given by using the discrete parabola {(t, t 2 ) : t ∈ Z p } in Z p × Z p . For a survey of other constructions of Sidon sets, see . Let N be a large integer. Give examples of finite non-empty sets A, B of integers such that \|A\| = \|B\| = N and σ [A], σ [B] ≤ 2, but σ [A ∪ B] ≥ N . This example shows that doubling constants can behave very badly 2 under set union (see however Exercise 2.3.17). On the other hand, establish the inequality σ [A ∪ B] ≤ σ [A] + \|B\|; thus adding a small set to A will not significantly affect the doubling constant. Let N be a large integer. Give examples of finite non-empty sets A, B of integers such that \|A\| = \|B\| = N and σ [A], σ [B] ≤ 10, but σ [A ∩ B] ≥ 1 N 1/2 . (Hint: concatenate a Sidon set with an arithmetic progression.) 10 Compare this result against Exercise 2.2.6. This example shows that doubling constants can behave badly under set intersection (but see Exercise 2.4.7). Let A be an additive set in Z , and let π : Z → Z be a group homomorphism. Show by example that σ [π(A)] is not necessarily less than or equal to σ [A]. (Hint: this is surprisingly delicate. One way is to start with an additive set C in some additive group Z 0 with σ [C] > δ[C], and consider the additive set A := ((−C)n × {0} × G) ∪ (C n × X × {0}) in Z 0n × Z × G, where n ≥ 1 is large, G is a very large finite group, and X is a Sidon set of medium size in a group Z .) See however Exercise 2.3.8 and Exercise 6.5.17. Let A be an additive set in Z , and let G be a finite subgroup of Z . Show by example that σ [A + G] is not necessarily less than or equal to σ [A]. (Hint: use the previous exercise.) 2.3 Ruzsa distance and additive energy The doubling constant measures the amount of internal additive structure of a single additive set A. We now introduce two useful quantities measuring the amount 2 Sum set estimates 60 of common additive structure between two additive sets A, B – the Ruzsa distance and the additive energy. Definition 2.5 (Ruzsa distance) Let A and B be two additive sets with a common ambient group Z . We define the Ruzsa distance d(A, B) between these two sets to be the quantity d(A, B) := log \|A − B\| . \|A\|1/2 \|B\|1/2 Thus for instance d(A, A) = log δ[A]. We now justify the terminology “Ruzsa distance”. Lemma 2.6 (Ruzsa triangle inequality) The Ruzsa distance d(A, B) is non-negative, symmetric, and obeys the triangle inequality d(A, C) ≤ d(A, B) + d(B, C) for all additive sets A, B, C with common ambient group Z . Proof The non-negativity follows from (2.1). The symmetry follows since B − A = −(A − B). Now we prove the triangle inequality, which we can rewrite as \|A − C\| ≤ \|A − B\|\|B − C\| . \|B\| From the identity a − c = (a − b) + (b − c) we see that every element a − c in A − C has at least \|B\| distinct representations of the form x + y with (x, y) ∈ (A − B) × (B − C). The claim then follows. For an approximate version of this inequality in which one replaces complete difference sets with nearly complete difference sets (using at least 75% of the differences), see Exercise 2.5.4. The Ruzsa distance thus satisfies all the axioms of a metric except one; we do not have that d(A, A) = 0 for all sets A (also, we have d(G + x, G + y) = 0 whenever G + x, G + y are cosets of a group G). Indeed we have a precise characterization on when this Ruzsa distance vanishes: Proposition 2.7 Suppose that (A, Z ) is an additive set. Then the following are equivalent: r σ [A] = 1 (i.e. \|A + A\| = \|A\|); r δ[A] = 1 (i.e. \|A − A\| = \|A\|, or d(A, A) = 0); r d(A, B) = 0 for at least one additive set B; 2.3 Ruzsa distance and additive energy 61 r \|n A − m A\| = \|A\| for at least one pair of non-negative integers n, m with n + m ≥ 2; r \|n A − m A\| = \|A\| for all non-negative integers n, m; r A is a coset of a finite subgroup G of Z . Proof Apply Proposition 2.2 and the Ruzsa triangle inequality. Later on in this chapter we shall generalize this proposition to the case when the Ruzsa distance, difference constant, or doubling constant are a little larger than 0, 0, or 1 respectively, but still fairly small; see Proposition 2.26. Despite the non-vanishing of the distance d(A, A) in general, it is still a useful heuristic to view the Ruzsa distance as behaving like a metric1 . Now we relate the difference constant to the doubling constant. From the definition of Ruzsa distance and doubling constant we have the identity d(A, −A) = log σ [A]. (2.4) In particular, from Lemma 2.6 we have log δ[A] = d(A, A) ≤ 2 log σ [A] and hence we obtain the estimate δ[A] ≤ σ [A]2 or in other words that \|A − A\| ≤ estimate \|A+A\| \|A\| 2 (2.5) . A similar argument gives the more general \|B − B\| ≤ \|A + B\|2 \|A\| (2.6) for any two additive sets A, B with common ambient group Z . It turns out that we can conversely bound the doubling constant of a set by its difference constant; see (2.11) below. Having introduced the Ruzsa distance, we now turn to the closely related notion of additive energy E(A, B) between two additive sets. Definition 2.8 (Additive energy) If A and B are two additive sets with ambient group Z , we define the additive energy E(A, B) between A and B to be the quantity E(A, B) := \|{(a, a , b, b ) ∈ A × A × B × B : a + b = a + b }\|. 1 One could artificially convert the Ruzsa distance into a genuine metric by identifying A with A + x for all x, and redefining d(A, A) to be zero, or alternatively by introducing the metric space X := {A × { j} : A ⊆ Z ; 0 < \|A\| < ∞; j ∈ {1, 2}} – consisting of two copies of each finite non-empty subset of Z (again identifying A with its translations) – with the metric d X (A × { j}, B × {k}) defined to equal d(A, B) if A × { j} = B × {k} and equal to 0 otherwise. However there appears to be no significant advantage in working in such an artificial setting. 2 Sum set estimates 62 We observe the trivial bounds \|A\|\|B\| ≤ E(A, B) ≤ \|A\|\|B\| min(\|A\|\|B\|). (2.7) The lower bound follows since a + b = a + b whenever (a, b) = (a , b ). To see the upper bound, observe that if one fixes a, a , b, then b = a + a − b is completely determined, and hence E(A, B) ≤ \|A\|2 \|B\|. A similar argument gives E(A, B) ≤ \|A\|\|B\|2 . Note that Proposition 2.3 addresses the case when E(A, B) = \|A\|\|B\|. We will analyze the additive energy more comprehensively in Section 4.2, when we have developed the machinery of Fourier transforms, and in Section 2.5, when we have developed the Balog–Szemer´edi–Gowers theorem. For now we concentrate on the elementary properties of this energy. We first observe the symmetry property E(A, B) = E(B, A) and the translation invariance property E(A + x, B + y) = E(A, B) for all x, y ∈ Z . From the trivial observation a + b = a + b ⇐⇒ a − b = a − b we also see that E(A, B) = E(A, −B), and similarly if we reflect A to −A. The additive energy reflects the extent to which A intersects with translates of B or −B, as the following simple identities show: Lemma 2.9 Let A, B be additive sets with ambient group Z . Then we have the identities \|A\|\|B\| = \|A ∩ (x − B)\| = \|A ∩ (B + y)\| x∈A+B and E(A, B) = y∈A−B \|A ∩ (x − B)\|2 x∈A+B = \|A ∩ (B + y)\|2 y∈A−B = \|A ∩ (z + A)\|\|B ∩ (z + B)\|. z∈(A−A)∩(B−B) In particular, if we let r A+B (n) denote the number of representations of n as a + b for some a ∈ A and b ∈ B, and define r A−B (n) similarly, then we have \|A\|\|B\| = r A+B (n) = r A−B (n); E(A, B) = r A+B (n)2 = r A−B (n)2 . n n n n Proof A simple counting argument yields \|A\|\|B\| = \|{(a, b) ∈ A × B : a + b = x}\| = \|A ∩ (x − B)\|; x∈A+B x∈A+B 2.3 Ruzsa distance and additive energy 63 By replacing B with −B we similarly obtain \|A\|\|B\| = y∈A−B \|A ∩ (B + y)\|. This gives the first set of identities. For the second set we compute \|A ∩ (x − B)\|2 x∈A+B = \|{(a, b) ∈ A × B : a + b = x}\|2 x∈A+B = \|{(a, a , b, b ) ∈ A × A × B × B : a + b = a + b = x}\| x∈A+B = \|{(a, a , b, b ) ∈ A × A × B × B : a + b = a + b }\| = \|{(a, a , b, b ) ∈ A × A × B × B : a − b = a − b}\| = \|{(a, b ) ∈ A × B : a − b = a − b}\|2 y∈A−B = \|A ∩ (B + y)\|2 y∈A−B and \|A ∩ (z + A)\|\|B ∩ (z + B)\| z∈(A−A)∩(B−B) = \|{(a, a , b, b ) ∈ A × A × B × B : z = a − a = b − b}\| z∈(A−A)∩(B−B) = \|{(a, a , b, b ) ∈ A × A × B × B : a − a = b − b}\| = \|{(a, a , b, b ) ∈ A × A × B × B : a + b = a + b }\| and the claims follow from the definition of E(A, B). The last identity follows since r A+B (n) = \|A ∩ (n − B)\| and r A−B (n) = \|A ∩ (B + n)\|. As a consequence of this Lemma we have the following inequalities, which assert that pairs of sets with small Ruzsa distance have large additive energy, and pairs with large additive energy have large intersection (after translating and possibly reflecting one of the sets). Corollary 2.10 Let A, B be additive sets. Then there exists x ∈ A + B and y ∈ A − B such that \|A ∩ (x − B)\|, \|A ∩ (B + y)\| ≥ E(A, B) \|A\|\|B\| ≥ \|A\|\|B\| \|A ∓ B\| (2.8) for either choice of sign ±. In particular all of the above quantities are bounded by \|(A − A) ∩ (B − B)\|. Finally we have the Cauchy–Schwarz inequality E(A, B) ≤ E(A, A)1/2 E(B, B)1/2 . (2.9) 2 Sum set estimates 64 Proof From Lemma 2.9 and Cauchy–Schwarz we have E(A, B) \|A\|\|B\| ≥ . \|A\|\|B\| \|A ± B\| Also, from the last part of Lemma 2.9 we have E(A, B) ≤ \|A\|\|B\| max r A+B (x), \|A\|\|B\| max r A−B (y) x∈A+B y∈A−B which establishes (2.8). To bound \|A ∩ (x − B)\| and \|A ∩ (B + y)\|, observe that if z ∈ A ∩ (x − B), then A ∩ (x − B) ⊂ z + ((A − A) ∩ (B − B)), hence \|A ∩ (x − B)\| ≤ \|(A − A) ∩ (B − B)\|, and similarly \|A ∩ (B + y)\| ≤ \|(A − A) ∩ (B − B)\|. Finally, (2.9) follows from the formula E(A, B) = z∈(A−A)∩(B−B) \|A ∩ (z + A)\|\|B ∩ (z + B)\| from Lemma 2.9 and the Cauchy–Schwarz inequality. Another connection in a similar spirit is Lemma 2.11 Let A, B be additive sets. Then for any x ∈ A + B we have \|A ∩ 2 (x − B)\| ≤ \|A−B\| . \|A+B\| Proof (Lev Vsevolod, private communication) We can rewrite the inequality as \|{(a, b, c) ∈ A × B × (A + B) : a + b = x}\| ≤ \|(A − B) × (A − B)\|. Now for each (a, b, c) in the set on the left-hand side, we can write c = ac + bc for some ac ∈ A, bc ∈ B, and then form the pair (a − bc , ac − b) ∈ (A − B) × (A − B). Using the identity c = x − (a − bc ) + (ac − b) we can verify that this map is injective. The claim follows. Corollary 2.12 Let A, B be additive sets with ambient group Z . Then there exists x ∈ A + B such that \|A − B\|2 \|A − B\|2 \|A\|\|B\| \|A − B\|3 ≤ ≤ . \|A ∩ (x − B)\| E(A, B) \|A\|\|B\| (2.10) Furthermore we have d(A, −B) ≤ 3d(A, B). Proof The inequalities in (2.10) follow from (2.8), and the final inequality d(A, −B) ≤ 3d(A, B) then follows from Lemma 2.11 and the definition of Ruzsa distance. From (2.10) and and (2.5) we obtain the inequalities δ[A]1/2 ≤ σ [A] ≤ δ[A]3 (2.11) which were first observed in . Thus an additive set has small doubling constant if and only if its difference constant is small. It is not known whether the lower 2.3 Ruzsa distance and additive energy 65 bound is best possible. However, the upper bound can be improved to σ [A] ≤ δ[A]2 using Pl¨unnecke inequalities; see Exercise 6.5.15. We now show how the Ruzsa distance can be used to control iterated sum sets. We begin with a lemma which controls iterated sum sets of “most” of A + B. Lemma 2.13 Let A and B be additive sets in a common ambient group. Then there exists S ⊂ A + B such that \|{(a, b) ∈ A × B : a + b ∈ S}\| ≥ \|A\|\|B\|/2 (2.12) and such that \|A + B + nS\| ≤ 2n \|A + B\|2n+1 \|A\|n \|B\|n (2.13) for all integers n ≥ 0. Note that (2.12) gives a lower bound on \|S\|, namely \|S\| ≥ max(\|A\|, \|B\|)/2. Proof (2.14) If we define S to be the set of all x ∈ A + B such that \|{(a, b) ∈ A × B : a + b = x}\| ≥ \|A\|\|B\| 2\|A + B\| then we have \|{(a, b) ∈ A × B : a + b ∈ (A + B)\S}\| < \|A + B\| \|A\|\|B\| 2\|A + B\| which gives (2.12). Now we prove (2.13). A typical element of A + B + nS can be written as a0 + s1 + s2 + · · · + sn + bn+1 where a0 ∈ A, bn+1 ∈ B, and s1 , . . . , sn ∈ S. By definition of S, we can expand \|A\|\|B\| n this in at least ( 2\|A+B\| ) different ways as a0 + (b1 + a1 ) + (b2 + a2 ) + · · · + (bn + an ) + bn+1 where bi ∈ B, ai ∈ A, and bi + ai = si for all 1 ≤ i ≤ n. We regroup this as the sum of n + 1 elements from A + B, (a0 + b1 ) + (a1 + b2 ) + · · · + (an + bn+1 ) and observe that for fixed a0 , s1 , . . . , sn , bn+1 , the quantities a0 + b1 , a1 + b2 , . . . , an + bn+1 completely determine all the variables a0 , . . . , an , b1 , . . . , bn+1 . \|A\|\|B\| n Thus we have shown that every element of A + B + nS has at least ( 2\|A+B\| ) representations of the form t0 + · · · + tn where each ti ∈ A + B. The claim then follows. 2 Sum set estimates 66 This result can then be used, together with the Ruzsa triangle inequality, to deduce control on iterated sum sets of A and B; see Exercise 2.3.10. However we will pursue an approach that gives slightly better bounds in the next section (and an even better result will be developed in Section 6.5). Exercises 2.3.1 2.3.2 If φ : Z → Z is a surjective group homomorphism whose kernel ker(φ) := φ −1 ({0}) is finite, and A, B are additive sets in Z , show that d(φ −1 (A), φ −1 (B)) = d(A, B). Also show that d(A + x, B + y) = d(A, B) for any x, y ∈ Z . If A, B, C, D are additive sets in Z , show that d(A, B) − 1 log \|C\|\|D\| ≤ d(A + C, B + D) ≤ d(A, B) + log \|C − D\| 2 and d(A, B ∪ C) ≤ max(d(A, B), d(A, C)) + If A , B are additive sets in Z , show that 1 log 2. 2 d(A × A , B × B ) = d(A, B) + d(A , B ). 2.3.3 2.3.4 2.3.5 2.3.6 2.3.7 Let A, B be additive sets with common ambient group. Show that d(A, B) ≤ 12 log \|A\| + 12 log \|B\|, and that d(A, B) = 12 log \|A\| + 1 log \|B\| if and only if d(A, −B) = 12 log \|A\| + 12 log \|B\|. 2 Let A, B, C be additive sets in Z . Show that 1 \|B\| d(A, C) ≤ d(A, B) + log (2.15) 2 \|C\| whenever C ⊆ B; this shows that the Ruzsa distance d(A, B) is stable under refinement of one or both of the sets A, B. By combining this inequality with the triangle inequality d(A, −B) ≤ d(A, (x − A) ∩ B) + d((x − A) ∩ B, −B), give another proof of Lemma 2.11. Show that for any n ≥ 1, there exists an additive set A such that \|A\| = 4n , \|A + A\| = 10n , and \|2A − A\| = 28n . Thus it is not possible to obtain an estimate of the form \|2A − A\| = O(σ 2 [A]\|A\|). Let A, B be additive sets with common ambient group. Show that e−2d(A,B) \|A\| ≤ \|B\| ≤ e2d(A,B) \|A\|. Thus sets which are close in the Ruzsa distance are necessarily close in cardinality also. Of course the converse is far from true. Let A, B be additive sets with common ambient group Z . Show that d(A, B) = 0 if and only if A, B are cosets of the same finite subgroup G of Z . (We shall generalize this result later; see Proposition 2.27.) 2.3 Ruzsa distance and additive energy 2.3.8 2.3.9 2.3.10 67 Let A be an additive set in an additive group Z , and let G be a finite subgroup of Z . Show that σ [A + G] ≤ \|3A\| . (Hint: apply the Ruzsa tri\|A\| angle inequality to 2A, −A, and G.) Conclude that if π : Z → Z is a group homomorphism then σ [π(A)] ≤ \|3A\| . One cannot replace the \|A\| tripling constant \|3A\| with the doubling constant; see Exercise 2.2.10. See \|A\| however Exercise 6.5.17. Let K be a large integer, and let A = B = {e1 , . . . , e K } be the standard basis of Z K . Show that if S is any subset of A + B obeying (2.12) then \|A + B\|2n+1 \|A + B + nS\| = n \|A\|n \|B\|n where we are using the Landau notation (). This shows that Lemma 2.13 cannot be significantly improved (except possibly by improving the bound (2.14)). Let A, B be additive sets with common ambient group such that \|A + B\| ≤ K \|A\|1/2 \|B\|1/2 for some K ≥ 1. Using Lemma 2.13 and many applications of the Ruzsa triangle inequality, establish the estimate \|n 1 A − n 2 A + n 3 B − n 4 B\| = On 1 ,n 2 ,n 3 ,n 4 K On1 ,n2 ,n3 ,n4 (1) \|A\|1/2 \|B\|1/2 for all integers n 1 , n 2 , n 3 , n 4 . In particular, establish the bounds d(n 1 A − n 2 A + n 3 B − n 4 B, n 5 A − n 6 A + n 7 B − n 8 B) ≤ On 1 ,...,n 8 (1 + d(A, B)) 2.3.11 for all integers n 1 , . . . , n 8 . We shall improve this bound slightly in Corollary 2.23 and Corollary 2.24; see also Corollary 2.19 for the “tensor power trick” that can eliminate lower order terms such as the implicit constant preceding the K On1 ,n2 ,n3 ,n4 (1) factor. Let G and H be subgroups of Z . Show that d(G, H ) = log 2.3.12 \|G\|1/2 \|H \|1/2 \|G ∩ H \| Conclude that d(G, H ) = d(G, G + H ) + d(G + H, H ) = d(G, G ∩ H ) + d(G ∩ H, H ). Also, if K is another subgroup of Z , prove the contractivity properties d(G + K , H + K ) ≤ d(G, H ) and d(G ∩ K , H ∩ K ) ≤ d(G, H ). Note that the Ruzsa distance, when restricted to subgroups of Z , is indeed a genuine metric, thanks to Proposition 2.7. See also Exercises 2.4.7 and 2.4.8 below. Let A be an additive set. Show that σ [A ∪ (−A)] ≤ 2σ [A] + σ [A]2 . 68 2.3.13 2.3.14 2.3.15 2.3.16 2 Sum set estimates Thus a set with small doubling can be embedded in a symmetric set (i.e. a set B such that −B = B) with small doubling which has at most twice the cardinality. Let A be an additive set. Prove the inequalities \|A − A\| ≤ \|A + A\|3/2 and \|A + A\| ≤ \|A − A\|3/2 . (Hint: use (2.11), Corollary 2.12 and (2.1).) Let A be an additive set. Show that there exists an element x ∈ A − A such that the set F := A ∩ (x + A) has size \|F\| ≥ \|A\|/σ [A] and doubling constant σ [F] ≤ σ [A]2 . Thus every additive set A of small doubling contains a large symmetric subset F of small doubling, though the set F may be symmetric around a non-zero origin x/2. Let A, B be additive sets with common ambient group Z . Show that δ[A] ≤ e2d(A,B) and σ [A] ≤ e6d(A,B) . Thus only sets with small doubling constant can be close to other sets in the Ruzsa metric. (The 6 can be lowered to a 4, see Exercise 6.5.15.) Let A, B be additive sets with common ambient group Z . Show that σ [A ∪ B] ≤ ed(A,B) + 2e4d(A,B) . Thus a pair of sets which are close in the Ruzsa metric can be embedded in a slightly larger set with small doubling. In the converse direction, establish the estimate d(A, B) ≤ log σ [A ∪ B] + 2.3.17 1 \|A ∪ B\| 1 \|A ∪ B\| log + log . 2 \|A\| 2 \|B\| Let A, B be additive sets with common ambient group Z , such that σ [A], σ [B] ≤ K for some K ≥ 1, and such that A ∩ B is non-empty. Show that σ [A ∪ B] ≤ 2K + K 3 2.3.18 min(\|A\|, \|B\|) . \|A ∩ B\| Thus the union of sets with small doubling remains small doubling provided that those two sets had substantial intersection. , Let K ≥ 1, and let A1 , A2 , A3 be additive sets with common ambient group Z , such that \|A j ∩ A3 \| ≥ 1 \|A j \| and \|A j + A j \| ≤ K \|A j \| K for all j = 1, 2, 3. Prove that \|A1 + A2 \| ≤ K 6 \|A3 \|. Hint: use the triangle inequality d(A1 , −A2 ) ≤ d(A1 , −(A1 ∩ A3 )) + d(−(A1 ∩ A3 ), A2 ∩ A3 ) + d(A2 ∩ A3 , −A2 ) 2.4 Covering lemmas 2.3.19 2.3.20 69 Suppose that A and B are subgroups of Z , and let x = y = 0. Show that all the inequalities in (2.8) are in fact equalities. Let A, B, C be additive sets in an ambient group Z . Show that max(E(A, B), E(A, C)) ≤ E(A, B ∪ C)1/2 ≤ E(A, B)1/2 + E(A, C)1/2 . 2.3.21 2.3.22 2.3.23 2.3.24 (Hint: use Lemma 2.9 and the triangle inequality for the l 2 norm.) Let A, B, C be additive sets in an ambient group Z with \|A\| = \|B\| = \|C\| = N . Give examples of such sets where E(A, B) and E(A, C) are comparable to N 2 and E(B, C) is comparable to N 3 , or where E(A, B) and E(A, C) are comparable to N 3 and E(B, C) are comparable to N 2 . These examples show that there is no hope of any useful “triangle inequality” connecting E(A, B), E(B, C), and E(A, C). Suppose A, B are additive sets in an ambient group Z . Show that E(A, B) = \|A\|2 \|B\| holds if and only if \|A + B\| = \|B\|. One can thus use Proposition 2.2 to determine when the upper bound in (2.7) is obtained. Conclude in particular that E(A, B) = \|A\|3/2 \|B\|3/2 if and only if d(A, B) = 0, which in turn occurs if and only if A and B are cosets of the same finite group G. Give an example of an additive set A ⊂ Z of cardinality \|A\| = N such 1 1 that E(A, A) ≥ 100 N 3 but d(A, A) ≥ 100 log N . Compare this with (2.8) (and with Corollary 2.31 below). Let A be an additive set. Show that there exists a subset A of A of cardinality \|A \| ≥ 2σ1[A] \|A\| and an element a0 ∈ A such that \|(a + A) ∩ (a0 + A)\| ≥ 2σ1[A] \|A\| for all a ∈ A . (Hint: first obtain a lower bound for E(A, A).) 2.4 Covering lemmas We now describe some covering lemmas, which roughly speaking have the following flavor: if A and B have similar additive structure (for instance, if their Ruzsa distance is small) then one can cover A by a small number translates of B (or some modification of B). Lemma 2.14 (Ruzsa’s covering lemma) For any additive sets A, B with common ambient group Z , there exists an additive set X + ⊆ B with B ⊆ A − A + X +; \|X + \| ≤ \|A + B\| ; \|A\| \|A + X + \| = \|A\|\|X + \| 2 Sum set estimates 70 and similarly there exists an additive set X − ⊆ B with B ⊆ A − A + X −; \|X − \| ≤ \|A − B\| ; \|A\| \|A − X − \| = \|A\|\|X − \|. In particular, B can be covered by min( \|A+B\| , \|A−B\| ) translates of A − A. \|A\| \|A\| Remark 2.15 One useful side benefit of this covering lemma is that there exist at \|B\| least \|A−A\| disjoint translates A + b of A with b ∈ B, as can be seen by restricting b to X + . Proof It suffices to prove the claim concerning A + B, since the claim concerning A − B follows by replacing B with −B and X + with −X − (note that A − A is symmetric around the origin). Consider the family {A + b : b ∈ B} of translates of A by elements of B. All of these translates have volume \|A\| and are contained inside A + B. Thus if we take a maximal disjoint sub-family of these translates, i.e. {A + x : x ∈ X + } for some X + ⊆ B, then X + can have cardinality at most \|A+B\| . \|A\| Also we have \|A + X + \| = \|A\|\|X + \| by construction. Now for any element b ∈ B, we see that A + b cannot be disjoint from every member of {A + x : x ∈ X + } as this would contradict the maximality of X + . Thus A + b must intersect A + X + , which implies that b is in A − A + X + . Since b ∈ B was arbitrary, we thus have B ⊆ A − A + X + and the claim follows. Covering lemmas such as the one above are convenient for a number of reasons. Firstly, they allow for easy computation of iterated sum sets. For instance, if one knows that A+B ⊆ A+X then one can immediately deduce that A + n B ⊆ A + n X for all n ≥ 0. This is advantageous if X is substantially smaller than B. Also, a covering property such as A + B ⊆ A + X is preserved under Freiman homomorphisms, whereas bounds such as \|A + A\| ≤ K \|A\| are only preserved by Freiman isomorphisms (see Chapter 5, in particular Exercise 5.3.13). Remark 2.16 Observe that we are covering B by A − A rather than by A. This reflects the fact that A − A is a “smoother” set than A, and tends to contain fewer “holes” that would render it unsuitable for covering other sets. Later on we shall see that higher-order sum-difference sets such as 2A − 2A are even smoother, in that they tend to contain very large arithmetic progressions; see Section 4.7 and Chapter 12 for further discussion. 2.4 Covering lemmas 71 One can modify Ruzsa’s covering lemma in a number of ways. For instance, one can ensure the covering of B by translates of A − A has very high multiplicity (at the cost of increasing the number of covers by a factor of 2). Lemma 2.17 (Green–Ruzsa covering lemma) Let A and B be additive sets with common ambient group. Then there exists an additive set X ⊆ B with \|X \| ≤ 2 \|A+B\| − 1 such that for every y ∈ B there are at least \|A\|/2 triplets \|A\| (x, a, a ) ∈ X × A × A with x + a − a = y. More informally, A − A + X covers B with multiplicity at least \|A\|/2. Furthermore, we have B − B ⊆ A − A + X − X. Similar claims hold if \|A+B\| \|A\| is replaced by \|A−B\| . \|A\| Proof Again it suffices to prove the claim for \|A+B\| . We perform the following \|A\| algorithm. Initialize X to be the empty set, so that X + A − A is also the empty set. We now run the following loop. If we cannot find any element y in B which is “sufficiently disjoint from X + A − A” in the sense that \|(y + A) ∩ (X + A)\| ≤ \|A\|/2, we terminate the algorithm. Otherwise, if there is such an element y, we add it to X , and then repeat the algorithm. Every time we add an element to X , the size of \|X + A\| increases by at least \|A\|/2, by construction, and at the first stage it increases by \|A\|. However, X + A must always lie within the set B + A. Thus this algorithm terminates after at most 2\|A+B\| − 1 steps. \|A\| Now let y be any element of B. By construction, we have \|(y + A) ∩ (X + A)\| > \|A\|/2, and hence y has at least \|A\|/2 representations of the form x + a − a for some (x, a, a ) ∈ X × A × A , as desired. Finally, if y and y are two elements of B, then we have \|{a ∈ A : y + a ∈ X + A}\| = \|(y + A) ∩ (X + A)\| > \|A\|/2 and similarly we have \|{a ∈ A : y + a ∈ X + A}\| > \|A\|/2. Thus by the pigeonhole principle there exists a ∈ A such that y + a ∈ X + A and y + a ∈ X + A, thus y − y = (y + a) − (y + a) ∈ X + A − (X + A) = A − A + X − X . Since y, y ∈ B is arbitrary, we have B − B ⊆ A − A + X − X as claimed. In Section 5.4 we develop yet another covering lemma (Lemma 5.31), in which the covering set X is not arbitrary, but is in fact a cube. We now give an application of the Green–Ruzsa covering lemma, namely a variant of (2.6) which controls quadruple sums rather than double sums. 2 Sum set estimates 72 Proposition 2.18 Let A, B be additive sets in an ambient group Z . Then \|2B − 2B\| ≤ 16 \|A + B\|4 \|A − A\| . \|A\|4 Proof Applying the Green–Ruzsa covering lemma, we may find a set X of cardinality \|X \| ≤ 2 \|A+B\| such that A − A + X covers B with multiplicity at least \|A\| \|A\|/2. Now let z be any element of B − B. By definition, we have z = b1 − b2 for some b1 , b2 ∈ B. By construction of X , we can find at least \|A\|/2 triplets (x, a1 , a2 ) ∈ X × A × A such that b2 = x + a1 − a2 , and thus \|{(x, a1 , a2 ) ∈ X × A × A : z = b1 − a1 + a2 − x}\| ≥ \|A\|/2. Making the change of variables c := b1 + a2 ∈ A + B, we conclude that \|{(x, c, a1 ) ∈ X × (A + B) × A : z = c − a1 − x}\| ≥ \|A\|/2. Similarly, if z is another element of B − B, we have \|{(x , c , a1 ) ∈ X × (A + B) × A : z = c − a1 − x }\| ≥ \|A\|/2, and hence \|{(x, x , c, c ,a1 , a1 ) ∈ X × X × (A + B) × (A + B) × A × A : z = c − a1 − x, z = c − a1 − x }\| ≥ \|A\|2 /4. Now write d := a1 − a1 ∈ A − A, and observe that if z = c − a1 − x and z = c − a1 − x then z − z = c − c − d − x + x . Also, if one fixes z, z , c, c , d, x, x , then a1 and a1 are determined by the equations a1 = c − x − z, a1 = c − x − z . Thus we have \|{(x, x , c, c , d) ∈ X × X × (A + B) × (A + B) × (A − A) : z − z = c − c − d − x + x }\| ≥ \|A\|2 /4. Note that z − z is an arbitrary element of (B − B) − (B − B) = 2B − 2B. Thus we have shown that an arbitrary element of 2B − 2B has at least \|A\|2 /4 representations of the form c − c − d − x + x where (x, x , c, c , d) ∈ X × X × (A + B) × (A + B) × (A − A). The claim then follows since \|X \| ≤ 2 \|A+B\| . \|A\| We can eliminate the factor of 16 by the following elegant “tensor power trick” of Ruzsa : Corollary 2.19 Let A, B be additive sets in an ambient group Z . Then \|2B − 2B\| ≤ \|A + B\|4 \|A − A\| . \|A\|4 2.4 Covering lemmas 73 Proof Fix A, B, and let M be a large integer parameter. We consider the M-fold Cartesian product A⊕M := A × · · · × A, which is a subset of the additive group Z ⊕M := Z ⊕ · · · ⊕ Z ; similarly consider B ⊕M . Then one easily verifies 2B ⊕M − 2B ⊕M = (2B − 2B)⊕M ; A⊕M + B ⊕M = (A + B)⊕M ; A⊕M − A⊕M = (A − A)⊕M . Thus by applying Lemma 2.18 with A, B replaced by A⊕M , B ⊕M we obtain \|2B − 2B\| M ≤ 16 \|A + B\|4M \|A − A\| M . \|A\|4M Taking Mth roots of both sides and letting M → ∞, we obtain the result. Specializing Corollary 2.19 to the case B := −A, we obtain Corollary 2.20 Let A be an additive set. Then \|2A − 2A\| ≤ \|A − A\|5 \|A\|4 or, in other words, d(A − A, A − A) ≤ 4d(A, A). Remark 2.21 One can improve these estimates slightly by using the machinery of Pl¨unnecke inequalities; see Corollary 6.28. Combining Corollary 2.20 with the Ruzsa covering lemma (Lemma 2.14 with B = 2A − A) we obtain Corollary 2.22 For any additive set A, 2A − A can be covered by δ[A]5 translates of A − A. This then shows that 3A − A is covered by δ[A]5 translates of 2A − A, and hence by δ[A]10 translates of A − A. Continuing in this fashion, an easy induction then shows m A − n A can be covered by δ[A]5(m+n−2) translates of A − A (2.16) for all m, n ≥ 1. In particular we have \|m A − n A\| ≤ δ[A]5(m+n−1) \|A\| for all m, n ≥ 1. (2.17) From this (and the trivial estimates \|k A\| ≥ \|A\| for any k ≥ 1) we obtain Corollary 2.23 (Symmetric sum set estimates, preliminary version) Let A be an additive set. Then we have the estimates d(n 1 A − n 2 A, n 3 A − n 4 A) ≤ 5(n 1 + n 2 + n 3 + n 4 )d(A, A) 2 Sum set estimates 74 for any non-negative integers n 1 , n 2 , n 3 , n 4 . (The constant 5 is not best possible; we will improve it later.) Thus if A has small difference constant, then in fact all iterated sum sets of A are close to each other in the Ruzsa metric. Another consequence of the corollary is that σ [n 1 A − n 2 A] ≤ σ [A]10(n 1 +n 2 ) for all non-negative integers n 1 , n 2 . The factor of 10 is not best possible; we shall obtain improvements to this constant later when we develop the machinery of Pl¨unnecke inequalities in Section 6.5. However, the linear growth in n 1 and n 2 is necessary; see Exercise 2.4.9. By combining the above corollary with the Ruzsa triangle inequality one can obtain similar estimates for pairs of sets: Corollary 2.24 (Asymmetric sum set estimates, preliminary version) Let A, B be additive sets with common ambient group Z . Then we have the estimates d(n 1 A − n 2 A + n 3 B − n 4 B,n 5 A − n 6 A + n 7 B − n 8 B) = O((n 1 + · · · + n 8 )d(A, B)) for any n 1 , . . . , n 8 ∈ N. The proof is left as an exercise. We can use the above machinery to place additive sets with small difference or doubling constant inside a more structured set, namely an “approximate group”. Definition 2.25 (Approximate groups) Let K ≥ 1. An additive set H is said to be a K -approximate group if it is symmetric (so H = −H ), contains the origin, and H + H can be covered by at most K translates of H . Observe that a 1-approximate group is necessarily a finite group, and conversely every finite group is a 1-approximate group. We can summarize many of the preceding results by giving the following partial generalization of Proposition 2.7. Proposition 2.26 Let A be an additive set and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the jth property holds for some absolute constant C j , then the kth property will also hold for some absolute constant Ck depending on C j : (i) (ii) (iii) (iv) σ [A] ≤ K C1 (i.e. \|A + A\| ≤ K C1 \|A\|); δ[A] ≤ K C2 (equivalently, d(A, A) ≤ C2 log K or \|A − A\| ≤ K C2 \|A\|); d(A, B) ≤ C3 log K for at least one additive set B; \|n A − m A\| ≤ K C4 (n+m) \|A\| for all non-negative integers n, m; 2.4 Covering lemmas 75 (v) there exists a K C5 -approximate group H such that A ⊆ x + H for all x ∈ A, and furthermore \|A\| ≥ K −C5 \|H \|. Proof The equivalence of the first three properties follows from the Ruzsa triangle inequality and (2.11). The equivalence of the fourth property with (say) the second follows from Corollary 2.24. To see that the fifth property implies (say) the first, observe that if the former holds, then \|A + A\| ≤ \|H + H \| ≤ K C5 \|H \| ≤ K 2C5 \|A\|. To deduce the fifth from the fourth, take H = A − A and apply the Ruzsa covering lemma. Thus, in a qualitative sense, we have reduced the study of additive sets with small difference or doubling constant to the study of approximate groups, or precisely to the study of dense subsets of translates of approximate groups. This is a fairly satisfactory state of affairs, except for the fact that we do not have a good characterization of which sets are approximate groups. The well known structure theorem for finite groups (see Corollary 3.8 below) asserts that every finite group is the product of finite cyclic groups; we shall eventually be able to obtain a somewhat similar characterization of approximate groups, showing that they are efficiently contained in a generalized arithmetic progression. For some other properties of approximate groups, see the exercises below. There is an asymmetric counterpart to Proposition 2.26, whose proof we leave as an exercise. Proposition 2.27 Let A, B be additive sets in an ambient group Z , and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the jth property holds for some absolute constant C j , then the kth property will also hold for some absolute constant Ck depending on C j : d(A, B) ≤ C1 log K ; d(A, −B) ≤ C2 log K ; \|A + B\| ≤ K C3 min(\|A\|, \|B\|); \|A − B\| ≤ K C4 min(\|A\|, \|B\|); \|n 1 A − n 2 A + n 3 B − n 4 B\| ≤ K C5 (n 1 +n 2 +n 3 +n 4 ) \|A\| for all non-negative integers n 1 , n 2 , n 3 , n 4 ; (vi) σ [A], σ [B] ≤ K C6 , and there exists x ∈ Z such that \|A ∩ (B + x)\| ≥ K −C6 \|A\|1/2 \|B\|1/2 ; (vii) σ [A], σ [B] ≤ K C7 , and E(A, B) ≥ K −C7 \|A\|3/2 \|B\|3/2 ; (viii) there exists a K C8 -approximate group H such that A ⊆ H + a and B ⊆ H + b for all a ∈ A, b ∈ B, and furthermore \|A\|, \|B\| ≥ K −C8 \|H \|. (i) (ii) (iii) (iv) (v) Observe that Exercise 2.3.7 is essentially the K = 1 case of this Proposition. 2 Sum set estimates 76 Proposition 2.27 gives a satisfactory characterization of pairs of sets with small Ruzsa distance, in terms of approximate groups, provided that one is ready to lose some absolute constants in the exponents. Note however that it is restricted to treating those sets A, B which are comparable in magnitude up to powers of K (cf. Exercise 2.3.6). A partial analogue of this proposition exists in the case when A and B are very different in magnitude, but the theory here is not as satisfactory; see Section 2.6. Exercises 2.4.1 2.4.2 Let Z be a finite additive group, and let A be a random subset of Z such that the events a ∈ A are independent with probability 3/4 for all a ∈ Z . Show that with probability 1 − o\|Z \|→∞ (1), \|A\| > \|Z \|/2 (so in particular A + A = A − A = Z , by Exercise 2.1.6), but that it is not possible to 1 cover Z using fewer than 10 log \|Z \| translates of A. (Hint: if X is an 1 additive set with \|X \| ≤ 10 log \|Z \|, use Lemma 2.14 to find an additive set Y with \|Y \| = (\|Z \|/ log2 \|Z \|) such that the translates y − X are disjoint for all y ∈ Y . Compute the probability that A is disjoint from at least one of the sets y − X , and conclude an upper bound for the probability that A + X = Z . Now take the union bound over all choices of X .) This shows that we cannot replace A − A by A in Lemma 2.14 without admitting some sort of logarithmic loss. Let A be an additive set in a group Z , and let φ : Z → Z be a group homomorphism. Establish the inequalities \|A\| ≤ \|φ(A)\| sup \|A ∩ φ −1 (x)\| ≤ \|2A\|. x∈Z 2.4.3 2.4.4 2.4.5 (Hint: use the Ruzsa covering lemma to cover A by translates of a subset of φ −1 (0).) In particular equality is attained in both inequalities when A is the coset of a group. Prove Corollary 2.24. What value of the implicit constant in the O() notation do you get? Let A be an additive set such that \|2A − 2A\| < 2\|A\|. Conclude that A − A is a group. (Hint: use Lemma 2.14.) From this and Corollary 2.19 we see that if \|A − A\| < 21/5 \|A\|, then A − A is a group. The constant 21/5 can be improved to 32 ; see Exercise 2.6.5 below. Let G be group for some integer K ≥ 1. Show that a K -approximate \|nG\| ≤ K +n−1 \|G\| for all integers n ≥ 1. Conclude in particular the n bounds \|nG\| ≤ min(K n , n K −1 )\|G\| for all n ≥ 1; 2.4 Covering lemmas 2.4.6 thus the numbers \|nG\| grow exponentially in n for n ≤ K but settle down to become polynomial growth for n > K . In fact for any additive set, \|n A\| is a polynomial in n for sufficiently large n; see for a proof of this fact and some further discussion. Let A be an additive set with doubling constant σ [A] = K for some K ≥ 1. Show that \|n A\| ≤ min(K Cn , n K 2.4.7 77 C −1 )\|A\| for all n ≥ 1 and some absolute constant C > 0. (Note that if K is very close to 1, then one can use Exercise 2.4.4 to obtain a much stronger bound.) Let G be a K -approximate group in an ambient group Z , and let H be a K approximate group in Z . Show that G + H is a K K -approximate group. Show that 2G ∩ 2H is a (K K )3 -approximate group. (Hint: first show that (2G ∩ 2H ) − (2G ∩ 2H ) ⊂ (G + X ) ∩ (H + Y ) for some X, Y of cardinality at most K 3 and (K )3 respectively, and then show that each set of the form (G + x) ∩ (H + y) is contained in a translate of 2G ∩ 2H .) Modify Exercise 2.2.9 to show that this type of statement fails quite badly if the set 2G ∩ 2H is replaced by G ∩ H . Also, establish the cardinality bounds \|G\|\|H \| \|G\|\|H \| 1 ≤ \|2G ∩ 2H \| ≤ . 3 \|G + H \| (K K ) \|G + H \| (Hint: use (2.8) for the lower bound, and the Ruzsa triangle inequality for the upper bound.) Conclude the estimates d(G, H ) ≤ d(G, G + H ) + d(G + H, H ) ≤ d(G, H ) + log K K and d(G, H ) ≤ d(G, 2G ∩ 2H ) + d(2G ∩ 2H, H ) ≤ d(G, H ) + 3 log K K , 2.4.8 and compare this with Exercise 2.3.11. For each j = 1, 2, 3, let G j be a K j -approximate group in an ambient group Z . Using the Ruzsa triangle inequality, show that \|G 1 + G 2 + G 3 \| ≤ K 2 \|G 1 + G 2 \|\|G 2 + G 3 \| . \|G 2 \| Conclude that d(G 1 + G 2 , G 1 + G 2 + G 3 ) ≤ d(G 2 , G 2 + G 3 ) + log K 1 K 2 . 78 2 Sum set estimates Similarly for permutations. Conclude from this and the preceding exercise that d(G 1 , G 2 ) ≤ d(G 1 + G 3 , G 2 + G 3 ) + 2 log K 1 K 2 K 3 2.4.9 2.4.10 and compare this with Exercise 2.3.11. (A corresponding statement exists for intersections but is somewhat tricky to establish.) For any integers K , n 1 , n 2 ≥ 1, give an example of an additive set A with σ [A] = K and σ [n 1 A − n 2 A] = n 1 ,n 2 (K n 1 +n 2 ). Let A, B be additive sets in a common ambient group Z . Show that σ [A + B] ≤ (σ [A]σ [B])C where C ≥ 1 is an absolute constant. (Hint: use Proposition 2.26 to place A and B inside translates of approximate groups. To obtain lower bounds on \|A + B\|, use the inequality \|A + B\| ≥ 2.4.11 2.4.12 2.4.13 2.4.14 \|A\|\|B\| \|(A − A) ∩ (B − B)\| from (2.8).) Prove Proposition 2.4.11. (Hint: to construct the approximate group H , one possible choice is H = A − A + B − B.) Try to improve upon the constant 5 in (2.17), by using the Ruzsa triangle inequality instead of the Ruzsa covering lemma. This exercise demonstrates that the triangle inequality is slightly sharper than the covering lemma when one wants cardinality bounds, but the covering lemmas of course give much more information than just cardinality. Let A, B be additive sets in an ambient group Z , and let G be the group generated by A. Show that there exists an additive set B ∈ B such \| that B is contained in a coset of G, and such that \|A + B \| ≤ \|B \|A + B\|. \|B\| Let A, B, A , B be additive sets with common ambient group Z . Establish the inequality d(A + A , B + B ) = O(d(A, B) + d(A , B )). (Hint: argue as in Exercise 2.4.10.) Conclude that if φ : Z → Z is a group homomorphism, then d(φ(A), φ(B)) = O(d(A, B)). Thus group homomorphisms are “Lipschitz” with respect to the Ruzsa distance. 2.5 The Balog–Szemer´edi–Gowers theorem In the previous sections we have only considered complete sum sets A + B and complete difference sets A − B. In many applications one only controls a partial collection of sums and differences. Fortunately, there is a very useful tool, the Balog–Szemer´edi–Gowers theorem, which allows one to pass from control of partial sum and difference sets to control of complete sum and difference sets (after refining the sets slightly). We begin with some notation. 2.5 The Balog–Szemer´edi–Gowers theorem 79 Definition 2.28 (Partial sum sets) If A, B are additive sets with common ambient group Z , and G is a subset of A × B, we define the partial sum set G A + B := {a + b : (a, b) ∈ G} and the partial difference set G A − B := {a − b : (a, b) ∈ G}. One may like to think of G as a bipartite graph connecting A and B. Note that when G = A × B is complete, then the notion of partial sum set and partial difference set collapse to just the complete sum set and difference set. Partial sum sets and partial difference sets are not as nice to work with algebraically as complete sum sets. In particular, the above machinery of sum set estimates do not directly yield any conclusion if one only assumes that the cardiG nality \|A + B\| of a partial sum set is small. Note that even when G is very large, G it is possible for \|A + B\| to be small while \|A + B\| is large; see exercises. Fortunately, the Balog–Szemer´edi–Gowers theorem, which we will present shortly, does allow us to conclude information on complete sum sets from information on partial sum sets, if we are willing to refine A and B by a small factor (i.e. replace A and B by subsets A and B which are only slightly smaller than A and B). The first result in this direction was by Balog and Szemer´edi , using the regularity lemma. A different, more effective proof, was found by Gowers (with a slight refinement by Bourgain ), in particular with dependence of constants that are only polynomial in nature. Here we present a modern formulation of the theorem, following . Theorem 2.29 (Balog–Szemer´edi–Gowers theorem) Let A, B be additive sets in an ambient group Z , and let G ⊆ A × B be such that G \|G\| ≥ \|A\|\|B\|/K and \|A + B\| ≤ K \|A\|1/2 \|B\|1/2 for some K ≥ 1 and K > 0. Then there exists subsets A ⊆ A, B ⊆ B such that \|A\| \|A \| ≥ √ 4 2K \|B\| \|B \| ≥ 4K \|A + B \| ≤ 212 K 4 (K )3 \|A\|1/2 \|B\|1/2 . In particular we have d(A , −B ) ≤ 5 log K + 3 log K + O(1). (2.18) (2.19) (2.20) 2 Sum set estimates 80 The proof of this theorem is graph-theoretical. It is elementary, but a little lengthy and so we postpone it to Section 6.4. One can of course combine this theorem with Corollary 2.24 and Proposition 2.26 to gain more information on the iterated sum and difference sets of A and B . It is likely that the factor of 212 K 4 (K )3 in (2.20) can be improved. However, the bounds (2.18), (2.19) cannot be significantly improved; see exercises. To apply the Balog–Szemer´edi–Gowers theorem, it is convenient to introduce the following lemma connecting large additive energy to small partial sum sets or small partial difference sets. Lemma 2.30 Let A, B be additive sets in an ambient group Z , and let G be a non-empty subset of A × B. Then E(A, B) ≥ Conversely, if E(A, B) ≥ \|A\| G ⊆ A × B such that 3/2 \|G\| ≥ \|A\|\|B\|/2K ; \|G\|2 G \|G\|2 , G \|A + B\| \|A − B\| . \|B\|3/2 /K for some K ≥ 1, then there exists G \|A + B\| ≤ 2K \|A\|1/2 \|B\|1/2 . and similarly there exists H ⊆ A × B such that \|H \| ≥ \|A\|\|B\|/2K ; Proof Observe that H \|A − B\| ≤ 2K \|A\|1/2 \|B\|1/2 . \|{(a, b) ∈ G : a + b = x}\| = \|G\| G x∈A+B and hence by Cauchy–Schwarz \|{(a, b) ∈ G : a + b = x}\|2 ≥ \|G\|2 G x∈A+B G \|A + B\| . But the left-hand side is equal to \|{(a, a , b, b ) ∈ A × A × B × B : a + b = a + b ; (a, b), (a , b ) ∈ G}\| G which was less than E(A, B). This proves that E(A, B) ≥ \|G\|2 /\|A + B\|; using the G symmetry E(A, B) = E(A, −B) we thus also obtain E(A, B) ≥ \|G\|2 /\|A − B\|. Now assume E(A, B) ≥ \|A\|3/2 \|B\|3/2 /K . Then by Lemma 2.9 we have x∈A+B \|A ∩ (x − B)\|2 ≥ \|A\|3/2 \|B\|3/2 . K 2.5 The Balog–Szemer´edi–Gowers theorem 81 If we set S := {x ∈ A + B : \|A ∩ (x − B)\| ≥ \|A\|1/2 \|B\|1/2 /2K }, we then have (by Lemma 2.9 again) \|A\|3/2 \|B\|3/2 \|A\|\|B\|\|A\|1/2 \|B\|1/2 \|A\|3/2 \|B\|3/2 \|A ∩ (x − B)\|2 ≥ − = . K 2K 2K x∈S Now observe from Lemma 2.9 again that \|S\|\|A\|1/2 \|B\|1/2 \|A ∩ (x − B)\| ≤ \|A\|\|B\| ≤ 2K x∈S and hence \|S\| ≤ 2K \|A\|1/2 \|B\|1/2 . G Now let G := {(a, b) ∈ A × B : a + b ∈ S}, then clearly A + B ⊆ S and hence G \|A + B\| ≤ 2K \|A\|1/2 \|B\|1/2 . Furthermore we have \|G\| = \|{(a, b) ∈ A × B : a + b = x}\| x∈S = \|A ∩ (x − B)\| x∈S ≥ \|A ∩ (x − B)\|2 \|A\|1/2 \|x − B\|1/2 x∈S ≥ \|A\|3/2 \|B\|3/2 /2K \|A\|1/2 \|B\|1/2 = \|A\|\|B\|/2K . This gives the desired set G. The construction of H follows by using the symmetry E(A, B) = E(A, −B). Combining this Lemma with the Balog–Szemer´edi–Gowers theorem, we can obtain a characterization of pairs of sets with large additive energy. Theorem 2.31 (Balog–Szemer´edi–Gowers theorem, alternative version) Let A, B be additive sets in an ambient group Z , and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the jth property holds for some absolute constant C j , then the kth property will also hold for some absolute constant Ck depending on C j : (i) E(A, B) ≥ K −C1 \|A\|3/2 \|B\|3/2 ; (ii) there exists G ⊂ A × B such that \|G\| ≥ K −C2 \|A\|\|B\| and G \|A + B\| ≤ K C2 \|A\|1/2 \|B\|1/2 ; 2 Sum set estimates 82 (iii) there exists G ⊂ A × B such that \|G\| ≥ K −C3 \|A\|\|B\| and G \|A − B\| ≤ K C3 \|A\|1/2 \|B\|1/2 ; (iv) there exists subsets A ⊆ A, B ⊆ B with \|A \| ≥ K −C4 \|A\|, \|B \| ≥ K −C4 \|B\|, and d(A , B ) ≤ C4 log K ; (v) there exists a K C5 -approximate group H and x, y ∈ Z such that \|A ∩ (H + x)\|, \|B ∩ (H + y)\| ≥ K −C5 \|H \| and \|A\|, \|B\| ≤ K C5 \|H \|. We leave the proof of this theorem to the exercises. Theorem 2.31 should be compared with Exercise 2.3.22, which is the K = 1 case of this Theorem. As with Proposition 2.27, this Theorem is restricted to sets A, B which are close in cardinality (see exercises). We shall address the question of sets A, B of widely differing cardinalities in the next section. Exercises 2.5.1 2.5.2 Let A, B be additive sets with common ambient group Z such that E(A, B) ≥ K −1 \|A\|3/2 \|B\|3/2 . Show that K −2 \|A\| ≤ \|B\| ≤ K 2 \|A\|, and show by means of an example that these bounds cannot be improved. Give an example of an additive set A ⊂ Z of cardinality N , and a set G 2.5.3 G ⊂ A × A of cardinality N 2 /4, such that \|A + A\| ≤ N but \|A + A\| ≥ N 2 /8. (Hint: concatenate a Sidon set with an arithmetic progression.) Let N K 1 be large integers, with N a multiple of K . Give an example of sets A, B ⊂ Z of cardinality \|A\| = \|B\| = N and a subset G ⊂ G 2.5.4 A × B of cardinality \|G\| = \|A\|\|B\|/K with the property that \|A + B\| ≤ 2N , but such that \|A + B \| ≥ N 2 /K 2 whenever A ⊂ A and B ⊂ B is such that \|A \| ≥ 2\|A\|/K . (Hint: take B to be a long progression, and take A to be a short progression concatenated with some generic integers.) This shows that the conditions (2.18), (2.19) in Theorem 2.29 cannot be significantly improved. Let A, B, C be additive sets in an ambient group Z , let 0 < ε < 1/4, and let G ⊂ A × B, H ⊂ B × C be such that \|G\| ≥ (1 − ε)\|A\|\|B\| and \|H \| ≥ (1 − ε)\|B\|\|C\|. Show that there exists subsets A ⊆ A and C ⊆ C with \|A \| ≥ (1 − ε 1/2 )\|A\| and \|C \| ≥ (1 − ε 1/2 )\|C\| such that \|A − C \| ≤ G H \|A − B\|\|B − C\|/(1 − 2ε 1/2 )\|B\|. (Hint: show that at most ε 1/2 \|B\| elements of B have a G-degree of less than (1 − ε1/2 )\|A\|, and similarly at most ε1/2 \|B\| elements have a H -degree of less than (1 − ε 1/2 )\|C\|.) This result is can be used as a substitute for the Balog–Szemer´edi–Gowers theorem in the case when the graph G is extremely dense; it has the advantage that it does not require A, B, C to be comparable in size and 2.6 Symmetry sets and imbalanced partial sum sets 2.5.5 2.5.6 83 it does not lose any constants in the limit ε → 0; indeed it collapses to Ruzsa’s triangle inequality in that limit. Prove Theorem 2.31. (Hint: for K large, e.g. K ≥ 1.1, one can use the Balog–Szemer´edi–Gowers theorem and Proposition 2.27. For K small, e.g. 1 ≤ K < 1.1, one can use Exercise 2.5.4 as a substitute for the Balog– Szemer´edi–Gowers theorem.) Let A, B be additive sets with common ambient group such that G \|A\| = \|B\| = N and \|A + A\| ≤ K N . Suppose also that \|A + B\| ≤ K N , where G ⊂ A × B is a bipartite graph such that every element of B is connected to at least K −1 N elements of A. Show that \|A + B\| ≤ K O(1) N and \|B + B\| ≤ K O(1) N . (Hint: write the elements of A + B in the form G x − y + z where x ∈ A + A, y ∈ A + A, and z ∈ A + B.) 2.5.7 G Let A be an additive set such that \|A + A\| ≤ K \|A\|, where G ⊂ A × A is such that every element of A is connected via G to at least K −1 \|A\| elements of A. Show that one can partition A into O(K O(1) ) subsets A1 , . . . , Am such that \|Ai + Ai \| = O(K O(1) \|A\|) for each 1 ≤ i ≤ m. (Hint: use the Balog–Szemer´edi–Gowers theorem and an iteration argument to obtain most of the subsets, and then Exercise 2.5.6 to deal with the remainder.) 2.6 Symmetry sets and imbalanced partial sum sets The Balog–Szemer´edi–Gowers theorem is a very powerful tool when studying two additive sets A, B with additive energy E(A, B) close to \|A\|3/2 \|B\|3/2 ; however from (2.7) we see that this situation only occurs when \|A\| and \|B\| are comparable in size. This leaves open the question of what happens in the case \|A\| \|B\| (say) and E(A, B) is close to the upper bound of \|A\|\|B\|2 given by (2.7). A special sub-case of this (thanks to (2.8)) is the case when \|A + B\| or \|A − B\| is comparable to \|A\|. Note that Proposition 2.2 already gives an answer to this question in the extreme case when \|A + B\| = \|A\| or \|A − B\| = \|A\| (or equivalently if E(A, B) = \|A\|\|B\|2 ; see Exercise 2.3.22). However, an example of Ruzsa shows that things become bad when \|A\| and \|B\| are very widely separated; see the exercises. If however we are prepared to endure logarithmic-type losses in the ratio \|A\|/\|B\| (or more precisely losses of the form (\|A\|/\|B\|)ε where ε can be chosen to be small), then one can recover a reasonable theory. In analogy with Proposition 2.2, one expects that if \|A + B\| is comparable to \|A\|, or if E(A, B) is close to \|A\|\|B\|2 , then there should be an approximate group H such that A is approximately the 84 2 Sum set estimates union of translates of H , and B is approximately contained in a single translate of H . To achieve this will be the main objective of this section. In the extreme case when \|A + B\| = \|A\| or E(A, B) = \|A\|\|B\|2 , the approximate group H was in fact an exact group and in the proof of Proposition 2.2 it was constructed as the symmetry group Sym1 (A) of the larger additive set A. In the general case this symmetry group is likely to be trivial. However, a more general notion is still useful. Definition 2.32 (Symmetry sets) Let (A, Z ) be an additive set. For any nonnegative real number α ≥ 0, define the symmetry set Symα (A) ⊆ Z at threshold α to be the set Symα (A) := {h ∈ Z : \|A ∩ (A + h)\| ≥ α\|A\|}. Note that Sym1 (A) = {h ∈ Z : A + h = A} is the same symmetry group applied in the proof of Proposition 2.2. The other symmetry sets are not groups in general, but nevertheless they are still symmetric (so −Symα (A) = Symα (A)) and contain the origin, and they obey the nesting property Symα (A) ⊆ Symβ (A) for α ≥ β. It is also clear that Symα (A) ⊆ A − A for all 0 < α ≤ 1. Note that as Symα (A) is empty for α > 1 and equal to all of Z for α ≤ 0, we shall mostly restrict ourselves to the non-trivial region where 0 < α ≤ 1. We now relate the size of these symmetry sets to the additive energy. From Lemma 2.9 we have E(A, A) = \|A ∩ (A + h)\|2 h∈A−A and hence for any 0 < α ≤ 1 and the crude bounds \|A ∩ (A + h)\| ≤ \|A\| when h ∈ Symα (A) and \|A ∩ (A + h)\| ≤ α\|A\| when h ∈ Symα (A), we have α 2 \|A\|2 \|Symα (A)\| ≤ E(A, A) ≤ α 2 \|A\|2 \|A − A\| + \|A\|2 \|Symα (A)\|, which indicates that Symα (A) should be large whenever the energy is large. In particular, from (2.7) we have \|Symα (A)\| ≤ \|A\|/α 2 . (2.21) Now let A, B be additive sets in an additive group Z . From Lemma 2.9 again, we have E(A, B) = \|A ∩ (A + b − b )\| b,b ∈B and hence for any 0 < α ≤ 1 we have E(A, B) ≤ \|B\|2 α\|A\| + \|A\|\|{(b, b ) ∈ B : b − b ∈ Symα (A)}\|. 2.6 Symmetry sets and imbalanced partial sum sets 85 In particular, if E(A, B) ≥ 2α\|A\|\|B\|2 , then we conclude that there is a set G ⊂ B × B of cardinality \|G\| ≥ α\|B\|2 such that G B − B ⊆ Symα (A). (2.22) At first glance it seems that one may now be able to apply the symmetric Balog– Szemer´edi–Gowers theorem. However, the fact that A is much larger than B means G that B − B may be much larger than B (compare (2.22) to (2.21)). To get around this difficulty we need to iterate this construction, and exploit the fact that Symα (A) behaves like a group. This is already clear when α = 1, when Sym1 (A) is indeed a genuine group; the following lemma shows that this behavior persists in an approximate sense for α less than 1. Lemma 2.33 Let A be an additive set. Then we have Sym1−ε (A) + Sym1−ε (A) ⊆ Sym1−ε−ε (A) (2.23) whenever ε, ε > 0. Furthermore, if 0 < α ≤ 1 and S ⊆ Symα (A) is a non-empty set, then there exists a set G ⊆ \|S\|2 with \|G\| ≥ α 2 \|S\|2 /2 (2.24) such that G S − S ⊆ Symα2 /2 (A). (2.25) Proof To verify the first claim, observe that if x ∈ Sym1−ε (A) and y ∈ Sym1−ε (A) then \|(A + x)\A\| = \|A\| − \|A ∩ (A + x)\| ≤ ε\|A\| and \|(A + x)(A + x + y)\| = \|A\| − \|A ∩ (A + y)\| ≤ ε \|A\|, and hence \|A ∩ (A + x + y)\| ≥ \|(A + x) ∩ A ∩ (A + x + y)\| ≥ (1 − ε − ε )\|A\| which proves (2.23). Now we prove the second claim. By definition of S, we see that for each x ∈ S there exist at least α\|A\| elements a ∈ A such that a + x ∈ A. Summing this over all x we see that \|{x ∈ S : a + x ∈ A}\| ≥ α\|A\|\|S\|. a∈A 2 Sum set estimates 86 Applying Cauchy–Schwarz we conclude that \|{a ∈ A : a + x, a + y ∈ A}\| = \|{x ∈ S : a + x ∈ A}\|2 ≥ α 2 \|A\|\|S\|2 . x,y∈S×S a∈A If we set G ⊆ S × S to be all the pairs (x, y) such that \|{a ∈ A : a + x, a + y ∈ A}\| ≥ α 2 \|A\|/2 then we have \|A\|\|G\| ≥ \|{a ∈ A : a + x, a + y ∈ A}\| ≥ α 2 \|A\|\|S\|2 − (x,y)∈G α 2 \|A\| 2 \|S\| 2 which gives (2.24). Also, if (x, y) ∈ G then \|A ∩ (A + x − y)\| ≥ α 2 \|A\|/2 by definition of G, which gives (2.25). Before we proceed with the main theorem, we need a technical lemma that G uniformizes the size of the fibers {(a, a ) ∈ G : a − a = x} of A − A. Lemma 2.34 (Dyadic pigeonhole principle) Let A be an additive set, and let G G ⊂ A × A be such that \|G\| ≥ α\|A\|2 and \|A − A\| ≤ L\|A\| for some 0 < α < 1 and L ≥ 1. Then there exists a subset G of G with α 2 \|G \| = \|A\| 1 + log α1 + log L and \|{(a, a ) ∈ G : a − a = x}\| ≥ \|G \| G 2\|A − A\| G for all x ∈ A − A. It is important to note that the dependence on L only enters in a logarithmic manner. Proof Let D be the set of all x such that 2 ˜ : a − a = x}\| ≥ α\|A\| = α \|A\| \|{(a, a ) ∈ G 2L\|A\| 2L ˜ to be the pairs (a, a ) in (thus D is the set of “popular differences”) and set G G ˜ ≤ α \|A\|\|A − A\| ≤ α\|A\|2 /2, and G such that a − a ∈ D. Then we have \|G\G\| 2L ˜ ≥ α\|A\|2 /2. On the other hand, we have the crude upper bound hence \|G\| ˜ : a − a = x}\| ≤ \|{(a, a ) ∈ G \|{a ∈ A : a = x + a }\| ≤ \|A\|. a ∈A 2.6 Symmetry sets and imbalanced partial sum sets 87 α ˜ = Thus if we let M be the least integer such that 2−M < 2L , we can partition G ˜ G 1 ∪ · · · ∪ G M where G m := {(a, a ) ∈ G : a − a ∈ Dm } and ˜ G ˜ : a − a = x}\| ≤ 2−m+1 \|A\|}. Dm := {x ∈ A − A : 2−m \|A\| < \|{(a, a ) ∈ G By the pigeonhole principle, there exists 1 ≤ m ≤ M such that \|G m \| ≥ 1 α \|A\|2 . \|G\| ≥ M C 1 + log α1 + log L By the definition of Dm , we have \|G m \| −m+1 2 \|A\| ≤ \|Dm \| ≤ \|G m \| ; 2−m \|A\| Gm since Dm = A − A, we thus see that \|{(a, a ) ∈ G : a − a = x}\| ≥ 2−m \|A\| ≥ Gm \|G \| G 2\|A − A\| for all x ∈ A − A. The claim then follows by setting G := G m . Now we give the main theorem of this section. Theorem 2.35 (Asymmetric Balog–Szemer´edi–Gowers theorem) Let A, B be additive sets in an additive group Z such that E(A, B) ≥ 2α\|A\|\|B\|2 and \|A\| ≤ L\|B\| for some L ≥ 1 and 0 < α ≤ 1. Let ε > 0. Then there exists a Oε (α −Oε (1) L ε )-approximate group H in Z , an additive set X in Z of cardinality \|X \| = Oε (α −Oε (1) L ε \|A\|/\|H \|) such that \|A ∩ (X + H )\| = ε (α Oε (1) L −ε \|A\|), and an x ∈ Z such that \|B ∩ (x + H )\| = ε (α Oε (1) L ε \|B\|). Observe in the converse direction that if the conclusions of this theorem are true, then E(A, B) = ε (α Oε (1) L −O(ε) \|A\|\|B\|2 ) (Exercise 2.6.3 at the end of this section). Thus this theorem is sharp up to polynomial losses in α and L ε , where ε can be made arbitrary small; the example in Exercise 2.6.1 can be adapted to show that this loss is necessary (Exercise 2.6.2). Proof A direct application of Theorem 2.31 will lose far too many powers of L. The trick is to embed B in a long increasing sequence of sets B0 , B1 , B2 , . . . , with each B j being (roughly speaking) a partial difference set of the previous one, and use the pigeonhole principle to show that at some stage the ratio \|B j+1 \|/\|B j \| is bounded by a small power of L. One can then apply Theorem 2.31 with acceptable losses and conclude the theorem. (This method of proof is inspired by a similar argument in .) 2 Sum set estimates 88 We turn to the details. It will be convenient to use a variant of the Landau O() and () notation which can absorb factors of α and log L (which we think of as being relatively close to 1). If X, Y are non-negative quantities and j is a parameter, ˜ j (Y ) or Y = ˜ j (X ) if one has an estimate of the form let us say that X = O X ≤ C( j)α −C( j) Y logC( j) L for some C( j) > 0 depending only on j. Let J = J (ε) 1 be a large integer to be chosen later. Let 1 > α1 > · · · > α J +1 > 0 be the sequence defined recursively by α1 := α and α j+1 := α 2j /2 for ˜ j (1).We claim that we can find a all 1 ≤ j ≤ J . From induction we see that α j = sequence B0 , B1 , . . . , B J , B J +1 of additive sets in Z with the following properties. r B = B, and for all 1 ≤ j ≤ J + 1 we have 0 B j ⊆ Symα j (A). (2.26) r For all 0 ≤ j ≤ J + 1, we have ˜ α −2 j L\|B\| ≥ \|B j \| = j (\|B\|). (2.27) r For all 0 ≤ j ≤ J , there exists G ⊆ B × B such that j j j ˜ j (\|B j \|2 ) \|G j \| = (2.28) and Gj B j+1 = B j − B j . Furthermore, for all x ∈ B j+1 we have (2.29) ˜j \|{(b, b ) ∈ G j : b − b = x}\| = \|B j \|2 . \|B j+1 \| (2.30) We construct the B j as follows. We set B0 := B. From (2.22) followed by G0 Lemma 2.34 we can construct G 0 ⊆ B0 × B0 and B1 := B0 − B0 obeying (2.26), G0 (2.28), (2.29), (2.30). Since each element in B0 − B0 can be represented as a difference of a pair in G in at most \|B0 \| ways, we have G0 ˜ j (\|B\|), \|B1 \| = \|B0 − B0 \| ≥ \|G 0 \|/\|B0 \| = which is the lower bound in (2.27); the upper bound follows from (2.26) and (2.21). Next, suppose inductively that B j ⊆ Symα j (A) has already been chosen for some 1 ≤ j ≤ J . Applying Lemma 2.33 (with S := B j ) followed by Lemma 2.34, and using the cardinality bounds already obtained in (2.27) and the construction Gj α 2j+1 := α 2j /2 of the α j , we can thus find G j ⊆ B j × B j and B j+1 := B j − B j 2.6 Symmetry sets and imbalanced partial sum sets 89 obeying (2.26), (2.28), (2.29), (2.30). This closes the induction and so we can construct the B j for all 0 ≤ j ≤ J + 1, and similarly obtain the G j for all 1 ≤ j ≤ J. Now for the crucial step (which explains why we iterated the above procedure so many times). From (2.27) and the pigeonhole principle, there exists 1 ≤ j ≤ J such that ˜ J L O(1/J ) \|B j \| ; \|B j+1 \| = O the point is that we have managed to replace L by the substantially smaller quantity L O(1/J ) . If we now apply (2.29), (2.28), and Theorem 2.31, we can thus find a ˜ J (L O(1/J ) )-approximate group H of cardinality O ˜ J L O(1/J ) \|B j \| \|H \| = O (2.31) and an x j ∈ Z such that ˜ J L −C0 /J \|B j \| \|B j ∩ (H + x j )\| = (2.32) for some absolute constant C0 . It remains to relate H to B and to A. We begin with B. From (2.32) and (2.30) (with j replaced by j − 1) we have ˜ J L −C0 /J \|B j−1 \|2 , \|{(b, b ) ∈ G j−1 : b − b ∈ B j ∩ (H + x j )}\| = so in particular ˜ J L −C0 /J \|B j−1 \|2 . \|{(b, b ) ∈ B j−1 × B j−1 : b ∈ H + x j + b }\| = Thus by the pigeonhole principle, there exists a b such that ˜ J L −C0 /J \|B j−1 \| . \|{b ∈ B j−1 : b − b ∈ H + x j + b }\| = Thus if we set x j−1 := x j + b then we have ˜ J L −C0 /J \|B j−1 \| . \|B j−1 ∩ (H + x j−1 )\| = (2.33) We now repeat this argument with j replaced by j − 1 and (2.32) replaced by (2.33). Iterating this at most J times, we eventually locate an x = x0 ∈ Z such that ˜ J L −C0 /J \|B\| , \|B ∩ (H + x)\| = which gives the desired control on B if J is sufficiently large depending on ε. It remains to control A. From (2.32), (2.31) and (2.26) we have ˜ J L −O(1/J ) \|H \| \|{y ∈ H + x j : y ∈ Symα j (A)}\| = and thus by definition of Symα j (A) and α j ˜ J L −O(1/J ) \|H \|\|A\| . \|{(a, y) ∈ A × (H + x j ) : a + y ∈ A}\| = 2 Sum set estimates 90 We rewrite this as ˜ J L −O(1/J ) \|H \|\|A\| . \|A ∩ (H + x)\| = x∈x j +A We can therefore find a subset X 0 of x j + A with ˜ J L −O(1/J ) \|A\| \|X 0 \| = such that (2.34) ˜ J L −O(1/J ) \|H \| for all x ∈ X 0 . \|A ∩ (H + x)\| = Now we use an argument similar to that used to prove Ruzsa’s covering lemma (Lemma 2.14). Let X be a subset of X 0 such that the sets {H + x : x ∈ X } are all disjoint, and which is maximal with respect to set inclusion. Then we have ˜ J L −O(1/J ) \|H \|\|X \| . \|A ∩ (H + X )\| = \|A ∩ (H + x)\| = (2.35) x∈X On the other hand, if y ∈ X 0 , then by maximality of X there exists x ∈ X such that x + H intersects y + H . In other words, X 0 is covered by X + H − H , and ˜ O(1/J ) )-approximate group) hence (since H is a O(L ˜ \|X \|L O(1/J ) \|H \| . \|X 0 \| ≤ \|X \|\|H − H \| = O (2.36) Combining (2.34), (2.35), (2.36) we see that X obeys all the desired properties, if J is chosen sufficiently small depending on ε. The above theorem can also be put in a form resembling Theorem 2.29: Corollary 2.36 Let A, B be additive sets with common ambient group such that E(A, B) ≥ 2α\|A\|\|B\|2 and \|A\| ≤ L\|B\| for some L ≥ 1 and 0 < α ≤ 1. Let ε > 0. Then there exists subsets A ⊆ A and B ⊆ B such that \|A \| = ε α Oε (1) L −ε \|A\| \|B \| = ε α Oε (1) L −ε \|B\| n+m \|A + n B − m B \| = Oε α −Oε (1) L ε \|A\| for all integers n, m ≥ 0. Proof Apply Theorem 2.35 and set A := A ∩ (X + H ) and B := B ∩ (x + H ). Because of (2.8), the above results give some partial results concerning the situation when \|A + B\| ≤ K \|A\| and \|A\| is much larger than \|B\|, but these results will be rather weak. We will give a better result concerning this problem in Section 6.5, once we develop the Pl¨unnecke inequalities. 2.6 Symmetry sets and imbalanced partial sum sets 91 Exercises 2.6.1 Let n be a large integer, and let Z := Z2n . Let A be the additive set A := {(x1 , x2 , . . . , x2n ) ∈ Z2n : x1 + · · · + x2n = n; x1 , . . . , x2n ≥ 0} 2.6.2 2.6.3 2.6.4 and let B := {e1 , . . . , e2n }. Show that \|B\| = 2n, that \|A\| = (27/4)n+o(1) , that \|A + B\| = O(\|A\|), but that \|A − B\| ≥ n\|A\|. (You may find Stirling’s formula (1.52) to be useful.) Modify Exercise 2.6.1 to show that one cannot take ε = 0 in Theorem 2.35. Let A, B be additive sets and let ε > 0, 0 < α < 1, and L ≥ 1 be such that the conclusions of Theorem 2.35 are satisfied. Conclude that E(A, B) = ε (α Oε (1) L −O(ε) \|A\|\|B\|2 ). Let A be an additive set. By modifying the proof of Lemma 2.13, establish the inequality \|A − A + nSymα (A)\| ≤ 2.6.5 2.6.6 2.6.7 δ[A]n+1 \|A\| αn for all integers n ≥ 0 and all 0 < α < 1. Let A be an additive set such that A − A is not a group. Show that there exists h ∈ A − A such that 1 ≤ \|A ∩ (A + h)\| ≤ \|A\|/2. (Hint: argue by contradiction, and analyze Symα (A) for some α slightly greater than 1/2.) Conclude in particular that if \|A − A\| < 32 \|A\|, then A − A is a group. Note that the example A = {0, 1} ⊂ Z shows that the constant 32 cannot be improved; one can also make this example larger, for instance by taking the Cartesian product of {0, 1} with a finite group. For a more refined estimate on A − A, see Theorem 5.5 and Corollary 5.6. Let A, B be additive sets with common ambient group such that \|A + B\| ≤ K \|A\| and \|A\| ≤ L\|B\| for some K , L ≥ 1. Let ε > 0. Show that there exists a Oε (K Oε (1) L ε )-approximate group H such that B is contained in a translate of H , and that A is contained in at most Oε (K Oε (1) L ε \|A\|/\|H \|) translates of H ; compare this with Proposition 2.2. (Hint: Apply Theorem 2.35 and the Ruzsa covering lemma (Lemma 2.14).) Let A be an additive set, and let B be a subset of A such that \|B\| ≥ (1 − ε)\|A\| for some 0 < ε < 1. Prove that Symα/(1−ε) (B) ⊆ Symα (A) ⊆ Sym(α−2ε)/(1−ε) (B) for every α ∈ R. 2 Sum set estimates 92 2.6.8 Let A be an additive set. Refine (2.21) slightly to \|Symα (A)\| ≤ 1 + 2.6.9 \|A\|(\|A\| − 1) for all α > 0. α Let A, B be additive sets in Z, such that B consists entirely of positive numbers. Show that there exists b ∈ B such that \|A ∩ (A + b)\| Then there exists subsets A ⊆ A, B ⊆ B such that \|A\| \|A \| ≥ √ 4 2K \|B\| \|B \| ≥ 4K \|A · B \| ≤ 212 K 4 (K )3 \|A\|1/2 \|B\|1/2 . (2.37) (2.38) (2.39) In particular we have d(A , B −1 ) ≤ 5 log K + 3 log K + O(1). Define the multiplicative energy E(A, B) between two multiplicative sets A, B with common ambient group to be E(A, B) := \|{(a, a , b, b ) ∈ A × A × B × B : ab = a b }\|. (2.40) A significant difficulty here is that E(A, B) obeys far fewer symmetries in the non-commutative case than in the commutative case; indeed, the only symmetry available is that E(A, B) = E(B −1 , A−1 ). However in the case when B = A−1 we have a crucial additional identity E(A, A−1 ) = E(A−1 , A) (see exercises), which can be thought of as a very weak, restricted form of commutativity. The following variant of Lemma 2.30 holds, with basically the same proof. Lemma 2.45 Let A, B be multiplicative sets in an ambient group Z , and let G be a non-empty subset of A × B. Then E(A, B) ≥ \|G\|2 G \|A · B\| . 2.7 Non-commutative analogues 97 Conversely, if E(A, B) ≥ \|A\|3/2 \|B\|3/2 /K for some K ≥ 1, then there exists G ⊆ A × B such that \|G\| ≥ \|A\|\|B\|/2K ; G \|A · B\| ≤ 2K \|A\|1/2 \|B\|1/2 . Finally, notice that by the triangle inequality d(A , A ) ≤ d(A , B −1 ) + d(B −1 , A ) = 2d(A , B −1 ), which means that if d(A , B −1 ) is small, then d(A , A ) is also small. From here, we can use the same arguments for the commutative case to deduce Corollary 2.46 Let A, B be multiplicative sets in an ambient group Z such that E(A, B) ≥ \|A\|3/2 \|B\|3/2 /K for some K > 1. Then there exists a subset A ⊂ A such that \|A \| = (K −O(1) \|A\|) and \|A · (A )−1 \| = O(K O(1) \|A\|) for some absolute constant C. Combining this with the identity E(A, A−1 ) = E(A−1 , A) we obtain the following weak commutativity property between A and A−1 : Corollary 2.47 Let A be a multiplicative set such that \|A · A\| ≤ K \|A\| for some K ≥ 1. Then there exists a subset A ⊂ A such that \|A \| = (K −O(1) \|A\|) and \|A · (A )−1 \| = O(K O(1) \|A\|). It is now not too hard to obtain the following theorem. Theorem 2.48 Let A, B be multiplicative sets in a group G, and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the jth property holds for some absolute constant C j , then the kth property will also hold for some absolute constant Ck depending on C j : (i) E(A, B) ≥ C1−1 K −C1 \|A\|3/2 \|B\|3/2 ; (ii) there exists a subset G ⊂ A · B with \|G\| ≥ C2−1 K −C2 \|A\|\|B\| such that G \|A · B\| ≤ C2 K C2 \|A\|1/2 \|B\|1/2 ; (iii) there exists a C3 K C3 -approximate group H and x, y ∈ G such that \|H \| ≤ C3 K C3 \|A\|1/2 \|B\|1/2 and \|A ∩ (x · H )\|, \|B ∩ (H · y)\| ≥ C3−1 K −C3 \|H \|. We leave the proofs of these statements to the exercises. Despite these characterizations, there is much left to be done in the study of product sets in noncommutative groups. For instance we do not currently have a satisfactory version of Freiman’s theorem in general. However there has been some progress in the case of very small doubling and also in certain special groups such as S L 2 (Z) or free groups; see for instance , . 2 Sum set estimates 98 Exercises 2.7.1 2.7.2 2.7.3 2.7.4 2.7.5 2.7.6 2.7.7 Prove a multiplicative version of Lemma 2.1. Prove a multiplicative version of Lemma 2.6. Prove Proposition 2.38. Let (A, G) be a multiplicative set. Prove that \|A · A\| = \|A\| if and only if A is a normal coset of H , i.e. A = x · H = H · x for some x ∈ N (H ). Let A be a symmetric multiplicative set, so A = A−1 , and let σn [A] denote the n-fold doubling numbers \|An \|/\|A\|. Using the Ruzsa triangle inequality, show that σm+n−2 [A] ≤ σm [A]σn [A] for all m, n ≥ 2. Let A and B be multiplicative sets. Establish the identities E(A, B) = E(B −1 , A−1 ) and E(A, A−1 ) = E(A−1 , A), and the inequal2 \|B\|2 ity E(A, B) ≥ \|A\| . \|A·B\| Let A, B, C be additive sets in an ambient group Z , let 0 < ε < 1/4, and let G ⊂ A × B −1 , H ⊂ B × C −1 be such that \|G\| ≥ (1 − ε)\|A\|\|B\| and \|H \| ≥ (1 − ε)\|B\|\|C\|. By modifying the solution of Exercise 2.5.4, show that there exists subsets A ⊆ A and C ⊆ C with \|A \| ≥ (1 − ε 1/2 )\|A\| G 2.7.8 −1 H −1 · B \|\|B · C \| . and \|C \| ≥ (1 − ε 1/2 )\|C\| such that \|A · (C )−1 \| ≤ \|A(1−2ε 1/2 )\|B\| −1 −1 Let A be a multiplicative set such that \|A · A \| ≤ K \|A\| and \|A · A\| ≤ ˜ ≥ \|A\|/2K K \|A\|. Show that there exists a subset A˜ of A such that \| A\| and n+1 \| A˜ · A˜ −1 · . . . · A˜ (−1) \| ≤ 2n−2 K 2n−3 \|A\| 2.7.9 2.7.10 2.7.11 2.7.12 for all n ≥ 2, where the product consists of n factors alternating between A˜ and A˜ −1 . If A and B are multiplicative sets in a group G, show that there exist sets X 1 , X 2 ⊆ A such that \|X 1 \| ≤ \|A·B\| , \|X 2 \| ≤ \|B·A\| , and A ⊆ X 1 · B · B −1 \|B\| \|B\| and A ⊆ B −1 · B · X 2 , by modifying the proof of Lemma 2.14. Prove Lemma 2.41. Show that the direct analogue of Proposition 2.18 fails in the noncommutative case, even when A = B = A−1 . Let A, B be multiplicative sets in an ambient group G, and let A˜ be the set \|A\|\|B\|2 −1 ˜ A := a ∈ A : \|{(a , b, b ) ∈ A × B × B : a = a b b }\| ≥ . 2\|A · B\| Establish the bounds ˜ ≥ \| A\| \|A\|2 2\|A · B\| 2.8 Elementary sum-product estimates 99 and −1 \|A · B\| \|A \|A · A˜ −1 · A˜ · A−1 \| ≤ 4 \|A\|4 4 · A\| . Compare this against Exercise 2.7.11. Hint: if x := a1 a2−1 a3 a4−1 be a typ2 ical element of A · A˜ −1 · A˜ · A−1 , obtain at least ( \|A\|\|B\| )2 representations 2\|A·B\| of the form x = a1 b2 −1 [(a2 )−1 a3 ]b3 [a4 b2 ]−1 2.7.13 where a1 b2 , a4 b2 ∈ A · B, b2 , b3 ∈ B, and (a2 )−1 a3 ∈ A−1 · A. Prove Theorem 2.48. 2.8 Elementary sum-product estimates We now discuss some results concerning the sum set and product set of a subset A of a commutative ring Z , thus combining both the additive and multiplicative theory of the preceding sections (but keeping the multiplication commutative, for simplicity). The question here is to analyze the extent to which a set A can be approximately closed under addition and multiplication simultaneously. Of course, one way that this can happen is if A is a subring of Z ; it appears that up to trivial changes (such as removing some elements, adding a small number of new elements, or dilating the set), this is essentially the only such example, although we currently only have a satisfactory and complete formalization of this principle when Z is a field (Theorem 2.55). In some ways the theory here is in fact easier than the sum set theory, because one can exploit two rather different structures arising from the smallness of A + A and the smallness of A · A to obtain a conclusion. As in the rest of this chapter, our discussion is for general fields, with a particular emphasis on the finite field Z p . We remark that for the field R much better results are known, see Sections 8.3, 8.5. In this section Z will always denote a commutative ring, and Z ∗ will denote the elements of Z which are not zero-divisors; these form a multiplicative cancellative commutative monoid in Z . The situation is significantly better understood in the case that Z is a field (see in particular Theorem 2.55 below); in such cases we shall emphasize this by writing the field as F instead of Z , and F × instead of F ∗ = F{0} to emphasize that F × is now a multiplicative group. A fundamental concept in the field setting is that of a quotient set, which is the arithmetic equivalent of the concept of a quotient field of a division ring. 2 Sum set estimates 100 Definition 2.49 (Quotient set) Let A be a finite subset of a field F such that \|A\| ≥ 2. Then the quotient set Q[A] of A is defined to be A− A a−b Q[A] := := : a, b, c, d ∈ A; c = d . (A − A)\0 c−d We also set Q[A]× := Q[A]\0 to be the invertible elements in Q[A]. Observe that Q[A] contains both 0 and 1, and is symmetric under both additive and multiplicative inversion, thus Q[A] = −Q[A] and Q[A]× = (Q[A]× )−1 . It is also invariant under translations and dilations of A, thus Q[A] = Q[A + x] = Q[λ · A] for all x ∈ F and λ ∈ F × . Geometrically, Q[A] can be viewed as the set of slopes of lines connecting points in A × A. The relevance of the quotient set to sum-product estimates lies in the trivial but fundamental observation: Lemma 2.50 Let A be a finite subset of a field F such that \|A\| ≥ 2, and let x ∈ F. Then \|A + x · A\| = \|A\|2 if and only if x ∈ Q[A]. Proof We have \|A + x · A\| = \|A\|2 if and only if the map (a, b) → a + xb is injective on A × A, which is true if and only if a + xb = c + xd for all distinct (a, b), (c, d) ∈ A × A, which after some algebra is equivalent to asserting that x ∈ Q[A]. This has an immediate corollary: Corollary 2.51 If A is a subset of a finite field F such that \|A\| > \|F\|1/2 , then Q[A] = F. Note that the condition \|A\| > \|F\|1/2 is absolutely sharp, as can be seen by considering the case when A is a subfield of F of index 2. Lemma 2.50 has another important consequence: it gives a criterion under which Q[A] is a subfield of F. Corollary 2.52 Let A be a finite subfield of a field F such that \|A\| ≥ 2 and \|A + Q[A] · Q[A] · A\|, \|A + (Q[A] + Q[A]) · A\| < \|A\|2 . Then Q[A] is a subfield of F. This corollary may be compared with Exercise 2.6.5. Proof From Lemma 2.50 and the hypotheses we see that Q[A] · Q[A] ⊆ Q[A] and Q[A] + Q[A] ⊆ Q[A]. In particular Q[A]× · Q[A]× = Q[A]× . Since Q[A] is finite and contains 0, 1, we see from Proposition 2.7 that Q[A] is an additive group, and similarly from the multiplicative version of this Proposition we see that Q[A]× is a multiplicative group. The claim follows. 2.8 Elementary sum-product estimates 101 In order to use this corollary, one needs to control rational expressions of A such as A + Q[A] · Q[A] · A. In analogy with sum set estimates such as Corollary 2.23, one might first expect that once \|A + A\| ≤ K \|A\| and \|A · A\| ≤ K \|A\|, then all polynomial or rational expressions of A are controlled in cardinality by C K C \|A\|. This however is not the case, even if one normalizes A to contain 0 and 1. To see this, consider A = G ∪ {x} where G is a subfield of F and x ∈ G. Then one easily verifies \|A + A\|, \|A · A\| < 2\|A\| but \|A · A + A · A\| ≥ (\|A\| − 1)2 , since A · A + A · A contains G + x · G, which has size \|G\|2 by Lemma 2.50. This example is similar to one appearing in the preceding section, and it is resolved in a similar way, namely by passing from A to a subset of A. Lemma 2.53 (Katz–Tao lemma) , Let Z be a commutative ring, and let A ⊆ Z ∗ be a finite non-empty subset such that \|A + A\| ≤ K \|A\| and \|A · A\| ≤ K \|A\| for some K ≥ 1. Then there exists a subset A of A such that \|A \| ≥ \|A\|/2K − 1 and \|A · A − A · A \| = O(K O(1) \|A \|). Note that this lemma works in arbitrary commutative rings, not just in fields. The requirement that none of the elements of A be zero-divisors is not serious in the case of a field, since one can simply remove the origin 0 from A if necessary, but is a non-trivial requirement in other commutative rings. Proof We use an argument from . We may assume that A > 10K (for instance) since the claim is trivial otherwise. Consider the dilates {a · A : a ∈ A} of A. Since a ∈ Z ∗ , a · A has the same cardinality as A. In particular we have 1a·A (x) = \|A\|2 . x∈A·A a∈A Since \|A · A\| ≤ K \|A\|, we may apply Cauchy–Schwarz and conclude 2 1a·A (x) ≥ \|A\|3 /K . x∈A·A We rearrange this as a∈A \|(a · A) ∩ (b · A)\| ≥ \|A\|3 /K . a,b∈A By the pigeonhole principle we can thus find a b ∈ A such that \|(a · A) ∩ (b · A)\| ≥ \|A\|2 /K . a∈A Fix this b. Setting A to be the set of all a ∈ A such that \|(a · A) ∩ (b · A)\| ≥ \|A\|/2K 2 Sum set estimates 102 we conclude that \|(a · A) ∩ (b · A)\| ≥ \|A\|2 /2K a∈A and hence \|A \| ≥ \|A\|/2K . By shrinking A by one if necessary we may assume b ∈ A . Now recall the Ruzsa distance d(A, B) := log \|A\|\|A−B\| 1/2 \|B\|1/2 , and observe that d(a · A, a · B) = d(A, B) whenever a is not a zero-divisor. Then d(A, A) ≤ 2d(A, −A) = 2 log K , and hence d(a · A, a · A) = d(b · A, b · A) = d(A, A) ≤ 2 log K for all a ∈ A . Since (a · A) ∩ (b · A) is a large subset of a · A and b · A, one can compute d(a · A, a · A ∩ b · A), d(b · A, a · A ∩ b · A) = O(1 + log K ) and hence by the Ruzsa triangle inequality d(a · A, b · A) = O(1 + log K ) for all a ∈ A . (2.41) Dilating this, we obtain d(a1 a2 · A, ba2 · A), d(ba2 · A, b2 · A) = O(1 + log K ) for all a1 , a2 ∈ A and hence by the Ruzsa triangle inequality d(a1 a2 · A, b2 · A) = O(1 + log K ) for all a1 , a2 ∈ A . (2.42) To proceed further we need to “invert” elements in A. For any a ∈ A let aˆ := ∗ a ∈A{a} a ∈ Z . By dilating (2.41) (with a replaced by a3 ) by a1 a2 a ∈A{a3 ,b} a for a1 , a2 , a3 ∈ A , we obtain d(a1 a2 bˆ · A, a1 a2 aˆ 3 · A) = O(1 + log K ) for all a1 , a2 , a3 ∈ A . Meanwhile, from dilating (2.42) we have d(a1 a2 bˆ · A, b2 bˆ · A) = O(1 + log K ) for all a1 , a2 , a3 ∈ A . Applying the Ruzsa triangle inequality, we thus have d(a1 a2 aˆ 3 · A, a1 a2 aˆ 3 · A) = O(1 + log K ) for all a1 , a2 , a3 , a1 , a2 , a3 ∈ A and hence \|a1 a2 aˆ 3 · A − a1 a2 aˆ 3 · A\| = O(K O(1) )\|A\|. Therefore we have x,y∈A ·A · Aˆ \|x · A − y · A\| = O(K O(1) )\|A\|\|A · A · Aˆ \|2 , 2.8 Elementary sum-product estimates 103 where Aˆ := {aˆ : a ∈ A }. But since \|A · A\| ≤ K \|A\| and \|A \| ≥ \|A\|/2K − 1, we see from the multiplicative version of sum set estimates (working in the formal multiplicative group generated by the cancellative commutative monoid Z ∗ ) that \|A · A · Aˆ \| = O(K O(1) \|A\|). We thus have \|x · A − y · A\| ≤ O(K O(1) \|A \|3 ). x,y∈A ·A · Aˆ We rewrite the left-hand side as \|{(x, y) : ∃ a, b ∈ A such that z = xa − yb}\|. z∈Z Write ω := a∈A a, and observe that whenever a1 , a2 , a3 , a4 ∈ A , the number ω(a1 a2 − a3 a4 ) has at least \|A \|2 representations of the form xa − yb with x, y ∈ A · A · Aˆ and a, b ∈ A , with (x, y) distinct, thanks to the identity ˆ ˆ − (a3 a4 b)b. ω(a1 a2 − a3 a4 ) = (a1 a2 a)a Thus \|ω · (A · A − A · A )\| = O(K O(1) \|A \|) and the claim follows since ω ∈ Z ∗ . A modification of the above argument also gives the following statement, which can be viewed as a variant of Corollary 2.23 for the sum-product setting; we leave the proof to Exercise 2.8.1. Lemma 2.54 Let Z be a commutative ring, and let A ⊆ Z ∗ be a finite nonempty set such that \|A · A − A · A\| ≤ K \|A\|. Then we have \|Ak − Ak \| ≤ K O(k) \|A\| for all k ≥ 1, where Ak = A · . . . · A is the k-fold product set of A. We can now classify those finite subsets of fields with small additive doubling and multiplicative doubling constant, up to polynomial losses: Theorem 2.55 (Freiman theorem for sum-products) Let A be a finite nonempty subset of a field F, and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the jth property holds for some absolute constant C j , then the kth property will also hold for some absolute constant Ck depending on C j : (i) \|A + A\| ≤ C1 K C1 \|A\| and \|A · A\| ≤ C1 K C1 \|A\|; (ii) either \|A\| ≤ C2 K C2 , or else there exists a subfield G of F, a non-zero element x ∈ F, and a set X in F such that \|G\| ≤ C2 K C2 \|A\|, \|X \| ≤ C2 K C2 , and A ⊆ x · G ∪ X . This is a slight strengthening of a result in , . 2 Sum set estimates 104 Proof We shall only show the forward implication, leaving the easy backward implication to Exercise 2.8.2. By relabeling C1 K C1 as K , we may thus assume that \|A + A\| ≤ K \|A\| and \|A · A\| ≤ K \|A\|. We may assume that \|A\| ≥ C0 K C0 for some large absolute constant C0 , since the claim is trivial otherwise. We may also remove 0 from A without any difficulty, thus we may assume A ⊆ F ∗ . Applying Lemma 2.53 and Lemma 2.54, we may find a subset A of A with \|A \| = (K −O(1) \|A\|) and \|(A )k − (A )k \| = O(K ) O(k) \|A \| for all k ≥ 1. By Corollary 2.23 this implies that \|n(A )k − m(A )k \| ≤ O(K ) Ok,n,m (1) \|A \| for all n, k, m ≥ 1. (2.43) Dilating A with a non-zero factor if necessary, we may assume 1 ∈ A (noting that the hypothesis and conclusion of the theorem are invariant under such dilations). We may now add 0 back to A and A without affecting (2.43). Now we apply Corollary 2.52. Let D := (A − A ){0} and G := Q[A ] = (A − A )/D. Using lowest common denominators, we observe that (A · D · D − (A − A ) · (A − A ) · A ) (4(A )3 − 4(A )3 ) ⊆ . D2 D2 on the other hand, from (2.43) we have A +G · G · A ⊆ \|(4(A )3 − 4(A )3 ) · D 2 \| = O(K O(1) \|A \|), so by the multiplicative version of Corollary 2.12 we see that \|A + G · G · A \| = O(K O(1) \|A \|) < \|A \|2 if C0 is sufficiently large. A similar argument gives \|A + (G + G) · A \| = O(K O(1) \|A \|) < \|A \|2 . Applying Corollary 2.52 we see that G is in fact a field. Now let x be a non-zero element of A , and let y be an element of A . Then (a − y)/x ∈ Q[A ] = G for all a ∈ A , thus A ⊆ x · G + y. Thus x · G + y = A + x · G ⊆ A + A · Q[A ] and hence (x · G + y)2 ⊆ (A + A · Q[A ])2 . But an argument using (2.43) and Corollary 2.12 as before gives \|(A + A · Q[A ])2 \| = O(K O(1) \|A \|) ≤ O(K O(1) \|G\|). Direct computation shows that \|(x · G + y)2 \| ≥ \|G\|2 unless y ∈ x · G. Thus (if C0 is sufficiently large) we can take y ∈ x · G. Because A contains 1, we thus have A ⊆ G. 2.8 Elementary sum-product estimates 105 Since \|A + A \| ≤ K \|A\| = O(K O(1) \|A \|), we may apply Ruzsa’s covering lemma (Lemma 2.14) and cover A by O(K O(1) ) translates of A − A , and hence by O(K O(1) ) translates of G. A similar argument using the multiplicative version of this lemma (and temporarily removing the non-invertible 0 element from A if necessary) covers A by O(K C ) dilates of G. On the other hand, we have \|(G · x) ∩ (G + y)\| ≤ 1 whenever x = 1. Thus we have \|A\G\| = O(K O(1) ), and the claim follows. This theorem implies that at least one of A + A or A · A is large if A does not intersect with a subfield of F: Corollary 2.56 (Sum-product estimate) , Let A be a finite non-empty subset of a field F, and suppose that K ≥ 1 is such that there is no finite subfield G of F of cardinality \|G\| ≤ K \|A\| and no x ∈ F such that \|A(x · G)\| ≤ K . Then we have either \|A\| = O(K O(1) ) or \|A + A\| + \|A · A\| = (K c \|A\|) for some absolute constant c > 0. Remark 2.57 In the particular case when F has no finite subfields we thus obtain \|A + A\| + \|A · A\| = (\|A\|1+ε ) for some absolute constant ε > 0; this result was first obtained (when F = R) by Erd˝os and Szemer´edi . In the setting of the real line it is was in fact conjectured in that one can take ε arbitrarily close to 1 in the above estimate. For the most recent value of ε, see Theorem 8.15. In the particular case of the field F = F p of prime order, which has no subfields other than {1} and F p , one obtains Corollary 2.58 (Sum-product estimate for F p ) , Let A be a non-empty subset of F p . Then \|A + A\| + \|A · A\| = (min(\|A\|, \|F p \|/\|A\|)c \|A\|) for some absolute constant c > 0. If H is any non-empty subset of F p , then we have k H k + k H k , k H k · k H k ⊂ 2 k H k for all k ≥ 2. Thus we have 2 2 \|k 2 H k \| = (min(\|k H k \|, p/\|k H k \|)c \|k H k \|) for some absolute constant c > 0. We can iterate this estimate (starting with k = 2 and squaring repeatedly) to establish Corollary 2.59 Let H be any non-empty subset of F p , and let A, δ > 0. Then there exists an integer k = k(A, δ) ≥ 1 such that \|k H k \| = A,δ (min(\|H \| A , p 1−δ )). 2 Sum set estimates 106 We leave the proof of this corollary as an exercise. By using Lemma 4.10 from Chapter 4 one can in fact set δ = 0 here, though we will not need this fact here. In the special case when H is a multiplicative subgroup of F p , we have H k = H , and hence Corollary 2.59 gives \|k H \| = A,δ (min(\|H \| A , p 1−δ )). Thus multiplicative subgroups have rather rapid additive expansion. It turns out that one can do something similar for approximate groups: Theorem 2.60 Let H be a non-empty subset of F p such that \|H 2 \| ≤ K \|H \|, and let A, δ > 0. Then there exists an integer k = k(A, δ) ≥ 1 such that \|k H \| = A,δ (K −O A,δ (1) min(\|H \| A , p 1−δ )). This result can be deduced from Corollary 2.59 and the following proposition; we leave the precise deduction as an exercise. Proposition 2.61 Let F be an arbitrary field, and let H ⊂ F × be a finite non-empty subset of invertible field elements such that \|H 2 \| ≤ K \|H \| for some K ≥ 1. Let k ≥ 1 and L ≥ 1 be such that k H obeys the following “additive non-expansion” property: we have \|2k H \| ≤ L\|k H \| for any subset H of H 1 of cardinality \|H \| ≥ 2K \|H \|. Then there exists a subset H of H of cardinality 1 \|H \| ≥ 2K \|H \| such that \| j(H ) j \| = O j ((1 + log \|H \|) j K O( j ) L O( j ) \|k H \|) 2 2 2 for all j ≥ 1. Proof From the multiplicative version of Exercise 2.3.24 we can find H ⊂ H 1 1 with \|H \| ≥ 2K \|H \| and h 0 ∈ H such that \|(h · H ) ∩ (h 0 · H )\| ≥ 2K \|H \| for all h ∈ H . By dilation we may normalize h 0 = 1. From the additive non-expansion property we conclude that \|2k H \| ≤ L\|k((h · H ) ∩ H )\| ≤ L\|Ah \| for all h ∈ H , where Ah := k(h · H ) ∩ k H . Since \|k H + Ah \| ≤ \|2k H \|; \|k(h · H ) + Ah \| ≤ \|2k(h · H )\| = \|2k H \| we thus obtain the Ruzsa distance estimates d(k H, −Ah ), d(k(h · H ), −Ah ) ≤ log L and hence by the triangle inequality d(k H, k(h · H )) ≤ 2 log L . (2.44) 2.8 Elementary sum-product estimates 107 Now we turn to controlling j(H ) j for some j. We first observe that \|(H )2 \| ≤ \|H 2 \| ≤ K \|H \| ≤ 2K 2 \|H \| and thus by the multiplicative analog of Exercise 2.3.10 we have \|(H )2 · (H )−1 \| = O K O(1) \|H \| . We can then apply the multiplicative version of Exercise 1.1.8 to obtain a set X ⊂ (H )2 · (H )−1 of cardinality \|X \| = O(K O(1) (1 + log \|H \|)) such that (H )2 ⊂ X · H , and thus (H ) j ⊂ X j−1 · H . Thus by the pigeonhole principle we can bound \| j(H ) j \| ≤ \| j(X j−1 H )\| ≤ \|X \| j( j−1) \|x1 · H + · · · + x j · H \| for some x1 , . . . , x j ∈ X j−1 ; it thus suffices to show that 2 \|x1 · H + · · · + x j · H \| = O j L O( j ) \|k H \| . Since x H is contained in a translate of k(x H ), we have the somewhat crude estimate \|x1 · H + · · · + x j · H \| ≤ \| j B\| where B := k(x1 · H ) ∪ · · · ∪ k(x j · H ). But the xi are all products of O( j) elements from H and (H )−1 . From repeated application of (2.44) and the triangle inequality we conclude that d(k(xi · H ), k(xi · H )) ≤ O( j log L) for all 1 ≤ i, i ≤ j and hence d(B, B) ≤ O( j log L) + O(log j). 2 From Exercise 2.3.10 we conclude that \| j B\| = O j (L O( j ) \|B\|), and the claim follows. By combining Corollary 2.60 with the asymmetric Balog–Szemer´edi–Gowers theorem, we can show that multiplicative subgroups of F p cannot have high additive energy: Corollary 2.62 Let H be a multiplicative subgroup of F p such that \|H \| ≥ p δ for some 0 < δ ≤ 1. Then there exists an ε = ε(δ) > 0, depending only on δ, such that E(A, H ) ≤ p −ε \|A\|\|H \|2 for all A ⊆ F p with 1 ≤ \|A\| ≤ p 1−δ , if p is sufficiently large and depending on δ. 108 2 Sum set estimates Proof Let ε = ε (δ) > 0 be a small number to be chosen later, and let ε = ε(ε , δ ) > 0 be an even smaller number to be chosen later. Suppose for contradiction that there existed a set A such that E(A, H ) ≥ p −ε \|A\|\|H \|2 . Applying Corollary 2.36 (with L := p and ε replaced by ε ) we can find (if ε is sufficiently small and depending on ε ) a subset H of H with cardinality \|H \| = ε p −ε /2 \|H \| such that \|k H \| ≤ \|A + k H \| = Oε ,k p kε /2 \|A\| for all k. Since H is a multiplicative subgroup, we see that \|H · H \| ≤ \|H 2 \| = \|H \| = Oε p ε /2 \|H \| . Since \|H \| ≥ p δ , we also see (if ε is sufficiently small depending on δ) that \|H \| A ≥ p 1−δ/2 for some A depending only on δ. We can thus apply Corollary 2.60 (with δ replaced by δ/2) and conclude that for a sufficiently large k depending on δ we have \|k H \| = ε ,δ p 1−δ/2−Oδ (ε ) . This gives a contradiction if ε is sufficiently small and depending on δ, and p is sufficiently large. We shall apply this to exponential sums over multiplicative subgroups; see Theorem 4.41. For a variant of this estimate, see Lemma 9.44. It seems of interest to obtain estimates of this type for more general commutative rings, and possibly even to non-commutative rings by combining these arguments with those in the preceding section. In this direction, Bourgain has established Theorem 2.63 Let p be a large prime, and let A be a subset of the commutative ring F p × F p (endowed with the product structure (a, b) · (c, d) = (ac, bd)) be such that \|A\| ≥ p δ and \|A + A\|, \|A · A\| ≤ p ε \|A\| for some δ, ε > 0. Then there exists a set G of F p × F p such that \|G\| ≤ p Oδ (ε) \|A\| and \|A ∩ G\| ≥ p −Oδ (ε) \|A\|, where G is one of the following objects: r r r r the whole space G = F p × F p ; a horizontal line G = F p × {a} for some a ∈ F p ; a vertical line G = {a} × F p for some a ∈ F p ; a line G = {(x, ax) : x ∈ F p } for some a ∈ F p× . We sketch a proof of this proposition in the exercises. This is not as complete a characterization of sets with small sum-product as Theorem 2.55 – in particular, it does not address the case of very small A – but is already sufficient to control 2.8 Elementary sum-product estimates 109 a number of exponential sums of importance in number theory and cryptography. See , . The problem of obtaining good sum-product estimates when the ambient commutative ring is the integers Z = Z has attracted a lot of interest. In this case it has been conjectured by Erd˝os and Szemer´edi that \|k A\| + \|Ak \| = k,ε (\|A\|k−ε ) (2.45) for all ε > 0, all k ≥ 2 and all additive sets A ⊂ Z. Even the k = 2 case is open (and considered very difficult); this k = 2 case has currently been verified for all 8 ε > 11 , see Theorem 8.15. In another direction towards (2.45), a recent result of Bourgain and Chang has shown that for every m > 1 there exists an integer k = k(m) ≥ 1 such that \|k A\| + \|Ak \| = m (\|A\|m ) (2.46) for all additive sets A ⊂ Z. This last result is rather deep, in particular using an intricate “induction on scales” argument, coupled with some quantitative Freimantype theorems. Exercises 2.8.1 2.8.2 2.8.3 2.8.4 Modify the proof of Lemma 2.53 to prove Lemma 2.54. (Hint: first use multiple applications of the triangle inequality to obtain control on ˆ \|x · A − y · A\| for all x, y ∈ Ak · A.) Prove the remaining implication in Theorem 2.55. Deduce Corollary 2.56 and Corollary 2.58 from Theorem 2.55. , Let A, A , B be non-empty subsets of a field F such that 0 ∈ B. Using the first moment method, show that there exists ξ ∈ B such that E(A, ξ · A ) ≤ \|A\|2 \|A \|2 + \|A\|\|A \| \|B\| and conclude from (2.8) that \|A + ξ · A \| ≥ 2.8.5 \|A\|\|A \|\|B\| . \|A\|\|A \| + \|B\| Let A be a subset of a finite field F such that \|A\| > \|F\|1/2 . Show that \|(A − A) · A + (A − A) · A\| ≥ supx∈F \|A + x · A\| ≥ \|F\| and then con2 clude that F = (A − A) · A + (A − A) · A + (A − A) · A + (A − A) · A. (Hints: the first inequality follows easily from Corollary 2.51. For the second inequality, use Exercise 2.8.4.) 110 2.8.6 2.8.7 2.8.8 2.8.9 2.8.10 2.8.11 2.8.12 2.8.13 2.8.14 2 Sum set estimates (Croot, personal communication) Let A be a subset of a finite field F such that \|A\| > \|F\|1/k for some integer k ≥ 2. Show that \|Q[A]\| ≥ \|F\|1/(k−1) ; this clearly generalizes Corollary 2.51. (Hint: exploit the fact that the maps (a1 , . . . , ak ) → x1 a1 + · · · + xk ak fail to be injective for arbitrary x1 , . . . , xk ∈ F.) Let A be a subset of a field F such that \|A\| ≥ \|F\|ε for some ε > 0. Show that there exists an integer k = k(ε) > 1 depending only on ε such that k(Ak ) − k(Ak ) = G for some subfield G of F. (Use Exercise 2.8.5 or Lemma 4.10.) Let F p be a field of prime order p and Z = F p × F p . Let A ⊆ Z be such that \|A ∩ ({a} × F p )\| ≥ p δ and \|A ∩ ({b} × F p )\| ≥ p δ for some 0 < δ < 1 and a, b ∈ F p . Show that for some k = k(δ) > 0 we have k(Ak ) − k(Ak ) = Z . (Hint: use Exercise 2.8.7.) Let F p , Z , be as in Exercise 2.8.8, and let π1 : Z → F p , π2 : Z → F p be the coordinate projections. Suppose that A ⊆ Z is such that \|π1 (A)\|, \|π2 (A)\| ≥ p δ for some 0 < δ < 1 and such that at least one of π1 , π2 is not injective. Show that for some k = k(δ) > 0 we have k(Ak ) − k(Ak ) = Z . (Hint: by Exercise 2.8.8 it suffices to find some k such that k (Ak ) − k (Ak ) contains a large intersection with either a horizontal line or a vertical line.) Let F p , Z , π1 , π2 be as in Exercises 2.8.8, 2.8.9. Suppose that A ⊆ Z is such that \|π1 (A)\|, \|π2 (A)\| ≥ p δ for some 0 < δ < 1. Show that either A is contained in a line {(x, ax) : x ∈ F p } for some a ∈ F p× , or else k(Ak ) − k(Ak ) = Z for some k = k(δ) > 0. (Hint: by Exercise 2.8.7 one can reduce to the case where π1 (A) = π2 (A) = F p . Now divide into two cases depending on whether π1 or π2 is injective on 2A − 2A or not.) Use Exercise 2.8.10 and Lemmas 2.53, 2.54 to deduce Theorem 2.63. (You will have to take a small amount of care concerning the zero-divisors {0} × F p ∪ F p × {0}.) Let Z be a commutative ring, and A1 , A2 , A3 , A4 be subsets of Z × such that \|A1 \| = \|A2 \| = \|A3 \| = \|A4 \| = N and \|A1 · A2 − A3 · A4 \| ≤ K N . Show that \|A j · A j − A j · A j \| ≤ K O(1) N for all j = 1, 2, 3, 4. This lemma allows one to extend several of the above results to the setting where the single set A is replaced by a number of sets of comparable cardinality. Prove Corollary 2.59. Use Corollary 2.59 and Proposition 2.61 to prove Theorem 2.60. (Hint: start with k equal to a large power of 2, and set L equal to a small power of \|H \|. If the hypotheses of Proposition 2.61 are satisfied, then one can lower bound \|k H \| by \| j(H ) j \|, which can be controlled using 2.8 Elementary sum-product estimates 2.8.15 2.8.16 111 Corollary 2.59. If not, we can lower bound \|2k H \| by L\|k H \| for some large subset H of H ; now replace k by k/2 and H by H and argue as before. Continuing this process, one eventually obtains a good lower bound on \|k H \| or \|2k H \|, either by combining Proposition 2.61 with Corollary 2.59, or by accumulating enough powers of L.) Prove the following variant of Corollary 2.62: for any δ > 0 there exists ε > 0 such that whenever H, A are subsets of F p with \|H \| ≥ p δ , \|H · H \| ≤ p ε \|H \|, and 1 < \|A\| < p 1−δ , then E(A, H ) = Oδ ( p −ε \|A\|\|H \|2 ). In particular we have \|A + H \| = δ ( p ε \|H \|). Let A be an additive set in F p such that \|A\| < p 1−δ for some δ > 0. Show that there exists an ε > 0 depending on δ such that \|{(a, b, c, d, e, f ) ∈ A6 : ab + c = de + f }\| = Oε,δ (\|A\|5−ε ). (Hint: use the Balog–Szemer´edi–Gowers theorem in both the additive and multiplicative forms, together with Corollary 2.58.) This estimate is used in to show that iterations of the map X → X 1 · X 2 + X 3 on random variables in F p (where X 1 , X 2 , X 3 are independent trials of X ) converge in a certain sense to the uniform distribution, which has applications to random number generation. 3 Additive geometry In Chapter 2 we studied the elementary theory of sum sets A + B for general subsets A, B of an arbitrary additive group Z . In order to progress further with this theory, it is important first to understand an important subclass of such sets, namely those with a strong geometric and additive structure. Examples include (generalized) arithmetic progressions, convex sets, lattices, and finite subgroups. We will term the study of such sets (for want of a better name) additive geometry; this includes in particular the classical convex geometry of Minkowski (also known as geometry of numbers). Our aim here is to classify these sets and to understand the relationship between their geometrical structure, their dimension (or rank), their size (or volume, or measure), and their behavior under addition or subtraction. Despite looking rather different at first glance, it will transpire that progressions, lattices, groups, and convex bodies are all related to each other, both in a rigorous sense and also on the level of heuristic analogy. For instance, progressions and lattices play a similar role in arithmetic combinatorics that balls and subspaces play in the theory of normed vector spaces. In later sections, by combining methods of additive geometry, sum set estimates, Fourier analysis, and Freiman homomorphisms, we will be able to prove Freiman’s theorem, which shows that all sets with small doubling constant can be efficiently approximated by progressions and similarly structured sets. Closely related to all of these additive geometric sets are Bohr sets, which are in many ways the dual object to progressions, but we shall postpone the discussion of these sets (and their relationship with progressions) in Section 4.4, once we have introduced the Fourier transform. 112 3.1 Additive groups 113 3.1 Additive groups We first review the theory of additive groups, which we introduced in Definition 0.1, obtaining in particular the classification theorem for finitely generated additive groups (Corollary 3.9). This is a fundamental result in additive group theory, but it will also motivate similar results concerning other additively structured sets such as progressions, Bohr sets, and the intersection of convex sets and lattices. Typical examples of additive groups include the integers Z, the reals R, the lattices Zd , the Euclidean spaces Rd , the torus groups Rd /Zd , and the cyclic groups Z N := Z/N · Z. Note that the direct sum Z ⊕ Z of two additive groups is again an additive group. We now make an important distinction between torsion groups and torsion-free groups. Definition 3.1 (Torsion) If Z is an additive group and x ∈ Z , we let ord(x) be the least integer n ≥ 1 such that n · x = 0, or ord(x) = +∞ if no such integer exists. We say that Z is a torsion group if ord(x) is finite for all x ∈ Z , and we say that it is an r -torsion group for some r ≥ 1 if ord(x) divides r for all x ∈ Z . We say that Z is torsion-free if ord(x) = +∞ for all x ∈ Z . Examples 3.2 The groups Z, R, Zd , Rd are torsion-free, whereas any finite group such as Z N is a torsion group. A group homomorphism φ : Z → Z between two additive groups Z , Z is any map which preserves addition, negation, and zero (thus φ(x + y) = φ(x) + φ(y), φ(−x) = −φ(x), and φ(0) = 0 for all x, y ∈ Z ). If φ is also invertible, then the inverse φ −1 is automatically a group homomorphism, and we say that φ is an group isomorphism, and Z and Z are group isomorphic. Since all of our notions here shall be defined in terms of the addition, negation, and zero operations, they will all be preserved by group isomorphism, and so we will treat group isomorphic groups to be essentially equivalent. Later on we shall develop a weaker notion of Freiman homomorphism and Freiman isomorphism which is more suitable for the study of “approximate groups” (sets that are “almost” closed under addition); see Section 5.3. If G is a subgroup of an additive group Z , then we can form the quotient group Z /G := {x + G : x ∈ Z } formed by taking all the cosets of G; this is easily verified to be a group (though it is no longer a subgroup of Z ). For instance, the cyclic group Z N = Z/(N · Z) is the quotient of the integers Z by the subgroup N · Z. Observe that the map π : Z → Z /G defined by π (x) := x + G is a surjective homomorphism. 114 3 Additive geometry The sumset G + H and intersection G ∩ H of two subgroups are still subgroups. Indeed, the arbitrary intersection of a family of subgroups is still a subgroup. Hence, given any subset X of Z , we can define the span X of Z to be the smallest subgroup of Z which contains X ; equivalently, X is the space of all finite Z-linear combinations of elements of X . Thus for instance if x ∈ Z , then x is a group with cardinality ord(x). We say that an additive group Z is finitely generated if it can be written as the span Z = X of some finite set X . Clearly, every additive set X is contained in at least one finitely generated group, namely X . Thus in the theory of additive sets one can usually reduce to the case when the ambient group Z is finitely generated (though it is sometimes convenient to work in some selected non-finitely generated additive groups, such as Q, R, or Rd ). In Corollary 3.9, we shall completely classify all finitely generated additive groups up to isomorphism. Let v = (v1 , . . . , vd ) ∈ Z d denote a d-tuple of elements in Z . We can rewrite the span v := {v1 , . . . , vd } of this d-tuple in the following manner. For any element n = (n 1 , . . . , n d ) ∈ Zd we define the dot product n · v in the usual manner as n · v := n 1 v1 + · · · + n d vd . The map n → n · v is then a homomorphism from Zd to Z , and its image Zd · v is precisely the span of v: v = Zd · v. The notion of a progression, introduced in Definition 0.2, is a truncated version of the concept of a span, in which the infinite lattice Zd is replaced instead by a box. Alternatively, one can think of lattices as infinite progressions. 3.1.1 Lattices We now study a special type of additive group, namely the lattices in Euclidean space. Definition 3.3 (Lattices) A lattice in Rd is any additive subgroup of the Euclidean space Rd which is discrete (i.e. every point in is isolated). We define the rank k of to be the dimension of the linear space spanned by the elements of , thus 0 ≤ k ≤ d. If k = d, we say that has full rank. If is another lattice in Rd which is contained in , we say that is a sub-lattice of . Thus for instance Zd is a lattice of full rank in Rd . More generally, a typical example of a lattice of rank k is the set Zk · v, where v = (v1 , . . . , vk ) is a collection of linearly independent vectors in Rd for some 0 ≤ k ≤ d. In fact, this is the only possible type of lattice, as we shall see in Lemma 3.4. We observe that if 3.1 Additive groups 115 T : Rd → Rd is an invertible linear transformation on Rd , and is a lattice, then T () is also a lattice with the same rank as . If is a lattice, then the quotient space Rd / is a smooth manifold with a natural Lebesgue (or Haar) measure induced from Rd . If has full rank, it is easy to see that Rd / is also compact, and thus has a volume mes(Rd / ), which we refer to as the covolume of . Next, we classify all lattices in Rd . Call a vector v in irreducible if v/n ∈ for any integer n ≥ 2. Lemma 3.4 (Fundamental theorem of lattices) If is a lattice in Rd of rank k, then there exist linearly independent vectors v1 , . . . , vk in Rd such that = Zk · v. In particular every lattice of rank k is finitely generated and is isomorphic (via an invertible linear transformation from the linear span of to Rd ) to the standard lattice Zk . Furthermore, if w is an irreducible vector in , we may choose the above representation = Zk · v so that v1 = w. Proof We first observe that we may assume that the vectors in span Rd , else we could pass from Rd to a smaller vector space and continue the argument. In other words, we may assume that the rank k of is equal to d. We may also clearly assume that d ≥ 1, since the d = 0 case is vacuously true. Observe that contains at least one irreducible vector w, since one can start with any non-zero vector v in and take w to be the smallest vector of the form v/n (such a vector must exist since is discrete). Now let w be an irreducible vector. By the full rank assumption, we can find d linearly independent vectors v1 , . . . , vd in with v1 = w, so in particular the volume \|v1 ∧ · · · ∧ vd \| of the parallelepiped spanned by v1 , . . . , vd is strictly positive. Since contains Zd · (v1 , . . . , vd ), we obtain an upper bound for the covolume: \|v1 ∧ · · · ∧ vd \| ≥ mes(Rd / ). We now use the method of descent. If Zd · (v1 , . . . , vd ) is equal to then we are d done. Otherwise, the half-open parallelepiped { i=1 ti vi : 0 ≤ ti < 1} generated by the vectors v1 , . . . , vd , being a fundamental domain of Zd · (v1 , . . . , vd ), must d contain a non-zero lattice point x in . Write x = i=1 ti vi ; note that at least one of t2 , . . . , td must be non-zero otherwise we would have tw ∈ for some 0 < t < 1, which (by the Euclidean algorithm) contradicts the irreducibility of w. By permuting the indices 1, . . . , d if necessary we may assume that td > 0. We may also assume that td ≤ 1/2 since we could replace w by v1 + · · · + vd − x otherwise. Then the volume \|v1 ∧ · · · ∧ vd−1 ∧ x\| is at most half that of \|v1 ∧ · · · ∧ vd \|, but is still non-zero. We thus replace vd by x and repeat the above argument. Because of our absolute lower bound on the volume of parallelepipeds, this argument must eventually terminate, at which point we have found the desired 116 3 Additive geometry presentation for . Note that this procedure will never alter v1 and hence v1 is equal to w as desired. Corollary 3.5 (Splitting lemma) Let be a lattice of rank k, and let v be an irreducible vector in . Then there exists a sub-lattice of of rank k − 1 such that is the direct sum of Z · v and , i.e. = Z · v + and Z · v ∩ = {0}. Proof Apply Lemma 3.4 with v1 := v, and set := Zk−1 · (v2 , . . . , vk ); the claim then follows from the linear independence of v1 , . . . , vk . Corollary 3.6 (Fundamental theorem of finitely-generated torsion-free additive groups) Let Z be a finitely generated torsion-free additive group. Then Z is isomorphic to Zd for some d ≥ 0. Proof We shall use the homomorphism theorems (Exercise 3.1.1). Since Z is finitely generated, we may find elements v1 , . . . , vn in Z such that Zn · (v1 , . . . , vn ) = Z . Now let be the set {n ∈ Zn : n · (v1 , . . . , vn ) = 0}; then is a sub-lattice of Zn and Z is isomorphic to Zn / . In particular, Zn / is torsion-free. We shall show that this implies that Zn / is isomorphic to some Zd , as desired. We induce on n, the case n = 0 being trivial. If = {0} we are done, so suppose contains a non-zero vector v ∈ , which we may assume without loss of generality to be irreducible in . It is also irreducible in Zn , for if v = m · w for some w ∈ Zd and m > 1, then w + would be a non-zero element of Zn / such that m · (w + ) = 0 + , contradicting the torsion-free assumption. By Lemma 3.4 or Corollary 3.5, this implies that Zn /(Z · w) is isomorphic to Zn−1 . Since Zn / is isomorphic to (Zn /(Z · w))/(/(Z · w)), the claim then follows from the induction hypothesis. 3.1.2 Quotients of lattices Let G be a finitely generated additive group generated by d elements v1 , . . . , vd ∈ G. If we write v := (v1 , . . . , vd ), and let ⊂ Zd be the lattice := {n ∈ Zd : n · v = 0}, it is easy to see that G is isomorphic to the quotient Zd / . Thus it is of interest to understand the quotient of two lattices. A basic tool for doing so is Theorem 3.7 (Smith normal form) Let and be two lattices of full rank in Rd such that is a sub-lattice of . Then there exist linearly independent vectors v1 , . . . , vd in such that = Zd · (v1 , . . . , vd ) and = Zd · (N1 v1 , . . . , Nd vd ), 3.1 Additive groups 117 where 1 ≤ N1 ≤ · · · ≤ Nd are positive integers such that N j divides N j+1 for all j = 1, . . . , d − 1. Note that by applying an invertible linear transformation one can set (v1 , . . . , vd ) equal to the standard basis (e1 , . . . , ed ), so that becomes just the standard lattice Zd , while is the sub-lattice of Zd of vectors whose jth coordinate is a multiple of N j for j = 1, . . . , d. Proof We induce on d. For d = 0 the statement is vacuously true, so suppose d ≥ 1 and the claim has already been proven for d − 1. Given any non-zero vector v ∈ , define the index of v to be the largest positive integer n such that v/n ∈ ; note that the index is finite since is discrete. Note that the index of v is n if and only if v = nw for some irreducible vector w in . Since has full rank, it contains non-zero vectors, each of which has an index. Let N1 denote the minimum index of all such vectors. By the well-ordering principle, this index is attained, and thus there exists an irreducible vector v1 ∈ such that N1 v1 ∈ . Using Lemma 3.4, we may apply an invertible linear transformation to map to Zd , in such a way that v1 is now equal to the standard basis vector e1 . Now let (n 1 , . . . , n d ) be any vector in . Observe that n 1 , . . . , n d are integers; furthermore, n 1 must be a multiple of N1 , otherwise by subtracting a multiple of N1 e1 we could ensure that \|n 1 \| < N1 , which contradicts the definition of N1 as the minimal index of . Thus we may factorize = N1 Z · e1 + , where is some sub-lattice of Zd−1 (which we think of as the span of e2 , . . . , ed ). Note that if x ∈ , then (N1 , x) ∈ , and hence (since (N1 , x) must have index at least N1 ), x must be a multiple of N1 . Thus actually lies in N1 · Zd−1 , and we may therefore write = N1 (Z · e1 + ) for some sub-lattice of Zd−1 . Note that must have rank d − 1 since has rank d. We now invoke the inductive hypothesis, and, by applying an invertible linear transformation to Zd−1 if necessary, we may assume that = {(n 2 M2 , . . . , n d Md ) : n 2 , . . . , n d ∈ Z} for some 1 ≤ M2 ≤ · · · ≤ Md such that M j divides M j+1 for all j = 2, . . . , d − 1. The claim follows by setting N j := N1 M j for j = 2, . . . , d. We can now obtain the well-known classification of finite and finitely generated additive groups: Corollary 3.8 (Fundamental theorem of finite additive groups) Every finite additive group G is isomorphic to the direct sum of a finite number of cyclic groups Z N = Z/(N · Z). 3 Additive geometry 118 Proof Let g1 , . . . , gd be a finite set of generators for G. Then the map φ : Zd → G defined by φ(n) := n · (g1 , . . . , gd ) is a surjection, and thus G is isomorphic to Zd /φ −1 (0), which is a subgroup of Rd /φ −1 (0). The kernel φ −1 (0) is clearly a lattice of some rank 0 ≤ k ≤ d, and hence by Lemma 3.4 is generated by k linearly independent vectors v1 , . . . , vk in Zd . Observe that we must have full rank k = d, otherwise Zd /φ −1 (0) (and hence G) will be infinite. Using the Smith normal form, we can after applying an isomorphism write φ −1 (0) as the lattice generated by N1 e1 , . . . , Nd ed for some integers N1 , . . . , Nd ≥ 1; this makes G isomorphic to G ≡ Z/N1 Z ⊕ · · · ⊕ Z/Nd Z, as desired (indeed we even obtain a normal form in which N j divides N j+1 for j = 1, . . . , d − 1). Corollary 3.9 (Fundamental theorem of finitely generated additive groups) Every finitely generated additive group G is isomorphic to the direct sum of a finite number of cyclic groups Z/(N · Z), and a lattice Zd for some d ≥ 0. ˜ := {x ∈ G : nx = 0 for some n > 0} be the torsion group of G; Proof Let G ˜ is the direct sum of cyclic groups. The quotient group then by Corollary 3.8, G ˜ is torsion-free and is thus isomorphic to Zd for some d ≥ 0 by Corollary 3.6. G/G If we let e˜ 1 , . . . , e˜ d be arbitrary representatives in G of the standard basis e1 , . . . , ed ˜ and Z · e˜ 1 , . . . , Z · e˜ d , and the of Zd , we thus see that G is the direct sum of G claim follows. Exercises 3.1.1 3.1.2 3.1.3 3.1.4 (Homomorphism theorems) If φ : Z → Z is a homomorphism between groups, show that the range φ(Z ) is a group which is isomorphic to the quotient group Z /φ −1 (0). If G, H are subgroups of Z , show that (G + H )/G is isomorphic to H/(G ∩ H ). If furthermore G ⊂ H , show that H/G is a subgroup of Z /G and that (Z /G)/(H/G) is isomorphic to Z /H . If G is a subgroup of Z , show that (Z ⊕ Z )/(G ⊕ G ) is isomorphic to (Z /G) ⊕ (Z /G ). (Cauchy’s theorem) Show that if G is a subgroup of a finite additive group Z , then \|Z /G\| = \|Z \|/\|G\| (and in particular \|G\| must divide \|Z \|). By considering the groups x for various x ∈ Z , conclude that every finite additive group Z is an \|Z \|-torsion group; in particular, ord(x) divides \|Z \| for all x ∈ Z . Show that if x is any element of a additive group Z , then the group x = Z · x has cardinality ord(x). More generally, if v = (v1 , . . . , vd ) ∈ Z d , show that the group Zd · v has cardinality at most ord(v1 ) · · · ord(vd ), but at least as large as the least common multiple of ord(v j ). Let Z be an additive group. Show that Z is an N -torsion group if and only if for every x ∈ Z , the torsion of x is a divisor of N . Show that Z is 3.2 Progressions 3.1.5 3.1.6 3.1.7 3.1.8 119 torsion-free if and only if Z contains no finite subgroups other than the trivial subgroup {0}. Let Z = Z 1 ⊕ Z 2 be a direct sum of additive groups and r ≥ 1. Show that Z is torsion-free (resp. r -torsion) if and only if Z 1 and Z 2 are torsion-free (resp. r -torsion). Prove that Q and R are not finitely generated. If x, y are elements of an additive group Z with finite order, show that x + y also has finite order, and that ord(x + y) divides the least common multiple of ord(x) and ord(y). Conclude that the set tor(Z ) := {x ∈ Z : ord(x) < ∞} is a torsion group; we refer to it as the torsion subgroup of Z . It is clearly the largest subgroup of Z which is a torsion group. Show that the quotient group Z /tor(Z ) is torsion-free, and is in fact the largest quotient which is torsion-free (in the sense that all other torsion-free quotients are quotients of Z /tor(Z )). Show that Corollary 3.5 fails whenever v is not irreducible. 3.2 Progressions We now study a basic example of an additive set, namely that of a generalized arithmetic progression (or progression for short), as defined in Definition 0.2. These will be model examples of additive sets with large amounts of additive structure; they can be viewed as a hybrid between a lattice and a convex set. (For a more quantitative realization of this heuristic, see Lemma 3.36 below.) Note that progressions with the same set of basis vectors add very easily (a + [0, N ] · v) + (a + [0, N ] · v) = (a + a ) + [0, N + N ] · v (3.1) (so in particular the rank and basis vectors do not change), whereas progressions with different basis vectors add via the formula (a + [0, N ] · v) + (a + [0, N ] · v ) = (a + a ) + [0, N ⊕ N ] · (v ⊕ v ). (3.2) Note the progression on the right-hand side of (3.2) is likely to be highly improper if v and v share some basis vectors in common. Also one can replace the box [0, N ] by another one and also obtain a progression: a + [N , M] · v = (a + N · v) + [0, M − N ] · v. Similarly if one uses boxes such as [N , M), etc. In particular, the negation of a progression is also a progression: −(a + [0, N ] · v) = (−a) + [0, N ] · (−v) = (−a − N · v) + [0, N ] · v. (3.3) 3 Additive geometry 120 From this and (3.2) we see that the sum or difference of two progressions is again a progression. Finally, we make the easy observation that the Cartesian product of two progressions is again a progression. We now show that, up to errors of O(1)d , that progressions of rank d are essentially closed under addition. Lemma 3.10 Let P = a + [0, N ] · v be a progression of rank d in an additive group Z ; we do not require that P be proper (see Definition 0.2). Then for any integers n < m and any b ∈ Z , we can cover b + [n N , m N ] · v by (m − n)d translates of P. In particular for any n, m ≥ 0 with (n, m) = (0, 0), we can cover n P − m P by (n + m)d translates of P, and in particular \|n P − m P\| ≤ (n + m)d \|P\|. Furthermore, n P − m P is also a progression of rank d and volume at most vol(n P − m P) ≤ (n + m)d vol(P). Proof The first claim is clear since [n · N , m · N ] · v = [0, N ] · v + [(n, . . . , n), (m, . . . , m)] · (N1 v1 , . . . , Nd vd ). From (3.1) we have n P − m P = (na − ma − m N · v) + [0, (n + m)N ] · v from which the remaining claims follow. From this lemma we see in particular that if P is a symmetric progression of rank d and contains the origin (e.g. if P = [−N , N ] · v), then P is a 2d -approximate group in the sense of Definition 2.25. Indeed one can think of (symmetric) progressions of small rank as substitutes for subgroups in torsion-free settings (since torsion-free groups cannot contain finite subgroups). They also are the arithmetic analogue of boxes (or more generally, parallelepipeds) in Euclidean space, and in fact many of the results from real-variable harmonic analysis regarding covering by boxes (in physical space, Fourier space, or both) will have analogues for progressions. In the special case when the rank d is equal to 1, a generalized arithmetic progression is the same as an ordinary arithmetic progression (or arithmetic progression for short) P = a + [0, N ] · v = {a + nv : 0 ≤ n ≤ N } with base point a ∈ Z , basis vector or step v ∈ Z , and length N + 1. Note again that the cardinality of P may be less than N + 1 if P is not proper, though in a torsion-free group this is only possible if the step v is zero. 3.2 Progressions 121 We record a trivial lemma that asserts that the sum set of a progression and a small set can be contained (somewhat inefficiently) in another progression. Lemma 3.11 If P is a progression of rank d, and P + w1 , . . . , P + w K are translates of P, then all the translates P + w1 , . . . , P + w K can be contained inside a single progression of rank d + K − 1 and volume 2 K −1 vol(P). Proof Write P = a + [0, N ] · v. By translation invariance we may set w K = 0. Then the claim follows by using the progression a + [0, N ] · v + [0, 1] K −1 · (w1 , . . . , w K −1 ). Thus if one adds a small number of elements to an progression, one can still place the combined set inside a progression of slightly larger rank and volume, although the volume can grow exponentially in \|A\|. This is unavoidable: see Exercise 3.2.2. Because of this exponential loss, it is sometimes better not to invoke this lemma, and deal with multiple shifts of a single progression rather than trying to contain everything inside a single progression. Note that we have not guaranteed that the progressions in Lemma 3.11 are proper; we will return to this point in Section 3.6. Exercises 3.2.1 3.2.2 3.2.3 3.2.4 3.2.5 Let N = (N1 , . . . , Nd ) be a collection of non-negative integers. Show that every proper ordinary arithmetic progression of length (N1 + 1) · · · (Nd + 1) is equal (as a set) to a proper generalized arithmetic progression of dimension N . (This example shows that the rank of a progression cannot be uniquely determined from the set of its elements, even if we restrict the progression to be proper.) Let K ≥ 1 and d ≥ 0 be integers, and P = a + [0, N ] · v be a rank d progression in an additive group Z for some basis vectors v = (v1 , . . . , vd ), and let X = {e1 , . . . , e K } be a set of K elements in Z . Suppose that the elements v1 , . . . , vd , e1 , . . . , e K are linearly independent over Z. Show that any progression which contains P + X must necessarily have rank at least d + K − 1 and volume at least 2 K −1 vol(P), which shows that Lemma 3.11 is sharp. Show that in a torsion-free additive group, the intersection of two ordinary arithmetic progressions is again an ordinary arithmetic progression. What happens if the torsion-free hypothesis is removed? What happens if one or both of the progressions is allowed to have rank greater than one? Show that every finite additive group is also a proper progression. Let P be a progression of rank d. Show that P contains an arithmetic progression Q with \|Q\| ≥ \|P\|1/d , and furthermore that Q is proper if P is, and Q can be chosen to be symmetric around the origin if P is. 122 3.2.6 3.2.7 3 Additive geometry Let P be a proper progression of rank d, and let A be a subset of P such that \|A\| ≤ ε\|P\| for some 0 < ε < 1. Show that P\A contains a proper progression Q of rank d with \|Q\| ≥ C −d /ε for some absolute constant C. Let A be an additive set in an ambient group Z , and let v ∈ Z . Show that \|(A + v)\A\| ≤ 1 if and only if A is equal to a proper arithmetic progression of step v, union a finite (possibly zero) number of translates of the group v . In particular, if \|A\| < ord(v), then \|(A + v)\A\| > 0, and \|(A + v)\A\| = 1 if and only if A is a proper arithmetic progression of step v. 3.3 Convex bodies We now review some of the theory of convex bodies in Rd , which are in some sense the continuous analogue of generalized arithmetic progressions. This is of course a vast field, and we shall restrict ourselves with just a small sample of results, relating to the additive theory of such sets, to covering lemmas, and the relationship between addition and volume. We shall use mes(A) to denote the volume of a set A in Rd ; to avoid issues with measurability we shall mostly concern ourselves with bounded open sets A. If A ∈ Rd and λ ∈ R, we use λ · A to denote the dilation λ · A := {λx : x ∈ A}. Observe that mes(λA) = \|λ\|d mes(A). Recall that a set A in Rd is convex if we have (1 − θ)x + θ y ∈ A whenever x, y ∈ A and 0 ≤ θ ≤ 1; equivalently, a set is convex if and only if a · A + b · A = (a + b) · A for all real a, b ≥ 0 (Exercise 3.3.3). In particular we have n A = \|n\| · A for any integer A. We call A a convex body if it is convex, open, non-empty, and bounded. In particular we see that if A is a convex body, then mes(A + A) = mes(2 · A) = 2d mes(A), (3.4) so convex bodies have small doubling constant. As for A − A, we can use Lemma 3.12 For any bounded open subsets A, B, C of Rd (not necessarily convex), we have mes(A − C)mes(B) ≤ mes(A − B)mes(B − C). 3.3 Convex bodies 123 This is proven by modifying the proof of Lemma 2.6 appropriately and is left as an exercise. From this Lemma (with A = C and B = −A) and (3.4) we obtain mes(A − A) ≤ 4d mes(A); (3.5) compare these bounds with Lemma 3.10. For a slight refinement of (3.5), see Exercise 3.4.6. In the converse direction, the Brunn–Minkowski inequality (Theorem 3.16 below) will give mes(A − A) ≥ 2d mes(A). Call a convex body A symmetric if A = −A; thus for us symmetry will always be with respect to the origin. The following theorem of John essentially classifies all convex bodies (symmetric and non-symmetric) up to a (dimension-dependent) constant factor. Theorem 3.13 (John’s theorem) Let A be a convex body in Rd . Then there exists an invertible linear transformation T : Rd → Rd on Rd and a point x0 ∈ A such that Bd ⊆ T (A − x0 ) ⊆ d · Bd , where Bd is the unit ball {(x1 , . . . , xd ) ∈ Rd : x12 + · · · + xd2 < 1}. If A is symmetric, then we can improve these inclusions to √ Bd ⊆ T (A) ⊆ d · Bd . √ The constants d and d are sharp; see the exercises. Proof We will use a variational argument. Define an ellipsoid to be any set E of the form E = L(Bd ) + x0 , where Bd is the unit ball, x0 ∈ Rd , and L is a (possibly degenerate) linear transformation in Rd ; we allow the ellipsoid to be degenerate for compactness reasons. Since A is open and bounded, it is easy to see that the set of all ellipsoids E contained in A is a compact set (with respect to the usual topology on L and x0 ). Also the volume of the ellipsoid E is mes(E) = \| det(L)\|, whch is clearly a continuous function of E. Thus there exists an ellipsoid E = L(Bd ) + x0 in A which maximizes the volume mes(E); since A is open, this volume is nonzero, and hence L is invertible. By applying L −1 if necessary (observing that the conclusion of the lemma is invariant under invertible linear transformations) we may thus assume that E is a translate E = Bd + y0 of the unit ball, where y0 = L −1 (x0 ). Let us now restrict to the case where A is symmetric. Observe that if A contains Bd + y0 then it also contains Bd − y0 by symmetry, and hence contains Bd , which is in the convex hull of Bd + y0 and Bd − y0 . To√ conclude the proof of the lemma in this case we need to show that √ A is contained in d · Bd . Suppose for contradiction that A was not contained in d · Bd ; without loss of generality (and using√the hypothesis that A is open) we may then suppose that r e1 ∈ A for some r > d, 124 3 Additive geometry where e1 is the first basis vector. Observe now from elementary geometry that if ω is any point on the boundary of Bd making an angle ∠(ω, e1 ) < arctan(r 2 − 1), then the line segment connecting ω to r e1 is disjoint from (and not tangent to) Bd , and, since Bd and r e1 both lie in the convex set A, we thus see that ω√ also lies in the open set A. By symmetry, the same is true if ∠(ω, −e1 ) < arctan( r 2 − 1). We now perturb the ball Bd by an epsilon. Now let δ > 0 be a small number, let ε > 0 be an even smaller one, and consider the ellipsoid L ε,δ (Bd ), where √ L ε,δ (x1 , . . . , xd ) := ((1 + ( d − 1 + δ)ε)x1 , (1 − ε)x2 , . . . , (1 − ε)xd )). When ε = 0, L ε,δ (Bd ) is just Bd . Now consider √ how L ε,δ (Bd ) evolves in ε. The determinant of this transformation is (1 + ( d − 1 + δ)ε)(1 − ε)d−1 , which has a positive ε-derivative at ε = 0. Thus L ε,δ (Bd ) has larger volume than B for sufficiently small ε (depending on δ). Now we check which points on the surface of L ε,δ (Bd ) expand away from the origin, and which ones contract. A simple computation shows that for any ω = (ω1 , . . . , ωd ) on the boundary of Bd , the derivative d L ε,δ (ω)2 , ε=0 dε where (y1 , . . . , yd )2 := y12 + · · · + yd2 , is negative unless (d − 1 + δ)ω12 − ω22 − · · · − ωd2 ≥ 0, or in other words that √ ∠(ω, ±e1 ) ≤ arctan( d − 1 + δ). But if δ is small enough depending on r , this region is contained entirely within the interior of A by the previous discussion. Thus for ε small enough L ε,δ (Bd ) is completely contained inside A but has larger volume, contradicting the maximality of Bd , and we are done. Now suppose that A is not symmetric. In this case we may translate so that y0 = 0. Thus again we have Bd ⊆ A, and the task is to show that A ⊆ n · Bd . Suppose again for contradiction that r e1 ∈ A for some r > d; again this means that every point ω in the boundary of Bd with ∠(ω, e1 ) < arctan(r 2 − 1) will lie in the interior of A. Now let δ, ε > 0 and consider the ellipsoid L ε,δ (x1 , . . . , xd ) + (d − 1 + δ)εe1 ; again, this ellipsoid has larger volume than Bd if ε is sufficiently small. Also, we see that d L ε,δ (ω) + (d − 1 + δ)εe1 2 ε=0 dε 3.3 Convex bodies 125 is negative unless (d − 1 + δ)ω12 + (d − 1 + δ)ω1 − ω22 − · · · − ωd2 ≥ 0, which can be rewritten (using ω = 1) as ((d + δ)ω1 − 1)(ω1 + 1) ≥ 0, or equivalently ∠(ω, e1 ) ≤ arctan( (d + δ)2 − 1). We now argue as in the symmetric case to obtain again the desired contradiction, if δ is chosen so that d + δ < r . As a corollary of Theorem 3.13 we see that if A is a convex body, we can cover A + A or A − A by a relatively small number of copies of A: A ± A can be covered by O(d)d translates of A. (3.6) This follows immediately from the geometric observation that d · Bd + d · Bd = 2d · Bd can be covered by O(d)d translates of Bd . If A is symmetric, we can improve this somewhat. In the special case when A is a cube or a box, it is clear that A ± A can be covered by 2d translates of A (cf. Lemma 3.10), but one cannot hope for this in general; for instance if A is a disk in R2 then one needs six copies of A to cover A ± A. In the general case, we will need the following continuous version of Lemma 3.14 (Ruzsa’s covering lemma) , For any bounded subsets A, B of Rd with positive measure (not necessarily convex), we can cover B by at most min( mes(A+B) , mes(A−B) ) translates of A − A. mes(A) mes(A) The proof of this lemma is nearly identical to that of Lemma 2.14 and is left as an exercise. As a consequence we can improve (3.6) for symmetric convex bodies: Corollary 3.15 Let A ⊂ Rd be a convex body, and let λ, μ > 0 be real. Then λ · A can be covered by at most (λ + 1)d translates of A − A, and λ · A − μ · A can be covered by (2 max(λ, μ) + 1)d translates of A − A. If A is symmetric, then λ · A can be also covered by (2λ + 1)d translates of A. Proof The first claim follows from Lemma 3.14 since mes(λ · A + A) = (λ + 1)d mes(A). To prove the second claim, we may take λ ≥ μ. The first claim implies that 2λ · A can be covered by (2λ + 1)d translates of A − A = 2 · A, and the third claim follows by rescaling by 1/2. Finally, the second claim follows by applying the third claim to A − A. 3 Additive geometry 126 Observe that all the bounds obtained here tend to be exponential in d or worse. Thus when using the theory of convex bodies to obtain explicit estimates, it is often important to keep the dimension d as low as possible, even at the cost of making some other parameters larger than would otherwise be necessary. See for further discussion of sum set and covering estimates for convex bodies. We have not yet seen what happens to the sum or difference of two unrelated convex bodies A and B. The relationship here is given by the Brunn–Minkowski inequality, which we turn to next. Exercises 3.3.1 3.3.2 3.3.3 3.3.4 3.3.5 Prove Lemma 3.12. Prove Lemma 3.14. Verify that the two definitions of convexity given are indeed equivalent. Let A be an open bounded subset of Rd . Show that A is convex if and only if 2A = 2 · A, and that A is convex and symmetric if and only if 2A = −2 · A. ∞ For any s > 0 let (s) := 0 e−x x s−1 d x denote the Gamma function. Show that (s + 1) = s(s) for all s > 0, that (d) = (d − 1)! for all √ d ≥ 1, that (1/2) = π, and we have the Stirling formula log (s) = s log s − s + O(log s) 3.3.6 for all large s. (Hint: use (1.52) and the monotonicity ofthe function.) 2 Let Bd be the unit ball in Rd . By evaluating the integral Rd e−π \|x\| d x in both Cartesian and polar coordinates, and using the preceding exercise, establish the volume formula mes(Bd ) = 3.3.7 3.3.8 3.3.9 (3.7) (3/2)d 2d = (2π e + o(1))d/2 d −d/2 . (d/2 + 1) (3.8) Let Od be the octahedron given by the convex hull of ±e1 , . . . , ±ed in Rd . Show that mes(Od ) = 2d /d! = (2e + o(1))d d −d . Thus in large dimension the octahedron becomes considerably smaller than the circumscribing ball Bd which contains it, which in turn is considerably smaller than the circumscribing√cube. Show that the constants d and d in Theorem 3.13 cannot be improved. (For the non-symmetric case, take A to be a d-simplex (the convex hull of d points in Rd ); for the symmetric case, take A to be a cube.) If A and A are two symmetric convex bodies in Rd , show that there exists an invertible linear transformation T : Rd → Rd such that A ⊆ T (A ) ⊂ d · A. 3.4 The Brunn–Minkowski inequality 3.3.10 State and prove a similar result in the case when A and A are not necessarily symmetric. Let A, B be open bounded sets. Show that mes((A − A) ∩ (B − B)) ≥ 3.3.11 127 mes(A)mes(B) mes(A ± B) for either choice of sign ±, by developing a continuous analogue of the arguments used to prove (2.8). (Alternatively, one can try to discretize A and B to replace them with finite sets, and then use (2.8) directly.) Let A be a symmetric convex body in Rd , which contains the ball ρ · B of radius ρ > 0 centered at the origin. Let V be any r -dimensional d! subspace of Rd . Show that mesr (A ∩ V ) ≤ r !(2ρ) d−r mesd (A), where mesr denotes r -dimensional measure. (Hint: first show that if r < d, then there exists an r + 1-dimensional space V1 containing V such that mesr +1 (A ∩ V1 ) ≥ r2ρ mesr (A ∩ V ). Then continue inductively.) +1 3.4 The Brunn–Minkowski inequality The purpose of this section is to prove the following lower bound for the volume mes(A + B) of a sum set. Theorem 3.16 (Brunn–Minkowski inequality) If A and B are non-empty bounded open subsets of Rd , then mes(A + B)1/d ≥ mes(A)1/d + mes(B)1/d . This inequality is sharp (Exercise 3.4.2). The theorem also applies if A and B are merely measurable (as opposed to being bounded and open), though one must then also assume that A + B is measurable; we will not prove this here. In general, there is no upper bound for mes(A + B); consider for instance the case when A is the x-axis and B is the y-axis in R2 , then A, B both have measure zero but A + B is all of R2 . One can easily modify this example to show that there is no upper bound for mes(A + B) in terms of mes(A) and mes(B) when A, B are bounded open sets. See for a thorough survey of the Brunn–Minkowski inequality and related topics. To prove this theorem, it suffices to prove the following dimension-independent version: Theorem 3.17 If A and B are non-empty bounded open subsets of Rd , and 0 < θ < 1, then mes((1 − θ ) · A + θ · B) ≥ mes(A)1−θ mes(B)θ . 3 Additive geometry 128 To see why Theorem 3.17 implies the Brunn–Minkowski inequality, apply Theorem 3.17 with A and B replaced by mes(A)−1/d · A and mes(B)−1/d · B to obtain 1−θ θ mes · A + · B ≥1 mes(A)1/d mes(B)1/d for any 0 < θ < 1. Setting θ := mes(B)1/d mes(A)1/d + mes(B)1/d we obtain the result. Conversely, one can easily deduce Theorem 3.17 from the Brunn–Minkowski inequality (Exercise 3.4.1). It remains to prove Theorem 3.17. We begin by first proving Lemma 3.18 (One-dimensional Brunn–Minkowski inequality) If A and B are non-empty bounded open subsets of R, then mes(A + B) ≥ mes(A) + mes(B). Proof The hypotheses and conclusion of this lemma are invariant under independent translations of A and B, so we can assume that sup(A) = 0 and inf(B) = 0, hence in particular A and B are disjoint. But then we see that A + B contains both A and B separately, and we are done. Using this Lemma, we deduce Proposition 3.19 (One-dimensional Pr´ekopa–Leindler inequality) Let 0 < θ < 1, and let f, g, h : R → [0, ∞) be lower semi-continuous, compactly supported non-negative functions on R such that h((1 − θ )x + θ y) ≥ f (x)1−θ g(y)θ for all x, y ∈ R. Then we have 1−θ θ h≥ f g . R R R Proof By multiplying f, g, h by appropriate positive constants we may normalize supx f (x) = sup y f (y) = 1. Let 1 > λ > 0 be arbitrary. Observe that if f (x) > λ and g(y) > λ, then by hypothesis h((1 − θ)x + θ y) > λ. Thus we have {z ∈ R : h(z) > λ} ⊆ (1 − θ ) · {x ∈ R : f (x) > λ} + θ · {y ∈ R : g(y) > λ}. Since f, g, h are lower semi-continuous and compactly supported, all the sets above are open and bounded, hence by Lemma 3.18 mes({z ∈ R : h(z) > λ}) ≥ (1 − θ)mes({x ∈ R : f (x) > λ}) + θmes({y ∈ R : g(y) > λ}). 3.4 The Brunn–Minkowski inequality 129 Integrating this for λ ∈ [0, ∞) and using Fubini’s theorem (cf. (1.6)), the claim follows from the arithmetic mean–geometric mean inequality. Now we iterate this to higher dimensions. Proposition 3.20 (Higher-dimensional Pr´ekopa–Leindler inequality) Let 0 < θ < 1, d ≥ 1, and let f, g, h : Rd → [0, ∞) be lower semi-continuous, compactly supported non-negative functions on Rd such that h((1 − θ )x + θ y) ≥ f (x)1−θ g(y)θ for all x, y ∈ Rd . Then we have 1−θ θ h≥ f g . R R R Proof We induce on d. When d = 1 this is just Proposition 3.19. Now assume inductively that d > 1 and the claim has already been proven for all smaller dimensions d. Define the one-dimensional function h d : R → [0, ∞) by h d (xd ) := h(x1 , . . . , xd ) d x1 · · · d xd−1 , Rd−1 and similarly define f d , gd . One can easily check that (using Fatou’s lemma) that these functions are lower semi-continuous and compactly supported. Also, applying the inductive hypothesis at dimension d − 1 we see that h d ((1 − θ)xd + θ yd ) ≥ f d (yd )1−θ gd (yd )θ for all xd , yd ∈ R. If we then apply the one-dimensional Pr´ekopa–Leindler inequality, we obtain the desired result. If we apply Proposition 3.20 with f := 1 A , g := 1 B , and h := 1(1−θ )A+θ B we obtain Theorem 3.17, and the Brunn–Minkowski inequality follows. Exercises 3.4.1 3.4.2 3.4.3 Show that Theorem 3.16 implies Theorem 3.17. Show that equality in Theorem 3.17 can occur when A is convex, and B = λ · A + x0 for some λ, x0 ∈ Rn . Conversely, if A and B are nonempty bounded open subsets of Rd , show that the preceding situation is in fact the only case in which equality can be attained. (The case when A and B are merely measurable is a bit trickier, and is of course only true up to sets of measure zero; see for further discussion). Let A be a convex body in Rd . Using Theorem 3.17, show that the cross-sectional areas f (xd ) := mes({x ∈ Rd−1 : (x , xd ) ∈ A}) are a 3 Additive geometry 130 3.4.4 3.4.5 3.4.6 log-concave function of xd , i.e. f ((1 − λ)xd + λyd ) ≥ f (xd )1−λ f (yd )λ for all 0 ≤ λ ≤ 1 and xd , yd ∈ R; this is known as Brunn’s inequality. Let A be a bounded open set with smooth boundary ∂ A, and let B be a ball with the same volume as A. Prove the isoperimetric inequality mes(∂ A) ≥ mes(∂ B). (Hint: Use the Brunn–Minkowski inequality to estimate mes(A+ε·B)−mes(A) for ε > 0 small, and then let ε → 0.) ε Let A, B be symmetric convex bodies in Rd . Show by examples that there is no upper bound for mes(A + B) in terms of mes(A), mes(B), amd d alone, except in the d = 1 case. However, by using Lemma 3.12, show that mes(A + B) ≤ 4d mes(A)mes(B) . mes(A∩B) Let A be a convex body. Use the Brunn’s inequality to show that mes(A ∩ (x + A)) ≥ (1 − r )n mes(A) whenever 0 ≤ r ≤ 1 and x ∈ r · (A − A). Conclude that mes(A)2 = mes(A ∩ (x + A)) d x ≥ A−A 1 n(1 − r )n−1 mes(A)mes(r · (A − A)) dr 0 1 = 2n mes(A)mes(A − A) n 3.4.7 3.4.8 3.4.9 whence one obtains the Rogers–Shepard inequality mes(A − A) ≤ 2n mes(A). Show that this inequality is sharp when A is a simplex. Use n Stirling’s formula to compare this inequality with (3.5). Let A, B be additive sets in Zd . Use the Brunn–Minkowski inequality to show that \|A + B + {0, 1}d \| ≥ 2d min(\|A\|, \|B\|). (Hint: consider A + [0, 1]d and B + [0, 1]d .) Let A, B be additive sets in Rd . Show that \|A + B + {0, 1}d \| ≥ 2d min(\|A\|, \|B\|). (Hint: partition Rd into cosets of Zd , locate the coset with the largest intersection with A or B, and apply the preceding exercise.) Let A be an open bounded set in Rd . Show that mes(A + A) ≥ 2d mes(A), with equality if and only if A is convex. (Hint: A + A contains 2 · A.) 3.5 Intersecting a convex set with a lattice In previous sections we have studied lattices, which are discrete but unbounded, and convex sets, which are bounded but continuous. We now study the intersection B ∩ of a convex set B and a lattice in a Euclidean space Rd , which is then necessarily 3.5 Intersecting a convex set with a lattice 131 a finite set. A model example of such set is the discrete box [0, N ) for some N = (N1 , . . . , Nd ), which is the intersection of the convex body {(x1 , . . . , xd ) : −1 < xi < Ni for all 1 ≤ i ≤ d} with the Euclidean lattice Zd . One of the main objectives of this section shall to show a “discrete John’s lemma” which shows that all intersections B ∩ can be approximated in a certain sense by a discrete box. We begin with some elementary estimates. Lemma 3.21 Let be a lattice in Rd . If A ⊂ Rd is an arbitrary bounded set and P ⊂ Rd is a finite non-empty set, then \|A ∩ ( + P)\| ≤ \|(A − A) ∩ ( + P − P)\|. (3.9) If B is a symmetric convex body, then (k · B) ∩ can be covered by (4k + 1)d translates of B ∩ (3.10) for all k ≥ 1. If furthermore is a sub-lattice of of finite index \|/ \|, then we have \|B ∩ \| ≤ \|B ∩ \| ≤ 9d \|/ \|\|B ∩ \|. (3.11) Proof We first prove (3.9). We may of course assume that A ∩ ( + P) contains at least one element a. But then A ∩ ( + P) ⊆ ((A − A) ∩ ( + P − P)) + a, and the claim follows. Now we prove (3.10). The lower bound is trivial, so it suffices to prove the upper bound. By the preceding argument we can cover \|( 12 · B + x) ∩ \| by a translate of B ∩ for any x ∈ Rd . But by Corollary 3.15 we can cover k · B by (4k + 1)d translates of 12 · B, and the claim (3.10) follows. Finally, we prove (3.11). The lower bound is trivial. For the upper bound, observe that is the union of \|/ \| translates of , so it suffices to show that \|B ∩ ( + x)\| ≤ 9d \|B ∩ \| for all x ∈ Rd . But by (3.9) and (3.10) we have \|B ∩ ( + x)\| ≤ \|(2 · B) ∩ \| ≤ 9d \|B ∩ \| as desired. Next, we recall a result of Gauss concerning the intersection of a large convex body with a lattice of full rank. Lemma 3.22 Let ⊂ Rd be a lattice of full rank, let v1 , . . . , vd ∈ be a set of generators for , and let B be a convex body in Rd . Then for large R > 0, we have \|(R · B) ∩ \| = (R d + O,B,d (R d−1 )) mes(B) . \|v1 ∧ · · · ∧ vd \| Here \|v1 ∧ · · · ∧ vd \| denotes the volume of the parallelepiped with edges v1 , . . . , vd . 3 Additive geometry 132 Proof We use a “volume-packing argument”. Since has full rank, v1 , . . . , vd are linearly independent. By applying an invertible linear transformation we may assume that v1 , . . . , vd is just the standard basis e1 , . . . , ed , so that = Zd . Now let Q be the unit cube centered at the origin. Observe that the sets {x + Q : x ∈ (R · B) ∩ Zd } are disjoint up to sets of measure zero, and their union differs from √ R · B only in the d-neighborhood of the surface of R · B, which has volume O,B,d (R d−1 ). The claim follows. Remark 3.23 The task of improving the error term O,B,d (R d−1 ) for various lattices and convex bodies (e.g. Gauss’ circle problem) is a deep and important problem in number theory and harmonic analysis, but we will not discuss this issue in this book; our only concern is that the error term is strictly lower order than the main term. If is a lattice, we define a fundamental parallelepiped for to be any parallelepiped whose edges v1 , . . . , vd generate . From the above lemma we conclude that all fundamental parallelepipeds have the same volume; indeed this volume is nothing more than the covolume mes(Rd / ) of . Thus for instance mes(Rd /Zd ) = 1. By another volume-packing argument we can establish mes(Rd / )\|/ \| = mes(Rd / ) (3.12) whenever ⊆ ⊂ Rd are two lattices of full rank; see the exercises. In particular we see that the quotient group \|/ \| is finite. Yet another volume-packing argument gives the following continuous and periodic analogue of (2.8). Lemma 3.24 (Volume-packing lemma) Let ⊂ Rd be a lattice of full rank, let V be a bounded open subset of Rd , and let P be a finite non-empty set in Rd . Then \|(V − V ) ∩ ( + P − P)\| ≥ mes(V )\|P\| . mes(Rd / ) In particular, we have \|(V − V ) ∩ \| ≥ mes(V ) . mes(Rd / ) Proof Let B be the unit ball on Rd , and let R > 0 be a large number. Consider the integral of the function f (x) := 1V +y+ p (x). y∈∩(R·B) p∈P 3.5 Intersecting a convex set with a lattice 133 On the one hand we can compute this integral using Lemma 3.22 as f (x) d x = mes(V + y) Rd y∈∩(R·B) p∈P = \| ∩ (R · B)\|\|P\|\|mes(V )\| mes(B)mes(V ) = (R d + O,B,d (R d−1 ))\|P\| mes(Rd / ) On the other hand, from (3.9) we have f (x) ≤ \|(x − V ) ∩ ( + P − P)\| ≤ \|(V − V ) ∩ ( + P − P)\|. Furthermore, f (x) is only non-zero when x lies in R · B + V + P ⊂ (R + OV,P (1)) · B, which has volume R d + OV,P,d (R d−1 ). Thus f (x) d x ≤ \|(V − V ) ∩ ( + P − P)\|R d + OV,P,d (R d−1 ). Rd Combining these inequalities, dividing by R d , and taking limits as R → ∞, we obtain the result. To see the utility of this lemma, let us pause to establish the following classical result in number theory, which we will need later in this book. Let xR/Z denote the distance from x to the nearest integer. Corollary 3.25 (Kronecker approximation theorem) Let α1 , . . . , αd be real numbers, and let 0 < θ1 , . . . , θd ≤ 1/2. Then for any N > 0, we have \|{n ∈ (−N , N ) : nα j R/Z < θ j for all j = 1, . . . , d}\| ≥ N θ1 · · · θd . In particular, if N θ1 · · · θd ≤ 1, then there exists an integer 0 < n < N such that nα j R/Z ≤ θ j for all j = 1, . . . , d. Proof Apply Lemma 3.24 with := Zd , V := {(t1 , . . . , td ) + Zd : 0 < t j < θ j for all 1 ≤ j ≤ d}, and P equal to the arithmetic progression P = [0, N ) · (α1 , . . . , αd ) in Rd . Even when B is symmetric, it is possible for \|B ∩ \| to be extremely large commes(B) 2 2 pared with 2d mes(R < d / ) ; consider for instance := Z and B := {(x, y) : −1/N 2 x < 1/N ; −N < y < N }. However, if B ∩ has full rank, then we can complement the lower bound (3.14) with an upper bound: Lemma 3.26 Let be a lattice of full rank in Rd , and let B be a symmetric convex body in Rd such that the vectors in B ∩ linearly span Rd . Then \|B ∩ \| ≤ 3d d!mes(B) . 2d mes(Rd / ) (3.13) 134 3 Additive geometry This bound is with a factor of 3d /(2d + 1) of being sharp, as can be seen by the example where = Zd and B is (a slight enlargement of) the octahedron with vertices ±e1 , . . . , ±ed . Indeed this example motivates the volume-packing argument used in the proof. Proof By hypothesis, B ∩ contains a d-tuple (v1 , . . . , vd ) of linearly independent vectors. Since B ∩ is finite, we can choose v1 , . . . , vd in order to mind imize the volume mes(O) = 2d! \|v1 ∧ · · · ∧ vd \| of the octahedron with vertices ±v1 , . . . , ±vd . Since B is symmetric and convex, we see that O ⊆ B. Also O does not contain any elements of other than v1 , . . . , vd , since otherwise one could replace one of v1 , . . . , vd with this element and reduce the volume of O, a contradiction. Thus we see that the sets {x + 12 · O : x ∈ B ∩ } are all disjoint and are contained in B + 12 · O ⊆ 32 · B. Thus mes 32 · B 3d d! 1 = d \|B ∩ \| ≤ mes(B). 2 \|v1 ∧ · · · ∧ vd \| mes 2 · O Since \|v1 ∧ · · · ∧ vd \| ≥ mes(Rd / ), the claim follows. A special case of the volume-packing lemma gives Lemma 3.27 (Blichtfeld’s lemma) Let ⊂ Rd be a lattice of full rank, and let V be an open set in Rd such that mes(V ) > mes(Rd / ). Then there exists distinct x, y ∈ V such that x − y ∈ . Now let us apply Lemma 3.24 to the case V = 12 · B and P = {0}, where B is a symmetric convex body; we obtain the lower bound \|B ∩ \| ≥ mes(B) , 2d mes(Rd / ) (3.14) which is the classical Minkowski’s first theorem. The assumption of symmetry is essential. Consider for instance := Z2 and a convex set of the form B := {(x, y) : 1/3 < x < 2/3; −N < y < N } for arbitrarily large N . Theorem 3.28 (Minkowski’s first theorem) Let be a lattice of full rank, and let B be a symmetric convex body such that mes(B) ≥ 2d mes(Rd / ). Then the closure of B must contain at least one non-zero element of (in fact it contains at least two, by symmetry). If we have strict inequality, mes(B) > 2d mes(Rd / ), then we can replace the closure of B with the interior of B in the above statement. Proof Apply (3.14) to (1 + )B and let go to zero. The constant in Minkowski’s first theorem is sharp. We may apply an invertible linear transformation to set := Zd , and then the example of the cube A := 3.5 Intersecting a convex set with a lattice 135 {(t1 , . . . , td ) : −1 < t j < 1 for all j = 1, . . . , d} shows that the constant 2d cannot be improved. Nevertheless, it is possible to improve Minkowski’s first theorem by generalizing it to a “multiparameter” version as follows. Definition 3.29 (Successive minima) Let be a lattice in Rd of rank k, and let B be a convex body in Rd . We define the successive minima λ j = λ j (B, ) for 1 ≤ j ≤ k of B with respect to as λ j := inf{λ > 0 : λ · B contains k linearly independent elements of }. Note that 0 < λ1 ≤ · · · ≤ λk < ∞. Thus, for instance, if = Zd and B is the box B := {(t1 , . . . , td ) : \|t j \| < a j for all j = 1, . . . , d} for some a1 ≥ a2 ≥ · · · ≥ ad > 0, then λ j = 1/a j for j = 1, . . . , d. Note that the assumption that has rank k ensures that the λ j are both finite and non-zero. Theorem 3.30 (Minkowski’s second theorem) Let be a lattice of full rank in Rd , and let B be an symmetric convex body in Rd , with successive minima 0 < λ1 ≤ · · · ≤ λd . Then there exists d linearly independent vectors v1 , . . . , vd ∈ with the following properties: r for each 1 ≤ j ≤ d, v lies in the boundary of λ · B, but λ · B itself does not j j j contain any vectors in outside of the span of v1 , . . . , v j−1 ; r the octahedron with vertices ±v contains no elements of in its interior, j other than the origin; r we have 2d \|/(Zd · (v1 , . . . , vd ))\| λ1 · · · λd mes(B) ≤ ≤ 2d ; d! mes(Rd / ) (3.15) in particular, the sub-lattice Zd · (v1 , . . . , vd ) of has bounded index: \|/(Zd · (v1 , . . . , vd ))\| ≤ d!. (3.16) One can state (3.15) rather crudely as λ1 · · · λd mes(B) = d O(d) mes(Rd / ) thus relating the successive minima to the volume of the body B and the covolume of the lattice . Note that if B contains no non-zero elements of then λ j ≥ 1 for all j, so Minkowski’s second theorem implies Minkowski’s first theorem. Conversely, we shall see from the proof that Minkowski’s second theorem can be obtained from Minkowski’s first theorem by a non-isotropic dilation. The basis v1 , . . . , vd is 136 3 Additive geometry sometimes referred to as a directional basis for A with respect to , although one should caution that this basis does not quite generate (the index in (3.16) is bounded but not necessarily equal to 1). Proof By definition of λ1 , we may find a vector v1 ∈ such that v1 lies in the closure of λ1 · B, but that λ · B contains no non-zero elements of for any λ ≤ λ1 . By definition of λ2 , we can then find a vector v2 ∈ , linearly independent from v1 , such that v2 lies in the closure of λ2 B, but that λ · B contains no elements of outside of the span of v1 for any λ ≤ λ2 . Continuing inductively we can eventually find a linearly independent set v1 , . . . , vd in such that v j lies in the boundary of λ j · B, but λ j · A itself does not contain any vectors in outside of the span of v1 , . . . , v j−1 , for all 1 ≤ j ≤ n. The set v1 , . . . , vd is a basis of Rd ; by applying an invertible linear transformation we may assume it is the standard basis e1 , . . . , ed (this changes both B and , but one may easily verify that the conclusion of the theorem remains unchanged). In particular this forces to contain Zd , hence by (3.12) mes(Rd / ) = mes(Rd /Zd )/\|/Zd \| = 1/\|/Zd \| ≤ 1. (3.17) Let O d be the open octahedron whose vertices are ±e1 , . . . , ±ed . We need to verify that O d contains no lattice points from other than the origin. Suppose for contradiction that O d ∩ contained w = t1 e1 + · · · + t j e j where 1 ≤ j ≤ d and t j = 0. Then (1 + ε)w would be a linear combination of ±e1 , . . . , ±e j for some ε > 0. All of these points lie in the closure of λ j · B, hence w lies in the interior of λ j · B, but does not lie in the span of e1 , . . . , e j−1 . But this contradicts the construction of v j = e j . Hence O d ∩ = {0}. Next, observe that ±v j = ±e j lies on the boundary of λ j · B for each 1 ≤ j ≤ d. Thus B contains the open octahedron whose vertices are ±e1 /λ1 , . . . , ±ed /λd . d This octahedron is easily verified to have volume d!λ21 ···λd ; indeed one can rescale to the case when all the λ j are equal to 1, and then one can decompose the octahedron into 2d simplices, each of which has volume 1/d!. This establishes the lower bound in (3.15). Now we establish the upper bound in (3.15). We need the following lemma. Lemma 3.31 (Squeezing lemma) Let K be a symmetric convex body in Rd , let A be an open subset of K , let V be a k-dimensional subspace of Rd , and let 0 < θ ≤ 1. Then there exists an open subset A of K such that mes(A ) = θ k mes(A) and (A − A ) ∩ V ⊆ θ · (A − A) ∩ V . Note that we do not assume any convexity on A or A . Indeed the squeezing operation we define in the proof below does not preserve the convexity of A. 3.5 Intersecting a convex set with a lattice 137 Proof Without loss of generality we may take V = Rk , and write Rd = Rk × Rd−k . Let π : Rd → Rd−k be the orthogonal projection map, which restricts to a map π : K → π (K ). Let f : π (K ) → K be any continuous right-inverse of π ; thus for instance f (y) could be the center of mass of π −1 (y). A point w ∈ K can be written as w = (x, y), using the decomposition Rd = k R × Rd−k . Consider the map which maps w = (x, y) to θw + (1 − θ ) f (y) and set A = (A). Since both w and f (y) belong to K and K is convex, it follows that A is an open subset of K . Furthermore, the second coordinate of (w) is y as is that of f (y). By applying Cavalieri’s principle (or Fubini’s theorem) we see that mes(A ) = θ k mes(A) (the map contracts A by a factor θ with respect to V = Rk ). Consider a point v = (w) − (w ), where w = (x, y), w = (x , y ) are points from A. If v ∈ V , then the second coordinate of v is zero, which means y = y . Then by the definition of , v = θ(w − w ). Thus v ∈ θ · (A − A), concluding the proof of Lemma 3.31. We apply the squeezing lemma iteratively, starting with A0 := open sets A1 , . . . , Ad−1 ⊆ A0 such that λj j mes(A j ) = mes(A j−1 ) λ j+1 λd 2 · B, to create and (A j − A j ) ∩ R j ⊆ λj · (A j−1 − A j−1 ) ∩ R j λ j+1 for all 1 ≤ j ≤ d − 1, where R j is the span of e1 , . . . , e j . In every application of the squeezing lemma, A0 plays the role of the mother set K . Using the definition of A0 , it is easy to check that mes(Ad−1 ) = λ1 · · · λd 2−d mes(B). (3.18) Furthermore, by induction one can show (Ad−1 − Ad−1 ) ∩ R j ⊆ λj · (A j−1 − A j−1 ) ∩ R j . λd On the other hand, A j−1 ⊂ A0 = (λd /2) · B. Since B is symmetric, λd · B = λd · B. It follows that 2 λd 2 ·B− (Ad−1 − Ad−1 ) ∩ R j ⊂ λ j · B ∩ R j for all 1 ≤ j ≤ d. By the definition of the successive minima, λ j · B ∩ R j does not contain any lattice point in , except for those in R j−1 . This implies that Ad−1 − Ad−1 does 138 3 Additive geometry not contain any point in other than the origin. Applying Blichtfeld’s lemma, we conclude that mes(Ad−1 ) ≤ mes(Rd / ), which when combined with (3.18) gives the upper bound in (3.15). We now give several applications of this theorem. First we “factorize” a convex body B as the finitely overlapping sum of a subset of and and a dilate of a small convex body B , up to some scaling factors of O(d) O(1) : Lemma 3.32 Let B be a symmetric convex body in Rd , and let be a lattice in Rd . Then there exists a symmetric convex body B ⊆ B such that B contains no nonzero elements of , and such that B ⊆ O(d 3/2 ) · B + ((O(d 3/2 ) · B) ∩ . In particular, the projection of B in Rd / is contained in the projection of O(d 3/2 ) · B . Furthermore, we have the bounds mes(B) mes(B) ≤ mes(B ) ≤ O(1)d . O(d)5d/2 \|B ∩ \| \|B ∩ \| (3.19) Proof By using John’s√theorem and an invertible linear transformation we may assume that Bd ⊆ B ⊆ d · Bd , where Bd is the unit ball. We may assume that the vectors in B ∩ generate , since otherwise we could replace by the lattice generated by B ∩ . Let us temporarily assume that has full rank, and thus that the linear span of B ∩ is Rd . Thus if we let λ1 ≤ · · · ≤ λd be the successive minima of B, then we have λ j ≤ 1 for all j. Now we take a directional basis v1 , . . . , vd of , and let B be the open octahedron with vertices ±v j ; this octahedron then contains no non-zero elements of , and is also contained in B (since ±v j /λ j already lies on the boundary of B). Observe that d · B contains a parallelepiped with edges v1 , . . . , vd , and hence d · B + = Rd . Thus B ⊆ d · B + ((B − d · B ) ∩ ) ⊆ d · B + (((d + 1) · B) ∩ ) as desired (with about d 1/2 room to spare). In particular we have mes(B) ≤ mes(d · B )\|(d + 1) · B ∩ \| ≤ (d(4d + 5))d mes(B )\|B ∩ \| thanks to (3.10); this proves the lower bound in (3.19) (with a factor of d d/2 to spare). Conversely, the sets {x + 12 · B : x ∈ B ∩ } are disjoint (since B contains no non-zero elements of ) and contained in 2 · B, hence 1 \|B ∩ \|mes · B ≤ mes(2 · B) 2 3.5 Intersecting a convex set with a lattice 139 which gives the upper bound in (3.19). This concludes the proof when has full rank. Now suppose that has rank r < d, then after a rotation we may assume that is contained in Rr × {0} ⊂ Rr × Rd−r . The point is that the behavior in the d − r dimensions orthogonal to Rr is rather trivial and can be easily dealt with as follows. Let B˜ ⊂ Rr be the intersection of B with Rr × {0}, identifying Rr × {0} with Rr in the usual manner. Then by John’s theorem we have the inclusions √ 1 ˜ · ( B × Bd−r ) ⊆ B ⊆ d · ( B˜ × Bd−r ). 2 ˜ and then defining Applying the previous arguments to B˜ to obtain a set B˜ ⊆ B, 1 ˜ B := 2 · ( B × Bd−r ), we can verify the claim in this case (losing some additional factors of d 1/2 and d d/2 ); we omit the details. In this theorem, we did not use the full strength of Minkowski’s second theorem (in particular we did not need the upper bound). The notion of a directional vector is, however, useful. As another consequence of Minkowski’s second theorem, we show how to find large proper progressions inside sets of the form B ∩ . Lemma 3.33 Let B be a convex symmetric body in Rd , and let be a lattice in Rd . Then there exists a proper progression P in B ∩ of rank at most d such that \|P\| ≥ O(d)−7d/2 \|B ∩ \|. Proof Applying John’s theorem (Theorem 3.13) and (3.10) followed by a linear transformation, we may reduce to the case where B is the unit ball B = Bd in Rd , provided that we also reduce the 7d/2 exponent to 3d. We may assume that B ∩ spans Rd , since otherwise we may restrict B to the linear span of B ∩ , which is then isomorphic to a Euclidean space of some lower dimension. In particular this means has full rank, and that the successive minima 0 < λ1 ≤ · · · ≤ λd of B with respect to cannot exceed 1. Let v1 , . . . , vd ∈ ∩ B be the corresponding directional basis. Let Q denote the parallelepiped Q := {t1 v1 + · · · + td vd : 0 ≤ t j < 1/2 for all j ∈ [1, d]}. By (3.16), Since each translate of Q − Q is a fundamental domain for Zd · (v1 , . . . , vd ), it contains at most d! elements of . By Lemma 2.14, we can cover B by at most mes(B+Q) translates of Q − Q, and thus mes(Q) \|B\| ≤ d! mes(B + Q) . mes(Q) 3 Additive geometry 140 Since the v1 , . . . , vd lie in the unit ball B, we see that Q ⊆ d2 · B and hence B + Q ⊆ ( d2 + 1) · B. Crudely bounding d! = O(d d ), we thus conclude that \|B ∩ \| ≤ O(d)2d /mes(Q). From (3.15) we have λ1 · · · λd ≤ O(1)d mes(Zd / ) ≤ O(1)d mes(Q) and thus \|B ∩ \| ≤ O(d)2d /λ1 · · · λd . The claim now follows by setting P := [−N , N ] · (v1 , . . . , vd ), where N j := 1/2dλ j for j ∈ [1, d]; note that one can easily verify that P is contained in B ∩ . We now give an alternative approach that gives results similar to Lemma 3.33. We first need a lemma to modify the directional basis given by Minkowski’s second theorem (which only spans a sub-lattice of , see (3.16)) into a genuine basis. Theorem 3.34 (Mahler’s theorem) Let be a lattice of full rank in Rd , and let B be an symmetric convex body in Rd , with successive minima 0 < λ1 ≤ · · · ≤ λd . Let v1 , . . . , vd be a directional basis for . Then there exists a basis w1 , . . . , wd of such that w1 lies in the closure of λ1 · B, and wi lies in the closure of iλ2i · B for all 2 ≤ i ≤ d. Furthermore, if Vi is the linear span of v1 , . . . , vi , then w1 , . . . , wi forms a basis for ∩ Vi . The basis w1 , . . . , wd is sometimes known as a Mahler basis for . Proof We choose w1 := v1 ; clearly w1 forms a basis for ∩ V1 . Now suppose inductively that 2 ≤ i ≤ d and w1 , . . . , wi−1 have already been chosen with the desired properties. The lattice ∩ Vi has one higher rank than ∩ Vi−1 and hence there exists a vector wi in ∩ (Vi \Vi−1 ) which, together with ∩ Vi−1 , generates ∩ Vi ; in particular, w1 , . . . , wi will generate ∩ Vi . Since v1 , . . . , vi linearly span Vi , we may write wi = t1 v1 + · · · + ti−1 vi−1 + ti vi for some real numbers t1 , . . . , ti with ti = 0. Since vi lies in ∩ Vi−1 + W , we must have ti = ±1/n for some integer n. If \|ti \| = 1, then ∩ Vi is generated by ∩ Vi−1 and vi , and we can take wi := vi . Thus we may assume \|ti \| ≤ 1/2. Also, by subtracting integer multiples of v1 , . . . , vi−1 from wi if necessary (which will not affect the fact that ∩ Vi is generated by ∩ Vi−1 and wi ) we may assume that \|t j \| ≤ 1/2 for all 1 ≤ j < i. But since each v j lies in the closure of λ j · B and hence λi · B, we conclude by convexity that wi lies in the closure of iλ2i · B, and so we can continue the iterative construction. Setting i = d we obtain the remaining claims in the theorem. As an application we give 3.5 Intersecting a convex set with a lattice 141 Corollary 3.35 Let be a lattice of full rank in Rd . Then there exists linearly independent vectors w1 , . . . , wd ∈ which generate , and such that mes(Rd / ) = \|w1 ∧ · · · ∧ wd \| ≥ (d −3d/2 )\|w1 \| · · · \|wd \|. (3.20) Proof Let w1 , . . . , wd be a Mahler basis for with respect to the unit ball B, and let λ1 , . . . , λd be the successive minima. Then by Theorem 3.34 we have \|w1 \| · · · \|wd \| ≤ λ1 d iλi i=2 2 . Applying (3.15) we obtain \|w1 \| · · · \|wd \| ≤ 2d! mes(Rd / ). mes(B) On the other hand, from (3.8) we have mes(B) = (3/2)d 2d = (2πe + o(1))d/2 d −d/2 . (d/2 + 1) Crudely bounding d! = O(d d ), the claim follows. As a consequence, we can give a “discrete John’s theorem” to characterize the intersection of a convex symmetric body with a lattice. Lemma 3.36 (Discrete John’s theorem) Let B be a convex symmetric body in Rd , and let be a lattice in Rd of rank r . Then there exists a r -tuple w = (w1 , . . . , wr ) ∈ r of linearly independent vectors in and and a r -tuple N = (N1 , . . . , Nr ) of positive integers such that (r −2r · B) ∩ ⊆ (−N , N ) · w ⊆ B ∩ ⊆ (−r 2r N , r 2r N ) · w. Notice that the fact (−N , N ) · w ⊆ B ∩ is similar to the conclusion of Lemma 3.33. However, the generalized arithmetic progression in Lemma 3.33 has higher density. Proof We first observe, using John’s theorem and an invertible linear transformation, that we may assume without loss of generality that Bd ⊆ B ⊆ d · Bd , where Bd is the unit ball in Rd . We may assume that has full rank r = d, for if r < d then we may simply restrict B to the linear span of , which can then be identified with Rr . We may assume d ≥ 2 since the claim is easy otherwise. Now let w = (w1 , . . . , wd ) be as in Lemma 3.35. For each j, let L j be the least integer greater than 1/d\|w j \|. Then from the triangle inequality we see that \|l1 w1 + · · · + ld wd \| < 1 whenever \|l j \| < L j , and so (−L , L) · w is contained in Bd and hence in B. 3 Additive geometry 142 Now let x ∈ B ∩ . Since w generates , we have x = l1 w1 + · · · + ld wd for some integers l1 , . . . , ld ; since B ⊆ d · Bd , we have \|x\| ≤ d. Applying Cramer’s rule to solve for l1 , . . . , ld and (3.20), we have \|l j \| = = \|x ∧ w1 · · · w j−1 ∧ w j+1 ∧ wd \| \|x\|\|w1 \| · · · \|wd \| ≤ \|w1 ∧ · · · ∧ wd \| \|w j \|\|w1 ∧ · · · ∧ wd \| \|x\| mes(Rd / ) 2d · d! ≤ , \|w j \| \|w j \| which is certainly at most d 2d L j . It follows that x ∈ (−d 2d L , d 2d L) · w, which is what we wanted to prove. A more-or-less identical argument gives the inclusion (d −2d · B) ∩ ⊆ (−L , L) · w. It would be of interest to see if the constant r 2r could be significantly improved here, for instance to e O(r ) or even r O(1) . Progress on this issue may well have applications to improvements for Freiman’s theorem (see Chapter 5), which can be viewed as a variant of the above theorem in which the set B ∩ is replaced by a more general set of small doubling. Exercises 3.5.1 3.5.2 3.5.3 3.5.4 3.5.5 3.5.6 3.5.7 Prove (3.12). Let α be an irrational number, and let I be any open interval in R. Show that Z · α and I + Z have non-empty intersection. (In other words, the integer multiples of α are dense in R/Z.) Let be a lattice in Rd , and let A be a convex body (possibly asymmetric). Show that σ [A ∩ ] ≤ O(1)d . Let v1 , . . . , vd be any vectors in a lattice ⊂ Rd of full rank. Show that \|v1 ∧ · · · ∧ vd \| is an integer multiple of the covolume mes(Rd / ). Let be a lattice of full rank in Rd , let B be a symmetric convex body, and let v1 , . . . , vd be a directional basis with successive minima λ1 ≤ · · · ≤ λd . Let O be the open octahedron with vertices ±v j /λ j . Show that O ⊆ B ⊆ O(d)d · O. Thus Minkowski’s second theorem can be used to give a rather weak version of John’s theorem. Let be a lattice of full rank in Rd , let B be a symmetric convex body, and let λ1 ≤ · · · ≤ λd be the successive minima of B. Establish the bounds 1 1 −O(d) O(d) O(d) ≤ \|B ∩ \| ≤ O(d) . max 1, max 1, λi λi 1≤i≤d 1≤i≤d (3.21) Generalize Lemma 3.32 and Lemma 3.36 to the case when B is an asymmetric convex body. 3.6 Progressions and proper progressions 3.5.8 Let A be a bounded open subset of Rd , and let B, C be open subsets of A. Prove that mes((B − B) ∩ (C − C)) ≥ 3.5.9 143 mes(B)mes(C)mes(A) . mes(A − B)mes(A − C) (Hint: use the volume-packing argument to locate a large set of the form (x + B) ∩ (y + C) where x ∈ A − B and y ∈ A − C.) Let B the the unit ball in R5 , and let be the lattice generated by the five basis vectors e1 , . . . , e5 and by 12 (e1 + · · · + e5 ). Show that in this case the directional basis for does not actually generate . 3.6 Progressions and proper progressions In this section we work in a fixed additive group Z , which may or may not be torsion-free. Recall from Definition 0.2 that a progression P = a + [0, N ] · v is proper if the map n → n · v is injective on [0, N ]. Not all progressions are proper; however it turns out that, just as John’s theorem (Theorem 3.13) shows that all convex sets are in some sense comparable to ellipsoids, all progressions are comparable to proper progressions. This is most obvious in the rank 1 case, in which every arithmetic progression is equal (as a set) to a proper arithmetic progression: Lemma 3.37 Let a + [0, N ] · v be an arithmetic progression in an additive group Z . Then there exists an n > 0 such that a + [0, n) · v is a proper arithmetic progression and a + [0, n) · v = a + [0, N ] · v. Proof If a + [0, N ] · v is already proper, then we are done. Otherwise, there exist distinct n 1 , n 2 ∈ [0, N ] such that a + n 1 · v = a + n 2 · v. In particular, there exists n ∈ [1, N ] such that n · v = 0. Let n be the least integer in [1, N ] with this property. Then a + [0, n) · v is necessarily proper, and by the Euclidean algorithm it is clear that a + [0, n) · v = a + [0, N ] · v. We now consider the higher rank case; as with John’s theorem, the constants will deteriorate worse than exponentially in d. We first show the easier of the two containments, namely that every progression contains a large proper progression of equal or lesser rank. Theorem 3.38 Let P be a progression of rank d in an additive group Z . Then P contains a proper progression of rank at most d and volume at least O(d)−5d \|P\|. 3 Additive geometry 144 Remark 3.39 For a result of similar flavor (but proven by completely different methods), see Theorem 4.42 below. Note that the d = 1 case already follows from Lemma 3.37 (with a constant of 1 instead of O(d)−5d ). Proof The idea is to pass to a convex body, apply Lemma 3.32 to obtain a “proper” subset of this body, and then use Lemma 3.33 to pass back to a progression. By translating and enlarging P slightly we may assume P = [−N , N ] · v. We may assume that none of the components N j of N are equal to 0 or 1, since otherwise we could refine P by at worst a factor of 3d to eliminate those dimensions. Now consider the set := {n ∈ Zd : n · v = 0}, which is clearly a sub-lattice of Zd , and let A be the symmetric convex box A := {(x1 , . . . , xd ) ∈ Rd : −N j ≤ x j ≤ N j for all 1 ≤ j ≤ d}. By Lemma 3.32, we may find a symmetric convex subset A of A such that A − A is disjoint from − {0}, and such that A ⊂ O(d)3/2 · A + for some x ∈ Rd . From Corollary 3.15, we thus see that A can be covered by O(d)3d/2 translates of 12 · A + . Since [−N , N ] = A ∩ Zd and ⊆ Zd , we conclude that [−N , N ] can be covered by O(d)3d/2 sets of the form [( 12 · A + x) ∩ Zd ] + . Taking inner products with v, we conclude that P = [−N , N ] · v can be covered by O(d)3d/2 sets of the form [( 12 · A + x) ∩ Zd ] · v. By the pigeonhole principle, there must thus exist an x such that 3d/2 1 1 d \|P\| 2 · A +x ∩Z ≥ d and hence by (3.9) 3d/2 1 \|A ∩ Z \| ≥ \|P\|. d d We now apply Lemma 3.33 to find a proper progression P˜ ⊆ A ∩ Zd ⊆ [0, N ] of rank at most d such that 5d 1 −7d/2 d ˜ \| P\| ≥ O(d) \|A ∩ Z \| ≥ \|P\|. d The set P˜ · v is then clearly a progression of rank at most d contained in P; it is ˜ The proper since A − A is disjoint from − {0}, so in particular \| P˜ · v\| = \| P\|). claim follows. Now we show the more difficult containment, that every progression can be contained inside a proper progression of equal or lesser rank, but somewhat larger volume. 3.6 Progressions and proper progressions 145 Theorem 3.40 Let P be a progression of rank d in an additive group Z . Then P is contained in a proper progression Q of rank at most d and volume at most 3 d C0 d \|P\| for some absolute constant C0 > 0. Also, Q is contained in a translate of 2 d C0 d P. If d ≥ 2 and P is not proper, then Q can be chosen to have rank at most d − 1. Finally, if Z is torsion-free and P is symmetric, then one can ensure that Q is symmetric also. Remark 3.41 Theorems of this type first appeared in the literature in , and later in some unpublished work of Gowers–Walters and Ruzsa. The version we give here is taken from . 3 Comparison with Theorem 3.38 suggests that the factor d C0 d is probably not best possible, but we do not know what the correct constant here should be. This theorem can be thought of as the analogue of Corollary 3.8 or Corollary 3.9, but for progressions rather than finitely generated additive groups. Proof This claim is analogous to the basic linear algebra statement that every linear space spanned by d vectors is equal to a linear space with a basis of at most d vectors. Recall that the proof of that fact proceeds by a descent argument, showing that if the d spanning vectors were linearly dependent, then one could exploit that dependence to “drop rank” and span the same linear space with d − 1 vectors. Our proof of Theorem 3.40 shall be based on a similar strategy. We shall work only in the case when Z is torsion-free; the general case is proven similarly but contains a few additional technicalities, and we leave it as an exercise (Exercise 3.6.3). We induce on d. When d = 1 the claim follows from Lemma 3.37. Now suppose inductively that d ≥ 2, and the claim has already been proven for d − 1 (for arbitrary groups Z and arbitrary progressions P). Let P = a + [0, N ] · v be a progression in Z of rank d, where N = (N1 , . . . , Nd ) and v = (v1 , . . . , vd ); we may translate P so that the base point a equals 0. If P is proper, then we are done. Similarly, if one of the N j is equal to zero, then we are done by induction hypothesis. Suppose instead that P is not proper and all the N j are at least 1; then there exist distinct n, n ∈ [0, N ] such that n · v = n · v. If we then let 0 ⊆ Zd denote the lattice {m ∈ Zd : m · v = 0}, then we see that 0 ∩ [−N , N ] contains at least one non-zero element, namely n − n. Let m = (m 1 , . . . , m d ) be a non-zero element of 0 ∩ [−N , N ], thus m 1 · v1 + · · · + m d · vd = 0. (3.22) We may assume without loss of generality that m is irreducible in 0 . Since Z is torsion-free, this also implies that m is irreducible in Zd (i.e. that the m 1 , . . . , m d have no common divisor) unless Z is torsion-free. The strategy shall be to contain 3 Additive geometry 146 P inside a progression Q of rank d − 1 and size 2 \|Q\| ≤ d O(d ) \|P\|, (3.23) such that Q is contained in a translate of d O(d) P. If we can achieve this, then by the induction hypothesis we can contain Q inside a proper progression R of rank at most d − 1 and cardinality 3 2 \|R\| ≤ (d − 1)C0 (d−1) (O(d)) O(d ) \|P\| 2 and which is contained in a translate of d C0 (d−1) d O(d) P. If C0 is sufficiently large, we will have completed the induction. It remains to cover P by a progression of rank at most d − 1 with the bound (3.23) and contained in a translate of d O(d) P. Observe that m lies in [−N , N ], so the rational numbers m 1 /N1 , . . . , m d /Nd lie between −1 and 1. Without loss of generality we may assume that m d /Nd has the largest magnitude, thus \|m d \|/Nd ≥ \|m j \|/N j (3.24) for all 1 ≤ j ≤ d. By replacing vd with −vd if necessary, we may also assume that m d is positive. To exploit the cancellation in (3.22) we introduce the rational vector q ∈ 1 · Zd−1 by the formula md m1 m d−1 q := − , . . . , − . md md Since m is irreducible in Zd , we see, for any integer n, that n · q lies in Zd−1 if and only if n is a multiple of m d , because (m 1 , . . . , m d ) is irreducible in Zd . Next, let ⊂ Rd−1 denote the lattice := Zd−1 + Z · q. Since q is rational, this is indeed a lattice; since it contains Zd−1 , it is certainly full rank. We define the homomorphism f : → Z by the formula f ((n 1 , . . . , n d−1 ) + n d q) := (n 1 , . . . , n d ) · v; the condition (3.22) ensures that this homomorphism is indeed well defined, in the sense that different representations v = (n 1 , . . . , n d−1 ) + n d q of the same vector v ∈ give the same value of f (v). We also let B ⊆ Rd−1 denote the convex symmetric body B := {(t1 , . . . , td−1 ) ∈ Rd−1 : −3N j < t j < 3N j for all 1 ≤ j ≤ d − 1}. We now claim the inclusions P ⊆ f (B ∩ ) ⊆ 5P − 5P. 3.6 Progressions and proper progressions 147 To see the first inclusion, let n · v ∈ P for some n ∈ [0, N ], then we have n · v = f ((n 1 , . . . , n d−1 ) + n d q); from (3.24) we see that the jth coefficient of (n 1 , . . . , n d−1 ) + n d q has magnitude at most 3N j , and thus n · v lies in f (B ∩ ) as claimed. To see the second inclusion, let (n 1 , . . . , n d−1 ) + n d q be an element of B ∩ . By subtracting if necessary an integer multiple of m d from n d (and thus adding integer multiples of m 1 , . . . , m d−1 to n 1 , . . . , n d−1 ) we may assume that \|n d \| ≤ \|m d \|/2. By (3.24) and the definition of B, this forces \|n j \| ≤ 5N j for all 1 ≤ j ≤ d, and hence f ((n 1 , . . . , n d−1 ) + n d q) = (n 1 , . . . , n d ) · v ⊆ [−5N , 5N ] · v = 5P − 5P. Next, we apply Theorem 3.36 to find vectors w1 , . . . , wd−1 ∈ and M1 , . . . , Md−1 such that (−M, M) · w ⊆ B ∩ ⊆ (−d O(d) M, d O(d) M) · w. Applying the homomorphism f , we obtain (−M, M) · f (w) ⊆ f (B ∩ ) ⊆ (−d O(d) M, d O(d) M) · f (w) where f (w) := ( f (w1 ), . . . , f (wd−1 ). Observe that (−d O(d) M, d O(d) M) · f (w) is a progression of rank d − 1 which contains f (B ∩ ) and hence contains P. Furthermore, by two applications of Lemma 3.10 we have 2 \|(−d O(d) M, d O(d) M) · f (w)\| ≤ (O(d)) O(d ) \| f (B ∩ )\| 2 ≤ (O(d)) O(d ) \|5P − 5P\| 2 ≤ (O(d)) O(d ) O(1)d \|P\| which proves (3.23). Also, since (−M, M) · f (w) is contained in f (B ∩ ), which is contained in 5P − 5P, which is a translate of 10P, we see that (−d O(d) M, d O(d) M) · f (w) is contained in a translate of d O(d) P. This completes the induction and proves the theorem. When P is symmetric, one can easily modify the above argument to ensure that all progressions in the above construction are also symmetric; we leave this modification to the interested reader. Exercises 3.6.1 Let P = a + [0, N ] · v be a progression of rank d in some additive group Z , and let := {n ∈ Zd : n · v = 0} be the associated sub-lattice of Zd . Prove the inequalities \|[0, N ]\| \|[0, N ]\| ≤ \|[−N , N ] ∩ \| ≤ 3d . \|P\| \|P\| Thus the ratio between the volume and cardinality of a progression P is essentially controlled by the quantity \|[−N , N ] ∩ \|. (Hints: for the lower 148 3.6.2 3.6.3 3.6.4 3.6.5 3.6.6 3 Additive geometry bound, first use Cauchy–Schwarz to obtain a lower bound for {(n, n ) ∈ [0, N ] : n · v = n · v}. For the upper bound, consider the multiplicity of the map f : [−N , 2N ] → Z defined by f (n) := n · v.) Let [0, N ] be a box in Zd , and let be a sub-lattice of Zd . Show that \|[−k N , k N ] ∩ \| ≤ (2k)d \|[−N , N ] ∩ \| for all integers k ≥ 1. Prove Theorem 3.40 in the case when Z is not necessarily torsion-free. (The main new difficulty is that the vector m is not always irreducible in Zd ; in such a case one will have to “quotient out” a finite cyclic group from P before proceeding with the rest of the argument. However, this will only introduce additional factors of C d into the inductive bound (3.23), which is acceptable.) Note that the second part of the Theorem does not extend to the torsion case, as can already be seen by considering P = Z = Z2 . Prove an extension of Theorem 3.40 in the torsion-free case in which one requires that k Q is also proper for some fixed constant k ≥ 1 (of course, the bounds on Q will depend on k). Note that the torsion-free hypothesis is now essential, as can be seen by considering the case when P = [1, N ] · 1 in Z N . Let N1 , N2 , a1 , a2 be positive integers such that 0 < a2 < N1 /5 and 0 < a1 < N2 /5, and a1 , a2 are coprime. Use the Chinese remainder theorem to show the inclusion 1 4 (a1 N1 + a2 N2 ), (a1 N1 + a2 N2 ) ⊆ [0, (N1 , N2 )] · (a1 , a2 ). 5 5 Conclude that if P is any progression of rank 2 in the integers of dimensions N1 , N2 and steps v1 , v2 with 0 < v2 < N1 /5 and 0 < v1 < N2 /5, then P contains a proper arithmetic progression of length 3(N1 v1 + N2 v2 )/5gcd(v1 , v2 ) and spacing gcd(v1 , v2 ). Let A be an additive set in an ambient group Z . Show that there exists d = O(log \|A\|) and distinct elements v1 , . . . , vd ∈ A such that the cube [0, 1]d · (v1 , . . . , vd ) has cardinality at least 14 \|A\|. (Hint: Using (2.21), show that if S is any additive set in Z such that \|S\| < 14 \|A\|, then there exists a ∈ A such that \|S ∪ (S + a)\| ≥ 32 \|S\|. Then use the greedy algorithm.) 4 Fourier-analytic methods In Chapter 1 we have already seen the power of the probabilistic method in additive combinatorics, in which one understands the additive structure of a random object by means of computing various averages or moments of that object. In this chapter we develop an equally powerful tool, that of Fourier analysis. This is another way of computing averages and moments of additively structured objects; it is similar to the probabilistic method but with an important new ingredient, namely that the quantities being averaged are now “twisted” or “modulated” by some very special complex-valued phase functions known as characters. This gives rise to the concept of a Fourier coefficient of a set or function, which measures the bias that object has with respect to a certain character. These coefficients serve two major purposes in this theory. Firstly, one can exploit the orthogonality between different characters to obtain non-trivial bounds on these coefficients; this orthogonality plays a role somewhat similar to the role of independence in probability theory. Secondly, Fourier coefficients are very good at controlling the operation of convolution, which is the analog of the sum set operation, but for functions instead of sets. Because of this, the Fourier transform is ideal for studying certain arithmetic quantities, most notably the additive energy introduced in Definition 2.8. Using Fourier analysis, one can essentially divide additive sets A into two classes. At one extreme are the pseudo-random sets, whose Fourier transform is very small (except at 0); we shall introduce the linear bias Au and the ( p) constants to measure this pseudo-randomness. Such sets are very “mixing” with respect to set addition (or to locating progressions of length three), and as the terminology implies, they behave more or less like random sets. At the other extreme are the almost periodic sets, which include arithmetic progressions, Bohr sets, and other sets with small doubling constant or large additive energy. The behavior of these sets with respect to set addition or progressions of length three is almost completely described by a rather small spectrum Specα (A), defined as the set of frequencies where the Fourier transform of 1 A is large. We shall rely on this 149 150 4 Fourier-analytic methods dichotomy between randomness and structure in a number of ways, most strikingly in proving Roth’s celebrated theorem (which we discuss in Chapter 10) that subsets of integers of positive upper density contain infinitely many progressions of length 3. (Progressions of higher length cannot be treated by linear Fourier techniques, requiring either higher order Fourier analysis or other approaches; see Chapter 11.) Fourier analysis can be performed on any additive group Z (and even on nonabelian groups). However, we shall only need this transform on finite groups, where the theory is slightly simpler technically. Thus we shall restrict our attention exclusively to the finite case. The cases Z = Z, Z = R/Z, and Z = R are also of importance to additive combinatorics (in particular leading to the Hardy– Littlewood circle method in analytic number theory), but it turns out that the finite Fourier theory forms an acceptable substitute for these infinite Fourier theories in our applications. 4.1 Basic theory Let Z be a finite additive group (for instance, Z could be a cyclic group Z = Z N ). In this section we recall the basic theory of the finite Fourier transform on such groups. Fourier analysis relies on the duality between a group Z and its Pontryagin dual Zˆ , which can be defined as the space of homomorphisms from Z to the circle group R/Z. In the case of finite groups, it turns out that a group Z and its Pontryagin dual Zˆ are always isomorphic, and so it shall be convenient to identify the two in order to simplify the theory slightly. This can be done by means of a non-degenerate bilinear form: Definition 4.1 (Bilinear forms) A bilinear form on an additive group Z is a map (ξ, x) → ξ · x from Z × Z to R/Z, which is a homomorphism in each of the variables ξ, x separately. We say that the form is non-degenerate if for every nonzero ξ the map x → ξ · x is not identically zero, and similarly for every non-zero x the map ξ → ξ · x is not identically zero. We say the form is symmetric if ξ · x = x · ξ. Examples 4.2 If Z is a cyclic group Z N then the bilinear form x · ξ := xξ/N is symmetric and non-degenerate. If Z is a standard vector space F n over a finite field F, then the bilinear form (x1 , . . . , xn ) · (ξ1 , . . . , ξn ) := φ(x1 ξ1 + · · · + xn ξn ) is symmetric and non-degenerate whenever φ : F → R/Z is any non-trivial homomorphism from F to R/Z (e.g. if F = Z p we can take φ(x) := x/ p). This particular choice has the useful additional property that aξ · x = ξ · ax for all a ∈ F and x, ξ ∈ Z . 4.1 Basic theory 151 Lemma 4.3 (Existence of bilinear forms) Every finite additive group Z has at least one non-degenerate symmetric bilinear form. Proof From Corollary 3.8 we know that every finite additive group is the direct sum of cyclic groups. We have already seen in Example 4.2 that each cyclic group has a symmetric non-degenerate bilinear form. Finally, observe that if Z 1 and Z 2 have symmetric non-degenerate bilinear forms, then the direct sum Z 1 ⊕ Z 2 also has a symmetric non-degenerate bilinear form, defined by (ξ1 , ξ2 ) · (x1 , x2 ) := ξ1 · x1 + ξ2 · x2 . The claim follows. Remark 4.4 A given additive group Z generally has multiple bilinear forms (see Exercise 4.1.10), but from the point of view of Fourier analysis they are all equivalent1 . The symmetry property has some minor aesthetic advantages but is not essential to the Fourier theory, as the physical space variable and the frequency space variable usually play completely different roles. Henceforth we fix a finite additive group Z , equipped with a non-degenerate symmetric bilinear form ξ · x; in practice we shall usually use one of the two examples from Example 4.2. To perform Fourier analysis, it will be convenient to adopt the following “ergodic” notation. Let C Z denote the space of all complex-valued functions f : Z → C. If f ∈ C Z , we define the mean or expectation of f to be the quantity 1 E Z ( f ) = Ex∈Z f (x) := f (x). \|Z \| x∈Z Similarly, if A ⊆ Z , we define the density or probability of A as P Z (A) = Px∈Z (x ∈ A) := E Z (1 A ) = \|A\| . \|Z \| We can generalize this notation to other finite non-empty domains than Z , thus 1 for instance Ex∈A,y∈B f (x, y) := \|A\|\|B\| x∈A,y∈B f (x, y). This notation not only suggests the connections between Fourier analysis, ergodic theory, and probability, but is also useful in concealing from view a number of normalizing powers of \|Z \| which would otherwise clutter the estimates. Generally, we shall use this ergodic notation for the physical variable, but use the discrete notation ξ ∈Z f (ξ ) and \|A\| (without the normalizing \|Z \| factor) for the frequency variable. We shall also rely 1 One way of viewing this is that the identification between Zˆ and Z is non-canonical, and one should really be placing the frequency variable in Zˆ instead of Z . This is ultimately the more correct viewpoint; however since we shall usually be working in very concrete situations such as cyclic groups Z N , where one does have a standard identification, we have chosen to rely on the bilinear form approach here rather than the abstract approach. 4 Fourier-analytic methods 152 heavily on the exponential map e : R/Z → C, defined by e(θ ) := e2πiθ . (4.1) The following two orthogonality properties form the foundation for Fourier analysis. Lemma 4.5 (Orthogonality properties) For any ξ, ξ ∈ Z we have Ex∈Z e(ξ · x)e(ξ · x) = I(ξ = ξ ) and for any x, x ∈ Z we have e(ξ · x)e(ξ · x ) = \|Z \|I(x = x ). ξ ∈Z Proof We prove the first identity only, as the second is similar. Since e(ξ · x)e(ξ · x) = e((ξ − ξ ) · x), it will suffice to show the claim in the ξ = 0 case, i.e. it suffices to show Ex∈Z e(ξ · x) = I(ξ = 0). This is clear in the case ξ = 0. If ξ = 0, then by non-degeneracy there exists h ∈ Z such that e(ξ · h) = 1. Shifting x by h we then have Ex∈Z e(ξ · x) = Ex∈Z e(ξ · (x + h)) = e(ξ · h)Ex∈Z e(ξ · x) and hence Ex∈Z e(ξ · x) = 0 = I(ξ = 0) as desired. For every ξ ∈ Z , we can define the associated character eξ ∈ C Z by eξ (x) := e(ξ · x). The above lemma then shows that the eξ are an orthonormal system in C Z , with respect to the complex Hilbert space structure f, g C Z := E Z ( f g) = Ex∈Z f (x)g(x). Since the number \|Z \| of characters equals the dimension \|Z \| of the space, we see that this system is in fact a complete orthonormal system. This motivates Definition 4.6 (Fourier transform) If f ∈ C Z , we define the Fourier transform fˆ ∈ C Z by the formula fˆ(ξ ) := f, eξ C Z = Ex∈Z f (x)e(ξ · x). We refer to fˆ(ξ ) as the Fourier coefficient of f at the frequency (or mode) ξ . Since the eξ are a complete orthonormal basis, we have the Parseval identity 1/2 2 1/2 2 (E Z \| f \| ) = \| fˆ(ξ )\| (4.2) ξ ∈Z 4.1 Basic theory the Plancherel theorem f, g C Z = fˆ(ξ )gˆ (ξ ) 153 (4.3) ξ ∈Z and the Fourier inversion formula f = fˆ(ξ )eξ . (4.4) ξ ∈Z In particular we see that two functions are equal if and only if their Fourier coefficients match at every frequency. In other words, the Fourier transform is a bijection from C Z to C Z (in fact it is a unitary isometry, thanks to (4.2), (4.3)). From Lemma 4.5 we see that the Fourier coefficients of a character eξ are just a Kronecker delta function: eξ (ξ ) = I(ξ = ξ ). ˆ ) = I(ξ = 0). In particular 1(ξ A special role in the additive theory of the Fourier transform is played by the zero frequency ξ = 0. This is because the zero Fourier coefficient is same concept as expectation: fˆ(0) = f, 1 C Z = E Z ( f ). (4.5) If S is any subset of Z , define the orthogonal complement S ⊥ ⊆ Z of S to be the set S ⊥ := {ξ ∈ Z : ξ · x = 0 for all x ∈ S}. One can easily verify that S ⊥ is a subgroup of Z . Also one has the pleasant identity 1 G = P Z (G)1G ⊥ (4.6) whenever G is a subgroup; see Exercise 4.1.6. Applying (4.2) we see in particular that \|G\|\|G ⊥ \| = \|Z \|. (4.7) We now introduce the fundamental notion of convolution, which links the Fourier transform to the theory of sum sets. Definition 4.7 (Convolution) If f, g ∈ L 2 (Z ) are random variables, we define their convolution f ∗ g to be the random variable f ∗ g(x) = E y∈Z f (x − y)g(y) = E y∈Z f (y)g(x − y). We also define the support supp( f ) of f to be the set supp( f ) = { f = 0} = {x ∈ Z : f (x) = 0}. 4 Fourier-analytic methods 154 The significance of convolution to sum sets lies in the obvious inclusion supp( f ∗ g) ⊆ supp( f ) + supp(g) and particularly in the identity A + B = supp(1 A ∗ 1 B ). Indeed we have the more precise statement 1 A ∗ 1 B (x) := P Z (A ∩ (x − B)). (4.8) The relevance of the Fourier transform to convolution lies in the easily verified identity f ∗ g = fˆ · gˆ (4.9) Applying (4.9) at the zero frequency we have the basic formula E Z ( f ∗ g) = (E Z f ) · (E Z g). (4.10) In particular, if f or g has mean zero, then so does f ∗ g. As one consequence of (4.9) we see that convolution is bilinear, symmetric, and associative. We also have a dual version of (4.9), namely the formula f g(ξ ) = fˆ(η)gˆ (ξ − η) (4.11) η∈Z which converts pointwise product back to convolution; we leave the verification of these identities as an exercise. In the exercises below, Z is a fixed finite additive group, with a fixed symmetric non-degenerate bilinear form ·. Exercises 4.1.1 4.1.2 4.1.3 4.1.4 Let Zˆ be the additive group consisting of all the homomorphisms from Z to R/Z. Show that the identification of a frequency ξ ∈ Z with the homomorphism x → ξ · x gives an isomorphism from Z to Zˆ . Define a character to be any map χ : Z → C with χ (0) = 1 and χ (x + y) = χ (x)χ (y) for all x, y ∈ Z . Show that the set of all characters is precisely {eξ : ξ ∈ Z }. Show that for any ξ ∈ Z , eξ takes values in the \|Z \|th roots of unity. Define a linear phase function to be any map φ : Z → R/Z with the property that φ(x + h 1 +h 2 )−φ(x + h 1 )−φ(x + h 2 )+φ(x) = 0 for all x, h 1 , h 2 ∈ Z . 4.1 Basic theory 4.1.5 4.1.6 4.1.7 155 Show that φ : Z → R/Z is a linear phase function if and only if there exists ξ ∈ Z and c ∈ R/Z such that φ(x) = ξ · x + c for all c. (The map φ is also a Freiman homomorphism of order 2; see Definition 5.21.) Let x be an element of Z chosen uniformly at random. Show that the random variables {eξ (x) : ξ ∈ Z } are pairwise independent, and have variance 1 and mean zero for ξ = 0, and variance 0 and mean 1 for ξ = 0. Use this and (1.9), (4.4) to give an alternative proof of (4.2). Prove (4.6). Let f : Z → C. If H is a subgroup of Z , and g := f 1 H , show that gˆ (ξ ) = Eη∈H ⊥ fˆ(ξ + η) for all ξ ∈ Z and conclude in particular the Poisson summation formula Ex∈H f (x) = Eξ ∈H ⊥ fˆ(ξ ). In the converse direction, if h = f ∗ of H , i.e. 1 1 P Z (H ) H is the average of f on cosets h(x) := E y∈H f (x + y), 4.1.8 4.1.9 4.1.10 4.1.11 4.1.12 4.1.13 show that hˆ = fˆ · 1 H ⊥ . If φ : Z → Z is a group isomorphism of Z , then there exists a unique group isomorphism φ † : Z → Z , called the adjoint of φ, such that ξ · φ(x) = φ † (ξ ) · x for all x, ξ ∈ Z . Furthermore if g(x) = f (φ(x)) for all x ∈ Z then gˆ (x) = f ((φ † )−1 (x)) for all x ∈ Z . If φ : Z → Z and ψ : Z → Z are group isomorphisms, show that (φ ◦ ψ)† = ψ † ◦ φ † . Let • : Z × Z → R/Z and •˜ : Z × Z → C be two non-degenerate symmetric bilinear forms on a finite additive group Z . Show that there exists a self-adjoint group isomorphism φ : Z → Z such that ξ •˜ x = ξ • φ(x) = φ † (ξ ) • x for all x, ξ ∈ Z . This shows that all Fourier transforms are equivalent up to isomorphisms of either the x or ξ variable. Prove (4.9) and (4.11). Let x be an element of Z chosen uniformly at random, and let ξ1 , . . . , ξn ∈ Z . Show that the random variables eξ1 (x), . . . , eξn (x) are jointly independent if and only if the group ξ1 , . . . , ξn generated by ξ1 , . . . , ξn has order ord(ξ1 ) . . . ord(ξn ). Let G, H be two subgroups of Z . Show that (G + H )⊥ = G ⊥ ∩ H ⊥ , (G ∩ H )⊥ = G ⊥ + H ⊥ , and d(G ⊥ , H ⊥ ) = d(G, H ), where d is the Ruzsa distance defined in Definition 2.5. This may help explain the symmetric nature of G + H and G ∩ H in the estimates in Exercise 2.3.11. 4 Fourier-analytic methods 156 4.1.14 4.1.15 4.1.16 Let G, H be two subgroups of Z and let x be an element of Z chosen randomly. Show that the indicators I(x ∈ G) and I(x ∈ H ) have nonnegative correlation, i.e. Cov(I(x ∈ G), I(x ∈ H )) ≥ 0; establish this both by Fourier-analytic means and by direct computation. Show that equality occurs if and only if G + H = Z . Show that for any subgroup G of Z , we have (G ⊥ )⊥ = G, and for any random variable f , we have fˆ(x) = \|Z \|−1 f (−x). More generally, for any A ⊂ Z , we have A = (A⊥ )⊥ , where A is the group generated by A. If Z and Z are finite groups, formulate a rigorous version of the statement that the Fourier transform on Z × Z is the composition of the Fourier transform on Z and the Fourier transform on Z . 4.2 L p theory We now turn to the analytic theory of the Fourier transform and of convolutions, starting with the L p theory, and then apply it to the problem of locating arithmetic progressions inside sum sets. If f ∈ C Z and 0 < p < ∞, we define the L p (Z ) norm of f to be the quantity f L p (Z ) := (E Z \| f \| p )1/ p = (Ex∈Z \| f (x)\| p )1/ p . Thus for instance f L 2 (Z ) is just the Hilbert space magnitude of f . We also define f L ∞ (Z ) = sup \| f (x)\|. x∈Z Similarly we define 1/ p f l p (Z ) := \| f (ξ )\| p ξ ∈Z for 0 < p < ∞ and f l ∞ (Z ) := sup \| f (ξ )\|. ξ ∈Z p We have the following two basic L estimates on the Fourier transform and on convolution. Theorem 4.8 Let f, g : Z → C be functions on an additive group Z . Then for any 1 ≤ p ≤ 2 we have the Hausdorff–Young inequality fˆl p (Z ) ≤ f L p (Z ) (4.12) 4.2 L p theory 157 where the dual exponent p to p is defined by 1p + p1 = 1. Also, whenever 1 ≤ p, q, r ≤ ∞ are such that 1p + q1 = r1 + 1, we have the Young inequality f ∗ g L r (Z ) ≤ f L p (Z ) g L q (Z ) . (4.13) Both inequalities follow easily from Riesz–Thorin complex interpolation theorem. With this theorem, one only needs to verify the extremal (and easy) cases. The Riesz–Thorin theorem, however, is beyond the scope of this book. On the other hand, one can also have an elementary proof, using combinatorial arguments (see Exercise 4.2.3). Recall the additive energy E(A, B) between two additive sets A, B in Z , defined in Definition 2.8. From that definition one can easily check that E(A, B) = \|Z \|3 1 A ∗ 1 B 2L 2 (Z ) . By (4.2) and (4.9) we obtain the fundamental identity E(A, B) = \|Z \|3 E(1 A , 1 B ) = \|Z \|3 \|1ˆ A (ξ )\|2 \|1ˆ B (ξ )\|2 . (4.14) ξ ∈Z This formula may illuminate some of the properties of the additive energy that were obtained in Section 2.3, such as the symmetries E(A, B) = E(B, A) = E(A, −B) and the Cauchy–Schwarz inequality (2.9); see Exercise 4.2.7. For the purposes of additive combinatorics, the Fourier transform is most useful when applied to characteristic functions f = 1 A , and in this case one can say quite a bit about the Fourier transform and its relation to the additive energy E(A, A). Lemma 4.9 Let A be a subset of a finite additive group Z , and let 1A : Z → C be the Fourier transform of the characteristic function of A. Then we have the identities: 1A l ∞ (Z ) = sup \|1A (ξ )\| = 1A (0) = P Z (A); ξ ∈Z 1A l22 (Z ) = 1A l44 (Z ) \|1A (ξ )\|2 = P Z (A); (4.15) (4.16) ξ ∈Z 1A (ξ ) = 1A (−ξ ); E(A, A) = \|1A (ξ )\|4 = ; \|Z \|3 ξ ∈Z 1A (ξ ) = 1A (η)1A (ξ − η). (4.17) (4.18) (4.19) η∈Z This lemma follows easily from the estimates that have already been established; see Exercise 4.2.4. 4 Fourier-analytic methods 158 We now present a simple application of the Fourier transform in the setting of a finite field F. Lemma 4.10 Let F be a finite field, and let A be a subset of F{0} such that \|A\| > \|F\|3/4 . Then 3(A · A) = A · A + A · A + A · A = F. Proof We give F a symmetric non-degenerate bilinear form of the type in Example 4.2. Let f : F → R denote the non-negative function f := Ea∈A 1a·A . Observe that supp( f ) = A · A and fˆ(0) = E F f = P F (A). Taking Fourier transforms we obtain fˆ(ξ ) = Ea∈A 1A (ξ/a) for any ξ ∈ F. If ξ = 0, then we observe that the frequencies ξ/a are all distinct as a varies. Using Cauchy–Schwarz and then (4.16), we then obtain \| fˆ(ξ )\| ≤ 1 \|A\|1/2 P F (A)1/2 = 1/\|F\|1/2 for ξ = 0. \|A\| Now let x ∈ F be arbitrary. We use (4.4) and (4.9) to compute f ∗ f ∗ f (x) = Re f ∗ f ∗ f (x) = Re fˆ(ξ )3 e(ξ · x) ξ ∈F ≥ Re fˆ(0)3 − \| fˆ(ξ )\|3 ξ ∈F{0} ≥ P F (A) − 3 \|F\|−1/2 \| fˆ(ξ )\|2 ξ ∈F = P F (A)3 − \|F\|−1/2 P F (A) >0 since P F (A) > \|F\|−1/4 by hypothesis. Since supp( f ∗ f ∗ f ) = 3(A · A) and x was arbitrary, we are done. Remark 4.11 Lemma 4.10 is a simple example of a sum-product estimate – an assertion that a combination of a sum and product of a set A is necessarily much larger than A itself. It can be viewed as a quantitative reflection of the fact that a set A of cardinality greater than \|F\|3/4 has difficulty behaving like a subfield of F. It should be compared with the results in Section 2.8. 4.2 L p theory 159 Exercises 4.2.1 Let 1 ≤ p < ∞. By exploiting the convexity of the function x → \|x\| p , establish the convexity of the set { f ∈ C Z : f L p (Z ) ≤ 1}, and conclude the triangle inequality f + g L p (Z ) ≤ f L p (Z ) + g L p (Z ) . 4.2.2 Argue similarly for the p = ∞ case and with L p replaced by l p . Let 1 < p < ∞, and let p the dual exponent, thus 1/ p + 1/ p = 1. By exploiting the convexity of the function x → e x , establish the preliminary inequality Ex∈Z \| f (x)\|\|g(x)\| ≤ 1 whenever f L p (Z ) , g L p (Z ) ≤ 1, and then conclude H¨older’s inequality f g L r (Z ) ≤ f L p (Z ) g L q (Z ) 4.2.3 whenever 0 < p, q, r ≤ ∞ are such that 1p + q1 = r1 . Similarly with the L p norms replaced by l p norms. The purpose of this exercise is to give a proof of Theorem 4.8 that does not require complex interpolation. First use (4.2), the trivial bound fˆl ∞ (Z) ≤ f L 1 (Z) , (4.20) and H¨older’s inequality to establish the weaker estimate fˆl p (Z ) = O p ( f L p (Z ) ) whenever f ∈ C Z is supported on a set A and obeys an estimate of the form \| f (x)\| = (λ) for all x ∈ A and some threshold λ. Then, prove the even weaker estimate fˆl p (Z ) = O p ( f L p (Z ) log(1 + \|Z \|)) 4.2.4 4.2.5 4.2.6 for arbitrary f ∈ C Z by applying the previous inequality to a dyadic decomposition of f , followed by the triangle inequality. Finally, remove the O p (log(1 + \|Z \|)) factor to establish (4.12) by replacing Z with a large power Z M of Z , and similarly replacing f with a large tensor power (as in Corollary 2.19) and letting M → ∞. Argue similarly to establish (4.13). Prove Lemma 4.9. Let A be an additive set in a finite additive group Z . Show that 1ˆ A is real-valued if and only if A is symmetric. (Law of large numbers for finite groups) Let f : Z → R≥0 be such that E Z f = 1 and f (0) = 0, and let H be the subgroup of Z generated by supp( f ). Show that \| fˆ(ξ )\| ≤ 1, with equality if and only if ξ ∈ H ⊥ . 160 4.2.7 4.2.8 4.2.9 4 Fourier-analytic methods Next, define the iterated convolutions f (n) for n = 1, 2, . . . inductively by f (1) := f and f (n+1) := f ∗ f (n) , and show that limn→∞ f (n) = 1 1 . What can happen when the hypothesis f (0) = 0 is dropped? P Z (H ) H Use Fourier-analytic methods to give another proof of Corollary 2.10. Use Fourier-analytic methods to give another proof of Proposition 2.7. Let f be a random variable which is not identically zero. By using (4.2) and (4.20), establish the uncertainty principle \|supp( f )\|\|supp( fˆ)\| ≥ \|Z \|. 4.2.10 (4.21) Prove that equality occurs if and only if f (x) = ce(ξ · x)1 H +x0 (x) for some complex number c ∈ C, some subgroup H of Z , and some ξ, x0 ∈ Z . This inequality can be improved for certain groups Z : see Theorem 9.52. Let f ∈ C Z be normalized so that f 2L 2 (Z ) = ξ ∈Z \| fˆ(ξ )\|2 = 1. By differentiating the Hausdorff–Young inequality in p, establish the entropy uncertainty principle 1 1 Ex∈Z \| f (x)\|2 log + \| fˆ(ξ )\|2 log ≥ 0, 2 ˆ \| f (x)\| \| f (ξ )\|2 ξ ∈Z where we adopt the convention that 0 log 10 = 0. (Hint: differentiate the Hausdorff–Young inequality in p at p = 2, using the fact that equality holds at that endpoint.) Using Jensen’s inequality, show that this inequality implies (4.21). 4.3 Linear bias One common way to apply the Fourier transform to the theory of sum sets or to arithmetic progressions is to introduce the notion of Fourier bias of that set (also known as linear bias or pseudo-randomness). Roughly speaking, this notion separates sets into two extremes, ones which are highly uniform (and behave like random sets, especially with regard to iterated sum sets), and ones which are highly non-uniform (and behave like arithmetic progressions). Definition 4.12 (Fourier bias) Let Z be a finite additive group. If A is a subset of Z , we define the Fourier bias Au of the set A to be the quantity Au := sup \|1ˆ A (ξ )\|. ξ ∈Z {0} This quantity is always non-negative, with Au = 0 if and only if A is equal to Z or the empty set (Exercise 4.3.1). It obeys the symmetries Au = − Au = A + hu = Z \Au for any h ∈ Z (Exercise 4.3.2). We warn that this quantity 4.3 Linear bias 161 is not monotone: A ⊆ B does not imply Au ≤ Bu . However, the Fourier bias does obey a triangle inequality (Exercise 4.3.3). The Fourier bias Au can be as large as the density P Z (A), but is usually smaller (Exercise 4.3.4). Sets A with Fourier bias less than α are sometimes called α-uniform or α-pseudo-random; sets with small Fourier bias are called linearly uniform, Gowers uniform of order 1, or pseudo-random. The connection between Fourier bias and sum sets can be described by the following lemma. Lemma 4.13 (Uniformity implies large sum sets) Let n ≥ 3, and let A1 , . . . , An be additive sets in a finite additive group Z . Then for any x ∈ Z we have 1 \|{(a , . . . , a ) ∈ A ×· · ·× A : x = a + · · · +a }\| − P (A ) · · · P (A ) n 1 n 1 n Z 1 Z n \|Z \|n−1 1 ≤ A1 u · · · An−2 u P Z (An−1 )1/2 P Z (An )1/2 . In particular, if we have A1 u · · · An−2 u < P Z (A1 ) · · · P Z (An−2 )P Z (An−1 )1/2 P Z (An )1/2 then A1 + · · · + An = Z . Of course, a similar result is true if we permute the A1 , . . . , An . Note that the quantity P Z (A1 ) · · · P Z (An ) is the quantity one would expect for 1 \|{(a1 , . . . , an ) ∈ A1 × · · · × An : x = a1 + · · · + an }\| if the events a1 ∈ \|Z \|n−1 A1 , . . . , an ∈ An were jointly independent conditioning on x = a1 + · · · + an . This may help explain why uniformity is sometimes referred to as pseudorandomness. Proof By (4.9), the function 1 A1 ∗ · · · ∗ 1 An has Fourier transform 1 A1 · · · 1 An . Applying the Fourier inversion formula (4.4), (4.15), the Cauchy–Schwarz inequality and (4.16) we thus see that 1 A1 ∗ · · · ∗ 1 An (x) = Re1 A1 ∗ · · · ∗ 1 An (x) 1 = Re A1 (ξ ) · · · 1 An (ξ )e(x · ξ ) ξ ∈Z ≥ 1 A1 (0) · · · 1 An (0) − \|1 A1 (ξ )\| · · · \|1 An (ξ )\| ξ ∈Z {0} ≥ P Z (A1 ) · · · P Z (An ) − A1 u · · · An−2 u \|1 An−1 (ξ )\|\|1 An (ξ )\| ξ ∈Z ≥ P Z (A1 ) · · · P Z (An ) − A1 u · · · An−2 u 1 An−1 l 2 (Z ) 1 An (ξ )l 2 (Z ) = P Z (A1 ) · · · P Z (An ) − A1 u · · · An−2 u P Z (An−1 )1/2 P Z (An )1/2 . 162 4 Fourier-analytic methods A similar argument gives 1 A1 ∗ · · · ∗ 1 An (x) ≤ P Z (A1 ) · · · P Z (An ) + A1 u · · · An−2 u P Z (An−1 )1/2 P Z (An )1/2 . Since by definition of convolution 1 A1 ∗ · · · ∗ 1 An (x) = \|Z \|1−n \|{(a1 , . . . , an ) ∈ A1 × · · · × An : x = a1 + · · · + an }\|, and the lemma follows. We now give an application of the above machinery to the finite field Waring problem. We first need a standard lemma. Lemma 4.14 (Gauss sum estimate) Let F be a finite field of odd order, and let 1 A := F ∧ 2 = {a 2 : a ∈ F} be the set of squares in F. Then Au ≤ 2\|F\| + 2\|F\|1 1/2 . Proof Let ξ ∈ F\0. Since every non-zero element in A has exactly two representations of the form a 2 , we have 1 1 1 1ˆ A (ξ ) = + e(−ξ · x) = e(−ξ · a 2 ). \|F\| x∈A 2\|F\| 2\|F\| a∈F On the other hand, we may square 2 2 2 2 e(−ξ · a ) = e(ξ · a ) = e(ξ · (a 2 − b2 )) a∈F a∈F a,b∈F = e(ξ · (a 2 − (a + h)2 )) a,h∈F = h∈F e(−ξ · h 2 ) e(ξ · 2ah). a∈F If h = 0, then 2h = 0, and = c∈F e(ξ · c) = 0 thanks to a∈F e(ξ · 2ah) Lemma 4.5. On the other hand, if h = 0, then a∈F e(ξ · 2ah) = \|F\|. We conclude that \| a∈F e(ξ · a 2 )\|2 = \|F\|, and the claim follows. By combining this lemma with Lemma 4.13, one immediately obtains Corollary 4.15 Let F be a finite field of odd order, and let A = F ∧ 2 be the set of squares in F. Then k A = F for all k ≥ 3. Indeed, for any x ∈ F, the number of representations of x as a sum x = a1 + · · · + ak with a1 , . . . , ak ∈ F is (21−k + Ok (\|F\|−(k−2)/2 ))\|F\|k−1 . We leave the verification of this corollary as an exercise. It shows that the sum sets k A are more or less uniformly distributed for k ≥ 3. Note that when k = 2, one can still prove that 2A = F, but the sum sets can be quite irregular; for instance, if −1 is not a square in F, then 0 only has one representation as the sum of two elements in F. 4.3 Linear bias 163 We now present a lemma which asserts, roughly speaking, that if B is a randomly-chosen subset of A, then Bu is approximately equal to \|B\| Au ; thus \|A\| the Fourier bias decreases proportionally when passing to random subsets. Lemma 4.16 Let A be an additive set in a finite additive group Z , and let 0 < τ ≤ 1. Let B be a random subset of A defined by letting the events a ∈ B be independent with probability τ . Then for any λ > 0 we have √ 2 P(\|Bu − τ Au \| ≥ λσ ) ≤ 4\|Z \| max e−λ /8 , e−λσ/2 2 , where σ 2 := \|A\|τ (1 − τ )/\|Z \|2 . The lemma is an easy consequence of Chernoff’s inequality and is left as an exercise. Applying it with λ = C log1/2 \|Z \| for some large C, and assuming \|A\|τ (1 − τ ) log \|Z \|, we see in particular that P Bu = τ Au + O σ log1/2 \|Z \| = 1 − O(\|Z \|−100 ) (for instance). In particular if we set A = Z then we have Bu = 1/2 τ Z + O(τ (1 − τ ) log\|Z \|\|Z \| ) with high probability; thus random subsets of Z tend to be extremely uniform. Note that P Z (B) ≈ τ with high probability, thanks to Corollary 1.10. A major application of Fourier bias is in the study of arithmetic progressions of length 3. We will study this application in detail in Chapter 10. Exercises 4.3.1 4.3.2 4.3.3 4.3.4 4.3.5 Let A be a subset of a finite additive group Z . Show that Au = 0 if and only if A = Z or A = ∅. Let A be a subset of a finite additive group Z . Show that Au = − Au = T h Au = Z \Au for any h ∈ Z . More generally, if φ : Z → Z is any isomorphism from one additive group to another, show that φ(A)u = Au . In a similar spirit, show that the Fourier bias of a set A does not depend on the choice of symmetric non-degenerate bilinear form. Let A, B be disjoint subsets of a finite additive group Z . Show that \|Au − Bu \| ≤ A ∪ Bu ≤ Au + Bu . Let A be an additive set in a finite additive group Z . Show that Au ≤ P Z (A), with equality if and only if A is contained in a coset of a proper subgroup of Z . Let A and A be subsets of finite additive groups Z and Z respectively. Show that A × A u = Au A u . 4 Fourier-analytic methods 164 4.3.6 4.3.7 Let A be a subset of a finite additive group Z . Show that Au = supφ \| 1 A , e(φ) C Z \|, where φ : Z → R/Z ranges over all non-constant linear phase functions (as defined in Exercise 4.1.4). Let A, B be additive sets in a finite additive group Z . Show that E(A, B) ≤ \|A\|2 \|B\|2 + \|Z \|2 A2u \|B\|. \|Z \| Using (2.8), conclude that if Au ≤ αP Z (A), then 1 1 \|A + B\| ≥ min \|Z \|, 2 \|B\| . 2 α 4.3.8 Thus α-uniform sets tend to expand sum sets by a factor of roughly α −2 (unless this is impossible due to the trivial bound \|A + B\| ≤ \|Z \|). Let A be an additive set in a finite additive group Z . Show that A4u ≤ 4.3.9 4.3.10 4.3.11 4.3.12 (4.22) 1 E(A, A) − P Z (A)4 ≤ A2u P Z (A). \|Z \|3 (4.23) Thus uniform sets have additive energy E(A, A) close to the minimal value of P Z (A)4 \|Z \|3 , and vice versa. Let A be an additive set in a finite additive group Z , and let n ≥ 3 be 1 an integer. Using Lemma 4.13, show that if n A = Z , then P Z (A)1+ n−2 ≤ Au ≤ P Z (A). This estimate is especially useful when n is very large, as it shows that 1 A has a very large non-trivial Fourier coefficient. Prove Corollary 4.15. Also show the identity A · 2A = A and conclude that 2A = F (using the fact that 3A = F to show that 2A = A). Use Chernoff’s inequality (in the form of Exercise 1.3.4) to prove Lemma 4.16. Let A, B be additive sets in a finite additive group Z . Use Lemma 4.13 to establish the inequality Su ≥ P Z (A)1/2 P Z (B)1/2 P Z (S) 4.3.13 whenever S is disjoint from A + B. In particular, this inequality holds when S = Z (A + B). This shows that complements of sum sets are “hereditarily non-uniform”. Let A be a subset of a cyclic group Z p of prime order. Show that for any arithmetic progression P in Z p , we have the uniform distribution estimate 1 PZ p (A ∩ P) = PZ p (A)PZ p (P) + O(ε) + O log Au ε for any 0 < ε ≤ 1. (Hint: apply a change of variables to make P = [−N , N ] for some N . Approximate the indicator 1 P by something a bit 4.4 Bohr sets 4.3.14 165 smoother (smoothed out at scale εp) and then compute the Fourier expansion. Apply Plancherel’s theorem (4.3) with this smoothed out function and 1 A − P(A).) This inequality is a crude form of the famous Erd˝os– Tur´an inequality in discrepancy theory, and is related to the Weyl criterion for uniform distribution modulo one. Let A = Z2p be the set of squares in a cyclic group of prime order. Show that for any arithmetic progression P in Z p , we have 1 √ \|P\| + O( p log p). 2 (Hint: use Lemma 4.14 and the preceding exercise.) This is a special case of the Polya–Vinogradov inequality from analytic number theory. Let F be a finite field, let Z be a vector space over F, and let M : Z → Z be a linear transformation. Show that if dim F (Z ) ≥ 3, then there exists a non-zero x ∈ Z such that M x · x = 0. (Hint: reduce to the case when M has full rank, and then modify Lemma 4.14. One can also solve this problem by purely algebraic methods.) Let W be a vector space over a finite field F of odd order, and let M : W → W be a linear transformation. Show that there exists a subspace U of W with dimension dim F (U ) ≥ 12 dim F (W ) − 32 such that M is null on U , i.e. M x · y = 0 for all x, y ∈ U . (Hint: take a maximal space U which is null with respect to M. If the orthogonal complement U ⊥ := {y ∈ W : M x · y = 0 for all x ∈ U } is at least three dimensions larger than U , then use the previous lemma.) For a purely algebraic proof of this fact, see Exercise 9.4.11. \|A ∩ P\| = 4.3.15 4.3.16 4.4 Bohr sets In many applications of the Fourier-analytic method, one starts with some additive set A and concludes some information about the Fourier transform 1ˆ A of A (for instance, one may obtain some bound on the Fourier bias Au ). One would then like to pass from this back to some new combinatorial information on the original set A. For some special groups (e.g. finite field geometries F pn ) one can do this quite directly (see for instance Lemma 10.15). However, to convert Fourier information on general groups to combinatorialinformation we need the notion of a Bohr set (also known as Bohr neighborhoods in the literature). We first define a “norm” θ R/Z on the circle group by defining θ + ZR/Z = \|θ\| whenever −1/2 < θ ≤ 1/2; in other words, θ R/Z is the distance from θ (or more precisely, any representative of the coset θ ) to the integers. We observe the elementary bounds 4θR/Z ≤ \|e(θ ) − 1\| ≤ 2π θ R/Z (4.24) 4 Fourier-analytic methods 166 which follow from elementary trigonometry and the observation that the sinc function sin(x)/x varies between 1 and 2/π when \|x\| ≤ π/2. Definition 4.17 (Bohr set) Let S ⊂ Z be a set of frequencies, and let ρ > 0. We define the Bohr set Bohr(S, ρ) = Bohr Z (S, ρ) as Bohr(S, ρ) := x ∈ Z : sup ξ · xR/Z < ρ . ξ ∈S We refer to S as the frequency set of the Bohr set, and ρ as the radius. The quantity \|S\| is known as the rank of the Bohr set. Remark 4.18 Note that if Z is a vector space over a finite field F, then every subspace of Z can be viewed as a Bohr set (with radius O(1/\|F\|), and rank equal to the codimension). Thus Bohr sets can be viewed as a generalization of subspaces. Note that most finite groups Z tend to have very few actual subgroups (the extreme case being the cyclic groups Z p of prime order), so it is convenient to be able to rely on the much larger class of Bohr sets as a substitute. Remark 4.19 One way to think of Bohr sets is to consider the embedding of Z into the complex vector space C S (and in particular to the standard unit torus inside C S ) by the multiplicative map x → (e(ξ · x))ξ ∈S . A Bohr set is thus the inverse image of a cube. Observe that the R/Z norm is symmetric and subadditive; − xR/Z = xR/Z and x + yR/Z ≤ xR/Z + yR/Z . From this we see that the Bohr sets Bohr(S, ρ) are symmetric, decreasing in S, and increasing in ρ (and fill out the whole space Z once ρ > 1/2); they are always unions of cosets of S ⊥ , and if ρ is sufficiently small they consist entirely of S ⊥ . One can also easily verify the intersection property Bohr(S, ρ) ∩ Bohr(S , ρ) = Bohr(S ∪ S , ρ) and the addition property Bohr(S, ρ) + Bohr(S, ρ ) ⊆ Bohr(S, ρ + ρ ). In particular we have kBohr(S, ρ) ⊆ Bohr(S, kρ) for any k ≥ 1. Next, we establish some bounds for the size of Bohr sets. Lemma 4.20 (Size bounds) If S ⊂ Z and ρ > 0, then we have the lower bound P Z (Bohr(S, ρ)) ≥ ρ \|S\| (4.25) 4.4 Bohr sets 167 and we have the doubling estimate P Z (Bohr(S, 2ρ)) ≤ 4\|S\| P Z (Bohr(S, ρ)). (4.26) This lemma should be compared with the Kronecker approximation theorem (Corollary 3.25); indeed the two results are very closely related. Proof For each ξ ∈ S let θξ be an element of R/Z chosen independently and uniformly at random. For any x ∈ Z , one can easily verify that P Z (ξ · x − θξ R/Z < ρ/2 for all ξ ∈ S) = ρ \|S\| . Summing this over all x ∈ Z using linearity of expectation (1.4), we conclude E\|{x ∈ Z : ξ · x − θξ R/Z < ρ/2 for all ξ ∈ S}\| ≥ ρ \|S\| \|Z \| and thus there exists a choice of θξ such that \|{x ∈ Z : ξ · x − θξ R/Z < ρ/2 for all ξ ∈ S}\| ≥ ρ \|S\| \|Z \|. (4.27) Now observe from the triangle inequality that if x, x lie in the above set, then x − x lies in Bohr(S, ρ). The claim (4.25) follows. Now we prove (4.26). By a limiting argument we may replace 2ρ by 2ρ − ε on the left-hand side for some small ε > 0. Observe that we can cover the interval {θ ∈ R/Z : θR/Z < 2ρ − ε} by four intervals of the form {θ ∈ R/Z : θ − θ0 R/Z < ρ/2}. We can thus can cover Bohr(S, 2ρ) by 4\|S\| sets of the type appearing in the left-hand side of (4.27). The claim follows by arguing as before. We have already mentioned that subspaces of a vector space are one example of a Bohr set. Progressions can form another example; for instance intervals such as (−N , N ) in a cyclic group Z M can easily be seen to be a Bohr set of rank 1. We can combine these two examples by introducing the concept of a coset progression. Definition 4.21 (Coset progressions) A coset progression in an additive group Z is any set of the form P + H where P is a progression and H is a finite subgroup of Z . We say that the coset progression P + H is proper if P is proper and \|P + H \| = \|P\|\|H \| (i.e. all the sums in P + H are distinct). We say that a coset progression P + H has rank d if the component P has rank d. We say that P + H is symmetric if P has the form P = (−N , N ) · v. Of course, Corollary 3.8 shows that every coset progression can also be viewed as an ordinary progression, but possibly of much larger rank. If however Z is a cyclic group of prime order, then H will either be trivial or equal to the whole space, and will thus increase the rank by at most 1. Indeed we can view vector 168 4 Fourier-analytic methods spaces over small finite fields on the one hand, and cyclic groups of prime order on the other, as the two extremes of additive behavior for finite groups Z . Now we relate Bohr sets of rank d with coset progressions of rank d. Lemma 4.22 (Bohr sets contain large coset progressions) Let Bohr(S, ρ) be a Bohr set of rank d in Z with 0 < ρ < 12 . Then there exists a proper symmetric coset progression P + H of rank 0 ≤ d ≤ d, obeying the inclusions Bohr(S, d −2d ρ) ⊆ P + H ⊆ Bohr(S, ρ). (4.28) In particular, from Lemma 4.20 we have P Z (P + H ) ≥ ρ d d −4d . 