Calculus II. Practice Problems, Methods, and Solutions 9783031453526, 9783031453533

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Mehdi Rahmani-Andebili

Calculus II

Practice Problems, Methods, and Solutions

Calculus II

Mehdi Rahmani-Andebili

Calculus II Practice Problems, Methods, and Solutions

Mehdi Rahmani-Andebili Electrical Engineering Department Arkansas Tech University Russellville, AR, USA

ISBN 978-3-031-45352-6 ISBN 978-3-031-45353-3 https://doi.org/10.1007/978-3-031-45353-3

(eBook)

# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

Preface

Calculus is one of the most important courses of many majors, including engineering and science, and even some non-engineering majors like economics and business, which is taught in three successive courses at universities and colleges worldwide. Moreover, in many universities and colleges, a precalculus course is mandatory for under-prepared students as the prerequisite course of Calculus 1. Unfortunately, some students do not have a solid background and knowledge in math and calculus when they start their education in universities or colleges. This issue prevents them from learning calculus-based courses such as physics and engineering. Sometimes, the problem escalates, so they give up and leave the university. Based on my real professorship experience, students do not have a serious issue comprehending physics and engineering courses. In fact, it is the lack of enough knowledge of calculus that hinder them from understanding those courses. Therefore, a series of calculus textbooks covering Precalculus, Calculus 1, Calculus 2, and Calculus 3 have been prepared to help students succeed in their major. The subjects of the calculus series books are as follows.

Precalculus: Practice Problems, Methods, and Solution • • • • • • •

Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities Systems of Equations Quadratic Equations Functions, Algebra of Functions, and Inverse Functions Factorization of Polynomials Trigonometric and Inverse Trigonometric Functions Arithmetic and Geometric Sequences

Calculus 1: Practice Problems, Methods, and Solution • • • • •

Characteristics of Functions Trigonometric Equations and Identities Limits and Continuities Derivatives and Their Applications Definite and Indefinite Integrals

Calculus 2: Practice Problems, Methods, and Solution • • • •

Applications of Integration Sequences and Series and Their Applications Polar Coordinate System Complex Numbers v

vi

Preface

Calculus 3: Practice Problems, Methods, and Solution • • • • • •

Linear Algebra and Analytical Geometry Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System Multivariable Functions Double Integrals and their Applications Triple Integrals and their Applications Line Integrals and their Applications

The textbooks include basic and advanced calculus problems with very detailed problem solutions. They can be used as practicing study guides by students and as supplementary teaching sources by instructors. Since the problems have very detailed solutions, the textbooks are helpful for under-prepared students. In addition, they are beneficial for knowledgeable students because they include advanced problems. In preparing the problems and solutions, care has been taken to use methods typically found in the primary instructor-recommended textbooks. By considering this key point, the textbooks are in the direction of instructors’ lectures, and the instructors will not see any untaught and unusual problem solutions in their students’ answer sheets. To help students study in the most efficient way, the problems have been categorized into nine different levels. In this regard, for each problem, a difficulty level (easy, normal, or hard) and a calculation amount (small, normal, or large) have been assigned. Moreover, problems have been ordered in each chapter from the easiest problem with the smallest calculations to the most difficult problems with the largest ones. Therefore, students are suggested to start studying the textbooks from the easiest problems and continue practicing until they reach the normal and then the hardest ones. This classification can also help instructors choose their desirable problems to conduct a quiz or a test. Moreover, the classification of computation amount can help students manage their time during future exams, and instructors assign appropriate problems based on the exam duration. Russellville, AR, USA

Mehdi Rahmani-Andebili

The Other Works Published by the Author

The author has already published the books and textbooks below with Springer Nature. Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2023. Calculus III – Practice Problems, Methods, and Solutions, Springer Nature, 2023. Calculus II – Practice Problems, Methods, and Solutions, Springer Nature, 2023. Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2023. Planning and Operation of Electric Vehicles in Smart Grid, Springer Nature, 2023. Applications of Artificial Intelligence in Planning and Operation of Smart Grid, Springer Nature, 2022. AC Electric Machines- Practice Problems, Methods, and Solutions, Springer Nature, 2022. DC Electric Machines, Electromechanical Energy Conversion Principles, and Magnetic Circuit Analysis- Practice Problems, Methods, and Solutions, Springer Nature, 2022. Differential Equations- Practice Problems, Methods, and Solutions, Springer Nature, 2022. Feedback Control Systems Analysis and Design- Practice Problems, Methods, and Solutions, Springer Nature, 2022. Power System Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2022. Advanced Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2022. Design, Control, and Operation of Microgrids in Smart Grids, Springer Nature, 2021. Applications of Fuzzy Logic in Planning and Operation of Smart Grids, Springer Nature, 2021. Operation of Smart Homes, Springer Nature, 2021. AC Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2021. Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

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viii

The Other Works Published by the Author

Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. DC Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2020. Planning and Operation of Plug-in Electric Vehicles: Technical, Geographical, and Social Aspects, Springer Nature, 2019.

Contents

1

Problems: Applications of integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Mean Value of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Surface Area Bounded by Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Volume Resulted from Rotation of an Enclosed Region . . . . . . . . . . . . . . . . 1.4 Arc Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Surface Area of a Solid of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Center of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 2 7 10 12 13

2

Solutions of Problems: Applications of Integration . . . . . . . . . . . . . . . . . . . . . . 2.1 Mean Value of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Surface Area Bounded by Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Volume Resulted from Rotation of an Enclosed Region . . . . . . . . . . . . . . . . 2.4 Arc Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Surface Area of a Solid of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Center of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15 15 16 28 37 45 49

3

Problems: Sequences and Series and Their Applications . . . . . . . . . . . . . . . . . .

53

4

Solutions of Problems: Sequences and Series and Their Applications . . . . . . . .

59

5

Problems: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73

6

Solutions of Problems: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . .

77

7

Problems: Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89

8

Solutions of Problems: Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . .

95

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

ix

1

Problems: Applications of integration

Abstract

In this chapter, the basic and advanced problems related to the applications of integration are presented. The subjects include mean value of a function, surface area bounded by curves, volume resulted from rotation of an enclosed region, arc length of a curve, surface area of a solid of revolution, and center of gravity. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations.

1.1

Mean Value of a Function

1.1. For the range of 2 ≤ x ≤ 5, calculate the mean value of the following function [1–3]. y ¼ ax þ b Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 5 1) a þ 3b 2 7 2) a þ 3b 2 5 3) a þ b 2 7 4) a þ b 2 1.2. Consider the functions of f (x) ¼ 2x and g (x) ¼ 3x2 - 2x. Calculate the value of λ if the mean value of the functions in the range of [1, λ] is the same. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 1þ 5 1) 2 p 1þ 3 2) 2 p 5 3) 2 p 3 3 4) 2

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_1

1

2

1.2

1 Problems: Applications of integration

Surface Area Bounded by Curves

1.3. Calculate the surface area enclosed between the curves of y ¼ 2x2 - 2x and y ¼ x2. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 3 2 2) 3 4 3) 3 7 4) 3 1.4. Calculate the surface area bounded by the functions of y (x) ¼ sin x and y (x) ¼ cos x for x ¼ Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 2 p 2) 2 p 3) 2 2 p 4) 3 2 1.5. Calculate the surface area enclosed between the curves of y ¼ x2and y ¼ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 2 1) 3 2) 1 1 3) 3 1 4) 6

p

x.

1.6. Calculate the surface area enclosed between the curve of y ¼ x3 + 2x2 + x and x-axis. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 12 1 2) 10 1 3) 9 1 4) 7 1.7. Calculate the surface area enclosed between the curve of y ¼ x2 + 1 and the line of y ¼ 2. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 3 2 2) 3

π 5π . , 4 4

1.2 Surface Area Bounded by Curves

3

3) 1 4 4) 3 1.8. Calculate the surface area restricted by the curve of y2(x) ¼ x2 - x4. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 2 2 3) 3 4 4) 3 1.9. Calculate the surface area of the shaded region shown in Fig. 1.1. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 8 - 2π π 2) 8 2 π 3) 4 4 4) 2π - 4

Figure 1.1 The graph of problem 1.9

1.10. Calculate the surface area restricted by the following function, above the line of y(x) ¼ 0, and the right-hand side of x ¼ 1. yð xÞ ¼

6 ð2x þ 1Þðx þ 2Þ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 2 2) ln 2 3) 2 ln 2 4) 1

4

1 Problems: Applications of integration

p p 1.11. Calculate the surface area enclosed between the curve with the function below and x-axis in the range of - 2, 2 . y¼

1 2 þ x2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p π 2 1) 4 π 2) 4p π 2 3) 2 π 4) 2 1.12. Calculate the surface area restricted by the function below and x-axis in the domain of [1, e2]. ln x yð xÞ ¼ p x Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 5 2) 4 3) 3 4) 2 1.13. Calculate the surface area of the shaded region shown in Fig. 1.2. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 3 2 2) 3 2 1 3) ln 2 3 2 4) - ln 2 3

Figure 1.2 The graph of problem 1.13

1.2 Surface Area Bounded by Curves

5

1.14. Calculate the surface area of the shaded region shown in Fig. 1.3. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) π π 2) 2 π 1 3) 2 2 π 1 4) þ 2 2

Figure 1.3 The graph of problem 1.14

π 1.15. Determine the value of parameter of c, where 0 < c < , if the surface area restricted by the function of y (x) ¼ cos x, 2 the function of y (x) ¼ cos (x - c), and the line of x ¼ 0 is equal to the surface area bounded by the function of y (x) ¼ cos (x - c), the line of x ¼ π, and the line of y (x) ¼ 0. ○ Easy ● Normal ○ Hard Difficulty level Calculation amount ○ Small ○ Normal ● Large π 1) 5 π 2) 3 π 3) 4 π 4) 6 p 1.16. Determine the surface area bounded by the function of yðxÞ ¼ x ln x, x-axis, and the lines with the equations of x ¼ 1 and x ¼ 2. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 2 ln 2 2 2) 2 ln 2 - 1 3 3) ln 2 þ 4 3 4) ln 2 8 1.17. Estimate the surface area of the shaded region shown in Fig. 1.4. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) ln2

6

1 Problems: Applications of integration

3) cos2 4) 2 ln 2 - cos 2

Figure 1.4 The graph of problem 1.17

1.18. Calculate the surface area of the shaded region shown in Fig. 1.5. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 12 2) 14 3) 16 4) 18

Figure 1.5 The graph of problem 1.18

1.3

Volume Resulted from Rotation of an Enclosed Region

1.19. Calculate the volume resulted from the rotation of the function of yðxÞ ¼ Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π 1) 3 2π 2) 3

p

x2 - x3 around x-axis for x ¼ [-1, 1].

