www.pdfnotes.co 's a n h Kris TEXT BOOK on C alculus (For B.A. and B.Sc. Ist year students of All Colleges affiliated to universities in Uttar Pradesh) As per U.P. UNIFIED Syllabus (w.e.f. 2011-2012) By A. R. Vasishtha Retired Head, Dep’t. of Mathematics Meerut College, Meerut (U.P.) KRISHNA Prakashan Media (P) Ltd. KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India www.pdfnotes.co Jai Shri Radhey Shyam Dedicated to Lord Krishna Authors & Publishers www.pdfnotes.co Preface This book on CALCULUS has been specially written according to the latest Unified Syllabus to meet the requirements of the B.A. and B.Sc. Part-I Students of all Universities in Uttar Pradesh. The subject matter has been discussed in such a simple way that the students will find no difficulty to understand it. The proofs of various theorems and examples have been given with minute details. Each chapter of this book contains complete theory and a fairly large number of solved examples. Sufficient problems have also been selected from various university examination papers. At the end of each chapter an exercise containing objective questions has been given. We have tried our best to keep the book free from misprints. The authors shall be grateful to the readers who point out errors and omissions which, inspite of all care, might have been there. The authors, in general, hope that the present book will be warmly received by the students and teachers. We shall indeed be very thankful to our colleagues for their recommending this book to their students. The authors wish to express their thanks to Mr. S.K. Rastogi, Managing Director, Mr. Sugam Rastogi, Executive Director, Mrs. Kanupriya Rastogi Director and entire team of KRISHNA Prakashan Media (P) Ltd., Meerut for bringing out this book in the present nice form. The authors will feel amply rewarded if the book serves the purpose for which it is meant. Suggestions for the improvement of the book are always welcome. Preface to the Revised Edition The authors feel great pleasure in presenting the thoroughly revised edition of the book CALCULUS and wish to record thanks to the teachers and students for their warm reception to the previous edition. The present edition has been specially designed, made up-to-date and well organised in a systematic order according to the latest syllabus. The authors have always endeavoured to keep the text update in the best interests of the students community- a gesture which the authors hope would be appreciated by the students and teachers alike. Suggestions for the improvement of the book will be thankfully received. — Authors www.pdfnotes.co Syllabus Calculus U.P. UNIFIED (w.e.f. 2011-12) B.A./B.Sc. Paper-II M.M. : 33 / 65 Differential Calculus Unit-1: ε -δ definition of the limit of a function, Continuous functions and classification of discontinuities, Differentiability, Chain rule of Differentiability, Rolle’s theorem, First and second mean value theorems, Taylor’s theorems with Lagrange’s and Cauchy’s forms of remainder, Successive differentiation and Leibnitz’s theorem. Unit-2: Expansion of functions (in Taylor’s and Maclaurin’s series), Indeterminate forms, Partial differentiation and Euler’s theorem, Jacobians. Unit-3: Maxima and Minima (for functions of two variables), Tangents and normals (polar form only), Curvature, Envelopes and evolutes. Unit-4(a): Asymptotes, Tests for concavity and convexity, Points of inflexion, Multiple points, Tracing of curves in Cartesian and Polar coordinates. Integral Calculus Unit- 4(b): Reduction formulae, Beta and Gamma functions. Unit-5: Quadrature. Rectification. Volumes and surfaces of solids of revolution, Pappus theorem, Double and triple integrals, Change of order of integration, Dirichlet’s and Liouville’s integral formulae. www.pdfnotes.co B rief C ontents DIFFERENTIAL CALCULUS.......................D-01— D-330 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. Limits and Continuity.....................................................................D-03 — D-40 Differentiability ..................................................................................D-41 — D-84 Differentiation .................................................................................D-85 — D-100 Successive Differentiation .........................................................D-101 — D-120 Expansions Of Functions ............................................................D-121 — D-134 Indeterminate Forms ...................................................................D-135 — D-152 Partial Differentiation .................................................................D-153 — D-168 Jacobians ...........................................................................................D-169 — D-184 Maxima And Minima Of Functions of Two Independent Variables ....................................................D-185 — D-196 Tangents And Normals ...............................................................D-197 — D-218 Curvature ..........................................................................................D-219 — D-240 Envelopes, Evolutes And Involutes ......................................D-241 — D-262 Asymptotes ......................................................................................D-263 — D-288 Singular Points: Curve Tracing ...................................................D-289 — D-330 INTEGRAL CALCULUS ................................I-01—I-198 1. Reduction Formulae (For Trigonometric Functions) .................I-03 — I-30 2. Reduction Formulae Continued (For Irrational Algebraic and Transcendental Functions) ......... I-31 — I-44 3. Beta and Gamma Functions ............................................................... I-45 — I-74 4. Multiple Integrals ....................................................................................I-75 — I-104 5. Dirichlet's and Liouville's Integrals...............................................I-105 — I-120 6. Areas of Curves ......................................................................................I-121 — I-144 7. Rectification (Lengths of Arcs and Intrinsic Equations of Plane Curves) ......................................................................................... I-145 — I-166 8. Volumes and Surfaces of Solids of Revolution ......................I-167 — I-198 www.pdfnotes.co SECTION A DIFFERENTIAL CALCULUS C hapters 1. Limits and Continuity 1. 1. Differentiability 2. 1. Differentiation 3. 1. Successive Differentiation 4. 1. Expansions of Functions 5. 1. Indeterminate Forms 6. 1. Partial Differentiation 7. 1. Jacobians 8. 9. Maxima And Minima of Functions of Two Independent Variables 1. Tangents and Normals 10. 1. Curvature 11. 12. Envelopes, Evolutes and Involutes 1. Asymptotes 13. 1. Singular Points : Curve Tracing 14. 1.1 Definitions Constant : A symbol wh ich r et ains t h e sam e value t h roughout a set of mathematical operations is called a constant. A variable is a quantity, or a symbol representing a number, which is capable of assuming different values. A continuous variable is one which can take all the numerical values between two given numbers. An independent variable is one which may take up any arbitrary value that may be assigned to it. A dependent variable is a symbol which can assume its value as a result of some other variable taking some assigned value. Domain of a Variable : If we give the independent variable x only those values which lie between x = a and x = b, then all these numerical values taken collectively will be called domain or interval of the variable. The domain is said to be closed if a and b are included in it and is denoted by the symbol [a, b]. An open domain is denoted by ]a, b[ or by (a, b). Similarly the symbols [a, b [ and ] a, b] stand for semi-open domains. These semi-open domians are also denoted by [a, b) and (a, b] respectively. Function : If y depends upon x in such a manner that for every value of x in its domain of variation there corresponds a definite (i.e., a unique) value of y, then y is said D-4 DIFFERENTIAL CALCULUS to be a single-valued function of x and is denoted by y = f (x), f denoting the kind of dependence or relationship that exists between x and y. This relationship is often called functional relation and f (x1), f (x2), … , f (xr) are called functional values of f (x) for x = x1, x2 , …, xr respectively. Note : The essential thing about the definition of a function is that for each value of x there must correspond a definite value of f (x). We must be in possession of a set of rules which determine for each value of x in a certain interval, a definite value of the funct ion. These rules may t ake t he shape of a single compact formula such as f (x) = sin x or a number of such formulae that apply to different parts of the domain of x, for example f (x) = sin x for 0 ≤ x ≤ π ⁄ 2⎫ …(1) f (x) = x for π ⁄ 2 < x < π ⎪⎬ ⋅ ⎪ ⎭ f (x) = cos x for x ≥ π. In the first case f (x) = sin x is defined for values of x in any interval. In the second case f (x) given by (1) is defined in the interval [0, ∞ [. The above definition of a function of x brings about (1) idea of the dependence of the function on x (2) idea of definiteness of the values of the function for each value of x (3) idea of single valuedness of the function (4) idea of the domain of the variable x. We are accust omed t o t hink t hat every funct ion is capable of graphical representation. Majority of functions are certainly capable of graphical representation but there are some functions which cannot be represented by a graph. The function defined as follows is such a function : f (x) = 0 when x is rational, f (x) = 1 when x is irrational. Set-theoretic definition of a function : Let A and B be two given sets. Suppose there exists a correspondence denoted by f, which associates to each member of A a unique member of B. Then f is called a function or a mapping from A to B. The mapping f of A to B is denoted by f : A → B. The set A is called the domain of the function f, and B is called the co-domain of f. The element y ∈ B which the mapping f associates to an element x ∈ A is denoted by f (x) and is called the f-image of x or the value of the function f for x. E ach element of A has a unique image and each element of B need not appear as the image of an element in A . We define the range of f to consist of those elements in B which appear as the image of at least one element in A . Equality of two functions : Two functions f and g of A → B are said to be equal if and only if f (x) = g (x) — V x ∈ A and we write f = g. For two unequal mappings from A to B, there must exist at least one element x ∈ A such that f (x) ≠ g (x). Constant function : A function f : A → B is called a constant function if the same element b ∈ B is assigned to every element in A . Real valued function : If both A and B are the sets of real numbers, then f : A → B is called a real valued function of a real variable. Single-valued and multiple-valued functions : If y has only one definite value when a definite value is given to x then y is called a single-valued function of x. When y has more than one value for a value of x, it is called a multiple-valued function of x. Odd and Even functions : A function is said to be odd if it changes sign when the sign of the variable is changed i.e., if f (− x) = − f (x). A function is said to be even if its sign does not change when the sign of the variable is changed i.e., if f (− x) = f (x). D-5 LIMITS AND CONTINUITY Bounded and unbounded functions : If for all values of x in a given interval, f (x) is never greater than some fixed number M, the number M is said to be an upper bound for f in that interval, whereas if f (x) is never less than some number m then m is called a lower bound for f in that interval. If both upper and lower bounds of a function are finite, the function is said to be bounded otherwise it is said to be unbounded. By a supremum of f in an interval we mean the least of all the upper bounds of f in that interval. Similarly an infimum of f is the greatest of all the lower bounds of f in the interval. A rational integral function, or a polynomial, is a function of the form a0 x n + a1 x n − 1 + … + a n − 1 x + an where a0 , a1,…, an are constants and n is a positive integer or zero. A rational function is defined as the quotient of one polynomial by another. For example, 7x + 4 is a rational function. 2 2x + 3x + 6 An algebraical function is a function which can be expressed as the root of an equation of the form yn + A1 yn − 1 + A2 yn − 2 + … + A n − 1 y + An = 0 where A 1, A 2 ,… , A n are rational functions of x. In particular a rational function is also algebraical. A transcendental function is a function which is not algebraical. Trigonometrical, exponential and logarithmic functions are examples of transcendental functions. Monotonic functions : Th e fu n ct io n y = f (x) i s s a i d t o b e monotonically increasing if corresponding to an increase in the value of x in a certain interval I in which the function f (x) is defined, the value of y never decreases i.e., x > x ⇒ f (x ) ≥ f (x ) — V x , x ∈ I. 1 2 1 2 1 2 Similarly the function f (x) is monotonically decreasing if V x , x ∈ I. x > x ⇒ f (x ) ≤ f (x ) — 1 2 1 2 1 2 Also f is said to be strictly increasing iff x1 > x2 ⇒ f (x1) > f (x2) and strictly decreasing iff x1 > x2 ⇒ f (x1) < f (x2). The function f defined by f (x) = sin x is monotonically increasing in the interval 0≤ x≤ 1 2 π and monotonically decreasing in the interval 1 2 π ≤ x ≤ π. Explicit and implicit functions : A function is said to be explicit when expressed directly in terms of the independent variable or variables e.g., y = sin− 1 x + log x. If the function cannot be expressed directly in terms of the independent variable or variables, the function is said to be implicit e.g., the equation x y + y x = ab expresses y as an implicit function of x. Sum, Difference, Product and Quotient of two functions : Let f, g be two functions with domains D1 and D2 . If D = D1 ∩ D2 , then D is common to the domains of f and g. V x ∈ D. The sum function f + g is defined as ( f + g) (x) = f (x) + g (x) — If c ∈ R, the function cf is defined as (cf ) (x) = c f (x) — V x ∈ D 1. V x ∈ D. The difference function f − g is defined as ( f − g) (x) = f (x) − g (x) — D-6 DIFFERENTIAL CALCULUS The product function fg is defined as ( fg) (x) = f (x) g (x) — V x ∈ D. The reciprocal function 1/g of the function g is defined as 1 — ⎛ 1⎞ V x ∈ D2 and g (x) ≠ 0. ⎜ ⎟ (x) = ⎝ g⎠ g (x) f (x) — ⎛ f ⎞ V x ∈ D and g (x) ≠ 0. The quotient function f / g is defined as ⎜ ⎟ (x) = g (x) ⎝ g⎠ 1.2 Limits Consider the function y = (x2 − 1) ⁄ (x − 1). The value of this function at x = 1 is of the form 0 ⁄ 0 which is meaningless. In this case we cannot divide the numerator by the denominator since x − 1 is zero. Now suppose x is not actually equal to 1 but very nearly equal to 1. Then x − 1 is not equal to zero. H ence in this case we can divide the numerator by the denominator. x2 − 1 = x + 1. x− 1 If x is little greater than 1, then the value of y will be greater than 2 and as x gets nearer to 1, y comes nearer to 2. Now the difference between y and 2 is ∴ x2 − 2x + 1 (x − 1) 2 x2 − 1 − 2= = = x − 1. x− 1 x− 1 x− 1 This difference (x − 1) can be made as small as we please by letting x tend to 1. Thus we see that when x has a fixed value 1, the value of y is meaningless but when x tends to 1, y tends to 2 and we say that the limit of y is 2 when x tends to 1. Thus we write as lim ⎡ 2 ⎤ x → 1 ⎣ (x − 1) ⁄ (x − 1) ⎦ = 2. (Bundelkhand 2006; Purvanchal 10; Kashi 14) Definition of limit : L et f be a function defined on some neighbourhood of a point a except possibly at a itself. Then a real number l is said to be the limit of f as x approaches a if for any arbitrarily chosen positive number ε, however small but not zero, there exists a corresponding num ber δ greater than zero such that ⏐ f (x) − l⏐ < ε for all values of x for which 0 < ⏐ x − a⏐ < δ, where ⏐ x − a⏐ m eans the absolute value of x − a without any regard to sign. lim In symbols, we then write x → a f (x) = l. We have to negate the above definition in order to show that f does not approach l as x approaches a. If it is not true that for every ε > 0, there is some δ > 0 such that 0 < ⏐ x − a⏐ < δ ⇒ ⏐ f (x) − l⏐ < ε, then there must exist an ε > 0, such that for every δ > 0, there is some x for which 0 < ⏐ x − a⏐ < δ but ⏐ f (x) − l⏐ <| ε. This means that in order to show that f does not approach l as x approaches a, it is sufficient to produce an ε > 0 such that for each δ > 0 there is some x satisfying 0 < ⏐ x − a⏐ < δ and ⏐ f (x) − l⏐ ≥ ε. D-7 LIMITS AND CONTINUITY lim Note 1 : It is not at all necessary for x → a f (x) to exist that f be defined at x = a. It is enough that for some δ > 0, f be defined whenever 0 < ⏐ x − a⏐ < δ. Note 2 : If N be a neigh bourhood of a, t hen N ~ {a} is called a deleted neighbourhood of a. Note 3 : If a function f has a finite limit at a point a, then by the definition of the limit of a function a deleted neighbourhood of a exists on which f is bounded. Now we shall prove a theorem which is the foundation on which the definition of limit rests. If this theorem were not true, the definition of limit would have been useless. lim lim lim Theorem : If x → a f (x) = l , and x → a f (x) = m , then l = m i.e., if x → a f (x) exists, then it is unique. Proof : Suppose, if possible, l ≠ m. Let us take ε = 12 ⏐ l − m⏐ . Then ε > 0. lim Since x → a f (x) = l, for a given ε > 0, there exists δ 1 > 0 such that …(1) ⏐ f (x) − l⏐ < ε whenever 0 < ⏐ x − a⏐ < δ 1. lim Again since x → a f (x) = m, for a given ε > 0, there exists δ 2 > 0 such that …(2) ⏐ f (x) − m⏐ < ε whenever 0 < ⏐ x − a⏐ < δ 2 . If we ch o o se δ = m i n . {δ 1, δ 2}, t h e n 0 < ⏐ x − a⏐ < δ imp lies t h at bo t h 0 < ⏐ x − a⏐ < δ 1 and 0 < ⏐ x − a⏐ < δ 2 hold, and hence, we have i.e., ⏐ f (x) − l⏐ < ε and ⏐ f (x) − m⏐ < ε whenever 0 < ⏐ x − a⏐ < δ. This implies that if 0 < ⏐ x − a⏐ < δ, then ⏐ l − m⏐ = ⏐ { f (x) − m} − { f (x) − l}⏐ ≤ ⏐ f (x) − m⏐ + ⏐ f (x) − l⏐ < ε + ε = 2 ε = ⏐ l − m⏐ ⏐ l − m⏐ < ⏐ l − m⏐ , which is absurd and so our assumption is wrong. lim H ence, l = m i.e., x → a f (x) is unique. 1.3 Algebra of Limits Now we shall give some theorems on limits of functions which are similar to those of limits of sequences. lim Theorem 1 : If x → a f (x) = l ≠ 0 , then there exist num bers k > 0 and δ > 0 such that ⏐ f (x)⏐ > k whenever 0 < 1 lim A lso then x → a = f (x) Proof : Let ε = 1 2 ⏐ x − a⏐ < δ. 1 ⋅ l ⏐ l⏐ . Then ε > 0, because l ≠ 0. lim Since x → a f (x) = l, therefore, given ε > 0, there exists δ > 0 such that ⏐ f (x) − l⏐ < ε, whenever 0 < ⏐ x − a⏐ < δ. …(1) Now ⏐ l⏐ = ⏐ l − f(x) + f (x)⏐ ≤ ⏐ l − f (x)⏐ + ⏐ f (x)⏐ < ε + ⏐ f (x)⏐ , whenever 0 < ⏐ x − a⏐ < δ, from (1). D-8 DIFFERENTIAL CALCULUS ∴ Whenever 0 < ⏐ x − a⏐ < δ, we have ⏐ f (x)⏐ > ⏐ l⏐ − ε = ⏐ l⏐ − 12 ⏐ l⏐ = Thus taking k = 1 2 1 2 ⏐ l⏐ > 0. …(2) ⏐ l⏐ > 0, we get ⏐ f (x)⏐ > k whenever 0 < ⏐ x − a⏐ < δ. This proves the first part of the theorem. 1 1 lim Second part : Now to prove that x → a = ⋅ f (x) l ⏐ l − f (x)⏐ 1 ⎪ ⎪ l − f (x) ⎪ ⎪ 1 − ⎪= ⎪ ⋅ …(3) We have ⎪ ⎪= l ⎪ ⎪ l . f (x) ⎪ ⏐ l⏐ . ⏐ f (x)⏐ ⎪ f (x) By first part of this theorem there exist numbers k > 0 and δ 1 > 0 such that ⏐ f (x)⏐ > k i.e., 1 1 < whenever 0 < ⏐ x − a⏐ < δ 1. ⏐ f (x)⏐ k …(4) Let ε′ > 0 be given. lim Since x → a f (x) = l, therefore, given ε′ > 0, there exists δ 2 > 0 such that ⏐ f (x) − l⏐ < k ⏐ l⏐ ε′ whenever 0 < ⏐ x − a⏐ < δ 2 . Let δ = min. {δ 1, δ 2}. Then from (3), (4) and (5), we have 1⎪ 1 1 ⎪ 1 − ⎪< ⋅ k ⏐ l⏐ ε′ ⋅ whenever 0 < ⏐ x − a⏐ < δ. ⎪ l ⎪ ⏐ l⏐ k ⎪ f (x) Thus for given ε′ ⎪ 1 − ⎪ ⎪ f (x) = ε′. > 0, there exists δ > 0 such that 1⎪ ⎪ < ε′ whenever 0 < ⏐ x − a⏐ < δ. l⎪ 1 1 lim H ence, x → 0 = ⋅ f (x) l Theorem 2 : The lim it of a sum is equal to the sum of the limits. lim lim Proof : Let x → a f (x) = l and x → a g (x) = m. lim x → a {( f + g) (x)} = l + m. We have to show that Let ε > 0 be given. lim Since x → a f (x) = l, therefore, there exists δ 1 > 0 such that ⏐ f (x) − l⏐ < 1 2 ε whenever 0 < ⏐ x − a⏐ < δ 1. lim Again since x → a g (x) = m, therefore, there exists δ 2 > 0 such that ⏐ g (x) − m⏐ < 1 2 ε whenever 0 < ⏐ x − a⏐ < δ 2 . If we take δ = min. {δ 1, δ 2}, then 0 < ⏐ x − a⏐ < δ ⇒ both 0 < ⏐ x − a⏐ < δ 1 and 0 < ⏐ x − a⏐ < δ 2 hold, and consequently if 0 < ⏐ x − a⏐ < δ, then both ⏐ f (x) − l⏐ < 12 ε and ⏐ g (x) − m⏐ < 1 2 ε are true. Now if 0 < ⏐ x − a⏐ < δ, then ⏐ ( f + g) (x) − (l + m)⏐ = ⏐ f (x) − l + g (x) − m⏐ ...(5) D-9 LIMITS AND CONTINUITY ≤ ⏐ f (x) − l⏐ + ⏐ g (x) − m⏐ < 1 2 ε+ 1 2 ε = ε. Thus ⏐ ( f + g) (x) − (l + m)⏐ < ε whenever 0 < ⏐ x − a⏐ < δ. lim lim ∴ x → a ( f + g) (x) exists and x → a ( f + g) (x) = l + m. The above result can be extended to any finite number of functions. lim In the same way, we can prove that x → a (f − g) (x) = l − m. Theorem 3 : The lim it of a product is equal to the product of the limits. Proof : Using the notations of theorem 2, we have to prove that lim ( fg ) (x) = lm. x →a Let ε > 0 be given. Now ⏐ ( fg) (x) − lm⏐ = ⏐ f (x) g (x) − lg (x) + lg (x) − lm⏐ ≤ ⏐ f (x) g (x) − lg (x)⏐ + ⏐ lg (x) − lm⏐ = ⏐ g (x)⏐ ⏐ f (x) − l⏐ + ⏐ l⏐ ⏐ g (x) − m⏐ . …(1) lim Since x → a g (x) = m, therefore g (x) is bounded in some deleted neighbourhood o f x = a. H en ce t h e re exist s k > 0 a n d δ 1 > 0 su ch t h a t ⏐ g (x)⏐ ≤ k wh en ever 0 < ⏐ x − a⏐ < δ 1. lim lim Since x → a f (x) = l and x → a g (x) = m, therefore, corresponding to any given ε > 0, we can find positive numbers δ 2 and δ 3 such that ε ⏐ f (x) − l⏐ < whenever 0 < ⏐ x − a⏐ < δ 2 2k ε and ⏐ g (x) − m⏐ < whenever 0 < ⏐ x − a⏐ < δ 3 . 2 (⏐ l⏐ + 1) If we take δ = min. {δ 1, δ 2 , δ 3}, then from (1), we get ε ε + ⏐ l⏐ ⋅ ⏐ ( fg) (x) − lm⏐ < k ⋅ 2k 2 (⏐ l⏐ + 1) ⏐ l⏐ ε ε ⎡. . ⎤ + < 1⎥ ⎢ . 2 2 ⏐ l⏐ + 1 ⎣ ⎦ = ε whenever 0 < ⏐ x − a⏐ < δ. T h u s fo r ε > 0, we h a ve δ > 0 s u c h t h a t ⏐ ( fg) (x) − lm⏐ < ε wh en ever 0 < ⏐ x − a⏐ < δ. lim lim lim ∴ x → a ( fg) (x) = x → a f (x) g (x) exists and x → a ( fg) (x) = lm. < The above theorem can evidently be extended to any finite number of functions. Theorem 4 : The limit of a quotient is equal to the quotient of the lim its provided the limit of the denom inator is not zero. lim lim Proof : Let x → a f (x) = l and x → a g (x) = m ≠ 0. Now f (x) ⎫ ⎧ f (x) l ⎪ ⎪ ⎧ f (x) l ⎫⎪ ⎪ f (x) − ⎪ = ⎪⎨ − − ⎬⎪ ⎪ ⎬+ ⎨ m ⎪ ⎪ ⎩ g (x) m ⎭ ⎩ m m ⎭⎪ ⎪ g (x) 1 ⎪ f (x) ⎪ {m − g (x)} + {f (x) − l}⎪ m ⎪ m g (x) ⎪ = ⎪ D-10 DIFFERENTIAL CALCULUS ≤ ⏐ f (x)⏐ 1 ⏐ m − g (x)⏐ + ⏐ f (x) − l⏐ ⏐ m⏐ ⏐ g (x)⏐ ⏐ m⏐ …(1) lim S i n c e x → a f (x) = l, t h e r e fo r e t h e r e e xi st s a d e l e t e d n e i gh b o u r h o o d ] a − δ 1, a + δ 1 [ − {a} of the point x = a in which the function f is bounded. Let K > 0 be such that ⏐ f (x)⏐ ≤ K whenever 0 < ⏐ x − a⏐ < δ 1 . lim Again since g (x) ≠ 0 for all x in the domain of g and x → a g (x) = m ≠ 0, therefore there exist numbers k > 0 and δ 2 > 0 such that 1 1 < whenever 0 < ⏐ x − a⏐ < δ 2 . ⏐ g (x)⏐ > k i.e., k ⏐ g (x)⏐ [See theorem 1 of article 1.3] Let δ′ = min (δ 1, δ 2). The inequality (1) can then be written as l⎪ K 1 ⎪ f (x) − ⎪≤ ⏐ m − g (x)⏐ + ⏐ f (x) − l⏐ , ...(2) ⎪ ⎪ g (x) m ⎪ k ⏐ m⏐ ⏐ m⏐ for all x such that 0 < ⏐ x − a⏐ < δ′. Now take any given ε > 0. lim lim Since x → a f (x) = l and x → a g (x) = m, we can find positive numbers δ 3 and δ 4 such that ε ⏐ f (x) − l⏐ < ⏐ m⏐ ⋅ whenever 0 < ⏐ x − a⏐ < δ 3 2 k ⏐ m⏐ ε and ⏐ g (x) − m⏐ < ⋅ whenever 0 < ⏐ x − a⏐ < δ 4. K 2 Take δ = min {δ′, δ 3 , δ 4}. Then from (2), we get ε l⎪ ε ⎪ f (x) − ⎪ < + = ε, whenever 0 < ⏐ x − a⏐ < δ. ⎪ m⎪ 2 2 ⎪ g (x) l lim f (x) lim f (x) lim ⎛ f ⎞ x → a g (x) exists and x → a g (x) = x → a ⎜⎝ g ⎟⎠ (x) = m , if m ≠ 0. Alternative Proof : 1 1 lim Since m ≠ 0, therefore, by theorem 1 of § 3, x → a exists and equals ⋅ g (x) m ∴ 1 ⎫ lim ⎛ f ⎞ lim ⎧ x → a ⎜⎝ g ⎟⎠ (x) = x → a ⎨⎩ f (x) ⋅ g (x) ⎬⎭ 1 ⎫ ⎧ lim ⎫ ⎧ lim = ⎨ x → a f (x) ⎬ ⎨ x → a [By theorem 3 of article 1.3] ⎬ g (x) ⎭ ⎩ ⎭⎩ 1 l = l⋅ = ⋅ m m Theorem 5 : L et f be defined on D and let f (x) ≥ 0 for all x ∈ D. lim lim If x → a f (x) exists, then x → a f (x) ≥ 0. Now D-11 LIMITS AND CONTINUITY lim Proof : Suppose that x → a f (x) = l and l is negative. Taking ε = − 1 2 l, we can find a positive number δ > 0 such that ⏐ f (x) − l⏐ < − 1 2 l whenever 0 < ⏐ x − a⏐ < δ. It gives that 3l l < f (x) < < 0 whenever 0 < ⏐ x − a⏐ < δ. 2 2 This is a contradiction since we are given that f (x) ≥ 0 for all x ∈ D. H ence l cannot be negative. lim Consequently x → a f (x) ≥ 0. Corollary : L et f be defined on D and let f (x) > 0 for all x ∈ D. lim lim If x → a f (x) exists, then x → a f (x) ≥ 0. Proof : Since f (x) > 0 ⇒ f (x) ≥ 0, therefore now we can apply theorem 5 of article 1.3. Theorem 6 : L et f and g be defined on D and let f (x) ≥ g (x) for all x ∈ D. Then lim lim x → a f (x) ≥ x → a g (x), provided these limits exist. Proof : Let lim lim x → a f (x) = l, x → a g (x) = m. Let us define a function h by h (x) = f (x) − g (x) — V x ∈ D. Then, we have — (i) h (x) ≥ 0 V x ∈ D. lim lim (ii) x → a h (x) exsits and x → a h (x) = l − m. lim (iii) x → a h (x) ≥ 0, by theorem 5 of article 1.3. i.e., Thus, from (ii) and (iii), we get l − m ≥ 0 i.e., l ≥ m, lim lim x → a f (x) ≥ x → a g (x). lim lim Corollary : L et f (x) > g (x) for all x ∈ D. Then x → a f (x) ≥ x → a g (x) , provided these lim its exist. Theorem 7 : L et f, g and h be defined on D and let f (x) ≥ g (x) ≥ h (x) for all x. lim lim L et x → a f (x) = x → a h (x) . lim lim lim lim Then x → a g (x) exists, and x → a g (x) = x → a f (x) = x → a h (x). lim lim Proof : Let x → a f (x) = x → a h (x) = l. Then corresponding to any given ε > 0, we can find positive numbers δ 1 and δ 2 such that ⏐ f (x) − l⏐ < ε whenever 0 < ⏐ x − a⏐ < δ 1 i.e., l − ε < f (x) < l + ε whenever 0 < ⏐ x − a⏐ < δ 1 …(1) and l − ε < h (x) < l + ε whenever 0 < ⏐ x − a⏐ < δ 2. …(2) D-12 or DIFFERENTIAL CALCULUS Choosing δ to be smaller than δ 1 and δ 2 , we see from (1) and (2) that l − ε < h (x) ≤ g (x) ≤ f (x) < l + ε whenever 0 < ⏐ x − a⏐ < δ. Thus l − ε < g (x) < l + ε whenever 0 < ⏐ x − a⏐ < δ ⏐ g (x) − l⏐ < ε whenever 0 < ⏐ x − a⏐ < δ. lim lim H ence x → a g (x) exists and x → a g (x) = l. lim lim Theorem 8 : If x → a f (x) = l, then x → a ⏐ f (x)⏐ = ⏐ l⏐ . Proof : We have ⏐ f (x) − l⏐ ≥ ⏐ ⏐ f (x)⏐ − ⏐ l⏐ ⏐ , for all x. …(1) [... ⏐ p − q⏐ ≥ ⏐ ⏐ p⏐ − ⏐ q⏐ ⏐ ] Let ε > 0 be given. lim Since x → a f (x) = l, therefore, given ε > 0, there exists a number δ > 0 such that ⏐ f (x) − l⏐ < ε whenever 0 < ⏐ x − a⏐ < δ. …(2) From (1) and (2), we get ⏐⏐ f (x)⏐ − ⏐ l⏐⏐< ε whenever 0 < ⏐ x − a⏐ < δ. lim lim Consequently x → a ⏐ f (x)⏐ exists and x → a ⏐ f (x)⏐ = ⏐ l⏐ . Theorem 9 : I f t h e re i s a n u m b e r δ > 0 s u c h t h a t h (x) = 0 wh en ever lim 0 < ⏐ x − a⏐ < δ, then x → a h (x) = 0. Proof : For any ε > 0, the number δ > 0, given in the hypothesis of the theorem is such that h (x) = 0 whenever 0 < ⏐ x − a⏐ < δ or ⏐ h (x) − 0⏐ = 0 < ε whenever 0 < ⏐ x − a⏐ < δ. H ence lim x → a h (x) = 0. Corollary : I f t h e re i s a n u m b e r δ > 0 s u c h t h a t f (x) = g (x) wh en ever lim lim 0 < ⏐ x − a⏐ < δ , then x → a f (x) = x → a g (x). Proof : Let us define a function h by setting h (x) = f (x) − g (x) for all x. Then h (x) = 0 whenever 0 < ⏐ x − a⏐ < δ. Now, apply theorem 9 of article 1.3. Note : The above corollary has deep implications. It asserts that the concept of limit is a ‘local’ one. If two functions agree on some neighbourhood of a point a, then they cannot approach different limits as x approaches a. lim Theorem 10 : I f x → a f (x) = 0 a n d g (x) i s b o u n d ed in so m e d e le t ed lim neighbourhood of a, then x → a f (x) g (x) = 0. Proof : Since g (x) is bounded in some deleted neighbourhood of a, therefore there exist numbers k > 0 and δ 1 > 0 such that ⏐ g (x)⏐ ≤ k whenever 0 < ⏐ x − a⏐ < δ 1. ...(1) LIMITS AND CONTINUITY D-13 Now take any given ε > 0. lim Since x → a f (x) = 0, therefore there exists δ 2 > 0 such that ε ...(2) ⏐ f (x) − 0⏐ = ⏐ f (x)⏐ < whenever 0 < ⏐ x − a⏐ < δ 2 k Now take δ = min (δ 1, δ 2). Then for all x such that 0 < ⏐ x − a⏐ < δ, we have H ence ⏐ f (x) g (x) − 0⏐ = ⏐ f (x) g (x)⏐ = ⏐ f (x)⏐ . ⏐ g (x)⏐ ε < ⋅ k = ε , using (1) and (2). k lim x → a f (x) g (x) = 0. lim Illustration : We have x → 0 x sin (1 ⁄ x) = 0 lim because x → 0 x = 0 and ⏐ sin (1 ⁄ x)⏐ ≤ 1 for all x ≠ 0 i.e., sin (1 ⁄ x) is bounded in some deleted neighbourhood of zero. 1.4 Right Hand and Left Hand Limits Definition : (Right-hand limit) : A function f is said to approach l as x approaches a from right (or from above) if corresponding to an arbitrary positive num ber ε, there exists a positive num ber δ such that ⏐ f (x) − l⏐ < ε whenever a < x < a + δ. lim It is written as x → a + 0 f (x) = l or f (a + 0) = l. The working rule for finding the right hand : “Put a + h for x in f (x) where h is + ive and very very small and make h approach zero”. lim In short, we have f (a + 0) = h → 0 f (a + h). Definition : (Left-hand limit) : A function f is said to approach l as x approaches a from the left (or from below) if corresponding to an arbitrary positive number ε, there exists a positive num ber δ such that ⏐ f (x) − l⏐ < ε whenever a − δ < x < a. lim It is written as x → a − 0 f (x) = l or f (a − 0) = l. The working rule for finding the left hand : “Put a − h for x in f (x) where h is + ive and very very small and make h approach zero.” lim In this case, we have f (a − 0) = h → 0 f (a − h)⋅ Important Note : If both right hand limit and left hand limit of f as x → a, exist and are equal in value, their common value, evidently, will be the limit of f as x → a. If however, either or both of these limits do not exist, the limit of f as x → a does not exist. E ven if both these limits exist but are not equal in value then also the limit of f as x → a does not exist. D-14 DIFFERENTIAL CALCULUS 1.5 Limits as x → + ∞ (− ∞ ) Definition : A function f is said to approach l as x becom es positively infinite, if corresponding to each ε > 0, there exists δ > 0 such that ⏐ f (x) − l⏐ < ε whenever x ≥ δ. lim Then we write x → ∞ f (x) = l or f (x) → l as x → ∞ . Definition : A function f is said to approach l as x becom es negatively infinite, if corresponding to each ε > 0 there exists δ > 0 such that ⏐ f (x) − l⏐ < ε whenever x ≤ − δ. lim Then we write x → − ∞ f (x) = l f (x) → l as x → − ∞ . Note 1 : The results on the limits of sum, product and quotient of functions also hold good here provided that in these cases l + m, lm, l ⁄ m are defined. lim lim Note 2 : If x → ∞ f (x) = l exists, x → ∞ g (x) does not exist (as a finite real or lim number), even then x → ∞ f (x) g (x) can exist. Similar is the case as x → − ∞ . 1.6 Infinite Limits Definition : A fu n ctio n f is said to a pproach + ∞ as x approaches a, if co rrespo n din g t o a n y ε > 0, t h e re e xi s t s δ > 0 s u c h t h a t f (x) > ε whenever 0 < ⏐ x − a⏐ < δ. lim Then we write x → a f (x) = ∞ or f (x) tends to ∞ as x tends to a. Definition : A fu n ctio n f is said to a pproach − ∞ as x approaches a, if corresponding to any ε > 0 , there exists δ > 0 such that f (x) < − ε whenever 0 < ⏐ x − a⏐ < δ. lim Then we write x → a f (x) = − ∞ or f (x) tends to − ∞ as x tends to a. Example 1 : L et f be the function given by f (x) = lim Using (ε, δ) definition show that x → a f (x) = 2a. x2 − a2 , x ≠ a. x− a Solution : Let ε > 0 be given. In order to show that lim x → a f (x) = 2a, we have to show that for any given ε > 0, there exists a number δ > 0 such that ⏐ f (x) − 2a⏐ < ε whenever 0 < ⏐ x − a⏐ < δ. ⎪ x2 − a 2 ⎪ − 2a⎪ ⎪ x− a ⎪ If x ≠ a, then ⏐ f (x) − 2a⏐ = ⎪ = ⏐ x − a⏐ . = ⏐ (x + a) − 2a⏐ [... x ≠ a] D-15 LIMITS AND CONTINUITY ∴ ⏐ f (x) − 2a ⏐ < ε, if ⏐ x − a⏐ < ε. Choosing a number δ such that 0 < δ ≤ ε, we have ⏐ f (x) − 2a⏐ < ε whenever 0 < ⏐ x − a⏐ < δ. lim H ence x → a f (x) = 2a. 1⎞ lim ⎛ Example 2 : Using (ε, δ) definition show that x → 0 ⎜ x sin ⎟ = 0⋅ x⎠ ⎝ (Meerut 2012, 13; Rohilkhand 13B) 1⎞ lim ⎛ Solution : Let ε > 0 be given. In order to show that x → 0 ⎜ x sin ⎟ = 0, x⎠ ⎝ we have to show that for any given ε > 0, there exists a number δ > 0 such that 1 ⎪ ⎪ ⎪ x sin − 0⎪ < ε whenever 0 < ⏐ x − 0⏐ < δ. x ⎪ ⎪ 1 1⎪ 1⎪ ⎪ ⎪ ⎪ ⎪ Now ⎪ x sin − 0⎪ = ⏐ x ⏐ ⎪ sin ⎪ ≤ ⏐ x⏐ , because ⎪ sin ⎪ ≤ 1. x x⎪ x⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ∴ ⎪ x sin − 0⎪ < ε whenever ⏐ x⏐ < ε. x ⎪ ⎪ Choosing a number δ such that 0 < δ ≤ ε, we have 1 ⎪ ⎪ ⎪ x sin − 0⎪ < ε whenever 0 < ⏐ x⏐ < δ. x ⎪ ⎪ 1 lim H ence x → 0 x sin = 0. x Example 3 : Show by (ε, δ) m ethod that the function f, defined on R − {0} by f (x) = sin (1 ⁄ x) whenever x ≠ 0 , does not tend to 0 as x tends to 0. (Meerut 2013B) Solution : In order to show that sin (1 ⁄ x) does not tend to 0 as x tends to 0, take ε = 12 . By Archimedean property of real numbers for any δ > 0 there exists a positive integer n such that 1 1 i.e., δ > ⋅ nπ πδ 2 1 1 0< < < < δ. (4n + 1) π 2nπ nπ n> ∴ 2 ⋅ Then 0 < ⏐ x − 0⏐ < δ. (4n + 1) π Take x= Also, ⏐ sin (1 ⁄ x) − 0⏐ = ⏐ sin (2nπ + 1 2 π)⏐ = 1 > ε. 1 2 Thus we have shown that there exists an ε > 0, namely , such that for every ⎡ 2 , wh e r e n i s a p o sit i ve i n t e ge r su ch t h a t δ > 0 t h e r e i s a n x ⎢⎣ = (4n + 1) π ⎤ 2 < δ ⎥⎦ such that (4n + 1) π 0 < ⏐ x − 0⏐ < δ and ⏐ sin (1 ⁄ x) − 0⏐ > ε. H ence sin (1 ⁄ x) does not tend to 0 as x tends to 0. ⏐ x − 2⏐ lim does not exist. Example 4 : Show that x → 2 x− 2 D-16 DIFFERENTIAL CALCULUS Solution : Let f (x) = ⏐ x − 2⏐ ⁄ (x − 2)⋅ We have the right hand limit i.e., lim lim ⏐ 2 + h − 2⏐ f (2 + 0) = h → 0 f (2 + h) = h → 0 (2 + h − 2) lim h lim = h →0 = h →0 1 = 1 ; h and the left hand limit i.e., lim lim ⏐ 2 − h − 2⏐ f (2 − 0) = h → 0 f (2 − h) = h → 0 (2 − h − 2) h lim ⏐ − h⏐ lim lim = h →0 = h →0 = − 1 = − 1. − h − h h →0 lim ⏐ x − 2⏐ does not exist. Since f (2 + 0) ≠ f (2 − 0), hence x → 2 x− 2 Example 5 : Evaluate the following limits if they exist : (a) lim x2 + 3x + 2 ⋅ x →2 x− 2 Solution : H ere the right hand limit i.e., lim lim (2 + h) 2 + 3 (2 + h) + 2 f (2 + 0) = h → 0 f (2 + h) = h → 0 2+ h− 2 ⎞ lim 12 + 7h + h2 lim ⎛ 12 = h →0 = h →0 ⎜ + 7 + h⎟ = ∞ ; h ⎝ h ⎠ and the left hand limit i.e., lim lim (2 − h) 2 + 3 (2 − h) + 2 f (2 − 0) = h → 0 f (2 − h) = h → 0 2− h− 2 ⎞ lim 12 − 7h + h2 lim ⎛ 12 = h →0 = h →0 ⎜ − + 7 − h⎟ = − ∞ . − h h ⎝ ⎠ lim Since f (2 + 0) ≠ f (2 − 0), hence x → 2 f (x) does not exist. (b) lim 1⁄x x → 0 (1 + x) . Solution : H ere the right hand limit i.e., lim lim lim f (0 + 0) = h → 0 f (0 + h) = h → 0 f (h) = h → 0 (1 + h) 1 ⁄ h ⎡ ⎢ lim ⎢⎢⎢ = 1+ h → 0 ⎢⎣ lim ⎡ = ⎢1 + h →0 ⎣ 1 ⎛1 ⎞ ⎜ − 1⎟ ⎠ h ⎝h 1 ⎛1 ⎞ ⎛1 ⎞ ⎜ − 1⎟ ⎜ − 2⎟ ⎠ ⎝h ⎠ h ⎝h ⎤ ⎥ ⎥ 1 ⎥ ⋅h+ h2 + h3 + …⎥⎥ h 1.2 1.2.3 ⎦ 1.(1 − h) 1.(1 − h) (1 − 2h) 1 ⎤ + + + …⎥ 1! 2! 3! ⎦ 1 1 1 = 1+ + + + …∞ = e. 1! 2! 3! Similarly, the left hand limit i.e., lim lim lim f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) = h → 0 (1 − h) − 1 ⁄ h = e. D-17 LIMITS AND CONTINUITY Thus both f (0 + 0) and f (0 − 0) exist and are equal to e. lim H ence x → 0 (1 + x) 1 ⁄ x = e. (c) lim sin x x →0 x ⋅ Solution : Let H ere (Bundelkhand 2008; Kanpur 09) sin x ⋅ x f (x) = lim lim lim sin h f (0 + 0) = h → 0 f (0 + h) = f (h) = h →0 h →0 h h3 h5 + − … ⎞ h2 h4 3! 5! lim lim ⎛⎜ = = 1− + − …⎟ = 1. ⎠ h →0 h →0 ⎝ h 3! 5! lim lim f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) h− Similarly = lim sin (− h) lim sin h = h →0 = 1. h →0 − h h lim sin x f (0 + 0) = f (0 − 0) = 1, hence x → 0 = 1. x sin x lim (d) x →∞ x ⋅ lim sin x lim Solution : x → ∞ = y → 0 y sin (1 ⁄ y), putting x = 1 ⁄ y. x Let f ( y) = y sin (1 ⁄ y)⋅ lim We have, right hand limit i.e., f (0 + 0) = h → 0 f (0 + h) Since (Bundelkhand 2008) lim lim = h → 0 f (h) = h → 0 h sin (1 ⁄ h) = 0 × a finite quantity lying between − 1 and 1 = 0. lim Similarly, left hand limit i.e., f (0 − 0) = h → 0 f (0 − h) lim lim lim = h → 0 f (− h) = h → 0 (− h) sin (− 1 ⁄ h) = h → 0 h sin (1 ⁄ h) = 0. Since i.e., (e) lim f (0 + 0) = f (0 − 0) = 0, therefore y → 0 y sin (1 ⁄ y) = 0 lim sin x x → ∞ x = 0. 1 lim x → 0 sin x ⋅ Solution : Let f (x) = sin (1 ⁄ x). H ere 1 lim lim lim f (0 + 0) = h → 0 f (0 + h) = h → 0 f (h) = h → 0 sin ⋅ h As h → 0, the value of sin (1 ⁄ h) oscillates between + 1 and − 1, passing through zero and intermediate values an infinite number of times. Hence there is no definite D-18 DIFFERENTIAL CALCULUS number l to which sin (1 ⁄ h) tends as h tends to zero. Therefore the right hand limit f (0 + 0) does not exist. lim Similarly the left hand limit f (0 − 0) also does not exist. Thus x → 0 sin (1 ⁄ x) does not exist. (f) lim a x − 1 ⋅ x →0 x ax− 1 x 1 + x log a + (x2 ⁄ 2 !) (log a) 2 + … − 1 = x x [log a + 12 x (log a) 2 + …] = ⋅ x h ⎡ ⎤ h ⎢ log a + (log a) 2 + …⎥ ⎣ ⎦ 2 lim lim = log a. f (0 + 0) = h → 0 f (h) = h → 0 h Solution : Let H ere (Meerut 2003; Kanpur 10) f (x) = Also lim lim f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) = log a. Since lim a x − 1 f (0 + 0) = f (0 − 0) = log a, therefore x → 0 = log a. x lim 1 (g) x → 0 ⋅ e1 ⁄ x. x Solution : Let Then Also Since f (x) = 1 1⁄x .e . x lim lim lim 1 f (0 + 0) = h → 0 f (0 + h) = h → 0 f (h) = h → 0 e1 ⁄ h h = ∞ , since both 1 ⁄ h and e1 ⁄ h tend to ∞ as h → 0. 1 lim lim lim f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) = h → 0 − e− 1 ⁄ h h − 1 − 1 lim lim = h →0 1 ⁄ h = h →0 1 1 1 ⎛ ⎞ he ⋅ + …⎟ h ⎜1 + + h 2 ! h2 ⎝ ⎠ − 1 lim = 0. = h →0 h + 1 + (1 ⁄ 2h) + … lim 1 f (0 + 0) ≠ f (0 − 0) , therefore x → 0 e1 ⁄ x does not exist. x (h) lim (1 + x) n − 1 ⋅ x →0 x Let f (x) = Then lim lim (1 + h) n − 1 f (0 + 0) = h → 0 f (h) = h → 0 h (1 + x) n − 1 ⋅ x D-19 LIMITS AND CONTINUITY lim = h →0 lim = h →0 1 + nh + ⎡ ⎣ h ⎢n + n (n − 1) 2 h + …− 1 2! h n (n − 1) ⎤ h + …⎥ 2! ⎦ h n (n − 1) ⎤ lim ⎡ = h →0 ⎢ n + h + …⎥ = n. 2! ⎣ ⎦ Also lim lim f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) lim (1 − h) n − 1 = h →0 − h n (n − 1) (− h) 2 + … − 1 2! lim = n. = h →0 − h lim Since f (0 + 0) = f (0 − 0) = n, therefore x → 0 f (x) = n. 1 + n (− h) + (i) lim x m − a m x →a x − a ⋅ Solution : Let Then f (x) = xm − a m ⋅ x− a lim lim (a + h) m − am f (a + 0) = h → 0 f (a + h) = h → 0 a+ h− a lim = h →0 ⎡⎛ ⎣⎝ am ⎢ ⎜ 1 + h⎞ m ⎤ ⎟ − 1⎥ ⎦ a⎠ h am ⎡ 2 ⎤ lim ⎢ 1 + m ⋅ h + m (m − 1) h + … − 1⎥ = h →0 ⎢ ⎥ 2 ⎥ a 2! h ⎢⎣ a ⎦ m (m − 1) h m ⎡m ⎤ lim = h → 0 am ⎢ + . 2 + …⎥ = am ⋅ = mam − 1. 2 a a ⎣ ⎦ a Also lim lim (a − h) m − am f (a − 0) = h → 0 f (a − h) = h → 0 = mam − 1. a− h− a Since lim f (a + 0) = f (a − 0) = mam − 1, hence x → a f (x) = m am − 1. Example 6 : Find the right hand and the left hand limits in the following cases and discuss the existence of the lim it in each case : lim 2 x2 − 8 lim e1 ⁄ x − 1 (i) x → 2 ; (ii) x → 0 1 ⁄ x ; x− 2 e + 1 (Meerut 2003; Kanpur 11; Rohilkhand 14) D-20 DIFFERENTIAL CALCULUS lim (iii) x → 0 f (x) where f (x) is defined as f (x) = x, when x > 0 ; f (x) = 0, when x = 0 ; f (x) = − x, when x < 0. (Purvanchal 2008) Solution : (i) Let f (x) = 2x2 − 8 ⋅ x− 2 lim lim 2 (2 + h) 2 − 8 We have f (2 + 0) = h → 0 f (2 + h) = h → 0 2+ h− 2 lim 2(4 + 4h + h2) − 8 lim 8h + 2h2 = h →0 = h →0 h h h (8+ 2h) lim lim = h → 0 (8 + 2h) = 8. = h →0 h Again lim lim 2 (2 − h) 2 − 8 f (2 − 0) = h → 0 f (2 − h) = h → 0 2− h− 2 lim 2 (4 − 4h + h2) − 8 lim − 8h + 2h2 = h →0 = h →0 − h − h − h (8 − 2h) lim lim = h →0 = h → 0 (8 −. 2h) = 8 − h Since (ii) Let lim 2x2 − 8 exists and is equal to 8. f (2 + 0) = f (2 − 0) = 8, therefore x → 2 x− 2 f (x) = e1 ⁄ x − 1 ⋅ e1 ⁄ x + 1 H ere the right hand limit, i.e., lim lim lim e1 ⁄ h − 1 f (0 + 0) = h → 0 f (0 + h) = h → 0 f (h) = h → 0 1 ⁄ h + 1 e 1 ⁄ h 1 ⁄ h lim e [1 − (1 ⁄ e ] = h →0 1 ⁄ h = 1. [1 + (1 ⁄ e1 ⁄ h )] e Again the left hand limit, i.e., lim lim lim e− 1 ⁄ h − 1 f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) = h → 0 − 1 ⁄ h e + 1 1 ⁄ h (1 ) ⁄e − 1 0− 1 lim = h →0 = = − 1. (1 ⁄ e1 ⁄ h ) + 1 0 + 1 lim e1 ⁄ x − 1 f (0 + 0) ≠ f (0 − 0), hence x → 0 1 ⁄ x does not exist. e + 1 (iii) We have the right hand limit i.e., f (0 + 0) lim = h → 0 f (0 + h), where h is + ive but sufficiently small Since lim lim = h → 0 f (h) = h → 0 h, [... h > 0 and f (x) = x if x > 0] = 0. Also, the left hand limit, i.e., f (0 − 0) lim = h → 0 f (0 − h), where h is + ive but sufficiently small D-21 LIMITS AND CONTINUITY lim lim = h → 0 f (− h) = h → 0 − (− h), [... − h < 0 and f (x) = − x if x < 0] lim = h → 0 h = 0. Thus both the limits f (0 + 0) and f (0 − 0) exist and are equal to zero. lim H ence x → 0 f (x) exists and is equal to zero. Example 7 : L et ⎧ x if x is rational ⎩ − x if is irrational. f (x) = ⎨ lim Show that x → a f (x) exists only when a = 0. (Purvanchal 2007) Solution : Case I : If a is a non-zero rational number. lim In this case f (a − 0) = h → 0 f (a − h) lim = h → 0 (a − h) or lim h → 0 − (a − h), according as (a − h) is rational or irrational = a or − a i.e., is not unique. f (a − 0) does not exist. lim x → a f (x) does not exist. ∴ ∴ lim lim Case II : If a = 0. In this case f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) lim = h → 0 (− h) or lim h → 0 h, according as − h is rational or irrational = 0. Again lim lim f (0 + 0) = h → 0 f (0 + h) = h → 0 f (h) lim = h →0 h or lim h → 0 (− h), according as h is rational or irrational = 0. lim Since f (0 + 0) = f (0 − 0) = 0, hence x → 0 f (x) exists and is equal to zero. Case III : If a is an irrational number. lim In this case f (a − 0) = h → 0 f (a − h) lim = h → 0 (a − h) ∴ ∴ or lim h → 0 − (a − h), according as (a − h) is rational or irrational = a or − a i.e., is not unique. f (a − 0) does not exist. lim x → a f (x) does not exist. lim Thus we see that x → a f (x) exists only when a = 0. D-22 DIFFERENTIAL CALCULUS Example 8 : Discuss the existence of the lim it of the function f defined as f (x) = 1, if x < 1; f (x) = 2 − x, if 1 < x < 2; f (x) = 2, if x ≥ 2 at x = 1 and x = 2. Solution : At x = 1. We have lim f (1 + 0) = h → 0 f (1 + h), where h is + ive and sufficiently small lim lim = h → 0 [2 − (1 + h)] = h → 0 (1 − h) = 1; lim lim f (1 − 0) = h → 0 f (1 − h) = h → 0 (1) = 1. and Since lim f (1 + 0) = f (1 − 0) = 1, hence x → 1 f (x) exists and is equal to 1. lim lim At x = 2. We have f (2+ 0) = h → 0 f (2 + h) = h → 0 (2) = 2; lim lim lim f (2 − 0) = h → 0 f (2 − h) = h → 0 [2 − (2 − h)] = h → 0 h = 0. and Since lim f (2 + 0) ≠ f (2 − 0), hence x → 2 f (x) does not exist. lim Example 9 : If x → a f (x) = ± ∞ , then Solution : Let lim x → a f (x) = + ∞ . 1 lim x → a f (x) = 0. Let ε > 0 be given. If ε1 = 1 ⁄ ε, then ε1 > 0. lim Since x → a f (x) = ∞ , therefore for ε1 > 0, there exists δ > 0 such that f (x) > ε1 whenever 0 < ⏐ x − a ⏐ < δ 1 1 < whenever 0 < ⏐ x − a ⏐ < δ f (x) ε1 1 i.e., 0< < ε whenever 0 < ⏐ x − a ⏐ < δ [... ε = 1 ⁄ ε1] f (x) 1 < ε whenever 0 < ⏐ x − a ⏐ < δ i.e., − ε< f (x) ⎪ 1 ⎪ − 0⎪ < ε whenever 0 < ⏐ x − a ⏐ < δ. i.e, ⎪ ⎪ f (x) ⎪ 1 lim ∴ x → a f (x) = 0⋅ Similarly it can be proved that 1 lim lim x → a f (x) = 0 when x → a f (x) = − ∞ . sin [x] Example 10 : If f (x) = , [x] ≠ 0 and f (x) = 0, [x] = 0, [x] lim where [x] denotes the greatest integer less than or equal to x, then find x → 0 f (x). i.e., (Kanpur 2010) D-23 LIMITS AND CONTINUITY lim lim Solution : H ere f (0 + 0) = h → 0 f (0 + h) = h → 0 f (h) lim = h →0 0 [... [h] = 0] = 0. Also lim lim f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) lim sin [− h] , = h →0 [ [ ... [− h] = − 1 ≠ 0 ] − h] lim sin (− 1) sin (− 1) = = sin 1 ≠ 0. = h →0 (− 1) (− 1) lim Since f (0 + 0) ≠ f (0 − 0), therefore x → 0 f (x) does not exist. 1. lim U sing definition of limit, show that x → 0 f (x) = 1 where 2. if x ≠ 0 if x = 0. ⎩ ⎧ 2, if x is irrational If f is defined on R as f (x) = ⎨ ⎩ 1, if x is rational, lim prove that x → a f (x) does not exist for any a ∈ R. ⎧ ⎪ 1 + x2 f (x) = ⎨⎪ 3. 4. 0 ⎧ 0, if x is irrational If f is defined on R as f (x) = ⎨ ⎩ 1, if x is rational, lim prove that x → a f (x) does not exist for any a ∈ R. If x → 0, then does the limit of the following function f exist or not ? f (x) = x, when x < 0 ; f (x) = 1, when x = 0 ; f (x) = x2, when x > 0. 5. 6. 7. 8. 9. 2x − 1 ax− 1 lim lim = log a to find x → 0 ⋅ U se the formula x → 0 x (1 + x) 1 ⁄ 2 − 1 If f (x) = e− 1 ⁄ x, show that at x = 0, the right hand limit is zero while the left hand limit is + ∞ , and thus there is no limit of the function at x = 0. lim G ive an example to show that x → a f (x) may exist even when the function is not defined for x = a. ⎧ x, 0≤ x< 1 Let f (x) = ⎨⎩ 3 − x, 1 ≤ x ≤ 2. lim Show that x → 1 + f (x) = 2. Does the limit of f (x) at x = 1 exist ? G ive reasons for your answer. x − ⏐ x⏐ lim E valuate x → 0 ⋅ x (Meerut 2001) D-24 DIFFERENTIAL CALCULUS ⏐ sin x⏐ lim 10. E valuate x → 0 ⋅ x lim 11. E valuate x → 0 e1 ⁄ x e1 ⁄ x + 1 ⋅ (Avadh 2010) 12. If f (x) = a0 x n + a1 x n − 1 + a2 x n − 2 + … + an , then prove that lim x → a f (x) = f (a). (Garhwal 2011) 4. lim Yes; x → 0 f (x) = 0. 9. R ight hand limit is 0 and left hand limit is 2 and so the limit does not exist. 5. 2 log 2. 8. Does not exist. 10. Does not exist because the right hand limit is 1 and the left hand limit is − 1. 11. The limit does not exist because the right hand limit is 1 and the left hand limit is 0. 1.7 Continuity (Purvanchal 2010, 11; Avadh 14) The intuitive concept of continuity is derived from geometrical considerations. If the graph of the function y = f (x) is a continuous curve, it is natural to call the function continuous. This requires that there should be no sudden changes in the value of the function. A small change in x should produce only a small change in y. Moreover for the graph to be a continuous running curve, it should possess a definite direction at each point. But the continuity as defined in pure analysis is quite distinct from the intuitive or the geometrical concept of the term. Sometimes drawing a graph is difficult. We now give the arithmetical definition of continuity given by Cauchy. Cauchy’s definition of continuity : A real valued function f defined on an open interval I is said to be continuous at a ∈ I iff for any arbitrarily chosen positive number ε , however sm all, we can find a corresponding num ber δ > 0 such that ⏐ f (x) − f (a)⏐ < ε whenever ⏐ x − a⏐ < δ. …(1) (Bundelkhand 2010; Kanpur 11) that We say that f is a continuous function if it is continuous at every x ∈ I. In other words, f is continuous at a if for any given ε > 0, we can find a δ > 0 such ⏐ x − a⏐ < δ ⇒ ⏐ f (x) − f (a)⏐ < ε. This means that the function f will be continuous at x = a if the difference between f (a) and the value of f (x) at any point in the interval ]a − δ, a + δ[ can be made less than a pre-assigned positive number ε. Note that we choose δ after we have chosen ε. D-25 LIMITS AND CONTINUITY Geometrical Interpretation of Continuity : A geometrical interpretation of the above definition is immediate. Corresponding to any pre-assigned positive number ε, we can determine an interval of width 2δ about t he p o in t x = a (see the figure) such t hat for any point x lyin g in t he in terval ]a − δ, a + δ[, f (x) is confined to lie between f (a) − ε and f (a) + ε . The inequality (1) may be written in the form of an equality as f (x) = f (a) + η, where ⏐ η ⏐ < ε. Note 1 : For a function f (x) to be continuous at x = a, it is necessary that lim x → a f (x) must exist. Note 2 : The function must be defined at the point of continuity. Note 3 : The value of δ depends upon the values of ε and a. Note 4 : The interval I may be of any one of the forms : ]a, b[ , ] − ∞ , b [ , ]a, ∞ [ , ] − ∞ , ∞ [. An alternative definition of continuity : A function f is said to be continuous at lim a ∈ I i f f x → a f (x) exists, is finite and is equal to f (a) otherwise the function is discontinuous at x = a. This definition of continuity follows immediately from the definition of limit and th e definition of continuity. Thus a function f is said to be continuous at a, if f (a + 0) = f (a − 0) = f (a). This is a working formula for testing the continuity of a (Bundelkhand 2008, 10; Kashi 12) function at a given point. n n − 1+ …+ a Important Remark : I f f (x) = a0 x + a1 x n − 1 x + an i s a polynomial in x of degree n, then by the above definition it can be easily seen that f (x) is continuous for all x ∈ R. If c be any real number, then lim lim n n− 1+ …+ a x → c f (x) = x → c {a0 x + a1 x n − 1 x + an} lim lim lim lim = a0 x → c x n + a1 x → c x n − 1 + … + a n − 1 x → c x + x → c an = a0 c n + a1 c n − 1 + … + a n − 1 c + an = f (c). ⎡ ⎢ ⎣ ... lim x = c⎤⎥ x →c ⎦ D-26 DIFFERENTIAL CALCULUS lim Since x → c f (x) = f (c), therefore f (x) is continuous at x = c. Thus f (x) is continuous at every real number c and so f (x) is continuous for all x ∈ R. Thus remember that a polynomial function f (x) is always continuous at each point of its domain. Continuity from left and continuity from right : L et f be a function defined on an open interval I and let a ∈ I. We say that f is lim continuous from the left at a if x → a − 0 f (x) exists and is equal to f (a). Sim ilarly f is said lim to be continuous from the right at a if x → a + 0 f (x) exists and is equal to f (a). From the above definitions it is clear that for a function f to be continuous at a, it is necessary as well as sufficient that f be continuous from the left as well as from the right at a. Continuous function : A function f is said to be a continuous function if it is continuous at each point of its domain. Continuity in an open interval : A function f is said to be continuous in the open interval ]a, b[ if it is continuous at each point of the interval. (Bundelkhand 2009) Continuity in a closed interval : Let f be a function defined on the closed interval [a, b]. We say that f is continuous at a if it is continuous from the right at a and also that f is continuous at b if it is continuous from the left at b. Further, f is said to be continuous on the closed interval [a, b], if (i) it is continuous from the right at a, (ii) continuous from the left at b and (iii) continuous on the open interval ]a, b[. Thus if a function f is defined on the closed interval [a, b], then (i) it is continuous at the left end point a if f (a) = f (a + 0) lim i.e., f (a) = x → a + 0 f (x) i.e., (ii) it is continuous at the right end point b if f (b) = f (b − 0) lim f (b) = x → b − 0 f (x) and (iii) it is continuous at an interior point c of [a, b] i.e., at c ∈ ] a, b [ if lim lim f (c − 0) = f (c) = f (c + 0) i.e., if x → c − 0 f (x) = f (c) = x → c + 0 f (x). 1.8 Discontinuity Definition : If a function is not continuous at a point, then it is said to be discontinuous at that point and the point is called a point of discontinuity of this function. Types of discontinuity : (Avadh 2014) (i) Removable discontinuity : (Meerut 2011) lim A function f is said to have a rem ovable discontinuity at a point a if f (x) x →a exists but is not equal to f (a) i.e., if f (a + 0) = f (a − 0) ≠ f (a). The function can be made continuous by defining it in such a way that D-27 LIMITS AND CONTINUITY lim x → a f (x) = f (a). (Meerut 2010B) (ii) Discontinuity of the first kind or ordinary discontinuity : A function f is said to have a discontinuity of the first kind or ordinary discontinuity at a if f (a + 0) and f (a − 0) both exist but are not equal. The point a is said to be a point of discontinuity from the left or right according as f (a − 0) ≠ f (a) = f (a + 0) or f (a − 0) = f (a) ≠ f (a + 0). (iii) Discontinuity of the second kind : A funct ion f i s sa i d t o h a ve a discontinuity of the second kind, at a if none of the limits f (a + 0) and f (a − 0) exist. The point a is said to be a point of discontinuity of the second kind from the left or right according as f (a − 0) or f (a + 0) does not exist. (Meerut 2003, 10B) (iv) Mixed discontinuity : (Meerut 2012B) A function f is said to have a m ixed discontinuity at a, if f has a discontinuity of second kind on one side of a and on the other side a discontinuity of first kind or may be continuous. (v) Infinite discontinuity : A function f is said to have an infinite discontinuity at a if f (a + 0) or f (a − 0) is + ∞ or − ∞ . O bviously, if f has a discontinuity at a and is unbounded in every neighbourhood of a, then f is said to have an infinite discontinuity at a. 1.9 Jump of a Function at a Point If both f (a + 0) and f (a − 0) exist, then the jump in the function at a is defined as the non-negative difference f (a + 0) ~ f (a − 0). A function having a finite number of jumps in a given interval is called piecewise continuous or sectionally continuous. Example 1 : Test the following functions for continuity : (i) f (x) = x sin (1 ⁄ x), x ≠ 0, f (0) = 0 at x = 0. (Kanpur 2005; Avadh 08; Meerut 09B; Purvanchal 09; Kashi 12; Rohilkhand 14) A lso draw the graph of the function. (ii) f (x) = 21 ⁄ x when x ≠ 0, f (0) = 0 at x = 0. (iii) f (x) = 1 ⁄ (1 − e− 1 ⁄ x), x ≠ 0, f (0) = 0 at x = 0. lim Solution : (i) H ere f (0 + 0) = h → 0 f (0 + h), h > 0 1 lim lim = h → 0 f (h) = h → 0 h sin = 0. h [See theorem 10 of article 1.3] ⎡ . . lim 1 ⎢ . h → 0 h = 0 and ⎪⎪ sin ⎪⎪ ≤ 1 for all h ≠ 0 i.e., sin (1 ⁄ h) is bounded ⎣ h⎪ ⎪ ⎤ in some deleted neighbourhood of zero ⎥⎦ lim Similarly f (0 − 0) = h → 0 f (0 − h), h > 0 1 ⎛ 1 ⎞ lim lim lim = h → 0 f (− h) = h → 0 (− h) sin ⎜ ⎟ = h → 0 h sin = 0, as before. h ⎝ − h⎠ Also f (0) = 0. D-28 x = 0. DIFFERENTIAL CALCULUS Thus f (0 − 0) = f (0) = f (0 + 0). ∴ t he fun ctio n f (x) is con tinu ous at To draw the graph of the function we put y = f (x). So the graph of the function is the curve y = x sin (1 ⁄ x), x ≠ 0 and y = 0 when x = 0. If we put − x in place of x, the equation of this curve does not change and so this curve is symm et rical about t he y-axis a n d it is sufficient to draw the graph when x > 0. Also ⏐ f (x)⏐ = ⏐ x sin (1 ⁄ x)⏐ = ⏐ x⏐ ⋅ ⏐ sin (1 ⁄ x)⏐ ≤ ⏐ x⏐ . [... ⏐ sin (1 ⁄ x)⏐ ≤ 1] ∴ for all x the curve y = x sin (1 ⁄ x) lies between the lines y = x and y = − x. E xcluding origin the curve meets the y-axis at the points where 1 1 1 1 , 1 , sin = 0 i.e., where = π, 2π, 3π, … i.e., where x = π , …. x x 2π 3π 1 π 5π , 9π , 1 … Also y = x at the points where sin = 1 i.e., = , x 2 2 2 x 2 2 , 2 , i.e., x= π , … 5π 9π 1 3π , 7π , 1 … and y = − x at the points where sin = − 1 i.e., = x 2 2 x 2 , 2 , i.e., x= …. 3π 7π 1 1⎞ ⎛ 1 ⎞ 1 1 1 dy ⎛ = sin + x ⎜ cos ⎟ ⎜ − 2 ⎟ = sin − cos ⋅ We have x x⎠ ⎝ x ⎠ x x x dx ⎝ So at the points where sin (1 ⁄ x) = 1, we have cos (1 ⁄ x) = 0 and dy ⁄ dx = 1 i.e., at these points the curve touches the straight line y = x. Similarly at the points where sin (1 ⁄ x) = − 1, the curve touches the straight line y = − x. 1 lim [Form ∞ × 0] Also x → ∞ x sin x ⎡ sin (1 ⁄ x) 0⎤ lim ⎢ Form ⎥ = x →∞ ⎣ 0⎦ 1⁄x sin θ , 1 lim = θ →0 θ putting x = θ so that θ → 0 as x → ∞ = 1. Thus y → 1 as x → ∞ and so the straight line y = 1 is an asymptote of the curve. Although the function is continuous at the origin, yet the graph of the function in the vicinity of the origin cannot be drawn, since the function oscillates infinitely often in any interval containing the origin. F rom t he graph it is clear that the function makes an infinite number of oscillations in the neighbourhood of x = 0. The oscillations, however, go on diminishing in length as x → 0. LIMITS AND CONTINUITY D-29 Note 1 : If we are to check the continuity of f (x) at any point x = c, where 1 1 lim lim c ≠ 0, t hen we see that x → c f (x) = x → c x sin = c sin = f (c) a nd so f (x) is c x continuous at x = c. Thus f (x) is continuous for all x ∈ R i.e., f (x) is continuous on the whole real line. Note 2 : If we take f (0) = 2, the function becomes discontinuous at x = 0 and has a removable discontinuity at x = 0. (ii) H ere lim lim f (0 + 0) = h → 0 f (0 + h) = h → 0 21 ⁄ h = 2∞ = ∞ , lim lim f (0 − 0) = h → 0 f (0 − h) = h → 0 2− 1 ⁄ h = 2− ∞ = 0, and f (0) = 0. Since f (0 + 0) ≠ f (0 − 0), therefore the function is discontinuous at the origin. It has an infinite discontinuity there. 1 lim lim = 1, (iii) H ere f (0 + 0) = h → 0 f (0 + h) = h → 0 1 − e− 1 ⁄ h 1 lim lim = 0. f (0 − 0) = h → 0 f (0 − h) = h → 0 1 − e1 ⁄ h Si n ce f ( 0 + 0) ≠ f (0 − 0), h e n c e f (x) i s d isco n t in u o u s a t x = 0 a n d h a s discontinuity of the first kind. This function has a jump of one unit at 0 since f (0 + 0) − f (0 − 0) = 1. Example 2 : Consider the function f defined by f (x) = x − [ x ], where x is a positive variable and [ x ] deontes the integral part of x and show that it is discontinuous for integral values of x and continuous for all others. Draw its graph. Solution : From the definition of the function f (x), we have f (x) = x − (n − 1) for n − 1 < x < n, f (x) = 0 for x = n, f (x) = x − n for n < x < n + 1, where n is an integer. We shall test the function f (x) for continuity at x = n. We have f (n) = 0; lim lim f (n + 0) = h → 0 f (n + h) = h → 0 {(n + h) − n} [... n < n + h < n + 1] lim = h →0 h = 0 ; and lim lim f (n− 0) = h → 0 f (n− h) = h → 0 {(n− h)− (n− 1)} [... n − 1 < n − h < n] lim = h → 0 (1 − h) = 1; Since f (n + 0) ≠ f (n − 0), the function f (x) is discontinuous at x = n. Thus f (x) is discontinuous for all integral values of x. It is obviously continuous for all other values of x. Since x is a positive variable, putting n = 1, 2, 3, 4, 5, … we see that the graph of f (x) consists of the following straight lines : y = x when 0 < x < 1, y = 0 when x = 1 y = x − 1 when 1 < x < 2, y = 0 when x = 2 y = x − 2 when 2 < x < 3, y = 0 when x = 3 D-30 y = x − 3 when 3 < x < 4, The graph of the function thus obtained is shown by thick lines from x = 0 to x = 4. From the graph it is evident that : (i) T h e fu n c t i o n is discontinuous for all integral values of x but continuous for other values of x. (ii) The function is bounded be t wee n 0 an d 1 in every domain which includes an integer. (iii) T h e l o we r b o u n d 0 i s attained but the upper bound 1 is not attained since f (x) ≠ 1 for any value of x. DIFFERENTIAL CALCULUS y = 0 when x = 4 and so on. Example 3 : Show that the function f (x) = [x] + [− x] has rem ovable discontinuity for integral values of x. (Kanpur 2009) Solution : We observe that f (x) = 0, when x is an integer and f (x) = − 1, when x is not an integer. Hence if n is any integer, we have f (n − 0) = f (n + 0) = − 1 and f (n) = 0. So the function f (x) has a removable discontinuity at x = n, where n is an integer. Example 4 : L et y = E (x) , where E (x) denotes the integral part of x. Prove that the function is discontinuous where x has an integral value. A lso draw the graph. Solution : From the definition of E (x), we have E (x) = n − 1 for n − 1 ≤ x < n, E (x) = n for n ≤ x < n + 1 E (x) = n + 1 for n + 1 ≤ x < n + 2, and so on where n is an integer. We consider x = n. Then E (n) = n, E (n − 0) = n − 1 and E (n + 0) = n. Since E (n + 0) ≠ E (n − 0), the function E (x) is discontinuous at x = n i.e.,when x has an integral value. E vidently it is continuous for all other values of x. To draw the graph, we put n = …, − 4, − 3, − 2, − 1, 0, 1, 2, 3, 4, … , so that y = − 4, when − 4 ≤ x < − 3, y = − 3, when − 3 ≤ x < − 2, y = − 2, when − 2 ≤ x < − 1, y = − 1, when − 1 ≤ x < 0, y = 0, when 0≤ x< 1 y = 1, when 1≤ x< 2 y = 2, when 2≤ x< 3 y = 3, when 3≤ x< 4 y = 4, when 4 ≤ x < 5 and so on. The graph is shown by thick lines. D-31 LIMITS AND CONTINUITY Example 5 : Show that the function φ defined as ⎧0 for x = 0 ⎪ ⎪ 1 − x for 0 < x < 1 ⎪2 2 ⎪1 for x = 12 φ (x) = ⎨ 2 ⎪3 ⎪ 2 − x for 12 < x < 1 ⎪ ⎪1 for x = 1 ⎩ has three points of discontinuity which your are required to find. A lso draw the graph of the (Rohilkhand 2009; Avadh 10, 13) function. Solution : H ere the domain of the function φ (x) is the closed interval [0, 1]. When 0 < x < 12 , φ (x) = 12 − x which is a polynomial in x of degree 1. We know that a polynomial function is continuous at each point of its domain and so φ (x) is continuous at each point of the open interval 0 < x < 12 ⋅ Again when 12 < x < 1, φ (x) = 32 − x which is also a polynomial in x and so φ (x) is also continuous at each point of the open interval 12 < x < 1. 1 2 Now it remains to test the function φ (x) for continuity at x = 0, and 1. For x = 0, we have φ (0) = 0, lim lim lim φ (0 + 0) = h → 0 φ (0 + h) = h → 0 φ (h) = h → 0 ⎛⎜ 12 − h⎞⎟ = 12 ⋅ ⎝ ⎠ Since φ (0) ≠ φ (0 + 0), the function φ (x) is discontinuous at x = 0 and the discontinuity is ordinary. (ii) For x = 12 , we have φ ⎛⎜ 12 ⎞⎟ = 12 , (i) ⎝ ⎠ ⎞⎤ lim lim ⎡ 1 ⎛ 1 φ ⎛⎜ 12 − 0⎞⎟ = h → 0 φ ⎛⎜ 12 − h⎞⎟ = h → 0 ⎢ − ⎜ − h⎟ ⎥ , ⎡⎢ Note that 0 < ⎝ ⎠ ⎝ ⎠ ⎣2 ⎝2 ⎠⎦ ⎣ lim = h → 0 h = 0. 1 2 − h < 12 ⎤⎥ ⎦ Since φ ⎛⎜ 12 − 0⎞⎟ ≠ φ ⎛⎜ 12 ⎞⎟ , the function φ (x) is discontinuous from the left at ⎝ ⎠ ⎝ ⎠ x = 1 ⁄ 2. lim Again φ ( 12 + 0) = h → 0 φ ( 12 + h), h > 0 D-32 DIFFERENTIAL CALCULUS lim = h → 0 ⎡⎢ 32 − ⎛⎜ 12 + h⎞⎟ ⎤⎥ ⎣ ⎝ ⎠⎦ lim = h → 0 (1 − h) = 1 ≠ φ ⎛⎜ 12 ⎞⎟ = ⎝ ⎠ ⎡ ... ⎢ ⎣ 1 2 1 2 < 1 2 + h < 1⎤⎥ ⎦ ⋅ Thus the function φ (x) is discontinuous from the right also at x = 12 ⋅ In this way φ (x) has discontinuity of the first kind i.e., ordinary discontinuity at x = 12 a n d t h e j u m p o f t h e fu n c t i o n a t x = 1 ⁄ 2 i s φ ⎛⎜ 12 + 0⎞⎟ − φ ⎛⎜ 12 − 0⎞⎟ i.e., ⎝ ⎠ ⎝ ⎠ 1 − 0 i.e., 1. (iii) For x = 1, we have φ (1) = 1, lim φ (1 − 0) = h → 0 φ (1 − h) lim = h → 0 [(3 ⁄ 2) − (1 − h)], ⎡ Note that ⎢ ⎣ 1 2 < 1 − h < 1⎤⎥ lim = h → 0 ⎛⎜ 12 + h⎞⎟ = 12 ⋅ ⎝ ⎠ Sin ce φ (1) ≠ φ (1 − 0), φ (x) is discon tinu ous at x = 1 and the discont inuity is ordinary. H ence the function φ (x) has three points of discontinuity at x = 0, 12 and 1. The graph of the function consists of the point (0, 0); t he segment of the line y = 12 − x, 0 < x < 12 ; the point ⎛⎜ 12 , 12 ⎞⎟ ; the segment of the line ⎝ ⎠ y = 32 − x, 12 < x < 1; and the point (1, 1). Thus the graph is as shown in the figure. From the graph we observe that the function is discontinuous at x = 0, 12 and 1. Example 6 : Determ ine the values of a, b, c for which the function f (x) = ⎧ sin (a + 1) x + ⎪ x ⎪ ⎪ ⎪ c ⎨ ⎪ ⎪ ⎪ (x + bx2) 1 ⁄ 2 − ⎪ ⎩ bx3 ⁄ 2 is continuous at x = 0. sin x for x < 0 for x = 0 x1 ⁄ 2 for x > 0 lim lim (h + bh2) 1 ⁄ 2 − h1 ⁄ 2 Solution : H ere f (0 + 0) = h → 0 f (0 + h) = h → 0 bh3 ⁄ 2 1 1⁄2 − 1 lim (1 + bh) lim {1 + 2 bh + …} − 1 1 , = h →0 = h →0 = 2 bh bh which is independent of b and so b may have any real value except 0. lim lim sin (a + 1) (− h) + sin (− h) Again f (0 − 0) = h → 0 f (0 − h) = h → 0 (− h) 2 sin ⎛⎜ 12 a + 1⎞⎟ h cos (ah ⁄ 2) ⎝ ⎠ lim sin (a + 1) h + sin h lim = h →0 = h →0 h h ⎦ D-33 LIMITS AND CONTINUITY i.e., lim sin {(a + 2) ⁄ 2} h = h →0 (a + 2) cos (ah ⁄ 2) = a + 2. {(a + 2) ⁄ 2} h For continuity at x = 0, we have f (0 + 0) = f (0 − 0) = f (0) 1 = a + 2 = c. ∴ c = 12 and a = − 32 ⋅ 2 Example 7 : A function f (x) is defined as follows : ⎧ (x2 ⁄ a) − a, when x < a ⎪ f (x) = ⎨ when x = a 0, ⎪ ⎩ a − (a 2 ⁄ x), when x > a. Prove that the function f (x) is continuous at x = a. (Bundelkhand 2007; Avadh 09; Rohilkhand 13) a2 ⎤ lim lim ⎡ ⎥, Solution : We have f (a + 0) = h → 0 f (a + h) = h → 0 ⎢ a − ⎣ (a + h) ⎦ [... f (x) = a − (a2 ⁄ x) for x > a] = [a − (a2 ⁄ a)] = a − a = 0 ; 2 ⎤ lim lim ⎡ (a − h) f (a − 0) = h → 0 f (a − h) = h → 0 ⎢ − a⎥ , [... f (x) = (x2 ⁄ a) − a for x < a] ⎣ ⎦ a = [(a2 ⁄ a) − a] = a − a = 0. Also, we have f (a) = 0. Since f (a + 0) = f (a − 0) = f (a), therefore f (x) is continuous at x = a. Example 8 : Exam ine the function defined below for continuity at x = a : ⎛ 1 ⎞ 1 f (x) = cosec ⎜ ⎟ ,x≠ a ⎝ x − a⎠ x− a f (x) = 0, x = a. (Avadh 2004) Solution : We have lim f (a + 0) = h → 0 f (a + h) 1 1 1 lim lim = h →0 cosec = a+ h− a a + h − a h → 0 h sin (1 ⁄ h) = + ∞ , since h sin (1 ⁄ h) → 0 as h → 0. 1 1 ⎛ ⎞ lim lim f (a − 0) = h → 0 f (a − h) = h → 0 cosec ⎜ ⎟ a− h− a ⎝ a − h − a⎠ 1 1 ⎡1 ⎤ lim lim = h →0 − ⎢ ⋅ ⎥ = h →0 h sin (1 ⁄ h) ⎣ h sin {− (1 ⁄ h)} ⎦ = + ∞, since h sin (1 ⁄ h) → 0 as h → 0. Also, we have f (a) = 0. Since f (a + 0) = f (a − 0) ≠ f (a), the function f (x) is discontinuous at x = a, having an infinite discontinuity of the second kind. Example 9 : Examine the function defined below for continuity at x = 0 : sin2 ax for x ≠ 0, f (x) = 1 for x = 0. x2 Solution : We have f (0) = 1 ; lim lim lim sin2 ah f (0 + 0) = h → 0 f (0 + h) = h → 0 f (h) = h → 0 h2 f (x) = (Meerut 2010) D-34 DIFFERENTIAL CALCULUS lim ⎛ sin ah ⎞ 2 2 = h →0 ⎜ ⎟ . a = 1 . a2 = a2 ; ⎝ ah ⎠ and lim lim lim sin2 (− ah) f (0 − 0) = h → 0 f (0 − h) = h → 0 f (− h) = h → 0 (− h) 2 lim sin2 ah = a 2. = h →0 h2 Now f (x) is continuous at x = 0 iff f (0 + 0) = f (0 − 0) = f (0). H ence f (x) is discontinuous at x = 0 unless a = 1. Example 10 : A function f (x) is defined as follows : f (x) = 1 + x if x ≤ 2 and f (x) = 5 − x if x ≥ 2. Is the function continuous at x = 2 ? (Meerut 2002, 06) Solution : H ere f (2) = 1 + 2 or 5 − 2 = 3 ; lim f (2 + 0) = h → 0 f (2 + h), where h is + ive and sufficiently small lim = h → 0 [5 − (2 + h)], [... 