2 (4.29) Furthermore we have H = S ⊥ . Proof Let φ : Z → (R/Z) S be the group homomorphism φ(x) := (ξ · x)ξ ∈S . Observe that φ(Z ) is a finite subgroup of the torus (R/Z) S , and that Bohr(S, ρ) contains the inverse image of the cube Q := {(yξ )ξ ∈S ∈ R S : \|yξ \| ≤ ρ} ⊂ R S (which we identify with its projection in (R/Z) S ) under φ. Let ⊆ R S be the lattice φ(Z ) + Z S . Though it is a slight abuse of notation, we consider φ(Z ) ∩ Q to be the same as ∩ Q. Applying Lemma 3.36, we can find a progression P˜ := (−L , L) · w for some linearly independent w1 , . . . , wd ⊆ with 0 ≤ d ≤ d such that ∩ d −2d · Q ⊆ P˜ ⊆ ∩ Q. Since the w j are independent, P˜ is necessarily proper. The claim now follows by setting v j to be an arbitrary element of φ −1 (w j ) for each 1 ≤ j ≤ d , and setting H equal to the kernel of φ, which is of course just S ⊥ . In the case of a cyclic group, we can dispense with the group H and sharpen the constants somewhat (though at the cost of losing the first inclusion in (4.28)): Proposition 4.23 Let Z = Z N be a cyclic group, and let Bohr(S, ρ) be a Bohr set of rank d with 0 < ρ < 12 . Then Bohr(S, ρ) contains a symmetric proper progression P of rank d and cardinality \|P\| ≥ ρd N. dd Furthermore we may choose P to be symmetric (i.e. P = −P). Proof The main tool here will be Minkowski’s second theorem. We use the standard bilinear form ξ · x = ξ x/N , and write S = (ξ1 , . . . , ξd ). Let α ∈ Rd be 4.4 Bohr sets 169 the vector α := ( ξN1 , . . . , ξNd ), and let be the lattice Z · α + Zd ; this clearly has full rank, and by (3.12) mes(Rd / ) = mes(Rd /Zd )/\|/Zd \| ≥ 1/N . Let Q be the cube Q := {(x1 , . . . , xd ) ∈ Rd : \|x j \| < ρ for all 1 ≤ j ≤ n}, and let 0 < λ1 ≤ · · · ≤ λd be the succesive minima of Q with respect to , with a corresponding directional basis v1 , . . . , vd ∈ as given by Theorem 3.30. In particular we see that every coordinate of v j has magnitude at most λ j ρ. Let 1 ≤ j ≤ d be arbitrary. Since v j ∈ , we see from the definition of that there exists w j ∈ Z N such that v j ∈ αw j + Zd . In particular we see that ξi · w j R/Z ≤ λ j ρ for all 1 ≤ i, j ≤ d. Set w := (w1 , . . . , wd ). Now we let M j := dλ1 j , and let M := (M1 , . . . , Md ); we now claim that the progression P := (−M, M) · w is proper and lies in Bohr(S, ρ) (it is clearly symmetric). Let us first verify that P ⊆ Bohr(S, ρ). If n = (n 1 , . . . , n d ) ∈ (−M, M), then for any 1 ≤ j ≤ d we have ξ j · (n · w)R/Z ≤ d \|n j \|ξ j · w j R/Z 0 2r1 μ{y : x − r < y < x + r }. (It can be verified that Mμ is a measurable function.) Using the Vitali covering lemma, establish the distributional inequality mes({x : Mμ(x) ≥ λ}) ≤ 2 μ(R). λ 172 4 Fourier-analytic methods 4.5 ( p) constants, Bh [g] sets, and dissociated sets In Section 4.3 we discussed one Fourier-analytic characteristic of an additive set A in a finite additive group Z , namely its linear bias. In this section we discuss a rather different characteristic, namely the ( p) constants of a set S of frequencies. These constants measure how “dissociated” or “Sidon-like” a set1 S is; in more practical terms, the ( p) constants quantify the independence of the characters associated to S in a certain L p (Z ) sense. These constants can be used to obtain precise control on the arithmetic structure of S, for instance in controlling iterated sum sets of S. One feature of these constants is that they are stable under passage to subsets, thus ( p) constants will also control iterated sum sets of subsets S of S. This stability (which is not present in the Fourier bias, unless one takes random subsets as in Lemma 4.16) is useful for a number of applications. We begin with the formal definition of the ( p) constants. Definition 4.26 (( p) constants) Let S be an additive set in a finite2 additive group Z , and let 2 ≤ p ≤ ∞. We define the ( p) constant of S, denoted S( p) , to be the best constant such that the inequality c(ξ )e(ξ · x) ≤ S( p) cl 2 (S) (4.30) ξ ∈S L p (Z ) holds for all sequences c : S → C of complex numbers. One can easily establish the bound S( p) ≤ \|S\|1/2−1/ p , (4.31) for 2 ≤ p ≤ ∞, with equality at the endpoints p = 2, ∞; see Exercise 4.5.6. This exercise indicates that largeness of ( p) constants is correlated to strong additive structure of S. At the other extreme, we now show that smallness of ( p) constants is correlated to strong lack of additive structure of S. Definition 4.27 (Bh sets) Let h ≥ 2. A non-empty subset S of an additive group Z is a Bh set if for any ξ1 , . . . , ξh , η1 , . . . , ηh ∈ S, one has ξ1 + · · · + ξh = η1 + · · · + ηh if and only if (ξ1 , . . . , ξh ) is a permutation of (η1 , . . . , ηh ). We say S is a Sidon set if it is a B2 set. These sets are the g = 1 version of the Bh [g] sets, encountered in Section 1.7.1; Sidon sets were also briefly mentioned in Section 2.2. Note that we do not bother with the notion of a B1 set, since every set is trivially a B1 set. 1 2 Here, we use “Sidon set” to denote a set whose pairwise sums are all disjoint. There is another, more Fourier-analytic, notion of a Sidon set related to ( p) constants which we will not discuss here. One can also define the concept of a ( p) constant for subsets of the integers, or more general additive groups, but we will not need to do so in this book. 4.5 ( p) constants, Bh [g] sets, and dissociated sets 173 Example 4.28 For any M > 1, the set S := {0} ∪ (M ∧ N) = {0, 1, M, M 2 , . . .} is a Bh set in Z if and only if h < M. In particular, the powers of 2 form a Sidon set. One can of course truncate these examples to finite additive groups such as Z N ; note that any non-empty subset of a Bh set is also a Bh set. Proposition 4.29 Let S be a non-empty subset of a finite additive group Z . Then we have 1 1/4 S(4) ≥ 2 − , (4.32) \|S\| with equality holding if and only if S is a Sidon set. More generally, if h ≥ 1, then there exists a number 1 ≤ α(h, \|S\|) < (h!)1/2h depending on h and \|S\| such that S(2h) = α(h, \|S\|) when S is a Bh set, and S(2h) > α(h, \|S\|) otherwise. Proof We first prove (4.32). By testing (4.31) with cξ identically equal to 1, it will suffice to show that 4 1 e(x, ξ ) ≥ 2 − \|S\|2 . ξ ∈S \|S\| L 4 (Z ) The left-hand side can be expanded as Ex∈Z e((ξ1 + ξ2 − η1 − η2 ) · x). ξ1 ,ξ2 ,η1 ,η2 ∈S By Lemma 4.5 this simplifies to \|{ξ1 , ξ2 , η1 , η2 ∈ S : ξ1 + ξ2 = η1 + η2 }\|. Clearly ξ1 + ξ2 will equal η1 + η2 when (ξ1 , ξ2 ) is a permutation of (η1 , η2 ), so this expression is at least as large as 1 1 = \|S\|(\|S\| − 1)2 + \|S\| = 2 − \|S\|2 \|S\| ξ1 ,ξ2 ,η1 ,η2 ∈S:{ξ1 ,ξ2 }={η1 ,η2 } as claimed. Note that this argument also shows that the inequality in (4.32) is strict if S is not a Sidon set, since then we have additional terms coming from pairs (ξ1 , ξ2 ) and (η1 , η2 ) which are not permutations of each other. Now suppose that S is a Sidon set. To prove equality in (4.32) it suffices to show that 4 1 cξ e(x, ξ ) ≤2− ξ ∈S \|S\| 4 L (Z ) 4 Fourier-analytic methods 174 assuming the normalization ξ ∈S \|cξ \|2 = 1. The left-hand side can be expanded as cξ1 cξ2 cη1 cη2 Ex∈Z e((ξ1 + ξ2 − η1 − η2 ) · x) ξ1 ,ξ2 ,η1 ,η2 ∈S which as before simplifies to cξ1 cξ2 cη1 cη2 . ξ1 ,ξ2 ,η1 ,η2 ∈S:ξ1 +ξ2 =η1 +η2 Since S is a Sidon set, (η1 , η2 ) must be a permutation of (η1 , η2 ). Splitting into the cases ξ1 = ξ2 and ξ1 = ξ2 , we can thus rewrite the previous expression as cξ 4H + 2 \|cξ1 \|2 \|cξ2 \|2 ξ ∈S which by the normalization ξ1 ,ξ2 ∈S:ξ1 =ξ2 ξ ∈S \|cξ \|2 = 1 can be written as 2− \|cξ \|4 . ξ ∈S 2 But from Cauchy–Schwarz and the normalization ξ ∈S \|cξ \| = 1 we have 4 ξ ∈S \|cξ \| ≥ 1/\|S\|, and the claim follows. The general case h ≥ 2 is similar but is left to Exercise 4.5.9. Another quantification of the heuristic that large ( p) constants corresponds to strong additive structure is given by Lemma 4.30 Let S be a non-empty subset of a finite additive group Z , and let h ≥ 1. Then we have \|h 1 S − h 2 S\| ≥ \|S\|h S2h (2h) whenever h 1 , h 2 ≥ 0 are such that h 1 + h 2 = h. In particular we have \|h S\| ≥ \|S\|h . S2h (2h) Remark 4.31 This lemma shows that if S has a small (2h) constant, then not only do the sum sets h S become very large, but so do the sum sets h S of all subsets S of S, thanks to the monotonicity of ( p) constants. The converse statement is also true up to logarithmic factors; see exercises. Thus (2h) constants measure the failure of S, or any of its subsets, to have good closure properties under h-fold sums. 4.5 ( p) constants, Bh [g] sets, and dissociated sets Proof 175 From (4.30) with p := 2h, and cξ set identically equal to 1, we have 2h h e(ξ · x) ≤ S2h (2h) \|S\| . ξ ∈S L 2h (Z ) The left-hand side is equal to h 1 h 2 2 e(ξ · x) e(ξ · x) ξ ∈S 2 ξ ∈−S L (Z ) since e(x, −ξ ) is the conjugate of e(x, ξ ). We can expand h 1 h 2 e(ξ · x) e(x, ξ ) = rh 1 ,h 2 (ξ )e(ξ · x) ξ ∈S ξ ∈−S ξ ∈S where rh 1 ,h 2 is the counting function rh 1 ,h 2 (ξ ) : = \|{(ξ1 , . . . , ξh 1 , ξ1 , . . . , ξh 2 ) ∈ S h 1 +h 2 : ξ = ξ1 + · · · + ξh 1 − ξ1 − · · · − ξh }\|. By (4.2) we thus have ξ ∈S h rh 1 ,h 2 (ξ )2 ≤ S2h (2h) \|S\| . On the other hand, the function rh 1 ,h 2 is supported in h 1 S − h 2 S, so by Cauchy– Schwarz h rh 1 ,h 2 (ξ ) ≤ \|h 1 S − h 2 S\|1/2 S(2h) \|S\|h/2 . ξ ∈S But from the definition of rh 1 ,h 2 we have rh 1 ,h 2 (ξ ) = \|S h 1 +h 2 \| = \|S\|h 1 +h 2 ξ ∈S The claim follows. We now investigate the ( p) constants of Sidon-like sets as p → ∞. Definition 4.32 An additive set S with cardinality \|S\| = d is said to be dissociated if the cube [0, 1]d · S is proper, or in other words, the 2d subset sums F S(S) := ξ:S ⊆S ξ ∈S are all distinct. 4 Fourier-analytic methods 176 This should be compared with the concept of a Sidon set, which is a set S of cardinality d whose d(d+1) pairwise sums {ξ1 + ξ2 : ξ1 , ξ2 ∈ S} are all distinct 2 (except for the trivial identification ξ1 + ξ2 = ξ2 + ξ1 ). A good example of a dissociated set is the set of powers of 2: S = {1, 2, . . . , 2n } in any cyclic group Z/N Z with N ≥ 2n+1 . Observe that if S is a dissociated set of cardinality d, and v is a non-zero element of [−1, 1]d , then v · S = 0 (since otherwise we could find two disjoint sets S1 , S2 in S, corresponding to where the components of v are +1 or −1, such that ξ ∈S1 ξ = ξ ∈S2 ξ ). Dissociativity is the Fourier analog of joint independence. It leads to the following Fourier-analytic analog of Chernoff’s inequality: Lemma 4.33 (Rudin’s inequality) If S is dissociated, then we have 2 Ex∈Z exp σ Re c(ξ )e(ξ · x) ≤ eσ /2 (4.33) ξ ∈S whenever cl 2 (S) ≤ 1 and σ ≥ 0. We also have the distributional estimates 2 Px∈Z (4.34) c(ξ )e(ξ · x) ≥ λ = Oε e−λ /(4+ε) ξ ∈S for every ε > 0, and the ( p) estimate √ S( p) = O( p) (4.35) for all 2 ≤ p < ∞. √ Note that when p = 2h then (h!)1/2h is comparable to p by Stirling’s formula (1.52), and hence so (4.35) and shows that dissociated sets are comparable in (2h) constant to B2h sets for any given h (if S is sufficiently large). This also shows that the bounds in the above lemma cannot be significantly improved except in the constants, even if one imposes even more additive independence conditions on S. Proof Write c(ξ ) = \|c(ξ )\|e(θξ ) for some phase θξ ∈ R/Z. We begin by observing the inequality et x ≤ cosh(x) + t sinh(x) for all x ≥ 0 and −1 ≤ t ≤ 1, which is simply a consequence of the convexity of et x as a function of t. In particular we see that exp(σ Rec(ξ )e(x, ξ )) ≤ cosh(σ \|c(ξ )\|) + sinh(σ \|c(ξ )\|)Re e(ξ · x + θξ ), 4.5 ( p) constants, Bh [g] sets, and dissociated sets 177 which upon multiplying and taking expectations becomes Ex∈Z exp σ Rec(ξ )e(x, ξ ) ξ ∈S ≤ Ex∈Z 1 sinh(σ \|c(ξ )\|)e(ξ · x + θξ ) 2 1 + sinh(σ \|c(ξ )\|)e(−ξ · x − θξ ) . 2 (cosh(σ \|c(ξ )\|) + ξ ∈S Now we multiply the product out and inspect its behavior in x. We obtain a large number of terms (3\|S\| , to be exact) that are of the form e((v · S) · ξ ), for some v ∈ [−1, 1]\|S\| , times some constant independent of x, where we select some enumeration S = (ξ1 , . . . , ξ\|S\| ) of S. There is one constant term, namely ξ ∈S cosh(σ \|c(ξ )\|), but all the others have a non-zero frequency vector v · S because S is dissociated, and thus integrate out to zero by the Fourier inversion formula. Thus we have Ex∈Z exp σ Re c(ξ )e(ξ · x) ≤ cosh(σ \|c(ξ )\|), ξ ∈S ξ ∈S and the claim (4.33) then follows from the elementary inequality cosh(x) ≤ e x /2 (which follows by comparing Taylor series). From Markov’s inequality we thus obtain 2 Px∈Z Re c(ξ )e(ξ · x) ≥ λ ≤ eσ /λ e−σ λ 2 ξ ∈S for every λ ≥ 0; choosing σ := λ/2, we obtain 2 Px∈Z Re c(ξ )e(ξ · x) ≥ λ ≤ e−λ /4 . ξ ∈S Replacing λ by (1 − ε)λ and rotating c(ξ ) by an arbitrary angle e(θ ), we obtain 2 2 Px∈Z Re e(θ ) c(ξ )e(ξ · x) ≥ (1 − ε)λ ≤ e−λ (1−ε )/4 . ξ ∈S If take the union of these estimates as eiθ varies over a finite number of angles (depending on ε) we obtain (4.34). To obtain (4.35), we observe from the identity p ∞ p−1 c(ξ )e(ξ · x) ≤p λ Px∈Z c(ξ )e(ξ · x) ≥ λ dλ ξ ∈S ξ ∈S 0 L p (Z ) 4 Fourier-analytic methods 178 and (4.34) (with ε = 1, say) that p c(ξ )e(ξ · x) ξ ∈S L p (Z ) =O p ∞ λ p−1 e−λ 2 /5 dλ . 0 To estimate the integral, we observe from elementary calculus that the integrand √ 2 λ p−1 e−λ /5 is bounded by O( p) p/2 for λ = O( p), and then decays exponentially √ for λ p. From this we can easily bound the integrand by p O(1) O( p) p/2 , and the claim follows (note that p 1/ p is bounded by e). In the next few sections we shall use Rudin’s inequality to obtain structural control on various sets of frequencies. Exercises 4.5.1 4.5.2 4.5.3 Show that the ( p) constant of a set S does not depend on the choice of bilinear form used to define the Fourier transform, and is also invariant under translations or isomorphisms of the set S. For any 2 ≤ p ≤ ∞ and any disjoint S1 , S2 , show the triangle inequality S( p) ≤ S1 ( p) + S2 ( p) whenever S ⊆ S1 ∪ S2 . Let ε be the uniform distribution on {−1, 1}, and let ε1 , . . . , ε N be independent trials of ε. If c1 , . . . , c N are arbitrary complex numbers and 2 ≤ p < ∞, prove Bernstein’s inequality p 1/ p 1/2 N N 2 \|c j \| ≤E εjcj j=1 j=1 ⎛ 1/2 ⎞ N √ ⎠. ≤ O⎝ p \|c j \|2 j=1 (Hint: for the lower bound, compute the p = 2 moment. For the upper bound, modify the proof of Lemma 4.33; alternatively, apply Lemma 4.33 to the group Z = Z2N , where S is the standard basis for Z2N .) Conclude that if f 1 , . . . , f N are any complex-valued functions on Z , then we have Khintchine’s inequality ⎛ p ⎞1/ p 1/2 N N 2 ⎠ ⎝ \| f \| ≤ E ε f j j j j=1 j=1 p L p (Z ) L (Z ) ⎞ ⎛ 1/2 N √ ⎠. ≤ O⎝ p \| f j \|2 p j=1 L (Z ) 4.5 ( p) constants, Bh [g] sets, and dissociated sets 4.5.4 179 Let f : Z 1 × Z 2 → C be a function on two variables in two nonempty finite sets Z 1 , Z 2 , and let 2 ≤ p < ∞. Establish the Minkowski inequality 1/ p 1/2 E y∈Z 2 (Ex∈Z 1 \| f (x, y)\|2 ) p/2 ≤ Ex∈Z 1 (E y∈Z 2 \| f (x, y)\| p )2/ p (4.36) (Hint: use the triangle inequality for the L p/2 norm.) Conclude that S( p) is the best constant such that 1/2 2 c(ξ )e(x, ξ ) ≤ S( p) c(ξ ) H ξ ∈S ξ ∈S H L p (Z ) 4.5.5 for all finite-dimensional Hilbert spaces H and all sequences (c(ξ ))ξ ∈S taking values in H . Using this, conclude that S1 × S2 ( p) = S1 ( p) S2 ( p) whenever S1 , S2 are additive sets in finite additive groups Z 1 , Z 2 and 2 ≤ p ≤ ∞. , Let n ≥ 1 be an integer, let Z := Zn2 . For ξ = (ξ1 , . . . , ξn ) ∈ Zn2 , let \|ξ \| denote the number of coefficients ξ1 , . . . , ξn which are equal to one. Establish the Bonami–Beckner inequality \|ξ \| ε c(ξ ) ≤ cl 2 (Z ) ξ ∈Z 2 L 1+1/ε (Z ) 4.5.6 4.5.7 4.5.8 4.5.9 4.5.10 for all 0 < ε < 1 and all c ∈ l 2 (Z). (Hint: first establish this by hand for n = 1, and then exploit (4.36) to obtain the general case.) Conclude in particular that if Sk := {ξ ∈ Zn2 : \|ξ \| = k}, then Sk ( p) ≤ ( p − 1)k/2 for all 2 < p < ∞. Let 2 ≤ p ≤ ∞, and let S be a non-empty subset of Z . Prove (4.31). (Hint: use the Hausdorff–Young inequality.) If 2 < p < ∞, show that equality occurs if and only if S is a translate of a subgroup of Z . (You may need Exercise 4.2.9.) Let S be an additive set in a finite additive group. Show that S( p) ≥ min 1, \|Z \|−1/ p \|S\|1/2 for all 2 ≤ p < ∞. It turns out that these bounds are essentially sharp for randomly chosen sets S in Z of a fixed cardinality: see . Let S be a Bh set in a finite additive group Z . Show that \|S\| ≤ \|Z \|1/ h . Complete the proof of Proposition 4.29. Let S be an additive subset of Z . Show that E(S, S) ≤ S4(4) \|S\|2 ; thus the additive energy of an additive set is controlled by its (4) constant. 180 4.5.11 4 Fourier-analytic methods Let S be an additive set, and let h ≥ 1. Suppose that A > 0 is a constant such that \|S \|h \|h S \| ≥ 2h A for all non-empty subsets S of S. Show that S(2h) = O(A(1 + log \|S\|)); 4.5.12 thus Lemma 4.30 can be reversed after conceding a factor of a logarithm. (Hint: first verify the estimate (4.30) when c is a characteristic function by reversing the proof of Lemma 4.30. For general c, decompose c into at most O(1 + log \|S\|) functions which are comparable to constant multiples of characteristic functions, by partitioning the range of c using powers of 2, and discarding those values of c smaller than (say) \|S\|−100 cl 2 .) Show that S( p) is the best constant such that fˆl 2 (S) ≤ S( p) f L p (Z ) for all random variables f , where p is the dual exponent to p, thus 1/ p + 1/ p = 1. Next, write \|S\| fˆl22 (Z ) = e(ξ · (x − y)) f 2L 2 (Z ) + Ex,y∈Z f (x) f (y)I(x = y) \|Z \| ξ ∈S and observe the inequalities e(ξ · (x − y)) ≤ f L 2 (Z ) g L 2 (Z ) Ex,y∈Z f (x)g(y)I(x = y) ξ ∈S and Ex,y∈Z f (x)g(y)I(x = y) e(ξ · (x − y)) ≤ \|Z \|Su f L 1 (Z ) g L 1 (Z ) . ξ ∈S Using Riesz–Thorin interpolation (or arguing as in Exercise 4.2.3) conclude that e(ξ · (x − y)) Ex,y∈Z f (x)g(y)I(x = y) ξ ∈S ≤ (\|Z \|Su )1−2/ p f L p (Z ) g L p (Z ) . From this, conclude the Tomas–Stein inequality S2( p) ≤ \|S\|\|Z \|− p + (Su \|Z \|)1− p 2 2 (compare with (4.31)). Thus, Fourier-uniform sets tend to have fairly small ( p) constants. See also Lemma 10.22. 4.6 The spectrum of an additive set 181 4.6 The spectrum of an additive set We now use Fourier analysis to investigate the spectral properties of additive sets A which have high additive energy E(A, A); examples of such sets include sets with small sum set \|A + A\| or small difference set \|A − A\| (cf. (2.8)). One can already conclude from estimates such as (4.23) that such sets must be highly nonuniform, i.e. 1 A contains non-trivial Fourier coefficients. However, this by itself is not the strongest Fourier-analytic statement one can say about such sets. In order to proceed further it is convenient to introduce the notion of the α-spectrum of a set. Definition 4.34 (Spectrum) Let A be an additive set in a finite additive group Z with a non-degenerate symmetric bilinear form · and let α ∈ R be a parameter. We define the α-spectrum Specα (A) ⊆ Z to be the set Specα (A) := {ξ ∈ Z : \|1A (ξ )\| ≥ αP Z (A)}. One could define this spectrum without the assistance of the bilinear form ·, but then it would be a subset of the Pontryagin dual group Zˆ rather than Z . From Lemma 4.9 we see that the sets Specα (A) are symmetric, decreasing in α, empty for α > 1, contain the origin for α ≤ 1, and are the whole space Z whenever α ≤ 0. Thus the spectrum is really only an interesting concept when 0 < α ≤ 1. In the extreme case α = 1 the spectrum becomes a group, see Exercise 4.6.2. From (4.16) (and Markov’s inequality) we observe the upper bound \|Specα (A)\| ≤ α −2 /P Z (A) (4.37) on the cardinality of the α-spectrum. In fact we can use Rudin’s inequality to obtain a more precise structural statement, in which the polynomial loss in P Z (A) is replaced with a logarithmic loss. To prove this statement, we first need an easy lemma (cf. Corollary 1.42). Lemma 4.35 (Cube covering lemma) Let S be an additive set in an ambient group Z , and let d ≥ 1 be an integer. Then we can partition S = D1 ∪ · · · ∪ Dk ∪ R where D1 , . . . , Dk are disjoint dissociated subsets of S of cardinality d + 1, and the remainder set R is contained in a cube [−1, 1]d · (η1 , . . . , ηd ) for some η1 , . . . , ηd ∈ Z . Proof We use the greedy algorithm. We initially set k = 0. If we can find a dissociated subset D of S of cardinality d + 1, we remove it from S and add it to the collection D1 , . . . , Dk , thus incrementing k + 1. We continue in this manner until we are left with a remainder R where all dissociated subsets of S have cardinality d or less. Let {η1 , . . . , ηd } be a dissociated subset of R with maximal cardinality; thus d ≤ d. Observe that if R contained an element ξ which was not contained in [−1, 1]d · (η1 , . . . , ηd ), then {η1 , . . . , ηd , ξ } would be dissociated, 182 4 Fourier-analytic methods so contradicting maximality of d . Thus we have R ⊆ [−1, 1]d · (η1 , . . . , ηd ), and the claim follows (padding out the progression with some dummy elements ηd +1 , . . . , ηd if necessary). Lemma 4.36 (Fourier concentration lemma) Let A be an additive set in a finite additive group Z , and let 0 < α ≤ 1. Then there exist d = O(α −2 (1 + log P Z1(A) )) and frequencies η1 , . . . , ηd ∈ Z such that Specα (A) ⊆ [−1, 1]d · (η1 , . . . , ηd ). This result is essentially sharp in a number of ways; see . It will suffice to show that for each phase θ ∈ R/Z, the set α Sθ := ξ ∈ Z : Re e(θ )1A (ξ ) ≥ P Z (A) 2 can be contained in a progression of the desired form, since from Definition 4.34 we see that Specα (A) is contained in the union of a bounded number of the Sθ , and we can simply add all the progressions together (here the fact that we have α/2 instead of α in the definition of Sθ is critical). Fix θ. By Lemma 4.35, it will suffice to show that 1 −2 \|S \| ≤ Cα 1 + log P Z (A) Proof for all dissociated sets S in Sθ . But if S ∈ Sθ , then by definition of Sθ α Re e(θ ) 1A (ξ )1 S (ξ ) ≥ P Z (A)\|S \|. 2 ξ ∈Z Let f (x) := \|S 1\|1/2 ξ ∈S e(x, ξ ) be the normalized inverse Fourier transform of 1 S ; then by (4.3) the left-hand side is equal to Re e(θ )\|S \|1/2 E Z 1 A f . Thus we have α E Z 1 A \| f \| ≥ P Z (A)\|S \|1/2 . 2 The left-hand side can be rewritten as ∞ EZ 1 A\| f \| = Px∈Z (x ∈ A; \| f (x)\| ≥ λ) dλ, 0 cf. (1.6). To bound Px∈Z (x ∈ A; \| f (x)\| ≥ λ), we can either use the trivial bound 2 of P Z (A) or use (4.34) to obtain a bound of Ce−λ /5 (for instance). Thus we have λ α 2 min P Z (A), Ce−λ /5 dλ ≥ P Z (A)\|S \|1/2 . 2 0 The left-hand side is at most CP Z (A)(1 + log1/2 1 ), P Z (A) and the claim follows. 4.6 The spectrum of an additive set 183 The above lemma suggests that the spectrum has some additive structure. This is confirmed by the following closure properties of the α-spectrum under addition: Lemma 4.37 Let A be an additive set in an finite additive group Z , and let ε, ε > 0. Then we have Spec1−ε (A) + Spec1−ε (A) ⊆ Spec1−2(ε+ε ) . (4.38) In a similar spirit, for any 0 < α ≤ 1 and for any non-empty S ⊆ Specα (A) we have 2 (ξ1 , ξ2 ) ∈ S × S : ξ1 − ξ2 ∈ Spec 2 (A) ≥ α \|S\|2 (4.39) α /2 2 See Exercise 4.6.2 for the ε = 0 case of this lemma. This lemma should be compared with Lemma 2.33. Indeed there is a strong analogy between the spectra Specα (A) and the symmetry sets Symα (A), which are heuristically dual to each other. Proof We first prove (4.38). Let ξ ∈ Spec1−ε and ξ ∈ Spec1−ε , then there exists phases θ, θ ∈ R/Z such that Re Ex∈Z e(ξ · x + θ )1 A (x) ≥ (1 − ε)P Z (A); Re Ex∈Z e(ξ · x + θ )1 A (x) ≥ (1 − ε )P Z (A). Since Re Ex∈Z 1 A = P Z (A), we thus have Re Ex∈Z [2e(ξ · x + θ ) + 2e(ξ · x + θ ) − 3]1 A (x) ≥ (1 − 2(ε + ε ))P Z (A). To conclude that ξ + ξ ∈ Spec1−2(ε+ε ) (A), it will thus suffice to establish the pointwise estimate Re [2e(ξ · x + θ ) + 2e(ξ · x + θ ) − 3] ≤ Re ei(θ +θ ) e(x, ξ + ξ ) . Writing e(ξ · x + θ) = eiβ and e(ξ · x + θ ) = eiβ for some −π/2 ≤ β, β ≤ −π/2, we reduce to showing 2 cos(β) + 2 cos(β ) − 3 ≥ cos(β + β ). But by the convexity of cos between −π/2 and π/2, we have β + β 2 cos(β) + 2 cos(β ) − 3 ≥ 4 cos −3 2 2 β + β 2 β + β = 2 cos − 1 − 2 1 − cos 2 2 ≥ cos(β + β ) as desired. 4 Fourier-analytic methods 184 Now we prove (4.39), which is due to Bourgain . Set a(ξ ) := sgn(1ˆ A (ξ )) for ξ ∈ S; thus Ex∈Z a(ξ )e(ξ · x)1 A (x) = \|1ˆ A (ξ )\| ≥ αP Z (A)\|S\|. ξ ∈S ξ ∈S Applying Cauchy–Schwarz, we conclude 2 Ex∈Z a(ξ )e(ξ · x) 1 A (x) ≥ α 2 P Z (A)\|S\|2 . ξ ∈S But the left-hand side can be rearranged as a(ξ1 )a(ξ2 )1A (ξ1 − ξ2 ), ξ1 ,ξ2 ∈S so by the triangle inequality we have \|1A (ξ1 − ξ2 )\| ≥ α 2 \|S\|2 . ξ1 ,ξ2 ∈S In particular (cf. Exercise 1.1.4) \|1A (ξ1 − ξ2 )\| ≥ α 2 /2\|S\|2 ξ1 ,ξ2 ∈S:ξ1 −ξ2 ∈Specα2 /2 (A) and (4.39) follows. We now show that small sum sets force large spectra (cf. Exercise 4.3.9, or Exercise 4.6.3 below). Lemma 4.38 Let A be an additive set in an finite additive group Z , and let 0 < α ≤ 1. For any integers n, m ≥ 0 with (n, m) = (0, 0), we have the lower bound on sum sets \|A\| \|n A − m A\| ≥ . \|Specα (A)\|P Z (A) + α 2(n+m)−2 Proof We may take n, m ≥ 0. Consider the function f = 1 A ∗ · · · ∗ 1 A ∗ 1−A ∗ · · · ∗ 1−A formed by convolving n copies of A and m copies of −A. Then f is non-negative and supported on n A − m A, and thus E Z f ≤ P Z (n A − m A)1/2 (E Z \| f \|2 )1/2 . From (4.10) we have E Z f = P Z (A)n+m . From (4.9) and (4.17) we have fˆ = m n 1A 1A . Combining these inequalities with (4.2) we see that \|Z \|P Z (A)2(n+m) \|n A − m A\| ≥ . 2(n+m) ξ ∈Z \|1 A (ξ )\| 4.6 The spectrum of an additive set But \|1A (ξ )\|2(n+m) ≤ ξ ∈Z 185 P Z (A)2(n+m) ξ ∈Specα (A) + α 2(n+m)−2 P Z (A)2(n+m)−2 \|1A (ξ )\|2 ξ ∈Specα (A) ≤ P Z (A)2(n+m) \|Specα (A)\| + α 2(n+m)−2 P Z (A)2(n+m)−1 and the claim follows. Now we consider the following inverse-type question: if A has additive structure in the sense that its energy E(A, A) is large or its difference set \|A − A\| is small, is it possible to approximate A (or a closely related set) by a Bohr set? We give two results of this type, one which places a relatively large Bohr set inside 2A − 2A, and another which places A − A inside a relatively small Bohr set. We begin with the former result, the main idea of which dates back to Bogolyubov. Proposition 4.39 Let 0 < α ≤ 1, and let A be an additive set in a finite additive group Z such that E(A, A) ≥ 4α 2 \|A\|3 . Then we have the inclusion 1 Bohr Specα (A), ⊆ 2A − 2A. (4.40) 6 Proof Let x be any element of the Bohr set Bohr(Specα (A), 16 ), thus Re e(ξ · x) > 1 for all ξ ∈ Specα (A). To show that x ∈ 2A − 2A, it would suffice to show that 2 1 A ∗ 1 A ∗ 1−A ∗ 1−A (x) = 0. But from (4.4), (4.9), (4.17) we have 1 A ∗ 1 A ∗ 1−A ∗ 1−A (x) = \|1ˆ A (ξ )\|4 e(ξ · x). ξ ∈Z Now take real parts of both sides and use the hypothesis on x to obtain 1 A ∗ 1 A ∗ 1−A ∗ 1−A (x) = \|1ˆ A (ξ )\|4 Re e(ξ · x) + ξ ∈Specα (A) ≥ \|1ˆ A (ξ )\|4 Re e(x, ξ ) ξ ∈Specα (A) 1 \|1ˆ A (ξ )\|4 − \|1ˆ A (ξ )\|4 2 ξ ∈Spec (A) ξ ∈Spec (A) α α 1 ˆ 3 = \|1 A (ξ )\|4 − \|1ˆ A (ξ )\|4 2 ξ ∈Z 2 ξ ∈Spec (A) α 1 E(A, A) 3 2 ≥ − α P Z (A)2 \|1ˆ A (ξ )\|2 2 \|Z \|3 2 ξ ∈Z 1 E(A, A) 3 2 − α P Z (A)3 2 \|Z \|3 2 >0 ≥ as desired, where we have used the hypothesis on α in the last step. 4 Fourier-analytic methods 186 Now we give a converse inclusion, which applies to sets of small difference constant δ[A] but requires the spectral threshold to be very large. Proposition 4.40 Let K ≥ 1. If A is an additive set in a finite additive group Z such that \|A − A\| ≤ K \|A\| (i.e. δ[A] ≤ K ) and 0 < ε < 1, then √ A − A ⊆ Bohr(Spec1−ε (A − A), 8εK ). Proof Let x, y ∈ A and ξ ∈ Spec1−ε (A − A). Then there exists a phase θ ∈ R/Z such that Re e(ξ · x + θ ) ≥ (1 − ε)\|A − A\| z∈A−A and hence (1 − Re e(ξ · x + θ)) ≤ ε\|A − A\| ≤ εK \|A\|. z∈A−A Since the summand is non-negative, and A − A contains both x − a and y − a, we thus have \|1 − Re e(ξ · (x − a) + θ )\| ≤ εK \|A\| a∈A and hence by Cauchy–Schwarz \|1 − Re e(ξ · (x − a) + θ )\|1/2 ≤ ε 1/2 K 1/2 \|A\|. a∈A From the elementary identity \|1 − e(α)\| = we conclude that √ 2\|1 − Re e(α)\|1/2 \|1 − e(ξ · (x − a) + θ )\| ≤ √ 2ε 1/2 K 1/2 \|A\|. a∈A Similarly for x replaced by y. By the triangle inequality we conclude that √ \|e(ξ · (y − a) + θ) − e(ξ · (x − a) + θ)\| ≤ 22ε 1/2 K 1/2 \|A\|. a∈A But the left-hand side is just \|A\|e(ξ · (x − y)); thus √ \|e(ξ · (x − y)) − 1\| ≤ 8εK . Since ξ ∈ Spec1−ε (A − A) was arbitrary, the claim follows from (4.24). In the next chapter we apply these propositions, together with the additive geometry results from Chapter 3, to obtain Freiman-type theorems in finite additive 4.6 The spectrum of an additive set 187 groups. For now, we shall give one striking application of the above machinery, namely the following Gauss sum estimate of Bourgain and Konyagin: Theorem 4.41 Let F = F p be a finite field of prime order, and let H be a multiplicative subgroup of F such that \|H \| ≥ p δ for some 0 < δ < 1. Then, if p is sufficiently large depending on δ, we have H u ≤ p −ε for some ε = ε(δ) > 0. In other words, we have sup e(xξ ) ≤ p −ε \|H \|. ξ ∈Z p \0 x∈H Proof We may use the standard bilinear form ξ · x = xξ/ p. Since h · H = H for all h ∈ H , we easily verify that 1ˆ H (h −1 ξ ) = 1ˆ H (ξ ) for all h ∈ H and ξ ∈ Z . This implies in particular that Specα (H ) = H · Specα (H ). Thus each Specα (H ) consists of multiplicative cosets of H , together with the origin 0. We use an iteration and pigeonhole argument, similar to that used to prove Theorem 2.35. Let J = J (δ) ≥ 1 be a large integer to be chosen later, and let ε = ε(J, δ) > 0 be a small number also to be chosen later. Define the sequence 1 > α1 > · · · > α J +1 > 0 by setting α1 := p −ε and α j+1 := α 2j /2. Suppose for contradiction that H u > p −ε ; then Specα1 (H ) contains a non-zero element, and hence by the preceding discussion \|Specα1 (H )\| ≥ \|H \| + 1 ≥ p δ + 1. Since Specα j (H ) is increasing in j, we see from the pigeonhole principle that there exists 1 ≤ j ≤ J such that \|Specα j+1 (H )\| ≤ p 1/J \|Specα j (H )\|. On the other hand, from Lemma 4.37 we have \|{(ξ1 , ξ2 ) ∈ Specα j (H ) × Specα j (H ) : ξ1 −ξ2 ∈ Specα j+1 (A)}\| ≥ α 2j 2 \|Specα j (H )\|2 . Applying Cauchy–Schwarz or Lemma 2.30 we conclude that E(Specα j (H ), Specα j (H )) = J p −O J (ε)−O(1/J ) \|Specα j (H )\|3 . If we let A := Specα j (H ){0}, we thus obtain E(A, A) = J p −O J (ε)−O(1/J ) \|A\|3 since \|A\| ≥ p δ , J is large enough depending on δ, and ε small enough depending on J , δ. But A is a union of cosets x · H of H for various x ∈ F p {0}. Applying Exercise 2.3.20 E(A, x · H ) = J p −O J (ε)−O(1/J ) \|A\|\|H \|2 . 4 Fourier-analytic methods 188 Dilating this by x −1 we obtain E(x −1 · A, H ) = J p −O J (ε)−O(1/J ) \|A\|\|H \|2 . But this will contradict Corollary 2.62 if J is sufficiently large depending on δ, and ε sufficiently small. In this result was extended (using slightly different arguments) to the case where H was not a multiplicative subgroup, but merely had small multiplicative doubling, for instance \|H · H \| ≤ p ε \|H \|. In the result was further extended to the case where the field F p was replaced by a commutative ring such as F p × F p (with Theorem 2.63 playing a key role in the latter result). This yields some estimates on exponential sums related to the Diffie–Hellman distribution and to Mordell sums; see , for further discussion. Exercises 4.6.1 4.6.2 4.6.3 Let A be an additive set in a finite additive group Z and let α ∈ R. Show that A, −A, and T h A all have the same spectrum for any h ∈ Z ; thus Specα (A) = Specα (−A) = Spec(T h A). If φ : Z → Z is a group isomorphism of Z , show that Specα (φ(A)) = φ † (Specα (A)), where φ † is the adjoint of φ, defined in Exercise 4.1.8. Let A be an additive set in Z . Show that the spectrum Spec1 (A) is a group and is in fact equal to (A − A)⊥ , the orthogonal complement of the group generated by A − A. Also, recall that Sym0 (A) := {h ∈ A : A + h = A} is the symmetry group of A; show that the orthogonal complement Sym0 (A)⊥ of this group is the smallest group which contains the Specα (A) for all α > 0. Let A be an additive set in an finite additive group Z , and let 0 < α ≤ 1. Establish the inequalities α 4 \|Specα (A)\|P Z (A) ≤ 4.6.4 4.6.5 E(A, A) ≤ \|Specα (A)\|P Z (A) + α 2 . \|A\|3 Thus, large energy forces large spectrum (and conversely). Let 0 < α ≤ 1, and let A, B be additive sets in Z with \|A\| = \|B\| = N 2 and E(A, B) ≥ 4α 2 N 3 . Show that \|Specα (A) ∩ Specα (B)\| ≥ 2αN\|Z \| . Thus pairs of sets with large additive energy must necessary have a large amount of shared spectrum. If A is an additive set in a finite additive group Z , and A is an additive set in a finite additive group Z , show that Specα (A) × Specβ (A ) ⊆ Specαβ (A × A ) for all 0 < α, β ≤ 1, where we give Z × Z the bilinear form induced from Z and Z . 4.7 Progressions in sum sets 4.6.6 4.6.7 189 Show that Theorem 4.41 implies Corollary 2.62. (Hint: use (4.14).) Let S be a subset of a finite additive group Z , and let 0 < ρ < 1/4. Show that if A is any additive set in Bohr(S, ρ), then S ⊆ Speccos(πρ) (A). This can be viewed as a kind of converse to Proposition 4.39. 4.7 Progressions in sum sets A cornerstone of additive combinatorics is Szemer´edi’s theorem. One form of this theorem states that if A is a subset of the interval [1, N ] with positive density α, then A contains an arithmetic progression of length f (N , α), where f tends to infinity as N does and α is fixed. In Chapters 10 and 11, we will discuss this result in more detail, but let us mention here that f tends to infinity very slowly as a function of N . In this section, we are going to show that if we replace the additive set A by a larger set, such as A + B, A + A + A, or 2A − 2A, then one can locate significantly larger progressions inside these sets by taking advantage of the existence of functions supported on those sets with good Fourier transform, namely 1 A ∗ 1 B , 1 A ∗ 1 A ∗ 1 A and 1 A ∗ 1 A ∗ 1−A ∗ 1−A . To illustrate this, we begin with a theorem of Chang (based on earlier work of Ruzsa ) which demonstrates the existence of a large generalized progression inside 2A − 2A; this theorem will be a key ingredient in one of the formulations of Freiman’s theorem (see Theorem 5.30). Theorem 4.42 (Chang’s theorem) Let K , N ≥ 1. Let A be an additive set in a cyclic group Z = Z N such that E(A, A) ≥ \|A\|3 /K . Then there exists a proper progression P ⊆ 2A − 2A of rank at most O(K (1 + log P Z1(A) )) and size 1 \|P\| ≥ O K 1 + log P Z (A) −O(K (1+log P 1(A) )) Z N. (4.41) Furthermore we may choose P to be symmetric (−P = P). Note from (2.8) that the hypothesis E(A, A) ≥ \|A\|3 /K will be obeyed if \|A + A\| ≤ K \|A\| or \|A − A\| ≤ K \|A\|; thus this theorem covers the case of sets with small doubling constant or small Ruzsa diameter. Alternatively, from the trivial bound E(A, A) ≥ \|A\|2 we see this hypothesis is always satisfied with K = 1/P Z (A), but this is costly as the dependence of (4.41) on K is exponential. On the other hand, if A has small doubling then this theorem can be applied efficiently even when A is a rather sparse subset of Z . 4 Fourier-analytic methods 190 Proof Set α := 1/2K 1/2 . By Proposition 4.39, we have 1 Bohr Specα (A), ⊆ 2A − 2A. 2 On the other hand, from Lemma 4.36 we can find a set S := {η1 , . . . , ηd } of frequencies with 1 1 −2 d = \|S\| = O α 1 + log = O K 1 + log P Z (A) P Z (A) such that Specα (A) ⊆ [−1, 1]d · (η1 , . . . , ηd ). This implies (from the triangle inequality) that 1 1 Bohr S, ⊆ Bohr Specα (A), . 6d 6 Applying Proposition 4.23 we see that Bohr(S, 6d1 ) contains a proper symmetric progression of rank d and cardinality −O(K (1+log P 1(A) )) Z (1/6d)d 1 \|P\| ≥ N ≥ O K 1 + log N d d P Z (A) and the claim follows. In the proof of the above theorem (or more precisely, in the proof of Proposition 4.39) one took advantage of the fact that 1 A ∗ 1 A ∗ 1−A ∗ 1−A had positive Fourier coefficients \|1A (ξ )\|4 . However, it turns out that with a slight modification to the argument one does not need positivity of the Fourier coefficients, and in fact one only needs three summands instead of four: Theorem 4.43 Let K , N ≥ 1. Let A1 , A2 , A3 be additive sets in Z N such that \|A1 \| = \|A2 \| = \|A3 \| and \|A1 + A2 + A3 \| ≤ K \|A1 \|. Then there exists a 1 proper progression P ⊆ A1 + A2 + A3 of rank at most O(K 2 (1 + log P Z (A )) and 1) size \|P\| ≥ O K 1 + log 1 P Z (A1 ) −O(K 2 (1+log P 1 )) Z (A1 ) N. (4.42) One can of course generalize the hypotheses to deal with sets A1 , A2 , A3 of differing cardinalities, but the statement of the theorem becomes a little messier and we do not pursue it here. 4.7 Progressions in sum sets 191 Proof We adapt some arguments of . We consider the non-negative function f := 1 A1 ∗ 1 A2 ∗ 1 A3 . From (4.10) we have E Z f = P Z (A1 )3 . On the other hand, we have P Z (supp( f )) = P Z (A1 + A2 + A3 ) = K P Z (A1 ). By the pigeonhole principle, we can thus find an element x0 ∈ A1 + A2 + A3 such that f (x0 ) ≥ P Z (A1 )2 /K . By translating one of the A j , if necessary, we may assume x0 = 0, thus f (0) ≥ P Z (A1 )2 /K . Next, we observe from (4.9) that fˆ(ξ ) = 1 A1 (ξ )1 A2 (ξ )1 A3 (ξ ). From (4.4), Cauchy–Schwarz, (4.16) and (4.24) we thus have for any x ∈ Z \| f (x) − f (0)\| = (ξ ) 1 (ξ ) 1 (ξ )(e(ξ · x) − 1) 1 A2 A3 ξ ∈Z A1 ≤ \|1 A1 (ξ )\|\|1 A2 (ξ )\|\|1 A3 (ξ )\|\|e(ξ · x) − 1\| ξ ∈Z ≤ sup \|1 A1 (ξ )\|\|e(ξ · x) − 1\| 1 A2 L 2 (Z ) 1 A3 (ξ ) L 2 (Z ) ξ ∈Z = P Z (A1 ) sup \|1 A1 (ξ )\|\|e(ξ · x) − 1\| ξ ∈Z ≤ 2π P Z (A1 ) sup \|1 A1 (ξ )\|ξ · xR/Z . ξ ∈Z Combining this with our bound on f (0) and the support of f , we see that x ∈ Z : sup \|1 (ξ )\| ξ · x < P (A )/2π K ⊆ A1 + A2 + A3 . A1 R/Z Z 1 ξ ∈Z Since \|1 A1 (ξ )\| ξ · xR/Z < P Z (A1 )/2π K whenever ξ ∈ Spec1/2π K (A1 ), we obtain x∈Z: sup \|1 A1 (ξ )\| ξ · xR/Z < P Z (A1 )/2π K ⊆ A1 + A2 + A3 . ξ ∈Spec1/2π K (A1 ) Moreover, as \|1 A1 (ξ )\| ≤ P Z (A1 ) for all non-zero ξ , we obtain Bohr(Spec1/2π K (A1 ), 1/2π K ) ⊆ A1 + A2 + A3 1 (for instance). But by Lemma 4.36 we can find d = O(K 2 (1 + log P Z (A )) and 1) frequencies S := {η1 , . . . , ηd } ⊂ Z such that Spec1/2π K (A1 ) ⊆ [−1, 1]d · (η1 , . . . , ηd ) and hence by the triangle inequality Bohr(S, 1/2πd K ) ⊆ Bohr(Spec1/2π K (A1 ), 1/2π K ) ⊆ A1 + A2 + A3 . 4 Fourier-analytic methods 192 Applying Proposition 4.23, we can locate a proper progression P in Bohr(S, 1/2π d K ) of rank d and cardinality at least (1/2d K )d 2 N ≥ (C K (1 + log(1/P Z (A1 ))))−C K (1+log(1/P Z (A1 ))) N dd and the claim follows. \|P\| ≥ The above arguments relied crucially on having three or more summands; roughly speaking, two of the summands were treated by Plancherel’s theorem, leaving at least one other summand to be free to exploit the smallness of its Fourier coefficients outside of its spectrum. They break down quite significantly for sums of two sets1 . Nevertheless, it is still possible to obtain some relatively large progressions in a set of the form A + B, because the function 1 A ∗ 1 B still has l 1 type control on the Fourier coefficients. We follow the arguments of Bourgain . We first give a convenient criterion for establishing the existence of progressions. Lemma 4.44 (Almost periodicity implies long progressions) Let f : Z → R+ be a non-negative random variable on an additive group Z , let J ≥ 1 be an integer, and suppose that r ∈ Z is such that E Z max \|T jr f − f \| < E Z f, 1≤ j≤J where T f (x) := f (x − jr ) is the shift of f by jr . Then supp( f ) contains an arithmetic progression a + [0, J ] · r of length J + 1 and spacing r . jr Proof By the pigeonhole principle, there exists x ∈ Z such that max \|T jr f (x) − f (x)\| < f (x) 1≤ j≤J and hence f (x − jr ) = T jr f (x) > 0 for all 0 ≤ j ≤ J . The claim follows. To apply this lemma, we need to estimate expressions of the form E Z max1≤ j≤J \|T jr f − f \|. This can be done easily if f has Fourier transform in a dissociated set: Lemma 4.45 Let S ⊆ Z be a dissociated set, and let f be a random variable such that supp( fˆ) ⊆ S. Then for any non-empty set of shifts H ⊂ Z we have max \|T h f \| 2 = O(1 + log \|H \|)1/2 f L 2 (Z ) . h∈H 1 L (Z ) There is a similarity with the Goldbach conjectures. The weak conjecture – every large odd number is the sum of three primes – has been solved by Fourier methods, whereas the strong conjecture – every large even number is the sum of two primes – is still open, and probably not amenable to a purely Fourier-analytic method. 4.7 Progressions in sum sets Proof 193 Let p > 2 be a large exponent to be chosen later. Then max \|T h f \| 2 ≤ max \|T h f \| p h∈H h∈H L (Z ) L (Z ) !1/ p h p ≤ \|T f \| p L (Z ) h∈H = 1/ p T h f L p (Z ) h∈H ≤ \|H \|1/ p f L p (Z ) ≤ \|H \|1/ p S( p) f L 2 (Z ) √ = O \|H \|1/ p p f L 2 (Z ) by Rudin’s inequality (Lemma 4.33). The claim now follows by setting p := O(1 + log \|H \|). By combining this lemma with Lemma 4.35, we can obtain an estimate when supp( fˆ) is not dissociated, but fˆ is uniform in size: Lemma 4.46 Let f be a random variable, and let J, d > 1. Suppose that there exists an integer m such that 2m ≤ \| fˆ(ξ )\| ≤ 2m+1 for all ξ ∈ supp( fˆ). Then one can find a set S ⊂ Z of cardinality \|S\| = d such that such that ⎛ ⎞ " log J E Z max \|T jr f − f \| = O ⎝ \| fˆ(ξ )\| + J d max η · r R/Z ⎠ η∈S 1≤ j∈J d ˆ ξ ∈supp( f ) for all r ∈ Z . Proof Applying Lemma 4.35, we may write supp( fˆ) = D1 ∪ · · · ∪ Dk ∪ R where D1 , . . . , Dk are disjoint dissociated sets of cardinality d + 1, and R ⊆ [−1, 1]d · (η1 , . . . , ηd ) for some S = {η1 , . . . , ηd } ⊂ Z . Using the Fourier transform, we may then split f = f D1 + · · · + f Dk + f R accordingly. From Lemma 4.45 we have, for any 1 ≤ i ≤ k, E Z max \|T jr f Di − f Di \| ≤ 2 max \|T jr f Di \| 2 1≤ j∈J 0≤ j∈J L (Z ) 1/2 ≤ O log J f Di L 2 (Z ) ⎛ 1/2 ⎞ ⎠ = O ⎝log1/2 J \| fˆ(ξ )\|2 " ≤O ξ ∈Di log J ˆ \| f (ξ )\| D ξ ∈Di 194 4 Fourier-analytic methods thanks to the uniformity assumption 2m ≤ \| fˆ(ξ )\| ≤ 2m+1 . Also, we have from the triangle inequality, (4.24) and the hypothesis on R jr max \|T f R − f R \| ˆ(ξ )\| × \|e(x + jr, ξ ) − e(ξ · x)\| ≤ max \| f 1≤ j∈J 1 1≤ j≤J ξ ∈R L (Z ) L 1 (Z ) ≤ \| fˆ(ξ )\| max \|e( jr, ξ ) − 1\| ξ ∈R ≤ 2π J d 1≤ j≤J ;ξ ∈R ˆ \| f (ξ )\| max η · r R/Z . η∈S ξ ∈R Summing these estimates using the triangle inequality, the claim follows. Now we can prove Bourgain’s theorem. Theorem 4.47 Let N ≥ 1 be a prime number, and let A, B be additive sets in log N )3 Z N such that \|A\|, \|B\| ≥ δ N for some C (loglog < δ ≤ 1 for some large absolute N constant C > 1. Then A + B contains a proper arithmetic progression of length at least exp((δ log N )1/3 ). Proof We may assume N to be large. By removing elements from A and B and increasing δ if necessary we may assume P Z (A) = P Z (B) = δ. Set f := 1 A ∗ 1 B , and let exp((δ log N )1/3 ) ≤ J < N be chosen later: thus supp( f ) = A + B and E Z f = P Z (A)P Z (B) = δ 2 ; note also that J 1/δ. By Lemma 4.44, it suffices to show that E Z max \|T jr f − f \| < δ 2 1≤ j≤J for some non-zero r . The Fourier coefficients fˆ of f cannot exceed fˆ(0) = E Z f = δ. Furthermore we have by Cauchy–Schwarz and (4.16) \| fˆ(ξ )\| = \|1A (ξ )\| \| 1 B (ξ )\| ξ ∈Z ξ ∈Z ≤ 1A l 2 (Z ) 1 B l 2 (Z ) = P Z (A) 1/2 P Z (B) 1/2 = δ. To exploit this, we let M ≥ 1 be chosen later and partition # Z= m ∪ err 0≤m Report "Additive Combinatorics 9780521853866, 9780511245305, 0511245300, 0521853869" × Close Submit Contact information Michael Browner info@dokumen.pub Address: 1918 St.Regis, Dorval, Quebec, H9P 1H6, Canada. 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15455	https://en.wikipedia.org/wiki/Legendre%27s_equation	Jump to content Search Contents (Top) 1 References Legendre's equation Català Deutsch Français עברית Русский Svenska Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Special diophantine equation involving squares For the equation satisfied by Legendre polynomials, see Legendre's differential equation. In mathematics, Legendre's equation is a Diophantine equation of the form: The equation is named for Adrien-Marie Legendre who proved it in 1785 that it is solvable in integers x, y, z, not all zero, if and only if −bc, −ca and −ab are quadratic residues modulo a, b and c, respectively, where a, b, c are nonzero, square-free, pairwise relatively prime integers and also not all positive or all negative. References [edit] L. E. Dickson, History of the Theory of Numbers. Vol.II: Diophantine Analysis, Chelsea Publishing, 1971, ISBN 0-8284-0086-5. Chap.XIII, p. 422. J.E. Cremona and D. Rusin, "Efficient solution of rational conics", Math. Comp., 72 (2003) pp. 1417-1441. \| \| \| This number theory-related article is a stub. You can help Wikipedia by expanding it. \| v t e Retrieved from " Categories: Diophantine equations Number theory stubs Hidden categories: Articles with short description Short description is different from Wikidata All stub articles Legendre's equation Add topic
15456	https://arxiv.org/html/2509.14138v1	SeqVLA: Sequential Task Execution for Long-Horizon Manipulation with Completion-Aware Vision-Language-Action Model Back to arXiv Back to arXiv This is experimental HTML to improve accessibility. We invite you to report rendering errors. Use Alt+Y to toggle on accessible reporting links and Alt+Shift+Y to toggle off. Learn more about this project and help improve conversions. Why HTML?Report IssueBack to AbstractDownload PDF Table of Contents Abstract I INTRODUCTION II Related Work II-A Vision-Language-Action Models II-B The π 0\pi_{0} Model III Methods III-A Problem Formulation III-B VLA Architecture with Completion Detection Head III-C Subtask Execution IV Experiments IV-A Experimental Setup IV-B Dataset construction IV-C Finetuning Strategy IV-D Long-horizon task results V CONCLUSIONS and FUTURE WORK V-A Conclusion V-B Future Work References No License arXiv:2509.14138v1 [cs.RO] 17 Sep 2025 SeqVLA: Sequential Task Execution for Long-Horizon Manipulation with Completion-Aware Vision-Language-Action Model Report issue for preceding element Ran Yang 1⋆, Zijian An 2⋆, Lifeng Zhou 2, and Yiming Feng 1†1 Ran Yang and Yiming Feng are with Virginia Seafood Agricultural Research and Extension Center, and Department of Biological Systems Engineering, Virginia Tech, 15 Rudd Ln, Hampton, VA, 23669, USA {ryang17,yimingfeng}@vt.edu 2 Zijian An and Lifeng Zhou are with the Department of Electrical and Computer Engineering, Drexel University, 3141 Chestnut St, Philadelphia, PA 19104, USA {za382, lz457}@drexel.edu⋆\star Equally contributed†\dagger Corresponding author Report issue for preceding element Abstract Report issue for preceding element Long-horizon robotic manipulation tasks require executing multiple interdependent subtasks in strict sequence, where errors in detecting subtask completion can cascade into downstream failures. Existing Vision-Language-Action (VLA) models such as π 0\pi_{0} excel at continuous low-level control but lack an internal signal for identifying when a subtask has finished, making them brittle in sequential settings. We propose SeqVLA, a completion-aware extension of π 0\pi_{0} that augments the base architecture with a lightweight detection head perceiving whether the current subtask is complete. This dual-head design enables SeqVLA not only to generate manipulation actions but also to autonomously trigger transitions between subtasks. We investigate four finetuning strategies that vary in how the action and detection heads are optimized (joint vs. sequential finetuning) and how pretrained knowledge is preserved (full finetuning vs. frozen backbone). Experiments are performed on two multi-stage tasks: salad packing with seven distinct subtasks and candy packing with four distinct subtasks. Results show that SeqVLA significantly outperforms the baseline π 0\pi_{0} and other strong baselines in overall success rate. In particular, joint finetuning with an unfrozen backbone yields the most decisive and statistically reliable completion predictions, eliminating sequence-related failures and enabling robust long-horizon execution. Our results highlight the importance of coupling action generation with subtask-aware detection for scalable sequential manipulation. Report issue for preceding element I INTRODUCTION Report issue for preceding element Robotic manipulation in complex tasks often involves executing long-horizon tasks that require precise temporal sequencing. Such sequential dependencies arise in many real-world applications: when preparing a meal, a robot must first retrieve a plate, then place utensils, followed by arranging food items in a specific order; in kit assembly, it must insert components into a fixture in the correct sequence, where skipping or misordering steps leads to failure; in laboratory automation, liquid handling tasks must follow an exact pipetting and dispensing order to preserve chemical validity. These settings exhibit sparse supervision, implicit subgoal transitions, and strong error propagation across stages. A key challenge is enabling manipulation policies to detect subtask completion and determine when and how to transition to the next stage [1, 2]. Naive end-to-end models often struggle under such constraints, lacking a reliable internal signal to segment the task or recover from temporal drift. Report issue for preceding element Vision-Language-Action (VLA) models are based on large models pretrained on vast robotic datasets and then fine-tuned on domain-specific robotic data to ground instructions in perception and control. They have shown strong capability in mapping rich multimodal context to action sequences with semantic alignment, enabling general-purpose control in a variety of single-stage settings [3, 4, 5, 6, 7, 8, 9, 10]. These models have been used to support robots in performing practical behaviors such as folding clothes, placing objects, and other structured manipulation tasks. However, their vanilla formulation lacks an internal notion of subtask completion, making them brittle when naively chained for long-horizon, sequential objectives: they either switch prematurely, linger redundantly, or propagate early-stage mistakes into downstream failure or perform in the wrong order [11, 12]. Report issue for preceding element Classical approaches to long-horizon control often impose structure to handle sequencing. Modern methods handle long horizons using temporal abstraction and hierarchy. Instead of fixed options, they learn hierarchical world models and use goal-conditioned planners to operate at multiple timescales [13, 14]. Related methods aim to determine when a subtask or goal is completed. Learned termination critics, progress monitors, and vision-language success detectors have been used to give adaptive stop signals or catch failures in multi-stage policies [15, 16]. Despite these advances, termination signals are usually kept separate from action generation, and hierarchical or planning systems still depend on external controllers or hand-specified intermediate goals to sequence stages. As a result, a gap remains: how to endow a VLA generation model with a tightly coupled, learned signal that detects when each one is complete, allowing it to autonomously trigger the next stage and self-pace the chaining of dependent subtasks without external control. Report issue for preceding element In this paper, we bridge this gap with Sequential-VLA (SeqVLA), which is built upon the π 0\pi_{0} framework and features a Completion Detection Head. Given the same visual, linguistic, and action-history context, this head predicts whether the current subtask has been successfully achieved. We explore finetuning strategies by freezing different sections of π 0\pi_{0} to identify which components need adaptation for reliable sequential execution. We then evaluate the proposed model on two distinct scenarios, demonstrating substantial improvements in overall success rate compared to the original π 0\pi_{0} and several strong baselines. Our contributions are: Report issue for preceding element •We integrate a learned task completion detection head into the π 0\pi_{0} model, enabling the model to infer subtask completion from multi-modal context. Report issue for preceding element •We identify the most effective finetuning strategy for the augmented VLA by freezing different sections of π 0\pi_{0}, demonstrating how to adapt the model to generate correct subtasks in sequence reliably. Report issue for preceding element •We evaluate the resulting framework in two real-world sequential scenarios and compare it with the strong baseline, demonstrating that our approach yields substantially superior task-level performance. Report issue for preceding element II Related Work Report issue for preceding element II-A Vision-Language-Action Models Report issue for preceding element Vision-Language-Action (VLA) models aim to unify visual perception, language understanding, and continuous robot control within a single generative framework. Instead of decoupling language grounding, visual recognition, and action synthesis into separate modules, VLA systems directly learn the conditional distribution. Report issue for preceding element p(𝐚 1:T∣𝐨 1:T,𝐥),p(\mathbf{a}{1:T}\mid\mathbf{o}{1:T},\mathbf{l}), over action sequences 𝐚 1:T\mathbf{a}{1:T}, conditioned on image and state observations 𝐨 1:T\mathbf{o}{1:T} and a natural language prompt 𝐥\mathbf{l}. This joint modeling approach enables better semantic alignment between high-level instructions and low-level behaviors, particularly in open-world or zero-shot settings. Report issue for preceding element Most VLA systems employ a vision-language encoder backbone, such as BLIP or SigLIP , to embed both the language command and visual observations into a shared representation space. The output of the VLM is then passed to an action-generation module, which produces either continuous control commands or discrete action tokens depending on the model architecture. Report issue for preceding element The core distinction between different VLAs lies in how they generate action sequences. One class of models, exemplified by SayCan and PaLM-E , decomposes tasks into high-level symbolic skills and uses large language models to plan over these skills before grounding them into low-level controllers. This symbolic planning approach facilitates semantic reasoning, but often relies on predefined skill libraries and structured environments. A second category of methods, such as RT-1 , RT-2 , and OpenVLA , formulates action prediction as a token-by-token autoregressive generation task. These models discretize continuous actions into token sequences and train transformers to decode them from vision-language context. A third class of approaches, including VIMA , CoT-VLA [10, 9], and SmoVLA , leverages latent trajectory modeling, contrastive learning, or smooth policy regularization in the latent space. These models typically embed demonstrations into compact latent codes and generate behavior via infilling or trajectory decoding. Report issue for preceding element II-B The π 0\pi_{0} Model Report issue for preceding element The π 0\pi_{0} model is a representative VLA framework that emphasizes continuous high-frequency robotic control, with a particular focus on robustness to horizon length and ambiguity. Unlike models that decode action sequences token-by-token or operate in a latent trajectory space, π 0\pi_{0} directly learns a continuous function π θ(𝐚 t∣𝐳 t)\pi_{\theta}(\mathbf{a}{t}\mid\mathbf{z}{t}) to predict actions based on a shared observation-language context 𝐳 t\mathbf{z}_{t}. Report issue for preceding element The core of π 0\pi_{0} is a flow matching strategy for trajectory supervision. During finetuning, the model first samples a noisy initial action 𝐚 t(0)\mathbf{a}{t}^{(0)} and a ground-truth action 𝐚 t(1)\mathbf{a}{t}^{(1)} from human demonstration. A random scalar τ∈[0,1]\tau\in[0,1] is drawn to interpolate between them via a straight-line path: Report issue for preceding element 𝐚 t(τ)=(1−τ)⋅𝐚 t(0)+τ⋅𝐚 t(1).\mathbf{a}{t}^{(\tau)}=(1-\tau)\cdot\mathbf{a}{t}^{(0)}+\tau\cdot\mathbf{a}_{t}^{(1)}. The model is trained to predict the velocity vector pointing from 𝐚 t(0)\mathbf{a}{t}^{(0)} to 𝐚 t(1)\mathbf{a}{t}^{(1)} using the intermediate 𝐚 t(τ)\mathbf{a}_{t}^{(\tau)} and conditioning context, resulting in the following supervised loss: Report issue for preceding element ℒ flow=𝔼 τ,𝐚 t(0),𝐚 t(1)[‖π θ(𝐚 t(τ),τ,𝐳 t)−(𝐚 t(1)−𝐚 t(0))‖2 2].\mathcal{L}{\text{flow}}=\mathbb{E}{\tau,\mathbf{a}{t}^{(0)},\mathbf{a}{t}^{(1)}}\left[\left\\|\pi_{\theta}(\mathbf{a}{t}^{(\tau)},\tau,\mathbf{z}{t})-(\mathbf{a}{t}^{(1)}-\mathbf{a}{t}^{(0)})\right\\|_{2}^{2}\right]. This formulation allows the policy to match the target velocity field over the entire trajectory manifold rather than only at the final endpoints, offering smoother control and better resilience to noise or partial observability. Report issue for preceding element Compared to token-based autoregressive models, π 0\pi_{0} benefits from continuous-time reasoning and variable rollout lengths. It does not require action discretization or fixed-length decoding, which helps mitigate error accumulation and temporal drift across long sequences. The flow-based formulation also supports iterative refinement: even if the intermediate predictions are ambiguous, the model can gradually converge toward the goal through continuous updates. These design choices make π 0\pi_{0} particularly suitable for robotic manipulation tasks that demand both temporal smoothness and precise low-level control, providing a compelling advantage over symbolic or token-based VLAs that often struggle with continuity, compounding errors, or rigid horizon constraints.1 1 1 At the time our experiments were conducted, the latest variant π 0.5\pi_{0.5} had been announced but was not yet publicly released. Accordingly, we base our implementation on the most recent available model, π 0\pi_{0}. Report issue for preceding element However, despite these advantages, π 0\pi_{0} operates uniformly over entire action segments through flow matching on observation-language context and lacks an explicit mechanism for identifying or sequencing subtasks. Specifically, the action finetuning is solely based on 𝐳 t\mathbf{z}{t} without awareness of task completion status during sequential execution. This limitation poses challenges in structured long-horizon manipulation, where the agent must not only produce smooth low-level motion but also recognize when a subtask is complete and transition to the next. To address this, we extend π 0\pi{0} with a dedicated completion detection head that monitors subtask status, and a chaining mechanism that enables self-paced progression across sequential stages. Report issue for preceding element III Methods Report issue for preceding element Figure 1: Architecture of SeqVLA. The inputs are the same as π 0\pi_{0}. The completion detection head takes the feature layer from the Action Expert as input to generate the task completion probability. Report issue for preceding element III-A Problem Formulation Report issue for preceding element Real-world manipulation tasks, such as food handling or assembly operations, usually require sequential execution of multiple subtasks. We fomulate such long-horizon tasks as a sequential order 𝒯={𝒯 1,𝒯 2,…,𝒯 n}\mathcal{T}={\mathcal{T}{1},\mathcal{T}{2},...,\mathcal{T}{n}} where the completion of task 𝒯 i\mathcal{T}{i} is the prerequisite of the initialization of task 𝒯 i+1\mathcal{T}_{i+1}. Challenges arise when using existing VLA models, as they may struggle to execute subtasks in the correct order, especially when visual scenes appear similar across different subtask stages. For example, a partially filled food container would resemble a nearly filled one before or after loading the last food type. SeqVLA addresses the task sequence confusion issue by learning not only the action to execute but also when each subtask is completed, which enables the proper execution of long-horizon tasks that consist of multiple subtasks. Report issue for preceding element III-B VLA Architecture with Completion Detection Head Report issue for preceding element SeqVLA extends a standard VLA architecture with a lightweight completion detection head, as illustrated in Figure1. The model takes multi-view visual observations from three cameras, the current robot joint states, and a language prompt as inputs, and jointly produces two outputs: continuous robot control actions for the current subtask and a binary classification score indicating whether the subtask has been completed. The completion signal then triggers a transition to the next subtask in the sequence, enabling structured long-horizon execution. We implement SeqVLA on top of π 0\pi_{0}, which provides a strong foundation through its SigLIP vision encoder, Gemma-2B language backbone, and Gemma-300M action expert. By leveraging π 0\pi_{0}’s extensive pretraining across diverse robotic tasks and systems, SeqVLA inherits robust low-level control and mistake recovery capabilities. At the same time, our additional completion detection mechanism equips it with the ability to reason over subtask boundaries and enforce correct sequencing. SeqVLA extends the base π 0\pi_{0} model with a parallel completion detection output that shares the exact feature representations from the action expert layer. The original action head retains the function of predicting a 14-DoF output for manipulating a dual-arm Aloha. Our add-on task-tracking head performs binary classification to determine the completion status of the ongoing subtask. This feature-sharing architecture guarantees computational efficiency during inference while leveraging the information extracted by the powerful pre-trained backbone model. Report issue for preceding element Completion detection head, shown in Figure 1, is a binary classifier that transforms the feature representations from the action expert into a probability, which we denote as task state indicator p p, Report issue for preceding element p=σ(W⋅F+b),p=\sigma(\textbf{W}\cdot\textbf{F}+b), where W∈ℝ 1024,b∈ℝ\textbf{W}\in\mathbb{R}^{1024},b\in\mathbb{R} are learnable parameters, σ\sigma is sigmoid activation function σ(x)=1/(1+e−x)\sigma(x)=1/(1+e^{-x}), and 𝐅∈ℝ 1024\mathbf{F}\in\mathbb{R}^{1024} is the generated feature layer from the action expert. We measure the loss of task completion term by binary cross-entropy loss : Report issue for preceding element L completion(y,y^)=−[y⋅log(y^)+(1−y)⋅log(1−y^)].L_{\text{completion}}(y,\hat{y})=-[y\cdot log(\hat{y})+(1-y)\cdot log(1-\hat{y})]. Accordingly, the loss function is given by L total=L action+λ⋅L completion L_{\text{total}}=L_{\text{action}}+\lambda\cdot L_{\text{completion}} where L action L_{\text{action}} is the standard π o\pi_{o} action prediction loss from the 3M action expert. We set λ=0.1\lambda=0.1 as the weight for the completion loss contribution. Report issue for preceding element III-C Subtask Execution Report issue for preceding element SeqVLA offers a simple yet effective solution that leverages the binary head output to manage subtask transitions and maintain the accuracy of the task sequence. It operates by processing each inference to obtain both an action chunk for execution and a completion assessment for transition decisions. At each inference step, SeqVLA produces two outputs: an action chunk and an execution indicator p∈[0,1]p\in[0,1], which represents the probability that the agent should continue the current subtask (p p close to 1) or terminate it and transition to the next (p p close to 0). When a subtask completion signal is detected (p<τ p<\tau), the indicator would initialize a transition process: 1) stop execution of the robot action immediately, 2) send the robot to home pose for consistent task initialization, and 3) begin the following subtask by switching the task prompt. A threshold τ=0.2\tau=0.2 is selected to trigger task transitions based on our empirical results.2 2 2 In experiments, it is observed that when the threshold is set too low (e.g., 0.1), the policy becomes overly conservative and rarely issues a stop signal, making it difficult to terminate at the correct time. Conversely, when the threshold is set too high, the policy becomes overly sensitive to minor variations in the scale readings, which often leads to premature stopping and fluctuating behavior. Report issue for preceding element IV Experiments Report issue for preceding element IV-A Experimental Setup Report issue for preceding element Our experiments feature a bimanual 14-DoF Aloha robot, equipped with three cameras for collecting RGB images from central top, left wrist, and right wrist views during teleoperation, as shown in Figure˜2 and Figure˜3. We collected two types of teleoperation demonstrations: individual subtask demonstrations for finetuning SeqVLA, and complete long-horizon sequential demonstrations for finetuning the baseline π 0\pi_{0}. Such a data collection strategy enables the precise labeling of subtask completion status for SeqVLA, while also allowing π 0\pi_{0} to perform the same long-horizon tasks for fair comparison. We present experiments on two different long-horizon tasks, each consisting of multiple subtasks, as illustrated in Figure˜4(a). The detailed setup for the two experiments is illustrated in Figure˜4(b) and Figure˜4(c). We have attached a video of our experiments to the supplementary material. Report issue for preceding element Salad Packing: this environment is set up to mimic a food service scenario, where the robot is assigned to load the foodstuffs in the designated order (spinach, coleslaw, meatball, chicken, tomato, and sauce cup) into the food container and then close it. The left arm handles loading tasks for spinach and coleslaw, while the right arm picks up and loads the other items. Container closing requires bimanual coordination for lid placement and securing. This task assesses the ability to maintain the correct sequential order across diverse subtasks involving different objects and manipulations. Report issue for preceding element Candy Packing: this task examines scenarios where certain subtasks are executed multiple times within a single long-horizon sequence. The robot picks gummies (left arm), places Kinder chocolates twice consecutively (left arm), adds Snickers bars twice (right arm), and finishes with lollipop placement (right arm). This task examines the capability to handle repeated subtask execution within a long-horizon sequence. Report issue for preceding element Figure 2: Experiment setup. The three red boxes denote the cameras: two are mounted on the gripper to monitor the conditions at the gripper’s end, and one is positioned in front of the robot to provide a global view. The two green boxes indicate the robotic arms responsible for grasping objects. The two blue boxes correspond to the kinesthetic teaching arms, which are used to guide the motion of the front robotic arms during data collection for finetuning. Report issue for preceding element (a)Camera views in salad task Report issue for preceding element (b)Camera views in candy task. Report issue for preceding element Figure 3: Camera view on two tasks, with high camera (left), left camera (middle), and right camera (right). Report issue for preceding element (a)Subtask sequences for long-horizon salad and candy packing tasks. Report issue for preceding element (b)Task1: Salad packing Report issue for preceding element (c)Task2: Candy packing Report issue for preceding element Figure 4: Task setup. (a) expected long-horizon task sequences for the Salad and Candy tasks, showing the sequential order of all subtasks. (b),(c) visualization of the experimental setup for the two tasks, where the green boxes mark the robotic arm. Report issue for preceding element IV-B Dataset construction Report issue for preceding element We collect two complementary types of demonstration data. The first type is subtask-level demonstrations, where each subtask in Figure˜4(a) is paired with a unique language prompt. For example, S-1 S\text{-}1 is prompted as “Pick up the spinach,” while C-2 C\text{-}2 is prompted as “Pick up the kinder.” These demonstrations are used to fine-tune SeqVLA to ground prompts into subtask-specific behaviors and to learn subtask completion detection. The second type is long-horizon demonstrations, where entire salad-packing and candy-packing tasks are executed following the designed subtask sequences. These complete trajectories are used to finetune the baseline VLA π 0\pi_{0} for end-to-end long-horizon execution. Report issue for preceding element All subtask-level demonstrations are manually annotated with binary labels to indicate whether each frame still contributes to the execution of the subtask. Frames labeled as 1 correspond to movements that are actively advancing the subtask. In contrast, frames labeled as 0 correspond to non-contributory actions, such as holding the arm stationary after an object has already been placed in the container. Consequently, the label sequence follows patterns such as 1111⋯\cdots 1100⋯\cdots 00), where 1s indicate the execution phase and 0s indicate the post-completion phase. In practice, the transition to label 1 is determined by strict criteria: the target object must have been successfully placed in the container, the gripper must be fully released, and there should be no contact between the gripper and the container. These conditions ensure that subsequent reset motions, such as returning the arm to the home pose, do not interfere with the current task state. Report issue for preceding element IV-C Finetuning Strategy Report issue for preceding element Integrating a completion detection head into a pre-trained VLA model raises the question of how to fine-tune the dual-head architecture best so that the model can adapt to predicting both actions and task completion without erasing its pre-trained competencies. To integrate the additional completion-detection head into the pre-trained π 0\pi_{0} backbone, we consider two orthogonal design dimensions in the finetuning procedure as illustrated in Figure˜5. The first dimension concerns how to optimize the dual outputs, balancing action prediction with task-completion classification. The second dimension addresses how to preserve the generalizable knowledge obtained through π 0\pi_{0}’s pretraining while enabling adaptation to our tasks. By combining choices along these two dimensions, we arrive at four distinct finetuning configurations for SeqVLA, which we evaluate systematically. Report issue for preceding element We first examine different approaches for finetuning the dual outputs of SeqVLA. Since the model must simultaneously predict low-level manipulation actions and high-level task completion signals, a natural option is to fine-tune both heads jointly, allowing the two objectives to inform each other during optimization. However, joint optimization may also introduce conflicts, especially if the classification head is immature in early finetuning and destabilizes action learning. To address this, we also consider a sequential strategy in which the action head and backbone are trained first to ensure stable control, and the classification head is introduced afterward to learn on top of a well-initialized representation. The following outlines these two types of finetuning in more detail. Report issue for preceding element •Joint finetuning: Action and classification heads are trained simultaneously for coupled learning of manipulation and task completion detection. Report issue for preceding element •Sequential finetuning: Action head is trained first together with the entire backbone, with the classification head frozen; then the entire backbone with the action head is frozen to fine-tune the completion classification head. Report issue for preceding element In addition, we explore different strategies for preserving the knowledge obtained through π 0\pi_{0}’s pretraining while adapting to our tasks. Full fine-tuning allows all parameters, including the pre-trained vision-language backbone, to update, thereby providing maximum flexibility for domain-specific adaptation. However, such aggressive updating risks overwriting the generalizable representations that π 0\pi_{0} has acquired from large-scale pretraining. As a contrasting strategy, we also consider freezing the pre-trained backbone so that only the action expert and the two prediction heads are updated. This approach helps safeguard the original vision-language grounding ability of π 0\pi_{0}, but may limit task-specific adaptation capacity. Report issue for preceding element •Full finetuning strategy: All parameters, including the pre-trained VLM backbone, are allowed to update during finetuning for domain-specific learning on our tasks. Report issue for preceding element •Frozen backbone strategy: Pre-trained VLM backbone is fixed during finetuning while only the action expert, as well as the two prediction heads, are trained, for preserving original vision-language knowledge obtained via pre-finetuning. Report issue for preceding element Combining the two dimensions discussed above, we obtain four distinct finetuning configurations for SeqVLA: Report issue for preceding element •SeqVLA-J (Joint finetuning, No Freezing). Report issue for preceding element •SeqVLA-JF (Joint finetuning, Freezing). Report issue for preceding element •SeqVLA-S (Sequential finetuning, No Freezing). Report issue for preceding element •SeqVLA-SF (Sequential finetuning, FreezingSS). Report issue for preceding element (a)SeqVLA-J Report issue for preceding element (b)SeqVLA-JF Report issue for preceding element (c)SeqVLA-S Report issue for preceding element (d)SeqVLA-SF Report issue for preceding element Figure 5: Four finetuning strategies. Report issue for preceding element Figure 6: Finetuning strategies comparison. Average success rate for all subtasks within the two long-horizon tasks. Report issue for preceding element We instantiate four SeqVLA models corresponding to the finetuning configurations, and evaluate each model on the two long-horizon tasks. For every configuration, we compute the execution success rate over all subtasks in the salad-packing and candy-packing sequences, as summarized in Figure˜6. The results show that configurations with an unfrozen backbone, regardless of whether trained jointly or sequentially, consistently achieve higher success rates across subtasks. This finding highlights the importance of allowing the pre-trained backbone to adapt during finetuning, as freezing it restricts the model’s ability to transfer pre-trained representations to our domain-specific tasks. Report issue for preceding element A further comparison of the completion-detection head is conducted on SeqVLA-J and SeqVLA-S, which are trained without freezing the backbone, as frozen-backbone variants already underperform substantially and are less informative for analyzing detection behavior. As shown in Figure˜7, which displays the raw binary classification outputs, SeqVLA-J produces predictions that are both more decisive and more consistent: the output values remain much closer to 1 during task execution and drop clearly after subtask completion. In contrast, SeqVLA-S exhibits a more dispersed distribution with larger variance, often producing intermediate confidence scores that blur the boundary between execution and completion. Report issue for preceding element TABLE I: Classification Confidence Comparison \| Task \| SeqVLA-S Entropy \| SeqVLA-J Entropy \| KS \| \| Salad Tasks \| \| Spinach \| 1.32 \| 0.76 \| 0.77 \| \| Cole Slaw \| 1.28 \| 0.70 \| 0.79 \| \| Meatball \| 0.79 \| 0.71 \| 0.57 \| \| Chicken \| 1.13 \| 0.81 \| 0.78 \| \| Tomato \| 1.47 \| 0.99 \| 0.69 \| \| Sauce \| 1.51 \| 0.57 \| 0.81 \| \| Container \| 1.19 \| 0.48 \| 0.79 \| \| Overall \| 1.35 \| 0.76 \| 0.75 \| \| Candy Tasks \| \| Gummies \| 0.69 \| 0.64 \| 0.52 \| \| Kindar \| 1.10 \| 0.79 \| 0.72 \| \| Snicker \| 1.78 \| 0.67 \| 0.85 \| \| Lollipop \| 0.83 \| 0.67 \| 0.80 \| \| Overall \| 1.11 \| 0.79 \| 0.72 \| Note: All comparisons show statistical significance (p< 0.001). Report issue for preceding element Report issue for preceding element We quantify these differences in Table˜I using the entropy of the classification outputs as a measure of confidence. Across all subtasks, SeqVLA-J consistently yields lower entropy values than SeqVLA-S, indicating that joint finetuning leads to sharper and more confident detection decisions. To further assess statistical reliability, we compute the Kolmogorov–Smirnov (KS) statistic between the output distributions of execution and completion phases. SeqVLA-J achieves substantially higher KS values, with all p-values below 0.001, confirming that its classification head separates the two phases more distinctly and with a significant margin. Report issue for preceding element (a)Salad task Report issue for preceding element (b)Candy task Report issue for preceding element Figure 7: Task completion prediction comparison between sequential finetuning (a. SeqVLA-S) and joint finetuning (b. SeqVLA-J) strategies. Left panels show scatter plots of predicted task completion confidence for individual subtasks, where points closer to 1 indicate higher confidence in continuing execution. Right panels show success rate distributions across different confidence thresholds for (a) the Salad task with seven subtasks and (b) the Candy task with four subtasks. Overall, models trained with joint optimization (SeqVLA-J) achieve higher confidence and success rates than those trained sequentially (SeqVLA-S). Report issue for preceding element Taken together, these results demonstrate that joint finetuning enables the completion-detection head to leverage shared representations with the action head, leading to more confident, statistically reliable, and practically useful subtask completion signals. In contrast, sequential finetuning tends to decouple the learning of detection from action execution, resulting in noisier predictions that may undermine reliable stage transitions in long-horizon tasks. Report issue for preceding element Report issue for preceding element Report issue for preceding element Figure 8: Task execution record of baseline π 0\pi_{0} policy. Report issue for preceding element Report issue for preceding element Report issue for preceding element Figure 9: Task execution record of the SeqVLA-J policy. Report issue for preceding element Figure 10: Success rate on long-horizon task. Report issue for preceding element IV-D Long-horizon task results Report issue for preceding element We evaluate the performance of the model trained with SeqVLA-J, i.e., the best model, on the salad and candy tasks. In addition, we also finetuned the baseline π 0\pi_{0} model to perform the salad and candy long-horizon tasks without subtask decomposition. The baseline model for each task is finetuned on complete sequential demonstrations (following sequences in Figure˜4(a)), and only executes the action chunks during inference for the entire workflow, without a specific subtask monitoring scenario. Report issue for preceding element As shown inFigure˜8 andFigure˜9, the finetuned π 0\pi_{0} model achieved a comparable success rate as the SeqVLA trained on individual subtask demonstrations for both salad and candy tasks. However, the raw π 0\pi_{0} model struggles to maintain the proper task sequences, frequently executing subtasks out of sequence or repeating completed subtasks. In contrast, SeqVLA-J with a subtask completion detection head exhibits superior performance in executing long-horizon tasks (as shown in Figure˜9), addressing the critical issue of sequential task management while maintaining overall success rates (as illustrated in Figure˜10). Although both models experience failure in some subtasks due to manipulation challenges, SeqVLA-J eliminates sequence-related failures. The dual-head architecture ensures that when failures occur, they stem from genuine manipulation difficulties rather than incorrect task ordering, making the behavior more predictable during execution. Report issue for preceding element From the success rates of the two long-horizon tasks, we observe that SeqVLA can handle long-horizon tasks that the original π 0\pi_{0} VLA model performs poorly on. The highly accurate performance of SeqVLA is based on π 0\pi_{0}’s sensitivity to prompts. In particular, during finetuning, we collected 50 sets of data for each subtask and then combined them into a unified dataset. For the salad packing task, which consists of seven subtasks, the training set contained a total of 350 episodes. For the candy packing task, which consists of four distinct subtasks, the training set contained 200 episodes. We then finetuned π 0\pi_{0} on these aggregated datasets, enabling the model to accurately infer from the input prompt which subtask it should execute at a given stage. However, as illustrated in Figure 8, when deployed for long-horizon tasks, the raw π 0\pi_{0} lacks the ability to sustain long-horizon task execution, particularly for functions with strict sequential requirements. π 0\pi_{0} often exhibits failures such as repeated actions or incorrect ordering. In contrast, SeqVLA mitigates these issues by incorporating a completion detection mechanism that monitors subtask status and ensures proper sequential execution, thereby enabling π 0\pi_{0} to accurately execute long-horizon tasks that require structured coordination across multiple subtasks. Report issue for preceding element V CONCLUSIONS and FUTURE WORK Report issue for preceding element V-A Conclusion Report issue for preceding element This work addresses the challenge of long-horizon sequential manipulation, where existing VLA models, such as π 0\pi_{0}, lack explicit signals for subtask completion. We proposed SeqVLA, which augments π 0\pi_{0} with a lightweight completion detection head to generate actions and subtask transitions jointly. By evaluating four finetuning strategies that balance dual-head optimization and pre-trained knowledge preservation, we showed that SeqVLA achieves higher success rates than the baseline across salad- and candy-packing benchmarks. Notably, joint finetuning with an unfrozen backbone yielded the most confident and reliable completion predictions. These results highlight that coupling action generation with explicit subtask-aware detection is a promising direction for scaling VLA models to complex sequential manipulation tasks. Report issue for preceding element V-B Future Work Report issue for preceding element Several promising directions emerge from this work that warrant further investigation. A natural extension is to move beyond linear task sequences toward hierarchical structures with conditional branching and parallel execution. Instead of a binary detection head, future models could employ multi-level state detection to track progress across subtasks and dependencies, enabling more sophisticated workflows such as multi-dish preparation or dual-arm coordination, where subtasks may run concurrently or adapt their order to environmental conditions. Report issue for preceding element A further extension is to explore human–robot collaboration. In many realistic scenarios, robots will work alongside human operators on long-horizon tasks. Extending our framework to such settings would require mechanisms for effective communication, shared understanding of subtask states, and smooth handover of responsibilities between human and robotic execution. Report issue for preceding element Finally, real-world deployment beyond laboratory conditions presents its own challenges, including safety constraints, fault tolerance, and integration into existing workflows. Robust error recovery mechanisms, particularly for unexpected subtask failures, will be essential to ensure reliable operation in environments such as restaurants, homes, or industrial facilities. Addressing these issues will be critical to bridging the gap between controlled experiments and practical, large-scale deployment. Report issue for preceding element References Report issue for preceding element ↑ Oliver Kroemer, Scott Niekum, and George Konidaris. A review of robot learning for manipulation: Challenges. Representations, and Algorithms, page 82, 2021. ↑ Mostafa Hussein and Momotaz Begum. Detecting incorrect visual demonstrations for improved policy learning. In Conference on Robot Learning, pages 1817–1827. PMLR, 2023. ↑ Kevin Black, Noah Brown, Danny Driess, Adnan Esmail, Michael Equi, Chelsea Finn, Niccolo Fusai, Lachy Groom, Karol Hausman, Brian Ichter, et al. π\pi 0: A vision-language-action flow model for general robot control. corr, abs/2410.24164, 2024. doi: 10.48550. arXiv preprint ARXIV.2410.24164. ↑ Danny Driess, Fei Xia, Mehdi SM Sajjadi, Corey Lynch, Aakanksha Chowdhery, Ayzaan Wahid, Jonathan Tompson, Quan Vuong, Tianhe Yu, Wenlong Huang, et al. Palm-e: An embodied multimodal language model. 2023. ↑ Michael Ahn, Anthony Brohan, Noah Brown, Yevgen Chebotar, Omar Cortes, Byron David, Chelsea Finn, Chuyuan Fu, Keerthana Gopalakrishnan, Karol Hausman, et al. Do as i can, not as i say: Grounding language in robotic affordances. arXiv preprint arXiv:2204.01691, 2022. ↑ Anthony Brohan, Noah Brown, Justice Carbajal, Yevgen Chebotar, Joseph Dabis, Chelsea Finn, Keerthana Gopalakrishnan, Karol Hausman, Alex Herzog, Jasmine Hsu, et al. Rt-1: Robotics transformer for real-world control at scale. arXiv preprint arXiv:2212.06817, 2022. ↑ Brianna Zitkovich, Tianhe Yu, Sichun Xu, Peng Xu, Ted Xiao, Fei Xia, Jialin Wu, Paul Wohlhart, Stefan Welker, Ayzaan Wahid, et al. Rt-2: Vision-language-action models transfer web knowledge to robotic control. In Conference on Robot Learning, pages 2165–2183. PMLR, 2023. ↑ Yunfan Jiang, Agrim Gupta, Zichen Zhang, Guanzhi Wang, Yongqiang Dou, Yanjun Chen, Li Fei-Fei, Anima Anandkumar, Yuke Zhu, and Linxi Fan. Vima: General robot manipulation with multimodal prompts. arXiv preprint arXiv:2210.03094, 2(3):6, 2022. ↑ Tony Z Zhao, Vikash Kumar, Sergey Levine, and Chelsea Finn. Learning fine-grained bimanual manipulation with low-cost hardware. arXiv preprint arXiv:2304.13705, 2023. ↑ Qingqing Zhao, Yao Lu, Moo Jin Kim, Zipeng Fu, Zhuoyang Zhang, Yecheng Wu, Zhaoshuo Li, Qianli Ma, Song Han, Chelsea Finn, et al. Cot-vla: Visual chain-of-thought reasoning for vision-language-action models. In Proceedings of the Computer Vision and Pattern Recognition Conference, pages 1702–1713, 2025. ↑ Christoph Willibald and Dongheui Lee. Hierarchical task decomposition for execution monitoring and error recovery: Understanding the rationale behind task demonstrations. The International Journal of Robotics Research, page 02783649251352112, 2025. ↑ Tao Huang, Kai Chen, Wang Wei, Jianan Li, Yonghao Long, and Qi Dou. Value-informed skill chaining for policy learning of long-horizon tasks with surgical robot. In 2023 IEEE/RSJ International Conference on Intelligent Robots and Systems (IROS), pages 8495–8501. IEEE, 2023. ↑ Robin Schiewer, Anand Subramoney, and Laurenz Wiskott. Exploring the limits of hierarchical world models in reinforcement learning. Scientific Reports, 14(1):26856, 2024. ↑ Mianchu Wang, Rui Yang, Xi Chen, Hao Sun, Meng Fang, and Giovanni Montana. Goplan: Goal-conditioned offline reinforcement learning by planning with learned models. arXiv preprint arXiv:2310.20025, 2023. ↑ Jiafei Duan, Wilbert Pumacay, Nishanth Kumar, Yi Ru Wang, Shulin Tian, Wentao Yuan, Ranjay Krishna, Dieter Fox, Ajay Mandlekar, and Yijie Guo. Aha: A vision-language-model for detecting and reasoning over failures in robotic manipulation. arXiv preprint arXiv:2410.00371, 2024. ↑ Qiao Gu, Yuanliang Ju, Shengxiang Sun, Igor Gilitschenski, Haruki Nishimura, Masha Itkina, and Florian Shkurti. Safe: Multitask failure detection for vision-language-action models. arXiv preprint arXiv:2506.09937, 2025. ↑ Junnan Li, Dongxu Li, Caiming Xiong, and Steven Hoi. Blip: Bootstrapping language-image pre-training for unified vision-language understanding and generation. In International conference on machine learning, pages 12888–12900. PMLR, 2022. ↑ Xiaohua Zhai, Basil Mustafa, Alexander Kolesnikov, and Lucas Beyer. Sigmoid loss for language image pre-training. In Proceedings of the IEEE/CVF international conference on computer vision, pages 11975–11986, 2023. ↑ Moo Jin Kim, Karl Pertsch, Siddharth Karamcheti, Ted Xiao, Ashwin Balakrishna, Suraj Nair, Rafael Rafailov, et al. Openvla: An open-source vision-language-action model. arXiv preprint arXiv:2406.09246, 2024. ↑ Mustafa Shukor, Dana Aubakirova, Francesco Capuano, Pepijn Kooijmans, Steven Palma, Adil Zouitine, Michel Aractingi, Caroline Pascal, Martino Russi, Andres Marafioti, et al. Smolvla: A vision-language-action model for affordable and efficient robotics. arXiv preprint arXiv:2506.01844, 2025. ↑ Physical Intelligence, Kevin Black, Noah Brown, James Darpinian, Karan Dhabalia, Danny Driess, Adnan Esmail, Michael Equi, Chelsea Finn, Niccolo Fusai, Manuel Y. Galliker, Dibya Ghosh, Lachy Groom, Karol Hausman, Brian Ichter, Szymon Jakubczak, Tim Jones, Liyiming Ke, Devin LeBlanc, Sergey Levine, Adrian Li-Bell, Mohith Mothukuri, Suraj Nair, Karl Pertsch, Allen Z. Ren, Lucy Xiaoyang Shi, Laura Smith, Jost Tobias Springenberg, Kyle Stachowicz, James Tanner, Quan Vuong, Homer Walke, Anna Walling, Haohuan Wang, Lili Yu, and Ury Zhilinsky. π 0.5\pi_{0.5}: a vision-language-action model with open-world generalization, 2025. ↑ Zipeng Fu, Tony Z Zhao, and Chelsea Finn. Mobile aloha: Learning bimanual mobile manipulation with low-cost whole-body teleoperation. arXiv preprint arXiv:2401.02117, 2024. ↑ Anqi Mao, Mehryar Mohri, and Yutao Zhong. Cross-entropy loss functions: Theoretical analysis and applications. In International conference on Machine learning, pages 23803–23828. pmlr, 2023. ↑ William H Press. Numerical recipes 3rd edition: The art of scientific computing. Cambridge university press, 2007. 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15457	https://glencoe.mheducation.com/sites/0078805732/student_view0/chapter6/lesson1/	Slope of a Line and Slope-Intercept Form MathMatters 3: An Integrated Program ==================================== Lesson 1: Slope of a Line and Slope-Intercept Form -------------------------------------------------- In this Lesson: Extra Examples Self-Check Quizzes Extra Examples Self-Check Quizzes Search Search for: Site Preferences(Log out) This form changes settings for this website only. To make changes to your user profile instead, please click here. Send mail as: Instructor email: TA email: Other email: "Floating" navigation? Drawer speed: Teacher Log In Log in here to access teaching material for this site. Username: Password: Online Student Edition Math Modeling with Applications Multilingual eGlossary Textbook Resources Chapter Themes Chapter Test Extra Examples MathWorks Concepts in Motion Personal Tutor Self-Check Quizzes Standardized Test Practice Vocabulary Review Chapter Activities Chapter Themes Chapter Test Standardized Test Practice MathWorks Lesson Resources Extra Examples Self-Check Quizzes Teacher Educational Partners Math Modeling with Applications Meet the Authors National Resources Professional Development State Resources Teaching Today TechConnect Student Online Student Edition Math Modeling with Applications Multilingual eGlossary Textbook Resources Chapter Themes Chapter Test Extra Examples MathWorks Concepts in Motion Personal Tutor Self-Check Quizzes Standardized Test Practice Vocabulary Review Chapter Activities Student Chapter Themes Chapter Test Standardized Test Practice MathWorks Lesson Resources Student Extra Examples Self-Check Quizzes Home>Student Center>Chapter 6>Lesson 1 Mathematics HomeProduct InfoSite MapContact Us Please read our Terms of Use and Privacy Policy before you explore our Web site. To report a technical problem with this Web site, please contact the Web Producer.
15458	https://www.grc.nasa.gov/www/k-12/BGP/airprop.html	Air Properties Definitions + Text Only Site + Non-Flash Version + Contact Glenn Earth's atmosphere is composed of air. Air is a mixture of gases, 78% nitrogen and 21% oxygen with traces of water vapor, carbon dioxide, argon, and various other components. We usually model air as a uniform (no variation or fluctuation) gas with properties that are averaged from all the individual components. Any gas has certain properties that we can detect with our senses. The values and relations of the properties define the state of the gas. On this slide you will find typical values of the properties of air at sea level static conditions for a standard day. We are all aware that pressure and temperature of the air depend on your location on the earth and the season of the year. And while it is hotter in some seasons than others, pressure and temperature change day to day, hour to hour, sometimes even minute to minute during severe weather. The values presented on the slide are simply average values used by engineers to design machines. That's why they are called standard values. We also know that all of the state-of-the-gas variables will change with altitude, which is why the typical values are given at sea level, static conditions. Because the gravity of the Earth holds the atmosphere to the surface, as altitude increases, air density, pressure, and temperature (for lower altitudes) decrease. At the edge of space, the density is almost zero. The variation of the air from the standard can be very important since it affects flow parameters like the speed of sound. A gas is composed of a large number of molecules which are in constant, random motion. The sum of the mass of all the molecules is equal to the mass of the gas. A gas occupies some volume in three dimensional space. For a given pressure and temperature, the volume depends directly on the amount of gas. Since the mass and volume are directly related, we can express both the mass and volume by a single variable. When a gas is moving, it is convenient to use the density of a gas, which is the mass divided by the volume the gas occupies. The sea level standard value of air density r is r = 1.229 kilograms/cubic meters = .00237 slug/cubic feet When working with a static or unmoving gas, it is more convenient to use specific volume, which is the volume divided by the mass. The sea level standard value of specific volume v is v = .814 cubic meters/kilogram = 422 cubic feet/slug The pressure of a gas equals the perpendicular force exerted by the gas divided by the surface area on which the force is exerted. The sea level standard value of air pressure p is p = 101.3 kilo Newtons/square meter = 14.7 pounds/square inch The temperature of a gas is a measure of the kinetic energy of the molecules of the gas. The sea level standard value of air temperature T is T = 15 degrees C = 59 degrees Fahrenheit A gas can exert a tangential (shearing) force on a surface, which acts like friction between solid surfaces. This "sticky" property of the gas is called the viscosity and it plays a large role in aerodynamic drag. The sea level standard value of air viscosity mu is mu = 1.73 x 10^-5 Newton-second/square meters = 3.62 x 10^-7 pound-second/square feet The density (specific volume), pressure, and temperature of a gas are related to each other through the equation of state. The state of a gas can be changed by external processes, and the reaction of the gas can be predicted using the laws of thermodynamics. A fundamental understanding of thermodynamics is very important in describing the operation of propulsion systems. Activities: Guided Tours Standard Atmosphere Model: Navigation .. Beginner's Guide Home Page + Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility CertificationEditor: Tom Benson NASA Official: Tom Benson Last Updated: May 13 2021 + Contact Glenn
15459	https://math.stackexchange.com/questions/2436782/computing-f0-given-other-information	algebra precalculus - Computing $f(0)$ given other information - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Computing f(0)f(0) given other information Ask Question Asked 8 years ago Modified8 years ago Viewed 38 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I encountered this problem today. f(x)f(x) is a quadratic polynomial with real coefficients. f(2)=0 f(2)=0 f(f(x))=0 f(f(x))=0 has only 1 1 real root at x=5 x=5 Compute f(0)f(0). I attempted to answer this question by writing f(x)f(x) in vertex form and using bullet 1 1. f(x)=a(x−h)2+k=a(x−2)[x−(2 h−2)]+k f(x)=a(x−h)2+k=a(x−2)[x−(2 h−2)]+k I don't know how to use the other bullet in the solution. I would like to know how to proceed from here. (The answer is −32 9−32 9) algebra-precalculus functions polynomials quadratics Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Sep 20, 2017 at 0:47 user343705 user343705 0 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. f(x)=a(x−2)(x−b)f∘f(x)=a(a(x−2)(x−b)−2)(a(x−2)(x−b)−b)f(x)=a(x−2)(x−b)f∘f(x)=a(a(x−2)(x−b)−2)(a(x−2)(x−b)−b) Since f∘f(x)f∘f(x) is a quartic (degree 4) equation and it has only one real root. 5 5 must be a root of multiplicity. Which means a(x−2)(x−b)−2=a(x−5)2 a(x−2)(x−b)−2=a(x−5)2 or a(x−2)(x−b)−b=a(x−5)2 a(x−2)(x−b)−b=a(x−5)2 Suppose: a(x−2)(x−b)−2=a(x 2−10 x+25)b=8 a(x 2−10 x+16−2 a)a=−2 9 a(x−2)(x−b)−2=a(x 2−10 x+25)b=8 a(x 2−10 x+16−2 a)a=−2 9 In which case, the other factor equals −2 9(x−2)(x−8)−8−2 9(x−5)2−6−2 9(x−2)(x−8)−8−2 9(x−5)2−6 which has no real roots. However if we solve... a(x−2)(x−b)−b=a(x 2−10 x+25)a(x−2)(x−b)−b=a(x 2−10 x+25) we would find the other factor equals: −8 9(x−5)2+6−8 9(x−5)2+6 and there would still be real roots. f(x)=−2 9(x−2)(x−8)f(x)=−2 9(x−2)(x−8)f(0)=−2 9(16)f(0)=−2 9(16) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 20, 2017 at 1:57 Doug MDoug M 58.8k 4 4 gold badges 35 35 silver badges 69 69 bronze badges Add a comment\| You must log in to answer this question. 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15460	https://www.convertunits.com/from/J/to/erg	Convert J to erg - Conversion of Measurement Units Convert joule to erg Please enable Javascript to use the unit converter. Note you can turn off most ads here: \| \| \| --- \| \| \| J \| \| \| erg \| \| \| More information from the unit converter How many J in 1 erg? The answer is 1.0E-7. We assume you are converting between joule and erg. You can view more details on each measurement unit: J or erg The SI derived unit for energy is the joule. 1 joule is equal to 10000000 erg. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between joules and ergs. Type in your own numbers in the form to convert the units! Quick conversion chart of J to erg 1 J to erg = 10000000 erg 2 J to erg = 20000000 erg 3 J to erg = 30000000 erg 4 J to erg = 40000000 erg 5 J to erg = 50000000 erg 6 J to erg = 60000000 erg 7 J to erg = 70000000 erg 8 J to erg = 80000000 erg 9 J to erg = 90000000 erg 10 J to erg = 100000000 erg Want other units? You can do the reverse unit conversion from erg to J, or enter any two units below: Enter two units to convert \| \| \| --- \| \| From: \| \| \| To: \| \| \| \| \| Common energy conversions Definition: Joule The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889). Definition: Erg An erg is the unit of energy and mechanical work in the centimetre-gram-second (CGS) system of units, symbol "erg". Its name is derived from the Greek word meaning "work". Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
15461	https://www.facebook.com/groups/ceboardexam/posts/2006282286467747/	CE Board Exam Study Group \| A differential manometer shown is measuring the difference in pressure two water pipes \| Facebook Log In Log In Forgot Account? CE Board Exam Study Group \| A differential manometer shown is measuring the difference in pressure two water pipes ================================================================================================================== ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- · 351.9K members Join group CE Board Exam Study Group Anonymous participant · September 24, 2024 · A differential manometer shown is measuring the difference in pressure two water pipes. The indicating liquid is mercury (specific gravity = 13.6), h1 is 675 mm, hm1 is 225 mm, and hm2 is 300 mm. What is the pressure differential between the two pipes. Please pahelp po mga Engineers. Thank you in advance All reactions: 2 7 comments Like Comment Share All comments Rafael Miguel Mazo 1y 7 View all 6 replies See more on Facebook See more on Facebook Email or phone number Password Log In Forgot password? or Create new account
15462	https://dokumen.pub/fluid-mechanics-and-hydraulic-machines-problems-and-solutions-1nbsped-9780070699809-0070699801.html	Fluid mechanics and hydraulic machines : problems and solutions [1 ed.] 9780070699809, 0070699801 - DOKUMEN.PUB Anmelden Registrierung Deutsch English Español Português Français Dom Najlepsze kategorie CAREER & MONEY PERSONAL GROWTH POLITICS & CURRENT AFFAIRS SCIENCE & TECH HEALTH & FITNESS LIFESTYLE ENTERTAINMENT BIOGRAPHIES & HISTORY FICTION Najlepsze historie Najlepsze historie Dodaj historię Moje historie Home Fluid mechanics and hydraulic machines : problems and solutions [1 ed.] 9780070699809, 0070699801 Fluid mechanics and hydraulic machines : problems and solutions [1 ed.] 9780070699809, 0070699801 52,252 6,116 63MB English Pages Year 2011 Report DMCA / Copyright DOWNLOAD FILE Polecaj historie ###### Fluid Mechanics: With Problems and Solutions, and an Aerodynamics Laboratory 112 38 8MB Read more ###### A textbook of fluid mechanics and hydraulic machines 9899107446, 9911310888, 8121916666 494 101 14MB Read more ###### A Textbook of Fluid Mechanics and Hydraulic Machines [5 ed.] 9899107446, 9911310888, 8121916666 15,059 1,489 17MB Read more ###### A Text Book of Fluid Mechanics and Hydraulic Machines [9 ed.] 8170083117, 9788170083115 16,523 1,890 30MB Read more ###### Engineering Mechanics: Problems and Solutions 1108411622, 9781108411622 This comprehensive and self-contained textbook will help students in acquiring an understanding of fundamental concepts 65,375 4,481 31MB Read more ###### Numerical Fluid Mechanics I Exercises with Solutions 134 25 3MB Read more ###### Introduction to Fluid Mechanics and Fluid Machines [2 ed.] 0070667624, 9780070667624 5,204 525 48MB Read more ###### Introduction To Fluid Mechanics And Fluid Machines [3 ed.] 0071329196, 9780071329194 12,634 1,342 9MB Read more ###### Hydraulic Machines 9781259006845 7,991 743 28MB Read more ###### Problems and Solutions on Thermodynamics and Statistical Mechanics 9810200560, 9789810200565 The material for these volumes has been selected from the past twenty years' examination questions for graduate stu 4,415 672 20MB Read more Author / Uploaded K. Subramanya _Table of contents : Cover Contents Properties of Fluids Introduction 1.1 Density, Specific Volume and Specific Weight 1.2 Pressure 1.3 Shear Stress and Viscosity 1.4 Surface Tension 1.5 Compressibility 1.6 Vapour Pressure Worked Examples Problems Objective Questions Fluid Statics Introduction 2.1 Pressure in a Static Fluid 2.2 Forces on Plane Surfaces 2.3 Forces on Curved Surfaces 2.4 Buoyancy 2.5 Rigid Body Motion Worked Examples Problems Objective Questions Fluid Flow Kinematics Introduction 3.1 Classification of Flow 3.2 Streamline 3.3 Acceleration 3.4 Continuity Equation 3.5 Rotational and Irrotational Motion 3.6 Stream Function 3.7 Potential Function 3.8 Relation between y and f for 2-Dimensional Flow 3.9 Some Common Formulae in Cylindrical Co-ordinates 3.10 Elementary Inviscid Plane Flows Worked Examples Problems Objective Questions Energy Equation and Its Applications Introduction 4.1 Bernoulli Equation 4.2 Practical Applications of Bernoulli Equation 4.3 Energy Equation 4.4 Power Worked Examples Problems Objective Questions Momentum Equation and its Applications Introduction 5.1 Linear Momentum Equation 5.2 The Moment of Momentum Equation Worked Examples Problems Objective Questions Dimensional Analysis and Similitude Introduction 6.1 Common Variables in Fluid Flow 6.2 Dimensional Homogeneity 6.3 Dimensional Analysis 6.4 Similitude 6.5 Important Dimensionless Flow Parameters 6.6 Model Scales 6.7 Distorted models Worked Examples Problems Objective Questions Laminar Flow Introduction 7.1 Basic Equations 7.2 Creeping Motion 7.3 Lubrication 7.4 Viscometers 7.5 Internal and External Flows Worked Examples Problems Objective Questions Boundary Layer Concepts Introduction 8.1 Boundary Conditions 8.2 Laminar Boundary Layer over a Flat Plate 8.3 Karman Momentum Integral Formulation 8.4 Boundary Conditions for a Proper f(h) 8.5 Transition from Laminar Boundary Layer 8.6 Turbulent Boundary Layer 8.7 Laminar Sublayer 8.8 Establishment of Flow in a Pipe 8.9 Boundary Layer Separation Worked Examples Problems Objective Questions Drag and Lift on Immersed Bodies Introduction 9.1 Drag 9.2 Lift Worked Examples Problems Objective Questions Turbulent Pipe Flow Introduction 10.1 Characteristics of Turbulence Flows 10.2 Turbulent Pipe Flow 10.3 Commercial Pipes 10.4 Aging of Pipes 10.5 Simple Pipeline Design Problems 10.6 Velocity Distribution in the Neighbourhood of Flat Surfaces Worked Examples Problems Objective Questions Pipe Flow Systems Introduction 11.1 Minor Losses 11.2 Simple Pipe Problems 11.3 Pipe Network 11.4 Miscellaneous Problems Worked Examples Problems Objective Questions Flow in Open Channels Introduction 12.1 Classification 12.2 Uniform Flow 12.3 Rapidly Varied Flow 12.4 Gradually Varied Flow 12.5 Flow Measurement Worked Examples Problems Objective Questions Flow Measurement Introduction 13.1 Orifices 13.2 Orifice Meter 13.3 Venturimeter 13.4 Pitot Tube 13.5 Weirs 13.6 Rotameter 13.7 Measurement of Turbulence Worked Examples Problems Objective Questions Unsteady Flow Introduction 14.1 Surges in Open Channels 14.2 Water Hammer 14.3 Establishment of Flow 14.4 Surge Tanks Worked Examples Problems Objective Questions Compressible Flow Introduction 15.1 Thermodynamic Principles 15.2 Basic Definitions 15.3 Basic Equations for Compressible Fluid Flow 15.4 Application of Energy Equation 15.5 Sonic Velocity 15.6 Flow in a Nozzle 15.7 Normal Shock Wave Worked Examples Problems Objective Questions Hydraulic Machines Introduction 16.1 Turbines 16.2 Rotodynamic Pumps 16.3 Reciprocating Pump 16.4 Miscellaneous Hydraulic Machinery and Devices Worked Examples Problems Objective Type Questions Additional Objective Questions on Hydraulic Machines Appendices Appendix A Multiple Choice Objective Questions Appendix B1 Answers to Objective Questions in Chapter 1 through 16 Appendix B1 Answers to Additional Objective Questions in Hydraulic Machines Appendix B2 Answers to Multiple Choice Questions of Appendix-A Index_ Citation preview Fluid Mechanics And Hydraulic Machines Problems and Solutions About the author K Subramanya is a retired Professor of Civil Engineering at the Indian Institute of Technology, Kanpur, UP, India. He obtained his Bachelor’s degree in Civil Engineering from Mysore University and Master’s degree from the University of Madras. Further, he obtained another Master’s degree and a PhD degree from the University of Alberta, Edmonton, Canada. He has taught at IIT Kanpur for over 30 years and has extensive teaching experience in the areas of Fluid Mechanics, Open Channel Hydraulics and Hydrology. During his tenure at IIT Kanpur, Dr Subramanya worked for a short while as visiting faculty at Asian Institute of Technology, Bangkok. He has authored several successful books for McGraw-Hill Education India. Besides the current book his other books include Flow in Open channels (3rd Edition, TMH, 2009) and Engineering Hydrology (3rd Edition, TMH, 2008). Dr Subramanya has published over eighty technical papers in national and international journals and conferences. He is a Fellow of the Institution of Engineers (India); Fellow of Indian Society for Hydraulics, Member of Indian Society for Technical Education and member of Indian Water Resources Association. Currently, he resides in Bangalore and is active as a practicing consultant in Water Resources Engineering. He can be contacted at subramanyak1@gmail.com Fluid Mechanics And Hydraulic Machines Problems and Solutions K Subramanya Retired Professor Department of Civil Engineering Indian Institute of Technology, Kanpur Tata McGraw Hill Education Private Limited NEW DELHI New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto Tata McGraw-Hill Published by the Tata McGraw Hill Education Private Limited, 7 West Patel Nagar, New Delhi 110 008. Fluid Mechanics and Hydraulic Machines: Problems and Solution, 1/e Copyright © 2011 by Tata McGraw Hill Education Private Limited. 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Ltd, A-102/4, Phase II, Okhla Industrial Area, New Delhi-110 020 Cover Printer: Sai Printo Pack RZZCRRAZDAAZC The McGraw-Hill Companies Dedicated to My Parents Contents Preface xiii Properties of Fluids Introduction 1 1.1 Density, Speciﬁc Volume and Speciﬁc Weight 1.2 Pressure 3 1.3 Shear Stress and Viscosity 3 1.4 Surface Tension 5 1.5 Compressibility 6 1.6 Vapour Pressure 8 Worked Examples 8 Problems 22 Objective Questions 25 Fluid Statics 1 2 30 Introduction 30 2.1 Pressure in a Static Fluid 30 2.2 Forces on Plane Surfaces 33 2.3 Forces on Curved Surfaces 34 2.4 Buoyancy 36 2.5 Rigid Body Motion 36 Worked Examples 38 Problems 84 Objective Questions 96 Fluid Flow Kinematics Introduction 105 3.1 Classiﬁcation of Flow 105 3.2 Streamline 106 3.3 Acceleration 106 3.4 Continuity Equation 107 105 viii Contents 3.5 3.6 3.7 3.8 3.9 3.10 Rotational and Irrotational Motion 107 Stream Function 109 Potential Function 109 Relation between y and f for 2-Dimensional Flow 109 Some Common Formulae in Cylindrical Co-ordinates 110 Elementary Inviscid Plane Flows 110 Worked Examples 112 Problems 130 Objective Questions 135 Energy Equation and Its Applications Introduction 141 4.1 Bernoulli Equation 141 4.2 Practical Applications of Bernoulli Equation 4.3 Energy Equation 142 4.4 Power 143 Worked Examples 144 Problems 162 Objective Questions 168 141 142 Momentum Equation and its Applications Introduction 172 5.1 Linear Momentum Equation 172 5.2 The Moment of Momentum Equation Worked Examples 177 Problems 193 Objective Questions 197 172 175 Dimensional Analysis and Similitude Introduction 201 6.1 Common Variables in Fluid Flow 201 6.2 Dimensional Homogeneity 201 6.3 Dimensional Analysis 201 6.4 Similitude 203 6.5 Important Dimensionless Flow Parameters 203 6.6 Model Scales 204 6.7 Distorted models 204 Worked Examples 206 Problems 220 Objective Questions 225 201 ix Contents Laminar Flow 228 Introduction 228 7.1 Basic Equations 228 7.2 Creeping Motion 232 7.3 Lubrication 233 7.4 Viscometers 233 7.5 Internal and External Flows 233 Worked Examples 233 Problems 247 Objective Questions 250 Boundary Layer Concepts 253 Introduction 253 8.1 Boundary Conditions 254 8.2 Laminar Boundary Layer over a Flat Plate 254 8.3 Karman Momentum Integral Formulation 255 8.4 Boundary Conditions for a Proper f(h) 256 8.5 Transition from Laminar Boundary Layer 256 8.6 Turbulent Boundary Layer 256 8.7 Laminar Sublayer 257 8.8 Establishment of Flow in a Pipe 258 8.9 Boundary Layer Separation 258 Worked Examples 259 Problems 271 Objective Questions 274 Drag and Lift on Immersed Bodies 278 Introduction 278 9.1 Drag 278 9.2 Lift 281 Worked Examples 283 Problems 292 Objective Questions 295 Turbulent Pipe Flow Introduction 298 10.1 Characteristics of Turbulence Flows 298 10.2 Turbulent Pipe Flow 300 10.3 Commercial Pipes 302 10.4 Aging of Pipes 303 298 x Contents 10.5 Simple Pipeline Design Problems 304 10.6 Velocity Distribution in the Neighbourhood of Flat Surfaces Worked Examples 306 Problems 317 Objective Questions 320 Pipe Flow Systems 305 324 Introduction 324 11.1 Minor Losses 324 11.2 Simple Pipe Problems 326 11.3 Pipe Network 330 11.4 Miscellaneous Problems 331 Worked Examples 331 Problems 360 Objective Questions 365 Flow in Open Channels 370 Introduction 370 12.1 Classiﬁcation 370 12.2 Uniform Flow 371 12.3 Rapidly Varied Flow 375 12.4 Gradually Varied Flow 377 12.5 Flow Measurement 379 Worked Examples 379 Problems 400 Objective Questions 408 Flow Measurement Introduction 414 13.1 Oriﬁces 414 13.2 Oriﬁce Meter 416 13.3 Venturimeter 417 13.4 Pitot Tube 418 13.5 Weirs 419 13.6 Rotameter 421 13.7 Measurement of Turbulence 422 Worked Examples 423 Problems 444 Objective Questions 450 414 xi Contents Unsteady Flow 456 Introduction 456 14.1 Surges in Open Channels 456 14.2 Water Hammer 458 14.3 Establishment of Flow 460 14.4 Surge Tanks 461 Worked Examples 462 Problems 475 Objective Questions 478 Compressible Flow Introduction 482 15.1 Thermodynamic Principles 483 15.2 Basic Deﬁnitions 483 15.3 Basic Equations for Compressible Fluid Flow 15.4 Application of Energy Equation 485 15.5 Sonic Velocity 485 15.6 Flow in a Nozzle 486 15.7 Normal Shock Wave 491 Worked Examples 492 Problems 509 Objective Questions 512 482 485 Hydraulic Machines 517 Introduction 517 16.1 Turbines 517 16.2 Rotodynamic Pumps 522 16.3 Reciprocating Pump 528 16.4 Miscellaneous Hydraulic Machinery and Devices 533 Worked Examples 540 Problems 579 Objective Type Questions 588 Additional Objective Questions on Hydraulic Machines 594 Appendices Appendix A 598 Multiple Choice Objective Questions 598 Appendix B1 Answers to Objective Questions in Chapter 1 through 16 620 Appendix B1 Answers to Additional Objective Questions in Hydraulic Machines Appendix B2 Answers to Multiple Choice Questions of Appendix-A Index 621 622 623 Preface Fluid mechanics is an important constituent of the undergraduate syllabi of a very large number of engineering disciplines. The subject of ﬂuid mechanics is given considerable importance in Mechanical, Civil and Chemical engineering programmes at the core as well as at professional levels. While problem-solving is an important aspect of learning ﬂuid mechanics, a typical textbook in this subject does not provide adequate space for this aspect due to various constraints. Also, in the teaching schedule, adequate time and facilities for tutorials are available in very few institutions. As such, students have to make their own arrangements for learning problem solving, information manipulation and processing skills. The common perception is that students ﬁnd the subject of ﬂuid mechanics difﬁcult to cope with in their studies. This book, primarily based on my earlier books Theory and Applications in Fluid Mechanics (TMH, 1993) and 1000 Solved Problems in Fluid mechanics (TMH, 2005) is designed as an essential and compatible supplement to any to good textbook in ﬂuid mechanics with the needs of the undergraduate engineering students in mind. It meets the requirements of undergraduate ﬁrst and second courses in the subject; speciﬁcally, the requirements of Mechanical engineering and Civil engineering students in the area of ﬂuid mechanics study. A typical undergraduate syllabus in Fluid Mechanics is covered in sixteen chapters. In each of the chapters, an outline of the basic theoretical considerations and application methodologies is given. This is followed by a large set of carefully chosen and graded worked examples covering all the sub-areas of the chapter theme. A set of tested practice problems with answers is provided in each chapter to help students hone up their skills through practice. Further, a set of objective questions, with answers provided at the end of the book (in Appendix B), is included in each chapter. This will help the students have a quick review of the chapter and also aid them in thorough understanding of the concepts. The contents of each chapter are so designed as to help the user in all aspects of the subject matter, viz, theory, application and information processing. At the end of the book, three kinds of multiple-choice objective question sets are provided in Appendix A. These sets cut across various chapters and are especially of immense use to those preparing for national level competitive examinations. Worked examples, practice problems and objective questions are graded in three levels (simple, medium and difﬁcult) and designated by the markings of , and respectively. These markings are provided at the beginning of each item. This will be of particular use to teachers in selecting problems for class work, assignments, quizzes and examinations. Students would also ﬁnd this classiﬁcation useful in planning their preparation for various examinations and, particularly, the national-level competitive examinations. The contents of the book, which cover essentially all the important normally accepted basic areas of ﬂuid mechanics, are presented in simple, lucid style. A total of 1941 items consisting of worked examples, practice problems and objective questions with answers to the above are provided in the book. In addition to students taking formal courses in ﬂuid mechanics offered in university engineering colleges, the book is xiv Preface useful to students appearing in AMIE examinations. Candidates taking competitive examinations like Central Engineering Services examinations, Central Civil Services examinations and GATE will ﬁnd this book useful in their preparations related to the topic of ﬂuid mechanics. I would like express my sincere thanks to all those who have directly and indirectly helped me in bringing out this revised edition. In this regard, the reviewers of the book deserve a special mention. Shaligram Tiwari Indian Institute of Technology (IIT) Madras, Chennai, Tamil Nadu R B Anand National Institute of Technology (NIT), Tiruchirapalli, Tamil Nadu S Jayaraj National Institute of Technology (NIT), Calicut, Kerala S Suresh Sona College of Technology, Salem, Tamil Nadu T P Ashok Babu National Institute of Technology Karnataka (NITK), Surathkal M V Ramamurthy University College of Engineering, Osmania University, Hyderabad, Andhra Pradesh A K Mishra Harcourt Butler Technological Institute (HBTI), Kanpur, Uttar Pradesh Amarnath Mullick National Institute of Technology (NIT), Durgapur, West Bengal Dr. Debasish Roy Jadavpur University, Kolkata, West Bengal Dipankar Bhanja Dr B C Roy Engineering College, Durgapur, West Bengal D C Mahale SSVPS’s B S Deore College of Engineering, Dhule, Maharashtra S N Londhe Vishvakarma Institute of Information Technology Pune, Maharashtra S S Maghrabi Rizvi College of Engineering, Mumbai, Maharashtra Comments and suggestions for further improvement of the book would be greatly appreciated. I could be contacted at the following e-mail address: subramanyak1@gmail.com K SUBRAMANYA Publisher’s Note Tata McGraw-Hill invites comments, suggestions and feedback from readers, all of which can be sent to tmh.mechfeedback@gmail.com mentioning the title and author’s name in the subject line. Piracyrelated issues can also be reported. Properties of Fluids Concept Review Introduction 1 Continuum In engineering problems dealing with ﬂuids, one generally deals with dimensions that are very large compared to molecular sizes. The space between the molecules is not considered and the ﬂuid properties are considered to vary continuously in space. The density of ﬂuid is thus a point function. This method of considering ﬂuid as a continuous mass is stated as continuum principle. Except in dealing with rariﬁed gases, all normal ﬂuid mechanics analysis deals with ﬂuid as a continuum. Units In ﬂuid mechanics four fundamental dimensions namely mass, length, time and temperature are involved. These days the SI units are adopted in describing the various parameters of ﬂuid ﬂow. In this system the fundamental units are Mass [M] Length [L] Time [T] Temperature [q] kilogram metre second Kelvin (for thermodynamic calculations) or °Celsius kg m s K °C Based upon these fundamental units, a number of derived units are developed. The most commonly used derived units is the unit of force which is newton (N). A newton of force corresponds to an acceleration of 1 m/s2 of a mass of 1 kg. The commonly used derived terms and their units are listed in Table 1.1. 2 Fluid Mechanics and Hydraulic Machines Commonly used Derived Terms in Fluid Mechanics Derived term Area Volume Velocity Acceleration Force Pressure (or stress) Energy (or work) Power Dimension SI unit 2 Abbreviation 2 (L ) (L3) (LT –1) (LT –2) (MLT –2) (ML–1T –2) m m3 m/s m/s2 N N/m2 (ML2T –2) N.m (ML2T –3) J/s of mercury) the density of water is 998 kg/m3. Thus, the speciﬁc weight of water at 20°C temperature and 1 atmospheric pressure (known as NTP = normal temperature and pressure) is g = rg = 998 ¥ 9.81 = 9790 N/m3 = 9.79 kN/m3 Pascal Pa = N/m2 Joule J = N.M. Watt W = J/s Relative Density (RD) of a ﬂuid is the ratio of its density to that of a standard reference ﬂuid, water (for liquids) and air (for gasses). In engineering practice, the term speciﬁc gravity (SG or S) is used synonymously with the term relative density. Thus RDliquid = (SGliquid) kg/m3 (ML–3) (ML–1T –1) kg/m.s Pa.s (= N.s/m2) N/m Surface tension (MT –2) Density Viscosity 1.1 DENSITY, SPECIFIC VOLUME AND SPECIFIC WEIGHT 1.1.1 Density The density r of a ﬂuid is its mass per unit volume. The units are kg/m3. In general, the density of a ﬂuid depends upon the temperature and pressure. For incompressible ﬂuids (liquids), the variation of density with pressure is however small. = Weight The speciﬁc weight g of a ﬂuid is its weight per unit volume. Thus, 998 (kg/m3) RDgas = (SGgas) = Density of gas (kg/m3) 1.205 (kg/m3) For example if the relative density of a liquid is 0.85, it means that its density is 0.850 ¥ 998 = 848.3 kg/m3. Commonly used values of approximate speciﬁc gravities in ﬂuid ﬂow calculations are 1.0 for water and 13.6 for mercury. When no other information is available, the following values corresponding to NTP (20°C temperature and one atmospheric pressure) are used: Item The reciprocal of mass density is known as speciﬁc volume. It represents volume per unit mass of the ﬂuid and has units of m3/kg. Density of liquid (kg/m3) Water Air 3 Density r 998 kg/m Speciﬁc gravity 1.00 (= Relative density) Speciﬁc weight g 9790 N/m3 (= 9.79 kN/m3) 1.205 kg/m3 1.0 11.82 N/m3 Unless otherwise stated, the above values are used for r and g (for water and air) in this book. g = rg in units of N/m2 The standard value of acceleration due to gravity g is 9.086 m2/s and is usually taken as 9.81 m2/s. At 20°C temperature and one atmospheric pressure (760 mm [Note In approximate/quick calculations, for water r = 1000 kg/m3 and g = 9.8 or 10.0 kN/m3 are used ] 3 Properties of Fluids 1.2 PRESSURE and p p (N/m 2 ) = rg g (N/m3) In such cases, h is called the pressure head. h (meters of ﬂuid) = Units Pressure is the compressive stress on the ﬂuid and is given by Force F for uniform pressure. Area A dF for variable pressure. p= dA p= The units of pressure are N/m2 = Pa. (Pa is the abbreviation for pascal) 1 Pa = 1 pascal = 1 N/m2 1 kPa = 1 kilo pascals = 1000 N/m2 Bar is a unit extensively used in meteorology and in calculations involving atmosphere and high pressures. Here, 1 bar = 105 Pa = 100 kN/m2 atmospheric pressure at sea level which is 101,325 kN/m2. The pressure of 101,325 N/m2 = 101.325 kPa is called one atmosphere and is denoted by 1 atm. deﬁned by IUPAC, is air pressure at 0°C (= 273.16 K = 32°F) at 1 atmospheric pressure (= 1 atm = 101.325 N/m2 = 101.325 kPa = 760 mm of mercury = 10.336 m of water) For example, (i) A pressure head of 5.0 m of water is equivalent to a pressure of 5.0 ¥ 9790 = 48950 Pa = 48.95 kPa. (ii) Similarly, a pressure of 4.0 kPa is equivalent to a pressure head h of mercury where 4000 = 0.03004 m = 30.04 mm of h= 13.6 ¥ 9790 mercury. 1.3 SHEAR STRESS AND VISCOSITY While the pressure, a normal stress, is encountered in both ﬂuid static and dynamic conditions the shear stress (t) is encountered only in real ﬂuids and also only when they are in motion. The units of shear stress is N/m2 and is designated in Pa or kPa depending on the magnitude. 1.3.2 Viscosity Dynamic Viscosity Viscosity is the resisting property of a ﬂuid to shearing force. The shear stress t is related to the deformation rate in most of the commonly occurring ﬂuids by the Newton’s law of viscosity, as t =m a standard commonly used in engineering practice and refers to 20°C temperature and 1 atmospheric pressure (1 atm = 101.325 kPa). of the height of an equivalent column of a ﬂuid of density r. Thus p = rgh = g h (1.1) du dy (1.2) du = velocity gradient in the Y direction and dy m = coefﬁcient of viscosity, which is a ﬂuid property. where The ﬂuids which obey Newton’s law of viscosity are known as Newtonian ﬂuids. Most of the common liquids like water, kerosene, petrol, ethanol, benzene, 4 Fluid Mechanics and Hydraulic Machines Glycerin and mercury are Newtonian. Further, all gases are Newtonian. The coefﬁcient of viscosity, m, is also known variously as the coefﬁcient of dynamic viscosity, absolute viscosity or simply as viscosity. It has the units t N/m 2 = m= = N.s/m2 = Pa.s Ê du ˆ Ê m/s ˆ ÁË m ˜¯ ÁË dy ˜¯ Sometimes, the coefﬁcient of dynamic viscosity m is designated by a unit poise (abbreviated as P) or as centipoises (abbreviated as CP) where 1 poise = 1 = 1 centipoise = gm dyne . second =1 cm . second cm 2 10-5 -2 2 N.s (10 ) m 2 = 1 Pa.s 10 1 1 poise = Pa.s 100 1000 The coefﬁcient of viscosity m depends upon the temperature. Generally, for liquids the value of m decreases with an increase in temperature, and for gases, the value of m increases with an increase in the temperature. Kinematic Viscosity The ratio of dynamic viscosity to the density of the ﬂuid is known as kinematic viscosity. This term is designated by the Greek Ê 2ˆ letter n (nu) and has the dimensions Á L ˜ as shown ËT ¯ below: m N.s/m 2 kg . m -1 s -1 = = = m2/s. 3 -3 r kg/m kg . m Sometimes, the kinematic viscosity n is designated by a unit stoke or as centistoke where n= 1 stoke = 1 cm 2 m2 = 1(10–2)2 second s = 10 – 4 m2/s 1 stoke = 10 –6 m2/s 1 centistoke = 100 Table 1.2 gives the dynamic and kinematic viscosities of some commonly used ﬂuids at 20°C and 1 atm pressure. While most of the common ﬂuids like water, air, petrol, ethanol and benzene follow Newton’s law of viscosity as given by Eq. (1.2), there exists a large number of ﬂuids which do not follow this linear relationship between the shear stress t and du . Such ﬂuids which do the rate of deformation, dy not obey Newton’s law of viscosity are known as Non-Newtonian ﬂuids. Typical examples of nonNewtonian ﬂuids are blood, suspension of corn starch in water, paint, slurries, pastes and polymer solutions. In the non-Newtonian ﬂuids, such as the ones mentioned above, the relationship between rate du , and the shear stress t can in of deformation, dy general be expressed as a power law relation like Ê du ˆ t = mÁ ˜ Ë dy ¯ n (1.3) In this, m is known as consistency index and the power n is the ﬂow index. n < 1, the ﬂuid is known as nonNewtonian pseudoplastic ﬂuid. Gelatine, milk and blood are typical examples of pseudoplastic ﬂuids. n > 1, the ﬂuid is known as nonNewtonian dilatant ﬂuid. Starch suspension, sugar solution and high-concentration sand suspension are typical examples of dilatant ﬂuids. n = 1 represents a Newtonian ﬂuid, with m = m. du is t and dy 5 Properties of Fluids known as rheological behavior and Fig. 1.1 Shear stress Elastic solid is a schematic representation of rheological classiﬁcation of ﬂuids. and Bingham plastic stic pla e Ps tension s. The most common interfaces and values of s, for clean surface at 20°C, are o ud Newtonian fluid Dilatanat fluid Ideal fluid Shear rate Fig. 1.1 In Fig. 1.1, the x-axis also represents a Newtonian ﬂuid with m = 0, that is a ﬂuid with zero viscosity. Such ﬂuid called an ideal ﬂuid or inviscid ﬂuid. du is zero for all t, the situation represents When dy an elastic solid. Some non-Newtonian ﬂuids can be modeled as Ê du ˆ t = t y + mpÁ ˜ Ë dy ¯ (1.4) Such ﬂuids which require a yield stress t y for the ﬂow to be established, are known as Bingham plastic. While the above non-Newtonian ﬂuids are time independent, there exist some non-Newtonian ﬂuids which are time dependent, that is the shear stress and corresponding deformation rate are functions of time. Further classiﬁcation of such time-dependent non-Newtonian ﬂuids are beyond the scope of this book. 1.4 SURFACE TENSION A liquid forms an interface with a second liquid or gas. The surface energy per unit area of interface is known as surface tension or coefﬁcient of surface s = 0.073 N/m for s = 0.480 N/m for air-water interface. air-mercury interface. Note that the surface tension s has the dimension of force/unit length (N/m). When a liquid interface interacts with a solid surface, a contact angle q is formed. For water-clean glass surface q ª 0° and for mercury-clean glass q ª 130°. Due to surface tension, pressure changes occur across a curved interface. The pressure difference between inside and outside of a curved surface Dp is related to the radius of curvature R and surface tension s as (1) For the interior of a liquid cylinder Dp = s R (1.5) (2) For a spherical droplet Dp = 2s R (1.6) (3) A soap bubble has two surfaces and the pressure difference is given by Dp = 4s R (1.7) Thus, the pressure inside a droplet or a soap bubble will be higher than the surrounding atmosphere. The pressure inside will be higher, the smaller the size of the droplet or bubble. Liquids have both cohesion and adhesion, which are forms of molecular attraction. Capillarity, the rise (or fall) of liquid in small-diameter tubes is due to this attraction. Liquids, such as water, which wet a surface cause capillary rise. In nonwetting liquids (e.g. mercury) capillary depression is caused. 6 Fluid Mechanics and Hydraulic Machines For a cylindrical glass tube the capillary rise (or depression) h (Fig. 1.2) is given by h= q g R s where 2s cos q gR (1.8) = contact angle, = unit weight of the liquid (= rg), = radius of curvature of the glass tube = coefﬁcient of surface tension. [Note Capillary rise is usually measured to the bottom of the meniscus] R s q 8312 (1.10) M where M = molecular weight of gas. For air M = 28.97, giving R air = 287 m2/(s2K). For any gas R= Another fundamental equation of a perfect gas between two state points is p1 p = 2 r1n r2n where p is absolute pressure. (1.11) (i) If the process is isothermal (i.e. at constant temperature), n = 1.0 (ii) If the process is adiabatic (i.e. without heat transfer) and without friction (isentropic) n=k where k = = ratio of speciﬁc heat at constant cv pressure (cp) and that at constant volume (cv). For air and diatomic gases k ª 1.4. Values of k for various gases are given in Table 1.2(b). Combining Eq. 1.9 and Eq. 1.11 h Fig. 1.2 cp Capillary Rise For clean glass and water q can be assumed to be zero. For clean mercury–air–glass interface, q = 130°. T2 Ê p2 ˆ = T1 ÁË p1 ˜¯ ( k -1) / k 1.5 COMPRESSIBILITY bc Gasses are highly compressible and their relationship between pressure, temperature and volume is expressed by perfect gas equation p = pv = RT r where p r v T = absolute pressure = mass density = speciﬁc volume = 1/r = absolute temperature in Kelvin = (273 + °C) Ê m2 ˆ R = characteristic gas constant = Á 2 ˜ Ë s K¯ (1.12) (1.9) Compressibility of a ﬂuid refers to its ability to change its volume and density when subjected to pressure. The coefﬁcient of compressibility b c is deﬁned as the relative change of volume (or density) per unit pressure and is represented as bc = - d ~/ ~ dp (1.13) where dp = change in pressure and d ~ = change in volume ~ of the ﬂuid. The negative sign indicates a decrease in volume ~ with increase in pressure. In practice, however, the reciprocal of compressibility, known as the bulk modulus of elasticity is extensively used to characterize compressibility effects. 7 Properties of Fluids Properties of Some Common Fluids at 20°C and 1 atm Pressure Fluids Density r(kg/m3) (a) Liquids Water Sea water Petrol Kerosene Glycerine Mercury SAE 10 oil SAE 30 oil Castor oil 998 1025 680 804 1260 13550 917 917 960 Dynamic viscosity m (Ns/m2) 1.00 ¥ 10–3 1.07 ¥ 10–3 2.92 ¥ 10–4 1.92 ¥ 10–3 1.49 1.56 ¥ 10–3 1.04 ¥ 10–1 2.90 ¥ 10–1 9.80 ¥ 10–1 r (kg/m3) (b) Gases Air Carbon dioxide Hydrogen Nitrogen Methane 0.747 K 1 dp = bc Ê d ~ ˆ ÁË - ~ ˜¯ Bulk modulus K (N/m2) 1.00 ¥ 10–6 1.04 ¥ 10–6 4.29 ¥ 10–7 2.39 ¥ 10–4 1.18 ¥ 10–3 1.15 ¥ 10–7 1.13 ¥ 10–4 3.16 ¥ 10–4 1.02 ¥ 10–3 7.28 ¥ 10–2 7.28 ¥ 10–2 2.16 ¥ 10–2 2.80 ¥ 10–2 6.33 ¥ 10–2 4.84 ¥ 10–1 3.60 ¥ 10–2 3.50 ¥ 10–2 3.92 ¥ 10–2 2.19 ¥ 109 2.28 ¥ 109 9.58 ¥ 108 1.43 ¥ 109 4.34 ¥ 109 2.55 ¥ 1010 1.31 ¥ 109 1.38 ¥ 109 1.44 ¥ 109 m (Ns/m2) n (m2/s) 1.494 ¥ 10–5 0.804 ¥ 10–5 10.714 ¥ 10–5 1.517 ¥ 10–5 2.000 ¥ 10–5 1.504 ¥ 10–5 1.352 ¥ 10–5 Speciﬁc heat ratio, k = cp/cn 1.40 1.28 1.40 1.40 1.30 1.40 1.33 where k = cp /cv = ratio of speciﬁc heat at constant pressure to that at constant volume and p = absolute pressure. The bulk modulus of elasticity K is deﬁned as K= Surface tension s (N/m) 1.80 ¥ 10–5 1.48 ¥ 10–5 0.90 ¥ 10–5 1.76 ¥ 10–5 1.34 ¥ 10–5 ¥ 10–5 1.01 ¥ 10–5 1.205 1.840 0.084 1.160 0.668 Water vapour Kinematic viscosity n (m2/s) (1.14) K represents the compressive stress per unit Ê d~ ˆ volumetric strain. Since Á is dimensionless, the Ë ~ ˜¯ dimensions of K is that of pressure p, viz., N/m2 = Pa. For a perfect gas, K = p for isothermal process, and K = k p for isentropic process. C Sound is propagaed in ﬂuid due to compressibility of the medium, and the speed of sound C is given by C= K r (1.15) where K = bulk modulus of elasticity of the medium and r = mass density of the ﬂuid. 8 Fluid Mechanics and Hydraulic Machines 1.6 VAPOUR PRESSURE Vapour pressure is the pressure at which a liquid boils and is in equilibrium with its own vapour. In many liquid ﬂow situations, such as in hydraulic machines and in ﬂow through constricted passages, a low pressure approaching vapour pressure of the liquid may occur. When this happens, the liquid ﬂashes into vapour, forming a rapidly expanding cavity. This phenomenon, known as cavitation, has serious implications on the operating performance of hydraulic machines and passages of high-speed ﬂows, (see Chapter 16, Sec. 16.2.6 for further details). Vapour pressure of a liquid depends upon temperature and increases with it. At 20°C, water has a vapour pressure (pv) of 2.34 kPa (i.e. vapour pressure head = pv = 0.24 m). g Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples (ii) If g2 = 18.0 m/s2 1.1 W2 = mg2 = 50.99 ¥ 18.0 = 917.8 N 2 g (a) What is its mass? 1.2 2 2 2 ? 2 Solution: Let W = weight of the liquid and m = its mass. (a) W = mg 500 = m ¥ 9.806 500 = 50.99 kg m= 9.806 (b) The mass of the ﬂuid remains constant regardless of its location. Hence m = 50.99 kg at all locations. (i) If g1 = 3.5 m/s2 W1 = mg1 = 50.99 ¥ 3.5 = 178.46 N ? Solution: The mass of the body, m= weight 400 = = 40.791 kg gravity 9.806 This mass is constant and does not change with location. Hence, when a force F is applied, by Newton’s second law, F = ma or acceleration a = F/m which is independent of g. Hence, both on the earth, 9 Properties of Fluids as well as on moon, 800 = 19.612 m/s2 a= 40.791 1.5 1.3 Solution: Solution: (1) Mass density of petrol (23.7/ 9.81) 3.0 = 0.805 kg/litre = 805 kg/m3 rp = mass/volume = Mass density of water = rw = 998 kg/m3 Speciﬁc gravity of petrol = 805/998 = 0.807 (2) Speciﬁc weight of petrol = weight per unit volume (i) Unit weight g = rg = (rwater ¥ RD) ¥ g Taking r water at 20° as 998 kg/m3, g = 998 ¥ 0.8 ¥ 9.81 = 7832.3 N/m3 = 7.832 kN/m3 (ii) Dynamic viscosity m = vr v = 2.3 centistoke = 2.3 ¥ 10– 6 m2/s r = 998 ¥ 0.8 = 798.4 kg/m3 m = 2.3 ¥ 10–6 ¥ 798.4 = 1.836 ¥ 10–3 Pa.s. 23.7 = 7.9 N/litre = 3.0 = 7.9 kN/m3 (3) Speciﬁc volume = volume per unit mass 1 1 = = rp 805 –3 1.6 V 3 = 1.242 ¥ 10 m /kg 3 cm 1.4 t Fig. 1.3 Solution: p = g h = (0.80 ¥ 9.79) ¥ 25 100 = 1.958 kPa If hm is the equivalent column of mercury, hm ¥ 13.6 ¥ 9.79 = 1.958 1.958 = 0.0147 m hm = (13.6 ¥ 9.79) = 1.47 cm Solution: Since the gap between the plates is very small, a linear variation of velocity can be assumed. du V 1.50 = = = 500(s–1) dy h 3 ¥ 10-3 t = Shear stress on the bottom plate du = 0.2 ¥ 500 = 100 N/m2 =m dy 1.7 u y – y2 y£2m 10 Fluid Mechanics and Hydraulic Machines u y 2 y y Solution: Given t u = 4y – y2 du = 4 – 2y Therefore dy du Shear stress t = = m (4 – 2y) dy At y = 0, t 0 = 4m = 4 ¥ 1.5 = 6.0 Pa.s At y = 2.0 m, t 2 = m(4 – 4) = 0 V q W Fig. 1.4 Solution: Given 1.8 y £ section at (i) y W = 90 N V = terminal velocity q = 30° p y) u y y Solution: Given, 5 Pa.s m = 5 poise = 10 Since u = 5.0 sin (5p y) du = 5.0 ¥ 5p cos (5py) dy du 5 Shear stress t = m = ¥ 25p cos (5p y) dy 10 = 12.5p cos (5p y) (i) At y = 0, t = 12.5 p cos (0) = 12.5p = 39.27 N/m2 (ii) At y = 0.05 m, t = 12.5p cos (5p ¥ 0.05) At the terminal velocity, the sum of the forces acting on the block in the direction of its motion is zero. Hence W sin q – tA = 0 where t = shear stress on the block and A = area of the block. du V =m t =m dy h where h = thickness of oil ﬁlm 8 m = 8 poise = Pa.s = 0.8 Pa.s 10 h = 3 mm = 3 ¥ 10–3 m A = 0.3 m2 Substituting the various values in the above equation, = 12.5p ¥ 0.707 = 27.76 N/m2 90 sin 30° – (iii) At y = 0.10 m, t = 12.5p cos (5p ¥ 0.1) = 0 1.9 \ (0.8 V ) 3 ¥ 10-3 ¥ (0.3) = 0 V = 45 80 = 0.5625 m/s 11 Properties of Fluids 1.10 h m1 and m2 N = 240 RPM Clearence h = 2.5 mm h y m1 50 cm Solution: Let y be the distance of the thin ﬂat plate from the top ﬂat surface (Fig. 1.5) and V = velocity of the thin plate. Thin flat plate h V m2 (h – y ) Fig. 1.5 du V = m1 dy y Since the gap in the bottom portion = h – y V Shear stress, t 2 = m2 (h - y ) Force on both sides of the plate Shear stress on the top portion t 1 = m1 Èm m2 ˘ F = A(t1 + t2 ) = VA Í 1 + ˙ Î y h - y˚ where A = area of the thin plate. dF =0 For F to be minimum dy m m2 - 21 + =0 y (h - y )2 y2 or 1.11 Êm ˆ = Á 1˜ Ë m2 ¯ (h - y )2 y = (h - y ) m1 / m 2 90 mm 95 mm Fig. 1.6 Solution: V = Circumferential velocity of the shaft 2pN ¥r = wr = 60 = 2 p ¥ 240 Ê 0.090 ˆ ¥Á Ë 2 ˜¯ 60 = 1.131 m/s 95 - 90 = 2.5 mm 2 Assuming linear variation of velocity across the gap, du V = Velocity gradient dr h 1.131 = = 452.4 s–1 -3 2.5 ¥ 10 m = 2.0 poise = 0.2 Pa.s Clearance h = 12 Fluid Mechanics and Hydraulic Machines du dr = 0.2 ¥ 452.4 = 90.48 Pa Shear force Fs = t ¥ 2 pr ¥ L 0.09 ˆ Ê = 90.48 ¥ Á 2 p ¥ ¥ 0.50 Ë 2 ˜¯ Shear stress on the shaft t = m = 12.791 N T = Fs r Torque = 12.791 ¥ Solution: The thickness of the glycerin layer is same on either side of the plate. t = thickness of glycerin layer = (15 – 3)/2 = 6.0 mm mV t Fs = Total shear force (considering both sides of the plate) Shear stress on one side of the plate = t = 0.09 = 0.6756 N.m 2 2pN ◊T 60 2 p ¥ 240 = ¥ 0.6756 = 14.5 W 60 Here, Power required = P = 2AmV t A = area of plate = 0.8 ¥ 0.8 = 0.64 m2 = 2 At = Weight of steel plate = Ws = 110 N Volume of the plate = 0.64 ¥ 0.003 = 0.00192 m3 2 ¥ 0.64 ¥ 1.5 ¥ 0.15 0.006 = 48 Shear force Fs = 1.12 rg 2 3 and m = Up thrust on submerged plate = gg ¥ (volume of the plate) Wu = (1260 ¥ 9.81) ¥ 0.00192 = 23.73 N Effective weight of plate = We = Ws – Wu = 110 – 23.73 = 86.27 N Total force required to pull the plate = F = Fs + We = 48 + 86.27 = 134.27 N 1.13 V L Fs Gap = t D Fig. 1.7 Example 1.12 Fig. 1.8 Example 1.13 13 Properties of Fluids Solution: At terminal velocity Shear force = Submerged weight of the sleeve. Vm t In the given set up L, D, m and t are invariant. F = constant Hence, V F1 F F Thus = 2 , giving F2 = 1 V2 V1 V2 V1 F = L ¥ (pD) ¥ Force 1250 ¥ 1.8 = 1500 N 1.5 = \ (2prL) ¥ t = Ws 0.03 10 Ê 4ˆ ÁË 2 p ¥ 2 ¥ 100 ¥ 3 ¥ 10 ˜¯ V = 7.5 282.74 V = 7.5 V = 0.02653 m/s = 2.66 cm/s 1.15 1.14 2 Solution: m = nr = 3.7 ¥ 10–4 ¥ 0.85 ¥ 998 = 0.3139 Pa.s 10 cm Shear stress t = m 3 cm V du V =m dy h = 0.3139 ¥ 25 ¥ 3.3 ¥ 523.1 100 = 1356 N = 1.356 kN =p¥ V Fig. 1.9 m = 6 poise = 0.6 Pa.s h = 0.02 mm = 0.02 ¥ 10–3 m (25.018 - 25.00)/(2 ¥ 102 ) = 523.1 N/m2 Frictional resistance Fs = At Fs = (pDL) ¥ t Clearence 0.02 mm Solution: Given 0.15 1.16 R m w Let V = Velocity of the sleeve when sliding down. du V =m dr h V = (0.6) ¥ = 3 ¥ 104 V 2 ¥ 10-5 Shear stress t = m Solution: Consider an element of disc of width dr at a radial distance r. Velocity at this radius = V = rw. Assuming linear variation of velocity with depth in the gap h, 14 Fluid Mechanics and Hydraulic Machines w w 2.5 cm R h r Stationary dr Fig. 1.10 Example 1.16 9.75 cm 10.00 cm V m = wr Shear stress t =m h h Viscous torque on the element Fig. 1.11 Velocity gradient m w r (2prdr) r h mw = ◊ 2pr3dr h dT = Total torque T= Ú R 0 dT = Ú R 0 du V 0.4712 = = 377 = dr h 0.00125 du = 377m dr Shear force Fs = t ¥ area mw 2p r 3dr h 0.1 2.5 ˆ Ê = 377m Á 2 p ¥ ¥ Ë 2 100 ˜¯ R Example 1.17 Shear stress t = m È mw r4 ˘ = Í ◊ 2p ˙ 4 ˙˚ ÍÎ h 0 1 T= R4 2 h or Clearence = h = 2.961m Torque = Fs ¥ lever arm = Fs r T = 2.961 m ¥ (0.1) 2 = 0.14805 m But Torque T = 1.2 N.m 1.17 Therefore 0.14805 m = 1.2 \ m = 120 0.14805 = 8.106 Pa.s Solution: Tangential velocity pD N p ¥ 0.10 ¥ 90 = V= 60 60 = 0.4712 m/s (10.00 - 9.75) Radial clearance h = cm 2 = 0.125 cm or 0.00125 m 1.18 r w 2q m T h 15 Properties of Fluids w Fluid r0 Values in Units ∂u ∂y dr t Oil h r 2q B ∂u ∂y C ∂u ∂y ds t=1 q ds = dr /sin q 2 3 t (a) (b) Fig. 1.12 Example 1.18 Solution: At any radius r £ r0, u = wr Shear stress on the inclined wall du t =m dy V wr =m h h Considering an elemental area (2pr ds) dr = 2pr ◊ sin q d (torque) = dT = r d (force) =m = rt 2pr◊ = rm = m Torque T = Ú r0 dT = 0 T= 1.19 ∂u ∂y dr sin q wr dr 2pr h sin q 2 pw 1 ◊ r 3 dr h sin q 2pw m h sin q 2h sin Ú r0 r 3 dr 0 r04 A to D t E ∂u ∂y t Solution: Towards classiﬁcation of the ﬂuids, plot ∂u plotted on X-axis against t the data given as ∂y plotted along y-axis. The plots are shown in Fig. 1.13 1. Fluid-A shows a linear increase of shear stress with shear rate starting from the origin (0, 0) and hence is a Newtonian ﬂuid. The slope of the line (2 units in this case) is its coefﬁcient of viscosity. 2. Fluid-B has a linear shear stress vs shear ∂u = 0, the shear rate behaviour, however at ∂y stress is 1.0 units, indicating an yield stress (ty). Hence this ﬂuid is classiﬁed as Bingham plastic. Fluid-C, shows shear stress decreasing with increase in shear rate. Hence it is a shear thinning ﬂuid and as such is classiﬁed as non-Newtonian and is sub-classiﬁed as Pseudoplastic. Fluid-D, shows shear stress increasing with increase in shear rate. Hence it is a shear thickening ﬂuid and as such is classiﬁed 16 Fluid Mechanics and Hydraulic Machines Rheological behaviour of fluids 6.00 Shear stress 5.00 Fluid - C Fluid - D Fluid - B Fluid - E Fluid - A B A 5.00 D 5.00 5.00 C 5.00 E 5.00 0.5 0 1 1.5 2 2.5 Shear rate Fig. 1.13 as non-Newtonian and is sub-classiﬁed as Dilatant Fluid-E, shows shear stress is zero for all shear rates. Hence, it is a Newtonian ﬂuid with zero viscosity. As such, it is classiﬁed as ideal ﬂuid (also known as inviscid ﬂuid). Solution: (i) For a spherical raindrop 2s 2 ¥ 0.073 = = 116.8 Pa 1 ˆ R Ê 2.5 ¥ ÁË 2 1000 ˜¯ Dp = (ii) For a circular cylindrical jet of liquid 1.20 Dp = Solution: An air bubble has only one surface. Hence 2s 2 ¥ 0.073 = Dp= R Ê 0.01ˆ -3 ÁË 2 ˜¯ ¥ 10 2 = 29200 N/m = 29.2 kPa Example 1.19 1.22 Solution: Hence Dp = 1.21 s 0.073 = = 41.7 Pa 1 ˆ R Ê 3.5 ¥ ÁË ˜ 2 1000 ¯ In a soap bubble, there are two interfaces. 4s 4 ¥ 0.088 = R Ê3 -2 ˆ ÁË 2 ¥ 10 ˜¯ = 23.47 N/m2 above atmospheric pressure s 1.23 17 Properties of Fluids 2s 1/1000 s = 0.0376 N/m 75.2 = Solution: Let h = difference in water levels in the two limbs. By assuming the angle of contact q = 0°, 1.25 Ê 2s 2s ˆ Dp = rgh = Á Ë R1 R2 ˜¯ 3 and a Ê 1 1 ˆ h ¥ 998 ¥ 9.81 = 2 ¥ 0.073 Á ˜ Ë 3 ¥ 10-3 8 ¥ 10-3 ¯ h = 3.1 ¥ 10–3 m = 3.1 mm 1.24 Solution: 2s cos q (1) gR Here, q = 0 and hence cos q = 1.0 h = 2.0 mm = 0.002 m; s = 0.06 N/m g = rg = 1530 ¥ 9.81 = 15009 N/m3 2s cos q 2 ¥ 0.06 = From Eq. 1, R = gh 15009 ¥ 0.002 Capillary rise h = 2 Solution: Pressure inside the bubble = 200 N/m2 1.5 ¥ 0.85 ¥ 9790 Pressure outside the bubble = 100 = 124.8 N/m2 Dp = 200.0 – 124.8 = 75.2 N/m2 2s Dp = R = 3.9976 ¥ 10 –3 m = say 4.0 mm Required diameter of the tube = 8.0 mm 1.26 3 Air 2 mm dia Solution: The liquid in the tube rises (or falls) due to capillarity. The capillary rise (or fall) 1.5 cm Bubble = 2 mm dia Fig. 1.14 2s cos q gR 3 R = mm = 1.5 ¥ 10–3 m 2 q = 130°, s = 0.48 N/m g = rg = 13.6 ¥ 103 ¥ 9.81 h = Example 1.24 Here 18 Fluid Mechanics and Hydraulic Machines \ h= 2 ¥ 0.48 ¥ cos 130∞ 3 = -3 (13.6 ¥ 10 ¥ 9.81) ¥ (1.5 ¥ 10 ) = – 3.08 ¥ 10–3 m = –3.08 mm Therefore, there is a capillary depression of 3.08 mm. 2 ¥ 0.073 (9.81 ¥ 998) ¥ (2.5 ¥ 10-6 ) = 5.95 m 1.29 s D 1.27 h Plate 3 Liquid h Solution: Here, q d1 d2 g = 0 and hence cos q = 1.0 = 1.0 mm, R1 = 0.0005 m; = 2.0 mm, R2 = 0.001 m; = rg = 800 ¥ 9.81 = 7848 N/m3 2s cos q 2s = h1 = g R1 7848 ¥ 0.0005 = 0.50968s 2s cos q 2s Similarly, h2 = = g R2 7848 ¥ 0.001 = 0.25484s 1.3 (h1 – h2) = = [0.50968s – 0.25484s] 100 0.013 = 0.2548s and s = 0.051 N/m 1.28 Diameter D Plate (a) F p0 s h p1 s Diameter D (b) Fig. 1.15 Example 1.29 Solution: Let the pressure difference between the ambient and that in the ﬂuid within the plate gap be Dp. Then Dp = p1 – p0 Balancing the forces in the free body (Fig. 1.15(b)) pDh (Dp) = 2s (pD) 2s h Force required to pull the plates apart or Solution: Diameter of pores = 2R = 0.005 mm R = 0.00025 mm = 2.5 ¥ 10–6 m 2s by assuming q = 0° Dh = gR Dp = Ê pD2 ˆ F= Á (Dp) Ë 4 ˜¯ = 2pD 2s p Ê Dˆ = Á ˜ (sD) 4h 2 Ë h¯ 19 Properties of Fluids p = 120 ¥ 103 Pa. (abs) T = 273 + 60 = 333 K p 120 ¥ 103 Density r = = RT 287 ¥ 333 = 1.256 kg/m3 2, M = 44 8312 Gas constant R = = 189 44 120 ¥ 103 Density r = (189)(333) 1.30 Solution: The soap bubble has two interfaces. Work done = Surface tension ¥ total surface area 2 Ê 12 ˆ = 0.040 ¥ 4 p Á ¥ 10-2 ˜ ¥ 2 Ë 2 ¯ = 36.2 ¥ 10– 4 N.m 1.31 = 1.907 kg/m3 Solution: Two water surfaces at the ring resist the lifting force. Referring to Fig. 1.16, by assuming d 1 and B = 0 (b) n = 1 and B = 0 (c) n > 1 and B π 0 (d) n < 1 and B = 0 1.19 The following shear stress – shear rate relationship was obtained for a ﬂuid: du/dy (units) t (units) 0 0 1 6 3 18 5 30 The ﬂuid is classiﬁed as (a) Bingham plastic (b) Dilatant (c) Newtonian (d) Ideal 1.20 The dimensions of the coefﬁcient of dynamic viscosity in [M, L, T] notation system are (a) M L–1 T (b) M L–1 T –1 –1 (c) M L T (d) M L T –1 27 Properties of Fluids 1.21 Poise is a unit of (a) dynamic viscosity (b) kinematic viscosity (c) vapour pressure (d) surface tension 1.22 If the unit of dynamic viscosity of a ﬂuid is stated as Poise, one unit of poise is equivalent to (a) 1/10 pa.s (b) 10 pa.s 1 –4 2 dyne.s/cm2 (c) 10 m /s (d) 100 1.23 The dimensions of the coefﬁcient of dynamic viscosity in [F, L, T] notation system are (a) F T L–2 (b) F L–1 T–1 2 –1 (c) F L T (d) F T–2 L 1.24 A Newtonian ﬂuid ﬁlls the clearance between a shaft and a sleeve. When a force of 800 N is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead, the shaft would move with a speed of (a) 1.5 cm/s (b) 13.5 cm/s (c) 0.5 cm/s (d) 4.5 cm/s 1.25 Typical example of a non-Newtonian ﬂuid of pseudoplastic veriety is (a) water (b) air (c) blood (d) printing ink 1.26 The kinematic viscosity n is related to the dynamic viscosity m and density r as n = (a) m/r (b) mr (c) r/m (d) m/rg 1.27 The unit of dynamic viscosity of a ﬂuid is (a) m2/s (b) N.s/m2 (c) Pa.s/m2 (d) kg.s/m 1.28 A ﬂow of a viscous ﬂuid with m = 1.0 Ns/ m2 has a velocity distribution given by u = 0.90 y – y2. The shear stress at y = 0.45 m is (a) 0.90 N/m2 (b) (c) zero (d) – 0.90 N/m2 1.29 A perfect gas (a) is same as an ideal ﬂuid (b) satisﬁes pv k = constant (c) has zero velocity (d) satisﬁes p/r = RT relation 1.30 A perfect gas (a) is a perfect ﬂuid (b) does not have viscosity (c) is incompressible (d) does not really exist 1.31 In an isentropic process (a) pv = constant (b) p/T = constant (c) p/v k = constant (d) pv k = constant 1.32 The bulk modulus of elasticity for a liquid, K (a) is a function of both temperature and pressure (b) at any given temperature decreases continuously with pressure (c) at any pressure increases continuously with temperature (d) is a constant 1.33 In a sample of water an increase of pressure by 18 MN/m2 caused 1% reduction in the volume. The bulk modulus of elasticity of this sample, in MN/m2, is (a) 1.80 (b) 180 (c) 1800 (d) 0.18 1.34 Broadly speaking, water is (a) 10 times more compressible than steel (b) 80 times more compressible than steel (c) 80 times less compressible than steel (d) 800 times less compressible than steel 1.35 The bulk modulus of elasticity K for a gas at constant temperature is (a) p/r (b) rT (c) rRT (d) p 1.36 The bulk modulus of elasticity for a gas undergoing adiabatic process pv k = constant is, 28 Fluid Mechanics and Hydraulic Machines 1.37 1.38 1.39 1.40 1.41 (a) p/r (b) p (c) kp (d) constant Kerosene is known to have a bulk modulus of elasticity K = 1.43 ¥ 109 N/m2 and a relative density of 0.806. The speed of sound in kerosene, in m/s, is (a) 1333 (b) 1075 (c) 1197 (d) 184 The bulk modulus of elasticity K of (a) a gas is larger than that of a solid (b) a liquid is smaller than that of a gas (c) a liquid is larger than that of a solid (d) a solid is larger than that of a liquid What is the dimension of bulk modulus of elasticity? (a) M L2 T –2 (b) M L–1 T –1 –2 –2 (c) M L T (d) M L –1 T –2 Which of the following is the correct expression for the bulk modulus of elasticity of a ﬂuid? dr dp (b) r (a) r dp dr dr dp (c) (d) rd p r dr Which of the following is the correct expression for the velocity of sound in a ﬂuid? dp dr (b) (a) dr dp (c) dr r dp (d) r dp dr 1.42 The dimension of surface tension is (a) N/m2 (b) J/m (c) J/m2 (d) N/m 1.43 If the capillary rise of water in a 2 mm diameter tube is 1.5 cm, the height of capillary rise in a 0.5 mm diameter tube, in cm, will be (a) 10.0 (b) 1.5 (c) 6.0 (d) 24.0 1.44 If the surface tension of water-air interface is 0.073 N/m, the gauge pressure inside a rain drop of 1 mm diameter is, (a) 146.0 N/m2 (b) 0.146 N/m2 2 (c) 73.0 N/m (d) 292.0 N/m2 1.45 The excess pressure (above atmospheric) inside a soap bubble of diameter 1 cm, by assuming the surface tension of soap solution to be 0.04 N/m is, (a) 32.0 N/m2 (b) 16.0 N/m2 (c) 160.0 N/m2 (d) 0.32 N/m2 1.46 The capillary rise in a 3 mm tube immersed in a liquid is 15 mm. If another tube of diameter 4 mm is immersed in the same liquid the capillary rise would be (a) 11.25 mm (b) 20.00 mm (b) 8.44 mm (d) 26.67 mm 1.47 The pressure difference between the inside and outside of a rain drop of diameter d is equal to 2s s (b) (a) d d s 4s (c) (d) 2d d where s = surface tension for air water interface. 1.48 The capillary rise of water at 20°C in a clean glass tube of 1.0 mm diameter tube is about (a) 15 mm (b) 50 mm (c) 25 mm (d) 30 mm 1.49 An apparatus produces water droplets of size 70 mm. If the coefﬁcient of surface tension of water in air is 0.07 N/m, the excess pressure in these droplets, in kPa, is, (a) 5.6 (b) 4.0 (c) 8.0 (d) 13.2 1.50 If the coefﬁcient of surface tension of water in air is 0.07 N/m, the diameter of a tube that can be used to keep the capillary height between 1.80 cm to 2.00 cm is, 29 Properties of Fluids (a) 1.65 mm (b) 3.33 cm (c) 1.65 cm (d) 1.40 cm 1.51 At a liquid-air-solid interface the contact angle q measured in the liquid is less than 90°. The liquid is, (a) wetting (b) non-wetting (c) ideal (d) does not form a stable bubble 1.52 If s Lg is surface tension at liquid-gas interface, sgs is surface tension at gassolid interface, and sLs is surface tension at liquid-solid interface, a small drop of liquid when dropped on to a solid surface will remain in equilibrium without spreading if (a) \| sLg – sLs \| > sgs (b) \| sgs – sLs \| > sLg (c) \| sLg – sLs \| < sgs (d) \| sgs – sLs \| < sLg 1.53 The predominant ﬂuid property associated with cavitation phenomenon is (a) surface tension (b) vapour pressure (c) mass density (d) bulk modulus of elasticity 1.54 At 20°C, pure water will have a vapour pressure, in kPa, of about (a) 0.5 (b) 2.34 (c) 101.3 (d) 8.67 1.55 At 100°C, at sea level, pure water will have a vapour pressure, in kPa, of about (a) 0.50 (b) 2.3 (c) 10.1 (d) 101.3 Fluid Statics Concept Review 2 Introduction 2.1 PRESSURE IN A STATIC FLUID Z Direction The basic law relating to the pressure (normal stresses) in a static ﬂuid is Pascal’s law which states that the pressure at a point in a ﬂuid at rest is same in all directions. For incompressible ﬂuids (i.e., for liquids and such of the gas ﬂow situations where compressibility effects can be ignored), the variation of pressure in vertical direction in a static ﬂuid is given by dp = -g (2.1) dz (p2 – p1) = g (Z1 – Z2) (2.1-a) ( p + g Z ) = ( p1 + g Z1 ) = ( p2 + g Z 2 ) = Constant (2.2) where g = Speciﬁc weight of the ﬂuid and Z = Vertical distance measured from a datum (positive upward). 1 (Z1 – Z2) Z1 2 Z2 Datum Fig. 2.1 At a free surface the pressure is atmospheric. If h is the depth below the free surface of a point M, the absolute pressure at M (Fig. 2.2) is pm (abs) = g h + patm If the pressure in excess of atmosphere is recorded then pm (abs) - patm = pm = g h (2.3) 31 Fluid Statics Patmos absolute pressures. Absolute pressures cannot be negative. h Pm (abs) = g h + Patm M Fig. 2.2 [Note: That h is measured positive downwards from the liquid surface]. The pressure pm is then called gauge pressure. The linear variation of pressure with depth below the free surface is known as hydrostatic pressure distribution. The variation of gauge pressure in a liquid below the free surface is shown in Fig. 2.3. From this, p1 = g h1 and p2 = g h2, or ( p2 - p1 ) = g ( h2 - h1 ) h1 h1 h2 g h1 h 1 h2 2 g h2 gh Fig. 2.3 Note that in the above the atmospheric pressure was assumed as the datum, i.e., reference with a zero value. Different references can be taken and depending upon the reference pressures we have the following: Absolute pressure is the pressure measured above the absolute zero, a thermodynamic concept. Absolute pressures are designated with (abs) following the symbol or numeral to distinguish from other forms. Thus, for example, 12.0 kPa (abs); pm (abs) are Gauge pressure is the pressure measured with respect to local atmospheric pressure. Gauge pressures are extensively used in engineering practice and as such are indicated with a symbol or a numeral without any other explanatory notation, e.g. 14.0 kPa, – 3.2 kPa, pm are gauge pressures. Note that gauge pressures can be positive or negative. Negative gauge pressures are also called vacuum pressures. It is seen that Absolute pressure = (Local atmospheric pressure) + (gauge pressure) Pressure has the dimension of [Force/Area] = [F L–2 ] and is usually expressed in pascals Pa (= N/m3); kilo pascals kPa (= 103 N/m2); height h of a column of a ﬂuid of speciﬁc weight g , in bars (= 105 Pa) or atmospheres (= number of standard atmospheric pressure value). The pressures are commonly indicated as gauge pressures and unless a pressure is speciﬁcally marked absolute the pressure is treated as gauge pressure. The atmosphere, however, is an exception and is an absolute pressure unit. Gauge pressures are commonly measured by a Bourdon gauge. Differences in pressures are measured by manometers. Local atmospheric pressure (i.e. the absolute pressure of the atmosphere at a place) is measured by a mercury barometer. The local atmospheric pressure varies with the elevation above mean sea level and local meteorological conditions. For engineering application, a standard atmospheric pressure at mean sea level at 15°C is often used. The value of this standard atmospheric pressure (called 1 atmosphere) is 1 atm = 10.336 m of water = 760 mm of mercury = 101.325 kPa = 10132.5 mbar Aneroid barometer is another instrument commonly used to measure local atmospheric pressure. 32 Fluid Mechanics and Hydraulic Machines 2.1.1 Aerostatics (2) Non-Isothermal Atmosphere The variation of pressure in the earth’s atmosphere is of importance in many aspects of engineering. The study of atmosphere in its state of static equilibrium is known as aerostatics. It is generally observed that from sea level up to an elevation of about 11,000 m the temperature varies linearly with the elevation. This region is know as troposphere. Beyond 11,000 m up to 24,000 m the region is known as stratosphere and the temperature is found to be approximately constant at 216.5 K in this region. Three approaches used in aerostatics studies are given below. It is usual to consider that in troposphere the temperature decreases linearly with elevation as Ú (2.6) Depending upon the process involved, i.e., isothermal, constant temperature lapse rate or adiabatic, the corresponding variation of pressure with Z can be determined. (1) Isothermal Process In an isothermal process, T = T0 = constant. p Since r= RT dp p using Eq. 2.5, = dZ RT0 2 dp g 2 = dZ 1 p RT0 1 Ú For standard atmosphere, a = 6.5 K/km and at sea level, tempereture T0 = 285 K and density r 0 = 101.325 kg/m3. p p = RT R(T0 - a Z ) Substituting in Eq. (2.5) dp pg = – rg = dz R(T0 - a Z ) dp g dZ = p R(T0 - a Z ) On integration g T - aZ Ê pˆ ln Á ˜ = ln 0 Ra T0 Ë p0 ¯ aZ ˆ Ê pˆ Ê ÁË p ˜¯ = ÁË1 - T ˜¯ 0 0 (2.8) For the case of adiabatic process (zero heat transfer), if there is no friction (isentropic) p = constant = Cs (2.9) rk where k = adiabatic constant for the gas. Combining with perfect gas law (Eq. 2.4) we get r k -1 (2.7) g /Ra (3) Adiabatic Process T Ú p2 È - g ( Z 2 - Z1 ) ˘ = exp Í ˙ p1 RT0 Î ˚ T = Temperature at an elevation Z above sea level a = a constant known as lapse ratet From Eq. (2.5) r = For a compressible ﬂuid, the density changes with pressure and temperature. For a perfect gas p = rRT (2.4) where p = absolute pressure r = mass density T = absolute temperature (in Kelvin), R = gas constant dp = – rg (2.5) Since dz = - g r dZ where T0 = Absolute temperature at sea level (that is at Z = 0) Variation of Pressure with Elevation Density–Pressure Relationship in Compressible Fluids Ú dp T = T0 – a Z = constant; and by using (Eq. 2.5), on integration = constant T Ê k -1ˆ Á ˜ rË k ¯ 33 Fluid Statics Substituting in Eq. (2.6) and on simpliﬁcation Ê k ˆ È ( k - 1) p2 Ê r ˆ ˘ÁË k -1˜¯ g ( Z 2 - Z1) Á 1 ˜ ˙ (2.10) = Í1 p1 k Ë p1 ¯ ˚ Î Ê k ˆ p È ( k - 1) ( Z 2 - Z1) ˘ÁË k -1˜¯ or 2 = Í1 (2.11) g p1 k RT1 ˙˚ Î The variation of the temperature with Z in adiabatic process is given by T2 È ( k - 1) ( Z 2 - Z1) ˘ = Í1 g (2.12) T1 k RT1 ˙˚ Î The rate of variation of the temperature with dT is known as lapse rate (L) and for the elevation dZ atmosphere having adiabatic process it is given by È g Ê k - 1ˆ ˘ dT L= = Í- Á ˜˙ dZ Î R Ë k ¯˚ (2.13) of liquid. Figure 2.4 shows some commonly used forms of manometers. In a manometer the basic point to remember is that, in a continuous mass of the same static ﬂuid, the pressure at the points in any horizontal plane will be the same. 2.2 FORCES ON PLANE SURFACES An important problem in the design of hydraulic structures and other structures which interact with ﬂuids is the computation of hydrostatic forces on plane surfaces. Computations of magnitude and point of application of hydrostatic forces on plane surfaces are described. 2.2.1 Magnitude of Force on a Plane When a plane area is immersed in a static liquid with its plane making an angle q with the free liquid surface (Fig. 2.5) the total hydrostatic force on one side of the area is 2.1.2 Manometers For a ﬂuid at rest it is seen from Eq. 2.1(a) that (p2 – p1) = g (Z2 – Z1). The pressure difference between two points can thus be measured by a static column of a liquid. A manometer is a device to determine the pressure in a ﬂuid by balancing it against a column F = g hA where g = speciﬁc weight of the liquid h = depth of the centre of gravity of the area below the free surface A = area of the immersed plane. patm = 0 (Z2 – Z1) 1 1 RD = S2 y1 RD = Sp 0 y2 RD = S1 0 y RD = Sm p1 = g [Sm y2 – Spy1] 0 0 RD = Sm ( p1 – p2) = g y(Sm – S1) + g S2(Z2 – Z1) (a) Open manometer Fig. 2.4 (2.14) (b) Differential manometer Manometers 34 Fluid Mechanics and Hydraulic Machines xp = x + q x F = g hA h I xy (2.16) Ay g where Ixy = product of inertia Ê = Ë Ú xy d Aˆ of the ¯ A yp Y¢ y cp G O x Y¢ xp area about axis GY ¢, passing through the centre of gravity of the area and parallel to OY and OX. When either of the centroidal axes x = x or y = y is an axis of symmetry, Ixy = 0 and x p = x . Properties of some commonly encountered simple geometrical shapes are collated in Table 2.1 g 2.3 FORCES ON CURVED SURFACES Y Fig. 2.5 Centre of Pressure It may be noted that the force F is independent of the angle of inclination q so long as the depth of the centroid h is unchanged. 2.2.2 Centre of Pressure The point of application of the force F on the submerged area is called the centre of pressure. Considering the line of intersection of the plane area with the liquid surface (Line OX) as the reference axis, the centre of pressure is located along the plane at yp = y + I gg Ay (2.15) where Igg = moment of inertia about an axis parallel to OX and passing through the centre of gravity of the area y = location of the centre of gravity with respect to the axis OX A = area of the plane area Note that the distances y are measured along the plane from the axis OX. The lateral position of the centre of pressure with respect to any axis OY perpendicular to OX and lying in the plane of the lamina is When the ﬂuid static force on a curved submerged surface is desired, it is convenient to consider the horizontal and vertical components of the force separately. 2.3.1 Horizontal Component The horizontal component of hydrostatic force in any chosen direction on any area (plane or curved) is equal to the projection of the area on a vertical plane normal to the chosen direction. The horizontal force acts through the centre of pressure of the vertical projection. 2.3.2 Vertical Component The vertical component of the hydrostatic force on any surface (plane or curved) is equal to the weight of volume of liquid extending above the surface of the object to the level of the free surface. This vertical component passes through the centre of gravity of the volume considered. The volume and the free surface can be real or imaginary. 2.3.3 Tensile Stress in a Pipe or Shell In a circular pipe subjected to high pressure, the pressure centre can be taken to be at the centre of the pipe. The tensile circumferential stress (hoop stress) in a pipe wall subjected to an internal pressure of p (Fig. 2.6) is 35 Fluid Statics Table 2.1 Properties of areas Sketch Area Location of centroid I or Ic b Rectangle Ic Triangle bh yc = h 2 Ic = bh3 12 bh 2 yc = h 3 Ic = bh3 36 pD 2 4 yc = D 2 Ic = pD 4 64 yc pD 2 8 yc = 4r 3p I= pD 4 128 h pbh 4 yc = h 2 Ic = pbh3 64 yc pbh 4 yc = 4h 3p I= pbh3 16 yc 2bh 3 xc = I= 2bh3 7 h yc h Ic y c b Circle Ic D yc r Semicircle I D b Ellipse Ic Semiellipse I yc h b b h Parabola 3b 8 3b yc = 5 I xc I = Moment of inertia about indicated axis Ic = Moment of inertia about indicated axis passing throuth the centre of gravity of the area hoop stress s h = t where T F = pD = 2T (per unit length) Fig. 2.6 T pD 2t (2.17) D = diameter of the pipe t = thickness of pipe. This formula assumes t/D < 0.1 and hence is based on thin cylinder theory. If the ends of a cylinder are closed and the cylinder has a ﬂuid under pressure, a longitudinal stress s L is 36 Fluid Mechanics and Hydraulic Machines produced in the cylinder. This stress is given by sL = 1 pD sh = 2 4t (2.18) M For thin spherical shells the tensile stress is pD ss = 4t G W (2.19) Fb Fb B G B B¢ W 2.4 BUOYANCY When a body is submerged or ﬂoating in a static ﬂuid the resultant force exerted on it by the ﬂuid is called buoyancy force. This buoyancy force is always vertically upwards, and has the following characteristics. 1. The buoyancy force is equal to the weight of the ﬂuid displaced by the solid body. 2. The buoyancy force acts through the centre of gravity of the displaced volume, called the centre of buoyancy. 3. A ﬂoating body displaces a volume of ﬂuid whose weight is equal to the weight of the body. (a) (b) Stable equilibrium (M is above G) Fig. 2.7 In this equation I = Moment of inertia of the water line area about an axis through the centre of the area and perpendicular to the axis of tilt (longitudinal axis). BG = Vertical distance between the centre of gravity and centre of buoyancy. V = Volume of the ﬂuid displaced by the body. If M coincides with G, MG is zero, the body is said to be in neutral equilibrium. 2.4.1 Stability A submerged body is stable if the centre of gravity of the body lies below the centre of buoyancy. For a ﬂoating body the stability depends upon the type of couple that is formed for small angular displacements. For a body shown in Fig. 2.7(a) the centre of gravity is G and the centre of buoyancy is B. Initially it is stable with G above B. Figure 2.7(b) shows the same body with a small displacement. If B¢ is the new centre of buoyancy a vertical from B¢ intersects the line of symmetry through G at M. M is known as the meta centre. If M is above G, then MG the metacentric height is positive and the equilibrium is stable. If M is below G, MG is negative and equilibrium is unstable. The metacentric height MG is independent of magnitude of angular rotation (so long as it is small) and is given by MG = I – BG V (2.20) 2.5 RIGID BODY MOTION When a ﬂuid mass in a container is subjected to a motion such that there is no relative motion between the particles, such a motion is known as rigid body motion. The motion can be either translation or rotation at constant acceleration or a combination of both. As there is no relative motion there is no shear stress in such a motion and the pressure distribution is similar to that in ﬂuids at rest, of course modiﬁed by the combined action of gravity and ﬂuid acceleration. 2.5.1 Translation If a container with a ﬂuid is given a translation (a linear motion) with a uniform acceleration the piezometric head will have a gradient in the direction of motion. 37 Fluid Statics If the motion is in the x-direction with a constant acceleration ax then a dh (2.21) = tan q = x g dx where h = (p/g + z) = piezometric head above datum q = Inclination of hydraulic grade line. = Inclination of water surface, measured clockwise with respect to the x-direction. Thus, if a vessel containing a liquid is given an acceleration ax in x-direction (Fig. 2.8) the surface will back up against the farthest side, i.e., it will have increasing depth in (– x) direction. Z In vertical acceleration the liquid suffers an apparent gravity equal to (g + az ). If the acceleration is as in any direction s, then the com-ponents ax and az in x- and z-directions are considered. The ﬂuid surface will now have an inclination tan q given by - 2.5.2 ax dh = tan q = (g + az ) dx Rigid Body Rotation When a vessel containing a liquid with a free surface is rotated about an axis, the free surface will be a paraboloid of revolution given by X y= q h (2.25) w = angular velocity y = height of the free surface above the vertex at a radial distance r from the axis. At any two points r1 and r2 from the axis g Fig. 2.8 w2 2 ( r2 - r12 ) 2g Since w r = V = tangential velocity. (y2 – y1) = Dy = difference in the liquid surface elevation between the points 2 and 1 (Fig. 2.9) (y2 – y1) = If a closed tank without a free surface is involved, an imaginary free surface equivalent to the piezometric head line can be considered. This piezometric head line will be inclined to the x-direction such that (2.22) It follows from the above that if acceleration is solely in the vertical direction (+ z direction) then ax = 0 and tan q = 0. This means that the liquid surface will remain horizontal. However, the pressure ph at any depth h below the free surface will now be Ê a ˆ ph = g h Á1 + z ˜ g¯ Ë w 2r2 2g where ax tan q = ax / g (2.24) (2.23) In this az = vertical acceleration in + z direction (if the acceleration is vertically downwards, az is taken as negative). 2 1 h2 y x A r Datum w Fig. 2.9 38 Fluid Mechanics and Hydraulic Machines V22 V12 = D (V 2/2g) 2g 2g = difference in the velocity head at these two points The pressure distribution in any vertical line at a radial distance r will, however, remain hydrostatic. pA + zA for all values of A on this At point 2, h2 = g vertical line. = If the free surface does not exist, the piezometric head will follow the relation for y (Eq. 2.22) as: ( h - h0 ) = w 2r2 2g where h = piezometric head above a datum at any radial distance r from the axis h0 = value of h at r = 0, i.e. on the axis w = angular velocity. Êp ˆ The piezometric head h = Á + z ˜ will vary with Ëg ¯ r as a paraboloid of revolution and this surface can be considered as an imaginary liquid surface. The volume of a paraboloid of revolution is one half the volume of the circumscribing cylinder. (2.26) Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples A. Measurement of Pressure column of Solution: 5 100 = 367.13 N/m2 (iii) For mercury column: g = 13.6 ¥ 9790 = 133144 N/m3 2 p = 133144 ¥ 100 = 2662.9 N/m2 p = 7342.5 ¥ 2.1 Pressure p = g h (i) For water column: g = 9790 N/m3 10 p = 9790 ¥ 100 = 979 N/m2 (ii) For the oil column:g = 0.75 ¥ 9790 = 7342.5 N/m3 2.2 39 Fluid Statics p1 = 0 Solution: For the liquid: g = 0.85 ¥ 9790 = 8321.5 N/m3 Atmospheric pressure = 0.750 ¥ 13.6 ¥ 9790 = 99858 N/m2 Absolute pressure in the liquid at 5.0 m depth = pressure due to 5 m of liquid + Atmospheric pressure = (5 ¥ 8321.5) + 99858 = 141465.5 N/m2 = 141.466 kPa 2.3 A hydraulic press has a ram of 150 mm and a plunger of 20 mm diameter. Find the force required on the plunger to lift a weight of 40 kN. If the plunger has a stroke of 0.40 m and makes 30 strokes per minute, determine the rate at which the weight is lifted per minute and the power required by the plunger. Assume no losses whatsoever. Solution: Let F = force on the plunger. Since the pressure in the ﬂuid is same at the plunger and at the weight: F 40 = p p Ê Ê 2ˆ 2ˆ ÁË (0.15) ˜¯ ÁË (0.02) ˜¯ 4 4 F = 0.711 kN Total length of stroke in one minute = 0.40 ¥ 30 = 12.0 m Distance travelled by weight/minute = 12 ¥ (0.02) 2 (0.15) 2 0.2133 ¥ 40 60 = 0.142 kW = 142 W h1 = 2 m 2.4 A 6 m deep tank contains 4 m of water and 2 m of oil of relative density 0.88. Determine the pressure at bottom of the tank (Refer Fig 2.10). Oil RD = 0.8 2m Water 4m 2 p2 h2 = 4 m p3 3 Fig. 2.10 Solution: First determine the pressure at the oilwater interface. p2 = p1 + pressure due to 2 m of oil = p1 + (g 0 ¥ 2) p1 = 0, g 0 = 0.88 ¥ 9790 = 8615.2 N/m3 \ p2 = 8615.2 (2) = 17230.4 N/m2 For water g w = 9790 N/m3 p3 = p2 + pressure due to 4 m of water = 17230.4 + (9790 ¥ 4) = 56390.4 N/m2 = 56.390 kPa 2.5 the pressure at an elevation of 2500 m above sea level. The mass density and atmospheric pressure at an elevation of 500 m above sea level are known to be 1.1677 kg/m3 and 95480 Pa. What is the density of air at that level? Solution: For an isothermal atmosphere, the pressure distribution is given by = 0.2133 m/minute Power required = 1 Since È g ( Z 2 - Z1) ˘ p2 = exp Í˙ p1 RT0 Î ˚ p1 = RT0 r1 È r g ( Z 2 - Z1) ˘ p2 = exp Í- 1 ˙ p1 p1 Î ˚ È (1.1677 ¥ 9.81) ¥ 2000 ˘ p2 = 95480 exp Í˙ 95480 Î ˚ = 75111 Pa p2 p = 1 r2 r1 40 Fluid Mechanics and Hydraulic Machines \ r2 = p2 r1 p1 3 75111 ¥ 1.1677 = 0.9186 kg/m3 = 95480 Solution: p0 = r 0 RT0 101.3 ¥ 103 = 291.9, R = 1.205 ¥ 288 a = 0.000635 K/m g 9.81 = = 5.17 Ra 291.9 ¥ 0.0065 2.6 pressure at an elevation of 1000 m above sea level 3 Solution: For an adiabatic atmosphere È Ê k - 1ˆ g ( Z 2 p2 = Í1 - Á ˜ p1 RT1 Î Ë k ¯ p1 = RT1 r1 since ˙ ˚ È Ê k - 1ˆ (gr1) ( Z 2 - Z1) ˘ p2 = Í1 - Á ˙ ˜ p1 p1 Î Ë k ¯ ˚ k is assumed as equal to 1.4. Hence k 1.4 = = 3.5 ( k - 1) 0.4 È Ê 1.4 - 1ˆ p2 = 89,890 Í1 - Á ˜ Î Ë 1.4 ¯ p1 ( k / k - 1) 2.8 A and B A 10 cm 3.5 Oil RD = 0.70 B Oil RD = 0.8 13 cm 8 cm 1/ k X X r1 Ê 65446 ˆ = Á Ë 89890 ˜¯ = 89,890 [1 – 0.086683]3.5 = 65446 Pa = 65.446 kPa p = 2k r2 Êp ˆ r2 = Á 2 ˜ Ë p1 ¯ \ g/Ra p È az˘ = Í1 p0 T0 ˙˚ Î p = (1 – 0.226)5.17 = 0.2665 p0 p = 0.2665 ¥ 101.3 = 27 kPa From Eq. (2.4) (9.81 ¥ 1.1120) (3500 - 1000) ˘ ˙ 89, 890 ˚ r1k az 0.0065 ¥ 10, 000 65 = = 0.226 = T0 288 288 k Z1) ˘ ( k - 1) Mercury RD = 13.6 1 / 1.4 3 (1.1120) = 0.88665 kg/m 2.7 2 Fig. 2.11 Solution: Equating the pressures on both the limbs at the horizontal plane X X (unit weight of water is taken as 9790 N/m3): 41 Fluid Statics pA + [0.10 + 0.13 + 0.08] ¥ g A = pB + 0.13 g B + 0.08 gm gA = 0.70 ¥ 9790 = 6853 N/m3 gB = 0.80 ¥ 9790 = 7832 N/m3 gm = 13.6 ¥ 9790 = 133144 N/m3 ( pA – pB) = – 0.31 ¥ 6853 + 0.13 ¥ 7832 + 0.08 ¥ 133144 = 9545 Pa = 9.545 kPa 2.9 For the manometer shown in Fig. 2.12 calculate the pressure between points M and N. difference 2.10 Find the difference in the pressure at the the Fig. 2.13. Solution: The pressures at the level of H2 above the ﬂoor of the tanks are equal. Let p1 = pressure on the ﬂoor of Tank 1. p2 = pressure on the ﬂoor of Tank 2. Hence p1 – r1 gH1 – (H2 – H1)r 3g = p2 – r2 gH2 (p2 – p1) = (r2 gH2) – (r1gH1 + (H2 – H1)r3g) (p2 – p1) = Oil of RD = 0.83 p3 X X 3.5 cm Water 6.0 cm p1 Water M 12 cm H2 H1 1 p2 2 Fig. 2.13 N Fig. 2.12 Solution: Equating the pressures at both the limbs along the horizontal plane XX pm – gw ◊ (0.06 + 0.035) = pN – gw (0.12 + 0.06) – g0 (0.035) gw = unit wt. of water = 9790 N/m3 (assumed) g0 = unit wt. of oil = 0.83 ¥ 9790 = 8125.7 N/m3 \ (pM – pN) = 9790 ¥ 0.095 – 9790 ¥ 0.18 – 8125.7 ¥ 0.035 2 = – 1116.5 N/m = – 1.1165 kPa Pressure at N is larger than at M by 1.117 kPa 2.11 the other (Fig. 2.14). A bourdon gauge M mm of mercury. Calculate the absolute pressure recorded at M and N in of mercury. Solution: A bourdon gauge records the gauge pressure relative to the pressure of the medium surrounding the tube. Local atmospheric pressure is measured by the aneroid barometer. In the present case local atmospheric pressure outside the gauge N = 750 mm. N M Fig. 2.14 42 Fluid Mechanics and Hydraulic Machines Hence absolute pressure at N = Ê ˆ 35 ¥ 1000˜ pN (abs) = 750 + Á Ë 13.6 ¥ 9.79 ¯ = 1012.9 mm of mercury (abs) The gauge M reads relative to its surrounding pressure of 1012.9 mm of mercury (abs). Hence, Ê 20 ¥ 1000 ˆ pM (abs) = 1012.9 + Á Ë 13.6 ¥ 9.79 ˜¯ = 1163.1 mm of mercury (abs) \ pA = (0.10 ¥ 133144) – 0.50 ¥ 9790 – 1.5 ¥ 7342.5 – 0.1 ¥ 9790 = – 3573.4 Pa = – 3.573 kPa 2.13 (pM – pN M N Oil of RD = 0.8 15 cm 20 cm 2.12 contains air at a pressure pA pA 15 cm 12 cm 16 cm A Open tube X 200 cm Oil RD = 0.75 150 cm Air Water Mercury Fig. 2.16 Solution: Equating the horizontal plane X-X X Mercury Fig. 2.15 Solution: Considering the pressure at the horizontal plane X–X: pA + 1.5 ¥ g 0 + (2.0 – 1.5) gw + 0.10 gw = 0.10 gm go = speciﬁc wt. of oil = 0.75 ¥ 9790 = 7342.5 N/m3 gw = speciﬁc wt. of water = 9790 N/m3 g m = speciﬁc wt. of mercury = 13.6 ¥ 9790 = 133144 N/m3 pressure across the pM + 0.20 go + 0.16 go = pN + 0.15 go + 0.15 gm – 0.12 go + 0.12 gm + 0.04 gm go = speciﬁc weight of oil = 0.8 ¥ 9790 = 7832 N/m3 gm = speciﬁc weight of mercury = 13.6 ¥ 9790 = 133144 N/m3 (pM – pN) = – 0.36 ¥ 7832 + 0.15 ¥ 7832 – 0.12 ¥ 0.7832 + (0.15 + 0.12 + 0.04) ¥ 133144 = – 2584.6 + 41274.6 = 38690 Pa = 38.690 kPa 10 cm X X 2.14 43 Fluid Statics p2 Hence Dc = Minimum diameter of cistern = d/ 0.005 = 14.142 d p1 (b) If H Zero level for (p1 – p2) = 0 DH O a = 1/500 Ac DH 1 = H 500 then Percentage error in DH 1 ¥ 100 = ¥ 100 = 0.2% H= H 500 O 2.15 M and N Mercury Fig. 2.17 M N Example 2.14 M and N H Solution: Given Refer to Fig. 2.18 pm = 10 kPa (vaccum). pn = 20 kPa (gauge). Take gw = 9.79 kPa. Let x be distance from the centerline of the pipes to the top of mercury column in limb connected to of (p2 – p1 H M Solution: The column height is measured with reference to zero level corresponding to (p1 – p2) = 0. When a pressure differential is applied the mercury in the cistern will go down by DH and the true column height is (DH + H). (a) It is required that H + DH £ 1.005 H DH £ 0.005 H DH £ 0.005 H Since (Area of cistern) ¥ DH = (area of tube) ¥ H For minimum area of cistern Ac; DH Ac = aH and Ê d ˆ DH a = 0.005 = = Á ˜ H Ac Ë Dc ¯ 2 N x Water Water h A A Mercury Fig. 2.18 44 Fluid Mechanics and Hydraulic Machines point M (see Fig. 2.18). Further let H = difference in the heights of the mercury columns. It is required to ﬁnd h. Equating the pressures in the two limbs at the plane AA. pm + g w x + (13.6 h)gw = pn + xg w + hgw (pn – pm) = (13.6 – 1) hgw ( pn - pm ) h = 12.6 ¥ 9.79 ( 20.0 - ( - 10.0)) = 123.354 = 0.2432 m Equating pressure at the level of mercury in the reservoir p + (0.30 ¥ 0.9 ¥ 9.79) = (0.50 ¥ 13.6 ¥ 9.79) Hence p = (0.50 ¥ 13.6 ¥ 9.79) – (0.30 ¥ 0.9 ¥ 9.79) = 66.572 – 2.643 = 63.93 kN/m2 [Note: Since the actual level of the mercury in the reservoir is used in the calculations, the area of the reservoir is of no consequence.] 2.17 A manometer is made of a tube of uniform 2 Hence difference in the height of mercury in the two limbs = 24.32 cm 2.16 3 Solution: Figure 2.20 is a schematic representation of the manometer set up. Initial liquid level Solution: Let p = pressure in the pipe. patm L2 h2 h1 Interface position L1 Water Angle = 30° Sp. Gravity = 1.25 Fig. 2.20 30 cm 50 cm Pressure = p SG = 0.9 SG = 13.6 Fig. 2.19 The total change in the inclined limb is made up of two parts: (1) length L1 by which the original meniscus moved down to the ﬁnal position, and (2) length L2 by which the ﬁnal water surface is positioned with respect to the initial liquid level. The corresponding changes in the vertical limb are h1 and h2 as shown in Fig. 2.20 7.5 = 15 cm Now (L1 + L2) = 0.5 45 Fluid Statics Considering the pressure exerted by this column, (L1 + L2)rg sin 30° = (r ¥ 1.25) g(h1 + h2) (i) 15 ¥ 0.5 = 6.0 cm (h1 + h2) = 1.25 Considering the extra volume of 15 cm3 of water added in the tube of area A = 0.5 cm2 15 ¥ A = L 2 A + h2A. But L 2 = 15 – L1 and h2 = 6.0 – h1 Thus 15 ¥ A = (15 – L1) A + (6.0 – h1)A (ii) Noting that L1 = 2h1, Eq. 2 can be written as 15 = (15 – 2h1) + (6.0 – h1) 3h1 = 21 – 15 = 6.0 and h1 = 2.0 cm h2 = rise in the level of the meniscus in the vertical limb = 6.0 – h1 = 6.0 – 2.0 = 4.0 cm 2.18 a densities r1, r2 and r3 r3 r1 < r2 r2 > (r3 + 2r1) 3 3 Solution: Referring to Fig. 2.21, let E, F and G be the interfaces. Let E A = x. Then DE = DA – EA = (a – x) Total tubing length = 4a 4 Length of each liquid = a 3 1 4 ˆ Ê ÁË Check: FB + BA + AE = 3 a - x + a + x = 3 a ˜¯ At the Interface F: The pressure balance is Pressures of (Column DG + Column GC) = Pressure of column AB Ê1 Ê2 ˆ ˆ r1g Á a + x˜ + r3 g Á a - x˜ = r2ga Ë3 Ë3 ¯ ¯ 1 x(r3 – r1) = a (2r3 + r1 – 3r2) 3 1 x = a (2r3 + r1 – 3r2)/(r3 – r1) 3 1 It is known that x > 0 and also x < a 3 a Hence 0 (1.2 + 2 ¥ 10) 3 > 1.0667 1 r2 < (2r3 + r1) 3 1 < (2 ¥ 1.2 + 1.0) 3 < 1.1333 1.0667 < r2 < 1.1333 r2 > B. Forces On Plane Surfaces 2.19 Y 0. 6 m X G C 2 0. 1. m m Y 1 6 Fig. 2.22 h = depth of CG of the plate = 0.7 + 0.6 sin 45° = 1.1243 m Total pressure force 1 1 (2r3 + r1) > r2 > (r3 + 2r1) 3 3 (b) r1 = 1.0, r3 = 1.2 1 h F = g Ah = 9790 ¥ (0.6 ¥ 1.2) ¥ 1.1243 = 7924.9 N = 7.925 kN Centre of pressure: Because of symmetry xp = x , i.e. C lies on the axis Y1Y1 passing through the CG of the area. I gg yp = y + Ay y = 1.1243 h = = 1.59 m sin 45∞ 1/ 2 bd 3 0.6 ¥ (1.2)3 = = 0.0864 12 12 0.0864 yp = 1.59 + (0.6 ¥ 1. 2) ¥ (1. 59) = 1.665 m [yp is measured along the plane of the area from the axis OX] Igg = 2.20 h is immersed Solution: Referring to Fig. 2.22 47 Fluid Statics Similarly for yp: g 2 bh yp = 3 Solution: Total force F = gAh 2h g Ê1 ˆ = bh2 = g Á bh˜ ¥ Ë2 ¯ 3 3 To determine the centre of pressure, consider the axis OX and OY as shown in Fig. 2.23. For an element of width x and height dy at a depth y, b x h = or y = x h y b h dy = dx b O X Ú h 0 g y ( x d y) y = = gb h yp = 3 h 4 Ú y3 d y = Úg y 2 Ê by ˆ ÁË h ˜¯ d y g b h4 h 4 2.21 A circular disc of diameter D is immersed r Solution: Figure 2.24 is the deﬁnition sketch of the problem. In this C = Centre of pressure and G = centre of gravity. y yp h h yp x dy G xp cp D C b Y Fig. 2.24 Fig. 2.23 Example 2.20 Taking moments of force on the element about OY and intergrating h x F ◊ xp = g y ( x d x) 0 2 b h h x g 2 bh xp = g x x dx 0 b b 2 3 Since the plane is vertical y = h Centre of pressure yp = h + Ú Ú = Ú 2 b gh 0 2b 2 2 2 gh b 8 3 xp = b 8 = h = I GG Ah D 2 Ê pD 4 ˆ IGG = Á Ë 64 ˜¯ and x 3 dx Hence yp = D Ê pD 4 ˆ Ê 1 ˆ Ê 1 ˆ ¥ +Á ˜ + 2 Ë 64 ¯ ÁË pD 2 / 4 ˜¯ ÁË ( D / 2) ˜¯ ÈD D˘ Ê 5 ˆ yp = Í + ˙ = Á D˜ 8˚ Ë8 ¯ Î2 48 Fluid Mechanics and Hydraulic Machines 2.22 Solution: 2.23 Refer to Fig. 2.25. h = 4.0 m 3.0 m r = 1.0 m 4r 3p CG 1.0 m 2.0 m Fig. 2.26 Fig. 2.25 (i) Force exerted on one face F = g Ah Outer diameter D = 2.0 m Inner diameter d = 1.0 m p F = 9.79 ¥ ((2.0)2 – (1.0)2) ¥ 3.0 4 = 69.20 kN (ii) Centre of pressure Due to symmetry the centre of pressure lies on the vertical axis passing through the centre of the circular areas. I yp = y + GG Ay Since the areas are immersed vertically yp = h = 3.0 m p A = ((2.0)2 – (1.0)2) = 2.3562 m2 4 p p (D4 – d 4) = ((2.0)4 – (1.0)4) IGG = 64 64 = 0.7363 yp = 3.0 + 0.7363 = 3.104 m 2.3562 ¥ 3.0 Example 2.23 Solution: For the given semicircular lamina: and Area A = p r 2 = p = 3.142 m2 and r = 1.0 m h = 4.0 Ê 4r ˆ Ê 4ˆ h = h - Á ˜ = 4.0 - Á ˜ = 3.576 m Ë 3p ¯ Ë 3p ¯ Hence force on one side of the plate = g Ah = 9.79 ¥ 3.142 ¥ 3.576 = 110 kN 2.24 Solution: Referring to Fig. 2.27, g = 0.80 ¥ 9790 = 7832 N/m3 h = 1.5 m F = Force on one side of the plate = g Ah Èp ˘ = 7832 ¥ Í ◊ (0.75) 2 ˙ ¥ 1.5 4 Î ˚ = 5190 N 49 Fluid Statics Referring to Fig. 2.28 30° 1.5 m yp 1.0 m yp h Y1 1.5 m G C G C Y1 5 0.7 Fig. 2.27 2.0 m m Fig. 2.28 Example 2.24 By symmetry xp = x , i.e., the centre of pressure lies on the Y1Y1 axis through the CG of the area. To ﬁnd yp: I yp = y + gg Ay y = 1.5/sin 30° = 3.0 m p 4 p ¥ (0.75)4 Igg = D = 64 64 p (0.75) 4 64 yp = 3.00 + p ¥ (0.75) 2 ¥ (3.0) 4 = 3.0117 m 2.25 By symmetry xp = x , i.e., the centre of pressure lies on the vertical axis passing through the CG of the plate. I yp = y + GG Ay Since the plane is vertical, y = h = 2.0 m 1 A = ¥ 2 ¥ 1.5 = 1.5 m2 2 1 IGG = ¥ bh3 36 1 = ¥ 2.0 (1.5)3 36 = 0.1875 m4 0.1875 yp = 2.0 + 1.5 ¥ 2.0 = 2.0625 m Solution: Total pressure force F = g Ah Ê1 ˆ = 9790 ¥ Á ¥ 2 ¥ 1. 5˜ Ë2 ¯ = 29370 N 2 Ê ˆ ÁË1 + 3 ¥ 1. 5˜¯ 2.26 Solution: g oil g water p0 pi = 0.9 ¥ 9790 = 8811 N = 9790 N = pressure at the top = 0 = pressure at the interface = 8811 ¥ 0.9 = 7930 N/m2 50 Fluid Mechanics and Hydraulic Machines pb = pressure on the bottom = 7930 + 9790 ¥ 0.6 = 13804 N/m2 F1 = Force on the top 0.9 m of a side. 1 = ¥ 7930 ¥ (0.9 ¥ 1.5) = 5353 N 2 Ê2 ˆ acting at Á ¥ 0.9˜ = 0.6 m below the surface. Ë3 ¯ F2 = Part of force on the bottom 0.6 m of a side = 7930 ¥ 0.6 ¥ 1.5 = 7137 N 0.6 ˆ Ê acting at Á 0.9 + = 1.2 m below the surface. Ë 2 ˜¯ F3 = remaining part of force on the bottom 0.6 m of a side 1 = ¥ (13804 - 7930) ¥ 0.6 ¥ 1.5 2 1 = ¥ 5874 ¥ 0.6 ¥ 1.5 = 2643 N 2 2 Ê ˆ acting at Á 0.9 + ¥ 0.6˜ = 1.3 m below the surface. Ë ¯ 3 Total force = F1 + F2 + F3 = 15133 N = 15.133 kN Centre of pressure = yp (5353 ¥ 0.6) + (7137 ¥ 1.2) + ( 2643 ¥ 1.3) 15133 = 1.005 m By symmetry x = xp, i.e., the centre of pressure acts on the vertical passing through the CG of the side. = 2.27 Solution: Consider unit width of a side wall. The pressure distribution is as in Fig. 2.29. Oil r2 h2 = 1.2 m Water r1 h1 = 0.80 m 2.0 m F2 g h2 F11 F12 g h2 g h1 Fig. 2.29 Consider a vertical strip of unit width in one of the sides. Depth of water in the tank = h1 = 0.8 m Speciﬁc weight of water = g 1 = 9.79 kN Depth of oil in the tank = h2 = 1.2 m Speciﬁc weight of oil = g 2 = 0.85 ¥ 9.79 = 8.3215 kN F2 = Pressure force of oil on the element 1 1 ¥ 8.3215 ¥ (1.2)2 = 5.991 kN = g h 22 = 2 2 2 = 0.8 m from This force acts at a depth of 1.2 ¥ 3 the free surface of the oil. Water pressure distribution is trapezoidal in shape with g h 2 on the top and (g h2 + g h1) on the bottom as shown in the ﬁgure. The force due to water pressure can be considered to be in two parts F11 and F12 as below: F11 = Part pressure force due to water on the element = g h2 h1 = 8.3215 ¥ 1.2 ¥ 0.8 = 7.989 kN 51 Fluid Statics 0.8 ˆ Ê acting at a depth of Á1.2 + ˜ = 1.60 m Ë 2 ¯ from the free oil surface. F12 = Part pressure force due to water on the 1 1 g h 21 = ¥ 9.79 ¥ (0.8)2 element = 2 2 0.8 ˆ Ê = 3.133 kN acting at Á 2.0 ˜ = 1.733 m Ë 3 ¯ from the free surface. The various forces on the element are summarized below: Magnitude Lever arm (kN) (m) Force Description F2 F11 F12 2 (1/2) ¥ (8.3215) ¥ (1.2) 5.991 (8.3215 ¥ 1.2 ¥ 0.8) 7.989 (1/2) ¥ (9.79) ¥ (0.8)2 3.1328 Total Force on the element 17.113 0.80 1.6 1.733 3.5 m Gate Pivot 1.6 m y Moment Fig. 2.30 (kN.m) yp = depth of centre of pressure below the water I surface = h + GG Ah 4.793 12.782 5.429 23.004 Total Force on the side = 17.113 ¥ 2.0 = 34.226 kN The vertical location of the centre of pressure is found by taking moments of all pressure forces about the water surface and dividing it by the sum of all forces. Hence the depth of centre of pressure below free surface of oil = 23.004/17.113 = 1.3443 m. By symmetry, the centre of pressure acts on the vertical centre line of the plane. Considering the width of 2 m of the side of the tank, total force due to pressure of the ﬂuids on the side = FTotal = (2 ¥ 17.113) = 34.226 kN. Water h = depth of CG of the plate from the water surface = (3.5 – 0.8) = 2.7 m (1.0) ¥ (1.6)3 = 0.3413 m4 IGG = 12 Depth of centre of pressure = (0.3413) = 2.779 m yp = 2.7 + (1 ¥ 1.6) ¥ ( 2.7) y = 3.50 – 2.779 = 0.721 m. 2.29 board AB pivoted at C of C above B A 2.28 A Pivot y Solution: The gate will open just when the centre of pressure coincides with the location of the pivot. By symmetry, the centre of pressure lies on the vertical sxis of symmetry. Consider unit width of the gate. C 4.5 m B Fig. 2.31 Example 2.29 52 Fluid Mechanics and Hydraulic Machines Solution: For critical stability, the centre of pressure must be at C. Thus h = height of C above B. 4.5 = 1.5 m = 3 2.31 2.30 h Solution: From the Fig. 2.32 h1 = depth of M below free surface = 2.0 m h2 = depth of N below free surface = 2.0 + 1.5 sin 45° = 3.06 m. hCG = 2.0 + 0.75 sin 45° = 2.53 m y = 2.53/sin 45° = 3.578 m The force on the gate MN can be considered to be made up of two parts [see Fig. 2.24(b)]. B B Solution: Consider unit width of the ﬂash board 4 Total force F = 9.790 ¥ ¥ ( 4 ¥ 1) 2 = 78.32 kN/m acting at 4/3 m above B. Let RB and RC be the reactions at B and C respectively. Taking moments of forces about C, F1 = force due to uniform pressure pa acting uniformly over the surface. = pa A = (10 ¥ 1.5 ¥ 0.6) = 9 kN 4ˆ Ê 78.32 ¥ Á1.5 - ˜ - RB ¥ 1.5 = 0 Ë 3¯ RB = 8.7 kN/m RC = F1 – RB = 78.32 – 8.70 = 69.62 kN/m acting at CG of the area hcg = 2.53 m i.e y = 2.53/sin 45° = 3.578 m along OM axis. F2 = Force due to hydrostatic pressure of liquid = g Ah = 9.79 ¥ (1.5 ¥ 0.6) ¥ 2.53 = 22.29 kN 1 gh 1 F m h1 2.0 m (b) 2. 0 45° m M Actual water surface 2.0 m 5 1. h2 1.025 m O gh 2 F 2 p a pA = 10 kN O Imaginary water surface O m Water 0. 6 m 0. 6 M N Water 45° (c) (a) Fig. 2.32 Example 2.31 N 53 Fluid Statics This force F2 acts at yc1, where I gg yc1 = y + Ay 0.6 ¥ (1.5)3 / 12 = 3.578 + (0.6 ¥ 1.5) (3.578) = 3.6308 m hc = vertical depth of this yc1 = 3.6308 ¥ sin 45° = 2.5674 m Total force F = F1 + F2 = 9 + 22.29 = 31.29 kN This force acts at yc. Taking moments about CG F ¥ (yc – y ) = F2 ¥ (yc1 – y ) 22.29 ¥ (3.6308 – 3.578) yc – y = 31.29 = 0.0376 m yc = 3.578 + 0.0376 = 3.616 m Depth of centre of pressure from the water surface = h p = 3.616 sin 45° = 2.557 m Alternative method: Consider the equivalent column of water corresponding to a pressure of 10 kN 10.0 = 1.0215 m (See Fig. 2.32(c)) he = 9.79 Now an imaginary water surface at a height he above the actual water surface be considered as reference surface. Hence force on the gate F = g Ah 15 Ê ˆ = 9.79 ¥ (0.6 ¥ 1.5) ¥ Á 3.0215 + sin 45∞˜ Ë ¯ 2 = 31.29 kN y p¢ = location of centre of pressure on the O¢M axis I = y ¢ + GG A y¢ Ê 3.0215 ˆ + 0.75˜ = 5.023 y¢ = Á Ë sin 45∞ ¯ I GG (0.6 ¥ (1.5 )3) / 12 = 0.0373 = A y¢ (0.6 ¥ 1.5 ) (5.023 ) y p¢ = 5.023 + 0.0373 = 5.060 m yp = distance of centre of pressure from actual water surface = 5.060 – (1.0215/sin 45°) = 3.616 m Depth of centre of pressure from actual water surface = 3.616 sin 45° = 2.557 m 2.32 H H Oil RD = 0.8 Oil pressure Hinge Air 1.5 m 30 kPa Gate: 0.6 m wide (a) Air pressure F2 F1 Gate (b) Fig. 2.33 Solution: Force due to oil F1 = g 0 Ah goil = 0.8 ¥ 9.79 = 7.832 kN/m3 A = (0.6 ¥ 1.5) = 0.9 m2 Ê 1. 5 ˆ h = (H – 1.5) + Á Ë 2 ˜¯ F1 = 7.832 ¥ 0.9 ¥ {(H – 1.5) + 0.75} = 7.0488 H – 5.2866 Centre of pressure of force F1 I gg hc1 = h + Ah 0.6 ¥ (1. 5 )3 / 12 = (H – 0.75) + (0.6 ¥ 1. 5 ) ( H - 0.75 ) 0.1875 ( H - 0.75 ) (2) Force due to air pressure F2 = pA = 30 ¥ (0.6 ¥ 1.5) = 27 kN = (H – 0.75) + 54 Fluid Mechanics and Hydraulic Machines Centre of pressure of this force (below the oil surface) hc2 = h = (H – 0.75) Taking the moments about the hinge: 5.0 = 10.0 m, y = 10.5093 m sin 30∞ h = depth of CG below water surface = 5.0 + OG sin 30° = 5.0 + 0.5093 sin 30° = 5.2546 m Force on the gate (normal to the plane of the gate) Since F1 ¥ [hc1 – (H – 1.5)] = F2 ¥ [hc2 – (H – 1.5)] ¥ {7.0488 H – 5.2866} Ï 0.1875 ¸ ¥ Ì( H - 0.75) - ( H - 1. 5) + ˝ ( H - 0.75) ˛ Ó = F = g Ah = 9.79 2.33 ( p ¥ 1. 2 2 ) ¥ 5.2546 2 = 116.36 kN Location of centre of pressure on OB I I (OG ) 2 yp = y + gg = y + 0102 Ay Ay y = 27 ¥ {(H – 0.75) – (H – 1.5)]} Ï 0.1875 ¸ {7.0488 H – 5.2866} Ì0.75 + ˝ ( H - 0.75) ˛ Ó = 27 ¥ 0.75 On solving by trial and error H = 4.33 m y0 = Ê pD 4 ˆ 1 1 (0.5093) 2 ¥ yp = 10.5093 + Á ˜ 10.5093 Ë 128 ¯ Ê pD 2 ˆ 10.5093 Á 8 ˜ ¯ Ë Solution: In the Fig. 2.34, O1 O2 B is the semicircular gate. 4R 4 ¥ 1. 2 = = 0.5093 m 3p 3p O 2 y – y0 = OG = Substituting D = 2.5 m, yp = 10.5218 m OP = yp – yo = 0.5218 m Let Fm = minimum force required to lift the gate. By taking moments about O, = W ¥ OG cos 30° = F ¥ OP – Fm ¥ OB 12 ¥ 0.5093 ¥ 0.866 = 116.36 ¥ 0.5218 – 1.2 Fm 5.293 = 60.717 – 1.2 Fm Fm = 46.19 kN m G O 1.2 2.34 1 B Hinge Fm O y O G Solution: Consider 1 m length of piling F1 = force due to salt water = g 1 A1 h1 P B 30° F 5.0 m W Fig. 2.34 hc1 Ê 3. 5 ˆ = (1.035 ¥ 9.79) (3.5 ¥ 1.0) Á Ë 2 ˜¯ = 62.062 kN = Centre of pressure of this force F1 I gg = h1 + A1 h1 55 Fluid Statics Sheet pilling N Salt water RD = 1.035 3.5 m 2.35 Fresh water 2.5 m M (a) 0.6 m 0.9 m h1 F1 F2 M g h1 h2 A A 0.6 m g h2 B (b) 1.2 m B Fig. 2.35 Fig. 2.36 5 1. 0 ¥ (3 . 5)3 / 12 + = 2 Ê 3. 5 ˆ (3 . 5 ¥ 1. 0) Á Ë 2 ˜¯ = 1.75 + 0.5833 = 2.333 m. Lever arm a1 = 3.5 – 2.333 = 1.167 m F2 = Force due to fresh water = g 2 A2 h 2 Ê 2.5ˆ = (9.79) (2.5 ¥ 1) Á Ë 2 ˜¯ = 30.594 kN I gg hc2 = h 2 + A2 h2 2.5 [1. 0 ¥ ( 2 . 5)3 ] / 12 + = 2 ( 2 . 5 ¥ 1. 0 ) ( 2 . 5 / 2) = 1.667 m. Lever arm a2 = 2.5 – 1.667 = 0.833 m Net moment about M = M1 = F1a1 – F2 a2 (Taking clockwise as positive) M = (62.062 ¥ 1.167) – (30.594 ¥ 0.833) = 72.43 – 25.48 = 46.95 kN.m (clockwise) Solution: Force on the bottom due to water F1 = g Ah = 9.79 ¥ (1.2 ¥ 2.0) ¥ 1.5 = 35.24 kN Force on the surface A A = 9.79 ¥ (0.6 ¥ 2.0) ¥ 0.9 = 10.57 kN When suspended from top the stress on the side walls s = 35.24 (1.2 + 1.2 + 2.0 + 2.0) ¥ 8 / 1000 = 688.3 kPa When supported from bottom the stress on the side walls 10.57 s = (1.2 + 1.2 + 2.0 + 2.0) ¥ 8 / 1000 = 206.4 kPa 2.36 56 Fluid Mechanics and Hydraulic Machines G X g h = 1.0536 m a a 0.0108 (0.36) (1.0243) = 1.0243 + 0.6 m G X G g C. Forces on Curved Surfaces h a = 0.6 m 2.37 b (b) (a) Fig. 2.37 Example 2.36 Solution: Area A = a 2 = 0.36 m2 a 0.6 = 0.6 + h = 0.6 + 2 2 = 1.0243 m Consider two triangular laminas of base b and height h, as shown in Fig. 2.37(b). Considering the axis XX Ixx Solution: Referring to Fig. 2.38 and considering 1 m width, FH = horizontal force = g (Projected area of the cylinder MSN on a vertical M¢N¢ ) ¥ h Ê 2.5ˆ = 30.59 kN = 9.79 ¥ (2.5 ¥ 1) ◊ Á Ë 2 ˜¯ Fv2 ÏÔ 1 Ê bh ˆ h2 ¸Ô = Ì bh3 + Á ˜ ˝¥2 Ë 2 ¯ 9 Ô˛ ÓÔ 36 = 1 bh3 bh3 ¥ 2 = 12 6 N¢ S FH N T Fv1 O 2.5 m For the present case Ixx = Igg b = 2a/ 2 Igg = = and h = a / 2 1 2a a3 1 4 ◊ = a 6 12 2 2 2 1 (0.6) 4 = 0.0108 12 Force on one side of the lamina F = g Ah = 9.79 ¥ (0.36) (1.0243) = 3.61 kN Centre of pressure: By symmetry xp = 0 I gg hp = yp = h + Ah M¢ M Fig. 2.38 Vertical force FV = FV1 on MS – FV2 on SN = g (volume NMST – volume NST) 2 ÔÏÊ 1 p ¥ ( 2 . 5) ˆ Ê 2 . 5 2 . 5 ˆ Ô¸ = g ÌÁ ¥ ˜ + ÁË 2 ¥ 2 ˜¯ ˝ ¥ 1 4 4 ¯ ˛Ô ÓÔË ÏÔÊ 2 . 5 2 . 5 ˆ Ê p ¥ ( 2 . 5) 2 ˆ ¸Ô ¥ – g ÌÁ ˝ ¥1 2 ˜¯ ÁË 4 ¥ 4 ˜¯ Ô˛ ÔÓË 2 1 p = g ¥ ¥ ( 2.5) 2 ¥ 1 = 24.03 kN 2 4 Hence FH = 30.59 kN FV = 24.03 kN 57 Fluid Statics 2.38 S 2.39 T 1.2 m 2.0 m FH Water N 5 1. FR q FV N Fv2 1.5 m m M S R O 1.8 m wide X (a) 2.4 m yR M a O FH N (b) Fv1 Fig. 2.39 Solution: Horizontal component of the force FH = g (Projected area) ¥ h 1. 5 ˆ Ê = 9.79 ¥ (1.5 ¥ 1.8) ¥ Á 2.0 + Ë 2 ˜¯ = 72.69 kN Vertical component of the force FV = g (volume TNMS) È = 9.79 Í(1.5 ¥ 1.8) ¥ ( 2.0 + 1.5) Î 1 ˘ - ¥ p ¥ (1.5) 2 ¥ 1.8 ˙ 4 ˚ = 61.37 kN Resultant force FR = = FH2 + FV2 (72.69) 2 + (61.37) 2 = 95.13 kN If q is the inclination of FR to vertical F 95.13 = 1.55 tan q = H = FV 61.37 q = 57.173° The resultant passes through O. yR = vertical height of resultant on the gate. = R cos q = 1.5 cos (57.173°) = 0.813 m 3.6 m Q R 1.2 m M Fv x Fig. 2.40 Solution: Vertical component: FV = volume contained in (OMN + ONRS) = FV1 + FV2 Èp ˘ FV = Í ¥ (1.2) 2 ¥ 3.0 ¥ 9.79˙ + (2.4 ¥ 1.2 4 Î ˚ ¥ 3.0 ¥ 9.79) = 33.22 + 84.59 = 117.81 kN Ê 4 ˆ ¥ 1.2˜ = 0.5093 m from OM FV1 acts at Á Ë 3p ¯ FV2 acts at 0.6 m from OM. By taking moments of vertical forces about O, FV ◊ x = (33.22 ¥ 0.5093) + (84.59) ¥ 0.6 67.67 = 0.574 m x= 117.81 Horizontal component FH = Force on the projection of curved face MN on a vertical plane 58 Fluid Mechanics and Hydraulic Machines = g Ah = 9.79 ¥ (1.2 ¥ 3) (2.4 + 0.6) = 105.73 kN I gg Depth of centre of pressure of FH = hc = h + Ah 1 ¥ [3 ¥ (1.2)3 ] = 3.0 + 12 (3 ¥ 1.2) ¥ 3 = 3.04 m Resultant force on the curved face MN = Fv2 3.0 m Water M FV2 + FH2 O S Fv1 (117.81) 2 + (105.73) 2 = 158.3 kN If a is the inclination of R to the horizontal F 117.81 = 1.114 tan a = V = FH 105.73 a = 48° 5¢ 36≤ The resultant passes through the centre of curvature O. Q 1.2 m R= T N Fig. 2.41 Example 2.40 2.41 2.40 Solution: Referring to Fig. 2.42, by symmetry the net horizontal force = 0 R Solution: S Referring to Fig. 2.41 1 4 ¥ ¥ p (1.2)3 = 35.44 kN 2 3 Horizontal force FH = g Ah where A = projected area of the hemisphere = pR2 FH = 9.79 ¥ p ¥ (1.2)2 ¥ 3.0 = 132.87 kN = 9.79 ¥ 3.0 m P R Vertical force FV = FV1 – FV2 = weight of volume of water NSTQ – weight of volume of water MSTQ. = weight of water contained by the hemisphere MSN 1 Ê4 ˆ = g ◊ ◊ Á p R3 ˜ ¯ 2 Ë3 M O N 1.2 m dia Fig. 2.42 Vertical force FV = weight of ﬂuid above the hemisphere MPN 1 4 È ˘ = g ÍpR 2 H - ◊ pR3 ˙ 2 3 Î ˚ 59 Fluid Statics If R is inclined at an angle a to the horizontal F 10.99 = 0.14032 tan a = V = FH 78.32 a = 7.988° 2 È ˘ = 9.79 Íp ¥ (0.6) 2 ¥ 3 . 0 - p (0.6)3 ˙ 3 Î ˚ = 28.79 kN Resultant force is the same as the vertical force FV = 28.79 kN acting vertically at the centre of the hemisphere. Since the surface of the gate is cylindrical, the resultant R passes through the centre of curvature O. 2.42 2.43 C1 N 5. 0 4.0 m R m Oil RD = 0.9 Ra q q S d 2.50 m C FH1 C2 Liquid M: RD = 0.80 FH2 A Fig. 2.44 Fig. 2.43 Solution: Consider a gate width of 1 m sin q = 2/5, q = 23.578°, 2q = 47.156° FH = horizontal component of force on the gate = g (projected area) ¥ h 4 = 78.32 kN = 9.79 ¥ (4 ¥ 1) ¥ 2 FV = Vertical component of the force on the gate = weight of water displaced by the gate = g [sector OMSN – triangle OMN] 1 È 47.156 ˘ ¥ p ¥ 52 - ¥ 4 ¥ 5 cos 23.578∞˙ 360 2 Î ˚ =9.79 Í = 10.99 kN Resultant force on the gate = R = Fv2 R a D O a Fv1 O M R= B FH2 + F V2 (78.32) 2 + (10.99) 2 = 79.08 kN Solution: Consider 1 m length of gate Force due to oil: Horizontal force FH1 = go (Projected area of ACB) ¥ h Ê 2.5ˆ = 0.9 ¥ 9.79 ¥ (2.5 ¥ 1) ¥ Á Ë 2 ˜¯ = 27.53 kN 2.5 This force acts at a height of = 0.833 m 3 above level of A. Vertical force FV1 = [weight of volume of oil ACC1B – weight of volume of oil BCC1] = weight of volume ACB 1 pD 2 = go ¥ ¥ ¥ 1.0 2 4 1 p = 0.9 ¥ 9.79 ¥ ¥ ¥ ( 2 . 5) 2 ¥ 1.0 2 4 = 21.63 kN 60 Fluid Mechanics and Hydraulic Machines 1˘ Èp = 0.8 ¥ 9.79 ¥ Í ¥ ( 2.5) 2 ¥ ˙ ¥ 1.0 4˚ Î4 = 9.61 kN 4 Ê Dˆ This force acts at a distance of 3 p ÁË 2 ˜¯ 4 ¥ 2.5 = = 0.531 m from the vertical at A. 3p ¥ 2 Resultant force: R H = horizontal component = FH1 – FH2 = 27.53 – 6.12 = 21.41 kN RV = vertical component = FV1 + FV2 = 21.63 + 9.61 = 31.24 kN Resultant force F = RH2 + RV2 2 = ( 21.41) + (31.24) = 37.87 kN 2 Let a = inclination of the resultant to the horizontal. Then R 31.24 = 1.459 tan a = V = RH 21.41 a = 55° 34¢ 32≤ The resultant acts normal to the cylindrical surface and passes through the centre of curvature O of the cylindrical surface. ¥ 2.44 ¥ ¥ Solution: First convert the base pressure of 60 kPa into equivalent column of oil (ho) 60 = go ¥ ho ho = 60 = 6.81 m (9.79 ¥ 0.9) An imaginary free surface of oil at an elevation of 6.81 m above the base AB can be imagined [see Fig. 2.45 (b)]. F G 1.6 m Imaginary water surface 6.81 m Oil RD = 0.90 2.4 m Fv D 8m Vertical force FV2 = go ¥ [volume AOD] C 0. 4( D / 2 ) This force acts at a distance of 3p 4 ¥ 2.5 / 2) = 3p = 0.531 m from vertical at A. Force due to liquid M: Horizontal force FH2 (Right to left) = g L (projected area) ¥ h 1. 25 = 0.80 ¥ 9.79 ¥ (1.25 ¥ 1) ¥ 2 = 6.12 kN 1. 25 = 0.417 m above level of This force acts at 3 A. O E 60 kPa A B 1.6 m (b) (a) Fig. 2.45 By symmetry, horizontal force on the cylindrical cover is zero. The vertical force FV = go [volume of prism COEGF – curved volume CDE] = (0.9 ¥ 9.79) È Ê p (1.6) 2 ˆ˘ ¥ 3˜ ˙ ¥ Í(1.6 ¥ 3.0) (6.81 - 2.4) - Á ¥ 4 Ë2 ¯ ˙˚ ÍÎ = 8.811[21.168 – 3.016] = 159.9 kN This force acts vertically at the centre of the tank along OD. 61 Fluid Statics = 8.126 (37.131 – 7.069) = 244.28 kN This force is shared by six bolts. 244.28 Tensile force on each bolt = kN 6 = 40.71 kN 2.45 Solution: First consider equivalent of pressure PA in terms of oil column. PA = go hA 50 = (0.83 ¥ 9.79) ◊ hA hA = 6.153 m of oil The imaginary oil surface at an elevation of hA = 6.153 m is now considered. [Fig. 2.46(b)]. Above the base plane MN of the dome, the elevation of the imaginary oil surface is = 6.153 – 0.9 = 5.253 m By symmetry, there is no horizontal force on the dome. Imaginary oil surface N¢ 5.253 m M¢ Hemisphere D 6.153 m Oil RD = 0.83 O 0.9 m 1.5 m A 0.6 m 50 kPa M O N 2.46 3 Solution: Refer to Fig. 2.47 consider 1 m length of the weir. The force acting on the weir are: (i) weight of masonry W = W1 + W2 + W3 (ii) vertical force due to water on the upstream slope, FV1. (iii) vertical force due to tailwater on the downstream slope, FV2. Fv1 2.0 m 0.6 m A C 3 m Dia (a) (b) Fig. 2.46 Example 2.45 3ˆ ˘ ÈÊ p (3.0) 2 ˆ Ê 1 4 Ê 3.0 ˆ = (0.83 ¥ 9.79) ÍÁ ¥ 5.253˜ - Á ¥ p Á ˜˙ Ë ¯˜ 4 ¯ Ë2 3 2 ¯ ˙˚ Fv2 W2 5.0 m The vertical force FV = weight of oil above the dome surface up to the imaginary oil surface = weight of volume MDN N¢ M¢. ÍË Î D FH1 W1 W3 FH2 2.0 m A B 0.5 m 3.75 m 2.0 m 6.25 m Fig. 2.47 Example 2.46 62 Fluid Mechanics and Hydraulic Machines Table 2.1 Magnitude of the Forces and Moments and of the Lever Arm about B-Example 2.36 Force Description (0.5 ¥ 5) ¥ W2 (2.0 ¥ 5.0) ¥ 1.0 ¥ 22 1 (3.75 ¥ 5) ¥ ¥ 1 ¥ 22 2 1 (0.5 ¥ 5) ¥ ¥ 1 ¥ 9.79 2 1 (1.5 ¥ 2.0) ¥ ¥ 1 ¥ 9.79 2 5 (5 ¥ 1) ¥ ¥ 9.79 2 2 (2 ¥ 1) ¥ ¥ 9.79 2 FV1 FV2 FH1 FH2 Vert. force (kN) 1 ¥ 1 ¥ 22 2 W1 W3 Horiz. force (kN) Sum 27.5 5.917 162.7 220.0 4.75 1045.0 206.25 2.50 515.6 12.24 6.083 74.5 14.69 0.50 7.3 122.38 1.667 – 19.58 0.667 102.8 (iv) horizontal water force on the upstream side, FH1. (v) horizontal water force on the downstream side, FH2. = 491.6 kN 204.0 s 1, 2 = = 1818.2 SV b 6eˆ Ê ÁË1 ± b ˜¯ 480.7 Ê 6 ¥ 0.233 ˆ 1± 6.25 ÁË 6.25 ˜¯ s1 = maximum stress = 94.1 kPa s2 = minimum stress = 59.7 kPa SV 2 + S H 2 ( 480.7) 2 + (102.8) 2 13.1 As b = base width = 6.25 m, the Eccentricity 6.25 b e=x– = 3.358 – = 0.233 m 2 2 Maximum and minimum stresses S V = Sum of vertical forces = 480.7 kN S H = Sum of horizontal forces = 102.8 kN = 204.0 480.7 The magnitudes of these forces, their distance from the toe of the weir (edge B) and the moments of these forces about B are tabulated as in Table 2.1. R = Resultant = Lever arm Moment Moment about B (Clockwise) (Anticlockwise) (m) (kN.m) (kN.m) 2.47 If q is the inclination of the resultant to horizontal SV 480.7 = = 4.676, and q = 77.93° SH 102.8 S M = 1818.2 – 204 = 1614.2 kN.m tan q = x = distance of point of action of the resultant from B SM 1614.2 = = 3.358 = SV 480.7 Solution: Hoop stress (circumferential tensile stress) in a cylinder s = pD 2t 63 Fluid Statics If s = fa = allowable stress in the material. Minimum thickness t required is pD t= 2 fa 1000 ¥ 2.5 = 0.01042 m = 2 ¥ (120 ¥ 1000) = 10.42 mm A thickness of 10.5 mm can therefore be used. [Note: The relative density of the ﬂuid does not have any role when once the pressure inside the pipe is stipulated.] Upstream 120° CL RP F 30° (a) 3.0 m B P q O Solution: Let F = resultant force due to water on each gate acting at the centre of each gate, point O R = resultant reaction at the hinges P = reaction between two gates at the common contact area, acting perpendicular to contact area, in this case normal to the centre line The reaction R must pass through C (see Fig. 2.48), the point of intersection of F and P, due to equilibrium of the system. In triangle ACB, because CO is the perpendicular bisector of AB, –CBA = –CAB = q \ R =P F Also P =R= 2sin q [Note: It is assumed that F, P and R are in the same plane.] Bottom hinge 3.0 m C F 0.8 m q A 2.48 The gates of a 8 m wide lock include an angle of 120° in the closed position. Each gate is held on by two hinges one placed at 0.8 m and another at 6.0 m from the bottom of the lock. If the water levels are 9.0 m and 3.0 m on the upstream and downstream respectively, determine (i) the resultant force due to water pressure and (ii) magnitudes of reaction at the hinges. Top hinge 9.0 m R Downstream 8.0 m (b) Fig. 2.48 (c) Lock Gates, Example 2.48 B = width of each gate = AB 4 = 4.62 m = cos 30∞ On the upstream: 9 F1 = 9.79 ¥ (4.62 ¥ 9) ¥ = 1831.8 kN 2 H acts at 1 = 3.0 m from bottom of the gate on the 3 downstream. 3 = 203.5 kN F2 = 9.79(4.62 ¥ 3) ¥ 2 acts at 1.0 m from the bottom. (i) Net water force F = F1 – F2 = 1831.8 – 203.5 = 1628.3 kN acts at h from base where F ¥ h = F1h1 – F2h2 1628.3 ¥ h = (1831.8 ¥ 3) – (203.5 ¥ 1) h = 3.25 m from the bottom F 1628.3 = = 1628.3 kN (ii) P = R = 2 sin 30∞ 2 sin 30∞ Let R T = reaction at the top hinge. RB = reaction at the bottom hinge. RT + RB = 1628.3 64 Fluid Mechanics and Hydraulic Machines Also by taking moments about the bottom hinge RT (6.0 – 0.8) 5.2 RT RT RB = R(3.25 – 0.8) = 1628.3 ¥ 2.45 = 767.2 kN = (R – RT) = 861.1 kN V1 = V2 + V3 Since the block is in equilibrium, Consider the weight of components of the block and hence the two displaced liquids to obtain r1V1g = r2V2 g + r3V3 g D. Buoyancy 2.49 Solution: Let V2 = volume of the block in upper liquid (Liquid-2.) Thus r1V1 = r2 V2 + r3V3 R r1 or rw V3 V1 r1 > rw Solution: Let V be the volume of the sphere. 4p 3 R r1g Weight of sphere = W = 3 4p 3 R r w g. Buoyant force = Fb = 3 = Since r1 > rw, W > Rb. The tension in the string, T, compensates the difference between the weight and the buoyant force and hence W = T + Fb Thus r1V1 - r3V3 r2 V3 V3 = = ( r r V2 + V3 1V1 3V3 ) + V3 r2 r2V3 = Ê r - r2 ˆ r1V1 - Á 3 Ë r2 ˜¯ V2 = 1 È r1 V1 ˘ È ( r3 - r2 ) ˘ Ír V ˙ - Í r ˙ 2 Î 2 3˚ Î ˚ r1 Ê r3 - r2 ˆ V3 =1 r2 ÁË r2 ˜¯ V1 Ê r1 ˆ ÁË r - 1˜¯ V3 2 = On simpliﬁng, = V1 Ê r3 ˆ ÁË r - 1˜¯ 2 T= 2.50 r1 r2 and r2 V2 is 2.51 section of 300 mm ¥ V1 V3 (r1 - r2) = V1 (r3 - r2) ¥ r2 Liquid-2 Block r1 r3 Fig. 2.49 Liquid-3 Solution: In this case buoyant force = weight of the ﬂoating body. Let A = cross-section of the ﬂoating body and L = Length of the lead prism. 65 Fluid Statics Hence gA[L + 2.0] = g A[(0.7 ¥ 2.5) + (12 ¥ L)] L + 2.0 = (0.7 ¥ 2.5) + 12 L L = (0.25/11) = 0.2272 m = 227.2 mm 2.52 Solution: Refer to Fig. 2.50. The tension in the chain indicates that the buoyant force is larger than the weight of the object in the air. For the lower sphere: Buoyant force Fb = 1.767 ¥ 9.79 = 17.3 kN Tension T = W – Fb – 20.0 – 17.3 = 2.7 kN Upper sphere: Buoyant force = F b¢ = W + T = 4.0 + 2.7 = 6.7 kN If the sphere is completely submerged the buoyant force would have been = Fb = 1.767 ¥ 9.79 = 17.3 kN. Since only 6.7 kN of buoyant force is being exerted. Percentage of volume above water (17.3 - 6.7) = 0.617 = 61.7%. = 17.3 2.54 Fb W T Fig. 2.50 Hence or T = Fb – W W = Fb – T 3 4 Ê 1.5 ˆ = p Á ˜ ¥ 9.79 – 5.30 3 Ë 2 ¯ = 17.30 – 5.30 = 12.0 kN 2.53 Solution: Let weight in air = W and volume = V Weight in a liquid of RD = S; WS = W – g S V \ WS1 = W – g S 1 V WS2 = W – g S2 V g V (S2 – S1) = WS2 – WS1 W - WS1 V = S2 g ( S2 - S1) = 20 - 10 9.79 ¥ 103 ¥ (1.2 - 0.8) = 2.5536 ¥ 10–3 m3 = 2.5536 L Weight in air WS = WS1 + g S1 V = 20 + 9790 ¥ 0.8 ¥ 2.5536 ¥ 10–3 = 40 N 2.55 D W stem h S Solution: Volume of the two spheres 3 4 Ê 1.5 ˆ V = ¥ Á ˜ = 1.767 m3 3 Ë 2 ¯ Solution: Refer to Fig. 2.51. 66 Fluid Mechanics and Hydraulic Machines RD = 1.0 h RD = S (S < 1) (ii) h2 = distance between markings of RD of 1.0 and 1.05. 0.2 Ê 1 ˆ = ¥Á - 1˜ Ê p Ê 5 ˆ 2 ˆ Ë 1.05 ¯ 9790 Á ¥ Á ˜ ˜ Ë 4 Ë 1000 ¯ ¯ = – 0.04954 = – 49.54 mm h2 will be below the marking corresponding to relative density of 1.0. Fig. 2.51 2.57 sectional area 15 cm ¥ Let V = submerged volume when immersed in water (RD = 1.0). W=g V When immersed in a liquid with RD = S where S < 1.0 Solution: 10 cm Ê ˆ pD 2 W = ÁV + ¥ h˜ ¥ g ◊ S 4 Ë ¯ pD 2 \ gV = g V S+ hg ◊S 4 or y N x W Ê1 ˆ Ê1 ˆ h= ÁË S - 1˜¯ = ( D /4) ÁË S - 1˜¯ 2 ( pD /4) Original liquid surface N M V 8.0 m Note that when S > 1, h will be negative, i.e. the markings will be below the mark corresponding to RD = 1.0. Refer to Fig. 2.52 2.56 Solution: (i) h1 = distance between marking of RD of 1.0 and 0.95. 0.2 Ê 1 ˆ h1 = ÁË 0.95 - 1˜¯ 2 Êp Ê 5 ˆ ˆ 9790 Á ¥ Á ˜ ˜ Ë 4 Ë 1000 ¯ ¯ = 0.05476 m = 54.76 mm A B 15 cm Fig. 2.52 Let x = depth to which the bottom of the cube falls below original liquid surface (cm) y = height of the rise of liquid above the original liquid surface (cm) x + y = depth of submergence of the cube (cm) Volume M (10 ¥ 10 ◊ x) x W W = Volume N = (152 – 102) ◊ y = 1.25 y = weight of cube = 5 N = buoyant force 67 Fluid Statics = (10 ¥ 10) ◊ ( x + y ) 106 5 = 0.7832 (x + y) ¥ 0.8 ¥ 9790 5 0.7832 y = 2.8374 cm x = 3.5467 cm Elevation of: bottom of cube above plane AB = (8.00 – 3.547) = 4.453 cm liquid surface above plane AB = (8.00 + 2.837) = 10.837 cm 2.25 y = (p2 – pa) ¥ Area of can = weight of the can 6.0 = 191 Pa p2 – pa = Êp 2ˆ ÁË 4 ¥ (0.2) ˜¯ p2 = 191 + pa From isothermal relationship paVa = p2V2 = constant V2 = pa 100000 Va = ¥ (A ¥ 0.40) p2 (100000 + 191) È 100000 ˘ hs A = Va – V2 = (A ¥ 0.40) Í1 ˙ Î 100191 ˚ hs = 0.40 ¥ 1.9064 ¥ 10–3 = 7.626 ¥ 10– 4 m = 0.0763 cm L = 40 – 1.950 – 0.076 = 37.974 cm 2.58 The top of the can is L = 37.974 cm above the water surface in the tank. 2. The water surface in the can is 1.95 cm below the water surface in the tank. 2.59 20 cm Dia L p2 40 cm Hb Water F 9.1 m E = 10 m pa Fig. 2.53 Solution: Buoyant weight = weight of can p ¥ (0.2)2 ¥ Hb = 6 9790 ¥ 4 \ Hb = 0.0195 m = 1.95 cm Also if p2 = pressure inside the can (abs) and pa = atmospheric pressure (abs) pa Air hs pc Hb e p2 h (a) (b) Fig. 2.54 68 Fluid Mechanics and Hydraulic Machines Case 1: Let pa = atmospheric pressure pc = pressure inside the container (pc – pa) ¥ area of the container = weight of the container. But (pc – pa) = g Hb = 9790 ¥ 0.10 = 979 N/m2. Hence W = weight of the container = 979 ¥ 1.5 = 1468.5 N = 1.469 kN Mass of container = 1468 . 5 9 . 81 External force F = Fb – W = 6.825 – 1.469 = 5.356 kN 2.60 Solution: Referring to Fig. 2.55. Case 2: When the container is completely immersed to a depth E below the water surface. e = height of air column h = total height of the container From isothermal consideration paVa = p2V2 where p = pressure and V = volume. pa ◊ (Area) ¥ h = p2 ¥ (area) ¥ e e p = a \ h p2 Also from pressure consideration p2 = pa + g (E – h + e) Here pa = 100 kPa, h = 0.9 m E = 10 m and g = 9.79 kN 100 e = Hence p2 0.9 90 or p2 e = 90 or p2 = e 90 = p2 = 100 + 9.79(10 – 0.9 + e) e 90 = [100 + 89.09 + 9.79 e] e 9.79 e2 + 189.09 e – 90 = 0 Solving, e = 0.4648 m Buoyant force Fb = g e (Area) = 9.79 ¥ 0.4648 ¥ 1.5 = 6.8258 kN For equilibrium Fb = weight + external force =W+F G B 12 cm 15 cm = 149.7 kg O 20 cm O¢ 40 cm O¢¢ 20 cm Fig. 2.55 Weight of body = weight of displaced volume of water = (12 ¥ 20 ¥ 40) ¥ 9.79 ¥ 10–3 N (1 cm3 of water weighs 9.79 ¥ 10–3 N) = 93.98 N OB = height of centre of buoyancy above the base of the block 12 = 6 cm = 2 OG = height of centre of gravity of the block above O 69 Fluid Statics 15 = 7.5 cm 2 If M is the metacentre I BM = V I = moment of inertia of the water line area 40 ¥ ( 20)3 about O¢O≤ = 12 V = volume of ﬂuid displaced by the body = (20 ¥ 12 ¥ 40) = 40 ¥ ( 20)3 / 12 = 2.778 cm ( 40 ¥ 20 ¥ 12) MG = BM – BG = BM – (OG – OB) = 2.778 – (7.5 – 6.0) = 1.278 cm Since M is above G, the body is in stable equilibrium. \ BM = vertical axis of symmetry and is equal to half the draft. Hence AB = 1.0 m G = centre of gravity. From given data AG = 2.0 + 1.5 = 3.5 m M = meta centre È(15 ¥ (8)3 )/12˘˚ = 2.667 m MB = I / V = Î (15 ¥ 8 ¥ 2) AM = (1.0 + 2.667) = 3.667 m It is seen that AM is larger than AG, that is M is above G and hence the barge is stable. 