1.3 Volume Resulted from Rotation of an Enclosed Region

7

3) π 3π 4) 2 1.20. Calculate the volume resulted from the rotation of the surface area around x-axis enclosed between one period of the curve of y ¼ sin (x) and x-axis. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π 2 2) 2π 2 π2 3) 2 π2 4) 4 1.21. Calculate the volume resulted from the rotation of the function of y(x) ¼ cos x around x-axis for x ¼ 0, Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π2 1) 4 π2 2) 8 π 3) 8 π 4) 4

π . 2

1.22. The volume resulted from the rotation of the surface area restricted by the function of y(x) ¼ e-x, the coordinate axes, and x ¼ b, where b > 0, around x-axis is V. Calculate the value of lim V. b→1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π 1) 4 π 2) 2 3) π 4) 1 1 1.23. The surface area restricted by the function of yðxÞ ¼ p , the coordinate axes, and x ¼ 1 is called S. Moreover, the x volume resulted from the rotation of the surface area around x-axis is called V. Which of the following options is correct? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) S ¼ 2, V ¼ 8π 2) S ¼ 4, V ¼ 8π 3) S ¼ 2, V ¼ 1 4) S ¼ 1 , V ¼ 1 1.24. Calculate the volume resulted from the rotation of the surface area around y-axis enclosed between the curve of 1 y ¼ 1 - x2 and x-axis. 4 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large

8

1 Problems: Applications of integration

1) 2) 3) 4)

π 2π 3π 4π

1.25. Calculate the volume resulted from the rotation of the surface area around x-axis enclosed between the curve with the π π function below, x-axis, x ¼ , and x ¼ . 4 2 y¼

1 sin 2 ðxÞ

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) π 2π 2) 3 3) 2π 4π 4) 3 1.26. The surface area confined by the function below, x-axis, and two lines with the equations of x ¼

π π and x ¼ are 2 6

rotated around x-axis. Calculate the resultant volume. p yðxÞ ¼

cos 3 x sin 2 x

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 7π 1) 3 5π 2) 3 4π 3) 3 2π 4) 3 1.27. Calculate the volume resulted from the rotation of the surface area bounded by the function of y (x) ¼ x e x and the lines with the equations of x ¼ 1 and y (x) ¼ 0 around x-axis. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large π 2 1) e -1 2 π 2 2) e -2 2 π 2 3) e -1 4 π 2 4) e þ1 4 1

1

1

1.28. Calculate the volume resulted from the rotation of the surface area restricted by the relation of y2 ðxÞ ¼ a2 - x2 and x- and y- axes around x-axis. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large

1.4 Arc Length of a Curve

9

1) 5πa3 1 2) πa3 2 1 3 3) πa 12 1 3 4) πa 15 1.29. Calculate the volume resulted from the rotation of the function of yðxÞ ¼ e - x Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large π 1 1) 5 1 - e - 2π π 1 2) 5 1 þ e - 2π π 3) 1 þ e - 2π 5 π 4) 1 - e - 2π 5

p

sin x around x-axis for 0 ≤ x ≤ π.

1.30. Calculate the volume created by the rotation of the shaded region (see Fig. 1.6) around x-axis. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 4π 1) 15 π 2) 2 14π 3) 3 3π 4) 5

Figure 1.6 The graph of problem 1.30

1.4

Arc Length of a Curve xp

1.31. Calculate the arc length of the function of f ðxÞ ¼

t 4 - 1dt for the interval of x ¼ [1, 3].

1

10

1 Problems: Applications of integration

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 19 1) 3 25 2) 3 26 3) 3 28 4) 3 1.32. Calculate the arc length of the parametric relation below for the interval of t ¼ [0, 4]. xðt Þ ¼ et cos t yðt Þ ¼ et sin t Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 1) 2 e4 - 1 p 2) 2 e4 þ 1 3) 2(e4 - 1) 4) 2(e4 + 1) 1.33. Calculate the arc length of the following parametric relation for the interval of 0 ≤ t ≤ 1. xðt Þ ¼ et ðcos t þ sin t Þ yðt Þ ¼ et ðcos t - sin t Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) e - 1 e-1 2) 2 3) 4(e - 1) 4) 2(e - 1) 1 1 1.34. Calculate the arc length of the function of yðxÞ ¼ x2 - ln x for the interval of x ¼ [1, 2]. 2 4 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 3 1 1) þ ln 2 2 2 3 1 2) þ ln 2 2 4 3 1 3) þ ln 2 4 2 3 1 4) þ ln 2 4 4 1 x4 for the interval of x ¼ [1, 2]. þ 8 4x2 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

1.35. Calculate the arc length of the function of yðxÞ ¼

1.4 Arc Length of a Curve

11

1) 15 2) 21 25 3) 3 33 4) 16 1.36. Calculate the arc length of the function of f (x) ¼ ln (sec x) for the interval of x ¼ 0, Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large p 1) ln 2 p 2) ln 2 - 2 p 3) ln 2 - 1 p 4) ln 1 þ 2 x

1.37. Calculate the arc length of the function of f ðxÞ ¼

π . 4

coshðt Þdt for the interval of x ¼ [0, 2].

0

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large p 1 1) 2 e e 1 2) 2 e e p 1 3) 2 e þ e 1 4) 2 e þ e p 1 p 2 x x - 1 - ln x þ x2 - 1 2 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 2 3 2) 2 p 3) 1 þ 2 4) 2

1.38. Calculate the arc length of the function of f ðxÞ ¼

1.39. Calculate the arc length of the function of f ðxÞ ¼ ln Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1 1) ln e e 1 2) ln e þ e 1 3) ln e2 - 2 e 1 4) ln e2 þ 2 e

ex þ 1 ex - 1

for the interval of x ¼ [1, 2].

for the interval of 0 ≤ x ≤ 2.

12

1.5

1 Problems: Applications of integration

Surface Area of a Solid of Revolution

1.40. The function of y (x) ¼ cosh x for the interval of 0 ≤ x ≤ 2 is rotated around x-axis. Calculate the surface area of the solid. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large π 1) ð2 þ cosh 2Þ 2 π 2) ð2 þ sinh 2Þ 2 π 3) ð4 þ cosh 4Þ 2 π 4) ð4 þ sinh 4Þ 2 1.41. The function of 3y(x) - x3 ¼ 0 is rotated around x-axis. Calculate the surface area of the solid of revolution. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large π p 1) 2 2-1 9 π p 2-1 2) 9 p π 3) 2-1 3 π p 4) 2 2-1 3 1.42. The function of y (x) ¼ x2 for the interval of y ≤ 2 is rotated around y-axis. Calculate the surface area of the solid of revolution. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 8π 1) 3 13π 2) 3 11π 3) 3 10π 4) 3 x for the interval of 0 ≤ x ≤ 5 is rotated around x-axis. What is the ratio of the volume 6 of the solid of revolution to its surface area. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 3 2) 2 3 3) 2 2π 4) 3

1.43. The function of f ðxÞ ¼ 6 cosh

1.6 Center of Gravity

13 x

1.44. The function of f ðxÞ ¼

sinh t 2 dt for the interval of 0 ≤ x ≤

p

ln 1396 is rotated around y-axis. Calculate the surface

0

area of the solid of revolution. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 2π 1396 1396 π 1 2) 1396 4 1396 1 3) π 1396 1396 1 π 4) 1396 1396 2

1.6

Center of Gravity

1.45. Determine the center of gravity of a surface from x-axis which is restricted by the function of f (x) ¼ 1 - x2 and x-axis. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 4 1) 15 4 2) 5 3 3) 5 2 4) 5 1.46. Determine the center of gravity of a surface from y-axis which is restricted by the function of f (x) ¼ 1 - x2 and x-axis. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 4 2) 5 3 3) 5 2 4) 5 1.47. Determine the center of gravity of the parametric curve below from x-axis. xðt Þ ¼ t þ sin t yðt Þ ¼ 1 - cos t Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) 1 1 2) 3 2 3) 3 3 4) 4

,

0≤t≤π

14

1 Problems: Applications of integration

References 1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

2

Solutions of Problems: Applications of Integration

Abstract

In this chapter, the problems of the first chapter are fully solved, in detail, step-by-step, and with different methods.

2.1

Mean Value of a Function

2.1. As we know, the average value of a function can be determined as follows [1–3]: f ave ¼

1 b-a

b

f ðxÞdx

a

In addition, from list of integral of functions, we know that: xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. f ave ¼

1 5-2

5

ðax þ bÞdx ¼

2

5 1 25a 1 a 2 4a ¼ x þ bx þ 5b - 2b 2 3 2 3 2 2

7 ⟹ f ave ¼ a þ b 2 Choice (4) is the answer. 2.2. As we know, the average value of a function can be determined as follows: 1 b-a

b

f ðxÞdx

a

Based on the problem, we know that: f ave ¼ gave Now, the problem can be solved as follows.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_2

15

16

2 λ

1 λ-1

1 λ-1

2x dx ¼

1

⟹ x2

λ

Solutions of Problems: Applications of Integration

3x2 - 2x dx

1

λ λ ¼ x 3 - x2 1 1

⟹ λ2 - 1 ¼ λ3 - λ2 - 0 ⟹ λ3 - 2λ2 þ 1 ¼ 0 ⟹ ð λ - 1 Þ λ2 - λ - 1 ¼ 0 p p 1- 5 1 þ 5 ⟹λ¼ , ,1 2 2 p 5

However, just 1þ2

is acceptable because the others are not within the range. p 1þ 5 ⟹λ¼ 2

Choice (1) is the answer. In this problem, the rule below was used. xn dx ¼

2.2

1 nþ1 x þc nþ1

Surface Area Bounded by Curves

2.3. First, we need to find the intersection points of the curves as follows: 2x2 - 2x ¼ x2 ⟹ x2 - 2x ¼ 0 ⟹ x ¼ 0, 2 Then: S¼

x2

ðy2 - y1 Þdx

x1 2

⟹S¼ 0

⟹S¼ -

- x2 þ 2x dx

0

x3 þ x2 3

2 23 ¼ - þ 2 2 - ð 0 þ 0Þ 3 0 ⟹S¼

Choice (3) is the answer.

2

x2 - 2x2 þ 2x dx ¼

4 3

2.2 Surface Area Bounded by Curves

17

In this problem, the rule below was used. xn dx ¼

1 nþ1 þc x nþ1

2.4. The problem can be solved as follows. x2

S¼

ðy2 - y1 Þdx

x1

)S¼

5π 4 π 4

ðsin x - cos xÞdx

5π 5π 5π π π ) S ¼ ð- cos x - sin xÞ 4 ¼ - cos - - cos - sin ¼ - sin π 4 4 4 4 4

p

p p p 2 2 2 2 þ - 2 2 2 2

p )S¼2 2 Choice (3) is the answer. In this problem, the rules below were used. sin xdx ¼ - cos x þ c cos xdx ¼ sin x þ c p 5π 2 cos ¼4 2 sin

p 5π 2 ¼4 2

cos

sin

p π 2 ¼ 4 2 p π 2 ¼ 4 2

2.5. First, we need to find the intersection points of the curves as follows: y1 ¼ x 2 p y2 ¼ x ⟹ y2 ¼ y1 ⟹

p

x ¼ x2 ⟹

p

p x x x - 1 ¼ 0 ⟹ x ¼ 0, 1

18

2 x2

S¼

Solutions of Problems: Applications of Integration

ðy2 - y1 Þdx

x1 1

⟹S¼

p

x - x2 dx

0

⟹S¼

2 32 x3 x 3 3 ⟹S¼

1 2 1 ¼ 0 3 3 1 3

Choice (3) is the answer. In this problem, the rule below was used. xn dx ¼

1 nþ1 þc x nþ1

2.6. First, we need to find the intersection points of the curves as follows: y2 ¼ y1 ⟹ x3 þ 2x2 þ x ¼ 0 ⟹ x x2 þ 2x þ 1 ¼ xðx þ 1Þ2 ¼ 0 ⟹ x ¼ 0, - 1, - 1 S¼

x2

ðy2 - y1 Þdx

x1

⟹S¼ ⟹S¼

x4 2 3 x2 þ x þ 4 3 2

0

x3 þ 2x2 þ x dx

-1

0 1 2 1 3-8 þ 6 ¼ 0¼ - þ -1 4 3 2 12

The surface area must be a positive quantity. Therefore, S¼

1 12

Choice (1) is the answer. In this problem, the rule below was used. xn dx ¼

1 nþ1 þc x nþ1

2.7. First, we need to find the intersection points of the curves as follows: y1 ¼ x2 þ 1 y2 ¼ 2

¼ -

1 12

2.2 Surface Area Bounded by Curves

19

⟹ x2 þ 1 ¼ 2 ⟹ x2 ¼ 1 ⟹ x ¼ ± 1 Then: x2

S¼

ðy2 - y1 Þdx

x1

⟹S¼

1 -1

1

2 - x2 þ 1 dx ¼ 2

1 - x2 dx

0

⟹ S ¼ 2 x-

x3 3

1 1 ¼ 2 1-0 0 3

⟹S¼

4 3

Choice (4) is the answer. In this problem, the rules below were used. xn dx ¼ a -a

a

f ðxÞdx ¼ 2

1 nþ1 þc x nþ1

f ðxÞdx ,

if f ðxÞ is an even function

0

2.8. As can be noticed from the function of y2 ¼ x2 - x4, the function is even with respect to both x and y. Therefore, we can calculate the surface area bounded in the first quadrant and then multiply its value by 4. To determine the range of x, we need to solve the equation of y ¼ 0 or quadrant gives x ¼ 0, 1. S¼

x2

p

p x2 - x4 ¼ x 1 - x2 ¼ 0 that for the first

ðy2 - y1 Þdx

x1 1

)S¼4

p x 1 - x2 dx

0

By defining a new variable, we have: 1 1 - x2 ¼ u ) - 2xdx ¼ du ) xdx ¼ - du 2 )S¼4 )S¼4 -

-

4 1 32 u ¼ - 1 - x2 3 3 )S¼

Choice (4) is the answer.