2 + h > 2 and f (x) = 5 − x if x > 2] lim = h → 0 (3 − h) = 3 ; lim and f (2 − 0) = h → 0 f (2 − h), where h is + ive and sufficiently small lim = h → 0 [1 + (2 − h)], [... 2 − h < 2 and f (x) = 1 + x if x < 2] lim = h → 0 (3 − h) = 3. Thus f (2 + 0) = f (2 − 0) = f (2). H ence the function f (x) is continuous at x = 2. Example 11 : Discuss the continuity of the function f (x) defined as follows: f (x) = x2 for x < − 2, f (x) = 4 for − 2 ≤ x ≤ 2, f (x) = x2 for x > 2. Solution : We shall test the continuity of f (x) only at the points x = − 2 and 2. O bviously it is continuous at all other points. At x = − 2. We have f (− 2) = 4; lim lim f (− 2 + 0) = h → 0 f (− 2 + h) = h → 0 4 = 4 ; lim lim f (− 2 − 0) = h → 0 f (− 2 − h) = h → 0 (− 2 − h) 2, [... − 2 − h < − 2] = 4. Since f (− 2 + 0) = f (− 2 − 0) = f (− 2), the function is continuous at x = − 2. At x = 2. We have f (2) = 4 ; lim lim f (2 + 0) = h → 0 f (2 + h) = h → 0 (2 + h) 2 = 4 ; lim lim f (2 − 0) = h → 0 f (2 − h) = h → 0 4 = 4. Since f (2 + 0) = f (2 − 0) = f (2), the function is continuous at x = 2. D-35 LIMITS AND CONTINUITY 1.10 Algebra Of Continuous Functions Theorem 1 : L et f and g be defined on an interval I. If f and g are continuous at a ∈ I, then f + g is also continuous at a. Theorem 2 : L et f and g be defined on an interval I. If f and g are continuous at a ∈ I, then fg is continuous at a. Theorem 3 : If f is continuous at a point a and c ∈ R , then cf is continuous at a. Theorem 4 : L et f and g be defined on an interval I, and let g (a) ≠ 0. If f and g are continuous at a ∈ I, then f ⁄ g is continuous at a . Theorem 5 : If f is continuous at a then ⏐ f ⏐ is also continuous at a. Note : The converse is not true. For example, if f (x) = − 1, for x < a and f (x) = 1 for x ≥ a then lim lim x → a ⏐ f (x)⏐ = 1 = ⏐ f (a)⏐ , but x → a f (x) does not exist. Thus ⏐ f ⏐ is continuous at a while f is not continuous at a. 1. 2. Discuss the continuity and discontinuity of the following functions : (ii) f (x) = x + x − 1. (i) f (x) = x3 − 3x. − 1 ⁄ x (iii) f (x) = e . (iv) f (x) = sin x. (v) f (x) = cos (1 ⁄ x) when x ≠ 0, f (0) = 0. (vi) f (x) = sin (1 ⁄ x) when x ≠ 0, and f (0) = 0. (Lucknow 2011) sin x (vii) f (x) = when x ≠ 0 and f (0) = 1. (Kanpur 2007; Avadh 08) x e1 ⁄ x − 1 (viii) f (x) = 1 ⁄ x when x ≠ 0 and f (0) = 1. (Meerut 2004B) e + 1 e1 ⁄ x (ix) f (x) = when x ≠ 0, f (0) = 0. (Bundelkhand 2011) 1 + e1 ⁄ x xe1 ⁄ x (x) f (x) = + sin (1 ⁄ x) when x ≠ 0, f (0) = 0. 1 + e1 ⁄ x (xi) f (x) = sin x cos (1 ⁄ x) when x ≠ 0, f (0) = 0. (i) Discuss the continuity of f (x) at x = 0, if f (x) = ⎧ ⎪ ⎪ ⎪1 ⎨ ⎪ ⎪ ⎪ ⎩ 2 e1/x 2 − e1/x 1 , when x ≠ 0 , when x = 0 (Meerut 2008) 3. 1 1 , (ii) If f (x) = sin find f (a + 0) and f (a − 0). x− a x− a Is the function continuous at x = a? Find out the points of discontinuity of the following functions : (i) f (x) = (2 + e1 ⁄ x ) − 1 + cos e1 ⁄ x for x ≠ 0, f (0) = 0. (ii) f (x) = 1 ⁄ 2n for 1 ⁄ 2n + 1 < x ≤ 1 ⁄ 2n , n = 0, 1, 2, … and f (0) = 0. D-36 4. 5. DIFFERENTIAL CALCULUS 1 1 sin for x ≠ 0 and f (0) = 0, show that f (x) is finite for every value of x x x in the interval [− 1, 1] but is not bounded. Determine the points of discontinuity of the function if any. ⎧ x, if x is rational A function f defined on [0, 1] is given by f (x) = ⎨ ⎩ 1 − x, if x is irrational. If f (x) = Show that f takes every value between 0 and 1 (both inclusive), but it is continuous only at the point x = 1 2 ⋅ (Rohilkhand 2012B) 6. Prove that the function f defined by ⎧1 , ⎪ 2 if x is rational f (x) = ⎨ 1 ⎪ , if x is irrational ⎩3 is discontinuous everywhere. 7. (i) Show that the function f defined by f (x) = 8. 9. xe1 ⁄ x , x ≠ 0, f (0) = 1 1 + e1 ⁄ x is not continuous at x = 0 and also show how the discontinuity can be (Meerut 2011; Rohilkhand 06) removed. (ii) Show that the function f (x) = 3x2 + 2x − 1 is continuous for x = 2. (iii) S h o w t h a t t h e fu n ct i o n f (x) = (1 + 2x) 1 ⁄ x, x ≠ 0, f (x) = e2, x = 0 i s continuous at x = 0. E xamine the continuity of the function ⎧ − x2 if x ≤ 0 ⎪ ⎪ 5x − 4 if 0 < x ≤ 1 f (x) = ⎨ 2 ⎪ 4 x − 3x if 1 < x < 2 ⎪ 3x + 4 if x ≥ 2 ⎩ at x = 0, 1 and 2. (Meerut 2004, 06B, 07B; Avadh 06; Purvanchal 06, 10) (i) Show that the function e1 ⁄ x − 1 , f (x) = 1 ⁄ x x ≠ 0 and f (0) = 0 e + 1 is discontinuous at x = 0. Show that the following function is continuous at x = 0. f (x) = 10. sin− 1 x , x ≠ 0, f (0) = 1. x Discuss the continuity of the function f (x) = 1 , when x ≠ 0 and f (0) = 0 1 − e1 ⁄ x for all values of x. 11. Prove that the function f (x) = except x = 0. 12. (Rohilkhand 2010B) ⏐ x⏐ for x ≠ 0, f (0) = 0 is continuous at all points x (Meerut 2009; Kanpur 08, 09) e1 ⁄ x sin (1 ⁄ x) Test the continuity of the function f (x) at x = 0 if f (x) = and f (0) = 0. 1 + e1 ⁄ x ,x≠ 0 (Meerut 2005) D-37 LIMITS AND CONTINUITY 13. E xamine the following function for continuity at x = 0 and x = 1 : f (x) = ⎧x2 ⎪ ⎪1 ⎨ ⎪1 ⎪ ⎩x for x ≤ 0 for 0 < x ≤ 1 for x > 1. (Meerut 2001, 03, 04B, 05) 14. Discuss the continuity of the following function at x = 0 : ⎧ cos x , x ≥ 0 f (x) = ⎨⎩ − cos x , x < 0. 15. Test the continuity of the following functions at x = 0 : (i) f (x) = x cos (1 ⁄ x), when x ≠ 0, f (0) = 0. (Meerut 2007) (ii) f (x) = x log x, for x > 0, f (0) = 0. Discuss the nature of discontinuity at x = 0 of the function f (x) = [ x ] − [− x] where [ x ] denotes the integral part of x. 16. 17. 18. 19. Discuss the continuity of f (x) = (1 ⁄ x) cos (1 ⁄ x). G ive an example of each of the following types of functions : (i) The function which possesses a limit at x = 1 but is not defined at x = 1. (ii) The function which is neither defined at x = 1 nor has a limit at x = 1. (iii) The function which is defined at two points but is nevertheless discontinuous at both the points. In the closed interval [− 1, 1] let f be defined by f (x) = x2 sin (1 ⁄ x2) for x ≠ 0 and f (0) = 0. In the given interval (i) Is the function bounded ? (ii) Is it continuous ? 1. 2. (i) Continuous for all x. (ii) Discontinuous at x = 0. (iii) Discontinuous at x = 0. (iv) Continuous for all x. (v) Discontinuous at x = 0. (vi) Discontinuous at 0. (vii) Continuous for all x. (viii) Discontinuous at 0. (ix) Discontinuous at 0. (x) Discontinuous at 0. (xi) Continuous for all x. (ii) No, it has a discontinuity of second kind. H ere both f (a + 0) and f (a − 0) do not exist. 4. (i) Discontinuous at x = 0. (ii) Discontinuous at x = 1 ⁄ 2n , n = 1, 2, 3, … Discontinuous at 0. 8. Continuous at x = 1, 2 and discontinuous at x = 0. 3. 10. Discontinuous only at x = 0 and the discontinuity is ordinary. 12. Discontinuity of the second kind at x = 0. 13. Discontinuous at x = 0 and continuous at x = 1. 14. 15. Discontinuous at x = 0. (i) Continuous. (ii) Continuous. D-38 DIFFERENTIAL CALCULUS 16. 17. Discontinuity of the first kind. Continuous for all x, except at x = 0 where it has discontinuity of the second kind. 18. (i) f (x) = x2 for x > 1, f (x) = x3 for x < 1. (ii) f (x) = − x2 for x < 1, f (x) = x2 for x > 1. (iii) f (x) = 0 for x ≤ 0, f (x) = 19. (i) 3 2 − x for 0 < x ≤ Yes; (ii) 1 2 , f (x) = 3 + x for x > 1 ⋅ 2 2 Yes. Fill in the Blanks: 1. Fill in the blanks “……”, so that the following statem ents are com plete and correct. lim A function f (x) is continuous at a point x = a if f (x) = …… . x →a (Bundelkhand 2008) x2 − 1 = …… . x− 1 2. lim x →1 3. lim sin (x ⁄ 4) = …… . x →0 x 4. lim x 3 − 1 = …… . x →1 x2 − 1 5. lim (1 + x) 1 ⁄ x = …… . x →0 6. lim ax − 1 = …… . x →0 x 7. If f (x) = x − [x], where [x] denotes the greatest integer less than or equal to x , then f (x) = …… , for 3 < x < 4. 8. ⎧ x , 0≤ x< 1 Let f (x) = ⎨⎩ 3 − x , 1 ≤ x ≤ 2. Then 9. lim x → 1− f (x) = …… . ⎧1 , x< 1 ⎪ Let f (x) = ⎨⎪ 2 − x , 1 ≤ x < 2 ⎩2 , x ≥ 2. 3 2 Then (i) f ( ) = …… (ii) lim x → 1+ f (x) = …… and (iii) lim x → 2− f (x) = …… (Meerut 2003) 10. lim | sin x | = …… . x →0 − x 11. A function f (x) has a removable discontinuity at x = a if is not equal to …… . lim f (x) exists but x →a D-39 LIMITS AND CONTINUITY 12. 13. sin x is …… . x ⎧ sin x , x≠ 0 ⎪ The domain of the function f (x) = ⎨⎪ x ⎩1 , x= 0 is …… . The domain of the function f (x) = Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 14. 15. 16. 17. 18. lim | x | x →0 x (a) 1 is equal to (c) 2 (d) The limit does not exist (c) –1 (d) 2 lim | x − 2| is equal to x → 2+ x − 2 (a) –1 (b) 1 (c) 2 (d) –2 lim | x − 3| is equal to x → 3− x − 3 (a) –1 (b) 3 (c) –3 (d) 1 (Meerut 2003; Rohilkhand 14) lim e 1 ⁄ x − 1 is equal to x → 0+ e 1 ⁄ x + 1 (a) –1 (b) 1 (c) 0 (d) 2 lim x →0 (a) 0 ex − x (b) –1 1 is equal to (b) 1 True or False: Write ‘T’ for true and ‘F’ for false statement. 19. If f (x) = then 20. 21. 22. ⎧ ⎪ ⎨ ⎪ ⎩ x , when x < 0 1 , when x = 0 x 2, when x > 0, lim f (x) = 0. x →0 ⎧ sin x , x≠ 0 ⎪ The function f (x) = ⎨⎪ x ⎩2 , x= 0 is continuous at x = 0. ⎧ sin x , x ≥ 0 − sin x , x < 0 is continuous at x = 0. The function f (x) = ⎨⎩ For lim f (x) to exist, the function f (x) must be defined at x = a. x →a D-40 23. 24. 25. 26. DIFFERENTIAL CALCULUS ⎧ x cos (1 ⁄ x), x ≠ 0 The function f (x) = ⎨⎩ 0 , x= 0 is discontinuous at x = 0. If a function f is continuous at a, then | f | is also continuous at a. ⎧ sin x , x≠ 0 ⎪ The function f (x) = ⎨⎪ x ⎩ 1 , x= 0 is continuous at x = 0. ⎧1 , x< 1 ⎪ The function f (x) = ⎨⎪ 2 − x , 1 ≤ x < 2 ⎩2 , x≥ 2 is discontinuous at x = 1. 27. lim sin x = 1. x →∞ x 28. lim sin 3x = 1. x →0 x 29. lim sin 2 x = 2. x →0 x 30. lim sin x 1 = ⋅ x →0 2 x 2 1. f (a). 2. 2. 3. 1 ⋅ 4 6. log e a. 7. x − 3. 8. 1. (iii) 0. 10. − 1. 11. f (a). 14. 19. 24. 29. (d). T. T. T. 15. 20. 25. 30. (b). F. T. T. 16. 21. 26. (b). T. F. 9. 12. 17. 22. 27. (i) 1 2 (ii) 1 R − {0}. 13. (a). 18. F. 23. F. 28. R. (b). F. F. 4. 3 ⋅ 2 5. e. 2.1 Definitions Derivative at a point : (Bundelkhand 2010; Purvanchal 11) L et I denote the open interval ]a, b[ in R and let x0 ∈ I. Then a function f : I → R is said to be differentiable (or derivable) at x0 iff lim f (x) − f (x0) lim f (x0 + h) − f (x0) or equivalently x → x h →0 h x − x0 0 exists finitely and this lim it, if it exists finitely, is called the differential coefficient or derivative of f with respect to x at x = x0 . It is denoted by f ′ (x0) or by D f (x0). Progressive and regressive derivatives : The progressive derivative of f at x = x0 is given by lim f (x0 + h) − f (x0) , h > 0. h →0 h It is also called the right hand differential coefficient of f at x = x0 and is denoted by R f ′ (x0) or by f ′ (x0 + 0). The regressive derivative of f at x = x0 is given by D-42 DIFFERENTIAL CALCULUS lim f (x0 − h) − f (x0) , h > 0. h →0 − h It is also called the left hand differential coefficient of f at x = x0 and is denoted by L f ′ (x0) or by f ′ (x0 − 0). It is obvious that f is derivable at x0 iff L f ′ (x0) and R f ′ (x0) both exist and are equal. Remark : If f (x) = a0 x n + a1 x n − 1 + … + an − 1 x + an is a polynomial in x of degree n, then f (x) is differentiable at every point a of R. (Meerut 2003; Purvanchal 11) Differentiability in an interval : Open interval ]a, b[ : A function f : ]a, b [→ R is said to be differentiable in ]a, b[ iff it is differentiable at every point of ]a, b[. Closed interval [a, b] : A function f : [a, b] → R is said to be differentiable in [a, b] iff R f ′ (a) exists, L f ′ (b) exists and f is differentiable at every point of ]a, b[. Derivative of a function : Let f be a function whose domain is an interval I. If I1 be the set of all those points x of I at which f is differentiable i.e., f ′ (x) exists and if I1 ≠ ∅, we get another function f ′ with domain I1. It is called the first derivative of f (or simply the derivative of f ). Similarly 2nd, 3rd, …, nth derivatives of f are defined and are denoted by f ′′, f ′′′, …, f (n) respectively. Note : The derivative of a function at a point and the derivative of a function are two different but related concepts. The derivative of f at a point a is a number while the derivative of f is a function. H owever, very often the term derivative of f is used to denote both number and function and it is left to the context to distinguish what is intended. An alternate definition of differentiability : Let f be a function defined on an interval I and let a be an interior point of I. Then, by the definition of f ′ (a), assuming it to exist, we have lim f (x) − f (a) f ′ (a) = x →a x− a i.e., f ′(a) exists if for a given ε > 0, there exists a δ > 0 such that ⎪ ⎪ f (x) − f (a) − f ′ (a) ⎪ < ε whenever 0 < ⏐ x − a⏐ < δ ⎪ ⎪ ⎪ x− a or equivalently f (x) − f (a) x ∈ ] a− δ, a + δ [⇒ f ′(a) − ε < < f ′ (a) + ε. x− a 2.2 Geometrical Meaning of a Derivative We take two neighbouring points P [a, f (a)] and Q [a + h, f (a + h)] on the curve y = f (x). Let the chord PQ and the tangent at P meet the x-axis in L and T respectively. Let ∠ Q L X = α and ∠ PTX = ψ. Draw PN and QM ⊥ to OX and PH ⊥ to QM. Then P H = NM = OM − ON = a + h − a = h, and QH = QM − MH = QM − PN = f (a + h) − f (a). QH f (a + h) − f (a) ∴ tan α = = ⋅ …(1) PH h DIFFERENTIABILITY D-43 As h → 0, the point Q moving a lo n g t h e cu rve ap p ro ach es t h e point P, the chord PQ approaches the tangent line TP as its limiting position and the angle α approaches the angle ψ. H ence taking limits as h → 0, the equation (1) gives tan ψ = f ′ (a). H ence f ′ (a) is the tangent of the angle which the tangent line to t h e cu r ve y = f (x) a t t h e p o i n t P [a, f (a)] makes with x-axis. 2.3 A Necessary Condition for the Existence of a Finite Derivative Theorem : Continuity is a necessary but not a sufficient condition for the existence (Meerut 2010, 10B, 11; Kanpur 07, 12; Avadh 10; Kashi 14) of a finite derivative. lim f (x) − f (x0) Proof : Let f be differentiable at x0 . Then x → x exists and equals x − x0 0 f (x) − f (x0) f ′ (x0). Now, we can write f (x) − f (x0) = (x − x0), if x ≠ x0 . x − x0 Taking limits as x → x0 , we get ⎫ lim ⎡⎢ f (x) − f (x ) ⎤⎥ lim ⎧ f (x) − f (x0) 0 ⎦ = x →x ⎪ (x − x0) ⎪⎬ x → x0 ⎣ 0⎨ ⎪ ⎪ x − x0 ⎩ ⎭ lim f (x) − f (x0) lim = x →x ⋅ x → x (x − x0) = f ′ (x0) . 0 = 0, x − x0 0 0 so that lim x → x0 f (x) = f (x0). H ence f is co n t in u o u s at x0 . Thus cont inuity is a necessary condition for differentiability but it is not a sufficient condition for the existence of a finite derivative. The following example illustrates this fact : Let f (x) = x sin (1 ⁄ x), x ≠ 0 and f (0) = 0. This function is continuous at x = 0 but not differentiable at x = 0. 1 lim lim Since f (x) = x sin = 0 = f (0), t h e r e fo r e t h e fu n ct i o n f (x) i s x →0 x →0 x continuous at x = 0. lim f (0 + h) − f (0) lim f (h) − f (0) Now R f ′ (0) = = h →0 h →0 h h h sin (1 ⁄ h) − 0 1 lim lim = = sin , h →0 h →0 h h which does not exist. Similarly L f ′ (0) does not exist. Thus f (x) is not differentiable at x = 0, though it is continuous there. D-44 DIFFERENTIAL CALCULUS 2.4 Algebra of Derivatives Now we shall establish some fundamental theorems regarding the differentiability of the sum, product and quotient of differentiable functions. Theorem 1 : If a function f is differentiable at a point x0 and c is any real num ber, then the function c f is also differentiable at x0 and (c f ) ′ (x0) = c f ′ (x0). Proof : By the definition of f ′ (x0), we have lim f (x) − f (x0) = f ′(x0). x → x0 x − x0 Now lim c f (x) − c f (x0) lim (c f ) (x) − (c f ) (x0) = x →x x → x0 x − x0 x − x0 0 f (x) − f (x0) ⎫ lim ⎧ ⎪ = x → x ⎪⎨ c ⋅ ⎬ ⎪ x − x0 ⎪⎭ 0⎩ lim f (x) − f (x0) = c ⋅ x →x = c f ′ (x0). x − x0 0 H ence cf is differentiable at x0 and (c f ) ′ (x0) = c f ′ (x0). Theorem 2 : L et f and g be defined on an interval I. If f and g are differentiable at x0 ∈ I, then so also is f + g and ( f + g)′ (x0) = f ′ (x0) + g′ (x0). Proof : Since f and g are differentiable at x0, therefore lim f (x) − f (x0) = f ′ (x0), x → x0 x− x …(1) 0 lim g (x) − g (x0) = g′ (x0). x → x0 x − x0 and Now …(2) lim ( f + g) (x) − ( f + g) (x0) x → x0 x − x0 lim [ f (x) + g (x)] − [ f (x0) + g (x0)] = x →x x − x0 0 lim ⎡ f (x) − f (x0) g (x) − g (x0) ⎤⎥ = x → x ⎢⎢ + ⎥ x − x0 x − x0 ⎥⎦ 0 ⎢⎣ lim f (x) − f (x0) lim g (x) − g (x0) , = x →x + x →x x − x0 x − x0 0 0 as the limit of a sum is equal to the sum of the limits = f ′ (x0) + g′ (x0), using (1) and (2). H ence f + g is differentiable at x0 and ( f + g)′ (x0) = f ′ (x0) + g′ (x0). Theorem 3 : L et f and g be defined on an interval I. If f and g are differentiable at x0 ∈ I, then so also is fg and ( fg)′ (x0) = f ′ (x0) g (x0) + f (x0) g′ (x0). D-45 DIFFERENTIABILITY Proof : Since f and g are differentiable at x0 , we have lim f (x) − f (x0) = f ′ (x0) x → x0 x− x …(1) 0 lim g (x) − g (x0) = g ′ (x0) x → x0 x − x0 and Now …(2) lim ( fg) (x) − ( fg) (x0) lim f (x) g (x) − f (x0) g (x0) = x →x x → x0 x − x0 x − x0 0 lim f (x) g (x) − f (x0) g (x) + f (x0) g (x) − f (x0) g (x0) = x →x x− x 0 0 g (x) − g (x0) ⎤ lim ⎡ f (x) − f (x0) ⎥ = x → x ⎢⎢ ⋅ g (x) + f (x0) ⎥ x − x0 x − x0 ⎥⎦ 0 ⎢⎣ lim f (x) − f (x0) lim lim g (x) − g (x0) = x →x . x → x g (x) + f (x0) . x → x x − x0 x − x0 0 0 0 = f ′ (x0) g (x0) + f (x0) g′ (x0), lim using (1), (2) and the fact that x → x g (x) = g (x0). 0 Note that g (x) is differentiable at x = x0 implies that g (x) is continuous at x0 and so lim x → x0 g (x) = g (x0). H ence fg is differentiable at x0 and ( fg)′ (x0) = f ′ (x0 ) g (x0) + f (x0) g′ (x0). Theorem 4 : If f is differentiable at x0 and f (x0) ≠ 0 , then the function 1 ⁄ f is differentiable at x0 and (1 ⁄ f )′ (x0) = − f ′ (x0) ⁄ { f (x0)}2. lim f (x) − f (x0) = f ′ (x0). x → x0 x − x0 Proof : Since f is differentiable at x0 , therefore …(1) Since f is differentiable at x0 , it is continuous at x0 , therefore lim x → x0 f (x) = f (x0) ≠ 0. …(2) Also, since f (x0) ≠ 0, hence, f (x0) ≠ 0 in some neighbourhood N of x0 . Now, we have for x ∈ N 1 1 − lim ⎧ f (x) − f (x0) lim f (x) f (x0) 1 1 ⎫⎪ = x → x ⎪⎨ − ⋅ ⋅ ⎬ x → x0 ⎪ x− x )⎪ f (x) f (x x − x 0 0 ⎩ 0 lim f (x) − f (x0) lim 1 1 = − x →x ⋅ x →x ⋅ f (x) f (x x − x 0 0 0 0) 0 ⎭ D-46 DIFFERENTIAL CALCULUS = − f ′ (x0) ⋅ 1 1 , ⋅ using (1) and (2) f (x0) f (x0) = − f ′ (x0) ⁄ { f (x0)}2. H ence 1 ⁄ f is differentiable at x0 and (1 ⁄ f )′ (x0) = − f ′ (x0) ⁄ { f (x0)}2. Theorem 5 : L et f and g be defined on an interval I. If f and g are differentiable at x0 ∈ I, and g (x0) ≠ 0, then the function f ⁄ g is differentiable at x0 and [g (x0) f ′ (x0) − f (x0) g′(x0)] ⎛⎜ f ⎞⎟ ′ ⎜⎜ ⎟⎟ (x0) = ⋅ [g (x0)]2 ⎝g⎠ Proof : Use theorems 3 and 4 of article 2.4. 2.5 The Chain Rule of Differentiability Theorem. L et f and g be functions such that the range of f is contained in the domain of g. If f is differentiable at x0 and g is differentiable at f (x0), then g ο f is differentiable at x0 , and ( g ο f )′ (x0) = g′ ( f (x0)). f ′ (x0). Proof. Let y = f (x) and y0 = f (x0). Since f is differentiable at x0 , we have lim f (x) − f (x0) = f ′ (x0) x → x0 x− x 0 f (x) − f (x0) = (x − x0) [ f ′ (x0) + λ (x)] or …(1) where λ (x) → 0 as x → x0. Further since g is differentiable at y0 , we have lim g( y) − g ( y0) = g′ ( y0) y → y0 y− y 0 g ( y) − g ( y0) = ( y − y0) [ g′ ( y0) + μ ( y)] where μ ( y) → 0 as y → y0 . or Now …(2) ( g ο f ) (x) − ( g ο f ) (x0) = g ( f (x)) − g ( f (x0)) = g ( y) − g ( y0) = ( y − y0) [ g′ ( y0) + μ ( y)], by (2) = [ f (x) − f (x0)] [ g′ ( y0) + μ ( y)] = (x − x0) [ f ′ (x0) + λ (x)] [ g′ ( y0) + μ ( y)], by (1). Thus if x ≠ x0, then ( g ο f ) (x) − ( g ο f ) (x0) x − x0 = [ g′ ( y0) + μ ( y)] . [ f ′ (x0) + λ(x)]. …(3) Also f bein g differen tiable at x0 , is co n tinuous at x0 and hence as x → x0 , f (x) → f (x0) i.e., y → y0. D-47 DIFFERENTIABILITY Consequently μ ( y) → 0 as x → x0 and λ (x) → 0 as x → x0 . Taking the limits as x → x0 , we get from (3) lim ( g ο f ) (x) − ( g ο f ) (x0) = g′ ( y0) . f ′ (x0). x → x0 x − x0 H ence the function gof is differentiable at x0 and ( gof )′ (x0) = g′ ( f (x0)) f ′ (x0). 2.6 Derivative of the Inverse Function Theorem : If f be a continuous one-to-one function defined on an interval and let f be differentiable at x0 , with f ′ (x0) ≠ 0, then the inverse of the function f is differentiable at f (x0) and its derivative at f (x0) is 1 ⁄ f ′ (x0). Proof : Before proving the theorem we remind that if the domain of f be X and its range be Y, then the inverse function g of f usually denoted by f − 1 is the function with domain Y and range X such that f (x) = y ⇔ g (y) = x. Also g exists if f is one-one. Let y = f (x) and y0 = f (x0). Since f is differentiable at x0 , we have lim f (x) − f (x0) = f ′ (x0) x → x0 x − x0 f (x) − f (x0) = (x − x0) [ f ′ (x0) + λ(x)] or ...(1) where λ (x) → 0 as x → x0. Further, we have g ( y) − g ( y0) = x − x0 , by definition of g. ∴ g ( y) − g ( y0) y − y0 = x − x0 y − y0 = x − x0 f (x) − f (x0) = 1 , by (1). f ′ (x0) + λ (x) It can be easily seen that if y → y0 , then x → x0 . In fact, f is continuous at x0 implies that g = f − 1 is continuous at f (x0) = y0 and consequently g ( y) → g ( y0) as y → y0 i.e., x → x0 as y → y0 , so that λ (x) → 0 as y → y0 . ∴ or lim g ( y) − g ( y0) lim 1 1 = y →y = y → y0 y − y0 f ′ (x0) 0 f ′ (x0) + λ (x) 1 1 g′ ( y0) = or g′ ( f (x0)) = ⋅ f ′ (x0) f ′ (x0) Example 1 : Prove that the function f (x) = ⏐ x⏐ is continuous at x = 0, but not differentiable at x = 0 where ⏐ x⏐ m eans the num erical value or the absolute value of x. (Bundelkhand 2008; Rohilkhand 07; Avadh 11; Meerut 13B) A lso draw the graph of the function. Solution : We have f (0) = ⏐ 0⏐ = 0, lim lim lim lim f (0 + 0) = f (0 + h) = f (h) = ⏐ h⏐ = h= 0 h →0 h →0 h →0 h →0 D-48 DIFFERENTIAL CALCULUS lim lim lim lim f (0 − h) = f (− h) = ⏐ − h⏐ = h = 0. h →0 h →0 h →0 h →0 ∴ f (0) = f (0 + 0) = f (0 − 0). H ence f (x) is continuous at x = 0. lim f (0 + h) − f (0) lim f (h) − f (0) Also, we have Rf ′ (0) = = h →0 h →0 h h ⏐ ⏐ h − 0 lim lim h = = , (h being positive) h →0 h →0 h h lim = 1 = 1, h →0 lim f (− h) − f (0) lim f (0 − h) − f (0) = and L f ′ (0) = h →0 h →0 − h − h ⏐ ⏐ − h − 0 lim lim h , (h being positive) = = h →0 h →0 − h − h lim = − 1 = − 1. h →0 Since R f ′ (0) ≠ L f ′ (0), the function f (x) is not differentiable at x = 0. To draw the graph of the function f (x) = ⏐ x⏐ . ⎧ x, x≥ 0 We have f (x) = ⎨ ⎩ − x, x ≤ 0. and f (0 − 0) = Let y = f (x). Then the graph of the function consists of the following straight lines : y = x, x≥ 0 y = − x, x ≤ 0. The graph is as shown in the figure. From the graph we observe that the function is continuous at the point O i.e., at the point x = 0 but it is not differentiable at this point. The tangent to the curve at the point O from the right is the straight line OA and from the left is the straight line OB. Thus the tangent to the curve at O does not exist and so the function is not differentiable at O. Example 2 : Show that the function f (x) = ⏐ x⏐ + ⏐ x − 1⏐ is not differentiable at x = 0 and x = 1. (Meerut 2005B, 08; Kashi 14) Solution : We first observe that if x < 0, then ⏐ x⏐ = − x and ⏐ x − 1⏐ = ⏐ 1 − x⏐ = 1 − x ; if 0 ≤ x ≤ 1, then ⏐ x⏐ = x and ⏐ x − 1⏐ = ⏐ 1 − x⏐ = 1 − x ; and if x > 1, then ⏐ x⏐ = x and ⏐ x − 1⏐ = x − 1. ∴ the function f (x) is given by ⎧ 1 − 2 x , if x < 0 ⎪ f (x) = ⎨1, if 0 ≤ x ≤ 1 ⎪ ⎩2 x − 1, if x > 1. D-49 DIFFERENTIABILITY At x = 0. We have = R f ′ (0) = lim f (0 + h) − f (0) h →0 h lim f (h) − f (0) lim 1 − 1 = , as f (x) = 1 if 0 ≤ x ≤ 1 h →0 h →0 h h lim 0 = 0, h →0 lim f (0 − h) − f (0) L f ′ (0) = h →0 − h = and lim [1 − 2 (− h)] − 1 lim f (− h) − f (0) = h →0 h →0 − h − h [... f (x) = 1 − 2x, if x < 0] lim 2h lim = = − 2 = − 2⋅ h →0 − h h →0 = ... R f ′ (0) ≠ L f ′ (0), so the given function is not differentiable at x = 0. At x = 1. We have lim f (1 + h) − f (1) lim [2 (1 + h) − 1] − 1 = R f ′ (1) = h →0 h →0 h h lim 2 + 2h − 1 − 1 lim 2h lim = = 2 = 2, h →0 h →0 h h →0 h lim f (1 − h) − f (1) lim 1 − 1 lim L f ′ (1) = = = 0 = 0. h →0 h →0 − h h →0 − h = and ... R f ′ (1) ≠ L f ′(1), so the given function f (x) is not differentiable at x = 1. Example 3 : L et f (x) be an even function. If f ′ (0) exists, find its value. Solution : f (x) is an even function, so f (− x) = f (x) — V x. f ′ (0) exists ⇒ R f ′ (0) = L f ′ (0) = f ′ (0). lim f (h) − f (0) Now f ′ (0) = R f ′ (0) = ,h> 0 h →0 h lim f (− h) − f (0) [... f (− x) = f (x)] = h →0 h lim f (− h) − f (0) = − L f ′ (0) = − f ′ (0). = − h →0 − h ∴ 2 f ′ (0) = 0 ⇒ f ′ (0) = 0. ⎧ − 1, − 2 ≤ x ≤ 0 Example 4 : L et f (x) = ⎨ 0 < x ≤ 2, and ⎩ x − 1, g (x) = f (⏐ x⏐ ) + ⏐ f (x)⏐ . Test the differentiability of g (x) in ]− 2, 2[. Solution : When − 2 ≤ x ≤ 0, ⏐ x⏐ = − x and when 0 < x ≤ 2 , ⏐ x⏐ = x. Now − 2 ≤ x ≤ 0 ⇒ ⏐ x⏐ = − x ⇒ f (⏐ x⏐ ) = f (− x) = − x − 1 . [... 0 < − x ≤ 2] So we have ⎧ x − 1, ⎩ − x − 1, f (⏐ x⏐ ) = ⎨ 0< x≤ 2 − 2≤ x≤ 0 D-50 and DIFFERENTIAL CALCULUS − 2≤ x≤ 0 1, ⎧ ⏐ f (x)⏐ = ⎪⎨ − x + 1, 0 < x ≤ 1 ⎪ ⎩ x − 1, 1 < x ≤ 2⋅ − 2≤ x≤ 0 ⎧ − x, ∴ g (x) = f (⏐ x⏐ ) + ⏐ f (x)⏐ = ⎪⎨ 0, 0< x≤ 1 ⎪ ⎩ 2x − 2, 1 < x ≤ 2. — We see that g (x) is differentiable V x ∈ ] − 2, 2 [, except possibly at x = 0 and 1. L g′ (0) = lim g (0 − h) − g (0) lim g (− h) − g (0) = h →0 h →0 − h − h = R g′ (0) = lim h − 0 = − 1, h →0 − h lim g (0 + h) − g (0) lim 0 − 0 = = 0. h →0 h →0 h h Since L g′ (0) ≠ R g′ (0), g (x) is not differentiable at x = 0. Again R g′ (1) = lim g (1 + h) − g (1) lim 2 (1 + h) − 2 − 0 = = 2, h →0 h →0 h h L g′ (1) = lim g (1 − h) − g (1) lim 0 − 0 = = 0 ≠ R g′ (1). h →0 h →0 − h − h ∴ g is not differentiable at x = 1. Example 5 : Suppose the function f satisfies the conditions : (i) f (x + y) = f (x) f ( y) — V x, y (ii) f (x) = 1 + x g (x) where lim g (x) = 1. x →0 Show that the derivative f ′ (x) exists and f ′ (x) = f (x) for all x. Solution : Putting δx for y in the first condition, we have f (x + δx) = f (x) f (δ x). Then f (x + δ x) − f (x) = f (x) f (δ x) − f (x) f (x + δx) − f (x) f (x) [ f (δx) − 1] or = δx δx f (x) δx g (δx) = , by given condition (ii) δx = f (x) g (δ x). lim lim f (x+ δx) − f (x) = f (x) g (δ x) = f (x) . 1. ∴ δx → 0 δx → 0 δx lim ⎡ .. ⎤ g (δx) = 1⎥ ⎢ . δ x →0 ⎣ ⎦ ∴ f ′ (x) = f (x). Since f (x) exists, f ′ (x) also exists. Example 6 : Show that the function f given by f (x) = x tan− 1 (1 ⁄ x) for x ≠ 0 and f (0) = 0 is continuous but not differentiable at x = 0. (Purvanchal 2008; Meerut 13) 1 lim lim − 1 Solution : Since f (x) = x tan = 0 = f (0), therefore the function x →0 x →0 x f is continuous at x = 0. lim f (0 + h) − f (0) lim f (h) − f (0) = Now R f ′ (0) = h →0 h →0 h h D-51 DIFFERENTIABILITY π ⎛ 1⎞ lim h tan − 1 (1 ⁄ h) − 0 lim = tan − 1 ⎜ ⎟ = tan − 1 ∞ = h →0 h → 0 ⎝ h⎠ 2 h lim f (0 − h) − f (0) lim f (− h) − f (0) L f ′ (0) = = h →0 h →0 − h − h = and lim − h tan− 1 (− 1 ⁄ h) − 0 h →0 − h π ⎛ 1⎞ lim = tan− 1 ⎜ − ⎟ = − tan− 1 ∞ = − ⋅ h →0 2 ⎝ h⎠ Since R f ′ (0) ≠ L f ′ (0), f is not differentiable at x = 0. = Example 7 : Investigate the following function from the point of view of its differentiability. Does the differential coefficient of the function exist at x = 0 and x = 1 ? − x ⎧ x< 0 if ⎪ if 0 ≤ x ≤ 1 x2 f (x) = ⎨ ⎪ 3 x > 1. ⎩ x − x + 1 if (Meerut 2006) Solution : We check the function f (x) for differentiability at x = 0 and x = 1 only. For other values of x, obviously f (x) is differentiable because it is a polynomial function. It can be seen that f (x) is continuous at x = 0 and x = 1. lim f (0 − h) − f (0) lim − (0 − h) − 0 Now L f ′ (0) = = = − 1 h →0 h →0 − h − h R f ′ (0) = and lim f (0 + h) − f (0) lim (0 + h) 2 − 0 = = 0. h →0 h →0 h h ... L f ′ (0) ≠ R f ′ (0), the function is not differentiable at x = 0. Again L f ′ (1) = = and R f ′ (1) = lim (1 − h) 2 − 1 lim f (1 − h) − f (1) = h →0 h →0 − h − h lim − 2h + h2 lim = (2 − h) = 2 h →0 h →0 − h lim f (1 + h) − f (1) lim (1 + h) 3 − (1 + h) + 1 − 1 = h →0 h →0 h h lim 2h + 3h2 + h3 lim = (2 + 3h + h2) = 2 = L f ′ (1). h →0 h →0 h H ence f ′ (1) exists i.e., the function is differentiable at x = 1. = x− 1 ⎧ , when x ≠ 1 ⎪ Example 8 : Find f ′ (1) if f (x) = ⎨ 2x2 − 7x + 5 ⎪ ⎩ − 1 ⁄ 3, when x = 1. lim f (1 + h) − f (1) Solution : We have f ′ (1) = h →0 h ⎛ 1⎞ ⎤ 1+ h− 1 lim ⎡ = − ⎜− ⎟ h h → 0 ⎢⎣ 2 (1 + h) 2 − 7 (1 + h) + 5 ⎝ 3 ⎠ ⎥⎦ / = lim 3h + 2 (1 + h) 2 − 7 (1 + h) + 5 h → 0 3h [2 (1 + h) 2 − 7 (1 + h) + 5] = 2h2 2 2 lim lim = = − ⋅ h → 0 3h (− 3h + 2h2) h → 0 − 9 + 6h 9 D-52 DIFFERENTIAL CALCULUS Example 9 : T est the continuity and differentiability in − ∞ < x < ∞ , of the following function : f (x) = 1 in − ∞ < x < 0 = 1 + sin x ⎛ ⎝ = 2 + ⎜x − 0≤ x< in 1 2 ⎞2 ⎠ π⎟ 1 2 in 1 2 π π ≤ x < ∞. (Avadh 2009) Solution : We shall test f (x) for continuity and differentiability at x = 0 and π ⁄ 2. It is obviously continuous as well as differentiable at all other points. (i) Continuity and differentiability of f (x) at x = 0. We have f (0) = 1 + sin 0 = 1; lim lim lim f (0 + 0) = f (0 + h) = f (h) = (1 + sin h) = 1; h →0 h →0 h →0 lim lim lim f (0 − h) = f (− h) = 1 = 1. h →0 h →0 h →0 f (0 − 0) = and Since f (0) = f (0 + 0) = f (0 − 0), f (x) is continuous at x = 0. Now R f ′ (0) = lim f (0 + h) − f (0) lim f (h) − f (0) = h →0 h →0 h h lim (1 + sin h) − (1 + sin 0) lim sin h = = 1; h →0 h →0 h h = lim f (0 − h) − f(0) lim f (− h) − f (0) = h →0 h →0 − h − h L f ′ (0) = and lim 1 − (1 + sin 0) lim 0 lim = = 0 = 0. h →0 h →0 − h h →0 − h Since R f ′ (0) ≠ L f ′ (0), f (x) is not differentiable at x = 0. = (ii) Continuity and differentiability of f (x) at x = ⎛1 ⎞ ⎝2 ⎠ ⎛1 ⎝2 f ⎜ π⎟ = 2 + ⎜ π − We have 1 2 1 2 π. ⎞2 ⎠ π⎟ = 2 ; ⎛1 ⎝2 ⎞ ⎠ 2 ⎞ ⎧⎛ 1 ⎞ lim ⎛ 1 lim ⎡⎢ 1 ⎫ ⎢ 2 + ⎨⎜ π + h⎟ − π ⎬ f ⎜ π + h⎟ = 2 ⎭ h →0 ⎝ 2 h → 0 ⎢⎣ ⎠ ⎠ ⎩⎝ 2 lim = (2 + h2) = 2 ; h →0 ⎛1 ⎝2 ⎞ ⎠ ⎞ ⎛1 ⎞⎤ lim ⎛ 1 lim ⎡ ⎢ 1 + sin ⎜ π − h ⎟ ⎥ f ⎜ π − h⎟ = h →0 ⎝ 2 h →0 ⎣ ⎠ ⎝2 ⎠⎦ lim = (1 + cos h) = 1 + 1 = 2 . h →0 f ⎜ π + 0⎟ = and f ⎜ π − 0⎟ = ⎛1 ⎞ ⎝2 ⎠ ⎛1 ⎝2 ⎞ ⎠ ⎛1 ⎝2 ⎞ ⎠ Since f ⎜ π ⎟ = f ⎜ π + 0⎟ = f ⎜ π − 0⎟ , f is continuous at x = ⎛1 ⎞ ⎛1 ⎞ f ⎜ π + h⎟ − f ⎜ π ⎟ 2 ⎝ ⎠ ⎝2 ⎠ lim ⎛1 ⎞ Now R f ′ ⎜ π ⎟ = h →0 ⎝2 ⎠ = 1 2 π. h 2 2 ⎧1 ⎛1 ⎡ ⎡ 1 ⎫ ⎤ 1 ⎞ ⎤ ⎢ 2 + ⎨ π + h − π⎬ ⎥ − ⎢ 2 + ⎜ π − π⎟ ⎥ 2 ⎭ ⎦ 2 ⎠ ⎦ ⎣ ⎩2 ⎝2 lim ⎣ h →0 h ⎤ ⎥ ⎥ ⎥ ⎦ D-53 DIFFERENTIABILITY lim 2 + h2 − 2 lim = h= 0; h →0 h →0 h = ⎛1 ⎞ lim L f ′ ⎜ π⎟ = h →0 ⎝2 ⎠ and ⎛1 ⎝2 ⎞ ⎠ ⎛1 ⎞ ⎝2 ⎠ f ⎜ π − h⎟ − f ⎜ π ⎟ − h ⎛1 ⎝2 ⎞ ⎠ 1 + sin ⎜ π − h⎟ − 2 = lim h →0 − h = lim − 1 + cos h lim 1 − cos h lim 2 sin2 (h ⁄ 2) = = h →0 h →0 h →0 h − h h = ⎤ lim ⎡ sin (h ⁄ 2) . sin (h ⁄ 2) ⎥ = 1 × 0 = 0. ⎢ h →0 ⎣ h ⁄ 2 ⎦ Since R f ′ (0) = L f ′ (0), f (x) is differentiable at x = 1 2 π. Example 10 : If f (x) = x2 sin (1 ⁄ x) , for x ≠ 0 and f (0) = 0 , then show that f (x) is continuous and differentiable everywhere and that f ′ (0) = 0. A lso show that the function f ′ (x) has a discontinuity of second kind at the origin. (Meerut 2006B; Kanpur 14) 1 1 lim lim 2 2 Solution : We have f (0 + 0) = (0 + h) sin = h sin = 0; h →0 0 + h h →0 h lim lim lim f (0 − 0) = f (0 − h) = f (− h) = (− h) 2 sin (− 1 ⁄ h) h →0 h →0 h →0 = − 1 lim 2 h sin = 0. h →0 h ... f (0 + 0) = f (0 − 0) = f (0), so the function is continuous at x = 0. lim f (0 + h) − f (0) lim f (h) − f (0) = Now R f ′ (0) = h →0 h →0 h h = and L f ′ (0) = 1 lim h2 sin (1 ⁄ h) − 0 lim = h sin = 0; h →0 h →0 h h lim f (0 − h) − f (0) lim f (− h) − f (0) = h →0 h →0 − h − h 1 lim (− h) 2 sin (− 1 ⁄ h) − 0 lim = h sin = 0. h →0 h →0 h − h Thus R f ′ (0) = L f ′ (0) implies that f (x) is differentiable at x = 0 and f ′(0) = 0. For all other values of x, f (x) is easily seen to be continuous and differentiable. 1 1 Now f ′ (x) = 2 x sin − cos at x ≠ 0 and f ′ (0) = 0. x x lim lim ∴ f ′ (0 + 0) = h → 0 f ′ (0 + h) = h → 0 f ′ (h) = 1 1⎞ lim ⎛ ⎜ 2 h sin − cos ⎟ , which does not exist. h →0 ⎝ h h⎠ Similarly it can be shown that f ′ (0 − 0) does not exist. H ence f ′ is discontinuous at the origin. Since both the limits f ′ (0 − 0) and f ′ (0 + 0) do not exist, therefore the discontinuity is of the second kind. Example 11 : A function f is defined by f (x) = x p cos (1 ⁄ x), x ≠ 0; f (0) = 0. What conditions should be im posed on p so that f m ay be (i) continuous at x = 0 (ii) differentiable at x = 0 ? Solution : We have = D-54 DIFFERENTIAL CALCULUS lim lim ⎡ f (0 + h) = (0 + h) p cos {1 ⁄ (0 + h)}⎤⎦ h →0 h →0 ⎣ lim p = h cos (1 ⁄ h) h →0 lim lim ⎡ (0 − h) p cos {1 ⁄ (0 − h)}⎤⎦ and f (0 − 0) = f (0 − h) = h →0 h →0 ⎣ lim = (− h) p cos (1 ⁄ h). h →0 Now if the function f (x) is to be continuous at x = 0, then f (0 + 0) = f (0) = 0 = f (0 − 0) i.e., the limits given in (1) and (2) must both tend to zero. This is possible only if p > 0, which is the required condition. lim f (0 + h) − f (0) lim f (h) − f (0) Now R f ′ (0) = = h →0 h →0 h h f (0 + 0) = 1 lim h p cos (1 ⁄ h) − 0 lim = h p − 1 cos h →0 h →0 h h = ...(1) …(2) …(3) lim f (0 − h) − f (0) lim (− h) p cos (− 1 ⁄ h) − 0 = h →0 h →0 − h − h lim = − (− 1) p h p − 1 cos (1 ⁄ h). …(4) h →0 Now if f ′ (x) exists at x = 0 then we must have R f ′ (0) = L f ′ (0) and this is possible only if p − 1 > 0 i.e., p > 1 which gives Rf ′ (0) = 0 = L f ′ (0). H ence in order that f is differentiable at x = 0, p must be greater than 1. Example 12 : For a real num ber y, let [ y] denote the greatest integer less than or tan (π [ x − π]) , show that f ′ (x) exists for all x. equal to y. Then if f (x) = 1 + [ x]2 Solution : From the definition of [ y], we see that [x − π] is an integer for all values of x. Then π (x − π) is an integral multiple of π and so tan (π [x − π]) = 0 — V x. Since [x] is an integer so 1 + [x]2 ≠ 0 for any x. Thus f (x) = 0 for all x i.e., f (x) is a constant function and so it is continuous and differentiable i.e., f ′ (x) exists — V x and is equal to zero. Exa mple 13 : D e t e rm i n e t h e s e t o f a l l p o i n t s w h e re t h e f u n c t i o n f (x) = x ⁄ (1 + ⏐ x⏐ ) is differentiable. Solution : Since ⏐ x⏐ = x, x > 0, ⏐ x⏐ = − x, x < 0, ⏐ x⏐ = 0, x = 0, x , x , x < 0; f (x) = 0, x = 0; f (x) = x > 0. ∴ f (x) = 1+ x 1− x h lim lim lim We have f (0 + 0) = f (0 + h) = f (h) = = 0; h →0 h →0 h →0 1 + h L f ′ (0) = and f (0 − 0) = Since lim lim lim − h f (0 − h) = f (− h) = = 0. h →0 h →0 h →0 1 + h f (0 + 0) = f (0) = f (0 − 0) = 0 so the function is continuous at x = 0. Further L f ′ (0) = lim f (0 − h) − f (0) lim f (− h) − f (0) = h →0 h →0 − h − h D-55 DIFFERENTIABILITY = 1 lim [ − h ⁄ (1 + h)] − 0 lim = = 1; h →0 h →0 1 + h − h lim f (0 + h) − f (0) lim [h ⁄ (1 + h)] − 0 = = 1. h →0 h →0 h h Since L f ′ (0) = R f ′ (0), so the function is differentiable at x = 0. It is obviously differentiable for all other real values of x. H ence it is differentiable in the interval ] − ∞ , ∞ [. Example 14 : L et f (x) = √(x) {1 + x sin (1 ⁄ x)} for x > 0, f (0) = 0, f (x) = − √(− x) {1 + x sin (1 ⁄ x)} for x < 0. Show that f ′ (x) exists everywhere and is finite except at x = 0 where its value is + ∞ . Solution : We have lim f (0 + h) − f (0) lim f (h) − f (0) = R f ′ (0) = h →0 h →0 h h lim (√h) {1 + h sin (1 ⁄ h)} − 0 = h →0 h 1 ⎤ lim ⎡ = + (√h) sin (1 ⁄ h) ⎥ = ∞ + 0 = ∞ ⎢ h → 0 ⎣ √h ⎦ lim f (0 − h) − f (0) lim f (− h) − f (0) = and L f ′ (0) = h →0 h →0 − h − h 1 ⎤ ⎡ − √[− (− h)] ⎢ 1 + (− h) sin ⎥ − 0 − h⎦ ⎣ lim = h →0 − h 1⎤ lim ⎡ 1 = + (√h) sin ⎥ = . ∞ + 0 = ∞ ⎢ h →0 ⎣ √ h h⎦ Since R f ′ (0) = L f ′ (0) = ∞ , ∴ f ′ (0) = ∞ . 3 1 1 1 1 + √x sin − cos for x > 0 We have f ′ (x) = x √x x 2√x 2 3 1 1 1 1 + √(− x) sin − cos for x < 0. and f ′ (x) = x √(− x) x 2 √(− x) 2 H ence f ′ (a) is finite for all a ≠ 0. R f ′ (0) = Example 15 : Draw the graph of the function y = ⏐ x − 1 ⏐ + ⏐ x − 2 ⏐ in the interval [0, 3] and discuss the continuity and differentiability of the function in this interval. (Meerut 2007B, 09; Garhwal 08) Solution : From the given definition of the function, we have y = 1 − x + 2 − x = 3 − 2x when x ≤ 1 y= x− 1 + 2 − x= 1 when 1 ≤ x ≤ 2 y = x − 1 + x − 2 = 2x − 3 when x ≥ 2 . Thus the graph consists of the segments of the three straight lines y = 3 − 2x, y = 1 and y = 2x − 3 co r r e sp o n din g t o t h e in t e r va ls [0, 1], [1, 2], [2, 3] respectively. The graph of the function for the interval [0, 3] is as given in the figure. The graph shows that the function is continuous throughout the interval but is not differentiable at D-56 DIFFERENTIAL CALCULUS x = 1, 2 because the slopes at these points are different on the left and right hand sides. To test it analytically, we write y = f (x). Then f (x) = 3 − 2x when x ≤ 1 = 1 when 1 ≤ x ≤ 2 = 2x − 3 when x ≥ 2 . This function is obviously continuous and differentiable at all points of the interval [0, 3] except possibly at x = 1 and at x = 2 . At x = 1, we have f (1) = 1; lim lim f (1 − 0) = [3 − 2 (1 − h)] = 1; f (1 + 0) = (1) = 1. h →0 h →0 Since f (1 − 0) = f (1 + 0) = f (1), f is continuous at x = 1. lim f (1 + h) − f (1) lim 1 − 1 = = 0 Again R f ′ (1) = h →0 h →0 h h lim f (1 − h) − f (1) lim 3 − 2 (1 − h) − 1 and L f ′ (1) = = = − 2. h →0 h →0 − h − h Since R f ′ (1) ≠ L f ′ (1), f is not differentiable at x = 1. At x = 2 , we have f (2) = 1; lim lim f (2 − 0) = (1) = 1 ; f (2 + 0) = [2 (2 + h) − 3] = 1. h →0 h →0 Since f (2 − 0) = f (2 + 0) = f (2), f is continuous at x = 2. lim 2 (2 + h) − 3 − 1 lim f (2 + h) − f (2) = = 2 Again Rf ′ (2) = h → 0 h →0 h h lim f (2 − h) − f (2) lim 1 − 1 and L f ′ (2) = = = 0⋅ h →0 h →0 − h − h Since Rf ′ (2) ≠ L f ′ (2), f is not differentiable at x = 2 . Example 16 : Show that the function f : R → R defined by ⎡ ⎣ f (x) = x ⎢ 1 + 1 3 ⎤ ⎦ sin log x2⎥ , x ≠ 0 and f (0) = 0 is everywhere continuous but has no differential coefficient at the origin. (Garhwal 2009) Solution : O bviously the function f (x) is continuous at every point of R except possibly at x = 0. We test at x = 0. G iven f (0) = 0. ⎧ ⎫⎤ lim lim ⎡ 1 f (0 + 0) = f (0 + h) = ⎢ (0 + h) ⎨ 1 + sin log (0 + h) 2⎬ ⎥ 3 h →0 h →0 ⎣ ⎩ ⎭⎦ lim = [h + (h ⁄ 3) sin log h2] = 0 + 0 × a finite quantity = 0. h →0 [... sin log h2 oscillates between − 1 and + 1 as h → 0] Similarly we can show that f (0 − 0) = 0. H ence f is continuous at x = 0. ⎧ ⎩ (0 + h) ⎨ 1 + Now R f ′ (0) = lim h →0 1 3 ⎫ ⎭ sin log (0 + h) 2⎬ − 0 h D-57 DIFFERENTIABILITY = ⎫ lim ⎧ 1 ⎨ 1 + sin log h 2⎬ , which does not exist since sin log h 2 3 h →0 ⎩ ⎭ oscillates between − 1 and 1 as h → 0. ⎧ ⎩ (0 − h) ⎨ 1 + 1 3 ⎫ ⎭ sin log (0 − h) 2⎬ − 0 lim h →0 − h ⎡ ⎤ lim 1 ⎢ 1 + sin log h 2⎥ , which does not exist as above. = 3 h →0 ⎣ ⎦ H ence f has no differential coefficient at x = 0. e1 ⁄ x − e− 1 ⁄ x , Example 17 : L et f (x) = x 1 ⁄ x x ≠ 0; f (0) = 0. e + e− 1 ⁄ x Show that f (x) is continuous but not derivable at x = 0. L f ′ (0) = (Meerut 2005; Purvanchal 07, 14; Kanpur 08; Bundelkhand 14) Solution : We have f (0) = 0; f (0 + 0) = lim lim lim f (0 + h) = f (h) = h h →0 h →0 h →0 e1 ⁄ h − e1 ⁄ h + e− 1 ⁄ h e− 1 ⁄ h 1 − e− 2 / h , lim h dividing the Nr. and Dr. by e1 ⁄ h h → 0 1 + e− 2 / h 1− 0 = 0× = 0 × 1 = 0; 1+ 0 lim lim f (0 − 0) = f (0 − h) = f (− h) h →0 h →0 = and = e1 ⁄ − h − e− 1 ⁄ − h e− 1 ⁄ h − e1 ⁄ h lim lim − h 1⁄− h = − h − 1⁄h − 1 ⁄ − h h →0 h → 0 e + e e + e1 ⁄ h e− 2 / h − 1 0− 1 lim − h − 2/h = 0× = 0. h →0 0+ 1 + 1 e Since f (0 + 0) = f (0 − 0) = f (0), the function is continuous at x = 0. lim f (0 + h) − f (0) lim f (h) − f (0) Now R f ′ (0) = = h →0 h →0 h h = = and L f ′ (0) = = ⎤ lim ⎡⎢ e1 ⁄ h − e− 1 ⁄ h − 0⎥⎥ ⎢h h → 0 ⎢⎣ e1 ⁄ h + e− 1 ⁄ h ⎥ ⎦ / h = hlim→ 0 11 −+ ee − 2/h − 2/h = 1− 0 = 1, 1+ 0 lim f (0 − h) − f (0) lim f (− h) − f (0) = h →0 h →0 − h − h ⎤ e− 1 ⁄ h − e1 ⁄ h lim ⎡⎢ ⎥ − 0 ⎢ (− h) − 1 ⁄ h ⎥ ⎥ h → 0 ⎢⎣ e + e1 ⁄ h ⎦ / (− h) e− 2 / h − 1 0 − 1 = = − 1. e− 2 / h + 1 0 + 1 Since R f ′ (0) ≠ L f ′ (0), the function is not derivable at x = 0. = lim h →0 2 Example 18 : L et f (x) = e− 1 ⁄ x sin (1 ⁄ x) when x ≠ 0 and f (0) = 0. Show that at every point f has a differential coefficient and this is continuous at x = 0. Solution : We test differentiability at x = 0. D-58 DIFFERENTIAL CALCULUS 2 lim e− 1 ⁄ h sin (1 ⁄ h) − 0 lim sin (1 ⁄ h) = h →0 h → 0 he1 ⁄ h 2 h sin (1 ⁄ h) sin (1 ⁄ h) lim lim = = h → 0 ⎧⎪ h →0 1 1 1 1 1 ⎫⎪ + …⎬ ⋅ 3+ … h ⎨1 + 2 + h+ + 4 2 ! h ⎪⎩ ⎪ h 2!h h ⎭ a finite quantity lying between − 1 and + 1 = = 0. ∞ Similarly L f ′ (0) = 0. Since R f ′ (0) = L f ′ (0) = 0, hence the function f (x) is differentiable at x = 0 and f ′ (0) = 0. If x is any point other than zero, then R f ′ (0) = 2 2 f ′ (x) = (2 ⁄ x3) e− 1 ⁄ x sin (1 ⁄ x) − (1 ⁄ x2) e− 1 ⁄ x cos (1 ⁄ x) 2 Now …(1) = {(2 ⁄ x) sin (1 ⁄ x) − cos (1 ⁄ x)} (1 ⁄ x2) (1 ⁄ e1 ⁄ x ) 1 1⎞ 1 lim lim ⎛ 2 f ′ (0 + 0) = f ′ (0 + h) = ⎜ sin − cos ⎟ ⋅ h →0 h →0 ⎝ h h h ⎠ h2 e1 ⁄ h 2 lim ⎛⎜ 2 sin (1 ⁄ h) cos (1 ⁄ h) ⎞⎟ = − 2 ⎟ h → 0 ⎜⎜ h3 e1 ⁄ h 2 h2 e1 ⁄ h ⎟ ⎝ ⎠ 2 sin (1 ⁄ h) cos (1 ⁄ h) ⎤ lim ⎡ = − ⎢ ⎥ h →0 ⎢ 3 ⎛ 1 1 1 1 ⎞ ⎛ ⎞⎥ 2 ⎢ h ⎜1 + 2 + h + … 1 + + + … ⎟ ⎜ ⎟⎥ ⎣ ⎝ ⎠ ⎝ ⎠⎦ h 2 !. h4 h2 2 !. h4 some finite quantity some finite quantity = − = 0⋅ ∞ ∞ Similarly f ′ (0 − 0) = 0. H ence f ′ is continuous at x = 0. 1. 2. 3. 4. 5. Show that the function f (x) = ⏐ x − 1⏐ is not differentiable at x = 1. (a) If f (x) = x ⁄ (1 + e1 ⁄ x), x ≠ 0, f (0) = 0, show that f is continuous at x = 0, but f ′(0) does not exist. (Purvanchal 2014) xe1 ⁄ x (b) If f (x) = for x ≠ 0 and f (0) = 0, show that f (x) is continuous at 1 + e1 ⁄ x x = 0, but f ′ (0) does not exist. A function φ is defined as follows : φ (x) = − x for x ≤ 0, φ (x) = x for x ≥ 0. T e st t h e ch a r a ct e r o f t h e fu n ct io n a t x = 0 a s r e ga r ds co n t in u it y a n d differentiability. Show that the function f defined on R by f (x) = ⏐ x − 1⏐ + 2 ⏐ x − 2⏐ + 3 ⏐ x − 3⏐ is continuous but not differentiable at the points 1, 2 and 3. (Bundelkhand 2009) Show that the function f (x) = x, 0< x≤ 1 = x − 1, 1< x≤ 2 has no derivative at x = 1. DIFFERENTIABILITY 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. D-59 Show that the function f (x) = x2 − 1, x ≥ 1 = 1 − x, x < 1 has no derivative at x = 1. The following limits are derivatives of certain functions at a certain point. Determine these functions and the points. lim log x − log 2 lim √(a + h) − √(a) (i) ⋅ (ii) ⋅ x →2 h →0 x− 2 h Let f (x) = x2 sin (x− 4 ⁄ 3) except when x = 0 and f (0) = 0. Prove that f (x) has zero as a derivative at x = 0. A function φ (x) is defined as follows : φ (x) = 1 + x if x ≤ 2 φ (x) = 5 − x if x > 2 . Test the character of the function at x = 2 as regards its continuity and differen tiability. E xamine the following curve for continuity and differentiability at x = 0 and x= 1 : for x ≤ 0 y = x2 y= 1 for 0 < x ≤ 1 y = 1 ⁄ x for x > 1. (Meerut 2004B, 09B) Also draw the graph of the function. A function f (x) is defined as follows : f (x) = 1 + x for x ≤ 0, f (x) = x for 0 < x < 1, f (x) = 2 − x for 1 ≤ x ≤ 2 , f (x) = 3 x − x2 for x > 2 . Discuss the continuity of f (x) and the existense of f ′ (x) at x = 0, 1 and 2. Discuss the continuity and differentiability of the following function : f (x) = x2 for x < − 2 f (x) = 4 for − 2 ≤ x ≤ 2 for x > 2 . f (x) = x2 Also draw the graph. (Meerut 2007, 10B) A function f (x) is defined as follows : f (x) = x for 0 ≤ x ≤ 1 f (x) = 2 − x for x ≥ 1. Test t he character of t he function at x = 1 as regards t he con tinu it y and differentiability. (Meerut 2003) E xamine the function defined by f (x) = x2 cos (e1 ⁄ x), x ≠ 0, f (0) = 0 with regard to (i) continuity (ii) differentiability in the interval ]− 1, 1[ . (a) Define continuity and differentiability of a function at a given point. If a function possesses a finite differential coefficient at a point, show that it is continuous at this point. Is the converse true ? Give example in support of your answer. (b) What do you understand by the derivative of a real valued function at a point ? Show that f (x) = x sin (1 ⁄ x), x ≠ 0, f (0) = 0 is not derivable at x = 0. D-60 DIFFERENTIAL CALCULUS (c) Prove that if a function f (x) possesses a finite derivative in a closed interval [a, b], then f (x) is continuous in [a, b]. 12. Continuous at x = 0 but not differentiable at x = 0. (i) The function is log x and the point is x = 2. (ii) The function is √x and the point is x = a. Continuous but not differentiable at x = 2. Discontinuous and non-differentiable at x = 0, continuous and non-differentiable at x = 1. D isco n t in uo u s an d n on -differen t iable at x = 0, 2 a nd co nt in uo u s bu t n ot differentiable at x = 1. Continuous but not differentiable at x = − 2, 2. 13. 14. Continuous but non-differentiable at x = 1. Continuous and differentiable throughout R. 3. 7. 9. 10. 11. 2.7 Rolle’s Theorem If a function f (x) is such that (i) f (x) is continuous in the closed interval [a, b], (ii) f ′ (x) exists for every point in the open interval ]a, b[, (iii) f (a) = f (b), then there exists at least one value of x, say c, where a < c < b, such that f ′ (c) = 0. (Lucknow 2007; Purvanchal 07; Kanpur 08; Meerut 12B; Avadh 14; Kashi 13,14 ) Proof : Since f is continuous on [a, b], it is bounded on [a, b]. Let M and m be the supremum and infimum of f respectively in the closed interval [a, b]. Now either M = m or M ≠ m. If M = m, then f is a constant function over [a, b] and consequently f ′ (x) = 0 for all x in [a, b]. H ence the theorem is proved in this case. If M ≠ m, then at least one of the numbers M and m must be different from the equal values f (a) and f (b). For the sake of definiteness, let M ≠ f (a). Since every continuous function on a closed interval attains its supremum, t herefore, there exist s a real number c i n [a, b] su ch t hat f (c) = M. Also, since f (a) ≠ M ≠ f (b), t herefore, c is different from both a a n d b. This implies that c ∈ ] a, b [. Now f (c) is the supremum of f on [a, b], therefore, f (x) ≤ f (c) — V x in [a, b]. …(1) In particular, for all positive real numbers h such that c − h lies in [a, b], f (c − h) ≤ f (c) f (c − h) − f (c) ⇒ ≥ 0. …(2) − h Since f ′ (x) exists at each point of ] a, b [, and hence, in particular f ′ (c) exists, so taking limit as h → 0, (2) gives L f ′ (c) ≥ 0. …(3) DIFFERENTIABILITY have D-61 Similarly, from (1), for all positive real numbers h such that c + h lies in [a, b], we f (c + h) ≤ f (c). By the same argument as above, we get Rf ′ (c) ≤ 0. …(4) Since f ′ (c) exists, hence, L f ′ (c) = f ′ (c) = R f ′ (c). …(5) From (3), (4) and (5) we conclude that f ′ (c) = 0. In the same manner we can consider the case M = f (a) ≠ m. Note 1 : There may be more than one point like c at which f ′ (x) vanishes. Note 2 : R olle’s theorem will not hold good (i) if f (x) is discontinuous at some point in the interval a ≤ x ≤ b, or (ii) if f ′ (x) does not exist at some point in the interval a < x < b, or (iii) if f (a) ≠ f (b). Note 3 : It can be seen that the conditions of R olle’s theorem are not necessary fo r f ′ (x) t o va n ish a t so m e p o i n t in ] a, b [. F o r e xa m p le , f (x) = cos (1 ⁄ x) i s discontinuous at x = 0 in the interval [ − 1, 2] but f ′ (x) vanishes at an infinite number of points in the interval. Geometrical interpretation of Rolle’s Theorem : Suppose the function f (x) is not constant and satisfies the conditions of R olle’s theorem in t h e in t e r val [a, b]. T hen it s ge omet rical interpretation is that the curve representing the graph of the function f must have a tangent parallel to x-axis, at least at one point between a and b. Algebraical interpretation of Rolle’s Theorem : R olle’s theorem leads to a very important result in the theory of equations, when f (a) = f (b) = 0 and f : [a, b] → R is a polynomial function f (x). H ere a and b are the roots of the equation f (x) = 0. Since a polynomial function f (x) is continuous and differentiable at every point of its domain and we have taken f (a) = f (b), therefore, all the three conditions of R olle’s theorem are satisfied and consequently there exists a point c ∈ ] a, b [ such that f ′ (c) = 0 i.e., if a and b are any two roots of the polynom ial equation f (x) = 0, then there exists at least one root of the equation f ′ (x) = 0 which lies between a and b. Example 1 : Discuss the applicability of Rolle’s theorem for f (x) = 2 + (x − 1) 2 ⁄ 3 in the interval [0, 2]. (Meerut 2012) 2 ⁄ 3 Solution : We have f (x) = 2 + (x − 1) . H ere f (0) = 3 = f (2), which shows that the third condition of R olle’s theorem is satisfied. Since f (x) is an algebraic function of x, it is continuous in the closed interval [0, 2]. Thus the first condition of R olle’s theorem is satisfied. D-62 DIFFERENTIAL CALCULUS Now f ′ (x) = 2 ⋅ 3 [1 ⁄ (x − 1) 1 ⁄ 3]. We see that for x = 1, f ′ (x) is not finite while x = 1 is a point of the open interval 0 < x < 2. Thus the second condition of R olle’s theorem is not satisfied. H e n c e t h e R o l l e ’s t h e o r e m i s n o t a p p l i c a b l e fo r t h e fu n c t i o n f (x) = 2 + (x − 1) 2 ⁄ 3 in the interval [0, 2]. Example 2 : Discuss the applicability of Rolle’s theorem in the interval [− 1, 1] to the function f (x) = ⏐ x⏐ . Solution : G iven f (x) = ⏐ x⏐ . H ere f (− 1) = ⏐ − 1⏐ = 1, f (1) = ⏐ 1⏐ = 1, so that f (− 1) = f (1). Further the function f (x) is continuous throughout the closed interval [− 1, 1] but it is not differentiable at x = 0 which is a point of the open interval ] − 1, 1 [. Thus the second condition of R olle’s theorem is not satisfied. Hence the R olle’s theorem is not applicable here. Example 3 : A re the conditions of Rolle’s theorem satisfied in the case of the following functions ? (ii) f (x) = cos (1 ⁄ x) in − 1 ≤ x ≤ 1, (i) f (x) = x2 in 2 ≤ x ≤ 3, (iii) f (x) = tan x in 0 ≤ x ≤ π. Solution : (i) The function f (x) = x2 is continuous and differentiable in the interval [2, 3]. Also f (2) = 4 and f (3) = 9, so that f (2) ≠ f (3). Thus the first two conditions of R olle’s theorem are satisfied and the third condition is not satisfied. (ii) The function f (x) = cos (1 ⁄ x) is discontinuous at x = 0 and consequently is not differentiable there. Thus the first two conditions of R olle’s theorem are not satisfied. H ere f (− 1) = cos (− 1) = cos 1 and f (1) = cos 1. Thus f (− 1) = f (1) i.e., the third condition is satisfied. (iii) The function f (x) = tan x is not continuous at x = π ⁄ 2 and consequently is not differentiable there. Thus the first two conditions of R olle’s theorem are not satisfied here. Further f (0) = tan 0 = 0 and f (π) = tan π = 0. Thus f (0) = f (π) i.e., the third condition is satisfied. ⎡ x2 + ab ⎤ ⎥ Example 4 : Discuss the applicability of Rolle’s theorem to f (x) = log ⎢⎢⎢ ⎥, ⎥ (a + b) x ⎣ ⎦ in the interval [a, b] , 0 < a < b. (Rohilkhand 2014) ⎡ a 2 + ab ⎤ ⎥ = log 1 = 0, Solution : H ere f (a) = log ⎢ ⎣ (a + b) a ⎦ ⎡ b2 + ab ⎤ ⎥ ⎥ = log 1 = 0. f (b) = log ⎢⎢⎢ ⎥ ⎣ (a + b) b ⎦ and f (a) = f (b) = 0. lim f (x + h) − f (x) Also R f ′ (x) = h →0 h Thus = ⎧ (x + h) 2 + ab ⎫ ⎧ x2 + ab ⎫ ⎤ lim 1 ⎡⎢ ⎪ ⎪ ⎪ ⎪⎥ ⎢ log ⎨ ⎬ − log ⎨ ⎬⎥ ⎢ ⎪ ⎪ ⎪ (a + b) x⎪ ⎥ h →0 h ⎣ ⎩ (a + b) (x + h) ⎭ ⎩ ⎭⎦ = (x2 + 2 x h + h2 + ab) (a + b) x⎤⎥ lim 1 ⎡⎢ log ⎥ h → 0 h ⎢⎢⎣ (a + b) (x + h) (x2 + ab) ⎥⎦ D-63 DIFFERENTIABILITY = ⎧ (x2 + 2 x h + h 2 + ab) x lim 1 ⎡⎢ ⎪ × ⎢ log ⎨ 2 + ab ⎪ h → 0 h ⎢⎣ x + x ⎩ = ⎧ ⎧ 2 x h + h2 ⎫⎪ h ⎫ ⎤⎥ lim 1 ⎡⎢ log ⎪⎨ 1 + 2 ⎬ − log ⎨ 1 + ⎬ ⎥ ⎪ ⎪ h → 0 h ⎢⎢⎣ ⎩ x ⎭ ⎥⎦ x + ab ⎭ ⎩ = ⎤ lim 1 ⎡⎢ 2 x h + h2 h − + …⎥⎥ , ⎥ h → 0 h ⎢⎢⎣ x2 + ab x ⎦ ⎫ ⎤⎥ ⎪⎥ ⎬ h ⎪⎭ ⎥⎦ …(1) ⎡. . ⎤ 1 ⎢ . log (1 + y) = y − y2 + …⎥ 2 ⎣ ⎦ = 2x 1 − ⋅ x2 + ab x Again L f ′ (x) = = lim ⎡ f (x − h) − f (x) ⎤ ⎢ ⎥ h →0 ⎣ − h ⎦ 1 lim h → 0 (− h) ⎡ − 2hx + h 2 ⎤ (− h) ⎢ − + …⎥⎥ , replacing h by − h in (1) ⎢ 2 ⎢ x + ab ⎥ x ⎣ ⎦ 2x 1 − ⋅ + ab x Since R f ′ (x) = L f ′ (x), f (x) is differentiable for all values of x in [a, b]. This implies that f (x) is also continuous for all values of x in [a, b]. Thus all the three conditions of R olle’s theorem are satisfied. Hence f ′ (x) = 0 for at least one value of x in the open interval ]a, b[. 2x 1 Now f ′ (x) = 0 ⇒ 2 − = 0 or 2 x2 − (x2 + ab) = 0 x + ab x = x2 or x2 = ab or x = √(ab), which being the geometric mean of a and b lies in the open interval ] a, b[. H ence the R olle’s theorem is verified. Remark : In this question to find f ′ (x), we can also proceed as follows : We have f (x) = log (x2 + ab) − log (a + b) − log x. 1 2x − ⋅ ∴ f ′ (x) = 2 x + ab x O bviously f ′ (x) exists for all values of x in [a, b] . Example 5 : Verify Rolle’s theorem in the case of the functions (i) f (x) = 2x3 + x2 − 4x − 2, (ii) f (x) = sin x in [0, π], (iii) f (x) = (x − a) m (x − b) n , where m and n are + ive integers, and x lies in the interval [a, b]. Solution : (i) Since f (x) is a rational integral function of x, therefore, it is continuous and differentiable for all real values of x. Thus the first two conditions of R olle’s theorem are satisfied in any interval. H ere f (x) = 0 gives 2x3 + x2 − 4x − 2 = 0 (x2 − 2) (2x + 1) = 0 i.e., x = ± √2 , − or Thus ⎛ f (√2) = f (− √2) = f ⎜ − ⎝ 1⎞ ⎟ 2⎠ = 0. 1 ⋅ 2 D-64 DIFFERENTIAL CALCULUS If we take the interval [− √2 , √2], then all the three conditions of R olle’s theorem are satisfied in this interval. Consequently there is at least one value of x in the open interval ] − √2 , √2[ for which f ′ (x) = 0. f ′ (x) = 0 ⇒ 6 x2 + 2 x − 4 = 0 ⇒ 3x2 + x − 2 = 0 or (3x − 2) (x + 1) = 0 or x = − 1, 2 ⁄ 3 i.e., f ′ (− 1) = f ′ (2 ⁄ 3) = 0. Since both the points x = − 1 and x = 2 ⁄ 3 lie in the open interval ] − √2, √2[, R olle’s theorem is verified. (ii) The function f (x) = sin x is continuous and differentiable in [0, π]. Also f (0) = 0 = f (π). Thus all t he three condt ions of R olle’s theorem are satisfied. H ence f ′ (x) = 0 for at least one value of x in the open interval ]0, π [. 3π 5π π ,± ,…. Now f ′ (x) = 0 ⇒ cos x = 0 ⇒ x = ± , ± 2 2 2 Since x = π ⁄ 2 lies in the open interval ]0, π[, the R olle’s theorem is verified. (iii) We have f (x) = (x − a) m (x − b) n . As m and n are positive integers, (x − a) m and (x − b) n are polynomials in x on being expanded by binomial theorem. H ence f (x) is also a polynomial in x. Consequently f (x) is continuous and differentiable in the closed interval [a, b]. Also f (a) = f (b) = 0. Thus all the three conditions of R olle’s theorem are satisfied so that there is at least one value of x in the open interval ]a, b[ where f ′ (x) = 0. Now f ′ (x) = (x − a) m . n (x − b) n − 1 + m (x − a) m − 1 (x − b) n . Solving the equation f ′ (x) = 0, we get x = a, b, (na + mb) / (m + n). O ut of these values the value (na + mb) ⁄ (m + n) is a point which lies in the open interval ]a, b[, since it divides the interval ]a, b[ internally in the ratio m : n. H ence the R olle’s theorem is verified. Example 6 : Verify Rolle’s theorem for Now f (x) = x (x + 3) e− x ⁄ 2 in [− 3, 0]. Solution : We have f (x) = x (x + 3) e− x ⁄ 2. ∴ ⎛ 1⎞ f ′ (x) = (2x + 3) e− x ⁄ 2 + (x2 + 3x) e− x ⁄ 2. ⎜ − ⎟ ⎡ ⎢ ⎣ = e− x ⁄ 2 2 x + 3 − 1 2 ⎤ ⎥ ⎦ (x2 + 3x) = − ⎝ 2⎠ 1⎛ 2 x 2⎝ − x − 6⎞⎠ e− x ⁄ 2, which exists for every value of x in the interval [− 3, 0]. H ence f (x) is differentiable and so also continuous in the interval [− 3, 0]. Also f (− 3) = f (0) = 0. Thus all the three conditions of R olle’s theorem are satisfied. So f ′ (x) = 0 for at least one value of x lying in the open interval ]− 3, 0[. Now f ′ (x) = 0 ⇒ − 1 2 (x2 − x − 6) e− x ⁄ 2 = 0 or x2 − x − 6 = 0 (x − 3) (x + 2) = 0 or x = 3, − 2 . Since the value x = − 2 lies in the open interval ] − 3, 0[, the R olle’s theorem is verified. or 1. (i) State R olle’s theorem. (ii)Verify R olle’s theorem when (Kanpur 2005) f (x) = e x sin x, a = 0, b = π. D-65 DIFFERENTIABILITY 2. 3. Verify R olle’s theorem for the following functions : (i) f (x) = (x − 4) 5 (x − 3) 4 in the interval [3, 4]. (ii) f (x) = x3 − 6 x2 + 11 x − 6. (iii) f (x) = x3 − 4 x in [− 2, 2]. (iv) f (x) = e x (sin x − cos x) in [π ⁄ 4, 5π ⁄ 4]. (v) f (x) = 10x − x2 in [0, 10]. Discuss the applicability of R olle’s theorem to the function (Meerut 2013B) (Kanpur 2006) f (x) = x2 + 1, when 0 ≤ x ≤ 1 = 3 − x, when 1 < x ≤ 2 . 4. 5. 6. 7. Show that between any two roots of e x cos x = 1 there exists at least one root of e x sin x − 1 = 0. St ate and prove R olle’s th eorem. In terp ret it geometrically. Verify R olle’s theorem for the function f (x) = x2 in [− 1, 1]. Verify the truth of R olle’s theorem for the function f (x) = x2 − 3x + 2 on the interval [1, 2]. Does the function f (x) = ⏐ x − 2⏐ satisfy the conditions of R olle’s theorem in the interval [1, 3]. Justify your answer with correct reasoning. 8. The function f is defined in [0, 1] as : f (x) = 1 = 2 9. 10. 3. 7. for 0 ≤ x < for 1 2 1 2 ≤ x ≤ 1. Show that f (x) satisfies none of the conditions of R olle’s theorem, yet f ′ (x) = 0 for many points in [0, 1]. If a + b + c = 0, then show that the quadratic equation 3ax2 + 2bx + c = 0 has at least one root in ] 0, 1 [ . a0 a1 a2 an − 1 Let + + + …+ + an = 0. Show that there exists at least n+ 1 n n− 1 2 one real x between 0 and 1 such that a0 x n + a1 x n − 1 + … + an = 0. The given function is not differentiable at x = 1 and so R olle’s theorem is not applicable to the given function in the interval [0, 2]. The function does not satisfy the third condition that f (x) must be differentiable in the open interval ] 1, 3 [ . 2.8 Lagrange’s Mean Value Theorem Theorem : If a function f (x) is (i) continuous in a closed interval [a, b], and (ii) differentiable in the open interval ]a, b [ i.e., a < x < b, then there exists at least one value ‘c’ of x lying in the open interval ]a, b[ such that f (b) − f (a) = f ′(c). b− a (Kanpur 2011; Rohilkhand 12, 12B; Meerut 12; Kashi 14; Avadh 07, 12, 14) Proof : Consider the function φ (x) defined by φ (x) = f (x) + Ax, …(1) D-66 DIFFERENTIAL CALCULUS where A is a constant to be chosen such that φ (a) = φ (b) f (b) − f (a) ⋅ …(2) i.e., f (a) + Aa = f (b) + Ab or A = − b− a (i) Now the function f is given to be continuous on [a, b] and the mapping x → Ax is continuous on [a, b], therefore φ is continuous on [a, b]. (ii) Also, since f is given to be differentiable on ]a, b[ and the mapping x → Ax is differentiable on ]a, b[, therefore, φ is differentiable on ]a, b[. (iii) By our choice of A, we have φ (a) = φ (b). From (i), (ii) and (iii), we find that φ satisfies all the conditions of R olle’s theorem on [a, b]. Hence there exists at least one point, say x = c, of the open interval ]a, b[, such that φ ′ (c) = 0. But φ ′ (x) = f ′ (x) + A, from (1). ∴ φ ′ (c) = 0 ⇒ f ′ (c) + A = 0 f (b) − f (a) or f ′ (c) = − A = , from (2). b− a This proves the theorem. It is usually known as the ‘First Mean Value Theorem of Differential Calculus’. Another form of Lagrange’s mean value theorem. If in the above theorem, we take b = a + h, then a number c, lying between a and b can be written as c = a + θh, where θ is some real number such that 0 < θ < 1. Now Lagrange’s theorem can be stated as follows : If f be defined and continuous on [a, a + h] and differentiable on ]a, a + h[, then there exists a point c = a + θh (0 < θ < 1) in the open interval ]a, a + h[ such that f (a + h) − f (a) = f ′ (a + θh) h or f (a + h) − f (a) = hf ′ (a + θh). Geometrical interpretation of the mean value theorem. L e t y = f (x) a n d le t ACB be t h e gra p h of y = f (x) in [a, b]. The coordinates of the point A are (a, f (a)) and those of B are (b, f (b)). If the chord A B makes an angle α with the x-axis, then f (b) − f (a) tan α = b− a = f ′ (c), by Lagrange’s mean value theorem where a < c < b. Thus Lagrange’s mean value theorem says that there is some point c in ]a, b[ such that the tangent to the curve at this point is parallel to the chord joining the points on the graph with abscissae a and b. 2.9 Important Deductions from the Mean Value Theorem Theorem 1 : If a function f is continuous on [a, b] , differentiable on ]a, b[ and if f ′ (x) = 0 for all x in ]a, b[ , then f (x) has a constant value throughout [a, b]. Proof : Let c be any point of ]a, b]. Then the function f is continuous on [a, c] and differentiable on ]a, c[. Thus f satisfies all the conditions of Lagrange’s mean value DIFFERENTIABILITY D-67 t h eo rem o n [a, c]. Consequently t here exists a real number d bet ween a a n d c i.e., a < d < c such that f (c) − f (a) = (c − a) f ′ (d). But by hypothesis f ′ (x) = 0 throughout the interval ]a, b[, therefore, in particular f ′ (d) = 0 and hence f (c) − f (a) = 0 or f (c) = f (a). Since c is any point of ]a, b], t herefo re, it gives t hat f (x) = f (a) — V x in ]a, b]. Th u s f (x) has a const ant value throughout [a, b]. Theorem 2 : If f (x) and φ (x) are functions continuous on [a, b] and differentiable on ]a, b[ and if f ′ (x) = φ ′ (x) throughout the interval ]a, b[, then f (x) and φ (x) differ only by a constant. Proof : Consider t he fu nction F (x) = f (x) − φ (x). Throughout t he interval ]a, b[, we have F ′ (x) = f ′ (x) − φ ′ (x) = 0, because f ′ (x) = φ ′ (x). Consequently, from theorem 1, we get F (x) = constant or f (x) − φ (x) = constant. Theorem 3 : If f ′ (x) = k for each point x of [a, b], k being a constant, then f (x) = k x + C — V x ∈ [a, b] , where C is a constant. Proof : Consider the interval [a, x] such that [a, x] lies in the interval [a, b] i.e., [a, x] ⊂ [a, b]. Since f ′ (x) exists — V x ∈ [a, b], f is differentiable on [a, b] and hence on [a, x] and consequently continuous on [a, x]. Thus f satisfies all the conditions of Lagrange’s mean value theorem on [a, x] and hence there is a point c ∈ ]a, x[ such that f (x) − f (a) = (x − a) f ′ (c). But by hypothesis f ′ (x) = k — V x ∈ [a, b], therefore, in particular f ′ (c) = k as a < c < x < b i.e., a < c < b. H ence f (x) − f (a) = (x − a) k or f (x) = k x + f (a) − ak or f (x) = k x + C where C = f (a) − ak is a constant. Theorem 4 : If f is continuous on [a, b] and f ′ (x) ≥ 0 in ]a, b[ , then f is increasing in [a, b]. Proof : Let x1 and x2 be any two distinct points of [a, b] such that x1 < x2 . Then f satisfies the conditions of the Lagrange’s mean value theorem in [x1, x2]. Consequently there exists a number c such that x1 < c < x2 , and f (x2) − f (x1) = (x2 − x1) f ′ (c). V x ∈ ]a, b[ and c is a point of Now x2 − x1 > 0 and f ′ (c) ≥ 0 (as f ′ (x) ≥ 0 — ]a, b[), therefore f (x2) − f (x1) ≥ 0 i.e., f (x1) ≤ f (x2). V x1, x2 ∈ [a, b]. Thus x1 < x2 ⇒ f (x1) ≤ f (x2) — H ence f is an increasing function in the interval [a, b]. Similarly, we can prove that if f ′ (x) ≤ 0 in ]a, b[, then f is decreasing in [a, b]. Corollary : If f is continuous on [a, b] , then f is strictly increasing or strictly decreasing on [a, b] according as f ′ (x) > 0 or < 0 in ]a, b[⋅ D-68 DIFFERENTIAL CALCULUS 2.10 Cauchy’s Mean Value Theorem If two functions f (x) and g (x) are (i) continuous in a closed interval [a, b] , (ii) differentiable in the open interval ]a, b[ , and (iii) g′ (x) ≠ 0 for any point of the open interval ]a, b[, then there exists at least one value c of x in the open interval ]a, b[, such that f (b) − f (a) f ′ (c) , = a < c < b. g (b) − g (a) g′ (c) (Kanpur 2007; Avadh 12; Rohilkhand 14) Proof : F i r st we o b se r ve t h a t a s a co n s e q u e n c e o f c o n d i t i o n ( i i i ) , g (b) − g (a) ≠ 0. For if g (b) − g (a) = 0 i.e., g (b) = g (a), then the function g (x) satisfies all the conditions of R olle’s theorem in [a, b] and consequently there is some x in ]a, b[ for which g′ (x) = 0, thus contradicting the hypothesis that g ′ (x) ≠ 0 for any point of ]a, b[. Now consider the function F (x) defined on [a, b], by setting F (x) = f (x) + Ag (x), …(1) where A is a constant to be chosen such that F (a) = F (b) i.e., f (a) + Ag (a) = f (b) + Ag (b) f (b) − f (a) or − A= ⋅ …(2) g (b) − g (a) Since g (b) − g (a) ≠ 0, therefore A is a definite real number. (i) Now f and g are continuous on [a, b], therefore, F is also continuous on [a, b]. (ii) Again, since f a n d g are different iable o n ]a, b[, t h erefo re F is also differentiable on ]a, b[. (iii) By our choice of A, F (a) = F (b). Thus the function F (x) satisfies the conditions of R olle’s theorem in the interval [a, b]. Consequently there exists, at least one value, say c, of x in the open interval ]a, b[ such that F ′ (c) = 0. But F ′ (x) = f ′ (x) + Ag′ (x), from (1). ∴ F ′ (c) = 0 ⇒ f ′ (c) + Ag′ (c) = 0 f ′ (c) ⋅ …(3) or − A= g′ (c) f (b) − f (a) f ′ (c) From (2) and (3), we get = ⋅ g (b) − g (a) g′ (c) Another form : If b = a + h, then a + θh = a when θ = 0 and a + θ h = b when θ = 1. Therefore, if 0 < θ < 1, then a + θh means some value between a and b. So putting b = a + h and c = a + θ h, the result of the above theorem can be written as f ′ (a + θh) , f(a + h) − f (a) = 0 < θ < 1. g′ (a + θh) g (a + h) − g (a) Note 1 : If we take g (x) = x for all x in [a, b], then Cauchy’s mean value theorem gives Lagrange’s mean value t heorem as a part icular case. F or g (x) = x means g (b) = b, g (a) = a, g′ (x) = 1 and so g′ (c) = 1. Putting these values in Cauchy’s mean value theorem, we get Lagrange’s mean value theorem. Thus Cauchy’s mean value theorem is more general than Lagrange’s mean value theorem. D-69 DIFFERENTIABILITY Note 2 : Ca uchy’s mean value t heorem cannot be obt ained by applying Lagrange’s mean value theorem to the functions f and g. For applying Lagrange’s mean value theorem to f (x) and g (x) separately, we get f (b) − f (a) = (b − a) f ′ (c1), where a < c1 < b g (b) − g (a) = (b − a) g ′ (c2), where a < c2 < b. and Dividing, we have f ′ (c1) f (b) − f (a) = ⋅ g′ (c2) g (b) − g (a) Note that here c1 is not necessarily equal to c2 . Example 1 : I f f (x) = (x − 1) (x − 2) (x − 3) a n d a = 0, b = 4, find ‘c’ using (Lucknow 2007; Rohilkhand 14) L agrange’s m ean value theorem . Solution : We have ∴ ∴ f (x) = f (a) = f (b) − b− (x − 1) (x − 2) (x − 3) = x3 − 6x2 + 11x − 6. f (0) = − 6 and f (b) = f (4) = 6. f (a) 6 − (− 6) 12 = = = 3⋅ 4− 0 4 a Also f ′ (x) = 3x2 − 12x + 11 gives f ′ (c) = 3c2 − 12c + 11. Putting these values in Lagrange’s mean value theorem f (b) − f (a) = f ′ (c), (a < c < b), we get b− a 3 = 3c2 − 12c + 11 or 3c2 − 12c + 8 = 0 2 √3 12 ± √ (144 − 96) = 2± ⋅ or c= 3 6 As both of these values of c lie in the open interval ]0, 4[, hence both of these are the required values of c. Example 2 : L et f : [0, 1] → R be defined by f (x) = (x − 1) 2 + 2 — V x ∈ [0, 1]. Find the equation of the tangent to the graph of this curve which is parallel to the chord joining the points (0, 3) and (1, 2) of the curve. Solution : Since f (x) is a polynomial function, therefore it is continuous on [0, 1] and differentiable in ]0, 1[. H ence, by Lagrange’s mean value theorem, there is some c ∈ ]0, 1[ such that 2− 3 f (1) − f (0) = f ′ (c) or = f ′ (c) or − 1 = f ′ (c). 