2.62 section, of sides a and b L a b a 2.61 b y M G a B O Solution: Fig. 2.56 Consider the schematic layout shown in Fig. 2.57 M Length = 15.0 m 2.0 CG B A 8.0 Fig. 2.56 In the Fig. 2.55 A = a point on the bottom of the barge lying on the vertical axis of symmetry B = Centre of buoyancy. Since the submerged volume is a rectangular prism B lies on the OG = height of centre of gravity above the base. = a/2 Weight of the block = Sg (L ba) = buoyant force = g bLy where y = depth of immersion. y = Sa OB = height of centre of buoyancy above the base. y Sa = = 2 2 If M is the metacentre 1 ( L b3 ) I b2 12 BM = = = V 12 Sa (bL) ( Sa) 70 Fluid Mechanics and Hydraulic Machines Sa b2 + 2 12 Sa MG = metacentric height = OM – OG Sa b2 a = + 2 12 Sa 2 b2 a - (1 – S) = 12 Sa 2 For stability MG > 0 b2 a > (1 – S) Hence 12 Sa 2 b2 > 6 S(1 – S) a2 b > 6 S (1 - S ) or a \ where g = unit weight of water. OM = OB + BM = 2.63 pD 2 H g Sc 12 WL = weight of liquid displaced Wc = pD12 y p D2 3 g SL = y g SL 12 12 H 2 But weight of cone = weight of liquid displaced, i.e. = Wc = WL pD 2 H g Sc pD 2 ◊ y3 g SL = 2 12 12 H ÊS ˆ ÊS ˆ y3 = H 3 Á c ˜ or y = H Á c ˜ Ë SL ¯ Ë SL ¯ Let B be the centre of buoyancy. D and 3 3 ÊS ˆ OB = y = H Á c ˜ 4 4 Ë SL ¯ H SL Sc H2 2.65 O D D Fig. 2.59 Solution: Let g = Speciﬁc weight of water. Weight of cylinder = weight of water displaced. If A = Cross-sectional area of cylinder, A = pD2/4 y = Depth of immersion of the cylinder. Weight of cylinder = AH g S = weight of displaced volume of water = g yA Hence y = SH SH OB = height of the centre of buoyancy = 2 OG = height of the centre of gravity of the H cylinder = 2 If M is the metacentre, I pD 4 / 64 1 D2 BM = = = ¥ V 16 SH pD 2 ¥ SH 4 SH 1 D2 OM = OB + BM = + ¥ 2 16 SH For stable equilibrium with the axis vertical OM > OG H G B y O Fig. 2.60 Solution: Let O be point of intersection of the vertical axis and the bottom place surface of the cylinder when ﬂoating in water with a depth of immersion of y. If S is the speciﬁc gravity of the cylinder material, p Weight of the cylinder = (D2) H(Sg w) 4 p 2 ((0.3) ) ¥ 0.15 ¥ 0.9 ¥ 9790 = 93.42 N 4 This is equal to the buoyant force p Fb = 1.03 ¥ 9790 ¥ (0.3)2 ¥ y = 712.8y 4 93.42 = 0.131 m = 13.10 cm Hence y = 712.8 72 Fluid Mechanics and Hydraulic Machines The centre of buoyancy above A A as datum = OB = 0.1310/2 = 0.0655 m Height of centre of gravity G above the bottom plane = OG = 0.15/2 = 0.075 m Height BG = 0.075 – 0.0655 = 0.0095 m I – (BG) Metacentric height GM = V Ê pD 2 ˆ 2 ÁË ˜ I 64 ¯ = D = V 16 y Ê pD 2 ˆ y˜ ÁË 4 ¯ = (0.3) 2 = 0.0429 16 ¥ 0.131 0.6 m 0.4 m H = 1.0 m G Distance GM = 0.0429 – 0.0095 = 0.0334 m Thus the metacentric height is positive, i.e., the metacenter is above the center of gravity. As such, the cylinder will be ﬂoating in stable equilibrium. B A Water surface M 2.66 h A Fig. 2.61 Metacentric height I – (BG) V p I = (D 14 – D24) = 0.0051 4 GM = Solution: Let AA be the bottom plane surface of the cylinder when ﬂoating in water with a depth of immersion of h. Volume displaced by the hollow cylinder = weight of the cylinder p = (D12 – D 22) hgrw = 700 4 p = ((0.6)2 – (0.4)2) ¥ h ¥ 1.0 ¥ 9.81 ¥ 998 4 = 700 Hence h = 0.455 m. The centre of buoyancy above AA as datum = AB = 0.455/2 = 0.2275 m Height of centre of gravity G above place A A = AG = 1.0/2 = 0.50 m Height BG = 0.50 – 0.2275 = 0.2725 m V = volume of the cylinder = (D 12 – D 22) h = p 4 700 700 = rw g 998 ¥ 9.81 = 0.0715 m3 0.0051 – (0.272) Distance GM = 0.0715 = – 0.20 m Thus the metacentric height is negative, i.e., the metacenter is below the centre of gravity. As such, the cylinder will be in unstable equilibrium. Hence the hollow cylinder will not be able to ﬂoat in water with its axis vertical. 73 Fluid Statics 3 H = 0.75 ¥ 20 = 15.000 4 d = diameter at water surface = 2y tan q = 22.280 cm I pd 4 / 64 = BM = V 1 Ê pd 2 ˆ ◊y 3 ÁË 4 ˜¯ 2.67 OG = D d A M G B = B H = 5.014 cm OM = OB + BM = 13.925 + 5.014 = 18.939 cm OG = 15.000 MG = (18.939 – 15.000) = 3.939 cm (i.e. M is above G by 3.939 cm) Hence the cone is under stable equilibrium. y q Fig. 2.62 Solution: For the cone E. Rigid Body Motion D = 24 cm H = 20 cm S = 0.8 Let q = Semi-vertex angle 12 = 0.6, q = 30.96° tan q = 20 Diameter of cone at water surface d = 2y tan q 1 pD 2 weight of cone = ¥ ¥ H ¥g S 3 4 = weight of water displaced 1 pd 2 yg = 3 4 \ D2HS = d2y Ê Dˆ 4 y3 Á ˜ Ë 24 ¯ 2 Ê D ˆ = D2HS = 4y3 tan2 q = 4 y 3 Á Ë 2 H ˜¯ 2.68 2 q , ax O x 0.9 m N M 5.0 m 2 y3 = H3S or y = H(S)1/3 = HS1/3 y = 20 ¥ (0.8)1/3 = 18.566 cm If B is the centre of buoyancy OB = 3 Ê d2 ˆ 3 = y tan2 q Á ˜ 16 Ë y ¯ 4 3 y = 0.75 ¥ 18.566 = 13.925 cm 4 Fig. 2.63 Solution: The new oil surface will be inclined at an angle q given by a 3.0 tan q = x = g 9.81 = 0.3058 74 Fluid Mechanics and Hydraulic Machines q = 17° (a) The depth of oil at the front edge M is 5 ¥ tan q = 0.135 m hM = 0.9 – 2 The depth of oil at the rear edge N is 5 tan q = 1.665 m hN = 0.9 + 2 (b) Pressure at M = pM = gs hM = gwater ¥ S ¥ hM = 9.79 ¥ 0.9 ¥ 0.135 = 1.189 kN/m2 Pressure at N = pN = gs hN = 9.79 ¥ 0.9 ¥ 1.665 = 14.67 kN/m2 2.69 2 Solution: Refer to Fig. 2.64(a). Inclination of the oil surface If there were no spill, the oil surface would swing about an axis at O. Piezometric head at 5 ¥ tan q = 2.0468 m N = h N¢ = 0.9 + 2 Since this is larger than the depth of the tank there will be a spill of the oil. The new oil surface will have a depth hN = depth of tank = 2.0 at N and a slope of q. X intercept of the surface at the bottom 2.0 2.0 = = 4.36 tan q 0.4587 AB is the new oil surface [Fig. 2.64(b)]. Volume of oil = DANB ¥ width of tank 1 ¥ 2.0 ¥ 4.36 ¥ 3.0 = 13.08 m3 = 2 Original volume of oil = 0.9 ¥ 5.0 ¥ 3.0 = 13.50 m3 Spill of oil = 13.50 – 13.08 = 0.42 m3 2.70 a 4.5 = 0.4587 tan q = x = g 9.81 A¢ A q 2.0 m Solution: The oil surface will be inclined at an angle q to the horizontal where a tan q = x g 10.0 = = 1.0194 9.81 In Fig. 2.65, ROS is the original water surface and MON is the new surface at an acceleration of ax. The surface tilts around O. ax O 0.9 m M N M¢ 5.0 m (a) A 2 4.5 m/s 2.0 m M B M O R S q 5.0 m r = 1.2 m RD = 0.8 (b) Fig. 2.64 N T Fig. 2.65 1.2 m N ax 75 Fluid Statics The maximum pressure acts on the boundary point where the depth (measured normal to the free surface) is maximum. In this case the maximum depth is OT = radius = 1.2 m Hence Ê pˆ ÁË g ˜¯ = 1.2 m pm = 9.79 ¥ 0.8 ¥ 1.2 = 9.4 kN 2.71 2 Solution: In Fig. 2.66, RS is the original water surface. After the acceleration of ax = 2.4 m/s2, the water surface slope is a 2.4 = 0.2446 tan q = x = g 9.81 q = 13.75° x P M A q 0.6 m R 1.2 m 1 1 xyB = x2 B tan q 2 2 1 2 7.2 = x ¥ 2 ¥ 0.2446 2 7.2 2 = 29.436 x = 0.2446 x = 5.4255 m and y = 1.327 m Hence CN = depth of water in the front = 1.80 – 1.327 = 0.473 m AM = 6.0 – 5.4255 = 0.5745 m AP = AM tan q = 0.5745 ¥ 0.2446 = 0.1405 m The pressure proﬁle on the top is represented by the triangle APM extending over the width. Pressure force on the top and V= Ê1 ˆ Ft = Á ¥ AP ¥ AM ¥ Breadth˜ g Ë2 ¯ 1 ¥ 0.1405 ¥ 0.5745 ¥ 2.0 ¥ 9.79 2 Ft = 0.790 kN AM = The force acts vertically upwards at 3 0.1915 m from A at the mid-width section. = D Air S Water y = x tan q [Note: In this case the free surface does not tilt at the mid length. As there is no spill that volume of water and air volume are conserved.] y N ax B C 2.72 6.0 m Fig. 2.66 As there is no spill of water, the air space will remain same as at start. Air space volume, V = 0.6 ¥ 6.0 ¥ 2 = 7.2 m3 Let MN be the new water surface at an inclination of q to the horizontal. If MD = x and DN = y, B = breadth of the tank. Solution: Refer to Fig. 2.67 76 Fluid Mechanics and Hydraulic Machines M (c) Since the pressure distribution is hydrostatic in any vertical direction and the hydraulic grade line is inclined at q to horizontal (line MN) the lines of equal pressure will be parallel to MN, as shown in Fig. 2.67. Hydraulic grade line q Lines of equal pressure N D A hc = h d Closed tank oil RD = 0.81 3m 2.73 ha = h b ax B C a Datum 10.0 m Fig. 2.67 It is required to ﬁnd: (a) pa – pd and (b) pb – pa and (c) Lines of equal pressure 1.2 m At an acceleration of ax let MN be the hydraulic grade line. Its inclination tan q = Opening D hd - ha a = x = 0.25 L g E But and Zd = Za Hence (pd – pa) = gs ¥ 2.5 A 0.8 m C B 6.2 m (a) M Atmospheric pressure at E E D Hydraulic grade line = 9.79 ¥ 0.81 ¥ 2.5 Hence A F = 19.83 kPa (b) Along BA the hydraulic grade line is constant. 1.5 m 5.0 m F 2.3 m (a) (hd – ha) = 0.25 ¥ 10 = 2.5 m Êp ˆ Êp ˆ (hd – ha) = Á d + Zd ˜ - Á a + Za ˜ Ë gs ¯ Ë gs ¯ ax C ax B h hb = ha h Ê pb ˆ Ê pa ˆ ÁË g + Zd ˜¯ - ÁË g + Za ˜¯ = 0 s s (pb – pa) = gs (Za – Zb) = 9.79 ¥ 0.81 ¥ 3.0 = 23.79 kPa N Datum (b) Fig. 2.68 77 Fluid Statics Solution: At the acceleration ax the hydraulic grade line will be inclined at q, given by a tan q = x g Since pe = pressure at E = atmospheric pressure the hydraulic grade line will pass through E as shown in Fig. 2.68(b) by the line MEN. Then above an arbitrary datum: he = hf and ha = hb Also (he – ha) = AF ◊ tan q = 5 tan q At the onset of cavitation at A, pa = pv = vapour pressure. Considering absolute pressures, \ = (10.0 – 0.5) + (1.5) = 11.0 m But (he – ha) = 5 tan q a 11.0 = 2.2 Hence tan q = x = g 5 or ax = 2.2 g = 2.2 ¥ 9.81 = 21.585 m/s2 (b) Pressure at B: Êp ˆ Êp ˆ (hb – ha) = Á b + Z b ˜ - Á a + Za ˜ = 0 Ë g ¯ Ë g ¯ pb p = a + (Za – Zb) = 0.5 + 0.8 = 1.3 m g g pb = 1.3 ¥ 9.79 = 12.727 kPa (abs) Pressure at F: Êp ˆ Êp ˆ (he – hf) = Á e + Ze ˜ - Á f + Zf ˜ = 0 Ë g ¯ Ë g ¯ pf p = e + (Ze – Zf) = 10.0 + 1.5 g g = 11.5 m pf = 11.5 ¥ 9.79 = 112.585 kPa (abs) Pressure at D: (hd – he) = 1.2 tan q = 1.2 ¥ 2.2 = 2.64 m pd p = e + (Ze – Zd) + 2.64 g g = 10.0 + 0 + 2.64 = 12.64 m pd = 9.79 ¥ 12.64 = 123.75 kPa (abs) Pressure at C: Êp ˆ Êp ˆ (hd – hc) = Á d + Zd ˜ - Á c - Zc ˜ = 0 Ë g ¯ Ë g ¯ pc pd = + (Zd – Zc) g g = 12.64 + 2.03 = 14.94 m pc = 9.79 ¥ 14.94 = 146.26 kPa (abs) Êp ˆ Êp ˆ (he – ha) = Á e + Ze ˜ - Á a + Za ˜ Ë g ¯ Ë g ¯ Êp p ˆ = Á e - a ˜ + (Ze – Za) g ¯ Ë g Êp ˆ Êp ˆ (hd – he) = Á d + Zd ˜ - Á e + Ze ˜ Ë g ¯ Ë g ¯ Also 2.74 2 Solution: Refer to Fig. 2.69 Resolving the acceleration as into x- and z-components: ax = 2 cos 30° = 1.732 m/s2 az = – 2 sin 30° = – 1.00 m/s2 Z A q C a X O q 1.2 m F N 1.5 m M 3.0 m (a) D 30° (b) Fig. 2.69 78 Fluid Mechanics and Hydraulic Machines Water surface slope (q measured clockwise from x-direction) ax 1.732 = = 0.1966 tan q = ( az + g ) ( - 1 + 9 . 81) or q = 11.12° Piezometric head at the rear 3. 0 hN = 1.2 + ¥ tan q 2 = 1.2 + 0.2949 = 1.495 m Piezometric head at the front 3. 0 tan q = 0.905 m hM = 1.2 – 2 Pressures at N and M: pN = 1.495 ¥ 9.79 = 14.64 kPa pM = 0.905 ¥ 9.79 = 8.86 kPa 2.75 2 Pressure force on a side wall per metre width 1. 5 FH = 0.7452 ¥ 9.79 ¥ 1.5 ¥ 2 = 8.207 kN 2.76 Solution: At the maximum permissible angular velocity the water surface will just touch the top rim of the cylinder. Referring to Fig. 2.70, let C represent the original water level at the axis. As the water surface is a paraboloid of revolution, rise of water surface at the edge above C = distance OC = y/2 where y = water surface elevation at the outer edge above the vertex 0. B A 0.5 m C Solution: (a) When the acceleration is upwards az = + 2.5 m/s2 Pressure at any depth h below the free surface Ê a ˆ p = g h Á1 + z ˜ g¯ Ë (b) When the acceleration is downwards az = – 2.5 m/s2 Pressure at any depth h below free surface Ê a ˆ p = g h Á1 + z ˜ = 0.7452 g h g¯ Ë 2.0 m O 1.5 m N S M w Ê 2.5 ˆ = g h Á1 + = 1.2548 g h 9 . 81˜¯ Ë Pressure force on a side wall per metre width 1. 5 FH = 1.2548 ¥ 9.79 ¥ 1.5 ¥ 2 = 13.82 kN y 1.0 m Fig. 2.70 Hence y = 2(2.0 – 1.5) = 1.0 m In the present case w 2r2 y = 2g w 2 ¥ (0 . 5) 2 2 ¥ 9 . 81 2 ¥ 9 . 81 w2 = = 78.48 (0 . 5 )2 1.0 = 79 Fluid Statics w = 8.858 2 pN when N = rotations/min w = 60 60 ¥ 8 . 858 N= = 84.6 rpm 2p But 2.77 Solution: Referring to Fig. 2.71, (a) If N = rotations/min Angular velocity w = 2 pN 60 In the present case 2 p ¥ 180 = 18.85 rad./s w= 60 20 cm w 2r2 (18.85) 2 (0.10) 2 = 2g 2 ¥ 9.81 = 0.181 m = 18.1 cm Rise at ends = y/2 above original level C. hN = piezometric head at a radial distance 10 cm on the base, i.e. point N. = original depth + y/2 18.10 = 39.05 cm = 30.00 + 2 hs = piezometric head at the centre of the base = hN – y = 39.05 – 18.10 = 20.95 cm Pressure at N = pN = g h N = 9.79 ¥ 0.3905 = 3.823 kPa Pressure at S = ps = g hs = 9.79 ¥ 0.2095 = 2.051 kPa (b) In this case y = 20 cm, or y = 40 cm 2 hs = 10 cm and hN = 50 cm y= y= y/2 C y or y/2 2g y r = 28.014 rad./s 60 ¥ 28.014 2p = 267.5 rpm = N = S w 2.78 Fig. 2.71 If y = water surface elevation above the vertex at a radial distance r, at the radial distance r = 10 cm 2 ¥ 9.81 ¥ 0.4 0.10 N = revolutions/min = 30 cm O w= w 2r2 2g Solution: (a) When N = 180 rpm 60w 2p 80 Fluid Mechanics and Hydraulic Machines Volume of the paraboloid AOB of height 40.75 cm 1 ¥ (volume of enclosing cylinder) = 2 = Volume of cylinder of height 20.375 cm Hence total volume of water inside the cylinder = Volume corresponding to a depth of (9.25 + 20.375) = 29.625 cm p ¥ (0.3)2 ¥ 0.29625 = 4 = 20.9 ¥ 10–3 m3 = 20.9 L p ¥ (0.3)2 ¥ 0.50 Original volume = 4 = 0.0353 m3 = 35.3 L Volume of water spilled = 35.3 – 20.9 = 14.4 L (b) When N = 240 rpm 2p ¥ 180 = 18.85 rad./s 60 At the wall of the cylinder, r = 15 cm = 0.15 m w 2r2 (18.85) 2 = ¥ (0.15)2 ymax = 2g 2 ¥ 9.81 = 0.4075 m = 4.075 cm The position of the water surface in the cylinder will be as shown in Fig. 2.72 (a). The water surface will start from the rim of the cylinder and extend downwards as a paraboloid to its vertex at O. OS = The depth at the vertex = hs = 50 – 40.75 = 9.25 cm w= B 20.375 cm A 50 cm 2p ¥ 240 = 25.13 rad./s 60 w= O 9.25 cm At r = 0.15 m, w 2r2 ( 25.13 ¥ 0.15) 2 = 2g 2 ¥ 9 . 81 = 0.724 m = 72.4 cm S 30 cm ymax = 180 RPM (a) A B 50 cm 50 cm x N R2 S R1 M 22.4 cm As this value is larger than 50 cm it means that the theoretical paraboloid will extend below the base thereby leaving a part of the bottom uncovered by water. The paraboloid will start from the rim and extend up to its vertex O which is 72.4 cm below it as shown in Fig. 2.72(b). Let x = radius of the exposed portion of the bottom of the cylinder x = SR1 = SR2 ymax r2 = 2 OS x O 240 RPM (b) Fig. 2.72 Example 2.78 \ x = r OS / ymax = 15 ¥ 22.4 = 8.34 cm 72.4 81 Fluid Statics Volume of water spilled = Volume of the paraboloid AOB – Volume of the paraboloid R1OR2 = p r2 p x2 ¥ ymax – ¥ (OS) 2 2 p Èp ˘ = Í ¥ (0.15)2 ¥ 0.724 - ¥ (0.0834)2 ¥ (0.224)˙ 2 Î2 ˚ 1 [0.05118 – 0.004895] 2 = 0.0232 m3 = 23.2 L 60 ¥ 2.0238 2p = 19.326 rpm Hence, RPM = N = 2.80 = Solution: Referring to Fig. 2.74, let, at the top, centre point be O. Excess pressure due to rotation at the centre point =0 Hence in the plane AOB pressure head at any radial distance r is p w 2r2 = g 2g 2.79 Solution: Refer to Fig. 2.73, CL R1 = 2 m Ú R=h H = R1 = 2 m h Considering a thin annular ring of radius r and thickness d r, Pressure force on the top lid R g w 2r2 FT = 2 p r dr p ◊ 2p r ◊ dr = 0 2g g w 2p 4 R = 240 RPM 4g 45° 45° A Ú O B 1 2 1 p r H = p R12 3 3 = height of water before rotation 1 = p h3 3 = R/(21/3) = 0.7937 = 2.0 m, h = 1.5874 m. H = 30 cm Fig. 2.73 Volume of cone = If h 1 p R13 (1 / 2) 3 \ h Since R On rotation, (w2R12 )/2g = (H – h) = (2.0 – 1.5874) = 0.4126 m 2 ¥ 9.81 ¥ 0.4126 w = ( 2.0) 2 = 2.0238 2 pN and w = 60 S R = 10 cm M dr r Fig. 2.74 82 Fluid Mechanics and Hydraulic Machines In the present case, 2 pN 2 p ¥ 240 = = 25.13 rad./s w = 60 60 R = 0.10 m 9.79 ¥ ( 25.13) 2 ¥ p \ ¥ (0.10)4 FT = 4 ¥ 9.81 = 0.0495 kN = 49.5 N Force on the bottom = FT + g H (Area of the base) = 49.5 + 9790 ¥ 0.30 ¥ p (0.10)2 = 141.8 N Ê 1 w 2r2 ˆ r = exp Á ˜ r0 Ë 2 RT ¯ \ But r p = and hence r0 p0 r ˆ p = exp 2RT ˜¯ p0 2.82 2.81 w p = rRT, Ê w 2r 2 ˆ Á 2RT ˜ Ë ¯ p p0 p0 Solution: The Euler equation relating the pressure gradient in normal direction is dp dp = = ra dn dr But a = w2 r and hence dp = rw 2 r dr From equation of state p = rRT dp = RT dr Ê w 2r ˆ dr = rÁ ˜ dr Ë RT ¯ \ Ê w 2r ˆ dr = Á ˜ dr r Ë RT ¯ 1 w 2r2 +C 2 RT r = r0, r = r0 and p = p0 On integration ln r = when Solution: Case (a): 2 pN 2 p ¥ 120 = = 12.57 rad./s 60 60 In this case because of the symmetry the limbs AN and BM will be having a liquid column of 25 cm as at start. The points A and B being free surfaces will have atmospheric pressure. An imaginary free liquid surface will extend from A and B as a paraboloid with the vertex at 0. Thus the pressures on the limb MN will decrease towards the centre to have a least value at S, the midpoint of the bottom limb. w= \ BM = hM = 25.0 cm AN = hN = 25.0 cm w 2r2 (12.57) 2 ¥ (0.15) 2 = 2g 2 ¥ 9.81 = 0.181 m = 18.1 cm Hence hs = (25.0 – 18.1) = 6.9 cm Pressures at M, N and S are: pM = g hM = 9.79 ¥ 1.25 ¥ 0.25 = 3.06 kPa pN = g hN = 9.79 ¥ 1.25 ¥ 0.25 = 3.06 kPa pS = g hS = 9.79 ¥ 1.25 ¥ 0.069 = 0.844 kPa (BM – OS) = y = Case (b): When N = 240 rpm 83 Fluid Statics 15 cm 15 cm w A Imaginary liquid surface B 25 cm O N S Note: The location of R1 can be calculated by considering points O and R1 as M (a) 15 cm M = pressure at N. = pressure due to a column of 25 cm of liquid = 3.06 kPa Then it decreases to atmospheric pressure at R1 and R2, the point of intersection of the imaginary free surface with the limb MN. At S, pressure ps = – 47.4 cm of liquid = – 0.474 ¥ (1.25 ¥ 9.79) = – 5.8 kPa (gauge) 15 cm y = 0.474 m = A w 2 x2 2g 2.83 B 25 cm N R2 S R1 M Imaginary liquid surface Solution: O 240 RPM (b) Fig. 2.75 2 pN 2 p ¥ 60 = 60 60 = 6.283 rad./s The liquid levels in both the limbs will be part of a paraboloid with vertex at point 0 [see Fig. 2.76(b)] w = Example 2.82 y = 2 p ¥ 240 = 25.13 rad./s 60 w 2r2 ( 25.13) 2 ¥ (0.15) 2 (BM – OS) = y = = 2g 2 ¥ 9.81 = 0.724 m = 72.4 cm Since BM = 25 cm, OS = – 72.4 + 25 = – 47.4 cm (i.e. O is below the point S at a depth of 47.4 cm) This signiﬁes negative pressure at S. The free surface will be as in Fig. 2.75(b). As before, the pressure at w = \ (y1 – y2) = w 2r2 2g w2 2 (r 1 – r 22) 2g (6.283) 2 [(0.30)2 – (0.15)2] 2 ¥ 9.81 = 0.136 m = 13.6 cm = w 2 r12 (6.283) 2 = ¥ (0.30)2 2g 2 ¥ 9.81 = 0.181 m = 18.1 cm y2 = 18.1 – 13.6 = 4.5 cm y1 = 84 Fluid Mechanics and Hydraulic Machines 30 cm 15 cm Liquid RD = 1.25 N M S 60 RPM (a) Imaginary liquid surface A y1 B O N M S r1 y2 r2 There is another relationship relating to total volume of liquid in the tube. The horizontal portions MN is common. The sum of two vertical columns of liquid before rotation = (15 + 15) = 30 cm Also y1 + y2 = 18.1 + 4.5 = 22.6 cm The difference (30.0 – 22.6) = 7.4 cm is equally distributed in the two limbs to get the column heights in the right and left limbs as: 7.4 = 3.7 cm Pressure head hS = 2 7.4 Column NA = hN = 18.1 + = 21.8 cm 2 7.4 Column MB = hM = 4.5 + = 8.2 cm 2 pN = pressure at N = 1.25 ¥ 9790 ¥ 0.218 = 2668 Pa pM = pressure at M = 1.25 ¥ 9790 ¥ 0.082 = 1003 Pa w (b) Fig. 2.76 Example 2.83 Problems A. Measurement of Pressure tank is now carried to a higher elevation where the atmospheric pressure is 70.11 kPa, what reading will be indicated by the gauge? (Ans. p (gauge) = 60.37 kPa) 2.1 Calculate the pressure in pascals corresponding to (a) 8 cm column of a liquid of relative density 0.80 (b) 6 cm column of mercury (c) 2.0 m column of water (Ans. (a) 626.6 Pa, (b) 7988.6 Pa, (c) 19.58 kPa) 2.2 A bourdon gauge connected to a closed tank indicates 35 kPa at a place where the atmospheric pressure is 95.48 kPa. If the 2.3 A hydraulic press has a ram of 150 mm and a plunger of 20 mm diameter. Find the force required on the plunger to lift a weight of 40 kN. If the plunger has a stroke of 0.40 m and makes 30 strokes per minute, determine the rate at which the weight is lifted per minute and the power required by the plunger. 85 Fluid Statics 2.4 2.5 2.6 2.7 (Ans. F = 0.711 kN, rate = 0.2133 m/min, P = 142 W) An open tank contains water to a depth of 2.5 m and an oil of relative density 1.25 to a depth of 1.5 m. Determine the pressure at (a) the water surface (b) at the oil-water interface (c) at a depth of 3.5 m below the free surface and (d) at the bottom of the tank. (Ans. (a) p1 = 0, (b) p2 = 24.475 kPa (c) p3 = 36.713 kPa, (d) p4 = 41.831 kPa) Calculate the pressure and density of air at an elevation of 4500 m above sea level if the atmospheric pressure and density at sea level are 101.35 kPa and 1.2255 kg/m3. Assume isothermal process. (Ans. p2 = 59.429 kPa, r2 = 0.7186 kg/m3) Assuming adiabatic process exists, determine the pressure and density of air at an elevation of 500 m above sea level given that the atmospheric pressure and density at an elevation of 3000 m are 70.107 kPa and 0.9092 kg/m3 respectively. Assume k = 1.4. (Ans. p2 = 95.054 kPa, r2 = 1.1187 kg/m3) For the system shown in Fig. 2.77 calculate the air pressure pA to make the pressure at N one third of that at M. (Ans. pA = 0.2937 kPa) 2.8 For the system shown in Fig. 2.78 determine the pressures at A and B. B Oil RD = 0.8 Atmosphere 60 cm Air A 8 cm Oil RD = 0.8 RD = 0.8 Oil Fig. 2.78 Problem 2.8 (Ans. pA = 626.6 Pa (gauge), pB = – 4072.6 Pa (gauge)) 2.9 If the pipe in Fig. 2.79 contains water and there is no ﬂow, calculate the value of the manometer reading h. (Ans. h = 0) CL B 0m 5. Horizontal 30° A h=? pA A Air Mercury Fig. 2.79 60 cm Water Mercury 10 cm N Fig. 2.77 M Problem 2.10 2.10 For the manometer arrangement of Problem 2.9 (Fig. 2.79) there is a ﬂow of water from A towards B. If the manometer reading is h = 5 cm, calculate the pressure difference (pA – pB). (Ans. (pA – pB) = 30.643 kPa) 86 Fluid Mechanics and Hydraulic Machines 2.11 Two points M and N in a pipe system (Fig. 2.80) are connected a mercury differential manometer. The connecting tube is ﬁlled with oil of relative density 0.85. The high pressure point M is 0.80 m above point N. If the mercury column reading is 8 cm (as shown in the ﬁgure), what is the pressure difference between M and N in (i) kPa and (ii) metres of water. (Ans. (i) (pM – pN) = 3.33 kPa; (ii) hw = 0.34 m) M 0.8 m N Oil RD = 0.85 Oil RD = 0.85 8 cm and B if the atmospheric pressure is 95.48 kPa. (Ans. pA (abs) = 28.91 kPa (abs), pB (abs) = 102.137 kPa (abs)) 2.13 Calculate the pressure and density of air at an elevation of 5000 m above mean sea level. The atmospheric pressure and temperature at sea level are 101.35 kPa and 15°C respectively. The temperature lapse rate is 0.007 K/m. The density of air at sea level is 1.2255 kg/m3. (Ans. p2 = 52.854 kPa (abs), r2 = 0.7413 kg/m3) 2.14 For the manometer set-up shown in Fig. 2.82, ﬁnd the difference in pressure (pA – pB) when (a) H1 = 0.3 m and H2 = 0.05 m. (b) H1 = 0.05 m and H2 = 0.30 m. Take gw = 9.79 kN/m2. (Ans. (a) (pA – pB) = – 0.4895 kPa. (pB is greater than pA.) (b) (pA – pB) = + 0.4895 kPa. (pA is greater than pB.)) Fig. 2.80 Pressure = PA Pressure = PB B A 2.12 A tube ﬁlled with mercury is placed in a bowl of mercury as in Fig. 2.81. The tube is closed at point A and the other end is open. Calculate the absolute pressures at points A H1 Specific weight = 0.8 gw A Specific weight = 0.6 gw H2 Mercury 50 cm Water = gw Fig. 2.82 B Fig. 2.81 5 cm Problem 2.14 2.15 A U-tube manometer is used to measure the pressure of water in a pipe line which is in excess of the atmospheric pressure. The right limb of the manometer contains 87 Fluid Statics mercury and is open to atmosphere. The contact between the water and the mercury is in the left limb. Determine the pressure of water in the pipe line, if the difference in level of mercury in the limbs of the U-tube is 100 mm and the free surface of the mercury is in level with the center of the pipe. If the pressure of water in the pipe is reduced to 9.79 kPa, calculate the new difference in level of mercury. Sketch the arrangements in both cases. (Ans. (i) 13.314 kPa (ii) 79.4 mm) 2.16 A certain ﬂuid of speciﬁc gravity 0.8 ﬂows upwards through a vertical pipe. A and B are two points on the pipe, B being 0.3 m higher than A. A U-tube mercury manometer is connected at points A and B. If the difference in pressure between A and B is 5 kPa, ﬁnd the difference in the heights of the mercury columns in the manometer. (Ans. h = 21.1 mm) 2.17 Considering the standard temperature and pressure at sea level as 288 K and 101.32 kN/m2 respectively, ﬁnd the atmospheric pressure at a height of 5 km above sea level by taking in to account the linear temperature lapse rate as 6.35 K/km. Standard density of air at sea level is 1.205 kg/m3. (Ans. p = 54.6 kPa) B. Forces of Plane Surfaces 2.18 A vertical rectangular gate 2.0 m wide and 2.5 m high is subjected to water pressure on one side, the water surface being at the top of the gate. The gate is hinged at the bottom and is held by a horizontal chain at the top. Calculate the tension in the chain. (Ans. F = 20.39 kN) 2.19 An annular ring cut in a sheet metal has 1.5 m outer diameter and 1.0 m inner diameter. It is inserted vertically in a liquid of relative density 0.90 with its centre 1.75 m below the surface. Calculate the total force on one side of this ring and the location of the centre of pressure. (Ans. F = 15.138 kN; hp = 1.866 m) 2.20 A stone masonry wall 6 m high and 2.5 m wide retains water to a height of 5 m on one side. If the speciﬁc weight of masonry is 25 kN/m3, ﬁnd the direction and magnitude of the resultant force per unit width of wall. Also ﬁnd the point of action of the resultant force on the base of the wall. What are the maximum and minimum stresses at the base of the wall? (Ans. R = 394.5 kN, q = 71.9° to the horizontal, x = 1.794 m from the upstream, s max = 3.46 Pa, smin = – 46 kPa) 2.21 A right angled triangular plate of base b and height h is immersed in water vertically with its base horizontal and coincident with the water surface. Calculate the total pressure force on one side of the plate and the coordinates of the centre of pressure. Ê g bh2 b hˆ Á Ans. F = 6 , xp = 4 , yp = 2 ˜ Ë ¯ 2.22 A vertical gate of width 2.0 m and height 2.5 m controls a sluice opening in a dam. The top of the gate is 10 m below the water surface. If the gate weights 80 kN, ﬁnd the vertical force required to raise the gate. The coefﬁcient of friction between, the gate and the guides can be assumed to be 0.25. (Neglect buoyancy effect on the gate) (Ans. F = 218 kN) 2.23 Some regular geometrical laminae are shown in Fig. 2.83. These laminae are immersed vertically with their upper edges or top most points on the water surface as indicated in the ﬁgures. Derive the expression for the centre of pressure as indicated against each lamina in Table 2.2. 88 Fluid Mechanics and Hydraulic Machines Table 2.2 Location of Centre of Lamina Description Depth of centre of pressure hp Fig. 2.83(i) Rectangle Fig. 2.83(ii) Trapezoid Fig. 2.83(iii) Triangle with base on the free surface Fig. 2.83(iv) Triangle with vertex on the surface; base is horizontal Fig. 2.83(v) Circle Fig. 2.83(vi) Semicircle with diameter on the free surface 3 pD 32 Fig. 2.83(vii) Quadrant of circle with one of the bounding radii on the free surface 3 pD 32 2 h 3 h ( a + 3b) 2 ( a + 2b ) h 2 3 h 4 5 D 8 h 2.24 A thin plate in the form of a rhombus is immersed in water with a vertex on the free surface and the diagonal through that vertex h b (i) b (ii) b a h (iii) h D b (iv) (v) D/2 D D/2 (vi) Fig. 2.83 (vii) Problem 2.23 vertical. Show that the centre of pressure is 5 located at a depth of hp = h where h = 7 length of the vertical diagonal. 2.25 Calculate the force F required to hold the hinged door in Fig. 2.84 in closed position. The door is a 0.5 m square. An air pressure of 30 kPa acts over the water surface. (Ans. F = 7.217 kN) 2.26 A vertical gate 5 m ¥ 2.5 m in size and weighing 0.5 tonnes slides along guides ﬁtted on the side walls of an overﬂow spillway at its crest. The coefﬁcient of friction between the gate and the guide is 0.25. What force will be exerted at the hoisting mechanism to lift the gate when the head of water over the crest is 2.0 m? (Ans. F = 3 tonnes) 89 Fluid Statics Air 30 kPa x = 0.3 m y = 0.6 m H 2.5 m Water x G y Hinge 2m 0.5 m Door Door 0.5 m F 0.5 m Fig. 2.84 Problem 2.25 Fig. 2.86 2.27 The counterweight pivot gate shown in Fig. 2.85 controls the ﬂow from a tank. The gate is rectangular and is 3 m ¥ 2 m. Determine the value of the counterweight W such that the upstream water can be 1.5 m deep. (Ans. W = 84.8 kN) Problem 2.28 2.29 A tank shown in Fig. 2.87 has an isosceles triangular gate hinged at the edge M. Find the horizontal force at M required to keep the gate closed. The weight of the gate is too small and can be neglected. (Ans. F = 13.06 kN) M 0 m 1.50 m 2. 2.0 m Pivot M M te Ga Hinge N 0. 6 m 2.0 m 60° F N Fig. 2.87 Fig. 2.85 Problem 2.27 2.28 The gate shown in Fig. 2.86 weighs 10 kN per metre width. Its centre of gravity is 0.3 m from the vertical face and 0.6 m above the lower horizontal face. Find the height H at which the gate will just topple about the hinge. (Ans. H = 3.38 m) Problem 2.29 2.30 The rectangular gate AB in Fig. 2.88 is 2.0 m high and 1.5 m wide. Calculate the net hydrostatic force on the gate. (Ans. F1 – F2 = 37.83 kN) 2.31 A circular plate 3 m in diameter is submerged in water so that the greatest and least depths below the free surface are 2.0 m and 1.0 m respectively. Find (a) the total pressure force on one side of 90 Fluid Mechanics and Hydraulic Machines 1.6 m 0.6 m A Oil (RD = 0.88) Water 2.0 m B Fig. 2.88 Gate Problem 2.30 the plate and (b) the position of the centre of pressure. (Ans. F = 103.77 kN, hp = 1.5416 m) 2.32 For the ﬂash board shown in Fig. 2.89 ﬁnd the depth of water H and the compressive force on the strut per metre length of the ﬂash board when the board is about to trip. (Hint: At the time of tipping the centre of pressure coincides with the hinge.) (Ans. H = 1.559 m; F = 13.73 kN) Flash board m Hinge 0.6 H Fig. 2.89 vertical side facing the water. The height of the dam is 10 m and the base width is 8 m. Assuming a speciﬁc weight of 25 kN/m3 for the concrete, calculate the resultant force per unit length of dam and its point of action on the ﬂoor of the dam when it retains 10 m of water. What are the minimum and maximum vertical stresses on the base of the dam? (Ans. R = 1113.3 kN, q = 26.08°, s1 = 153.1 kPa, s2 = 96.9 kPa) 2.35 A tank contains a suspension to a height H. The suspension can be considered to be a liquid whose density increases linearly, from r0 at the surface, towards the bottom. Show that the pressure at any depth y is given by K 2 p = r0 g y + g y + p0 2 where p0 = pressure at the free surface and K = rate of increase of the density with depth. Show that the force per unit width on a vertical side of the tank is given by Strut Problem 2.32 2.33 A square aperture in the vertical side of a tank has one diagonal vertical and is completely covered by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are 2 m long and the tank contains a liquid of relative density 1.15. The centre of aperture is 1.5 m below the free surface. Calculate the thrust exerted on the plate by the liquid and the position of the centre of pressure. (Ans. F = 33.78 kN, hp = 1.611 m) 2.34 A concrete gravity dam is in the form of a right angled triangle in section with the È r gH 2 K ˘ + gH 3 + p0 H ˙ F= Í 0 6 ÍÎ 2 ˙˚ C. Forces On Curved Surfaces 2.36 A circular cylinder of 1.8 m diameter and 2.0 m long is used for water level control in a tank. If at a given instance it retains water as shown in Fig. 2.90 determine the reaction at the joint A. (Ans. RH = 31.72 kN, R = 61.96 kN Rv = 53.23 kN, tan q = 0.596 where q = inclination of R to the vertical. MA = 36.0 kN.m (clockwise)) 2.37 A cylinder of diameter 0.6 m is located in water as shown in Fig. 2.91. The cylinder and the wall are smooth. If the length of the cylinder is 1.5 m, ﬁnd (i) its weight, (ii) 91 Fluid Statics 2.39 For a cylindrical gate 4 m long shown in Fig. 2.93 calculate the resultant force due to ﬂuid pressure. (Ans. R = 219.64 kN, a = inclination of R to the horizontal = 40.34°, the resultant passes through the centre of the gate.) Water A 1.80 m O Fig. 2.90 Problem 2.36 Oil 1.50 m RD = 0.80 the resultant force exerted by the wall on the cylinder and (iii) the resultant moment about the centre of the cylinder due to water forces on the cylinder. (Ans. 4410 kN, 661.5 kN, zero) Water 1.50 m Fig. 2.93 r= 0. 3 m 2.40 For the container shown in Fig. 2.94 estimate the resultant force on the hemispherical bottom. (Ans. RV = 52.46 kN, RH = 0 (due to symmetry)) Water Fig. 2.91 Problem 2.39 Air 15 kPa Problem 2.37 2.38 A radial gate retains 5 m of water above the crest of a dam as shown in Fig. 2.92. Find the resultant water force on the gate per metre length. (Ans. R = 124.4 kN, a = inclination of R to the horizontal = 10.235°. The resultant passes through the centre O.) Oil RD = 0.75 3.0 m Hemisphere 5.0 Water q 5.0 m 5.0 m M Fig. 2.92 1.5 m m Problem 2.38 Fig. 2.94 O Problem 2.40 2.41 A cylindrical barrier of diameter 2.0 m and width 3.0 m retains water on both sides with water surface elevations of 1.5 m and 0.5 m above the lowest point 92 Fluid Mechanics and Hydraulic Machines of the barrier, respectively. Calculate the resultant hydrostatic force on the barrier. (Ans. R = 54.69 kN; a = inclinations of R to the horizontal = 57.5°) 2.42 A 1.8 m diameter cylindrical tank is laid with its axis horizontal on a level ground. Each of its ends are closed by a hemispherical dome. The tank contains oil of relative density 0.9 under pressure. If a pressure gauge on the top of the tank reads 22 kPa, calculate the resultant force on the hemispherical end. (Ans. R = 54.69; a = inclination of R to the horizontal = 10°, the resultant passes through the centre of the hemisphere) 2.43 A cylindrical tank of diameter 2.5 m is founded with its axis horizontal. The tank contains water in the bottom half and an oil of relative density 0.82 on the top half with a common interface. The ﬂuids are under pressure and a pressure gauge on the top of the cylinder reads 15 kPa. Calculate the force per metre length on the upper half of the cylindrical tank. (Ans. FV = 42.9 kN/m length (upward)) 2.44 A conical valve of weight 2.5 kN deeps water from ﬂowing out of a tank as shown in Fig. 2.90. It is held in position by a counter weight of 10 kN connected by a string passing over frictionless pullies as in Fig. 2.95. Find the maximum height H 2.45 2.46 2.47 2.48 2.49 Water H 1.0 m 10 kN 60° Weight Cone 2.5 kN Fig. 2.95 Problem 2.44 of water in the tank, which will make the device to function without any leak. (Ans. H = 1.064 m) A rectangular water tank has its vertical sides joined to the horizontal bottom through a smooth curve which can be approximated to a quadrant of a circle of radius 1.6 m. If the tank contains 2.0 m of water and 2.5 m of oil relative density 0.8 over it, calculate the resultant force per metre length of the curved surface. (Ans. R = 76.1 kN; a = inclination of R to the horizontal = 48.8°, the resultant passes through the centre of the arc.) A cylinder of 0.8 m diameter is made of 5 mm thick plates. If the maximum permissible stress in the plates is 100 MPa, calculate the maximum permissible pressure inside the cylinder. (Ans. s = 1.25 MPa) A pressure vessel in the form of cylinder of 1.5 m diameter is to have a ﬂuid under a pressure of 1.2 MPa. If the allowable stress in the material of the walls is 140 MPa ﬁnd the minimum thickness of the walls. (Ans. t = 6.5 mm) A spherical vessel is made up of a material of permissible stress of 150 MPa. The diameter of the vessel is 15 m and the thickness of the sides is 5 mm. Find the maximum permissible pressure in this vessel. (Ans. p = 2.0 MPa) A pair of lock gates each of 3 m width make an angle of 120° when closed. The gates are supported by two hinges at 0.5 m and 4.5 m above the bottom. Determine the reaction force on each hinge when the lock has 4.8 m and 1.5 m of water on upstream and downstream respectively. (Ans. R top = 93.0 kN, Rbottom = 212.3 kN) 93 Fluid Statics D. Buoyancy 2.50 A closed cylindrical tank of diameter 2.0 m, height 1.2 m and weighing 20 kN is ﬂoating with its axis vertical in sea water (relative density = 1.025). (a) Find the depth of the cylinder below the water surface. (b) What would be the depth of immersion if an additional load of 5.0 kN is added at the top? (Ans. (a) h1 = 0.6344 m, (b) h2 = 0.793 m) 2.51 A metal sphere of volume Vm = 1 m3, relative density Sm = 2 and fully immersed in water is attached by a ﬂexible wire to a buoy of volume V b = 1 m3 and relative density Sb = 0.1 (See Fig. 2.96). Calculate the tension T in the wire and volume of the buoy that is submerged. (Ans. T = 1.96 kN, Vb = 0.2 m3) Vb Vm Fig. 2.96 Problem 2.51 2.52 An iceberg has a speciﬁc weight of 9.0 kN/m3 and ﬂoats in sea water of speciﬁc weight 10.05 kN/m3. What percentage of the total volume of the iceberg will be above the sea water surface? (Ans. 10.5%) 2.53 A hydrometer is to be so built that the mark corresponding to relative density of 1.0 and 1.5 are 8 cm apart on a 5 mm diameter stem. How far from the mark of 1.0 will the relative density mark of 1.25 be located? (Ans. h = 4.8 cm) 2.54 A hydrometer has a 6 mm diameter stem. The distance between markings of relative density 1.0 and 0.90 is 10 cm. Determine the weight of the hydrometer. (Ans. W = 0.249 N) 2.55 A 10 cm cube of steel (relative density = 7.85) is to ﬂoat on mercury in a container of square cross section, with a clearance of 1 cm all round. Find the weight of mercury (relative density = 13.6) required. (Ans. W = 53 N) 2.56 An open cylindrical bucket 30 cm diameter and 50 cm long whose wall thickness and weight can be considered negligible is forced open end ﬁrst into water until its lower edge is 10 m below the water surface. What force will be required to maintain this position? Assume that the trapped air undergoes any change under isothermal conditions. Atmospheric pressure = 100 kPa, temperature of water = 20°C. (Ans. F = 177 N) 2.57 A pontoon of rectangular cross sectional area is 7.0 m long, 3.0 m wide and 1.5 m high. The depth of submergence of the pontoon is 0.9 m and its centre of gravity is 0.7 m above its bottom. Determine its metacentric height. (Ans. MG = 0.583 m) 2.58 A solid cylinder of diameter 30 cm and height 15 cm ﬂoats with its axis vertical in sea water (rel. den. = 1.03). If the relative density of the cylinder material is 0.9, examine the stability of the cylinder. (Ans. MG = 3.34 cm; stable) 2.59 A cube of side a ﬂoats with one of its axes vertical in a liquid of relative density SL. If the relative density of the cube material is Sc, ﬁnd the condition for the metacentric height to be zero. Ê ˆ SL ÁË Ans. S = 1.268 or 4.732˜¯ c 94 Fluid Mechanics and Hydraulic Machines 2.60 A solid cube of sides 0.5 m is made of a material of relative density 0.5. The cube ﬂoats in a liquid of relative density 0.95 with two of its faces horizontal. Examine its stability. (Ans. MG = – 0.0393 m; unstable) 2.61 A rectangular barge of width b and depth of submergence H has its centre of gravity at the waterline. Find the metacentric height and the value of the ratio b/H for which the barge is stable. 2.66 2.67 Ê ˆ b2 H - , for stability b /H ≥ 6 ˜ Ans . MG = Á 12 H 2 Ë ¯ 2.62 A solid cone with an apex angle of 60° and relative density Sc is ﬂoating in mercury (rel. den. = 13.6) with its vertex downwards and axis vertical. Determine the range of Sc over which the cone is in stable equilibrium. (Ans. Sc > 5.738) 2.63 A cone of relative density 0.8 is to ﬂoat in water with its axis vertical and vertex downwards. Find the least apex angle of the cone for stable equilibrium. (Ans. q = 31° 03¢ 33≤) 2.68 E. Rigid Body Motion 2.64 An open tank is 7 m long, 2 m wide and 1.5 m deep. It contains oil of relative density 0.8 to a depth of 1.0 m. If the tank is given a horizontal acceleration at a constant value of 2.5 m/s2 along its length, calculate the amount of oil spill. What are the pressures on the bottom of the tank at its front and rear end? (Ans. Spill = 5.171 m3, p front = 0 (atmospheric), p rear = 14.685 kPa) 2.65 An open tank 5 m long, 3 m wide and 2 m deep contains 1.5 m water. What is the maximum horizontal acceleration that can be given to the tank without causing spill over, and what is the alignment of the tank 2.69 2.70 with respect to the movement? (Ans. ax = 3.27 along the width) An open tank 3.0 m long, 2.0 m wide and 2.0 m deep contains water to a depth of 0.9 m. What minimum horizontal acceleration along the length should be given to have zero depth of the water along the front edge of the tank? (Ans. ax = 5.886 m/s2) A 2.5 m long open tank is mounted on a carriage which moves up a plane inclined at 35° to the horizontal at an acceleration of 2.0 m/s2. What is the slope of the water surface? If the tank is 1.2 m deep and initially contains water to a height of 0.6 m, what would be the depths at the forward and rear edges of the tank? (Ans. q = 8.5° with lower depth in the front, yfront = 0.413 m, yrear = 0.787 m) A 30 cm diameter cylinder contains oil of speciﬁc weight 7.5 kN/m3 to a height of 120 cm. Calculate the force on the bottom of the tank when the tank undergoes an acceleration of 3.5 m/s2 (a) vertically downwards and (b) vertically upwards. (Ans. (a) FV = 0.863 kN, (b) FV = 0.409 kN) An open tank containing water slides down an inclined plane without friction. Show that the water surface will be (a) parallel to the plane if the acceleration is equal to the component of g along the inclined plane, (b) horizontal if the velocity of slide is constant. A closed tank shown in Fig. 2.97 is given an acceleration of 3.6 m/s2, the accelerations vector being inclined at an angle of 35° to the horizontal. Calculate the pressures at point M, N and R. Ans. (PM = 10.406 kPa, PN = 25.73 kPa; PR = 3.703 kPa) 95 Fluid Statics R Z 0.5 m 2.0 m N a= 35° Oil (RD = 0.9) 10 m M 2 s 3.6 m/ x 30 cm Width = 2.0 m 50 cm Fig. 2.97 Problem 2.70 2.71 An open cylindrical tank of 30 cm diameter is 40 cm high. The tank is ﬁlled with water to a depth of 30 cm. If the tank is rotated about the vertical axis of the cylinder, ﬁnd the maximum speed of rotation that does not cause any spill of the liquid. (Ans. N = 126 rpm) 2.72 A 40 cm diameter cylindrical tank is 35 cm high and is open at the top. Initially it contains water to a depth of 20 cm. If the tank is rotated about its vertical axis at 120 rpm, calculate the amount of liquid spilled out. (Ans. Vs = 0.34 L) 2.73 If the tank in Problem 2.73 is rotated at 150 rpm, calculate the volume of water spilled out. (Ans. Vs = 7.01 L) 2.74 A closed cylinder 40 cm in diameter and 40 cm in height is ﬁlled with oil of relative density 0.80. If the cylinder is rotated about its vertical axis at a speed of 200 rpm, calculate the thrust of oil on top as well as bottom of the cylinder. (Ans. FT = 440 N, FB = 834 N) 2.75 A hemispherical bowl of radius 1.0 m is full of water and is to be rotated about its vertical axis at 30 rpm. Estimate the amount of water that will overﬂow. (Ans. 0.790 m3/s) 2.76 A U-tube has a liquid of relative density 0.85 in its limbs to a height of 50 cm above the horizontal limb of 30 cm length (Fig. 2.98). What will be the difference in N S 10 cm M 20 cm 180 rpm Axis Fig. 2.98 Problem 2.76 and 2.77 elevation of the two free surfaces when the tube is rotated about a vertical axis 10 cm from one leg and 20 cm from the other, at 180 rpm? (Ans. (y2 – y1) = 54.3 cm) 2.77 For the U-tube of Problem 2.76, under the given conditions, determine the pressures at points N, S and M. (Ans. pN = 1.79 kPa, pS = 0.372 kPa, pM = 6.042 kPa) 2.78 A 20 cm high cylinder open at one end contains 10 cm of water and its top is connected to a 1.0 m lever arm. If the cylinder is rotated in the vertical plane at an angular velocity of 5 rad./s, calculate the pressure at the bottom of the cylinder when (a) it is at its highest point and (b) at its lowest point in the rotation. (Use an average radial distance of 1.15 m in calculating accelerations). (Ans. (a) pb = 1890 Pa, (b) pb = 3848 Pa) 2.79 At what angular velocity must a U-tube, of 30 cm horizontal limb length and ﬁlled with water to a depth of 30 cm in the vertical limbs, be rotated about a vertical axis at mid distance from the vertical limbs, to cause cavitation at the point of intersection 96 Fluid Mechanics and Hydraulic Machines of the axis with the horizontal limb? The vapour pressure of water can be taken as 2.50 kPa (abs) and the atmospheric pressure as 95.5 kPa (abs). The density of water is 998 kg/m3. (Ans. N = 869 rpm) (Hint: Cavitation in water occurs when the local pressure reaches the vapour pressure) 2.80 For the U-tube containing mercury, shown in Fig. 2.99 what speed of rotation causes the differences in limb heights as indicated? (Ans. N = 48.8 rpm) 50 cm 25 cm 25 cm N rpm Axis Fig. 2.99 Problem 2.80 Objective Questions 2.1 For a ﬂuid at rest (a) the shear stress depends upon the coefﬁcient of viscosity (b) the shear stress is maximum on a plane inclined at 45° to horizontal (c) the shear stress is zero (d) the shear stress is zero only on horizontal planes 2.2 If a Mohr circle is drawn for a ﬂuid element inside a ﬂuid body at rest, it would be (a) a circle not touching the origin (b) a circle touching the origin (c) a point on the normal stress axis (d) a point on the shear stress axis 2.3 Indicate the incorrect answer: Hydrostatic pressure variation implies that (a) the pressure varies linearly with depth (b) the piezometric head [p/g + Z] in constant (c) the density of the ﬂuid in constant (d) pressure varies linearly with distance 2.4 Normal stresses are of the same magnitude in all directions at a point in a ﬂuid (a) only when the ﬂuid is frictionless (b) only when the ﬂuid is at rest (c) only when there is no shear stress (d) in all cases of ﬂuid motion 2.5 The basic differential equation for the variation of pressure p in a static ﬂuid with vertical distance y (measured upwards) is (a) dp = – g dy (b) dy = – g dp (c) dp = – rdy (d) dp = – dy 2.6 In an isothermal atmosphere the pressure (a) decreases linearly with elevation (b) decreases exponentially with elevation (c) increases logarithmically with elevation (d) varies inversely as the density 2.7 The piezometric head in a static liquid (a) remains constant only on a horizontal plane (b) increases linearly with depth below a free surface (c) remains constant at all points in the ﬂuid (d) decreases linearly with depth below the free surface 97 Fluid Statics 2.8 Identify the correct statement: (a) Local atmospheric pressure is always less than the standard atmospheric pressure. (b) Local atmospheric pressure depends only on the elevation of the place. (c) A barometer reads the difference between the local and standard atmospheric pressure. (d) Standard atmospheric pressure is 760 mm of mercury. 2.9 Aneroid barometer measures (a) local atmospheric pressure (b) standard atmospheric pressure (c) gauge pressure (d) difference between the standard and local atmospheric pressures 2.10 Bourdon gauge measures (a) absolute pressure (b) gauge pressure (c) local atmospheric pressure (d) standard atmospheric pressure 2.11 A barometer at a given location, (a) shows the local atmospheric pressure which is invariant with time (b) always shows the local atmospheric pressure which may change with time (c) shows the standard atmospheric pressure, if it is of aneroid type (d) shows the local temperature if it is of mercury column type 2.12 In a mercury column-type barometer the correct local atmospheric pressure is obtained by adding correction due to vapour pressure of mercury as follows: Ha = (a) H0 – hv (b) H0 + hv (c) H0/hv (d) hv – H0 where Ha = correct local pressure in mm of mercury, H0 = observed barometer reading in mm of mercury and hv = vapour pressure of mercury in mm. 2.13 When the barometer reads 740.0 mm of mercury, a pressure of 10 kPa suction at that location is equivalent to (a) 10.02 m of water (abs) (b) 9.87 m of water (abs) (c) 88.53 kPa (abs) (d) 0.043 kPa (abs) 2.14 The standard sea-level atmospheric pressure is equivalent to (a) 10.0 m of fresh water of r = 998 kg/m3 (b) 10.1 m of salt water of r = 1025 kg/m3 (c) 12.5 m of kerosene of r = 800 kg/m3 (d) 6.4 m of carbon tetrachloride of r = 1590 kg/m3 2.15 The standard atmospheric pressure is 760 mm of mercury. At a certain location the barometer reads 710 mm of mercury. At this place an absolute pressure of 360 mm of mercury corresponds to a gauge pressure in mm of mercury. (a) 400 mm of vacuum (b) 350 mm of vacuum (c) 360 mm of vacuum (d) 710 mm 2.16 The standard atmospheric pressure is 101.32 kPa. The local atmospheric pressure at a location was 91.52 kPa. If a pressure is recorded as 22.48 kPa (gauge), it is equivalent to (a) 123.80 kPa (abs) (b) 88.84 kPa (abs) (c) 114.00 kPa (abs) (d) 69.04 kPa (abs) 2.17 A U-tube manometer measures (a) absolute pressure at a point (b) local atmospheric pressure (c) difference in total energy between two points (d) difference in pressure between two points 98 Fluid Mechanics and Hydraulic Machines 2.18 In the setup shown in Fig. 2.100 assuming the speciﬁc weight of water as 10 kN/m3, the pressure difference between the two points A and B will be (a) 10 N/m3 (b) –10 N/m3 3 (c) 20 N/m (d) –20 N/m3 Oil of Sp. Gr =0.98 50 cm 50 cm Water Sp. Gr = 1.0 A Fig. 2.100 B Question 2.18 2.19 An inclined manometer contains a liquid of relative density 0.8 and has an inclination of 30° to the horizontal. For a certain pressure the column length was 10 cm. If there is an uncertainty of 1° in the measurement of the angle of inclination, the calculated pressure would have an uncertainty of (a) 1% (b) 0.28% (c) 1.75% (d) 3.33% 2.20 A U-tube open at both ends and made of 8 mm diameter glass tube has mercury in the bottom to a height of 10 cm above the horizontal limb. If 19 cc of water is added to one of the limbs, the difference in mercury levels at equilibrium is (a) 3.0 cm (b) 2.8 cm (c) 1.0 cm (d) zero 2.21 For a submerged plane in a liquid, the resultant hydrostatic force F on one side of the plane is related to area A, centroidal depth h , depth of the centre of pressure hcp and depth of bottom edge hb as F = (a) g Ahcp (b) g A h (c) g Ahb (d) Ah /g 2.22 For an inclined plane submerged in a liquid the centre of ﬂuid pressure on one side of the plane will be (a) above the top edge of the area (b) vertically below the centre of gravity (c) below the centre of gravity (d) in the same horizontal plane as the centre of gravity 2.23 A rectangular water tank, full to the brim, has its length, breadth and height in the ratio 2 : 1 : 2. The ratio of hydrostatic forces on the bottom to that on any larger vertical surface of the tank is (a) 1/2 (b) 1 (c) 2 (d) 4 2.24 The depth of centre of pressure for a rectangular lamina of height h and width b immersed vertically in water with its longest edge vertical and top edge touching the water surface is (a) b/3 (b) h/4 (c) 2h/3 (d) 3h/2 2.25 An equilateral triangular plate is immersed in water as shown in Fig. 2.101. The centre of pressure below the water surface is at a depth of (a) 3h/4 (b) h/3 (c) 2h/3 (d) h/2 Plate h Fig. 2.101 Question 2.25 2.26 When the water surface coincides with the top edge of a rectangular gate 2.5 m wide ¥ 3 m deep, the depth of centre of pressure is (a) 1.0 m (b) 1.5 m (c) 2.0 m (d) 2.5 m 99 Fluid Statics 2.27 A circular plate of diameter D is submerged in water vertically so that the topmost point is just at the water surface. The centre of pressure of the plate will be below the water surface at a depth of (a) 5D/8 (b) 11D/16 (c) 2D/3 (d) 3D/4 2.28 The tank in Fig. 2.102 discharges water at constant rate for all water levels above the air inlet R. The height above datum to which water would rise in the manometer tubes M and N respectively, are (a) (60 cm, 20 cm) (b) (40 cm, 40 cm) (c) (20 cm, 20 cm) (d) (20 cm, 60 cm) 2.31 2.32 Open to atmosphere M N 40 cm 2.33 R 20 cm Datum Fig. 2.102 Question 2.28 2.29 A cylindrical tank of 2 m diameter is laid with its axis horizontal and is ﬁlled with water just to its top. The force on one of its end plate in kN, is (a) 123.0 (b) 61.51 (c) 30.76 (d) 19.58 2.30 A rectangular plate 0.75 m ¥ 2.4 m is immersed in a liquid of relative density 0.85 with its 0.75 m side horizontal and just at the water surface. If the plane of the plate makes an angle of 60° with the horizontal, 2.34 the pressure force on one side of the plate, in kN, is (a) 15.6 (b) 7.8 (c) 24.0 (d) 18.0 A circular annular plate bounded by two concentric circles of diameter 1.2 m and 0.8 m is immersed in water with its plane making an angle of 45° with the horizontal. If the centre of the circles is 1.625 m below the free surface, the total pressure force on one side of the plate in kN, is (a) 7.07 (b) 10.0 (c) 14.14 (d) 18.0 A rectangular plate 30 cm ¥ 50 cm is immersed vertically, in water with its longer side vertical. The total force on one side of the plate is estimated as 17.6 kN. If the plate is turned in the vertical plane at its centre of gravity by 90° and if all other factors remain the same, the total force on one side of the plate would now be (a) 8.8 kN (b) 15.6 kN (c) 17.6 kN (d) 19.6 kN A stationary liquid is so stratiﬁed that its density is r0(1 + h) at a depth h below the free surface. At a depth h in this liquid the pressure p in excess of r0 gh is (a) r0 gh (b) r0gh2 3 (c) r0 gh /3 (d) r0gh2/2 A curved surface is submerged in a static liquid. The horizontal component of pressure force on it is equal to (a) the pressure force on a horizontal projection of the surface (b) product of the surface are and pressure at the centre of gravity (c) pressure force on a vertical projection of the surface (d) weight of the liquid contained between the curved surface and the liquid surface 100 Fluid Mechanics and Hydraulic Machines 2.35 A hollow hemispherical object of diameter D was immersed in water with its plane surface coinciding with the free surface. The vertical component of force on the curved surface is given by Fv = 3 1 g p D3 g p D3 (a) (b) 8 12 1 g p D3 (c) (d) zero 24 2.36 A cylindrical gate of 2.0 m diameter is holding water on one side as shown in the Fig. 2.103. The resultant vertical component of force of water per meter with of gate, by taking g = 10 kN/m for water, is in kN/m (a) 15.71 (b) 31.42 (c) 20.0 (d) zero 2.40 2.41 2.42 Water 2m Fig. 2.103 Question 2.36 2.37 A 2.0 m diameter penstock pipe carries water under a pressure head of 100 m. If the wall thickness is 7.5 mm, the tensile stress in the pipe wall, in MPa, is (a) 65.3 (b) 130.5 (c) 231.0 (d) 1305.0 2.38 The centre of buoyancy of a submerged body (a) coincides with the centre of gravity of the body (b) coincides with the centroid of the displaced volume of the ﬂuid (c) is always below the centre of gravity of the body (d) is always above the centroid of the displaced volume of liquid 2.39 An object weighing 100 N in air was found to weigh 75 N when fully submerged in 2.43 2.44 2.45 water. The relative density of the object is (a) 4.0 (b) 4.5 (c) 2.5 (d) 1.25 An iceberg has 12% of its volume projecting above the surface of the sea. If the density of sea water is 1025 kg/m3, the density of the iceberg is (a) 878 kg/m3 (b) 1000 kg/m3 3 (c) 1148 kg/m (d) 902 kg/m3 An object weighs 50 N in water and 80 N in an oil of relative density 0.80. Its volume in litres is (a) 15.3 (b) 60.0 (c) 30.6 (d) 50.0 A ﬂoating body displaces (a) a volume of liquid equal in magnitude to its own volume (b) a volume of liquid equal to its own submerged weight (c) a weight of liquid equal in magnitude to its own weight (d) a weight of liquid which depends upon the volume of the container A metal block is thrown into a deep lake. As it sinks deeper in water, the buoyant force acting on it (a) increases (b) remains the same (c) decreases (d) ﬁrst increases and then decreases When a block of ice ﬂoating on water in a container melts, the level of water in the container (a) rises (b) ﬁrst falls and then rises (c) remains the same (d) falls When a ship enters sea from a river one can expect it (a) to rise a little (b) to sink a little 101 Fluid Statics 2.46 2.47 2.48 2.49 (c) to remain at the same level of draft (d) to rise or fall depending on whether it is of wood or steel Buoyant force is the (a) lateral force acting on a submerged body (b) resultant force acting on a submerged body (c) resultant hydrostatic force on a body due to ﬂuid surrounding it (d) the resultant force due to water on a body A right circular wooden cone (Speciﬁc gravity = 0.8) with a base diameter of 0.6 m and hight of 0.8 m ﬂoats in water such that its axis is vertical and apex is downward. The immersed depth of cone is (a) 0.480 m (b) 0.533 m (c) 0.600 m (d) 0.743 m A hydrometer weighs 0.03 N and has a stem at the upper end. The stem is cylindrical and 3 mm in diameter. It will ﬂoat deeper in oil of speciﬁc gravity 0.75 than in alcohol of speciﬁc gravity 0.8 by how much amount? (a) 10.7 mm (b) 43.3 mm (c) 33 mm (d) 36 mm A wooden rectangular block of length L is made to ﬂoat in water with its axis vertical. The center of gravity of the ﬂoating body is 0.15 L above the center of buoyancy. What is the speciﬁc gravity of the wooden block? (a) 0.6 (b) 0.65 (c) 0.7 (d) 0.75 h CG 0.15 L L B L/2 A Fig. 2.104 A Question 2.49 2.50 A 12 cm steel cube (RD = 7.6) is submerged in a two layered ﬂuid system. The bottom layer is mercury (rel. den = 13.6) and the top layer is kerosene. The height of the top surface of the cube above the interface of the two liquids is (a) 4.76 cm (b) 2.95 cm (c) 6.0 cm (d) zero 2.51 Two cubes of size 1.0 m sides, one of relative density = 0.60 and another of relative density = 1.15, are connected by a weightless wire and placed in a large tank of water. Under equilibrium the lighter cube will project above the water surface to a height of (a) 10 cm (b) zero (c) 50 cm (d) 25 cm 2.52 If B = centre of buoyancy, G = is the centre of gravity and M = metacentre, of a ﬂoating body, the body will be in stable equilibrium if (a) MG = 0 (b) M is below G (c) BG = 0 (d) M is above G 2.53 In a ﬂoating body I = moment of inertia of water line area about the longitudinal axis, V = volume of displaced ﬂuid, B = centre of pressure, G = centre of gravity and M = metacentre. For stable equilibrium of this body I + MG (a) BG = V I (b) MG = + BG V Ê I ˆ (c) MG = Á ˜ /BG ËV ¯ I (d) BG + MG = V 2.54 A body is ﬂoating in a liquid as shown in Fig. 2.105. The centre of buoyancy, centre of gravity and metacentre are labeled as B, G and M respectively. The body is (a) vertically stable 102 Fluid Mechanics and Hydraulic Machines G B Fig. 2.105 2.55 2.56 2.57 2.58 M Question 2.54 (b) vertically unstable (c) rotationally stable (d) rotationally unstable A large metacentric height in a vessel (a) improves stability and makes periodic time of oscillation longer (b) impairs stability and makes periodic time of oscillation shorter (c) has no effect on stability or periodic time of oscillation (d) improves stability and makes periodic time of oscillation shorter An open box of base 2 m ¥ 2 m contains a liquid of speciﬁc gravity 0.80 to a height of 2.5 m. When it is given an acceleration vertically upwards of 4.9 m/s2, the pressure on the base of the tank, in kPa, is (a) 9.8 (b) 36.8 (c) 19.6 (d) 29.4 A tank, open at top, contains a liquid with a relative density of 0.85 to a depth of 1.2 m. The acceleration which should be given to the tank to make the pressure at the bottom atmospheric is (a) 9.8 m/s2 vertically upward (b) 9.8 m/s2 vertically downward (c) 8.33 m/s2 vertically upward (d) 8.33 m/s2 vertically downward An open tank contains a liquid and is made to slide down an inclined plane with uniform velocity. The free surface of the liquid (a) will be parallel to the plane of the inclined plane (b) will be horizontal (c) will be inclined to the horizontal at an angle of tan–1 1/g (d) will be inclined to the horizontal at an angle which depends upon the slope of the inclined plane 2.59 An open tank containing a liquid slides down a frictionless inclined plane. The free surface of the liquid will be (a) inclined to the horizontal at an angle which depends upon the slope of the inclined plane (b) horizontal (c) parallel to the plane of the inclined plane (d) inclined at an angle of (tan–1 1/g) to the horizontal 2.60 A U-tube shown in Fig. 2.106 is closed at A and open at B. It is ﬁlled with water. The acceleration to the right needed to make the pressure at C atmospheric is 1 g (b) g (a) 2 3 (c) 2g (d) g 2 A 50 cm 50 cm 50 cm Fig. 2.106 C Question 2.60 2.61 An open cubical tank was initially ﬁlled with water. When the tank was accelerated 103 Fluid Statics 2.62 2.63 2.64 2.65 2.66 on a horizontal plane along one of its sides it was found that one third of the volume of water spilled out. The acceleration was 1 1 g (b) g (a) 3 2 2 g (d) g (c) 3 An open cylindrical vessel ﬁlled with a liquid is falling freely with an acceleration g. The absolute pressure at any point in the liquid is (a) zero (b) above atmospheric pressure (c) below atmospheric pressure (d) equal to atmospheric pressure Oil of density 800 kg/m3 stands within 0.8 m from the top of an open tank 2.2 m high, 3.0 m wide and 8 m long. The greatest acceleration that this tank can have without any spilling over is (a) 9.81 m/s2 (b) 0.98 m/s2 (c) 1.96 m/s2 (d) 5.23 m/s2 A liquid in a circular container is given a rigid body rotation about the axis of the cylinder. The piezometric line in a cross section is (a) a horizontal line (b) a circular arc (c) a parabola (d) a vertical line An open cylindrical tank with its axis vertical is 0.8 m high and is 0.8 m in diameter. It is ﬁlled with an oil of density 800 kg/m3 and is rotated at 120 rpm about the axis of the cylinder. The gauge pressure at the centre of the bottom of the tank is (a) –0.49 m of oil (b) zero (c) 1.29 m of oil (d) 6.27 kPa A liquid in an open right circular cylinder is given rigid body rotation about the axis of the cylinder. The pressure distribution (a) in any horizontal plane is uniform 2.67 2.68 2.69 2.70 2.71 (b) in any vertical plane is uniform (c) on the bottom of the tank is uniform (d) in any vertical plane is hydrostatic An open circular cylinder 1.2 m high is ﬁlled with a liquid to its top. The liquid is given a rigid body rotation about the axis of the cylinder and the pressure at the centre of the bottom is found to be 0.3 m of liquid. The ratio of the volume of liquid spilled out of the cylinder to the original volume is (a) 3/8 (b) 3/4 (c) 1/2 (d) 1/4 A liquid undergoing a rigid body rotation in a container is said to have (a) circulatory ﬂow (b) circulation (c) forced vortex motion (d) free vortex motion A 20 cm diameter open cylindrical container contains kerosene (relative density = 0.80) to a height of 20 cm. It is rotated about a vertical axis coinciding with the axis of the cylinder. If the bottom of the cylinder at the axis is just exposed, the speed of rotation is (a) 94.6 rpm (b) 267.5 rpm (c) 133.8 rpm (d) 535.0 rpm If a cylinder containing a liquid is rotated about a vertical axis coinciding with the axis of the cylinder, the pressure in a vertical (a) decreases as depth (b) increases as depth (c) decreases as square root of depth (d) increases as square root of depth A right circular cylinder is open at the top and ﬁlled with a liquid to its top level. It is rotated about its vertical axis at such a speed that half of the liquid spills out. The pressure at the point of intersection of the axis and the bottom is (a) same as before rotation (b) 1/2 the value before rotation 104 Fluid Mechanics and Hydraulic Machines (c) 1/4 the value before rotation (d) equal to the atmospheric pressure 2.72 For a U-tube containing water and rotating about an axis as in Fig. 2.107 the difference in water surface elevations (ha – hb) is (a) 6.04 cm (b) 1.51 cm (c) 3.02 cm (d) 24.15 cm A 10 cm 10 cm Axis 20 cm B Axis O 120 rpm ha Fig. 2.108 hb 20 cm 10 cm 60 rpm Fig. 2.107 Question 2.72 2.73 The U-tube shown in Fig. 2.108 contains water and is rotated at 120 rpm, about a vertical axis passing through the midpoint O of the horizontal limb. The pressure head at O is (a) 8.05 cm (b) 11.95 cm (c) 20.0 cm (d) 6.35 cm Question 2.73 2.74 A 25 cm long pipe closed at one end (A) is ﬁlled with water and then the other end (B) is capped. It is then placed in a horizontal position and rotated at 30 rad./s about a vertical axis passing through the capped end, (Fig. 2.109). The difference of pressure between the two ends is (a) 28.1 kPa (b) 5.3 kPa (c) 65.1 kPa (d) 113.2 kPa Axis 25 cm B A 30 rad/s Fig. 2.109 Question 2.74 Fluid Flow Kinematics Concept Review 3 Introduction stream line path line streak line time line 3.1 CLASSIFICATION OF FLOW (A) Steady Flow: Fluid ﬂow conditions at any point do not change with time. For example ∂V ∂p ∂r = 0, = 0, =0 ∂t ∂t ∂t In a steady ﬂow steam line, path line and streak line are identical. Flow parameters at any point ∂V π 0. change with time, e.g. ∂t Unsteady Flow: 106 Fluid Mechanics and Hydraulic Machines (B) Uniform Flow: The velocity vector V is identically same at all points at a given instant. The velocity vector V at any instant varies from point to point. Non-Uniform Flow: hq V + hV Stream line s 3.2 STREAMLINE In a ﬂuid ﬂow, a continuous line so drawn that it is tangential to the velocity vector at every point is known as a streamline. If the velocity vector V = iu + jv + kw then the differential equation of a streamline is given by dx dy dz = = u v w (3.1) 3.2.1 Stagnation Point A point of interest in the study of the kinematics of ﬂuid is the occurrence of points where the ﬂuid ﬂow stops. When a stationary body is immersed in a ﬂuid, the ﬂuid is brought to a stop at the nose of the body. Such a point where the ﬂuid ﬂow is brought to rest is known as the stagnation point. Thus, a stagnation point is deﬁned as a point in the ﬂow ﬁeld where the velocity is identically zero. This means that all the � components of the velocity vector V , viz., u, v, and w are identically zero at the stagnation point. Pitot tube (Sec. 13.5) which is used to measure the velocity in a ﬂuid ﬂow is an example where the properties of the stagnation point are made use. 3.3 ACCELERATION Acceleration is a vector. (i) In the natural co-ordinate system, viz., along and across a streamline (Fig. 3.1). dV and a = as2 + an2 a= dt In the tangential direction: as = ∂Vs ∂Vs + Vs ∂t ∂s (3.2) V hV hVs Fig. 3.1 In the normal direction ∂Vn Vs2 + (3.4) ∂t r where r = radius of curvature of the streamline at the point, Vs = tangential component of the velocity V and Vn = normal component of velocity generated due to change in direction. The terms ∂Vs/ ∂t and ∂Vn /∂t ∂V are called local accelerations. Also Vs s ∂s = tangential convective acceleration and V s2/r = normal convective acceleration. (ii) In Cartesian co-ordinates: an = V = iu + jv + kw Acceleration ax , ay and az in the x, y, z directions respectively are: ∂u ∂u ∂u ∂u +u +v +w (3.5) ax = ∂t ∂x ∂y ∂z ∂v ∂v ∂v ∂v +u +v +w (3.6) ay = ∂t ∂x ∂y ∂z az = ∂w ∂w ∂w ∂w +u +v +w (3.7) ∂t ∂x ∂y ∂z (iii) In two-dimensional (Fig. 3.2) polar co-ordinates ∂ vr ∂v v ∂ vr vq2 + vr r + q (3.8) ∂t ∂r r ∂q r ∂ vq ∂v v ∂ vq vr vq + vr q + q + (3.9) aq = ∂t ∂r r ∂q r ar = (3.3) hVn hq 107 Fluid Flow Kinematics V Y v Vq v V 2 P(r, q) r O 1 q Fig. 3.4 3.4.2 X In Differential Form Cartesian co-ordinates: Fig. 3.2 ∂ r ∂( r u) ∂( r v ) ∂( r w ) + + + = 0 (3.13) ∂t ∂x ∂y ∂z For incompressible ﬂuid (dr/d t) = 0) and hence Eq. 3.13 is simpliﬁed as 3.4 CONTINUITY EQUATION 3.4.1 In One-dimensional Analysis ∂u ∂ v ∂ w + + =0 ∂x ∂ y ∂z In steady ﬂow, mass rate of ﬂow into stream tube is equal to mass rate of ﬂow out of the tube r1 A1V1 = r2A2V2 (3.14) (3.10) (i) For incompressible ﬂuid, under steady ﬂow (Fig. 3.3). A1V1 = A2 V2 (3.11) 3.5 ROTATIONAL AND IRROTATIONAL MOTION Consider a rectangular ﬂuid element of sides dx and dy [(Fig. 3.5(a)]. Under the action of velocities acting on it let it undergo deformation as shown in Fig. 3.5(b) in a time dt. Stream tube V2 2 V1 Fig. 3.3 (ii) When there is a variation of velocity across the cross section of a conduit, for an incompressible ﬂuid discharge, (Fig. 3.4) Ú A1 vd A = ∂v ∂x g 2 = angular velocity of element AD = ∂u ∂y Considering the anticlockwise rotation as positive, the average of angular velocities of two mutually perpendicular elements is deﬁned as the rate of rotation. Thus rotation about z-axis 1 Q= g 1 = angular velocity of element AB = Ú A2 vd A (3.12) wz = 1 Ê ∂ v ∂u ˆ 2 ÁË ∂ x ∂ y ˜¯ (3.15) Thus for a three-dimensional ﬂuid element, three rotational components as given in the following are possible: 108 Fluid Mechanics and Hydraulic Machines u+ D ∂u dy ∂y v+ C ∂v dx ∂x u+ ∂u dy ∂y i.e. ∂ v ∂u ∂u ∂ w = 0, = 0; ∂x ∂y ∂z ∂x and ∂w ∂v =0 ∂y ∂ z dy v+ v A u Thus for a two-dimensional irrotational ﬂow ∂v dx ∂x wz = B dx Ê ∂ v ∂u ˆ ÁË ∂ x - ∂ y ˜¯ = 0 (a) or ∂u dydt ∂y 3.5.1 (3.17) Circulation In rotational ﬂuid motion, circulation is very useful concept. Circulation is deﬁned as the line integral of the tangential component of the velocity taken around a closed contour (Fig. 3.6). The limiting value of circulation divided by the area of the closed contour, as the area tends to zero, is the vorticity along an axis normal to the area. g2 g1 1 Ê ∂ v ∂u ˆ =0 2 ÁË ∂ x ∂ y ˜¯ ∂v dxdt ∂x (b) Fig. 3.5 Y C 1 Ê ∂ v ∂u ˆ ¸ About z axis, w z = Á Ô 2 Ë ∂ x ∂ y ˜¯ Ô 1 Ê ∂ u ∂ w ˆ ÔÔ About y axis, w y = Á ˝ 2 Ë ∂ z ∂ x ˜¯ Ô 1 Ê ∂ w ∂ v ˆ ÔÔ About x axis, w x = Á 2 Ë ∂ y ∂ z ˜¯ Ô˛ ∂u u + dy ∂y (3.16) Fluid motion with one or more of the terms w x, wy or wz different from zero is termed rotational motion. Twice the value of rotation about any axis is called as vorticity along that axis. Thus the equation Ê ∂ v ∂u ˆ . for vorticity along z-axis is z z = 2wz = Á Ë ∂ x ∂ y ˜¯ A ﬂow is said to be irrotational if all the components of rotation are zero, viz. wx = wy = wz = 0, v dy dx u G v+ ∂v dx ∂x ds a V X Fig. 3.6 Circulation Concept Circulation is taken as positive in anticlockwise direction. Referring to Fig. 3.6 � � G = V ◊ dS �Ú C = �Ú (udx + vdy + wdz) C For two-dimensional ﬂow G= �Ú V cos a ds C = �Ú (udx + vdy) C 109 Fluid Flow Kinematics G = area of closed curve C Vorticity along the axis perpendicular to the plane contaiining the closed curve C. 3.6 STREAM FUNCTION ∂y ∂y and v = – (3.18) ∂y ∂x The stream function y is deﬁned as above for two dimensional ﬂows only. ∂ v ∂u = 0 and hence, For an irrotational ﬂow, ∂x ∂ y u= ∂ 2y ∂ x2 In polar coordinates vr = 3.7 In a two-dimensional ﬂow consider two streamlines S1 and S2. The ﬂow rate (per unit depth) of an incompressible ﬂuid across the two streamlines is constant and is independent of the path, (path a or path b from A to B in Fig. 3.7). A stream function y is so deﬁned that it is constant along a streamline and the difference of ys for the two streamlines is equal to the ﬂow rate between them. Thus yB – yA = ﬂow rate between S1 and S2. The ﬂow from left to right is taken as positive, in the sign convention. The velocities u and v in x and s directions are given by – is satisﬁed by the stream function in irrotational ﬂow. Conversely, if y does not satisfy — 2y = 0, then the ﬂow is rotational. - ∂ 2y =0 ∂ y2 That is, the Laplace equation ∂ 2y ∂x 2 + ∂ 2y ∂ y2 =0 (3.19) B yB > yA a S2 A S1 Fig. 3.7 POTENTIAL FUNCTION In irrotational ﬂows, the velocity can be written as a gradient of a scalar function f called velocity potential. u= ∂f ∂f ∂f ,v= and w = ∂x ∂y ∂z (3.20) Considering the equation of continuity (Eq. 3.14) for an incompressible ﬂuid, ∂u ∂ v ∂ w + + =0 ∂x ∂y ∂z and substituting the expressions for u, v and w in terms of f, — 2f = ∂ 2f + ∂ 2f + ∂ 2f = 0 (3.21) ∂x ∂y ∂ z2 Thus the velocity potential satisﬁes the Laplace equation. Conversely, any function f which satisﬁes the Laplace equation is a possible irrotational ﬂuid ﬂow case. Lines of constant f are called equipotential lines and it can be shown that these lines will form orthogonal grids with y = constant lines. This fact is used in the construction of ﬂow nets for ﬂuid ﬂow analysis. 2 2 È Í Note : Some authors define f such that Î u=- b 1 ∂y ∂y and vq = – r ∂q ∂r ∂f ∂f ∂f ˘ ,v=and w = ˙ ∂x ∂y ∂z ˚ 3.8 RELATION BETWEEN y AND f FOR 2-DIMENSIONAL FLOW f exists for irrotational ﬂow only. 110 Fluid Mechanics and Hydraulic Machines ∂ x2 ∂ 2y ∂ x2 ∂f ∂y =∂y ∂x (3.23) + + ∂ 2f ∂ y2 ∂ 2y ∂ y2 =0 (3.21(a)) ELEMENTARY INVISCID PLANE FLOWS Since the Laplace equation is linear, several interesting potential ﬂow situations can be constructed by using elementary solutions and method of superposition. The basic ﬂow types are Uniform ﬂow, Source, Sink and Vortex. These are brieﬂy described below. 3.10.1 Uniform Flow A stream of constant velocity U in x-direction is shown in Fig. 3.8 and has y = Uy =0 (3.19) y = constant along a streamline. f = constant along an equipotential line which is normal to streamlines. y = U r sin q U (3.31) y f = U r cos q and (3.32) y = 3Uh h y = 2Uh h Equation of continuity: 1 ∂ 1 ∂ (rVr ) + (rVr ) + (rVq ) = 0 r ∂r r ∂q (3.24) For incompressible ﬂuid ﬂow: Laplace equation f = Ux In polar coordinates 3.9 SOME COMMON FORMULAE IN CYLINDRICAL CO-ORDINATES Vr ∂Vr 1 ∂Vq =0 + + r r ∂q ∂r 2. Stream function y: 1 ∂y Vr = r ∂q ∂y Vq = – ∂r 3. Potential function f: ∂f Vr = ∂r 1 ∂f Vq = r ∂q and f = 3Uh ∂ 2f (3.22) f = 2Uh v= 3.10 ∂f ∂y = ∂ x ∂y f = Uh u= (3.25) (3.26) (3.27) (3.28) (3.29) 1 ∂f ∂ 2f 1 ∂ 2f + 2 + 2 =0 r ∂r ∂r r ∂q 2 (3.30) y = Uh h 0 h Fig. 3.8 3.10.2 h h x Uniform Flow Line Source and Sink A two-dimensional ﬂow emanating from a point in the x-y plane and imagined to ﬂow uniformly in all directions is called a source. Since the twodimensional source is a line in the z-direction, it is known as a line source. The total ﬂow per unit time per unit length of the line source is called the strength m of the source. The velocity at a radial distance r from the source is m (3.33) vr = r The stream function y and the potential function f for such line source is given by y = mq and f = m ln r (3.34) 111 Fluid Flow Kinematics Figure 3.9 shows a source ﬂow. y= y= f= mp 2 3mp 4 Kp 2 f= y= mp 4 Kp 4 y = –K ln(r1) f = mln(r1) y = mp f=0 y=0 y = –K ln(r2) f = mln(r2) Fig. 3.11 Line Vortex 3.10.4 Two-Dimensional Doublet Fig. 3.9 Line Source A sink is a negative source. It is a line in z-direction into which ﬂuid ﬂows radially in x-y plane (Fig. 3.10). –mp y= 2 –3mp y= 4 The limiting case of a line source approaching a line sink of equal strength while keeping constant the product of their strength and the distance between them (l) is known as a two dimensional doublet. For a doublet. y doublet – –mp y= 4 (l y) 2 2 (x + y ) =– f doublet = f = mln(r1) y = –mp y=0 f = m ln(r2) l sin q r lx 2 2 (x + y ) (3.37) = l cos q (3.38) r Figure 3.12 shows the streamlines and equipotential lines in a doublet. y y = C3 Fig. 3.10 3.10.3 Streamlines Line Sink y = C2 Velocity potential lines Line Vortex Suppose we reverse the role of y and f in Fig. 3.9 yielding y = – K ln r and f = K q (3.35) from which we get vr = 0 and v0 = K/r representing a circulating ﬂow (Fig. 3.11). Such a ﬂow is known as line vortex and K in Eq. 3.35 is known as Vortex strength. The centre of the vortex is a singular point and the circulation G of the vortex around a circular path about the centre is given by G = 2pK (3.36) x Fig. 3.12 Doublet 112 Fluid Mechanics and Hydraulic Machines 3.10.5 Other Inviscid Flows Using the basic ﬂow elements described above various ﬂow situations can be created by the method of superposition. A few examples are given below in Table 3.1. Table 3.1 Some Ideal Fluid Flow Simulations Sl. No. Name 1. Rankine half body 2. Rankine oval 3. Circular cylinder 4. Rotating circular cylinder Combination (and Flow description) Equation of Stream function Source + uniform ﬂow [curved, roughly elliptical half body] Source + sink + Uniform ﬂow [cylindrical oval shaped body] y = Ur sin q + mq Uniform ﬂow + doublet [circular cylinder] Uniform ﬂow + doublet + vortex [rotating circular cylinder] y = Ur sin q – l sin q r y = Ur sin q – l sin q – K ln r r y = Ur sin q + m(q 1 – q2 ) Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples Constant emission velocity 3.1 Ve 10 cm V1 Porous pipe V2 Ve 2.0 m 1 Solution: A = area of pipe cross section 2 Fig. 3.13 113 Fluid Flow Kinematics p ¥ (0.1)2 = 7.854 ¥ 10–3 m2 4 Q1 = inlet discharge = V1 A = 2.0 ¥ 7.854 ¥ 10–3 = 0.01571 m3/s Q2 = outlet discharge = V2 A = 1.2 ¥ 7.854 ¥ 10–3 = 0.0094248 m3/s = (i) Q e = discharge emitted through walls of the porous pipe = Q1 – Q2 = 0.01571 – 0.0094248 = 0.0052852 m3/s (ii) Surface area of emission Ae = pDL = p ¥ 0.1 ¥ 2.0 = 0.6283 m2 Q Ve = Velocity of emission = e Ae = = 2um 1 when R È 1 1 ˘ = 2u m Í ˙ Î ( n + 1) ( n + 2) ˚ Ê 1 1 ˆ n = 1/5, V = 2u m Á Ë 1/ 5 + 1 1/ 5 + 2 ˜¯ = 2u m ¥ when 25 25 = um 66 33 Ê 1 1 ˆ n = 1/2, V = 2u m Á Ë 1/ 2 + 1 1/ 2 + 2 ˜¯ = 2u m ¥ 4/15 = 8/15 um rˆ Ê u = um Á - ˜ Ë R¯ r/ R=0 n r Ê rˆ Ê rˆ 1 - ˜ d Á ˜ = 2u m R ÁË R¯ Ë R¯ n+ 2 n +1 È 1 1 r Ê rˆ ˘ Ê rˆ ˙ 1 1 ¥ ÍÁ ˜ ( n + 1) R ÁË R ˜¯ ˙ ÍÎ ( n + 1) ( n + 2) Ë R ¯ ˚0 6.283 ¥ 10 -3 = 0.01 m/s 0.6283 3.2 Ú r / R =1 3.3 n um r=R n n pipe? Solution: Refer to Fig. 3.14 12 cm Dia dr R r 25 mm Vr Va Vr 30 cm Dia Fig. 3.15 Fig. 3.14 Solution: Discharge At the outlet Area If average velocity = V Then V◊ pR2 = Q = Ú R 0 V= Ú R 0 u ◊ 2p r ◊ d r n um ◊ 2 p r Ê rˆ ÁË1 - R ˜¯ d r 2 pR Q = 0.08 m3/s A = p ¥ 0.30 ¥ 0.025 ¥ 0.95 = 0.02238 m2/s Radial velocity at the edge of the impeller = Q/A = 0.08/0.02238 = 3.575 m/s 114 Fluid Mechanics and Hydraulic Machines At the inlet, Va . A a = 0.08 Y where Va is the axial velocity, at the inlet pipe and Aa is the area of the inlet. 0.08 = 7.074 m/s Va = p (0.12) 2 4 B A U0 U0 (h – d ) D E U0 h u BL U0 3.4 O y d y C X Impervious x Fig. 3.17 V r Moving plate yo Vo Vr r Fixed plate Fig. 3.16 Example 3.4 Solution: In time Dt, at a radial distance r, volume displaced by the upper plate = V0 ¥ p r 2 = Radial out ﬂow = Vr ¥ 2p r ¥ y0 where Vr is the ﬂuid velocity at a radial distance r. V p r2 V r = 0 Vr = 0 2 p r y0 2 y0 As the ﬂow is radially outwards, acceleration a r = Vr V02 r ∂Vr V r Ê V ˆ = 0 Á 0 ˜ = ∂r 2 y0 Ë 2 y0 ¯ 4 y02 rU0d – Ú d 0 ru dy = rQAB = mass rate across AB. rQAB = rU0d – r Ú d 0 3 Ê y ˆ Ê yˆ U 0 Á 2 - 2 Á ˜ + ( y / d ) 4 ˜ dy Ëd¯ Ë d ¯ È 2 d2 2 d4 1 d5 ˘ = rU0 Íd + 3 ˙ d 2 d 4 5 d 4 ˙˚ ÍÎ 3 2 1ˆ Ê rU0d = rU0d Á1 - 1 + - ˜ = Ë 10 4 5¯ = Mass rate of outﬂow across AB per unit width. 3.6 3.5 u = h – h + h4 U d x Solution: By continuity equation for a volume ABCO: Mass rate of ﬂow across AO – Mass rate across BC = Mass rate of ﬂow across AB. Since across section BC, u = U0 for y ≥ d, the mass rate across AD = rU0(h – d) = mass rate across BE = rU0(h – d). Hence, per unit width ∂Q ∂y +T ∂x ∂t h=y d T Q t 115 Fluid Flow Kinematics Solution: Refer to Fig. 3.18. ∂Q ∂y +T =0 ∂x ∂t or Water surface at t 3.7 W. S. at (t + ht) Q1 y Q2 hx 1 2 30 (a) 100 T dA = Tdy y ? 20 B dy Consider section 1 and 2 D x apart. Let Q2 > Q1 at any instant t. Then ∂Q . Dx Q2 – Q1 = ∂x In a time interval Dt, volume rate of excess outﬂow ∂Q D x Dt. If the top width of the over the inﬂow = ∂x channel is T at a depth y, the water surface will drop ∂y Dt and the decrease in storage between by Dy = ∂t sections 1 and 2 is ∂y Dt Dx ∂t By continuity consideration, ∂Q ∂y D x Dt = - T Dt Dx ∂x ∂t ? ? D Fig. 3.19(a) Fig. 3.18 ? ? 50 (b) F ? 70 A – DS = – DA ¥ Dx = – TD y Dx = - T A C (a) 80 E 90 Example 3.7 Solution: By continuity criterion the ﬂow entering into a node must be equal to the ﬂow going out of the node. Thus by considering ﬂow into a node as positive, the algebraic sum of discharges at a node is zero. Thus at node A: 100 – 70 – QAB = 0 or QAB = 30 and QAB is from A to B. At node D: 70 + 50 – QDC = 0 QDC = 120 and QDC is from D to C. At node C: 120 – 80 – QCB = 0 QCB = 40 and QCB from C to B. At node B: 30 + 40 – 30 – QBE – 20 = 0 QBE = 20 and QBE is from B to E. At node E: 80 + 20 – QEF – 90 = 0 QEF = 10 and QEF is from E to F. At node F: 20 + 10 – QF = 0 QF = discharge out of node F = 30. 116 Fluid Mechanics and Hydraulic Machines is the differential equation of the streamline. The distribution of discharges are as in Fig. 3.19(b). Ú 6 x dx = Ú y 30 A 100 30 B 20 F 3x 2 - 30 10 D 120 Fig. 3.19(b) C 80 (b) Answer E 90 3x 2 - y3 8 = 3 3 9x2 – y3 = 8 or Example 3.7-Answer It can be seen now that at each node the continuity equation is satisﬁed. Putting x = 1, y = 1 1 2 =2 3 3 Hence the equation of the required streamline is 40 50 dy + c c = 3- 20 70 y3 = c. 3 2 3.9 V xi + 4yj – zk 3.8 V xi – yj V = – y i – xj Solution: (i) u = 3x and v = – 3y The equation of a streamline in twodimensional ﬂow is dx dy = u v Here dx dy = . On integration 3x 3y 1 1 1 ln x = - ln y + ln c 3 3 3 where c = a constant ln xy = ln c or xy = c For the streamline passing through (1, 1), c = 1 and hence the required streamline equation is xy = 1. (ii) u = – y2 and v = – 6x dx dy - 2 =6x y Solution: The equation of the streamline is dx dy d z = = u v w Here u = 3x, v = 4y and w = – 7z d x dy dz Hence = =3x 4 y 7z Considering equations involving x and y, on integration 1 1 ln x = ln y + ln C1¢ 3 4 where C 1¢ = a constant or y = C1 x 4/3 (i) where C1 is another constant. Similarly, by considering equations with x and z and on integration 1 1 ln x = - ln z + ln C2¢ 3 7 where C2¢ = a constant C z = 7 2/ 3 (ii) x where C2 is another constant. 117 Fluid Flow Kinematics Putting the coordinates of the point M (1, 4, 5) 4 from Eq. (i) C1 = =4 (1) 4 / 3 3.11 u =x+y+ v =x–y– C2 = 5 ¥ 17/3 from Eq. (ii) The streamline passing through M is given by y = 4x4/3 and z = 5/x7/3 3.10 x y=– V V = 4xi z t t + xy i – yz – t j – yz + z tj – 4y Solution: (i) V = ui + vj + wk and A (x = 2, y = – 3, z = 1, t = 2) Here u = 10t + xy ﬁ ua = (10 ¥ 2) + (2 ¥ (– 3)) = 14 v = – yz – 10t ﬁ va = – (– 3 ¥ 1) – (10 ¥ 2) = – 17 w = – yz + z2/2 ﬁ wa = – (– 3 ¥ 1) + (1)2/2 = 3.5 Magnitude of velocity at A = VA = = (14) 2 + ( - 17) 2 + (3.5) 2 (ii) V = 4xi + (– 4y + 3t)j + (0)k At A (x = 2, y = – 3, z = 1, t = 2) u = 4x ﬁ ua = 4 ¥ 2 = 8 v = – 4y + 3t ﬁ va = (– 4) ¥ (– 3) + (3 ¥ 2) = 18 w =0 ﬁ wa = 0 Magnitude of velocity at A = = Solving for x and y, x = 0.5 and y = – 1.5. Thus the stagnation point occurs at (0.5, – 1.5). 3.12 L xˆ Ê V = 2t Á 1 ˜ Ë 2L ¯ V t ua2 + va2 + wa2 = 22.3 units VA = k Solution: At the stagnation point u = 0 and v = 0 Hence u = x + y + 1.0 = 0 and v =x–y–2=0 Thus x + y = –1 and x–y =2 ua2 + va2 + wa2 82 + 182 = 19.7 units 2 t x x L Solution: (i) Local acceleration = ∂V x ˆ Ê = 2 Á1 Ë ∂t 2 L ˜¯ 2 at t = 3 s and x = 0.5 m, Ê ∂V 0.5 ˆ = 2 Á1 ∂t 2 ¥ 0.8 ˜¯ Ë 2 = 0.945 m/s2 (ii) Convective acceleration = V 2 ∂V ∂x x ˆ x ˆÊ 1 ˆ Ê Ê = 2t Á1 ◊ 2t ◊ 2 Á 1 ˜ Ë Ë 2L ¯ 2 L ˜¯ ÁË 2 L ˜¯ = - x ˆ 4t 2 Ê 1Á Ë L 2 L ˜¯ 3 118 Fluid Mechanics and Hydraulic Machines At t = 3 s and x = 0.5 m Convective acceleration =– V = Velocity = Ê 0.5 ˆ ÁË1 - 2 ¥ 0.8 ˜¯ 4 ¥ 32 0.8 3 = – 14.623 m/s2 (iii) Total acceleration = (local + convective) acceleration = 0.945 – 14.623 = – 13.68 m/s2 3.13 Solution: Refer to Fig. 3.20. 1 L = 2.0 m X 20 cm ∂V = rate of increase of velocity ∂t ∂ (Q / A) 1 ∂Q = = ∂t Ax ∂t 1 = (0.050) = 0.1507 m/s2 0.3318 = local acceleration at XX ∂V (ii) Convective acceleration as = Vx x ∂x Q Q Vx = = p Ax (0.3 x + 0.2) 2 4 0.20 = p (0.3 x + 0.2) 2 4 = 0.2546/(0.3x + 0.20)2 (i) ∂Vx = (0.2546) (– 2) (0.3x + 0.20)–3 (0.3) ∂x = – 0.15276/(0.3x + 0.20)3 Hence convective acceleration 2 Vx as = Vx 80 cm X x Fig. 3.20 Diameter at section XX Ê D - D1 ˆ Dx = Á 2 ◊ x + D1 Ë L ˜¯ Ê 0.8 - 0.2 ˆ = Á ˜¯ ◊ x + 0.20 Ë 2 ∂Vx ∂x Ê - 0.15276 ˆ ¥Á ˜ (0.3 x + 0.2) Ë (0.3 x + 0.20)3 ¯ 0.03889 = (0.3 x + 0.2)5 At x = 1.5 m, convective acceleration 0.03889 = – 0.3352 m/s2 as = 5 (0.3 ¥ 1.5 + 0.2) (iii) Total acceleration = local acceleration + convective acceleration ∂Vx ∂V + Vx x = ∂t ∂t At x = 1.5 m, total acceleration = 0.1507 – 0.3352 = – 0.1845 m/s2 = = 0.3x + 0.20 At x = 1.5 m, D = 0.65 m A = area at section XX p = ¥ (0.65)2 = 0.3318 m2 4 Q 0.200 = 0.6027 m/s = A 0.3318 0.2546 2 119 Fluid Flow Kinematics ay = 4 + (5 ¥ 22) + (15 ¥ 3) = 69 3.14 a = u v x y– x– y– ∂u ∂u +v ∂x ∂y = (2x + 3y – 5) ¥ (2) + (5x – 2y – 9) ¥ (3) = (19x – 37) At point (1, 2) ax = (19 ¥ 1 – 37) = –18 units ∂v ∂v ay = u +v ∂x ∂y = (2x + 3y – 5) ¥ (5) + (5x – 2y – 9) ¥ (4) = (30x + 7y – 61) At point (1, 2), ay = (30 ¥ 1) + (7 ¥ 2) – 61 = 17 units 3.16 V xt + yz i ax = u Acceleration a = = ( - 18) 2 + (17) 2 613 = 24.76 units 3.15 u =t v = 4t y x t ax = ∂u ∂u ∂u +u +v ∂t ∂x ∂y = 2t + (t2 + 3y)(0) + (4t + 5x)(3) = 14t + 15x ∂v ∂v ∂v +u +v ∂t ∂x ∂y = 4 + (t2 + 3y)(5) + (4t + 5x) (0) ay = = 4 + 5t 2 + 15y At point (5, 3) ax = (14 ¥ 2) + (15 ¥ 5) = 103 t + xy j xy – xyz – tz k t Solution: (i) V = (6xt + y z 2 )i + (3t + xy 2 )j + (xy – 2xyz – 6tz)k = ui + vj + wk ∂u = 6t u = 6xt + yz2; ∂x ∂v v = 3t + xy2; = 2 xy ∂y ∂w w = xy – 2xyz – 6tz; = – 2xy – 6t ∂z Ê ∂u ∂ v ∂ w ˆ ÁË ∂ x + ∂ y + ∂ z ˜¯ = 6t + 2xy – 2xy – 6t = 0 Hence the continuity equation is satisﬁed. (ii) Acceleration a = a x i + ay j + azk ∂u ∂u ∂u ∂u +u +v +w ∂t ∂x ∂y ∂z = 6x + (6xt + yz 2 ) (6t) + (3t + xy2) (z 2) + (xy – 2xyz – 6tz) (2yz) At point A (1, 1, 1) and at t = 1, ax = 6 + (6 + 1) (6) + (3 + 1) (1) + (1 – 2 – 6)(2) = 38 units ∂v ∂v ∂v ∂v +u +v +w ay = ∂t ∂x ∂y ∂z = 3 + (6xt + yz2) (y2) + (3t + xy2) (2xy) + (xy – 2xyz – 6tz) (0) ax = Solution: (103) 2 + (69) 2 = 123.97 units Solution: ( ax ) 2 + ( a y ) 2 = 120 Fluid Mechanics and Hydraulic Machines At point A (1, 1, 1) and at t = 1 ay = 3 + (6 + 1)(1) + (3 + 1) (2) = 18 units Similarly ∂w ∂w ∂w ∂w +u +v +w az = ∂t ∂x ∂y ∂z = – 6 z + (6xt + yz 2 ) (y – 2yz) + (3t + xy 2) ¥ (x – 2xz) + (xy – 2xyz – 6tz) (– 2xy – 6t) At point A (1, 1, 1) and at t = 1 az = – 6 + (6 + 1) (1 – 2) + (3 + 1) (1 – 2) + (1 – 2 – 6) (– 2 – 6) = – 6 – 7 – 4 + 56 = 39 units Hence at A (1, 1, 1) and at t = 1, a = 38i + 18j + 39k ∂v - ( x 2 + y 2) + 2 y 2 = ∂y ( x 2 + y 2 )2 ∂u ( x 2 - y 2) = 2 ; ∂ x ( x + y 2 )2 ∂v (- x 2 + y 2) = ∂ y ( x 2 + y 2 )2 ∂u ∂ v x 2 - y 2 - x 2 + y 2 =0 + = ∂x ∂y ( x 2 + y 2 )2 Hence, the ﬂow is possible. \ 3.18 3.17 u v u = 4xy + y v xy x u x + y v = – 4xy u=–x x +y v=–y x +y Solution: For steady, incompressible ﬂow the following continuity equation must be satisﬁed: ∂u ∂ v + =0 ∂ x ∂y u = cx v = – cy u = A sin xy v = – A sin xy u x + zy y zy v = – xy 3 w = - z – xy – yz 2 u = – cx y v = c xy u=x+y v=x–y Solution: (a) u = cx; v = – cy (i) u = 4xy + y2; v = 6xy + 3x ∂u ∂v = 4y; = 6x ∂x ∂y ∂u ∂ v \ + = 4y + 6x π 0 ∂x ∂y Hence the ﬂow is not possible. ∂u ∂v = c; =–c ∂x ∂y ∂u ∂ v + =c–c=0 ∂x ∂y Hence, the continuity equation is satisﬁed. (b) u = – cx/y; v = c ln xy (ii) u = 2x2 + y2; v = – 4xy ∂u ∂v = 4x; = – 4x ∂x ∂y ∂u ∂ v + = 4x – 4x = 0 ∂ x ∂y Hence the ﬂow is possible. (iii) u = – x/(x2 + y2); v = – y/(x2 + y2) ∂u c ∂v = - c/ y; ◊ x = c/y = ∂x xy ∂y ∂u ∂ v + = – c/y + c/y = 0 ∂ x ∂y The continuity equation is satisﬁed. (c) u = A sin xy; v = – A sin xy ∂u ∂v = Ay cos xy; = – Ax cos xy ∂x ∂y ∂u - ( x 2 + y 2) + 2 x 2 = ; ∂x ( x 2 + y 2 )2 ∂u ∂ v π0 + ∂x ∂ y 121 Fluid Flow Kinematics Hence the continuity equation is not satisﬁed. (d) u = x + y ; v = x – y ∂u ∂v = 1; =–1 ∂x ∂y ∂u ∂ v + =1–1=0 ∂x ∂ y Hence the continuity equation is satisﬁed. ∂u = 4x (e) u = 2x2 + zy; ∂x v = – 2xy + 3y2 + 3zy; ∂v = – 2x + 9y 2 + 3z ∂y 3 w = - z 2 – 2xz – 6yz; 2 ∂w = – 3 z – 2x – 6y ∂z ∂u ∂ v ∂ w + + ∂x ∂y ∂z = 4x – 2x + 9y2 + 3z – 3z – 2x – 6y π0 The continuity equation is thus not satisﬁed. (c) u = – A ln (x/L) ∂u A 1 A ∂v = =- =∂x ( x / L) L x ∂y A v = d y = Ay/x + f (x) x (d) v = Axy Ú ∂v ∂u = Ax = – ∂y ∂x u = Ú – Ax dx = – A x2 + f (y) 2 3.20 u =ax+b y v =a x+b y a a b b so that the Hence 3.19 First, the continuity condition must be ∂u ∂ v + =0 ∂ x ∂y a1 + b2 = 0 i.e., u x +y v=? x u = - A ln L v=? u e v=? u=? xy v Solution: (a) u = A(x2 + y2) ∂u ∂v = 2Ax = – ∂x ∂y v= (b) u = Ae Solution: satisﬁed. x v= Ú - Ae d y = – Aex y + f (x) x …(i) Next the irrotational ﬂow condition must be satisﬁed. For this ∂ v ∂u = ∂ x ∂y a 2 = b1 ...(ii) The conditions given by equations (i) and (ii) must be satisﬁed to make the ﬂow ﬁeld a possible irrotational ﬂow ﬁeld. 3.21 u=x v Ú - 2Ax dy = – 2A xy + f (x) ∂u ∂v = Aex = – ∂x ∂y a1 = – b2 Solution: xy Refer to Fig. 3.21 122 Fluid Mechanics and Hydraulic Machines Y D (1,2) u C (2,2) 3 2 3 2 x - y 2 2 v=– x xy v = u=y A (1,1) Solution: The components of rotation about the various axes are: B (2,1) X wz = 1 Ê ∂ v ∂u ˆ 2 ÁË ∂ x ∂ y ˜¯ wx = 1 Ê ∂w ∂v ˆ 2 ÁË ∂ y ∂ z ˜¯ wy = 1 Ê ∂ y ∂w ˆ 2 ÁË ∂ z ∂ x ˜¯ Fig. 3.21 �Ú (udx + vdy + wdz) = Ú u dx + Ú G= + Ú x=2 y=2 y =1, x =1 x = 2 , y =1 vdy + Ú x =1 u dx y = 2, x = 2 y =1 v dy (i) u = xy3z; v = – y2 z2; w = yz 2 - x =1, y = 2 Along the line AB x=2 Ú x=2 È x3 ˘ È x3 ˘ 1 = [(8) - (1)] udx = Í ˙ =Í ˙ ÍÎ 3 ˙˚ y =1, x =1 ÍÎ 3 ˙˚ , x =1 3 y=2 y=2 = È - 2 y2 ˘ vd y = È - xy2 ˘ Î ˚ y =1, x = 2 Î ˚ , y =1 = [(– 8) – (–2)] = – 6.00 Along the line CD x =1 Ú x =1 È x3 ˘ È x3 ˘ 1 = Í ˙ = [(1) - (8)] ud x = Í ˙ Î 3 ˚ y = 2, x = 2 Î 3 ˚ , x = 2 3 = – 2.333 Along the line DA Ú vdy = ÈÎ x y 2˘ y =1 y =1 2 ˚ x =1, y = 2 = ÈÎ - y ˘˚ , y = 2 3.22 u = xy z v = – y z wx = 3 2 xy z 2 1 Ê ∂w ∂v ˆ 2 ÁË ∂ y ∂ z ˜¯ Ê 2 3 y2 z2 ˆ + 2 y2 z˜ Áz 2 Ë ¯ 1 Ê ∂u ∂ w ˆ 1 1 = (xy3 – 0) = xy 3 wy = Á ˜ 2 Ë ∂z ∂x ¯ 2 2 = 1 2 3 2 3 2 x - y 2 2 1 Ê ∂ v ∂u ˆ 1 = (3 x - 3 x ) = 0 wz = Á 2 Ë ∂ x ∂ y ˜¯ 2 (ii) u = 3xy; v = = [(–1) – (– 4)] = +3.00 Circulation G = 2.333 – 6.0 + 2.333 + 3.00 = 3.0 units 1 Ê ∂ v ∂u ˆ 1 = (0 – 3xy 2 z) 2 ÁË ∂ x ∂ y ˜¯ 2 = – = + 2.333 Along the line BC Ú wz = y3 z 2 2 As the ﬂow is two-dimensional in the x–y plane, wx = wy = 0 (iii) u = y2; v = – 3x wz = y 3 z2 w = yz 2 1 Ê ∂ v ∂u ˆ 1 = (– 3 – 2y) 2 ÁË ∂ x ∂ y ˜¯ 2 As the ﬂow is two-dimensional in x–y plane, wx = wy = 0. 123 Fluid Flow Kinematics ∂2 y 3.23 ∂ x2 x u = Ax + By y-component v y Solution: u = Ax2 + By ∂u ∂v = 2 Ax = By continuity equation ∂x ∂y v = – 2Axy + C Since v = 0 at y = 0, The constant C = 0. Hence v = – 2 Axy ∂v ∂u = - 2 Ay and =B ∂x ∂y ∂ v ∂u Since π 0, ∂x ∂y the ﬂow is not an irrotational ﬂow. (i) y = y2 – x2 ∂y ∂y = - 2 x and = 2y ∂x ∂y ∂2 y ∂ x2 ∂2 y ∂ y2 = 2 Ax 2 ; ∂2 f ∂2 y ∂ x2 ∂2 y Hence ∂x2 + = 0 and ∂2 y ∂ y2 ∂2 y ∂ y2 = - 2B; = - 2B π 0 Hence the stream function y = Ax – By2 does not represent an irrotational ﬂow. 3.25 xy v=a +x –y y = Ax – By Solution: A stream function y represents irrotational ﬂow if it satisﬁes Laplace equation. Hence + u y=y –x y = Ax y ∂2 f = 2A(x 2 + y2) π 0 ∂ x2 ∂ y2 Hence the stream function y = Ax 2 y2 does not represent an irrotational ﬂow. (iii) y = Ax – By2 ∂y ∂y = A and = - 2By ∂x ∂y Hence 3.24 ∂2 f = 2 Ay 2 and + = - 2 and ∂2 y ∂ y2 = 2; ∂2 y =0 ∂ x2 ∂ y2 Hence, the stream function y = y2 – x2 represents irrotational ﬂow. Solution: u = 2xy; v = a2 + x2 – y2 ∂u ∂v = 2 y; = - 2y ∂x ∂y ∂u ∂ v + = 2y – 2y = 0 ∂x ∂y The continuity equation for steady, incompressible ﬂow is satisﬁed. Hence the ﬂow is possible. The stream function y is related to u and v as ∂y = 2xy u = ∂y y = 2 2 (ii) y = Ax y ∂y ∂y = 2 Axy 2 and = 2 Ax 2 y ∂x ∂y ∂y = – y 2 – f ¢(x) = v = a 2 + x 2 – y2 ∂x f ¢(x) = – (a 2 + x2) Hence Ú 2xy dy = xy2 + f (x) 124 Fluid Mechanics and Hydraulic Machines x3 + constant 3 Thus the relevant stream function is y = xy 2 – a2x x3 + constant. – 3 f (x) = – a2 x – 3.26 u = – cx y xy v u x v=– y u=x+y v=x–y Solution: (i) u = – cx/y; v = c ln xy ∂u ∂v = – c/y; = c/y ∂x ∂y ∂u ∂ v + = – c/y + c/y = 0 ∂x ∂y The ﬂow is possible and y exists. ∂y = - cx / y ∂y y = – cx ln y + f (x) ∂y = c ln y – f ¢(x) = v = c ln xy ∂x = c ln x + c ln y Hence f ¢(x) = – c ln x u= Ú f (x) = - c ln x ◊ dx = – c(x ln x – x) + c2 where c2 = constant. Hence the stream function representing this ﬂow is y = – cx ln y – cx ln x + cx + c2 y = – cx ln xy + cx + c2 (ii) u = x + y; v = x – y ∂u ∂v = 1 and = –1 Therefore ∂x ∂y ∂u ∂ v + = 1 – 1 = 0. ∂x ∂y Hence the ﬂow is possible and y exists. ∂y =x+y ∂y y2 + f (x) y = xy + 2 ∂y = – y – f ¢(x) = v = x – y ∂x \ f ¢(x) = – x x2 and f (x) = - x dx = +c 2 where c = a constant. y2 x2 Hence +c y = xy + 2 2 1 \ y = ( y2 – x 2) + xy + c 2 (iii) u = 2cx, v = – 2cy ∂u ∂v = 2c and = – 2c ∂x ∂y ∂u ∂ v + = 2c – 2c = 0. ∂x ∂y Hence the ﬂow is possible and y exists. ∂y = 2cx u = ∂y y = 2cxy + f (x) ∂y = – 2cy – f ¢(x) = v = – 2cy ∂x Hence f ¢(x) = 0 and f(x) = c1 = a constant y = 2cxy + c1 u = Ú 3.27 y xy Solution: ∂y = 2x ∂y ∂y v= = – 2y ∂x At (2, 3), u = 2 ¥ 2 = 4 v =–2¥3=–6 (i) u = y x y–y 125 Fluid Flow Kinematics V = u2 + v2 = 4 2 + ( - 6) 2 tan 150° = - = 52 units – 0.5774 = – y/x y = 0.5774 x Substituting in Eq. (1) ∂y = 3x2 – 3y 2 (ii) u = ∂y ∂y = – 6xy ∂x At (2, 3), u = 3 ¥ (2)2 – 3(3)2 = – 15 v = – 6 ¥ 2 ¥ 3 = – 36 v= - V= 2 ( - 15) + ( - 36) 2 2 3 y ◊ 2 3 x (2) 4 ¥ 3 ¥ [( x 2 + (0.5774) 2 x 2 )] 4.0 = 16 x 2 = 4x = x = 1.00 y = 0.5774 The required point (x, y) is (1.00, 0.5774). = 39 units 3.28 y = 2 3x y 3.29 y a in Solution: The velocity vector is diagrammatically shown in Fig. 3.22 Ê a2 ˆ y = U Á1 - 2 ˜ r ¯ Ë r sin q U y = 2 3xy q ∂y = 2 3x u= ∂y Solution: ∂y v= = -2 3 y ∂x V= Vr = Ê 1 a2 ˆ 1 ∂y = U Á1 - 2 ˜ r ◊cosq Ë r r ∂q r ¯ Ê a2 ˆ = U Á1 - ˜ cosq Ë r2 ¯ u2 + v2 4 = 4(3) ( x 2 + y 2 ) v Also, tan q = u (1) At r = a and q = 90°, Vr = 0 ∂y = – U sin q Vq = – ∂r Ê 1 ˆˆ 2Ê ÁË1 - a ÁË - 2 ˜¯ ˜¯ r Ê a2 ˆ = – U sin q Á1 + 2 ˜ r ¯ Ë Y At r = a and q = 90°, Vq = – 2 U V 150° V (x, y) q u u = 4x v=– y y x v X Fig. 3.22 3.30 x y V at point y Solution: u = 4x3 and v = –12x 2y 126 Fluid Mechanics and Hydraulic Machines ∂y = u = 4x3 ∂y y = 4x 3y + f (y) ∂y = –12x 2y + f ¢(y) = v ∂x Since v = – 12x 2y, f ¢(y) = 0 Thus f (y) = constant and y = 4x3y + C. Since y = 0 at x = 0 and y = 0, C = 0 and y = 4x3y At the point (1, 2): u = 4x3 = 4 ¥ (1)3 = 4 units v = –12x 2y = –12 ¥ (1)2 ¥ (2) = – 24 units Comparing (i) and (ii) f ¢(y) = –5y f (y) = - 3.31 y = Ax + By Solution: (i) y = Ax + By ∂y ∂f =B= ∂y ∂x f = Bx + f (y) xy f y = 5xy ∂y ∂y = 5y and = 5x ∂x ∂y ∂ x2 ∂2 y + ∂2 y ∂ y2 =0 u= ∂2 y = 0 and hence the ﬂow is irrotational. ∂ x2 ∂ y2 To ﬁnd the potential function f. y = 5xy ∂y ∂f u= = 5x = ∂y ∂x 2 5x + f ( y) 2 ∂y v= = - 5y ∂x ∂f But by using f, v = = f ¢ ( y) . ∂y ∂y =-A ∂x ∂f = f ¢ ( y) But by using f, v = ∂y v = - …(i) …(ii) Comparing (i) and (ii) f ¢(y) = –A f (y) = –Ay + a constant Hence f = Bx – Ay + a constant (ii) y = xy Solution: = 0 and y = xy u = y = ∂2 y 5 2 ( x – y 2 ) + a constant 2 3.32 Velocity V = u 2 + v 2 = ( 4) 2 + ( - 24) 2 = 24.33 units y = 4x3y = 4 ¥ (1)3 ¥ (2) = 8 units f = Hence 5 y2 + a constant 2 ∂y ∂f =x= ∂y ∂x x2 + f ( y) 2 ∂y v= =-y ∂x ∂f But by using f, v = = f ¢ ( y) ∂y f= (i) ...(ii) Comparing (i) and (ii) f ¢(y) = –y f = f (y) = - …(i) …(ii) Hence y2 + a constant 2 Ê x2 y2 ˆ – f =Á ˜ + a constant Ë 2 2¯ 127 Fluid Flow Kinematics At point (1, 2) 3.33 f V 3 ˘ È y2 = Í 2 ¥ (1 ¥ 2) - (1 - 4) ˙ = 8.5 units 2 ˚ Î xy – x Solution: f = 2xy – x ∂f = 2y – 1 u= ∂x ∂f v= = 2x ∂y At point (2, 1): u = 2y – 1 = (2 ¥ 1) – 1 = 1 unit v = 2x = (2 ¥ 2) = 4 units Velocity V= u2 + v2 = (1) 2 + ( 4) 2 3.35 f x y f = 4x – y Solution: A valid potential function must satisfy the Laplace equation. (i) f = 2x + 5y = 4.12 units Flow rate between the stream lines passing through (1, 1) and (1, 2) = Dy = y2 – y1 = (8.5 – 2.0) = 6.5 units ∂f ∂f = 2 and =5 ∂x ∂y 3.34 f x – y xy ∂2 f ∂ x2 Hence + ∂2 f ∂ y2 = 0; ∂2 f =0 ∂ x2 ∂ y2 Hence f = 2x + 5y is a valid potential function. Solution: f = (x2 – y2) + 3xy ∂f ∂y = 2x + 3y = u = ∂x ∂y 3 2 y = 2xy + y + f (x) …(i) 2 ∂f ∂y = –2y + 3x = v = …(ii) ∂y ∂x ∂y And from (i) = –2y – f ¢(x) ∂x Thus f ¢(x) = –3x and hence 3 f (x) = - x 2 2 The required stream function is 3 y = 2xy - ( x 2 - y 2 ) 2 3 Ê ˆ At point (1, 1) y1 = ÁË 2 - (1 - 1)˜¯ = 2 units 2 ∂2 f = 0 and (ii) f = 4x2 – 5y2 ∂f ∂f = 8 x and = - 10 y ∂x ∂y ∂2 f ∂ x2 ∂2 f ∂2 f ∂2 f ∂ y2 = - 10 ; = -2 π 0 ∂x2 ∂ y2 Hence f = 4x2 – 5y2 is not a valid potential function. Hence + = 8 and 3.36 Solution: The differential 128 Fluid Mechanics and Hydraulic Machines (ii) f = 4(x 2 – y2) ∂y ∂y dx + dy dy = ∂x ∂y ∂f ∂y = 8x = ∂x ∂y Hence y = 8xy + f(x) ∂f ∂y = – 8y = v = ∂y ∂x = – 8y – f ¢(x) Hence f ¢(x) = 0 and f(x) = constant = c \ y = 8xy + c (iii) f = x + y + 3 ∂f ∂y =1= u = ∂x ∂y Hence y = y + f (x) u = Slope of stream line m1 = Ê ∂y ˆ ÁË ∂ x ˜¯ v Ê dy ˆ = =ÁË d x ˜¯ u Ê ˆ y ∂ y = const ÁË ∂ y ˜¯ Similarly the differential df = ∂f ∂f dx + dy ∂x ∂y Slope of equipotential line, m2 = Ê ∂f ˆ ÁË ∂ x ˜¯ ∂f ∂y =1= ∂y ∂x = – f ¢(x) f ¢(x) = – 1 and hence f (x) = – x Hence y = – x + y + c where c is a constant u Ê dy ˆ = = ÁË d x ˜¯ v Ê ˆ f ∂ f = const ÁË ∂ y ˜¯ Ê vˆ (m1 ◊ m2) = Á - ˜ Ë u¯ Since Ê uˆ ÁË v ˜¯ = - 1, the f = constant line and y = constant line are orthogonal to each other. 3.37 v = 3.38 f f f f=m f xy x –y x x y f xy f=x+y f x –y Solution: A valid potential function satisﬁes the Laplace equation. (i) f = A xy Solution: (i) f = 3xy ∂f ∂y = 3y = ∂x ∂y 3 2 = y + f(x) 2 ∂f ∂y = 3x = – = ∂y ∂x = – f ¢(x) = – 3x and hence f (x) = – 3/2x2 + c = –3/2(x2 – y2) + c where c is a constant u= Hence y v f ¢(x) \ y ∂f ∂f = A y; = Ax ∂x ∂y ∂2 f + ∂2 f =0+0=0 ∂ x2 ∂ y2 Hence f = Axy is a valid potential function. (ii) f = m ln x ∂f m ∂f = ; =0 ∂x x ∂y ∂2 f ∂ x2 =- m x2 ∂2 f ∂ x2 ; + ∂2 f ∂ y2 ∂2 f ∂x2 =0 = – m/x 2 π 0 129 Fluid Flow Kinematics Hence f = m ln x is not a valid potential function. (iii) f = A(x2 – y2) ∂f ∂f = 2Ax; = –2A y ∂x ∂y ∂2 f ∂ x2 ∂2 f = 2A; ∂2 f ∂ y2 = –2A ∂2 f = 2A – 2A = 0 ∂x2 ∂ y2 Hence f = A (x 2 – y2) is a valid potential function. (iv) f = A cos x ∂f ∂f = – A sin x; =0 ∂x ∂y ∂2 f ∂2 f = – A cos x π 0 ∂x2 ∂y2 Hence f = A cos x is not a valid potential function. + ∂2 y + ∂2 y = 2A – 2A = 0 ∂x2 ∂ y2 Hence y = A(x2 – y 2) represents a possible irrotational ﬂow ﬁeld. U cos q (iii) f = U r cos q + r Laplace equation is radial co-ordinates (r, q) 1 ∂f ∂ 2 f 1 ∂2 f + 2 + 2 =0 r ∂r ∂r r ∂q 2 ∂f U = U cos q – cos q ∂r r2 ∂ 2f 2U = + 3 cos q 2 ∂r r ∂f U = – Ur sin q – sin q ∂q r ∂2 f U = – Ur cos q – cos q 2 r ∂q L.H.S. of Laplace equation is 1 Ê 2U ˆ 1Ê U ˆ U cos q - 2 cos q ˜ + Á 3 cosq ˜ + 2 Á Ë ¯ Ë ¯ r r r r 3.39 f = Ur cos q + y Ê ˆ f = Á r - ˜ sin q Ë r¯ x –y U Ê ˆ ÁË - U r cos q - 2 cos q ˜¯ r U cos q r y = xy Solution: For an irrotational ﬂuid ﬂow phenomenon, f as well as y satisfy Laplace equation. (i) y = xy ∂y ∂y = y; =x ∂x ∂y ∂2 y ∂2 y = 0; =0 ∂ x2 ∂ y2 ∂2 y ∂2 y + =0 ∂x2 ∂x2 Hence y = xy represents a possible irrotational ﬂow. (ii) y = A(x2 – y 2) ∂y ∂y = 2Ax; = – 2Ay ∂x ∂y 2 1 1ˆ Ê1 1 = U cos q Á - 3 + 3 - - 3 ˜ Ër r r r ¯ r =0 The Laplace equation is satisﬁed and hence the given function f represents a possible irrotational ﬂow. 2ˆ Ê (iv) f = Á r - ˜ sin q Ë r¯ ∂f Ê 2ˆ = Á1 + 2 ˜ sin q Ë ∂r r ¯ ∂ 2f ∂r 2 = - 2 r3 sin q ∂f 2ˆ Ê = Á r - ˜ cos q Ë ∂q r¯ 130 Fluid Mechanics and Hydraulic Machines ∂2 f ∂q 2 f = f (q) 2ˆ Ê = - Á r - ˜ sin q Ë r¯ 1 ∂f 1 ∂y m = f ¢(q) = = r ∂q r ∂r r f ¢(q) = – m and hence f(q) = – m q + c where c = a constant. Hence, f = – m q + c. m cosq (ii) For f = r ∂f m cosq 1 ∂y vr = = = 2 ∂r r ∂q r ∂y m = cos q ∂q r m and y= sin q + f (r) r vq = The Laplace equation, in radial coordinates (r, q) is 1 ∂f ∂ 2 f 1 ∂ 2f + + 2 =0 2 r ∂r ∂r r ∂q 2 Substituting for L.H.S. terms 2 1 2˘ È1 2 = sin q Í + 3 - 3 - + 3 ˙ r r r ˚ r r Î Ê 2ˆ = sin q Á ˜ π 0 Ë r3 ¯ Hence, the given function does not represent any possible irrotational ﬂow. ∂y m = - 2 sin q – f ¢(r) ∂r r 1 ∂f = r ∂q vq = - 3.40 y=m r f= m cosq r = Solution: In radial co-ordinates, ∂f 1 ∂y = vr = ∂r r ∂q 1 ∂f ∂y vq = ◊ = r ∂q ∂r (i) For y = m ln r 1 ∂y ∂f =0= vr = r ∂q ∂r 1Ê m m ˆ - ◊ sinq ˜ = - 2 sin q ¯ r ÁË r r Thus f ¢(r) = - \ 2m sin q and r2 2m f (r) = sin q + c r m 2m sin q – sin q + c y = r r m sin q + c y =– r Problems 3.1 Water is pumped into the tank shown in Fig. 3.23 at 100 L/s. Both kerosene (relative) density = 0.8) and oil relative density = 0.90) are driven out. If 30 L/s oil is driven out estimate the volume of kerosene coming out of tank per second. (Ans. Q = 91.25 L/s) 3.2 A water tank has a 3 cm diameter inlet at A, a 4 cm diameter outlet at B and a 3 cm diameter controllable inlet at C, (Fig. 3.24). If the velocity of water at the inlet A is 2.0 m/s and the velocity of ﬂow going out at B = 1.85 m/s, what should be the velocity at the inlet at C to see that the water level in 131 Fluid Flow Kinematics r u um u1 Kerosene (RD = 0.8) 2 6 cm Dia Oil (RD = 0.9) 1 8 cm Dia 3 4 cm Dia r u um Water Fig. 3.23 Fig. 3.25 C Problem 3.3 3.4 A two-dimensional duct 10 cm high carries an incompressible ﬂuid ﬂow. At the entrance section 1 (Fig. 3.26), the velocity can be assumed to be as shown in the ﬁgure. 2 cm Tank A (Water) CL 6 cm um B u B 1 Fig. 3.24 the tank does not change? (Ans. V = 1.285 m/s) 3.3 Figure 3.25 shows a Y-junction of three pipes. The velocity distribution in pipe 1 is uniform while in pipes 2 and 3 it is given 1/ 7 rˆ Ê by u = um Á1 - ˜ . Pipes 1, 2 and 3 have Ë R¯ diameters 8 cm, 6 cm and 4 cm respectively. If the maximum velocity um in pipes 2 and 3 are 0.8 m/s and 0.6 m/s respectively, estimate the value of the uniform velocity u1 in pipe 1. (Ans. u1 = 0.49 m/s) 2 cm y 2 Fig. 3.26 After the ﬂow is fully developed, at a section 2, the velocity u at a distance y from the boundary is given by Êy u y2 ˆ = 4Á - 2 ˜ ËB B ¯ um where B = height of the duct and u m = velocity at the centreline of the duct. If the maximum velocity at entrance section 1 = 0.8 m/s, estimate the value of um at section 2. (Ans. um = 0.96 m/s) 3.5 Figure 3.27 shows a ﬂow of incompressible ﬂuid of density r ﬂowing past an 132 Fluid Mechanics and Hydraulic Machines B U U C 3.7 The velocity in a ﬂow is found to vary as vx 2 xˆ Ê = V0 / Á1 - ˜ . Determine the acceleration Ë L¯ u h at a point x = 0.5 m when L = 1.5 m and V0 y = 3.0 m/s. A D L Fig. 3.27 Problem 3.5 impervious plate AD. BC is an imaginary plane at a height h above the plate. The ﬂow approaching the plate is of uniform velocity U and the velocity proﬁle u = (y/h)1/7. at a section DC of plate is U Estimate the mass rate of outﬂow across BC (Ans. a = 71.40 m/s2) 3.8 A two-dimensional duct contains a straight sided contraction as in Fig. 3.29. The depth is constant at 30 cm. At a certain time an incompressible ﬂuid ﬂows in the duct at a rate of 0.3 m3/s and is decreasing at the rate of 0.1 m3/s per second. Estimate the acceleration at the section AA 20 cm from the inlet. (Ans. a = – 8.648 m/s2) per unit width of the plate. 60 cm 1 Ê ˆ ÁË Ans. qBC = rUh˜¯ 8 3.6 Water ﬂows in a pipe network shown in Fig. 3.28 Fill in the missing discharges so that continuity equation is satisﬁed. A 100 40 50 B 110 F ? ? D ? ? ? 25 C ? E 110 F 100 30 100 A 40 50 B 100 5 50 60 20 G 30 20 70 30 D 5 25 C 30 Answer Fig. 3.28 E 100 20 cm 20 cm A 2-D contraction Fig. 3.29 G ? 50 cm 100 5 50 A 3.9 A 90 cm diameter pipe is reduced to 30 cm diameter in a length of 1.5 m. Water ﬂows through this pipe at a rate of 280 L/s. If at an instant the discharge is found to decrease at the rate of 60 L/s per second, estimate the total acceleration at a distance of 75 cm from the 90 cm diameter inlet. (Ans. a = 1.0953 m/s2) 3.10 A 40 cm diameter pipe is reduced uniformly to 20 cm diameter in a length of 0.5 m. If the steady discharge of water through this pipe is 100 L/s, determine the acceleration at 10 cm from the 20 cm diameter end (i) when the ﬂow is steady and (ii) when the ﬂow is increasing at a rate of 600 L/s per minute. 133 Fluid Flow Kinematics (Ans. (i) a = 16.289 m/s2, (ii) a = 16.51 m/s2) 3.11 For the following ﬂows ﬁnd the equation of the streamline passing through the indicated point: (i) V = 2x i – yj – zk ..... passing through (1, 1, 1) (ii) V = 4yi – 3xj ...... passing through (– 1, 2) (iii) V = – xi + 2 yj + (3 – z)k ...... passing through (1, 1, 2) (Ans. x1/2 y = 1 and x1/2 z = 1, (ii) 3x 2 + 4y2 = 19, (iii) xy1/2 = 1 and x = 3 – z) 3.12 A steady, incompressible, two-dimensional velocity ﬁeld is given by u = x + 2y + 2.0 v = 2x – y – 3.5 Determine the location of the stagnation point, if it exists. (Ans. stagnation point = (1.0, – 1.5)) 3.13 The velocity in a ﬂuid ﬂow is expressed as V = (xy + 2zt)i + (2y2 + xyt)j + (12xy)k Determine ax, the x-component of the acceleration of the particle of ﬂuid, at the point (1, 1, 2) at t = 2.0 seconds. (Ans. ax = 65 units) 3.14 In a two-dimensional ﬂow, u = x + y and v = x2 – y. Calculate the circulation about a square shaped contour enclosed by lines joining the points (1, 1), (2, 1), (2, 2), (1, 2). (Ans. Circulation = 2.0 units) 3.15 Calculate the circulation per unit area around a closed contour not including the centre in the following types of vortex motion: (i) vs = cr and (ii) vs = c/r, where vs = tangential velocity at a radial distance r and c = a constant. (Ans. (i) G/A = 2C and (ii) G/A = 0) 3.16 Verify whether the following ﬂow ﬁelds are rotational. If so, determine the components of rotation about various axes. (i) u = xyz v = zx w= (ii) u = xy 1 v = ( x 2 - y 2) 2 1 2 yz – xy 2 1 Ê ÁË Ans. (i) Rotational, w z = 2 z (1 - x ), 1 1 w x = ( z 2 / 2 - 2 x ), w y = y ( x + 1) 2 2 ˆ (ii) Irrotational, w x = w y = 0˜ ¯ 3.17 For the following set of velocity components verify whether the continuity equation is satisﬁed. If so, determine the (i) vorticity vector and (ii) acceleration vector at point A (1, 1, 1) u = 2x 2 + 3y v = – 2xy + 3y3 + 3zy 3 w = – z 2 – 2xz – 9y2 z 2 (Ans. (i) z z = – (2y + 3); z x = – (18yz + 3y), z y = 2z, (ii) ax = 32, ay = – 7.5 and az = 93) 3.18 Check whether the following sets of velocity components satisfy the continuity equation of steady incompressible ﬂow. (i) u = 4x + 2y – 3 v = 2x + 4y + 3 (ii) u = 4xy + y2 v = 6xy + 3x + 2 (iii) u = 2x2 + y2 v = – 4xy (iv) u = x3 + y3 v = x – 3x2y C ( y 2 - x 2) (v) u = ( x 2 + y 2 )2 - 2C x y v= 2 ( x + y 2 )2 134 Fluid Mechanics and Hydraulic Machines (Ans. (i) not satisﬁed, (ii) not satisﬁed, (iii) satisﬁed (iv) satisﬁed, (v) not satisﬁed) 3.19 Calculate the unknown velocity component in the following so that the equation of continuity is satisﬁed. 2 (i) u = xy3 – x2y (ii) u = Aye x 3 v=? v=? 1 4 Ê 2 ÁË Ans. (i) v = - 6 y + xy + f (x ), ˆ Ay 2 x (ii) v = e + f (x)˜ 2 ¯ 3.20 For a two-dimensional incompressible ﬂow, the x component of velocity and boundary conditions are as follows: u = x2 y2 + 2xy and at y = 0, u = v = 0 Determine the y component of the velocity. 2 3 Ê 2ˆ ÁË Ans. v = - 3 xy - y ˜¯ 3.21 Given the components of a velocity vector in a three-dimensinal ﬂow, as below, determine the missing component. (i) u = 2x 2 + 2xy, v = 4yz, w = ? (ii) u = 3x2, v = 4yxz, w = ? (iii) u = x3 + y2 + 2 z2, v = – x 2y – yz – xy, w=? {(Ans. (i) w = – (4xz + 2yz + 2z 2) + f(x, y), (ii) w = – (6xz + 2xz 2) + f (x, y), z2 – xz + f (x, y)} (iii) w = 2x 2 z – 2 3.22 For the following sets of velocity components obtain the relevant stream functions. y3 + 2x – x2 y (i) u = 6y (ii) u = 3 x2 v = 6x v = xy2 – 2y – 3 (iii) u = – A ln x y + x2 v=A x Ê 2 2 ÁË Ans. (i) y = 3(y - x ) + C , y4 x2 y2 x3 + 2 xy + , 12 2 9 ˆ x3 (iii) y = - Ay ln x + C˜ 3 ¯ (ii) y = 3.23 The velocity components of a steady, twodimensional ﬂow of an ideal ﬂuid, are u = 2xy and v = a2 + x2 – y2. Show that the velocity potential exists and determine the same. (Ans. f = a 2y + x2y – y3/3 + constant) 3.24 If the velocity potential f = (x 3/3) – x2 – xy 2 + y2, determine the corresponding stream function. (Ans. y = x 2 y – 2xy – (y3/3) + constant) 3.25 Find the velocity potential if the velocity ﬁeld is x y ;v= 2 u= 2 2 x +y x + y2 1 Ê 2 2 ˆ ÁË Ans. f = 2 log ( x + y )˜¯ 3.26 Calculate the velocity at the point (3, 3) for the following stream function. (i) y = – x ln xy + x 1 2 2 (y – x ) + xy + 6 (ii) y = 2 (Ans. (i) V = 2.42 units, (ii) V = 6 units) 3.27 Given the following stream functions, determine the corresponding potential functions. Also, estimate the discharge, per unit depth, in the z direction passing between the streamlines through the points (1, 3) and (3, 3). (i) y = 3xy 3 (ii) y = (y2 – x 2 ) 2 135 Fluid Flow Kinematics 3 2 Ê 2 ÁË Ans. f = 2 ( x - y ) + c ; 18 units, ˆ (ii) f = 3 xy ; - 12 units˜ ¯ 3.32 If the stream function of a two-dimensional irrotational ﬂow is given as y = Axy + B(x2 – y2) in which A and B are constants, evaluate the potential function of this ﬂow. A 2 Ê ˆ 2 ÁË Ans. f = 2 ( x - y ) - 2Bxy + a constant˜¯ 3.28 Verify whether the following functions are valid potential functions. (i) f = y 3 – 3 x2y (ii) f = y 4 – 6x2 y2 (iii) f = x 2 – 3x2y (iv) f = x 3 – y3 (Ans. (i) Yes, (ii) No, (iii) No, (iv) No.) 3.29 Determine the corresponding conjugate functions relating to the following functions representing irrotational ﬂuid ﬂows. (i) f = Ur cos q (ii) y = – mq + k ln r (Ans. (i) y = U r sin q + C, (ii) f = – kq – m ln r) 3.30 Show that if two velocity potentials f1 and f2 have velocity components (u1, v1) and (u2, v2) respectively, then for a velocity potential f = (f1 + f2) the velocity components are ((u1 + u2),(v1 + v2)). 3.31 The velocity potential for a two-dimensional ﬂow is given by f = y2 – x2. Develop an expression for the stream function for this ﬂow and ﬁnd the ﬂow rate between the stream lines passing through points (0, 1) and (1, 0). (Ans. y = – 2xy + C, Flow rate = – 2 units) 3.33 A velocity potential for a two-dimensional ﬂow is given by f = x2 – y2 + y. Calculate (i) the stream function and (ii) the ﬂow rate between the streamlines passing through points (1, 1) and (1, 2) (Ans. y = 2xy – x, Flow rate = 2 units) 3.34 A steady, incompressible, two-dimensional velocity ﬁeld is given by y3 + 2x – x2y 3 x3 v = xy2 – 2y – 3 Does this represent an irrotational ﬂow? If so, determine the relevant potential function which can represent this ﬂow. u = Ê ÁË Ans. flow is irrotational; f= xy 3 x3 ˆ y + ( 2 x 2 - y 2) 3 3 ˜¯ Objective Questions 3.1 The ﬂow ﬁeld represented by the velocity vector V = axi + by2j + czt 2 k where a, b and c are constants, is (a) three-dimensional and unsteady (b) two-dimensional and steady (c) three-dimensional and steady (d) two-dimensional and unsteady 3.2 The ﬂow of a liquid at constant rate in a conically tapered pipe is classiﬁed as (a) steady, uniform ﬂow (b) steady, non-uniform ﬂow 136 Fluid Mechanics and Hydraulic Machines 3.3 3.4 3.5 3.6 3.7 (c) unsteady, uniform ﬂow (d) unsteady, non-uniform ﬂow A pathline is the (a) mean direction of a number of particles at the same instant of time. (b) instantaneous picture of positions of all particles in the ﬂow which passed a given point. (c) trace made by a single particle over a period of time (d) path traced by continuously injected tracer at a point A streamline is a line (a) which is normal to the velocity vector at every point (b) which represents lines of constant velocity potential (c) which is normal to the lines of constant stream function (d) which is tangential to the velocity vector everywhere at a given instant In a steady ﬂow (a) streamlines and pathlines are identical but are different from streaklines (b) streakline and pathlines are identical but are different from streamlines (c) streamline, streakline and pathline can all be different from each other (d) none of the above In two-dimensional ﬂow the equation of a streamline is given as dy dx = (a) u v dx dy (b) = u v dx dy (c) = u, =v dt dt u dy (d) = dx v The shape of the stream line passing through the origin in a ﬂow ﬁeld u = cos q ; v = sin q for a constant q, is determined by (a) y = x 3 (b) y = x cot2 q (c) y = x tan q (d) y = sin q 3.8 A velocity vector V in two-dimensional ﬂow is inclined at an angle q to the X-axis. The resulting acceleration vector a (a) will be always normal to V (b) will be always parallel to V (c) will have an inclination of (90 – q) to the y-axis (d) will have an inclination a to the X-axis which depends on the components of the acceleration. 3.9 A streamline is deﬁned in terms of stream function y and potential function f as (a) f = constant ∂f = constant ∂s ∂y (c) = constant ∂s (d) y = constant. (b) 3.10 If y = 2xy, the magnitude of the velocity vector at (2, – 2) is (a) 4 2 (b) 4 (c) – 8 (d) 2 3.11 A continuity equation for steady twodimensional compressible ﬂow is ∂u ∂v =0 +r ∂x ∂y ∂ u ∂v + =0 (b) ∂x ∂y (a) r ∂ ( r u) ∂ ( r v ) + =0 ∂x ∂y ∂r ∂r (d) u +v =0 ∂x ∂y 3.12 The continuity equation (c) ∂ u ∂ v ∂w + + =0 ∂x ∂ y ∂ z 137 Fluid Flow Kinematics 3.13 3.14 3.15 3.16 3.17 (a) is not valid for unsteady, incompressible ﬂuids. (b) is valid for incompressible ﬂuids whether the ﬂow is steady or unsteady (c) is valid for steady ﬂow, whether compressible or incompressible. (d) is valid for ideal ﬂuid ﬂow only. If u and v, the components of velocity in x and y directions respectively are given by u = ax + by and v = cx + dy then the condition to the satisﬁed is (a) a + c = 0 (b) b + d = 0 (c) a + b + c + d = 0 (d) a + d = 0 Which of the following can be a set of velocity components in a two-dimensional ﬂow? (a) u = x + y; v = x2 + y2 (b) u = x + y; v = x – y (c) u = x y; v = x/y (d) u = x 2 + y2; v = x2 – y2 A ﬂow has diverging straight streamlines. If the ﬂow is steady, the ﬂow (a) is a uniform ﬂow with local acceleration (b) has convective normal acceleration (c) has convective tangential acceleration (d) has convective normal as well as tangential accelerations. A ﬂow has parallel curved streamlines and is steady. This ﬂow has (a) tangential convective acceleration (b) local acceleration (e) normal convective as well as local accelera-tion (d) normal convective acceleration In a 2 m long tapered duct the area decreases as A = (0.4 – 0.1x) where x is distance in metres. At a given instant a discharge of 0.48 m3/s was ﬂowing in the duct and it was found to increase at a rate of 0.12 m3/s. The local acceleration at x = 0 in m/s2 is (a) 0.3 (b) 3.6 (c) 3.9 (d) – 0.30 3.18 In a two-dimensional ﬂow acceleration component in the X-direction is given by ax = ∂u ∂u ∂v (a) +u +v ∂t ∂x ∂y ∂u ∂u (b) u +v ∂x ∂y ∂u ∂u ∂u +u +v (c) ∂t ∂x ∂y ∂u ∂u ∂v +v +u ∂t ∂x ∂y 3.19 In a natural coordinate system the acceleration an in the normal direction when local and convective terms are present is given by an = ∂v ∂v v2 ∂v +v (b) (a) +v ∂t ∂t r ∂r (d) u d (v 2 / r ) ∂ vn v 2 + (d) dt ∂t r 3.20 The velocity of an incompressible ﬂuid is given by (c) V = (Px – Q) i + R y j + S t k m/s where P = 3 s–1, Q = 4 m/s, R = – 3 s–1. x and y are in m and t in s. The local and convective acceleration components at x = 1 m, y = 2 m and t = 5 s are respectively, (a) 5k and (–3i + 18j) m/s (b) zero and (–3i + 18j) m/s (c) 5k and (18i – 3j) m/s (d) – 5k and (–3i + 18j) m/s 3.21 A ﬂow is said to be rotational when (a) the streamlines are curved (b) a velocity gradient in the normal direction to ﬂow exists. (c) every ﬂuid element has ﬁnite angular velocity about its mass centre 138 Fluid Mechanics and Hydraulic Machines (d) every ﬂuid element has an angular velocity about a common axis 3.22 In three-dimensional motion of a ﬂuid the component of rotation about the X-axis is wx = (a) 1 Ê ∂w ∂v ˆ 1 Ê ∂u ∂ w ˆ - ˜ (b) Á 2 Ë ∂ y ∂z ¯ 2 ÁË ∂ z ∂ y ˜¯ (c) 1 2 Ê ∂ v ∂u ˆ 1 ÁË ∂ x - ∂ y ˜¯ (d) 2 3.28 Ê ∂v ∂w ˆ ÁË ∂ x - ∂ y ˜¯ 3.23 If w z = component of rotation of a ﬂuid about z-axis, the vorticity along that axis is usually deﬁned as z z = 1 wz (a) (b) 2wz 2 (c) ∂w z /∂x (d) w z dz Ú 3.24 A two-dimensional ﬂow in x–y plane is rotational if ∂u ∂ v ∂ u ∂u = (b) = (a) ∂x ∂ y ∂x ∂ y ∂v ∂v ∂ v ∂u (c) (d) = = ∂x ∂y ∂x ∂y 3.25 Vorticity in z-direction is given by È ∂ u ∂v ˘ + (a) Í ˙ Î ∂x ∂ y ˚ È ∂u ∂v ˘ (b) Í ˙ Î ∂x ∂ y ˚ È ∂ v ∂u ˘ (c) Í + ˙ Î∂x ∂y ˚ È ∂ v ∂u ˘ (d) Í ˙ Î∂x ∂y ˚ 3.29 3.30 3.31 3.32 3.26 If y2 and y1 are the values of stream function at points 2 and 1 respectively, the volume rate of ﬂow per unit depth across an element Ds connecting 2 and 1 is given by Dy (b) S Dy . Ds (a) Ds 1 (d) Dy (c) Dy 3.27 A two-dimensional ﬂow is described by velocity components u = 2 x and v = – 2y. 3.33 3.34 The discharge between points (1,1) and (2,2) is equal to (a) 9 units (b) 8 units (c) 7 units (d) 6 units If y is a stream function then the velocity u and v are given by (a) u = ∂y/∂y (b) u = – ∂y/ ∂y v = – ∂y/ ∂x v = – ∂y/∂x (c) u = ∂y/ ∂y (d) u = ∂y/ ∂x v = ∂y/∂x v = – d y/∂y The stream function in a two-dimensional ﬂow ﬁeld is given by y = x 2 – y2. The magnitude of the velocity at point (1, 1) is (a) 2 (b) 2 2 (c) 4 (d) 8 In a two-dimensional, incompressible ﬂow, if the ﬂuid velocity components are u = x – 4y and v = –y then the stream function y is given by (a) x 2 – xy + 2 y2 (b) 2x 2 + 2xy + y2 (c) 2x 2 + xy – 2y2 (d) 2x 2 – xy + 2y2 Stream function is deﬁned for (a) Flow of a perfect ﬂuid only (b) All 2–D incompressible ﬂows (c) All 3–D ﬂows (d) Irrotational ﬂows only Velocity potential exists for (a) Flow of a perfect ﬂuid only (b) Stready, irrotational ﬂow only (c) All irrotational ﬂows (d) All 3–D ﬂows A velocity potential exists (a) whenever the real ﬂuid ﬂow exists (b) when the ﬂow is real and rotational (c) when the ﬂow satisﬁes the conditions of irrotational motion. (d) when the ﬂow satisﬁes the equations of continuity. If f is a potential function in twodimensional ﬂow the velocity components u and v are deﬁned as 139 Fluid Flow Kinematics 3.35 3.36 3.37 3.38 3.39 3.40 (a) u = ∂f/∂y (b) u = ∂f/∂x v = – ∂f/∂x v = ∂f/∂y (c) u = ∂f/∂x (d) u = ∂f/∂y v = – ∂f/∂y v = ∂f/∂x The velocity potential function for a line source varies with radial distance r as (a) 1/r (b) 1/r 2 (c) r (d) ln r Lines of constant f (a) are parallel streamlines (b) are parallel to the streamlines (c) are normal to the streamlines (d) can intersect each other In 2-D radial co-ordinates the velocities vr and vq are expressed in terms of f as 1 ∂f ∂f (b) vr = (a) vr = r ∂q ∂r ∂f 1 ∂f vq = vq = ∂r r ∂q 1 ∂f ∂f (c) vr = (d) vr = r ∂q ∂r 1 ∂f ∂f vq = vq = r ∂q ∂r If f = 3 x y, the x and y components of velocity at the point (1, 3) will be (a) u = – 9, v = – 3 (b) u = – 3, v = – 9 (c) u = 9, v = – 3 (d) u = 9, v = 9 A stream function y = x3 – y3 is observed for a two-dimensional ﬂow ﬁeld. What is the magnitude of the velocity at point (1, – 1)? (a) 4.24 (b) 2.83 (c) 0 (d) – 2.83 Which one of the following stream functions y is a possible irrotational ﬂow ﬁeld? (a) y = y2 – x2 (b) y = Asin(xy) (c) y = Ax2y2 (d) y = Ax + By2 3.41 The stream function y of a ﬂow ﬁeld is given by the expression y = xy. The potential function relevant to this ﬂow is 1 2 (a) (b) 2xy ( x - y 2) 2 1 2 (c) (d) x2y + y2x ( x + y 2) 2 3.42 Which of the following stream functions is a possible irrotational ﬂow ﬁeld? (a) y = x2y (b) y = 2xy (c) y = Ax2y2 (b) y = Ax + By2 3.43 If for a ﬂow, a stream function y exists and satisﬁes the Laplace equation, then (a) the ﬂow is rotational (b) the ﬂow is irrotational but does not necessarily satisfy continuity equation (c) the ﬂow satisﬁes continuity equation but does not necessarily satisfy condition for irrotational ﬂow (d) the continuity equation is satisﬁed and the ﬂow is irrotational 3.44 If a stream function y exists it implies that (a) the function y represents a possible ﬂow ﬁeld (b) the ﬂow is irrotational (c) the ﬂow is steady, incompressible (d) the potential function also exists 3.45 The potential function exists for (a) irrotational motion of incompressible ﬂuids only (b) irrotatinal motion of ﬂuids whether compressible or incompressible (c) for two-dimensional irrotational ﬂow only (d) for steady ﬂows only 3.46 Cauchy-Reimann equations relating f and y are ∂f ∂f = ; ∂x ∂y ∂f ∂f = (b) ; ∂x ∂y (a) ∂y ∂y = ∂x ∂y ∂f ∂y = ∂y ∂x 140 Fluid Mechanics and Hydraulic Machines ∂f ∂y ∂f ∂y = ; = ∂x ∂x ∂y ∂y ∂f ∂f ∂y ∂y = = (d) ; ∂x ∂y ∂x ∂y 3.47 Indicate the incorrect statement: A ﬂow net (a) is applicable to irrotational ﬂuid ﬂow (b) for a given boundary is the same whether the ﬂow is in one direction or the other (c) for a given boundary is applicable to one chosen direction of ﬂow; if the ﬂow is reversed the ﬂow net will change (d) will be no constructed that the size of the mesh is inversely proportional to the local velocity 3.48 The velocity potential f at any point for a two-dimensional, steady, irrotational ﬂow in polar coordinates is given by l cosq f= r (c) This equation represents a (a) vortex (b) sink (c) source (d) doublet 3.49 The velocity potential f at any point for a two-dimensional, steady, irrotational ﬂow is given by f = K q. This equation represents a (a) vortex (b) sink (c) source (d) doublet 3.50 Inviscid, incompressible ﬂow about a stationary cylinder in uniform ﬂow can be simulated by superposition of uniform ﬂow and (a) a sink and a vortex (b) a doublet (c) a vortex (d) a doublet and a vortex Energy Equation and Its Applications Concept Review 4 Introduction 4.1 BERNOULLI EQUATION Euler equation: For the frictionless ﬂow along a streamline of an incompressible ﬂuid the relationship among the pressure, elevation and velocity is given by the Euler equation. ∂V 1 ∂ p ∂z ∂V + +g +V =0 ∂ t r ∂s ∂s ∂s (4.1) Bernoulli equation: Integration of the Euler equation for steady, incompressible ﬂuid ﬂow, without friction, yields the Bernoulli equation p V2 + + Z = constant = H g 2g (4.2) It can be shown that the Bernoulli equation is applicable across the streamlines also if the ﬂow is irrotational. In Eq. (4.2) the term V 2/2 g represents kinetic energy of the ﬂow per unit weight of the ﬂuid. Similarly, Z represents potential energy per unit weight. The term p/g represents ﬂow work, i.e. the work done by the ﬂuid on the surroundings. All the terms in Eq. 4.2 have unit of [L] = (N.m/N) of ﬂuid. The constant H is called the total energy. For any two points in a steady irrotational ﬂow ﬁeld of an ideal ﬂuid, ÊV 2 V 2 ˆ ( p1 - p2 ) + Á 1 - 2 ˜ + (Z1 – Z2) = H 1 – H2 = 0 g Ë 2g 2g ¯ (4.3) 142 Fluid Mechanics and Hydraulic Machines 4.2 PRACTICAL APPLICATIONS OF BERNOULLI EQUATION p2 V22 + + Z2 g 2g H E = HP = energy input per unit weight of ﬂuid per second by the pump H L = energy loss between points 1 and 2 H2 = In practical applications of Bernoulli equation the restriction of frictionless ﬂow is accommodated by introducing a loss of energy term and the restriction of irrotational ﬂow is waived in most of the cases. Equation 4.2 is used as a special case of the general energy equation. The general energy equation dealing with the conservation of energy is written for steady, incompressible ﬂuid ﬂow between two sections 1 and 2 as H 1 + HE – HL = H2 (4.3a) where H1 = total energy at section 1 H E = energy input to the system between sections 1 and 2 HL = energy loss due to friction, etc. between sections 1 and 2 H2 = total energy at section 2. Energy is transferred to the system as mechanical work done on the ﬂuid by a pump. Similarly, energy is extracted from the system by a turbine. For incompressible ﬂuid ﬂow all non-recoverable energy such as change of internal energy and heat transfer are usually clubbed under a common term energy loss. Thus for a ﬂuid ﬂow system shown in Fig. 4.1 the Bernoulli equation is 4.3 The general equation for conservation of energy for an incompressible ﬂuid ﬂow can be written as Ê p1 V12 ˆ Á g + 2g + Z1 ˜ + qw + H E Ë ¯ Êp ˆ V2 = Á 2 + 2 + Z 2 ˜ + (e2 - e1 ) 2g Ë g ¯ H1 = p V2 + + Z, then Eq. 4.4 If the total head H = g 2g is written as H 1 + HE – [(e2 – e1) – qw) = H2 The term: (e2 – e1) – qw = (reversible + irreversible) head V12 In incompressible ﬂuid ﬂow irreversible head is called head loss HL and represents energy loss per unit weight of ﬂuid due to friction and other causes. Thus for an incompressible ﬂuid p1 + + Z1 g 2g 2 1 Z2 Z1 È Head added due ˘ È Total head ˘ Í to a machine such ˙ - [ Head loss ] + ÍÎat section 1˙˚ ˙ Í as a pump ˚ Î È Total head ˘ =Í Îat section 2˙˚ P Datum Fig. 4.1 (4.4) where qw = heat added per unit weight of ﬂuid e1, e2 = internal energy per unit weight of ﬂuid at the respective states H E = external work done (i.e. shaft work added) on the ﬂuid per unit weight of ﬂuid from a device such as a pump. H1 + H E - H L = H 2 where ENERGY EQUATION or H1 + H E - H L = H 2 (4.3a) 143 Energy Equation and Its Applications When a pump is used H E = HP (a positive quantity), and when a turbine is used HE = H t (a negative quantity). 4.3.1 Hydraulic Grade Line A line joining the piezometric heads at various points in a ﬂow is known as the hydraulic grade line (HGL). p As the piezometric head h = + Z, the HGL g Ê p ˆ represents the variation of h Á = + Z ˜ measured Ë g ¯ section. The actual velocity distribution in the cross section may be non-uniform. Hence, the kinetic energy calculated by using V must be multiplied by a correction factor to obtain proper kinetic energy at the cross section due to non-uniform velocity distribution. Thus the velocity head in the Bernoulli equation will be a V2 where 2g a= above a datum, (Fig. 4.2). HL 2 V1 /2g Energy line 2 V 2/2g Hydraulic grade line p1/g p2/g 2 CL Z2 Z1 Datum 1 A Ú 4.3.2 Energy Line The total energy Êp ˆ V2 V2 H = Á + Z˜ + =h+ 2g 2g Ëg ¯ A line joining the elevation of total energy of a ﬂow measured above a datum is known as energy line, (Fig. 4.2). The energy line lies above the HGL by an amount of V2/2g. 4.3.3 Kinetic Energy Correction Factor, a In one-dimensional method of analysis, the average velocity V is used to represent the velocity at a cross (4.5) The term a is called the kinetic energy correction factor. For uniform velocity distribution a = 1.0 and in all other cases it will be greater than 1.0. Greater the non uniformity in velocity distribution larger will be the value of a. For laminar ﬂow through a pipe, a = 2.0 and for turbulent ﬂow through a pipe its value varies from 1.01 to 1.20. In the absence of speciﬁc information about the value of a, it is usual practice to assume its value as unity. 4.4 POWER In the case of work done over a ﬂuid the power input into the ﬂow is P = g Q Hm Fig. 4.2 3 Ê vˆ ÁË V ˜¯ dA where (4.6) g = unit weight of ﬂuid in N/m 3, Q = discharge in m3/s and H m = head added to the ﬂow, in m In a pump H m = Hp is positive. In a turbine Hm = H t is negative and power is extracted from the ﬂow. If hp = efﬁciency of the pump, the power input required at the pump is Pin = g Q Hm hp (4.7) In the case of a turbine, in h t is the efﬁciency of the turbine, power delivered by the turbine is Pout = g Q H m ht (4.8) 144 Fluid Mechanics and Hydraulic Machines Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples 3 4.1 2 107.00 m Solution: From given data: 800 ¥ 9.81 g = rg = = 7.848 kN/m3 1000 p1 36.0 = Z1 = 7.0 m and = 4.587 m 7.848 g The piezometric head at Section 1 = p h1 = 1 + Z1 = 4.587 + 7.0 = 11.587 m g The total head at Section 1 = p1 V2 V2 + Z1 + a1 1 = h1 + a1 1 g 2g 2g 2 Ê (1.8) ˆ H1 = 11.587 + (1.15) ¥ Á ˜ Ë 2 ¥ 9.81¯ = 11.587 + 0.19 H1 = = 11.777 m 4.2 30 cm Dia 1 15 cm Dia 100.00 m Datum A B Fig. 4.3 Solution: Q = 0.05 m3/s Let sufﬁxes 1 and 2 refer to sections A and B respectively. Q 0.05 V1 = = = 2.829 m/s p A1 ¥ (0.15) 2 4 Q 0.05 = V2 = = 0.7074 m/s p A2 ¥ (0.30) 2 4 145 Energy Equation and Its Applications g = unit weight of water = 998 ¥ 9.81/1000 = 9.79 kN/m3 (i) When the ﬂow is from A to B: Taking the atmospheric pressure as zero 1 Ê p1 V12 ˆ p2 V22 Z + + = + + Z 2 + HL 1˜ Ág 2g g 2g Ë ¯ 30 ( 2.829) 2 + + 100.00 9.79 2 ¥ 9.81 p (0.7074) 2 = 2 + + 107.00 + 2.0 2 ¥ 9.81 g 3.064 + 0.4080 + 100.00 p = 2 + 0.0255 + 107.0 + 2.0 g p2 = – 5.554 m (gauge) g p2 = – 5.554 ¥ 9.79 = – 54.37 kPa (gauge) (ii) When the ﬂow is from B to A: Taking atmospheric pressure as zero, Ê p2 V22 ˆ Ê p1 V12 ˆ = Z + + – H + + Z2 ˜ L 2˜ Á g Á 2g 2g Ë ¯ Ëg ¯ p2 (0.7074) 2 + + 107.00 – 2.00 2 ¥ 9.81 g 30 ( 2.829) 2 = + + 100.00 9.79 2 ¥ 9.81 p2 + 0.0255 + 107.00 – 2.00 g = 3.064 + 0.4080 + 100.00 p2/g = – 1.554 m (gauge) p2 = – 1.554 ¥ 9.79 = – 15.21 kPa (gauge) H 2 Fig. 4.4 Solution: pD12V1 pD22V2 = 4 4 Ê D2 ˆ Hence V2 = Á 12 ˜ V1 = 4V1. Ë D2 ¯ Further it is given that p1 = p2 and also (Z1 – Z2) = H Applying Bernoulli’s equation between sections 1 and 2, p1 V22 p V2 + Z1 + 1 = 2 + Z2 + 2g g 2g g On simplifying by substitution of given data, V12 V22 = (Z1 – Z2) + 2g 2g 16 or 15 V12 = 1.5 2g V1 = 1.962 = 1.40 m V2 = 4V1 = 5.6 m/s and 4.3 V12 V12 = 1.5 + 2g 2g 4.4 r 3 146 Fluid Mechanics and Hydraulic Machines Solution: p1 = 50 kPa, p2 = 25 kPa and 680 ¥ 9.81 = 6.67 kN/m2. 1000 p 50.0 From given data: 1 = = 7.496 m, 6.67 g p 25.0 Z1 = 0 and 2 = 6.67 g = 3.748 m g = rg = pD22V2 pD12V1 = 4 4 Ê D2 ˆ V2 = Á 12 ˜ V1 = 4V1. Ë D2 ¯ Hence, V22 V12 Solution: pDa2Va pDb2Vb = 4 4 2 Ê D2 ˆ Ê 10.0 ˆ V = 1.5625 Va. Vb = Á a ˜ Va = Á 2 Ë 8.0 ˜¯ a Ë Db ¯ Hence, Vb2 V2 V2 = (1.5625)2 a = 2.4414 a 2g 2g 2g Applying Bernoulli’s equation to sections A and B, = 16 2g 2g Applying Bernoulli’s equation to sections 1 and 2, p1 p V2 V2 + Z1 + 1 = 2 + Z2 + 2 2g 2g g g 7.496 + 0 + 15 Discharge 4.5 V12 V2 = 3.748 + 0 + 16 1 2g 2g V12 2g pa p V2 V2 + Za + 1 = b + Zb + b 2g 2g g g 25.00 V2 + 102.00 + a 9.79 2g = = 3.748 and V1 = 2.214 m/s pD12V1 Q= 4 p(0.2) 2 ¥ 2.214 = = 0.0696 m3/s 4 = 69.6 Litres/s (2.4414 – 1) 18.00 V2 + 102.500 + 2.4414 a 9.79 2g Va2 = 104.5536 – 104.3386 2g = 0.215 m Va2 0.215 = 0.14916 m = 1.4414 2g and Va = 1.727 m/s Discharge Q= = p Da2Va 4 p(0.1) 2 ¥ 1.727 = 0.01356 m3/s 4 = 13.56 litres/s = 13.56 ¥ 60 = 814 Litres/minute 147 Energy Equation and Its Applications 20.0 (1.25) 2 + + 12.00 9.79 2 ¥ 9.81 = 2.043 + 0.0796 + 12.00 = 14.123 m As H1 > H2, the ﬂow is from section 1 to 2. Head loss between the sections 1 and 2, i.e. HL = H1 – H2 = 15.187 – 14.123 = 1.064 m 4.6 H2 = Solution: 4.7 Refer to Fig. 4.5 2 V V EL:12.00 m 1 1 EL:10.00 m Datum H Oil (RD = 0.8) 25 mm Dia 100 mm Dia pipe Fig. 4.5 2 Discharge Q = A1V1 = A2V2 p Q= ¥ (0.25)2 ¥ 1.25 4 = 0.0614 m3/s = 61.4 L/s As A1 = A2, V1 = 1.25 m/s g = 998 ¥ 9.81/1000 = 9.79 kN/m3 Taking atmospheric pressure as zero, Total energy at section 1, p1 V12 + + Z1 g 2g 50.0 (1.25) 2 H1 = + + 10.00 9.79 2 ¥ 9.81 = 5.107 + 0.0796 + 10.00 = 15.187 m Total energy at section 2, H1 = H2 = p2 V22 + + Z2 g 2g 3 Fig. 4.6 Solution: By continuity criterion, p p V2 (D)2 = V3 (d)2 = Q 4 4 2 2 Êdˆ Ê 25 ˆ . V Velocity in the pipe V2 = V3 Á ˜ = Á Ë D¯ Ë 100 ˜¯ 3 1 = V3 16 Apply Bernoulli equation to points 1 and 3 with the centre line of the pipe as datum and atmospheric pressure as zero. The velocity at point 1 can be taken as zero. p1 V12 p V2 + + Z1 = 3 + 3 + Z3 + H L g 2g g 2g 0+0+H =0+ V32 V22 + 0 + 20 2g 2g 148 Fluid Mechanics and Hydraulic Machines H= = 1.07813 V32 2g Ê 2 ¥ 9.81 ¥ 4.0 ˆ As H = 4.0 m, V3 = Á Ë 1.07813 ˜¯ 2 1/ 2 = 8.5319 m/s p Discharge Q = ¥ (0.025)2 ¥ 8.5319 4 = 4.1881 ¥ 10–3 m3/s = 4.1881 L/s V2 = Velocity in the pipe 1 = ¥ 8.53188 = 0.5332 m/s 16 Loss of head in the pipe HL = 20 ¥ V22 2g (0.5332) 2 = 0.2899 m 2 ¥ 9.81 Applying Bernoulli equation to points 1 and 2. = 20 ¥ p2 V22 + + 0 + HL g 2g 0+0+H = p2 (0.5332) 2 + + 0 + 0.2899 2 ¥ 9.81 g 4.0 = p2 = 4.0 – 0.3044 = 3.6956 m g g = (0.8 ¥ 998 ¥ 9.81/1000) = 7.832 kN/m3 p 2 = pressure at the base of the nozzle = 3.6956 ¥ 7.832 = 28.944 kPa 4.8 h 15 cm 2 2 V32 Ê 1 ˆ V3 + 20 ¥ Á ˜ Ë 16 ¯ 2g 2g 80 cm Water 1 x h=? Flow 30 cm Mercury Fig. 4.7 Solution: Let S = Relative density of mercury. For the manometer: Considering the elevation of section 1 as datum p1 p + x + h = 2 + 0.8 + x + Sh g g Ê p1 p2 ˆ ÁË g - g ˜¯ – 0.8 = (S – 1)h = (13.6 – 1)h = 12.6 h (1) By continuity criterion, p Q = ¥ (0.30)2 ¥ V1 4 p = ¥ (0.15)2 ¥ V2 = 0.120 m3/s 4 V1 = 1.6977 m/s, V2 = 6.79 m/s By Bernoulli equation for points 1 and 2, p1 V12 p V2 + + Z1 = 2 + 2 + Z 2 g 2g g 2g 149 Energy Equation and Its Applications Solution: By continuity V1 D 12 = V2 D22 Ê p1 p2 ˆ V22 - V12 ÁË g - g ˜¯ + 0 – 0.8 = 2g (6.79) 2 - (1.6977) 2 = 2 ¥ 9.81 = 2.2034 4.9 2 = 0.49V2 g = 0.800 ¥ 9.81 = 7.848 kN/m3 800 = 0.8016 998 For manometer connected to 1 and 2: Relative density of kerosene = p1 p Ê 13.6 ˆ h1 + x + h1 + x = 2 + Á g g Ë 0.8016 ˜¯ (p1 – p2)/g = 15.966 ¥ 0.04 = 0.6386 m of kerosene. By applying Bernoulli equation to sections 1 and 3 r Ê 7.0 ˆ V1 = V2 (D2 /D1)2 = V2 Á Ë 10.0 ˜¯ Hence Ê p1 - p2 ˆ ÁË g ˜¯ – 0.8 = 12.6h = 2.2034 2.2034 Therefore h= = 0.175 m 12.6 = 17.5 cm Deﬂection: As indicated in Fig. 4.7, the manometer limb connected to section 1 will be having a smaller column of mercury than the other limb. 2, Z h D D3 D p1 V12 p V2 + + Z1 = 2 + 2 + Z2 g 2g g 2g D ( p1 - p2 ) V 2 - V12 = 2 g 2g As Z1= Z2, D = Q V Z3 V22 V22 [1 – (0.49)2] = 0.7599 2g 2g h Horizontal pipe Kerosene consideration, 1 3 2 4 5 Atmos Z0 D1 x D2 h1 D3 = D 4 Z3 D4 h4 Datum Fig. 4.8 Example 4.9 D5 150 Fluid Mechanics and Hydraulic Machines \ 0.7599 V22 = 0.6386 2g V2 = 4.06 m/s, V1 = 0.49 ¥ 4.06 = 1.9899 m/s 4.10 p ¥ (0.07)2 4 = 0.015625 m3/s = 15.625 L/s (i) Discharge Q = 4.06 ¥ 0.015625 = 7.9577 m/s p ¥ (0.05) 2 4 (iii) V4 = V1 = 1.9899 m/s (ii) Velocity V5 = By applying Bernoulli equation to sections 4 and 5, Solution: Consider points 1 and 2 at the surface of the oil in tank A and at the outlet as in Fig. 4.9. The velocity V1 can be assumed to be zero. Between points 1 and 2, by Bernoulli equation we have p4 V12 P V2 + + Z 4 = atm + 5 + Z5 g 2g g 2g Patm = 0, and Z5 = Z4 p4 = (V 25 – V 12 )/2g g (7.9577) 2 - (1.9899) 2 = 2 ¥ 9.81 = 3.0258 m Total energy at point 4: p4 V12 + + Z4 g 2g (1.9899) 2 = 3.0258 + + Z3 2 ¥ 9.81 As there is no loss of energy, H4 = Z 0 = 20.00 (1.9899) 2 Hence Z3 = 20.0 – 3.0258 – 2 ¥ 9.81 = 16.7724 m (iv) For the manometer at point 4 H4 = p4 Ê 13.60 ˆ = h4 Á - 1˜ = 3.0258 Ë 0.8016 ¯ g 3.0258 = 0.18946 m 15.97 = 0.189 m h4 = C EL: 5.50 m 1 EL: 4.00 m Tank-A EL: 1.00 m 2 V2 Fig. 4.9 p1 V12 p V2 + + Z1 = 2 + 2 + Z 2 + H L(1–2) g 2g g 2g 0 + 0 + 4.00 = 0 + V22 + 1.00 + (0.5 + 120) 2g V22 = 4.00 – 2.70 = 1.30 2g V2 = (2 ¥ 9.81 ¥ 1.30)1/2 = 5.05 m/s p ¥ (0.15)2 ¥ 5.05 4 = 0.0892 m3/s = 89.2 L/s (i) Discharge = Q = 151 Energy Equation and Its Applications (ii) Using sufﬁx 3 to denote the conditions at the summit C, by applying Bernoulli equation to points 1 and 3, p1 V1 2 p V2 + + Z1 = 3 + 3 + Z3 + H L(1–3) g 2g g 2g V3= V2 = 5.05 m/s as the pipe is of uniform cross section. p1 p V2 V2 + Z1 + 1 = 2 + Z2 + 2 2g 2g g g 10.3 + 0 + 0 = 0.2 + Hm + 1.274 where Hm is the maximum height of the summit above the tank water surface. Hm = 10.3 – 0.2 – 1.274 = 8.826 m \ 0 + 0 + 4.0 = p3 (5.05) 2 + + 5.50 + 0.5 2 ¥ 9.81 g p3 = 4.0 – 1.30 – 5.50 – 0.5 g = – 3.3 m p3 = – 3.3 ¥ (0.8 ¥ 998 ¥ 9.81)/1000 = – 25.85 kPa (gauge) 4.12 L Atmosphere 1 4.11 150 m A L=? Solution: Consider Section 1 at the tank water surface where p1 = 10.3 m (abs) and V1 = 0 g p1 p V2 V2 + Z1 + 1 = 2 + Z2 + 2 g 2g 2g g At the summit, Section 2, least pressure that can p occur = 2 = 0.2 m (abs) g Velocity head in the siphon pipe = V12 2g = (5.0) 2g 2 = 1.274 m By applying Bernoulli’s equation to Section 1 and Section 2: 2 Atmosphere V2 Fig. 4.10 Solution: pA = vapour pressure = 4.00 kPa (abs) p1 = p2 = atmospheric pressure = 95.48 kPa (abs) By applying Bernoulli equation to points 1 and A. p1 V1 2 p V2 + + Z1 = A + A + Z A g 2g g 2g VA2 Ê 95.48 ˆ Ê 4.00 ˆ + +0 ÁË 9.79 ˜¯ + 0 + 1.5 = Á 2g Ë 9.79 ˜¯ 152 Fluid Mechanics and Hydraulic Machines VA2 95.48 - 4.00 + 1.5 = 10.844 m = 2g 9.79 VA= (2 ¥ 9.81 ¥ 10.844)1/2 = 14.59 m/s 1 V2 = VA = 14.59/2 2 = 7.29 m/s \ Solution: p Discharge Q = 0.015 m3/s = ¥ (0.05)2 ¥ V2 4 V2 = 7.639 m/s By applying Bernoulli equation to points 1 and 2, p1 V1 2 + + Z1 g 2g p1 +0+0 g p1 g Hence, p1 By applying Bernoulli equation to points 1 and 2, with datum at point 2, p1 V1 2 p V2 + + Z1 = 2 + 2 + Z 2 g 2g g 2g (95.48) + 0 + (L + 1.50) 9.79 p2 V22 + + Z2 + HL g 2g (7.639) 2 =0+ + 3.00 + 1.5 2 ¥ 9.81 = 7.475 m and g = 9.79 kN/m3 = = 9.79 ¥ 7.475 = 73.18 kPa 4.14 V – V (7.29) 2 Ê 95.48 ˆ = Á + +0 2 ¥ 9.81 Ë 9.79 ˜¯ L = 2.709 – 1.50 = 1.21 m 4.13 p Solution: Refer to Fig. 4.12. D1 = 10 cm, A1 = EL: 3.00 m 2 Roof 5 cm Dia Air p ¥ (0.1)2 4 = 7.854 ¥ 10–3 m2 Energy line Piezometric headline 0.572 m 0.3605 m HL = 0.119 m 0.453 m 0.4398 m p1 1 EL: 0.00 m CL 20 cm 10 cm 1 Fig 4.11 2 Fig. 4.12 153 Energy Equation and Its Applications p ¥ (0.2)2 4 = 3.142 ¥ 10–2 m2 D2 = 20 cm, A 2 = Q = V1 A1 = V2 A2. 2 1 ÊD ˆ Hence V2 = Á 1 ˜ V1 = V1 4 Ë D2 ¯ Head loss at the expansion H L = (V1 – V2)2 /2g (V1 - V1 / 4) 2 9 V1 2 = 2g 16 2g By applying Bernoulli equation to sections 1 and 2 = p1 V1 2 p V2 + + Z1 = 2 + 2 + Z 2 + H L g 2g g 2g Ê p2 p1 ˆ V12 V22 ÁË g - g ˜¯ = 2g – 2g + Z1 – Z2 – H L 660 V2 = 1 (0.85 ¥ 9790) 2g 1 9ˆ Ê ÁË1 - 16 - 16 ˜¯ + 0 3 V12 8 2g 8 V 12 = ¥ 2 ¥ 9.81 ¥ 0.0793 3 = 4.1496 V1 = 2.037 m/s Discharge Q = A1V1 = 7.854 ¥ 10–3 ¥ 2.037 m3/s = 0.016 m3/s = 16.0 L/s V1 = 2.037 m/s, V12/2g = 0.2115 2.037 V2 = = 0.509 m/s, 4 V22 /2g = 0.0132 m (V1 – V2) = 1.528 m/s, (V - V ) 2 H L = 1 2 = 0.1190 m 2g p1 3000 = 0.3605 m = 0.85 ¥ 9790 g p2 = 0.3605 + 660/(0.85 ¥ 9790) g = 0.4398 m With centre line of pipe as datum, the elevations of energy and piezometric head are as follows: At Section 1: p1 V2 + 1 + Z1 g 2g = 0.3605 + 0.2115 + 0 = 0.572 m p Piezometric head = h1 = 1 + Z1 = 0.3605 m g At Section 2: p V2 Energy head = H2 = 2 + Z2 + 2 2g g = 0.4398 + 0.0132 = 0.453 m Piezometric head = h2 = 0.4398 m Energy loss = H1 – H 2 = 0.572 – 0.453 = 0.119 m = (V12 – V22)/2g The energy line and the piezometric head line (hydraulic grade line) are shown in Fig. 4.12. Energy head = H1 = = 4.15 3 Solution: In this case the sea around the torpedo is stationary and the torpedo is in motion. Hence, this is an unsteady ﬂow motion. However, this could be converted to equivalent steady motion by considering relative ﬂuid motion. The torpedo is considered to be stationary and the sea water is assumed to move with an approach velocity of 25 m/s (Fig. 4.13). The Relative flow of sea water po TORPEDO (Stationary) Vo Stagnation point Vs = 0, ps Fig. 4.13 Equivalent Steady Flow 154 Fluid Mechanics and Hydraulic Machines pressure at the nose of the torpedo is the stagnation pressure corresponding to this approach velocity. Referring to the Fig. 4.13 by using the sufﬁx O to denote the approach ﬂow and the sufﬁx s to denote the stagnation point conditions, the Bernoulli theorem applied to a streamline passing through O and S is po p V2 V2 + Zo + 0 = s + Zs + s 2g 2g g g Considering the horizontal plane passing through O and S as datum, in the present case po Zo = Zs = 0, and = 15.0 m. g Further, since S is a stagnation point Vs = 0. Thus ps p V2 ( 25) 2 = o + s = 15.0 + = 46.86 2 ¥ 9.81 2g g g m ps = 1025 ¥ 9.81 ¥ 46.86 = 471.2 KN/m2 1000 4.16 Total energy line 2 Vox/2g 2 2 + Voy Vox V 2/2g = Vox 2g Trajectory Vo 2 ym = Voy/2g y Voy q O Vox x Fig. 4.14 Voy 1 – g x2/(Vox)2 2 Vox g x2 y = x tan q – 2 2Vo cos 2 q Equation of the trajectory: y =x y = x tan 60° – 2 ¥ (6) 2 ¥ cos 2 60 (6) y = 1.732x – 0.545x2 Equation (6) is the equation of the trajectory. (ii) At the point of maximum elevation, Vy = 0 and hence 2 Vox2 V 2 Voy =H + ym = ox + 2g 2g 2g ym = or Solution: (i) In a free jet the pressure is atmospheric throughout the trajectory. Referring to Fig. 4.14. Vox = Vo cos q = constant = Vx (1) Voy = Vo sin q (2) x = Vox t (3) 1 2 y = Voy t – g t (4) 2 x t= and substituting in Eq. 4 Vox (9.81) x 2 (5) Voy2 2g In the present case Vo = 6.0 m/s and q = 60° Vox = 6.0 cos 60° = 3.0 m/s If d m is the diameter of the jet at the point of maximum elevation Êp ˆ Ê pˆ Ê 5 ˆ Vox Á dm2 ˜ = Vo Á ˜ Á Ë4 ¯ Ë 4 ¯ Ë 100 ˜¯ dm = 5 2 Ê V ˆ ÁË V ˜¯ = 5 ox 6.0 3.0 = 7.07 cm Voy = 6.0 sin 60° = 5.196 m/s 155 Energy Equation and Its Applications Maximum elevation of the jet = Voy2 (5.196) 2 = 1.376 m 2 ¥ 9.81 2g above the jet exit level (iii) For maximum horizontal distance at jet exit level: Putting y = 0 in the trajectory equation ym = = 1.732 x – 0.545 x 2 = 0 x (1.732 – 0.545x) = 0 x = 0 or x = 1.732/0.545 = 3.178 m Hence maximum horizontal distance at jet exit level is x m = 3.178 m. 4.17 x tan q – 7.848 ¥ 10–3 x2 sec2 q – 13.5 = 0 (2) The maximum value of x is obtained by differentiating with respect to q and putting dx/dq = 0. On differentiation with respect to q. Ê dx ˆ 2 –3 ÁË x sec q + tan q d q ˜¯ – 7.848 ¥ 10 Ê dx ˆ ¥ Á x 2 ◊ 2 sec 2 q ◊ tan q + 2 x sec 2 q ◊ ˜ = 0 dq ¯ Ë dx Putting =0 dq x sec2 q – 7.848 ¥ 10–3 (2x 2 sec2 q . tan q ) = 0 63.71 1 – 0.0157 x tan q = 0 or x = tanq Substituting this in Eq. 2, Solution: Refer to Fig. 4.15. Window B Vo A Eliminating t, the equation of the trajectory is g x2 y = x tan q – sec2 q (1) 2Vo2 Substituting y = 15 – 1.5 = 13.5 m, 9.81 13.5 = x tan q – x 2 sec2 q 2 2 ¥ ( 25) 15 m q 1.5 m x È (63.71) 2 ˘ sec 2 q ˙ – 13.5 = 0 63.71 – 7.848 ¥ 10–3 Í 2 ÍÎ tan q ˙˚ 31.855 – 13.5 = 0 63.71 – sin 2 q sin q = 0.7965 and q = 52.8° tan q = 1.3174 63.71 63.71 = x= = 48.36 m 1.3174 tanq 4.18 D Fig. 4.15 Let q V0 x y = angle of inclination of the nozzle. = 25 m/s = Vo cos q . t 1 = Vo sin q . t – gt 2 2 L H 156 Fluid Mechanics and Hydraulic Machines 3 Thus the results are For Oriﬁce A 1 H B 1 Rounded orifice For Pipe Velocity at 1 = 2g H 2g ( H + L) Velocity at 2 = 2g ( H + L) 2g ( H + L) Discharge Q = p 2 D 2g H 4 p 2 D 2g ( H + L) 4 L Pipe with rounded entry 2 Fig. 4.16 4.19 2 Example 4.18 Vr Rounded entry oriﬁce: Applying Bernoulli equation to a point on the water surface 3 and point 1. p1 V1a2 + +0 g 2g p As the oriﬁce discharges to atmosphere, 1 = 0 g and Vla = 2gH . 0+0+H= p 2 D 2g H . 4 At point 2, the pressure is atmospheric and hence by applying Bernoulli equation between points 3 and 2. The discharge Q a = V22a +0 2g Solution: For an irrotational vortex Vr = C and Bernoulli equation can be used across the streamlines also. Considering the duct to be in a horizontal plane, applying Bernoulli equation to points 1 and 2 shown in Fig. 4.17, p2 V22 p V2 + + Z 2 = 1 + 1 + Z1 g 2g g 2g As the discharge is Q a, the diameter at 2 will be smaller than D. Z2 = Z1 and V = C/r r (p2 – p1) = (V12 – V22 ) 2 r C2 Ê 1 1ˆ - 2˜ = Á 2 2 Ë r1 r2 ¯ Pipe: By applying Bernoulli equation between points 3 and 2. V 0 + 0 + (H + L) = 0 + or V2a = 2g ( H + L) V2 0 + 0 + (H + L) = 0 + 2 b + 0 2g or V2b = 2g ( H + L) As the pipe size is uniform from point 1 to 2, by continuity criterion V1b = V2b = 2g ( H + L) As 2 r 1 Fig. 4.17 O 157 Energy Equation and Its Applications Thus 998 ¥ C2 30 ¥ 10 = 2 3 È 1 1 ˘ Í ˙ 2 (0.65) 2 ˚ Î (0.4) = 1937.7 C2 C = 15.48 and C = 3.9348 Discharge per unit depth 2 q= Ú r2 v dr r1 r2 C dr = C ln r2/r1 r 0.65 = 3.9348 ln 0.40 = 1.910 m3/s per metre width = Ú r1 Va = 0.06 = 3.395 m/s, Va2 = 0.588 m 2g p ¥ (0.15) 2 4 0.06 V2 = 7.639 m/s, b = 2.975 m Vb = p 2g ¥ (0.10) 2 4 Power delivered by pump P = g Q H p = 10 kW 9.79 ¥ 0.060 ¥ Hp = 10 H p = head delivered by the pump = 17.02 m (i) By applying Bernoulli equation to points C and A, 0+0+3 = 4.20 pa V2 + a +0 g 2g pa = 3 – 0.588 = 2.412 m g pa = 9.79 ¥ 2.412 = 23.61 kPa By applying Bernoulli equation between C and B with level at A as datum: pc Vc2 p V2 + + Zc + H p = b + b + Zb g 2g g 2g 0 + 0 + 3.0 + 17.02 B 1.20 m C 10 cm Dia 3.0 m A P = pb + 2.975 + (3.0 + 1.2) g pb = 12.845 m g and p b = 9.79 ¥ 12.845 = 125.75 kPa (ii) When losses are considered: By applying Bernoulli equation between C and B with level at A as datum, pc Vc2 p V2 + + Zc + H p = b + b + Zb + HL g 2g g 2g 0 + 0 + 3.0 + 17.02 = 15 cm Dia Fig. 4.18 Solution: Discharge Q = 0.060 m3/s = AaVa = A bVb and pb + 2.975 g (3.0 + 1.2) + (2 ¥ 2.975) pb = 6.895 m g p b = 9.79 ¥ 6.895 = 67.50 kPa 158 Fluid Mechanics and Hydraulic Machines 4.21 – The elevation of point 4, the summit of the jet, is = 4.00 + 19.22 = 23.22 m (iii) To ﬁnd the power delivered by the pump: Apply Bernoulli equation to points 1 and 3: 0 + 0 + 0 + Hp = 0 + V32 /2g + 4.00 Hp = 19.22 + 4.00 = 23.22 m Solution: Applying Bernoulli equation to points 1 and 2 (Fig. 4.19). Power delivered by pump P = g QH p P = 9.79 ¥ 0.1525 ¥ 23.22 = 34.67 kW 4.22 4 h EL: 4.00 m 3 25 cm Dia Nozzle 2 EL: 2.50 m P 3 30 cm Dia EL: 0.00 m 1 Water Fig. 4.19 1 V1 2 + 2.5 2g V12 /2g = 0.492 m and V1 = 3.107 m/s 0 + 0 + 0 = (– 0.22 ¥ 13.6) + p ¥ (0.25)2 ¥ 3.107 4 = 0.1525 m3/s = 152.5 L/s 40 m 60 cm Dia (i) Discharge Q = A1V1 = (ii) V3 = V1 ¥ (D1/D3)2 = 3.107 ¥ (25/10)2 = 19.419 m/s (19 .419 ) 2 = = 19.22 m 2 ¥ 9.81 Hence, the height to which the jet will reach h = 19.22 m. V32 /2g T 2 45 cm Dia Fig. 4.20 Solution: (i) Water friction loss is neglected: Discharge Q = 0.8 m3/s. V2 = Velocity at outlet 0.8 = = 5.03 m/s. p ¥ (0.45) 2 4 159 Energy Equation and Its Applications Applying Bernoulli equations to points 1 and 2, 2 Solution: Q 0.5 = = 3.98 m/s p Aa ¥ (0.4) 2 4 0.5 Vb = = 1.768 m/s p ¥ (0.6) 2 4 Applying Bernoulli equation to points A and B: 2 p1 V1 V + + Z1 = H T + 2 + Z2 g 2g 2g Va = (5.03) 2 +0 2 ¥ 9.81 H T = Head extracted by the turbine = 40 – 1.29 = 38.71 m Power extracted by the turbine P = g Q HT = 9.79 ¥ 0.8 ¥ 38.71 = 303.2 kW (ii) When losses are included: By applying Bernoulli equation to points 1 and 2 0 + 0 + 40 = H T + (5.03) 2 + 10.0 + 0 2 ¥ 9.81 HT = 28.71 m Power extracted by the turbine = g QHT Power output of the turbine Pn = g QHT h where h = efﬁciency of the turbine \ Pn = 9.79 ¥ 0.8 ¥ 28.71 ¥ 0.85 = 191.1 kW pa Va2 p V2 + + Za = b + b + Z b + H t g 2g g 2g (3.98) 2 (1.768) 2 + 2.0 = – 4 + + 0 + Ht 2 ¥ 9.81 2 ¥ 9.81 H t = 32.807 + 3.841 = 36.65 m Power output P = g Q H t ¥ h = 9.79 ¥ 0.5 ¥ 36.65 ¥ 0.90 = 161.45 kW 30.0 + 0 + 0 + 40 = HT + 4.24 factor a r u u 4.23 Ê rˆ ÁË 1 - r ˜¯ 0 u u – r r Solution: Ê u rˆ (i) = Á1 - ˜ um r0 ¯ Ë 3 Refer to Fig. 4.22(a) The kinetic energy correction factor 1 a= 3 u 3 dA A V A where V = average velocity. – Ú 40 cm Dia Turbine A Q = p r 02 V = 2.0 m = 0 B 60 cm Dia Fig. 4.21 Ú r0 Ú r0 u2prd r 0 Ê rˆ um Á1 - ˜ 2prd r r0 ¯ Ë Ê r2 r2ˆ p r02 = 2pu m Á 0 - 0 ˜ = um 3¯ 3 Ë 2 1 V = um 3 160 Fluid Mechanics and Hydraulic Machines a = r0 um u r CL = (a) 1 V 3 p r02 Ú r0 0 Ê r2 ˆ 3 um Á1 - 2 ˜ 2pr dr r0 ¯ Ë 3 2 p um ( u m / 2) 3 p r02 ¥ Ú r0 0 Ê r3 r5 r7 ˆ Á r - 3 2 + 3 4 - 6 ˜ dr r0 r0 r0 ¯ Ë V 16 È 2 Ê 1 3 3 1 ˆ ˘ Ír0 Á - + - ˜ ˙ r02 Î Ë 2 4 6 8 ¯ ˚ a = 2.0 = um CL r r0 u (b) 4.25 Fig. 4.22 Now a = 1 V 3p r02 Ú r0 0 3 2 p um = 3 Ê rˆ u 3m Á1 - ˜ 2 prd r r0 ¯ Ë 3 Ú r0 Ê rˆ ÁË1 - r ˜¯ rd r 0 Ê1 ˆ 2 0 ÁË 3 um ˜¯ p r0 54 È Ê 1 3 3 1 ˆ ˘ = 2 Ír02 Á - + - ˜ ˙ r0 Î Ë 2 3 4 5 ¯ ˚ = (ii) 54 = 2.7 20 2 ÈÊ u rˆ ˘ = ÍÁ1 - ˜ ˙ um r0 ¯ ˙ ÍË Î ˚ Refer to Fig. 4.22(b) Q= = p r 20 V Ú r0 0 = Ú r0 3 correction factor a Solution: Distribution (a): u1 y. B Consider unit width of the conduit. Refer to Fig. 4.23(a). u Average velocity V = 1 2 B 3 1 a = 3 u dy V B 0 3 B Êu 1 ˆ 1 = ◊ y Á ˜¯ dy 3 0 Ë B Ê u1 ˆ ÁË 2 ˜¯ B u= Ú Ú = u2 pr dr 0 Ê r2 ˆ u m Á1 - 2 ˜ 2pr dr r0 ¯ Ë Ê r2 r2 ˆ p um r02 = 2 p um Á 0 - 0 ˜ = 4¯ 2 Ë 2 um V= 2 8 Ê B4 ˆ = 2.0 B ÁË 4 B3 ˜¯ Distribution (b): and u = constant for 0 £ y £ 2B/3 u = 0 for y > 2B/3. Refer to Fig. 4.23(b). Consider unit width of the conduit. 2 ˆ Ê Discharge q = V . B = Á u ¥ B˜ + 0 Ë 3 ¯ 161 Energy Equation and Its Applications EL: 25.00 m 1 D1 = 25 cm B u y u1 (a) EL: 20.00 m D2 = 35 cm 2 Fig. 4.24 B 2 B 3 y u 0.20 = 4.074 m/s p 2 ¥ (0.25) 4 0.20 V2 = = 2.079 m/s p ¥ (0.35) 2 4 V1 = (b) Fig. 4.23 2 u 3 Average velocity V = a= = È Í 3 V B Î 1 Ú 2/3B 0 1 ( 2 / 3 u )3 B u3d y + Ú B 2/3B ¥ u3 [y] 02/3 B 27 2 9 B= 8B 3 4 a = 2.25 = ˘ u 3d y˙ ˚ (V1 – V2) = 1.995 m/s a1 = 1.1 and a2 = 1.5 By applying Bernoulli equation to points 1 and 2, p1 V2 p V2 + a1 1 + Z1 = 2 + a 2 2 + Z2 + HL g 2g g 2g ( 4.074) 2 Ê 120 ˆ ÁË 9.79 ˜¯ + 1.1 ¥ 2 ¥ 9.81 + 25.00 4.26 = p2 ( 2.079) 2 1.2 ¥ (1.995) 2 + 1.5 ¥ + 20.00 + 2 ¥ 9.81 2 ¥ 9.81 g 12.257 + 0.931 + 25.00 = 3 V –V p2 + 0.330 + 20.00 + 0.243 g p2 = 38.188 – 20.573 = 17.615 m g p2 = 9.79 ¥ 17.615 = 172.45 kPa Solution: Q = 0.20 m3/s 162 Fluid Mechanics and Hydraulic Machines Problems 4.1 For the pipeﬂow system shown in Fig. 4.25 the following data are available: Item Diameter Elevation (m) Pressure Velocity Point 1 20 cm 103.00 55 kPa 2.5 m/s Point 2 30 cm 106.00 75 kPa tank. (Ans. Q = 22 L/s; h1 = 3.03 cm; V4 = 5.022 m/s; H = 1.285 m) H 1 2 3 4 Water h1 2 Fluid r = 800 kg/m h3 Fig. 4.26 3 1 Horizontal pipe Datum Fig. 4.25 Determine the direction of ﬂow and the loss of energy between these two points. (Ans. The ﬂow is from 2 to 1. HL = 5.292 m) 4.2 A 15 cm diameter pipe is reduced to 7.5 cm diameter through gradual contraction. At this contraction the difference between the piezometric heads at the main and the contracted section is 4 cm of mercury. By neglecting losses, calculate the discharge of water. (Ans. Q = 14.9 L/s) 4.3 For the ﬂow system shown in Fig. 4.26 diameter D1 = D3 = 20 cm, D2 = 10 cm, D4 = 7.5 cm. The ﬂuid in the manometer is mercury. The mercury water differential head h3 = 10 cm. Assuming zero energy loss, ﬁnd (i) the discharge, (ii) manometer differential head h1, (iii) velocity of ﬂow at section 4 and (iv) head of water H in the 4.4 Two points A and B are located in a long 20 cm diameter pipe. When a downstream valve is completely closed the difference in pressure between B and A, (pB – pA) = 100 kPa. When the valve is open and a discharge of 70 L/s of water is ﬂowing, (pA – pB) = 50 kPa. Calculate the head loss between A and B. (Ans. H L = 15.32 m) 4.5 A siphon consisting of a 3 cm diameter tube is used to drain water from a tank. The outlet end of the tube is 2.0 m below the water surface in the tank. Neglecting friction, calculate the discharge. If the summit of the siphon is 1.4 m above the water surface in the tank, estimate the pressure at the summit of the siphon. (Ans. Q = 4.428 L/s, p = – 33.29 kPa) 4.6 For the system shown in Fig. 4.27, ﬁnd the height h to which the jet from the nozzle would rise. If the nozzle and the pipe were to have diameters of 10 cm and 20 cm respectively, calculate the discharge and the velocity in the pipe. Neglect frictional 163 Energy Equation and Its Applications losses. (Ans. h = 2.361 m, Q = 53.5 L/s, Vp = 1.702 m/s) Air 10 kPa Oil RD = 0.85 0.40 m Water 0.50 m h=? 0.5 m Pipe Nozzle Fig. 4.27 4.7 A 10 cm long nozzle of exit diameter 10 cm is attached to a pipe of 30 cm diameter. The nozzle is vertical. If a water jet issuing out of the nozzle reaches a height of 4.5 m above the nozzle exit, calculate the discharge. Also, by assuming a head loss in the nozzle equal to 10% of the exit velocity head, calculate the pressure at the base of the nozzle. (Ans. Q = 73.8 L/s, p1 = 48.895 kPa) 4.8 Two sections A and B at the two ends of a transition in a pipe line carrying water have the following properties: Section Datum height (Z) Diameter Pressure Kinetic energy correction factor A 13.00 m 15 cm 10.00 kPa 1.5 4.9 A 25 cm diameter pipe carries oil of speciﬁc gravity 0.8 at the rate of 150 litres per second. At a point A, which is 3.5 m above the datum, the pressure is 19.62 kN/m2. Calculate the total energy at section A in meters of oil. (Ans. H = 6.476 m of oil) 4.10 When a body A moves through still water at a constant velocity of 4.5 m/s, the velocity of water at point M which is 0.8 m/s ahead of the nose of the body is found to be 3.0 m/s. What will be difference in pressure between the nose and the point 0.8 m ahead of it? (Ans. 1125 N/m2) 4.11 A nozzle at the end of a hose has a diameter of 5 cm. The inclination of the nozzle is at 45° to the horizontal and is directed upwards. A point on the jet axis is at a distance of 3 m from the nozzle and 2.0 m above it. Estimate the discharge from the nozzle. (Ans. Q = 18.45 L/s) 4.12 A ﬁre extinguishing service is trying to train a ﬁre hose at A on to a window B as shown in Fig. 4.28. The velocity of the jet is 30 m/s. Calculate the maximum distance x of the nozzle at which this could be achieved. What is the corresponding angle q of the nozzle? (Ans. xmax = 71.51 m, q = 52.06°) B 12.00 m 12 cm 7.00 kPa 1.05 Estimate the discharge in the pipe by assuming zero energy loss between the sections. (Ans. Q = 92.2 Litres/s) B A q x 2m Nozzle Fig. 4.28 164 Fluid Mechanics and Hydraulic Machines 4.13 Water ﬂows radially outwards between two horizontal circular plates of diameter 0.80 m. The plates are 3 cm apart and the ﬂow is supplied through a 20 cm diameter pipe at the centre of the plates (Fig. 4.29). If the discharge, from the gap between the plates to atmosphere is 40 L/s. Calculate the pressure at point A in the pipe 0.8 m above the plates. (Ans. pa = – 8.5 kPa) the pressure difference (p3 – p1). Given are A11 = 20 cm3, A3 = 40 cm2. r = 1000 kg (Ans. (p3 – p1) = 158.72 kPa) A11 V11 Flow A12 A3 V3 V12 3 1 Fig. 4.30 20 cm Dia 4.16 A 15 cm diameter pipe is expanded to 25 cm diameter suddenly at a section. The head loss at a sudden expansion from section 1 to 2 is given by hL = (V1 – V2)2 /2g. For a discharge of 45 L/s for a pipeline set-up shown in Fig. 4.31. Calculate the reading h of the mercurywater differential manometer. (Ans. h = 1.21 cm) A 0.8 m 3 cm 3 cm 0.80 m Dia Fig. 4.29 15 cm Problem 4.13 4.14 A conical tube is ﬁxed vertically with its larger diameter at the top and forms a part of a pipeline carrying kerosene (RD = 0.80). The velocity at the smaller end is 3.0 m/s and at the larger end it is 1.5 m/s. The tube is 2.0 m long. At the bottom of the tube the pressure is 50 kPa. The head loss in the tube can be assumed to be 0.35 times the difference in the velocity heads at the two ends. Estimate the pressure at the top of the tube when the ﬂow is upwards. (Ans. p2 = 36.09 kPa) 4.15 Two streams of water at the same pressure p1 but with velocities V11 = 25 m/s and V12 = 5 m/s enter a mixing chamber, and after complete mixing emerge as a single stream with uniform properties [See Fig. 4.30]. If no loss of any kind occurs in the ﬂow, determine the exit velocity V3 and 1 50 cm Water 2 h 25 cm Mercury Fig. 4.31 Problem 4.16 4.17 For the ﬂow situation shown in Fig. 4.32 calculate the discharge of oil (RD = 0.75) when the oil-mercury differential manometer 165 Energy Equation and Its Applications reading h = 10 cm. Neglect all losses in the ﬂow system. (Ans. Q = 47 L/s) 2u1 10 cm B u y 2 u1 (a) 0.60 m Oil RD = 0.75 u1 u1 1 B h B/3 20 cm u1 Mercury Fig. 4.32 (b) 4.18 Show that the average velocity V and the kinetic energy correction factor a for the following velocity distribution in a pipe of radius r0: Ê rˆ u = Á1 - ˜ um r Ë 0¯ B/2 2 um V= (1 + m)( 2 + m) (1 + m)3 ( 2 + m)3 4 (1 + 3 m) ( 2 + 3 m) In the above um = maximum velocity and u = velocity at any radius r, and m = a coefﬁcient. 4.19 Find the kinetic energy correction factor a for the velocity distributions in a twodimensional duct, as shown in Fig. 4.33(a), (b) and (c). 10 2 È ˘ Í Ans. (a ) a = 9 , ( b) a = 4 3 , (c) a = 1.543˙ Î ˚ u y y2 y u – 2 um = 4 B B (c) are given by a= um B m Fig. 4.33 Problem 4.19 4.20 A conical pipe has diameters 0.40 m and 0.80 m at its two ends. The smaller end is 2 m above the larger end. For a ﬂow of 0.30 m3/s of water the pressure at the lower end is 10 kPa. Assuming a head loss of 2 m and kinetic energy correction factor a = 1.1 and 1.5 at the smaller and larger ends respectively, estimate the pressure at the smaller end. (Ans. p1 = 7.144 kPa) 166 Fluid Mechanics and Hydraulic Machines 4.21 A pipeline has the following data at its two sections A and B: Item Section A Diameter 30 cm Elevation (m) 10.000 Pressure 40.0 kPa Kinetic energy correction factor, a 1.08 Section B 45 cm 16.000 30 kPa 1.25 Assume a head loss equal to 20 times the velocity head at A and calculate the discharge of water through this pipeline when ﬂowing from B to A. (Ans. Qa = 147.5 L/s) 4.22 A pump has a 30 cm diameter suction pipe and 25 cm diameter delivery pipe. When 220 L/s of water was being pumped, the pressure on the suction side of the pump was 4 m of vacuum and on the delivery side the pressure was 100 kPa. Assuming an efﬁciency of 50% for the pump-motor set, estimate the electrical power consumed. (Ans. P = 63.51 kW) 4.23 A mercury-water differential manometer connected to the 15 cm diameter suction pipe and 12 cm delivery pipe of a pump shows a deﬂection of 40 cm. The centerlines of the suction and delivery pipes are at the same level. If the pump is discharging 70 L/s of water, estimate the head developed by the pump. (Ans. H p = 6.193 m) 4.24 A pump draws from a sump whose water surface is 1.5 m below the centre of pump and discharges it freely to atmosphere at 1.2 m above the pump centreline. The suction and delivery pipes are 20 cm and 25 cm in diameter respectively. If the pressure at the suction end of the pump is (–2.0) cm of mercury (gauge) determine the discharge and power imparted by the pump. Neglect all losses. (Ans. Q = 153.7 L/s, P = 4.81 kW) 4.25 Determine the shaft power for a 70% efﬁcient pump to discharge 1.5 m3/min of oil of RD = 0.90 from a tank with oil surface elevation 100.00 m to another with oil surface elevation at 120.00 m. The pipeline is of 15 cm diameter. The headloss in the pipe can be taken to be 10 times the velocity head in the pipeline. (Ans. Ps = 6.615 kW) 4.26 For a hydraulic machine shown in Fig. 4.34 the following data are available: A x x M B Fig. 4.34 Flow Diameters Elevation (m) Pressures Discharge Problem 4.26 : : : : From A to B at A: 20 cm; at B : 30 cm at A: 105.00; at B: 100.00 at A: 100 kPa; at B: 200 kPa : 200 L/s of water. Is this machine a pump or a turbine? Calculate the power input or output depending on whether it is pump or a turbine. (Ans. The machine is a pump. P = 6.965 kW) 4.27 A 20 cm diameter pipe leading water from a reservoir ends in a nozzle of exit diameter of 10 cm at elevation 90.00 m. The water surface in the reservoir is at elevation 100.00 m (Fig. 4.35). The energy loss in the system can be assumed as 12 times the velocity head in the pipe. Calculate the 167 Energy Equation and Its Applications EL: 100.00 m commencement of the draft tube, next to the turbine, which is 3.50 m above the tailwater, (Fig. 4.36) (Ans. P = 405 kW; Pb = – 34.79 kPa) Nozzle: 10 cm Dia 20 cm Dia pipe Fig. 4.35 EL: 90.00 m Turbine Problem 4.27 T discharge. If the discharge is to be increased by 75%, calculate the power input through a pump introduced in the pipeline at the base of the nozzle. (Ans. Q0 = 83.16 L/s; P = 12.236 kW) 4.28 A turbine discharges 2.0 m3/s of water into a vertical draft tube as shown in Fig. 4.36. The diameter of the tube is 0.8 m at A. If the head loss in the draft tube can be assumed as 1.5 times the velocity head at A, estimate the pressure at A. (Ans. pa = – 30.32 kPa) 4.29 A reaction turbine has a supply pipe of 0.80 m diameter and a draft tube with a diameter of 1.2 m at the turbine and expanding gradually downwards. The tailwater surface is 4.0 m below the centreline of the supply pipe at the turbine. For a discharge of 1.1 m3/s, the pressure head just upstream of the turbine is 40 m. Estimate the power output of the turbine by assuming 85% efﬁciency. Also, determine the pressure at the A Dia. Da = 0.8 m 3.5 m B 1.2 m Fig. 4.36 4.30 A pipeline delivering water from a reservoir is shown in Fig. 4.37. A pump M adds energy to the ﬂow and 45 L/s of water is discharged to atmosphere at the outlet. Calculate the power delivered by the pump. Assume the head loss in the pipe as two times the velocity head at the suction side and 10 times the velocity head in the delivery pipe. Draw a neat sketch showing energy line and hydraulic grade lines. (Ans. P = 5.218 kW) EL: 108.00 EL: 100.00 15 cm Dia P 20 cm Dia Pump: EL: 103.00 Fig. 4.37 168 Fluid Mechanics and Hydraulic Machines Objective Questions 4.1 The Bernoulli equation is written with usual notation as p/g + V 2/2g + Z = constant. In this equation each of the terms represents (a) energy in kg.m/kg mass of ﬂuid (b) energy in N.m/kg mass of ﬂuid (c) energy in N.m/N weight of ﬂuid (d) power in kW/kg mass of ﬂuid 4.2 Bernoulli equation is applicable between any two points (a) in any rotational ﬂow of an incompressible ﬂuid (b) in any type of irrotational ﬂow of a ﬂuid (c) in steady rotational ﬂow of an incompressible ﬂuid (d) in steady, irrotational ﬂow of an incompressible ﬂuid 4.3 The piezometric head of a ﬂow is (a) the sum of the velocity head and datum head (b) the sum of the pressure head and datum head (c) the sum of the pressure head and velocity head. (d) the sum of the velocity head, pressure head and datum head. 4.4 In a ﬂow of a real ﬂuid with no addition of energy (a) the energy line will be horizontal or sloping upward in the direction of ﬂow. (b) the energy line can never be horizontal or slopping upward in the direction of the ﬂow. (c) the piezometric line can never be horizontal or sloping downward in the direction of the ﬂow. (d) the centre line of the pipe can never be above the energy line 4.5 The total head in a ﬂow is the sum of (a) piezometric head and datum head (b) piezometric head and pressure head (c) piezometric head and velocity head (d) piezometric head, velocity head and datum head. 4.6 The difference between the total head line and the hydraulic grade line represents (a) the velocity head (b) the piezometric head (c) the pressure head (d) the elevation head 4.7 In a pipeline the hydraulic grade line is above the pipe centre line in the longitudinal section at point A and below the pipe centre line at another point B. From this it can be inferred that (a) vacuum pressures prevail at B (b) vacuum pressures prevail at A (c) the ﬂow is from A to B (d) the ﬂow is from B to A 4.8 Sections A and B in a pipeline (shown schematically in Figure 4.38 given below) are at the same elevation of 2.5 m above datum. A valve lies in-between A and B. The ﬂow parameters at A are: velocity head of 0.5 m and the pressure head of 2.5 m. The valve loss is 0.2 m. The piezometric head at B is (a) 5.5 m (b) 5.3 m (c) 5.0 m (d) 4.8 m Valve A B Fig. 4.38 169 Energy Equation and Its Applications 4.9 The dimensions of Kinetic energy correction factor a is (a) M0 L T0 (b) M0 L1 T–2 0 0 0 (c) M L T (d) M1 L2 T–2 4.10 The kinetic energy correction factor a is deﬁned as a = 1 1 (a) v3 dA (b) v3 dA 3 3 A AV Ú 1 1 v dA (d) v3 dA AV 3 AV 3 where V = average velocity in the cross section. In a two-dimensional duct ﬂow, air ﬂows in the bottom half of the duct with uniform velocity and there is no ﬂow in the upper half. The value of the kinetic energy correction factor a for this ﬂow is (a) 2.0 (b) 2 ¼ (c) 4.0 (d) 3.0 A 15 cm diameter pipe carries a ﬂow of 70 L/s of an oil (RD = 0.75). At a section 12 cm above the datum the pressure is vacuum of 2 cm of mercury. If the kinetic energy correction factor a for this section is 1.1, the total head at the section in metres of oil is (a) 0.648 (b) 0.728 (c) 0.557 (d) 0.637 A 20 cm diameter horizontal pipe is attached to a tank containing water. The water level in the tank is 7 m above the pipe outlet and the pipe discharges into the atmosphere. Assuming a total loss of 3 m in the pipe and the kinetic energy correction factor a of the jet issuing from the pipe to be 1.20, the discharge in the pipe is L/s is (a) 254 (b) 278 (c) 368 (d) 305 Water ﬂows steadily down a vertical pipe of constant cross section. Neglecting friction, according to Bernoulli’s equation, (c) 4.11 4.12 4.13 4.14 Ú Ú Ú (a) pressure is constant along the length of the pipe (b) velocity decreases with height (c) pressure decreases with height (d) pressure increases with height 4.15 In the siphon shown in Fig. 4.39 assuming ideal ﬂow, pressure pB (a) = pA (b) < pA (c) > pA (d) = pC pA pB pC Fig. 4.39 Question 4.15 4.16 A siphon used to empty a tank consists essentially of a pipe with its summit 0.5 m above the water surface of the tank and its outlet at 2.0 m below the summit. Neglecting friction and other losses, the velocity in the siphon is (a) 4.4 m/s (b) 5.4 m/s (c) 3.1 m/s (d) 3.8 m/s 4.17 Figure 4.40 shows a tank being emptied by a pipe of length L. If the friction of the pipe is neglected, the pressure at a point A at the pipe inlet would (a) increase if the length L is increased (b) be constant and equal to gL (c) remain constant at g H for all lengths L (d) decrease with an increase in L 170 Fluid Mechanics and Hydraulic Machines H A 4.22 L Fig. 4.40 Question 4.17 4.18 In a ﬂuid ﬂow, point A is at a higher elevation than point B. The head loss between these points is HL. The total heads at A and B are H a and Hb respectively. The ﬂow will take place (a) from A to B if H a + HL = Hb (b) from B to A if Ha + HL = Hb (c) always from A to B (d) from B to A if Hb + HL = Ha 4.19 In a siphon the summit is 4 m above the water level in the reservoir from which the ﬂow is being discharged out. If the head loss from the inlet of the siphon to the summit is 2 m and the velocity head at the summit is 0.5 m the pressure at the summit is (a) – 63.64 kPa (b) – 9.0 m of water (c) 6.5 m of water (abs) (d) – 39.16 kPa 4.20 A liquid jet issues out from a nozzle inclined at an angle of 60° to the horizontal and directed upwards. If the velocity of the jet at the nozzle is 18 m/s the maximum vertical distance attained by the jet, measured above the point of exit from the nozzle is (a) 14.30 m (b) 16.51 m (c) 4.12 m (d) 12.39 m 4.21 A nozzle emits a 5 cm diameter liquid jet at 20 m/s in to air at an angle of elevation of 4.23 4.24 4.25 4.26 30° to horizontal. At a point of maximum elevation, if the jet is assumed to be unbroken throughout, the diameter of the jet is (a) 5.373 cm (b) 5.000 cm (c) 4.653 cm (d) 2.582 cm A water jet with a velocity of 20 m/s is directed upwards at an angle of 45° to the horizontal. If air resistance is neglected, it will reach a maximum elevation at a horizontal distance x from the nozzle. The value of x in metres is (a) 20.38 (b) 40.77 (c) 10.19 (d) 14.41 A nozzle directs a liquid jet at an angle of elevation of 45°. The hydraulic grade line for the jet (a) coincides with the centre line of the jet (b) will be horizontal at the level of the jet (c) will be horizontal at the level of the energy line (d) coincides with the energy line A pump delivers 50 L/s of water and delivers 7.5 kW of power to the system. The head developed by the pump is (a) 7.5 m (b) 5.0 m (c) 1.53 m (d) 15.32 m In a hydro-project a turbine has a head of 50 m. The discharge in the feeding penstock is 3.0 m3/s. If a head loss of 5 m takes place due to losses, and a power of 1000 kW is extracted, the residual head downstream of the turbine is (a) 5.0 m (b) 10.95 m (c) 15.95 m (d) 20.95 m In a turbine having a ﬂow of 1.2 m3/s the net head is 120 m. If the efﬁciency of the turbine is 90% the shaft power developed, in kW, is (a) 1440 (b) 160 (c) 1566 (d) 1269 171 Energy Equation and Its Applications 4.27 In a two-dimensional irrotational vortex ﬂow the velocity V at a radial distance r from the centre is V = C/r where C = constant. The difference in pressure between any two points 1 and 2 is given by (p1 – p2) = Ê 1 1ˆ r 2 (a) (b) 2r Á 2 – 2 ˜ (r 2 – r 12) 2 r1 ¯ Ë r2 (c) r Ê 1 1ˆ – 2˜ Á 2 2 Ë r2 r1 ¯ (d) r Ê 1 1ˆ – 2˜ Á 2 2 Ë r1 r2 ¯ 4.28 At a distance of 10 cm from the axis of a whirlpool in an ideal liquid, the velocity is 5 m/s. At a radius of 30 cm the depression of the free surface below the surface of the liquid at a very large distance is (a) 7.98 cm (b) 3.33 cm (c) 14.16 cm (d) 21.37 cm 4.29 In a ﬂow net for the ﬂow in a twodimensional constriction, the size of the mesh in the uniform ﬂow upstream of the constriction is 4 mm and at a point B in the constriction the mesh surrounding the point has a size of 3 mm. With the usual notations the pressure coefﬁcient Cp = Ê1 ˆ (p b – p0)/ Á r U 02 ˜ at point B is Ë2 ¯ (a) 7/9 (b) 4/9 (c) –1/3 (d) –7/9 4.30 In an irrotational ﬂow past a body the free stream velocity and pressure are V0 and p0 respectively. The stagnation pressure ps is given by ps = r r (a) p0 + V02 (b) p0 – V02 2 2 r 2 r Ê ˆ 2 (c) p0 / Á V0 ˜ (d) V0 2 Ë2 ¯ Momentum Equation and Its Applications Concept Review 5 Introduction The momentum principle is derived from Newton’s second law and is considered under two categories as (i) Linear momentum equation and (ii) Moment of momentum equation. Both the equations are applicable to a control volume and are vector equations. A control volume 5.1 LINEAR MOMENTUM EQUATION This equation states that the vector sum of all external forces acting on a control volume in a ﬂuid ﬂow equals the time rate of change of linear momentum vector of the ﬂuid mass in the control volume. The external forces are of two kinds, viz. boundary (surface) forces and body forces. Boundary forces consist of 1. Pressure intensities acting normal to a boundary, Fp, and 2. Shear stresses acting tangential to a boundary, Fs. Body forces are those that depend upon the mass of the ﬂuid in the control volume, for example weight, Fb. The linear momentum equation in a general ﬂow can be written for any direction x as Â F =F x px Fsx + Fbx = ∂ ( M x )cv + M xout - M xin ∂t (5.1) where Mx = momentum ﬂux in x-direction = rQVx. Sufﬁxes out represent the ﬂux going out of the control volume and in represent the ﬂux coming into the control volume. Fpx, Fsx and Fbx represent x-component of pressure force, shear force and body force respectively acting on the control volume surface. ∂ (Mx)cv = rate of change of x-momentum ∂t within the control volume. This component is zero in a steady ﬂow. 173 Momentum Equation and Its Applications Thus for a steady ﬂow, in the x-direction, Fpx + Fsx + Fbx = (Mx)out – (Mx)in = (rQVx)out – (rQVx)in (5.2) Similar momentum equations are applicable to other coordinate directions, y and z also. 5.1.1 Application to One-dimensional Flow Momentum Correction Factor In one-dimensional ﬂow analysis the ﬂow characteristic in one major direction, say longitudinal axis direction, is considered and the variation in other directions neglected. Thus, for example, in the two-dimensional transition shown in Fig. 5.1, the velocity distribution of u with y is accounted for by taking average 1 u d y and V is used in the analysis. velocity V = B The discharge Q = VA. Ú A momentum correction factor 1 u dA AÚ 2 (5.3) V is used to account for the variation of the velocity across the area in the calculation of the momentum ﬂux. Thus the momentum ﬂux at section 1 is M1 = b1rQV1 (5.4a) and the momentum ﬂux at section 2 is M2 = b2rQV2 (5.4b) For uniform velocity distribution b = 1 and for all other cases b > 1.0. In laminar ﬂow through a circular tube, b = 1.33 and for turbulent ﬂow through pipes b = 2 b ª 1.05. By deﬁnition b depends upon the nature of the velocity distribution; larger the non-uniformity, greater will be the value of b. If no other information is given, it is usual practice to assume b = 1.0: Control Volume In the application of the linear momentum equation the control volume can be assumed arbitrarily. It is usual practice to draw a control volume in such a way that (Fig. 5.2): (i) Its boundaries are normal to the direction of ﬂow at inlets and outlets. (ii) It is inside the ﬂow boundary and has the same alignment as the ﬂow boundary. (iii) Wherever the magnitudes of the boundary forces (due to pressure and shear stresses) are not known, their resultant is taken as a reaction force R (with components, Rx, Ry and Rz) on the control volume. This reaction R is the Force acting on the ﬂuid in the control volume due to reaction from the boundary. The Force F of the ﬂuid on the boundary will be equal and opposite to the reaction R. Rx a R Ry Y b1, V1, r1 X q 1 Control volume V1 u V2 Y 2 X B Fig. 5.2 2 1 Fig. 5.1 b2, V2, r2 Reaction of the Boundary, R As indicated above, the reaction of the boundary R, with component Rx and Ry is the force exerted by the boundary on the ﬂuid. In most of the applications, R is an unknown to be determined. As such, Rx and Ry are assumed to 174 Fluid Mechanics and Hydraulic Machines act in chosen directions and the momentum equation written. Upon solving for Rx and Ry, depending upon the sign of the answer, the assumption is corrected, if need be. Thus, Rx and Ry can be assumed to be in positive or negative direction of x and y respectively and upon solving, the ﬁnal answer will emerge out with the proper direction of the reaction force, R. Also, R = R x2 + R y2 and u as in Fig. 5.3. V2 = vr + u The relative velocity is always assumed to leave the blade tangentially. Hence, the momentum equation can be applied to the relative velocities. q Ry Rx vr (5.5) V2 u CV and its inclination a to x-axis is tan a = (5.8) Py u vr Px (5.6) vr = relative velocity vr When b at a section is given, the momentum ﬂux past the section in the chosen x-direction is given by M x = b r QVx In Fig. 5.2, the directions are: at 1, in x-direction: at 1, in y-direction: at 2, in x-direction: at 2, in y-direction: Discharge Mx1 My1 Mx2 My2 Absolute velocity (5.7) momentum ﬂux in various = b1rQV1 =0 = b2rQV2 cos q = b2rQV2 sin q Q = AV = A1 V1 = A2 V2 5.1.2 Forces on Moving Blades A major application of the momentum equation relates to impact of liquid jets on blades. Figure 5.3 shows a liquid jet of velocity V impacting on a curved blade moving at a velocity u. The static pressure is atmospheric everywhere. Relative velocity of water entering the blade = vr = V1 – u, where V1 = absolute velocity of the jet. If there is no friction, the relative velocity will remain constant all over the blade. At the exist of the blade, the relative velocity Vr2 = Vr = V1 – u. The absolute velocity V2 is obtained as vector sum of Vr u V1 Fig. 5.3 If Px is the reaction of the blade on the ﬂuid in the control volume. 0 – Px = rQr (– vr cos q – vr) (5.8) 2 0 – Px = – r Avr (cos q + 1) Px = r A (V1 – u)2 (1 + cos q) (5.9) Force on the blade = + Fx = \|Px \| in the positive x-direction Power developed = Fxu (5.10) If a series of vanes are so arranged on a wheel that the entire jet is intercepted by one blade or other, the discharge to be used in Eq. (5.8) is the actual discharge of the jet Q instead of Qr. This principle is used in pelton turbines. In reaction turbines, the pressure on the blade is not atmospheric and the velocity triangles have to be written for both inlet and outlet of the blades. Details about turbines and pumps are presented in Chapter 16. 5.1.3 Momentum Equation for Steady Flow For a control volume lying in a horizontal plane, shown in Fig. 5.2, the linear momentum equation for steady ﬂow is written as outlined below. 175 Momentum Equation and Its Applications Let Rx along positive x-direction and R y in negative y-direction be the reaction of the boundary on the ﬂuid of the control volume (cv). Then in x-direction: ÈThe resultant of all forces ˘ Èx-Momentum flux ˘ Í ˙ =Í ˙ Îon cv in x-direction ˚ Îgoing out of cv ˚ Èx-Momentum flux ˘ – Í ˙ Îgoing into cv ˚ Thus p1A1 – p2A2 cos q + Rx = Mx2 – Mx1 = (b2rQV2 cos q – b1rQV1) (5.11) Similarly in y-direction, 0 – p2A2 sin q – Ry = My2 – My1 = b2 rQ V2 sin q (5.12) For any direction, that does not lie in a horizontal plane, the component of the body force (weight of ﬂuid in cv) should be suitably included among the forces on cv. In the solution of Eqs 5.11 and 5.12 often, depending upon the data, the continuity equation. A1V1 = A2V2 (5.13) and the Bernoulli equation 2 2 p2 a 2V 2 p1 a1V1 + + Z2 + + Z1 = rg 2g rg 2g (5.14) will have to be used. 5.2 THE MOMENT OF MOMENTUM EQUATION The moment of momentum equation is based on Newton’s second law applied to a rotating ﬂuid mass system. Moment of momentum about an axis is known as angular momentum. The moment of a force about a point is torque. The moment of momentum principle states that in a rotating system the torque exerted by the resultant force on the body with respect to an axis is equal to the time rate of change of angular momentum. In a steady ﬂow rotating system, i.e. when the rotating speed is constant, ÈTorque exerted Í Íon the fluid by the ÍÎ rotating element ˘ ˙ = ˙ ˙˚ ÈAngular momentum ˘ ÈAngular momentum ˘ Í ˙ – Í ˙ Í of fluid leaving ˙ Íof fluid entering ˙ ÍÎ out of cv ˙˚ ÍÎthe cv ˙˚ T = rQ [(S Vur)out – (S Vur)in] (5.15) where Q = discharge, Vu = tangential component of absolute velocity, r = moment arm of Vu, out and in denote items leaving or entering a control volume (cv) respectively. Equation (5.15) ﬁnds considerable application in the analysis of roto dynamic machines, viz., turbines, pumps, propellors, etc. details of which can be had in Chapter 16. In the following section, the details of reaction with rotation with a typical application to a lawn sprinkler is given. 5.2.1 Reaction with Rotation The reaction of ﬂuid discharging in a rotating system would generate force and hence a torque. This is clearly illustrated in a rotating arm of a lawn sprinkler. Figure 5.4 shows a ﬂuid entering the arm of a sprinkler normal to the arm at 1 and discharging at the outlet 2. If Q = discharge and a = area of the outlets (two in number here) then Q/2a = v2 = relative velocity of exit. Also b = angle of the jet with the direction of rotation and u2 = velocity of the arm = rw where w = angular velocity. The absolute velocity is V2. Its tangential component = Vu2 = V2 cos a. It is the change in the absolute velocity that causes change in angular momentum. From the velocity triangle (in vector notation) V2 – u2 = v2 = relative velocity (5.16) i.e., u2 + v2 cos b = Vu2 = V2 cos a Force exerted by the ﬂuid on the system, F = – rQ(Vu2) = – rQ (u2 + v2 cos b) 176 Fluid Mechanics and Hydraulic Machines Vu2 = V2 cos a = 0 and a = inclination of the absolute velocity with the tangential direction = 90°. (c) For a sprinkler with b = 180°: T = – rQr (w r – v2) (5.20a) and when T = 0 (frictionless system) w = v2/r (5.20b) (d) The torque required to hold the sprinkler in ﬁxed position (i.e. to prevent it from rotating) is T for w = 0, i.e. T0 = rQrv2 cos b (5.21) The retarding torque (due to bearing friction, etc.) Also, is T = – rQr (u2 + v2 cos b) (5.17) where u2 = w r2. (a) Thus when the retarding torque = T, the angular velocity w is v2 cos b T r r Q r2 w =– (5.18) (b) The maximum speed of a sprinkler (run away speed) is when T = 0 and hence v2 cos b r wmax = – (5.19) w b u2 V2 1 V2 a u2 r b V2 = absolute velocity b v2 Tangential velocity = u2 a V2 Vu2 b Vu2 (a) (b) v2 v2 = velocity of jet relative to tangential velocity Fig. 5.4 Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult 177 Momentum Equation and Its Applications Worked Examples V1 A: Linear Momentum t 5.1 A 7.5 cm diameter water jet having a n velocity of 12 m/s impinges on a plane, smooth plate at an angle of 60° to the normal to the plate. What will be the impact force when (i) the plate is stationary and (ii) when moving in the direction of the jet at 6 m/s. Estimate the work done per unit time on the plate in each case. Rn V1 X q D Solution: Consider the normal and tangential directions and a control volume as in Fig. 5.5. Let Rn be the normal reaction on the ﬂuid in the control volume. Consider the normal direction n. The pressure in the jet is atmospheric. CV V1 Fig. 5.5 (i) When the plate is stationary: 0 – Rn = rQ (0 – V1 cos q) Rn = rQV1 cos q Èp ˘ = 998 ¥ Í ¥ (0.075) 2 ¥ 12˙ Î4 ˚ ¥ 12 cos 60° = 317.45 N The normal force of the jet on plate is Fn = 317.45 N in the positive n direction (opposite to Rn). (ii) When the plate moves in the x-direction with u = 6 m/s. Considering normal direction and relative velocities: – Rn = rQr (0 – V1r cos q) Rn = r AV 21r cos q V1r = 12.0 – 6.0 = 6.0 m/s p Rn = 998 ¥ ¥ (0.075)2 ¥ 62 ¥ cos 60° 4 = 79.36 N The normal force of the water jet on the plate Fn will be equal and opposite to Rn. Hence Fn = 79.36 N and acts in positive n direction. Fn Jet Impingement on a Plate Work done W = Fn ¥ un = Fn cos 60° ¥ u In case (i): u = 0, W = 0 (ii): u = 6 m/s, W = 79.36 ¥ 0.5 ¥ 6.0 = 238.0 N.m/s 5.2 A jet of water 6 cm in diameter has a velocity of 15 m/s and impinges normally on a vertical stationary plate. jet impingement. (b) What is the work done in this instance? Solution: Consider the control volume as shown in Fig 5.6. The pressure in the jet is atmospheric at the boundaries of the control volume. Since the jet impinges normally on the vertical plate and spreads radially, the force on the plate is in the X-direction. By assuming the plate to be frictionless there is no force on the plate in the radial (Y)-direction. Let Rx = reaction of the plate in the X-direction acting on the ﬂuid in the control volume. The force on the plate is equal and opposite to Rx. 178 Fluid Mechanics and Hydraulic Machines Rn = rq1Vn = 998 ¥ (1.0) ¥ 8.67 = 8652.7 N = 8.653 kN = Fn (along the outward normal as indicated in Fig. 5.7) \ V2 Y Rx V1 q3V3 Fx Control volume Control volume t-direction V3 By steady state momentum equation in Xdirection applied to the control volume: S (forces in x-direction) = (Momentum ﬂux)out – (Momentum ﬂux)in 0 – Rx = rQ (0 – V1) \ Rx = rQV1 = Fx in X-direction. p (0.06)2 ¥ 20 ¥ 20 = 1128.7 N Fx = 998 ¥ 4 = 1.1287 kN (Fx is in positive x-direction) Since the Plate is stationary, no work is done by the force on the plate. 5.3 A two-dimensional jet of water of thickness 10 cm and issuing with a velocity of 10 m/s strikes a stationary plate at an angle of 30° to the normal of the plate. (a) Calculate the force on the plate. (b) Estimate the discharge of the two streams that move on the plate on either side of the impact zone. Solution: Consider unit width of the jet. In the normal direction, Rn = reaction of the plate acting on the control volume enclosing the ﬂuid. Pressure is atmospheric throughout. Vn = V1 cos 30° = 10.0 ¥ 0.867 = 8.67 m/s, and q1 = V1h1 = 10.0 ¥ 0.10 = 1.0 m3/s/m By momentum equation in the normal direction; 0 – Rn = rq1 (0 – Vn) n-direction Angle = 30° to normal Fig. 5.6 Rn Normal En q 1, V 1 2-D Jet V3q3 Fig. 5.7 In the transverse direction, the pressure is atmospheric at the control volume surfaces, and by assuming no friction, the force in the transverse direction is zero. Hence, the force on the plate due to impingement of jet is = Fn = 8.653 kN along the outward normal. (b) Since there is no force on the boundary in the transverse direction, the momentum ﬂux is preserved. Also due to absence of friction V1 = V2 = V3 (i) Thus the momentum equation in the transverse direction is written as: rq1V1 sin q = rq2V2 – rq3V3 (ii) Using the identity of Eq.(i), Eq.(ii) becomes q1 sin q = q2 – q3 (iii) Further by continuity q1 = q2 + q3 (iv) Simplifying (iii) and (iv) q2(1 – sin q) = q3 (1 + sin q) 179 Momentum Equation and Its Applications q2 1 + sin q = 1 - sin q q3 1 + sin 30∞ = 3.0 = 1 - sin 30∞ Since q1 = 1.0 m3/s/m, q2 = 0.75 m3/s/m and q3 = 0.25 m3/s/m Force on the plate = Fx = Rx in opposite direction = 282.15 N (b) Work done per second = P = Fxu = 282.15 ¥ 8.0 = 2257.2 W = 2.26 kW (c) Efﬁciency of power transfer by jet 5.4 A 5 cm diameter jet of water having a impingement = velocity of 20 m/s impinges normally on the jet with a velocity of 8.0 m/s. Find (a) force on the plate, (b) work done per second and (c) Kinetic energy ﬂux of the jet = r (AV1) Solution: In this case the relative velocity of the jet with respect to the plate is to be considered, (Fig. 5.8). As such, relative velocity of the jet Vr = (V – u) = 20 – 8 = 12 m/s Velocity of plate u V12 2 Êp ˆ ( 20) 2 KE = 998 ¥ Á ¥ (0.05) 2 ¥ 20˜ ¥ Ë4 ¯ 2 = 7838.3 W 2257.2 Efﬁciency h = = 0.288 = 28.8% 7838.3 V–u 5.5 A liquid jet of area A and velocity V strikes amoving vertical plate normally.The plate is moving with a velocity u in the direction of the jet velocity. Determine the power transmitted y transfer. Obtain the ratio u/V V–u Rx V x Fx u Control volume Work done per second Kinetic Energy of the Jet V–u Relative steady flow Fig. 5.8 Quantity of ﬂuid mass that strikes the plate per second = relative discharge = p p Qr = (D)2 (V – u) = (0.05)2 (20 – 8) 4 4 = 0.02356 m3/s Reaction of plate = Rx = rQrVr = 998 ¥ (0.02356) ¥ 12 = 282.15 N impingement. Solution: Consider relative motion of the jet with respect to the plate. Relative velocity of the jet Vr = (V – u) Quantity of ﬂuid mass that strikes the plate per second = relative discharge = Qr = A(V – u) Force on the plate = rQrVr = rA (V – u)2 Work done per second = P = Fxu = rA (V – u)2u V2 2 Efﬁciency of power transfer by jet impingement = Worke done per sencond h= Kinetic Energy of the net Kinetic energy ﬂux of the jet = r(AV) 180 Fluid Mechanics and Hydraulic Machines h= r A (V - u ) 2 u uˆ Ê uˆ Ê = 2 Á ˜ Á1 - ˜ ËV ¯ Ë V ¯ 2 r AV 3 / 2 u Putting = e, h = 2e(1 – e)2 and in this V expression the limits of e are, e ≥ 0 and e < 1.0. For maximum efﬁciency dh = 0. Thus de ( d e (1 - e ) ( d e - 2e 2 + e 3 de de ) 2 ) = 0. = 0. Thus D2 = D = 0.0424 = 0.0036 m2 Ê pˆ ÁË ˜¯ ¥ 15.0 4 0.0036 = 0.06 m = 6 cm 5.7 A undershot water wheel consists of a wheel. A free jet of water impinges on the blade normally and it can be assumed that at any time one blade or other is in front of the jet. Obtain V to the tangential velocity of the wheel blade u Leading to 3e 2 – 4e + 1 = 0 1 . Since 3 e = 1 is not feasible and hence for maximum efﬁciency 1 e= . 3 Substituting this in the expression for h the maximum efﬁciency is obtained as power from the jet to the wheel. The roots to this equation are e = 1 and e = hmax. = 2 ¥ 2 1 Ê 8 1ˆ ¥ Á1 - ˜ = = 29.6% 3 Ë 27 3¯ 5.6 A circular jet of water emanating from a nozzle strikes a stationary vertical plate N. If the discharge from the nozzle is measured as 42.4 liters/s estimate the diameter of the Solution: Force on the stationary plate due to jet impingement F = rQV p Here Q = d 2V = 0.0424 m3/s 4 and F = 635 N Thus F = 635 = 998 ¥ 0.0424 ¥ V V = 15.0 m/s. Solution: Since at any time one blade or other is in front of the jet the discharge intercepted by the wheel is Q = AV where A is the area of the jet. Relative velocity of the jet Vr = (V – u) Force on the plate = rQVr = rAV (V – u) Work done per second = P = Fxu = rAVu (V – u) V2 2 Efﬁciency of power transfer by jet impingement = Work done per second h = Kinetic Energy of the Jet Kinetic energy ﬂux of the jet = r(AV) h = r AVu (V - u ) 3 uˆ Ê uˆ Ê = 2 Á ˜ Á1 - ˜ ËV ¯ Ë V ¯ r AV / 2 u Putting = e, h = 2e (1 – e) and in this expression V the limits of e are, e ≥ 0 and e < 1.0. For maximum d (e (1 - e ) ) dh efﬁciency = 0. Thus = 0. de de 1 Leading to –2e + 1 = 0 i.e., e = 2 Substituting this in the expression for h the maximum efﬁciency is obtained as hmax = 2 ¥ 1ˆ 1 Ê 1 = 50% ¥ Á1 - ˜ = Ë ¯ 2 2 2 181 Momentum Equation and Its Applications 5.8 A vertical jet is issuing upwards from a nozzle with a velocity of 11 m/s. The r = 800 kg/m3 bearing a total load of 400 N is supported only by the impact of the Jet (Fig. 5.9). Determine the equilibrium height of the plate above the nozzle 400 N By momentum equation in Y-direction, R y = Reaction force on the cv in (– Y)-direction = weight supported by the jet = 400 N Hence 0 – Ry = 0 – rQV2 Ry = 400 = 800 ¥ 0.0486 ¥ [(11)2 – 2 ¥ 9.81 ¥ h]1/2 121 – 19.62 h = 105.86 h = 0.0772 m The equilibrium height of the plate is 0.772 m 5.9 A jet of oil (RD = 0.80) issues from nozzle of 15 cm diameter with a velocity of 2 Y CV Calculate the force required to hold the cone in position. h X 1 Fig. 5.9 Solution: Consider a control volume as shown in Fig. 5.10. Let Rx = reaction of the cone on the ﬂuid in the control volume. The pressure is everywhere atmospheric. As the cone is smooth, by neglecting friction the velocity of the sheet of water over the cone is V everywhere. The inclination of this velocity V to x-axis is 90/2 = 45°. p ¥ (0.15)2 = 0.01767 m2, 4 r = 0.8 ¥ 998 = 798.4 kg/m3 Q = AV = 0.01767 ¥ 12 = 0.2121 m3/s Solution: A= Q = discharge = A1 V1 p = ¥ (0.075)2 ¥ 11.0 4 = 0.0486 m3/s Since the jet is issuing into atmosphere p1 = p2 = 0. Let h = equilibrium height of the plate above 1 V2 = velocity of the plate just before impact. By Bernoulli equation between points 1 and 2, 0+ Hence or V12 V22 +0 =0+ +h 2g 2g V Y 45° V 45° (V12 - 2 gh) Cone CV V 22 = V 12 – 2gh V2 = X Rx V Fig. 5.10 182 Fluid Mechanics and Hydraulic Machines By momentum equation in X-direction: 0 – Rx = rQ (V cos 45° – V) Rx = rQV (1 – cos 45°) = 798.4 ¥ 0.2121 ¥ 12 ¥ (1 – cos 45°) = 595 N By symmetry Ry = 0 Hence the resultant reaction force on the ﬂuid R = Rx. Thus the force required to hold the cone in position is F = R = Rx = 595 N along (– X)-direction. V2 30° Y 2 X CV V1 Rx 1 (a) 5.10 A 10 cm diameter jet of water strikes a curved vane with a velocity of 25 m/s. The inlet angle of the vane is zero and the outlet angle is 30°. Determine the resultant force on the vane (a) When the vane is stationary and (b) When the vane is moving in the direction of the jet at 10 m/s velocity. (a) When the vane is stationary: Let Rx and Ry, as shown in Fig. 5.11(a), be the reaction forces on the ﬂuid in the control volume. The pressure is atmospheric at inlet and outlet. By momentum equation in the X-direction, – Rx = rQ (– V2 cos 30° – V1) V2 = V1 = 25 m/s as there are no losses. p ¥ (0.10)2 4 ¥ 25 (– 25 cos 30° – 25) = – 195.95 ¥ (46.65) Rx = 9141.5 N By momentum equation in the Y-direction, Ry = r Q (V2 sin 30° – 0) – Rx = 998 ¥ p ¥ (0.10)2 ¥ 25 ¥ 25 sin 30° 4 = 195.95 ¥ 25 ¥ 0.5 = 2449.5 N Resultant Reaction = 998 ¥ R= (9141.5) 2 + ( 2449.5) 2 = 9464 N acting at an angle a to the (– X)-direction given by Ry v – u = 15 m/s 2 30° CV u v – u = 15 m/s R a Ry Rx 1 (b) Fig. 5.11 Ry 2449.5 = 15° 9151.5 Rx The force on the blade is equal and opposite to R. Hence F = 9464 N and acts at (360 – 15) = 345° to the x-direction. (b) When the vane moves at 10 m/s in the X-direction. The situation is shown in Fig. 5.8(b). The relative velocity entering the vane Vr1 = the relative velocity leaving the vane Vr1 = Vr2 = V1 – u = 25 – 10 = 15 m/s The relative discharge acting on the vane = Qr a = tan–1 = tan–1 = A ¥ (V – u) = p ¥ (0.10)2 ¥ 15 4 = 0.1178 m3/s By momentum equation in x-direction, – Rx = rQr (– Vr2 cos 30° – Vr1) 183 Momentum Equation and Its Applications = 998 ¥ 0.1178 (– 15 cos 30° – 15) Rx = 3290.7 N By momentum equation in Y-direction, Ry = rQr (Vr2 sin 30° – 0) = 998 ¥ 0.1178 ¥ (15 sin 30°) Ry = 881.7 N Resultant reaction R on the water in the control volume: R= (3290.7) 2 + (881.7) 2 = 3406.7 N Inclination of R with (– X)-direction –1 a = tan Ry Rx 881.7 = 15° 3290.7 The resultant force on the vane will be equal and opposite to R. Hence R = 3406.7 N and inclined at (360° – 15°) = 345° to the X-direction. Vu2 u a2 5.11 A jet of water with a velocity of 30 m/s impinges on a moving vane of velocity 12 m/s at 30° to the direction of motion. The vane angle at the outlet is 18°. Find (i) the blade angle at inlet so that the water enters without shock. (ii) the work done on the vane per unit weight of water per second entering the vane, and Solution: Here the jet velocity = absolute velocity = V1 = 30 m/s Velocity of blade = u = 12 m/s The inlet and outlet velocity triangles are shown in Fig. 5.12. (i) Consider the inlet velocity triangle. V1 = 30 m/s, Vf1 = V1 sin 30° = 30 ¥ 0.5 = 15 m/s Vu1 = V1 cos 30° = 30 ¥ 0.866 = 25.98 m/s Vf2 Outlet u X Direction of motion Inlet 30° V1 Vr1 30° u Vf1 a1 Vu1 a = tan–1 V2 Vr2 Fig. 5.12 Jet Impingement on a Moving Vane If a1 = angle of the relative velocity with the direction of the vane = vane angle at inlet. Vf 1 15 = (Vu1 - u ) ( 25.98 - 12.0) = 1.073 a1 = 47° tan a1 = The relative velocity Vr1 = Vr2 = Vf1 sin 47∞ = 20.505 m/s From the outlet velocity triangle, a2 = 18° Vu2 = Vr2 cos a2 – u2 As u2 = u1 = 12 m/s Vu2 = 20.505 cos 18° – 12.0 = 7.50 m/s (in the – x-direction) (ii) Considering the direction of motion, x, force exerted by the jet in the x-direction Fx = rQr [Vu2 + Vu1] Work done/second P = Fxu = rQ r (Vu2 + Vu1) u Work done/second/unit weight of liquid P = gQ 184 Fluid Mechanics and Hydraulic Machines = Head extracted = plane, 1 (Vu2 + Vu1) u g 1 (7.50 + 25.98) ¥ 12 = 40.95 m 9.81 (iii) Initial energy head of the jet = Velocity head = V 12/2g = (30)2/2 ¥ 9.81 = 45.87 m = 5.12 A discharge of 0.06 m3 a horizontal bend as shown in Fig. 5.1. Calculate the force on the bolts in section 1. CV 5 cm Dia V2 Rx 2 Y X 15 cm Dia 1 Fig. 5.13 The control volume is shown in dotted lines. The Reaction on the control volume ﬂuid is shown as Rx in positive x-direction. Discharge p (D2)2 V2 = 0.06 m3/s 4 0.06 V2 = = 30.56 m/s p 2 ¥ (0.05) 4 V1 = V2 (D2/D1)2 Q= 2 p ¥ (0.15)2 + Rx 4 = 998 ¥ 0.06 ¥ (30.56 + 3.395) Rx = 8132 + 2033 = 10165 N The force F exerted by the ﬂuid on the pipe, and hence on the bolts in Section 1, is equal and opposite to Rx. Thus F = 10165 N and acts to the left, i.e., in the negative x-direction, as a pull (tension) on the joint. – (460.2 ¥ 103) ¥ Head extracted Efﬁciency = Intial energy head = 40.95/45.87 = 89.27% (30.56) 2 P (3.395) 2 … (p2 = atmospheric) = 1+ 2 ¥ 9.81 g 2 ¥ 9.81 P1 = 47.59 – 0.59 = 47.00 m g p1 = 47.00 ¥ 9.79 = 460.2 kPa By momentum equation in the x-direction, – p1A1 + Rx – 0 = rQ [V2 – (– V1)] 0+ Ê 5ˆ = 30.56 ¥ Á ˜ = 3.395 m/s Ë 15 ¯ By applying Bernoulli equation to sections 2 and 1, by assuming the bend to be in the horizontal 5.13 A 30 cm diameter pipe is bifurcated into two nozzles at a Y-junction as shown in Fig. 5.14. The nozzles discharge to atmosphere and have a velocity of 10 m/s each. The junction is in a horizontal plane and the friction can be neglected. Determine the magnitude and direction of the resultant force on the Y-junction. Solution: Consider the control volume and reaction forces on the ﬂuid in the control volume as in Fig. 5.14. p A1 = ¥ (0.30)2 = 0.07069 m2 4 p A2 = ¥ (0.075)2 = 0.004418 m2, V2 = 10 m/s 4 p A3 = ¥ (0.10)2 = 0.007854 m2, V3 = 10 m/s 4 Q = A1 V1 = Q1 + Q2 Q2 = A2V2 = (0.004418 ¥ 10) = 0.04418 m3/s Q3 = A3V3 = (0.007854 ¥ 10) = 0.07854 m3/s Q = Q1 + Q2 = 0.1227 m3/s V1 = 0.1227/0.07069 = 1.736 m/s 185 Momentum Equation and Its Applications Inclination q of R with the (– x) direction F Rx R q CV 7.5 cm dia 10 m/s Ry 2 Y 25° 30 cm dia X 35° 1 10 m/s 3 Ry 263 = 5.79° 2592 Rx Force F on the body of the junction exerted by the ﬂuid will be equal and opposite to R. Hence F = 2605 N and will be directed at 5.79° to the positive x-direction. q = tan–1 = tan–1 5.14 A reducer bend having an outlet diameter of 15 cm discharges freely. The bend, connected to a pipe of 20 cm diameter, 10 cm dia Fig. 5.14 By applying Bernoulli theorem to sections 2 and 1 P1 V12 P2 V22 + + = g 2g g 2g (10) 2 p1 (1.736) 2 + = 2 ¥ 9 .81 g 2 ¥ 9 .81 p1/g = 5.0968 – 0.1536 = 4.943 m p1 = 9790 ¥ 4.943 = 48394 Pa Applying momentum equation in the x-direction: p1A1 – Rx = r[Q2 V2 cos 25° + Q3V3 cos 35° – Q1V1] (48394 ¥ 0.07069) – Rx = 998 [0.04418 ¥ 10 ¥ cos 25° + 0.07854 ¥ 10 ¥ cos 35° – 0.1227 ¥ 1.736] 3421 – Rx = 998 (0.4004 + 0.6434 – 0.2130) Rx = 3421 – 829 = 2592 N By applying momentum equation in the y-direction: 0 – Ry = r (Q2V2 sin 25° – Q3V3 sin 35°) = 998 (0.04418 ¥ 10 ¥ sin 25° – 0.07854 ¥ 10 ¥ sin 35°) = 998 (0.1867 – 0.4505) = – 263 Ry = 263 N The reaction R on the ﬂuid in the control volume is 0+ 2 2 R = R x + Ry = = 2605 N ( 2592) 2 + ( 263) 2 plane. Determine the magnitude and direction of force on the anchor block supporting the pipe when a discharge of 0.3 m3/s passes through the pipe. Solution: Consider the control volume as shown by the dotted line in Fig. 5.15 (a). At section 2: p2 = 0 = atmospheric pressure. 0 .3 = 16.98 m/s p 2 ¥ (0 .15) 4 2 2 Ê 15 ˆ ÊD ˆ At section 1: V1 = V2 Á 2 ˜ = 16.98 ¥ Á ˜ Ë 20 ¯ Ë D1 ¯ = 9.55 m/s By applying Bernoulli equation to sections 2 and 1, V2 = 0 + (16.98)2/(2 ¥ 9.81) = p1 + (9.55)2/(2 ¥ 9.81) g V2 P2 = 0 R Ry 2 R q Y 60° V1 X p1 Y Ry R q Rx CV F 1 (a) (b) Fig. 5.15 X 186 Fluid Mechanics and Hydraulic Machines p1 = 14.689 – 4.648 = 10.041 m g p1= 9790 ¥ 10.041 = 98301 Pa Let Rx and R y be the reaction of the pipe on the ﬂuid in the control volume in (– x) and y-directions, respectively [Fig. E-5.12 (b)]. By applying momentum equation in x-direction), p1A1 – Rx = rQ(V2 cos 60° – V1) p ¥ (0.2)2 – Rx 4 = 998 ¥ 0.3 ¥ (16.98 cos 60° – 9.55) Rx = 3088.2 + 317.4 = 3405.6 N By momentum equation in y-direction, 0 + Ry = rQ(V2 sin 60° – 0) Ry = 998 ¥ 0.3 ¥ 16.98 ¥ 0.866 = 4402.7 N 0.25 mf 2 CV 2.0 m Ry W 98301 ¥ Resultant R= = R 2x + R 2y (3405.6) 2 + ( 44027) 2 = 5564.3 N Inclined at an angle q such that tan q = Ry Rx 4402 .7 = 52.28° 3405.6 The force F on the pipe is equal and opposite to R and hence F = 5564 N inclined at (360 – q) = 307.72° to positive x-axis (see Fig. 5.15 (b)]. q = tan–1 5.15 nozzle of inlet and outlet diameters 0.5 m and 0.25 m respectively as shown in Fig. 5.13. The pressure at section 1 is 15 kPa and 3 /s. Find (i) the velocities at section 1 and 2, (ii) pressure at section 2 and (iii) total force acting on the walls of the nozzle [Neglect frictional resistance] Solution: Refer to Fig. 5.16. A1 = (p /4) (0.5)2 = 0.1964 m2 A2 = (p/4) (0.25)2 = 0.0491 m2 V1 = 0.5/0.1964 = 2.546 m/s 1 Flow Y 0.5 mf X Fig. 5.16 V2 = 0.5/0.04951 = 10.185 m/s By applying Bernoulli’s theorem to sections 1 and 2 V12 V 22 p1 p + Z1 + = 2 + Z2 + g 2g g 2g (10.185) 2 15.0 ( 2 . 546) 2 p +0+ = 2 + 2.0 + g 2 ¥ 9.81 9.79 2 ¥ 9.81 p2 1.532 + 0.3304 = + 2.0 + 5.287 g p2 = – 5.425 m g and p2= – 53.11 kPa (gauge) Now, consider a control volume encompassing section 1 and 2 shown in Fig. 5.16. W = weight of water in the control volume which is in the shape of frustum of cone = g (p /3)h [r 12 + r1r2 + r 22 ] = 9.79 (p/3) ¥ 2.0 [(0.25)2 + (0.25) (0.125) + (0.125)2] = 2.243 kN Applying linear momentum equation in the vertical direction to cv with Ry = reaction force on the water in the control volume. S (Forces) Y-direction = – W + Ry + p1A1 – p2 A2 = r Q(V2 – V1) 187 Momentum Equation and Its Applications – 2.243 + Ry + (15.0 ¥ 0.1964) – (– 53.11 ¥ 0.0491) = 0.998 ¥ 0.5 ¥ (10.185 – 2.546) Ry – 2.243 + 2.945 + 2.607 = 3.812 Ry = 0.503 kN F= Net force on the nozzle walls is equal and opposite to Ry. Hence F = 0.503 kN acting vertically downwards B dh 2 and limits are when y = 0, h = 0 B ,h=1 and when y = 2 dy = 2 b = u 2 dy BV um2 V 2 = 2 1 um2 V 1 Ú (1 - h) 2 2 dh 0 Ú (1 - 2h + h ) dh 2 0 1 um2 È h3 ˘ h - h2 + ˙ 2 Í 3 ˙˚ V ÍÎ 0 um2 1 = 2 ¥ 3 V 4V 2 um = 2V, b = = 1.333 3V 2 = 5.16 The velocity distribution in a two- dimensional duct is shown in Fig. 5.17. Calculate the value of the momentum correction factor b. Since um B/2 0 b = [Note: Remember to include the component of body force (i.e. weight of ﬂuid in the control volume)in the application of linear momentum equation in a direction that does not lie in a horizontal plane.] Ú 5.17 The velocity distribution in a pipe of radius r0 is given by y B u Ê rˆ = Á1 – ˜ um r0 ¯ Ë B/2 u um m with m < 1. Determine the momentum correction factor b and calculate the value of b for m = 1/ 7. Fig. 5.17 Solution: Average velocity Solution: Average velocity 2 or Ú V = B/2 udy 2 = B B/2 y 0 ) dy um (1 V= 0 B/2 B 2 ÈÊ B ˆ Ê u ˆ Ê B 2 ˆ ˘ Ê um ˆ V = ÍÁ um ˜ - Á m ˜ Á ˙ = 2 ¯ Ë B / 2 ¯ Ë 8 ˜¯ ˙˚ ÁË 2 ˜¯ B ÍÎË um = 2V 2 b= = Ú B/2 Ú V = u 2 dy Ú B/2 u2m (1 - BV 0 y 2 Putting h = , dh = dy B/2 B 2u m r02 Ú r0 0 Ú r0 0 u ◊ 2p r dr y 2 ) dy B/2 m Ê rˆ r Á1 - ˜ dr r0 ¯ Ë r02 Ê rˆ – 1- ˜ Á r0 ¯ ( m + 1) Ë BV 2 2 p r02 2u m È r02 Ê rˆ 1- ˜ = 2 Í Á m 2 r ( + ) Ë r0 ÍÎ 0¯ 0 2 1 = V = m + 1˘ 2um È r02 r2 ˘ - 0 ˙ 2 Ím +1 m + 2 ˙˚ r0 ÍÎ 2um ( m + 1) ( m + 2) m+2 r0 ˙ ˙ ˚0 188 Fluid Mechanics and Hydraulic Machines Momentum correction factor b= b= = 1 V 2 Ú p r02 2 u 2 2p rdr 0 Ú V 2 r02 r0 r0 0 For the hose: rˆ 2 Ê um ◊ Á1 - ˜ r0 ¯ Ë = = For ◊ rdr 2m + 1 ˘ r0 ˙ ˙ ˚0 2 È 2um r02 r02 ˘ Í ˙ V 2 r02 ÍÎ 2m + 1 2m + 2 ˙˚ 2 2um 2 V Substituting for V, b= 2m 2 m+ 2 2 È Ê 2um r02 rˆ Í 1 r0 ˜¯ V 2 r02 Í ( 2m + 2) ÁË Î r02 Ê rˆ 1- ˜ Á r0 ¯ ( 2m + 1) Ë = V22 ( 27.85) 2 = = 39.53 m 2 ¥ 9.81 2g ◊ 1 ( 2m + 1) ( 2m + 2) 2 2um ( m + 1) 2 ( m + 2) 2 2 4 um 1 ◊ 2 ( m + 1) ( 2m + 1) 1 ( m + 2)2 ( m + 1) 4 (2m + 1) m = 1/7, 1 ( 2 + 1/ 7) 2 (1 + 1/ 7) 4 1 + 2/ 7 225 ¥ 8 = = 1.02 4 ¥ 9 ¥ 49 b= 5.18 A 40 mm nozzle connected to a 120 mm hose ejects a horizontal jet of water to atmosphere. If the discharge from the nozzle is measured as 35 liters/s, estimate the force on Solution: Refer to Fig. 5.18 For the nozzle: p2 = patmosphere = 0 0.035 p (0.04) 2 4 2 ¥ 27.85 = 3.094 m/s; V12 2g = (3.094) 2 = 0.488 m 2 ¥ 9.81 By Bernoulli equation applied to section 1 and 2 p1 V2 V2 +0+ 1 =0+0+ 2 g 2g 2g p1 + 0 + 0.488 = 0 + 0 + 39.53; g p1 = 39.04 m; g p1 = 9.79 ¥ 39.04 = 382.2 kN/m2 Applying linear momentum equation in the x-direction to control volume (shown in Fig. 5.18) Taking Rx = reaction of the boundary on the ﬂuid in the control volume S Forces = [(Momentum going out) – (Momentum ﬂux coming in)] P1 A1 – Rx – p2 A2 = rQ (V2 – V1) Êp ˆ 382.2 ¥ Á (0.12) 2 ˜ – Rx – 0 Ë4 ¯ V2 = 2 Ê 4ˆ ÊD ˆ V1 = Á 2 ˜ V2 = Á ˜ Ë 12 ¯ Ë D1 ¯ = 27.85 m/s; Ê 998 ˆ = Á ¥ 0.035 ¥ (27.85 – 3.09) Ë 1000 ˜¯ 4.3225 –Rx = 0.8649. Thus Rx = 5.187 kN. The force on the ﬂange connection, being equal and opposite to Rx = 5.187 kN, acts in the positive X-direction. When the hose is held ﬁrmly, the ﬂange bolts will be in tension, with a total force of 5.187 kN. 189 Momentum Equation and Its Applications 1 2 p2 = Atmospheric pressure y Rx p 1A1 p2A2 x Fx Control volume 1 D1 = 120 mm Fig. 5.18 = reduces the diameter to 15 cm without Neglect any losses in the transition. Solution: Consider the control volume as shown by the dotted line in Fig. 5.19. R R y V1 Y Rx p 1 A1 1 p 2 A2 CV V2 X 2 Fig. 5.19 R is the reaction on the ﬂuid in the control volume. p ¥ (0.30)2 = 0.07069 m2 4 p A2 = ¥ (0.15)2 = 0.01767 m2 4 V2 = 6.0 m/s, V1 = V2 (D2 /D1)2 A1 = D2 = 40 mm Example 5.18 5.19 A transition in a 30 cm diameter pipe transition is horizontal. Velocity measurements indicated the momentum correction factor b and kinetic energy correction factor a in the 30 cm pipe to be 1.30 and 1.90 respectively. Corresponding values of b and a for the 15 cm pipe are 1.05 and 1.15 respectively. If the pressure and the mean velocity in the 15 cm pipe at the end of the transition are to be 15 kPa and 6.0 m/s respectively, calculate the resultant z 1 V2 = 1.5 m/s 4 Q = A2 V2 = 0.01767 ¥ 6.0 = 0.106 m3/s p2 = 15 kPa, Z2 = Z1 By applying Bernoulli equation to sections 1 and 2, V12 V 22 p1 p + a1 + Z1 = 2 + a 2 + Z2 g 2g g 2g p1 1. 90 ¥ (1. 5) 2 15.0 1.15 ¥ (6.0) 2 + = + g 2 ¥ 9 .81 9.79 2 ¥ 9 .81 p1 = 1.532 + 2.110 – 0.218 = 3.424 g p1 = 9.79 ¥ 3.424 = 33.521 kPa There is no change in momentum ﬂux in y-direction. Hence R y = 0. By momentum equation in x-direction: p1 A1 – Rx – p2A2 = rQ(b2V2 – b1V1) (33521 ¥ 0.07069) – Rx – (15000 ¥ 0.01767) = 998 ¥ 0.106 ¥ (1.05 ¥ 6.00 – 1.30 ¥ 1.5) 2370 – Rx – 265 = 460 Rx = 1645 N = Reaction force on the ﬂuid in the control volume. Hence the resultant force on the transition which is equal and opposite to Rx is F = 1645 N in the positive x-direction. 5.20 A sluice gate in an open channel is shown in Fig. 5.20(a). Estimate the force on the unit width of the gate. Neglect frictional force on the channel bottom. 190 Fluid Mechanics and Hydraulic Machines 1 1 ¥ 9.79 ¥ (2.8)2 – ¥ 9.79 ¥ (0.3)2 – Rx 2 2 998 = (2.1) (7.0 – 0.75) 1000 38.38 – 0.44 – Rx = 13.10 Rx = 24.84 kN Gate 0.75 m/s 2.8 m 0.30 m The force FG on the gate due to water is equal and opposite to Rx. Thus FG = 24.84 kN and acts in the positive x-direction. (a) CV Y 1 Rx X FG 2 F1 F2 1 V2 2 (b) Fig. 5.20 Sluice Gate Solution: Consider a unit width of the gate and the control volume (cv) as shown in Fig. 5.20(b). The frictional force on the channel bottom is neglected. The forces on the control volume surfaces are: F1 = pressure force on the section 1 = g y 12 /2 (by assuming hydrostatic pressure distribution) F2 = pressure force on the section 2 = g y22 /2 Rx = reaction of the gate on the water in the control volume acting in the (– x)-direction. Here y1 = 2.8 m, y2 = 0.3 m, V1 = 0.75 m/s q = discharge per unit width of channel = y1V1 = y2V2 = 0.75 ¥ 2.8 = 2.1 m3/s/m V2 = q/y2 = 2.1/0.3 = 7.0 m/s From momentum equation to the control volume in the x-direction. F1 – F2 – Rx = rQ(V2 – V1) 5.21 A jet of water with velocity V1 and area of cross section A1 enters a stream of slow moving water in a pipe of area A2 and velocity V2. The two streams enter with the same pressure p1 in the pipe the stream emerges as a single stream with velocity V3 and pressure p2 (see Fig. 5.21) Assuming no losses in the pipe, determine (p2 – p1) for V1 = 20 m/s and V2 = 10 m/s, A1 = 0.01 m2, A2 = 0.02 m2 and density of water r = 1000 kg/m3. Solution: Consider the control volume as shown in Fig. 5.21. The pressure at section A is p1 and at B it is p2. There are no frictional losses and hence no additional force on cv. Pressures p1 and p2 act over the cross sectional area A2. Q1 = V1A1 = 20 ¥ 0.01 = 02 m3/s Q2 = V2(A2 – A1) = 10 ¥ (0.02 – 0.01) = 0.1 m3/s Q = Q1 + Q2 = total ﬂow = 0.2 + 0.1 = 0.3 m3/s V3 = Q/A2 = 0.3/0.02 = 15 m/s By momentum equation in the x-direction: p1A2 – p2A2 = Mout – Min = r(Q ◊V3 – Q1V1 – Q2V2) (p1 – p2) ¥ 0.02 = 1000 [(0.3 ¥ 15) – (0.2 ¥ 20) – (0.1 ¥ 10)] = –500 (p2 – p1) = 500 = 25000 Pa = 25 kPa 0.02 191 Momentum Equation and Its Applications V2 A1 A2 p1A2 Y = V1 p2A2 V3 X A2 V2 A 1 È 2˘ Í( p1 - p2 ) - 3 r U 0 ˙ Î ˚ The wall drag F on the pipe is equal and opposite p D2 È 1 ˘ ( p1 - p2 ) - r U 20 ˙ and 4 ÍÎ 3 ˚ acts in the (+ x) direction. to Rx. Thus F = CV Fig. 5.21 pD 2 4 B Example 5.21 B: Moment of Momentum 5.22 5.23 Figure 5.23 shows a lawn sprinkler with two jets, each located at 30 cm from the centre. The jets are of 1 cm diameter. Assuming factor at section 2 is 4/3, show that the wall drag F is given by discharge of 2.5 L/s. pD 2 Ê 1 2ˆ F= ÁË p1 - p2 - r U 0 ˜¯ 4 3 v2 u2 30 cm w CV 2 30 cm r U0 D 1 u x u2 v2 Fig. 5.23 Rx (reaction force on the fluid) (Drag on the wall) F 2 1 Fig. 5.22 Solution: Let Example 5.22 Solution: Consider a control volume encompassing section 1 and 2 as shown in Fig. 5.19. Let Rx = reaction force on the ﬂuid in the control volume (cv). Average velocity at section 1 = V1 = U0 = V2 = average velocity at section 2. At section 1, the velocity distribution is uniform. Hence, b1 = 1.0. It is given b2 = 4/3. By momentum equation in x-direction. S (Forces)x = – Rx + (p1 – p2) A = rQ [b2U0 – b1U0] 4ˆ Ê Rx = (p1 – p2) A + rAU 02 Á1 - ˜ Ë 3¯ Example 5.23 w = angular velocity u2 = w r a = p ¥ (1)2 = 0.7854 cm2 4 Q = 2500 cm3/s 2500 = 1519.5 cm/s 2 ¥ 0.7854 = 15.20 m/s = relative velocity of jet. T = – rQ r (u2 – v2) = 0 v2 = u2 = w r = 15.20 m/s v2 = Torque or w = 15.20 = 50.67 rad./s 0. 30 192 Fluid Mechanics and Hydraulic Machines Speed of rotation per minute, w N= ¥ 60 2p 50.67 \ N= ¥ 60 = 483.8 rpm 2p Solution: a = area of the jet = = 0.5027 cm2 Qn = discharge from each nozzle 1200 = 600 cm3/s 2 v2 = v3 relative velocity of jets 5.24 Find the torque required to hold the = Solution: When the sprinkler is stationary, w = 0 and torque T0 = rQrv2 = 2 .5 ¥ 0.30 ¥ 15.20 1000 = 11.38 N.m 5.25 what will be its steady rotation rate if it has a retarding friction torque of 1.5 N.m? Solution: T = – rQr (u2 – v2) 2 .5 ¥ 0.30 ¥ (w r – 15.20) 1000 = – 0.7485 (w ¥ 0.3 – 15.20) 1.5 = –998 ¥ 1 Ê 1. 5 ˆ 15.20 = 43.99 Á 0.3 Ë 0.7485 ˜¯ 43.99 N= ¥ 60 = 420 rpm 2p w = Torque (a) For zero friction T = 0. Hence r2 (w r2 + v2 cos b) + r3 (wr3 + v3 cos b) = 0 0.2 (0.2 + 11.94 cos 120°) + 0.4 (0.4w + 11.94 cos 120°) = 0 0.04w – 1.194 + 0.16w – 2.388 = 0 w = 17.91 rad./s 17.91 ¥ 60 = 171 rpm 2p (b) When the sprinkler is stationary, w = 0 Torque required to hold the sprinkler stationary = T0 = – rQ n (r2v2 cos b + r3v3 cos b) = – rQn v2 cos b (r2 + r3) or 5.26 Figure 5.24 shows a lawn sprinkler with unequal arms. The jets issuing out of the sprinkler are of 0.8 cm in diameter and the total discharge is 1.2 L/s. (a) Assuming zero friction, determine the rotational speed of the sprinkler. (b) What torque would be required to hold the sprinkler stationary? w 3 2 v2 60° 20 cm 1 Fig. 5.24 60° Example 5.26 v3 N = 0.6 ¥ 11.94 ¥ (0.5) (0.2 + 0.4) 1000 = 2.145 N.m T0 = 998 ¥ 40 cm 1200 2 ¥ 0.5027 = 1194 cm/s = 11.94 m/s b = Inclination of v2 with the positive direction sprinkler arm = 180 – 60° = 120° r2 = 0.20 m and r3 = 0.40 m T = – rQ n (r2(u2 + v2 cos b ) + r3 (u3 + v3 cos b)] = 998 ¥ p ¥ (0.8)2 4 5.27 A sprinkler with unequal arms and jets of area 0.8 cm2 facing in the same direction assembly normal to the rotating arm. (i) Assuming the frictional resistance to be zero, calculate its speed of rotation, (ii) What torque is required to hold it from rotating? 193 Momentum Equation and Its Applications v2 v1 30 cm 40 cm 3 2 1 w u2 u3 Fig. 5.25 Example 5.27 (Note that u2 and v2 are in the same direction) Absolute velocity: V2 = v2 + wr2 = 9.375 + 0.3 w V3 = v3 – wr3 = 9.375 – 0.4 w (i) Torque on the arm = T0 = – rQ n[0 – (r3 V3 – r2 V2)] For zero frictional resistance, T = 0 and hence r3V3 = r2V2 0.4 (9.375 – 0.4 w) = 0.3 (9.375 + 0.3 w) Solution: Q n = discharge from each nozzle 0.9375 = 3.75 rad./s 0.25 3.75 ¥ 60 N = 2p = 35.81 rpm w= 5 = 0.75 l/s 2 1. 5 ¥ 1000 v= = 937.5 cm/s = 9.375 m/s 0.8 ¥ 2 Let w = angular velocity of the arm. Designating the jet at shorter arm by 2 and at the longer arm by 3, relative velocities v2 = v3 = 9.375 m/s Tangential velocity u2 = w r2 = 0.3w u3 = wr3 = 0.4w or = (ii) When the arm is stationary, w = 0 Torque T0 = – rQ n (– r3V3 + r2V2) T0 = – rQn(– v3 r3 + v2r2) = 998 ¥ 0.75 ¥ 9.375 ¥ (0.4 – 0.3) 1000 = 0.702 N.m Problems Linear Momentum 5.1 A two-dimensional jet of water impinges on a plane at an angle q to the normal to the plane. If the jet splits into two streams of discharges in the ratio 1 : 2, calculate the angle q. (Ans. q = 19.47°) 5.2 A two-dimensional jet of liquid issuing from a long slot strikes a plate at an angle of 60° with the plate. This causes the ﬂow to divide into two parts q1 and q2 on either side of the impact zone. Calculate the ratio of q1/q2. (Ans. q1/q2 = 3.0) 5.3 A 15 cm diameter jet of water with a velocity of 15 m/s strikes a plane normally. If the plate is moving with a velocity of 6 m/s in the direction of the jet calculate the work done per second on the plate and the efﬁciency (h) of energy transfer. (Ans. P = 8571 N.m/s, h = 28.8%) 5.4 A 15 cm diameter jet of oil (RD = 0.8) strikes a stationary ﬂat plate at an angle of 35° to the normal. Calculate the force exerted on the plate when the velocity of the jet is 16 m/s. (Ans. F = 2958.7 N) 194 Fluid Mechanics and Hydraulic Machines 5.5 A 20 cm diameter jet of oil (RD = 0.9) strikes a ﬂat plate at an angle of 25° to the normal. The plate is moving at a velocity of 3 m/s in the direction of the jet. Calculate the absolute velocity of the jet if the resultant force exerted on the plate is 2500 N. (Ans. V = 12.89 m/s) 5.6 A vertical jet of oil (r = 900 kg/m3) issues out of a 10 cm diameter nozzle at a velocity of 15 m/s. The jet is directed upwards and is deﬂected by a horizontal ﬁxed plate kept at a height of 3.0 m above the nozzle exit. Estimate the force of impact of the jet on the plate. (Ans. F = 1367 N, upwards) 5.7 A 20 cm long nozzle having an outlet of 5 cm diameter is attached to a 10 cm diameter pipe and is directed vertically downwards. The pressure in the pipe at the base of the nozzle is 10 kPa. The jet is discharged to atmosphere and strikes a horizontal plate at a depth of 1.5 m below the nozzle exit. Calculate the force on the plate due to impact of the jet. (Ans. Fy = 69.2 N) 5.8 A tank shown in Fig. 5.26 has a nozzle of exit diameter D1 at a depth H1 below the free surface. At the side opposite to that of nozzle 1, another nozzle is proposed at a depth H1/2. What should be the diameter D2 in terms of D1 so that the net horizontal force on the tank is zero? (Ans. D2 = 2 .D1) H1/2 D2 = ? H1 D1 Fig. 5.26 5.9 A stationary curved plate deﬂects a 10 cm diameter water jet through an angle of 120° in the horizontal plane. Calculate the force required to hold the plate in position of the velocity of the jet is 15 m/s. (Ans. F = 3054 N at 30° to – ve x-direction) 5.10 A jet of water 15 cm in diameter strikes a curved blade at 20 m/s velocity. The inlet angle and the outlet angles of the vane are 0° and 45° respectively. Determine the resultant force exerted on the blade when (i) the jet is stationary and (ii) the blade moves against the direction of the water at 5 m/s. Neglect friction along the blade. (Ans. (i) F = 13035 N inclined at 337.5° to x-direction. (ii) F = 20367 N inclined at 337.5° to x-direction) 5.11 A 10 cm diameter jet having a velocity of 18 m/s impinges on a moving vane of velocity 8 m/s at 25° to the direction of motion. The vane angle at the outlet is 30°. Find the (i) blade angle at inlet so that the water enters without shock, (ii) the component of force exerted by the jet on the vane in the direction of motion of the vane. (Ans. (i) a1 = 42.46°, (ii) FX = 1597 N in x-direction) 5.12 A jet of oil (RD = 0.90) issues out horizontally from a nozzle and has a size of 12 cm diameter. A 60° vertex angle cone with its vertex pointing to the jet and its axis aligned in the direction of ﬂow deﬂects the jet. If a force of 600 N is required to hold the cone in position, calculate the velocity of the jet. (Ans. V = 21 m/s) 5.13 A reducer bend has an inlet of 30 cm diameter and an outlet of 15 cm diameter. The outlet is turned at a deﬂection angle of 45° and the bend lies in a horizontal plane. Water ﬂows through the bend at a rate of 90 L/s with a pressure of 20 kPa 195 Momentum Equation and Its Applications at the outlet end. Determine the magnitude and direction of the force required to keep the bend in position. (Ans. F = 1900.5 N inclined at 197.53° to x-direction.) 5.14 A 20 cm diameter pipe has a 90° bend (to the right) in the horizontal plane. When a discharge of 150 L/s of oil (RD = 0.8) is sent in this pipe, the pressure at the beginning of the bend is found to be 0.5 m of oil. Estimate the resultant force exerted by the oil on the bend. (Ans. F = 1612.5 N inclined at 41.9° to x-direction.) 5.15 A reducer bend at the end of a pipe having inlet diameter of 30 cm and inlet axis horizontal is turned through 60° and discharges freely with an exit diameter of 20 cm. The bend lie in a vertical plane and the centre of the exit section is 60 cm higher than the axis at the inlet. The water inside the bend weighs 700 N. If the exit velocity is 10 m/s, determine the force required to hold the bend in position. (Ans. R = 4593 N at q = 131.95° to x-direction.) 5.16 Water ﬂows through a 180° vertical reducing bend shown in Fig. 5.27. The pressure at the inlet pipe is 20 kPa, and the discharge is 0.4 m3/s. If the bend volume is 0.8 m3, calculate the force required to hold the bend in place. 