1 12 u du 2

4 3

3 2

1 0

¼ -

4 ð 0 - 1Þ 3

20

2

Solutions of Problems: Applications of Integration

2.9. The problem can be solved as follows. S¼

ðy1 ðxÞ - y2 ðxÞÞdx 1

)S¼2

4-

0

) S ¼ 2ð4x - 4 arctan xÞ

1 0

4 dx 1 þ x2

¼ 2 4-4

π - 0 ¼ 2ð 4 - π Þ 4

) S ¼ 8 - 2π Choice (1) is the answer.

Figure 2.1 The graph of problem 2.9

In this problem, the rules below were used. adx ¼ ax þ c 1 dx ¼ arctan x þ c 1 þ x2 a -a

a

f ðxÞdx ¼ 2

f ðxÞdx ,

if f ðxÞ is an even function

0

arc tan 1 ¼

π 4

arc tan 0 ¼ 0

2.10. The problem can be solved as follows. S¼

ðy1 ðxÞ - y2 ðxÞÞdx

2.2 Surface Area Bounded by Curves

21 1

)S¼ 1

1

6 - 0 dx ¼ ð2x þ 1Þðx þ 2Þ

) S ¼ ð2lnð2x þ 1Þ - 2 lnðx þ 2ÞÞ

1 1

4 2 dx 2x þ 1 x þ 2

1

¼ 2 ln

2x þ 1 1 ¼ 2½ln 2 - ln 1 xþ2 1

) S ¼ 2 ln 2 Choice (3) is the answer. In this problem, the rules below were used. 1 du ¼ ln u þ c u ln 1 ¼ 0

2.11. The problem can be solved as follows. S¼ p

ðy1 ðxÞ - y2 ðxÞÞdx p 2

2

1 )S¼ p 2 þ x2 dx ¼ 2 - 2

)S¼

p

0

p

1 dx ¼ 2 2 þ x2

1 2 1þ

px 2

p

2 p π ¼ 2 -0 4 0

x 2 arc tan p 2 ⟹S¼

0

2

p π 2 4

Choice (1) is the answer. In this problem, the rules below were used. 1 x 2 a

1þ a -a

a

f ðxÞdx ¼ 2

dx ¼ a arc tan

f ðxÞdx ,

x a

þc

if f ðxÞ is an even function

0

arc tan 1 ¼

π 4

arc tan 0 ¼ 0

2

dx

22

2

Solutions of Problems: Applications of Integration

2.12. From list of integral of functions or by using the method of integration by parts, we know that: ln u du ¼ u ln u - u þ c The problem can be solved by defining a new variable as follows. x ¼ t2 )

p

dx x ¼ t ) p ¼ 2dt x

ðy1 ðxÞ - y2 ðxÞÞdx

S¼

e2

)S¼ 1 t2

)S¼

ln x p - 0 dx x t2

ln t 2 ð2dt Þ ¼ 4

t1

ln tdt ¼ 4ðtln t - t Þ

t1

)S¼4

p

xln

p

p e2 x - x 1

) S ¼ 4½ðeln e - eÞ - ð1ln 1 - 1Þ )S¼4 Choice (2) is the answer. In this problem, the rules below were used. ln ab ¼ b ln a ln e ¼ 1 ln 1 ¼ 0 2.13. From list of integral of functions, we know that: ax dx ¼ xn dx ¼

ax þc ln a

1 nþ1 þc x nþ1

The problem can be solved as follows. S¼

ðy1 ðxÞ - y2 ðxÞÞdx

t2 t1

2.2 Surface Area Bounded by Curves

23 1

)S¼ 0

)S¼

1

2x - 1 - x2 dx ¼

2x - 1 þ x2 dx

0

x3 2x -x þ ln 2 3

1

¼

0

)S¼

2 1 1 -1 þ ln 2 3 ln 2

2 1 ln 2 3

Choice (3) is the answer.

Figure 2.2 The graph of problem 2.13

2.14. The problem can be solved as follows. S¼

ðy1 ðxÞ - y2 ðxÞÞdx 1

)S¼ 0

)S¼

2arctan x -

x2 2

1 0

2 - x dx 1 þ x2

¼ 2arctan 1 -

)S¼

1 - ð2arctan 0 - 0Þ 2

π 1 2 2

Choice (3) is the answer.

Figure 2.3 The graph of problem 2.14

24

2

Solutions of Problems: Applications of Integration

In this problem, the rules below were used. 1 dx ¼ arctan x þ c 1 þ x2 xn dx ¼

1 nþ1 þc x nþ1

arc tan 1 ¼

π 4

arc tan 0 ¼ 0 2.15. First, we need to determine the points of intersection of two graphs related to the first surface area, that is, y1(x) ¼ cos x and y2(x) ¼ cos (x - c). y1 ðxÞ ¼ y2 ðxÞ ) cosðx - cÞ ¼ cos x ) ð x - cÞ ¼ ± x ) x - c ¼ - x ) x ¼

c 2

Also, from the problem, we have x ¼ 0. Thus, the problem can be solved as follows. S1 ¼

ðy1 ðxÞ - y2 ðxÞÞdx c 2

) S1 ¼

½cos x - cosðx - cÞdx

0 c

) S1 ¼ ½sin x - sinðx - cÞ02 ¼ sin

c c c - sin - ð0 - sinð- cÞÞ ¼ 2 sin - sin c 2 2 2

Likewise, we need to determine the points of intersection of two graphs related to the second surface area, that is, y3(x) ¼ 0 and y4 ¼ cos (x - c). y3 ðxÞ ¼ y4 ðxÞ ) cosðx - cÞ ¼ 0 ) x-c ¼

π π )x¼cþ 2 2

Also, from the problem, we have x ¼ π. Therefore: S2 ¼

ðy3 ðxÞ - y4 ðxÞÞdx π

S2 ¼

cþπ2

½0 - cosðx - cÞdx

) S2 ¼ - ½sinðx - cÞπcþπ ¼ - sinðπ - cÞ þ sin 2

Based on the problem, we have: S1 ¼ S2

π ¼ - sin c þ 1 2

2.2 Surface Area Bounded by Curves

25

) 2 sin

c c 1 c π - sin c ¼ 1 - sin c ) sin ¼ ) ¼ 2 2 2 2 6 )c¼

π 3

Choice (2) is the answer. In this problem, the rules below were used. cos x ¼ sin x þ c cosðx þ aÞ ¼ sinðx þ aÞ þ c sinðπ - cÞ ¼ sinðcÞ sin

for 0 < c
0, b > 0 an þ bn

57

58

3 Problems: Sequences and Series and Their Applications

3.18. Which one of the following series is convergent? S1 ¼

1

n-1 n

n

nþ1 n

n

nþ1 n

n2

n¼1

S2 ¼

1 n¼1

S3 ¼

1 n¼1

S4 ¼

1 n¼1

1 p n n

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) S1 2) S2 3) S3 4) S4

References 1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

4

Solutions of Problems: Sequences and Series and Their Applications

Abstract

In this chapter, the problems of the third chapter are fully solved, in detail, step-by-step, and with different methods.

4.1. From the concept of growth rate, for n ⟶ 1, a, b > 1 and k > 0, we know that the order of growth rate is as follows [1–3]. log a n < nk < bn < n! < nn Thus, for this problem, we have: 2n 1 1 ¼  lim ¼0 n⟶1 ðn þ 2Þ! n⟶1 ðn þ 2Þ! 1! lim

2n ¼0 n⟶1 ðn þ 2Þ!

) lim Choice (1) is the answer. 4.2. The problem can be solved as follows.

lim

n⟶1

8

8 n! n! þ 2n8 þ ln n ¼ lim 1 ¼ 1  lim n n → þ1 n! n → þ1 n! þ 5 þ 4n

) lim

n⟶1

8

n! þ 2n8 þ ln n ¼1 n! þ 5n þ 4n

Choice (2) is the answer. In this problem, the rule below was used. Based on the concept of growth rate, for n ⟶ 1, a, b > 1 and k > 0, the order of growth rate is as follows. log a n < nk < bn < n! < nn

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_4

59

60

4

Solutions of Problems: Sequences and Series and Their Applications

4.3. From the concept of equivalent functions, we know that for a > 0 and an even k: k

lim

n⟶1

ank þ bnk - 1 þ . . . ¼ lim

p k

a nþ

b ka

a nþ

b ka

n⟶1

And for an odd k, we have: lim

k

n⟶1

p k

ank þ bnk - 1 þ . . . ¼ lim

n⟶1

Moreover, based on growth rate, we know that: lim a1 nk1 þ a2 nk2  a1 nk1

if k 1 > k 2 > 0

n⟶1

Based on the information given in the problem, we have: p

p p n n n an ¼ p þp þ⋯þp 3 3 3 4n þ sin n þ 1 4n þ sin n þ 2 4n þ sin n þ n ) lim an  lim n→1

n→1

p p p n n n p þp þ ⋯p 4n3 4n3 4n3

¼ lim

) lim an ¼ lim n  n→1

n→1

n→1

1 1 1 þ þ⋯þ 2n 2n 2n

1 2n

1 2

) lim an ¼ n⟶1

Choice (1) is the answer. 4.4. Based on the problem, we need to determine the coefficient of x3 in the Maclaurin expansion of the function below based on the following information. f ð 0Þ ¼ 1 f 0 ðxÞ ¼ 1 þ ðf ðxÞÞ10 From Maclaurin series or Maclaurin expansion, we know that f f ð x Þ ¼ f ð 0Þ þ f 0 ð 0Þ

000

ð0Þ 3! is

the coefficient of x3 as can be seen in the following.

x x2 x3 xn þ f 00 ð0Þ þ þf 000 ð0Þ þ . . . þ f ðnÞ ð0Þ þ . . . 2! 3! n! 1!