1 1− 0 Now f ′ (x) = 2 (x − 1) gives f ′ (c) = 2 (c − 1). Thus 2 (c − 1) = − 1 i.e., c = ∴ f (c) = 9 4 1 2 ⋅ , so that the point of contact of the tangent is ⎛⎜ 1 , 9 ⎞⎟ and its slope is f ′ (c) = − 1. Hence the equation of the required tangent is y− 9 4 ⎛ ⎝ = − 1 ⎜x − 1⎞ ⎟ 2⎠ or 4 x + 4y = 11. ⎝ 2 4⎠ D-70 DIFFERENTIAL CALCULUS Example 3 : Com pute the value of θ in the first m ean value theorem f (x + h) = f (x) + hf ′ (x + θh), if f (x) = ax2 + bx + c. Solution : H ere f (x) = ax2 + bx + c. f (x + h) = a (x + h) 2 + b (x+ h) + c, f ′ (x) = 2ax + b, f ′ (x + θh) = 2a (x + θh) + b. Substituting all these values in the Lagrange’s mean value theorem, we get ∴ a (x + h) 2 + b (x + h) + c = ax2 + bx + c + h [2a (x + θh) + b] …(1) The relation (1) being identically true for all values of x, hence when x → 0, we have ah2 + bh + c = c + h [2aθ h + b] ah2 = 2aθh2 or θ = 1 ⁄ 2. Example 4 : A function f (x) is continuous in the closed interval [0, 1] and differentiable in the open interval ]0, 1[ , prove that f ′ (x1) = f (1) − f (0), where 0 < x1 < 1. or Solution : H ere a = 0, b = 1 so that f (b) − f (a) f (1) − f (0) = = f (1) − f (0). 1− 0 b− a If we take c = x1, and substitute these values in the result of Lagrange’s mean value theorem, we get f (1) − f (0) = f ′ (x1) where 0 < x1 < 1. This is a particular case of Lagrange’s mean value theorem. Students can give an independent proof of this. Example 5 : Separate the intervals in which the polynom ial 2x3 − 15x2 + 36x + 1 is increasing or decreasing. Solution : We have f (x) = 2x3 − 15x2 + 36x + 1. f ′ (x) = 6x2 − 30x + 36 = 6 (x − 2) (x − 3). f ′ (x) > 0 for x < 2 or for x > 3, f ′ (x) < 0 for 2 < x < 3, and f ′ (x) = 0 for x = 2, 3. Thus f ′ (x) is + ive in the intervals ] − ∞ , 2[ and ]3, ∞ [ and negative in the interval ∴ Now ]2, 3[. H ence f is mo no to nically increasing in the intervals ] − ∞ , 2 ], [ 3, ∞ [ an d monotonically decreasing in the interval [2, 3]. x (Bundelkhand 2011) < log (1 + x) < x for x > 0. Example 6 : Show that 1+ x x Solution : Let f (x) = log (1 + x) − ⋅ ∴ f (0) = 0. 1+ x 1 . (1 + x) − x . 1 1 1 1 x Then f ′ (x) = − = − = ⋅ 2 2 1+ x 1 + x (1 + x) (1 + x) (1 + x) 2 We observe that f ′ (x) > 0 for x > 0. H ence f (x) is monotonically increasing in the interval [0, ∞ [. Therefore x ⎤ ⎡ f (x) > f (0) for x > 0 i.e., ⎢ log (1 + x) − ⎥ > 0 for x > 0 1 + x⎦ ⎣ x i.e., log (1 + x) > for x > 0. …(1) 1+ x D-71 DIFFERENTIABILITY Again, let φ (x) = x − log (1 + x). ∴ φ (0) = 0. x 1 = ⋅ Then φ ′ (x) = 1 − 1+ x 1+ x We observe that φ ′ (x) > 0 for x > 0. Hence φ (x) is monotonically increasing in the interval [0, ∞ [. Therefore φ (x) > φ (0) for x > 0 i.e., [x − log (1 + x)] > 0 for x > 0 i.e., x > log (1 + x) for x > 0. From (1) and (2), we get x < log (1 + x) < x for x > 0. 1+ x …(2) Example 7 : Verify Cauchy’s m ean value theorem for the functions x2 and x3 in the interval [1, 2]. (Avadh 2013) Solution : Let f (x) = x2 and g (x) = x3. Then f (x) and g (x) are continuous in the cl o se d i n t e r va l [1, 2] a n d d i ffe r e n t i a b l e i n t h e o p e n i n t e r va l ]1, 2[. A lso g′ (x) = 3x2 ≠ 0 for any point in the open interval ] 1, 2 [. H ence by Cauchy’s mean value theorem there exists at least one real number c in the open interval ]1, 2[, such that f ′ (c) f (2) − f (1) = ⋅ …(1) g (2) − g (1) g ′ (c) Now f (2) − f (1) 4− 1 3 = = ⋅ g (2) − g (1) 8 − 1 7 Also f ′ (x) = 2x, g′ (x) = 3x2. f ′ (c) 2c 2 3 2 14 ∴ = = ⋅ Putting these values in (1), we get = or c = which g′ (c) 7 3c 9 3c2 3c lies in the open interval ]1, 2[. H ence Cauchy’s mean value theorem is verified. Example 8 : If in the Cauchy’s m ean value theorem , we write f (x) = e x and g (x) = e− x, show that ‘c’ is the arithm etic mean between a and b. f (b) − f (a) eb − ea Solution : H ere = − b = − ea eb = − ea + b, g (b) − g (a) e − e− a f ′ (x) f ′ (c) ex ec and = so that = = − e2c. g′ (x) g′ (c) − e− x − e− c Putting these values in Cauchy’s mean value theorem, we get 1 − ea + b = − e2c or 2c = a + b or c = (a + b). 2 and if Thus c is the arithmetic mean between a and b. Example 9 : If in the Cauchy’s m ean value theorem, we write (i) f (x) = √x and g (x) = 1 ⁄ √x, then c is the geom etric m ean between a and b, (Rohilkhand 2014) (ii) f (x) = 1 ⁄ x2 and g (x) = 1 ⁄ x, then c is the harm onic m ean between a and b. (Bundelkhand 2005) Solution : (i) f (b) − f (a) √ b − √a H ere = = − √(ab), g (b) − g (a) (1 ⁄ √b) − (1 ⁄ √a) 1 and x− 1 ⁄ 2 f ′(c) f ′ (x) c− 1 ⁄ 2 2 = so that = − − 3 ⁄ 2 = − c. 1 − 3⁄2 g′(c) g′ (x) c − x 2 D-72 DIFFERENTIAL CALCULUS Putting these values in Cauchy’s mean value theorem, we get − √(ab) = − c or c = √(ab) i.e., c is the geometric mean between a and b. (ii) H ere (1 ⁄ b2) − (1 ⁄ a2) a + b f (b) − f (a) = = ab (1 ⁄ b) − (1 ⁄ a) g (b) − g (a) f ′ (x) − 2x− 3 f ′ (c) − 2 c− 3 2 = so that = = ⋅ g′ (x) g′(c) c − x− 2 − c− 2 Putting these values in Cauchy’s mean value theorem, we get a+ b 2 2ab = or c = ab c a+ b i.e., c is the harmonic mean between a and b. and 1. State Lagrange’s mean value theorem. Test if Lagrange’s mean value theorem holds for the function f (x) = ⏐ x⏐ in the interval [− 1, 1]. (Kanpur 2010; Rohilkhand 13B) If f (x) = 1 ⁄ x in [− 1, 1], will the Lagrange’s mean value theorem be applicable to f (x) ? (Meerut 2012B) 3. Verify Lagrange’s mean value theorem for the function f : [− 1, 1] → R given by f (x) = x3. 2. 4. Find ‘c’ of the mean value theorem, if f (x) = x (x − 1) (x − 2) ; a = 0, b = 5. 6. 7. 8. 9. 10. 11. 1 ⋅ 2 Find ‘c’ of mean value theorem when (ii) f (x) = 2 x2 + 3 x + 4 in [1, 2] (i) f (x) = x3 − 3x − 2 in [− 2 , 3] (iii) f (x) = x (x − 1) in [1, 2] (Meerut 2013B) ⎛ 11 13 ⎞ 2 , ⎟⋅ (iv) f (x) = x − 3x − 1 in ⎜⎝ − 7 7⎠ State the conditions for the validity of the formula f (x + h) = f (x) + h f ′ (x + θh) and investigate how far these conditions are satisfied and whether the result is true, when f (x) = x sin (1 ⁄ x) (being defined to be zero at x = 0) and x < 0 < x + h. (a) Show that x3 − 3x2 + 3x + 2 is monotonically increasing in every interval. 2x is increasing when x > 0. (b) Show that log (1 + x) − 2+ x Determine the intervals in which the function (x4 + 6 x3 + 17x2 + 32 x + 32) e− x is increasing or decreasing. U se the function f (x) = x1 ⁄ x, x > 0 to determine the bigger of the two numbers e π and π e. Show that the set of all x for which log (1 + x) ≤ x is equal to [0, ∞ [. U se Lagrange’s mean value theorem to prove that 1 + x < e x < 1 + xe x — V x > 0. D-73 DIFFERENTIABILITY 12. 13. 14. 15. 1. 2. 5. 6. 8. 9. 14. If a = − 1, b ≥ 1 and f (x) = 1 ⁄ ⏐ x⏐ , show that the conditions of Lagrange’s mean value theorem are not satisfied in the interval [a, b] , but the conclusion of the theorem is true if and only if b > 1 + √2 . (Kanpur 2007) State Cauchy’s mean value theorem. Verify Cauchy’s mean value theorem for f (x) = sin x, g (x) = cos x in [− π ⁄ 2, 0]. If f (x) = x2, g (x) = cos x, then find the point c ∈ ] 0, π ⁄ 2 [ which gives the result of Cauchy’s mean value theorem in the interval [0, π ⁄ 2] for the functions f (x) and g (x). U se Cauchy’s mean value theorem to show that sin α − sin β π = cot θ, where 0 < α < θ < β < ⋅ cos β − cos α 2 Th e mea n value t h eore m doe s not hold since t he given funct ion is not differentiable at x = 0. √21 ⋅ not applicable. 4. 1 − 6 (i) ± √(7 ⁄ 3). (ii) 3 ⁄ 2. (iii) 3 ⁄ 2. (iv) 1 ⁄ 7. Con dition of differen tiabilit y is n ot satisfied in x < 0 < x + h sin ce f (x) i s non-differentiable at x = 0. Increasing in the intervals [− 2, − 1] and [0, 1] and decreasing in the intervals ] − ∞ , − 2], [− 1, 0] and [1, ∞ [ . e π is bigger than π e. R oot of the equation sin c − (8c ⁄ π 2) = 0 in the open interval ] π ⁄ 6, π ⁄ 2 [. 2.11 Taylor’s Theorem with Lagrange’s Form of Remainder after n Terms If f (x) is a single-valued function of x such that (i) all the derivatives of f (x) upto (n − 1)th are continuous in a ≤ x ≤ a + h, and (ii) f (n) (x) exists in a < x < a + h, then f (a + h) = f (a) + h f ′ (a) + h2 f ′′ (a) + … 2! hn (n) hn − 1 (n − 1) f (a) + f (a + θh), where 0 < θ < 1. (n − 1) ! n! Proof : Consider the function φ defined by + (a + h − x) 2 f ′′ (x) + … 2! (a + h − x) n − 1 (n − 1) A f (x) + (a + h − x) n , + (n − 1) ! n! where A is a constant to be suitably chosen. We choose A such that φ (a) = φ (a + h). φ (x) = f (x) + (a + h − x) f ′ (x) + D-74 DIFFERENTIAL CALCULUS Now φ (a) = f (a) + h f ′ (a) + and φ (a + h) = f (a + h). H ence A is given by h2 hn − 1 (n − 1) A n f ′′ (a) + … + f (a) + h , n! 2! (n − 1) ! f (a + h) = f (a) + hf ′ (a) + h2 f ′′ (a) + … 2! + hn − 1 (n − 1) hn f (a) + A. …(1) (n − 1) ! n! Now, by hypothesis, all the functions f (x), f ′ (x), f ′′ (x),……, f (n − 1) (x) are continuous in the closed interval [a, a + h] and differentiable in the open interval ]a, a + h[. Further (a + h − x), (a + h − x) 2 ⁄ 2 !, …, (a + h − x) n ⁄ n !, all being polynomials, are continuous in the closed interval [a, a + h] and differentiable in the open interval ]a, a + h[. Also A is a constant. ∴ φ (x) is continuous in the closed interval [a, a + h] and differentiable in the open interval ]a, a+ h[. By our choice of A , φ (a) = φ (a + h). Hence φ (x) satisfies all the conditions of R olle’s theorem. Consequently φ ′ (a + θh) = 0, where 0 < θ < 1. Now φ ′ (x) = f ′ (x) − f ′ (x) + (a + h − x) f ′′ (x) − (a + h − x) f ′′ (x) + …+ (a + h − x) n − 1 (n) A f (x) − (a + h − x) n − 1 (n − 1) ! (n − 1) ! (a + h − x) n − 1 (n) [ f (x) − A] , (n − 1) ! since other terms cancel in pairs. φ ′ (a + θh) = 0 gives = ∴ [a + h − (a + θh)]n − 1 (n − 1) ! [ f (n) (a + θh) − A] = 0 or hn − 1 (1 − θ) n − 1 (n) [ f (a + θh) − A] = 0 (n − 1) ! or f (n) (a + θh) − A = 0 A = f (n) (a + θh). [... h ≠ 0, (1 − θ) ≠ 0 as 0 < θ < 1] or Putting this value of A in (1), we get f (a + h) = f (a) + h f ′ (a) + h2 f ′′ (a) + … 2! hn (n) hn − 1 f (n − 1) (a) + f (a + θh). (n − 1) ! n! This is Taylor’s development of f (a + h) in ascending integral powers of h. The hn (n) (n + 1)th term f (a + θh) is called Lagrange’s form of remainder after n terms in n! Taylor’s expansion of f (a + h). + D-75 DIFFERENTIABILITY Note : If we take n = 1, we see that Lagrange’s mean value theorem is a particular case of the above theorem. Corollary : (Maclaurin’s development) : If we take the interval [0, x] instead of [a, a + h], so that changing a to 0 and h to x in Taylor’s theorem, we get f (x) = f (0) + x f ′(0) + xn − 1 xn (n) x2 f ′′ (0) + …… + f (n − 1) (0) + f (θx), 2! (n − 1) ! n! which is known as Maclaurin’s theorem or Maclaurin’s development of f (x) in the x n (n) f (θx) after n terms. interval [0, x] with Lagrange’s form of remainder n! 2.12 Taylor’s Theorem with Cauchy’s Form of Remainder If f (x) is a single-valued function of x such that (i) all the derivatives of f (x) upto (n − 1)th are continuous in a ≤ x ≤ a + h, and (ii) f (n) (x) exists in a < x < a + h, then f (a + h) = f (a) + h f ′ (a) + h2 f ′′ (a) + … 2! hn − 1 (n − 1) hn f (a) + (1 − θ) n − 1 f (n) (a + θh), where 0 < θ < 1. (n − 1) ! (n − 1) ! Proof : Consider the function φ defined by + φ (x) = f (x) + (a + h − x) f ′ (x) + (a + h − x) 2 f ′′ (x) + … 2! (a + h − x) n − 1 (n − 1) f (x) + (a + h − x) A, (n − 1) ! where A is a constant to be suitably chosen. We choose A such that φ (a) = φ (a + h). hn − 1 (n − 1) h2 f ′′ (a) + …+ f (a) + h A, Now φ (a) = f (a) + h f ′ (a) + 2! (n − 1) ! and φ (a + h) = f (a + h). H ence A is given by + h2 hn − 1 (n − 1) f ′′ (a) + … + f (a) + hA. …(1) 2! (n − 1) ! As explained earlier in article 2.11, it can be easily seen that φ (x) satisfies all the conditions of R olle’s theorem. Consequently φ ′ (a + θh) = 0, where 0 < θ < 1. f (a + h) = f (a) + h f ′ (a) + Now φ ′ (x) = ∴ or (a + h − x) n − 1 (n) f (x) − A, since other terms cancel in pairs. (n − 1) ! φ ′ (a + θh) = 0 gives [a + h − (a + θh)]n − 1 (n − 1) ! hn − 1 (1 − θ) n − 1 f (n) (a + θh). (n − 1) ! Putting this value of A in (1), we get A= f (n) (a + θh) − A = 0 D-76 DIFFERENTIAL CALCULUS f (a + h) = f (a) + h f ′ (a) + h2 hn − 1 f ′′ (a) + …+ f (n − 1) (a) 2! (n − 1) ! + hn (1 − θ) n − 1 f (n) (a + θh). (n − 1) ! hn (1 − θ) n − 1 f (n) (a + θh) is called Cauchy’s form of (n − 1) ! remainder after n terms in the Taylor’s expansion of f (a + h) in ascending integral powers of h. The (n + 1)th term Corollary : (Maclaurin’s development with Cauchy’s form of remainder) : If we change a to 0 and h to x in the above result, we get f (x) = f (0) + x f ′(0) + x2 xn − 1 f ′′ (0) + … + f (n − 1) (0) 2! (n − 1) ! xn (1 − θ) n − 1 f (n) (θx), (n − 1) ! which is Maclaurin’s theorem with Cauchy’s form of remainder. The (n + 1)th term xn (1 − θ) n − 1 f (n) (θ x) is known as Cauchy’s form of remainder after n terms in (n − 1) ! Maclaurin’s development of f (x) in the interval [0, x]. + 2.13 Expansions of Some Basic Functions (i) Expansion of e x : Let f (x) = e x. V n ∈ N,— V x ∈ R so that Then f (n) (x) = e x — f (n) (0) = e0 = 1 — V n ∈ N. Now Maclaurin’s expansion of f (x) with Lagrange’s form of remainder is f (x) = f (0) + x f ′ (0) + where Now Rn = x n − 1 (n − 1) x2 f ′′(0) + …+ f (0) + R n 2! (n − 1) ! x n (n) f (θx), 0 < θ < 1. n! lim lim x n (n) lim x n θ x R = f (θx) = e n →∞ n n →∞ n ! n →∞ n ! = eθx lim x n = e θ x × 0 = 0. n →∞ n ! ⎡⎢ . . lim x n ⎤ ⎢⎢ . = 0⎥⎥⎥ n → ∞ n! ⎣ ⎦ Thus f (n) (x) exists in [0, x] for each n ∈ N and R n → 0 as n → ∞ i.e., all the conditions of Maclaurin’s series expansion are satisfied. H ence — V x ∈ R the expansion of e x is e x = 1+ x + (ii) Let x2 x3 xn + + …+ + … 2! 3! n! Expansion of sin x : f (x) = sin x. D-77 DIFFERENTIABILITY 1 2 ⎛1 ⎞ sin⎜ n π ⎟ ⎝2 ⎠ f (n) (x) = sin (x + n π) — V n ∈ N ,— V x∈ R so that f (n) (0) = — V n∈ N or f (n) (0) = ⎨ Then ⎧ 0, if n is even , ⎪ ⎪ (− 1) (n − 1) / 2 , if n is odd. ⎩ The Maclaurin’s expansion of f (x) with Lagrange’s form of remainder is x2 x n − 1 (n − 1) f ′′ (0) + … + f (0) + R n 2! (n − 1) ! f (x) = f (0) + xf ′(0) + Rn = where n π⎞ , x n (n) xn ⎛ f (θx) = sin ⎜ θx + ⎟ 0 < θ < 1. 2 ⎠ n! n! ⎝ ⎪ xn n π ⎞ ⎪⎪ ⎛ Now ⏐ R n ⏐ = ⎪⎪⎪ sin ⎜ θx + ⎟ ⎪= 2 ⎠ ⎪⎪ n ! ⎝ ⎪ lim | Rn | ≤ n →∞ ∴ ⎪⎪ x n ⎪⎪ ⎪ n π ⎞ ⎪ ⎪⎪ x n ⎪⎪ ⎛ ⎪⎪ ⎪⎪ ⎪ sin ⎜ θx + ⎟⎪ ≤ ⎪ ⎪ ⋅ 2 ⎠ ⎪ ⎪⎪ n ! ⎪⎪ ⎝ ⎪ n !⎪ ⎪ ⎡⎢ . . lim x n ⎤ lim ⎪⎪ x n ⎪⎪ ⎪⎪ ⎪⎪ = 0 ⎢⎢ . = 0⎥⎥⎥ n → ∞ n →∞ ⎪n ! ⎪ n ! ⎣ ⎦ lim R = 0. n →∞ n Thus all the conditions of Maclaurin’s series expansion are satisfied. H ence the expansion of sin x is or sin x = x − (iii) x5 x7 x2 n − 1 x3 + − + … + (− 1) n − 1 + … 3! 5! 7! (2n − 1) ! Expansion of cos x : Proceed as above. In this case we get cos x = 1 − x4 x6 x2 + − + … 2! 4! 6! Expansion of log e ( 1+ x ) : (iv) f (x) = log e ( 1+ x ), Let (− 1 < x ≤ 1). V n ∈ N, f (n) (x) = (− 1) n − 1(n − 1) !(1 + x) − n — (n) n − 1 (n − 1) ! — V n∈ N. so that f (0) = (− 1) The Maclaurin’s expansion of f (x) with Lagrange’s form of remainder is Then f (x) = f (0) + x f ′(0) + where Rn = x2 x n − 1 (n − 1) f ′′(0)+ … + f (0) + R n 2! (n − 1) ! x n (n) f (θx) , 0 < θ < 1 n! 1 ⎛ x ⎞n x n (− 1) n − 1(n − 1) ! = (− 1) n − 1 . ⎜ ⎟ ⋅ n n ⎝ 1 + θx⎠ n! (1 + θx) In order to show that R n → 0 as n → ∞ , we consider two cases : = Case I : 0 ≤ x ≤ 1 . Since 0 ≤ x ≤ 1 and 0 < θ < 1, therefore x < 1 + θx and hence lim (− 1) n − 1 lim ⎛ x ⎞ n = 0. ⎜ ⎟ = 0. Also n →∞ n → ∞ ⎝ 1 + θ x⎠ n D-78 DIFFERENTIAL CALCULUS Thus, in this case, lim R n = 0. n →∞ Case II : − 1 < x < 0 . In this case it will not be convenient to show that Lagrange’s form of remainder R n → 0 as n → ∞ because x ⁄ (1 + θx) may not be numerically less than unity. Therefore we use the Cauchy’s form of remainder. We have Rn = = xn (1 − θ) n − 1f (n) (θx) (n − 1) ! xn (1 − θ) n − 1(− 1) n − 1(n − 1) ! (1 + θx) − n (n − 1) ! = (− 1) n − 1 xn ⎛ 1− θ⎞n− 1 1 , ⎟ ⎜ ⋅ 1 + θx 1 + θx ⎠ ⎝ (0 < θ < 1). Now as 0 < θ < 1 and − 1 < x < 0 , we have 1− θ lim ⎛ 1 − θ ⎞ n − 1 < 1, so that = 0. 0< ⎜ ⎟ n → ∞ ⎝ 1 + θx ⎠ 1 + θx lim Also, x n = 0 as − 1 < x < 0. n →∞ Thus, in this case also R n → 0 as n → ∞ . H ence, f (x) satisfies all the conditions of Maclaurin’s series expansion for − 1 < x ≤ 1. Therefore, for − 1< x ≤ 1, we get log e (1 + x) = x − (v) Let x2 x3 x4 xn + − + … + (− 1) n − 1 + … 2 3 4 n Expansion of (1 + x)m : f (x) = (1 + x) m , — V x∈ R. V n∈ N. Then f (n) (x) = m(m − 1) (m − 2) … (m − n + 1) (1 + x) m − n — Now we consider two cases : Case I : If m is a positive integer. In this case, we notice that for n > m , f (n) (x) = 0. So all the terms after the (m + 1)th term vanish and so the expansion consists of finite number of terms in the form x2 x m (m) f ′′(0) + … + f (0). 2! m! Case II : If m is a fraction or a negative integer. In this case, let ⏐ x ⏐ < 1. We have f (n) (x) = m (m − 1) (m − 2) … (m − n + 1) (1 + x) m − n , x ≠ − 1. H ere, we use Maclaurin’s expansion with Cauchy’s form of remainder. Thus, we f (x) = f (0) + x f ′(0) + have f (x) = f (0) + x f ′(0) + x n − 1 (n − 1) x2 f ′′(0) + … + f (0) + R n 2! (n − 1) ! D-79 DIFFERENTIABILITY Rn = where = ∴ xn (1 − θ) n − 1f (n) (θ x), 0 < θ < 1 (n − 1) ! xn (1 − θ) n − 1. m(m − 1) … (m − n + 1) (1 + θ x) m − n (n − 1) ! ⎛ 1− θ ⎞n− 1 m(m − 1) … (m − n + 1) ⎟ = ⎜ . (1 + θ x) m − 1 ⋅ ⋅ x n. (n − 1)! ⎝ 1 + θ x⎠ an + 1 m − n m(m − 1) … (m − n + 1) ⋅ x n , then = ⋅ x If an = (n − 1)! n an (on simplification) a ⎛ ⎞ lim n + 1 lim m = ⎜ − 1⎟ x = (0 − 1) x = − x. n → ∞ an n →∞ ⎝ n ⎠ This gives lim a = 0, since ⏐ − x⏐ = ⏐ x⏐ < 1. n →∞ n Further 0 < θ < 1 ⇒ 0 < 1 − θ < 1 + θ x n− 1 1− θ lim ⎛ 1 − θ ⎞ ⎟ ⎜ < 1 which gives = 0. n → ∞ ⎝ 1 + θ x⎠ 1 + θx lim H ence in this case R = 0. Thus f (x) satisfies the conditions of Maclaurin’s n →∞ n series expansion. Therefore for − 1 < x < 1 , we have so that (1 + x) m = 1 + mx + m(m − 1) m(m − 1)(m − 2) ⋅ x2 + ⋅ x3 + … 2! 3! Example 1 : If f (x) = f (0) + x f ′ (0) + x2 f ′′ (θx), 2! find the value of θ as x → 1, f (x) being (1 − x) 5 ⁄ 2. Solution : H ere f (x) = (1 − x) 5 ⁄ 2. ∴ f ′ (x) = − 5 2 (1 − x) 3 ⁄ 2 and f ′′ (x) = 15 (1 − x) 1 ⁄ 2. 4 , f ′′ (θ x) = 15 (1 − θ x) 1 ⁄ 2. 4 Putting these values in (1), we get Thus f (0) = 1, f ′(0) = − 5 2 5 x2 15 x+ × (1 − θ x) 1 ⁄ 2. 2! 2 4 Therefore as x → 1, we have 1 15 5 ⋅ (1 − θ) 1 ⁄ 2 0= 1− + 2 2! 4 4 16 9 or (1 − θ) 1 ⁄ 2 = or (1 − θ) = or θ= ⋅ 5 25 25 (1 − x) 5 ⁄ 2 = 1 − …(1) D-80 DIFFERENTIAL CALCULUS Example 2 : Show that the num ber θ which occurs in the Taylor’s theorem with L agrange’s form of rem ainder after n terms approaches the lim it 1 ⁄ (n + 1) as h → 0 provided that f (n + 1) (x) is continuous and different from zero at x = a. Solution : Applying Taylor’s theorem with Lagrange’s form of remainder after n terms and (n + 1) terms successively, we get for θ, θ′ ∈ ]0, 1[, and f (a + h) = f (a) + h f ′ (a) + … + hn − 1 (n − 1) hn (n) f (a) + f (a + θ h), (n − 1) ! n! f (a + h) = f (a) + h f ′ (a) + … + hn (n) hn + 1 (n + 1) f (a) + f (a + θ′ h). n! (n + 1) ! Subtracting these, we have hn f (n) (a) hn + 1 (n + 1) hn (n) + f (a + θ′h) = f (a + θh) n! (n + 1) ! n! h f (n + 1) (a + θ′h) f (n) (a + θh) − f (n) (a) = n+ 1 or …(1) Applying Lagrange’s mean value theorem to the function f (n) (x) in the interval [a, a + θh], we get f (n) (a + θh) − f (n) (a) = θhf (n + 1) (a + θθ′′h), 0 < θ′′ < 1. From (1) and (2), we have h θhf (n + 1) (a + θθ′′h) = f (n + 1) (a + θ′h) n+ 1 θ= or …(2) f (n + 1) (a + θ′h) 1 ⋅ n + 1 f (n + 1) (a + θθ′′h) 1 1 f (n + 1) (a) lim = , provided f (n + 1) (a) ≠ 0. θ= h →0 n + 1 f (n + 1) (a) n + 1 Example 3 : Assuming the derivatives which occur are continuous, apply the m ean value theorem to prove that φ ′ (x) = F ′ { f (x)} f ′ (x) where φ (x) = F { f (x)}. ∴ Solution : Let f (x) = t so that φ (x) = F (t). lim F { f (x + h)} − F { f (x)} lim φ (x + h) − φ (x) = h →0 h →0 h h F { f (x) + h f ′ (x + θ h)} − F{ f (x)} lim 1 = , (0 < θ1 < 1) h →0 h [... f (x + h) = f (x) + h f ′ (x + θ1 h), by mean value theorem] Now φ ′ (x) = lim F (t + H) − F (t) , where H = h f ′ (x + θ1h) h →0 h lim HF ′ (t + θ2 H) = h →0 h . . [ . F(t + H) = F (t) + HF ′ (t + θ2 H), by mean value theorem] = lim h f ′ (x + θ1 h) F ′ [t + θ2 h f ′(x + θ1 h)] h →0 h = f ′ (x) F ′ (t) = F ′ { f (x)} f ′ (x). Note : This example gives an alternative proof of the chain rule. = D-81 DIFFERENTIABILITY 1. 2. 3. h2 f ′′ (x + θh), 2! find the value of θ as x → a, f (x) being (x − a) 5 ⁄ 2. If f (x + h) = f (x) + h f ′ (x) + h2 f ′′ (x + θh), 0 < θ < 1, and 2! (ii) f (x) = x3. (i) f (x) = ax3 + bx2 + c x + d Show that ‘θ’ (which occurs in the Lagrange’s mean value theorem) approaches 1 the limit as ‘h ’ approaches zero provided that f ′′ (a) is not zero. It is assumed Find θ, if f (x + h) = f (x) + hf ′ (x) + 2 4. 1. that f ′′ (x) is continuous. Show that the number θ which occurs in the Taylor’s theorem with Lagrange’s form of remainder after n terms approaches the limit 1 ⁄ (n + 1) as h → 0 provided that f (n + 1) (x) is continuous and different from zero at x = a. θ= 64 ⋅ 225 2. (i) θ = 1 3 ⋅ (ii) θ= 1 3 ⋅ Fill in the Blanks: 1. 2. 3. 4. 5. 6. Fill in the blanks “……” so that the following statem ents are com plete and correct. A function f (x) is said to be differentiable at x = a if lim f (x) − … exists. x →a x − a The right hand derivative of f (x) at x = a is given by lim f (a + h) − f (a) , h > 0, h →0 … provided the limit exists. The left hand derivative of f (x) at x = a is given by lim f (a − h) − f (a) , h > 0, h →0 … provided the limit exists. A function f : ]a, b[ → R is said to be differentiable in ]a, b[ if and only if it is differentiable at every point in … . If a function f (x) is differentiable at x = a, then f ′(a) is the tangent of the angle which the tangent line to the curve y = f (x) at the point P (a, f (a)) makes with …. Continuity is a necessary but not a … condition for the existence of a finite derivative. D-82 7. 8. 9. 10. 11. 12. DIFFERENTIAL CALCULUS The function f (x) = | x | is differentiable at every point of R except at x = … . If a function f (x) is such that (i) f (x) is continuous in the closed interval [a, b], (ii) f ′(x) exists for every point in the open interval ]a, b[ , (iii) f (a) = f (b), then there exists at least one value of x, say c , where a < c < b, such that f ′(c) = 0. The above theorem is known as … . If a function f (x) is (i) continuous in the closed interval [a, b], and (ii) differentiable in the open interval ]a, b[ i.e., a < x < b, then there exists at least one value ‘c ’ of x lying in the open interval ]a, b[ such that f (b) − f (a) = …. b− a If two functions f (x) and g (x) are (i) continuous in a closed interval [a, b] (ii) differentiable in the open interval ]a, b[ , and (iii) g′(x) ≠ 0 for any point of the open interval ]a, b[, then there exists at least one value c of x in the open interval ]a, b[, such that f (b) − f (a) f ′(c) = ⋅ g′(c) … If f is continuous in [a, b] and f ′(x) ≥ 0 in ]a, b[, then f is … in [a, b]. If f (x) = sin x, then lim f (x + h) − f (x) = …. h →0 h Multiple Choice Questions: 13. 14. 15. 16. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). The function f (x) = | x − 1| is not differentiable at (a) x = 0 (b) x = − 1 (c) x = 1 (d) x = 2 The function f (x) = | x + 3| is not differentiable at (a) x = 3 (b) x = − 3 (c) x = 0 (d) x = 1 A function f (x) is differentiable at x = a if (a) R f ′(a) = L f ′(a) (b) R f ′(a) = 0 (c) L f ′(a) = 0 (d) R f ′(a) ≠ L f ′(a) A function φ (x) is defined as follows : φ (x) = 1 + x if x ≤ 2 φ (x) = 5 − x if x > 2 . Then (a) φ (x) is continuous but not differentiable at x = 2 (b) φ (x) is differentiable at every point of R (c) φ (x) is neither continuous nor differentiable at x = 2 (d) φ (x) is differentiable at x = 2 but is not continuous at x = 2 . D-83 DIFFERENTIABILITY 17. 18. 19. O ut of the following four functions tell the function for which the conditions of R olle’s theorem are satisfied. (a) f (x) = | x | in [− 1, 1] (b) f (x) = x2 in 2 ≤ x ≤ 3 (c) f (x) = sin x in [0, π] (d) f (x) = tan x in 0 ≤ x ≤ π. The function f (x) = sin x is increasing in the interval ⎡ π⎤ (a) [0, π] (b) ⎢ 0, ⎥ ⎣ 2⎦ ⎡ π 3π ⎤ ⎡π ⎤ (c) ⎢⎣ , ⎥⎦ (d) ⎢ , π ⎥ ⎣ 4 4 2 ⎦ The value of ‘c ’ of Lagrange’s mean value theorem for f (x) = x (x − 1) in [1, 2] is given by (a) c = (c) c = 20. 5 4 7 4 (b) c= (d) c= 3 2 11 6 The value of ‘c ’ of R olle’s theorem for the function f (x) = e x sin x in [0, π] is given by 3π (a) c = 4 π (b) c = 4 π (c) c = 2 5π (d) c = 6 True or False: Write ‘T ’ for true and ‘F ’ for false statem ent. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. If a function f (x) is continuous at x = a, it must also be differentiable at x = a. If a function f (x) is differentiable at x = a, it must be continuous at x = a. If a function f (x) is differentiable at x = a, it may or may not be continuous at x = a. The function f (x) = | x | is differentiable at every point of R. R olle’s theorem is applicable for f (x) = sin x in [0, 2 π]. R olle’s theorem is applicable for f (x) = | x| in [− 1, 1]. Lagrange’s mean value theorem is applicable for f (x) = | x| in [− 1, 1]. ⎡ π π⎤ The function f (x) = sin x is increasing in ⎢ − , ⎥ ⋅ ⎣ 2 2⎦ If a + b + c = 0, then the quadratic equation 3ax2 + 2 bx + c = 0 has no root in ]0, 1[. If f is continuous on [a, b] and f ′(x) ≤ 0 in ]a, b[, then f is increasing in [a, b]. The function f (x) = 2 x3 − 15 x2 + 36 x + 1 is decreasing in the interval [2 , 3]. Let f (x) = | x| + | x − 1| . Then Rf ′(0) = 0. R olle’s theorem is not applicable for the function f (x) = x (x + 2) e − x ⁄ 2 in [− 2 , 0]. D-84 34. DIFFERENTIAL CALCULUS The value of ‘c ’ of Lagrange’s mean value theorem for the function f (x) = 2 x2 + 3 x + 4 in [1, 2] is given by c = 5 ⋅ 4 lim f (x + h) − f (x) = n x n− 1. h →0 h 35. If f (x) = x n , then 36. If f (x) = cos x, then 37. lim f (x) − f (x0) = e x. If f (x) = e x, then x → x x − x0 0 1. 6. 10. 15. 20. 25. 30. 35. f (a). sufficient. g (b) − g (a). (a). (a). T. F. T. lim f (x) − f (a) = − sin a. x →a x− a 2. h. 7. 0. 11. increasing. 16. (a). 21. F. 26. F. 31. T. 36. T. 3. − h. 4. ]a, b[. 8. R olle’s theorem. 12. cos x. 13. (c). 17. (c). 18. (b). 22. T. 23. F. 27. F. 28. T. 32. T. 33. F. 37. F. 5. 9. 14. 19. 24. 29. 34. the x-axis. f ′(c). (b). (b). F. F. F. 3.1 Increments In differential calculus we use the word ‘Increment’ to denote a small change (increase or decrease) in the value of any variable. Thus if x be a variable, then a small change in the value of x is called the increment in x and we usually denote it by δx which is read as ‘delta x’. It should be noted that δx does not mean δ multiplied by x. It represents a single quantity which stands for the increment in x. Sometimes we also use the single letters h, k etc. to denote increments. Now suppose y = f (x) is a function of the variable x. Let δy denote the increment in y corresponding to an increment δx in x. Then y + δy = f (x + δx). Therefore δy = f (x + δx) − f (x). δy f (x + δx) − f (x) = The quotient δx δx is called the average rate of change of y with respect to x in the interval (x, x + δx). 3.2 The Differential Coefficient Definition : The differential coefficient of a function y = f (x) -with respect to x is defined as D-86 DIFFERENTIAL CALCULUS f (x + δx) − f (x) δy lim lim δx → 0 δx = δx → 0 δx provided the lim it exists. The differential coefficient is also called the derivative, or the derived function. The differential coefficient of y = f (x) with respect to x may be denoted by any of the symbols d d dy y, , y′, Dy, f (x), f ′ (x), Df (x). dx dx dx The process of finding the differential coefficient is called differentiation. The differential coefficient (dy ⁄ dx) is also called the instantaneous rate of change (or simply, the rate of change) of y with respect to x. The Differential Coefficient at a Point : If y = f (x) is a function of x, then lim f (a + h) − f (a) h →0 h is called the differential coefficient of f (x) for x = a, provided the above limit exists. It ⎛ dy⎞ is denoted by ⎜ ⎟ , ( y′) a , or f ′ (a). It gives us the rate of change of y with respect ⎝ dx⎠ x = a to x at x = a. 3.3 Differential Coefficient of x n (n being Real Number) Let y = x n . Then y + δy = (x + δx) n . Therefore δy = (x + δx) n − x n . δy (x + δx) n − x n = ⋅ δx δx Taking limit when δx → 0, we get ∴ δy dy lim lim (x + δx) n − x n = δx → 0 = δx → 0 δx dx δx lim = δx → 0 δx ⎞ n + ⎟ − xn x⎠ ⎝ ⎛ xn ⎜1 δx lim = δx → 0 n ⎡⎛ ⎤ δx ⎞ xn ⎢ ⎜1 + ⎟ − 1⎥ x⎠ ⎣⎝ ⎦ δx ⎡ ⎤ n (n − 1) ⎛ δx ⎞ ⎛ δx ⎞ xn ⎢ 1 + n ⎜ ⎟ + ⎜ ⎟ + … − 1⎥ ⎝ ⎠ ⎝ ⎠ 2 ! x x ⎣ ⎦ 2 lim = δx → 0 lim = δx → 0 δx [E xpanding by binomial theorem since δx ⁄ x is numerically less than unity, δx being numerically small] ⎡ ⎤ δx n (n − 1) ⎛ δx ⎞ 2 ⎛ ⎞ xn ⎢ n ⎜ ⎟ + ⎜ ⎟ + .....⎥ 2! ⎝ x⎠ ⎣ ⎝ x⎠ ⎦ δx n (n − 1) δx n ⎡n ⎤ lim = δx → 0 x n ⎢ + + ......⎥ = x n . = nx n − 1. 2! x ⎣x ⎦ x2 D-87 DIFFERENTIATION d n x = nx n − 1. dx d 4 (x ) = 4x4 − 1 = 4x3. Illustration 1 : dx d ⎛⎜ 1 ⎞⎟ d − 1⁄3 1 1 Illustration 2 : = (x ) = − x− 4⁄3 = − ⋅ dx ⎜⎜ x1 ⁄ 3 ⎟⎟ dx 3 3x 4 ⁄ 3 ⎝ ⎠ Thus 3.4 Differential Coefficient of sin x We have Thus sin (x + δx) − sin x d lim sin x = δx → 0 , by definition dx δx δx ⎞ δx ⎛ δx 2 cos ⎜ x + ⎟ sin sin 2⎠ 2 ⎝ δx⎞ 2 ⎛ lim lim = δx → 0 = δx → 0 cos ⎜ x + ⎟ ⎝ 2 ⎠ δx δx 2 lim sin (δx ⁄ 2) = cos x, since δx → 0 = 1. δx ⁄ 2 d sin x = cos x. dx Similarly, it can be shown that d cos x = − sin x. dx lim sin x Note : x → 0 = 1 is true only when x is expressed in terms of radians. In x case x is given in terms of degrees, it should be first expressed in terms of radians before applying the above results. 3.5 Differential Coefficient of e x We have d x lim e x + δx − e x e = δx → 0 , by definition δx dx lim e x eδx − e x lim e x (e δx − 1) = δx → 0 = δx → 0 δx δx 2 3 ⎡ ⎤ δx (δx) (δx) ex ⎢ 1 + + + + ..... − 1⎥ 1! 2! 3! ⎣ ⎦ lim = δx → 0 δx 2 3 ⎡ ⎤ δx (δx) (δx) ex ⎢ + + + ....⎥ ⎣1 ! ⎦ 2! 3! lim = δx → 0 δx Thus ⎡ ⎤ δx (δx) 2 lim = δx → 0 e x ⎢ 1 + + + …⎥ = e x. ⎣ ⎦ 2! 3! d x (e ) = e x. dx D-88 DIFFERENTIAL CALCULUS 3.6 Differential Coefficient of log e x We have log (x + δx) − log x d lim log x = δx → 0 , by definition dx δx δx ⎞ ⎛ x + δx ⎞ ⎛ log ⎜ 1 + log ⎜ ⎟ ⎟ ⎝ x ⎠ ⎝ x⎠ lim lim = δx → 0 = δx → 0 δx δx δx (δx) 2 (δx) 3 − + − ..... x 2x2 3x3 lim = δx → 0 δx (the expansion is justified since δx ⁄ x is numerically less than unity, δx being numerically small) Thus lim = δx → 0 d 1 (log e x) = ⋅ dx x 2 ⎡1 ⎤ ⎢ − δx + (δx) − ......⎥ = 1 ⋅ ⎢ ⎥ 2 3 ⎢x ⎥ x 2x 3x ⎣ ⎦ 3.7 Differential Coefficient of a Constant Let f (x) = c, where c is a constant. d lim f (x + δx) − f (x) f (x) = δx → 0 , by definition Then δx dx lim c − c , since f (x) = c for every value of x = δx → 0 δx 0 lim = δx → 0 = 0. δx Thus, the differential coefficient of a constant is zero. 3.8 Differential Coefficient of the Product of a Constant and a Function Let c be a constant and f (x) be a function of x. Then by definition d lim c f (x + δx) − c f (x) {c f (x)} = δx → 0 δx dx f (x + δx) − f (x) lim = δx → 0 c ⋅ δx d lim f (x + δx) − f (x) = c f (x). = c ⋅ δx → 0 dx δx Thus the differential coefficient of the product of a constant and a function is equal to the product of the constant and the differential coefficient of the function. d d x Illustration 1 : (4e x) = 4. e = 4e x. dx dx d ⎛⎜ 1 ⎞⎟ 1 d ⎛⎜ 1 ⎞⎟ 1 d − 4⁄3 Illustration 2 : = = x dx ⎜⎜ 2x4 ⁄ 3 ⎟⎟ 2 dx ⎜⎜ x4 ⁄ 3 ⎟⎟ 2 dx ⎝ ⎠ ⎝ ⎠ D-89 DIFFERENTIATION = 1 ⎛ 4⎞ − 7 ⁄ 3 2 = − ⋅ ⎜− ⎟ x 2 ⎝ 3⎠ 3x7 ⁄ 3 3.9 Differential Coefficient of log a x We have log a x = log e x log a e = log a e log e x, where loga e is simply a constant. d 1 1 d (log a x) = log a e (log e x) = (log a e) . = ⋅ log a e ∴ dx x x dx 1 , since log e . log a = 1. = a e x log e a Thus d 1 1 (log a x) = = ⋅ dx x log e a x log a 3.10 Differential Coefficient of the Sum of Two Functions f (x) = f1 (x) + f2 (x). Then f (x + δx) = f1 (x + δx) + f2 (x + δx). Therefore, by definition d lim f (x + δx) − f (x) f (x) = δx → 0 δx dx { f (x + δx) + f2 (x + δx)} − { f1 (x) + f2 (x)} 1 lim = δx → 0 δx Let lim { f1 (x + δx) − f1 (x)} + { f2 (x + δx) − f2 (x)} = δx → 0 δx ⎧ ⎫ lim ⎪ f1 (x + δx) − f1 (x) f2 (x + δx) − f2 (x) ⎪⎬ = δx → 0 ⎨⎪ + ⎪ δx δx ⎩ ⎭ lim f1 (x + δx) − f1 (x) lim f2 (x + δx) − f2 (x) = δx → 0 + δx → 0 δx δx d d = f (x) + f (x). dx 1 dx 2 Thus, the differential coefficient of the sum of two functions is equal to the sum of their differential coefficients. This theorem can be extended for the sum of any number of functions. Thus d d d d { f (x) + f2 (x) + .... + fn (x)} = f (x) + f (x) + .... + f (x) . dx 1 dx 2 dx n dx 1 d d d d 3 Illustration 1 : (7e x + 4 log x + x3) = 7e x + 4 log x + x dx dx dx dx Illustration 2 : = 7e x + 4 . (1 ⁄ x) + 3x2. d d d (8 x3 − sin x) = 8 x3 − sin x = 24 x2 − cos x. dx dx dx D-90 DIFFERENTIAL CALCULUS 3.11 Differential Coefficient of the Product of Two Functions Let f (x) = f1 (x) f2 (x). Then f (x + δx) = f1 (x + δx) f2 (x + δx). Therefore, by definition d lim f (x + δx) − f (x) f (x) = δx → 0 dx δx lim f1 (x + δx) f2 (x + δx) − f1 (x) f2 (x) = δx → 0 δx lim = δx → 0 f1 (x + δx) f2 (x + δx) − f1 (x + δx) f2 (x) + f1 (x + δx) f2 (x) − f1 (x) f2 (x) δx (by adding and subtracting the term f1 (x + δx) f2 (x) in the numerator) lim f1 (x + δx) { f2 (x + δx) − f2 (x)} + f2 (x) { f1 (x + δx) − f1 (x)} = δx → 0 δx ⎡ f2 (x + δx) − f2 (x) ⎤⎥ lim = δx → 0 f1 (x + δx) ⎢⎢⎢ ⎥⎥ δx ⎣ ⎦ ⎡ f1 (x + δx) − f1 (x) ⎤⎥ lim ⎥⎥ + δx → 0 f2 (x) ⎢⎢⎢ δx ⎣ ⎦ d d = f1 (x) f (x) + f2 (x) f (x). dx 2 dx 1 Thus, the differential coefficient of the product of two functions = first function × differential coefficient of the second + second function × differential coefficient of the first. Illustration 1 : d x d d (e cos x) = e x cos x + cos x e x dx dx dx Illustration 2 : = − e x . sin x + cos x . e x = e x (cos x − sin x). d 3 d d (x log x) = x3 (log x) + log x x3 dx dx dx = x3 . (1 ⁄ x) + (log x) . 3x2 = x2 (1 + 3 log x). 3.12 Differential Coefficient of the Quotient of Two Functions f1 (x) Let f (x) = Then f (x + δx) = f2 (x) ⋅ f1 (x + δx) f2 (x + δx) Therefore, by definition ⋅ D-91 DIFFERENTIATION f1 (x + δx) f (x + δx) − f (x) d lim lim f (x) = δx → 0 = δx → 0 dx δx f1 (x) f1 (x + δx) − − f2 (x + δx) f2 (x + δx) lim = δx → 0 δx f2 (x + δx) f1 (x) f2 (x) + − f1 (x) f2 (x) δx f1 (x) f2 (x + δx) f1 (x) ⎛ ⎞ ⎜ by adding and subtracting the term in the numerator⎟⎟ ⎜ ⎜ ⎟ f2 (x + δx) ⎝ ⎠ ⎧ f (x + δx) − f (x) ⎫ ⎪ 2 ⎪ 1 2 ⎬ { f (x + δx) − f1 (x)} − f1 (x) ⎨ ⎪ f2 (x) f2 (x + δx) ⎪ f2 (x + δx) 1 ⎩ ⎭ lim = δx → 0 δx ⎧ f1 (x + δx) − f1 (x) ⎫⎪ 1 lim ⎪ ⎬ = δx → 0 ⎨ ⋅ ⎪ f2 (x + δx) ⎪ δx ⎩ ⎭ ⎧ f1 (x) f2 (x + δx) − f2 (x) ⎫⎪ lim ⎪ ⎬ − δx → 0 ⎨ ⋅ ⎪ f2 (x) f2 (x + δx) ⎪ δx ⎩ ⎭ f1 (x) d d 1 ⋅ f1 (x) − ⋅ f (x) f2 (x) dx f2 (x) f2 (x) dx 2 d d f (x) − f1 (x) . f (x) f2 (x) . dx 1 dx 2 ⋅ = [ f2 (x)]2 = ⎧ ⎫ d ⎪ f1 (x) ⎪ ⎨ ⎬= Thus, dx ⎪ f2 (x) ⎪ ⎩ ⎭ (Diff. coeff. of Numerator) × (Denominator) − (Numerator) × (Diff. Coeff. of Denominator) ⋅ Square of the Denominator 3.13 Differential Coefficient of tan x We have, = Thus, d d ⎛ sin x ⎞ tan x = ⎜ ⎟ = dx dx ⎝ cos x ⎠ d ⎛ d ⎞ sin x⎟ cos x − sin x cos x ⎜ dx ⎝ dx ⎠ cos2 x , by article 3.12 cos x . cos x − sin x . (− sin x) cos2 x + sin 2 x 1 = = = sec 2 x. 2 2 cos x cos2 x cos x d tan x = sec 2 x . dx Similarly, we can show that d cot x = − cosec 2 x . dx D-92 DIFFERENTIAL CALCULUS 3.14 Differential Coefficient of cosec x d d ⎛ 1 ⎞ cosec x = ⎜ ⎟ dx dx ⎝ sin x⎠ d ⎛ d ⎞ 1⎟ . sin x − 1 . (sin x) ⎜ dx ⎝ dx ⎠ 0 − cos x = = − cosec x cot x. = 2 sin x sin2 x d cosec x = − cosec x cot x . Thus, dx d Similarly, we can show that sec x = sec x tan x . dx We have, 3.15 Differential Coefficient of a Function of a Function Consider the function log sin x. H ere log (sin x) is a function of sin x whereas sin x is itself a function of x. Thus we have case of a function of a function. Let y = f {φ (x)}. Put t = φ (x). Then t + δt = φ (x + δx). As δx → 0, δt also → 0. dy lim δy lim ⎛ δy δt ⎞ We have = δx → 0 = δx → 0 ⎜ . ⎟ dx δx ⎝ δt δx⎠ δy⎞ ⎛ lim δt ⎞ ⎛ lim = ⎜ δx → 0 ⎟ .⎜ ⎟ δt ⎠ ⎝ δx → 0 δx⎠ ⎝ δy⎞ ⎛ lim δt ⎞ ⎛ lim = ⎜ δt → 0 ⎟ . ⎜ δx → 0 ⎟ , since δt → 0 when δx → 0 ⎝ δt ⎠ ⎝ δx⎠ dy dt = ⋅ ⋅ dt dx Thus, if y is a function of t and t is a function of x, then y is also a function of x and we have dy dy dt = ⋅ ⋅ dx dt dx Similarly, if y is a function of u, u is a function of v and v is a function of x,then y is also a function of x and we have, dy dy du dv = ⋅ ⋅ ⋅ dx du dv dx Example 1 : Find the differential coefficient of sin 2x. Solution : Put 2x = t. d d d ⎛d ⎞ dt sin 2x = sin t = ⎜ sin t⎟ . = cos t . (2x) Then dx dx dx ⎝ dt ⎠ dx = (cos t) . 2 = 2 cos t = 2 cos 2x. D-93 DIFFERENTIATION Example 2 : Find the differential coefficient of tan3 x. Solution : Put tan x = t. d 3 ⎛ d 3⎞ dt d d tan3 x = t = ⎜ t ⎟⋅ = 3t2 ⋅ (tan x) Then dx dx dx ⎝ dt ⎠ dx = 3t2 . sec 2 x = 3 tan2 x sec 2 x. Example 3 : Find the differential coefficient of log sin x. Solution : We have d 1 d ⎛ ⎞ d log sin x = ⎜ log sin x⎟ . (sin x) = . cos x = cot x. dx sin x ⎝ d (sin x) ⎠ dx 3.16 Differential Coefficient of a x We have ∴ x a x = e log a = e x log a. d x d d a = (e x log a) = e x log a . (x log a) dx dx dx = (log a) . e x log a = a x log a. Thus, d x a = a x log a. dx 3.17 Differential Coefficient of sin− 1 x Let Then sin− 1 x = y. x = sin y. Differentiating both sides with respect to x, we get 1 = or d (sin y) dx d sin y . dy dy or 1 = cos y ⋅ ⋅ dy dx dx 1 1 1 dy = = = ⋅ ∴ dx cos y √(1 − sin2 y) √(1 − x2) d 1 Thus, sin− 1 x = ⋅ dx √(1 − x2) Similarly, we can prove the following other results for inverse circular functions : 1 d 1 d 1 d cos− 1 x = − ; tan− 1 x = ; cot− 1 x = − ; 2 2 dx dx dx √(1 − x ) 1+ x 1 + x2 d 1 d 1 sec− 1 x = ; cosec− 1 x = − ⋅ 2 dx dx x√(x − 1) x √ (x2 − 1) 1= 3.18 Inverse Functions Let y = f (x) be a function of x. If when solved for x, this relation can be written as x = f − 1 ( y), then f − 1 is called the inverse function of the function f. H ere f − 1 should be regarded as one symbol like F, or φ , or g. D-94 DIFFERENTIAL CALCULUS In the relation y = f (x), x is regarded as the independent variable, while in the relation x = f − 1 ( y), y is the independent variable. dy dx By differentiation, we get and respectively. dx dy The relation between these two differential coefficients can be obtained as follows : Let δx and δy be the corresponding increments in x and y respectively. Then we have δy δx ⋅ = 1. δx δy Taking limit of both sides when δx → 0, we get ⎛ δy δx⎞ lim δx → 0 ⎜⎝ δx ⋅ δy⎟⎠ = 1 δy⎞ ⎛ lim δx⎞ ⎛ lim or ⎜ δx → 0 ⎟ ⎜ ⎟ = 1 ⎝ δx⎠ ⎝ δx → 0 δy⎠ δy⎞ ⎛ lim δx⎞ ⎛ lim ⎜ δx → 0 ⎟ .⎜ ⎟ = 1, since δy → 0 as δx → 0 ⎝ δx⎠ ⎝ δy → 0 δy⎠ or or Thus dy ⋅ dx dy ⋅ dx dx = 1 dy dx = 1 dy dy 1 = ⋅ dx dx ⁄ dy dy 1 i.e., = ⋅ dx dx ⁄ dy or 3.19 Logarithmic Differentiation Whenever we are required to differentiate a function of x in which a function of x is raised to a power which itself is a function of x, neither the formula for a x nor that for x n is applicable. In such cases we first take logarithm of the function and then differentiate. This process is called logarithmic differentiation. It is also helpful in the cases where we are to differentiate a function which consists of the product or the quotient of a number of functions. Example 1 : Find the differential coefficient of (sin − 1 x) log x. Solution : Let y = (sin − 1 x) log x. Taking logarithm of both sides, we have log y = (log x). log sin − 1 x. Differentiating both sides with respect to x, we get 1 dy 1 1 1 = (log x) ⋅ ⋅ + ⋅ log sin − 1 x. − 1 2 y dx x sin x √(1 − x ) ∴ ⎡ log x log sin − 1 x⎤⎥ dy = y ⎢⎢ + ⎥ ⎢ (sin − 1 x) √(1 − x2) ⎥ x dx ⎣ ⎦ ⎡ = (sin − 1 x) log x ⎢⎢ log x + ⎢ (sin − 1 x) √(1 − x2) ⎣ log sin − 1 x⎤⎥ ⎥ ⋅ ⎥ x ⎦ DIFFERENTIATION D-95 3.20 Differential Coefficient of the Product of any Number of Functions Let Then ∴ y = f1 (x) f2 (x) f3 (x) … fn (x). log y = log f1 (x) + log f2 (x) + … + log fn (x). fn ′ (x) ⎤ 1 dy ⎡⎢ f1′ (x) f2 ′ (x) ⎥ ⋅ = ⎢ + + …+ ⎥ y dx ⎢⎣ f1 (x) f2 (x) fn (x) ⎥⎦ f2′ (x) fn ′ (x) ⎤ ⎡ f1′ (x) dy ⎥ = y ⎢⎢ + + …+ ⎥ ⎢ f1 (x) (x) dx f fn (x) ⎥⎦ 2 ⎣ dy or = f1′ (x) f2 (x) f3 (x)… fn (x) + f1 (x) f2′ (x) f3 (x)…fn (x) + dx … + f1 (x) f2 (x)… fn − 1 (x) fn ′ (x). Thus to differentiate the product of any num ber of functions multiply the differential coefficient of each function taken separately by the product of all the rem aining functions and then add up the results. ∴ 3.21 Implicit Functions If y is a function of x given by a relation of the type y = f (x), then y is said to be an explicit function of x. O n the other hand, if the relation between x and y is given by an equation involving both x and y , then y is said to be an implicit function of x. If we are given y implicitly in terms of x, we can find dy ⁄ dx without first expressing y explicitly in terms of x. Thinking of y as a function of x, we differentiate both sides of the given equation with respect to x and then solve the resulting relation for dy ⁄ dx. Example 1 : Find dy/dx when ax2 + 2hxy + by2 + 2g x + 2f y + c = 0. Solution : R egarding y as a function of x, differentiating both sides of the given equation with respect to x, we get dy dy dy = 0. 2ax + 2hy + 2hx + 2by + 2g + 2f dx dx dx dy Therefore (2hx + 2by + 2f ) = − (2ax + 2hy + 2g) dx dy ax + hy + g or = − ⋅ dx hx + by + f dy (cos x)…to inf Example 2 : Find if y = (cos x) (cos x) . dx Solution : From the given expression for y, it follows that y = (cos x) y or log y = y log cos x. Now differentiating both sides with respect to x, we get 1 dy dy 1 = log cos x + y. (− sin x) y dx dx cos x D-96 DIFFERENTIAL CALCULUS dy ⎡ 1 ⎤ ⎢ − log cos x⎥ = − y tan x ⎦ dx ⎣ y or or dy ( y2 tan x) = − ⋅ dx 1 − y log cos x 3.22 Parametric Equations If x and y are both expressed in terms of a third variable, say t, then t is usually called a param eter. In the case of parametric equations we can always find dy ⁄ dx, without first eliminating the parameter. Thus, if the parametric equations are x = φ ( t ), y = ψ ( t ), then dy dy dt = ⋅ dx dt dx dy dy dx or = ⋅ dx dt dt Example : If x = a ⎛⎜ cos t + log tan 12 t ⎞⎟ , y = a sin t, find dy ⁄ dx. ⎝ ⎠ ⎧ 1 t ⎞ 1⎫ dx ⎛ = a ⎨ − sin t + ⋅ ⎜ sec2 ⎟ ⋅ ⎬ Solution : H ere tan (t 2 ⁄ 2) dt ⎝ ⎠ 2⎭ ⎩ / ⎧ ⎨ ⎪ ⎩ = a ⎪ − sin t + Also Now ⎛ 1 − sin2 t ⎞ cos2 t ⎫ ⎟ = a ⋅ ⎪= a⎜ 1 1 ⎝ sin t ⎠ sin t 2 sin t cos t ⎬⎪ 2 2 ⎭ 1 dy ⁄ dt = a cos t. dy dy ⁄ dt a cos t = = = tan t. dx dx ⁄ dt (a cos2 t) ⁄ sin t 3.23 Differentiation of a Function with Respect to a Function Suppose we are to find the differential coefficient of the function u = f (x) with respect to another function, say, v = φ (x). It means we are to find du ⁄ dv, where u and v are both given in terms of a third du du ⁄ dx , = variable x. Therefore, as in the case of parametric equations, we have dv dv ⁄ dx d f (x) d φ (x) d f (x) = ⋅ i.e., dx d φ (x) dx / Example : Differentiate x sin − x sin − 1 1 x with respect to sin − 1 x. and v = sin − 1 x. Solution : Let u = Then log u = sin − 1 x. log x. 1 1 1 du = log x + sin − 1 x ∴ x u dx √(1 − x2) or Again x − 1 ⎡ log x du sin − 1 x⎤⎥ = x sin x ⎢⎢ + ⎥ ⋅ ⎢ √(1 − x2) x ⎥⎦ dx ⎣ 1 dy = ⋅ dx √(1 − x2) D-97 DIFFERENTIATION x sin Now − 1 ⎡ log x sin − 1 x ⎤⎥ + ⎢ ⎥ ⎢ √(1 − x2) x ⎥⎦ ⎣ x⎢ du du ⁄ dx = = dv dv ⁄ dx = x sin 1 √(1 − x2) − 1 ⎡ x log x + √(1 − x2) sin − 1 x⎤ ⎥ ⋅ x ⎦ ⎣ x⎢ 3.24 Trigonometrical Transformations So m e t i m e s a fu n ct io n ca n be e a sily diffe r e n t ia t e d a ft e r m a kin g so m e trigonometrical transformation. Following formulae of trigonometry are of frequent use in such cases : (ii) 1 − cos x = 2 sin2 (x ⁄ 2), (i) 1 + cos x = 2 cos2 (x ⁄ 2), 2 tan (x ⁄ 2) , 2 tan (x ⁄ 2) , (iv) sin x = (iii) tan x = 2 1 − tan (x ⁄ 2) 1 + tan2 (x ⁄ 2) (v) cos x = 1 − tan2 (x ⁄ 2) , 1 + tan2 (x ⁄ 2) x+ y , (vi) tan− 1 x + tan− 1 y = tan− 1 1 − xy x− y , (vii) tan− 1 x − tan− 1 y = tan− 1 1 + xy 3x − x3 , (ix) 3 tan− 1 x = tan− 1 1 − 3x2 (viii) 2 tan− 1 x = tan− 1 2x , 1 − x2 (x) sin 3x = 3 sin x − 4 sin3 x. (xi) cos 3x = 4 cos3 x − 3 cos x. a− x with respect to x. Example 1 : Differentiate tan− 1 1 + ax a− x ⋅ Solution : Let y = tan− 1 1 + ax Then y = tan− 1 a − tan− 1 x. dy 1 , ∴ = 0− since tan − 1 a is constant dx 1 + x2 = − 1 ⁄ (1 + x2). Example 2 : Differentiate tan − 1 [{√(1 + x2) − 1} ⁄ x] with respect to tan − 1 x. √(1 + x2) − 1 and v = tan − 1 x. Solution : Let u = tan − 1 x Then to find du ⁄ dv. Since v = tan − 1 x, therefore x = tan v. ∴ √(1 + tan2 v) − 1 sec v − 1 = tan − 1 u = tan − 1 tan v tan v D-98 DIFFERENTIAL CALCULUS 1 2 sin2 v 1 − cos v 2 1 1 = tan − 1 = tan − 1 = tan − 1 tan v = v. 1 1 2 2 sin v 2 sin v cos v 2 2 H ence du ⁄ dv = 1 ⁄ 2. √(1 + x2) + tan− 1 √(1 + x2) − √(1 + x2) + √(1 − y = tan− 1 √(1 + x2) − √(1 − Example 3 : Differentiate Solution : Let √(1 − √(1 − x2) ⋅ x2) x2) ⋅ x2) Put x2 = cos 2 θ. Then √(1 + cos 2 θ) + √(1 − cos 2θ) y = tan − 1 √(1 + cos 2θ) − √(1 − cos 2 θ) cos θ + sin θ 1 + tan θ = tan − 1 = tan − 1 cos θ − sin θ 1 − tan θ ⎛ π ⎞ π = tan − 1 tan ⎜ + θ⎟ = + .θ 4 ⎝4 ⎠ 1 4 1 2 cos − 1 x2. ∴ y= H ence dy 1 x 1 = − ⋅ 2x = − ⋅ 2 √(1 − x4) dx √(1 − x4) π+ 3.25 Hyperbolic Functions We define the hyperbolic functions as follows : ex − e− x , ex + e− x , sinh x = 2 2 cosh x , sinh x , coth x = tanh x = sinh x cosh x 1 , 1 sech x = cosech x = ⋅ cosh x sinh x Relations between different hyperbolic functions. cosh x = We have, cosh2 x − sinh2 x = = 1 4 (e2x + cosh2 2+ 1 4 e− 2x) 1 (e x + e− x) 2 − (e x − e− x) 2 − 4 1 4 (e2x − 1 2 + e− 2x) = (2 + 2) = 1. 4 sinh2 Thus, x− x = 1. Similarly, we can establish the following other relations for hyperbolic functions : cosh 2x = cosh2 x + sinh2 x , sinh 2 x = 2 sinh x cosh x cosh 2 x = 2 cosh2 x − 1, cosh 2 x = 1 + 2 sinh2 x, sech2 x = 1 − tanh2 x , cosech2 x = coth2 x − 1. Note : In order to remember the relations for hyperbolic functions it should be noted that they can be obtained from the corresponding relations for circular functions simply by changing them to hyperbolic functions and also by changing the sign of the term which contains the product of two sines. D-99 DIFFERENTIATION Differential Coefficients of Hyperbolic Functions : d ⎛⎜ e x + e− d ⎜ (cosh x) = 2 dx ⎜⎝ dx We have x⎞ 1 x ⎟⎟ − x ⎟⎠ = 2 (e − e ) = sinh x. d cosh x = sinh x . dx d sinh x = cosh x. Similarly, dx (cosh x) (cosh x) − (sinh x) (sinh x) d d ⎛ sinh x ⎞ Again ⎟ = (tanh x)= ⎜ dx dx ⎝ cosh x⎠ cosh2 x Thus, = cosh2 x − sinh2 x 1 = = sech2 x. 2 cosh x cosh2 x d tanh x = sech 2 x . dx d Similarly, coth x = − cosech 2 x , dx d cosech x = − cosech x coth x , dx d and sech x = − sech x tanh x . dx Thus, 3.26 Inverse Hyperbolic Functions and their Derivatives If sinh y = x, then we write y = sinh− 1 x. Similarly, we can define cosh− 1 x, x and other inverse hyperbolic functions. Logarithmic values of inverse hyperbolic functions. sech− 1 Let y = cosh− 1 x, then cosh y = x. ∴ sinh y = √(cosh2 y − 1) = √(x2 − 1). But e y = cosh y + sinh y = x + √(x2 − 1). ∴ y = log [x + √(x2 − 1)] Similarly, cosh− 1 x = log [x + √(x2 − 1)]. 1+ x 1 sinh− 1 x = log [x + √(x2 + 1)] and tanh− 1 x = log ⋅ 2 1− x i.e., Now, in order to find out the differential coefficient of cosh− 1 x , we have d d 1 cosh− 1 x = log [x + √(x2 − 1)] = ⋅ 2 dx dx √(x − 1) 1 d cosh− 1 x = ⋅ 2 dx √(x − 1) d 1 , Similarly, sinh− 1 x = 2 dx √(x + 1) 1 d tanh− 1 x = ⋅ dx 1 − x2 Thus, D-100 DIFFERENTIAL CALCULUS 3.27 List of Standard Results to be Committed to Memory d n x = nx n − 1 dx d x e = ex dx d 1 log e x = dx x 1 d log a x = log a e x dx d x x a = a log e a dx d sin x = cos x dx d cos x = − sin x dx d tan x = sec 2 x dx d cot x = − cosec 2 x dx d tanh x = sech2 x . dx d coth x = − cosech2 x dx d sech x = − sech x tanh x dx d cosech x = − cosech x coth x dx d 1 sin − 1 x = dx √(1 − x2) d 1 cos − 1 x = − dx √(1 − x2) d 1 tan − 1 x = dx 1 + x2 d 1 cot − 1 x = − dx 1 + x2 d 1 sec − 1 x = dx x √(x2 − 1) d sec x = sec x tan x dx d cosec x = − cosec x cot x dx d sinh x = cosh x dx d cosh x = sinh x dx d 1 cosec − 1 x = − 2 dx x √(x − 1) d 1 − 1 sinh x= dx √(x2 + 1) d 1 cosh − 1 x = dx √(x2 − 1) d 1 tanh − 1 x = ⋅ dx 1 − x2 4.1 Successive Differential Coefficients Let y = f (x) be a differentiable function of x ; then its differential coefficient is called the first differential coefficient of y. If the first differential coefficient dy dx dy is dx d ⎛ dy⎞ ⎜ ⎟ is called the second differential dx ⎝ dx⎠ d2 y d2 y coefficient of y and is denoted by 2 ⋅ Similarly, the differential coefficient of 2 is dx dx d3 y called the third differential coefficient of y and is written as 3 ⋅ In general, the nth dx n d y differential coefficient of y is denoted by n ⋅ dx If y = f (x) be a function of x, then the various ways of writing the successive differential coefficients of y are as follows : differentiable, then its differential coefficient i.e., dy , d2 y , d3 y , … , dn y , … ; D y, D2 y, D3 y, …, D n y, … dx dx2 dx3 dxn D-102 DIFFERENTIAL CALCULUS y1, y2 , y3 , …, yn ,… ; y′, y′′, y′′′, …, y(n) ,… f ′ (x), f ′′ (x), f ′′′ (x), …, f (n) (x), …. If y = f (x) be a function of x, then the nth differential coefficient of yr is the (n + r) th differential coefficient of y D n yr = D n + r y = yn + r . In particular, D n y2 = D n + 2 y = yn + 2 . The value of the nth differential coefficient of y = f (x) at x = a is denoted by ( yn ) x = a or by ( yn ) a , or by f (n) (a). It should be noted that the differential coefficient of a given order at a point can exist only when the function and all derivatives of lower order are differentiable at the point. i.e., Example 1 : Find the second differential coefficient of e3x sin 4x. Solution : Let y = e3x sin 4 x. dy = 3e3x sin 4 x + 4e3x cos 4 x = e3x (3 sin 4 x + 4 cos 4 x). Then dx ∴ d 3x d ⎛ dy⎞ d2 y = ⎜ ⎟ = {e (3 sin 4 x + 4 cos 4 x)} 2 dx dx dx ⎝ ⎠ dx = 3e3 x (3 sin 4 x + 4 cos 4 x) + e3 x (12 cos 4 x − 16 sin 4x) = e3 x (24 cos 4x − 7 sin 4x). Example 2 : If y = (sin− 1 x) 2, prove that (1 − x2) y2 − xy1 − 2 = 0. Solution : We have y = (sin− 1 x) 2. 2 sin− 1 x ⋅ Differentiating both sides with respect to x, we get y1 = √(1 − x2) Squaring both sides, we get (1 − x2) y12 = 4 (sin− 1 x) 2 or (1 − x2) y12 − 4y = 0, since y = (sin− 1 x) 2. Differentiating again, we get (1 − x2) 2y1 y2 − 2 xy12 − 4y1 = 0. Since 2y1 ≠ 0, therefore cancelling 2y1, we get (1 − x2) y2 − xy1 − 2 = 0. Example 3 : If x = a (cos θ + θ sin θ) , y = a (sin θ − θ cos θ) , find Solution : We have x = a (cos θ + θ sin θ). dx ∴ = a (− sin θ + sin θ + θ cos θ) = a θ cos θ. dθ Also y = a (sin θ − θ cos θ). dy = a (cos θ − cos θ + θ sin θ) = a θ sin θ. ∴ dθ dy d y ⁄ d θ aθ sin θ Now = = = tan θ. dx d x ⁄ d θ aθ cos θ d2 y ⋅ dx2 D-103 SUCCESSIVE DIFFERENTIATION ∴ ⎡ d ⎤ dθ d d2 y d ⎛ dy⎞ = ⎜ ⎟ = (tan θ) = ⎢ (tan θ) ⎥ 2 dx dx dx dθ ⎝ ⎠ dx ⎣ ⎦ dx 1 1 sec3 θ = sec2 θ ⋅ = ⋅ aθ cos θ a θ Example 4 : If p2 = a2 cos2 θ + b2 sin2 θ, prove that p + or d2 p a 2 b2 = ⋅ p3 dθ2 Solution : We have p2 = a2 cos2 θ + b2 sin2 θ. Differentiating both sides of (1) w.r.t. ‘θ’, we have dp 2p = − 2a2 cos θ sin θ + 2b2 sin θ cos θ dθ dp = (b2 − a2) sin θ cos θ. . p dθ Now differentiating both sides of (2) w.r.t. ‘θ’, we have d2 p ⎛ dp⎞ 2 p 2 + ⎜ ⎟ = (b2 − a2) (cos2 θ − sin2 θ) ⎝ dθ ⎠ dθ …(1) …(2) = (b2 cos2 θ + a2 sin2 θ) − (a2 cos2 θ + b2 sin2 θ) = (b2 cos2 θ + a2 sin2 θ) − p2. ∴ p d2 [From (1)] ⎛ dp ⎞ 2 p + p2 = (b2 cos2 θ + a2 sin2 θ) − ⎜ ⎟ ⎝ dθ ⎠ dθ2 = (b2 cos2 θ + a2 sin2 θ) − (b2 − a2) 2 sin2 θ cos2 θ , p2 substituting for dp ⁄ dθ from (2) 1 2 2 [ p (b cos2 θ + a2 sin2 θ) − (b2 − a2) 2 sin2 θ cos2 θ] p2 1 = 2 [(a2 cos2 θ + b2 sin2 θ) (b2 cos2 θ + a2 sin2 θ) p = − (b2 − a2) 2 sin2 θ cos2 θ] = 1 2 2 [a b cos4 θ + a2 b2 sin4 θ + 2a2 b2 sin2 θ cos2 θ] p2 2 2 a 2 b2 2 θ + sin2 θ) 2 = a b ⋅ (cos p2 p2 2 2 2 d p a b p 2 + p2 = ⋅ dθ p2 = Thus Dividing both sides by p, we have 1. a 2 b2 d2 p + p = ⋅ p3 dθ2 If x = a (t − sin t ) and y = a (1 + cos t ), prove that d2 y 1 ⎛ t⎞ = cosec 4 ⎜ ⎟ ⋅ 4a ⎝ 2⎠ dx2 D-104 2. DIFFERENTIAL CALCULUS (i) If y = A sin mx + B cos mx, prove that y2 + m 2 y = 0. (ii) If y = A eax + Be− ax, show that y2 − a2 y = 0. 3. If y = eax cos bx, prove that y2 − 2ay1 + (a2 + b2) y = 0. Also prove that yn + 1 = 2ayn − (a2 + b2) yn− 1 . 4.2 nth Differential Coefficients of Some Standard Functions (i) If y = eax + b, then y1 = aeax + b, y2 = a2 eax + b, y3 = a3 eax + b, ……and so on. In general yn = an eax + b. Thus, D n e ax + b = (Bundelkhand 2005; Agra 07; Rohilkhand 11B) a n e ax + b. (ii) If y = ax, then y1 = (log a) ax, y2 = (log a) 2 ax, etc. In general yn = (log a) n ax. Thus D n a x = (log a) n a x. (Meerut 2001; Bundelkhand 05; Agra 07) (iii) If y = (ax + b) m , then y1 = m a (ax + b) m − 1, y2 = m (m − 1) a2 (ax + b) m − 2, y3 = m (m − 1) (m − 2) a3 (ax + b) m − 3, etc. In general yn = m (m − 1) (m − 2) …{m − (n − 1)}an (ax + b) m − n . Thus, D n (ax + b) m = m (m − 1) (m − 2) … (m − n + 1) a n (ax + b) m − n. If m is a positive integer we can write the above result in a compact form by using the factorial notation. Thus, in this case D n (ax + b) m m (m− 1) (m − 2) … (m − n + 1) (m − n) (m − n − 1)…2.1 = ⋅ an (ax + b) m − n (m − n) (m − n − 1)…2.1 m! = an (ax + b) m − n . (m − n) ! If m is a negative integer, say m = − p, where p is a positive integer, then D n (ax + b) − p = (− p) (− p − 1) (− p − 2) … {− p − (n − 1)} an (ax + b) − p − n = (− 1) n p ( p + 1) ( p + 2) … ( p + n − 1) an (ax + b) − p − n = (− 1) n ( p + n − 1)! n a (ax + b) − p − n . ( p − 1)! Note : If m is a positive integer, the m th differential coefficient of (ax + b) m is const ant. Therefore t he (m + 1) th and all t h e high er differen tial co efficien ts of (ax + b) m will be zero. (iv) If y = (ax + b) − 1, then (Agra 2007) y1 = (− 1) a (ax + b) − 2, y2 = (− 1) (− 2) a2 (ax + b) − 3, y3 = (− 1) (− 2) (− 3) a3 (ax + b) − 4, etc. In general, yn = (− 1) (− 2) (− 3)…(− n) an (ax + b) − (n + 1) . Thus, D n (ax + b) − 1 = (− 1) n n ! a n (ax + b) − n − 1. D-105 SUCCESSIVE DIFFERENTIATION (v) If y = log (ax + b), then y1 = a = a (ax + b) − 1, ax + b y2 = a2 (− 1) (ax + b) − 2, y3 = a3 (− 1) (− 2) (ax + b) − 3, etc. In general, yn = an (− 1) (− 2)…{− (n − 1)} (ax + b) − n . (− 1) n − 1 (n − 1) ! a n ⋅ (ax + b) n Thus, D n log (ax + b) = (vi) If y = cos (ax + b), then ⎛ ⎝ y1 = − a sin (ax + b) = a cos ⎜ ax + b + (Meerut 2003) π⎞ ⎟ 2⎠ π⎞ π⎞ ⎛ ⎟ , y = a 3 cos ⎜ ax + b + 3 . ⎟ , etc. 2⎠ 3 2⎠ ⎝ nπ ⎛ ⎞ In general, yn = an cos ⎜ ax + b + ⎟ ⋅ 2⎠ ⎝ n π⎞ ⎛ Thus, D n cos (ax + b) = a n cos ⎜ ax + b + ⎟ . 2 ⎠ ⎝ n π⎞ ⎛ (vii) Similarly, D n sin (ax + b) = a n sin ⎜ ax + b + ⎟ . ⎝ 2 ⎠ y2 = a2 cos ⎛⎜ ax + b + 2 ⋅ ⎝ (Gorakhpur 2005; Bundelkhand 07; Kumaun 08) (viii) If y = e ax sin (bx + c), then y1 = ae ax sin (bx + c) + be ax cos (bx + c) = e ax {a sin (bx + c) + b cos (bx + c)}. Putting a = r cos φ , b = r sin φ , so that r2 = a2 + b2 and φ = tan− 1 (b ⁄ a), we get y1 = re ax sin (bx + c + φ ). Similarly y2 = r2 e ax sin (bx + c + 2φ ), etc. In general, yn = r n e ax sin (bx + c + n φ ). Thus, D n { e ax sin (bx + c)} = r n e ax sin (bx + c + n φ ), where, r = (a2 + b2) 1 ⁄ 2, and φ = tan− 1 (b ⁄ a). (ix) Similarly, D n { e ax cos (bx + c)} = r n e ax cos (bx + c + nφ ) where r = (a2 + b2) 1 ⁄ 2, and φ = tan− 1 (b ⁄ a). 4.3 Decomposition in a Sum All the standard results obtained in article 4.2 should be committed to memory. In order to find the nth differential coefficient of any other function, it will be often necessary to express that function as the sum or difference of suitable functions with the help of some algebraic or trigonometrical transformations as discussed below. D-106 DIFFERENTIAL CALCULUS 4.4 Use of Partial Fractions In order to find the nth differential coefficient of a fraction in which numerator and denominator are both rational, integral algebraic functions, we should resolve the fraction into partial fractions after breaking its denominator into linear factors, real or imaginary. In case we get imaginary factors in the denominator we shall make use of De-Moivre’s Theorem of trigonometry in order to simplify the result. Example 1 : Find the nth differential coefficient of x2 ⋅ (x − a) (x − b) (Rohilkhand 2014) ⋅ (x − a) (x − b) Since the given fraction is not a proper one, therefore we should first divide the numerator by the denominator before resolving it into partial fractions. H ere we observe orally that the quotient will be 1. So let A B x2 ≡ 1+ + ⋅ (x − a) (x − b) x− a x− b Clearing the fractions, we get x2 ≡ (x − a) (x − b) + A (x − b) + B (x − a). Solution : Let y= x2 Putting x = a, we get A = a2 ⁄ (a − b) and putting x = b, we get B = b2 ⁄ (b − a). H ence y= 1 + a2 b2 + (a − b) (x − a) (b − a) (x − b) a2 b2 (x − a) − 1 − (x − b) − 1. (a − b) (a − b) Therefore differentiating both sides n times, we get = 1+ a2 b2 (− 1) n n ! (x − a) − n − 1 − (− 1) n n ! (x − b) − n − 1 (a − b) (a − b) ⎤ (− 1) n n ! ⎡⎢ b2 a2 ⎥⋅ = − ⎢ ⎥ n + 1 n + 1 ⎥ (a − b) ⎢⎣ (x − a) (x − b) ⎦ yn = Example 2 : Find the nth differential coefficient of (i) tan − 1 (x ⁄ a). (Meerut 2001, 05B, 09; Purvanchal 10, 14; Avadh 13) (ii) tan − 1 x. x a a = ⋅ Solution:(i) If y = tan− 1 , then y1 = 2 a x + a2 (x + ia) (x − ia) A B a ≡ + ⋅ Now let (x + ia) (x − ia) x + ia x − ia Clearing the fractions, we get a ≡ A (x − ia) + B (x + ia). Putting x = ia, we get B = 1 ⁄ 2i and putting x = − ia, we get A = − 1 ⁄ 2i. ∴ y1 = 1 ⎡ 1 ⎡ 1 1 ⎤ − (x − ia) − 1 − (x + ia) − 1⎤⎦ ⋅ ⎥ = ⎢ 2i⎣ 2 i ⎣ x − ia x + ia ⎦ D-107 SUCCESSIVE DIFFERENTIATION Now differentiating both sides (n − 1) times, we get 1 ⎡ (− 1) n − 1 (n − 1) ! (x − ia) − n − (− 1) n − 1 (n − 1) ! (x + ia) − n ⎤⎦ yn = 2i⎣ (− 1) n − 1 (n − 1) ! ⎡ − n − (x + ia) − n ⎤ . ⎦ ⎣ (x − ia) 2i = Put x = r cos φ and a = r sin φ . Then yn = = (− 1) n − 1 (n − 1) ! ⎡ − n − n − r− n (cos φ + i sin φ ) − n ⎤ ⎦ ⎣ r (cos φ − i sin φ ) 2i (− 1) n − 1 (n − 1) ! − n ⎡ r ⋅ ⎣ (cos nφ + i sin n φ ) − (cos nφ − i sin n φ ) ⎤⎦ 2i (− 1) n − 1 (n − 1) ! − n r . 2i sin nφ = (− 1) n − 1 (n − 1) ! r− n sin nφ 2i a ⎛ a ⎞− n = (− 1) n − 1 (n − 1) ! ⎜ sin nφ , since r = ⎟ sin φ ⎝ sin φ ⎠ = = (− 1) n − 1 (n − 1) ! a− n sin n φ sin n φ , where φ = tan− 1 (a ⁄ x). (ii) Proceeding as in part (i), we get Dn (tan− 1 x) = (− 1) n − 1 (n − 1) ! sinn φ sin n φ , where φ = tan− 1 (1/x). 4.5 Use of Trigonometrical Transformations Su p p o se we a r e t o fi n d t h e nth differe nt ia l coe fficient of t h e funct ion x cosn x, where m and n are positive integers. With the help of trigonometry, we express this function as the sum of sines or cosines of multiples of x and then we apply standard results. sinm Example 1 : Find the nth differential coefficient of sin2 x sin 2x. Solution : Let y = sin2 x sin 2x. Then y= = 1 2 1 2 (1 − cos 2x) sin 2x, since 2 sin2 x = 1 − cos 2x sin 2x − 1 2 sin 2 x cos 2 x = 1 2 sin 2 x − 1 4 sin 4 x . Now differentiating both sides n times, we have n π⎞ n π⎞ 1 1 ⎛ ⎛ yn = ⋅ 2n sin ⎜ 2 x + ⎟ − . 4n sin ⎜ 4 x + ⎟ ⎝ ⎝ 2 2 ⎠ 4 2 ⎠ = n π⎞ n π⎞ ⎤ 1⎡ n ⎛ ⎛ ⎢ 2.2 sin ⎜ 2x + ⎟ − 4n sin ⎜ 4 x + ⎟⎥⋅ ⎝ ⎝ 4⎣ 2 ⎠ 2 ⎠⎦ Example 2 : Find the nth differential coefficient of eax cos2 x sin x. Solution : Let y = eax cos2 x sin x. Then y= 1 ax e (1 2 + cos 2 x) sin x = 1 ax e sin 2 x+ 1 ax e cos 2x sin 2 x D-108 DIFFERENTIAL CALCULUS = 1 ax e sin 2 = 1 ax 1 1 e sin x + eax sin 3x − eax sin 2 4 4 1 ax ax [e sin x + e sin 3x]. 4 x+ 1 ax e [sin 2 3x − sin x] as 2 cos A sin B = sin (A + B) − sin (A − B) = x Now differentiating both sides n times, we have yn = 1⎡ ⎢ (1 4⎣ 1⎞ ⎛ + a2) n ⁄ 2 eax sin ⎜ x + n tan− 1 ⎟ ⎝ a⎠ 3⎞ ⎤ ⎛ + (9 + a2) n ⁄ 2 eax sin ⎜ 3x + n tan− 1 ⎟ ⎥ ⎝ a⎠ ⎦ 1 ⎛ ⎞ 1 ax⎡ = e ⎢ (1 + a2) n ⁄ 2 sin ⎜ x + n tan− 1 ⎟ 4 a⎠ ⎝ ⎣ 3⎞ ⎤ ⎛ + (9 + a2) n ⁄ 2 sin ⎜ 3x + n tan− 1 ⎟ ⎥ ⋅ a⎠ ⎦ ⎝ th 5 3 Example 3 : Find the n differential coefficient of sin x cos x. Solution : Let z = cos x + i sin x, then z− 1 = (cos x + i sin x) − 1 = cos x − i sin x. Therefore z + z− 1 = 2 cos x and z − z− 1 = 2i sin x. Also by De-Moivre’s theorem, z m = cos mx + i sin mx, z− m = cos m x − i sin mx. Therefore z m + z− m = 2 cos mx and z m − z− m = 2i sin mx. Now (2i sin x) 5 (2 cos x) 3 = (z − z− 1) 5 (z + z− 1) 3 = (z8 − z− 8) − 2 (z6 − z− 6) − 2 (z4 − z− 4) + 6 (z2 − z− 2) = 2i sin 8x − 2 (2i sin 6x) − 2 (2i sin 4x) + 6 (2i sin 2x). Therefore sin5 x cos3 x = 2− 7 [sin 8x − 2 sin 6x − 2 sin 4x + 6 sin 2x]. H ence D n (sin5 x cos3 x) ⎡ nπ ⎞ n π⎞ ⎛ ⎛ = 2− 7 ⎢ 8n sin ⎜ 8x + ⎟ − 2. 6n sin ⎜ 6x + ⎟ 2 2 ⎠ ⎝ ⎠ ⎝ ⎣ nπ ⎞ n π⎞⎤ ⎛ ⎛ − 2 . 4n sin ⎜ 4x + ⎟ + 6.2n sin ⎜ 2x + ⎟⎥ ⋅ ⎝ ⎝ 2⎠ 2 ⎠⎦ 1. Find the nth differential coefficients of : (i) log [(ax + b) (cx + d)]. (ii) cos 2x cos 3x (Bundelkhand 2001; Kashi 11) (iii) cos x cos 2x cos 3x. (iv) cos4 x. 2. 3. (i) cos2 x sin3 x. (ii) eax cos3 bx. (iii) ea x sin bx cos cx. 1 (i) 1 − 5x + 6x2 (iv) e2 x sin3 x. 1 (ii) 2 x − a2 D-109 SUCCESSIVE DIFFERENTIATION (iii) 4. 5. 6. x2 ⋅ (x + 2) (2x + 3) x4 (x − 1) (x − 2) ⎧1 + (i) tan− 1 ⎨ ⎩1 − (iv) x ⋅ (x − a) (x − b) (x − c) , n ≥ 3. x⎫ ⎬. x⎭ (Agra 2014) ⎧ 2x ⎫ ⎪. (ii) tan− 1 ⎪⎨ ⎬ ⎪ 1 − x2 ⎪ ⎩ ⎭ (Purvanchal 2011) ⎧ ⎫ ⎪ √(1 + x2) − 1 ⎪ ⎬ , show that yn = 1 (− 1) n − 1 (n − 1) ! sin n θ sin n θ, If y = tan− 1 ⎨⎪ ⎪ 2 x ⎩ ⎭ θ = cot− 1 x. x , prove that y = (− 1) n n ! a− n − 1 sinn + 1 φ cos (n + 1) φ , 7. If y = 2 n (x + a2) where where 8. 9. 10. φ = tan− 1 (a ⁄ x). (Kanpur 2009) If y = sin mx + cos mx, prove that yn = mn [1 + (− 1) n sin nth 2mx]1 ⁄ 2. (Meerut 2000, 09B) 3 2 x ⁄ (x − 1) for x = 0 is differential coefficient of Prove that the value of the zero if n is even, and is − n ! if n is odd and greater than 1. If y = (tan− 1 x) 2, prove that (x2 + 1) 2 y2 + 2 x (x2 + 1) y1 − 2 = 0. 1. (i) (− 1) n − 1 (n − 1)! {a n (ax + b) − n + c n (cx + d ) − n }. 1 1 ⎞⎫ ⎛ ⎞ ⎛ 1⎧ (ii) ⎨⎩ 5n cos ⎜ 5x + n π ⎟ + cos ⎜ x + nπ ⎟ ⎬⎭ ⋅ ⎝ ⎠ ⎝ 2 2 ⎠ 2 1⎧ n 1 1 ⎞ 1 ⎛ ⎞ ⎛ ⎛ ⎞⎫ n (iii) ⎨ 2 cos ⎜ 2x + n π ⎟ + 4 cos ⎜ 4x + nπ ⎟ + 6 n cos ⎜ 6 x + n π ⎟ ⎬ . 4⎩ 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎭ ⎧ ⎫ 1 ⎞ 1 ⎞ ⎛ ⎛ 1 (iv) ⎨ 4n cos ⎜ 4x + nπ ⎟ + 2n + 2 cos ⎜ 2x + nπ ⎟ ⎬ . ⎝ ⎝ 2 ⎠ 2 ⎠⎭ 8⎩ 2 (i) 1 ⎧ 1 1 1 ⎞⎫ ⎛ ⎞ ⎛ ⎞ ⎛ ⎨ 2 sin ⎜ x + n π ⎟ + 3n sin ⎜ 3x + n π ⎟ − 5n sin ⎜ 5x + nπ ⎟ ⎬ . 16 ⎩ 2 2 2 ⎠⎭ ⎝ ⎠ ⎝ ⎠ ⎝ (ii) (1 ⁄ 4) (a2 + 9b2) n ⁄ 2. eax cos {3bx + n tan− 1 (3b ⁄ a)} + (3 ⁄ 4) (a2 + b2) n ⁄ 2 eax cos {bx + n tan− 1 (b ⁄ a)}. 1 1 (iii) r n eax sin {(b + c) x + nφ } + r1n eax sin {(b − c) x + nψ}, 2 2 where r2 = a2 + (b + c) 2, φ = tan− 1 {(b + c) ⁄ a}, r12 = a2 + (b − c) 2, 3. ψ = tan− 1 {(b − c) ⁄ a}. 3 1 (iv) . 5n ⁄ 2 . e2x sin [x + n tan− 1 (1 ⁄ 2)] − ⋅ (13) n ⁄ 2 . e2x [sin 2x + n tan− 1 (3 ⁄ 2)]. 4 4 (i) (− 1) n n ! [2n + 1 (2x − 1) − n − 1 − 3n + 1 (3x − 1) − n − 1]. (ii) (1 ⁄ 2a) n ! (− 1) n {(x − a) − n − 1 − (x + a) − n − 1}. D-110 DIFFERENTIAL CALCULUS ⎡ ⎢ ⎤ 4 9. 2n − 1 ⎥ ⎥. − (x + 2) n + 1 ⎥⎦ ⎣ (2x + 3) n + 1 a b ⎧ (iv) (− 1) n n ! ⎪⎨ + ⎪⎩ (a − b) (a − c) (x − a) n + 1 (b − c) (b − a) (x − b) n + 1 c + (c − a) (c − b) (x − c) n + (iii) (− 1) n n ! ⎢⎢ 4. 5. (− 1) n n ! {16 (x − 2) − n − 1 − (x − 1) − n − 1}. (i) (− 1) n − 1 (n − 1) ! sin n φ sin nφ , where φ = tan− 1 (1 ⁄ x). ⎫ ⎪. 