1 Y 30 cm 30 cm X 20 cm 2 Fig. 5.27 (The bend is in a vertical plane). (Ans. F = 10854 N at 136.19° to x-direction) 5.17 Water ﬂows through the Y-joint as shown in Fig. 5.28. Find the horizontal and vertical components of the force acting on the joint because of the ﬂow of water. Neglect energy loss and body forces. [Take r = 1000 kg/m3] (Ans. Fx = 27.5 kN in (– x) direction, Fy = 16.65 kN in (– y) direction) 2 1 A2 = 0.1 m V1 = 10 m/s 2 p1 = 150 kN/m 120° 2 90° 3 2 2 A2 = 0.1 m p3 = ? V3 = ? Fig. 5.28 A2 = 0.1 m p2 = ? V2 = ? Example 5.17 5.18 0.28 m3/s of water ﬂows up a vertical enlargement, one meter high, 30 cm in diameter at the bottom and 60 cm diameter at the top. Calculate the magnitude and direction of force exerted by the ﬂow on the enlargement. (Ans. F = 21.30 kN acting vertically downwards) 5.19 A sluice gate in a rectangular channel carrying water is so opened to create depths of ﬂow of 1.70 m and 0.25 m on the upstream and downstream of the gate respectively. The discharge intensity is 1.30 m3/s per metre width and the ﬂow from the 196 Fluid Mechanics and Hydraulic Machines sluice gate is in free ﬂow mode. Estimate the force per metre width on the gate. (Ans. F = 8.09 kN) 5.20 For the velocity distribution in a two dimensional duct, shown in Fig. 5.29 (a) and (b) calculate the value of the momentum correction factor. (Ans. (a) 1.333, (b) 1.5) 2u1 B u y u1 with the tangential direction of rotation. The ﬂow enters the sprinkler normally at the centre. The nozzle ends lie on a diameter of 0.8 m and the discharge of water entering the sprinkler is measured as 2.0 L/s. Calculate the speed of rotation by (i) assuming zero frictional resistance (ii) assuming the torque due to friction as 2.0 N.m (Ans. (i) N = 117 rpm, (ii) N = 57.2 rpm) 5.23 A four-arm lawn sprinkler, shown in Fig. 5.30 has provision for admitting water at the axis of rotation at a rate of 1.5 L/s. calculate the steady rotational speed by neglecting friction. (Ans. N = 237.5 rpm) (a) u 2/3B r = 30 cm B B 3 (b) Fig. 5.29 w Problem 5.20 5.21 A 20 cm diameter pipe has a nozzle with 10 cm diameter exit attached to it. A jet of water with 15 m/s velocity is discharged, horizontally, from this nozzle into the atmosphere. The momentum correction factor b for the pipe ﬂow is 1.25 and 1.0 for the jet. The kinetic energy correction factor a for the pipe ﬂow is 1.75 and it is 1.0 for the jet. Calculate the force on the nozzle at its junction with the pipe. (Ans. F = 1929 N in x-direction) Nozzle dia: 8 mm Fig. 5.30 5.24 For the water sprinkler, shown in Fig. 5.31 b = 158°, radius r = 0.50 m, jet diameter = 1.1 cm. If the speed of rotation is 240 rpm calculate the discharge supplied to the sprinkler. Assume zero frictional losses. (Ans. Q = 2.758 L/s) w Moment of Momentum 5.22 A lawn sprinkler has 1.5 cm diameter nozzles at the two ends of a rotating arm. The nozzles make an angle of 150° Problem 5.23 b 1 2 u Fig. 5.31 Problem 5.24 197 Momentum Equation and Its Applications 5.25 In Problem 5.31, what external torque is required to make the sprinkler stationary? (Ans. T0 = 17.3 N.m) 5.26 A lawn sprinkler has two jets of 1.5 cm diameter on a rotating arm which can describe a circle of 60 cm diameter (Fig. 5.32). If the rotating speed is 210 rpm and the discharge is 3.0 L/s. Calculate the torque due to friction at the axis of rotation. (Ans. T = 1.698 N.m) 5.27 A lawn sprinkler has unequal arms and has 1.25 cm nozzles at the ends as shown in Fig. 5.33. If the discharge through the sprinkler is 2.8 L/s calculate the constant speed of rotation (i) by neglecting friction and (ii) by assuming frictional torque as 6 Nm. (Ans. (i) N = 512.7 rpm, (ii) N = 30.22 rpm) 25 cm 15 cm w 60 cm w 40 cm Fig. 5.33 30 cm Fig. 5.32 Problem 5.27 Problem 5.26 Objective Questions Linear Momentum 5.1 The linear momentum equation is based on (a) Newton’s law of viscosity (b) Newton’s ﬁrst law (c) Newton’s second law (d) Newton’s third law 5.2 A control volume is (a) the volume of ﬂuid ﬂowing per unit of time. (b) a volume ﬁxed in space. (c) the volume in which a control device is situated. (d) the volume of the ﬂuid controlling device. 5.3 The linear momentum equation is (a) a scaler relation (b) an approximate relation for engineering analysis (c) a relation applicable to incompressible ﬂuids only (d) a vector relation 5.4 In steady, incompressible, ﬂuid ﬂow with uniform velocity distribution, the momentum ﬂux in a given x-direction past a given section is expressed as Mx = (a) rQV (b) rV2/2 (c) rQVx (d) Q2/A 5.5 The linear momentum equation applied to a control volume in a ﬂow through a nozzle yielded the resultant reaction force R on the ﬂuid in the control volume. The force required to keep the nozzle in position is (a) the same as in magnitude and direction. (b) equal to R but opposite in direction (c) equal to the x-component of R (d) equal to R minus the friction force 198 Fluid Mechanics and Hydraulic Machines 5.6 A jet of oil (RD = 0.8) has an area of 0.02 m2 and a velocity of 10 m/s. If it strikes a plate normally, the force exerted on the plate is (a) 1597 N (b) 1996 N (c) 15665 N (d) 19581 N 5.7 A water jet has an area of 0.03 m3 and impinges normally on a plate. If a force of 1 kN is produced as a result of this impact, the velocity of the jet, in m/s, is (a) 15 (b) 33.4 (c) 3.4 (d) 5.78 5.8 A water Jet 0.015 m2 in area has a velocity of 15 m/s. If this jet impinges normally on a plate which is moving at a velocity of 5 m/s in the direction of the jet, the force on the plate due to this impact is (a) 3368 N (b) 2246 N (c) 1497 N (d) 14686 N 5.9 Uniform ﬂow of a real ﬂuid takes place in a horizontal pipe of diameter D. If p1 and p2 are the pressures at the upstream and downstream sections of a stretch of length L off the pipe, the boundary shear stress t0 could be expressed by the momentum equation as t0 = (a) (p1 – p2) pD2/4L (b) (p1 – p2) D/4L (c) (p1 – p2) 4L/D (d) (p1 – p2) D/rgL 5.10 A ﬁre house has a nozzle attached to it and the nozzle discharges a jet of water into the atmosphere at 20 m/s. This places the joint of the nozzle. (a) in compression (b) in tension (c) in a state of zero stress (d) in bending stress 5.11 A two-dimensional jet strikes a ﬁxed twodimensional plane at 45° to the normal to the plane. This causes the jet to split into two streams whose discharges are in the ratio (a) 1.0 (b) 2.41 (c) 5.83 (d) 1.414 5.12 A jet of water with a velocity of 20 m/s impinges on a single vane at 5.0 m/s in the direction of the jet and transmits a power P1. If the same jet drives a series of similar vanes mounted on a wheel under similar velocity conditions, the power transmitted is P2. The ratio of P1 and P2 is (a) 0.25 (b) 0.33 (c) 0.50 (d) 0.75 5.13 When a steady two-dimensional jet of water impinges on a stationary inclined plate and if the ﬂuid friction is neglected, the resultant force on the plate (a) is tangential to the surface (b) is normal to the surface (c) is in the direction of jet ﬂow (d) is normal to the direction of the jet 5.14 A symmetrical stationary vane experiences a force F = 100 N in a ﬂow as shown in Fig. 5.34. The mass rate of ﬂow of water is 5 kg/s with a velocity V = 20 m/s without any friction. The angle a of the vane is (a) zero (b) 30° (c) 45° (d) 60° V a a F = 100 N V Fig. 5.34 5.15 For turbulent ﬂow in a long straight reach of a pipe carrying a ﬂuid, the momentum 199 Momentum Equation and Its Applications correction factor b can be expected to be in the range (a) 2.0 – 4.0 (b) 1.7 to 2.3 (c) 1.70 to 1.30 (d) 1.01 to 1.10 5.16 The dimensions of Momentum correction factor b is (a) M0 L T0 (b) M0 L1 T –2 0 0 0 (c) M L T (d) M1 L2 T–2 5.17 The momentum correction factor is given by b = (a) (c) 1 3 V 1 A2 AÚ Ú u 3 d A (b) A A2 d u (d) 1 VA 1 Ú V D 0.75 V 2. 1.25 V D 0.75 V u dA A Ú 0.5 V u2 d A V 2A A 5.18 A nozzle discharging under a head H has an area a and a discharge coefﬁcient Cd = 1.0. A vertical plate is acted upon by the ﬂuid force Fj when held across the free jet and by the ﬂuid force Fn when held against the nozzle to stop the ﬂow. The ratio Fj /Fn is (a) 1/2 (b) 1 A 1. (c) 2 (d) 2 5.19 Consider the following velocity proﬁles in a pipeline (Fig. 5.35) Among these proﬁles the momentum correction factor would be (a) least in 4 (b) highest in 1 (c) more in 3 than that for 2 (d) the same in 1, 2, 3 and 4 5.20 The velocity distribution over one half of a cross section is uniform and is zero over the remaining half. The momentum correction factor for this cross section is (a) 2.0 (b) 4.0 (c) 1.0 (d) 3.0 5.21 A jet strikes a stationary plate normally with a velocity of 8 m/s and the plate suffers a force of 120 N. The power obtained, in kW is 3. 1.5 V D 0.5 V 4. 2V Fig. 5.35 D Question 5.19 (a) 0.96 (b) 9.4 (c) zero (d) 958 5.22 If all the energy in a jet of velocity 20 m/s issuing as a jet of 15 cm diameter could be extracted, the power available, in kW, is (a) 70.54 (b) 7.21 (c) 20.39 (d) 705 Moment of Momentum 5.23 The moment of momentum principle states that in a rotating system (a) the resultant force exerted by the ﬂuid on the body is equal to the rate of change of angular momentum (b) the torque exerted by the resultant force is equal to the time rate of change of angular momentum 200 Fluid Mechanics and Hydraulic Machines (c) the torque exerted by the resultant force is equal to the time rate change of linear momentum (d) the angular moment is conserved 5.24 A lawn sprinkler has two nozzles of area 0.75 cm2 each, on either side of an arm capable of rotating about its midpoint. A discharge of 1.5 L/s is introduced at its axis to be discharged out at its nozzles, and the sprinkler arm rotates at a constant speed. The jet issuing from the nozzle will have (a) an absolute velocity of 10.0 m/s (b) an absolute velocity of 20.0 m/s (c) a relative velocity of 10.0 m/s (d) a relative velocity of 20.0 m/s 5.25 Figure 5.36 shows a rotating water sprinkler. The area of the nozzle at each end of the arm is 1.2 cm2. If the discharge ﬂowing out of the sprinkler is 2.4 L/s and the angular velocity is 7.69 rad/s, the absolute velocity of water emanating from A (at the end of the shorter arm), in m/s, is (a) 7.69 (b) 8.462 (c) 10.00 (d) 11.538 30 cm 20 cm w A B Fig. 5.36 Question 5.25 5.26 For the sprinkler of Question 5.25 the absolute velocity of water issuing out from nozzle B (at the end of the longer arm), in m/s, is (a) 10.0 (b) 7.693 (c) 12.307 (d) zero 5.27 Consider the lawn sprinkler shown in Fig. 5.37 with v2 = relative velocity and u2 = tangential velocity of the sprinkler jet. For a frictionless system, the angular velocity is given by (a) w = v2/r (b) w = (v2 cos b + u2)/r (c) w = v2 sin b/r (d) w = v2 cos b/r v2 r r b b w u2 v2 Fig. 5.37 Question 5.27 5.28 For a lawn sprinkler shown in Fig. 5.38, if T = torque due to friction at bearings, etc., and Q = discharge, then the angular velocity w is related to relative velocity v2 as (a) w = (c) w = v2 r T rQr 2 v2 T r rQr 2 v2 T + (d) w = r rQr 2 (b) w = 5.29 A sprinkler, such as shown in Fig. 5.38 is frictionless and requires a torque of 10 N.m to keep it stationary when a discharge of 1.5 L/s is passing through the system. If the discharge is doubled, the torque required to keep the sprinkler stationary, in N.m, is (a) 40 (b) 20 (c) 10 (d) 5 v2 u2 r w r u2 v2 Fig. 5.38 Question 5.29 Dimensional Analysis and Similitude Concept Review 6 Introduction Dimensional analysis is a type of compacting technique to reduce the number of variables to be studied in an experimental investigation of a physical phenomenon. All physical phenomena are expressible in terms of a set of basic or fundamental M, length L, time T and temperature q in what is known as the MLT q system. Sometimes the force F is used in place of M to get F, L, T and q as the basic dimensions in what is known as the engineering system. 6.1 COMMON VARIABLES IN FLUID FLOW The commonly occurring variables in ﬂuid mechanics together with their common notations and their dimensions are given in Table 6.1. This table will be very helpful in performing dimensional analysis and as such warrants careful study. 6.2 DIMENSIONAL HOMOGENEITY An equation which expresses the proper relationship between the variables in a physical phenomenon will be dimensionally homogeneous. This means that each of the additive terms in an equation should have the same dimension. The principle is useful in checking the proper form of the equation, and in converting the equations having dimensional constants from one system to another (Example 6.1) 6.3 DIMENSIONAL ANALYSIS In the dimensional analysis of a physical phenomenon the relationship between the dependent and independent variables is studied in terms of their basic dimensions to obtain the information about the functional relationship between the dimensionless parameters that control the phenomenon. There are several methods of reducing the number of 202 Fluid Mechanics and Hydraulic Machines Table 6.1 Common Variables in Fluid Flow Quantity Length Area Volume Angle Angular velocity Frequency Discharge Velocity Mass density Dynamic viscosity Kinematic viscosity Surface tension Volume modulus of elasticity Speciﬁc weight Relative density Force Moment, Torque Momentum Work, Energy Power Rotation Strain Strain rate Stress, Pressure Temperature Speciﬁc heat Thermal conductivity Notation expressed as Dimensions (MLT q) System (FLT q) System L A V a w f Q U.V. r m n s K L L2 L3 M0 L0 T0 T–1 T–1 L3 T–1 LT–1 ML–3 ML–1 T–1 L2 T–1 MT–2 ML–1 T–2 L L2 L3 F0 L0 T0 T–1 T–1 L3 T–1 LT–1 FL–4 T2 FL–2 T L2 T–1 FL–1 FL–2 g RD F M, T M W, E P rpm – – p T cp, cv k ML–2 T–2 M0 L0 T0 ML T–2 ML2 T–2 MLT–1 ML2 T–2 ML2 T–3 T–1 M0 L0 T0 T–1 ML–1 T–2 q L2 T–2 q –1 MLT–3 q –1 FL–3 F0 L0 T0 F FL FT FL FLT–1 T–1 F0 L0 T0 T–1 FL–2 q L2 T–2 q –1 FT–1 q–1 dimensional variables into a smaller number of dimensionless parameters. Two of the commonly used methods are the (i) Raleigh’s method, and (ii) Buckingham Pi theorem method. 6.3.1 Raleigh’s Method If A1 is a dependent variable and A2, A3, º An are independent variables in a phenomenon, A1 is A1 = k A2a A3b A4c º An (6.1) where k is a dimensionless constant. The dimensions of each of the quantities A1, A2, A3, º An are written and the sum of exponents of each of M, L, and T on both sides are equated. Solution of the equations on simpliﬁcation yields dimensionless groups controlling the phenomenon. Example 6.2, 6.3 and 6.4 illustrate this method. While this method is simple for a small number of parameters, it becomes rather cumbersome when a large number of parameters are involved. 6.3.2 Buckingham Pi Theorem The Buckingham Pi theorem states that if there are m primary dimensions involved in the n variables controlling a physical phenomenon, then the phenomenon can be described by (n – m) independent dimensionless groups (known as p s). The word Pi here refers to a product of variables and the Greek letter p is used to indicate these products, for example p1, p 2 .... In the application of this method, m number of repeating variables are selected and dimensionless groups obtained by each one of the remaining variables one at a time. Raleigh’s method is used in this part of the operation. Examples 6.5 to 6.12 illustrate the use of Buckingham Pi theorem. This method is also known as the method of repeating variables. Care is needed in selecting the repeating variables. (i) They must have amongst themselves all the basic dimensions involved in the problem. (ii) The dependent variable must not be chosen as a repeating variable. (iii) Usually a length parameter (such as a diameter D or head over a weir, H ); a typical velocity V and the ﬂuid density r are convenient set of repeating variables. 203 Dimensional Analysis and Similitude 6.4.3 6.4 SIMILITUDE In hydraulic and aeronautical engineering valuable results are obtained at a relatively small cost by performing tests on small scale models of full size systems (prototypes). Similarity laws help us interpret the results of model studies. Similitude, the relation between model and a prototype, is classiﬁed into three kinds as follows: 6.4.1 Geometric Similarity If the ratios of corresponding lengths in a model and the prototype are the same, the model is said to be geometrically similar model. In such models if Two systems are dynamically similar, if geometric and kinematic similarities exist and further the ratios of all corresponding forces in the two systems are the same. If forces due to Gravity = FG Viscosity = Fv Elasticity = FE Surface tension = FT Inertia = FI and sufﬁxes m and p stand for model and prototype respectively, strict dynamic similarity means L model L = m = Lr L prototype Lp then and Area model A ( L )2 = m = m 2 = L 2r Area prototype Ap ( Lp ) FGm Fvm FEm FT m FIm = = = = = constant FGp Fvp FEp FT p FIp (6.2) ( Volume) model V ( L )3 = m = m 3 = L r3 (6.3) ( Volume) prototype Vp ( Lp ) 6.4.2 Kinematic Similarity Kinematic similarity means geometric similarity and in addition the ratio of velocities at all corresponding points in the ﬂow is the same. If L r = Lm = length ratio, and if Lp tm L = tr = r tm Vr (ii) acceleration ratio = (6.4) am V2 L = a r = r = 2r ap Lr Tr (6.5) (iii) discharge ratio = Qm L3 = Qr = r Qp Tr From the above the following relationships can be derived. (Inertia force) p (Inertia force) m = ( Viscous force) m ( Viscous force) p = Constant 1 (Inertia force) p (Inertia force) m = (Gravity force) m (Gravity force) p = Constant 2 and so on for all the forces. 6.5 ÊV ˆ ( Velocity ) model = Á m ˜ = Vr then ( Velocity) prototype Ë Vp ¯ (i) time ratio = Dynamic Similarity (6.6) IMPORTANT DIMENSIONLESS FLOW PARAMETERS The following dimensionless parameters representing ratios of forces per unit volume are of great signiﬁcance in the analysis of ﬂuid ﬂow: Inertial force (1) Reynolds number = Re = Viscous force rVL VL = m v For dynamic similarity where viscous forces are predominant. = Ê VL ˆ Ê VL ˆ ÁË v ˜¯ = Re m = Re p = Á ˜ Ë v ¯p m 204 Fluid Mechanics and Hydraulic Machines (Inertia force)1/ 2 (2) Froude number = Fr = (Gravity force) V = Ê V ˆ Ê V ˆ Á ˜ = (Fr)m = (Fr)p = Á Ë gL ¯ m Ë g L ˜¯ p (3) Mach number = 1/ 2 (Inertial force) (Compressibility force)1/ 2 V = E/r = V C where C = velocity of sound in the medium. Where compressibility effects predominate, for dynamic similarity, we have the relations: ÊV ˆ ÊV ˆ ÁË C ˜¯ = Mm = Mp = Á ˜ ËC¯p m (4) Weber number = W= (Inertial force) rV 2 L = (Surface tension ) s (Inertial force)1/ 2 ( Pressure force)1/ 2 = V Dp / r 6.6 MODEL SCALES Fluid ﬂow models are usually designed to account for one most dominant force, and occasionally for two dominant forces. Thus, if the dominant force is the gravity force, then the Froude number must be the same in the model and prototype. Thus, Vm g Lm Table 6.2 Similitude Scale Ratios Scale Ratios for Laws of Parameter Dimension Froude Reynolds L L2 L3 Lr L2r L3r Lr L2 L3r LT–1 (Lr)1/2 (m r /Lr rr) Time T (Lr)1/2 (L2r rr/mr) Acceleration LT–2 1 (m 2r /rr2 L3r) Discharge L3T (Lr)5/3 (Lr m r /r r) M (L3r r r) (L3r r r) Force MLT–2 (L3r r r) (m2r/r r) Pressure ML–1 T–2 (Lr r r ) (m2r /L2r rr) (L7/2 r r r) (Lr4r r) (L7/2 r rr) (L2r mr) Geometric Length Area Volume Kinematic Velocity Dynamic Mass Momentum (5) Euler number = E= = Lr From this other scales ratios, for such Froude law modelling are developed as shown in Table 6.2. gL For dynamic similarity where gravity forces are predominant. M= Vm = Vr = Vp 1/ 2 Vp g Lp As g is the same for both model and prototype, –1 MLT Energy of work ML2T–2 Power 2 –3 ML T (Lrm 2r /rr) (m r3/L r r r2) Similarly, if viscous forces are predominant the Reynolds number (VL/n ) must be the same in the model and prototype. The scale ratios for Reynolds number modeling are also shown in Table 6.2. Similarly appropriate scales can be developed for Mach law and Weber number law. Some commonly used applications of Froude number similarity and Reynolds number similarity in model studies are indicated in Table 6.3 6.7 DISTORTED MODELS Hydraulic modeling of rivers, harbors, estuaries, etc. which have longitudinal slope and large areal spread, pose many practical problems if strict 205 Dimensional Analysis and Similitude Table 6.3 Common Applications of Froude Number and Reynolds Number Similarities in Model Studies Froude number similarity is used when there is dominant action of gravity. Typically in Surface wave action as in (i) Motion of ships, boats, (ii) Break waters and harbours Reynolds number similarity is used when there is dominant action of viscous forces. Typically in Completely submerged ﬂow as in (i) Air planes, (ii) Torpedos Free surface ﬂow as in Flow in canals, streams and Completely enclosed ﬂow rivers as in Flow through pipes, ﬂow past plates, ﬂuid drag and lift on body shapes (cars, trains, parachutes, etc.) Hydraulic structures with free surface ﬂow as in Spillways, stilling basins, hydraulic jumps, weirs and notches Settling of particles and creeping ﬂow Structures subjected to free surface ﬂow action as in Wave and water ﬂow forces Flow in ﬂow meters in pipes as in in bridge piers, off-shore structures, jetties and piers Fluid ﬂow machines as in similitude is attempted. To overcome most of these problem, they are usually modeled by using distorted scales. The vertical ﬂow dimension (viz. depth) is used to simulate Froude’s law while the other two dimensions (viz. length and width) are scaled to suit available space. Thus, vertical scales of 1/100 and horizontal scales of 1/200 to 1/1000 are common. In such distorted models, Horizontal scale is Lr = length ratio = (Lm /L p) = width ratio = (Bm /Bp) Vertical scale is h r = depth ratio = (ym/yp) In the above prototype and model quantities are denoted by sufﬁxes p and m respectively. Based on these two scales, various scaling ratios for physical parameters are obtained as follows: Cross sectional area ratio = (area in model)/(area in prototype) = (By)m /(By)p = Lr hr Froude number ratio = (Fm/Fp) = 1 Hence (Vm /V p ) 2 ( ym / y p ) Thus the velocity ratio = Vr2 . hr Viscous ﬂow as in Venturi meters, Oriﬁce meters, etc. Fans, blowers, propellers, pumps and turbines Vr = (Vm/Vp) = hr Discharge ratio Qr = (area ratio) (velocity ratio) = Lr h 1.5 r Slope ratio Sr = Sm /Sp = hr/L r. Time ratio Tr = Lr /Vr = L r / hr Roughness Ratio By Manning’s formula, considering the channel to be wide, 1 2/3 1/2 Vr = h r Sr nr Substituting the various ratios derived above 1 2/3 hr = hr ( hr / Lr ) nr Manning’s Hence, nr = hr2 / 3 L1r/ 2 Similarly, scale ratio for any other physical quantity or ﬂow phenomenon can be derived. Similitude scale ratios of some commonly used parameters in distorted scale models are listed in Table 6.3. 206 Fluid Mechanics and Hydraulic Machines Table 6.4 Similitude Scale Ratios in Distorted Models Parameter Symbol Table 6.4 Contd. Parameter Scale ratio Symbol Scale ratio Length L Lr Time T L r / hr Width Depth B y Lr hr Area of ﬂow A L r hr Acceleration Discharge Slope a Q S hr/Lr Lr h 1.5 r hr /Lr Volume V L2r hr Manning’s roughness n Velocity V hr hr2 / 3 L1r/ 2 Reynolds number Re h 1.5 r Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples Dimensional Homogeneity Dimensional Analysis: Raleigh Method 6.1 V in V= R 1 2/3 1/2 S0 R n S0 n of n Solution: 6.2 D R 2 / 3 S 01/ 2 V Writing the dimensions of the terms on the right hand side and noting that S0 does not have any dimensions, n= [n] = (L) 2/3 -1 ( LT ) = [L–1/3 T] r V m for F Solution: Since Rewriting the equation explicitly for n, F FD = fn (D, V, r, m) FD = KD aV br cmd where K is a dimensionless constant. Using F, L, T as primary units, F = [L]a (LT–1]b [FL–4 T2]c [FL–2 T]d Equating the powers of F, L and T on both sides: For F: 1 =c+d (1) for L: 0 = a + b – 4c – 2d (2) for T: 0 = – b + 2c + d (3) 207 Dimensional Analysis and Similitude Since there are 3 equations and 4 unknowns, three variables can be expressed in terms of the fourth. from (1) c =1–d from (3) b = 2c + d = 2 – d from (2) a = – b + 4c + 2d = –(2 – d) + 4 – 4d + 2d = 2 – d \ F = K . D2–d . V 2– d . r1–d . md Ê m ˆ = KrD 2V 2 Á Ë VD r ˜¯ d Solution: F = fn (L, V, D, r, m, E) Hence F = KLaV bD crdm eE f Using the MLT system of basic units, [MLT –2] = [L]a [LT –1]b [L]c [ML–3]d ˆ FD = rD V fn VD ˜¯ 2 length L, velocity V, diameter D properties like density r, dynamic viscosity m and bulk modulus of elasticity E. Derive an expression for F. [ML–1 T –1] e [ML–1 T –2] f 2 6.3 pressure Dp is a function of the pipe length L, its diameter D V and the dynamic viscosity m. Using Raleigh’s method, develop an expression for Dp. Solution: Dp = fn (L, D, V, m) This can be written in terms of a dimensionless constant K as Dp = KLa D b V c m d Using the MLT system of basic units Equating powers of M: 1 =d+e+f (1) L: 1 = a + b + c – 3d – e – f (2) T: –2 = – b – e – 2f (3) Note in this case there are 6 variables and 3 equations. Values of three unknowns are expressed in term of other three (viz. c, e, and f ). From 1: d = – e – f + 1 From 3: b = 2 – e – 2f From 2: a = 1 – b – c + 3d + e + f = 1 – 2 + e + 2f – c – 3e – 3f + 3 + e + f =2–c–e \ F = KL2– c–e . V 2– e – 2f . Dc . r–e–f +1 . me . E f [ML–1 T –2] = [L]a [L]b [LT –1]c [ML–1 T–1]d F = KrV L Equating the Powers of M: d = 1 Powers of T: –2 = – c – d = – c – 1; c = 1 Powers of L: –1 = a + b + c – d = a + b + 1 – 1; a = – 1 – b F = V 2 L2 Dp = KL–1–b D bVm \ ÊVmˆ Ê Dˆ = KÁ Ë L ˜¯ ÁË L ˜¯ \ Dp = b Ê Dˆ V ◊ fn Á ˜ ËL¯ L 6.4 The drag force F on a body in supersonic 2 2. 1 ÈD˘ ÍL˙ Î ˚ c È m ˘ ◊Í ˙ Î rVL ˚ e È E ˘ Í 2˙ ÍÎ rV ˙˚ f D Ê ˆ Ê E ,Á , L Ë VL ˜¯ ÁË V 2 [Note: It is very important not to make a mistake in writing the basic dimensions of the variables at the beginning of the problem. As such it is essential to know the dimensions of parameters given in Table 6.1 thoroughly]. Dimensional Analysis: Buckingham Pi Theorem 6.5 The resistance force F of a ship is a function of its length L, velocity V, 208 Fluid Mechanics and Hydraulic Machines like density r and viscosity m. Write this relationship in a dimensionless form. F Hence 2 2 rV L F = [Fr, Re] V 2 L2 Fr = Froude number = V / g L Re = Reynold’s number = rVL/m. Solution: F = fn (L, V, g, r, m) For using Buckingham Pi theorem, list the dimensions of each variable. where and F L V g r m [MLT –2] [L] [LT–1] [LT–2] [ML–3] [ML–1 T–1] There are a total of six variables (n = 6) and 3 primary dimensions. Hence m = 3. As such there are n – m = 3 dimensionless Pi terms. Using L, V and r as repeating variables: p1 = [M 0 L0 T0] = FLa Vbr c or [M0L0T0] = [MLT–2] [L]a [LT–1]b [ML–3]c Hence by equating powers of M, L and T 1+c =0 1 + a + b – 3c = 0 –2 – b = 0 c = – 1, b = – 2, a = – 2 F p1 = rV 2 L2 p 2 term: p2 = gLa V b r c or [M0 L0T0] c 1+a+b –2–b a \ p3 term: = [LT–2] [L]a [LT–1]b [ML–3]c =0 =0 =0 = 1, b = – 2, c = 0 gL p2 = 2 V p3 = mLa V b r c [M0L0T0] 1+c –1 + a + b – 3c –1 – b \ c = [ML–1 T–1] [L]a [LT–1]b [ML–3]c =0 =0 =0 = –1, b = –1, a = –1 m p3 = LVr È gL m ˘ = fn Í 2 , ˙ Î V rVL ˚ 6.6 The discharge Q over a small rectangular weir is known to depend upon the head H over the weir, the weir height P, gravity g, width of the weir L r, dynamic viscosity m and surface tension s. Express the relationship between the variables in dimensionless form. Solution: Q = fn (H, P, g, L, r, m, s) For using Buckingham Pi theorem list the dimensions of each variable. Q 3 H –1 P g –2 L [L T ] [L] [L] [LT ] [L] r m –3 s –1 –1 [ML ] [ML T ] [MT–2] In this case, the number of variables n = 8 number of primary variables m = 3 Hence number of dimensionless terms = 8 – 3 =5 Selecting r, H and g as the repeating variables p1 = Q H agbr c [M0L0T0] = [L3 T –1] [L]a [LT–2]b [ML–3]c Equating exponents of M, L, T on both sides, c =0 3 + a + b – 3c = 0 –1 – 2b = 0 \ c = 0, b = – 1/2 and a = – 5/2 Q Hence p1 = 5 / 2 1/ 2 H g = PH a g b r c [M L T ] = [L] [L]a [LT–2]b [ML–3]c By inspection b = 0, c = 0 and a = – 1 \ p 2 = P/H p 2 term: p2 0 0 0 209 Dimensional Analysis and Similitude = LH agb r c [M L T ] = [L] [L]a [LT –2]b [ML–3]c By inspection, here also b = 0, c = 0 and a = – 1 p 3 = L/H p 4 term: p4 = mHag brc p 3 term: p3 0 0 0 [M0L0T0] = [ML–1 T–1] [L]a [LT–2]b [ML–3]c By inspection, we have 1+c =0 –1 + a + b – 3c = 0 –1 – 2b = 0 1 \ c = – 1, b = 2 3 and a = 1 – b + 3c = 2 m p 4 = 3 / 2 1/ 2 H ◊g r [M0L0T0] 1+c a + b – 3c –2 – 2b c a = [MT–2] [L]a [LT–2]b [ML–3]c =0 =0 =0 = – 1, b = – 1 = 3c – b = – 2 s p5 = 2 H ◊g◊ r Thus Q \ V rs –1 m rf D –3 –1 gH 5/ 2 = fn P L , , , H H H 3/2 g1/2 gH 2 V = fn [rs, D, r f, m, g] m –1 –1 s [MT–2] and 3 basic dimensions, m = 3 Hence there are (6 – 3) = 3 dimensionless terms. Select D, rf and g as the repeating variables 1 term: p1 = VD ar bf gc Following the usual procedure [M0L0T0] b 1 + a – 3b + c –1 – 2c Hence b \ = [LT–1] [L]a [ML–3]b [LT–2]c =0 =0 =0 = 0, c = – 1/2, a = – 1/2 p1 = V gD p 2 = rsDarfbgc II term: By inspection, it is easy to see p 2 = rs rf III term: p 3 = mD arfbgc In this case, we have [M0L0T0] 1+b –1 + a – 3b + c –1 – 2c = [ML–1 T–1] [L] a [ML–3]b [LT–2]c =c =0 =0 1 b = –1, c = - and a = 3b – c + 1 2 3 = 2 m p 3 = 3 / 2 1/ 2 D rf g \ Hence Solution: –2 There are six variables, n = 6 6.7 A small sphere of density rs and diameter D settles at a terminal velocity V in a liquid of density rf and dynamic viscosity m. Gravity g is known to be a parameter. Express the functional relationships between these variables in a dimensionless form. g –1 [LT ] [ML ] [L] [ML ] [ML T ] [LT ] [ML T ] –3 p5 = sH ag brc p 5 term: and List the dimensions of each variable as follows: V gD = È Í Î s f , ˘ ˙ D g D f ˚ 210 Fluid Mechanics and Hydraulic Machines 6.8 The pressure drop Dp generated by a pump of a given geometry is known to depend upon the impeller diameter D, the Q, the rotational speed N r and viscosity m. Obtain the dimensionless form of the functional relationship. Solution: Dp = fn (D, N, Q, r, m) List the dimensions of each variable as: Dp D N Q r m m As usual, [M0L0T0] 1+c –1 + a – 3c –1 – b or c = [ML–1T–1] [L]a [T–1]b [ML–3]c =0 =0 =0 = –1, b = – 1, a = – 2 m \ p3 = 2 D Nr Thus the functional relationship can be expressed as Ê Q = fn Á , ND Ë ND3 p s 2 [ML–1 T–2] [L] [T–1] [L3 T–1] [ML–3] [ML–1T–1] [ML–1T–1] [MT–2] There are a total of six variables: n = 6 Number of primary dimensions m = 3 Hence there are (6 – 3) = 3 dimensionless Pi terms. Select D, N and r as repeating variables. I term: p1 = DpD aN brc The dimensional equation will be [M0L0T0] = [ML–1 T–2] [L]a [T–1]b [ML–3]c from which we get 1+c =0 –1 + a – 3c = 0 –2 – b = 0 or c = – 1, b = – 2 and a = – 2 Dp \ p1 = 2 2 D N r II term: ˆ ND ¯ 2˜ 6.9 The time period T of water surface waves is known to depend on the wave length D r, l acceleration due to gravity g and surface tension s. Obtain the dimensionless form of the functional relationship. Solution: T = fn (l, D, r, g, s) List the dimensions of each variable as follows: T l D r s g m [T] [L] [L] [ML–3] [LT –2] [MT –2] [ML–1T–1] s [MT–2] There are a total of six variables; n = 6 Number of primary dimensions; m = 3 Hence there are (6 – 3) = 3 dimensionless Pi terms. Select l, g, and r as repeating variables. p2 = QDaNbrc Equating dimensions on either side [M0L0T0] = [L3T–1] [L]a [T–1]b [ML–3]c we have c =0 3 + a – 3c = 0 –1 – b = 0 or a = –3, b = –1, c = 0 Q \ p2 = 3 D N III term: p3 = mDaNbrc 2 I term: p1 = Tlagbrc [M0L0T0] 1 – 2b c a + b – 3c \ = [T] [L]a [LT–2]b [ML–3]c =0 \ b = 1/2, =0 = 0; a = –1/2 p1 = T g l II term: p2 = Dlagb rc 211 Dimensional Analysis and Similitude By inspection it is easy to see that p2 = Thus D l b =– 2 –1 – a = 0 a =– 1 È v ˘ p2 = Í ˙ Îw D2 ˚ Thus III term: a b c p3 = sl g r Hence Again [M0L0T0] 1+c –2 – 2b a + b – 3c = [MT–2] [L]a [LT–2]b [ML–3]c =0 \ c = –1 = 0; b = –1 = 0; a = –2 s p3 = 2 l gr Hence T g ÈD = fn Í , Î 6.10 2 ˘ g ˙˚ h w p3: [M0L0T0] = [L3T –1] [T–1]a [L]b By equating the powers of M, L and T 3 + b = 0, Thus b = –3 –1 – a = 0 Thus a = –1 È Q ˘ Hence p2 = Í ˙ Î w D3 ˚ Q ˘ È v Hence h = fn Í , ˙ 2 Î w D w D3 ˚ which can also be written as Èw D2 Q ˘ h = fÍ , ˙ w D3 ˚ Î v D h Q È w D2 Q ˘ h= fÍ , ˙ w D3 ˚ Î v 6.11 T of a D m N Solution: r Ê m ˆ T = D5N 2 rf Á 2 ˜ Ë D Nr ¯ h = fn(v, w, D, Q) Listing the dimensions of each variable: Solution: h v w M0 L0T0 L2T –1 T –1 D Q L L3 T– There are ﬁve variables; n = 5 Two basic dimensions; m = 2 Hence number of dimensionless terms = 3. For dimensional analysis, select w and D as repeating variables. I term p1: Since h is dimensionless, p1 = h II term p2: [M 0 L0 T0] = [L2T–1[T –1]a [L]b By equating the powers of M, L and T 2 + b = 0, h = fn(D, N, m, r) Listing the dimensions of each variable: T D ML2T –2 L N m r T –1 ML–1T –1 ML–3 There are ﬁve variables; n = 5 Two basic dimensions; m = 3 Hence number of dimensionless terms = 2. For dimensional analysis, select w and D as repeating variables. I term p1: p1 = TDaNbrc [M 0L0 T 0] = [ML2 T –2] [L]a [T–1]b [ML–3]c 212 Fluid Mechanics and Hydraulic Machines By equating the powers of M, L and T 1 + c = 0, Thus c = –1 2+a+c =0 Thus a = –5 There are six variables; n = 6 Two basic dimensions; m = 3 Hence number of dimensionless terms = 3. For dimensional analysis, select r, g and H as repeating variables. ˘ È T p1 = Í 2 5˙ Î rN D ˚ Hence I term p1: [M 0L0T0] = [LT –1] [ML–3]a [LT–2]b [L]c p2 = mDa N b rc 0 0 0 –1 –1 a –1 a –3 c p2: [M L T ] = [ML T ] [L] [T ] [ML ] By equating the powers of M, L and T 1 + c = 0, Thus c = –1 –1 + a –3c = 0 Thus a = 3c + 1 = –2 –1 – b = 0 Thus b = –1 È m ˘ Hence p2 = Í 2 ˙ Î rD N ˚ ˘ È T È m ˘ = fn Í Í 2 5˙ 2 ˙ r N D ˚ Î Î rD N ˚ Hence T = D5N 2 r or D2 N 6.12 Using Buckingham’s p theorem, show that the velocity V mass density r and dynamic viscosity m, through D under a head H is given by m ˘ ÈD V = 2gH f Í , ˙ Î H r VH ˚ In this expression g is the acceleration due to gravity. Solution: V = fn(r,m D, H) Listing the dimensions of each variable: V r g m M0L0T0 ML–3 LT–2 ML–1T–1 p1 = Vra g b H c D H L L By equating the powers of M, L and T a =0 1 – 3a + b + c = 0 –1 –2b = 0 Thus b = –1/2 and from (i) 1 –1/2 + c = 0 giving c = –1/2 Hence …(i) È V ˘ p1 = Í ˙ Î gH ˚ II Term p2: p2 = m ra gb Hc [M0L0T0] = [ML–1T–1] [ML–3]a [LT–2]b [L]c By equating the powers of M, L and T 1 + a = 0 giving a = –1 –1 – 3a + b + c = 0 …(ii) –1 – 2b = 0 giving b = –1/2 and from (ii) –1 + 3 – (1/2) + c = 0 giving c = – 3/2 Hence m ˘ È p2 = Í 1/ 2 3 / 2 ˙ r g H ˚ Î III Term p3: p3 = D r a g b H c It is obvious the third term is (D/H), both the parameters having the dimensions of length. D H Thus from the above three Pi terms Hence p3 = ÈÊ ˆ Ê Dˆ˘ m È V ˘ = fn , Í Á ˜˙ Í ˙ 1/ 2 3 / 2 ˜ Á ÍÎË r g H ¯ Ë H ¯ ˙˚ Î gH ˚ …(iii) To get (iii) in to the form needed, consider p2 being multiplied and divided by V, to get 213 Dimensional Analysis and Similitude ÈÊ ˆ V˘ m p2 = ÍÁ ˙ 1/ 2 3 / 2 ˜ ÍÎË r g H ¯ V ˙˚ = Thus m ¥ rVH V gH = = m ¥ p1 rVH Work ratio = Energy ratio = V= 2gH . 1 VH = , D H Power ratio = rr L3r m r2 r 2r L2r Ê m2 L ˆ = Á r r ˜ = (n2rrLr) Ë rr ¯ Pm = Pr = (Force ¥ velocity)r Pp = (rL2V3)r Similitude Em = Er Ep = (Force ¥ distance)r = (rL3V2)r ÈÊ m ˆ Ê D ˆ ˘ È V ˘ ˙ = f ÍÁË rVH ˜¯ , ÁË H ˜¯ ˙ Í ÍÎ ˙˚ Î gH ˚ and hence Ê m 2r ˆ v 2r rr = = Á ˜ rr2 L2r Ë rr L2r ¯ L2r rr m 2r 6.13 Pr = in a model which is to be constructed by using Reynolds model law. Find the expressions for model to prototype ratios of velocity, discharge, pressure, work and power. Solution: Using the subscripts m for model, p for prototype and r for the ratio of model to prototype: Ê rVL ˆ Ê rVL ˆ In Reynolds law Re = Á = Á ˜ Ë m ¯ m Ë m ˜¯ p L Lr = m , then Lp Let Velocity ratio Êv ˆ Vm m = Vr = r = Á r ˜ Vp rr Lr Ë Lr ¯ where vr = ratio of the kinematic viscosities = Discharge ratio Pressure ratio mr rr Qm = Qr = Vr L2r Qp pm pp m L = r r = (vr Lr) rr = pr Nothing that Force μ rL2 V2 Ê Force ˆ Ê r L2V 2 ˆ pr = Á = (rrVr2) =Á ˜ ˜ 2 Ë ¯ Ë Area ¯ r L r rr L3r m 3r r 3r L3r Ê m3 ˆ = Á 2r ˜ = Ë r r Lr ¯ 2 r r Lr [Note: A full list of the various ratios by Reynolds model law is given in Table 6.2.] Froude Model Law 6.14 of velocity, discharge, force, work and power in terms of the length scale. Solution: In Froude model law the model and prototype Froude numbers are the same. Hence Vm Froude number (Fr)m = g Lm Vp = (Fr)p = g Lp The gravity g is same for both model and prototype. Hence if the length ratio Lm/Lp = Lr and rm/rp = rr Velocity ratio Vr = Lr Qr = (Velocity ¥ area)r = VrL2r = (Lr5/2) Force ratio Fr = (rL2V2)r = (rrLr3) Work ratio = Energy ratio = Er = (Force ¥ distance)r = (r rLr4) Discharge ratio 214 Fluid Mechanics and Hydraulic Machines Power ratio = (Force ¥ velocity)r = Pr 7/2 = rrLr3L1/2 r = (rrLr ) Frm = [Note: A full list of the various ratios by Froude model law is given in Table 6.2.] Vm = Vp In the present case, and 12 hours to occur in the prototype, how long should it take in the model? Solution: Since this is the case of a free surface phenomenon affected by gravity, Froude model law is appropriate. Vp Vm Frm = = = Frp g Lm g Lp \ If Lr = Discharge ratio = Qr = VrL2r = L5/2 r Qm 125 Ê 1 ˆ 2.5 1 , = As Lr = =Á ˜ Qp Qp Ë 50 ¯ 50 Prototype discharge Time ratio Tr = 2.5 gL p Lm Lp Vm = 0.81 m/s, Lm = 1.0 m, Lp = 64.0 m Vr = 0.81 = Vp 1.0 1 = 64 8 Vp = 0.81 ¥ 8 = 6.48 m/s 6.17 A model boat, 1/100 size of its prototype has 0.12 N of resistance when simulating a speed of 5 m/s of the prototype. Water is the Solution: The resistance offered at the free surface is the signiﬁcant force and as such Froude model law is appropriate. Frm = 3 ¥ 1.25 = 22097 m /s Lr L = r = Vr Lr If Lr tm tm = = 1/ 50 t p 12 12 = 1.697 hours tm = 50 Vp = Frp = resistance in the prototype? [The frictional forces can be neglected]. Lm , Lp Qp = (50) gLm Vr = 6.15 A 1:50 spillway model has a discharge of 1.25 m3/s. What is the corresponding Vm gLm = Vp g Lp Lm ; Vr = Lp = Frp Lr and (Force) m = rrLr2Vr2 = rrLr3 (Force) p Since the same ﬂuid is used in the model and prototype, rm = rp and rr = 1. Hence, Fm = Lr3 Fp 6.16 A 1.0 m long model of a ship is towed in a towing tank at a speed of 81 cm/s. To what speed of the ship of 64 m long does this correspond? Solution: For ship models the resistance offered at the free surface is the signiﬁcant force and as such for dynamic similarity the Froude number should be same for the model and the prototype. Lr = Vm Ê 1 ˆ Fp = Prototype force = (0.12)/ Á Ë 100 ˜¯ = 120000 N = 120 kN 3 6.18 A model of an open channel is built to a scale of 1/100. If the model has a Manning’s n = 0.013, to what value of prototype 215 Dimensional Analysis and Similitude Solution: Froude model law is applicable here Vm gLm If = gLp Lm = Lr, Lp Vm = Vp Vr = Lr By Manning formula 1 ◊ R2/3 S 01/2 V= n The dimension of R = [L] S0 = [M0 L0 T0] As nr = R r2 / 3 Vr nm = nr = np \ Qr = V rLr = L3/2 r Lr 3/2 qp = qm/L3/2 r = 0.20 ¥ (20) qr = Vp = 17.89 m3/s/m (iii) pressure ratio pr = (Lrrr) Assume rm = rp, i.e. rr = 1.0 Hence pr = Lr \ pp = pm/Lr = 5 ¥ 20 = 100 cm of mercury (iv) Power ratio = (Energy loss/second)r = [Lr7/2 rr] As rr = 1.0 (assumed), Pr = Lr7/2 Pm = Pp ◊ Lr7/2 L2r / 3 Lr np = nm/L1/6 r = Ê 1ˆ = 1000 ¥ Á ˜ Ë 20 ¯ = L1/6 r 0.013 (1/100)1/ 6 = 0.028 7/ 2 = 0.028 W Reynolds Model Law 6.20 Oil of density 917 kg/m3 and dynamic 6.19 Estimate for a 1/20 model of a spillway (i) the prototype velocity corresponding to a model velocity of 1.5 m/s (ii) the prototype discharge per unit width corresponding to a model discharge per unit width of 0.2 m3/s per metre (iii) the pressure head in the prototype corresponding to a model pressure head of 5 cm of mercury at a point (iv) The energy dissipated per second in the model corresponding to a prototype value of 1 kW. Solution: For dynamic similarity Froude number must be the same in the model and prototype. If Lr is the length ratio, then V (i) Vr = m = Lr Vp Vp = Vm / Lr = 1.5 20 = 6.71 m/s (ii) ratio of discharge per unit width = qr = (Q / L) m (Q / L) p diameter 15 cm at a velocity of 2.0 m/s. What dynamically similar? The density and viscosity of water can be taken as 998 kg/m3 and 1.31 ¥ 10–3 Pa.s respectively. Solution: Reynolds similarity law is applicable. Vp dp V d (Re)m = m m = (Re)p = vm vp Vm vr mr \ Vr = = = Vp Lr Lr rr Vm m 1 = m Vp m p Ê Lm ˆ Ê rm ˆ ÁL ˜Ár ˜ Ë p¯Ë p¯ Referring to oil with a subscript p and water with a sufﬁx m Vm 1.31 ¥ 10 -3 1 = ¥ Vp 0.29 Ê 1.0 ˆ Ê 998 ˆ ÁË 15.0 ˜¯ ÁË 917 ˜¯ = 0.0623 216 Fluid Mechanics and Hydraulic Machines Vm = Velocity of water ﬂow = Vp ¥ 0.0623 = 2 ¥ 0.0623 = 0.1246 m/s Solution: The Reynolds model law is applicable. (Re)m = 6.21 A 1 : 6 scale model of a passenger car is Lm/Lp = Lr = 1/6 mr Vr = Lr rr tested in a wind tunnel. The prototype velocity is 60 km/h. If the model drag is 250 N what is the drag and the power required to overcome the drag in the prototype? The air in the model and prototype can be assumed to have the same properties. Solution: \ Reynolds similarity law is applicable. Vp Lp V L (Re)m = m m = (Re)p = vm vp vr Vr = Lr Pressure ratio Force ratio mr = If rr = 1, and mr = 1, \ \ \ Fm = 1.0 Fp m r Lr rr 1 ¥ 10 -3 = 9.615 ¥ 10–3 0.104 998 = 1.109 900 Lr = 1/6 (9.615 ¥ 10 -3 ) 2 pr = = 3 ¥ 10–3 2 Ê 1ˆ ÁË 6 ˜¯ ¥ 1.109 p 450 pp = m = = 150,000 Pa pr 3 ¥ 10 -3 = 150 kPa Fp = 250 N (same as in the model) Power to overcome drag in the prototype: Qr = Ê 60 ¥ 103 ˆ Pp = Fp ◊ Vp = 250 ¥ Á ˜ = 4167 W Ë 3600 ¯ = 4.167 kW L2r rr rr = m r2 Fm = Fp rr m r2 In the present problem: Vm = Vp/Lr = 60 ¥ 6 = 360 km/h = 100 m/s pr = rrVr2 = and discharge ratio Qr = Vr ◊ L2r = vr = 1 (i.e. rm = rp, mm = mp) Vr = 1/Lr, If then rmVm Lm rpVp Lp = mm mp of 0.9 m/s is to be estimated by model studies. A 1 : 6 scale model using water is used. If the pressure drop in the model is 450 Pa, what will be the prototype pressure drop? If the prototype discharge is 200 L/s what is the model discharge? The following data are relevant: Item Density Viscosity Qm = Qp ¥ 1.445 ¥ 10–3 = 200 ¥ 1.445 ¥ 10–3 = 0.289 L/s 6.22 Prototype 900 kg/m3 0.104 Pa.s Model 998 kg/m3 1 ¥ 10 – 3 Pa.s (9.615 ¥ 10 -3 ) ¥ 1/ 6 = 1.445 ¥ 10–3 1.109 6.23 An underwater device is 1.5 m long, and is to move at 3.5 m/s. A geometrically similar model 30 cm long is tested in a variable pressure wind tunnel at a speed of 35 m/s. Calculate the pressure of air in the model. If the model exhibits a drag force 40 N, calculate the prototype drag force. [Assume rwater = 998 kg/m3, rair at standard atmospheric pressure = 1.17 kg/m3, mair = 1.90 ¥ 10 – 5 Pa.s, at local atmospheric pressure and mwater = 1.0 ¥ 10 – 3 Pa.s]. 217 Dimensional Analysis and Similitude Solution: Reynolds model law is applicable. Hence r V L (Re)m = m m m = (Re)p = mm \ If Lm r = Lr, m = rr and Lp rp rpVp Lp mp mm = mr mp Vr = 35 = 10 3.5 mr = Hence Hence rr = 1.90 ¥ 105 1 ¥ 10 -3 transport an oil of relative density 0.9 and kinematic viscosity = 3 ¥ 10 –2 stoke at a rate of 3.0 m3/s. If a 15 cm diameter pipe with water at 20°C (v = 0.01 stoke) is used to model Solution: The Reynolds number must be the same in the model and prototype for similar pipe ﬂows. In the present case 0.30 1 = 1.50 5 6.24 A pipe of diameter 1.5 m is required to the model. Vm mr = Vr = Vp rr Lr Lr = Vp Dp Vm Dm vm vp D v Vm = Vp p m Dm vp 3.0 = 1.6977 m/s p pD / 4 ¥ (1.5) 2 4 1.5 0.01 Vm = 1.6977 ¥ ¥ = 5.659 m/s 0.15 0.03 p Qm = discharge in the model = D m2 ¥ Vm 4 p ¥ (0.15)2 ¥ (5.659) = 0.1 m3/s = 4 Vp = = 1.9 ¥ 10–2 mr 1.9 ¥ 10 -2 = = 0.0095 Vr Lr 10 ¥ 1/ 5 rair = rm = 998 ¥ 0.0095 = 9.481 This is about 8 times larger than the density at atmospheric pressure. Since at constant temperature, by the equation of state p/r = constant. Hence (pressure) model rmodel = (Atmospheric pressure) ratmospheric pmodel 9.481 ¥ pa = 8.103 pa = 1.17 = 8.103 times local atmospheric pressure Force ratio = Fr = rrV r2 L2r = m r2 (1.9 ¥ 10 -2 ) 2 = rr 0.0095 = 0.038 Fm 40 Fp = = = 1053 N 0.038 0.038 = 1.053 kN = Q 2 = 6.25 A 1/10 model of an airplane is tested in a variable density wind tunnel. The atmospheric conditions The pressure used in the wind tunnel is 10 times the atmospheric pressure. Calculate the velocity of air in the model. To what prototype value would a measured drag of 500 N in the model correspond? If some vortices are shed at a frequency of 25 Hz in the model, what would be the corresponding prototype frequency? Solution: The Reynolds number in the model and the prototype must be the same. (Re)m = Here rV L rmVm Lm = (Re)p = p p p mm mr Lm = Lr = 1/10. Lp 218 Fluid Mechanics and Hydraulic Machines Since pressure does not affect the viscosity appreciably mm = mp and hence mr = 1.0. Further, at constant temperature p/r = constant. p r Pressure ratio m = 10 = m = rr pa rp \ Vm = Vp ¥ mr = rr Lr 400 1 10 ¥ 10 = 400 km/h Model velocity is the same as prototype velocity, i.e. Vr = 1.0 (Lr)3/2 = \ Hence 1 (10) 2 ¥ 10 ¥ (1) = 1 10 Fm = 500 ¥ 10 = 5000 N Fr T 1 V Frequency ratio fr = p = = r Tr Lr Tm Fp = f 1 = 10 fr = m = fp 1/10 Hence, f 25 fp = m = = 2.5 Hz 10 10 6.26 Obtain an expression for the scale of a In this case, g Lm = Vp g Lp Vr = L1/2 r Also by Reynolds model law, rpVp Lp rm Vm Lm = mm mp \ Lr = = (nr)2/3 È mr L1r/ 2 ˘ Í ˙ =1 Î sr ˚ Solution: With the sufﬁx r denoting the ratio of model to prototype, the following values of dimensionless numbers are to be satisﬁed to meet dynamic similarity requirements of different forces: For surface tension force: Weber number ratio = Wr = rr Vr2 Lr = 1 …(i) sr For Viscous force: Reynolds number ratio rr Vr Lr =1 mr For gravity force: Froude number ratio = = (Re)r = (Fr)r = Vr Lr =1 rr Lr Vr2 = 1 or sr s rr Lr = 2r Vr m From (ii) rr Lr = r Vr rr m Hence from (iv) and (v) = r or 2 Vr Vr …(ii) …(iii) From (i) Vm i.e. 2/3 and surface tension are equally important in a model, show that for dvnamic similarity the relationship between viscosity ratio mr, surface tension ratio sr and model scale ration Lr is given by model, which has to satisfy both Froude’s model law and Reynolds model law. Solution: Êm ˆ scale = Lr = Á r ˜ Ë rr ¯ 6.27 If acceleration due to gravity, viscosity Force ratio = Fr = Lr2 rrV2r = mr = nr rr mr mr = 1/ 2 Vr rr ( Lr ) ◊ rr m r Vr =1 sr …(iv) …(v) …(vi) 219 Dimensional Analysis and Similitude By Froude law relation (eq. iii) Vr = L1/2 r Substituting in Eq. (vi) For the prototype, by Froude’s law, the wave resistance (Fw)p is given by m r L1r/ 2 =1 sr ( Fw ) p ( Fw ) m Drag Components in Ship Model Froude number = 6.28 A 1 : 25 scale model of a ship has a = (Fw)r = r rVr2 Lr2 Vm g Lm 2 submerged surface area of 6 m , a length of 5 m and experiences a total drag of 25 N when towed through water with a velocity of 1.2 m/s. Estimate the total drag on the prototype when cruising at the corresponding speed. The skin friction force can be estimated by Fs = Cf ArV2 Cf = 0.0735/(Re)1/5. Assume mwater = 1 ¥ 10 –3 Pa.s and rwater = 1030 kg/ m3 for both model and the prototype. Solution: The total drag is the sum of the wave resistance and the skin friction. The Froude criterion of similarity is used for the gravity-affected component of the drag, namely the wave resistance. The skin friction is estimated by the formula separately. For the model: rVL Reynolds number Re = m = 1030 ¥ 1.2 ¥ 5 1 ¥ 10 -3 = 6.18 ¥ 106 0.0735 Cf = (6.18 ¥ 106 )1/ 5 = 3.222 ¥ 10–3 Skin friction resistance Fs = Cf ArV2/2 = (3.222 ¥ 10–3) ¥ 6 ¥ 1030 ¥ (1.2)2/2 = 14.34 N Total model drag = 25.00 N Hence, the model wave resistance = 25.00 – 14.34 (Fw)m = 10.66 N The corresponding wave resistance in the prototype is calculated by Froude’s law of similarity. = (Fr)m = Vp g Lp = (Fr)p Vr = Lr and (Fw)r = rrLr3 In the present case, rr = 1.0 (Fw)r = Lr3 (F ) 10.66 = 166.6 ¥ 103 N \ (Fw)p = w3 m = Lr (1/ 25)3 = 166.6 kN Skin friction for the prototype: Prototype Reynolds number rp Vp Lp = = (Re)p mp L Since Vp = Vm/ Lr and Lp = m Lr Also rp = rm and mp = mm (Re)p = (Re)m 1 L3r / 2 = (25)3/2 ¥ 6.18 ¥ 106 = 7.73 ¥ 108 0.0735 Cf = (7.73 ¥ 108 )1/ 5 = 1.2266 ¥ 10–3 Skin friction resistance (Fs)p = (Cf ArV2/2)p = (1.2266 ¥ 10–3) ¥ {6 ¥ (252)} ¥ 1030 ¥ [(1.2) ¥ 25 ]2/2 = 85279 N = 85.3 kN Total prototype resistance = (Fw)p + (Fs)p = 166.6 + 85.3 = 251.9 kN 220 Fluid Mechanics and Hydraulic Machines Distorted Models Time ratio 6.29 A proposed model of a river stretch of 15 km is to have a horizontal scale of 1/200 and vertical scale of 1/40. If the normal discharge, width and depth of the river are 152 m3/s, 90 m and 2 m respectively, estimate the corresponding model quantities. What value of Manning’s roughness n is to be provided in the model to represent a prototype roughness value of 0.025? Solution: Horizontal scale = Lr = 1/200 Vertical scale = hr = 1/40 1. Discharge Qm = Qp (Lr) (hr)1.5 = 152 ¥ (1/200) (1/40)1.5 = 0.03 m3/s 2. Depth ym = yp (hr) = 2.0/40 = 0.05 m 3. Width Bm = Bp (Lr) = 90/200 = 0.045 m 4. Manning’s nm = np(hr)2/3/(Lr)1/2 = (0.025) (1/40)2/3/(1/200)1/2 = 0.03 [Note that the model has to be rougher than the prototype] 6.30 In a tidal model, the horizontal scale ratio is 1/500. The vertical scale is 1/50. What model period would correspond to a prototype period of 12 hours? Solution: Tr = Lr / hr = 1/ 500 1/ 50 = 0.01414 Model period Tm = TpTr = (12 ¥ 60 ¥ 60) ¥ 0.01414 = 610 s = 10 minutes 10 seconds 6.31 For a river model of horizontal scale 1/250 and vertical scale 1/25, estimate the model value corresponding to a prototype slope of 0.0002. If the model velocity and discharges are 0.50 m/s and 0.02 m3/s respectively, estimate the corresponding prototype velocity and discharge values. Solution: Lr = 1/250 and hr = 1/25 (1) Slope ratio Sr = Sm/Sp = hr/Lr = (1/25)/(1/250) = 10 Sm = Sp Sr = 0.0002 ¥ 10 = 0.002 (2) Velocity ratio = Vr = (Vm/Vp) = hr = (1/25)0.5 = 1/5 Vp = Vm/Vr = 0.50/(1/5) = 2.5 m/s (3) Discharge ratio Qr = Qm/Qp = Lr hr1.5 = (1/250) (1/25)1.5 = 1/31250 Qp = Qm/Qp = 0.02/(1/31250) = 625 m3/s [Note that the model has to be steeper than the prototype.] Lr = 1/500 and hr = 1/50 Problems Dimensional Analysis 6.1 The Chezy formula for velocity V in an open channel is given by V= C RS0 where R = hydraulic radius, S0 = longitudinal slope of the channel and C = Chezy coefﬁcient. Find the dimensions of C. 221 Dimensional Analysis and Similitude 6.2 The capillary rise h of a ﬂuid of density r and surface tension s in a tube of diameter D depends upon the contact angle q and gravity g. Obtain an expression for h by Raleigh’s method. Ê Ê wD m P K ˆˆ = fn , , Á Ans. Á ˜˜ rV 3 D 2 Ë V rVD rV 2 ¯ ¯ Ë 6.7 The head loss hL due to ﬂuid friction in a pipe is known to depend on the diameter D, length L and roughness magnitude e of the pipe; the velocity of ﬂow V, the gravity g; and ﬂuid density r and viscosity m. Derive an expression for hL in dimensionless form. Ê Ê s ˆˆ h , q˜˜ Á Ans. = fn Á 2 D Ë r gD ¯¯ Ë 6.3 The critical depth yc in a trapezoidal channel depends upon the discharge Q, the side slope of the channel m, the bottom width B and the gravity g. Obtain an expression for yc by Raleigh’s method. Ê Ê L e gD m ˆ ˆ hL Á Ans. D = fn Á D , D , 2 , rVD ˜ ˜ Ë ¯¯ V Ë 6.8 In laminar ﬂow through a tube the discharge Q is a function of diameter D, the ﬂuid viscosity m and the pressure gradient dp/dx. Obtain an expression for Q in a dimensionless form. Ê Ê yc Q ˆˆ = fn Á m, 2 Á Ans. ˜˜ ÁË B Ë B gB ¯ ˜¯ Ê ˆ Ê Qm ˆ Á Ans. Á 4 dp ˜ = Constant ˜ Á ˜ ÁË D ˜ Ë ¯ dx ¯ 6.4 The stagnation pressure ps in an air ﬂow depends upon the static pressure p0, the velocity V of the free stream and density r of the air. Derive a dimensionless expression for ps. Ê Ê r ˆˆ ps = fn Á 0 2 ˜ ˜ Á Ans. p0 Ë rV ¯ ¯ Ë 6.5 The discharge Q over a V-shaped notch is known to depend on the angle q of the notch, the head H of the water surface, the velocity of approach V0 and the acceleration due to gravity g. Determine the dimensionless form of the discharge equation. Ê Ê Q V = fn Á q , 0 Á Ans. 2 ÁË gH H gH Ë 6.10 The shear stress t0 at the bed of a rough channel depends upon the depth of ﬂow y, velocity of the ﬂuid V, roughness height e of the bed and ﬂuid density r and viscosity m. Derive an expression for t0 in dimensionless form. ˆˆ ˜ ˜˜ ¯¯ Ê Ê m e ˆˆ t0 = fn Á , Á Ans. ˜ 2 Ë rVy y ˜¯ ¯ rV Ë 6.6 The power P required to drive a propeller is known to depend on the diameter D and angular velocity w of the propeller; the density r, viscosity m and bulk modulus of elasticity K of the ﬂuid; and the free stream velocity V. Derive the functional relationship for P in a dimensionless form. 6.9 The ﬂow velocity u very near a rotating disk depends on the angular velocity w of the disk, the radial distance r, vertical distance z and kinematic viscosity of the ﬂuid v. Derive a relation for u in a dimensionless form. Ê Ê z w r2 ˆ ˆ Ê m ˆ Á Ans. Á ˜ = fn Á , ˜˜ Ë wr¯ Ë r v ¯¯ Ë 6.11 The shear stress t0 on the walls of a triangular channel depends upon the vertex angle q, depth of ﬂow y, longitudinal slope S, density r and acceleration due to 222 Fluid Mechanics and Hydraulic Machines gravity g. Obtain an expression for t0 in dimensionless form. Ê ˆ t0 ÁË Ans. r g y = fn [q , S ]˜¯ Ê Ê gD B d ˆˆ m Ê nD ˆ , , , ˜˜ = fn Á Á Ans. Á ˜ ÁË Ë V ¯ rVD D D ¯ ˜¯ Ë V 6.16 The discharge Q from a centrifugal pump is dependent upon the pump speed N (rpm), diameter of the impeller D, head H, acceleration due to gravity g, density of he ﬂuid r and viscosity m. Derive an expression for Q in dimensionless form. 6.12 The terminal velocity of descent V of a hemispherical parachute is found to depend on its diameter D, weight W, acceleration due to gravity g, density of air ra and viscosity of air m. Obtain an expression for V in dimensionless form. Ê Ê W ˆˆ V m = fn Á , Á Ans. ˜˜ 3 ÁË gD Ë r D g r D gD ¯ ˜¯ Ê Ê N D H m D2 ˆ ˆ Q = fn Á , , Á Ans. 2 ˜˜ ÁË D r Q ¯ ˜¯ g D gD Ë 6.17 Obtain an expression for the thrust (F) developed by a propeller which depends upon the angular velocity w, approach velocity V. dynamic viscosity m, density r, propeller diameter D and compressibility of the medium measured by the local velocity of sound C. Ê 2 2 È V Dw rVD ˘ ˆ Á Ans. F = r D V f Í C , V , m ˙˜ Ë Î ˚¯ 6.13 The lift force F on an airfoil is a function of the angle of attack, a, velocity of ﬂow V, chord length C, span L, density r, viscosity m, and bulk modulus of elasticity E. Obtain the dimensionless form of the functional relationship. Ê Ê rVC V r L ˆ ˆ F = fn Á , , , a˜˜ Á Ans. 2 2 ÁË E C ¯ ˜¯ rV C Ë m 6.14 The variables controlling the motion of a ﬂoating vessel through water are the drag force F, speed V, length L, acceleration due to gravity g, ﬂuid density r and viscosity m. Derive an expression for F by dimensional analysis. Ê Ê gL m ˆ ˆ F = fn Á , Á Ans. ˜˜ 2 2 ÁË rVL ¯ ˜¯ rV L Ë V 6.15 In the study of the vortex shedding phenomenon due to a bluff body in an open channel ﬂow the following parameters are found to be important: velocity of ﬂow = V, depth of ﬂow = D, density of ﬂuid = r, acceleration due to gravity = g, dynamic viscosity = m, width of body = B, thickness of body = d and frequency of vortex shedding = n. Obtain the dimensionless parameters governing the phenomenon. Similitude 6.18 A spillway model is constructed on a scale of 1 : 25. Calculate: (a) the prototype discharge corresponding to a model discharge of 0.12 m3/s; (b) the model velocity corresponding to a prototype velocity of 3.5 m/s; (c) the discharge per metre width in the prototype when the model discharge is 0.15 m3/s and the length of the spillway model is 40 cm. (Ans. (a) Qp = 375 m3/s, (b) Vm = 0.7 m/s, (c) qp = 46.88 m3/s per metre length) 6.19 A 1 : 36 model of a spillway crest records an acceleration of 1.5 m/s2, a velocity of 0.5 m/s and a force of 0.30 N at a certain area of the model. What would be the values of the corresponding parameters in the prototype? (Ans. ap = 1.5 m/s2, Vp = 3.0 m/s, Fp = 14.0 kN) 223 Dimensional Analysis and Similitude 6.20 A concrete open channel has Manning’s n = 0.014. A 1/64 model of this channel is needed. Find the value of n for the model. Comment on the result. (Ans. nm = 0.007. It is not possible to get such a low value of n. It is better to go for a bigger model.) 6.21 A geometrically similar model of a spillway built to 1/50 scale is tested. The discharge and velocity of ﬂow over the model were measured as 2.5 m3/s and 1.5 m/s respectively. Find the corresponding discharge and velocity of ﬂow in the prototype. (Ans. Qp = 44194 m3/s, Vp = 10.61 m/s) 6.22 Model tests are to be conducted on a seawall constructed on a scale of 1 : 25. If the wave period in the prototype is 12 seconds what should be the corresponding wave period in the model? To what prototype force per metre length of wall would a model value of 200 N/m correspond? Assume rmodel = rprototype. (Ans. Tm = 2.4 s, Fp = 125 kN/m) 6.23 A gravity fed lock in a navigational channel is to be studied with a 1/75 scale model. (a) If the model lock ﬁlls up in 1.15 minutes estimate the corresponding time for the prototype. (b) If the pressure at a point in the model is 0.5 kPa, what is the corresponding pressure in the prototype? (Ans. Tp = 9.96 min, pp = 37.5 kPa) 6.24 An offshore platform is known to encounter waves 4.5 m high at a frequency of 0.15 Hz and a steady current of 1.5 m/s. If a 1/30 model of the platform is to be built, determine the values of the above parameters in that model. (Ans. hm = 0.15 m, fm = 0.82 Hz, Vm = 0.274 m/s) 6.25 The resistance offered to the movement of a 2.0 m long ship model in a towing tank full of fresh water while moving with a speed of 1.5 m/s was 450 N. (i) If the prototype is 60 m in length what will be the corresponding speed? (ii) What would be the force required to drive at a corresponding speed, a prototype of 80 m length in sea water of relative density 1.025? (Ans. Vp = 8.22 m/s, Fp = 29520 kN) 6.26 The drag of a submerged mine due to ocean currents is studied in a wind tunnel on a scale of 1 : 5. What wind velocity is needed to simulate a 2.0 m/s current? What prototype value would correspond to a model resistance of 10 N? [rair = 1.23 kg/m3, rwater = 998 kg/m3, mair = 1.71 ¥ 10–5 Pa.s, mwater = 1.0 ¥ 10–3 Pa.s]. (Ans. Vm = 139 m/s, Fp = 42.1 N) 6.27 Characteristics of a small underwater craft are studied under dynamic similarity conditions in a variable density wind tunnel on a model scale of 1 : 12. What prototype speed and power are indicated by model values of 100 m/s of velocity and 30 N of drag force? The model is operated at 8 atm pressure. [Take rair at atmospheric pressure = 0.95 kg/m3, mair = 2.17 ¥ 10–5 Pa.s, rwater = 998 kg/m3 and mwater = 1 ¥ 10–3 Pa.s]. (Ans. Vp = 2.93 m/s, (Power)p = 1422 W) [Hint: rm = 8 ¥ ratmos = 8 ¥ 0.95 kg/m3] 6.28 If a 1.0 m diameter pipe carrying air at a velocity of 3.8 m/s is to be modelled for dynamic similarity by a water pipe of diameter 10 cm, what would be the velocity of water? (Ans. Vm = 1.335 m/s) 6.29 A component of a airplane of length 3.0 m is tested in a variable density wind tunnel. The model has a length of 60 cm. If the 224 Fluid Mechanics and Hydraulic Machines model has the same speed as the prototype, what pressure in the wind tunnel, relative to local atmospheric pressure (pa), is needed? 6.35 The pressure drop in an air duct depends on the length and diameter of the duct, the mass density, viscosity of the ﬂuid and the velocity of the ﬂow. Obtain an expression for the pressure drop in dimensionless form. Estimate the pressure drop in a 20 m long air duct if a model of the duct operating with water produces a pressure drop of 10 kN/m2 over 10 m length. The scale ratio is 1 : 50. rwater = 1000 kg/m3 rair = 1.2 kg/m3 mwater = 0.001 N.s/m2 mair = 0.0002 N.s/m2 Ê ˆ pm ÁË Ans. p = 5˜¯ a 6.30 A ﬂowmeter to be installed in a 0.80 m pipe is to be tested in the laboratory by using a 1/4 model. If the same ﬂuid as in the prototype is used in the dynamically similar model what would be the prototype pressure drop corresponding to a model value of 6 kPa? If the model discharge is 0.2 m3/s, what would be the prototype discharge? (Ans. D pp = 375 Pa, Qp = 0.8 m3/s) 6.31 A venturimeter ﬁxed in a 60 cm pipe is being modelled to scale of 1 : 5 in a model using air as the working ﬂuid. If rair = 1.1 kg/m3, rwater = 998 kg/m3, mair = 1.95 ¥ 10–5 Pa.s, mwater = 1 ¥ 10–3 Pa.s, estimate the prototype discharge corresponding to a model discharge of 5 m3/s. (Ans. Qp = 1.413 m3/s) 6.32 In a ﬂow condition where both viscous and gravity forces dominate, both the Froude number and the Reynolds number are kept the same in the model and prototype. If the ratio of the kinematic viscosity of model to that of the prototype are is 0.0894, determine the model scale. (Ans. Lm/Lp = 1/5) 6.33 Obtain expressions for the velocity ratio and force ratio similitude for a model which obeys Mach’s law of similarity. (Ans. Vr = Cr = K r / rr , Fr = L2rKr) 6.34 If a model is so constructed as to have the same Weber number in the model and prototype, obtain expressions for the velocity ratio and force ratio similitude. Ê ˆ Ê sr ˆ Á Ans. Vr = Á L r ˜ , Fr = s r Lr ˜ Ë r r¯ Ë ¯ Ê ˆ Ê m Dp Lˆ = fn Á , ˜ ; D pp = 133.3 N/ m 2 ˜ Á Ans. 2 Ë rVD D ¯ rV Ë ¯ 6.36 A 1/20 model of a ship having a submerged surface area of 5 m2 and length of 8 m has a total drag of 20 N when towed through sea water at a velocity’ of 1.5 m/s. Calculate the total drag on the prototype when moving at the corresponding speed. The skin friction can be estimated by Fs = Cf ArV2/2 where 0.0735 . [Take m = 1.07 the coefﬁcient Cf = ( Re)1/ 5 ¥ 10–3 Pa.s and r = 1025 kg/m3]. (Ans. Total drag = 82.17 kN) 6.37 An ocean-going vessel has a length of 80 m and a submerged surface area of 1200 m2. Its cruising speed is 30 km/h. A 1/20 model of this vessel is tested in a towing tank using fresh water and Froude law of similarity. A total drag force of 18.5 N was measured in the model at a velocity corresponding to the cruising speed. Calculate the total drag on the prototype at the cruising speed. The skin friction is estimated by Fs = CfArV2/2 where Cf = 0.074 ( Re)1/ 5 . 225 Dimensional Analysis and Similitude Use for fresh water: r = 998 kg/m3 m = 0.001 Pa.s and for sea water r = 1025 kg/m3 m = 1.07 ¥ 10–3 Pa.s (Ans. Total drag = 73.36 kN) 6.38 A model propeller with diameter d = 0.5 m is tested in water tunnel at a speed of n = 360 rpm when the ﬂow velocity V = 2.5 m/s. The model produces a thrust of 250 N at a torque of 22 Nm. The prototype has a diameter of 4 m and operates at 100 rpm in water. If the signiﬁcant non-dimensional group is (V/nd) only, determine (a) the ﬂow velocity, (b) thrust produced and (c) the applied torque for the prototype. (Ans. (a) Vp = 0.556 m/s, (b) Fp = 460.8 kN, (c) Tp = 324.4 kN.m) Distorted Models 6.39 A distorted model of a river has a horizontal scale of 1/750. The slope in the model is 0.0025. If the prototype slope is 0.0001, estimate the prototype discharge corresponding to a model discharge of 0.01 m3/s. (Ans. Qp = 1232 m3/s) 6.40 A model of a river has a horizontal scale of 1/500 and vertical scale of 1/50. Fill up the following table pertaining to the above model—prototype pair. SI. No. 1 2 3 4 5 6 7 8 Parameter Prototype Value Model Value Cross-section area Longitudinal slope Manning’s roughness coefﬁcient Wave period Volume of water Width Velocity Depth 100 m2 0.0001 ? ? ? 0.05 1 hour ? ? 1.5 m/s 2.5 m ? 0.5 m3 0. 2 m ? ? (Ans. (1) 0.04 m2, (2) 0.001, (3) 0.030, (4) 50.9 s, (5) 6.25 Mm3, (6) 100 m, (7) 0.353 m/s, (8) 0.05 m) 6.41 A river carries a discharge of 16,000 m3/s of water at a depth of 8.0 and slope of 0.0025 when its width is 400 m. Above 15 km reach of this river is to be reproduced in the laboratory where 30 m long space is available. Determine the appropriate horizontal and vertical scales for this model. Also, determine the roughness scale and the model slope. (Ans. Lr = 1/500, hr = 1/30, nr = 2.316 and Sm = 0.04167) Objective Questions Dimensional Analysis 6.1 Which of the following is a dimensionless number: (a) Manning’s coefﬁcient n (b) Pipe friction factor f (c) Chezy coefﬁcient C (d) Hazen-William coefﬁcient CH 6.2 The dimensions of volume modulus of elasticity K are (a) FL–1 T–2 (b) FL–2 T –2 (c) FL (d) FL–4 T–2 6.3 The dimensions of speciﬁc heat (cv or cp) are (a) L2 q–2 (b) L2 T–2 q–1 (c) FT–1 q–1 (d) L2q–1 226 Fluid Mechanics and Hydraulic Machines 6.4 The Euler number En is written as En = (a) V/ K / r 2 (b) rV L/s (c) Vr/ Dp (d) V/ Dp / p 6.5 Assuming the thrust T of a propeller depends upon the diameter D, speed of advance V, angular velocity w, dynamic viscosity m, and density r, which of the following dimensionless parameters can be derived by dimensional analysis? T VDm (a) (b) 2 2 r rD V Dw VDr (c) (d) V m Select the correct answer using the codes given below: (a) 1, 2 and 3 (b) 2, 3 and 4 (c) 1, 3 and 4 (d) 1, 2 and 4 6.6 A dimensionless combination of pressure drop Dp, dynamic viscosity m, velocity V and length L is Dp Dpm L (b) (a) mVL V DpL Dpm (c) (d) mV V 2L Similitude 6.9 The variables controlling the motion of a ﬂoating vessel through water are the drag force (F), the speed (V), length (L), density r and dynamic viscosity m of water and gravitational constant (g). If the nondimensional groups are Reynolds number (Re), Weber number (We), Prndtl number (Pr), and Froude number (Fr), the expression for drag force F is given by F (a) = f n(Re) rV 2 L2 F (b) = f n(Re, Pr) rV 2 L2 F (c) = f n (Fr, We) rV 2 L2 F (d) = fn(Fr, Re) rV 2 L2 6.10 The time scale ratio for a model based on Froude law criterion in terms of length scale ratio Lr is (a) Lr (b) Lr (c) 1/ Lr (d) L1.5 r 6.11 If Froude law of similitude exists between a model and a prototype, then the force ratio Fr = 6.7 Dynamic similarity exists when the model and prototype have the same (a) length scale ratio and time scale ratio (b) length scale ratio and velocity scale ratio (c) length scale ratio, time scale ratio and velocity scale ratio (d) length scale ratio, velocity scale ratio and force scale ratio 6.8 Both Reynolds number and Froude number assume signiﬁcance in one of the following example: (a) Motion of submarine at large depths (b) Motion of ship in deep seas (c) Cruising of a missile in air (d) Flow over a spillway (a) L3r Lr3rr (b) Lrrr (c) (d) L3r rr–1 6.12 In a model experiment with a weir, if the dimensions of the model weir are reduced by a factor k, the ﬂow rate through the model weir is the following fraction of the ﬂow rate through the prototype: (a) k5/2 (b) k2 (c) 1 (d) k–2 6.13 In the Froude law of similitude the acceleration ratio ar = (a) L2r (b) 1.0 (c) 1/Lr (d) Lr–3/2 6.14 If Reynolds law of similitude exists between a model and a prototype, then the force ratio Fr = 227 Dimensional Analysis and Similitude (a) Lr3rr Lr mr rr–1 6.15 6.16 6.17 6.18 6.19 6.20 (b) mr2rr–1 Lr–2mr rr–1 (c) (d) In Reynolds law of similitude the discharge ratio Qr = (a) Lrmr/rr (b) Lr3rr 5/2 (c) Lr (d) Lrrr/mr A hydraulic model of a spillway is constructed with a scale 1 : 16. If the prototype discharge is 2048 m3/s, then the corresponding discharge in m3/s for which the model should be tested is (a) 1 (b) 2 (c) 4 (d) 8 In the model of a highway bridge constructed to a scale of 1:25, the force of water on the pier was measured as 5 N. The force on the prototype pier will, approximately, be (a) 15.6 kN (b) 25.3 kN (c) 78.1 kN (d) 90.5 kN A ship whose full length is 100 m is to travel at 10 m/s. For dynamic similarity, with what speed should a 1:25 model of the ship be towed? (a) 2 m/s (b) 10 m/s (c) 4 m/s (d) 0.4 m/s A model test is to be conducted in a water tunnel using a 1:20 model of a submarine which is used to travel at a speed of 12 km/h deep under the sea. The water temperature in the tunnel is so maintained, that its kinematic viscosity is half that of the sea water. At what speed is the model test to be conducted? (a) 12 km/h (b) 240 km/h (c) 24 km/h (d) 120 km/h The fall velocity of a sand grain in water is to be modeled by using particles of the same relative density as sand and a liquid whose kinematic viscosity is 100 times larger than that of water. The diameters of the particles in the model that will have the same fall velocity as the prototype will be (a) 10 times smaller 6.21 6.22 6.23 6.24 (b) 10 times larger (b) 100 times smaller (d) 100 times larger In a model built on Froude law of similarity a phenomenon lasts for 20 min. If the model scale is 1/25, the duration of the phenomenon in the prototype, in minutes, is (a) 50 (b) 100 (c) 2500 (d) 4 If the same ﬂuid is used both in the model and prototype, and if it is desired to have equal Reynolds number and Froude number in the model and prototype, the scale of the model is (a) Vr (b) Vr1/2 (c) 1.0 (d) Vr–1/2 where Vr = velocity ratio. A harbor model has a horizontal scale of 1/150 and a vertical scale of 1/75. The interval between successive daily high tides in the model will be nearly (a) 90 min (b) 40 min (c) 15 min (d) 5 hours In the distorted model of a river, the horizontal and vertical scales are Lr and hr respectively. The discharge ratio will be 2 (a) L1/2 r hr (b) Lrhr3/2 (c) Lr2h1/2 (d) Lr3 hr1/2 r 6.25 A river model is constructed to a horizontal scale of 1 : 1000 and a vertical scale of 1 : 100. If the model discharge were 0.1 m3/s, then the discharge in the prototype would be (a) 103 m3/s (b) 104 m3/s (c) 105 m3/s (d) 102 m3/s 6.26 A model of a weir made to a horizontal scale of 1/40 and vertical scale of 1/9 passes a discharge of 1 Litre/s. The corresponding discharge in the prototype would be (a) 10.8 Lps (b) 108 Lps (c) 1080 Lps (d) 10800 Lps Laminar Flow Concept Review 7 Introduction Critical Reynolds Number Critical Reynolds Number Re 7.1 BASIC EQUATIONS The basic equations which govern the motion of incompressible viscous ﬂuid in laminar motion are called as Navier–Stokes equations. In Cartesian coordinates, for two-dimensional ﬂow, these are: Ê ∂2 u ∂2 u ˆ Ê ∂u ∂u ∂u ˆ ∂p rÁ +u +v ˜ = X + mÁ 2 + 2˜ ∂x ∂y ¯ ∂x Ë ∂t ∂y ¯ Ë ∂x (7.1) ÊVDˆ n ˜¯ crit = Á Ë Ê ∂2 v ∂2 v ˆ Ê ∂v ∂v ∂v ˆ ∂p rÁ +u + v ˜ =Y +mÁ 2 + 2˜ ∂x ∂y¯ ∂y Ë ∂t ∂y ¯ Ë ∂x (7.2) The continuity equation is ∂u ∂ v (7.3) + =0 ∂x ∂y These equations can be solved exactly for only a few simple ﬂow situations. 229 Laminar Flow An important result that can be obtained from the above for the two-dimensional, steady, uniform ﬂows in the X-direction is ∂p ∂t = ∂x ∂y Velocity t0 = or R 8 mV D (7.8) r R (7.9) For a horizontal pipe, for two sections 1 and 2 distance L apart, For inclined pipes, replace Ê dpˆ Ê d ˆ ÁË - d x ˜¯ by ÁË - ds ( p + g Z )˜¯ t d hˆ Ê i.e. by Á - g where h = p/g + Z = piezometric Ë d s ˜¯ head. t0 V = um/2 Laminar Flow in a Circular Conduit Velocity distribution: Ê dpˆ ÁË - d x ˜¯ Dp Ê d p ˆ Ê p1 - p2 ˆ =ÁË - d x ˜¯ = ÁË L ˜¯ L Shear stress R r R 2 Ê dpˆ Pressure gradient Á - ˜ : Ë dx ¯ t0 um Fig. 7.1 (7.7) Variation of the shear stress: t = t 0 Consider a horizontal circular pipe carrying an incompressible ﬂuid in laminar motion, as illustrated in Fig. 7.1. The following relationships for the velocity distribution, shear stress and its distribution and for the head loss have been established analytically. u um Ê 1 d p ˆ 2 = R 2 ÁË 8m d x ˜¯ Shear stress at the boundary: t0 = 7.1.1 Flow in Circular Conduits r V= (7.4) which states that in steady uniform ﬂow the pressure gradient depends upon the existence of viscous shear stress and its variation across the ﬂow. D Mean velocity: Ê 1 d pˆ (R 2 - r 2 ) u = ÁË 4 m d x ˜¯ Ê p1 ˆ Êp ˆ + Z1 ˜ - Á 2 + Z 2 ˜ Á Ê d hˆ h - h2 Ëg ¯ Ë g ¯ = Here Á - ˜ = 1 Ë ds ¯ L L Dh = L (7.5) Head Loss, hf Maximum velocity: Hence Ê 1 d pˆ 2 um = Á R Ë 4 m d x ˜¯ È Ê r ˆ2˘ u = u m Í1 - Á ˜ ˙ ÍÎ Ë R ¯ ˙˚ Designating hf = – Dh = head loss in a length L (7.6) dh ds = hf L Note that for a uniform ﬂow the velocity is same all along the length and hence the energy loss = head loss = drop in piezometric head. 230 Fluid Mechanics and Hydraulic Machines In general, the variation of the head loss h f due to uniform laminar ﬂow in a length L of a pipe of diameter D is given by, hf = Re = (7.10) g D2 7.1.2 For uniform laminar ﬂow between two stationary parallel plates separated by a distance B, (Fig. 7.2) an exact solution of the Navier–Stokes equations yields: 2v 3 m t0 y t B/2 CL Power, P Power required to overcome a head H is P = g QH (7.12) 128 m Q 2L pD 4 (7.13) Friction Factor, f It is usual to designate the frictional resistance to ﬂow in a pipe by Darcy–Weisbach equation as 2 f LV 2gD where f = friction factor. For laminar ﬂow 32 mV L f LV 2 = hf = 2g D g D2 32 mV L 2 g D ◊ Hence f= g D 2 LV 2 64m 64 = = rVD Re vm v B Shear stress y Hence, in laminar ﬂow the power required to overcome frictional resistance in a pipe of length L and diameter D, carrying a discharge Q of a ﬂuid of speciﬁc weight g and viscosity m is P = g Qhf = VD = Reynolds number n (7.11) g pD 4 64 Re Flow Between Two Stationary Parallel Plates V= 128m QL hf = f = where 32 mVL This equation is known as Hagen–Poiseuille equation. Since the mean velocity Q V= where Q = discharge p 2 D 4 hf = or (7.14) Velocity Fig. 7.2 t0 Laminar Flow Between Stationary Parallel Plates Velocity distribution Ê 1 dp ˆ n = Á(By – y2) Ë 2m d x ˜¯ È Ê y ˆ Ê y ˆ2˘ ˙ = n m Í2 Á Í Ë B/ 2 ˜¯ ÁË B/ 2 ˜¯ ˙ Î ˚ (7.16) (7.16(a)) Maximum velocity Ê 1 dp ˆ 2 B nm = Á Ë 8m d x ˜¯ (7.17) Average velocity V = 2 Ê d p ˆ B2 vm = Á - ˜ Ë d x ¯ 12m 3 (7.18) Shear stress at the boundary 6mV Ê dp ˆ B t0 = Á - ˜ = Ë dx¯ 2 B (7.15) (7.19) 231 Laminar Flow Variation of the shear stress hf 3mV = S0 = sin q = L g d2 Ê dp ˆ Ê B ˆ t = Á - ˜ Á - y˜ Ë dx¯ Ë 2 ¯ and where Ê y ˆ = t 0 Á1 for y < B/2 B/ 2 ˜¯ Ë (7.20(a)) Ê y ˆ - 1˜ for y > B/2 t = t0 Á Ë B/ 2 ¯ (7.20(b)) The head loss hf in a length L is hf = 12 mVL (7.21) g B2 [Note: As in laminar pipe ﬂow case, for inclined ﬂow between two stationary parallel plates use Ê d ( p + g Z)ˆ ÁË ˜¯ in place of ds formulae.] Ê dp ˆ ÁË - d x ˜¯ in the various 7.1.4 S0 = slope of the inclined plane = slope of the liquid surface = slope of the hydraulic grade line Coutte Flow The ﬂow between two stationary parallel plates of Sec. 7.1.2 is a special case of a general ﬂow situation representing ﬂow under pressure gradient in the gap between two parallel plates, with one of the plates moving relative to the other. This general ﬂow, schematically represented in Fig. 7.4 is called General Coutte ﬂow. In Fig. 7.4, U = velocity of the top plate, u = velocity at a distance y from the bottom ﬁxed plate, B = gap between the two plates. 7.1.3 Viscous Flow with a Free Surface When a viscous uniform ﬂow takes place in laminar regime down an inclined plane with a free surface (Fig. 7.3), the ﬂow is similar to ﬂow between two parallel plates. Here the depth of ﬂow d = B/2 = half the spacing between the plates. On this basis the various parameters of the ﬂow, viz. the velocity distribution and shear stress distribution can be estimated (see Example 7.20). For the head loss equation, s y (7.22) Moving plate U U Y B u y x Fixed plate Fig. 7.4 General Coutte Flow between Two Parallel Plates The solution of two-dimensional Navier – Stokes equation for the boundary conditions represented in Fig. 7.4 yields u d u = Uy By dp Ê yˆ 1- ˜ Á B 2 m dx Ë B¯ (7.23) q Ê dp ˆ In this equation Á - ˜ = pressure gradient in the Ë dx ¯ Fig. 7.3 direction of ﬂow. Using the nondimensionl pressure 232 Fluid Mechanics and Hydraulic Machines U 0.8 –3 –2 Moving plate 0.6 –1 = 1 P 0.4 2 3 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 u/U Fig. 7.5 Velocity Distribution in Coutte Flow. B 2 y Ê dp ˆ Á - ˜ the velocity distribution 2mU Ë dx ¯ of Equation 7.23 can be represented as gradient P = u y yÊ yˆ = - P Á1 - ˜ U B BË B¯ Fig. 7.6 shows the variation of velocity and shear stress across the gap between the plates in a simple Coutte ﬂow. Velocity distribution U (7.24) u Èy ˘ The variation of = fn Í , P ˙ is shown in U ÎB ˚ u y with for various Fig. 7.5 as the variation of U B values of P. For non-horizontal Coutte ﬂow, the pressure p is to be replaced by piezometic head h as Êp ˆ p Æ g Á + Z˜ = g h Ëg ¯ Thus for inclined Coutte ﬂow p d( + Z ) Uy By yˆ g Ê g u= ÁË1 - B ˜¯ B 2m dx du U = and hence the dy B mU du shear stress t = m is constant all = B dy across the gap. The velocity gradient is 0.2 –0.4 –0.2 u y y = or u = U , i.e., the U B B velocity varies linearly from zero at the ﬁxed boundary to U at the moving boundary. In plain Coutte ﬂow, 0 y B 1.0 (7.25) Moving plate t0 t u B Shear stress distribution y t0 Fixed plate Fig. 7.6 7.2 Velocity and Shear stress distribution in CREEPING MOTION Very slow motion of an object in an inﬁnite expanse of a viscous ﬂuid is known as creeping motion. For the case of a sphere of diameter D moving with a velocity V0 in a viscous ﬂuid, the creeping motion occurs at the Reynolds number V0 D (7.26) £ 1.0 v Through an analytical procedure Stokes has shown that the net longitudinal force F exerted upon the sphere is (7.27) F = 3p D mV0 Re = U = 0, It is easy to see that we get the case of ﬂow between two ﬁxed parallel plates (known as 2-D Poisuille ﬂow) discussed in Sec. 7.1.2. Plain Coutte Flow The particular case of Ê dp ˆ Coutte ﬂow with Á - ˜ = 0 is know as Ë dx ¯ Simple or Plain Coutte Flow. This equation, known as Stokes Equation, ﬁnds application in the determination of the fall velocity of small particles [For details see Chapter 9]. 233 Laminar Flow 7.3 LUBRICATION 7.4 VISCOMETERS Whenever there is relative motion of two surfaces in contact there exists friction and consequent loss of energy. In machine elements having moving parts, the friction is considerably reduced through application of lubrication and use of bearings. There are a wide variety of bearings in use and the mechanics of commonly used bearings can be modeled through laminar ﬂow in passages of simple geometries. Examples of common bearings that can be analyzed by simple laminar ﬂow concepts include journal bearing, conical bearing, collar bearing, pedestal bearing and slipper bearings. Few examples to illustrate the analysis procedure are given in the example set that follows. In mechanics of ﬂow related to lubrication, it is always assumed that the ﬂow is laminar. A viscometer is a device for determining the viscosity of a liquid. Many of these instruments use laminar ﬂow situations to estimate the viscosity of the liquid. The capillary tube viscometer utilizes the Hagen-Poiseuille equation to estimate the coefﬁcient of viscosity m of the liquid. 7.5 INTERNAL AND EXTERNAL FLOWS It may be realized that the examples considered in this chapter are ﬂows bounded by walls. Such ﬂows are known as internal ﬂows. If the ﬂows are not bounded by walls such ﬂows are known as external ﬂows. Both laminar ﬂows and turbulent ﬂows exist as internal or external ﬂows. While this chapter dealt with internal ﬂows, the next chapter, viz. Chapter 8, deals with external ﬂows, both viscous and turbulent. Also, Chapter 9 deals with external ﬂows while chapter 10 deals with internal ﬂows. Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples 7.1 2prhxt 2 pr .p pr r R Solution: (i) Consider a cylindrical element of ﬂuid as shown in Fig. 7.7. Since the ﬂow is steady and hx Fig. 7.7 2 p+ dp hx dx 234 Fluid Mechanics and Hydraulic Machines uniform, there is no acceleration and, as such, the sum of horizontal forces on the element must be zero. Hence Ê Ë pr2p – pr 2 Á p + R D p ˆ D x ˜ – 2prDxt = 0 ¯ x t Note that the shear stress t at a radial distance r acts over the surface of the cylindrical element, the surface area of which is (2pr ◊ Dx). dp r Simplifying t= dx 2 dp R At the boundary r = R, t = t0 = dx 2 At the centre line r = 0 tc = 0 The variation of the shear stress t with r is shown in Fig. 7.1. 7.2 0 q Z1 2 L dp Dp Dp dp is constant. Hence = = over (ii) d x D x D L dx a length L. 2t 0 L 4t 0 L The pressure drop (– DP) = = R D 1 Z2 Datum Fig. 7.8 But L sin q = (Z1 – Z2). Hence on simplifying Ê p1 ˆ Ê p2 ˆ 2t 0 L 4t 0 L ÁË g + Z 1˜¯ - ÁË g + Z 2 ˜¯ = h f = g R = g D The shear stress t0 is expressed as t0 = f rV 2 8 where f = Darcy–Weisbach friction factor \ f hf = 2 ¥ f 1 f rV 2 ¥ L ¥ g D /2 8 f LV 2 2g D This is known as Darcy–Weisbach equation and is valid for both laminar and turbulent ﬂows. In laminar ﬂow by Hagen–Poiseuille equation 32 mVL hf = g D2 Thus, f LV 2 32 mVL = 2g D g D2 By solving for f: hf = Solution: Considering two sections 1 and 2, distance L apart, as in Fig. 7.8, V12 V 22 p1 p + Z1 + = 2 + Z2 + + hf g 2g g 2g As the ﬂow is uniform V1= V2 = V Êp ˆ Ê p1 ˆ Ê p2 ˆ ÁË g + Z 1˜¯ - ÁË g + Z 2 ˜¯ = hf = - D ÁË g + Z ˜¯ Considering the force balance between sections 1 and 2 pD2 pD2 - p2 + g ( p R 2 ) L sin q - t 0 2 p R L = 0 p1 4 4 (i) f= 64 64 = rVD /m Re 235 Laminar Flow f rV 2 8 64 1 ◊ ◊ rV 2 = rVD /m 8 8mV t0 = D t0 = (ii) Shear stress 7.3 Solution: Ê g d hˆ For a laminar ﬂow u = Á (R2 – r2) ˜ Ë 4m d s ¯ = K (R2 – r2) At R = 0.06 m and r = 0.02 m, u = 0.6 m/s 0.6 = K (0.062 – 0.022) K = 187.5 Ê g d hˆ 2 (a) Maximum velocity um = Á R Ë 4 m d s ˜¯ = KR2 = 187.5 ¥ (0.06)2 = 0.675 m/s u = m = 0.3375 m/s 2 p Q = ¥ (0.12)2 ¥ (0.3375) 4 = 3.817 ¥ 10–3 m2/s = 3.817 L/s Solution: The velocity distribution in laminar ﬂow in a circular tube is given by Ê g d hˆ 2 2 v = Á(R – r ) Ë 4 m d s ˜¯ Average velocity 1 V= = pR 2 v ( 2 p r ) dr Ê R4 R4 ˆ Á 2 - 4 ˜ Ë ¯ r = R2 2 R d hˆ Ê ÁË - g d s ˜¯ = 0.707 R 2 Distance from the boundary y = R – r = 0.293 R 7.4 (c) Discharge d hˆ Ê ÁË - g d s ˜¯ 1 d hˆ Ê 2 2 ÁË - g d s ˜¯ = (R – r ) ◊ 4m r2 = or 0 V =v R2 8m i.e. R 2 Ê g d hˆ Á˜ R2 Ë 4m d s ¯ R2 = 8m When Ú (b) Mean velocity 7.5 Solution: u = K(R2 – r2) Ê 1 dp ˆ where K = Á for horizontal pipes Ë 4m dx ˜¯ and Ê g d ( p /g + Z)ˆ K = Á˜¯ for inclined pipes ds Ë 4m Average velocity V = = 1 2 pR 2pK pR 2 Ú R 0 Ú R u 2p r d r (R 2r – r3) dr 0 2 K È R4 R4 ˘ K 2 R ˙ = 2 Í 2 4 2 R ÍÎ ˙˚ Kinetic energy correction factor 1 u3 d A a = 3 V A = Ú 236 Fluid Mechanics and Hydraulic Machines = = = 1 3 Ê K 2ˆ 2 ÁË 2 R ˜¯ pR 16 8 R 16 R 8 Ú Ú R Ú R Pressure drop per unit length of pipe. (–D p)1 = (32 ¥ 0.08 ¥ 0.7 ¥ 1.0)/(0.1)2 = 179.2 N/m2 K 3 (R2 – r2)3 2prdr 0 Alternate method for Part (b) (R2 – r2)3 r d r 4t 0 L D [See Examples 7.1 and 7.2] Pressure drop per unit length of pipe, (– D p)1 = (4 ¥ 4.48 ¥ 1)/0.01 = 179.2 N/m2 0 Pressure drop (–Dp) = R [R 6 r – r7 – 3R4r3 + 3R 2 r5]dr 0 È1 1 3 3˘ = 16 Í - - + ˙ Î2 8 4 6˚ È12 - 3 - 18 + 12 ˘ = 16 Í ˙ 24 Î ˚ 16 ¥ 3 = 2.0 a= 24 7.7 7.6 f ur 2 u r R ] R r Solution: m = 1.5 poise = 0.15 Pa.s r = 0.85 ¥ 998 = 848.3 kg/m3 Solution: 2 u = 1.4 [1 – (r/R) ] (a) t = m du du = -m dr dy (Distance y is from the boundary and = R – r; dy = – dr) t = 2m (1.4)r/R2 At the boundary, r = R and 2 ¥ 0.08 ¥ 1.4 t0 = 0.05 = 4.48 N/m2 (b) Maximum velocity um occurs at r = 0 Hence um = 1.4 m/s Mean velocity V = um/2 = 0.7 m/s Pressure drop (–D p) = (32 mVL)/D2 (a) Wall shear stress R Êgh ˆ t0 = Á f ˜ 2Ë L ¯ 0.30 Ê 848.3 ¥ 9.81 ¥ 20 ˆ ˜¯ 2 ÁË 3000 = 8.32 Pa (b) Shear stress t at r= 10 cm t 8.32 ¥ 0.10 t = 0r = R 0.15 = 5.548 Pa (c) If the ﬂow is laminar 32 mV L hf = g D2 hf g D 2 V = L 32 m = 237 Laminar Flow 20 848.3 ¥ 9.81 ¥ (0.30) 2 ¥ 3000 32 ¥ 0.15 = 1.04 m/s = Power Reynolds number V Dr 1.04 ¥ 0.30 ¥ 848.3 = Re = m 0.15 = 1764.5 < 2000 The assumption of laminar ﬂow is therefore correct. 64 Re = 64/1764.5 = 0.03627 p ¥ (0.1) 2 ¥ 5.0 4 = 0.03927 m3/s P = (1260 ¥ 9.81) ¥ 0.03927 ¥ 23.3 = 11309.8 W = 11.31 kW Discharge Q = AV = 7.9 r 3 ¥ m In laminar ﬂow f = 7.8 m Solution: The maximum Reynolds number = Recrit r = 2000 = 3 VD n 8 ¥ 10 - 2 1 ¥ 950 0.15 = 1.123 m/s 32 mV L hf = g D2 V = 2000 Head loss Reynolds number rV D 1260 ¥ 5.0 ¥ 0.10 = Re = m 1.50 = 420 (a) As this value is less than 2000, the ﬂow is laminar. In laminar ﬂow in a conduit 8mV D 8 ¥ 1.50 ¥ 5.0 = = 600 Pa 0.10 t0 = = 32 ¥ 8 ¥ 10 -2 ¥ 1.123 ¥ 200 (950 ¥ 9.81) (0.15) 2 = 2.742 m = maximum difference in oil surface elevations 7.10 2 (b) In laminar ﬂow the head loss hf = = 32 mV L g D2 32 ¥ 1.50 ¥ 5.0 ¥ 12 (1260 ¥ 9.81) (0.1) 2 (c) Power expended P = g Qh f Solution: = 23.3 m P = Power spent in ﬂuid friction = 5.4 ¥ 0.6 = 3.24 kW = 128mQ 2 L pD 4 238 Fluid Mechanics and Hydraulic Machines Therefore 3240 = Pressure drop 128 ¥ 0.1 ¥ Q 2 ¥ 1000 p ¥ (0.075) 4 – Dp = Q2 = 2.51611 ¥ 10–5 Q = 0.00502 m3/s = 0.00502 ¥ 1000 ¥ 60 = 301.2 L/min 0.00502 Q Velocity V= = p p ¥ (0.075) 2 ¥ D2 4 4 = 1.136 m/s V Dr Reynolds number Re = m 1.136 ¥ 0.075 ¥ (0.90 ¥ 998) = 0.10 = 32 mVL D2 32 ¥ 0.097 ¥ 0.4725 ¥ 10.0 (0.1)2 = 1467 Pa Reynolds number of the ﬂow rVD 898.2 ¥ 0.4725 ¥ 0.1 = Re = m 0.097 = 437.5 < 2000 Hence the ﬂow is laminar. 7.12 = 765 7.11 Solution: Solution: Given data: m = 0.97 poise = 0.097 N.m/s r = 0.9 ¥ 998 = 898.2 kg/m3 rQ = mass rate of ﬂow = 100 = 3.333 kg/s 30 Q = volume rate of ﬂow = 3.333 988.2 = 3.711 ¥ 10–3 m3/s = 3.711 Liters/s p 2 0.1) (0.1)2 ( 4 = 0.007854 m2 Average velocity Area of ﬂow = A = V= 3.711 ¥ 10 - 3 = 0.4725 m/s 0.007854 m r D Q = 2.5 poise = 0.25 Pa.s = 0.9 ¥ 998 = 898.2 kg/m3 = 100 mm = 0.1 m = 2L/s = 0.002 m3/s 0.002 V = = 0.2546 m/s p 2 ¥ (0.1) 4 (a) Reynolds number rVD Re = m 898.2 ¥ 0.2546 ¥ 0.1 0.25 = 91.49 As Re < 2000, the ﬂow is laminar. (b) Head loss due to friction: 32 mVL hf = g D2 = 239 Laminar Flow = 32 ¥ 0.25 ¥ 0.2546 ¥ 500 2 (898.2 ¥ 9.81) (0.1) = 11.55 m If A is the pump end and B is the outlet, then pA pB V2 V2 +0+ + 20 + + hf = g 2g g 2g But pB = 0 = atmospheric, as the outlet is free. Also hf = 11.55 m pA = 20.0 + 11.55 = 31.55 m g pA = 31.55 ¥ (898.2 ¥ 9.81) = 278000 Pa = 278 kPa (c) Power required for pumping the ﬂuid P = g QH where H = overall head = static head + friction head = 31.55 m P = (898.2 ¥ 9.81) ¥ 0.002 ¥ 31.55 = 556 W As the overall efﬁciency of the pump set is 65% Power input required = Pi = 0.01417 = 0.8017 m/s p ¥ (0.15) 2 4 Dp 95, 000 Head loss = = = 10.56 m g 917 ¥ 9.81 Assuming the ﬂow as laminar V = hf = 10.56 = 32 mV L g D2 32 ¥ m ¥ 0.8017 ¥ 800 (917 ¥ 9.81) ¥ (0.15) 2 = 101.4 m 10.56 m = = 0.1041 Pa.s 101.4 Reynolds number V Dr 0.8017 ¥ 0.15 ¥ 917 = Re = m 0.104 = 1059 As this value of Re is less than 2000, the ﬂow is laminar as assumed initially. 7.14 3 556 = 855.4 W 0.65 3 7.13 A 80 cm Oil D = 2 cm L = 70 cm Solution: Q = 850 L/min = = 0.01417 m3/s 850 1000 ¥ 60 B Q Fig. 7.9 240 Fluid Mechanics and Hydraulic Machines Solution: m = 1.5 poise = 0.15 Pa.s 7.15 A By energy equation between sections A and B, D L Vb2 VA2 pb pa + Zb + + hf + Za + = g 2g g 2g H where hf = head lost in friction. As the tank is large VA = 0, p a = pb = 0 = atmospheric pressure Za – Zb = 1.50 m 1 H1 dh V2 hf = 1.5 – b 2g Assuming laminar ﬂow in the pipe, hf = \ 1.5 – H2 H2 Length = L Diameter = D h 32 mVb L 2 g D2 Datum Vb2 32 ¥ 0.15 ¥ 0.70 Vb = 2 ¥ 9.81 (920 ¥ 9.81) ¥ (0.02) 2 (Note the length of the tube is 70 cm). V2 = 0.9307 Vb 1.5 – b 19.62 Solving for Vb, Vb = 1.49 m/s Reynolds number Re = V Dr m 1.49 ¥ 0.02 ¥ 920 0.15 = 182.8 As this value of Re is less than 2000, the ﬂow is laminar as assumed. Discharge = Q = AV p ¥ (0.02)2 ¥ 1.49 = 4 = 4.681 ¥ 10–4 m3/s = 28.086 L/min Q Fig. 7.10 Solution: Let at any instant t the liquid surface be at an elevation h above the datum drawn at the level of the outlet. In a time dt the liquid surface will drop by a height dh. By neglecting (1) the velocity of the free surface in the tank, (2) velocity head at the outlet, and (3) all other minor losses, the total head loss due to friction hf = h 128 m QL 32 mV L = Hence hf = h = 2 p D 4g gD By continuity – Adh = Q dt p D 4g h dt = K h dt = 128 m L where K = - p D 4g 128 m L A dh = dt K h 241 Laminar Flow Integrating Ú A dt = T = K 0 Hence H2 dh H A ln 1 = H1 h K H2 H1 128 m AL ◊ ln T = H2 p D 4g T Ú 7.17 A F r m N d v= 7.16 N pFd 4 I 28 m A2L L ¥ n Solution: m2 Reynolds number VD = 1200 Re = n VD = 1200 ¥ 1.92 ¥ 10–3 = 2.304 Se = Energy gradient Solution: Consider one tubular opening which acts as pipe of diameter d and length L for the ﬂow of oil. Pressure difference across the pipe (i) ÏÔÊ p ˆ Êp ˆ ¸Ô = ÌÁ 1 + Z1 ˜ - Á 2 + Z 2 ˜ ˝ L ¯ Ë g ¯ Ô˛ ÔÓË g When pressure is constant p1 = p2 Also for a vertical pipe (Z1 – Z2) = L (– Dp) Hence V = 32 mV hf = L g D2 Ê pd2 ˆ pd4 F = = VÁ ˜ 128 m L A Ë 4 ¯ 32nV Since there are N tubular openings, total discharge Q = NQ1 gD2 In the present case 32nV gD2 V or D 2 Rate of descent of the piston = =1 g 9.81 = 32n 32 ¥ 1.92 ¥ 10 -3 = 159.67 v = = (ii) = From (i) and (ii) V D2 F d2 ◊ A 32 m L Q1 = discharge through one tubular opening For laminar ﬂow = F 32 mVL = A d2 where V = average laminar ﬂow velocity in the tubular opening. Se = (Z1 – Z2)/(Z1 – Z2) = 1.0 Sf = = 1 159.67 = 69.3 (VD ) = 3 = 2.304 D D = 0.243 m Q pd4 F Ê N ˆ NQ1 = = A 128 m L A ÁË A ˜¯ A p d 4 NF 128 m L A2 N Fd 4 Hence v = 128 A2 L 242 Fluid Mechanics and Hydraulic Machines 7.18 F 7.19 3 –3 m 2 f 0.5 F 10 Solution: The ﬂow can be considered as laminar ﬂow between two parallel plates of spacing B = 0.12 mm. f L = 0.50 m h f = head loss = 5.0 m 12 mVL = g B2 40 mm Fig. 7.11 Solution: Given data: m = 10–3 N ◊m/s, Diameter D = 0.5 mm = 0.5 ¥ 10–3 m, L = 40 ¥ 10–3 m Discharge Q = 500 ¥ 10–9 m3/s, Velocity V= 5.0 = 12 ¥ (998 ¥ 0.01 ¥ 10 -4 ) ¥ V ¥ 0.50 9790 ¥ (0.12 ¥ 10 -3 ) 2 = 42.475 V V = 0.1177 m/s Discharge per metre width = q = 1 ¥ 0.1177 ¥ 0.12 ¥ 10–3 = 1.4126 ¥ 10–5 m3/s q = 0.848 L/min per metre width of crack 500 ¥ 10 - 9 Êpˆ 2 -6 ÁË 4 ˜¯ (0.5) ¥ 10 = 2.546 m/s 7.20 Pressure difference across the needle F 32 mVL = A D2 A = area of the piston (–Dp) = where (– Dp) = ( ) ( 32 ¥ 10 - 3 ¥ 2.546 ¥ 40 ¥ 10 - 3 (0.5) 2 ¥ 10 - 6 = 13035 Pa ) Solution: Consider the laminar ﬂow between two inclined plates spaced B apart and inclined at q to the horizontal, as shown in Fig. 7.12. The velocity u at any y from the boundary is. Force F = (– Dp)A Êpˆ = (13035.5) ¥ Á ˜ (10)2 ¥ 10–6 Ë 4¯ = 1.023 N u = 1 2m È d ( p + g Z )˘ 2 Í˙ (By – y ) dl Î ˚ Ê g ÁÁ = 2m Á Á Ë Êp ˆˆ d Á + Z˜ ˜ Ëg ¯ ˜ (By – y2) ˜ dl ˜ ¯ 243 Laminar Flow Using result in Eq. 1, t0 = B um u 3mV d 7.21 s y t0 d q Fig. 7.12 For a free surface ﬂow of depth d, the maximum velocity will be at y = d. Hence the depth d can be considered as d = B/2. Also on the free surface the pressure is atmospheric and hydraulic grade line coincides with the free surface. Êp ˆ - d Á + Z˜ hf Ëg ¯ = Thus = sin q = S0 L dl g sin q (2dy – y2) Thus u= 2m or if s = depth below the surface is used, s = d – y and (2dy – y2) = (d2 – s2) g (d2 – s2) sin q Thus u= 2m Maximum velocity g 2 d sin q = Surface velocity um = 2m g È1 d 2 ˘ sin q Í Mean velocity = ( d - s 2 ) ds˙ 2m Îd 0 ˚ g 2 d sin q V = 3m Discharge per unit width g 3 q = Vd = d sin q (1) 3m Shear stress on the bed t0 = g d sin q Ú Solution: (a) For two-dimensional laminar ﬂow between parallel plates 3 um = maximum velocity = V 2 3 = ¥ 1.40 = 2.10 m/s 2 (b) Since Ê d p ˆ B2 V = Á- ˜ Ë d x ¯ 12m 12 mV 12 ¥ 0.105 ¥ 1.40 Ê dp ˆ ÁË - d x ˜¯ = B 2 = (0.012) 2 = 12250 Boundary shear stress Ê dp ˆ B t0 = Á - ˜ Ë dx ¯ 2 = 12250 ¥ 0.012 2 = 73.5 Pa (c) Shear stress t at any y from the boundary Ê dp ˆ t = Á- ˜ Ë dx ¯ ÊB ˆ ÁË 2 - y ˜¯ At y = 0.002 m: Velocity Ê 0.012 ˆ - 0.002˜ t = (12250) Á Ë 2 ¯ = 49 Pa 1 Ê dp ˆ (By – y2) v = 2m ÁË d x ˜¯ 244 Fluid Mechanics and Hydraulic Machines 1 ¥ 12250 2 ¥ 0.105 ¥ [0.012 ¥ 0.002 – (0.002)2] v = 1.167 m/s 12 mVL (d) Head loss h f = g B2 12 ¥ 0.105 ¥ 1.4 ¥ 25 = (0.92 ¥ 998 ¥ 9.81) (0.012) 2 = 34.0 m = 7.23 a Solution: Let B = thickness of the gap between the two ﬁxed parallel plates. Then the velocity v at a distance y from the boundary is given by Ê 1 dp ˆ v = Á(By – y2) Ë 2m dx ˜¯ 7.22 Ê 1 dp ˆ v = K (By – y2) where K = Á Ë 2m dx ˜¯ Average velocity 2 B B Ú ( ) 1 1 K By - y 2 dy vdy = V = B B Ú 0 0 Solution: Given data: Clearance Diameter m B D g Head difference = 0.02 N.m/s, = 0.2 mm = 0.0002 m, = 200 mm = 0.2 m, = 0.9 ¥ 9790 = 8811 N/m3 = hf = 10.0 m, Length L = 0.5 m, 12 mVL 12 ¥ 0.02 ¥ V ¥ 0.5 = hf = 2 g B2 8811 ¥ (0.02) = 10.0 10.0 V= = 0.0294 m/s 340.5 Equivalent width of plate = pD = p ¥ 0.2 = 0.6283 m Leakage Q = V ¥ Area of ﬂow = 0.0294 ¥ (0.6283 ¥ 0.0002) = 3.6944 ¥ 10–6 m3/s = 0.369 ¥ 10–6 Liters/sec B˘ È 1 Í Ê y 2 y3 ˆ ˙ B2 KÁB = - ˜ = K BÍ Ë 2 3¯ ˙ 6 0 ˚ Î Kinetic energy correction factor a = = = = B 1 V 3 v BÚ 3 dy 0 B 1 Ê B2 ˆ ÁK 6 ˜ ¯ Ë K B 216 B 7 7 3 dy 0 B (6 )3 3 Ú ( K ( By - y ) 2 3 Ú (K 3 ) 3 ( By - y 2 ) dy 0 B Ú (B y 3 3 0 216 È B 7 B 7 3 B 7 3 B 7 ˘ + Í ˙ 7 5 6 ˚˙ B 7 ÍÎ 4 È1 1 3 3˘ = 216 Í - - + ˙ Î4 7 5 6˚ = = 54 = 1.543 35 ) y 6 - 3 B 2 y 4 + 3 By 5 dy 245 Laminar Flow Oil of thickness = h 7.24 Shaft RPM = N 3 r L Journal bearing Diameter = D Fig. 7.13 dp = 0 and U = 4.0 m/s, B = 2.00 mm dx = 0.002 m; m = 0.5 Pa.s (i) Since the ﬂow is laminar, the velocity U du = distribution is linear and B dx Hence the shear stress at the boundaries (i.e, at both top and bottom plates) 4.0 U t0 = m = 1000 N/m2 = (0.50) ¥ 0.002 B U = 2.0 m/s (ii) Mean velocity V = 2 Discharge per unit width 4 ¥ 0.002 = 0.004 m3/s/m q = VB = 2 Given: 7.25 2 pmNr V = 60 h h Shear force on the shaft surface 2 pmNr Fs = t ¥ 2pr ¥ L = ¥ 2pr ¥ L 60 h Shear stress = t = m Solution: This is a case of plain Coutte ﬂow. Torque T = Fs ¥ r = Power lost P =T¥w 4p 2 mNLr 3 60 h 4p 2 m NL r 3 2p N ¥ 60 h 60 3 2 3 8p mN L r = 3600 h = Since r = D , power P = 2 3 N 2 LD 3 3600 h 7.26 L D h Solution: p 3 m N 2LD 3 P= 3600h Solution: Refer to the deﬁnition sketch Fig. 7.13 2p N r. Tangential velocity V = wr = 60 Assuming linear variation of velocity in the gap, Refer to the deﬁnition sketch Fig. 7.14. 2p N . w = angular velocity of the shaft = 60 V = Velocity at radius r = w r By assuming the velocity variation to be linear across the oil ﬁlm, shear stress mwr V t = m = h h 246 Fluid Mechanics and Hydraulic Machines R = 0.20 m N = 4.5 RPM Shaft h = 0.0015 m Solution: Collar bearings are employed to take the axial thrust of a rotating shaft. The collar arrangement of Fig. 7.15 shows the collar separated from the bearing surface by an oil ﬁlm of very small thickness, h. Linear variation of velocity across the gap is assumed. Oil r dr Foot step bearing Fig. 7.14 Example 7.26 D1 Consider an area element of shaft of radius r and width dr. d A = 2pr ◊dr Viscous torque on the element mw r = dT = t (dA)r = (2p r dr ) r h 2pmw 3 r dr = h Shaft D2 Collar ( ) Total torque T = R R 0 0 Ê 2pmw ˆ 3 r dr h ˜¯ Ú dT = Ú ÁË mpw 4 R T= 2h 2p N 2p ¥ 1200 In the present case w = = 60 60 = 125.67 rad/s h = 2.5 mm = 0.0015 m; R = 200/2 = 100 mm = 0.1 m mp (125.67) T= (0.10)4 2 ¥ 0.0015 = 4.5 4.5 = 13.16m 4.5 = 0.3419 Pa.s m= 13.16 Oil Collar bearing Fig. 7.15 The shear stress on a circular element of radius r and thickness dr is wr t = m h Viscous torque on the element = dT = t (dA)r mw r 2pmw 3 2p r dr ) r = r dr = ( h h ( R2 Total torque T = Ú dT R1 R2 7.27 Example 7.27 = R1 = Ê 2pmw ˆ 3 r dr h ˜¯ Ú ÁË mpw È 4 R - R24 ˘ ˚ 2h Î 1 ) 247 Laminar Flow In the present problem 2p N 2p ¥ 500 = 60 60 = 52.36 rad/s w= h = 1.2 mm = 0.0012 m; R1 = 225/2 = 112.5 mm = 0.1125 m R2 = 175/2 = 87.5 mm = 0.0875 m, m = 0.5 Pa.s 0.5 ¥ p ¥ (52.36 ) È(0.1125) 4 - (0.0875) 4 ˘ T= Î ˚ 2 ¥ 0.0012 T = 3.48 N◊m Power lost = P = Tw = 3.48 ¥ 52.36 = 182.2 W Solution: Given data: m = 2 N.m/s, Clearance B = 100 mm = 0.1 m, (i) For laminar ﬂow between two parallel plates Average velocity 2 2 vm = ¥ 1.5 = 1.0 m/s 3 3 Discharge; V = q = BV = 0.1 ¥ 1.0 = 0.10 m3/s/m width (ii) Boundary shear stress t0 = 6mV 6 ¥ 2.0 ¥ 1.0 = = 120 N/m2 B 0.10 Ê dp ˆ B (iii) Since t0 = Á - ˜ Ë dx ¯ 2 7.28 2t 0 2 ¥ 120 Ê dp ˆ ÁË - dx ˜¯ = B = 0.1 2 = 2400 Pa/m Problems 7.1 An oil of dynamic viscosity 0.008 Pa.s and relative density 0.86 is to ﬂow in a 6 cm diameter pipe. What is the maximum discharge that can be achieved while maintaining laminar ﬂow? If crude oil (RD = 0.925 and dynamic viscosity = 0.09 Pa.s) is used instead, at the same velocity, would the ﬂow be laminar? (Ans. Q = 0.878 L/s; ﬂow would still be laminar) 7.2 An oil of relative density 0.92 and dynamic viscosity 0.082 Pa.s ﬂows in an 80 mm diameter pipe. In a distance of 20 m the ﬂow has a head loss of 2 m. Calculate (i) the mean velocity, (ii) discharge, (iii) velocity and shear stress at a radial distance of 38 mm from the pipe axis and (iv) boundary shear stress. (Ans. (i) V = 2.197 m/s. (ii) Q = 11.04 L/s; (iii) u = 0.4284 m/s; t = 17.114 Pa (iv) t0 = 18.02 Pa) 7.3 It is required to maintain a shear stress of 3 Pa at the wall when water (n = 1 ¥ 10–6 m2/s) ﬂows with a head loss of 10 cm in 1000 m. What diameter pipe would achieve this? Is this applicable only to laminar ﬂow? (Ans. D = 12.26 cm; this is true for all regimes of ﬂow.) 248 Fluid Mechanics and Hydraulic Machines 7.4 With laminar ﬂow in a circular pipe, at what radial distance from the centre line does the local velocity equal one third the maximum velocity? (Ans. r = 0.8165 R) 7.5 With laminar ﬂow between two ﬂat plates, at what distance from the centre line does the local velocity equal the mean velocity? y¢ Ê ˆ ÁË Ans. B = 0.2887 on either side of centre line˜¯ 7.6 Show that the momentum correction factor for laminar ﬂow in a circular tube is 1.33. 7.7 Determine the kinetic energy correction factor and momentum correction factor for laminar ﬂow between two ﬁxed parallel plates. (Ans. a = 1.543, b = 1.20) 7.8 What power will be required per kilometre length of a pipeline to overcome viscous resistance to the ﬂow of an oil of viscosity 2.0 poises through a horizontal 10 cm diameter pipe at the rate of 200 L/min? Find the Reynolds number of the ﬂow if the relative density of the oil is 0.92. (Ans. Re = 194.8; P = 0.905 kW) 7.9 An oil of relative density 0.90 ﬂows at a rate of 10.0 L/s through a horizontal pipe of 7.5 cm diameter. The pressure drop over a length of 300 m of pipe is found to be 40 N/cm2. Estimate the viscosity of the oil. What is the Reynolds number of the ﬂow? (Ans. m = 0.1035 Pa.s, Re = 1473) 7.10 A capillary tube of 1.5 mm diameter and 15 cm long is connected horizontally to a tank 6 cm in diameter. The tank contains oil up to a height of 9.0 cm above the axis of the capillary tube. When the oil is allowed to discharge through the capillary to atmosphere, it takes 10 min to discharge 85 cm3 of oil. Estimate the kinematic viscosity of the oil. (Ans. n = 1.77 ¥ 10–6 m2/s) 7.11 A capillary viscometer has a capillary of diameter 2.0 mm and length 50 cm. A liquid of density 850 kg/m3 and dynamic viscosity 0.5 poise is sent through the capillary under a constant pressure difference of 6 kPa. Find the time taken to collect 50 cm3 of liquid at the capillary outlet. (Ans. T = 8 min 50.5 s) 7.12 A liquid of dynamic viscosity 0.07 Pa.s and relative density 0.86 ﬂows through an inclined pipe of 2 cm diameter. A discharge of 13 L/min is to be sent through the pipe in such a manner that the pressure along the length is constant. Find the required inclination of the pipe. hf ˆ Ê ÁË Hint : For constant pressure the slope = sin q = ˜¯ L (Ans. q = 27° 18¢ 17≤) 7.13 Calculate the least diameter of a pipe to carry 10 L/s of an oil of density 900 kg/m3 and dynamic viscosity 1.5 poise with a permissible energy gradient in laminar ﬂow regime of 0.03. (Ans. D = 12.3 cm) 7.14 An oil of relative density 0.92 and dynamic viscosity 0.9 poise ﬂows though a 10 cm diameter pipe 30 m long. Determine the largest ﬂow that can be passed through this pipe while maintaining laminar regime. What is the head loss between the two ends of the pipe under this ﬂow? (Ans. Q = 15.4 L/s and hL = 1.88 m) 7.15 A liquid of relative density 0.85 ﬂows in a pipe 8 cm is diameter with a velocity of 0.8 m/s in laminar regime. If two pressure gauges located at the ends of a 12 m long horizontal stretch of pipe record a pressure difference of 60 kPa, estimate the (i) viscosity of the ﬂuid and (ii) Reynolds number of the ﬂow. (Ans. m = 1.25 Pa.s; Re = 43.4) 249 Laminar Flow 7.16 For a steady fully developed laminar ﬂow of an oil of density rf through two pipes in series as shown in Fig. 7.16 ﬁnd the ratio h1/h2 of the manometer ﬂuid deﬂections Consider only friction losses in the pipes. 4 Ê h1 Ê D2 ˆ Ê L1 ˆ ˆ Á Ans. ˜ = h2 ÁË D1 ˜¯ ÁË L2 ˜¯ ˜¯ ÁË L2 L1 Flow f D1 h1 f D2 rm h2 rm Fig. 7.16 7.17 A pump delivers a lubricating oil of relative density 0.90 and dynamic viscosity 0.85 poise through 25 m of 30 mm pipe to a tank whose oil surface is 10 m higher than the oil surface of the supply tank. The pump efﬁciency is 70%. (a) Estimate the power input required to pump oil at a rate of 250 L/min (b) What power input is required to pump the liquid at a Reynolds number of 2000 in this system? (Ans. (a) P = 3.18 kW; (b) P = 3.6 kW) 7.18 A 5 cm diameter pipe carries lubricating oil of relative density 0.92 and dynamic viscosity 2.0 poise in a vertical pipe. Two pressure gauges are connected 25 m apart. The upper gauge records 220 kPa and the lower gauge records 350 kPa. Find the direction and rate of ﬂow. (Ans. Flow is upwards: Q = 175 L/min) 7.19 A ﬂow of 60 L/s per metre width of glycerine of relative density 1.25 and dynamic viscosity 1.5 Pa.s takes place between two parallel plates having a gap of 25 mm between them. Calculate the (i) maximum velocity, (ii) boundary shear stress and (iii) energy gradient. (Ans. um = 3.6 m/s; t0 = 864 Pa and h f /L = 5.648) 7.20 A masonry wall of a water tank is 0.90 m thick. At the bottom a crack of thickness 0.3 mm and 60 cm wide has developed and the crack extends to the entire thickness of the wall. If the tank contains 4 m of water above the crack and the other end of the crack is at atmospheric pressure, estimate the leakage volume per day from the crack (n = 1 centistoke.) (Ans. Q = 5.09 m3/day) 7.21 An oil having a viscosity of 0.098 N.s/m2 and a relative density of 1.59 ﬂows through a horizontal pipe of 5 cm diameter with a pressure drop of 0.3 N/cm2 per metre length of pipe. Determine the (i) rate of ﬂow, (i) shear stress at the pipe wall and (iii) power required for 100 m of pipe to maintain the ﬂow. (Ans. (i) Q = 4.696 L/s; (ii) t0 = 37.51 Pa; (iii) P = 1.41 kW) 7.22 A ﬂuid ﬁlm of RD = 0.90 and thickness 2.0 mm ﬂows down a vertical surface at a surface velocity of 0.45 m/s. Estimate the (i) discharge in cm3/s/cm width, (ii) viscosity of the ﬂuid and (iii) the boundary shear stress. (Ans. (i) 6 cm3/s/cm; (ii) m = 0.0392 Pa.s; (iii) t0 = 17.62 Pa) 7.23 A ﬁlm of liquid moves down a plane inclined at 60° to the horizontal. The surface velocity of the ﬁlm, of thickness 3 mm, is found to be 3.2 cm/s. If the dynamic viscosity of the liquid is 1.5 Pa.s, calculate the (i) boundary shear stress and (ii) speciﬁc weight of the liquid. (Ans. (i) t0 = 31.95 Pa; (ii) g = 12317 N/m3) 250 Fluid Mechanics and Hydraulic Machines 7.24 Show that the momentum correction factor for (i) laminar ﬂow in a circular tube is 1.33 (ii) laminar ﬂow between two parallel ﬁxed plates is 1.20 7.25 A vertical shaft has a hemispherical bottom of radius R which rotates inside a bearing of identical shape and an all round clearance h at its end. An oil of viscosity m is maintained in the bearing. Show that the viscous torque in the shaft when its rotating with angular velocity w is given by 4pmw 4 R T= 3h 7.26 A 90 mm diameter shaft rotates at 1200 rpm in a 100 mm long journal bearing of 90.5 mm internal diameter. The annular space. In the bearing is ﬁlled with lubricating oil having a viscosity of 0.12 Pa.s. Estimate the power dissipated as heat. [Ans: P = 434 W] Objective Questions 7.1 The equations of motion for laminar ﬂow of a real ﬂuid are known as (a) Euler’s equations (b) Bernoulli equation (c) Navier–Stokes equation (d) Hagen–Poiseuille equation 7.2 In a two-dimensional, steady, horizontal, uniform laminar ﬂow the shear gradient in the normal direction is equal to (a) the velocity gradient in the normal direction. (b) the velocity gradient in the longitudinal direction (c) the pressure gradient in the normal direction. (d) the pressure gradient in the direction of ﬂow. 7.3 An oil of kinematic viscosity 0.25 stokes ﬂows through a pipe of diameter 10 cm. The ﬂow is critical at a velocity of (a) 7.2 m/s (b) 5.0 m/s (c) 0.5 m/s (d) 0.72 m/s 7.4 Air (r = 1.2 kg/m3 and m = 1.80 ¥ 10–5 Pa.s) ﬂows with a velocity of 20 m/s in a 10 cm diameter pipe. If the friction factor f = 0.02, the shear stress at the wall is (a) Zero (b) 153.6 Pa (c) 2.4 Pa (d) 1.2 Pa 7.5 In a steady ﬂow of an oil in a pipe in laminar regime the shear stress is (a) constant across the pipe (b) maximum at the centre and decreases parabolically towards the sides (c) zero at the boundary and increases linearly towards the centre (d) zero at the centre and increases linearly towards the boundary. 7.6 Oil of viscosity 1.5 Pa.s and relative density 0.9 ﬂows through a circular pipe a diameter 5 cm with a mean velocity of 1.2 m/s. The shear stress at the wall in Pa is (a) 360 (b) 288 (c) 180 (d) 144 7.7 In a circular pipe of certain length carrying oil at a Reynolds number 100, it is proposed to triple the discharge. If the viscosity remains unchanged, the power input will have to be (a) decreased to 1/3 its original value (b) increased by 100% (c) increased to 3 times the original value (d) increased to 9 times its original value 251 Laminar Flow 7.8 A liquid ﬂowing in a pipe has a head loss of 2 m in a pipe length of 10 m. The Reynolds number of the ﬂow is 100. If the ﬂow rate is doubled and all other ﬂuid properties remain the same, the head loss in m is (a) 0.5 (b) 8.0 (c) 4.0 (d) 2.0 7.9 The pressure drop in an 8 cm horizontal pipe is 75 kPa in a distance of 15 m. The shear stress at the pipe wall, in kPa is (a) 0.2 (b) 2.0 (c) 5.0 (d) 0.4 7.10 The Reynolds number for ﬂow of an oil in a certain pipe is 640. The Darcy–Weisbach friction factor f for this ﬂow is (a) 0.02 (b) 0.01 (c) 0.1 (d) 0.064 7.11 The minimum value of friction factor f that can occur in laminar ﬂow through a circular pipe is (a) 0.025 (b) Zero (c) 0.064 (d) 0.032 7.12 A 20 cm diameter pipe carries a ﬂuid of relative density 0.9. If the boundary shear stress in the pipe is 0.50 Pa, the head loss in a length of 100 m of the pipe line is (a) 11.35 m (b) 4.54 m (c) 0.36 m (d) 9.08 m 7.13 The friction factor f in a laminar pipe ﬂow was found to be 0.04. The Reynolds number of the ﬂow was (a) 2000 (b) 1000 (c) 800 (d) 1600 7.14 In a laminar ﬂow through a circular pipe of diameter 20 cm, the maximum velocity is found to be 1 m/s. The velocity at a radial distance of 5 cm from the axis of the pipe will be (a) 0.25 m/s (b) 0.50 m/s (c) 0.75 m/s (d) 0.10 m/s 7.15 In a circular tube of radius R carrying a laminar ﬂow, the ratio of average velocity 7.16 7.17 7.18 7.19 7.20 7.21 7.22 to the maximum velocity in the conduit is (a) 0.50 (b) 1.00 (c) 0.67 (d) 0.33 When water passes through a given pipw at mean velocity V, the ﬂow is found to change from laminar to turbulent regime. If another ﬂuid of speciﬁc gravity 0.8 and of coefﬁcient of viscosity 20% that of water is passed through the same pipw, the transition to turbulent ﬂow is expected at a velocity of (a) 2V (b) V (c) V/2 (d) V/4 The momentum correction factor b for laminar ﬂow through a circular pipe is (a) 1.5 (b) 2.0 (c) 1.67 (d) 1.33 The kinetic energy correction factor a for laminar ﬂow through a circular pipe is (a) 1.54 (b) 2.0 (c) 1.67 (d) 2.33 In a uniform laminar ﬂow through a twodimensional passage, the ratio of maximum velocity to the average velocity is (a) 2.0 (b) 1.67 (c) 1.5 (d) 1.33 A laminar motion between two vertical parallel plates occurs in such a manner that the hydraulic grade line is vertical. This indicates that (a) the viscosity is negligibly small (b) the ﬂow is with a free surface (c) the pressure is same at all the sections (d) the ﬂow has stopped A wall shear stress of 28 Pa exists in a laminar ﬂow in an 8 cm diameter pipe. At a radial distance of 3 cm from the axis, the shear stress, in Pa, is (a) 21.0 (b) 28.0 (c) 7.8 (d) 12.25 If the maximum velocity in a laminar ﬂow through two parallel static plates is 9 m/s, 252 Fluid Mechanics and Hydraulic Machines 7.23 7.24 7.25 7.26 7.27 7.28 then the average velocity of the ﬂow will be (a) 3.0 m/s (b) 4.5 m/s (c) 6.0 m/s (d) 7.5 m/s In a laminar ﬂow between two parallel plates with a separation distance of 6 mm, the centre line velocity is 1.8 m/s. The velocity at a distance of 1 mm from the boundary is (a) 0.15 m/s (b) 1.0 m/s (c) 0.55 m/s (d) 0.75 m/s In laminar ﬂow between two ﬁxed parallel plates, the shear stress is (a) constant across the passage (b) maximum at centre and zero at the boundary (c) zero all through the passage (d) maximum at the boundary and zero at the centre. A ﬂuid (RD = 0.9 and m = 1.2 Pa.s) ﬂows in laminar regime between two parallel plates ﬁxed 3 cm apart. If the discharge is 600 cm3/s/cm width of plate, the shear stress on the boundary, in Pa, is (a) 800 (b) 640 (c) 480 (d) 240 In the laminar ﬂow of a liquid down an inclined plane, the surface velocity is found to be 30 cm/s. The average velocity of the ﬂow, in cm/s, is (a) 20 (b) 30 (c) 15 (d) 10 In laminar ﬂow between two parallel plates, the slope of hydraulic grade line was found to be 0.05. If the discharge remains the same but the viscosity increases by 50%, the value of the new slope of hydraulic grade line will be (a) larger by 25% (b) smaller by 50% (c) larger by 100% (d) larger by 50% The velocity proﬁle for laminar ﬂow of water between two parallel plates shown in Fig. 7.17 is given as u = 0.01 [1 – 1000 y2] y u 2.0 cm x Fig. 7.17 where u is in m/s and y is in m. The viscosity of water can be assumed to be 10–3 Ns/m2. The shear stress on each plate will be (a) 2.0 N/m2 (b) 0.002 N/m2 2 (c) 0.04 N/m (d) 0.004 N/m2 7.29 The creeping motion obeys Stokes law up to a critical Reynolds number of value (a) 0.001 (b) 1.0 (c) 100 (d) 2000 7.30 In Coutte ﬂow with zero pressure gradient the shear stress t0 at the boundary is given by UB mU (b) t0 = (a) t0 = m B mB B (c) t0 = (d) t0 = U m where B = gap between the plates 7.31 In plain Coutte ﬂow (i.e., with zero pressure gradient) if B = gap between the plates, then the discharge per unit witdh of the plates is given by UB UB (b) q = (a) q = m 2 (c) q = UB 4 (d) q = 2UB Boundary Layer Concepts Concept Review 8 Introduction A boundary layer u y U in Y U U U Y U U u = 0.99 U u = 99 U Y u u y y d u d d d y d x Laminar flow region Fig. 8.1 Transition zone Turbulent flow region (a) Nominal and displacement thicknesses Fig. 8.2 Boundary Layer Growth As u Ux Rex = n u (b) Relative magnitudes of d, d and q Boundary Layer Thickness d U y q d U v 254 Fluid Mechanics and Hydraulic Machines 5 Rex = 5 ¥ turbulent boundary layer y u Nominal thickness d x U y u U Displacement thickness d d = Ú d 0 uˆ Ê ÁË 1 - ˜¯ d y U Momentum thickness q q= Ú d 0 u Ê uˆ Á 1 - ˜¯ d y U Ë U Energy thickness d δ d = Shape factor H u ∫0 U () 2 ⎛ u ⎞ 1 − dy ⎜⎝ U ⎟⎠ H = d /q 8.1 BOUNDARY CONDITIONS For a laminar boundary layer, the boundary conditions are: 1. At the wall y = 0, u = 0 and v = 0. 2. At the outer edge y = d, u = U. 3. Shear stress at the wall, t0 = m ∂u ∂y Êdp ˆ ÁË d x is + ve˜¯ in which U = f (x). These are beyond the scope of this book. The boundary layer thickness d and the local shear stress t0 are functions of x. 8.2 y=0 The ﬂow over a ﬂat plate which is described in this section is a particular case in which U = constant dp or the pressure gradient is zero. This case is also dx known as zero pressure gradient ﬂow. There are situations in which the pressure Êdp ˆ gradient can be favourable Á is - ve˜ or adverse Ë dx ¯ LAMINAR BOUNDARY LAYER OVER A FLAT PLATE For laminar ﬂow over a ﬂat plate, Blasius solved the basic boundary layer equations and obtained analytical solution which have been veriﬁed experimentally to be remarkably accurate. The classic Blasius solution for laminar bound-ary layer are: 5.0 d = x Rex (8.1) 255 Boundary Layer Concepts where From the boundary conditions for a laminar boundary layer, Ux n Rex = rU 2 2 where Cf = local shear stress coefﬁcient. t0 = Cf By deﬁning Cf = we have (8.3) Rex If the total drag force on one side of a plate of length L and width B is deﬁned as, FD = B then CDf = where ReL = Ú 0 t 0 d x = CDf ( L ◊ B) rU 2 1.328 ReL 2 (8.4) UL and CDf = total frictional drag n coefﬁcient. 8.3 KARMAN MOMENTUM INTEGRAL FORMULATION Putting This is an approximate but simple method of solving boundary layer equations. By the application of momentum principle to a steady boundary layer over a ﬂat plate it can be shown that, rU 2 Putting = 1 Cf 2 = ∂ Ê ∂ x ÁË mU b (8.7) d Equating the two expressions for t0, dd mU b rU 2 a = dx d m 1 Êbˆ or ddd = dx r U ÁË a ˜¯ Integrating with the boundary condition (x = 0, d = 0) x d = 2( b / a ) Rex Ú d 0 u Ê u ˆ ˆ ∂q dy = 1U ÁË U ˜¯ ˜¯ ∂ x (8.5) u = f (h) where h = y/d, U 1 2 ∂d f (h) (1 – f (h)) dh t0 = r U ∂x 0 then Ú 1 f (h) (1 – f (h)) dh = a 0 2 t0 = r U a ∂d ∂x (8.6) 2( b / a ) d = x (8.8) Rex Substituting the value of d in Eq. (8.7) for t0, and simplifying, t0 1 (8.9) Rex The drag force on one side of the plate, for a plate of unit width is Cf = Ú Let df (h) dh h = 0 = b t0 = or t0 mU È df (h) ˘ d ÍÎ d h ˙˚h = 0 = 0.664 L Ê ∂u ˆ t0 = m Á ˜ Ë ∂y ¯ y = 0 (8.2) FD = rU 2 / 2 Ú L 0 = 2 ab t 0 dx FD/(0.5 rU 2L) = CDf = 2ab / ReL (8.10) Example 8.5 illustrates in detail the use of this method for a speciﬁc f(h). In Table 8.1, some of the commonly adopted forms of u/U = f (h) and the corresponding boundary layer parameters d, Cf, CDf, obtained by using the Karman momentum integral equations, are given. After 256 Fluid Mechanics and Hydraulic Machines Table 8.1 (y/d = h) u/U = f(h) (d/x) Rex Exact (Blasius) (d/x) Rex Cf Rex CDf ReL 5.00 1.729 0.664 1.328 5.84 1.752 0.686 1.372 4.64 1.740 0.626 1.292 2h – h2 5.48 1.826 0.730 1.460 sin Ê p hˆ Á ˜ 4.80 1.741 0.654 1.308 h 3.46 1.730 0.577 1.154 3 2h – 2h + h 4 3 1 h - h3 2 2 Ë2 ¯ studying the Example 8.5, the reader is advised to derive all the elements listed in Table 8.1 as a good exercise. onwards the ﬂow of the boundary layer will be turbulent. 8.6 TURBULENT BOUNDARY LAYER 8.4 BOUNDARY CONDITIONS FOR A PROPER f(h) A proper function u/U = f(h) must satisfy the following essential and desirable boundary conditions: Essential At the wall, y = 0; u = 0 At y = d; u = U Desirable 2 ∂ u/ ∂y2 = 0 ∂u/ ∂y = ∂2u/ ∂y2 = 0 8.5 TRANSITION FROM LAMINAR BOUNDARY LAYER As the ﬂow passes down the plate, i.e. as Rex increases the boundary layer thickness increases and soon it becomes unstable. Turbulence persists and grows in the boundary layer at higher values of Rex. It is generally believed that the transition from laminar to turbulent boundary layer takes place between Rex = 1.3 ¥ 105 and 4 ¥ 106, with the mean value of Rex = (Rex)crit = 5 ¥ 105 taken as the commonly accepted critical Reynolds number. In a ﬂow past a long plate, the initial part in the boundary layer up to xcrit will be laminar and then The turbulent boundary layer will have much more steeper velocity gradients at the boundary than the laminar boundary layer. The velocity distribution is logarithmic and could be conveniently expressed in the form of a power law, u/U = (y/d)1/n over a range of Reynolds number. The power n can be 5 to 10, depending on the Reynolds number range. Next to the boundary, in a turbulent boundary layer over a smooth bed, there exists a thin layer called as laminar sublayer. For Rex between 5 ¥ 106 and 2 ¥ 107 the velocity distribution can be expressed by the 1/7 power law, u/U = (y/d)1/7. The turbulent boundary layer characteristics found by experiments and analytical calculations, to be valid for 5 ¥ 105 < Rex < 2 ¥ 107, are d/ x = 0.377/Re 1x/ 5 (8.11) Cf = 0.059/Re 1x/ 5 (8.12) CDf = 0.074/Re 1L/ 5 (8.13) The above formulae assume the boundary layer to 257 Boundary Layer Concepts be turbulent from x = 0. To account for initial laminar boundary layer, CDf can be calculated by CDf = (0.074/Re 1L/ 5) - (1700 /ReL ) If correction for initial laminar boundary layer is applied then, CDf = 0.455 /(log ReL ) 2.58 - 1700 /ReL (8.14) For higher Reynolds numbers (107 < Rex < 109), the logarithmic form of velocity distribution in the turbulent boundary layer is more appropriate. The boundary layer shear coefﬁcients are expressed by the following formulae given by Schlichting: Cf = 0.370 /(log Rex ) 2.58 (8.15) CDf = 0.455 /(log ReL ) 2.58 (8.16) (8.18) The term 1700/ReL is so small that omitting it does not cause any appreciable error. In Fig. 8.3, the values of CDf are plotted against the Reynolds number ReL as obtained from various equations, e.g. 8.4, 8.13, 8.14 and 8.16 for various regimes of ﬂow. 8.7 LAMINAR SUBLAYER The boundary layer thickness d is estimated by The laminar sublayer is usually very thin and its thickness d ¢ is found by experiments to be d/ x = 0.22 /Re 1x/ 6 d ¢ = 11.6 n /u (8.17) (8.19) 0.007 Equation 8.13 Equation 8.16 0.006 r Tu nt le bu 0.005 B. 0.004 Cdf L. Blasius Equ. (8.4) (Exact) Equation (8.14) 0.003 Tran s ition 0.002 La m ina rB .L 0.001 10 5 10 6 . 10 7 10 UL ReL = u Fig. 8.3 8 10 9 10 10 258 Fluid Mechanics and Hydraulic Machines where u = t 0 /r = shear velocity. CDf = 1/ (1.89 + 1.62 log ( L /e )) 2.5 If the roughness magnitude of a surface e is very small compared to d ¢, i.e. e 0.25 (a) Smooth boundary d¢ Datum e/d¢ > 6 (b) Rough boundary Smooth and Rough Surfaces If the laminar sublayer thickness d ¢ is very small compared to roughness height e, (i.e. e >> d ¢), in such ﬂows viscous effects are not important and the boundary is said to be hydrodynamically rough. Usually e/d¢ > 6 is taken as the criterion for hydrodynamically rough boundaries. In the region 0.25 < e/d ¢ < 6, the boundary is in the transition regime and both viscosity and roughness control the ﬂow. Rough ﬂat plate: For ﬂow on a completely rough ﬂat plate the local friction coefﬁcient Cf and total drag coefﬁcient CDf are given by Cf = 1/ ( 2.87 + 1.58 log ( x /e )) ESTABLISHMENT OF FLOW IN A PIPE When a ﬂow enters a pipe from a reservoir a boundary layer forms in the pipe at the entrance. The thickness of the boundary layer in the radial direction grows along the length of the pipe till it merges at the centre line at a distance Le known as entrance length. The ﬂow is uniform beyond Le. The establishment length in the laminar and turbulent ﬂow is given by the following formulae: In laminar ﬂow: Le /D = 0.07 Re (8.22) In turbulent ﬂow: Le /D = 50 (8.23) 8.9 Laminar sublayer Fig. 8.4 8.8 5 (8.20) (8.21) BOUNDARY LAYER SEPARATION 8.9.1 Separation Phenomenon The ﬂow past a ﬂat plate held parallel to the ﬂow is a case of boundary layer with zero pressure gradient. Flows in converging boundaries are examples of favourable pressure gradient and ﬂows in diverging conduits or diverging boundaries are examples of adverse pressure gradient ﬂows. In adverse pressure gradient boundary layer ﬂow the boundary layer may at some section leave the boundary. This is called as separation and downstream of the separation section turbulent y y y u s S = Separation point Separation streamline Wake region Negative velocity Note: du = 0 dy s Fig. 8.5 Separation of Boundary Layer 259 Boundary Layer Concepts eddies exist and this disturbed region is called as a wake (Fig. 8.5). Separation can take place in both laminar and turbulent boundary layers. The location of the separation section on the surface of a body and the size of the wake have important bearing on the total drag force experienced by the body. At the separation point, the shear stress is zero and the velocity gradient ∂u/∂ y = 0. Figure. 8.6 shows some commonly used boundary layer control methods. 1 8.9.2 Control of Separation Separation of ﬂow from the boundary leads to inefﬁciency of the ﬂow unit. In the lifting surfaces such as aerofoils, it may cause reduction of lift and even stalling. Diffusers, conduit transitions, pump and turbine blades and aerofoils are some common ﬂow units where separation may impair the performance. Common procedures to control separation are based on the following methodologies: 2 3 Fig. 8.6 Different arrangements for boundary Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples 8.1 u y = d U Solution: (i) Displacement thickness The displacement thickness d is given by d d = d Ê Ê u ˆˆ Ú ÁË1 - ÁË U ˜¯ ˜¯ 0 dy 260 Fluid Mechanics and Hydraulic Machines u y Given = ◊ . Hence d = U d d Ú 0 Ê Ê yˆˆ ÁË1 - ÁË d ˜¯ ˜¯ dy Putting h = Also when y = 0, h = 0 and when y = d, h = 1. y Putting h = , dy = d dh. d Also when y = 0, h = 0 and when y = d, h = 1. 1 1 Now Ú 1 Ê 1 1ˆ d = d Á - ˜ = Ë 2 4¯ 4 Ú d = d 1 – 1 d = 2 2 8.2 (ii) Momentum thickness The momentum thickness q is given by d q= uÊ u 1 3˘ È3 = Í h- h ˙ U 2 2 Î ˚ y u = . Hence q = d U d yÊ Ê yˆˆ Ú d ÁË1 - ÁË d ˜¯ ˜¯ dy 0 1 Ú q = d h h(1 – h) dh 0 1 È h 2 h3 ˘ =d Í - ˙ 3 ˙˚ ÍÎ 2 0 Ê 1 1ˆ q = dÁ - ˜ = Ë 2 3¯ 6 (iii) Energy Thickness The energy thickness d is given by d d 2 u Ê Ê uˆ ˆ = Á1 ˜ dy U Ë ÁË U ˜¯ ¯ 0 Ú u y Given = . U d Solution: The displacement thickness d is given by d d = Hence d 2 y Ê Ê yˆ ˆ 1 = Á ˜ dy d Ë ÁË d ˜¯ ¯ 0 Ú Ê Ê u ˆˆ Ú ÁË1 - ÁË U ˜¯ ˜¯ dy 0 u 1 ˘ È3 = Í h - h3 ˙ . U 2 2 ˚ Î 1 Noting that dh = dy. d y = 0, h = 0 and when y = d, h = 1, When It is given: 1 3 1 ˆ Ê d = d Á1 - h + h3 ˜ dh Ë 2 2 ¯ Ú 0 1 Now È 3h 2 1 4 ˘ + h ˙ d = d Íh 4 2 ˙˚ ÍÎ 0 d = 3d 8 Momentum thickness The momentum thickness q is given by d d y d d y Putting h = , dy = d dh. d Also when y = 0, h = 0 and when y = d, h = 1. Now h= Ê u ˆˆ Ú U ÁË1 - ÁË U ˜¯ ˜¯ dy 0 Given 1 È h2 h4 ˘ Now d = d h (1 – h2) dh = d Í ˙ 4 ˚˙ ÍÎ 2 0 0 È h2 ˘ d = d (1 - h) dh = d Íh ˙ 2 ˙˚ ÍÎ 0 0 y , dy = d dh. d q = uÊ Ê u ˆˆ Ú U ÁË1 - ÁË U ˜¯ ˜¯ dy 0 261 Boundary Layer Concepts Given y u = . Hence q = d U For given Ú 0 y Ê Ê yˆˆ 1dy d ÁË ÁË d ˜¯ ˜¯ u 1 ˘ È3 = Í h - h3 ˙ U 3 ˚ Î2 1 Now d q =d Ê 3h Ú ÁË 2 0 - 1 3ˆ Ê 3h 1 3 ˆ h 1+ h ˜ dh 2 ˜¯ ÁË 2 2 ¯ Ê y 1 Ê y ˆ 2ˆ = 2Á - Á ˜ ˜ Ëd 2Ëd ¯ ¯ y Substituting h = , dy = d dh, d u = (2h – h2) U Displacement thickness d d = Ê 3h 9 2 3 4 1 3 ˆ - h + h - h Á 2 4 4 2 ˜ q =d Á ˜ dh 3 4 1 6 ˜ Á + h h – 0Á ˜¯ 1/ 4 Ë 4 1 Ú 0 Ú Substituting uˆ Ê ÁË1 - U ˜¯ dy = d 1 Ú (1 - h) dh u = (2h – h2) U 1 Ú 2 d = d (1 - 2h + h ) dh 3 1 3 3˘ È3 3 - + - ˙ q =d Í - + 4 4 20 8 20 28 Î ˚ 0 1 È h3 ˘ d d = d Íh - h 2 + ˙ = d 3 ˙˚ ÍÎ 3 0 Momentum thickness 39 q= 280 8.3 d - q = uÊ uˆ Ú U ÁË1 - U ˜¯ dy. 0 Substituting yˆ Ê t = t0 Á - ˜ Ë d¯ u = (2h – h2) U 1 Ú q = d ( 2h - h 2 ) ◊ (1 – 2h + h2) dh 0 1 d Ú 2 3 4 q = d ( 2h - 5h + 4h - h ) dh Solution: 0 1 du yˆ Ê t = t0 Á 1 - ˜ = m dy Ë ¯ d t0 Ê y2 ˆ d y◊ Á md Ë 2 ˜¯ Hence u= At y = d, u = U and hence t Ê d 2 ˆ t0d = U = 0 Ád 2 md Ë 2 ˜¯ 2m t0 Ê y2 ˆ u 2m d y ¥ = md ÁË 2 ˜¯ U t0d È 2 5 4h 4 h5 ˘ q = d Íh 2 - h3 + ˙ = 15 3 4 5 ÍÎ ˚˙ 0 8.4 u/U H y/d) Solution: The shape factor H = d /q Putting u/U = f(h) = h1/7 where h = y/d d = d Ú 1 0 (1–h1/7)dh 262 Fluid Mechanics and Hydraulic Machines 1 7 È ˘ = d Íh - h8 / 7 ˙ = d/8 8 Î ˚0 q=d Ú 1 Hence all essential and one desirable boundary conditions are satisﬁed. (b) u = U [1 + (y/d) – 2(y/d)2] h1/7(1 – h1/7)dh ∂u È1 4y ˘ = U Í - 2˙ ∂y Îd d ˚ 0 1 7 7 È7 ˘ = d Í h8 / 7 - h 9 / 7 ˙ = d 72 9 Î8 ˚0 = U [– 4/d 2 ] ∂ y2 1 Ê 72 ˆ = 1.286 8 ÁË 7 ˜¯ H = d /q = ∂2u At y = 0, u = U π 0 ∂2u = – 4U /d 2 π 0 ∂ y2 8.5 At y = d, u = 0 π U ∂u = –3U/d π 0 ∂y Êp y ˆ u/U = sin Á Ë d ˜¯ u/U h ∂2u h ∂y2 h = y/d None of the essential and desirable boundary conditions is satisﬁed. Hence this is not a proper velocity distribution in a laminar boundary layer. Solution: The requisite boundary conditions are: Essential Desirable 2 At y = 0, u = 0, At y = d, u = U, ∂ u =0 ∂y2 ∂u ∂2u = =0 ∂y ∂ y2 Ê py ˆ (a) u/U = sin Á ˜ Ë 2d ¯ ∂2u ∂y 2 = - U p2 4d 2 Ê py ˆ sin Á ˜ Ë 2d ¯ At y = 0, u = U sin (0) = 0 2 ∂ u ∂y2 =– Up 2 4d 2 At y = d, u = U sin (p/2) = U (0) = 0 ∂u Ê Up ˆ = Á cos (p/2) = 0 Ë 2d ˜¯ ∂y 2 ∂ u ∂y2 =– Up 2 4d 2 sin (p/2) π 0 8.6 u/U = a c ∂u Up Ê py ˆ = cos Á ˜ ∂y 2d Ë 2d ¯ = – 4U/d 2 π 0 bh h = y/d d ch dh3 a, b, Solution: Let u/U = f (h) The requisite boundary conditions are: At y = 0, i.e. h = 0, u = 0 i.e. f(h) = 0 ∂2u = 0 i.e. f ≤(h) = 0 ∂y2 u = U i.e. f(h) = 1 At y = d, i.e. h = 1, ∂u = 0 i.e. f ¢(h) = 0 ∂y ∂2u ∂y2 = 0 i.e. f ≤(h) = 0 263 Boundary Layer Concepts f(h) = a + bh + ch 2 + dh3 f ¢(h) = b + 2ch + 3dh 2 f ≤(h) = 2c + 6dh Here Therefore at h = 0, f(h) = 0 f ≤(h) = 0 at h = 1, f(h) = 1 i.e. a i.e. c i.e. a + b + c + d i.e. b+d i.e. b + 2c + 3d i.e. b + 3d f ¢(h) = 0 Solving and =0 =0 =1 =1 =0 =0 dd 37 rU 2 (i) dx 315 From the boundary conditions for a laminar boundary layer Therefore, Solution: y d )3 h = y/d ∂ Ê t0 = rU ∂x ÁË Ú d 0 y d)4 dd dx Ú 1 0 f (h) (1 – f(h)) dh Since at \ 1 0 = x2 1260 m x = 34.05 ( rU x /m ) 37 rU d 5.835 where Rex = pU x/m = x Re x d2 = 2 mU ( rUx /m ) 5.835 x m1/ 2U 3 / 2 r1/ 2 t0 = 2mU/d = f (h) (1 – f(h))dh Ú 630 m x + constant 37 rU x = 0, d = 0, constant = 0 Shear stress: Substituting the value of d in the second expression for t0, But, Ú 1 0 (2h – 2h3 + h4) (1 – 2h – 2h3 + h4)dh (ii) d 2/2 = u = f(h) = 2h – 2h3 + h4 U Substituting, 2 t 0 = rU t0 = 2m U/d Equating the two expressions for t0, 2mU dd 37 rU 2 = d dx 315 630 m dx d dd = 37 rU On integration, uÊ uˆ ˆ 1 - ˜ dy ˜ Á Ë U U¯ ¯ put y/d = h, dy = d dh and the limits of h are 0 and 1. f(h) = 2h – 2h2 + h4 È df (h) ˘ = [2 – 4h + 4h 3]h = 0 = 2 Í dh ˙ Î ˚h = 0 By Karman momentum integral equation 2 t0 = Ê ∂u ˆ uU È df (h) ˘ t0 = m Á ˜ = Ë ∂y¯ y = 0 d ÍÎ d h ˙˚h = 0 Since where (2h – 4h2 – 2h3 + 9h4 – 4h5 + 4h6 = 37/315 \ y d 0 – 4h7 – h8)dh 8.7 u U Ú 1 1 d= 2 3 1 u = h - h3 U 2 2 Thus 3 b= 2 = = 0.3428 \ t0 = 0.6855 x1/ 2 rU 2 Ê m ˆ 2 ÁË rUx ˜¯ 1/ 2 264 Fluid Mechanics and Hydraulic Machines t0 i.e. ( rU 2 / 2) = Cf = Substituting the given expression of 0.6855 Re x Force on one side, FD: Consider a plate of unit width and length L. FD = Ú L 0 t 0 dx = rU 2 2 Ú L 0 0.6855 ( rU x / m )1/ 2 t0 = 1 rU 2 L 2 = CDf = 1.3710 ( rU L /m )1/ 2 = Re L ( rU / 2) u where È3 Ê y ˆ 1 Ê y ˆ3˘ u = Í Á ˜ - Á ˜ ˙◊ U ÍÎ 2 Ë d ¯ 2 Ë d ¯ ˙˚ d = = 3 13U ◊ 2 280 vx m1/ 2U 3 / 2 r1/ 2 x1/ 2 rU 2 Ê m ˆ = 0.646 2 ÁË rUx ˜¯ 1.371 t0 8.8 13U 280 vx t0 = 0.323 2 mU 3 ◊ 2 dx rU 2 0.6855 = (2L1/2) 2 ( rU /m )1/ 2 FD 280 v x in the above, 13U d = = Cf = Rex = 1/ 2 0.646 Rex rUx m 8.9 280v x 13U Cf x=L Solution: Solution: Putting y h = , the velocity distribution is d 1 ˘ È3 u = Í h - h3 ˙ = f(h). 2 ˚ U Î2 t0 = UL 6.0 ¥ 0.45 = n 0.9 ¥ 10 -4 = 3.0 ¥ 104 ReL = (Note: 1 stoke = 10–4 m2/s) Since ReL is less than Re(crit) = 5.0 ¥ 105, the boundary layer is laminar. Using the Blasius’ results, Also the boundary conditions are y = 0, h = 0 and y = d, h = 1 Also dy = d dh Ê du ˆ mU d f (h) t0 = m Á ˜ = d dh Ë dy ¯ y = 0 Reynolds number at the trailing edge is (i) Boundary layer thickness: h=0 3 mU mU È 3 1 ˘ - ◊ 3h 2 ˙ = ◊ Í 2 d d Î2 2 ˚h = 0 d = x 5.0 Rex At the trailing edge x = L = 0.45 m 5.0 ¥ 0.45 5.0 ¥ 0.45 = ReL 3 ¥ 10 4 = 0.01299 m = 1.3 cm dL = 265 Boundary Layer Concepts (ii) Shear stress at the trailing edge, tL tL = rU 2 rU Ê 0.664 ˆ Á ˜ 2 Ë ReL ¯ 2 2 Cf (L) = 0.925 ¥ 1000 0.664 ¥ (6 ) 2 ¥ = 2 3 ¥ 10 4 = 63.8 N/m2 (iii) Drag on one side of the plate FD = CDf (L◊B) rU 2/2 whereCDf = CDf = Cf = local friction coefﬁcient = 1.875 ¥ 10 t0m = Cf 1.328 4 = 7.667 ¥ 10–3 4 Rex = 4.849 ¥ 10–3 FD = (7.667 ¥ 10 ) ¥ (0.45 ¥ 0.15) 0.925 ¥ 1000 ¥ 62 2 = 8.617 N On both sides of the plate drag force F2D = 2 ¥ FD = 2 ¥ 8.617 = 17.23 N ¥ rU 2 2 = 4.849 ¥ 10–3 ¥ –3 0.664 0.664 Shear stress ReL 3 ¥ 10 1.875 ¥ 10 4 = 0.0456 m = 4.56 cm = 1.328 5.0 ¥ 1.25 d m = d at midpoint = 0.8 ¥ 1000 ¥ (1.5)2 2 = 4.364 N/m2 (b) At the trailing edge, x = L = 2.5 m 1.5 ¥ 2.5 = 3.75 ¥ 104 10 -4 ReL < ReL(crit) = 5 ¥ 105 The boundary layer is laminar ReL = 8.10 dL = 5.0 L ReL = 5.0 ¥ 2.5 3.75 ¥ 10 4 = 0.0645 m = 6.45 cm v –4 Solution: (a) At the centre of the plate x = 1.25 m U = 1.5 m/s Rex = 1.5 ¥ 1.25 10 -4 = 1.875 ¥ 104 This is less than Re(crit) = 5 ¥ 105 and hence the boundary layer is laminar. d = x 5.0 Rex Cf = 0.664 = ReL 0.664 3.75 ¥ 10 4 = 3.429 ¥ 10–3 and shear stress at trailing edge t0L is t0L = Cf rU 2 2 = (3.429 ¥ 10–3) ¥ (0.8 ¥ 1000)(1.5) 2 2 = 3.086 N/m2 (c) Total force (on both sides of the plate) F = CDf ¥ (area) ¥ rU 2/2 266 Fluid Mechanics and Hydraulic Machines CDf = 1.328 ReL 1.328 = 3.75 ¥ 10 8.12 4 = 6.858 ¥ 10–3 F = (6.858 ¥ 10–3) (2 ¥ 2.5 ¥ 2.0) ¥ (0.8 ¥ 1000) ¥ (1.5) 2 2 = 61.72 N Power required to tow the plate P =F¥U = 61.72 ¥ 1.5 = 92.58 W Solution: Let x be the distance from the leading edge such that the drag force in distance x is half of the total drag force. FDx = 8.11 L is ? L 3 rair ¥ nair –5 Solution: The maximum length of plate corresponds to the critical Reynolds number which can be taken as, ReL(crit) UL = = 5 ¥ 105 n 3.0 ¥ L = 5 ¥ 10 1.45 ¥ 10 -5 L = 2.417 m For this length of plate in a laminar sublayer Thus, CDf 1.328 = = ReL = 1.878 ¥ 10 Drag force on one side of the plate FD = CDf ¥ area ¥ Ê Lˆ \ Á ˜ Ë x¯ 1/ 2 ◊ Ê xˆ ÁË L ˜¯ x 1 = L 2 1/ 2 = 1 2 i.e. x= u Êyˆ = Á ˜ Ëd¯ U 5 ¥ 105 1/ 7 t0 –3 n ˆ rU ÊÁ Ë Ud ˜¯ t0 rU 2 2 2 ¥ (3.0) 2 1 L. 4 8.13 1.328 = 1.878 ¥ 10–3 ¥ (1.5 ¥ 2.417) ¥ = 0.03677 N rU 2 2 rU 2 FDL = CDfL (B L) 2 CDf x x FD x 1 = ◊ = \ FD L 2 CDf L L 1.328 But from CDfx = (Ux / n ) 1.328 CDfL = (UL / n ) = CDfx (Bx) or 5 1 FDL 2 2 C Solution: Putting d ReL h = y/d, u/U = f(h) = h1/7 Cf 1/ 4 Rex 267 Boundary Layer Concepts (i) By Karman momentum integral equation (see Example 8.7) dd t0 = rU dx 2 = rU 2 dd dx (ii) Force on one side of a plate of unit width and length L 1 Ú f(h)[1 – f(h)]dh L FD = 0 1 Ú Ú t 0 dx = 0 h1/7(1 – h1/7)dh 0 dd 7 = rU 2 dx 72 It is given that FD (1) Ê n ˆ t0 = 0.0228 rU2 Á Ë U d ˜¯ rU 2 2 1/ 4 Ênˆ d 5/4 = 0.2931 Á ˜ ËU ¯ = CDf = CDf = (2) Equating (1) and (2), Ênˆ d1/4dd = 0.2345 Á ˜ ËU ¯ On integrating Re1L/ 5 Ú 0 0.0555 1 dx Ê Ux ˆ 5 ÁË n ˜¯ where ReL = UL n 8.14 1/ 4 dx 1/ 4 x+C Ênˆ t0 = 0.0228rU 2 ÁË ˜¯ U ¥ (3) (0.375 x )1/ 4 1/ 5 Ê v ˆ = 0.02775rU 2 Á ˜ Ë Ux ¯ –1/5 ¥ Reynolds number UL 2 ¥ 20 = 4 ¥ 107 = n 1 ¥ 10 -6 The boundary layer is turbulent at the trailing edge. Ê Ux ˆ ÁË n ˜¯ Local shear stress coefﬁcient (0.002775) ¥ 2 t0 Cf = = Re 1x/ 5 rU 2 2 ) Solution: n ReL = 1/ 4 1 3 r d 0.375 = Thus 1/5 x Ê Ux ˆ ÁË n ˜¯ Substituting Eq. (3) in Eq. (2), = 0.0555 Rex L 4 0.0555 Ê 5 5 ˆ L Á ˜ 1 Á4 ˜¯ Ë (U / v ) 5 0.0694 Using the boundary condition d = 0 when x = 0, C = 0 ( rU 2 2 1 / 20 (4) (i) Taking the critical Reynolds number Re(crit) = 5 ¥ 105, Ux crit 2 ¥ x crit = 5 ¥ 105 = n 1 ¥ 10 -6 xcrit = 0.25 m = 25 cm Laminar boundary layer exists in the ﬁrst 25 cm of the plate. (ii) At xcrit, d = xcrit 5.0 Re( crit ) 268 Fluid Mechanics and Hydraulic Machines The boundary layer thickness at the edge of the laminar boundary layer dc = 5.0 ¥ 0.25 5 Reynolds number, UL 1.75 ¥ 5 = n 1.475 ¥ 10 -5 = 5.932 ¥ 105 ReL = = 1.768 ¥ 10–3m 5 ¥ 10 = 1.768 mm At the trailing edge, as the Reynolds number is > 107, the thickness of the turbulent boundary layer is obtained by putting x = L = 20 m in the equation (i) For laminar boundary layer Drag coefﬁcient CDf = 0.370 (log Rex )2 . 58 Here x = L = 20 m and Rex = ReL = 4 ¥ 107 Cf = 0.370 = 1.9742 ¥ 10–3 (log 4 ¥ 107 ) 2. 50 rU 2 to = Cf 2 = 1.9742 ¥ 10–3 ¥ rU 2 2 = 1.724 ¥ 10–3 ¥ (1.8 ¥ 5.0) ¥ 1.22 ¥ (1 .75) 2 2 = 0.029 N (ii) For a completely turbulent boundary layer Since ReL is between 5 ¥ 105 and 107, the 1/7th power law-is applicable. Thus the drag coefﬁcient CDf = 0.074 998 ¥ 22 2 Re1L/ 5 = 0.074 (5.932 ¥ 105 )1/ 5 = 5.183 ¥ 10–3 Drag force on one side of the plate rU 2 2 = 5.183 ¥ 10–3 ¥ (1.8 ¥ 5.0) FD = CDf ¥ area ¥ 8.15 3 5.932 ¥ 105 FD = CDf ¥ (area) ¥ = 3.94 Pa ReL 1.328 = = 1.724 ¥ 10–3 0.22 dt = x ( Re x )1/ 6 0.22 ¥ 20 dt = = 0.238 m ( 4 ¥ 107 )1/ 6 = 23.8 cm (iii) The shear stress coefﬁcient in the turbulent boundary with Re > 107 is Cf = 1.328 ¥ –4 ¥ 1.22 ¥ (1.75) 2 2 = 0.0871 N 8.16 3 Solution: –4 –5 m = 1.8 ¥ 10 poise = 1.8 ¥ 10 Pa.s v = m/r = 1.8 ¥ 10–5/1.22 = 1.475 ¥ 10–5 m2/s ¥ —5 269 Boundary Layer Concepts Solution: Wind velocity Reynolds number 20 ¥ 1000 = 5.56 m/s 3600 Reynolds number U= UL 3.0 ¥ 3.0 = = 9 ¥ 10–6 n 1.0 ¥ 106 The boundary layer is turbulent and the drag coefﬁcient appropriate to this ReL = 9 ¥ 106 is ReL = UL 5.56 ¥ 6 = 2.22 ¥ 106 = n 1.5 ¥ 10 -5 The boundary layer is turbulent and the appropriate drag coefﬁcient is the one corresponding to the power law, viz. ReL = CDf = = 0.074 1700 Re1L/ 5 Re L 0.074 6 1/ 5 - CDf = = = 0.003215 ¥ (6.0 ¥ 1.0) ¥ 1.2 ¥ (5.56) 2 2 = 0.3578 N The turbulent boundary layer thickness at the trailing edge is given by putting x = L = 6.0 m in 6 1/ 5 FDf = CDf ¥ area ¥ 6 rU 2 FD = CDf ¥ area ¥ 2 1700 Re1L/ 5 Re L 0.074 1700 - (9.00 ¥ 10 ) 9.00 ¥ 106 = 2.82 ¥ 10–3 Drag force due to skin friction 1700 ( 2.22 ¥ 10 ) 2.22 ¥ 10 = 0.003215 Drag force on one side of the plate per unit metre width: 0.074 rU 2 2 = 2.82 ¥ 10–3 ¥ 3.5 ¥ 998 ¥ (3) 2 2 = 44.3 N Total measured drag = Skin friction drag + wave drag \ 70.0 = 44.3 + FDW Wave drag FDW = 70.0 – 44.3 = 25.7 N 8.18 0.377 d = x Re1L/ 5 d =L¥ 0.377 Re1L/ 5 = 6.0 ¥ 0.377 ( 2.22 ¥ 106 )1/ 5 = 0.1217 m = 12.17 cm Solution: Consider a distance x from the leading edge. The total drag force on the plate of length x on one side of it is 8.17 rU 2 2 Similarly, the drag force on one side of a plate of length L is FDx = CDfx (Bx) r 3 n ¥ Solution: The wetted surface of the model is considered as an equivalent ﬂat plate 3.0 m long and having a surface area of 3.50 m2 FDL = CDfL (BL) rU 2 2 270 Fluid Mechanics and Hydraulic Machines FDx CDfx x = FDL CDfL L However, as the boundary layer is turbulent, the plate is smooth and Rex ª 106, and Thus and CDfx CDfL CDfx /CDfL FDx/FDL = 0.074/Rex1/5 = 0.074/ReL1/5 = (L/x)1/5 = (L/x)1/5 (x/L) rU 2 2 = 0.007096 ¥ (2 ¥ 2 ¥ 10) FDf = CDf ¥ area ¥ ¥ Power 2 If x = L, as in the present case, FDx = drag 3 on the ﬁrst 2/3 of the plate = F1 and Drag force on both sides of the plate 1020 ¥ (5) 2 2 = 3619 N P = FD ¥ U = 3619 ¥ 5 = 18095 W = 18.1 kW 8.20 F1/FDL = (3/2)1/5 (2/3) = 0.723 F2 = drag on the rear 1/3 of the plate = FDL – F1 F2/FDL = 1 – (F1/FDL ) = 1 – 0.723 = 0.277 Thus F1/F2 = F1 /FDL = 0.723/0.277 = 2.61 F2 /FDL 8.19 3 r Solution: m Reynolds number rUL 1020 ¥ 5.0 ¥ 10.0 = m 0.0018 7 = 2.83 ¥ 10 The boundary layer is turbulent. Relative roughness L/e = 10.0/0.005 = 2000 The drag coefﬁcient for fully rough turbulent boundary layer ﬂow is ReL = CDf = = 1 Lˆ Ê ÁË1.89 + 1.62 log e ˜¯ 1 [r ¥ m Frm = Up Um = Frp = g Lm g Lp Um = Up ( Lm /Lp ) = 10.0 ¥ 1 / 30 = 1.826 m/s 1 = 3.3333 m 30 2 = 0.007096 –3 Solution: The total resistance of the model and the prototype consists of two parts: Surface drag and wave drag. The wave drag follows the Froude law of similarity. The surface drag being dependent on Reynolds number is not modelled. It has to be estimated separately for the model as well the prototype. Let subscripts m and p stand for the model and prototype respectively. Froude number Lm = 100 ¥ 5 (1.89 + 1.62 log 2000) 2. 5 ? 3 È1˘ Am = 1600 ¥ Í ˙ = 1.778 m2 Î 30 ˚ 271 Boundary Layer Concepts Reynolds number: ReLp = rU p Lp = m = 9.58 ¥ 108 ReLm = 1025 ¥ 10 ¥ 100 1.07 ¥ 10 -3 \ 1025 ¥ 1.826 ¥ 3.333 1.07 ¥ 10 -3 = 5.83 ¥ 106 The surface resistance in the model Fsm is ﬁrst calculated. Since ReLm = 5.83 ¥ 106 the boundary layer is turbulent and CDfm This Fwm is modelled by Froude’s law. Hence (Wave drag)p = Fwp = (Fwm)/(rrL3r) Since rm = rp’ rr = 1.0 Fwp = = 294705 N (1 / 30)3 = 294.7 kN = Prototype wave drag For the prototype, the surface drag Fsp is 2 9 Since ReLp = 9.58 ¥ 10 0.074 1700 = 1/ 5 Re ReLm Lm 0.074 1700 = 6 1/ 5 (5.83 ¥ 10 ) 5.83 ¥ 106 CDfp = = 0.455 (log ReLp) 2. 58 - 1700 ReLp 0.455 8 2. 58 (log 9.58 ¥ 10 ) - 1700 9.58 ¥ 108 = 1.577 ¥ 10–3 rU m2 = CDfm ¥ (area)m ¥ 2 = 2.99 ¥ 10–3 ¥ 1.778 ¥ rU p2 Fsp = CDfp ¥ (area)p ¥ = 2.99 ¥ 10–3 Fsm 10.915 1025 ¥ (10) 2 2 = 129,314 N = 129.3 kN Total prototype drag Fp = Fwp + Fsp = 294.7 + 129.3 = 424.0 kN \ 1025 ¥ (1.826) 2 2 = 9.085 N (Wave drag)m = (Total measured drag)m – (surface drag)m Fwm = 20.0 – 9.085 = 10.915 N Fsp = 1.577 ¥ 10–3 ¥ 1600 ¥ Problems 8.1 Calculate the displacement thickness and momentum thickness in terms of d, the nominal boundary layer thickness, for the following velocity distributions. (a) u/U = 2h – 2h3 + h4 u (b) = h1/7 U y where h = d (Ans. (a) d = 0.3d; q = 0.117d (b) d = 0.125d; q = 0.0977d) 8.2 For the velocity proﬁle u/U = sin (py/2d), calculate the shape factor H. How does it compare with the shape factor of the laminar boundary layer obtained by the exact solution of Blasius? Ê d Á Ans. H = q = 2.662; Ë ˆ H Blasius = 2.604˜ ¯ 8.3 If the velocity distribution in a turbulent u boundary layer is expressed as = U 272 Fluid Mechanics and Hydraulic Machines 8.4 8.5 8.6 8.7 8.8 8.9 (y/d)1/m, show that d/d = 1/(m + 1) and q/d = m/(m + 1) (m + 2). For the laminar boundary layer over a ﬂat plate the velocity distribution is given by u/U = a + bh + ch2, where h = y/d. Determine the coefﬁcients a, b and c. (Ans. a = 0, b = 2, c = –1; u/U = 2h – h2) For a laminar boundary layer on a ﬂat plate, derive the expressions for d, Cf and CDf when u/U = f(h) given by (a) f(h) = 3h/2 – h3/2 (b) f(h) = 2h – h2 (c) f(h) = sin (ph/2) (d) f(h) = h (Ans. See Table 8.1) A laminar boundary layer on a ﬂat plate has a velocity distribution given by u/U = 2h – 2h3 + h4 where h = y/d. At a location the boundary layer thickness is 1.6 cm and the free stream velocity is 1.25 m/s. If m = 1.136 ¥ 10–5 Pa.s, calculate the shear stress at that location. (Ans. t0 = 1.775 ¥ 10–3 Pa) A thin plate 2 m ¥ 2 m is placed edgewise in a ﬂow of oil. Calculate the boundary layer thickness and the shear stress at the trailing edge when the free stream velocity is 2 m/s. (RD of oil = 0.85 and v = 10–5 m2/s). (Ans. t0L = 1.785 Pa) A sharp edged ﬂat plate 1.5 m along the direction of ﬂow and 3 m across is placed parallel to the ﬂow of air. Find the drag on one side of the plate. Also ﬁnd d, d and q at the trailing edge for ﬂow of air at 3 m/s past the plate. (r = 1.23 kg/m3 and v = 1.45 ¥ 10–5 m2/s) (Ans. d = 0.01346 m, d = 0.023 m, q = 1.788 ¥ 10–3 m, FD = 0.0594 N) If the boundary layer over a ﬂat plate, kept parallel to the ﬂow, is laminar ﬁnd the ratio of the skin friction drags on the front half of the plate to the rear half. Ê ˆ F1 ÁË Ans. F = 2.414˜¯ 2 8.10 A smooth ﬂat plate 1.5 m wide and 2 m long at a uniform velocity of 2 m/s. Find (a) the extent of the laminar boundary layer on the plate, (b) the thickness of the boundary layer at the edge of the laminar boundary layer and at the trailing edge and (c) the shear stress at the trailing edge. (r = 998 kg/m3, v = 1 ¥ 10–6 m2/s). (Ans. xcrit = 25 cm, dc = 1.768 mm, dL = 3.6 cm, t0 = 5.631 Pa) 8.11 A thin plate is moving in a direction parallel to its length in still air at a velocity of 4.0 m/s. The length of the plate is 0.5 m and width is 0.6 m. Taking vair = 1.5 ¥ 10–5 m2/s and rair = 1.25 kg/m3, calculate (a) the boundary layer thickness at the end of the plate, (b) shear stress at 20 cm from the leading edge and (c) drag force on one side of the plate. (Ans. d = 6.847 mm, t0 = 0.02875 Pa, FD = 0.0109 N) 8.12 Find the ratio of friction drags on the front half and rear half of a plate kept in a stream at zero angle of incidence. Assume the boundary layer to be turbulent over whole plate and the Reynolds number to be of the order of 106. Ê ˆ F1 ÁË Ans. F = 1.349˜¯ 2 8.13 A smooth ﬂat plate is kept at zero angle of incidence in a stream and the Reynolds number in terms of the length of the plate is of the order of 106. At what fraction of the total length, measured from the leading edge, would the drag force on the front portion would be equal to half of the 273 Boundary Layer Concepts total drag force on the plate? Assume the boundary layer to be turbulent over the whole plate. x Ê ˆ ÁË Ans. L = 0.420˜¯ 8.14 Wind at 100 km/h blows along a long ﬂat surface 50 m long. Estimate (i) the thickness of the boundary layer at distances of 5 m and 50 m from the leading edge, and (ii) shear stress at 5 m and 50 m from the leading edge. [r = 1.22 kg/m3 and v = 1.5 ¥ 10–5 m2/s] (Ans. d5 = 7.62 cm, d50 = 51.7 cm; t05 = 1.123 Pa, t050 = 0.8236 Pa) 8.15 A barge has a rectangular bottom 30 m long and 10 m wide. Calculate (i) the frictional force on the bottom when the barge moves at a velocity of 1.5 m/s, and (ii) the thickness of the boundary layer and the shear stress at the trailing edge. (r = 998 kg/m3 and v = 1 ¥ 10–6 m2/s). (Ans. (i) FD = 722 N; (ii) d = 33.3 cm, t0 = 1.952 Pa) 8.16 A train is 250 m long and its surface area of top, sides and bottom add up to 15 m2 per metre length of the train. If the train moves at a speed of 120 km/h, calculate the power required to overcome surface resistance. The surfaces can be assumed to be smooth. Take rair = 1.2 kg/m3 and mair = 1.80 ¥ 10–5 Pa.s (Ans. P = 140.7 kW) 8.17 The length of a submarine is 80 m and its surface area is 3000 m2. If the submarine is moving with a velocity of 4 m/s, determine the frictional drag, considering (a) the boundary layer to be turbulent over the entire surface (b) the turbulent boundary layer is preceded by a laminar boundary layer (c) the surface of the submarine is rough with roughness height e = 3 mm. [m = 1.0 ¥ 10–3 Pa.s and r = 1040 kg/m3] (Ans. (a) FD = 45.13 kN, (b) FD = 45.003 kN, (c) FD = 101.02 kN) 8.18 In a ﬂow over a smooth ﬂat plate the Reynolds number at the trailing edge is 106. If the critical Reynolds number is 5 ¥ 105, what fraction of the total frictional force occurs in the laminar boundary layer? (Ans. Flam = 31.6% of the total drag) 8.19 An air stream ﬂows over a smooth ﬂat plate with a terminal Reynolds number of 106. The boundary layer can be assumed to be turbulent over the entire plate. If the length of the plate is increased by 10%, keeping all other factors same, what is the percentage change in the (a) total drag coefﬁcient and (b) total drag force? (Ans. (a) 1.9% decrease in CDf; (b) 7.92% increase in FD) 8.20 In a turbulent boundary layer over a ﬂat plate it is found that u/U = (y/d )1/7 and Cf = 0.02 Re d–1/6. By using Karman momentum integral equation obtain expressions (i) for d and Cf in terms of Rex and (ii) for CDf in terms of ReL [Red = U d/n, Rex = U x/n, and ReL = UL/n] Ê 0.027 0.031 Á Ans. Cf = 1/ 6 , CDf = 1/ 6 , Re x Ë Re L 0.16 ˆ d = 1/ 7 ˜ x Re x ¯ 8.21 A 1/25 model of an ocean-going ship was towed in a towing tank containing fresh water. The model had the same Froude number as the prototype. The model had a length of 3 m, wetted surface area of 2 m2 and was towed at a speed to reproduce the prototype speed of 9 m/s. What is the 274 Fluid Mechanics and Hydraulic Machines prototype drag corresponding to a measured total drag of 16 N in the model? [For sea water: rs = 1025 kg/m3 and ms = 1.07 ¥ 10–3 Pa.s rw = 998 kg/m3 mw = 1.00 ¥ 10–3 Pa.s]. (Ans. FDp = 186.1 kN) For fresh water and Objective Questions 8.1 The nominal distance of a boundary layer is deﬁned as the distance from the wall to a point (a) where the velocity is 99% less than the asymptotic limit (b) where the velocity ceases to be laminar (c) where the velocity is within 90% of the asymptotic limit (d) where the velocity is 99% of its asymptotic limit 8.2 The displacement thickness of a boundary layer is (a) The distance to the point where u/U0 = 0.99 (b) The distance where u = u where u = shear velocity (c) The distance by which the main ﬂow is to be shifted from the boundary to maintain the continuity equation (d) One half of the actual thickness of the boundary layer 8.3 In a two-dimensional boundary layer over a ﬂat surface (a) The longitudinal pressure gradient is important and transverse pressure gradient can be neglected (b) The transverse pressure gradient is important and longitudinal pressure gradient can be neglected (c) Both the longitudinal and transverse pressure gradients can be neglected (d) Both the longitudinal and transverse pressure gradients are important 8.4 The displacement thickness d of a boundary layer is deﬁned as d = (a) Ú d Ú d Ú dÊ Ú dÊ 0 (b) u U uˆ Ê ÁË1 - U ˜¯ dy u /U dy 0 (c) 0 (d) 0 uˆ ÁË1 - U ˜¯ dy 2 uˆ ÁË1 - U ˜¯ dy 8.5 A laminar boundary layer has a velocity distribution given by u/U = y/d. The displacement thickness d for this boundary layer is (a) d (b) d/2 (c) d/4 (d) d /6 8.6 If the velocity distribution in a laminar boundary layer can be assumed as u/U = y/d the ratio of momentum thickness q to nominal thickness d is given by q/d = (a) 1/2 (b) 1/3 (c) 1/6 (d) 1.25 8.7 The following boundary conditions exist at the wall (y = 0) in a boundary layer. (a) u = U (b) dp/dx = – ve (c) t0 = 0 (d) u = 0, v = 0 8.8 If the velocity distribution in a laminar boundary layer over a ﬂat plate is to be expressed as u/U = sin (Ap y/d) where d = thickness of the boundary layer, the appropriate value of A is 275 Boundary Layer Concepts 8.9 8.10 8.11 8.12 8.13 (a) 1.0 (b) – 1/2 (c) 1/2 (d) 2 In a boundary layer developed along the ﬂow, the pressure decreases along the downstream direction. The boundary layer thickness would (a) tend to decrease along the ﬂow (b) remain constant (c) increase rapidly along the ﬂow (d) increase gradually along the ﬂow What is the ratio of displacement thickness to momentum thickness for linear velocity distribution in a laminar boundary layer along a ﬂat plate? (a) 1.5 (b) 2.0 (c) 2.5 (d) 3.0 If the velocity distribution in a turbulent boundary layer is u/U = (y/d )1/10 the displacement thickness d is given by d/d = (a) 0.3 (b) 0.125 (c) 0.091 (d) 0.0758 In a boundary layer at a certain location the boundary layer thickness d = 0.5 cm, displacement thickness d = 0.15 cm and the momentum thickness q = 0.0585 cm. The shape factor H at this location is (a) 2.564 (b) 0.39 (c) 3.333 (d) 8.547 In a laminar boundary layer the shear stress t0 at a location x is given by t0 = (a) m (c) m ∂u ∂x ∂u ∂y (b) m y=0 (d) m y =d 8.14 In a laminar boundary thickness varies with distance x as (a) x–1/2 (b) (c) x (d) ∂u ∂y y=0 ∂u ∂y y=x layer the nominal the longitudinal x–1/5 x1/2 8.15 In a laminar boundary layer over a ﬂat plate the ratio of shear stresses t1 and t2 at two sections 1 and 2 at distances from the leading edge such that x2 = 5 x1, is given by t1/t2 = 1 5 (c) 1.0 (d) 5.0 If d 1 and d2 denote boundary layer thicknesses at a point distance x from the leading edge when the Reynolds numbers are 100 and 256 respectively, then the ratio of d1/d2 will be (a) 0.625 (b) 1.6 (c) 2.56 (d) 4.90 The mean drag coefﬁcient CDf for a laminar boundary layer over a ﬂat plate was found to be 0.015. If all other ﬂow factors remain the same and the length of the plate is decreased to 1/4 of its original value, the drag coefﬁcient CDf would be equal to (a) 0.015 (b) 0.060 (c) 0.030 (d) 0.0075 In the case of ﬂow over a ﬂat plate the growth of the boundary layer d/x (a) decreases with an increase in the kinematic viscosity (b) increases with an increase in the free stream velocity (c) decreases with an increase in the free stream velocity only if the boundary layer is laminar (d) increases with an increase in the kinematic viscosity in both laminar and turbulent boundary layers. A ﬂuid with kinematic viscosity n ﬂows in laminar stage along a ﬂat plate with free stream velocity of U. At a distance x from the leading edge, the Reynolds Ux number is given as Re = . The v thickness of the boundary layer at x will be (a) 8.16 8.17 8.18 8.19 5 (b) 276 Fluid Mechanics and Hydraulic Machines proportional to 1/2 8.20 8.21 8.22 8.23 8.24 –1/2 (a) x Re (b) x Re (c) Re1/2/x (d) Re–1/2/x A ﬂat plate with a sharp leading edge is placed along a free stream of ﬂuid ﬂow. The local Reynolds number at 3 cm from the leading edge is 105. What is the thickness of the boundary layer? (a) 0.47 mm (b) 0.35 mm (c) 0.23 mm (d) 0.12 mm If the velocity u in a turbulent boundary layer varies as y1/7, the growth of the boundary layer thickness d/x varies as (a) Re x–1/5 (b) Re x–1/2 –4/5 (c) Re x (d) Re x–1 The rate of growth of the boundary layer thickness with the longitudinal distance on a ﬂat plate (a) is faster when the boundary layer is laminar than when it is turbulent (b) is the same whether the boundary layer is laminar or turbulent (c) is faster in a turbulent boundary layer when compared to that in a laminar boundary layer (d) depends only on the aspect ratio of the plate In a turbulent boundary layer on a ﬂat plate the velocity distribution is given by the 1/7th power law. If the boundary layer over the whole plate is turbulent, the ratio of the shear stresses t1 and t2 at two sections 1 and 2 at distances from the leading edge x2 = 5 x1 is given by t1/t2 = (a) 1.379 (b) 1.258 (c) 2.236 (d) 1.308 The ratio of the coefﬁcient of friction drag in laminar boundary layer compared to that in turbulent boundary layer is proportional to (a) R 1/2 (b) R 1/5 L L 3/10 (c) R L (d) R 3/10 L 8.25 The laminar sublayer exists (a) only in laminar boundary layers (b) in all turbulent boundary layers (c) only in smooth turbulent boundary layers (d) only in rough fully developed turbulent boundary layers 8.26 On a ﬂat plate, point A is at the mid-section and point B is at the trailing edge. The shear stresses tA at A and tB at B are such that (a) tA > tB (b) tB > tA (c) tA = tB (d) tA = tB if the boundary layer is laminar 8.27 The laminar sublayer is (a) a boundary layer that occurs before the formation of a regular laminar boundary layer (b) is the ﬁrst 1/8 of a laminar boundary layer, next to a boundary (c) is a region where the wall roughness predominates (d) is a region next to the wall where the laminar motion persists while the rest of the ﬂow is turbulent 8.28 In a hydrodynamically smooth surface the roughness magnitude e and laminar sublayer thickness d ¢ are related as (a) e /d ¢ > 1.0 (b) e /d ¢ < 0.25 (c) e /d ¢ ≥ 6.0 (d) e /d ¢ = 1/30 8.29 The thickness of laminar sublayer d ¢ is given by (a) 11.6 u/n (b) u/(11.6n) (c) 11.6 n/u (d) n/u 8.30 In a boundary layer ﬂow the parameter u e/n was equal to 1.5. The boundary can be classiﬁed hydrodynamically as (a) rough (b) in transition from rough to smooth (c) in transition from smooth to rough (d) smooth 277 Boundary Layer Concepts [Here u = shear velocity, e = equivalent roughness] 8.31 In a boundary layer ﬂow the parameter u e/n was equal to 12. The boundary can be classiﬁed hydrodynamically as (a) smooth (b) rough (c) in transition (d) unstable 8.32 A turbulent boundary layer occurs over a ﬂat plate practically right from the leading edge. (a) it is possible for the surface to be hydrodynamically rough upstream and yet hydrodynamically smooth downstream. (b) it is possible for the surface to be hydrodynamically smooth upstream and yet hydrodynamically rough downstream. (c) if the ﬂow is rough on the upstream it will have to be rough on the downstream also. will behave as rough plate everywhere. 8.33 The separation of boundary layer takes place when the pressure gradient is (a) negative (b) positive (c) zero (d) constant 8.34 At the point of separation (a) velocity is negative (b) shear stress is zero (c) shear stress is maximum (d) pressure gradient is zero 8.35 Separation of boundary layer takes place when (a) (∂ u/∂y)y = 0 > 0 (b) (∂2u/∂ y2)y = 0 > 0 (c) (∂ u/∂y)y = 0 = 0 (d) (∂ u/∂y)y = d > 0 8.36 The separation of a boundary layer occurs when (a) the ﬂow is accelerated past a boundary (b) the boundary layer comes to rest (c) any adverse pressure is encountered (d) the ﬂuid is ideal 8.37 It is required to have laminar ﬂow in a 6 cm diameter pipe at a Reynolds number of 1500. The entrance length required for fully developed laminar ﬂow to exist in the pipe is (a) 105 cm (b) 630 cm (c) 257 cm (d) 900 cm 8.38 An 8 cm diameter pipe is to carry water at a Reynolds number of 105. The entrance length required for the establishment of turbulent ﬂow is about (a) 4.0 m (b) 560 cm (c) 10.0 m (d) 8.7 m 8.39 In the ﬂow of a ﬂuid past a circular cylinder the angle q from the front stagnation point to the location at which separation takes place is about (a) 82° if the boundary layer is laminar (b) 82° if the boundary layer is turbulent (c) 110° if the boundary layer is laminar (d) 90° whether the boundary layer is laminar or turbulent. Drag and Lift on Immersed Bodies Concept Review 9 Introduction Lift (lateral force) Resultant force ds p t 0 V0 Drag V0 drag lift 9.1 DRAG The total drag (sometimes called as proﬁle drag) of a body is made up of two parts: frictional drag and pressure drag (or form drag), depending on whether the shear stresses or pressure differences cause it. The relative proportion of these two parts depends upon the shape of the body and ﬂow condition. In practical applications, it is the total drag rather than its constituent components that is important. Hence Fig. 9.1 the total drag FD on an immersed body in a relative free stream velocity Vo is expressed as FD = CD A where rV 02 2 (9.1) CD = total drag coefﬁcient A = a characteristic area of the body In general CD = fn (geometry, Reynolds number, Froude number, Mach number). 279 For incompressible ﬂow and in the absence of free surface effects C D = f n (geometry, Re) The variation of the drag coefﬁcient for certain interesting and commonly used objects is discussed in the following sub-sections. The characteristic area A is one of the following, as per convention: 9.1.1 (1) A = frontal area = projected area on a plane normal to V0. This is used for blunt (bluff) shaped objects such as spheres, cylinders, cars, trains, projectiles and missiles. (2) A = planform area = area of the body as seen from above. This is used for thin, ﬂat surfaces where frictional forces are predominant as for example: ﬂat plate ﬂow, airplane wings and hydrofoils. (3) A = wetted area. Customarily this is used for water crafts such as boats, barges and ships. Sphere The variation of the drag coefﬁcient CD with VD Reynolds number Re = 0 where D = diameter of v the sphere, is illustrated in Fig. 9.2. The variation is studied in three regimes: For Re < 1.0, the ﬂuid motion is known as creeping motion. The total drag is given by Stokes’ law, as (i) Very Small Reynolds Numbers Writing FD = 3pD m V0 (9.2) Ê pD 2 ˆ rV02 FD = CD Á ˜ Ë 4 ¯ 2 (9.3) 2 10 8 6 4 2 10 8 6 4 2 Disk Stokes law: CD = 24/Re V CD 1 8 6 4 V D Sphere 2 10 D –1 8 6 4 2 –2 10 –1 4 6 8 2 ´ 10 1 2 4 6 8 10 2 4 6 8 10 2 2 4 68 10 3 2 4 68 Reynolds number, Re = VD/n Fig. 9.2 D 10 4 4 6 8 10 5 2 4 68 10 6 280 Fluid Mechanics and Hydraulic Machines CD = 24 r 24 = V0 D /m Re (9.4) Stokes’ law is valid for Re £ 1.0 and its application to determine fall velocities of small particles is given in Examples 9.1 to 9.4. 1 £ Re < 2 ¥ 105 � In this regime, increasing Reynolds number reﬂects decreasing role of viscosity and CD drops from a value of 24 at Re = 1 to about 0.4 at Re ª 104. Then onwards, CD is essentially constant, and at around Re = 105, CD = 0.5. The region 104 < Re < 2 ¥ � � 105 marks laminar boundary layer ﬂow. In the range 1 < Re < 100, CD can be approximated by An interesting feature of ﬂow past twodimensional bodies is the formation and alternate release of vortices behind the cylinder for Re > 30. This vortex shedding leads to lateral vibration of two-dimensional bodies. The frequency of vortex shedding n is expressed as Strouhal number (ii) Reynolds Number CD = 24 Ê 3 ˆ 1+ R e˜ Á Ë ¯ Re 16 1/ 2 (9.5) (iii) Transition Region (Re ª 2 ¥ 105) Around Re = 2 ¥ 105 the laminar boundary layer undergoes transition to turbulent boundary layer. Consequently, the boundary layer separation line which occurred at around 85° angle shifts downstream to an angular position of about 120° causing a reduction in wake area. This is reﬂected in a rapid reduction in the value of CD from 0.5 to 0.2. The critical Reynolds number depends upon boundary roughness and free stream turbulence, and an average value of 2 ¥ 105 is generally adopted. (iv) Reynolds Number 5 Re > 2 ¥ 10 The boundary layer is turbulent and the drag coefﬁcient assumes an essentially constant value of 0.2. S= (9.6) For cylinders, the value of Strouhal number S can be estimated by the formula 20 ˆ Ê S = 0.20 Á1 Ë Re ˜¯ for 250 < Re < 2 ¥ 10–5. 9.1.3 (9.7) Miscellaneous Bodies The drag coefﬁcients of a variety of commonly used three-dimensional and two-dimensional body shapes are given in Table 9.1. Table 9.1 Form of body D L/D Hemisphere: Hollow upstream Hollow downstream 3. Ellipsoid (1:2, major axis along ﬂow) 4. Circular cylinder (axis along ﬂow) Circular cylinder (axis normal to ﬂow) Re 5 Sphere 9.1.2 Cylinder A cylinder is a typical example of a two-dimensional bluff body. The nature of variation of CD with Reynolds number Re is essentially similar to that of the sphere. The values of CD are, however, different and depend upon L/D also in addition to Re. For L/D ª , CD = 1.20 in laminar boundary layer region (Re ª 105) and it drops to 0.33 at Re > 5 ¥ 105. nD V0 CD 10 > 3 ¥ 105 0.50 0.20 103 > 103 1.33 0.34 2 ¥ 105 0.07 0 1 2 4 7 103 1.12 0.91 0.85 0.87 0.99 1 5 20 105 5 0.63 0.74 0.90 1.20 0.35 > 5 ¥ 105 0.33 (Contd.) 281 Table 9.1 (Contd. For incompressible ﬂow Form of body 6. Rectangular plate (L = length, D = width) L/D Re CD 1 5 20 103 1.16 1.20 1.50 1.90 1.12 103 Circular disk CL (or CD ) = f n (a , Rec ) where a = angle of attack and VC Rec = Reynolds number = 0 where C = Chord n length. 9.2.1 9.2 LIFT The lift force FL, which occurs normal to the direction of relative motion V0 is expressed as FL = CL A rV 02 CL = Lift coefﬁcient and A = a characteristic area. For lifting bodies such as airfoils, hydrofoil vanes, the practice is to have large surface areas with small thickness and large chord lengths (Fig. 9.3). For such lifting bodies the characteristic area A in the deﬁnition of the lift coefﬁcient is the planform area A = Ap = CL. Note that the drag coefﬁcients for lifting bodies area also deﬁned in terms of Ap. Vc = tangential velocity on the circumference of the cylinder = wr The circulation where FL Ap = planform area = CL Chordline FD a C Chor L an sp d The lift force FL = LrV0G Expressing FL = CL = r (Area) where or V 02 2 , Area = LD, LrV0 G = CLr LD CL = 2G/ V0 D = V 02 2 2pDVc 2pVc = DV0 V0 2 sin q = - Fig. 9.3 Thus FL Lift coefficient CL = Ap ◊ ( rV 02 / 2) (for a lifting body) G = 2prVc (9.12) The production of lift due to a rotating body is known as Magnus effect. The resulting streamline pattern round such a rotating cylinder is shown in Fig. 9.4 (b, c and d). It will be noticed from Fig. 9.4(b) that the stagnation points S1 and S2 are not at 0 and 180° to ﬂow direction, but are at angle q > 180°. The value of q is given by Midline (camber line) a = angle of attack Rotating Cylinders For a cylinder of radius r rotating with an angular velocity w in an ideal ﬂuid ﬂowing with a velocity V0: (9.8) 2 (9.11) or (9.9) FD Drag coefficient CD = (9.10) (for a lifting body) Ap ◊ ( rV 02 / 2) Vc V0 Ê 1 Vc ˆ q = sin -1 Á Ë 2 V0 ˜¯ Thus the two stagnation points merge and occur at q = 270° when Vc = 2V0, (Fig. 9.4(c)). For Vc > 2V0, the stagnation point is removed from the cylinder 282 Fluid Mechanics and Hydraulic Machines w a q S (a) Flow past a non-rotating cylinder (ideal fluid flow) (c) Single stagnation point (V c = 2 V 0 ) FL w w q S a S2 S1 (b) Flow past a rotating cylinder (Vc < 2V0) (d) Stagnation point away from the cylinder (Vc > 2V0) Fig. 9.4 surface and a ring of ﬂuid is dragged around the cylinder, (Fig. 9.4(d)). The above are theoretical results. In practice, due to viscosity, there will be considerable variation in CL and q values. 9.2.2 Lift and Stall in an Aerofoil In a lifting surface such as an aerofoil, at small angles of attack the rounded leading edge prevents ﬂow separation. However, the sharp trailing edge causes ﬂow separation. This separation at the tail generates a vortex which in turn causes realignment of stream lines resulting in lift. The lift increases with the angle of attack up to a limit of about 15°–20°. At this limit, the ﬂow separates completely from the leading edge and the aerofoil is said to be stalled. At the stalling point, the lift drops off markedly and the drag increases very signiﬁcantly making the aerofoil incapable of ﬂying (Fig. 9.5). 283 Stall CL Angle of attack Fig. 9.5 A L Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples Drag 9.2 ¥ 9.1 D D 3 F r 4 < Re £ 3 ¥ Re > 3 ¥ D D 3 r m Solution: FD = CD A rV 2 2 = 1.2 ¥ (0.5 ¥ 1.2) ¥ = 2250 N = 2.25 kN ¥ – Solution: 1000 ¥ ( 2.5) 2 2 Velocity 100 ¥ 1000 3600 = 27.78 m/s V = 284 Fluid Mechanics and Hydraulic Machines Buoyant force Fb Reynolds number Re = rVD 1.2 ¥ 27.78 ¥ 0.0713 = m 1.8 ¥ 10 -5 Drag 5 = 1.32 Re < 3 ¥ 105, CD = 0.5 Since rs Drag force on the ball = FD = CD A rV 2 2 Vot rf p 1.2 ¥ ( 27.78) 2 = 0.5 ¥ ¥ (0.0713)2 ¥ 4 2 = 0.924 N Weight Fig. 9.6 9.3 D r 3 m ¥ – Solution: rV 2 2 For a parachute with terminal fall velocity the CD of a hemisphere with concave frontal surface is appropriate. Hence take CD = 1.33. The total weight is balanced by the drag. FD = CDA Ê pˆ W = FD = 1.33 ¥ Á ˜ Ë 4¯ (7.0)2 ¥ 1.2 ¥ (5.0) 2 2 drag (see Fig. 9.6), i.e. p 3 p ◊D rs g – D3r f g = FD 6 6 (a) If the Reynolds number is very small (Re < 1) Stokes’ law is applicable and FD = 3pDmV0t. where V0 t = terminal velocity. p 3 D (rs – rf) g = 3pDmV0t 6 D 2 g( rs - rf ) V0t = m 18 D 2 (g s - g f ) 18 m where g f = rf g = speciﬁc weight of the liquid and gs = r sg = speciﬁc weight of the sphere. = Hence terminal velocity V0t = D 2 (g s - g f ) 18 m and this is valid for V0t D £ 1.0 n = 767.8 N = Say 767 kN 9.4 D r r (b) In general FD = CD Ar f where D Solution: At terminal velocity, since there is no change in the velocity, the net force on the body is zero. Hence, (weight of cylinder – buoyant force) = Re = V 02t 2 CD = drag coefﬁcient and A = frontal area pD 2 = for a sphere. 4 Ê pD 2 ˆ V 02t p Hence D3 (rs – r f) g = CD Á r f ˜ 6 2 Ë 4 ¯ 285 1/2 and on solving D = 9.5 1 D2 (gs – g f ) 18 m Substituting for D, V0t = 3 r [ m rf V0 t Also the fall velocity È 4 gD Ê rs ˆ˘ - 1˜ ˙ V0t = Í Á ¯ ˙˚ ÍÎ 3 C D Ë rf 4 D Re > 3 ¥ n ¥ < Re < 3 ¥ –6 D V0t = 1 m2 (gs – g f ) 18 rf2V 02t m V 03t = ˆ 1 Ê m ˆÊgs - 1˜ g 18 ÁË rf ˜¯ ÁË g f ¯ D Solution: Assume Re > 3 ¥ 105. Then CD = 0.20. At terminal velocity V0t, submerged weight = Drag rf V 02t p 3 D (gs – g f ) = FD = CDA 6 2 3 p D (g s - g f ) V 02t = 3 (p D 2 /4) rf CD = Ê 1 ¥ 10 -3 ˆ Ê 2.65 ¥ 998 ˆ 1 ¥ 9.81 ¥ Á - 1˜ ˜Á ¯ 18 998 Ë 998 ¯ Ë = 9.01 ¥ 10–7 V0t = 9.66 ¥ 10–3 m/s = 9.66 mm/s = ˆ 4 D Ê rs g - 1˜ 3 CD ÁË rf ¯ = 4 0.2 Ê 998 ¥ 2.8 ˆ ¥ ¥ 9.81 Á - 1˜ Ë 998 ¯ 3 0.2 D = V 02t = 23.544 m 1 ¥ 10 -3 = rf V0 t 998 ¥ 9.66 ¥ 10 -3 = 1.0373 ¥ 10–4 m = 0.1037 mm 9.7 V0t = 4.852 m/s V0 D 4.852 ¥ 0.2 = n 1 ¥ 10 -6 = 9.70 ¥ 105 > 3 ¥ 105 Hence the initial assumption is correct and V0t = 4.852 m/s. 3 Re = Solution: Assuming the validity of Stokes’ law, Fall velocity V0t = 9.6 1 D 2 (g s - g f ) 18 m (i) Using Eq. (i) (r 3 m ¥ –3 Solution: Stokes’ law is valid up to Re = 1.0. For maximum size particles that obey Stokes’ law, r V ◊D (Re)max = f 0 t = 1.0 m m = = D 2 (g s - g f ) 18V0 t ( 2 ¥ 10 -3 ) 2 ( 2.60 ¥ 998 ¥ 9.81 - 917 ¥ 9.81) 18 ¥ 0.0125 = 0.293 Pa.s 286 Fluid Mechanics and Hydraulic Machines Reynolds number 9.9 rV0t D Re = m = 917 ¥ 0.0125 ¥ 2 ¥ 10 -3 0.293 = 0.078 < 1.0 Hence the assumed Stokes’ law is valid. 3 r 9.8 Solution: V0 = 60 km/h = 16.67 m/s 4 D (a) FD = Drag force = CD A ¥ Re > 3 ¥ D (16.67) 2 = 93.33 N 2 Power = FD ¥ V0 = 93.33 ¥ 16.67 = 1555.6 W = 1.556 kW – ¥ 3 r Solution: Assume Re > 3 ¥ 105 and hence CD = 0.2. (b) Power = CD A Ê pD (rair – r helium) g Á ˜ = W + FD Ë 6 ¯ FD = CD A rV 2 2 Ê pD 3 ˆ (1.2 – 0.2) ¥ 9.81 ¥ Á ˜ Ë 6 ¯ rV 03 = 1555.6 2 1555.6 ¥ 2 V 03 = = 5401.4 0.3 ¥ 1.6 ¥ 1.2 V0 = 17.545 m/s = 63.16 km/h 3ˆ But 2 = 0.35 ¥ 1.6 ¥ 1.2 ¥ 3 r m rV 02 9.10 Ê (3.0) 2 ˆ pD 2 = 210 + Á 0.2 ¥ ¥ 1.2 ¥ 2 ˜¯ 4 Ë 5.1365 D3 = 210 + 0.8482 D2 By trial and error, Diameter of the balloon = D = 3.5 m V A Reynolds number rVD 1.2 ¥ 3.0 ¥ 3.5 Re = = 7 ¥ 105 = m 1.8 ¥ 10 -5 Corresponding CD = 0.2. Hence the assumption is OK. [ D D r Solution: 3 D Refer Fig. 9.7. FD = Drag force on a cup = CD A rV 02 2 287 10 cm V0 V0 Wind Wind N M 10 cm 50 cms 45 cm 12 cms diameter Fig. 9.8 Fig. 9.7 Since both the cups are subjected to same V0, torque about the vertical axis, T = (FD1 – FD2) ¥ r = (CD1 – CD2) A where r = lever arm rV 02 2 ¥r p ¥ (0.10) 2 4 1.2 0.45 ¥ V 02 ¥ 2 2 = 1.04968 ¥ 10–3 V 02 = (1.33 – 0.34) ¥ (a) When V0 = 45 km/h = 12.5 m/s Torque = T = 1.04968 ¥ 10–3 ¥ (12.5)2 = 0.164 N.m (b) When V0 = 60 km/s = 16.67 m/s T = 1.04968 ¥ 10–3 ¥ (16.67)2 = 0.2916 N.m 9.11 p D2 r (V0 – 0.25 w)2 4 2 p D2 r (V0 + 0.25 w)2 FDM = 0.34 ¥ 4 2 Torque = (FDN – FDM) ¥ r = 0 FDN = FDM (as r π 0) 1.33 (V0 – 0.25 w)2 = 0.34 (V0 + 0.25 w)2 FDN = 1.33 ¥ i.e. \ Ê 1.33 ˆ ÁË 0.34 ˜¯ D r 3 Solution: Let w = steady angular velocity in a wind velocity of V0. For constant w, the net torque about the axis of rotation is zero. V0 = 60 km/h = 16.67 m/s v = velocity of the cup = wr = 0.25 w Relative velocity of the concave cup N = V0 – 0.25w Relative velocity of the convex cup M = V0 + 0.25w CD for N = 1.33 CD for M = 0.34 Drag force 1/ 2 (V0 – 0.25 w) = V0 + 0.25w 0.9778 V0 = 0.7445 w w = 1.31346 V0 = 1.31346 ¥ 16.667 = 21.89 rad/s 60w Rotations per minute, rpm = p ( 2r ) 60 ¥ 21.89 = p ¥ 0.50 = 836 rpm 288 Fluid Mechanics and Hydraulic Machines [Note: Notice the difference in calculating the drag in this and the previous problem. Here the cups are rotating and hence the relative velocity is used. In Example 9.7 the cups are held stationary and as such the relative velocity is equal to the velocity of wind.] [ 9.13 D ¥ 3 r ¥ n – Solution: 9.12 3 [r [ 4 D Re ¥ ¥ – < Re £ ¥ n D D Solution: V0 = 80 km/h = 22.22 m/s VD 22.22 ¥ 0.05 = 7.4 ¥ 104 Re = 0 = -5 v 1.5 ¥ 10 Hence CD = 1.20 (i) Drag force for unit length of cable: (A = LD) F D = CD A rV02 2 V0 = 80 km/h = 22.22 m/s Reynolds number VD 22.22 ¥ 2.5 Re = 0 = v 1.5 ¥ 10 -5 6 = 3.7 ¥ 10 This is larger than the critical Reynolds number and, as such, CD = 0.33. Force on the chimney rV 2 FD = CD ◊ (LD) 0 2 1.2 ¥ ( 22.22) 2 = 0.33 ¥ (50 ¥ 2.5) ¥ 2 = 12,222 N = 12.222 kN Bending moment at the base L 50 = 12.222 ¥ M0 = F = 305.6 kN.m 2 2 9.14 3 1.2 ¥ (22.22)2 2 = 17.78 N/metre length of cable (ii) Strouhal number = 1.2 ¥ (1.0 ¥ 0.05) ¥ 3 20 ˆ Ê S = 0.20 Á1 Ë Re ˜¯ Ê 20 ˆ = 0.20 Á1 ˜ ª 0.20 7.4 ¥ 10 4 ¯ Ë Since S= nD = 0.20 V0 V0 22.22 = 0.20 ¥ D 0.05 = 88.88, say 89.0. Frequency of vertex shedding n = 0.20 Thus, n = 89 Hz [n ¥ D 4 – ¥ £ Re 3 ¥ D Solution: When the balloon is rising: (a) Assume Re > 3 ¥ 105 so that CD = 0.2 Buoyant force = weight + drag 2 (rair – rHe) g rV 0 p 3 D = 135 + CD A 2 6 289 (1.22 – 0.22) ¥ 9.8 ¥ q = 28.89° = 28° 53¢ 30≤ p ¥ D3 6 Êp ˆ 1.22 ¥ 22 = 135 + 0.2 ¥ Á D 2 ˜ ¥ Ë4 ¯ 2 3 3.672 3.672 = sin q 0.48313 = 7.6 N Tension in the cable = 2 5.1365 D = 135 + 0.3833 D By trial and error D = 3.0 m 2.0 ¥ 3.0 Re = 1.5 ¥ 10 -5 = 4 ¥ 105 > Re (critical) Hence assumed CD = 0.2 is correct. (b) When the Balloon is tethered Referring to Fig. 9.9, V0 = 10 km/h = 2.78 m/s Let q = Inclination of the cable to the horizontal. rV 02 T cos q = Drag force = CD A 2 p T sin q = (rair – rHe) g D 3 – W 6 where T = tension in the cable. D = 3.0 m. Assume CD = 0.2 T sin q = 5.136 D3 – 135 = 3.672 ( 2.78) Êp ˆ T cos q = 0.2 ¥ Á ¥ 32 ˜ ¥ 1.22 ¥ Ë4 ¯ 2 = 6.654 3.672 tan q = = 0.5518 6.654 V0 = 2.78 m/s FD 9.15 3 [r < Re £ 3 ¥ 4 ¥ – Re ≥ 3 ¥ n D D Solution: Neglecting the weight and drag of the string, and referring to Fig. 9.10, T cos q = FD = CDA rV 02 2 T sin q = Wnet \ tan q = Re = As Wnet CD ArV 02 / 2 V0 D 25 ¥ 0.03 = 53571 = n 1.4 ¥ 10 -5 104 < Re < 3 ¥ 105; CD = 0.5 2 Wnet = W = (2.5 ¥ 9790) ¥ p ¥ (0.03)3 6 = 0.346 N p ( 25) 2 ¥ (0.03)2 ¥ 1.25 ¥ 4 2 = 0.138 N 0.346 tan q = = 2.506 0.138 FD = 0.5 ¥ q = tan–1 (2.506) = 68° 14¢ 51≤ w T q Tension in the string = T = = Wnet sin q 0.346 sin (68.246)0 = 0.3725 N Fig. 9.9 290 Fluid Mechanics and Hydraulic Machines Solution: q Velocity V0 = T 200 ¥ 1000 = 55.56 m/s 3600 Lift force on the plane 3 cm Diameter q = FL = CL A FD rV02 2 25000 = CL ¥ 25 ¥ V0 = 25 m/s 1.2 ¥ (55.56) 2 2 CL = 0.54 Wnet 9.18 Fig. 9.10 9.16 Solution: Solution: 2 rV 2 rV 3 Power P = FDV = CDA 2 Let sufﬁxes 1 and 2 refer to the two systems with the same power. V0 = 30 km/h = Drag force = FD = CD A rV13 rV 3 = CD2A 2 2 2 = 0.85 CD1, 30 ¥ 103 = 8.33 m/s 3600 FD = Drag force = CD A CD2 Ê 1 ˆ V2 = Á Ë 0.85 ˜¯ 1/ 3 V1 = 1.05567 V1 Lift = 20790 N = 20.79 kN FL = Lift force = CL A rV 02 2 = 0.60 ¥ (2.0 ¥ 1.5) ¥ 998 ¥ F = 9.17 = r 3 (8.33) 2 2 (8.33) 2 2 = 62370 N = 62.37 kN Resultant force i.e., 5.57% increase in the velocity. 2 = 0.20 ¥ (2.0 ¥ 1.5) ¥ 998 ¥ P = CD1A If rV 02 F D2 + F L2 ( 20.79) 2 ¥ (62.37) 2 = 65.74 kN Power required to tow the plate P = FD ¥ V0 = 20.79 ¥ 8.33 kW = 173.2 kW 291 Drag and Lift on Immersed Bodies Considering the vertical component of T 9.19 A kite is in the form of a rectangular airfoil with a chord length of 60 cm and a width of 45 cm and weights 0.8 N. It is maintained at an angle of 10° to horizontal and the string makes an angle of 30° to the vertical. If the wind speed is 15 km/h and CD is 0.25 (Fig. 9.11) estimate the tension in the string and the rair = 1.2 kg/m3] FL (1.406) cos 30° + 0.8 = CL ¥ (0.6 ¥ 0.45) ¥ 2.0176 = 2.81295 CL CL = 0.717 [Note the planform area A = 0.6 ¥ 0.45 has been used in the calculation of drag and lift forces, as per the convention relating to lifting surfaces.] 9.20 Experiments were conducted in a FD 15 km/h wind tunnel with a wind speed of 10° 30° 1.2 ¥ ( 4.167) 2 2 m wide. The density of air is 1.20 kg/m3. The Kite W lift and drag are 0.75 and 0.15 respectively. Determine (i) the lift force, (ii) drag force, (iii) resultant force, (iv) direction of resultant force, and (v) power expended in overcoming resistance of the plate. T Fig. 9.11 Solution: Solution: V0 = 15 km/h = 4.167 m/s rV 02 Drag force = FD = CD A 2 rV 02 Lift force = FL = CL A 2 Weight of kite = W Resolving the tension T into components in horizontal and vertical directions: T cos 30° + W = FL T sin 30° = FD \ T= FD sin 30∞ 0.25 ¥ (0.60 ¥ 0.45) ¥ 1.2 ¥ ( 4.167) sin 30∞ ¥ 2 Tension in the string T = 1.406 N = 2 V0 = 50 km/h = 13.89 m/s A = planform area = 2 ¥ 1.2 = 2.4 m2 (i) Lift force FL = CLA rV 02 2 = 0.75 ¥ (2.0 ¥ 1.2) ¥ 1.2 ¥ (13.89) 2 2 = 208.3 N (ii) Drag force FD = CD A rV 02 2 = 0.15 ¥ (2.0 ¥ 1.2) ¥ = 41.67 N (iii) Resultant force F = F L2 + F D2 1.2 ¥ (13.89) 2 2 292 Fluid Mechanics and Hydraulic Machines = (i) Circulation G = 2prVc = 2 ¥ p ¥ 0.6 ¥ 13.19 = 49.73 m2/s (ii) Lift force FL = LrV0G = 9.0 ¥ 1.2 ¥ 10.0 ¥ 49.73 = 5371 N = 5.371 kN Lift coefﬁcient V CL = 2p c V0 = 2p ¥ 1.319 = 8.29 (iii) Stagnation point location ( 208.3) 2 + ( 41.67) 2 = 212.43 N (iv) Inclination of F with free stream = q tan q = FL/FD = 5 q = tan–1 5 = 78° 42¢ (v) Power expended P = FD ¥ V 0 = 41.67 ¥ 13.89 = 578.8 W 9.21 Ê 1 Vc ˆ q = sin–1 Á Ë 2 V0 ˜¯ 3 r Ê 1.319 ˆ = sin–1 Á Ë 2 ˜¯ Solution: Vc = tangential velocity due to rotation = – (41.26)° = 318.74° (Stagnation point S2) and q = 180 + 41.26° = 221.26° (Stagnation point S1) (See Fig. 9.4 (b) for direction of measurement of q). pDN p ¥ 1.2 ¥ 210 = = 13.19 m/s 60 60 V0 = 10 m/s Vc 13.19 = = 1.319 V0 10.0 = Problems [Air: ra = 1.20 kg/m3, ma = 1.80 ¥ 10–5 Ns/m2 Water: rw = 998 kg/m3, mw = 1.0 ¥ 10–3 Ns/m2] (Ans. ReD = 267, CD = 0.679) 9.1 Find the terminal settling velocity of a 0.8 mm diameter sediment particle (RD = 2.65) in water (n = 1.03 ¥ 10–6 m2/s). The drag coefﬁcient for 1.0 < Re < 200 can be taken as 3 Ê ˆ CD = 24 Á1 + Re˜ Ë 16 ¯ 1/ 2 /Re (Ans. V0t = 12.7 cm/s) 9.2 If a raindrop of 1 mm diameter has a terminal velocity of 4.0 m/s, (1) what is the Reynolds number of the ﬂow? (2) What is the magnitude of the drag coefﬁcient CD? 9.3 Determine the largest diameter and the corresponding terminal velocity of particle that will obey Stokes’ law in the following cases: (a) Polysterene (RD = 1.05) spheres settling in air (ra = 1.2 kg/m3, na = 1.5 ¥ 10–5 m2/s) (b) Polysterene (RD = 1.05) spheres settling in water (nw = 1 ¥ 10–6 m2/s) 293 (c) Aluminium (RD = 2.8) spheres settling in glycerene (rg = 1260 kg/m3, mg = 1.49 Pa.s) (Ans. (a) Dm = 0.078 mm; V0t = 0.1925 m/s, (b) Dm = 0.332 mm; V0t = 3 mm/s (c) Dm = 1.282 cm; V0t = 9.22 cm/s) The parachute is to be designed for a total load of 1.0 kN. Estimate the minimum size of the parachute. [ra = 1.2 kg/m3; CD for a hemisphere = 0.34 and 1.33 depending on a convex or concave frontal surface to the ﬂow]. 9.4 A 2 mm sphere made of stainless steel (RD = 7.8) is observed to have a fall velocity of 9.5 mm/s in a liquid of density 1260 kg/m3. Estimate the kinematic viscosity of the liquid. (Ans. n = 1.188 ¥ 10–3 m2/s) (Ans. D = 6.5 m) 9.9 A parachute is to lower a box of machinery weighting 1450 N. The vertical component of landing velocity is not to exceed 4.0 m/s. How many parachutes of 6 m diameter are to be used for the purpose? Assume CD = 1.33 for the parachutes. [ra = 1.22 kg/m3]. (Ans. 4 parachutes) 9.10 A 1940 model car has a drag coefﬁcient CD = 0.95. (a) What is the aerodynamic drag on this car when it travels at 30 km/h. Assume frontal area of 1.8 m2. (b) What is the power spent in overcoming this drag? [ra = 1.22 kg/m3]. (Ans. FD = 72.44 N; P = 603.6 W) 9.5 Determine the terminal velocity at which an air bubble of 3 mm diameter will rise in SAE 30 oil. What is the corresponding Reynolds number? [rf = 917 kg/m3 and mf = 0.29 Pa.s] (Ans. V0t = 1.55 cm/s; Re = 0.147) 9.6 A hydrogen balloon is 2.0 m in diameter and contains hydrogen of density rh = 0.066 kg/m3. The balloon is found to rise at a terminal speed in air of density 0.95 kg/m3 when the combined weight of empty balloon and payload is 24.5 N. Estimate the terminal speed. [na = 2.3 ¥ 10–5 m2/s. CD = 0.5 for 104 £ Re £ 3 ¥ 105 and CD = 0.2 for Re > 3 ¥ 105] (Ans. V0t = 6.3 m/s) 9.7 Calculate the weight of a ball of diameter 15 cm which is just supported by a vertical air stream of velocity 10 m/s, ra = 1.25 kg/m3 and na = 1.5 stoke. The variation of CD with Reynolds number Re is as follows: Re CD 104 0.4 105 0.5 3 ¥ 105 0.2 (Ans. W = 0.4418 N) 9.8 It is desired to provide a para–trooper with a hemispherical parachute which will provide a terminal fall velocity no greater than that caused by a jump from a 2 m high wall. 9.11 A radio antenna having a diameter of 6 mm extends 1.5 m upwards into air in the front part of a car. If the car moves at 100 km/h, estimate the bending moment at the base of the antenna. [ra = 1.2 kg/m3. CD = 1.2]. (Ans. M = 0.375 Nm) 9.12 A sphere of 4 cm diameter made of aluminium (RD = 2.8) is attached to a string and suspended from the roof of a wind tunnel test section. If an air stream of 30 m/s ﬂows past the sphere, ﬁnd the inclination of the string and the tension in the string. [ra = 1.2 kg/m3, na = 1.5 ¥ 10–5 m2/s. CD = 0.5 for 104 < Re £ 3 ¥ 105 and CD = 0.2 for Re > 3 ¥ 105]. Neglect the drag of the string. (Ans. T = 0.979 N; q = 69° 43¢ 39≤) 9.13 A wind anemometer has two 6 cm diameter hemispherical cups on a horizontal rod, the centre to centre distance between the cups being 30 cm. The arrangement rotates about a vertical axis at the middle 294 Fluid Mechanics and Hydraulic Machines of the rod. If the frictional resistance at the middle of the horizontal bearing can be neglected, compute the speed of rotation of the anemometer in a wind of 10 km/h [CD of hemispherical cups = 1.33 and 0.34. ra = 1.2 kg/m3]. (Ans. N = 387 rpm) the kite? (Ans. FL/FD = 0.822) 9.19 A small airplane has a weight of 10 kN, a wing area of 25 m2 and a take off speed of 100 km/h. The lift and drag coefﬁcients of the wing for small angles of attack can be approximated as CL = 0.11 a and CD = 0.002 + 0.0025 a where a = angle of attack in degrees. Find the angle of attack needed and the power required at the take off? [ra = 1.2 kg/m3]. (Ans. a = 7.85°; P = 6.96 kW) 9.14 A horizontal pipeline 15 cm diameter crosses a deep canal at mid-depth. If the velocity of ﬂow in the canal is 3.0 m/s, estimate the frequency of vortex shedding from the pipe. [n = 1.0 ¥ 10–6 m2/s]. (Ans. n = 4.2 Hz) 9.15 At what wind velocity would an overhead transmission wire of 1.5 cm diameter attain a vibration of frequency equal to 100 Hz? [na = 1.5 ¥ 10–5 m2/s]. (Ans. V0 = 25.7 km/h) 9.16 An advertisement hoarding has a height of 3.0 m and a length of 10 m. If a gale with wind velocity 90 km/h is expected, what would be the force on the hoarding if the wind blows normal to it? [na = 1.2 ¥ 10–5 m2/s]. The variation of CD with L/D is as follows: L/D CD 1.0 1.16 5 1.20 20 1.5 9.20 An airfoil has a planform area of 10 m2 and travels at 200 km/h in air [ra = 1.2 kg/m3]. If the lift and the drag coefﬁcients at the particular angle of attack are 0.8 and 0.005 respectively, calculate (a) the lift force (b) drag force and (c) the resultant force. (Ans. (a) FL = 14.815 kN; (b) FD = 0.0926 kN; (c) FR = 14.82 kN) 9.21 A 2.0 m diameter cylinder is 10 m long and rotates at 300 rpm about its axis which is normal to an air stream of velocity 20 m/s. Calculate the (a) theoretical lift force per unit length (b) the position of stagnation points and (c) actual lift, drag and resultant force on the cylinder. [For actual lift and drag take CL = 3.40 and CD = 0.65 rair = 1.25 kg/m3]. (Ans. (a) FL = 4.935 kN/m; 1.9 (Ans. FD = 13.309 kN) 9.17 A kite 0.8 m ¥ 0.8 m weighing 4 N assumes an angle of 12° to the horizontal. The string attached to the kite makes an angle of 45° to the horizontal. The pull on the string is 25 N when the wind is blowing at a speed of 30 km/h. Find the coefﬁcients of lift and drag? [ra = 1.2 kg/m3]. (Ans. CL = 0.813; CD = 0.663) 9.18 A kite having a weight of 0.5 N soars at an angle to the horizontal. The string holding the kite makes an angle of 35° to the horizontal and has a tension of 5 N. Calculate the ratio of lift to drag force on (b) q1 = – 51.77°, q2 = 231.77°; (c) FaL = 17 kN, FaD = 3.25 kN, a = 79.18°) 9.22 A cylinder of 1.5 m diameter and 6 m long rotates inside a stream of water. The velocity of water is 2 m/s and the rotational speed is such that a double stagnation point is formed. Find the theoretical lift force and the corresponding lift coefﬁcient. (Ans. FL = 225.7 kN; CL = 12.57) 295 Drag and Lift on Immersed Bodies Objective Questions 9.1 The drag force on a body (a) is the net frictional force on the body (b) is the net pressure force on the body in the direction of the relative velocity. (c) is the component of the resultant force in the direction of the relative velocity. (d) is the component of the resultant force in a direction perpendicular to the direction of gravity 9.2 The lift force on a body (a) is due to buoyant force (b) is always in the direction of the gravity (c) is the component of the resultant force in a vertical direction (d) is the component of the resultant force in a direction normal to relative velocity 9.3 In calculating the drag force using CD the area used is (a) always the frontal area (b) the planform area when the body is ﬂat like an airfoil (c) the planform area when the body is bluff like a sphere (d) always the planform area 9.4 Pressure drag results due to (a) formation of wake (b) turbulence in the wake (c) existence of stagnation point in the front of a body (d) high Reynolds numbers 9.5 In calculating the lift force (a) always the frontal area is used (b) always the planform area is used (c) planform area is used if the body is a lifting surface (d) actual surface area of the body is used 9.6 If a streamlined body A and a sphere B both having the same maximum cross sectional area are subjected to laminar ﬂow past them, then (a) the body drag on A will be smaller than on B (b) the total drag on A will be larger than on B (c) the total drag on both A and B will be identical (d) initially A will have larger drag but later on, both the objects will experience the same drag 9.7 A streamlined body with a round nose and a tapering back is generally best suited for (a) creeping motion (b) turbulent sub-sonic ﬂow (c) supersonic ﬂow (d) laminar ﬂow with low Reynolds number 9.8 For a sphere falling at terminal velocity in the Stokes’ law range, the drag coefﬁcient CD is given by (a) 24 Re (b) 64/Re (c) 24/Re (d) 24 (1 + 3/16 Re)/Re where Re is Reynolds number of the ﬂow. 9.9 Stokes’ law is valid up to a maximum Reynolds number of (a) 0.1 (b) 1.0 (c) 2000 (d) 5 ¥ 105 9.10 A very tiny sphere is settling down in a viscous liquid at a Reynolds number of 0.2. Its drag coefﬁcient is (a) 320 (b) 120 (c) 80 (d) 32 296 Fluid Mechanics and Hydraulic Machines 9.11 Body M has twice the weight, twice the projected area and twice the drag coefﬁcient of body N. The terminal velocity of body M in air would be x times that of N, where x is (a) 8 (c) (b) 2 2 (d) 1/ 2 9.12 For a spherical sand particle in Stokes’ law range the fall velocity V0 is related to the diameter D such that (a) V0 increases as D (b) V0 varies inversely as D (c) V0 varies as D2 (d) V0 varies inversely as D2 9.13 For a solid sphere falling under gravity at terminal velocity in a ﬂuid (a) buoyant force = drag (b) weight of the body = buoyant force (c) weight of the sphere = buoyant force + drag (d) drag = weight 9.14 The drag coefﬁcient of a cylinder at small Reynolds number (Re < 10) (a) increases with increase in the Reynolds number (b) decreases with increase in the Reynolds number (c) is essentially constant (d) increases with the Reynolds number in the range 0 < Re < 1 and then decreases for higher Reynolds numbers 9.15 When compared to a streamlined body, a bluff body will have (a) more pressure drag but less friction drag (b) more pressure drag and more friction drag (c) less pressure drag and less friction drag (d) less pressure drag but more friction drag 9.16 A very long circular cylinder at a Reynolds number of Re > 5 ¥ 105 will have a drag coefﬁcient CD (a) 1.20 (b) 0.50 (c) 0.33 (d) 0.20 9.17 At a Reynolds number Re > 103, one can expect a long rectangular plate held normal to the ﬂow to have a drag coefﬁcient CD (a) 0.50 (b) 2.40 (c) 1.00 (d) 1.90 9.18 In the case of ﬂow past a circular cylinder at the critical Reynolds number, the drag coefﬁcient drops from about (a) 0.03 to 0.015 (b) 0.6 to 0.3 (c) 1.20 to 0.33 (d) 2.0 to 1.0 9.19 In the case of ﬂow past a sphere at the critical Reynolds number, the drag coefﬁcient drops from about (a) 1.2 to 0.5 (b) 0.5 to 0.2 (c) 0.2 to 0.07 (d) 2.1 to 1.2 9.20 The drag coefﬁcient CD of a sphere undergoes a sudden drop in its value at the critical Reynolds number of about. (a) 1.0 (b) 2000 5 (c) 2 ¥ 10 (d) 5 ¥ 106 9.21 The critical Reynolds number at which there is a sudden drop in the CD value for a cylinder is about (a) 2000 (b) 5 ¥ 104 5 (c) 5 ¥ 10 (d) 5 ¥ 106 9.22 The Karman vortex trail occurs (a) in all shapes and at all Reynolds numbers (b) in two-dimensional body shapes and in a range of Reynolds numbers (c) in two-dimensional bodies and at all Reynolds numbers > 30 (d) in circular cylinder only 9.23 The Strouhal number S is deﬁned as S = (a) V0/nd (b) V0 d/g (c) nd2/g (d) nd/V0 297 where V0 = free stream velocity and d = diameter of the cylinder and n = vortex shedding frequency 9.24 For circular cylinders the Strouhal number S (a) decreases very slowly with Re. (b) varies as Re1/4 (c) increases linearly with Re (d) is essentially constant at a value of 0.12 9.25 When a cylinder rotates in a ﬂuid (a) only one stagnation point is possible (b) always two stagnation points occur (c) no stagnation point is formed (d) either two or one stagnation point is formed depending upon the ratio of free stream and rotational velocity 9.26 When an airfoil reaches the stall angle (a) the drag coefﬁcient is zero (b) the lift decreases rapidly for any further increase in the angle of attack (c) the drag decreases rapidly for any further increase in the angle of attack (d) the lift coefﬁcient is zero. 9.27 The overall drag of an aircraft of weight W and wing area S is given by CD = a + bC 2L where a and b are constants. The maximum drag in horizontal ﬂight will be (a) 6W ab (b) 4W ab (c) 2W ab (d) W ab 9.28 The optimum efﬁciency of a lifting vane is limited by the (a) onset of stall (b) separation from the leading edge (c) separation from the leading edge (d) more rapid increase in CD than in CL 9.29 A streamlined body is one for which (a) the skin friction is zero (b) the skin friction is minimum (c) the thickness of the body is minimum (d) the separation point occurs on the far downstream part of the body Turbulent pipe 10 10.1 CHARACTERISTICS OF TURBULENCE FLOWS Turbulence is the breakdown of orderly laminar ﬂow in to a state of random ﬂuctuations of velocity. The source of turbulence is the formation of eddies at the shear layer formed either at the boundary or at the layer of separation at the surfaces of discontinuity in the ﬂow. If the turbulence is generated at the wall as in internal ﬂows it is known as wall turbulence and those developed in external ﬂows, away from any boundary, such as in free jets, is known as free turbulence. property such as a velocity is considered to be made up of a mean value and a ﬂuctuating Introduction component. Thus the velocity components are u = u + u¢, v = v + v¢, w = w + w¢ where u = 1 T T Ú u dt 0 etc for v and w. It is obvious that u ¢ = v ¢ = w ¢ =0 ﬂuctuations is an important statistical property of turbulence. Thus for x-component rms = Similarly È1 u¢ = Í ÍT Î 2 v ¢ 2 and 1/ 2 ˘ u ¢ dt ˙ ˙ 0 ˚ T Ú 2 w ¢ 2 are deﬁned. These 299 Turbulent Pipe Flow rms values are measures of average values of turbulence intensities in x, y and z-directions. ﬂow is written for the mean motion as ∂u ∂v ∂w + + = 0 and it should satisfy the ∂x ∂x ∂x continuity condition for the ﬂuctuations as expressed as 1 1 (u ¢ 2 + v ¢ 2 + w ¢ 2 ) V 3 where V is the mean velocity of ﬂow given by I= V= 1 2 (u + v 2 + w 2 ) 3 ∂u ¢ ∂v ¢ ∂w ¢ + + = 0. ∂x ∂x ∂x 10.1.1 Shear Stress In turbulent ﬂow the shear stress t t is expressed as unit of mass is deﬁned as 1 KE per unit mass = (u ¢ 2 + v ¢ 2 + w ¢ 2 ) 2 turbulent ﬂuctuations u¢, v¢ and w¢. These are represented as, for example, u ¢ v ¢ = 1 T T Ú u¢v¢dt . Similarly for v ¢ w ¢, w ¢ u ¢ du du du = ( m + h) +h dy dy dy where m = dynamic viscosity and h = eddy viscosity which is not a ﬂuid property but depends upon turbulence conditions of the ﬂow. Different models are proposed for the estimation of the turbulent shear du dy Prandtl’s model assumes stress t t = h 0 and so on. These correlations of ﬂuctuations of velocities cause additional tangential stresses and normal stresses due to momentum exchange and could be represented in a compact form as - r u¢2 r u ¢v ¢ r u ¢w ¢ r v ¢u ¢ r v¢ 2 r v¢ w ¢ 2 Ê du ˆ t t = rl 2 Á ˜ Ë dy ¯ or r w ¢u¢ - r w ¢ v ¢ - r w ¢ 2 2 t t = t lam + t turb = m 2 Ê du ˆ h = rl 2 Á ˜ Ë dy ¯ where mixing length 2 In this ( - ru ¢ ) , ( - rv ¢ ) and ( - rw ¢ ) are normal stresses on planes normal to x, y and z directions respectively. The remaining are stresses on appropriate ( - ru ¢ v ¢ ) is the turbulent shear stress on xy plane. Obviously ( - ru ¢ v ¢ ) = ( - rv ¢ u ¢ ) and so on. These turbulent shear stresses play a very important role in the ﬂow mechanism and energy losses of turbulent ﬂows. l = ky in which k Karman’s model assumes the mixing length to be l=k and du / dy (d 2 u / dy 2 ) Ê du ˆ t t = rl 2 Á ˜ Ë dy ¯ 2 300 Fluid Mechanics and Hydraulic Machines 10.1.2 Turbulent Flow Near a Wall Three important regions are to be noted: tlam predominates where both tlam and tturb are important Overlap region u yu = u v and the thickness of laminar sublayer d ¢ is taken as 11.6 v d¢ = u 10.2 TURBULENT PIPE FLOW 10.2.1 and tturb predominates Critical Reynolds Number VD of a pipe ﬂow v exceeds a critical value, the ﬂow becomes turbulent. Even though the critical Reynolds number can assume a value within a range depending upon many ﬂow parameters, for practical purposes Recrit = 2000 is usually adopted. When the Reynolds number Re = Y y = d (x) t (x, y) 10.2.2 t turb equilibrium boundary layer ﬂow. Here boundary layer thickness d = r0 = D/2 = constant. The shear stress varies linearly with the distance from the boundary to become zero at the centre t lam t0 (a) Y Outer turbulent layer Over lap region u(x, y) Laminar sublayer (b) Fig. 10.1 t = t 0 (1 - y0 / r0 ) y = d (x) U(x) y Pipe Flow where r0 = radius of the pipe. In a horizontal pipe of diameter D carrying a p2 – p in a length L resistance to the difference in pressure forces. p – p2 or pD 2 = t0 pDL 4 t0 p – p2 D Ê D p◊Dˆ = ÁË L L ˜¯ Turbulent Flow Near a Wall The shear stress at the wall t0 is an important ﬂow parameter. t 0 / r = u is called shear velocity. In the laminar sublayer f rV 2 4 2 where f = Darcy–Weisbach friction factor, Designating t0 = t 0 / r = u = V f /8 301 Turbulent Pipe Flow p – p2 Dp f L rV 2 D 2 Dp Thus written as Ê p1 ˆ Ê p2 ˆ f LV 2 Z Z h + + = = f 1 2 ÁË g ˜¯ ÁË g ˜¯ 2g D Here hf = drop in piezometric head in a distance L as Darcy–Weisbach formula. 10.2.3 Hydrodynamically Smooth and Rough Pipes On the basis of relative magnitudes of surface protrusions e and thickness of laminar sublayer d¢, a pipe surface is classiﬁed as follows: e/d ¢ £ i.e ue £ 3.0 n e d¢ i.e. ue 3 < < 70 n Re f e e r0 = = d ¢ r0 d ¢ ( r0 / e ) (65.6) Hence e d¢ e ≥ d¢ 10.2.4 Re f ( r0 / e ) Re f ≥ ( r0 / e ) Velocity Distribution (a) Local Velocity diameter D r0 = D at any distance y from the boundary, the local velocity u is given by the following expressions: u yu = 5.75 log + 5.5 u n u y = 5.75 log + 8.5 e u (b) Mean Velocity The mean velocity V in a pipe of radius r0 is obtained by integrating the velocity u over the entire area of the pipe and dividing by pr02. This gives smooth pipes: V ru = 5.75 log 0 + 1.75 u n rough pipes e/d¢ i.e. ue ≥ 70 n It may be noted that the laminar sublayer thickness d ¢ could be expressed in terms of pipe radius, as 11.6n 11.6n 65.6 d¢ = = = D r0 r0 u Re f V f /8 2 V r = 5.75 log 0 + 4.75 u e both smooth and rough boundaries, the maximum velocity um is at y = r0 and is given by (c) Maximum Velocity um u - um y = 5.75 log u r0 302 Fluid Mechanics and Hydraulic Machines (d) Relations Among V, u, um, u and f: (ii) (i) In terms of mean velocity V = [Q/(pD2/4)], the velocity u and the maximum velocity um are given for both smooth and rough pipes as: f = 0.0032 + (ii) In terms of centreline velocity umax, the mean velocity V and f are related by the expression. umax = 1.43 f + 1 V umax - V = 3.75 u [From Eq. 10.14] Up to Re £ 2000 64 Re (b) Smooth Turbulent Flow (i) f = 0.316 Re1/ 4 (10.26) Ê ˆ Re f £ 17˜ Á When ( r0 / e ) Ë ¯ ...Blasius equation valid for 104 < Re < 105 Re0.237 = 1.8 log Re - 1.5186 f (10.29) (10.30) 1 f for = 2 log e ≥ 6.0 d¢ r0 + 1.74 e i.e. (10.31) Re f ≥ 400 ( r0 / e ) (c) Transitional regime in turbulent ﬂow: Re f r 1 - 2 log 0 = 2 log - 0.8 e ( r0 / e ) f (10.32) valid for uniform sand grain roughness only. and the variation of the friction factor f is as follows: f = 1 0.221 to get approximate but adequate value of f. (10.25) The energy loss due to friction in pipe ﬂow is expressed by Darcy–Weisbach formula as (a) Laminar Flow (10.28) (c) Rough Turbulent Flow Frictional Resistance f LV 2 2gD or (10.24) This equation is obtained from Eq. (10.23(b)) after making a small change in the coefﬁcients to ﬁt the experimental data better. (iii) In terms of umax, V and u are related as hf = = 2 log Re f - 0.80 f /8 u -V y = 2 log + 1.32 (10.23b) r V f 0 10.2.5 f for all Reynolds numbers in the turbulent range in smooth pipes. For explicit relationship of f one can use either u -V y = 5.75 log + 3.75 (10.23a) u r0 Replacing u = V 1 (10.27) 10.3 COMMERCIAL PIPES The rough turbulent ﬂow Eqs. 10.31 and 10.32 mentioned in Sub-section 10.2.5 were based on Nikuradse’s classical experiments on uniform sand grain roughness. However, in commercial pipes the roughness magnitude, shape and distribution vary widely. To overcome the difﬁculty of describing all these parameters the concept of equivalent sand grain roughness is used. The roughness of a uniform sand grain coated pipe of the same size as the given commercial pipe which gives the same value of f in the completely rough turbulent regime is termed as the equivalent sand grain roughness, and also as effective roughness (es). Some of the usual ranges of equivalent sand grain roughness of commercial pipes are given in Table 10.1. 303 Turbulent Pipe Flow Table 10.1 Values of es for Commercial Pipes Pipe material Some ¥ Common es (mm) ¥ and es/D values within ± (2) Swamee and Jain Equation piping plastic, fibre glass 1 f 21.25 ˆ Êe = 1.14 - 2 log Á s + Ë D Re 0.9 ˜¯ £ Re £ es/D f –2 . This (3) Haaland Equation 10.3.1 È 6.9 Ê e / D ˆ 1.11 ˘ 1 = - 1.8 log Í +Á s ˜ ˙ f ÍÎ Re Ë 3.7 ¯ ˙˚ Colebrook Equation empirical relationship amongst f, Re and r0/es covering the smooth pipe, transition and rough pipe turbulent ﬂow regimes, as – 2 log f r0 es Ê r0 / e s ˆ ˜ Á1 + 18.7 Re f ¯ Ë which simpliﬁes to Èe 1 9.35 = 1.14 - 2 log Í s + f ÍÎ D Re f f [Note: as Stanton diagram as it is believed that Stanton 10.3.2 ˘ ˙ ˙˚ which is asymptotic to smooth pipe and rough pipe relationships of f Re, r0/es is a graphical plot of this relationship. available for easy and explicit solution of f. Three of these are given below. In the following Re = Reynolds number = VD/n. (1) Moody Equation 1/ 3 ˘ È Ê e s 106 ˆ ˙ Í f = 0.0055 1 + Á 20000 + Í Ë D Re ˜¯ ˙ Î ˚ Moody Diagram It is a chart showing the variation of f es/D Re, calculations and ﬁnds a large number of applications. It can be used for non-circular conduits and also for open channels, by replacing D Rh where Rh is diagram. 10.4 AGING OF PIPES characteristics due to continuous usage. The increase es with age is usually taken to be linear as e s = e s0 + a t 304 Transition zone 0.08 Critical zone 0.1 0.09 Laminar flow Fluid Mechanics and Hydraulic Machines Complete turbulence, rough pipe 0.05 0.04 0.07 0.06 0.03 0.05 0.02 0.03 0.004 0.025 0.002 0.02 0.015 0.001 0.0008 0.0006 0.0004 Laminar flow f = 64/Re Relative roughness Friction factor f 0.01 0.008 0.006 es D 0.015 0.04 0.0002 0.0001 Smooth pipe 0.00005 0.01 es/D = 0.000005 es/D = 0.000001 0.009 0.008 10 3 3 4 5 6 8 10 4 3 4 5 6 8 10 5 3 4 5 6 8 10 6 3 4 5 6 8 10 0.00001 7 3 4 5 6 10 8 VD Reynolds number, Re = n Fig. 10.2 Moody Diagram where t is the number of years of use after es0 was recorded. 10.5 SIMPLE PIPELINE DESIGN PROBLEMS hf = f LV 2 2g D The friction factor f When the pipe friction is the only loss, the variables in a pipeline ﬂow are: Q, L, D, hf, es and n diagram. Out of these es, n and L are known or can be determined. Thus one has the following three types of problems: This is a straight forward problem. I II III Q, L, D, n, es hf, L, D, n, es Q, hf, L, n, es hf Q D In each of these types of problems the head loss hf is found by the Darcy–Weisbach formula Type I Problem Type II Problem Here es/D Re f = To ﬁnd hf To ﬁnd Q Re VD Ê 2ghf D ˆ v ÁË V 2 L ˜¯ f is given by 1/ 2 = D 3 / 2 Ê 2ghf ˆ v ÁË L ˜¯ 1/ 2 Using these two parameters, f is found from the 305 Turbulent Pipe Flow Knowing f and hf, V Type III Problem Q are determined. u u y = 5.75 log + 5.5 u n To ﬁnd D u y v well beyond the laminar sublayer. valid in a region f LV 2 f LQ 2 = 2g D 2g( p / 4) 2 D 5 D =C f hf = where C = Re = 8 LQ u y v given by 2 hf gp2 VD QD C = 2 = n D Ê p 2ˆ ÁË 4 D ˜¯ n u u y = u n u y Ê ˆ ÁË 70 > v > 5.0˜¯ an approximate relation is Ê Qˆ C2 = Á Ë pn ˜¯ where Thus by using D =C f and Re = C2/D D is solved by trial and error. The procedure involves the following steps: u u y = 11.5 log - 3.0 n u u y = 5.75 log + 8.5 es u f. D Re es/D. where es f for Re and es/D f till satisfactory value of f is obtained. 10.6 VELOCITY DISTRIBUTION IN THE NEIGHBOURHOOD OF FLAT SURFACES The turbulent ﬂow velocity distribution in pipe ﬂow is essentially applicable to describe the velocity distribution near boundaries other than pipes also. Thus, and rough surfaces in the turbulent ﬂow zone u y = 5.75 log u v¢ where y¢ = a parameter that depends upon laminar sublayer thickness d ¢ and roughness magnitude es es d¢, y¢ = es 30 306 Fluid Mechanics and Hydraulic Machines Y and for es u y¢ = d¢, d¢ 0.108 n = 107 u proﬁles over ﬂat surfaces, e.g. wind blowing over a ground, ﬂow over a ﬂat plate and other similar turbulent boundary layer ﬂows. y Transition & Laminar sublayer y¢ Fig. 10.3 s Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Simple Difﬁcult 1.847 L/min 10.1 f = 64 64 = Re 2000 u = V Solution: The largest discharge corresponds to the critical Reynolds number. VD Re crit = 2000 = v 2000 ¥ 0.0098 ¥ 10 -4 V= 0.02 Q= p 2 ¥ ¥ ¥ m /s ¥ f /8 = 0.098 0.032 8 ¥ t0 = ru2 = 0.0383 Pa 10.2 r ¥ ¥ 2 307 Turbulent Pipe Flow Solution: For both smooth and rough turbulent ﬂows u -V y = 5.75 log + 3.75 u r0 (a) When u = V y r0 y log r0 y r0 y 5.75 log or (b) When u = V 2 = – 3.75 = - 3.75 – 0.65217 5.75 = 0.22275 = 0.22275 r0 and u = 1 V 20 Ê 0.5 V - V ˆ y ÁË 1/ 20 V ˜¯ = 5.75 log r + 3.75 0 y r0 y log r0 y r0 y 5.75 log or = – 10.0 – 3.75 = – 13.75 The maximum velocity um is given by um = 1.43 f + 1 V = 1.43 0.02 + 1 = 1.20223 um = 1.20223 V (b) Here y = distance from the wall = r0 – r y = r0 – 0.3 r0 = 0.7 r0 y = 0.7 r0 u -V y = 5.75 log + 3.75 u r0 = 5.75 log (0.7) + 3.75 = 2.8593 But u = 0.05 V u 1 \ – = 2.8593 0.05V 0.05 u = 1.14297 or V 10.4 r0 = – 2.391 u u = 0.00406 y/r0 = 0.00406 r0 10.3 Solution: r0 f u V. r0 Solution: (a) In turbulent ﬂow the shear velocity u = V =V y f /8 0.02 = 0.05 V 8 For smooth pipes, u yu = 5.75 log + 5.5 u v At the centreline where y = r0, u = um, so that um ru = 5.75 log 0 + 5.5 u v um - u r0 = 5.75 log (1) \ u y In rough pipes: u y = 5.75 log + 8.5 u e At y = r0, u = um um r \ = 5.75 log 0 + 8.5 u e 308 Fluid Mechanics and Hydraulic Machines um - u r = 5.75 log 0 u y (2) The ﬂow is in the turbulence mode and the Blasius formula is applicable. It is seen that Eqs (1) and (2) are identical. Hence, the same equation is valid for both smooth and rough pipe ﬂow. 10.5 A 30 cm diameter pipe conveys water in f= 0.316 (Re)1/ 4 Loss of head = 0.316 (1.2 ¥ 10 4 )1/ 4 = 0.0299 0.0299 ¥ 100 ¥ ( 2.5) 2 = 4.762 m (of crude oil) 2 ¥ 9.81 ¥ 0.2 Discharge hf = Ê pˆ Q = Á ˜ ¥ (0.2)2 ¥ (2.5) = 0.07854 m3/s Ë 4¯ Power P = g Qhf = (0.9 ¥ 9.79) ¥ 0.07854 ¥ 4.762 = 3.30 kW Solution: Maximum velocity um is related to the mean velocity in both smooth and rough turbulent ﬂows as um - V u Since = 3.75. f = 0.02, f 0.02 =V 8 8 = 0.05 V. (um – V) = 3.75 u = (3.75) ¥ 0.05 V = 0.1875 V um = 3.57 = 1.1875 V 10.7 ¥ 10 –6 m Shear velocity u = V Hence V= Discharge Q = 3.57 = 3.00 m/s 1.1875 p ¥ (0.30)2 ¥ 3.00 = 0.212 m3/s 4 10.6 Solution: Reynolds number Re = = = 0.0032 + = 1.44 ¥ 105. 0.221 (Re) 0.237 0.221 (1.44 ¥ 105 )0.237 Loss of head in 1000 m of pipe Solution: VD Reynolds number Re = v 2.5 ¥ 0.2 = 1.2 ¥ 104. = 0.40 ¥ 10 -4 -6 2.5 ¥ 10 Since the Reynolds number is > 105, the Blasius formula is not applicable. Using the explicit relationship for f in the smooth turbulent ﬂow regime (Eq. 10.29), f = 0.0032 + n VD v 1.2 ¥ 0.3 = hf = = 0.01644 fLV 2 2 gD 0.01644 ¥ 1000 ¥ (1.2) 2 2 ¥ 9.81 ¥ 0.3 = 4.02 m = 309 Turbulent Pipe Flow 10.8 es r es 0.80 = 0.175 d¢ – ¥ m f transition regime. 10.10 - Solution: V= Q 0.30 = A Ê pˆ 2 ÁË 4 ˜¯ ¥ (0.3) –6 Solution: f Shear velocity u = V ¥ n begins at 0.0097 8 Re f ( r0 / e ) um is related to the mean velocity in both smooth and rough turbulent ﬂows as um - V Ê 0.10 ˆ ¥ Á Ë 0.0002 ˜¯ The friction factor in smooth-turbulent ﬂow is f Re u = 2.0 log Re f f um = V um u ¥ \ Re ¥ ¥ VD V ¥ 0.2 ¥ = v 1 ¥ 10 -6 6 ¥ 10 4 ¥ 10 -6 V = = 0.3 m/s 0.2 Hence the velocity at the upper limit of smooth um = 4.8 m/s 10.9 f d ¢ is 65.6 d¢ = r0 Re f Re f ( r0 / e ) Re ¥ f 65.6 d¢ = 5 0.05 10 0.035 d¢ ¥ m = 0.175 mm f 0.1 = 200,000 0.0002 = 2.0 log r0 e 310 Fluid Mechanics and Hydraulic Machines f Re f ¥ f 0.1 ˆ Ê = 2.0 Á log Ë 0.0002 ˜¯ 10.12 ¥ n (r = Re = 200,000 ¥ – f ¥ Re n Therefore, V= D 1.428 ¥ 106 ¥ 10 -6 = 0.2 = 7.14 m/s The ﬂow will be at fully-rough turbulent regime at V Solution: Reynolds number VD 8.0 ¥ 0.25 ¥ = v 1.5 ¥ 10 -5 The friction factor f in fully rough-turbulent ﬂow Re = is f 10.11 = 2.0 log r0 e Ê 0.125 ˆ = 2.0 log Á ˜ Ë 0.50 ¥ 10 -3 ¯ f Shear stress at the boundary Solution: um - u u r0 y Ê 4.50 - 4.20 ˆ ÁË ˜¯ u u = 0.30 (0.30 - 0.10) u The mean velocity V is related to um as t0 = t 0 / r = V0 rV02 8 = 0.2285 Pa 1.22 ¥ (8.0) 2 ¥ 0.0234 8 10.13 4.50 - V 0.2963 V = = f e um - V u The discharge Q = f f /8 n pD 2 V 4 –6 ¥ Solution: p ¥ (0.6) 2 ¥ 4 3 = 0.958 m /s = 2 log f r0 e 1 r = 2 log 0 0.028 e 311 Turbulent Pipe Flow 2 log r0 e r0 e 0.150 e = 131.24 e = 1.14 mm u = and ¥ t0 /r = V ¥ t0 /r m u es 1 ◊ y es 2 ¥ 0.028 / 8 1.5 0.07 es es = 42.79 Pa y = r0 – r The shear stress varies linearly with y with zero at the centre and as such 0.10 ˆ Ê ÁË1 - 0.15 ˜¯ y/r0 = 14.26 Pa u -V u and 0.025 es ¥ = 0.141 mm u u y/es r0 es is used. u u y e y e Ê 0.075 ˆ Á ˜ Ë 1.411 ¥ 10 -4 ¯ V 0.07 V e ¥ y e du = u dy Discharge Q = V ¥ D2 p u = V f /8 Ê 0.07 ˆ Êu ˆ f = 8Á ˜ = 8Á Ë 1.4295 ˜¯ ËV ¯ = 0.0192 r 2 ¥ /s = 25.3 L/s 2 10.14 p ¥ u/y ¥ = 5.177 s–1 m y/r0 Hence V/u u/y u u t0 = u2 r t 0 = 4.493 Pa 2 t = t0 y es f /8 t0 r u u du dy 2 – [Note: f could also have been evaluated by the Solution: friction factor formula for rough turbulent ﬂow: r = 2.0 log 0 es f 312 Fluid Mechanics and Hydraulic Machines Solution: 10.15 Re = r e f Solution: The ﬂow is rough-turbulent, and as such u y u e u2 - u1 y2 Hence u y1 u2 y u1 u u1 u u1 u and Thus Reynolds number rVD 804 ¥ 3.0 ¥ 0.20 = m 1.92 ¥ 10 -3 the ﬂow will be in smooth-turbulent regime. Hence the friction factor f for Re as 0.221 f Re0.237 0.221 ( 2.513 ¥ 105 )0.237 u and y2 Since the pipe will behave as smooth, at the limiting value 0.03 0.015 Re f ( r0 / e ) 2.513 ¥ 105 ¥ 0.0148 r0 = 17 e 0.015 e 0.10 ¥ 1798 = 0.0556 mm e = 0.015 e or e ¥ Relative roughness r0 0.05 = 113.5 = e 4.41 ¥ 10 -4 f is = 2 log r0 e 10.17 f = 0.0292 10.16 r m m [Note: of energy when it is in the smooth pipe regime. In other words, for a given Reynolds number the lowest value of f occurs in the smooth pipe given by f ¥ ¥ – Solution: = 2 log f r0 e 313 Turbulent Pipe Flow = 2 log 0.05 0.0002 f and 1 f2 f2 P = g Qhf 2 But \ hf = 0.150 e0 e0 = 2 log 0.075 0.0002 f LQ 2 1 2 P Ratio of powers 1 for same Q, g and L P2 = 5 0.0234 Ê 15 ˆ ¥ 0.0211 ÁË 10 ˜¯ = 2.0 log r0 e1 = 2.0 log 0.15 e1 0.025 Ê pˆ 2g Á ˜ D 5 Ë 4¯ Ê f ˆÊ D ˆ P1 = Á 1˜Á 2˜ P2 Ë f 2 ¯ Ë D1 ¯ e t and f f LV = 2gD ¥ m = 0.324 mm 5 0.15 e1 e ¥ m = 0.765 mm The roughness magnitude can be taken to increase linearly with age as e = e0 + at e – e0 at (0.765 - 0.324) a = 10 log consumed. Hence t e2 = e0 + at Cost of pumping in 10 cm pipe = 8.42 Cost of pumping in 15 cm pipe 10.18 ¥ The friction factor f2 will be 1 0.15 = 2 log 0.0014265 f2 f2 = 0.0299 Solution: Initially t = 0 years: roughness = e0 and 1 r = 2.0 log 0 e0 f0 1 0.150 = 2.0 log e0 0.02 log 0.150 e0 10.19 e e = n Solution: ¥ –6 Before the lining: 4.0 V = p ¥ (1.5) 2 4 314 Fluid Mechanics and Hydraulic Machines Reynolds number Re = VD v = hf2 = 2.264 ¥ 1.5 1.0 ¥ 10 -6 ¥ 15 ¥ 10 e s1 = 1.5 D1 -3 Saving in head: hs = h – hf2 Saving in power Ps = g Qhs ¥ ¥ = 162.5 kW –2 Ê e s1 21.25 ˆ ˜ ÁD + Ë 1 Re10.9 ¯ f 0.0132 ¥ 1000 ¥ ( 2.325) 2 2 ¥ 9.81 ¥ 1.48 10.20 e Ê -2 ˆ 21.25 Á10 + 6 0.9 ˜ (3.395 ¥ 10 ) ¯ Ë or f Solution: Relative roughness 0.15 ¥ 10 -3 es = 0.25 D ¥ D2 V2 = 4.0 p ¥ (1.48) 2 4 D 3 / 2 Ê 2gh f ˆ = v ÁË L ˜¯ ¥ ¥ f = Re Ê ˆ 21.25 -4 Á1.351 ¥ 10 + 6 0.9 ˜ (3.44 ¥ 10 ) ¯ Ë f2 2 Head loss hf = h = f LV 2gD 0.0379 ¥ 1000 ¥ ( 2.264) 2 2 ¥ 9.81 ¥ 1.50 1/ 2 1/ 2 hf /L (0.25)3 / 2 1 ¥ 10 -6 Ê e s 2 21.25 ˆ ˜ ÁD + Ë 2 Re20.9 ¯ 1 f2 ¥ Ê VD ˆ Ê 2 gh f D ˆ f = Á Ë v ˜¯ ÁË V 2 L ˜¯ Re Reynolds number V D 2.325 ¥ 1.48 Re2 = 2 2 = v 1 ¥ 10 -6 0.2 ¥ 10 -3 es/D2 = 1.48 f2 is given by –6 ¥ n or ¥ f Ê es 9.35 ˆ ˜ Á + Ë D Re f ¯ f 9.35 ˆ Ê -4 ÁË 6 ¥ 10 + 87545 ˜¯ f Re 0.01806 = \ ¥ V V ¥ (0.25) 1 ¥ 10 -6 ¥ 315 Turbulent Pipe Flow p Discharge Q = 2 ¥ ¥ Ê e s 21.25 ˆ ÁË D + ˜ Re0.9 ¯ /s = 127.9 L/s f Ê 21.25 ˆ -4 Á 2.864 ¥ 10 ¥ ˜ (759665)0.9 ¯ Ë ¥ P = g Qhf ¥ f ¥ f 3.13 kW 2nd trial: Using f D ¥ Re es/D ¥ 10.21 e –6 ¥ f f This is practically the same as the assumed value and as such no further trials are necessary. The D = 38.3 cm. In practice, the next larger standard size would be used. Solution: hf = f LV 2 8 LQ 2 f = 2gD p 2g D 5 È 8 ¥ 1500 ¥ (0.250) 2 ˘ f Í ˙ 5 p 2 ¥ 9.81 ÍÎ ˙˚ D f \ Re = 10.22 n ¥ –6 D f D Reynolds number Re = Ê 21.25 ˆ -4 Á 3.133 ¥ 10 ¥ ˜ (831070)0.9 ¯ Ë VD Ê 4Q ˆ 1 = Á v Ë pv ˜¯ D 4 ¥ 0.25 p ¥ 1 ¥ 10 -6 1 D ¥ f D È Ê 6 ˆ 1/ 3 ˘ Í1 + Á 2000 e s + 10 ˜ ˙ Í Ë D Re ¯ ˙ Î ˚ Solution: The Reynolds number 1st trial f D Re es/D = Re = ¥ 0.12 ¥ 10 -3 0.419 ¥ es VD 1¥ D = v 1 ¥ 10 -6 D ¥ m, so 316 Fluid Mechanics and Hydraulic Machines 1/ 3 ˘ È Ê -3 6 ˆ Í1 + 2000 ¥ 0.45 ¥ 10 + 10 ˙ ˜ Í ÁË D 106 D ¯ ˙ Î ˚ f f È Ê 0.9 1 ˆ 1/ 3 ˘ Í1 + Á + ˜ ˙ D¯ ˙ ÍÎ Ë D ˚ ¥ ÏÔ ¸Ô 21.25 -4 Ì1.625 ¥ 10 ¥ 5 0.9 ˝ ( 4.10 ¥ 10 ) ˛Ô ÓÔ /D 2 f LV 2gD 10 f 100 ¥ 1 = ¥ 100 D 2 ¥ 9.81 or Head loss hf = f LV 2 2gDh 0.0154 ¥ 200 ¥ ( 20) 2 2 ¥ 9.81 ¥ 0.3077 = 204 m (of air column) = Dp = pressure loss = rghf ¥ ¥ 2.40 kPa ¥ 10.23 r n f hf = head loss = D f The value of D is obtained by trial and error from f corresponding D = 0.678 m. In practice, the next larger standard size would be used. ÏÔ e s 21.25 ¸Ô + Ì 0.9 ˝ ÓÔ Dh ( Reh ) ˛Ô ¥ = – = e Solution: Dh Dh Area Perimeter ¥ hydraulic radius Rh 2 ¥ 10.24 Solution: ¥ 0.1 1.3 VDh 20 ¥ 0.3077 Reh = = v 1.5 ¥ 10 -5 ¥ es/Dh = 1.2 ¥ 9.81 ¥ ( 20 ¥ 0.4 ¥ 0.25) ¥ 204 0.60 8.0 kW ¥ ¥ Dh g Qhf h 0.05 ¥ 10 0.3077 -3 ¥ u u y y2 u2 - u1 u 6.0 - 5.0 u u y/y¢ u u2 y2/y 10.0 5.0 317 Turbulent Pipe Flow y y is given by y3 y1 u3 - u1 u u3 - 5.0 0.5777 u 10.1 Which of the following two pipes carrying water has larger shear stress at the wall? What are the magnitudes of these stresses? m and n ¥ ¥ show that um - V V Ê yˆ f log Á ˜ Ë R¯ velocity to shear velocity. If the mean velocity is 2.0 m/s, estimate the value of the centreline velocity. r Ê ˆ um ÁË Ans. u = 21.93; um = 2.452 m/s˜¯ Ans. t0 10.2 In a turbulent ﬂow through a pipe of radius r0 at what radial distance would the local 7.585 m/s 10.5 In a turbulent ﬂow the friction factor f = m2 t0 Ê 30 ˆ ÁË 5 ˜¯ 10.6 The centreline velocity in a pipe of 20 cm the ﬂow to be in the fully developed roughturbulent regime, ﬁnd the discharge in the pipe. The effective roughness of the pipe maximum velocity? Ê ˆ r ÁË Ans. r = 0.9914˜¯ 0 Ans. Q 10.7 10.3 In a turbulent ﬂow in a pipe the friction factor f the ﬂuid in the pipe. Ans. Q 10.8 In a fully developed rough-turbulent ﬂow velocity. um V u V 10.4 If the velocity distribution in a pipe of radius R is given by Ans. u -V u y R grain roughness of the pipe. Ans. es 10.9 In a turbulent ﬂow in a pipe the velocity velocity. Determine the ratio of average 318 Fluid Mechanics and Hydraulic Machines velocity to the centreline velocity and the value of the friction factor f. Ê ˆ V ÁË Ans. u = 0.783; f = 0.0376˜¯ m 10.10 A 30 cm pipeline carrying water has a centreline velocity of 2.0 m/s. If the velocity at mid-radius is 1.6 m/s, estimate the discharge. (Ans. Q = 80 L/s) 10.11 The centreline velocity in a smooth 10 cm pipe carrying water (n = 1 ¥ 10–6 m2s) is found to be 3.5 m/s. Determine the discharge and the friction factor f. The ﬂow can be assumed to be hydrodynamically smooth. {Hint: Find u by trial and error method} (Ans. Q = 23.7 L/s; f = 0.0143) 10.12 A 9.0 cm diameter pipeline carries 10 L/s of oil (n = 2.5 ¥ 10–6 m2/s). Calculate (i) the friction factor f, (ii) shear stress at the boundary, (iii) shear stress and velocity at a radial distance of 3.0 cm from the pipe axis and (iv) the thickness of laminar sublayer. Assume the pipe to be hydrodynamically smooth. (roil = 800 kg/m3). (Ans. (i) f = 0.0205; (ii) t0 = 5.063 Pa; (iii) t = 3.375 Pa, u = 1.663 m/s; (iv) d ¢ = 0.365 mm) 10.13 A pipe 30 cm in diameter carries water (n = 0.9 ¥ 10–6 m2/s). If the velocities at the centreline and at a radial distance of 8 cm from the pipe axis are 5.0 m/s and 4.7 m/s respectively, calculate the discharge in the pipe. Identify the pipe ﬂow regime if the equivalent sand grain roughness is 0.3 mm. (Ans. Q = 311.6 L/s; the ﬂow is in transition between hydrodynamically rough and smooth boundary turbulent ﬂow.) 10.14 A 60 cm diameter pipe is to carry 1.2 m3/s of oil (n = 2.4 ¥ 10–6 m2/s). Estimate the maximum height of surface roughness that will have no effect on the resistance. (Ans. emax = 0.045 mm) 10.15 A 30 cm diameter pipe is known to have roughness elements of equivalent height 0.25 mm. What is the maximum velocity of ﬂow of oil (n = 2.4 ¥ 10–6 m2/s) at which the pipe will behave as a smooth pipe? (Ans. V = 0.589 m/s) 10.16 Galvanized iron pipes can be assumed to have equivalent roughness magnitude of 0.15 mm. What minimum size of galvanized iron pipe will be hydrodynamically smooth at a Reynolds number of 2 ¥ 105? (Ans. D = 0.439 m) 10.17 A 10 cm diameter pipe is to carry water (n = 1 ¥ 10–6 m2/s). The equivalent sand grain roughness of the pipe of 0.5 mm. What is the minimum discharge at which the pipe will have fully developed roughturbulent ﬂow? (Ans. Q = 18.0 L/s) 10.18 Water (n = 1.0 ¥ 10–6 m2/s) ﬂows in a 35 cm diameter pipe in fully roughturbulent regime. If at a radial distance of 15 cm from the pipe surface the velocity is 2.5 m/s and the velocity gradient is 7 ¥ 10–4 s–1 determine the discharge in the pipe. (Ans. Q = 175 L/s) 10.19 Two pipes A and B have the same diameter and carry the same discharge of different ﬂuids. The Reynolds number in pipe A is 1000 while that in B is 50,000. Determine (i) the ratio of maximum velocities in pipes A and B, (ii) the ratio of energy gradients in pipes A and B. Ê umA = 1.677; ÁË Ans. (i) u mB ˆ (ii) ( hf / L) A /( hf / L) B = 3.033 ˜ ¯ 319 Turbulent Pipe Flow m2/s. The 10.20 to reservoir B. Two alternative pipes are friction factor may be used. has an absolute roughness magnitude of has an absolute roughness magnitude of the same discharge of water through these two pipes. The ﬂow can be assumed to be fully developed rough-turbulent. Ê e s 21.25 ˆ Á D + 0.9 ˜ Re ¯ Ë Ans. Rough hf Smooth: hf f 10.24 smooth pipe through which air is supplied. If the Reynolds number of the ﬂow is 2¥ of the pipe. Express your answer in cms of water column. [rair ,n ¥ m2 Ê ˆ C1 ÁË Ans. C = 0.47˜¯ 2 10.21 In a 20 cm diameter pipe carrying water the Ans. hf water 10.25 What is the head loss per metre of pipe roughness of the pipe. The absolute roughness magnitude at this stage is ¥ effective The pipe can be assumed to be in roughturbulent ﬂow regime. Ans. roughness Ans. f r 10.26 magnitude as hf and m 10.22 ¥ es n ¥ m2 2.0 m/s. The surface irregularities are estimated as e0 estimated that the rate of growth of surface Ans. hf r 10.27 many years would this ﬂow cease to be in smooth pipe regime? Ans. T P and m ¥ es 10.23 a distance of 200 m. Ans. Q 10.28 How much head loss would be expected if this pipe were smooth? Take kinematic P es n ¥ m2 with a maximum allowable pressure drop 320 Fluid Mechanics and Hydraulic Machines maximum discharge is possi ble under these conditions? 10.30 Wind velocities over a ﬂat land are Ans. Q Estimate the wind velocity at a height es 10.29 ¥ m2/s. If the head loss is not to exceed fully developed rough-turbulent ﬂow, determine the effective roughness height of the surface. Ans. u es Ans. D 10.1 The intensity of turbulence refers to u¢ and v¢ 10.6 In a turbulent ﬂow in a pipe the shear stress is per unit mass. linearly towards the wall. velocity ﬂuctuations es logarithmically towards the wall. - centreline and the wall. 10.2 The turbulent shear stress in xy plane is given by ru ¢ 2 r u¢ v¢ r linearly to a zero value at the centre. 10.7 Water of kinematic viscosity v u ¢v ¢ r u¢ v¢ pipe. The critical ﬂow in this pipe would correspond to a discharge of approximately 10.3 the turbulent shear stress t = rl2 u/dy 2 rl u/dy r2 l2 u/dy 2 rl u/dy 2 10.4 The turbulent intensity u ¢ 2 in a turbulent pipe ﬂow is maximum at 10.8 In a turbulent ﬂow, u, v and w are time averaged velocity components. The ﬂuctuating components are u¢, v¢ and w¢ respectively. The average kinetic energy per unit mass is given by T 1 u ¢ v ¢ dt T Ú 0 l is 10.5 k y2 k log y ky k/y 1 (u ¢ 2 + v ¢ 2 + w ¢ 2 ) 2 1 2 (u + v 2 + w 2 ) 3 321 Turbulent Pipe Flow 1 V 10.14 1 (u ¢ 2 + v ¢ 2 + w ¢ 2 ) 3 ﬂow regimen. Estimate the value of the friction factor f if the diameter of the pipe 10.9 Shear stress in turbulent ﬂow is due to roughness of the pipe e direction of ﬂow 10.15 In a turbulent ﬂow through a pipe the direction of ﬂow as well as transverse to it friction factor f = 0.02. The mean velocity of the ﬂow in m/s is 10.10 In a horizontal pipe ﬂow if Dp is the difference in pressure between two sections distance L apart in a pipe of diameter D then t0 = Dp D L 1 D pLV 2 8 10.16 In a turbulent ﬂow through pipes the mean velocity V and the centreline velocity um Dp L D are related as D p/8 um = V f 10.11 f 10.17 Shear velocity is f wall. laminar sublayer 10.12 In a turbulent pipe ﬂow, the term Re f r0/es d¢/r0 es/d¢ r0/d¢ d¢/es roughness 10.18 where d¢ = thickness of laminar sublayer and es 10.13 In a turbulent ﬂow through smooth pipe of radius r0 the velocity distribution plotted against y/r0 of the Reynolds number f in laminar and turbulent ﬂow in a pipe varies as Re and Re– respectively. If V is the average velocity, the pressure drop in a horizontal pipe for laminar and turbulent ﬂow respectively will be proportional to and V2 and V 10.19 . If the friction decrease in Reynolds number sublayer, in mm, is the Reynolds number increase in Reynolds number 322 Fluid Mechanics and Hydraulic Machines 10.20 10.26 In a turbulent pipe ﬂow the boundary is hydrodynamically smooth if es/d ¢ £ es/d ¢ d ¢/es £ es/d ¢ ≥ sand grain roughness to the thickness of e/d¢ can be classiﬁed as f in turbulent ﬂow through pipes relates f to the Reynolds number Re as f = Re Re Re Re 10.28 In a fully rough-turbulent pipe ﬂow, the friction factor f is Re and es/D Re only es/D only Re and es/D 10.27 ﬂow smooth to rough regime 10.21 In a turbulent ﬂow through a pipe of radius r0, the radial distance at which the local r0 r0 r0 r0 10.22 In a pipe ﬂow the shear velocity u is related to friction factor f and mean velocity V as u/V = f /8 where Re = Reynolds number and es/D = relative roughness. 