Therefore: f 0 ðxÞ ¼ 1 þ ðf ðxÞÞ10 ) f 00 ðxÞ ¼ 10f 0 ðxÞ  ðf ðxÞÞ9 ) f 000 ðxÞ ¼ 10f 00 ðxÞ  ðf ðxÞÞ9 þ 90ðf 0ðxÞÞ2  ðf ðxÞÞ8 For x ¼ 0: f 0 ð0Þ ¼ 1 þ ðf ð0ÞÞ10 ¼ 1 þ 1 ¼ 2

4

Solutions of Problems: Sequences and Series and Their Applications

61

f 00 ð0Þ ¼ 10f 0 ð0Þ  ðf ð0ÞÞ9 ¼ 10  2  19 ¼ 20 f 000 ð0Þ ¼ 10f 00 ð0Þ  ðf ð0ÞÞ9 þ 90ðf 0 ð0ÞÞ  ðf ð0ÞÞ8 ¼ 10  20  19 þ 90ð2Þ2  ð1Þ8 2

) f 000 ð0Þ ¼ 380 ) coefficient of x3 ¼

f 000 ð0Þ 380 190 ¼ ¼ 3! 6 3

Choice (2) is the answer. 4.5. First, we need to find a general term for the sequence as follows. nþ1

1, -

nð- 1Þ 2 3 -4 5 , , , , ... ¼ 2 4 8 16 2n - 1

From the concept of growth rate, for n ⟶ 1, a, b > 1 and k > 0, we know that the order of growth rate is as follows. log a n < nk < bn < n! < nn Hence: nð- 1Þnþ1 1 1 1 ¼ lim n - 1 ¼ 1 - 1 ¼ n-1 n⟶1 n⟶1 1 2 2 2 lim

nð- 1Þnþ1 ¼0 n⟶1 2n - 1

) lim Choice (1) is the answer.

4.6. Based on the information given in the problem, we have: S¼

þ1 n¼1

1 þ cos n n2

As we know, the series below, called P-series, is convergent if P > 1; otherwise, it is divergent. 1 n¼1

1 nP

Based on the above-mentioned rule, the series below is convergent. S0 ¼

1 n¼1

2 n2

Since S0 is convergent and the relation below is held for every term, the series of S is convergent. 2 1 þ cos n ≤ 2 n n2 Choice (1) is the answer.

62

4

Solutions of Problems: Sequences and Series and Their Applications

In this problem, the theorems below were used. Theorem: The P-series, presented below, is convergent for P > 1 and it is divergent for P ≤ 1. 1 n¼1

1 nP 1 n¼1 an

Theorem: Suppose that for every term an ≤ bn. Then, if 1 1 n¼1 bn is convergent, n¼1 an will be convergent.

is divergent,

1 n¼1 bn

will be divergent as well. Also, if

4.7. Based on growth rate, we know that: lim a1 nk1 þ a2 nk2  a1 nk1

if k 1 > k 2 > 0

n⟶1

Therefore: S¼

þ1 n¼1

10n2 þ 9n þ 8  12n3 þ 11n2 þ 10n þ 9

þ1 n¼1

10n2 5 ¼ 12n3 6

þ1 n¼1

1 n

As we know, harmonic series, shown below, is divergent. þ1 n¼1

1 n

Hence, the series of S has a similar behavior, and consequently it is divergent as well. Choice (2) is the answer. In this problem, the theorem below was used. Theorem: Harmonic series is a divergent series presented in the following. þ1 n¼1

1 n

4.8. Based on the problem, we have: S¼

1 n¼1

)S¼

1 n¼1

p

p n þ 1- n p n2 þ n

p p n þ 1- n ¼ p p n nþ1

1 n¼1

1 1 p -p n nþ1

As can be noticed, the series is a telescoping series. Therefore: 1 1 )S¼p -p ¼ 1-0 1þ1 1 )S¼1 Choice (2) is the answer.

4

Solutions of Problems: Sequences and Series and Their Applications

63

In this problem, the rule below was used. Telescoping series, presented below, is a series in which pairs of consecutive terms cancel each other so that only the initial and final terms are left. N

ðan - anþ1 Þ ¼ a1 - aNþ1 n¼1

4.9. Based on the problem, we have: 1

S¼

n1390 xn n¼1

To determine the convergence radius (R) and convergence range of a power series in the form of use two methods below.

1 n n¼1 an ðx - cÞ ,

1 ¼ lim n jan j R n→1 1 a ¼ lim nþ1 R n → 1 an The convergence range can be determined as follows. j xj ≤ R ) - R ≤ x ≤ R For this problem, we can use the second method as follows. ðn þ 1Þ1390 1 a n1390  lim ¼1 ¼ lim nþ1 ¼ lim n→1 n → 1 n1390 R n → 1 an n1390 )R¼1 ) j xj ≤ 1 ) - 1 ≤ x ≤ 1 It should be noted that for x ¼ ±1, the series do not have the necessary convergence criterion. Hence: -1 0, the order of growth rate is as follows. log a n < nk < bn < n! < nn 4.11. Based on the problem, we have: S¼

1 n¼1

ðn!Þxn ð n þ 1Þ n

4

Solutions of Problems: Sequences and Series and Their Applications

65 1

To determine the convergence radius (R) and convergence range of a power series in the form of

an ðx - cÞn , we can

n¼1

use two methods below. 1 ¼ lim n jan j R n→1 1 a ¼ lim nþ1 R n → 1 an The convergence range can be determined as follows. j xj ≤ R ) - R ≤ x ≤ R For this problem, we can use the first method as follows. p n 1 n! n! n ¼ lim ¼ lim n n → þ1 n þ 1 R n → þ1 ðn þ 1Þ Based on Stirling’s approximation, we have: n

)

1 1 n e ¼ lim ¼ lim R n → þ1 n þ 1 e n → þ1 n þ 1 )

1 1 ¼ R e

)R¼e Choice (2) is the answer. In this problem, the rule below was used. Stirling’s approximation: lim

n → þ1

p n

n! ¼ lim

n

n → þ1 e

4.12. Based on the problem, we need to determine the Taylor series of the function below around x ¼ 2. f ðxÞ ¼

1 5-x

The Taylor series or Taylor expansion of the function of f (x) around x ¼ a can be calculate as follows. f ð x Þ ¼ f ð a Þ þ f 0 ð aÞ

ð x - aÞ ð x - aÞ 2 ð x - aÞ n þ f 00 ðaÞ þ . . . þ f ð n Þ ð aÞ þ ... 1! 2! n!

Moreover, if in the Taylor series a ¼ 0, the series is called Maclaurin series or Maclaurin expansion.

66

4

f ð x Þ ¼ f ð 0Þ þ f 0 ð 0Þ

Solutions of Problems: Sequences and Series and Their Applications

x x2 xn þ f 00 ð0Þ þ . . . þ f ðnÞ ð0Þ þ . . . 1! 2! n!

However, the problem can be easily solved without using the direct formula of Taylor series as follows. f ðxÞ ¼

1 -1 -1 ¼ ¼ 5 - x x - 5 ðx - 2Þ - 3

) f ðxÞ ¼

-1 1 1 ¼ 3 1 - x -3 2 - 3 1 - x -3 2

As we know: 1 ¼ 1 þ x þ x2 þ ⋯ ¼ 1-x

1

xn n¼0

Therefore: ) f ðxÞ ¼

1 3

1 n¼0

x-2 3

n

2

¼

1 ðx - 2Þ ðx - 2Þ þ þ⋯ þ 3 33 32

Choice (1) is the answer. 4.13. From the concept of equivalent functions, we know that for a > 0 and even k: k

lim

n⟶1

ank þ bnk - 1 þ . . . ¼ lim

p k

n⟶1

a nþ

b ka

a nþ

b ka

And for odd k, we have: lim

k

n⟶1

ank þ bnk - 1 þ . . . ¼ lim

p k

n⟶1

Moreover, based on growth rate, we know that: lim a1 nk1 þ a2 nk2  a1 nk1

if k 1 > k 2 > 0

n⟶1

Therefore, for this problem, we have: lim an ¼ lim

n⟶1

n⟶1

p 3

p 3 n2 - n3 þ n ¼ lim - n3 - n2 þ n  lim - n þ n⟶1

n⟶1

) lim an ¼ n⟶1

1 þn 3ð- 1Þ

¼ lim

1 3

p p p p p n2 þ 1 - n n- n n p p lim bn ¼ lim p ¼ lim 4 n ¼ 4 1 p  nlim p  nlim 4 4 → 1 → 1 n → 1 n⟶1 n⟶1 4 n3 þ n þ 3 3 n n n þ n ) lim bn ¼ 1 n⟶1

Choice (3) is the answer.

n⟶1

1 3

4

Solutions of Problems: Sequences and Series and Their Applications

67

4.14. Based on the information given in the problem, we have: an ¼

1 þ 12 þ 13 þ ⋯ þ n13 ln n

1 þ 12 þ 13 þ ⋯ þ n13 1 ¼ lim  n → þ1 n → þ1 ln n ln n

ð1Þ n3

lim an ¼ lim

n → þ1

k¼1

1 k

ð2Þ

From the concept of equivalent functions, we know that if n → + 1: n k¼1

1  ln n k

ð3Þ

1  ln n3 k

ð4Þ

Hence: n3 k¼1

Solving (2) and (4): lim an  lim

n→1

n → þ1

ln n3 3 ln n ¼ lim ¼ lim 3 n → 1 ln n n→1 ln n

) lim an ¼ 3 n→1

Choice (3) is the answer. In this problem, the rule below was used. ln ab ¼ b ln a 4.15. Based on the information given in the problem, we have: an ¼ p

1 1 1 þp þ ⋯p n2 þ 2 n2 þ n n2 þ 1

Each term of the sequence is equal or larger than pn12 þn and equal or smaller than pn12 þ1. In other words: p

1 1 ≤ an ≤ p , 8n E ℕ n2 þ n n2 þ 1

Therefore, the value of lim an is larger than pnn2 þn and smaller than pnn2 þ1. In other words: n → þ1

lim p

n → þ1

n n ≤ lim an ≤ lim p 2 n → þ1 n → þ1 þn n þ1

n2

Based on the concept of equivalent functions, we have: lim

n → þ1

n n ≤ lim an ≤ lim n → þ1 n → þ1 jnj n þ 12

68

4

Solutions of Problems: Sequences and Series and Their Applications

) 1 ≤ lim an ≤ 1 n → þ1

Thus, based on Sandwich theorem, we have: lim an ¼ 1

n → þ1

Choice (3) is the answer. In this problem, the rules below were used. From the concept of equivalent functions, we know that for a > 0 and an even k: k

lim

n⟶1

ank þ bnk - 1 þ . . . ¼ lim

p k

a nþ

b ka

a nþ

b ka

n⟶1

And for an odd k, we have: lim

n⟶1

k

ank þ bnk - 1 þ . . . ¼ lim

p k

n⟶1

Theorem: Sandwich theorem states that if we have: bn ≤ an ≤ cn lim bn ¼ L

n → þ1

lim cn ¼ L

n → þ1

Then: lim an ¼ L

n → þ1

4.16. Based on the problem, we have: S¼

1 1 1 1 þ þ þ þ⋯ 1 1þ2 1þ2þ3 1þ2þ3þ4 )S¼

1

1

nðnþ1Þ n¼1 2

)S¼2

1 n¼1

1

¼

n¼1

2 nð n þ 1 Þ

1 1 n nþ1

As can be noticed, the series is a telescoping series. Therefore: S ¼ 2 1-

1 ¼ 2 ð 1 - 0Þ 1þ1 )S¼2

Choice (1) is the answer.