1⎬ ⎪ ⎭ (ii) 2 (− 1) n − 1 (n − 1) ! sin n φ sin nφ , where φ = tan− 1 (1 ⁄ x). 4.6 Leibnitz’s Theorem This theorem is useful for finding the nth differential coefficient of the product of two functions. The statement of this theorem is as follows : If u and v are any two functions of x such that all their desired differential coefficients exist, then the nth differential coefficient of their product is given by D n (uv) = (D n u). v + n C1 D n − 1 u. Dv + n C2 D n − 2 u. D2 v + … … + n Cr D n − r u. D rv + … + u D n v. (Meerut 2004, 05BP, 08, 09; Bundelkhand 05; Agra 07; Kumaun 08; Purvanchal 11; Kashi 13) Proof : We shall prove this theorem by mathematical induction. By actual differentiation, we have D (uv) = (Du) . v + u. Dv. …(1) From (1) we see that the theorem is true for n = 1. Now suppose that the theorem is true for a particular value of n. Then we have D n (uv) = (D n u) v + n C1 D n − 1 u Dv + n C2 D n − 2 u D2 v + … … + n Cr D n − r u D r v + n Cr + 1 D n − r − 1 u D r + 1 v + … + u. D n v. …(2) Differentiating both sides of (2) with respect to x, we have D n + 1 (uv) = {(D n + 1 u). v + D n u Dv} + {n C1 D n u. Dv + n C1 D n − 1 u D2 v} + {n C2 D n − 1 u. D2 v + n C2 D n − 2 u. D 3 v} + … … + {n Cr D n − r + 1 u D r v + n Cr D n − r u D r + 1v} + {n Cr + 1 D n − r u. D r + 1 v + n Cr + 1 D n − r − 1 u D r + 2 v}+ … … + {Du D n v + u D n + 1 v}. R earranging the terms, we have D n + 1 (uv) = (D n + 1 u). v + (1 + n C1) (D n u Dv) + ( n C1 + n C2) (D n − 1 u D2 v) + … + ( n Cr + n Cr + 1) (D n − r u. D r + 1 v) + … + u D n + 1v. But we know that n Cr + n Cr + 1 = n + 1Cr + 1. Therefore n C0 + n C1 = n + 1 C1 where n C0 = 1, n C1 + n C2 = n + 1C2 , etc. H ence (3) becomes …(3) D-111 SUCCESSIVE DIFFERENTIATION D n + 1 (uv) = (D n + 1 u). v + n + 1C1 D n u. Dv + n + 1C2 D n − 1 u. D2 v + … + n + 1 Cr + 1 D n − r u. D r + 1v + … + u. D n + 1 v. …(4) The result (4) shows that if the theorem is true for any particular value of n, it is also true for the next value of n. But we have already seen that the theorem is true for n = 1. H ence it must be true for n = 2 and so for n = 3 ; and so on. Therefore the theorem is true for every positive integral value of n. Note : While applying Leibnitz’s theorem if we see that one of the two functions is such that all its differential coefficients after a certain stage become zero then we should take that function as the second function. Example 1 : Find the nth differential coefficient of x3 cos x. (Meerut 2010) Solution : Since the fourt h and higher derivatives of x3 will become zero, therefore for the sake of convenience we should choose x3 as the second function. Applying Leibnitz’s theorem, we have D n [(cos x) . x3] = (D n cos x) . x3 + n C1 (D n − 1 cos x) . (Dx3) + n C2 (D n − 2 cos x) (D2 x3) + n C3 (D n − 3 cos x) (D3 x3), since all other terms become zero ⎛ n π⎞ 3 ⎧ π⎫ ⎟ . x + n cos ⎨ x + (n − 1) ⎬ . 3x2 = cos ⎜ x + 2 ⎠ 2⎭ ⎝ ⎩ + n (n − 1) ⎧ π⎫ n (n − 1) (n − 2) ⎧ π⎫ cos ⎨ x + (n − 2) ⎬ . 6 x + cos ⎨ x + (n − 3) ⎬ . 6 1⋅ 2 2 1 2⎭ ⋅ 2 ⋅ 3 ⎩ ⎩ ⎭ ⎛ ⎛ nπ ⎞ nπ ⎞ = x3 cos ⎜ x + ⎟ + 3x2 . n sin ⎜ x + ⎟ 2⎠ 2⎠ ⎝ ⎝ ⎛ ⎛ nπ ⎞ nπ ⎞ ⎟ − n (n − 1) (n − 2) sin ⎜ x + ⎟ − 3n (n − 1) x cos ⎜ x + 2⎠ 2⎠ ⎝ ⎝ ⎛ ⎛ nπ ⎞ nπ ⎞ ⎟ + ⎡⎣ 3x2 n − n (n − 1) (n − 2) ⎤⎦ sin ⎜ x + ⎟ ⋅ = ⎡⎣ x3 − 3n (n − 1) x⎤⎦ cos ⎜ x + 2⎠ 2⎠ ⎝ ⎝ Example 2 : Find the nth differential coefficient of x n − 1 log x. (Meerut 2010B) Solution : Let Then y1 = xn − 1 y = x n − 1 log x. . (1 ⁄ x) + (n − 1) …(1) . x n − 2. Multiplying both sides by x, we have or log x. x y1 = x n − 1 + (n − 1) x n − 1 log x x y1 = x n − 1 + (n − 1) y . …(2) [... from (1) , y = x n − 1 log x] Differentiating both sides of (2), (n − 1) times, we have D n − 1 (y1 x) = D n − 1 x n − 1 + (n − 1) D n − 1 y or (D n − 1 y1). x + n − 1C1 (D n − 2 y1). 1 = (n − 1) ! + (n − 1) yn − 1 or x yn + (n − 1) yn − 1 = (n − 1) !+ (n − 1) yn − 1 D-112 or DIFFERENTIAL CALCULUS x yn = (n − 1) ! or yn = (n − 1) ! ⁄ x. H ence D n (x n − 1 log x) = (n − 1) ! ⁄ x. Example 3 : If y = a cos (log x) + b sin (log x) , show that x2 y2 + xy1 + y = 0, (Bundelkhand 2006, 11, 12; Avadh 08; Kashi 12; Meerut 13) or x2 yn + 2 + (2n + 1) xyn + 1 + (n2 + 1) yn = 0. Solution : We have y = a cos (log x) + b sin (log x). Differentiating both sides with respect to x, we have a b y1 = − sin (log x) + cos (log x) x x xy1 = − a sin (log x) + b cos (log x). or Differentiating both sides again with respect to x, we have a b xy2 + y1 = − cos (log x) − sin (log x) x x 2 x y2 + xy1 = − [a cos (log x) + b sin (log x)] and or x2 y2 + xy1 = − y or x2 y2 + xy1 + y = 0. Differentiating both sides of this equation n times by Leibnitz’s theorem, we get D n (x2 y2) + D n (xy1) + D n ( y) = 0 or (D n y2) x2 + n C1 (D n − 1 y2) . (Dx2) + n C2 . (D n − 2 y2) . (D2 x2) + (D n y1) . x + n C1 (D n − 1 y1) . (Dx) + D n y = 0 or x2 yn + 2 + 2 xn yn + 1 + or x2 yn + 2 + (2n + 1) xyn + 1 + (n2 + 1) yn = 0. Example 4 : If y = ea sin (1 − x2) Solution : We have Therefore y1 = − ea sin − 1 x, show that yn + 2 − (2n + 1) x yn + 1 − (n2 + a2) yn = 0. y= 1 n (n − 1) . 2yn + xyn + 1 + nyn + yn = 0 2 (Garhwal 2000, 01; Gorakhpur 05; Rohilkhand 05, 08; Agra 06, 08; Purvanchal 07) − 1 a sin x e . x . a ⁄ √(1 − x2) or − 1 y1 . √(1 − x2) = aea sin x = ay, or y12 (1 − x2) = a2 y2. [replacing ea sin − 1 x by y] ...(1) Differentiating (1) w.r.t. ‘x’, we have 2 y1 y2 (1 − x2) + y12 (− 2 x) = 2 a2 y y1 or 2 y1 [ y2 (1 − x2) − y1 x− a2 y] = 0. Cancelling 2y1, since 2y1 ≠ 0, we get y2 (1− x2) − y1 x − a2 y = 0. ...(2) Differentiating (2) n times by Leibnitz’s theorem, we have D n [ y2 (1 − x2)] − D n ( y1x) − a2 D n y = 0 (Bundelkhand 2007) D-113 SUCCESSIVE DIFFERENTIATION n (n − 1) ⎡ ⎤ yn (− 2) ⎥ ⎢ yn + 2 . (1 − x2) + n yn + 1 (− 2x) + 2! ⎣ ⎦ or − [ yn + 1 x + n yn . 1] − a2 yn = 0 (1 − x2) yn + 2 − (2n + 1) x yn + 1 − (n2 + a2) yn = 0. or Example 5 : If y 1 ⁄ m + y − (x2 Solution : 1⁄m = 2 x, prove that − 1) yn + 2 + (2n + 1) x yn + 1 + (n2 − m 2) yn = 0. (Agra 2002; Meerut 04, 12B; Rohilkhand 06, 09B, 10B, 11; Purvanchal 06) We have y1 ⁄ m + y− 1 ⁄ m = 2 x. Multiplying both sides by y1 ⁄ m , we get y2 ⁄ m + 1 = 2 x y1 ⁄ m or y2 ⁄ m − 2 x y1 ⁄ m + 1 = 0. 2 x ± √(4x2 − 4) = x ± √(x2 − 1) or y = [x ± √(x2 − 1)]m . ...(1) 2 x ⎧ ⎫⎪ ∴ y1 = m[x ± √(x2 − 1)]m − 1 ⎪⎨ 1 ± ⎬ 2 ⎪⎩ √(x − 1) ⎪⎭ my my , from (1). [x ± √(x2 − 1)]m = ± = ± 2 √(x2 − 1) √(x − 1) Squaring both sides, we get ∴ y1 ⁄ m = y12 (x2 − 1) = m 2 y2. Differentiating again, we get 2 y1 y2 (x2 − 1) + 2 x y12 = 2 m 2 y y1 or 2 y1 [ y2 (x2 − 1) + x y1 − m 2 y] = 0 or y2 (x2 − 1) + x y1 − m 2 y = 0, since 2 y1 ≠ 0. Differentiating (2) n times by Leibnitz’s theorem, we get ...(2) D n { y2 (x2 − 1)} + D n ( y1 x) − m 2 D n y = 0 or yn + 2 . (x2 − 1) + n C1 yn + 1 2 x + n C2 yn . 2 + yn + 1 . x + n C1 yn . 1 − m 2 y n = 0 or (x2 − 1) yn + 2 + (2n + 1) x yn + 1 + (n2 − m 2) yn = 0. Example 6 : If hence show that In = n ! In = dn (x n log x) , prove that dx n In = n In − 1 + (n − 1) !; (Meerut 2004B; Agra 06; Gorakhpur 06; Rohilkhand 08; Avadh 11) ⎛ 1 1 1⎞ ⎜ log x + 1 + + + … + ⎟⎠ ⋅ ⎝ 2 3 n Solution : We have, In = = dn n dn − 1 ⎡ d n ⎤ [x log x] = (x log x) ⎥ ⎢ n n − 1 ⎣ dx ⎦ dx dx dn − 1 ⎡ n − 1 1⎤ log x + x n ⋅ ⎥ ⎢ nx x⎦ dx n − 1 ⎣ n− 1 dn − 1 n − 1 log x)+ d (x (x n − 1) dx n − 1 dx n − 1 ! = n In − 1 + (n − 1) Proved. = n We have just proved that In = n In − 1 + (n − 1) !. ...(1) D-114 DIFFERENTIAL CALCULUS In − 1 1 + ⋅ ...(2) n ! (n − 1) ! n Changing n to n − 1 in the above relation (2), we have In − 1 In − 2 1 = + ⋅ (n − 1) ! (n − 2) ! n − 1 In − 1 Putting this value of in (2), we have (n − 1) ! In − 2 In 1 1 = + + ⋅ n ! (n − 2) ! n − 1 n Thus making repeated use of the reduction formula (2), we ultimately have I1 In 1 1 1 = + + + …+ ⋅ n n! 1! 2 3 d 1 But I1 = (x log x) = x ⋅ + log x = log x+ 1. dx x In 1 1 1 = log x + 1 + + + … + ∴ 2 3 n n! 1 1 1⎞ ⎛ In = n ! ⎜ log x + 1 + + + … + ⎟ ⋅ 2 3 n⎠ ⎝ Dividing both sides by n !, we have or 1. 2. State Leibnitz’s theorem. Find the 4th In = (Meerut 2005B, 08, 11; Bundelkhand 08; Agra 08) differential coefficients of x3 log x ; x2 sin 3x ; e2 x sin2x. Find the nth differential coefficients of : 3. (i) x2 e− x. 4. 1 1 If y = x2 e x, show that yn = n (n − 1) y2 − n (n − 2) y1 + (n − 1) (n − 2) y. 2 2 (ii) x3 log x. (iii) e x log x. (iv) x2 tan− 1 x. (Bundelkhand 2008) 5. Prove that the nth differential coefficient of xn (1 − x) n is equal to ⎧ ⎫ n2 (n − 1) 2 n2 x x2 n ! (1 − x) n ⎪⎨ 1 − 2 + − …⎪⎬ . 2 2 2 ⎪ ⎪ 1 1− x 1 ⋅ 2 (1 − x) ⎩ ⎭ ⎡ ⎤ n ! x n − r⎥ . ⎢ Hint. D r x n = ⎣ ⎦ (n − r) ! (Rohilkhand 2007; Kanpur 08) (− 1) n (n !) ⎡ d n ⎛ log x ⎞ 1⎤ 1 1 ⎢ log x − 1 − − − … − ⎥ . = ⎜ ⎟ n n + 1 2 3 n x ⎝ ⎠ ⎣ ⎦ x dx 6. Prove that 7. If y = x n log x , prove that x yn+ 1 = n ! . (Meerut 2001; Bundelkhand 09; Rohilkhand 11B) ⎛ 1 1 1⎞ H ence show that In = n ! ⎜ log x + 1 + + + … + ⎟ . 2 3 n⎠ ⎝ 8. By forming in two different ways the nth derivative of x2n , prove that D-115 SUCCESSIVE DIFFERENTIATION 1+ (2 n) ! n2 n2 (n − 1) 2 n2 (n − 1) 2 (n − 2) 2 + + + …= . 2 2 2 2 2 2 1 1 ⋅ 2 1 ⋅ 2 ⋅ 3 (n !) 2 [Hint. Find the nth derivative of x n . x n and of x2n and equate]. 9. ⎛ sin x ⎞ ⎛ ⎞ ⎛ ⎞⎫ 1 1 ⎧ ⎟ = ⎨ P sin ⎜ x + nπ ⎟ + Q cos ⎜ x + nπ ⎟ ⎬ Prove that D n ⎜ 2 2 ⎝ ⎠ ⎝ ⎠⎭ ⎩ ⎝ x ⎠ /x n + 1, where P = x n − n (n − 1) x n − 2 + n (n − 1) (n − 2) (n − 3) x n − 4 − …, and Q = nx n − 1 − n (n − 1) (n − 2) x n − 3 + … 10. − 1 If y = etan x, prove that (1 + x2) yn + 2 + [2 (n + 1) x − 1] yn + 1 + n (n + 1) yn = 0. (Avadh 2010; Kanpur 14) 11. 12. yn + 2 + (2n + 1) xyn + 1 + (n2 + 1) yn = 0. If y = (sin− 1 x) 2, prove that (1 − x2) y2 − x y1 − 2 = 0, If y = cos (log x), prove that and 13. x2 (1 − x2) yn + 2 − x (2n + 1) yn + 1 − n2 yn = 0. (Meerut 2002; Agra 08) If y = (x2 − 1) n , prove that (x2 − 1) yn + 2 + 2 x yn + 1 − n (n + 1) yn = 0. (Meerut 2008; Rohilkhand 06, 11B; Kashi 13) dPn ⎫⎪ ⎧ 2 − 1) n , show that d ⎪ (x ⎨⎪ (1 − x2) ⎬ + n (n + 1) Pn = 0. dx ⎩ dx ⎪⎭ dx n ⎛ y⎞ ⎛ x⎞ n If cos− 1 ⎜ ⎟ = log ⎜ ⎟ , prove that x2 yn + 2 + (2n + 1) xyn + 1 + 2n2 yn = 0. ⎝ b⎠ ⎝ n⎠ H ence if Pn = 14. 15. If y = and dn (Meerut 2006B; Rohilkhand 13; Purvanchal 14) [ x + √(1 + prove that (1 + x2) y2 + x y1 − m 2 y = 0 (1 + x2) yn + 2 + (2n + 1) x yn + 1 + (n2 − m 2) yn = 0. x2)]m , (Kanpur 2006; Avadh 09; Bundelkhand 14) 16. 17. x2)}]2, If y = [log {x + √(1 + prove that 2 (1 + x ) yn + 2 + (2n + 1) x yn + 1 + n2 yn = 0. If y = (Agra 2005; Purvanchal 09) sin− 1 x , prove that (1 − x2) yn + 1 − (2n + 1) x yn − n2 yn − 1 = 0. √(1 − x2) (Meerut 2007B; Kanpur 10) 2. 3. (6 ⁄ x); 33 (3x2 − 4) sin 3x − 63 x cos 3x ; − 64e2x sin 2x. (i) (− 1) n e− x [x2 − 2nx + n (n − 1)]. (ii) (− 1) n (n − 4) ! 6 x− n + 3. (iii) e x [log x + n C1 x− 1 − n C2 x− 2 + n C3 2 ! x− 3 + … + (− 1) n − 1 (n − 1) ! x− n ]. (iv) (− 1) n − 1 (n − 3) ! {(n − 1) (n − 2) x2 sin n φ sin nφ − n C1 2x (n − 2) sin n − 1 φ sin (n − 1) φ + 2 . n C2 sin n − 2 φ sin (n − 2) φ }, where φ = tan− 1 (1 ⁄ x). D-116 DIFFERENTIAL CALCULUS 4.7 nth Differential Coefficient for x = 0 Sometimes we are required to find the nth differential coefficient of y for x = 0 i.e. ( yn ) 0. This may be done even though we may not be able to find the nth differential coefficient in a compact form for the general value of x. The method will be clear from the following example : Example 1 : If y = sin (m sin− 1 x) , find ( yn ) 0 . (Meerut 2000, 03) Solution : We have y = sin (m sin− 1 x). Differentiating both sides with respect to x, we get m ⋅ y1 = cos (m sin− 1 x) ⋅ √(1 − x2) ...(1) …(2) Squaring both sides of (2) and multiplying by (1 − x2), we get (1 − x2) y12 = m 2 cos2 (m sin− 1 x) or (1 − x2) y12 = m 2 [1 − sin2 (m sin− 1 x)] or (1 − x2) y12 = m 2 (1 − y2) or [since y = sin (m sin − 1 x)] (1 − x2) y12 + m 2 y2 − m 2 = 0. Differentiating both sides of (3) with respect to x, we get …(3) (1 − x2) 2y1 y2 − 2 x y12 + 2m 2 yy1 = 0. Cancelling 2y1, since 2y1 ≠ 0, we get (1 − x2) y2 − xy1 + m 2 y = 0. Differentiating both sides of (4) n times by Leibnitz’s theorem, we get or …(4) (1 − x2) yn + 2 + n C1 yn + 1 (− 2 x) + n C2 yn (− 2) − x yn + 1 − n C1 yn + m 2 yn = 0 (1 − x2) yn + 2 − (2n + 1) xyn + 1 − (n2 − m 2) yn = 0. …(5) (Meerut 2000) Putting x = 0 in (1), we get ( y) 0 = 0. Putting x = 0 in (2), we get ( y1) 0 = m. Putting x = 0 in (4), we get ( y2) 0 + m 2 ( y) 0 = 0 i.e., ( y2) 0 = 0. Also putting x = 0 in (5), we get ( yn + 2) 0 = (n2 − m 2) ( yn ) 0…(6) Putting n − 2 in place of n in (6), we get ( yn ) 0 = {(n − 2) 2 − m 2} ( yn − 2) 0 = {(n − 2) 2 − m 2} {(n − 4) 2 − m 2}( yn − 4) 0. [Since from (6), we have ( yn − 2) 0 = {(n − 4) 2 − m 2} ( yn − 4) 0]. Now there arise two cases. Case I : When n is even. ( yn ) 0 = {(n− 2) 2 − m 2}{(n− 4) 2 − m 2} {(n− 6) 2 − m 2}… {42 − m 2} {22 − m 2}( y2) 0 = 0 , since ( y2) 0 = 0. D-117 SUCCESSIVE DIFFERENTIATION Case II : When n is odd. ( yn ) 0 = {(n − 2) 2 − m 2} {(n − 4) 2 − m 2} {(n − 6) 2 − m 2}… {32 − m 2} {12 − m 2} ( y1) 0 = {(n − 2) 2 − m 2} {(n − 4) 2 − m 2} {(n − 6) 2 − m 2}… {32 − m 2} {12 − m 2} m. 1. If y = sin− 1 x, prove that (1 − x2) yn + 2 − (2n + 1) xyn + 1 − n2 yn = 0, and hence find the vlaue of ( yn ) 0 . (Agra 2005; Bundelkhand 11) 2. Find ( yn ) 0 , when y = log [ x + √(1 + x2)]. 3. If y = [log {x + √(1 + x2)}]2, prove that ( yn + 2) 0 = − n2 ( yn ) 0 , hence find (yn ) 0. 4. 5. dn + 2 y + (2n + 1) x dx n + 2 H ence find, at x = 0, the value of (dn y ⁄ dx n ). If y = [x + √(1 + x2)]m , find ( yn ) x = 0 . If y = (sinh− 1 x) 2, prove that (1 + x2) (Meerut 2005, 09B) n + 1y d dn y + n2 n = 0. n + 1 dx dx (Meerut 2006, 07, 09; Bundelkhand 2001) 6. 7. If y = cos (m sin − 1 x), find ( yn ) 0 . ⎛1 ⎞ − 1 If x = sin ⎜ log y ⎟ or if y = ea sin x, prove that ⎝a ⎠ (1 − x2) y2 − x y1 − a2 y = 0, x2) (1 − yn + 2 − x (2n + 1) yn + 1 − and hence find the value of ( yn ) 0. (n2 + (Bundelkhand 2007) a 2) yn = 0, (Rohilkhand 2005, 08; Agra 06, 08; Gorakhpur 05) 8. 9. If y = − ea cos 1 x, prove that (1 − x2) yn + 2 − (2n + 1) xyn + 1 − (n2 + a2) yn = 0. H ence find the value of yn for x = 0. (Meerut 2001; Purvanchal 14) − 1 2 If y = tan x, prove that (1 + x ) y2 + 2 x y1 = 0, and hence find the value of all the derivatives of y with respect to x, when x = 0. Also show that ( yn ) 0 is 0, (n − 1) ! or − (n − 1) ! according as n is of the form 2p, 4p + 1 or 4p + 3 respectively. 1. 0 when n is even, and (n − 2) 2 (n − 4) 2…… 52. 32. 12, when n is odd. 2. 0 when n is even, and (− 1) (n − 1) ⁄ 2 (n − 2) 2 (n − 4) 2 … 32 . 12, when n is odd. D-118 3. 4. 5. DIFFERENTIAL CALCULUS 0 when n is odd, and (− 1) (n − 2) ⁄ 2 (n − 2) 2 (n − 4) 2 (n − 6) 2 … 42 . 22. 2 , when n is even. 0 when n is odd, and (− 1) (n − 2) ⁄ 2 (n − 2) 2 (n − 4) 2 (n − 6) 2… 42.22.2 when n is even. {m 2 − (n − 2) 2} {m 2 − (n − 4) 2}… (m 2 − 12) m, n odd; {m 2 − (n − 2) 2} {m 2 − (n − 4) 2}…(m 2 − 22) m 2, n even. 6. 0 when n is odd, and − {(n − 2) 2 − m 2} {(n − 4) 2 − m 2}…(22 − m 2) m 2, when n is even. 7. {(n − 2) 2 + a2} {(n − 4) 2 + a2}… (32 + a2) (12 + a2) a, n odd; {(n − 2) 2 + a2} {(n − 4) 2 + a2}… (42 + a2) (22 + a2) a2, n even. 8. − {(n − 2) 2 + a2} {(n − 42) + a2}… (32 + a2) (12 + a2) aea π ⁄ 2, n odd; {(n − 2) 2 + a2}{(n − 4) 2 + a2}…(42 + a2) (22 + a2) a2 ea π ⁄ 2, n even. 9. 0 when n is even and (− 1) (n − 1) ⁄ 2 (n − 1) ! when n is odd. Fill in the Blanks: Fill in the blanks “……”, so that the following statem ents are com plete and correct. 1. If y = sin (ax + b), then D n sin (ax + b) = …… . 2. If y = (ax + b) − 1, then D n (a x + b) − 1 = …… . 3. The nth differential coefficient of e x sin2 x = …… . 4. If y = a cos (log x) + b sin (log x), then x 2 y 2 + x y1 = …… . sin− 1 x , If y = then (1 − x 2) yn + 1 − (2 n + 1) x yn = …… . ⎯⎯⎯⎯⎯ √ 1 − x⎯2 5. 6. If y = e− x, then D n e− x = …… . (Agra 2006) Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 7. If y = log x , then D n log x is (− 1) n (n − 1) ! xn (− 1) n − 1 (n − 1) ! (b) xn (− 1) n − 1 n ! (c) xn (− 1) n − 1 (n − 1) ! (d) xn+ 1 (a) D-119 SUCCESSIVE DIFFERENTIATION 8. If x = a (cos θ + θ sin θ), y = a (sin θ − θ cos θ), then (a) (b) (c) (d) 9. 10. d 2y is d x2 1 sec 3 θ a θ a sec 3 θ θ θ sec 3 θ a a θ sec 3 θ By Leibnitz’s theorem we find the nth differential coefficient of the …… of two functions. (a) sum (b) difference (c) product (d) quotient If y = e tan − 1 x, then (1 + x 2) yn + 2 + [2 (n + 1) x − 1] yn+ 1 = …… . (a) − n (n − 1) yn n (b) (n + 1) yn 2 n (c) − (n − 1) yn 2 (d) − n (n + 1) yn True or False. Write ‘T’ for true and ‘F’ for false statement. 11. If y = f (x), then the nth differential coefficient of yr is the (n + r) th differential coefficient of y. 12. If y = e a x sin (b x + c), then D n {e a x sin (b x + c)} = r n e a x sin {b x + c + (n + 1) φ }, where r = (a2 + b 2) 1 ⁄ 2 and φ = tan − 1 (b ⁄ a). 13. While applying Leibnitz’s theorem if we observe that one of the two functions is such that all its differential coefficients after a certain stage become zero, then we should take that function as second function. D-120 DIFFERENTIAL CALCULUS 1. ⎛ n π ⎞⎟ ⋅ an sin ⎜ a x + b + ⎝ 2 ⎠ 2. (− 1) n n ! an (a x + b) − n − 1. 3. 1 2 6. 10. [e x − (5) n ⁄ 2 e x cos (2x + n tan − 1 2)]. (− 1) n e− x. (d). 7. (b). 11. T. 4. − y. 8. (a). 12. F. 5. n2 yn − 1. 9. (c). 13. T. 5.1 Accurate Statement of Taylor’s Theorem If f (x) is a single-valued function of x such that (i) all the derivatives of f (x) upto (n − 1) th are continuous in a ≤ x ≤ a + h , and (ii) f (n) (x) exists in a < x < a + h, then h2 f ′′ (a) + … 2! hn (n) hn − 1 (n − 1) f (a) + f (a + θh), where 0 < θ < 1. + (n − 1) ! n! f (a + h) = f (a) + hf ′ (a) + Taylor’s Series : (Meerut 2009B, 10B; Kashi 11, 13) Su ppose f (x) possesses continuous derivat ives of all orders in the interval [a, a + h]. Then for every positive integral value of n, we have h2 f ′′ (a) + … f (a + h) = f (a) + h f ′ (a) + 2! hn − 1 (n − 1) + f (a) + R n , …(1) (n − 1) ! hn (n) f (a + θ h), (0 < θ < 1). n! Suppose R n → 0, as n → ∞ . Then taking limits of both sides of (1) when n → ∞ , where we get Rn = D-122 DIFFERENTIAL CALCULUS h2 hn (n) f ′′(a) + … + f (a) + … …(2) 2! n! The series given in (2) is known as Taylor’s infinite series for the expansion of f (a + h) as a power series in h . f (a + h) = f (a) + h f ′ (a) + 5.2 Maclaurin’s Series (Rohilkhand 2009B; Kashi 12) Suppose f (x) possesses continuous derivatives of all orders in the interval [0, x]. Then for every positive integral value of n, we have x2 x n − 1 (n − 1) f (x) = f (0) + x f ′ (0) + f ′′ (0) + … + f (0) + R n , … ...(1) 2! (n − 1) ! x n (n) f (θ x), (0 < θ < 1). where Rn = n! Suppose R n → 0, as n → ∞ . Then taking limits of both sides of (1) when n → ∞ , we get x2 x n (n) f ′′ (0) + … + f (0) + … …(2) 2! n! The series given in (2) is known as Maclaurin’s infinite series for the expansion of f (x) as a power series in x. Maclaurin’s series is a particular case of Taylor’s series. If in Taylor’s series we put a = 0 and h = x, we get Maclaurin’s series. Maclaurin’s expansion of f (x) fails if any of the functions f (x), f ′ (x), f ′′ (x), …, becomes infinite or discontinuous at any point of the interval [0, x] or if R n does not f (x) = f (0) + x f ′ (0) + tend to zero as n → ∞ . 5.3 Formal Expansions of Functions We have seen that for the validity of the expansion of a function f (x) as an infinite Maclaurin’s series, it is necessary that R n → 0 as n → ∞ . But to examine the behaviour of R n as n → ∞ is not an easy job because in many cases it is not possible to find a general expression for the nth derivative of the function to be expanded. So in this chapter we shall simply obtain form al expansion of a function f (x) without showing that R n → 0 as n → ∞ . Such an expansion will not give us any idea of the range of values of x for which the expansion is valid. To obtain such an expansion of f (x) we have only to calculate the values of its derivatives for x = 0 and substitute them in the infinite Maclaurin’s series x x n (n) f (x) = f (0) + f ′ (0) + … + f (0) + … n! 1! F or t he convenience of t he st udents we shall now give formal proofs of Maclaurin’s and Taylor’s theorems without bothering about the nature of R n as n → ∞ . (Bundelkhand 2006; Kashi 12, 13; Purvanchal 14) Maclaurin’s Theorem : L et f (x) be a function of x which possesses continuous derivatives of all orders in the interval [0, x]. A ssum ing that f (x) can be expanded as an infinite power series in x, we have f (x) = f (0) + x2 x n (n) x f ′ (0) + f ′′ (0) + … + f (0) + … 2! n! 1! Proof : Suppose f (x) = A 0 + A 1 x + A 2 x2 + A 3 x3 + … …(1) EXPANSIONS OF FUNCTIONS D-123 Let the expansion (1) be differentiable term by term any number of times. Then by successive differentiation, we have f ′ (x) = A 1 + 2A 2 x + 3 A 3 x2 + 4A 4 x3 + …, f ′′ (x) = 2 . 1 A 2 + 3 . 2 A 3 x + 4 . 3 A 4 x2 + …, f ′′′ (x) = 3 . 2 . 1 A 3 + 4 . 3 . 2 A 4 x + …, and so on. Putting x = 0 in each of these relations, we get f (0) = A 0 , f ′ (0) = A 1, f ′′ (0) = 2! A 2 , f ′′′ (0) = 3! A 3 , … . Substituting these values of A 0 , A 1 , A 2 , … in (1), we get x2 x n (n) f ′′ (0) + … + f (0) + … 2! n! This is Maclaurin’s Theorem. If we denote f (x) by y, then Maclaurin’s theorem can also be written in the following way : f (x) = f (0) + x f ′ (0) + x x2 x3 xn ( y1) 0 + ( y2) 0 + ( y3) 0 + … + (y ) + …. y = ( y) 0 + 1! 2! 3! n! n 0 Taylor’s Theorem : (Bundelkhand 2005; Avadh 09, 10, 14; Kashi 11, 13, 14) L et f (x) be a function of x which possesses, continuous derivatives of all orders in the interval [a, a + h]. A ssum ing that f (a + h) can be expanded as an infinite power series in h, we have h2 hn (n) f (a + h) = f (a) + hf ′ (a) + f ′′(a) + … + f (a) + … 2! n! …(1) Proof : Suppose f (a + h) = A 0 + A 1 h + A 2 h2 + A 3 h3 + … Let the expansion (1) be differentiable term by term any number of times w.r.t. ‘h’. Then by successive differentiation w.r.t. ‘h’, we have f ′ (a + h) = A 1 + 2 A 2 h + 3A 3 h2 + …, f ′′ (a + h) = 2 . 1 A 2 + 3 . 2 A 3 h + …, f ′′′ (a + h) = 3 . 2 . 1 A 3 + …, and so on. Putting h = 0 in each of the above relations, we get f (a) = A 0 , f ′ (a) = A 1, f ′′ (a) = 2! A 2 , f ′′′ (a) = 3! A 3 , and so on. 1 1 A 0 = f (a), A 1 = f ′ (a), A 2 = f ′′ (a), A 3 = f ′′′ (a), and so on. 2! 3! Substituting these values of A 0 , A 1 , A 2 , A 3 , … in (1), we get ∴ h2 hn (n) f ′′ (a) + … + f (a) + … 2! n! This is Taylor’s theorem. Another useful form is obtained on replacing h by (x − a). Thus f (a + h) = f (a) + h f ′ (a) + (x − a) n (n) (x − a) 2 f ′′ (a) + … … + f (a) + …, 2! n! which is an expansion of f (x) as a power series in (x − a). Note : If we expand f (x + h), by Taylor’s theorem, as a power series in h, then the result is as follows : hn (n) h2 f ′′ (x) + … + f (x) + … f (x + h) = f (x) + hf ′ (x) + 2! n! f (x) = f (a) + (x − a) f ′ (a) + D-124 DIFFERENTIAL CALCULUS Example 1 : Expand e x in ascending powers of x. (Bundelkhand 2008) x (n) x Solution : Let f (x) = e . Then f (0) = 1, f (x) = e so that f (n) (0) = 1, where n = 1, 2, 3, 4, … Substituting these values in Maclaurin’s series f (x) = f (0) + xf ′ (0) + x3 x2 f ′′(0) + f ′′′ (0) + …, we get 2! 3! x2 x3 x4 xn + + + …+ + … 2! 3! 4! n! This is known as Exponential series. Example 2 : Expand (1 + x) n in ascending powers of x. Solution : Let f (x) = (1 + x) n , so that f (0) = 1. We have f (m) (x) = n (n − 1) …(n − m + 1) (1 + x) n − m . ex = 1 + x + ∴ f (m) (0) = n (n − 1)…(n − m + 1). Putting m = 1, 2, 3, …, we have f ′ (0) = n, f ′′ (0) = n (n − 1), f ′′′ (0) = n (n − 1) (n − 2), and so on. Substituting these values in Maclaurin’s series for f (x), we get n (n − 1) 2 n (n − 1) …(n − m + 1) m x + …… + x + … (1 + x) n = 1 + nx + 2! m! This is known as Binomial series. If n is a positive integer, the series will consist of (n + 1) terms. Example 3 : Expand sin x. Solution : Let f (x) = sin x. Then f (0) = 0 f ′ (x) = cos x, f ′ (0) = 1 f ′′ (x) = − sin x, f ′′ (0) = 0 f ′′′ (x) = − cos x, f ′′′ (0) = − 1. 1 ⎛ ⎞ (n) In general f (x) = sin ⎜ x + nπ ⎟ so that 2 ⎝ ⎠ 1 (n) f (0) = sin nπ = 0 when n = 2m and 2 (Kashi 2012) = (− 1) m when n = 2m + 1. H ence substituting these values in Maclaurin’s series, we get sin x = 0 + x . 1 + 0 + x2m + 1 x3 (− 1) + 0 + … + 0 + (− 1) m + … 3! (2m + 1)! x3 x5 x2m + 1 + − … + (− 1) m + …. 3! 5! (2m + 1) ! This is known as Sine series. Similarly we may obtain Cosine series : x2m x2 x4 + − … + (− 1) m + …. cos x = 1 − 2! 4! (2m) ! = x− Example 4 : Expand log (1 + x) by Maclaurin’s theorem . Solution : Let f (x) = log (1 + x). (Kanpur 2006) (Meerut 2003, 11; Agra 05) D-125 EXPANSIONS OF FUNCTIONS Then f (0) = log 1 = 0, f (n) (x) = (− 1) n − 1 (n − 1)! (x + 1) n so that f (n) (0) = (− 1) n − 1 (n − 1)!, where n = 1, 2, 3, 4, … . Now by Maclaurin’s theorem, x2 f (x) = f (0) + x f ′ (0) + f ′′ (0) + … . 2! Substituting the values of f (0), f ′ (0), f ′′ (0), etc., we get log (1 + x) = 0 + x − x3 x4 x2 ⋅ 1! + ⋅ 2! − ⋅ 3! + … 2! 3! 4! xn (− 1) n − 1 (n − 1)! + … n! xn x2 x3 x4 + − + … + (− 1) n −. 1 + … = x− 2 3 4 n Example 5 : A pply Maclaurin’s theorem to find the expansion in ascending powers of x of loge (1 + e x) to the term s containing x4. (Kanpur 11; Rohilkhand 12) …+ Solution : Let y = log e (1 + e x). Now y1 = (1 + e x) − 1 ex 1 1 1 = = 1− so that ( y1) 0 = 1 − = , 2 2 1 + ex 1 + ex 1 + ex y2 = 0 + so that Then ( y) 0 = log e (1 + e0) = log e 2. ex ex 1 = . = y1 (1 − y1) = y1 − y12, (1 + ex) 2 1 + e x 1 + e x ⎛ ⎞2 1 1 1 ( y2) 0 = ( y1) 0 − [( y1) 0]2 = − ⎜⎝ ⎟⎠ = , 2 2 4 1 1 1 y3 = y2 − 2y1 y2 so that ( y3) 0 = − 2 ⋅ ⋅ = 0, 4 2 4 ⎛ 1⎞ 2 1 y4 = y3 − 2y22 − 2y1 y3 so that ( y4) 0 = 0 − 2 ⋅ ⎜ ⎟ − 0 = − , and so on. 8 ⎝ 4⎠ Now by Maclaurin’s theorem, we have x2 x3 x4 y = ( y) 0 + x ( y1) 0 + ( y2) 0 + ( y3) 0 + (y ) + …. 2! 3! 4! 4 0 1 x2 1 x3 x4 ⎛ 1 ⎞ + ⋅ + ⋅ 0+ ⋅ ⎜− ⎟ + … 2 2! 4 3! 4! ⎝ 8 ⎠ x4 x x2 − + …. = log 2 + + 8 192 2 Example 6 : Expand log {x + √(1 + x2)} in ascending powers of x and find the general term . ∴ log (1 + e x) = log 2 + x ⋅ Solution : Let or y = log {x + √(1 + x2)}. ⎧ ⎫ 1 2x 1 ⎪= ⋅ ⎪⎨ 1 + ⎬ 2 2 x + √(1 + x ) ⎪⎩ 2 √(1 + x ) ⎪⎭ √(1 + x2) Then y1 = ∴ y12 (1 + x2) − 1 = 0. Differentiating again, we get (1 + x2) 2 y1 y2 + 2x y12 = 0 2y1 [(1 + x2) y2 + xy1] = 0 …(1) …(2) D-126 DIFFERENTIAL CALCULUS or (1 + x2) y2 + xy1 = 0, since 2y1 ≠ 0. or …(3) Now differentiating (3) n times by Leibnitz’s theorem, we get n (n − 1) yn . 2 + yn + 1 . x + n . yn . 1 = 0 (1 + x2) yn + 2 + n . yn + 1 . 2x + 1.2 …(4) (1 + x2) yn + 2 + (2n + 1) xyn + 1 + n2 yn = 0. Putting x = 0 in (1), (2), (3) and (4), we get ( y) 0 = 0, ( y1) 0 = 1, ( y2) 0 = 0, and ( yn + 2) 0 = − n2 ( yn ) 0. Now putting n = 1, 3, 5, … in (5), we get …(5) ( y3) 0 = − 12 ( y1) 0 = − 12, ( y5) 0 = (− 32) ( y3) 0 = (− 32) (− 12) = 32.12, ( y7) 0 = (− 52) ( y5) 0 = (− 52) (− 32) (− 12) = − 52 . 32 . 12, and so on. Putting n − 2 in place of n in (5), we get ( yn ) 0 = {− (n − 2) 2} ( yn − 2) 0 = {− (n − 2) 2} {− (n − 4) 2} ( yn − 4) 0 . [... replacing n by n − 2 in (6), we have ( y n − 2) 0 …(6) = − (n − 4) 2 ( yn − 4) 0] Thus if n is odd, we have ( yn ) 0 = {− (n − 2) 2} {− (n − 4) 2}… (− 52) (− 32) (− 12) . 1 = (− 1) (n − 1) ⁄ 2 (n − 2) 2 (n − 4) 2 …52 . 32 . 12. Again, putting n = 2, 4, 6,…in (5), we get …(7) ( y4) 0 = − 22. ( y2) 0 = 0, ( y6) 0 = − 42 . ( y4) 0 = 0, and so on. Thus, if n is even, we have ( yn ) 0 = 0. Now by Maclaurin’s theorem, we have x x2 x3 y = ( y) 0 + ( y1) 0 + ( y2) 0 + (y ) + …. 2! 3! 3 0 1! ∴ x2 x3 x4 .0+ ⋅ (− 12) + ⋅ 0 log {x + √(1 + x2)} = 0 + x . 1 + 2! 3! 4! x5 2 2 x6 x7 + (3 ⋅ 1 ) + ⋅ 0+ (− 52 ⋅ 32 ⋅ 12) + … 5! 6! 7! = x− The general term = x5 2 2 x7 x3 ⋅ 12 + (3 ⋅ 1 ) − . (52 ⋅ 32 ⋅ 12) + … 3! 5! 7! xn ( y ) , where ( yn ) 0 is given by (7) when n is odd and n! n 0 ( yn ) 0 = 0 , when n is even. Putting 2n − 1 in place of n in (7), we find that ( y2n − 1) 0 = (− 1) n − 1 (2n − 3) 2 (2n − 5) 2 …52. 32. 12. D-127 EXPANSIONS OF FUNCTIONS H ence log {x + √(1 + x2)} = x − 12 ⋅ x3 x5 x7 + 12 ⋅ 32 ⋅ − 12 ⋅ 32 ⋅ 52 ⋅ + … 3! 5! 7! + (− 1) n − 1 12 ⋅ 32 ⋅ 52 … (2n − 3) 2 ⋅ x2n − 1 + …. (2n − 1)! Example 7 : If y = sin− 1 x = a0 + a1 x + a2 x2 + … , prove that (n + 1) (n + 2) an + 2 = n2 an . (Meerut 2010B; Kumaun 08) Solution : Let y = sin− 1 x 1 Then y1 = ⋅ √(1 − x2) ∴ or . …(1) …(2) y12 (1 − x2) − 1 = 0. Differentiating again, we get (1 − x2) 2y1 y2 − 2 xy12 = 0 2y1 [(1 − x2) y2 − x y1] = 0 (1 − x2) y2 − xy1 = 0, 2y1 ≠ 0. or since …(3) Now differentiating (3) n times by Leibnitz’s theorem, we get n (n − 1) yn (− 2) − yn + 1 . x − nyn . 1 = 0 (1 − x2) yn + 2 + n . yn + 1 . (− 2x) + 1.2 or (1 − x2) yn + 2 − (2n + 1) x yn + 1 − n2 yn = 0 Putting x = 0 in (4), we get ( yn + 2) 0 = n2 ( yn ) 0. By Maclaurin’s theorem, we have x x2 x3 xn y = ( y) 0 + ( y1) 0 + ( y2) 0 + ( y3) 0 + … + (y ) + … 1! 2! 3! n! n 0 Also we are given that …(4) …(5) y = sin− 1 x = a0 + a1 x + a2 x2 + … + an x n + … . E quating the coefficients of x n in the two expansions for y, we get an = ∴ H ence ( yn ) 0 n! ⋅ an + 2 ( yn + 2) 0 ( yn + 2) 0 n! 1 = ⋅ = ⋅ (n + 2) (n + 1) an (n + 2)! ( yn ) 0 ( yn ) 0 (y ) n2 , substituting for n + 2 0 from (5). = (n + 2) (n + 1) ( yn ) 0 (n + 1) (n + 2) an + 2 = n2 an . ⎛ 1 ⎞⎟ π by using Taylor’s series. 2 ⎠ (Meerut 2005; Rohilkhand 06, 10; Agra 06) Example 8 : Expand sin x in powers of ⎜⎝ x − Solution : Let f (x) = sin x. We want to expand f (x) in powers of x − ⎡1 ⎣2 We can write f (x) = f ⎢ π + ⎛ ⎜ ⎝ x− 1 2 ⎞ ⎟ ⎠ ⎤ π ⎥. ⎦ 1 2 π. D-128 DIFFERENTIAL CALCULUS ⎡1 ⎣2 Now expanding f ⎢ π + we get ⎡1 ⎛ ⎜ ⎝ 1 2 x− ⎛ ⎞ ⎤ ⎦ ⎛ ⎝ π ⎟⎠ ⎥ by Taylor’s theorem in powers of ⎜ x − 1 ⎞⎤ ⎛1 ⎞ ⎛ 1 ⎞ 1 2 ⎞ ⎠ π⎟ , ⎛1 ⎞ f (x) = f ⎢⎣ π + ⎜ x − π ⎟ ⎥⎦ = f ⎜ π ⎟ + ⎜ x − π ⎟ f ′ ⎜ π ⎟ 2 2 ⎠ 2 ⎠ ⎝ ⎝2 ⎠ ⎝ ⎝2 ⎠ + 2 3 ⎛1 ⎞ ⎛1 ⎞ 1 ⎛ 1 ⎛ 1 ⎞ 1 ⎞ ⎜ x − π ⎟ f ′′ ⎜ π ⎟ + ⎜ x − π ⎟ f ′′′ ⎜ π ⎟ + … …(1) 2 ⎠ 2 ⎠ ⎝2 ⎠ ⎝2 ⎠ 2!⎝ 3!⎝ 1 ⎛1 ⎞ f (x) = sin x. Therefore f ⎜ π ⎟ = sin π = 1, 2 ⎝2 ⎠ Now ⎛1 ⎞ 1 2 ⎝2 ⎠ ⎛1 ⎞ 1 f ′′ (x) = − sin x so that f ′′ ⎜ π ⎟ = − sin π = − 1, 2 ⎝2 ⎠ ⎛1 ⎞ 1 f ′′′ (x) = − cos x so that f ′′′ ⎜ π ⎟ = − cos π = 0, 2 ⎝2 ⎠ ⎛ ⎞ 1 1 f i v (x) = sin x so that f i v ⎜ π ⎟ = sin π = 1, etc. 2 ⎝2 ⎠ f ′ (x) = cos x giving f ′ ⎜ π ⎟ = cos π = 0, Substituting these values in (1), we get ⎛ ⎝ sin x = 1 + ⎜ x − 1 2 ⎞ ⎠ π⎟ . 0 + 2 3 1 ⎛ 1 ⎛ 1 ⎞ 1 ⎞ ⎜ x − π ⎟ . (− 1) + ⎜ x − π⎟ . 0 2 ⎠ 2 ⎠ 2! ⎝ 3! ⎝ + 4 1 ⎛ 1 ⎞ ⎜ x − π⎟ . 1 + … 2 4!⎝ ⎠ 2 4 1 ⎛ 1 ⎛ 1 ⎞ 1 ⎞ ⎜ x − 2 π⎟ + ⎜ x − 2 π⎟ − … . 4!⎝ 2!⎝ ⎠ ⎠ Example 9 : Expand log sin (x + h) in powers of h by Taylor’s theorem . = 1− (Purvanchal 2006; Meerut 10; Bundelkhand 09; Kashi 14) Solution : Let f (x + h) = log sin (x + h). Then by Taylor’s theorem, we have log sin (x + h) = f (x + h) h2 h3 h f ′ (x) + f ′′ (x) + f ′′′ (x) + … . 2! 3! 1! f (x + h) = log sin (x + h). f (x) = log sin x , f ′ (x) = (1 ⁄ sin x) . cos x = cot x , = f (x) + Now ∴ f ′′ (x) = − cosec 2 x , f ′′′ (x) = 2 cosec x cosec x cot x = 2 cosec 2 x cot x . … … … … … … … … … H ence, log sin (x + h) = log sin x + h cot x − 1 2 h2 cosec 2 x + 1 3 h3 cosec 2 x cot x + … D-129 EXPANSIONS OF FUNCTIONS Ex. 10. Use Taylor’s theorem to prove that sin θ sin 2 θ − (h sin θ) 2 tan− 1 (x + h) = tan− 1 x + h sin θ 1 2 sin 3 θ sin n θ 3 n − 1 + (h sin θ) + … + (− 1) (h sin θ) n + … 3 n θ = cot − 1 x. (Gorakhpur 2005; Agra 07; Rohilkhand 08B, 09B; where Kashi 11; Avadh 13) Solution : Let Then or y = f (x) = tan− 1 x. y1 = 1 1 1 ⎡ 1 1 = = ⎢ − 2 (x + i) (x − i) 2i x − i x + 1+ x ⎣ y1 = 1 [(x − i) − 1 − (x + i) − 1]. 2i ⎤ ⎥ i⎦ …(1) Differentiating (1), (n − 1) times, we get 1 yn = [( − 1) n − 1 (n − 1)! (x − i) − n − (− 1) n − 1 (n − 1)! (x + i) − n ] 2i yn = or (− 1) n − 1 (n − 1)! [(x − i ) − n − (x + i) − n ]. 2! …(2) Now put x = r cos θ, 1 = r sin θ in (2). Then yn = (− 1) n − 1 (n − 1)! − n .r [(cos θ − i sin θ) − n − (cos θ + i sin θ) − n ] 2i (− 1) n − 1 (n − 1)! − n r [(cos n θ + i sin n θ) − (cos n θ − i sin n θ)], 2i by De Moivre’s theorem n − 1 (n − 1)! − n (− 1) r . 2i sin n θ = 2i = (− 1) n − 1 (n − 1)! sinn θ sin n θ. [... r− 1 = 1 ⁄ r = sin θ] = H ence f (n) (x) = (− 1) n − 1 (n − 1) ! sinn θ sin n θ, where cot θ = x, i.e., θ = cot− 1 x. Putting n = 1, 2, 3,…, we get f ′ (x) = sin θ . sin θ, f ′′ (x) = − sin2 θ sin 2θ, f ′′′ (x) = 2! sin3 θ sin 3 θ, and so on. Substituting these values in Taylor’s series h2 h3 hn (n) f ′′ (x) + f ′′′ (x) + … + f (x) + …, we get 2! 3! n! 2 1 (x + h) = tan− 1 x + h sin θ . sin θ − h sin2 θ sin 2 θ 2! hn h3 2! sin3 θ sin 3 θ − … + (− 1) n − 1 (n − 1)! sin n θ sin n θ + … + 3! n! 1 (x + h) = tan− 1 x + h sin θ. sin θ − (h sin θ) 2 sin 2 θ 1 2 sin 3 θ sin n θ 3 n + (h sin θ) . + … + (− 1) − 1 (h sin θ) n . + … 3 n f (x + h) = f (x) + hf ′ (x) + tan− or tan− D-130 1. (i) (i) DIFFERENTIAL CALCULUS State Maclaurin’s theorem. State Taylor’s theorem. (Meerut 2000; Bundelkhand 01, 08, 11; Agra 07) (Bundelkhand 2006, 08, 11) Expand the following functions by Maclaurin’s theorem : 2. (i) a x (ii) tan x (Meerut 2012B) (Kanpur 2014) e x cos x (iii) (iv) tan− 1 x (v) sec x. 3. (Bundelkhand 2001) (i) O btain by Maclaurin’s theorem the first five terms in the expansion of esin x. (Bundelkhand 2007) (ii) E xpand by Maclaurin’s theorem ex 1 + ex as far as the term x3. (Meerut 2006B) (iii) O btain by Maclaurin’s theorem the first five terms in the expansion of log (1 + sin x). (Meerut 2007) (iv) Find the first three terms in the expansion in the powers of x of log (1 + tan x) . (Rohilkhand 2011B) 4. (i) Apply Maclaurin’s theorem to prove that log sec x = 1 2 x 2 + 1 4 x 12 + 1 6 x 45 + …. (Bundelkhand 2011; Rohilkhand 13) 2x2 4x3 (ii) U se Maclaurin’s formula to show that e x sec x = 1 + x + + + … 2! 3! (Meerut 2004; Rohilkhand 08B) 5. (iii) Expand sinh x cos x to fifth powers of x. Show that (i) e x cos x = 1 + x − 2 x3 22 x4 22 x5 23 x7 − − + + … 3! 4! 5! 7! ⎛ 1 ⎞ 2n ⁄ 2 n x + …. + cos ⎜ nπ ⎟ ⋅ n! ⎝4 ⎠ (Bundelkhand 2014; Agra 14) 2 3 22 5 ⎛ 1 ⎞ 2n ⁄ 2 n (ii) e x sin x = x + x2 + x − x − …+ sin ⎜ nπ ⎟ x + … 5! 3! ⎝4 ⎠ n! (Meerut 2003; Gorakhpur 06) 6. Apply Maclaurin’s theorem to prove that 3 a 2 b − b3 3 (i) e ax sin bx = bx + abx2 + x + … 3! …+ ⎛ (a2 + b2) n ⁄ 2 n b⎞ x sin ⎜ n tan− 1 ⎟ + … a⎠ n! ⎝ D-131 EXPANSIONS OF FUNCTIONS 7. a2 − b2 2 a (a2 − 3b2) 3 (ii) e ax cos bx = 1 + ax + x + x + … 3! 2 ⎛ (a2 + b2) n ⁄ 2 n b⎞ x cos ⎜ n tan− 1 ⎟ + … . + a⎠ n! ⎝ 2 3 x x Show that e x cos α cos (x sin α) = 1 + x cos α + cos 2α + cos 3α + … 2! 3! (Rohilkhand 2007; Avadh 11) 8. 9. sin− 1 (i) E xpand (x + h) in powers of x as far as the term x3. [Hint. Use Taylor’s series] x x2 x3 (ii) Prove that log (x + h) = log h + − + − … 2 h 2h 3h3 ⎛ 1 ⎞ (i) E xpand tan− 1 x in powers of ⎜ x − π ⎟ . 4 ⎠ ⎝ [Hint. Let f (x) = tan− 1 x. We can write ⎡1 ⎛ 1 ⎞⎤ f (x) = f ⎢⎣ π + ⎜ x − π ⎟ ⎥⎦ . Now apply Taylor’s theorem] 4 4 ⎠ ⎝ ⎞ ⎛1 (ii) E xpand sin ⎜ π + θ⎟ in powers of θ. ⎠ ⎝4 (iii) E xpand 2x3 + 7x2 + x − 1 in powers of x − 2. (Meerut 2004B, 05B; Gorakhpur 06; Rohilkhand 09; Purvanchal 11; Kashi 11) (iv) Write the value of α , if by Taylor’s theorem 2x3 + 7x2 + x − 1 = α + 53 (x − 2) + 19 (x − 2) 2 + 2 (x − 2) 3. (Meerut 2001) (v) E xpand log sin x in powers of (x − a). (Meerut 2001, 06; Rohilkhand 07B; Avadh 10) 10. − 1 (i) If y = e a sin x, show that (1 − x2) yn + 2 − (2n + 1) xyn + 1 − (n2 + a2) yn = 0. H ence by Maclaurin’s theorem, show that a2 2 a (12 + a2) 3 x + x + … 3! 2! 1 2 Also deduce that e θ = 1 + sin θ + sin2 θ + sin3 θ + … 2! 3! − 1 e a sin x= 1 + ax + (ii) If y = sin (m sin− 1 x) , then show that d2 y dy − x + m 2 y = 0. dx dx2 H ence or otherwise expand sin m θ in powers of sin θ. (1 − x2) (iii) If y = sin log (x2 + 2x + 1), prove that (x + 1) 2 yn + 2 + (2n + 1) (x + 1) yn + 1 + (n2 + 4) yn = 0. H ence or otherwise expand y in ascending powers of x as far as x6. (Kumaun 2008) D-132 11. 12. 13. DIFFERENTIAL CALCULUS By Maclaurin’s theorem or otherwise find the expansion of y = sin (e x − 1) upto and including the term in x4. Find also the first two non-vanishing terms in the expansion of x as a series of ascending powers of y. E xpand log {1 − log (1 − x)} in powers of x by Maclaurin’s theorem as far as the (Avadh 2009) term x3. x By substituting for x deduce the expansion of log {1 + log (1 + x)} as far as 1+ x the term in x3. 1 1 If y = sin − 1 x ⁄ √(1 − x2) when − 1 < x < 1, and − π < sin − 1 x < π, prove that dn + 1 2 n d − 1 dn 2 y y y − (2n + 1) x n − n2 n − 1 = 0. (Meerut 2007B) d xn + 1 dx dx Assuming that y can be expanded in ascending powers of x in the form (1 − x2) a0 + a1 x + a2 x2 + … + prove that (n + 1) an + 1 = nan − expansion. 14. 15. If y = e m tan − 1 an x n + …, 1, and hence obtain the general term of the x= a0 + a1 x + a2 x2 + … + an x n + … , prove that (n + 1) an + 1 + (n − 1) an − 1 = m an . (Purvanchal 2007) Prove that f (m x) = f (x) + (m − 1) x f ′(x) + (m − 1) 2 2 (m − 1) 3 3 x f ′′ (x) + x f ′′′ (x) + … 2! 3! (Meerut 2001; Agra 07; Rohilkhand 13) 16. Prove that ⎛ x2 ⎞ f ′′(x) x x2 ⎟ = f (x) − (i) f ⎜ f ′ (x) + − … 2 ⎝ 1 + x⎠ 1+ x (1 + x) 2 ! (ii) f (x) = f (0) + x f ′ (x) − x2 2! f ′′ (x) + x3 3! x x2 ⎞ ⎛ ⎟ = f ⎜x − ⎝ 1 + x⎠ 1+ ⎝ Now apply Taylor’s theorem. (ii) We have f (0) = f (x − x). Apply Taylor’s theorem and transpose the [ Hint. (i) 2. ⎛ Write f ⎜ (i) 1 + x log a + f ′′′ (x) − … ⎞ ⎟. x⎠ terms to get the result.] (x log a) 2 (x log a) n + …+ + …. 2! n! 2 5 x3 + x + …. 3 15 x2 x3 11 x4 x5 − − − + …. (iii) 1 + x + 2 3 24 5 (ii) x + (Rohilkhand 2008) D-133 EXPANSIONS OF FUNCTIONS x3 x5 x7 x2 n − 1 + − + … + (− 1) n − 1 + …. 3 5 7 (2n − 1) x2 5x4 61 x6 (v) 1 + + + + …. 2! 4! 6! x3 1 x2 x4 1 1 − + …. (ii) + x − + …. (i) 1 + x + 2 4 2 8 8 3! x4 x5 x2 x3 1 2 + − + − …. (iv) x − x2 + x3 + … . (iii) x − 2 3 2 6 12 24 3 5 2x 4x (iii) x − − + …. 3! 5! x2 h (1 − h2) − 3 ⁄ 2 sin− 1 h + x (1 − h2) − 1 ⁄ 2 + 2! x3 + {(1 − h2) − 5 ⁄ 2 (1 + 2h2)}+ … . 3! (iv) x − 3. 4. 8. 9. 2 ⎛ 1 ⎞ (i) tan− 1 (π ⁄ 4) + ⎜ x − π ⎟ ⎝ 1 ⎛⎜ 1+ θ− √2 ⎝ (iii) 45 + 53 (x − (iv) 45. (ii) 4 ⎛ 1 ⎞ (1 + π 2 ⁄ 16) − π ⎜ x − π ⎟ / {4 (1 + π 2 ⁄ 16) 2} + … . 4 ⎠ ⎠ / ⎝ ⎞ θ2 θ3 θ4 θ5 − + + − …⎟ ⋅ 2! 3! 4! 5! ⎠ 2) + 19 (x − 2) 2 + 2 (x − 2) 3. (x − a) 2 (x − a) 3 cosec 2 a + 2 cosec 2 a cot a + … . 2! 3! m (12 − m 2) (32 − m 2) m (12 − m 2) sin3 θ + sin5 θ + … . (ii) sin m θ = m sin θ + 3! 5! 3 5 3 2 (iii) y = 2x − x2 − x3 + x4 − x5 + x6 + … . 3 2 3 2 y2 x2 5 x4 x+ − + … ., y − + … 2! 24 2 x3 7 x3 x+ + … ., x − x2 + + … 6 6 2m (2m − 2) (2m − 4)…2 ⋅ a2 m = 0, a2m + 1 = (2m + 1) (2m − 1) …3 (v) log sin a + (x − a) cot a − 10. 11. 12. 13. Fill in the Blanks: Fill in the blanks “……”, so that the following statem ents are com plete and correct. 1. 2. By Maclaurin’s theorem expansion of sin− 1x is …… . By Maclaurin’s theorem y = log (sec x + tan x) = x + x3 x5 + + …… , then ( y3) 0 = …… . 6 24 D-134 3. DIFFERENTIAL CALCULUS By Maclaurin’s theorem 2n ⁄ 2 sin (n π ⁄ 4 ) n 2 3 x + …… + x + …… , y = e x sin x = x + x2 + 3! n! then ( y3) 0 = …… . (Meerut 2001) 4. By Taylor’s theorem the expansion of log (x + h) in ascending powers of x is …… . Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 5. 6. By Taylor’s theorem the second term in the expansion of log sin x in the powers of (x − a) is (a) (x − a) cot a (b) (x − b) cosec a (c) (x − a) cot a cosec a (d) (x − a) 2 cot a By Maclaurin’s theorem the second term in the expansion of 1 x (a) x (b) 48 4 1 3 (c) 0 (d) − x 48 (Rohilkhand 2005) x e ⁄ (1 + e x) is (Rohilkhand 2007) True or False: Write ‘T’ for true and ‘F’ for false statement. 7. 8. The function log x does not possess Maclaurin’s series expansion because it is not defined at x = 0. If in Taylor’s series we put a = 0 and h = x, we get Maclaurin’s series. 9. Is this f (x) = f (0) + x f ′ (0) + x2 x3 f ′′ (0) + f ′′′ (0) + …… Taylor’s Theorem. 2! 3! (Agra 2005, 06) 12 . x 3 32 . 12 . x 5 52 . 32 . 12 . x 7 + + + …… . 3! 5! 7! 1. x+ 2. 1. 3. 2. 4. log h + 5. 8. (a). T. 6. 9. (b). F. 7. T. x x2 x3 − + − …… . h 2 h2 3 h3 6.1 Indeterminate Forms The form 0 ⁄ 0 has got no definite value. For if we write 0 ⁄ 0 = y, then the equation 0y = 0 reduces to an identity in y, i.e., it is true for all values of y. We cannot cancel 0 from both sides. Therefore the form 0 ⁄ 0 is meaningless. lim lim Now suppose x → a φ (x) = 0 and x → a ψ (x) = 0. lim x → a φ (x) lim φ (x) = Then we cannot write x → a lim ψ (x) x → a ψ (x) φ (x) takes the form 0 ⁄ 0 which is meaningless. It, however, ψ (x) lim φ (x) takes the form 0 ⁄ 0, then the limit itself does not exist. does not mean that if x → a ψ (x) lim 2 2 x → a (x − a ) lim x2 − a2 For example, x → a takes the form 0 ⁄ 0 if we write it as ⋅ x− a lim x → a (x − a) lim because in that case x → a lim (x − a) (x + a) lim lim x2 − a2 = x →a = x → a (x + a) = 2a But, we have x → a x− a (x − a) and thus the limit exists. D-136 DIFFERENTIAL CALCULUS H ere it should not be confused that we have made an attempt to find the value of 0/0. We have simply evaluated the limit of a function which is the quotient of two functions such that if we take their limits separately, then the combination takes the form 0/0. The form 0/0 is an indeterminate form. It has no definite value. The other indeterminate forms are ∞ ⁄ ∞ , ∞ − ∞ , 0 × ∞ , 1∞ , 00, ∞ 0. In this chapter we shall discuss methods which enable us to evaluate the limits of indeterminate forms. 6.2 The Form 0/0 Suppose φ (x) and ψ (x) are functions which can be expanded by Taylor’s theorem in the neighbourhood of x = a. A lso let φ (a) = 0, and ψ (a) = 0. Then φ ′ (x) lim lim φ (x) x → a ψ (x) = x → a ψ ′ (x) ⋅ lim φ (x) We have, by Taylor’s theorem, x → a ψ (x) (x − a) 2 φ ′′ (a) + … + R 1 2! lim , = x →a (x − a) 2 ψ (a) + (x − a) ψ ′ (a) + ψ ′′ (a) + … + R 2 2! φ (a) + (x − a) φ ′ (a) + (x − a) n (n) φ {a + θ1 (x − a)}, 0 < θ1 < 1, n! (x − a) n (n) and R2 = ψ {a + θ2 (x − a)}, 0 < θ2 < 1. n! But, by hypothesis, φ (a) = 0 and ψ (a) = 0. (x − a) 2 φ ′′ (a) + … + R 1 (x − a) φ ′ (a) + φ (x) 2! lim lim Therefore, x → a = x →a . ψ (x) (x − a) 2 (x − a) ψ ′ (a) + ψ ′′ (a) + … + R 2 2! Dividing the numerator and denominator by x − a, we have 1 ⎧ 1 ⎫ φ ′ (a) + (x − a) ⎨ φ ′′ (a) + (x − a) φ ′′′ (a) + …⎬ 3 ! 2 ! ⎩ ⎭ lim lim φ (x) x → a ψ (x) = x → a 1 ⎧ 1 ⎫ ψ ′ (a) + (x − a) ⎨ ψ ′′ (a) + (x − a) ψ ′′′ (a) + …⎬ 3! ⎩2 ! ⎭ φ ′ (a) , = if φ ′ (a) and ψ ′ (a) are not both zero ψ ′ (a) lim φ ′ (x) ⋅ = x →a ψ ′ (x) This proves the theorem which is generally known as L ’ Hospital’s Rule. I t ca n b e e a s i l y s e e n t h a t i f φ ′ (a), φ ′′ (a),……, φ (n − 1) (a) a n d ψ ′ (a), ψ ′′ (a),…, ψ (n − 1) (a) are all zero, but φ (n) (a) and ψ(n) (a) are not both zero, then where R1 = lim φ (x) lim φ (n) (x) ⋅ x → a ψ (x) = x → a ψ (n) (x) D-137 INDETERMINATE FORMS The theorem of this article is true even if x tends to ∞ or − ∞ instead of a, i.e., if lim lim x → ∞ φ (x) = 0, x → ∞ ψ (x) = 0, then φ ′ (x) lim lim φ (x) x → ∞ ψ (x) = x → ∞ ψ ′ (x) ⋅ Writing x = 1 ⁄ y, we have as x → ∞ , y → 0. φ (1 ⁄ y) lim φ ′ (1 ⁄ y) y− 2 , = y →0 ψ (1 ⁄ y) ψ ′ (1 ⁄ y) y− 2 by L’H ospital’s rule φ ′ (x) lim lim φ ′ (1 ⁄ y) = ⋅ = y →0 ψ ′ (1 ⁄ y) x → ∞ ψ ′ (x) Note 1 : L’Hospital’s rule implies φ (x) φ (x) lim lim x → a + ψ (x) = x → a − ψ (x) ⋅ φ (x) Note 2 : While applying L’H ospital’s rule we are not to differentiate by the ψ (x) rule for finding the differential coefficient of the quotient of two functions. But we are to differentiate the numerator and denominator separately. Note 3 : Important. Befo r e a p p lyin g L ’H o sp it a l’s r u le we mu st sat isfy ourselves that the form is 0/0. Sometimes it happens that at some stage the resulting function is not indeterminate of the type 0/0 and we still apply L’H ospital’s rule which is not justified in that case. This is a fairly common error. ∴ lim lim φ (x) x → ∞ ψ (x) = y → 0 lim e x − e− x − 2 log (1 + x) Example 1 : Evaluate x → 0 . x sin x lim e x − e− x − 2 log (1 + x) Solution : We have x → 0 x sin x 2 e x + e− x − (1 + x) lim = x →0 sin x + x cos x 2 e x − e− x + 1− 1+ 2 2 (1 + x) 2 lim = x →0 = = = 1. cos x + cos x − x sin x 1 + 1 − 0 2 lim Example 2 : Evaluate x → 0 lim Solution : We have x → 0 lim = x →0 1 3 x 6 x5 sin x − x + cos x − 1 + 5 x4 sin x − x + x5 1 2 x 2 1 3 x 6 [form 0/0] [form 0/0] ⋅ [form 0/0] [form 0/0] D-138 DIFFERENTIAL CALCULUS lim − sin x + x = x →0 20 x3 lim − cos x + 1 = x →0 60 x2 lim sin x = x →0 120 x 1 lim cos x = x →0 = ⋅ 120 120 [form 0/0] [form 0/0] [form 0/0] 6.3 Algebraic Methods In many cases the limits are easily obtained by the use of well known algebraic and trigonometrical expansions. We can also make use of some well known limits in order to solve the problems or to shorten the work. The following expansions should be remembered : (i) (1 − x) − 1 = 1 + x + x2 + x3 + …ad. inf., | x | < 1. x2 x3 x4 + − + …ad. inf., | x | < 1. 2 3 4 x3 x5 x7 (iii) sin x = x − + − + …ad. inf. 3! 5! 7! (ii) log (1 + x) = x − x2 + 2! x3 + (v) sinh x = x + 3! (iv) cos x = 1 − (vi) cosh x = 1 + x4 − 4! x5 + 5! x6 + …ad. inf. 6! x7 + …ad. inf. 7! x2 x4 x6 + + + …ad. inf. 2! 4! 6! 2 5 x3 + x + …ad. inf. 15 3 x x2 x3 (viii) e x = 1 + + + + …ad. inf. 1! 2! 3! (vii) tan x = x + x3 x5 x7 + 32 . 12 ⋅ + 52 . 32 . 12 ⋅ + …… . 3! 5! 7! The following values of logarithms to the base e should also be remembered : log 1 = 0, log e = 1, log ∞ = ∞ , log 0 = − ∞ . (ix) sin− 1 x = x + 12 ⋅ lim x cos x − log (1 + x) ⋅ Example 1 : Evaluate x → 0 x2 lim x cos x − log (1 + x) Solution : We have x → 0 x2 2 4 ⎛ ⎞ ⎛ ⎞ x x x2 x3 x ⎜1 − + − …⎟ − ⎜ x − + − …⎟ ⎝ ⎠ ⎝ ⎠ 2 ! 4 ! 2 3 lim = x →0 2 x (Meerut 2001) D-139 INDETERMINATE FORMS x2 5 3 − x + … 6 lim 2 = x →0 x2 ⎞ lim ⎛ 1 5 1 = x → 0 ⎜ − x + terms containing higher powers of x⎟ = . ⎝2 ⎠ 6 2 x1 ⁄ 2 tan x ⋅ (Kanpur 2007; Rohilkhand 05) (e x − 1) 3 ⁄ 2 Solution : H ere it should be noted that we cannot apply H ospital’s rule since x1 ⁄ 2 cannot be expanded by Taylor’s theorem in the neighbourhood of x = 0. H owever, we can get the result by the use of algebraic methods. We have thus, lim Example 2 : Evaluate x → 0 lim x →0 x1 ⁄ 2 tan x (e x − 1) 3 ⁄ 2 lim = x →0 x1 ⁄ 2 tan x ⎛ ⎞ x x2 ⎜1 + + + … − 1⎟ ⎝ ⎠ 1! 2! lim = x →0 3⁄2 x1 ⁄ 2 tan x x ⎛ ⎞3⁄2 + …⎟ x3 ⁄ 2 ⎜ 1 + 2! ⎝ ⎠ 1 lim sin x = x →0 ⋅ ⋅ x cos x ⎛ 1 x ⎞3⁄2 + …⎟ ⎜1 + 2 ! ⎝ ⎠ lim sin x lim = 1, since x → 0 = 1 and x → 0 cos x = 1. x lim (1 + x) 1 ⁄ x − e Example 3 : Evaluate x → 0 ⋅ x (Rohilkhand 2013; Kashi 13) 0 lim (1 + x) 1 ⁄ x − e Solution : H ere x → 0 is of the form because 0 x lim 1⁄x x → 0 (1 + x) = e. To evaluate the given limit first we shall obtain an expansion for (1 + x) 1 ⁄ x in ascending powers of x. Let y = (1 + x) 1 ⁄ x . Then ⎞ 1 1⎛ x2 x3 x x2 log (1 + x) = ⎜ x − + − …⎟ = 1 − + − … ⎠ x x⎝ 2 2 3 3 = 1 + z , where z = − (x ⁄ 2) + (x2 ⁄ 3) − … log y = ∴ ⎛ ⎞ z z2 y = e1 + z = e . e z = e . ⎜ 1 + + + …⎟ ⎝ ⎠ 1! 2! 2 2 ⎡ ⎛ x ⎞ ⎞2 ⎤ x x x 1⎛ = e ⎢1 + ⎜− + − …⎟ + ⎜ − + − …⎟ + …⎥ 2⎝ ⎝ 2 ⎠ ⎠ ⎣ ⎦ 3 3 2 x x2 1 2 ⎤ + + x + terms containing powers of x higher than 3 ⎥ 2 8 3 ⎦ 1 11 2 ⎡ ⎤ = e ⎢1 − x + x + …⎥ . 2 24 ⎣ ⎦ ⎡ = e ⎢1 − ⎣ Now lim (1 + x) 1 ⁄ x − e lim = x →0 x →0 x ⎡ ⎣ e ⎢1 − 1 2 x+ 11 2 x 24 x ⎤ ⎦ + …⎥ − e D-140 DIFFERENTIAL CALCULUS lim = x →0 1. 2. 3. ⎡ ⎣ e ⎢− 1 11 2 ⎤ x+ x + …⎥ 2 24 ⎦ x State L’ H ospital’s rule. Evaluate the following limits : lim sin x ⋅ (i) x → 0 x lim x − sin x ⋅ (ii) x → 0 x3 lim e x − 1 (iii) x → 0 ⋅ x lim 1 − cos x ⋅ (iv) x → 0 x2 ⎡ 1 ⎤ 11 1 lim = x →0 e ⎢ − + x + …⎥ = − e. ⎣ 2 ⎦ 24 2 (Meerut 2012B) lim x5 − 2x3 − 4 x2 + 9x − 4 (i) x → 1 ⋅ x4 − 2x3 + 2x − 1 lim a x − b x ⋅ (ii) x → 0 x (Agra 2003) lim log (1 − x2) (iii) x → 0 ⋅ log cos x 4. lim xe x − log (1 + x) (iv) x → 0 ⋅ x2 lim x − sin x ⋅ (i) x → 0 tan3 x lim sin 2x + 2 sin2 x − 2 sin x (ii) x → 0 ⋅ cos x − cos2 x lim (1 + x) n − 1 . (iii) x → 0 x lim log x (iv) x → 1 ⋅ x− 1 5. (Garhwal 2001) lim e x − esin x (i) x → 0 ⋅ x − sin x lim {1 − √(1 − x2)} ⋅ (ii) x → 0 x2 lim tan x − x (iii) x → 0 ⋅ x2 tan x lim tan x − x (iv) x → 0 ⋅ x − sin x (Avadh 2014) D-141 INDETERMINATE FORMS 6. lim {cosh x + log (1 − x) − 1 + x} (i) x → 0 ⋅ x2 lim 5 sin x − 7 sin 2x + 3 sin 3x ⋅ (ii) x → 0 tan x − x lim (iii) x → 0 7. ⎛ 1 − x⎞ ⎟ ⎝ e ⎠ (Meerut 2001) e x + log ⎜ tan x − x ⋅ lim x √ (3x − 2x4) − x6 ⁄ 5 ⋅ (iv) x → 1 1 − x2 ⁄ 3 cos x lim (i) x → π ⁄ 2 ⋅ 1 x− π 2 lim a x − x a (ii) x → a x ⋅ x − aa lim (iii) x → 0 (1 + x) 1 ⁄ x − e + lim sin (iv) x → 0 1 2 ex x2 x . sin− 1 x− x6 x2 ⋅ (Avadh 2006; Purvanchal 14) ⋅ lim x2 + 2 cos x − 2 ⋅ (i) x → 0 x sin3 x lim e x − e x cos x (ii) x → 0 ⋅ x − sin x lim cosh x − cos x ⋅ (iii) x → 0 x sin x lim sin 2x + a sin x (iv) x → 0 ⋅ x2 lim sin2 x − x2 9. (i) x → 0 ⋅ x4 lim sin x sin − 1 x ⋅ (ii) x → 0 x2 8. 10. 11. 12. 2. 3. (Meerut 2012) lim x (1 + a cos x) − b sin x , may be Find the values of a and b in order that x → 0 x3 equal to 1. (Meerut 2013B) lim ae x − b cos x + ce− x Find the values of a, b, c so that x → 0 = 2. x sin x lim x(a + b cos x) − c sin x Find the value of a, b and c so that x → 0 = 1. x5 (i) 1. (i) 4. (ii) 1 ⁄ 6. (ii) log (a ⁄ b). (iii) 1. (iii) 2. (iv) 1 ⁄ 2. (iv) 3 ⁄ 2. D-142 4. 5. DIFFERENTIAL CALCULUS (i) 1 ⁄ 6. (ii) 4. (i) 1. (ii) (i) 0. 7. (i) − 1. ⋅ 1 ⋅ 3 (iv) 1. (iii) 1 ⁄ 3. (iv) 2. 1 ⋅ 2 (iv) 81 ⁄ 20. (iv) 1 ⁄ 18. (iii) − log a − 1 ⋅ log a + 1 (i) 1 ⁄ 12. (ii) 3. (iv) Infinite if a ≠ − 2 and 0 if a = − 2. 9. (i) − 10. 11. 12. (iii) n. (ii) − 15. 6. 8. 1 2 (ii) (iii) 11e ⁄ 24. (iii) 1. (ii) 1. a = − 5 ⁄ 2, b = − 3 ⁄ 2. a = 1, b = 2, c = 1. a = 120, b = 60, c = 180. 6.4 The Form ∞ ∞ lim lim Suppose x → a φ (x) = ∞ and x → a ψ (x) = ∞ . lim φ (x) lim φ ′ (x) Then x → a = x →a ⋅ ψ (x) ψ ′ (x) lim 1 ⁄ ψ (x) lim φ (x) We have x → a ψ (x) = x → a 1 ⁄ φ (x) − ψ ′ (x) [ψ (x)]2 lim = x →a − φ ′ (x) [φ (x)]2 lim ⎡ ψ′ (x) ⎧ φ (x) ⎫ 2⎤ .⎨ = x →a ⎢ ⎬ ⎥ . ⎣ φ ′ (x) ⎩ ψ (x) ⎭ ⎦ φ (x) ⎫ 2 lim φ (x) lim ψ ′ (x) ⎧ Thus, x → a ψ (x) = x → a φ ′ (x) . ⎪⎨⎪ lim ψ (x) ⎪⎬⎪ . x →a ⎩ ⎭ [Form 0/0] [by L’H ospital’s rule] …(1) lim φ (x) Now suppose x → a = λ. …(2) ψ (x) Then three cases arise. Case I : λ is neither zero nor infinite. In this case dividing both sides of (1) by λ2, we get lim ψ ′ (x) lim φ ′ (x) or λ = x → a ⋅ λ− 1 = x → a φ ′ (x) ψ ′ (x) Case II : λ = 0. In this case adding 1 to each side of equation (2), we get ⎫ lim ⎧ φ (x) lim φ (x) + 1 = x →a ⎨ + 1⎬ λ + 1 = x →a ψ (x) ⎩ ψ (x) ⎭ lim φ (x) + ψ (x) lim φ ′ (x) + ψ ′ (x) = x →a = x →a ψ (x) ψ ′ (x) ∞ ⎧ ⎫ ⎨ by case I, since form is ∞ and λ + 1≠ 0⎬ ⎩ ⎭ D-143 INDETERMINATE FORMS lim φ ′ (x) = x →a + 1. ψ ′ (x) lim φ ′ (x) ⋅ Therefore λ = x →a ψ ′ (x) Case III : λ = ∞ . In this case, we have 1 lim ψ (x) lim ψ ′ (x) = = ⋅ [by case II] lim φ (x) x → a φ (x) x → a φ ′ (x) x → a ψ (x) lim φ ′ (x) lim φ (x) = ⋅ Therefore x → a ψ (x) x → a ψ ′ (x) lim lim H ence in every case in which x → a φ (x) = ∞ and x → a ψ (x) = ∞ , we get lim φ ′ (x) lim φ (x) x → a ψ (x) = x → a ψ ′ (x) ⋅ Note 1 : By writing x = 1 ⁄ y, we can show as in § 2 that the proposition of this article is also true when x → ∞ or − ∞ in place of a. Note 2 : O bviously the proposition of this article is true when one or both the limits are − ∞ . Important : We have seen that in both cases when the form is ∞ ⁄ ∞ or 0 ⁄ 0 the ru le o f evaluat ing t h e limit by differen tiating the numerat or and denominat or separately holds good. Also we can easily convert the form ∞ ⁄ ∞ to the form 0 ⁄ 0 and vice-versa. Therefore at every stage we should note carefully that which form will be more suitable to evaluate the limit most quickly. Moreover in some cases it will be necessary t o convert the form ∞ ⁄ ∞ t o t h e fo r m 0 ⁄ 0, ot herwise t he process of differentiating the numerator and the denominator would never terminate. lim log x ⋅ Example 1 : Evaluate x → 0 cot x lim log x Solution : We have, x → 0 cot x ⁄ x 1 lim = x →0 − cosec 2 x (Agra 2014; Purvanchal 14) lim − sin2 x = x →0 x lim − 2 sin x cos x − 2 × 0 × 1 = = 0. = x →0 1 1 lim log sin 2x Example 2 : Evaluate x → 0 ⋅ log sin x lim log sin 2x Solution : We have x → 0 log sin x (2 ⁄ sin 2x) . cos 2x lim lim 2 cot 2x = x →0 ⋅ = x →0 (1 ⁄ sin x) . cos x cot x [form ∞ ⁄ ∞ ] [form ∞ ⁄ ∞ ] [form 0/0] (Agra 2001) [form ∞ ⁄ ∞ ] [form ∞ ⁄ ∞ ] D-144 DIFFERENTIAL CALCULUS lim − 4 cosec2 2x = x →0 − cosec2 x [form ∞ ⁄ ∞ ] lim 4 sin2 x = x →0 sin2 2x 4 sin2 x (2 sin x cos x) 2 1 lim = 1. = x →0 cos2 x lim log log (1 − x2) Example 3 : Evaluate x → 0 ⋅ log log cos x [form 0/0] lim = x →0 (Kumaun 2000; Avadh 13) lim log log (1 − x2) Solution : We have, x → 0 log log cos x 1 1 ⋅ ⋅ (− 2 x) lim log (1 − x2) 1 − x2 = x →0 ⋅ 1 1 ⋅ ⋅ (− sin x) log cos x cos x x cos x log cos x lim = 2 x →0 sin x . (1 − x2) log (1 − x2) x cos x log cos x lim lim lim = 2 x →0 ⋅ ⋅ sin x x → 0 1 − x2 x → 0 log (1 − x2) log cos x lim = 2 × 1 × 1 × x →0 log (1 − x2) 1 ⋅ (− sin x) lim cos x = 2 x →0 1 ⋅ (− 2x) 1 − x2 = 2× 1 lim ⎛ sin x 1 − x2 ⎞ ⋅ ⋅ ⎜ ⎟ cos x ⎠ 2 x →0 ⎝ x = 1. 1. 2. Evaluate the following limits : x lim (i) x → ∞ x ⋅ e lim e x (ii) x → ∞ 3 ⋅ x lim log (1 − x) ⋅ (i) x → 1 cot π x lim log x , (ii) x → ∞ a > 1. ax [form ∞ ⁄ ∞ ] [form 0/0] D-145 INDETERMINATE FORMS log (x − a) ⋅ log (e x − ea) lim sec π x ⋅ (ii) x → 1 2 tan 3 π x 3. lim (i) x → a 4. lim (i) x → ∞ ⎧⎪ (log x) 3 ⎫⎪ ⎨⎪ ⎬⋅ x ⎪⎭ ⎩ x (log x) 3 ⋅ 1 + x + x2 1 lim (i) x → ∞ x tan ⋅ x lim (ii) x → ∞ 5. lim log tan 2x (ii) x → 0 ⋅ log tan 3x 6. 7. lim (i) x → π ⁄ 2 ⎛ ⎝ log ⎜ x − 1 2 tan x (Bundelkhand 2001) ⎞ ⎠ π⎟ ⋅ lim c {e1 ⁄ (x − a) − 1} ⋅ (ii) x → a {e1 ⁄ (x − a) + 1} 1 ⎫⎪ lim ⎧ 2 (i) x → 1 ⎪⎨ 2 − ⋅ x − 1 ⎬⎪⎭ ⎪⎩ x − 1 (Garhwal 2002) lim (ii) x → π ⁄ 2 (sec x − tan x). 1. 3. 5. (i) 0. (i) 1. (i) 1 7. (i) − (ii) ∞ . (ii) 3 (ii) 1. 1 2 ⋅ 2. 4. 6. (i) 0. (i) 0 (i) 0. (ii) 0. (ii) 0. (ii) c. (ii) 0. 6.5 The Form ∞ − ∞ This form can be easily reduced to the form ∞ 0 or ∞ ⋅ 0 lim lim Suppose x → a φ (x) = ∞ and x → a ψ (x) = ∞ . Then lim x → a {φ (x) − ψ (x)} 1 ⎫ lim ⎧ 1 = x →a ⎨ − ⎬ 1 ⁄ ψ (x) ⎭ ⎩ 1 ⁄ φ (x) 1 1 − 0 lim ψ (x) φ (x) = x →a , which is of the form ⋅ 1 0 φ (x) . ψ (x) [form ∞ − ∞ ] D-146 DIFFERENTIAL CALCULUS 1 ⎞ ⎟. 1 − sin x⎠ 1 ⎞ lim ⎛ Solution : We have, x → π ⁄ 2 ⎜ sec x − ⎟ 1 − sin x ⎠ ⎝ ⎛ lim Example 1 : Evaluate x → π ⁄ 2 ⎜ sec x − ⎝ 1 ⎞ lim ⎛ 1 − = x →π ⁄ 2 ⎜ ⎟ ⎝ cos x 1 − sin x⎠ [form ∞ − ∞ ] [form ∞ − ∞ ] 1 − sin x − cos x lim = x →π ⁄ 2 [form 0/0] cos x − cos x sin x − cos x + sin x − 0+ 1 lim = = ∞. = x →π ⁄ 2 2 2 − 1+ 1− 0 − sin x + sin x − cos x 1 ⎞ lim ⎛ 1 Example 2 : Evaluate x → 0 ⎜ 2 − ⎟ . (Kanpur 2011) ⎝x sin2 x⎠ 1 ⎞ lim ⎛ 1 Solution : We have, x → 0 ⎜ 2 − [form ∞ − ∞ ] ⎟ sin2 x ⎠ ⎝x lim sin2 x − x2 = x →0 x2 sin2 x lim = x →0 [form 0/0] ⎛ ⎞2 x3 ⎜x − + …⎟ − x2 ⎝ ⎠ 3! ⎛ x2 ⎜ x − ⎝ ⎞2 x3 + …⎟ ⎠ 3! 2 x4 − + terms containing higher powers of x 3! lim = x →0 x4 + terms containing higher powers of x 2 − + terms containing powers of x only in the numerator 3! lim = x →0 1 + terms containing powers of x only in the numerator − 2 1 = = − ⋅ 3! 3 6.6 The Form 0 − ∞ This form can be easily reduced to the form lim Suppose x → a φ (x) = 0 and Then lim x → a φ (x). ψ (x) lim = x →a φ (x) 1 ψ (x) ∞ 0 or to the form ∞ ⋅ 0 lim x → a ψ (x) = ∞ . [form 0 × ∞ ] [form 0/0] D-147 INDETERMINATE FORMS ψ (x) [form ∞ ⁄ ∞ ] 1 φ (x) We shall reduce the form 0 × ∞ to the form 0 ⁄ 0 or ∞ ⁄ ∞ according to our convenience. lim = x →a lim Example 1 : Evaluate x → 0 x log sin x. lim Solution : We have, x → 0 x log sin x lim log sin x = x →0 1⁄x lim (1 ⁄ sin x) . cos x = x →0 − 1 ⁄ x2 lim − x2 cos x = x →0 sin x [form 0 × ∞ ] [form ∞ ⁄ ∞ ] [form ∞ ⁄ ∞ ] [form 0/0] lim x2 sin x − 2x cos x = 0. = x →0 cos x 1. 2. x ⎞ lim ⎛ 1 − (i) x → 1 ⎜ ⎟ ⋅ log x⎠ ⎝ log x ⎛ cot x − 1 ⎞ x ⎟⎟ lim ⎜⎜ (ii) x → 0 ⎜ ⎟ ⋅ x ⎝ ⎠ lim ⎛ cosec x − cot x⎞ (i) x → 0 ⎜ ⎟ ⋅ ⎝ ⎠ x ⎛⎜ 1 1⎞ − ⎟⎟ ⋅ ⎜⎜ x x ⎟⎠ − 1 e ⎝ π lim ⎛ ⎞ (i) x → π ⁄ 2 ⎜ x tan x − sec x⎟ ⋅ 2 ⎝ ⎠ lim (ii) x → 0 3. lim (ii) x → 0 x log x. 4. lim (i) x → 0 sin x. log x. lim (ii) x → ∞ x (a1 ⁄ x − 1). 5. a lim (i) x → ∞ 2 x sin x ⋅ 2 lim (ii) x → π ⁄ 2 (1 − sin x) tan x. (Agra 2003) D-148 DIFFERENTIAL CALCULUS 6. lim m n x → 0 x (log x) , where m and n are positive integers. 1. (i) − 1. (ii) − 3. 5. (i) − 1. (i) a. (ii) 0. (ii) 0. 1 3 ⋅ 2. (i) 1 2 ⋅ 4. (i) 0. 6. 0. (ii) − 1 2 ⋅ (ii) log a. 6.7 The Forms 1∞ , 00, ∞ 0 lim Suppose x → a {φ (x)}ψ (x) takes any one of these three form s. Then let lim y = x → a {φ (x)}ψ (x) . lim Taking logarithm of both sides, we get log y = x → a ψ (x). log φ (x). Now in any of the above three cases, log y takes the form 0 × ∞ which can be evaluated by the process of aricle 6.6. 2 lim Example 1 : Evaluate x → 0 (cos x) cot x.; 2 lim Solution : Let y = x → 0 (cos x) cot x. ∴ lim log y = x → 0 cot2 x. log cos x (Kumaun 2001; Bundelkhand 14) [form 1∞ ] [form ∞ × 0] lim log cos x = x →0 [form 0/0] tan2 x lim (1 ⁄ cos x) . (− sin x) (by L’H ospital’s rule) = x →0 2 tan x sec 2 x − tan x 1 1 lim lim = x →0 = − ⋅ = x →0 2 − 2 sec 2 x 2 tan x sec 2 x ∴ y = e− 1 ⁄ 2. a⎞ x lim ⎛ Example 2 : Evaluate x → ∞ ⎜ 1 + ⎟ ⋅ x⎠ ⎝ a⎞ x lim ⎛ Solution : Let y = x → ∞ ⎜ 1 + ⎟ ⋅ x⎠ ⎝ ∴ a⎞ ⎫ ⎛ lim ⎧ log y = x → ∞ ⎨ x log ⎜ 1 + ⎟ ⎬ x⎠ ⎭ ⎩ ⎝ lim log (1 + a ⁄ x) = x →∞ 1⁄x [form 1∞ ] [form ∞ × 0] [form 0/0] D-149 INDETERMINATE FORMS 1 ⋅ (− a ⁄ x2) a lim 1 + a ⁄ x lim = x →∞ = x →∞ = a. 2 1 + a⁄x − 1⁄x y = e a. Therefore, 6.8 Compound Forms Suppose a function is the product of two or more factors the limit of each of which can be easily found. Then the limit of the entire function will be equal to the product of the limits of the factors provided that the product is not in itself an indeterminate form. A similar rule is applicable in the case of a sum, difference, quotient or power. 1 ⎤ lim ⎡ 1 Example 1 : Evaluate x → 0 ⎢ − 2 log (1 + x) ⎥ . ⎣x ⎦ x 1 ⎤ lim ⎡ 1 lim x − log (1 + x) Solution : We have, x → 0 ⎢ − 2 log (1 + x) ⎥ = x → 0 [form 0/0] x ⎣ ⎦ x x2 1 1− 1+ x , lim by L’Hospital’s rule [The form is again 0/0] = x →0 2x 1 1 lim (1 + x) 2 = x →0 = ⋅ 2 2 2 lim ⎛ tan x⎞ 1 ⁄ x . Example 2 : Evaluate x → 0 ⎜ ⎟ ⎝ x ⎠ (Meerut 2000; Garhwal 01; Kumaun 02; Agra 03; Kashi 14; Purvanchal 14) 2 lim ⎛ tan x ⎞ 1 ⁄ x Solution : Let y = x → 0 ⎜ ⋅ ⎟ ⎝ x ⎠ ∴ ⎡1 ⎛ ⎞⎤ x3 2 5 ⎛ tan x⎞ lim 1 lim 1 + x + …⎟ ⎥ log y = x → 0 2 log ⎜ ⎟ = x → 0 2 log ⎢ ⎜ x + ⎣x ⎝ ⎠⎦ 15 3 ⎝ x ⎠ x x ⎡ ⎛ x2 ⎞⎤ 2 4 lim 1 = x → 0 2 log ⎢ 1 + ⎜ + x + …⎟ ⎥ ⎣ ⎝ ⎠⎦ 15 3 x lim = x →0 ⎛ x2 ⎞ ⎞2 2 4 1 ⎛ x2 2 4 ⎜ + x + …⎟ − ⎜ + x + …⎟ + … ⎝3 ⎠ ⎠ 15 2⎝3 15 x2 x2 3 lim = x →0 ⎡1 ⎤ ⎢ + terms containing powers of x only in the numerator⎥ ⎣3 ⎦ = ∴ + terms containing higher powers of x lim = x →0 y= 1 ⋅ 3 e1 ⁄ 3. x2 D-150 1. DIFFERENTIAL CALCULUS lim (i) x → 0 x x . (Agra 2002; Kanpur 04) lim (ii) x → 0 (cos x) 1 ⁄ x. 2. 2 lim (i) x → 0 (cos x) 1 ⁄ x . (Garhwal 2003) lim (ii) x → π ⁄ 2 (sin x) tan x. 3. lim (i) x → π ⁄ 2 (sec x) cot x. lim (ii) x → π ⁄ 4 (tan x) tan 2x. 4. 5. x ⎞ tan (π x ⁄ 2a) lim ⎛ (i) x → a ⎜ 2 − ⎟ a⎠ ⎝ lim (ii) x → 1 (1 − x2) 1 ⁄ log (1 − x) . (Rohilkhand 2012) lim (i) x → 1 x1 ⁄ (1 − x) . lim (ii) x → ∞ (a0 x m + a1 x m − 1 + … + am ) 1 ⁄ x. 6. lim ⎛ tan x⎞ 1 ⁄ x (i) x → 0 ⎜ ⎟ . ⎝ x ⎠ (Garhwal 2001, 03) 3 lim ⎛ tan x⎞ 1 ⁄ x (ii) x → 0 ⎜ . ⎟ ⎝ x ⎠ 2 7. lim ⎛ sinh x⎞ 1 ⁄ x (i) x → 0 ⎜ . ⎟ ⎝ x ⎠ (Kumaun 2008) 2 lim ⎛ sin x⎞ 1 ⁄ x (ii) x → 0 ⎜ . ⎟ ⎝ x ⎠ 8. 9. 10. (Kumaun 2003) 2 ⎧⎪ 2 (cosh x − 1) ⎫⎪ 1 ⁄ x . ⎨⎪ ⎬⎪ x2 ⎩ ⎭ lim ⎧ log x⎫ 1 ⁄ x (ii) x → ∞ ⎨ ⎬ . ⎩ x ⎭ lim (i) x → 0 1⁄x π lim (i) x → ∞ ⎛⎜ − tan− 1 x⎞⎟ . ⎝2 ⎠ lim (ii) x → 0 (cosec x) 1 ⁄ log x. πx lim x → 1 (1 − x) tan 2 ⋅ D-151 INDETERMINATE FORMS 1. 3. 5. 7. 9. (i) (i) (i) (i) (i) 1. 1. 1 ⁄ e. e1 ⁄ 6. 1. (ii) (ii) (ii) (ii) (ii) 1. 1⁄e . 1. e − 1 ⁄ 6. 1 ⁄ e. 2. (i) e− 1 ⁄ 2. 4. (i) e2 ⁄ π . 6. (i) 1. 8. (i) e1 ⁄ 12. 10. 2 ⁄ π. (ii) (ii) (ii) (ii) 1. e. ∞. 1. Fill in the Blanks: 1. Fill in the blanks “……”, so that the following statem ents are com plete and correct. sin a x lim = …… . x → 0 sin b x log (1 + k x 2) = …… . 1 − cos x x →0 x2 + 2 x = …… . 3. lim 2 x → ∞ 5 − 3x ⎛ π⎞ 4. lim ⎜⎝ sec ⎟⎠ ⋅ log x = …… . 2 x x →1 2. lim 5. The value of lim 6. The value of lim x − tan x = …… . x3 x →0 (Agra 2002) ax− bx = …… . x x →0 (Agra 2003) Multiple Choice Questions: 7. 8. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1 − cos x The value of lim is 3x2 x →0 2 (a) 0 (b) 3 1 (Garhwal 2002) (c) (d) 1 6 Which of the following is not an indeterminate form ? ∞ (a) ∞ (b) 0 × ∞ (c) 10 9. (d) 00 log tan x is log x x →0 The value of the lim (a) 0 (c) –1 (b) 1 (d) None of these D-152 10. 11. DIFFERENTIAL CALCULUS Which of the following is an indeterminate form ? (a) ∞ + ∞ (b) ∞ × ∞ ∞ (c) 1 (d) 0∞ x a − 1 − x log e a The value of lim is x →0 x2 (log e a) (a) (log e a) 2 (b) 2 (c) a − log e a 12. The value of lim x →1 (d) (Garhwal 2001) 2 log x is x− 1 (a) − 1 13. (log e a) 2 (b) ∞ log e x , a > 1 is The value of lim x →∞ ax 1 (a) (b) a log e a (c) 1 (d) 0 (c) 1 (d) 0 (Garhwal 2003) True or False: Write ‘T’ for true and ‘F’ for false statement. 14. 15. 16. 17. 1. 4. 7. 10. 13. 16. While applying L’ H ospital’s rule to evaluate lim x →a f (x) , 0 if the form is , we φ (x) 0 are to differentiate f (x) ⁄ φ (x) as a fraction. ∞ 0 The indeterminate form ∞ can be easily converted to the form and vice-versa. 0 f (x) f ′ (x) lim lim lim lim If f (x) = ∞ and φ (x) = ∞ , then = ⋅ x →a x →a x → a φ (x) x → a φ ′ (x) ex (1 + x) 1 ⁄ x − e + ∞ 2 lim is of the form ⋅ (Meerut 2001) 2 x →a 0 x a ⋅ b 2 π⋅ (c). (c). (d) T. 2. 2 k . 1 ⋅ 3 (c). (d). F. F. 5. − 8. 11. 14. 17. 3. − 6. 1 ⋅ 3 log 9. (b). 12. (c). 15. T. a ⋅ b 7.1 Functions of Several Independent Variables So far we have considered functions of one independent variable only. H owever, in practice, we often come across functions of more than one independent variable. For example, the area of a rectangle depends upon two independent variables, namely the length and the breadth. Similarly, the volume of a rectangular parallelopiped depends upon three independent variables, namely the length, the breadth and the height. There are a number of differences between the calculus of one and of two variables. Fortunately the calculus of functions of three or more variables differs only slightly from that of functions of two variables. The study here will be limited largely to functions of two variables. Definition : L et z be a sym bol which has one definite value for every pair of values of x and y. Then z is called a function of the two independent variables x and y, and is usually written as z = f (x, y). A function of x and y is also written as φ (x, y) or ψ(x, y) etc. (Kanpur 2014) A similar definition can be given for functions of more than two independent variables. If to each point (x, y), of a part of the xy-plane is assigned a unique real number z, even then z is said to be given as a function, z = f (x, y), of the independent variables x and y. The locus of all points (x, y, z) satisfying z = f (x, y) is a surface in ordinary space. D-154 DIFFERENTIAL CALCULUS 7.2 Continuity of a Function of Two Variables A function f (x, y) is said to have a limit A as x → a and y → b if for any arbitrarily chosen positive number ε, however small (but not zero), there exists a corresponding number δ > 0 such that ⏐ f (x, y) − A⏐ < ε, for all values of x and y satisfying 0 < √{(x − a) 2 + ( y − b) 2} < δ. H ere 0 < √{(x − a) 2 + (y − b) 2} < δ defines a deleted neighbourhood of (a, b), namely all points except (a, b) lying within a circle of radius δ and centre (a, b). A function f (x, y) is said to be continuous at (a, b) provided f (a, b) is defined and lim x → a, y → b f (x, y) = f (a, b). 