10.29 8/ f 8g/ f f 10.23 The Darcy–Weisbach friction factor f is related to boundary shear t0 as f = 8 rV 2 t0 t0 8 rV 2 rV 2 8t 0 8t 0 rV 2 friction factor f es/r0 only Re and es/r0 only Re only where Re = Reynolds number and es/r0 = relative roughness. 10.30 In a fully developed rough-turbulent regime in pipe ﬂow, same friction factor 10.24 f the roughness projections the Reynolds number 10.25 In laminar ﬂow through a pipe the Darcy– Weisbach friction factor f is given by f = 64 Re 16 Re 24 Re 3 Re 16 the relative roughness es/r0 10.31 transitional regime, if the friction factor f is and relative roughness 323 Turbulent Pipe Flow sistance at low Reynolds number and offers less resistance at high Reynolds number 10.35 f for turbulent ﬂow are based on number and relative roughness 10.32 es/D. Here es refers to roughness coated to a pipe - grain roughness cial pipes commercial pipes 10.33 Two pipes are identical in diameter and carry the same ﬂuid. One of the pipes is rough and the other has a very smooth inside surface. If both the pipes have the same friction factor, when carrying the same discharge, then 10.36 friction is for 10.37 used for smaller than the laminar sublayer from smooth to rough ﬂow only 10.34 discharge, ﬂuid properties and roughness of the boundary between a circular sectional conduit, resistance resistance cal resistance References Fluid Mechanics, th Edition, Tata commercial pipes 10.38 With increasing aging of pipes, the proportion between maximum velocity and the mean velocity in turbulent ﬂow Pipe Flow Systems Concept Review 11 Introduction Pipeline systems used in water distribution; industrial application and in many engineering systems may range from simple arrangement to extremely complex ones. This chapter deals with the basic elements of a pipeline and methods of evaluating or including their effects in computations concerning pipe systems. In general, the friction factor f is a function of Reynolds number and relative roughness. The method of evaluating f is described in Chapter 10. However, for purposes of simplifying computations the value of f is assumed to be constant and known in the examples that follow in this chapter. The basic equation used in the calculation of head loss hf in a pipe is the Darcy–Weisbach equation hf = where f LV 2 2g D (11.1) L = length of a pipe of diameter D; V = mean velocity in the pipe; f = friction factor = fn (Reynolds number, relative roughness) in general. hf = Sf represents the energy slope L In long pipelines hf forms a very large part of the total loss. The ratio 11.1 MINOR LOSSES In pipeline systems there will be a large number of pipe ﬁttings such as bends, elbows, joints, valves and transitions. These ﬁttings cause localised energy losses due to their shape and these losses are classiﬁed as minor losses. Table 11.1 lists some important minor losses. Minor losses are usually neglected as insigniﬁcant if they are less than 5% of the frictional losses. 325 Pipe Flow Systems Table 11.1 Minor Losses in Pipeline Systems Situation Head loss = hL 1. Sudden expansion (V1 - V2) 2 2g 2. Sudden contraction 3. Square edged entrance to a pipe 0.5 4. Exit from a pipe V2 2g 5. Conical expansion k (V1 - V2) 2 2g k = fn (q, D2/D1) 6. Bends k V2 2g k = fn (q, R/D) 7. Pipe ﬁttings k V12/2g Values of k given in Table 11.2. 8. Nozzle Ê 1 ˆ 2 Á 2 - 1˜ V /2g ËCv ¯ Cv = coeff. of velocity of the nozzle Table 11.2 8. Globe valve (fully open) Angle valve (fully open) Gate valve (fully open) Gate valve (half open) Standard tee Standard elbow Sharp 90° bend Sharp 90° bend with vanes 11.1.1 Expansion from section 1 to 2 2 2 Ê 1 ˆ V2 ÁË C - 1˜¯ 2 g c Cc = coefﬁcient of contraction V2 = velocity in contracted section V2 2g Loss at Pipe Fittings (hL = kV2/2g)— Average Values of k Fitting Remarks/Explanation k 10.0 5.0 0.19 5.6 1.8 0.90 1.20 0.20 Equivalent Length Equivalent length Le of a minor loss is that length of the pipe which will have the same head loss for the same discharge. Square edged entrance from a reservoir Thus if the head loss in pipe ﬁttings is expressed as V2 then the length Le of a pipe of diameter D 2g and friction factor f which has the same velocity V is called its equivalent length if hL = k V2 f LeV 2 = 2g D 2g The equivalent length Le will then be k Le = 11.1.2 kD f (11.2) (11.3) Equivalent Pipes (i) A pipe with length L1, diameter, D1 and friction factor f1 will be equivalent to another pipe of corresponding parameters L2, D2 and f2, if 326 Fluid Mechanics and Hydraulic Machines f1L1 (11.4) D 52 (ii) If a set of pipes described by (L1, D1, f1), (L2, D2, f2)... are connected in series then an equivalent pipe (L e, De, fe) is related as fe Le = De5 f1L1 D15 + f 2 L2 D 52 + f 3 L3 D 53 +º 1/ 2 Ê D15 ˆ =Á ˜ Ë f1 L1 ¯ 1/ 2 Ê D 52 ˆ +Á ˜ Ë f 2 L2 ¯ 1/ 2 Ê D 53 ˆ +Á ˜ Ë f3 L3 ¯ Also, since the velocity V is same at sections 1 and 2. Ê p1 ˆ Ê p2 ˆ ÁË g + Z1 ˜¯ - ÁË g + Z 2 ˜¯ = H L (11.9) (11.5) 11.2.1 (iii) If a set of pipes (L1, D1, f1), (L2, D2, f2), (L3, D3, f3)... are connected in parallel between two points, then an equivalent pipe (L e, D e, fe) is related as Ê D 5e ˆ Á ˜ Ë f e Le ¯ H1 = H 2 + H L Then f 2 L2 = D15 Siphon A siphon is a situation where the centreline of the pipe is above the hydraulic grade line (Fig. 11.1 and 11.2). 2 1/ 2 +º (11.6) Siphon 11.2 SIMPLE PIPE PROBLEMS If a pipeline has a large number of minor head losses, the total head loss HL = f LV 2 Ê +Á 2g D Ë ˆ V2 n Â k ˜¯ 2g i 3 Z3 Datum Fig. 11.1 Power required to pump the ﬂuid over a length L, 2 p1 V + Z1 + g 2g H2 = total energy at the end of the pipe (after a length L from section 1) = Energy line HG line p2 V2 + Z2 + g 2g Siphon Tank A (11.8) By energy consideration, if H1 = total energy at beginning of the pipe = Siphon (11.7) 1 P = g QH L line Z1 V2 V2 +… HL = h f + k1 + k2 2g 2g or HG Z2 Pipe Fig. 11.2 Tank B Siphon In a siphon that portion of the pipeline which is above the hydraulic grade line experiences negative pressures. In Fig. 11.1, by taking atmospheric pressure as datum (zero). Z1 + 0 + 0 = p2 V + Z2 + 2 + HL(1 – 2) g 2g = Z3 + HL(1 – 3) 327 Pipe Flow Systems where HL(1 – 2) = total head loss between sections 1 and 2 = (friction loss + minor losses) between sections 1 and 2. Ê V2ˆ Á S k 2g ˜ represents the total minor losses in each Ë ¯ pipe. Also by continuity, since the same discharge passes through all the pipes, H L (1-3) = Z1 - Z3 It is seen that Q1 = Q2 = Q3 and p2 = ( Z1 - Z 2 ) - H L (1- 2) g (11.10) 11.2.2 Pipes in Parallel A combination of two or more pipes connected between two points so that the discharge divides at the ﬁrst junction and rejoins at the next is known as pipes in parallel (Fig. 11.4). Here the head loss between the two junctions (M and N in Fig. 11.4) is same for all the pipes. Energy line M HL HA 1 Q1 2 Q2 Q HGL HB 1 2 Q3 N Q 3 B 3 (11.12) The solution of pipes in series is generally relatively straight forward and simple. 11.2.3 Pipes in Series When two or more pipes of different diameters or roughnesses are so connected that the full discharge of the ﬂuid from one ﬂows into the others serially, the system represents a series pipeline. Figure 11.3 shows a typical set of pipes in series. The head losses are cumulative. A V1D12 = V2 D 22 = V3 D 23 i.e. (gauge pressure) Fig. 11.4 Pipes in Parallel Datum Fig. 11.3 Thus as total discharge Pipes in Series Q = Q1 + Q2 + Q3 The total head H at A an B are related as: Head loss HA - H B = H L where HL = Sum of energy losses in pipes 1, 2 and 3 Ê V2 f LV 2 ˆ Ê V2 f L V2ˆ = Á (S k ) 1 + 1 1 1 ˜ + Á (S k ) 2 + 2 2 2 ˜ 2g 2 g D1 ¯ Ë 2g 2g D2 ¯ Ë Ê V 33 Ë 2g Á (S k ) + f3 L3V 32 ˆ ˜ 2gD3 ¯ Êp ˆ Êp ˆ H L1 = H L 2 = H L3 = Á M + Z M ˜ - Á N + Z N ˜ g g Ë ¯ Ë ¯ = hM - hN (11.13) It is usual to consider minor losses as equivalent lengths in parallel pipe ﬂow problems and as such (11.11) H L =hf = fLV 2 2gD 328 Fluid Mechanics and Hydraulic Machines Two types of problems can be recognised: (1) Given piezometric heads at M and N (hM and hN) to ﬁnd Q1, Q2, ... etc. This is a straight forward problem, especially if f of each pipe is known. (2) Given total discharge Q, to determine the discharge division (i.e. to ﬁnd Q1, Q2, ... etc.). Three solution procedures are available to solve this problem 1. Exact method 2. Equivalent Pipe method 3. Trial and Error method Let there be N pipes in parallel. (1) Exact Method Put hf = r1Q 12 = r2Q 22 = … = riQ i2 = … = rnQ n2 In this ri = 8 fi Li p 2 g D 5i Ê fi LiV i2 ˆ obtained from h = Á ˜ f 2g Di ¯ Ë Q2 = r1 Q1 r2 Q3 = r1 Q1 r3 ..................... Qi = r1 Q1 ri (11.15) By continuity, total discharge ÂQ i 1 Substituting Qi from Eq. (11.15) È Q = Q1 Í1 + ÍÎ r1 + r2 È = Q1 r1 Í ÍÎ N Â 1 r1 +º+ r3 1 ri ˘ ˙ ˙˚ r1 +º+ ri Â 1 (11.16) Knowing Q1, other discharges Qi can be found from Eq. 11.15. In addition, from Eq. 11.14 h f can be found. This is an alternative exact method. In this an equivalent pipe is used as a replacement of the set of parallel pipes. Examples 11.14 and 11.16 illustrate the use of this method. (2) Equivalent Pipe Method (3) Trial and Error Method (i) Assume a trial discharge Q1¢ in pipe 1, (Fig. 11.4). f L1V1¢ 2 2 g D1 (iii) Using hf¢ as common head ﬁnd Q¢2 and Q 3¢. (iv) Find Q¢ = (Q 1¢ + Q 2¢ + Q 3¢) (v) Assume the correct discharge is split among the pipes in the same ratio as Q 1¢ : Q2¢ : Q 3¢, thus Q1 = K Q 1¢, Q2 = K Q 2¢ and Q3 = K Q 3¢ Ê Qˆ where K = Á ˜ Ë Q¢ ¯ (vi) Check the accuracy of this assumption by calculating h f 1, h f2 and h f3. If the difference is greater than 5% repeat step (iv) onwards. The Example 11.13 illustrates this method. 11.2.4 n Q = Q1 + Q2 + Q3 + ... + Qi + ... + Qn = N ˘ ˙ 1 ˙ ˙ 1 ˙ ri ˙˚ (ii) Using Q 1¢ ﬁnd hf¢ = (11.14) Thus È Í Q Í Thus Q1 = Í r1 Í Í Î r1 ˘ ˙ rN ˙˚ Branching Pipes In this at a junction three or more pipes lead off to different terminals. A typical branched pipe problem is the three-reservoir problem (Fig. 11.5). Here at the junction J the continuity equation must be satisﬁed, i.e. the ﬂow into the junction = ﬂow out of the junction. Three-Reservoir Problem Referring to Fig. 11.5, three reservoirs A, B and C are connected by three pipes 1, 2 and 3 at a common 329 Pipe Flow Systems junction J. The three-reservoir problem is about ﬁnding the direction and magnitude of the discharges in the three pipes when the geometric characteristics of the pipes and the water surface elevations of the three reservoirs are known. Two solution procedures viz. (i) Exact method and (ii) Trial and error method are in use. In this type of problem the calculations are greatly simpliﬁed by putting h = rQ2 where r = 8 fl p 2g D 5 between the junction j and the reservoirs A, B and C successively. Ha = Hj + r1Q 12 Hb = Hj – r2Q 22 Hc = Hj – r3Q 32 Put Q3 = mQ2 resulting in Q1 = (1 + m)Q2 (11.17) H (Ha – Hb) = [r1 (1 + m)2 + r2]Q 22 (Hb – Hc) = [r3m2 – r2]Q 22 (11.18) . Dividing Eq. 11.17 by 11.18 (1) Exact Method Let Hj = piezometric head at the r1 (1 + m 2 ) + r2 Ha - H b = h¢ (11.19) H b - Hc r3 m - r2 Equation 11.19 is a quadratic equation in m. Considering the positive root as the relevant value, Q2 is found by applying this value of m in Eq. 11.17 or Eq. 11.18 and hence values of Q3 and Q1. Hj is found by energy relationship between a reservoir and the junction, e.g., Ha = Hj + r1Q 12. Example 11.18 illustrates the use of the exact method for Type-1 ﬂow. junction and Ha, Hb and Hc are the piezometric heads at the highest reservoir (A), intermediate reservoir (B) and lowest reservoir (C) respectively. 2 (1) To determine the direction of ﬂow: Let Ha - H b r1 = R1 = h¢ and H b - Hc r2 h¢ > R1, the ﬂow is of Type-1 with Hj > Hb and Q1 = Q2 + Q3. h¢ < R1, the ﬂow is of Type-2 with Hj < Hb and Q1 + Q2 = Q3. h¢ = R1, the ﬂow is of Type-3 with Hj = Hb and Q1 = Q3. This is a rare situation and the problem degenerates to a case of two pipes in series. = (3) Solution Procedure for Type-2 ﬂow: The energy relationships for this type of ﬂow are: Ha = Hj + r1Q 12 (2) Solution Procedure for Type-1 ﬂow: By the application of Energy equation Hb = Hj + r2Q 22 Hc = Hj – r3Q 32 Piezometer at J EL or HGL A B 1 2 Za = Ha 3 J Zb = Hb Hj Zj Zc = Hc C Datum Fig. 11.5 Three Reservoir Problem J = Junction 1,2,3 = Pipes A,B,C = Reservoirs 330 Fluid Mechanics and Hydraulic Machines Put Q3 = mQ2 resulting in Q1 = (m – 1)Q2 (Ha – Hb) = [r1 (m – 1)2 – r2]Q 22 (11.20) (Hb – Hc) = [r3m2 + r2]Q 22 (11.21) Dividing Eq. 11.20 by 11.21 r1 ( m - 12 ) - r2 2 r3 m + r2 H - Hb = a = h¢(11.22) H b - Hc This quadratic equation in m has two positive roots. Selecting the root where m > 1 as relevant (as m < 1 gives negative Q1 values) the discharge Q2 is found from Eq. 11.20 or Eq. 11.21 and thence Q3 and Q1. The head at the junction, Hj, is found from energy relationship between a reservoir and the junction J. Example 11.19 illustrates the use of the exact method for Type-2 ﬂow. (2) Trial and Error Method (i) Assume a trial value of Hj. The ﬁrst trial Hj may be taken around the average value of the lowest and highest reservoir levels. (ii) For each Hj calculate Qi in each pipeline with positive sign if it is towards the junction and negative sign if away from the junction. Find DQ = S Qi and also ﬁnd S \|Q/h f \|. (iii) The additive correction, to be added to the assumed value of Hj for purposes of next trial, is DHj = 2D Q S \| Q /hf \| (11.23) (iv) New Hj for next trial is Hj = previous Hj + DHj. (v) Continue till DQ is very small. [Usually two iterations would sufﬁce.] 11.3 PIPE NETWORK An interconnected system of pipes is called a pipe network. The ﬂow to an outlet may come from different pipes. Figure 11.6 shows a typical network. In a network: (1) Flow into each junction must be equal to ﬂow out of each junction. (2) Algebraic sum of head losses round each loop must be zero. Qa r1 A r3 r2 r7 Qc Qb B C r8 F r5 Qf r4 r9 E D r6 Fig. 11.6 Typical Pipe Network Generally, the solution of real pipe networks, as used in engineering practice, is difﬁcult and needs the help of digital computers. Here, for illustration purposes simpliﬁed situation is used. The head loss in each pipe is expressed as hf = rQn. The coefﬁcient r depends upon pipe length, diameter and friction factor. For turbulent ﬂow n is of the order of 2. 11.3.1 Hardy–Cross Method of Analysis (i) In this method a trial distribution of discharges is made, arbitrarily but in such a way as to satisfy continuity at each node. (ii) Head loss in each pipe is calculated as h f = rQn = r Q \|Q n–1\| and also the quantity (rn \| Qn–1\|) is calculated for each pipe. (iii) For each loop, the quantity DQ = S r Qn S rn \| Q n-1 \| (11.24) is calculated. This represents the correction to the assumed discharge. (iv) Value of DQ is calculated for all loops. (v) Corrections are now applied to each pipe in a loop and to all loops. Clockwise direction is considered positive. (vi) The procedure is repeated till DQ is very small. 331 Pipe Flow Systems Examples 11.22, 11.23 and 11.24 illustrate the Hardy–Cross procedure of network analysis. where Cout = average (mixed) contaminant concentration going out Qi = inﬂow rates and Ci = contaminant concentrations of inﬂows. Example 11.25 illustrates the method of estimating the propagation of contaminants to various nodes in a steady state network. 11.3.2 11.4 [Note: The directional sense is important. Flows in a loop which are clockwise are taken positive. Positive DQ represents a correction to be added to clockwise (positive) ﬂows. Some pipes which are common to two loops may get two corrections.] Contaminant Propagation If a contaminant enters a steady state pipe network shown in Fig. 11.6 at any node (say node A) with a concentration Ca, its propagation through the network can be determined by simple mass balance, i.e. continuity relationship extended to include the contaminant. It is assumed that complete mixing takes place at a node and the contaminant is conservative, i.e. does not die off. Then at any node Cout = S QiCi S Qi (11.25) MISCELLANEOUS PROBLEMS A large number of variations in the basic situation described above are possible. Some of the common types of pipe ﬂow problems of interest are: (i) Pipelines with pumps/turbines. (ii) Nozzle at the end of a pipe. (iii) Nonuniform ﬂow in a pipe, e.g. tapering pipe or pipe with gradually decreasing discharge. (iv) Combination of series and parallel pipes. (v) Non-circular conduits These are illustrated through carefully chosen examples. Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples 11.1 A horizontal pipeline, 50 m long, is connected to a reservoir at one end and discharges freely in to the atmosphere at the reservoir the pipe has a diameter of 15 cm and it has a square entrance at the reservoir. The 30 cm. The junction of the two pipes is in the form of a sudden expansion. The 15 cm pipe has a gate valve (k the height of the water surface in the tank is 10 m above the centerline of the pipe, estimate 332 Fluid Mechanics and Hydraulic Machines the discharge in the pipe by considering the Darcy–Weisbach friction factor f = 0.02 for both the pipes. [Include all minor losses in the calculations]. Solution: Refer to Fig. 11.7. The pipe system has following (a) minor losses: (1) Square entrance with loss of head = 0.5 Here H = 10.0 m, D1 = 0.15 m, D2 = 0.30 m h12 2 ˆ V 2 ÊÊ D ˆ = 2 Á Á 2 ˜ - 1˜ ˜¯ 2 g ÁË Ë D1 ¯ 2g = (2) Valve with loss of head =k V12 V12 = 0.2 2g 2g È (V - V2 ) 2 f L2 V22 V22 ˘ + Í 1 + + ˙ (i) 2g 2 gD2 2 g ˙˚ ÍÎ 2 V22 2g From Eq. (i) h12 = È V2 f L1 V22 ˘ V12 1 Í ˙ 0 . 5 + 0 . 2 + H= 2g 2g 2 gD1 ˙ ÍÎ ˚ 2 2 ˆ V22 Ê Ê 0.30 ˆ - 1˜ ÁÁ ˜ 2 g Ë Ë 0.15 ¯ ¯ =9 (3) Sudden expansion with loss of head (V1 - V2 ) 2 2g (b) Next there exists friction loss hf in pipes 1 and 2 given by f LV 2 hf = 2 gD (c) Further there exists an exit velocity head at the V2 end of pipe 2, of magnitude 2 2g Now writing the energy equation 2 Since V1 D12 = V2 D 22 h12 V12 (V - V2 ) 2 ˆ V2 ÊV = 2 Á 1 - 1˜ = 1 2g 2 g Ë V2 ¯ H = 10.0 = V12 È 0.02 ¥ 25 ˘ 0.5 + 0.2 + Í 2g Î 0.15 ˙˚ + = 4.03 V22 2g 0.02 ¥ 25 ˘ È Í9 + 0.30 + 1˙ Î ˚ V2 V12 + 11.667 2 2g 2g Substituting V1 = V2 D22 D12 2 Ê 0.30 ˆ = V2 Á = 4V2 Ë 0.15 ˜¯ H = 10.0 = [((4)2 ¥ 4.03) + 11.667] V22 2g 10 . 0 ¥ 19 .62 = 2.5766 and V 22 = 76.147 V2 = 1.605 m/s = 76.147 10.0 m Pipe 1 Pipe 2 Square Valve Sudden enlargement entrance Fig. 11.7 Example 11.1 V2 Exit velocity V22 2g Discharge Ê pˆ Q = Á ˜ ¥ (0.30)2 ¥ 1.605 Ë 4¯ = 0.1135 m3/s. 333 Pipe Flow Systems 11.2 A 6 cm diameter pipe has a discharge of 450 11.3 A horizontal pipe of diameter D1 has a sudden expansion to a diameter D . At what ratio D1/D would the differential pressure on either side of the expansion be maximum? What is the corresponding loss of head and differential pressure head? L/min. At a section the pipe has a sudden expansion to a size of 9 cm diameter. If the kN/m , calculate the pressure just after the expansion. Assume the pipe to be horizontal at the expansion region. Solution: 450 = 7.50 L/s = 0.0075 m3/s 60 Discharge Q = Solution: (a) In a sudden expansion, the loss of head (V1 - V2 ) 2 2g hL = V1 = Velocity before expansion For an expansion in a horizontal pipe 0.0075 = p ¥ (0.06) 2 4 = 2.653 m/s 2 2 p1 V1 p2 V 2 (V1 - V2 ) 2 + = + + g 2g g 2g 2g Dp Ê p2 p1 ˆ ÁË g - g ˜¯ = g V2 = Velocity downstream of the expansion 2 ÊD ˆ = V1 Á 1 ˜ = 2.653 Ë D2 ¯ 2 Ê 0.06 ˆ ÁË 0.09 ˜¯ = 1.179 m/s Loss of head at sudden expansion HL = = (V1 - V2 ) 2 2g 2g - V 22 2g - (V1 - V2) 2 2g By continuity V1D 12 = V2D 22 ÊD ˆ V2 = V1 Á 1 ˜ = V1x2 Ë D2 ¯ ( 2.653 - 1.179) 2 = 0.111 m 2 ¥ 9.81 where 2 2 p2 V 2 p1 V1 + + Z2 + H L + + Z1 = g 2g g 2g \ Z1 = Z2 Ê D1 ˆ x= Á Ë D2 ˜¯ V12 V12 Dp 4 = (1 – x ) – (1 – x2)2 2g g 2g 2 V12 [1 – x 4 – (1 – x2 )2] 2g For maximum pressure differential = 2 2 p1 V1 V 2 p2 + - HL = g 2g 2g g = V12 2 By energy equation: As = 2 20.0 ( 2.653) (1.179) + - 0.111 9.79 2 ¥ 9.81 2 ¥ 9.81 = 2.043 + 0.359 – 0.071 – 0.111 p2 = 2.22 m g p2 = 2.22 ¥ 9.79 = 21.73 kPa d (D p / g ) =0 dx \ – 4x3 – 2 (1 – x 2) (– 2x) = 0 – 2x3 + 2x – 2x3 = 0 or i.e. (2x2 – 1) = 0 D2 = or x = 2 D1 1 2 334 Fluid Mechanics and Hydraulic Machines (b) Head loss = V12 2g V1 = V2 2 2 (1 – x ) D12 2 V2 Ê 0.30 ˆ = V2 Á = ˜ 4 Ë 0.60 ¯ Substituting these values in Eq. (i) Ê p1 p2 ˆ V22 È 1˘ ÁË g - g ˜¯ = 2.043 = 2 g Í1 + 0.29 - 16 ˙ Î ˚ Differential pressure head = D22 V12 1 V22 = 16 2 g 2g 2 2 2 1 ˆ V1 Ê 1 V1 = Á1 - ˜ = Ë 2 ¯ 2g 4 2g = V1 D21 = V2 D 22, Since (V1 - V2) 2 hL = 2g V12 Dp = (1 – x 4) – hL g 2g = 1.2275 2 V12 Ê 1 1ˆ 1 V1 1 = ◊ 2 g ÁË 4 4 ˜¯ 2 2g V 22 = V22 2g 2.043 ¥ 19.62 = 32.655 and 1.2275 V2 = 5.71 m/s 11.4 When a sudden contraction from 60 cm diameter to 30 cm diameter is introduced in a horizontal pipeline, the pressure drops from 100 kPa at the upstream of the contraction to 80 kPa on the downstream. Ê pˆ Q = Á ˜ ¥ (0.30)2 ¥ 5.71 Ë 4¯ Discharge = 0.404 m3/s Loss of head in the contraction = 0.29 ¥ (5.71) 2 V22 = 19.62 2g = 0.482 m of head due to contraction. hLc = 0.29 Solution: Designating a section upstream of contraction as 1 and the one on the downstream as 2, the loss of head at sudden contraction 2 hLc Ê 1 ˆ = Á - 1˜ . Ë Cc ¯ 2g V22 D1 = 0.60 m, D2 = 0.30 m and Cc = 0.65. Here 2 V22 Ê 1 ˆ V2 hLc = Á . - 1˜ 2 = 0.29 2g Ë 0.65 ¯ 2 g Applying energy equation to sections 1 and 2 11.5 Three pipes with diameter, length and friction factor values of (D1, L1, f1 ); (D , L , f ) and (D3, L3, f3) are connected in parallel between two points A and B in a pipeline. If an equivalent pipe (De, Le, fe) is to replace the set of parallel pipes, obtain an expression for estimating the equivalent pipe parameters. Hence p1 V12 p V2 + = 2 + 2 + hlC g 2g g 2g Ê p1 p2 ˆ V22 V2 = + hlC – 1 ÁË g ˜ g ¯ 2g 2g Ê p1 p2 ˆ Ê 100 - 80 ˆ ÁË g - g ˜¯ = ÁË 9.79 ˜¯ = 2.043 m Solution: Since there is a common head drop for the parallel pipes f1L1V12 f L V2 f L V2 = 2 2 2 = 3 3 3 2g D1 2g D2 2g D3 The sum of the ﬂows = total discharge or Q = Q1 + Q2 + Q3 hf = …(i) From Eq. (1) Ê D ˆ V1 = Á 1 ˜ Ë f1L1¯ 1/ 2 2 g ◊ hf (1) (2) 335 Pipe Flow Systems p Ê D15 ˆ 4 ÁË f1L1˜¯ 2 g ◊ hf p Ê D 25 ˆ Q2 = Á 4 Ë f 2 L 2 ˜¯ 2 g ◊ hf ˆ pÊ Á 4 Ë f3 L3 ˜¯ 2 g ◊ hf Q1 = Similarly Q3 = (i) fe Le D 5e Le D 5e D 35 Le 5 p 2 D e Ve 4 (ii) p Ê D e5 ˆ 4 ÁË fe Le ˜¯ 1/ 2 2 g ◊ hf 1/2 Ê D5 ˆ = Á 1 ˜ Ë f1 L1 ¯ 1/2 Ê D5 ˆ +Á 2 ˜ Ë f2 L 2 ¯ 1/2 Ê D5 ˆ +Á 3 ˜ Ë f3 L 3 ¯ = L1 D15 + 1800 = 3600 D 5e = 5 1800 (0.5) D 35 L2 + D 52 L3 D 35 1200 + 5 + 1200 (0.4) 5 + 5 + 600 (0.3)5 600 (0.3)5 2 2 Ê f L ˆ V3 Ê f L ˆ Ve = Á 3 3˜ = Á e e˜ Ë D3 ¯ 2 g Ë De ¯ 2 g 1/2 (3) [Note: For the equivalent pipe, out of the 3 variables De, Le and fe, given any two, the third one can be solved by Eq. (3).] f 3 L3 + 2 2 Ê f L ˆ V1 Ê f L ˆ V2 \ Á 1 1˜ = Á 2 2˜ Ë D1 ¯ 2 g Ë D2 ¯ 2 g Thus from Eq. (2) Ê De5 ˆ Á f L ˜ Ë e e¯ D 52 De = 38.57 cm (iii) Three pipes in parallel: Q0 = Q1 + Q2 + Q3 and h f1 = h f 2 = h f 3 = h fe f L V2 hf = e e e De 2 g Q= f 2 L2 + (0.4) (0.5) (0.4) Le = 4318.2 m and the head loss or D15 fe = f1 = f2 = f3 Also for the equivalent pipe Q= f1L1 = 11.6 A compound piping system consists of 600 m of 30 cm diameter pipes of the same material connected in series. (i) What is the equivalent length of a 40 cm pipe of the same material? (ii) What is the equivalent size of a pipe 3600 m long? (iii) If the three pipes are in parallel, what is the equivalent length of a 50 cm pipe? Solution: It is assumed that f is same for all the pipes. f1 = f2 = f3 = fe But \ V2 = V1 L1 D2 ◊ = L2 D1 = 1.0954 1800 ¥ 0.40 1200 ¥ 0.50 Similarly V3 = V1 1800 ¥ 0.3 = 1.3416 600 ¥ 0.5 The total discharges Q0 is given by Ê pˆ Q0 = Á ˜ [(0.5)2 V1 + (0.4)2 V2 + (0.3)2 V3] Ë 4¯ p = V1 [ 0.25 + (0.16) (1.095) 4 + (0.09) (1.3416)] Êp ˆ = (0.546) Á V1 ˜ Ë4 ¯ For an equivalent pipe Qe = Q0, the diameter De = 0.50 m and length = Le 336 Fluid Mechanics and Hydraulic Machines Since h f 1 = h f e Ve = V1 f1L1V12 f e LeVe2 = 2 gD1 2 gDe 1800 ¥ 0.5 Le ¥ 0.5 L1 De = Le D1 In the present case L1 = Le and f1 = fe 1800 /Le = p ˆ Ê Q0 = Á 0.546 ¥ V1˜ = Qe Ë 4 ¯ Also by continuity Since p = ¥ (0.5)2 Ve 4 Q e = Q0 \ Ve 0.546 = = 2.184 = V1 0.250 Le = 377.37 m Also Ê D e5 ˆ Á f L ˜ Ë e e¯ 1/ 2 Here Ê 5ˆ = Á D1 ˜ Ë f1L1 ¯ 1/ 2 Ê D5 ˆ +Á 2 ˜ Ë f 2 L2 ¯ 1/ 2 1800 Le 1/ 2 Ê D5 ˆ +Á 3 ˜ Ë f 3 L3 ¯ = 9.1004 ¥ 10–3 ( L1e/ 2 ) Le = 377.37 m 11.7 For the distribution main of a city water supply a 36 cm diameter pipe is required. As pipes above 30 cm are not available, it is decided to lay two parallel mains of the same diameter. assuming the friction factor of all the pipes to be same, determine the diameter of the parallel mains. Solution: Let sufﬁx 1 denote the 36 cm diameter pipe and with sufﬁx e denoting the equivalent pipe, (i.e., the two parallel pipes) (0.36) 2 = 15.432 D 2e Ve Ve2 (15.432 De2 ) 2Ve2 = De (0.36) +º 1/ 2 1/ 2 1/ 2 ˘ ÈÊ 5ˆ 5ˆ 5ˆ Ê Ê = ÍÁ (0.5) ˜ + Á (0.40) ˜ + Á (0.30) ˜ ˙ ÍË 1800 ¯ Ë 1200 ¯ Ë 600 ¯ ˙ Î ˚ De2 Substituting this value of V1 in Eq. (i) 1/ 2 = [(4.1667 ¥ 10–3) + (2.9212 ¥ 10–3) + (2.0125 ¥ 10–3)] 0.17678 p p V1 D 12 = 2 ¥ Ve D e2 4 4 V1 = 2Ve De = 0.50 m, f1 = f2 = f3 = fe È (0.5)5 ˘ Í ˙ ÍÎ Le ˙˚ …(i) p p V1 (0.36)2 = 2 ¥ Ve De2 4 4 Alternatively By the equivalent pipe formula for parallel pipes. Ve2 V12 = De D1 Hence D e5 = 1.5117 ¥ 10–3, and the size of the set two parallel pipes De = 0.2728 m = 27.28 cm The next higher size of standard pipe would be used. 11.8 Two reservoirs with a difference in water surface elevation of 10 m are connected by a pipeline ABC which consists of two pipes of AB and BC joined in series. Pipe AB is 10 cm in f an f = 0.018. The junctions with the reservoirs and between the pipes are abrupt. (a) Calculate the discharge. (b) What difference in reservoir elevations is necessary to have a discharge of 15 L/s? [Include all minor losses]. Solution: Use sufﬁx 1 for pipe AB and sufﬁx 2 for pipe BC. (i) Entrance loss hL1 = 0.5 V 12 /2g 337 Pipe Flow Systems (ii) Loss at sudden expansion = heL = 2 (V1 - V2) 2 2g V12 Ê V2 ˆ = 1- ˜ V1 ¯ 2 g ÁË 2 = 0.39063 V1 HL = (5.8714 + 0.4292) Since p 2 p D 1 V1 = D 22 V2 = Q 4 4 ÊD ˆ V2 = Á 1˜ V1 Ë D2 ¯ = 6.300 heL Ê Ê D ˆ 2ˆ V 2 1 = Á1 - Á 1 ˜ ˜ ÁË Ë D2 ¯ ˜¯ 2 g 2 Ê Ê 0.10 ˆ 2 ˆ V12 = Á1 - Á ˜ ˜ Ë Ë 0.16 ¯ ¯ 2 g V12 = 0.3714 2g V 22 2g (iv) Friction loss 2 f1L1 V1 D1 2 g 2 V12 0.02 ¥ 20 V1 = 4.0 = 0.10 2 g 2g 2 f L V2 = 2 2 D2 2 g 2 0.018 ¥ 25 V 2 = 0.16 2g V 22 = 2.8125 2g hf1 = hf2 Total loss HL HL = 0.5 V12 2g 0.3714 + 4.0 HL = 5.8714 V12 2g V12 2g V12 2g 2.8125 + pipes through division at M to rejoin at N (Fig. 11.8). Estimate the division of discharge in the two pipes. f1 = 0.018 2g 1 Q1 6 cm, 1000 m Q 2g V 22 11.9 A pipe 6 cm in diameter, 1000 m long and with f = 0.018 is connected in parallel between two points M and N with another pipe 8 cm diameter, 800 m long and having f V 22 2.8125 2g V12 (b) When Q = 15 L/s = 0.015 m3/s 0.015 0.015 V1 = = p 0 . 007854 ¥ (0.1) 2 4 = 1.91 m/s V 12 = 3.648, HL = 6.300 V 12/2g HL = 1.171 m (iii) Loss at exit he = V12 2g (a) When HL = 10 m, Ê 10 ¥ 2 ¥ 9.81ˆ V 12 = Á ˜¯ = 31.14; Ë 6.300 V1 = 5.582 m/s V2 = 2.18 m/s p and Q = ¥ (0.10)2 ¥ 5.582 4 = 0.0438 m3/s = 43.8 L/s 2 2 \ 2 ÊD ˆ Ê 0.10 ˆ But V2 = Á 1 ˜ V1 = Á V Ë 0.16 ˜¯ 1 Ë D2 ¯ Q N M V 22 2g 8 cm, 800 m f2 = 0.020 2 Q2 Fig. 11.8 338 Fluid Mechanics and Hydraulic Machines Solution: By continuity consideration, A1V1 = A2V2 = Q = Total discharge p p (0.06)2 V1 + (0.08)2 V2 = 0.020 4 4 11.10 Three pipes are connected in parallel between two reservoirs A and B. The details of the pipes are: Pipe Diameter Length f V1 + 1.7778 V2 = 7.074 (1) By considering the head loss between M and N f1L1V12 = 2 g D1 f 2 L2V 22 If the difference in the water level elevations of 2 g D2 0.018 ¥ 1000 2 0.02 ¥ 800 2 V1 = V2 0.06 0.08 2 V 12 = V 22 3 V1 = 0.8165 V2 or in each pipe. (2) Substituting this in Eq. (1) and and Solution: The head loss h f = 12.0 m is common to all the three pipes. f1 L1 V12 Hence h f = 12.0 = D1 2 g f 2 L2 V 22 f3 L3 V32 = D2 2 g D3 2 g Discharge calculations in the three pipes are shown in the following Table: = 0.8165 V2 = 1.7778 V2 = 7.074 V2 = 2.727 m/s p Q2 = ¥ (0.08)2 ¥ 2.727 4 = 0.0137 m3/s = 13.7 L/s V1 = 0.8165 ¥ 2.727 = 2.227 m/s Q1 = 0.0063 m3/s = 6.3 L/s Table 11.3 Table of Example 11.10 Since 8 fi Li p 2 g D i5 3 Diameter D (m) 0.10 0.15 0.12 0.00785 0.01767 .01131 220 96 158.3 0.05455 0.1250 0.0758 1.034 1.566 1.219 0.0081 0.0277 0.0138 Ê p 2ˆ ÁË 4 D ˜¯ = 0.0826 f L/D5. Pipe f L/D5 r 1 2 23,148,000 4,882,800 1912000 403320 hf = 2 Area (m ) h f = ri Q 2i = r1 Q 12 1 2 Alternate Solution: Put Pipe No. ( fL ) D ÊV2 ˆ hf Á 2g ˜ = ( fL / D ) Ë ¯ = r2Q 22 0.5 Q2 = (r1/r2) Q1 = (403320/1912000)0.5 Ql = 0.4593 Q1 Q = Q1 + Q2 = 1.4593 Q1 = 0.02 Q1 = 0.0137 m3/s and Q2 = 0.02 – 0.0137 = 0.0063 m3/s. V (m/s) 3 Q (m /s) 11.11 Two pipes A and B are connected in parallel between two points. Pipe A is 150 m long and has a diameter of 15 cm. Pipe B the pipes have the same friction factor of 0.018. 339 Pipe Flow Systems A partially closed valve in pipe A causes the discharge in the two pipes to be the same (Fig. All other minor losses can be neglected. f = 0.018 A Valve D h1 to h . Neglect h . If D1 minor losses and assume the friction factor f to be constant and to have the same value for both the pipes. Solution: f LV 2 2g D Head loss hf = 150 m, 15 cm dia f LQ 2 = Êp ˆ 2gD Á D 2 ˜ Ë4 ¯ 100 m, 12 cm dia B f = 0.018 p p ¥ (0.15)2 Va = ¥ (0.12)2 Vb 4 4 2 Ê 0.15 ˆ Vb = Á Va = 1.5625 Va Ë 0.12 ˜¯ Let the loss in the valve be KL V a2 /2g. Head losses in both pipes are the same. Hence fa LaVa2 V2 f L V2 + KL a = b b b 2 g Da 2g 2 g Db h1 = Ê Q1 ˆ ÁË Q ˜¯ 2 or or p 2g D15 ÊD ˆ = Á 1˜ Ë D2 ¯ Also p 2 g D5 or = 8 f LQ22 p 2g D25 5 5/ 2 or ÊD ˆ Q1 = Q2 Á 1 ˜ Ë D2 ¯ 5/ 2 Q1 + Q2 = Q ÈÊ D ˆ 5 / 2 ˘ Q2 ÍÁ 1 ˜ + 1˙ = Q ÍË D2 ¯ ˙ Î ˚ 5/2 Putting D1 = 2D2, Q = (2 + 1) Q2 = 6.65685 Q2 h1 = = 8 f LQ2 2 p g D25 (6.65685) 2 1 8 f LQ2 ( 44.313) p 2g D25 Case II: h2 = 11.12 Two pipes each of length L and diameters D1 and D are arranged in parallel; the loss of head when a total quantity of water Q h1. If the pipes are arranged in series and the same quantity of water, Q 2 8 f LQ12 Ê Q1 ˆ Ê D1 ˆ ÁË Q ˜¯ = ÁË D ˜¯ 2 2 0.018 ¥ 150 V a2 V2 ¥ + KL a 0.15 2g 2g 0.018 ¥ 100 ¥ (1.5625) 2 Va2 = ¥ 0.12 2g 18 + KL = 36.62 KL = 18.62 8 f LQ 2 Case I: Fig. 11.9 Solution: Since the discharges are same in both the pipes: Aa Va = Ab Vb 2 = 8 f LQ2 p 2g D15 + 8 f LQ2 p 2g D25 = 8 f LQ2 Ê 1 1 ˆ + Á ˜ p 2g Ë D15 D 52 ¯ = 5 ˆ 8 f LQ2 Ê Ê D2 ˆ 1 + ˜ Á Á ˜ ˜¯ p 2g D25 ÁË Ë D1 ¯ (1) 340 Fluid Mechanics and Hydraulic Machines for D1 = 2D2; h2 = = (1.03125) 5 ¸Ô 8 f LQ2 ÏÔÊ 1 ˆ + 1˝ Ì ˜ 2 5 Á p g D2 ÔÓË 2 ¯ Ô˛ Also 8 f LQ2 (2) p 2g D25 From Eqs (1) and (2), 11.13 Three pipes with details as the following, are connected in parallel between two points. Pipe Diameter Length 0.02 ¥ 800 ¥ V 32 = 12.91 2 ¥ 9.81 ¥ 0.15 V 32 = 2.375 and Discharge h1 1 1 ¥ = 0.02188 = h2 ( 44.313) (1.03125) f3 L3 V32 = 12.91 2g D3 f When a total discharge of 0.30 m3 through the system, calculate the distribution of the discharge and the head loss between the junctions. p ¥ (0.15)2 ¥ 1.541 4 = 0.0272 m3/s Q3 = Total discharge SQi = 0.0500 + 0.1453 + 0.0272 = 0.2225 m3/s But the actual total discharge is 0.30 m3/s The discharges in each pipe is now corrected by 0.30 multiplying the trial discharges by the ratio 0.2225 = 1.3483. Hence corrected discharges in each pipe are: Pipe 1 = Q1 = 0.05 ¥ 1.3483 = 0.0674 m3/s Pipe 2 = Q2 = 0.1453 ¥ 1.3483 = 0.1959 m3/s Pipe 3 = Q3 = 0.0272 ¥ 1.3483 = 0.0367 m3/s Solution: A trial and error procedure is adopted. Total 3 1st trial: Assume a trial discharge Q1 = 0.05 m /s in pipe 1. Èp ˘ V1 = 0.05 Í ¥ (0.2) 2 ˙ = 1.592 m/s Î4 ˚ hf = f1 L1 V12 0.02 ¥ 1000 ¥ (1.592) 2 = 2 ¥ 9.81 ¥ 0.20 2 g D1 = 12.91 m Using this h f: f 2 L2 V22 = 12.91 2g D2 0.015 ¥ 1200 ¥ V 22 = 12.91 2 ¥ 9.81 ¥ 0.3 V 22 = 4.22157 V2 = 2.055 m/s p and discharge Q2 = ¥ (0.3)2 ¥ 2.055 4 = 0.1453 m3/s V3 = 1.541 m/s 0.3000 m3/s 2nd trial: Head loss between the junctions by considering pipe 1: f1L1V12 0.02 ¥ 1000 (0.0674) 2 ¥ = 2 ¥ 9.81 ¥ 0.20 Ê p ˆ 2 2g D 4 ÁË 4 ˜¯ (0.2) = 23.46 m Q2 and Q3 are found by using this h f. By considering pipe 2: hf = f 2 L2 V 22 0.015 ¥ 1200 ¥ V22 ¥ 2 ¥ 9.81 ¥ 0.3 2g D 2 = 23.46 m V 22 = 7.6713 or V2 = 2.7697 m/s p Discharge Q2 = ¥ (0.3)2 ¥ 2.7697 4 = 0.1958 m3/s hf = 341 Pipe Flow Systems f3 L3 V32 In pipe 3 h f = 2g D3 0.02 ¥ 800 ¥ V 32 = 2 ¥ 9.81 ¥ 0.15 = 23.46 m V 32 = 4.3151 or V3 = 2.0773 m/s p Discharge Q3 = ¥ (0.15)2 ¥ 2.0773 4 = 0.0367 m3/s Total discharge = S Qi = 0.0674 + 0.1958 + 0.0367 = 0.2999 m3/s As this discharge is essentially the same as the given discharge, no more trials are needed. Hence Q1 = 0.0674 m3/s, Q2 = 0.1958 m3/s, Q3 = 0.0367 m3/s and hf = 23.46 m. Q = Q1 + Q2 + Q3 = [1 + 2.9048 + 0.5446] Q1 0.30 = 4.4494 Q1 Q1 = 0.0674 m3/s Q2 = 2.9048 ¥ 0.0674 = 0.1959 m3/s and Q3 = 0.30 – 0.0674 – 0.1959 = 0.0366 m3/s 11.14 branch off from a point A in a pipeline 600 m long. The total head at A is 30 m. A short is discharged into atmosphere through it (Fig. 11.10). Assuming f the total discharge and division of discharge in A [Note: If the total discharge at the end of second trial differed substantially the discharges would have to be once again readjusted by the ratio Qactual/(SQi)trial and the distribution checked by a 3rd trial and so on. Usually 2 trials would be sufﬁcient.] Alternative Solution: Put Since 10 cm dia, 600 m 8 cm dia 2 Fig. 11.10 p 2g D 5i = 0.0826 f L/Ds. 5 Pipe 1 2 3 10 cm dia, 400 m f L/D 62500 7407 210700 = r2Q 22 = r3Q 32 (r1/r2)0.5 Q1 = (5164/612)0.5 HB = Head at B = r 5164 612 1741 HA – HB = 30.0 – V 32 2g V 32 2g Consider an equivalent pipe De = 0.08 m and fe = 0.02 to replace the parallel pipes 1 and 2. Then Ê D 5e ˆ Á f L ˜ Ë e e¯ r1Q 12 hf = Q2 = Q1 = 2.9048 Q1 Q3 = (r1/r3)0.5 Q1 = (5164/1741)0.5 Q1 = 0.5446 Q1 Example 11.14 As the 8 cm pipe is short, the friction loss in it can be neglected. 8 fi L i h f = riQ i2 where ri = To atmos B 1 Ha = 30 m Since 1/ 2 Ê D5 ˆ = Á 1 ˜ Ë f1L1 ¯ 1/ 2 Ê D5 ˆ +Á 2 ˜ Ë f 2 L2 ¯ 1/ 2 fe = f1 = f2 and D1 = D2 = 0.10 m (0.08)5 / 2 L1e/ 2 È 1 1 ˘ = (0.1)5/2 Í + ˙ 400 600 Î ˚ L 1/2 e = 6.3026 and Le = 39.72 m 342 Fluid Mechanics and Hydraulic Machines As De = 0.08 m, velocity in this pipe = V3 = Ve \ Head loss HA – HB = 30 – = V 32 f L V2 = e e e De 2 g 2g 0.02 ¥ 39.72 V 32 V 32 = 9.93 0.08 2g 2g V 32 30 = 2.745 = 2g 10.93 V3 = 7.338 m/s p Q= ¥ (0.08)2 ¥ 7.338 4 = 0.0369 m3/s HA – HB = 30 – 2.745 = 27.255 m = h f 1 = h f2 f1L1 V12 f 2 L 2 V 22 0.02 ¥ 600 V 22 ¥ = 27.255 = D2 2 g 0.1 2g V2 = 2.111 m/s p Q2 = ¥ (0.1)2 ¥ 2.111 4 = 0.0166 m3/s Check: Q1 + Q2 = 0.0203 + 0.0166 = 0.0369 = Q 15 m A 11.15 A city water supply main is 1000 m long two reservoirs with a head difference of 15 m. adding another pipe of the same diameter as the main from the upstream reservoir in parallel and joining it to the main at a suitable junction. Estimate the length of the additional pipe and the head at the junction relative to the downstream reservoir water surface. Assume the lengths of L1 J L3 B L2 C Fig. 11.11 (a) Before introduction of the pipe: V12 0.02 ¥ 400 ¥ = = 27.255 D1 2 g 0.1 2g V1 = 2.585 m/s p Q1 = ¥ (0.1)2 ¥ 2.585 4 = 0.0203 m3/s \ the two parallel pipes to be same. Also, all pipes can be assumed to have the same friction factor f. [Refer Fig. 11.11]. hf = where f L0 Q 02 f L0V 02 = 2g D 2 g D 5 ( p /4) 2 = KL0Q02 K = f /[2gD5 (p/4) 2 ] L0 = 1000 m, Q0 = 0.1 m3/s, h f = 15 m 15 = 1.5 K = 1000 ¥ (0.1) 2 (b) After the introduction of the parallel pipe: Fig. 11.11 shows the schematic layout. For pipes AJ and BJ f LV 2 f L V2 hf1 = 1 1 1 = 2 2 2 2g D1 2g D 2 Since f1 = f2, L1 = L2, D1 = D2 = D V1 = V2 \ Q1 = Q2 Hence Q3 = Q1 + Q2 = 2Q1 Head loss between the two reservoirs f L1V 12 f L3V 32 + HL = 15 m = 2g D 2g D Since L1 + L3 = 1000 m Here f L1V 12 f (1000 - L1) 2 + V 3 = 15 2g D 2g D In terms of the new total discharge Q3 343 Pipe Flow Systems L2, D2, V2, Q2 1 2 Q3 f L1 f (1000 - L1) Q 32 ¥ 4 2 + = 15 2 2g Ê p ˆ Ê pˆ 5 5 2g Á ˜ D ÁË 4 ˜¯ D Ë 4¯ Putting, as before, f =K 2 Ê pˆ 5 2g Á ˜ D Ë 4¯ ÊL ˆ K Á 1 + 1000 - L1 ˜ Q 32 = 15 Ë 4 ¯ 3 ˆ Ê K Á1000 - L1 ˜ Q 32 = 15 Ë 4 ¯ = K ¥ 1000 ¥ Q 02 2 ÊQ ˆ 3 ˆ Ê Hence Á1000 - L1 ˜ = Á 0 ˜ ¥ 1000 Ë ¯ 4 Ë Q3 ¯ But Q3= 1.30 Q0, and as such 1000 – L1, D1 V1, Q1 Fig. 11.12 15 = p ¥ (0.5)2 ¥ 1.918 4 = 0.3766 m3/s Discharge Q0 = Case II: L1 = 1000 m D1 = 0.50 m = K (455.6) (0.13)2 = 1.5 ¥ (455.6) (0.13)2 = 11.55 m 11.16 A pipeline carrying water has a diameter the delivery another pipeline of the same the second half of its length. Find the increase in discharge if the total head loss in both the cases is 15 m. Assume f Solution: Refer to Fig. 11.12 Case I: hf = f LV02 2g D0 L2 = 1000 m D2 = 0.50 m Total head loss h f = 15 m = h f1 + hf2 3 L1 = 591.7 4 L1 = L2 = 544.4 m f L3V32 f (1000 - 544.4) 2 Q3 = 2 2gD Êpˆ 5 2gÁ ˜ D Ë 4¯ 0.02 ¥ 2000 ¥ V02 = 4.0775 V 02 2 ¥ 9.81 ¥ 0.5 V 02 = 3.679 or V0 = 1.918 m/s h f = 15 = f1L1V12 f L V2 + 2 2 2 2g D1 2g D2 0.02 ¥ 1000 0.02 ¥ 1000 V 12 + V 22 2 ¥ 9.81 ¥ 0.5 2 ¥ 9.81 ¥ 0.5 h f = 2.039 (V 12 + V 22 ) = 15 m Head at the junction, Hj (above the downstream reservoir water surface elevation): Hj = h f3 = head loss in pipe 3 = L2, D2, V2, Q2 = Since the discharge Q1 = 2Q2 D 12 V1 = 2D 22 V2 Since D1 = D2, V1 = 2V2 \ h f = 2.039 (4V 22 + V22 ) = 15 V22 = 1.4715 or V2 = 1.213 m/s p Q1 = 2Q2 = 2 ¥ ¥ (0.5)2 ¥ 1.213 4 = 0.4763 m3/s Increase in discharge = Q1 – Q0 = 0.4763 – 0.3766 = 0.0997 m3/s 0.0997 or ¥ 100 = 26.48% 0.3766 Alternative method: For the set of two parallel pipes in the second half, consider an equivalent pipe of 0.5 m diameter and f = 0.02. 344 Fluid Mechanics and Hydraulic Machines The equivalent pipe (De, fe, Le) is related to the two parallel pipes it replaces as ÈÊ 5 ˆ 1/ 2 Ê 5 ˆ 1/ 2 ˘ D D = ÍÁ 1 ˜ + Á 2 ˜ ˙ ÍË f1L1 ¯ f Ë 2 L2 ¯ ˙ Î ˚ Here D1 = 0.50 m D2 = 0.50 m De = 0.50 m L1 = 1000 m L2 = 1000 m Le = ? f1 = 0.02 f2 = 0.02 fe = 0.02 Ê De5 ˆ Á f L ˜ Ë e e¯ 1/ 2 È (0.5)5 ˘ Í ˙ ÍÎ 0.02 ¥ Le ˙˚ 1 1/ 2 hf = 15.0 = = = Ê 0.02 ¥ 1250 ˆ 2 hf = Á V = 15.0 m Ë 2 ¥ 9.81 ¥ 0.5 ˜¯ 0.025 ¥ (120 + 680)V 2 2 ¥ 9.81 ¥ 0.15 (ii) Applying the energy equation between the water surface at station A and the summit station S, by considering the absolute pressures Za + V2 = 5.886 and V = 2.426 m/s p Discharge Q = ¥ (0.5)2 ¥ 2.426 4 = 0.4763 m3/s pa p V2 + 0 = Zs + s + s g g 2g 100.00 + 10.0 = 105.00 + 11.17 The pipeline connecting Tanks A and ps = (4.887 ¥ 9.79) = 47.85 kPa. (absoute) B passes over a high ground S. The elevations of water levels in the tanks and the centerline of the pipe at S are as follows: Station A (Water surface elevation) 100.00 m Station B (water surface elevation) 85.00 m Station S (Centerline of pipe at S) 105.00 m The pipeline connecting A to S is 15 cm in tank B it is of 15 cm diameter and is 680 m long. estimate the (i) discharge and (ii) pressure at the summit station S. [Take atmospheric pressure as 10.0 m (abs).] ps Vs2 + g 2g ps V2 = 5.0 – s g 2g È (1.486) 2 ˘ = Í5.0 ˙ 2 ¥ 9.81 ˙˚ ÍÎ = 4.887 m (absolute) Increase in discharge = 0.4763 – 0.3766 = 0.0997 m3/s = 26.48% f lV 2 2 gD 15.0 = 6.795 V 2 V = 1.486 m/s Discharge p Q = ¥ (0.15)2 ¥ (1.486) 4 = 0.0263 m3/s 1/ 2 ˘ È ÔÏ (0.5)5 Ô¸ ˙ Í = 2¥Ì ˝ Í ÔÓ 0.02 ¥ 1000 Ô˛ ˙ Î ˚ 2 31.623 and Le = 250 m Now the total length of pipe = L = 1000 + 250 = 1250 m L1/2 e Solution: (i) Total loss of head = 100.0 – 85.0 = 15.0 m 11.18 For the branching system shown in Fig. 11.13, calculate the discharge in each pipe. Take f Pipe 1 Dia 15 cm Length (m) 350 (Neglect minor losses.) Connectivity AJ 345 Pipe Flow Systems EL: 126.00 m Pipe r hf (m) A 1 1 7617 126.00 – 116.54 = 9.46 + 0.0352 35.2 2 33051 116.54 – 109.00 = 7.54 – 0.0151 –15.1 3 41313 116.54 – 100.00 = 16.54 – 0.020 – 20.0 EL: 109.00 m B EL: 100.00 m J As the error in the discharge is 0.1 L/s (0.0001 m /s) no further iteration is necessary and the ﬁnalised discharges are 3 Fig. 11.13 Solution: Putting h f = r Q2, 8f L 8 ¥ 0.02 L r= 2 5 = 2 p gD p ¥ 9.81 D 5 L = 1.6525 ¥ 10–3 5 D A trial and error solution method is adopted. Flow away from the junction is taken as negative. For ﬁrst trial assume Hj = elevation of hydraulic grade line at the junction J = 114.00 m. Q1 = 35.20 L/s Q2 = –15.15 L/s Q3 = –20.05 L/s Alternate solution procedure (Exact method) h f = rQ2 Putting r= Estimated h f Q = hf / r (m) (m3/s) Q \|Q/h f\| (L/s) (Q in L/s) 1 7617 126.0 – 114.0 = 12.0 + 0.0397 +39.7 2 33051 114.0 – 109.0 = 5.0 –0.0123 –12.3 3 41313 114.0 – 100.0 = 14.0 –0.0184 –18.4 3.31 2.46 1.31 DQ = 9.0 S\|Q/h f \| = 7.08 2 ¥ DQ 2 ¥ 9.0 DHj = = S \| Q / hf\| 7.08 = 2.54 m Hj for next trial = 114.00 + 2.54 = 116.54 m Second trial Hj = 116.54 m 8fl 2 Pipe 1 2 3 p gD 5 = 8 ¥ 0.02 2 p ¥ 9.81 L = 1.6525 ¥ 10–3 5 D Hj = 114.0 m r (towards reservoir B) (towards reservoir C) The elevation of the piezometric head at junction Hj = 116.54 m. First trial Pipe 3.72 2.00 1.21 DQ = 0.1 S\|Q/hf\| = 6.93 2 3 C Q Q \|Q/hr\| (m3/s) (L/s)(Q in L/s) L(m) 350 200 250 D(m) 0.15 0.10 0.10 ¥ L D5 r 7617 33051 41313 Let Hj = piezometric head at the junction J. Similarly Ha, Hb and Hc are piezometric heads at reservoirs A, B and C respectively. h¢ = Ha - Hb 126.00 - 109.00 = = 1.89 Hb - Hc 109.00 - 100.00 r1 7617 = = 0.184 < h¢ r3 41313 Since h¢ > R1, the ﬂow is of Type-1, i.e. Hj > Hb and Q1 = Q2 + Q3. By the application of Energy equation between the junction J and the reservoirs A, B and C successively. R1 = 346 Fluid Mechanics and Hydraulic Machines Ha = Hj + r1 Q 12 Hb = Hj – r2 Q 22 Hc = Hj – r3 Q 32 \ (Ha – Hb) = r1Q 12 + r2Q 22 = 126.00 – 109.00 = 17 (Hb – Hc) = r3 Q 32 – r2 Q 22 = 109.00 – 100.00 =9 Substituting the values of r1, r2 and r3 7617Q 12 + 33051Q 22 = 17 41313Q 32 11.19 A water supply system consists of three reservoirs A, B and C connected to a common junction J as shown in Fig. 11.14. EL: 100.00 A EL: 98.00 B 1 2 33051Q 22 – =9 Put Q3 = mQ2 resulting in Q1 = (1 + m) Q2. Substituting in the above equations Q 22 [7617 (1 + m)2 + 33051] = 17 Q 22 [41313 m2 – 33051] = 9 3 (1) (2) EL: 90.00 C Dividing (1) by (2) 7617(1 + m) 2 + 33051 41313m 2 - 33051 (1 + 2m + m 2 ) + 4.339 5.424 m 2 - 4.339 Fig. 11.14 Example 11.19 = 1.889 Pipe Dia Length f Connectivity = 1.889 On simplifying m2 – 0.216 m – 1.464 = 0 \ 0.216 ± (0.216) 2 + 4 ¥ 1.464 = 1.323 m= 2 [Note: Only positive root is relevant.] Hence Q2 Q3 Q1 Hj Solution: Putting h f = r Q2 r= Substituting in Eq. (2) Q 22 Calculate the discharge in each pipe and the piezometric head at the junction. 2 5 p gD = 34.0 fL 2 = [41313 (1.323) – 33051] = 9 = 0.0151 m3/s = mQ2 = 1.323 (0.0151) = 0.200 m3/s = (1 + m) Q2 = 2.323 (0.151) = 0.0351 m3/s = piezometric head at the junction = Hb + r2 Q 22 = 109.000 + 33051 (0.0151)2 = 116.54 m 8f l Pipe 1 2 3 L(m) 200 125 250 = 8 2 p ¥ 9.81 f 0.02 0.016 0.016 ¥ fL (0.30)5 r 136 68 136 Let Hj = piezometric head at the junction J. Similarly Ha, Hb and Hc are piezometric heads at reservoirs A, B and C respectively. 347 Pipe Flow Systems Ha - Hb 100.00 - 98.00 = = 0.25 Hb - Hc 98.00 - 90.00 r 136 R1 = 1 = = 1.0 > h¢ r3 136 Hf = piezometric head at the junction = Hc + r3Q 32 = 90.00 + 136 (0.234)2 = 97.447 m h¢ = Since h¢ < R1, the ﬂow is of Type-2, i.e., Hj < Hb and Q1 + Q2 = Q3 The energy relationships for this type of ﬂow are Hb = Hj + r2Q 22 Hc = Hj – r3Q 32 (Ha – Hb) = r1Q 12 – r2Q 22 = 100.00 – 98.00 = 2.0 (Hb – Hc) = r2Q 22 + r3Q 32 = 98.00 – 90.00 = 8 Substituting the value of r1, r2 and r3 136Q 12 – 68Q 22 = 2 68Q 22 136Q3 = 8 Put Q3 = mQ2 resulting in Q1 = (m – 1) Q2. Substituting in the above equations Q 22 [136 (m – 1]2 – 68] = 2 (1) 2 2 Q 2 [68 + 136 m ] = 8 (2) Dividing (1) by (2) 136( m - 1) 2 - 68 68 + 136m 2 2( m - 1) 2 - 1 m= pipe branching into 60 m long 10 cm diameter 60 m branch. The water surface in the tank is at f pipes, estimate the discharges in the two outlets. Solution: Figure 11.15 shows the schematic layout of the tanks and pipelines. According to the given data: Pipe 1 2 3 Diameter 20 cm 10 cm 10 cm Length 100 m 60 m 120 m 2.667 ± ( 2.667) 2 - 4 ¥ 0.5 = 2.46 2 Outlet Junction J Atmosphere Atmosphere [This problem is a variation of the three reservoir problem and the same technique is adopted for solution]. = 0.25 = 0.25 1 + 2m 2 On simplifying m2 – 2.667 m + 0.5 = 0 \ 11.20 A water tank discharges to the atmosphere elevations above datum of the outlets are Ha = Hj + r1Q 12 \ Ê 8f L ˆ f1L1 V12 ◊ = Á 2 5 ˜ Q 12 D1 2 g Ë p gD ¯ For pipe 1: h f1 = r1Q 12 = Ê 8f L ˆ 8 ¥ 0.02 ¥ 100 r1 = Á 2 5 ˜ = 2 p ¥ 9.81 ¥ (0.2)5 Ë p gD ¯ = 516.4 EL: 125.00 m [Note: Only positive root is relevant as m < 1 gives negative values of Q1.] Substituting in Eq. (2) Q 22 Hence Q2 Q3 Q1 = [68 – 138 (2.46)2] = 8 = 0.095 m3/s = mQ2 = 2.46 (0.095) = 0.234 m3/s = (m – 1) Q2 = 1.46 (0.095) = 0.139 m3/s J 2 1 B EL: 105.00 m To atmos 3 C Fig. 11.15 EL: 100.00 m 348 Fluid Mechanics and Hydraulic Machines Ê f 2 L2 ˆ V2 + 1˜ 2 For pipe 2: h f 2 = r2Q 22 = Á Ë D2 ¯ 2g r2 = Ê f L ˆ 8 Á 2 2 + 1˜ Ë D2 ¯ = p 4 gD 4 r2 = 10742 For pipe 3: h f 3 = r3 = r3Q 32 = DHj = – p2 g D 4 Ê 0.02 ¥ 120 ˆ + 1˜ 8Á Ë ¯ 0.1 p 2 ¥ g ¥ (0.1) 4 No. hf (m) Q= hf / r Q (m3/s) \|Q/hf\| (L/s)(Q in L/s) 22.0 2.4 1.5 2 ¥ SQ 2 ¥ 17.5 = 25.9 S \|Q / hf \| = 1.35 m New Hj = 121.0 + 1.35 = 122.35 m for the next trial. Second trial Hj = 122.35 m (m) hf / r Q 3 (m /s) \|Q/hf\| (L/s) (Q in L/s) 1 516.4 125.0 – 122.35 = 2.65 +0.0716 +71.6 2 10742 122.35 – 105.0 = 17.35 –0.0402 –40.2 3 20657 122.35 – 100.0 = 22.35 –0.0329 –32.9 hf / r Q 3 (m /s) \|Q/hf\| (L/s) (Q in L/s) 26.5 2.3 1.5 S = – 0.0 S = 30.3 DHj = Q= Q= 1 516.4 125.0 – 122.25 = 2.75 +0.073 73.0 2 10742 122.25 – 105.0 = 17.25 –0.040 –40.0 3 20657 122.25 – 100.0 = 22.25 –0.033 –33.0 S = +17.5 S = 25.9 hf hf = 20657 1 516.4 125.0 – 121.0 = 4.0 +0.0880 +88.0 2 10742 121.0 – 105.0 = 16.0 –0.0386 –38.6 3 20657 121.0 – 100.0 = 21.0 –0.0319 –31.9 Pipe r Hj = 122.25 (m) Q 32 First trial Let Hj = 121.0 m r Third trial Pipe r Ê f L ˆ 8 Á 3 3 + 1˜ Ë D3 ¯ A trial and error procedure with an assumption of the energy head at the junction J is adopted. Pipe New Hj for the 3rd trial = 122.35 – 0.10 m = 122.25 m Ê 0.02 ¥ 60 ˆ + 1˜ 8Á Ë ¯ 0.1 p 2 ¥ g ¥ (0.1) 4 2 ¥ 1.5 = – 0.1 m 30.8 27.0 2.3 1.5 S = –1.5 S = 30.8 DHj = 2 ¥ ( 0.0) = – 0.00 30.3 As the error in DQ is nil no further iteration is needed. Hence and Hj Q1 Q2 Q3 = 122.25 m above datum. = 73.0 L/s = 40.0 L/s = 33.0 L/s Alternative Precedure: Exact Method Obviously the ﬂow is of Type-1, i.e. Hj > Hb and Q1 = Q2 + Q3. By the application of Energy equation between the junction J and the reservoir A, and outlets B and C successively Ha = Hj + r1Q 12 Hb = Hj – r2 Q 22 Hc = Hj – r3 Q 32 \ (Ha – Hb) = r1Q 12 + r2Q 22 = 125.00 – 105.00 = 20 (Hb – Hc) = r3 Q 23 – r2 Q 22 = 105.00 – 100.00 = 5 Substituting the values of r1, r2 and r3 516.4Q 21 + 10742Q 22 = 20 20657Q 32 – 10742Q 22 = 5 Put Q3 = mQ2 resulting in Q1 = (1 + m) Q2. 349 Pipe Flow Systems Substituting in the above equations Q 22 EL: 80.00 m 2 [516.4 (1 + m) + 10742] = 20 (1) Q 22 [20657 m2 – 10742] = 5 (2) A J Dividing (1) by (2) 3 561.4(1 + m) 2 + 10742 = 4.0 20657m 2 - 10742 0.25 m /s EL: 70.00 m B (1 + 2m + m 2 ) + 20.80 =4 40.0 m 2 - 20.80 On simplifying m2 – 0.01258 m – 0.660 = 0 EL: 60.00 m 2 \ m= 0.01258 ± (0.01258) + 4 ¥ 0.660 = 0.819 2 C Fig. 11.16 [Note: Only positive root is relevant.] Substituting in Eq. (2) Q 22 = [20657 (0.819)2 – 10742] = 5 Q2 = 0.04 m3/s = 40 L/s Q3 = m Q2 = 0.819 (0.04) = 0.0328 m3/s = 32.8 L/s Q1 = (1 + m) Q2 = 1.819 (0.0.04) = 0.0728 m3/s = 72.8 L/s Hence hf = rQ2 Pipe Diameter Length f Feeding to reservoir elevation 8f L where r = p 2 gD 5 A trial and error solution is adopted. Flow away from the junction is marked negative. First trial Let Hj = Elevation of hydraulic grade line at J = 150.0 m Pipe 11.21 A water supply main trifurcates at a junction point J into three branches each feeding a separate reservoir. The details of the pipes and the reservoir are as follows: Example 11.21 JA JB JC r hf Q = hf / r Ê 8f L ˆ Á= 2 5 ˜ Ë p gD ¯ (m) (m3/s) 10328 150 – 80 = 70 – 0.0823 12910 150 – 70 = 80 – 0.0787 15493 150 – 60 = 90 – 0.0762 Q \| Q/hf \| (L/s) (Q in L/s) –82.3 –78.7 –76.2 1.18 0.98 0.85 S = –237.2 S = 3.01 SQ = Inﬂow + S Outﬂow = 250 – 237.2 = 12.8 L/s DHj = Correction to assumed Hj = 3 /s, deter mine the delivery into each reservoir. [Refer Fig. 11.16]. 2 ¥ 12.8 = 8.50 m 3.01 New Hj = 158.5 m = 2 ¥ SQ S\| Q / hf \| 350 Fluid Mechanics and Hydraulic Machines Second trial Solution: The basic premises to remember are: Hj = 158.5 m Pipe r hf Q= hf / r 3 (m) (m /s) Q \| Q/hf \| (L/s) (Q in L/s) JA 10328 158.5 – 80 = 78.5 –0.0872 –87.2 JB 12910 158.5 – 70 = 88.5 –0.0828 –82.8 JC 15493 158.5 – 60 = 98.5 –0.0797 –79.7 1.11 0.94 0.81 S = –249.7 S = 2.86 (i) ﬂow into each junction of a network must be equal to ﬂow out of that junction (ii) Algebraic sum of head losses round each closed loop must be zero. Using these: By considering the ﬂow in to the node as positive and ﬂow out of the node as negative, Discharges: For Node D: QD + Q3 + Q4 = 0 100 – 40 – Q4 = 0 Hence Q4 = 60 (from D towards A) SQ = 250 – 249.7 = + 0.3 L/s 0.3 ¥ 2 = + 0.2 m 2.86 As DQ is very small no further iterations are necessary and DQ is distributed equally to the 3 pipes. DHj = + Hence QJA = 87.3 L/s QJB = 82.9 L/s QJC = 79.8 L/s The elevation of the hydraulic gradient line at the junction is 158.5 + DHj = 158.7 m. 11.22 which Q and hf refer to discharges and head losses respectively. Determine the head losses and discharges indicated by a question mark, for this pipe network. QA = 20 A For Node C: Q3 – QC + Q5 + Q2 = 0 40 – 30 + 10 + Q2 = 0 Hence Q2 = –20 (from C towards B) For Node B: Q1 + QB + Q2 = 0 30 + QB + 20 = 0 Hence Q4 = –50 (Out of B) The discharges in all the nodes and the lines are shown in Fig. 11.18 QA = 20 A QB = ? Q1 = 30; hf 1 = 60 Q2 = ? hf 2 = 40 D QD = 100 D QD = 100 Q3 = 40 hf 3 = 120 Fig. 11.17 QB = 50 Q1 = 30; hf 1 = 60 Q3 = 40 hf 3 = 120 Fig. 11.18 C B Q5 = 10 hf 5 = 20 Q4 = 60 hf 4 = 100 B Q5 = ? hf 5 = ? Q4 = ? hf 4 = ? For Node A: – QA – Q1 + Q5 + Q4 = 0 – 20 – 30 + Q5 + 60 = 0 Hence Q5 = –10 (from A towards C) QC = 30 Head Loss: Consider the loop ABC: hAB + hBC + hCA = 0 Q2 = 20 hf 2 = 40 C QC = 30 351 Pipe Flow Systems 60 + 40 + hCA = 0 hCA = –20 (Drop from A to C) that is hAC = 20 Consider the loop ADC: hAD + hDC + hCA = 0 hAD + 120 – 20 = 0 hAD = – 100 that is hDA = 100 (Drop from D to A) B r 11.23 For the network given below the dis- charges at the nodes are known. Verify whether the following suggested distribution of discharges in the pipelines of the network as given below in the Table, is satisfactory. If not, adjust the distribution. The head loss in a pipe is given by hf = rQ . The values of r for various pipes are indicated in the Fig. An accuracy of 0.5 unit of discharge is adequate. Line Suggested Discharge (Units) direction is assumed positive clockwise. The Hardy– Cross method is used for checking the correctness of the suggested distribution and to determine corrections if needed. The calculations are made in tabular form and are given in Table. 11.4 given below. The suggested distribution is found to be satisfactory at the given level of accuracy of 0.5 units of discharges. AB BC CA BE ED DC 60 19 40 41 16 34 100 E r=4 25 =1 r=4 A r=3 r=2 r=5 C D 50 25 Fig. 11.19 Example 11.23 11.24 the head loss is given by hf = rQ . The values of r for each pipe, and the discharge into or out of various nodes are shown in the sketch. The discharges are in an arbitrary unit. Obtain the distribution of discharge in the network. Solution: The Suggested discharges satisfy continuity at each node, For the loops, the ﬂow Table 11.4 Loop ABC Line AB BC CA Loop BCDE 2 \|2r Q\| rQ 1 ¥ 602 = 3600 4 ¥ 192 = 1444 –3 ¥ 402 = –4800 2 ¥ 1 ¥ 60 = 120 2 ¥ 4 ¥ 19 = 152 2 ¥ 3 ¥ 40 = 240 S = –244 DQ = – 244 512 S = 512 = – 0.48 ª0 Satisfactory to an accuracy of 0.5 units of Q. Line BC CD DE EB rQ 2 – 4 ¥ 192 –5 ¥ 342 + 2 ¥ 162 +4 ¥ 412 S \|2r Q\| = –1444 = –5780 = +512 = + 6724 = 12 DQ = – 12 884 2 ¥ 4 ¥ 19 2 ¥ 5 ¥ 34 2 ¥ 2 ¥ 16 2 ¥ 4 ¥ 41 S = – 0.01 ª0 Satisfactory to an accuracy of 0.5 units of Q. = 152 = 340 = 64 = 328 = 884 352 Fluid Mechanics and Hydraulic Machines A 20 r=7 D Loop ADC 15 rQ2 Line r=8 r=6 45 B r=4 (G) (Given) Fig. 11.20 40 C – 6 ¥ 5 = – 150 7 ¥ 102 = +700 8 ¥ 52 = +200 CD DA AC r=6 Example 11.24 Solution: Flow direction is assumed positive clockwise for all loops. A ﬁrst trial set of discharges is selected to satisfy continuity at each node [Fig. 11.20(a)), (i.e. ﬂow into a node = ﬂow out of the node). The Hardy–Cross method is used to ﬁnd the corrections DQ. \| 2rQ \| 2 2 ¥ 6 ¥ 5 = 60 2 ¥ 7 ¥ 10 = 140 2 ¥ 8 ¥ 5 = 80 S = 750 Correction DQ for loop ADC 750 DQ = – 280 DQ = – 3 S = 280 The corrected ﬂows are as in Fig. 11.20(b) Note that the line AC has two corrections. A 20 7 D 15 6 8 19 10 A 20 D 15 45 5 26 B Fig. 11.20(b) B 30 (a) C 40 Second trial Loop ABC AB BC CA First Trial Loop ABC AB CB CA rQ2 –6 ¥ 152 = –1350 4 ¥ 302 = +3600 –8 ¥ 52 = –200 S = 2050 Correction DQ for loop ABC 2050 DQ = – 500 DQ = – 4 rQ2 Line Fig. 11.20(a) Line 40 (b) 5 15 45 C –6 ¥ 19 = –2166 +4 ¥ 262 = + 2704 –8 ¥ 62 = –288 \|2rQ\| 2 ¥ 6 ¥ 19 = 228 2 ¥ 4 ¥ 26 = 208 2 ¥ 8 ¥ 6 = 96 S = 250 250 DQ = – = – 0.5 532 \| 2rQ \| 2 ¥ 6 ¥ 15 = 180 2 ¥ 4 ¥ 30 = 240 2 ¥ 8 ¥ 5 = 80 S = 500 2 S = 532 Loop ADC Line CD DA AC rQ 2 2 –6 ¥ 8 = – 384 7 ¥ 72 = +343 8 ¥ 62 = +288 \|2rQ\| 2 ¥ 6 ¥ 8 = 96 2 ¥ 7 ¥ 7 = 98 2 ¥ 8 ¥ 6 = 96 S = +247 247 = –1.0 DQ = – 290 S = +290 353 Pipe Flow Systems The corrections of the second trial are applied to get a distribution as shown in Fig. 11.20(c) 5.97 A 20 D 15 5.62 A 20 D 6.0 9.03 15 19.65 5.5 9.0 45 45 B C 25.5 (c) 40 40 C Fig. 11.20(d) Fig. 11.20(c) 11.25 values of discharges (in arbitrary units) at the nodes and in each pipe are shown in Third trial Loop ABC rQ2 Line AB BC CA 25.35 Final (d) B 19.5 \|2rQ\| –6 ¥ 19.52 = –2281.5 +4 ¥ 25.52 = 2601.0 –8 ¥ 5.52 = –242.0 2 ¥ 6 ¥ 19.5 = 234 2 ¥ 4 ¥ 25.5 = 204 2 ¥ 8 ¥ 5.5 = 88 S = 77.5 77.5 DQ = – 526 = –0.15 S = 526 concentration of Ca enters the steady state contaminant where the water leaves the system at points B, E and F. Water entering the network at D contains no contaminant. Assume perfect mixing at nodes. 30 A 100 30 B 20 F 30 20 Loop ADC Line CD DA AC rQ2 2 – 6 ¥ 9 = – 486 7 ¥ 62 = + 252 8 ¥ 5.52 = + 242 S =8 8 DQ = – 280 = –0.03 2 ¥ 6 ¥ 9 = 108 2 ¥ 7 ¥ 6 = 84 2 ¥ 8 ¥ 5.5 = 88 S = 280 The corrections are applied and further trials are discontinued by taking this level of distribution to be satisfactory. The calculations can be continued if further accuracy is needed. The ﬁnal distribution is, therefore, as shown in Fig. 11.20(d). 40 70 \|2rQ\| 10 50 D 120 Fig. 11.21 80 C E 90 Example 11.25 Solution: Node A. From an inspection of Fig. 11.21, it is obvious that CAB = CAD = Ca 70 ¥ C AD (70 + 50) 70 Ca = 120 = 0.583Ca Node D: Concentration at D = CDC = CCB = CCE = 0.583Ca 354 Fluid Mechanics and Hydraulic Machines Node B: Concentration at 30 ¥ C AB + 40 ¥ CCB (30 + 40) 30Ca + 40(0.583)Ca = 70 = 0.762Ca = CBE = 0.762Ca Solution: For a given discharge Q, the head loss due to friction B= CBF Contaminant concentration in ﬂow leaving B = 0.762 Ca Node E: Concentration at hf = where 20 ¥ C BE + 80 ¥ CCE ( 20 + 80) 20(0.762Ca ) + 80(0.583)Ca = 100 = 0.619 Ca Contaminant concentration in ﬂow leaving E = 0.619 Ca Node F: Concentration at 20 ¥ C BF + 10 ¥ C EF ( 20 + 10) 20(0.762Ca ) + 10(0.619Ca ) = 30 = 0.7143 Ca Contaminant concentration in ﬂow leaving F = 0.7143 Ca Check: Contaminant entering per second = 100 Ca + 50 ¥ (0) = 100 Ca units Contaminant leaving per second = 30 ¥ (0.762 Ca) + 30 ¥ (0.7143 Ca) + 90 ¥ (0.619 Ca) = 100 Ca units fL 2gD ( A2 ) in which A = area of the conduit. Power P = g Q(H – hf) = g Q(H – rQ2) dP =0 dQ For maximum power E= F= r= f LV 2 = r Q2 2gD dP = g H – g (3rQ2) = 0 dQ H – 3hf = 0 H hf = 3 11.27 diameter D carrying water. Show that for maximum kinetic energy to be supplied by the nozzle, the diameter of the nozzle d is given by Ê D5 ˆ d= Á ˜ Ë 2f L ¯ where f = friction factor and L = length of the pipe. Solution: Let n = velocity in the nozzle and V = velocity in the pipe. The total head H = velocity head at the nozzle + head lost in friction H = 11.26 Power is transmitted through a pipeline connected to a reservoir. Show that for a given total head H, the power transmitted is maximum when the loss of head due to friction hf = H . (Minor losses can be neglected) 3 1/ 4 By continuity Hence v2 f LV 2 + 2g 2g D p 2 p d v = D2 V 4 4 Ê d2 ˆ V = vÁ 2˜ ËD ¯ (1) 355 Pipe Flow Systems 4 v2 Ê fL Ê d ˆ ˆ Á1 + ˜ =H Á ˜ D Ë D¯ ¯ 2g Ë \ for purposes of power generation. The pipe is 1000 m long and has a head of 100 m at the inlet. If a nozzle, discharging into atmosphere v2 H (2) = 4 2g Ê fLÊdˆ ˆ Á1 + ˜ D ÁË D ˜¯ ¯ Ë P = gQ Power of jet the nozzle for maximum transmission of power. What is the magnitude of the maximum power? [Assume f = 0.019]. v2 2g Solution: For the condition of maximum power, the diameter of the nozzle v2 f L V2 =H– D 2g 2g From Eq. (1) 11.28 A 50 cm diameter pipe conveys water Ê D5 ˆ d = Á ˜ Ë 2 f L¯ 4 È f L Ê d ˆ v2 ˘ ÍH ◊ ◊ ˙ D ÁË D ˜¯ 2g ˙ ÍÎ ˚ dP For maximum power =0 dv p Thus P= g d 2 v 4 f L Ê d ˆ v2 =0 ◊ D ÁË D ˜¯ 2g H–3 v2 = 2g fLÊdˆ D ÁË D ˜¯ But by Eq. (2), H 100 = = 33.33 m 3 3 f LV 2 0.019 ¥ 1000 hf = ¥ V2 = 2 ¥ 9.81 ¥ 0.5 2g D = 33.33 V = Velocity in the pipe = 4.148 m/s 3 or (3) 4 v2 = 2g 1+ fLÊdˆ D ÁË D ˜¯ p ¥ (0.5)2 ¥ 4.148 4 = 0.8146 m3/s Maximum Power Pm = g Q(H – hf) 2 = 9.79 ¥ 0.8146 ¥ ¥ 100 3 = 531.6 kW Discharge Q = H Hence 4 H 1+ fL Êdˆ ◊ D ÁË D ˜¯ 4 4 È fLÊdˆ ˘ ˙ = H Í1 + D ÁË D ˜¯ ˙ ÍÎ ˚ fLÊdˆ D ÁË D ˜¯ 1/ 4 Head loss hf = H 3 Ê ˆ (0.5)5 = Á ˜ Ë 2 ¥ 0.019 ¥ 1000 ¯ = 16.93 cm 4 i.e. 1/ 4 4 =3 d4 = fLÊdˆ D ÁË D ˜¯ D5 2f L Ê D5 ˆ d = Á ˜ Ë 2 f L¯ 1/4 4 11.29 A pipeline 60 cm in diameter takes off from a reservoir whose water surface elevation is 150 m above datum. The pipe is 5000 m long and is laid completely at the datum withdrawn by a series of pipes at a uniform rate of 0.088 m3/s per 300 m. Find the pressure at the end of the pipeline. Assume f to have a dead end. Solution: First an expression for head loss in a pipe having a uniform withdrawal of q m3/s per metre length is derived. 356 Fluid Mechanics and Hydraulic Machines Consider a section at a distance x from the start of the uniform withdrawal at q per metre length (see Fig. 11.22). 3 ¥ p ¥ 9.81 ¥ (0.6)5 1 = [(0.352)3 – 0] ( 2.933 ¥ 10 -4 ) Dia = D q L0 Residual head at the dead end = 150.0 – 11.053 = 138.947 m above datum. Fig. 11.22 Discharge Qx = Q0 – qx In a small distance dx, f LV 2 dhf = = 2g D dx = 8f 2 p gD L0 hf = Ú dh f = 0 2 a length of 10 m. If the Darcy–Weisbach friction 2 Ê pˆ 2g ¥ Á ˜ ¥ D 5 Ë 4¯ -8 f 2 3p g D 5 5 pipe, determine the head loss in friction when Solution: Consider a stretch of length dx at a distance x from the 20 cm diameter end, (Fig. 11.23). 1 [(Q0 – qx)3]0L0 q 1 [Q03 – (Q0 – qL0)3] q 8f = 11.30 f ( Q 0 - q x ) 2 (Q0 – qx)2 dx 5 3p g D In the present problem, 1200 ¥ (0.088) = 0.352 m3/s 300 0.088 q = = 2.933 ¥ 10–4 m3/s/m 300 L0 = 1200 m HL = total head lost = (head lost in ﬁrst (5000 – 1200) m with a discharge Q0 = 0.352 m3/s) + head lost in 1200 with a uniform withdrawal of q. = hf1 + hf2 x 20 cm 0.02 ¥ 3800 ¥ (0.352) 2 8f 2 1 5 3p g D q 10 cm D 1 2 Fig. 11.23 dhfx = f dxV 2 = 2g D f Q 2 dx 2 Ê pˆ 2g Á ˜ D 5 Ë 4¯ where D = diameter at the section (x). 2 Ê pˆ 2 ¥ 9.81 ¥ Á ˜ ¥ (0.6)5 Ë 4¯ = 10.00 m hf2 = dx 10 cm Q0 = hf1 = 2 = 1.053 m Total head loss = 10.00 + 1.053 = 11.053 m Q0 x 8 ¥ 0.02 = [Q03 – (Q0 – q L0)3] ( 20 - 10) ˘ 1 È D = Í20 x ˙ cm = (20 – x ) m 10 100 Î ˚ Hence dhfx = 0.008263 fQ2 (10)10 dx ( 20 - x )5 357 Pipe Flow Systems = 0.08263 ¥ 0.02 ¥ (0.050)2 ¥ 1010 ( 20 - x ) Total head loss Ú ( 20 - x ) 11.32 For the pumping set-up shown in Fig. 5 and (b) the pressure at the suction side of the pump. Include minor losses. Atmospheric pressure head = 10.0 m. dx = 41313 hf = dx 5 10 0 dhfx = 41313 Ú 10 EL: 110.00 m –5 (20 – x) dx Pump Q 0 10 È ˘ 1 = 41313 Í 4˙ Î 4( 20 - x ) ˚ 0 41313 È 1 1 ˘ = Í ˙ = 0.968 m 4 4 Î (10) ( 20) 4 ˚ P C L/s EL: 100.00 m EL: 95.00 m A Fig. 11.24 0 =2 Example 11.32 11.31 Kerosene (r = 804 kg/m3) is pumped from tank M to tank N through a 100 m long, 5 cm diameter pipe. The total minor losses can static lift is 5.0 m, calculate the power required to pump 300 liters of kerosene per minute. (Take Solution: 0.3 = 0.005 m3/s 60 0.005 V = = 2.546 m/s Ê pˆ 2 ¥ ( 0 . 05 ) ÁË 4 ˜¯ Discharge Q = Velocity Head loss Pipe Diameter 2ˆ Ê 0.02 ¥ (100 + 25) ¥ ( 2.546) = Á ˜ 2 ¥ 9.81 ¥ 0.05 Ë ¯ = 16.525 m Static head = 5.00 m Total head = H = (16.525 + 5.00) = 21.525 m Power required = P = g QH = = (804 ¥ 9.81) ¥ (0.005) ¥ (21.525) = 849 W = 0.85 kW f Solution: Static head Hs = 110.00 – 95.00 = 15 m V1 = Velocity in the suction pipe = Q A1 0.02 = 1.132 m/s p ¥ (0.15) 2 4 V2 = Velocity in the delivery pipe V1 = ÊD ˆ = V1 Á 1 ˜ Ë D2 ¯ Ê f ( L + Le) V 2 ˆ h1 = Á ˜ 2gD Ë ¯ Length 2 Ê 15 ˆ = 1.132 Á ˜ Ë 12 ¯ 2 = 1.769 m/s hf1 = head loss in suction pipe = 2 (1.132) 0.02 ¥ 20 ¥ 2 ¥ 9.81 0.15 = 2.667 ¥ 0.0653 = 0.174 m = f1L1 V12 ◊ D1 2g 358 Fluid Mechanics and Hydraulic Machines hL1 = Inlet loss = 0.5 V12 (1.132) 2 = 0.5 ¥ 2 ¥ 9.81 2g = 0.5 ¥ 0.0653 = 0.033 m hf2 = friction loss in the delivery pipe = f 2 L2 V22 D2 2g (1.769) 2 0.02 ¥ 300 ¥ = 50 ¥ 0.1595 2 ¥ 9.81 0.12 = 7.975 m = hL2 = Loss at exit = (1.769) 2 V22 = 2 ¥ 9.81 2g 11.33 A pump delivers water from a tank A (water surface elevation = 100.00 m) to tank B (water surface elevation = 150.00 m). The suction pipe is 50 m long (f cm in diameter. The delivery pipe is 900 m long (f discharge relationships for the pump is given by Q , calculate the discharge in the Hp = 80 – pipeline and the power delivered by the pump. (Neglect minor losses). [In the pump equation Hp is in metres and Q in m3/s]. Solution: Consider the schematic sketch, Fig. 11.25 Here, D1 = 0.30 m L1 = 50.0 m f1 = 0.025 = 0.160 m Total loss = 0.174 + 0.033 + 7.975 + 0.160 = 8.342 m Head delivered by pump Hf = Static head + losses = 15.0 + 8.342 = 23.342 m Power delivered = g QHf = 9.79 ¥ 0.02 ¥ 23.342 kW = 4.57 kW (b) By energy equation between reservoir A and the pump P: Let ps = Pressure at the suction side of the pump. 95.00 + 0 + 0 = 100.00 + ps V2 + 1 g 2g D2 = 0.20 m L2 = 900 m f2 = 0.020 EL: 150.00 m B D2, L2, f2 D1, L1, f1 P EL: 100.00 m A losses in the suction pipe = 100 + ps + 0.065 + 0.174 + 0.033 g ps = – 5.272 m (gauge) g = 10.0 – 5.272 = 4.728 m (abs) Ps = 4.728 ¥ 9.79 = 46.29 kPa (abs) Fig. 11.25 Example 11.33 Suction pipe: Head loss = hL1 = = 4.167 f1L1V12 0.025 ¥ 50 V12 = 0.30 2g 2g D1 V12 m 2g 359 Pipe Flow Systems Delivery pipe: f L V2 0.020 ¥ 900 V22 Head loss = hL2 = 2 2 2 = 0.20 2g 2g D2 = 90 V22 m 2g V12 V2 + 90 2 2g 2g 2 2 By continuity V1 (0.30) = V2 (0.20) Total head loss = HL = 41.167 11.34 A Closed loop air drying system has a conduit of rectangular section, 0.50 m ¥ 0.30 m, which has a total length of 85 m. The minor losses can be accounted for by an equivalent length of 15 m. What power is required to circulate air through this loop at f r = 3 . 2 Ê 0.20 ˆ V1 = V2 Á = 0.444 V2 Ë 0.30 ˜¯ 0.30 m V12 V2 = 0.1975 2 2g 2g 0.50 m V2 HL = [(4.167 ¥ 0.1975) + 90] 2 2g = 90.82 V22 m 2g Static head = 150.0 – 100 = 50.0 m 2g Q2 Èp 2˘ Í 4 ¥ (0.2) ˙ Î ˚ 2 ¥ 1 2 ¥ 9.81 (1) Hp = 50 + 4690 Q2 But by the given pump performance relation Hp = 80 – 7000 Q2 \ Area (0.5 ¥ 0.3) = Wetted perimeter 2 ¥ (0.5 + 0.3) = 0.09375 m V22 = 50 + 90.82 ¥ Solution: Hydraulic radius of the conduit = Rh = Hp = Head delivered by pump = Static head + friction head = 50 + 90.82 Fig. 11.26 80 – 7000 Q2 = 50 + 4690 Q2 30 Q2 = = 2566 ¥ 10–3 11690 Q = 0.0506 m3/s = 50.6 L/s Hp = 50 + 4690 (0.0506)2 = 62.04 m Power delivered by the pump P = g QHp = 9.79 ¥ 0.0506 ¥ 62.04 = 30.73 kW For non-circular conduits, the hydraulic radius Rh is used in the Darcy–Weisbach equation and related calculations by replacing D in the circular pipe equations by (4 Rh). Hence D = 4 Rh = 4 ¥ 0.09375 = 0.375 m Head loss in friction (including minor losses) = hf = f ( L + Le )V 2 0.02 ¥ (85 + 15) ¥ ( 25) 2 = 2 ¥ 9.81 ¥ 0.375 2gD = 169.9 m (of air column.) Velocity head = ha = V2 ( 25) 2 = 2g 2 ¥ 9.81 = 31.85 m (or air column.) Total head = H = 169.9 + 31.85 = 201.8 m Power required = P = g QH = (1.20 ¥ 9.81) ¥ (0.5 ¥ 0.3 ¥ 25) ¥ (201.8) = 8908 W = 8.91 kW 360 Fluid Mechanics and Hydraulic Machines Problems 11.1 A tank discharges water through a horizontal pipe into atmosphere. The pipe is 2.5 m long and contains a gate valve of K = 0.2 for fully open conditions. Calculate the discharge in the pipe for a head of 3.0 m in the tank when (a) the pipe is of 8 cm diameter with rounded entrance (K = 0.05). (b) the pipe is of 10 cm diameter with square entrance. (Assume f = 0.02 for both cases and the valve to be fully open). (Ans. (a) Q1 = 28.2 L/s; (b) Q2 = 40.6 L/s) 11.2 An 8 cm diameter pipe carrying water has an abrupt expansion 12 cm diameter at a section. If a differential mercury manometer connected to upstream and downstream sections of the expansion indicate a gauge reading of 2.0 cm, estimate the discharge in the pipe. (Ans. Q = 15.9 L/s) 11.3 When a sudden contraction from 50 cm diameter to 25 cm diameter is introduced in a horizontal pipeline the pressure changes from 105 kPa to 69 kPa. Assuming a coefﬁcient of contraction of 0.65, calculate the ﬂowrate. Following this contraction if there is a sudden enlargement to 50 cm and if the pressure in the 25 diameter section is 69 kPa, what is the pressure in the 50 cm section? (Ans. Q = 0.376 m3/s; p3 = 80 kPa) 11.4 A 30 cm diameter pipe is required for a town’s water supply. As pipes of this diameter were not available in the market, it was decided to lay two parallel pipes of equal diameter. Find he diameter of the parallel pipes. Assume f is same for all the pipes. (Ans. De = 22.74 cm; the next higher standard size pipe would be used.) 11.5 Two pipes, each of 10 cm diameter and length 100 m, are connected in parallel, between two points. Calculate the (a) equivalent length of a single pipe of 10 cm diameter. (b) equivalent size (diameter) of a single pipe of length 100 m. (Assume f is same for all the pipes). (Ans. (a) Le = 25 m; (b) De = 0.132 m) 11.6 Three pipes A, B and C with details as given in the following list are connected in series: Pipe A B C Length 60 m 80 m 100 m Diameter 10.0 cm 8.0 cm 6.0 cm f 0.018 0.020 0.020 Calculate (a) the size of a pipe of length 125 m and f = 0.020, equivalent to the pipeline ABC. (b) the length of an 8 cm diameter (f = 0.015) pipe equivalent to the pipeline ABC. (Ans. (a) De = 6.0 cm; (b) Le = 692 m) 11.7 If n number of pipes of different diameters but of same length L are connected in parallel between two points, obtain an expression for the size of an equivalent single pipe of length L. All the n pipes and the equivalent pipe can be assumed to have the same friction factor f. Ê È Á Ans. D = D Í e 1 Á Í Î Ë n Â 1 Ê Di ˆ ÁË D ˜¯ 1 5/ 2 ˘ ˙ ˙ ˚ 2/5ˆ ˜ ˜ ¯ 361 Pipe Flow Systems 11.8 Two pipes A and B with details as below are connected in parallel between two points: Pipe A B Length 150 m 200 m Diameter 10 cm 8 cm f 0.02 0.015 Calculate the size of an equivalent pipe of length 175 m and having a friction factor of 0.015. (Ans. De = 11.67 cm) 11.9 Two reservoirs are connected by a pipeline consisting of two pipes in series; one of 15 cm diameter and 6 m long and another of 22.5 cm diameter and 15 m long. If the difference in water levels of the reservoirs is 6.0 m, calculate the discharge by considering all losses. Assume f = 0.020 for both pipes. [ Assume square entrance and exit and sudden expansion at the junction of the two pipes] (Ans. Q = 133.3 L/s) 11.10 Three pipes: 300 m long of 30 cm diameter, 150 m long of 20 cm diameter and 200 m long of 25 cm diameter are connected in series in the same order as indicated above between a high level reservoir and a low level reservoir. The friction factor f for the pipes are: 0.018, 0.02 and 0.019 respectively. Determine the rate of ﬂow for a difference in elevation of 15 m between the two reservoirs. Account for all losses. Contractions and expansions are sudden. (Assume k for contraction = 0.30). (Ans. Q = 106.3 L/s) 11.11 Two pipes 1 and 2 are connected in parallel between points M and N. The details of the pipes are: Pipe Length Diameter f 1 75 m 8 cm 0.018 2 150 m 12 cm 0.020 Sk 15 7.5 (Sk = sum of minor loss coefﬁcients in a pipe). The pipes are horizontal and the pressure difference between M and N is kN/m2. Determine the discharge of water in each pipe. (Ans. Q1 = 6.3 L/s, Q2 = 14.0 L/s) 11.12 A 30 cm pipeline is 750 m long and connects two reservoirs A and B. The elevation of the water surface in the upper reservoir A is 122.50 m. At a certain point, distance 500 m from A on the pipeline, the elevation of the centreline of the pipe is 122.00 m. If cavitation is expected at a pressure of 21 kPa (abs) determine the lowest elevation of the water surface in the reservoir B that is admissible. Neglect minor losses. [Assume f = 0.02 and atmospheric pressure = 10.3 m (abs)]. (Ans. Elevation of B = 109.895 m) 11.13 The reservoir M with its water surface at an elevation of 120.00 m above datum is connected to the reservoir N, whose water surface is at the elevation 100.00, through a pipeline system. Two pipes, AC (8 cm diameter and 100 m long) and BC (10 cm diameter and 800 m long) take off from the reservoir M and joint at a junction C. From C a pipe CD (12 cm diameter and 1200 m long) is connected a reservoir N. Assuming f = 0.02 for all the pipes and neglecting minor losses, estimate the discharges in all the pipes and the head at the junction. (Ans. Q1 = 7.976 L/s, Q2, = 4.084 L/s and Q3 = 12.060 L/s; Hc = 111.59 m above datum) 11.14 Two reservoirs having a difference in water surface elevation of 12 m are connected by a pipeline system. This pipeline consists of a 30 cm diameter 1000 m long pipe leading from the higher reservoir to a junction from which two parallel pipes, each of 20 cm diameter and 800 m long, connect the lower reservoir. Assuming f = 0.020 for the 362 Fluid Mechanics and Hydraulic Machines 30 cm pipe and f = 0.015 for the 20 cm pipe, estimate the total discharge transferred to the lower reservoir? (Ans. Q = 90.8 L/s) 11.15 Two reservoirs with 15 m difference in their water levels are connected by 300 mm diameter pipeline 3000 m long. Calculate the discharge. If a parallel pipeline of 300 mm diameter is attached to the last 1500 m length of the existing pipe, determine the modiﬁed discharge. [Assume f = 0.02 for all pipes] (Ans: Qo = 0.0857 m3/s, Qnew = 0.1085 m3/s) 11.16 Three pipes whose data are given below are connected in parallel between two reservoirs A and B. If a total discharge of 50 L/s of water is transmitted from A to B, (a) estimate the discharge in each pipe. (b) What is the difference in the water surface elevations of reservoirs A and B? Pipe Length Diameter f 1 2 3 200 m 250 m 300 m 9 cm 10 cm 12 cm 0.020 0.020 0.018 (Ans. Q1 = 12.7 L/s, Q2 = 14.8 L/s and Q3 = 22.5 L/s; h f = 9.028 m) 11.17 Two pipes: L1 = 400 m, D1 = 30.0 cm and L2 = 500 m and D2 = 20 cm are connected in parallel between two reservoirs. If f1 = 0.020 and f2 = 0.015, what difference in reservoir water surface elevations will produce a total ﬂow of 0.25 m3/s from one reservoir to another? (Ans. DH = 9.0 m) 11.18 Three reservoirs A, B and C are interconnected as shown in Fig. 11.27. Determine the distribution of discharge in the pipes. Pipe 1 2 3 Length 1500 m 2000 m 2000 m Diameter 30 cm 25 cm 30 cm f 0.015 0.018 0019 A EL: 100.00 m EL: 90.00 m 1 B 2 J 3 EL: 82.00 m C Fig. 11.27 Problem 11.18 (Ans. Hj = 91.36 m; Q1 = 106.3 L/s; Q2 = 21.1 L/s and Q3 = 85.2 L/s) 11.19 The water levels in two reservoirs A and B are 65 m and 50 m above datum. The reservoirs are connected by two pipes to a common junction J, where the elevation of the hydraulic grade line is 45 m above datum. The junction J is also connected to another reservoir C. Estimate the discharge in the pipes and the elevation of the water surface in the reservoir C. The pipe data are: Pipe AJ BJ JC Diameter Length 40 cm 1000 m 30 cm 750 m 60 cm 850 m f 0.018 0.020 0.022 (Ans. Hc = 41.11 m; Q1 = 371 L/s from A; Q2 = 99 L/s from B; Q3 = 470 L/s to C) 11.20 Water ﬂows from the reservoir through a pipe of 0.15 m diameter and 180 m long to a point 13.5 m below the surface of the reservoir. Here it branches into two pipes, each of 0.1 m diameter, one of which is 48 m long discharging to atmosphere 18 m below the reservoir level and the other 60 m long discharging to atmosphere 24 m below the reservoir level. Assuming a constant friction factor f = 0.032, calculate the discharge from each pipe. 363 Pipe Flow Systems Neglect any losses at the junction. (Ans. Q1 = 0.0453 m3/s. Q2 = 0.0193 m3/s and Q 3 = 0.0260 m3/s) 11.21 A pipe network is shown in Fig. 11.28 in which Q and h f refer to discharges and head loss, respectively. Determine the head losses and discharges indicated by question marks for this pipe network. B 100 42.1 57.9 20.6 20 C A 17.3 30 32.7 D 100 A 50 Q=? hf = 35 25 Fig. 11.29(a) B Q = 40 hf = 16 Q =? =? hf Q = 40 hf = 9 C D Q=? hf = ? Fig. 11.28 75 Related to Problem 11.21 (Ans. QDB = 10, QDC = 35, QAB = 60 hf (DC)= 28, hf (DB)= 19, 11.22 Verify whether the suggested discharges in various pipelines in the following network (Fig. 11.29) are proper. If not, determine the proper distribution. The head loss in each pipe is given by hf = rQ2. B r=2 20 units An accuracy of 0.1 unit of discharge is adequate. 11.23 For the following pipe network (Fig. 11.30) obtain the discharge distribution. The head loss is given by hf = rQ2. An accuracy of 0.01 units of discharge is sufﬁcient. r=1 10 units Fig. 11.30 10.00 A A 7 units r=3 C r=4 C B r=2 100 units r=1 Answers to Problem 11.22 3 units Related to Problem 11.23 5.97 B 7.00 1.03 30 units 4.03 r=1 r=5 C D 3.00 50 units Fig. 11.29 Fig. 11.30(a) Answers to Problem 11.23 Related to Problem 11.22 Line AB BC CD DA AC Suggested 58 42 32 18 20 discharge (units) 11.24 If a conservative contaminant enters the steady state network of Fig. 11.31 at node 364 Fluid Mechanics and Hydraulic Machines 50 100 40 A ? B 50 ? 40 20 D ? 20 G S C F 100 ? 30 ? E ? Fig. 11.31 Problem 11.24 D with a concentration Cd, and at the node E with a concentration 0.5 Cd, ﬁnd the concentration of contaminant in the ﬂows leaving the nodes E and F. Water entering the nodes A and B contains no contaminant. Assume perfect mixing at the nodes. The discharges entering the nodes and in the pipes are shown in the ﬁgure in arbitrary units. (Ans: Cf = 0.0893 Cd; Ce = 0.2608 Cd) m and at the upstream and of the pipe it is 200 m. If f = 0.02, estimate the discharge and the velocity in the pipe. What is the diameter of the pipe? (Ans. Q = 75.66 L/s, V = 2.973 m/s; D = 18.0 cm) 11.27 The supply of water from a reservoir to a nozzle situated 108 m below the reservoir free surface is through a 300 mm diameter pipe of 600 m length. The friction factor of the pipe is f = 0.02. Find the greatest possible power of the issuing jet. (Ans. Pm = 209.4 kW) 11.28 Water ﬂows in to junction J from reservoirs A and B through connecting pipes, the head loss through these being 10.0 Q A2 and 4.0 Q B2. The water level elevations at the reservoirs at A and B are 25.9 m and 18.0 m, respectively. The inﬂow at C is discharged in to atmosphere. The head loss through pipe JC is 1.0 Q C2 . The gauge pressure at J is 9.0 m. What is the residual gauge pressure of the outﬂow at C? A, 25.9 m B, 18.0 m 2 2 10.0 Q A [Note: The discharges in the pipes and at the nodes are indicated by numerals in the Figure. The arrows indicate the direction of ﬂow.] 4.0 Q B J, 9.0 m 11.25 A pipe 10 cm in diameter and 400 m long has a nozzle ﬁxed at the discharge end. The nozzle discharges at atmospheric pressure. If the pipe is horizontal and the head at the pipe inlet is 45 m, determine the maximum power that could be transmitted. What would be the corresponding size of the nozzle? (Assume f = 0.02). (Ans. Pmax = 4.42 kW; d = 28.1 mm) 11.26 A 1000 m long pipe supplies water to a turbine which develops a brake power of 100 kW. The efﬁciency of the turbine is 90%. The head at the turbine inlet is 150 2 1.0 Q C C Fig. 11.32 Problem 11.28 (Ans. Hc = 1.16 m) 11.29 Calculate the power required to pump oil (RD = 0.85) at a rate of 20 L/s from a storage sump to an elevated tank. The static lift is 20 m and the pipe is 10 cm in diameter and 250 m long. Assume f = 0.02. (Include entrance and exit losses). (Ans. P = 6.16 kW) 365 Pipe Flow Systems 11.30 A ﬁre brigade pump delivers water through a hose of 200 m length and 8 cm diameter to a nozzle which produces a 2 cm diameter jet. The Cv of the nozzle is 0.985 and the pipe friction factor f = 0.025. The end of the nozzle is 25 m above the water level in the supply tank. Calculate the power delivered by the pump when the nozzle velocity is 64 m/s. (Ans. P = 57.3 kW) 11.31 A 25 cm pipe 2000 m long (f = 0.02) discharges freely into air at an elevation of 8 m below the water surfaces of the supply tank. It is required to increase the discharge by 75% by inserting a pump in the pipeline. If the pump has an efﬁciency of 60%, calculate the power to be supplied to the pump. (Ans. P = 22.8 kW) 11.32 A pump with discharge characteristics Hp = 80 – 2000 Q 2 where Hp is in metres and Q is in m3/s is used to pump gasoline (r = 680 kg/m3) in a 500 m long pipe of 30 cm diameter (f = 0.02). The static lift is 10 m. What ﬂow rate will result? What is the power expended on the ﬂuid? (Neglect minor losses). (Ans. Q = 173 L/s; P = 23.2 kW) 11.33 Two reservoirs are connected by 150 m of horizontal 25 cm diameter pipe with f = 0.025. Midway in the pipe is a turbine. If the difference in reservoir elevations is 25 m and the change in head at the turbine is 18 m, estimate the rate of ﬂow. (Include minor losses). (Ans. Q = 141.6 L/s). 11.34 A horizontal duct in an air conditioning system is rectangular in cross section with a width of 50 cm and height of 40 cm. If the pressure difference between two sections 50 m apart is 3 mm of mercury, calculate the volumetric rate of ﬂow of air in the duct. (Take rair = 1.20 kg/m3 and the friction factor f = 0.017. Consider the ﬂow to be incompressible) (Ans. 3.73 m3/s) Objective Questions 11.1 Hydraulic grade line for ﬂow in a pipe of constant diameter is (a) always above the centreline of the pipe (b) always above the energy grade line (c) always sloping downwards in the direction of the ﬂow (d) coincides with the pipe centreline 11.2 The head loss in a sudden expansion from 6 cm diameter pipe to 12 cm diameter pipe, in terms of velocity V1 in the 6 cm pipe is (a) 15 V12 16 2g (b) 3 V12 4 2g 1 V12 9 V12 (d) 4 2g 16 2g 11.3 The head loss caused due to sudden expansion from area A1 to area A2 causing velocity to change from V1 to V2 is (c) 2 Ê A ˆ V2 (a) Á1 - 1 ˜ 1 A2 ¯ 2g Ë 2 Ê A ˆ V2 (b) Á1 - 1 ˜ 2 A2 ¯ 2g Ë 366 Fluid Mechanics and Hydraulic Machines 11.9 In a 15 cm pipe line, the minor losses add V2 up to 15 . The length of a pipe of 15 2g 2 Ê A ˆ V2 (c) Á1 - 2 ˜ 2 A1 ¯ 2g Ë Ê A1 ˆ V12 (d) Á1 - ˜ A2 ¯ 2g Ë cm diameter (f = 0.02) equivalent to this loss is (a) 75 m (c) 7.5 m 11.4 In a sudden contraction, the velocity head changes from 0.5 m to 1.25 m. The coefﬁcient of contraction is 0.66. The head loss in this contraction is (a) 0.133 m (b) 0.332 m (c) 0.644 m (d) 0.750 m 11.5 The minor loss due to sudden contraction is due to (a) ﬂow contraction (b) expansion of ﬂow after sudden contraction (c) boundary friction (d) cavitation 11.10 Two identical pipes of length L, diameter D and friction factor f, are connected in series between two reservoirs. The size of a pipe of length L and of the same friction factor f, equivalent to the above pipeline, is (a) 0.5 D (b) 0.87 D (c) 1.15 D (d) 1.40 D 11.11 Two pipe systems in series are said to be equivalent when (a) the average diameter in both systems is the same (b) the average friction factor in both systems is the same (c) total length of the pipe is the same in both the systems. (d) the discharge under the same head is the same in both systems. 11.6 A pipe has a well rounded entrance from a reservoir. If the head loss at the entrance is V2 expressed as K , the value of K would 2g be about (a) 0.02 (c) 0.5 (b) 0.2 (d) 1.0 11.7 The loss at the exit of a submerged pipe in a reservoir is V2 V2 (a) 0.5 (b) 2g 2g V2 (d) negligibly small 2g 11.8 Minor losses in a pipe ﬂow are those losses (a) which are insigniﬁcantly small (b) which can be neglected always (c) caused by local disturbance due to pipe ﬁttings (d) caused by frictional resistance (b) 200 m (d) 150 m 11.12 Two identical pipes of length L, diameter D and friction factor f, are connected in parallel between two reservoirs. The size of a pipe of length L and of same friction factor f, equivalent to the above pipes, is (a) 0.5 D (b) 0.87 D (c) 1.40 D (d) 2.0 D (c) 0.1 11.13 Two identical pipes of length L, diameter D and friction factor f, are connected in parallel between two points. The length of a single pipe of diameter D and the same friction factor f, equivalent to the above pair, is (a) 2L (b) L 2 367 Pipe Flow Systems L L (c) (d) 2 4 11.14 In using Darcy–Weisbach equation for ﬂow in a pipe, the friction factor is misjudged by + 25%. The resulting error in the estimated discharge Q is (a) +25% (b) –16.67% (c) –5% (d) –12.5% 11.15 A pipeline connecting two reservoirs has its diameter reduced by 10% over a length of time due to chemical deposit action. If the friction factor remains unaltered, for a given head difference in the reservoirs this would reﬂect in a reduction in discharge of (a) 10% (b) 14.6% (c) 23.2% (d) 31.6% 11.16 Two pipelines of equal length and diameter of 20 cm and 30 cm respectively are connected in parallel between two reservoirs. If the friction factor f is the same for both the pipes, the ratio of the discharges in the smaller to the larger size of the pipe is (a) 0.363 (b) 0.444 (c) 0.667 (d) 0.137 11.19 A nozzle with a coefﬁcient of velocity Cv = 0.95 is attached at the end of a pipe. If the head loss in the nozzle is K(V 2/2g), where V = velocity of jet issuing from the nozzle, the value of K for this nozzle is (a) 0.905 (b) 0.108 (c) 0.053 (d) 0.0028 11.20 Two reservoirs are connected by two pipes A and B of identical friction factor and length, in series. If the diameter of A is 30% larger than that of B the ratio of the head loss in A to that in B is (a) 0.77 (b) 0.59 (c) 0.50 (d) 0.27 11.21 Two reservoirs connected by a pipe of friction factor f has a difference in water surface elevation of H. If H and f were to remain constant, to increase the ﬂow by 100%, one would need to increase the cross sectional area of ﬂow by (a) 74% (b) 100% (c) 50% (d) 32% 11.22 The discharge Q in a pipe of known f is estimated by using the head loss h f in a length L and diameter D. If an error of 1% is involved in the measurement of D, the corresponding error in the estimation of Q is (a) 2.5% (b) 1.0% (c) 0.4% (d) 5% 11.17 Three pipes are connected in series. Then (a) the head loss in each pipe is the same (b) the total discharge is the sum of the discharge in the individual pipes (c) the discharge through each pipe is the same (d) the Reynolds number for each pipe is the same 11.18 For maximum transmission of power through a pipeline with a total head H, the head loss due to friction h f is given by hf = (a) H/3 (c) H/2 2 H 3 (d) 0.1 H (b) 11.23 A 12 cm diameter straight pipe is laid at a uniform downgrade and the. ﬂow rate is maintained such that the velocity head in the pipe is 0.5 m. If the pressure is observed to be uniform along the length when the down-slope of the pipe is 1 in 10, what is the friction factor for the ﬂow? (a) 0.012 (b) 0.024 (c) 0.042 (d) 0.050 11.24 The velocities and the corresponding ﬂow areas of branches labeled (1), (2), (3), (4), 368 Fluid Mechanics and Hydraulic Machines and (5) for a given pipe system shown in the ﬁgure are given in the following table: Pipe 1 Velocity 5.0 (cm/s) Area (cm2) 4.0 2 3 4 5 6.0 V3 4.0 V5 5.0 2.0 10.0 8.0 The velocity V5 would be (a) 2.2 cm/s (b) 5.0 cm/s (c) 7.5 cm/s (d) 10.0 cm/s (2) (5) (1) (3) 11.27 Two tanks are connected in parallel by two pipes A and B of identical friction factors and lengths. If the size of pipe A is double than that pipe B, then their discharges will be in the ratio of (a) 2 (b) 4 (c) 5.66 (d) 32 11.28 Consider the following conditions for the pipe network shown in the Fig. 11.34 (1) Q1 = Q3 (2) Q2 = Q1 + Q3 (3) hf1 = hf3 (4) hf1 = hf2 = hf3 Which of these conditions must be satisﬁed by this pipe network? (4) Fig. 11.33 Question 11.24 11.25 A pipe is connected in series to another pipe whose diameter is twice and the length is 32 times that of the ﬁrst. The ratio of the frictional head losses for the ﬁrst pipe to that of the second pipe, by assuming both the pipes to have same frictional coefﬁcient f, is (a) 8 (b) 4 (c) 2 (d) 1 11.26 Three pipes A, B, C have the following basic geometries: Pipe Diameter Length A D L B D/2 L C 2D 4L If these three pipes are connected in series, by assuming the value of f to be the same for all the three pipes, the equivalent length of a pipe of diameter D, in terms of length L, is 1 1 (a) 5 L (b) 4 L 8 8 1 (c) 265 L (d) 33 L 8 [The friction factor f can be assumed to be the same for the pipes A, B, C and the equivalent pipe] A Pipe-3 Junction B Pipe-1 Pipe-2 Fig. 11.34 Question 11.28 (a) 1 and 3 (b) 2 and 3 (c) 1 and 4 (d) 2 and 4 11.29 Velocity of air passing through a rectangular duct and a circular duct is the same. Which one of the following is the correct expression for the equivalent diameter of the circular duct in respect of a rectangular duct for the same pressure loss per unit length? In the following a and b are the length and breadth of the rectangular duct cross section. a+b 2ab (a) (b) ab a+b 2a 2b (c) (d) a-b a+b 11.30 In a pipeline design the usual practice is to assume that due to aging (a) the effective roughness increases linearly with time 369 Pipe Flow Systems (b) the friction factor increases linearly with time (c) the pipe becomes smoother with time (d) the friction factor decreases linearly with time 11.31 A rectangular conduit 0.8 m ¥ 0.4 m carries air (kinematic viscosity = 1.5 ¥ 10–5 m2/s) at a velocity of 3 m/s. The Reynolds number of the ﬂow for calculation of friction factor f is (a) 8 ¥ 104 (b) 1.07 ¥ 105 5 (c) 1.6 ¥ 10 (d) 6 ¥ 104 11.32 In a pipe network (a) the algebraic sum of discharges around each elementary circuit must be zero (b) the head at each node must be the same (c) the algebraic sum of the piezometric head drops around each elementary circuit is zero (d) the piezometric head loss in each line of a circuit is the same Flow in open Channels Concept Review 12 12.1 Introduction Flow in a conduit with a free surface is known as . Flows in irrigation channels, streams and rivers, navigation channels, drainage channels and bottom slope and the component of the weight of the liquid along the slope acts as CLASSIFICATION 12.1.1 Open channel ﬂows with no addition or withdrawal of ﬂow along the channel are classiﬁed for identiﬁcation and analysis as shown in the following chart. Open channel flow Steady Uniform Non-uniform Gradually varied Rapidly varied The depth, slope and velocity remain constant along the channel. The water surface slope S w, the slope of the energy line Sf, and the bed slope S0, will all be equal to each other. 12.1.2 Unsteady Gradually varied Rapidly varied Uniform Flow Rapidly Varied Flow The ﬂow is non-uniform and the change in depth takes place rapidly. The frictional loses are relatively unimportant. The hydraulic jump is a typical example of the rapidly varied ﬂow. 371 12.1.3 Gradually Varied Flow The ﬂow is non-uniform and the change in depth takes place gradually. The curvatures of the water surface are small and the pressure can be assumed to be hydrostatic. The frictional resistance at the boundary plays an important role in this phenomenon. The backwater curve produced by an obstruction to ﬂow such as a weir is a typical example. 12.2 of the boundary surface and has the dimensions of [L–1/3 T ]. It varies from 0.013 for a smooth ﬂoat ﬁnished concrete surface to 0.025 for irregular channels in excavated rock. Natural streams and rivers may have a high value of n ranging from 0.020 to as high as 0.13. popular equation, known as Chezy formula, is Chezy Formula Another V = C RS0 UNIFORM FLOW 12.2.1 The average shear stress t 0 at the boundary of a channel in uniform ﬂow (Fig. 12.1) is given by t 0 = g RS0 Resistance Equation The Darcy–Weisbach resistance equation, developed for pipe ﬂow, could be used for representing open channel resistance as Darcy–Weisbach V = 8g/ f where g = unit weight of water Area of flow wetted perimeter A (12.3) ÈÊ 4 RV ˆ Ê 4 R ˆ ˘ f = fn ÍÁ Re = , ˙ v ˜¯ ÁË e s ˜¯ ˙˚ ÍÎË (12.4) where e S = equivalent sand grain roughness of the surface. y 12.2.3 P Fig. 12.1 Uniform Flow Resistance Formulas: Manning’s Formula The most widely used resistance equation for uniform ﬂow, known as Manning’s formula, is 1 V = R 2 / 3 S01/ 2 n RS0 where f = Darcy–Weisbach friction factor the variation of which is given by the Moody Diagram as = A/P S0 = bottom slope of the channel 12.2.2 (12.2) where C = Chezy coefﬁcient. Boundary Shear R = hydraulic radius = resistance (12.1) where V = average velocity of ﬂow and n = a roughness coefﬁcient known as Manning’s roughness coefﬁcient. This coefﬁcient is essentially a function Relationship between n, C and f The Manning’s n, Chezy’s C and Darcy–Weisbach’s f are related as 1 1/6 C = 8g/ f = (12.5) R n Ê n2 ˆ Thus f = Á 1/ 3 ˜ (8g) (12.6) ËR ¯ and n= R1/ 6 C and n = R1/ 6 8g f (12.7) It should be noted that Manning’s formula is applicable to fully developed rough turbulent ﬂow regimes only. 372 Fluid Mechanics and Hydraulic Machines 12.2.4 Uniform Flow Computation The discharge Q = AV and thus Q= 1 AR 2 / 3 S01/ 2 n (12.8) Usually the uniform ﬂow parameters are designated with a subscript 0. For channel cross sections whose top width is constant (e.g. rectangular) or increases with depth. (e.g. triangular, trapezoidal, parabolic) there is only one depth at which a given discharge will ﬂow as uniform ﬂow in a channel of known slope S0. Such a depth is known as normal depth, y0. This is a very important parameter in all open channel ﬂow computations. Uniform ﬂow computations are relatively simple. The available relations are: (a) Manning’s formula (In this chapter Manning’s formula is used in all cases unless speciﬁcally mentioned otherwise). (b) Continuity equation and (c) Geometry of the channel. The following ﬁve types of basic problems can arise: Problem type 1 2 3 4 5 Given y0 , n, S 0, Geometric elements (GE) Q, y0, n, GE Q, y0. S0, GE Q. n, S0, GE Q. y0. n. S0 Required Method of solution Q and V Direct (explicit) Direct Direct Trial and error Trial and error S0 n y0 GE and discharge can reach a maximum value at ﬁnite values of depth. dQ =0 (12.9) For maximum discharge dy d ( AR 2 / 3 ) = 0 dy i.e. Similarly for maximum velocity dV = 0 (12.10) dy d ( R) = 0 dy i.e. Channel Section In uniform ﬂow 1 1 1 AR 2/3 S01/2 = A5/3 2 / 3 S 01/2 n n p For a given roughness coefﬁcient n, slope S 0 and area of channel A, a minimum perimeter channel section represents the hydraulically efﬁcient section that will convey maximum discharge. Such a channel is called as the efﬁcient or best section. For an efﬁcient section, A = constant and dP/dy = 0. Analysing various channel sections it can be shown that Q= (a) Of all the various possible open-channel sections, the semi-circular shape has the least amount of perimeter for a given area. (b) For any other cross section shape, by using the subscript ‘e’ to denote the hydraulically efﬁcient section value, (i) For a rectangular section: Typical geometric elements of common open channel cross sections are shown in Table 12.1. 12.2.5 Maximum Velocity and Discharge Channels with closing top, i.e. those in which the top width decreases with depth in a certain range, do have an interesting property in which the velocity and Be = 2 ye (12.11) Re = ye /2 (12.12) (ii) For a triangular section the vertex angle: 2q e = 90∞ (12.13) 373 Table 12.1 Shape Rectangle (a) Triangle (b) Trapezoidal (c) 1 y y 1 Area, A 2q my2 By q m q m B Top width, T B 2my By m B + 2y 2 2 m +1 B + 2y m + 1 m = tan q m = cot q Other relationships For wide rectangular channel, i.e. (y0/B £ 0.02), R=y 2 y (iii) For a trapezoidal section of side slope m horizontal: 1 vertical B + 2y Be = 2 ye ( 1 + m 2 - m) (12.14) Ae = ( 2 1 + m 2 - m) ye2 (12.15) Re = ye /2 (12.16) Proportions of some common most efﬁcient sections are shown in Table 12.2 Energy The total energy of a channel section referred to the channel bed as the datum is 2 8 2 m +1 (2q – sin 2q) Dq B + 2my D sin q ( B + my ) y D ( 2q - Sin 2q ) q 8 2 Ê Ë cos q = Á 1 - 2 yˆ ˜ D¯ known as the speciﬁc energy, E. Thus E = y+a If the side slope can also be varied, the optimum 1 value of m = mem = gives the most efﬁcient 3 section. (a) D (B + my)y m +1 y 2q B + 2y 2y D B Wetted perimeter, P Hydraulic radius, R Circular (d) V2 2g (12.17) where a = kinetic energy correction factor, usually taken as unity, when no other information on it is available. (b) Alternate Depths For a given discharge Q in a channel, there will be two depths for a given speciﬁc energy E. These two depths are known as the alternate depths, (Fig. 12.2). (c) Critical Depth The depth at which the speciﬁc energy is the minimum for a given discharge is known as the critical depth. It can also be shown that the discharge for a given speciﬁc energy is maximum at the critical depth. Using subscript ‘c’ to denote the critical ﬂow condition, at the critical depth yc 374 Fluid Mechanics and Hydraulic Machines Table 12.2 Sl. No. Channel 1 Area Shape (A em) Rectangle 2 y 2em Bottom Hydraulic Top Width Width Perimeter (Pem) (Bem) Radius (Rem) (Tem) Wetted 4 yem 2 yem (Half square) 2 3 y 2em Trapezoidal 1 (Half regular hexagon; m = 2 2 3 yem 3 yem yem Qn 8 / 3 1/ 2 yem S0 = Kem 2 yem 2 1.260 4 yem yem 1.091 3 2 ) 3 p 3 Circular (semi-circular) 4 Triangle (Vertex angle = 90°) 2 y 2em y1, y2 = Alternate depths Depth, y p yem D = 2 yem 2 3 y em – nt Q = 2 yem 2yem 0.9895 2yem 0.500 (Q / Ac ) 2 = F c2 = 1. g ( Ac /Tc ) co At any other depth y, the Froude number is given y > yc = Subcritical flow y < yc = Supercritical flow y = yc = Critical flow by F= V (12.19) g ( A/T ) [Note: In this chapter the notation for the Froude number is F instead of Fr as used in Chapter 6. This is a commonly used notation in open cannel ﬂow.] y2 yc yem 2 2 and hence ta ns yc = Critical depth y1 y 2em Ec E Specific energy, v2 E=y+ 2g 12.2.8 Using Eq. (12.18) simple explicit expressions for critical depth for rectangular and triangular channels are obtained. Fig. 12.2 Q 2 Ac3 = Tc g Calculation of Critical Depth Rectangular Channel For a rectangular channel (12.18) (Fig. 12.3), A = By and T = B where Tc = top width at critical depth. This relationship is used to deﬁne the Froude number F for open channels. At critical depth Fc = 1 Hence from Eq. (12.18) Q 2Tc3 gA 3c = Vc2 g y 3c =1 375 T=B 2 my 1 y y m B q Fig. 12.4 Fig. 12.3 The speciﬁc energy at critical depth Vc2 or 2g = yc 2 (12.20) Ec = yc + The speciﬁc energy at critical depth V2 3 Ec = yc + c = yc 2 2g = yc + (12.21) Note that Eq. (12.21) is independent of the width of the channel. Q If q = = discharge per unit width of the channel B from Eq. (12.18) i.e., 1/ 3 (12.22) A = y, by Eq. (12.19), the Froude T number for a rectangular channel will be deﬁned Also, since as F= V gy (12.23) Triangular Channel For a triangular channel (Fig. 12.4) having side slopes of m horizontal to 1 vertical, A = m y2 and T = 2m y. Hence from Eq. (12.18) Q2 A3 m3 yc6 m 2 yc5 = = c = Tc g 2myc 2 1/ 5 Hence È 2Q 2 ˘ ye = Í 2 ˙ ÍÎ gm ˙˚ (12.24) 2 gAc2 = yc + Ec = 1.25yc F= Ê q2 ˆ yc = Á ˜ Ë g¯ Q2 m 2 yc5 4 m 2 yc4 (12.25) Note that Eq, (12.25) is independent of the side A y slope m of the channel. Also, since = , by T 2 Eq. (12.19), the Froude number for a triangular channel will be deﬁned as q2 = yc3 g that is Vc2 2g V 2 gy (12.26) For other sections, a trial and error procedure is required for the evaluation of yc. For details regarding the critical ﬂow computation and use of speciﬁc energy relationship for solving problems relating to open channel transitions, the reader is advised to refer to Ref. 12.1. 12.3 12.3.1 RAPIDLY VARIED FLOW Hydraulic Jump Hydraulic jump is a case of rapidly varied ﬂow. This phenomenon occurs when a super-critical ﬂow stream tries to reach its alternate depth in sub-critical mode. In the process it loses substantial energy and falls short of the alternate depth. The depths on either side of the jump are known as sequent depths (Fig. 12.5). Depending upon the Froude number of the supercritical stream, hydraulic jumps are classiﬁed 376 Fluid Mechanics and Hydraulic Machines Energy line Eq. 12.27 can be rearranged as E1 Ê Q2 ˆ P + M = g Á Ay + = constant g A ˜¯ Ë E1 y2 E2 V1, F1 V2, F2 y1 Horizontal 1 2 (a) A1 y1 + Q2 Q2 = A2 y 2 + g A1 g A2 This is a general expression to determine the sequent depths, y1 and y2 in a hydraulic jump in horizontal frictionless channels of any shape. The energy loss EL in the jump is EL = E1 - E2 y2 y CG y y1 (12.29) The computation usually involves trial and error procedure Hydraulic Jumps in Rectangular Channels For a Area A (b) Fig. 12.5 (12.28) Hydraulic Jump hydraulic jump in a horizontal, frictionless rectangular channel, Equation 12.28 can be simpliﬁed to obtain sequent depth ratio y2/y1 as y2 1 = [- 1 + 1 + 8 F12 ] (12.30) y1 2 in to ﬁve categories as below: Sl. No Classiﬁcation Froude number range 1 2 3 4 5 Undular jump Weak Jump Oscillating jump “Steady” jump Strong or Choppy jump 1.0 < F1 £ 1.7 1.7 < F1 £ 2.5 2.5 < F1 £ 4.5 4.5 < F1 £ 9.0 F1 > 9.0 Consider a horizontal, frictionless channel of any arbitrary shape (Fig. 12.5) By applying momentum equation in the ﬂow direction to a control volume encompassing sections 1 and 2 P1 – P2 = M2 – M1 (12.27) where P = pressure force on a section = g A y M = momentum ﬂux passing a section = r Q V = r Q 2/A In the above A = cross sectional area and y = depth of center of gravity of the area from the water surface. where F1 = V1 = Froude number of the approach- g y1 ing ﬂow. In terms of the Froude number of the sub critical ﬂow, F2, the sequent depth ratio can be expressed as y1 1 = È- 1 + 1 + 8 F22 ˘ ˙˚ y2 2 ÍÎ (12.30-a) The energy loss in a hydraulic jump occurring in a rectangular channel is E L = E1 - E2 = ( y2 - y1 )3 4 y1 y2 (12.31) The length of the jump is Lf = 6.1y2 for F1 > 4.5. The power dissipated in the jump is P = g QE L (12.32) 377 12.4 12.4.1 GRADUALLY VARIED FLOW Basic Equation In gradually varied ﬂow the energy slope Sf, the water surface slope Sw, and the bed slope So are all different. At any depth y the energy slope Sf is assumed to be given by Manning’s formula. 1 2/3 1/2 R Sf n V= or Sf = n2V 2 R4 / 3 = n2 Q 2 (12.33) A2 R 4 / 3 The basic differential equation governing the gradually varied ﬂow is S - Sf dy (12.34) = o dx Q 2T 1g A3 For a given channel, when Q, n and S0 are ﬁxed, the normal depth y0 and critical depth yc are ﬁxed depths. Depending upon the relative values of y0 and yc, the channels are divided into ﬁve categories as shown in Table 12.3. Further, based on the relative positions of the actual depth y, normal depth y0 and critical depth yc, the possible gradually varied ﬂow (GVF) proﬁles are grouped into twelve types as shown in Table 12.4, and also in Fig. 12.6. Table 12.4 Channel Condition Type Mild Slope y > y0 > yc y0 > y > yc y0 > yc > y y > yc > y0 yc > y > y0 yc > y0 > y y > (yc = y0) y < (yc = y0) y > yc y < yc y > yc y < yc M1 M2 M3 S1 S2 S3 C1 C3 H2 H3 A2 A3 Steep Slope This could also be written in terms of speciﬁc energy E as dE (12.35) = So – Sf dx Using Manning’s formula Eq. 12.35 can be expressed in terms of yo and yc for a wide rectangular channel as dy = dx Êy ˆ 1- Á o ˜ Ë yc ¯ Critical Slope Horizontal bed Adverse Slope 3.33 Êy ˆ 1- Á c ˜ Ë y¯ (12.36) 3 There are a host of methods for computing the GVF proﬁles. The direct step method is a simple procedure suitable for use in prismatic channels. The basic equation Table 12.3 Sl. No. Channel category Symbol 1 2 3 4 Mild slope Steep slope Critical slope Horizontal bed M S C H 5 Adverse slope A Characteristic condition y0 yc yc S0 y0 S0 yc > y0 = y0 =0 = yc, the channel is a mild slope channel for this discharge. If y is the depth of ﬂow: For M1 curve y > 0.536 m M2 curve 0.536 m > y > 0.382 m M3 curve y < 0.382 m NDL y01 = 1.350 m CDL yc = 0.618 m S2 CDL NDL y02 = 0.913 m Steep Fig. 12.23 Example 12.38 12.39 3 12.38 A wide rectangular channel has a n 3 grade of the channel: S Solution: In gradually varied ﬂow the Manning’s formula is written for any section as 1 V = R 2/3 S f1/2 n where S f = energy slope at that section. S Solution: Discharge intensity q = 1.5 m3/s/m Critical depth 2 yc = (q /g) 1/3 Ê (1.5) 2 ˆ = Á ˜ Ë 9.81 ¯ 1/ 3 = 0.612 m Normal depth y0: For a wide rectangular channel R = y0 1 q = y0 y02/3 S 01/2 n È nq ˘ y0 = Í ˙ ÍÎ S0 ˙˚ yc (m) 0.612 3/ 5 Slope 0.0004 0.016 y01 y02 (m) (m) 1.197 0.396 È 0.018 ¥ 1.5 ˘ = Í ˙ S0 ÍÎ ˙˚ 3/ 5 y0 1.197 m 0.396 m Type of grade change Mild to Steep Hence Sf = n2V 2 R4 / 3 Average energy slope between two sections S +S Sf = f 1 f 2 2 The calculations are shown in Table 12.5. Table 12.5 Energy Slope Calculation Property A P R V Sf Section M Section N 2 3.0 ¥ 1.4 = 4.2 m 3.0 ¥ 1.05 = 3.15 m2 3.0 + (2 ¥ 1.4) = 5.8 m 3.0 + (2 ¥ 1.05) = 5.10 m 4.2/5.8 = 0.724 m 3.15/5.10 = 0.6176 m 8.0/4.2 = 1.9048 m/s 8.0/3.15 = 2.5397 m/s (0.018) 2 ¥ (1.9048) 2 (0.724) 4 / 3 (0.018) 2 ¥ ( 2.5397) 2 (0.6176) 4 / 3 = 1.8077 ¥ 10–3 = 3.973 ¥ 10–3 397 Flow in Open Channels 1.8077 ¥ 10 -3 + 3.9730 ¥ 10 -3 2 = 2.890 ¥ 10–3 Sf = 12.40 A rectangular channel has a bed width = 4.0 m, bottom slope = 0.0004 and Manning’s n in this channel is 2.0 m. If the channel empties into a pool at the down stream end and the pool elevation is 0.60 m higher than the canal bed elevation at the downstream end, calculate the Solution: For uniform ﬂow, y0 = 2.0 m A0 = 4.0 ¥ 2.0 = 8.0 m2 8.0 R0 = = 1.0 m ( 4.0 + 2 ¥ 2.0) 1 Q1 = Q0 = A0R02/3 S 01/2 n 1 = ¥ 8.0 ¥ (1.0)2/3 ¥ (0.0004)1/2 0.02 = 8.0 m3/s For critical depth, Q 8.0 q= = = 2.0 m3/s/m B 4.0 Critical depth yc = (q 2/g)1/3 = (22/9.81)1/3 = 0.742 m Since y0 > yc, the channel slope is mild. Since the downstream pool elevation is 0.6 m above the canal invert at that section, and yc = 0.742 m, the critical depth will be the downstream control. The water surface proﬁle will be an M2 curve extending from y0 = 2.0 m to yc = 0.742 m at the downstream end. The direct step method with 4 steps is used. The calculations start with y = yc = 0.742 m and end at y = 0.99 y0 = 1.98 m. The intermediate water surface depths are chosen as 1.00 m, 1.40 m and 1.80 m, The calculations are performed in a tabular manner and is shown in Table 12.6. The table is self explanatory. The distance of the water surface in each reach (step) Dx is obtained as DE Dx = S 0 - Sf 12.41 A rectangular channel (n = 0.017) is 3.0 m wide and is laid on a bottom slope of 0.0009. It carries a discharge of 10 m3/s and A is 2.5 m, calculate the distance to the section B Solution: Consider one step with depths of 2.5 m and 2.7 m on its either ends. The distance of the water surface in the reach (step) Dx is obtained as DE Dx = . The calculations are performed in a S0 - Sf tabular fashion and is shown in Table 12.7. which is self explanatory. The explanation of each column is same as in Table 12.6 The distance between the two sections is found as 566 m. 12.42 A sluice gate discharges a stream of depth 0.15 m at the vena contracta. The channel can be taken as a wide rectangular m3 of 0.25 m estimate the distance from the toe of n = 0.015) Solution: Consider two steps with depths of 0.15 m, 0.22 m and 0.25 m forming the ends of the reaches. The distance of the water surface in the (step) Dx is DE obtained as D x = . The calculations are S0 - Sf performed in a tabular fashion and is shown in Table 12.8, which is self explanatory. The explanation of each column is same as in Table 12.6 The distance between the two depths 0.15 m and 0.25 m is found as 22 m. I. Broad Crested Weir 12.43 a broad crested weir. y 4.000 5.600 7.200 7.920 1.00 1.40 1.80 1.98 3 5 2.0320 1.8629 1.5040 1.2039 1.1129 (m) E 0.1691 0.3589 0.3001 0.0917 (m) –3 8 DE = E2 – E1 Col. 6 Col. 11 Col. 10 V2 (Col. 4) 2 +y= + Col. 1 2g 2 ¥ 9.81 E= Col. 5 Col. 9 Q 8.0 = A Col. 2 V= Col. 4 Col. 8 Col. 7 4.7079 ¥ 10 –4 7.9411 ¥ 10–4 19.024 ¥ 10–4 –4 9 –4 –2389 –910.7 –199.7 –21.5 3521 1132 221.2 21.5 0 (m) x (m) 11 10 Dx DE Col. 6 = S0 - Sf Col. 9 x = S Dx Dx = S0 – Sf = 0.0004 – Col. 8 n2V 2 (0.02) 2 ¥ (Col. 4) 2 = 4/3 R (Col. 3) 4 / 3 1 (Sf2 + Sf1) Sf = 2 Sf = –0.7079 ¥ 10 –4 –3.9411 ¥ 10–4 –15.024 ¥ 10–4 –42.673 ¥ 10 S0 – Sf yc = 0.742 m y0 = 2.0 m 46.673 ¥ 10 Sf A Col. 2 = P (4 + 2 ¥ (Col.1)) 4.1086 ¥ 10–4 5.3071 ¥ 10–4 1.0575 ¥ 10–3 2.7473 ¥ 10–3 6.5872 ¥ 10 7 Sf 6 DE A = By = 4.0 ¥ Col. 1 1.0101 1.1111 1.4285 2.000 2.695 (m/s) V 4 S0 = 0.0004 B = 4.0 m R= 0.9950 0.9474 0.8235 0.6666 0.5412 (m) R n = 0.02 Q = 8.0 m3/s 2 Col. 3 Col. 2 2.968 0.742 Note: (m ) (m) 2 2 A 1 Table 12.6 398 Fluid Mechanics and Hydraulic Machines 399 Table 12.7 Q = 10.0 m3/s, n = 0.017, S0 = 0.0009 1 y 2 A 3 R 4 V 5 E 6 DE 2.5 7.5 0.83 1.33 2.59 2.7 10.8 1.15 1.39 2.80 7 Sf 8 Sf 9 S0 – Sf 10 Dx 11 x 566 566 6.552 ¥ 10–4 0.2077 4.104 ¥ 10–4 0 5.33 ¥ 10–4 3.67 ¥ 10–4 Table 12.8 q = 1.4 m3/s/m, Wide Channel, n = 0.015, S0 = 0 = horizontal y V E DE Sf Sf S0 – S f 0.15 0.22 0.25 9.33 6.36 5.60 4.590 2.284 1.848 2.306 0.436 0.2459 0.0686 0.0448 0.1573 0.0567 – 0.1573 – 0.0563 Solution: Referring to Fig. 12.24, 2 V0 /2g H H1 Ideal discharge of weir. Energy line Dx x –14.7 –7.7 0 15 22 Q t = LVc yc where L = length 2 ( 2 / 3 g) LH 3/2 3 Putting g = 9.81, Q t = 1.705 LH3/2 When the weir occupies the full width of the channel L = B = width of channel. To account for energy losses a coefﬁcient of discharge Cd is introduced as = yc Q = Q tCd = 1.705 Cd LH 3/2 Fig. 12.24 V02 where V0 is 2g the velocity of approach. It is usual to neglect the velocity of approach and in that case It is to be noted that H = H1 + The critical depth yc occurs over the weir crest. Assuming no loss of energy and considering the elevation of the weir crest as datum the energy equation is H1 + Thus V02 V2 = H = yc + c 2g 2g Ê 3 V2ˆ yc Á∵ yc = c ˜ = 2 g ¯ Ë yc = 2 H and 3 Vc = g yc Q = 1.705 Cd LH 13/2 Usually Cd is about 0.85 for a weir with square entrance. For weirs with rounded entrance Cd is about 0.98. 12.44 A broad crested weir spanning the full width of a 2.0 m wide channel is 1.5 m required to pass a discharge of 3.0 m3/s? 400 Fluid Mechanics and Hydraulic Machines Solution: For a broad-crested weir, Q = 1.705 Cd LH 3/2 where H = H1 + V02/2g V0 = Velocity of approach Assume Cd = 0.85 as the weir has a square entrance. B = L = 2.0 m Q = 3.0 = 1.705 ¥ 0.85 ¥ 2.0 ¥ H 3/2 V0 = Q 3.0 = B ( H1 + P ) 2.0 (1.023 + 1.5) = 0.594 m/s V02 (0.594) 2 = = 0.018 m 2 ¥ 9.81 2g New Ê V2ˆ H = 1.023 m = Á H1 + 0 ˜ 2g ¯ Ë H1 = 1.023 – 0.018 = 1.005 m Q 3.0 V0 = = B ( H1 + P ) 2.0 (1.005 + 1.5) = 0.599 m/s V02 = 0.0183 ª 0.018 as obtained earlier 2g A trial and Error method in used to ﬁnd H1 To ﬁnd V0: Assume H1 = H = 1.023 m. Then if P = height of the weir, (Fig. 12.24) Hence H1 = head over the broad-crested weir = 1.005 m Note An exhaustive presentation of worked examples, practice problems and objective questions covering practically all aspects of Open Channel Flow is available in Flow in Open Channels By Subramanya, K. Tata McGraw Hill Education Private Ltd., New Delhi, 3rd Edition, 3rd Reprint, 2009. Problems A. Uniform Flow 12.1 A trapezoidal channel has a bottom width of 3.5 m and side slopes of 1.5 horizontal : 1 vertical. The channel has a longitudinal slope of 1/4000 and carries a certain discharge at a depth of 1.70 m. Calculate the average shear stress on the boundary. (Ans. t 0 = 2.614 Pa) 12.2 Show by using Manning’s formula that the average boundary shear stress in an open channel is given by t0 = g n2V 2 R1/ 3 12.3 Show that the Chezy coefﬁcient C, Manning’s coefﬁcient n and Darcy– Weisbach friction factor f are related as 401 f= 8g = 12.8 Figure 12.26 shows the situation in a uniform ﬂow in a wide rectangular channel. Calculate the longitudinal slope if n = 0.02. (Ans. S0 = 3.96 ¥ 10–3) 8gn2 C2 R1/ 3 12.4 A trapezoidal channel has a bed width of 2.0 m, side slope of 1.25 horizontal : 1 vertical and carries a discharge of 9.00 m3/s at a depth of 2.0 m. Calculate the average velocity and bed slope of the channel. [Assume Manning’s coefﬁcient n = 0.015]. (Ans. V = 1.0 m/s; S0 = 2.0534 ¥ 10–4) 0.20 m 0.30 m 12.5 In a trapezoidal channel of bottom width 3.0 m and side slopes 2 horizontal : 1 vertical, the depth of ﬂow is 1.4 m. If the channel has a Manning’s coefﬁcient n = 0.015, estimate the values of Chezy coefﬁcient C and Darcy–Weisbach friction factor f. Flow Fig. 12.26 12.9 A rectangular channel 3.0 m wide had a badly damaged lining whose Manning’s n was estimated as 0.025. The lining was repaired and it now has an n value of 0.014. If the depth of ﬂow remains the same at 1.30 m as before the repair, estimate the new discharge and percentage increase in discharge as a result of repair. (Ans. C = 65.2; f = 0.0184) 12.6 An open channel of trapezoidal section has a base width of 2.5 m and sides inclined at 60° to the horizontal. The bed slope is 1 in 500. It is found that when the discharge is 1.5 m3/s the normal depth is 0.5 m. Using Manning’s formula calculate the discharge when the normal depth is 0.70 m. (Ans. Q = 2.594 m3/s) 12.7 A trapezoidal channel with cross section as in Fig. 12.25 carries a discharge of 10 m3/s at a depth of 1.5 m under uniform ﬂow conditions. The longitudinal slope of the channel bed is 0.001. Compute the average shear stress in N/ m2 on the boundary. Also compute the value of Manning’s n. (Ans. t 0 = 9.134 Pa; n = 0.025) 1.5 m 30° 30° 3.0 m Fig. 12.25 1.50 m (Ans. Q2 = 4.894 m3/s; 78.5% increase) 12.10 What diameter of a semicircular channel will have the same discharge as a rectangular channel of width 2.0 m and depth 1.2 m? Assume the bed slope and Manning’s n are the same for both the channels. (Ans. D = 2.396 m) 12.11 For a channel shown in Fig. 12.27 the bed slope is 1 in 1000 and Manning’s n = 0.014. Calculate the discharge. (Ans. Q = 1.898 m3/s) 12.12 A commonly used lined canal section is shown in Fig. 12.28. It consists of a triangular section with side slopes of m horizontal : 1 vertical. Further, it is rounded off at the bottom by a radius equal to full supply depth. 402 Fluid Mechanics and Hydraulic Machines 1.50 m q 0. 75 Fig. 12.27 r0 1.10 m For one such channel, commonly known as the standard lined triangular section, the full supply depth is 2.0 m, m = 1.5 and bed slope = 1/4000. Manning’s n = 0.015. Determine the full supply discharge. (Ans. Q = 8.804 m3/s) B. Normal Depth 12.13 A wide rectangular channel has a slope of 0.0004 and its Manning’s roughness coefﬁcient is 0.02. If a discharge intensity of 2.54 m3/s per metre width is to be passed in this channel, estimate the normal depth. (Ans. y0 = 1.75 m) 12.14 A triangular channel has a vertex angle of 75° and a longitudinal slope of 0.001. If Manning’s n = 0.015, estimate the normal depth for a discharge of 250 L/s in this channel. (Ans. y0 = 0.668 m) 12.15 A 1.5 m wide rectangular channel carries a discharge of 300 L/s. This longitudinal slope of the channel is 0.00016. If Manning’s n = 0.018, estimate the normal depth. (Ans. y0 = 0.595 m) [Hint: A trial and error procedure will be needed to estimate the normal depth in this case.] q m r0 = y 0 m Problem 12.11 q 2q 1 Fig. 12.28 (Ans. y0 = 2.218 m) 12.16 A wide rectangular channel is to carry a discharge of 2.5 m3/s/metre width at a Froude number of 0.5. Assuming Manning’s n = 0.015, calculate the (a) normal depth and (b) required bed slope (Ans. (a) y0 = 1.6815 m; (b) S0 = 4.973 ¥ 10–4;) 12.17 Show that the normal depth in a triangular channel of side slopes m horizontal: 1 vertical is given by È Qn ˘ y0 = 1.1892 Í ˙ ÍÎ S0 ˙˚ 3/8 1/ 8 È m2 + 1˘ Í 5 ˙ ÍÎ m ˙˚ C. Maximum Velocity and Discharge 12.18 Water ﬂows in a triangular duct resting on one of its sides. The duct is in the shape of an isosceles triangle of bed width B and sides making an angle 45° with the bed. The water surface is below the vertex and the bed is horizontal laterally. Determine the relationship between the depth of ﬂow y and the bed width B for maximum velocity condition. (Ans. y = 0.338 B) 12.19 A circular channel of diameter 0.6 m is laid on a slope of 1 in 3000. Calculate the maximum discharge it can convey as an open channel. (Manning’s n = 0.018). (Ans. Qm = 0.0871 m3/s) 403 12.20 A rectangular channel (Manning’s n = 0.020) is 5.0 m wide, 0.9 m deep and has a slope of 1 in 1600. If the channel had been designed to be of efﬁcient rectangular section, for the same wetted perimeter what additional discharge would it carry? (Ans. DQ = 2.211 m3/s; 51.76% increase) 12.21 The following Table 12.9 lists some typical problems relating to efﬁcient trapezoidal channels. Out of the nine variables listed in the Table, four are given in each problem. You are to determine the others and ﬁll in the blanks in the table. Note that when m = 0 the channel is rectangular and when Be = 0 the channel is triangular. 12.22 Determine the dimensions of a concrete lined (n = 0.014) trapezoidal channel of most efﬁcient proportions to carry a discharge of 10.0 m3/s. The bed slope of the channel is 0.005. (Ans. yem = 1.250 m; Bem = 1.444 m; m = 0.5773) 12.23 Determine the efﬁcient section and bed slope of a trapezoidal channel (n = 0.025) designed to carry 15 m3/s of ﬂow. To prevent scouring the velocity is to be 1.0 m/s and the side slopes of the channel are 1 vertical to 2 horizontal. (Ans. ye = 2.463 m; Be = 1.163 m; S0 = 4.735 ¥ 10–4) 12.24 What should be the dimensions of the efﬁcient trapezoidal section of side slopes Table 12.9 Prob. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) Q Be ye (m3/s) (m) (m) 15 1.2 1.5 6.0 3.0 m 2.0 1.5 1.5 1.0 1.5 1.0 0 V Pe Ae (m/s) (m) (m2) 1.5 12.0 5.0 0 n 0.015 0.020 0.020 0.015 0.018 0.014 0.015 0.015 S0 0.0004 0.0009 0.0006 0.0004 0.001 0.0004 0.0008 0.0008 Answers to Problem 12.21 Prob. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) Q Be ye (m3/s) (m) (m) 15.0 1.256 30.94 5.87 3.79 8.01 6.0 3.0 0.991 0.84 1.726 1.370 0.727 1.50 1.190 0 2.10 0.20 2.85 1.654 1.20 1.811 2.380 1.544 m 2.0 1.5 1.5 1.0 1.5 1.0 0 1.0 V Pe Ae (m/s) (m) (m2) 10.383 0.842 12.0 6.047 5.053 6.621 9.522 4.366 10.90 0.837 19.952 5.0 3.032 5.994 2.833 2.832 1.376 1.50 1.551 1.175 1.250 1.337 2.118 1.059 n 0.015 0.020 0.020 0.015 0.018 0.014 0.015 0.015 S0 0.0004 0.0009 0.0006 0.0004 0.001 0.0004 0.0008 0.0008 404 Fluid Mechanics and Hydraulic Machines 1.5 horizontal to 1 vertical, if the design discharge is 12 m3/s and the channel slope is 0.0005? (Take n = 0.02). (Ans. ye 2.19 m; Be = 1.326 m) 12.25 What are the dimensions of an efﬁcient rectangular brick channel (n = 0.015) designed to carry 6.0 m3/s of water with a bed slope of 0.002? How much additional discharge is carried by a semicircular channel of the same area? (Ans. ys = 1.192 m; Be = 2.384 m; Qs = 6.5033 m3/s; DQ = 0.5033 m3/s) 12.26 A trapezoidal channel is 6.0 m wide and has a side slope of 0.5 horizontal : 1 vertical. If the bed slope of the channel is 0.0004 and Manning’s n = 0.02, ﬁnd the discharge which can make the channel a hydraulically efﬁcient section. (Ans. Q = 73.9 m3/s) 12.27 A circular channel 1.20 m in diameter is laid on a slope of 0.008. The Manning’s n for the channel can be taken as 0.018. Calculate the speciﬁc energy when the depth of ﬂow is 0.60 m. (Ans. E = 0.853 m) 12.28 A rectangular channel 1.5 m wide carries a discharge of 9.0 m3/s at a depth of 2.0 m. Calculate the (a) speciﬁc energy and (b) depth alternate to the given depth of 2.0 m. (Ans. E = 2.4587 m; y2 = 1.214 m) 12.29 For a constant speciﬁc energy of 3.0 m, what maximum ﬂow may occur in a rectangular channel of 4.5 m bed width? (Ans. Qm = 39.87 m3/s) 12.30 What is the lowest possible speciﬁc energy for a water discharge of 12 m3/s to ﬂow through a 3.0 m wide rectangular channel? (Ans. E min = 1.766 m) 12.31 A rectangular channel 12.0 m wide carries 200 m3/s. Find the critical depth and critical velocity. What slope will produce this critical velocity in the channel if n = 0.02? (Ans. S0 = 4.682 ¥ 10–3; yc = 3.048 m; Vc = 5.468 m/s) 12.32 What is the maximum discharge that may be carried by a 3.2 m wide rectangular channel at a speciﬁc energy of 1.8 m? (Ans. Qm = 13.18 m3/s) 12.33 Fill in the missing data in Table 12.10 connected with critical depth computation in a rectangular channel. Table 12.10 Prob. (i) (ii) (iii) (iv) c in a Q B yc Ec (m3/s) (m) (m) (m) 2.6 0.70 0.40 10.48 6.0 2.0 5.0 1.2 Answers to Problem 12.33 Prob. (i) (ii) (iii) (iv) Q B yc Ec (m3/s) (m) (m) (m) 5.15 10.48 6.0 14.0 2.6 4.0 2.0 5.0 0.40 0.70 0.917 0.80 0.60 1.05 1.376 1.2 12.34 In a rectangular channel the speciﬁc energy is 2.0 m and one of the alternate depths is 0.5 m. Calculate the (a) other alternate depth, (b) critical depth and (c) discharge intensity. (Ans. (a) y1 = 1.896 m; (b) yc = 1.333 m; (c) q = 4.822 m3/s/m) 12.35 Calculate the discharge corresponding to a critical depth of 1.5 m in a 405 (a) rectangular channel of width 1.5 m; (b) triangular channel of vertex angle 120°; (c) trapezoidal channel of bed width = 6.0 m and side slopes 1.5 horizontal : 1 vertical. (Ans. (a) Q = 8.631 m3/s; (b) Q = 10.571 m3/s; (c) Q = 42.08 m3/s) 12.36 A triangular channel has an apex angle of 60° and carries a ﬂow with a velocity of 2.0 m/s and depth of 1.25 m (a) Is the ﬂow sub-critical or super critical? (b) What is the critical depth? (c) What is the speciﬁc energy? (Ans: (a) Sub-critical, (b) yc = 1.148 m, (c) E = 1.454 m) 12.37 A triangular channel with a vertex angle of 120° carries a discharge of 2.0 m3/s. Find the critical depth and the speciﬁc energy corresponding to critical ﬂow. (Ans. yc = 0.771 m; Ec = 0.964 m) F. Transitions 12.38 A horizontal rectangular channel 3.0 m wide is narrowed to 1.5 m width to cause critical ﬂow in the contracted section. If the depth in the contracted section is 0.8 m, calculate the discharge in the channel and the possible depths of ﬂow and corresponding Froude numbers in the 3.0 m wide section. Neglect energy losses in the transition. (Ans. Q = 3.362 m3/s; y1 = 1.153 m; y2 = 0.261 m; F1 = 0.289; F2 = 2.683) 12.39 A 2.0 m wide rectangular channel has a ﬂow with a velocity of 1.3 m/s and depth of 1.2 m. A smooth hump is to be built at a section to cause critical ﬂow. Calculate the minimum height of the hump required to achieve this. (Ans. D z = 0.344 m) 12.40 A discharge of 10.0 m3/s ﬂows at a depth of 2.0 m in a rectangular channel 4.0 m wide. At a section the width is reduced to 3.5 m. It is desired to provide a smooth hump at this contracted section to achieve critical depth without affecting the upstream depth. Estimate the height of the hump required. Neglect frictional losses. (Ans. Dz = 0.669 m) 12.41 A 3.0 m wide rectangular channel carries a discharge of 10.0 m3/s at a depth of 1.35 m. If at a section a smooth hump of height 0.25 m is built, estimate the depth of ﬂow over the hump and upstream of it by neglecting frictional losses. (Ans. y2 = yc = 1.042 m; y1¢ = 1.59 m) 12.42 A discharge of 15 m3/s ﬂows through a rectangular channel 3.0 m wide. The depth of ﬂow is 2.0 m. A smooth hump of 0.10 m is built at a section. Also, at that section the bottom width is reduced to 2.8 m. What are the depths of water at the contracted section and upstream of it? Neglect frictional losses. (Ans. y2 = 1.73 m; y1 = 2.00 m) 12.43 Water ﬂows in a rectangular channel at a depth of 1.5 m. A smooth hump 30 cm high in the bed produces a drop of 15 cm in the water surface elevation without affecting the upstream depth. Neglecting losses, calculate the rate of ﬂow per metre width. (Ans. q = 2.52 m3/s/m) G. Hydraulic Jump 12.44 Water is being discharged from under a sluice gate at a rate of 18 m3/s in a 3.0 m wide rectangular channel. A hydraulic jump is found to occur at a section where the depth of ﬂow is 0.50 m. Determine the 406 Fluid Mechanics and Hydraulic Machines (a) depth after the jump, (b) energy dissipated in the jump. (Ans. (a) y2 = 3.59 m; (b) EL = 4.11 m) 12.45 A hydraulic jump occurs in a horizontal rectangular channel with sequent depths of 0.70 m and 4.2 m. Calculate the rate of ﬂow per unit width, energy loss and the initial Froude number. (Ans. q = 8.4 m3/s/m; EL = 3.65 m; F1 = 4.58) 12.46 A hydraulic jump occurs in a horizontal rectangular channel at an initial Froude number of 10.0. What percentage of initial energy is lost in this jump? (Ans. EL/E1 = 72.7%) 12.47 A hydraulic jump formed below a sluice gate in a horizontal rectangular channel has a depth of 0.60 m before the jump and an initial speciﬁc energy of 10.0 m. Find the sequent depth and percentage of initial energy lost in the jump. the Froude number of the supercritical approach ﬂow. (Ans. F1 = 4.883) 12.49 A hydraulic jump in a rectangular channel has the Froude number at the beginning of the jump F1 = 5.0. Find the Froude number F2 at the end of the jump. (Ans. F2 = 0.2956) 12.50 Complete the following Table 12.11 relating to elements of a hydraulic jump in a horizontal rectangular channel. 12.51 A rectangular channel carrying a supercritical stream is to be provided with a hydraulic jump type of energy dissipator. If it is desired to have an energy loss of 5 m in the jump when the inlet Froude number is 8.5, determine the sequent depths. (Ans. y1 = 0.198 m; y2 = 2.277 m) (Ans. y2 = 4.459 m; EL/E1 = 53.7%) 12.48 The Froude number of the sub-critical ﬂow after a hydraulic jump in a horizontal rectangular channel is 0.3. Esti mate 12.52 At the foot of a 30 m wide spillway in a dam where the discharge velocity is 28.2 m/s and the depth is 0.96 m, a hydraulic jump is formed on a horizontal apron. Calculate the height of the jump and the total power dissipated in the jump. (Ans. H j = 11.04 m; P = 232.17 MW) Table 12.11 Prob. V1 y1 q (m/s) (m) (m3/s/m) (i) (ii) (iii) F1 y2 V2 (m) (m/s) 2.5 0.6 F2 EL EL/E1 (m) 12.0 9.0 0.20 8.0 F2 EL EL/E1 (m) % Answers to Problem 12.50 Prob. V1 y1 No. (m/s) (m) (i) (ii) (iii) 15.03 21.02 14.55 0.16 0.0714 0.2237 q F1 3 (m /s/m) 2.405 1.50 3.255 12.0 25.12 9.82 y2 V2 (m) (m/s) 2.636 2.5 3.00 0.9126 0.6 1.085 0.1794 0.1212 0.2 9.0 20.06 8.0 77.1 88.8 72.63 407 12.53 A hydraulic jump in a 1.5 m wide horizontal rectangular channel was used for estimating the ﬂow. If the ﬂow depths before and after the jump are 0.3 m and 2.1 m respectively in that channel, estimate the discharge. (Ans. Q = 4.085 m3/s) 12.54 In a hydraulic jump on a horizontal rectangular channel the depth and Froude number before the jump are 0.20 m and 9.0 respectively. Estimate the energy loss and speciﬁc head at the end of the jump. (Ans. EL = 5.798 m; E2 = 2.5 m) 12.55 A hydraulic jump takes place in a horizontal, triangular channel having side slopes of 1.5 H : IV. The depths before and after the jumps are 0.30 m and 1.20 m respectively. Estimate (i) the ﬂow rate, (ii) Froude numbers at the beginning of the jump and (iii) energy loss in the jump. (Ans. Q = 1.096 m3/s, F1 = 6.693, and EL = 2.447 m) H. Gradually Varied Flow 12.56 A wide rectangular channel (n = 0.015) carries a ﬂow of 2.5 m3/s per metre width. The bed slope of the channel is 0.005. Determine whether the channel slope is mild, steep or critical. (Ans. Steep) 12.57 Sketch the possible GVF proﬁles in the following break in grades. The ﬂow is from left to right. (a) Horizontal channel to steep. (b) Mild to milder. (Ans. (a) H2, S2; (b) M1; on mild channel) 12.58 Sketch the possible gradually varied ﬂow proﬁles in the following serial arrangement of channels and controls. The ﬂow is from left to right. Mild-sluice gate-steep (Ans. (a) M1; S3;) 12.59 Water is ponded up to a depth of 5 m just upstream of a weir in a wide channel. Estimate the depth of ﬂow 1.5 km upstream of the weir, given q = 2 m3/s/m, Manning’s n = 0.018 and S0 = 0.001. Take two steps. In Direct Step method. (Ans. 3.51 m) 12.60 At a certain section M in rectangular channel of bed width 2 m the depth of ﬂow is 1.20 m. When the ﬂow rate is 6.0 m3/s, estimate the distance from M to another section N where the depth is 1.40 m. The bed slope is 0.0020 and Manning’s n = 0.015. Take two steps. in Direct Step method. (Ans. 245 m) 12.61 A rectangular brick-lined channel (n = 0.016) has a bed width of 4.0 m and a longitudinal slope of 0.0009. At a certain discharge the normal depth was 2.0 m. In a reach where the ﬂow was non-uniform the depth of ﬂow at a section A was 2.6 m. Calculate the depth at a section B, 500 m downstream of A, by using (a) one step only and (b) two steps. (Ans. (a) 2.85 m; (b) 2.87 m) 12.62 A weir is 3.0 m long and has a head of 1.2 m over the weir crest. The height of the weir crest above the channel bed is 0.8 m. Neglecting the velocity of approach, estimate the discharge if the weir is (a) a broad crested weir with well rounded entrance. (b) a broad crested weir with square entrance. (c) a sharp crested weir. (Ans. (i) Q = 6.59 m3/s; 3 (ii) Q = 5.715 m /s; (iii) Q = 8.513 m3/s) 12.63 A broad crested weir is 2.5 m long and passes a discharge of 3.5 m3/s under a head of 0.9 m. Neglecting the velocity of approach, estimate its coefﬁcient of discharge. (Ans. Cd = 0.962 m) 408 Fluid Mechanics and Hydraulic Machines Objective Questions 12.1 Uniform ﬂow in a channel is characterized by the following statement: (a) Total energy remains constant along the channel. (b) Gradient of the total energy is parallel to the channel bed (c) Speciﬁc energy decreases along the channel. (d) Total energy line either rises or falls depending upon the Froude number. 12.2 Uniform ﬂow in an open channel exists when the ﬂow is steady and the (a) channel is frictionless (b) channel is non-prismatic (c) channel is prismatic (d) channel is prismatic and the depth of ﬂow is constant along the channel. 12.3 In deﬁning a Froude number applicable to channels of any shape, the length parameter used is the (a) depth of ﬂow (b) hydraulic radius (c) ratio of area to top width (d) wetted perimeter 12.4 The ﬂow can be uniform in (a) a non-prismatic channel (b) a wide rectangular channel (c) a horizontal trapezoidal channel (d) a frictionless rectangular channel 12.5 A rectangular channel has its width reduced from 6.0 m to 4.0 m at a transition. If the depth of ﬂow upstream of the contraction is 1.2 m, the change in the bottom elevation at the transition required to cause zero change in the water surface elevation is (a) 0.60 m drop (b) 0.60 m rise (c) 0.30 m drop (d) 0.30 m rise 12.6 The term alternate depths in open channel ﬂow is used to designate the depths (a) at the beginning and end of a hydraulic jump (b) having the same kinetic energy for a given discharge (c) having the same speciﬁc energy for a given discharge (d) at the beginning and end of a gradually varied ﬂow proﬁle 12.7 Which of the following conditions is the chief characteristic of critical ﬂow? Q 2T QT 2 = 1 (b) =1 (a) g A3 g A2 (c) Q2R =1 Q 2T 2 =1 g A3 g A3 12.8 If the alternate depths for certain ﬂow in a rectangular channel are 0.5 m and 3.0 m respectively, the critical depth for this channel is (a) 1.087 m (b) 1.333 m (c) 1.500 m (d) 3.500 m 12.9 While determining the critical depth applicable to channels of any shape, the length parameter used along with average velocity is the (a) ratio of area to wetted perimeter (b) wetted perimeter (c) depth of ﬂow (d) ratio of area to top width 12.10 For a triangular channel having side slopes of 2 horizontal : 1 vertical, the Froude number is given by F = (a) V/ g y (b) 2V/ g y (c) V/ 2g y (d) (d) V/ g( y / 2) 409 12.11 In a rectangular channel if the critical depth is 2.0 m, the speciﬁc energy at critical depth is (a) 3.0 m (b) 1.5 m (c) 2.0 m (d) 2.5 m 12.12 In a rectangular channel the depth of ﬂow is 1.6 m and the speciﬁc energy at that section is 2.7 m. The ﬂow is (a) sub-critical (b) supercritical (c) critical (d) not possible 12.13 For a uniform ﬂow with a depth of 0.6 m and Froude number of 2.0 in a rectangular channel, the speciﬁc energy will be (a) 2.4 m (b) 0.8 m (c) 2.6 m (d) 1.8 m 12.14 A rectangular channel carries a uniform ﬂow with a Froude number of 2.83. The ratio of critical depth to normal depth of this ﬂow is (a) 1.68 (b) 2.83 (c) 2.00 (d) 4.75 12.15 In a triangular channel with side slopes of 2.0 horizontal : 1 vertical, the critical depth is 2.8 m. The speciﬁc energy at critical depth is (a) 3.5 m (b) 3.0 m (c) 4.2 m (d) 3.72 m 12.16 For a given discharge in a channel at critical depth, (a) the total energy is minimum (b) the total energy is maximum (c) the speciﬁc energy is maximum (d) the speciﬁc energy is minimum. 12.17 At critical depth, (a) the discharge is minimum for a given speciﬁc energy (b) the discharge is maximum for a given speciﬁc force (c) the discharge is minimum for a given speciﬁc force (d) the discharge is maximum for a given speciﬁc energy 12.18 The speciﬁc energy Ec in a critical ﬂow at a depth yc occurring in a triangular channel is given by Ec = (a) 1.25 yc (b) 1.50 yc (c) 1.75 yc (d) 2.5 yc 12.19 For a given discharge in a channel the critical depth is a function of (a) slope of the channel (b) roughness of the channel (c) geometry of the channel (d) viscosity of the liquid 12.20 If the Froude number characterising the ﬂow in an open channel is less than unity, an increase in the channel width at a transition causes the water surface elevation to (a) remain unchanged (b) decrease (c) increase (d) form ripples 12.21 In a supercritical ﬂow in a rectangular channel, a smooth expansion changes the width from B1 to B2. This causes the water surface elevation after the expansion to (a) increase (b) decrease (c) remain unchanged (d) increase or decrease depending upon the channel roughness 12.22 In subcritical ﬂow in a channel, Dzm is the minimum height of a smooth hump that can be installed to cause critical ﬂow over the hump. If the hump of height Dz > Dzm is installed, then the ﬂow over the hump will be (a) subcritical (b) supercritical (c) critical and the upstream water surface will rise (d) critical and a lowering of the upstream water surface will occur 410 Fluid Mechanics and Hydraulic Machines 12.23 For a given discharge in a horizontal frictionless channel two depths may have the same speciﬁc force. These two depths are known as (a) speciﬁc depths (b) sequent depths (c) alternate depths (d) normal and critical depths 12.24 For ﬂow under a sluice gate where the upstream depth is 1.2 m and the depth at the vena contracta is 0.3 m, the discharge per metre width would be nearly. (a) 0.36 m3/s (b) 1.25 m3/s 3 (c) 1.45 m /s (d) 4.0 m3/s Uniform Flow: Resistance and Computation 12.25 A rectangular channel 3 m wide is laid on a slope of 0.0002. The average boundary shear stress for depth of ﬂow of 1.5 m is nearly (a) 0.90 N/m2 (b) 0.45 N/m2 2 (c) 0.30 N/m (d) 0.15 N/m2 12.26 In a wide rectangular channel the full supply depth is 1.52 m. If 50% of the full supply discharge is ﬂowing in this channel, the depth of ﬂow will be (a) 0.76 m (b) 0.90 m (c) 1.00 m (d) 0.43 m 12.27 The dimensions of Manning’s roughness coefﬁcient n are (a) L1/2 T–1 (b) L–1/3T (c) M0L0T0 (d) L 12.28 The dimensions of Chezy coefﬁcient C are (a) L–1/3 T (b) M 0L0T0 1/2 –1 (c) L T (d) LT–1 12.29 Manning’s roughness coefﬁcient n is related to Darcy–Weisbach friction factor f as È f R1/ 3 ˘ (a) n = Í ˙ ÍÎ 8 g ˙˚ (b) n = R 2/3/ 8g f (c) n = [8f R 1/3/g)1/2 (d) n = 8g/ f R 1/6 12.30 The Chezy coefﬁcient C and Manning’s n are related as 1 (a) C = n1/3 R1/6 (b) C = R1/6 n n1/ 6 (c) C = (d) n = C R1/6 R 12.31 A rectangular channel, 2.0 m wide has a bed slope of 1/800. Taking Chezy coefﬁcient as 60, the discharge in the channel at a depth of ﬂow of 1.0 m is (a) 1.0 m3/s (b) 1.5 m3/s 3 (c) 2.0 m /s (d) 3.0 m3/s 12.32 In a wide rectangular channel, an increase in a normal depth by 20% correspond to an increase in discharge by about (a) 13% (b) 25% (c) 36% (d) 48% 12.33 For a hydraulically efﬁcient rectangular channel of bed width 4.0 m, the depth of ﬂow is (a) 4.0 m (b) 8.0 m (c) 1.0 m (d) 2.0 m 12.34 For a hydraulically efﬁcient triangular section the hydraulic radius R = (a) 2 2 y (b) y/2 2 (c) y/2 (d) y 12.35 In a hydraulically efﬁcient circular channel the ratio of the hydraulic radius to the diameter of the channel is (a) 1.0 (b) 0.5 (c) 0.25 (d) 0.125 12.36 A hydraulically most efﬁcient trapezoidal channel section carries water at the optimal depth of 0.72 m Chezy coefﬁcient is 75 and the longitudinal slope is 1 in 2500. What is the discharge through the channel? 411 (a) 0.808 m3/s (b) 1.14 m3/s 3 (c) 0.900 m /s (d) 0.090 m3/s 12.37 In a hydraulically most efﬁcient trapezoidal channel section the hydraulic radius R = (a) y/2 (b) y 4 y 3 12.38 In a hydraulically most efﬁcient trapezoidal channel section the ratio of the bed width to depth is (a) 0.50 (b) 0.707 (c) 0.866 (d) 1.155 12.39 At the same mean velocity, the ratio of head loss per unit length for a sewer pipe running full to that for the same pipe ﬂowing half full would be (a) 2.0 (b) 1.67 (c) 1.0 (d) 0.67 (c) y/2 2 (d) Hydraulic Jump 12.40 The sequent depth ratio of a hydraulic jump in a rectangular channel is 16.48. The Froude number at the beginning of the jump is (a) 5.0 (b) 8.0 (c) 10.0 (d) 12.0 12.41 The Froude number at the end of a hydraulic jump in a rectangular channel is 0.25. The sequent depth ratio of this jump is (a) 2.5 (b) 5.2 (c) 8.9 (d) 9.8 12.42 The type of jump that forms when the initial Froude number lies between 2.5 and 4.5 is known as (a) weak jump (b) steady jump (c) Undular jump (d) Oscillating jump 12.43 In a horizontal rectangular channel a hydraulic jump with a sequent depth ratio of 5.0 is formed. This jump can be classiﬁed as (a) weak jump (b) oscillating jump (c) strong jump (d) steady jump 12.44 The sequent depths in a hydraulic jump formed in a 4.0 m wide rectangular channel are 0.2 m and 1.0 m. The discharge in the channel, in m3/s, is (a) 5.00 (b) 1.12 (c) 2.17 (d) 4.34 12.45 The sequent depths in a hydraulic jump formed in a horizontal rectangular channel are 0.2 m and 2.0 m. The length of the jump is about (a) 50 m (b) 12 m (c) 8 m (d) 2 m 12.46 In a hydraulic jump occurring in a horizontal rectangular channel the sequent depths are 0.25 m and 1.25 m. The energy loss in this jump is (a) 0.8 m (b) 1.0 m (c) 1.25 m (d) 1.50 m 12.47 The discharge per metre width at the foot of a spillway is 10 m 3/s at a velocity of 20 m/s. A perfect free jump will occur at the foot of the spillway when the tail water depth is nearly (a) 4.5 m (c) 6.50 m (b) 5.00 m (d) 8.50 m Gradually Varied Flow 12.48 The differential equation of the gradually varied ﬂow can be written by using Manning’s formula for the case of a wide dy rectangular channel as = dx 412 Fluid Mechanics and Hydraulic Machines (a) S0 (b) S0 (c) S0 (d) S0 1 - ( y0 / y )3.33 1 - ( yc / y )3 1 - ( yc / y ) 12.55 3.33 1 - ( y0 / y )3 1 - ( y0 / y )3 1 - ( yc / y )3 1 - ( y0 / yc )3 1 - ( yc / y ) 12.56 12.57 3.33 12.49 If E = speciﬁc energy at a section in a gradually varied ﬂow, then dE/dx = 12.50 12.51 12.52 12.53 12.54 (a) S0 + S f (b) S0 – Sf (c) S f – S0 (d) Sf /S0 – 1 where S f = energy slope and S0 = bed slope. If in a gradually varied ﬂow dy/dx is positive, then dE/dx (a) is always negative (b) is always positive (c) is positive if y/yc > 1 (d) is negative if y > yc A 3 m wide rectangular channel ﬂowing at its normal depth of 0.8 m carries a discharge of 9.5 m3/s. The channel slope is (a) steep (b) critical (c) mild (d) none of the above In an M1 type of gradually varied ﬂow proﬁle (a) y0 > y > yc (b) y0 > yc > y (c) y > y0 > yc (d) yc > y0 > y In an M2 type of gradually varied ﬂow proﬁle (a) y0 > y > yc (b) y > y0 > yc (c) y0 > yc > y (d) yc > y > y0 The ﬂow will be in supercritical state in the following proﬁles (a) M3, S3 and M1 (b) M2, S1 and M3 12.58 (c) S2, S3 and M3 (d) S1, S2 and S3 Which of the following is the correct representation of sequence of surface proﬁles if the channel slope changes from Mild to Steep? (a) M1, S1 (b) M1, S2 (c) M2, S3 (d) M2, S2 A wide rectangular channel carries a ﬂow of 2.96 m3/s per metre width. The bed slope of the channel is 1.0 ¥ 10–4 and Manning’s n = 0.021. If at a section the depth of ﬂow is 1.5 m the energy slope at that section is (a) 0.01 (b) 0.00228 (c) 0.0009 (d) 0.001 A 2 m wide rectangular channel ﬂowing at its normal depth of 1.2 m carries a discharge of 6.0 m3/s. If at a section, the depth of ﬂow is 1.10 m, the water surface at that location is a part of the gradually varied ﬂow of type (a) S2 (b) M2 (c) M3 (d) S3 The ﬂow in a long 4.0 m wide rectangular channel is 8.0 m3/s. The normal depth of ﬂow is 1.5 m. If, at a certain section A, the depth of ﬂow in the channel is 1.0 m, the depth of ﬂow at a section downstream of A would be (a) > 1.0 m (b) < 1.0 m (c) = 1.0 m (d) £ 1.0 m Broad-crested Weir 12.59 The discharge Q over a broad-crested weir of length L is often expressed as Q = 1.705 Cd LH3/2. In this expression H is the difference in elevation between (a) the upstream energy line and the crest (b) the upstream water surface and the crest 413 (c) the upstream water surface and the upstream bed (d) the upstream energy line and the downstream energy line 12.60 The modular limit of a broad-crested weir is about (a) 15% (b) 35% (c) 67% (d) 90% 12.61 If the Cd and length L of a rectangular notch and a broad-crested weir are the same, then for the same head on both of References Subramanya, K., Flow in Open Channels, Tata McGraw Hill Education Private Ltd., New Delhi, 3rd Edition, 3rd Reprint, 2009. these (a) the broad-crested weir passed 73% more discharge than the rectangular notch (b) the rectangular notch passes 57.7% less discharge (c) the rectangular notch passes 73% more discharge than the broadcrested weir (d) both notch and the broad-crested weir pass equal discharges. Flow Measurement Concept Review 13 Introduction - 13.1 ORIFICES An oriﬁce is an opening in a ﬂuid container. It is an important ﬂow element which ﬁnds application in diverse ﬂuid ﬂow situa tions including ﬂuid ﬂow measurement and control. Figure 13.1 shows the trajectory from a circular oriﬁce of diameter d and area a in a tank containing a liquid to a height H above the centreline of the oriﬁce. If d 3H1, H1/P < 1.0. For a triangular weir with a central angle q (Fig. 13.11), the discharge under a head H1 is given by Q = Cd 8 15 2 g tan q ◊ H15 / 2 2 (13.21) (13.19) where P = height of the weir crest above the channel bottom. This relationship for Cd is valid for H1/P £ 5. Contracted Weirs When the length of the weir L is less than the width of the channel B, (Fig. 13.10) the weir is known as a contracted weir. Due to the presence of the end contractions the effective length of the weir will be smaller than L. The discharge equation for contracted rectangular weirs ﬂowing free is given by Francis formula H1 q B Triangular Weir The coefﬁcient of discharge Cd is, in general, a function of q and has a value around 0.58. L End contraction H1 Weir crest P B Contracted Weir The ﬂow from a trapezoidal weir (Fig. 13.12) of side slope m horizontal : 1 vertical is considered as a combination of ﬂow from a suppressed rectangular weir of length L = L1, and from a triangular weir with a central angle of 2q, where tan q = m. (See Example 13.41) However, an exception to the above rule is for a weir with side slope value of 1 horizontal: 4 vertical; the discharge is calculated by using the 421 Flow Measurement q H1 q L1 B Trapezoidal Weir suppressed weir formula, viz. Q= 2 Cdc 2 g BH13 / 2 3 (13.22) A current meter measures point velocity in the cross section of an open channel. It is a mechanical device, consisting of a rotating element (cup assembly/ propeller) the rotational speed of which, when immersed in the ﬂow, is a measure of the velocity at the meter. Figure 13.14 shows a horizontal axis propeller type current meter. This type of meter comes in a wide variety of size with propellers diameters in the range 6 cm – 15 cm to register velocities in the range 0.15 – 4.0 m/s. A current meter is so designed that its rotational speed varies linearly with the stream velocity. where the coefﬁcient of discharge Cdc is a constant and has a value of about 0.63. This weir is known as Cipolletti weir. If the tailwater level in a weir is above the weir crest, the weir will be functioning under submerged ﬂow mode (Fig. 13.13). Hoisting & electrical connection Propeller Sounding weight Fin for stabilization Horizontal Axis Current Meter 13.6 ROTAMETER H1 H2 P Submerged Weir Flow The discharge over the weir Qs is estimated by the Villemonte formula È Ê H ˆn ˘ Qs = Q1 Í1 - Á 1 ˜ ˙ Í Ë H2 ¯ ˙ Î ˚ 0.385 (13.23) where Q1 = free-ﬂow discharge under the head H1, n = exponent in the head-discharge relationship for the weir (For rectangular weir n = 1.5, and for triangular weir n = 2.5), H2 = downstream water surface elevation measured above the weir crest, and H1 = upstream head. A rotameter is a device to measure ﬂow rate of ﬂuid in a pipe. A rotameter consists of a transparent vertical tapering tube with a ﬂoat in it (Fig. 13.15). Fluid entering from the bottom of the tube will raise the ﬂoat increasing the annular area between the tube and the ﬂoat. An equilibrium position is reached at which the upward force of the ﬂuid on the ﬂoat is balanced by the weight of the ﬂoat. The position of the ﬂoat is a measure of the discharge, the greater the ﬂow the higher the position of the ﬂoat in the tube. The ﬂoat is so shaped that it will rotate in the ﬂow (hence the name) and maintains its position on the axis of the tube. The shapes of the tube and ﬂoat are so adjusted to get a linear discharge scale, which is etched on the tube. In practice, it is necessary to install the rotameter in a vertical position with the ﬂow entering from bottom position. 422 Fluid Mechanics and Hydraulic Machines Out Tapered glass metering tube Float Float stopper In Rotameter Hot-wire anemometer is an instrument used for measuring velocity of ﬂow and turbulent properties of a ﬂow of gas or air. This instrument consists of a probe for data acquisition and an electronics unit for signal processing of the probe output. The probe consists essentially of a very thin platinum or nickel wire of size of about 5 ¥ 10–3 mm diameter and of length of about 5 mm. The wire is mounted on the ends of two pointed prongs and is introduced into the ﬂow ﬁeld so that the ﬂow is normal to the wire. A small electric current is passed through the wire to heat it. As the gas ﬂow passes past the wire, the hot wire is cooled. The amount of heat lost form the wire to the gas ﬂow is a function of the velocity of the ﬂow. The heat transfer alters the resistance of the wire. This change in the resistance is appropriately processed in the electronics of the instrument. Through calibration the output is made proportional to the velocity. Since the response of the instrument is very fast, turbulent ﬂuctuations of the velocity could be measured with sufﬁcient accuracy. Hot wire anemometer is the basic instrument in experimental turbulent ﬂow studies. Laser Doppler velocimetry is a technique for measuring the velocity of ﬂuids with high accuracy. In this technique, two beams of collimated, monochromatic, and coherent laser light are crossed in the ﬂow of the ﬂuid being measured. The two beams are usually obtained by splitting a single beam, thus ensuring coherency between the two. The two beams at the intersection volume interfere and generate a set of straight fringes. Particles passing through the fringes reﬂect light into a photo detector, and since the fringe spacing is known (from calibration), the velocity can be calculated. The normal impurities present in the liquids serve as source for the necessary particles for ﬂow measurement. In gasses, however, sometimes they have to be seeded. Laser Doppler Anemometry (LDA) is ideal for nonintrusive 1D, 2D and 3D point measurement of velocity and turbulence distribution in both free ﬂows and internal ﬂows. Other advantages of LDA are: (i) high spatial resolution of the ﬂow ﬁeld, and (ii) velocity data are independent of the thermodynamic properties of the ﬂuid. Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult 423 Flow Measurement Worked Examples (iii) The trajectory of the jet from the vena contracta is x Cv = 4yH 13.1 Solution: Ideal velocity V = 2g H = V2 Head loss = hL = H – a 2g 2.0 – Va2 = 0.2 2g Va = 2 ¥ 9.81 ¥ ( 2.0 - 0.2) = 5.943 m/s V (i) Coefﬁcient of velocity Cv = a V 5.943 = = 0.949 6.264 Coefﬁcient of discharge Cd = Cc ◊ Cv = 0.63 ¥ 0.949 = 0.598 (ii) Discharge through the oriﬁce Q = Cda x = Cv for y = 0.50 m, x = 0.949 = 1.898 m 4 ¥ 0.50 ¥ 2.0 13.2 2 g ¥ 2.0 = 6.264 m/s \ 4yH or Solution: Here H = 5.5 m, x = 1.5 m and y = 0.12 m. x 1.5 Cv = = 4yH 4 ¥ 0.12 ¥ 5.5 = 0.923 Q = Cd a 2g H Êp ˆ 3 ¥ 10–3 = Cd Á ¥ (0.025) 2 ˜ ¥ Ë4 ¯ Since Cd = 0.588 Cd = Cv Cc Cc = 0.588 = 0.637 0.923 13.3 Cv Ha 2g H Êp ˆ = 0.598 ¥ Á ¥ (0.04) 2 ˜ Ë4 ¯ ¥ (2 ¥ 9.81 ¥ 2.0)1/2 = 4.707 ¥ 10–3 m3/s = 4.707 L/s 2 ¥ 9.81 ¥ 5.5 H Ha H 424 Fluid Mechanics and Hydraulic Machines Solution: Refer to Fig. 13.16. = Vj = velocity of jet = Cv Ha A = 0.98 Hb 2 ¥ 9.81 ¥ 5.426 Discharge Q = A1V1 = ajVj p ¥ (0.10)2 ¥ 2.5 = aj ¥ 10.11 4 ya yb p (Dj)2 4 Dj = 0.0497 m = 4.97 cm aj = 1.942 ¥ 10–3 m2 = x O Head loss The trajectory is given by x2 = 4yHC v2 At the point of intersection, (1) Ê 1 ˆ Vj2 HL = Á 1 ˜ 2g Ë Cv2 ¯ Ê 1 ˆ (10.11) 2 ¥ = Á 1 ˜ 2 ¥ 9.81 Ë (0.98) 2 ¯ xa2 = 4ya HaCv2 = xb2 = 4ybHbCv2 Also 2g H = 10.11 m/s B ya H = b yb Ha Hence 50 ( 2.5) 2 + = 5.426 m 9.79 2 ¥ 9.81 = 0.215 m (2) 13.5 Ha + ya = Hb + yb ya – yb = Hb – Ha Solving for ya from Eqs. (2) and (3) ya = Hb and yb = Ha (3) Substituting in Eq. (1), x2 = 4ya Ha Cv2 = 4yb HbCv2 \ x= 4 H a H bC v2 = 2Cv 13.4 Ha Hb Solution: In Fig. 13.17, let the sufﬁx j refer to the actual jet at the vena contracta. Thus aj = area of the jet at the vena contracta and vj = velocity of the jet at that section. It is given that vj = 25 m/s 10 cm Vena contracta V1 vj A1 aj Solution: By taking Z1 = 0 Total head H= 3 cm dia p1 V2 + 1 g 2g 1 J 425 Flow Measurement aj = Cc ¥ (area of the nozzle end) p = 0.80 ¥ ¥ (0.03)2 = 5.655 ¥ 10–4 m2 4 Discharge Q = ajvj = 5.655 ¥ 10–4 ¥ 25 = 0.01414 m3/s By continuity, Q = ajvj = Cc avj = A1V1 vj = 0.96 ¥ and 2 ¥ 9.81 ¥ 34.565 = 25.0 m/s] 13.6 vena contracta 2 Ê aˆ Ê 3ˆ V1 = Cc Á ˜ vj = 0.8 ¥ Á ˜ ¥ 25 Ë A1 ¯ Ë 10 ¯ = 1.8 m/s Energy loss in the nozzle Ê 1 ˆ v 2j = HL = Á 2 - 1˜ Ë Cv ¯ 2g By applying the Bernoulli theorem to section 1 and J \ Ê p1 v 2j V12 ˆ p0 = + Z + + + Z – H L 0 1 Ág 2 g ˜¯ g 2g Ë p0 = 0 = atmospheric pressure g head, Z1 = Z0 v 2j Ê p1 V12 ˆ Ê 1 ˆ v 2j – = 1 + Ág Á 2 ˜ 2g 2g 2 g ˜¯ ËCv ¯ Ë v 2j 1 p1 V12 = – g 2 g C 2v 2g È ( 25) 2 ˘ 1 =Í - (1.8) 2 ˙ 2 2 ¥ 9.81 ÍÎ (0.96) ˙˚ = 34.40 m p1 = 34.40 ¥ 9.79 = 336.8 kPa [Check: You can check your calculations by noting that vj = Cv 2g H (1.8) 2 2 ¥ 9.81 H = 34.565 m, = 34.40 + \ and H= p1 V2 + 1 g 2g m r D D2 K VD v 4 2¥ 4 4¥ 4 ¥ 4 K Solution: Discharge Q = K0 ◊ A2 2g D H A2 = Area of the oriﬁce = = 7.854 ¥ 10–3 m2 DH = 0.5 m, p ¥ (0.10)2 4 2g D H = 2 ¥ 9.81 ¥ 0.5 = 3.132 m/s As K0 is not known to start with, a trial and error method is adopted. Let K0 = 0.62 Q = 0.62 ¥ (7.854 ¥ 10–3) ¥ 3.132 = 0.01525 m3/s Q 0.01525 = Velocity V = 2 Êp ( pD1 /4) 2ˆ ÁË 4 ¥ 0.20 ˜¯ = 0.485 m/s Reynolds number VD1 0.485 ¥ 0.20 = Re = = 96806 v (0.001/ 998) = 105 For this value of Re, the ﬂow coefﬁcient K0 = 0.62 which is the same as the assumed value. Hence no more iterations are required. The discharge Q = 0.01525 m3/s = 15.25 L/s. 426 Fluid Mechanics and Hydraulic Machines 13.7 H Solution: Discharge Q = Kf A2 2 M R N area = a C 2 2g D H Q = 0.002 m3/s, Kf = 0.99 p A2 = area of the nozzle = ¥ (0.03)2 4 = 7.0686 ¥ 10–4 0.002 = 0.99 ¥ 7.0686 ¥ 10–4 2 ¥ 9.81 ¥ D H DH = piezometric head difference across the nozzle = 0.4163 m Êp p ˆ DH = Á 1 - 2 ˜ + (Z1 – Z2) g ¯ Ëg For a horizontal nozzle Z1 – Z2 = 0. Hence D p = (p1 – p2) = g DH = 9.79 ¥ 0.70 ¥ 0.4163 = 2.853 kPa C 13.8 tube is essentially due to sudden expansion from the contracted section C-C to full pipe section (as at 2-2). Hence, Head loss HL = (Vc – V2)2/2g But Q = Vcac = V2a Also ac = Cc a V Thus Vc = 2 Cc 2 HL = V22 Ê 1 ˆ ˆ V22 Ê 1 = 1 ÁË 0.62 - 1˜¯ Á ˜ 2 g 2 g Ë Cc ¯ = 0.376 V22 2g Applying Bernoulli equation between points M and N and, if Ha = atmospheric pressure head, Ha + H = Ha + 1.376 Solution: A mouthpiece is a short tube ﬁtted in place of an oriﬁce. The ﬂow in a mouthpiece ﬁrst contracts to a vena contracta as in an oriﬁce (Section C-C in Fig. 13.18) and then it expands to ﬁll the tube of area a. The loss of head in the 2 V22 + HL 2g V22 = H or V2 = 0.853 2g H 2g V 0.853 Vc = 2 = 2g H C2 0.62 = 1.376 2g H Since the velocity in the tube is the highest at the vena contracta, the minimum pressure occurs at section c-c. By applying Bernoulli equation to N and R. 427 Flow Measurement Ha + Êp ˆ V22 V2 + HL = Á c ˜ + c 2g 2g Ë g ¯ abc p V2 Ha – c = 2 g 2g or Ê 1 ˆ ÁË 0.62 ˜¯ = 1.225 2 H1 = 3.0 m dh V2 – 1.376 2 2g 2g Ê H ˆ = 1.225 ¥ Á Ë 1.376 ˜¯ = 0.890 H Hence – A dh = Cd a T= dt = Ú H2 Adh H1 Cd a 2 g h1/ 2 p ¥ (2)2 4 p a = area of oriﬁce = ¥ (0.1)2 4 2 A Ê 2 ˆ = Á ˜ = 400 Ë 0.1¯ a Cd = 0.63, H1 = 3.0 m, and H2 = 2.0 T = time to lower the water surface from 3.0 m to 2.0 m 2 ¥ 400 = ( 3 - 2) 0.63 ¥ 2 ¥ 9.81 A = area of tank = \ (9.855 - 0.409) = 10.61 m 0.890 Thus cavitation will take place in the tube at a head H = 10.61 m. H= Ú 2g H dt Here A is constant with respect to h and as such 2A T = ( H1 - H 2 ) Cd a 2 g 0.409 = 9.855 – 0.890 H or area = a vj Ê pc ˆ = (Ha – 0.890 H) (absolute) ÁË g ˜¯ abs Êp ˆ [The gauge Á c ˜ will be negative by an extent Ë g ¯ of 0.89 H.] (ii) For cavitation to take place the local pressure must reach the vapour pressure. Hence, the cavitation conditions. pc = pv = 4.0 kPa (abs) pa 96.48 Also, = 9.855 m = Ha = 9.79 g Ê pc ˆ 4.0 = 0.409 m = ÁË g ˜¯ 9.79 abs \ H2 = 2.0 m Water h V22 = 91.1 s 13.9 13.10 R L a C Solution: Refer to Fig. 13.19. Let A = area of the tank at height h. Let the water level fall by dh in time dt. Then H to H2 428 Fluid Mechanics and Hydraulic Machines Solution: Figure 13.20 shows a cross section of the tank. At any instant t, let the water surface be at a height h above the oriﬁce. The top width of water surface b = 2x = 2 ¥ =2 2 (OB) - (OA ) ( R 2 - ( R - h) 2 = 2 13.11 R a Cd 2 ( 2 R h - h2 ) T= Area of the water surface Solution: Refer to Fig. 13.21 which is a deﬁnition sketch of the problem. Consider the water surface to be at an elevation h above the oriﬁce at any instant t from start of the draining process. Initially (at t = 0) the water level was at a height h = R above the oriﬁce. It is required to ﬁnd the time t = T at which the height h = 0. Let x = radius of the water surface at a height h above the oriﬁce. From Fig. 13.1.11, ( 2 R h - h2 ) A = L ¥ b = 2L H1 O R A x H2 14 pR5/ 2 15 Cda 2g B h OC2 - OB 2 x = BC = But Orifice of area = a OB = R – h and OC = R. Hence If the water surface drops by an amount dh in time dt, –A dh = Cd a T= Ú =– =– 2g h dt R 2 - ( R - h) 2 = 2 Rh - h2 Area of the water surface = A = p x2 = p(2Rh – h2) If the water surface drops by an amount dh in time dt, –Adh = Cd a 2gh dt T Adh - p( 2 Rh - h2 )dh = Cda 2g Cda 2g p Putting K = Cda 2g dt = – K(2 Rh – h2)dh Integrating between the limits t = 0, h = R and t = T, h = 0 dt = dt 0 Ú x= H2 1 H1 Cd a 2 g 2L Cd a 2 g Ú H2 H1 ◊ 2 L 2 R h - h2 h dh 2R - h d h 2L 2 = ¥ [(2R – H2)3/2 – (2R – H1)3/2] 3 Cd a 2 g L 4 T= ◊ [(2R – H2)3/2 – (2R – H1)3/2] 3 Cd a 2g T T= Ú 0 dt = – K 0 Ú (2Rh – h ) dh 2 R R T =K È4 Ú (2Rh – h ) dh = K ÍÎ 3 R 2 R 14 T = KR5/2. 15 5/ 2 - 2 5/ 2 ˘ R ˙ 5 ˚ 429 Flow Measurement Ê Aˆ dh = dH Á1 + 1 ˜ A2 ¯ Ë Hemisphere with radius = R O B –A1 dH = Cd a C x dh h T= Orifice of diameter = d T = Flow from orifice Ú or dH = dh (1 + A1/ A2) 2g H dt T dt = – 0 Ú A1h-1/ 2dh H2 H1 Cd a (1 + A1/ A2) 2 A1 Cd a (1 + A1/ A2) 2g 2g (H11/2 – H21/2) A = A1 + A2 = 6 m2 A1/A2 = 2 \ A1 = 4 m2 and A2 = 2 m2 p a = ¥ (0.06)2 = 2.827 ¥ 10–3 m2 4 H1 = 2.0 m, H2 = 1.5 m Substituting in the expression for T Here Substituting for K, T = 14 pR5 / 2 15 Cd a 2g 13.12 T = 2 ¥ 4 ¥ ( 2 - 1.5 ) 0.63 ¥ ( 2.827 ¥ 10 -3 ) ¥ (1 + 2) ¥ 2 ¥ 9.81 = 64 s ¥ 13.13 C = Solution: Referring to Fig. 13.22, Area = A dH h dH. Orifice, area = a A1 A2 Solution: The discharge in the venturimeter Q = Area = A2 Let at any instant of time the difference in head is h, and let dH be the drop in tank A1. New head difference A h – dh = h – dH – dH ◊ 1 A2 Cd A2 1 - ( D2 /D1 ) 4 2g D h The differential manometer reading y will give directly the difference in piezometric head Dh as ÊS ˆ Dh = y Á m - 1˜ Ë Sp ¯ Here Sm = 13.6, Sp = 0.80, y = 0.4 m 430 Fluid Mechanics and Hydraulic Machines where HL = head loss between 1 and 2. But Oil RD = 0.8 Ê p1 ˆ Ê p2 ˆ ÁË g + Z1 ˜¯ – ÁË g + Z 2 ˜¯ = Dh Also Hence 2 1 Ê 13.6 ˆ Dh = 0.4 Á - 1 = 6.4 m of oil Ë 0.80 ˜¯ Cd = 1.0 as the losses are neglected. p A2 = (0.2)2 = 0.031416 4 D2 20 = = 0.5, 1 - (0.5) 4 D1 40 = 0.96825 Substituting the various values, 0.031416 Q= ¥ 2 ¥ 9.81 ¥ 6.4 0.96825 \ Q = 0.3636 m3/s = 363.6 L/s of oil –C V2 = Q A2 Q2 Ê 1 1 ˆ - 2 ˜ = Dh – HL Á 2 2 g Ë A2 A1 ¯ 2 Q 2 1 Ê Ê A2 ˆ ˆ 1 Á ˜ = Dh – HL 2 g A22 ÁË ÁË A1 ˜¯ ˜¯ Q2 or 2g A22 2 C Dh Solution: If sufﬁxes 1 and 2 refer to the inlet and the throat respectively, p1 V2 p V2 + 1 + Z1 = 2 + 2 + Z2 + HL g 2g g 2g = C d2 (1 - ( D2 /D1 ) 4 Dh (1) (2) Substituting from (2) in (1) Cd2 Dh = Dh – HL HL = (1 – Cd2) Dh or [Note: This is a very useful result and is to be kept in mind while solving venturimeter problems. Many problems involving loss of energy in the inlet portion of the venturimeter, (for instance Worked Examples, 13.15,13.19 and 13.22) can be solved very easily through use of this result.] 13.14 H and 4 Q 2 1 Ê Ê D2 ˆ ˆ 1 Á ˜ = Dh – HL 2 g A22 ÁË ÁË D1 ˜¯ ˜¯ But the venturimeter ﬂow equation Cd A2 Q = 2g D h 1 - ( D2 /D1 ) 4 Mercury Dh Q A1 i.e. y = 0.4 m V1 = 13.15 431 Flow Measurement Solution: (i) The venturimeter discharge is given by Cd A2 Q= 2g D h 1 - ( D2 /D1 ) 4 13.16 ¥ The differential manometer reading y is related to Dh as ÊS ˆ Dh = y Á m - 1˜ Ë Sp ¯ In this case, Ê 13.6 ˆ Dh = 0.20 Á - 1˜ = 2.52 m Ë 1.0 ¯ p 2 ¥ (0.05) = 1.9635 ¥ 10–3 m2 A2 = 4 D2 5 = = 0.4, D1 12.5 ÊD ˆ 1- Á 2 ˜ Ë D1 ¯ V1 = Velocity in the pipe = = Q A1 0.01343 = 1.094 m/s p ¥ (0.125) 2 4 V12 = 0.061 m 2g HLi = head loss in the converging cone. = (1 – Cd2) Dh = [1 – (0.96)2] ¥ 2.52 = 0.1976 m of water HLd = head loss in the diverging cone V12 = 10 ¥ 0.061 = 0.61 m of water 2g Total head loss in the meter HL = HLi + HLd = 0.1976 + 0.61 = 0.8076 m of water = 10 Solution: The discharge in the venturimeter is Q = Cd A2 2g D h 1 - ( D2 /D1 ) 4 For a differential manometer, ÊS ˆ Dh = y Á m - 1˜ Ë Sp ¯ 4 = 0.9871 0.96 ¥ 0.019635 Q= 2 ¥ 9.81 ¥ 2.52 0.9871 = 0.01343 m3/s = 13.43 L/s (ii) C In the present case, Sm = 13.6, Sp = 0.9 Ê 13.6 ˆ \ Dh = y Á - 1˜ = 14.11 y Ë 0.9 ¯ p 2 ¥ (0.1) = 7.854 ¥ 10–3 m2 A2 = 4 D2 = 0.5, 1 - ( D2 /D1 ) 4 = 0.96825 D1 Q = 0.99 ¥ 7.854 ¥ 10 -3 ¥ 0.96825 Q = 0.1336 2 ¥ 9.81 ¥ 14.11 y y (i) When y = 9 cm = 0.09 m, Q = 0.1336 = 0.040 m3/s = 40.0 L/s (ii) When Q = 50 L/s = 0.050 m3/s 0.050 = 0.1336 y y = 0.14 m = 14 cm 13.17 ¥ 0.09 432 Fluid Mechanics and Hydraulic Machines Solution: Cavitation occurs when the local pressure reaches the vapour pressure. Hence, for maximum discharge the pressure at throat p2 = pv = 4.0 kPa (abs) Inlet pressure (abs) = 10 kPa (gauge) + atmospheric pressure p1 = 10.0 + 96.0 = 106.0 kPa (abs) For a horizontal venturimeter Z1 = Z2 Êp ˆ Êp ˆ p - p2 Dh = Á 1 + Z1 ˜ – Á 2 + Z 2 ˜ = 1 g Ëg ¯ Ë g ¯ \ = difference in piezometric heads. 4 V22t Ê Ê D2 ˆ ˆ Á1 ˜ = Dh 2 g ÁË ÁË D1 ˜¯ ˜¯ 1 - ( D2 /D1 ) 4 Here 1 - ( D2 /D1 ) Q= 4 2g D h p ¥ (0.05)2 4 = 1.9635 ¥ 10–3 m2 50 = = 0.5, 100 = 1 - (0.5) 0.95 ¥ 1.9635 ¥ 10 0.96825 -3 4 = 0.96825 Cd A2 Q = 1 - ( D2 /D1 ) 4 \ ¥ 2g D h ¥ 2g D h = V2 A2 V2 = actual velocity at section 2 = Cd 2g D h 1 - ( D2 /D1 ) 4 = CdV2t Hence here Cd = Cv = coefﬁcient of velocity Head loss ¥ HLi = 2 ¥ 9.81 ¥ 10.419 p1 V2 p V2 + Z1 + 1 – 2 – Z2 – 2 g 2g g 2g = Dh + = 0.0275 m3/s = 27.5 L/s (maximum discharge) 1 - ( D2 /D1 ) 4 Now let us consider the actual ﬂow: The actual discharge Cd = 0.95, A2 = D2 D1 \ ¥ 1 V2t = or 106.0 - 4.0 = 10.419 m 9.79 The venturimeter discharge equation is Cd A2 V1D12 = V2D22 Also = Q= Ê p1 ˆ p2 ÁË g + Z1 - g - Z 2 ˜¯ = Dh But V12 V2 – 2 2g 2g Substituting Dh = [1 – (D2/D1)4] 13.18 and as 4 Ê 1 ˆÊ Ê D ˆ ˆ V2 H = Á 2 - 1˜ Á 1 - Á 2 ˜ ˜ 2 Ë D1 ¯ ˜¯ 2 g Ë Cd ¯ ÁË Solution: First consider the situation without losses V22t p1 V2 p + 1 + Z1 = 2 + + Z2 g 2g g 2g where V2t = theoretical velocity at the throat V12 V2 Ê D ˆ = 2 Á 2˜ 2g 2g Ë D1 ¯ HLi = È V22 Í 1 2 g Í Cd2 Î 1 V22 Cd2 2 g 4 ÏÔ Ê D ˆ 4 ¸Ô Ê D ˆ 4 ˘ 2 2 ˙ Ì1 - Á ˜¯ ˝ + ÁË D ˜¯ - 1 ˙ D Ë 1 1 ÓÔ ˛Ô ˚ 4 Ê 1 ˆ ÏÔ Ê D ˆ ¸Ô V 2 HLi = Á 2 - 1˜ Ì1 - Á 2 ˜ ˝ 2 Ë Cd ¯ ÔÓ Ë D1 ¯ ˛Ô 2g 433 Flow Measurement ¥ 13.19 The discharge Q by the venturimeter equation is Cd A2 Q = 1 - ( D2 /D1 ) C 4 ¥ 2g D h Cd = 0.98, A2 = p ¥ (0.15)2 = 0.01767 m2 4 D2 15 = = 0.5, D1 30 1 - ( D2 /D1 ) 4 = 0.96825 0.98 ¥ 0.01767 19.62 ¥ 3.78 0.96825 = 0.154 m3/s = 154 L/s (ii) DZ = Z2 – Z1 = difference in elevation between the throat and the inlet = 0.45 sin 30° = 0.225 m \ Q = 2 1 30° DZ 0.45 m x Êp ˆ Êp ˆ Since Á 1 + Z1 ˜ – Á 2 + Z 2 ˜ = Dh = 3.78 m Ëg ¯ Ë g ¯ 0.3 m Z Datum Mercury p2 p = 1 – (Z2 – Z1) – Dh g g 50 – 0.225 – 3.78 = 1.102 m 9.79 p2 = 1.102 ¥ 9.79 = 10.79 kPa (iii) Head loss in the converging section = HLi Solution: Êp ˆ Êp ˆ (i) Á 1 + Z1 ˜ – Á 2 + Z 2 ˜ = D h Ëg ¯ Ë g ¯ From differential manometer ÊS ˆ Dh = y Á m - 1˜ Ë Sp ¯ Ê 13.6 ˆ = 0.3 Á - 1˜ = 3.78 m Ë 1 ¯ Ê 1 ˆ = Á 2 - 1˜ Ë Cd ¯ Ê Ê D ˆ4ˆ V 2 Á1 - Á 2 ˜ ˜ 2 ÁË Ë D1 ¯ ˜¯ 2 g 4 Ê 1 ˆ ÏÔ Ê 15 ˆ ¸Ô = Á 1 1 ˜¯ Ì ÁË 30 ˜¯ ˝ Ë 0.982 ÓÔ ˛Ô 2 1 Ê 0.154 ˆ ÁË 0.01767 ˜¯ 2 ¥ 9.81 = 0.0387 ¥ 3.871 = 0.15 m Alternatively: HL = (1 – Cd2) Dh = (1 – 0.982) (3.78) = 0.15 m 434 Fluid Mechanics and Hydraulic Machines 13.20 13.21 V22 2g Solution: (a) HLi = Head loss at inlet = (1 – Cd2) Dh = 0.03 Dh \ Cd = 0.985 From the inverted differential manometer Solution: For the inverted differential U-tube manometer, Dh = difference in piezometric heads Ê 0.75 ˆ S ˆ Ê = y Á1 - m ˜ = 30 Á1 = 7.5 cm Ë 1.0 ˜¯ Sp ¯ Ë = 0.075 m The loss of head between inlet and throat: Ê S ˆ Dh = y Á1 - m ˜ Sp ¯ Ë 4 Ê 1 ˆ Ê Ê D2 ˆ ˆ V22 HLi = Á 1 1 Á ˜ Á ÁË D ˜¯ ˜˜ 2 g Ë Cd2 ¯Ë 1 ¯ 0.6 ˆ Ê = 0.15 Á1 = 0.06 m Ë 1.0 ˜¯ Ê 1 ˆ È Ê 0.1 ˆ 4 ˘ V22 V22 = Á 1 ˜ Í1 - ÁË 0.2 ˜¯ ˙ 2 g 2g Ë Cd2 ¯ ÍÎ ˙˚ Ê 1 ˆ Á 2 - 1˜ = 0.10667 Ë Cd ¯ By the venturimeter equation Q= Cd A2 1 - ( D2 /D1 ) 4 0.1 2g D h Cd = 0.985, p A2 = ¥ (0.1)2 = 7.854 ¥ 10–3 m2 4 D2 10 = = 0.5, 1 - ( D2 /D1 ) 4 = 0.96825 D1 20 Hence 0.985 ¥ 7.854 ¥ 10 -3 Q= ¥ 2 ¥ 9.81 ¥ 0.06 0.96825 = 8.667 ¥ 10–3 m3/s = 8.67 L/s (b) Head loss in the inlet section HLi = 0.03 ¥ Dh = 0.03 ¥ 0.06 = 1.8 ¥ 10–3 m = 1.8 mm of water V2 Cd = 0.95 HLi = (1 – Cd2) Dh Also, \ V22 = [1 – (0.95)2] (0.075) 2g = 7.313 ¥ 10–3 2 V2 = 0.07313 and V2 = 1.198 m/s 2g Discharge Q = A2V2 p = ¥ (0.1)2 ¥ 1.198 4 = 9.41 ¥ 10–3 m3/s = 9.41 L/s 0.1 435 Flow Measurement 13.22 C Solution: In the venturimeter equation Cd A2 Q = ÊD ˆ 1- Á 2 ˜ Ë D1 ¯ 4 2gDh Given data are: Cd = 0.984 and Solution: Difference in the piezometric head between the inlet and the throat, by taking the throat as the datum ÈS ˘ È13.6 ˘ Dh = y Í m - 1˙ = 0.50 ¥ Í - 1˙ = 6.3 m S Î 1.0 ˚ ÍÎ p ˙˚ Êp ˆ Êp ˆ Dh = Á 1 + z1 ˜ - Á 2 + z2 ˜ Ëg ¯ Ë g ¯ Q = 0.02 m3/s. Substituting these in the equation Ê 150 ˆ Ê ˆ 90 + 0.10˜ - Á = Á + 0˜ Ë 0.9 ¥ 9.79 ¯ Ë 0.9 ¥ 9.79 ¯ 0.02 = Êp ˆ 0.984 ¥ Á D22 ˜ Ë4 ¯ ÊD ˆ 1- Á 2 ˜ Ë D1 ¯ = 6.91 m Head loss HL = (1 – C 2d) Dh = 0.02 Dh (1 – C 2d) = 0.02 and Cd = 0.99 The discharge Q by venturimeter equation is Q= Cd A2 ÊD ˆ 1- Á 2 ˜ Ë D1 ¯ Q= 4 2gDh Ê pˆ 0.99 ¥ Á ˜ ¥ (0.10) 2 Ë 4¯ Ê Ê 0.1 ˆ 4 ˆ Á1 - ÁË ˜ ˜ Ë 0.3 ¯ ¯ Dr = Putting 0.02 = (1 - ( D ) ) 4 2 ¥ 9.81 ¥ 6.3 r On simplifying 0.0004 ¥ (1 – D 4r) = 0.0073826 D 4r (1 – D 4r) = 18.456 D 4r 2g ¥ (6.91) Dr = 0.476 and D2 = 0.0476 m = 4.76 cm 13.23 D2 D2 = = 10 D2 D1 0.10 Êp ˆ 0.984 ¥ Á ( Dr2 ) ¥ (0.1) 2 ˜ Ë4 ¯ = 0.091 m3/s = 91 litres/s 2 ¥ 9.81 ¥ 6.3 4 13.24 436 Fluid Mechanics and Hydraulic Machines Solution: VPR = 200 – (–60) = 260 km/h = 72.2 m/s In a pitot tube (assuming C = 1) Êp 0.05 ¥ 9790 p ˆ Dh = Á s - o ˜ = g ¯ (1.2 ¥ 9.81) Ëg = 41.58 m of air column. 2 VPR Dp = g 2g V = C 2gDh = 0.98 ¥ 2 ¥ 9.81 ¥ 41.58 = 28.0 m/s Dp = \ 13.25 Dp = or 1.20 ¥ (72.2) 2 2 rVPR 2 2 = 3127.7 Pa Dp = Ps – p0 = Differential pressure intensity in the instrument = 3.128 kPa. 13.27 r – Solution: For the differential manometer, Solution: ÊS ˆ Ê 13.6 ˆ Dh = y Á m - 1˜ = 0.04 Á -1 Ë 0.85 ˜¯ Ë Sp ¯ Dh = = 0.6 m of oil For the pitot tube V0 = C Ê p - p0 ˆ ps p – 0 = Á s ˜¯ g g Ë g V0 = C 2g D h = 0.99 2 ¥ 9.81 ¥ 0.6 Velocity at M = 3.397 m/s In a pitot-static tube, Ê p - p0 ˆ 2g Á s ˜¯ = C Ë g Ê p - p0 ˆ 2Á s Ë r ˜¯ where ps = stagnation pressure and p0 = static pressure. 13.26 V0 = 0.98 2 [3.0 - ( -3.0)] ¥ 1000 1.20 = 98 m/s 13.28 rair = 60 km/h 200 km/h Vwind Vplane 2 260 km/h Vplane (relative) Solution: Solution: Relative velocity of plane with respect to wind. p0 p 10 ¥ 13.6 = – 1.36 m = static = – g g 100 437 Flow Measurement p ps 1 ¥ 10 4 = 1.021 m = stagnation = 9790 g g Ê p - p0 ˆ Dh = Á s ˜¯ = [1.021 – (–1.36)] Ë g = 2.381 m For the pitot tube V = C 158.1 1.2 Coefﬁcient of the pitot tube C = 0.955 15.495 = C ◊ vm = Centreline velocity = C 2¥ 13.30 h h2 2g D h A A2 2 ¥ 9.81 ¥ 2.381 = 0.98 2 ( Dps /r ) = 6.698 m/s Mean velocity in the pipe = 0.85 ¥ vm V = 0.85 ¥ 6.698 = 5.693 m/s p Discharge Q = ¥ (0.3)2 ¥ (5.693) 4 = 0.402 m3/s A1 Density = pa A2 13.29 h1 rair h2 Cv vena contracta Solution: For the oriﬁce ( D p) r Dp = Pchamber – Patmospheric = 150 Pa 2 g D h0 = Cv V = Cv Here 2 150 = 15.495 m/s 1.2 For the inclined manometer 38 Dhm = y sin 30° = ¥ sin 30° 1000 = 0.019 m of liquid V = 0.98 2¥ Dpmanometer = pstagnation – patmospheric = D ps = 0.019 ¥ 0.85 ¥ 9790 = 158.1 Pa pb For pitot tube: ps + h1 (rag) = p2 + h1 (rbg) For piezometric tubes p1 + h2 (rag) = p2 + h2 rbg \ (ps – p1) + rag (h1 – h2) = (h1 – h2) rbg (ps – p1) = (h1 – h2) (rb – ra)g Ê ps - p1 ˆ V12 = ÁË r g ˜¯ 2g a Êr ˆ = (h1 – h2) Á b - 1˜ Ë ra ¯ From piezometer tappings (p1 – p2) = h2g (rb – ra) Ê rb ˆ Ê p1 - p2 ˆ ÁË r g ˜¯ = h2 ÁË r - 1˜¯ a a 438 Fluid Mechanics and Hydraulic Machines By Bernoulli equation: 13.32 p1 V2 p V2 + 1 = 2 + 2 ra g 2g ra g 2g 2 ˘ Ê p1 - p2 ˆ V22 - V12 V12 ÈÊ V2 ˆ Í ˙ = = 1 ÁË r g ˜¯ 2g 2 g ÍÁË V1 ˜¯ ˙ a Î ˚ = V12 2g ÈÊ A ˆ 2 ˘ ÍÁ 1 ˜ - 1˙ ÍË A2 ¯ ˙ Î ˚ Solution: By the weir formula 2 Q = Cd 2g LH13/2 3 2 0.025 = Cd 2 ¥ 9.81 ¥ 0.40 ¥ (0.10)3/2 3 = 0.03735 Cd Cd = 0.669 Êr ˆ Êr ˆ h2 Á b - 1˜ = (h1 – h2) Á b - 1˜ (4 – 1) Ë ra ¯ Ë ra ¯ \ Ê h1 ˆ ÁË h - 1˜¯ 3 = 1 2 h1 1 4 = +1= h2 3 3 h1 4 = h2 3 13.31 Solution: Here H1 = 0.35 m, P = 0.70 m, L = 2.5 m By Rehbock formula Cd = 0.611 + 0.075 H1/P 0.35 ˆ Ê = 0.611 + Á 0.075 ¥ = 0.649 Ë 0.70 ˜¯ The discharge 2 Q = Cd 2g LH13/2 3 2 = ¥ 0.649 ¥ 2 ¥ 9.81 ¥ 2.5 ¥ (0.35)3/2 3 = 0.9925 m3/s Q = 992.5 L/s 13.33 - Solution: Here P = 10 cm = 0.1 m and L = 0.80 m Also H1 + P = Channel height – free broad = 0.75 – 0.15 = 0.60 m \ H1 = 0.60 – 0.10 = 0.50 m 0.50 H1/P = = 5.0 0.10 This is at the limit of the applicability of Rehbock’s formula for Cd. \ Cd = 0.611 + 0.075 (H1/P) = 0.611 + 0.075 (5.0) = 0.986 The discharge over the weir, by the weir formula, is 2 Q = Cd 2g LH13/2 3 2 = ¥ 0.986 ¥ 2 ¥ 9.81 ¥ 0.80 ¥ (0.50)3/2 3 = 0.8235 m3/s 439 Flow Measurement Cd = 0.611 + 0.075 ¥ 0.3793 = 0.639 No further trials are necessary. Thus the weir height P = 1.0875 m. 13.34 13.35 Solution: Refer to Fig. 13.27. H1 C 1.50 m P Solution: By Rehbock formula, H1 P Ê 0.5 ˆ = 0.611 + 0.075 Á Ë 0.75 ˜¯ = 0.661 Cd = 0.611 + 0.075 In this problem since both H1 and P are unknowns, a trial and error procedure is used. First assume a value of Cd. I trial: Assume Cd = 0.64 2 Q = Cd 2g LH13/2 3 2 1.5 = ¥ 0.64 ¥ 2 ¥ 9.81 3 ¥ 3.0 ¥ H13/2 3/2 H1 = 0.2646, H1 = 0.412 m P = 1.50 – 0.412 = 1.088 m By Rehbock formula H1 P Ê 0.412 ˆ = 0.611 + 0.075 ¥ Á Ë 1.088 ˜¯ = 0.639 Cd = 0.611 + 0.075 2nd trial: Use Cd = 0.639 in the 2nd trial for the calculation of discharge by the weir formula. 2 Q = 1.50 = ¥ 0.639 ¥ 2 ¥ 9.81 ¥ 3.0 ¥ (H1)3/2 3 H13/2 = 0.2650, H1 = 0.4125 m P = 1.50 – 0.4125 = 1.0875 m H1/P = 0.3793 and substituting this in the Rehbock formula, (a) For a suppressed weir, 2 Q = Cd 2g LH13/2 3 2 = ¥ 0.661 ¥ 2 ¥ 9.81 ¥ 1.5 ¥ (0.5)3/2 3 = 1.035 m3/s (b) For a contracted weir, Le = effective length of weir = L – 0.1 ¥ 2 ¥ H = 1.5 – (0.1 ¥ 2 ¥ 0.5) = 1.4 m 2 Q = ¥ Cd 2g Le H13/2 3 2 = ¥ 0.661 ¥ 2 ¥ 9.81 ¥ 1.4 3 ¥ (0.5)3/2 3 = 0.966 m /s 13.36 C 440 Fluid Mechanics and Hydraulic Machines Solution: Effective crest length = Le = L – 0.1 nH1 Here crest length L = 2.5 – 2 ¥ 0.15 = 2.20 m n = number of end contractions = 2 + (2 ¥ 2) = 6 Le = 2.20 – (0.1 ¥ 6 ¥ 0.7) = 1.78 m The discharge, from Francis formula by neglecting the velocity of approach, 2 Cd 2g LeH13/2 3 2 = ¥ 0.62 ¥ 2 ¥ 9.81 ¥ 1.78 ¥ (0.7)3/2 3 = 1.91 m3/s Q= 13.37 C Solution: This is the case of a contracted weir. Here L = 1.0 m, P = 0.60 m, H1 = 0.30 m. n = number of end contractions = 2. The discharge is estimated by the Francis formula Q= 2 Cd 3 2g (L – 0.1 nH1) ÈÊ 2ˆ Í H1 + V0 ÍÁË 2 g ˜¯ Î 3/ 2 Ê V02 ˆ -Á ˜ Ë 2g ¯ Second trial: Using the above Q, the velocity of approach Q Q 0.2828 V0 = = = A B (H + P) 2.0 ¥ 0.9 = 0.157 m/s V02 = 0.001258 2g Revised discharge Q = Q2 = 1.721 [(0.30 + 0.001258)3/2 – (0.001258)3/2] = 0.2845 m3/s Third trial: Using the above Q, revised 0.2845 V0 = = 0.15805 m/s 2 ¥ 0.9 V02 = 0.001273 2g Revised discharge Q3 = 1.721 ¥ [(0.3 + 0.001273)3/2 – (0.001273)3/2] 3 = 0.2845 m /s This is the same as in the previous trial. Thus no more trials are needed. \ Discharge in channel Q = 0.2845 m3/s. 13.38 3/ 2 ˘ ˙ ˙ ˚ Since the discharge is involved in both sides of this equation, a trial and error method is adopted. Since (V02/2g) is usually a very small quantity as a ﬁrst trial, Q is calculated by assuming V0 = 0. C First trial. Ï2 Q1 = Ì ¥ 0.62 ¥ Ó3 ¸ 2 ¥ 9.81 ¥ (1.0 - 0.2 ¥ 0.3) ˝ ˛ (0.30)3/2 = 1.721 (0.30)3/2 = 0.2828 m3/s Solution: For a contracted weir, by neglecting the velocity of approach, Q = 2 Cd 3 2g LeH13/2 441 Flow Measurement where Le = effective length. Considering Le = constant dQ dH = 1.5 1 Q H1 Considering the accuracies desired 0.0005 0.005 = 1.5 ¥ Hm Hm = minimum head desired = 0.15 m = 15 cm. Thus H1 must be greater than or equal to 0.15 m. By the discharge equation, for the smallest discharge at the maximum Le, Q = 0.100 = 2 ¥ 0.62 ¥ 3 Here dH = 0.001 m and H1 = 0.437 m dQ 5 0.001 = ¥ = 5.72 ¥ 10–3 Q 2 0.437 = 0.572% Since Q = 100 L/s, dQ = 0.572 L/s Possible error in discharge = ± 0.572 L/s 13.40 2 ¥ 9.81 Le (0.15)3/2 0.100 = 0.94 m 0.106 Le should be less than or equal to 0.94 m Since Le = L – 0.2 H1 L = length of the weir £ (0.94 + (0.2 ¥ 0.15) Maximum length of weir = 0.97 m. Le = dQ 5 dH1 = Q 2 H1 \ 13.39 vertex angle q Solution: For a V-notch, 8 q Q = Cd 2g tan H15/2 15 2 q = K tan H15/2 where K = a constant. 2 dQ 5 dH1 = Q 2 H1 At H = 0.25 cm, dQ 5 1 = ¥ dH1 = 10 dH1 Q 2 0.25 dQ 1 Ê 2 qˆ = KH15/2 ◊ ◊ Á sec ˜ dq 2¯ 2 Ë C Solution: For a triangular notch, the discharge Q is given by 8 q Q= Cd 2g tan H15/2 15 2 Ï8 ¸ 0.100 = Ì ¥ 0.58 ¥ 19.62 ¥ tan 30∞˝ H15/2 Ó15 ˛ 5/2 0.100 = 0.79107 H1 H1 = 0.437 m Writing the discharge equation as Q = KH15/2 5 dQ = KH13/2 dH1 2 dQ sec 2 (q /2) dq = Q 2 tan (q /2) 1 rad 57.296 dQ 1 1 dq = Q 2 cos 2 ( 45∞) tan ( 45∞) At q = 90° and dq = 1° = = If 1 57.296 dQ is to be the same in both cases Q 10 dH1 = 1 57.296 442 Fluid Mechanics and Hydraulic Machines dH1 = 1.745 ¥ 10–3 m = 1.745 mm = required error in measurement of H1. 13.41 C q H1 q I m L Trapezoidal Notch when H1 = 1.2 m, 2.0 = 1.8308 ¥ (1.2)3/2 (L + (0.8 ¥ 1.2 ¥ tan q)] Simplifying L + 0.96 tan q = 0.8310 (i) When H1 = 1.2/2 = 0.6 m, 0.6 = 1.8308 ¥ (0.6)3/2 (L + (0.8 ¥ 0.6 ¥ tan q)] Simplifying L + 0.48 tan q = 0.7051 Subtracting Eq. (ii) from Equation (i) (ii) 0.48 tan q = (0.8310 – 0.7051) = 0.1259 or tan q = 0.2623 q = 14.7° From Eq. (ii) L = 0.7051 – 0.1259 = 0.5792 m 13.43 Solution: tan q = m = 0.5 The discharge Q for a head H1 is 2 4 Ê ˆ Cd 2g H13/2 Á L + H1 tanq ˜ Ë ¯ 3 5 2 = ¥ 0.63 ¥ 2g ¥ (0.50)3/2 3 Ï Ê4 ˆ¸ Ì0.75 + Á ¥ 0.5 ¥ 0.5˜ ˝ Ë5 ¯˛ Ó Q= = 0.65774 (0.95) = 0.625 m3/s 13.42 C Solution: The discharge over a Cipolletti weir is calculated by using the suppressed weir formula. 2 Q = Cd 2g LH13/2 3 2 = ¥ 0.63 ¥ 2 ¥ 9.81 ¥ 0.50 ¥ (0.25)3/2 3 = 0.1163 m3/s = 116.3 L/s 13.44 C C = Solution: For a trapezoidal notch, 2 4 Ê ˆ Cd 2g H13/2 Á L + H1 tanq ˜ Ë ¯ 3 5 2 = ¥ 0.62 ¥ 2 ¥ 9.81 H13/2 (L + 08 H1 tan q) 3 = 1.8308 H13/2 (L + 0.8 H1 tan q) Q= Solution: This is a case of submerged ﬂow. H1 = 75 – 30 = 45 cm = 0.45 m H2 = 50 – 30 = 20 cm = 0.20 m Q1 = free ﬂow mode discharge under H1 8 q = ¥ Cd 2g tan H15/2 15 2 443 Flow Measurement 8 ¥ 0.6 ¥ 2 ¥ 9.81 tan (37.5°) ¥ (0.45)5/2 15 = 0.1477 m3/s By the Villemonte equation = È Ê H ˆn˘ Qs = Q1 Í1 - Á 2 ˜ ˙ Í Ë H1 ¯ ˙ Î ˚ 0.385 or T = 5 ◊ 4 Ê 1 1 ˆ - 3/2 ˜ Á 3/2 q H1 ¯ 2 g tan Ë H2 2 A Cd 2 ¥ 13.46 in For a triangular notch, n = 5/2. È Ê 0.20 ˆ 2.5 ˘ Qs = 0.1477 Í1 - Á ˜ ˙ ÍÎ Ë 0.45 ¯ ˙˚ 0.385 C = 0.1477 [1 – 0.1317)0.385 = 0.140 m3/s The discharge over the notch is 140 L/s. 13.45 A elevation H H2 Solution: Let the water surface be at an elevation h above the vertex at any instant t, and in time dt let it drop by dh. By continuity, the volume of outﬂow –A dh = Q ◊ dt = 8 Cd 15 2g tan Solution: Let, at any instant of time, the water level be at a height h above the weir crest. In time dt, volume of water outﬂow is 2 –A dh = Q dt = Cd 2g Lh3/2 dt 3 A dh dt = – 2 Cd 2 g Lh3 / 2 3 T = q 5/2 h dt 2 = where q = angle of the V-notch. dt = Ú T 0 -A dt = T = Ú H2 -A –5/2 (h dh) 8 q Cd 2 g tan 15 2 2 -A T= ¥ (H2–3/2 – H1–3/2) q 3 8 / 15 Cd 2 g tan 2 H1 dt = 0 Ú H1 H2 A 2 Cd 3 Ê Á 2g L Ë 2A 2 Cd 3 1 H2 (h–3/2) dh 2g L - 1 ˆ ˜ H1 ¯ Here T = 30 min = 1800 s H1 = 1.60 m, H2 = 10 m dh q h5 / 2 2 g tan 2 8 Cd 15 Ú T L = Ê Á1 Ê 2ˆ Ë 0 735 19 62 1800 ¥ ( . ) . ¥ ÁË 3 ˜¯ = 15.0 m 2 ¥ 1.4 ¥ 105 1 ˆ ˜ 1.6 ¯ 444 Fluid Mechanics and Hydraulic Machines Problems 13.1 A closed tank A contains 3.0 m depth of water and an air space at 15 kPa pressure. A 5 cm diameter oriﬁce at the bottom of the tank discharges the water to a tank B containing pressurised air at 25 kPa. If the coefﬁcient of discharge of the oriﬁce is 0.61, calculate the discharge of water from tank A. (Ans. Q = 7.46 L/s) 13.2 A 5 cm diameter oriﬁce discharges 7.75 L/s of water under a head of 2.0 m. A ﬂat plate held normal to the jet just downstream of the vena contracta experiences a force of 4.5 N. Find he values of Cc, Cv and Cd of the oriﬁce. (Ans. Cc = 0.678, Cv = 0.929 and Cd = 0.630) 13.3 A 25 mm diameter nozzle discharges 0.76 m3/min of water when the head is 60 m. The diameter of the jet is 22.5 mm. Determine (i) the values of the coefﬁcients Cc, Cv and Cd and (ii) the loss of head due to ﬂuid resistance in the nozzle. (Ans. Cc = 0.810, Cv = 0.930 and Cd = 0.750; HL = 8.106 m) 13.4 Water discharges at the rate of 100 L/s through a 12 cm diameter vertical sharp edged oriﬁce placed under a constant head of 10 m. A point on the jet, issuing into atmosphere, has coordinates measured from the vena contracta of 4.5 m horizontal and 0.55 m vertical. Find the coefﬁcients Cv, Cc and Cd. (Ans. Cv = 0.959, Cc = 0.658 and Cd = 0.631) 13.5 A 12 cm ¥ 3 cm nozzle is attached to the end of a 12 cm diameter water pipe. The pressure at the base of the nozzle is 200 kPa. If the coefﬁcient, of velocity is 0.96 and the coefﬁcient of contraction is 0.90, determine the discharge of the jet. (Ans. Q = 12.246 L/s) 13.6 A closed tank contains kerosene (RD = 0.8) to a depth of 2.5 m. The top portion of the tank contains air under a pressure of 25 kPa. If a sharp edged circular oriﬁce of diameter 3 cm (Cd = 0.61) is provided at the bottom of the tank, estimate the discharge through the oriﬁce. (Ans. Q = 4.294 L/s) 13.7 A jet issues in an upward trajectory out of an oriﬁce located on the side of a tank which has an inclination of a with the horizontal. If the head of the liquid in the tank over the centre of the oriﬁce is H, show that the maximum elevation of the jet Y and its horizontal distance X from the vena contracta are given by X = Cv2 H sin2 a and Y = Cv2 H cos2 a 13.8 A 9 cm diameter base and 4.5 cm diameter tip nozzle has a pressure of 60 kPa at the base. If the coefﬁcients of velocity and contraction of the nozzle are 0.95 and 0.85 respectively, determine the jet velocity and power of the jet of water. (Ans. Vj = 10.636 m/s and P = 0.813 kW) 13.9 A nozzle 5 cm in diameter is attached to a 10 cm diameter pipe. The pressure at the base of the nozzle is 8 N/cm2. If the coefﬁcient of contraction = 0.99 and the coefﬁcient of velocity = 0.98, ﬁnd the following: (a) the velocity of the jet, (b) discharge, (c) power available from the jet, (d) efﬁciency of the nozzle and (e) height to which the jet will rise if directed 445 Flow Measurement 13.10 13.11 13.12 13.13 13.14 vertically upwards. (Ans. (a) Vj = 12.8 m/s; (b) Q = 24.88 L/s (c) P = 2.034 kW; (d) h = 96.04%; (e) hm = 8.35 m) Determine the discharge from a 10 cm diameter mouthpiece ﬁtted to the side of a tank containing 3 m of water above the centreline of the mouthpiece. The coefﬁcient of contraction for the mouthpiece can be taken as 0.65. (Ans. Q = 53.1 L/s) A rectangular tank of cross section 0.9 m ¥ 1.2 m is 3.0 m high. At 20 cm from the bottom, an 8 cm oriﬁce with Cd = 0.65 is provided. If the tank is full when the oriﬁce is opened ﬁnd the time taken to lower the water surface by 1.0 m. (Ans. T = 49.5 s) A vertical prismatic tank of cross section area 1.2 m2 has a 4 cm diameter oriﬁce at the bottom. If it takes 60 s to lower the water surface elevation from 1.2 m to 1.0 m above the oriﬁce, ﬁnd the coefﬁcient of discharge of the oriﬁce. (Ans. Cd = 0.686) A cylindrical water tank 0.6 m in diameter has its axis vertical and is to be provided with an oriﬁce at its bottom. If the Cd of the oriﬁce is 0.65, what size of oriﬁce is needed to lower the water surface elevation, measured above the base of the tank, from 1.5 m to 0.8 m in 30 s? (Ans. d = 5.25 cm) A cylindrical tank with its axis horizontal is 2.5 m in diameter and is 4.0 m long and contains oil. Estimate the time required to lower the oil surface in the tank from 2.0 m to 1.20 m above the bottom of the tank, through a 12 cm sharp edged oriﬁce of discharge coefﬁcient 0.63 situated in the bottom of the tank. (Ans. T = 191 s) 13.15 Two water tanks of area 2 m2 and 0.8 m2 have a common partition wall. A circular oriﬁce of diameter 10 cm allows ﬂow of water between the tanks. If initially the level of the water in the larger tank is 1.5 m above that in the other, determine the time required for the difference in water surface elevations to reduce to 0.5 m. The coefﬁcient of discharge of the oriﬁce can be taken as 0.7. (Ans. T = 24.3 s) 13.16 A hemispherical tank 3 m in diameter has an oriﬁce 15 cm diameter at the bottom. Assuming Cd = 0.62, ﬁnd the time required to lower the level of the water surface from 2.0 m to 1.2 m above the oriﬁce. (Ans. T = 90.37 s) 13.17 A vertical prismatic tank of cross sectional area A has a steady inﬂow of Q into the tank, while an oriﬁce of area a at the bottom is discharging the ﬂow freely. Show that the time taken to lower or raise the water surface between heights H1 and H2 above the oriﬁce is 2A T=– 2 ¥ K È ˘ Ê Q - K H2 ˆ ÍQ In Á ˜ + K ( H 2 - H1 ) ˙ ÍÎ ˙˚ Ë Q - K H1 ¯ where K = Cd a 2g . 13.18 A vertical venturimeter 15 cm ¥ 10 cm installed in a pipe carrying water downwards shows the same pressure at the inlet and at the throat. The throat is 25 cm below the inlet. If Cd of the meter is 0.95, calculate the discharge in the pipe. (Ans. Q = 18.4 L/s) 13.19 A horizontal venturimeter 9 cm ¥ 4 cm is installed in a 9 cm water pipe. A differential 446 Fluid Mechanics and Hydraulic Machines water-mercury manometer reads 30 cm. If the coefﬁcient Cd of the meter is 0.96, estimate the (i) discharge in the pipe and (ii) head loss in the converging section of the meter. (Ans. Q = 10.6 L/s, HL = 0.296 m) 13.20 A venturimeter is used for the measurement of discharge of water in a horizontal pipeline. The upstream diameter is 300 mm, the throat is of 150 mm diameter and the difference in pressure between the inlet and throat is 3 m head of water. If the loss of head through the converging section of the meter is one-eighth of the throat velocity head, calculate the discharge in the pipe. (Ans. Q = 131.5 L/s) 13.21 Oil of relative density 0.9 ﬂows in a 10 cm pipe. A 10 cm ¥ 5 cm venturimeter (Cd = 0.97) in the pipe exhibits a reading of 12 cm in a mercury-oil differential manometer. Calculate the head loss in the inlet to throat region and estimate the discharge. (Ans. HL = 0.10 m, Q = 11.33 L/s) 13.22 Crude oil (relative density = 0.85) ﬂows upwards at a volumetric rate of 60 L/s through a vertical venturimeter with an inlet diameter of 200 mm and a throat diameter of 100 mm. The coefﬁcient, of discharge of the venturimeter is 0.98. The vertical distance between the pressure tappings is 300 mm. (i) Determine the difference in the readings of pressure gauges connected to the inlet and throat sections. (ii) If a differential U-tube mercury-oil manometer is used to connect the two tappings determine the manometer reading. (Ans. (i) D p = 26.66 kPa; (ii) y = 19.36 cm) 13.23 A 20 cm ¥ 10 cm venturimeter with Cd = 0.96 carries 30 L/s of water. A differential gauge has an indicator liquid M and the manometer reading is 1.16 m. What is the relative density of the manometer liquid M? (Ans. Sm = 1.75) 13.24 A 15 cm ¥ 7.5 cm venturimeter has Cd = 0.96. If the loss of head between inlet and throat is 0.78 cm calculate the discharge. (Ans. Q = 6.12 L/s) 13.25 A 20 cm ¥ 10 cm venturimeter (Cd = 0.96) carries water. A differential mercury water manometer shows a reading of 10 cm. (i) Calculate the discharge. (ii) If the loss of head in the expansion part of the meter is taken as eight times the velocity head in the pipe, calculate the total head loss due to the venturimeter. (Ans. (i) Q = 38.7 L/s; (ii) HL = 0.7178 m) 13.26 A 20 cm ¥ 15 cm venturimeter is ﬁxed vertically in a pipe carrying water. The throat is 45 cm above the inlet. If the inlet pressure is 103 kPa (abs), what is the maximum discharge that can be passed through the pipe without cavitation occurring in the meter? The vapour pressure of water = 3.5 kPa (abs). The Cd of the meter is 0.95. (Ans. Qmax = 280 L/s) 13.27 A 15 cm diameter pipe has a discharge of 60 L/s of water and the pressure head at a section is 5 m (gauge). If a horizon tal venturimeter is to be installed at this section, what is the minimum diameter to be adopted to ensure that the pressure head at the throat does not go below 2.3 m (abs)? The atmospheric pressure is 10.3 m. Assume Cd = 0.95. (Ans. D2 = 7.19 cm) 447 Flow Measurement 13.28 A venturimeter with a throat diameter of 12 cm is inserted in a pipe of 20 cm diameter. The pipe carries a liquid of relative density 0.65. The differential mercuryliquid U-tube manometer connected to the throat and the inlet records a readings of 4.0 cm. If the head loss between the inlet and the throat is 8% of the velocity head in the throat, estimate (a) discharge and (b) the coefﬁcient of discharge of the venturimeter. (Ans. (a) Q = 45.9 L/s; (b) Cd = 0.957) Oil RD = 0.85 A 2 cm Mercury 13.29 A pitot tube is inserted into an air stream at 95.0 kPa (abs). If a mercury-air manometer indicates a positive gauge pressure with a reading of 20 mm, calculate the velocity of air. rair = 1.22 kg/m3 and Patmos = 100 kPa. Assume the pitot tube coefﬁcient C = 0.98. (Ans. V = 109.8 m/s) 13.30 The velocity of an oil ﬂow (RD = 0.90) was measured by a pitot-static tube. The instrument had a coefﬁcient of 0.98. Calculate the velocity corresponding to a mercury-oil differential manometer, connected to the pitot-static tube, reading of 6 cm. (Ans. V0 = 4.0 m/s). 13.31 For the pitot tube shown in Fig. 13.29 determine the velocity at point A. Assume the instrument coefﬁcient to be 0.99. (Ans. V0 = 2.426 m/s) 13.32 For the ﬂow of water in a frictionless uniform pipe, a pitot tube was arranged on the centreline as shown in Fig. 13.30. Calculate he centreline velocity in the pipe by assuming the instrument coefﬁcient to be 1.0. (Ans. V0 = 3.851 m/s) 13.33 A pitot tube is placed on the centreline of a pipe carrying kerosene (RD = 0.8) as Flow 1 15 cm 2 Water 6 cm CL Mercury shown in Fig. 13.31. Neglecting frictional losses in the pipe, estimate the centreline velocity. [Assume C = 0.99]. (Ans. V0 = 3.07 m/s) 13.34 Kerosene (RD = 0.81) is ﬂowing in a pipe and the static pressure at a station is 3 kPa. If the stagnation pressure of a pitot tube (C = 0.99) inserted at the centreline at 448 Fluid Mechanics and Hydraulic Machines CL will be the stagnation pressure in kPa (abs)? Atmospheric pressure = 101 kPa. (Ans. Pat = 74.81 kPa (abs)) 2 Kerosene RD = 0.8 10 cm 1 3 cm Flow Mercury that section indicates 4.0 kPa, what is the centreline velocity at that section? (Ans. V0 = 1.56 m/s) 13.35 A pitot-static tube (C = 0.97) is connected to an inverted U-tube manometer containing an oil of relative density 0.82. If the pitot tube is to measure the velocity of water up to 0.6 m/s, what would be the largest manometer reading? (Ans. y = 10.8 cm) 13.36 A pipe is known to be conveying water at a centreline velocity of 2.52 m/s. A pitot-static tube is inserted at the centreline and is connected to a differential carbon tetrachloride (RD = 1.60)-water differential manometer. If the manometer reading is 55 cm, what is the value of the instrument coefﬁcient? (Ans. C = 0.99) 13.37 At the summit of a siphon in a water pipe the centreline velocity of ﬂow is 2.5 m/s and the pressure is 3 m of water (vacuum). If a pitot tube (C = 0.99) is inserted into the pipe centreline at this section, what 13.38 A suppressed rectangular weir in a 1.5 m wide rectangular channel is located so that its crest is 0.30 m above the bed. Estimate the discharge over the weir for a head of 0.30 m over the weir. (Ans. Q = 0.499 m3/s) 13.39 A sharp crested rectangular weir is 2.0 m long. Calculate the height of the weir to pass a ﬂow of 1.10 m3/s while maintaining an upstream depth of 1.20 m. (Ans. P = 0.767 m) 13.40 The head on a sharp crested rectangular suppressed weir 1.2 m long and 0.9 m high is 10 cm. Calculate the discharge and the velocity of approach. (Ans. Q = 69.4 L/s, Va = 0.0578 m/s) 13.41 What is the coefﬁcient of discharge of a suppressed rectangular weir of 1.5 m length which passes 1.12 m3/s under a head of 0.50 m? (Ans. Cd = 0.715) 13.42 A 20 cm high sharp crested rectangular weir plate is installed at the end of a 1.2 m wide rectangular channel. The weir plate spans the full width of the channel. What maximum discharge can be passed if the side walls of the channel are 0.90 m high and the minimum speciﬁed free board is 0.20 m? (Ans. Qm = 1.0 m3/s) 13.43 A sharp crested suppressed rectangular weir of height 0.6 m is used to measure the discharge in a 1.8 m wide rectangular channel. At a certain discharge the head over the weir was recorded as 0.6 m by a point gauge. It was however found later that the point gauge had a zero error and 449 Flow Measurement 13.44 13.45 13.46 13.47 13.48 was recording heads 2 cm too small. Determine the percentage error in the estimated discharge corresponding to an observed head of 0.6 m. (Ans. Error = 5.14%) An overﬂow weir is 16 m long between the abutments. There are three piers of 0.30 m thick on the weir. Estimate the discharge for a head of 0.5 m over the weir. Assume Cd = 0.63 and the velocity of approach to be zero. (Ans. Q = 9.67 m3/s) A rectangular channel 1.8 m wide has, at its end, a sharp crested rectangular weir 1.2 m long and 0.50 m high. Calculate the discharge in the channel when the head recorded over the weir is 0.25 m, Cd of the weir = 0.62. (Ans. Q = 0.266 m3/s) Find the discharge in a triangular notch of vertex angle 30° corresponding to a head of 35 cm. What is the head corresponding to a discharge of 50 L/s? Take Cd = 0.60. (Ans. Q = 27.5 L/s; H1 = 44.4 cm) A right angled triangular notch is used for measuring the discharge in a laboratory ﬂume. The coefﬁcient of discharge of the notch is 0.59. If the heads can be measured with an accuracy of 2 mm, ﬁnd the head measured above the level of the vertex of the notch and the likely error in the calculated discharge of 60 L/s. (Ans. H1 = 28.4 cm; error in Q = ± 1.056 L/s) It is desired to measure a discharge varying from 50 L/s to 150 L/s with an accuracy of 1% throughout the range. The depth can be measured with an accuracy of 1 mm. What is the maximum permissible vertex angle of a V-notch that will satisfy this condition? Take Cd = 0.60. (Ans. Q £ 96° 55¢ 12≤) 13.49 A 60° triangular weir has a coefﬁcient of discharge of 0.59. If the accuracy of measuring the angle is 1°, estimate the likely error in the estimated discharge when the head over the vertex of the weir is 40 cm. (Ans. Likely error = 1.64 L/s) 13.50 Estimate the head over a Cipolletti weir of base width 0.90 m required to pass a discharge of 600 L/s. Assume Cd = 0.63. (Ans. H1 = 50.45 cm) 13.51 A trapezoidal notch is to be designed to pass a discharge of 1.0 m3/s at a head of 0.8 m over the crest and 0.50 m3/s at a head of 0.51 m. Assuming Cd = 0.7, calculate the base width and side slope of the notch. (Ans. L = 0.643 m, q = 2.96°) 13.52 A suppressed rectangular weir is ﬁtted in a 2.5 m channel. Calculate the discharge over the weir when the water surface elevations on the upstream and downstream of the weir, measured from the weir crest, are respectively 0.60 m and 0.30 m. A constant value of Cd = 0.62 can be assumed. (Ans. Qs = 1.798 m3/s) 13.53 A triangular notch in a channel is known to have carried 60 L/s under a head of 30 cm when ﬂowing free. If it is submerged with water surface elevations at 50 cm and 30 cm above the vertex on the upstream and downstream of the notch respectively, estimate the discharge in the channel. (Ans. Qs = 190 L/s) 13.54 The discharge over a weir could be expressed as Q = KH n where H is the head over the weir and K and n are constants. It was found that the discharge was 7.48 m3/s and 3.84 m3/s when the head H was 2.0 m and 1.25 m, respectively. Calculate (a) the discharge over the weir for a head 450 Fluid Mechanics and Hydraulic Machines of 1.50 m and (b) head required to pass a discharge of 2.5 m3/s. (Ans. (a) Q = 4.973 m3/s; (b) H = 0.9237 m) 13.55 A sharp crested rectangular suppressed weir 90 cm long and a 90° V-notch are placed in the same vertical face of a tank. The vertex of the notch is 20 cm below the crest of the weir. Assuming Cd to be the same for both the notch and the weir, ﬁnd: (a) the head over the V-notch when the discharges in the notch and the weir are identical, and (b) the head above the vertex of the V-notch when the rectangular weir discharges its greatest excess amount over and above the discharge of the V-notch. (Ans. (a) H = 0.645 m; (b) H = 0.535 m) 13.56 Estimate the time required to lower the water level from an elevation of 13.00 m to an elevation of 12.00 m in a reservoir through a ﬂow over a 75° V-notch whose vertex is at elevation 10.00 m. The tank has a constant cross sectional area of 6000 m2. (The Cd of the notch = 0.60). (Ans. T = 9 min 52.5 s) 13.57 Find the time required to lower the water surface of a rectangular tank of cross sectional area 5000 m2 from an elevation of 10.20 m to 9.20 m. A rectangular suppressed weir of length 2.0 m with its crest at elevation 7.7 m above the datum is used for lowering the water surface (Assume Cd = 0.62). (Ans. T = 8 min 22.6 s) Objective Questions 13.1 A 20 cm diameter oriﬁce discharging from a tank issues out a jet of 15.75 cm diameter at the vena contracta. The coefﬁcient of contraction is (a) 0.520 (b) 0.620 (c) 0.790 (d) 0.887 13.2 The loss of head HL in an oriﬁce discharging under a head H is (a) H (Cv – 1) (b) H (1 – Cv) È 1 ˘ (d) Í 2 - 1˙ H ÍÎ Cv ˙˚ 13.3 If a tank discharges water from an oriﬁce under variable head h, the water surface will be lowered at constant velocity, if the surface area of the tank varies as 1 (a) h (b) h (c) H (1 – Cv)2 1 (d) h h 13.4 A 10 cm diameter Borda’s internal mouthpiece running free, discharges 24 L/s under a head of 2.0 m. The coefﬁcient of velocity Cv is (a) 0.5 (b) 0.948 (c) 0.976 (d) 0.995 13.5 The Cd of an oriﬁce is always (a) greater than Cc (b) equal to Cv (c) equal to Cc (d) less than Cc 13.6 An oriﬁce is discharging under a head of 1.25 m of water. A pitot tube kept at its centre line at the vena contracta indicates a head of 1.20 m of water. (c) 451 Flow Measurement 13.7 13.8 13.9 13.10 13.11 The coefﬁcient of velocity of the oriﬁce is (a) 0.990 (b) 0.980 (c) 0.965 (d) 0.960 The head loss at an oriﬁce (Cv = 0.98) discharging under a head of 2.0 m is (a) 0.02 m (b) 0.04 m (c) 0.06 m (d) 0.08 m The velocity of efﬂux from an oriﬁce is observed to be 3.1 m/s. If the Cv of the oriﬁce is 0.98, the head loss at the oriﬁce is (a) 0.441 m (b) 0.009 m (c) 0.015 m (d) 0.020 m The water level in a cylindrical tank with its axis vertical was lowered form an elevation of 4 m to 2.0 m in 200 s by discharging the contents through a oriﬁce at the bottom of the tank. If the diameter of the oriﬁce is doubled, the time required for the above lowering of the water surface would be (a) 50 s (b) 100 s (c) 12.5 s (d) 200 s In submerged oriﬁce ﬂow, the discharge is proportional to (a) square root of the upstream head H1 (b) square root of the downstream head H2 (c) square root of the difference between upstream and downstream heads, (H1 – H2). (d) square of the upstream head H1. An oriﬁce meter consists of an oriﬁce of diameter d in a pipe of diameter D. In general, the Cd of the oriﬁce meter is (a) a function of d/D only (b) a function of Reynolds number only (c) independent of d/D and Reynolds number. (d) a function of d/D and Reynolds number 13.12 For a given Reynolds number, the Cd of an oriﬁce meter of oriﬁce diameter d in a pipe of diameter D (a) increase with an increase in d/D (b) decreases with increase in d/D (c) independent of d/D (d) increases with d/D up to d/D = 0.5. 13.13 An oriﬁce meter with d/D = 0.5 and Reynolds number = 106 is expected to have Cd of about (a) 0.45 (b) 0.61 (c) 0.80 (d) 0.95 13.14 The percentage error in the estimation of the discharge due to an error of 2% in the measurement of the reading of a differential manometer connected to an oriﬁce meter is (a) 4% (b) 2% (c) 1% (d) 0.5% 13.15 A standard, long radius, ﬂow nozzle has a discharge coefﬁcient Cd larger than that of a corresponding standard venturimeter. The non-recoverable energy loss under identical condition is (a) larger in the ﬂow nozzle (b) larger in the venturimeter (c) essentially same in both the meters (d) larger in the ﬂow nozzle upto a critical Reynolds number and then onwards it is smaller. 13.16 In a standard oriﬁce meter. (a) the bevel of the plate is on the upstream (b) the bevel angle is 45° to 60° (c) the non-recoverable energy loss is independent of the location of the pressure taps (d) the Cd value is independent of the location of the pressure taps. 13.17 A standard venturimeter of diameter ratio 0.5, at a Rey nolds number of 10 6 has a 452 Fluid Mechanics and Hydraulic Machines 13.18 13.19 13.20 13.21 coefﬁcient of discharge of about (a) 0.65 (b) 0.75 (c) 0.87 (d) 0.97 The discharge coefﬁcient of a standard venturi-meter can be expressed, in genial as Cd = (a) fn (Re) (b) fn (b) (c) fn (Re, b) (d) a constant for all Re and b. where Re = Reynolds number and b = ratio of throat to inlet diameter. A venturimeter has a differential mercury water manometer connected to its inlet and throat. The gauge reading y of the manometer for a given discharge in the pipe (a) depends on the orientation of the venturimeter (b) is independent of the orientation of the venturi-meter (c) varies as the slope of the venturimeter with respect to the horizontal (d) depends on whether the manometer is above or below the pipe centreline. A venturimeter has a Cd = 0.95. For a differential head of 2.8 m across the inlet and the throat, the loss of head between the inlet and throat is (a) 0.273 m (b) 0.140 m (c) 0.302 m (d) 0.95 m A venturimeter with a throat diameter of 6 cm is connected to a pipe of 10 cm diameter. The Cd of the instrument is 0.95. For this meter, the head loss from the inlet to throat section can be expressed as KV2/2g where V = velocity at the throat and K = (a) 0.0975 (b) 0.0105 (c) 0.014 (d) 0.108 13.22 The value of the coefﬁcient of discharge of an oriﬁce meter of d/D = 0.5 lies in the range (a) 0.95 to 0.98 (b) 0.70 to 0.80 (c) 0.81 to 0.94 (d) 0.60 to 0.62 13.23 In a venturimeter, the coefﬁcient of discharge of the meter Cd is related to the head loss between the inlet and the throat as (a) (1 – C 2d) Dh (b) (1 – Cd)2 Dh 2 (c) (1 – Cd Dh) 13.24 13.25 13.26 13.27 2 Ê 1 ˆ (d) Á - 1˜ Dh Ë Cd ¯ In the above Dh is the difference of piezometric heads at the inlet and the throat. The stagnation pressure in front of an object in a ﬂuid ﬂow is equal to (a) static pressure (b) dynamic pressure (c) sum of the static and dynamic pressures (d) piezometric head A static tube is used to measure (a) the velocity (b) undisturbed ﬂuid pressure (c) the total head (d) datum head The pitot-static tube measures (a) the dynamic pressure (b) the static pressure (c) the total head (d) the difference in static and dynamic pressures To measure static pressure in a pipe, one uses a pressure gauge connected to a (a) pitot tube (b) venturimeter (c) oriﬁce meter (d) piezometer tapping 453 Flow Measurement 13.28 A Pitot-static tube indicates a differential head of 0.75 m of water between its two openings when inserted in a stream of water. If the coefﬁcient of the tube is 0.99, the velocity in m/s, at the location of the tube is (a) 4.43 13.29 13.30 13.31 13.32 13.33 (b) 0.78 (c) 3.84 (d) 3.80 A Pitot tube (coefﬁcient = 1.0) is used to measure the velocity of air mass density 1.2 kg/m3. If the head difference in a vertical U-tube ﬁtted with water is 12 mm, then the velocity of air in m/s is (a) 10 (b) 14 (c) 17 (d) 20 A hot-wire anemometer is used to measure essentially the (a) wind speed over ground surfaces (b) turbulent velocity ﬂuctuations in a ﬂow (c) shear stress on a boundary (d) drag force on a body A laser-doppler anemometer (LDA) is a device to measure (a) the turbulent velocity ﬂuctuations in a ﬂow (b) shear stress at a boundary (c) drag force on an airfoil (d) surface tension of a ﬂuid Which one of the following is measured by a rotameter? (a) Viscosity of ﬂuids (b) Velocity of ﬂow (c) Discharge of a ﬂow (d) Rotational speed of a wind anemometer The instrument preferred for highly ﬂuctuating velocities in air ﬂow is (a) pitot-static tube (b) propeller-type anemometer (c) three-cup anemometer (d) laser doppler anemometer 13.34 The coefﬁcient of discharge of a suppressed rectangular weir at the limit of application of Rehbock formula is (a) 0.786 (c) 0.886 (c) 0.986 (d) 1.06 13.35 The approximate discharge over a 4 m long rectangular suppressed weir with head over the crest as 0.36 m is (a) 0.39 m3/s (b) 2.4 m3/s 3 (c) 0.8 m /s (d) 1.6 m3/s 13.36 The discharge over a suppressed sharp crested rectangular weir is given by Q = 2 Cd 2g LH 13/2. Here H1 is 3 (a) the difference in elevation between the upstream water surface and the energy line (b) the difference in elevation between the energy line and the weir crest (c) the difference in elevation of the water surface and the weir crest (d) the difference in elevation between the water surface and the bottom of the channel. 13.37 For a suppressed rectangular weir an arrangement for aeration of nappe is necessary. (a) to maintain water quality (b) to prevent submergence of the weir (c) to have the highest value of Cd (d) to have a constant head-discharge relationship which is independent of time. 13.38 In a triangular notch there is an error of 4% in observing the head. The error in the computed discharge is (a) 4% (b) 10% (c) 6% (d) 2.5% 454 Fluid Mechanics and Hydraulic Machines 13.39 In a suppressed rectangular weir the computed discharge was found to be 3% in excess of the actual discharge. If this discrepancy was due to an error in reading the head, the measured head was (a) 3% excess (b) 2% less (c) 2% excess (d) 1.2% excess 13.40 In a 90° triangular notch, for a given head, the error in the estimated discharge due to a 2% error in the measurement of the vertex angle is (a) p% (b) 5.0% (c) 3.0% (d) p/2 % 13.41 A 1.5 m long, suppressed rectangular weir had a head of water of 62 cm. However, it was measured as 60.0 cm and used in computation. The percentage error in computed ﬂow is (a) 3.3 (b) 5.0 (c) 6.7 (d) 2.0 13.42 The discharge in a triangular notch and a rectangular suppressed weir both having the same head and Cd are identical when the ratio of the water surface width in the V-notch to the length of the rectangular weir is 1 (a) 1 (b) 1 2 7 1 (c) 1 (d) 2 8 2 13.43 An overﬂow spillway is 10.0 m long between two square abutments and has two piers of 0.25 m width on its crest when the head over the weir is 0.60 cm. The effective length of the spillway for calculation of the discharge by the weir formula is (a) 9.14 m (b) 9.50 m (c) 9.26 m (d) 9.40 m 13.44 A separate arrangement for aeration of nappe is desired in the following: (a) triangular weir 13.45 13.46 13.47 13.48 13.49 (b) trapezoidal weir (c) contracted rectangular weir (d) none of the above A suppressed rectangular weir and a triangular V-notch are installed in a large tank such that the vertex of the notch and the weir are at the same level. If Cd of both the weir and the notch is same. (a) the discharge from the V-notch will be larger for all heads. (b) the discharge from the rectangular weir will be larger for all heads (c) the rectangular weir will have higher discharge from zero value to a critical head and then onwards it will be smaller than that of the V-notch. (d) the triangular notch will have higher discharge from zero value to a critical head beyond which it will be smaller than that of the rectangular weir. The discharge over a 90°V-notch is written as Q = 1.37 H5/2, where Q is in m3/s and H is in metres. The Cd of this notch is (a) 0.611 (b) 0.580 (c) 0.464 (d) 0.710 The discharge over a Cipolletti weir of base width B is written as Q = KH3/2, where Q is in m3/s and H in metres. If the Cd = 0.62, the coefﬁcient K = (a) 1.86 B (b) 1.83 B (c) 1.46 B (d) 1.52 B A Cippolletti weir discharges water with the head of water above the crest being 250 mm. if the heat due to velocity of approach is 0.01 m, what will be the excess percentage of discharge, as compared to when not so corrected? (a) 3.2% (b) 4.2% (c) 5.3% (d) 6.3% A Cipolletti weir has a side slope of (a) 1 vertical : 4 horizontal 455 Flow Measurement (b) 1 vertical : 2 horizontal (c) 1 horizontal : 4 vertical (d) 1 horizontal : 2 vertical 13.50 A submerged weir is one in which the water level on the downstream of the weir is (a) just at the crest level (b) below the crest level (c) is above the crest level (d) at the same elevation as the upstream water surface. 13.51 In a rectangular suppressed weir the tailwater head is 30% of the upstream head both measured above the crest. The submerged ﬂow discharge is x per cent of the free ﬂow discharge at the same upstream head, where x is (a) 98.1% (b) 70% (c) 93.3% (d) 87.7% 13.52 While conducting ﬂow measurement using a rectangular notch, an error of 2% in head over the notch and an error of –3% in the length of the notch occurred. The percentage error in the computed discharge would be (a) +6% (b) –1% (c) –2.5% (d) zero Unsteady Flow Concept Review 14 14.1 SURGES IN OPEN CHANNELS Whenever there is a sudden change in the discharge or depth or both in an open channel, a rapidly varied unsteady phenomenon, known as surge, develops. Such situations occur during sudden operation of a control gate. A surge producing an increase in depth is known as positive surge and the one which causes a decrease in the depth is known as negative surge. Positive surges have steep fronts, more like a hydraulic jump, and the shape Introduction of the wave does not change during its translation. They are also known as moving hydraulic jumps. These are relatively easy to analyse than negative surges. In this section only positive surges are considered. 14.1.1 Positive Surge Moving Downstream Figure 14.1 shows a horizontal, frictionless rectangular channel in which a positive surge is moving downstream. Sufﬁx 1 refers to the 457 Unsteady Flow Vw V2 Vw y2 y1 V1 y1 Positive surge moving downstream (a) (Vw – V2) y2 Positive surge moving upstream (a) Vw Vw Vw y1 y1 (Vw – V1) y2 (V1 + Vw) Fig. 14.2 Fig. 14.1 conditions before the arrival of the surge and sufﬁx 2 refers to a section after the passage of the surge. The absolute velocity of the surge is Vw and is assumed constant. The unsteady ﬂow situation can be simulated to an equivalent steady state ﬂow by superimposing a velocity (–Vw), (directed to the left in Fig. 14.1 (a)) at all sections [Fig. 14.1(b)]. The conditions are now similar to that of a hydraulic jump with approach velocity of (Vw – V1) and approach depth y1. The depth after the surge (jump) is y2 and the corresponding velocity is (Vw – V2). The continuity equation is, A1 (Vw – V1) = A2 (Vw – V2) (14.1) By considering unit width of the rectangular channel V1 ) = y2 (Vw Vw (V2 + Vw) Simulated steady flow (b) Simulated steady flow (b) y1 (Vw y2 V2 V1 V2 ) (14.2) By application of momentum equation ˆ (Vw - V1 ) 2 1 y2 Ê y2 = + 1˜ Á g y1 2 y1 Ë y1 ¯ (14.3) From Eqs. 14.2 and 14.3 two out of the ﬁve variables V1, V2, y1, y2 and Vw can be evaluated if the other three are given. 14.1.2 Positive Surge Moving Upstream Figure 14.2(a) shows a positive surge moving upstream. This kind of surge occurs on the upstream of a sluice gate when the gate is closed suddenly and in the phenomenon of tidal bores. The unsteady ﬂow situation is converted to a simulated steady ﬂow by superposition of a velocity Vw directed downstream [to the right in Fig. 14.2(b)]. As before, sufﬁxes 1 and 2 refer to the conditions at sections of the channel before and after the passage of the surge. The continuity equation is A1 (Vw + V1 ) = A2 (Vw + V2 ) (14.4) For unit width of the rectangular channel, Eq. 14.4 becomes y1 (Vw + V1 ) = y2 (Vw + V2 ) (14.4a) From Fig. 14.2(b) it is easy to see that the ﬂow is similar to that of a hydraulic jump with initial velocity of (Vw + V1) and initial depth y1. The ﬁnal velocity is (Vw + V2) and the depth after the surge is y2. By momentum equation (Vw + V1 ) 2 1 Ê y2 ˆ Ê y ˆ = Á ˜ Á1 + 2 ˜ 2 Ë y1 ¯ Ë y1 ¯ g y1 (14.5) 458 Fluid Mechanics and Hydraulic Machines From Eqs. 14.4a and 14.5 two of the ﬁve variables y1, y2, V1, V2 and Vw can be determined if three other variables are given. It is to be remembered that, in the real ﬂow, Vw is directed upstream (to the left in the ﬁgure). 14.1.3 Other Forms of Equations Equations 14.3 and 14.5, being symmetrical, could also be expressed as follows to suit the problem at hand: Alternative form of Eq. 14.3: (Vw - V2 ) 2 1 y1 Ê y ˆ = 1+ 1 ˜ 2 y2 ÁË y2 ¯ g y2 Alternative form of Eq. 14.5 (Vw + V2 ) 2 1 y1 Ê y ˆ = 1+ 1 ˜ Á 2 y2 Ë y2 ¯ g y2 14.1.4 14.2 (14.5a) (14.6) WATER HAMMER When a liquid ﬂow in a long pipeline is reduced suddenly, due to compressibility of the liquid, the sudden change in momentum would cause a pressure surge to develop. This pressure moves through the pipe at high speed and undergoes reﬂection at the ends. The phenomenon is known as Water hammer and is of importance in all major pipeline designs. 14.2.1 (14.7) K /r where K = bulk modulus of the liquid in Pa. and r = mass density of the liquid in kg/m3. If the pipe material is elastic, the velocity of propagation will be less than that given by Eq. 14.7 and depends upon the diameter D, thickness t of the pipe and the modulus of elasticity E of the pipe material. The velocity of the pressure wave in an elastic pipe is given by 1/ 2 C= The velocity of the surge relative to the initial ﬂow velocity in the channel is known as celerity of the surge, Cs. Thus for the surge moving downstream, Cs = Vw – V1 and for the surge moving upstream Cs = Vw + V1. From Eqs. 14.3 and 14.5 it is seen that in both cases 1 y2 g ( y1 + y2 ) 2 y1 C= (14.3a) Celerity of the Surge Cs = sound in an inﬁnite expansion of the medium, and is given by Velocity of Pressure Wave If the pipe is rigid, the pressure wave will travel in the medium at a velocity C equal to the velocity of È ˘ 1 K /r Í ˙ Î1 + ( D / t ) ( K / E ) ˚ (14.7-a) For water, normally K / r is about 1400 m/s and velocity of pressure wave C by Eq. 14.7-a for normal dimensions would be between 900 and 1200 m/s. 14.2.2 Rapid Closure Consider a pipe of length L leading from a reservoir and terminating in a valve at its downstream end. When the valve is instantaneously closed a pressure of magnitude ph is formed and moves up with a velocity C. The wave undergoes reﬂections at the reservoir end as well as at the valve. For the case of a frictionless ﬂow, the various stages of the pressure wave are shown in Fig. 14.3 (a and b). Figure 14.4 shows the water hammer pressure ph as a function of time at various locations of the pipe. It is seen that at all locations the time period for a complete cycle is 4L/C. The time T0 = 2L/C is called as critical time. If the time of closure T of a valve is such that T < T0, the pressure head at the valve will be same as that for instantaneous closure. As such, the time of closure T £ 2L/C is known as rapid closure. If x0 = length of the pipe having peak pressure in a closure time T < 2L/C, the distance x0 is given by x0 = L - CT 2 (14.8) 459 Unsteady Flow L V0 t=0 Valve L Reservoir A C L/2 (1) +ph B M 2L/C 2L/C At (B) C time V0 t = L/2C (2) C t = L/C –ph 2L/C +p h At (A) (3) 2L/C At (M) O –ph C V0 t = 3/2C (4) Fig. 14.4 (a) Fig. 14.3(a) Various Stages of a Water Hammer Pressure Wave L t = 2L/C t = 5L/2C 2L/C 4L/C 6L/C 8L/C time Variation of Water Hammer Pressure with Time Thus it is seen that in instantaneous closure (T = 0), x0 = L, i.e. the entire pipe will have peak pressure. C V0 V0 O (5) C (6) Water Hammer Pressure Let the velocity of ﬂow in a pipe be changed from V1 to V2 rapidly. The resulting water hammer pressure ph is given in pascals by ph = –rC DV where DV = (V2 – V1) (14.9) Thus, if the ﬂow in a pipe is completely stopped, DV = –V1 and the water hammer pressure is, in pascals, C ph = r C V1 t = 3L/C (for complete closure) (14.10) (7) 14.2.3 t = 7L/2C C V0 (8) V0 t = 4L/C C (9) (b) Fig. 14.3(b) Various Stages of a Water Hammer Pressure Wave (a) Slow Closure If the time of closure T > To (critical time) then the closure is known as slow closure. If T is slightly larger than To (say 1 < T/To £ 1.5), compressibility effects are important but the peak water hammer pressure will be less than that in rapid closure. An approximate value of peak water hammer pressure can be obtained by phs To = phr T (for 1 < T/To £ 1.5) (14.11-a) 460 Fluid Mechanics and Hydraulic Machines where T0 = 2L/C = critical time. T = actual time of slow closure, (T > T0) Phs = peak water hammer pressure in slow closure Phr = peak water hammer pressure in rapid closure [Eq. 14.9 or Eq. 14.10 as is the case]. H Pipe of length, L Valve (a) (b) Very Slow Closure If the actual valve closure time is many times greater than To, the compressibility effects are no longer important and the pressure rise is due to change in momentum and is given by pvsc = 14.2.4 r LV T (14.11-b) time t (b) Fig. 14.5 Design of Pipe Thickness The water hammer pressure ph will be over and above the steady state pressure in the pipe, ps (referred commonly as static pressure). Hence, the pipe will have to withstand a total pressure pt given by pt = ps + ph (14.12) The stress s in the pipe wall is given by the thin cylinder formula pD s= t (14.13) 2t where t = thickness of the pipe wall and D = diameter of the pipe. For design, s should be less than the working stress s w of the pipe material. Thus the minimum thickness of a pipewall is tm = ( ps + ph ) D pt D = 2s w 2s w (14.14) For steel, the normal working stress is of the order of 0.1 kN/mm2 (= 100 MPa). 14.3 V/V0 ESTABLISHMENT OF FLOW Consider a pipe leading from a large tank. A valve at the downstream end controls the ﬂow (Fig. 14.5). If the valve is opened suddenly, it takes a while for the Establishment of Flow ﬂow to be fully established. By considering the ﬂuid as incompressible and the pipe to be rigid, the time for ﬂow establishment is obtained by integration of the Euler equation. Let V = velocity at any time t V0 = ﬁnal steady state velocity. f = Darcy–Weisbach friction factor k1 = sum of minor loss coefﬁcients for the pipe H = static head at the outlet. Ê fL ˆ V2 Total head loss = Á + k1 ˜ Ë D ¯ 2g =k Then V0 = V2 2g where k= fL + k1 D 2gH (1 + k ) If V = 0 at t = 0, the time taken to attain a velocity V after sudden opening of the valve, is given by t= L (1 + V /V0 ) ln (1 + k ) V0 (1 - V /V0 ) (14.15) It is seen that V approaches V0 asymptotically (Fig. 14.5(b)). For a frictionless pipe, k = 0. 461 Unsteady Flow 14.4 SURGE TANKS In many hydropower projects, large penstocks carry considerable quantity of water from a reservoir to the turbines. When there is a sudden drop in the load at the generator, the ﬂow in to the turbines is reduced suddenly leading to water hammer situation. Similarly, when there is sudden increase in the discharge requirement at the turbines, a sudden opening of the control valves may cause negative pressures. To overcome these problems it is a common practice to install Surge tanks in such systems. A simple Surge tank is essentially a large cylindrical vertical tank of sufﬁcient height connected to the penstock. Figure 14.6 is a deﬁnition sketch of a surge tank installation. Reservoir Surge tank h1 h2 B High pressure penstock Power house Low pressure penstock A Tail race Valve Fig. 14.6 In this ﬁgure, a set of large pipes, made up of two parts A and B, connects a reservoir to a turbine and generator set. Pipeline B of length Lb is the low pressure penstock taking off from the reservoir and pipe A of length La is the high pressure penstock connected to the turbine. A surge tank is provided near the turbine valve. Generally Pipeline B will be considerably longer than Pipeline A. In this ﬁgure h1 represents the static level corresponding to no ﬂow in the pipelines. When there is uniform ﬂow with a velocity V0 in the pipe B, due to friction, the hydraulic grade line will be as shown in the ﬁgure and the piezometric head at the surge tank is h2. When there is sudden closure of the valve at the turbine, water hammer pressures are created in pipeline A. The free surface of the surge tank acts as a reservoir type end condition for the water hammer process in the pipe line and the effect of the water hammer is conﬁned to the penstock A only. Water from pipeline A rushes in to the surge tank and the water surface in the surge tank rises above the level h2. After the momentum change has dissipated the water surface moves down and due to inertia goes lower than the original level h2 and thus a mass oscillation is set up in the tank. Due to friction this oscillation will be dampened and die down eventually. A new stable level h2 will be eventually established. If it were not for the surge tank, the entire pipe length (La + Lb) would have been affected by the water hammer effect. A similar, but converse situation exists at the sudden load acceptance and opening of the valve at the turbine. In this case the water ﬂows out quickly out of the surge tank and the water surface in the surge tank drops down below the level h2, and undergo mass oscillation. The new stable level h2 will be eventually established due to ﬂow from the reservoir. Thus the surge tank in a hydropower system as above, (i) helps reduce the length of the pipe affected by water hammer effects due to valve operations, (ii) provides a source for storage of water rejected by the turbine due to sudden valve closure and (iii) provides a source of water to meet sudden demands created by the turbine due to sudden valve opening. Surge tanks are usually open at the top and of sufﬁcient height so that they do not overﬂow. The above is a brief description of a simple surge tank. Many designs, which are the variants of the basic simple surge tank, such as (i) restricted entry surge tank, and (ii) differential surge tank, are in use with speciﬁc advantages. 