4

Solutions of Problems: Sequences and Series and Their Applications

69

In this problem, the rules below were used. 1 þ 2 þ 3 þ ... þ n ¼

nð n þ 1 Þ 2

Telescoping series, presented below, is a series in which pairs of consecutive terms cancel each other so that only the initial and final terms are left. N

ðan - anþ1 Þ ¼ a1 - aNþ1 n¼1

4.17. A sequence in the form of an ¼ {a1, a2, . . .} is convergent if its limit when n → 1 is a unique finite number ( lim an); n → þ1

otherwise, the sequence is divergent. Choice (1): Based on Stirling’s approximation, we have: lim

p n

n → þ1

n! ¼ lim

n

n → þ1 e

) lim

p n

n → þ1

n! ¼ 1

Hence, the sequence is divergent. Choice (2): Each term of the sequence is equal or larger than n2nþn and equal or smaller than n2nþ1. In other words: n n n n n ≤ þ þ⋯þ 2 ≤ , 8nEℕ n2 þ n n2 þ 1 n2 þ 2 n þ n n2 þ 1 2

2

Therefore, the value of lim an is larger than n2nþn and smaller than n2nþ1. In other words: n → þ1

lim

n → þ1 n2

n2 n2 ≤ lim an ≤ lim 2 n → þ1 n þ 1 þ n n → þ1 ) 1 ≤ lim an ≤ 1 n → þ1

Therefore, based on Sandwich theorem: lim an ¼ 1

n → þ1

Hence, the sequence is convergent. Choice (3): Based on growth rate, we know that: lim a1 n þ a2 n  a1 n

if a1 > a2 > 0

n⟶1

Thus: lim

n → þ1

p n

3n þ 2n  lim

n → þ1

p n

3n ¼ lim 3 ¼ 3 n → þ1

70

4

Solutions of Problems: Sequences and Series and Their Applications

) lim an ¼ 3 n → þ1

Thus, the sequence is convergent. Choice (4): Based on growth rate, we know that: lim a1 n þ a2 n  a1 n

n⟶1

if a1 > a2 > 0

Now, if a > b: an - bn an ¼1 n  lim n n → þ1 a þ b n → þ1 an lim

) lim an ¼ 1 n → þ1

If b > a: an - bn - bn n  lim n ¼ -1 n n → þ1 a þ b n → þ1 b lim

) lim an ¼ - 1 n → þ1

Therefore, in any condition, the sequence is convergent. Choice (1) is the answer. In this problem, the rule below was used. Theorem: Sandwich theorem states that if we have: bn ≤ an ≤ cn lim bn ¼ L

n → þ1

lim cn ¼ L

n → þ1

Then: lim an ¼ L

n → þ1

4.18. The problem can be solved by using the following theorem. Theorem: It states that the necessary, but not enough, criterion for a series in the form of Sn ¼

þ1

an to be convergent is n¼1

that the limit of its general term must be zero when n → 1. In other words: lim an ¼ 0

n → þ1

Choice (1): lim

n→1

n-1 n

n

¼ lim 1 n→1

1 n

n

¼ e-1 ≠ 0

References

71

Choice (2): nþ1 n

lim

n→1

n

¼ lim 1 þ n→1

n

1 n

¼ e≠0

Choice (3): lim

n→1

nþ1 n

n2

¼ lim

n→1

nþ1 n

n

n

¼ lim

1þ

n→1

1 n

n

n

¼ e1 ¼ 1 ≠ 0

Choice (4): lim

n→1 n

1 1 1 p ¼ lim 3 ¼ ¼0 n n → 1 n2 1

Thus, the series of S4 has the necessary criterion of convergence. On the other hand, we know that P-series, shown below, is convergent for P > 1. 1 n¼1

1 nP

For Choice (4), P ¼ 1.5 > 1; therefore, the series is convergent. Choice (4) is the answer. In this problem, the rule below was used. lim 1 þ

n→1

a bn

cn

ac

¼ eb

References 1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

5

Problems: Polar Coordinate System

Abstract

In this chapter, the basic and advanced problems concerned with polar coordinate system are presented. The subjects include tangent line on a curve, radius of curve, polar equation, spiral, transferring from cartesian coordinate to polar coordinate and vice versa, and curve types such as ellipse, straight line, parabola, hyperbola. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. p 1 3 5.1. Express the cartesian position of the point of ,2 2 Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π 1) 2, 3 π 2) 1, 6 π 3) 1, 6 π 4) 1, 3

in polar coordinate system (r, θ) [1–3].

p 5.2. Express the cartesian position of the point of - 3, 3 in polar coordinate system (r, θ). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 5π 1) 2 3, 6 p 5π 2) 3 3, 6 p π 3) 2 3, 6 p π 4) 2 3, 3 5.3. Express the polar position of the point of r ¼ 1, θ ¼ 0 in cartesian coordinate system (x, y). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) (1, 1) 2) (1, 0) # The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_5

73

74

5

Problems: Polar Coordinate System

3) (0, 1) 4) (0, 0) 5.4. Express the polar position of the point of r ¼

p

2, θ ¼ -

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p p 2, 2 1) p p 2 2 2) , 2 2 3) (1, -1) 4) (1, 1)

π in cartesian coordinate system (x, y). 4

5.5. Calculate the amount of angle between the tangent line and the radius of the curve of r ¼ θ2 + 1 at θ ¼ 1. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large π 1) 6 π 2) 4 π 3) 3 π 4) 2 π 5.6. Calculate the amount of angle between the tangent line and the radius of the curve of r ¼ 2 + 2 sin θ at the point of (3, ). 6 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large π 1) 6 π 2) 4 π 3) 3 π 4) 2 5.7. Determine the polar equation of a circle with the center on positive side of x-axis and radius of two while passing from the origin. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) r ¼ 4 cos θ 2) r ¼ 2 cos θ 3) r ¼ 4 sin θ 4) r ¼ 2 sin θ θ

5.8. Calculate the surface area of the spiral of r ¼ e4π from θ ¼ 0 to θ ¼ 2π. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) π(e - 1) p 2) 2π e - 1 3) 2π(e - 1) p 4) 4π e - 1

5

Problems: Polar Coordinate System

75 1

5.9. Calculate the surface area that the curve of r ¼ ð4 - 4 sin θ cos θÞ2 separates from the first quadrant. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) π - 1 2) π 3) π + 1 4) π - 2 5.10. What is the curve type of the polar function of r ¼ cot θ csc θ. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) Ellipse 2) Straight line 3) Parabola 4) Hyperbola 5.11. Determine the curve type of the polar function below. r¼

5 3 cos θ þ 2 sin θ

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) Straight line 2) Ellipse 3) Parabola 4) Hyperbola 5.12. Express the function of cos 2θ ¼ 1 in cartesian coordinate system. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) x - y ¼ 0 2) x + y ¼ 0 3) x ¼ 0 4) y ¼ 0 5.13. Express the function below in cartesian coordinate system. x ¼ cos 2 θ y ¼ sin θ cos θ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1 2 þ y2 ¼ 1) x 4 2 2 1 1 2) x2 þ y ¼ 2 4 2 1 1 3) x þ þ y2 ¼ 2 4 1 2 1 2 1 þ y¼ 4) x 2 2 4

76

5

Problems: Polar Coordinate System

5.14. Calculate the amount of angle between the curves of r ¼ 2(1 + sin θ) and r ¼ 3(1 - sin θ) at the intersection point. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large π 1) 2 π 2) 4 2 3) arctan 3 1 4) arctan 7 5.15. Determine the curve type of the polar function below. r¼

4 2 - cos θ

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) Ellipse 2) Parabola 3) Hyperbola 4) Two crossing straight lines 5.16. Determine the number of intersection points of the following polar functions. r¼ r¼ -

cos θ sin 2 θ

3 cos θ þ 4 sin θ

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) 1 2) 2 3) 1 4) 0

References 1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

6

Solutions of Problems: Polar Coordinate System

Abstract

In this chapter, the problems of the fifth chapter are fully solved, in detail, step-by-step, and with different methods. 6.1. Based on the information given in the problem, we have [1–3]: p 1 3 ,2 2

ðx, yÞ ¼

The relations below are used to transfer from cartesian coordinate to polar coordinate. r¼ θ ¼ tan - 1

x2 þ y2

y x

if x > 0, y > 0

θ ¼ π - tan - 1

y x

if x < 0, y > 0

θ ¼ π þ tan - 1

y x

if x < 0, y < 0

θ ¼ - tan - 1

y x

if x > 0, y < 0

Therefore, for this problem, we have:

r¼

1 2

2

p þ p

θ ¼ - tan - 1

3 2 1 2

3 2

2

1 3 þ ¼1 4 4

¼

p π ¼ - tan - 1 3 ¼ 3

) ðr, θÞ ¼ 1, -

π 3

Choice (4) is the answer. In this problem, the rule below was used.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_6

77

78

6 Solutions of Problems: Polar Coordinate System

p π tan - 1 3 ¼ 3 6.2. Based on the information given in the problem, we have: p ðx, yÞ ¼ - 3, 3 The relations below are used to transfer from cartesian coordinate to polar coordinate. x 2 þ y2

r¼ y x

θ ¼ tan - 1

if x > 0, y > 0

θ ¼ π - tan - 1

y x

if x < 0, y > 0

θ ¼ π þ tan - 1

y x

if x < 0, y < 0

θ ¼ - tan - 1

y x

if x > 0, y < 0

Therefore, for this problem, we have: p 3

ð- 3Þ2 þ

r¼

2

¼

p

p 9þ3¼2 3

p θ ¼ π - tan

-1

3 π 5π ¼ π- ¼ 6 6 3 p 5π 2 3, 6

) ðr, θÞ ¼ Choice (1) is the answer. In this problem, the rule below was used. p tan

-1

3 π ¼ 3 6

6.3. Based on the information given in the problem, we have: ðr, θÞ ¼ ð1, 0Þ The relations below are used to transfer from polar coordinate to cartesian coordinate. x ¼ r cos θ y ¼ r sin θ Therefore, for this problem, we have:

6

Solutions of Problems: Polar Coordinate System

79

x ¼ 1 cos 0 ¼ 1 y ¼ 1 sin 0 ¼ 0 ) ðx, yÞ ¼ ð1, 0Þ Choice (2) is the answer. In this problem, the rules below were used. cos 0 ¼ 1 sin 0 ¼ 0 6.4. Based on the information given in the problem, we have: p

ðr, θÞ ¼

2, -

π 4

The relations below are used to transfer from polar coordinate to cartesian coordinate. x ¼ r cos θ y ¼ r sin θ Therefore, for this problem, we have: x¼ y¼

p

2 cos -

π ¼1 4

p π 2 sin ¼ -1 4

) ðx, yÞ ¼ ð1, - 1Þ Choice (3) is the answer. In this problem, the rules below were used. cos sin -

p π 2 ¼ 4 2

p π 2 ¼ 4 2

6.5. The amount of angle between the tangent line on a polar curve and the radius of the curve can be calculated as follows. tan α ¼

f ðθ Þ r ðθ Þ ¼ f 0 ðθÞ drðθÞ

ð1Þ

dθ

Based on the information given in the problem, we have: r ¼ θ2 þ 1

ð2Þ

80

6 Solutions of Problems: Polar Coordinate System

θ¼1

ð3Þ

Solving (1)–(3): tan α ¼

θ2 þ 1 ¼1 2θ θ ¼ 1

) α ¼ tan - 1 ð1Þ ¼

π 4

Choice (2) is the answer. In this problem, the rule below was used. π 4

tan - 1 ð1Þ ¼ 6.6. Based on the information given in the problem, we have:

r ¼ 2 þ 2 sin θ ðr, θÞ ¼ 3,

ð1Þ

π 6

ð2Þ

The amount of angle between the tangent line on a polar curve and the radius of the curve can be calculated as follows. tan α ¼

f ðθ Þ r ðθ Þ ¼ f 0 ðθÞ drðθÞ

ð3Þ

dθ

Solving (1)–(3):

tan α ¼

2 þ 2 sin θ 2 cos θ

ðr, θÞ ¼ 3,

π 6

) α ¼ tan - 1

p 2 þ 2 sin π6 3 ¼p ¼ 3 2 cos π6 3

¼

p

3 ¼

Choice (3) is the answer. In this problem, the rules below were used. d sin x ¼ cos x dx p π 3 cos ¼ 6 2 sin tan - 1