7.3 Partial Differential Coefficients Suppose z = f (x, y) is a function of two independent variables x and y. Since x and y are independent, we may (i) allow x to vary while y is kept fixed, (ii) allow y to vary while x is kept fixed, (iii) allow x and y to vary simultaneously. In the first two cases, z is practically a function of a single variable and we can differentiate it in accordance with the usual rules. The partial differential coefficient of z = f (x, y) with respect to x is the ordinary differential coefficient of f (x, y) with respect to x when y is regarded as a constant. It is usually written as ∂f ∂z or or fx . ∂x ∂x ∂f lim f (x + δx, y) − f (x, y) , = δx → 0 provided the limit exists. Thus, δx ∂x Similarly, the partial differential coefficient of z = f (x, y) with respect to y is the ordinary differential coefficient of f (x, y) with respect to y when x is kept as constant. It is written as ∂z ∂f or or fy . ∂y ∂y In a similar manner, if z = f (x1, x2 , … , xn ) be a function of n independent variables x1, x2 , …, xn , then the partial differential coefficient of z with respect to x1, is the ordinary differential coefficient of z with respect to x1, when all the variables except x1 are regarded as constants. We shall write it as ∂f ∂z or ⋅ ∂x1 ∂x1 7.4 Partial Differential Coefficients of Higher Orders ∂z of z = f (x, y) may again be differentiated ∂x partially with respect to x and to y, thus giving the second partial differential coefficients The partial differential coefficient D-155 PARTIAL DIFFERENTIATION ∂ ⎛ ∂z ⎞ ∂2 z = fx x = and ⎜ ⎟ 2 ∂x ⎝ ∂x⎠ ∂x ∂z may be obtained Similarly, from ∂y ∂2 z ∂ ⎛ ∂z ⎞ = fx y = ⎜ ⎟ ∂x ⎝ ∂y⎠ ∂x ∂y ∂2 z ∂ ⎛ ∂z ⎞ = fy x = ⎜ ⎟ ⋅ ∂y ⎝ ∂x⎠ ∂y ∂x ∂2 z ∂ ⎛ ∂z ⎞ = fy y = ⎜ ⎟ ⋅ ∂y ⎝ ∂y⎠ ∂y2 If z = f (x, y) and its partial derivatives are continuous (as is true in all ordinary cases), the order of differentiation is immaterial, that is, and ∂2 z ∂2 z = ⋅ ∂x ∂y ∂y ∂x ∂2 u ∂2 u and ⋅ ∂y ∂x ∂x ∂y u = ax2 + 2hxy + by2. Example 1 : If u = ax2 + 2hxy + by2, find Solution : We have, ∂u = 2ax + 2hy. ∴ ∂x H ence ∂2 u ∂ ⎛ ∂u ⎞ ∂ = (2ax + 2hy) ⎜ ⎟ = ∂y ∂y ∂x ∂y ⎝ ∂x ⎠ = 2h. ∂u = 2hx + 2by . ∂y Again ∴ (treating y as constant) (treating x as constant) (treating x as constant) ∂ ⎛ ∂u ⎞ ∂ ∂2 u = (2hx + 2by) ⎜ ⎟ = ∂x ∂y ∂x ⎝ ∂y ⎠ ∂x = 2h. ∂2 (treating y as constant) ∂2 u u = ⋅ ∂x ∂y ∂y ∂x ∂u ∂u ⎛ y⎞ + y = 0. Example 2 : If u = f ⎜ ⎟ , show that x ∂x ∂y ⎝ x⎠ ⎛ y⎞ Solution : We have, u = f ⎜ ⎟ ⋅ ⎝ x⎠ y⎞ ∂u ⎧ ⎛ y⎞ ⎫ ⎛ ∴ = ⎨f ′ ⎜ ⎟ ⎬ ⎜ − ⎟ ⋅ ∂x ⎩ ⎝ x⎠ ⎭ ⎝ x2 ⎠ H ere we note that ∴ Again ∴ ∂u = − ∂x ∂u ⎧ = ⎨f′ ∂y ⎩ x y y ⎛ y⎞ f′ ⎜ ⎟ ⋅ x ⎝ x⎠ ⎛ y⎞ ⎫ ⎛ 1 ⎞ ⎜ ⎟⎬ ⎜ ⎟ ⋅ ⎝ x⎠ ⎭ ⎝ x ⎠ (treating y as constant) …(1) (treating x as constant) ∂u y ⎛ y⎞ = f′ ⎜ ⎟ ⋅ ∂y x ⎝ x⎠ Adding (1) and (2), we have x (Agra 2003) …(2) ∂u ∂u + y = 0. ∂y ∂x D-156 DIFFERENTIAL CALCULUS Example 3 : If u = log (x3 + y3 + z3 − 3 xyz) , show that ∂ ∂ ⎞2 9 ⎛ ∂ + + ⋅ ⎜ ⎟ u= − ∂y ∂z ⎠ ⎝ ∂x (x + y + z) 2 (Kanpur 2007; Purvanchal 07; Rohilkhand 12; Avadh 13; Kashi 14) z3 − 3xyz). Solution : We have, u = log (x3 + y3 + Differentiating partially with respect to ∂u 1 = (3x2 − ∂x x3 + y3 + z3 − 3xyz or x, we have 3yz) 3 (x2 − yz) ∂u = 3 ⋅ ∂x x + y3 + z3 − 3xyz Similarly, by symmetry, we have ∂u 3 (y2 − zx) , = 3 ∂y x + y3 + z3 − 3xyz and 3 (z2 − xy) ∂u = 3 ⋅ ∂z x + y3 + z3 − 3xyz Adding (1), (2) and (3), we have ∂u ∂u ∂u 3 (x2 + y2 + z2 − yz − zx − xy) + + = ∂x ∂y ∂z x3 + y3 + z3 − 3xyz …(1) …(2) …(3) 3 (x2 + y2 + z2 − yz − zx − xy) 3 = ⋅ x+ y+ z (x + y + z) (x2 + y2 + z2 − yz − zx − xy) ∂ ∂ ⎞2 ∂ ∂ ⎞ ⎛ ∂u ∂u ∂u ⎞ ⎛ ∂ ⎛ ∂ + + + + + + ⎟ ⎜ ⎟ u= ⎜ ⎟ ⎜ ∂y ∂z ⎠ ∂y ∂z ⎠ ⎝ ∂x ∂y ∂z ⎠ ⎝ ∂x ⎝ ∂x ∂ ∂⎞ ⎛ ∂ 3 ⎞ = ⎛⎜ + + ⎟ ⎟ ⎜ ∂y ∂z ⎠ ⎝ x + y + z ⎠ ⎝ ∂x ∂ ⎛ ∂ ⎛ 1 1 1 ⎡ ∂ ⎛ ⎞ ⎞ ⎞⎤ = 3⎢ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ∂y ⎝ x + y + z ⎠ ∂z ⎝ x + y + z ⎠ ⎦ ⎣ ∂x ⎝ x + y + z ⎠ 1 1 1 9 ⎡ ⎤ = 3 ⎢− − − ⋅ ⎥ = − ⎣ (x + y + z) 2 (x + y + z) 2 (x + y + z) 2 ⎦ (x + y + z) 2 = Now Example 4 : If x = r cos θ, y = r sin θ, show that ∂r ∂x , ∂x ∂θ , = = r ∂x ∂r r∂θ ∂x ∂2 θ ∂2 θ + = 0. ∂x2 ∂y2 and ∂θ ⋅ ∂x Solution : We have x = r cos θ. ∂x = cos θ. ∴ ∂r A lso find the value of Also we have, r2 = x2 + y2. ∂r ∴ 2 r = 2x ∂x (Garhwal 2002) (regarding θ as constant) (regarding y as constant) D-157 PARTIAL DIFFERENTIATION or Thus Again, ∴ ∂r x r cos θ = = = cos θ. ∂x r r ∂r ∂x = ⋅ ∂x ∂r ∂x ∂θ = − r sin θ. 1 ∂x = − sin θ. r ∂θ (regarding r as constant) Also we have, θ = tan− 1 ( y ⁄ x). y⎞ y ∂θ r sin θ sin θ 1 ⎛ ∴ = = − = − ⋅ ⎜− ⎟ = − 2 2 2 ∂x r y2 ⎝ x2 ⎠ x + y r 1+ x2 ∂θ ∴ r = − sin θ. ∂x ∂θ ∂x = r ⋅ H ence ∂x r∂θ Finally, we have θ = tan− 1 ( y ⁄ x). ∴ ∂θ y = − 2 ⋅ ∂x (x + y2) ∴ ∂2 θ 2xy = 2 ⋅ (x + y2) 2 ∂x2 …(1) Also ∂θ = ∂y ⎛⎜ ∴ ∂2 θ − 2 xy = 2 ⋅ ∂y2 (x + y2) 2 …(2) 1 y2 ⎞ ⎟ ⎜ 1 + 2⎟ ⎜ x ⎟⎠ ⎝ ⋅ 1 x = ⋅ x x2 + y2 ∂2 θ ∂2 θ + = 0. ∂x2 ∂y2 Example 5 : If u = (1 − 2 xy + y2) − 1 ⁄ 2, prove that ∂u ⎫ ∂ ⎧ 2 ∂u ⎫ ∂ ⎧ ⎨ (1 − x2) ⎬+ ⎨y ⎬ = 0. ∂x ⎭ ∂y ⎩ ∂y ⎭ ∂x ⎩ Adding (1) and (2), we get Solution : H ere u = (1 − 2xy + y2) − 1 ⁄ 2. ∂u 1 ∴ = − (1 − 2xy + y2) − 3 ⁄ 2 (− 2y) = yu3, 2 ∂x ∂u 1 = − (1 − 2xy + y2) − 3 ⁄ 2 . (− 2x + 2y) = (x − y) u3. and 2 ∂y ∂u ⎫ ∂ ⎧ ∂ ⎧ Now ⎨ (1 − x2) ⎬= ⎨ (1 − x2) . yu 3⎫⎬⎭ ∂x ⎭ ∂x ⎩ ∂x ⎩ ∂u = y (− 2x) u3 + y (1 − x2) . 3u2 ∂x = − 2xy u3 + 3y (1 − x2) u2 yu3 Also = − 2 x y u3 + 3y2 u5 (1 − x2). ∂ ⎧ 2 ∂u ⎫ ∂ ⎧ 2 ∂ {(y2x − y3) u3} ⎨y ⎬= ⎨⎩ y (x − y) u 3⎫⎬⎭ = ∂y ⎩ ∂y ⎭ ∂y ∂y ∂u = (2xy − 3y2) u3 + ( y2 x − y3) . 3u2 ∂y = (2 x y − 3y2) u3 + y2 (x − y) . 3u2 . (x − y) u3 …(1) D-158 DIFFERENTIAL CALCULUS = (2 x y − 3y2) u3 + y2 (x − y) 2 . 3u5 = 2 x y u3 + 3y2 u5 [(x − y) 2 − u− 2] [... u− 2 = 1 − 2xy + y2] = 2 x y u3 + 3y2 u5 [(x − y) 2 − (1 − 2xy + y2)], …(2) = 2 x y u3 + 3y2 u5 [x2 − 1] = 2 x y u3 − 3y2 u5 (1 − x2). Adding (1) and (2), we have ∂ ⎧ ∂u ⎫ ∂ ⎧ 2 ∂u ⎫ ⎨ (1 − x2) ⎬+ ⎨y ⎬ = 0. ∂x ⎩ ∂x ⎭ ∂y ⎩ ∂y ⎭ 2 ∂θ 1 ∂ ⎛ 2 ∂θ ⎞ Example 6 : If θ = t n e− r ⁄ 4t, what value of n will m ake 2 ? ⎜r ⎟ = ∂r ∂r ∂t ⎝ ⎠ r (Garhwal 2003; Lucknow 11) 2 ∂θ r ⎛ 2 r⎞ Solution : We have = t n . e− r ⁄ 4t . ⎜ − ⎟ = − ⎝ ∂r 2 4t ⎠ 2 ∂θ 1 ∴ r2 = − r3 t n − 1 e− r ⁄ 4t. 2 ∂r ∂ ⎛ 2 ∂θ⎞ 3 r 2 n − 1 − r 2 ⁄ 4t 1 3 n − t e − r t ∴ ⎜r ⎟ = − 2 2 ∂r ⎝ ∂r ⎠ = − 2 t n − 1 e− r ⁄ 4t. 1 e− r 2 ⁄ 4t . ⎛⎜ − ⎝ 2 r⎞ ⎟ 4t ⎠ 2 3 2 n − 1 − r 2 ⁄ 4t 1 r t e + r 4 t n −. 2 e− r ⁄ 4t . 2 4 2 2 1 ∂ ⎛ 2 ∂θ⎞ 3 1 ⎜r ⎟ = − 2 t n − 1 e− r ⁄ 4t + 4 r 2 t n − 2 e− r ⁄ 4t. r 2 ∂r ⎝ ∂r ⎠ 2 2 ∂θ r2 = nt n − 1 e− r ⁄ 4t + t n e− r ⁄ 4t . 2 ∂t 4t ∴ Also 2 2 1 = nt n− 1 e− r ⁄ 4t + r 2 t n − 2 e− r ⁄ 4t . 4 ∂θ 1 ∂ ⎛ 2 ∂θ⎞ ⎜r ⎟ = ∂t r 2 ∂r ⎝ ∂r ⎠ Now 2 2 2 3 n − 1 − r 2 ⁄ 4t 1 1 t e + r 2 t n − 2 e− r ⁄ 4t = nt n − 1 e− r ⁄ 4t + r 2 t n − 2 e− r ⁄ 4t 2 4 4 2 2 3 − t − n − 1 e− r ⁄ 4t = nt n − 1 e− r ⁄ 4t, for all possible values of r and t 2 3 n= − ⋅ 2 ⇒ − ⇒ ⇒ y2 ∂u ∂u x2 and when u = 2 + 2 − 1. ∂y ∂x a b 1. Find 2. Prove that ∂2 u ∂2 u = in each of the following cases : ∂x ∂y ∂y ∂x (i) u = x4 + x2 y2 + y4. (ii) u = log tan (y ⁄ x). ⎧ x2 + y2 ⎫⎪ (iii) u = log ⎪⎨⎪ ⎬⋅ ⎩ x + y ⎪⎭ (iv) u = x y. D-159 PARTIAL DIFFERENTIATION 3. 4. ∂u ∂u If u = sin − 1 (x ⁄ y) + tan− 1 ( y ⁄ x), show that x + y = 0. ∂x ∂y ⎛⎜ y ⎞⎟ ∂u ∂u + y ⋅ If u = x y f ⎜⎜ ⎟⎟ , then write the value of the expression x ⎝ x⎠ ∂y ∂x (Meerut 2001; Kanpur 07) 5. If z = f (x + ay) + φ (x − ay), prove that ∂2 z ∂2 z = a2 2 ⋅ 2 ∂y ∂x (Bundelkhand 2001; Kanpur 05) 6. If u = f (r), where r = √(x 2 + y 2), show that ∂2 u ∂x 2 + ∂2 u 1 = f ′′(r) + f ′(r). r ∂y 2 (Meerut 2001; Avadh 04) ∂2 x2 z − y2 = 2 ⋅ ∂x∂y x + y2 7. If z = x2 tan− 1 ( y ⁄ x) − y2 tan− 1 (x ⁄ y), prove that 8. ∂2 u ∂2 u = 0. (i) If u = 2 (ax + by) 2 − (x2 + y2) and a2 + b2 = 1, prove that 2 + ∂x ∂y2 (ii) If u = e x(x cos y − y sin y), then show that 9. ∂2u ∂2u + = 0. ∂x2 ∂y2 (Garhwal 2003) ∂z ⎞ 2 ∂z ∂z ⎞ ⎛ ∂z ⎛ If z (x + y) = x2 + y2, show that ⎜ − − ⎟ = 4 ⎜1 − ⎟ . ⎝ ∂x ⎝ ∂y⎠ ∂x ∂y⎠ (Bundelkhand 2011; Kashi 13; Avadh 14) 10. ∂3 u If u = e x y z, show that = (1 + 3xyz + x2 y2 z2) e x y z. ∂x ∂y ∂z 11. ∂2 z If x x y y z z = c, show that at x = y = z, = − (x log e x) − 1. ∂x ∂y (Kanpur 2011; Kashi 12; Rohilkhand 13) (Rohilkhand 2013; Bundelkhand 14; Purvanchal 14) 12. Show that u u + = 0 when ∂x2 ∂y2 u = e m y cos m x. (ii) log (x2 (iv) u= u= + tan− 1 e x (x cos y − (x2 y2 (Agra 2014) y2). ( y ⁄ x). y sin y ) (Garhwal 2003) z2) − 1 ⁄ 2, If V = + + show that ∂V ∂V ∂V + y + z = − V. (i) x ∂y ∂z ∂x (ii) 14. ∂2 (i) (iii) 13. ∂2 ∂2 V ∂2 V ∂2 V + + = 0. ∂x2 ∂y2 ∂z2 If u = tan− 1 2 xy 1 , show that ∂ u = ⋅ 2 2 ∂x ∂y (1 + x2 + y2) 3 ⁄ 2 √(1 + x + y ) (Kumaun 2008) D-160 15. DIFFERENTIAL CALCULUS x2 y2 z2 If 2 + 2 + 2 = 1, prove that a + u b + u c + u ⎛ ∂u ⎞ 2 ⎛ ∂u ⎞ 2 (Rohilkhand 2011B, 12B) ⎛ ∂u ⎞ 2 ∂u ∂u ⎞ ⎛ ∂u + y + z ⎟. ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 2 ⎜x ∂y ∂z ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂z ⎠ ⎝ ∂x 16. If x = r cos θ, y = r sin θ, prove that (i) (∂r ⁄ ∂ x) 2 + (∂ r ⁄ ∂ y) 2 = 1. (ii) 17. 1. ∂2 r ∂2 r 1 ⎧ ⎛ ∂ r⎞ 2 ⎛ ∂r⎞ 2⎫ + = ⎨⎜ ⎟ + ⎜ ⎟ ⎬ ⋅ r ⎩ ⎝ ∂x ⎠ ⎝ ∂y⎠ ⎭ ∂x2 ∂y2 (Lucknow 2007, 11) ∂u ∂u ∂u (i) If u = x2 y + y2 z + z2 x, show that + + = (x + y + z) 2. ∂x ∂y ∂z ⎪ 1 1 1⎪ ⎪ ⎪ ∂u ∂u ∂y + + = 0. (ii) If u = ⎪ x y z ⎪ , show that ∂y ∂z ∂x ⎪ 2 2 2⎪ (Rohilkhand 2013B) ⎪x y z ⎪ 2x , 2y ⋅ a 2 b2 4. 2 u. 7.5 Homogeneous Functions An expression in x and y in which every term is of the same degree is called a homogeneous function of x and y. Consider the function defined by f (x, y) = a0 x n + a1 x n − 1 y + a2 x n − 2 y2 + …… + an − 1 xy n − 1 + an y n . …(1) In this function every term is of degree n. Therefore it is a homogeneous function of x and y of degree n. Moreover, (1) may be written as ⎧ ⎛ y⎞ ⎛ y⎞ 2 ⎛ y⎞ n ⎫ f (x, y) = x n ⎨ a0 + a1 ⎜ ⎟ + a2 ⎜ ⎟ + … + an ⎜ ⎟ ⎬ ⎩ ⎝ x⎠ ⎝ x⎠ ⎝ x⎠ ⎭ n or f (x, y) = x F (y ⁄ x), where F ( y ⁄ x) is some function of y ⁄ x. Thus a homogeneous function of x and y of degree n may be put in the form x n F ( y ⁄ x). Therefore we give the general definition of a homogeneous function as follows : x n F ( y ⁄ x) is called a hom ogeneous function of x and y of degree n, whatever the function F may be. Similarly, y n F (x ⁄ y) is also a homogeneous function of x and y of degree n. Thus x3 sin ( y ⁄ x) is a homogeneous function of x and y of degree 3. Similarly, 2 y cos (x ⁄ y) is a homogeneous function of x and y of degree 2. In general, if the function f (x1, x2 , …, xp) of the p variables x1, x2 ,… , xp can be put in the form xp ⎞ ⎛ x1 x2 xr n F ⎜⎜ , , …, ⎟⎟ , ⎜ xr xr xr ⎟⎠ ⎝ then f (x1, x2 , … , xp) is called a homogeneous function of x1, x2 , … , xp of degree n. D-161 PARTIAL DIFFERENTIATION 7.6 Euler’s Theorem on Homogeneous Functions If u be a hom ogeneous function of x and y of degree n, then x ∂u ∂u + y = nu. ∂x ∂y (Meerut 2000; Gorakhpur 06; Kashi 11, 13, 14) Proof : Since u is a homogeneous function of x and y of degree n, therefore u may be put in the form u = x n F ( y ⁄ x). Differentiating (1) partially with respect to x, we get y⎞ ∂u ⎛ y⎞ ⎧ ⎛ y⎞ ⎫ ⎛ = nx n − 1 F ⎜ ⎟ + x n ⎨ F ′ ⎜ ⎟ ⎬ ⎜ − 2 ⎟ ⋅ ∂x ⎝ x⎠ ⎩ ⎝ x⎠ ⎭ ⎝ x ⎠ …(1) ∂u ⎛ y⎞ ⎛ y⎞ = nx n F ⎜ ⎟ − y x n − 1 F ′ ⎜ ⎟ ⋅ …(2) ∂x ⎝ x⎠ ⎝ x⎠ Again differentiating (1) partially with respect to y, we get ∂u ⎧ ⎛ y⎞ ⎫ ⎛ 1 ⎞ = x n ⎨F ′ ⎜ ⎟ ⎬ ⎜ ⎟ ⋅ ∂y ⎩ ⎝ x⎠ ⎭ ⎝ x ⎠ ⎛ y⎞ ∂u ...(3) = yx n − 1 F ′ ⎜ ⎟ ⋅ ∴ y ∂y ⎝ x⎠ Adding (2) and (3), we get ∂u ∂u ⎛ y⎞ x + y = nx n F ⎜ ⎟ = nu. ∂x ∂y ⎝ x⎠ ∂u ∂u + y = nu. H ence x ∂y ∂x This proves the theorem. In general, if u be a homogeneous function of x1, x2 , … , xm of degree n, then ∂u ∂u ∂u x1 + x2 + … + xm = nu. ∂x1 ∂xm ∂x2 ∴ x The proof is similar to that of two variables. Example 1 : Verify Euler’s theorem for the function u = (x1 ⁄ 4 + (x1 ⁄ 5 + y1 ⁄ 4) ⋅ y1 ⁄ 5) (Rohilkhand 2014) Solution : H ere we see that u is a homogeneous function of x and y of degree 1 ⁄ 4 − 1 ⁄ 5 i.e., 1 ⁄ 20. Therefore in order to verify Euler’s theorem we are to show that ∂u ∂u 1 x + y = u. ∂x ∂y 20 We have, log u = log (x1 ⁄ 4 + y1 ⁄ 4) − log (x1 ⁄ 5 + y1 ⁄ 5). Differentiating (1) partially with respect to x, we have ⎛ ⎞ ⎛ 1 1 1 ∂u 1 1 − 4 ⁄ 5⎞ ⎜ ⎟ . = ⋅ ⎜ x− 3 ⁄ 4⎟⎠ − 1 ⁄ 5 x ⎝ 5 1 ⁄ 5 u ∂x x1 ⁄ 4 + y1 ⁄ 4 ⎝ 4 ⎠ x + y ∴ ⎡1 ∂u x− 3 ⁄ 4 1 x− 4 ⁄ 5 ⎤⎥ = u ⎢⎢ 1 ⁄ 4 − ⎥. ⎢4 x ∂x + y1 ⁄ 4 5 x1 ⁄ 5 + y1 ⁄ 5 ⎥⎦ ⎣ …(1) D-162 ∴ DIFFERENTIAL CALCULUS x ⎡1 ⎤ ∂u x1 ⁄ 4 1 x1 ⁄ 5 ⎥. = u ⎢⎢ 1 ⁄ 4 − ⎥ 1 ⁄ 4 1 ⁄ 5 1 ⁄ 5 ⎢4 x ⎥ ∂x 5 + y x + y ⎣ ⎦ …(2) Again differentiating (1) partially with respect to y, we get 1 y− 4 ⁄ 5 ⎤⎥ 1 ∂u ⎡⎢ 1 y− 3 ⁄ 4 = ⎢ 1⁄4 − ⎥. 1 ⁄ 4 5 x1 ⁄ 5 + y1 ⁄ 5 ⎥⎦ u ∂y ⎢⎣ 4 x + y ∴ y ⎡1 ⎤ ∂u y1 ⁄ 4 1 y1 ⁄ 5 ⎥. = u ⎢⎢ 1 ⁄ 4 − ⎥ 1 ⁄ 4 1 ⁄ 5 1 ⁄ 5 ⎢4 x ⎥ ∂y 5 + y x + y ⎣ ⎦ …(3) Adding (2) and (3), we get x ⎡1 ∂u ∂u + y = u ⎢⎢ ⎢4 ∂y ∂x ⎣ x1 ⁄ 4 + x1 ⁄ 4 + y1 ⁄ 4 1 x1 ⁄ 5 + − y1 ⁄ 4 5 x1 ⁄ 5 + ⎡1 y1 ⁄ 5 ⎤⎥ 1 1⎤ u. ⎥ = u ⎢ 4 − 5⎥ = 1 ⁄ 5 ⎥ 20 ⎣ ⎦ y ⎦ This verifies Euler’s theorem. ⎧ x2 + y2 ⎫⎪ ∂u ∂u Example 2 : If u = sin− 1 ⎪⎨⎪ + y = tan u. ⎬⎪ , show that x x + y ∂x ∂y ⎩ ⎭ (Garhwal 2002; Kumaun 08; Avadh 12) x2 + y2 Solution : We have, sin u = . x+ y x2 + y2 Let v = ⋅ Then v is a homogeneous function of x and y of degree 1. x+ y ∂v ∂v + y = v. …(1) Therefore by E uler’s theorem, we have x ∂y ∂x Now v = sin u. ∂u ∂v ∂u ∂v = cos u and = cos u ⋅ ∴ ∂x ∂y ∂y ∂x Putting these values in (1), we get ∂u ∂u x cos u + y cos u = v ∂x ∂y This proves the result. 1. 2. ⎡. . x2 + y2 ⎤ ⎢ . v = sin u = ⎥ ⎣ x+ y ⎦ ∂u ⎞ v sin u ⎛ ∂u + y ⎟= = ⎜x ⎝ ∂x ∂y ⎠ cos u cos u or = tan u . State E uler’s theorem on homogeneous functions. Verify Euler’s theorem in the following cases : (Bundelkhand 2001) (iii) u = a x y + b y z + c z x. x (x3 − y3) ⋅ x3 + y3 (iv) u = x n sin (y ⁄ x). (v) u = x n log (y ⁄ x). (vi) u = 1 ⁄ √(x2 + y2). (i) u = ax2 + 2hxy + by2. (ii) u = D-163 PARTIAL DIFFERENTIATION 3. ⎛ x3 + y3 ⎞ ∂u ∂u ⎟ , prove that x (i) If u = tan− 1 ⎜ + y = sin 2u. ⎝ x− y ⎠ ∂x ∂y (Rohilkhand 2012B) (ii) If u = 4. sin− 1 √x − √y , ∂u ∂u show that x + y = 0. √x + √y ∂x ∂y (Gorakhpur 2005; Garhwal 02) x+ y , ∂u ∂u 1 (i) If u = sin − 1 show that x + y = tan u. √x + √y ∂x ∂y 2 ⎛ x+ y ⎞ ∂u ∂u 1 (ii) If u = cos− 1 ⎜ ⎟ , show that x + y + cot u = 0. ∂x ⎯ √ ⎯x + ⎯ √ y ∂y 2 ⎠ ⎝ (Rohilkhand 2013B) 5. x3 + y3 ∂u ∂u (i) If u = log , show that x + y = 2. ∂x ∂y x+ y (ii) If u = log x4 + y4 ∂u ∂u , show that x + y = 3. x+ y ∂x ∂y (iii) If u = log 6. 7. ∂u ∂u x2 + y2 , prove that x + y = 1. ∂x ∂y x+ y (Kanpur 2006; Avadh 13) U se E uler’s theorem on homogeneous functions to show that if u = tan− 1 ( y ⁄ x), ∂u ∂u + y = 0. then x ∂y ∂x If u be a homogeneous function of x and y of degree n, show that ∂2 u ∂2 u ∂u (i) x 2 + y = (n − 1) ⋅ ∂x ∂x ∂y ∂x (ii) x ∂2 u ∂u ∂2 u + y 2 = (n − 1) ⋅ ∂y ∂x ∂y ∂y ∂2 u ∂2 u ∂2 u (iii) x2 2 + 2xy + y2 2 = n (n − 1) u. ∂x ∂y ∂x ∂y ∂2 u ∂2 u ∂2 u xy , show that x2 2 + 2 x y + y2 2 = 0. ∂x ∂y x+ y ∂x ∂y 8. If u = 9. ∂2 u ∂2 u ∂2 u + y2 2 = 0. If u = x φ ( y ⁄ x) + ψ ( y ⁄ x), prove that x2 2 + 2xy ∂x ∂y ∂x ∂y (Kanpur 2008) 7.7 Total Differential Coefficient If u = f (x, y) where x = φ 1 (t) and y = φ 2 (t), then x and y are not independent variables. Substituting the values of x and y in u, we can express u as a function of the single variable t and we can find the ordinary differential coefficient du ⁄ dt. To distinguish du ⁄ dt from the partial differential coefficients ∂u ⁄ ∂x and ∂u ⁄ ∂y, we shall call du ⁄ dt as the total differential coefficient. We shall now obtain a formula which will enable us to find du ⁄ dt without first expressing u in terms of t only. Suppose δx, δy and δu are the increments in x, y and u respectively corresponding to an increment δt in t. Then u + δu = f (x + δx, y + δy). D-164 DIFFERENTIAL CALCULUS ∴ δu = f (x + δx, y + δy) − f (x, y). δu f (x + δx, y + δy) − f (x, y) = ∴ δt δt { f (x + δx, y + δy) − f (x, y + δy)} + { f (x, y + δy) − f (x, y)} = δt [adding and subtracting the term f (x, y + δy) in the numerator] f (x + δx, y + δy) − f (x, y + δy) f (x, y + δy) − f (x, y) + = δt δt f (x + δx, y + δy) − f (x, y + δy) δx f (x, y + δy) − f (x, y) δy = ⋅ + ⋅ ⋅ δx δt δy δt δu du lim Now = δt → 0 dt δt lim f (x + δx, y + δy) − f (x, y + δy) δx = δt → 0 ⋅ δx δt lim f (x, y + δy) − f (x, y) δy + δt → 0 ⋅ ⋅ …(1) δy δt Now δx and δy also tend to zero when δt tends to zero. So we have ∂f ∂u , lim f (x + δx, y + δy) − f (x, y + δy) = = δx → 0 ∂x ∂x δx because while x becomes x + δx , y + δy remains unchanged. ∂f ∂u lim f (x, y + δy) − f (x, y) Similarly, δy → 0 = = ⋅ δy ∂y ∂y lim δy dy lim δx dx = and δt → 0 = ⋅ Also δt → 0 dt δt dt δt du ∂u dx ∂u dy = ⋅ + ⋅ ⋅ Therefore (1) gives ∂x dt ∂y dt dt Similarly, if u = f (x1, x2 , … , xm ) and x1, x2 , … , xm are all functions of t, we can prove that ∂u dx1 ∂u dx2 ∂u dxm du = ⋅ + ⋅ + …+ ⋅ ⋅ ∂x1 dt ∂x2 dt ∂xm dt dt 7.8 First Differential Coefficient of an Implicit Function Suppose u = f (x, y), where y = φ (x). Then supposing t to be the same as x in the formula of article 7.7, we get du ∂u dx ∂u dy du ∂u ∂u dy = ⋅ + ⋅ or = + ⋅ ⋅ dx ∂x dx ∂y dx dx ∂x ∂y dx Now suppose we are given an implicit relation between x and y of the form u ≡ f (x, y) = c, where c is a constant and y is a function of x. Then, we have du ⁄ dx = 0 ∂u ∂u dy ∂u ⁄ ∂x ∂f ⁄ ∂x dy dy or + = 0 or = − or = − ⋅ ∂x ∂u ⁄ ∂y ∂f ⁄ ∂y ∂y dx dx dx D-165 PARTIAL DIFFERENTIATION Example 1 : If (tan x) y + ( y) cot x = a, find the value of dy ⁄ dx . (Rohilkhand 2014) Solution : Let f (x, y) ≡ (tan x) y + ( y) cot x = a. Then, we have ∂f / ∂x dy = − ⋅ dx ∂f / ∂y …(1) ∂f = y (tan x) y − 1 sec 2 x + ycot x . log y. (− cosec 2 x) ∂x ∂f = (tan x) y log tan x + (cot x) . ycot x − 1 . and ∂y Therefore (1) gives y (tan x) y − 1 . sec 2 x − ycot x . log y. cosec 2 x dy = − ⋅ dx (tan x) y log tan x + cot x. ycot x − 1 Now Example 2 : If u = x2 y, where x2 + xy + y2 = 1, find du ⁄ dx. du ∂u ∂u dy = + ⋅ ⋅ Solution : We have, ∂x ∂y dx dx ∂u ∂u Now = 2xy and = x2. ∂x ∂y …(1) f (x, y) ≡ x2 + xy + y2 = 1. ∂f ⁄ ∂x 2x + y dy = − = − ⋅ Then ∂f ⁄ ∂y x + 2y dx So putting the values in (1), we get Let 2x + y ⎞ x2 (2x + y) du ⎛ = 2xy + x2 . ⎜ − ⋅ ⎟ = 2 xy − dx x + 2y ⎝ x + 2 y⎠ ∂u ∂u ∂u + + = 0. Example 3 : If u = f ( y − z , z − x , x − y), prove that ∂x ∂y ∂z (Kanpur 2009; Avadh 10) Solution : We have u = f ( y − z , z − x , x − y). Let y− z = A , z − x= B, and x− y= C. Then u = f (A, B, C ) where A, B and C are functions of x, y and z. ∂u ∂A ∂u ∂B ∂u ∂C ∂u = ⋅ + ⋅ + ⋅ Now ∂B ∂x ∂C ∂x ∂x ∂A ∂x ∂u ∂u ∂u ∂u ∂u = ⋅ (0) + ⋅ (− 1) + ⋅ (1) = − + ⋅ ∂A ∂B ∂C ∂B ∂C ∂u ∂A ∂u ∂B ∂u ∂C ∂u = ⋅ + ⋅ + ⋅ Similarly ∂B ∂y ∂C ∂y ∂y ∂A ∂y ∂u ∂u ∂u ∂u ∂u = ⋅ (1) + ⋅ (0) + ⋅ (− 1) = − ∂A ∂B ∂C ∂A ∂C ∂u ∂u ∂A ∂u ∂B ∂u ∂C and = ⋅ + ⋅ + ⋅ ∂z ∂A ∂z ∂B ∂z ∂C ∂z ∂u ∂u ∂u ∂u ∂u = ⋅ (− 1) + ⋅ (1) + ⋅ (0) = − + ⋅ ∂A ∂B ∂C ∂A ∂B ...(1) ...(2) ...(3) D-166 DIFFERENTIAL CALCULUS Adding (1), (2) and (3), we have 1. ∂u ∂u ∂u + + = 0. ∂x ∂y ∂z Find dy ⁄ dx in the following : (i) x y + y x = ab. (ii) ax2 + 2hxy + by2 = 1. (Kashi 2013) y2) 2. dy √(1 − If √(1 − x2) + √(1 − y2) = a (x − y), prove that = ⋅ dx √(1 − x2) 3. Find du ⁄ dx if u = sin (x2 + y2), where a2 x2 + b2 y2 = c2. 4. If u = x4 y5, where x = t 2 and y = t 3, find du ⁄ dt. ∂f ∂φ dz ∂f ∂φ ⋅ ⋅ = ⋅ ⋅ If f (x, y) = 0, φ ( y, z) = 0, show that ∂y ∂z dx ∂x ∂y 5. 6. I f u = √(x2 + y2) a n d x3 + y3 + 3axy = 5a2, fi n d t h e va l u e o f du ⁄ dx wh e n x = a, y = a. (ax + hy) { y x y − 1 + y x log y} , (ii) − ⋅ (hx + by) {x y log x + xy x − 1} ⎛ a2 ⎞ 3. 2 x {cos (x2 + y2)} ⎜⎜ 1 − 2 ⎟⎟ ⋅ 4. ⎜ b ⎟⎠ ⎝ 6. 0. 1. (Lucknow 2009) (i) − 23t 22. Fill in the Blanks: 1. 2. 3. 4. 5. Fill in the blanks “……”, so that the following statem ents are com plete and correct. ∂2 u ∂2 u = …… . If u = e my cos mx , then 2 + ∂x ∂y 2 x2 + y2 , then ∂u = …… . ∂x x+ y An expression in which every term is of the same degree is called a …… function. du If u = f (x, y), where x = φ (t) and y = ψ (t), then = …… . dt If u (x , y) is a homogeneous function of x and y of degree n, then ∂ ∂ x (ux ) + y (ux ) = …… , ∂y ∂x ∂u where ux = ⋅ ∂x If u = tan− 1 D-167 PARTIAL DIFFERENTIATION 6. If u = f( y + ax) + φ ( y − ax), then 7. If u = f( y / x), then, x ∂2u a 2∂2u − = ...... . ∂x2 ∂y 2 (Agra 2002) ∂u ∂u + y = ...... . ∂x ∂y (Agra 2003) Multiple Choice Questions: 8. 9. 10. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). ⎛ ∂2 z ⎞ ⎛ ∂2 z ⎞ ⎜ ⎟ ⎜ ⎟ If z = tan ( y + a x)+ ( y − a x) 3 ⁄ 2, then ⎜⎜ 2 ⎟⎟ − a2 ⎜⎜ 2 ⎟⎟ is equal to ⎝ ∂x ⎠ ⎝ ∂y ⎠ (a) 0 (b) ( y − a x) (c) 1 (d) sec ( y + a x) ∂u ∂u If u is a homogeneous function of x and y of degree n, then x + y is equal to ∂x ∂y (a) n (b) n u n (c) (d) u u ∂u ∂u ∂u 1 + y + z is equal to If = √(x 2 + y 2 + z 2), then x ∂x ∂y ∂z u (a) u (b) − u (c) u2 11. If x and y are connected by an equation of the form f (x , y) = 0, then ∂f ⁄ ∂x ∂f ⁄ ∂y ∂f ⁄ ∂y (c) ∂f ⁄ ∂x ∂f ⁄ ∂x ∂f ⁄ ∂y ∂f ⁄ ∂x (d) − ∂f ⁄ ∂y ∂θ If x = r cos θ, y = r sin θ, then the value of is ∂x sin θ r (b) (a) − sin θ r sin θ 1 (c) (d) − r r sin θ (a) 12. 13. (d) 0 dy dx is (b) (− 1) n If f(x, y) be a homogeneous function of x and y of degree n, then ∂f ∂f ∂f ∂f (a) x + y = f (b) y + x = f ∂x ∂y ∂x ∂y ∂f ∂f ∂f ∂f (d) x + y = nf (c) y + x = nf ∂y ∂x ∂y ∂x (Garhwal 2001) (Garhwal 2002) (Garhwal 2003) True or False: 14. 15. Write ‘T’ for true and ‘F’ for false statement. I f u = f (x , y) and its part ial derivat ives are continuous, then the order of differentiation is immaterial. ∂u ∂u If u is a homogeneous function of x and y of degree n, then and are also ∂x ∂y homogeneous functions of x and y each being of degree n. D-168 16. DIFFERENTIAL CALCULUS If f (x , y) = 0 be an implicit function of x and y and p= d2y dx 2 1. 0. 3. homogeneous. 5. (n − 1) ux . 0. (b). (d). (d). F. 7. 9. 11. 13. 15. ∂f , ∂2 f , ∂2 f ∂2 f , ∂f , q= r= s= and t = then 2 ∂y ∂x ∂y ∂x ∂x ∂y 2 = − (q 2 r − 2 pqs + p 2 t) ⁄ q 3. (x 2 + 2 xy − y 2) ⋅ {(x + y) 2 + (x 2 + y 2) 2} ∂u dx ∂u dy 4. ⋅ + ⋅ ⋅ ∂x dt ∂y dt 6. 0. 2. 8. 10. 12. 14. 16. (a). (b). (a). T. T. 8.1 Jacobian (Kanpur 2014) Definition : If u1, u2 , … , un are functions of n independent variables x1, x2 , ..., xn then the determinant ∂u1 ∂u1 ∂u … …… 1 ⎪ ⎪ ⎪ ∂x ∂x2 ∂xn ⎪⎪ ⎪ 1 ⎪ ⎪ ⎪ ∂u 2 ∂u2 ∂u2 ⎪⎪ ⎪ … … … ⎪ ⎪ ⎪ ∂x1 ∂x2 ∂xn ⎪ ⎪ ⎪ ⎪ …………… ……… ⎪ ⎪ ⎪ ⎪ ∂un ∂un ⎪⎪ ⎪ ∂u n … …… ⎪ ⎪ ⎪ ∂x ∂x2 ∂xn ⎪ 1 is called the Jacobian of u1, u2 , … , un with respect to x1, x2 , … , xn and is denoted either ∂ (u1, u2 , … , un ) or by J (u1, u2 , … , un ) . The second notation is used when there is ∂ (x1, x2 , … , xn ) no doubt as regards the independent variables. Thus if u and v are functions of two independent variables x and y, we have ∂u ⎪ ⎪ ∂u ∂ (u, v) ⎪⎪ ∂x ∂y ⎪⎪ = ⎪ ⎪ = J (u, v) . ∂v ⎪ ⎪ ∂v ∂ (x, y) ⎪ ⎪ ⎪ ∂x ∂y ⎪ by D-170 DIFFERENTIAL CALCULUS Similarly if u, v and w are functions of three independent variables x, y and z, we have ∂u ∂u ∂u ⎪ ∂x ∂y ∂z ⎪⎪ ∂v ∂v ∂v ⎪⎪ ⎪ = J (u, v, w). ∂x ∂y ∂z ⎪⎪ ∂w ∂w ∂w ⎪⎪ ∂x ∂y ∂z ⎪ Note : If the functions u1, u2 , … , un of n independent variables x1, x2 , ..., xn are ⎪ ⎪ ⎪ ∂ (u, v, w) ⎪⎪ = ⎪ ⎪ ∂ (x, y, z) ⎪ ⎪ ⎪ ⎪ of the following forms, u1 = f1 (x1), u2 = f2 (x1, x2), … , un = fn (x1, x2 , … , xn ), then ⎪ ∂u 1 ⎪ ⎪ 0 0 … 0 ⎪⎪ ⎪ ∂x ⎪ 1 ⎪ ⎪ ⎪ ⎪ ∂u ⎪ ∂u 2 2 ⎪ ∂u1 ∂u2 ∂u ∂ (u1, u2 , … , un ) ⎪⎪ ⎪ 0 … 0 …… n , ⎪ = = ⎪⎪ ∂x1 ∂x2 ⎪ ∂x1 ∂x2 ∂xn ∂ (x1, x2 , … , xn ) ⎪ ⎪ ⎪… … … … … ⎪⎪ ⎪ ⎪ ∂u ∂un ∂un ∂un ⎪⎪ ⎪ n ⎪ ⎪ … ⎪ ∂x ∂x2 ∂x3 ∂xn ⎪ 1 i.e., in such cases the Jacobian reduces to the principal diagonal term of the determinant. Example 1 : If x = r sin θ cos φ , y = r sin θ sin φ , z = r cos θ, show that ∂ (x, y, z) = r2 sin θ. ∂ (r, θ, φ ) (Meerut 2003; Garhwal 02; Rohilkhand 12; Avadh 07, 12) Solution : We have ∂x ∂x ⎪ ⎪ ∂r ∂θ ⎪ ⎪ ∂y ∂ (x, y, z) ∂y = ⎪⎪ ∂θ ∂ (r, θ, φ ) ⎪⎪ ∂r ⎪ ∂z ∂z ⎪ ⎪ ∂θ ∂r ∂x ∂φ ∂y ∂φ ∂z ∂φ ⎪ ⎪ ⎪ ⎪ sin θ cos φ ⎪ ⎪ = ⎪ sin θ sin φ ⎪ ⎪ ⎪ ⎪ ⎪ cos θ ⎪ ⎪ ⎪ ⎪ r cos θ cos φ r cos θ sin φ − r sin θ − r sin θ sin φ r sin θ cos φ 0 ⎪ ⎪ ⎪ ⎪ ⎪ = cos θ (r2 sin θ cos θ cos2 φ + r2 sin θ cos θ sin2 φ ) + r sin θ (r sin2 θ cos2 φ + r sin2 θ sin2 φ ) , expanding the determinant along the third row = r2 sin θ cos2 θ + r2 sin3 θ = r2 sin θ (cos2 θ + sin2 θ) = r2 sin θ. Example 2 : Find the Jacobian ∂ (x, y, z) being given ∂ (r, θ, φ ) x = r cos θ cos φ , y = r sin θ √(1 − m 2 sin2 φ ) , z = r sin φ √(1 − n2 sin2 θ) , where m 2 + n2 = 1. D-171 JACOBIANS Solution : H ere x2 + y2 + z2 = r2 cos2 θ cos2 φ + r2 sin2 θ − r2 m 2 sin2 θ sin2 φ + r2 sin2 φ − r2 n2 sin2 φ sin2 θ [ ... m 2 + n2 = 1] = r2 (cos2 θ cos2 φ + sin2 θ + sin2 φ − sin2 θ sin2 φ ) = r2 (cos2 θ cos2 φ + sin2 θ + sin2 φ cos2 θ) = r2 (sin2 θ + cos2 θ) = r2. ∂x ∂y ∂z ∂x ∂y ∂z ⎫ ∴ x + y + z = r;x ∂θ + y ∂θ + z ∂θ = 0 ⎪⎪ ∂r ∂r ∂r ⎬ ⎪ ∂y ∂z ∂x and ⎪ x ∂φ + y ∂φ + z ∂φ = 0. ⎭ ∂x ∂x ∂x ∂x ∂x ⎪ ⎪ x ∂x x ∂θ x ∂φ ∂θ ∂φ ⎪⎪ ⎪ ∂r ⎪ ∂r ⎪ ⎪ ⎪ ⎪ ∂y ∂y ∂y ⎪⎪ 1 ⎪⎪ ∂y ∂y ∂y Now J (x, y, z) = ⎪⎪ ∂θ ∂φ ⎪⎪ = x ⎪⎪ ∂r ∂θ ∂φ ⎪ ∂r ⎪ ⎪ ⎪ ⎪ ∂z ⎪ ∂z ∂z ∂z ∂z ∂z ⎪⎪ ⎪ ⎪ ⎪ ∂r ⎪ ⎪ ∂r ∂θ ∂φ ∂θ ∂φ r 0 0 ⎪ ⎪ ⎪ ∂y ⎪ ∂y ∂y ⎪ by adding y R + zR to R 1 ⎪⎪ 2 3 1 = ⎪ ∂r ∂θ ∂φ ⎪⎪ , ⎪ and using the relations (1) x ⎪⎪ ⎪ ∂z ∂z ⎪ ⎪ ∂z ⎪ ∂r ∂θ ∂φ ⎪ ∂y ⎪ ⎪ ∂y ∂z ∂y ⎫ r ⎪⎪ ∂θ ∂φ ⎪⎪ r ⎧ ∂y ∂z = ⎪ − ∂θ ∂φ ⎬ ⎪= ⎨ ∂z ⎪ x ⎩ ∂θ ∂φ ⎭ x ⎪⎪ ∂z ⎪ ⎪ ∂θ ∂φ ⎪ = ...(1) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ r ⎧⎪ 2 2 2 2 ⎨ r cos θ √(1 − m sin φ ) . r cos φ √(1 − n sin θ) x ⎪⎩ − = r3 cos θ cos φ x = r3 cos θ cos φ ⋅ r cos θ cos φ r sin φ . n2 sin θ cos θ r sin θ . m 2 sin φ cos φ ⎫⎪ ⋅ ⎬ √(1 − n2 sin2 θ) √(1 − m 2 sin2 φ ) ⎪⎭ ⎡ (1 − m 2 sin 2 φ ) (1 − n 2 sin 2 θ) − n 2 m 2 sin 2 θ sin 2 φ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ √[(1 − n2 sin2 θ) (1 − m 2 sin2 φ )] ⎣ ⎦ ⎡ 1 − m 2 sin2 φ − n 2 sin2 θ + m 2 n 2 sin2 φ sin2 θ − m 2 n 2 sin2 θ sin2 φ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ √[(1 − n2 sin2 θ) (1 − m 2 sin2 φ )] ⎣ ⎦ r2 (m 2 cos2 φ + n2 cos2 θ) ⋅ [ ... m 2 + n2 = 1] √[(1 − n2 sin2 θ) (1 − m 2 sin2 φ )] Example 3 : If y1 = r sin θ1 sin θ2 , y2 = r sin θ1 cos θ2 , y3 = r cos θ1 sin θ3 , = y4 = r cos θ1 cos θ3 , find the value of the Jacobian ∂ ( y1, y2 , y3 , y4) ∂ (r, θ1, θ2 , θ3) ⋅ Solution : Squaring and adding the given relations, we have y12 + y22 + y32 + y42 = r2 . D-172 ∴ and DIFFERENTIAL CALCULUS y1 y1 ∂y1 ∂r ∂y1 ∂θr + y2 + y2 ∂y2 ∂r ∂y2 ∂θr + y3 + y3 ∂y3 ∂r ∂y3 ∂θr + y4 + y4 ∂y4 ∂r ∂y4 ∂θr = r = 0, r = 1, 2, 3. ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ...(1) Also y32 + y42 = r2 cos2 θ1, so that ∂y3 y3 ∂θ + 1 ∂y3 y3 ∂θ + r ∂y4 y4 ∂θ = − r2 cos θ1 sin θ1 ; 1 ∂y4 y4 ∂θ = 0, r = 2, 3. r Now the required Jacobian ∂y ∂y1 ⎪ 1 ⎪ ∂r ∂θ 1 ⎪ ⎪ ∂y2 ⎪ ∂y2 ⎪ ⎪ ∂r ∂θ1 J = ⎪⎪ ∂y3 ⎪ ∂y3 ⎪ ⎪ ∂r ∂θ ⎪ 1 ⎪ ⎪ ∂y4 ∂y4 ⎪ ⎪ ∂θ1 ∂r ∂y1 ∂θ2 ∂y2 ∂θ2 ∂y3 ∂θ2 ∂y4 ∂θ2 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ...(2) ∂y1 ⎪ ∂θ3 ⎪⎪ ∂y2 ⎪⎪ ⎪ ∂θ3 ⎪⎪ ⎪⋅ ∂y3 ⎪ ⎪ ∂θ3 ⎪⎪ ⎪ ∂y4 ⎪ ⎪ ∂θ3 ⎪ O perating y1R 1 + ( y2 R 2 + y3 R 3 + y4 R 4) , and using the results (1), we get ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪⎪ J= ⎪ y1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = r 0 0 ∂y2 ∂y2 ∂θ1 ∂y3 ∂θ1 ∂y4 ∂θ1 ∂y2 ∂θ2 ∂y3 ∂θ2 ∂y4 ∂θ2 ∂r ∂y3 ∂r ∂y4 ∂r ⎪ ∂y2 ⎪ ⎪ ∂θ1 0 ⎪⎪ ∂y2 ∂θ3 ∂y3 ∂θ3 ∂y4 ∂θ3 r ⎪⎪ 2 ⎪ − r cos θ1 sin θ1 y1 y3 ⎪ ⎪ ∂y ⎪ 4 ⎪ ⎪ ∂θ1 ∂y ⎪ ⎪ 2 ⎪ ⎪ ∂θ ⎪ ⎪ 1 ⎪ ⎪ ⎪ r ⎪⎪ ∂y3 ⎪ ⎪ ⎪ = ⎪ y1 ⎪ ∂θ1 ⎪ ⎪ ⎪ ∂y ⎪ ⎪ 4 ⎪ ⎪ ⎪ ⎪ ∂θ1 ⎪ ⎪ ⎪ ∂y2 ∂θ2 0 ∂y4 ∂θ2 ∂y2 ∂θ2 ∂y3 ∂θ2 ∂y4 ∂θ2 ∂y2 ∂θ3 ∂y3 ∂θ3 ∂y4 ∂θ3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂y2 ⎪ ⎪ ∂θ3 ⎪⎪ ⎪ 0 ⎪, ⎪ ∂y4 ⎪ ⎪ ∂θ3 ⎪⎪ adding y4 R 3 to y3 R 2 and using the results (2) ∂y4 ∂y2 ⎤ ⎡ ∂y2 ∂y4 r ⋅ r2 cos θ1 sin θ1 ⎢⎢ ∂θ ⋅ ∂θ − ∂θ ⋅ ∂θ ⎥⎥ ⎢ y1 y3 2 3 2 3 ⎥⎦ ⎣ r3 cos θ1 sin θ1 = [(− r sin θ1 sin θ2) (− r cos θ1 sin θ3) − 0] y1 y3 = = r5 sin2 θ1 cos2 θ1 sin θ2 sin θ3 r2 sin θ1 cos θ1 sin θ2 sin θ3 = r3 sin θ1 cos θ1 . D-173 JACOBIANS 1. 2. If x = r cos θ, y = r sin θ, show that ∂ (x, y) (Kanpur 2005; Meerut 13B; Kashi 13) (i) = r, ∂ (r, θ) ∂ (r, θ) 1 (Meerut 2003) (ii) = ⋅ ∂ (x, y) r If x = u (1 + v), y = v (1 + u), find the Jacobian of x, y with respect to u, v. (Meerut 2013) 3. If x = c cos u cosh v, y = c sin u sinh v, prove that ∂ (x, y) 1 = c2 (cos 2u − cosh 2v) . ∂ (u, v) 2 y2 ,v= x2 + 2x y2 (Rohilkhand 2013) , find ∂ (u, v) ⋅ ∂ (x, y) 4. If u = 5. If u1 = x2 x3 ⁄ x1 , u2 = x3 x1 ⁄ x2 , u3 = x1 x2 ⁄ x3 , prove that J (u1, u2 , u3) = 4 . 2x (Meerut 2012) (Bundelkhand 2014; Purvanchal 14) c2 sin 2 6. If x = sin θ √(1 − 7. [(1 − c2) cos2 θ + c2 cos2 φ ] ∂ (x, y) = − sin φ ⋅ ∂ (θ, φ ) √(1 − c2 sin2 φ ) If y1 = 1 − x1, y2 = x1 (1 − x2), y3 = x1 x2 (1 − x3), … , yn = x1 x2 … xn − 1 (1 − xn ), prove that J ( y1, y2 , ..., yn ) = (− 1) n x1n − 1 x2n − 2 … xn − 1 . 8. φ ) , y = cos θ cos φ , then show that If y1 = cos x1 , y2 = sin x1 cos x2 , y3 = sin x1 sin x2 cos x3 , … , yn = sin x1 sin x2 sin x3 … sin xn − 1 cos xn , find the Jacobian of y1 , y2 , … , yn with respect to x1 , x2 , … , xn . 2. 1 + u + v. 8. 1) n (− sin n 4. − y ⁄ 2 x . x1 sin n − 1 x2 … sin xn . 8.2 Case of Functions of Functions (Chain Rule) We shall establish the formula for two variables and the result can be easily extended to any number of variables. Theorem : If u1, u2 are functions of y1, y2 and y1, y2 are functions of x1, x2 , then ∂ (u1, u2) ∂ (x1, x2) = ∂ (u1, u2) ∂ ( y1, y2) ⋅ ∂ ( y1, y2) ∂ (x1, x2) ⋅ (Kumaun 2003) D-174 DIFFERENTIAL CALCULUS Proof : We have ∂u1 ∂u1 = ∂x1 ∂y1 ∂u2 ∂u2 = ∂x1 ∂y1 ∂y1 ∂x1 ∂y1 ∂x1 + + ∂u1 ∂y2 ∂y2 ∂x1 ∂u2 ∂y2 ∂y2 ∂x1 ⎪ ⎪ ∂ (u1, u2) ∂ ( y1, y2) ⎪⎪ ⋅ = ⎪ Now ∂ ( y1, y2) ∂ (x1, x2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = ⎪⎪ ⎪ ⎪ ⎪ ⎪ ∂u1 ∂y1 ∂y1 ∂x1 ∂u2 ∂y1 ∂y1 ∂x1 + + ∂u1 ∂y1 ∂u2 ∂y1 , , ∂u1 ∂x2 ∂u2 ∂x2 ∂u1 ∂y1 = ∂y1 ∂x2 ∂u2 ∂y1 = ∂y1 ∂x2 ∂u1 ⎪ ⎪ ∂y2 ⎪⎪ ⎪× ∂u2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂y2 ⎪⎪ ∂u1 ∂y2 ∂u1 ∂y1 ∂y2 ∂x1 ∂u2 ∂y2 ∂y1 ∂x2 ∂u2 ∂y1 ∂y2 ∂x1 ∂y1 ∂x2 ∂y1 ∂x1 ∂y2 ∂x1 + + ⎫ , ⎪ ⎪ ∂x2 ⎬ ∂y2 ⎪ ⋅ ⎪ ∂x2 ⎭ ∂u1 ∂y2 ∂y2 ∂u2 ∂y2 ...(1) ∂y1 ⎪ ⎪ ∂x2 ⎪⎪ ⎪ ∂y2 ⎪ ⎪ ∂x2 ⎪⎪ ∂u1 ∂y2 ⎪ ⎪ ∂y2 ∂x2 ⎪⎪ ⎪, ∂u2 ∂y2 ⎪ + ⎪ + ∂y2 ∂x2 ⎪⎪ applying row-by-column multiplication rule ⎪ ⎪ ⎪ = ⎪⎪ ⎪ ⎪ ⎪ ⎪ = ∂u1 ∂x1 ∂u2 ∂x1 ∂u1 ⎪ ⎪ ∂x2 ⎪⎪ ⎪ , using the relations (1) ∂u2 ⎪ ∂ (u1, u2) ∂ (x1, x2) ⎪ ∂x2 ⎪⎪ ⋅ Note : T h e a b o ve fo r m u l a r e s e m b l e s ve r y m u c h wi t h t h e fo r m u l a d f d f dt , = ⋅ for the derivative of the function of a function. dx dt dx Generalization of the above formula : I f u1, u2 , ..., un a re f u n ct io n s o f y1, y2 , … , yn and y1, y2 , ..., yn are functions of x1, x2 , ..., xn , then ∂ (u1, u2 , …, un ) ∂ (x1, x2 , ..., xn ) = ∂ (u1, u2 , ..., un ) ∂ ( y1, y2 , ..., yn ) ⋅ ∂ ( y1, y2 , ..., yn ) ∂ (x1, x2 , ..., xn ) ⋅ The proof may be easily extended as in the case of two variables and has been left as an exercise for the students. 8.3 Jacobian of Implicit Functions Theorem : S uppose u1, u2 , ..., un instead of being given explicitly in term s of x1, x2 , ..., xn are connected with them by equations such as F1 (u1, u2 , ..., un , x1, x2 , ..., xn ) = 0, F2 (u1, u2 , ..., un , x1, x2 , ..., xn ) = 0, … … … … … … Fn (u1, u2 , ..., un , x1, x2 , ..., xn ) = 0. D-175 JACOBIANS ∂ ( F1, F2 , ..., Fn ) Then, we have ∂ (u1, u2 , ..., un ) ∂ (x1, x2 , ..., xn ) = (− 1) n ∂ (x1, x2 , ..., xn ) ∂ ( F1, F2 , ..., Fn ) ⋅ ∂ (u1, u2 , ..., un ) Proof : Here also we shall establish the result for two variables and the proof can be extended easily for n variables. The students should themselves write the proof for n variables on the basis of the proof given below for two variables. In the case of two variables, the connecting relations are F1 (u1, u2 , x1, x2) = 0, ⎫ ⎪ ...(1) F2 (u1, u2, x1, x2) = 0. ⎬⎪⎭ From relations (1), we have by differentiation ⎫ ∂F1 ∂F1 ∂u1 ∂F1 ∂u2 + + = 0, ⎪ ⎪ ∂x1 ∂u1 ∂x1 ∂u2 ∂x1 ⎪ ∂F1 ∂F1 ∂u1 ∂F1 ∂u2 ⎪ + + = 0, ⎪ ∂x2 ∂u1 ∂x2 ∂u2 ∂x2 ⎪ ⎬ ∂F2 ∂F2 ∂u1 ∂F2 ∂u2 ⎪ + + = 0, ⎪ ∂x1 ∂u1 ∂x1 ∂u2 ∂x1 ⎪ ⎪ ∂F2 ∂F2 ∂u1 ∂F2 ∂u2 ⎪ + + = 0. ⎪ ∂x2 ∂u1 ∂x2 ∂u2 ∂x2 ⎭ ∂F ∂F ⎪ 1 1⎪ ⎪ ∂u 1 ⎪ ⎪ ⎪ ⎪ ∂ ( F1, F2 ) ∂ (u1, u2) ⎪⎪ ∂u1 ∂u2 ⎪ ⎪⎪ ∂x1 Now ⋅ = ⎪ ⎪× ⎪ ⎪ ∂F2 ∂F2 ⎪ ⎪ ∂u2 ∂ (u1, u2) ∂ (x1, x2) ⎪ ⎪ ⎪ ∂u 1 ⎪ ⎪ ⎪ = ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂x1 ∂u2 ⎪⎪ ∂F1 ∂u1 + ∂u1 ∂x1 ∂F2 ∂u1 + ∂u1 ∂x1 ∂F1 ∂u2 ∂u2 ∂x1 ∂F2 ∂u2 ∂u2 ∂x1 ...(2) ∂u1 ⎪ ⎪ ∂x2 ⎪⎪ ⎪ ∂u2 ⎪ ⎪ ∂x2 ⎪⎪ ∂F1 ∂u1 + ∂u1 ∂x2 ∂F2 ∂u1 + ∂u1 ∂x2 ∂F1 ∂u2 ⎪ ⎪ ∂u2 ∂x2 ⎪⎪ ⎪, ∂F2 ∂u2 ⎪ ⎪ ∂u2 ∂x2 ⎪⎪ applying row-by-column multiplication rule ⎪ − ∂F1 ⎪ ⎪ ∂x 1 = ⎪⎪ ⎪ − ∂F2 ⎪ ⎪ ⎪ ∂x1 = (− 1) 2 − ∂F1 ⎪ ⎪ ∂x2 ⎪⎪ ⎪ , using the relations (2) − ∂F2 ⎪ ⎪ ∂x2 ⎪⎪ ∂ ( F1, F2) ∂ (x1, x2) ⋅ Accordingly, we have ∂ ( F1, F2) ∂ (u1, u2) ∂ (x1, x2) = (− 1) 2 ∂ (x1, x2) ∂ ( F1, F2) ∂ (u1, u2) ⋅ D-176 DIFFERENTIAL CALCULUS Example 1 : Prove that ∂ (u, v) ∂ (x, y) × = 1. ∂ (x, y) ∂ (u, v) (Bundelkhand 2011; Kanpur 08; Meerut 12; Avadh 13) Solution : Let u = f1 (x, y), v = f2 (x, y) . ...(1) O bviously x and y can also be expressed as functions of u and v . Differentiating relations (1) partially with respect to u and v, we get ∂u ∂x ∂u ∂y , ∂u ∂x ∂u ∂y , ⎫ ⎪ 1= + 0= + ∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v ⎪ ...(2) ⎬ ∂v ∂x ∂v ∂y ∂v ∂x ∂v ∂y , + 1= + ⋅ ⎪⎪ 0= ∂x ∂v ∂y ∂v ∂x ∂u ∂y ∂u ⎭ ∂u ∂u ∂x ∂x ⎪ ⎪ ⎪ ⎪ ⎪ ∂x ∂ (u, v) ∂ (x, y) ∂y ⎪⎪ ⎪⎪ ∂u ∂v ⎪⎪ ⎪ Now × = ⎪ ⎪× ⎪ ⎪ ∂v ⎪ ⎪ ∂y ∂y ⎪ ∂ (x, y) ∂ (u, v) ⎪ ∂v ⎪ ⎪ ⎪ ⎪ ⎪ ∂x ∂y ⎪ ⎪ ∂u ∂v ⎪ ∂u ∂x ∂u ∂y ⎪ ⎪ ∂u ∂x + ∂u ∂y + ⎪ ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v ⎪⎪ , ⎪ = ⎪ ⎪ ∂v ∂y ∂v ∂x ∂v ∂y ⎪ ⎪ ∂v ∂x ⎪ ⎪ + + ⎪ ∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v ⎪ ⎪1 = ⎪ ⎪0 applying row-by-column multiplication 0⎪ ⎪ , using the relations (2) 1⎪ = 1. Example 2 : If u3 + v + w = x + y2 + z2, u + v3 + w = x2 + y + z2, u + v + w3 = x2 + y2 + z, prove that ∂ (u, v, w) 1 − 4 (x y + yz + z x) + 16 xyz = ⋅ ∂ (x, y, z) 2 − 3 (u2 + v2 + w2) + 27 u2 v2 w2 Solution : The given relations can be written as F1 ≡ u3 + v + w − x − y2 − z2 = 0, F2 ≡ u + v3 + w − x2 − y − z2 = 0, F3 ≡ u + v + w3 − x2 − y2 − z = 0. Now H ere / ∂ ( F1, F2 , F3) ∂ ( F1, F2 , F3) ∂ (u, v, w) = (− 1) 3 ⋅ ∂ (x, y, z) ∂ (u, v, w) ∂ (x, y, z) − 2y − 2z ⎪ ∂ ( F1, F2 , F3) ⎪⎪ − 1 = ⎪ − 2x − 1 − 2z ⎪⎪ ⎪ ⎪ ∂ (x, y, z) ⎪ − 2x − 2y − 1 ⎪ = − 1 (1 − 4yz) + 2x (2y − 4yz) − 2x (4yz − 2z) = − 1 + 4 ( yz + zx + xy) − 16xyz . And ∂ ( F1, F2 , F3) ∂ (u, v, w) ⎪ 3u 2 ⎪ = ⎪⎪ 1 ⎪ ⎪ 1 ⎪ 1 3v2 1 1 ⎪⎪ 1 ⎪⎪⎪ 3w2 ⎪⎪ ...(1) D-177 JACOBIANS = 3u2 (9v2 w2 − 1) − 1 (3w2 − 1) + 1 . (1 − 3v2) = 2 − 3 (u2 + v2 + w2) + 27u2 v2 w2. ∂ (u, v, w) 1 − 4 ( yz + zx + xy) + 16 xyz ∴ From (1), = ⋅ ∂ (x, y, z) 2 − 3 (u2 + v2 + w2) + 27u2 v2 w2 1. If u3 + v3 = x + y, u2 + v2 = x3 + y3, show that ∂ (u, v) 1 y2 − x2 = ⋅ 2 uv (u − v) ∂ (x, y) 2. If x + y + z = u, y + z = uv, z = uvw, show that ∂ (x, y, z) = u2 v. ∂ (u, v, w) (Rohilkhand 2005; Kashi 14) 3. 4. 1 1 1 1, 2 = + + w = x2 + y2 + z2, prove that v x y z ∂ (u, v, w) v ( y − z) (z − x) (x − y) (x + y + z) = − ⋅ ∂ (x, y, z) 3u2 w ( yz + zx + xy) If u3 = xyz , u3 + w3 u2 v2 w2 x3 y3 = x + y + z, + + = + + ∂ (u, v, w) ( y − z) (z − x) (x − y) then prove that = ⋅ ∂ (x, y, z) (u − v) (v − w) (w − u) If + v3 z3, (Rohilkhand 2012B) u + v + w = x2 + y2 + z2, (Purvanchal 2007; Kanpur 12) 5. 6. Compute the Jacobian ∂ (u, v) where ∂ (r, θ) u = 2 x y, v = x2 − y2, x = r cos θ, y = r sin θ. I f u1 = x1 + x2 + x3 + x4 , u1u2 = x2 + x3 + x4 , u1 u2 u3 = x3 + x4 , u1 u2 u3 u4 = x4 , show that ∂ (x1, x2 , x3 , x4) = u 13 u 22 u 3 . ∂ (u1, u2 , u3, u4) 7. G iven y1 (x1 − x2) = 0, y2 (x12 + x1 x2 + x22) = 0, show that ∂ ( y1, y2) x1 + x2 = 3y1 y2 3 ⋅ ∂ (x1, x2) x1 − x23 8. If u = x (1 − r2) − 1 ⁄ 2, v = y (1 − r2) − 1 ⁄ 2, w = z (1 − r2) − 1 ⁄ 2, ∂ (u, v, w) where r2 = x2 + y2 + z2, show that = (1 − r2) − 5 ⁄ 2 . ∂ (x, y, z) (a) Find the Jacobian of y1 , y2 , y3 , ..., yn , being given 9. y1 = x1 (1 − x2 ), y2 = x1 x2 (1 − x3), … , yn − 1 = x1 x2 … xn − 1 (1 − xn ) , yn = x1 x2 x3 … xn . (b) If y1 = r cos θ1, y2 = r sin θ1 cos θ2 , y3 = r sin θ1 sin θ2 cos θ3 , .... yn − 1 = r sin θ1 sin θ2 … sin θn − 2 cos θn − 1 and yn = r sin θ1 sin θ2 … sin θn − 1 , D-178 DIFFERENTIAL CALCULUS prove that ∂ ( y1 , y2 , … , yn ) ∂ (r, θ1 , ..., θn − 1) 10. 11. = r n − 1 sin n − 2 θ1 sin n − 3 θ2 … sin θn − 2 . If λ , μ , ν are the roots of the equation in k , y x z + + = 1, a+ k b+ k c+ k prove that ( μ − ν) (ν − λ) (λ − μ) ∂ (x, y, z) = − ⋅ (b − c) (c − a) (a − b) ∂ (λ , μ , ν) The roots of the equation in λ , ( λ − x) 3 + ( λ − y) 3 + ( λ − z) 3 = 0 are u, v, w. Prove that ∂ (u, v, w) ( y − z) (z − x) (x − y) = − 2 ⋅ ∂ (x, y, z) (v − w) (w − u) (u − v) (Kumaun 2008; Kanpur 10) 12. If x, y, z are connected by a functional relation f (x, y, z) = 0, show that ∂ ( y, z) ⎛ ∂y⎞ = ⎜ ⎟ ∂ (x, z) ⎝ ∂x⎠ z = const. 13. (i) Prove that 5. ∂ (u, v, w) ∂ (x, y, z) × = 1. ∂ (x, y, z) ∂ (u, v, w) ∂ ( y1, y2 , ..., yn ) ∂ (x1, x2 , ..., xn ) (ii) Prove that ⋅ = 1. ∂ (x1, x2 , ..., xn ) ∂ ( y1, y2 , ..., yn ) − 4 r3 . 9. (a) (Kanpur 2009, 11) x1n − 1 x2n − 2 … xn − 1 . 8.4 Necessary and Sufficient Condition for a Jacobian to Vanish Theorem : L et u1, u2 , ..., un be functions of n independent variables x1, x2 , ..., xn . In order that these n functions m ay not be independent, i.e., there m ay exist between these n functions a relation F (u1, u2 , ..., un ) = 0, ...(1) ∂ (u1, u2 , ..., un ) should vanish identically. ∂ (x1, x2 , .., xn ) Proof : The condition is necessary i.e., if there exists between u1, u2 , ..., un a it is necessary and sufficient that the Jacobian relation F (u1, u2 , ..., un ) = 0 their Jacobian is necessarily zero. Differentiating (1) partially with respect to x1, x2 , ..., xn , we get ∂F ∂u1 ∂F ∂u2 … ∂F ∂un + + + = 0, ∂u1 ∂x1 ∂u2 ∂x1 ∂un ∂x1 ...(1) D-179 JACOBIANS ∂F ∂u2 … ∂F ∂un ∂F ∂u1 + + + = 0, ∂u2 ∂x2 ∂un ∂x2 ∂u1 ∂x2 ............................................................. ∂F ∂u2 … ∂F ∂un ∂F ∂u1 + + + = 0. ∂u2 ∂xn ∂un ∂xn ∂u1 ∂xn ∂F , ∂F , … , ∂F E liminating from these equations, we get ∂u1 ∂u2 ∂un ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂u1 ∂u2 ∂x1 ∂u1 ∂x1 ∂u2 ∂x2 ∂x2 … … ∂u1 ∂u2 ∂xn ∂xn … … … … ∂un ⎪⎪ ∂x1 ⎪⎪ ⎪ ∂un ⎪ ⎪ ∂x2 ⎪⎪ ⎪ = 0 or ⎪ … ⎪⎪ ⎪ ∂un ⎪⎪ ⎪ ∂xn ⎪ ⎪ ∂ (u1, u2 , ..., un ) ∂ (x1, x2 , ..., xn ) = 0. The condition is sufficient, i.e., if the Jacobian J (u1, u2 , ..., un ) is zero, then there must exist a relation between u1, u2 , ..., un . Th e equ at io ns co nn ectin g th e fun ct io ns u1, u2 , ..., un a nd t h e var ia bles x1, x2 , ..., xn are always capable of being put into the following form : φ 1 (x1, x2 , ..., xn , u1) = 0 φ 2 (x2 , x3 , ..., xn , u1, u2) = 0 .............................................. φ r (xr , xr + 1 , ..., xn , u1, u2 , ..., ur) = 0 ............................................... φ n (xn , u1, u2 , ..., un ) = 0. Then, we have ∂ (φ 1, φ 2 , ..., φ n ) J= ∂ (u1, u2 , ..., un ) ∂ (x1, x2 , ..., xn ) = (− 1) n ∂ (x1, x2 , ..., xn ) ∂ (φ 1, φ 2 , ..., φ n ) ∂ (u1, u2 , ..., un ) ∂φ 1 ∂φ 2 ∂φ … n ∂x1 ∂x2 ∂xn . = (− 1) n ∂φ ∂φ 1 2 … ∂φ n ∂u1 ∂u2 ∂un Now, if J = 0, we have ∂φ 1 ∂φ 2 ∂φ ∂φ … r… n = 0 ⋅ ∂x1 ∂x2 ∂xr ∂xn [See note after article 8.1] D-180 i.e., ∂φ r ∂xr DIFFERENTIAL CALCULUS = 0 for some value of r between 1 and n . H ence, for that particular value of r the function φ r must not contain xr ; and accordingly the corresponding equation is of the form φ r (xr + 1, ..., xn , u1, u2 , ..., ur) = 0. Consequently between this and the remaining equations φ r + 1 = 0, φ r + 2 = 0, ..., φ n = 0, the variables xr + 1, xr + 2 , … , xn can be eliminated so as to give a final equation between u1, u2 , ..., un alone. H ence the theorem is established. Example 1 : Show that the functions u = x + y− z, v = x − y + z, w = x2 + y2 + z2 − 2yz are not independent of one another. A lso find the relation between them. (Garhwal 2000) Solution : H ere 1 ∂ (u, v, w) ⎪⎪ = ⎪ 1 ⎪ ∂ (x, y, z) ⎪ 2x ⎪ 1 = ⎪⎪ 1 ⎪ ⎪ 2x 1 − 1 2 ( y − z) 1 − 1 2 ( y − z) − 1 ⎪ ⎪ 1 ⎪ ⎪ 2 (z − y) ⎪ 0⎪ 0⎪⎪ , adding C2 to C3 ⎪ 0⎪ = 0. Since the Jacobian is zero, the functions are not independent. Now u + v = 2 x and u − v = 2 ( y − z) . Therefore (u + v) 2 + (u − v) 2 = 4 (x2 + y2 + z2 − 2yz) = 4w. This is the required relation between u, v, w. Example 2 : Show that ax2 + 2hxy + by2 and Ax2 + 2Hxy + By2 are independent unless a h b = = ⋅ A H B or or Solution : Let u = ax2 + 2hxy + by2, v = A x2 + 2Hxy + By2. If the functions u, v are not independent, then ∂ (u, v) = 0 ∂ (x, y) ∂u ⎪ ⎪ ∂u ⎪ ∂x ∂y ⎪⎪ ⎪ ⎪ ⎪= 0 ∂v ⎪ ⎪ ∂v ⎪ ⎪ ⎪ ∂x ∂y ⎪ ⎪ 2 (ax + hy) ⎪ ⎪ 2 (A x + Hy) 2 (hx + by) ⎪ ⎪= 0 2 (Hx + By) ⎪ D-181 JACOBIANS or (ax + hy) (Hx + By) − (hx + by) (A x + Hy) = 0 or (aH − A h) x2 + (aB − A b) xy + (Bh − bH) y2 = 0. Since the variables x, y are independent, the coefficients of x2 and y2 in the above equation must be separately zero. Hence, we have aH − A h = 0 and Bh − bH = 0 h b a = = ⋅ whence H B A ( μ − ν) (ν − λ) ( λ − μ) ∂ (x, y, z) = − ⋅ (b − c) (c − a) (a − b) ∂ (λ, μ, ν) 1. If u = x2 + y2 + z2, v = x + y + z , w = x y + y z + z x , show that the Jacobian ∂ (u, v, w) vanishes identically. Also find the relation ∂ (x, y, z) between u, v and w. 2. (Avadh 2014) ∂ (u, v) If u = (x + y) ⁄ (1 − xy) and v = tan− 1 x + tan− 1 y, find ⋅ ∂ (x, y) Are u and v functionally related ? If so, find the relationship. 3. 4. 5. If the functions u, v, w of three independent variables x, y, z are not independent, prove that the Jacobian of u, v, w with respect to x, y, z vanishes. Show that the functions u = 3x + 2y − z, v = x − 2y + z and w = x (x + 2y − z) are not independent and find the relation between them. Show that the functions u = x + y+ z, v = xy + yz + zx , w = x3 + y3 + z3 − 3xyz are not independent. Find the relation between them. 6. 7. 8. 9. (Meerut 2013B) If u = x + 2y + z, v = x − 2y + 3z and w = 2xy − xz + 4yz − 2 z2, show that they are not independent. Find the relation between u, v and w. y+ z , y (x + y + z) , x+ y, v= w= show that u, v, w are If u = x xz z not independent and find the relation between them. If u = x + y + z + t , v = x + y − z − t , w = xy − zt , r = x2 + y2 − z2 − t2, show ∂ (u, v, w, r) = 0 and hence find a relation between u, v, w and r . that ∂ (x, y, z, t ) 1 , If f (0) = 0 and f ′ (x) = prove without using the method of integration, 1 + x2 that ⎛ x+ y⎞ ⎟ ⋅ ⎝ 1 − xy⎠ f (x) + f ( y) = f ⎜ (Meerut 2012B, 13) D-182 1. 5. 8. DIFFERENTIAL CALCULUS v2 = u + 2w. u3 = 3uv + w . uv = r + 2w . 2. u = tan v . 6. u2 − v2 = 4w . 4. u2 − v2 = 8w . 7. uv = w + 1 . Fill in the Blanks: Fill in the blanks “……” so that the following statem ents are com plete and correct. 1. If u and v are functions of two independent variables x and y, then ⎪ ∂u ⎪ ⎪ …⎪ ⎪ ∂ (u, v) ⎪⎪ ∂x ⎪⋅ = ⎪ ⎪ ∂v ∂v ∂ (x, y) ⎪ ⎪ ⎪ ∂x ∂y ⎪ 2. If x = r cos θ, y = r sin θ, then 3. ∂ (u, v) ∂ (x, y) × = …. ∂ (x, y) ∂ (u, v) 4. If x = u (1 + v), y = v (1 + u), then ∂ (r, θ) = …. ∂ (x, y) ∂ (x, y) = …. ∂ (u, v) Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 5. If x = r cos θ, y = r sin θ, then ∂ (x, y) = r (a) ∂ (r, θ) (b) ∂ (x, y) 1 = ∂ (r, θ) r (c) ∂ (x, y) = r2 ∂ (r, θ) (d) ∂ (x, y) 1 = ∂ (r, θ) r 2 D-183 JACOBIANS 6. y2 , v= 2x ∂ (u, v) (a) = ∂ (x, y) ∂ (u, v) (b) = ∂ (x, y) ∂ (u, v) (c) = ∂ (x, y) ∂ (u, v) (d) = ∂ (x, y) If u = x2 + y2 , then 2x y 2x y − 2x 2x y 2x − y True or False: Write ‘T ’ for true and ‘F ’ for false statem ent. x x x x ∂ (u1, u2 , u3) , u = 3 1 , u = 1 2 , then = − 4. 2 3 x1 x2 x3 ∂ (x1, x2 , x3) x2 x3 7. If u1 = 8. If u1, u2 are functions of y1, y2 and y1, y2 are functions of x1, x2 , then ∂ (u1, u2) ∂ (x1, x2) 9. = ∂ (u1, u2) ∂ ( y1, y2) ⋅ ∂ ( y1, y2) ∂ (x1, x2) ⋅ If x = r sin θ cos φ , y = r sin θ sin φ , z = r cos θ, then ∂ (x, y, z) = r 2 sin θ. ∂ (r, θ, φ ) 10. If u, v and w are functions of three independent variables x, y and z , then ∂ (u, v, w) ∂ (x, y, z) ⋅ = 0. ∂ (x, y, z) ∂ (u, v, w) 11. If u1, u2 and u3 are functions of three independent variables x1, x2 and x3 connected by the equations F1 (u1, u2 , u3 , x1, x2 , x3) = 0, F2 (u1, u2 , u3 , x1, x2 , x3) = 0, and F3 (u1, u2 , u3 , x1, x2 , x3) = 0, ∂ (F1, F2 , F3) then ∂ (u1, u2 , u3) ∂ (x1, x2 , x3) = ∂ (x1, x2 , x3) ∂ (F1, F2 , F3) ⋅ ∂ (u1, u2 , u3) 12. T h e fu n c t i o n s u = x2 + y2 + z 2, independent of each other. 13. If x + y + z = u, y + z = v, z = uvw, ∂ (x, y, z) then = u2 v. ∂ (u, v, w) v = x + y+ z, w = xy + yz + zx a r e n o t D-184 14. 1. 5. 9. 13. DIFFERENTIAL CALCULUS If u, v, w are functions of three independent variables x, y, z , then u, v, w are not ∂ (u, v, w) ≠ 0. independent of each other if ∂ (x, y, z) ∂u ⋅ ∂y (a). T. T. 1 ⋅ r 6. (b). 10. F. 14. F. 2. 3. 1. 7. 11. F. F. 4. 1 + u + v. 8. T. 12. T. 9 M axima and M inima of F unctions of T wo I ndependent V ariables D-186 DIFFERENTIAL CALCULUS ∂2 f ∂2 f ⎞ 1 ⎛⎜ 2 ∂2 f + … + 2hk + k 2 2 ⎟⎟ ⎜h 2 ⎜ 2 ! ⎝ ∂x ∂x ∂y ∂y ⎟⎠ x = a, y= b ⎛ ∂f ⎞ ⎛ ∂f ⎞ ∴ f (a + h, b + k ) − f (a, b) = h ⎜ ⎟ x = a + k ⎜ ⎟ x = a ⎝ ∂x ⎠ ⎝ ∂y ⎠ y= b y= b ...(1) + terms of the second and higher orders in h and k . By taking h and k , sufficiently small, the first degree terms in h and k can be made to govern the sign of the right hand side and therefore of the left hand side of (1). Thus the sign of [ f (a + h , b + k) − f (a, b)] ⎡ ⎛ ∂f ⎞ ⎤ ⎛ ∂f ⎞ ⎥ + k⎜ ⎟ ...(2) = the sign of ⎢⎢ h ⎜ ⎟ ⎝ ∂y ⎠ x = a ⎥⎥ ⎢ ⎝ ∂x ⎠ x = a ⎣ ⎦ y= b y= b + ⎛ ∂f ⎞ ≠ 0, the right hand side of (2) changes sign ⎟ ⎝ ∂x ⎠ x = a Taking k = 0, we find that if ⎜ y= b when h changes sign. Therefore f (x, y) cannot have a maximum or minimum at x = a, ⎛ ∂f ⎞ ≠ 0. y = b if ⎜ ⎟ ⎝ ∂x ⎠ x = a y= b Similarly taking h = 0, we can see that f (x, y) cannot have a maximum or a ⎛ ∂f ⎞ minimum at x = a, y = b if ⎜ ⎟ ≠ 0. ⎝ ∂y ⎠ x = a y= b Thus a set of necessary conditions that f (x, y) should have a m axim um or a m inim um at x = a, y = b is that ⎛ ∂f ⎞ ⎛ ∂f ⎞ = 0 and ⎜ ⎟ = 0. ⎜ ⎟ ⎝ ∂x ⎠ x = a ⎝ ∂y ⎠ x = a y= b y= b The above conditions are necessary but not sufficient for the existence of maxima or minima. 9.3 Stationary , Extreme and Saddle Points (Kashi 2014) A point (a, b) is called a stationary point, if both the first order partial derivatives of the function f (x, y) vanish at that point. A stationary point which is either a maximum or a minimum is called an extrem e point and the value of the function at the point is called an extrem e value. A stationary point is not necessarily an extreme point. Thus a stationary point may be a maximum or a minimum or neither of these two. To decide whether a point is really an exreme point, a further investigation is necessary. Saddle Point : A stationary point which is neither a maximum nor a minimum is called a saddle point. MAXIMA AND MINIMA of Functions of Two Independent Variables D-187 9.4 Sufficient Condition for Maxima or Minima (Lucknow 2008) ⎛ ∂2 f Let r = ⎜⎜ 2 ⎜ ∂x ⎝ ⎞ ⎛ ∂2 f ⎞ ⎟ ⎟ ,s= ⎜ and t = ⎟ ⎟ x= a ⎝ ∂x ∂y ⎠ x = a ⎠ y= b y= b ⎛ ∂2 f ⎜ ⎜ 2 ⎜ ∂y ⎝ ⎞ ⎟ ⎟ ⎟ x= a ⎠ y = b. ⎛ ∂f ⎞ ⎛ ∂f ⎞ = 0 and ⎜ ⎟ = 0, i.e. if t he necessary condit ions for t he ⎟ ⎝ ∂x ⎠ x = a ⎝ ∂y ⎠ x = a If ⎜ y= b y= b existence of maxima or minima are satisfied, we have 1 f (a + h , b + k ) − f (a, b) = (rh2 + 2shk + tk 2) + R 3 , 2! where R 3 consists of terms of third and higher orders in h and k . For sufficiently small values of h and k , the sign of 1 2 (rh2 + 2shk + tk 2) + R 3 is the same as that of rh2 + 2shk + t k 2. Now the following three different cases arise : Case I : rt − s2 > 0. In this case obviously neither r nor t can be zero. Therefore we write 1 r h2 + 2 s h k + t k 2 = [r2 h2 + 2 s r h k + r t k 2 ] r 1 = [(rh + sk) 2 + (rt − s2) k 2 ]. r Since r t − s2 is positive, therefore (rh + sk) 2 + (r t − s2) k 2 is positive for all values of h and k except when rh + sk = 0, k = 0 i.e. when h = 0, k = 0 which is obviously not possible. Thus in this case the expression rh2 + 2 s hk + t k 2 will have the sam e sign for all values of h and k . This sign is determined by the sign of r . Thus f (x, y) will have a maximum or a minimum at x = a, y = b if r t > s2. Further f (x, y) is a maximum or a m inim um according as r is negative or positive. Case II : rt − s2 < 0. In this case if r ≠ 0, we can write 1 r h2 + 2 shk + t k 2 = [(rh + sk) 2 + (r t − s2) k 2 ] . r If k = 0, h ≠ 0, the sign of this expression will be the same as that of r . But if k ≠ 0, rh + sk = 0, the sign of this expression will be opposite to that of r since rt − s2 is negative. Thus in this case the expression rh2 + 2shk + tk 2 is not of invariable sign. A similar argument can be given if t ≠ 0. In case r = 0 as well as t = 0, we have rh2 + 2 shk + t k 2 = 2shk , which obviously does not keep the same sign for all values of h and k . T hus f (x, y) will have neither a m axim um nor a m inim um at x = a, y = b, if r t < s2. D-188 of DIFFERENTIAL CALCULUS Case III : rt − s2 = 0. If r ≠ 0, we can write 1 r h2 + 2 shk + t k 2 = [(rh + sk) 2 + (r t − s2) k 2 ] r 1 = (rh + sk) 2 . [ ... r t − s2 = 0] r This expression becomes zero when rh + sk = 0. Therefore the nature of the sign f (a + h, b + k ) − f (a, b) depends upon the consideration of R 3 . The case is, therefore, doubtful. In case r = 0, we must have s = 0, because of the condition r t − s2 = 0. ∴ rh2 + 2shk + t k 2 = t k 2, which is zero when k = 0 whatever h may be. The case is again doubtful. Thus, if r t − s2 = 0, the case is doubtful and further investigation is needed to determ ine whether f (x, y) is a m axim um or a m inim um at x = a, y = b, or not. 9.5 Working Rule for Maxima and Minima Suppose f (x, y) is a given function of x and y. Find ∂f ⁄ ∂x and ∂f ⁄ ∂y and solve the simultaneous equations ∂f ⁄ ∂x = 0 and ∂f ⁄ ∂y = 0. In order to solve these equations we may either eliminate one of the variables, or factorise the equations. In the latter case each factor of the first equation must be solved in conjunction with each factor of the second equation. Suppose solving these equations we get the pairs of values of x and y as (a1, b1), (a2 , b2) etc. Then all these pairs of roots will give stationary values of f (x, y) . To discuss the maximum or minimum at x = a1, y = b1, we should find ⎛ ∂2 u ⎞ ⎛ ∂2 u ⎞ ⎟ r = ⎜⎜ 2 ⎟⎟ ,s= ⎜ ,t= ⎜ ∂x ⎟ x = a 1 ⎝ ∂x ∂y⎠ x = a 1 ⎝ ⎠ y = b1 y = b1 ⎛ ∂2 u ⎞ ⎜ ⎟ ⎜ 2⎟ ⎜ ∂y ⎟ x = a 1 ⎝ ⎠ y = b1 . Then calculate r t − s2. If r t − s2 > 0 and r is negative, f (x, y) is maximum at x = a1 , y = b1. If r t − s2 > 0 and r is positive, f (x, y) is m inim um at x = a1, y = b1. If r t − s2 < 0, f (x, y) is neither maximum nor minimum at x = a1, y = b1 . In this case the function z = f (x , y) is stationary at x = a , y = b but the stationary value f (a , b) is neither maximum nor minimum. H ence, (x = a , y = b , z = (a , b)) is a saddle point of the surface z = f (x , y). If r t − s2 = 0, the case is doubtful and further investigation will be required to decide it. We shall leave this case. Example 1 : Discuss the maximum or m inim um values of u where u = 2 a2 x y − 3ax2 y − ay3 + x3 y + xy3. Solution : We have ∂u ⁄ ∂x = and ∂u ⁄ ∂y = 2 a2 x − 3ax2 2 a2 y − − 3ay2 6axy + 3x2 y + + x3 + 3xy2. (Avadh 2013) y3, MAXIMA AND MINIMA of Functions of Two Independent Variables Also D-189 r = ∂2 u ⁄ ∂x2 = 6ay + 6xy, s = ∂2 u ⁄ ∂x ∂y = 2a2 − 6ax + 3x2 + 3y2, and t = ∂2 u ⁄ ∂y2 = − 6ay + 6xy . For a maximum or minimum of u , we have ∂u ⁄ ∂x = 0 and ∂u ⁄ ∂y = 0. Thus, we have y (2 a2 − 6ax + and 3ax2 − 3ay2 + Therefore we have to consider 3x2 + y2) = 0 ⎫⎪ ⎬⋅ x3 + 3xy2 = 0 ⎪⎭ the pairs of equations, viz., y = 0 ⎫⎪ ⎬ 2 a2 x − 3ax2 − 3ay2 + x3 + 3xy2 = 0 ⎪⎭ 2 a2 x − and 2 a2 x − ...(1) 2 a2 − 6ax + 3x2 + y2 = 0 ⎫⎪ ⎬ 3ax2 − 3ay2 + x3 + 3xy2 = 0 ⎪⎭ ...(2) Putting y = 0 in the second equation of the pair (1), we get 2 a2 x − 3ax2 + x3 = 0 i.e., x (x2 − 3ax + 2a2) = 0 i.e., x (x − a) (x − 2a) = 0 i.e., x = 0, x = a, x = 2a. Thus the pair (1) gives the following values of x and y : x = 0, y = 0; x = a, y = 0; x = 2a, y = 0. Multiplying the first equation of the pair (2) by x and subtracting it from the second equation of the pair, we get 3ax2 − 3ay2 − 2x3 + 2xy2 = 0 or ∴ x= When x = 3 2 3 2 (x2 − y2) (3a − 2x) = 0. a and x = ± y . a , the first equation of the pair (2) gives y = ± 1 2 a. 1 2 When x = y, we have 2 a2 − 6ay + 4y2 = 0 i.e., y = a, a . Also when x = − y, we have 2 a2 + 6ay + 4y2 = 0 i.e., y = − a, − 1 2 a. Thus in all we get the following pairs of values of x and y which make the function u stationary : ⎛ 3 2 ⎞ 1 2 ⎛ 3 2 (0, 0), (a, 0), (2 a, 0), ⎜⎝ a, a⎟⎠ , ⎜⎝ a, − ⎛ 1 2 ⎞ 1 2 ⎛ 1 2 (a, a), ⎜⎝ a, a⎟⎠ , (a, − a), ⎜⎝ a, − 1 2 1 2 ⎞ a⎟⎠ ⎞ a⎟⎠ ⋅ For (0, 0), r = 0, s = 2a2, t = 0 so that r t − s2 is negative. Therefore we have neither a maximum nor a minimum of u at (0, 0). Similarly, we can show that u has neither a maximum nor a minimum at (a, 0), (2 a, 0), (a, a), (a, − a). For (3a ⁄ 2 , a ⁄ 2) , r= 3 2 a2 , s = 1 2 a 2, t = 3 2 a2 so that r t − s2 is positive. Since r is positive, therefore u has minimum at this point. Similarly, we can show that u has a maximum at ⎛ ⎜ ⎝ 1 2 a, − 1 2 ⎞ a⎟⎠ ⋅ D-190 DIFFERENTIAL CALCULUS For (3a ⁄ 2 , − a ⁄ 2) , r= − 3 2 a2 , s = − 1 2 a2 , t = − 3 2 a2 so that r t − s2 is positive. Since r is negative, therefore u has a maximum at this point. Similarly, we can show that u has a maximum at (a ⁄ 2, a ⁄ 2) . Example 2 (a) : Find the extrem e values of xy (a − x − y) . Solution : Let u = xy (a − x − y) . ∂u Then = ay − 2 xy − y2 ∂x ∂u = ax − x2 − 2 x y . and ∂y ∂2 u ∂2 u = − 2 y, s = = a − 2 x − 2 y, Also r= ∂y ∂x ∂x2 ∂2 u = − 2 x. and t= ∂y2 For a maximum or minimum of u , we have ∂u ⁄ ∂x = 0 and ∂u ⁄ ∂y = 0. Thus, we have ay − 2 x y − y2 = 0 ⎫⎪ ⎬⋅ ax − x2 − 2 x y = 0 ⎪⎭ These equations can be written as y (a − 2 x − y) = 0, x (a − x − 2 y) = 0, so that we have to consider the four pairs of equations, viz., y = 0, x = 0; a − 2 x − y = 0, x = 0; y = 0, a − x − 2 y = 0; a − 2 x − y = 0, a − x − 2 y = 0 . Solving these, we get the following pairs of values of x and y which make the function stationary : ⎛ 1 3 1 3 ⎞ (0, 0), (0, a), (a, 0), ⎜⎝ a, a⎟⎠ ⋅ For (0, 0), r = 0, s = a, t = 0 so that r t − s2 is negative. Therefore we have neither a maximum nor a minimum of u at (0, 0) . For (0, a), r = − 2 a, s = − a, t = 0 so that r t − s2 is negative. Therefore u has not an extreme value at (0, a) . Similarly, we may show that u has not an extreme value at (a, 0) . For ⎛ ⎜ ⎝ 1 3 1 3 ⎞ a, a⎟⎠ , r= − 2 3 a, s = − 1 3 a, t = − 2 3 a so that r t − s2 is positive. Therefore u has an extreme value at ⎛ ⎜ ⎝ 1 3 1 3 ⎞ a, a⎟⎠ and it will be a maximum or minimum according as, r is negative or positive i.e., according as, a is positive or negative. ∴ the extreme value of u is (1 ⁄ 27) a3. MAXIMA AND MINIMA of Functions of Two Independent Variables D-191 Example 3 : S h o w th a t th e dista n ce l of any point (x, y, z) on th e plane 2 x + 3y − z = 12 from the origin is given by l = √[x2 + y2 + (2 x + 3y − 12) 2] . Hence find the point on the plane that is nearest to the origin. Solution : I f l i s t h e d i st a n ce fr o m (0, 0, 0) o f a n y p o i n t (x, y, z) , t h e n l = √(x2 + y2 + z2) . I f t h e p o i n t (x, y, z) l i e s o n t h e p l a n e 2 x + 3y − z = 12 , then l = √[x2 + y2 + (2 x + 3y − 12) 2] . [... z = 2 x + 3y − 12 , from the equation of the plane]. ∴ l2 = x2 + y2 + (2 x + 3y − 12) 2 = 5x2 + 10 y2 + 12 x y − 48x − 72 y + 144 = u , say. Now l is maximum or minimum according as l2 i.e., u is maximum or minimum. For a maximum or minimum of u , we have ∂u = 10x + 12 y − 48 = 0 , ∂x ∂u and = 20y + 12 x − 72 = 0 . ∂y Solving these equations, we get x = 12 ⁄ 7 , and y = 18 ⁄ 7 . Also and r= t= ∂2 u ∂2 u = 10 , s = = 12 , 2 ∂x ∂y ∂x ∂2 u = 20 . ∂y2 ∴ r t − s2 = 10 × 20 − 122 = + ive. Since r t − s2 > 0 and r > 0 , therefore u is minimum and hence l is minimum when x = 12 ⁄ 7 and y = 18 ⁄ 7 . Putting these values of x and y in the equation of the plane, we get z = 2 .(12 ⁄ 7) + 3 . (18 ⁄ 7) − 12 = − 6 ⁄ 7 . Therefore the required point is (12 ⁄ 7, 18 ⁄ 7, − 6 ⁄ 7) . Example 4 : L ocate the stationary points of x4 + y4 − 2 x2 + 4 x y − 2 y2 a n d determ ine their nature. Solution : Let u = x4 + y4 − 2 x2 + 4 x y − 2 y2. ∂u Then = 4 x3 − 4 x + 4y ∂x ∂u and = 4y3 + 4 x − 4y . ∂y The stationary points are given by ∂u = 0 i.e., 4 x3 − 4 x + 4y = 0 , ...(1) ∂x ∂u and = 0 i.e., 4y3 + 4 x − 4y = 0 . ...(2) ∂y Now we shall find the points (x, y) satisfying the simultaneous equations (1) and (2). Adding (1) and (2), we get 4 x3 + 4y3 = 0 i.e., x3 + y3 = 0 D-192 i.e., ∴ or i.e., DIFFERENTIAL CALCULUS (x + y) (x2 − xy + y2) = 0 . either x + y = 0 , ...(3) x2 ...(4) y2 − xy + = 0. First we solve the simultaneous equations (1) and (3). From (3), we have y = − x . Putting y = − x in (1), we get 4 x3 − 8x = 0 i.e., x3 − 2 x = 0 i.e., x (x2 − 2) = 0 x = 0 , √2 or − √2 . The corresponding values of y are y = 0 , − √2 , √2 . Thus the points (0, 0) , (√2, − √2) and (− √2, √2) satisfy (1) and (2). If we solve the equations (1) and (4), we get (0, 0) as the only real solution. H ence the function u is stationary at the points (0, 0) , (√2, − √2) , (− √2, √2) . ∂2 u ∂2 u ∂2 u = 12 x2 − 4 , s = = 4, t = = 12 y2 − 4 . 2 ∂x ∂y ∂x ∂y2 At (0, 0) , r = − 4 , s = 4 , t = − 4 , so that r t − s2 = 16 − 16 = 0. Thus at the point (0, 0), the case is doubtful and further investigation is needed. At (√2, − √2), r = 20, s = 4, t = 20, so that r t − s2 = 400 − 16 = + ive. Therefore u has an extreme value at this point. Since r is positive, therefore u has a minimum at this point. At (− √2, √2) , r = 20 , s = 4 , t = 20 , so that rt − s2 is positive. Since r is positive, therefore u has a minimum at this point also. Note : We may tackle the doubtful case at the point (0, 0) by the following consideration : We have u = x4 + y4 − 2 (x − y) 2 . At the point (0, 0) , we have u = 0 . At the points in the neighbourhood of the point (0, 0) where x ≠ y , the value of u is approximately given by We have r = u = − 2 (x − y) 2 , [Neglecting the terms x4 + y4 because the numerical values of x and y are small]. Thus at such points u is –ive. Again at the points in the neighbourhood of the point (0, 0) , where x = y , we have u = x4 + y4 which is positive. Thus in the neighbourhood of the point (0, 0) , there are points at which u takes values less than its value at the point (0, 0) and there are points at which u takes values greater than its value at the point (0, 0) . H ence u cannot have a maximum or a minimum value at the point (0, 0) . Example 5 : Find the m inim um value of x2 + y2 + z2 when ax + by + cz = p . (Kashi 2014) Solution : Let u = x2 + y2 + z2 . H ere u is a function of three variables x, y and z . But we can eliminate one variable with the help of the given relation, viz., a x + by + cz = p . p − ax − by ⋅ From this relation, we have z = c MAXIMA AND MINIMA of Functions of Two Independent Variables D-193 Putting this value of z in the value of u , we get (p − ax − by) 2 , u = x2 + y2 + c2 where u has been expressed as a function of two independent variables x and y . ∂u 2a = 2 x − 2 ( p − ax − by) , We have ∂x c ∂u 2b and = 2 y − 2 ( p − ax − by) . ∂y c ∂u ∂u = 0 and = 0 , we get Solving ∂y ∂x ap bp , and y = x= 2 ⋅ 2 2 2 a + b + c a + b2 + c2 Again, we get r = ∂2 u 2 b2 = 2+ 2 ⋅ 2 ∂y c t= and ∴ ∂2 u ∂2 u 2 a2 2 ab = 2+ 2 ,s= = 2 , 2 ∂x ∂y ∂x c c ⎛ ⎛ a2 ⎞ ⎛ b2 ⎞ 4 a 2 b2 a 2 b2 ⎞ r t − s2 = 4 ⎜⎜ 1 + 2 ⎟⎟ ⎜⎜ 1 + 2 ⎟⎟ − = 4 ⎜⎜ 1 + 2 + 2 ⎟⎟ ⋅ 4 ⎜ ⎜ c ⎟⎠ ⎜⎝ c ⎟⎠ c c c ⎟⎠ ⎝ ⎝ Since r t − s2 is positive and r is also positive, therefore u is minimum for the values of x and y found above. p2 The minimum value of u , therefore, is 2 ⋅ (a + b2 + c2) 1. Discuss the maxima and minima of the following functions : (i) u = xy + a3 (1 ⁄ x + 1 ⁄ y) . 2. (ii) u = x3 y2 (1 − x − y) . (iii) u = x3 + y3 − 3axy . (iv) u = x2 + y2 + 6x + 12 . (v) u = x2 + y2 + 2 ⁄ x + 2 ⁄ y. (vi) u = y2 (i) u = 3x2 (ii) u = 2 x2 y + 3x2 + 4 xy + − (iii) u = 2 sin y2 1 2 + x2 + (Meerut 2003, 13; Kashi 11; Avadh 11; Purvanchal 14) (Bundelkhand 2011, 14; Rohilkhand 12; Avadh 12) (Meerut 2003, 12; Kanpur 07, 08) − (Meerut 2012B) x3. x3. y2 (Kashi 2013) (Meerut 2013B) + 2 y. 1 2 (x + y) cos (x − y) + cos (x + y) . (iv) u = sin x sin y sin (x + y) . (Bundelkhand 2011; Meerut 12B) [Hint : It is sufficient to consider the values of x and y between 0 and π since the function u is periodic with period π both for x and y ] (v) u = x2 y2 − 5x2 − 8 x y − 5y2. (Avadh 2006) D-194 DIFFERENTIAL CALCULUS (vi) u = x4 + 2 x2 y − x2 + 3y2. 3. 4. If u = a x3 y2 − x4 y2 − x3 y3, prove that x = a ⁄ 2 , y = a ⁄ 3 make u a maximum. Investigate the maxima and minima of 5. 2 (x − y) 2 − x4 − y4 , leaving aside any doubtful case that may arise. E xamine the following surface for high and low points : 6. 7. 8. 9. 10. 11. 1. (Kanpur 2009) z = x2 + xy + 3x + 2y + 5 . Find the saddle points of the surface if there are any. Find all the stationary points of the function x3 + 3x y2− 15x2 − 15y2 + 72 x , examining whether they are maxima or minima. E xamine for maximum and minimum values of the function z = x2 − 3x y + y2 + 2 x . Find the points (x, y) where the function xy (1 − x − y) is maximum or minimum. What is the maximum value of the function ? E xamine the function z = x2 y − y2 x − x + y for maxima and minima. Find the saddle points of the given surface. L e t f (x, y) = x2 − 2 x y + y2 + x3 − y3 + x5. S h o w t h a t f (x, y) h a s n e i t h e r a maximum nor a minimum at (0, 0) . Find a point within a triangle such that the sum of the squares of its distances from the three vertices is a minimum. (Kanpur 2010; Kumaun 08) (i) Minimum at x = y = a . (ii) Maximum at x = 1 2 ,y= 1⋅ 3 (iii) x = y = a gives a maximum or a minimum according as a is negative or positive. (iv) Minimum at (− 3, 0). (v) Minimum at (1, 1). 2, 4 (vi) Minimum at x = y= − ⋅ 3 2. 4. 5. 6. 7. 8. 3 (ii) No exreme value. (i) Maximum at (− 2, 0) . (iii) Maximum when x = y = n π + (− 1) n π ⁄ 6 and minimum when x = y = 2 n π − π ⁄ 2 . (iv) Maximum at x = y = π ⁄ 3 and minimum at x = y = 2 π ⁄ 3 . (v) Maximum at x = y = 0. (vi) Minimum when x = ± √3 ⁄ 2__, y = − 1 ⁄ 4 . Maximum when x = ± √2 , y = + √2 . No high and low points. The point (2, 1, 3) is a saddle point. Maximum at (4, 0) and minimum at (6, 0). The function z is stationary at the point (4 ⁄ 5, 6 ⁄ 5). But it is neither maximum nor minimum at this point. Maximum at the point ⎛ ⎜ ⎝ 1 3 ⎞ , 1 ⎟ ⋅ Max. value = 1 ⁄ 27 . ⎠ 3 MAXIMA AND MINIMA of Functions of Two Independent Variables 9. 11. D-195 The function is stationary at the points (1, 1) and (− 1, − 1), but it has no exreme value. The points (1, 1, 0) and (− 1, − 1, 0) are the two saddle points of the surface z = x2 y − y2 x − x + y. Centroid of the triangle. Fill in the Blanks: 1. 2. Fill in the blanks “……” so that the following statem ents are com plete and correct. Let f (x, y) be a function of two independent variables x and y. The necessary co ndit io n s fo r t he e xist e nce o f a m a xim u m o r a minimu m o f f (x, y) a t x = a, y = b are ∂f ∂f = 0, and = … at x = a, y = b. ∂y ∂x Let f (x, y) be a function of two independent variables x and y. If at the point (a, b), we have 2 ∂f ∂2 f ∂2 f ⎛ ∂2 f ⎞⎟ ∂f = 0, = 0, 2 ⋅ 2 − ⎜⎝ > 0 ∂y ∂x ∂y⎠ ∂x ∂x ∂y ∂2 f > 0, then f (x, y) is … at (a, b). ∂x2 Let f (x, y) be a function of two independent variables x and y. Let ∂2 f ∂2 f ∂2 f , s = and t = ⋅ r= ∂x ∂y ∂x2 ∂y2 If at the point (a, b), we have ∂f ∂f = 0, = 0, rt − s 2 > 0 and r < 0, then f (x, y) is … at (a, b). ∂x ∂y and 3. Multiple Choice Questions: 4. 5. 6. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). Let f (x, y) be a function of two independent variables x and y. Let ∂2 f ∂2 f ∂2 f , s = and t = ⋅ r= ∂x ∂y ∂x2 ∂y2 ∂f ∂f = 0 and = 0, then f (x, y) is maximum at (a, b) If at the point (a, b), we have ∂y ∂x if at (a, b) we have (b) rt − s 2 > 0 and r < 0 (a) rt − s 2 < 0 2 (c) rt − s > 0 and r > 0 (d) rt − s 2 < 0 and r < 0 The function u = x2 + y2 + 6x + 12 is minimum at (a) (3, 0) (b) (0, 3) (c) (− 3, 0) (d) (3, 3) The function u = 3x2 − y2 + x3 is maximum at (b) (2 , 0) (a) (− 2 , 0) (c) (0, − 2) (d) (0, 2) D-196 DIFFERENTIAL CALCULUS True or False: 7. Write ‘T ’ for true and ‘F ’ for false statem ent. Let f (x, y) be a function of two independent variables x and y. Let ∂2 f , ∂2 f ∂2 f s = and t = ⋅ ∂x ∂y ∂x2 ∂y2 ∂f ∂f = 0, = 0 and rt − s 2 < 0, then f (x, y) has an If at the point (a, b), we have ∂y ∂x extreme value at (a, b). Let f (x, y) be a function of two independent variables x and y. If at (a, b), we have ∂f ∂f = 0 and = 0, then f (x, y) must have a maximum or a minimum at (a, b). ∂y ∂x p2 The minimum value of x2 + y2 + z 2 when ax + by + c z = p is 2 ⋅ (a + b2 + c 2) r= 8. 9. ⎛ ⎜ ⎝ 1 2 ⎞ , 1⎟ ⋅ 10. The function xy (1 − x − y) has a maximum value at the point 11. 12. The function x3 + y3 + 3xy has a maximum value at the point (− 1, − 1). The minimum value of the function x3 + y3 − 6 x y is − 8. 1. 4. 7. 10. 0. (b). F. F. 2. 5. 8. 11. minimum. (c). F. T. 3. 6. 9. 12. 2⎠ maximum. (a). T. T. 10.1 Tangent and Normal to a Curve Tangent to a Curve : Let P be any given point on a curve and Q any other point on it in the neighbourhood of P. The point Q may be taken on either side of P. As Q tends to P, the straight line PQ, in general, tends to a definite straight line TP passing through P. This straight line is called the tangent to the curve at the point P. (Kanpur 2014) Normal to a Curve : The normal to a curve at any point P of it is the straight line which passes through that point and is perpendicular to the tangent to the curve at that point. 