462 Fluid Mechanics and Hydraulic Machines Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples A. Surges in Canals Vw - V1 14.1 g y1 = 1 2 1/ 2 È y2 Ê y2 ˆ˘ + 1˜ ˙ Í Á ¯˚ Î y1 Ë y1 1/ 2 È1 ˘ 9.81 ¥ 1.3 Í ¥ 2 ¥ ( 2 + 1) ˙ Î2 ˚ = 6.185 Vw = 8.185 m/s (in simulated ﬂow) The surge moves downstream with a velocity of 8.185 m/s. By continuity equation: y1 (Vw – V1) = y2 (Vw – V2) 1.3 (8.185 – 2.0) = 2.6 (8.185 – V2) V2 = 8.185 – 3.0925 = 5.093 m/s New discharge Q2 = By2 V2 = 2.0 ¥ 2.6 ¥ 5.093 = 26.484 m3/s Vw – 2.0 = Solution: Conditions of relative steady ﬂow are simulated by adding the velocity – Vw vectorially, i.e. to the left in Fig. 14.7. Vw V y2 2 y1 V1 (a) Vw (Vw – V2) y2 Vw y 1 (V – V ) w 1 (b) Simulated steady flow Fig. 14.7 14.2 Solution: Here y1 = 2.0 m and the surge height = 0.5 m. Hence y2 = 2.0 + 0.5 = 2.5 m Surge Moving Downstream Celerity Cs = Here y1 = 1.30 m, V1 = 2.0 m/s y2 = 2.60 m By the surge equation obtained from a combination of momentum and continuity equations: = 1 y2 g ( y1 + y2 ) 2 y1 1 2.5 ¥ 9.81 ¥ ( 2.0 + 2.5) 2 2.0 = ± 5.25 m/s 463 Unsteady Flow y2 -1 ± 1 + 4 ¥ 3.5384 = 1.446 = y1 2 y2 = 1.30 ¥ 1.446 = 1.88 m 6.175 V2 = – 4.0 = – 0.715 m/s 1.88 = 0.715 m/s in upstream direction. 14.3 Solution: Superimpose a velocity of Vw to the right as shown in Fig. 14.8 to simulate steady ﬂow situation. 14.4 Vw V1 y2 y1 V2 (a) Vw V1 + Vw Vw y1 y2 V2 + Vw Solution: Let Vw (directed upstream) to be velocity of the bore. Superimpose a velocity Vw directed to the right, to get simulated steady ﬂow as shown in Fig. 14.9. Vw (b) Simulated steady flow y1 = 2.0 m Fig. 14.8 y1 = 1.30 m, V1 = 0.75 m/s Vw = 4.0 m/s By continuity, y1 (V1 + Vw) = y2 (V2 + Vw) 1.3 ¥ (0.75 + 4.0) = y2 (V2 + 4.0) 6.175 V2 = – 4.0 y2 By the surge equation, 1/ 2 È 1 y2 Ê V1 + Vw y ˆ˘ = Í 1+ 2 ˜˙ Á g y1 y1 ¯ ˚ Î 2 y1 Ë (0.75 + 4.0) 2 1 y2 Ê y ˆ = 1+ 2 ˜ 9.81 ¥ 1.30 2 y1 ÁË y1 ¯ 2 Ê y2 ˆ Ê y2 ˆ ÁË y ˜¯ + ÁË y ˜¯ – 3.5384 = 0 1 1 V2 + V w y2 = 5.0 m V1 + V w Fig. 14.9 Here Vw Simulated Steady Flow 5.0 ¥ 1000 = 1.389 m/s 60 ¥ 60 y1 = 2.0 m, y2 = 5.0 m By continuity equation: y1 (V1 + Vw) = y2 (V2 + Vw) 2.0 ¥ (1.389 + Vw) = 5.0 (V2 + Vw) 1 V2 = (2.778 – 3 Vw) 5 By the surge equation ˆ (Vw + V1 ) 2 1 y2 Ê y2 = + 1˜ Á 2 y1 Ë y1 g y1 ¯ Here V1 = (Vw + 1.389) 2 1 5.0 Ê 5.0 ˆ = ¥ +1 9.81 ¥ 2.0 2 2.0 ÁË 2.0 ˜¯ (Vw + 1.389)2 = 85.838 Vw = 7.876 m/s 464 Fluid Mechanics and Hydraulic Machines The bore travels upstream with a velocity of 7.876 m/s. 1 Velocity V2 = (2.778 – 3 ¥ 7.876) 5 = – 4.17 m/s After the passage of the tidal bore the ﬂow in the river is upstream with a velocity of 4.17 m/s. The ﬂow in the river after the bore will be in the upstream direction with a velocity of 2.202 m/s. 14.6 14.5 Solution: Let Vw = velocity of surge wave. Superimpose a velocity Vw to the left (towards upstream) to simulate steady ﬂow as in Fig. 14.11. Solution: By superimposing a velocity Vw to the right, a steady ﬂow situation is simulated, as in Fig. 14.10. y2 = 3.5 m V1 + V w (a) Vw y2 Fig. 14.10 Here Vw = 6.0 m, y1 = 2.0 and y2 = 4.25 m By continuity equation y1 (V1 + Vw) = y2 (V2 + Vw) 2.0 (V1 + 6.0) = 4.25 (V2 + 6.0) V2 = 0.4706 V1 – 3.1765 By the surge equation, (Eq. 14.5), (V1 + Vw ) 2 1 y2 Ê y ˆ = 1+ 2 ˜ Á 2 y1 Ë y1 ¯ g y1 9.81 ¥ 2.0 y1 V1 Vw Simulated steady flow (V1 + 6.0)2 V2 Vw V + V 2 w y2 = 4.25 m Vw y1 = 2.0 m Vw = 8.0 m/s = 1 4.25 Ê 4.25 ˆ ¥ 1+ 2 2.0 ÁË 2.0 ˜¯ (V1 + 6)2 = 65.145 By taking the positive root V1 = 2.07 m/s V2 = 0.4706 ¥ 2.07 – 3.1765 = – 2.202 m/s Vw – V2 Vw – V1 y1 (b) Simulated steady flow Fig. 14.11 Here y2 = 3.5 m, V2 = 4.5 m/s, and Vw = 8.0 m/s By the continuity equation, y1 (Vw – V1) = y2 (Vw – V2) y1 (8.0 – V1) = 3.5 (8.0 – 4.5) 12.25 V1 = 8.0 – y1 By the surge equation, (Eq. 14.3), (Vw - V1 ) 2 1 y2 Ê y ˆ = 1+ 2 ˜ Á 2 y1 Ë y1 ¯ g y1 Ê 12.25 ˆ ÁË 8.0 - 8.0 + y ˜¯ 1 9.81 y1 2 = 1 3.5 Ê 3.5 ˆ 1+ Á 2 y1 Ë y1 ˜¯ 465 Unsteady Flow 15.297 = y13 1.75 Ê 3.5 ˆ 1+ Á y1 Ë y1 ˜¯ y12 = 3.5 y1 – 8.741 = 0 - 3.5 ± (3.5) 2 + 4 ¥ 8.741 2 y1 = 1.686 m y1 = 12.25 = 0.734 m/s 1.686 Alternative method: Writing the alternative form of the surge equation Eq. (14.3a) V1 = 8.0 – Here V1 = 2.0 m/s Vw = 4.5 m/s y2 = 3.6 m By continuity equation: y1 (Vw + V1) y1 (4.5 + 2.0) 6.5 y1 V2 By the surge equation: (Vw + V1 ) 2 1 y2 Ê y ˆ = 1+ 2 ˜ Á 2 y1 Ë y1 ¯ g y1 (Vw - V2 ) 2 1Ê y ˆÊ y ˆ = Á 1 ˜ Á1 + 1 ˜ 2 Ë y2 ¯ Ë y2 ¯ g y2 (8.0 - 4.5) 2 1Ê y ˆÊ y ˆ = Á 1 ˜ Á1 + 1 ˜ 9.81 ¥ 3.5 2 Ë 3.5 ¯ Ë 3.5 ¯ y2 y1 + 1 – 2.4975 = 0 3.5 y12 + 3.5 y1 – 8.741 = 0 Solving y1 = 1.686 m 14.7 ( 4.5 + 2.0) 2 1 Ê 3.6 ˆ = Á 9.81 y1 2 Ë y1 ˜¯ Vw y1 Fig. 14.12 Vw 14.8 Solution: The absolute velocity of the surge is Vw in the downstream direction. By superposing a velocity Vw in the opposite direction (i.e. to the left) a steady ﬂow is simulated as in Fig. 14.13. Vw + V2 y2 = 3.6 m Vw (Vw – V2) y 2 Vw + V1 Simulated Steady Flow Ê 3.6 ˆ ÁË1 + y ˜¯ 1 Ê 3.6 ˆ ÁË1 + y ˜¯ = 2.3927 1 y1 = 2.585 m Discharge before the passage of the surge = Q1 = B y1V1 = 2.0 ¥ 2.585 ¥ 2.0 = 10.34 m3/s V2 = Velocity after the passage of the surge = 1.806 y1 – 4.5 = (1.806 ¥ 2.585) – 4.5 = 0.168 m/s Solution: Let Vw (directed to the left) be the velocity of the surge wave. Superimpose a velocity Vw directed to the right, to obtain a simulated steady ﬂow, as in Fig. 14.12. = y2 (Vw + V2) = 3.6 (4.5 + V2) = 16.2 + 3.6 V2 = 1.806 y1 – 4.5 Vw y1 (Vw – V1) Original downstream Fig. 14.13 466 Fluid Mechanics and Hydraulic Machines y1 = 0.8 m 1.6 V1 = = 2.0 m/s 0.8 3.2 V2 y2 = 2 ¥ 1.6 = 3.2 and V2 = y2 By continuity equation: y1 (Vw – V1) = y2 (Vw – V2) 0.8 (Vw – 2.0) = Vw y2 – 3.2 Vw (y2 – 0.8) = 1.6 Vw = 1.6/(y2 – 0.8) From the surge equation, (Eq. 14.3). Gate Here (Vw - V1 ) 2 1 y2 Ê y ˆ = 1+ 2 ˜ Á 2 y1 Ë y1 ¯ g y1 Ê 1.6 ˆ ÁË y - 0.8 - 2.0˜¯ 2 9.81 ¥ 0.8 y1 V1 V =0 y2 2 (a) Vw (V1 + Vw) y1 Vw y2 Vw 2 1 y2 Ê y ˆ = 1+ 2 ˜ Á 2 0.8 Ë 0.8 ¯ 2 Ê 3.2 - 2.0 y2 ˆ ÁË y - 0.8 ˜¯ = 6.131 y2 (0.8 + y2) 2 By trial and error, y2 = 1.088 m 3.2 3.2 V2 = = = 2.941 m/s y2 1.088 1.6 Vw = = 5.556 m/s (1.088 - 0.8) Vw = + 5.556 m/s in simulated ﬂow. Hence the surge moves downstream with a velocity of 5.556 m/s. Vw 14.9 (b) Simulated steady flow Fig. 14.14 By continuity equation, y1 (V1 + Vw) = y2 (V2 + Vw) 0.8 (0.9 + Vw) = y2 (0 + Vw) 0.72 Vw = ( y2 - 0.8) By the surge equation, (Eq. 14.5), (V1 + Vw ) 2 1 y2 Ê y ˆ = 1+ 2 ˜ Á 2 y1 Ë y1 ¯ g y1 ÏÔ Ê 0.72 ˆ ¸Ô Ì0.9 + Á ˝ Ë y2 - 0.8 ˜¯ Ô˛ ÔÓ = 9.81 ¥ 0.8 ¥ Ê 0.9 y2 ˆ ÁË y - 0.8 ˜¯ 2 Solution: Let Vw = velocity of the surge. Superimpose a velocity Vw to the right to simulate the steady ﬂow conditions as in Fig. 14.14. Here V2 = 0 y1 = 0.8 m V1 = 0.9 m/s 2 1 y2 Ê y ˆ 1+ 2 ˜ ¥ 2 0.8 ÁË 0.8 ¯ 2 = 6.13125 y2 (0.8 + y2) By trial and error y2 = 1.075 m. 0.72 Velocity of surge Vw = (1.075 - 0.8) = 2.618 m/s in simulated ﬂow. The surge moves upstream with a velocity of 2.618 m/s. 467 Unsteady Flow B. Water Hammer 14.10 elasticity K in MPa ¥ ¥ ¥ 2 ¥ K Solution: Here Solution: Velocity of pressure surge = sonic velocity = C= 14.12 Change in pressure Dph = –rCDV C = Velocity of pressure surge = K /r 2.22 ¥ 10 998 9 = 1491.5 m/s (c) for gasolene (RD = 0.68) Ê 9.58 ¥ 108 ˆ C = Á ˜ Ë 0.68 ¥ 998 ¯ E 14.13 K 1/ 2 = 1484.7 m/s V2 – V1 = DV = 1.0 – 40 = – 3.0 m/s Dph = 998 ¥ 1484.7 ¥ 3 = 4.445 ¥ 106 Pa = 4.445 MPa 1/ 2 = 1370.7 m/s 1/ 2 Also (b) for crude oil (RD = 0.8) Ê 1.50 ¥ 109 ˆ C= Á ˜ Ë 0.8 ¥ 998 ¯ (as the pipe is rigid) Ê 2.2 ¥ 109 ˆ C= Á ˜ Ë 998 ¯ (a) for water C= K /r ¥ ¥ = 1188 m/s 14.11 ¥ E Solution: The velocity of pressure wave ¥ K C = Solution: The velocity of pressure wave C in a elastic pipe carrying a liquid is C= Ê ˆ 1 K /r Á Ë 1 + ( D / t ) ( K / E ) ˜¯ Ê 1.43 ¥ 109 ˆ È = Á ˜Í Ë 0.8 ¥ 998 ¯ Í Ê 250 ˆ Í1 + ÁË 2 ˜¯ ÍÎ 1/ 2 Ê 2.0 ¥ 109 ˆ = Á ˜ Ë 998 ¯ 1/ 2 1/ 2 ¥ 1/ 2 1/ 2 1 ˘ Ê 1.43 ¥ 109 ˆ ˙ ˙ Á ˜ ÁË 2.07 ¥ 1011 ˜¯ ˙ ˙˚ = 1338.3 ¥ 0.7325 = 980.4 m/s Ê ˆ 1 K /r Á Ë 1 + ( D / t ) ( K / E ) ˜¯ È ˘ 1 Í 9 11 ˙ Î1 + (90 /1.25) ( 2.0 ¥ 10 /1.0 ¥ 10 ) ˚ = 1415.6 ¥ 0.64 = 906.2 m/s Pressure rise Dph = – rCDV D ph C DV Water hammer head = h h = =– g g 468 Fluid Mechanics and Hydraulic Machines Here DV = 0 – V = – 2.60 m/s. Hence, 906.2 ¥ 2.60 hw = = 240.2 m 9.81 14.15 ¥ 14.14 ¥ ¥ 2 Solution: Velocity of pressure wave, 2 ¥ 1/ 2 C = È ˘ 1 K /r Í ˙ Î1 + ( D / t ) ( K / E ) ˚ Ê 1.43 ¥ 109 ˆ =Á ˜ Ë 0.8 ¥ 998 ¯ Solution: (i) By neglecting the elasticity of the pipe material, velocity of pressure wave C= K /r Ê 2.1 ¥ 103 ¥ 106 ˆ =Á ˜ 998 Ë ¯ 1/ 2 = 1450.6 m/s Water hammer pressure rise Dph = –rC DV = – 998 ¥ 1450.6 ¥ (0 – 2.1) = 3.04 MPa (ii) By considering the elasticity of the pipe material: 1/ 2 ¥ 1/ 2 1 ˘ È ˙ Í Ê 800 ˆ Í1 + Á (1.43 ¥ 109 / 2.10 ¥ 1011 ) ˙ ˜ ˙˚ ÍÎ Ë 80 ¯ = 1295 m/s Critical time T0 = 2L/C = (2 ¥ 1000)/1295 = 1.544 s Hence, maximum time for a sudden closure is 1.544 s. 14.16 1/ 2 C= È ˘ 1 K /r Í ˙ 1 + ( D / t ) ( K / E ) Î ˚ È 1/ 2 ˘ 9 11 ˙ Î1 + (60 /1.2) ( 2.1 ¥ 10 / 2.1 ¥ 10 ) ˚ = 1450.6 Í 1 = 1184.4 m/s Water hammer pressure rise, Dph = –rC DV Dph = –998 ¥ 1184.4 ¥ (0 – 2.1) = 2.48 MPa Solution: 2 ¥ 100 = 0.14 s. 1430 (i) For instantaneous closure, water hammer pressure ph = – rC DV 0.5 Here – DV = V = Ê pˆ 2 ÁË 4 ˜¯ ¥ (0.5) = 0.2546 m/s Critical time To = 2 L/C = 469 Unsteady Flow ph = 998 ¥ 1430 ¥ 0.2546 = 363350 Pa = 363.35 kPa (ii) Closure in T = 1s is a very slow closure as T/To = (1/0.14) ª 7 Hence, at the valve, the pressure rise is approximately rLV 998 ¥ 100 ¥ 0.2546 Pvsc = = T 1 = 25409 Pa = 25.41 kPa Hence is approximately To 1.48 (ph)rapid = ¥ 3.235 T 2.00 = 2.394 MPa (c) When T = 0.8 s, as T < T0, it is a rapid closure and ph is the same as in case (a) viz ph = 3.235 MPa phs = 14.18 14.17 2 ¥ E K ¥ E Solution: Velocity of pressure wave, K 2 ¥ 1/ 2 C= Solution: Velocity of pressure wave 1/ 2 C= È ˘ 1 K /r Í ˙ Î1 + ( D / t ) ( K / E ) ˚ Ê 2.11 ¥ 109 ˆ C= Á 998 ˜¯ Ë ¥ È ˘ 1 K /r Í ˙ 1 + ( D / t ) ( K / E ) Î ˚ Ê 2.10 ¥ 109 ˆ = Á 998 ˜¯ Ë 1/ 2 1/ 2 1/ 2 ¥ 1/ 2 1 ˘ È ˙ Í Ê 60 ˆ Í1 + Á ˜ ( 2.11 ¥ 109 /1.04 ¥ 1011 ) ˙ ˙˚ ÍÎ Ë 1.5 ¯ = 1454 ¥ 0.743 = 1080.4 m/s Critical time T0 = 2L/C = 2 ¥ 800/1080.4 = 1.48 s (a) Closure in T = 1.25 s is rapid closure as T < T0. Hence, water hammer pressure ph = – rC DV = – 998 ¥ 1080.4 ¥ (– 3.0) Pa = 3.235 MPa (b) When T = 3.0 s, as T/T0 = (2.0/1.48) = 1.35 it is a slow closure. The pressure rise at the valve 1 ˘ È ¥ Í ˙ Ê 400 ˆ 9 11 ˙ Í1 + Á ( . / . ) ¥ ¥ 2 10 10 2 11 10 ˜ ˙˚ ÍÎ Ë 10 ¯ = 1450.6 ¥ 0.8457 = 1227 m/s Critical time for rapid closure T0 = 2L/C = (2 ¥ 2500)/1227 = 4.076 s Actual time of closure T = 5.0 and T/T0 = 5.0/4.076 = 1.227 it is a slow closure case. For a rapid closure: phr = water hammer pressure = – rC DV = – 998 ¥ 1227 ¥ (0.5 – 3.0) p hr = 3.061 ¥ 106 Pa = 3061 kPa At slow closure, pressure rise: T0 4.076 p hr = ¥ 3061 T 5.0 = 2495 kPa phs = 470 Fluid Mechanics and Hydraulic Machines Total pressure at the valve end pt = pstatic + phs = (9.79 ¥ 250) + 2495 = 4943 kPa Solution: Ê 1.956 ¥ 109 ˆ K /r = Á ˜ 998 Ë ¯ C = 14.19 Critical time T0 = Solution: As the pipe is rigid, velocity of pressure wave C= Ê 2.2 ¥ 109 ˆ K /r = Á ˜ Ë 998 ¯ 1/ 2 = 1484.7 m/s 700 m 2L 2 ¥ 3500 = = 4.715 s C 1484.7 CT 2 = 3500 – 1400 m Valve V B ph 0 1 2 3 4 5 1484.7 ¥ 4.0 = 530.6 m 2 K 6 7 8 9 10 11 12 13 –ph Time in seconds (a) Fig. 14.15(a) 14.20 ¥ 700 m Lake L C As T = time of closure = 4.0 s, the closure is rapid. Hence the water hammer pressure ph = rCV0 = 998 ¥ 1484.7 ¥ 0.8 Pa = 1.185 MPa Length of the pipe, from the valve end, affected by this peak pressure, during the closure time. x0 = L – 2L 2 ¥ 2800 = =4s C 1400 (i) Figure 14.15(a) shows the variation of the water hammer pressure ph, at the valve V with the static pressure at V as datum. Critical time T0 = = 1400 m/s Period of pressure ﬂuctuation: 2T0 = 8 s Water hammer pressure ph = rCV0 = 998 ¥ 1400 ¥ 2.0 Pa = 2.794 MPa ¥ K 1/ 2 Variation of ph at Valve V Time taken for ph to reach the lake = 2800/1400 = 2 s Time taken for the reﬂected wave to reach valve V = 2 s Hence from 0 to 4 s, the pressure at V will be + ph and from 4 to 8 s, it will be – ph and from 8 to 12 s, it will be + ph and so on. Note the period of the wave is 8 s. (ii) Figure 14.15(b) shows the variation of the water hammer pressure ph, at mid point B. The static pressure at B is taken as datum. 471 Unsteady Flow Time for ph to reach the lake = 0.5 s. Time for reﬂected pressure to reach C = 0.5 s. Hence, pressure at C will remain zero (relative to static pressure at C) from 2.5 s to 5.5 s. From 5.5 s to 6.5 s the pressure at C will be –ph From 6.5 to 9.5 s the pressure at C will be zero From 9.5 s to 10.5 s, it will be + ph, and so on. Note the period of pressure ﬂuctuation at C = 8 s. +ph 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 –ph Time in seconds Fig. 14.15(b) Variation of ph of mid-point B Time for the pressure wave to reach B = 1400/1400 = 1 s Time for ph to reach the lake = 1 s Time for the reﬂected wave to reach = 1 s Hence, duration of max pressure ph at B is 2 s, i.e. from t = 1.0 s to 3.0 s. Time for the reﬂected wave to reach the valve = 1s Time for the –ph wave to reach point B = 1 s Hence from 3.0 s to 5.0 s the pressure at B will remain at zero (relative to the static pressure). From 5.0 s to 7.0 s it will register – ph. From 7.0 s to 9.0 s the pressure B will be 0. From 9.0 s to 11.0 s it will be + ph and so on. The period of pressure ﬂuctuation is 8 s. (iii) Fig. 14.15(c) shows the variation of the water hammer pressure ph, at point C. The static pressure at C is taken at datum. Time for pressure ph to reach C = 2100/1400 = 1.5 s –ph 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 –ph Time in seconds Fig. 14.15(c) Variation of ph at point C 14.21 2 E 2 K 2 Solution: Assume the pipe to be rigid. The velocity of pressure wave for this case C = Ê 2.10 ¥ 109 ˆ K /r = Á 998 ˜¯ Ë 1/ 2 = 1450.6 m/s Velocity of ﬂow = V0 = Q 1.40 = = 1.238 m/s A Ê pˆ 2 ( 1 . 20 ) ÁË 4 ˜¯ Water hammer pressure ph = rCV0 = 998 ¥ 1450 ¥ 1.238 ¥ 10–3 kPa = 1792 kPa Total pressure pt = Static pressure + water hammer pressure = (9.79 ¥ 300) + 1792 pt = 4729 kPa 472 Fluid Mechanics and Hydraulic Machines Minimum thickness of pipe required pD tm = t where sw = working stress 2s w = (4729 ¥ 1.2)/(2 ¥ 0.1 ¥ 106) = 0.02837 m = 28.3 mm When the elasticity of the pipe is considered, the thickness required will be slightly less than 28 mm. A trial and error approach is adopted. s = stress in the pipe wall = pt D 2t 4428 ¥ 1.20 = 98405 kPa < sw 2 ¥ 0.027 Hence the adopted thickness t = 27 mm is satisfactory. = 14.22 Trial 1: Assume t = 26 mm D/t = 1200 K 2.1 1 = 46.15 and = = 26 E 210 100 Velocity of pressure, wave, 1/ 2 È ˘ 1 C = 1450.6 ¥ Í ˙ Î1 + ( D / t ) ( K / E ) ˚ 1/ 2 È ˘ 1 = 1450.6 ¥ Í ˙ Î1 + ( 46.15) (1/100) ˚ = 1200 m/s Water hammer pressure ph = rCV0 ph = 998 ¥ 1200 ¥ 1.238 ¥ 10–3 kPa = 1483 kPa ps = Static pressure = 9.79 ¥ 300 = 2937 kPa pt = total pressure = ps + ph = 4420 kPa tm = minimum thickness needed = = 4420 ¥ 1.2 2 ¥ 0.1 ¥ 106 pt D 2s w = 0.02652 m = 26.5 mm ¥ Solution: Velocity of pressure wave, 1/ 2 C = È ˘ 1 K /r Í ˙ Î1 + ( D / t ) ( K / E ) ˚ 1/ 2 Ê 1.52 ¥ 109 ˆ È 1 ˘ =Á ˜ Í ˙ Ë 0.8 ¥ 998 ¯ Í1 + Ê 200 ˆ (1520 / 207000) ˙ ÁË 10 ˜¯ ˙˚ ÍÎ = 1380 ¥ 0.9338 = 1289 m/s Velocity of ﬂow = V0 = 0.040 Ê pˆ 2 ÁË 4 ˜¯ ¥ (0.2) = 1.273 m/s. Hence adopt 27 mm thickness pipe. As a check: D/t = 1200/27 = 44.44 1/ 2 È ˘ 1 C = 1450.6 ¥ Í ˙ 1 44 44 1 100 + ( . ) ( / ) Î ˚ = 1207 m/s ph = 998 ¥ 1207 ¥ 1.238 ¥ 10–3 kPa = 1491 kPa ps = 2937 kPa pt = 4428 kPa Critical closure time T0 = 2L/C = (2 ¥ 1000)/1289 = 1.55 s Actual closure time T = 1.25 s < T0 Hence the closure is rapid. Water hammer pressure ph = rCV0 ph = (998 ¥ 0.8) ¥ 1289 ¥ 1.273 Pa = 1310 kPa 473 Unsteady Flow Additional stress due to water hammer: p D sa = h 2t 1310 ¥ 0.20 sa = = 13100 kPa 2 ¥ 0.01 = 0.0131 kN/mm2 pD 2t 2341 ¥ 300 = 2¥3 = 1.17 ¥ 105 kPa As the working stress sw = 1.0 ¥ 105 kPa is lower than this value, the valve must be closed with a time T > T0 to have slow closure. Maximum allowable total pressure is therefore Stress in the pipe s = 14.23 s w 2t 1.0 ¥ 105 ¥ 2 ¥ 3 = = 2000 kPa D 300 Allowable water hammer pressure in slow closure p hw = 2000 – pstatic = 2000 – 1284 = 716 kPa In slow closure with time T, pt = 2 K E ¥ ¥ ˆ T phw T Ê = 0 Á for 1 << 1.5˜ ~ T0 ph T Ë ¯ Solution: Velocity of pressure wave, 1/ 2 C= È ˘ 1 K /r Í ˙ Î1 + ( D / t ) ( K / E ) ˚ Ê 1 ¥ 109 ˆ = Á ˜ Ë 0.82 ¥ 998 ¯ \ 1/ 2 ¥ 1/ 2 1 ˘ È ˙ Í Ê 300 ˆ 9 11 Í1 + Á (1 ¥ 10 / 2.14 ¥ 10 ) ˙ ˜ ˙˚ ÍÎ Ë 3 ¯ = 1105.4 ¥ 0.8255 = 912.6 m/s Velocity of ﬂow V0 = T = Time of slow closure to attain phw 2.19 ¥ 1057 = T0 (ph /phw) = = 3.233 716 Since T/T0 = 3.233/2.19 = 1.476 < 1.5 the assumption of slow closure is O.K. Hence, minimum time of closure Tm = 3.23 s. C. Establishment of Flow 14.24 0.100 = 1.415 m/s Ê pˆ 2 ( 0 . 3 ) ÁË 4 ˜¯ Critical time of closure T0 = 2L/C = (2 ¥ 1000)/912.6 = 2.19 s At rapid closure, at T £ T0, Water hammer pressure ph = rCV0 ph = (0.82 ¥ 998) ¥ 912.6 ¥ 1.415 Pa = 1057 kPa Static pressure ps = 160 ¥ (0.82 ¥ 9.79) = 1284 kPa Total pressure p = 1284 + 1057 = 2341 kPa V2 2g f Solution: Friction loss hf = f LV 2 V2 = kf 2g D 2g kf = 0.02 ¥ 1500 fL = = 200 D 0.15 Minor loss coefﬁcient = k1 = 5.0 Total has coefﬁcient = k = k1 + kf = 205 474 Fluid Mechanics and Hydraulic Machines For steady ﬂow: V0 = 2gH /(1 + k ) 1/ 2 È 2 ¥ 9.81 ¥ 20 ˘ = Í ˙ Î (1 + 205) ˚ = 1.380 m/s During the establishment of ﬂow, t= = (ii) When Q/Q0 = 0.95, V/V0 = 95 1 + 0.95 t = 75.72 ln = 277.4 s 1 - 0.95 14.26 L V +V ln 0 (1 + k ) V0 V0 - V 1500 1 + (V /V0 ) ln (1 + 205) ¥ 1.380 1 - (V /V0 ) = 5.276 ln 1 + (V /V0 ) 1 - (V /V0 ) f Solution: Refer to the schematic layout shown in Fig. 14.16. (i) When Q/Q0 = 0.50, V/V0 = 0.5 ln Ê 1 + 0.5 ˆ 1 + (V /V0 ) = ln Á = 1.0986 1 - (V /V0 ) Ë 1 - 0.5 ˜¯ t = 5.276 ¥ 1.0986 = 5.8 s (ii) When Q/Q0 = 0.95, V/V0 = 0.95 ln 2 Pipe, D1 V2 /2g Ê 1 + 0.95 ˆ 1 + (V /V0 ) = ln Á = 3.6636 1 - (V /V0 ) Ë 1 - 0.95 ˜¯ D2 Nozzle t = 5.276 ¥ 3.6636 = 19.33 s hf TE Line Fig. 14.16 14.25 Let Solution: If friction is neglected, k = 0 and the time of ﬂow establishment t= Here L V +V ln 0 V0 V0 - V V0 = 2gH = (2 ¥ 9.81 ¥ 20)1/2 = 19.81 m/s 1500 1 + (V /V0 ) t= ln 19.81 1 - (V /V0 ) = 75.72 ln V1 = ultimate velocity in the pipe. V2 = ultimate velocity in the nozzle. p p By continuity V1 D12 = V2 D22 4 4 V1 = V2(D2 /D1)2 V22 = 40.0 m, 2g hence 2 ¥ 9.81 ¥ 40 = 28.0 m/s Ê 10 ˆ V1 = 28.0 ¥ Á ˜ Ë 40 ¯ 1 + (V /V0 ) 1 - (V /V0 ) (i) When Q/Q0 = 0.5, V/V0 = 0.5 1 + 0.5 t = 75.72 ln = 83.19 s 1 - 0.5 V2 = Head loss h f1 2 = 1.75 m/s fL V2 = (V12/2g) = k 1 D1 2g 475 Unsteady Flow fL 0.02 ¥ 2000 = D1 0.40 = 100 For 98% ultimate ﬂow establishment, time required k= t = = L 1 + V /V1 ln (1 + k ) V1 1 - V /V1 2000 1 + 0.98 ln (1 + 100) ¥ 1.75 1 - 0.98 = 11.315 ln 99 = 52 s Problems Surges in Canals 14.1 In a tidal river the depth and velocity of ﬂow are 0.90 m and 1.25 m/s respectively. Due to tidal action a tidal bore of height 1.2 m is observed to travel upstream. Estimate the speed of the bore and the speed of ﬂow after the passage of the bore. 14.2 14.3 14.4 14.5 (Ans. Vw = 4.61 m/s (travels upstream) V2 = –2.1 m/s) In a rectangular channel a positive surge of velocity 6.0 m/s was seen moving down the stream. If the depth and velocity after the passage of the surge are 2.5 m and 5.0 m/s respectively, estimate the height of the surge. (Ans. Height of surge = 2.31 m) For a positive surge in a horizontal rectangular channel ﬁll in the blanks in Table 14.1. The normal downstream direction is taken as positive. A tidal bore moves up a wide river with a velocity of 6.0 m/s. If the river had a depth of 1.8 m and velocity of 0.85 m/s before the passing of the bore, estimate the height of the bore and velocity of ﬂow in the river after the passage of the bore. (Ans. Height of bore = 1.546 m, V2 = –2.315 m/s) A rectangular channel 3.0 m wide is conveying 15.0 m3/s of discharge at a depth of 1.50 m. If a downstream sluice gate is Table 14.1 Data on Positive Surge in Rectangular Channel Sl. No. y1 (m) y2 (m) V1 (m/s) V2 (m/s) Vw (m/s) (a) (b) (c) (d) (e) 1.2 º º 1.75 º 3.6 4.0 3.4 º 1.80 0.8 º º 0.70 2.80 º 5.3 –0.25 º º º 9.2 – 5.0 –5.50 –3.50 Answers to Question 14.5 Sl. No. y1 (m) y2 (m) V1 (m/s) V2 (m/s) Vw (m/s) (a) 1.2 3.6 0.8 6.403¸ ˝ - 4.803˛ 9.204 ¸ ˝ - 7.604 ˛ (b) (c) (d) (e) 2.05 2.60 1.75 0.515 4.0 3.4 4.388 1.8 1.59 1.211 0.70 2.8 5.3 –0.25 –2.607 –1.698 9.2 –5.00 –5.50 –3.50 abruptly lowered to reduce the discharge by 60%, estimate the characteristics of the resulting surge. (Ans. y2 = 2.64 m, Vw = 2.632 m/s (travels upstream)) 14.6 A rectangular channel carries a ﬂow with a velocity of 0.65 m/s and depth of 1.40 m. If the discharge in the channel is abruptly increased three-fold by a sudden lifting 476 Fluid Mechanics and Hydraulic Machines of a gate on the upstream, estimate the velocity and height of the resulting surge. (Ans. y2 – y1 = 0.36 m, Vw = 5.06 m/s in downstream direction) [Note: Problems 14.5 and 14.6 require trial and error procedure.] Water Hammer 14.7 Calculate the velocity of propagation of pressure wave in the following cases of ﬂow through pipes: ﬂow is suddenly stopped by a valve at the downstream end of the pipe. (Ans. s = 24.77 MPa) 14.10 A valve is closed in 5.5 s at the end of a 4000 m long pipe carrying oil of density 917 kg/m3 at a velocity of 2.0 m/s. Assuming the pipe to be rigid, estimate the peak pressure developed by this closure and the length of the pipe subjected to this peak pressure at the end of the closure time. For the oil K = 1.38 ¥ 109 Pa. (Ans. x0 = 626.4 m; ph = 2250 kPa) Case Liquid Density kg/m3 Bulk modulus of elasticity (K) MPa Pipe dia mm (i) (ii) (iii) (iv) (v) Water Water water Sea water Mercury 998 998 998 1025 13550 2190 2190 2190 2280 25500 300 800 450 300 10 Pipe thickness mm 8.0 10.0 ———— ———— ———— Material Steel Cast iron Rigid pipe Rigid pipe Rigid pipe Modulus of elasticity (E) MPa 2.1 ¥ 105 1.0 ¥ 105 ———— ———— ———— (Ans. (i) 1256 m/s (ii) 893 m/s (iii) 1481.3 m/s (iv) 1491.4 m/s (v) 1371.8 m/s) 14.8 A valve at the downstream end of a 1.20 m diameter steel penstock carrying water is operated suddenly so as to reduce the ﬂow from 3.0 m3/s to 0.5 m3/s. Estimate the maximum water hammer pressure rise at the valve if the pipe thickness is (i) 7.5 mm and (ii) 12 mm. [For water: K = 2200 MPa; for steel: E : 2.10 ¥ 105 MPa]. (Ans. (i) Dph = 2.264 MPa; (ii) Dph = 2.589 MPa) 14.9 A brass pipe (E = 0.8 ¥ 1011 Pa) is 5 cm in diameter and 600 m long. A liquid of density 800 kg/m3 and bulk modulus of elasticity 9.0 ¥ 108 Pa is conveyed through this pipe at a rate of 450 L/min. The thickness of the pipe wall is 3 mm. Estimate the hoop stress in the pipe wall due to water hammer pressure when the 14.11 A cast iron pipe 30 cm in diameter and 8 mm thick is 1500 m long. The pipe is to convey 200 L/s of water. (a) Estimate the maximum time of closure of a valve at the downstream end that would be reckoned as rapid closure. (b) What is the peak water hammer pressure produced by rapid closure. (c) What is the length of the pipe subjected to peak water hammer pressure at the and of the valve closure time if the time of closure is 2.0 s? [For Water : K = 2200 MPa; for cast iron: E = 80 ¥ 109 Pa] (Ans. (a) 2.88 s, (b) ph = 2.942 MPa, (c) x0 = 458 m) 14.12 A 20 cm steel pipe is 1500 m long and conveys 50 L/s of water with a static head of 200 m at the downstream end of the pipe. If a valve at the downstream end is closed in 3 s, estimate the stress in the 477 Unsteady Flow pipe wall at the valve. The pipe thickness is 6 mm. [For water: K = 2.20 ¥ 109 Pa; for steel: E = 2.11 ¥ 1011 Pa] (Ans. s = 0.059 kN/mm2) 14.13 A 2300 m long pipeline leading from a large tank has a diameter of 15 cm and thickness of 2.8 mm. When a discharge of 2200 L/min of water was ﬂowing the valve was suddenly closed completely. Sketch the variation of the water hammer pressure with time at (i) the valve end, (ii) a distance of 575 m from the valve and (iii) 57.5 m from the upstream tank. [Take K = 2.0 ¥ 109 Pa for water and E = 2.08 ¥ 1011 Pa for steel.] (Ans. Fig. 14.17) ph + 0 2 4 6 8 10 1214 1618 20 22 – Time seconds Case (i) ph + 12.5 15.5 4.5 7.5 0.523.5 6 8 10 12 14 16 18 20 22 8.5 – 11.5 Time in seconds Case (ii) ph + 2.05 6.05 10.05 01.955.95 9.95 – Time in seconds Case (iii) Fig. 14.17 Answer to Problem 14.13 14.14 A steel pipeline 0.50 m in diameter and 3.0 km long discharge 250 L/s of water freely at its lower end under a head of 160 m. The thickness of the pipewall is 6 mm. A valve at the downstream end is used to regulate the ﬂow. If the working stress in the steel is 0.11 kN/mm2, calculate the minimum time of complete closure of the valve. [For water, K = 2.2 ¥ 109 Pa and for steel, E = 2.1 ¥ 1011 Pa.] (Ans. Tmin = 7.09 s) 14.15 A steel pipe 45 cm in diameter conveys water at a velocity of 0.90 m/s. The pipe is 2000 m long and has a static head of 150 m at a valve provided at tits downstream end. The valve can be expected to be closed rapidly and the pipe thickness has to be designed to include the contingency. Estimate the minimum thickness of the pipewall to the nearest millimetre. [For steel: E = 2.21 ¥ 1011 Pa and safe working stress s w = 0.12 kN/mm2. For water: K = 2.05 ¥ 109 Pa.] (Ans. t = 5 mm) 14.16 A steel pipe 3000 m long has a diameter of 0.90 m. It is a carry 0.6 m3/s of oil of density 800 kg/m3 with a static pressure of 900 kPa at the outlet valve. A valve at the outlet is designed to close completely in 6 s. Estimate the minimum thickness of pipe, to the nearest mm, required at the valve. [For steel E = 2.05 ¥ 1011 Pa and working stress s w = 0.125 kN/mm2; for oil K = 1500 MPa.] (Ans. t = 6 mm) Establishment of Flow 14.17 Two reservoirs with a constant difference of 10 m in their water surface elevation are connected by a 15 cm diameter pipe of length 400 m and f = 0.025. The minor 478 Fluid Mechanics and Hydraulic Machines loses in the pipe can be taken as 15 times the velocity head in the pipe. If a valve controlling the ﬂow is suddenly opened, (a) estimate the time for 95% of ultimate ﬂow to be established and (b) ﬁnd the ﬂow at the end of 10 s from the start of the valve operation. (Ans. (a) 11.51 s (b) 92% of ultimate ﬂow) 14.18 A turbine is supplied with water through a 30 cm diameter pipe leading 1500 m from reservoir. The friction factor f for the pipe can be taken as 0.015 and all other losses can be neglected. If the rate of ﬂow during normal operation is 140 L/s, (a) estimate the minimum time required for the turbine the reach 98% of its capacity from fully shut off condition. (b) What is the corresponding time if the pipe can be assumed to be frictionless? (Ans. (a) 45.8 s (b) 3479 s) 14.19 A pipe 3000 m long and of 0.45 m diameter leads from a reservoir of water surface elevation of 120.00 m. The outlet of the valve is at an elevation of 100.00 m and discharges to atmosphere. If minor losses are 16 V 2/ 2g where V = velocity in the pipe and f = 0.018, (a) estimate the time, after sudden opening of the outlet valve, for the ﬂow to attain 90% of the ultimate value and (b) what is the velocity in the pipe after 20 s? (Ans. (a) 38.1 s (b) V = 1.097 m/s) 14.20 A 7.5 m diameter pipeline is 350 m long and discharges water from a large tank through a 3.0 cm nozzle into atmosphere at the outlet. If a valve at the outlet is suddenly opened and if the ultimate velocity head at the nozzle is 12 m, (i) estimate the time required for the ﬂow to attain 95% of its ultimate value. Assume f = 0.020 and minor losses = 15 V2/2g where V = velocity in the pipe. (ii) What would be the corresponding time if the friction and minor losses are neglected? (Ans. (i) 4.82 s (ii) 522.3 s) Objective Questions Surges in Canals 14.1 In a rectangular channel of depth 1.2 m and velocity 2.0 m/s, an elementary wave travelling upstream will have an absolute velocity of (a) 5.43 m/s (b) 3.43 m/s (c) 1.43 m/s (d) 2.0 m/s 14.2 A sluice gate controlling ﬂow in a canal is suddenly lowered by an amount to cause partial closure. This will produce. (a) a negative wave on the upstream (b) a positive surge on the downstream (c) a positive wave on the upstream (d) a standing wave on the downstream 14.3 A positive surge travels upstream in a canal with an absolute velocity Vw. With sufﬁxes 1 and 2 referring to sections upstream and downstream of the surge respectively, the continuity equation is written as (a) A1 V1 = A2 V2 (b) A1 (V1 + Vw) = A2 (V2 – Vw) (c) A2 (Vw – V2) = A1 (Vw + V1) (d) A1 (V1 + Vw) = A2 (V2 + Vw) 14.4 In a rectangular channel carrying a ﬂow with a depth of 1.2 m and velocity of 2.0 m/s, a gate on the downstream is suddenly closed. If a positive surge of speed 3.75 479 Unsteady Flow m/s travelling upstream is produced, the height of the surge is (a) 1.5 m (b) 0.25 m (c) 2.3 m (d) 0.8 m 14.10 14.11 Water Hammer 14.5 If the bulk modulus of water is 1.96 ¥ 109 N/m2 the water hammer wave velocity through a rigid pipe is (a) 448 m/s (b) 996 m/s (c) 4390 m/s (d) 1401 m/s 14.6 The velocity of pressure wave in a rigid pipe carrying a ﬂuid of density r and viscosity m varies as (a) r (b) r (c) r/m (d) 1/ r 14.7 In a pipe ﬂow the ratio of the bulk modulus of the ﬂuid and the coefﬁcient of elasticity of the pipe material is 1/100. The thickness of the pipe wall is 1/50 of the diameter of the pipe. The ratio of the velocity of propagation of a pressure wave C in this pipe to the velocity of sound in an inﬁnite medium of the ﬂuid C1 is given by C/C1 = (a) 0.667 (b) 0.816 (c) 0.500 (d) 0.732 14.8 The velocity of a pressure wave in water of inﬁnite extent is 1440 m/s, For a pipe with diameter = 40 cm, thickness = 4 mm, with E of the pipe material = 2.1 ¥ 1011 Pa and K for water = 2.1 ¥ 109 Pa, the velocity of propagation of water hammer pressure wave, in m/s, is (a) 1440 (b) 720 (c) 2036 (d) 1018 14.9 A pipe 1000 m long conveys a ﬂuid whose velocity of propagation of pressure wave is 1000 m/s. After a sudden closure of the downstream end valve, the peak water hammer pressure will exist for a 14.12 14.13 14.14 duration of (a) 2s (b) 1s (c) 4s (d) 0.5s A penstock is 3000 m long. Pressure wave travels in it with a velocity of 1500 m/s. If the gates of the turbine are closed uniformly and completely in 4 s, then the closure is called (a) rapid (b) slow (c) sudden (d) ultra-rapid A penstock is 2000 m long and the velocity of pressure wave in it is 1000 m/s. Water hammer pressure head for instantaneous closure of valve at the downstream end of the pipe is 60 m. If the valve is closed in 4 s, then the peak water hammer pressure is equal to (a) 15 m (b) 30 m (c) 60 m (d) 120 m In a pipe 2000 m long carrying oil, the velocity of propagation of the pressure wave is 1000 m/s. A valve at the downstream end is closed suddenly. At the midpoint of the pipeline, the peak water hammer pressure will exist for a duration of (a) 1.0 s (b) 4.0 s (c) 3.0 s (d) 2.0 s A valve at the downstream end of 2500 m long pipe is closed in 4 s. If the velocity of propagation of the pressure wave in this pipe is 1000 m/s, the length of the pipe subjected to peak water hammer pressure at the end of the closure time is (a) 1000 m (b) 500 m (c) 2000 m (d) 2500 m In an 800 m long pipe the velocity of propagation of pressure wave C = 960 m/s. If the peak water hammer pressure due to sudden closure of the ﬂow at the downstream end is 900 kPa, the peak water hammer pressure due to closure of 480 Fluid Mechanics and Hydraulic Machines 14.15 14.16 14.17 14.18 14.19 the same valve in 2.0 s is (a) 900 kPa (b) 750 kPa (c) 1080 kPa (d) 821 kPa A long pipeline 3840 m in length carries water. It has a velocity of propagation of pressure wave C = 960 m/s. When a valve at the downstream end of the pipe was closed in T seconds, 960 m of the pipeline from the valve end was subjected to peak pressure at the end of the closure time. The time of closure T is (a) 5s (b) 6s (c) 8s (d) 12s In a pipeline 1920 m long the velocity of propagation of pressure wave is 960 m/s. If a rapid closure of a downstream end valve is desired the largest time of closure is (a) 2s (b) 4s (c) 6s (d) 8s In a 2000 m long pipeline the velocity of pressure wave is 1000 m/s. If a ﬂow of water with a velocity of 0.8 m/s in this pipe is suddenly stopped completely in 1.5 s the resulting peak water hammer pressure in kPa is (a) 798.4 (b) 1197.6 (c) 532.2 (d) 81.5 A 3150 m long pipeline has a velocity of propagation of pressure wave C = 1050 m/s. If a ﬂow of water with a velocity of 1.2 m/s is stopped completely by a downstream valve in 8 s, the approximate peak water hammer pressure, in kPa, is (a) 1257 (b) 1656 (c) 943 (d) 1050 In a long pipeline at a downstream valve the normal static pressure head is 140 m. If the valve is suddenly closed, the expected water hammer pressure head is 80 m. If the pipe thickness is to be designed for this contingency, it 14.20 14.21 14.22 14.23 14.24 will have to be designed to withstand a pressure head of (a) 60 m (b) 80 m (c) 140 m (d) 220 m A surge tank is provided in hydropower schemes to (a) strengthen the penstocks (b) reduce water hammer pressure (c) reduce frictional losses in the system (d) increase the net head A downstream end valve in a long pipe connected to a water tank is suddenly opened. If t0 is the time to reach 95% of the ultimate ﬂow by neglecting friction and other losses and t1 is the corresponding time obtained by including friction and other losses, then (a) t1 > t0 (b) t1 = t0 (c) t1 ≥ t0 (d) t1 < t0 Indicate the incorrect statement: The time of establishment of 95% of the ultimate ﬂow due to sudden opening of a downstream valve in a pipeline (a) will increase if the pipe length is increased (b) will increase if the pipe friction is reduced (c) will decrease if the head of the reservoir is increased (d) will decrease if the pipe diameter is increased A pipe of length L leads from a large reservoir. When a downstream valve is suddenly opened, the time to attain 25% of the ultimate ﬂow is estimated as 2.18 s. Then, the time to attain 75% of the ﬂow would be. (a) 6.54 s (b) 3.78 s (c) 15.15 s (d) 8.30 s In a pipe leading from a reservoir the time for the establishment of ﬂow due to sudden opening of a downstream valve is 481 Unsteady Flow calculated. If the friction and other minor losses are neglected the time to attain 50% of the ultimate ﬂow is 1215 s. The corresponding time when there is a head loss, expressed as equal to k times the velocity head is 15 s. The value of the loss coefﬁcient k is (a) 81 (b) 80 (c) 9 (d) 10 14.25 The correct sequence, in the direction of ﬂow of water, for installation in a hydropower plant is (a) reservoir, surge tank, turbine, high pressure penstock (b) reservoir, penstock, surge tank, turbine, (c) reservoir, high pressure penstock, turbine, surge tank (d) reservoir, surge tank, high pressure penstock, turbine 14.26 Consider the following statements: A surge tank provided on the penstock connected to a water turbine (1) helps reducing the water hammer effect (2) stores extra water when needed (3) provides increased demand of water when needed Which of these statements are correct? (a) 1 and 3 (b) 2 and 3 (c) 1 and 2 (d) 1, 2 and 3 14.27 The function of a surge tank is to (a) avoid reversal of ﬂow (b) reduce the water hammer effect in the pipeline (c) prevent occurrence of mass oscillation of water (d) smoothen the ﬂow. Compressible 15 Introduction M = V/C M 483 Compressible Flow 15.1 THERMODYNAMIC PRINCIPLES The basic thermodynamic principles and properties of gases used in the brief treatment of this chapter are as follows: p = pv = RT r (15.1) where p = absolute pressure r = density (mass per unit volume) v = speciﬁc volume (= 1/r = volume per unit mass) R = gas constant T = absolute temperature in Kelvin. [Note: It is important to note that only absolute values are used in compressible ﬂuid ﬂow. Absolute values are measured above absolute zero. Gauge pressures and vacuum pressures must be ﬁrst converted to absolute values. For temperatures t°C = 273 + t Kelvin = T K. It is a convention not to indicate degree (°) in Kelvin. It is simply T kelvin or T K. (like for example 293 K).] = 8314 J/( kg ◊ K ) Mg p r R = cp - cv R k -1 k = pv k = constant (15.4) p1 = p2 r 2k ÊT ˆ Ê p ˆ and Á 2 ˜ = Á 2 ˜ Ë T1 ¯ Ë p1 ¯ ( k - 1) / k (15.5) (15.2) 15.2 BASIC DEFINITIONS Internal energy, u, is the energy of unit mass of ﬂuid due to molecular activity. Change of internal energy (15.3) where cp = speciﬁc heat at constant pressure and cv = speciﬁc heat at constant volume. The ratio cp /cv = k is an important thermodynamic property of a gas. For air and other diatomic gases k = 1.4. From Eq. 15.3 cv = 287 = 0.0596 kcal /(kg ◊K) 4812 Thus by Eqs 15.4 and 15.1, between two sections 1 and 2, r1k where Mg = molecular weight of gas. For air Mg = 28.97 and R = 287 J/(kg◊K). Further (15.4b) Table 15.1 gives the values of Mg, r, R, cv, cp and k for some common gases. (2) Another fundamental equation for a perfect gas is The gas constant R is given by R= kR k -1 Note that the units of R, cp and cv are J/(kg◊K). Since 1 Joule (J) = N◊m = (kg ◊m/s2) (m) the unit J/(kg ◊K) = m2/(s2 ◊K). If R is to be expressed in heat units, the relevant conversion is 1 kcal = 4812 J. Thus for air R = 287 J/(kg ◊K) (1) The gas is assumed to be a perfect gas. Equation of State: cp = and (15.4a) u2 - u1 = cv (T2 - T1 ) Thus Ê ∂u ˆ cv = Á Ë ∂T ˜¯ v = constant (15.6) (15.7) Enthalpy, h The energy possessed by a unit mass of a gas by virtue of its absolute temperature under which it exists is known as enthalpy. It is represented as a sum of pressure per unit mass (p/r) and internal energy 484 Fluid Mechanics and Hydraulic Machines Table 15.1 Properties of Common Gases at 1 atm. and 20°C Gas Mol. Weight Mg r kg/m3 CO2 29 44 1.205 1.84 CO 28 1.16 O2 N2 Cl2 CH4 He H2 32 28 71 16 4 2 1.33 1.16 2.946 0.668 0.166 0.0839 Formula Air Carbon dioxide Carbon monoxide Oxygen Nitrogen Chlorine Methane Helium Hydrogen per unit mass (u) as h = u + p /r (15.8) For a perfect gas the change in enthalpy Dh = cpDT (15.9) Ê ∂h ˆ cp = Á Ë ∂T ˜¯ p = constant Thus (15.10) Entropy, s Entropy is deﬁned as a measure of the availability of energy for conversion into mechanical work. The entropy change ds for a perfect gas is dp r dh = cp dT and Tds = dh – Putting Ú 2 ds = 1 Ú 2 1 cp s2 - s1 = cp ln (15.11) Ú 2 1 T2 p - R ln 2 T1 p1 cp J/(kg ◊ K) cV J/(kg◊ K) k 287 188 1003 858 716 670 1.40 1.28 297 1040 743 1.40 260 297 117 520 2077 4120 909 1040 461 2250 5220 14450 649 743 344 1730 3143 10330 1.40 1.40 1.34 1.30 1.66 1.40 A ﬂow in which entropy does not change is known as isentropic ﬂow, i.e. s2 – s1 = 0 For an isentropic ﬂow ÊT ˆ R p ln Á 2 ˜ = ln 2 T c p1 Ë 1¯ p p 2 Ê T2 ˆ = p1 ÁË T1 ˜¯ k /( k - 1) Êr ˆ = Á 2˜ Ër ¯ k (15.13) 1 Isothermal process is a process in which the temperature is constant p = constant r (15.14) Adiabatic process is a process in which there is no heat transfer to or from a gas. A frictionless adiabatic process is isentropic, and for such a process [from Eq. (15.13)], rT = p/R dT -R T R J/(kg ◊ K) dp p (15.12) p rk = constant (15.15) 485 Compressible Flow 15.3 BASIC EQUATIONS FOR COMPRESSIBLE FLUID FLOW (i) Continuity: In expansion, r2 < r1 and work done by gas is positive. If w is negative, as in compression, it means that work is done on the gas and heat is rejected. Mass rate of ﬂow (15.16) m� = r AV = constant In differential form d A d r dV + + =0 A r V p V2 +u+ + g Z = constant r 2 (15.17) h+ V2 = constant 2 (15.18) cp T + V2 = constant 2 (15.19) (15.20) 15.4 APPLICATION OF ENERGY EQUATION If a gas is expanded to perform work, then per unit mass, by energy equation Heat absorbed by gas = Work done by gas + Increase in internal energy. q = w + De (15.21) Isothermal process If the process is from state 1 to state 2 then in isothermal conditions, T1 = T2. Hence there will be no change in internal energy, De = 0. Thus Work done by gas per unit mass, w = Heat absorbed by gas per unit mass, q w = (p1/r1) ◊ ln Ú 2 pd v 1 r1 Êr ˆ = RT1 ln Á 1 ˜ r2 Ë r2 ¯ (15.22) Ú 2 pd v 1 ( p1 /r1) - ( p2 /r2 ) R (T1 - T2) = ( k - 1) ( k - 1) = cv (T1 – T2 ) (15.23) Note that in isentropic expansion there will be a decrease in internal energy and in isentropic compression there will be an increase in internal energy. w = 15.5 Ê k ˆ p V2 ÁË k - 1˜¯ r + 2 = constant Also the work done/kg by gas = showing that the work done in expansion is at the expense of the internal energy of the gas. Work done/kg by gas = In gas ﬂow, the Z terms are negligible and hence, i.e. w = - D u = (e1 - e2 ) (15.16a) (ii) Energy equation for isentropic ﬂow: If no heat is added and no mechanical work is done, i.e. Isentropic process: In an isentropic process q = heat absorbed by unit mass of gas = 0. Hence, SONIC VELOCITY The sonic velocity, i.e. the velocity of sound is the speed of propagation of a pressure wave in the medium and is given for an isentropic ﬂow as C = k p/r = kRT (15.24) The ratio of the velocity of ﬂow to the sonic speed is known as Mach number M. Thus M = V /C (15.25) In supersonic ﬂow, if a point source of disturbance is present, it creates a conical space in which all the disturbances are piled up. The boundary of the cone is a shock where there is a sudden change in the ﬂuid properties like density and pressure. Figure 15.1 is a schematic diagram of a supersonic ﬂow moving past a disturbance. OAA¢ is the trace of a cone of disturbance, and this conical zone is known as Mach cone. The semi vertex angle AOB = a is the Mach angle. The Mach angle a is given by 486 Fluid Mechanics and Hydraulic Machines 15.5.2 Effect of Area Variation For Steady, one-dimensional, isentropic ﬂow the Euler equation is A VdV + C V>C O B a Ma ch line (Sh d A d r dV + + =0 A r V and noting that C 2 = dp/dr and M = V/C the following important relation between variation of area and Mach number is obtained. A¢ k li ne ) Fig. 15.1 sina = 1 M When the ﬂow velocity V = 0, the corresponding values of other parameters, viz. pressure, temperature and density are known as stagnation values and are designated with sufﬁx 0. Thus stagnation pressure = p0, stagnation temperature = T0 and stagnation density is r0. Stagnation values are important reference parameters in gas dynamics. The energy equation [Eq. 15.19] becomes and hence V12 = c pT0 2 (as V0 = 0) V12 = 2cp (T0 - T1) (15.27) The ratios of pressure p1, density r1, and temperature T1 with their corresponding stagnation values p0, r0 and T0 are expressed in terms of Mach number M1 (= V1/C ) and k as follows: T0 k - 1 2ˆ Ê = Á1 + M1 ˜ Ë ¯ 2 T1 (15.28) k /( k - 1) p0 ÊT ˆ k - 1 2ˆ = ÊÁ1 + = Á 0˜ M1 ˜ p1 Ë ¯ 2 Ë T1 ¯ 1/( k - 1) r0 ÊT ˆ k - 1 2ˆ = ÊÁ1 + = Á 0˜ M1 ˜ r1 Ë ¯ 2 Ë T1 ¯ dV d A Ê 1 ˆ dp = =Á ˜ 2 V A Ë M - 1¯ rV 2 (15.26) Stagnation Values cpT1 + (15.31) Combining with the continuity equation [Eq. 15.16 (a)] oc 15.5.1 dp =0 r k /( k - 1) (15.29) 1 /( k - 1) (15.30) (15.32) An analysis of this equation indicates the qualitative relationship between M, duct geometry and other ﬂow parameters as in Table 15.2. 15.6 15.6.1 FLOW IN A NOZZLE Discharging from a Tank Consider a tank containing a gas at pressure p0, temperature T0 and density r0 discharging through a converging nozzle. Since the velocity in the tank is zero for all practical purposes, the pressure p0 temperature T0 and density r0 are all stagnation values. Let the pressure outside the nozzle (ambient pressure) be p2 (Fig. 15.3). Let p1, V1, r1 and T1 are the values of pressure, velocity, density and temperature at the throat (exit) of the nozzle. These values, which depend on the nature of ﬂow are controlled by a critical pressure ratio p1 /p0 given by the following relationship. p1 È 2 ˘ = Í ˙ p0 Î ( k + 1) ˚ k /( k - 1) = 0.528 when k = 1.4 (15.33) 487 Compressible Flow Table 15.2 Compressible Flow in Converging and Diverging Ducts Geometry Subsonic M1 dV > 0, dp < 0 Along the ﬂow: dV < 0, dp > 0 Along the ﬂow: subsonic nozzle. supersonic diffuser. Converging, dA < 0 Fig. 15.2(a) Coverging Duct Subsonic M1 dV > 0, and dp < 0 Along the ﬂow: dV < 0, dp > 0 Along the ﬂow: subsonic diffuser supersonic nozzle. Diverging, dA < 0 Fig. 15.2(b) Diverging Duct (Note the opposing behaviour of subsonic and supersonic ﬂows in a given geometry.) Tank p0 r0 T0 V1 = 0 Fig. 15.3 Ambient p2 1 Nozzle Discharging from a Tank This critical pressure p 1 corresponds to occurrence of sonic ﬂow at the throat. Two types of ﬂow are possible. (1) When p2 > p1: Subsonic ﬂow prevails through the nozzle and the nozzle exit pressure p1 = p2. The relationship of the various ﬂow parameters will be as follows: When p2 > p1, p1 = p2 and p0 È k - 1 2˘ = Í1 + M1 ˙ p1 2 Î ˚ Ê k ˆ ËÁ k - 1¯˜ (15.34) 488 Fluid Mechanics and Hydraulic Machines T0 k - 1 2ˆ Ê = Á1 + M1 ˜ Ë ¯ 2 T1 r0 k - 1 2ˆ Ê M1 ˜ = Á1 + Ë ¯ 2 r1 The exit velocity (15.35) k /( k - 1) (15.36) m� = A2 r1 kRT ( k - 1) / k ˘ Ê k ˆ p0 È Ê p1 ˆ Í ˙ 2Á 1 Ë k - 1˜¯ r0 Í ÁË p0 ˜¯ ˙ Î ˚ (15.37) Mass rate of ﬂow . m = A1r1V1 (15.38) V1 = m� = A1r1 ( k - 1) / k ˘ Ê k ˆ p0 È Ê p1 ˆ Í ˙ 2Á 1 Ë k - 1˜¯ r0 Í ÁË p0 ˜¯ ˙ Î ˚ (15.38a) (2) When p 1 ≥ p2: Sonic conditions (M1 = 1.0) prevail at the nozzle exit and p1 = p 1. The mass rate of ﬂow . m = A1r 1 V (15.39) where the parameters with superscript represent critical condition. This mass rate . m is the maximum mass rate of ﬂow that the nozzle can discharge under the given upstream conditions. The values of critical ﬂow parameters are p1 Ê 2 ˆ = Á p0 Ë k + 1˜¯ k /( k - 1) = 0.528 for k = 1.4 (15.40) Ê 2 ˆ T1 = Á = 0.833 T0 Ë k + 1˜¯ for k = 1.4 (15.41) r1 Ê 2 ˆ = Á r0 Ë k + 1˜¯ 1 /( k - 1) = 0.634 for k = 1.4 (15.42) V 1 = C 1 = kRT 1 p1 < 0.528 critical ﬂow p2 prevails in the nozzle and the maximum mass rate will pass through the nozzle. Also Thus for air, when (15.43) (15.44) The condition of occurrence of critical ﬂow at the nozzle exit is known as choking condition. 15.6.2 Converging–Diverging Nozzle In a converging nozzle the subsonic ﬂow is similar to that from a reservoir where the gas is initially at rest. The exit area is the minimal cross section area. The maximum mass discharge corresponds to the choking condition at which situation the Mach number is unity. If supersonic ﬂow at exit is desired the Mach one ﬂow will have to be suitably expanded to achieve the desired exit Mach number. A converging–diverging nozzle does this. The ﬂow conditions at the exit of a converging–diverging nozzle depend upon the back pressure that exists at the exit section. To appreciate this aspect, consider the ﬂow through a converging–diverging nozzle for different back pressures as in Fig. 15.4. In the converging–diverging nozzle shown in the Fig. 15.4, sufﬁx 0 denotes the stagnation conditions which is prevalent at the inlet and pb is the back pressure at the exit of the divergent part of the nozzle. The ﬂow pattern is analyzed for various values of pb indicated by letters A, B, C, D, E and F in Fig. 15.4, each representing a speciﬁc case. 1. Case A: When p0 = pb, there is no ﬂow in the system 2. Case B: When p0 > pb > pc (where pc = back pressure at Case C depicted in Fig. 15.4): Subsonic ﬂow occurs in the converging nozzle as well as in the diverging nozzle. The ﬂow is accelerated in the converging nozzle and decelerated in the diverging part. The diverging nozzle acts as a diffuser. 489 Compressible Flow Inlet Exit Throat Pe P0 Vi @ 0 Pb x P Pb A B P0 C D P PA PB PC PD Sonic flow at throat Shock in nozzle 0 Inlet Throat Shock in nozzle M PE PF E, F Subsonic flow at nozzle exit (no shock) Subsonic flow at nozzle exit (shock in nozzle) Supersonic flow at nozzle exit (no shock in nozzle) x Exit E, F Sonic flow at throat I D C 0 Inlet Fig. 15.4 B A Throat Exit x Characteristics of Convergent–Divergent Nozzle (Ref. 15.5) Case C: When p0 > (pb = pc): Critical condition is reached at the throat section. Mach number at the throat is unity. The ﬂow is still subcritical in the diverging nozzle. p0 > (pb > pE), (where pE = back pressure at Case E depicted in the ﬁgure and is the design pressure value of the nozzle): The ﬂow is accelerated in the diverging nozzle and the supersonic ﬂow may have a normal shock at some section in the divergent nozzle. The ﬂow is no more isentropic beyond the shock. The ﬂow beyond the shock will be subsonic. The throat will continute to be in critical state. 490 Fluid Mechanics and Hydraulic Machines Case E: When p0 > (pb = pE): The ﬂow is supersonic throughout the diverging nozzle with no shock anywhere in it. The throat will continue to be in critical state. This is the design condition. 6. Case F: When p0 > Pb and (pb < pE): Supersonic ﬂow exists in the divergent section, but expansion shocks occur outside the nozzle exit. Flow is not isentropic beyond exit. Thus the only relevant conditions for analysis of convergent divergent nozzles are: (i) Case E where the design backup pressure exists and supersonic ﬂow prevails all over the divergent nozzle. Further, the ﬂow is isentropic throughout. Here, critical condition prevails at the throat and thus M = 1 and p, T and r occur at the throat. (ii) Case B where p0 > pb > pc the divergent nozzle acts as a diffuser and the ﬂow is subsonic in the divergent portion. The throat is not in critical state. (iii) Case C where p0 > (pb = pc): The divergent nozzle acts as a diffuser and the ﬂow is subsonic in the divergent portion. However, the throat is in critical state with Mthroat = 1. In cases E and C, the throat is under critical condition and the isentropic ﬂow prevails all over from throat to the exit. For these two cases, the throat area and any area in the nozzle are uniquely related to the local Mach number as Ê k +1 ˆ Á ˜ ÈÊ 2 ˆ Ê ( k - 1) ˆ 2 ˘Ë 2( k - 1) ¯ 1 + M Í ˙ Á ˜Á A 2 ˜¯ ÎË k + 1¯ Ë ˚ (15.45) Eq. 15.45 for k = 1.4 simpliﬁes as 3 1 A È(1 + 0.2 M 2 ) ˘ º (15.45-a) = Î ˚ 1.728 M A In this A throat area and A = area of the diverging nozzle where the Mach number is M. Note that Eq. 15.45 is applicable only when the throat is under critical condition and isentropic ﬂow prevails all over the divergent section. A = 1 M The convergent–divergent nozzle with back pressure equal the design state is a well used method of creating supersonic stream of gas. Such nozzles are standard components in rocket engines and supersonic aircrafts. The convergent–divergent nozzle was ﬁrst developed by a Swedish engineer Carl G.B. de Laval and in honor of him convergent– divergent nozzles are often called Laval nozzles. In the analysis of Laval nozzle ﬂow, equations 15.34 through 15.44 are used along with appropriate boundary conditions. Examples 15.26 through 15.29 illustrate the analysis procedure. 15.6.3 Compressibility Effects on Pitot Static Tube In incompressible ﬂow, the Pitot-static tube measures the difference between the stagnation pressure p0 and the static pressure p1, that is (p0 – p1) of the ﬂow. When the compressibility effects are ignored ( p0 - p1 ) = 1. (15.46) 1 2 rV1 2 However, when the compressibility effects are included, for an isentropic process, from Eq. (15.34) Ê k ˆ p0 È ( k - 1) 2 ˘ÁË k - 1˜¯ = Í1 + M ˙ 2 p1 Î ˚ In this sufﬁx 0 denotes stagnation values and the sufﬁx 1 denotes local values. For subsonic ﬂow, this equation could be expanded by Binomial theorem for (M2 < 1), as È ˘ p0 - p1 M 2 (2 - k ) 4 = Í1 + + M +º˙ (15.47) 1 2 4 24 ÍÎ ˙˚ rV1 2 It is seen from the comparison of Eq. (15.47) with Eq. (15.46) that the pressure difference (p0 – p1) for a given ﬂow becomes large as the value of Mach number in subsonic ﬂow increases. The factor È ˘ M 2 (2 - k ) 4 + M +º˙ Í1 + 4 24 ÍÎ ˙˚ is called as the compressibility correction factor (CCF). Usually the ﬁrst two terms are sufﬁcient and 491 Compressible Flow Iso-energetic T01 = T02 Normal shock p2 r2 M1 > 1 V2 T2 Isentropic u/s s = s1 Isentropic d/s s = s2 Fig. 15.5 as such the common practice is to write È M2 ˘ CCF = Í1 + (15.48) ˙ 4 ˙˚ ÍÎ Thus in subsonic ﬂow, the Pitot–static tube can be used to measure the velocity of ﬂow adequately by using the relation 2 1 ( p0 - p1 ) rÈ M2 ˘ ˙ Í1 + 4 ˙˚ ÍÎ Perfect gas relationship: (15.50) ÈÊ p ˆ Ê r ˆ k ˘ s2 - s1 = cv ln ÍÁ 2 ˜ Á 1 ˜ ˙ ÍË p1 ¯ Ë r2 ¯ ˙ Î ˚ (a) Mach number relation, M 22 = (15.51) Note that this relationship is valid for subsonic ﬂow only. 15.7 (15.52) The relationships between the various ﬂuid and ﬂow parameters on either side of the shock are obtained as If the instrument has a Cd value less than unity, then Eq. (15.49) is modiﬁed to accommodate Cd as V1 = Cd p1 p2 = r1T1 r 2T2 It is found that the entropy changes across the shock and the change is given by (15.49) The local velocity of ﬂow is given by V1 = V 22 V 12 + h2 + h1 = 2 2 = h 0 = constant = cpT0 (by considering no heat transfer and no work done) Energy: Normal Shock Weve È p0 - p1 M2 ˘ = Í1 + ˙ 1 4 ˙˚ ÎÍ rV12 2 shock. Such a shock wave is known as normal shock wave. This shock wave will have very little thickness and is analogous to the hydraulic jump that takes place in an open channel ﬂow. Considering a section 1 on the upstream and a section 2 on the downstream of the shock and noting that A1 = A2 = A (as the thickness of the shock is very small) the controlling equations for adiabatic ﬂow are . Continuity: m = r1V1 = r 2V2 Momentum: p2 – p1 = r 1V 21 – r 2 V22 NORMAL SHOCK WAVE In a supersonic stream, under certain conditions, a shock wave normal to the ﬂow direction may occur and the ﬂow may change into subsonic state after the and M12 = 2 + ( k - 1) M 12 2 kM12 - ( k - 1) 2 + ( k - 1) M 22 2 kM 22 - ( k - 1) (15.53) (15.54) (b) Pressure ratio, p2 1 + kM12 2 kM12 - ( k - 1) = = p1 1 + kM 22 ( k + 1) (15.55) 492 Fluid Mechanics and Hydraulic Machines Note the following characteristics of a normal shock: M12 ( k + 1) r2 = r1 2 + M12 ( k - 1) (15.56) downstream ﬂow is subsonic. V2 1 2 + M12 ( k - 1) = = V1 ( r2 /r1) M12 ( k + 1) the same across the shock and hence all over the ﬂow. (15.57) (e) Temperature: Stagnation temperature constant across the shock. Thus, density decrease with M1 in the same ratio, viz. is r02 p02 = r01 p01 (15.58) T01 = T02 (f) Stagnation pressure ratio and density ratio, p02 r02 È ( k + 1) M 12 ˘ = =Í ˙ p01 r01 ÍÎ 2 + ( k - 1) M 12 ˙˚ k /( k - 1) consequent decrease in stagnation pressure and stagnation density across the shock. ¥ 1 /( k - 1) È ˘ k +1 Í ˙ 2 ÍÎ 2 kM 1 - ( k - 1) ˙˚ (15.59) Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Simple Medium Worked Examples A. Thermodynamic Properties For carbon dioxide, Mg = 44 8314 = 189 N◊m/kg◊K) 44 Mg = 32 Hence 15.1 R For oxygen, Solution: Gas constant R = 8314 32 = 260 N◊m/(kg ◊K) Hence 8314 Molecular weight of gas 8314 = Mg R = R= 15.2 k ◊ ◊ cp cv for air in 493 Compressible Flow Solution: Gas constant R= = 4373,226 J = 4373 kJ 8314 8314 = Mg 28.96 = 287 J/(kg ◊ K) 15.4 - k 1.4 R= ¥ 287 k -1 (1.4 - 1.0) = 1004.5 J/(kg ◊ K) cp = 287 R cv = = = 717.5 J/ (kg ◊ K) k - 1 (1.4 - 1.0) In heat units, 1 kcal = 4187 J 1004. 5 = 0.240 Kcal/(kg◊ K) 4187 717. 5 cv = = 0.171 kcal/(kg ◊K) 4187 \ cp = 15.3 ◊ cv ◊K) (1) Change in internal energy per unit mass, Du = cv (T2 – T1) = 670 ¥ 50 = 33,500 J/kg Total change in internal energy = mDu = 101.91 ¥ 33, 500 = 3414,880 J = 3415 kJ (2) Change in enthalpy per unit mass, Dh = cp(T2 – T1) = 858 ¥ 50 = 42,900 J/kg Total change in enthalpy = mDh = 101.91 ¥ 42,900 cp pv k = constant For a perfect gas, pv = RT pv k = (pv) (v k–1) = (RT) v k–1 R = constant, T v k–1 = constant Since T1v1k–1 = T2v 2k–1 T1 = 25 + 273 = 298 K v1/v2 = 1.0/(0.5) = 2.0 1000 9.81 = 101.94 kg T1 = 273 + 15 = 288 K T2 = 273 + 65 = 338 K (T2 – T1) = (338 – 288) = 50 K cp 909 = = 1.4 649 cv m = mass of gas = 5.0 kg For an isentropic process, k = Solution: m = Mass of 1000 N of CO2 = ◊ Solution: Hence cp cv ◊ k -1 (1) (2) Since Êv ˆ T2 = T1 Á 1 ˜ = 298 ¥ (2)(1.4 –1.0) Ë v2 ¯ = 393.2 K = 120.2°C p2 v2 p1v1 = T2 T1 p2 = p1 (v1/v2) (T2/T1) Ê 393.2 ˆ = 100 (2.0) Á Ë 298 ˜¯ = 263.9 kPa (abs) (3) Change in internal energy per unit mass, Du = cv (T2 – T1) = 649 (393.2 – 298) = 61784.8 J/kg Total change of internal energy = Work required 494 Fluid Mechanics and Hydraulic Machines 1 kJ 1000 = 308.9 kJ = Work done on the gas (4) Change in enthalpy per kg: = mDu = 5.0 ¥ 61784.8 ¥ Dh = cp (T2 – T1) = 909 ¥ (393.2 – 298) = 86537 J/kg Total change of enthalpy = mDh = 5.0 ¥ 86537 ¥ = 432.7 kJ 1 kJ 1000 Solution: T1 = 273 + 250 = 523 K = T2 1000 ¥ 1000 p1 = 287 ¥ 523 RT = 6.662 kg/m3 Initial density r1 = Initial volume V1 = m/r = 3.0 = 0.450 m3 6.662 (i) In an isothermal process p1 p2 = as T1 = T2 r1 r2 p2 200 r1 = p2 = ¥ 6.662 1000 r2 15.5 = 1.3324 kg/m3 Final volume R= ◊ k Solution: R 260 = = 650 J/(kg ◊ K) k - 1 (1.4 - 1.0) and T2 = 85 + 273 T1 = 27 + 273 = 300 K = 358K p1 = 150 kPa (abs) and p2 = 450 kPa (abs) Change in entropy per kg: cv = ÈÊ T ˆ k Ê p ˆ k -1 ˘ s2 – s1 = cv ln ÍÁ 2 ˜ Á 1 ˜ ˙ ÍË T1 ¯ Ë p2 ¯ ˙ Î ˚ ÈÊ 358 ˆ 1.4 Ê 150 ˆ (1.4 -1.0 ) ˘ ˙ = 650 ln ÍÁ ˜ Á ˜ ÍÎË 300 ¯ Ë 450 ¯ ˙˚ = –124.8 J/K per kg Total change in entropy = m(s2 – s1) = 7 ¥ (– 124.8) = – 873.6 J/K 15.6 p Work done 6.662 1.3324 = 724700 J = 724.7 kN◊m Final temperature T2 = 532 K = 250°C (ii) In isentropic expansion = 3.0 ¥ 287 ¥ 523 ln p1 r1k or = p2 r2k Êp ˆ r2 = Á 2 ˜ Ë p1 ¯ k R 1/ k Ê 200 ˆ r1 = Á Ë 1000 ˜¯ 1 / 1.4 ¥ 6.662 = 2.1103 kg/m3 Final volume 3.0 = 1.42 m3 2.1103 Final temperature V2 = Êp ˆ T2 = Á 2 ˜ Ë p1 ¯ T ◊ 3.0 = 2.252 m3 1.3324 r W = m RT1 ln 1 r2 V2 = ( k -1) / k Ê 200 ˆ = Á Ë 1000 ˜¯ = 57.2°C T1 0.4 / 1.4 ¥ 523 = 330.2 K 495 Compressible Flow Work done per unit mass = cp (T1 – T2) R (T1 – T2) k -1 287 = (523 – 330.2) = 138334 J/kg 0.4 W = Total work done = m cp(T1 – T2) = 3.0 ¥ 138334 = 415002 J = 415 kJ = 415 kN◊m 15.9 = Solution: Stagnation pressure ratio = k /( k - 1) È ( k - 1) M12 ˘ p0 = Í1 + ˙ p1 2 ÍÎ ˙˚ Here p1 = 35 kPa, p0 = 65.4 kPa, k = 1.4. Substituting these, the pressure ratio = B. Speed of Sound & Mach Number 15.7 ◊ R È (1.4 - 1) M12 ˘ 70 = Í1 + ˙ 35.0 2 ÍÎ ˙˚ 2 3.5 2.0 = [1 + 0.2M 1] k Solution: C= [1 + 0.2M 21] = 1.219 k = 1.40, R = 260 J/kg ◊K T = 273 + 25 = 298 K and C= kRT and hence 1.4 ¥ 260 ¥ 298 = 329.4 m/s M 21 = 1.095 M1 = 1.0465 Temperature = –38 + 273 = 235 K C = 15.8 R Solution: Temperature = –50°C = –50 + 273 = 223 K C= = Mach Number 15.10 kRT 1.4 ¥ 287 ¥ 223 = 299.3 m/s V M= = 2.0 C V = MC = 2.0 ¥ 299.3 = 598.6 m/s 598.6 ¥ 3600 1000 = 2155 km/hour = kRT = 1.4 ¥ 287 ¥ 235 = 307.3 m/s V = MC = 1.0465 ¥ 307.3. = 321.6 m/s M k (1.4 / 0.4 ) Solution: C= = Speed of place k ( p / r) 1.4 ¥ ( 20000 / 0.32) = 295.8 m/s V = MC = 1.80 ¥ 295.8. = 532.44 m/s 532.44 ¥ 3600 = 1000 = 1916.8 km/h 496 Fluid Mechanics and Hydraulic Machines 15.11 k R ◊ R k Solution: Solution: Temperature = – 40°C = –40 + 273 = 233 K. C = kRT = 1.4 ¥ 287 ¥ 233 = 306 m/s 2160 ¥ 1000 V= = 600 m/s 3600 V 600 Mach number M = = = 1.96 C 306 Mach angle a is given by the relation 1 C sin a = = M V In the present case, 306 sin a = = 0.51 giving 600 a = 30.66° V1t A B a 1000 m 15.12 F – k Fig. 15.6 1 1 = M 2.2 a = 27.036° R sin a = Solution: Temperature = – 30°C = –30 + 273 = 243 K. C = kRT = 1.4 ¥ 287 ¥ 243 = 312.5 m/s 1 C If Mach angle = a, then sin a = = M V In the present case a = 40°. Hence 312.5 sin 40° = 0.643 = V 312, 5 Speed of plane = V = = 486.2 m/s 0.643 C = kRT = 1.4 ¥ 287 ¥ ( 273 + 22) = 344.3 m/s V C M = 344.3 ¥ 2.2 = 757.4 m/s Considering the Fig. 15.6, when the plane is at B the sonic boom reaches F. If angle ABF = a then 15.13 - Vt 1000 AF = tan( 27.036∞) tana = 1959.6 m t = time elapsed in the boom reaching the AB = AB V 1959.6 = = 2.59 s 757.4 point F = 15.14 497 Compressible Flow V 337.8 Speed of plane V = 1.867 ¥ 337.8 = 630.7 m/s 1.867 = ◊ R 60 ¥ 60 1000 = 2271 km/h = 630.7 ¥ Solution: As the value of k is not given, k = 1.4 for air is assumed. Sonic velocity kRT C= C. Stagnation Properties 1.4 ¥ 287 ¥ ( 273 + 11) = 15.15 = 337.8 m/s In Fig 15.7, B is the location of the plane when the sonic boom is heard at 0. ˆ =a A BO where sin a = 1 M Ct. ˆ = (90 – a) AOD \ AD Ct = = sin (90 – a) AO (AO) = cos a = (1 – sin2 a)1/2 p r k R ◊ Solution: (1) Stagnation pressure p0 is given by p0 È k -1 2˘ = Í1 + M1 ˙ 2 p Î ˚ k /( k -1) 1.4 /(1.4 -1) 1 ˆ Ê = Á1 ˜ Ë M2¯ 1/ 2 È (1.4 - 1.0) 2 ˘ 70 M1 ˙ = Í1 + 2 40 Î ˚ 1 ˆ 337.8 ¥ 7.5 Ê = Á1 - 2 ˜ Ë 3000 M ¯ 1 1 - 2 = (0.8445)2 M M = 1.867 1/ 2 1.5909 = [1+ 0.2 M 21]3.5 M 21 = 0.7093 and M1 = 0.8422 Sonic speed Mach number M = C = = 312.7 m/s Mach number M1 = V1/C V C V1t A B a ct 3000 m D O = observer Fig. 15.7 k p/r = (1.4 ¥ 44000)/ 0.63 Plane V1 = C M1 = 312.7 ¥ 0.8422 = 263.4 m/s (2) Temperature of the atmosphere, T1 = p/rR Ê 44000 ˆ T1 = Á = 243.35 K Ë 0.63 ¥ 287 ˜¯ Stagnation temperature T0 is given by T0 È k -1 2˘ = Í1 + M1 ˙ T1 2 Î ˚ 498 Fluid Mechanics and Hydraulic Machines È Ê 0.4 ˆ T0 2˘ = Í1 + Á ˜¯ (0.8422) ˙ Ë 2 243.35 Î ˚ = 1.1419 T0 = 277.87 K = 4.87°C 15.16 1.4 /(1.4 -1.0 ) È (1.4 - 1.0) ˘ p0 (1.7505) 2 ˙ = Í1 + 2 ˚ 28 . 5 Î = (1.61285)3.5 = 5.328 p0 = 151.85 kPa (abs) (3) Stagnation density r0 p0 151850 = RT0 287 ¥ 364.8 = 1.450 kg/m3 r0 = k ◊ R 15.17 Solution: Sonic speed C= = k p/r : k 1.4 ¥ 28500 0.439 = 301.5 m/s 1900 ¥ 1000 = 527.8 m/s 3600 527.8 Mach number M = V/C = = 1.7505 301.5 For the atmosphere: p1 = 28500 Pa (abs) r1 = 0.439 kg/m3 Speed of plane V = Temperature T1 = p1/r1R = 28500 0.439 ¥ 287 = 226.2 K (1) Stagnation temperature T0 T0 È k - 1 2˘ M1 ˙ = Í1 + T1 2 Î ˚ È (1.4 - 1.0) ˘ T0 ¥ (1.7505) 2 ˙ = Í1 + 2 Î ˚ 226.2 = 1.61285 T0 = 364.8 K = 91.8°C (2) Stagnation pressure p0 p0 È k - 1 2˘ M1 ˙ = Í1 + 2 p1 Î ˚ k /( k -1) R ◊ Solution: Speed of sound in CO2 =C= kRT = 1.28 ¥ 188 ¥ ( 273 + 30) = 270 m/s Velocity V1 = 150 m/s Mach number 150 = 0.5555 270 Stagnation temperature T0: M1 = V1/C = È Ê k - 1ˆ 2 ˘ T0 = Í1 + ÁË 2 ˜¯ M 1 ˙ T1 Î ˚ È (1.28 - 1) ˘ T0 (0.5555) 2 ˙ = Í1 + 2 ( 273 + 30) Î ˚ = 1.0432 T0 = 316 K = 43°C Stagnation pressure p0: k /( k -1) p0 È k -1 2˘ M1 ˙ = Í1 + 2 p1 Î ˚ 1.28 /(1.28 -1.0 ) p0 È (1.28 - 1.00) ˘ (0.5555) 2 ˙ = Í1 + 2 500 Î ˚ = (1.0432)4.57143 = 1.2133 p0 = 606.65 kPa (abs) 499 Compressible Flow Solution: 15.18 Referring to Fig. 15.8. T1 = 273 – 17 = 256 K 1 cp 2 T1 p1 [Stagnation conditions exist at 2] T2 p2 = p 0 r1 V1 r2 V2 = 0 Fig. 15.8 Solution: Given Sonic speed C1 = 800 ¥ 1000 = 222.2 m/s 3600 T = 217 K and p = 12.06 kN/m2 The maximum possible temperature and pressure on the airplane skin correspond to stagnation temperature and pressure respectively. = V = 800 km/h = (i) T0 = T1 + k /( k -1) p0 k - 1 2ˆ Ê = Á1 + M1 ˜ p1 Ë ¯ 2 Ê 1.4 - 1.0 2 ˆ M1 ˜ = Á1 + Ë ¯ 2 ( k / k - 1) and 75 = (1+ 0.2 M 21 )3.5 50 M 21 = 0.6141 M1 = 0.7837 Since M1 = 1.4 / 0.4 Ê 241.56 ˆ = 1.206 ¥ 104 ¥ Á Ë 217 ˜¯ = 12060 ¥ 1.4554 = 17552 N/m2 = 17.552 kN/m2 C = kRT = 1.4 ¥ 287 ¥ 217 = 295.28 m/s Mach number of the ﬂight = V 222.2 M= = = 0.753 C 295.28 1.4 ¥ 287 ¥ 256 = 320.72 m/s The stagnation pressure p0 is related to free stream pressure p1 as V02 ( 222.2) 2 = 217 + = 241.56 K 2 ¥ 1005 2c p ÊT ˆ (ii) p0 = p1 Á 0 ˜ Ë T1 ¯ kRT1 1.4 /(1.4 -1) V1 , the speed of plane V1 is C1 V1 = C1 M1 = 320.72 ¥ 0.7837 = 251.3 m/s 15.20 D. Pitot-Static Tube 15.19 k R k R ◊ Solution: (i) ◊K)] T0 = 273 + 30 = 303 K 500 Fluid Mechanics and Hydraulic Machines For isentropic ﬂow with subscript zero denoting stagnation values Ê p ˆ T1 = T0 Á 1 ˜ Ë p0 ¯ = ( k -1) / k Ê 50 ˆ = 303 Á ˜ Ë 95 ¯ (1.4 -1) / 1.4 Here = 252.2 K kR V12 (T – T1) = cp (T0 – T1) = ( k - 1) 0 2 2 kR V 21 = (T0 – T1) ( k - 1) 2 ¥ 1.4 ¥ 287 = (303 – 252.2) (1.4 - 1.0) = 102057.2 V1 = 319.5 m/s (ii) When compressibility effects are ignored: = 1.092 kg/m 1.4 /( 0.4 ) 1.8 = [1 + 0.2M 21]3.5 [1 + 0.2M 21] = 1.1829 and hence M 21 = 0.9143 M1 = 0.9562 Temperature = 15°C = 15 + 273 = 288 K C = kRT 1.4 ¥ 287 ¥ 288 = 340 m/s V = MC = 0.9562 ¥ 340 = 325.1 m/s (b) By considering the ﬂow as incompressible: = ( p0 - p) =1 1 2 r1V 1 2 2( p0 - p) \ V 21 = r1 2 ¥ (95 - 50) ¥ 1000 1.092 = 82418 V1 = 287.1 m/s = È (1.4 - 1) M12 ˘ 18 = Í1 + ˙ 2 10 ÍÎ ˙˚ Hence 95000 287 ¥ 303 3 k /( k - 1) È ( k - 1) M12 ˘ = Í1 + ˙ 2 ÍÎ ˙˚ po = stagnation pressure = 10.0 + 8.0 = 18.0 kPa p1 = static pressure = 10.0 kPa. Also r1 = r0 = p0/RT0 = p0 p1 r = p1 10.0 ¥ 1000 = RT 287 ¥ 288 = 0.121 kg /m3 V = 15.21 = 2Dp / r 2 ¥ (8 ¥ 1000) / 0.121 = 363.6 m/s E. Flow Through a Nozzle k 15.22 R Solution: (a) compressible ﬂow: Stagnation pressure ratio k R ◊ 501 Compressible Flow Solution: Inside the tank stagnation conditions prevail. Hence V0 = 0, and T0 = 273 + 35 = 308 K p0 = 250 kPa (abs) p1 95 = = 0.6333 p0 150 As this is larger than the critical value of 0.528 the ﬂow in the nozzle throat will be subsonic. At the nozzle throat (exit): V1 = 200 m/s V 12 150000 287 ¥ 313 = 1.6698 kg/m3 r0 = p0 /RT0 = = 2cp (T0 – T1) cp = kR 1.4 = ¥ 287 = 1004.5 k -1 0.4 r0 Êp ˆ = Á 0˜ r1 Ë p1 ¯ 2002 = 2009 (308 – T1) T1 = 288 K The Mach number M1 at the exit is given by 1/ k Ê p ˆ r1 = r0 Á 1 ˜ Ë p0 ¯ \ 1/k = 1.6698 (0.6333)1/1.4 = 1.205 kg/m3 T0 k - 1 2ˆ Ê = Á1 + M1 ˜ Ë ¯ 2 T1 V1 1/ 2 ( k -1) / k ˘ ¸ Ï Ê k ˆ p0 È Ê p1 ˆ Ô Í ˙ Ô˝ V1 = Ì2 Á 1 ˜¯ r Í ÁË p ˜¯ 1 k Ë ˙Ô 0 0 ÔÓ Î ˚˛ 1/ 2 Ï Ê 1.4 ˆ Ê 150, 000 ˆ 0.4 / 1.4 ¸ 2 1 0 6333 [ ( . ) ] =Ì Á ˝ ˜Á ˜ Ó Ë 0.4 ¯ Ë 1.6698 ¯ ˛ 308 = 1 + 0.2 M 21 288 M 21 = 0.3472 or M1 = 0.589 15.23 = {628.818 (1 – 0.87765)}1/2 = 2774 m/s k R ◊ A1 = area of nozzle exit = = 1.2566 ¥ 10–3 m2 Mass rate of ﬂow . m = r1 A1V1 = 1.205 ¥ 1.2566 ¥ 10–3 ¥ 277.4 = 0.42 kg/s Solution: Referring to Fig. 15.9, Tank p0 = 95 kpa (abs) T0 p2 = 95 kpa (abs) 1 r0 V0 = 0 p ¥ (0.04)2 4 15.24 p1 V1 T1 Fig. 15.9 k Temperature inside the tank = T0 = 273 + 40 = 313 K R ◊K)] 502 Fluid Mechanics and Hydraulic Machines Solution: Inside the tank T0 = 273 + 60 = 333 K p0 = 200,000 Pa (abs) r0 = [k = 1.66 and R = 2077 J/(kg◊K)] from a container tank to a receiver through a 2.0 cm convergent nozzle. The container has a pressure of 300 kPa (abs) and temperature of – 10°C. What 200, 000 p0 = 287 ¥ 333 RT0 = 2.093 kg/m3 Pressure ratio p1/p0 = 98/200 = 0.49 this arrangement? What is the corresponding maximum pressure in the receiver tank? p1 = 0.528 p0 chocking condition will prevail. The ﬂow at the nozzle throat (exit) will be critical, i.e. M1 = 1.0. At critical condition As this is less than the critical value of T 1 = 0.833 and hence T0 T 1 = 0.833 ¥ 333 = 277.4 K C 1 = Sonic velocity at throat = = kRT1 1.4 ¥ 287 ¥ 277.4 = 333.8 m/s A1 = area of throat = p ¥ (0.03)2 4 = 7.0686 ¥ 10–4 m2 r1 = 0.634 r0 \ 15.25 It is proposed to discharge helium p1 = 0.634 ¥ 2.093 = 1.327 kg/ m3 Mass rate of ﬂow under choked condition � = r 1 A1C 1 = m = 1.327 ¥ 7.0686 ¥ 10–4 ¥ 333.8 = 0.313 kg/s Pressure at the nozzle throat = p1 p 1 = 0.528 p0 = 0.528 ¥ 200 = 105.6 kPa (abs) [Note: When the chocking condition prevails, in a converging nozzle the pressure at the throat will be p1. The receiver pressure p2 will be either equal to or less than p1.] Referring to Fig 15.10, Solution: T0 = 273 – 10 = 263 K p0 = 300,000 Pa (abs) r0 = 300, 000 p0 = 2077 ¥ 263 RT0 = 0.549 kg/m3 Under conditions of maximum ﬂow, critical conditions will prevail at the nozzle throat. At critical condition M1 = M 1 = 1.0 Ê 2 ˆ p1 =Á Ë k + 1˜¯ p0 Hence C 1 = V 1 k /( k -1) Ê 2 ˆ =Á Ë 1 + 1.66 ˜¯ 1.66 /(1.66 -1) = (0.7519)2.515 = 0.488 p 1 = 0.488 ¥ 300 = 146.43 kPa (abs) The receiver pressure p2 should be less than or equal to p 1 = 146.43 kPa (abs) for maximum rate of ﬂow condition. Hence maximum downstream receiver pressure possible = p 1 = 146.43 kPa (abs) Container tank p0 V0 = 0 T0 Area = A1 1 r0 p1 ³ p 2 V1 T1 r1 Fig. 15.10 Receiver p2 503 Compressible Flow Solution: Given Ê 2 ˆ T1 = Á = (0.7519) T0 Ë k + 1˜¯ T 1 = 263 ¥ 0.7519 = 197.7 K Ê 2 ˆ r1 = Á r0 Ë k + 1˜¯ 1 /( k -1) 1/0.66 = (0.7519) = 0.649 For isentropic ﬂow in the divergent nozzle, considering the throat and the exit section, since k = 1.4 r1 = 0.549 ¥ 0.649 = 0.3565 kg/m3 V 1 = C 1 = Speed of sound = Ae A kRT1 Ae = 1.66 ¥ 2077 ¥ 197.7 = 825.6 m/s Area of nozzle A 15.26 A convergent–divergent nozzle is fed from a tank containing air at pressure of 600 kPa (abs). Calculate the back pressure required to cause Mach number of 2.0 at the exit. k = 1.4. Solution: = (1 + 0.2(2.0)2)3.5 = 7.824 pe = p0 600 = 76.68 kPa(abs) = 7.824 7.824 = 1 [(1 + 0.2 ¥ (2.5)2)]3 1.728 ¥ 2.5 k = 1.4, Hence p0 = p = 0.528 p0 p 280 = 530.3 kPa = 0.528 0.528 Applying the isentropic pressure relationship between the stragnation value and the pressure at exit p0 = (1 + 0.2 M2e)3.5 = [1 + 0.2 (2.5)2]3.5 = 17.086 pe 530.3 Exit pressure pe = = 31.04 kPa. 17.086 p0 = (1 + 0.2M 2e)3.5 pe 1 [(1 + 0.2 M 2e)]3 1.728 M At the throat for Maximum mass rate of ﬂow under the ﬂow conditions prevailing . m� = mmax = r 1 A1V 1 = 0.3565 ¥ 3.1416 ¥ 10–4 ¥ 825.6 = 0.0925 kg/s = = 2.637 Exit area = Ae = 2.637 A = 2.637 ¥ 8.0 = 21.094 cm2 p ¥ (0.02)2 4 = 3.1416 ¥ 10–4 m2 A1 = E. Convergent–Divergent Nozzle A = 8.0 cm2 and p = 280 kPa, Me = 2.5 15.28 A tank contains air at –5°C under a pressure of 303 kPa (abs). A convergent– divergent nozzle of exit diameter 5 cm is designed to discharge the air to the ambient atmosphere of pressure 101 kN/m2. Calculate the (i) Mach at the exit and (iii) mass discharge rate. (Assume k = 1.4 and R = 287 J/kg.K.) 15.27 area is 8 cm2 and throat pressure is 280 kPa. Estimate the exit pressure and exit area if the exit Mach number 2.5. (Take k = 1.4) Solution: The one dimensional isentropic compressible ﬂow functions are used between the inlet and the exit. 504 Fluid Mechanics and Hydraulic Machines Sufﬁx 0 refers to the stagnation value, which is the same as the values at the inlet tank. Sufﬁx e refers to the exit section. k /( k - 1) p0 ( k - 1) 2 ˆ Ê = Á1 + M ˜ pe Ë ¯ 2 velocity by considering the case of sonic velocity at the throat and diverging section acting as a nozzle. compressible functions for an ideal gas. The superscript indicates the critical values at the = (1 + 0.2M2)3.5 303 = (1 + 0.2M2)3.5 101 1 + (0.2M2) = 1.369, M2 = 1.8486 M = 1.36 Inlet temperature = T0 = – 5 + 273 = 268 K = T0 ( k - 1) 2 ˆ Ê M ˜ = (1 + 0.2M2) = Á1 + Ë ¯ 2 Te T0 Te = ( k - 1) 2 ˆ Ê ÁË1 + 2 M ˜¯ 268 = 195.6 K = 1 + 0.2 ¥ (1.36) 2 ( ) 101 ¥ 1000 pe = = 1.8 kg/m3 287 ¥ 195.6 RTe Velocity of sound at exit temperature re = Ce = kRTe = 1.4 ¥ 287 ¥ 195.6 = 280.3 m/s V Mach Number M = = 1.36 C Velocity V = MC = 1.36 ¥ 280.3 = 381.3 m/s Mass rate of ﬂow = . m = re VeAe = 1.734 ¥ 381.3 ¥ 1.964 ¥ 10–3 = 1.30 kg/s 15.29 A convergent–divergent nozzle has an throat area. of 950 kPa and a stagnation temperature of 350 K. The throat area is 490 mm2. Determine the temperature, (iv) exit Mach number and (v) exit M 1.00 2.197 A/A 1.00 2.00 p/p0 0.528 0.0939 T/T0 0.8333 0.5089 Solution: Sufﬁx 0 refer to stagnation values at the inlet, sufﬁx 2 to the conditions at the exit and the superscript refer to the throat section. Given: A2/A = 2.0, p0 = 950 kPa, T0 = 350 K. Referring to the given Table at A2/A = 2.0, M2 = 2.197, p2/p0 = 0.0939, T2/T0 = 0.5089. p2 = 0.0939 ¥ p0 = 0.0939 ¥ 950 = 89.205 kPa. T2 = 0.5089 ¥ T0 = 0.5089 ¥ 350 = 178.115 K Velocity of sound at exit C2 = kRT2 = 1.4 ¥ 287 ¥ 178.115 = 267.52 m/s Velocity at exit V2 = M2C2 = 2.197 ¥ 267.52 = 587.74 m/s . Mass ﬂow rate = m = rVA = r2V2 A2 Referring to the given Table, for M = M = 1.0, p/p0 = 0.528, T/T0 = 0.8333 p = 0.528 ¥ p0 = 0.528 ¥ 950 = 501.6 kPa. T = 0.833 ¥ T0 = 0.833 ¥ 350 = 291.55 K 501.6 p r = = 287 ¥ 291.55 RT = 0.005995 kg/m3 V = C = kRT 1.4 ¥ 287 ¥ 291.55 = 342.26 m/s = 505 Compressible Flow Mass ﬂow rate = . m = r VA = 0.005995 ¥ 342.26 ¥ (490 ¥ 10–6) = 1.005 ¥ 10–3 kg/s Alternatively p2 89.205 r2 = = RT2 287 ¥ 178.115 = 0.001745 A2 = 2 ¥ 490 = 980 mm2 . m = r2V2A2 = 0.001745 ¥ 587.74 ¥ (980 ¥ 10–6) = 1.005 ¥ 10–3 kg/s = 1.4 ¥ 287 ¥ 343 = 371.24 m/s Mach number M1 = V1 /C1 = 100 371.24 = 0.2694 Stagnation pressure = p01 k - 1 2ˆ Ê p01 = p1 Á1 + M1 ˜ Ë ¯ 2 k /( k -1) = 200 [1 + 0.2 ¥ (0.2694)2 ]3.5 = 210.3 kPa (abs) = p02 At Section 2: V2 = 250 m/s M2 = 0.70 Stagnation pressure = p02 F. Energy Equation in Isentropic Flow (iii) At Section 1, V1 = 100 m/s Sonic speed = C1 = kRT 1 15.30 k /( k -1) k p02 k - 1 2ˆ Ê = Á1 + M2˜ Ë ¯ 2 p2 p02 = [1 + 0.2 ¥ (0.7)2 ]3.5 kPa (abs) p2 = 1.3871 R ◊K)] Solution: T1 = 273 + 70 = 343 K V1 = 100 m/s and V2 = 250 m/s p1 = 200 kPa (abs) cp = p2 = k 1.4 R= ¥ 287 = 1004.5 k -1 0.4 (i) V 22 – V 12 = 2cp(T1 – T2) (250)2 – (100)2 = 2 ¥ 1004.5 (343 – T2) T2 = 316.87 K = 43.87°C Sonic speed at section 2 C2 = kRT2 1.4 ¥ 287 ¥ 316.87 = 356.82 m/s (ii) Mach number at Section 2 C2 = V2 C2 250 = = 0.70 356.82 M2 = 210.3 = 151.6 kPa (abs) 1 . 3871 r1 = 200, 000 p1 = = 2.0317 kg/m3 287 ¥ 343 RT1 r2 = p2 151600 = = 1.667 kg/m3 RT2 287 ¥ 316.87 [Note: 1. Check on calculation — r01 = Stagnation density at 1 1/( k -1) k - 1 2ˆ Ê = r1 Á1 + M1 ˜ Ë ¯ 2 = 2.0317 [1 + 0.2 ¥ (0.2694)2]2.5 = 2.106 Ê ˆ 2 r02 = r2 Á1 + M 22˜ k -1 Ë ¯ 1/( k -1) 506 Fluid Mechanics and Hydraulic Machines 2 + 0.4 ¥ ( 2.5) ( 2 ¥ 1.4) ¥ ( 2.5) - 0.4 M 22 = 0.4545 and M2 = 0.674 Writing the pressure ratio in terms of M1 and M2 = 1.667[1 + 0.2 ¥ (0.700)2]2.5 = 2.106 r02 = r01 = 2.106. 2. In isentropic ﬂow the stagnation pressure p0, stagnation temperature T0 and stagnation density r0 remain constant throughout the ﬂow.] = p2 1 + kM12 1 + (1.4 ¥ 2.5) = = 3 p1 1 ¥ (1.4 ¥ 0.4545) 1 + kM 2 = 2.75 G. Normal Shock 15.31 15.33 k k M Solution: M 22 = (0.4)2 = and 2 + ( k - 1) M 12 2 kM 12 ( k - 1) 2.0 + (1.32 - 1.0) M 12 V2 V1 r2/r1 = 3.0 r2 1 + b ( p2 / p1) = r1 b + ( p2 / p1) M 12 2 ¥ 1.32 ¥ - (1.32 - 1.0) 0.4224 M 21 – 0.0512 = 2.0 + 0.32 M 21 M 21 = 20.03 M1 = 4.476 15.32 Solution: Given k + 1 1.66 + 1.0 = = 4.03 k - 1 1.66 - 1.0 1 + 4.03 ( p2 / p1) 3.0 = 4.03 + ( p2 / p1) (4.03 – 3.0) (p2 /p1) = 12.09 – 1.0 (p2 /p1) = 10.77 = pressure ratio Also the pressure ratio where b = Ê p2 ˆ 2 kM 12 - ( k - 1) = Á ˜ = ( k + 1) Ë p1 ¯ Solution: Considering the velocity ratio equation for a normal shock, V2 2 + 0.4 M12 = 0.5 = V1 2.4 M12 1.2M 12 = 2.0 + 0.4M12 M12 = 2.5 and M1 = 1.581 Considering the relationship between the two Mach numbers M1 and M2, M22 = 2 + 0.4 M12 ( 2 ¥ 1.4) M12 - 0.4 2 ¥ 1.66 ¥ M 12 - (0.66) 2.66 M 21 = 8.8278 and M1 = 2.971 Upstream Mach number M1 = 2.971 10.77 = V2 1 = 1/(r2/r1) = = 0.333 V1 3 M 22 = 2 + ( k - 1) M 12 2 kM 12 - ( k - 1) 507 Compressible Flow Temperature after the shock, 2 + (1.66 - 1.0)( 2.971) 2 = 2 ¥ 1.66 ¥ ( 2.971) 2 - (1.66 - 1.0) = 7.826/28.645 = 0.2732 M2 = 0.523 T2 = 483.2 ¥ 1000 3.815 ¥ 287 = 441.3 K = 168.3°C T2 = 15.34 k p2 r2 R 15.35 ◊K)] R k = Solution: ◊K)] R r1 = p1 100, 000 = RT1 287 ¥ ( 273 - 20) Solution: = 1.3772 kg/m3 Sonic speed C1 = = 1.4 ¥ 287 ¥ 253 = 318.83 m/s V1 660 = = 2.07 C1 318.83 2 kM 12 - ( k - 1) = k +1 Mach number M1 = Pressure ratio p2 p1 2 = 250, 000 p2 = 287 ¥ ( 273 + 100) RT2 = 2.335 kg/m3 r2 = kRT1 2 ¥ 1.4 ¥ ( 2.07) - (1.4 - 1.0) (1.4 + 1.0) = 4.832 p2 = 100 ¥ 4.832 = 483.2 kPa (abs) Sonic speed C2 = 1 (V2 /V1) = 1.3772/0.361 = 3.815 kg/m3 r2 = r1 1.4 ¥ 287 ¥ 373 = 387.1 m/s Mach number after the shock M2 = V2 /C2 180.0 = 0.465 387.1 From the Mach number relation M2 = M 21 = 1) M 12 (k + 2.0 V2 = V1 ( k + 1) M 12 (1.4 - 1.0) ( 2.07) 2 + 2.0 = (1.4 + 1.0) ( 2.07) 2 = 0.361 V2 = 660 ¥ 0.361 = 238.4 m/s r1A1V1 = r2A2V2 and since A2 = A1 r1V1 = r2V2 kRT2 = = 2 + ( k - 1) M 22 2 kM 22 - ( k - 1) 2 + 0.4 ¥ (0.465) 2 2 ¥ 1.4 ¥ (0.465) 2 - 0.4 M 21 = 10.1567 M1 = 3.187 Pressure ratio 2 p2 1 + kM1 = p1 1 + kM 22 = 1 + 1.4 (3.187) 2 1 + 1.4 (0.465) 2 = 11.684 p1 = 250/11.684 = 21.4 kPa (abs) 508 Fluid Mechanics and Hydraulic Machines Normal shock V2 ( k - 1) M 12 + 2 = V1 ( k + 1) M 12 = 0.4 ¥ (3.187) 2 + 2 ( 2.4) (3.187) 2 p2 = static pressure = 0.2487 M1 > 1 p1 V1 p02 V1 = 180/0.2487 = 723.8 m/s r1: r1 = M2 < 1 r2V2 = 2.335 ¥ 0.2487 V1 Pitot static tube = 0.5807 kg/m3 21.4 ¥ 1000 p1 = 287 ¥ 0.5807 R r1 = 128.4 K = – 144.6°C Temperature: T1 = Fig. 15.11 For the normal shock from the Mach number relation 15.36 M 21 = = k R ◊ Solution: Referring to Fig 15.11 for the pitot tube, p02 = 220 kPa (abs) p2 = 170 kPa (abs) Stagnation pressure ratio k /( k -1) p02 k - 1 2ˆ Ê = Á1 + M2 ˜ p2 Ë ¯ 2 As k = 1.4, p02 220 = = (1 + 0.2 M 22 )3.5 p2 170 M 22 = 0.382 and M 2 = 0.618 2 + ( k - 1) M 22 2 kM 22 - ( k - 1) 2 + 0.4 ¥ (0.618) 2 = 3.216 2 ¥ 1.4 ¥ (0.618) 2 - 0.4 M1 = 1.793 Since stagnation temperature T0 = 350 K T0 k -1 2 =1+ M1 T1 2 = 1 + 0.2 ¥ (1.793)2 = 1.643 T1 = 350/1.643 = 213 K Sonic speed at T1 = C1 = C1 = kRT1 1.4 ¥ 287 ¥ 213 = 292.55 m/s V1 = C1 M1 = 292.55 ¥ 1.793 = 524.5 m/s 509 Compressible Flow Problems 15.1 Calculate the value of R for chlorine, helium and hydrogen. Express R in both (J/(kg◊K)) and (kcal/kg◊K) units. (Ans.) Gas R R (J/kg◊K) (kcal/kg ◊K) Cl2 117 0.0243 He 2079 0.4319 H2 4157 0.8639 15.2 A mass of 4 kg of oxygen (k = 1.4) has its temperature decreased from 85°C to 10°C at constant volume conditions. Calculate cp, cv and work involved. (Ans. cp = 909.3 J/(kg◊K); cv = 649.5 J/ (kg ◊K); Work W = 194.85 kJ (work is done by the gas) 15.3 Show that for a perfect gas cp = cv + R. 15.4 A mass of 5 kg of a gas of molecular weight 44 has a work of 100.5 kJ done on it at constant volume. This causes the temperature of the gas to rise by 30°C. Calculate R, cv, cp and k of the gas. (Ans. R = 189 J/(kg◊K); cv = 670; cp = 859; k 1.282) 15.5 Six kilograms of oxygen at 200 kPa (abs) and 10°C, in a container, is expanded isentropically to 120 kPa (abs). Find the ﬁnal temperature and the work involved. (k = 1.4). (Ans. T2 = – 28.4°C; W = work done by the gas = 149.6 kJ) 15.6 Carbon dioxide of mass 2 kg is expanded isothermally from 400 kPa (abs) to 100 done by the gas, the initial volume and ﬁnal volume, Take k = 1.30 and R = 189 J/ (kg◊K) (Ans. W = 137.8 kJ, V1 = 0.2485 m3; V2 = 0.994 m3) 15.7 Calculate the speed of sound wave in the following ﬂuids: No. Gas R Temp. (J/kg ◊K) (°C) 4. Air CO2 H2 N2O No. Liquid Bulk modulus K (N/m2) Density (kg/m3) 7. Water Gasoline Mercury 2.19 ¥ 109 9.58 ¥ 108 2.55 ¥ 1010 998 680 13550 287 189 4124 189 20 20 20 20 k 1.4 1.3 1.41 1.31 Answers to Problem 15.7 No. Fluid 7. Air CO2 H2 N2O Water Gasoline Mercury C(m/s) 343.1 268.3 1305 269.3 1481 1187 1372 15.8 An airplane is ﬂying at a Mach number of 1.8 in an atmosphere where the pressure is 14 kPa (abs) and density is 0.225 kg/m3. Calculate the speed of the plane. (Ans. V = 531 m/s) 15.9 A rocket travels at 1800 km/h in air of pressure 35.6 kPa (abs) and temperature –37°C. Find the Mach number and Mach angle. [Take k = 1.5 and R = 287 J/(kg◊K)]. (Ans. M = 1.624, a = 38°) 510 Fluid Mechanics and Hydraulic Machines 15.10 A supersonic plane ﬂies at an altitude of 2500 m and 6.5 s after it has passed over the head of an observer on the ground, the sonic boom is heard. Calculate the speed of the plane and its Mach number. The average temperature of the atmosphere can be assumed to be 5°C. Take R = 287 J/(kg ◊K). (Ans. M = 2.02 and V = 675.2 m/s) 15.11 A bullet ﬁred from a gun creates a Mach angle of 30° in still air. If the air temperature is 15°C, calculate the velocity of the bullet. Take k = 1.4 and R = 287 J/kg.K. (Ans: 680 m/s) 15.12 The Concorde airplane ﬂies at a Mach number of 2.2. If it ﬂies in a standard atmosphere at 15000 m altitude where the pressure is 12 kPa (abs) and density is 0.1935 kg/m3, calculate the pressure, temperature and density at a stagnation point. [Take R = 287 J/(kg ◊K) and k = 1.4] (Ans. T0 = 152°C, p0 = 128.3 kPa (abs); r0 = 1.052 kg/m3) 15.13 A conduit conveys air at a Mach number of 0.70. At a certain section the static pressure is 30 kPa (abs) and the temperature is 25°C. (i) Calculate the stagnation temperature and pressure (ii) If the stagnation temperature is 90°C, what would be the Mach number of the ﬂow? (Take k = 1.4.) (Ans. (i) T0 = 59.7°C, p0 = 41.61 kPa (abs); (ii) M1 = 0.995) 15.14 Estimate the maximum velocity of air at 35°C so that the stagnation point temperature is less than 40°C. [Take k = 1.4 and R = 287 J/(kg ◊K).] (Ans. V1 = 100.2 m/s) 15.15 A pitot-static tube mounted on an airplane in ﬂight records a stagnation pressure of 90 kPa (abs) and a static pressure of 70 kPa (abs). If the density 15.16 15.17 15.18 15.19 15.20 of air in the atmosphere at the level is 0.80 kg/m3, calculate the velocity of the airplane. (Take k = 1.4). (Ans. V1 = 213.5 m/s) A pitot-static tube to be used in an airplane for on-ﬂight measurement was calibrated on ground by using air of density 1.20 kg/m3. A measurement taken at an altitude of 3000 m where the ambient pressure and density were 70 kPa (abs) and 0.91 kg/m3 respectively, indicated a velocity of 200 m/s by using the above ground level calibration. Estimate the true speed of the airplane [Take k = 1.4 and R = 287 J/(kg ◊K)]. (Ans. V1 = 217.5 m/s) Show that in an isentropic ﬂow of a gas in a duct, if a pitot-static tube measures the stagnation pressure p0, stagnation temperature T0 and static pressure p1, the velocity of ﬂow V1 can be calculated by the relation. V1 = {2c p T 0 [1 – (p1/p0)(k–1)/k]}1/2 Air ﬂow in a duct can be considered to be isentropic. At section 1, the velocity, pressure and temperature are 125 m/s, 200 kPa (abs) and 300 K respectively. If at a downstream section the velocity is 220 m/s, calculate the (i) Mach number, temperature and pressure at section 2 and (ii) the densities at sections 1 and 2. (Ans. (i) M1 = 0.360, T2 = 283.7 K, p2 = 164.4 kPa (abs) (ii) r1 = 2.3229 kg/m3, p2 = 2.0196 kg/m3) Air at 200 kPa (abs) and 27°C is expanded isentropically. What is the maximum possible attainable speed? Take k = 1.4 and R = 287 J/(kg◊K). (Ans. Vm = 776.3 m/s) Oxygen [k = 1.4 and R = 260 J/(kg ◊K)] is contained in a tank at 150 kPa (abs) and 20°C. If it is expanded isentropically to attain a Mach number of unity, what is 511 Compressible Flow the sonic speed and temperature at that section? (Ans. C1 = 298.6 m/s, T 1 = – 29°C) 15.21 Air at 40°C and pressure of 300 kPa (abs) ﬂows from a large tank through a converging nozzle. If the Mach number at the outlet of the nozzle is 0.5, calculate the velocity, pressure, temperature and density at the nozzle exit. (Take k = 1.4 and R = 287 J/(kg ◊K). (Ans. V1 = 173.04 m/s; p1 = 252.9 kPa (abs); T1 = 298.1 K; r1 = 2.956 kg/m3) 15.22 A tank contains nitrogen [ k = 1.4 and R = 297 J/(kg ◊K)] at 225 kPa (abs) and 50°C. A convergent nozzle of exit area 0.05 m2 is used to exhaust this gas to an ambient pressure of 100 kPa (abs). Calculate the mass rate of ﬂow and the pressure at the nozzle throat. . (Ans. m = 26.81 kg/s, p1 = 118.8 kPa (abs)) 15.23 Oxygen ﬂows steadily from a reservoir at –10°C and 265 kPa (abs) through a convergent nozzle of exit diameter of 10 cm into another large tank where the pressure is 180 kPa (abs). Calculate the mass rate of ﬂow and Mach number at the nozzle exit. Assume isentropic ﬂow with k = 1.4 and R = 260 J/(kg◊K). . (Ans. m = 5.17 kg/s, M1 = 0.915) 15.24 Chlorine [k = 1.34 and R = 117 J/(kg◊K)] is stored in a tank at 300 kPa (abs) and 5°C. If a convergent nozzle of exit area 0.08 m2 discharges the gas to another tank having a pressure of 150 kPa (abs), calculate the mass rate of ﬂow. What is the pressure at the nozzle throat? . (Ans. m = 89.72 kg/s, p 1 = 161.6 kPa (abs)) 15.25 Air form tank A at 80°C ﬂows through a convergent nozzle of 8 cm diameter isentropically into a receiving atmosphere of pressure 100 kPa (abs). Estimate the mass ﬂow rate when the pressure inside the tank A is (a) 200 kPa (abs), (b) 300 kPa (abs) and (c) 400 kPa (abs). Take k = 1.4 and R = 287 J/(kg ◊K). . (Ans. (a) m = 0.2162 kg/s; . . (b) m = 0.3243 kg/s; (c) m = 0.4324 kg/s) 15.26 Air with pressure p0 and temperature T0 in a tank is discharged through a convergent nozzle into a receiver with ambient pressure p1. If p1 is kept constant, show that the mass rate of ﬂow increases linearly with p0 when p0 > 1.893 p1. 15.27 Air at 10 bar and 500 K stagnation condition ﬂows through a convergent– divergent nozzle. The area at the nozzle exit is 0.25 ¥ 10–4 m2. The pressure at exit volume and mean ﬂow rate through the nozzle. (Take k = 1.4 and R = 287 J/kg.K) (Ans: 608.4 m/s, 0.453 m3/kg, 0.0335 kg/s) 15.28 Air enters a convergent–divergent nozzle with an inlet pressure of 750 k Pa, temperature of 25°C and very low velocity. If the exit Mach number is 2.3, calculate the ratio of throat area to exit area, exit pressure and exit temperature. Assume isentropic ﬂow and k = 1.4. (Ans: Ae/A = 2.193, Pe = 59.98 kPa(abs), Te = 144.8 K) 15.29 At section 1, upstream of the throat of a convergent–divergent nozzle the properties of an isentropic ﬂow are V1 = 250 m/s, T1 = 320 K and p1 = 800 kPa. Calculate the (i) exit temperature, (ii) exit pressure and (iii) Mach number at section 1. The exit Mach number is 2.3. (Take Cp = 1005 J/kg.K) (Ans: Te = 144.8 K, pe = 59.98 kPa, M1 = 6056) 15.30 In a supersonic stream of air a pitot-static tube creates a stagnation pressure of 50 kPa (abs) and a stagnation temperature 512 Fluid Mechanics and Hydraulic Machines of 410 K. A normal shock occurs in front of the tube. If the static pressure is 35 kPa (abs), estimate the velocity of the supersonic stream. Take k = 1.4 and R = 287 J/(kg◊K). (Ans. V1 = 485.9 m/s) 15.31 In a supersonic stream of air at 600 m/s and a temperature of 250 K, a normal shock wave occur in front of a body. If the static pressure of the supersonic stream is 20 kPa (abs), calculate (i) the pressure after the shock, (ii) the temperature and velocity after the shock. Take k = 1.4 and R = 287 J/(kg ◊K). (Ans. (i) p2 = 80.3 kPa (abs); (ii) T2 = 127.5°C, V2 = 239.5 m/s) 15.32 Air at 5°C is moving at Mach number 2.5. If a normal shock wave occurs, estimate the ratios of pressure, velocity, temperature and density across the shock. Assume k = 1.4 and R = 287 J/(kg ◊K). (Ans. p2 /p1 = 7.125, V2/V1 = 0.3, T2/T1 = 2.1375, r2/r1 = 3.333) 15.33 If for a normal shock occurring in a supersonic stream of air at Mach number of 1.5, the upstream stagnation pressure is 210 kPa (abs), calculate the stagnation pressure after the shock. (k = 1.5). (Ans. p02 = 195.24 kPa (abs)) 15.34 In a normal shock wave in air the Mach number after the shock is 0.60. If the temperature and pressure after the shock are 150°C and 360 kPa (abs) respectively, determine the Mach number, pressure, temperature and density of the supersonic stream before the shock. Take k = 1.4 and R = 287 J/(kg ◊K). (Ans. M1 = 1.878; r1 = 91.19 kPa (abs); V1 = 614 m/s; r1 = 1.195 kg/m3; T1 = – 7.1°C) 15.35 The velocity ratio V2 /V1 across a normal shock wave in air (k = 1.4) is 0.5. Estimate corresponding pressure ratio. What are the Mach numbers upstream and downstream the shock? (Ans. p2 /p1 = 2.75, M1 = 1.581, M2 = 0.674) Objective Questions 15.1 A gas has a molecular weight of 44. The gas constant R for the gas, in J/(kg◊K), (a) 0.045 (b) 189 (c) 1130 (d) 1854 15.2 A gas has a molecular weight of 16 and has a cv = 1730 J/(kg◊K). The value of the speciﬁc heat ratio k for this gas is (a) 1.30 (b) 1.40 (c) 1.65 (d) 1.21 15.3 The speciﬁc heat ratio k is given by the following expression 1 1 - (R /cv ) (c) 1 + (R/cv) (d) 1 + (cv /R) 15.4 A gas of 2 kg with cp = 5220 J/(kg◊K) and cv = 3143 J/(kg◊K) has its temperature raised by 20°C isentropically. The change in internal energy is (a) 62.86 kJ (b) 313.2 kJ (c) 156.60 kJ (d) 125.72 kJ 15.5 A gas of 3 kg with cv = 745 J/(kg◊K) and k = 1.40 has its temperature raised by 30°C isentropically. The change in enthalpy is (a) 1 + (cp/R) (b) 513 Compressible Flow (a) dr + d(rV 2 ) = 0 (a) 67.0 kJ (b) 94.5 kJ (c) 31.5 kJ (d) 22.3 kJ 15.6 An isentropic process is (a) adiabatic and irreversible (b) adiabatic and frictionless (c) reversible and isothermal (d) any adiabatic process 15.7 The speed of sound in air varies as (a) 15.8 15.9 15.10 15.11 15.12 T (b) r (c) 1/ p (d) p where T = absolute temperature, p = density and p = absolute pressure. In an atmosphere where pressure p = 16.5 kN/m2 (abs), density r = 0.265 kg/m3 and k = 1.4, the speed of sound in the medium is (a) 9.34 m/s (b) 78.2 m/s (c) 295 m/s (d) 334 m/s An aircraft moves at 1580 km/h in an atmosphere where the temperature is –60°C. If k = 1.4 and R = 287 J/(kg ◊K) the Mach number of the plane is (a) 0.67 (b) 1.50 (c) 2.10 (d) 5.4 In an atmosphere the speed of sound is 300 m/s. If a plane travels at 1620 km/h in this atmosphere the Mach angle is (a) 30.5° (b) 56.3° (c) 10.7° (d) 41.8° In a standard atmosphere the temperature is 15°C at sea level and –56°C at an altitude of 20 km. A supersonic plane has the same speed at the sea level as well as at an altitude of 20 km. If its Mach number is 1.5 at sea level, its Mach number at an altitude of 20 km in standard atmosphere is (a) 2.80 (b) 1.30 (c) 1.99 (d) 1.73 The differential equation for energy for reversible adiabatic ﬂow may take the form (b) VdV + C 2 dr =0 r dr =0 r dr (d) 2VdV + =0 r (c) VdV + 15.13 Considering that p/p = constant, expresses an isentropic process, which one of the following is NOT a representation of speed of sound? (a) k gRT (b) dp dr (c) k/r (d) p/r 15.14 The maximum Mach number for which the ﬂow of air can be considered incompressible within 1% error is (a) 0.1 (b) 0.2 (c) 0.4 (d) 0.6 15.15 An airplane is cruising at a speed of 800 km/h at an altitude where the air temperature is 0°C. The ﬂight Mach number at this speed is nearly (a) 1.33 (b) 0.67 (c) 0.25 (d) 2.4 15.16 In the ﬂow of air (k = 1.4) in a duct the ambient temperature is 30°C and the stagnant temperature is measured as 59.7°C. The Mach number of the ﬂow is (a) 0.50 (b) 2.22 (c) 1.59 (d) 0.70 15.17 Air (k = 1.4) ﬂows at a Mach number of 1.5 in a duct. If the ambient temperature is 7°C, the stagn ation temperature is (a) 133°C (c) 406°C (b) 10.5°C (d) 256°C 15.18 If air [k = 1.4 and R = 287 J/(kg◊K)] at 19°C is expanded isentropically, the maximum velocity that can be achieved is 514 Fluid Mechanics and Hydraulic Machines 15.19 15.20 15.21 15.22 15.23 (a) 342.5 m/s (b) 766 m/s (c) 1000 m/s (d) 518 m/s For isentropic ﬂow of air (k = 1.4), the stagnation temperature T0 and the temperature T at any Mach number are related as T0/T = (a) (1 + 0.2 M2) (b) (1 + 0.2 M2)2.5 (c) (1 + 0.2 M2)3.5 (d) (1 + 0.2 M2)–1/2 For air (k = 1.4) the critical pressure ratio p1/p0 for isentropic ﬂow is (a) 0.833 (b) 0.728 (c) 0.628 (d) 0.528 For helium (k = 1.66) the critical pressure ratio p 1/p0 is (a) 0.728 (b) 0.528 (c) 0.488 (d) 0.833 For N2O the critical temperature ratio is 0.866 and the critical density ratio is 0.628. Its critical pressure ratio is (a) 0.544 (b) 0.725 (c) 0.737 (d) 0.528 In isentropic ﬂow of a perfect gas (a) the velocity always decreases in conduits of increasing area. (b) the velocity is always critical at the throat of a nozzle. (c) if the ﬂow is subsonic ﬂow through a convergent nozzle, the maximum velocity is always at the throat. (d) if the ﬂow is in a convergent-divergent nozzle, the maximum velocity is always at the throat. 15.24 A gas is being discharged isentropically from a tank of pressure p1(abs) through a converging nozzle in a receiving chamber of pressure p2 (abs) with critical ﬂow condition in the nozzle exit. If p2 is kept constant and p1 is doubled, the mass rate of ﬂow through the nozzle will (a) not change (b) be doubled (c) be halved (d) increase by 2 times 15.25 In isentropic ﬂow between two points (a) stagnation pressure and stagnation temperature may vary (b) the stagnation pressure decreases in the direction of the ﬂow (c) the stagnation temperature and stagnation pressure decrease with increase in the velocity (d) the stagnation temperature and stagnation pressure remain constant 15.26 In ﬂow through a convergent nozzle, the ratio of back pressure to the inlet pressure is given by pb È 2 ˘ = Í ˙ p1 Î k + 1˚ k /( k + 1) If the back pressure is lower than pB given in the above equation, then (a) Pressure in the nozzle is supersonic (b) A shock wave exists inside the nozzle (c) The gasses expand outside the nozzle and a shock wave appears outside the nozzle. (d) A shock wave appears at the nozzle exit. 15.27 The stagnation temperature of an isentropic ﬂow of air (k = 1.4) is 400 K, if the temperature is 200 K, then the Mach number of the ﬂow will be (a) 1.046 (b) 1.264 (c) 2.236 (d) 3.211 15.28 Air from a reservoir is to be passed through a supersonic nozzle so that the jet will have a Mach number of 2.0. If the static temperature of the jet is not to be less than 27°C, the minimum temperature of the air in the reservoir should be 515 Compressible Flow 15.29 15.30 15.31 15.32 15.33 15.34 (a) 48.6°C (b) 167°C (c) 267°C (d) 367°C In an isentropic ﬂow of air (k = 1.4) the stagnation temperature T0 = 300 K. If at a section the temperature is 166.7 K, the Mach number of the ﬂow is (a) 0.56 (b) 1.80 (c) 2.0 (d) 4.0 Which of the following is analogous to normal shock wave? (a) An elementary wave in a still liquid (b) Hydraulic jump (c) Flow of liquid in an expanding nozzle (d) Subcritical ﬂow in a rough channel In a normal shock taking place in a gas (a) the velocity, pressure and density increase across the shock (b) the entropy remains constant (c) the entropy decreases across the shock (d) the entropy increases across the shock In a normal shock in a gas (a) the upstream ﬂow is supersonic (b) the upstream ﬂow is subsonic (c) the downstream ﬂow is sonic (d) the downstream ﬂow as well as the upstream ﬂow is supersonic In a normal shock in a gas (a) the stagnation pressure remains the same on both sides of the shock (b) the stagnation density remains the same on both sides of the shock (c) the stagnation temperature remains the same on both sides of the shock (d) the Mach number remains the same on both sides of the shock A normal shock wave in a gas (a) is reversible (b) is isentropic throughout 15.35 15.36 15.37 15.38 (c) is irreversible (d) causes an increase in Mach number In a normal shock occurring in air (k = 1.4), if the upstream Mach number is 3.52, the Mach number after the shock is (a) 0.61 (b) 0.45 (c) 0.13 (d) 0.28 In a normal shock wave in a gas with k = 1.4, one of the Mach numbers is 0.5. The other Mach number is (a) 2.65 upstream of the shock (b) 0.06 downstream of the shock (c) 0.02 upstream of the shock (d) 3.75 upstream of the shock In a normal shock wave in air (k = 1.4), the density ratio across the shock r2 /r1 is 3.0. The corresponding pressure ratio p2/p1 is (a) 3.0 (b) 0.61 (c) 5.70 (d) 1.5 In a compressible ﬂow the area of ﬂow = A, velocity V and mass density = r. At a particular section, the differential form of continuity equation is d A dV d r + = V r A dA dV d r (b) =– + A V r d A dV d r (c) = A V r (a) d A dV d r = A V r 15.39 In a supersonic ﬂow, a diverging passage results in (a) increase in velocity and pressure (b) pressure, density and temperature increase (c) velocity, pressure and density increase (d) pressure, density and momentum ﬂux increase (d) 516 Fluid Mechanics and Hydraulic Machines References Aerodynamics of Subsonic Flight, Fundamentals of Compressible Flow The Dynamics and Thermodynamics of Compressible Fluid Flow 1954 Fluid Mechanics, Sixth Ed., Tata McGraw-Hill, Special India Ed., 2008 Fluid Mechanics, Tata McGraw-Hill Ed., 2006 HYDRAULIC MACHINES Concept Review 16 16.1 TURBINES 16.1.1 Francis Turbine Basic Equations Figure 16.1(a) shows the schematic sketch of a Francis turbine and Fig. 16.1(b) shows the inward ﬂow through a radial turbine runner, typically a Francis turbine. The following notations are employed: Sufﬁx 1 is used for inﬂow and sufﬁx 2 for outﬂow conditions. Further, V = absolute velocity of the ﬂuid. Introduction v = relative velocity of ﬂuid with respect to the blade. 2prN u = peripheral velocity of the blade = . 60 r = radius N = revolutions per minute. a = angle made by the absolute velocity vector V with the positive direction of the peripheral velocity u. b = angle made by the relative velocity vector v with the negative direction of the peripheral velocity u; known as blade angle. 518 Fluid Mechanics and Hydraulic Machines b¢ = Vu = = a= Vf = = b= Main shaft Pivot Scroll casing Guide vane or wicket gate Shroud ring Scroll casing Runner vane Draft tube Tail race 180 – b. tangential component of absolute velocity V V cos a, known as swirl velocity. guide vane angle radial component of absolute velocity V V sin a, known as ﬂow velocity. width of ﬂow passage. Discharge By continuity From penstock Scroll casing Q = 2pr1b1V1 sin a1 = 2pr2b2V2 sin a2 Q = 2 p r1 b1Vf 1 = 2 p r2 b2 Vf 2 or (16.1) Torque Torque on the axis of the runner, T = rQ(r1 V1cos a1 – r2 V2 cos a2) Guide vane or wicket gate T = rQ ( rV 1 u1 - r2 Vu 2 ) (16.2) Power Power delivered to the runner, P = T w = rQ (u1Vu1 - u2 Vu 2 ) Fig. 16.1(a) Schematic Sketch of a Francis Turbine Head If He = head utilised by the turbine P = rQHe = rQ(u1Vu1 – u2 Vu2) He = Vu1 u1 a1 v1 u1Vu1 - u2Vu 2 g (16.4) In a reaction turbine the net available head H is the difference between the energy levels just upstream of the turbine and that at the tail race. Net Available Head, H b1 b (16.3) Vf1 V1 hH b2 Vf2 b2 u 2 V2 a2 Vu2 Thus r1 v2 r2 Head extracted H = e Net available head H hH = (u1Vu1 - u2Vu 2 ) gH (16.5) hm hm = Fig. 16.1(b) hH = Inward Flow through a Turbine = power available at the shaft power exerted by the water on the rotor brake power (brake power + power used up in mechanical friction) (16.6) 519 Hydraulic Machines h0 Power delivered to the shaft h0 = power available in water = (brake power)/g QH (16.7) h0 = hm ◊ hH Thus (16.8) s The speciﬁc speed Ns, in SI units, is deﬁned for turbines as N P Ns = (16.9) H 5/ 4 where P = power in kW, H = net available head in metres and N = speed of rotation in rpm. The speciﬁc speed of Francis turbine runners ranges from 40 to 420 but more commonly it is in the range 75 to 305. Note that the dimensions of Ns are [F1/2 L–3/4 T–3/2] but, as a convention, the units are not written. As the speciﬁc speed is a dimensional quantity its value for a given machine depends on the system of units adopted. In this book the SI units are used. A dimensionless form of speciﬁc speed, known as shape number, is sometimes used. It is written as Shape number = Dimensionless speciﬁc speed Symbolically Sp = N P 1/ 2 r (16.10) (g H )5 / 4 where u1 = pD1 N in which N = rpm and D1 = 60 diameter at the inlet. Speed factor f is also sometimes called as Speed ratio. The value of f for Francis turbine ranges from 0.70 to 0.85. Flow Ratio, y y= The ﬂow ratio y is deﬁned as Vf1 flow velocity at inlet = 2g H 2g H (16.12) For Francis turbines y ranges from 0.15 to 0.35. 16.1.2 Propeller and Kaplan Turbines These turbines are of the axial ﬂow type. A propeller turbine with an adjustable blade is known Kaplan turbine. Figure 16.2 shows a propeller turbine in a schematic form. The following are some special features of these turbines: (1) The peripheral velocity at inlet and outlet are the same p D0 N Hence u1 = u2 = (16.13) 60 (2) The ﬂow velocity remains unchanged from inlet to outlet. General Features of a Francis Runner The relative velocity must be tangential to the blade at the entry to cause the least amount of disturbance to the ﬂow. Also, for maximum efﬁciency the ﬂuid should leave the outlet with zero swirl velocity (i.e. Vu2 = 0 which means a2 = 90°). In solving problems, if no mention is made about the discharge at outlet, the radial discharge (i.e. a2 = 90°) can be assumed as this is a common practice in design. Speed Factor, f f= Blade Db The speed factor f is deﬁned as peripheral velocity of a rotating element 2g H = Guide vane Hub or boss Do u1 2g H (16.11) Runner Fig. 16.2 Schematic Sketch of a Propeller Turbine 520 Fluid Mechanics and Hydraulic Machines Hence, Vf1 = Vf2 (16.14) (3) It is usual to assume the relative velocities to remain unchanged, i.e. v1 = v2 (16.15) p (D 2o – D b2 ) 4 (16.16) where Do = outside diameter of the runner and Db = diameter of the hub (or boss). (5) The ﬂow leaves the runner radially, i.e. a 2 = 90°. N P (6) The speciﬁc speed Ns = for propeller H 5/ 4 type turbines ranges from 380 to 950. The speed factor f for Kaplan turbines ranges from 1.40 to 2.0. N (4) The inlet area = outlet area = 16.1.3 Impulse Turbine Figure 16.3 shows the schematic deﬁnition sketch of an impulse turbine. The net available head H for an impulse turbine is the energy head available at the base of the nozzle. As in a reaction turbine Power P = g QHh0 (16.17) where ho= overall efﬁciency = hH hm. A small part of the head H is lost in friction in the nozzle, a portion is expended in bucket friction and in the kinetic energy carried away by the water leaving the buckets. A part of the energy developed is lost in mechanical friction b V1 u = pND Nozzle, dia = d Fig. 16.3 and windage losses. Figure 16.4 shows the inlet and outlet velocity vectors in an impulse turbine bucket. Here, the sufﬁxes 1 and 2 denote the inlet and outlet conditions respectively. u = u1 = u2 = peripheral velocity of the wheel. V1 = jet velocity = absolute velocity of impingement. = Cv 2gH (16.18) where Cv = coefﬁcient of velocity with a value of about 0.95 to 0.98 v1 = (V1 – u) = relative velocity at inlet. (16.19) v2 = relative velocity at outlet = V2 – u2 = kv1 where k = a coefﬁcient to account for losses in the wheel. When k = 1, v2 = v1. Vu1 (V 1 – u ) u Outlet b V1 v2 Inlet b a2 b¢ u Fig. 16.4 Velocity Vectors in Impulse Turbine V2 521 Hydraulic Machines b = bucket angle = deﬂection angle of the jet and is usually an obtuse angle. b¢ = (180 – b). By momentum equation, the force exerted on the bucket in x-direction (i.e., in the initial direction of the jet). Fx = r Q(V1 – u) (1 + k cos b¢) (16.20) Power transmitted to the bucket = power extracted. P = r Qu (V1 - u ) (1 + k cos b ¢ ) (16.21) Head extracted, He = 1 u (V1 - u ) (1 + k cos b ¢ ) (16.22) g H e u (V1 - u ) (1 + k cos b ¢ ) = H gH Hydraulic efﬁciency hH is also sometimes called as Wheel efﬁciency or Blade efﬁciency. Mechanical efﬁciency, hH = power available at shaft r Qu (V1 - u ) (1 + k cos b ¢ ) Speed ratio, f: The ratio of peripheral velocity u to ideal velocity Similitude in Turbines Scale models are often used in designing and other studies relating to turbines. Geometric similarity is a basic requirement. Kinematic similarity is assured by having geometrically similar velocity vector diagrams. It is usual to neglect viscous effects in the model studies. The model and prototype characteristic relationships are usually expressed in terms of the following relationships between the variables: N pDp N m Dm = (16.24) Hm Hp Qm N m Dm3 Hydraulic efﬁciency hH = 16.1.4 2gH is known as speed ratio or speed factor, f. Pm N m3 Dm5 = = Qp N p D 3p Pp N 3p D 5p f = u / 2g H (16.23) For most efﬁcient operations f depends upon the speciﬁc speed to some extent and is found to be in the range 0.43 to 0.47. Speciﬁc speed, Ns = N P H 5/ 4 For multiple jet impulse turbines, the speciﬁc speed is based on the brake power per jet. The impulse turbines have speciﬁc speeds ranging from 8 to 30 and attain best efﬁciency at a value of Ns around 17. (16.26) It is seen from the above that the speciﬁc speed Ns is same for both the model and prototype. Also for a given diameter ratio N μ H1/2 Q μ H1/2 P μ H3/2 This fact is expressed in terms of unit quantities. The unit speed Nu is deﬁned as the speed of a geometrically similar turbine working under a head of 1 m. Nu = N / H Thus (16.25) (16.27) The unit discharge Qu = ﬂow rate in a geometrically similar turbine working under a head of 1 m Qu = Q / H (16.28) Unit Power Pu = Power generated in a geometrically similar turbine working under a head of 1 m. Pu = P / H 3 / 2 (16.29) These relationships are useful in studying the performance of a turbine under varying heads. 522 Fluid Mechanics and Hydraulic Machines 16.1.5 Typical Characteristics of Common Turbines 16.2 Table 16.1 gives the ranges of various characteristics of commonly used turbines. Table 16.1 Ranges of Characteristics of Common Turbines Characteristics Pelton Francis Kaplan & Propeller Head H(m) 100–1760 30–450 1.5–75 Speed N(rpm) 75–1000 70–1000 70–600 Speciﬁc speed Ns 8–30 40–420 380–950 105 600 125 Max. power Pm (MW) Typical inlet and outlet velocity triangles of the different kinds of turbines are shown in Fig. 16.5. ROTODYNAMIC PUMPS A rotodynamic pump consists essentially of a rotating element inside a casing. The rotating element is called an impeller. The ﬂuid enters the casing at its centre and ﬂows outward by the action of the rotating impeller and is discharged around the circumference of the casing. During this process the ﬂuid receives energy. Basically three kinds of rotodynamic pumps are recognised. They are: (i) centrifugal pumps, (ii) mixed ﬂow pumps, and (iii) axial ﬂow pumps. This classiﬁcation is based on the direction of ﬂow of the ﬂuid in the impeller. Figure 16.6 shows a typical centrifugal pump and a typical casing. In a pump the mechanical energy through the shaft and impeller is converted to ﬂuid energy. The difference between the total energy heads at the intake and discharge ﬂanges of the pump is denoted as net head H developed in the pump. The intake end is commonly called the suction end and the discharge end as the delivery end. Denoting these by sufﬁxes s and d respectively (see Fig. 16.7) the net head H = Hd – Hs Net Head Developed, H 16.1.6 Draft Tube Reaction turbines work under pressure and hence the turbine system consisting of the runner assembly and the volute casing are completely enclosed. The inlet and outlet will be through closed pipes ﬂowing full. Any kinetic energy at the point of discharge of water to the tail race is a waste of energy so far as the turbine is concerned. By minimizing the kinetic energy at the outlet, the efﬁciency of the system can be improved. Towards this, a diverging tube that connects the outlet of the runner to the tail race, called a draft tube, is used in reaction turbines. Thus, a draft tube is a conduit attachment to the turbine exit to achieve the following beneﬁts: (i) To enable the turbine to be set up at an elevation higher than the tail water level (ii) To utilize a major part of the kinetic energy of the water exiting the turbine Analysis of ﬂow through draft tube consists essentially of application of Bernoulli equation to the inlet and outlet ends of the tube along with appropriate boundary conditions. Examples 16.14 through 16.15 illustrate this aspect. or Ê pd Vd2 ˆ Ê ps Vs2 ˆ Z Z + + + + d s H = ÁË g 2g ˜¯ ÁË g 2g ˜¯ (16.30) Velocity Triangles Velocity triangles at inlet and outlet of an impeller are shown in Fig. 16.8 (a), (b) and (c). Three kinds of vane conﬁgurations viz. (i) backward curved vanes (ii) radial vane and (iii) forward curved vanes are shown. The backward curved vane pumps are the most common. Unless the data are explicitly clear about any other types, the backward curved blade with (b1 and b2) < 90∞ is assumed in all problems. Notations The following notations are used in connection with ﬂow in pump impellers: 523 Hydraulic Machines u1 a1 a1 b1 u1 u1 a1 b1 v1 V1 v1 V1 V1 v2 v2 V2 a2 b2 b2 v2 u2 Impulse turbine, b1 > 90° (i) b2 Medium francis turbine, b1 = 90° (iii) u2 u1 u1 a1 a1 b1 v2 a2 b1 v1 V1 v2 u2 v1 V1 V2 a2 b2 V2 u2 Fast francis turbine, b1 < 90° (iv) Propeller turbine, b1 < 90° (v) Fig. 16.5 a2 u2 V2 a2 Slow francis turbine, b1 > 90° (ii) b2 b1 = 90° v1 Different Kinds of Turbines V2 524 Fluid Mechanics and Hydraulic Machines and Vu = V cos a = whirl velocity. The discharge Q itself is given in terms of ﬂow velocity Vf = V sin a as Q = pD2b2Vf2 = pD1b1 Vf1 (16.32) Casing Impeller where b = width of the impeller at the given radius. Power P The power transmitted by the impeller to water is P = Tw = r Q (u2Vu2 – u1Vu1) = g QH¢ Volute Fig. 16.6 Centrifugal Pump D S pd ps P Suction Vs Delivery z d , Vd zs Fig. 16.7 Inlet V1 = v1 = u1 = a1 = absolute velocity relative velocity peripheral velocity inclination of V1 with u1 direction = direction of absolute velocity at inlet b1 = inclination of v1 with u1 direction = blade angle at inlet Outlet V2 = v2 = u2 = a2 = absolute velocity relative velocity peripheral velocity inclination of V2 with u2 direction = direction of absolute velocity at exit b2 = inclination of v2 with u2 direction = blade angle at exit Note that all angles are measured with respect to positive direction of u. If some angles are obtuse their complementary angles are used with a prime. For example: b1¢ = 180 – b1 and b2¢ = 180 – b2. The sufﬁxes 1 and 2 are used for inlet and outlet conditions respectively. In Eq. 16.33 H¢ = theoretical head 1 = (u V – u1Vu1) g 2 u2 (16.34) This is also known as manometric head. Now H¢ = head transferred from the shaft = head supplied to machine. If H = actual head delivered to water (= net head developed), then H¢ = H + hL where hL = losses. The losses are composed of shock losses at entrance and exit to the blades, frictional losses in blade passages and circulation in the passages. hH hH = H gH h = = 1 – 1 (16.35) H¢ u2Vu 2 - u1Vu1 H¢ For radial entry, i.e., Vu1 = 0, h H = gH u2Vu 2 (16.36) The hydraulic efﬁciency is also known as manometric efﬁciency. hv hv = Q Q + QL (16.37) where Q = discharge actually delivered and QL = leakage of discharge. Torque The torque acting on the ﬂuid is T = r Q (r2V2 cos a2 – r1V1 cos a1) = r Q (r2Vu2 – r1Vu1) (16.31) where Q = discharge (16.33) hm hm = brake power - power loss due to friction brake power 525 Hydraulic Machines u2 = Vu2 u2 Vf2 = v2 a2 Vf2 a2 b2 = 90° V u2 V2 V2 b2 v2 V1 b v1 v1 V1 u1 b1 b1 u1 (a) Backward curved vane b2 < 90° (fast speed) (b) Radial vane b2 = 90° (medium speed) Vu2 u2 b2 Vf2 V1 v2 b1 V2 v1 u1 (c) Forward curved blade b2 > 90° (slow speed) Fig. 16.8 a2 526 Fluid Mechanics and Hydraulic Machines = hm ( BP) - Pf ( BP ) (16.38) h0 power delivered to the fluid h0 = power put into the shaft (BP) h0 = 16.2.2 g QH = hv hm hH ( BP ) Slow speed 10 – 30 Median speed 30 – 50 High speed 50 – 80 80 – 200 200 – 300 Radial ﬂow Pumps (16.39) Basic Features of Rotodynamic Pumps Mixed ﬂow pumps Axial ﬂow pumps Minimum Speed When the pump is switched At the design value of discharge, the ﬂow is assumed to enter the impeller radially. This implies a1 = 90° and Vf1 = V1. If this information is not explicitly speciﬁed, it can be assumed in the solution of problems. The inlet velocity diagram will be as in Fig. 16.9. The manometric head Radial Entry H¢ = For double suction pumps the speciﬁc speed Ns is based on one half of the total capacity of the pump. The typical ranges of Ns of commonly used pump types are as follows: on, the ﬂow will take place only when the rise in pressure due to impeller action is large enough to overcome manometer head. The pressure head created by centrifugal action on the rotating liquid is (u22 – u12)/2g. Thus the ﬂow will commence only if (u22 - u12 )/ 2g ≥ H ¢ (16.42) where H¢ = manometric head given by Eq. 16.34. Noting that u2 = p D2N/60 and u1 = p D1N/60 the minimum speed (in rpm) to start the pump is u2 Vu2 g N min = 60 p [ D22 - D12 ]1/ 2 2g H ¢ (16.43) v1 V1 = Vf1 16.2.3 a1 = 90° b1 u1 Fig. 16.9 Radial Entry s For a pump, the speciﬁc speed Ns is deﬁned in SI units as N Q (16.40) H 3/ 4 where N = rotational speed in rpm for maximum efﬁciency, Q = discharge in m3/s, and H = net head developed in m. Note that the speciﬁc speed Ns has the dimensions of [L3/4 T–3/2]. The non-dimensional form of speciﬁc speed for pumps, known as shape number is N Q (16.41) Sq = (g H )3 / 4 Ns = Similarity Laws In the testing of pumps one is interested in the operation of pumps at a certain speed as determined by the motor to which it is coupled. Hence, the rotative speed N, diameter D are the basic repeating variables in the similarity relations. The similitude laws for discharge, head and power are expressed in terms of N and D. For homologous pumps, by using the sufﬁxes m and p to designate the model and prototype respectively, we have the following relations: Hp Hm = 2 2 (16.44-a) 2 2 Dm N m Dp N p Qm 3 N m Dm Pm 5 3 g m Dm Nm = = Qp N p Dp3 Pp g m Dp5 N p3 (16.44-b) (16.44-c) 527 Hydraulic Machines Note that only the power relation contains the ﬂuid property term, g. From the above relationships, it follows that for two homologous pumps the speciﬁc speed Ns will be the same. Thus between a geometric model and its prototype, ÊN Q ˆ Nsm = Á m 3/ 4 m ˜ = Nsp = Ë Hm ¯ 16.2.4 Qt = Qa + Qb H = Ha = H b Where ˆ ˜ (16.45) ˜¯ Pumps in Parallel If two similar pumps A and B are connected in parallel (Fig. 16.10) the combined discharge Qt will be the sum of the individual discharges Qa and Qb, i.e. Qt = Qa + Qb (16.46) Qb P P B Pumps Connected in Parallel The head H will however be the same in both the pumps and will also be the net head of the combined discharge. Thus, H = Ha = Hb (16.47) 16.2.5 Pumps in Series If two similar pumps 1 and 2 are connected in series (Fig. 16.11) the discharge will not change and the heads will be added up. Thus the total net head and the discharge P Ht = H1 + H2 (16.48) Q = Q1 = Q2 (16.49) Q 1 Fig. 16.11 H = head developed by the pump NPSH = Net positive suction head = A Fig. 16.10 Cavitation When the absolute local pressure at any point in a conduit carrying a liquid approaches the vapour pv of the liquid, the dissolved gases and liquid vapour emerge out of the liquid as bubbles. These bubbles may travel to regions of higher pressure and collapse. At the point of bubble collapse the boundary gets damaged. The phenomenon of formation, travel and collapse of vapour bubbles is known as cavitation. In pumps the suction end of the pump and blade passages are susceptible to cavitation. Cavitation in pumps causes reduction of the efﬁciency and often causes damage to the material of the pump assembly. An important factor in the pump operation is the avoidance of cavitation. A cavitation parameter s (also known as Thoma number) is deﬁned as ( NPSH ) s = (16.50) H Ê N p Qp Á ÁË H p3/ 4 Qa 16.2.6 P 2 Q H = H1 + H 2 Pumps in Series ( patm ) abs p – Zs – hL – v (16.51) g g in which ( patm ) abs = Pressure head (absolute) acting g upon the liquid surface at the sump, (normally atmospheric pressure) Zs = elevation of pump above the liquid surface in the sump. (If the pump is set below the liquid level in the sump then Zs would be negative) hL = head loss in the suction pipe assembly pv = Vapour pressure head of the liquid at the g prevailing temperature Minimum NPSH for a pump is usually speciﬁed by the manufacturer. Depending upon the pump, a critical cavitation number (= critical Thoma number) sc is identiﬁed. Any value of s < sc will result in cavitation and cause severe reduction in the pump efﬁciency. Hence for operational purposes it should be seen that 528 Fluid Mechanics and Hydraulic Machines s ≥ sc i.e., (NPSH) ≥ sc H Hence, the minimum NPSH = sc H (16.53) RECIPROCATING PUMP 16.3.1 Hd Introduction A reciprocating pump is a positive displacement pump. In this a piston moving inside a cylinder draws in the ﬂuid by suction and discharges it to the delivery pipe by pushing it bodily by the action of the piston. The piston gets its reciprocating motion through a crank and connecting mechanism that are connected to a prime mover. One-way valves provided in the suction and delivery sides of the piston help in pumping action. Figure 16.12(a) shows schematically a single acting reciprocating pump. In the single acting pump shown in Fig. 16.12(a) the suction stroke (crank angle 0° – 180°) gets the liquid from the sump to the cylinder and the delivery stroke (crank angle 180° – 360°) drives the liquid out of the cylinder. The suction and delivery strokes take place alternatively. The variation of the discharge with the crank angle is shown in Fig. 16.12(b). In a double acting reciprocating pump, the suction and delivery strokes occur simultaneously. Figures 16.13(a) and (b) show a schematic sketch of a double acting reciprocating pump and the variation of discharge with crank angle in such a pump. Multi-cylinder pumping arrangements are employed to get more steady ﬂow in the delivery pipe. In these the cranks in a common drive provide the necessary phase shift and the outlets from a set of multi-cylinders are connected to a common delivery pipe. Figures 16.14(a, b and c) show two-throw and three-throw pumps and the discharge performance of a three–throw pump respectively. 16.3.2 Receiving tank Notations A = Cross-sectional area of piston dp = Diameter of piston L = Length of stroke = twice the crank radius Delivery pipe Connecting rod Delivery valve Crank q Piston Suction Hs valve Suction pipe Sump (a) Suction stroke Discharge 16.3 (16.52) 0 60° 120° Delivery stroke 180° 240° Crank angle (b) 300° 360° Fig. 16.12 Single Acting Reciprocating Pump (a) Schematic Sketch (b) Variation of Discharge with Crank Angle r N Hd dd Ld Hs ds Ls Qt P 16.3.3 = crank radius = Revolutions per minute of the crank = Static delivery head = diameter of delivery pipe = Length of delivery pipe = Static suction head = Diameter of suction pipe = Length of suction pipe = Theoretical discharge = Power Discharge Theoretical discharge for single acting pump ALN Qt = (16.54) 60 529 Hydraulic Machines Delivery q = wt = 2 pN t 60 where D1 S1 D2 Piston S2 To connecting rod and crank Suction (a) Delivery D1 Discharge Delivery D2 O 60° 120° 180° 240° 300° Crank angle (b) 360° Fig. 16.13 Double Acting Reciprocating Pump (c) Schematic Sketch (d) Variation of Discharge with Crank Angle For a double acting pump 2 ALN (16.55) 60 Volumetric efﬁciency (percentage) Actual discharge he = ¥ 100 Theoretical discharge Qt = = Qt ¥ 100 Qa (16.56) Ê Q - Qa ˆ Slip = Á t ¥ 100 = 100 – he (16.56a) Ë Qt ˜¯ Coefﬁcient of discharge Qa = Qt 16.3.4 (16.57) Simple Harmonic Motion The motion of the piston in the cylinder is treated as a simple harmonic motion. The crank rotates with an angular velocity w radians/second. Then in a time t reckoned from inner dead centre N = rotational speed in rpm, r = crank radius = stroke/2 = L/2 Displacement x = r (1 – cos q) dx Velocity of piston v = = w r sin q = wr sin wt dt If A = area of piston and A1 = area of a pipe (suction or delivery pipe) Velocity of water in the pipe at any instant A v1 = w r sin q (16.58) A1 Maximum velocity in the pipe A v1m = wr A1 Time averaged velocity per cycle A wr V1 = (16.59) A1 p Acceleration of piston dv p = w 2 r cos wt =a= dt = w2 r cos q Acceleration of ﬂuid in the pipe (suction or delivery) at any instant A 2 = a1 = w r cos q (16.60) A1 Maximum acceleration in the pipe = a1m = A 2 w r A1 16.3.5 Acceleration Head The acceleration of ﬂuid in the pipe requires a force F given by Ê Aˆ F = rA1L1 Á ˜ w2r cos q Ë A1 ¯ The pressure head caused by this force F is Ha1 = F L A 2 = 1 w r cos q (16.61) A1rg g A1 Out 2 1 Piston In 1 Crankshaft set at 180° to each other Out 3 Crankshaft set at 120° to one another 3 Piston In (a) 120° (b) Resultant Cylinder 2 Cylinder 3 Discharge Cylinder 1 O Fig. 16.14 60° 120° 180° 240° Crank angle (c) 300° 360° 60° Multi Throw Reciprocating Pump (a) Two-throw Pump (b) Three-throw Pump (c) Variation of Discharge in a Three-throw Pump Fluid Mechanics and Hydraulic Machines 2 531 Hydraulic Machines Ha1 = is known as acceleration or inertial head. This is an additional head that is required to be developed by the pump. By replacing the sufﬁx 1 denoting any pipe by the speciﬁc sufﬁxes s to denote suction and d to denote delivery pipes F Ls A 2 Has = = w r cos q As rg g As Maximum Acceleration head in suction pipe is at q = 0 and is L A 2 H as m = s (16.62) w r g As For delivery pipe: F L A 2 Ha d = = d w r cos q Ad rg g Ad Maximum acceleration head in delivery pipe is at q = 0 and is Ld A 2 H adm = (16.63) w r g Ad Table 16.1 Friction Darcy–Weisbach formula with friction factor f is used to estimate the friction losses hfs and hfd in suction pipe and delivery pipe respectively. (i) For Suction Pipe 2 (16.64) ˆ fLs Ê A rw ˜ 2g ds ÁË As ¯ Beginning of Storke (q = 0) Mid-stroke (q = 90°) End of stroke (q = 180°) Total head in Delivery Pipe – (Has+ Hs + hfs) (absolute) (Had + Hd + hfd ) (absolute) hfs = 0 hfd = 0 Has = 0 Had = 0 hfs = 0 hfd = 0 Indicator Diagram Delivery hfd 5 2 2 (16.65) (ii) For Delivery Pipe 2 h fd ˆ fLd Ê A flVd2 = = 2g d ÁË A rw sin q ˜¯ (16.66) 2 g dd d d Maximum h fd is at (q = p/2) and hence h fdm = ˆ fLd Ê A rw ˜ Á 2g ds Ë Ad ¯ 4 Had Had Pressure head ˆ 1 fLs Ê A rw ˜ Á 3 g ds Ë As ¯ Htd = Hatmos An indicator diagram is a plot showing the variation of pressure in the cylinder at various stages of the strokes. The area of the indicator diagram represents the work done by the pump. Figure 16.15 is an indicator diagram showing effect of acceleration and friction in a single stage reciprocating pump. h fsa = Average hfs = (2/3) hfsm = (16.67) Total head in Suction Pipe Hts = H atmos Maximum hfs is at (q = p/2) and hence h fsm = 2 6 Hd O O 3 Has Hatmo Has 1 2 2 h fs Suction Stroke length Fig. 16.15 Indicator diagram Head (abs) flVs2 ˆ fLs Ê A rw sin q ˜ = 2g ds 2g ds ÁË As ¯ ˆ 1 fLd Ê A rw ˜ 3 g ds ÁË Ad ¯ Combined Effect of Acceleration and Item 16.3.7 Friction Head hfs = = Atmospheric 16.3.6 h fda = Average hfd = (2/3) hfdm 532 Fluid Mechanics and Hydraulic Machines against friction in suction pipe against friction in delivery pipe If L = length of the stroke, and noting that the mean ordinate of a parabola is equal to 2/3 of the maximum ordinate Total work done in one complete cycle of the crank = area of the indicator diagram = (Hs L + (2/3) h fs L + HdL + (2/3) hfd L) = L (Hs + h fsa + Hd + hfda) Total work done per second = Power expended g AN P= (Area of indicator diagram) 60 g ALN = (Hs + h fsa + Hd + hfda) (16.68a) 60 For a double acting pump 2 g ALN P= (Hs + h fsa + Hd + hfda) (16.68b) 60 The safe speed of the pump is decided of the following considerations: Ld = L¢d + Lda L¢d Air vessel Lda Pump Lsa Air vessel Ls = L¢s + Lsa L¢s Sump head should be more than the minimum head at which cavitation (Separation of vapour) can occur i.e., At (q = 0°) Hts = Hatmo– (Has + Hs) > Hv (abs) (16.69a) should be more than the minimum head at which cavitation (Separation of vapour) can occur At (q = 180°) Htd = Hatmo + (Had + Hd) > Hv (abs) (16.69b) 16.3.8 Air Vessel An air vessel is a large closed chamber ﬁtted to a reciprocating pump to eliminate pulsations of pressure and discharge in suction and delivery pipes. These are ﬁtted in either/both delivery and suction Fig. 16.16 Schematic Arrangement of Air Vessels sides and as close to the pump cylinder as possible. Figure 16.16 shows a schematic arrangement of two air vessels in a pump system. The chief advantages of air vessels in a reciprocating pump are (i) Reduces cavitation possibility (ii) For a given minimum pressure head, the pump can run at higher speed (iii) Suction pipe length can be increased (i) Almost constant delivery discharge is obtained (ii) Reduction in friction loss and hence saving in power Referring to Fig. 16.16, the acceleration heads are conﬁned to the lenghts Lsa and Lda only. Beyond the 533 Hydraulic Machines air vessels, the velocity is constant at Vs and Vd in the suction pipe of length L¢a and delivery pipe of length L¢d respectively. Hd + In the suction pipe Ê A ˆ Ê LN ˆ Ê A ˆ rw Qt = Á ˜Á ˜ = As Ë As ¯ Ë 60 ¯ ÁË As ˜¯ p Similarly in delivery pipe Ê A ˆ Ê LN ˆ Ê A ˆ rw Qt Vd = = Á ˜Á = Á ˜ ˜ Ad Ë Ad ¯ Ë 60 ¯ Ë Ad ¯ p Vs = If Qi = instantaneous discharge, then in any pipe (suction or delivery), Qi = A1v1 = A1wr sin q Discharge in air vessel 1ˆ Ê Qav = Qt – Qi = A1rw Á sin q - ˜ (16.70a) Ë p¯ For a double acting pump, 2ˆ Ê Qav = Qt – Qi = A1rw Á sin q - ˜ (16.70b) Ë p¯ (In the above equations, the sufﬁx 1 should be replaced by sufﬁx s for suction pipe and by sufﬁx d for delivery pipe). Qav is positive, then the ﬂow is into the air vessel. Qav is negative, then the ﬂow is out of the air vessel. acting pump, sin q = 1/p or q = 18.56° or 161.44°. double acting pump, sin q = 2/p or q = 39.54° or 140.46°. 16.3.9 Work Done Per Second (Power Consumed) Work done per second (Power consumed) with air vessels in the suction and delivery pipes is Pt = 2 ˆ g Qt È f ( Ls - Lsa ) 2 2 1 fLsa Ê A ÍHs + Vs + r w ˜¯ + 1000 Í 2g ds 3 g ds ÁË As Î 2 ˆ ˘ f ( Ld - Lda ) 2 2 1 fLda Ê A ˙ w Vd + r ˜¯ ˙ g dd 3 2g dd ÁË Ad ˚ (16.71) In this Qt = Theoretical discharge = ALN and 60 2 pN . 60 If h is the efﬁciency of the pumping system, then the power (in kW) to be supplied is P (16.72) P= t h w= 16.4 MISCELLANEOUS HYDRAULIC MACHINERY AND DEVICES 16.4.1 Introduction While the turbines and pumps described in the previous sections form important hydraulic machines, there are a host of other hydraulic machines and devices that are used as components of a ﬂuid system. A few of these devices and a few interesting types of pumps are brieﬂy described in this section. The devices described below are used to assist in power transmission and include 1. 2. 3. 4. 5. Hydraulic press Hydraulic accumulators Hydraulic intensiﬁers Fluid coupling Fluid torque converter The pumps described include (i) hydraulic ram, (ii) gear pump, and (iii) jet pump. 16.4.2 Hydraulic Systems A hydraulic system is a circuit in which power and forces are transmitted through a liquid. These can be classiﬁed in to two categories as (1) hydrostatic systems, and (ii) hydro dynamic systems Hydrostatic systems In these systems, the power and force are transmitted primarily by the ﬂuid static pressure. The motivating force in such systems is 534 Fluid Mechanics and Hydraulic Machines a change in pressure whereas the velocity of the ﬂuid usually remains constant. There is no energy transfer as kinetic energy to and from the ﬂuid. The hydraulic press, hydraulic accumulator and hydraulic intensiﬁer are examples of hydrostatic systems. Hydrodynamic system In this form of power transmission, energy is transferred by a change in velocity or kinetic energy. The change in the pressure of the working ﬂuid is generally of no consequence. Fluid coupling and torque converter are typical examples of this type of system. 16.4.3 Based on this principle, working devices have been developed and the hydraulic presses have been used as a standard industry device for providing large compressive forces. A schematic hydraulic press is shown in Fig. 16.18. The plunger/ram is activated by a pump providing the hydraulic pressure and causes a displacement of a movable platform. The material placed between the stationary platform and movable platform undergo high compressive forces. Common industrial use of hydraulic press include, Compression forming, blanking and forming and punching. Hydraulic Press Stationary platform Hydraulic press is a device where a smaller force is used to provide a much higher force for purposes of providing compressive or lifting force. To understand the principle of hydraulic press, consider two interconnected piston-cylinder units as in Fig. 16.17. Fixed frame Fm Movable platform Ram N Piston From supply M Liquid Fig. 16.17 Fig. 16.18 Basics of Hydraulic Press The cylinders M and N contain an incompressible ﬂuid. Force Fm applied on piston at M transforms in to pressure p = Fm/A1 where A1 is the area of cylinder M. By Pascal’s law the pressure is common to all points in the ﬂuid and hence it will also be acting on the bottom of piston/ram of cylinder N. The force applied to the bottom of the piston of the larger cylinder N having an area A2 is then F A Fn = pA2 = m 2 A1 (16.73) If the volume of the ﬂuid is held constant, the displacement of the larger piston, relative to the smaller piston, will be proportionately smaller. Hydraulic Press Similar to the principle of a hydraulic press, the arrangement of a ram working in a cylinder under hydraulic pressure created by a pump ﬁnds applications in many industrial applications. These include the hydraulic jack, hydraulic lift and hydraulic crane. 16.4.4 Hydraulic Accumulator Hydraulic accumulator is a device, akin to that of a storage battery, to store hydraulic liquid under pressure when not required by the system. Normally, pumps in a hydraulic system work continuously and when they are idling due to no load, hydraulic accumulators serve the purpose of storing the ﬂuid under pressure to be released as and when required. 535 Hydraulic Machines A hydraulic accumulator consists essentially of a high pressure cylinder in which a ram can move. Figure 16.19 is a schematic representation of a simple hydraulic accumulator. When the load on the pump in a hydraulic system is reduced or rejected, the high pressure liquid ﬂows into the accumulator and causes the ram to move. A resistance load on the movement of the ram is provided either through a dead weight or in the form of a compression spring. This arrangement keeps the liquid in the accumulator under required pressure. When needed, this pressure liquid from the accumulator is released through the out let to the desired hydraulic machine/device. The release of the pressure liquid causes the ram to descend towards its original position. Weight when the ram falls uniformly through a distance of L in time t, P= WL t …(16.75) A hydraulic intensiﬁer is a component that converts the low pressure form a cylinder into high pressure in a smaller cylinder. Intensiﬁers consist essentially of two different-sized cylinders connected by a common piston. Intensiﬁers operate on the ratio-of-areas principle in interconnected cylinders. A common rod connects the pistons of two cylinders of different bore, Fig. 16.20. Lower-pressure ﬂuid, acting on the larger piston, exerts, a force that is transferred mechanically by the rod to the smaller piston. The smaller piston generates a higher pressure in the ﬂuid in its bore: the pressure ratio is inversely proportioned to the areas ratio. RAM Liquid under pressure Inlet Fig. 16.19 High pressure discharge Low pressure hydraulic liquid Suction for intensifier (low pressure) Outlet Hydraulic Accumulator The commonly used types of accumulators can be classiﬁed as (i) raised weight type, (ii) spring type, (iii) metal bellows type, and (iv) compressed gas type. Figure 16.19 shows the raised weight type of accumulator. In this the pressure p of the liquid in the accumulator is related to the weight W on the ram of diameter D by the simple relation W= To reservoir p D2 p 4 Fig. 16.20 Let sufﬁxes 1 and 2 denote the larger and smaller cylinders respectively. For small cylinder velocities, that is when there is no acceleration head, p p p1 D12 = p2 D22 4 4 2 …(16.74) If L = Stroke of the ram = length of the ram movement, the power supplied by the accumulator (16.76) p2= p1 Ê D1 ˆ ÁË D ˜¯ 2 Thus the input pressure is increased by a factor which is equal to the square of the ratio of the inlet to 536 Fluid Mechanics and Hydraulic Machines outlet cylinder diameters. The ratio p2/p1 is called as intensiﬁcation ratio. Another form of intensiﬁer is the coaxial type. (Fig. 16.21). The arrangement consists of a ﬁxed cylinder, a movable hollow cylindrical ram and a ﬁxed ram. Initially the low pressure liquid enters through the ﬁxed ram and enters in to the hollow of the movable ram. This pushes the movable ram outwards till it reaches its full stroke length. Now the low pressure inlet valve is closed and the low pressure liquid supply is let in to the ﬁxed cylinder which pushes the movable ram (down in the Figure) causing the liquid in it to escape through the ﬁxed ram to the outlet. This action causes the out ﬂowing liquid to have a higher pressure. If p1 = pressure of low pressure supply A1 = inside cross sectional area of ﬁxed (larger) cylinder of diameter D1 A2 = inside cross sectional area of movable ram of inside diameter D2 Low pressure supply Fixed cylinder Movable ram Low pressure inlet valve Fixed ram High pressure outlet valve High pressure out flow Fig. 16.21 Delivery pressure at the outlet ÊAˆ ÊD ˆ p2 = p1 Á 1 ˜ = p1 Á 1 ˜ Ë A2 ¯ Ë D2 ¯ 2 The above relationship assumes that there is no frictional loss in the movement of the ram. If, however, there is frictional effect amounting to e % at each of the packings of the ram, from equilibrium considerations of the moving ram at any position, p2 = p2 A1 Ê e ˆ 1Á Ë 100 ˜¯ A2 2 (16.77) Hydraulic intensiﬁers are simple, rugged, and reliable ﬂuid power components. 16.4.6 Fluid Coupling The hydraulic coupling (also generally called as Fluid coupling) is the simplest means of transmitting torque hydraulically. It can be deﬁned as a device in which a ﬂuid, usually oil, transmits torque from one shaft to another, producing an equal torque in the other shaft. A ﬂuid coupling is widely used to transfer rotating power from a prime mover, such as an internal combustion engine or electric motor, to a rotating driven load. The ﬂuid coupling consists basically of two elements: a centrifugal pump or impeller connected to the driving shaft, and a turbine wheel or runner on the output (driven) shaft. There is no mechanical connection between both shafts. Figure 16.22 represents a schematic view of a ﬂuid coupling. The device consists of two split toroidal grooved discs, each facing the other with a small clearance between them. Radial blades are provided across the grooves to divide them in to curved cells. The hollow space in the impeller and the runner is ﬁlled with oil. When the driving member or impeller, begins to rotate (as the engine is started and runs), the oil is set into motion. The vanes in the driving member start to carry the oil round with them. As the oil is spun round, it is thrown outward or away from the shaft, by centrifugal force. However, since the oil is being carried round with the rotating driving 537 Hydraulic Machines Pump impeller Turbine runner w1 Driving shaft Fig. 16.22 w2 Driven shaft Fluid coupling member, it is thrown into the driven member. The oil thus strikes the vanes of the driven member at an angle, thereby imparting torqre to the driven member due to transfer of kinetic energy. The inlet angles of the runners are set such that the ﬂow from the impeller enters without any shock. The oil in the turbine moves towards the shaft of the wheel from where it ﬂows to the pump to complete a closed ﬂuid circuit. Since the rate of ﬂow and change in the velocity vector are numerically same in both the impeller and the runner, the torque of the input and output shafts of a ﬂuid coupling are identical at all speeds, if friction and other losses are neglected. If the speeds of both the impeller and the runner are same there would be no ﬂow. However, due to ﬂuid friction and turbulence effects, the angular velocity of the driven shaft w2 will be a little less than the angular velocity of the driving shaft w1. The Ê w - w1 ˆ ratio Á 2 is known as slip and the slip is Ë w1 ˜¯ generally very small being about 3% at peak speed. The necessary reduction of the speed of the driven shaft thus maintains continuous ﬂow of oil from the impeller to the runner. Thus, the power loss is very small and the torque ratio between the output and input shaft is very near unity. The condition when the speed of the driving and driven shafts are the same is known as stall. The operational characteristics of a ﬂuid coupling show that for a given slip the input torque increases as the cube of the driving shaft speed. Hence, the power transmitted also varies with the cube of the speed for a given slip. In ﬂuid coupling, low viscosity ﬂuids are generally preferred and the physical properties, like density and viscosity, of the oil determine the operation characteristics of the coupling. For example, increasing the density of ﬂuid increases the torque that can be transmitted at a given speed. Fluid couplings are vary widely used and their application includes Diesel locomotives, Automobiles, Aviation and Marine machinery and Agricultural machinery. 16.4.7 Torque Converter A torque converter is a modiﬁed form of ﬂuid coupling. Like a ﬂuid coupling, the torque converter normally takes the place of a mechanical clutch, allowing the load to be separated from the power source. Unlike a ﬂuid coupling, however, a torque converter is able to multiply torque when there is a substantial difference between input and output rotational speed, thus providing the equivalent of a reduction gear. A torque converter consists of (i) a pump impeller connected mechanically to the driving shaft, (ii) turbine runner connected to the driven shaft, and (iii) a stator, usually known as a reaction member, positioned in the middle of the ﬂow from the impeller and the runner. The stator has guide vanes which change the direction to the liquid impinging on the runner and thus causes the torque delivered to the driven shaft to be higher than that of the driving shaft. Figure 16.23 is a schematic sketch of a simple ﬂuid torque converter. In a ﬂuid coupling, under conditions of high slippage the ﬂuid ﬂow returning from the turbine to the pump opposes the direction of pump rotation. This leads to a signiﬁcant loss of energy. Under the 538 Fluid Mechanics and Hydraulic Machines 16.4.8 Stationary guide vane Turbine runner Pump impeller w1 w2 Hydraulic ram is a pumping device that utilizes the principle of water hammer to lift small quantities of water to a higher level through use of large quantities of water at lower head. Fig. 16.24 shows a schematic layout of the hydraulic ram set-up. The source is a tank or a stream. A pipeline connects the source to the ram located at a lower level. The pumping device consists of a chamber C, set of valves V1, and V2. Driven shaft Driving shaft Supply source Fig. 16.23 Hydraulic Ram Torque Converter same condition in a torque converter, the returning ﬂuid will be redirected by the stator so that it aids the rotation of the pump, instead of impeding it. This leads to recovery of much of the energy in the returning ﬂuid and addition to the energy being supplied by the pump itself. This action causes a substantial increase in the mass of ﬂuid being directed to the turbine, producing an increase in the output torque. Unlike the radially straight blades used in a ﬂuid coupling, the turbine and stator of a torque converter use angled and curved blades. The shape of the blades is important as even minor variations can result in signiﬁcant changes in the performance of the device. Efﬁciency curve of a typical torque converter indicates that the converter attains its maximum efﬁciency at a speed ratio of about 0.5 and at higher speed ratios the efﬁciency drops. There have been many advances over the basic features given above and torque converters are used extensively in (i) Automatic transmissions on automobiles, such as cars, buses and light trucks, (ii) marine propulsion systems, and (iii) industrial power transmission. H1 Air vessel Waste Valve V2 Delivery pipe H2 Valve V1 C G Inlet valve Chamber Fig.16.24 Hydraulic Ram Operation To start with, the gate valve G in the pipe is opened. Water rushes down the pipe in to the chamber and some water passes out through the waste valve V2 which is open. The buildup of the dynamic pressure within the chamber C at the seat of the valve V2 causes the valve to shut down suddenly. This sudden closure and consequent sudden change of momentum causes a pressure build up in the chamber. The excess pressure causes the valve V1 to open and the water rushes to the air chamber. The compression of the air in the air chamber causes the water to pass out through the delivery pipe to the reservoir at its outlet. The dissipation of the pressure wave caused by the momentum change causes the valve V1 connecting the air chamber to close and also the waste valve V2 to open. The process repeats. The valves V1 and V2 are one way valves and act due to self weight or due to spring loading. It is usual to provide an air bleed valve, called snifting valve, to prevent formation of negative pressures in the chamber due to air entrainment action of the ﬂows in the chamber. 539 Hydraulic Machines Let hfs and hfd be the frictional head losses in the supply and delivery pipes respectively. Further, H1 = height of water surface in the source above the chamber of the ram H2 = Height of delivery reservoir above the chamber Q1 = discharge delivered by the ram Q2 = discharge wasted by the ram Qs = discharge supplied to the ram = (Q1 + Q2) Discharge line From sump Connected to motor The Work done per second = gQ1 (H2 + hfd) Work done in supplying the ﬂow to the ram = g Qs(H1 – hfs) = g (Q1 + Q2)(H1 – hfs) Efﬁciency of the ram = Q1 ( H 2 + hfd ) h= (Q1 + Q2 ) ( H1 - hfs ) If frictional losses are neglected Q1H 2 h= (Q1 + Q2 ) H1 (16.78) Fig. 16.25 (17.78-a) In general, a ram can pump approximately one tenth of the received water volume to a height of about ten times greater than the intake. A hydraulic ram pump is useful where the water source ﬂows constantly and the usable fall from the water source to the pump location is at least 1.0 m 16.4.9 Nozzle Throat Impeller Jet Pump The efﬁciency of the pump is deﬁned as h= Qs ( H s + H d ) Qn ( H n - H d ) …(16.79) where Qn = discharge through the nozzle, Qs = discharge through the suction pipe, Hn = pressure head applied at the nozzle, Hs = suction head, Hd = delivery head. Generally the efﬁciency of the jet pump is low, being less than 50% Jet Pump A jet pump is a combination of a normal centrifugal pump and a jet device at the suction end. When the pump is started a part of the water from delivery side of the pump is diverted in to a nozzle. Water under high pressure is forced through this nozzle in to the throat of a venturimeter shaped converging pipe-diffuser section of a pipe located in the suction side of the pump assembly. The negative pressure caused by the jet ﬂow causes the water to be sucked up from the sump and delivers it to the pump. This causes the suction head of the pump assembly to be larger. Fig. 16.25 is a schematic sketch of a jet pump assembly. As much as 5 to 6 m of suction lift can be obtained by this device. 16.4.10 Gear Pump Gear pumps are positive displacement pumps. Here a pair of identical spur gears meshed inside a closed casing, rotate in opposite directions as shown in Fig. 16.26. The casing has an inlet and outlet for liquid to be pumped. The teeth of the gear have a perfect meshing that prevents any leakage and also aids in pushing the ﬂuid forward. The rotation of the gears causes the liquid in the casing to be bodily pushed continuously. There is no dynamic action of imparting pressure or kinetic energy as in other rotodynamic pumps. The ﬂow is continuous, uniform and very high pressures can be achieved. 540 Fluid Mechanics and Hydraulic Machines Outlet Inlet Fig. 16.26 Gear Pump Gear pumps ﬁnd application in lubrication of internal combustion engines and in hydraulic control of machines. The actual discharge from the pump is given by the following empirical formula: Q = 0.95 hpc (D – c)LN/60 m3/s (16.80) where D = outside diameter of the gears c = centre to centre distance between the axis of the gears L = axial length teeth N = revolutions per minute h = volumetric efﬁciency of the pump Gradation of Numericals All the Worked Examples, Objective Questions and Problems have been graded in three levels—Simple, Medium and Difﬁcult. The markings for these are given below. Simple Medium Difﬁcult Worked Examples u1 A. Francis Turbine a1 16.1 b1 = 90° v1 = Vf1 V1 Inlet Solution: Referring to Fig. 16.27, pD1N p ¥ 1.2 ¥ 250 u1 = = 15.7 m/s = 60 60 As b1 = 90°, and v1 = Vf1 = 3.5 m/s tan a1 = v1 3.5 = = 0.2228 u1 15.7 Fig. 16.27 (i) a = 12.56° V1 = absolute velocity at entry = = 3.5 = 16.093 m/s sin (12.56∞) v1 sina 541 Hydraulic Machines (ii) Discharge Q = pD1b1 Vf1 = p ¥ 1.2 ¥ 0.25 ¥ 3.5 = 3.299 m2/s (iii) At outlet Q = pD2b2 Vf 2 3.299 = p ¥ 0.6 ¥ 0.35 ¥ Vf2 Vf2 = velocity of ﬂow at outlet = 5.0 m/s 16.2 Flow velocity at inlet 8.0 = 6.366 m/s p ¥ 0.4 Velocity of ﬂow at outlet Vf1 = D1 ¥ 6.366 D2 2.0 = ¥ 6.366 1.2 = 10.61 m/s As the vanes are radial at the inlet, v1 = Vf1 = 6.366 m/s. From the inlet velocity triangle, Vf2 = Vf1 6.366 = = 0.243 u1 26.18 \ a1 = 13.67° = Inlet guide vane angle At the outlet, a2 = 90° tan a1 = Solution: Referring to Fig. 16.28, u1 Vf 2 10.61 = = 0.675 u2 15.708 b2 = 34.04° = outlet blade angle tan b2 = a1 b1 = 90° v1 = Vf1 V1 \ 16.3 Inlet u2 a2 = 90° b2 V2 = Vf2 v2 Outlet Fig. 16.28 Peripheral velocity pD1N p ¥ 2.0 ¥ 250 = 60 60 = 26.18 m/s u1 = D2 1.2 u1 = ¥ 26.18 D1 2.0 = 15.708 m/s At the outlet u2 = Discharge Q = pD1bVf1 8.0 = p ¥ 2.0 ¥ 0.2 ¥ Vf1 Solution: Assuming constant ﬂow velocity and radial discharge at outlet Vf1 = Vf 2 = V2 Flow ratio = y = Vf 1 = 0.15 2gH 542 Fluid Mechanics and Hydraulic Machines Vf1 = Vf2 = 0.15 ¥ Vu 2 ¥ 9.81 ¥ 70 u1 = 5.56 m/s Overall efﬁciency h0 = hm ¥ hH = 0.84 ¥ 0.95 = 0.798 Now power developed P = h0g QH. Hence discharge, Q = a1 Since B1 D1 B1 Also since D2 Peripheral velocity v1 V1 P h0g H Vf 1 (a) Inlet 370 ¥ 10 0.798 ¥ 9790 ¥ 70 = 0.677 m3/s = (peripheral area) ¥ Vf1 = (1 – 0.05) ¥ pD1B1 ¥ 5.56 0.677 = = 0.0408 p ¥ 0.95 ¥ 5.56 = 0.1 D1, D 12 = 0.408 and = 0.639 m = 0.1 D1 = 0.064 m = 0.5 D1, D2 = 0.319 m a2 = D1B1 b 1¢ b1 3 But discharge 0.677 1 pD1N p ¥ 0.639 ¥ 750 = 60 60 = 25.09 m/s pD2 N p ¥ 0.319 ¥ 750 u2 = = 60 60 = 12.55 m/s uV Hydraulic efﬁciency hH = 1 u1 = 0.95 gH Hence, swirl velocity at entry u1 = Vf = Vf 1 2 u2 b2 v2 (b) Outlet Fig. 16.29 Example 16.3 Form inlet velocity triangle at inlet: b1 = 180 – b 1¢ Vf 1 5.56 tan b 1¢ = = Vu1 - u1 ( 26.0 - 25.09) = 6.129 b 1¢ = 80.73° and b1 = 180 – b 1¢ = 99.27° (c) From outlet velocity triangle V 5.56 tan b 2 = f 2 = = 0.443 u2 12.547 b2 = 23.90° 16.4 hH gH 0.95 ¥ 9.81 ¥ 70 = u1 25.09 = 26.0 m/s Vu1 = (a) Guide vane angle at inlet, a1 From inlet velocity triangle at inlet: V 5.56 tan a1 = f 1 = = 0.2138 Vu1 25.093 a1 = 12.07° (b) Blade angle at inlet, b1 Since Vul > u1, angle b1 is obtuse and the velocity triangle at inlet will be as shown in Fig. 16.19. Solution: Given: Q = 12.0 m3/s, 543 Hydraulic Machines P = 13000 kW, N = 450.0 rpm, Vf1 = 10.0 m/s, D1 = 1.5 m pD1N 60 p ¥ 450 ¥ 1.5 u1 = 60 = 35.34 m/s Power produced P = rQu1 Vu1 Swirl velocity at entry Hence, effective head He = At entrance Peripheral velocity u1 = Vul = P rQu1 13 ¥ 106 998 ¥ 12 ¥ 35.34 = 30.716 m/s = (i) Guide vane angle at inlet, a1 From inlet velocity triangle at inlet: V 10.0 tan a1 = f 1 = = 0.3256 Vu1 30.716 a1 = 18° (ii) Blade angle at inlet, b1 Since Vul < u1, angle b1 is acute and the velocity triangle at inlet will be as shown in Fig. 16.30 u1 Vu 1 a1 b1 Vf V1 Fig. 16.30 1 v1 Example 16.4 From inlet velocity triangle at inlet, 10.0 Vf 1 tan b1 = = (35.34 - 30.716) u1 - Vu1 = 2.163 b1 = 65.18° Now power developed by reaction of water ﬂow P = gQHe. where He = effective head. P 13 ¥ 106 = = 110.66 m 9790 ¥ 12 gQ (Note: This is also equal to Ê u1 - Vu1 ˆ and is ÁË g ˜¯ sometimes known Euler head) Applying energy equation to the entrance to the runner (section 1) and exit from the runner (section 2): H1 = H2 + He + HL where H1 = Total energy head at Section 1 È p1 ˘ V12 = Í + Z1 ˙ + Îg ˚ 2g H2 = Total energy head at Section 2 Èp ˘ V2 = Í 2 + Z2 ˙ + 2 Îg ˚ 2g He = Effective head and HL = Energy head lost in the runner V1 = absolute velocity at Section 1 = Vf21 + Vu21 = (10.0) 2 + (30.716) 2 = 32.30 m/s V12 (32.30) 2 = = 53.17 m 2g 2 ¥ 9.81 V2 = absolute velocity at Section 2 = ﬂow velocity = Vf 2 = Vf1 = 10.0 m/s V22 (10.00) 2 = = 5.10 m 2g 2 ¥ 9.81 From given data, difference in piezometric heads between section 1 and 2 = È p1 ˘ È p2 ˘ + Z 2 ˙ = 70.0 m Í + Z1 ˙ - Í Îg ˚ Îg ˚ Energy equation is now HL = [H1 – H2] – He ÏÔ È p ˘ Èp ˘ ¸Ô ÏÔV 2 V 2 ¸Ô = Ì Í 1 + Z1 ˙ - Í 2 + Z 2 ˙ ˝ + Ì 1 - 2 ˝ – He ÔÓ Î g ˚ Îg ˚ Ô˛ ÔÓ 2g 2g Ô˛ HL = 70.0 + (53.17 – 5.10) – 110.66 = 7.41 m 544 Fluid Mechanics and Hydraulic Machines From the inlet velocity triangle Vf1 3.537 tan a1 = = = 0.4159 Vu1 8.505 a1 = 22.58° 16.5 16.6 b Vu1 u1 a1 b¢1 b1 v1 Vf1 Solution: Refer to Fig. 16.32. V1 u1 A V2 = Vf2 60° b1 Inlet v1 Vf1 V1 u2 90° b¢1 B 20° a1 a2 Vu1 C b2 v2 u2 a2 Fig. 16.31 Example 16.5 V2 Solution: pD1N p ¥ 3.0 ¥ 200 = 60 60 = 31.42 m/s Discharge Q = pD1 bVf1 30 = p ¥ 3.0 ¥ 0.9 ¥ Vf1 Vf1 = velocity of ﬂow at inlet = 3.537 m/s Power P = rQ(u1 Vu1 – u2 Vu2) Since the outlet ﬂow is radial a2 = 90° and Vu2 = 0 P = rQ u1Vu1 8000 ¥ 103 = 998 ¥ 30.0 ¥ 31.42 ¥ Vu1 Vu1 = 8.505 m/s b2 90° Outlet v2 u1 = Fig. 16.32 Example 16.6 At the inlet a1 = 20°, b 1¢ = 60°, b = 120° pD1N p ¥ 1.2 ¥ 450 u1 = = 60 60 = 28.27 m/s ˆ = 40° Form DABC, ACB u1 V1 = sin 40∞ sin 120∞ 545 Hydraulic Machines V1 = 28.27 ¥ sin 120∞ = 38.09 m/s sin 40∞ Velocity of ﬂow Vf1 = V1 sin 20° = 38.09 sin 20° = 13.03 m/s Velocity of whirl Vu1 = 38.09 cos 20° = 35.79 m/s Discharge Q = area ¥ Vf1 = 0.4 ¥ 13.03 = 5.212 m3/s Power developed P = r Q(u1 Vu1 – u2 Vu2) But Vu2 = velocity of whirl at outlet = 0 (i) Hence P = r Qu 1 Vu1 = 998 ¥ 5.212 ¥ 28.27 ¥ 35.79 = 5263 ¥ 103 W = 5263 kW (ii) Hydraulic Efﬁciency 28.27 ¥ 35.79 uV h H = 1 u1 = 9.81 ¥ 115 gH = 0.897 = 89.7% 16.7 Vu1 A a1 u1 a1 = 30°, b1 = 120°, b 1¢ = 60° ˆ = 90° In triangle ABC, ACB V1 = u1 cos 30° = 0.866 u1 Vu1 = V1 cos 30° = (0.866)2 u1 = 0.75 u1 As the outﬂow is radial, Vu2 = 0 u1Vu1 Head extracted He = = hH . H g 0.75 u12 = = 0.88 ¥ 15.0 9.81 = 13.2 m u1 = 13.14 m/s If D1 = Diameter of the runner at the inlet pD1N u1 = 60 p ¥ D1 ¥ 1000 13.14 = 60 (i) D1 = 0.25 m = 25 cm (ii) Discharge Q = p D1bVf1 0.25 D b = 1 = = 0.0625 m 4 4 Vf1 = V1 sin 30° = 0.866 ¥ 13.14 ¥ 0.5 = 5.69 m/s Q = p ¥ 0.25 ¥ 0.0625 ¥ 5.69 = 0.279 m3/s Available power = g QHe Developed power = ho g QH where h o = overall efﬁciency P = h o g QH = 0.85 ¥ (9.81 ¥ 998) ¥ 0.279 ¥ 15 = 34864 W = 34.864 kW B 30° b1 Vf1 120° Inlet v1 V1 C Fig. 16.33 Solution: Example 16.7 16.8 546 Fluid Mechanics and Hydraulic Machines Vf1 = tan b1 = tan 80° = 5.671 u1 - Vu1 u1 – Vu1 = 0.1763 Vf1 u1 = Vu1 + 0.1763 Vf1 = (2.4751 + 0.1763) Vf1 u1 = 2.6514 Vf1 On the outlet side: D2 0.45 u2 = u1 = u1 = 0.75 u1 D1 0.60 = 0.75 ¥ 2.6514 Vf1 = 1.9885 Vf1 Also Solution: Refer to Fig. 16.34. u1 Vu1 A a1 B 22° b1 = 80° Since tan 25° = v1 V1 Inlet Vf2 u2 + Vu2 Vf1 (∵Vf1 = Vf2) 1.9885 Vf1 + Vu2 Vu2 = 0.156 Vf1 V Vf1 tan a ¢2 = f2 = Vu2 0.156 Vf1 tan a ¢2 = 6.41, a ¢2 = 81.13° or a2 = 98.87° Actual head extracted: u1Vu1 - u2Vu2 He = g 0.4663 = C v2 b2 V2 Vf2 Outlet a2 25° u2 Fig. 16.34 Vu2 Example 16.8 Given data: a1 = 22°, b1 = 80°, D1 = 0.60 m b2 = 25°, D2 = 0.45 m, Ht = 60 m Discharge Q = p D1b1(1 – 0.05) Vf1 = p D2b2(1 – 0.05) Vf2 Vf1 D1b1 0.60 0.06 = = ¥ = 1.0 Vf2 D2b2 0.45 0.08 Hence Vf1 = Vf2 ˆ = 78° At the inlet from D ABC, ACB V1 u1 = sin 80∞ sin 78∞ V1 = 1.0068 u1 Vf1 tan a1 = = tan 22° = 0.404 Vu1 Vf1 = 0.404 Vu1 i.e. Vu1 = 2.4751 Vf1 È ( 2.6514 ¥ 2.4751) - (1.9885 ¥ 0.156) ˘ 2 =Í ˙ Vf 1 9.81 Î ˚ = 0.6373 V 2f1 Since the hydraulic efﬁciency H h H = e = 0.90 H He = 0.6373 V2f1 = 0.90 ¥ 60 Vf1 = 9.205 m/s p ¥ 0.60 ¥ N u1 = 2.6514 ¥ 9.205 = 60 N = 776.9 rpm Discharge Q = (1 – 0.05) ¥ p D1b1Vf1 = 0.95 ¥ p ¥ 0.6 ¥ 0.06 ¥ 9.205 = 0.989 m3/s Power delivered = g QH h m = (9.81 ¥ 998) ¥ 0.989 ¥ (0.90 ¥ 60) ¥ 0.95 547 Hydraulic Machines u1Vu1 = V 2f1 cot a1 (cot a1 + cot b1) = 497 ¥ 103 W = 497 kW u1Vu1 Vf12 + g 2g 1 gH = [2 V 2f1 cot a1 (cot a1 + cot b1) + V 2f1] 2 1 2 = V f1 [2 cot a1 (cot a1 + cot b1) + 1] 2 uV hH = 1 u1 gH H = 16.9 Show that for a Francis turbine the È h H = Í1 Î ˘ 1 ˙ 1 + 2 cot a1(cot a1 + cot b1) ˚ a1 = b1 = Solution: The velocity of ﬂow is constant. Hence Vf1 = Vf2. Since the outlet ﬂow is radial V2 = V f2. From the inlet velocity triangle (see Fig. 16.35) Vf12 ( 2 cot a1 ) (cot a1 + cot b1 ) Vf12 [2 cot a1 (cot a1 + cot b1 ) + 1] È ˘ 1 hH = 1 – Í ˙ 1 2 + cot a (cot a + cot b ) 1 1 1 ˚ Î 16.10 A Francis turbine works under a head of 3 u1 Vu1 a1 b1 Vf1 V1 Fig. 16.35 v1 Inlet Velocity Triangle-Example 16.9 Vf1 = tan a1, Vu1 hence Vu1 = Vf1 cot a1 Vf1 = tan b1, hence u1 – Vu1 = Vf1 cot b1 u1 - Vu1 u1 = Vu1 + Vf1 cot b1 = Vf1 (cot a1 + cot b1) Head extracted uV V2 V2 He = 1 u1 = H – 2 = H – f 1 g 2g 2g where H = net available head. Also \ h H = hydraulic efﬁciency = u1Vu1 / g u1Vu1 = u V gH 1 u1 + V22 / 2g g 2gH 2gH where H Solution: In Fig. 16.36 showing the velocity triangle at the inlet, b1 is acute. Given: Q = 10.0 m3/s and H = 30.0 m. At inlet, Peripheral velocity = u1 = 0.9 2gH = 0.9 2 ¥ 9.81 ¥ 30 = 21.83 m/s Velocity of ﬂow = Vf1 = 0.3 2gH = 0.3 2 ¥ 9.81 ¥ 30 = 7.278 m/s (i) Power developed = P = h0g Q H = 0.8 ¥ 9790 ¥ 10 ¥ 30 = 2349600 W = 2349.6 kW 548 Fluid Mechanics and Hydraulic Machines discharge = (peripheral area) ¥ Vf1 = pD1 B1 ¥ Vf1 10.0 = p ¥ 1.39 ¥ B1 ¥ 7.278, giving B1 = 0.3146 m = 31.46 cm Hydraulic efﬁciency u1Vu1 hH = = 0.90 gH Swirl velocity at entry But, hH gH 0.9 ¥ 9.81 ¥ 30 = u1 21.83 = 12.133 m/s (ii) Guid vane angle at inlet, a1 From inlet velocity triangle at inlet: Vf 1 7.278 tan a1 = = = 0.5998 Vu1 12.133 a1 = 30.98° (iii) Blade angle at inlet, b1 Since Vul < u1, angle b1 is acute and the velocity triangle at inlet will be as shown in Fig. 16.36 Vu1 = u1 Vu1 a1 b1 Vf1 V1 Fig. 16.36 v1 (v) u1 = N P H 5/ 4 = 300 ¥ 2349.6 (30)5 / 4 16.11 Solution: u1 = 2.0 2g H Here H = H t = 30 m Hence u1 = 2.0 ¥ 2 ¥ 9.81 ¥ 30 = 48.52 m/s Diameter of boss D b = 0.35 D Flow ratio = Vf1/ 2gH = 0.65 Speed ratio f = = 15.77 m/s = 207 pD1N 60 pD1 ¥ 300 or 60 21.83 ¥ 60 D1 = =1.39 m p ¥ 300 21.83 = Vf1 = 0.65 2 ¥ 9.81 ¥ 30 Example 16.10 From inlet velocity triangle at inlet: Vf 1 7.278 tan b1 = = u1 - Vu1 ( 21.83 - 12.133) = 0.750 b1 = 36.88° (iv) Speciﬁc speed Ns = B. Kaplan Turbine Power P = g QH h0 15000 ¥ 103 = (9.81 ¥ 998) ¥ Q ¥ 30 ¥ 0.90 Q = 56.745 m3/s p But discharge Q = (D2 – D 2b ) ¥ Vf1 4 p 56.745 = [D2 – (0.35 D)2] ¥ 15.77 4 = 10.868 D2 D = 2.285 m D b = 0.35 ¥ 2.285 = 0.80 m If speed = N rpm, pD N u1 = 60 p ¥ 2.285 ¥ N 48.52 = 60 N = 405.5 rpm 549 Hydraulic Machines (iii) Speciﬁc speed Ns = Here H 5/ 4 H = 30 m, P = 15000 kW, N = 405.5 rpm Ns = N P 405.5 ¥ 15000 (30)5 / 4 = 707.4 But u1 = pD N p¥4¥N = 60 60 \ N = 60 ¥ 38.72 = 184.9 rpm p¥4 Speciﬁc speed Ns = N P 5/ 4 H = 485 = 184.9 ¥ 10955 (19.1)5 / 4 16.12 16.13 Solution: Power Solution: Discharge = Q = p (D 2 – D2b ) Vf1 4 p 70 = [(4)2 – (1.2)2]Vf1 4 Velocity of ﬂow = Vf1 = Vf2 = 6.1213 m/s Let H = Net available head on the turbine. Since at the outlet V2 = Vf2 = 6.1213 m/s, by energy equation V22 = Head extracted = He = h H H 2g = 0.9 H (6.1213) 2 \ 0.1 H = 2 ¥ 9.81 H = 19.1 m Power developed P = g QH ¥ h H ¥ hm = 9.79 ¥ 70 ¥ 19.1 ¥ 0.9 ¥ 0.93 = 10955 kW u1 Speed ratio = = 2.0 2g H H– u1 = 2.0 2 ¥ 9.81 ¥ 19.1 = 38.72 m/s P = ho g Q H 20,000 = 0.85 ¥ 9.79 ¥ Q ¥ 35 20.000 = 68.67 m3/s 0.85 ¥ 9.79 ¥ 35 p Q = (D 2 – D2b) Vf1 4 p 68.67 = {(2.5)2 – (0.85)2} Vf1 4 = 4.3413 Vf1 Flow velocity at inlet 68.67 Vf1 = = 15.82 m/s 4.3413 Peripheral velocity at inlet Discharge Q = pD N p ¥ 2.5 ¥ 420 = 60 60 = 54.98 m/s Hydraulic efﬁciency u1 = hH = Vu1 u1 gH Vu1 (54.98) 9.81 ¥ 35 Whirl velocity at inlet Vu1 = 5.495 m/s 0.88 = 550 Fluid Mechanics and Hydraulic Machines Since Vu1 < u1, the inlet velocity triangle is as in Fig. 16.37. Let a1 = Inlet ﬂow angle V 15.82 tan a1 = f1 = = 2.879 Vu1 5.495 a1 = 70.84° Let b1 = Inlet blade angle 15.82 b1 = Vf1 /(u1 – Vu1) = (54.98 - 5.495) = 0.320 b 1 = 17.73° tan Vu1 a1 v1 Inlet Velocity Triangle-Example 16.13 Draft Tube Loss of head HL (V ) = 0.35 ¥ 2 2 2 2g p1 V12 p V2 + + Z1 = 2 + 2 + Z2 + HL g 2g g 2g b1 Vf1 Fig. 16.37 V22 (3.82) 2 = = 0.744 m 2 ¥ 9.81 2g = 0.35 ¥ 0.744 = 0.260 m By Bernoulli equation applied to sections 1 and 2, u1 V1 12.0 = 6.79 m/s 1.767 12.0 = 3.82 m/s V2 = 3.142 V12 (6.79)2 = = 2.351 m 2g 2 ¥ 9.81 V1 = 16.14 (a) Take the bottom of the draft tube as datum (b) Let the depth of draft tube below tail water level = y, (Ref. Fig. 16.38). (c) Taking atmospheric pressure head = 10.3 m of p water, 2 = 10.3 + y g The energy equation now reads as p1 + 2.351 + (7.0 + y) g = (10.3 + y) + 0.744 + 0 + 0.260 p1 = 10.3 + 0.744 + 0.260 – 2.351 – 7.0 g = 1.953 m (abs) Turbine 1 1 1.5 m dia V1 Draft tube 7.0 m Solution: Discharge Q = 12 m3/s p (1.5)2 = 1.767 m2 A1 = 4 p A2 = (2.0)2 = 3.142 m2 4 Tail race level y V2 Datum 2 2 2.0 m dia Fig. 16.38 Draft tube set up of Example 16.14 551 Hydraulic Machines (ii) Efﬁciency of the draft tube = Loss of head HL hd = 1 – Ê V12 V22 ˆ Á 2g - 2g ˜ ¯ Ë 0.260 =1– = 0.838 ( 2.351 - 0.744) = 83.8% HL (V ) = 0.15 ¥ 2 1 2 2g = 0.15 ¥ 7.339 = 1.101 m Considering the Bernoulli equation between sections 1 and 2 p1 V2 + 1 + Z1 g 2g p V2 = 2 + 2 + Z2 + HL g 2g p1 + 7.339 + (2.0 + y) g = (10.3 + y) + 0.106 + 0 + 1.101 p1 = 12.80 – 7.096 = 2.168 m (abs) g 16.15 (ii) Power wasted to the tail race = PL = 12 m/s = 1.5 m2 = 12.5 m2 = 12 ¥ 1.5 = 18.0 m3/s 18.0 V2 = = 1.44 m/s 12.5 V12 (12.00) 2 = = 7.339 m 2 ¥ 9.81 2g V22 (1.44) 2 = = 0.106 m 2 ¥ 9.81 2g C. Pelton Turbine 2.0 Tail race y Datum 2 Fig. 16.39 2 1.101 = 0.848 (7.33 - 0.106) = 84.8% =1– Turbine Elbow type draft tube 2 2 2g = 9.79 ¥ 18.0 ¥ 0.106 = 18.68 kW (iii) Efﬁciency of the draft tube = HL hd = 1 – Ê V12 V22 ˆ Á 2g - 2 g ˜ ¯ Ë Solution: Velocity V1 A1 A2 Q 1 (V ) = gQ Draft Tube set up in Example 16.15 16.16 552 Fluid Mechanics and Hydraulic Machines Solution: V1 = Cv H = 300 m, D = 2.5 m, d = 0.20 m Cv = 0.98, k = 0.95, b¢ = 180 – b = 15° Here V1 = jet velocity = Cv b = 165°, b¢ = 180 – 165 = 15°, k = 1.0 (assumed) Power P = r Qu (V1 – u) (1 + cos b¢) = 998 ¥ 0.8 ¥ 14.0 ¥ (29.27 – 14.0) ¥ (1 + cos 15°) = 335550 W = 335.6 kW Power delivered to shaft = 335.6 ¥ h m = 335.6 ¥ 0.95 = 318.77 kW Overall efﬁciency 2gH = 75.186 m/s Discharge Q = p (d) 2 V1 4 p = ¥ (0.2)2 (75.186) = 2.362 m3/s 4 u = p ¥ 2.5 ¥ 300 = 39.27 m/s 60 He = head extracted 1 = u (V1 – u) (1 + k cos b¢) g = 1 ¥ 39.27 ¥ (75.186 – 39.27) 9.81 ¥ (1 + 0.95 cos 15°) = 275.7 m Hydraulic efﬁciency = h H = 275.7 = 0.919 300 Power developed P = g QH h o = (hH . hm) g QH = 0.919 ¥ 0.95 ¥ 9.79 ¥ 2.362 ¥ 300 kW = 6057 kW Speciﬁc speed (per jet) Ns = N P H 5/ 4 16.17 = 300 ¥ 6057 (300) 5/ 4 = 18.7 2 ¥ 9.81 ¥ 45 = 29.27 m 2 ¥ 9.81 ¥ 300 = 0.98 2gH = 0.985 h0 = = power delivered to shaft g QH 318.77 = 0.905 (9.81 ¥ 998) ¥ 0.8 ¥ 45 1000 16.18 C Solution: Net available Head Power per jet Speciﬁc speed H = 400 (1 – 0.05) = 380 m = 500 kW = N P H 5/ 4 [Note: For multiple jet Pelton wheels the speciﬁc speed is based on brake power per jet.] 14 = Solution: u = 14.0 m/s N 500 = 0.01333 N (380)5 / 4 Rotational speed N = 1050 rpm V1 = Cv 2gH 553 Hydraulic Machines 2 ¥ 9.81 ¥ 380 = 84.62 m/s u = 0.46 2g H = 0.98 Speed ratio f = u = 0.46 ¥ 19.62 ¥ 380 = 39.72 m/s pD N p ¥ D ¥ 1050 = 60 60 D = 0.722 m = mean diameter of bucket circle Power developed = 1000 kW = g Q H ho 1000 = 9.79 ¥ Q ¥ 380 ¥ 0.85 = 3162.17 Q Q = 0.3162 m3/s As there are two jets of diameter d, p 2¥ ¥ d 2 ¥ 84.62 = 0.3162 4 d = 0.04877 m = 4.88 cm u = 39.72 = Let a 2¢ be the direction of the absolute velocity V2 with the peripheral velocity. Vf2 = V2 sin a 2¢ = (V1 – u) sin b¢ = (96 – 44) sin 10° = 9.03 Vu2 = V2 cos a 2 = (V1 – u) cos b¢ – u = (96 – 44) cos 10° – 44 = 7.21 tan a 2¢ = 9.03 = 1.252, a 2¢ = 51.39° 7.21 9.03 V2 = = 11.555 m/s sin 51.39∞ (11.555) 2 V22 = = 6.81 m 2 ¥ 9.81 2g 16.20 C k 1 C 2 h k b¢ 16.19 b¢ b Solution: Let the net head at the base of the nozzle = H. Velocity of jet V1 = Cv Solution: From the outlet velocity triangle (see Fig. 16.40), b¢ = 180 – b = 10° v2 = (V1 – u) V2 b a2 b¢ u2 = u Fig. 16.40 2gH Ê V12 ˆ H = Á 2 ˜ Ë Cv ◊ 2g ¯ Head extracted in the turbine 1 (V1 – u) . u (1 + k cos b¢) He = g Hydraulic efﬁciency He u Ê uˆ 1 - ˜ (1 + k cos b¢) hH = = 2 Cv2 Á H V1 Ë V1 ¯ Vf2 At outlet where a¢2 Vu2 Outlet Velocity Triangle-Example 16.19 = 2C v2 e (1 – e) (1 + k cos b¢) e = u/V1 For maximum h H, (1 – 2e) = 0 dhH =0 de or e = 1/2 554 Fluid Mechanics and Hydraulic Machines Maximum value of hydraulic efﬁciency (h H)Max = = 2C 2v 1ˆ 1 Ê ÁË1 - 2 ˜¯ (1 + k cos b¢) 2 16.22 h Solution: 16.21 H = 500 m V1 = Cv 2 ¥ 9.81 ¥ 500 = 97.06 m/s p 2 p Discharge Q = d V1 = ¥ (0.18) 2 ¥ 97.06 4 4 = 2.47 m3/s Power developed P = ho g Q H = 0.85 ¥ 9.79 ¥ 2.47 ¥ 500 = 10,277 kW Speciﬁc speed C Solution: 3000 = 1500 kW 2 P = ho g Q H 1500 = 0.90 ¥ 9.79 ¥ Q ¥ 270 Q = 0.6305 m3/s Power per wheel = Velocity of the jet = V1 = Cv 2gH 2 ¥ 9.81 ¥ 270 = 69.14 m/s p For the nozzle: Q = ¥ d 2 ¥ V1 4 p 0.6305 = ¥ d2 ¥ 69.14 4 d = diameter of the nozzle = 0.1078 m = 10.78 cm Peripheral velocity of the bucket u = pD N 60 p ¥ 1 . 5 ¥ 400 u= = 31.42 m/s 60 u 31.42 Speciﬁc ratio f = = 2g H 2 ¥ 9.81 ¥ 270 = 0.95 = 0.432 Speciﬁc speed Ns = = N P H 5/ 4 400 ¥ 1500 ( 270) 5/ 4 = 14.15 2gH = 0.98 f Power f C 1 C 2v (1 + k cos b¢) 2 Ns = N P H 5/ 4 = 420 10, 277 (500)5 / 4 = 18 16.23 Solution: Since there are two jets, for each jet: Power P = 7500/2 = 3750 kW. H = 400 m, k = 0.85, Cv = 0.98 and j = 0.47. u (i) Speed ratio j = = 0.47, 2gH Hence u = 0.47 ¥ 2 ¥ 9.81 ¥ 400 = 41.64 m/s 555 Hydraulic Machines P = 3750 = h0 g Q H Q= 3750 0.80 ¥ 9.79 ¥ 400 = 1.197 m3/s. Total discharge through the turbine = 2 Q = 2.394 m3/s (ii) V1 = 0.47 1 ¥ 495 = 165 m 3 Net head at turbine = 495 m – 165 = 330 m Loss of head at penstock = 2gH = 0.98 ¥ 2 ¥ 9.81 ¥ 400 (i) Power: P = h0 g QH = 0.85 ¥ 9.79 ¥ 2.0 ¥ 330 = 5492 kW 2gH = 0.98 ¥ = 78.86 m/s (ii) V1 = Cv = 86.82 m/s Discharge Q = 1.197 = p d 2 V1 4 1 . 197 Hence d2 = = 0.01755, p ¥ 86.82 4 u =f 2 ¥ 9.81 ¥ 330 = 36.21 m/s Head Extracted He = Diameter of each jet = d = 0.1325 m = 13.25 cm 1 u (V1 – u) (1 + k cos b ¢) g 1 ¥ 36.21 ¥ (78.86 – 36.21) 9.81 (1 + (0.95 ¥ cos15º) = 301.89 m = (iii) Total force exerted by each jet in tangential direction Fx1 = rQ(V1 – u)(1 + k cos b ¢) b¢ = (180 – b) = 15°. Cosb¢ = cos 15° = 0.9659 998 Fx1 = ¥ 1.197 ¥ (86.82 – 41.64) 1000 ¥ (1 + (0.85 ¥ 0.9695) = 98.286 kN 2gH = 0.45 ¥ 2 ¥ 9.81 ¥ 330 301.89 = 0.915 hH = H e = 330 H D. Similitude in Turbines 16.25 Total tangential force on the wheel = FxT = 2Fx1 = 2 ¥ 98.286 = 196.57 kN 16.24 C K Solution: Gross head = 495 m Solution: Power P = h o g QH For the prototype: 6750 = 0.82 ¥ (9.81 ¥ 998) ¥ Q ¥ 45 1000 = 361.265 Q Discharge Qp = 18.685 m3/s Using the sufﬁxes m and p to denote model and prototype parameters respectively 556 Fluid Mechanics and Hydraulic Machines 1 Dm = = Scale ratio 8 Dp Hence Dp = 3.0 m, Dm = 3.0 = 0.375 m 8 Hm = 9 = 1 Also Hp = 45 m, Hm = 9.0 m, Hp 45 5 N m Dm N p Dp = Speed: Hm Hp Nsm As \ Nm = speed of the model ˆ Ê Hm ˆ ˜ ˜Á m ¯ Ë Hp ¯ Ê Dp = NP Á D Ë Ê 1ˆ = 300 (8) Á ˜ Ë 5¯ Discharge: Qm 3 N m Dm = = N p Pp H p5 / 4 = = 1073.3 9.433 (9)5 / 4 300 6760 ( 45)5 / 4 16.26 = 1073.3 rpm Qp Solution: For geometrically similar turbines, the unit speed N Nu = H N1 N2 = \ H1 H2 N p Dp3 3 Ê 1073.3 ˆ Ê 1 ˆ = 18.685 Á Ë 300 ˜¯ ÁË 8 ˜¯ 3 3 Nm Dm 5/ 4 Hm = 211.5 rpm 1/ 2 Ê N m ˆ Ê Dm ˆ = Qp Á N ˜ Á D ˜ Ë p¯Ë p¯ Pm N m Pm = 211.5 rpm It is seen that Nsm = Nsp, as expected. This is a check on the calculations. 1/ 2 Qm = Model discharge Power: Nsp = N2 = N 1 3 Pm = Model power 3 3 5 Ê 1073.3 ˆ Ê 1 ˆ = 6750 Á Ë 300 ˜¯ ÁË 8 ˜¯ = 9.433 kW 5 Speciﬁc speed: Since the model and the prototype are similar we expect them to have the same speciﬁc speed. 18 / 30 = 77.46 rpm P Unit power Pu = 3 / 2 H P1 P = 32/ 2 H13 / 2 H2 3/ 2 ÊH ˆ P2 = P1 ¥ Á 2 ˜ Ë H1 ¯ 3/ 2 Ê 18 ˆ = 8000 ¥ Á ˜ = 3718 kW Ë 30 ¯ = 0.13056 m3/s Pp = 3 5 N p Dp Ê N m ˆ Ê Dm ˆ = Pp Á N ˜ Á D ˜ Ë p¯ Ë p¯ H 2 / H1 = 100 16.27 557 Hydraulic Machines Solution: Power developed P = h o g QH (9.81 ¥ 998) 6750 = 0.85 ¥ ¥ Q ¥ 45 1000 Discharge Q = 18.03 m3/s. For calculating N, Q and P at H = 60 m the unit relationships are used. N Nu = H N1 N2 = \ H1 H2 or N2 = N1 H 2 / H1 = 300 Dm 1 = , Dp 5 Hm 25 = Head ratio Hp 49 For a 1/5 Model: Speed: \ 60 / 45 H 2 / H1 = 18.03 60 / 45 25 49 3 Ê 1 ˆ Ê 892.9 ˆ = 43.82 ¥ Á ˜ Á Ë 5 ¯ Ë 250 ˜¯ P = 1.252 m3/s H 3/ 2 ÊH ˆ P2 = P1 Á 2 ˜ Ë H1 ¯ 3/ 2 Ê 60 ˆ = 6750 Á ˜ Ë 45 ¯ Model discharge is 1.252 m3/s Power developed by the model = Pm = h0 g Qm Hm = 0.88 ¥ 9.79 ¥ 1.252 ¥ 25 = 269.7 kW 3/ 2 = 10,392 kW Hm Hp 3 = 20.82 m /s \ Hp ÊD ˆ ÊN ˆ Qm = Qp Á m ˜ Á m ˜ Ë Dp ¯ Ë N p ¯ 3 Pu = N p Dp Ê 5ˆ = 250 ¥ Á ˜ Ë 1¯ = 892.9 rpm Hence speed of model = 852.9 rpm Qp Qm = Discharge: 3 N m Dm N p Dp3 Q H Q2 = Q1 Hm = Ê Dp ˆ Nm = Np Á Ë Dm ˜¯ = 346.4 rpm Qu = N m Dm 16.28 E: General 16.29 Solution: For Prototype, Power Pp = h0g Qp Hp = 18500 kW Discharge Q p = Pp h0g H p = = 43.82 m3/s 18500 0.88 ¥ 9.79 ¥ 49 Solution: Power P = h0g Q H = 0.9 ¥ 9.79 ¥ 300 ¥ 35 = 92516 kW 558 Fluid Mechanics and Hydraulic Machines For a Speciﬁc speed of 400, power produced per machine P1 is given by: 400 = 16.31 150 p1 (35)5 / 4 2 Ê 400 ˆ P1 = (35)5/2 ¥ Á = 51536 kW Ë 150 ˜¯ and Number of turbines required n = (92516)/51536 = 2 E. Rotodynamic Pumps 16.30 Solution: Given: H = 10.0 m; N = 1000 rpm b 2 = 30°; D2 = 0.30 m b = 0.05 m; h H = 0.95 u2 = tangential velocity of the impeller at the outlet pD2 N p ¥ 0.30 ¥ 1000 = = 15.708 m/s = 60 60 Manometric efﬁciency Solution: Refer to Fig. 16.41. hH = u2 b2 a2 90° 0.95 = v2 V2 At outlet Fig. 16.41 Example 16.30 From the outlet velocity triangle Vf2 = v2 and V2 cos a2 = Vu2 = u2 u2 = p D 2 N/60 = p ¥ 0.30 ¥ 1450/60 = 22.78 m/s Hence, manometric efﬁciency hH = 0.82 = gH gH = 2 u2Vu2 u2 gH (The oulﬂow is assumed to be u2Vu2 radial.) 9.81 ¥ 10.0 15.708 ¥ Vu2 Vu2 = 6.574 m/s From the outlet velocity triangle (Fig. 16.42), since b2 < 90° Vf2 tan b2 = u2 - Vu 2 Vf2 tan 30° = 15.708 - 6.574 Vf2 = 5.274 m/s u2 b2 = 30° Vf2 v2 9.81 ¥ H ( 22.78) 2 H = Net head developed = 43.36 m Vu2 (u2 – Vu2) At the outlet: Fig. 16.42 Example 16.31 V2 559 Hydraulic Machines Discharge Q = p D2b2V f2 = p ¥ 0.30 ¥ 0.05 ¥ 5.274 = 0.249 m3/s = 249 L/s 16.33 16.32 Solution: Solution: At the outlet pD2 N 60 p ¥ 0.25 ¥ 1450 = 60 = 18.98 m/s Assuming radial ﬂow at the inlet, the manometric efﬁciency gH hH = u2Vu2 9.81 ¥ 15 0.80 = 18.98 ¥ Vu2 Swirl velocity at outlet Vu2 = 9.69 m/s The discharge Q = p D2b2Vf2 0.100 = p ¥ 0.25 ¥ 0.06 ¥ Vf2 Velocity of ﬂow at outlet Vf2 = 2.122 m/s From the outlet velocity triangle (Fig. 16.43) 2.122 Vf2 tan b2 = = 0.2284 = (18.98 - 9.69) u2 - Vu2 Peripheral velocity u2 = Since Since D2 = 0.80 m, Q = 1.10 m3/s, H = 70 m, N = 1000 rpm, B2 = 0.08 m, h h = 0.82. Leakage loss 4% Qth = (1.10 ¥ 1.04) = 1.144 m3/s pDN p ¥ 0.8 ¥ 1000 u2 = = = 41.89 m/s 60 60 Qth = p D2 B2 Vf2 1.144 Vf2 = = 5.69 m/s p ¥ 0.8 ¥ 0.08 gH , hH = u2Vu 2 9.81 ¥ 70 Vu2 = = 20.0 m/s 41.89 ¥ 0.82 From velocity triangle at the outlet Fig. 16.44 u2 Vn2 b2 a2 v2 Vf 2 V2 Outlet b2 = tan–1 0.2284 = 12.87° u2 Vu2 (u2 – Vu2) V1 = V f 1 Inlet b2 Vf2 v2 V2 Fig. 16.44 tan b2 = Fig. 16.43 Velocity Triangle of OutletExample 16.32 a1 v1 b1 u1 Velocity Triangles-Example 16.33 5.69 Vf 2 = 0.26 = ( 41.89 - 20.0) (u2 - Vu 2 ) b2 = 14.57° 560 Fluid Mechanics and Hydraulic Machines Power required: From velocity triangle at the outlet Vf 2 4.244 = ( 22 - 13.94) (u2 - Vu 2 ) = 0.527 b2 = 27.77° È (Vu 2 u2 ) ˘ g Qth ˙ + 10.0 P= Í g Î ˚ Ê 20.0 ¥ 41.89 ˆ = Á ˜¯ ¥ 9.79 ¥ 1.144 + (10.0) Ë 9.81 = 966.5 kW tan b2 = 956.5 = 0.99 Mechanical efﬁciency hmec = 966.5 16.35 Overall efﬁciency h0 = hmec ¥ hH = 0.99 ¥ 0.82 = 0.812 h0 = 81.2% 16.34 Solution: p D2 N 60 p ¥ 0.30 ¥ 1200 = 60 = 18.85 m/s Vf2 = 2.0 m/s and b2 = 30° From the outlet velocity triangle [Fig. 16.46(a)] At the outlet u2 = Solution: Let b2 be the blade angel at he outlet. pDN p ¥ 0.3 ¥ 1400 u2 = = = 22.0 m/s 60 60 gH , Since hH = u2Vu 2 9.81 ¥ 25 Vu2 = = 13.94 m/s 22.0 ¥ 0.80 Since Q = p D2 B2Vf2 0.2 Vf 2 = = 4.244 m/s p ¥ 0.3 ¥ 0.05 u2 Vu2 b2 = 30° a2 Vf2 v2 V2 u2 (a) Outlet Vu2 b2 a2 v2 Vf 2 V2 V1 V1 = Vf1 Outlet a1 Fig. 16. 45 Outlet Velocity Triangle-Example 16.34 b1 90° (b) Inlet Fig. 16.46 u1 Example 16.35 561 Hydraulic Machines Vf2 u2 - Vu2 2.0 tan 30° = 18.85 - Vu2 tan b2 = 18.85 – Vu2 = 3.464; and hence Vu2 = 15.386 m/s (i) V2 = absolute velocity at outlet If = 2 Vu2 + Vf22 = (15.386) 2 + ( 2.0) 2 = 15.515 m/s a2 = Inclination of V2 to tangential direction at outlet V 2.0 tan a2 = f2 = = 0.13 Vu2 15.386 a2 = 7.4° (ii) Manometric efﬁciency gH hH = u2Vu2 Solution: layout. Figure 16.47(a) shows the schematic Delivery pipe Static lift = 40 m P Pump Suction pipe Head developed u2Vu2 g 0.85 ¥ 18.85 ¥ 15.386 = 9.81 = 25.13 m (iii) From the inlet velocity diagram [Fig. 16.46(b)] V1 = Vf1 = Vf2 = 2.0 m/s u1 = peripheral velocity p ¥ 0.15 ¥ 1200 = 60 = 9.425 m/s The inlet blade angle b1 is given by V 2.0 tan b1 = f1 = = 0.2122 u1 9.425 b1 = 11.98° = H = hH 16.36 (a) Schematic layout u2 Vu2 (u2 – Vu2) b2 = 20° Vf2 v2 V2 (b) Outlet velocity triangle Fig. 16.47 Net head Example 16.36 H = Static lift + friction loss = 40.0 + 2.0 + 6.0 = 48 m u2 = peripheral velocity at outlet u2 = p ¥ 0.5 ¥ 1200 = 31.42 m/s 60 562 Fluid Mechanics and Hydraulic Machines Assume the ﬂow to be radial at the inlet. gH Manometric efﬁciency h H = u2Vu2 0.85 = ps p = atm – 5.308 g g = 5.308 m vacuum patm Assuming = 10.35 m g ps = 10.35 – 5.308 g = 5.042 m of water (absolute) ps = 5.042 ¥ 9.79 = 49.36 kN/m2 (abs) 9.81 ¥ 48.0 31.42 ¥ Vu2 Vu2 = 17.63 m/s Blade angle b2 = 20° From the outlet velocity triangle [Fig. 16.47(b)] Vf2 tan b2 = u2 - Vu2 tan 20° = Vf2 = 0.3639 31.42 - 17.63 Vf2 = 5.02 m/s Discharge Q = p D2b2Vf2 = p ¥ 0.5 ¥ 0.03 ¥ 5.02 = 0.2366 m3/s Vd = Velocity in delivery pipe = Vs = Velocity in suction pipe = Q 2 p ¥ D /4 = 0.2366 p ¥ (0.35) 2 4 = 2.459 m/s 2 Vd V2 ( 2.459) 2 = s = = 0.308 m 2 ¥ 9.81 2g 2g Let the pressure on delivery side = pd 16.37 Solution: Manometric efﬁciency, h H = 0.85 = 9.81 ¥ 16 u2Vu2 (1) u2Vu2 = 184.66 From the outlet velocity triangle (Fig. 16.48) pd Vd2 + = Hd + HLd g 2g u2 Vu2 Vd2 pd = Hd + HLd – g 2g = 37 + 6 – 0.308 = 42.69 m (9.81 ¥ 998) Pd = 42.69 ¥ 1000 = 418 kN/m2 (gauge) Let the pressure on the suction side be ps and the atmospheric pressure be patm. Then gH u2Vu2 patm p V2 = Hs + HLs + s + s g g 2g ps =3+2+ + 0.308 g a2 b2 = 35° Vf2 V2 v2 Fig. 16.48 Outlet Velocity Triangle-Example 16.37 Vf2 u2 - Vu2 1.50 = = 2.142 tan 35∞ tan 35° = u2 – Vu2 563 Hydraulic Machines Vu2 = (u2 – 2.142) (2) Substituting for Vu2 in Eq. 1, (u2 – 2.142) u2 = 184.66 u 22 – 2.142 u2 – 184.66 = 0 Taking the positive root u2 = 14.702 Vu2 = 12.560 m/s pD2 N Since u2 = 60 p ¥ D2 ¥ 1000 14.702 = 60 D2 = 0.280 m Discharge Q = p D2 b2Vf2 0.080 = p ¥ 0.280 ¥ b2 ¥ 1.50 b2 = 0.061 m Width of impeller at outlet = 6.1 cm tan b ¢2 = tan 30° = Vf2 4.244 = (Vu 2 - u2 ( 22 - 13.94) = 0.5774 2.0 Vu2 – u2 = = 3.464 0.5774 giving Vu2 = u2 + 3.464 hm = gH 9.81 ¥ 6.0 = = 0.70 Vu 2 ◊u2 (u2 + 3.464) ¥ u2 u 22 + 3.464u2 – 58.86 = 0 This gives u2 = 6.133 m/s pD2 N Also u2 = = 6.133 60 6.133 ¥ 60 = 0.234 m D2 = 500 ¥ p Discharge 16.38 Q = pD2 B2 Vf2 0.090 = p ¥ 0.234 ¥ B2 ¥ 2.0 B2 = Width of the impeller = 0.0612 m 16.39 Solution: Vf1 = Vf2 = 2.0 m/s. Also, Blade angle b2 = 150°. Hence, b ¢2 = 180 – 150 = 30° From the outlet velocity triangle (Fig. 16.49), Vu2 (Vu2 – u2) b 2¢ u2 a2 b2 = 150° V2 Outlet Fig. 16.49 Z Z u Z (u22 - u12 ) 2g - (v 22 - v12 ) 2g v p Solution: For a rotodynamic pump the net head developed H is given by the manometric efﬁciency gH hH = (u2Vu2 - u1Vu1 ) For no-loss situation h H = 1.0. Hence v2 Vf 2 (p2 - p1) g Outlet Velocity Triangle-Example 16.38 H = 1 (u V – u V ) g 2 u2 1 u1 From the velocity vector diagram (Fig. 16.50), absolute velocity V, peripheral velocity u and relative velocity v are related as 564 Fluid Mechanics and Hydraulic Machines Vu p2 - p1 u2 - u2 = 2 1 g 2g pD1N p ¥ 0.2 ¥ 600 u2 = = 60 60 = 6.283 m/s a b v Z2 = Z1 Assuming u V pD2 N p ¥ 0.1 ¥ 600 = 60 60 = 3.142 m/s u1 = Fig. 16.50 Example 16.39 V 2 + u2 – 2uV cos a = v 2 p2 - p1 = Difference in pressure head g across the periphery. 1 = [(6.283)2 – (3.142)2] 2 ¥ 9.81 = 1.51 m 1 uV cos a = uVu = (V 2 + u2 – v2) 2 Hence 1 [(V 22 – V 12 ) + (u22 – u 21) – (v 22 – v 21)] 2g By Bernoulli equation, between a point on the inlet and a point on the outlet, H= 16.41 D D Ê p2 V22 ˆ Ê p1 V12 ˆ Z Z + + + + 2 1 H= Á g 2g ˜¯ ÁË g 2g ˜¯ Ë H N Hence Ê p2 p1 ˆ (V22 - V12 ) ÁË g - g ˜¯ + (Z2 – Z1) = H – 2g ( u22 - u12 ) ( v22 - v12 ) 2g 2g = increase in piezometric head = 16.40 D Solution: For a pump to just start pumping, if there are no loses, the centrifugal head must be equal to the actual lift. Hence under ideal conditions u22 - u12 2g Allowing for manometric efﬁciency, if the net head delivered by the pump is H, then Hi = H/hH Hi = Hence H = 1 Ê u22 - u12 ˆ hH ÁË 2g ˜¯ But u2 = pD2 N 60 H = 1 p 2 D12 N 2 hH 2g(60) 2 Solution: Increase in piezometric head (u 2 - u12 ) ( v22 - v12 ) ( p2 - p1 ) + (Z2 – Z1) = 2 g 2g 2g When the outlet is shut off, the ﬂow is zero and hence v1 = v2 = 0, 81.7 H N and pD1N 60 È Ê D ˆ2˘ Í1 - Á 2 ˜ ˙ Í Ë D1 ¯ ˙ Î ˚ u1 = 565 Hydraulic Machines Since D2/D1 = 0.5 2 H= D12 ¥ 2 1 p ¥ N ¥ (1 – 0.25) 0.70 3600 ¥ 2 ¥ 9.81 D12 = 6679.4 or D1 = 81.7 = 0.3935 = 39.35 % (c) Minimum starting speed, N, in rpm is given by H N 2 H N Least diameter of the impeller D1 = 81.7 Manometric efﬁciency gH 9.81 ¥ 8.5 = = u2Vu2 15.71 ¥ 13.49 h H = 0.70 and p 2 N m2 (60) 2 H /N p 2 N m2 (60) 16.42 2 [D 22 – D 21 ] = 2gH [(0.5)2 – (0.25)2] = 2 ¥ 9.81 ¥ 8.5 Nm = 570 rpm u2 Vu2 a2 b2 Vf2 v2 Solution: (a) Figure 16.51 show the velocity triangles at the inlet and oulet 8000 ¥ 10 Vf1 = Vf2 = = 2.22 m/s 60 ¥ 0.06 pD1N p ¥ 0.25 ¥ 600 u1 = = 7.85 m/s = 60 60 pD2 N p ¥ 0.5 ¥ 600 u2 = = 15.71 m/s = 60 60 From the inlet triangle, V 2.22 tan b1 = f1 = = 0.2828 u1 7.85 V1 = Vf1 Inlet a1 Fig 16.51 16.43 b1 = vane angle at inlet = 15°47° (b) From outlet velocity triangle, Vf 2 tan b 2 = tan 45° = (u2 - Vu 2 ) 15.71 – Vu2 = 2.22 Vu2 = 15.71 – 2.22 = 13.49 m/s v1 b1 u1 Velocity Triangles-Example 16.42 G. Similitude in Pumps Outlet V2 Solution: BP1 = Power required = g QH ho 566 Fluid Mechanics and Hydraulic Machines 9.79 ¥ 0.350 ¥ 8.0 = 39.16 kW 0.70 = Ns = Speciﬁc speed = Solution: For homologous pumps the speciﬁc speed is the same. N Q H Hence 3/ 4 2000 0.35 NsA = Q1 N1D13 = 20.18 = Q2 = N 2 D23 ÊN ˆ Q2 = Q1 Á 2 ˜ Ë N1 ¯ Ê D2 ˆ ÁË D ˜¯ 1 D 2 = D1 hence Ê 2500 ˆ Q2 = 0.350 Á (1) Ë 2000 ˜¯ D 3B = Hence = 0.4375 m /s (ii) N12 D12 = N B QB 2 D2 ˆ H2 = Á H1 Ë N1 ˜¯ ÁË D1 ˜¯ = H B3/ 4 = 20.18 600 0.3 H B3/ 4 328.63 H B3/ 4 QB . N A . 3 DA QA NB Ê 0.3 ˆ DB = Á Ë 0.4 ˜¯ H2 N 22 D22 2 Ê N2 ˆ Ê 503 / 4 Ê QA ˆ Ê QB ˆ = Á ˜ Á 3 ˜ Ë N A DA ¯ Ë N B DB3 ¯ 3 H1 600 0.4 HB = 41.28 m From the similarity relation 3 Here = H A3/ 4 For pump B, NsB = NsA = 20.18 = 248.7 (8)3 / 4 At higher speed: For homologous conditions (i) N A QA 1/ 3 ¥ (0.5) = 0.454 m 16.45 2 Ê 2500 ˆ = Á (1) (8) = 12.5 m Ë 2000 ˜¯ (iii) BP2 = brake power required g Q2 H 2 = ho 9.79 ¥ 0.4375 ¥ 12.5 0.70 = 76.48 kW = Solution: In this case, D1 = D2. Q1 Q2 3 = N1D1 N 2 D23 3 16.44 Hence ÊQ ˆ Ê D ˆ N2 = Á 2 ˜ Á 1 ˜ N 1 Ë Q1 ¯ Ë D2 ¯ H1 D12 N12 Ê 0.60 ˆ = Á (1)3 1200 = 1440 rpm Ë 0.50 ˜¯ H = 222 D2 N 2 567 Hydraulic Machines 2 ÊN ˆ ÊD ˆ H2 = Á 2 ˜ Á 2 ˜ Ë N1 ¯ Ë D1 ¯ \ 2 2 Ê 1440 ˆ = Á ¥ 20 = 28.8 m Ë 1200 ˜¯ At shut off head: H 1 = 30 m, discharge Q 1 = 0 and N1 = 1200 rpm. When shut off head H 2 = 22 m H 2 H1 2 2 = D2 N 2 D12 N12 Ê H ˆ Ê D ˆ2 N 22 = Á 2 ˜ Á 1 ˜ N12 Ë H1 ¯ Ë D2 ¯ H 2 / H1 = 1200 = 1401 rpm N 2 = N1 60 15173.6 = 3.954 ¥ 10–3 Q = 0.0629 m3/s = 62.9 L/s H = Hp = 100 – 6000 ¥ (0.0629)2 = 76.27 m Q2 = H1 Net head of pump 16.47 Q H h 30 / 22 G. Pump Characteristics 16.46 Solution: Plots of Q vs H and Q vs h are plotted on the same graph sheet as shown in Fig. 16.52 from which the BEP is h max = 73.5 with Q = 30.3 L/s and H = 30 m. f f Q H 40 H Q vs h Q BEP 75 70 35 Solution: Pipe system— (1) Static head Hs = 40 m (2) Friction head HL = hL1 + hL2 HL = 65 60 H(m) 30 f1L1V12 f L V2 + 2 2 2 2g D1 2g D1 55 25 50 45 2 0.02 ¥ 50 Ï p ¸ ¥ (0.25) 2 ˝ ÌQ 4 2 ¥ 9.81 ¥ 0.25 Ó ˛ 2 0.022 ¥ 1600 p ¸ Ï + ¥ ÌQ ¥ (0.20) 2 ˝ 4 2 ¥ 9.81 ¥ 0.20 Ó ˛ 2 = (84.6 + 9089) Q = 9173.6 Q2 At equilibrium, system head = pump head. Hence 40 + 9173.6 Q2 = 100 – 6000 Q2 = h Q vs H 20 40 35 15 0 5 Fig. 16.52 10 15 20 25 30 35 40 Q = discharge in litres/s 45 30 50 Pump Characteristics-Example 16.47 The speciﬁc speed Ns is calculated for the Best Efﬁciency Point (Design Point). 568 Fluid Mechanics and Hydraulic Machines Ns = N Q = 1500 0.0303 Table 16.2(a) Pumps A and B in Parallel = 20.37 H 3/ 4 (30)3 / 4 By extrapolating the Q vs H curve, the shut off head which corresponds to Q = 0 is found to be 35.5 m. Head 45 43 40 38 29 23 16 16.48 (i) Pumps in parallel The individual headdischarge curves for pump A and B are plotted in Fig. 16.53(a). When the pumps are connected in parallel, the discharge gets added up for a given head. The head-discharge data are obtained as in Table 16.2(a) and plotted in Fig. 16.53(a). 0 0 0 0.1 0.2 0.3 0.4 Head H (m) 30 A B 20 0 0.1 Fig. 16.53(a) 0.2 0.3 0.4 0.5 0.6 Discharge Q (m3/s) 0.7 0.8 Pumps in Parallel-Example 16.48 0 0.1 0.17 0.20 0.29 0.34 0.38 0 0.1 0.17 0.30 0.49 0.64 0.78 Table 16.2(b) Pumps A and B in Series Discharge Q (m3/s) 0 0.1 0.2 0.25 0.30 0.35 Pumps A II B curve - C 40 Combined discharge Q1 = Qa + Qb (m3/s) Values of Qb indicated by an asterisk () are interpolated values of curve B from Fig. 16.53(a). Values of Qt and corresponding H are plotted in Fig. 16.53 (a) to get curve C which is the head-discharge relationship for pumps A and B operating in parallel. (ii) Pumps in series When the pumps are connected in series the heads of individual pumps for a given discharge get added up. Thus the combined head H t = Ha + Hb for a given discharge. Table 16.2(b) is prepared from the individual H-Q relationship shown in Fig. 16.53(a). 50 10 Discharge Discharge from A alone from B alone Qa Qb 3 (m /s) (m3/s) Head(m) Ha (for pump A alone) Head (m) Hb (for pump B alone) Combined head Ht (m) (H t = Ha + Hb) 40 38 29 23 16 10 45 43 38 34 28 22 85 81 67 57 44 32 Interpolated values from curves in Fig. 16.53(a). The variation of H t vs Q is shown plotted as curve D in Fig. 16.53(b) and represents the head-discharge curve for pumps A and B in series. 569 Hydraulic Machines 100 Head H (m) 80 16.50 Pumps A and B in series Curve-D 60 40 20 0 0.1 0.2 0.3 0.4 0.5 Discharge Q (m3/s) Fig. 16.53(b) Pumps in Series-Example 16.48 Solution: NPSH = ( patm )abs p – Zs – hL – v g g ( patm )abs = Atmospheric pressure (absolute) g 98.00 = 10.01 m = 9.79 Zs = elevation of pump above the reservoir water surface = 1.50 m hL = head loss in the suction pipe = 1.75 m pv = vapour pressure head g 2.30 = 0.24 = 9.79 NPSH = 10.01 – 1.50 – 1.75 – 0.24 = 6.57 m Since (NPSH = 6.57 m) is greater than [(NPSH)min = 6.30 m)] cavitation should not be a problem for the installation. where 16.49 s Solution: (i) By Eq. 16.53 sc = ( NPSH) min H ( NPSH) min 30 Minimum NPSH = 3.6 m 0.12 = (p ) p NPSH = atm abs - v – Zs – hL g g (ii) I. Reciprocating Pumps 16.51 where Zs = elevation of the pump above the sump water surface. (Zs)max corresponds to sc. Hence ( patm )abs pv – hL – (NPSH)min g g 96.0 3.0 = – 0.3 – 3.6 = 5.6 m 9.79 9.79 (Zs)max = Solution: Theoretical discharge Q t = ALN 60 570 Fluid Mechanics and Hydraulic Machines p (0.15)2 = 0.01767 m2 4 L = 2r = 2 ¥ 0.15 = 0.30 m; N = 60 rpm 0.01767 ¥ 0.30 ¥ 60 Qt = 60 = 0.0053 m3/s = 318 l/min Qa = actual discharge = 310 l/min (318 - 310) ¥ 100 Slip = = 2.52 % 318 Coefﬁcient of discharge = 310/318 = 0.975 Total head = Ht = (Hs + Hd ) = 15 m Power = Pt = rQa H t = 9790 ¥ 0.0053 ¥ 15 = 778 W = 0.778 kW Total head = Ht = (Hs + hfs) + (Hd + hfd ) = 80 + 2 + 18 = 100 m Power = P = g QaHt/h = (9790 ¥ 0.00833 ¥ 100)/0.90 = 9061 W = 9.06 kW A= 16.52 16.53 Solution: At incipient cavitation in the delivery pipe Hatmo + Hd – Had = Hv 9.75 + 40 – Had = 2.75 Had = 47 m At the end of the delivery stroke Had = Here and Solution: p (0.2)2 = 0.3142 m2, L = 0.4 m, 4 Qa = (50000/(60 ¥ 1000) = 0.0833 m3/s For a three-throw pump A= Qt = (6.28 ¥ 10 -4 N - 0.0833) 6.28 ¥ 10 -4 N N = 135.4 rpm Ld = 45.0 m, (A/Ad) = (20/10)2 = 4, r = 0.40 m 2pN = 0.1047 N 60 45 Hd = 47 = ¥ 4 ¥ (0.1047 N)2 ¥ 0.40 9.81 w= Thus 3 ALN 60 0.3142 ¥ 0.40 ¥ N Qt = 60 = 6.283 ¥ 10–4 N Q - Qa Since, Slip = 2%, t = 0.02 Qt Ld Ê A ˆ 2 wr g ÁË Ad ˜¯ = 0.0805 N 2 N = 24.17 rpm = 0.02 16.54 571 Hydraulic Machines Solution: pipe Hasm = Ls Ê A ˆ 2 wr g ÁË As ˜¯ 2 p ¥ 30 = 3.141 rad/s, r = 0.40/2 = 0.20 m 60 (A/As ) = (20/10)2 = 4, Ls = 5.0 m 5.0 Hasm = ¥ 4 ¥ (3.141)2 ¥ 0.20 = 4.024 m 9.81 At limiting condition for a suction pipe Hasm + Hv + Hs = Hatmo 4.24 + 2.5 + Hs = 10.0 Hence Hs = suction lift = 3.476 m w = Solution: Maximum acceleration Ls Ê A ˆ 2 Hasm = wr g ÁË As ˜¯ 2pN = 0.1047 N, r = 0.30/2 = 0.15 m 60 (A/As) = (15/5)2 = 9 5.0 Hasm = ¥ 9 ¥ (0.1047 N)2 ¥ 0.15 9.81 = 0.007543 N2 At limiting condition for a suction pipe Hasm + Hv + Hs = Hatmo Hasm + 2.0 + 2.5 = 10.0 Hence Hasm = 5.5 = 0.007543 N2 N = 27.0 rpm w= Maximum acceleration head in suction 16.56 16.55 Solution: 2 p ¥ 40 = 4.189 rad/s, r = 0.40/2 = 0.20 m 60 2 (A/As) = (25/15) = 2.778, (A/Ad) = (25/20)2 = 1.5625, Calculations of Velocity and acceleration are as in table: w = Item Maximum velocity Suction Pipe Ê Aˆ Vsm = Á ˜ wr Ë As ¯ = 2.778 ¥ 4.189 ¥ 0.20 = 2.327 m/s Maximum acceleration Ê Aˆ asm = Á ˜ w 2r Ë As ¯ = 2.778 ¥ (4.189)2 ¥ 0.20 = 9.749 m/s2 Delivery pipe Ê Aˆ Vdm = Á wr Ë Ad ˜¯ = 1.5625 ¥ 4.189 ¥ 0.20 = 1.309 m/s Ê Aˆ adm = Á w 2r Ë Ad ˜¯ = 1.5625 ¥ 4.189)2 ¥ 0.20 = 5.484 m/s2 572 Fluid Mechanics and Hydraulic Machines q Condition 16.57 Beginning Mid End Has (m) 0° 3.0182 90° 0 180° –3.0182 hfs (m) Pressure head Hts (abs) 0 0.302 0 2.982 m 5.698 m 9.018 m Delivery Side: Acceleration head: Had = Ld Ê A ˆ 2 w r cos q g ÁË Ad ˜¯ 25 ¥ 4 ¥ (3.1412)2 ¥ 0.125 ¥ cos q 9.81 = 1.2576 cos q Friction head: = 2 f hfd Solution: 2 p ¥ 30 = 3.1412 rad/s, r = 0.25/2 = 0.125 m 60 (A/As) = (10/5)2 = 4, (A/Ad ) = (10/5)2 = 4, w= Suction side: = 1.2576 sin2 q Pressure head on the piston Htd = Hatmo + (Hd + Had + hfd ) = 10.0 + (16.0 + Had + hfd) = 26.0 + (Had + hfd) m (abs) Acceleration head: Ls Ê A ˆ 2 w r cos q g ÁË As ˜¯ 6.0 = ¥ 4 ¥ (3.1412)2 ¥ 0.125 ¥ cos q 9.81 Has = = 3.0182 cos q Friction head: hfs = = ˆ fLs Ê A w r sin q ˜ 2g ds ÁË As ¯ ˆ fLd Ê A = w r sin q ˜ 2g dd ÁË Ad ¯ 0.02 ¥ 25 (4 ¥ 3.1412 ¥ 0.125 ¥ sin q)2 = 2 ¥ 9.81 ¥ 0.05 Condition Beginning 2 0.02 ¥ 6 (4 ¥ 3.1412 ¥ 0.125 ¥ sin q)2 2 ¥ 9.81 ¥ 0.05 = 0.302 sin2 q Pressure head on the piston Hts = Hatmo– (Hs + Has + hfs) = 10.0 – (4.0 + Has + hfs) = 6.0 – (Has + hfs) m (abs) Mid End 16.58 q Had (m) hfd (m) Pressure Pressure head head Htd Htd (abs) (gauge) 0° 12.576 0 38.576 m 28.576 m 90° 0 1.258 27.258 m 17.258 m 180° –12.576 0 13.424 m 3.424 m 573 Hydraulic Machines Solution: Crank radius r = L/2 = 0.4/2 = 0.20 m Since there are two strokes per revolution, N = 50/2 = 25 rpm 2 p ¥ 25 w = 60 = 2.618 rad/s, (A/As) = (d/ds)2 With No Air vessel: hfd = Ls Ê A ˆ 2 wr g ÁË As ˜¯ 2 5.0 Ê ˆ ¥ (2.618)2 ¥ 0.20 ¥ d 9.81 ÁË ds ˜¯ = 0.699 (d/ds)2 At limiting condition for a suction pipe Hasm + Hv + Hs = Hatmo 0.699 (d/ds)2 + 2.5 + 3.0 = 10.0 (d/ds)2 = 6.438 and as such (d/ds) = 2.537 Since d = 25 cm, ds = 9.86 cm Hasm = 2 Maximum head loss occurs when q = 0° hfdm = maximum head loss 2 = ˆ fLd Ê A rw ˜ Á 2g dd Ë Ad ¯ = fLd Ê A L 2 p N ˆ 2g dd ÁË Ad 2 60 ˜¯ = fLd Ê A LN ˆ p 2g dd ÁË Ad 60 ˜¯ Maximum acceleration head in suction pipe Hasm = ˆ fLd Ê A rw sin q ˜ Á 2g dd Ë Ad ¯ 2 2 = p2 hfdl Time averaged head loss hfd2 = (2/3) hfdm = 2 2 p hfdl 3 Work done per stroke P2 = W hfd2 = 2 2 p P1 3 Saving in work due to air vessel Ê 2 2ˆ ÁË 3 p ˜¯ - 1 P2 - P1 = = 0.848 Ê 2 2ˆ 2 p ÁË 3 ˜¯ 16.59 Solution: (a) Single Acting Pump With Air Vessel: When an air vessel is provided very near to the cylinder in the delivery pipe the ﬂow in the pipe is steady. The average velocity in the delivery pipe. h fdl = head loss in the delivery pipe fLdVd2 fLd Ê A LN ˆ = 2 g dd 2g dd ÁË Ad 60 ˜¯ Ê A ˆ LN Vd = Q t/Ad = 2 Á ˜ . Ë Ad ¯ 60 hfd3 = head loss in the delivery pipe = Ê A ˆ LN Vd = Qt /Ad = Á ˜ . Ë Ad ¯ 60 = Percentage savings in work done per stroke = 84.8 % (b) Double Acting Pump 2 If W = weight of water pumped per stroke, Work done per stroke P1 = W h fdl fLdVd2 4 fLd Ê A LN ˆ = 2g dd 2g dd ÁË Ad 60 ˜¯ 2 If W = weight of water pumped per stroke, Work done per stroke P3 = W hfd3 With No Air vessel hfd = ˆ fLd Ê A rw sin q ˜ 2g dd ÁË Ad ¯ 2 574 Fluid Mechanics and Hydraulic Machines Maximum head loss occurs when q = 0º hfdm = maximum head loss 2 = ˆ fLd Ê A rw ˜ Á 2g dd Ë Ad ¯ = fLd Ê A L 2 p N ˆ 2g dd ÁË Ad 2 60 ˜¯ fLd Ê A LN ˆ p = 2g dd ÁË Ad 60 ˜¯ 2 2 p2 hfd3 3 4 2 p2 P 3 4 3 Savings in work due to air vessel Ê 2 2 1ˆ ÁË 3 p 4 ˜¯ - 1 P4 - P3 = = P4 Ê 2 2 1ˆ ÁË 3 p 4 ˜¯ = 0.392 Percentage savings in work done per stroke = 39.2% 16.60 f Case-1: No Air Vessel At q = 0°, hfs = 0 and Has = Hasm Hasm = Ls Ê A ˆ 2 wr g ÁË As ˜¯ 6.0 ¥ 4 ¥ (0.1047 N)2 ¥ 0.225 9.81 = 0.00634 N2 At limiting condition for a suction pipe Hasm + Hv + Hs = Hatmo Hasm + 3.0 + 2.5 = 10.0 Hence Hasm = 4.5 = 0.006034 N 2 N = 27.3 rpm = Work done per stroke P4 = W hfd3 = 2p ¥ N = 0.1047 N rad/s, r = 0.45/2 60 = 0.225 m (A/As) = (30/15)2 = 4, w= 2 = (p2/4) hfd3 Time averaged head loss hfd4 = (2/3) hfdm = Solution: Case-2: With Air Vessel When an air vessel is ﬁtted in the suction pipe at 2.0 m from the cylinder. Acceleration pressure head is conﬁned to a 2.0 m length next to the cylinder. Friction loss in remaining 4.0 m of suction pipe is constant over time, as the ﬂow is steady. Hasm = Ls Ê A ˆ 2 w r g ÁË As ˜¯ 2.0 ¥ 4 ¥ (0.1047 N)2 ¥ 0.225 9.81 = 0.002011 N 2 Vs = average steady velocity in pipe below air vessel Ê A ˆ wr 1 = Á ˜ ¥ 0.1047 N ¥ 0.225 = Ë As ¯ p p = = 0.03 N fLs hfs = V2 2g ds s 0.02 ¥ 40 = (0.03 N)2 2 ¥ 9.81 ¥ 0.15 = 2.446 10–5 N 2 575 Hydraulic Machines At limiting condition for a suction pipe Hasm + Hv + h fs = Hatmo 0.002011 N 2 + 2.5 + 3.0 + 2.446 ¥ 10–5 N 2 = 10.0 N 2 = 4.5/2.0355 = 2210.8 N = 47 rpm ALN Discharge Q t = 60 Ratio of discharge Q2/Q1 = N2/N1 = 47.0/27.3 = 1.722 Percentage change in discharge = 72.2% increase after ﬁtting the air vessel. L¢d = length of delivery pipe after air vessel = 19.0 m At the beginning of the stroke: Friction head is conﬁned only to length Lda Acceleration head: Lda Ê A ˆ 2 wr g ÁË Ad ˜¯ 1.0 = ¥ 4 ¥ (4.189)2 ¥ 0.15 9.81 = 1.073 m Friction head: fLd¢ hfdl = Vd2 2 g dd 0.02 ¥ 19 = ¥ (0.8)2 = 0.165 m 2 ¥ 9.81 ¥ 0.075 Had = 16.61 Pressure head on the piston Htd = Hd + Had + hfdl = 15.0 + 1.073 + 0.165 = 16.238 m (gauge) At the middle of the stroke: Acceleration head Had = 0 Friction head hfd = hfd1 + hfd2 hfd1 = same as at the beginning of the stroke = 0.165 m hfd2 = additional friction head in pipe of length Lda f Solution: p (0.15)2 = 0.0177 m2, L = 0.3 m, 4 For a single acting pump ALN 0.0177 ¥ 0.3 ¥ 40 Qt = = 60 60 A= hfd2 = = 0.00354 m3/s 2 0.02 ¥ 1 ¥ (4 ¥ 4.189 ¥ 0.15)2 2 ¥ 9.81 ¥ 0.075 = 0.086 m Pressure head on the piston Htd = Hd + h fd1 + hfd2 = 15.0 + 0.165 + 0.086 = 15.251 m (gauge) = 2 p ¥ 40 w = = 4.189 rad/s, r = 0.30/2 60 = 0.150 m (A/Ad) = (15/7.5)2 = 4, Ad = (0.01777/4) = 0.004425 m3/s Vd = steady velocity after the air vessel = Qt/Ad = 0.00354/0.004425 = 0.8 m/s Lda = length of delivery pipe before air vessel = 1.0 m ˆ fLda Ê A w r˜ Á 2g dd Ë Ad ¯ 16.62 576 Fluid Mechanics and Hydraulic Machines 16.63 f Solution: Without Air Vessel p A= (0.15)2 = 0.0177 m2, 4 For a single acting pump L = 0.3 m, ALN 0.0177 ¥ 0.3 ¥ 60 = 60 60 3 = 0.00531 m /s 2 p ¥ 60 w = = 6.283 rad/s, r = 0.30/2 60 = 0.150 m Maximum friction head Qt = hfdm = ˆ fLd Ê A w r˜ 2g dd ÁË Ad ¯ 2 0.02 ¥ 30 ¥ (4 ¥ 6.283 ¥ 0.15)2 2 ¥ 9.81 ¥ 0.075 = 5.795 m Time averaged friction head = hfda = (2/3) hfdm = (2/3) ¥ 5.795 = 3.863 m With Air Vessel: The velocity is steady in the delivery pipe with an average value of Vd = Qt /Ad = 0.00531/0.004425 = 1.2 m/s Friction head: fL hfdl = V2 2g dd d 0.02 ¥ 30 = ¥ (1.2)2 = 0.587 m 2 ¥ 9.81 ¥ 0.075 Friction power saved due to air vessel in delivery pipe, in kW, g Qt Pf = (hfda – hfdl) 1000 = (9790/1000) ¥ 0.00531 ¥ (3.863 – 0.587) = 0.17 kW Solution: p (0.2)2 = 0.0314 m2, L = 0.4 m, 4 2 p ¥ 90 w= = 9.425 rad/s, r = 0.40/2 60 = 0.20 m (A/As ) = (20/10)2 = 4, As = A/4 = 0.007854 m2 Since the pump is double acting 2 ALN 2 ¥ 0.0314 ¥ 0.4 ¥ 90 Qt = = 60 60 = 0.03768 m3/s vs = Instantaneous velocity of unsteady ﬂow in suction pipe between air vessel and the cylinder A = wr sin q = 4 ¥ 9.425 ¥ 0.20 sin q As A = = 7.54 sin q Discharge in this part of suction pipe at any instant Qs = vs As = 7.54 sin q ¥ 0.007854 = 0.05922 sin q Discharge into/out of air vessel: If Qt < Qs, then ﬂow goes out of the air vessel If Qt > Qs, then ﬂow goes into the air vessel Crank angle q Qt m3/s Qs m3/s Remarks 45° 0.03768 0.04187 Flow going out of air vessel = 0.00419 m3/s 150° 0.03768 0.02961 Flow going into air vessel = 0.00807 m3/s 577 Hydraulic Machines Power required in kW g Qt P= (Hs + hfs + Hd + hfd) 1000 ¥ h 16.64 9790 ¥ 0.003534 (4.0 + 0.097 + 2.5 + 0.364) 1000 ¥ 0.8 = 0.30 1 kW = 16.65 f f Solution: p (0.15)2 = 0.01767 m2, L = 0.2 m, 4 2 p ¥ 60 w = = 6.283 rad/s, r = 0.20/2 = 0.10 m 60 (A/As) = (15/7.5)2 = 4, Ad = As = A/4 = 0.004179 m2 ALN 0.01767 ¥ 0.2 ¥ 60 Qt = = 60 60 = 0.003534 m3/s A = Since the air vessels are very near to the cylinder, it is assumed that the entire suction an delivery pipes will have steady ﬂow. Vs = Vd = velocity in suction and delivery pipes = 0.003534/0.004179 = 0.8457 m/s Hence fLs 0.025 ¥ 8 hfs = ¥ (0.8457)2 V2 = 2 ¥ 9.81 ¥ 0.075 2g ds s = 0.097 m hfd = fLd 0.025 ¥ 30 ¥ (0.8457)2 V2 = 2g dd d 2 ¥ 9.81 ¥ 0.075 = 0.364 m Solution: At the press: Force to be provided by the ﬂuid pressure = 500 + 25 = 525 kN Fluid pressure at the base of the ram = 525 = 7427 kN/m2 Ê pˆ 2 ÁË 4 ˜¯ ¥ (0.30) Area of the plunger = Ê pˆ Ap = Á ˜ ¥ (0.10)2 = 0.007854 m2 Ë 4¯ Force required on the plunger (without frictional losses) F1 = 7427 ¥ 0.007854 = 58.33 kN Frictional resistance in plunger assembly = 0.01 ¥ 58.33 = 0.5833 kN Total force required at the plunger = Fp = (58.33 + 0.583) = 58.91 kN 16.66 578 Fluid Mechanics and Hydraulic Machines Solution: Effective load = 800 – (0.02 ¥ 800) = 784 kN Power supplied by the accumulator = WL 784 ¥ 5.0 = = 32.67 kW t 120 Pressure at the accumulator = 784 = 11091.3 kN/m2 p= Ê pˆ 2 ÁË 4 ˜¯ ¥ (0.30) Discharge Q = 15 liter/s Power supplied by the pump = pQ P= Solution: From equilibrium considerations of the moving ram at any position, p2 A1 e ˆ Ê p1 A1 Á1 = ˜ e ˆ Ê Ë 100 ¯ ÁË1 - 100 ˜¯ p2 = Ê 15 ˆ = 11091.3 ¥ Á = 166.37 kW Ë 1000 ˜¯ 16.67 2 2 ˆ Ê 50 ˆ Ê p2 = 200 ¥ Á ˜ Á1 Ë 10 ¯ Ë 100 ˜¯ = 960.4 kPa Total power supplied by the hydraulic system = 32.67 + 166.67 m = 199.04 kW p2 A1 Ê e ˆ 1Á Ë 100 ˜¯ A2 2 16.69 l l Solution: Here, Hn = 13.0 m, (Hd + Hs) = (11.0 + 3.0) = 14.0 m, Q (H + Hd ) Efﬁciency of the jet pump h = s s Qn ( H n - H d ) Solution: Effective load = 350 – (0.02 ¥ 350) = 343 kN Power supplied by the accumulator = P = WL t 343 ¥ 3.0 = 11.43 kW 90 The displacement of the accumulator is the volume displaced by the ram in one full stroke. h= = (15.0 - 2.5)(14.0) = 35.0% ( 2.5)(13.0 - 11.0) 16.70 p ¥ (0.250)2 ¥ 3.03 4 = 0.147 m3 = 147.3 liters Displacement = 16.68 0.005 0.007854 579 Hydraulic Machines fLV 2 2gDd h 0.02 ¥ 100 ¥ (0.6367)2 2 ¥ 9.81 ¥ 0.10 h 16.71 Q1 ( H 2 + hfd ) (Q1 + Q2 )( H1 - hfs ) f Solution: H1 H2 Q1 Q2 Qs = 4.0 m, hfs = 0 = 15.0, hfd = 1.5 m = 2.0 liters/s = 0.002 m3/s = 15.0 liters/s = 0.015 m3/s = discharge supplied to the ram = (0.002 + 0.015) = 0.017 m3/s Solution: = 4.0 m = 18.0, Q1 = 5.0 liters/s = 0.005 m3/s = 75.0 liters/s = 0.075 m3/s = discharge supplied to the ram = (0.005 + 0.075) = 0.08 m3/s Ad = area of delivery pipe p = (0.1)2 = 0.007854 m2 4 Vd = velocity in the delivery H1 H2 Q2 Qs Efﬁciency of the ram =h= h= Q1 ( H 2 + hfd ) (Q1 + Q2 )( H1 - hfs ) 0.002 ¥ (15 + 1.5) = 0.485 0.017 ¥ 4 h= = 48.5% 0.005 ¥ (18 + 0.413) 0.08 ¥ 4 = 0.288 = 28.8% Problems A. Turbines Reaction Turbines Velocity Triangle Relationships 16.1 An inward ﬂow reaction turbine has a runner of outer diameter 1.2 m and inner diameter 0.6 m. The blades occupy 5% of the peripheral area and the widths of the blades are 25 cm and 30 cm at the inlet and outlet respectively. If a discharge of 3.0 m3/s enters radially, determine the ﬂow velocities at the inlet and outlet of the runner. (Ans. Vf1 = 3.35 m/s; Vf2 = 5.58 m/s) 16.2 A Francis turbine has a wheel of outer diameter = 1.25 m and inner diameter = 0.6 m. The runner blades are radial at inlet and the discharge is radial at outlet. If the ﬂow enters the vanes at 10° and the velocity of ﬂow is 3.0 m/s calculate the speed of the runner and the vane angle at the outlet. (Ans. N = 260 rpm; b2 = 20.18°) 16.3 A Francis turbine has wheel diameters of 0.9 m at entrance and 0.45 m at exit. The runner blades are radial at entrance and the guide vanes are at 12°. The head developed in the turbine is 25 m. Assuming 580 Fluid Mechanics and Hydraulic Machines that the ﬂow leaves the turbine radially, calculate the speed of rotation and the blade angle at the exit. (Ans. N = 332.3 rpm; b2 = 23.03°) 16.4 A Francis turbine having an overall efﬁciency of 76% is to produce 105 kW of power under a head of 12 m and speed of 150 rpm. The peripheral velocity at inlet is 10 m/s and the velocity of ﬂow at inlet is 5.0 m/s. Assuming the hydraulic losses as 20% of available energy, calculate the (i) guide vane angle, (b) wheel blade angle at inlet and (c) width of wheel at inlet. (Ans. a1 = 12.07°, b1 = 99.27°, b2 = 23.90°) 16.5 An inward ﬂow reaction turbine develops 1750 kW at 750 rpm under a net head of 100 m. The guide vanes are at an angle of 15° with tangent at the inlet. The breadth of the blade at inlet is 0.1 times the inlet diameter. The blade thickness blocks 5% of the inlet area. The hydraulic efﬁciency of the wheel is 88% and the overall efﬁciency is 84%. Determine the (a) wheel diameter at inlet and (b) blade angle at the inlet. (Ans. D1 = 0.519 m; b1 = 152.7°) 16.6 In an outward ﬂow reaction turbine rotating at 300 rpm the inner and outer diameters are 1.50 m and 1.85 m respectively. The wheel has 32 vanes 15 mm thick at inlet and 30 mm thick at the outlet. The breadth of the passage is 25 cm throughout. The net head available is 40 m. The discharge is 7.5 m3/s and takes place to atmosphere. Determine the (i) blade angles at the inlet and outlet and (ii) power developed. (The ﬂow leaves the outlet radially). {Hint: V2 = Vf2 and head extracted He = u1 Vu1/g = Ht V22 V2 = H – f 2 .} 2g 2g (Ans. (i) b1 = 42.56°; b2 = 12.01°; (ii) P = 2794 kW) 16.7 An inward ﬂow reaction turbine works under a total head of 25 m. The outer and inner diameters of the runners are 0.70 m and 0.35 m respectively. The vane tip is radial at the inlet and the ﬂow leaves the turbine radially. The guide vane angle is 12°. Calculate the speed of the runner and the vane angle at exit, if the velocity of ﬂow at the exit of the draft tube is 4.0 m/s. Assume the velocity of ﬂow to be constant and the hydraulic efﬁciency to be 0.90. {Hint: Head extracted = Net available head-velocity head at the end of the draft tube.} (Ans. N = 420.2 rmp; b2 = 23.03°) 16.8 An inward ﬂow reaction turbine has a guide vane angle of 20° and a vane angle at inlet of 30°. Determine the hydraulic efﬁciency of the turbine. Assume the outﬂow to be radial and the velocity of ﬂow to be constant. {Hint: Refer to Example 16.7.} (Ans. h H = 96 %) 16.9 An inward ﬂow radial turbine works under a head of 30 m and discharges 10 m3/s. The speed of the runner is 300 rpm. At inlet tip of runner vane, the peripheral velocity of wheel is 0.9 2gH and radial velocity of ﬂow is 0.3 2gH where H is the head on the turbine. If the overall efﬁciency and hydraulic efﬁciency of the turbine are 80% and 90% respectively, (i) determine the power developed in kW, (ii) diameter and width of runner at inlet, (iii) guide blade angle at inlet, (iv) inlet angle of runner vane and (v) diameter of runner at outlet. [Assume radial ﬂow at outlet] (Ans. 2352 kW, 1.391 m, 30° 58¢ and 1.323. m) 581 Hydraulic Machines 16.10 An inward ﬂow reaction turbine has a setting in the tailwater without a draft tube. Show that for radial inlet and outlet vanes and for constant velocity of ﬂow, the hydraulic efﬁciency hH is given by hH = 2 cot 2 a1 1 + 2 cot 2 a1 where a1 = guide vane angle. {Hint: Refer to Example 16.7.} 16.11 A Kaplan turbine runner has an outer diameter of 4.5 m and an inner diameter of 2.0 m and develops 20,000 kW of power while runner at 140 rpm. The head is known to be 20 m. Assuming a hydraulic efﬁciency of 85% ﬁnd the discharge through the turbine, the blade angle at the inlet and guide vane angle at inlet. (Ans. Q = 120.17 m3/s: a1 = 59.3°, b1 = 161.03°) 16.12 A Kaplan turbnine develops 2300 kW at an available head of 32 m. The speed ratio and ﬂow ratios are respectively 2.1 and 0.62. The diameter of the boss is 1/3 the diameter of the runner. Assuming an overall efﬁciency of 88%, estimate the diameter, speed of rotation and speciﬁc speed of the runner. (Ans. D = 0.877 m; N = 1146 rpm; Ns = 722) 16.13 A Kaplan turbine has speed ratio of 2.0 and a speciﬁc speed of 450. Determine the diameter of the propeller in order that it will develop 10,100 kW under a head of 20 m. (Ans. D = 4.0 m) 16.14 A Kaplan turbine develops 1 MW under a head of 4.5 m. For speed ratio f = 1.8, ﬂow ratio y = 0.5, boss diameter = 0.35 times the outer diameter and overall efﬁciency of 90%, ﬁnd the diameter, speed of rotation and speciﬁc speed of the runner. (Ans. D = 2.79 m, N = 115.8 rpm, Ns = 558.7) 16.15 A Kaplan turbine has a runner diameter to hub diameter of 3. The speed ratio is 1.61. If this turbine produces 6500 kW of power at a head of 15 m and speed of 150 rpm, calculate the (i) speciﬁc speed, (ii) ﬂow ratio and (iii) discharge. (Ans. Ns = 410, Q = 48.11 m3/s, y = 0.324) Draft Tube 16.16 A turbine has an exit velocity of 10 m/s and is provided with a straight conical draft tube. The velocity head at the exit of the draft tube is 1.0 m and loss of head in the draft tube is 1.5 m. To avoid cavitation, the minimum pressure head in the turbine is set at 2.0 m (abs). Taking atmospheric pressure as 10.3 m of water, estimate the maximum height of setting of the turbine above the tail water level. (Ans. hs = 5.704 m) Impulse Turbines 16.17 A Pelton wheel has a bucket diameter of 90 cm and has one jet of diameter 8 cm with a coefﬁcient of velocity of 0.97. The wheel has a speed ratio of 0.45 and a blade angle of 170°. The blade friction coefﬁcient is 0.93. If the net head available on the wheel is 500 m, calculate (i) the power extracted and (ii) speciﬁc speed of the wheel. (Ans. (i) P = 2120 kW; (ii) Ns = 18.4) 16.18 A Pelton wheel has a diameter of 1.25 m and operates under a net available head of 200 m. The other characteristics of the installation are: Cv = 0.98, N = 250 rpm, blade angle b = 162°, blade friction coefﬁcient k = 0.96, diameter of the jet d = 10 cm and mechanical efﬁciency h m = 0.96. Determine the (i) power delivered to the shaft, (ii) hydraulic efﬁciency and (iii) speciﬁc speed. 582 Fluid Mechanics and Hydraulic Machines 16.19 16.20 16.21 16.22 16.23 16.24 (Ans. (i) P = 650.8 kW; (ii) h H = 71.83%; (iii) Ns = 8.48) A Pelton wheel has a mean bucket speed of 10 m/s with a jet of water ﬂowing at a rate of 0.7 m3/s under a head of 30 m. The bucket deﬂects the jet through an angle of 160°. Assuming Cv = 0.98, calculate the power and overall efﬁciency of the turbine. (Ans. P = 186.7 kW, ho = 0.908) Find the hydraulic efﬁciency of an impulse turbine for which: coefﬁcient of velocity Cv = 0.98, bucket angle b = 165°, speed factor f = 0.46 and ratio of outlet relative velocity to inlet relative velocity in a bucket k = 0.99. (Ans. h H = 0.936) In a Pelton wheel the overal efﬁciency is 0.85, Cv = 0.95, speed factor f = 0.46 and the ratio of wheel diameter to jet diameter is 12. Calculate the speciﬁc speed of the wheel. (Ans. Ns = 17.0) Show that with usual notations the speciﬁc speed of an impulse turbine can be written as Êdˆ Ns = 493.7 f Á ˜ hoCv Ë D¯ The water jet in a Pelton wheel is 8 cm in diameter and has a velocity of 93 m/s. The rotational speed of the wheel is 600 rpm and the deﬂection angle of the jet is 170°. If the speed ratio f = 0.47, determine the (i) diameter of the wheel and (ii) power developed. (Assume Cv = 1.0). (Ans. (i) D = 1.39 m; (ii) P = 1995 kW) A Pelton turbine has two jets of diameter 15 cm each and develops 6500 kW of power. If the net available head on the turbine is 300 m determine the overall efﬁciency of the turbine. Also, if the rotatinoal speed is 375 rpm, calculate the speciﬁc speed. Take Cv = 0.98. (Ans. Ns = 17.12; ho = 0.833) 16.25 The water available to a power house is 3.0 m3/s and the total head from the reservoir to the nozzle is 250 m. There are three pelton wheels of two jets each. All the six jets have the same diameter and are supplied with water from a single pipe of length 500 m. The efﬁciency of power transmission through the pipe is 90% and the overall efﬁciency of the turbie is 85%. The Cv for each nozzle is 0.95 and the Darcy–Weisbach friction factor f = 0.02. Determine (i) the power developed, (ii) diameter of the jet and (iii) diameter of the pipe. (Ans. (i) P = 5617 kW; (ii) d = 10 cm; (iii) D1 = 0.785 m) 16.26 A double overhung impulse turbine installation is to develop 15000 kW at 260 rpm under a net head of 350 m. (a) Determine the speciﬁc speed and wheel pitch diameter (b) What would these values be when (i) a single wheel with a single nozzle is used and (ii) a single wheel with four nozzles are used? (Assume a velocity ratio of 0.46 and Cv = 1.0). (Ans. (a) Ns = 14.87; D = 2.8 m; (b) (i) Ns = 21.03; D = 2.8 m; (ii) Ns = 10.52; D = 2.8 m) 16.27 In a hydroelectric station, water is available at the rate of 175 m3/s under a head of 18 m. The turbines run at a speed of 150 rpm whith overall efﬁcicney of 82%. Find the number of turbines required if they have the maximum speciﬁc speed of 460 rpm. (Ans: 2 Nos) Similarity Relationships in Turbines 16.28 A turbine is to operate under a head of 25 m at a speed of 300 rpm. The discharge 583 Hydraulic Machines is 12 m3/s. Assuming an efﬁciency of 0.85 calculate the power developed. What would be the speciﬁc speed, power, discharge and rotational speed at a head of 15 m? (Ans. (i) Ns = 268.1; P1 = 2496.5 kW; (ii) Ns = 268.1; P2 = 1160 kW; Q2 = 9.30 m3/s; N2 = 232.4 rpm) 16.29 A turbine develops 150 kW while running at 120 rpm under a head of 10.0 m. The diameter of the runner is 1.5 m. A 1 : 3 scale model of this turbine is tested under a head of 3.0 m. Determine the speed and power developed in the model. Assuming an overall efﬁciency of 0.90 for both the model and prototype calculate the discharges in the model and prototype. (Ans. (i) Nm = 197.18 rpm; Pm = 2.739 kW; (ii) Qp = 1.7024 m3/s; Qm = 0.1036 m3/s) 16.30 A 1 : 5 model of a turbine develops 2.0 kW of power at 400 rpm under a head of 3.0 m. What is the speciﬁc speed of the runner? Assuming an overall efﬁciency of 0.85 for both the model and prototype, calculate the rotational speed, power and discharge of the prototype when working under a head of 20 m. (Ans. Ns = 143.3; Np = 206.6 rpm; Pp = 861.2 kW; Q p = 5.165 m3/s) 16.31 In a small hydro development a Kaplan turbine runs under a head of 2.1 m. It has a runner of 3.5 m diameter and develops 600 kW at 80 rpm. Assuming an overall efﬁciency of 80%, estimate the discharge and speciﬁc speed of the machine. If a 1.5 m diameter homologous turbine is to be tested at a head of 3.0 m, What are the rotational speed, discharge and power of that unit? (Ans. Qp = 36.48 m3/s; Ns = 775; Nm = 156.2 rpm; Qm = 5.607 m3/s; Pm = 64.6 kW) 16.32 A turbine is to operate under a head of 25 m at 200 rpm. The discharge is 9.0 m3/s. If the eﬁciency is 90%, determine the performance of the turbine under a head of 20 m. (Ans. N2 = 178.9 rpm; Q2 = 8.05 m3/s; P2 = 1418.6 kW) B. Rotodynamic Pumps Velocity Triangle Relationships 16.33 A centrifugal pump having an overall efﬁciency of 70% delivers 1500 L/min of water against a static head of 20 m. The suction and delivery pipes are of 20 cm diameter and has a combined total length of 1000 m. Assuming f = 0.02 estimate the power input required. (Ans. Ps = 8.122 kW) 16.34 A centrifugal pump runs at 800 rpm and delivers 5000 L/min against a head of 7 m. The impeller has an outer diameter of 25 cm and a width of 5 cm at the outlet. If the backward curved vane at the outlet makes an angle of 45°, determine the manometric efﬁciency. What is the speciﬁc speed of the pump? (Ans. h H = 78.5%; Ns = 53.65) 16.35 A centrifugal pump has an impeller of diameter 30 cm whose width at exit is 6.0 cm. The velocity of ﬂow through the impeller is constant at 3.0 m./s. The impeller vanes are radial at the outer periphery. If the rotational speed is 1000 rpm and the manometric efﬁciency is 80%, calculate (i) the head produced and (ii) discharge. (Ans. (i) H = 20.12 m; (ii) Q = 169.6 L/s) 16.36 A centrifugal pump has vanes which are radial at the outer periphery. The impeller has an outer diameter of 20 cm and a width of 3 cm at that diameter. If the discharge is 1800 L/min and the net head produced is 3.5 m, calculate the 584 Fluid Mechanics and Hydraulic Machines 16.37 (i) (ii) 16.38 16.39 16.40 (i) rotational speed of the impeller and (ii) magnitude and direction of absolute velocity at exit. Manometric efﬁciency can be assumed as 0.85. (Ans. N = 607 rpm; (ii) V2 = 6.552 m/s, a2 = 14.06°) A centrifugal pump impeller is 40 cm in outer diameter and 2.5 cm wide at the exit, and its blade angle is 30°. When run at a speed of 2100 rpm, the ﬂow rate through the pump is 80 L/s. Calculate the radial, relative and absolute velocities at the impeller exit. If there is no inlet whirl, what would be the head added to the water by the impeller? (Ans. (i) 2.55 m/s, 5.10 m/s and 39.62 m/s. (ii) 177.18 m) A centrifugal pump running at 1200 rpm delivers water at a net head of 10.0 m. At the outlet of the impeller the vane angle is 30° with the peripheral velocity. The impeller has an outer diameter of 25 cm and a width of 5 cm at the outlet. Estimate the discharge (Assume manometric efﬁciency = 85%). (Ans. Q = 190 L/s) A centrifugal pump impeller rotates at a speed of 1000 rpm. The external and internal diameters of the impeller are 0.40 m and 0.20 m respectively. The vanes make an angle of 35° with the tangential direction of rotation at the outlet. If the radial velocity of ﬂow through the impeller is constant at 1.75 m/s, ﬁnd the (i) angle of vanes at the inlet, (ii) absolute velocity and its direction at outlet and (iii) net head of the pump. (Assume manometric efﬁciency = 80%) (Ans. (i) b1 = 9.5°; (ii) V2 = 18.52 m/s, a2 = 5.4°; (iii) H = 31.5 m) A centrifugal pump with impeller diameter of 50 cm at outlet and 25 cm at inlet runs at a speed of 1000 rpm. The discharge is 150 L/s. The width is 8 cm at the inlet and at the outlet it is 6 cm. The vanes are curved back and make an angle of 25° with the peripheral velocity direction at the outlet. Assuming a manometric efﬁciency of 0.85 and mechanical efﬁciency of 0.80, calculate (i) the net head produced by the pump and (ii) brake power input to the pump. (iii) What is the speciﬁc speed of the pump? (Ans. (i) H = 51.65 m; (ii) BP = 111.5 kW; (iii) Ns = 20.1) 16.41 A centrifugal pump has an impeller of outer diameter 35 cm and an inside diameter of 20 cm. The impeller vanes make angles of 35° and 30° with the direction of the peripheral velocity at the outlet and inlet respectively. The rotational speed of the impeller is 1000 rpm. Assuming the ﬂow velocity to remain constant throghout the impeller, estimate the net head developed by the pump. Assume manometric efﬁciency as 0.80. (Ans. H = 21.06 m) General Characteristics 16.42 The impeller of a centrifugal pump has an outer diameter of 50 cm and inner diameter of 25 cm . If the discharge pipe is closed and the pump is full of water, what would be the difference in pressure between the outer and inner periphery when the impeller rotates at 600 rpm? (Ans. H = 9.43) 16.43 A centrifugal pump with impeller of outer diameter 45 cm and inner diameter 25 cm, is required to develop a net head of 20 m. Find the lowest speed to start pumping. (Ans. N = 1011 rpm) 16.44 A pump has a head-discharge characteristic given by Hp = 35 – 2200 Q2 where Hp = head developed by the pump in metres 585 Hydraulic Machines and Q = corresponding discharge in m3/s. The pump is to deliver a discharge against a static head of 12 m. The suction pipe is 15 cm in diameter, 20 m long and has an f value of 0.018. The delivery pipe is 20 cm in diameter, 400 m long and has an f value of 0.0. Calculate the head and discharge delivered by the pump. If the overall efﬁciency of the pump is 0.70, calculate the driving power supplied to the pump. (Ans. H = 24.14 m; Q = 70 L/s; BP = 23.63 kW) 16.45 (a) A pump delivers water at a net head of 45 m. The atmospheric pressure is 90 kPa (abs) and the vapour pressure is 4.24 kPa (abs). If the head lost in the intake pipe due to friction is 0.76 m, calculate the maximum allowable elevation above the sump water level at which the pump can be located. The critical cavitation number for the pump can be taken as 0.15. (b) In another location, with all other factors remaining the same except the net head of the pump which is now 80 m, estimate the maximum allowable elevation of the pump above the sump water level. (Ans. (a) (Z c)m = 1.25 m; (b) (Zc ) m = – 4.0 m (i.e. 4 m below the sump water level.) 