π 1 ¼ 6 2 p

3 ¼

π 3

π 3

6

Solutions of Problems: Polar Coordinate System

81

6.7. The polar equations of a circle with the center on positive and negative sides of x-axis and radius of “a” while passing from the origin are as follows, respectively. r ¼ 2a cos θ r ¼ - 2a cos θ Moreover, the polar equations of a circle with the center on positive and negative sides of y-axis and radius of “a” while passing from the origin are as follows, respectively. r ¼ 2a sin θ r ¼ - 2a sin θ Thus, for this problem, we have: r ¼ 4 cos θ Choice (1) is the answer. 6.8. The surface area of a spiral in the form of r ¼ f(θ) from θ1 to θ2 can be calculated as follows. 1 2

S¼

θ2

r 2 dθ

θ1

Based on the information given in the problem, we have: θ

r ¼ e4π

ð1Þ

0 ≤ θ ≤ 2π

ð2Þ

For this problem, we have: 1 2

S¼ θ

) S ¼ πe2π

2π

θ

e2π dθ 0

2π ¼ π e1 - e0 0

) S ¼ π ðe - 1Þ Choice (1) is the answer. In this problem, the rule below was used. 1 eax dx ¼ eax þ c a

82

6 Solutions of Problems: Polar Coordinate System

6.9. The surface area of a spiral in the form of r ¼ f(θ) from θ1 to θ2 can be calculated as follows. S¼

θ2

1 2

r 2 dθ

θ1

Based on the information given in the problem, we have: 1

r ¼ ð4 - 4 sin θ cos θÞ2 0≤θ≤

ð1Þ

π 2

ð2Þ

Thus, for this problem, we have: 1 )S¼ 2 )S¼

θ2 θ1

π 2

1 r dθ ¼ 2 2

ð4 - 4 sin θ cos θÞdθ

0

π 1 1 4θ - 2 sin 2 θ 2 ¼ 2 2 0

4

π π - ð 0 - 0Þ - 2 sin 2 2 2

) S ¼ π-1 Choice (1) is the answer. In this problem, the rule below was used. un du ¼

1 unþ1 þ c nþ1

6.10. Based on the information given in the problem, we have: r ¼ cot θ csc θ

ð1Þ

It can be simplified as follows. r¼

cos θ sin θ

1 cos θ ¼ sin θ sin 2 θ

ð2Þ

By transferring from polar coordinate to cartesian coordinate, we have: y ¼ r sin θ

ð3Þ

x ¼ r cos θ

ð4Þ

x2 þ y2 ¼ r 2

ð5Þ

Solving (2)–(5):

)

x2 þ y 2 ¼

x þ y2

x2

2

y x2 þ y 2

)

y2 ¼ x2 þ y2

x þ y2

x2

6

Solutions of Problems: Polar Coordinate System

83

) y2 ¼ x It is the equation of a parabola. Choice (3) is the answer. In general, the equation of a parabola is as follows. ðx - x0 Þ2 ¼ 4aðy - y0 Þ ðy - y0 Þ2 ¼ 4aðx - x0 Þ In this problem, the rules below were used. cot θ ¼

cos θ sin θ

cscθ ¼

1 sin θ

6.11. Based on the information given in the problem, we have: r¼

5 3 cos θ þ 2 sin θ

) 3r cos θ þ 2r sin θ ¼ 5

ð1Þ ð2Þ

By transferring from polar coordinate to cartesian coordinate, we have: y ¼ r sin θ

ð3Þ

x ¼ r cos θ

ð4Þ

x2 þ y2 ¼ r 2

ð5Þ

Solving (2)–(5): 3x þ 2y ¼ 5 It is the equation of a straight line. Choice (1) is the answer. In general, the equation of a straight line is as follows. ax þ by ¼ c 6.12. Based on the information given in the problem, we have: cos 2θ ¼ 1

ð1Þ

) cos 2 θ - sin 2 θ ¼ 1

ð2Þ

By transferring from polar coordinate to cartesian coordinate, we have:

84

6 Solutions of Problems: Polar Coordinate System

y ¼ r sin θ

ð3Þ

x ¼ r cos θ

ð4Þ

x2 þ y 2 ¼ r 2

ð5Þ

Solving (2)–(5): 2

x x2 þ y 2

2

y x2 þ y2

-

¼ 1 ) x2 - y2 ¼ x2 þ y2 ) 2y2 ¼ 0 )y¼0

Choice (4) is the answer. In this problem, the rule below was used. cos 2θ ¼ cos 2 θ - sin 2 θ 6.13. Based on the information given in the problem, we have: x ¼ cos 2 θ

ð1Þ

y ¼ sin θ cos θ By transferring from polar coordinate to cartesian coordinate, we have: y ¼ r sin θ

ð2Þ

x ¼ r cos θ

ð3Þ

x2 þ y2 ¼ r 2

ð4Þ

Solving (1)–(4): 2

x¼ y¼

x 2 x þ y2 y x2 þ y 2

) x x2 þ y 2

x2 þ y2 xy y¼ 2 x þ y2 x¼

x2

From (5) or (6), we have: x 2 þ y2 ¼ x ) x2 - x þ y2 ¼ 0 ) xChoice (1) is the answer.

1 2

2

-

1 1 þ y2 ¼ 0 ) x 4 2

2

þ y2 ¼

1 4

ð 5Þ ð 6Þ

6

Solutions of Problems: Polar Coordinate System

85

6.14. Suppose α1 is the angle between the tangent line and the radius of the curve of r ¼ f1(θ) at θ0 and α2 is the angle between the tangent line and the radius of the curve of r ¼ f2(θ) at θ0. The acute or straight angle between the two polar curves at the intersection point (θ ¼ θ0) can be calculated as follows. tan ψ ¼

tan α1 - tan α2 1 þ tan α1 tan α2

ð1Þ

Moreover, the amount of angle between the tangent line on a polar curve and the radius of the curve can be calculated as follows. tan α ¼

f ðθ Þ r ðθ Þ ¼ f 0 ðθÞ drðθÞ

ð2Þ

dθ

Based on the information given in the problem, we have: r ¼ 2ð1 þ sin θÞ

ð1Þ

r ¼ 3ð1 - sin θÞ

ð2Þ

First, we need to find the intersection point of the polar curves as follows. 2ð1 þ sin θÞ ¼ 3ð1 - sin θÞ ) 5 sin θ ¼ 1 ) sin θ ¼

p 1 24 ) cos θ ¼ 5 5

Then: p f 2 ðθÞ 2ð1 þ sin θÞ 1 þ sin θ 1 þ 15 6 6 ¼ ¼ tan α1 ¼ 0 ¼ p ¼p ¼ 24 2 cos θ cos θ 2 f 2 ðθ Þ 24 5

tan α2 ¼

p 1 - 15 f 2 ðθÞ 3ð1 - sin θÞ 1 - sin θ 4 - 6 p p ¼ ¼ ¼ ¼ ¼ - 3 cos θ - cos θ 3 f 02 ðθÞ - 524 - 24

Therefore:

) tan ψ ¼

p 6 2

1þ

p

6 2

p 6 3

p

¼

p - 6 3

) ψ ¼ tan - 1 ð1Þ ¼ Choice (1) is the answer. In this problem, the rules below were used. sin 2 x þ cos 2 x ¼ 1 d sin x ¼ cos x dx tan - 1 ð1Þ ¼

π 2

6 6

1-1 π 2

¼1

86

6 Solutions of Problems: Polar Coordinate System

6.15. Based on the information given in the problem, we have: r¼

4 2 - cos θ

ð1Þ

) 2r - r cos θ ¼ 4

ð2Þ

By transferring from polar coordinate to cartesian coordinate, we have: y ¼ r sin θ

ð3Þ

x ¼ r cos θ

ð4Þ

x2 þ y 2 ¼ r 2

ð5Þ

Solving (2)–(5): 2

x2 þ y2 - x ¼ 4 ) 2

x2 þ y2 ¼ x þ 4

) 4 x2 þ y2 ¼ x2 þ 8x þ 16 ) 3x2 - 8x þ 4y2 ¼ 16 ) 3 x2 -

8 16 16 4 þ 4y2 ¼ 16 þ xþ ) 3 x3 9 3 3 )

4 2 3 8 2 3

x-

þ

ð y - 0Þ 2 8 p 2 3

2

2

þ 4y2 ¼

64 3

¼1

It is the equation of an ellipse. Choice (1) is the answer. In general, the equation of an ellipse is as follows. ðx - x0 Þ2 ðy - y0 Þ2 þ ¼1 a2 b2 6.16. Based on the information given in the problem, we have: r¼ r¼ -

cos θ sin 2 θ

3 cos θ þ 4 sin θ

ð1Þ ð2Þ

The problem can be easily solved by transferring from polar coordinate to cartesian coordinate as follows:

Solving (1) and (3)–(5):

y ¼ r sin θ

ð3Þ

x ¼ r cos θ

ð4Þ

x2 þ y2 ¼ r 2

ð5Þ

References

87

p x2

þ

y2

¼

x x2 þy2

py x2 þy2

2

)

y2 ¼ x2 þ y2

x ) y2 ¼ x x2 þ y 2

ð6Þ

Solving (2) and (3)–(5): x2 þ y2 ¼ -

p

x x2 þy2

3 þ 4p

y

) x þ 4y ¼ - 3

x2 þy2

Solving (6) and (7): y2 þ 4y ¼ - 3 ) y2 þ 4y þ 3 ¼ 0 p - 4 ± 42 - 4  1  3 - 4 ± 2 )y¼ ¼ 2 2 ) y ¼ - 1, - 3 Choice (2) is the answer. In this problem, the rule below was used. ax2 þ bx þ c ¼ 0 p - b ± b2 - 4ac )x¼ 2a

References 1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

ð7Þ

7

Problems: Complex numbers

Abstract

In this chapter, the basic and advanced problems of complex numbers are presented. The subjects include the operations on complex numbers as well as on functions in complex form. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 7.1. Calculate the value of e0i. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) -1 4) 1 7.2. Calculate the value of eπi [1–3]. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) -1 4) 1 7.3. Calculate the complex conjugate of i. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 - i 2) 0 3) 1 + i 4) -i 7.4. Which one of the relations below is wrong? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) ða - ibÞ ¼ a þ ib 2)

z1 z2

¼

z1 z2

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_7

89

90

7 Problems: Complex numbers

3) (a + ib)(a - ib) ¼ a2 - b2 p 4) i ¼ - 1 7.5. Which one of the relations below is wrong? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

j a þ ib j ¼ a2 þ b2 Re(z1z2) ¼ x1x2 - y1y2 Im(z1z2) ¼ x1x2 + y1y2 eix ¼ cos x + i sin x

7.6. Which one of the relations below is wrong? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large jz j z 1) 1 ¼ 1 eiðθ1 - θ2 Þ z2 jz2 j 1 1 - iθ2 ¼ e 2) z2 jz2 j 3) z1 z2 ¼ jz1 jjz2 jeiðθ1 þθ2 Þ 4) z1 þ z2 ¼ ðjz1 j þ jz2 jÞeiðθ1 þθ2 Þ 7.7. Calculate the value of zz if: z¼

ð1 þ 2iÞð1 þ 3iÞð1 þ 4iÞ ð2 - 3iÞð2 - 4iÞ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 86 1) 25 96 2) 25 85 3) 26 95 4) 26 7.8. Calculate the value of ii. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

2

eπ π e-2 Ln i Ln π

7.9. Present the equation below in complex form. x2 - y 2 ¼ 1 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

7

Problems: Complex numbers

1) 2) 3) 4)