10.2 Equation of the Tangent (Cartesian Co-ordinates) Let y = f (x) be the cartesian equation of a curve. Let P be any given point (x, y) on this curve. Take a point Q (x + δx, y + δy) on this curve in the neighbourhood of P. D-198 DIFFERENTIAL CALCULUS If (X, Y ) are current co-ordinates of any point on the chord PQ, then the equation of the chord PQ is (y + δy) − y δy Y − y= (X − x) or Y − y= (X − x). …(1) (x + δx) − x δx Now, as Q tends to P, δx → 0 and chord PQ tends to the tangent at P. Therefore the equation (1) tends to the equation δy dy⎤ ⎡. . dy lim (X − x). = Y − y= ⎢ . δx → 0 ⎥. ⎣ δx dx ⎦ dx Hence the equation of the tangent to the curve y = f (x) at the point (x, y) is dy (X − x). Y− y= dx Note 1 : If we are to find the equation of the tangent to the curve y = f (x) at the point (x1, y1) on it, we should first find the value of dy ⁄ dx of the curve at the point (x1, y1). The equation of the tangent at the point (x1, y1) will then be ⎛ dy⎞ (x − x1), ⎟ ⎝ dx⎠ (x , y ) 1 1 y − y1 = ⎜ where (x, y) are the current co-ordinates of any point on the tangent. Note 2 : If the equations of the curve be given in parametric cartesian form say x = f ( t ) and y = φ ( t ), then dy dy ⁄ dt φ ′ ( t ) = = ⋅ dx dx ⁄ dt f′ (t) H ence the equation of the tangent at any point ‘t’ on the curve is given by φ′ ( t ) Y− φ (t)= [X − f ( t )]. f ′ (t ) 10.3 Geometrical Meaning of dy ⁄ dx Let P be any given point (x, y) on the curve y = f (x). Suppose the positive direction of the tangent at P is that in which x increases. Let ψ be the angle which the positive direction of the tangent at P makes with the positive direction of the axis of x. The equation of the tangent at P is dy Y − y= (X − x) dx ⎛ dy⎞ ⎛ dy⎞ or Y = ⎜ ⎟ X + ⎜y− x ⎟ ⋅ …(1) dx⎠ ⎝ dx⎠ ⎝ This equation is of the form Y = mX + c, …(2) which is the equation of the straight line whose gradient is m i.e., the line makes an angle, with the positive direction of x-axis, whose tangent is m . Therefore comparing (1) and (2), we get dy = tan ψ. dx Thus the differential coefficient dy ⁄ dx at any point (x, y) on the curve y = f (x) is equal to the tangent of the angle which the positive direction of the tangent at P to the curve m akes with the positive direction of the axis of x. TANGENTS AND NORMALS D-199 10.4 Tangents Parallel to the Co-ordinate Axes If the tangent at any point is parallel to the axis of x, then ψ = 0 i.e., tan ψ = 0 and so we have dy ⁄ dx = 0 at that point. O n the other hand if a tangent is parallel to the axis of y or perpendicular to the axis of x, then ψ = π ⁄ 2 i.e., tan ψ = tan (π ⁄ 2) = ∞ and so we have dy ⁄ dx = ∞ or dx ⁄ dy = 0 at that point. 10.5 Equation of the Normal Let P be any given point (x, y) on the curve y = f (x). The equation of the tangent at P is ⎛ ⎛ dy⎞ dy⎞ dy (X − x) or Y = ⎜ ⎟ X + ⎜ y − x ⎟ ⋅ Y − y= dx ⎠ dx ⎝ ⎝ dx ⎠ Therefore the gradient of the tangent at P is dy ⁄ dx. If m be the gradient of the normal at P, then dy 1 dx m. = − 1 or m = − = − ⋅ dx dy ⁄ dx dy H ence the equation of the norm al to the curve at P is dx dy Y− y= − (X − x) or (Y − y) + (X − x) = 0. dy dx Important : If the equation of a curve is given in the form f (x, y) = 0, then ∂f ⁄ ∂x dy = − ⋅ ∂f ⁄ ∂y dx Example 1 : Find the equations of the tangent and the norm al at any point (x, y) of ym xm the curve m + m = 1. a b ym xm Solution : Let f (x, y) ≡ m + m − 1 = 0. a b ∂f ∂f my m − 1 mx m − 1 Then = and = ⋅ ∂x ∂y am bm ∂f ⁄ ∂x b m xm − 1 dy = − = − m m − 1. ∴ ∂f ⁄ ∂y dx a y Hence the equation of the tangent at (x, y) is b m xm − 1 (X − x) am y m − 1 ym − 1 xm − 1 (Y − y) = − (X − x) m b am ym Y ym − 1 X xm − 1 xm + = m+ m⋅ m m b a a b Y − y= − i.e., i.e., D-200 DIFFERENTIAL CALCULUS But the point (x, y) lies on the given curve. ym xm + m = 1. Therefore m a b X xm − 1 Y ym − 1 + = 1. am bm dx (X− x) Also, the equation of the normal at (x, y) is Y − y = − dy X− x Y− y a m ym − 1 i.e., Y − y = m m − 1 (X − x) i.e., = . b m xm − 1 a m ym − 1 b x Example 2 : Find the equation of the tangent at the point ‘t’ to the cycloid x = a (t + sin t ) , y = a (1 − cos t ) . (Agra 2014) Solution : We have dx ⁄ dt = a (1 + cos t), dy ⁄ dt = a sin t. 2 sin t ⁄ 2 cos t ⁄ 2 dy dy ⁄ dt a sin t t ∴ = = = = tan ⋅ 2 dx dx ⁄ dt a (1 + cos t ) 2 2 cos t ⁄ 2 H ence the equation of the tangent at the point ‘t’ is t y − a (1 − cos t ) = tan [ x − a (t + sin t )] 2 t t t i.e., y − 2a sin2 = (x − at ) tan − a sin t. tan 2 2 2 t t t i.e., y − 2a sin2 = (x − at ) tan − 2a sin2 2 2 2 t i.e., y = (x − at ) tan , 2 where (x, y) are the current co-ordinates of any point on the tangent. H ence the equation of the tangent at (x, y) is Note : If t h e e qu a t io n s o f a cu r ve a r e give n in t h e p a r a m e t r ic fo r m x = f (t ), y = φ (t ), then by the point t we mean the point whose co-ordinates are x = f ( t ) and y = φ ( t ). 10.6 Angle of Intersection The angle of intersection of two curves is defined as the angle between their tangents at their point of intersection. In order to determine the angles of intersection of two given curves f (x, y) = 0, …(1) and φ (x, y) = 0 …(2) we should first solve the equations (1) and (2) simultaneously to get the points of intersection of (1) and (2). If (x1, y1) is one of the points of intersection, then to find the angle of intersection at (x1, y1), we should find the slopes m 1 and m 2 of the tangents of the two curves at the point (x1, y1). ⎛ dy⎞ We have, m 1 = ⎜ ⎟ at (x1, y1) of the curve (1) ⎝ dx⎠ and ⎛ dy⎞ m 2 = ⎜ ⎟ at (x1, y1) of the curve (2). ⎝ dx⎠ D-201 TANGENTS AND NORMALS If m 1 = m 2 , the angle of intersection is 0o . If m 1 = ∞ , m 2 = 0, the angle of intersection is 90o . If m 1 m 2 = − 1, again the angle of intersection is 90o and we say that the two curves intersect orthogonally. In all other cases, the acute angle between the tangents is equal to ⎪ m − m ⎪ 1 2 ⎪ tan− 1 ⎪⎪ ⎪. ⎪ 1 + m 1 m 2⎪ ⎪ ⎪ 10.7 Length of Cartesian Tangent, Normal, Subtangent and Subnormal The lentgh of the tangent of a curve at one of its points is defined as the length of the portion of the tangent between its point of contact and the x-axis. The length of the projection of this segment on the x-axis is called the length of the subtangent. Similarly the length of the norm al is defined a s t he lengt h of t he p ort ion of t he n orma l between the point of contact of the tangent and the x-axis. The length of the projection of this segment on the x-axis is called the length of the subnormal. Let P be any point (x, y) on the curve y = f (x). Suppose tangent and the normal at P meet the x-axis in T and N respectively. Let PS be the ordinate of the point P. Then PS = y. I f ψ b e t h e a n g l e wh i c h t h e t a n g e n t a t P m a k e s wi t h x-a xis, t hen ∠ PTS = ∠ SPN = ψ and tan ψ = dy ⁄ dx. Length of tangent = PT = y cosec ψ = y√(1 + cot2 ψ) = y Length of sub-tangent = TS = y cot ψ = y √ 1+ ⎧ ⎨ ⎩ ⎛ ⎜ ⎝ dx⎞⎟ 2 ⎫⎬ . dy⎠ ⎭ y dx = ⋅ dy dy ⁄ dx Length of normal = PN = y sec ψ = y √ (1 + tan2 ψ) = y Length of subnormal √ 1+ ⎧ ⎨ ⎩ ⎛ dy⎞ 2⎫⎬ ⎜ ⎟ ⎭. ⎝ dx⎠ dy ⋅ dx Intercepts made by the tangent on the coordinate axes. dy (X − x). The equation of the tangent at P (x, y) is Y − y = dx = SN = y tan ψ = y D-202 DIFFERENTIAL CALCULUS This meets OX , where Y = 0 i.e., where y dy (X − x) or X = x − ⋅ 0 − y= dy ⁄ dx dx H ence the length of the intercept OT that the tangent cuts off from the x-axis is y x− ⋅ (dy ⁄ dx) Again the tangent meets y-axis, where X = 0, i.e., where dy dy Y − y= (0 − x) or Y = y − x ⋅ dx dx dy H ence the intercept OT′, made by the tangent on y-axis is y − x ⋅ dx 10.8 Polar Co-ordinates Besides the cartesian system of co-ordinates, there are other systems also for representing the position of a point in a plane. Polar system, which is one of them, will be described here. In polar system, we start with a fixed half line OX, called the initial line and a fixed point O on it, called the pole. If P is any point in the plane, the distance OP= r is called the radius vector of the point P and ∠ XOP = θ is called the vectorial angle of P. Also (r, θ) are called polar coordinates of the point P. The line OP is called the revolving line. For any point P (r, θ) the angle θ is taken to be positive when measured in the anti-clockwise direction from the initial line and negative when measured in the clockwise direction from the initial line. The radius vector r is considered to be positive when measured away from O in the direction of the line governing the vectorial angle θ. If for any point P (r, θ), r is negative, we first draw through O a line making an angle θ with the initial line. Producing this line backwards through O, we mark a point P on it such that OP = ⏐ r⏐ ; P is then the required point (r, θ). Thus in polar coordinates both r and θ are capable of varying in the interval (− ∞ , ∞ ). To each ordered pair (r, θ) of real numbers there corresponds one and only one point. But the converse is not true. For example, in the figure the point P whose polar coordinates are (r, θ) may also be described as (r , θ + 2π), (− r, θ + π) etc. In particular, the polar coordinates of pole may be given as (0, θ) where θ is perfectly arbitrary. It is usual to regard θ as the independent va r ia bl e a n d t h e cu r ve wh o se e qu a t io n is r = f (θ) or F (r, θ) = 0 consists of the totality of distinct points (r, θ) which satisfy the equation. Positive and Negative radii vectors : If we measu re t he dist an ce r alo n g th e revolving line in the direction in which the line projects from the pole O then the radius vector r is + ive and if it is measured in the opposite direction then the radius vector r is negative. TANGENTS AND NORMALS D-203 If the distance r is measured along OP in the direction of OP and OP1 = r, then P1 will be called (r, θ). But if r is measured in the opposite direction, then the point P3 obtained is called (− r, θ). H ere the line OP3 can be said to make an angle π + θ with the initial line. In this case the distance r measured along OP till P3 will be called + ive and point P3 will be said to be the point (r, π + θ). If now we measure a distance r in the opposite direction of OP3, then we get the point P1 which we shall call (− r, π + θ). Positive and Negative Vectorial Angles : If the revolving line makes an angle θ in the anticlockwise direction with the initial line, then the vectorial angle is said to be positive and if it makes an angle θ in the clockwise direction, then the vectorial angle is said to be negative. Thus, if OP4 = r, then the point P4 is said to be (r, − θ). Similarly if OP2 = r, then the point P2 is said to be (− r, − θ). Relation between Cartesian and polar coordinates : Take the pole O as origin, the initial line as the positive direction of x-axis, and π with OX in the anti-clockwise direction as the the line through O making angle 2 positive direction of y-axis. Suppose (r, θ) are the polar and (x, y) are the cartesian coordinates of any point P. D raw PM perpendicular to OX . Then OM = x and MP = y. From Δ OPM, we have x = r cos θ, …(1) y = r sin θ. …(2) Squaring and adding (1) and (2), we get x2 + y2 = r2 and dividing (2) by (1), we get y y tan θ = or θ = tan− 1 ⋅ x x Exercise : Plot the positions of the points whose polar coordinates are π⎞ ⎛ π⎞ ⎛ π⎞ ⎛ 3 π⎞ ⎛ π⎞ ⎛ ⎟ , (1, − π), (1, π) . ⎜ − 2, ⎟ , ⎜ 2, − ⎟ , (3, π), (1, 0), ⎜ 2, ⎟ , ⎜ 2, − ⎟ , ⎜ 1, 2 4 4⎠ ⎝ 4 ⎠ 2 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎝ 10.9 Angle Between Radius Vector and Tangent Let P be any point (r , θ) on the curve r = f (θ). The line T P T′ is tangent to this curve at P. We denote by φ the angle which the positive direction of the tangent at P (the direction in which θ increases) makes with the positive direction of the radius vector OP (the direction in which r increases). Let Q be any other point (r + δr, θ + δ θ) on the curve in the neighbourhood of P. Draw QM perpendicular to OP. As Q → P, we have δθ → 0, the chord PQ →tangent at P and the angle QPM → φ . QM Lim Lim Lim QM = δθ → 0 Thus tan φ = δθ → 0 tan ∠ QPM = δθ → 0 OM − OP PM D-204 DIFFERENTIAL CALCULUS (r + δr) sin δθ Lim = δθ → 0 (r + δr) cos δθ − r ⎛ ⎞ δθ3 (r + δr) ⎜ δθ − + …⎟ 3! ⎝ ⎠ Lim Lim r δθ , = δθ → 0 = δθ → 0 2 ⎞ ⎛ δr δθ + …⎟ − r (r + δr) ⎜ 1 − 2! ⎠ ⎝ neglecting small quantities of the second and higher order dθ = r ⋅ dr dθ Hence, tan φ = r (Meerut 1990, 96) dr 1 dr or cot φ = ⋅ r dθ Note 1 : The angle φ is taken to be positive if it is measured in the anti-clockwise direction. Note 2 : From the figure, we have an important relation ψ = θ + φ . Note 3 : If the equation of a curve is given in the form r = f (θ) and we are to find the value of φ , then differentiating with respect to θ after taking logarithm of both sides, we shall at once get cot φ . 10.10 Angle of Intersection of Two Curves Let r = f (θ) and r = F (θ) be the polar equations of two curves and P be one of their points of intersection. The two curves have a common radius vector at P. Suppose φ 1 is the angle which the tangent to the first curve at P makes with the radius vector of P and φ 2 is the angle which the tangent to the second curve at P makes with the radius vector of P. Then the acute angle of intersection of the two curves at P is obviously = φ 1~ φ 2 i.e., ⏐ φ 1 − φ 2⏐ . If tan φ 1 = n1 and tan φ 2 = n2, then the angle of intersection is ⎛ tan φ − tan φ ⎞ n1 − n2 1 2 ⎟ . φ 1 − φ 2 = tan− 1 tan (φ 1 − φ 2) = tan− 1 ⎜⎜ ⎟ = tan− 1 1 + tan φ tan φ 1 + n1 n2 ⎜ 1 2 ⎟⎠ ⎝ n1 − n2 If is positive, we shall get acute angle of intersection at P and if 1 + n1 n2 n1 − n2 is negative we shall get the obtuse angle of intersection at P. 1 + n1 n2 In p a r t icu la r , t h e t wo cu r ves in t e r sect o rt h o go n a lly if n1 n2 = − 1, i.e., tan φ 1 . tan φ 2 = − 1. Example 1 : Show that the parabolas r = a ⁄ (1 + cos θ) and r = b ⁄ (1 − cos θ) (Meerut 2011; Avadh 13; Rohilkhand 14) intersect orthogonally. D-205 TANGENTS AND NORMALS Solution : The equations of the curves are …(1) r = a ⁄ (1 + cos θ) and r = b ⁄ (1 − cos θ). …(2) Taking logarithm of both sides of (1), we get log r = log a − log (1 + cos θ). Differentiating with respect to θ, we get 1 1 2 sin θ cos θ (− sin θ) 1 dr 2 2 1 = − = = tan θ. 1 2 2 r dθ 1 + cos θ 2 cos θ 2 ∴ cot φ H ence φ1 = 1 = tan θ = 2 1 1 π − θ. 2 2 1 ⎞ ⎛1 cot ⎜ π − θ⎟ 2 ⎠ ⎝2 or φ= 1 2 1 2 π− θ. Again taking logarithm of both sides of (2), we get log r = log b − log (1 − cos θ). Differentiating with respect to θ, we get 1 1 2 sin θ cos θ sin θ 1 dr 2 2 1 = − = − = − cot θ. 1 2 1 − cos θ r dθ 2 sin2 θ 2 ∴ H ence ⎛ 1 ⎞ cot φ = − cot θ = cot ⎜ π − θ⎟ 2 ⎠ ⎝ 1 φ 2 = π − 2 θ. 1 2 or φ= π− 1 2 θ. 1 ⎞ 1 ⎞ 1 ⎛ ⎛1 Therefore, angle of intersection = φ 1 ~ φ 2 = ⎜⎝ π − 2 θ⎟⎠ − ⎜⎝ 2 π − 2 θ⎟⎠ = 2 π . Thus the two curves intersect orthogonally. r2 Example 2 : Find the angle of intersection of the curves r2 = 16 sin 2 θ and sin 2 θ = 4. Solution : The given curves are r2 = 16 sin 2 θ, …(1) and …(2) r2 sin 2 θ = 4. From (1), 2 log r = log 16 + log sin 2θ . cos 2θ 2 dr Therefore = 2 or cot φ 1 = (1 ⁄ r) (dr ⁄ dθ) = cot 2 θ. Thus φ 1 = 2θ . r dθ sin 2θ From (2), 2 log r + log sin 2 θ = log 4. 1 dr 2 dr 2 cos 2θ + = 0 or cot φ 2 = = − cot 2 θ = cot (π − 2 θ) . Therefore sin 2θ r dθ r dθ Thus φ2 = π − 2 θ . Now the angle of intersection of (1) and (2) = φ 1 ~ φ 2 = (π − 2θ) − 2θ = π − 4 θ, where θ is to be found at the point where (1) and (2) intersect. E liminating r between (1) and (2), we get sin2 2θ = sin 2 θ = ± 1 2 ⋅ 1 4 ⋅ Therefore D-206 But sin 2 θ = − and (2). Now sin 2 θ = DIFFERENTIAL CALCULUS 1 2 is inadmissible because it gives imaginary values of r from (1) 1 2 gives 2 θ = 1 6 π or θ = π ⁄ 12. H en ce, t h e an gle o f in t ersect io n o f (1) an d (2) at t he p o in t θ = π ⁄ 12 i s π − 4 (π ⁄ 12) i.e., 2π ⁄ 3 . 2. Show that in the equiangular spiral r = ae θ cot α the tangent is inclined at a constant angle α to the radius vector. (Avadh 2014) Find the angle at which the radius vector cuts the curve l ⁄ r = 1 + e cos θ. 3. Find the angle φ for the curve aθ = (r2 − a2) 1 ⁄ 2 − a cos− 1 (a ⁄ r). 4. If φ be the angle between tangent to a curve and the radius vector drawn from the origin of coordinates to the point of contact, prove that ⎛ dy ⎞ ⎛ dy⎞ tan φ = ⎜ x − y⎟ / ⎜ x + y ⎟ ⋅ dx⎠ ⎝ dx ⎠ ⎝ y⎤ dy ⎡ and tan θ = ⎥ . ⎢ Hint . We have ψ = θ + φ , tan ψ = dx x⎦ ⎣ 5. Prove 6. Show that the spirals r n = an cos nθ and r n = bn sin nθ intersect orthogonally. 7. Show that the cardioids r = a(1 + cos θ) and r = b (1 − cos θ) intersect at right (Meerut 2000; Kanpur 07, 11) angles. Show that the curves r = a (1 + sin θ) and r = a (1 − sin θ) cut orthogonally. Find the angle of intersection of the curves r = sin θ + cos θ and r = 2 sin θ. Show that the curves r = 2 sin θ and r = 2 cos θ intersect at right angles. Find the angle between the tangent and the radius vector in the case of the curve r n = an sec (nθ + α), an d p ro ve t h at t h is cu rve is in t ersect ed by t h e cu rve r n = bn sec (n θ + β) at an angle which is independent of a and b. Find the angle of intersection between the pair of curves r = 6 cos θ and r = 2 (1 + cos θ). 1. (Rohilkhand 2013) 2 ⎛ du ⎞ 1 1 = u2 + ⎜⎝ ⎟⎠ , where u = and p is the lenght of perpendicular form 2 θ r d p pole to the tangent of the curve at any point p (r , θ). (Bundelkhand 2001) (Kumaun 2008) 8. 9. 10. 11. 12. 2. 9. 12. tan− 1 {(1 + e cos θ) ⁄ (e sin θ)}. π ⁄ 4. (π ⁄ 6). 3. 11. cos− 1 (a ⁄ r). (π ⁄ 2) − (nθ + α). D-207 TANGENTS AND NORMALS 10.11 Lengths of Polar Sub-tangent and Polar Sub-normal (Kashi 2013) Let P be any point (r, θ) on a given curve. Suppose the tangent and normal at P meet the straight line through the pole O perpendicular to the radius vector OP in T and N respectively. Then OT and ON are respectively called the polar subtangent and the polar sub-normal at P. We have ∠ OPT = φ and ∠ ONP = φ . dθ dθ From Δ PTO, OT = OP tan φ = r . r = r2 ⋅ dr dr dθ ⋅ H ence Polar sub-tangent = r2 dr 1 dr dr Again from Δ PON, ON = OP cot φ = r ⋅ = ⋅ r dθ dθ dr ⋅ H ence Polar sub-normal = dθ Example : Show that in the curve r = aθ the polar sub-normal is constant and in the curve r θ = a the polar sub-tangent is constant. Solution : From the curve r = aθ, we have dr ⁄ dθ = a. ∴ Polar sub-normal = dr ⁄ dθ = a, which is a constant i.e., independent of r and θ. Again, from the curve r θ = a or r = a ⁄ θ, we have dθ θ2 = − ⋅ dr a dθ a2 ⎛ − θ2 ⎞ ⎟ = − a, which is constant. = ⋅ ⎜ ∴ Polar sub-tangent = r2 dr θ2 ⎝ a ⎠ dr = − a ⁄ θ2 dθ or 10.12 Length of the Perpendicular from Pole to the Tangent From the pole O draw OT perpendicular to the tangent at any point P (r , θ) on the curve r = f (θ). Let O T = p. Thus, p is the length of the perpendicular from pole to the tangent. We have ∠ OPT = φ . From the right angle triangle OTP, we have OT = OP sin φ or p = r sin φ . ...(1) O ften we require the value of p in terms of r and θ only. For this we shall substitute the value of φ in (1) from the equation cot φ = 1 dr ⋅ r dθ 1 1 1 Thus from (1), we have 2 = 2 cosec 2 φ = 2 (1 + cot2 φ ) r r p 2 ⎡ 1 1 1 ⎛ dr ⎞ 2⎤ 1 ⎛ dr ⎞ = 2 ⎢1+ 2 ⎜ ⎟ ⎥ = 2 + 4 ⎜ ⎟ ⋅ r ⎣ r r ⎝ dθ⎠ ⎦ r ⎝ dθ⎠ D-208 H ence DIFFERENTIAL CALCULUS 2 1 1 1 ⎛ dr ⎞ ⎟ = + ⎜ ⋅ p2 r2 r4 ⎝ dθ⎠ …(2) (Meerut 2003) 1 Sometimes, we write u = ⋅ r Then du 1 dr = − 2 ⋅ dθ r dθ ∴ 2 1 1 ⎛ du ⎞ = + ⎜ ⎟ p2 r2 ⎝ dθ ⎠ ⎛ ⎞2 1 2 + ⎜ du⎟ ⋅ = u …(3) p2 ⎝ dθ ⎠ Note : The results (1), (2) and (3) are very important and should be committed to memory. or 10.13 Pedal Equation (Meerut 2009) The relation between p and r for a given curve is called its pedal equation where r is the radius vector of any point on the curve and p is the length of the perpendicular from pole to the tangent at that point. Case I : To form the pedal equation of a curve whose cartesian equation is given. Let f (x, y) = 0, …(1) be the cartesian equation of the given curve. The equation of the tangent at (x, y) to the curve (1) is ⎞ dy dy ⎛ dy Y − y= (X − x), or Y − X + ⎜ x − y⎟ = 0. dx dx ⎝ dx ⎠ If p be the length of perpendicular from (0, 0) to this tangent, we have dy x − y dx ⋅ p= ⎧ ⎛ dy⎞ 2⎫ ⎨1 + ⎜ ⎟ ⎬ ⎝ dx⎠ ⎭ ⎩ √ …(2) …(3) Also r2 = x2 + y2. E liminating x and y between (1), (2) and (3), we get a relation between p and r i.e., the required pedal equation of the given curve. Case II : To form the pedal equation of a curve whose polar equation is given. Let f (r, θ) = 0 …(1) be the polar equation of the given curve. We have p = r sin φ , …(2) 1 dr and cot φ = ⋅ …(3) r dθ E liminating θ and φ between (1), (2) and (3), we obtain the required pedal equation of the given curve. Important : Sometimes we do not get the value of φ from equation (3) in a convenient form. In that case instead of using the relations (2) and (3), we can use the single relation TANGENTS AND NORMALS 2 1 1 ⎛ dr ⎞ 1 ⎟ = + ⎜ . p2 r2 r4 ⎝ dθ⎠ D-209 …(4) E liminating θ between (1) and (4), we obtain the required pedal equation of the curve. y2 x2 Example 1 : Show that the pedal equation of the ellipse 2 + 2 = 1, is a b 2 1 1 1 r = + − ⋅ (Meerut 2010B; Avadh 13) p2 a 2 b2 a 2 b2 x2 y2 + = 1. …(1) a 2 b2 Th e co-ordinates (x, y) o f an y p o in t P on ( 1) ma y be t aken as x = a cos t, y = b sin t. dy dy b cos t dx = − a sin t, = b cos t. ∴ = − . ∴ dt dx a sin t dt H ence the equation of the tangent to the ellipse at the point ‘t’ is b cos t Y − b sin t = − (X − a cos t) a sin t or ab − b cos t . X − a sin t . Y = 0. …(2) Therefore p = the length of perpendicular from (0, 0) to (2) ab = √(a2 sin2 t + b2 cos2 t ) Solution : The equation of the curve is i.e., Also a2 sin2 t + b2 cos2 t 1 = ⋅ …(3) p2 a 2 b2 r2 = x2 + y2 = a2 cos2 t + b2 sin2 t = a2 (1 − sin2 t ) + b2 (1 − cos2 t ) …(4) = a2 + b2 − a2 sin2 t − b2 cos2 t . E liminating t between (3) and (4), we obtain the pedal equation of the given curve. From (4), we get a2 sin2 t + b2 cos2 t = (a2 + b2) − r2. (a2 + b2) − r2 1 1 1 r2 1 = or = 2+ 2− 2 2, H ence (3) gives 2 2 2 2 p a b a b a b p which is the required pedal equation of the ellipse. Example 2 : Show that the pedal equation of the parabola y2 = 4a (x + a) is 2 p = ar. (Meerut 2010) Solution : Differentiating the equation of the curve, we get dy dy 2a 2y = 4a, or = ⋅ dx dx y Therefore the equation of the tangent at (x, y) is or (2a ⁄ y) X − Y + y − (2a ⁄ y) x = 0. Y − y = (2a ⁄ y) (X − x) ∴ p= y − (2ax ⁄ y) y2 − 2ax = 2 2 √(1 + 4a ⁄ y ) √(y2 + 4a2) D-210 DIFFERENTIAL CALCULUS 4a (x + a) − 2ax 2ax + 4a2 = 2 √[4a (x + a) + 4a ] √ [4a (x + 2a)] 2a (x + 2a) = = √[a (x + 2a)] . √[4a (x + 2a)] = r2 = x2 + y2 = x2 + 4a (x + a) = (x + 2a) 2. Also ∴ r = (x + 2a). ∴ p2 = a (x + 2a) or p2 = ar, which is the required pedal equation of the given curve. Example 3 : Form the pedal equation of the sine spiral r n = an sin n θ. …(1) Solution : The curve is r n = an sin n θ. Taking logarithm of both sides, we get n log r = n log a + log sin n θ. Differentiating with respect to θ, we obtain cos nθ n dr 1 dr = n = n cot nθ. ∴ cot φ = = cot nθ. r dθ sin nθ r dθ Therefore φ = nθ. Now p = r sin φ . ∴ p = r sin nθ. …(2) E liminating θ between (1) and (2), we obtain the required pedal equation of the given curve. From (2), we have sin n θ = p ⁄ r. Putting this value in (i), we obtain r n = an ( p ⁄ r) or pan = r n + 1, which is the required pedal equation. 1. 2. Find the polar subtangent of the ellipse l ⁄ r = 1 + e cos θ. For the parabola 2a ⁄ r = 1 − cos θ, prove that (i) φ= π− 1 2 θ, (ii) 1 2 p = a cosec θ, (iii) p2 = ar, (iv) the polar subtangent = 2a cosec θ . 3. (Purvanchal 2014) For the cardioid r = a (1 − cos θ), prove that (i) φ= 1 2 θ, (Meerut 2001) 1 2 (ii) p = 2a sin3 θ, (iii) the pedal equation is 2ap2 = r 3, 4. (iv) the polar subtangent = 2a Show that the pedal equation sin 2 (Meerut 2001; Rohilkhand 12B) (θ ⁄ 2) tan (θ ⁄ 2). (i) of the lemniscate r 2 = a2 cos 2 θ is r 3 = a2 p, (ii) of the hyperbola r 2 cos 2 θ = a2 is pr = a2, (iii) of the cosine spiral r n = an cos n θ is pan = r n + 1, (iv) of the curve r = a θ is p2 = r 4 ⁄ (r2 + a2) . D-211 TANGENTS AND NORMALS l = 1 + e cos θ is r 5. Show that the pedal equation of the conic 6. 1 = p2 Show that the 1 = p2 7. Show that the pedal equation of the cardioid r = a (1 + cos θ) is r3 = 2ap2. 8. 9. Show that the pedal equation of the astroid x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3 is r2 = a2 − 3p2. Show that the locus of the extremity of the polar subnormal of the curve ⎞ 1 ⎛⎜ 2l − 1 + e2⎟⎠ ⋅ ⎝ 2 r l pedal equation of the spiral r = a sech n θ is of the form A + B. r2 ⎛ ⎝ r = f (θ) is r = f ′ ⎜ θ − 1 2 ⎞ ⎠ π⎟ ⋅ (Gorakhpur 2006) H ence show that the locus of the extremity of the polar subnormal of the equiangular spiral r = aemθ is another equiangular spiral. 10. 1. Prove that the normal at any point (r, θ) to the curve r n = an cos n θ makes an angle (n + 1) θ with the initial line. l ⁄ e sin θ. 10.14 Differential Coefficient of Arc Length (Cartesian Formula) Let s denote the length of the arc A P of the curve y = f (x) measured from some fixed point A on it to any other point P (x, y). Then s is obviously some function of x and we want to find ds ⁄ dx. Take a point Q (x + δx, y + δy) on the curve in the neighbourhood of P such that arc AQ = s + δs. Then arc PQ = δs. Also δx → 0 as Q → P. From the right angled triangle PSQ, we have chord (PQ) 2 = PS 2 + SQ 2 = (δx) 2 + (δy) 2. …(1) Dividing (1) throughout by (δx) 2, we get ⎛ δy⎞ 2 ⎛ δy⎞ 2 ⎛ chord PQ ⎞ 2 ⎛ chord PQ ⎞ 2 ⎛ arc PQ ⎞ 2 ⎜ ⎟ = 1 + ⎜ ⎟ or ⎜ ⎟ .⎜ ⎟ = 1+ ⎜ ⎟ δx ⎝ arc PQ ⎠ ⎝ δx ⎠ ⎝ ⎠ ⎝ δx⎠ ⎝ δx ⎠ or or 2 ⎛ δy⎞ 2 ⎛ chord PQ ⎞ 2 ⎛ δs ⎞ ⎟ .⎜ ⎟ = 1+ ⎜ ⎟ . ⎜ ⎝ arc PQ ⎠ ⎝ δx⎠ ⎝ δx⎠ Taking limit of both sides as Q → P, we get 2 2 ⎛ δy⎞ 2⎤ lim ⎛ chord PQ ⎞ lim ⎛ δs ⎞ lim ⎡ ⎜ ⎟ . ⎜ ⎟ Q → P arc PQ δx → 0 δx = δx → 0 ⎢ 1 + ⎜⎝ δx⎟⎠ ⎥ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 2 2 ⎛ dy⎞ ⎡ . . lim chord PQ ⎤ ⎛ ds ⎞ ⎢ . Q →P = 1⎥ . ⎜ ⎟ = 1+ ⎜ ⎟ ⋅ dx arc PQ dx ⎝ ⎠ ⎣ ⎦ ⎝ ⎠ D-212 DIFFERENTIAL CALCULUS Thus ds = ± dx ⎧ √ ⎨⎩ 1 + ⎛ dy⎞ 2⎫ , ⎜ ⎟ ⎬ ⎝ dx⎠ ⎭ where plus or minus sign is to be taken before the radical sign according as s increases as x increases or decreases. H ence if s increases as x increases, we have √ ⎧ ⎛ dy⎞ 2⎫ ds ⎨1 + ⎜ ⎟ ⎬⋅ = ⎝ dx⎠ ⎭ ⎩ dx Corollary 1 : If x = f (y) be the equation of the curve, then ‘s’ is obviously a function of y. In this case dividing (1) throughout by (δy) 2 and proceeding to limits, we get √ ⎧ ⎛ dx⎞ 2⎫ ds ⎨1 + ⎜ ⎟ ⎬ , = ± ⎝ dy⎠ ⎭ ⎩ dy where plus or minus sign is to be taken before the radical sign according as s increases as y increases or decreases. Corollary 2 : If the equations of the curve be given in the parametric form x = f1 ( t ) and y = f2 ( t ), then ‘s’ is evidently a function of t. In this case dividing (1) throughout by (δt ) 2 and proceeding to limits, we have √ ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2⎫ ds ⎨⎜ ⎟ + ⎜ ⎟ ⎬, = ± (Param etric formula) ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ dt where plus or minus sign is to be taken before the radical sign according as s increases as t increases or decreases. 10.15 cos ψ = dy dx and sin ψ = ds ds We know that tan ψ = dy ⁄ dx. If s increases as x increases, we have ⎧ ds ⎛ dy⎞ 2⎫ ⎨ 1 + ⎜ ⎟ ⎬ = √(1 + tan2 ψ) = sec ψ. = dx ⎝ dx⎠ ⎭ ⎩ √ dx = cos ψ. ds Again if s increases as y increases, we have ds ⎧ ⎛ dx⎞ 2⎫ = ⎨ 1 + ⎜ ⎟ ⎬ = √(1 + cot2 ψ) = cosec ψ. ⎩ ⎝ dy⎠ ⎭ dy ∴ √ dy = sin ψ. ds Important : We can remember these results very easily with the help of the adjoining hypothetical figure. ∴ D-213 TANGENTS AND NORMALS 10.16 Differential Coefficient of Arc Length (Polar Formula) To prove that ds = dθ ⎧ 2 √ ⎨⎩ r + ⎛ dr ⎞ 2⎫ ⎜ ⎟ ⎬ for the curve r = f (θ). ⎝ dθ ⎠ ⎭ Let s denote the length of the arc A P measured from some fixed point A on the curve r = f (θ) to any other point P (r, θ). Then s is a function of θ. Ta ke a p oint Q (r + δr, θ + δθ) o n t h e cu rve in t h e neighbourhood of P such that arc AQ = s + δs. Then arc PQ = δs. Also δθ → 0 and δr → 0 as Q → P. From the triangle OPQ, we have (chord PQ) 2 = OP2 + OQ 2 − 2OP. OQ cos ∠ QOP or (chord PQ) 2 = r2 + (r + δr) 2 − 2r (r + δr) cos δθ or (chord PQ) 2 = (δr) 2 + 2rδr (1 − cos δθ) + 2r2 (1 − cos δθ) or (chord PQ) 2 = (δr) 2 + 2rδr (1 − cos δθ) + 2r2 (1 − cos δθ). Dividing by (δθ) 2, we get ⎛ δθ ⎞ 2 ⎛ δθ ⎞ 2 ⎜ sin ⎟ ⎜ sin ⎟ ⎛ δr ⎞ ⎛ chord PQ ⎞ ⎛ δs ⎞ 2⎟ 2⎟ ⎜ ⎜ ⎜ ⎟ .⎜ ⎟ = ⎜ ⎟ + r.⎜ ⎟ . δr + r2 ⎜ ⎟ ⋅ ⎝ δθ⎠ ⎜ δθ ⎟ ⎜ δθ ⎟ ⎝ arc PQ ⎠ ⎝ δθ⎠ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 2 Taking limits of both sides as Q → P, we get ⎛ ds ⎞ 2 ⎛ dr ⎞ 2 ⎡ δθ δθ⎞ lim ⎛ ⎜ ⎟ = ⎜ ⎟ + r . 1 . 0 + r2 . 1 . ⎢ ... δθ → 0 ⎜ sin / 2 ⎟⎠ = 1, dθ 2 dθ ⎝ ⎠ ⎣ ⎝ ⎠ ⎝ ⎤ dr lim chord PQ lim δr δθ → 0 δθ = dθ and Q → P arc PQ = 1⎥⎦ ∴ ⎛ dr ⎞ 2 ⎛ ds ⎞ 2 2 ⎜ ⎟ = r + ⎜ ⎟ ⋅ ⎝ dθ⎠ ⎝ dθ⎠ √ ⎫ ⎧ ⎛ ⎞2 ⎪ ⎪ 2 ds ⎨ r + ⎜ dr ⎟ ⎬ , where plus or minus sign is to be taken before the = ± ⎪ ⎝ dθ⎠ ⎪⎭ dθ ⎩ radical sign according as s increases or decreases as θ increases. H ence if s increases as θ increases, we have Thus √ ⎧ dr ⎞⎟ 2⎫ ds ⎨ r2 + ⎛⎜ ⎬. = ⎩ dθ ⎝ dθ⎠ ⎭ Corollary : If θ = f (r) be the equation of the curve, then ‘s’ is a function of r and we have ⎧ 2 ⎛ dr ⎞ 2⎫ ds ds dθ dθ = = ⎨r + ⎜ ⎟ ⎬ dr dr dθ dr ⎝ dθ⎠ ⎭ ⎩ or ds = dr √ √ ⎧ ⎨1 ⎩ ⎛ dθ⎞ + r2 ⎜⎝ ⎟⎠ ⎬⎭ . dr 2⎫ D-214 10.17 DIFFERENTIAL CALCULUS cos φ = dθ dr and sin φ = r ds ds We know that tan φ = r (dθ ⁄ dr). 1 1 = = ∴ cos φ = 2 √(sec φ ) √(1 + tan2 φ ) 1 √ ⎧ ⎛ dθ ⎞ 2⎫ ⎨ 1 + r2 ⎜ ⎟ ⎬ ⎝ dr ⎠ ⎭ ⎩ dr ⁄ dθ dr , 1 = = = ds ⁄ dθ ds ⎛ dθ ⎞ ⎧ 2 ⎛ dr ⎞ 2⎫ ⎜ ⎟ ⎨r + ⎜ ⎟ ⎬ ⎝ dr ⎠ ⎩ ⎝ dθ ⎠ ⎭ s being measured in such a way that s increases as θ increases. dr ⋅ Thus cos φ = ds dθ dθ dr ⋅ = r ⋅ Also sin φ = tan φ cos φ = r ds dr ds dθ H ence sin φ = r ⋅ ds Important : We can remember these results very easily with the help of the adjoining hypothetical figure. √ Example 1 : For the curve y = a log sec (x ⁄ a) , prove that Solution : We have y = a log sec (x ⁄ a). Differentiating with respect to x, we get ⎛ x⎞ 1 ⎛ x⎞ dy 1 = a⋅ sec ⎜ ⎟ tan ⎜ ⎟ ⋅ dx sec (x ⁄ a) ⎝ a⎠ a ⎝ a⎠ ⎛ x⎞ = tan ⎜ ⎟ . ⎝ a⎠ Now ⎧ ⎛ dy⎞ 2⎫ √ ⎨⎩ 1 + ⎜⎝ dx⎟⎠ ⎬⎭ = √ 1 + tan ds = dx ⎧ ⎨ ⎩ ⎛ x⎞ ds = sec ⎜ ⎟ . dx ⎝ a⎠ 2 ⎛⎜ x ⎞⎟ ⎫ ⎬ ⎝ a⎠ ⎭ ⎛ x⎞ = sec ⎜ ⎟ ⋅ ⎝ a⎠ Example 2 : For the ellipse x = a cos t, y = b sin t, prove that ds ⁄ dt = a (1 − e2 cos2 t) 1 ⁄ 2. dy dx = − a sin t and = b cos t. Solution : H ere dt dt ds ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2⎫ Now = ⎨⎜ ⎟ + ⎜ ⎟ ⎬ ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ dt √ = √(a2 sin2 t + b2 cos2 t ) = √{a2 sin2 t + a2 (1 − e2) cos2 t} [... b2 = a2 (1 − e2)] D-215 TANGENTS AND NORMALS = a√(sin2 t + cos2 t − e2 cos2 t ) = a√(1 − e2 cos2 t ). Example 3 : Show that for the curve rm = am cos mθ, i.e., ds am = m− 1⋅ dθ r Solution : We have r m = am cos mθ. Differentiating logarithmically, we obtain m sin mθ m dr = − , r dθ cos mθ dr = − r tan mθ. dθ ds ⎧ 2 ⎛ dr ⎞ 2⎫ Now = ⎨r + ⎜ ⎟ ⎬ dθ ⎩ ⎝ dθ⎠ ⎭ √ = √(r2 + r2 tan2 mθ) = r sec mθ = 1. r am r am am r = m = m = m − 1⋅ cos mθ a cos mθ r r Calculate ds ⁄ dx for the following curves: (i) y2 = 4ax; (ii) y = a cosh (x ⁄ a); (iii) x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3. 2. Calculate ds ⁄ dt for the following curves : (i) y = a (1 − cos t ), x = a (t + sin t ) . (ii) x = a cos3 t, y = a sin3 t . (iii) x = 2 sin t, y = cos 2t. 3. Calculate ds ⁄ d θ for the following curves : (i) r = log sin 3θ; (ii) r = a (1 − cos θ). 4. For the curve r = aeθ cot α, prove that s ⁄ r = constant, s being measured from the pole. 5. In any curve, prove that ds r2 , = dθ p r ds = ⋅ (ii) dr √(r2 − p2) (i) d2 r 6. For the curve r n = an cos nθ, prove that a2 n 2 + nr 2 n − 1 = 0. ds D-216 DIFFERENTIAL CALCULUS 7. For the cycloid x = a (1 − cos t), y = a (t + sin t), find ds (i) dt ds (ii) dx ds (iii) ⋅ dy 1. (i) (1 + a ⁄ x) 1 ⁄ 2, (ii) cosh (x ⁄ a), (iii) (a ⁄ x) 1 ⁄ 3. 2. (i) 2a cos (t ⁄ 2), (ii) 3a cos t sin t, (iii) 2 cos t √ (1 + 4 sin2 t ). 3. (i) √(r2 + 9 cot2 3 θ), (ii) 2a sin (θ ⁄ 2). 7. t , 2 t , (ii) cosec 2 t (iii) sec ⋅ 2 (i) 2a cos Fill in the Blanks: Fill in the blanks “……”, so that the following statem ents are com plete and correct. 1. 2. 3. 4. 5. If φ is the angle between the radius vector and the tangent of a curve then tan φ = …… . ds = …… . For the curve r = f (θ), dθ 2a For the parabola = 1 − cos θ, φ = …… . r ds = …… . For the cycloid x = a (1 − cos t), y = a (t + sin t), we have dt ds For the curve r 2 = a2 cos 2 θ, the value of is …… . dθ D-217 TANGENTS AND NORMALS 6. dθ 7 , = at a point on the curve r = f (θ), then at that point polar subnormal 3 dr (Meerut 2001) is …… . If Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 7. For the cardioid r = a (1 − cos θ), the value of φ is θ (a) θ (b) 2 θ (c) − (d) − θ 2 8. Two curves cut orthogonally if tan φ 1 . tan φ 2 is equal to (a) 1 (c) –1 9. 10. (b) 0 (d) None of these For the curve r = f (θ), the value of cos φ is dθ ds (b) r (a) r ds dθ ds dr (c) (d) dr ds For any curve r = f (θ), the value of ds is dθ (a) r2 p (b) (c) r p (d) p r2 p r True or False: Write ‘T’ for true and ‘F’ for false statement. 11. If p be the length of perpendicular drawn from the pole O to tangent at any point P (r , θ) on the curve r = f (θ), then 2 12. 1 1 ⎛ dr ⎞ 1 = 2 + 4 ⎜⎝ ⎟⎠ ⋅ 2 r r dθ p The relation between p and r for a given curve is called its polar equation. 13. ⎛ dr ⎞ 2 ⎛ d θ ⎞ 2 For the curve r = f (θ), we have ⎜⎝ ⎟⎠ + ⎜⎝ r ⎟⎠ = 1. ds ds 14. p = r sin θ is the pedal equation of some curve. D-218 DIFFERENTIAL CALCULUS dθ ⋅ dr 1. r 4. 2 a cos 7. 10. 13. (b). (a). T. 2. 1 t. 2 √ a2 ⋅ r 8. (c). 11. T. 14. F. 5. ⎧ ⎪ 2 ⎨r ⎪ ⎩ 2⎫ ⎛ dr ⎞ ⎪ + ⎜⎝ ⎟⎠ ⎬⎪ ⋅ dθ ⎭ 3. π − 3 ⋅ 7 9. (d). 12. F. 6. 1 θ. 2 11.1 Meaning of Curvature In the adjoining figure we see that the curve bends more sharply at the point P than at the point Q. We express this feeling by saying that the curve has a greater curvature at P than at Q. H owever, in order to get a quantitative estimate of curvature, we should give a mathematical definition of curvature which should be in agreement with our intuitive notion of curvature. 11.2 Definition of Curvature (Purvanchal 2010) Let P and Q be two neighbouring points on a curve, ψ and ψ + δψ the angles which the tangents at P and Q make with the x-axis. Let A be any fixed point on the curve. Let arc AP = s, arc AQ = s + δs, so that arc PQ = δs. The symbol, δψ denotes the angle through which the tangent turns as a point moves along the curve from P to Q through a distance δs. The angle δψ is called the contingence of the arc PQ. O bviously, δψ will be large or small, as compared with δs, depending on the degree of sharpness of D-220 DIFFERENTIAL CALCULUS the bend of the arc PQ. This suggests us to make the following definitions : (i) δψ is defined to be total curvature of the arc PQ ; δψ (ii) the ratio is defined to be the average curvature of the arc PQ ; δs dψ lim δψ i.e., ⋅ (iii) the curvature of the curve at P is defined to be Q → P ds δs dψ is a m athem atical m easure for the curvature of curve at any point P. Thus ds 11.3 Radius of Curvature (Meerut 2010) Let P be a given point on a given curve, and Q any other point on it in the neighbourhood of P. Let N be the point of intersection of the normals at P and Q. Suppose N tends to definite position C as Q tends to P, whether from the right or from the left. Then C is called the centre of curvature of the curve at P. The distance CP is called the radius of curvature of the curve at P and is usually denoted by the G reek letter ρ. The circle with its centre at C and radius CP is called the circle of curvature at P. Any chord drawn through P, of the circle of curvature at P, is called a chord of curvature at P. 11.4 Intrinsic Formula for the Radius of Curvature (Kashi 2011) The relation between s and ψ for any curve is called its intrinsic equation. Let P be a given point on t h e cu rve s = f (ψ), a n d Q a p o i n t o n it in t h e neighbourhood of P. Let ψ and ψ + δψ be the angles which tangents at P and Q make with the x-axis. Let A be any fixed point on the curve. Let arc AP = s, arc AQ = s + δs, so that arc PQ = δs. Let R be the point of intersection of the tangents at P and Q and N be the point of intersection of the n or ma ls at t h ese t wo p o in t s. Su p p o se N → C a s Q → P. lim Then the radius of curvature at P = ρ = Q → P PN. We have ∠ PNQ = ∠ TRT ′ = δψ. Now from the triangle PNQ, we have PN chord PQ chord PQ = = ⋅ sin N Q P sin PNQ sin δψ δψ chord PQ δs chord PQ sin NQP = ⋅ δψ ⋅ ⋅ sin NQP. ∴ PN = δs sin δψ sin δψ Now as Q → P, we have δψ → 0, δs → 0, chord PQ → tangent at P, π QN → normal at P and consequently ∠ NQP → ⋅ 2 D-221 CURVATURE Lim Therefore ρ = Q → P PN ⎛ Lim chord PQ ⎞ ⎛ Lim δs ⎞ ⎛ Lim δψ ⎞ = ⎜ Q →P ⎟ ⋅ ⎜ δψ → 0 δψ⎟ ⋅ ⎜ δψ → 0 ⎟ arc PQ sin δψ⎠ ⎠ ⎝ ⎝ ⎠ ⎝ ⎞ ⎛ Lim ⋅ ⎜ Q → P sin NQP⎟ ⎝ ⎠ ⎛ ds ⎞ π ds ⋅ = 1 ⋅ ⎜ ⎟ ⋅ 1. sin = 2 dψ ⎝ dψ⎠ ds H ence, ρ = ⋅ dψ Corollary : The curvature of the curve at any point P is by definition, equal to dψ . H ence the curvature of the curve at any point is equal to the reciprocal of the radius ds 1 of curvature at that point i.e., curvature = ρ⋅ Example : Find the radius of curvature for the curve whose intrinsic equation is ⎛ π ψ⎞ s = a log tan ⎜ + ⎟ ⋅ 2⎠ ⎝4 ψ⎞ 1 1 ds 2⎛π = a Solution : We have ρ = ⎛ π ψ ⎞ sec ⎜⎝ 4 + 2 ⎟⎠ ⋅ 2 dψ tan ⎜ + ⎟ 2⎠ ⎝4 a a a = ⎞ = cos ψ = a sec ψ. ⎛ π ψ⎞ = ⎛ π ψ⎞ ⎛π 2 sin ⎜ + ⎟ cos ⎜ + ⎟ sin ⎜ + ψ⎟ 2⎠ 2⎠ ⎠ ⎝4 ⎝4 ⎝2 11.5 Cartesian Formula for Radius of Curvature (Gorakhpur 2005) Let the equation of the curve be y = f (x). dy We know that = tan ψ. dx Differentiating with respect to x, we get dψ d ψ ds d2 y = sec 2 ψ ⋅ = sec 2 ψ ⋅ ⋅ dx ds dx dx2 ∴ dψ = ds But we have d2 y dx2 sec 2 ψ ds = dx ds = dψ √ ⋅ ds dx ⎧ ⎛ dy⎞ 2⎫ ⎨1 + ⎜ ⎟ ⎬. ⎝ dx⎠ ⎭ ⎩ 3⁄2 ⎧ ⎛ dy⎞ 2⎫ ⎨1 + ⎜ ⎟ ⎬ ⎝ dx⎠ ⎭ ⎩ ∴ (1 + y12) 3 ⁄ 2 (1 + tan2 ψ) ⋅ d2 y dx2 ds dx ⋅ ⋅ (Bundelkhand 2008) y2 d2 y 2 dx Note 1 : The radius of curvature ρ can come out to be positive or negative. If in the relation H ence ρ= = ds = dψ D-222 DIFFERENTIAL CALCULUS ds = dx ⎧ √ ⎨⎩ 1 + ⎛ dy⎞ 2⎫ ⎜ ⎟ ⎬, ⎝ dx⎠ ⎭ we take the positive sign before the radical, the value of ρ is positive or negative d2 y according as 2 is positive or negative. H owever, if we define the radius of curvature dx in such a way that it is to be always positive, then we should ignore the sign whenever we get a negative value for ρ. Note 2 : Sin ce t h e radiu s o f cu rvat u re is a len gt h t herefo re it s valu e is independent of the choice of x-axis and y-axis. H ence interchanging x and y, we obtain ρ= 3⁄2 ⎧ ⎛ dx⎞ 2⎫ ⎨1 + ⎜ ⎟ ⎬ ⎝ dy ⎠ ⎭ ⎩ ⋅ d2 x dy2 This formula is specially useful when (dy ⁄ dx) is infinite i.e., when the tangent is perpendicular to x-axis. Example 1 : Fin d t h e cu rv a t ure a t t h e po in t (3a ⁄ 2, 3a ⁄ 2) o f t h e c u rv e x3 + y3 = 3axy. (Meerut 2010; Agra 05; Kashi 14; Avadh 14) or Solution : The curve is x3 + y3 = 3axy. Differentiating with respect to x, we get dy dy = 3ay + 3ax 3x2 + 3y2 dx dx dy dy x2 + y2 = ay + ax ⋅ dx dx ⎡ dy⎤ dy x2 − ay ∴ = ⋅ ∴ ⎢ ⎥ = − 1. dx ax − y2 ⎣ dx⎦ ⎛ 3 a , 3 a⎞ ⎜2 ⎝ 2 …(1) …(2) ⎟ ⎠ Again, differentiating (2), with respect to x, we get ⎡ dy⎤ 2 d2 y dy dy d2 y + a + ax 2 2x + 2y ⎢ ⎥ + y2 2 = a dx dx dx dx ⎣ dx⎦ or ⎛ dy⎞ 2 d2 y dy (ax − y2) 2 = 2x + 2y ⎜ ⎟ − 2a ⋅ dx ⎝ dx⎠ dx ⎡ dy⎤ 3a 3a ,y= and ⎢ ⎥ = − 1 in (3), we get 2 2 ⎣ dx ⎦ (3a ⁄ 2 , 3a ⁄ 2) ⎡ d2 y⎤ 32 1 ⎢ ⎥ = − ⋅ ⋅ ⎢ 2⎥ 3 a ⎢ dx ⎥ ⎣ ⎦ (3a ⁄ 2 , 3a ⁄ 2) ⎛ 3a , 3a ⎞ H ence the radius of curvature ρ at ⎜ ⎟ ⎝ 2 2⎠ Putting x = …(3) D-223 CURVATURE ⎡ ⎧⎨ 1 + ⎢ ⎢⎩ ⎢ = ⎢ ⎢ ⎢ ⎢ ⎣ ⎛ dy⎞ 2⎫ 3 ⁄ 2 ⎤ ⎜ ⎟ ⎬ ⎥ ⎝ dx⎠ ⎭ ⎥ d2 y dx2 (1 + 1) 3 ⁄ 2 ⎥ 3a = = − ⋅ ⎥ 32 1 8√2 ⎥ − . ⎥ 3 a ⎥ ⎦ (3a ⁄ 2 , 3a ⁄ 2) ⎛ 3a 3a ⎞ 8 √2 1 ⋅ ∴ Curvature at ⎜ , ⎟ = ρ = − 2 3a 2 ⎠ ⎝ ⎛ 3a , 3a ⎞ 8 √2 If we ignore the negative sign, the value of curvature at ⎜ ⎟ = ⋅ 3a ⎝ 2 2⎠ Example 2 : If a curve is defined by the equations x = f ( t ) and y = φ ( t ) , prove (x ′ 2 + y ′ 2) 3 ⁄ 2 , that the radius of curvature ρ is equal to where accents (i.e., dashes) denote x ′ y ′′ − y ′ x ′′ differentiation with respect to t. y′ dy dy ⁄ dt φ ′ ( t ) = = = ⋅ Solution : We have x′ dx dx ⁄ dt f ′ ( t ) ⎧ d ⎛ y ′ ⎞ ⎫ dt d2 y d ⎛ y′⎞ = ⎜ ⎟ = ⎨ ⎜ ⎟⎬ ⋅ ⋅ 2 dx x ′ dx ⎩ dt ⎝ x ′ ⎠ ⎭ dx ⎝ ⎠ Also H ence ρ= = y ′′ x ′ − x ′′ y ′ 1 ⋅ x′ x ′2 = y ′′ x ′ − x ′′ y ′ ⋅ x ′3 ⎧ ⎛ dy⎞ 2⎫ 3 ⁄ 2 ⎨1 + ⎜ ⎟ ⎬ ⎩ ⎝ dx ⎠ ⎭ d2 y = dx2 ⎡ . . dx dt 1⎤ = x ′ and = ⎥ ⎢ . dt dx x ′⎦ ⎣ 2 ⎫3 ⁄ 2 ⎧ ⎪1 + y ′ ⎪ ⎨ ⎬ ⎪ x ′ 2 ⎪⎭ ⎩ y ′′ x ′ − x ′′ y ′ x ′3 = (x ′ 2 + y ′ 2) 3 ⁄ 2 ⋅ y ′′ x ′ − x ′′ y ′ Example 3 : In th e cyclo id x = a (t + sin t ), y = a (1 − cos t ), prove that 1 2 ρ = 4a cos t. (Bundelkhand 2007; 12, 14) dx dy Solution : H ere = a (1 + cos t ) and = a sin t. dt dt ∴ 2 sin t ⁄ 2 cos t ⁄ 2 dy dy ⁄ dt a sin t t = = = = tan ⋅ 2 dx dx ⁄ dt a (1 + cos t ) 2 2 cos t ⁄ 2 Also d ⎛ d ⎛ dy⎞ d2 y = ⎜ ⎟ = ⎜ tan 2 dx ⎝ dx ⎝ dx⎠ dx = H ence 1 t⎞ t dt ⎟ = ⋅ sec2 ⋅ 2 2⎠ 2 dx 1 t 1 1 t 1 1 t sec 2 ⋅ = sec 2 ⋅ = sec4 ⋅ 2 2 a (1 + cos t) 2 2 2a cos2 t ⁄ 2 4a 2 ρ= 3⁄2 ⎧ t⎫ ⎨ 1 + tan2 ⎬ 2⎭ ⎩ t 1 sec4 2 4a 4a sec3 = sec4 t 2 t 2 1 2 = 4a cos t ⋅ D-224 DIFFERENTIAL CALCULUS Example 4 : If CP, CD be a pair of conjugate sem i-diam eters of an ellipse, prove that the radius of curvature at P is CD3 ⁄ ab, a and b being the lengths of the semi-axes of the ellipse. (Meerut 2001, 05B; Rohilkhand 11) Solution : (Note. Two perpendicular diameters are called conjugate diameters) x2 y2 Let CP and CD be a pair of conjugate semi-diameters of the ellipse 2 + 2 = 1, a b the centre C is origin. Let ‘t’ be the eccentric angle of the point P. Then the co-ordinates of P are x = a cos t, y = b sin t. The eccentric angle of D will be t + 1 2 π, so that the co-ordinates of D are ⎛1 ⎞ ⎡ ⎛1 ⎞⎤ ⎢ a cos ⎜ 2 π + t⎟ , b sin ⎜ 2 π + t ⎟ ⎥ i.e., (− a sin t, b cos t ). ⎝ ⎠ ⎣ ⎝ ⎠⎦ Now for the point P we have x = a cos t and y = b sin t. dy dx = − a sin t, = b cos t. ∴ dt dt dy b H ence = − cot t. dx a 2 ⎞ d y d ⎛ b d ⎛ dy⎞ Also = ⎜ ⎟ = ⎜ − cot t ⎟ 2 dx ⎝ a dx ⎝ dx⎠ ⎠ dx ⎧d ⎛ b ⎞ ⎫ dt cot t ⎟ ⎬ ⋅ ⎜− ⎩ dt ⎝ a ⎠ ⎭ dx = ⎨ b ⎛b ⎞ ⎛ 1 ⎞ cosec2 t ⎟ ⋅ ⎜ − cosec t ⎟ = − 2 cosec3 t ⋅ ⎝a ⎠ ⎝ a ⎠ a = ⎜ ∴ R adius of curvature of the point ‘t’ = ρ= 3⁄2 ⎧ b2 cos2 t ⎫⎪ ⎪ ⎨1 + ⎬ ⎪ a2 sin2 t ⎪⎭ ⎩ − b cosec3 t a2 = − (a2 sin2 t + b2 cos2 t ) 3 ⁄ 2 ⋅ ab Neglecting the negative sign, we have ρ = (a2 sin2 t + b2 cos2 t ) 3 ⁄ 2 ⋅ ab Now CD = √ {(− a sin t − 0) 2 + (b cos t − 0) 2} = (a2 sin2 t + b2 cos 2 t) 1 ⁄ 2 ⋅ ∴ CD3 (a2 sin2 t + b2 cos 2 t ) 3 ⁄ 2 = = ρ. ab ab ax , if ρ is the radius of curvature at any point a+ x (x, y), show that (2ρ ⁄ a) 2 ⁄ 3 = ( y ⁄ x) 2 + (x ⁄ y) 2. Example 5 : For the curve y = (Kumaun 2008; Rohilkhand 10B, 13, 14; Avadh 10, 13) Solution : We have y = ∴ ax a+ x (a + x) − x dy a2 = a = = a2 (a + x) − 2, 2 dx (a + x) (a + x) 2 …(1) D-225 CURVATURE − 2a2 − 2a2 d2 y d ⎛ dy⎞ 2y3 = ⎜ ⎟ = − 2a2 (a + x) − 3 = = = − ⋅ 2 3 3 dx ⎝ dx⎠ (a + x) (ax ⁄ y) ax3 dx and Now ⎛ dy⎞ 2 a4 a4 y4 x4 + y4 = 1+ = 1+ 4= ⋅ 1+ ⎜ ⎟ = 1+ 4 4 (a + x) (ax ⁄ y) x x4 ⎝ dx⎠ ∴ ρ= [1 + (dy ⁄ dx) 2]3 ⁄ 2 d2 y ⁄ dx2 = [(x4 + y4) ⁄ x4]3 ⁄ 2 (2y3 ⁄ ax3) , (negative sign being neglected) = H ence y4) 3 ⁄ 2 (x4 + a a = 2 2 x6 ( y3 ⁄ x3) (x4 y4) 3 ⁄ 2 + x3 y3 ⋅ 2 ⎛ 2ρ⎞ 2 ⁄ 3 x4 + y4 ⎛ y⎞ 2 x4 y4 x2 y2 ⎛ x⎞ ⎜ ⎟ ⎟ = = + = + = ⎜ + ⎜ ⎟ ⋅ ⎝ a⎠ ⎝ x⎠ x2 y2 x2 y2 x2 y2 y2 x2 ⎝ y⎠ 11.6 Radius of Curvature at the Origin (Another Method) R adius of curvature at the origin can be found by substituting x = 0, y = 0 in the value of ρ obtained from the formula of §11.5. H ere we shall give an alternative method. Since the curve passes through the origin, therefore ( y) 0 = 0 i.e., the value of y at x = 0 is 0. Let ⎛ dy⎞ ⎜ ⎟ = ( y1) 0 = p and ⎝ dx⎠ (0, 0) ⎛ d2 y ⎞ ⎟ ⎜ = ( y2) 0 = q . ⎜ 2⎟ ⎜ dx ⎟ ⎠ (0, 0) ⎝ (1 + p2) 3 ⁄ 2 ⋅ q To get the values of p and q, we know by Maclaurin’s theorem that ( y2) 0 y = ( y) 0 + ( y1) 0 x + x2 + … 2! Then ρ (at origin) = …(1) …(2) Since the curve passes through origin, therefore (2) becomes y = px + 1 2 q x2 + …… Thus to get the values of p and q, we should get from the equation of the curve an expansion for y in ascending powers of x by algebraic or trigonometric methods. The coefficient of x in this expansion will be equal to p and the coefficient of x2 will be equal to 1 2 q. Putting the values of p and q in (1), we shall get ρ at origin. 11.7 Newton’s Method for Radius of Curvature at the Origin Suppose a curve passes through the origin and the x-axis is tangent to the curve at origin. ⎛ dy⎞ i.e, ( y1) 0 = 0. Then ( y) 0 = 0 and ⎜ ⎟ ⎝ dx ⎠ (0, 0) Therefore in this case by Maclaurin’s expansion, we have D-226 DIFFERENTIAL CALCULUS y = 0 + 0. x + where q 2 ( y3) 0 3 .x + x + … 2 3! …(1) ( y2) 0 = q. 2 2y 2 Multiplying (1) by 2 , we get 2 = q + ( y ) x+ … 3! 3 0 x x Lim 2y Taking limit as x → 0 of both sides of (2), we get x → 0 2 = q. x (1 + 0) 3 ⁄ 2 1 Lim x2 Also in this case ρ at origin = = = x →0 ⋅ q q 2y Therefore when x-axis is tangent to the curve at the origin, Lim ⎛ x2 ⎞ ρ (at origin) = x → 0 ⎜ ⎟ ⋅ ⎝ 2y⎠ ...(2) Similarly it can be shown that if y-axis is tangent to the curve at the origin, then Lim y2 ⋅ ρ (at origin) = x → 0 2x These two formulae are known as Newton’s formulae. Example 1 : Find the radius of curvature at origin for the curve x3 + y3 − 2x2 + 6y = 0. Solution : The curve passes through origin. Equating to zero the lowest degree terms we get y = 0, i.e., x-axis as tangent to the curve at origin lim x2 ∴ By Newton’s method ρ (at origin) = x → 0 ⋅ 2y Dividing by 2y, the equation of the curve can be written as x⋅ x2 x2 1 2 + y − 2. + 3 = 0. 2y 2 2y lim x2 Taking limit as x → 0, y → 0 and x → 0 = ρ, we get 2y 0ρ + 0 − 2ρ + 3 = 0 i.e., ρ = 3 ⁄ 2. Example 2 : Show that the radii of curvature of the curve y2 = x2 (a + x) ⁄ (a − x) at the origin are ± a √2. (Gorakhpur 2006) Solution : The curve passes through the origin and the tangents at origin are y2 = x2 i.e., y = ± x. Thus neither of the coordinate axes is tangent at the origin. Therefore we cannot apply Newton’s method. But the equation of the curve can be written as ± x (a + x) 1 ⁄ 2 ⎛ x⎞ 1 ⁄ 2 ⎛ x⎞ − 1 ⁄ 2 y= or y = ± x ⎜ 1 + ⎟ ⎜1 − ⎟ 1 ⁄ 2 ⎝ ⎝ a⎠ a⎠ (a − x) or or 1 x 1 x ⎫⎧ ⎫ + ..……⎬ ⎨ 1 + + ……..⎬ 2a 2a ⎭⎩ ⎭ expanding by Binomial Theorem x ⎡ ⎤ y = ± x ⎢ 1 + + ……⎥ . a ⎣ ⎦ ⎧ ⎩ y = ± x ⎨1 + D-227 CURVATURE Comparing this equation with the equation x2 2 + …, we get p = 1, q = a 2 (1 + p2) 3 ⁄ 2 ⋅ But ρ at origin = q y = px + q or p = − 1, q = − 2 ⋅ a (1 + 1) 3 ⁄ 2 2, ρ at origin = = a √2. 2 a a (1 + 1) 3 ⁄ 2 2 Also when p = − 1, q = − , ρ at origin = = − a√2. a − 2⁄a Example 3 : Find the radii of curvature at the origin for the curve y2 − 3xy − 4 x2 + x3 + x4 y + y5 = 0. Solution : The curve passes through the origin and the tangents at origin are y2 − 3xy − 4x2 = 0. Thus n eith er of the co-ordinate axes is tangent at the origin. Therefore Newton’s method cannot be applied. Also we cannot put the equation of the qx2 curve in the form y = px + + …… 2 qx2 + … for y in the equation of the curve, we get the H ence substituting px + 2 identity, ⎛ ⎞2 ⎛ ⎞ qx2 qx2 ⎜ px + + ……⎟ − 3x ⎜ px + + ……⎟ ⎝ ⎠ ⎝ ⎠ 2 2 ⎛ ⎞ qx2 − 4x2 + x3 + x4 ⎜ px + + ……⎟ + … = 0 . 2 ⎝ ⎠ ∴ When p = 1, q = E quating to zero the coefficients of x2 and x3, we get 3q + 1 = 0. p2 − 3p − 4 = 0 and pq − 2 Solving these we get p = 4, − 1. When p = 4, q = − 2 ⁄ 5 and when p = − 1, q = 2 ⁄ 5. Now ∴ ρ (at origin) = (1 + p2) 3 ⁄ 2 ⋅ q When p = 4, q = 2 ⁄ 5, ρ at origin = and when p = − 1, q = 2 ⁄ 5, ρ at origin = 1. (1 + 16) 3 ⁄ 2 85√17 = − 2 − 2⁄5 (1 + 1) 3 ⁄ 2 = 5√2. 2⁄5 Find the radius of curvature at the point (s, ψ) on the following curves : (i) s = c tan ψ (Catenary) 1 6 (ii) s = 8a sin2 ψ (Cardioid) (iii) s = 4a sin ψ (Cycloid) (Bundelkhand 2001; Rohilkhand 08; Kashi 11) D-228 2. DIFFERENTIAL CALCULUS (iv) s = c log sec ψ ( Tractrix). (Kashi 2012) Find the radius of curvature at the point (x, y) on the following curves : (i) a2 y = x3 − a3 (iii) xy = c2 (v) y = 1 2 (ii) y2 = 4ax (iv) ay2 = x3 a (e x ⁄ a + e− x ⁄ a) (Agra 2007) (vi) y = c log sec (x ⁄ c) (vii) x1 ⁄ 2 + y1 ⁄ 2 = a1 ⁄ 2. (viii) x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3 (ix) x m + y m = 1. 3. (Kanpur 2007; Purvanchal 09) (Rohilkhand 2009B; Kashi 12) (i) Find the radius of curvature of the curve y = e x, at the point where it crosses (Agra 2014) the y-axis. ⎛ 1 1⎞ (ii) Find the radius of curvature of the curve √x + √y = 1 at the point ⎜ , ⎟ . ⎝4 4. 5. 4⎠ 5√5 ⋅ (i) Prove that at the point x = π of the curve y = 4 sin x − sin 2x, ρ = 4 ⎛ π ψ⎞ (ii) Prove that for the curve s = a log cot ⎜ − ⎟ + a sin ψ sec2 ψ, ρ = 2a sec 3 ψ ; 2⎠ ⎝4 2 1 d y ⋅ and hence that 2 = 2a dx 1 2 In the curve y = ae x ⁄ a, prove that ρ = a sec 2 θ cosec θ, where θ = tan− 1 ( y ⁄ a). 6. Show that the radius of curvature at a point (a cos3 θ, a sin3 θ) on the curve x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3 is 3a sin θ cos θ. 7. Prove that for the ellipse x2 ⁄ a2 + y2 ⁄ b2 (Meerut 2000, 05; Kashi 13) = 1, a 2 b2 , p being the perpendicular from the centre upon the tangent at (x, y). ρ= p3 (Meerut 2002, 04B, 07; Avadh 05, 09) 8. 9. x2 ⁄ a2 y2 ⁄ b2 In the ellipse + = 1, show that the radius of curvature at an end of the major axis is equal to the semi-latus rectum of the ellipse. If ρ and ρ′ be the radii of curvature at the extremities of two conjugate diameters of an ellipse, prove that ( ρ2 ⁄ 3 + ρ′ 2 ⁄ 3) (ab) 2 ⁄ 3 = a2 + b2. (Meerut 2001, 03, 04, 06, 11; Bundelkhand 06; Kanpur 11; Rohilkhand 13B; Kashi 14) 11. Prove t hat if ρ be the radius of curvature at any point P on the parabola y2 = 4ax and S be its focus, then ρ2 varies as (SP) 3. If the co-ordinates of a point on a curve be given by the equations 12. x = c sin 2 θ (1 + cos 2 θ), y = c cos 2 θ (1 − cos 2θ), show that the radius of curvature at the point is 4c cos 3 θ. If the co-ordinates of a point on a curve be given by the equations 10. x = a sin t − b sin (at ⁄ b), y = a cos t − b cos (at ⁄ b), a− b 4ab sin t. show that the radius of curvature at the point is 2b a+ b D-229 CURVATURE 13. Prove that for the curve s = ae x ⁄ a, aρ = s (s2 − a2) 1 ⁄ 2. ⎛ y⎞ ⎜1− ⎟ ⋅ Show that for the curve s2 = 8ay, ρ = 4a 2a ⎠ ⎝ √ (Kanpur 2009) 14. Find the radius of curvature at the origin of the following curves : (ii) y = x3 + 5x2 + 6x. (i) y = x4 − 4x3 − 18x2. 15. Show that the radii of curvature of the curve a (y2 − x2) = x3 at the origin are ± 2a√2. 1. (i) c sec 2 ψ 2. (a4 (i) (ii) 4 3 a sin 9x4) 3 ⁄ 2 + 6a4x 1 (4a + 9x) 3 ⁄ 2 x1 ⁄ 2 6a 2 (x + y) 3 ⁄ 2 (vii) √a (iv) 3. (i) √8. 1 3 ψ (iii) 4a cos ψ (iv) c tan ψ. (x2 + y2) 3 ⁄ 2 2c2 (ii) ⎛ 2 ⎞ ⎜ ⎟ (x + a) 3 ⁄ 2 ⎝ √a ⎠ (iii) (v) y2 ⁄ a (vi) c sec (x ⁄ c) (viii) 3a1 ⁄ 3 x1 ⁄ 3 y1 ⁄ 3 (ii) 1 ⋅ √2 14. (ix) (x 2 m − 2 + y 2 m − 2) 3 ⁄ 2 ⋅ (1 − m) x m − 2 y m − 2 (i) 1 ⁄ 36, (ii) 37 √(37) ⁄ 10. 11.8 Pedal Formula for Radius of Curvature We have the relation ψ = θ + φ , …(1) as is obvious from the adjoining figure. Differentiating (1) w.r.t. s, we get dψ dθ dφ dr dψ dθ dφ = + or = + ⋅ ds ds ds ds dr ds ds dφ 1 1 or ρ = r sin φ + cos φ dr ⎡.. dθ dr⎤ ds , ⎢ . ρ= sin φ = r and cos φ = ⎥ ds ds ⎦ dψ ⎣ or H ence dφ ⎞ 1 d 1 dp 1 1⎛ ρ = r ⎜⎝ sin φ + r cos φ dr ⎟⎠ = r dr (r sin φ ) = r dr ⋅ dr ρ= r ⋅ dp [... p = r sin φ ] (Meerut 2003, 07B) 11.9 Polar Formula for Radius of Curvature 2 1 1 1 ⎛ dr ⎞ We know that 2 = 2 + 4 ⎜ ⎟ ⋅ p r r ⎝ dθ⎠ …(1) D-230 DIFFERENTIAL CALCULUS Differentiating (1) with respect to r, we get − 2 2 dp 2 4 ⎛ dr ⎞ 1 ⎧ d ⎛ dr ⎞ 2⎫ ⎟ = − − ⎜ + 4⎨ ⎜ ⎟ ⎬ 3 3 5 dr dθ p r r ⎝ ⎠ r ⎩ dr ⎝ dθ⎠ ⎭ 2 4 ⎛ dr ⎞ 1 ⎧ d ⎛ dr ⎞ 2⎫ d θ 2 = − 3− 5⎜ ⎟ + 4⎨ ⎜ ⎟ ⎬⋅ dθ r ⎝ ⎠ r ⎩ dθ ⎝ dθ⎠ ⎭ dr r 2 ⎛ dr ⎞ d2 r dθ 2 4 ⎛ dr ⎞ 1 = − 3− 5⎜ ⎟ + 4⋅ 2⎜ ⎟ ⋅ ⋅ r r ⎝ dθ⎠ r ⎝ dθ⎠ dθ2 dr 2 4 ⎛ dr ⎞ 2 d2 r 2 = − 3− 5⎜ ⎟ + 4 2 r ⎝ dθ⎠ r dθ r ⎧ ⎫ 1 dp 1 ⎪2 d2 r⎪ ⎛ dr ⎞ 2 = ⎨r + 2 ⎜ ⎟ − r . ⎬ ⎝ dθ⎠ p3 dr r5 ⎪⎩ dθ2 ⎪⎭ 1 r⋅ 3 dr p Therefore ρ= r = ⋅ dp d2 r⎫⎪ 1 ⎧⎪ 2 ⎛ dr ⎞ 2 r + 2 − r ⎜ ⎟ ⎨ ⎬ ⎝ dθ ⎠ r5 ⎪⎩ dθ2 ⎪⎭ ∴ But from (i), 1 = p3 ρ= H ence ⎧1 ⎫3 ⁄ 2 1 ⎧ 2 ⎛ dr ⎞ 2⎫ 3 ⁄ 2 ⎪ + 1 ⎛⎜ dr ⎞⎟ 2⎪ = ⎨r + ⎜ ⎟ ⎬ . ⎨ 2 ⎬ ⎪r ⎝ dθ⎠ ⎭ r4 ⎝ dθ⎠ ⎪⎭ r6 ⎩ ⎩ 3⁄2 1 ⎧ ⎛ dr ⎞ 2⎫ r6 ⋅ 6 ⎨ r2 + ⎜ ⎟ ⎬ ⎝ dθ ⎠ ⎭ r ⎩ ⎛ dr ⎞ 2 d2 r + 2⎜ ⎟ − r 2 dθ ⎝ dθ⎠ ⎧ 2 ⎛ dr ⎞ 2⎫ 3 ⁄ 2 ⎨r + ⎜ ⎟ ⎬ ⎝ dθ⎠ ⎭ ⎩ Therefore, ρ= ⋅ 2 2r ⎞ ⎛ dr d r2 + 2 ⎜ ⎟ − r 2 dθ ⎝ dθ ⎠ 1 1 Corollary : If we put u = or r = , then u r ⋅ r2 2 dr 1 du d2 r 2 ⎛ du ⎞ 1 d2 u ⎟ = − 2 and = ⎜ − ⋅ dθ dθ2 u3 ⎝ dθ ⎠ u dθ u2 dθ2 Putting these values in the polar formula for ρ, we get ρ= 2 ⎫3 ⁄ 2 ⎧1 ⎪ + u′ ⎪ ⎨ 2 ⎬ ⎪u u4 ⎪⎭ ⎩ = (u2 + u′ 2) 3 ⁄ 2 , u3 (u + u′′) 2u′ 2 2u′ 2 u′′ 1 + − + 3 2 u u4 u4 u where dashes denote differentiation with respect to θ. Note : We see that the pedal formula for ρ is simpler than the polar formula. Therefore in case the equation of the curve is given in polar form, it is often convenient to change it first to pedal equation and then to find ρ with the help of the pedal formula. D-231 CURVATURE 2 √(2ar). 3 (Purvanchal 2006, 11; Rohilkhand 09B, 10; Agra 14) Example 1 : Show that for the cardioid r = a (1 + cos θ), ρ = Solution : The curve is r = a (1 + cos θ). d2 r dr = − a sin θ and = − a cos θ. ∴ dθ dθ2 Now {r2 + (dr ⁄ dθ) 2}3 ⁄ 2 ρ= 2 r + 2 (dr ⁄ dθ) 2 − r (d2 r ⁄ dθ2) {a2 (1 + cos θ ) 2 + a2 sin2 θ}3 ⁄ 2 = 2 a (1 + cos θ) 2 + 2 (− a sin θ) 2 − a (1 + cos θ) (− a cos θ) = = 3⁄2 ⎛ 2 41 2 21 21 ⎞ ⎜ 4a cos 2 θ + 4a cos 2 θ sin 2 θ⎟ ⎝ ⎠ a2 + 2a2 (cos2 θ + sin2 θ) + 3a2 cos θ 1 2 1 2 1 2 (4a2 cos2 θ) 3 ⁄ 2 [cos2 θ + sin2 θ]3 ⁄ 2 3a2 (1 + cos θ) = 1 2 1 2 2 6a cos 2 8a3 cos3 θ ⎡ 4a ⎤ 1 = ⎢ ⎥ cos θ. 2 3 θ ⎣ ⎦ 1 2 But r = a (1 + cos θ) = 2a cos2 θ . ∴ cos θ = √(r ⁄ 2a). H ence ρ= 1 2 4a 3 √ ⎡ ⎢ ⎣ r ⎤⎥ = (2 ⁄ 3) √(2ar). 2a ⎦ Note : We could have solved this problem more easily by changing the equation of the curve to pedal form. Example 2 : Show that in the rectangular hyperbola r2 cos 2 θ = a2, ρ = r3 ⁄ a2. or …(1) Solution : The curve is r2 cos 2 θ = a2. Taking logarithm of both sides of (1), we get 2 log r + log cos 2θ = 2 log a. Differentiating with respect to θ, we get 1 2 dr + (− 2 sin 2θ) = 0 r dθ cos 2θ ⎡π ⎤ 1 dr = cot φ = tan 2θ = cot ⎢ − 2θ⎥ . r dθ ⎣2 ⎦ π ∴ φ = − 2 θ. 2 ⎡π ⎤ p = r sin φ = r sin ⎢ − 2θ⎥ = r cos 2θ. ⎣2 ⎦ H ence the pedal equation of the curve is Now a2 p = r. 2 r ∴ dp a2 = − 2⋅ dr r or p= a2 ⋅ r But cos 2θ = a2 ⋅ r2 D-232 H ence DIFFERENTIAL CALCULUS ρ= r dr r3 = − 2⋅ dp a r3 ⋅ a2 Example 3 : Find the radius of curvature at the point ( p, r) on the ellipse Neglecting the negative sign, we have ρ = 1 1 1 r2 = 2 + 2 − 2 2⋅ 2 p a b a b Solution : Differentiating the given equation with respect to r, we get 2r 2 dp = − 2 2⋅ ∴ a b p3 dr a 2 b2 a 2 b2 dr = r⋅ = ⋅ ρ= r dp r p3 p3 − H ence dp r p3 = 2 2⋅ dr a b 11.10 Tangential Polar Formula for Radius of Curvature A relation between p and ψ holding for every point of a curve, is called tangential polar equation of the curve. Thus the tangential polar equation of the curve is of the form p = f (ψ). dp dp dr ds dp We have = ⋅ ⋅ = cos φ . ρ dψ dr ds dψ dr ⎡ . . dr dr ds dr ⎤ dp cos φ . r = cos φ and ρ = = r ⎥ = ⎢ . dp dψ dp ⎦ dr ⎣ ds = r cos φ . Also p = r sin φ . ∴ p2 ⎛ dp ⎞ 2 + ⎜ ⎟ = r2 (sin2 φ + cos2 φ ) or ⎝ dψ⎠ Differentiating (1) with respect to p, we get ⎛ dp ⎞ d2 p dψ dr 2p + 2 ⎜ ⎟ ⋅ ⋅ = 2r dp ⎝ dψ⎠ dψ2 dp H ence ρ= p+ p2 or ⎛ dp ⎞ 2 + ⎜ ⎟ = r2. ⎝ dψ ⎠ r dr d2 p = p+ ⋅ dp dψ2 d2 p ⋅ dψ2 Example 1 : Show that for the epi-cycloid p = a sin bψ, ρ varies as p. dp d2 p Solution : We have = ab cos bψ and = − ab2 sin bψ = − b2 p. dψ dψ2 d2 p = p − b2 p = (1 − b2) p. dψ2 H ence ρ varies as p. ∴ ρ= p+ …(1) D-233 CURVATURE 1. 2. 3. 4. 5. Find the radius of curvature at the point ( p, r) on the following curves : (i) p2 = ar (parabola). (iii) a2 p = r3 (Lemniscate). (ii) 2ap2 = r3 (cardioid). r4 (iv) p2 = 2 ⋅ (r + a2) ⎛ dφ ⎞ r Prove that for any curve ρ = sin φ ⎜ 1 + ⎟ , where ρ is the radius of curvature dθ ⎠ ⎝ dθ (Gorakhpur 2005) and tan φ = r ⋅ dr In the curve p = r n + 1 ⁄ an , show that the radius of curvature varies inversely as the (n − 1) th power of the radius vector. Find the radius of curvature at the point (r, θ) on each of the following curves : (i) r = a cos θ. (Kanpur 2006) (ii) r (1 + cos θ) = 2a. (iii) r n = an cos n θ. (Rohilkhand 2005) (iv) r n = an sin n θ (Agra 2006; Rohilkhand 12; Avadh 12) (v) r = a (1 − cos θ). (Avadh 2010) (vi) r2 = a2 cos 2θ. Forming the pedal equation of the curve a θ = a− 1 (r2 − a2) 1 ⁄ 2 − cos− 1 ⎛⎜ ⎞⎟ , show that ρ = √(r2 − a2). ⎝ r⎠ (Meerut 2006B, 08; Rohilkhand 06; Kashi 11) 6. 1 For the rectangular hyperbola xy = c2, prove that ρ = r3 ⁄ c2, r being the central 2 radius vector of the point considered. 7. 8. Show that at any point on the equiangular spiral r = aeθ cot α, ρ = r cosec α, and that it subtends a right angle at the pole. If ρ1, ρ2 be radii of curvature at the extremities of any chord of the cardioid r = a (1 + cos θ), which passes through the pole, then show that ρ12 + ρ22 = 16a2 ⁄ 9. 9. 10. 1. Show that the radius of curvature at any point on the curve r = a (1 ± cos θ) varies as square root of the radius vector. Find the radius of curvature of the cardioid r = a (1 − cos θ) at the pole (origin). (i) 2p3 a2 (ii) 2 3 √(2ar) (iv) (r2 + a2) 3 ⁄ 2 / (r2 + 2a2). 4. (Kanpur 2008) (i) a ⁄ 2 (ii) 2√(r3 ⁄ a) (iii) a2 ⁄ 3r a 2 b2 ⋅ p3 an (iii) (n + 1) r n − 1 2. D-234 (iv) 10. DIFFERENTIAL CALCULUS an (n + 1) r n − 1 (v) 2 3 √(2ar) (vi) a2 ⋅ 3r 0. 11.11 Co-ordinates of Centre of Curvature (Meerut 2008) Let the equation of the curve be y = f (x). Let P be the given point (x, y) on this curve and Q the point (x + δx, y + δy) in the neighbourhood of P. (See the fig. of article 11.3). Let N be the point of intersection of the normals at P and Q. As Q → P, suppose N → C. Then C is the centre of curvature of P. Suppose co-ordinates of C are (α, β). dy From the equation of the curve, we have = f ′ (x) = φ (x), say. dx The equation of normal at P is (Y − y) φ (x) + (X − x) = 0. ...(1) The equation of normal at Q is …(2) {Y − ( y + δy)} φ (x + δx) + {X − (x + δx)} = 0. Subtracting (1) from (2), we get (Y − y) {φ (x + δx) − φ (x)} − φ (x + δx) δy − δx = 0. Dividing by δx, we get ⎧ φ (x + δx) − φ (x) ⎫ δy (Y − y) ⎨ ⎬ − φ (x + δx) − 1 = 0. δx δx ⎩ ⎭ …(3) The value of Y obtained from this equation will give us the y co-ordinate of the point of intersection of (1) and (2). Now as Q → P, δx → 0 and Y obtained from (3)→ β. Therefore taking limit of (3) as δx → 0, we get lim ⎧ φ (x + δx) − φ (x) ⎫ ⎬ ( β − y) δx → 0 ⎨ δx ⎭ ⎩ lim lim δy − 1= 0 − δx → 0 φ (x + δ x) . δx → 0 δx or ( β − y) d dy φ (x) − φ (x) ⋅ − 1= 0 dx dx or ( β − y) dy dy d ⎛ dy⎞ ⎜ ⎟ − ⋅ − 1 = 0. dx dx dx ⎝ dx⎠ or ( β − y) ⎫ d2 y ⎧ ⎛ dy⎞ 2 − ⎨ ⎜ ⎟ + 1⎬ = 0. 2 dx ⎝ ⎠ ⎩ ⎭ dx ⎛ ∴ β= y+ ⎞2 dy 1 + ⎜⎝ ⎟⎠ dx d2 y dx2 ⋅ ⎡.. dy⎤ ⎥ ⎢ . φ (x) = dx⎦ ⎣ D-235 CURVATURE Also (α, β) lies on (1). Therefore, we get dy + (α − x) = 0 ( β − y) dx i.e., ∴ ⎛ dy⎞ 2 1+ ⎜ ⎟ ⎝ dx⎠ dy dy (α − x) = − ( β − y) = − ⋅ dx dx d2 y dx2 2 ⎛ dy⎞ ⎫ dy ⎧ ⎨ 1 + ⎜⎝ ⎟⎠ ⎬ dx ⎭ dx ⎩ ⋅ α= x− d2 y dx2 11.12 Evolute of a Curve The locus of the centre of curvature of a curve is called its evolute. 11.13 Equation of the Circle of Curvature If (α, β) be the co-ordinates of the centre of curvature and ρ the radius of curvature at any point (x, y) on a curve, then the equation of the circle of curvature at that point is (X − α) 2 + (Y − β) 2 = ρ2. Example 1 : Find the co-ordinates of the centre of curvature for the point (x, y) on the parabola y2 = 4ax. A lso find the equation of the evolute of the parabola. ⎛ a⎞ dy dy 2a ⎜ ⎟⋅ Solution : H ere 2y = 4a, i.e., = = a1 ⁄ 2 x− 1 ⁄ 2 = ⎝ x⎠ dx dx y d2 √ 1 1⁄2 − 3⁄2 a x . 2 If (α, β) be the centre of curvature for the point (x, y), then ⎛ a⎞ ⎧ dy ⎧ a⎫ ⎛ dy⎞ 2⎫ ⎜ ⎟ ⎨1 + ⎬ ⎨1 + ⎜ ⎟ ⎬ ⎝ ⎠ ⎝ dx ⎠ ⎭ dx ⎩ x⎭ x ⎩ α= x− = x− = x + 2x (1 + a ⁄ x)⋅ 2 ⎛ a⎞ d y 11 ⎜ ⎟ − ⎝ x⎠ 2 x dx2 ∴ α = 3x + 2a, ...(1) 2 ⎛ dy⎞ ⎛ a⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ ⎝ dx⎠ ⎝ x⎠ and β= y+ = y+ d2 y 11 ⎛ a⎞ − ⎜ ⎟ ⎝ x⎠ 2 x dx2 ∴ y dx2 = − √ √ √ D-236 ∴ DIFFERENTIAL CALCULUS = 2a1 ⁄ 2 x1 ⁄ 2 − 2a− 1 ⁄ 2 x3 ⁄ 2 (1 + a ⁄ x) = 2a1 ⁄ 2 x1 ⁄ 2 − 2a− 1 ⁄ 2 x3 ⁄ 2 − 2a1 ⁄ 2 x1 ⁄.2 ⎛ x⎞ ⎜ ⎟ ⋅ β = − 2x ⎝ a⎠ √ …(2) ⎛ ⎜ Therefore the required centre of curvature is ⎜⎜ (3x + 2a), − 2x ⎝ √ ⎞ ⎛ x⎞ ⎟ ⎜ ⎟ ⎟⎟ ⎝ a⎠ ⎠ ⋅ Evolute of the parabola : Let us eliminate x between (1) and (2). From (2), we get β2 = 4 x3 a or x= From (i), we get x3 = aβ2 ⋅ 4 α − 2a ⋅ 3 ⎛ α − 2a ⎞ 3 aβ2 ⎜ ⎟ = or 27 aβ2 = 4 (α − 2a) 3. 4 3 ⎠ ⎝ H ence the locus of (α, β) is 27 ay2 = 4 (x − 2a) 3, which is the evolute of parabola. Example 2 : Prove that the co-ordinates (α, β) of the centre of curvature at any point (x, y) can be expressed in the form dy dx α= x− and β = y + ⋅ dψ dψ Solution : L e t C (α, β) b e t h e ce n t r e o f curvature of the point P (x, y) on the curve y = f (x). The line TP is tangent to the curve at P and obviously PC is normal at P to the curve, since the centre of curvature is a point on the normal. ρ = radius of curvature at P = PC and CN is the o r d in a t e o f C. D r a w PM p e rp e n dicu la r t o CN . O bviously ∴ ∠ CPM = Then π − ψ. α = x − PM = x − ρ sin ψ = x− = x− Also 1 2 β= y+ = y+ = y+ dy ds ⋅ ds dψ dy ⋅ dψ CM = y + ρ cos ψ dx ds . ds dψ dx ⋅ dψ ⎡.. ⎤ ds dy and = sin ψ⎥ ⎢ . ρ= dψ ds ⎣ ⎦ ⎡.. dx⎤ ⎥ ⎢ . cos ψ = ds ⎦ ⎣ 11.14 Chord of Curvature through the Origin (Pole) Let C be the centre of curvature at the point P on any given curve. Then CP = ρ = radius of curvature at P. O is the pole. Join OP to meet the circle of curvature in E. Then PE is the chord of curvature through the origin. D-237 CURVATURE PD is the diameter of the circle of curvature. We have PD = 2ρ and ∠ PED = 90o , being an angle in a semi-circle. Also PD is normal to the curve at P. ∴ ∠ EPD = 1 2 π − φ. H ence from the right-angled triangle PED, we have ⎞ ⎛1 PE = PD cos ⎜ π − φ ⎟ = 2ρ sin φ . ⎠ ⎝2 ∴ chord of curvature through pole = 2ρ sin φ . Corollary : Chord of curvature perpendicular to the Radius Vector : Suppose a line through P, perpendicular to the radius vector OP, meets the circle of curvature in F. Then PF is the chord of curvature perpendicular to the radius vector. ⎞ ⎛1 We have PF = ED = 2ρ sin ⎜ π − φ ⎟ = 2ρ cos φ . ⎠ ⎝2 ∴ Chord of curvature perpendicular to the radius vector = 2ρ cos φ . 11.15 Chord of Curvature Parallel to the Axes (i) Chord of Curvature Parallel to the x-axis : Let C be the centre of curvature at the point P on any given curve. Suppose a line through P, drawn parallel to the axis of x, meets the circle of curvature in E. Then PE is the chord of curvature parallel to x-axis. PD is the diameter of the circle of curvature. We have PC = 2ρ and ∠ PED = 90o . O bviously ∠ EPD = 1 2 π − ψ. H ence from the right angled triangle PED, we have ⎞ ⎛1 PE = 2ρ cos ⎜ π − ψ⎟ = 2ρ sin ψ. ⎠ ⎝2 ∴ Chord of curvature parallel to x-axis = 2ρ sin ψ. (ii) Chord of Curvature Parallel to y-axis : Draw a line through P, parallel to y-axis, to meet the circle of curvature in F. Then PF = chord of curvature parallel to y-axis ⎛π ⎞ = ED = 2ρ sin ⎜ − ψ⎟ = 2ρ cos ψ. ⎝2 ⎠ rn Example 1 : Show that the chord of curvature through the pole of the curve = an cos n θ is 2r ⁄ (n + 1). (Purvanchal 2014) Solution : The curve is r n = an cos n θ. Taking logarithm of both sides, we get n log r = n log a + log cos n θ. …(1) D-238 DIFFERENTIAL CALCULUS Differentiating with respect to θ, we get ⎛ 1 2 n dr n = − sin n θ r dθ cos nθ ⎞ cot φ = − tan n θ = cot ⎜⎝ π + n θ⎟⎠ ⋅ i.e., ∴ φ= 1 2 π + n θ. ⎛ ⎞ 1 2 Now p = r sin φ = r sin ⎜⎝ π + n θ⎟⎠ = r cos n θ. Therefore the pedal equation of the given curve is p = r n + 1 ⁄ an . ∴ dp (n + 1) rn = ⋅ dr an an dr = ⋅ dp (n + 1) r n − 1 H ence the chord of curvature through the pole ⎞ ⎛1 = 2ρ sin φ = 2ρ sin ⎜ π + n θ⎟ = 2ρ cos n θ ⎠ ⎝2 Also ρ= r rn 2r an ⋅ = ⋅ (n + 1) r n − 1 an (n + 1) Example 2 : In the curve y = a log sec (x ⁄ a) , prove that the chord of curvature parallel to the axis of y is of constant length. (Rohilkhand 2009) Solution : Differentiating the equation of the curve with respect to x, we get 1 dy = a. . sec (x ⁄ a) tan (x ⁄ a) (1 ⁄ a) = tan (x ⁄ a). sec (x ⁄ a) dx = 2⋅ d2 y = (1 ⁄ a) sec 2 (x ⁄ a). dx2 Chord of curvature parallel to y-axis 2ρ 2ρ = = 2ρ cos ψ = sec ψ ⎧ ⎛ dy⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ ⎩ ⎝ dx ⎠ ⎭ ∴ √ = 2⋅ = 2⋅ = 2⋅ ⎧ ⎛ dy⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ ⎩ ⎝ dx⎠ ⎭ d2 y dx2 ⎧ ⎛ dy⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ ⎩ ⎝ dx ⎠ ⎭ 3⁄2 ⋅ √ 1 ⎧ ⎛ dy⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ ⎩ ⎝ dx⎠ ⎭ d2 y dx2 ⎛ x⎞ ⎜ 1 + tan 2 ⎟ ⎝ a⎠ 1 ⎛ x⎞ sec2 ⎜ ⎟ ⎝ a⎠ a = 2a, which is constant. D-239 CURVATURE 1. 2. 3. In the parabola x2 = 4ay, prove that the co-ordinates of the centre of curvature ⎛ 3x2 ⎞⎟ x3 , 2a + are ⎜⎜ − ⎟ ⋅ 2 4a ⎟ ⎜ 4a ⎠ ⎝ In the catenary y = c cosh (x ⁄ c), show that the centre of curvature (α, β) is given by α = x − y {( y2 ⁄ c2) − 1)}1 ⁄ 2, β = 2y. For the curve a2 y = x3, show that the centre of curvature (α, β) is given by 9x4 ⎫⎪ 5x3 a2 x ⎧⎪ + ⋅ ⎨1 − 4 ⎬ , β = 2 ⎪⎩ a ⎪⎭ 2a2 6 x Show that the centre of curvature (α, β) at the point determined by t on the ellipse a 2 − b2 a 2 − b2 x = a cos t, y = b sin t, is given by α = cos3 t, β = − sin3 t. a b α= 4. Also show that the evolute of the ellipse is (ax) 2 ⁄ 3 + (by) 2 ⁄ 3 = (a2 − b2) 2 ⁄ 3. 5. Prove that the centre of curvature (α, β) for the curve x = 3t, y = t 2 − 6 is α = − 6. 7. 8. 4 3 t , 3 10. ⋅ Show that in any curve the chord of curvature perpendicular to the radius vector is 2ρ √(r2 − p2) ⁄ r. Show that the chord of curvature through the pole of the equiangular spiral r = ae m θ is 2r. Show that the chord of curvature, through the pole, for the cardioid r = a (1 + cos θ) is 9. 3 2 β = 3t 2 − 4 3 r. Show t hat t he circle of curvature at the point (am 2, 2am) of the parabola y2 = 4ax has for its equation x2 + y2 − 6am 2 x − 4ax + 4am 3 y − 3a2 m 4 = 0. If Cx and Cy be the chords of curvature parallel to the axes at any point of the curve 1 1 1 y = ae x ⁄ a, prove that 2 + = ⋅ 2 2aC Cx Cy x (Agra 2007; Rohilkhand 07; Purvanchal 07) Fill in the Blanks: 1. 2. 3. 4. 