16.46 The head-discharge characteristic of a centrifugal pump is given below. The pump delivers water through a 20 cm diameter, 2000 m long pipe. The friction factor of the pipe f = 0.02. Neglecting minor losses, determine the discharge in the pipe for a static lift of 12 m. Discharge (L/s) 0 10 20 30 40 50 Head (m) 26 25.5 24.5 22.5 18.5 12.5 (Ans. H = 22.0 m and Q = 32 L/s) Similarity Relations for Pumps 16.47 A centrifugal pump with 30 cm diameter impeller was found to be most efﬁcient when discharging 0.20 m 3/s of water at 1200 rpm against a head of 15 m of water. A similar pump is required to deliver 2.0 m3/s at 1000 rpm. Calculate the dimater of the impeller and the head that could be developed by this pump. (Ans. D2 = 0.687 m; H2 = 54.6 m) 16.48 A centrifugal pump of 25 cm diameter runs at 1450 rpm and delivers 0.3 m3/s against a head of 12 m. Calculate the speciﬁc speed of the pump. A similar pump with half the size is to run at a head of 10 m. Find the working speed, discharge and power required. The efﬁciencies of the pumps can be assumed to be 75%. (Ans. Ns = 123; N2 = 4833 rpm; Q2 = 0.0205 m3/s; BP2 = 2.68 kW) 16.49 A centrifugal pump with an impeller diameter of 20 m discharges 120 L/s at 1200 rpm and 10 m head. What is its speciﬁc speed? If a homologous pump produces 240 L/s at 20 m head, determine its size and speed of rotation. (Ans. Ns = 73.9; D2 = 25.94 cm; N2 = 1427 rpm) 16.50 A centrifugal pump running at 750 rpm discharges water at 0.1 m3/s against a head of 10 m at its best efﬁciency. A second pump of the same homologous series, when working at 500 rpm, is to deliver water at 0.05 m3/s at its best efﬁciency. What will be the design head of the second pump and what is the scale ratio between the ﬁrst and the second? (Ans. H1 = 3.67 m, D1/D2 = 1.1) 16.51 A pump is installed at a height of 5.0 m above the water level in the sump. Frictional loss in the suction side is 0.6 m. If the atmospheric pressure is 10.3 and the 586 Fluid Mechanics and Hydraulic Machines vapor pressure is 0.43 m (abs) estimate the NPSH for this pump installation. (Ans. 4.3 m) C. Reciprocating Pumps —Basic Relationships 16.52 A single acting reciprocating pump has the following characteristics: Piston diameter = 30 cm Stroke = 50 cm Speed = 40 rpm Total lift = 25 m If the discharge delivered by the pump at the outlet is 1380 litres/minute, calculate the slip, coefﬁcient of discharge and theoretical power in kW required to drive the pump. (Ans. Slip = 2.34%, Coefﬁcient of discharge = 0.9766, Power = 5.78 kW) 16.53 A double acting reciprocating pump, running at 40 rpm is discharging 1.0 m3/min. The pump has a stroke of 40 cm and the diameter of the piston is 20 cm. The delivery and suction heads are 20 m and 5 m respectively. Find the slip of the pump and the power required to drive the pump. (Ans. Slip = 0.53%, Power = 4.08 kW.) 16.54 A plunger is ﬁtted to a vertical pipe ﬁlled with water. The lower end of the pipe is submerged in a sump. If the plunger is drawn up with an acceleration of 5.0 m/s2, ﬁnd the maximum height above the sump level at which the plunger will work without separation. Assume atmospheric pressure = 10.0 m water (abs) and separation occurs at 2.0 m water (abs). Take acceleration due to gravity as 10 m/s2. (Ans: Ls = 5.33 m) 16.55 A single acting reciprocating pump has the following data: Piston diameter = 10 cm Stroke = 30 cm Suction head = 4.0 m Diameter of suction pipe = 7.5 cm Suction pipe length = 4.0 m Assuming atmospheric pressure = 10.0 m water (abs) and cavitation occurs at 2.5 m water (abs) determine the maximum speed at which the pump can be run without cavitation. Assume Frictional losses = 1.0 m. (Ans: N = 45.8 rpm) 16.56 A single acting reciprocating pump has the following characteristics: Cylinder diameter = 22.5 cm Stroke length = 45 cm Suction head = 4.5 m Diameter of suction pipe = 22.5 cm Suction pipe length = 20.0 m Assuming atmospheric pressure = 10.0 m water (abs) and cavitation occurs at 2.0 m water (abs) determine the maximum speed at which the pump can be run without cavitation. (Ans. N = 26.36 rpm) 16.57 A double acting reciprocating pump has a 20 cm cylinder with a stroke of 20 cm. The suction pipe is 20 cm diameter and 15 m long. If the speed of the pump is 60 rpm, determine the maximum suction lift if separation occurs at 2.5 m water (abs). Assume atmospheric pressure = 10.0 m water (abs). Neglect frictional losses. (Ans: Hs = 1.457 m) 16.58 A double acting reciprocating pump has a cylinder of diameter 20 cm and stroke of 30 cm. The piston makes 30 strokes/ minute. Estimate the maximum velocity and acceleration in the suction pipe of diameter 20 cm and delivery pipe of diameter 25 cm. (Ans: Vsm = 0.471 m/s, asm = 1.480 m/s2: Vdm = 0.301 m/s, adm = 0.947 m/s2) 587 Hydraulic Machines 16.59 A single acting reciprocating pump has a stroke length of 15 cm. The suction pipe is 7 m long. The water level in the sump is 2.5 m below the cylinder. The diameters of suction pipe and the plunger are 7.5 cm and 10.0 cm respectively. If the speed of the pump is 75 rpm, determine the pressure head on the piston at the beginning, mid and end of the suction stroke. Take Darcy– Weisbach friction factor f = 0.02. (Ans. Hts = – 8.369 m (gauge), –2.604 m (gauge), and +3.868 m (gauge)) 16.60 A single acting reciprocating pump has the following data: Cylinder diameter = 35 cm Stroke = 35 cm Static suction head = 3.0 m Diameter of suction pipe = 20.0 cm Suction pipe length = 6.0 m Crank speed = 20 rpm Delivery pipe diameter = 20.0 cm Length of delivery pipe = 25.0 m Static delivery head = 20.0 m Estimate the power required to drive the pump. Take Darcy–Wesbach friction factor f = 0.02 and pump efﬁciency = 0.9. (Ans. P = 1.7 kW) Air Vessels 16.61 A single acting reciprocating pump has a stroke length of 37.5 and a cylinder of diameter 22.5 cm. The suction pipe is 12 m long and has a diameter of 15 cm. The water level in the sump is 3.0 m below the level of the cylinder. If the speed of the pump is 20 rpm, determine the pressure head on the piston at the beginning of the suction stroke (1) when no air vessel is ﬁtted and (2) when an air vessel is ﬁtted to the suction pipe at the level of the cylinder and at a distance of 1.5 m from the cylinder. Take Darcy–Weisbach friction factor f = 0.02. (Ans. (1) Hts = 5.263 m water (vacuum); (2) Hts = 3.289 m water (vacuum) 16.62 A single acting reciprocating pump has an air vessel in the delivery side ﬁtted very close to the cylinder. The cylinder has a diameter of 30 cm and a stroke length of 45 cm. The delivey pipe is 40 m long and has a diameter of 20 cm. The speed of the pump is 60 rpm. Determine the power saved by the air vessel in overcoming friction in the delivery pipe. Take Darcy– Weisbach friction factor f = 0.03. (Ans: Power savings = 5.45 kW) 16.63 A single acting reciprocating pump has a stroke length of 40 cm and a cylinder diameter of 25 cm. The delivery pipe is 20 m long and has a diameter of 15 cm. A large diameter air vessel is ﬁtted to the delivery pipe. For a crank speed of 40 rpm, determine the quantity of water going in or coming out of the air vessel when the crank angle is (i) 15° (ii) 90° and (iii) 120°. Also, determine the crank angle at which there is no ﬂow into or out of the air vessel. (Ans. (i) 0.0025 m3/s into the air vessel, (ii) 0.028 m3/s goes out of the air vessel (iii) 0.016 m3/s goes out of the air vessel; q = 18, 58° or 161.41°) 16.64 A double acting reciprocating pump has the following data: Cylinder diameter = 10 cm Stroke length = 20 cm Static suction head = 3.0 m Diameter of suction pipe = 8.0 cm Suction pipe length = 8.0 m Crank speed = 40 rpm Delivery pipe diameter = 8.0 m Length of delivery pipe = 15.0 m Static delivery head = 15.0 m Darcy–Weisbach friction factor f = 0.03. Estimate the power required to drive the 588 Fluid Mechanics and Hydraulic Machines pump by assuming a pump efﬁciency of 90%. (Ans: P = 4.97 kW) D. Miscellaneous Devices 16.65 A hydraulic press has a ram of 25 cm diameter and a plunger of 25 mm diameter. The plunger has a stroke of 25 cm and makes 30 strokes per minute. If a weight of 40 kN is to be lifted by the press ﬁnd the force on the plunger and rate of rise of the weight. (Ans. 7.5 cm/min) 16.66 A hydraulic ram receives 100 liter/s of water from a source under a head of 5.0 m and delivers 10.0 liter/s to a reservoir 20 m above the ram. The delivery pipe is 50 m long and has a diameter of 100 mm. The supply pipe is 15 m long and is 200 mm in diameter. Assuming a friction factor f = 0.02 for both the pipes, estimate the efﬁciency of the ram. (Ans. h = 52%) Objective Questions Turbines 16.1 A hydraulic turbine has a discharge of 3 m3/s when operating under a head of 15 m and a speed of 500 rpm. If it is to operate under 12 m of head, the rotational speed will be (a) 600 rpm (b) 559 rpm (c) 447 rpm (d) 400 rpm 16.2 A turbine has a discharge of 3 m3/s when operating under a head of 14 m and a speed of 400 rpm. If it is to operate under a head of 18 m, the discharge, in m3/s, will be (a) 3.86 (b) 3.40 (c) 2.65 (d) 2.23 16.3 A turbine works at 30 m head and 400 rpm speed. Its 1:2 scale model to be tested at a head of 30 m should have a rotational speed of (a) 800 rpm (b) 566 rpm (c) 400 rpm (d) 200 rpm 16.4 The unit speed Nu of a turbine of rotational speed N and head H is equal to (a) N H (b) N/ H (c) H /N (d) HN 16.5 The unit power Pu of a turbine developing a power P under a head H is equal to (a) P H 5/ 2 (c) P H3/2 (b) P/ H (d) P H 3/ 2 16.6 For a 1 : m scale model of a turbine the speciﬁc speed of the model Nsm is related to the prototype speciﬁc speed Nsp as Nsm = (a) Nsp/m (b) mNsp (c) Nsp (d) (Nsp)1/m 16.7 A turbine works under a head of 20 m, has a speed of 375 rpm and develops 400 kW of power. Its speciﬁc speed is (a) 375 (b) 83 (c) 177 (d) 1677 16.8 The cavitation parameter s for hydraulic machines is deﬁned as s = p - patm p - pv (b) (a) rV 2 / 2 rV 2 / 2 589 Hydraulic Machines (c) 16.9 16.10 16.11 16.12 16.13 patm - pv (d) p - pv patm - pv rV 2 / 2 where p = absolute pressure at the point of interest, pv = vapour pressure of liquid and V = reference velocity, patm = atmospheric pressure. The speciﬁc speed for a turbine has the dimensions of (a) T–1 (b) dimensionless (c) F1/2 L–3/4 T–3/2 (d) F1/2 L–5/2 T–3/2 In all reaction turbines, maximum efﬁciency is obtained, if (a) the guide vane angle is 90° (b) the blade angle of the runners is 90° at the inlet (c) the blade angle of the runners is 90° at the outlet (d) the angle of the absolute velocity vector at the outlet is 90° [Note: All angles are measured with respect to the direction of the peripheral velocity.] Two hydraulic turbines are similar and homologous when they are geometrically similar and have (a) the same Thoma number (b) the same Froude number (c) the same speciﬁc speed (d) the same rotational speed A turbine develops 2515 kW at 240 rpm. The torque in the shaft is (a) 400 kN.m (b) 3335 kN.m (c) 1000 kN.m (d) 100 kN.m The moment of momentum of water in a turbine is reduced by 15915 N.m in a turbine rotating at 600 rpm. The power developed, in kW, is (a) 1000 (b) 1500 (c) 2000 (d) 5000 16.14 A turbine develops 500 kW power under a net head of 30 m. If the overall efﬁciency of the turbine is 0.83, the discharge through the turbine, in m3/s, is (a) 20.5 (b) 2.05 (c) 1.41 (d) 1.51 16.15 A reaction turbine has a discharge of 30 m3/s passing through it under a net head of 10 m. If the overall efﬁciency is 0.851 the power developed, in kW, is (a) 3400 (b) 3450 (c) 2940 (d) 2500 16.16 Pelton wheels are available in the following range of speciﬁc speeds (Ns) (a) 8–30 (b) 40–420 (c) 380–950 (d) 4–70 16.17 Francis turbines are available in the following range of speciﬁc speeds (Ns) (a) 8–30 (b) 40–420 (c) 380–950 (d) 10–300 16.18 The speed factor f of a Pelton turbine for most efﬁcient operation is in the range (a) 1.4–2.0 (b) 0.70–0.85 (c) 0.43–0.47 (d) 0.1–0.3 16.19 In all reaction turbines, for maximum efﬁciency (a) the velocity of swirl at entrance must be zero (b) the velocity of ﬂow at oultet must be zero (c) the velocity of swirl at outlet must be zero (d) the velocity of ﬂow at entrance must be zero 16.20 The net available head H in a Pelton turbine installation is the (a) kinetic energy of the jet issuing from the nozzle (b) difference in elevation between forebay water level and the nozzle outlet 590 Fluid Mechanics and Hydraulic Machines 16.21 16.22 16.23 16.24 16.25 (c) head at the base of the nozzle (d) difference in level between water levels at the forebay and the tailwater level In Pelton turbines the ﬂow from the nozzle is controlled by a needle which works inside the nozzle. The typical value of the coefﬁcient of velocity Cv for needle nozzles, for partly open conditions, is (a) 0.61 (b) 0.95 (c) 0.75 (d) 0.10 In a general sense, water turbines may be put in the following decreasing order of speciﬁc speeds, as: (a) propeller turbine, Francis turbine and Pelton wheel (b) Pelton wheel, Francis turbine, Kaplan turbine (c) Kaplan turbine, impulse turbine, Francis turbine (d) Francis turbine, Kaplan turbine, Pelton wheel To generate 8100 kW under a head of 81 m while working at a speed of 540 rpm, the turbine of choice will be (a) Pelton (b) Kaplan (c) Bulb (d) Francis A turbine develops 7500 kW of power when rotating at 135 rpm under a head of 25 m. The normal speed under a head of 20 m would be (a) 150.9 rpm (b) 108 rpm (c) 120.7 rpm (d) 168.8 rpm A Francis turbine has runner of outer and inner diameters 4 m and 2 m respectively. The breadth at inlet and also at outlet is 0.8 m and the velocity of ﬂow is constant at 3.0 m/s. The discharge through the turbine, in m3/s, is (a) 28.28 (b) 37.70 (c) 30.16 (d) 15.08 16.26 In a Kaplan turbine the diameter of the runner is 2.25 m. The boss is having a diameter one third of the runner. If the velocity of ﬂow and the peripheral velocity are 16 m/s and 48 m/s respectively, the discharge through the runner, in m3/s, is (a) 254.47 (b) 63.62 (c) 169.6 (d) 56.55 16.27 In a Kaplan turbine working under a head of 35 m, the speed ratio is 2.0. If the outer diameter of the runner is 2.0 m and the hub diameter is 0.6 m, the rotational speed, in rpm, is (a) 500 (b) 125 (c) 150 (d) 250 16.28 In a Francis turbine the discharge leaves the runner radially at the exit. For this turbine (a) the blade tip is radial at the outlet (b) the blade tip is radial at the inlet (c) the guide vane angle is 90° (d) the absolute velocity is radial at the outlet 16.29 In a Francis turbine, the runner blades are radial at the entrance and the discharge is radial at the outlet. For this turbine (a) the relative velocity is radial at the outlet (b) the absolute velocity is radial at the outlet (c) the guide vane angle is 90° (d) the velocity of ﬂow is constant 16.30 A turbine with a speciﬁc speed of 400 is connected to a generator on the same shaft. The generator has 10 pairs of poles and produces an AC current with a frequency of 50 Hz. The synchronous speed, in rpm, is (a) 300 (b) 400 (c) 500 (d) 600 16.31 A turbine has a runner of outer diameter 0.56 m and works under a head of 30 m. 591 Hydraulic Machines If the rotational speed of the runner is 620 rpm the speed factor f for this runner is (a) 0.46 (b) 2.0 (c) 1.33 (d) 0.75 16.32 A turbine works at 20 m head and 500 rpm speed. Its 1 : 2 scale model is to be tested at a head of 20 m. The model should have a rotational speed of (a) 1000 rpm (b) 250 rpm (c) 500 rpm (d) 2000 rpm 16.37 16.38 Centrifugal Pumps 16.33 In a centrifugal pump the inlet angle will be designed to have (a) relative velocity vector in the radial direction (b) absolute velocity vector in the radial direction (c) velocity of ﬂow to be zero (d) peripheral velocity to be zero 16.34 A fast centrifugal pump impeller will have (a) forward facing blades (b) radial blades (c) backward facing blades (d) propeller type blades 16.35 In a centrifugal pump with redial entry of liquid, the manometric efﬁciency h H is given by hH = u2Vu2 gH H (c) g u2Vu2 (a) gH u2Vu2 2g H (d) u2Vu2 16.39 16.40 16.41 (b) where H = net head, u2 = peripheral velocity at outlet and Vu2 = velocity of swirl at the outlet. 16.36 In a centrifugal pump of 0.50 m diameter, running at 1500 rpm the absolute velocity at exit is 8.0 m/s and makes an angle of 45° with the tangential direction. With a hydraulic efﬁciency of 80% the net head produced is 16.42 16.43 (a) 18.1 m (b) 28.3 m (c) 9.1 m (d) 14.2 m A centrifugal pump has outer and inner diameters of 0.30 m and 0.15 m respectively. When running at 1450 rpm, with the discharge pipe completely shut off, the head produced is (a) 26.4 m (b) 6.6 m (c) 15.7 m (d) 19.8 m The speciﬁc speed of a pump has the dimensions of (a) L3/4 T–3/2 (b) M0 L0 T0 (c) M–1/2 L1/2 T–1/4 (d) L3/4 T–1/2 A pump is required to deliver 150 L/s at a head of 45 m when running at 1750 rpm. The speciﬁc speed of the pump is (a) 17.4 (b) 39 (c) 89 (d) 1233 A pump running at 1400 rpm is required to deliver 360 L/S of water at a head of 16 m. The pump of choice will be (a) normal speed, radial type (b) low speed, radial type (c) mixed ﬂow type (d) axial ﬂow type Typical range of speciﬁc speeds for axial ﬂow pumps is (a) 380–950 (b) 80–200 (c) 10–100 (d) 200–300 A pump delivers 100 L/s at a head of 25 m when running at 1450 rpm. A homologous pump to deliver 100 L/s at 12.5 m head will have a speciﬁc speed of (a) 1025 (b) 69 (c) 41 (d) 1450 A similar model of a pump is built to a ratio of 1 : 2. If the model pumps the same ﬂuid as the prototype at the same rotative speed, the ratio of model power to prototype power input is 592 Fluid Mechanics and Hydraulic Machines (a) 1/2 2 (b) 1/32 (c) 1/8 (d) 1/16 2 16.44 If two pumps identical in all respects and each capable of delivering a discharge Q against a head H are connected in parallel, the resulting discharge is (a) 2Q against a head 2H (b) 2Q against a head H (c) Q against a head 2H (d) 2Q against a head 2 H 16.45 If two pumps identical in all respects and each capable of delivering a discharge Q against a head H are connected in series, the resulting discharge is (a) 2Q against a head 2H (b) 2Q against a head H (c) Q against a head 2H 16.46 16.47 16.48 16.49 (d) Q against a head 2 H The maximum permissible suction lift for centrifugal pump in prectice (at sea level and 30° C) is (a) 12 m (b) 10 m (c) 6 m (d) 3 m Cavitation parameter s deﬁned in terms of NPSH (net positive suction head) and net head H as, s = H NPSH (a) (b) NPSH H H NPSH (c) (d) NPSH H In a pump installation the local atmospheric pressure is 9.8 m of water, vapour pressure head is 0.4 m (abs), height of the pump above sump water level is 5 m. For a head loss in the suction side of 0.6 m, the NPSH is (a) 3.8 m (b) 5.0 m (c) 4.0 m (d) 15.8 m Water is to be lifted by a net head of 180 m. Identical pumps with speciﬁc speed of 30 and rotative speed of 1450 rpm capable of discharging 200 L/s are available. The number of pumps required is (a) 1 (b) 2 (c) 3 (d) 4 16.50 Water is pumped through a pipeline to a height of 10 m at the rate of 0.1 m3/s. Frictional and other minor losses are 5 m. Pumping power required, in kW, is (a) 14.7 (c) 20.0 (b) 9.80 (d) 13.3 Reciprocating Pumps 16.51 In a reciprocating pump without air vessel, the friction head in the delivery pipe is maximum at the crank angle q = (a) 0° (b) 90° (c) 135° (d) 180° 16.52 In a reciprocating pump without air vessels, the acceleration head in the suction pipe is maximum at the crank angle q = (a) 0° (b) 90° (c) 135° (d) 180° 16.53 In a single acting reciprocating pump without air vessel, the average velocity in the delivery pipe is given by Vs = Ê Aˆ (a) Á ˜ wr Ë As ¯ Ê A ˆ wr (b) Á ˜ Ë As ¯ p Ê Aˆ (c) Á ˜ pw r Ë As ¯ Ê A ˆ pw r (d) Á ˜ Ë As ¯ 2 where w = angular velocity of the crank of radius = r 16.54 In a single reciprocating pump without air vessel, the ratio of the average friction head to the maximum fraction head in the delivery pipe is given by hfda/hfdm = (a) 1/2 (b) 1/3 (c) 2/3 (d) 3/2 593 Hydraulic Machines 16.55 The indicator diagram of a reciprocating pump is a plot of (a) work done vs stroke length (b) acceleration head vs stroke length (c) angular displacement vs stroke length (d) pressure head vs stroke length 16.56 An air vessel in the delivery side of a reciprocating pump (a) maintains steady discharge output (b) prevents cavitation in the system (c) enables suction head to be increased (d) enables the pump to run at higher speed Miscellaneous Devices 16.57 A hydraulic coupling (a) connects two shafts rotating at about the same speed. (b) connects two shafts running at different speeds (c) is used to segment the torque to the driven shaft (d) is used to connect the centrifugal pump to an electric motor for efﬁcient operation 16.58 In a jet pump (a) kinetic energy of the high velocity stream is converted in to potential energy (b) energy of high pressure stream is converted into kinetic energy of low pressure stream (c) high velocity stream aided by the geometry creats a suction pressure which results in increased suction head. (d) potential energy of the ﬂuid is converted in to kinetic energy thereby increasing the delivery head. 16.59 A hydraulic accumulator is a device to store (a) sufﬁcient quantity of water to compensate for the change in discharge (b) sufﬁcient energy to drive the machine when normal source does not function (c) sufﬁcient energy in case of machines which work intermittently to supplement the discharge form the normal source. (d) liquid which would otherwise have gone to waste. 16.60 Fluid ﬂow machines use the principle of either (i) supplying energy to the ﬂuid, or (ii) extracting energy from the ﬂuid. Some ﬂuid ﬂow machines are a combination of both (i) and (ii) listed above. They are classiﬁed as (a) compressors (b) hydraulic turbines (c) torque converters (d) wind mills 16.61 In a ﬂuid coupling, the torque transmitted is 50 kNm when the speed of driven and driving shafts are 900 rpm and 720 rpm respectively. The efﬁciency of this ﬂuid coupling is (a) 20% (b) 25% (c) 80% (d) 90% 16.62 A hydraulic coupling transmits 1 kW of power at an input speed of 200 rpm with a slip of 2%. If the input speed is changed to 400 rpm, the power transmitted with the same slip is about (a) 2 kW (b) 1/2 kW (c) 4 kW (d) 8 kW 594 Fluid Mechanics and Hydraulic Machines Additional objective questions on hydraulic machines A1. Which of the following pairs or formulas represent the speciﬁc speeds of turbine and pump respectively? (notations have their usual meaning.) 1/ 2 NQ (a) H 3/ 4 NQ1/ 2 (b) 3/ 4 H NP1/ 2 (c) H 3/ 4 NP1/ 2 (d) H 3/ 4 and and and and NP The angle a of the vane is (a) zero, (c) 45° 1/ 2 H 5/ 4 NP1/ 2 H A4. Efﬁciency of Pelton wheel shall be maximum if the ratio of jet velocity to tangential velocity of the wheel is (a) 1/2 (c) 2 3/ 4 A5. The maximum efﬁciency in case of a Pelton wheel with k = 1 and angle of deﬂection of the jet = (180 – b) is H 5/ 4 NQ1/ 2 1 + cos b 1 - cos b (b) 2 2 cos b 1 + cos b (c) (d) 2 4 A6. If H is the head available to a turbine, the power, speed and discharge are respectively, proportional to, (a) H 5/ 4 A2. Consider the speciﬁc speed ranges of the following types of turbines: 1. Francis 2. Kaplan 3. Pelton The sequence of their speciﬁc speed in increasing order is (a) 1, 2, 3 (b) 3, 1, 2 (c) 3, 2, 1 (d) 2, 3, 1 A3. A symmetrical stationary vane experiences a force F of 200 N as shown in Fig. 16.54 when the mass ﬂow rate of war over the vane is 20 kg/s with a velocity V of 10 m/s without friction. V (a) H 1/2, H1/2, H3/2 (b) H3/2, H1/2, H1/2 (c) H1/2, H3/2, H1/2 (d) H1/2, H1/2, H A7. In utilizing the scale models in the designing of turbo-machines, which of the following relationships must be satisﬁed? (a) (b) Force F (c) Angle q (d) Fig. 16.54 (b) 1 (d) 4 NQ1/ 2 V (b) 30° (d) 60° H ND = constant; 3 Q HD 2 Q N D2 = constant; 2 H N 3D = constant = constant P H = constant; 2 2 = constant QH N D NQ1/ 2 H 3/ 2 = constant; NP1/ 2 H 3/ 4 = constant 595 Hydraulic Machines Head Kaplan Francis Specific speed (a) Head Francis Kaplan Specific speed Head (b) Kaplan Francis Specific speed (c) Francis Head A8. A draft tube is used in a reaction turbine to (a) guide water downstream without splashing (b) convert residual pressure energy in to kinetic energy (c) convert residual kinetic energy in to pressure energy (d) to streamline the ﬂow in the tailrace A9. Two Pelton wheels A and B have the same speciﬁc speed and are working under the same head. Wheel A produces 400 kW at 1000 rpm. If wheel B produces 100 kW then its rpm is (a) 4000 (b) 2000 (c) 1500 (d) 1250 A10. Which of the following graphs correctly represents the relationship between head and speciﬁc speed for Kaplan and Francis turbines? Kaplan Specific speed (d) Fig. 16.55 A11. What is the range of speed ratio j for Kaplan turbine for its most efﬁcient operation? (a) 0.10 to 0.30 (b) 0.43 to 0.65 (c) 0.85 to 1.20 (d) 1.40 to 2.00 A12. An impulse turbine operating with a single nozzle has a speciﬁc speed of 10. What will be the appoximate speciﬁc speed of the turbine if the turbine is operating with one more additional nozzle of the same size? (a) 10 (b) 12 (c) 14 (d) 20 A13. A Francis turbine is coupled to an alternator to generate electricity with a frequency of 50 Hz. If the alternator has 12 poles, then the turbine should be regulated to run at which one of the following speeds? (a) 250 rpm (b) 500 rpm (c) 600 rpm (d) 1000 rpm A14. A Pelton wheel with a single jet rotates at 600 rpm. The velocity of the jet from the nozzle is 100 m/s. If the ratio of the vane velocity to jet velocity is 0.44, what is the diameter of the Pelton wheel? (a) 0.7 m (b) 1.4 m (c) 2.1 m (d) 2.8 m A15. Which of the following advantages is/ are possessed by a Kaplan turbine over a Francis turbine? (1) Low frictional losses (2) Part load efﬁciency is considerably high (3) More compact and smaller in size 596 Fluid Mechanics and Hydraulic Machines A16. A17. A18. A19. A20. A21. A22. The correct answer is (a) only 1 (b) 1 and 2 (c) 2 and 3 (d) 1, 2 and 3 The use of a draft tube in a reaction type turbine helps to (a) prevent air from entering (b) increases the ﬂow rate (c) convert the kinetic energy to pressure energy (d) convert the kinetic energy to potential energy Which of the following is correct? If the number of jets is a Pelton turbine is n then the speciﬁc speed is proportional to (a) n2 (b) n 1/2 (c) n (d) n0 Kaplan turbine is (a) A low head axial ﬂow turbine (b) a high head mixed ﬂow turbine (c) an outward ﬂow reaction turbine (d) an inward ﬂow impulse turbine Cavitation in a turbine is most likely to occur at the turbine (a) turbine entry (b) draft tube exit (c) rotor entrance (d) rotor exit Consider the following data for the performance of a centrifugal pump: Speed = 1200 rpm, ﬂow rate = 30 L/s, head = 20 m and power = 5 kW. If the speed is increased to 1500 rpm, the power will be nearly equal to (a) 6.5 kW (b) 8.7 kW (c) 9.8 kW (d) 10.9 kW A centrifugal pump gives maximum efﬁciency when the blades are (a) bent forward (b) bent backward (c) straight (d) wave shaped A pump running at 1000 rpm consumes 1 kW of power and generates head of 10 m of water. When it is operated at 2000 rpm, and if the overall efﬁciency remains the same, A23. A24. A25. A26. A27. its power consumption and head generated would be about (a) 4 kW and 50 m of water (b) 6 kW and 20 m of water (c) 3 kW and 30 m of water (d) 8 kW and 40 m of water A pump and its (1/4) scale model are being compared. If the ratio of the heads is 5: 1, then the ratio of the power consumed by the prototype and the model is (a) 100 (b) 3.2 (c) 179 (d) 12.8 Which one of the following statements relevant to the speciﬁc speed of a centrifugal pump? (a) Head developed is unity and discharge is unity. (b) Head developed is unity and power absorbed is unity. (c) Discharge is unity and power absorbed is unity. (d) Each of head developed, power absorbed and discharge is equal to unity. A centrifugal pump needs 1000 W of power when operating at 1500 rpm. What is the power requirement if the speed of the pump is increased to 3000 rpm? (a) 2000 W (b) 4000 W (c) 6500 W (d) 8000 W Which one of the following is a correct statement for a given centrifugal pump? (a) The discharge varies directly as the speed. (b) The head varies inversely as the speed. (c) The power varies as the square of the speed. (d) The discharge varies as the square of the speed. A centrifugal pump is started with its delivery valve kept 597 Hydraulic Machines (a) fully open (b) fully shut (c) partially open (d) 50% open A28. Consider the following pumps: (1) Centrifugal pump–single stage (2) Centrifugal pump–Multi stage (3) Reciprocating pump (4) Jet Pump The pump (s) which could be used to lift water through a suction head of 12 m from a well would include (a) 2 alone (b) 1, 3 and 4 (c) 4 alonge (d) 1 and 3 A29. The shape number in the case of pumps refer to (a) ratio of sizes of suction pipe and delivery pipes (b) ratio of diameter and thickness of impeller (c) non-dimensional form of speciﬁc speed (d) non-dimensional cavitation number A30. A fast centrifugal pump will have the following feature: (a) Forward facing blades (b) Radial blades (c) Backward facing blades (d) Propeller type blades A31. In an M L T system, what is the dimension of the speciﬁc speed of a rotodynamic pump? (a) LT3/2 (b) L–3/4T3/2 1/2 1/4 –5/2 (c) M L T (d) L3/4 T–3/2 [Hint: Ns = N Q ] H 3/ 4 A32. In positive displacement pumps, the slip can sometimes be negative when the actual discharge is greater than the theoretical discharge. This happens in (a) small suction pipes coupled with low delivery head (b) small suction pipes coupled with medium delivery head (c) long suction pipes coupled with low delivery head (d) long suction pipes coupled with medium delivery head A33. An air vessel is used in a reciprocating pump (a) to obtain continuous supply of the ﬂuid at uniform rate (b) to reduce suction head (c) to increase delivery head (d) to reduce cavitation Appendix-A Multiple Choice Objective Questions Series–1 1. Match List I with List II and select the correct answer using the code given below the lists. List I (Description) List II (Property of Fluid) A. Property which explains the spherical shape of the drop of a liquid B. Property which explains the phenomenon of cavitation in a ﬂuid ﬂow C. Property which explains the rise of sap in a tree D. Property which explains the ﬂow of a jet of oil in an unbroken stream Viscosity 2. Surface Tension 3. Compressibility 4. Vapour pressure 5. Capillarity (a) (b) (c) (d) A 1 2 4 1 B 2 4 2 2 C 4 5 5 3 D 5 1 1 4 Figure M.02 shows the rheological behaviour of four types of viscous ﬂuids. With reference to this ﬁgure, match the following two lists. List I (Curve in Fig. M 02) A. B. C. D. Curve A Curve B Curve C Curve D D Shear stress Codes: C B List II (Classiﬁcation) 1. 2. 3. 4. Dilatant Ideal Bingham Plastic Pseudo Plastic Newtonian A Shear strain rate Fig. M.02 Question No. 2 599 Appendix–A Codes: A 2 1 3 3 (a) (b) (c) (d) B 4 4 2 1 C 3 3 1 4 D 1 2 4 2 Match List I with List II and select the correct answer using the code given below the lists. List I A. B. C. D. List II Submerged body Floating body Metacentric height Buoyancy Codes: A 5 1 5 1 (a) (b) (c) (d) B 4 2 4 2 C 2 4 2 5 Force on a curved surface Moment of inertia Force acting vertically up Metacentre Centre of Buoyancy D 1 3 3 3 Consider the following statements: 1. 2. 3. 4. In a source, equipotential lines are circles. Flownet is a representation of two-dimensional irrotational ﬂow of incompressible ﬂuid. Boundary acts as limiting equipotential lines in a ﬂownet. In uniform ﬂow region, streamlines will be parallel and equidistant. Of these statements, (a) 1, 2, and 3 are correct (c) 2, 3, and 4 are correct (b) 1, 2, and 4 are correct (d) 1, 3, and 4 are correct Match List I with List II and select the correct answer using the code given below the lists. List I List II A. Rotational ﬂow A ﬂuid motion in which streamlines are concentric circles. 2. The ﬂuid particles moving in concentric circles may not rotate about their mass centres. 3. The ﬂuid particles moving in concentric circles may rotate about their mass centres. 4. Flow near a curved boundary B. Vortex ﬂow C. Free vortex D. Forced vortex Codes: (a) (b) (c) (d) A 4 3 1 4 B 2 1 3 1 C 3 2 2 2 D 1 4 4 3 600 Fluid Mechanics and Hydraulic Machines Match List I with List II and select the correct answer using the code given below the lists. List I (Phenomenon) A. B. C. D. List II (Condition) Rotational ﬂow Irrotational ﬂow Singularities Streamline spacing Codes: A 3 4 4 3 (a) (b) (c) (d) B 4 3 3 4 C 1 1 2 2 Velocity is zero or inﬁnite. Proportional to velocity Vorticity is zero Vorticity exists D 2 2 1 1 Match List I with List II and select the correct answer using the code given below the lists. List I List II A. Stream lines B. Streak lines C. Path lines Tracing of motion of any one ﬂuid particle 2. Tracing of motion of different ﬂuid particles 3. Identiﬁcation of location of number of ﬂuid particles 4. Orthogonal to streak lines 5. Location of equal piezometric heads. D. Equipotential lines Codes: (a) (b) (c) (d) A 2 3 1 2 B 3 2 2 3 C 4 1 4 1 D 5 4 3 5 Which of the following pairs are correctly matched? 1. 2. 3. 4. Piezometric head Dynamic head Stagnation head Total head … … … … Sum of datum head and pressure head Sum of datum head and velocity head Sum of piezometric head and velocity head Sum of piezometric head and dynamic head Select the correct answer using the codes given below: Codes: (a) 1, 2 and 3 (c) 1, 2 and 4 (b) 1, 3 and 4 (d) 2, 3 and 4 p V2 = constant, will be valid in the whole ﬂow +Z+ 9. Under which of the following conditions will the equation 2g g ﬁeld? 1. Flow is rotational 2. Flow is irrotational 3. Flow is incompressible 601 Appendix–A Flow is steady 5. Flow is laminar Select the correct answer using the codes given below. Codes: (a) 1, 3 and 5 (c) 1, 3 and 4 (b) 2, 4 and 5 (d) 2, 3 and 4 Match List I (phenomenon) with List II (equation/concept involved) and select the correct answer using the code given below the lists. List I List II A. Force developed in a bend B. Pitot-static tube C. Flow through smaller passage produces higher velocity D. Vortex ﬂow Codes: (a) (b) (c) (d) A 3 3 2 2 B 2 2 3 3 C 4 1 4 1 Continuity equation 2. Energy equation 3. Momentum equation 4. Moment of momentum equation D 1 4 1 4 Match List I with List II and select the correct answer using the code given below the lists. List I List II A. Moment of momentum equation B. Bernoulli equation C. Euler’s equation D. Hagen–Poiseuille equation Codes: (a) (b) (c) (d) A 2 3 2 3 B 3 2 3 2 C 4 1 1 4 Equation to ﬁnd energy loss in a pipeline having laminar ﬂow 2. Equation of motion for one-diamensional steady ﬂow of ideal incompressible ﬂuid. 3. Equation based on conservation of momentum principle applicable to circulatory ﬂow. 4. Three-dimensional equation of motion based on principle of conservation of momentum for ideal and incompressible ﬂuid. D 1 4 4 1 Match List I (units) with List II (dimensions) and select the correct answer using the code given below the lists. List I A. B. C. D. Pressure Power Reynolds number Speciﬁc weight List II 1. 2. 3. 4. M0 L0 T0 ML–2 T –2 ML2 T –3 ML–1 T–2 602 Fluid Mechanics and Hydraulic Machines Codes: (a) (b) (c) (d) A 3 4 4 3 B 4 3 3 4 C 2 2 1 1 D 1 1 2 2 Match List I with List II and select the correct answer using the code given below the lists. List I (Fluid property) A. B. C. D. List II (Flow phenomenon) Compressibility Gravity Viscosity Vapour pressure Codes: (a) (b) (c) (d) A 4 4 3 3 B 3 3 4 4 C 2 1 1 2 Flow of real ﬂuid past a tiny sphere Cavitation Hydraulic jump Flight of supersonic aircraft D 1 2 2 1 Match List I (Phenomenon) with List II (Tool for dynamic similarity) and select the correct answer using the code given below the lists. List I A. B. C. D. List II Flow in open channel Flow in pipes Movement of bullet in air Capillary ﬂow Codes: (a) (b) (c) (d) A 1 1 2 1 B 2 2 3 2 C 3 3 4 4 5. Froude number Reynolds number Weber number Mach number Thoma number D 4 5 5 3 Match List I with List II and select the correct answer using the code given below the lists. List I (Non-Dimensional numbers) A. B. C. D. Mach number Thoma number Reynolds number Weber number List II (Application) 1. 2. 3. 4. 5. Waves in an ocean Launching of rockets Cavitation phenomenon Capillary ﬂow in soil Motion of a submarine 603 Appendix–A Codes: (a) (b) (c) (d) A 1 2 3 2 B 3 3 2 1 C 5 5 4 3 D 2 4 1 4 Match List I with List II and select the correct answer using the case given below the lists. List I (Fluid problems) A. B. C. D. List II (Model laws) Flow over a spillway dam Flow through a butterﬂy valve Rise of moisture in the stem of a plant Water hammer in a penstock Codes: (a) (b) (c) (d) A 2 4 2 5 B 4 2 3 1 C 5 3 1 4 5. Euler number Froude number Mach number Renynolds number Weber number D 3 1 4 2 In a distorted model of a free surface ﬂow phenomenon Lr = horizontal scale and hr = vertical scale. Match the following two lists pertaining to the model. List I List II A. Slope ratio 1. B. Discharge ratio C. Time scale D. Manning’s roughness ratio 4. L–1/2 hr3/2 r –1/2 Lr h r Lr h r3/2 L–1 r hr Codes: (a) (b) (c) (d) A 1 4 3 4 B 4 2 4 3 C 3 1 2 2 D 2 3 1 1 Consider the following statements regarding the laminar ﬂow through a circular pipe: 1. 2. 3. 4. 5. The friction factor is constant The friction factor depends upon the pipe roughness The friction factor varies inversely with the Reynolds number The velocity distribution is parabolic The pressure drop varies directly with the square of mean velocity of these statements (a) 1 and 4 are correct (c) 2 and 4 are correct (b) 3 and 5 are correct (d) 3 and 4 are correct 604 Fluid Mechanics and Hydraulic Machines Match List I with List II and select the correct answer using the code given below the lists. List I List II A. Boundary layer thickness B. Displacement thickness C. Laminar boundary layer thickness D. Turbulent boundary layer thickness Codes: (a) (b) (c) (d) A 1 1 5 2 B 2 2 4 1 C 4 5 2 5 Distance from the boundary where velocity is 99% of uniform velocity 2. Distance from the boundary by which the main ﬂow can be assumed to be shifted 3. Varies as x4/5 4. Varies as x(–1/2) 5. Varies as x (–1/2) D 3 3 1 3 Match List I (Typical occurrence) with List II (Relevant ﬂow condition) and select the correct answer using the code given below the lists. List I (Typical occurrence) A. B. C. D. List II (Relevant ﬂow condition) Cavitation Separation Stagnation point Wake Codes: (a) (b) (c) (d) A 4 2 4 2 B 2 4 2 4 C 3 3 1 1 Absence of ﬂuid velocity Fluid pressure reduces to vapour pressure Bluff body in ﬂow Adverse pressure gradient in widening boundaries of ﬂow D 1 1 3 3 Match List I with List II and select the correct answer using the code given below the lists. List I (Body) A. B. C. D. List II (Drag coeﬁcient) Aerofoil (Re > 107) Sphere (104 < Re < 105) Flat plate perpendicular to ﬂow Droplet (Re ª 0.1) Codes: (a) (b) (c) (d) A 1 4 4 4 B 2 3 3 1 C 3 1 2 3 D 4 2 1 2 4. 240 1.90 0.50 0.003 605 Appendix–A Match List I with List II and select the correct answer using the code given below the lists. List I (Object in a ﬂuid) A. B. C. D. List II (Drag coefﬁcient) Submarine Parachute Aircraft wing (air foil) Smoke stack (Chimney) (Reynolds number = 105) Codes: A 1 3 1 3 (a) (b) (c) (d) B 3 1 3 1 C 2 4 4 2 4. 1.33 0.10 0.15 1.00 D 4 2 2 4 Match List I with List II and select the correct answer using the code given below the lists. List I (Nature of ﬂow) List II (Friction factor equation) A. Turbulent ﬂow with rough boundary f = 64/Re B. Turbulent ﬂow with smooth boundary f = C. Turbulent ﬂow with Re < 105 3. D. Laminar ﬂow Codes: (a) (b) (c) (d) A 4 3 3 4 B 3 4 4 3 C 2 2 1 1 0.316 R 0e .25 1 f 1 f = 2 log r0 + 1.74 e = 2 log Re f – 0.8 D 1 1 2 2 Consider the following statements: 1. Pipe network analysis is normally necessary in analyzing ﬂow in pipes at city water systems. 2. Hardy–Cross method of solving pipe network is a method of successive approximations and is not a direct method. 3. The network must satisfy the momentum equation because the ﬂow in each pipe satisﬁes the head loss equation. 4. Principle of continuity is satisﬁed in a pipe network. Of these statements, (a) 1, 2 and 3 are correct (c) 1, 3 and 4 are correct (b) 2, 3 and 4 are correct (d) 1, 2 and 4 are correct 606 Fluid Mechanics and Hydraulic Machines Consider the following statements: 1. Flow is established in a pipe when the boundary layer thickness is equal to the radius of the pipe. 2. For laminar ﬂow, the friction factor in Darcy–Weisbach equation varies inversely as the Reynolds number. 3. For turbulent ﬂow, the friction factor in Darcy–Weisbach equation varies inversely as the square of Reynolds number. 4. When the boundary is rough, the friction factor varies with the relative roughness of the pipe. Of these statements, (a) 1, 2 and 3 are correct (c) 2, 3 and 4 are correct (b) 1, 2 and 4 are correct (d) 1, 3 and 4 are correct V2 . Match List I (Value of K) with List II (Pipe 26. The loss of head at various pipe ﬁttings (hL) is given by hL = K 2g ﬁtting) given below: List I A. B. C. D. List II 0.40 0.90 1.50 10.0 Codes: (a) (b) (c) (d) A 4 2 2 3 B 2 3 3 2 C 3 1 4 1 Pump foot valve Standard 45° elbow Standard 90° elbow Globe valve (Fully open) D 1 4 1 4 Match List I (Surface proﬁle) with List II (Description of the proﬁle and select the correct answer using the code given below the lists. List I A. M2 B. S3 C. C1 D. A3 Codes: (a) (b) (c) (d) A 2 2 3 3 B 4 1 4 1 List II 1. Convex upward; asymptotic to horizontal at downstream end; depth increasing in downstream direction. 2. Concave downwards; upstream asymptotic to normal depth with depth decreasing in downstream direction. 3. Depth increasing downstream and meeting at an angle to CDL; a curve with an inﬂexion point 4. Convex upward and depth increasing in ﬂow direction; asymptotic to NDL at downstream. C 1 4 1 4 D 3 3 2 2 607 Appendix–A Match List I with List II and select the correct answer using the code given below the lists. List I (Hydraulic structure) A. B. C. D. List II (Water surface proﬁle at the structure) On a ﬂat topped broad crested weir Immediately below a sluice gate on a level apron Behind the weir on an alluvial river In a chute spillway Codes: A 5 4 2 5 (a) (b) (c) (d) B 6 6 5 4 C 1 2 6 1 6. M1 M2 S1 S2 H2 H3 D 4 1 4 3 Match List I with List II and select the correct answer using the code given below the lists. (yo = Normal depth, yc = critical depth and y = depth in gradually varied ﬂow) A. B. C. D. Codes: (a) (b) (c) (d) List I List II y c > yo > y yo > y > yc y > yc > yo y > y o > yc 5. A 2 3 5 3 B 4 4 4 1 C 1 5 1 4 S2 M1 S3 M2 S1 D 3 2 3 2 Match List I with List II and select the correct answer using the code given below the lists. List I A. B. C. D. List II Speciﬁc force Speciﬁc energy Hydraulic jump Darcy–Weisbach equation Codes: (a) (b) (c) (d) A 4 4 3 3 B 3 3 4 4 C 2 1 1 2 D 1 2 2 1 Head loss due to friction Rapidly varied ﬂow Alternate depths Conjugate depths 608 Fluid Mechanics and Hydraulic Machines Match List I with List II and select the correct answer using the code given below the lists. List I (Devices) A. B. C. D. List II (Use) Hydrometer Pitot–Static tube Oriﬁce meter Engler’s apparatus Codes: (a) (b) (c) (d) A 1 3 3 3 B 4 1 2 2 C 2 2 1 4 Measures kinematic viscosity Measures ﬂow velocity Measures speciﬁc gravity Measures volumetric ﬂow rate D 3 4 4 1 Match List I with List II and select the correct answer using the code given below the lists. List I A. B. C. D. Codes: (a) (b) (c) (d) List II Venturimeter Current meter Piezometer Barometer A 2 1 1 2 B 1 2 2 1 C 4 3 4 3 Flow rate Flow velocity Gauge pressure Atmospheric Pressure D 3 4 3 4 Match List I (Name of instrument) with List II (Variable measured) and select the correct answer using the code given below the lists. List I A. B. C. D. Codes: (a) (b) (c) (d) List II Hot-wire anemometer Oriﬁce meter Pitot tube Preston tube A 2 5 2 5 B 3 2 5 2 C 4 3 1 3 D 1 4 3 1 Boundary shear stress Discharge Velocity Pressure Turbulence 609 Appendix–A Match List I with List II and select the correct answer using the code given below the lists. List I (Shape of the weir) A. B. C. D. List II (Value of the exponent in the equation Q = KH n) Rectangular weir Triangular weir Sutro weir Parabolic weir Codes: A 4 2 3 3 (a) (b) (c) (d) B 2 3 4 4 C 1 4 2 1 1.0 2.0 1.5 2.5 D 3 1 1 2 Which of the following statements are true in relation to water hammer phenomenon? 1. 2. 3. 4. 5. It causes surface erosion in pipes. The pressure rise is given by r CV for sudden closure of valve. It is accompanied by serious cavitation. The volume modulus of ﬂuid is the relevant ﬂuid property. It is governed by the Reynolds number of the ﬂow. Select the correct answer using the codes given below. Codes: (a) 3 and 5 (c) 2 and 5 (b) 2 and 4 (d) 1, 2, 3 and 4 Match List I with List II and select the correct answer using the code given below the lists. List I List II A. Positive surge traveling upstream B. Positive surge traveling downstream C. Negative surge traveling upstream D. Negative surge traveling downstream Codes: (a) (b) (c) (d) A 1 1 3 3 B 4 2 4 4 C 3 3 1 2 D 2 4 2 1 Occurs on upstream of gate that is partly closed suddenly. 2. Occurs on downstream of gate that is partly closed suddenly. 3. Occurs on upstream of gate that is opened suddenly. 4. Occurs on the downstream that is opened suddenly. 610 Fluid Mechanics and Hydraulic Machines Match List I with List II and select the correct answer using the code given below the lists. List I (Type of turbines) A. B. C. D. List II (Ranges of speciﬁc speeds in SI units) Francis Kaplan Pelton (one jet) Pelton (two jets) Codes: (a) (b) (c) (d) A 3 4 3 4 B 4 3 4 3 C 2 2 1 1 8–30 26–40 40–420 380–950 D 1 1 2 2 Match List I (outlet vane angle) with List II (curves lebeled 1, 2, and 3 in the given ﬁgure) for a pump and select the correct answer using the code given below: List I List II Head H (A) b 2 < 90∞ (B) b 2 = 90∞ (C) b 2 > 90∞ 3 2 1 Codes: (a) (b) (c) (d) A 1 1 2 3 B 2 3 1 2 C 3 2 3 1 Discharge Q Fig. M.38 Question No. 38 Match List I (Turbine) with List II (Head and discharge status) relating to turbines and select the correct answer using the code given below: List I (Turbine) List II (Head and discharge status) A. Pelton wheel (Single jet) B. Francis Turbine C. Kaplan Turbine Codes: (a) (b) (c) (d) A 1 1 4 4 B 2 3 1 3 C 3 4 3 2 4. Medium discharge, Low head High discharge, Low head Medium discharge, Medium head Low discharge, High head 611 Appendix–A Series-2 40. Which of the following will be satisﬁed by irrotational ﬂow of an imcompressible ﬂuid? 1. ∂u ∂v ∂w + + =0 ∂ x ∂y ∂ z 2. ∂u ∂v ∂u ∂ w ∂w ∂v + = + = + =0 ∂y ∂x ∂z ∂x ∂y ∂z 3. ∂2u ∂2 v ∂2 w + + =0 ∂ x 2 ∂ y2 ∂ z2 4. ∂v ∂u = ; ∂x ∂y ∂u ∂w = ; ∂z ∂x ∂v ∂w = ∂z ∂y Select the correct answer from the codes given below: (a) 3 and 4 (b) 1 and 2 (c) 1 and 3 (d) 1 and 4 Which of the following conditions will be satisﬁed by two-dimensional steady, irrotational ﬂow? (a) ∂u ∂v + =0 ∂y ∂x (b) ∂v ∂u =0 ∂ x ∂y (c) ∂u ∂v + =0 ∂ x ∂y (d) ∂u ∂v =0 ∂ x ∂y Select the correct answer using the codes given below: Codes: (a) 1 and 2 (c) 2 and 4 (b) 1 and 3 (d) 2 and 3 Assuming the thrust T of a propeller depends upon the diameter D, speed of advance V, angular velocity w, dynamic visocsity m, and density r, which of the following dimensionless parameters can be derived by dimensional analysis? T VDm (2) r r D 2V 2 Select the correct answer using the codes given below: (1) (a) 1, 2 and 3 (b) 2, 3 and 4 (3) Dw V (c) 1, 3 and 4 (4) VDr m (d) 1, 2 and 4 Which of the following statements are correct in respect of steady laminar ﬂow through a circular pipe? 1. 2. 3. 4. Shear stress is zero at the centre Discharge varies directly with the viscosity of the ﬂuid Velocity is maximum at the centre Hydraulic gradient varies directly with the velocity Select the correct answer using the codes given below: Codes: (a) 1, 2 and 4 (c) 1 and 3 (b) 1, 3 and 4 (d) 3 and 4 Given that, as the ﬂow takes place between two parallel plates, the velocity midway between the plates is 2.0 m/s, the Reynolds number is 1200 and the distance between the plates is 10 cm, which of the following statements are true? 612 Fluid Mechanics and Hydraulic Machines 4. The velocity at the boundary is 1 m/s The rate of ﬂow is 0.13 m3/s per metre width The ﬂow is turbulent The energy correction factor a is 2.0. Select the correct answer using the codes given below: Codes: (a) 2 and 3 (c) 1 and 3 (b) only 2 (d) 1, 2 and 4 Consider the following statements: In a laminar ﬂow occurring in a circular conduit 1. 2. 3. 4. The ﬂow is rotational Loss of head is proportional to the square of velocity Loss of head is proportional to ﬁrst power of viscosity Other quantities remaining same, increase in diameter increases the Reynolds number. Of these statements: (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2, 3 and 4 are correct Consider the following conditions for the pipe network shown is Fig. M-44 1. Q1 = Q2 hf 1 = hf 3 Q2 = Q1 + Q3 = hf3 hf 1 = hf2 3 A 1 B 2 Which of these conditions must be satisﬁed by the pipe system? (a) 1 and 3 (b) 2 and 3 47. The following parametes relate to ﬂow in a penstock: 1. 2. 3. 4. (d) 1, 2 and 4 are correct Fig. M-44 (c) 2 and 4 Question No. 46 (d) 1 and 4 water level in a reservoir density of water elasticity of water roughness of pipe Pressure rise due to water hammer is a penstock depends upon (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 Consider the follwing statements: In a supersonic ﬂow through a converging duct, along the ﬂow the 1. 2. 3. 4. velocity decreases density increases pressures decreases temperature decreases (d) 1, 2, 3 and 4 613 Appendix–A Of these, the correct statements are: (a) 3 and 4 (b) 1 and 2 (c) 1, 3 and 4 (d) 3 and 4 Consider the following statements relating to compressible ﬂow 1. In a contracting conduit, sonic velocity is reached in the minimum cross sectional area. 2. In a contracting conduit, supersonic velocities are reached only in the expanding section downstream from the minimum section. 3. An expanding shock wave is physically possible and can be exhibited. 4. In the Laval nozzle, it is possible to obtain a velocity greater than the velocity of sound in the expanding section of the nozzle. Of these statements (a) 1, 2 and 3 are correct (c) 2, 3 and 4 are correct (b) 1, 2 and 4 are correct (d) 1, 3 and 4 are correct Consider the following statements: 1. 2. 3. 4. Pumps in series operation allow the head to increase. Pumps in series operation increase the ﬂow rate. Pumps is parallel operation increase the ﬂow rate. Pumps is parallel operation allow the head to increase. Of these statements, (a) 1 and 3 are correct (c) 3 and 4 are correct (b) 2 and 4 are correct (d) 1 and 2 are correct Consider the following types of turbines: 1. 2. 3. 4. Francis turbine Pelton wheel with two or more jets Pelton wheel with a single jet Kaplan turbine The correct sequence of these turbines in increasing order of their speciﬁc speeds is (a) 2, 3, 1, 4 (b) 3, 2, 1, 4 (c) 2, 3, 4, 1 (d) 3, 2, 4, 1 Consider the following statements regarding the volute casing of a centrifugal pump: 1. Loss of head due to change in the velocity is eliminated. 2. Efﬁciency of the pump is increased. 3. Water from the periphery of the impeller is collected and transmitted to the pump at constant velocity. Out of the above, the following statements are correct: (a) 1, 2 and 3 (b) 1 and 2 Consider the following statement: (c) 2 and 3 If pump NPSH requirements are not satisﬁed, then Sufﬁcient head will not be developed to raise water (d) 1 and 3 614 Fluid Mechanics and Hydraulic Machines Efﬁciency will be lowered 3. Very low discharge will be delivered 4. Cavitation will result. Of these statements (a) 1, 2 and 3 are correct (c) 2, 3 and 4 are correct (b) 1 and 4 are correct (d) All 4 statements are correct Consider the speciﬁc speed ranges of the following types of turbines: 1. Francis Kaplan Pelton The sequence of their speciﬁc speed in increasing order is (a) 1, 2, 3 (b) 3, 1, 2 (c) 3, 2, 1 (d) 2, 3, 1 Consider the following statements pertaining to a centrifugal pump: 1. 2. 3. 4. The manometric head is the head developed by the pump The suction pipe has, generally, a larger diameter as compared to the discharge pipe. The suction pipe is provided with a foot valve and a strainer. The delivery pipe is provided with a foot valve and a strainer Of these statements (a) 1, 2, 3 and 4 are correct (c) 2 and 3 are correct (b) 1 and 2 are correct (d) 1 and 3 are correct Consider the following statements about Kaplan turbine: 1. It is a reaction turbine 2. It is a mixed ﬂow turbine 3. It has adjustable blades Out of the above statements (a) 1, 2 and 3 are correct (c) 1 and 3 are correct (b) 2 and 3 are correct (d) 1 and 2 are correct Consider the following statements relating to speciﬁc speeds in turbo machines: 1. Speciﬁc speed varies with the shape of the runner and other parts of the machine 2. Machines with higher speciﬁc speeds are limited to lower speeds 3. Speciﬁc speed is dimensionless and is independent of the variation of type of ﬂuid used. Out of the above statements (a) 1, 2 and 3 are correct (c) 2 and 3 are correct (b) 1 and 2 are correct (d) 1 and 3 are correct The volute casing of a hydraulic turbine has the following features: 1. Eliminating loss of head due to change in velocity after exit from impeller 2. Directs the ﬂow towards the delivery pipe 615 Appendix–A Converts part of the velocity head to pressure head 4. Given a constant velocity of ﬂow Out of the above statements (a) 1, 2 and 4 are correct (c) 1 and 4 are correct (b) 2 and 3 are correct (d) 2 and 4 are correct Consider the following pumps: 1. 2. 3. 4. Centrifugal pump–single stage Centrifugal pump–Multi stage Reciprocating pump Jet Pump The pump (s) which could be used to lift water through a suction head of 12m from a well would include (a) 2 alone (b) 1, 3 and 4 (c) 4 alone (d) 1 and 3 Consider the following statements: An air vessel is ﬁtted on the suction side of a reciprocating pump to 1. 2. 3. 4. achieve higher speed without separation reduce work in overcoming frictional resistance avoid excessive vibration have uniform discharge Which of these statements are correct? (a) 1, 2 and 4 (b) 1 and 2 only (c) 3 and 4 only (d) 2, 3 and 4 Which of the following advantages is/are possessed by a Kaplan turbine over a Francis turbine? 1. Low frictional losses 2. Part load efﬁciency is considerably high. 3. More compact and smaller in size The correct answer is (a) only 1 (b) 1 and 2 (c) 2 and 3 (d) 1, 2 and 3 Consider the following statements regarding a torque converter: 1. 2. 3. 4. Its maximum efﬁciency is less than that of ﬂuid coupling It has two runners and a set of stationary vanes intercepted between them. It has only one runner and a stationary wheel. the ratio of secondary to primary torque is zero for the zero value of angular veloctiy of the secondary. Which of the following is correct? (a) 1 and 2 (b) 3 and 4 (c) 1 and 4 (d) 2 and 4 616 Fluid Mechanics and Hydraulic Machines Series-3 Directions: The following items consist of two statements labeled Assertion A and Reason R. You are to examine each statement carefully and decide if the Assertion A and Reason R are individually true and if so, whether the Reason R is a correct explanation of Assertion A. Select your answers to these items using the codes given below. Codes: (a) (b) (c) (d) Both A and R are true and R is the correct explanation of A Both A and R are true but R is not a correct explanation of A A is true but R is false A is false but R is true Assertion (A): Reason (R): The kinematic viscosity of both air and water decreases as the temperature increases. The kinematic viscosity of liquids and gases at a given pressure is a function of temperature. Assertion (A): The relative movement of two blocks of wood joined with hot glue requires greater and greater effort as the glue dries up. Viscosity of liquids varies inversely with temperature. Reason (R): 65. Assertion (A): Reason (R): For submerged bodies, static forces in the lateral direction are balanced. For submerged bodies static forces in the vertical directions are not balanced. Assertion (A): The following potential function in two-dimensional ﬂow ﬁeld represents rotational ﬂow j = 2x2 – 3y 2 Reason (R): Assertion (A): Reason (R): 68. Assertion (A): Reason (R): 69. Assertion (A): Reason (R): 70. Assertion (A): Reason (R): For the given function ∂u ∂v + π0 ∂x ∂y In Raleigh’s method of dimensional analysis, the dependent variable is written as the function of independent variables. In Raleigh’s method, when the number of independent variables exceeds three, the exponents of non-repeating variables are expressed as exponents of repeating variables. In dimensional analysis, dimensionless numbers can be expressed as ratios of forces acting on the system. In dimensional analysis, Mach number is the ratio of inertia force to elastic force. If laminar ﬂow of oil between two points of a given pipeline is doubled, then the power consumption is increased to four times the original power. In laminar ﬂow through circular pipes, head loss varies directly as the discharge. In the pipe ﬂow, the upper critical Reynolds number is not well deﬁned. Flow changes from transition to full turbulence depending upon the external disturbance. 617 Appendix–A Assertion (A): Reason (R): Assertion (A): Flow in boundary layer is always laminar. In turbulent ﬂow on a smooth boundary, a laminar sublayer still exists within the boundary layer. l V2 , for laminar ﬂow through a pipe of diameter d, the term V d 2g (mean velocity) is given by In the equation hf = f V= Reason (R): Assertion (A): Reason (R): ( p1 - p2 )r 2 8m l The term f (friction factor) in the above equation equals 64/Re, where Re is the Reynolds number. Water ﬂows through a pipe connecting two reservoirs. The line joining the water surface levels in the reservoirs is the hydraulic grade line. There will be no negative pressure anywhere in the pipeline, as long as the pipeline lies below the hydraulic gradient. Assertion (A): Reason (R): Energy is lost in a sudden contraction in a pipeline. If the ﬂow is reversed, energy is gained at the transition which acts as an expansion. Assertion (A): A loss of head at a sudden contraction in a pipe is smaller than that at a sudden expansion. When the ﬂow contracts, it tends to become irrotational. Reason (R): 76. Assertion (A): Reason (R): Loss of head at a sudden expansion in a pipe is greater than that at a sudden contraction. Flow in a sudden expansion tends to be irrotational Assertion (A): In an open channel ﬂow, the maximum velocity ﬁlament does not occur on the free surface. There is wind drag on the free surface of an open channel. Reason (R): 78. Assertion (A): Reason (R): A minor change in the speciﬁc energy at or close to critical state will cause a major change in depth. Froude number being equal to unity characterizes a critical state of ﬂow. Assertion (A): Reason (R): At the critical state of ﬂow, the speciﬁc force is a minimum for the given discharge. For a minimum value of the speciﬁc force, the ﬁrst derivative of force with respect to depth should be unity. Assertion (A): A hydraulic jump cannot be expected in a long steep slope (fed by a large reservoir) when it is followed by a short stretch of adverse slope terminating in a deep and wide reservoir. The terminal depth in such an adverse slope reach will be critical and the ﬂow in the steep slope may be nearly at or at normal depth. Reason (R): Assertion (A): In a running irrigation canal, a regulating gate is partially closed suddenly. Such movement gives rise to a positive surge travelling downstream and a negative surge travelling upstream of the gate. 618 Fluid Mechanics and Hydraulic Machines Reason (R): Assertion (A): Reason (R): The sudden change in the discharge gives rise to changes in water elevations both upstream and downstream of the gate and develops in the form of an unsteady, rapidly verying ﬂow which propagates in the form of a surge wave moving on either side of the gate. The lower critical Reynolds number for pipe ﬂow is well deﬁned at 2000 while the upper critical Reynolds number has been found to exceed even 40,000. When the Reynolds number increases above that of the lower critical value, the transition tends to depend on external disturbance. Assertion (A): Reason (R): Surges can be positive or negative. Negative surges may occur when the gate at the head of a channel is suddenly opened or when the gate at tail end of a channel is suddenly closed. Assertion (A): The speciﬁc speed of a Pelton wheel is generally much less than that of a reaction turbine. Pelton wheels generally use more than one nozzle and the speciﬁc speed is deﬁned for power developed per nozzle. Reason (R): Assertion (A): Reason (R): Inlet side of a pump is less susceptible to cavitation damage. Cavitation occurs when the velocity is high and pressure is low. Assertion (A): Reason (R): The inlet velocity triangle for a Pelton wheel is a straight line. For a Pelton turbine, the vane angle at inlet is 180°. Assertion (A): By providing air vessels on the suction and delivery sides of a reciprocating pump, it is possible to increase the delivery head of the pump. The air vessel eliminates the acceleration head and makes discharge uniform. Reason (R): 88. Assertion: Reason (R): Reﬂux valve is provided on the suction side of a centrifugal pump. Reﬂux valve prevents backﬂow from the line into the pump and consequent damage to the pump set. Assertion A: In the boundary layer concept, the shear stress at the outer edge of the layer is considered to be zero. Local velocity is almost equal to velocity in the potential core. Reason R: 90. Assertion A: Reason R: A normal shock wave can occur at any section in a convergent divergent nozzle A normal shock wave occurs only when the ﬂow of the ﬂuid is supersonic and subsequent ﬂow after the shock is subsonic. Assertion A: The centre of pressure for a vertical surface submerged in a liquid lies above the centroid (centre of gravity) of the vertical surface. The distance of the centre of pressure from the free surface of the liquid for a vertical surface submerged in a liquid is independent of the density of the liquid. Reason R: Assertion A: Reason R: The power transmitted through a pipe is maximum when the loss of head due to friction is equal to one-third of total head at the inlet. Velocity is maximum when the friction loss is one-third of total head at the inlet 619 Appendix–A Assertion A: Reason R: Assertion A: Reason R: 95. Assertion A: Reason R: Assertion A: Reason R: Runaway speed of a turbine is the speed under maximum head at full gate opening when the load is disconnected suddenly. The various rotating components of a turbine are designed to remain safe at the runaway speed. The buoyant force of a ﬂoating body passes through the centroid of the displaced volume. The force of buoyancy is vertical force and is equal to the weight of the ﬂuid displaced. A circular plate is immersed in a liquid touching the free surface and the plane makes an angle q with the free surface. With different values of q, the position of centre of pressure will be different. Since the centre of pressure is dependent on the second moment of the area, with different values of q the second moment of the area for the circular plate will change. For the same power, the rotor of an impulse turbine need not be as large as that of a reaction turbine. In case of reaction turbine, water has to be admitted to the runner around its entire circumference. Assertion A: Reason R: For higher speciﬁc speeds, radial ﬂow pumps have the creates efﬁciency. Pumps having larger discharge under smaller head have higher speciﬁc speed. Assertion A: The volute casing of a centrifugal pump helps in creating the high velocity head necessary for enabling water ﬂow upwards to a higher level. The water ﬂows through a diverging passage in the volute chamber. Reason R: 99. Assertion A: Reason R: Assertion A: Reason R: Assertion A: Reason R: Aircraft wings are slotted to control separation of boundary layer especially at large angles of attack. This helps to increase the lift and the aircraft can take off from and land on short runways. Reynolds number must be the same for model and prototype immersed in subsonic ﬂows. Equality of Reynolds number for the model and prototype satisﬁes the dynamic similarity criteria towards viscosity. The speciﬁc speed of a Pelton turbine is low. Pelton turbine works under high head and handles low discharges. 620 Fluid Mechanics and Hydraulic Machines Appendix-B1 Answers to Objective Questions in Chapters 1 through 16 Chapter Chapter-1 Chapter-2 Chapter-3 Chapter-4 Chapter-5 Chapter-6 Chpater-7 Chapter 8 Chapter-9 Question Nos 0 1 2 3 4 5 6 7 8 9 0 10 20 30 40 50 b b d b c d b a d b b b a a a d d d a a c c b c d d b d b b b c d a d b b a c a a d d a a a b c d d c c d d b c b b b a d c d a c c b a c b c c c d d a b b a c b d c b d b d d b d d b c d c b c c c c c a d b c b a a b d b a d b b c c a c d d c d a b d b a d a d b d b c a a c a c c b b c d d b b d d c d c a d d c d c a a c d d b a a c c a d d a b a a b c d c b b a b d a d d d b c c a d b d c c c b c b b d a c a c d b b b b a c d c d b b a d d c a c a a b b b c d d d d c c d b b a b c a d d c c c d a d a c c d b d c d d d c b d a d d d c b b a c b a d c a c c d c a c c d d a b a b b c c d d b a b b a b c c a c a a b b c d c d b b d c d b a c c d b c a b b c c b d a d b c a d 0 10 20 30 40 50 60 70 0 10 20 30 40 50 0 10 20 30 0 10 20 0 10 20 0 10 20 0 10 20 30 0 10 20 b b a d b b b a a c a b d d a b a a a d 621 Appendix–B Chapter Chapter-10 Chapter-11 Chapter-12 Chapter-13 Chapter-14 Chapter 15 Chapter 16 Question Nos 0 1 2 3 4 5 6 7 8 9 10 20 30 0 10 20 30 0 10 20 30 40 50 60 0 10 20 30 40 50 0 10 20 0 10 20 30 0 10 20 30 40 50 60 a b c b c b c d a b b a b d c a c b d a a b c a c d b d c b c c b d d b c c a a d c a c d b c d d c d c b a d b a d a b d a b a c b c d c a b a a d b b d b c c a c d c a b c d c d b d d a b c b b d c a a d c b d c d c a a d a d b a c b a a c d a b d c a c b a c a d c d d c d d a c a a b d a a c a d a b a d a c c c c d d a b d c b b a c b d d d d b d c d b c a b b c c d b d c a c d a a d b b b d b b d b b c b b c c b c d b b b b d b d d c b c d d a d b d c a c a d a c a b d b a a c c c b a d d a d c a d c b a d b c d a a c b a c c c c b b c c b d a d c b d c c c c b a c b b d d b d c a c a c a b a Answers to Additional Objective Questions in Hydraulic Machines Question Nos. 0 A0 Chapter-16 1 2 3 4 5 6 7 8 9 d c d c b b c c b A10 b d c b b d c c a d A20 c b d c a d a b c c A30 c d c a 622 Fluid Mechanics and Hydraulic Machines Appendix-B2 Answers to Multiple Choice Questions of Appendix-A Series Question Nos. 0 0 Series-1 Series-2 1 2 3 4 5 6 7 8 9 b b c b b b d b d 10 b d c b d b a d d b 20 d c d b b b b a a b 30 a d b d d b a c a d 40 d d c b b d b d b b 50 a b a b b c c b b c 60 a d a d a b d c b a 70 a d d b c b c b b a 80 a d a c b d a d d a 90 100 d a d b c c b a b d d c 60 Series-3 Index A Absolute pressure 31, 39 Acceleration 106, 119, 529, 531, 572, 574, 575 Acceleration Head 529 acceleration vector 119, 133, 136 adiabatic 6, 20, 24, 27 adiabatic atmosphere 40 adiabatic constant 32 Adiabatic process 484 Adiabatic Process 32 adverse pressure gradient 258 aeration 419, 453, 454 aerofoils, 259 Aerostatics 32 Aging of Pipes 303 Air 2, 7, 17, 19, 23, 24 air bleed valve, 538 airfoils, hydrofoil vanes 281 air vessel 532, 533, 573, 574, 575, 576, 587, 592, 597 Air Vessel 532, 573, 574, 576 Air Vessels 532, 587 A laser-doppler anemometer 453 alternate depth 375, 404 alternate depths, 373 Aneroid barometer 31, 97 angular momentum. 175 atmospheres 31 atmospheric pressure 31, 39, 40, 41, 67, 68, 75, 76, 77, 82, 83, 84, 85, 86, 87, 96, 97, 103, 104 Atmospheric Pressure 3 axial ﬂow pumps 522, 591 axial ﬂow pumps. 522 Axial ﬂow pumps 526 B backward curved vanes 522 backwater curve 371 barometer 31, 38, 41, 97 bearings 233, 246 bed slope 370, 377, 379, 401, 402, 403, 404, 407, 410, 412 Bends 325 Bernoulli equation 141, 142, 143, 147, 148, 149, 150, 151, 152, 153, 156, 157, 158, 159, 161, 168, 522, 550, 551, 564 Bernoulli equation: 141 Best Efﬁciency Point 567 best section. 372 Bingham 5, 15, 26 Bingham plastic 5, 15, 26 blade angle 517, 524, 541, 544, 547, 549, 550, 553, 559, 560, 561, 580, 581, 584, 589 Blade efﬁciency. 521 Blasius equation 302, 322 Blasius solution 254 bluff body. 280 Body forces 172 Borda’s internal mouthpiece 450 boss 520, 548, 581, 590 624 Index Boundary forces 172 boundary layer 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277 Boundary layer Separation 258 Boundary Shear 371 Bourdon gauge 31, 97 Branching Pipes 328 broad-crested weir. 379 Broad Crested Weir 379, 397, 399 Buckingham Pi theorem 202, 208, 211 bulk modulus of elasticity 6, 7, 20, 21, 24, 27, 28, 29 Bulk Modulus of Elasticity: K 7 Buoyancy 36, 64, 93 buoyancy force 36 C Calculation of Critical Depth 374 Capillarity 5 capillary rise 5, 6, 17, 18, 24, 28 capillary viscometer 248 Cauchy-Reimann equations 139 cavitation 426, 427, 431, 446, 527, 532, 569, 570, 571, 572, 581, 585, 586, 588, 593, 596, 597 cavitation? 151 Cavitation 527, 592, 595 celerity 458, 462 Celerity 458, 462 centipoises 4 centre of buoyancy 36, 68, 69, 70, 71, 72, 73, 100, 101 centre of gravity 33, 34, 35, 36, 47, 68, 69, 71, 72, 89, 93, 94, 98, 99, 100, 101 Centre of Pressure 34 centrifugal pump 522, 536, 539, 558, 559, 560, 561, 562, 563, 564, 565, 566, 567, 583, 584, 585, 591, 592, 593, 595, 596 centrifugal pumps, 522 Centrifugal Pumps 591 Chezy coefﬁcient. 371 Chezy formula 371, 381, 384 choking 488 Choppy jump 376 Cipolletti weir 421, 442, 449, 454 circulation 108, 111, 121, 133, 281, 292, 524 Circulation 108, 122, 133, 292 Circulation Concept 108 Classiﬁcation 370, 376, 377 closed loop 350 Coefﬁcient of contraction 414, 416 Coefﬁcient of discharge 415, 416, 423, 529, 570 coefﬁcient of dynamic viscosity, absolute viscosity 4 coefﬁcient of surface tension 5, 6, 16, 28 Coefﬁcient of velocity 415, 423 Colebrook Equation 303 collar bearing 233, 246 collar bearing, pedestal bearing 233 commercial pipes 302, 303, 323 Commercial Pipes 302, 303 components of rotation 108, 122, 133 Compressibility 6 Compressibility Coefﬁcient: bc 6 compressibility correction factor (CCF). 490 compressibility effects 482, 490, 499, 500 conical bearing 233 Conical expansion 325 conjugate functions 130, 135 conservative, 331 conservative contaminant 353, 363 consistency index 4 Contaminant Propagation 331 continuity equation 228, 299 Continuity Equation 107 Continuum 1 continuum principle 1 Contracted weirs 420 Control of Separation 259 control volume 172, 173, 174, 175, 177, 178, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 197, 376 Control Volume 173 Converging–Diverging Nozzle 488 Coutte Flow 231, 232 Coverging Duct 487 crank angle 528, 576, 587, 592 creeping motion. 279 Creeping Motion 232 625 Index critical depth. 373, 374, 387, 390 Critical Depth 373, 374, 386, 404, 408 critical Reynolds number. 256 Critical Reynolds Number 228, 300 critical time. 458, 460 cup anemometer 286, 287 Current Meter 421 Curved Surfaces 56, 90 Cylinder 280 Cylindrical Co-ordinates 110 D Darcy–Weisbach equation 230, 234 Darcy–Weisbach formula. 301 Darcy–Weisbach friction factor, 300 delivery end. 522 density 1, 2, 3, 4, 6, 7, 9, 10, 13, 17, 18, 19, 22, 23, 24, 26, 27, 28, 29 diatomic gases 483 differential manometer 164, 166 differential surge tank, 461 diffuser. 487, 488 dilatant ﬂuid 4, 26 Dimensional analysis 201 Dimensional Homogeneity 201, 206 dimensional variables 202 dimensionless constant. 202, 206 dimensionless parameters 201, 202, 203, 211, 222, 226 direct step method 377, 378, 397 Direct step method 397 discharge division 328 Displacement thickness, 254 distorted models, 205 Distorted models 204 Distorted Models 206, 220, 225 distribution. 31 Diverging Duct 487 double stagnation point 294 doublet. 111 Doublet 111 draft tube 159, 167 Draft Tube 522, 550, 551, 581 drag. 278, 284 Drag 278, 280, 283, 284, 285, 286, 287, 288, 289, 290, 291 drag coefﬁcient 278, 279, 280, 284, 286, 290, 292, 293, 295, 296, 297 drowned 419 Dynamic Similarity 203 Dynamic Viscosity 3 E eddies 298 eddy viscosity 299 effective roughness 302, 305, 315, 317, 319, 320 Efﬁcient Channel Sections 403 efﬁcient trapezoidal channels. 403 elastic pipe 458, 467 end contractions 420, 440, 443 energy equation 141, 142 Energy Equation 141, 142 energy line 141, 143, 152, 153, 167, 168, 170 energy loss. 142 energy slope 324 Energy thickness, d 254 engineering system 201 Enthalpy 483 entrance length. 258 Entropy 484, 492 equation of continuity 109, 120, 121, 134 equation of continuity: 120 equilibrium boundary layer ﬂow. 300 equipotential lines 109, 111 Equivalent Length 325 equivalent pipe 326, 328, 334, 335, 336, 341, 343, 344, 360, 361, 368 Equivalent Pipes 325 equivalent sand grain roughness 302, 303, 305, 309, 310, 311, 317, 318, 319, 321, 322 equivalent steady motion 153 equivalent steady state 457 establishment of ﬂow 456, 474, 480 Establishment of Flow 258, 460, 473, 477 Euler 543 Euler equation: 141 Euler equation. 460 Euler number 204, 226 626 Index Euler’s equations 250 external cylindrical mouthpiece 426 external ﬂows 233, 298 fundamental dimensions. 201 Fxu 180 G F fall velocity 232 fast centrifugal pump 591, 596 favourable pressure gradient 258 ﬂash board 51, 52, 90 ﬂow coefﬁcient, 416 Flow in a nozzle 486 ﬂow index 4 Flow Measurement 379 ﬂow nets 109 Flow Nozzle 416 ﬂow properties 414 Flow Ratio, 519 Fluid Coupling 536 Fluid injection 259 Fluid mechanics 1 ﬂuid system 533 Force acting on the ﬂuid 173 Force on a Plane 33 Forces on curved surfaces 34 form drag 278 forward curved vanes 522 Francis formula 420, 440 Francis Turbine 517, 518, 540 free ﬂow. 419 free ﬂows 422 free jets 298 free stream velocity 125 free surface 30, 31, 33, 34, 37, 38, 48, 50, 51, 52, 60, 75, 77, 78, 83, 85, 87, 88, 89, 90, 96, 99, 100, 102 free turbulence. 298 Frictional Resistance 302 Friction Factor 230 Friction Head 531 frontal area 279, 284, 286, 293, 295 Froude Model Law 213 Froude number 204, 205, 208, 213, 214, 215, 218, 219, 224, 226, 227, 374, 375, 376, 387, 391, 392, 393, 394, 395, 402, 406, 407, 408, 409, 411 gas constant 483, 492, 512 Gas constant 492, 493 gauge pressure. 31 Gear Pump 539, 540 General Coutte ﬂow 231, 232 Geometric Similarity 203 gradually varied ﬂow 377, 396, 407, 408, 411, 412 Gradually Varied Flow 371, 377, 378, 395, 407, 411 Guide vane wicket gate 518 GVF Proﬁles 377 H Haaland equation 303 Hagen-Poiseuille equation 233 Hagen–Poiseuille equation. 230 Hardy–Cross Method 330 head loss 142, 147, 162, 163, 164, 165, 166, 167, 170 Head Loss, 229 homologous conditions 556, 565, 566 homologous pump 585, 591 homologous pumps 526, 527, 566 homologous pumps, 526 homologous turbine 583 hoop stress 34 Hoop stress 62 hot-wire anemometer 453 Hot-Wire Anemometer 422 hub 520, 549, 581, 590 hydraulic accumulator 534, 535, 593 Hydraulic Accumulator 534, 535 hydraulically efﬁcient 372, 384, 404, 410 Hydraulically Efﬁcient Channel 384 Hydraulically Efﬁcient Channel Section 372 hydraulic coupling 536, 593 hydraulic crane. 534 Hydraulic efﬁciency 521, 541, 542, 548, 549, 552, 553 627 Index Hydraulic Efﬁciency, 518, 524 hydraulic grade line 37, 76, 77, 143, 153, 168, 170, 231, 243, 251, 252 hydraulic gradient 324, 350 Hydraulic Intensiﬁer 535, 536 hydraulic jack, 534 hydraulic jump 370, 376, 392, 393, 394, 395, 397, 405, 406, 407, 408, 411 Hydraulic jump 375 Hydraulic Jump 375, 376, 392, 405, 411 hydraulic lift 534 hydraulic press 39, 84, 534, 577, 587 Hydraulic Press 534 hydraulic radius 371, 381, 384, 408, 410, 411 Hydraulic Ram 538 Hydraulic Systems 533 Hydrodynamic 534 hydrodynamically rough. 258 Hydrodynamically smooth: 301 hydrodynamically smooth. Roughness 258 hydro dynamic systems 533 hydrofoils. 279 hydrostatic. 371 hydrostatic pressure 31, 52 hydrostatic systems, 533 Hypersonic 482 I ideal ﬂuid 5, 16, 25, 26, 27 ideal ﬂuid, 141 ideal ﬂuid ﬂow. 253 impeller. 522, 563 Impulse Turbine 520 Impulse Turbines 581 incompressible ﬂuids 418 independent variables 201, 202 indicator diagram 531, 532, 592 Indicator Diagram 531 inlet head loss 417, 434 integrated properties 414 intensiﬁcation ratio. 536 intensity of turbulence 299, 320 Internal energy, 483 internal ﬂows 233, 298 internal ﬂows. 422 inviscid ﬂuid 5, 16 irrotational 108, 109, 110, 121, 123, 126, 127, 129, 130, 135, 138, 139, 140 irrotational ﬂow 108, 109, 110, 121, 123, 127, 129, 130, 135, 138, 139, 140 irrotational ﬂow: 123, 129 Irrotational Motion 107 irrotational vortex 156, 171 isentropic 6, 7, 19, 20, 24, 27, 32 isentropic ﬂow 484, 485, 486, 490, 500, 503, 505, 506, 510, 511, 514, 515 isentropic process 7, 19, 27 Isentropic process: 485 isothermal 6, 7, 19, 20, 24 isothermal atmosphere 39, 96 isothermal process 7, 19, 24 Isothermal process 32, 484, 485 J jet pump 533, 539, 578, 593 Jet Pump 539, 596 journal bearing 233, 245, 250 K Kaplan turbine. 519 Kaplan Turbine 548 Karman momentum integral equation 263, 266, 267, 273 Karman momentum integral equations, 255 Karman’s model 299 Kinematic Similarity 203 Kinematic Viscosity 4 kinetic energy correction factor 141, 143, 144, 159, 160, 165, 169, 235, 244, 251 kinetic energy correction factor. 143 kinetic energy correction factor a 159, 160, 165, 169 L laminae 87 laminar boundary layer. 256, 262 Laminar boundary layer 254, 267 laminar ﬂow 228, 230, 233, 234, 235, 236, 237, 240, 241, 242, 243, 247, 248, 249, 250, 251, 252 628 Index laminar sublayer 256, 257, 258, 266, 276 Laminar sublayer 257, 300 Laplace equation 109, 110, 123, 127, 128, 129, 130, 139 Laplace equation. 109, 123, 127, 128, 129 lapse rate 32, 33, 40, 86, 87 lapse ratet 32 Laser Doppler Anemometer 422 Laser Doppler velocimetry 422 Laval nozzles. 490 lawn sprinkler 175, 191, 192, 196, 197, 200 leakage of discharge. 524 lift. 278, 282 Lift 278, 281, 282, 290, 291, 292 Lift coefﬁcient 281, 292 lift force 281, 291, 292, 294, 295 lifting bodies 281 lifting surfaces 259 Lifting Vane 281 Linear Momentum 172, 177, 193, 197 linear momentum equation 172, 173, 174, 186, 187, 188, 197 Linear Momentum Equation 172 Line Sink 111 line source. 110 Line Source 110, 111 Line Source and Sink 110 Line Vortex 111 local accelerations 106 local properties 414 Local velocity 301 lock 63, 92 longitudinal stress 35 Loss of Head 415 Lubrication 233 M Mach angle. 485, 496, 509 Mach cone. 485 Mach number 204, 482, 485, 486, 488, 489, 490, 491, 495, 496, 497, 498, 499, 500, 501, 503, 504, 505, 506, 507, 508, 509, 510, 511, 512, 513, 514, 515 Mach Number 495, 504 Mach one ﬂow 488 Magnus effect. 281 Manning’s formula 371, 372, 377, 380, 381, 382, 383, 384, 386, 396, 400, 401, 411 Manning’s roughness coefﬁcient. 371 Manning’s roughness ratio 205 Manometers 33 manometric efﬁciency. 524, 583 manometric head. 524 mass oscillation 456, 461, 481 maximum discharge 372, 383, 384, 402, 404 maximum velocity 372, 383, 402 Maximum velocity 301, 308, 309 Maximum velocity: 229 Mean velocity 301 Measurement of Pressure 38, 84 Mechanical Efﬁciency, 518, 524 mechanical work 484, 485 Mercury 7 meta centre 36, 69 metacentric height 36, 70, 72, 93, 94, 102 method 110, 112 method of repeating variables. 202 method of superposition 110, 112 mild slope 396 Minimum NPSH 527 Minimum Speed 526 Minor Losses 324, 325 Miscellaneous Devices 587, 593 mixed ﬂow pumps, 522 Mixed ﬂow pumps 526 mixing length 299, 320, 321 model 203, 204, 205, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 224, 225, 226, 227 Model Scales 204 model slope. 225 modular limit 413 Mohr circle 96 molecular weight 483, 492, 509, 512 moment of inertia 34, 69, 101 Moment of inertia 35 Moment of Momentum 175, 191, 196, 199 moment of momentum equation 175 momentum correction factor 173, 187, 189, 191, 196, 198, 199, 248, 250, 251 momentum equation 376 629 Index momentum ﬂux 172, 173, 174, 178, 189, 197 Momentum thickness 254, 260, 261 Moody Diagram 303, 304, 371 Moody equation 303 most efﬁcient section. 373 mouthpiece 426, 445, 450 moving hydraulic jumps 456 N Navier–Stokes equations. 228 negative surge 456 neutral equilibrium. 36 newton 1 Newtonian ﬂuids 3, 4, 5, 25 Newton’s law of viscosity 3, 4, 25 Newton’s second law 172, 175, 197 Nominal thickness, 254 non-circular conduits 303, 316 non-horizontal Coutte 232 Non-Isothermal Atmosphere 32 Non-Newtonian Fluids 4 Non-uniform ﬂow: 106 normal convective acceleration 106, 137 normal depth 372, 377, 381, 382, 385, 397, 401, 402, 407, 409, 410, 412 normal shock 489, 491, 492, 506, 507, 508, 512, 515 Normal shock wave 491 normal stresses 299 notches. 419 Nozzle 325, 331 NPSH 527, 528, 569, 585, 592 NPSH. 569 O one-dimensional ﬂow 173 open channel ﬂow 370, 372, 379, 408 Oriﬁce Meter 416 oriﬁces 414, 415, 423 Oriﬁces 414, 423, 444 Oscillating jump 376, 411 Other Inviscid Flows 112 Outer layer 300 Overall Efﬁciency, 519, 526 Overlap region 300 P paraboloid of revolution 37, 38, 78 parallel 326, 327, 328, 331, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 360, 361, 362, 366, 367, 368 Parallel Plates 230, 231 pascal 3, 25 pascals 31, 84 Pascal’s law 30, 534 path line 105 pathline 136 Pelton Turbine 551 penstocks 461, 480 perfect gas, 7 perfect gas. 483 perfect gas equation 6 phase shift 528 piezometer. 419 piezometric head 36, 37, 38, 79, 96, 143, 144, 152, 153, 168 piezometric head. 229 piezometric head line 152, 153 Pipe Fittings 325 Pipe Flow 300 Pipeline systems 324 Pipe Network 330 Pipes in Parallel 327 Pipes in Series 327 pitot-static tube 419, 436, 447, 448, 452, 453 Pitot Static tube 490 pitot tube 414, 419, 436, 437, 447, 448, 450, 452 Pitot tube 106 Pitot Tube 418, 435, 447 Plain Coutte Flow 232 plastic. 5, 15 poise 4, 10, 11, 13, 22, 23, 27 positive displacement machines. 517 positive surge 456, 457, 462, 464, 465, 475, 478 Positive Surge 456, 457, 475 Potential Function 109 Power 143, 157, 158, 159 power transmission 533, 534, 538, 582 Prandtl’s model 299 630 Index Pressure 30, 31, 32, 34, 38, 41, 45, 50, 74, 75, 77, 78, 79, 81, 84 pressure coefﬁcient 419 pressure force, shear force 172 Pressure gradient 229 Pressure Wave 458, 459 primary dimensions 202, 208, 210 Prndtl number 226 product of inertia 34 proﬁle drag 278 proﬁles 377, 396, 407, 412 Propeller and kaplan turbines 519 propeller turbine 519, 590 Properties of areas 35 prototype 203, 204, 205, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 224, 225, 226, 227 prototype. 204, 213, 214, 215, 219, 220, 223, 224, 225 pseudoplastic ﬂuid 4, 26 pump 517, 522, 526, 527, 528, 529, 531, 532, 533, 534, 535, 536, 537, 538, 539, 540, 558, 559, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569, 570, 571, 572, 573, 574, 575, 576, 577, 578, 579, 583, 584, 585, 586, 587, 591, 592, 593, 595, 596, 597 Pump Characteristics 567 pumps 517, 522, 526, 527, 528, 533, 534, 539, 540, 566, 568, 577, 578, 585, 591, 592, 596, 597 Pumps in parallel 568 Pumps in Parallel 527, 568 Pumps in series 568 Pumps in Series 527, 569 R R 250 Radial Entry 526 Radial ﬂow 526 radial vane 522 Raleigh’s method, 202, 207 Rankine half body 112 Rankine oval 112 rapid closure. 458, 459, 476 Rapid Closure 458 rapidly varied ﬂow. 370, 375 Rapidly Varied Flow 370, 375 Reaction of the Boundary 173 Reaction Turbines 579 Reaction with Rotation 175 Reciprocating pump 528, 596 Reciprocating pumps, 517 Reciprocating Pumps 569, 585, 592 Rehbock formula 420, 438, 439, 453 Relative Density 2 relative ﬂuid motion. 153 relative velocity 174, 175, 179, 182, 183, 191, 192, 200 restricted entry surge tank, 461 retarding torque 176 Reynolds Model Law 215 Reynolds number 203, 204, 205, 206, 217, 218, 219, 224, 226, 227 Rigid body motion 36 Rigid Body Rotation 37 rms value 298 rotameter 421, 453 Rotameter 421, 422 Rotating circular cylinder 112 Rotating Cylinders 281 rotating system 175, 199 rotation 107, 108, 122, 133, 138 rotational 107, 108, 109, 133, 137, 138, 139 rotational motion 108 roto dynamic machines 175 Rotodynamic pumps 522 Rotodynamic Pumps 526, 558, 583 rough boundary 301 Rough ﬂat plate: 258 roughness scale 225 Rough turbulent ﬂow 302 run away speed 176 S safe speed 532 Schlichting: 257 Scroll casing 518 section 372, 373, 376, 378, 381, 384, 385, 386, 389, 390, 391, 392, 396, 397, 401, 402, 403, 404, 405, 407, 409, 410, 411, 412 separation 258, 259, 277 Separation 532, 572 631 Index Separation Phenomenon 258 separation point 259 sequent depths 375, 376, 392, 406, 410, 411 series 326, 327, 329, 331, 335, 336, 339, 355, 360, 361, 366, 367, 368 shape factor 254, 261, 271, 275 Shape factor, 254 shape number 519, 526, 596 shape number, 519 sharp crested 419, 438, 440, 448, 449, 450, 453 shear coefﬁcients 257 shear intensity 263 Shear stress 229, 230, 232, 235, 236, 243, 245 Shear Stress 3, 299 shear velocity 300, 306, 307, 317, 322 shear velocity. 258 Ship Model 219 shock 482, 485, 489, 490, 491, 492, 506, 507, 508, 512, 514, 515 shock wave 482, 491, 506, 507, 512, 514, 515 Similarity laws 203 Similarity Relations 585 Similarity Relationships 582 Similitude 201, 203, 204, 205, 206, 213, 222, 226 Similitude in Pumps 565 Similitude in Turbines 521, 555 Similitude Scale Ratios 204, 206 simple Coutte ﬂow. 232 Simple Harmonic Motion 529 sink 111, 112, 140 Siphon 326 Skin friction drag 269 slip 537, 569, 570, 586, 593, 597 Slip 529, 570, 586 slipper bearings 233 Slow Closure 459, 460 smooth boundary 301, 318, 322 Smooth turbulent ﬂow 302 snifting 538 soap bubble 5, 16, 19, 28 sonic velocity 467, 468 Sonic velocity 485, 497, 502 Speciﬁc energy 373, 386, 387, 391, 408 Speciﬁc Energy 373, 374, 386, 404, 408 Speciﬁc Energy and Critical Depth 373, 386, 404, 408 Speciﬁc Gravity 2 speciﬁc heat 483, 512 Speciﬁc Speed 519, 526 Speciﬁc Speed, 519, 526 speciﬁc volume 483, 511 Speciﬁc Volume 2 Speciﬁc weight 2, 9 speed factor 519, 520, 521, 582, 589, 590 Speed Factor, 519 speed of sound 7, 20, 28 Speed of Sound 495 Speed ratio 519, 521, 548, 549, 552, 553, 554 Speed ratio. 519 Sphere 279, 280 square entrance 331, 360, 361 Stability 36 stable 36, 69, 70, 71, 72, 73, 93, 94, 101, 102 stagnation density 486, 492, 506, 515 stagnation point 418 Stagnation point 292 Stagnation Point 106 stagnation pressure 221 Stagnation pressure 492, 495, 497, 498, 500, 505, 508 Stagnation Properties 497 stagnation temperature 486, 492, 497, 498, 499, 504, 506, 508, 510, 511, 513, 514, 515 Stagnation Values 486 stall. 537 Stall 282, 283 stalled. 282 stalling. 259 stalling point 282 standard lined triangular section 385, 402 standard pipe size 305 Stanton diagram 303 static ﬂuid 30, 33, 36, 96 static pressure 460, 470, 471, 477, 480 static pressure. 419, 436 Steady ﬂow: 105 “Steady” jump 376 Stokes Equation 232 Stokes’ law, 279, 285 stratosphere 32 streak line 105 stream ﬂow 419 632 Index Stream Function 109 stream line 105, 128, 136 streamline 106, 109, 110, 116, 117, 133, 136 Streamline 106 streamlined body 295, 296, 297 Streamlining 259 strength 110, 111 stresses 299, 317, 320 Strouhal number 280, 288, 296, 297 sub-critical 375, 405, 406, 409 submerged ﬂow. 419, 442 Submerged Flow 421 submerged oriﬁce ﬂow, 451 Subsonic 482, 487, 488, 516 subsonic diffuser 487 subsonic ﬂow 418 subsonic nozzle. 487 suction end 522, 527, 539 Suction of ﬂuid 259 Sudden contraction 325 sudden expansion 152, 164 Sudden expansion 325, 332 super-critical ﬂow 375 Supersonic 482, 487, 490 supersonic aircrafts. 490 supersonic diffuser. 487 supersonic ﬂow 482, 485, 488, 489, 490, 508, 515 supersonic nozzle. 487 suppressed rectangular weir 419, 420, 438, 448, 449, 453, 454 Suppressed weir 419 Surface drag 270 Surface in transition: 301 surface tension 5, 6, 16, 17, 18, 19, 23, 24, 26, 27, 28, 29 Surface Tension 5 surges 456 Surges 456, 462, 475, 478 SURGE TANKS 461 Swamee and Jain equation 303, 314 synchronous speed 590 T tangential 299 tangential convective acceleration 106, 137 tapering draft tube 159 temperature lapse rate 32, 40, 86, 87 tensile stress 36, 62, 100 Tensile Stress 34 terminal velocity 10, 13, 284, 285, 286, 292, 293, 295, 296 The capillary tube viscometer 233 The Moody diagram 303 Thermodynamic principles 483 Thoma number 527, 589 three-reservoir problem 328, 329 tidal bore 463, 464, 475 tidal bores. 457 tidal model, 220 time line 105 torque 524, 533, 534, 536, 537, 538, 589, 593 torque. 175 Torque 518, 524, 537, 538 Torque Converter 537, 538 total drag 219, 224, 278, 279, 295 total drag coefﬁcient 278 total drag force 255, 259, 266, 269, 273 total energy. 141 towing tank. 270 trajectory 414, 415, 423, 424, 444 Transition 256 Transitional regime 302 Transitions 389, 405 Translation 36 Transonic ﬂow 482 Trapezoidal channel 380 Trapezoidal Weirs 420 Triangular channel 382 Triangular Weir 420 troposphere 32 turbine 517, 518, 519, 520, 521, 522, 536, 537, 538, 540, 541, 542, 544, 545, 546, 547, 548, 549, 550, 551, 552, 553, 554, 555, 556, 557, 579, 580, 581, 582, 583, 588, 589, 590, 593, 594, 595 Turbines 517, 521, 522, 523, 555, 579, 581, 582, 588 Turbo-machines 517 Turbulence 298 TURBULENCE 422 633 Index turbulence intensities 299 turbulence property 298 turbulent boundary layer 254, 256, 257, 266, 268, 269, 270, 271, 273, 275, 276, 277 turbulent boundary layer. 254 turbulent eddies 258 Turbulent ﬂow. 298 Turbulent pipe Flow 300 turbulent shear stress 299, 320 two-dimensional ﬂow 108, 109, 110, 116, 119, 123, 125, 127, 133, 134, 135, 136, 137, 138, 139 U undershot water wheel 180 Undular jump 376, 411 Uniform ﬂow: 106 Uniform Flow 110, 370, 371, 372, 379, 400, 408, 410 Uniform Flow Computation 372 uniform sand grain roughness. 302 unit discharge 521 Unit Power 521 unit quantities 521 Units 1, 15 unit speed 521, 556, 588 unstable 36, 72, 94, 102 Unsteady ﬂow 105 Unsteady ﬂow: 105 unsteady ﬂows 456 V vacuum pressures. 31 vapor pressure head 151 Vapour Pressure 8 vectors 105, 117 velocity distribution; 173 Velocity distribution: 229 Velocity Distribution 301, 305 velocity potential. 109 Velocity Triangles 522, 559, 565 vena contracta. 414, 424, 450 venturimeter 414, 417, 418, 429, 430, 431, 432, 433, 434, 435, 445, 446, 447, 451, 452 Venturimeter 417, 418, 429, 445 Very Slow Closure 460 Villemonte equation 443 Villemonte formula 421 Viscometers 233 viscosity 3, 4, 5, 7, 9, 10, 11, 13, 14, 15, 16, 22, 23, 25, 26, 27 Viscosity 2, 3, 4 viscous forces 228 Volumetric efﬁciency 529 Volumetric Efﬁciency, 524 volute casing 522 vortex 111, 112, 133, 140 vortex shedding 222, 280, 288, 294, 297 Vortex strength. 111 vorticity 108, 133, 138 vorticity vector 133 W wake 259 wake area 280 wall turbulence 298 Water 7 water hammer 456, 458, 459, 460, 461, 467, 468, 469, 470, 471, 473, 476, 477, 479, 480, 481, 538 Water Hammer 458, 459, 467, 476, 479 water surface slope 370, 377 Water vapour 7 wave drag 269, 270, 271 wave drag. 269, 270 Weak Jump 376 Weber number 204, 218, 224, 226 weirs 414, 419, 420 Weirs 419, 420, 438, 448 wetted area. 279 Wheel efﬁciency 521 whirl velocity. 524 wicket gates 544 Y y 260 Z zero pressure gradient. 258 zero pressure gradient ﬂow. 254 Report "Fluid mechanics and hydraulic machines : problems and solutions [1 ed.] 9780070699809, 0070699801" × Close Submit Contact information Michael Browner info@dokumen.pub Address: 1918 St.Regis, Dorval, Quebec, H9P 1H6, Canada. 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15463	http://www.cad-conference.net/files/CAD24/CAD24_98-103.pdf	98 Title: Least Squares Fitting on a Segment of Ellipse and Its Application on Road Curvature Estimation Authors: Md Hashim F.I., farahiah@student.usm.my, Universiti Sains Malaysia Ramli A., alaramli@usm.my, Universiti Sains Malaysia Keywords: Least Squares Fitting, Curve Fitting, Parametric Equation, Ellipse, Minimization DOI: 10.14733/cadconfP.2024.98-103 Introduction: The tting of geometric features such as circles and ellipses to given points is applicable in various elds . Ellipse, for example, may represent many real-world situations, such as the orbits of the planets, satellites, and comets. These conic sections are also relevant in science, astronomy, and variety of engineering applications . In real-world practice, least squares tting is used in various manufacturing industries, software applications , and also signicant in pattern recognition . A circle, an ellipse, a parabola or a hyperbola are curves that are referred to as conic sections. In this paper, elliptical arcs are preferred over general conic arcs as ellipses are frequently encountered shapes found as a component in various objects and have been proven to be benecial in elds such as Computer-Aided Design (CAD), computer graphics and computer vision . For instance, we are interested in tting data that form an arc or a segment of an ellipse. Hence, we apply an ellipse tting rather than other curve ttings. In addition, ellipse ranks among the most prevalent geometric shapes found in the real world . In this paper, we are interested to t parametric curves in the least squares sense. The best tting curve to a given set of points is obtained by minimizing the sum of the square errors of the points from the curve . This mathematical procedure is called the least squares tting. Gander et al. , Watson , and Pilu et al. apply least squares tting based on data that forms entire circles and ellipses but here we only discuss a part of an ellipse or a circle that represents an arc or a curve. While the parameter z for a full ellipse ranges from 0 to 2π radians, the challenge with data that form a partial ellipse is to determine where it lies in the parameter z. We will determine the parameter z that best ts the data points presumed to be part of an ellipse. Then, we demonstrate how this parameter can be utilized to estimate road curvature and compare it with the experimental approach employed by engineers. A simple algorithm that uses a numerical optimization technique is introduced to obtain the best curve tting for any given set of data, specically data that forms a segment of an ellipse. This technique is known as simulated annealing. There are extensive usage of simulated annealing in real-life applications in which the primary benet of it lies in its simplicity . In road curvature estimation, people may relate road curvature with other benets. For example, Persyn et al. consider various factors related to road characteristics referring to OpenStreetMap (OSM) to estimate the expenses associated with road transportation. The factors include the existence of roundabouts and trac lights, the surface material and the curvature of the road. Good estimation of transportation costs is very helpful for trip planning, Proceedings of CAD'24, Eger, Hungary, July 1-3, 2024, 98-103 © 2024 U-turn Press LLC, 99 consequently may result in better fuel consumption and less pollution to the environment. The general parametric conic arcs introduced by outline the process of creating conic blending arcs by utilizing a unied rational parametric representation that merges the distinct cases of blending parallel and non-parallel edges based on given constraints such that it requires the arc to maintain a specied distance from a line, point, or a circle. Otherwise, the arc intersects a circle or line at a specied angle. In this paper, instead of interpolating points, we used least squares tting to approximate points for any given data. Fitting an Ellipse in Parametric Form: Generally, in order to t an ellipse in parametric form, we follow Späth and consider the equations: x(z) = a + p cos(z) y(z) = b + q sin(z) (2.1) where (a, b) is the center of the ellipse, p is the radius along the x-axis, q is the radius along the y-axis and parameter z lies between 0 to 2π radians. The function to be minimized is: S(a, b) = n X i=0 (xi −a −p cos(zi))2 + (yi −b −q sin(zi))2, (2.2) where zi is a parametrized value that lies between 0 to 2π radians and (xi, yi) are data points. Späth utilizes (2.1) to t an ellipse based on data forming a complete elliptical shape. However, we consider the case where collected data are from a segment and not the entire ellipse. The issue is dealt with in ; however, we use a parametrization approach in minimization as described in the next section. In this case, we aim to minimize (2.2) where zi is a parametrized value that lies between θ1 and θ2. Parameter zi should cover a certain part of an ellipse. For instance, if the data is half of an ellipse that forms the upper half of ellipse, zi should cover from 0 to π radians. If let's say the data forms the bottom half of the ellipse, then zi can be from π to 2π radians. The range can be determined through observation; nonetheless, we will select the optimal values for θ1 and θ2 by optimizing the equation (2.2). The values of a, b, p and q can be solved by dierentiating (2.2) with respect to each parameter and equate it to 0: δS δa = 0 , δS δb = 0 , δS δp = 0 , δS δq = 0. (2.3) The presence of parameter zi and the uncertainty regarding its interval render the problem dicult to solve. Therefore, we will determine the values of a, b, p, and q using simulated annealing, a method that will be further elaborated in the next section. Minimizing the Error Distances: To establish the minimization process using simulated annealing, we start by discussing about parameter z in the ellipse function. The parameter zi can be computed by: zi = θ1 + (i −1)h, (2.4) where i = 1, 2, ..., n and θ1 denotes the start of the interval while h is dened as the step size for parameter zi and it is assigned arbitrarily. For instance, if h = 0.1, then z = {θ1, θ1 + 0.1, θ1 + 0.2, θ1 + 0.3, ...}. By choosing any value from θ1 to θ2 from this range, where θ2 = θ1 + (n −1)h, we can see the pattern of the error distance to be either Proceedings of CAD'24, Eger, Hungary, July 1-3, 2024, 98-103 © 2024 U-turn Press LLC, 100 increasing or decreasing. To minimize (2.2), we employ a numerical approach to evaluate the value of S(a, b). We aim to nd the values of θ1 and h that will minimize the function. The error distance can be obtained by using: d = n X i=0 \|(Xi, Yi) −(xi, yi)\|, (2.5) where (Xi, Yi) are the points on the estimated curve and (xi, yi) are the original data points. Hence the minimum value of (2.5) is the solution to the minimization problem. We perform the minimization by using a numerical optimization technique called simulated annealing which is available as a built-in function on Mathematica. The purpose is to nd the optimum values of z, and parameters a, b, p, q, θ1 and h in order to obtain the best curve tting of an ellipse. The next part of the algorithm is to input data i.e. number of observations, n, coordinates (x1, y1), (x2, y2),..., (xn, yn), v1, and v2. Then, minimizing (2.2) using simulated annealing subject to constraint 0 ≤θ1 ≤2π and v1 ≤h ≤v2, whereas v1 and v2 are the minimum and maximum step sizes respectively. For our experiment, we let v1 and v2 ranges from 0.1 to 0.5 where v1 < v2. Meanwhile, the output obtained are the parameters a, b, p, and q for the equation of ellipse, the minimum error distance, and the values of θ1 and h. The Application of Ellipse Fitting on Road Curvature Estimation: After the minimization procedure, during which we obtained the best-tting ellipse for data forming a segment of an ellipse, we now aim to apply the proposed algorithm to t data from small segments of roads, particularly those with curvy shapes. Few points will be taken along the desired segment of a road. Hence, tting an ellipse on the segment of a road will allow us to calculate the radius of curvature for each point precisely based on its coordinate on the road. For the radius of curvature, we compare our approach to Luo et al. . In their paper, radius of curvature was calculated at 9 dierent test sites, chosen from highway ramps and eld measurement were used to conduct the validation tests. Besides, this paper uses roadway centerline to measure the radius of curvature and curve length. The road coordinates are determined by referencing to Test Site 3 as stated in . Ellipses will be constructed, and the radius of curvature can be calculated by using (2.6): R = [(x′)2 + (y′)2] 3 2 \|x′y” −y′x”\| (2.6) where x(z) and y(z) are from Equation (2.1). Test site 3 as shown in Fig. 1(a) is located in Interstate 35 (I-35) in Kansas, United States that begins at 39◦02'20.43" N, 94◦40'26.76" W and ends at 39◦02'29.26" N, 94◦40'22.67" W with the length of 324 m. The radius of curvature obtained from eld measurement is 104.15 m . Coordinates of 9 points along Test Site 3 from Google Maps were chosen and presented in Table 1, the best curve from minimization procedure is tted as shown in Fig. 1(b) and Fig. 1(c) displayed the tting of a full ellipse. We can observe that the centre of the ellipse segment is in between point 4 and point 5 and the radius of curvature in between those points lies between 123.686 m and 94.6137 m. The radius of curvature obtained from our proposed algorithm is nearly equal to the radius of curvature found by Luo et al. which is 104.15 m. We do not provide an exact comparison as we are uncertain of which specic point is referenced in . The method used in our paper shows a high similarity to the eld measurement in which radius of curvature obtained from the least squares tting on a segment of an ellipse is found to be approximately equal to the radius of curvature obtained by Luo et al. . . Proceedings of CAD'24, Eger, Hungary, July 1-3, 2024, 98-103 © 2024 U-turn Press LLC, 101 Fig. 1: Test Site 3: (a) Location on Google Maps, (b) 9 points chosen along Test Site 3, and (c) Fitting of a full ellipse on Test Site 3. Points Coordinates Radius of curvature (m) 1 (39.039408, -94.672916) 293.312 2 (39.039537, -94.672637) 213.213 3 (39.039712, -94.672455) 157.344 4 (39.039871, -94.672347) 123.686 5 (39.040104, -94.672268) 94.6137 6 (39.040312, -94.672229) 85.0239 7 (39.040496, -94.672251) 89.7698 8 (39.040704, -94.672315) 110.282 9 (39.040944, -94.672433) 153.020 Tab. 1: The coordinates and radius of curvature for 9 points taken along Test Site 3. Conclusions: In this paper, least squares tting is applied to obtain the best curve tting to the given data that form a segment of ellipse by minimizing the sum of square errors using simulated annealing. It can be observed that the solution to the minimization problem approximates the data closely by the ellipse. In addition, we t the data of a small segment of a road to obtain its curvature at any specic point on the road. For perspective, this can be extended in future research for travel time prediction in or for the purpose of road safety in . The positive aspect of our approach lies in its cost eciency as we rely on the readily available GPS data. Generally, if a set of a parametric data is assumed to behave in ellipse shape, we should be able to perform least squares tting using the proposed algorithm. A few segment of roads have been tested by using this approach and the results demonstrated were proven reliable by the proposed algorithm in obtaining the radius of curvature. Hence, it can be utilized in real-life applications. Acknowledgement: This research is supported by the Ministry of Higher Education Malaysia through Fundamental Research Grant Scheme (FRGS/1/2021/STG06/USM/02/6). The authors would also like to express their gratitude to the School of Mathematical Sciences, Universiti Sains Malaysia for supporting this work. Md Hashim F.I. , Ramli A. , Proceedings of CAD'24, Eger, Hungary, July 1-3, 2024, 98-103 © 2024 U-turn Press LLC, 102 References: Ahn, S.J.; Rauh, W.: Geometric least squares tting of circle and ellipse, Inter-national journal of pattern recognition and articial intelligence, 13(07), 1999, 987-996. Ahn, S. J.; Rauh, W.; Warnecke, H. J.: Least-squares orthogonal distances tting of cir-cle, sphere, ellipse, hyperbola, and parabola, Pattern Recognition, 34(12), 2001, 2283-2303. Chaudhuri, D.: A simple least squares method for tting of ellipses and circles depends on border points of a two-tone image and their 3-D extensions., Pattern Recognition Letters, 31(9), 2010, 818-829. Chernov, N.; Lesort, C.: Least squares tting of circles, Journal of Mathematical Imaging and Vision, 23(3), 2005, 239-252. Delahaye, D.; Chaimatanan, S.; Mongeau, M.: Simulated annealing: From basics to applications, Handbook of metaheuristics, 2019, 1-35. Delhi Babu, R.; Vaithyasubramanian, S.; Sundararajan, R.; Kirubhashankar, C. K.; Vengatakrish-nan, K.; PMSS, C.: Integrating Pedagogical Approaches in the Study of Conic Sections Using Dif-ferential Equation and Analysis via Bayesian Inference, Engineering Proceedings, 59(1), 2023, 93. Fudos, I.; Homann, C.M.: Constraint-based parametric conics for CAD, Computer-Aided Design, 28(2), 1996, 91-100. Gander, W.; Golub, G.H.; Strebel, R.: Least-squares tting of circles and ellipses, BIT Numerical Mathematics, 34(4), 1994, 558-578. Luo, W.; Li, L.: Automatic geometry measurement for curved ramps using inertial mea-surement unit and 3D LiDAR system, Automation in Construction, 94, 2018, 214-232. Misro, M.Y.; Ramli, A.; Ali, J.M.: Approximating maximum speed on road from curvature informa-tion of Bézier curve, World Academy of Science, Engineering and Technology, International Journal of Mathematical, Computational, Physical, Electrical and Computer Engineering, 9(12), 2015, 705-712. Persyn, D.; Díaz-Lanchas, J.; Barbero, J.: Estimating road transport costs be-tween and within European Union regions, Transport Policy, 124, 2022, 33-42. Pilu, M.; Fitzgibbon, A.W.; Fisher, R.B.: Ellipse-specic direct least-square tting, IEEE, 1996, 599-602. Rosin, P.L.: A survey and comparison of traditional piecewise circular approxi-mations to the ellipse, Computer Aided Geometric Design, 16(4), 1999, 269-286. Späth, H.: Least-squares tting of parametric curves with a linear function of several variables as argument, Mathematical Communications, 3(1), 1998, 61-66. Tamin, O.; Ikram, B.; Amri Ramli, A.L.; Moung, E.G.; Chin Pei Yee, C.: Travel-time estimation by cubic Hermite curve, Information, 13(7), 2022, 307. Wang, T.; Shi, Z.; Yu, B.: A parameterized geometric tting method for ellipse, Pattern Recognition, 116, 2021. Wang, Z.; Chen, D.; Gong, J.; Wang, C.: Fast high-precision ellipse detection method, Pattern Recognition, 111, 2021. Proceedings of CAD'24, Eger, Hungary, July 1-3, 2024, 98-103 © 2024 U-turn Press LLC, 103 Watson, G.: Least squares tting of circles and ellipses to measured data, BIT Numerical Mathe-matics, 39(1), 1999, 176-191. Proceedings of CAD'24, Eger, Hungary, July 1-3, 2024, 98-103 © 2024 U-turn Press LLC,
15464	https://math.stackexchange.com/questions/2795944/logarithmic-equation-log-3x3-x-log-3x2-1	algebra precalculus - Logarithmic equation $\log_{3x}(3/x)+(\log_3(x))^2=1$. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Logarithmic equation log 3 x(3/x)+(log 3(x))2=1 log 3 x⁡(3/x)+(log 3⁡(x))2=1. Ask Question Asked 7 years, 4 months ago Modified7 years, 4 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. The equation is log 3 x 3 x+log 2 3 x=1 log 3 x⁡3 x+log 3 2⁡x=1 I tried to solve it like this log 3 x x−1 3−1+log 2 3 x=1 log 3 x⁡x−1 3−1+log 3 2⁡x=1 log 3 x(x 3)−1+log 2 3 x=log 3 3 log 3 x⁡(x 3)−1+log 3 2⁡x=log 3⁡3 −log 3 x x 3+log 2 3 x=log 3 3−log 3 x⁡x 3+log 3 2⁡x=log 3⁡3 −log 3 x x−log 3 x 3+log 2 3 x=log 3 3−log 3 x⁡x−log 3 x⁡3+log 3 2⁡x=log 3⁡3 I don't have any other ideas. algebra-precalculus logarithms Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited May 25, 2018 at 20:55 nonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges asked May 25, 2018 at 17:22 HanlonHanlon 1,899 1 1 gold badge 19 19 silver badges 29 29 bronze badges Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Hint: Express everything in base 3 3 and make some substitution like t=log 3 x t=log 3⁡x. Also use addition theorem for log. For an example: log 3 x x=log 3 x log 3(3 x)=log 3 x log 3 3−log 3 x=t 1−t log 3 x⁡x=log 3⁡x log 3⁡(3 x)=log 3⁡x log 3⁡3−log 3⁡x=t 1−t So, you got 1−t 1+t=1−t 2=(1−t)(1+t)1−t 1+t=1−t 2=(1−t)(1+t) If we multiply by 1+t 1+t we get after some rearrangement t 3+t 2−2 t=0 t 3+t 2−2 t=0. This we can factor: t(t 2+t−2)=t(t+2)(t−1)=0 t(t 2+t−2)=t(t+2)(t−1)=0, so t∈{−2,0,1}t∈{−2,0,1}. Can you finish now? Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited May 26, 2018 at 10:49 answered May 25, 2018 at 17:25 nonusernonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges 0 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Hint: writing your equation in the form log 3 x 3−log 3 x x+(ln(x)ln(3))2=1 log 3 x⁡3−log 3 x⁡x+(ln⁡(x)ln⁡(3))2=1 And this is ln(3)ln(3)+ln(x)−ln(x)ln(3)+ln(x)+(ln(x))2(ln(3))2=1 ln⁡(3)ln⁡(3)+ln⁡(x)−ln⁡(x)ln⁡(3)+ln⁡(x)+(ln⁡(x))2(ln⁡(3))2=1 Can you finish? Substitute t=ln(x)t=ln⁡(x) Simplifying and factorizing we get t(3 ln(3)+t)(ln(3)−t)=0 t(3 ln⁡(3)+t)(ln⁡(3)−t)=0 Multiplying by ln(3)+t,ln 2(3)ln⁡(3)+t,ln 2⁡(3) we get (ln(3)−t)ln 2(3)+t 2(ln(3)+t)=ln 2(3)(ln(3)+t)(ln⁡(3)−t)ln 2⁡(3)+t 2(ln⁡(3)+t)=ln 2⁡(3)(ln⁡(3)+t) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited May 25, 2018 at 18:14 answered May 25, 2018 at 17:28 Dr. Sonnhard GraubnerDr. Sonnhard Graubner 97.5k 4 4 gold badges 42 42 silver badges 80 80 bronze badges 2 I don't think I can.Hanlon –Hanlon 2018-05-25 17:45:49 +00:00 Commented May 25, 2018 at 17:45 How did you get that? I get ln 3−t ln 3+t+t 2 ln 2 3=1 ln⁡3−t ln⁡3+t+t 2 ln 2⁡3=1.Hanlon –Hanlon 2018-05-25 17:51:56 +00:00 Commented May 25, 2018 at 17:51 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Perhaps it is simpler to rewrite 1=log 3 x(3/x)+(log 3(x))2=ln 3−ln x ln 3+ln x+(ln x ln 3)2 1=log 3 x⁡(3/x)+(log 3⁡(x))2=ln⁡3−ln⁡x ln⁡3+ln⁡x+(ln⁡x ln⁡3)2 and substitute u=ln x u=ln⁡x. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered May 25, 2018 at 17:28 gt6989bgt6989b 55k 3 3 gold badges 40 40 silver badges 75 75 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. HINT We have log 3 x 3 x+log 2 3 x=1 log 3 x⁡3 x+log 3 2⁡x=1 log 3 x 3−log 3 x x+log 2 3 x=1 log 3 x⁡3−log 3 x⁡x+log 3 2⁡x=1 log 3 x 3 x−2 log 3 x x+log 2 3 x=1 log 3 x⁡3 x−2 log 3 x⁡x+log 3 2⁡x=1 1−2 log 3 x x+log 2 3 x=1 1−2 log 3 x⁡x+log 3 2⁡x=1 −2 log 3 x x+(log 3 x)2=0−2 log 3 x⁡x+(log 3⁡x)2=0 −2 1 log x 3 x+(1 log x 3)2=0−2 1 log x⁡3 x+(1 log x⁡3)2=0 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered May 25, 2018 at 17:30 useruser 164k 14 14 gold badges 84 84 silver badges 157 157 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus logarithms See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 1Need help solving a logarithmic equation 3Solve for x x the equation x log 10 x=x 3 100 x log 10⁡x=x 3 100 0Logarithmic equation log 2 4 x−log 2 x+4=0 log 4 2⁡x−log 2⁡x+4=0 2log y z(x 2+4 4 y z√)+log z x(y 2+4 4 z x√)+log x y(z 2+4 4 x y√)=0 log y z⁡(x 2+4 4 y z)+log z x⁡(y 2+4 4 z x)+log x y⁡(z 2+4 4 x y)=0 0What is the value of (log 21(3))2+log 21(147)log 21(1323)(log 21⁡(3))2+log 21⁡(147)log 21⁡(1323)? 4Is there any way to solve log 2 10(x)+log 10(x)log 10(2)−log 10(5)=0 log 10 2⁡(x)+log 10⁡(x)log 10⁡(2)−log 10⁡(5)=0? 0Solving a logarithmic equation with different logarithmic exponents. 3Unable to solve a logarithmic equation Hot Network Questions "Unexpected"-type comic story. Aboard a space ark/colony ship. 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15465	https://vuir.vu.edu.au/18138/1/mon-element-exp.pdf	Monotonicity and Logarithmic Convexity for a Class of Elementary Functions Involving the Exponential Function This is the Published version of the following publication Qi, Feng (2006) Monotonicity and Logarithmic Convexity for a Class of Elementary Functions Involving the Exponential Function. Research report collection, 9 (2). The publisher’s official version can be found at Note that access to this version may require subscription. Downloaded from VU Research Repository MONOTONICITY AND LOGARITHMIC CONVEXITY FOR A CLASS OF ELEMENTARY FUNCTIONS INVOLVING THE EXPONENTIAL FUNCTION FENG QI Abstract. In this paper, the monotonicity and logarithmically convexity of the function e−αt−e−βt 1−e−t are obtained, where t ∈R and α and β are real numbers such that α ̸= β, (α, β) ̸= (0, 1) and (α, β) ̸= (1, 0). 1. Introduction For real numbers α and β with α ̸= β, (α, β) ̸= (0, 1) and (α, β) ̸= (1, 0) and for t ∈R, let qα,β(t) =    e−αt −e−βt 1 −e−t , t ̸= 0, β −α, t = 0. (1) In order to obtain the best bounds in Gautschi-Kershaw’s inequalities, it was proved in that the function qα,β(t) is logarithmically convex in (0, ∞) and loga-rithmically concave in (−∞, 0) if β −α > 1 and is logarithmically concave in (0, ∞) and logarithmically convex in (−∞, 0) if 0 < β −α < 1. When ones study the logarithmically completely monotonic property of some functions involving Euler’s gamma Γ function, the psi function ψ and the polygamma functions ψ(i) for i ∈N, the elementary function qα,β(t) is encountered now and then. The so-called logarithmically completely monotonic function on an interval I ⊂R is a positive function f which has derivatives of all orders on I and whose logarithm ln f satisﬁes 0 ≤(−1)kln f(x) < ∞for k ∈N on I. The set of the logarithmically completely monotonic functions on I is denoted by L[I]. For more information on the class L[I], please refer to [1, 2, 3, 5, 6, 7, 8, 9] and the references therein. The ﬁrst aim of this paper is to research the monotonicity of the function qα,β(t). The ﬁrst main result of ours is the following Theorem 1 or Corollary 1. Theorem 1. The following conclusions present the monotonic properties of qα,β(t). (1) The function qα,β(t) is increasing in (0, ∞) if either 1 ≥α + β > 2α + 1 or 1 ≤α + β < 2α < α + β + 1 holds. (2) The function qα,β(t) is decreasing in (0, ∞) if either 1 ≥α + β > 2β + 1 or 1 ≤α + β < 2β < α + β + 1 is valid. 2000 Mathematics Subject Classiﬁcation. Primary 26A48, 26A51; Secondary 26A09, 33B15. Key words and phrases. monotonicity, logarithmically convex function, elementary function, exponential function. The author was supported in part by the Science Foundation of the Project for Fostering Innovation Talents at Universities of Henan Province, China. This paper was typeset using A MS-L A T EX. 1 2 F. QI (3) The function qα,β(t) is increasing in (−∞, 0) if either 2α > α + β + 1 ≥2 or α + β < 2β < α + β + 1 ≤2 validates. (4) The function qα,β(t) is decreasing in (−∞, 0) if either 2β > α + β + 1 ≥2 or α + β < 2α < α + β + 1 ≤2 sounds. (5) The function qα,β(t) is increasing in (−∞, ∞) if and only if one of the following conditions holds: (a) α = β + 1 > 1, (b) α > β + 1 ≥1, (c) β = α + 1 < 1, (d) 1 ≥β > α + 1, (e) α < β < α + 1 ≤1, (f) β + 1 ≤α + β < 2α < α + β + 1. (6) The function qα,β(t) is decreasing in (−∞, ∞) if and only if one of the following conditions holds: (a) β = α + 1 > 1, (b) β > α + 1 ≥1, (c) β < α < β + 1 ≤1, (d) 1 > α = β + 1, (e) 1 ≥α > β + 1, (f) α + 1 ≤α + β < 2β < α + β + 1. Remark 1. The (α, β)-domain where the function qα,β(t) is monotonic in Theorem 1 can be described respectively by Figure 1 to Figure 6 below. -α 1 O 6 β 1 @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ β = α β = α + 1 β = α −1 β = 1 −α Figure 1. (α, β)-domain where the function qα,β(t) is increasing in (0, ∞) in Theorem 1 MONOTONICITY AND LOGARITHMIC CONVEXITY FOR A CLASS OF FUNCTIONS 3 - α 1 O 6 β 1 @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ β = α β = α + 1 β = α −1 β = 1 −α Figure 2. (α, β)-domain where the function qα,β(t) is decreasing in (0, ∞) in Theorem 1 Remark 2. Note that the (α, β)-domain where the function qα,β(t) is increasing (or decreasing) in (0, ∞) (or in (−∞, 0)) is an union where the function qα,β(t) increases (or decreases) in either (0, ∞) (or (−∞, 0)) or (−∞, ∞). Therefore, Theorem 1 can be restated as the following Corollary 1. Corollary 1. The following conclusions describe the monotonic properties of qα,β(t). (1) The function qα,β(t) is increasing in (−∞, ∞) if and only if (α, β) ∈ {(α, β) : α > β ≥0, α ≥1} ∪{(α, β) : α < β ≤0} ∪{(α, β) : α ≤ β −1, 0 ≤β ≤1} \ {(1, 0), (0, 1)}. (2) The function qα,β(t) is decreasing in (−∞, ∞) if and only if (α, β) ∈ {(α, β) : β > α ≥0, β ≥1} ∪{(α, β) : β < α ≤0} ∪{(α, β) : β ≤ α −1, 0 ≤α ≤1} \ {(1, 0), (0, 1)}. (3) The function qα,β(t) is increasing in (0, ∞) if and only if (α, β) ∈{(α, β) : α > β ≥1 2} ∪{(α, β) : α ≥1 −β, 0 ≤β < 1 2} ∪{(α, β) : α + 1 ≤β ≤ 1 −α, α < 0} ∪{(α, β) : β −1 ≤α < β ≤0} \ {(1, 0)}. (4) The function qα,β(t) is decreasing in (0, ∞) if and only if (α, β) ∈{(α, β) : β ≥1−α, 1 2 > α ≥0}∪{(α, β) : β > α ≥1 2}∪{(α, β) : β < α ≤0}∪{(α, β) : β ≤α −1, 0 ≤α ≤1} ∪{(α, β) : 1 ≤α ≤1 −β} \ {(1, 0), (0, 1)}. (5) The function qα,β(t) is increasing in (−∞, 0) if and only if (α, β) ∈{(α, β) : 1 −α ≤β < α, α ≥1} ∪{(α, β) : α < β ≤1, α ≤0} ∪{(α, β) : α < β ≤ 1 −α, 0 ≤α < 1 2} \ {(1, 0), (0, 1)}. 4 F. QI -α 1 O 6 β 1 @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ β = α β = 1 −α β = α −1 β = α + 1 Figure 3. (α, β)-domain where the function qα,β(t) is increasing in (−∞, 0) in Theorem 1 (6) The function qα,β(t) is decreasing in (−∞, 0) if and only if (α, β) ∈{(α, β) : 1 −β ≤α < β, β ≥1} ∪{(α, β) : β < α ≤1 2} ∪{(α, β) : β ≤1 −α, 1 2 < α ≤ 1} \ {(1, 0), (0, 1)}. Remark 3. The corresponding (α, β)-domains where the function qα,β(t) is mono-tonic in Corollary 1 can be described respectively by Figure 5 to Figure 10 below. The second aim of this paper is to reconsider the logarithmically convexity of the function qα,β(t) by a very simpler approach than that in . The second main result of ours is the following Theorem 2. Theorem 2. The function qα,β(t) in (−∞, ∞) is logarithmically convex if β−α > 1 and logarithmically concave if 0 < β −α < 1. Remark 4. Theorem 2 shows that the logarithmically convexity and logarithmically concavity in the interval (−∞, 0) of qα,β(t) presented in and mentioned at the beginning of this paper are wrong. However, this does not aﬀect the correctness of the main results established in , since the wrong properties about qα,β(t) in the interval (−∞, 0) are unuseful there luckily. Remark 5. Recall that a r-times diﬀerentiable function f(x) > 0 is said to be r-log-convex (or r-log-concave) on an interval I with r ≥2 if and only if ln f(x) exists and ln f(x) ≥0 (or ln f(x) ≤0) on I. In , the following conclusions are obtained: If 1 > β −α > 0, then qα,β(t) is 3-log-convex in (0, ∞) and 3-log-concave in (−∞, 0); if β −α > 1, then qα,β(t) is 3-log-concave in (0, ∞) and 3-log-convex in (−∞, 0). MONOTONICITY AND LOGARITHMIC CONVEXITY FOR A CLASS OF FUNCTIONS 5 -α 1 O 6 β 1 @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ β = α β = 1 −α β = α −1 β = α + 1 Figure 4. (α, β)-domain where the function qα,β(t) is decreasing in (−∞, 0) in Theorem 1 2. Proofs of theorems Proof of Theorem 1. It is clear that the function qα,β(t) can be rewritten as qα,β(t) = sinh (β−α)t 2 sinh t 2 exp (1 −α −β)t 2 ≜pα,β t 2 . (2) If α = β + 1, then qα,β(t) = −e−βt is increasing for β > 0 and decreasing for β < 0 in (−∞, ∞). If α = β −1, then qα,β(t) = e−αt is decreasing for α > 0 and increasing for α < 0 in (−∞, ∞). For \|α −β\| ̸= 1, direct diﬀerentiation shows p′ α,β(t) = sinh((β −α)t) sinh t e(1−α−β)tϕα,β(t), where ϕα,β(t) = (β −α) coth((β −α)t) −coth t −α −β + 1 (3) and ϕ′ α,β(t) = 1 sinh t 2 − β −α sinh((β −α)t) 2 = 1 t2 t sinh t 2 − (β −α)t sinh((β −α)t) 2 . (4) 6 F. QI -α 1 O 6 β 1 β = α β = α −1 β = α + 1 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` Figure 5. (α, β)-domain where the function qα,β(t) is increasing in (−∞, ∞) in Theorem 1 and Corollary 1 Since ϕ′ α,β(t) = ϕ′ α,β(−t) and the function t sinh t > 0 is decreasing in (0, ∞) and increasing in (−∞, 0), then ϕ′ α,β(t) ≥0 for \|α −β\| > 1 and ϕ′ α,β(t) ≤0 for 0 < \|α −β\| < 1 in (−∞, ∞). This means that the function ϕα,β(t) is increasing for \|α −β\| > 1 and decreasing for 0 < \|α −β\| < 1 in (−∞, ∞). It is not diﬃcult to obtain limt→−∞ϕα,β(t) = 2 −α −β −\|α −β\|, limt→0 ϕα,β(t) = 1 −α −β and limt→∞ϕα,β(t) = \|α −β\| −α −β. 1. If β > α + 1, then β −α > 0, \|α −β\| > 1, limt→−∞ϕα,β(t) = 2(1 −β) and limt→∞ϕα,β(t) = −2α. Further, if α ≥0, then ϕα,β(t) < 0 and p′ α,β(t) < 0 in (−∞, ∞), and then pα,β(t) and qα,β(t) are decreasing in (−∞, ∞). Therefore, for β > α + 1 ≥1, the function qα,β(t) is decreasing in (−∞, ∞). If β > α + 1 and β ≤1, then limt→−∞ϕα,β(t) ≥0, ϕα,β(t) > 0 and p′ α,β(t) > 0 in (−∞, ∞), and then pα,β(t) and qα,β(t) are increasing in (−∞, ∞). Hence, for 1 ≥β > α + 1, the function qα,β(t) is increasing in (−∞, ∞). If β > α+1 and α+β ≤1, then limt→0 ϕα,β(t) ≥0, ϕα,β(t) > 0 and p′ α,β(t) > 0 in (0, ∞), and then pα,β(t) and qα,β(t) are increasing in (0, ∞). Consequently, for 2α + 1 < α + β ≤1, the function qα,β(t) is increasing in (0, ∞). If β > α+1 and α+β ≥1, then limt→0 ϕα,β(t) ≤0, ϕα,β(t) < 0 and p′ α,β(t) < 0 in (−∞, 0), and then pα,β(t) and qα,β(t) are decreasing in (−∞, 0). Therefore, for 2β > α + β + 1 ≥2, the function qα,β(t) is decreasing in (−∞, 0). 2. If α < β < α + 1, then β −α > 0 and \|α −β\| < 1. Further, if α ≤0, then ϕα,β(t) > 0 and p′ α,β(t) > 0 in (−∞, ∞), and then pα,β(t) and qα,β(t) are increasing in (−∞, ∞). Accordingly, for α < β < α + 1 ≤1, the function qα,β(t) is increasing in (−∞, ∞). MONOTONICITY AND LOGARITHMIC CONVEXITY FOR A CLASS OF FUNCTIONS 7 -α 1 O 6 β 1 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` β = α β = α −1 β = α + 1 ``````````````````````````````````````````````````````````` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` Figure 6. (α, β)-domain where the function qα,β(t) is decreasing in (−∞, ∞) in Theorem 1 and Corollary 1 If α < β < α + 1 and β ≥1, then limt→−∞ϕα,β(t) ≤0, ϕα,β(t) < 0 and p′ α,β(t) < 0 in (−∞, ∞), and then pα,β(t) and qα,β(t) are decreasing in (−∞, ∞). Therefore, for α + 1 ≤α + β < 2β < α + β + 1, the function qα,β(t) is decreasing in (−∞, ∞). If α < β < α + 1 and α + β ≤1, then limt→0 ϕα,β(t) ≥0, ϕα,β(t) > 0 and p′ α,β(t) > 0 in (−∞, 0), and then pα,β(t) and qα,β(t) are increasing in (−∞, 0). As a result, for α + β < 2β < α + β + 1 ≤2, the function qα,β(t) is increasing in (−∞, 0). If α < β < α + 1 and α + β ≥1, then limt→0 ϕα,β(t) ≤0, ϕα,β(t) < 0 and p′ α,β(t) < 0 in (0, ∞), and then pα,β(t) and qα,β(t) are decreasing in (0, ∞). Con-sequently, for 1 ≤α + β < 2β < α + β + 1, the function qα,β(t) is decreasing in (0, ∞). 3. If α > β + 1, then β −α < 0, \|α −β\| > 1, limt→−∞ϕα,β(t) = 2(1 −α) and limt→∞ϕα,β(t) = −2β. Further, if β ≥0, then ϕα,β(t) < 0 and p′ α,β(t) > 0 in (−∞, ∞), and then pα,β(t) and qα,β(t) are increasing in (−∞, ∞). Therefore, for α > β + 1 ≥1, the function qα,β(t) is increasing in (−∞, ∞). 8 F. QI -α 1 O 6 β 1 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` β = α β = 1 −α β = α −1 @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` β = α + 1 Figure 7. (α, β)-domain where the function qα,β(t) is increasing in (0, ∞) in Theorem 1 If α > β + 1 and α ≤1, then limt→−∞ϕα,β(t) ≥0, ϕα,β(t) > 0 and p′ α,β(t) < 0 in (−∞, ∞), and then pα,β(t) and qα,β(t) are decreasing in (−∞, ∞). Hence, for 1 ≥α > β + 1, the function qα,β(t) is decreasing in (−∞, ∞). If α > β +1 and α+β ≤1, then limt→0 ϕα,β(t) ≥0, ϕα,β(t) > 0 and p′ α,β(t) < 0 in (0, ∞), and then pα,β(t) and qα,β(t) are decreasing in (0, ∞). Accordingly, for 1 ≥α + β > 2β + 1, the function qα,β(t) is decreasing in (0, ∞). If α > β +1 and α+β ≥1, then limt→0 ϕα,β(t) ≤0, ϕα,β(t) < 0 and p′ α,β(t) > 0 in (−∞, 0), and then pα,β(t) and qα,β(t) are increasing in (−∞, 0). Hence, for 2α > α + β + 1 ≥2, the function qα,β(t) is increasing in (−∞, 0). 4. If β < α < β + 1, then β −α < 0 and \|α −β\| < 1. Further, if β ≤0, then ϕα,β(t) > 0 and p′ α,β(t) < 0 in (−∞, ∞), and then pα,β(t) and qα,β(t) are decreasing in (−∞, ∞). Therefore, for β < α < β + 1 ≤1, the function qα,β(t) is decreasing in (−∞, ∞). If β < α < β + 1 and α ≥1, then limt→−∞ϕα,β(t) ≤0, ϕα,β(t) < 0 and p′ α,β(t) > 0 in (−∞, ∞), and then pα,β(t) and qα,β(t) are increasing in (−∞, ∞). Accordingly, for β + 1 ≤α + β < 2α < α + β + 1, the function qα,β(t) is increasing in (−∞, ∞). If β < α < β + 1 and α + β ≤1, then limt→0 ϕα,β(t) ≥0, ϕα,β(t) > 0 and p′ α,β(t) < 0 in (−∞, 0), and then pα,β(t) and qα,β(t) are decreasing in (−∞, 0). Consequently, for α + β < 2α < α + β + 1 ≤2, the function qα,β(t) is decreasing in (−∞, 0). If β < α < β + 1 and α + β ≥1, then limt→0 ϕα,β(t) ≤0, ϕα,β(t) < 0 and p′ α,β(t) > 0 in (0, ∞), and then pα,β(t) and qα,β(t) are increasing in (0, ∞). As a MONOTONICITY AND LOGARITHMIC CONVEXITY FOR A CLASS OF FUNCTIONS 9 -α 1 O 6 β 1 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` β = α β = 1 −α β = α −1 β = α + 1 @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ Figure 8. (α, β)-domain where the function qα,β(t) is decreasing in (0, ∞) in Theorem 1 result, for 1 ≤α + β < 2α < α + β + 1, the function qα,β(t) is increasing in (0, ∞). The proof of Theorem 1 is complete. □ Proof of Theorem 2. For β > α, the functions qα,β(t) and pα,β(t), related by (2), are positive. Taking logarithm of pα,β(t) and diﬀerentiating yields ln pα,β(t) = ln sinh((β −α)t) −ln sinh t + (1 −α −β)t, [ln pα,β(t)]′ = (β −α) coth((β −α)t) −coth t −α −β + 1 = ϕα,β(t), where ϕα,β(t) is deﬁned by (3). By the same argument as in the proof of Theorem 1 on page 5, it is easy to see that ϕ′ α,β(t) = [ln pα,β(t)]′′ ≥0 for β −α > 1 and ϕ′ α,β(t) = [pα,β(t)]′′ ≤0 for 0 < β −α < 1 in (−∞, ∞). This means that the function pα,β(t) = qα,β(2t) is logarithmically convex for β −α > 1 and logarithmically concave for 0 < β −α < 1 in the whole axis (−∞, ∞). The proof of Theorem 2 is complete. □ References C. Berg, Integral representation of some functions related to the gamma function, Mediterr. J. Math. 1 (2004), no. 4, 433–439. 10 F. QI -α 1 O 6 β 1 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` β = α β = α −1 β = α + 1 @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ β = 1 −α Figure 9. (α, β)-domain where the function qα,β(t) is increasing in (−∞, 0) in Theorem 1 A. Z. Grinshpan and M. E. H. Ismail, Completely monotonic functions involving the gamma and q-gamma functions, Proc. Amer. Math. Soc. 134 (2006), no. 4, 1153–1160. F. Qi, Certain logarithmically N-alternating monotonic functions involving gamma and q-gamma functions, RGMIA Res. Rep. Coll. 8 (2005), no. 3, Art. 5. Available online at F. Qi, Three-log-convexity for a class of elementary functions involving exponential function, J. Math. Anal. Approx. Theory 1 (2006), no. 2, in press. F. Qi and Ch.-P. Chen, A complete monotonicity property of the gamma function, J. Math. Anal. Appl. 296 (2004), no. 2, 603–607. F. Qi and B.-N. Guo, Complete monotonicities of functions involving the gamma and digamma functions, RGMIA Res. Rep. Coll. 7 (2004), no. 1, Art. 8, 63–72. Available online at http: //rgmia.vu.edu.au/v7n1.html. F. Qi, B.-N. Guo, and Ch.-P. Chen, Some completely monotonic functions involving the gamma and polygamma functions, RGMIA Res. Rep. Coll. 7 (2004), no. 1, Art. 5, 31–36. Available online at F. Qi, B.-N. Guo, and Ch.-P. Chen, Some completely monotonic functions involving the gamma and polygamma functions, J. Austral. Math. Soc. 80 (2006), 81–88. F. Qi, B.-N. Guo, and Ch.-P. Chen, The best bounds in Gautschi-Kershaw inequalities, Math. Inequal. Appl. (2006), in press. RGMIA Res. Rep. Coll. 8 (2005), no. 2, Art. 17. Available online at (F. Qi) Research Institute of Mathematical Inequality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China E-mail address: qifeng@hpu.edu.cn, fengqi618@member.ams.org, qifeng618@hotmail.com, qifeng618@msn.com, 316020821@qq.com URL: MONOTONICITY AND LOGARITHMIC CONVEXITY FOR A CLASS OF FUNCTIONS 11 -α 1 O 6 β 1 β = α β = α −1 β = α + 1 ``````````````````````````````````````````````````````````` @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ β = 1 −α Figure 10. (α, β)-domain where the function qα,β(t) is decreasing in (−∞, 0) in Theorem 1
15466	https://byjus.com/physics/gold-leaf-electroscope/	Electric charge is the basic physical property of matter that causes it to experience a force when kept in an electric or magnetic field. An electric charge is associated with an electric field, and the moving electric charge generates a magnetic field. Electric charges are of two types: Positive and Negative, commonly carried by charge carriers protons and electrons. In this article, let us familiarise ourselves with a device called Gold Leaf Electroscope, which is used to detect the charge in a body. \| \| \| Table of Contents What Is a Gold Leaf Electroscope? Construction of Gold Leaf Electroscope Applications of Gold Leaf Electroscope + Detect Charge + Identification of the Nature of the Charge + Identification of Body as a Conductor or an Insulator Frequently Asked Questions – FAQs \| What Is a Gold Leaf Electroscope? A gold-leaf electroscope is defined as A type of electroscope that consists of two gold leaves and is used for detecting the electrical charge of the body and for the classification of its polarity. There are two types of electroscope: Gold leaf electroscope Pith ball electroscope Abraham Bennet invented the gold-leaf electroscope in the year 1787. It is said that the gold leaf electroscope is more sensitive than the pith ball electroscope. Read More: Types of Electroscope Construction of Gold Leaf Electroscope The gold leaf electroscope is a sensitive electroscope type that is used for detecting charges. It consists of a brass rod with a brass disk at the top, and at the bottom, there are two thin gold leaves in the form of foils. In order to keep the rod in place, the rod travels through the insulator. The charges move from the disk to the leaves through the rod. At the lower portion of the jar, a thin aluminium foil is connected. The aluminium foil is grounded with the help of a copper wire so that the leaves are protected from external electrical disruptions. Applications of Gold Leaf Electroscope The following are the applications of gold leaf electroscope: Detect charge Identification of the nature of the charge Identification of the body as a conductor or an insulator Detect Charge For the detection of charge, the object that needs to be tested is touched with the metal cap. If the leaves diverge, the body is said to be charged, and if there is no change in the leaves of the electroscope, then the body is uncharged. Identification of the Nature of the Charge To identify the nature of the charge, let’s consider an example. A positively charged body is brought near the metal cap. Then an unknown body is brought near the metal cap. If the leaves diverge further, we can conclude that the unknown body has a positive charge. If the leaves come closer to each other, then the charge of the unknown body is negative. Identification of Body as a Conductor or an Insulator To identify if a body is a conductor or an insulator, two gold leaf electroscopes are taken. One gold leaf electroscope is charged so that the leaves will diverge. Then the other gold leaf electroscope is connected to the first one. If the leaves of the other electroscope diverge, then the body is said to be a conductor, and if there is no change in the leaves, the body is said to be an insulator. Frequently Asked Questions – FAQs Q1 What is the name of the first electroscope? The name of the first electroscope is Versorium. This was a pivoted needle-like electroscope invented by William Gilbert, who was a British physicist, in the year 1600. Q2 Why is a gold leaf electroscope enclosed in a glass case? A gold leaf electroscope is enclosed in a glass case so that the gold leaves are protected from the air. This also helps in capturing the charge leak through the air so that the sensitivity of the instrument can be increased. Q3 What does the degree of divergence of gold leaves mean in a gold leaf electroscope? The degree of divergence of gold leaves in a gold leaf electroscope is an indicator of the amount of charge that is transferred to the gold leaves. Q4 Why do the gold leaves in an electroscope fold back when touched with hands? The gold leaves in an electroscope fold back when they are touched with hands because the charge is earthed. Q5 What are the three methods for charging objects? The 3 methods for charging objects are friction, conduction, and induction. Stay tuned to BYJU’S to learn more about other concepts of Physics. Test your Knowledge on Gold Leaf Electroscope Q5 Put your understanding of this concept to test by answering a few MCQs. Click Start Quiz to begin! Select the correct answer and click on the "Finish" buttonCheck your score and explanations at the end of the quiz Congrats! Visit BYJU'S for all Physics related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
15467	https://stackoverflow.com/questions/20580028/flowchart-for-each-loop-loop-without-variable-increment	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Flowchart "for each" loop loop without variable increment Ask Question Asked Modified 1 year, 4 months ago Viewed 108k times 54 I'm designed a flowchart to describe at a high level how a programming process works. Part of this process involves looping through a set of items. I'm wondering if there is any standard or semi-standard way of representing a "for each" style loop in a flow chart, that does not involve making the iteration explicit with an iteration box like m = m + 1 (e.g. here). Most modern programming languages have some kind of "for each" construct for enumerating a set or sequence of items, without having to think about indices. I'm basically looking for a similar visual construct for a flow chart, to avoid wasting space with an explicit counter increment. flowchart Share Improve this question asked Dec 14, 2013 at 5:33 Bryce ThomasBryce Thomas 10.8k2626 gold badges7979 silver badges130130 bronze badges 2 2 I don't think there is a flowchart specifically designed for for..each loop since it was designed before such concept began. However, you could probably represent it same as the regular for loop, however instead of the standard increment say i=i+1, it would be Get the next Item of the Collection. Edper – Edper 2013-12-14 05:42:48 +00:00 Commented Dec 14, 2013 at 5:42 @Edper R is a syntactic clone of S. When you date its for construction you should back date it to the era when S was designed. IRTFM – IRTFM 2025-09-16 04:20:13 +00:00 Commented Sep 16 at 4:20 Add a comment \| 7 Answers 7 Reset to default 49 Sooo I know this topic is now 3 years old, but it might help others. I found a little trick to represent a "for each" loop in a UML flowchart. I don't think it is standart, though I find it quite instinctive. Here it is : Share Improve this answer edited Feb 27, 2020 at 22:57 Phil 166k2525 gold badges265265 silver badges269269 bronze badges answered Mar 23, 2017 at 11:00 user21715user21715 61155 silver badges22 bronze badges 2 Comments jfritz42 jfritz42 Too bad the image is not directly visible/embedded on Stackoverflow, but I like this answer best for its simplicity. Stewart Stewart I would imagine that, in any standard notation for this, the entry and exit of each loop iteration would be explicitly marked somehow. One idea I can see to do this is to add arrows connecting the "do this" box to the edge of the "for each element in array" box, maybe at the point where the arrows connected to that box meet it. 29 I had the same question and found the answer here. Share Improve this answer edited Sep 20, 2017 at 22:02 Nathan Smith 68111 gold badge1010 silver badges2424 bronze badges answered Jul 7, 2016 at 22:35 Ray HulhaRay Hulha 11.3k55 gold badges5757 silver badges5757 bronze badges 1 Comment Stewart Stewart That looks to me like an explicit variable increment, which is what the asker is trying to avoid. 3 Mendix is a rapid application development platform where most of the logic resides in so-called 'microflows', which are represented by flowcharts. Here is an example how it represents a 'for each' loop: It's similar to @user21715's answer, but it uses the same 'parameter' pentagon which is also used to denote input parameters for a flow, with a 'loop' icon. The small gray circle is the end of the loop iteration (similar to continue in most languages); they have an orange one for a break (a premature end of the loop). The first and last statements in a loop are recognizable by having no incoming resp. outgoing arrow. Share Improve this answer edited Oct 9, 2020 at 18:05 answered Jul 24, 2019 at 8:48 GlorfindelGlorfindel 22.8k1313 gold badges9696 silver badges124124 bronze badges Comments 1 Here's an example I found that seems pretty intuitive. I have no idea if this is a standard practice, but it looks good to me. Sorry about the resolution. The important parts are that the parallelogram on the left says "Listof numbers", the diamond in the middle says "For each", the arrow going down and to the right of the diamond says "Each Number", and the arrow going straight down from the diamond says "End of list" Share Improve this answer edited Nov 14, 2019 at 21:26 answered Jun 5, 2018 at 19:06 NateWNateW 99411 gold badge88 silver badges2828 bronze badges 3 Comments NateW NateW Hmm... Looks like the image I linked to has moved. Unfortunately, I don't really remember what it looked like. I'll see if I can find it again somewhere and if not, I'll delete my answer. NateW NateW The original link (svgur.com/s/zA) shows an X in a circle now. However, fortunately, it seems that google image search has cached the thumbnail of the original image, so I've just posted a screenshot of that thumbnail. It's lower resolution, but it gets the point across. Stewart Stewart What is the parallelogram on the right doing? Furthermore, why do this and two of the rectangles have two unlabelled outgoing arrows? 1 I prefer the "bread-loaf" convention, especially if one has to show lots of loops. Adjust wording as needed. I'm not sure if it's an official symbol anywhere, but it could be adopted as a shop convention regardless. Share Improve this answer edited Dec 21, 2023 at 20:49 answered Dec 20, 2023 at 22:33 FloverOweFloverOwe 37422 silver badges1212 bronze badges 2 Comments Stewart Stewart In your scheme, how do you represent multiple actions performed within a single foreach loop? FloverOwe FloverOwe To put multiple tasks inside the loop, one could stack in more adjacent rectangles below. Or just have a numbered or bulleted list. If and when it gets complex, such as conditionals within, then reverting to the original "long cut" may be in order. I've pondered other variants on the idea, but this is probably not the proper venue. 0 I've done that by setting an initial variable N=numberOfItems, and the rest is history ;) (i.e. a conventional "for" loop). Share Improve this answer answered Feb 27, 2014 at 2:29 antonimmoantonimmo 40355 silver badges88 bronze badges Comments 0 I really liked @FloverOwe 's suggestion, but, since it isn't a standard shape, I couldn't use it in Visio. I found that there are two standard shapes in Visio that I can add together and clearly indicate the concept of for each. I just rename the arrow to be what set is to be iterated over. Share Improve this answer answered May 20, 2024 at 12:33 cjbarthcjbarth 4,56577 gold badges4747 silver badges6767 bronze badges 1 Comment Stewart Stewart You can define your own shapes in Visio. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions flowchart See similar questions with these tags. The Overflow Blog The history and future of software development (part 1) Getting Backstage in front of a shifting dev experience Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure New and improved coding challenges New comment UI experiment graduation Policy: Generative AI (e.g., ChatGPT) is banned Linked 1 How to use a "Loop limit" symbol in a flowchart representing a simple R "for loop" Related 2 creating a recursive flowchart in visio with c# 0 Graphs in a loop in SPSS 0 Flowchart to loop an array 0 Generating a chart (python loop) 2 Showing nested for loops in a flowchart 0 R: Variable values in DiagrammeR flowchart How to increment the counter inside a Liquid for loop? 0 chart series data in first iteration of for loop vba 0 Creating a flowchart for an algorithm which receives values periodically 0 For Loop to build charts Hot Network Questions Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation What’s the usual way to apply for a Saudi business visa from the UAE? 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15468	https://www.khanacademy.org/python-program/testing/5702664022245376	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
15469	https://math.stackexchange.com/questions/1105602/system-of-differential-equations-pure-imaginary-eigenvalues-show-that-the-traj	System of differential equations, pure imaginary eigenvalues, show that the trajectory is an ellipse. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more System of differential equations, pure imaginary eigenvalues, show that the trajectory is an ellipse. Ask Question Asked 10 years, 8 months ago Modified10 years, 8 months ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I am stuck at the last part of a proof. When you have the system of equations: x′=A x x′=A x A=[a 11 a 21 a 12 a 22]A=[a 11 a 12 a 21 a 22] Show that when you have purely imaginary eigenvalues the trajectories in the phase plane x1 and x2 is an ellipse. First step When the eigenvalues are purely imaginary a 11+a 22=0 a 11+a 22=0 and a 11 a 22−a 12 a 21>0 a 11 a 22−a 12 a 21>0 this was easy to proof. Second step The trajectories can be written as the following expression: d y d x=a 21 x+a 22 y a 11 x+a 12 y d y d x=a 21 x+a 22 y a 11 x+a 12 y Now this expression is an exact differential equation when a 22=−a 11 a 22=−a 11 This was shown in the first step. So I could just fill this in and this equation is exact. So now I solved it giving the next equation: a 21 x 2+2 a 22 x y−a 12 y 2=c o n s t a n t a 21 x 2+2 a 22 x y−a 12 y 2=c o n s t a n t Now the last step is to show that this equation is an ellipse when the eigenvalues are purely imaginary. For some reason, I can't find a proof for this. I know the main form of an ellipse in any direction orientated is: A x 2−B x y+C y 2=c o n s t a n t A x 2−B x y+C y 2=c o n s t a n t Now, how can I proof that given that the eigenvalues of the matrix we started with are purely imaginary, we get an ellipse here ? Thank you very much. ordinary-differential-equations eigenvalues-eigenvectors Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 16, 2015 at 4:09 user856 asked Jan 15, 2015 at 17:32 ThomasThomas 97 1 1 silver badge 9 9 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. I recommend this site Essentially, the general equation for any conic section is A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 for some real numbers A,B,C,D,E,F A,B,C,D,E,F. However, not all choices of coefficients will yield a conic section, and from we have that the equation above is an ellipse precisely when the discriminant B 2−4 A C<0 B 2−4 A C<0. Proving this is probably fairly straightforward, albeit tedious. In the first link, the author proves that the discriminant is invariant under rotation of coordinates. In this problem you have, we have that det(λ I−A)=λ 2−t r a c e(A)λ+det(A),det(λ I−A)=λ 2−t r a c e(A)λ+det(A), and so in order to have purely imaginary coefficients, we need t r a c e(A)=0 t r a c e(A)=0 and det(A)=a 22 a 11−a 12 a 21>0 det(A)=a 22 a 11−a 12 a 21>0. From your equation for the trajectories: a 21 x 2+2 a 22 x y−a 12 y 2=c o n s t a n t,a 21 x 2+2 a 22 x y−a 12 y 2=c o n s t a n t, we see that the discriminant is (2 a 22)2−4 a 21(−a 12)=4 a 2 22+4 a 12 a 21=−4 a 22 a 11+4 a 12 a 22=−4 det(A).(2 a 22)2−4 a 21(−a 12)=4 a 22 2+4 a 12 a 21=−4 a 22 a 11+4 a 12 a 22=−4 det(A). Since det(A)>0 det(A)>0, the discriminant is negative, and hence the curve is an ellipse. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 16, 2015 at 3:40 Mark HubenthalMark Hubenthal 884 5 5 silver badges 8 8 bronze badges 1 Thank you, but in my course text of algebra, an ellipse was not defined with the D,E and F. So I wasn't paying attention to that.Thomas –Thomas 2015-01-16 11:37:14 +00:00 Commented Jan 16, 2015 at 11:37 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions ordinary-differential-equations eigenvalues-eigenvectors See similar questions with these tags. 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15470	https://www.cuemath.com/numbers/divisibility-rule-of-5/	LearnPracticeDownload Divisibility Rule of 5 According to the divisibility rule of 5 if the last digit of a number is either 0 or 5 then it is said to be divisible by 5. A number is said to be completely divisible by 5 if the remainder is zero and the quotient is a whole number. The divisibility rule of 5 is quite simple because we just need to check the last digit of the given number which should be either 5 or 0. If it satisfies this condition, then the number is considered to be divisible by 5. \| \| \| --- \| \| 1. \| What is the Divisibility Rule of 5? \| \| 2. \| Divisibility Rule of 5 for Large Numbers \| \| 3. \| Divisibility Rule of 5 and 10 \| \| 4. \| Divisibility Test of 5 and 6 \| \| 5. \| Divisibility Rule of 5 Examples \| \| 6. \| FAQs on Divisibility Rule of 5 \| What is the Divisibility Rule of 5? The divisibility rule of 5 states that if the digit on the units place, that is, the last digit of a given number is 5 or 0, then such a number is divisible by 5. For example, in 39865, the last digit is 5, hence, the number is completely divisible by 5. Similarly, in 3780, the last digit is 0, therefore, 3780 is considered to be divisible by 5. Divisibility Rule of 5 for Large Numbers Divisibility rules can easily check if a number can be exactly divided by another. It is specifically used for larger numbers, where the result can be known without performing the process of division. According to the divisibility rule of 5, numbers that are divisible by 5 should end with either 5 or 0. For example, in 347835, the digit on the units place is 5, hence, we know that the given number 347835 is divisible by 5. Let us take another example of 894720. In this case, the digit on the units place is 0, hence, the number is divisible by 5. Divisibility Rule of 5 and 10 As we discussed that the divisibility by 5 can easily be checked by observing the last digit of the given number which should either be 5 or 0. Similarly, the divisibility rule of 10 states that if the digit on the units place of a given number is 0, then the number is considered to be divisible by 10. For example, let us check if the number 73540 is divisible by 5 and 10. The last digit in the given number is 0, hence, 73540 is divisible by 5. The same condition is satisfied for the divisibility rule of 10 since the last digit is zero. Therefore, 73540 is divisible by 5 as well as 10. Let us take another example of the number 10105. In this case, the last digit is 5, therefore, it is divisible by 5, but not divisible by 10 since the last digit is not zero. Divisibility Test of 5 and 6 While the divisibility test of 5 is simple in which the digit on the units place is considered, the divisibility test of 6 is different. A number is divisible by 6 if it is a multiple of both 2 and 3. In other words, if the last digit of a number is even and the sum of all its digits is a multiple of 3, then the number is considered to be divisible by 6. Let us check if 61704 is divisible by 5 as well as 6. Since the last digit is neither 5 nor 0, the number is not divisible by 5. Now, let us check its divisibility by 6 with the help of the following steps: Step 1: Check if the last digit of the given number, 61704, is either 0 or a multiple of 2. In this case, it is 4, hence, it satisfies the condition of being a multiple of 2. Step 2: After this, calculate the sum of all digits of the given number. Here, 6 + 1 + 7 + 0 + 4 = 18. So, it satisfies the condition of being a multiple of 3, hence 18 is divisible by 3. Step 3: Since the given number 61704 is divisible by 2 and 3 both, it passes the divisibility test of 6. In other words, 61704 is divisible by 6. Numbers Divisible by 5 There is an infinite list of the numbers that can be completely divisible by 5. However, the simple rule of the divisibility by 5 says that if the last digit of the given number is either 0 or 5 then it is said to be divisible by 5. For example, starting from 5, 10, 15, 20, 25, 30, 35, and so on are a few numbers that are divisible by 5. These can also be called as the multiples of 5. ☛ Related Topics Divisibility Rule of 3 Divisibility Rule of 4 Divisibility Rule of 6 Divisibility Rule of 7 Divisibility Rule of 8 Divisibility Rule of 9 Divisibility Rule of 11 Divisibility Rule of 13 Read More Download FREE Study Materials Related Worksheet Divisibility Rules Worksheet Worksheet on Divisibility Rules Divisibility Rule of 5 Examples Example 1: Using the divisibility rule of 5, check whether 645 is divisible by 5 or not. Solution: The divisibility rule of 5 states that the last digit of the given number should be either 0 or 5. In 645, the last digit is 5. Hence, 645 is divisible by 5. 2. Example 2: Observe the following statements and write true or false using the divisibility rule of 5. a) 503 is divisible by 5. b) 16070 is divisible by 5. c) 43875 is not divisible by 5. Solution: As per the divisibility test of 5, the last digit of the given numbers should either be 0 or 5 to consider them to be divisible by 5. a.) False. In 503, the last digit is 3. Therefore, 503 is not divisible by 5. b.) True. In 16070, the last digit is 0. Therefore, 16070 is divisible by 5. c.) False. In 43875, the last digit is 5. Therefore, 43875 is divisible by 5. 3. Example 3: Check the divisibility of the following numbers by 5. a) 607 b) 6320 c) 21003 Solution: Using the divisibility rule of 5, let us check if the last digit of the given number is 0 or 5. a.) In 607, the last digit is neither 5 nor 0. Therefore, 607 is not divisible by 5. b.) In 6320, the last digit is 0. Therefore, 6320 is divisible by 5. c.) In 21003, the last digit is neither 5 nor 0. Therefore, 21003 is not divisible by 5. View Answer > Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class Practice Questions on Divisibility Rule of 5 Check Answer > FAQs on Divisibility Rule of 5 What is the Divisibility Rule of 5? According to the divisibility rule of 5, if the last digit of the given number is either 0 or 5, then the number is divisible by 5. For example, in 6735, the last digit is 5, therefore, the number 6735 is completely divisible by 5. Using the Divisibility Rule of 5, Check if 660 is Divisible by 5. The divisibility rule of 5 says that the last digit of the given number should be zero or 5. In this case, we can see that the last digit of 660 is 0. Therefore, 660 is divisible by 5. What is the Divisibility Rule of 5 and 6? The divisibility rule of 5 states that if the last digit of the given number is either 0 or 5, then the given number is divisible by 5. For example, in 615, the last digit is 5, therefore, 615 is said to be divisible by 5. The divisibility rule of 6 says that a number is divisible by 6 if it is divisible by 2 and 3 both. For example, in 312, the last digit is an even number which is divisible by 2, and the sum of the digits is 3 + 1 + 2 = 6, which is divisible by 3. Hence, it can be concluded that 312 is completely divisible by 6. Using the Divisibility Test of 5, Check if 250 is Divisible by 5. Using the divisibility test of 5, we can see that the last digit of 250 is 0. This means that 250 is divisible by 5. How do you know if a Big Number is Divisible by 5? It is easy to check the divisibility for smaller numbers. However, in order to check the divisibility for large numbers, we need to apply the given set of rules for the divisibility of a number. If the last digit of a given number is either 0 or 5, then the number is considered to be divisible by 5. For example, to check if 61,215 is divisible by 5, we check the last digit of the given number. Here, it is 5, therefore, we can say that 61,215 is divisible by 5. 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15471	https://openstax.org/books/precalculus-2e/pages/3-key-concepts	Skip to ContentGo to accessibility pageKeyboard shortcuts menu Precalculus 2e Key Concepts Precalculus 2eKey Concepts Search for key terms or text. Key Concepts ## 3.1 Complex Numbers The square root of any negative number can be written as a multiple of See Example 1. To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. See Example 2. Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See Example 3. Complex numbers can be multiplied and divided. To multiply complex numbers, distribute just as with polynomials. See Example 4, Example 5, and Example 8. To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator. See Example 6, Example 7, and Example 9. The powers of are cyclic, repeating every fourth one. See Example 10. ## 3.2 Quadratic Functions A polynomial function of degree two is called a quadratic function. The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down. The axis of symmetry is the vertical line passing through the vertex. The zeros, or intercepts, are the points at which the parabola crosses the axis. The intercept is the point at which the parabola crosses the axis. See Example 1, Example 7, and Example 8. Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph. See Example 2. The vertex can be found from an equation representing a quadratic function. See Example 3. The domain of a quadratic function is all real numbers. The range varies with the function. See Example 4. A quadratic function’s minimum or maximum value is given by the value of the vertex. The minimum or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue. See Example 5 and Example 6. Some quadratic equations must be solved by using the quadratic formula. See Example 9. The vertex and the intercepts can be identified and interpreted to solve real-world problems. See Example 10. ## 3.3 Power Functions and Polynomial Functions A power function is a variable base raised to a number power. See Example 1. The behavior of a graph as the input decreases beyond bound and increases beyond bound is called the end behavior. The end behavior depends on whether the power is even or odd. See Example 2 and Example 3. A polynomial function is the sum of terms, each of which consists of a transformed power function with positive whole number power. See Example 4. The degree of a polynomial function is the highest power of the variable that occurs in a polynomial. The term containing the highest power of the variable is called the leading term. The coefficient of the leading term is called the leading coefficient. See Example 5. The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function. See Example 6 and Example 7. A polynomial of degree will have at most x-intercepts and at most turning points. See Example 8, Example 9, Example 10, Example 11, and Example 12. ## 3.4 Graphs of Polynomial Functions Polynomial functions of degree 2 or more are smooth, continuous functions. See Example 1. To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero. See Example 2, Example 3, and Example 4. Another way to find the intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the axis. See Example 5. The multiplicity of a zero determines how the graph behaves at the intercepts. See Example 6. The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity. The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity. The end behavior of a polynomial function depends on the leading term. The graph of a polynomial function changes direction at its turning points. A polynomial function of degree has at most turning points. See Example 7. To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most turning points. See Example 8 and Example 10. Graphing a polynomial function helps to estimate local and global extremas. See Example 11. The Intermediate Value Theorem tells us that if have opposite signs, then there exists at least one value between and for which See Example 9. ## 3.5 Dividing Polynomials Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree. See Example 1 and Example 2. The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the quotient added to the remainder. Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form See Example 3, Example 4, and Example 5. Polynomial division can be used to solve application problems, including area and volume. See Example 6. ## 3.6 Zeros of Polynomial Functions To find determine the remainder of the polynomial when it is divided by See Example 1. is a zero of if and only if is a factor of See Example 2. Each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading coefficient. See Example 3 and Example 4. When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. Synthetic division can be used to find the zeros of a polynomial function. See Example 5. According to the Fundamental Theorem, every polynomial function has at least one complex zero. See Example 6. Every polynomial function with degree greater than 0 has at least one complex zero. Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form where is a complex number. See Example 7. The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer. The number of negative real zeros of a polynomial function is either the number of sign changes of or less than the number of sign changes by an even integer. See Example 8. Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division. See Example 9. ## 3.7 Rational Functions We can use arrow notation to describe local behavior and end behavior of the toolkit functions and See Example 1. A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote. See Example 2. Application problems involving rates and concentrations often involve rational functions. See Example 3. The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. See Example 4. The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero. See Example 5. A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero. See Example 6. A rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. See Example 7, Example 8, Example 9, and Example 10. Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior. See Example 11. If a rational function has x-intercepts at vertical asymptotes at and no then the function can be written in the form See Example 12. ## 3.8 Inverses and Radical Functions The inverse of a quadratic function is a square root function. If is the inverse of a function then is the inverse of the function See Example 1. While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible. See Example 2. To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one. See Example 3 and Example 4. When finding the inverse of a radical function, we need a restriction on the domain of the answer. See Example 5 and Example 7. Inverse and radical and functions can be used to solve application problems. See Example 6 and Example 8. ## 3.9 Modeling Using Variation A relationship where one quantity is a constant multiplied by another quantity is called direct variation. See Example 1. Two variables that are directly proportional to one another will have a constant ratio. A relationship where one quantity is a constant divided by another quantity is called inverse variation. See Example 2. Two variables that are inversely proportional to one another will have a constant multiple. See Example 3. In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation. See Example 4. 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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Chemistry: Atoms First 2e Summary Chemistry: Atoms First 2eSummary Contents Contents Highlights Table of contents Preface 1 Essential Ideas 2 Atoms, Molecules, and Ions 3 Electronic Structure and Periodic Properties of Elements 4 Chemical Bonding and Molecular Geometry 5 Advanced Theories of Bonding 6 Composition of Substances and Solutions 7 Stoichiometry of Chemical Reactions 8 Gases 9 Thermochemistry 10 Liquids and Solids 11 Solutions and Colloids 12 Thermodynamics 13 Fundamental Equilibrium Concepts 14 Acid-Base Equilibria 15 Equilibria of Other Reaction Classes 16 Electrochemistry 17 Kinetics 18 Representative Metals, Metalloids, and Nonmetals 19 Transition Metals and Coordination Chemistry Introduction 19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds 19.2 Coordination Chemistry of Transition Metals 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds Key Terms Summary Exercises 20 Nuclear Chemistry 21 Organic Chemistry A \| The Periodic Table B \| Essential Mathematics C \| Units and Conversion Factors D \| Fundamental Physical Constants E \| Water Properties F \| Composition of Commercial Acids and Bases G \| Standard Thermodynamic Properties for Selected Substances H \| Ionization Constants of Weak Acids I \| Ionization Constants of Weak Bases J \| Solubility Products K \| Formation Constants for Complex Ions L \| Standard Electrode (Half-Cell) Potentials M \| Half-Lives for Several Radioactive Isotopes Answer Key Index Search for key terms or text. Close 19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds ------------------------------------------------------------------------------------- The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce. Transition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity. 19.2 Coordination Chemistry of Transition Metals ------------------------------------------------ The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). Cis and trans configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use. 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds -------------------------------------------------------------------- Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal d orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting (Δ oct) depends on the nature of the ligands bonded to the metal. Strong-field ligands produce large splitting and favor low-spin complexes, in which the t 2 g orbitals are completely filled before any electrons occupy the e g orbitals. Weak-field ligands favor formation of high-spin complexes. The t 2 g and the e g orbitals are singly occupied before any are doubly occupied. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Paul Flowers, Edward J. Neth, William R. Robinson, PhD, Klaus Theopold, Richard Langley Publisher/website: OpenStax Book title: Chemistry: Atoms First 2e Publication date: Feb 14, 2019 Location: Houston, Texas Book URL: Section URL: © Jul 9, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . 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15473	https://pmc.ncbi.nlm.nih.gov/articles/PMC7393683/	Non—muscle-invasive Bladder Cancer: Overview and Contemporary Treatment Landscape of Neoadjuvant Chemoablative Therapies - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Search in PMC Search in PubMed View in NLM Catalog Add to search Non—muscle-invasive Bladder Cancer: Overview and Contemporary Treatment Landscape of Neoadjuvant Chemoablative Therapies Richard S Matulewicz Richard S Matulewicz, MD, MS 1 Department of Urology, NYU Grossman School of Medicine, NYU Langone Health, New York, NY 2 Department of Population Health, NYU Grossman School of Medicine, NYU Langone Health, New York, NY Find articles by Richard S Matulewicz 1,2, Gary D Steinberg Gary D Steinberg, MD 1 Department of Urology, NYU Grossman School of Medicine, NYU Langone Health, New York, NY Find articles by Gary D Steinberg 1 Author information Copyright and License information 1 Department of Urology, NYU Grossman School of Medicine, NYU Langone Health, New York, NY 2 Department of Population Health, NYU Grossman School of Medicine, NYU Langone Health, New York, NY © 2020 MedReviews®, LLC PMC Copyright notice PMCID: PMC7393683 PMID: 32760227 Abstract Non—muscle-invasive bladder cancer (NMIBC) is a heterogeneous subclassification of urothelial carcinoma with significant variation in individual risk of recurrence and progression to muscle-invasive disease. Risk stratification by American Urological Association (AUA) and European Association of Urology (EAU) guidelines or by using nomograms/risk calculators developed from clinical trial data can help inform patient treatment decisions but may not accurately classify all patients. Risk-adapted adjuvant (post—transurethral resection of bladder tumor [TURBT]) treatment strategies using intravesical therapies are an important means of balancing disease control with potential adverse effects. Adjuvant intravesical instillation with various chemotherapy agents and bacillus Calmette-Guérin (BCG) is well studied and associated with excellent outcomes for most patients. However, upwards of 40% of patients recur within 2 years and roughly 10% progress to muscle-invasive bladder cancer. Novel approaches and agents that aim to reduce the treatment burden associated with NMIBC are increasingly needed. We review the current landscape of NMIBC as it pertains to the use of and rationale for emerging neoadjuvant chemoablative therapies. Keywords: Non-muscle invasive bladder cancer, Urothelial carcinoma, Bacillus Calmette-Guérin (BCG), Mitomycin There are approximately 80,000 incident cases of bladder cancer each year in the United States, 90% of which are urothelial carcinoma.1 Median age at presentation is 70 years with male predominance. In the United States, chemical carcinogens from tobacco smoke and occupational exposures such as aromatic amines, polycyclic aromatic hydrocarbons (PAHs), and benzenes are the strongest established risk factors but the contributions of other environmental and genetic factors to the development of bladder cancer are increasingly being recognized.2 Patients with bladder cancer generally present with hematuria or voiding symptoms and are diagnosed by cystoscopy with biopsy and/or resection of the tumor(s) and cross-sectional imaging to evaluate for metastases. Bladder cancer is classified according to traditional American Joint Committee on Cancer (AJCC) TNM staging. In the absence of nodal (N stage) or distant metastases (M stage), depth of tumor invasion (T stage) is the most important determination to be made and can be dichotomized based on whether the tumor is invading into or beyond the muscularis propia (muscle-invasive bladder cancer, MIBC) or not (non-muscle-invasive bladder cancer, NMIBC). At presentation, 90% of bladder cancers are confined to the bladder wall layers and of all localized bladder cancers (≤T2), 75% are non-muscle invasive (Tis, Ta, T1).2 Tumors are further classified according to histological grade (low or high). Stratification of NMIBC into risk groups with similar rates of recurrence and progression help guide prognostication, counseling, and treatment recommendations. In this review, we describe the contemporary standards of diagnosis, risk stratification, and treatment of NMIBC with discussion of emerging neoadjuvant chemoablative treatment options. Evaluation of NMIBC Establishing an accurate diagnosis is critical to the proper treatment of bladder cancer. Diagnosis begins with a detailed history focusing on antecedent symptomatology, comorbidities, risk factors, and family history. Exploration of constitutional symptoms and a focused physical examination may uncover findings suggestive of advanced disease. Counseling patients on smoking cessation upon diagnosis can reduce the risk of perioperative complications and may reduce the risk of NMIBC recurrence.3 Office-based cystoscopy visually confirms the presence of bladder cancer which is followed by a transurethral resection of the bladder tumor (TURBT) under anesthesia. During the TURBT, description of tumor size, location, multiplicity, appearance, and associated mucosal changes is essential to staging. Resection of all visible tumor with safe sampling of the underlying detrusor muscle layer should be the goal of all TURBTs and is associated with improved histopathologic accuracy and improved oncologic outcomes. There are established diagnostic, prognostic, and therapeutic benefits to repeat TURBT and it is generally indicated for all high-grade tumors and especially for any initial pathology specimen devoid of detrusor muscle layer.4,5 The presence of metastatic or synchronous upper tract disease is evaluated using conventional cross-sectional imaging with delayed drainage films. CT or MRI urogram is the gold standard of evaluating the upper tracts concomitantly with the abdominopelvic viscera and lymph nodes. Chest imaging is also required to exclude metastatic disease in the lungs and is particularly important among smokers where a second primary malignancy may be present. Novel methods of visualization such as fluorescence cystoscopy with either 5-aminolaevulinic acid (5-ALA) or hexaminolevulinic acid-blue light cystoscopy (Cysview®, Photocure, Inc., Princeton, NJ; HAL-BLC) are increasingly being adopted. Compared with traditional white light cystoscopy (WLC), fluorescence cystoscopy has demonstrated improved sensitivity in detecting carcinoma in situ (CIS) and is able to detect additional primary tumors missed by WLC in roughly 20% of patients.6 Meta-analyses have also demonstrated improvements in recurrence but not progression or mortality rates when comparing HAL-BLC with WLC.7–9 Higher rates of false-positive results with BLC can lower specificity, whereas false-positive rates seen with traditional WLC have not been considerably different, historically.10 Narrow-band imaging is another enhanced cystoscopy technique to help improve the visual contrast between normal mucosal surfaces and tissue that is “hyper vascular” such as cancer, but evidence of effectiveness is limited.11 Urine cytology is a helpful adjunct to tumor visualization and biopsy because some CIS is difficult to visualize with standard WLC. Urine cytology is quite sensitive to high-grade tumors like CIS (0.80) but performs worse with low-grade tumors (0.04–0.30) and use is discouraged among those with low-risk NMIBC and normal cystoscopy.12 Urinary biomarkers continue to be studied as diagnostic adjuncts and as potential alternatives to cystoscopy for either bladder cancer diagnosis or surveillance.13 Six urinary biomarkers are currently approved by the US Food and Drug Administration (FDA), including those which assess nuclear matrix proteins (NMP22® BladderChek® Test, Abbott, Abbott Park, IL), bladder tumor antigen (BTA STAT & TRAK tests, Polymedco, Cortlandt Manor, NY), and those that utilize fluorescence in situ hybridization (UroVysion Bladder Cancer Kit [UroVysion Kit], Abbott) and fluorescent immunohistochemistry (ImmunoCyt/uCyt+ test, Scimedex, Denville, NJ). With sensitivities ranging from 0.57–0.82 and specificities between 0.74–0.88, few have proven accurate enough for routine clinical use in lieu of cystoscopy.14 There are histopathologic challenges to the diagnosis of NMIBC because of the known interobserver variability of pathologic evaluation. Therefore, it is incumbent on the urologist to provide the pathologists with the best possible specimen devoid of cautery or crush injury. A systematic and complete visual resection with either fractioned or en bloc removal that adequately samples the detrusor muscle is the most important step in the diagnosis and treatment algorithm for NMIBC. The presence or absence of variant histology (eg, micropapillary, plasmacytoid, sarcomatoid, etc), as well as factors such as concurrent CIS and lymphovascular invasion are important for risk stratification and may lead to consideration of upfront extirpative treatment with radical cystectomy rather than adjuvant intravesical therapies. Adjuvant intravesical treatments after TURBT have demonstrated reduction in the risk of recurrence and progression compared with TURBT alone. Although generally safe, the use of adjuvant intravesical therapy is based on a risk-adapted strategy that aims to balance reduction of recurrence and progression with side effects, especially systemic, from overtreatment. There are several available risk stratification methods using a variety of clinical and pathological factors that can be used to quantify risk of recurrence and progression. Based on each patients' cancer risk profile after TURBT and considered in the context of their comorbidities, a patient-centered discussion about adjuvant treatment options should be undertaken to guide next steps. Risk Stratification NMIBC is a widely heterogeneous subclassification of bladder cancer and treatment should be approached accordingly. Depending on tumor characteristics, the probability of recurrence for any given patient can range from 15% to 70% at 1 year.15 Several studies and subsequent meta-analyses have proven that adjuvant intravesical treatment after TURBT is an effective means of reducing the risk of recurrence and progression of NMIBC when compared with TURBT alone.16,17 Current adjuvant treatment strategies are risk adapted because not all intravesical therapy is similarly effective nor are all treatment options recommended across risk categories. There are several stratification tools available including nomograms that can be used to estimate individualized risk predictions for recurrence and progression. Post hoc analyses of patient-level data collected during large randomized, controlled trials (RCTs) conducted by the European Organization for Research and Treatment of Cancer (EORTC) and Club Urologico Espanol de Tratamiento Oncologico (CUETO) were used to generate these tools. Both the EORTC and CUETO groups' nomograms and risk models acknowledge that there are unique factors related to an increased risk of recurrence and/or progression in NMIBC (Table 1). As such, individual risks for recurrence and progression may be calculated from each. These nomograms have all been validated and avoid the somewhat arbitrary categorization of patients into broad risk groups. However, there are certain limiting factors with their use that are important to understand and consider. TABLE 1. Comparison of Risk Factors for Recurrence and Progression Used in CUETO and EORTC Nomograms \| \| EORTC \| CUETO \| :--- \| Number of tumors \| Gendera \| \| Tumor size \| Age \| \| Recurrent tumor \| Recurrent tumor \| \| T stage \| T stageb \| \| Concurrent CIS \| Association CIS \| \| WHO grade \| WHO grade \| Open in a new tab a Factor for recurrence only. b Factor for progression only. CIS, carcinoma in situ; CUETO, Club Urologico Espanol de Tratamiento Oncologico; EORTC, European Organization for Research and Treatment of Cancer; WHO, World Health Organization. The initial EORTC risk tables were based on patient-level data from 2596 patients with NMIBC, only 4.4% of which also had CIS.15 In this study, no patients received a second TURBT and 78% were treated primarily with a variety of intravesical chemotherapy regimens. Only 6.5% of patients received bacillus Calmette-Guérin (BCG) and none received maintenance treatments. The CUETO group developed a similar nomogram based on data from 1062 patients from 4 randomized trials, all of which were treated with BCG.18 Use of maintenance treatment in the CUETO cohort was limited to 5 to 6 months. In 2016, the EORTC subsequently produced a nomogram and additional risk groups specific to a combined cohort of patients from 2 RCTs assessing the benefit of 1 to 3 years of maintenance BCG among 1812 intermediateand high-risk patients, all without CIS.19 Risk estimates of early and later recurrence are given based on prior recurrence rate and number of tumors, with tumor grade as an additional prognostic factor for early recurrence. A summary and comparison of risk factors and predicted recurrence and progression rates can be found in Table 2 for the original EORTC risk tables and the CUETO nomogram. TABLE 2. Point Estimates and 95% Confidence Intervals for the 1- and 5-year Risk of Recurrence and Progression Based on the Initial EORTC Risk Groups and CUETO Nomograms for Respective NMIBC Risk Score Groups \| Risk Group \| Recurrence \| Progression \| :---: \| \| 1 y \| 5 y \| 1 y \| 5 y \| \| \| EORTC \| CUETO \| EORTC \| CUETO \| EORTC \| CUETO \| EORTC \| CUETO \| \| Lowest \| 15 (10–19) \| 8 (6–11) \| 31 (24–37) \| 13 (10–15) \| 0.2 (0–0.7) \| 1 (0–2) \| 0.8 (0–1.7) \| 4 (2–6) \| \| Second \| 24 (21–26) \| 12 (8–16) \| 46 (42–49) \| 22 (17–28) \| 1 (0.4–1.6) \| 3 (1–5) \| 6 (5–8) \| 12 (8–16) \| \| Third \| 38 (35–41) \| 25 (20–31) \| 62 (58–65) \| 40 (33–46) \| 5 (4–7) \| 6 (3–8) \| 17 (14–20) \| 21 (16–27) \| \| Highest \| 61 (55–67) \| 42 (28–56) \| 78 (73–84) \| 53 (38–67) \| 17 (10–24) \| 14 (7–21) \| 45 (35–55) \| 34 (23–44) \| Open in a new tab For CUETO, scores of 0–4, 5–6, 7–9, and 101 are used for quartiles scores for both recurrence and progression; for EORTC, recurrence scores for the quartiles correspond to 0, 1–4, 5–9, 10–17 points, respectively, while scores of 0, 2–6, 7–13, and 14–23 are used for progression. CUETO, Club Urologico Espanol de Tratamiento Oncologico; EORTC, European Organization for Research and Treatment of Cancer; NMIBC, non-muscle-invasive bladder cancer. As described, each nomogram was developed from pooled trial data with notable patient characteristics and specific nuances to the given treatment regimens; consideration of these factors must be given during use of each risk tool. All scoring models have been shown to overestimate recurrence and progression risk among patients treated according to contemporary standards.20–22 In addition, the historical variability of nomenclature and surveillance regimens inherently influences the estimated recurrence and progression rates. Therefore, it is not possible to use a single nomogram or scoring system that will be predictive for all contemporary patients with NMIBC. Indeed, each risk stratification system may be best applied only to patients whose tumor characteristics and treatment plans mirror the nomogram development cohort.23 Additionally, the American Urological Association (AUA) and the European Association of Urology (EAU) have produced evidence-based guidelines dividing NMIBC into three distinct risk categories. Comparison of evidence-based guideline recommendations from the AUA and EAU demonstrate good general agreement in how patients are categorized (Table 3). However, there are notable differences regarding how Ta high-grade (all are EAU high risk, primary Ta HG tumors ≤3 cm are AUA intermediate risk) and low-grade T1 (EAU high risk, AUA intermediate risk) are designated. Treatment recommendations are also well aligned aside from the AUA recommending 3 years of maintenance BCG for highrisk patients whereas the EUA suggests 1 to 3 years. During BCG shortages, it has been suggested that BCG be preferentially used for the initial induction treatment of high-risk disease. Recurrence and progression estimates at 6, 12, and 24 months for each risk category and treatment type have been developed for the AUA NMIBC risk guidelines (Table 4). TABLE 3. NMIBC Risk Stratification Groups, Criteria, and Initial Adjuvant Treatment Recommendations From AUA and EAU Guidelines \| Risk Category \| AUA \| EUA \| Treatment Recommendation \| :---: :---: \| \| Low \| Solitary LG Ta tumor, <3 cm PUNLMP \| Solitary LG Ta tumor, <3 cm PUNLMP \| Post-TURBT intravesical chemotherapy \| \| Intermediate \| Recurrence within 1 year, LG Ta \| All tumors not defined as low or high risk \| Post-TURBT intravesical chemotherapya + \| \| \| Solitary LG Ta, >3 cm \| \| Induction + maintenance BCG for 1 year \| \| \| LG Ta, multifocal \| \| or \| \| \| HG Ta, ≤3 cm LG T1 \| \| Induction + maintenance chemotherapy for 1 yearc \| \| High \| HG T1 \| Any T1 \| Induction 1 maintenance BCG for 1–3 yearsb \| \| \| Any recurrent, HG Ta \| Any HG tumor \| or \| \| \| HG Ta, >3 cm \| Any CIS \| Radical cystectomy \| \| \| Any CIS \| Multiple, recurrent, >3cm \| \| \| \| Any BCG failure in HG \| tumors \| \| \| \| Any variant histology \| T1HG with concurrent CIS \| \| \| \| Any LVI \| Variant histology \| \| \| \| Any HG prostatic urethral involvement \| Lymphovascular invasion \| \| Open in a new tab Special designation of “highest risk” and upfront radical cystectomy should be considered. a With an EORTC recurrence score <5; b EAU recommends 1–3 years, AUA recommends 3 years; c optimal schedule not known. AUA, American Urological Association; CIS, carcinoma in situ; EAU, European Association of Urology; HG, high grade (G2, all G3); LG, low grade (G1, some G2); LVI, lymphovascular invasion; NMIBC, non-muscle-invasive bladder cancer; PUNLMP, papillary urothelial neoplasm of low malignant potential, variant histology 5 micropapillary, plasmacytoid, sarcomatoid, etc. TABLE 4. Estimated Recurrence and Progression Rates for Each AUA NMIBC Risk Category at 6, 12, and 24 Months Based on Adjuvant Treatment Given \| Risk Category/Treatment \| Recurrence Rate (%) \| Progression Rate (%) \| :---: \| \| 6 mo \| 12 mo \| 24 mo \| 6 mo \| 12 mo \| 24 mo \| \| Low risk / TURBT alone \| 15 \| 20 \| 25 \| 0 \| 1 \| 1 \| \| Low risk / TURBT 1 perioperative chemo \| 5 \| 10 \| 15 \| 0 \| 1 \| 1 \| \| Intermediate risk / TURBT 1 chemo \| 20 \| 30 \| 40 \| 2 \| 3 \| 5 \| \| Intermediate risk / TURBT 1 BCG (induction and maintenance) \| 20 \| 25 \| 30 \| 2 \| 3 \| 4 \| \| High risk / TURBT 1 BCG (induction and maintenance) \| 20 \| 25 \| 30 \| 3 \| 5 \| 10 \| Open in a new tab Adapted from Kamat AM.29 Rates for all intermediate- and high-risk patients are based on BCG naive. AUA, American Urological Association; BCG, bacillus Calmette-Guérin; NMIBC, non-muscle-invasive bladder cancer; TURBT, transurethral resection of bladder tumor. Adjuvant treatment is any agent administered after complete TURBT aimed at reducing the risks of NMIBC recurrence and progression. Traditionally, only intravesical agents have been used in this setting, but emerging data on the use of systemic checkpoint inhibitor immunotherapies either as a monotherapy or combined with intravesical agents may radically change this treatment paradigm. Selecting an Adjuvant Therapy Types of Adjuvant Therapy Adjuvant treatment is any agent administered after complete TURBT aimed at reducing the risks of NMIBC recurrence and progression. Traditionally, only intravesical agents have been used in this setting, but emerging data on the use of systemic checkpoint inhibitor immunotherapies either as a monotherapy or combined with intravesical agents may radically change this treatment paradigm. Current adjuvant intravesical treatments can be defined by the timing and sequence of their administration: (a) perioperatively, usually within 24h of TURBT, (b) induction, as an initial course for that particular agent, or (c) maintenance, subsequent courses in the setting of no evidence of disease after TURBT and induction treatment. Standard intravesical therapies can also be categorized as either chemotherapy or immunotherapy agents. The most used contemporary chemotherapy agents are mitomycin C, gemcitabine, epirubicin, and docetaxel. These are given individually or in combination depending on the indication. Historically, agents such as doxorubicin, valrubicin, cisplatin, and thiotepa, among others, were also frequently studied and used. The gold standard intravesical immunotherapy is BCG, an attenuated mycobacterium strain first described for use in bladder cancer in 1976. Interferon (IFN) has also been studied, but contemporary use is limited. There are several emerging NMIBC therapies being developed and studied in the treatment-naive or treatment-unresponsive setting. Novel agents like oncolytic adenoviruses (CG0070), recombinant adenovirus interferon alpha2b with Syn3 (Instiladrin® [nadofaragene firadenovec], FKD Therapies Oy, Kuopio, Finland), modified herpes simplex virus (talimogene laherparepvec [TVEC]), and fusion proteins (Vicinium®, Sesen Bio, Inc., Cambridge, MA) are some examples. Additionally, as a result of KEYNOTE-057, the FDA recently approved the first systemic therapy, pembrolizumab, for the treatment of patients with BCGunresponsive CIS with or without papillary tumors who are ineligible for or have elected not to undergo cystectomy. Choosing an Adjuvant Treatment For patients with low-risk NMIBC, it is recommended that adjuvant therapy with a single instillation (SI) of perioperative chemotherapy be considered. It is critical that intravesical chemotherapy is not given in the setting of a bladder perforation because extravasation of chemotherapy into the peritoneum is a devastating complication. Therefore, a high-level of suspicion during extensive resections and completion of a cystogram prior to chemotherapy instillation is essential. The recommendation for a single dose of chemotherapy post-TURBT is supported by meta-analyses that demonstrate an approximately 15% reduction of recurrence with instillation of mitomycin or epirubicin compared with TURBT alone in low-risk disease.24 An alternative option is perioperative gemcitabine. A randomized trial of peri-TURBT gemcitabine versus saline demonstrated a reduction in 4-year recurrence rate from 54% to 34% with few adverse effects among 215 patients with low-grade NMIBC.25 Patients with intermediate-risk NMIBC comprise a heterogeneous subgroup with few proven treatment options and represent one of the greatest unmet needs in the treatment of NMIBC. SI perioperative intravesical chemotherapy can be considered but has only demonstrated efficacy among intermediate-risk patients with an EORTC risk score <5.24 Additional adjuvant intravesical chemotherapy or immunotherapy may be given in intermediate-risk patients for up to 1 year based on data from numerous studies of commonly used agents when compared with TURBT alone.12 However, individual treatment agent choice remains somewhat controversial and may be based on risk profile, prior intravesical treatment exposure, and tolerability. The International Bladder Cancer Group (IBCG) suggests further stratification of intermediate-risk patients based on number and size of tumors, recurrence and frequency of recurrence within 1 year, and previous treatment. For those patients without any of IBCG risk factors, SI is recommended; for those with 3 to 4 factors, BCG is recommended, and for those with 1 to 2 risk factors, BCG or chemotherapy may be used based on consideration of previously received therapies. These recommendations coincide with the principle that most patients with intermediate-risk NMIBC are at significantly greater risk of recurrence rather than progression. There is no well accepted schedule for the use of intravesical chemotherapy. Comparative studies between BCG and chemotherapy agents have generally included both intermediate- and high-risk patients. A meta-analysis of individual patient-level data (n = 2820) demonstrated risk of recurrence was reduced by 32% when BCG was given compared with mitomycin C (MMC) but minimal difference in risk of progression.26 RCTs and other analyses of pooled data suggest an improvement in progression rates and time to progression when BCG is used, with the strongest signal seen among those who received maintenance therapy.27 Duration of maintenance therapy has been suggested to be up to 3 years in high-risk patients based on data from SWOG 8507 and EORTC 30962. It is notable that no adjuvant treatment has been found to change the natural history of disease and improve overall mortality.28 “Early” radical cystectomy should be a consideration among patients at the very highest risk of progression to MIBC who are fit and willing to undergo surgery. Of note, comparing data from various historical studies is difficult due to differences in nomenclature, surveillance techniques, and definitions of recurrence. Strict nomenclature and use of standard endpoints have been suggested by expert consensus groups for trial use moving forward.29 Additionally, for the purposes of this review, we have primarily considered the management of index NMIBC cases naive to intravesical treatment. Special management considerations outside the scope of this review must be made for patients where initial adjuvant intravesical treatment, especially BCG, has failed. These management decisions should be based on what treatments they have been already been exposed to and the timing and aggressiveness of the recurrence. Rationale for Use of Neoadjuvant Treatments Upfront (neoadjuvant) chemoablative treatments like chemohyperthermia (hyperthermic intravesical chemotherapy (HIVECT), Combat Medical Ltd., UK) and reverse thermal gel formulated with MMC (UGN-102, UroGen Pharma, Inc., Princeton, NJ) are increasingly being explored as alternatives to repeat office or ambulatory procedures in patients with established, recurrent low-grade tumors. Chemoablation is a potentially attractive alternative to TURBT that may reduce treatment time, cost, and morbidity. Alternatively, others have also advocated for office-based fulguration of small, recurrent lowgrade tumors when feasible and tolerable as part of a risk-adapted strategy of treatment.30,31 Active surveillance has also been suggested as a safe and cost-effective alternative for highly selected patients.32 Upfront (neoadjuvant) chemoablative treatments like chemohyperthermia and reverse thermal gel formulated with MMC are increasingly being explored as alternatives to repeat office or ambulatory procedures in patients with established, recurrent low-grade tumors. HIVEC and UGN-102 Early Data HIVEC is neoadjuvant intravesical instillation of heated MMC delivered in a recirculated fashion by a proprietary system called the Bladder Recirculation System (BRS) from Combat Medical. Initial study of the system consisted of 8 courses (60 min dwell/recirculation time using 80 mg of MMC diluted in 50 mL of distilled water at 43±1 °C) in 24 patients with intermediate- or high-risk NMIBC delivered in the neoadjuvant setting with TURBT 2 weeks later.33 Of these, 15 patients (62.5%) had a complete response and 8 additional patients (33.3%) showed a partial response. The 4-year cumulative incidence of recurrence was 20.8% and the agent was well tolerated with 97% of patients completing all scheduled instillations. UGN-102 is a reverse thermal hydrogel that is formulated with MMC (UroGen Pharma). It is currently being studied (NCT03558503) as a chemoablative therapy in lieu of TURBT or surgical tumor destruction in patients with low-grade bladder cancer that meet 1 to 2 of 3 criteria for AUA Intermediate Risk NMIBC: multifocal, lesion >3 cm, and recurrence within 12 months. Because patients with low-grade intermediate-risk NMIBC are at low risk of progression (~5% at 2 y) and high recurrence risk (~40% at 2 y), the proposed benefit of this agent is two-fold: drug delivery will be improved by greater surface contact time and the morbidity and cost associated with TURBT will be reduced. The active study is a multicenter, single-arm Phase IIb that is treating patients with UGN-102 weekly for 6 weeks with the primary endpoint being a complete response at 3 months. A complete response was defined as negative cystoscopy, negative cytology, and negative biopsy if possible visual evidence of disease is noted. Secondary endpoints include 12-month durability and safety. Updated interim analysis of 63 patients (38 men, 25 women; age range, 33–96) demonstrated a 3-month complete response (CR) rate of 65% (41/63). Of those with a CR, 97% (31/32) patient and 85% (17/20) remained free of disease at 6 and 9 months, respectively.34 Mild or moderate adverse events were common, such as dysuria (38%) and hematuria (16%). Conclusions NMIBC is a diverse subclassification of urothelial bladder cancer with a wide variation in recurrence and progression rates. Risk-adapted treatment is essential to balance adverse effects of treatment with adequate disease control and there are many nomograms and strategies to assist, each with specific limitations. Standard treatment with intravesical chemotherapies and BCG is associated with good outcomes for most patients but novel agents and approaches will be necessary to continue to improve rates of durable cure and bladder preservation for those at highest risk of progression to MIBC. Main Points. Non-muscle-invasive bladder cancer is a diverse subclassification of urothelial bladder cancer with a wide variation in recurrence and progression rates. 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[Google Scholar] Articles from Reviews in Urology are provided here courtesy of MedReviews, LLC ACTIONS PDF (306.0 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Evaluation of NMIBC Risk Stratification Selecting an Adjuvant Therapy Rationale for Use of Neoadjuvant Treatments Conclusions References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15474	https://math.stackexchange.com/questions/4841668/solve-functional-equation-fx-fy-f-left-fracxy1-xy-right	calculus - Solve functional equation: $f(x) + f(y) = f\left(\frac{x+y}{1-xy}\right) $ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Solve functional equation: f(x)+f(y)=f(x+y 1−x y) Ask Question Asked 1 year, 8 months ago Modified1 year, 8 months ago Viewed 331 times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. The equation being : f(x)+f(y)=f(x+y 1−x y) The most obvious answer is, f(x)=C arctan x but there is another possible answer e.i f(x)=log(1−x 1+x) which i am unable to get by the process i have provided. First substituting: x=y=0⇒f(0)=0 then partially differentiating the original equation wrt to x and then y individually (provided x≠y), we'll get : f′(x)=f′(x+y 1−x y)1+y 2(1−x y)2−(i)f′(y)=f′(x+y 1−x y)1+x 2(1−x y)2−(i i) diving equations i by i i we get f′(x)f′(y)=1+y 2 1+x 2⇒f′(x)(1+x 2)=f′(y)(1+y 2) which is only possible if the above equation is equal to some constant c, since x≠y f′(x)(1+x 2)=C⇒f′(x)=C 1+x 2⇒f(x)=∫C 1+x 2=C arctan x+K since f(0)=0,k=0∴f(x)=C arctan x Also, when do i know that there is no more functions that satisfy the original equation ? calculus integration functions logarithms Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jan 11, 2024 at 7:46 gerw 33.5k 2 2 gold badges 30 30 silver badges 61 61 bronze badges asked Jan 9, 2024 at 18:19 Sh0unakSh0unak 75 5 5 bronze badges 7 Are you sure that f(x)=log(1−x)/(1+x) works? I checked the algebra, and I don't think it does. As a quick first check, the denominator on the LHS is (1+x)(1+y), while the denominator on the RHS is (1−x y)+(x+y). These are nowhere close to each other, so it's unlikely to cancel out easily.Calvin Lin –Calvin Lin 2024-01-09 18:41:00 +00:00 Commented Jan 9, 2024 at 18:41 1 @CalvinLin, I checked as well, and it works, at least if c=1.user1064129 –user1064129 2024-01-09 18:45:03 +00:00 Commented Jan 9, 2024 at 18:45 @Doug This is what I get in wolfram. We need x y(x+y)=0 for equality to hold. Alternatively, we can show that f(0.5)+f(0.5)≠f(4/3), because the LHS is defined, but the RHS is not defined.Calvin Lin –Calvin Lin 2024-01-09 18:48:57 +00:00 Commented Jan 9, 2024 at 18:48 2 So, I suppose that means that if x and y are imaginary, it should work, i.e. ln((1−i x)/(1+i x)), which is essentially another way of expressing arctangent.user1064129 –user1064129 2024-01-09 18:54:06 +00:00 Commented Jan 9, 2024 at 18:54 2 As for the original equation, I think it seems quite promising to substitute x=tan α and y=tan β and then use the formula for the tangent of sum (surely, this is motivated by the hypothetical answer).richrow –richrow 2024-01-09 18:59:07 +00:00 Commented Jan 9, 2024 at 18:59 \|Show 2 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. You should be able to say that there is only one differentiable solution (which you found) based on the uniqueness theorems for ODEs. You can also check that f is odd to conclude that the logarithm cannot be a solution. As far as non-smooth solutions, I'm not sure there is any general way to decide if there are any. But perhaps someone else knows more about this topic. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 9, 2024 at 19:18 user1064129 user1064129 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Choosing x=0 and y=0 gives f(0)+f(0)=f(0)⟹f(0)=0. Let z be any real number other than 0. 1/z is defined since z≠0. Substituting z and 1/z into the functional equation we get: f(z)+f(1/z)=f(z+1 z 0) f(z)=f(z+1 z 0)−f(1/z) but the right hand side is undefined because of the division by 0. So f(z) is undefined. In summary, f(x)=0 when x=0 but otherwise it is undefined. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 9, 2024 at 22:16 Will.Octagon.GibsonWill.Octagon.Gibson 1,540 6 6 silver badges 24 24 bronze badges 1 The solution is arctangent. There is no issue with letting f(±a/0)=f(±∞)=±π/2,a≠0.user1064129 –user1064129 2024-01-10 15:05:50 +00:00 Commented Jan 10, 2024 at 15:05 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus integration functions logarithms See similar questions with these tags. 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15475	https://byjus.com/chemistry/modern-periodic-table-modern-periodic-law/	What is Modern Periodic Law? The modern periodic table is developed after the periodic law and a periodic table given by Mendeleev. In the latter part of the 18th century, Mendeleev made his periodic table. Scientists did not know about the internal structure of the atom back then. The development of various atomic models and advances in quantum theory revealed that the atomic number is the most basic property of a chemical element. This led to the modification of Mendeleev’s periodic law, which is today called as modern periodic law. Check ⇒ History of Periodic Table Table of Contents Modern Periodic Law Definition Recommended Videos The Modern Periodic Table Classification of the Elements in the Periodic Table Frequently Asked Questions – FAQs Modern Periodic Law Definition The modern Periodic law can be stated as: “The physical and chemical properties of the elements are periodic functions of their atomic numbers”. The atomic number is equal to the number of electrons or protons in a neutral atom. After knowing the fundamental unit of elements, scientists now had a clear idea about quantum numbers and electronic configuration of elements in the periodic table. After knowing the periodic law, chemists noticed that there is an analogy between the 94 naturally occurring chemical elements. This analogy made people more curious about the chemistry of these elements. Scientists made various artificial elements. A new periodic table was developed based on the modern periodic law by modifying the Mendeleev’s periodic table. Recommended Videos The Modern Periodic Table The present form of a periodic table that is widely used across the globe is the long form of the periodic table. In this form of a periodic table, the horizontal rows are called periods and the vertical columns are known as the groups. Groups consist of elements that have similar outer shell electronic configuration in their atoms. Previously the groups were named as IA,…VIIIA, VIII, IB…VIIB and 0. But now they are named as 1, 2, 3…18. In the modern periodic table, periods are the seven horizontal rows. Principle quantum number ‘n’ decides the period of the element. Principle quantum number (n) is one of the four quantum numbers (n, l, m, and s). It tells us about the principle electron shell. For example, if n= 3, then it indicates the principle shell as 3. Classification of the Elements in the Periodic Table: Classification of the elements in the periodic table can be done in four ways on the basis of their electronic configurations: Noble gas elements: Elements of group 18 of the modern periodic table are considered a noble gas. The electronic configuration of the first element (helium) of this group is 1s2. Rest all the elements (neon, argon, krypton, xenon, and radon) have their outer shell electronic configuration is ns2np6. As the octet of these elements is complete, hence they are highly stable elements. Representative elements: S-lock and P-block elements come under the category of representative elements. Elements in groups 1 and 2 are known as the s – block elements (elements with 1s2and 2s2 outermost configuration). Group 13-17 are known as the p-block elements (outermost configuration varies from ns2np1 to ns2np5). Transition elements: Elements which belong to group 3 to 12 and have their outer shell electronic configuration as (n-1)d1-10ns1-2 are referred to as transition elements. These elements are also known as the d-block elements. Inner transition elements: Lanthanides and actinides series which fall at the bottom of the periodic table comes under the category of inner transition elements. In these elements the 4f and 5f orbitals are partially filled, rendering them special properties. Frequently Asked Questions on Modern Periodic Law Q1 Who gave the modern periodic law? In 1869, Dmitri Mendeleev and Lothar Meyer established the periodic law independently. The first periodic table was developed by Mendeleev and soon followed by Meyer. Each grouped the elements by their mass and proposed regularly reoccurring those properties. Q2 Why is the periodic law important? Periodic law is recognized as one of chemistry’s most important concepts. While dealing with the chemical elements, their properties, and their chemical reactions, each chemist makes use of Periodic Law, whether consciously or not. The development of the modern periodic table was driven by periodic law. Q3 How do periodic trends relate to periodic law? Periodic trends are common patterns in the periodic table showing us the various aspects of an element such as electronegativity, atomic radius, or ionizing power. The periodic law tells us that when grouped by atomic number, certain properties of elements occur periodically. Q4 Is atomic mass a periodic property? Generally, nuclear mass decreases from left to right and always increases from top to bottom. As the atomic number has been developed as the basis for organizing the elements on the periodic table, the atomic number will always increase from left to right and top to bottom. Q5 What is relative periodic property? In a given period, the valence shell electronic configuration of any two elements is not the same. Because of this, elements throughout time have different chemical properties with a periodic gradation from left to right for their physical properties. This is referred to as periodic property. This is just a brief description of the periodic table and the classification of elements. To know more about it, register with BYJU’S & download BYJU’S – the learning app. If you still have doubts regarding the periodic law and would like to learn about periodic table class 10, check out our Introduction to the Periodic Table for detailed clarification. Test Your Knowledge On Modern Periodic Table Modern Periodic Law! Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! 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15476	https://msp.org/involve/2009/2-2/involve-v2-n2-p07-s.pdf	inv lve a journal of mathematics mathematical sciences publishers 2009 Vol. 2, No. 2 On the existence of unbounded solutions for some rational equations Gabriel Lugo INVOLVE 2:2(2009) On the existence of unbounded solutions for some rational equations Gabriel Lugo (Communicated by Kenneth S. Berenhaut) We resolve several conjectures regarding the boundedness character of the ratio-nal difference equation xn = α + δxn−3 A + Bxn−1 + Cxn−2 + Exn−4 , n ∈N. We show that whenever parameters are nonnegative, A < δ, and C, E > 0, un-bounded solutions exist for some choice of nonnegative initial conditions. We also partly resolve a conjecture regarding the boundedness character of the ra-tional difference equation xn = xn−3 Bxn−1 + xn−4 , n ∈N. We show that whenever B > 25, unbounded solutions exist for some choice of nonnegative initial conditions. 1. Introduction Palladino [2009a] studies a trichotomy behavior of the k-th order rational difference equation with nonnegative parameters and nonnegative initial conditions, xn = α + Pk i=1 βixn−i A + Pk j=1 Bjxn−j , n ∈N. Palladino established that there is a trichotomy behavior which is dependent on the relation between A and Pk i=1 βi. In particular, in this paper, it was established that, under certain conditions, when A < Pk i=1 βi unbounded solutions exist. Here we will broaden that proof of unboundedness and show that when A < Pk i=1 βi unbounded solutions exist under different conditions. In Section 2 we present a MSC2000: 39A10, 39A11. Keywords: difference equation, periodic convergence, boundedness character, unbounded solutions, periodic behavior of solutions of rational difference equations, nonlinear difference equations of order greater than one, global asymptotic stability. 237 238 GABRIEL LUGO proof based on [Palladino 2009a, Section 5] which serves to generalize this work. An immediate consequence of this, as discussed later, will be to show that whenever parameters are nonnegative, A < δ, and C, E > 0, unbounded solutions exist for some choice of nonnegative initial conditions for the rational difference equation, xn = α + δxn−3 A + Bxn−1 + Cxn−2 + Exn−4 , n ∈N. This resolves the conjectures regarding boundedness character for equations 609, 611, 617, and 619 presented in [Camouzis and Ladas 2008]. In Section 3, we partially resolve Conjecture 2 in [Palladino 2009a]. We show that the rational difference equation xn = xn−3 Bxn−1 + xn−4 , n ∈N, has unbounded solutions whenever B > 25. In the process, we resolve the con-jecture in [Camouzis and Ladas 2008] regarding the boundedness character of equation 584. The proof here will use similar techniques to those presented in [Lugo and Palladino ≥2009]. 2. Preliminary results During this section we use the ideas of modulo classes. Let us introduce these ideas in the following remark. Remark 1. We say that a is congruent to b with modulus c and write a ≡b mod c if c \| a −b. It is well known that given z ∈Z, there exists a ∈{0, . . . , c−1} so that z ≡a mod c. We call such a the residue of z with respect to the modulus c, and write a = z mod c. Here we introduce a condition which allows us to construct unbounded solu-tions, namely Condition 1. Before doing so let us ﬁrst introduce some notation. Let us deﬁne the following sets of indices: Iβ = {i ∈{1, 2, . . . , k} \| βi > 0} and IB = { j ∈{1, 2, . . . , k} \| Bj > 0}. These sets are used extensively in [Palladino 2009b] when referring to the k-th order rational difference equation. Similarly we shall make extensive use of this notation. Condition 1. We say that Condition 1 is satisﬁed if, for some p ∈N, p \| gcd Iβ. We also must have disjoint sets B, L ⊂{0, . . . , p −1} with B ̸= ∅and with the following properties. (1) For all b ∈B, {(b −j) mod p : j ∈IB} ⊂L. (2) For all ℓ∈L, there exists j ∈IB so that (ℓ−j) mod p ∈B. UNBOUNDED SOLUTIONS FOR SOME RATIONAL EQUATIONS 239 We now present Theorem 1 which makes use of Condition 1. In the remainder of this section we will verify Condition 1 for a number of special cases of the fourth-order rational difference equation, thereby conﬁrming several conjectures in [Camouzis and Ladas 2008]. Theorem 1. Consider the k-th order rational difference equation xn = α + Pk i=1 βixn−i A + Pk j=1 Bjxn−j , n ∈N. (1) Assume nonnegative parameters and nonnegative initial conditions. Further as-sume that A < Pk i=1 βi and Pk i=1 βi > 0, and that Condition 1 is satisﬁed for Equation (1). Then unbounded solutions of Equation (1) exist for some initial conditions. Proof. By assumption, we may choose p ∈N and B, L ⊂{0, . . . , p −1} so that Condition 1 is satisﬁed. Choose initial conditions x−m where m ∈{0, . . . , k −1} so that the following holds. If (−m mod p) ∈B, then x−m > 2α Pk j=1 B j (min j∈IB B j)((Pk i=1 βi) −A) + Pk i=1 βi min j∈IB B j . If (−m mod p) ∈L, then x−m < (Pk i=1 βi) −A 2 Pk j=1 B j . Also assume x−m > 0 for all m ∈{0, . . . , k −1}. Under this choice of initial conditions our solution {xn} has the following prop-erties. (a) xn > 2α Pk j=1 B j (min j∈IB B j)((Pk i=1 βi) −A) + Pk i=1 βi min j∈IB B j whenever (n mod p) ∈B. (b) xn < (Pk i=1 βi) −A 2 Pk j=1 B j whenever (n mod p) ∈L. (c) xn > 0 for all n ∈N. We prove this using induction on n; our initial conditions provide the base case. Assume that the statement is true for all n ≤N −1. We show the statement for n = N. This induction proof has three cases. Let us begin by assuming (N mod p) ∈B. 240 GABRIEL LUGO Case (a). Condition 1(1) tells us that in this case {(N −j) mod p : j ∈IB} ⊂L. Hence xN−j < (Pk i=1 βi) −A 2 Pk j=1 B j for all j ∈IB. Since p \| gcd(Iβ), N mod p = (N −i) mod p for all i ∈Iβ.Thus for all i ∈Iβ, xN−i > 2α Pk j=1 B j (min j∈IB B j)((Pk i=1 βi) −A) + Pk i=1 βi min j∈IB B j . Hence xN = α + Pk i=1 βixN−i A + Pk j=1 B jxN−j ≥ Pk i=1 βi A + (Pk j=1 B j)(Pk i=1 βi) −A 2 Pk j=1 B j 2α Pk j=1 B j (min j∈IB B j)((Pk i=1 βi) −A) + Pk i=1 βi min j∈IB B j ≥ 2α Pk j=1 B j (min j∈IB B j)((Pk i=1 βi) −A) + Pk i=1 βi min j∈IB B j . This inequality is obtained by simply replacing the terms in the denominator with their upper bound, and replacing the terms in the numerator with their lower bound. This ﬁnishes case (a). Case (b). We now assume (N mod p) ∈L. Since p \| gcd(Iβ), we have N mod p = (N −i) mod p for all i ∈Iβ. Hence xN−i < (Pk i=1 βi) −A 2 Pk j=1 B j for all i ∈Iβ. Condition 1(2) guarantees that there exists j ∈IB so that xN−j > 2α Pk j=1 B j (min j∈IB B j)((Pk i=1 βi) −A) + Pk i=1 βi min j∈IB B j . Hence xN = α + Pk i=1 βixN−i A + Pk j=1 B jxN−j < α + (Pk i=1 βi)(Pk i=1 βi)−A 2 Pk j=1 B j (min j∈IB B j) 2α Pk j=1 B j (min j∈IB B j)((Pk i=1 βi)−A) + Pk i=1 βi min j∈IB B j UNBOUNDED SOLUTIONS FOR SOME RATIONAL EQUATIONS 241 = α + (Pk i=1 βi)(Pk i=1 βi)−A 2 Pk j=1 B j 2α Pk j=1 B j (Pk i=1 βi)−A + Pk i=1 βi = 2α Pk j=1 B j (Pk i=1 βi)−A + Pk i=1 βi 2α Pk j=1 B j (Pk i=1 βi)−A + Pk i=1 βi (Pk i=1 βi) −A 2 Pk j=1 B j = (Pk i=1 βi) −A 2 Pk j=1 B j . (2) This ﬁnishes case (b). Case (c). It is clear that if xn > 0 for n < N. Then xN > 0 so case (c) is trivial. We now use the facts we obtained from our induction to prove that a particular subsequence is unbounded. Take b ∈B. We now show that {xmp+b}∞ m=1 diverges to ∞. We explained earlier that xmp+b−j < (Pk i=1 βi) −A 2 Pk j=1 B j , since {(mp + b −j) mod p : j ∈IB} ⊂L. Hence, xmp+b = α + Pk i=1 βixmp+b−i A + Pk j=1 B jxmp+b−j > Pk i=1 βixmp+b−i A + (Pk j=1 B j) (Pk i=1 βi)−A 2 Pk j=1 B j ≥(Pk i=1 βi)(mini∈{1,...,⌊k/p⌋}(xmp+b−ip)) A + (Pk j=1 B j) (Pk i=1 βi)−A 2 Pk j=1 B j ≥ 2 Pk i=1 βi A + Pk i=1 βi min i∈{1,...,⌊k/p⌋}(xmp+b−ip), m ≥k. This is a difference inequality which holds for the subsequence {xmp+b} for m ≥k. We now rename this subsequence and apply the methods used in [Palladino 2008]. We set zm = xmp+b for m ∈N. Since we have just shown that {zm} satisﬁes the difference inequality zm ≥ 2 Pk i=1 βi A + Pk i=1 βi min i∈{1,...,⌊k/p⌋}(zm−i), m ≥k, 242 GABRIEL LUGO we can use the results of [Palladino 2008], particularly Theorem 3, to conclude that for m ≥k, min(zm−1, . . . , zm−⌊k/p⌋) ≥min(y⌊m−k ⌊k/p⌋⌋, . . . , ym−k), where {ym}∞ m=0 is a solution of the difference equation ym = 2 Pk i=1 βi A + Pk i=1 βi ym−1, m ∈N, (3) with y0 = min(zk−1, . . . , zk−⌊k/p⌋). Clearly every positive solution diverges to ∞for the simple difference equation (3), since A < Pk i=1 βi. Hence using the inequality we have obtained, {zm}∞ m=1 diverges to ∞. Hence with given initial conditions, there is a subsequence of our solution {xn}∞ n=1, namely {xmp+b}∞ m=1, which diverges to ∞. Hence our solution {xn}∞ n=1 is unbounded. So we have exhibited an unbounded solution whenever A < Pk i=1 βi. □ Corollary 1. Consider the fourth-order order rational difference equation xn = α + δxn−3 A + Bxn−1 + Cxn−2 + Exn−4 , n ∈N. (4) Assume nonnegative parameters and nonnegative initial conditions so that the de-nominator is nonvanishing. Further assume that δ, C, E > 0. (i) Whenever A > δ, the unique equilibrium is globally asymptotically stable. (ii) Whenever A = δ and α > 0, the unique equilibrium is globally asymptotically stable. (iii) Whenever A = δ and α = 0, every solution of Equation (4) converges to a periodic solution of period 3. (iv) Whenever A < δ, then Equation (4) has unbounded solutions for some choice of initial conditions. Proof. Cases (i), (ii), and (iii) were shown in [Palladino 2009b]. We now prove case (iv). Let us check Condition 1. Choose B = {0} and L = {1, 2}. Condition 1(1) is satisﬁed since for all b ∈B, namely b = 0, {(0−j) mod 3 : j ∈{2, 4}} = {(0 −j) mod 3 : j ∈{1, 2, 4}} = {1, 2}. Condition 1(2) is satisﬁed since for 1 ∈L, there exists 4 ∈IB so that (1 −4) mod 3 = −3 mod 3 = 0 ∈{0}. Also for 2 ∈L, there exists 2 ∈IB so that (2 −2) mod 3 = 0 mod 3 = 0 ∈{0}. Furthermore A < δ = k X i=1 βi and k X i=1 βi = δ > 0. UNBOUNDED SOLUTIONS FOR SOME RATIONAL EQUATIONS 243 Thus Theorem 1 applies and so in case (iv) Equation (4) has unbounded solutions for some choice of initial conditions. □ Notice that Corollary 1 resolves conjectures 609, 611, 617, and 619 in [Camouzis and Ladas 2008] regarding boundedness character. 3. The equation xn = xn−3 Bxn−1 + xn−4 In [Palladino 2009a] it is conjectured that the difference equation xn = xn−3 Bxn−1 + xn−4 , n ∈N, has unbounded solutions whenever B > 0. We show that whenever B > 25 un-bounded solutions exist for some choice of nonnegative initial conditions. This does not fully establish the conjecture in [Palladino 2009a]. It does however es-tablish the Conjecture 584 in [Camouzis and Ladas 2008]. We make use of the argument structure presented in Lemma 1 of [Lugo and Palladino ≥2009]. Let us repeat this lemma for the sake of the reader. Lemma 1. Let {xn}∞ n=1 be a sequence in [0, ∞). Suppose that there exists D > 1 and hypotheses H1, . . . , Hk so that for all n ∈N there exists pn ∈N so that the following holds. Whenever xn−i satisﬁes Hi for all i ∈{1, . . . , k}, then xn+pn−i satisﬁes Hi for all i ∈{1, . . . , k} and xn+pn−1 ≥Dxn−1. Further assume that for some N ∈N, xN−i satisﬁes Hi for all i ∈{1, . . . , k} and xN−1 > 0. Then {xn}∞ n=1 is unbounded. Particularly {xzm−1}∞ m=1 is a subsequence of {xn}∞ n=1 which diverges to ∞, where zm = zm−1 + pzm−1 and z0 = N. Proof. Let zm = zm−1+ pzm−1 and z0 = N. Using induction, we prove that given m ∈ N the following holds. xzm−1 ≥DmxN−1 and xzm−i satisﬁes Hi for all i ∈{1, . . . , k}. By assumption, xN−i satisﬁes Hi for all i ∈{1, . . . , k} and xN−1 ≥D0xN−1. This provides the base case. Assume xzm−1−i satisﬁes Hi for all i ∈{1, . . . , k} and xzm−1−1 ≥Dm−1xN−1. Using our earlier assumption this implies that there exists pzm−1 so that xzm−1+pzm−1−i satisﬁes Hi for all i ∈{1, . . . , k} and xzm−1+pzm−1−1 ≥ Dxzm−1−1 ≥(D)Dm−1xN−1 = DmxN−1. So we have shown that xzm−1 ≥DmxN−1 for all m ∈N. Hence the subsequence {xzm−1}∞ m=1 of {xn}∞ n=1 clearly diverges to ∞since D > 1. □ Theorem 2. Consider the fourth order rational difference equation, xn = xn−3 Bxn−1 + xn−4 , n ∈N. (5) Suppose B > 25. Then Equation (5) has unbounded solutions for some initial conditions. 244 GABRIEL LUGO Proof. We choose initial conditions so that x−2 > B, x−3 < 1 4, and one of the following holds: (1) x0 < 1 4B and x−1 < 1 B ; (2) 1 4B ≤x0 ≤2x−2 and x−1 < 1 B2x−2 ; (3) x0 > 2x−2 and x−1 < 1 B2x−2 . We show that there exists D = 2 so that for all n ∈N there exists pn ∈{2, 3, 5} so that the following holds. Whenever xn−3 > B, xn−4 < 1 4, and one of the following holds: (1) xn−1 < 1 4B and xn−2 < 1 B ; (2) 1 4B ≤xn−1 ≤2xn−3 and xn−2 < 1 B2xn−3 ; (3) xn−1 > 2xn−3 and xn−2 < 1 B2xn−3 ; then we have xn+pn−3 > Dxn−3 > B, xn+pn−4 < 1 4, and one of the following holds: (1) xn+pn−1 < 1 4B and xn+pn−2 < 1 B ; (2) 1 4B ≤xn+pn−1 ≤2xn+pn−3 and xn+pn−2 < 1 B2xn+pn−3 ; (3) xn+pn−1 > 2xn+pn−3 and xn+pn−2 < 1 B2xn+pn−3 . First assume xn−1 < 1 4B , xn−2 < 1 B , xn−3 > B, xn−4 < 1 4. In this case pn = 3. Since B > 25 we have xn+pn−4 = xn−1 < 1 4B < 1 4. Since xn−4 < 1 4 and xn−1 < 1 4B we have xn+pn−3 = xn = xn−3 Bxn−1 + xn−4 ≥ xn−3 2 max (Bxn−1, xn−4) > 2xn−3 > B. UNBOUNDED SOLUTIONS FOR SOME RATIONAL EQUATIONS 245 Since xn−2 < 1 B , xn+pn−2 = xn+1 = xn−2 Bxn + xn−3 ≤xn−2 Bxn < 1 B2xn < 1 B3 < 1 B . Hence regardless of the value of xn+pn−1 one of our requirements is satisﬁed. If xn+pn−1 < 1 4B then requirement (1) is satisﬁed. If 1 4B ≤xn+pn−1 ≤2xn+pn−3 then requirement (2) is satisﬁed. If xn+pn−1 >2xn+pn−3 then requirement (3) is satisﬁed. Next assume 1 4B ≤xn−1 ≤2xn−3, xn−2 < 1 B2xn−3 , xn−3 > B, xn−4 < 1 4. In this case pn = 5. Since B > 25 we have xn+pn−4 = xn+1 = xn−2 Bxn + xn−3 < xn−2 xn−3 < 1 B2x2 n−3 < 1 4. Since xn−2 < 1 B2xn−3 and B > 25 we have xn+pn−3 = xn+2 = xn−1 Bxn+1 + xn−2 ≥ xn−1 2 max (Bxn+1, xn−2) > xn−1 2 max 1 Bx2 n−3 , 1 B2xn−3 ≥B2xn−3 8B > 2xn−3 > B. Also notice that xn+pn−2 = xn+3 = xn Bxn+2 + xn−1 = xn−3 (Bxn+2 + xn−1)(Bxn−1 + xn−4) < xn−3 Bxn+2(Bxn−1+xn−4) < 8xn−3 B2xn−3(Bxn−1+xn−4) < 8 B3xn−1 < 25 B2 < 1 B . Notice that xn+pn−1 = xn+4 = xn+1 Bxn+3 + xn < 1 (B2x2 n−3)(Bxn+3 + xn) < 1 B2x2 n−3xn = Bxn−1 + xn−4 B2x3 n−3 < 2Bxn−3 + .25 B2x3 n−3 < 3 Bx2 n−3 < 1 4B . Hence requirement (1) is satisﬁed in this case. Finally assume xn−1 > 2xn−3, xn−2 < 1 B2xn−3 , xn−3 > B, xn−4 < 1 4. 246 GABRIEL LUGO In this case pn = 2. Immediately we have xn+pn−4 = xn−2 < 1 B2xn−3 < 1 4. Also by assumption, xn+pn−3 = xn−1 > 2xn−3 > B. Further since xn−1 > 2xn−3, xn+pn−2 = xn = xn−3 Bxn−1 + xn−4 < xn−3 Bxn−1 < 1 2B < 1 B . Furthermore xn+pn−1 = xn+1 = xn−2 Bxn + xn−3 < xn−2 xn−3 < 1 B2x2 n−3 < 1 4B . Hence requirement (1) is satisﬁed in this case, so after an application of Lemma 1 the proof is complete. □ 4. Conclusion As noted in the introduction, Theorem 2 partly resolves Conjecture 2 in [Palladino 2009a]; the latter, however, is only part of a larger conjecture, namely Conjecture 1 in the same reference. For convenience we restate this conjecture. Conjecture 1. Consider the k-th order rational difference equation xn = Pk i=1 βixn−i Pk j=1 B jxn−j , n ∈N. (6) Assume nonnegative parameters and nonnegative initial conditions so that the de-nominator is nonvanishing. Further assume that Pk i=1 βi > 0 and that there does not exist j ∈IB such that gcd(Iβ) \| j. Then unbounded solutions of Equation (6) exist for some initial conditions. It would be interesting to study this conjecture further utilizing techniques sim-ilar to that used in Theorem 2. 5. Acknowledgements I would like to thank Frank Palladino for his guidance, helpful discussions and his invaluable help with L A T EX. I would like to thank Gerry Ladas and the mem-bers of the MTH 691 class for providing a stimulating research environment for undergraduate students. UNBOUNDED SOLUTIONS FOR SOME RATIONAL EQUATIONS 247 References [Camouzis and Ladas 2008] E. Camouzis and G. Ladas, Dynamics of third-order rational difference equations with open problems and conjectures, Adv. Disc. Math. Appl. 5, Chapman & Hall/CRC, Boca Raton, FL, 2008. MR 2008h:39001 Zbl 1129.39002 [Lugo and Palladino ≥2009] G. Lugo and F. J. Palladino, “Unboundedness for some classes of rational difference equations”, Int. J. Difference Equ. To appear. [Palladino 2008] F. J. Palladino, “Difference inequalities,comparison tests,and some consequences”, Involve 1:1 (2008), 91–100. MR 2403068 [Palladino 2009a] F. J. Palladino, “On periodic trichotomies”, J. Difference Equa. Appl. 15 (2009). To appear. [Palladino 2009b] F. J. Palladino, “On the characterization of rational difference equations”, J. Dif-ference Equa. Appl. 15 (2009), 253–260. Received: 2008-12-08 Accepted: 2008-12-09 glugo@mail.uri.edu Department of Mathematics, University of Rhode Island, 5 Lippitt Road, Kingston, RI 02881, United States