91

ð zÞ 2 - z2 ¼ 2 z2 þ ðzÞ2 ¼ 2 z2 - ðzÞ2 ¼ 1 z2 þ ðzÞ2 ¼ 1

7.10. Calculate the value of following complex number if n E ℕ . ð1 þ iÞn ð1 - i Þn - 2 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2i n - 1 p 2) 2i2 p 3) 2i 4) i-2 7.11. Calculate the value of θ if the complex number below does not have a real part. 3 þ 2i sin θ 1 - 2i sin θ ○ Easy ● Normal ○ Hard Difficulty level Calculation amount ○ Small ● Normal ○ Large π 1) 6 π 2) 4 π 3) 3 π 4) 2 7.12. Calculate the value of following relation. p z ¼ ð1 - iÞð1 þ i 3Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p π π 1) 2 2 cos þ i sin 12 12 p π π - i sin 2) 2 2 cos 12 12 p 7π 7π þ i sin 3) 2 2 cos 12 12 p 7π 7π - i sin 4) 2 2 cos 12 12

92

7 Problems: Complex numbers

7.13. Calculate the value of following relation. ð1 þ iÞ15 ð1 - iÞ13 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2) -2 3) -3 4) 3 7.14. Calculate the value of relation below. p 3þi p 3-i

10

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p 1þi 3 1) 2 p -1 þ i 3 2) 2 p 1-i 3 3) 2 p -1-i 3 4) 2 7.15. Calculate the value of following relation. 1 þ cos

2π 2π þ i sin 3 3

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) -1 2) 1 3) -i 4) i 7.16. In the equation below, calculate the value of a. p eaþib ¼ 1 - i 3 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) ln 2 p 2) ln 2 3) 2 p 4) 2

120

7

Problems: Complex numbers

93

7.17. Calculate the value of relation below if zm ¼ cos

π π þ i sin m . 2m 2 1

∏ zm m¼1

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) -πi 2) πi 3) 1 4) -1 7.18. Which one of the following choices is correct if z1 and z2 are two non-zero complex numbers where: z1 - z2 ¼1 z1 þ z2 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) Re(z1z2) > 0 2) Re(z1z2) < 0 3) Re(z1z2) ¼ 0 4) Im(z1z2) > 0 7.19. Calculate the maximum value of z if the relation below is held. 6z - i ≤1 2 þ 3iz Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 5 1 2) 4 1 3) 3 1 4) 2 7.20. Calculate the value of the relation below in complex form. 1 þ sin θ þ i cos θ 1 þ sin θ þ i cos θ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) sin θ + i cos θ 2) sin θ - i cos θ 3) 1 - i θ θ 4) sin - i cos 2 2

94

7 Problems: Complex numbers

References 1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

8

Solutions of Problems: Complex Numbers

Abstract

In this chapter, the problems of the seventh chapter are fully solved, in detail, step-by-step, and with different methods. 8.1. As we know from Euler’ formula [1–3]: eθi ¼ cos θ þ i sin θ Therefore: e0i ¼ cos 0 þ i sin 0 ¼ 1 þ 0i e0i ¼ 1 Choice (2) is the answer. In this problem, the rules below were used. cos 0 ¼ 1 sin 0 ¼ 0 8.2. As we know from Euler’ formula: eθi ¼ cos θ þ i sin θ Therefore: eπi ¼ cos π þ i sin π ¼ - 1 þ 0i eπi ¼ - 1 Choice (3) is the answer. In this problem, the rules below were used. cos π ¼ - 1 sin π ¼ 0

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_8

95

96

8 Solutions of Problems: Complex Numbers

8.3. The complex conjugate of a complex number can be achieved by changing the sign of imaginary part of the complex number while the other parts of the complex number are left intact. In other words: ða þ ibÞ ¼ ða þ ibÞ ¼ a - ib Therefore: i ¼ i ¼ - i Choice (4) is the answer. 8.4. All the relations are correct except Choice (3). Its correct relation is as follows. ða þ ibÞða - ibÞ ¼ a2 - aib þ iba - ðibÞ2 ) ða þ ibÞða - ibÞ ¼ a2 þ b2 Choice (3) is the answer. In this problem, the rule below was used. i2 ¼ - 1 8.5. All the relations are correct except Choice (3). Its correct relation is as follows. z1 z2 ¼ ðx1 þ iy1 Þðx2 þ iy2 Þ ¼ x1 x2 þ x1 iy2 þ iy1 x2 þ i2 y1 y2 ) z1 z2 ¼ ðx1 x2 - y1 y2 Þ þ iðx1 y2 þ y1 x2 Þ ) Imðz1 z2 Þ ¼ x1 y2 þ y1 x2 Choice (3) is the answer. In this problem, the rule below was used. i2 ¼ - 1 8.6. All the relations are correct except Choice (4). Its correct relation is as follows. z1 þ z2 ¼ jz1 jeiθ1 þ jz2 jeiθ2 Choice (4) is the answer. 8.7. Based on the information given in the problem, we have: z¼

ð1 þ 2iÞð1 þ 3iÞð1 þ 4iÞ ð2 - 3iÞð2 - 4iÞ

)z¼

ð1 - 2iÞð1 - 3iÞð1 - 4iÞ ð2 þ 3iÞð2 þ 4iÞ

8

Solutions of Problems: Complex Numbers

97

) zz ¼

ð1 þ 4Þð1 þ 9Þð1 þ 16Þ 5  10  17 ¼ 13  20 ð4 þ 9Þð4 þ 16Þ ) zz ¼

85 26

Choice (3) is the answer. In this problem, the rules below were used. ða þ ibÞ ¼ a - ib z1 z2 z3 z4

¼

z1 z2 z3 z4

ða þ ibÞða - ibÞ ¼ a2 þ b2 8.8. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. i¼

p

- 11 0

1ei tan

¼ ei tan

-1

1

π

¼ e2 i

Then: π

ii ¼ e 2 i

i

π 2

π

¼ e2i ¼ e - 2

Choice (2) is the answer. In this problem, the rules below were used. a þ ib ¼

-1 b a2 þ b2 ei tan jaj

if a > 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπ - tan jajÞ

if a < 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπþ tan jajÞ

if a < 0, b < 0

a þ ib ¼

-1 b a2 þ b2 e - i tan jaj

eia

b

if a > 0, b < 0

¼ eiab

i2 ¼ - 1 8.9. Based on the information given in the problem, we have: x2 - y2 ¼ 1 As we know: z ¼ x þ iy ) z2 ¼ ðx þ iyÞ2 ¼ x2 - y2 þ i2xy

98

8 Solutions of Problems: Complex Numbers

z ¼ x - iy ) ðzÞ2 ¼ x2 - y2 - i2xy ) z2 þ ðzÞ2 ¼ 2 x2 - y2 Therefore: z2 þ ðzÞ2 ¼ 2 Choice (2) is the answer. In this problem, the rules below were used. i2 ¼ - 1 ða þ ibÞ ¼ a - ib 8.10. Based on the information given in the problem, we have: ð1 þ iÞn ð1 - i Þn - 2

ð1Þ

The problem can be solved as follows. )

ð1 þ iÞn ð1 þ i Þn - 2 1þi ¼ ð1 þ iÞ2 ¼ n-2 n-2 1-i ð1 - iÞ ð1 - i Þ

n-2

ð1 þ 2i - 1Þ ¼

1þi 1-i

n-2

ð2iÞ

ð2Þ

On the other hand: 1þi 1 þ i 1 þ i 1 þ 2i þ i2 2i ¼ ¼ ¼i  ¼ 1-i 1-i 1 þ i 2 1 - i2 Solving (1)–(3): ð1 þ iÞn ¼ in - 2 ð2iÞ ð1 - i Þn - 2 )

ð1 þ i Þn ¼ 2in - 1 ð1 - iÞn - 2

Choice (1) is the answer. In this problem, the rules below were used. i2 ¼ - 1 a þ ib a þ ib c - id ¼  c þ id c þ id c - id

ð3Þ

8

Solutions of Problems: Complex Numbers

99

8.11. Based on the information given in the problem, we know that: Re

3 þ 2i sin θ 1 - 2i sin θ

¼0

ð1Þ

The problem can be solved as follows. 3 þ 2i sin θ 3 þ 2i sin θ 1 þ 2i sin θ ¼  1 - 2i sin θ 1 - 2i sin θ 1 þ 2i sin θ ¼

3 - 4 sin 2 θ þ 8i sin θ 3 þ 6i sin θ þ 2i sin θ - 4 sin 2 θ ¼ 1 þ 4 sin 2 θ 1 þ 4 sin 2 θ ¼

3 - 4 sin 2 θ 8 sin θ þ i 1 þ 4 sin 2 θ 1 þ 4 sin 2 θ

ð2Þ

Solving (1) and (2): p 3 - 4 sin 2 θ 3 3 π 2 2 ¼ 0 ) 3 - 4 sin θ ¼ 0 ) sin θ ¼ ) sin θ ¼ ± )θ¼ ± 4 2 3 1 þ 4 sin 2 θ Choice (3) is the answer. In this problem, the rules below were used. a þ ib a þ ib c - id ¼  c þ id c þ id c - id i2 ¼ - 1 8.12. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. 1-i ¼ p 1þi 3¼

π

1 þ ð- 1Þ2 e - i4 ¼ ð 1Þ 2 þ

p

2

p

π

2e - i 4

π

π

3 ei3 ¼ 2ei3

Therefore: p - iπ p z ¼ ð1 - i Þ 1 þ i 3 ¼ 2e 4

p p π π π π 2ei3 ¼ 2 2eið3 - 4Þ ¼ 2 2ei12

p π π ) z ¼ 2 2 cos þ i sin 12 12 Choice (1) is the answer. In this problem, the rules below were used. a þ ib ¼

-1 b a2 þ b2 ei tan jaj

if a > 0, b > 0

100

8 Solutions of Problems: Complex Numbers

a þ ib ¼

-1 b a2 þ b2 eiðπ - tan jajÞ

if a < 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπþ tan jajÞ

if a < 0, b < 0

-1 b a2 þ b2 e - i tan jaj

a þ ib ¼

tan - 1 ð- 1Þ ¼ -

if a > 0, b < 0 π 4

p π tan - 1 3 ¼ 3 jz2 jeiθ2 ¼ jz1 jjz2 jeiðθ1 þθ2 Þ

z1 z2 ¼ jz1 jeiθ1

jzjeiθ ¼ jzjðcos θ þ i sin θÞ 8.13. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. p iπ 15 2e 4 ð1 þ iÞ15 ¼ p - iπ 13 13 ð1 - i Þ 2e 4 15π

¼

13π 2ei 4 ið15π 4 - ð - 4 ÞÞ 13π ¼ 2e e-i 4

28π

¼ 2ei 4 ¼ 2ei7π ¼ 2ðcos 7π þ i sin 7π Þ ¼ 2ðcos π þ i sin π Þ ¼ 2ð- 1 þ 0iÞ Thus: ð1 þ iÞ15 ¼ -2 ð1 - iÞ13 Choice (2) is the answer. In this problem, the rules below were used. a þ ib ¼

-1 b a2 þ b2 ei tan jaj

if a > 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπ - tan jajÞ

if a < 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπþ tan jajÞ

if a < 0, b < 0

a þ ib ¼

-1 b a2 þ b2 e - i tan jaj

tan - 1 1 ¼

if a > 0, b < 0

π 4

tan - 1 ð- 1Þ ¼ -

π 4

8

Solutions of Problems: Complex Numbers

101 b

eia

¼ eiab

z1 jz1 jeiθ1 jz1 j iðθ1 - θ2 Þ ¼ ¼ e z2 jz2 jeiθ2 jz2 j jzjeiθ ¼ jzjðcos θ þ i sin θÞ 8.14. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. p 3þi p 3-i