5. Fill in the blanks “……”, so that the following statem ents are com plete and correct. The relation between s and ψ for any curve is called its …… equation. By definition the curvature of the curve at any point P is equal to …… . For a curve y = f (x), we have ρ = …… . Intrinsic formula for the radius of curvature is …… . The radius of curvature at any point of the cycloid x = a (t + sin t), y = a (1 − cos t) is …… . D-240 6. DIFFERENTIAL CALCULUS For a curve defined by the equations x = f (t) and y = φ (t) the radius of curvature is …… . Multiple Choice Questions: 7. 8. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). If y-axis is the tangent to the given curve at the origin, then radius of curvature at the origin is equal to x2 y2 x2 y2 (b) Lim (c) Lim (d) Lim (a) Lim x →0 2 y x →0 2 x x →0 y x →0 x Pedal formula for radius of curvature is 1 dr dr (b) r r dp dp Chord of curvature parallel to y-axis is (a) 2 ρ sin φ (b) 2 ρ cos φ (a) 9. (c) 1 dp r dr (d) r (c) 2 ρ sin ψ dp dr (d) 2 ρ cos ψ True or False: 10. Write ‘T’ for true and ‘F’ for false statement. There is no difference between curvature of the circle and circle of curvature. (Meerut 2003) 11. The curvature of the curve at any point is equal to the reciprocal of the radius of curvature at that point. ⎧ ⎪ ⎨r2 ⎪ ⎩ ⎛ ⎞ 2 ⎫3 ⁄ 2 ⎪ 12. dr + ⎜⎝ ⎟⎠ ⎬⎪ dθ ⎭ ⋅ The polar formula for radius of curvature is : ρ = ⎛ d r⎞ 2 d2r 2 ⎜ ⎟ r + 2⎝ ⎠ − r 2 dθ dθ 13. The tangential polar formula for radius of curvature is ρ = p + 14. Pedal formula for radius of curvature is ρ = r 1. intrinsic. 5. 4a cos 7. 11. (b). T. 1 t. 2 2. dψ ⋅ ds 3. ⎡ ⎢ ⎢ ⎢1 ⎣ + d2p d ψ2 dr ⋅ dp ⎛ dy ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ dx ⎠ d2y dx 2 2 ⎤3⁄2 ⎥ ⎥ ⎥ ⎦ (Agra 2006) ⋅ 4. ds ⋅ dψ (x ′ 2 + y ′ 2) 3 ⁄ 2 , where x ′ y ′′ − y ′ x ′′ ≠ 0. x ′ y ′′ − y ′ x ′′ 8. (b). 9. (d). 10. F. 12. T. 13. T. 14. T. 6. ⋅ 12.1 One Parameter Family of Curves An equation of the form F (x, y, α) = 0 ...(1) in which α is a constant, represents a curve. If α is a parameter i.e., if α can take all real values, then (1) is the equation of a one parameter family of curves with param eter α . If we give different values to α we get different members of this family. O n any particular curve belonging to this family the value of α is constant but it changes from one curve to another. An equation of the type F (x, y, α, β) = 0 also defines a family of curves but in this case we have two parameters α and β . Illustrations : (i) The equation x cos α + y sin α = a determines a family of straight lines, and α is the parameter of this family. (ii) The equation y = m x − 2 am − am 3 determines a family of straight lines which are normals to the parabola y2 = 4ax . H ere m is the parameter. 12.2 Envelope of a One Parameter Family of Curves (Kashi 2011) Definition : Let F (x, y, α) = 0 be a family of curves, the parameter being α . D-242 DIFFERENTIAL CALCULUS Su p p o se P i s a p o i n t o f i n t e r s e c t i o n o f t wo m e m b e r s F (x, y, α) = 0 a n d F (x, y, α + δα) = 0 of t his family corresponding to t he paramet er values α a n d α + δα . As δα → 0 , let P tend to a definite point Q on the member α . The locus of Q (for varying values of α ) is called the envelope of the family. The above definition can be given in concise form as below : The envelope of a one param eter fam ily of curves is the locus of the limiting positions of the points of intersection of any two m em bers of the fam ily when one of them tends to coincide with the other which is kept fixed. Thus the envelope of a family of curves is the locus of the points of intersection of consecutive members of the family. 12.3 Method of Finding the Envelope Suppose F (x, y, α) = 0 ...(1) is the equation of a family of curves with parameter α . Consider the two members F (x, y, α) = 0 and F (x, y, α + δα) = 0 ...(2) of this family corresponding to the parameter values α and α + δα . The co-ordinates of the points of intersection of the curves (2) satisfy the equations F (x, y, α) = 0 , F (x, y, α + δα) − F (x, y, α) = 0 and therefore the equations F (x, y, α + δα) − F (x, y, α) F (x, y, α) = 0 , = 0. δα Taking limits as δα → 0, we see that the co-ordinates of the limiting positions of the points of intersection of the curves (2) satisfy the equations ...(3) ∂ F (x, y, α) = 0. F (x, y, α) = 0 and ∂α Thus, for all values of α , the co-ordinates of the points on the envelope satisfy the equations (3). Therefore eliminating α between the equations (3), we shall get the envelope of the family of curves (1). Working Rule : T h e eq u a t io n o f th e en v elo p e o f t h e fa m ily o f cu rv es F (x, y, α) = 0 where α is the param eter, is obtained by elim inating α between the equations F (x, y, α) = 0 ∂ F (x, y, α) and = 0. ∂α ∂ F (x, y, α) is the partial derivative of F (x, y, α) with respect to the parameter H ere ∂α α while x and y have been regarded as constants. Note : The equations x = φ (α) , y = ψ (α) obtained on solving F (x, y, α) = 0 and ∂F (x, y , α) = 0 are the parametric equations of the envelope, α being the parameter. ∂α Example 1 : Find the envelope of the fam ily of straight lines y = mx + (a ⁄ m) , the (Kanpur 2005) param eter being m . D-243 ENVELOPES, EVOLUTES AND INVOLUTES Solution : The equation of the given family of straight lines is y = m x + (a ⁄ m) , the parameter being m . Differentiating (1) partially with respect to m , we get ...(1) 0 = x − (a ⁄ m 2) or m = (a ⁄ x) 1 ⁄ 2 . E liminating m between (1) and (2), we get the required envelope. Putting m = (a ⁄ x) 1 ⁄ 2 in (1), we get y= x⋅ ...(2) x1 ⁄ 2 a1 ⁄ 2 + a ⋅ 1 ⁄ 2 = 2a1 ⁄ 2 . x1 ⁄ 2. 1 ⁄ 2 x a Squaring, we get y2 = 4ax , which is the required envelope. Example 2 : Find the envelope of the fam ily of straight lines by ax − = a 2 − b2 , cos θ sin θ where the parameter is θ . (Kashi 2013; Avadh 13) Solution : The equation of the given family of straight lines is by ax − = a2 − b2, θ being the parameter. ...(1) cos θ sin θ Differentiating (1) partially with respect to θ , we get ax sin θ by cos θ + = 0. ...(2) cos2 θ sin2 θ E liminating θ between (1) and (2), we get the required envelope. From (2), we get tan3 θ = − (by ⁄ ax) . or ∴ tan θ = − (by) 1 ⁄ 3 ⁄ (ax) 1 ⁄ 3. ∴ sin θ = (by) 1 ⁄ 3 (ax) 1 ⁄ 3 , cos θ = − 2 ⁄ 3 2 + (by) ] √[(ax) ⁄ 3 + (by) 2 ⁄ 3] √[(ax) 2 ⁄ 3 (by) 1 ⁄ 3 (ax) 1 ⁄ 3 , cos θ = ⋅ 2 ⁄ 3 2 + (by) ] √[(ax) ⁄ 3 + (by) 2 ⁄ 3] Substituting these values in (1), we get sin θ = − √[(ax) 2 ⁄ 3 ± [(ax) 2 ⁄ 3 + (by) 2 ⁄ 3] [(ax) 2 ⁄ 3 + (by) 2 ⁄ 3]1 ⁄ 2 = a2 − b2 or ± [(ax) 2 ⁄ 3 + (by) 2 ⁄ 3]3 ⁄ 2 = a2 − b2 i.e., (ax) 2 ⁄ 3 + (by) 2 ⁄ 3 = (a2 − b2) 2 ⁄ 3, which is the equation of the required envelope. 12.4 Envelope in Case the Equation of the Family of Curves is a Quadratic in the Parameter (Garhwal 2002) Let the equation of the family of curves be F (x, y, α) = 0 , the parameter being α . Suppose this equation can be arranged as a quadratic in α . Let this quadratic be Aα2 + Bα + C = 0 , where A , B and C are some functions of x and y . Differentiating (1) partially with respect to α , we get 2Aα + B = 0 . ...(1) ...(2) D-244 DIFFERENTIAL CALCULUS E liminating α between (1) and (2), we get the envelope. From (2), we have α = (− B ⁄ 2A) . Substituting this value of α in (1), we get B ⎞2 B⎞ ⎛ ⎛ A ⎜− ⎟ + B ⎜− ⎟ + C = 0 or B 2 − 4AC = 0 , ⎝ 2A ⎠ ⎝ 2A ⎠ which is the required equation of the envelope. Remember : The envelope of the fam ily of curves Aα2 + Bα + C = 0 , where A , B , C are functions of x and y , is B2 − 4AC = 0 . Example 1 : Find the envelope of the fam ily of straight lines y = m x + √(a2 m 2 + b2) , the param eter being m . (Garhwal 2002; Rohilkhand 14; Purvanchal 14) Solution : The equation of the given family of straight lines is y = m x + √(a2 m 2 + b2) or y − m x = √(a2 m 2 + b2) or ( y − m x) 2 = a2 m 2 + b2 or y2 − 2m xy + m 2 x2 − a2 m 2 − b2 = 0 ...(1) m 2 (x2 − a2) − 2xym + y2 − b2 = 0 . The equation (1) is a quadratic in the parameter m . So the required envelope is obtained by equating to zero the discriminant of (1). Hence the required envelope is (− 2xy) 2 − 4 (x2 − a2) ( y2 − b2) = 0 or or x2 y2 − (x2 − a2) ( y2 − b2) = 0 or x2 y2 − x2 y2 + x2 b2 + a2 y2 − a2 b2 = 0 or x2 b2 + a2 y2 = a2 b2 or y2 x2 + = 1 , which is an ellipse. a 2 b2 Example 2 : Find the envelope of the fam ily of straight lines x ⁄ a + y ⁄ b = 1 , where the two param eters a , b are connected by the relation ab = c2, c being a constant. (Kumaun 2002; Meerut 12) Solution : The equation of the given family of straight lines is x⁄ a + y⁄ b = 1, ...(1) where the parameters a , b are connected by the relation ab = c2. ...(2) We shall eliminate one parameter, say b . From (2), we have b = c2 ⁄ a . Putting the value of b in (1), we get y x ay x + = 1 or + 2 = 1 , ...(3) a c a c2 ⁄ a which is the equation of the given family and it contains only one parameter a . We can arrange (3) as a quadratic in a . Thus (3) can be written as ...(4) c2 x + a2 y = ac2 or a2 y − ac2 + c2 x = 0 . The equation (4) is a quadratic in the parameter a . So the required envelope is D-245 ENVELOPES, EVOLUTES AND INVOLUTES (− c2) 2 − 4yc2 x = 0 or or c4 − 4 xyc2 = 0 xy = c2 ⁄ 4 . 12.5 Geometrical Significance of the Envelope In general the envelope of a fam ily of curves touches each m em ber of the fam ily. Let the equation of the family of curves be ...(1) F (x, y, α) = 0 , α being the parameter. The envelope of (1) is obtained by eliminating α between (1) and ∂ F (x, y, α) = 0. ...(2) ∂α O bviously we can take (1) as the equation of the envelope provided we regard α as a function of x and y given by (2). Let (x, y) be a point common to the member ‘α’ of the family and the envelope. If ∂F ∂F , at the point (x, y) we do not have then at this point the slope of the tangent = 0= ∂x ∂y to the member ‘α’ of the family is ∂F ⁄ ∂x − ⋅ ...(3) ∂F ⁄ ∂y [R efer the chapter on partial differentiation] Also the slope of the tangent to the envelope at the point (x, y) is ∂F ∂F ∂α + ∂α ∂x ∂x ⋅ ...(4) − ∂F ∂F ∂α + ∂α ∂y ∂y Note that F (x, y, α) = 0 is also the equation of the envelope provided α is not a constant but is a function of x and y given by ∂F ⁄ ∂α = 0 . Since at every point of the envelope we have ∂F ⁄ ∂α = 0, therefore the two slopes given by (3) and (4) are the same. H ence the slopes of the tangents to the member of the family and the envelope at the common points are equal. This means that the curves of the family and the envelope have the same tangent at the points in common i.e., they touch each other at these points. E ach point on the envelope is a point on some curve of the family and each curve of the family has some point which is on the envelope. At these common points both touch each other. H ence, in general, the envelope of a fam ily of curves touches each curve of the fam ily and at each point is touched by som e m em ber of the fam ily. Note : If ∂F ⁄ ∂x and ∂F ⁄ ∂y are both zero for any point on the curve, the slopes of the tangents cannot be found from (3) and (4). In this case the above argument breaks down and the envelope may not touch a curve at the points where ∂F ⁄ ∂x = 0 = ∂F ⁄ ∂y i.e., at the singular points. If the given family of curves is a family of straight lines or a family of conics we have no singular points. H ence the envelope of a family of straight lines or of conics touches each member of the family at all their common points without exception. D-246 DIFFERENTIAL CALCULUS Example 1 : Find the envelope of the fam ily of straight lines x cos α + y sin α = a , the param eter being α , and interpret the result geom etrically. Solution : The equation of the given family of straight lines is x cos α + y sin α = a , the parameter being α . Differentiating (1) partially with respect to α , we get − x sin α + y cos α = 0 . ...(1) ...(2) E liminating α between (1) and (2), we get the envelope. So squaring and adding (1) and (2), we get (x cos α + y sin α) 2 + (− x sin α + y cos α) 2 = a2 or x2 (cos2 α + sin2 α) + y2 (sin2 α + cos2 α) = a2 or x2 + y2 = a2, which is the required envelope. Geometrical interpretation : x2 + y2 = a2 is the equation of a circle whose centre is origin and radius is a . This circle is the envelope of the family of straight lines x cos α + y sin α = a . So for each value of α , the straight line x cos α + y sin α = a t o uch es th e circle x2 + y2 = a2. Also th e circle x2 + y2 = a2 is touched at each point by some straight line belonging to the family x cos α + y sin α = a . Example 2 : Find the envelope of the fam ily of circles x2 + y2 − 2ax cos α − 2ay sin α = c2, where α is the parameter, and interpret the result. (Bundelkhand 2014) Solution : The equation of the given family of circles can be written as ...(1) 2ax cos α + 2ay sin α = x2 + y2 − c2. [Note that we have brought the terms containing cos α and sin α to one side and the rest of the terms to the other side]. Differentiating (1) partially with respect to α , we get − 2ax sin α + 2ay cos α = 0 . ...(2) Squaring and adding (1) and (2), we get 4a2 x2 + 4a2 y2 = (x2 + y2 − c2) 2 or (x2 + y2 − c2) 2 = 4a2 (x2 + y2) , which is the required envelope. Interpretation : The equation (3) can be written as (x2 + y2) 2 − 2 (2a2 + c2) (x2 + y2) + c4 = 0 . Solving it as a quadratic in (x2 + y2) , we get 2 (2a2 + c2) ± √{4 (2a2 + c2) 2 − 4c4} x2 + y2 = 2 ...(3) ENVELOPES, EVOLUTES AND INVOLUTES D-247 = 2a2 + c2 ± 2a √(c2 + a2) = (a2 + c2) ± 2a √(a2 + c2) + a2 = [√(a2 + c2) ± a]2. Therefore the required envelope consists of the two circles x2 + y2 = [√(a2 + c2) + a]2 and x2 + y2 = [√(a2 + c2) − a]2. These are the circles with centre at origin and radii √(a2 + c2) ± a . 1. 2. 3. 4. 5. 6. 7. 8. F in d t h e e n ve lo p e o f t h e st r a igh t lin e s (x ⁄ a) cos θ + ( y ⁄ b) sin θ = 1, the parameter being θ and interpret the result geometrically. (Kashi 2012) Find the envelope of the following families of straight lines : (i) y = m 2 x + (1 ⁄ m 2), the parameter being m . (ii) y = m x + a √(1 + m 2) , the parameter being m . (Kumaun 2000) (iii) y = m x + am 3, the parameter being m . (iv) y = m x + am p, the parameter being m . (v) x cosec θ − y cot θ = c , the parameter being θ. (vi) x cos3 α + y sin3 α = a , the parameter being α . Find the envelope of the family of circles x2 + y2 − 2ax cos α − 2ay sin α + c2 = 0, (a2 > c2) where α is the parameter, and interpret the result. Find the envelope of the following families of circles : (i) (x − α) 2 + y2 = 4α, α being the parameter. (ii) (x − α) 2 + ( y − α) 2 = 2α , α being the parameter. (Rohilkhand 2013) (iii) (x − c) 2 + y2 = R 2, where c is the parameter. (iv) y2 = m 2 (x − m), m being the parameter. (v) t x3 + t2 y = a, the parameter being t . x2 y2 (vi) 2 + 2 = 1, where α is the parameter. α k − α2 (Kanpur 2008) Find the envelope of the family of curves (a2 ⁄ x) cos θ − (b2 ⁄ y) sin θ = c2 ⁄ a, θ being the parameter. Find the envelope of the family of straight lines x cos n θ + y sin n θ = a, for different values of θ. Find the envelope of the ellipse x = a sin (θ − α), y = b cos θ, where α is the parameter. (Kanpur 2009) Projectiles are fired from a gun with a constant initial velocity v0 . Supposing the gun can be given any elevation and is kept always in the same vertical plane, what is the envelope of all possible trajectories, assuming their equation to be 1 gx2 y = x tan α − ? 2 2 v0 cos2 α D-248 9. DIFFERENTIAL CALCULUS Find the envelope of the straight lines x cos α + y sin α = l sin α cos α , where α is the parameter. Give the geometrical interpretation. (Rohilkhand 2012; Avadh 07, 12) 10. 11. 12. 13. 14. Find the envelope of the family of straight lines x cos mα + y sin mα = a (cos nα) m ⁄ n , where α is the parameter. [Hint. Change the equation to polar co-ordinates by substituting x = r cos θ, y = r sin θ.] Show that the radius of curvature of the envelope of the line x cos α + y sin α = f (α) is f (α) + f ′′ (α) . y x If x2 ⁄ 3 + y2 ⁄ 3 = k 2 ⁄ 3 is the envelope of the lines + = 1, then find the a b necessary relation between a, b and k . Find the envelope of the family of curves x2 sin α + y2 cos α = a2, where α is the parameter. Find the envelope of the family of curves ( y − c) 2 − 2 3 (x − c) 3 = 0, where c is the parameter. 1. 2. 3. 4. 5. 6. 8. 10. 12. 14. x2 ⁄ a2 + y2 ⁄ b2 = 1. E ach line of the given family is a tangent to the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1. (ii) x2 + y2 = a2 . (i) y2 = 4 x . (iv) ( p − 1) p − 1 x p + p p ay p − 1 = 0. (iii) 4 x3 + 27ay2 = 0. 2 2 2 (vi) a2 (x2 + y2) = x2 y2. (v) x − y = c . 2 2 2 2 2 2 2 (x + y + c ) = 4a (x + y ). Circles with centre at origin and radii a ± √(a2 − c2). (i) y2 − 4x − 4 = 0. (ii) (x + y + 1) 2 = 2 (x2 + y2). (iii) y = ± R . (iv) 4 x3 = 27y2. (v) x6 + 4ay = 0. (vi) x ± y = ± k . (a2 ⁄ x) 2 + (b2 ⁄ y) 2 = (c2 ⁄ a) 2. x2 ⁄ (2 − n) + y2 ⁄ (2 − n) = a2 ⁄ (2 − n) . 7. x= ± a. 9. x2 ⁄ 3 + y2 ⁄ 3 = l 2 ⁄ 3 . 1 y + ( g x2) ⁄ v02 = v02 ⁄ (2g) . 2 r n ⁄ (m − n) = an ⁄ (m − n) cos {nθ ⁄ (m (x, y) . a 2 + b2 = k 2 . x − y = 0, x − y = − n)}, where r, θ are the polar coordinates of 13. 2 ⋅ 9 x4 + y4 = a4. 12.6 Envelope of a Family of Curves whose Equation is not given in a Direct Form Sometimes, we are to find the envelope of a family of curves whose equation is not given in a direct form, but we are given a law in accordance with which the equation D-249 ENVELOPES, EVOLUTES AND INVOLUTES of any member of the family can be obtained. In such cases we should first find the equation of the family of curves in a proper form and then we should find the envelope. Example 1 : Find the envelope of the circles which pass through the origin and whose centres lie on the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 . Solution : Any point on the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 is (a cos θ, b sin θ) . Its distance from the origin is √(a2 cos2 θ + b2 sin2 θ) . Therefore the equation of the given family of circles is (x − a cos θ) 2 + ( y − b sin θ) 2 = a2 cos2 θ + b2 sin2 θ x2 + y2 − 2ax cos θ − 2by sin θ = 0 . ...(1) We are to find the envelope of the family of circles (1), where θ is the parameter. The equation (1) may be written as or 2ax cos θ + 2by sin θ = x2 + y2. Differentiating (2) partially with respect to θ , we get − 2ax sin θ + 2by cos θ = 0 . Squaring and adding (2) and (3), we get ...(2) ...(3) 4a2 x2 + 4b2 y2 = (x2 + y2) 2 or (x2 + y2) 2 = 4 (a2 x2 + b2 y2) , which is the required envelope. Example 2 : Find the envelope of the circles drawn on the radii vectors of the parabola y2 = 4ax as diam eter. Solution : Any point on the parabola y2 = 4ax is (at2, 2at) . E quation of the circle drawn on the line joining the origin (0, 0) to the point (at2, 2at) as diameter is (x − 0) (x − at2) + ( y − 0) ( y − 2at) = 0 x2 + y2 − axt2 − 2aty = 0 . ...(1) We are to find the envelope of the family of circles (1), where t is the parameter. Differentiating (1) partially with respect to t , we get 0 − 2axt − 2ay = 0 . ...(2) E liminating t between (1) and (2), we get the required envelope. From (2), we get t = − y⁄ x. Putting this value of t in (1), we get or x2 + y2 − ax .( y2 ⁄ x2) + 2ay .( y ⁄ x) = 0 x2 + y2 − (ay2 ⁄ x) + (2ay2 ⁄ x) = 0 or or x2 + y2 + (ay2 ⁄ x) = 0 ay2 + x (x2 + y2) = 0 , which is the required envelope. Example 3 : Find the envelope of the circles drawn on the radii-vectors of the curve r n = an cos nθ as diam eter. Solution : Let P be any point on the curve r n = an cos nθ . If α is the vectorial angle of P , then the radius vector OP is given by or (OP) n = an cos nα . ∴ OP = a (cos nα) 1 ⁄ n . D-250 DIFFERENTIAL CALCULUS Let Q be any point (r, θ) on the circle drawn on OP as diameter. From the right angled triangle OQP , we have OQ = OP cos ∠ POQ . ...(1) ∴ r = a (cos nα) 1 ⁄ n cos (θ − α) , is the equat ion of the circle drawn on OP as diameter. We are to find the envelope of the family of circles (1), where α is the parameter. Taking logarithm of both sides of (1), we get or or log r = log a + (1 ⁄ n) log cos nα + log cos (θ − α) . Differentiating (2) partially with respect to α , we get sin (θ − α) n 1 (− sin nα) + 0= 0+ ⋅ cos (θ − α) n cos nα tan nα = tan (θ − α) . ∴ nα = θ − α (taking principal value only) α = θ ⁄ (n + 1) . Substituting this value of α in (1), we get the required envelope. Thus from (1), [... θ − α = nα] r = a (cos nα) 1 ⁄ n cos nα or r = a (cos nα) 1 + 1 ⁄ n or ...(2) r = a (cos nα) (n + 1) ⁄ n or r n ⁄ (n + 1) = an ⁄ (n + 1) cos {nθ ⁄ (n + 1)}, which is the required envelope. Example 4 : Find the envelope of the straight lines drawn at right angles to the radii vectors of the cardioid r = a(1 + cos θ) through their extrem ities. Solution : Let P be any point on the cardioid r = a (1 + cos θ). If α is the vectorial angle of P, then the radius vector OP is given by OP = a (1 + cos α) . Let Q be any point on the straight line drawn through P and at right angles to OP . From the right angled triangle OPQ , we have OP = OQ cos ∠ POQ . ∴ a (1 + cos α) = r cos (θ − α) or 1 2 r cos (θ − α) = 2a cos2 α , ...(1) is the equation of the straight line drawn through P and at right angles to OP . We are to find the envelope of the family of straight lines (1), where α is the parameter. Taking logarithm of both sides of (1), we get ...(2) log r + log cos (θ − α) = log 2a + 2 log cos (α ⁄ 2) . Differentiating (2) partially with respect to α , we get ⎛ 1 2 1 ⎞ 1 ⎜ sin α⎟ ⋅ 0+ sin (θ − α) = 0 − 1 ⎝ cos (θ − α) 2 ⎠ 2 cos α 2 D-251 ENVELOPES, EVOLUTES AND INVOLUTES 1 2 tan (θ − α) = − tan α = tan (− or ∴ θ− α= nπ − 1 2 1 2 α) . α , where n is any integer. ∴ α = 2θ − 2n π . Substituting this value of α in (1), we get r cos {θ − (2θ − 2n π)} = 2a cos2 (θ − n π) or or pole. r cos (2n π − θ) = 2a cos2 θ r cos θ = 2a cos2 θ or r = 2a cos θ . Therefore the required envelope is r = 2a cos θ . It is a circle passing through the 12.7 Two Parameters Connected by a Relation Suppose the equation of the family of curves contains two parameters which are connected by a relation. We can find the envelope of this family by eliminating one parameter as we have done in one earlier example. But if the elimination of one parameter makes the subsequent process of finding the envelope difficult, we can proceed as in the following example. ym xm Example 1 : Find the envelope of the fam ily of curves m + m = 1 , where the a b param eters a and b are connected by the relation a p + b p = c p. Solution : The equation of the given family of curves is xm ym + = 1, ...(1) am bm where the parameters a and b are connected by the relation ...(2) a p + b p = c p. Since there is a relation between a and b , therefore we shall regard b as a function of a . Now we shall differentiate (1) and (2) with respect to a regarding x and y as constants and b as a function of a . From (1), we get x m ⁄ am + 1 m y m db db m xm = 0 i.e, = − m m+ 1⋅ ...(3) − m+ 1− m+ 1 da da a b y ⁄b Again from (2), we get pa p − 1 + pb p − 1 (db ⁄ da) = 0 i.e., db ⁄ da = − a p − 1 ⁄ b p − 1. ...(4) E quating the two values of (db ⁄ da) , we get ...(5) x m ⁄ am x m ⁄ am + 1 a p − 1 ap = or = p⋅ m m + 1 p − 1 m m y ⁄b b y ⁄b b E liminating a and b between (1), (2) and (5), we get the required envelope. From (5), we have x m ⁄ am y m ⁄ bm x m ⁄ a m + y m ⁄ bm 1 = = = p⋅ [ Note] p p c a b ap + bp D-252 ∴ or or or DIFFERENTIAL CALCULUS xm ⁄ a p + m = 1 ⁄ c p a p + m = xm c p a = (x m c p) 1 ⁄ ( p + m) 2 a p = (x m c p) p ⁄ ( p + m) = x m p ⁄ ( p + m) c p ⁄ ( p + m) . 2 Similarly b p = y mp ⁄ ( p + m) c p ⁄ ( p + m) . Substituting these values of a p and b p in (2), we get 2 c p ⁄ ( p + m) {x m p ⁄ ( p + m) + y m p ⁄ ( p + m) } = c p or 2 x m p ⁄ ( p + m) + y m p ⁄ ( p + m) = c p − p ⁄ ( p + m) or x m p ⁄ ( p + m) + y m p ⁄ ( p + m) = c m p ⁄ ( p + m) , which is the required envelope. 1. 2. 3. 4. Find the envelope of the circles drawn upon the radii vectors of the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 as diameter. Show that the envelope of the circles whose centres lie on the parabola y2 = 4ax and which pass through its vertex is the cissoid y2 (2 a + x) + x3 = 0. Show that the envelope of the circles whose centres lie on the rectangular hyperbola x2 − y2 = a2 and wh ich p ass through the origin is the lemniscate r2 = 4a2 cos 2θ. Show t hat t he en velo pe of th e circles described on t he cent ral radii of a rectangular hyperbola as diameters is a lemniscate r2 = a2 cos 2θ. 5. 6. x2 y2 Show that the envelope of the polars of points on the ellipse 2 + 2 = 1 with h k y2 h2 x2 k 2 y2 x2 + = 1. respect to the ellipse 2 + 2 = 1 is (Bundelkhand 2011) a b a4 b4 Show that the envelope of the straight line joining the extremities of a pair of conjugate diameters of the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 is the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 7. 8. 9. 1 ⋅ 2 Find the envelopes of circles described on the radii vectors of the following curves as diameters (ii) r3 = a3 cos 3θ, (iii) r cos n (θ ⁄ n) = a . (i) l ⁄ r = 1 + e cos θ, Find the envelopes of the straight lines drawn at right angles to the radii vectors of the following curves through their extremities : (i) r = ae θ cot α, (ii) r n = a n cos nθ, (iii) r = a + b cos θ. Find the envelope of the straight line x ⁄ a + y ⁄ b = 1, where the parameters a and b are connected by the following relations D-253 ENVELOPES, EVOLUTES AND INVOLUTES (i) an + bn = cn , c being a constant. (Garhwal 2003; Gorakhpur 06; Kashi 11; Meerut 13B) 11. (ii) am bn = c m + n , c being a constant. (iii) a + b = c, (Garhwal 2001, 03) (iv) a2 + b2 = c2, (v) ab = c2, c is a constant Prove that the envelope of the ellipses x2 ⁄ a2 + y2 ⁄ b2 = 1 having the sum of their semi-axes constant and equal to c, is the astroid x2 ⁄ 3 + y2 ⁄ 3 = c2 ⁄ 3. Find the envelope of the system of concentric and coaxial ellipses of constant area. 12. [Hint : Taking the common centre as origin and the common axes as coordinate axes, let the equation of the family of ellipses be x2 ⁄ a2 + y2 ⁄ b2 = 1. The area of an ellipse is πab. Since the ellipses are of constant area, so let ab = c2, where c is a constant. Now find the envelope.] Show that the envelope of the family of parabolas 10. (Gorakhpur 2005; Kanpur 10) (x ⁄ a) 1 ⁄ 2 + ( y ⁄ b) 1 ⁄ 2 = 1, 13. 1. 7. 8. 9. 11. under the condition ab = c2, is a hyperbola whose asymptotes coincide with the axes. (Garhwal 2000) A straight line of given length slides with its extremities on two fixed straight lines at right angles. Find the envelope of the circle drawn on the sliding line as diameter. (x2 + y2) 2 = a2 x2 + b2 y2. (i) r2 (e2 − 1) − 2 l e r cos θ + l2 = 0. (ii) r 3 ⁄ 4 = a3 ⁄ 4 cos (3θ ⁄ 4). (iii) r cosn − 1 {θ ⁄ (n − 1)} = a . (i) r sin α = ae(α − π ⁄ 2) cot α e θ cot α. (ii) r n ⁄ (1 − n) = a n ⁄ (1 − n) cos {nθ ⁄ (1 − n)} . (iii) r 2 − 2br cos θ + (b2 − a2) = 0. (i) x n ⁄ (n + 1) + y n ⁄ (n + 1) = c n ⁄ (n + 1). (ii) {(m + n) m + n x m y n } ⁄ m m nn = c m + n . (iii) x1 ⁄ 2 + y1 ⁄ 2 = c1 ⁄ 2. (iv) x2 ⁄ 3 + y2 ⁄ 3 = c2 ⁄ 3. 13. A circle. 4 x2 y2 = c4. 12.8 Evolute of a Curve (v) x y = c2 ⁄ 4. (Meerut 2012B) We define the evolute of a curve as the locus of the centre of curvature for that curve. Evolute as the envelope of the normals : The centre of curvature of a curve for a given point P on it is the limiting position of the intersection of the normal at P with the normal at any other consecutive point Q as Q → P . Therefore by the definition of D-254 DIFFERENTIAL CALCULUS envelope, the envelope of the normals to a curve is the evolute of that curve. Hence, the evolute of a curve is the envelope of the normals to that curve. Theorem : The norm al at any point of a curve is a tangent to its evolute touching at the corresponding centre of curvature. Proof : The co-ordinates (α, β) of the centre of curvature for any point P (x, y) on the given curve are given by α = x − ρ sin ψ ; β = y + ρ cos ψ . Differentiating these with respect to x, we get dψ dρ dα = 1 − ρ cos ψ − sin ψ dx dx dx dρ dρ ds dx d ψ = 1− ⋅ ⋅ − sin ψ = − sin ψ …(1) d ψ ds dx dx dx dψ dρ d β dy = − ρ sin ψ + cos ψ and dx dx dx dx dy dρ dρ ds dy d ψ …(2) = − ⋅ ⋅ + cos ψ = cos ψ ⋅ dx dx dx d ψ ds dx From (1) and (2), we have dβ = − cot ψ …(3) dα dβ is the slope of the tangent at But dα Q to the evolute and − cot ψ is the slope of the normal PQ at P to the given curve. These two slopes are equal, and Q is a common point on both the lines. H ence the tangent at Q to the evolute and the normal at P to the given curve coincide i.e., the normal at P to the given curve touches its evolute at the corresponding point. 12.9 Length of Arc of an Evolute The difference between the radii of curvature at any two points of a curve is equal to the length of the arc of the evolute between the two corresponding points. Let C (α, β) be the centre of curvature of the point P (x , y) on the given curve. Then CP = ρ . ...(1) Also α = x − ρ sin ψ , ...(2) β = y + ρ cos ψ . Let s be the arc length of the given curve measured from some fixed point A on it to the poin t (x, y) a n d s′ th e len gt h o f t he evolut e measured from some fixed point B on it to the point (α , β) . Differentiating both sides of (1) w.r.t. ‘s’, we have dα dx dρ dψ = − sin ψ − ρ cos ψ ds ds ds ds and D-255 ENVELOPES, EVOLUTES AND INVOLUTES dρ 1 sin ψ − ρ cos ψ ⋅ ρ ds = − (dρ ⁄ ds) sin ψ . Differentiating both sides of (2), w.r.t. ‘s’, we have dβ dy dρ dψ = + cos ψ − ρ sin ψ ds ds ds ds dρ 1 = sin ψ + cos ψ − ρ sin ψ ⋅ ρ ds = cos ψ − dψ 1 ⎤ ⎡ . . dx = cos ψ , = ρ⎥ ⎢ . ds ds ⎣ ⎦ ...(3) ⎡ . . dy ⎤ = sin ψ⎥ ⎢ . ds ⎣ ⎦ dρ cos ψ . ds Squaring and adding (3) and (4), we get ⎛ dα⎞ 2 ⎛ dβ⎞ 2 ⎛ dρ⎞ 2 ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ ⋅ ⎝ ds ⎠ ⎝ ds ⎠ ⎝ ds ⎠ = ∴ √ 2 ⎡⎢ ⎛ dα⎞ 2 ⎛ dβ⎞ ⎢⎢ ⎜ + ⎟ ⎜ ⎟ ⎝ ds ⎠ ⎣ ⎝ ds ⎠ ⎤⎥ dρ ⎥⎥ = ⋅ ds ⎦ ...(4) ...(5) Now s′ denotes the arc length of the locus of the point (α , β). R egarding α , β as the functions of the parameter s, we have ds ′ = ds √ 2 ⎡⎢ ⎛ dα⎞ 2 ⎛ dβ⎞ ⎢⎜ + ⎟ ⎜ ⎟ ⎢⎣ ⎝ ds ⎠ ⎝ ds ⎠ ⎤⎥ ⎥ ⎥⎦ ...(6) From (5) and (6), we have ds ′ ⁄ ds = dρ ⁄ ds . ∴ ds ′ = dρ ...(7) L e t Q 1 and Q 2 be t he point s on t he evolut e corresponding t o t he point s P1 and P2 on the given curve. Then integrating (7) between these points, we get ⎡ ⎤ Q 2 ⎡ ⎤ P2 = ⎢ ρ⎥ ⎢ s′ ⎥ ⎣ ⎦Q ⎣ ⎦P 1 i.e., 1 (s ′ at Q 2 ) − (s ′ at Q 1 ) = ( ρ at P2 ) − ( ρ at P1 ) i.e., the arc length of the evolute from Q 1 to Q 2 = the difference between the radii of curvature of the given curve at the points P1 and P2 . Example 1 : Find the evolute of the parabola y2 = 4ax . (Kanpur 2006; Avadh 08; Meerut 12B; Purvanchal 14) Solution : We know that the evolute of a curve is the envelope of the normals to that curve. E quation of any normal to the parabola y2 = 4ax is y = m x − 2am − am 3, where m is the parameter. So the envelope of (1) is the evolute of y2 = 4ax . Differentiating (1) partially with respect to m , we get 0 = x − 2a − 3am 2 i.e., m = {(x − 2a) ⁄ 3a}1 ⁄ 2. ...(1) D-256 DIFFERENTIAL CALCULUS Substituting this value of m in (1), we get x − 2a ⎤ ⎛ x − 2a ⎞ 1 ⁄ 2 ⎡ ⎟ ⎢ x − 2a − a ⋅ ⎥ ⎝ 3a ⎠ ⎣ 3a ⎦ y= ⎜ 2 2 (x − 2a) 3 ⁄ 2 ⎛ x − 2a ⎞ 1 ⁄ 2 (x − 2a) ⋅ = ⋅ ⎟ 3 √(3a) 3 ⎝ 3a ⎠ = ⎜ Squaring, we get 27ay2 = 4 (x − 2a) 3, which is the required evolute. Example 2 : Find the evolute of the hyperbola x2 ⁄ a2 − y2 ⁄ b2 = 1 . Soution : The given hyperbola is ...(1) x2 ⁄ a2 − y2 ⁄ b2 = 1 . The evolute of the hyperbola (1) is the envelope of the family of normals to the hyperbola (1). The coordinates (x, y) of any point P on the hyperbola (1) may be taken as x = a sec θ , y = b tan θ , where θ is the parameter. We have dx ⁄ dθ = a sec θ tan θ , dy ⁄ dθ = b sec 2 θ . ∴ slope of the normal to the hyperbola (1) at the point (a sec θ, b tan θ) dx ⁄ dθ dx = − = − dy ⁄ dθ dy a tan θ a sec θ tan θ = − ⋅ 2 b sec θ b sec θ ∴ E quation of the normal to the hyperbola (1) at the point (a sec θ, b tan θ) is a tan θ y − b tan θ = − (x − a sec θ) b sec θ = − ax tan θ + by sec θ = (a2 + b2) sec θ tan θ or ax cos θ + by cot θ = a2 + b2. ...(2) Now the evolute of the hyperbola (1) is the envelope of the family of straight lines (2), where θ is the parameter. Differentiating (2) partially with respect to θ , we get or − ax sin θ − by cosec 2 θ = 0 or ax sin θ = − by cosec 2 θ or sin3 θ = − ∴ by ax or cos θ = √(1 − sin2 θ) = = (by) 1 ⁄ 3 ⋅ (ax) 1 ⁄ 3 ⎡ (by) 2 ⁄ 3 ⎤⎥ ⎢ ⎢1 − ⎥ ⎢ (ax) 2 ⁄ 3 ⎥⎦ ⎣ sin θ = − √ [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3]1 ⁄ 2 (ax) 1 ⁄ 3 cos θ [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3]1 ⁄ 2 and cot θ = = − ⋅ sin θ (by) 1 ⁄ 3 Substituting the values of cos θ and cot θ in (2), the envelope of the family of straight lines (2) is [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3]1 ⁄ 2 [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3]1 ⁄ 2 ax ⋅ − by ⋅ = a 2 + b2 (ax) 1 ⁄ 3 (by) 1 ⁄ 3 D-257 ENVELOPES, EVOLUTES AND INVOLUTES or (ax) 2 ⁄ 3 [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3]1 ⁄ 2 − (by) 2 ⁄ 3 [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3]1 ⁄ 2 = a2 + b2 or [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3] [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3]1 ⁄ 2 = a2 + b2 or [(ax) 2 ⁄ 3 − (by) 2 ⁄ 3]3 ⁄ 2 = (a2 + b2) or (ax) 2 ⁄ 3 − (by) 2 ⁄ 3 = (a2 + b2) 2 ⁄ 3, which is the required evolute of the hyperbola (1). x2 ⁄ a2 Example 3 : S h o w t h a t t h e w h o le len gth o f th e evo lu te o f th e ellip se + y2 ⁄ b2 = 1 is 4 (a2 ⁄ b − b2 ⁄ a) . (Lucknow 2009, 10; Meerut 13) Solution : The given equation of the ellipse is x2 ⁄ a2 + y2 ⁄ b2 = 1 . Now ρ at the point (a cos t, b sin t) of (1) = [R efer the chapter on curvature] (a2 sin 2 t+ b2 cos2 But at the ends of major and minor axes t is equal to 0 and ∴ and ρ1 = ρ at the end of major axis = (b2) 3 ⁄ 2 ⁄ ab = 1 2 ...(1) t) 3 ⁄ 2 ⁄ ab . π respectively. b2 ⁄ a ρ2 = ρ at the end of minor axis = (a2) 3 ⁄ 2 ⁄ ab = a2 ⁄ b . Since the given ellipse is symmetrical about both the axes, therefore its evolute must also be symmetrical about both the axes. Hence the whole length of the evolute of the ellipse = 4 ( ρ2 − ρ1) = 4 (a2 ⁄ b − b2 ⁄ a) . 12.10 Evolute of Polar Curves In the case of polar curves there is no standard method of finding the evolute of a curve. H owever, if a curve is given in pedal form, we can easily find the pedal equation of its evolute by the method given below. Let the pedal equation of the given curve be p = f (r) . ...(1) Let C be the centre of curvature corresponding to the point P on the given curve. Then PC = ρ and PC is normal to t h e g i ve n c u r ve a t t h e p o i n t P . Corresponding to the point P on the given curve the point on the evolute is C . Since the evolute of a curve is the envelope of the normals of that curve, therefore the normal PC of the given curve is tangent to the evolute at the point C . If OT is the perpendicular from the pole O to the tangent to the given curve at the point P , then OT = p and OP = r . Draw OM perpendicular from the pole O to the tangent PC to the evolute at the point C . If OC = r ′ and OM = p′ , then the relation between p′ and r ′ will be the pedal equation of the evolute. From the Δ OPC , we have OC 2 = OP2 + PC 2 − 2OP . PC cos ∠ OPC D-258 DIFFERENTIAL CALCULUS r ′ 2 = r2 + ρ2 − 2r ρ sin φ i.e., r ′ 2 = r2 + ρ2 − 2ρp . [... p = r sin φ ] . ...(2) Now OTPM is a rectangle and so MP = OT = p . From the right angled triangle OMP , we have OM 2 = OP2 − MP2 i.e., p′ 2 = r 2 − p2. ...(3) dr Also ρ= r ⋅ ...(4) dp E liminating r , p and ρ between the equations (1) , (2) , (3) and (4) , we get the pedal equation of the evolute. i.e., Example 1 : Show that the evolute of an equiangular spiral is an equiangular spiral. (Rohilkhand 2012B) Solution : Let the pedal equation of the given equiangular spiral be p = r sin α . ...(1) We have dp ⁄ dr = sin α . dr 1 ∴ ρ= r = r⋅ = r cosec α . ...(2) dp sin α Corresponding to the point ( p, r) on the given curve (1), let the point on the evolute be ( p′, r ′) , the co-ordinates in each case being expressed in pedal form. Then r ′ 2 = r2 + ρ2 − 2ρp = r2 + r2 cosec 2 α − 2r cosec α . r sin α , [from (1) and (2)] = r2 cosec 2 α − r2 = r2 cot2 α . p′ 2 r2 p2 r2 = − = − Also Dividing (4) by (3), we get p′ 2 r ′2 = r2 sin2 α= r2 ...(3) cos2 α ...(4) r2 cos2 α = sin2 α . r 2 cot 2 α ∴ p′ 2 = r ′ 2 sin2 α or p′ = r ′ sin α . H ence the locus of the point ( p′, r ′) is p = r sin α . This is the pedal equation of the evolute and is an equiangular spiral. r= 1 3 Example 2 : Prove that the evolute of the cardioid r = a (1 + cos θ) is the cardioid 2 3 a (1 − cos θ) , the pole of the latter equation being at the point ( a, 0) . Solution : The given cardioid is r = a (1 + cos θ) . ...(1) Let O be the pole and OX the initial line. The cardioid (1) has been drawn in the figure. 2 ( 3 Take any point P (r, θ) on the given cardioid (1). Also let O′ be the given point a, 0) [this will be on the initial line because θ = 0]. D-259 ENVELOPES, EVOLUTES AND INVOLUTES Let C be the centre of curvature of the curve (1) corresponding to the point P . Then CP = ρ = radius of carvature of (1) at the point P . We have to find the locus of the point C w.r.t. O′ as pole. Let O′C = r1 and ∠ CO′X = θ1 . Draw CM and O′N perpendiculars from C and O′ respectively to OP . Let us first find the value of ρ . Taking logarithm of both sides of (1), we get log r = log a + log (1 + cos θ) . Differentiating w.r.t. ‘θ’, we get 1 cot φ = 1 − 2 sin θ cos θ − sin θ 1 dr 2 2 = = 1 2 r dθ 1 + cos θ 2 cos θ 2 1 2 = − tan θ = cot 1 2 1 2 1 ( 2 ∴ φ= Now p = r sin φ = r sin ( π + θ) . θ. 1 2 1 2 1 2 θ) = r cos θ . 1 2 r = 2 a cos2 θ = 2 a ( p ⁄ r) 2. From (1), ∴ π+ π+ 1 2 the pedal equation of the curve (1) is r3 = 2ap2. Differentiating (2) w.r.t. ‘p’, we get ...(2) 3r2 (dr ⁄ dp) = 4ap . ∴ ρ= r = 1⁄2 dr 4ap 4a ⎛ r3 ⎞ ⎜ ⎟ = = , dp 3r 3r ⎝ 2a ⎠ ⎡ .. r3 ⎤ ⎢ . from (2), p2 = ⎥ ⎣ 2a ⎦ 4a ⎛ r ⎞ 1 ⁄ 2 4a ⎧ a (1 + cos θ) ⎫ 1 ⁄ 2 = ⎜ ⎟ ⎨ ⎬ 3 ⎝ 2a ⎠ 3 ⎩ 2a ⎭ 4a = 3 4a = 3 ⎛ 1 ⎞1⁄2 ⎜ 2a cos2 θ ⎟ 2 ⎟ ⎜ ⎜ ⎟ 2a ⎝ ⎠ cos 1 θ. 2 …(3) D-260 DIFFERENTIAL CALCULUS 1 2 θ, ∴ ∠ CPM = 1 2 ∴ CM = PC sin θ = ρ sin θ = ∴ Also O′N = OO′ sin θ = Since φ = 1 2 π+ θ. 1 2 1 2 2 3 4 3 1 2 1 2 2 3 a cos θ sin θ = a sin θ . a sin θ . Thus CM = O′N and consequently O′N MC is a rectangle. Therefore O′C is parallel to OP , ...(4) i.e., θ1 = θ. Now r1 = O′C = N M = OP − ON − PM 1 2 = OP − OO′ cos θ − PC cos θ = r− 2 3 a cos θ − = a (1 + cos θ) − = = 1 3 1 3 4 1 1 a cos θ cos θ 3 2 2 2 2 a cos θ − a (1 3 3 + cos θ) a (3 + 3 cos θ − 2 cos θ − 2 − 2 cos θ) a (1 − cos θ) a (1 − cos θ1). [... θ1 = θ] E vo l u t e i s t h e l o cu s o f t h e ce n t r e o f cu r va t u r e . H e n ce ge n e r a li si n g 1 1 r1 = a (1 − cos θ1) , the required equation of the evolute is r = a (1 − cos θ) referred = 1 3 3 3 to O′ as pole. 12.11 Involutes Definition : If one curve is the evolute of another, then the latter is called an involute of the form er. Thus a curve C1 is an involute of a given curve C , if the curve C is the evolute of C1 . Theorem : Every curve has an infinite number of involutes. Let C be the given curve. Take an inextensible thin string and tie one end of this string to a fixed point, say A , of the curve C and then wrap the string round the convex side of C, keeping it taut all the while. Then any point on the string will describe an involute of C, since at each instant the free part of the string is a tangent to C whereas the direction of motion of the point on the string is at right angles to this tangent i.e., this tangent is along the normal to the locus of that point on the string. In the figure the locus of the point P1 on the string is the curve C1 which is an involute of C and the locus of the point P1′ on the string is the curve C2 which is also an involute of C . Similarly by taking some other point on the string we may obtain some other involute of C . S i n ce P1P1′ = P2 P2′ = P3 P3′ = P4 P4′ D-261 ENVELOPES, EVOLUTES AND INVOLUTES etc., therefore the curves C1 and C2 are at a constant distance along their common normal at their corresponding points. We have thus shown that there are an infinite number of involutes to the given curve C and all of them form a system of parallel curves. 1. Show that the equation of the normal to the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 at the point (a cos θ, b sin θ) is ax by − = a 2 − b2 . cos θ sin θ (Avadh 2006, 10) H ence find the evolute of the above ellipse. 2. Find the equation of the evolute of the parabola y2 = 2px . 3. Show that the evolute of the tractrix 1 2 x = a (cos t + log tan t ), y = a sin t , is the catenary y = a cosh (x ⁄ a) . 4. Prove that the evolute of the hyperbola 2 x y = a2 is (x + y) 2 ⁄ 3 − (x − y) 2 ⁄ 3 = 2 a2 ⁄ 3. 5. Prove that the evolute of the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 is the envelope of the family of ellipses given by a2 x2 sec 4 α + b2 y2 cosec 4 α = (a2 − b2) 2, α being the variable parameter. 6. Find the evolute of the curve x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3. 7. Show that the evolute of the curve whose pedal equation is r2 − a2 = m p2 is the curve whose pedal equation is r2 − (1 − m) a2 = mp2. 8. Prove that normals to a given curve are always tangent lines to its evolute. 1. (ax) 2 ⁄ 3 + (by) 2 ⁄ 3 = (a2 − b2) 2 ⁄ 3. 6. (x + y) 2 ⁄ 3 + (x − y) 2 ⁄ 3 = 2 a2 ⁄ 3. 2. 27py2 = 8 (x − p) 3. Fill in the Blanks: 1. Fill in the blanks “……” so that the following statem ents are com plete and correct. The equation of the envelope of the family of curves F (x , y , α) = 0 where α is the parameter, is obtained by eliminating α between the equations F (x , y , α) = 0 and ∂ F (x , y , α) = …… . ∂α D-262 2. 3. DIFFERENTIAL CALCULUS The envelope of the family of curves Aα2 + Bα + C = 0, wh ere A, B, C are functions of x and y is …… . The envelope of the family of curves t x2 + t 2 y = a , the parameter being t is … . Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 5. The envelope of the family of straight lines y = m x + a ⁄ m, the parameter being m is (a) y = 4 ax (b) y2 = 4 ax (c) y2 = ax (d) y2 = 2 ax. The envelope of the family of straight lines (x ⁄ a) cos θ + ( y ⁄ b) sin θ = 1, the parameter being θ is y2 y2 x2 x2 (a) + = 1 (b) + = 2 (c) x2 + y2 = a2 + b2 (d) a2 x2 + b2 y2 = 0. a 2 b2 a 2 b2 6. The envelope of the straight line y = m 2 x + 1 ⁄ m 2, the parameter being m is 4. (a) y2 = 4 m x 7. 8. (b) y2 = 2 mx (c) y2 = 4 x (d) y + x = m. Aα2 The envelope of t he family of curves + Bα + C = 0, where A, B, C are functions of x, y and α is a parameter is (Garhwal 2002) (a) A 2 = 4BC (b) B 2 = 4AC (c) A 2 + 4BC = 0 (d) A 2 − BC = 0 The locus of the centres of curvatures of all points of a given plane curve is called (a) radius of curvature (b) envelope (c) evolute (d) normal (Kumaun 2008) True or False: Write ‘T ’ for true and ‘F ’ for false statem ent. 9. In general, the envelope of a family of curves touches each curve of the family and at each point is touched by some member of the family. 10. 11. 12. 13. The envelope of x2 sin α + y2 cos α = a2, the parameter being α is x2 + y2 = a2. The evolute of a curve is the envelope of the tangents of that curve. If one curve is the evolute of another, then the latter is called an involute of the former. The evolute of an equiangular spiral is not an equiangular spiral. 1. 4. 7. 10. 13. 0. (b). (b). F. F. 2. 5. 8. 11. B 2 = 4 A C. (a). (c). F. 3. 6. 9. 12. x4 + 4 ay = 0. (c). T. T. 13.1 Asymptote Suppose a curve is not limited in extent i.e. it has some branch or branches which extend to infinity. Parabola and H yperbola are familiar curves of this type. Take a point on an infinite branch of such a curve and draw a tangent to the curve at this point. If the distance of the point of contact from the origin tends to infinity, the tangent itself, may or may not tend to a definite straight line. In case the tangent tends to a definite straight line, at a finite distance from the origin, it is called an asymptote of the curve. Thus we can define the asymptote of a curve as follows. Definition : A straight line at a finite distance from the origin to which a tangent to a curve tends, as the distance from the origin of the point of contact tends to infinity, is (Kanpur 2005, 07, 14; Kashi 11) called an asym ptote of the curve. The curve approaches the asymptote : R oughly speaking an asymptote is a tangent with its point of contact at a great distance from the origin. Therefore, when a point P on a curve tends to infinity, its perpendicular distance from the corresponding asymptote tends to zero. This property of an asymptote enables us to draw more accurately those curves which have asymptotes. We draw the asymptotes first, a n d t h e n t h e cu r ve, m a kin g it s bra n che s a pp r o ach t he corresponding asymptotes. D-264 DIFFERENTIAL CALCULUS Branch of a Curve : Suppose the equation of the curve is such that y has two or more values for every value of x. Corresponding to these distinct values of y we shall get different branches of the curve. It is just possible, that each branch may have its own separate asymptote. Therefore a curve may have m ore than one asym ptote. 13.2 Determination of Asymptotes Let the equation of the curve be f (x, y) = 0. …(1) We shall here consider the case of only those asymptotes which are not parallel to y-axis. We know that the equation of a straight line which is not parallel to y-axis is of the form y = mx + c. …(2) The abscissa, x, must tend to infinity as the point P (x, y) on the curve (1) tends to infinity along the line (2). The equation of the tangent to the curve (1) at the point P (x, y) is dy dy dy⎞ ⎛ Y − y= (X− x), or Y = X + ⎜y − x ⎟ ⋅ …(3) dx dx dx⎠ ⎝ dy dy and y − x As x → ∞ , dx dx must both tend to finite limits, in order that an asymptote might exist. Suppose the tangent (3) tends to the straight line (2) as x → ∞ . Then (2) is an asymptote of the curve (1). Also we have dy⎞ Lim dy Lim ⎛ m = x →∞ and c = x → ∞ ⎜ y − x ⎟ ⋅ dx dx⎠ ⎝ dy y− x dx Lim = 0 Since c is finite, therefore x → ∞ x Lim ⎛ y dy⎞ Lim y Lim dy i.e., i.e., x → ∞ ⎜⎝ x − dx⎟⎠ = 0 x → ∞ x = x → ∞ dx = m. Lim y Lim dy Therefore x → ∞ = x → ∞ = m. x dx dy⎞ Lim Lim dy Lim ⎛ = m. Also c = x → ∞ ⎜ y − x ⎟ = x → ∞ (y − m x), since x → ∞ dx⎠ dx ⎝ H ence, if y = mx + c is an asym ptote to the curve f (x, y) = 0, L im L im y m = x → ∞ and c = x → ∞ ( y − m x). x 13.3 The Asymptotes of the General Rational Algebraic Curve Let the equation to the curve be {a0 y n + a1 y n − 1 x + a2 y n − 2 x2 + … + an − 1 y x n − 1 + a n x n } + {b1 y n − 1 + b2 y n − 2 x + … + b n − 1 y x n − 2 + b n x n − 1} + {c2 y n − 2 + …} + … = 0, …(1) D-265 ASYMPTOTES x n φ n ( y ⁄ x) + x n − 1 φ n − 1 ( y ⁄ x) + x n − 2 φ n − 2 ( y ⁄ x) + … ⎛ y⎞ ⎛ y⎞ … + x φ 1 ⎜⎝ ⎟⎠ + φ 0 ⎜⎝ ⎟⎠ = 0, …(2) x x y ⎛ y⎞ where φ r ⎜ ⎟ is a polynomial in of degree r. x ⎝ x⎠ or Dividing (2) by x n , we get 1 1 ⎛ y⎞ φ n ( y ⁄ x) + φ n − 1 (y ⁄ x) + 2 φ n − 2 ⎜ ⎟ + …… x ⎝ x⎠ x 1 1 ⎛ y⎞ ⎛ y⎞ ………+ n − 1 φ 1 ⎜ ⎟ + n φ 0 ⎜ ⎟ = 0. …(3) ⎝ ⎠ ⎝ x⎠ x x x E xcluding at present the case of asymptotes parallel to the y-axis (i.e. excluding Lim the case in which x → ∞ ( y ⁄ x) is equal to ∞ ), (3) gives, on taking limits as x → ∞ , the equation φ n (m) = 0, ...(4) Lim ⎛ y⎞ m = x → ∞ ⎜ ⎟ = slope of an asymptote. ⎝ x⎠ The equation (4) is, in general, of degree n in m . Solving this equation, we shall get the slopes of the asymptotes. This equation will give us n values of m, corresponding to the n branches of the curve (1). H owever, some of the values of m may be equal, and this will be the case of parallel asymptotes. Since we are concerned only with real asymptotes, therefore we shall reject the imaginary roots of (4) if there are any. Now if y = m x + c is an asymptote of (1), then we know that corresponding to a specified value of m , we have Lim c = x → ∞ ( y − mx). where Therefore to determine the value of c corresponding to the value of m , we put y − m x = p in the equation of the curve, where p is a variable which → c as x → ∞ . So putting y = m x + p i.e., p y = m + in (2), we get x x p⎞ p⎞ p⎞ ⎛ ⎛ ⎛ xn φn ⎜ m + ⎟ + xn − 1 φn − 1 ⎜ m + ⎟ + xn − 2 φn − 2 ⎜ m + ⎟ + … x⎠ x⎠ x⎠ ⎝ ⎝ ⎝ p⎞ p⎞ ⎛ ⎛ … + x φ 1 ⎜ m + ⎟ + φ 0 ⎜ m + ⎟ = 0⋅ …(5) x⎠ x⎠ ⎝ ⎝ E xpanding each term of (5) by Taylor’s theorem, we get ⎡ ⎤ p p2 φ n ′′ (m) + …⎥⎥ x n ⎢⎢ φ n (m) + φ n ′ (m) + 2 ⎢ ⎥ x x 2! ⎣ ⎦ p ⎡ ⎤ + x n − 1 ⎢ φ n − 1 (m) + φ ′ n − 1 (m) + …⎥ x ⎣ ⎦ p ⎡ ⎤ + x n − 2 ⎢ φ n − 2 (m) + φ ′ n − 2 (m) + …⎥ + ……= 0. …(6) x ⎣ ⎦ D-266 DIFFERENTIAL CALCULUS Arranging terms in (6) according to descending powers of x, we get x n φ n (m) + x n − 1 [ pφ n ′ (m) + φ n − 1 (m)] ⎡ p2 ⎤ p + xn − 2 ⎢ φ ′′ (m) + φ′ (m) + φ n − 2 (m) ⎥ + …… = 0. …(7) ⎣2! n ⎦ 1! n− 1 n − 1 Putting φ n (m) = 0 in (7) and then dividing by x , we get [ p φ n ′ (m) + φ n − 1 (m)] p 1 ⎡ p2 ⎤ + ⎢ φ ′′ (m) + φ′ (m) + φ n − 2 (m) ⎥ + … = 0. x ⎣2 ! n 1! n− 1 ⎦ lim Taking limit as x → ∞ and remembering that x → ∞ p = c, we get …(8) …(9) c φ n′ (m) + φ n − 1 (m) = 0, which determines one value of c for each value of m found from (4). The asymptotes are then given by y = m x + c, where m is a root of (4) and the corresponding c is obtained from (9). Important : The polynomial φ n (m) is easily obtained by putting y = m and x = 1 in x n φ n (y ⁄ x) i.e. the nth degree terms in the equation of the curve. Similarly to obtain φ n − 1 (m), we should put y = m and x = 1 in the (n − 1) th degree terms in the equation of the curve. In general, to obtain φ r (m), we should put y = m and x = 1 in the rth degree terms in the equation of the curve. a 2 b2 Example 1 : Find the asymptotes of the curve 2 − 2 = 1⋅ x y (Bundelkhand 2006; Agra 07, 08; Kashi 14) Solution : The equation of the curve can be written as a2 y2 − b2 x2 = x2 y2 or x2 y2 − a2 y2 + b2 x2 = 0. Since the curve is of degree 4, therefore it cannot have more than four asymptotes. E quating to zero the coefficient of the highest power of y (i.e. o f y2), the asymptotes parallel to y-axis are given by x2 − a2 = 0 i.e. x = ± a. Also equating to zero the coefficient of the highest power of x (i.e. of x2), the asymptotes parallel to x-axis are given by y2 + b2 = 0, which gives two imaginary asymptotes. Thus all the four possible asymptotes of the curve have been found and the only real asymptotes are x = ± a. Example 2 : Find the asymptotes of the curve y2 (a2 − x2) = x4. (Meerut 2010) 2 2 4 2 2 Solution : The equation of the curve is y x + x − a y = 0. Since the curve is of degree 4, therefore it cannot have more than four asymptotes. E quating to zero the coefficient of the highest power of y (i.e., o f y2) the asymptotes parallel to y-axis are given by x2 − a2 = 0 i.e. x = ± a. The coefficient of the highest power x4 of x is merely a constant. H ence there is no asymptote parallel to x-axis. D-267 ASYMPTOTES To find the remaining oblique asymptotes, we put y = m and x = 1 in the highest i.e. four degree terms and we get φ 4 (m) = m 2 + 1. The roots of the equat ion φ 4 (m) = 0 are imaginary and consequently the corresponding asymptotes are imaginary. H ence the only real asymptotes of the curve are x = ± a. Find all the asymptotes of the following curves: 1. a2 ⁄ x2 + b2 ⁄ y2 = 1. 2. y2 3. xy2 = 4a2 4. x2 y2 a2 5. y2 6. x2 ⁄ a2 7. x2 y2 8. Find the asymptotes parallel to the axes of the curve (x2 − = (x2 a 2) = x. (Rohilkhand 2014) (2a − x). (x2 + y2). − a 2) − y2 ⁄ b2 − (Meerut 2007B; Purvanchal 14) x2 = y− x2 (x2 (Bundelkhand 2001, 05, 08) − 4a2). (Meerut 2003, 06; Agra 05; Rohilkhand 05, 06) = 1. xy2 (Gorakhpur 2005; Bundelkhand 11) + x + y + 1 = 0. (Meerut 2007) x2 y2 − x2 − y2 − x − y + 1 = 0. 9. (Bundelkhand 2001) Find the asymptotes of the curve x2 y2 − a2 (x2 + y2) − a3 (x + y) + a4 = 0. 1. 2. 3. 4. 5. 6. 7. 8. x = ± a, y = ± b. x = ± a, y = 0. x = 0. x = ± a, y = ± a . x = ± a, y = ± x. y x = ± ⋅ b a y = 0, y = 1, x = 0, x = 1. x = ± 1, y = ± 1. 9. (Meerut 2001) x = ± a , y = ± a. 13.4 Asymptotes Might Not Exist If one or more values of m obtained from φ n (m) = 0 are such that they make φ n ′ (m) = 0, but do not make φ n − 1 (m) zero, then the equation for determining the corresponding values of c becomes 0 . c + φ n − 1 (m) = 0. From this equation we get c = + ∞ or − ∞ and this corresponds to the case when the tangent goes farther and farther away from the origin as x → ∞ . Corresponding to such values of m , we shall get no asymptotes. D-268 DIFFERENTIAL CALCULUS 13.5 Two Parallel Asymptotes Suppose the equation (iv) i.e. φ n (m) = 0 of § 3 gives us two equal values of m. This repeated value of m will make φ n ′ (m) = 0. In case it does not make φ n − 1 (m) equal to zero, the asymptotes corresponding to it will not exist. If it also makes φ n − 1 (m) equal to zero, the equation from which c is usually determined reduces to the identity 0 . c + 0 = 0, and we cannot find the value of c in this way. To derermine c in this case, we put φ ′ n (m) = φ n − 1 (m) = 0 in equation (vii) of § 3 and we get ⎡ p2 ⎤ p φ ′′ n (m) + φ′ (m) + φ n − 2 (m) ⎥ + xn − 2 ⎢ ⎣2! ⎦ 1! n− 1 3 2 ⎡p ⎤ p p xn − 3 ⎢ φ ′′′ (m) + φ ′′ n − 1 (m) + φ′ (m) + φ n − 3 (m) ⎥ + …… = 0. ⎣3! n ⎦ 1! n− 2 2! Lim Dividing by x n − 2, taking limits as x → ∞ and remembering that x → ∞ p = c, we get c2 c φ ′′ n (m) + φ′ (m) + φ n − 2 (m) = 0. 2! 1! n− 1 This equation is quadratic in c. It will give us two values of c, say c1 and c2 corresponding to that repeated value of m . The two corresponding asymptotes will be y = m x + c1 and y = mx + c2 , which are obviously parallel. 13.6 Three Parallel Asymptotes If the equation φ n (m) = 0 gives us three equal values of m, then this repeated valu e o f m will ma ke φ ′ n (m) a n d φ ′′ n (m) equal to zero. F or the exist ence of corresponding asymptotes it must make φ n − 1 (m) equal to zero. If it also makes φ ′ n − 1 (m) and φ n − 2 (m) equal to zero, then the equation to determine c reduces to the identity 0 . c2 + 0 . c + 0 = 0, and we shall not be able to find the value of c in this way. S o p u t t i n g e a c h o f φ n (m), φ ′ n (m), φ ′′ n (m), φ n − 1 (m), φ ′ n − 1 (m) a n d φ n − 2 (m) equal to zero in equation (vii) of § 3 and dividing by x n − 3 and taking limit as x → ∞ , we get c2 c c3 φ ′′′ (m) + φ ′′ n − 1 (m) + φ′ (m) + φ n − 3 (m) = 0. 1 ! n− 2 3! n 2! This equation will give us three values of c corresponding to that repeated value of m and accordingly we shall get three parallel asymptotes. In a similar way we can discuss the case of more than three parallel asymptotes. D-269 ASYMPTOTES 13.7 Asymptotes Parallel to the Co-ordinate Axes (i) Asymptotes parallel to y-axis : Let x = k be an asymptote parallel to y-axis of the curve f (x, y) = 0. In this case, y, alone tends to infinity as a point P (x, y) on the curve tends to infinity along the line x = k. Also lim k = y → ∞ x, where (x, y) lies on the curve. Therefore to find the asymptotes parallel to y-axis, we find the definite value or values k 1, k 2 etc. to which x tends as y tends to infinity. Then the lines x = k 1, x = k 2 , etc. are the required asymptotes. Asymptotes parallel to y-axis of a rational algebraic curve. Let the equation of the curve, when arranged in descending powers of y, be …(1) y m φ (x) + y m − 1 φ 1 (x) + y m − 2 φ 2 (x) + … = 0, where φ (x), φ 1 (x), φ 2 (x) etc. are polynomials in x. Dividing the equation (1) by y m , we obtain 1 1 φ (x) + φ 1 (x) + 2 φ 2 (x) + … = 0. y y …(2) lim If x = k is an asymptote parallel to y-axis of (1), then k = y → ∞ x. Therefore taking limit of (2) as y → ∞ and remembering that x → k as y → ∞ , we get φ (k) = 0. Therefore k is a root of the equation φ (x) = 0. If k 1, k 2 etc., be the roots of φ (x) = 0, then the asymptotes of (1) parallel to y-axis are x = k 1, x = k 2 , etc. From algebra, we know that if k 1 is a root of φ (x) = 0, then x − k 1 must be a factor of φ (x). Also φ (x) is the coefficient of the highest power of y i.e. y m in the equation of the curve. H ence we have the following simple rule : The asym ptotes parallel to the axis of y are obtained by equating to zero the coefficient of the highest power of y in the equation of the curve. In case the coefficient of the highest power of y, is a constant or if its linear factors are all imaginary, there will be no asymptotes parallel to y-axis. (ii) Asymptotes parallel to x-axis : Proceeding as above, we have the following rule for finding asymptotes parallel to x-axis of a rational algebraic curve : The asymptotes parallel to the axis of x are obtained by equating to zero the co-efficient of the highest power of x, in the equation of the curve. In case the coefficient of highest power of x, is a constant or if its factors are all imaginary, there will be no asymptotes parallel to x-axis. 13.8 Total Number of Asymptotes of a Curve The num ber of asym ptotes, real or im aginary, of an algebraic curve of the nth degree cannot exceed n. The slopes of the asymptotes which are not parallel to y-axis are given as the roots of the equation φ n (m) = 0 which is of degree n at the most. If the equation of the curve D-270 DIFFERENTIAL CALCULUS possesses some asymptotes parallel to y-axis, then we can easily see that the degree of φ n (m) = 0 will be smaller than n by at least the same number. In general one value of m gives only one value of c. In case the equation for determining c is a quadratic, the equation φ n (m) = 0 has two equal roots. Similarly if the equation for determining c is cubic, the equation φ n (m) = 0 has three equal roots. H ence a curve of degree n cannot have m ore than n asymptotes. But the number of real asymptotes can be less than n. Some roots of the equation φ n (m) = 0 may come out to be imaginary or even corresponding to a real value of m the value of c may come out to be infinite. 13.9 Working Rule for Finding the Asymptotes of Rational Algebraic Curves (i) A curve of degree n cannot have more than n asymptotes real or imaginary. (ii) E quating to zero the coefficient of the highest power of y in the equation of the curve, we get asymptotes parallel to y-axis. Similarly equating to zero the coefficient of the highest power of x in the equation of the curve, we get asymptotes parallel to x-axis. If y = m x + c is an asymptote not parallel to y-axis, then the values of m and c are found as follows : (iii) Putting y = m and x = 1 in the highest i.e. nth degree terms in the equation of the curve, we get φ n (m). Solving the equation φ n (m) = 0, we get the slopes of the asymptotes. If some values of m are imaginary, we reject them. (iv) Corresponding to a value of m , the value of c is given by the equation c φ ′ n (m) + φ n − 1 (m) = 0, where φ n − 1 (m) is obtained by putting y = m and x = 1 in the (n − 1) th degree terms in the equation of the curve. The asymptotes corresponding to m = 0 are already found in (ii). So we need not find the value of c corresponding to m = 0. (v) If corresponding to two equal values of m , the equation for determining c, given in (iv) reduces to the identity 0 . c + 0 = 0, then the values of c are given by c2 c φ ′′ n (m) + φ′ (m) + φ n − 2 (m) = 0. 1! n− 1 2! (vi) Similarly, if three values of m are equal and the equation for determining c, given in (v), reduces to the identity 0 . c2 + 0 . c + 0 = 0, then the corresponding values of c are given by c3 c2 c φ ′′′ n (m) + φ ′′ n − 1 (m) + φ′ (m) + φ n − 3 (m) = 0. 1! n− 2 3! 2! Example 1 : Find the asymptotes of the curve y2 = 4 x. (Meerut 2010B; Kumaun 08) Solution : The equation of the curve is y2 − 4 x = 0. Putting y = m and x = 1 in the highest i.e. 2nd degree terms, we get φ 2 (m) = m 2. D-271 ASYMPTOTES Solving the equation φ 2 (m) = 0 i.e. m 2 = 0, we get m = 0, 0. Also putting y = m and x = 1 in the first degree terms, we get φ 1 (m) = − 4. Now c is given by the equation cφ ′ 2 (m) + φ 1 (m) = 0 i.e. 2mc − 4 = 0. If we put m = 0 in this equation, we get c = ∞ . Hence no asymptote exists. Example 2 : Find the asymptotes of the curve x3 + y3 − 3axy = 0. (Agra 2005; Bundelkhand 09; Meerut 12; Kashi 12; Avadh 13; Rohilkhand 14) Solution : O bviously there are no asymptotes parallel to the co-ordinate axes. Putting y = m and x = 1 in the highest i.e. third degree terms in the equation of the curve, we get φ 3 (m) = 1 + m 3. Solving the equation φ 3 (m) = 0, i.e. (1 + m 3) = 0, i.e. (m + 1) (m 2 − m + 1) = 0, we get m = − 1 as the only real root. The other two roots are imaginary. Again putting y = m and x = 1 in the second degree terms in the equation of the curve, we get φ 2 (m) = − 3am. Now c is given by c φ ′ 3 (m) + φ 2 (m) = 0, i.e. c (3m 2) − 3am = 0. Putting m = − 1, we get c = − a. H ence the only real asymptote of the curve is y = − x − a or y + x + a = 0. Example 3 : Find all the asymptotes of the curve 3x3 + 2x2 y − 7xy2 + 2y3 − 14xy + 7y2 + 4x + 5y = 0. Solution : O bviously there are no asymptotes parallel to the coordinate axes. Putting y = m and x = 1 in the highest i.e. third degree terms in the equation of the curve, we get φ 3 (m) = 3 + 2m − 7m 2 + 2m 3. The slopes of the asymptotes are given by φ 3 (m) = 2m 3 − 7m 2 + 2m + 3 = 0, or (m − 1) (2m + 1) (m − 3) = 0. ∴ m = 1, 3, − 1 ⋅ 2 Again putting y = m and x = 1 in the next highest i.e. second degree terms in the equation of the curve, we get φ 2 (m) = − 14m + 7m 2. Now c is given by c φ 3′ (m) + φ 2 (m) = 0, i.e., c (6m 2 − 14m + 2) + (7m 2 − 14m) = 0. When m = 1, c = − 7 ⁄ 6; when m = 3, c = − 3 ⁄ 2 and when m = − 1 2 ∴ The required asymptotes are y = x − 7 ⁄ 6 ; y = 3x − 3 ⁄ 2 and y = − i.e., , c = − 5 ⁄ 6. 1 2 x− 5⁄6 6y − 6x + 7 = 0; 2y − 6x + 3 = 0 and 2y + x + 5 ⁄ 3 = 0. Example 4 : Find all the asymptotes of the curve y3 − xy2 − x2 y + x3 + x2 − y2 − 1 = 0. Solution : Putting y = m and x = 1 in the highest i.e. third degree terms of the equation of the curve, we get φ 3 (m) = m 3 − m 2 − m + 1. D-272 DIFFERENTIAL CALCULUS The slopes of the asymptotes are given by φ 3 (m) = m 3 − m 2 − m + 1 = 0 or (m − 1) 2 (m + 1) = 0. ∴ m = 1, 1, − 1. Now putting y = m and x = 1 in the next highest i.e. second degree terms, we get φ 2 (m) = 1 − m 2. To determine c, we have c φ ′ 3 (m) + φ 2 (m) = 0, i.e. c (3m 2 − 2m − 1) + (1 − m 2) = 0. …(1) When m = − 1, we have c = 0 and the corresponding asymptote is y = − x + 0 i.e. y + x = 0. When m = 1, the equation (1) reduces to the identity c . 0 + 0 = 0 and we cannot determine c from it. In this case c is to be determined from the equation c2 c φ ′′ 3 (m) + φ ′ (m) + φ 1 (m) = 0. 1! 2 2! Putting y = m and x = 1 in the first degree terms in the equation of the curve, we get φ 1 (m) = 0, since there are no first degree terms. H ence for m = 1, c is to be given by, c2 (6m − 2) + c (− 2m) = 0 i.e. (3m − 1) c2 − 2mc = 0. 2 For m = 1, this becomes 2c2 − 2c = 0 i.e. c = 0 and 1. H ence y = x + 1 and y = x + 0 are two parallel asymptotes corresponding to the slope m = 1. ∴ The required asymptotes are y + x = 0, y − x = 0, y − x − 1 = 0. Example 5 : Find all the asymptotes of the curve (x + y) 2 (x + 2y + 2) = x + 9y + 2. Solution : The equation of the curve can be written as (x + y) 2 (x + 2y) + 2 (x + y) 2 − (x + 9y) − 2 = 0. H ere (Meerut 2011) φ 3 (m) = (1 + m) 2 (1 + 2m). The slopes of the asymptotes are given by φ 3 (m) = (1 + m) 2 (1 + 2m) = 0. i.e. 1 ⋅ 2 ∴ m = − 1, − 1, − Also φ 2 (m) = 2 (1 + m) 2. To determine c, we have c φ ′ 3 (m) + φ 2 (m) = 0, c {2 (1 + m) (1 + 2m) + 2 (1 + m) 2} + 2 (1 + m) 2 = 0. 1 When m = − , we have c = − 1 and the corresponding asymptote is …(1) 2 y= − 1 2 x− 1 i.e. 2y + x + 2 = 0. When m = − 1, the equation (i) reduces to the identity c . 0 + 0 = 0 and we cannot determine c from it. In this case c is to be determined from the equation c2 c φ ′′ 3 (m) + φ ′ (m) + φ 1 (m) = 0. 2! 1! 2 D-273 ASYMPTOTES Now φ 1 (m) = − (1 + 9m). H ence for m = − 1, c is given by c2 {2 (1 + 2m) + 4 (1 + m) + 4 (1 + m)} 2 + c {4 (1 + m)} − (1 + 9m) = 0 (6m + 5) c2 + 4 (1 + m) c − (1 + 9m) = 0. For m = − 1, this becomes − c2 + 8 = 0, i.e. c = ± 2 √2. H e n c e y = − x + 2 √2 a n d y = − x − 2 √2 a r e t wo p a r a l l e l a sym p t o t e s corresponding to the slope m = − 1. ∴ The required asymptotes are 2y + x + 2 = 0 and x + y ± 2 √2 = 0. i.e. Example 6 : Find the asymptotes of the curve x3 + 2x2 y + xy2 − x2 − xy + 2 = 0. (Meerut 2002; Kashi 12) Solution : The given curve is of degree 3. So it cannot have more than three asymptotes. E quating to zero the coefficient of the highest power of y (i.e., of y2), we get x = 0 as an asymptote parallel to y-axis. Also there is no asymptote parallel to x-axis because the coefficient of x2 is merely a constant. Now we proceed to find the remaining oblique asymptotes. Putting y = m and x = 1 in the third degree and second degree terms separately, we get φ 3 (m) = 1 + 2m + m 2, and φ 2 (m) = − 1 − m. The slopes of the asymptotes are given by the equation φ 3 (m) = 0 i.e., 1 + 2m + m 2 = 0 i.e., (1 + m) 2 = 0. ∴ m = − 1, − 1. To determine c, we have the equation c φ 3′ (m) + φ 2 (m) = 0 i.e., c (2 + 2m) − 1 − m = 0. …(1) For m = − 1, the equation (1) reduces to the identity c . 