15477	https://www.grandinetti.org/titrations	Solution Stoichiometry Now let's use molarity in some stoichiometric calculations. Calculate the minimum amount (in liters) of a 0.050 M BaCl2 solution that is required to precipitate all the SO42-(aq) in 0.10 liter of a 0.10 M Na2SO4 solution. Ba2+(aq) + SO42-(aq) → BaSO4 In order to use the chemical equation, we need to work in terms of moles. The concentration of SO42-(aq) is converted into moles by multiplying by the volume. Solution Stoichiometry - Calculate Moles or Mass:Quiz Solution Stoichiometry - Determine Volume:Quiz Solution Stoichiometry - Determine Molarity:Quiz Titrations A useful way to determine a solute's concentration in a solution is to react the solution with a solute in another solution of known concentration. This is known as a Titration. Titration: : Experiment which determines the concentration of a solute (reactant) using its reaction of known stoichiometry with another solution (reactant) of known concentration. For example, if I have a solution of sulfuric acid, H2SO4 (aq), but don't know its concentration, then I can react it with a NaOH solution of known concentration. 2 NaOH(aq) + H2SO4 (aq) → Na2SO4 (aq) + 2 H2O(l) In a titration, the titrant is added dropwise until the reaction is complete. Equivalence Point: : Point at which stoichiometrically equivalent quantities are brought together. 40.0 mL of 0.20 M NaOH is needed to neutralize (reach equivalence point) for 20.0 mL of H2SO4 solution. What is the concentration of H2SO4 solution? 2 NaOH + H2SO4 → Na2SO4 + 2 H2O How do we know when the reaction is complete? We add a tiny amount of indicator to the analyte that will change color when the solution has excess titrant (e.g., excess OH-). For example, phenolphthalein molecules are colorless in neutral and acidic solutions, but are reddish purple in basic (i.e., excess OH-) solutions. Indicator: : A material which (by changing colors or other means) signals that the equivalence point has been attained. End Point: : When the indicator indicates that the equivalence point has been reached (e.g., changes color). Indicators need to be chosen carefully so they don't change color too soon or too late. Homework from Chemisty, The Central Science, 10th Ed. 4.29, 4.31, 4.59, 4.61, 4.63, 4.65, 4.67 4.69, 4.71, 4.73, 4.77, 4.79, 4.81, 4.83, 4.85
15478	https://www.isa-afp.org/browser_info/current/AFP/Prime_Distribution_Elementary/outline.pdf	Elementary Facts About the Distribution of Primes Manuel Eberl March 17, 2025 Abstract This entry is a formalisation of Chapter 4 (and parts of Chapter 3) of Apostol’s Introduction to Analytic Number Theory. The main topics that are addressed are properties of the distribution of prime numbers that can be shown in an elementary way (i. e. without the Prime Number Theorem), the various equivalent forms of the PNT (which imply each other in elementary ways), and consequences that follow from the PNT in elementary ways. The latter include bounds for the number of distinct prime factors of n, the divisor function d(n), Euler’s totient function ϕ(n), and lcm(1, . . . , n). 1 Contents 1 Auxiliary material 4 1.1 Various facts about Dirichlet series . . . . . . . . . . . . . . . 6 1.2 Facts about prime-counting functions . . . . . . . . . . . . . . 6 1.3 Strengthening ‘Big-O’ bounds . . . . . . . . . . . . . . . . . . 7 2 Miscellaneous material 8 2.1 Generalised Dirichlet products . . . . . . . . . . . . . . . . . 8 2.2 Legendre’s identity . . . . . . . . . . . . . . . . . . . . . . . . 10 2.3 A weighted sum of the Möbius µ function . . . . . . . . . . . 11 3 The Prime ω function 11 4 The Primorial function 12 4.1 Deﬁnition and basic properties . . . . . . . . . . . . . . . . . 12 4.2 An alternative view on primorials . . . . . . . . . . . . . . . . 13 4.3 Maximal compositeness of primorials . . . . . . . . . . . . . . 15 5 The LCM of the ﬁrst n natural numbers 15 6 Shapiro’s Tauberian Theorem 16 6.1 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 6.2 Applications to the Chebyshev functions . . . . . . . . . . . . 17 7 Bounds on partial sums of the ζ function 18 8 The summatory Möbius µ function 20 9 Elementary bounds on π(x) and pn 22 9.1 Preliminary lemmas . . . . . . . . . . . . . . . . . . . . . . . 22 9.2 Lower bound for π(x) . . . . . . . . . . . . . . . . . . . . . . 23 9.3 Upper bound for ϑ(x) . . . . . . . . . . . . . . . . . . . . . . 23 9.4 Upper bound for π(x) . . . . . . . . . . . . . . . . . . . . . . 24 9.5 Bounds for pn . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 10 The asymptotics of the summatory divisor σ function 25 10.1 Case 1: α = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 10.2 Case 2: α > 0, α ̸= 1 . . . . . . . . . . . . . . . . . . . . . . . 26 10.3 Case 3: α < 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 11 Selberg’s asymptotic formula 27 2 12 Consequences of the Prime Number Theorem 28 12.1 Existence of primes in intervals . . . . . . . . . . . . . . . . . 29 12.2 The logarithm of the primorial . . . . . . . . . . . . . . . . . 29 12.3 Consequences of the asymptotics of ψ and ϑ . . . . . . . . . . 30 12.4 Bounds on the prime ω function . . . . . . . . . . . . . . . . 31 12.5 Bounds on the divisor function . . . . . . . . . . . . . . . . . 31 12.6 Mertens’ Third Theorem . . . . . . . . . . . . . . . . . . . . . 33 12.7 Bounds on Euler’s totient function . . . . . . . . . . . . . . . 34 3 1 Auxiliary material theory Prime-Distribution-Elementary-Library imports Zeta-Function.Zeta-Function Prime-Number-Theorem.Prime-Counting-Functions Stirling-Formula.Stirling-Formula begin lemma divisor-count-pos [intro]: n > 0 = ⇒divisor-count n > 0 ⟨proof ⟩ lemma divisor-count-eq-0-iﬀ[simp]: divisor-count n = 0 ← →n = 0 ⟨proof ⟩ lemma divisor-count-pos-iﬀ[simp]: divisor-count n > 0 ← →n > 0 ⟨proof ⟩ lemma smallest-prime-beyond-eval: prime n = ⇒smallest-prime-beyond n = n ¬prime n = ⇒smallest-prime-beyond n = smallest-prime-beyond (Suc n) ⟨proof ⟩ lemma nth-prime-numeral: nth-prime (numeral n) = smallest-prime-beyond (Suc (nth-prime (pred-numeral n))) ⟨proof ⟩ lemmas nth-prime-eval = smallest-prime-beyond-eval nth-prime-Suc nth-prime-numeral lemma nth-prime-1 [simp]: nth-prime (Suc 0) = 3 ⟨proof ⟩ lemma nth-prime-2 [simp]: nth-prime 2 = 5 ⟨proof ⟩ lemma nth-prime-3 [simp]: nth-prime 3 = 7 ⟨proof ⟩ lemma strict-mono-sequence-partition: assumes strict-mono (f :: nat ⇒′a :: {linorder, no-top}) assumes x ≥f 0 assumes ﬁlterlim f at-top at-top shows ∃k. x ∈{f k..<f (Suc k)} ⟨proof ⟩ lemma nth-prime-partition: assumes x ≥2 4 shows ∃k. x ∈{nth-prime k..<nth-prime (Suc k)} ⟨proof ⟩ lemma nth-prime-partition ′: assumes x ≥2 shows ∃k. x ∈{real (nth-prime k).. nth-prime k n < nth-prime (Suc k) shows ¬prime n ⟨proof ⟩ lemma nth-prime-partition ′′: includes prime-counting-syntax assumes x ≥(2 :: real) shows x ∈{real (nth-prime (nat ⌊π x⌋−1)).. 0 and s ̸= 1 shows hurwitz-zeta (a + real n) s = hurwitz-zeta a s −(P k 0 = ⇒real n ≤x = ⇒f n = f ′ n) = ⇒x = x ′ = ⇒sum-upto f x = sum-upto f ′ x ′ ⟨proof ⟩ lemma ﬁnite-primes-le: ﬁnite {p. prime p ∧real p ≤x} 5 ⟨proof ⟩ lemma frequently-ﬁltermap: frequently P (ﬁltermap f F) = frequently (λn. P (f n)) F ⟨proof ⟩ lemma frequently-mono-ﬁlter: frequently P F = ⇒F ≤F ′ = ⇒frequently P F ′ ⟨proof ⟩ lemma π-at-top: ﬁlterlim primes-pi at-top at-top ⟨proof ⟩ lemma sum-upto-ln-stirling-weak-bigo: (λx. sum-upto ln x −x ∗ln x + x) ∈O(ln) ⟨proof ⟩ 1.1 Various facts about Dirichlet series lemma fds-mangoldt ′: fds mangoldt = fds-zeta ∗fds-deriv (fds moebius-mu) ⟨proof ⟩ lemma sum-upto-divisor-sum1: sum-upto (λn. P d \| d dvd n. f d :: real) x = sum-upto (λn. f n ∗ﬂoor (x / n)) x ⟨proof ⟩ lemma sum-upto-divisor-sum2: sum-upto (λn. P d \| d dvd n. f d :: real) x = sum-upto (λn. sum-upto f (x / n)) x ⟨proof ⟩ lemma sum-upto-moebius-times-ﬂoor-linear: sum-upto (λn. moebius-mu n ∗⌊x / real n⌋) x = (if x ≥1 then 1 else 0) ⟨proof ⟩ lemma ln-fact-conv-sum-mangoldt: sum-upto (λn. mangoldt n ∗⌊x / real n⌋) x = ln (fact (nat ⌊x⌋)) ⟨proof ⟩ 1.2 Facts about prime-counting functions lemma abs-π [simp]: \|primes-pi x\| = primes-pi x ⟨proof ⟩ lemma π-less-self : includes prime-counting-syntax assumes x > 0 shows π x < x ⟨proof ⟩ 6 lemma π-le-self ′: includes prime-counting-syntax assumes x ≥1 shows π x ≤x −1 ⟨proof ⟩ lemma π-le-self : includes prime-counting-syntax assumes x ≥0 shows π x ≤x ⟨proof ⟩ 1.3 Strengthening ‘Big-O’ bounds The following two statements are crucial: They allow us to strengthen a ‘Big-O’ statement for n →∞or x →∞to a bound for all n ≥n0 or all x ≥x0 under some mild conditions. This allows us to use all the machinery of asymptotics in Isabelle and still get a bound that is applicable over the full domain of the function in the end. This is important because Newman often shows that f(x) ∈O(g(x)) and then writes X n≤x f(x n) = X n≤x O(g(x n)) which is not easy to justify otherwise. lemma natfun-bigoE: ﬁxes f :: nat ⇒-assumes bigo: f ∈O(g) and nz: Vn. n ≥n0 = ⇒g n ̸= 0 obtains c where c > 0 Vn. n ≥n0 = ⇒norm (f n) ≤c ∗norm (g n) ⟨proof ⟩ lemma bigoE-bounded-real-fun: ﬁxes f g :: real ⇒real assumes f ∈O(g) assumes Vx. x ≥x0 = ⇒\|g x\| ≥cg cg > 0 assumes Vb. b ≥x0 = ⇒bounded (f ‘ {x0..b}) shows ∃c>0. ∀x≥x0. \|f x\| ≤c ∗\|g x\| ⟨proof ⟩ lemma sum-upto-asymptotics-lift-nat-real-aux: ﬁxes f :: nat ⇒real and g :: real ⇒real assumes bigo: (λn. (P k=1..n. f k) −g (real n)) ∈O(λn. h (real n)) assumes g-bigo-self : (λn. g (real n) −g (real (Suc n))) ∈O(λn. h (real n)) assumes h-bigo-self : (λn. h (real n)) ∈O(λn. h (real (Suc n))) assumes h-pos: Vx. x ≥1 = ⇒h x > 0 assumes mono-g: mono-on {1..} g ∨mono-on {1..} (λx. −g x) assumes mono-h: mono-on {1..} h ∨mono-on {1..} (λx. −h x) shows ∃c>0. ∀x≥1. sum-upto f x −g x ≤c ∗h x 7 ⟨proof ⟩ lemma sum-upto-asymptotics-lift-nat-real: ﬁxes f :: nat ⇒real and g :: real ⇒real assumes bigo: (λn. (P k=1..n. f k) −g (real n)) ∈O(λn. h (real n)) assumes g-bigo-self : (λn. g (real n) −g (real (Suc n))) ∈O(λn. h (real n)) assumes h-bigo-self : (λn. h (real n)) ∈O(λn. h (real (Suc n))) assumes h-pos: Vx. x ≥1 = ⇒h x > 0 assumes mono-g: mono-on {1..} g ∨mono-on {1..} (λx. −g x) assumes mono-h: mono-on {1..} h ∨mono-on {1..} (λx. −h x) shows ∃c>0. ∀x≥1. \|sum-upto f x −g x\| ≤c ∗h x ⟨proof ⟩ lemma (in factorial-semiring) primepow-divisors-induct [case-names zero unit fac-tor]: assumes P 0 Vx. is-unit x = ⇒P x Vp k x. prime p = ⇒k > 0 = ⇒¬p dvd x = ⇒P x = ⇒P (p ^ k ∗x) shows P x ⟨proof ⟩ end 2 Miscellaneous material theory More-Dirichlet-Misc imports Prime-Distribution-Elementary-Library Prime-Number-Theorem.Prime-Counting-Functions begin 2.1 Generalised Dirichlet products deﬁnition dirichlet-prod ′ :: (nat ⇒′a :: comm-semiring-1) ⇒(real ⇒′a) ⇒real ⇒′a where dirichlet-prod ′ f g x = sum-upto (λm. f m ∗g (x / real m)) x lemma dirichlet-prod ′-one-left: dirichlet-prod ′ (λn. if n = 1 then 1 else 0) f x = (if x ≥1 then f x else 0) ⟨proof ⟩ lemma dirichlet-prod ′-cong: assumes Vn. n > 0 = ⇒real n ≤x = ⇒f n = f ′ n assumes Vy. y ≥1 = ⇒y ≤x = ⇒g y = g ′ y assumes x = x ′ shows dirichlet-prod ′ f g x = dirichlet-prod ′ f ′ g ′ x ′ ⟨proof ⟩ lemma dirichlet-prod ′-assoc: 8 dirichlet-prod ′ f (λy. dirichlet-prod ′ g h y) x = dirichlet-prod ′ (dirichlet-prod f g) h x ⟨proof ⟩ lemma dirichlet-prod ′-inversion1: assumes ∀x≥1. g x = dirichlet-prod ′ a f x x ≥1 dirichlet-prod a ainv = (λn. if n = 1 then 1 else 0) shows f x = dirichlet-prod ′ ainv g x ⟨proof ⟩ lemma dirichlet-prod ′-inversion2: assumes ∀x≥1. f x = dirichlet-prod ′ ainv g x x ≥1 dirichlet-prod a ainv = (λn. if n = 1 then 1 else 0) shows g x = dirichlet-prod ′ a f x ⟨proof ⟩ lemma dirichlet-prod ′-inversion: assumes dirichlet-prod a ainv = (λn. if n = 1 then 1 else 0) shows (∀x≥1. g x = dirichlet-prod ′ a f x) ← →(∀x≥1. f x = dirichlet-prod ′ ainv g x) ⟨proof ⟩ lemma dirichlet-prod ′-inversion ′: assumes a 1 ∗y = 1 deﬁnes ainv ≡dirichlet-inverse a y shows (∀x≥1. g x = dirichlet-prod ′ a f x) ← →(∀x≥1. f x = dirichlet-prod ′ ainv g x) ⟨proof ⟩ lemma dirichlet-prod ′-ﬂoor-conv-sum-upto: dirichlet-prod ′ f (λx. real-of-int (ﬂoor x)) x = sum-upto (λn. sum-upto f (x / n)) x ⟨proof ⟩ lemma (in completely-multiplicative-function) dirichlet-prod-self : dirichlet-prod f f n = f n ∗of-nat (divisor-count n) ⟨proof ⟩ lemma completely-multiplicative-imp-moebius-mu-inverse: ﬁxes f :: nat ⇒′a :: {comm-ring-1} assumes completely-multiplicative-function f shows dirichlet-prod f (λn. moebius-mu n ∗f n) n = (if n = 1 then 1 else 0) ⟨proof ⟩ lemma dirichlet-prod-inversion-completely-multiplicative: ﬁxes a :: nat ⇒′a :: comm-ring-1 9 assumes completely-multiplicative-function a shows (∀x≥1. g x = dirichlet-prod ′ a f x) ← → (∀x≥1. f x = dirichlet-prod ′ (λn. moebius-mu n ∗a n) g x) ⟨proof ⟩ lemma divisor-sigma-conv-dirichlet-prod: divisor-sigma x n = dirichlet-prod (λn. real n powr x) (λ-. 1) n ⟨proof ⟩ 2.2 Legendre’s identity deﬁnition legendre-aux :: real ⇒nat ⇒nat where legendre-aux x p = (if prime p then (P m \| m > 0 ∧real (p ^ m) ≤x. nat ⌊x / p ^ m⌋) else 0) lemma legendre-aux-not-prime [simp]: ¬prime p = ⇒legendre-aux x p = 0 ⟨proof ⟩ lemma legendre-aux-eq-0: assumes real p > x shows legendre-aux x p = 0 ⟨proof ⟩ lemma legendre-aux-posD: assumes legendre-aux x p > 0 shows prime p real p ≤x ⟨proof ⟩ lemma exponents-le-ﬁnite: assumes p > (1 :: nat) k > 0 shows ﬁnite {i. real (p ^ (k ∗i + l)) ≤x} ⟨proof ⟩ lemma ﬁnite-sum-legendre-aux: assumes prime p shows ﬁnite {m. m > 0 ∧real (p ^ m) ≤x} ⟨proof ⟩ lemma legendre-aux-set-eq: assumes prime p x ≥1 shows {m. m > 0 ∧real (p ^ m) ≤x} = {0<..nat ⌊log (real p) x⌋} ⟨proof ⟩ lemma legendre-aux-altdef1: legendre-aux x p = (if prime p ∧x ≥1 then (P m∈{0<..nat ⌊log (real p) x⌋}. nat ⌊x / p ^ m⌋) else 0) ⟨proof ⟩ lemma legendre-aux-altdef2: 10 assumes x ≥1 prime p real p ^ Suc k > x shows legendre-aux x p = (P m∈{0<..k}. nat ⌊x / p ^ m⌋) ⟨proof ⟩ theorem legendre-identity: sum-upto ln x = prime-sum-upto (λp. legendre-aux x p ∗ln p) x ⟨proof ⟩ lemma legendre-identity ′: fact (nat ⌊x⌋) = (Q p \| prime p ∧real p ≤x. p ^ legendre-aux x p) ⟨proof ⟩ 2.3 A weighted sum of the Möbius µ function context ﬁxes M :: real ⇒real deﬁnes M ≡(λx. sum-upto (λn. moebius-mu n / n) x) begin lemma abs-sum-upto-moebius-mu-over-n-less: assumes x: x ≥2 shows \|M x\| < 1 ⟨proof ⟩ lemma sum-upto-moebius-mu-over-n-eq: assumes x < 2 shows M x = (if x ≥1 then 1 else 0) ⟨proof ⟩ lemma abs-sum-upto-moebius-mu-over-n-le: \|M x\| ≤1 ⟨proof ⟩ end end 3 The Prime ω function theory Primes-Omega imports Dirichlet-Series.Dirichlet-Series Dirichlet-Series.Divisor-Count begin The prime ω function ω(n) counts the number of distinct prime factors of n. deﬁnition primes-omega :: nat ⇒nat where primes-omega n = card (prime-factors n) lemma primes-omega-prime [simp]: prime p = ⇒primes-omega p = 1 11 ⟨proof ⟩ lemma primes-omega-0 [simp]: primes-omega 0 = 0 ⟨proof ⟩ lemma primes-omega-1 [simp]: primes-omega 1 = 0 ⟨proof ⟩ lemma primes-omega-Suc-0 [simp]: primes-omega (Suc 0) = 0 ⟨proof ⟩ lemma primes-omega-power [simp]: n > 0 = ⇒primes-omega (x ^ n) = primes-omega x ⟨proof ⟩ lemma primes-omega-primepow [simp]: primepow n = ⇒primes-omega n = 1 ⟨proof ⟩ lemma primes-omega-eq-0-iﬀ: primes-omega n = 0 ← →n = 0 ∨n = 1 ⟨proof ⟩ lemma primes-omega-pos [simp, intro]: n > 1 = ⇒primes-omega n > 0 ⟨proof ⟩ lemma primes-omega-mult-coprime: assumes coprime x y x > 0 ∨y > 0 shows primes-omega (x ∗y) = primes-omega x + primes-omega y ⟨proof ⟩ lemma divisor-count-squarefree: assumes squarefree n n > 0 shows divisor-count n = 2 ^ primes-omega n ⟨proof ⟩ end 4 The Primorial function theory Primorial imports Prime-Distribution-Elementary-Library Primes-Omega begin 4.1 Deﬁnition and basic properties deﬁnition primorial :: real ⇒nat where primorial x = Q {p. prime p ∧real p ≤x} lemma primorial-mono: x ≤y = ⇒primorial x ≤primorial y ⟨proof ⟩ 12 lemma prime-factorization-primorial: prime-factorization (primorial x) = mset-set {p. prime p ∧real p ≤x} ⟨proof ⟩ lemma prime-factors-primorial [simp]: prime-factors (primorial x) = {p. prime p ∧real p ≤x} ⟨proof ⟩ lemma primorial-pos [simp, intro]: primorial x > 0 ⟨proof ⟩ lemma primorial-neq-zero [simp]: primorial x ̸= 0 ⟨proof ⟩ lemma of-nat-primes-omega-primorial [simp]: real (primes-omega (primorial x)) = primes-pi x ⟨proof ⟩ lemma primes-omega-primorial: primes-omega (primorial x) = nat ⌊primes-pi x⌋ ⟨proof ⟩ lemma prime-dvd-primorial-iﬀ: prime p = ⇒p dvd primorial x ← →p ≤x ⟨proof ⟩ lemma squarefree-primorial [intro]: squarefree (primorial x) ⟨proof ⟩ lemma primorial-ge: primorial x ≥2 powr primes-pi x ⟨proof ⟩ lemma primorial-at-top: ﬁlterlim primorial at-top at-top ⟨proof ⟩ lemma totient-primorial: real (totient (primorial x)) = real (primorial x) ∗(Q p \| prime p ∧real p ≤x. 1 −1 / real p) for x ⟨proof ⟩ lemma ln-primorial: ln (primorial x) = primes-theta x ⟨proof ⟩ lemma divisor-count-primorial: divisor-count (primorial x) = 2 powr primes-pi x ⟨proof ⟩ 4.2 An alternative view on primorials The following function is an alternative representation of primorials; instead of taking the product of all primes up to a given real bound x, it takes the 13 product of the ﬁrst k primes. This is sometimes more convenient. deﬁnition primorial ′ :: nat ⇒nat where primorial ′ n = (Q k 0 ⟨proof ⟩ lemma primorial ′-neq-0 [simp]: primorial ′ n ̸= 0 ⟨proof ⟩ lemma strict-mono-primorial ′: strict-mono primorial ′ ⟨proof ⟩ lemma prime-factorization-primorial ′: prime-factorization (primorial ′ k) = mset-set (nth-prime ‘ {..<k}) ⟨proof ⟩ lemma prime-factors-primorial ′: prime-factors (primorial ′ k) = nth-prime ‘ {..<k} ⟨proof ⟩ lemma primes-omega-primorial ′ [simp]: primes-omega (primorial ′ k) = k ⟨proof ⟩ lemma squarefree-primorial ′ [intro]: squarefree (primorial ′ x) ⟨proof ⟩ lemma divisor-count-primorial ′ [simp]: divisor-count (primorial ′ k) = 2 ^ k ⟨proof ⟩ lemma totient-primorial ′: totient (primorial ′ k) = primorial ′ k ∗(Q i 0 shows primorial ′ n = primorial (nth-prime (n −1)) ⟨proof ⟩ 14 4.3 Maximal compositeness of primorials Primorials are maximally composite, i. e. any number with k distinct prime factors is as least as big as the primorial with k distinct prime factors, and and number less than a primorial has strictly less prime factors. lemma nth-prime-le-prime-sequence: ﬁxes p :: nat ⇒nat assumes strict-mono-on {..<n} p and Vk. k < n = ⇒prime (p k) and k < n shows nth-prime k ≤p k ⟨proof ⟩ theorem primorial ′-primes-omega-le: ﬁxes n :: nat assumes n: n > 0 shows primorial ′ (primes-omega n) ≤n ⟨proof ⟩ lemma primes-omega-less-primes-omega-primorial: ﬁxes n :: nat assumes n: n > 0 and n < primorial x shows primes-omega n < primes-omega (primorial x) ⟨proof ⟩ lemma primes-omega-le-primes-omega-primorial: ﬁxes n :: nat assumes n ≤primorial x shows primes-omega n ≤primes-omega (primorial x) ⟨proof ⟩ end 5 The LCM of the ﬁrst n natural numbers theory Lcm-Nat-Upto imports Prime-Number-Theorem.Prime-Counting-Functions begin In this section, we examine Lcm {1..n}. In particular, we will show that it is equal to eψ(n) and thus (by the PNT) en+o(n). lemma multiplicity-Lcm: ﬁxes A :: ′a :: {semiring-Gcd, factorial-semiring-gcd} set assumes ﬁnite A A ̸= {} prime p 0 / ∈A shows multiplicity p (Lcm A) = Max (multiplicity p ‘ A) ⟨proof ⟩ The multiplicity of any prime p in Lcm {1..n} diﬀers from that in Lcm {1..n −1} iﬀn is a power of p, in which case it is greater by 1. lemma multiplicity-Lcm-atLeast1AtMost-Suc: 15 ﬁxes p n :: nat assumes p: prime p and n: n > 0 shows multiplicity p (Lcm {1..Suc n}) = (if ∃k. Suc n = p ^ k then 1 else 0) + multiplicity p (Lcm {1..n}) ⟨proof ⟩ Consequently, Lcm {1..n} diﬀers from Lcm {1..n −1} iﬀn is of the form pk for some prime p, in which case it is greater by a factor of p. lemma Lcm-atLeast1AtMost-Suc: Lcm {1..Suc n} = Lcm {1..n} ∗(if primepow (Suc n) then aprimedivisor (Suc n) else 1) ⟨proof ⟩ It follows by induction that Lcm {1..n} = eψ(n). lemma Lcm-atLeast1AtMost-conv-ψ: includes prime-counting-syntax shows real (Lcm {1..n}) = exp (ψ (real n)) ⟨proof ⟩ lemma Lcm-upto-real-conv-ψ: includes prime-counting-syntax shows real (Lcm {1..nat ⌊x⌋}) = exp (ψ x) ⟨proof ⟩ end 6 Shapiro’s Tauberian Theorem theory Shapiro-Tauberian imports More-Dirichlet-Misc Prime-Number-Theorem.Prime-Counting-Functions Prime-Distribution-Elementary-Library begin 6.1 Proof Given an arithmeticla function a(n), Shapiro’s Tauberian theorem relates the sum P n≤x a(n) to the weighted sums P n≤x a(n)⌊x n⌋and P n≤x a(n)/n. More precisely, it shows that if P n≤x a(n)⌊x n⌋= x ln x + O(x), then: • P n≤x a(n) n = ln x + O(1) • P n≤x a(n) ≤Bx for some constant B ≥0 and all x ≥0 • P n≤x a(n) ≥Cx for some constant C > 0 and all x ≥1/C 16 locale shapiro-tauberian = ﬁxes a :: nat ⇒real and A S T :: real ⇒real deﬁnes A ≡sum-upto (λn. a n / n) deﬁnes S ≡sum-upto a deﬁnes T ≡(λx. dirichlet-prod ′ a ﬂoor x) assumes a-nonneg: Vn. n > 0 = ⇒a n ≥0 assumes a-asymptotics: (λx. T x −x ∗ln x) ∈O(λx. x) begin lemma ﬁn: ﬁnite X if X ⊆{n. real n ≤x} for X x ⟨proof ⟩ lemma S-mono: S x ≤S y if x ≤y for x y ⟨proof ⟩ lemma split: ﬁxes f :: nat ⇒real assumes α ∈{0..1} shows sum-upto f x = sum-upto f (α∗x) + (P n \| n > 0 ∧real n ∈{α∗x<..x}. f n) ⟨proof ⟩ lemma S-diﬀ-T-diﬀ: S x −S (x / 2) ≤T x −2 ∗T (x / 2) ⟨proof ⟩ lemma shows diﬀ-bound-strong: ∃c≥0. ∀x≥0. x ∗A x −T x ∈{0..c∗x} and asymptotics: (λx. A x −ln x) ∈O(λ-. 1) and upper: ∃c≥0. ∀x≥0. S x ≤c ∗x and lower: ∃c>0. ∀x≥1/c. S x ≥c ∗x and bigtheta: S ∈Θ(λx. x) ⟨proof ⟩ end 6.2 Applications to the Chebyshev functions We can now apply Shapiro’s Tauberian theorem to ψ and ϑ. lemma dirichlet-prod-mangoldt1-ﬂoor-bigo: includes prime-counting-syntax shows (λx. dirichlet-prod ′ (λn. ind prime n ∗ln n) ﬂoor x −x ∗ln x) ∈O(λx. x) ⟨proof ⟩ lemma dirichlet-prod ′-mangoldt-ﬂoor-asymptotics: (λx. dirichlet-prod ′ mangoldt ﬂoor x −x ∗ln x + x) ∈O(ln) ⟨proof ⟩ 17 interpretation ψ: shapiro-tauberian mangoldt sum-upto (λn. mangoldt n / n) primes-psi dirichlet-prod ′ mangoldt ﬂoor ⟨proof ⟩ thm ψ.asymptotics ψ.upper ψ.lower interpretation ϑ: shapiro-tauberian λn. ind prime n ∗ln n sum-upto (λn. ind prime n ∗ln n / n) primes-theta dirichlet-prod ′ (λn. ind prime n ∗ln n) ﬂoor ⟨proof ⟩ thm ϑ.asymptotics ϑ.upper ϑ.lower lemma sum-upto-ψ-x-over-n-asymptotics: (λx. sum-upto (λn. primes-psi (x / n)) x −x ∗ln x + x) ∈O(ln) and sum-upto-ϑ-x-over-n-asymptotics: (λx. sum-upto (λn. primes-theta (x / n)) x −x ∗ln x) ∈O(λx. x) ⟨proof ⟩ end 7 Bounds on partial sums of the ζ function theory Partial-Zeta-Bounds imports Euler-MacLaurin.Euler-MacLaurin-Landau Zeta-Function.Zeta-Function Prime-Number-Theorem.Prime-Number-Theorem-Library Prime-Distribution-Elementary-Library begin We employ Euler–MacLaurin’s summation formula to obtain asymptotic estimates for the partial sums of the Riemann ζ(s) function for ﬁxed real a, i. e. the function f(n) = n X k=1 k−s . We distinguish various cases. The case s = 1 is simply the Harmonic num-bers and is treated apart from the others. lemma harm-asymp-equiv: sum-upto (λn. 1 / n) ∼[at-top] ln ⟨proof ⟩ lemma ﬁxes s :: real 18 assumes s: s > 0 s ̸= 1 shows zeta-partial-sum-bigo-pos: (λn. (P k=1..n. real k powr −s) −real n powr (1 −s) / (1 −s) −Re (zeta s)) ∈O(λx. real x powr −s) and zeta-partial-sum-bigo-pos ′: (λn. P k=1..n. real k powr −s) =o (λn. real n powr (1 −s) / (1 −s) + Re (zeta s)) +o O(λx. real x powr −s) ⟨proof ⟩ lemma zeta-tail-bigo: ﬁxes s :: real assumes s: s > 1 shows (λn. Re (hurwitz-zeta (real n + 1) s)) ∈O(λx. real x powr (1 −s)) ⟨proof ⟩ lemma zeta-tail-bigo ′: ﬁxes s :: real assumes s: s > 1 shows (λn. Re (hurwitz-zeta (real n) s)) ∈O(λx. real x powr (1 −s)) ⟨proof ⟩ lemma ﬁxes s :: real assumes s: s > 0 shows zeta-partial-sum-bigo-neg: (λn. (P i=1..n. real i powr s) −n powr (1 + s) / (1 + s)) ∈O(λn. n powr s) and zeta-partial-sum-bigo-neg ′: (λn. (P i=1..n. real i powr s)) =o (λn. n powr (1 + s) / (1 + s)) +o O(λn. n powr s) ⟨proof ⟩ lemma zeta-partial-sum-le-pos: assumes s > 0 s ̸= 1 deﬁnes z ≡Re (zeta (complex-of-real s)) shows ∃c>0. ∀x≥1. \|sum-upto (λn. n powr −s) x −(x powr (1−s) / (1−s) + z)\| ≤c ∗x powr −s ⟨proof ⟩ lemma zeta-partial-sum-le-pos ′: assumes s > 0 s ̸= 1 deﬁnes z ≡Re (zeta (complex-of-real s)) shows ∃c>0. ∀x≥1. \|sum-upto (λn. n powr −s) x −x powr (1−s) / (1−s)\| ≤c ⟨proof ⟩ lemma zeta-partial-sum-le-pos ′′: 19 assumes s > 0 s ̸= 1 shows ∃c>0. ∀x≥1. \|sum-upto (λn. n powr −s) x\| ≤c ∗x powr max 0 (1 − s) ⟨proof ⟩ lemma zeta-partial-sum-le-pos-bigo: assumes s > 0 s ̸= 1 shows (λx. sum-upto (λn. n powr −s) x) ∈O(λx. x powr max 0 (1 −s)) ⟨proof ⟩ lemma zeta-partial-sum-01-asymp-equiv: assumes s ∈{0<..<1} shows sum-upto (λn. n powr −s) ∼[at-top] (λx. x powr (1 −s) / (1 −s)) ⟨proof ⟩ lemma zeta-partial-sum-gt-1-asymp-equiv: ﬁxes s :: real assumes s > 1 deﬁnes ζ ≡Re (zeta s) shows sum-upto (λn. n powr −s) ∼[at-top] (λx. ζ) ⟨proof ⟩ lemma zeta-partial-sum-pos-bigtheta: assumes s > 0 s ̸= 1 shows sum-upto (λn. n powr −s) ∈Θ(λx. x powr max 0 (1 −s)) ⟨proof ⟩ lemma zeta-partial-sum-le-neg: assumes s > 0 shows ∃c>0. ∀x≥1. \|sum-upto (λn. n powr s) x −x powr (1 + s) / (1 + s)\| ≤c ∗x powr s ⟨proof ⟩ lemma zeta-partial-sum-neg-asymp-equiv: assumes s > 0 shows sum-upto (λn. n powr s) ∼[at-top] (λx. x powr (1 + s) / (1 + s)) ⟨proof ⟩ end 8 The summatory Möbius µ function theory Moebius-Mu-Sum imports More-Dirichlet-Misc Dirichlet-Series.Partial-Summation Prime-Number-Theorem.Prime-Counting-Functions Dirichlet-Series.Arithmetic-Summatory-Asymptotics Shapiro-Tauberian 20 Partial-Zeta-Bounds Prime-Number-Theorem.Prime-Number-Theorem-Library Prime-Distribution-Elementary-Library begin In this section, we shall examine the summatory Möbius µ function M(x) := P n≤x µ(n). The main result is that M(x) ∈o(x) is equivalent to the Prime Number Theorem. context includes prime-counting-syntax ﬁxes M H :: real ⇒real deﬁnes M ≡sum-upto moebius-mu deﬁnes H ≡sum-upto (λn. moebius-mu n ∗ln n) begin lemma sum-upto-moebius-mu-integral: x > 1 = ⇒((λt. M t / t) has-integral M x ∗ln x −H x) {1..x} and sum-upto-moebius-mu-integrable: a ≥1 = ⇒(λt. M t / t) integrable-on {a..b} ⟨proof ⟩ lemma sum-moebius-mu-bound: assumes x ≥0 shows \|M x\| ≤x ⟨proof ⟩ lemma sum-moebius-mu-aux1: (λx. M x / x −H x / (x ∗ln x)) ∈O(λx. 1 / ln x) ⟨proof ⟩ lemma sum-moebius-mu-aux2: ((λx. M x / x −H x / (x ∗ln x)) − − − →0) at-top ⟨proof ⟩ lemma sum-moebius-mu-ln-eq: H = (λx. −dirichlet-prod ′ moebius-mu ψ x) ⟨proof ⟩ theorem PNT-implies-sum-moebius-mu-sublinear: assumes ψ ∼[at-top] (λx. x) shows M ∈o(λx. x) ⟨proof ⟩ theorem sum-moebius-mu-sublinear-imp-PNT: assumes M ∈o(λx. x) shows ψ ∼[at-top] (λx. x) ⟨proof ⟩ We now turn to a related fact: For the weighted sum A(x) := P n≤x µ(n)/n, 21 the asymptotic relation A(x) ∈o(1) is also equivalent to the Prime Number Theorem. Like Apostol, we only show one direction, namely that A(x) ∈ o(1) implies the PNT. context ﬁxes A deﬁnes A ≡sum-upto (λn. moebius-mu n / n) begin lemma sum-upto-moebius-mu-integral ′: x > 1 = ⇒(A has-integral x ∗A x −M x) {1..x} and sum-upto-moebius-mu-integrable ′: a ≥1 = ⇒A integrable-on {a..b} ⟨proof ⟩ theorem sum-moebius-mu-div-n-smallo-imp-PNT: assumes smallo: A ∈o(λ-. 1) shows M ∈o(λx. x) and ψ ∼[at-top] (λx. x) ⟨proof ⟩ end end end 9 Elementary bounds on π(x) and pn theory Elementary-Prime-Bounds imports Prime-Number-Theorem.Prime-Counting-Functions Prime-Distribution-Elementary-Library More-Dirichlet-Misc begin In this section, we will follow Apostol and give elementary proofs of Chebyshev-type lower and upper bounds for π(x), i. e. c1x/ ln x < π(x) < c2x/ ln x. From this, similar bounds for pn follow as easy corollaries. 9.1 Preliminary lemmas The following two estimates relating the central Binomial coeﬃcient to pow-ers of 2 and 4 form the starting point for Apostol’s elementary bounds for π(x): lemma twopow-le-central-binomial: 2 ^ n ≤((2 ∗n) choose n) ⟨proof ⟩ lemma fourpow-gt-central-binomial: assumes n > 0 22 shows 4 ^ n > ((2 ∗n) choose n) ⟨proof ⟩ 9.2 Lower bound for π(x) context includes prime-counting-syntax ﬁxes S :: nat ⇒nat ⇒int deﬁnes S ≡(λn p. (P m∈{0<..nat ⌊log p (2∗n)⌋}. ⌊2∗n/p^m⌋−2 ∗⌊n/p^m⌋)) begin We now ﬁrst prove the bound π(x) ≥1 6x/ ln x for x ≥2. The constant could probably be improved for starting points greater than 2; this is true for most of the constants in this section. The ﬁrst step is to show a slightly stronger bound for even numbers, where the constant is 1 2 ln 2 ≈0.347: lemma ﬁxes n :: nat assumes n: n ≥1 shows π-bounds-aux: ln (fact (2 ∗n)) −2 ∗ln (fact n) = prime-sum-upto (λp. S n p ∗ln p) (2 ∗n) and π-lower-bound-ge-strong: π (2 ∗n) ≥ln 2 / 2 ∗(2 ∗n) / ln (2 ∗n) ⟨proof ⟩ lemma ln-2-ge-56-81: ln 2 ≥(56 / 81 :: real) ⟨proof ⟩ The bound for any real number x ≥2 follows fairly easily, although some ugly accounting for error terms has to be done. theorem π-lower-bound: ﬁxes x :: real assumes x: x ≥2 shows π x > (1 / 6) ∗(x / ln x) ⟨proof ⟩ lemma π-at-top: ﬁlterlim primes-pi at-top at-top ⟨proof ⟩ 9.3 Upper bound for ϑ(x) In this section, we prove a linear upper bound for ϑ. This is somewhat unnecessary because we already have a considerably better bound on ϑ(x) using a proof that has roughly the same complexity as this one and also only uses elementary means. Nevertheless, here is the proof from Apostol’s book; it is quite nice and it would be a shame not to formalise it. The idea is to ﬁrst show a bound for ϑ(2n) −ϑ(n) and then deduce one for ϑ(2n) from this by telescoping, which then yields one for general x by 23 monotonicity. lemma ϑ-double-less: ﬁxes n :: nat assumes n: n > 0 shows ϑ (2 ∗real n) −ϑ (real n) < real n ∗ln 4 ⟨proof ⟩ lemma ϑ-twopow-less: ϑ (2 ^ r) < 2 ^ (r + 1) ∗ln 2 ⟨proof ⟩ theorem ϑ-upper-bound-weak: ﬁxes n :: nat assumes n: n > 0 shows ϑ n < 4 ∗ln 2 ∗n ⟨proof ⟩ 9.4 Upper bound for π(x) We use our upper bound for ϑ(x) (the strong one, not the one from the previous section) to derive an upper bound for π(x). As a ﬁrst step, we show the following lemma about the global maximum of the function ln x/xc for c > 0: lemma π-upper-bound-aux: ﬁxes c :: real assumes c > 0 deﬁnes f ≡(λx. x powr (−c) ∗ln x) assumes x: x > 0 shows f x ≤1 / (c ∗exp 1) ⟨proof ⟩ Following Apostol, we ﬁrst show a generic bound depending on some real-valued parameter α: lemma π-upper-bound-strong: ﬁxes α :: real and n :: nat assumes n: n ≥2 and α: α ∈{0<..<1} shows π n < (1 / ((1 −α) ∗exp 1) + ln 4 / α) ∗n / ln n ⟨proof ⟩ The choice α := 2 3 then leads to the upper bound π(x) < cx/ ln x with c = 3(e−1 + ln 2) ≈3.183. This is considerably stronger than Apostol’s bound. theorem π-upper-bound: ﬁxes x :: real assumes x ≥2 shows π x < 3 ∗(exp (−1) + ln 2) ∗x / ln x ⟨proof ⟩ 24 corollary π-upper-bound ′: ﬁxes x :: real assumes x ≥2 shows π x < 443 / 139 ∗(x / ln x) ⟨proof ⟩ corollary π-upper-bound ′′: ﬁxes x :: real assumes x ≥2 shows π x < 4 ∗(x / ln x) ⟨proof ⟩ In particular, we have now shown a weak version of the Prime Number Theorem, namely that π(x) ∈Θ(x/ ln x): lemma π-bigtheta: π ∈Θ(λx. x / ln x) ⟨proof ⟩ 9.5 Bounds for pn By some rearrangements, the lower and upper bounds for π(x) give rise to analogous bounds for pn: lemma nth-prime-lower-bound-gen: assumes c: c > 0 and n: n > 0 assumes Vn. n ≥2 = ⇒π (real n) < (1 / c) ∗(real n / ln (real n)) shows nth-prime (n −1) ≥c ∗(real n ∗ln (real n)) ⟨proof ⟩ corollary nth-prime-lower-bound: n > 0 = ⇒nth-prime (n −1) ≥(139 / 443) ∗(n ∗ln n) ⟨proof ⟩ corollary nth-prime-upper-bound: assumes n: n > 0 shows nth-prime (n −1) < 12 ∗(n ∗ln n + n ∗ln (12 / exp 1)) ⟨proof ⟩ We can thus also conclude that pn ∼n ln n: corollary nth-prime-bigtheta: nth-prime ∈Θ(λn. n ∗ln n) ⟨proof ⟩ end end 10 The asymptotics of the summatory divisor σ function theory Summatory-Divisor-Sigma-Bounds 25 imports Partial-Zeta-Bounds More-Dirichlet-Misc begin In this section, we analyse the asymptotic behaviour of the summatory di-visor functions P n≤x σα(n) for real α. This essentially tells us what the average value of these functions is for large x. The case α = 0 is not treated here since σ0 is simply the divisor function, for which precise asymptotics are already available in the AFP. 10.1 Case 1: α = 1 If α = 1, σα(n) is simply the sum of all divisors of n. Here, the asymptotics is X n≤x σ1(n) = π2 12x2 + O(x ln x) . theorem summatory-divisor-sum-asymptotics: sum-upto divisor-sum =o (λx. pi2 / 12 ∗x ^ 2) +o O(λx. x ∗ln x) ⟨proof ⟩ 10.2 Case 2: α > 0, α ̸= 1 Next, we consider the case α > 0 and α ̸= 1. We then have: X n≤x σα(n) = ζ(α + 1) α + 1 xα+1 + O xmax(1,α) theorem summatory-divisor-sigma-asymptotics-pos: ﬁxes α :: real assumes α: α > 0 α ̸= 1 deﬁnes ζ ≡Re (zeta (α + 1)) shows sum-upto (divisor-sigma α) =o (λx. ζ / (α + 1) ∗x powr (α + 1)) +o O(λx. x powr max 1 α) ⟨proof ⟩ 10.3 Case 3: α < 0 Last, we consider the case of a negative exponent. We have for α > 0: X n≤x σ−α(n) = ζ(α + 1)x + O(R(x)) where R(x) = ln x if α = 1 and R(x) = xmax(0,1−α) otherwise. theorem summatory-divisor-sigma-asymptotics-neg: ﬁxes α :: real assumes α: α > 0 26 deﬁnes δ ≡max 0 (1 −α) deﬁnes ζ ≡Re (zeta (α + 1)) shows sum-upto (divisor-sigma (−α)) =o (if α = 1 then (λx. pi2/6 ∗x) +o O(ln) else (λx. ζ ∗x) +o O(λx. x powr δ)) ⟨proof ⟩ end 11 Selberg’s asymptotic formula theory Selberg-Asymptotic-Formula imports More-Dirichlet-Misc Prime-Number-Theorem.Prime-Counting-Functions Shapiro-Tauberian Euler-MacLaurin.Euler-MacLaurin-Landau Partial-Zeta-Bounds begin Following Apostol, we ﬁrst show an inversion formula: Consider a function f(x) for x ∈R> 0. Deﬁne g(x) := ln x · P n≤x f(x/n). Then: f(x) ln x + X n≤x Λ(n)f(x/n) = X n≤x µ(n)g(x/n) locale selberg-inversion = ﬁxes F G :: real ⇒′a :: {real-algebra-1, comm-ring-1} deﬁnes G ≡(λx. of-real (ln x) ∗sum-upto (λn. F (x / n)) x) begin lemma eq: assumes x ≥1 shows F x ∗of-real (ln x) + dirichlet-prod ′ mangoldt F x = dirichlet-prod ′ moebius-mu G x ⟨proof ⟩ end We can now show Selberg’s formula ψ(x) ln x + X n≤x Λ(n)ψ(x/n) = 2x ln x + O(x) . theorem selberg-asymptotic-formula: includes prime-counting-syntax shows (λx. ψ x ∗ln x + dirichlet-prod ′ mangoldt ψ x) =o (λx. 2 ∗x ∗ln x) +o O(λx. x) 27 ⟨proof ⟩ end 12 Consequences of the Prime Number Theorem theory PNT-Consequences imports Elementary-Prime-Bounds Prime-Number-Theorem.Mertens-Theorems Prime-Number-Theorem.Prime-Counting-Functions Moebius-Mu-Sum Lcm-Nat-Upto Primorial Primes-Omega begin In this section, we will deﬁne a locale that assumes the Prime Number Theorem in order to explore some of its elementary consequences. locale prime-number-theorem = assumes prime-number-theorem [asymp-equiv-intros]: π ∼[at-top] (λx. x / ln x) begin corollary ϑ-asymptotics [asymp-equiv-intros]: ϑ ∼[at-top] (λx. x) ⟨proof ⟩ corollary ψ-asymptotics [asymp-equiv-intros]: ψ ∼[at-top] (λx. x) ⟨proof ⟩ corollary ln-π-asymptotics [asymp-equiv-intros]: (λx. ln (π x)) ∼[at-top] ln ⟨proof ⟩ corollary π-ln-π-asymptotics: (λx. π x ∗ln (π x)) ∼[at-top] (λx. x) ⟨proof ⟩ corollary nth-prime-asymptotics [asymp-equiv-intros]: (λn. real (nth-prime n)) ∼[at-top] (λn. real n ∗ln (real n)) ⟨proof ⟩ corollary moebius-mu-smallo: sum-upto moebius-mu ∈o(λx. x) ⟨proof ⟩ lemma ln-ϑ-asymptotics: includes prime-counting-syntax shows (λx. ln (ϑ x) −ln x) ∈o(λ-. 1) ⟨proof ⟩ lemma ln-ϑ-asymp-equiv [asymp-equiv-intros]: 28 includes prime-counting-syntax shows (λx. ln (ϑ x)) ∼[at-top] ln ⟨proof ⟩ lemma ln-nth-prime-asymptotics: (λn. ln (nth-prime n) −(ln n + ln (ln n))) ∈o(λ-. 1) ⟨proof ⟩ lemma ln-nth-prime-asymp-equiv [asymp-equiv-intros]: (λn. ln (nth-prime n)) ∼[at-top] ln ⟨proof ⟩ The following versions use a little less notation. corollary prime-number-theorem ′: ((λx. π x / (x / ln x)) − − − →1) at-top ⟨proof ⟩ corollary prime-number-theorem ′′: (λx. card {p. prime p ∧real p ≤x}) ∼[at-top] (λx. x / ln x) ⟨proof ⟩ corollary prime-number-theorem ′′′: (λn. card {p. prime p ∧p ≤n}) ∼[at-top] (λn. real n / ln (real n)) ⟨proof ⟩ end 12.1 Existence of primes in intervals For ﬁxed ε, The interval (x; εx] contains a prime number for any suﬃciently large x. This proof was taken from A. J. Hildebrand’s lecture notes . lemma (in prime-number-theorem) prime-in-interval-exists: ﬁxes c :: real assumes c > 1 shows eventually (λx. ∃p. prime p ∧real p ∈{x<..c∗x}) at-top ⟨proof ⟩ The set of rationals whose numerator and denominator are primes is dense in R>0. lemma (in prime-number-theorem) prime-fractions-dense: ﬁxes α ε :: real assumes α > 0 and ε > 0 obtains p q :: nat where prime p and prime q and dist (real p / real q) α < ε ⟨proof ⟩ 12.2 The logarithm of the primorial The PNT directly implies the asymptotics of the logarithm of the primorial function: 29 context prime-number-theorem begin lemma ln-primorial-asymp-equiv [asymp-equiv-intros]: (λx. ln (primorial x)) ∼[at-top] (λx. x) ⟨proof ⟩ lemma ln-ln-primorial-asymp-equiv [asymp-equiv-intros]: (λx. ln (ln (primorial x))) ∼[at-top] (λx. ln x) ⟨proof ⟩ lemma ln-primorial ′-asymp-equiv [asymp-equiv-intros]: (λk. ln (primorial ′ k)) ∼[at-top] (λk. k ∗ln k) and ln-ln-primorial ′-asymp-equiv [asymp-equiv-intros]: (λk. ln (ln (primorial ′ k))) ∼[at-top] (λk. ln k) and ln-over-ln-ln-primorial ′-asymp-equiv: (λk. ln (primorial ′ k) / ln (ln (primorial ′ k))) ∼[at-top] (λk. k) ⟨proof ⟩ end 12.3 Consequences of the asymptotics of ψ and ϑ Next, we will show some consequences of ψ(x) ∼x and ϑ(x) ∼x. To this end, we ﬁrst show generically that any function g = ex+o(x) is o(cn) if c > e and ω(cn) if c < e. locale exp-asymp-equiv-linear = ﬁxes f g :: real ⇒real assumes f-asymp-equiv: f ∼[at-top] (λx. x) assumes g: eventually (λx. g x = exp (f x)) F begin lemma ﬁxes ε :: real assumes ε > 0 shows smallo: g ∈o(λx. exp ((1 + ε) ∗x)) and smallomega: g ∈ω(λx. exp ((1 −ε) ∗x)) ⟨proof ⟩ lemma smallo ′: ﬁxes c :: real assumes c > exp 1 shows g ∈o(λx. c powr x) ⟨proof ⟩ lemma smallomega ′: ﬁxes c :: real assumes c ∈{0<.. (3 :: real) shows ln x > 1 ⟨proof ⟩ lemma (in prime-number-theorem) primes-omega-primorial ′-asymp-equiv: (λk. primes-omega (primorial ′ k)) ∼[at-top] (λk. ln (primorial ′ k) / ln (ln (primorial ′ k))) ⟨proof ⟩ The number of distinct prime factors of n has maximal order ln n/ ln ln n: theorem (in prime-number-theorem) limsup-primes-omega: limsup (λn. primes-omega n / (ln n / ln (ln n))) = 1 ⟨proof ⟩ 12.5 Bounds on the divisor function In this section, we shall examine the growth of the divisor function σ0(n). In particular, we will show that σ0(n) < 2c ln n/ ln ln n for all suﬃciently large n if c > 1 and σ0(n) > 2c ln n/ ln ln n for inﬁnitely many n if c < 1. An equivalent statement is that ln(σ0(n)) has maximal order ln 2·ln n/ ln ln n. 31 Following Apostol’s somewhat didactic approach, we ﬁrst show a generic bounding lemma for σ0 that depends on some function f that we will specify later. lemma divisor-count-bound-gen: ﬁxes f :: nat ⇒real assumes eventually (λn. f n ≥2) at-top deﬁnes c ≡(8 / ln 2 :: real) deﬁnes g ≡(λn. (ln n + c ∗f n ∗ln (ln n)) / (ln (f n))) shows eventually (λn. divisor-count n < 2 powr g n) at-top ⟨proof ⟩ include prime-counting-syntax ⟨proof ⟩ Now, Apostol explains that one can choose f(n) := ln n/(ln ln n)2 to obtain the desired bound. proposition divisor-count-upper-bound: ﬁxes ε :: real assumes ε > 0 shows eventually (λn. divisor-count n < 2 powr ((1 + ε) ∗ln n / ln (ln n))) at-top ⟨proof ⟩ Next, we will examine the ‘worst case’. Since any prime factor of n with multiplicity k contributes a factor of k + 1, it is intuitively clear that σ0(n) is largest w. r. t. n if it is a product of small distinct primes. We show that indeed, if n := x# (where x# denotes the primorial), we have σ0(n) = 2π(x), which, by the Prime Number Theorem, indeed exceeds c ln n/ ln ln n. theorem (in prime-number-theorem) divisor-count-primorial-gt: assumes ε > 0 deﬁnes h ≡primorial shows eventually (λx. divisor-count (h x) > 2 powr ((1 −ε) ∗ln (h x) / ln (ln (h x)))) at-top ⟨proof ⟩ Since h(x) − →∞, this gives us our inﬁnitely many values of n that exceed the bound. corollary (in prime-number-theorem) divisor-count-lower-bound: assumes ε > 0 shows frequently (λn. divisor-count n > 2 powr ((1 −ε) ∗ln n / ln (ln n))) at-top ⟨proof ⟩ A diﬀerent formulation that is not quite as tedious to prove is this one: lemma (in prime-number-theorem) ln-divisor-count-primorial ′-asymp-equiv: (λk. ln (divisor-count (primorial ′ k))) ∼[at-top] (λk. ln 2 ∗ln (primorial ′ k) / ln (ln (primorial ′ k))) 32 ⟨proof ⟩ It follows that the maximal order of the divisor function is ln 2 · ln n/ ln ln n. theorem (in prime-number-theorem) limsup-divisor-count: limsup (λn. ln (divisor-count n) ∗ln (ln n) / ln n) = ln 2 ⟨proof ⟩ 12.6 Mertens’ Third Theorem In this section, we will show that Y p≤x 1 −1 p = C ln x + O 1 ln2 x with explicit bounds for the factor in the ‘Big-O’. Here, C is the following constant: deﬁnition third-mertens-const :: real where third-mertens-const = exp (−(P p::nat. if prime p then −ln (1 −1 / real p) −1 / real p else 0) − meissel-mertens) This constant is actually equal to e−γ where γ is the Euler–Mascheroni constant, but showing this is quite a bit of work, which we shall not do here. lemma third-mertens-const-pos: third-mertens-const > 0 ⟨proof ⟩ theorem deﬁnes C ≡third-mertens-const shows mertens-third-theorem-strong: eventually (λx. \|(Q p \| prime p ∧real p ≤x. 1 −1 / p) −C / ln x\| ≤ 10 ∗C / ln x ^ 2) at-top and mertens-third-theorem: (λx. (Q p \| prime p ∧real p ≤x. 1 −1 / p) −C / ln x) ∈O(λx. 1 / ln x ^ 2) ⟨proof ⟩ lemma mertens-third-theorem-asymp-equiv: (λx. (Q p \| prime p ∧real p ≤x. 1 −1 / real p)) ∼[at-top] (λx. third-mertens-const / ln x) ⟨proof ⟩ We now show an equivalent version where Q p≤x(1 −1/p) is replaced by Qk i=1(1 −1/pi): lemma mertens-third-convert: assumes n > 0 shows (Q k 0 deﬁnes C ≡third-mertens-const shows eventually (λn. totient n > (1 −ε) ∗C ∗n / ln (ln n)) at-top ⟨proof ⟩ include prime-counting-syntax ⟨proof ⟩ Next, we examine the ‘worst case’ of ϕ(n) where n is the primorial of x. In this case, we have ϕ(n) < cn/ ln ln n for any c > C for all suﬃciently large n. theorem (in prime-number-theorem) totient-primorial-less: ﬁxes ε :: real deﬁnes C ≡third-mertens-const and h ≡primorial assumes ε > 0 shows eventually (λx. totient (h x) < (1 + ε) ∗C ∗h x / ln (ln (h x))) at-top ⟨proof ⟩ It follows that inﬁnitely many values of n exceed cn/ ln(ln n) when c is chosen larger than C. corollary (in prime-number-theorem) totient-upper-bound: assumes ε > 0 deﬁnes C ≡third-mertens-const shows frequently (λn. totient n < (1 + ε) ∗C ∗n / ln (ln n)) at-top ⟨proof ⟩ Again, the following alternative formulation is somewhat nicer to prove: lemma (in prime-number-theorem) totient-primorial ′-asymp-equiv: (λk. totient (primorial ′ k)) ∼[at-top] (λk. third-mertens-const ∗primorial ′ k / ln k) ⟨proof ⟩ lemma (in prime-number-theorem) totient-primorial ′-asymp-equiv ′: (λk. totient (primorial ′ k)) ∼[at-top] (λk. third-mertens-const ∗primorial ′ k / ln (ln (primorial ′ k))) 34 ⟨proof ⟩ All in all, ϕ(n) has minimal order cn/ ln ln n: theorem (in prime-number-theorem) liminf-totient: liminf (λn. totient n ∗ln (ln n) / n) = third-mertens-const (is - = ereal ?c) ⟨proof ⟩ end References T. M. Apostol. Introduction to Analytic Number Theory. Undergraduate Texts in Mathematics. Springer-Verlag, 1976. A. Hildebrand. Introduction to Analytic Number Theory (lecture notes). h 35
15479	https://pmc.ncbi.nlm.nih.gov/articles/PMC4737182/	Novel nucleosomal particles containing core histones and linker DNA but no histone H1 - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Advanced Search Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Nucleic Acids Res . 2015 Sep 22;44(2):573–581. doi: 10.1093/nar/gkv943 Search in PMC Search in PubMed View in NLM Catalog Add to search Novel nucleosomal particles containing core histones and linker DNA but no histone H1 Hope A Cole Hope A Cole 1 Program in Genomics of Differentiation, Eunice Kennedy Shriver National Institute for Child Health and Human Development, National Institutes of Health, Bethesda, MD 20892, USA Find articles by Hope A Cole 1,†, Feng Cui Feng Cui 2 Thomas H. Gosnell School of Life Sciences, Rochester Institute of Technology, Rochester, NY, USA Find articles by Feng Cui 2,†, Josefina Ocampo Josefina Ocampo 1 Program in Genomics of Differentiation, Eunice Kennedy Shriver National Institute for Child Health and Human Development, National Institutes of Health, Bethesda, MD 20892, USA Find articles by Josefina Ocampo 1,†, Tara L Burke Tara L Burke 1 Program in Genomics of Differentiation, Eunice Kennedy Shriver National Institute for Child Health and Human Development, National Institutes of Health, Bethesda, MD 20892, USA Find articles by Tara L Burke 1, Tatiana Nikitina Tatiana Nikitina 3 National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA Find articles by Tatiana Nikitina 3, V Nagarajavel V Nagarajavel 1 Program in Genomics of Differentiation, Eunice Kennedy Shriver National Institute for Child Health and Human Development, National Institutes of Health, Bethesda, MD 20892, USA Find articles by V Nagarajavel 1, Naoe Kotomura Naoe Kotomura 1 Program in Genomics of Differentiation, Eunice Kennedy Shriver National Institute for Child Health and Human Development, National Institutes of Health, Bethesda, MD 20892, USA Find articles by Naoe Kotomura 1, Victor B Zhurkin Victor B Zhurkin 3 National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA Find articles by Victor B Zhurkin 3,, David J Clark David J Clark 1 Program in Genomics of Differentiation, Eunice Kennedy Shriver National Institute for Child Health and Human Development, National Institutes of Health, Bethesda, MD 20892, USA Find articles by David J Clark 1, Author information Article notes Copyright and License information 1 Program in Genomics of Differentiation, Eunice Kennedy Shriver National Institute for Child Health and Human Development, National Institutes of Health, Bethesda, MD 20892, USA 2 Thomas H. Gosnell School of Life Sciences, Rochester Institute of Technology, Rochester, NY, USA 3 National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA To whom correspondence should be addressed. Tel: +301 496 6966; Fax: +301 480 1907; Email: clarkda@mail.nih.gov Correspondence may also be addressed to Victor B. Zhurkin. Tel: +301 496 8913; Fax: +301 402 4724; Email: zhurkin@nih.gov † These authors contributed equally to the paper as the first authors. Accepted 2015 Sep 9; Revised 2015 Sep 4; Received 2015 Mar 25; Issue date 2016 Jan 29. Published by Oxford University Press on behalf of Nucleic Acids Research 2015. This work is written by (a) US Government employee(s) and is in the public domain in the US. PMC Copyright notice PMCID: PMC4737182 PMID: 26400169 See commentary "The proto-chromatosome: A fundamental subunit of chromatin?" in Nucleus, volume 7 on page 382. Abstract Eukaryotic chromosomal DNA is assembled into regularly spaced nucleosomes, which play a central role in gene regulation by determining accessibility of control regions. The nucleosome contains ∼147 bp of DNA wrapped ∼1.7 times around a central core histone octamer. The linker histone, H1, binds both to the nucleosome, sealing the DNA coils, and to the linker DNA between nucleosomes, directing chromatin folding. Micrococcal nuclease (MNase) digests the linker to yield the chromatosome, containing H1 and ∼160 bp, and then converts it to a core particle, containing ∼147 bp and no H1. Sequencing of nucleosomal DNA obtained after MNase digestion (MNase-seq) generates genome-wide nucleosome maps that are important for understanding gene regulation. We present an improved MNase-seq method involving simultaneous digestion with exonuclease III, which removes linker DNA. Remarkably, we discovered two novel intermediate particles containing 154 or 161 bp, corresponding to 7 bp protruding from one or both sides of the nucleosome core. These particles are detected in yeast lacking H1 and in H1-depleted mouse chromatin. They can be reconstituted in vitro using purified core histones and DNA. We propose that these ‘proto-chromatosomes’ are fundamental chromatin subunits, which include the H1 binding site and influence nucleosome spacing independently of H1. INTRODUCTION Eukaryotic DNA is organized into chromatin, which is essentially an array of regularly spaced nucleosomes that is occasionally interrupted by non-histone protein complexes bound at regulatory regions, such as promoters, enhancers and replication origins. The nucleosome core is composed of a core histone octamer around which is wrapped ∼147 bp of DNA in ∼1.7 superhelical turns (1–4). The linker histone (H1) binds externally to the nucleosome core, where it interacts with the linker DNA between nucleosomes to direct the folding of the chromatin fibre (5,6) and also seals the DNA turns at the nucleosomal DNA entry/exit points (7), forming a stem-loop structure (8,9). The nucleosome is a compact and stable structure capable of inhibiting transcription, DNA replication and repair. Accordingly, the cell contains various chromatin remodelling activities that can modify histones or move nucleosomes. Accurate maps of nucleosome positions in cells are therefore essential to an understanding of gene regulation and other processes. The DNA in the nucleosome core is strongly protected from digestion by micrococcal nuclease (MNase), which has a strong endonuclease activity and a relatively weak exonuclease activity (10). MNase cuts primarily in the linker DNA between nucleosome cores and then its exonuclease activity slowly trims the DNA ends from the initial cut sites. MNase digestion is temporarily impeded by H1, resulting in the transient formation of chromatosomes (∼160 bp), which contain the nucleosome core, H1 and some linker DNA (11). Eventually, MNase digests the linker DNA in the chromatosome, H1 dissociates, and the result is the nucleosome core particle (∼147 bp), which corresponds to the crystal structure (1–4). The protection of nucleosomal DNA from MNase digestion is made use of in MNase-seq, which involves massively parallel sequencing of mono-nucleosomal DNA (12). Paired-end sequencing represents a major advance in MNase-seq because it provides reads from both ends of each DNA molecule, such that the length of each molecule can be deduced after alignment to the genome sequence, resulting in more accurate position information (13–16). Ideally, MNase digestion would involve cutting the linker on both sides of the nucleosome core, followed by digestion of the protruding linker DNA, resulting in DNA fragments of 145–147 bp. Such fragments would provide a very accurate and precise map of nucleosome positions throughout the genome. In practice, paired-end sequencing has shown that nucleosomal DNA is heterologous in length (13–16), often with a significant fraction of nucleosomes containing DNA longer than 147 bp, primarily due to incomplete trimming by MNase. To solve this problem, we have augmented the slow exonuclease activity of MNase with E. coli exonuclease III (ExoIII). We have shown previously that this approach is effective on a reconstituted nucleosome in vitro (17). Here, we apply it to budding yeast chromatin. We observe a major, sharp, core particle peak at 145–147 bp, which provides very accurate nucleosome position data. We also observe clearly defined intermediate nucleosomal particles containing 154 bp or 161 bp. Surprisingly, these particles are not chromatosomes because they are observed in yeast lacking H1 and in mouse liver chromatin from which H1 has been removed. Furthermore, they can be reconstituted in vitro using just purified core histones and DNA. We propose that these novel particles, which we term ‘proto-chromatosomes’, play a critical role in nucleosome spacing by fixing the minimum distance between nucleosomes at 15 bp, independently of H1. MATERIALS AND METHODS Plasmids and strains p685 was constructed by inserting HHO1 (from −111 to +812 with respect to the start codon) as a Kpn I-Pme I fragment obtained by PCR into pNEB193 (New England Biolabs) cut with the same enzymes. p687 was constructed by replacing the Bam HI-Stu I HHO1 fragment in p685 with the Bam HI-Pme I URA3 fragment from pNEB-URA3 (18). pRS-ARG1-B (p688) was obtained by insertion of a PCR fragment containing ARG1 and flanking regions, from −1283 to +2936, relative to the start codon, at the Not I site in pRS414 (Stratagene), a CEN ARS TRP1 plasmid (ARG1 and TRP1 are transcribed in opposite directions). YVN381 (MATa can1–100 his3–11 leu2–3,112 lys2Δ trp1–1 ura3–1 hho1Δ::URA3) was constructed by transforming wild type strain JRY4012 (19) with a Kpn I-Pme I digest of p687. MNase-ExoIII digestion of yeast nuclei Yeast cells (250 ml culture) were grown to log phase (A 600 ∼0.8) and harvested. Nuclei were prepared (20) and resuspended in 5.2 ml Digestion Buffer (10 mM HEPES pH 7.5, 35 mM NaCl, 2 mM MgCl 2, 2 mM CaCl 2, 5 mM 2-mercaptoethanol, with protease inhibitors (Roche 04693159001)). Two series of six MNase digestions (15, 30, 60, 120, 240 and 480 Worthington units) were set up, with either 150 or 300 units of ExoIII (New England Biolabs). Nuclei (400 μl aliquots) were digested for 2 min at room temperature. The reaction was stopped by addition of Na-EDTA to 10 mM and SDS to 0.8%, mixing and incubating at room temperature for 5 min, before adding more SDS to a final concentration of 1.8%. The DNA was purified; 10% of each sample was analysed in an agarose gel. Appropriately digested samples were chosen for preparative gel electrophoresis. MNase-ExoIII digestion of mouse chromatin Livers were dissected from female NIH/S mice (E13.5) and stored at −80°C. All buffers contained 15 mM 2-mercaptoethanol and protease inhibitors as above. One liver (0.8 g) was disrupted in 6 ml Buffer A+ (0.34 M sucrose, 60 mM KCl, 15 mM NaCl, 15 mM TrisHCl pH 8.0, 0.5 mM spermidine–HCl, 0.15 mM spermine–HCl, 1 mM Na–EDTA) in a glass homogeniser on ice by 10 strokes with a teflon pestle and filtered through four layers of cheese cloth. The homogenate was adjusted to 12.5 ml Buffer A+ with 0.92 M sucrose in a 30-ml polycarbonate high-speed centrifuge tube and spun for 15 min at 2°C in a Sorvall SA600 rotor at 12 500 rpm. The pellet was resuspended in 5 ml Buffer A+ and spun for 5 min as above. The washed nuclei were resuspended in Buffer A+ to A 260 = 50 measured in 1 M NaOH (∼4 ml). A 1.4 ml-aliquot of nuclei in a microfuge tube was adjusted to 2 mM CaCl 2 ( = 1 mM excess over EDTA), warmed briefly to 37°C, and digested with MNase (20 Worthington units) for 3 min. The digestion was terminated by adding 0.1 M Na-EDTA pH 7.5 to a final concentration of 10 mM. The digested nuclei were recovered by a 5 min spin in a microfuge at 4°C and lysed by thorough resuspension in 1 ml 0.2 mM NaEDTA pH 7.5. The debris was removed by a 5 min spin. The chromatin concentration in the supernatant was measured by A 260 in 1% SDS. Two 400-μl aliquots of chromatin at ∼0.2 mg/ml were diluted with an equal vol. of 160 mM NaCl, 2 mM Na-phosphate pH 7.2, 0.4 mM Na-EDTA. To remove H1 (21), 200 μl AG50W-X2 resin (BioRad 142–1241; washed and equilibrated in 80 mM NaCl, 1 mM Na-phosphate pH 7.2, 0.2 mM Na-EDTA as described (22)) was added to one aliquot. Both aliquots were subjected to gentle rotation for 1 h at 4°C. The resin was allowed to settle and the H1-depleted chromatin was removed to a fresh tube on ice. The NaCl concentration was reduced to 50 mM by a 1.6-fold dilution in a pre-mixed buffer resulting in final concentrations of 10 mM HEPES pH 7.5, 2 mM MgCl 2, 2 mM CaCl 2. Histones in 100 μl aliquots were TCA-precipitated and analysed in a protein gel stained with Coomassie Blue. The chromatin was depleted of ∼85% of its H1 (estimated by protein gel analysis of H1 extracted by 5% perchloric acid). Six 100 μl-aliquots (∼5.5 μg) of native and H1-depleted chromatin were warmed briefly to 30°C and digested with 0, 0.1, 0.2, 0.4, 0.8 or 1.6 units MNase and 16.5 units ExoIII (all aliquots) for 3 min at 30°C. The digestion was stopped by adding 20 μl 50 mM Na-EDTA pH 7.5, 5% SDS. DNA was extracted as described above (nucleosomal DNA was not gel-purified). Reconstituted chromatin Reconstituted nucleosomes were made using recombinant yeast H3, H4, H2A and H2B purified from E. coli (23). Histone octamers were formed and purified by gel filtration (23). The octamer concentration was determined initially by amino acid composition analysis and then correlated with Bradford protein assay measurements using IgG as standard (BioRad): 1 mg/ml in the Bradford assay corresponds to an actual concentration of 0.36 mg/ml. pRS-ARG1-B was purified from E. coli lacking DNA methylases (dam- dcm-; New England Biolabs C2925). Briefly, 10 μg DNA and 4 μg octamer were mixed in 100 μl 2 M NaCl, 10 mM HEPES-K pH 7.5, 0.1 mM Na-EDTA and transferred to a Slide-A-Lyzer mini-dialysis unit with a molecular weight cut-off of 10 000 (ThermoScientific 69574). The solution was dialysed overnight at 4°C against 500 ml of 2 M NaCl in R-buffer (10 mM HEPES-K pH 7.5, 0.1 mM Na-EDTA, 0.05% NP40, 1 mM 2-mercaptoethanol) and then against R-buffer containing 1.5 M NaCl (2 h), 1 M NaCl (2 h), 0.75 M NaCl (3 h), and finally against 50 mM NaCl in R-buffer overnight. Alternatively, nucleosomes were reconstituted on pRS-ARG1-B at the same ratio using native chicken erythrocyte histones, as described (17). Reconstituted chromatin (1 μg) in 30 μl R-buffer containing 50 mM NaCl, 2 mM CaCl 2, 2 mM MgCl 2, was digested with 3.125 units ExoIII plus either 0.1, 0.125 or 0.2 units MNase for 3 min at 30°C. Reactions were stopped by addition of Na-EDTA pH 7.5–6.7 mM and cooling on ice. Libraries of gel-purified nucleosomal DNA were subjected to Illumina paired-end sequencing (20). The length of each sequenced DNA fragment was deduced after alignment of each pair of 50-nt reads to the yeast genome (SacCer2) or the mouse genome (mm10) using Bowtie 2 (24). A summary of the data is provided in Supplementary Table S1. RESULTS MNase–ExoIII digestion yields precisely trimmed core particles We added ExoIII to facilitate removal of linker DNA protruding from nucleosomes liberated from yeast nuclei by MNase. ExoIII is a partly processive 3′-to-5′ exonuclease which hydrolyses only one DNA strand, leaving a single-stranded 5′-overhang (25). Its activity is mildly sequence-dependent, in the order C > A, T > G (26). To remove the 5′ overhang, we initially included mung bean nuclease in the digestions, but it was found to be unnecessary, presumably because MNase digests single-stranded DNA much more rapidly than double-stranded DNA (10). Thus, the endonuclease activity of MNase cuts the linker DNA, ExoIII rapidly removes one strand of the linker until it is blocked by the histones, and then MNase destroys the single-stranded overhang (Figure 1A). Nuclei were digested at different MNase concentrations in the presence of a fixed concentration of ExoIII (Figure 1B) and paired-end sequencing libraries were prepared from gel-purified mono-nucleosomal DNA. Figure 1. Open in a new tab Simultaneous digestion of yeast chromatin with MNase and ExoIII results in a set of DNA fragments of discrete lengths, corresponding to the nucleosome core particle and its intermediates. (A) Digestion of chromatin with MNase and ExoIII. Ovals: nucleosome cores; black arrows: cut sites. (B) Agarose gel electrophoresis of chromatin digested with MNase and ExoIII. Markers: 1-kb ladder (M1); pBR322 Msp I digest (M2). (C) Length distribution histograms for nucleosome sequences obtained by paired-end sequencing of the mono-nucleosome band from wild type strain JRY4012: MNase-only. (D, E) Length histograms for MNase-ExoIII at two different MNase concentrations, corresponding to the mono-nucleosome bands in the gel. The length distributions for an MNase-only sample and two levels of MNase-ExoIII digestion are plotted as histograms (Figure 1C, D, E). The MNase-only sample has a typical length distribution with a fairly broad peak at 149 bp, corresponding to core particles, and a weak hump at ∼160 bp corresponding to particles with protruding linker DNA, as well as a set of poorly resolved peaks around 130 bp, corresponding to internally digested nucleosomes, sub-nucleosomes (i.e. nucleosomes lacking one or both H2A–H2B dimers; 27) and perhaps non-histone complexes (16). In the presence of ExoIII, much sharper peaks were observed in the length distribution, which shifted to slightly lower values with increased MNase digestion (Figure 1D, E). The major peak was at 147 bp or 146 bp, corresponding to precisely trimmed core particles. There were also minor peaks at sub-nucleosomal lengths of 134, 124, 114 and 104 bp, which show a 10 bp periodicity. This observation is consistent with internal digestion of nucleosomes by MNase and/or invasion of the nucleosome by ExoIII (28). Unexpectedly, there were also sharp peaks at 153/154 and 161 bp, apparently corresponding to defined nucleosomal particles containing more DNA than is present in the core particle. The periodicity in this case was 7 or 8 bp. The relative amounts of these particles decreased with MNase digestion, consistent with the hypothesis that they were eventually converted into core particles. Thus, the effect of ExoIII is to resolve and sharpen the peaks observed in the MNase-only digest; the resolution of the weak hump at ∼160 bp in the MNase-only digest into sharp peaks at 154 and 161 bp is particularly striking. Proto-chromatosomes The structural explanation for the pauses in digestion external to the nucleosome core is of interest. The most obvious possibility is that they represent chromatosomes: nucleosomes with additional linker DNA protected by the globular domain of histone H1, typically ∼160 bp (11,29). However, unlike higher organisms, yeast has much less H1 than nucleosomes (30), implying that it could not protect a large fraction of the total nucleosomes. Nevertheless, we tested the possibility that the 154-bp and 161-bp particles are chromatosomes by repeating the experiment using a strain in which the gene for H1 had been deleted (hho1Δ). The length distribution was very similar to that of wild type nucleosomes, yielding the same peaks, including those at 153/154 and 161 bp (Figure 2A). Since the 154-bp and 161-bp particles were observed in the absence of H1, they cannot be chromatosomes. Accordingly, we will refer to the particles that contain 154 bp and 161 bp as ‘proto-chromatosomes’, because they contain a precise amount of linker DNA but lack H1. Figure 2. Open in a new tab The 154-bp and 161-bp nucleosomal particles are present in yeast cells lacking linker histone and can be reconstituted in vitro using only core histones and DNA. (A) The 154-bp and 161-bp particles are not chromatosomes. Length distribution histograms for nucleosome sequences obtained by sequencing of mono-nucleosomes from hho1Δ cells. (B, C) The 154-bp and 161-bp particles can be formed using recombinant yeast core histones or native chicken erythrocyte core histones and plasmid DNA in vitro. Length distribution histograms for nucleosome sequences obtained at two different levels of MNase-ExoIII digestion. Proto-chromatosomes might result from protection of linker DNA by non-histone proteins, but this explanation is difficult to eliminate in vivo. Instead, we determined whether the 154-bp and 161-bp particles are observed after digestion of chromatin reconstituted in a purified system in vitro. We assembled nucleosomes on a large yeast plasmid using recombinant yeast histones and subjected them to MNase-ExoIII digestion (Figure 2B). Less digested mono-nucleosomes included a weak peak at 147 bp, corresponding to core particles, and stronger peaks at 155 bp and 162 bp, corresponding to proto-chromatosomes. More digested nucleosomes gave the expected sharp core particle peak at 145 bp and only a very minor peak at 153 bp. Some invasion of the nucleosomes was also observed (peaks at 133, 123 and 114, again with a ∼10 bp period). Clearly, the 162-bp and 155-bp particles were converted to core particles (145 bp) after extended digestion. We obtained the same result when native chicken erythrocyte core histones were used to reconstitute nucleosomes on the same plasmid, indicating that recombinant yeast histones are not atypical (Figure 2C). Thus, protection of the extra DNA in proto-chromatosomes requires only core histone-DNA interactions. We explored the relationship between native yeast proto-chromatosomes (152–155 bp) and core particles (144–147 bp) by comparing their sequences using cross-correlation analysis (31). The ends of the core particle and proto-chromatosome sequences were compared with one another: for every core particle sequence, the distance ‘d'’ of its right-hand end (with respect to the chromosome) to the first nucleotide of all proto-chromatosome sequences was summed. If the core particle and the proto-chromatosome have the same right-hand end, d = 0 (Figure 3A). A comparison of core particles (145–147 bp) and 152–155 bp proto-chromatosomes from wild-type cells (Figure 1E) revealed strong peaks at 0 and +8 bp, indicating that proto-chromatosomes tend to have one end that is identical to that of the corresponding core particle and a 7 or 8-bp extension at the other end (Figure 3B). That is, the extra 8 bp in the 154-bp particles are asymmetrically located on one side of the nucleosome. The additional correlation peaks at 19 and 30 bp etc. (Figure 3B), are shifted by 10 bp and represent the ends of rotationally related translational positions (31). A similar analysis of 161-bp proto-chromatosomes supports the proposal that, in most cases, the extra 15 bp in these particles correspond to 7 or 8 bp protruding from both sides of the nucleosome core (Figure 3C). There might be a fraction of asymmetric 161-bp particles, with 15 bp protruding from one side of the core, because the peak at 16–17 bp is quite wide, but such particles cannot be reliably distinguished from particles shifted translationally by 10 bp, which give rise to a peak at 19 bp, as observed for 154-bp particles (Figure 3B). The same results were obtained by cross-correlation analysis of core particles and proto-chromatosomes from less digested nucleosomes (Supplementary Figure S1). Figure 3. Open in a new tab Proto-chromatosomes have a 7 or 8-bp extension on one side (154-bp particles) or both sides (161-bp particles) of the nucleosome core in wild type cells. (A) Calculation of cross-correlation between the right-hand end of a core particle sequence (black line) and the right-hand end of a proto-chromatosome sequence (grey line). The right-hand end refers to the location of the sequence with respect to the chromosome. (B) Cross-correlation between nucleosome core particles (145–147 bp) and proto-chromatosomes (152–155 bp) (data in Figure 1E). (C) Cross-correlation between core particles (145–147 bp) and proto-chromatosomes (160–162 bp) (data in Figure 1E). In conclusion, MNase and ExoIII pause during digestion of the linker at a point ∼7 bp from the boundary of the nucleosome core, resulting in metastable proto-chromatosomes, which are eventually digested to core particles. This scenario implies that proto-chromatosomes have the same genomic distribution as core particles, and do not derive from specific regions of the genome. We confirmed that this was the case by comparing the nucleosome phasing patterns of core particles and proto-chromatosomes relative to the transcription start site (TSS) (12,32). Both core particles and proto-chromatosomes showed very similar phasing profiles (Supplementary Figure S2). Since genes are very close together in the budding yeast genome, most nucleosomes are represented by these phasing patterns. Proto-chromatosomes are present in mouse liver We determined whether proto-chromatosomes are present in the chromatin of higher organisms by MNase–ExoIII digestion of native and H1-depleted chromatin prepared from mouse liver. Nuclei were initially digested with MNase only to obtain long chromatin fragments. Some of this chromatin was depleted of H1 using a cation exchange resin; another aliquot was mock-treated (‘native’ chromatin). H1-depleted chromatin was digested more rapidly by MNase and ExoIII than native chromatin, consistent with protection of linker DNA by H1 (Figure 4A). Nucleosomal DNA was sequenced and aligned to the mouse genome to obtain the length distributions, which revealed the presence of the 154-bp and ∼161-bp peaks in both native and H1-depleted chromatin (Figure 4C). In native chromatin, the broad chromatosome peak at 163–166 bp is converted into 154 bp-particles and core particles (146 bp). Particles of the same size were observed in H1-depleted chromatin. The 154-bp and 161-bp particles were converted into a strong, sharp, core particle peak at 146 bp, consistent with conversion of proto-chromatosomes into core particles as digestion proceeds. Thus, the chomatosome is simply the proto-chromatosome with bound H1, which confers additional protection to the 7-bp extensions on either side of the classical core particle. Figure 4. Open in a new tab Proto-chromatosomes are present in mouse chromatin. (A) Agarose gel electrophoresis of DNA from native (N) and H1-depleted (D) mouse liver chromatin digested with MNase and ExoIII (lanes 1–6). ‘C’, long chromatin obtained by initial digestion of nuclei with MNase only. ‘M’, λ-Bst EII digest mixed with pBR322 Msp I digest (NEB). (B) Depletion of H1 shown by SDS-PAGE analysis of proteins in native (‘N’) and H1-depleted (‘D’) chromatin (M: marker). Stained with Coomassie blue. (C) Length histograms for the MNase-ExoIII digests of native (lanes N4, N5 and N6 in A) and H1-depleted chromatin (D3, D4 and D5). Enhanced WW and SS patterns in MNase–ExoIII yeast nucleosome sequences The initial aim of this study was to increase the accuracy of nucleosome position data by eliminating linker DNA protruding from mono-nucleosomes. This should improve alignment of nucleosomal DNA sequences and enhance the periodicities in the distributions of various dinucleotide and trinucleotide motifs relative to the nucleosome dyad (33). The strongest patterns are the 10 bp periodicities displayed by the distributions of A or T-containing dinucleotides (AT, TA, TT and AA = WW) and G or C-containing dinucleotides (GC, CG, GG and CC = SS), which are exactly out of phase with one another (33). The patterns have been attributed to the relative ease with which A-T base pairs can bend into the minor groove and the tendency for G-C base pairs to bend into the major groove, as the DNA is strongly bent around the core histone octamer (1,33). We compared the WW and SS patterns for MNase-only and MNase–ExoIII core particle DNA by aligning the sequences by their midpoints and calculating the frequencies of WW and SS dinucleotides within the nucleosome (Figure 5A) (34). The MNase-only data were restricted to the sequences in the major peak (147–152 bp), close to core particle length; likewise, the MNase–ExoIII data were restricted to the major peak (144–147 bp). In the case of the WW motif, both MNase-only and MNase–ExoIII core particles showed a clear ∼10 bp modulation in frequency across most of the nucleosome, with peaks at 7, 17, 27, 38, 48 and 59 bp from the boundary (i.e. 14, 25, 35, 46, 56 and 66 bp from the dyad). The amplitudes of all of these peaks were enhanced by ExoIII. In addition, two weak peaks at 64 and 70 bp in the MNase-only data, located near the dyad at 73 bp, were clearly enhanced by ExoIII. Although the peak at 70 bp is in phase with the other 10 bp-modulated peaks, the peak at 64 bp is exactly out of phase with them and therefore represents an anomaly, or ‘rogue’ peak, observed previously in chemical maps (35,36). Figure 5. Open in a new tab Enhanced dinucleotide periodicities and strand-specific preferences in MNase–ExoIII nucleosomes. (A) WW versus SS sequence patterns in MNase–ExoIII nucleosomes aligned at their dyads. WW versus SS patterns for MNase–ExoIII and MNase-only nucleosomes in wild type cells. (B) Single-nucleotide distributions in MNase–ExoIII and MNase-only nucleosomes (derived from the peaks in Figure 1C, E). The data were smoothed using a 3-bp running average. The amplitude of the SS signal was weaker than the WW signal, and is centered on lower dinucleotide frequencies because the yeast genome is AT-rich. SS dinucleotides showed a clear 10-bp modulation, with peaks at 12, 22, 34, 43 and 54 bp in both data sets (i.e. 19, 30, 39, 51 and 61 bp from the dyad), which are ∼5 bp out of phase with the WW peaks and enhanced by ExoIII. Additional peaks at 62 and 67 bp emerged in the MNase-ExoIII data, which flank the additional peaks observed in the WW profile. Thus there is a weak alternation between WW and SS with a ∼5 bp period close to the dyad (WW at 59, SS at 62, WW at 64, SS at 67, WW at 70 and SS at the dyad). At the nucleosome border (set at 0 in Figure 5), the MNase-only data show a large WW peak at −2 and similarly large trough at +2, which are resolved by ExoIII into two relatively small flanking peaks at −4 and +1 (Figure 5A). Neither of these peaks is in phase with the 10-bp WW period. Similarly, the large SS peak at +2 in the MNase-only data are absent from MNase–ExoIII nucleosomes, and the trough at −2 decreased in amplitude and shifted to −2, resulting in two new weak SS peaks at −1 and 4 bp in MNase–ExoIII nucleosomes, neither of which are in phase with the 10-bp signal. The MNase-only nucleosome boundary peak coincides with the MNase cut site and has therefore been attributed to its preference for AT-rich DNA (31,35), although this seems unlikely given its absence from MNase–ExoIII nucleosomes. In summary, the increased accuracy of nucleosome positions deduced from the MNase–ExoIII data results in a more accurate alignment of nucleosome sequences, enhancing the dinucleotide signals, such that strong peaks are sharpened and weak peaks are resolved. Strand-specific preferences for A versus T We also determined the distributions of all four mono-nucleotides within nucleosomes (Figure 5B). Intriguingly, all four nucleotides showed quite strong 10-bp periodicities, in line with those observed for the relevant dinucleotides, with A and T out of phase relative to G and C, although the amplitudes of the signals were not as strong as those observed for WW and SS. Surprisingly, the distributions of A and T are not identical, and neither are those of G and C. In the case of G and C, the differences are small, except near the dyad, where the weak G and C peaks are out of phase with one another; there is a G-peak at 63 that is not matched in the C profile, and a C peak that is stronger than the G peak at 67. These differences occur in the region next to the dyad, where the SS distribution shows a 5-bp period. The A and T distributions are more different from one another than the G and C distributions. The A profile is more periodic than that of T, such that A shows stronger peaks than T at 7, 27, 50, 59 and 64, essentially because the T peaks at these locations are split in two, suggesting that T tends to flank an A on both sides at these locations. Furthermore, the region near the dyad shows weak A and T peaks that are out of phase with one another, analogous to those observed for G and C in this region. The A peak at 64 is relatively strong and coincides with the rogue WW peak that is out of phase with the other WW peaks, indicating that WW at this location tends to be AA rather than TT (33). DISCUSSION We present a significant improvement on the MNase method for mapping nucleosomes, involving the combination of MNase and ExoIII for efficient removal of residual linker DNA from mono-nucleosomes. This results in a sharply defined peak of core particles (144–147 bp). More accurate nucleosome position data will facilitate nucleosome sequence analysis and may be critical in deciding whether a specific sequence, such as a transcription factor binding site, is located inside a nucleosome or in the more accessible linker DNA. The quality of MNase–ExoIII data is comparable with that of data obtained from the chemical mapping method, which involves sequencing DNA fragments generated by chemical nicking of DNA near the nucleosome dyad in cells carrying the S47C mutation in H4 (35,37). We find that dinucleotide periodicities are enhanced and that the WW signal at 64 is now apparent. Furthermore, the artifactual MNase-associated WW and SS peaks at the boundary of the nucleosome core are largely eliminated by ExoIII. The only significant discrepancy concerns the A peak located 3 bp from the dyad in chemically mapped nucleosomes (35). Given that this A-peak is far from the MNase cleavage sites, it may reflect a problem with the chemical cleavage method, perhaps involving the assignment of minor cleavage sites, or perhaps from an unexpected sequence preference of the DNA cleavage reagent and/or a structural alteration due to the H4-S47C mutation (38). We also observe a striking preference for A over T in the top strand at 7, 27, 50, 59 and 64 bp relative to the nucleosome boundary (Figure 5B), where A is located on the inside of the bend, just after the small kinks directed into the minor groove (Supplementary Figure S3). A differential distribution of A and T mono-nucleotides was reported previously, although the period was irregular, perhaps because the data were reduced to relatively few, idealized, nucleosome positions prior to analysis (39). The preference may reflect wedge formation by AA:TT dimeric steps, which occurs because the TT strand is slightly more extended than the AA strand (40), perhaps facilitating anisotropic bending (41). Alternatively, the fact that most arginine residues contacting DNA are closer to the bottom strand than the top strand may be important (42,43). However, neither of these explanations account for the opposite preference, for T over A, at 38 bp from the border. Thus, an additional factor is required to account completely for the A versus T strand asymmetry. We have identified clearly defined particles larger than the core particle, which have 7 or 8 bp projecting from one side (154-bp) or both sides (161-bp) of the nucleosome core. These proto-chromatosomes contain a similar amount of DNA to the chromatosome, but do not contain H1 and can be reconstituted in vitro using just DNA and core histones. Our observation is consistent with early papers showing that MNase digestion of H1-depleted chromatin yields a weak, transient band similar in size to the chromatosome (17,44–46). We propose that a steric block to MNase and ExoIII is located in the linker ∼7 bp from the nucleosome core (Figure 6A). In the case of the typical yeast linker of ∼15 bp, the putative steric blocks between two nucleosomes would coincide in the middle of the linker, as shown (Figure 6A). The ‘10 n +5’ rule for the linker length (35,45) predicts linker lengths of 5, 15, 25 bp, etc. If the linker is 25 bp then the steric blocks would be separated by 10 bp. We envisage that MNase cuts within one of the linkers, between the block and a core, resulting in one nucleosome with some protruding DNA that is trimmed away by MNase and ExoIII until it is stopped by the core particle boundary, and a second nucleosome which is trimmed until the exonucleases reach the block, resulting in a proto-chromatosome. Eventually, MNase cuts the proto-chromatosome between the block and the core, and the residual linker is digested, resulting in a core particle (Figure 6B). Figure 6. Open in a new tab Nucleosome core particles, proto-chromatosomes and linker length. (A) Provenance of proto-chromatosomes. The endonuclease activity of MNase (solid red arrows) cuts the linker DNA between nucleosome cores (grey ovals). ExoIII and the exonuclease activity of MNase (open red arrows) trim the linker, stopping either at the border of the nucleosome core, or at a putative block to digestion (black box), located in the linker ∼7 bp from the border of the nucleosome core. Another cut by MNase in the linker between the block and the core allows the exonucleases to remove the remaining linker and convert proto-chromatosomes to core particles. The typical linker length in yeast is 15 bp ( = 10 n +5; n = 1); in this case, the block due to each flanking nucleosome would be at the same location (in the middle of the linker, as shown). (B) Views of core particle and proto-chromatosomes from above, drawn approximately to scale, based on the nucleosome structure (1). The final 10 bp on each side of the nucleosome core are almost straight, projecting a short distance out of the particle. Proto-chromatosomes are shown with an extra 7 bp on one side (154 bp) or both sides (161 bp) with a continuing straight trajectory. (C) Potential role of the proto-chromatosome in determining linker length. The average nucleosome spacing is 161 bp (Supplementary Figure S4B). We propose that the proto-chromatosome fixes the minimum distance between nucleosomes at 15 bp, independently of H1. The 161-bp proto-chromatosome is significant for two reasons. Firstly, 161 bp is sufficient to complete two turns of DNA in the nucleosome. This observation might be coincidental, particularly in view of the tetra-nucleosome structure, which does not show additional DNA coiled around the nucleosome (47), but it is intriguing. Secondly, the average distance between MNase–ExoIII core particles determined by auto-correlation analysis of core particle sequences is also 161 bp (Supplementary Figure S4), suggesting that the proto-chromatosome might determine the distance of closest approach between two nucleosome cores, which corresponds to a minimum linker of ∼15 bp (Figure 6C) and may account for the ‘10 n +5’ rule, setting n> 0. What is the structural nature of the steric block? One possibility is a histone–DNA contact located ∼7 bp external to the nucleosome core, possibly involving the histone tail domains, since they interact with linker DNA and affect chromatin folding in vitro (48,49). In particular, it has been shown recently using ChIP-exo, a technique for mapping protein–DNA contacts genome-wide, that H3 makes a major contact with the linker in vivo, which is partly dependent on its N-tail domain (50). We are currently investigating whether the H3 tail domain is involved in proto-chromatosome formation. It should be noted that ChIP-exo works on a different principle to MNase–ExoIII: the locations of histone-specific cross-links on immuno-precipitated DNA are mapped using λ-exonuclease, which digests one strand until it is blocked by a histone–DNA cross-link. The method provides valuable information on individual histone occupancy at relatively high resolution. However, it cannot be used to obtain an accurate nucleosome dyad or boundary because the histone residue involved in each cross-link is unknown. Instead, the method provides an average dyad position, corresponding to the weighted average of the various rotationally related overlapping nucleosome positions within each nucleosome occupancy peak. Alternatively, the block might be due to a steric clash between the nucleosome core and the approaching MNase or ExoIII as it rotates around the DNA helix. The outer 10 bp on each side of the nucleosome core are almost straight and therefore deviate from the superhelix (Figure 6B) (1). If the linker maintains this straight trajectory in proto-chromatosomes, then a clash between ExoIII and the DNA at the dyad may occur (Supplementary Figure S5). On the other hand, MNase is smaller than ExoIII, reducing the probability of a clash, but it still detects the steric block, albeit less precisely. Irrespective of the structural origin of the steric block to exonuclease digestion, the binding of transcription factors to cognate sites located within 7 bp of the core (i.e. within the proto-chromatosome) may be inhibited, particularly if the rotational setting of the site is such that the factor must bind to the inner surface of the linker DNA. We propose that a critical function of the proto-chromatosome is to fix the linker length and geometry of the yeast chromatin fibre, resulting in compaction (51), even though it is H1-deficient (30). The proto-chromatosome would provide the correct DNA topography for H1 binding. In higher eukaryotes, longer linkers tend to de-stabilize the chromatin zigzag (52); this effect is offset by the binding of H1 (6), which forms a stem-loop structure in the linker (8,9). Nevertheless, linker histone is not essential for chromosome condensation or cell viability in vertebrates; complete loss of H1 results in chromatin with closely-spaced nucleosomes (53), as observed in yeast and in vitro (54). Thus, the proto-chromatosome discovered here may represent an essential chromatin subunit, influencing both H1 binding and nucleosome spacing. ACCESSION NUMBERS GEO database GSE65889. Supplementary Material SUPPLEMENTARY DATA supp_44_2_573__index.html (829B, html) Acknowledgments We thank the NHLBI Sequencing Core Facility (Yan Luo, Poching Lu, Yoshi Wakabayashi and Jun Zhu) and the Tufts University Core Facility for paired-end sequencing. We thank Rohinton Kamakaka for JRY4012, Will Huffman and Doug Fields for mouse livers, and Razvan Chereji for helpful comments on the manuscript. Author contributions: H.A.C. developed the method and performed experiments. F.C. analysed the data. T.L.B., J.O., T.N., V.N. and N.K. performed experiments. V.B.Z. and D.J.C. directed research and wrote the manuscript. Footnotes Present addresses: V. Nagarajavel, Genomic Medicine Institute, Lerner Research Institute NE-50, 9500 Euclid Avenue, Cleveland, OH 44195, USA. Naoe Kotomura, Department of Biochemistry, Fujita Health University School of Medicine, Toyoake, Aichi, Japan. Victor B. Zhurkin, Building 37, Room 3035A, National Cancer Institute, NIH, Bethesda, MD 20892, USA. SUPPLEMENTARY DATA Supplementary Data are available at NAR Online. FUNDING Intramural Research Program of the National Institutes of Health (NICHD and NCI). Funding for open access charge: Intramural Research Program of the National Institutes of Health (NICHD). Conflict of interest statement. None declared. REFERENCES 1.Luger K., Mäder A.W., Richmond R.K., Sargent D.F., Richmond T.J. Crystal structure of the nucleosome core particle at 2.8 Å resolution. Nature. 1997;389:251–260. doi: 10.1038/38444. [DOI] [PubMed] [Google Scholar] 2.White C.L., Suto R.K., Luger K. 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15480	https://math.stackexchange.com/questions/2872591/ratio-of-areas-determined-by-a-square-inscribed-in-the-corner-of-a-right-triangl	algebra precalculus - Ratio of areas determined by a square inscribed in the corner of a right triangle - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Ratio of areas determined by a square inscribed in the corner of a right triangle Ask Question Asked 7 years, 1 month ago Modified7 years, 1 month ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I’m having trouble working out how to algebraically get to the answer of this question. (See original image below.) A square is drawn in the corner of a right-angled triangle with side lengths a a, b b, and [hypotenuse] c c, as shown. Which expression gives the ratio of the unshaded area [inside the triangle, but outside the square] to the shaded area [of the square] in all cases? (A) 1:1 1:1 (B) c:(a+b)c:(a+b) (C) a b:c 2 a b:c 2 (D) (a+b)2:2 c 2(a+b)2:2 c 2 (E) c 2:2 a b c 2:2 a b Apparently the answer is c 2:2 a b c 2:2 a b (choice E), but how? Your help is greatly appreciated! Thank you in advance. (Please ignore the pen marks! They are incorrect assumptions a friend made on the diagram.) algebra-precalculus geometry triangles ratio Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Aug 5, 2018 at 3:39 Zelda_01Zelda_01 13 3 3 bronze badges 13 Do you see any similar triangles in the figure, and can you see how to use them to determine the length of the side of the square?Blue –Blue 2018-08-05 04:01:21 +00:00 Commented Aug 5, 2018 at 4:01 Hi, yeah all three triangles in the diagram are similar, but I don't know how to use them to determine the length of the side of the square. I ended up splitting a and b and calling them a1/a2 and b1/b2, but that made it confusing.Zelda_01 –Zelda_01 2018-08-05 04:03:52 +00:00 Commented Aug 5, 2018 at 4:03 Suppose the side of the square is s s. What are the sides of the two smaller triangles?Blue –Blue 2018-08-05 04:05:00 +00:00 Commented Aug 5, 2018 at 4:05 (a-s) and (b-s)? Or am I missing something easier?Zelda_01 –Zelda_01 2018-08-05 04:06:15 +00:00 Commented Aug 5, 2018 at 4:06 Good! Specifically, a−s a−s and b−s b−s are two of the legs. (I meant "legs", not "sides".) The two other legs are ... what? And what proportion links all four of these values?Blue –Blue 2018-08-05 04:08:03 +00:00 Commented Aug 5, 2018 at 4:08 \|Show 8 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. Let the side of the square is x x. Then, using similarity of triangles, b−x x=x a−x b−x x=x a−x or (b−x)(a−x)=x 2(b−x)(a−x)=x 2; a b−a x−b x+x 2=x 2 a b−a x−b x+x 2=x 2; a b=x(a+b)a b=x(a+b) or x=a b a+b x=a b a+b. Thus, the shaded area is (a b)2(a+b)2(a b)2(a+b)2. The unshaded area is 0.5 x(b−x)+0.5 x(a−x)=0.5(a b a+b(b−a b a+b)+a b a+b(a−a b a+b)=a b(b 2+a 2)2(a+b)2=a b c 2 2(a+b)2 0.5 x(b−x)+0.5 x(a−x)=0.5(a b a+b(b−a b a+b)+a b a+b(a−a b a+b)=a b(b 2+a 2)2(a+b)2=a b c 2 2(a+b)2 so the final ratio is 2 a b c 2 2 a b c 2 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 5, 2018 at 4:17 VasiliVasili 11.7k 1 1 gold badge 19 19 silver badges 31 31 bronze badges 2 Thanks for this, was just wondering how you went from working out both areas to then finding the final ratio? I haven't done ratios in a while sorry!Zelda_01 –Zelda_01 2018-08-05 04:42:59 +00:00 Commented Aug 5, 2018 at 4:42 Wait, nevermind I worked it out, thanks!Zelda_01 –Zelda_01 2018-08-05 04:49:36 +00:00 Commented Aug 5, 2018 at 4:49 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Let x x be the side length of the square. By similarity, The hypotenuse of the lower small right triangle is (c b)x(c b)x. The hypotenuse of the upper small right triangle is (c a)x(c a)x. Hence we get (c b)x+(c a)x=c(c b)x+(c a)x=c which yields x=a b a+b x=a b a+b If S,U S,U are the respective areas of the shaded and unshaded regions, then S=x 2 S=x 2 U=(1 2)a b−x 2 U=(1 2)a b−x 2 hence, the required ratio can be expressed as U S=(1 2)a b−x 2 x 2=(1 2)a b x 2−1=((1 2)a b)(1 x)2−1=((1 2)a b)(a+b a b)2−1=(a+b)2 2 a b−1=(a+b)2−2 a b 2 a b=a 2+b 2 2 a b=c 2 2 a b U S=(1 2)a b−x 2 x 2=(1 2)a b x 2−1=((1 2)a b)(1 x)2−1=((1 2)a b)(a+b a b)2−1=(a+b)2 2 a b−1=(a+b)2−2 a b 2 a b=a 2+b 2 2 a b=c 2 2 a b Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 5, 2018 at 4:51 quasiquasi 61.3k 3 3 gold badges 46 46 silver badges 105 105 bronze badges Add a comment\| You must log in to answer this question. 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15481	https://epubs.siam.org/doi/10.1137/0802001	Classification of Critical Stationary Points in Unconstrained Optimization \| SIAM Journal on Optimization Skip to main content Search Search This Journal This Journal Anywhere Books Journals Proceedings Quick Search in Journals Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Advanced Search 0 Register / Sign In Access via your Institution Skip main navigationClose Drawer Menu Open Drawer Menu Menu Journals SIAM Review Multiscale Modeling & Simulation SIAM Journal on Applied Algebra and Geometry SIAM Journal on Applied Dynamical Systems SIAM Journal on Applied Mathematics SIAM Journal on Computing SIAM Journal on Control and Optimization SIAM Journal on Discrete Mathematics SIAM Journal on Financial Mathematics SIAM Journal on Imaging Sciences SIAM Journal on Life Sciences SIAM Journal on Mathematical Analysis SIAM Journal on Mathematics of Data Science SIAM Journal on Matrix Analysis and Applications SIAM Journal on Numerical Analysis SIAM Journal on Optimization SIAM Journal on Scientific Computing SIAM/ASA Journal on Uncertainty Quantification Theory of Probability & Its Applications Locus E-books Bookstore Proceedings For Authors Journal Authors Book Authors ICM Authors For Librarians Collections Epidemiology Collection High Impact Article Collection JOIN SIAM HELP/CONTACT US Journal Home Current Issue All Issues About About this Journal Editorial Policy Editorial Board Instructions for Authors Instructions for Referees Submit Subscribe Share Share on Facebook X LinkedIn Email HomeSIAM Journal on OptimizationVol. 2, Iss. 1 (1992)10.1137/0802001 Next article Classification of Critical Stationary Points in Unconstrained Optimization Author: Stefan SchäfflerAuthors Info & Affiliations Get Access BibTeX Tools Add to favorites Download Citations Track Citations Contents ###### NEXT ARTICLE A Potential Reduction Algorithm Allowing Column Generation Next Abstract References Information & Authors Metrics & Citations Get Access References Figures Tables Media Share Abstract A stationary point of an unconstrained optimization problem is called critical if the Hessian matrix at this point is positive semidefinite. Such a point cannot be classified using second-order optimality conditions. In this paper the problem of classifying a critical stationary point of a given objective function is reduced to the application of higher-order optimality conditions for a special auxiliary function. MSC codes 65K05 90C30 Keywords optimality conditions critical stationary points positive-semidefinite Hessian Get full access to this article View all available purchase options and get full access to this article. Get Access Sign in as an individual or via your institution References 1. D. Castrigiano, S. Hayes-Widmann, Catastrophe Theory, Addison-Wesley, Reading, MA, 1991 Google Scholar 2. C. H. Edwards, Advanced calculus of several variables, Academic Press (a subsidiary of Harcourt Brace Jovanovich, Publishers), New York-London, 1973x+457 Google Scholar 3. Herbert Fischer, G. V. Milovanovic, Some aspects of automatic differentiation Numerical methods and approximation theory, III (Niš, 1987), Univ. Niš, Niš, 1988, 199–208, Yugoslavia Google Scholar 4. Herbert Fischer, Automatic differentiation of characterizing sequences, J. Comput. Appl. Math., 28 (1989), 181–185 Crossref Web of Science Google Scholar 5. H. Fischer, Automatic differentiation of implicitly defined functions, Tech. Report, 112A, Institut für Angewandte Mathematik und Statistik, Technische Universität München, München, Germany, 1989 Google Scholar 6. R. Fletcher, Practical methods of optimization. Vol. 1, John Wiley & Sons Ltd., Chichester, 1980viii+120 Google Scholar Show all references Information & Authors Information Authors Information Published In SIAM Journal on Optimization Volume 2 • Issue 1 • February 1992 Pages: 1 - 6 DOI: 10.1137/0802001 ISSN (online): 1095-7189 Copyright Copyright © 1992 Society for Industrial and Applied Mathematics. History Submitted: 25 June 1990 Accepted: 29 November 1990 Published online: 13 July 2006 MSC codes 65K05 90C30 Keywords optimality conditions critical stationary points positive-semidefinite Hessian Authors Affiliations Expand All Stefan Schäffler View all articles by this author Metrics & Citations Metrics Citations Metrics Metrics Downloads Citations No data available. 28 3 Total 6 Months 12 Months Total number of downloads Citations If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. Simply select your manager software from the list below and click Download. Format - [x] Direct import Cited By A growing string method for the reaction pathway defined by a Newton trajectory The Journal of Chemical Physics, Vol. 122, No. 17 \| 3 May 2005 Cross Ref Third and fourth–order optimality conditions in optimization Optimization, Vol. 33, No. 2 \| 1 Jan 1995 Cross Ref Critical stationary points and descent from saddlepoints in constrained optimization via implicit automatic differentiation Optimization, Vol. 27, No. 3 \| 1 Jan 1993 Cross Ref Figures Tables Media Share Share Copy the content Link Copy Link Copied! Copying failed. Share with email Email a colleague Share on social media FacebookX (formerly Twitter)LinkedInemail Get Access Get Access PurchaseSave for laterItem saved, go to cart Article Pay-Per-View $40.00 Add to cart Article Pay-Per-View Checkout Access via your Institution Questions about how to access this content? Contact SIAM at service@siam.org. References References 1. D. Castrigiano, S. Hayes-Widmann, Catastrophe Theory, Addison-Wesley, Reading, MA, 1991 Google Scholar 2. C. H. Edwards, Advanced calculus of several variables, Academic Press (a subsidiary of Harcourt Brace Jovanovich, Publishers), New York-London, 1973x+457 Google Scholar 3. Herbert Fischer, G. V. Milovanovic, Some aspects of automatic differentiation Numerical methods and approximation theory, III (Niš, 1987), Univ. Niš, Niš, 1988, 199–208, Yugoslavia Google Scholar 4. Herbert Fischer, Automatic differentiation of characterizing sequences, J. Comput. Appl. Math., 28 (1989), 181–185 Crossref Web of Science Google Scholar 5. H. Fischer, Automatic differentiation of implicitly defined functions, Tech. Report, 112A, Institut für Angewandte Mathematik und Statistik, Technische Universität München, München, Germany, 1989 Google Scholar 6. R. Fletcher, Practical methods of optimization. Vol. 1, John Wiley & Sons Ltd., Chichester, 1980viii+120 Google Scholar Recommended Content Linearly Constrained Non-Lipschitz Optimization for Image Restoration Wei Bian , Xiaojun Chen Abstract Nonsmooth nonconvex optimization models have been widely used in the restoration and reconstruction of real images. In this paper, we consider a linearly constrained optimization problem with a non-Lipschitz regularization term in the objective function which includes the $l_p$ norm ($0<p<1$) of the gradient of the underlying image in the $l_2$-$l_p$ problem as a special case. We prove that any cluster point of $\epsilon$ scaled first order stationary points satisfies a first order necessary condition for a local minimizer of the optimization problem as $\epsilon $ goes to $0$. We propose a smoothing quadratic regularization (SQR) method for solving the problem. At each iteration of the SQR algorithm, a new iterate is generated by solving a strongly convex quadratic problem with linear constraints. Moreover, we show that the SQR algorithm can find an $\epsilon$ scaled first order stationary point in at most $O(\epsilon^{-2})$ iterations from any starting point. Numerical examples are given to show good performance of the SQR algorithm for image restoration. Critical Imperfection of Symmetric Structures Kazuo Murota , Kiyohiro Ikeda Abstract A method is presented to determine the critical (most unfavorable) initial imperfection of structures of regular-polygonal symmetry (denoted by dihedral groups). A critical point of such structures is either simple or forced to be double by the symmetry, and the group representation theory is employed to deal with the degeneracy due to symmetry. The method is developed further to incorporate the symmetry of imperfections. In many cases of practical interest, the critical imperfection is given explicitly as the product of "imperfection sensitivity matrix" and the critical eigenvector. It is shown that the symmetry in the critical eigenvectors of the tangent-stiffness matrix is inherited to the symmetry in the critical imperfection mode. Worst-Case Complexity of Smoothing Quadratic Regularization Methods for Non-Lipschitzian Optimization Wei Bian , Xiaojun Chen Abstract In this paper, we propose a smoothing quadratic regularization (SQR) algorithm for solving a class of nonsmooth nonconvex, perhaps even non-Lipschitzian minimization problems, which has wide applications in statistics and sparse reconstruction. The proposed SQR algorithm is a first order method. At each iteration, the SQR algorithm solves a strongly convex quadratic minimization problem with a diagonal Hessian matrix, which has a simple closed-form solution that is inexpensive to calculate. We show that the worst-case complexity of reaching an $\epsilon$ scaled stationary point is $O(\epsilon^{-2})$. Moreover, if the objective function is locally Lipschitz continuous, the SQR algorithm with a slightly modified updating scheme for the smoothing parameter and iterate can obtain an $\epsilon$ Clarke stationary point in at most $O(\epsilon^{-3})$ iterations. Mountain Passes and Saddle Points James Bisgard Abstract Variational methods find solutions of equations by considering a solution as a critical point of an appropriately chosen function. Local minima and maxima are well-known types of critical points. We explore methods for finding critical points that are neither local maxima or minima, but instead are mountain passes or saddle points. Criteria for the existence of minima or maxima are well known, but those for mountain passes or saddle points are less well known. We give an accessible treatment of some criteria for the existence of such points (including the mountain pass lemma), as well as describe a method that could be used to find such points. Iteration Complexity of a Block Coordinate Gradient Descent Method for Convex Optimization Xiaoqin Hua , Nobuo Yamashita Abstract In this paper, we study the iteration complexity of a block coordinate gradient descent (BCGD) method with a cyclic rule for solving convex optimization problems. We propose a new Lipschitz continuity-like assumption and show that the iteration complexity for the proposed BCGD method can be improved to $O(\frac{\max{M, \;{L}}}{\varepsilon})$, where $M$ is the constant in the proposed assumption, ${L}$ is the usual Lipschitz constant for the gradient of the objective function, and $\varepsilon>0$ is the required precision. In addition, we analyze the relation between $M$ and ${L}$, and prove that, in the worst case, $M\leq \sqrt{N}{L}$, where $N$ is the number of blocks. 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15482	https://www.aafp.org/pubs/afp/issues/2007/1115/p1509.html	Scheduled maintenance is planned for September 26–29. You may experience brief interruptions during this time. LESLIE E. TINGLE, MD, DANIEL MOLINA, MD, AND CHARLES W. CALVERT, DO A more recent article on acute pericarditis is available. Am Fam Physician. 2007;76(10):1509-1514 Author disclosure: Nothing to disclose. Although acute pericarditis is most often associated with viral infection, it may also be caused by many diseases, drugs, invasive cardiothoracic procedures, and chest trauma. Diagnosing acute pericarditis is often a process of exclusion. A history of abrupt-onset chest pain, the presence of a pericardial friction rub, and changes on electrocardiography suggest acute pericarditis, as do PR-segment depression and upwardly concave ST-segment elevation. Although highly specific for pericarditis, the pericardial friction rub is often absent or transient. Auscultation during end expiration with the patient sitting up and leaning forward increases the likelihood of observing this physical finding. Echocardiography is recommended for most patients to confirm the diagnosis and to exclude tamponade. Outpatient management of select patients with acute pericarditis is an option. Complications may include pericardial effusion with tamponade, recurrence, and chronic constrictive pericarditis. Use of colchicine as an adjunct to conventional nonsteroidal anti-inflammatory drug therapy for acute viral pericarditis may hasten symptom resolution and reduce recurrences. Acute pericarditis is a common disease that must be considered in the differential diagnosis of chest pain in adults.1 The clinical syndrome of pericarditis results from inflammation of the pericardium, a fibrous sac that envelops the heart and the base of the great vessels. The pericardium has a visceral and a parietal layer, between which up to 50 mL of serous fluid is found in healthy patients.2 Pericarditis may present as an indolent process with no significant pain, as seen in patients with tuberculosis, or it may be heralded by the sudden onset of severe substernal chest pain in acute idiopathic or viral pericarditis. Although the incidence of acute pericarditis is unknown, up to 5 percent of visits to emergency departments for nonacute myocardial infarction (MI) chest pain may be related to pericarditis.3 Recognition of the clinical syndrome is important because it must be distinguished from acute coronary syndromes and pulmonary embolism. Further, the syndrome may lead to cardiac tamponade or constrictive pericarditis, or may be associated with underlying conditions (e.g., acquired immunodeficiency syndrome [AIDS], malignancy, MI, collagen vascular disease). \| Clinical recommendation \| Evidence rating \| References \| --- \| Echocardiography is recommended for patients with suspected pericardial disease, including effusion, constriction, or effusive-constrictive process. \| C \| 10, 25 \| \| Pericardiocentesis should be reserved to treat cardiac tamponade and suspected purulent pericarditis. \| C \| 10, 25 \| \| Colchicine should be considered as an adjunct to nonsteroidal anti-inflammatory drug therapy in patients with acute viral or idiopathic pericarditis. \| B \| 28, 29 \| \| Corticosteroid therapy alone should be avoided in patients with acute or idiopathic pericarditis. \| B \| 29 \| Etiology Although viral infection is the most common identifiable cause of acute pericarditis, the condition may be associated with many diseases.4 Nonviral causes of pericarditis include bacterial infection, MI, chest trauma, and neoplasm. Causes of pericarditis are listed in Table 1.5–7 \| \| \| --- \| \| Idiopathic (nonspecific, probably viral) \| \| \| Infectious causes \| \| \| \| Viruses: coxsackievirus A and B, hepatitis viruses, human immunodeficiency virus, influenza, measles virus, mumps virus, varicella virus \| \| \| Bacteria: gram-positive and gram-negative organisms; rarely, Mycobacterium tuberculosis \| \| \| Fungi (most common in immunocompromised patients): Blastomyces dermatitidis, Candida species, Histoplasma capsulatum \| \| \| Echinococcus granulosus \| \| Noninfectious causes \| \| \| \| Acute myocardial infarction \| \| \| Aortic dissection \| \| \| Renal failure† \| \| \| Malignancy: breast cancer, lung cancer, Hodgkin's disease, leukemia, lymphoma by local invasion \| \| \| Radiation therapy (usually for breast or lung cancer) \| \| \| Chest trauma \| \| \| Postpericardiotomy \| \| \| Cardiac procedures: catheterization, pacemaker placement, ablation \| \| \| Autoimmune disorders: mixed connective tissue disorder, hypothyroidism, inflammatory bowel disease, rheumatoid arthritis, scleroderma, spondyloarthropathies, systemic lupus erythematosus, Wegener's granulomatosis \| \| \| Sarcoidosis \| \| Medications \| \| \| \| Dantrolene (Dantrium), doxorubicin (Adriamycin), hydralazine (Apresoline; brand not available in the United States), isoniazid (INH), mesalamine (Rowasa), methysergide (Sansert; brand not available in United States), penicillin, phenytoin (Dilantin), procainamide (Procanbid), rifampin (Rifadin) \| Pericardial disease is the most common cardiovascular manifestation of AIDS, occurring in up to 20 percent of patients with human immunodeficiency virus infection/AIDS. Although the incidence of bacterial pericarditis is declining in developed countries, it has increased among patients with AIDS.8,9 Patients with AIDS and other immunocompromised persons are also at high risk for tuberculous and fungal pericarditis. The mortality rate for untreated tuberculous pericarditis approaches 85 percent. Tuberculous pericarditis often presents with subacute illness that includes fever, a large pericardial effusion, and tamponade. The diagnosis is confirmed by identification of Mycobacterium tuberculosis in pericardial fluid or tissue. Other diagnostic tools include polymerase chain reaction for DNA of mycobacteria, adenosine deaminase, and interferon-γ in pericardial fluid.10 Pericarditis also can arise as a complication of MI. Post–MI pericarditis may develop two to four days after an acute infarction and results from a reaction between the pericardium and the damaged adjacent myocardium. Dressler's syndrome is a post–MI phenomenon in which pericarditis develops weeks to months after an acute infarction; this syndrome is thought to reflect a late autoimmune reaction mediated by antibodies to circulating myocardial antigens.2 Pathophysiology The acute inflammatory response in pericarditis can produce either serous or purulent fluid, or a dense fibrinous material. In viral pericarditis, the pericardial fluid is most commonly serous, is of low volume, and resolves spontaneously. Neoplastic, tuberculous, and purulent pericarditis may be associated with large effusions that are exudative, hemorrhagic, and leukocyte filled.2,11 Gradual accumulation of large fluid volumes in the pericardium, even up to 250 mL, may not result in significant clinical signs.12 Although pericardial effusion may be absent in 60 percent of patients, large pericardial effusions may accumulate rapidly and impede diastolic filling of the right heart, resulting in cardiac tamponade and death.1,3,11 Cardiac tamponade associated with acute pericardial disease is more common in patients with neoplastic, tuberculous, and purulent pericarditis (60 percent) than in patients with acute viral pericarditis (14 percent).11,13 Prolonged pericarditis may result in persistent accumulation of pericardial fluid. In this case, the fluid constitutes to form a thick coating that surrounds the myocardium. The ultimate manifestation may be constrictive pericarditis.4 Diagnosis HISTORY AND PHYSICAL EXAMINATION Patients with acute pericarditis commonly report a prodrome of fever, malaise, and myalgias. The cardinal features of acute pericarditis are chest pain, pericardial friction rub, and gradual repolarization changes on electrocardiography (ECG).4,14,15 In patients with acute pericarditis, chest pain is abrupt in onset, pleuritic, and substernal or left precordial in location. It may radiate to the trapezius ridge, neck, arms, or jaw. The pain is relieved by leaning forward and is made worse by lying supine. The classic triphasic pericardial rub is best heard along the left sternal border with the patient sitting up and leaning forward (Table 2).7 The rightsholder did not grant rights to reproduce this item in electronic media. For the missing item, see the original print version of this publication. Although auscultation of a pericardial friction rub has high specificity (approaching 100 percent) for acute pericarditis, it has low sensitivity that varies with the frequency of auscultation.15,16 The pericardial rub is best auscultated with the diaphragm of the stethoscope over the left lower sternal border in end expiration with the patient leaning forward. It has a rasping or creaking sound similar to leather rubbing against leather. The classic pericardial rub is triphasic but occurs in only one half of patients with a rub, whereas the remainder of patients have a biphasic or monophasic rub. The three phases of the pericardial rub correspond to the movement of the heart against the pericardial sac during atrial systole, ventricular systole, and rapid ventricular filling. The rub of pericarditis may be transient, making it important for physicians to auscultate the heart repeatedly.4,5,15 The presence of a monophasic rub occurring with the respiratory cycle in the absence of diagnostic changes on ECG or elevated cardiac enzymes should alert physicians to the possibility of pleuritis.3 Physical findings that suggest acute cardiac tamponade include tachypnea, tachycardia, neck vein distention, hypotension, and inspiratory fall in arterial blood pressure. Echocardiographic evidence of diastolic chamber collapse of the right atrium, right ventricle, or both serves to confirm tamponade.4 Tuberculous pericarditis presents differently. A study of 233 patients found that fever, night sweats, weight loss, elevated serum globulin, and normal peripheral white blood cell count were independent predictors of tuberculous pericarditis and could be used in a clinical decision tool for making an accurate diagnosis.17 DIAGNOSTIC TESTS Superficial myocardial inflammation is believed to explain the four stages of ECG changes visible during acute pericarditis.18 These stages involve diffuse, upwardly concave ST-segment elevation; T-wave inversion; and PR-segment depression.19,20 The evolution of these ECG changes helps distinguish pericarditis from early repolarization and acute MI19 (Table 27 ). In stage I, which can last a few hours to several days, ST segments elevate, T waves remain upright, and PR segments are isoelectric or become depressed. After a few days, the ST and PR segments normalize, typifying stage II. In stage III, diffuse T-wave inversions remain after the ST segments have normalized. The ECG returns to normal in stage IV unless chronic pericarditis develops, leading to persistence of T-wave inversion.20 No reciprocal changes or Q waves are found in the 12-lead ECG during acute pericarditis, which is an important feature in distinguishing acute pericarditis from acute MI.13,21,22 The changes in stage I may be confused with findings of MI or early repolarization; old ECGs help differentiate among these conditions. It is important to remember that no PR-segment depression occurs in MI. The absence of PR-segment depression, however, does not rule out acute pericarditis because it may be found in about 25 percent of cases. In addition, the ST-segment elevation in acute infarction is upwardly convex in concordant leads, and Q waves often appear. The most reliable differential finding in the ECG is the ratio of the magnitude of the ST-segment elevation to the T-wave amplitude in the V6 lead; acute pericarditis is more likely when the ratio is greater than 0.25.19,23 Acute pericarditis is often associated with elevated markers of acute inflammation, including C-reactive protein, erythrocyte sedimentation rate, and leukocyte count. Markers of myocardial injury such as the MB isoenzyme of creatine kinase and cardiac troponins are often elevated. Troponin I elevation occurs in patients with ST-segment elevation and likely corresponds to epicardial cell damage. This type of cell damage in patients with acute pericarditis is seen in especially young patients and in those with recent infection. Family physicians should consider consultation with a cardiologist for patients with an atypical pericarditis presentation and elevated troponin I.24 In patients with suspected pericarditis, echocardiography helps detect pericardial effusion, cardiac tamponade, and underlying myocardial disease.10,25 When pericardial effusion is found in the setting of clinical pericarditis, the diagnosis is confirmed. However, results of echocardiography may be normal in patients with the clinical syndrome of pericarditis.10 Computed tomography (Figure 1) and magnetic resonance imaging are useful if the initial work-up for pericarditis is inconclusive.10 RECOMMENDED DIAGNOSTIC STRATEGY Most cases of acute pericarditis are idiopathic, and initial work-up should be limited to a history and physical examination, complete blood count, erythrocyte sedimentation rate, troponin I, serum chemistry, ECG, chest radiography, and echocardiography. For patients with tamponade, those with no known associated illness (Table 1),5–7 and those in whom pericardial disease does not improve within one week, antinuclear antibodies, rheumatoid factor, and mycobacterial studies (i.e., cultures of sputa and gastric aspirate) should be obtained.13 If the patient also has a pleural effusion, thoracentesis is recommended. The pleural fluid should be assessed for adenosine deaminase, cytology, and mycobacteria.13,25 Pericardiocentesis is only indicated to treat cardiac tamponade or when purulent pericarditis is suspected. If pericardiocentesis is ineffective or tamponade recurs, subxiphoid pericardial drainage and biopsy with histology and cultures are recommended.13 The diagnostic yield for pericardiocentesis and pericardial biopsy is about the same (19 percent).14 The yield increases to 34 percent when either procedure is performed therapeutically, as for patients with symptomatic cardiac tamponade. Some experts recommend that pericardiocentesis not be performed exclusively for diagnostic purposes unless purulent pericarditis is strongly suspected.14,26 Treatment The goal of treating acute viral pericarditis is to relieve pain and to prevent complications such as recurrence, tamponade, and chronic restrictive pericarditis. Because most episodes of acute pericarditis are uncomplicated, many patients may be managed in the outpatient setting.26,27 Indications for hospitalization of patients with acute pericarditis are listed in Table 3.26 When tamponade is identified, pericardiocentesis is the intervention of choice.13 If the patient has no poor prognostic indicators (e.g., body temperature > 100.4° F [38° C], findings of tamponade, history of immunocompromise or trauma) and the basic laboratory studies discussed earlier are reassuring, outpatient management is an option.24 \| \| \| Anticoagulation therapy \| \| Body temperature greater than 100.4° F (38° C) \| \| Echocardiographic findings of a large pericardial effusion \| \| Findings of cardiac tamponade (i.e., hypotension and neck vein distention) \| \| History of trauma and compromised immune system \| \| Myopericarditis \| \| Troponin I elevation \| Nonsteroidal anti-inflammatory drugs, including aspirin and ibuprofen (Motrin), have been used to relieve chest pain, fever, and friction rub in patients with acute pericarditis.10,24–26 Aspirin (up to 800 mg every six hours) and ibuprofen (300 to 800 mg every six hours) are conventional therapies.10 The authors of two recent randomized controlled trials examined the value of colchicine in the treatment of acute and recurrent pericarditis.28,29 Participants in each study received either conventional treatment with aspirin (800 mg orally every six or eight hours for seven to 10 days with gradual tapering over three to four weeks) or aspirin at the same dose combined with colchicine (1.0 to 2.0 mg for the first day, then 0.5 to 1.0 mg daily for three months). Patients intolerant of aspirin were given a tapering dose of prednisone over one month.29 The addition of colchicine reduced the recurrence rate of pericarditis at 18 months (32.3 percent for aspirin or prednisone alone, 10.7 percent for an anti-inflammatory plus colchicine; P = .004; number needed to treat [NNT] = 5). Addition of colchicine also decreased symptom persistence at 72 hours (11.7 percent versus 36.7 percent for aspirin or prednisone alone; P = .003; NNT = 4).29 Results were similar for patients with recurrent pericarditis.28 Use of corticosteroids alone was an independent risk factor for the subsequent recurrence of pericarditis in this study,29 although corticosteroids may be useful in refractory acute pericarditis. Side effects in the colchicine and noncolchicine groups were similar. This study was limited by an open-label design and the use of the subjective end points of symptom recurrence, but the results29 provide some evidence that colchicine should be considered as an adjunct to anti-inflammatory drugs, and that corticosteroids should be used with caution. Prognosis Generally, acute pericarditis is benign and self-limiting. Occasionally, acute pericarditis will be complicated by tamponade, constriction, or recurrence. Nearly 24 percent of patients with acute pericarditis will have recurrence. Most of these patients will have a single recurrence within the first weeks after the initial episode, and a minority may have repeated episodes for months or years. The first episode of viral pericarditis is usually the most severe, whereas recurrences are less severe and may present only with chest pain.16 Continue Reading Copyright © 2007 by the American Academy of Family Physicians. This content is owned by the AAFP. A person viewing it online may make one printout of the material and may use that printout only for his or her personal, non-commercial reference. This material may not otherwise be downloaded, copied, printed, stored, transmitted or reproduced in any medium, whether now known or later invented, except as authorized in writing by the AAFP. See permissions for copyright questions and/or permission requests. Copyright © 2025 American Academy of Family Physicians. All Rights Reserved.
15483	https://dev.to/phoenix_238501d86d417e/count-number-of-pairs-with-absolute-difference-k-3m6j	Forem Feed DEV Community A space to discuss and keep up software development and manage your software career Gamers Forem An inclusive community for gaming enthusiasts Future News and discussion of science and technology such as AI, VR, cryptocurrency, quantum computing, and more. Music Forem From composing and gigging to gear, hot music takes, and everything in between. DUMB DEV Community Memes and software development shitposting Vibe Coding Forem Discussing AI software development, and showing off what we're building. Popcorn Movies and TV Movie and TV enthusiasm, criticism and everything in-between. Design Community Web design, graphic design and everything in-between Maker Forem A community for makers, hobbyists, and professionals to discuss Arduino, Raspberry Pi, 3D printing, and much more. Open Forem A general discussion space for the Forem community. If it doesn't have a home elsewhere, it belongs here Scale Forem For engineers building software at scale. We discuss architecture, cloud-native, and SRE—the hard-won lessons you can't just Google Forem Core Discussing the core forem open source software project — features, bugs, performance, self-hosting. Security Forem Your central hub for all things security. From ethical hacking and CTFs to GRC and career development, for beginners and pros alike Crypto Forem A collaborative community for all things Crypto—from Bitcoin to protocol development and DeFi to NFTs and market analysis. DEV Community Posted on Sep 21, 2024 • Edited on Sep 23, 2024 Count Number of Pairs With Absolute Difference K Leetcode - 75 (4 Part Series) Problem Link - Count Pairs Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that \|nums[i] - nums[j]\| == k. The value of \|x\| is defined as: x if x >= 0. -x if x < 0. Example 1: Input: nums = [1,2,2,1], k = 1 Output: 4 Explanation: The pairs with an absolute difference of 1 are: Example 2: Input: nums = [1,3], k = 3 Output: 0 Explanation: There are no pairs with an absolute difference of 3. Naive Approach Code int countKDifference(vector<int>& nums, int k) { int count = 0; int n = nums.size(); for(int i=0;i<n-1;i++) { for(int j=i+1;j<n;j++) { if(abs(nums[i] - nums[j]) == k) { count++; } } } return count; } Optimal Approach Create a HashMap for Frequencies: Populate the HashMap: Find Valid Pairs: Avoid Double Counting Code int countKDifference(vector<int>& nums, int k) { int n=nums.size(); map<int,int> mp; int count=0; for(int i=0;i<n;i++){ mp[nums[i]]++; } for(int i=0;i<n;i++){ int t1=nums[i]-k; int t2=nums[i]+k; if(mp.find(t1)!=mp.end() ){ if(t1==nums[i]){ count=count+mp[t1]-1; }else{ count+=mp[t1]; } } if(mp.find(t2)!=mp.end()){ if(t2==nums[i]){ count=count+mp[t2]-1; }else{ count+=mp[t2]; } } } return count/2; } Note if we are aksed to calcuate distinct pairs, instead of adding frequencies of target element to the count variable, just add 1 Leetcode - 75 (4 Part Series) Top comments (0) Templates let you quickly answer FAQs or store snippets for re-use. Are you sure you want to hide this comment? It will become hidden in your post, but will still be visible via the comment's permalink. Hide child comments as well Confirm For further actions, you may consider blocking this person and/or reporting abuse More from Phoenix 💎 DEV Diamond Sponsors Thank you to our Diamond Sponsors for supporting the DEV Community Google AI is the official AI Model and Platform Partner of DEV Neon is the official database partner of DEV Algolia is the official search partner of DEV DEV Community — A space to discuss and keep up software development and manage your software career Built on Forem — the open source software that powers DEV and other inclusive communities. Made with love and Ruby on Rails. DEV Community © 2016 - 2025. We're a place where coders share, stay up-to-date and grow their careers.
15484	https://en.wikipedia.org/wiki/Non_sequitur_(literary_device)	Jump to content Non sequitur (literary device) Català Español Euskara Italiano Tagalog Edit links From Wikipedia, the free encyclopedia Conversational literary device Not to be confused with a non sequitur in the sense of a formal fallacy. A non sequitur (English: /nɒn ˈsɛkwɪtər/ non SEK-wit-ər, Classical Latin: [noːn ˈsɛkᶣɪtʊr]; "[it] does not follow") is a conversational literary device, often used for comedic purposes. It is something said that, because of its apparent lack of meaning relative to what preceded it, seems absurd to the point of being humorous or confusing. This use of the term is distinct from the non sequitur in logic, where it is a fallacy. Etymology [edit] The expression is Latin for "[it] does not follow". It comes from the words non meaning "not" and the verb sequi meaning "to follow". Usage [edit] A non sequitur can denote an abrupt, illogical, or unexpected turn in plot or dialogue by including a relatively inappropriate change in manner. A non sequitur joke sincerely has no explanation, but it reflects the idiosyncrasies, mental frames and alternative world of the particular comic persona. Comic artist Gary Larson's The Far Side cartoons are known for what Larson calls "absurd, almost non sequitur animal" characters, such as talking cows, to create a bizarre effect. He gives the example of a strip where "two cows in a field gaze toward burning Chicago, saying 'It seems that agent 6373 had accomplished her mission.'" See also [edit] Anacoluthon Anti-humor Dada Derailment (thought disorder) "Good day, fellow!" "Axe handle!" Gibberish Roger Irrelevant Surreal humour References [edit] ^ The Oxford Pocket Dictionary of Current English. Oxford University Press, 2009. ^ "Non Sequitur - Examples and Definition of Non Sequitur". Literary Devices. 2014-01-02. Retrieved 2021-08-31. ^ Merriam Webster's Online Dictionary. Archived 2012-02-18 at the Wayback Machine ^ "Definition of NON SEQUITUR". www.merriam-webster.com. Retrieved 2021-08-31. ^ Chambers, Robert (2010). Parody: The Art that Plays with Art. Peter Lang Publishers. p. 75. ISBN 978-1433108693. Retrieved 2014-09-17. Along with a rhythmic pattern, these jokes, however absurd they may be, build dual frames of reference, if not alternative worlds entirely reflecting the idiosyncrasies of the individual stand-up artist. ^ Harrington, Richard (16 June 1983). "The Bizarre Side". Washington Post. Retrieved 12 August 2020. Further reading [edit] The Koan: Texts and Contexts in Zen Buddhism. United Kingdom, Oxford University Press, 2000. Shabo, Magedah Rhetoric, Logic, and Argumentation: A Guide for Student Writers. United States, Prestwick House, 2010. External links [edit] Getting It: Human Event-Related Brain Response to Jokes in Good and Poor Comprehenders - "When asked to pick the punch-line of a joke from an array of choices, including straightforward endings, non sequitur endings, and the correct punch-line, RHD patients erred by picking non sequitur endings, indicating that they know surprise is necessary" Retrieved from " Categories: Humour Jokes Latin literary phrases Narratology Hidden categories: Webarchive template wayback links Articles with short description Short description matches Wikidata Pages with Classical Latin IPA
15485	https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem?srsltid=AfmBOoq3t07XNET8ppJ5UaAcKTn5k2Ko2sr3pDC9b3-NtCgRnfyvgUF6	Page Toolbox Search Angle Bisector Theorem \| \| \| This is an AoPSWiki Word of the Week for June 6-12 \| Contents Introduction & Formulas The Angle bisector theorem states that given triangle and angle bisector AD, where D is on side BC, then . It follows that . Likewise, the converse of this theorem holds as well. Further by combining with Stewart's theorem it can be shown that Proof By the Law of Sines on and , First, because is an angle bisector, we know that and thus , so the denominators are equal. Second, we observe that and . Therefore, , so the numerators are equal. It then follows that Examples & Problems See also Something appears to not have loaded correctly. Click to refresh.
15486	https://app.twinscience.com/en/twin-library/contents/677e1878ac2c3afd2dc98d80	Fractions: Intro \| Twin Science Educator Platform menu Home Classes Library Twin Library Lessons & Curricula My Library AI Tools Assistant Teacher Resources PD Resources Impact Project Teacher Community Coding Twin Code Lab Explore Blocks Twin Library Fractions: Intro How It Works Simulation Fractions: Intro Grades 1st - 2nd This interactive fractions game introduces parts of a whole to students aged 9-13. Teachers can use it to build foundational math skills with engaging, hands-on challenges. Assign to Class Introduction Explore the basics of fractions with this interactive simulation. Students can visualize fractions as parts of a whole and compare their sizes. By manipulating shapes and dividing them into equal parts, they develop a deeper understanding of numerators, denominators, and fraction equivalence. This tool makes learning fractions intuitive and fun by connecting abstract math concepts to everyday contexts. Let's Start Fractions: Intro by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC-BY 4.0 Curriculum & Lesson Library Home Classes Twin Library My Library AI Tools Assistant PD Resources Impact Project Teacher Community Twin Code Lab Explore Blocks Change Language
15487	https://or.stackexchange.com/questions/10837/compute-overlapping-time	mixed integer programming - Compute overlapping time - Operations Research Stack Exchange Join Operations Research By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Operations Research helpchat Operations Research Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Operations Research Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Compute overlapping time Ask Question Asked 2 years, 1 month ago Modified2 years, 1 month ago Viewed 232 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I am trying to solve an optimization problem in which there is a set of tasks, S S, where s i s i and e i e i are the starting and ending time of task i∈S i∈S. Each task i i must be done within its time own window [a i,b i][a i,b i]: a i≤s i,∀i∈S a i≤s i,∀i∈S e i≤b i,∀i∈S e i≤b i,∀i∈S Due to the presence of time windows, tasks could overlap. I have defined continuous variable t i j t i j to quantify the overlapping time of tasks i i and j j. The objective is to minimize the overlapping time: m i n∑i,j∈S t i j m i n∑i,j∈S t i j How can I define constraints to compute t i j t i j? mixed-integer-programming modeling constraint scheduling Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Aug 15, 2023 at 13:01 RobPratt 36.2k 2 2 gold badges 51 51 silver badges 94 94 bronze badges asked Aug 15, 2023 at 7:40 Christopher Expósito IzquierdoChristopher Expósito Izquierdo 45 2 2 bronze badges 1 Welcome to OR.SE. Do you have a single resource to process the tasks or there are multiple resources? Is it possible to have no overlap or this is mandatory?A.Omidi –A.Omidi 2023-08-15 19:50:31 +00:00 Commented Aug 15, 2023 at 19:50 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. First note that you can omit t i j t i j for any pair (i,j)(i,j) for which the time windows do not overlap because t i j=0 t i j=0 in that case. Tasks i i and j j overlap if and only if the later starting time max(s i,s j)max(s i,s j) precedes the earlier ending time min(e i,e j)min(e i,e j), so you want to enforce t i j≥max(min(e i,e j)−max(s i,s j),0),t i j≥max(min(e i,e j)−max(s i,s j),0), which can be linearized by introducing u i j u i j to represent max(s i,s j)max(s i,s j), v i j v i j to represent min(e i,e j)min(e i,e j), binary variables x i j x i j and y i j y i j, and linear constraints: 0≤u i j−s i 0≤u i j−s j(a j−b i)x i j≤v i j−e i(a i−b j)(1−x i j)≤v i j−e j t i j t i j≤(b j−a i)y i j≤(b i−a j)(1−y i j)≤0≤0≥v i j−u i j≥0(1)(2)(3)(4)(5)(6)(1)0≤u i j−s i≤(b j−a i)y i j(2)0≤u i j−s j≤(b i−a j)(1−y i j)(3)(a j−b i)x i j≤v i j−e i≤0(4)(a i−b j)(1−x i j)≤v i j−e j≤0(5)t i j≥v i j−u i j(6)t i j≥0 Constraints (1)(1) and (2)(2) enforce u i j=max(s i,s j)u i j=max(s i,s j). Constraints (3)(3) and (4)(4) enforce v i j=min(e i,e j)v i j=min(e i,e j). Constraints (5)(5) and (6)(6) enforce t i j≥max(v i j−u i j,0)t i j≥max(v i j−u i j,0). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Aug 15, 2023 at 20:34 answered Aug 15, 2023 at 13:01 RobPrattRobPratt 36.2k 2 2 gold badges 51 51 silver badges 94 94 bronze badges 3 Does this require the introduction of binary variables?prubin –prubin♦ 2023-08-15 15:49:57 +00:00 Commented Aug 15, 2023 at 15:49 @prubin Yes, it does.RobPratt –RobPratt 2023-08-15 16:02:30 +00:00 Commented Aug 15, 2023 at 16:02 @RobPratt thank you very much for your help. It works perfectly :)Christopher Expósito Izquierdo –Christopher Expósito Izquierdo 2023-08-16 10:12:29 +00:00 Commented Aug 16, 2023 at 10:12 Add a comment\| Your Answer Thanks for contributing an answer to Operations Research Stack Exchange! Please be sure to answer the question. Provide details and share your research! 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15488	https://www.scribd.com/document/523437711/LESSON-PLAN-IN-MATHEMATICS-9	Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Upload 100%(8)100% found this document useful (8 votes) 13K views5 pages Lesson Plan in Mathematics 9 (4a's METHOD) The document provides a lesson plan for teaching Grade 9 students how to solve quadratic equations by factoring. It includes learning objectives, content, procedures, examples, and exercises… Uploaded by MANDADO, ELLA V. You are on page 1/ 5 OBTTAM ^OLM CM ILXJBILXCDT 3 (8L–s IBXJAH) C. OBLPMCMG A@KBDXCUBT; Lt tjb bmh ae tjb dolss sbsscam, tjb Grlhb 3 stuhbmts iust b lob ta ha tjb eaooawcmg wctj lt oblst :>% obvbo ae lddurldy6 Hbecmb l zbra-prahudt prapbrty6 (`.) 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Afly, ear tjb cmeariltcam ae bvbryamb tjcs glib cs daipasb ae pcdturbs ar saib obttbrs lmh yau mbbh ta dai`cmb bldj pcbdb ta eari l warh ar pjrlsbs. Ha yau umhbrstlmh1 Sbs, il–li! Tcmdb yau jlvb l graup, tjb tbliiltbs dlm jbop tjraugj yaur graup djlts. Sau amoy jlvb <> sbdamhs ta lmswbr bldj puzzob. Wjbm tjb tcib cs up lmh yau hch mat lmswbr wctjcm tjb looattbh tcib, yaur lmswbr cs mat vloch lmyiarb. Cm lhhctcam, wb wcoo pcdf lmatjbr tbli lmh l polybr ta poly tjb glib. Lrb yau rblhy polybrs1 Sbs, wb lrb rblhy il–li! Afly, maw wb wcoo bgcm. Ear tjb ecrst raumh obt us sbb wja tjb oudfy polybr cs. (^olycmg tjb rlmhaiczbr) Afly wb jlvb erai graup QQ muibr QQ. Jbrb cs tjb pcdturb warh puzzob ear yau. (Plmhai stuhbmt wcoo poly) Ubry gaah polybr QQQQ. Sau jlvb blrmbh 2 pacmts ear jcttcmg tjb darrbdt lmswbr. Maw, wb wcoo pradbbh ta tjb mbxt. (Plmhaiczbr) Graup QQ mui`br QQ. Jbrb cs tjb pcdturb warh puzzob ear yau. (Plmhai stuhbmt wcoo poly) Sau gat tjb darrbdt lmswbr polybr QQQ. Sau losa glcm 2 pacmts ear tjcs raumh. Tcmdb tjb twa tblis jlvb lorblhy poly tjb glib. Obt–s gcvb l djlmdb ta tjb rbilcmcmg tbli. @ut stcoo obt aur rlmhaiczbr pcdf l polybr ear yaur tbli (raoocmg tjb rlmhaiczbr). Afly mui`br QQ jbrb cs tjb pcdturb warh puzzob ear yau. (Plmhai stuhbmt wcoo poly) Ubry gaah polybr QQQ. Saur tbli losa blrms 2 pacmts ear tjcs glib. Tbbis ocfb yau lrb maw rblhy ear aur mbxt tapcd, li C rcgjt1 Sbs il–li! OBTTAM ^PA^BP L. Ldtcvcty @bearb wb wcoo pradbbh ta aur prapbr hcsdusscam wb wcoo jlvb ecrst lm ldtcvcty. Sau wcoo lmswbr vcl Tadrltcvb Ttuhbmt. Sau lrb amoy gcvbm ecvb (2) icmutbs ta lmswbr tjcs. Xypb-cm ‟Tadrltcvb Ttuhbmt’ lmh tjbm fby -cm tjb qucz dahb ‟QQQQQQQQ’. Sau jlvb amoy l icmutb ta lmswbr bldj ctbi. Bldj ibi`br ae tjb graup sjauoh plrtcdcpltb lmh ilfb surb ta typb yaur daipobtb mlib. Lrb yau rblhy1 Sbs, il–li! Obt–s `bgcm! (Dahb cs prakbdtbh am tjb iamctar lmh/ar sjlrbh cm tjb Djlt @ax). (Ttuhbmt–s stlrts lmswbrcmg) @. Lmloyscs (Hcsdusscam) Download to read ad-free (Letbr 2 icmutbs) @lsb am tjb gcvbm ldtcvcty. Wjlt ha yau tjcmf cs aur obssam ear tjcs iarmcmg1 Il–li, aur obssam ear tjcs iarmcmg wcoo b laut saovcmg qulhrltcd bqultcam `y eldtarcmg. Sau gat ct rcgjt! Ear tahly–s tapcd, wb wcoo tldfob laut saovcmg qulhrltcd bqultcamy eldtarcmg. Ct cs lsbh am tjb zbra-prahudt prapbrty ae rblo muibrs. Wjlt cs tjb zbra-prahudt prapbrty1 Ce twa muibrs l lmh lrb iuotcpocbh tagbtjbr lmh tjb rbsuotcmg prahudt cs >, tjbm lt oblst amb ae tjb muibrs iustb >. ce l9 >, tjbm bctjbr l 9 > ar 9 > ar atj l 9 > lmh 9 >. Jbrb lrb tjb stbps jaw ta saovb qulhrltcd bqultcams `y eldtarcmg; Ttbps; <. Xrlmseari tjb qulhrltcd bqultcam cmta stlmhlrh eari lx 0 +x+d9> ce mbdbsslry. 0. Eldtar tjb qulhrltcd bxprbsscam. 4. Lppoy tjb zbra prahudt prapbrtyy sbttcmg bldj eldtar ae tjb qulhrltcd bxprbsscam bqulo ta zbra. 8. Taovb bldj rbsuotcmg bqultcam. 2. Djbdf tjb vloubs ae tjb vlrclob atlcmbh y sustctutcmg bldj cm tjb arcgcmlo bqultcam. Bxlipob <; Ecmh tjb saoutcams ae x 0 +3x9 -:. Taoutcam; x 0 +3x+:9-:+: trlmseari bqultcam cmta stlmhlrh eari x 0 3x+:9> (x+<)(x+:)9> Eldtar tjb qulhrltcd bxprbsscam x+<9>6 x+:9> Ttbp 4 x+<9> x9 -< x+:9> x9 -: Djbdfcmg; Xjb twa vloubs sltcsey tjb arcgcmlo bqultcam tjbrbearb x9-< lmh x9-: lrb tjb saoutcams ae tjb gcvbm qulhrltcd bqultcam. Bxlipob 0; Ecmh tjb raats ae tjb bqultcam Download to read ad-free 5x 0 +<:x9>. Taoutcam; 5x(x+4)9> Eldtar tjb bxprbsscam 5x 9 > x + 4 9 > Ttbp 0 Djbdfcmg; D. L`strldtcam (Gbmbrloczltcam) Taovcmg qulhrltcd bqultcam y eldtarcmg cslsbh am tjb zbra-prahudt prapbrty ae rblo mui`brs. Maw, wja dlm tboo ib tjb hbecmctcam ae l zbra-prahudt prapbrty1 Il–li, ce twa muibrs l lmh lrb iuotcpocbh tagbtjbr lmh tjb rbsuotcmg prahudt cs >, tjbm lt oblst amb ae tjb muibrs iustb >. Ce l9 >, tjbm bctjbr l 9 > ar 9 > ar atj l 9 > lmh 9 >. Ubry gaah! Jaw laut tjb stbps cm saovcmg qulhrltcd bqultcamy eldtarcmg1 Wja dlm bmuibrltb wjlt lrb tjasb1 Il–li tjb stbps cm saovcmg tjb qulhrltcd bqultcam `y eldtarcmg lrb; <. Xrlmseari tjb qulhrltcd bqultcam cmta stlmhlrh eari lx 0 +x+d9> ce mbdbsslry. 0. Eldtar tjb qulhrltcd bxprbsscam. 4. Lppoy tjb zbra prahudt prapbrtyy sbttcmg bldj eldtar ae tjb qulhrltcd bxprbsscam bqulo ta zbra. 8. Taovb bldj rbsuotcmg bqultcam. 2. Djbdf tjb vloubs ae tjb vlrclob atlcmbh y sustctutcmg bldj cm tjb arcgcmlo bqultcam. Hbecmctboy! Ct–s gaah ta fmaw tjlt yau rblooy ply lttbmtcam ta wjlt C jlvb hcsdussbh blrocbr. H. Lppocdltcam Lt tjcs iaibmt, obt us tbst yaur umhbrstlmhcmg. C jlvb prbplrbh saib bxbrdcsbs ear yau ta lmswbr. C wcoo kust eolsj tjb qubstcams am tjb sdrbbm lmh tjb sbobdtbh polybrs ae bldj tbli wcoo lmswbr tjraugj tjb djlt `ax. Lrb yau rblhy1 Sbs, il–li! Sau wcoo b gcvbm 0 icmutbs ta saovb bldj bqultcam. Lglcm, ce tjb sbobdtbh polybr dlmmat lmswbr wctjcm tjb looattbh tcib, jcs/jbr lmswbr cs lorblhy cmvloch lmh wb wcoo pcdf lmatjbr tbli lmh l polybr ta poly tjb glib. Ta, C wcoo maw poly tjb rlmhaiczbr ta gcvb us tjb ecrst oudfy polybr ear tjcs raumh. (^olycmg tjb rlmhaiczbr) Xjb oudfy polybr cs erai graup QQQ muibr QQQQ. 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15489	https://www.khanacademy.org/math/in-class-9-math-foundation/x6e1f683b39f990be:mensuration/x6e1f683b39f990be:3d-shapes/v/surface-area-of-a-box-cuboid-indian-accent	Surface area of a box (cuboid) (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever. Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. Skip to lesson content NCERT Math Class 9 (Bridge) Course: NCERT Math Class 9 (Bridge)>Unit 5 Lesson 2: 3D shapes Surface area of a box (cuboid) Volume of a rectangular prism Volume and surface area word problems Cylinder volume & surface area Volume of cylinders Math> NCERT Math Class 9 (Bridge)> Mensuration> 3D shapes © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Surface area of a box (cuboid) Google Classroom Microsoft Teams About About this video Surface area is total area on the surface of a three-dimensional shape. To find the surface area of a cuboid which has 6 rectangular faces, add the areas of all 6 faces. Or, you can label the length (l), width (w), and height (h) of the cuboid and use the formula: surface area (SA)=2lw+2lh+2hw.Created by Kalakrit (Dubbing). Skip to end of discussions Questions Tips & Thanks Want to join the conversation? 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15490	https://www.merriam-webster.com/thesaurus/prosperous	Definition of prosperous as in thriving marked by vigorous growth and well-being especially economically a prosperous business that will soon be expanding palmy halcyon rich well-to-do lush moneyed monied boomy opulent well-off substantial comfortable well-heeled Antonyms & Near Antonyms unsuccessful depressed failing unprosperous dying bankrupt struggling declining insolvent languishing floundering bankrupted 2 as in successful having attained a desired end or state of good fortune one of the most prosperous families in the community Synonyms & Similar Words successful thriving flourishing triumphant palmy promising in the clover going in clover growing booming robust coming roaring runaway gangbuster gangbusters boffo socko Antonyms & Near Antonyms unsuccessful inauspicious hopeless unpromising no-good failed failing futureless bankrupt collapsing washing-out folding destroyed slipping declining waning flopping ruined slumping wrecked flunking kaputt kerflooey kaput 3 as in lush growing thickly and vigorously our neighbor has a real green thumb and a yard full of healthy, prosperous plants Synonyms & Similar Words lush weedy rampant luxuriant dense thick lavish rank verdant overgrown tangled profuse close overrun Antonyms & Near Antonyms sparse dormant blighted stunted Example Sentences Examples are automatically compiled from online sources to show current usage. Read More Opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback. Recent Examples of prosperous Despite his prosperous posturing, Jake is scrambling just to pay the Rabbit’s vendors, let alone scrape together the funds to pursue a new opportunity at the Four Seasons’ iconic Pool Room. —Judy Berman, Time, 7 Sep. 2025 Every life saved strengthens the foundations for a more stable, secure, and prosperous global community. —Trickleup, Forbes.com, 1 Sep. 2025 So, after a prosperous run across multiple decades, CBS’ friendship ended with the SEC last year. —Steven Louis Goldstein, New York Times, 25 Aug. 2025 The route passed through different areas whose different atmospheres came and went, prosperous stretches giving way to rougher ones and then becoming prosperous again, like a perpetual argument or a struggle for victory. —Rachel Cusk, New Yorker, 24 Aug. 2025 See All Example Sentences for prosperous Recent Examples of Synonyms for prosperous thriving successful lush weedy wealthy rampant golden luxuriant Adjective The best and most enjoyable habit to maintain a healthy, thriving garden is a daily garden walk. — Andy Wilcox, Better Homes & Gardens, 11 Sep. 2025 In a world being reshaped by Gen Z, a thriving workplace needs more than technological agility. — Hanne Jesca Bax, Fortune, 11 Sep. 2025 Definition of thriving Adjective Nolan made a successful political comeback as then-President Barack Obama won a second term. — David Mark, The Washington Examiner, 12 Sep. 2025 In Season 1, viewers met exotic dancer Kimmie, a victim of abuse and an underground human trafficking operation, as her life deeply intersects with the Bellaires — a wealthy, powerful family running a successful hair care business in Chicago. — Taylor Ardrey, USA Today, 11 Sep. 2025 Definition of successful Adjective Then come the florals—lush, romantic, full of whimsy—before the icons land with striking confidence, like the bold centerpiece of the board. — Allure Editors, Allure, 10 Sep. 2025 Follow the guidelines recommended for your growing zone and type of grass to determine how and when to fertilize your lawn for lush grass growth. — Mary Marlowe Leverette, Southern Living, 10 Sep. 2025 Definition of lush Adjective For years, nearby residents came to this isolated corner of the lakefront to take walks and find comfort in the solitude and often unkept beach and weedy woodland behind it. — Michael McColly, Chicago Tribune, 24 Aug. 2025 Your neighbors probably won't be bothered by your weedy flower beds or rusty lawn furniture. — Brandee Gruener, Southern Living, 16 July 2025 Definition of weedy Adjective For perspective, the net worth of the seven wealthiest people in the world--all AI luminaries including Meta’s Mark Zuckerberg, Amazon’s Jeff Bezos, Google’s Larry Page and Sergey Brin and Nvidia’s Jensen Huang--exceeds $2 trillion, close to the GDP of Canada. — Martine Paris, Forbes.com, 12 Sep. 2025 In a recent interview with The School of Hard Knocks, a popular social-media channel known for interviewing wealthy entrepreneurs, Repole shared his contrarian view on entrepreneurship, emphasizing the brutal realities that most success stories gloss over. — Dave Smith, Fortune, 12 Sep. 2025 Definition of wealthy Adjective This rampant, unauthorized AI use degrades the educational experience of individual students who overly rely on the technology and those who wish to avoid using it. — Tyler Austin Harper, The Atlantic, 11 Sep. 2025 Speculation remains rampant that those three ACC schools will leave that league before the end of this decade for the SEC or Big Ten. — Fort Worth Star-Telegram, Fort Worth Star-Telegram, 11 Sep. 2025 Definition of rampant Adjective Lauren chose a golden, super-cinched Schiaparelli gown, while Jeff wore a classic tuxedo. — Anna Cafolla, Vogue, 13 Sep. 2025 The camera then pans to show the house opposite—where another golden is sitting in the exact same spot in its own yard, staring right back at Copper, but neither crossing the threshold on to the street. — Rachael O'Connor, MSNBC Newsweek, 12 Sep. 2025 Definition of golden Adjective Beyond it is a three-hectare haven of silvery olive trees, luxuriant Washingtonia palms, and lawns where peacocks wander freely. — Rebecca Ann Hughes, Forbes.com, 5 Aug. 2025 Meanwhile, a beard provided a simpler solution: a luxuriant facial covering requiring only the occasional scissor trim—no barber necessary. — Margaret Talbot, New Yorker, 21 July 2025 Definition of luxuriant Browse Nearby Words prosperity prosperous prosperousness See all Nearby Words Cite this Entry “Prosperous.” Merriam-Webster.com Thesaurus, Merriam-Webster, Accessed 16 Sep. 2025. Copy Citation Share More from Merriam-Webster on prosperous Nglish: Translation of prosperous for Spanish Speakers Last Updated: - Updated example sentences Love words? Need even more definitions? Subscribe to America's largest dictionary and get thousands more definitions and advanced search—ad free! Merriam-Webster unabridged More from Merriam-Webster ### Can you solve 4 words at once? Can you solve 4 words at once? Word of the Day debunk See Definitions and Examples » Get Word of the Day daily email! 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15491	https://artofproblemsolving.com/wiki/index.php/Power_Mean_Inequality?srsltid=AfmBOoolgFD-AxPHy-qpMt8sICLFrGqsLZKYDv8c78qw0dsyLifRil-E	Art of Problem Solving Power Mean Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Power Mean Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Power Mean Inequality The Power Mean Inequality is a generalized form of the multi-variable Arithmetic Mean-Geometric Mean Inequality. Inequality For positive real numbers and positive real weights with sum , the power mean with exponent , where , is defined by ( is the weighted geometric mean.) The Power Mean Inequality states that for all real numbers and , if . In particular, for nonzero and , and equal weights (i.e. ), if , then Considering the limiting behavior, we also have , and . The Power Mean Inequality follows from Jensen's Inequality. Proof We prove by cases: for for with Case 1: Note that As is concave, by Jensen's Inequality, the last inequality is true, proving . By replacing by , the last inequality implies as the inequality signs are flipped after multiplication by . Case 2: For , As the function is concave for all , by Jensen's Inequality, For , becomes convex as , so the inequality sign when applying Jensen's Inequality is flipped. Thus, the inequality sign in is flipped, but as , is a decreasing function, the inequality sign is flipped again after applying , resulting in as desired. Retrieved from " Categories: Algebra Inequalities Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15492	https://dk81.github.io/dkmathstats_site/math-soh-cah-toa.html	Right Angle Triangles & SOH CAH TOA dkmathstats Website Home About Math & Stats Pages R Programming Python Items Right Angle Triangles & SOH CAH TOA This topic is about right angle triangles and the memory aid SOH CAH TOA. This topic is typically found in (Canadian) high school mathematics. Table Of Contents Right Angle Triangles Finding Angles & Side Lengths SOH CAH TOA Memory Aid Examples Right Angle Triangles A right angle triangle has three sides and three angles with one of the three angles being a 90 degree (right) angle. The three angles inside a (right) triangle add up to 180 degrees. Also, the sides of the right angled triangle obey the formula c 2=a 2+b 2 c 2=a 2+b 2 (Pythagorean Theorem) where a a, b b and c c are side lengths of the triangle. (Recall that c c is the hypotenuse and the longest side of a right angle triangle. Here is an example of a right angled triangle: Image Source Image Source: Finding Angles and Side Lengths When dealing with triangles, we sometimes do not know all the angles and all the side lengths of a triangle. If we are given just enough information (i.e.2 angles and 1 side length), we can use the following formulas to determine the unknown angle(s) and the unknown side length(s). The sine of an angle θ θ (theta) is the ratio of the side length opposite to the angle θ θ over the length of the hypotenuse. The equation is as follows. sin(θ)=O p p o s i t e H y p o t e n u s e sin⁡(θ)=O p p o s i t e H y p o t e n u s e The cosine of an angle θ θ (theta) is the ratio of the side length which is adjacent to the angle θ θ over the length of the hypotenuse. Note that the adjacent side to the angle θ θ is NOT the hypotenuse. cos(θ)=A d j a c e n t H y p o t e n u s e cos⁡(θ)=A d j a c e n t H y p o t e n u s e The tangent of an angle θ θ (theta) is the ratio of the side length which is opposite to the angle θ θ over the side length which is adjacent to the angle θ θ. tan(θ)=O p p o s i t e A d j a c e n t tan⁡(θ)=O p p o s i t e A d j a c e n t SOH CAH TOA Memory Aid The three formulas can be hard to memorize and follow. A very neat memory aid is SOH CAH TOA. In SOH, the letter S refers to the sine function, the letter O refers to the side length opposite to the angle θ θ and H refers to the hypotenuse side length. The same logic applies to CAH and TOA. Here is a picture which provides a good summary. Source: Examples These examples will show how SOH CAH TOA can help in finding unknown side lengths and angles of right angled triangles. The pictures are my own and from my phone. Example One In the given picture, we have a right angle triangle with known side lengths and just the known right angle which is angle C. Referring to the angle θ θ we can determine the ratios using sine, cosine and tangent. sin(θ)=O p p o s i t e H y p o t e n u s e=6 10=0.6 sin⁡(θ)=O p p o s i t e H y p o t e n u s e=6 10=0.6 cos(θ)=A d j a c e n t H y p o t e n u s e=8 10=0.8 cos⁡(θ)=A d j a c e n t H y p o t e n u s e=8 10=0.8 tan(θ)=O p p o s i t e A d j a c e n t=6 8=0.75 tan⁡(θ)=O p p o s i t e A d j a c e n t=6 8=0.75 We have the ratios but not the angle θ θ. To find the angle θ θ we need to use the respective inverse trigonometric function. For example, the inverse sine function sin−1(x)sin−1⁡(x) would be used to find the angle θ θ given a (valid) numeric value x x. sin(θ)=0.6 sin⁡(θ)=0.6 sin−1(sin(θ))=sin−1(0.6)sin−1⁡(sin⁡(θ))=sin−1⁡(0.6) θ≈36.87∘θ≈36.87∘ cos(θ)=0.8 cos⁡(θ)=0.8 cos−1(cos(θ))=cos−1(0.8)cos−1⁡(cos⁡(θ))=cos−1⁡(0.8) θ≈36.87∘θ≈36.87∘ tan(θ)=0.75 tan⁡(θ)=0.75 tan−1(tan(θ))=tan−1(0.75)tan−1⁡(tan⁡(θ))=tan−1⁡(0.75) θ≈36.87∘θ≈36.87∘ Note that the inverse function of the original function is the argument inside the function. In this case, the inverse trigonometric function of the trigonometric function is the angle θ θ. Since the angle θ θ is found, we can now find the angle B. Angle B would be 180∘−36.87∘−90∘=53.13∘180∘−36.87∘−90∘=53.13∘. Example Two In this second example, we have only one known side length and know two of three angles in this right triangle. The side K L K L is known at 2 units with angle L at 35∘35∘ and angle K K being a right angle. The angle J J can be determined as 180∘−35∘−90∘=55∘180∘−35∘−90∘=55∘. One way of finding the side length x x is to use the tangent ratio. In this case, x x is adjacent to the angle 55∘55∘ and the side K L K L is the opposite side at length 2. Solving for x x gives us: tan(55∘)=O p p o s i t e A d j a c e n t tan⁡(55∘)=O p p o s i t e A d j a c e n t tan(55∘)=2 x tan⁡(55∘)=2 x x tan(55∘)=2 x tan⁡(55∘)=2 x=2 tan(55∘)x=2 tan⁡(55∘) x≈2÷1.428148 x≈2÷1.428148 x≈1.4004 x≈1.4004 Another way of solving for x x is using the tangent of 35∘35∘. The adjacent side of angle 35∘35∘ is K L K L with length 2 and the opposite side length is x x. tan(35∘)=O p p o s i t e A d j a c e n t tan⁡(35∘)=O p p o s i t e A d j a c e n t tan(35∘)=x 2 tan⁡(35∘)=x 2 2 tan(35∘)=x 2 tan⁡(35∘)=x 2(0.7002)≈x 2(0.7002)≈x 1.4004≈x 1.4004≈x Assuming the tangent formula is used correctly when solving for the unknown side length x x, the choice of (acute) angle does not matter too much. If we had wanted to find the length of the hypotenuse instead of x x, we would use the cosine of 35∘35∘.
15493	https://math.libretexts.org/Bookshelves/Precalculus/Corequisite_Companion_to_Precalculus_(Freidenreich)/1%3A_Simplifying_Expressions/1.01%3A_Exponent_Properties_and_More	1.1: Exponent Properties and More! - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 1: Simplifying Expressions Corequisite Companion to Precalculus (Freidenreich) { } { "1.01:_Exponent_Properties_and_More" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.02:_FOIL_Method_and_Special_Products" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.03:_Function_Notation_and_Simplify_Expressions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1:_Simplifying_Expressions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2:_Polynomials" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3:_Graphs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4:_Inequalities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5:_Rational_Expression" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6:_Radical_Expressions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7:_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8:_Problem_Solving" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Wed, 15 Sep 2021 03:19:19 GMT 1.1: Exponent Properties and More! 83107 83107 admin { } Anonymous Anonymous 2 false false [ "article:topic", "distributive property", "commutative property of addition", "scientific notation", "license:ccbysa", "showtoc:no", "Commutative Property of Multiplication", "licenseversion:40", "authorname:jfreidenreich", "Power of a Power Property", "Product of Powers Property", "Negative Exponent" ] [ "article:topic", "distributive property", "commutative property of addition", "scientific notation", "license:ccbysa", "showtoc:no", "Commutative Property of Multiplication", "licenseversion:40", "authorname:jfreidenreich", "Power of a Power Property", "Product of Powers Property", "Negative Exponent" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Bookshelves 3. Precalculus & Trigonometry 4. Corequisite Companion to Precalculus (Freidenreich) 5. 1: Simplifying Expressions 6. 1.1: Exponent Properties and More! Expand/collapse global location 1.1: Exponent Properties and More! Last updated Sep 15, 2021 Save as PDF 1: Simplifying Expressions 1.2: FOIL Method and Special Products Page ID 83107 Jennifer Freidenreich Diablo Valley College ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Property: Commutative Property of Addition 2. Property: Commutative Property of Multiplication 3. Example 1.1.1 4. Example 1.1.2 5. Note 6. Property: The Distributive Property 7. Example 1.1.3 8. The Product of Powers Property 9. The Power of a Power Property 10. Example 1.1.4 11. Note 12. The Power of a Product Property 13. Definition: Negative Exponents 14. Example 1.1.5 15. The Power of a Quotient Property 16. Example 1.1.6 17. The Quotient of Powers Property 18. Example 1.1.7 19. Try It! (Exercises) Figure 1.1.1 : “What do you remember most about your math classes?” Figure 1.1.2 : “All the rules and the x’s and y’s!” Folks remember, above all else, properties of math! Properties allow us to transform an expression or an equation into an equivalent form. We often need to use properties to move through a problem’s solution. This section introduces several properties. For all the properties in this section, the variables a, b, and c represent real numbers. Here are a few properties, followed by examples for their use: Property: Commutative Property of Addition (1.1.1)a+b=b+a The order of addition does not matter. Property: Commutative Property of Multiplication (1.1.2)a⁢b=b⁢a The order of multiplication does not matter. Example 1.1.1 Evaluate 40+−6+12+6+−40 Solution The commutative property of addition allows us to add in any order. \begin{array} &&40 + −40 + 6 + −6 + 12 &\text{Rearrange the order to group opposites.} \ &= 0 + 0 + 12 &\text{Opposites sum to zero.} \ &= 12 \end{array} Example 1.1.2 Why is 20% of 45 the same as 45% of 20? Solution Let’s verify the two expressions are equivalent: \begin{array} &&0.20 \cdot 45 = 9 &\text{(20\%) of (45) is (9).} \ &0.45 \cdot 20 = 9 &\text{(45\%) of (20) is (9).} \end{array} The commutative property of multiplication tells us why it’s true. \begin{array} &&20 \cdot 0.01 \cdot 45 = 45 \cdot 0.01 \cdot 20 &\text{The order of multiplication is just rearranged.} \end{array} The commutative property, as demonstrated, shows why the equivalency holds. Note Caution! The order of division and subtraction matters. Therefore, the commutative property only applies to multiplication and addition, NOT to division nor subtraction. Property: The Distributive Property (1.1.3)a⁢(b+c)=a⁢b+a⁢c The distributive property can be visualized as a rectangular area which is the sum of two smaller rectangular areas: Example 1.1.3 Simplify 6⁢(2⁢x+7) Solution \begin{array}&&6(2x + 7) = 6(2x) + 6(7) \ &= 12x + 42 \end{array} The Product of Powers Property (1.1.4)x a⋅x b=x a+b There are many power properties. Understanding why each property holds will help you decide whether to add or multiply exponents. Here’s why the Product Property holds: x⋅x⋅x⋅x⋅...⋅x⏟a number of x’s⋅x⋅x⋅x⋅x⋅...⋅x⏟b number of x’s⏟a+b number of x=x a⋅x b=x a+b The Power of a Power Property (1.1.5)(x a)b)=x a⁢b Again, understand why the property holds will help you master the rule: x a⋅x a⋅x a⋅x a⋅...⋅x a⏟b number of x a’s=x a+a+a+...+a⏟Product of Powers Property=x b⁢a⏟a+a+a+...+a=b⁢a⏟b number of a’s=x a⁢b Example 1.1.4 Simplify each of the following expressions. 5⁢y 8⋅y 4⋅y 9⁢(d 2)3 6⁢u⋅3⁢u 4+(u 3)2 Solution Product of Powers Property \begin{array} &&5y^8 \cdot y^4 \cdot y \ &= 5 \cdot y^{8+4+1} \ &= 5y^{13} \end{array} Power of a Power \begin{array} &&9(d^2)^3 \ &= 9d^{2 \cdot 3} \ &= 9d^6 \end{array} Both properties are used: \begin{array} &&6u \cdot 3u^4 + (u^3)^2 \ &= 6 \cdot 3 \cdot u^{1+4} \cdot u^{3 \cdot 2} \ &= 18u^5 + u^6 \end{array} Note The sum of powers can be simplified only if the powers are identical in base and exponent. For example, x 3+7⁢x 3=(1+7)⁢x 3=8⁢x 3. Simplify by combining like terms. Otherwise, the powers cannot be simplified. See solution to Example 4c. The Power of a Product Property (1.1.6)(x⁢y)a=x a⋅y a Each factor inside the parentheses is raised to the power indicated. Definition: Negative Exponents (1.1.7)b−n=(1 b)n=1 b n Create the reciprocal of base b, then raise b to the positive power, n. Think of the negative exponent as if it were flagging the power, letting you know that it is in the wrong place. Example 1.1.5 Simplify each of the following expressions. (3⁢n)4 2⁢(−4⁢x 3⁢y 2⁢z)2 (−2⁢u−3⁢w 5)−4 Solution \begin{array} &&(3n)^4 \ &=3^n \cdot 4^n \ &=81n^4 \end{array} \begin{array} &&2(−4x^3y^2z)^2 \ &=2(−4)^2 (x^3)^2(y^2)^2 (z)^2 \ &=2 \cdot 16 \cdot x^{3 \cdot 2} \cdot y^{2 \cdot 2} \cdot z^2 \ &= 32x^6y^4z^2 \end{array} \begin{array} &&(−2u^{−3}w^5 )^{−4} \ &= (-2)^{-4} \cdot (u^{-3})^{-4} \cdot (w^5)^{-4} \ &= \dfrac{1}{-2^4} \cdot u^{(-3)(-4)} \cdot w^{(5)(-4)} \ &=\dfrac{u^{12}}{16w^{20}} \end{array} The Power of a Quotient Property (1.1.8)(a b)n=a n b n The power is applied to both the numerator and the denominator. Example 1.1.6 Simplify (5⁢a−1⁢b 2 2⁢c 6)−2 Solution Let’s approach the solution like pealing an onion, beginning with the outer layer: \begin{array} &&=\left(\dfrac{2c^6}{5a^{-1}b^2}\right)^{2} &\text{The reciprocal of the fraction is taken. The power becomes positive.} \ &= \left(\dfrac{2ac^6}{5b^2}\right)^{2} &\text{The negative exponent flags the power. (\dfrac{1}{a^{-1}} = \dfrac{1}{\frac{1}{a}} = 1 \cdot a = a)} \ &= \left(\dfrac{(2ac^6)^2}{(5b^2)^2}\right) &\text{The Power of a Quotient Property is applied.} \ &=\dfrac{2^2 \cdot a^2 \cdot (c^6)^2}{5^2 \cdot (b^2)^2} &\text{The Power of a Product Property is applied.} \ &=\dfrac{4a^2c^{12}}{25b^4} &\text{The Power of a Power Property is applied. All is simplified.} \end{array} The Quotient of Powers Property (1.1.9)x a x b=x a−b The exponents are subtracted is due to the cancellation of common factors. For example, x 5 x 3=x⋅x⋅x⋅x⋅x x⋅x⋅x=x⋅x 1=x 2=x 5−3. Similarly, x 3 x 5=x⋅x⋅x x⋅x⋅x⋅x⋅x=1 x⋅x=x−2=x 3−5 Example 1.1.7 Simplify (4⁢p 3⁢q 2⁢p 2⁢q 5)2 Solution \begin{array} &&=\left( \dfrac{4}{2} \cdot p^{3-2} \cdot q^{1-5} \right)^2 &\text{The Quotient of Powers Property is applied.}\ &=(2pq^{-4})^2 \text{The exponents are simplified.} \ &=2^2 \cdot p^2 \cdot (q^{-4})^2 &\text{The Power of a Product Property is applied.} \ &=4p^2q^{(-4)(2)} &\text{The Power of a Power Property is applied.} \ &=4p^2q^{-8} &\text{The negative exponent flags the power (q^8).} \ &= \dfrac{4p^2}{q^8} &\text{The power (q^8) belongs in the denominator. All is simplified.} \end{array} Try It! (Exercises) Name the propertythe equation demonstrates. (5⋅7)2=5 2⋅7 2 5⁢(−20+2)=5⁢(−20)+5⁢(2) 7⁢(x+1)=(x+1)⋅7 6 12⋅6 3=6 15 8+(−9)=−9+8 4 30=(4 5)6 Evaluate the following sum. Rearrange the terms so a calculator isn’t necessary. 12+84+(−1)+(−84)+(−12)+10 (−1.9)+(−5.8)+13.2+1.9+5.8+(−1.2) For #3-11, use the distributive property to simplify. 10⁢(7⁢x+9) 2⁢y 2⁢(8+2⁢y) −7⁢p⁢(4⁢p 4+1) 3⁢b⁢[b 3+(−2)] 4⁢u⁢(5⁢u 2+9⁢u−1) n 3⁢m⁢(2⁢n 4⁢m−9⁢n⁢m 2+8) 1 2⁢(3 4−z 2) 2 5⁢q 2⁢(5⁢q+10) 7 8⁢t⁢(2⁢t 2−4⁢t+3) The commutative property applies to addition and multiplication. The commutative property does NOT apply to subtraction nor to division. The symbol ≠ means “does not equal.” Show that the two sides of the inequality produce different results. 6−2≠2−6 10÷5≠5÷10 For #13-15, use power properties to simplify. Answers should contain positive exponents only. (3⁢n)2+(5⁢n)⁢(6⁢n) (5⁢x 3⁢2⁢y)3 (−3⁢a 5⁢b−2⁢c)4 Scientific Notation, coupled with rules for exponents, eases the computations with very large and very small numbers. Use power properties to evaluate each of the following. Leave your answer in scientific notation. Leave negative powers of 10 as is for the final answer. (6.02⁢x⁢10 24)⁢(1.3⁢x⁢10 15) (3.14⁢x⁢10 10)⁢(2.65⁢x⁢10 18) (7.4⁢x⁢10−6)⁢(1.25⁢x⁢10−4) (3.8⁢x⁢10−16)⁢(2.4⁢x⁢10−8) For #17-25, use power properties to simplify. Answers should contain positive exponents only. Use fractions, not decimals. 12⁢w 6 4⁢w 5⁢u 2 10⁢u 3 (5 4)−1 (6⁢t 3)−2 (4⁢n 6 2⁢n)−1 (7⁢p⁢q 5 28⁢p 3⁢q)2 (12⁢x 10⁢y 3 24⁢x 9⁢y 8)3 (9⁢a⁢b 12⁢c 4 27⁢a 7⁢b⁢c 3)−4 (6⁢r 5⁢s−2 15⁢r−1⁢s)−3 For #26-35, simplify using properties. 2⁢(3⁢x+10)+5⁢(2⁢x−1) 4⁢a 3⁢(a+1)−2⁢a 2⁢(a 2−a) (8⁢y 2)⁢2−3⁢(7⁢y 3)⁢(2⁢y) (4⁢d)⁢(9⁢d 3)⁢(2⁢d 2)−(5⁢d 3)2 1 2⁢b 4⁢(6⁢b−2)−(3⁢b 2)2 3 4⁢u⁢v⁢(8⁢u−4⁢v)+3 2⁢u⁢(u⁢v+2⁢v u) p 2⁢q⁢(4⁢p+3⁢q)−(2⁢p⁢q)2 5⁢m⁢n 2⁢(m−n)−4⁢m 2⁢(n 2−1) 10⁢c 8⁢(2⁢c−3+c−4)+c−2⁢(c 6−12⁢c 7) 3⁢n 7⁢(3⁢n−2)−1+2⁢n 8⁢(n−1)+(−n 3)3 This page titled 1.1: Exponent Properties and More! is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jennifer Freidenreich. Back to top 1: Simplifying Expressions 1.2: FOIL Method and Special Products Was this article helpful? Yes No Recommended articles 2.2: The Commutative Property of Addition and Multiplication 1.1: Real Numbers - Algebra EssentialsIn this section, we will explore sets of numbers, calculations with different kinds of numbers, and the use of numbers in expressions. 1.2: Real Numbers - Algebra EssentialsIt is often said that mathematics is the language of science. If this is true, then the language of mathematics is numbers. The earliest use of number... 1.2: Real Numbers - Algebra EssentialsIt is often said that mathematics is the language of science. If this is true, then the language of mathematics is numbers. The earliest use of number... 1.2: Real Numbers - Algebra EssentialsIn this section, we will explore sets of numbers, calculations with different kinds of numbers, and the use of numbers in expressions. Article typeSection or PageAuthorJenny FreidenreichLicenseCC BY-SALicense Version4.0Show Page TOCno Tags commutative property of addition Commutative Property of Multiplication distributive property Negative Exponent Power of a Power Property Product of Powers Property scientific notation © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status×
15494	https://www.quora.com/How-do-you-convert-parametric-equations-to-standard-form-equations-motion-physics	Something went wrong. Wait a moment and try again. Standard Form Parametric Curves Motion (physics) Coordinate Plane Conversion (mathematics) Calculus (Mathematics) Mathematical Equations 5 How do you convert parametric equations to standard form equations (motion, physics)? Jeff Koch Physics / Math Professor · Author has 11.1K answers and 1.7M answer views · 3y If you have X and Y as functions of t, X = f(t) and Y = f(t) solve for t in terms of X with the first equation, then plug that expression in for each t in the second equation. You then have Y as a function of X, Y = f(x) Related questions How do you convert a parametric equation into a vector equation? How do you convert symmetric equations to parametric equations? What is the need for parametric equations? Is there a polar parametric equation? Could you explain the difference between vector equations, parametric equations, and Cartesian equations? David Joyce Professor Emeritus of Mathematics at Clark University · Author has 9.9K answers and 68.4M answer views · Updated 2y Related Does a set of parametric equations count as one equation, and if not, how do you convert it into one? A set of parametric equations is a single vector equation. Example Consider the parametric equations with parameter θ x=cosθ,y=sinθ which describe a point going around a unit circle. They correspond to the single vector equation (x,y)=(cosθ,sinθ). Sometimes, as in this example, you can take the two parametric equations and eliminate the parameter to get a single equation. Square them both and add them together to get x2+y2=1. Another example Here are the parametric equations for a helix in three-space: x=cosθ,y=sinθ,z=θ Those thr A set of parametric equations is a single vector equation. Example Consider the parametric equations with parameter θ x=cosθ,y=sinθ which describe a point going around a unit circle. They correspond to the single vector equation (x,y)=(cosθ,sinθ). Sometimes, as in this example, you can take the two parametric equations and eliminate the parameter to get a single equation. Square them both and add them together to get x2+y2=1. Another example Here are the parametric equations for a helix in three-space: x=cosθ,y=sinθ,z=θ Those three equations correspond to the single vector equation (x,y,z)=(cosθ,sinθ,θ). You can eliminate the parameter θ from the three equations to get two equations: x=cosz,y=sinz but that’s the best you can do—two ordinary equations. MSE M.S.c in Physics, University of the Witwatersrand (Graduated 2014) · Author has 220 answers and 724.7K answer views · 7y Related Could you explain the difference between vector equations, parametric equations, and Cartesian equations? Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · Updated 2y Related What is the definition of a parametric equation? How do you graph a parametric equation that crosses itself? I don’t think an actual “definition” would be of any actual use to you so I will just explain what parametric equations are. NOTE: you don’t just have one parametric equation. Suppose a stone is thrown up in the air and its x and y coordinates can be calculated at any time t seconds using: The graph has x and y axes and t seems to be an extra variable which is not mentioned on the graph with an x axis and a y axis! The parametric equations create the above function. “t” is called a PARAMETER. The parameter “t” is just a way of showing where the stone is at various times. The position of the stone is I don’t think an actual “definition” would be of any actual use to you so I will just explain what parametric equations are. NOTE: you don’t just have one parametric equation. Suppose a stone is thrown up in the air and its x and y coordinates can be calculated at any time t seconds using: The graph has x and y axes and t seems to be an extra variable which is not mentioned on the graph with an x axis and a y axis! The parametric equations create the above function. “t” is called a PARAMETER. The parameter “t” is just a way of showing where the stone is at various times. The position of the stone is shown by its x and y coordinates and there is no “t” axis on the graph. The parameter is just an extra variable. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Related questions How do you convert polar equations to parametric equations? Can each physical equation form a theory? How do I find a set of parametric equations from cartesian equations? What is the process for writing an equation in rectangular form from its parametric equations? Is there a general way to transform an implicit equation into parametric form? Martyn Hathaway BSc in Mathematics, University of Southampton (Graduated 1986) · Author has 4.7K answers and 6.7M answer views · 4y Related How do I do parametric equations on Desmos? Good question! As I didn’t know, I went to the Desmos site. Hang on, why didn’t you try this? For each curve, enter a point expressed in parametric form. For the ellipse x=4sin(t);y−cos(t), enter (4sin(t),cos(t)). When you enter this, Desmos adds a line underneath that seems to default to a parameter range of [0,1]. Amend this as required; for our ellipse, change this to [0,2π]. For the curve x=2t2;y=2−3t, enter (2t2,2−3t). Here, I retained the default range for the parameter. Good question! As I didn’t know, I went to the Desmos site. Hang on, why didn’t you try this? For each curve, enter a point expressed in parametric form. For the ellipse x=4sin(t);y−cos(t), enter (4sin(t),cos(t)). When you enter this, Desmos adds a line underneath that seems to default to a parameter range of [0,1]. Amend this as required; for our ellipse, change this to [0,2π]. For the curve x=2t2;y=2−3t, enter (2t2,2−3t). Here, I retained the default range for the parameter. Mr AB If you need an answer, you will get an answer · Author has 4K answers and 2.5M answer views · 1y Related Can you explain how to convert an equation into standard form and slope-intercept form? The conversion process is shown in... Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Jafar Mortadha BSc in Mechanical engineering · Author has 1.4K answers and 3.6M answer views · 8y Related How do you convert a parametric equation into a vector equation? If you have parametric equations, x=f(t), y=g(t), z=h(t) Then a vector equation is simply r(t)=xi+yj+zk r(t)=f(t)i+g(t)j+h(t)k The above form is a vector equation that describes any curve in 3D space. Nasir Haniff Lecturer in Math · 7y Related Does a set of parametric equations count as one equation, and if not, how do you convert it into one? A set of parametric equation is not ‘one equation’, but rather represents one object. (A curve, surface etc). In 2D a set of two parametric equation represents a curve and may be represented by one equation involving the coordinates - either Cartesian or polar or any other reasonable system. Just simply eliminate the parameter. Examples : x=acosθ;y=bsinθ represents an ellipse. Eliminate \theta to obtain x2a2+y2b2=1. But, in 3D a curve is represented by a set of three parametric equations. This can not be represented by a single equation ( involving only A set of parametric equation is not ‘one equation’, but rather represents one object. (A curve, surface etc). In 2D a set of two parametric equation represents a curve and may be represented by one equation involving the coordinates - either Cartesian or polar or any other reasonable system. Just simply eliminate the parameter. Examples : x=acosθ;y=bsinθ represents an ellipse. Eliminate \theta to obtain x2a2+y2b2=1. But, in 3D a curve is represented by a set of three parametric equations. This can not be represented by a single equation ( involving only the coordinate variables only). As an example: x=1+t;y=t;z=t represents a straight line. This can not be connected with one equation without t. In the meantime, a surface is a set of three parametric equations with ‘two’ independent parameters. By eliminating them, one may get a single equation. example: x=rsinθcosϕ;y=rsinθsinϕ;z=rcosθ represents a sphere. (r is constant. \theta, \phi are parameters) which has the single equation representation x2+y2+z2=r2 As a rule, in an n-dimensional coordinate system, a hyper-surface is represented by a set of parametric equation with can have a single equation representation. Anything less dimensional than n-1 can not have. Sponsored by Stake Stake: Online Casino games - Play & Win Online. Play the best online casino games, slots & live casino games! Unlock VIP bonuses, bet with crypto & win. Richard Elston B Sc in Mathematics & Computer Science, Imperial College London (Graduated 1970) · Author has 4.5K answers and 3.1M answer views · 2y Related What is the definition of parametric equations? How are they used to describe motion? What is the definition of parametric equations? I provide an intuitive understanding, rather than a formal definition: Suppose you have two related quantities. Some researchers may seek a single equation that connects them. But suppose that, instead, you write equations that relate each of the quantities to a third quantity. Then you have written parametric equations. How are they used to describe motion? Consider the motion of a projectile. Its trajectory is the relationship between its vertical displacement from its start point and its horizontal displacement from its start point. If you express What is the definition of parametric equations? I provide an intuitive understanding, rather than a formal definition: Suppose you have two related quantities. Some researchers may seek a single equation that connects them. But suppose that, instead, you write equations that relate each of the quantities to a third quantity. Then you have written parametric equations. How are they used to describe motion? Consider the motion of a projectile. Its trajectory is the relationship between its vertical displacement from its start point and its horizontal displacement from its start point. If you express both of these displacements as functions of time, you have described the motion parametrically. Vincent Middleton MAT in Mathematics, Georgia State University · Author has 269 answers and 483.8K answer views · 5y Related How can you convert parametric equations into Cartesian equations? Parametric equations substitute another parameter for the Cartesian coordinates x and y. This substituted parameter is often time (t) or an angle (θ). If the parameter is time, for instance, we can use two independent equations relating the parameter (t) to x and y to eliminate the parameter. Example: Convert the following parametric equation to Cartesian coordinates. (1) x = t^2 (2) y = 2t Use Equation (2) to define t in therms of y: y = 2t y/2 = t The substitute y/2 for t in Equation (1), x = t^2 x = (y/2)^2 x = (Y^2)/4 4x = y^2 2x^-1 = y y = plus or minus (2x^-1) This is the Cartesian equation of a Parametric equations substitute another parameter for the Cartesian coordinates x and y. This substituted parameter is often time (t) or an angle (θ). If the parameter is time, for instance, we can use two independent equations relating the parameter (t) to x and y to eliminate the parameter. Example: Convert the following parametric equation to Cartesian coordinates. (1) x = t^2 (2) y = 2t Use Equation (2) to define t in therms of y: y = 2t y/2 = t The substitute y/2 for t in Equation (1), x = t^2 x = (y/2)^2 x = (Y^2)/4 4x = y^2 2x^-1 = y y = plus or minus (2x^-1) This is the Cartesian equation of a parabola (quadratic) that goes to infinity if the positive and negative y-direction, and has a minimum at x = 0. If the parameter is θ, you can use a trigonometric relationship like to eliminate the parameter θ. Example: Convert the following parametric equation to Cartesian coordinates. (1) x = 3sinθ (2) y = 4cosθ We can use the trigonometric relationship sinθ^2 + cosθ^2 = 1, as follows: x = 3sinθ x/3 = sinθ (x/3)^2 = sinθ^2 y = 4cosθ y/4 = cosθ (y/4)^2 = cosθ^2 Substituting the values for sin^2 and cos^s into our trigonometric relationship sinθ^2 + cosθ^2 = 1 (x/3)^2 + (y/4)^2 = 1 This is the Cartesian equation of an ellipse. Nissim A. Author has 595 answers and 4.7M answer views · 10y Related How do I find the intercepts of a line if its equation is written in standard form? Set one of the values to 0 to find the intercept, because at the intercept one of the values is 0. Simple, right? In fact you can just look at a standard equation to find the intercept. An x-intercept is where the line intercepts (meets) the x-axis. y will be zero at the intercept. Similarly, A y-intercept is where the line intercepts (meets) the y-axis. x will be zero at the intercept. Take a look and try to guess the intercepts: Example 1 y = 6x + 3 That '3' is the y-intercept, so the coordinates for the y-intercept is (0,3) What's the x-intercept? Well, set y to 0! 0 = 6x + 3 x = -1/2 So, your Set one of the values to 0 to find the intercept, because at the intercept one of the values is 0. Simple, right? In fact you can just look at a standard equation to find the intercept. An x-intercept is where the line intercepts (meets) the x-axis. y will be zero at the intercept. Similarly, A y-intercept is where the line intercepts (meets) the y-axis. x will be zero at the intercept. Take a look and try to guess the intercepts: Example 1 y = 6x + 3 That '3' is the y-intercept, so the coordinates for the y-intercept is (0,3) What's the x-intercept? Well, set y to 0! 0 = 6x + 3 x = -1/2 So, your x-intercept's coordinates are (-0.5,0) and y's are (0,3) Example 2 y +2x = 8 Rearrange it in the form of y = mx + c y = -2x + 8 Your y-intercept is 8. So, so the coordinates for the y-intercept is (0,8) Similarly, for the x-intercept, 0 = 8 - 2x 2x = 8 x = 4 So, your x-intercept's coordinates are (4,0) and y's are (0,8) Example 3 2y = 4x + 8 What's the y coordinate for the y-intercept? If you say 8 then you'd be wrong! Why? Because the equation is not in standard form for it is '2' y and not just 'y'. To fix this, just do a simple division: 2y = 4x + 8 y = (4x + 8)/2 y = 2x + 4 Now we have a proper equation! Just like we did before, we can find the intercepts. The coordinates of the intercepts for x and y are (-2,0) and (0,4) respectively. Here are some more for your practice: 1) y - 2x = 5 2) 2y + 2x = 2 3) 5x + 4x + y = 10 Levie Bringmans Quant · Author has 88 answers and 146.3K answer views · 9y Related Could you explain the difference between vector equations, parametric equations, and Cartesian equations? I’ll use the equation of a plane in R3 as an example. The most general equation of a plane in cartesian form is ax+by+cz=0 This is just an algebraic equation. cartesian equations are just multivariate polynomials (not the other way around). If you would analyze the set of zeros of this equation and graph those zeros in R3, then you would get a plane. The vector equation of a plane is →x=→v0+s→v1+t→v2,s,t∈R \begin{bmatrix} x\y\z\end{bmatrix}=\begin{bmatrix} x_0\y_0\z_0\end{bmatrix}+s\begin{bmatrix} v_1\v_2\v_3\end{bmatrix}+t\begin{bmatrix} w_1\w_2\w_3\end I’ll use the equation of a plane in R3 as an example. The most general equation of a plane in cartesian form is ax+by+cz=0 This is just an algebraic equation. cartesian equations are just multivariate polynomials (not the other way around). If you would analyze the set of zeros of this equation and graph those zeros in R3, then you would get a plane. The vector equation of a plane is →x=→v0+s→v1+t→v2,s,t∈R ⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣x0y0z0⎤⎥⎦+s⎡⎢⎣v1v2v3⎤⎥⎦+t⎡⎢⎣w1w2w3⎤⎥⎦ This is just an equation involving vectors. Here →v0 is a point on the plane and →v1 and →v2 are direction vectors (two linearly independent vectors which lie in the plane). The second equation is just the vector equation expanded in matrix form using the coordinates of the vectors with respect to the standard basis of R3 (^i,^j,^k). The parametric equation of a plane is the following ⎧⎨⎩x=x0+sv1+tw1y=y0+sv2+tw2z=z0+sv3+tw3 It describes each coordinate as a function of two parameters s and t. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 8y Related What do you mean by parametric equation? Related questions How do you convert a parametric equation into a vector equation? How do you convert symmetric equations to parametric equations? What is the need for parametric equations? Is there a polar parametric equation? Could you explain the difference between vector equations, parametric equations, and Cartesian equations? How do you convert polar equations to parametric equations? Can each physical equation form a theory? How do I find a set of parametric equations from cartesian equations? What is the process for writing an equation in rectangular form from its parametric equations? Is there a general way to transform an implicit equation into parametric form? How do I describe the translation of a rule A (x; y) increase A' (x+10; y-8)? Is polar equation and parametric equation are same? How do I graph parametric equations? How can I find the parametric equations and standard equations of the line that contains the given points, has the given slope, or has the given direction vector, A (-1,2); B (5, 5)? What are parametric equations and why do we need to use them in physics? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15495	https://shop.readrelevant.com/products/vocabulary-in-context-worksheets-and-activities-for-grades-6-9	Vocabulary in Context Worksheets and Activities for grades 6-9 – Read Relevant Skip to content Just added to your cart Qty: View cart () Continue shopping Submit Close search Home RACE Strategy Reading Comprehension Writing Vocabulary Grammar Study Skills View All Emergency Sub Plans SearchCart 0 items Home RACE Strategy Reading Comprehension Writing Vocabulary Grammar Study Skills View All Emergency Sub Plans Vocabulary in Context Worksheets and Activities for grades 6-9 Regular price $5.00 Sale price $5.00 Regular price Sale Sold out Unit price/per Error Quantity must be 1 or more Add to cart More payment options This item is a recurring or deferred purchase. By continuing, I agree to the cancellation policy and authorize you to charge my payment method at the prices, frequency and dates listed on this page until my order is fulfilled or I cancel, if permitted. Adding product to your cart These context clue worksheets will help your students understand and use context clues to define unknown vocabulary words. This no-prep, printable packet shows examples of definition/explanationcontext clues, synonym/restatement context clues, antonym/contrast context clues, and inference context clues. Students will practice multiple choice as well as constructed response questions, all based on using context clues to determine a word's meaning. Great for progress monitoring and test-prep! This Download Contains: ♦ Types of Context Clues student guide with examples ♦ 14 pages of vocabulary activities ♦ Answer Key Teachers are Saying: "This year, especially with Covid, things feel very mundane. This was a nice way to break things up and play a "game" of sorts! I will be adding this into my regular curriculum every year! :)" Format:PDF Share Share on Facebook Tweet Tweet on Twitter Pin it Pin on Pinterest Frequently Bought Together Save money buying these products together ＋＋ Total price:$15.00 USD Add selected to cart [x] This item: Vocabulary in Context Worksheets and Activities for grades 6-9 $5.00 USD - [x] ### Figurative Language Sort Activity 100 Card Sorting Game Simile, Metap…$6.00 USD - [x] ### Grammar Worksheets : Practice Grammar activities worksheets$4.00 USD No reviews You may also like Grammar Worksheets : Practice Grammar activities worksheets Grammar Worksheets : Practice Grammar activities worksheets Regular price$4.00 Sale price$4.00 Regular price Unit price/per Sale Sold out Vocabulary in Context Worksheets and Activities for grades 4-6 Vocabulary in Context Worksheets and Activities for grades 4-6 Regular price$5.00 Sale price$5.00 Regular price Unit price/per Sale Sold out Types of Irony Sort : 50 Card Sorting Activity Types of Irony Sort : 50 Card Sorting Activity Regular price$4.00 Sale price$4.00 Regular price Unit price/per Sale Sold out Reading Comprehension and Writing with the RACE Strategy: Grades 4-6 Reading Comprehension and Writing with the RACE Strategy: Grades 4-6 Regular price$9.00 Sale price$9.00 Regular price Unit price/per Sale Sold out Quick links Search Newsletter Subscribe Payment methods Copyright © 2025 Read Relevant - Unlocking the joys of learning for little minds Payment methods Copyright © 2025 Read Relevant - Unlocking the joys of learning for little minds Use left/right arrows to navigate the slideshow or swipe left/right if using a mobile device Choosing a selection results in a full page refresh. Press the space key then arrow keys to make a selection. Opens in a new window. Opens external website. Opens external website in a new window.
15496	https://www.geeksforgeeks.org/competitive-programming/cses-solution-polygon-area/	CSES Solution - Polygon Area - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In DSA Tutorial Array Strings Linked List Stack Queue Tree Graph Searching Sorting Recursion Dynamic Programming Binary Tree Binary Search Tree Heap Hashing Sign In ▲ Open In App CSES Solution - Polygon Area Last Updated : 23 Jul, 2025 Comments Improve Suggest changes 1 Like Like Report Your task is to calculate the area of a given polygon. The polygon consists of n vertices(x1,y1),(x2,y2),.....(xn,yn). The vertices (xi,yi)and(xi+1,yi+1) are adjacent for i=1,2,....,n-1, and the vertices (x1,y1)and (xn,yn) are also adjacent. Example: Input:vertices[] = {{1, 1}, {4, 2}, {3, 5}, {1, 4}} Output:16 Input: vertices[] = {{1, 3}, {5, 6}, {2, 5}, {1, 4}} Output: 6 Approach: The idea is to useShoelace formula calculates the area of a polygon as half of the absolute difference between the sum of products of x-coordinates and y-coordinates of consecutive vertices. Shoelace formula Formula:A=1 2∣∑i=1 n−1(x i y i+1−x i+1 y i)+x n y 1−x 1 y n∣A = \frac{1}{2} \left\| \sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i) + x_n y_1 - x_1 y_n \right\| A=2 1∣∣∑i=1 n−1(x iy i+1−x i+1y i)+x ny 1−x 1y n∣∣ The formula involves summing the products of the x-coordinates of adjacent vertices and the y-coordinates of the next adjacent vertices, and then subtracting the products of the y-coordinates of adjacent vertices and the x-coordinates of the next adjacent vertices. The result is multiplied by 0.5 to get the area of the polygon. Steps-by-step approach: Calculate the area using the shoelace formula: Initialize a variable area to 0. Iterate over the vertices of the polygon. For each vertex i, calculate the product of the x-coordinate of vertexi and the y-coordinate of vertex i+1, and subtract the product of the y-coordinate of vertex i and the x-coordinate of vertex i+1. Add the result to the area variable. Take the absolute value of the area: The shoelace formula can result in a negative area if the polygon is oriented clockwise. To ensure that the area is always positive, take the absolute value of the area variable. Return the absolute value of the area variable. Below is the implementation of the above approach: C++ ```cpp include using namespace std; define ll long long define endl '\n' int main() { // Number of vertices in the polygon ll n = 4; // Array to store the vertices of the polygon pair<ll, ll> vertices[n] = { { 1, 3 }, { 5, 6 }, { 2, 5 }, { 1, 4 } }; // Variable to store the area of the polygon ll area = 0; // Calculating the area of the polygon using the // shoelace formula for (ll i = 0; i < n; i++) { area += (vertices[i].first vertices[(i + 1) % n].second - vertices[(i + 1) % n].first vertices[i].second); } // Printing the absolute value of the area cout << abs(area); return 0; } ``` include using namespace std;#define ll long long#define endl '\n'int main(){ // Number of vertices in the polygon ll n = 4; // Array to store the vertices of the polygon pair vertices[n] = { { 1, 3 }, { 5, 6 }, { 2, 5 }, { 1, 4 } }; // Variable to store the area of the polygon ll area = 0; // Calculating the area of the polygon using the // shoelace formula for (ll i = 0; i < n; i++) { area += (vertices[i].first vertices[(i + 1) % n].second - vertices[(i + 1) % n].first vertices[i].second); } // Printing the absolute value of the area cout << abs(area); return 0;} Java ```java import java.util.; public class Main { public static void main(String[] args) { // Number of vertices in the polygon int n = 4; // Array to store the vertices of the polygon int[][] vertices = { { 1, 3 }, { 5, 6 }, { 2, 5 }, { 1, 4 } }; // Variable to store the area of the polygon int area = 0; // Calculating the area of the polygon using the shoelace formula for (int i = 0; i < n; i++) { area += (vertices[i] vertices[(i + 1) % n] - vertices[(i + 1) % n] vertices[i]); } // Printing the absolute value of the area System.out.println(Math.abs(area)); } } Pythonpython Importing the math module for absolute function import math Number of vertices in the polygon n = 4 List to store the vertices of the polygon vertices = [(1, 3), (5, 6), (2, 5), (1, 4)] Variable to store the area of the polygon area = 0 Calculating the area of the polygon using the shoelace formula for i in range(n): area += (vertices[i] vertices[(i + 1) % n] - vertices[(i + 1) % n] vertices[i]) Printing the absolute value of the area print(math.fabs(area)) C#csharp using System; public class MainClass { public static void Main(string[] args) { // Number of vertices in the polygon int n = 4; // Array to store the vertices of the polygon Tuple<int, int>[] vertices = new Tuple<int, int>[] { Tuple.Create(1, 3), Tuple.Create(5, 6), Tuple.Create(2, 5), Tuple.Create(1, 4) }; // Variable to store the area of the polygon long area = 0; // Calculating the area of the polygon using the shoelace formula for (int i = 0; i < n; i++) { area += (vertices[i].Item1 vertices[(i + 1) % n].Item2) - (vertices[(i + 1) % n].Item1 vertices[i].Item2); } // Printing the absolute value of the area Console.WriteLine(Math.Abs(area)); } } JavaScriptjavascript // Number of vertices in the polygon let n = 4; // Array to store the vertices of the polygon let vertices = [ { x: 1, y: 3 }, { x: 5, y: 6 }, { x: 2, y: 5 }, { x: 1, y: 4 } ]; // Variable to store the area of the polygon let area = 0; // Calculating the area of the polygon using the // shoelace formula for (let i = 0; i < n; i++) { area += (vertices[i].x vertices[(i + 1) % n].y - vertices[(i + 1) % n].x vertices[i].y); } // Printing the absolute value of the area console.log(Math.abs(area)); ``` Output6 Time complexity: O(N^3), where N is the number of points. 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15497	https://www.reddit.com/r/PhysicsStudents/comments/yslq26/solutions_to_801sc_classical_mechanics_on_mit_ocw/	Solutions to 8.01SC Classical Mechanics on MIT OCW : r/PhysicsStudents Skip to main contentSolutions to 8.01SC Classical Mechanics on MIT OCW : r/PhysicsStudents Open menu Open navigationGo to Reddit Home r/PhysicsStudents A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to PhysicsStudents r/PhysicsStudents•3 yr. ago Groundbreaking-Call7 Solutions to 8.01SC Classical Mechanics on MIT OCW ~~ I made solutions to the course, so if anyone wants to take it, they can use these. Edit: sorry, I actually changed the "name" of the notion site, and forgot to update this - didn't even realize people saw this lol. This is the new link: Read more Share Related Answers Section Related Answers Resources for mastering quantum mechanics Balancing physics studies and personal life Best physics podcasts for students Essential skills for aspiring physicists How to find a good physics mentor New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of November 11, 2022 Reddit reReddit: Top posts of November 2022 Reddit reReddit: Top posts of 2022 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
15498	https://library.fiveable.me/key-terms/linear-algebra-and-differential-equations/positive-definiteness	Positive Definiteness - (Linear Algebra and Differential Equations) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms Linear Algebra and Differential Equations Positive Definiteness ➗linear algebra and differential equations review key term - Positive Definiteness Citation: MLA Definition Positive definiteness is a property of a symmetric matrix that indicates all its eigenvalues are positive, meaning it defines a quadratic form that is always greater than zero for non-zero vectors. This concept is crucial in determining the stability of systems, the geometry of inner product spaces, and ensuring unique solutions in optimization problems. 5 Must Know Facts For Your Next Test A matrix A is positive definite if for all non-zero vectors x, the expression $$x^T A x > 0$$ holds true. The leading principal minors of a positive definite matrix are all positive, providing a method for checking its definiteness. In optimization, positive definiteness ensures that a function has a unique minimum point, making it crucial in convex analysis. Positive definite matrices are always invertible, with their inverses also being positive definite. The concept of positive definiteness extends to forms defined on real vector spaces, influencing various mathematical fields such as statistics and control theory. Review Questions How does positive definiteness relate to the stability of systems in mathematical modeling? Positive definiteness is critical in assessing the stability of systems because it guarantees that energy or potential functions are minimized, which leads to stable equilibrium points. When analyzing dynamic systems, a positive definite matrix in the context of a Lyapunov function indicates that small perturbations will decay over time, ensuring that the system returns to equilibrium rather than diverging. This property is essential for control theory and stability analysis. Describe how one can determine if a given symmetric matrix is positive definite using its eigenvalues. To determine if a symmetric matrix is positive definite, one can calculate its eigenvalues. If all eigenvalues are positive, then the matrix is classified as positive definite. This relationship stems from the fact that for any non-zero vector x, the quadratic form associated with the matrix will yield a positive value, aligning with the definition of positive definiteness. Thus, eigenvalue analysis becomes an effective tool for classification. Evaluate the implications of using a positive definite matrix in optimization problems, particularly in gradient descent methods. Using a positive definite matrix in optimization problems greatly influences the convergence properties of algorithms like gradient descent. When the Hessian matrix of a function at a point is positive definite, it indicates that the function has a local minimum at that point and that moving along the direction of the negative gradient will lead toward this minimum efficiently. This characteristic ensures that the optimization process not only converges but does so without oscillating or diverging, providing reliable results in finding optimal solutions. Related terms Eigenvalues: Scalars associated with a square matrix, representing the factors by which the eigenvectors are scaled during linear transformations. Quadratic Form:A polynomial of degree two in a number of variables, typically expressed as $$Q(x) = x^T A x$$ where A is a symmetric matrix. Inner Product:A generalization of the dot product that defines a geometric structure on a vector space, allowing for the measurement of angles and lengths. 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15499	https://pmc.ncbi.nlm.nih.gov/articles/PMC5349625/	Peritoneal Dialysis Complicated by Pleuroperitoneal Communication and Hydrothorax - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Ochsner J . 2017 Spring;17(1):124–127. Search in PMC Search in PubMed View in NLM Catalog Add to search Peritoneal Dialysis Complicated by Pleuroperitoneal Communication and Hydrothorax Julian Yaxley Julian Yaxley, MBBS 1 Department of Internal Medicine, Redcliffe Hospital, Redcliffe, Queensland, Australia Find articles by Julian Yaxley 1, Kevin Twomey Kevin Twomey, MD 2 Department of Surgery, Royal Brisbane and Women's Hospital, Herston, Queensland, Australia Find articles by Kevin Twomey 2 Author information Copyright and License information 1 Department of Internal Medicine, Redcliffe Hospital, Redcliffe, Queensland, Australia 2 Department of Surgery, Royal Brisbane and Women's Hospital, Herston, Queensland, Australia ✉ Address correspondence to Julian Yaxley, MBBS, Department of Internal Medicine, Redcliffe Hospital, Anzac Ave., Redcliffe, Queensland, 4020, Australia. Tel: (+61) 42-080-8049. Email: julianyaxley@yahoo.com.au © Academic Division of Ochsner Clinic Foundation PMC Copyright notice PMCID: PMC5349625 PMID: 28331461 Abstract Background: Hydrothorax is an uncommon but well-recognized complication of peritoneal dialysis. It is a potentially serious condition that frequently requires cessation of peritoneal dialysis and permanent transition to hemodialysis. Hydrothorax is produced by movement of peritoneal dialysate through pleuroperitoneal fistulas. Pleural fluid analysis typically detects a high glucose concentration, and contrast imaging reveals tracer uptake transgressing the diaphragm. Experience with the treatment of hydrothorax related to peritoneal dialysis is limited. Case Report: We describe the case of a 54-year-old female on peritoneal dialysis for end-stage renal failure who developed a hydrothorax soon after beginning treatment. Conclusion: This case describes a classical presentation of hydrothorax in the context of peritoneal dialysis. Treatment is frequently unsuccessful. All clinicians prescribing peritoneal dialysis should be aware of this complication. Keywords:Ascitic fluid, hernias–diaphragmatic–congenital, hydrothorax, peritoneal dialysis INTRODUCTION Hydrothorax is an uncommon but well-recognized complication of peritoneal dialysis (PD). Hydrothorax is defined as a pleural effusion composed of serous fluid and is potentially life-threatening. The pathogenesis of hydrothorax in PD is incompletely understood. Typical signs include pleuritic chest pain and breathlessness. Development of a pleural effusion mandates immediate discontinuation of PD and precludes its resumption in many patients. Experience with patients on PD who develop hydrothorax is limited, and relatively few cases have been reported in the literature. This condition has no standard treatment, and practice is particularly varied among physicians unfamiliar with patients on dialysis. We describe a case of PD complicated by acute and significant hydrothorax and its subsequent management. CASE REPORT A 54-year-old female on PD presented to the emergency department in respiratory distress. She had begun continuous ambulatory peritoneal dialysis (CAPD) 6 weeks earlier for end-stage renal failure secondary to tuberous sclerosis. Her Tenckhoff catheter had been inserted surgically 4 weeks prior to the initiation of PD. The initial CAPD prescription comprised 3 daily exchanges of 2 liters of 2.5% dextrose-based dialysate. The patient developed progressive shortness of breath and right-sided pleuritic pain in the week prior to presentation that gradually worsened to the point of severe breathlessness. She also reported declining daily output from her Tenckhoff catheter during the previous 2 weeks. Her kidney failure was the result of focal segmental glomerulosclerosis and angiomyolipomas caused by tuberous sclerosis. She also suffered from cerebral, cutaneous, and skeletal involvement. Other significant medical history included right lobular carcinoma in situ with right total mastectomy, appendectomy, uterine fibroids requiring hysterectomy, and hemorrhoidectomy. The patient's medications included darbepoetin, lanthanum carbonate, oxcarbazepine, vigabatrin, and tamoxifen. On examination at arrival in the emergency department, the patient was in obvious respiratory distress. Her breathing was labored with a respiratory rate of 22 breaths per minute and an oxygen saturation of 89% on 2 liters of supplemental oxygen per minute, delivered via nasal cannula. Other vital signs were within normal limits. Reduced breath sounds and dullness to percussion were noted on the right side. Chest radiograph demonstrated a large right-sided pleural effusion (Figure 1). An intercostal catheter was inserted, and the patient's condition rapidly improved. Figure 1. Open in a new tab Large right-sided pleural effusion demonstrated on chest radiograph at the time of patient presentation. A total of 3.5 liters of pleural fluid was drained during the following days. A specimen of straw-colored fluid was collected and sent for analysis. Laboratory findings were consistent with a transudative effusion; no organisms were cultured and no malignant cells were visualized. The pleural fluid glucose concentration was 19.5 mmol/L compared with a serum glucose concentration of 6.8 mmol/L. Computed tomography (CT) with contrast administered through the Tenckhoff catheter into the peritoneal cavity failed to identify dispersal of contrast material into the right hemithorax. A nuclear isotope scan was subsequently done. Following administration of technetium 99m via the PD catheter, a high volume of radioactive dialysate was detected entering the right hemithorax (Figure 2). No tracer activity was seen in the left hemithorax. Figure 2. Open in a new tab Peritoneal scintigraphy demonstrating a high volume of radioactive dialysate in the right hemithorax with no abnormal tracer accumulation detected in the left hemithorax, thereby confirming a right pleuroperitoneal fistulous communication. The presence of a pleuroperitoneal fistula was suspected. After her effusion was drained, the patient's body weight was 5 kg heavier than her dry weight of 57.3 kg measured 1 month earlier. PD was withheld, and the patient transitioned to temporary hemodialysis. Video-assisted thoracoscopy 1 week later did not identify any pleuroperitoneal communication. Talc pleurodesis was performed under direct thoracoscopic vision. The patient recommenced PD 4 weeks after chemical pleurodesis. Her CAPD prescription was modified to 1.5 L of 1.5% dextrose-based dialysate solution exchanged 5 times daily. At 5-week follow-up after recommencing PD, the patient had no evidence of hydrothorax recurrence, and she continued to meet her targets of dialysis adequacy. DISCUSSION PD is a commonly used dialysis technique for patients with end-stage renal failure. PD is an effective treatment, but it is not without risk. The most frequent and important complication of PD is infection.1 Other common complications include catheter site leaks, catheter blockage, abdominal wall herniation, and intestinal perforation. Hydrothorax is a rare complication, caused by migration of fluid from the peritoneal cavity into the pleural space via pleuroperitoneal fistulas. These diaphragmatic defects are usually congenital and right-sided, explaining the predominance of right-sided effusions.2 Hydrothorax is estimated to occur in 1.6% of patients on PD.3 The majority of cases occur within 30 days of commencing PD, and up to 25% are asymptomatic.3,4 Hydrothorax should always be considered when patients on PD present with pleural effusions; however, the usual differential diagnoses for pleural effusions also apply. Typical features suggesting hydrothorax include dyspnea and pleuritic pain. Incomplete recovery of distilled fluid from the peritoneal catheter is an important clue to the diagnosis. Intraabdominal pressure is the predominant influence for hydrothorax but does not necessarily correlate with peritoneal dialysate volume.3,5,6 Hypertonic dialysate also increases risk7 because hyperosmolality produces an osmotically driven volume flux from tissues into the cavity, further increasing intraabdominal pressure.8 No single test definitively diagnoses hydrothorax. Diagnosis instead requires a combination of biochemical and radiographic findings. Analysis of the pleural fluid will identify a transudate with an elevated glucose concentration. The glucose concentration in dialysate and peritoneal fluid is higher than the serum glucose concentration. No consensus agreement on a diagnostic pleural glucose level has been reached, and its accuracy for this purpose has not been studied.9 Although the diagnostic contribution of effusate glucose is unreliable, it is generally accepted that pleural fluid with a glucose concentration >16.5 mmol/L or with a concentration greater than that of the serum concentration is consistent with hydrothorax.4,10 Pleural fluid glucose is influenced by previous dialysate composition, blood glucose, and the rate of pleuroperitoneal leakage. Interpretation in patients with diabetes is difficult and also in patients with small diaphragmatic defects in whom glucose absorption from the pleural cavity may be greater than in patients with larger defects.11 Various imaging studies can be helpful in identifying pleuroperitoneal communication. The most informative is peritoneal scintigraphy that has a sensitivity of approximately 50%.10 A radioisotope in the form of technetium 99m is instilled into the PD fluid, after which a series of images is obtained. Tracer uptake detected in the thoracic cavity confirms a communication between the peritoneum and the pleural space. Alternative techniques that are used less frequently are CT and magnetic resonance peritoneography. No current guidelines or standards of care for hydrothorax in patients on PD have been developed.12 A range of conservative and invasive management options has been reported in case series, but no comparative controlled trials have ever been undertaken to compare treatment strategies.2 The treatment of hydrothorax related to PD often has limited success, and most patients eventually require transition to permanent hemodialysis.2 Most authorities suggest beginning with a conservative approach. PD should first be withheld to permit spontaneous resolution of the hydrothorax and diaphragmatic connection. Peritoneal dialysate is hypothesized to act as a sclerosant and to spontaneously seal the pleuroperitoneal tracts in some patients.5 During this time, patients require conversion to temporary hemodialysis. After approximately 1 month, they may then be reintroduced to PD gradually with low-volume exchanges in a semi-upright position. Ultimately however, the success of this approach is suboptimal, and many patients experience hydrothorax recurrence.13,14 Removal of pleural fluid through insertion of an intercostal catheter should be restricted to patients in respiratory distress. Chemical pleurodesis is generally the next step if conservative measures fail. Because of the high failure rate associated with conservative management, many authorities advocate proceeding to pleurodesis at the outset.15 Chemical pleurodesis involves the instillation of irritants into the pleural cavity via an intercostal catheter with the intention of artificially obliterating the pleural space by inducing adherence of the pleura. Agents available for pleurodesis include talc, tetracycline, doxycycline, and autologous blood. No randomized study data support one agent over another, and their efficacy appears to be similar in small case series and observational studies.2 Following chemical pleurodesis, patients should wait at least 10 days before recommencing PD to allow time for sufficient scar formation over the defect.2,16 Most nephrologists defer reimplementation of PD for 2-6 weeks to improve the efficacy of pleurodesis.17 The likelihood of successfully resuming PD following chemical pleurodesis is approximately 50%.2,5,18 Surgical correction is the most efficacious treatment but also the most invasive. Several methods are available, including surgical pleurectomy, mechanical abrasion and pleurodesis, chemical pleurodesis, and diaphragmatic patching. These procedures may be performed through open surgery or video-assisted thoracoscopic surgery. Both surgical techniques are highly effective for this indication, and the likelihood of a patient being able to successfully reinstitute PD without hydrothorax recurrence exceeds 90%.2 Because of the invasive nature of these treatments, however, some patients do not elect surgery and instead convert permanently to hemodialysis if the conservative approaches fail. CONCLUSION Hydrothorax is an infrequent but well-described complication of PD. It is an important diagnosis that may require urgent treatment. No standard practice for this condition exists, and experience with investigations and management is limited. This case report demonstrates a classical presentation of hydrothorax and highlights its relevance for all clinicians encountering patients receiving PD. ACKNOWLEDGMENTS The authors have no financial or proprietary interest in the subject matter of this article. This article meets the Accreditation Council for Graduate Medical Education and the American Board of Medical Specialties Maintenance of Certification competencies for Patient Care and Medical Knowledge. REFERENCES Holley JL, Piraino BM. . Complications of peritoneal dialysis: diagnosis and management. Semin Dial. 1990. October; 3 4: 245- 248. 10.1111/j.1525-139X.1990.tb00057.x. [DOI] [Google Scholar] Chow KM, Szeto CC, Li PK. . Management options for hydrothorax complicating peritoneal dialysis. Semin Dial. 2003. Sep-Oct; 16 5: 389- 394. [DOI] [PubMed] [Google Scholar] Nomoto Y, Suga T, Nakajima K, et al. Acute hydrothorax in continuous ambulatory peritoneal dialysis—a collaborative study of 161 centers. Am J Nephrol. 1989; 9 5: 363- 367. [DOI] [PubMed] [Google Scholar] Bae EH, Kim CS, Choi JS, Kim SW. . Pleural effusion in a peritoneal dialysis patient. Chonnam Med J. 2011. April; 47 1: 43- 44. 10.4068/cmj.2011.47.1.43. 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