10

π

2ei6 π 2e - i6

¼

10

π

¼ ei 3

10

10π

¼ ei 3

¼ eið2πþ 3 Þ ¼ ei 3 4π

4π

p 4π 4π 1 3 ¼ cos þ i sin ¼ - -i 3 3 2 2 Therefore: p 3þi p 3-i

10

p -1-i 3 ¼ 2

Choice (4) is the answer. In this problem, the rules below were used. a þ ib ¼

-1 b a2 þ b2 ei tan jaj

if a > 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπ - tan jajÞ

if a < 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπþ tan jajÞ

if a < 0, b < 0

a þ ib ¼

-1 b a2 þ b2 e - i tan jaj

if a > 0, b < 0

p

3 π ¼ 3 6

tan - 1

p

3 3

tan - 1 -

¼ -

π 6

z1 jz1 jeiθ1 jz1 j iðθ1 - θ2 Þ ¼ ¼ e z2 jz2 jeiθ2 jz2 j eia

b

¼ eiab

eið2πþθÞ ¼ eiθ jzjeiθ ¼ jzjðcos θ þ i sin θÞ

102

8 Solutions of Problems: Complex Numbers

cos

sin

4π 1 ¼ 3 2

p 4π 3 ¼ 3 2

8.15. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. 1 þ cos

¼

120

2π 2π þ i sin 3 3

p 1 3 þi 2 2

120

π

¼ ei 3

¼

120

1-

p 1 3 þi 2 2

120

¼ ei40π ¼ eið202πþ0Þ

¼ cosð0Þ þ i sinð0Þ ¼ 1 þ 0i Therefore: 1 þ cos

2π 2π þ i sin 3 3

120

¼1

Choice (2) is the answer. In this problem, the rules below were used. cos

2π 1 ¼ 3 2

sin a þ ib ¼

p 2π 3 ¼ 3 2

-1 b a2 þ b2 ei tan jaj

if a > 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπ - tan jajÞ

if a < 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπþ tan jajÞ

if a < 0, b < 0

a þ ib ¼

-1 b a2 þ b2 e - i tan jaj

if a > 0, b < 0

p π tan - 1 3 ¼ 3 eia

b

¼ eiab

eið2πþθÞ ¼ eiθ jzjeiθ ¼ jzjðcos θ þ i sin θÞ cosð0Þ ¼ 1 sinð0Þ ¼ 0

8

Solutions of Problems: Complex Numbers

103

8.16. Based on the information given in the problem, we have: p eaþib ¼ 1 - i 3

ð1Þ

The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. p π 1 - i 3 ¼ 2e - 3i

ð2Þ

eaþib ¼ ea eib

ð3Þ

As we know:

Solving (1)–(3): π

ea eib ¼ 2e - 3i )

ea ¼ 2 e ¼e ib

) a ¼ ln 2, b ¼ -

- π3i

Choice (1) is the answer. In this problem, the rules below were used. -1 b a2 þ b2 ei tan jaj

a þ ib ¼

if a > 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπ - tan jajÞ

if a < 0, b > 0

a þ ib ¼

-1 b a2 þ b2 eiðπþ tan jajÞ

if a < 0, b < 0

a þ ib ¼

-1 b a2 þ b2 e - i tan jaj

if a > 0, b < 0

p π tan - 1 - 3 ¼ 3 eaþb ¼ ea eb 8.17. Based on the information given in the problem, we have: zm ¼ cos

π π þ i sin m 2m 2

Therefore: 1

1

m¼1

m¼1

∏ zm ¼ ∏ e2m i ¼ e2i e2ð2Þi e4ð2Þi ⋯ ¼ e2iþ2ð2iÞþ4ð2iÞþ⋯ π

π

1 π

1

1

m¼1

m¼1

) ∏ zm ¼ exp 1

1 π

π

1

π i 2m

¼ em¼1

1 π

π i 2m

1 π

¼e

πi 2 1-1 2

) ∏ zm ¼ eπi ¼ cos π þ i sin π ¼ - 1 þ 0i m¼1

π 3

104

8 Solutions of Problems: Complex Numbers 1

) ∏ zm ¼ - 1 m¼1

Choice (4) is the answer. In this problem, the rules below were used. 1

∏ zm ¼ z1 z2 . . . z1

m¼1

eix ¼ cos x þ i sin x jz1 jeiθ1 1

jz2 jeiθ2 ¼ jz1 jjz2 jeiðθ1 þθ2 Þ

am ¼

m¼1

a1 1-q

if jqj < 1

eπi ¼ - 1 8.18. Based on the information given in the problem, we have: z1 - z2 ¼1 z1 þ z2

ð1Þ

z1 ¼ x1 þ iy1

ð2Þ

z2 ¼ x2 þ iy2

ð3Þ

Let us assume:

Solving (1)–(3): ðx1 þ iy1 Þ - ðx2 - iy2 Þ ðx - x2 Þ þ iðy1 þ y2 Þ ¼1) 1 ¼1 ðx1 þ iy1 Þ þ ðx2 - iy2 Þ ðx1 þ x2 Þ þ iðy1 - y2 Þ )

jðx1 - x2 Þ þ iðy1 þ y2 Þj ¼1 jðx1 þ x2 Þ þ iðy1 - y2 Þj

) jðx1 - x2 Þ þ iðy1 þ y2 Þj ¼ jðx1 þ x2 Þ þ iðy1 - y2 Þj )

ð x1 - x2 Þ 2 þ ð y 1 þ y 2 Þ 2 ¼

ðx1 þ x2 Þ2 þ ðy1 - y2 Þ2

) ðx1 Þ2 þ ðx2 Þ2 - 2x1 x2 þ ðy1 Þ2 þ ðy2 Þ2 þ 2y1 y2 ¼ ðx1 Þ2 þ ðx2 Þ2 þ 2x1 x2 þ ðy1 Þ2 þ ðy2 Þ2 - 2y1 y2 ) - 4x1 x2 þ 4y1 y2 ¼ 0 ) x1 x2 - y1 y2 ¼ 0 ) Reðz1 z2 Þ ¼ 0 Choice (3) is the answer. In this problem, the rules below were used.

8

Solutions of Problems: Complex Numbers

105

ja þ ibj a þ ib ¼ c þ id jc þ id j ja þ ibj ¼

a2 þ b2

Reðz1 z2 Þ ¼ Reððx1 þ iy1 Þðx2 þ iy2 ÞÞ ¼ x1 x2 - y1 y2 i2 ¼ - 1 8.19. Based on the information given in the problem, we have: 6z - i ≤1 2 þ 3iz j6z - ij ≤ 1 ) j6ðx þ iyÞ - ij ≤ j2 þ 3iðx þ iyÞj j2 þ 3izj ) j6x þ ið6y - 1Þj ≤ jð2 - 3yÞ þ 3ixj )

ð6xÞ2 þ ð6y - 1Þ2 ≤

ð2 - 3yÞ2 þ ð3xÞ2

) 36x2 þ 36y2 - 12y þ 1 ≤ 4 - 12y þ 9y2 þ 9x2 27x2 þ 27y2 ≤ 3 ) x2 þ y2 ≤ ) jzj ≤

1 ) 9

1 3

) maxðjzjÞ ¼

1 3

Choice (3) is the answer. In this problem, the rules below were used. ja þ ibj a þ ib ¼ c þ id jc þ id j ja þ ibj ¼

a 2 þ b2

i2 ¼ - 1 8.20. Based on the information given in the problem, we have: 1 þ sin θ þ i cos θ 1 þ sin θ - i cos θ The problem can be solved as follows.

x2 þ y 2 ≤

1 3

106

8 Solutions of Problems: Complex Numbers

1 þ sin θ þ i cos θ 1 þ sin θ þ i cos θ 1 þ sin θ þ i cos θ ¼  1 þ sin θ - i cos θ 1 þ sin θ - i cos θ ð1 þ sin θÞ þ i cos θ ¼

1 þ sin 2 θ þ ðicos θÞ2 þ 2 sin θ þ 2 sin θi cos θ þ 2i cos θ ð1 þ sin θ þ i cos θÞ2 ¼ 2 1 þ sin 2 θ þ 2 sin θ þ cos 2 θ ð1 þ sin θÞ - i2 cos 2 θ ¼ ¼

1 þ sin 2 θ - cos 2 θ þ 2 sin θ þ 2i cos θð1 þ sin θÞ 2 þ 2 sin θ

sin 2 θ þ cos 2 θ þ sin 2 θ - cos 2 θ þ 2 sin θ þ 2i cos θð1 þ sin θÞ 2 þ 2 sin θ

¼

¼

2 sin 2 θ þ 2 sin θ þ 2i cos θð1 þ sin θÞ 2 þ 2 sin θ

¼

2 sin θð1 þ sin θÞ þ 2i cos θð1 þ sin θÞ 2ð1 þ sin θÞ

2ð1 þ sin θÞðsin θ þ i cos θÞ ¼ sin θ þ i cos θ 2ð1 þ sin θÞ

Therefore: 1 þ sin θ þ i cos θ ¼ sin θ þ i cos θ 1 þ sin θ - i cos θ Choice (1) is the answer. In this problem, the rules below were used. a þ ib a þ ib c - id ¼  c þ id c þ id c - id ða þ ibÞða - ibÞ ¼ a2 þ b2 ða þ b þ cÞ2 ¼ a2 þ b2 þ c2 þ 2ab þ 2bc þ 2ac sin 2 θ þ cos 2 θ ¼ 1

References 1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

Index

A Angle, 74, 76, 79, 80, 85 Applications of integration, v, 1–52 Arc length of a curve, 10–12, 37–45 Average value of a function, 15 C Cartesian coordinate system, 73–75 Center of gravity, 13–14, 49–52 Coefficient, 54, 60 Complex conjugate, 89, 96 Complex form, 90, 93 Complex numbers, 89–94, 96 Concept of equivalent functions, 60, 66–68 Concept of growth rate, 59, 61, 64 Consecutive terms, 63, 69 Convergence, 55, 63, 65, 71 Convergence radius, 55, 56, 63–65 Convergence range, 55, 63–65 Convergent, 54, 55, 58, 61, 62, 69–71 Curve type, 75, 76 D Defining a new variable, 19, 22, 46, 47 Divergent, 54, 55, 57, 61, 62, 69 E Ellipse, 75, 76, 86 Even function, 19–21, 50 G General term of a sequence, 54, 57

L Limit, 30, 53, 54, 56, 57, 69, 70 List of integral of functions, 15, 22, 25, 26, 28–30 M Maclaurin expansion, 54, 60, 65 Maclaurin series, 60, 65 Mean value of a function, 1, 15–16 N Necessary criterion of convergence, 71 O Odd function, 50 Order of growth rate, 59, 61, 64 P Parabola, 75, 76, 83 Parametric curve, 13, 38, 39 Parametric relation, 10 Polar coordinate system, 73–87 Polar curve, 79, 80, 85 Polar equation, 74, 81 Polar function, 75, 76 P-series, 61, 62, 71 Q Quadrant, 19, 75

H Harmonic series, 62 Hyperbola, 75, 76

R Radius of curve, 74, 79, 80, 85 Range of integration, 47 Real part, 91 Rotation of a surface area around x-axis, 28–30, 32–36, 47 Rotation of a surface area around y-axis, 8, 31

I Imaginary part, 96 Integration by parts, 22, 25, 33–35 Intersection point, 16–18, 76, 85 Intersection points of the curves, 16–18

S Sequences, 56, 57, 59–71 Series, 54–71 Spiral, 74, 81, 82 Stirling’s approximation, 64, 65, 69

# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3

107

108 Straight line, 75, 76, 83 Surface area, 2–9, 12, 13, 16–27, 30, 31, 45–50, 75, 81, 82 Surface area enclosed by curves, 31, 33 Surface area of a solid of revolution, 12–13, 45–49 T Tangent line, 74, 79, 80, 85 Taylor expansion, 65 Taylor series, 56, 65, 66

Index Telescoping series, 62, 63, 68, 69 Theorem, 62, 68–70 Transfer from cartesian coordinate to polar coordinate, 77, 78, 97, 99–103 Trigonometry, 26, 29 V Volume resulted from rotation of an enclosed region, 7–10, 28–37