0 + 0 = 0 and thus it fails to give c. In this case c is to be determined by the equation c2 c φ ′′ (m) + φ ′ (m) + φ 1 (m) = 0. 1! 2 2! 3 Now φ 3′′ (m) = 2, φ 2′ (m) = − 1, and φ 1 (m) = 0 because there are no first degree terms in the equation of the curve. So for m = − 1, c is to be given by 1 2 c . (2) 2 + c. (− 1) + 0 = 0 i.e., c2 − c = 0 i.e., c (c − 1) = 0. ∴ c = 0, 1. H ence y = − x + 0 and y = − x + 1 are two parallel asymptotes corresponding to the slope m = − 1. ∴ the required asymptotes are x = 0, x + y = 0 and x + y − 1 = 0. 1. Find all the asymptotes of the following curves : x3 + 2x2 y − xy2 − 2y3 + 4y2 + 2xy + y − 1 = 0. 2. 2x3 − x2 y − 2xy2 + y3 − 4 x2 + 8 xy − 4 x + 1 = 0. D-274 DIFFERENTIAL CALCULUS 3. x3 + 2x2 y − xy2 − 2y3 + xy − y2 − 1 = 0. 4. x2 y + xy2 + xy + y2 + 3x = 0. 5. y3 − 5xy2 + 8 x2 y − 4 x3 − 3y2 + 9xy − 6 x2 + 2y − 2x + 1 = 0. 6. 2 x (y − 3) 2 = 3y (x − 1) 2. 7. y2 (x − 2a) = x3 − a3. 8. y3 9. (x2 − y2) 2 − 4y2 + y = 0. − 2y2 x− yx2 + 2x3 (Bundelkhand 2008) + y2 − 6xy + 5x2 x3 + x2 y − xy2 − y3 − 3x − y − 1 = 0. 11. x2 y3 + x3 y2 = x3 + y3. 12. x3 − 2x2 y + xy2 + x2 − xy + 2 = 0. (Rohilkhand 2007) 13. (2x − 3y + 1) 2 (x + y) = 8x − 2y + 9. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. y = x + 1, y = − x + 1 and x + 2y = 0. y = − x + 2, y = x + 2, y = 2x − 4. 2y + x = 1, y = x, y + x + 1 = 0. x = − 1, y = 0, x + y = 0. y = x, y = 2x + 2, y = 2x + 1. x = 0, y = 0, 2y = 3x + 6. x = 2a, y = x + a, y = − x − a. y = x, y = 2x + 1, y = − x − 2. x + y = ± 1 and x − y = ± 1. y = x, y = − x + 1, y = − x − 1. y = ± 1, x = ± 1, y = − x. x = 0, y = x, y = x + 1. y − x = 0, 2y − x = 0, 2y − x − 1 = 0. y + x = 0, 3y − 2x − 3 = 0, 3y − 2x + 1 = 0. − + x2 (Kanpur 2014) 14. y+ 4y3 (Meerut 2001, 04, 06B; Gorakhpur 06) x3 − 8xy2 − 2y + 2x + 1 = 0. (Kanpur 2007) 10. 5x2 (Bundelkhand 2006) − 3xy + 13.10 Asymptotes by Expansion 2y2 − 1 = 0. (Kumaun 2008) To show that y = m x + c is an asym ptote of the curve B C A + 2+ 3+ … y = mx + c + x x x A B C Let the equation of the curve be y = m x + c + + 2+ 3+ … …(1) x x x B C A + 2 + 3 + ……is convergent for sufficiently large values of x. where the series x x x Differentiating (1), we have dy A 2B = m− 2− 3 − … dx x x D-275 ASYMPTOTES or or ∴ The equation of the tangent to (1) at (x, y) is A 2B ⎛ ⎞ Y − y = ⎜ m − 2 − 3 − …⎟ (X − x) ⎝ ⎠ x x A 2B A 2B ⎛ ⎞ ⎛ ⎞ Y = ⎜ m − 2 − 3 − …⎟ X + y − ⎜ m − 2 − 3 − …⎟ x x x x x ⎝ ⎠ ⎝ ⎠ ⎛ Y = ⎜m − ⎝ A 2B 2A 3B ⎞ − 3 − …⎟ X + c + + 2 + …, x ⎠ x2 x x …(2) substituting the value of y from (1). Suppose now x → ∞ . The equation (2) then tends to the equation Y = mX + c. H ence y = mx + c is the asymptote of the curve A B C y = mx + c + + + + …⋅ x x2 x3 y2 x2 Example : Find the asym ptotes of the hyperbola 2 − 2 = 1⋅ a b Solution : The equation of the hyperbola can be written as x2 y2 = − 1, b2 a 2 or b2 2 ⎛⎜ a2 ⎞ x ⎜ 1 − 2 ⎟⎟ , 2 ⎜ a x ⎟⎠ ⎝ or y2 = or y= ± b a ⎧ x ⎪⎨ 1 − ⎪ ⎩ y2 = or b2 2 (x − a2) a2 y= ± b a ⎛ x ⎜⎜ 1 − ⎜ ⎝ a2 ⎞⎟ ⎟ x2 ⎟⎠ 1⁄2 ⎫ 1 a2 1 a4 − + …⎪⎬ ⋅ ⎪ 2 x2 8 x4 ⎭ H ence by article 13.10, the asymptotes of the curve are y = ± b x. a 13.11 Alternative Methods of Finding Asymptotes of Algebraic Curves Theorem : The asymptotes of an algebraic curve are parallel to the lines obtained by equating to zero the linear factors of the highest degree term s in its equation. Let the equation of the curve be of degree n. Let y − m 1 x be a factor of the nth degree terms in the equation of the curve. Then obviously (m − m 1) is a factor of φ n (m). Therefore m 1 is a root of the equation φ n (m) = 0 and there is an asymptote parallel to the line y − m 1 x = 0. Conversely let m 1 be a root of the equation φ n (m) = 0, so that there is an asymptote parallel to the line y − m 1 x = 0. In this case m − m 1 must be a factor of φ n (m). Therefore ( y ⁄ x − m 1) must be a factor of φ n ( y ⁄ x). H ence ( y − m 1 x) must be a factor of x n φ n ( y ⁄ x) i.e. the highest degree terms in the equation of the curve. If the highest degree terms contain, x, as a factor, then after a little consideration it will be obvious that the curve will possess asymptotes parallel to x = 0 i.e. y-axis. H ence the theorem. D-276 DIFFERENTIAL CALCULUS We know that if y = m x + c is an oblique asymptote of the curve f (x, y) = 0, then Lim Lim y m = x → ∞ and c = x → ∞ , y ⁄ x → m ( y − m x). These facts together with the above x theorem enable us to find the asymptotes of algebraic curves very easily. The first step for this purpose is that we should collect the highest degree terms in the equation of the curve and resolve them into real linear factors. Then the following different cases may arise : Case I : Let y − m 1 x be a non-repeated factor of the highest i.e. nth degree terms in the equation of the curve. Then the equation to the curve can be written as ( y − m 1 x) Fn − 1 + Pn − 1 = 0, …(1) where Fn − 1 contains only terms of degree n − 1, and Pn − 1 contains terms of various degrees, none of which is of a degree higher than n − 1. O bviously y − m 1 x = c, where c is to be determined, is an asymptote of the curve. Lim Now c = x → ∞ , y ⁄ x → m 1 ( y − m 1x) where (x, y) lies on (1). Pn − 1 But when (x, y) lies on (1), y − m 1 x = − ⋅ Fn − 1 ∴ ⎛ − Pn − Lim c = x → ∞ , y ⁄ x → m 1 ⎜⎜ ⎜ F ⎝ 1 ⎞⎟ ⎟ ⋅ n − 1 ⎟⎠ ⎛ − Pn − Lim H ence y − m 1 x = x → ∞ , y ⁄ x → m 1 ⎜⎜ ⎜ F ⎝ 1 ⎞⎟ ⎟ is an asymptote of the curve. n − 1 ⎟⎠ Thus dividing (1) by Fn − 1 and taking limit as x → ∞ , y ⁄ x → m 1 we shall get an asymptote of (1). Similarly we can find asymptotes corresponding to other non-repeated linear factors. Example : Find all the asym ptotes of the curve (x2 − y2) (x + 2y + 1) + x + y + 1 = 0. Solution : The equation of the curve can be written as (x2 − y2) (x + 2y) + (x2 − y2) + x + y + 1 = 0 (x − y) (x + y) (x + 2y) = ( y2 − x2) − x − y − 1. The slope of the asymptote corresponding to the factor x − y is 1. Hence the asymptote corresponding to this factor is or ( y2 − x2) − x − y − 1 Lim x − y = x →∞ , y ⁄ x →1 (x + y) (x + 2y) ⎛ y2 ⎞ ⎜ − 1⎟ − 1 − y . 1 − 1 ⎜ 2 ⎟ ⎜x ⎟ x x x x2 ⎝ ⎠ Lim , = x →∞ , y ⁄ x →1 y⎞ ⎛ 2y⎞ ⎛ 1 + 1 + ⎜ ⎟ ⎜ ⎟ x⎠ ⎝ x⎠ ⎝ i.e., on dividing the numerator and denominator by x2 (1 − 1) 0 = = = 0, (1 + 1) (1 + 2) 6 x − y = 0 is one asymptote of the curve. Second asymptote corresponding to the factor x + y is D-277 ASYMPTOTES ( y2 − x2) − x − y − 1 Lim x + y = x →∞ , y ⁄ x →− 1 (x − y) (x + 2y) Lim = x →∞ , y ⁄ x →− 1 ⎡ y2 ⎤ ⎢ − 1⎥ − 1 − y ⋅ 1 − 1 ⎢ 2 ⎥ ⎢x ⎥ x x x x2 ⎣ ⎦ y⎤ ⎡ y⎤ ⎡ ⎢1 − ⎥ ⎢1 + 2 ⎥ ⎣ x⎦ ⎣ x⎦ (1 − 1) = 0, (1 + 1) (1 − 2) x + y = 0 is another asymptote of the curve. The third asymptote of the curve is = i.e. Lim x + 2y = x → ∞ , y ⁄ x → − 1 2 Lim = x →∞ , y ⁄ x →− 1 2 ( y2 − x2) + terms of degree lower than 2 (x − y) (x + y) ( y2 ⁄ x2 − 1) + terms which → 0 (1 − y ⁄ x) (1 + y ⁄ x) ⎛1 ⎞ 3 ⎜ − 1⎟ − ⎝4 ⎠ 4 = ⎛ = 3 = − 1. 1⎞ ⎛ 1⎞ ⎜1 + ⎟ ⎜1 − ⎟ 2⎠ ⎝ 2⎠ 4 ⎝ Therefore x + 2y + 1 = 0 is the third asymptote of the curve. Important : It should be noted that while taking limit we should reject all the terms in the numerator whose degree is lower than the degree of denominator. All such terms will tend to zero as x → ∞ . Case II : If ( y − m 1 x) 2 is a factor of the nth degree terms but ( y − m 1 x) is not a factor of the (n − 1) th degree terms, then φ n ′ (m 1) = 0 and φ n − 1 (m 1) ≠ 0. Therefore as in § 4, the asymptotes corresponding to the factor ( y − m 1 x) 2 will not exist. Therefore there will be no asym ptotes with slope m 1 if ( y − m 1 x) 2 is a factor of the nth degree terms and y − m 1 x is not a factor of the (n − 1) th degree term s. In case the equation of the curve does not contain terms of degree n − 1, we can add them with zero coefficient and obviously y − m 1 x can be taken as a factor of the (n − 1) th degree terms. Case III : Let the equation of the curve be of the form ( y − m 1 x) 2 Fn − 2 + ( y − m 1 x) G n − 2 + Pn − 2 = 0, …(1) where Fn − 2 and G n − 2 contain only terms of degree n − 2, and Pn − 2 contains terms of various degrees, none of which is of a degree higher than n − 2. Dividing (1) by Fn − 2 and taking limit as x → ∞ and y ⁄ x → m 1, we get an equation of the form c2 + Ac + B = 0, giving the values of c corresponding to the slope m 1. If c1 and c2 are the roots of this equation, then y − m 1 x = c1 and y − m 1 x = c2 will be the corresponding asymptotes. After a little consideration it will be obvious that the asymptotes corresponding to the factor ( y − m 1 x) 2 will be obtained by solving the quadratic ( y − m 1 x) 2 + A ( y − m 1 x) + B = 0. D-278 DIFFERENTIAL CALCULUS In a sim ilar way we can discuss t he ca se if t he nth de gre e term s contain ( y − m 1 x) 3 or a higher power of y − m 1 x as a factor. Example : Find the asym ptotes of the curve x2 (x2 − y2) (x − y) + 2x3 (x − y) − 4y3 = 0. Solution : The equation of the curve can be written as x2 (x − y) 2 (x + y) + 2x3 (x − y) − 4y3 = 0. The asymptotes corresponding to the factor (x − y) 2 are given by 2x3 Lim (x − y) 2 + (x − y) x → ∞ , y ⁄ x → 1 2 x (x + y) y3 Lim = 0 − 4 x →∞ , y ⁄ x →1 2 x (x + y) or 2 Lim (x − y) 2 + (x − y) x → ∞ , y ⁄ x → 1 (1 + y ⁄ x) ( y ⁄ x) 3 Lim = 0 − 4 x →∞ , y ⁄ x →1 (1 + y ⁄ x) or (x − y) 2 + 1 . (x − y) − i.e. (x − y) = 1 2 ⋅ 4 = 0 or − 1 ± √(1 + 8) 2 i.e. (x − y) 2 + (x − y) − 2 = 0 x − y = − 2 and x − y = 1. The asymptote corresponding to the factor x + y is 4y3 − 2x3 (x − y) Lim (x + y) = x → ∞ , y ⁄ x → − 1 x2 (x − y) 2 ⎛ y⎞ 3 1 ⎛ y⎞ 4 ⎜ ⎟ . − 2 ⎜1 − ⎟ ⎝ ⎠ ⎝ ⎠ x x x Lim = x →∞ , y ⁄ x →− 1 2 ⎛ y⎞ ⎜1 − ⎟ ⎝ x⎠ = − 4 ⁄ 4 = − 1 i.e., x + y + 1 = 0. Moreover the curve has two asymptotes parallel to y-axis and they can be obtained by equating to zero the coefficient of highest power of y i.e. y3 in the equation of the curve. So they are x2 − 4 = 0 i.e. x = ± 2. Case IV : Let the equation of the curve be of the form (ax + by + c) Pn − 1 + Q n − 1 = 0, ...(1) where Pn − 1 and Q n − 1 cont ain t erms none of which is of a higher degree than n − 1, and Pn − 1 contains at least one term of degree n − 1 so as to ensure that the equation (1) is of degree n. O bviously ax + by will be a factor of the nth degree terms in the equation of the curve. Now (1) can be written as (ax + by) Pn − 1 + cPn − 1 + Q n − 1 = 0. The asymptote of (1) corresponding to the factor ax + by (if it occurs as a non-repeated factor of the highest degree terms) is D-279 ASYMPTOTES Qn − ⎧ Lim (ax + by) + x → ∞ , y ⁄ x → − a ⁄ b ⎪⎨ c + ⎪ Pn − ⎩ or ⎛ Qn − Lim (ax + by + c) + x → ∞ , y ⁄ x → − a ⁄ b ⎜⎜ ⎜P 1⎫ ⎪ ⎬= 0 1⎪ ⎭ 1 ⎞⎟ ⎟ = 0⋅ ⎟ ⎝ n − 1⎠ Thus if the equation of a curve is given in the form (1), then there is no necessity of collecting separately the nth degree terms. A similar modification can be made in case III. Case V : Asymptotes by Inspection : If the equation of a curve of the nth degree can be put in the form Fn + P = 0, where Fn is of degree n (i.e., contains term s of degree n and m ay also contain term s of lower degrees) , and P is of degree n − 2 , or lower, and if Fn = 0 can be broken up into n linear factors which represent n straight lines no two of which are parallel or coincident then all the asym ptotes of the curve are given by equating to zero the linear factors of Fn . Let ax + by + c = 0 be a non-repeated factor of Fn . Then the equation of the curve can be written as (ax + by + c) Fn − 1 + P = 0, where Fn − 1 is of degree n − 1. The asymptote of the given curve parallel to the lim ⎛ P ax + by + c + x → ∞ , y ⁄ x → − a ⁄ b ⎜⎜ ⎜ Fn − ⎝ line ax + by + c = 0 is ⎞⎟ ⎟ = 0⋅ 1 ⎟⎠ …(1) Since Fn − 1 contains at least one term of degree n − 1 and P is of degree n − 2, or lower, therefore we shall have ⎛ P ⎞⎟ lim x → ∞ , y ⁄ x → − a ⁄ b ⎜⎜⎜ F ⎟⎟ = 0⋅ ⎝ n − 1⎠ Thus ax + by + c = 0 is an asymptote of the given curve. Example 1 : Find all the asymptotes of the curve (x − y − 1) 2 (x2 + y2 + 2) + 6 (x − y − 1) (xy + 7) − 8x2 − 2x − 1 = 0. Solution : The asymptotes parallel to the line x − y − 1 = 0 are xy + 7 lim (x − y − 1) 2 + 6 (x − y − 1) x → ∞ , y ⁄ x → 1 2 x + y2 + 2 or − 8x2 − 2x − 1 lim + x →∞ , y ⁄ x →1 = 0, x2 + y2 + 2 y 7 + 2 x x lim (x − y − 1) 2 + 6 (x − y − 1) x → ∞ , y ⁄ x → 1 2 ⎛ y⎞ 2 1+ ⎜ ⎟ + 2 ⎝ x⎠ x 1 2 − 8− − 2 x x lim + x →∞ , y ⁄ x →1 = 0 y2 2 1+ 2 + 2 x x D-280 DIFFERENTIAL CALCULUS (x − y − 1) 2 + 3 (x − y − 1) − 4 = 0, − 3 ± √(9 + 16) = 1, − 4. or (x − y − 1) = 2 Thus x − y − 2 = 0 and x − y + 3 = 0 are two parallel asymptotes of the given curve. Since the remaining linear factors of the fourth degree terms in the equation of the curve are imaginary, therefore the other two asymptotes are imaginary. Example 2 : Find all the asymptotes of the curve or (x2 − y2) (x + 2y + 1) + x + y + 1 = 0. Solution : The equation of the curve can be written as (x − y) (x + y) (x + 2y + 1) + x + y + 1 = 0. Since no two of the straight lines x − y = 0, x + y = 0 and x + 2y + 1 = 0 are parallel and x + y + 1 is of degree 1, therefore all the asymptotes of the curve are given by (x − y) (x + y) (x + 2y + 1) = 0. Find all the asymptotes of the following curves : 1. (x2 − y2) ( y2 − 4x2) − 6x3 + 5x2y + 3xy2 − 2y3 − x2 + 3xy − 1 = 0. 2. x ( y − x) 2 − 3y ( y − x) + 2x = 0. 3. ( y − x) ( y − 2x) 2 + ( y + 3x) ( y − 2x) + 2x + 2y − 1 = 0. 4. (x − 2y) 2 (x − y) − 4y (x − 2y) − (8x + 7y) = 0. 5. ( y − a) 2 (x2 − a2) = x4 + a4. 6. x ( y − 3) 3 = 4y (x − 1) 3. 7. (x − y) 2 (x2 + y2) − 10 (x − y) x2 + 12y2 + 2x + y = 0. 8. 9. x2 (x + y) (x − y) 2 + ax3 (x − y) − a2 y3 = 0. (x − y + 1) (x − y − 2) (x + y) = 8x − 1. 10. (Kanpur 2010; Meerut 12B; Bundelkhand 14; Purvanchal 14) xy (x2 − y2) (x2 − 4y2) + xy (x2 − y2) + x2 + y2 − 7 = 0. 7. y = x, y = 2x, y + x + 1 = 0, y + 2x + 1 = 0. 2. y = 2x − 2, y = 2x − 3, y − x = 4. 4. x = ± a, y = x + a, y = − x + a. 3 x = 0, y = 0, y = 2x + , 2y + 4x = 15. 2 x − y − 2 = 0, x − y − 3 = 0. 8. x = ± a, y = x + a, y = ± x − 1. 3. 5. 6. 9. 10. (Meerut 2005B; Bundelkhand 07) 1 2 x = 3, y − x = 1, y − x = 2. y = x + 4, x − 2y = 2 ± 3 √3. a. y + x = 0, y = x − 3 and y = x + 2. x = 0, y = 0, x − y = 0, x + y = 0, x − 2y = 0 and x + 2y = 0. D-281 ASYMPTOTES 13.12 Intersection of a Curve and its Asymptotes Let y = mx + c be an asymptote of the curve …(1) …(2) x n φ n ( y ⁄ x) + xn − 1 φ n − 1 ( y ⁄ x) + x n − 2 φ n − 2 ( y ⁄ x) + … = 0, which is of degree n. To find the points of intersection of (1) and (2), we should solve the two equations simultaneously. So eliminating y between (1) and (2), we get x n φ n (m + c ⁄ x) + x n − 1 φ n − 1 (m + c ⁄ x) + x n − 2 φ n − 2 (m + c ⁄ x) + … = 0. E xpanding each term by Taylor’s theorem and arranging the terms in descending powers of x, we get x n φ n (m) + [cφ ′ n (m) + φ n − 1 (m)] x n − 1 ⎡ c2 + ⎢ ⎣2! xn − 1 φ ′′ n (m) + ⎤ c φ′ (m) + φ n − 2 (m) ⎥ x n − 2 + … = 0. …(3) ⎦ 1! n− 1 Since y = mx + c is an asymptote of (2), therefore the coefficients of xn and are both zero in (3). H ence (3) reduces to ⎧⎪ c2 ⎫ c φ ′′ n (m) + φ ′ n − 1 (m) + φ n − 2 (m) ⎪⎬⎪ x n − 2 + … = 0, …(4) ⎨⎪ 1! ⎩2 ! ⎭ in which the coefficient of x n − 2 will be non-zero provided there is no other asymptote of the given curve parallel to y = m x + c. Now (4) gives us the abscissae of the points of intersection of (1) and (2). Since equation (4) is of degree n − 2 in x, therefore it will give n − 2 values of x. H ence, in general, any asymptote of a curve of the nth degree cuts the curve in (n − 2) points. Corollary 1 : The n asymptotes of a curve of the nth degree cut it in n (n − 2) points. Corollary 2 : If the equation of a curve of degree n can be put in the form Fn + P = 0, where P is of degree n − 2 at the m ost and Fn consists of n non-repeated linear factors, then the n (n − 2) points of intersection of the curve with its asymptotes lie on the curve P = 0. The asymptotes of the curve Fn + P = 0, are given by the equation Fn = 0. W e kn ow t h at if S = 0 a n d S′ = 0 represent two curves, then S − λS′ = 0 represents some curve through the points of intersection of S = 0 and S′ = 0. If we take λ = 1, then we see that (Fn + P) − Fn = 0 i.e. P = 0 is a curve passing through the points of intersection of Fn + P = 0 and Fn = 0. Thus, a curve of degree n − 2 , or less, can be m ade to pass through the n (n − 2) points of intersection of a curve of degree n with its n asym ptotes. Particular Cases : (i) If the given curve is of degree 3, then the 3 (3 − 2) i.e., 3 points of intersection of the curve and its asymptotes lie on a curve of degree 3 − 2 = 1 i.e., on a straight line. (ii) If the curve is of degree 4 then the 4 (4 − 2) i.e., 8 points of intersection of the curve and its asymptotes lie on a curve of degree 4 − 2 = 2 i.e. on a conic. D-282 DIFFERENTIAL CALCULUS Example 1 : Show that the asym ptotes of the curve 4 (x4 + y4) − 17x2 y2 − 4x (4y2 − x2) + 2 (x2 − 2) = 0 cut the curve in eight points which lie on the ellipse x2 + 4y2 = 4. Solution : The equation of the curve can be written as (Purvanchal 2007) (4 x4 + 4y4 − 17x2 y2) − 4 (4y2 x − x3) + 2x2 − 4 = 0. φ 4 (m) = 4 + H ere 4m 4 − …(1) 17m 2. The slopes of the asymptotes are given by the equation φ 4 (m) = 4m 4 − 17m 2 + 4 = 0 Therefore m = ± 1 2 i.e., (4m 2 − 1) (m 2 − 4) = 0. , ± 2. φ 3 (m) = − 4 (4m 2 − 1) and φ ′ 4 (m) = 16 m 3 − 34m. Also Now c is given by cφ ′ 4 (m) + φ 3 (m) = 0 i.e., c (16m 3 − 34m) − 4 (4m 2 − 1) = 0. 1 1 When m = , c = 0; when m = − , c = 0; when m = 2, c = 1; and when 2 2 m = − 2, c = − 1. Therefore the asymptotes are y= i.e., 1 2 x, y = − 1 2 x, y = 2x + 1 and y = − 2x − 1 2y − x = 0, 2y + x = 0, y − 2x − 1 = 0 and y + 2x + 1 = 0. The combined equation of the asymptotes is (2y − x) (2y + x) ( y − 2x − 1) ( y + 2x + 1) = 0 or (4y2 − x2) {( y2 − 4x2) − 4x − 1} = 0 or (4y2 − x2) ( y2 − 4x2) − 4x (4y2 − x2) − 4y2 + x2 = 0 …(2) 4y4 − 17x2 y2 + 4x4 − 4 (4y2 x − x3) − 4y2 + x2 = 0. Now each asymptote of (1) will cut it in 4 − 2 i.e. 2 points. Therefore the four asymptotes will cut it in 4 × 2 i.e. 8 points. Subtracting (2) from (1), we get 2x2 − 4 + 4y2 − x2 = 0 i.e. x2 + 4y2 = 4, which is the equation of an ellipse. H ence the eight points of intersection of (1) and (2) lie on the ellipse x2 + 4y2 = 4. or Example 2 : Find the equation of the cubic which has the sam e asym ptotes as the curve x3 − 6 x2 y + 11xy2 − 6y3 + x + y + 1 = 0 and which passes through the points (0, 0), (1, 0) and (0, 1). Solution : The equation of the given curve can be written as (x − y) (x − 2y) (x − 3y) + x + y + 1 = 0. …(1) By inspection, we find that x − y = 0, x − 2y = 0, and x − 3y = 0 are the asymptotes of (1). The combined equation of the asymptotes of (1) is F3 ≡ (x − y) (x − 2y) (x − 3y) = 0. D-283 ASYMPTOTES Since the points of intersection of a cubic curve with its asymptotes lie on a straight line, therefore the most general equation of the curve having F3 = 0 as its asymptotes is (x − y) (x − 2y) (x − 3y) + ax + by + c = 0 or x3 − 6x2 y + 11xy2 − 6y3 + ax + by + c = 0. If (2) passes through the points (0, 0), (1, 0) and (0, 1), then c = 0, 1 + a = 0 i.e., a = − 1 and − 6 + b = 0 i.e. b = 6. …(2) H ence the required curve is x3 − 6x2 y + 11xy2 − 6y3 − x + 6y = 0. 1. Find the asymptotes of the curve x2 y − xy2 + xy + y2 + x − y = 0 and show that they cut the curve again in three points which lie on the straight line x + y = 0. 2. Show that the asymptotes of the cubic x3 − 2y3 + xy (2x − y) + y (x − y) + 1 = 0 cut the curve in three points which lie on the straight line x − y + 1 = 0. (Kanpur 2009; Avadh 10) 3. Find the equation of the straight line on which lie the three points of intersection of the curve (x + a) y2 = ( y + b) x2 and its asymptotes. 4. Show that the eight points of intersection of the curve xy (x2 − y2) + x2 + y2 = a2 and its asymptotes lie on a circle whose centre is at the origin. 5. Show that the four asymptotes of the curve (x2 − y2) ( y2 − 4x2) + 6x3 − 5x2 y − 3xy2 + 2y3 − x2 + 3xy − 1 = 0 6. 7. 8. cut the curve in eight points which lie on the circle x2 + y2 = 1. Show that the eight points of intersection of the curve x4 − 5x2 y2 + 4y4 + x2 − y2 + x + y + 1 = 0 and its asymptotes lie on a rectangular hyperbola. Find the equation of the quartic curve which has x = 0, y = 0, y = x and y = − x for asymptotes and which passes through (a, b) and which cuts its asymptotes again in eight points that lie on a circle whose centre is origin and radius a. Find the equation of the cubic which has the same asymptotes as the curve x3 − 6 x2 y + 11xy2 − 6y3 + x + y + 1 = 0, and which touches the axis of y at the (Agra 2007) origin and passes through the point (3, 2). 1. 7. y = 0, x = 1, x − y + 2 = 0. 3. 2 2 2 2 2 2 2 bxy (x − y ) + a (b − a ) (x + y − a ) = 0. 8. x3 − 6x2 y + 11xy2 − 6y3 − x = 0. a2 ( y + b) = b2 (x + a). D-284 DIFFERENTIAL CALCULUS 13.13 Asymptotes to Non-Algebraic Curves Th e defin it ion o f t h e asym p t o t es h e lp s u s in finding t h e asympt o t es of non-algebraic curves as is clear from the following example. Example : Find the asym ptotes of the curve y = sec x. Solution : H ere dy ⁄ dx = sec x tan x. Therefore the tangent at (x, y) to the given curve is Y − sec x = sec x tan x (X − x) …(1) Y cos2 x − cos x = (X − x) sin x. Now as x → π ⁄ 2, y → ∞ and the distance of (x, y) from the origin tends to infinity. Therefore taking limit of (1) as x → π ⁄ 2, we get i.e., Y.0− 0= ⎛ ⎜ ⎝ X− 1 2 ⎞ π ⎟⎠ . 1 i.e., X= 1 2 π. This is one asymptote. The other asymptotes are X= − 1 2 π, ± 3 2 π,……. 13.14 Polar Curves Lemma : The polar equation of any line is p = r cos (θ − α) , where, p is the length of the perpendicular from the pole to the line and α is the angle which the perpendicular m akes with the initial line. Let O be the pole and OX the initial line. Let OM be the perpendicular from O to the given line. Then it is given that OM = p and ∠ XOM = α. Let P be any point (r, θ) on the given line. Then OP = r, ∠ XOP = θ and ∠ MOP = θ − α. From the right angled triangle OMP, we have OM = OP cos ∠ POM i.e. , p = r cos (θ − α), which is the required equation of the line. Asymptotes of Polar Curves : If α be a root of the equation f (θ) = 0 , then r sin (θ − α) = 1 ⁄ f ′ (α) is an asym ptote of the curve 1 ⁄ r = f (θ). Take any point P (r, θ) on the curve …(1) 1 ⁄ r = f (θ). Draw a line through the pole O perpendicular to the radius vector OP and meeting the tangent at P in T. dθ Then OT = polar subtangent of the curve at P = r2 ⋅ dr 1 dr But from (i), − 2 = f ′ (θ). r dθ dθ 1 = − = OT. ∴ r2 dr f ′ (θ) No w su p p o se θ t e n d s t o α. Si n ce f (α) = 0, therefore from (1), r → ∞ i.e. the distance of P from the pole tends to infinity. Also PT tends to the asymptote 1 ⎤ ⎡ and OT → ⎢ − ⎥ ⎣ f ′ (θ) ⎦ θ = α D-285 ASYMPTOTES 1 , if f ′ (α) ≠ 0. f ′ (α) Also OP and PT will tend to become parallel as is obvious from the dotted lines in the figure. Therefore the angle OTP will tend to a right angle and OT will tend to 1 ⋅ OM where OM is perpendicular to the asymptote. H ence OM = − f ′ (α) When θ = α, suppose OP tends to OP ′. Then ∠ XOP ′ = α. ⎛π ⎞ ∴ ∠MOX = − ⎜ − α⎟ , negative sign indicating that it has been measured ⎝2 ⎠ clockwise. ∴ The equation of the asymptote is π 1 r cos ⎡⎢ θ − ⎧⎨ − ⎛⎜ − α⎞⎟ ⎫⎬ ⎤⎥ = − f ′ (α) ⎣ ⎩ ⎝2 ⎠ ⎭⎦ i.e. OT → − i.e. r cos ⎛⎜ θ + i.e. − r sin (θ − α) = − ⎝ π 1 − α⎞⎟ = − 2 f ′ (α) ⎠ 1 f ′ (α) i.e. r sin (θ − α) = 1 ⋅ f ′ (α) 13.15 Working Rule for Finding the Asymptotes of Polar Curves 1 = f (θ). r (ii) Solve the equation f (θ) = 0. Let α, β, … be its roots. (iii) The asymptote corresponding to θ = α is 1 , where f ′ (α) = ⎡⎣ f ′ (θ) ⎤⎦ . r sin (θ − α) = θ= α f ′ (α) (i) Put the equation of curve in the form Example 1 : Find the asymptotes of the curve r = 2a ⁄ (1 − 2 cos θ). Solution : The equation to the curve can be written as 1 1 = (1 − 2 cos θ) = f (θ), say. r 2a Now f (θ) = 0 if 1 − 2 cos θ = 0 i.e 2 cos θ = 1 i.e. cos θ = θ = 2nπ ± i.e. Also ∴ ∴ π, where n is any integer 3 = α, say. 1 f ′ (θ) = (2 sin θ) = 2a 1 f ′ (α) = sin ⎛⎜ 2nπ ± a ⎝ 2a 1 = ± ⋅ √3 f ′ (α) 1 (sin θ). a π⎞ π √3 1 sin = ± ⋅ ⎟ = ± 3⎠ a 3 2a 1 2 D-286 or or i.e. i.e. DIFFERENTIAL CALCULUS H ence the asymptotes are given by π 2a r sin ⎧⎨ θ − ⎛⎜ 2nπ ± ⎞⎟ ⎫⎬ = ± 3⎠ ⎭ √3 ⎩ ⎝ π⎞ ⎛ ⎛ r sin θ cos ⎜ 2nπ ± ⎟ − r cos θ sin ⎜ 2nπ ± 3⎠ ⎝ ⎝ __ π π 2a (r sin θ) cos + r cos θ sin = ± 3 3 √3 π π 2a r sin θ cos − r cos θ sin = 3 3 √3 π π 2a and r sin θ cos + r cos θ sin = − 3 √3 3 π⎞ π⎞ 2a ⎛ ⎛ r sin ⎜ θ − ⎟ = and r sin ⎜ θ + ⎟ = − 3⎠ √3 3⎠ ⎝ ⎝ π⎞ 2a ⎟ = ± 3⎠ √3 2a ⋅ √3 Example 2 : Find the asymptotes of the curve r = a cosec θ + b. Solution : The equation of the curve can be written as sin θ 1 1 = = = f (θ), say. r a cosec θ + b a + b sin θ Now f (θ) = 0, if sin θ = 0 i.e. θ = nπ, where n is any integer = α, say. cos θ. (a + b sin θ) − sin θ. b cos θ Also f ′ (θ) = ⋅ (a + b sin θ) 2 cos nπ. (a + b sin nπ) − sin nπ . b cos nπ ∴ f ′ (α) = f ′ (nπ) = (a + b sin nπ) 2 a cos nπ 1 = = cos nπ. a a2 ∴ a 1 = ⋅ f ′ (α) cos nπ H ence the asymptotes are given by r sin (θ − nπ) = or (r sin θ cos nπ − r cos θ sin nπ) cos nπ = a or or r sin θ cos2 nπ = a r sin θ = a. a cos nπ 13.16 Circular Asymptotes Definition : L et the equation of a curve be r = f (θ). lim If θ → ∞ f (θ) = l, then the circle r = l is called the circular asymptote of the curve r = f (θ). θ Example 1 : Find the circular asym ptote of the curve r = a ⋅ ⋅ θ− 1 Solution : The circular asymptote is given by θ lim = a. r = a θ →∞ θ− 1 Thus r = a is the circular asymptote. D-287 ASYMPTOTES 2. 3. If α is a root of the equation f (θ) = 0, then write the equation of asymptote of the 1 polar curve = f (θ) corresponding to the root α. (Meerut 2001) r Find the asymptotes of the following curves : y = tan x. r sin mθ = a. (Meerut 2000) 4. r θ = a. 5. 6. 7. 2 ⁄ r = 1 + 2 sin θ. (i) r sin θ = 2 cos 2θ. (i) r cos θ = a sin θ. (ii) r sin θ = 2 cos θ. r = 4 (sec θ + tan θ). r sin 2θ = a. 1. 8. 9. (Meerut 2008, 12B) (ii) r sin θ = a cos 2θ. (Meerut 2004B) 10. r cos θ = 4 sin2 θ. 11. 12. r θ cos θ = a cos 2θ. r (eθ − 1) = a (eθ + 1). 2θ ⋅ r= sin θ Find the circular asymptotes of the following curves : 13. 14. r (eθ − 1) = a (eθ + 1). 15. r= (Meerut 2005) 3θ2 + 2θ + 1 ⋅ 2θ2 + θ + 1 1. r sin (θ − α) = 3. r sin ⎜ θ − 4. r sin θ = a. 6. 7. (i) (i) 8. r cos θ = 8. ⎛ ⎝ 1 ⋅ [ f ′(θ)]θ= α 2. x = ± π ⁄ 2, ± 3π ⁄ 2, …… . kπ ⎞ a , where k is any integer. ⎟ = m⎠ m cos kπ 5. r sin θ = 2. r cos θ = ± a. 10. r cos θ = 4. (ii) (ii) 9. 1 ⎤ π ⎥ = 2 ⁄ √3. 6 ⎦ r sin θ = a. r sin θ = ± 2. ⎡ ⎣ r sin ⎢ θ ± r sin θ = ± 11. r sin θ = a, r cos θ = 1 2 a, r cos θ = ± a 1⎞ ⎛ ⎜k + ⎟ π 2⎠ ⎝ 12. r sin θ = 2a. 13. r sin θ = 2kπ, k = ± 1, ± 2,…. . 14. r = a. 15. r= 3 ⋅ 2 1 2 a. , k is any integer. D-288 DIFFERENTIAL CALCULUS Fill in the Blanks: 2. Fill in the blanks “……”, so that the following statem ents are com plete and correct. The asymptotes parallel to the axis of y are obtained by equating to zero the coefficients of the …… power of y in the equation of the curve. (Agra 2008) A curve of degree n cannot have more than …… asymptotes. 3. The only asymptote of the curve x 3 + y 3 − 3a xy = 0 is …… . 4. If α be a root of the equation f (θ) = 0, then an asymptote of the curve 5. is …… . Circular asymptote of the curve r (θ2 + 1) = a θ2 − 1 is …… . 1. 1 = f (θ) r (Meerut 2001) Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 6. a2 b 2 The number of asymptotes of the curve 2 − 2 = 1 is x y (a) 2 (b) 3 (c) 4 (d) 1 y2 (x 2 (Bundelkhand 2006) a 2) 7. The asymptotes of the curve − = x , which are parallel to the x-axis are (a) x = ± a (b) y = ± a (c) y = 0, y = 0 (d) x = 0 8. The number of oblique asymptotes of the curve y 2 (x 2 − a2) = x 2 (x 2 − 4a2) is (a) 4 (b) 3 (c) 2 (d) None of these True or False: Write ‘T’ for true and ‘F’ for false statement. 9. 10. 11. 12. 13. The number of asymptotes, real or imaginary, of an algebraic curve of the nth degree cannot exceed n . The curve x 2 (x − y) 2 + a2 (x 2 − y 2) − a2 xy = 0 has no asymptotes parallel to y-axis. a has no asymptotes. The curve r = 1 − cos θ An asymptote is a tangent of the curve at infinity. An asymptote touch the curve at finite point. (Agra 2007) 1. highest. 2. n. 3. x + y + a = 0. 4. r sin (θ − α) = 5. 9. r = a. T. 6. (c). 10. F. 7. (c). 11. T. 8. (c). 12. T. 1 , f ′(α) ≠ 0. f ′(α) 13. F. 14.1 Concavity and Convexity (Meerut 2009) Let P be a given point on a curve and A B a given straight line which does not pass through P . Then the curve is said to be concave or convex at P with respect to A B , according as a sufficiently small arc of the curve containing P lies entirely within or without the acute angle formed by the tangent at P to the curve with the line A B . Thus in the figure 1 the curve at P is convex to A B , and in figure 2 it is concave to A B . The property of concavity or convexity of a curve at any point is not an inherent property of the curve. At a given point a curve may be concave with respect to some line, while at the same point it may be convex with respect to some other line. D-290 DIFFERENTIAL CALCULUS Concavity upwards and Concavity downwards : The curve shown in the Fig. 1 below is concave upwards and the curve shown in the Fig. 2 below is concave downwards. 14.2 Point of Inflexion (Avadh 2014) A point P on a curve is said to be a point of inflexion, if the curve is concave on one side and convex on the other side of P with respect to any line A B . Thus at a point of inflexion the curve changes its direction of bending from concavity to convexity or vice-versa. The two portions of the curve on the two sides of P lie on different sides of the tangent at P , i.e., the curve crosses the tangent at P . Thus a point where the curve crosses the tangent is a point of inflexion. Therefore the position of a point of inflexion of a curve will in no way depend on the choice of coordinate axes. In particular, the positions of x and y a xes m a y be int e r ch a n ge d wit h ou t a ffect ing t h e positions of the points of inflexion on the curve. Inflexional tangent. The tangent at a point of inflexion of a curve is called inflexional tangent. 14.3 Test of Concavity or Convexity We shall consider concavity and convexity with respect to the axis of x. Let the equation of the curve be y = f (x) and let P be the point (x, y) on this curve. Suppose the tangent at P is not parallel to y-axis so that at P the value of f ′(x) is finite. SINGULAR POINTS : CURVE TRACING D-291 Let Q be the point (x + h, y + k) on the curve in the neighbourhood of P. The point Q may be taken on either side of P. Suppose the ordinate QN of Q meets the tangent to the curve at P in Q ′. The equation of the tangent at P is Y − y = f ′(x) (X − x), …(1) where (X, Y ) are the current coordinates. Putting X = x + h in (1), we get N Q ′ − y = f ′(x) {x + h − x} …(2) or N Q ′ = f (x) + hf ′(x). . . [ . y = f (x)] Also from the equation of the curve, we get N Q = f (x + h) h2 1 f ′′(x) + … + h n− 1 f (n− 1) (x) 2! (n − 1) ! 1 n n h f (x + θh), …(3) + n! on expanding by Taylor’s theorem, where 0 < θ < 1. From (2) and (3), by subtraction, we get h2 h3 hn (n) f ′′(x) + f ′′′(x) + … + f (x + θh). …(4) NQ − NQ ′ = 2! 3! n! If f ′′(x) ≠ 0, then by taking h sufficiently small, the second degree terms in h on the R .H .S. of (4) can be made to govern its sign. Therefore (N Q − N Q ′) will be of the h2 f ′′(x). same sign as 2! h2 O bviously f ′′(x) will be of invariable sign whether h is positive or negative 2! i.e., whether Q lies to the right or the left of P. The curve at P will be convex with respect to the axis of x if N Q − N Q ′ is positive [See Fig. 1] and it will be concave at P with respect to the axis of x if NQ − NQ ′ is negative [See Fig. 2]. H ence, the curve is convex at P to the axis of x if f ′′(x) is positive, and concave if f ′′(x) is negative. We have drawn the above figures for the case when the curve is above the axis of x. If, however, the curve is below the axis of x, then N Q and N Q ′ are both negative, and N Q − N Q ′ = − {| N Q| − | N Q ′| }. H ence, in this case the curve at P is convex with respect to the axis of x if | NQ| − | NQ ′| is positive i.e., if N Q − N Q ′ is negative i.e., if f ′′(x) is negative. Similarly the curve at P will be concave with respect to the axis of x if f ′′(x) is positive. F rom t h e above discu ssion we observe that whether the curve lies below or above the axis of x, we have the following criterion for concavity or convexity at P with respect to the axis of x. = f (x) + hf ′(x) + D-292 DIFFERENTIAL CALCULUS A curve is convex or concave at P to the axis of x according as y d2 y dx2 is positive or negative at P. Test for concavity upwards or Concavity downwards : T h e cu rv e y = f (x) i s concave upwards i n [a, b] i . e. , w h e n a ≤ x ≤ b, if f ′′(x) > 0 — V x ∈ [a, b] and is concave downwards in [a, b] if f ′′(x) < 0 — V x ∈ [a, b]. Concavity and Convexity with respect to the axis of y : By considering y as the independent variable, we can easily show that a curve x = f ( y), at a given point P on it, is convex or concave to the axis of y according as d2 x x 2 is positive or negative at P. dy Example 1 : Show that the curve y = e x is everywhere concave upwards and the curve y = log x is everywhere concave downwards. Solution : First consider the curve y = e x. dy = e x and dx We have O bviously d2 y = e x. dx2 d2 y > 0, — V x ∈ R , where R is the set of real numbers. dx2 H ence , the curve y = e x is everywhere concave upwards. Now consider the curve y = log x, 0 < x < ∞ . dy 1 = dx x We have and d2 y 1 = − 2⋅ x dx2 d2 y < 0, — V x ∈ ]0, ∞ [. dx2 H ence, the curve y = log x is everywhere concave downwards. Example 2 : Find the intervals in which the curve y = e x (cos x + sin x) is concave upwards or downwards; x varying in the interval ]0, 2 π[. Solution : The given curve is y = e x (cos x + sin x). dy We have = e x (cos x + sin x) + e x (− sin x + cos x) = 2 e x cos x. dx O bviously, ∴ d2 y = 2 e x cos x − 2 e x sin x = 2 e x (cos x − sin x) dx2 ⎛ ⎛ π π⎞ π⎞ ⎯⎯2 . e x cos ⎜⎝ x + ⎟⎠ ⋅ = 2√ ⎯⎯2 . e x ⎜⎝ cos x cos − sin x sin ⎟⎠ = 2 √ 4 4 4 ⎤ 5π ⎡ ⎤ π⎡ , 2 π ⎢ , we have When x ∈ ⎥ 0, ⎢ or when x ∈ ⎥⎦ ⎣ ⎦ ⎣ 4 4 d2 y > 0. dx2 ⎤ 5π ⎡ ⎤ π⎡ , 2 π⎢ ⋅ H ence, the given curve is concave upwards in ⎥ 0, ⎢ and ⎥⎦ ⎣ ⎦ ⎣ 4 4 2 , 5π ⎡⎢ , we have d y < 0. ⎣ 4 4 dx2 ⎤π Again when x ∈ ⎥ ⎦ D-293 SINGULAR POINTS : CURVE TRACING ⎤π , 5π ⎡⎢ ⋅ 4 4⎣ Example 3 : Show that the sine curve y = sin x is everywhere concave with respect to the axis of x excluding the points where it meets the axis of x. Solution : The given curve is y = sin x. H ence, the given curve is concave downwards in ⎥ ⎦ d2 y dy = cos x and = − sin x. dx dx2 The function sin x is a periodic function with period 2 π. Hence, it is sufficient to consider the given curve in the interval [0, 2 π]. In the interval [0, 2 π], we have y = 0 when x = 0 or x = π or x = 2 π. d2 y When x ∈ ]0, π[, we have y > 0 and < 0. dx2 d2 y So y 2 < 0 when x ∈ ]0, π[. dx H ence, the curve y = sin x is concave to the axis of x in the interval ]0, π[. d2 y When x ∈ ]π, 2 π[, we have y < 0 and 2 > 0. dx d2 y So y 2 < 0 when x ∈ ]π, 2 π[. dx H ence, the curve y = sin x is concave to the axis of x in the interval ]π, 2 π[. Thus the curve y = sin x is everywhere concave with respect to the axis of x excluding the points where it meets the axis of x. We have 14.4 Test for Point of Inflexion (Avadh 2014) Let the equation of the curve be y = f (x) and let P be the point (x, y) on this curve. Suppose the tangent at P is not parallel to y-axis so that at P the value of f ′ (x) is finite. Let Q be the point (x + h, y + k) on the curve in the neighbourhood of P . The point Q may be taken on either side of P . Suppose the ordinate QN o f Q meets the tangent to the curve at P in Q′. The equation of the tangent at P is Y − y = f ′ (x) (X − x), ...(1) where (X, Y ) are the current coordinates. Putting X = x + h in (1), we get N Q′ − y = f ′ (x) {x + h − x} N Q′ = f (x) + hf ′ (x) , [... y = f (x)] ...(2) Also from the equation of the curve, we get N Q = f (x + h) h2 f ′′ (x) + … = f (x) + hf ′ (x) + 2! 1 n (n) 1 hn − 1 f (n − 1) (x) + h f (x + θh) , ...(3) + n! (n − 1) ! on expanding by Taylor’s theorem, if 0 < θ < 1 . From (2) and (3), by subtraction, we get or N Q − N Q′ = h2 h3 hn (n) f ′′ (x) + f ′′′ (x) + … + f (x + θh) . 2! 3! n! ...(4) D-294 DIFFERENTIAL CALCULUS If f ′′ (x) ≠ 0 , then by taking h sufficiently small, the second degree terms in h on the R .H.S. of (iv) can be made to govern its sign. Therefore (N Q − N Q′) will be of the same sign as (h2 ⁄ 2 !) f ′′ (x) . But (h2 ⁄ 2 !) f ′′ (x) will be of invariable sign whether h is positive or negative i.e., whether Q lies to the right or the left of P . Therefore on both sides of P the curve will be either concave or convex. H ence the necessary condition for the existence of a point of inflexion at P is that f ′′ (x) = 0 . Now if f ′′ (x) = 0 , we have from (iv) h4 iv hn (n) h3 f ′′′ (x) + f (x) + … + f (x + θh) . ...(5) N Q − N Q′ = 3! 4! n! If f ′′′ (x) ≠ 0 , then for sufficiently small values of h the sign of the right hand side of (5) is the same as that of (h3 ⁄ 3 !) f ′′′ (x) , which changes sign when h changes sign. Thus with respect to x-axis, the curve will be concave on one side of P and convex on the other side of P . So there will be a point of inflexion at P. H ence, there will be a point of inflexion at P, if d2 y ⁄ dx2 = 0 but d3 y ⁄ dx3 ≠ 0 . Generalisation : I f f ′′ (x) = f ′′′ (x) = f iv (x) = … = f (n − 1) (x) = 0, a n d (n) f (x) ≠ 0 , it is easy to see from the value of N Q − N Q′ , that there will be a point of inflexion if n is odd. If, however, n is even, the curve does not cross the tangent and so there will not be a point of inflexion at P . Such a point (if n is greater than 2) is called a point of undulation. To the eye a point of undulation appears just like an ordinary point. Corollary : The position of a point of inflexion is independent of the choice of coordinate axes. Therefore on interchanging x and y in the above results, we can say that, there will be a point of inflexion at P , if d2 x ⁄ dy2 = 0 , but d3 x ⁄ dy3 ≠ 0 . (Bundelkhand 2007) It will become necessary for us to use this criterion if the tangent at P is parallel to y-axis i.e., if dy ⁄ dx is infinite at P . It will also be useful where the equation of the curve is of the form x = f ( y) . 14.5 Concavity and Convexity for Polar Curves From the following figures it is obvious that if at any point P on the curve the perpendicular p drawn from the pole on the tangent increases as r increases, then the curve is concave at P to the pole. Thus, a curve is concave at P to the pole if dp ⁄ dr is positive there. Similarly, a curve is convex at P to the pole if dp ⁄ dr is negative there. If dp ⁄ dr is zero at P, positive for points on one side of P and negative for the points on the other side of P, there must be a point of inflexion at P. D-295 SINGULAR POINTS : CURVE TRACING But r dr = radius of curvature at P dp ⎧ ⎪ 2 ⎨r ⎪ ⎩ ⎛ 2 3⁄2 ⎞ ⎫⎪ dr + ⎜⎝ ⎟⎠ ⎬⎪ dθ ⎭ = ⎛ dr ⎞ 2 d2 r r2 + 2 ⎜⎝ ⎟⎠ − r 2 dθ dθ ⎧ ∴ dp = dr ⎛ dr ⎞ ⎪ r ⎨⎪ r2 + 2 ⎜⎝ ⎟⎠ dθ ⎩ 2 ⎧ ⎛ ⎞2 ⎪ 2 ⎨ r + ⎜ dr ⎟ ⎪ ⎝ dθ⎠ ⎩ 2 ⎛ dr ⎞ d2 r − r ⎫ d2 r⎪⎬ dθ2 ⎪⎭ ⎫3 ⁄ 2 ⎪ ⎬ ⎪ ⎭ Hence, if r2 + 2 ⎜⎝ ⎟⎠ − r 2 = 0 at P, there is, in general, a point of inflexion dθ dθ at P. Example 1 : Find the points of inflexion of the curve y = 3x4 − 4x3 + 1 . (Rohilkhand 2010, 11B; Avadh 10; Meerut 12) Solution : Differentiating the equation of the curve with respect to x , we get dy ⁄ dx = 12x3 − 12x2 and d2 y ⁄ dx2 = 36 x2 − 24x . For the points of inflexion, we must have d2 y ⁄ dx2 = 0 36 x2 − 24 x = 0 , i.e., x= 0 i.e., Now 2 3 or i.e., 12 x (3x − 2) = 0 , for the points of inflexion. d3 y ⁄ dx3 = 72 x − 24 . When x = 0, d3 y ⁄ dx3 ≠ 0, therefore x = 0 gives a point of inflexion. 2 2 Similarly, when x = , d3 y ⁄ dx3 ≠ 0 ; therefore x = also gives a point of inflexion. 3 3 From the equation of the curve, we have 11 , 2 y = 1 , when x = 0 and y = when x = ⋅ 3 27 ⎛ 2 11 ⎞ H ence (0, 1) and ⎜⎝ , ⎟⎠ are the required points of inflexion. 3 27 Important : Instead of finding d3 y ⁄ dx3 , we can use another criterion for points of inflexion. If d2 y ⁄ dx2 = 0 at x = a and the sign of d2 y ⁄ dx2 changes while x passes through a , then there will be a point of inflexion at x = a. Example 2 : Find the points of inflexion of the curve x = log ( y ⁄ x) . (Purvanchal 2011; Rohilkhand 14) Solution : The given curve is x = log ( y ⁄ x) or y ⁄ x = e x or y = x e x. Differentiating (1), we get dy ⁄ dx = x e x + e x = (x + 1) e x and d2 y ⁄ dx2 = e x + (x + 1) e x = e x (x + 2) . ...(1) D-296 DIFFERENTIAL CALCULUS For points of inflexion d2 y ⁄ dx2 = 0 . ∴ e x (x + 2) = 0 i.e., [... e x ≠ 0] x = − 2, Now d3 y ⁄ dx3 = e x + (x + 2) e x = e x (x + 3) ≠ 0 at x = − 2 . ∴ there is a point of inflexion at x = − 2 . From (1), when x = − 2 , y = − 2e− 2 = − 2 ⁄ e2. H ence the point of inflexion is (− 2, − 2 ⁄ e2) . Example 3 : Find the points of inflexion on the curve x = a (2θ − sin θ) , y = a (2 − cos θ) . Solution : Differentiating w.r.t. θ , we have dx ⁄ dθ = a (2 − cos θ) and dy ⁄ dθ = a sin θ . sin θ dy dy ⁄ dθ ∴ = = ⋅ dx dx ⁄ dθ 2 − cos θ And (Avadh 2013) d ⎛ sin θ ⎞ dθ d2 y = ⎜ ⎟ ⋅ dθ ⎝ 2 − cos θ⎠ dx dx2 (2 − cos θ) cos θ − sin θ (sin θ) 1 = ⋅ a (2 − cos θ) (2 − cos θ) 2 = 2 cos θ − (cos2 θ + sin2 θ) 2 cos θ − 1 = ⋅ 3 a (2 − cos θ) 3 a (2 − cos θ) Now for the points of inflexion, we must have d2 y ⁄ dx2 = 0 2 cos θ − 1 = 0 i.e., ∴ θ = 2n π ± 1 3 or cos θ = 1 2 1 3 = cos π . π , where n is any integer. Substituting the value of θ in the given equation of the curve we get the points of inflexion as 2π __ √3 ⎞ , 3a ⎤ ⎡ . . π⎞ √3 ⎤ ⎡ ⎛ ⎛ ⎛ π⎞ + ⎢ a ⎜ 4n π ± ⎟ ⎥ ⋅ ⎢ . sin ⎜ 2n π ± ⎟ = (− 1) 2n sin ⎜ ± ⎟ = ± ⎥⋅ 3 2⎠ 2⎦ ⎣ 3⎠ 2⎦ ⎣ ⎝ ⎝ ⎝ 3⎠ Example 4 : Find the ranges of values of x for which the curve y = x4 − 6 x3 + 12 x2 + 5x + 7 is concave upwards or downwards. A lso determine the points of inflexion. dy Solution : We have = 4 x3 − 18 x2 + 24 x + 5, dx d2 y = 12 x2 − 36 x + 24 = 12 (x − 1) (x − 2) dx2 (Purvanchal 2008) d3 y = 24 x − 36. dx3 and We have d2 y > 0, — V x ∈ ]− ∞ , 1[, dx2 d2 y d2 y < 0, — V x ∈ ]1, 2[ and 2 > 0, — V x ∈ ]2 , ∞ [. 2 dx dx H ence, the curve is concave upwards in the intervals ]− ∞ , 1[ and ]2 , ∞ [ and concave downwards in the interval ]1, 2[. D-297 SINGULAR POINTS : CURVE TRACING Now d2 y = 0 ⇒ (x − 1) (x − 2) = 0 ⇒ x = 1 or x = 2 . dx2 d2 y d3 y = − 12 ≠ 0 and at x = 2 , 2 = 12 ≠ 0. Thus, there are points of 3 dx dx inflexion at x = 1 and at x = 2 . When x = 1, we have y = 19 and when x = 2 , we have y = 33. H ence, (1, 19) and (2, 33) are the two points of inflexion on the curve. Example 5 : E xa m in e th e curve y = sin x for concavity upwards, concavity downwards and for points of inflexion in the interval [− 2 π, 2 π]. Solution : The given curve is y = sin x. At x = 1, We have d2 y d3 y dy = cos x, = − sin x and = − cos x. 2 dx dx dx3 We have, d2 y < 0, — V x ∈ ]− 2 π, − π[, dx2 d2 y d2 y > 0, — V x ∈ ]− π, 0[, < 0, — V x ∈ ]0, π[ 2 dx dx2 d2 y > 0, — V x ∈ ]π, 2 π[. dx2 H ence, the curve is concave downwards in the intervals ]− 2 π, − π[ and ]0, π[ and concave upwards in the intervals ]− π, 0[ and ]π, 2 π[. and Now or d2 y = 0 ⇒ sin x = 0 ⇒ x = − 2 π or x = − π or x = 0 dx2 x = π or x = 2 π. A t e ach o f t h e p o in t s x = − 2 π, x = − π, x = 0, x = π a n d x = 2 π, we h a ve d3 y ≠ 0. dx3 Thus there are points of inflexion at each of these points. Also, y = 0 at each of these points. H ence, the curve has points of inflexion at (− 2 π, 0), (− π, 0), (0, 0), (π, 0) and (2 π, 0). Example 6 : Find the points of inflexion on the curve r (θ2 − 1) = aθ2. (Meerut 2013) Solution : We have r = aθ2 ⁄ (θ2 − 1) . dr = a [(θ2 − 1) .2θ − θ2. 2θ] ⁄ (θ2 − 1) 2 = − 2aθ ⁄ (θ2 − 1) 2, ∴ dθ d2 r = − 2a [(θ2 − 1) 2. 1 − θ .2 (θ2 − 1) .2θ] ⁄ (θ2 − 1) 4 and dθ2 = 2a (3θ2 + 1) ⁄ (θ2 − 1) 3. We know that at the point of inflexion, the radius of curvature is infinite. H ence at the point of inflexion, we have r2 + 2 (dr ⁄ dθ) 2 − r (d2 r ⁄ dθ2) = 0 or a2 θ4 8a2 θ2 2a2 θ2 (3θ2 + 1) + − = 0 (θ2 − 1) 4 (θ2 − 1) 2 (θ2 − 1) 4 D-298 DIFFERENTIAL CALCULUS or a2 θ2 (θ2 − 3) (θ2 + 2) = 0 (θ2 − 1) 3 or θ2 (θ2 − 3) (θ2 + 2) = 0 θ2 = 0 , 3 , − 2 . ∴ R ejecting the values θ2 = − 2 and 0 we see that the points of inflexion are given by θ2 = 3 i.e., θ = ± √3 . 1. Show that the points of inflexion upon the curve x2 y = a2 (x − y) are given by x = 0, x = ± a √3 . (Meerut 2013B) 2. Find the points of inflexion of the curve y (a2 + x2) = x3. a2 3. Find the points of inflexion of the curve xy = 4. Find the points of inflexion of the curve x = (log y) 3. (Purvanchal 2010) log (y ⁄ a) . (Purvanchal 2009) 1) 4 (x − 2) 3. 5. Investigate the points of inflexion of the curve y = (x − 6. Show that every point in which the sine curve y = c sin (x ⁄ a) meets the axis of x is a point of inflexion. Show that points of inflexion of the curve y2 = (x − a) 2 (x − b) lie on the line 3x + a = 4b ; (Agra 2006; Avadh 11; Kashi 12) 7. (Agra 2014) 8. Find the points of inflexion on the curve y2 = x (x + 1) 2 and also obtain the equations of the inflexional tangents. 9. Show that origin is a point of inflexion of the curve am − 1. y = x m if m is odd and greater than 2. Show that the abscissae of the points of inflexion on the curve y2 = f (x) satisfy the equation [ f ′ (x)]2 = 2 f (x) . f ′′ (x). S h o w t h a t t h e l i n e j o i n i n g t h e p o i n t s o f i n fl e xi o n o f t h e c u r ve y2 (x − a) = x2 (x + a) subtends an angle of π ⁄ 3 at the origin. 1− x , Prove that the curve y = has three points of inflexion which lie in a 1 + x2 straight line. 10. 11. 12. − (x ⁄ a) 2 13. Show that the points of inflexion on the curve y = be x = ± a ⁄ √2 . 14. bθn Show that the points of inflexion of the curve r = r = b {− n (n + 2. ⎛ ⎝ (0, 0), ⎜ √3a , 1)}n ⁄ 2 . 3 √3 ⎞ , ⎛ − 3 √3a ⎞ a⎟ ⎜ − √3a , ⎟⋅ 4 ⎠ ⎝ 4 ⎠ are given by (Agra 2005) are given by D-299 SINGULAR POINTS : CURVE TRACING 3. 4. 5. 8. ⎛ ⎜ ⎝ 3 2 ⎞ ae− 3 ⁄ 2, ae3 ⁄ 2⎟⎠ ⋅ (0, 1), (8, e2) . Points of inflexion at x = 2 , (11 ± √2) ⁄ 7; point of undulation at x = 1. Points of inflexion are (1 ⁄ 3, ± 4 ⁄ 3√3) . Inflexional tangents are 9x ± 3√3y + 1 = 0. 14.6 Multiple Points (Meerut 2003) A point through which more than one branches of a curve pass is called a m ultiple point on the curve. A point on the curve is called a double point if two branches of the curve pass through it, a triple point if three branches pass through it. In general, if r branches pass through a point, it is called a multiple point of the r th order. 14.7 Singular Points A n unusual point on a curve is called a singular point. For example, at any point the tangent does not usually cross the curve. But at a point of inflexion the tangent crosses the curve and therefore it is a singular point. Similarly, through one point, usually one branch of the curve passes. But through a multiple point, more than one branches of the curve pass. Therefore multiple points are also singular points. 14.8 Classification of Double Points (i) Node : If the two branches through a double point on a curve are real and have different tangents there, then the double point is called a node. (Kumaun 2008; Kashi 11) (ii) Cusp : If the two branches through a double point on a curve are real and have coincident tangents there, then the double point is called a cusp. (Kashi 2011) (iii) Conjugate Point : If there are no real points on the curve in the neighbourhood of a point P on the curve, then P is called a conjugate point (or an isolated point). The process of finding the tangents usually gives imaginary tangents at such a point. D-300 DIFFERENTIAL CALCULUS Since through a double point two branches of the curve pass, therefore in the process of finding tangents at a double point we must get two tangents there, one for each branch. If the two tangents are real and distinct, the double point will be a node. If the two tangents are imaginary, the double point will be a conjugate point. If the two tangents are real and coincident, the double point may be a cusp or a conjugate point. The possibility of the point being a conjugate point in this case arises on account of the fact that sometimes imaginary expressions A ± iB become real by chance when B = 0 . In such cases the double point will be a cusp if there are other real points of the curve in its neighbourhood, otherwise it will be a conjugate point. 14.9 Species of Cusps We know that two branches of a curve have a common tangent at a cusp. A cusp is said to be single or double according as the curve lies entirely on one side of the common normal or on both sides. Also it is of the first or second species according as the two branches lie on opposite sides or on the sam e side of the common tangents. We have the following five different types of cusps : Single cusp of the first species as shown in Fig. 1. Single cusp of the second species as shown in Fig. 2. Double cusp of the first species as shown in Fig. 3. Double cusp of the second species as shown in Fig. 4. Double cusp with change of species as shown in Fig. 5. H ere the two branches lie on both the sides of the common normal but on one side they lie on the same and on the other on opposite sides of the common tangent. Such a point is called a point of oscul-inflexion. 14.10 Tangents at Origin In order to know the nature of a double point it is necessary to find the tangent or tangents there. Now we shall find a simple rule for writing down the tangent or tangents at the origin to rational algebraic curves. D-301 SINGULAR POINTS : CURVE TRACING If a curve passes through the origin and is given by a rational integral, algebraic equation, the equation to the tangent or tangents at the origin is obtained by equating to zero the lowest degree terms in the equation of the curve. Let the equation of the curve when arranged according to ascending powers of x and y be (a1 x + a2 y) + (b1 x2 + b2 xy + b3 y2) + (c1 x3 + c2 x2 y + …) + … = 0 , ...(1) where the constant term is absent since the curve passes through the origin. Let P (x, y) be any point on the curve. The slope of the chord OP is y ⁄ x . Therefore the equation to OP is Y = ( y ⁄ x) X , where (X, Y ) are current coordinates. As P → O i.e., as x → 0 and y → 0 , the chord OP tends to the tangent at O . E xcluding for the present the case when the tangent is the y-axis i.e., when ⎛ y⎞ ⎜ lim ⎜ ⎟⎟ x → 0 ⎜⎝ x⎟⎠ = ± ∞ , we have the equation of the tangent at O as ⎧ ⎛ ⎞ ⎫ ⎪ lim ⎜ y⎟ ⎪ ...(2) Y = ⎨⎪ x → 0 ⎜⎜ ⎟⎟ ⎬⎪ X . ⎩ ⎝ x⎠ ⎭ Case I : Let a2 ≠ 0 . Dividing (1) by x and taking limit as x→0 , we get ⎧ ⎪ lim a1 + a2 ⎨⎪ x → 0 ⎩ ⎛ y⎞ ⎜ ⎟ ⎜ ⎟ ⎜ x⎟ ⎝ ⎠ ⎫ ⎪ ⎬ = 0. ⎪ ⎭ ...(3) lim ⎛ y⎞ E liminating x → 0 ⎜ ⎟ bet ween (2) and (3), we get a1 X + a2 Y = 0 , as t h e ⎝ x⎠ equation of tangent at the origin to the curve (1). R eplacing the current coordinates X , Y by x, y this equation becomes a1 x + a2 y = 0 , ...(4) which is obviously the equation obtained by equating to zero the lowest degree terms in (1). If a2 = 0 , then a1 is also zero from (3), and we get the next case. Case II : Let a1 = 0 , a2 = 0 , but b2 and b3 are not both zero. Dividing (1) by x2 and taking limit as x→ 0 , we get lim ⎛ y⎞ lim ⎛ y⎞ 2 b1 + b2 x → 0 ⎜ ⎟ + b3 x → 0 ⎜ ⎟ = 0 , ⎝ x⎠ ⎝ x⎠ or b1 + b2 m + b3 m 2 = 0 , ...(5) lim ⎛ y⎞ x → 0 ⎜⎝ x⎟⎠ = m . E quation (5) is a quadratic in m , showing that there are two tangents at the origin in this case. Eliminating m between (2) and (5), we get where b1 x2 + b2 x y + b3 y2 = 0 , ...(6) as the equation of the tangents at the origin to (1) in this case. In equation (6), we have taken x, y as current coordinates. O bviously the equation (6) is obtained by equating to zero the lowest degree terms in the equation of the curve (1), where D-302 DIFFERENTIAL CALCULUS a1 = a2 = 0 . If b2 = b3 = 0 , then by (v), b1 = 0 . Case III : If a1 = a2 = b1 = b2 = b3 = 0 , we can show by the same process that the rule still holds; and so on. If tangent at the origin is the y-axis, we can easily show by supposing the axes of x and y to be interchanged for a moment, that the rule is still true. Hence the equation of the tangent or tangents at the origin is obtained by equating to zero the lowest degree terms in the equation of the curve. Corollary : If the origin is a double point on a curve, then the curve has two tangents at the origin. Therefore the equation of the curve should not contain the constant and the first degree terms and the second degree terms should be the lowest degree terms in the equation of the curve. Example 1 : Show that the origin is a node on the curve x3 + y3 − 3axy = 0. (Meerut 2003; Purvanchal 14) Solution : The curve passes through the origin as its equation does not contain the constant term. Also equating to zero the lowest degree terms in the equation of the curve, we get the equation to the tangents at origin as − 3axy = 0, i.e. x y = 0, i.e. x = 0, y = 0 are two real and distinct tangents at the origin. Therefore origin is a node. Example 2 : Show that the origin is a conjugate point on the curve a2 x2 + b2 y2 = (x2 + y2) 2 . Solution : O bviously the curve passes through the origin. The equation to the tangent at origin is a2 x2 + b2 y2 = 0, i.e., a x ± iby = 0. Thus there are two imaginary tangents at the origin. Therefore origin is a conjugate point. 14.11 Change of Origin Let (x, y) be the coordinates of a point P with r e fe r e n c e t o Ox and Oy a s c o o r d i n a t e a xe s. R eferred to Ox a n d Oy as co ordinate axes, let (h, k) be the coordinates of a point O′. Draw a line O′X parallel t o Ox an d a lin e O′Y parallel to Oy . Let (X , Y ) b e t h e co o r d i n a t e s o f P wi t h reference to O′X and O′Y as coordinate axes. O bviously, we have x = X + h and y = Y + k . Thus to obtain the equation of the curve referred to the point (h, k) as origin, the coordinate axes remaining parallel to their original directions, we should put X + h in place of x and Y + k in place of y in the equation of the curve, where X , Y are the current coordinates in the new equation. If in the new equation also we take x, y as the current coordinates, then in order to shift the origin to the point (h, k) , we should replace x by x + h and y by y + k in the given equation of the curve. SINGULAR POINTS : CURVE TRACING D-303 14.12 Tangents at the Point (h, k) to a Curve If we are to find the tangents at the point (h, k) to a curve, we should first shift the origin to the point (h, k) in the equation of the curve. Then the equation of the tangents at the new origin will be obtained by equating to zero the lowest degree terms in the new equation of the curve. Example : Show that the point (2 , 1) is a node on the curve (x − 2) 2 = y ( y − 1) 2. Solution : Shifting the origin to the point (2 , 1), the equation of the curve becomes {(x + 2) − 2}2 = ( y + 1) {( y + 1) − 1}2 x2 = y2 ( y + 1) . ...(1) E quating to zero the lowest degree terms in (1), the equation of the tangents at the new origin is i.e. x2 = y2, i.e. y = ± x . Thus there are two real and distinct tangents at the new origin. Therefore the new origin is a node. H ence there is a node at the point (2 , 1) on the given curve. 14.13 Position and Character of Double Points Let f (x, y) = 0 be any curve and P be any point (x, y) on it. The slope of the tangent at P is equal to dy ⁄ dx and it is given by the equation ∂f ⁄ ∂x ∂f ∂f dy dy = − or + ⋅ = 0 ...(1) dx ∂x ∂y dx ∂f ⁄ ∂y At a multiple point of a curve, the curve has at least two tangents and accordingly dy ⁄ dx must have at least two values at a multiple point. The equation (1) is of first degree in dy ⁄ dx . It can be satisfied for more than one value of dy ⁄ dx , if and only if, ∂f ∂f = 0, = 0. ∂x ∂y Therefore the necessary and sufficient conditions for any point (x, y) of the curve f (x, y) = 0 to be a m ultiple point are that ∂f ∂f = 0, = 0. ∂y ∂x H ence in order to find the multiple points of the curve f (x, y) = 0 , we should simultaneously solve the equations, ∂f ∂f = 0, = 0 , f (x, y) = 0 . ∂y ∂x Differentiating (1) with respect to x again, we get d ⎧ ∂f dy⎫ d ⎛ ∂f ⎞ ⎜ ⎟ + ⎨ ⎬= 0 dx ⎩ ∂y dx⎭ dx ⎝ ∂x⎠ or d ⎛ ∂f ⎞ d ⎛ ∂f ⎞ dy ∂f d2 y + ⋅ = 0 ⎜ ⎟ + ⎜ ⎟ ⋅ dx ⎝ ∂x⎠ dx ⎝ ∂y⎠ dx ∂y dx2 D-304 DIFFERENTIAL CALCULUS ⎡ ∂ ⎛ ∂f ⎞ ⎧ ∂ ⎛ ∂f ⎞ ⎫ dy⎤ ⎢ ⎜ ⎟ + ⎨ ⎜ ⎟⎬ ⎥ ⎣ ∂x ⎝ ∂x⎠ ⎩ ∂y ⎝ ∂x⎠ ⎭ dx⎦ or ∂f d2 y ⎡ ∂ ⎛ ∂f ⎞ ⎧ ∂ ⎛ ∂f ⎞ ⎫ dy⎤ dy + ⋅ = 0 ⎜ ⎟ + ⎨ ⎜ ⎟⎬ ⋅ ⎥ ⋅ ⎣ ∂x ⎝ ∂y⎠ ⎩ ∂y ⎝ ∂y⎠ ⎭ dx⎦ dx ∂y dx2 + ⎢ ∂2f ∂2 f dy ∂2 f dy ∂2 f ⎛ dy⎞ 2 ∂f d2 y + ⋅ + ⋅ + ⋅ = 0 ⎜ ⎟ + ∂y dx2 ∂x2 ∂x ∂y dx ∂x ∂y dx ∂y2 ⎝ dx⎠ or ∂2 f ∂2 f dy ∂2 f ⎛ dy⎞ 2 + 2⋅ ⋅ + ⋅ ⎜ ⎟ = 0, 2 ∂x ∂y dx ∂y2 ⎝ dx⎠ ∂x since at a multiple point ∂f ⁄ ∂y = 0 . or Therefore at the multiple point, the values of in dy are given by the quadratic dx dy , dx ∂2 f ⎛ dy⎞ ∂2 f ∂2 f ⎛ dy⎞ 2 + 2 + = 0. ⎜ ⎟ ⎜ ⎟ ∂x ∂y ⎝ dx⎠ ∂y2 ⎝ dx⎠ ∂x2 ...(2) ∂2 f , ∂2 f , ∂2 f are not all zero, the equation (2) will be a quadratic in dy ⁄ dx ∂x2 ∂x ∂y ∂y2 and the multiple point will be a double point. The two tangents will be real and distinct, coincident, or imaginary according as If ⎛ ∂2f ⎞ 2 ∂2 f ∂2 f ⎟ − 4 > , = or < 0 ⎝ ∂x ∂y⎠ ∂y2 ∂x2 4⎜ i.e., in general, the double point will be a node, cusp or conjugate point according as ⎛ ∂ 2 f⎞ ⎛ ∂ 2 f⎞ ⎛ ∂2 f ⎞ 2 ⎟ ⎜ ⎟⋅ ⎜ ⎟ > , = or < ⎜⎜ ⎟ ⎜ ⎟ (Meerut 2003) ⎝ ∂x ∂y⎠ ⎜ ∂x2 ⎟ ⎜ ∂y2 ⎟ ⎠ ⎝ ⎠ ⎝ ∂2 f ∂2 f ∂2 f = = = 0 , then the point (x, y) will be a multiple point of order ∂x2 ∂x ∂y ∂y2 higher than the second. If 14.14 Nature of a Cusp at the Origin Suppose the origin is a cusp. Then the curve will have two coincident tangents at the origin. Therefore the equation of the curve must be of the form (ax + by) 2 + terms of third and higher degrees = 0 . ...(1) The common tangent at the origin to the two branches of the curve is ax + by = 0 . ...(2) Let P be the perpendicular to (2) from any point (x, y) on (1) in the neighbourhood of the origin. Then ax + by , P= which is proportional to ax + by . Let us put √(a2 + b2) p = ax + by . ...(3) E liminate x or y (whichever is convenient) between (1) and (3). Suppose we eliminate y . Then we shall get a relation between p and x . Since p is small and also there SINGULAR POINTS : CURVE TRACING D-305 are only two branches of the curve (2) through the origin, therefore terms involving powers of p above the second will be neglected. Thus we shall get a quadratic in p of the form A p2 + Bp + C = 0 , ...(4) where A , B and C are some functions of x . Solving (4), we get ...(5) p = {− B ± √(B 2 − 4 A C )} ⁄ 2A . Also if p1, p2 are the roots of (4), we get p1 p2 = C ⁄ A . ...(6) The following different cases arise : (i) If for all values of x , positive or negative, provided they are numerically small, the values of p given by (5) are imaginary, the origin will be a conjugate point. (ii) If for all numerically small values of x , positive or negative, the values of p given by (5) are real, there will be a double cusp at the origin. (iii) If the reality of the values of p given by (5) depends on the sign of x , there will be a single cusp at the origin. (iv) If for numerically small values of x for which p is real, the sign of p1 p2 is positive, then p1 and p2 will be of the same sign. Therefore the two perpendiculars lie on the same side of the common tangent and there will be a cusp of the second species. If, on the other hand, the sign of p1 p2 is negative, then p1and p2 are of opposite signs. Therefore the two perpendiculars lie on opposite sides of the common tangent and there will be a cusp of the first species. Note : While investigating the sign of an expression for sufficiently small values of x , we should keep in mind only those terms which involve the lowest power of x . 14.15 Nature of a Cusp at any Point If there is a cusp at the point (h, k) , we should first shift the origin to (h, k) and then apply the methods given in article 14.14. Example 1 : Exam ine the nature of the origin on the curve (2x + y) 2 − 6xy (2x + y) − 7x3 = 0 . Solution : The tangents at the origin are (2x + y) 2 = 0 . Thus there are two coincident tangents at the origin. Therefore the origin may be a cusp or a conjugate point. Let p = 2x + y . Putting y = p − 2x in the equation of the curve, we get p2 − 6xp ( p − 2x) − 7x3 = 0 or p2 (1 − 6x) + 12x2 p − 7x3 = 0 . Let p1, p2 be the roots of (1). Then p= − 12x2 ± √{144x4 + 28x3 (1 − 6x)} 2 (1 − 6x) ...(1) D-306 DIFFERENTIAL CALCULUS p= i.e., − 6x2 ± √(7x3 − 6x4) , (1 − 6x) ...(2) 7x3 ⋅ ...(3) 1 − 6x From (2), we see that for sufficiently small positive values of x, p is real and for numerically small negative values of x, p is imaginary. Therefore, there is a single cusp at the origin. Also when x is + ive and very small, then from (3) we notice that p1 p2 is –ive. Therefore p1 and p2 are of opposite signs. H ence there is a single cusp of the first species p1 p2 = − and at the origin. Example 2 : Determ ine the existence and nature of the double points on the curve y2 = (x − 2) 2 (x − 1) . Solution : The equation of the given curve is (Meerut 2003) f (x, y) ≡ y2 − (x − 2) 2 (x − 1) = 0 . ...(1) We have ∂f ⁄ ∂x = − 2 (x − 2) (x − 1) − (x − 2) 2 = − (x − 2) {2 (x − 1) + (x − 2)} = − (x − 2) (3x − 4) and ∂f ⁄ ∂y = 2y . For double points, ∂f ⁄ ∂x = 0 , ∂f ⁄ ∂y = 0 and f (x, y) = 0 . H ere ∂f ⁄ ∂x = 0 gives (x − 2) (3x − 4) = 0 i.e., x = 2 , 4 ⁄ 3 ∂f ⁄ ∂y = 0 gives y = 0 . ∴ the possible double points are (2, 0) , (4 ⁄ 3, 0) . O ut of these only (2, 0) satisfies the equation of the curve. Therefore (2, 0) is the only double point on the given curve. Nature of the double point at (2, 0) : Shifting the origin to the point (2, 0) , the equation of the curve becomes and ...(2) y2 = (x + 2 − 2) 2 (x + 2 − 1) i.e., y2 = x2 (x + 1) E quating to zero the lowest degree terms in (2), the tangents at the new origin are y2 − x2 = 0 i.e., y2 = x2 i.e., y = ± x . Thus there are two real and distinct tangents at the new origin. Therefore the new origin is a node. H ence there is a node at the point (2, 0) on the given curve. Example 3 : Exam ine the nature of the double points of the curve 2 (x3 + y3) − 3 (3x2 + y2) + 12x = 4. Solution : The equation of the given curve is f (x, y) ≡ 2 (x3 + y3) − 3 (3x2 + y2) + 12x − 4 = 0 . We have ∂f = 6x2 − 18x + 12 and ∂x For the double points, H ere i.e., ∂f = 6y2 − 6y . ∂y ∂f ∂f = 0, = 0 and f (x, y) = 0 . ∂x ∂y ∂f = 0 gives 6x2 − 18x + 12 = 0 ∂x x2 − 3x + 2 = 0 i.e., (x − 1) (x − 2) = 0 i.e., x = 1 , 2 ...(1) D-307 SINGULAR POINTS : CURVE TRACING ∂f = 0 gives 6y2 − 6y = 0 i.e., y (y − 1) = 0 i.e., y = 0 , 1 . ∂y ∴ the possible double points are (1, 0) , (1, 1), (2, 0) and (2, 1) . O ut of these only (1, 1) and (2, 0) satisfy the equation of the curve. Therefore (1, 1) and (2, 0) are the only double points on the given curve. and Now ∂2 f ∂2 f ∂2 f = 12x − 18 , = 0 , = 12y − 6 . ∂x ∂y ∂x2 ∂y2 At the point (1, 1) , ∂2 f ∂2 f ∂2 f = − 6 , = 0 , = 6. ∂x ∂y ∂x2 ∂y2 ⎛ ∂2 f ⎞ 2 ∂2 f ∂2 f ⎟ = 0 and ⋅ = − 36 . ⎝ ∂x ∂y⎠ ∂x2 ∂y2 ∴ at the point (1, 1) , ⎜ ⎛ ∂2 f ⎞ 2 ∂2 f ∂2 f ⎟ > ⋅ ⋅ ⎝ ∂x ∂y⎠ ∂x2 ∂y2 Thus at the point (1, 1) , ⎜ Therefore there is a node at the point (1, 1) . ∂2 f ∂2 f ∂2 f = 0, 2 = − 6. At the point (2, 0) , 2 = 6 , ∂x ∂y ∂x ∂y ⎛ ∂2 f ⎞ 2 ∂2 f ∂2 f ⎟ > ⋅ ⋅ ⎝ ∂x ∂y⎠ ∂x2 ∂y2 ∴ at the point (2, 0) , ⎜ Thus there is a node at the point (2, 0) . Example 4 : Find the nature of the origin on the curve a4 y2 = x4 (x2 − a2) . (Meerut 2006B) Solution : The given curve is a4 y2 = x4 (x2 − a2) . ...(1) E quating to zero the lowest degree terms in the equation of the curve, we get the tangents at the origin as a4 y2 = 0 i.e., y = 0 , y = 0 are two real and coincident tangents at the origin. Thus the origin may be a cusp or a conjugate point. From (1), y = ± (x2 ⁄ a2) √(x2 − a2) . For small values of x ≠ 0 , + ive or –ive, (x2 − a2) is –ive i.e., y is imaginary. H ence no portion of the curve lies in the neighbourhood of the origin. H ence origin is a conjugate point and not a cusp. Example 5 : Show that the origin is a conjugate point on the curve x4 − a x2 y + a x y2 + a2 y2 = 0. Solution : E quating to zero, the lowest degree terms in the given curve, the tangents at the origin are given by a2 y2 = 0 i.e., y2 = 0 i.e., y = 0, y = 0. Thus there are two real and coincident tangents at the origin ∴ origin is either a cusp or a conjugate point. Now the equation of the given curve is ay2 (x + a) − ax2 y + x4 = 0. Solving it for y, we have a x2 ± √{a2 x4 − 4ax4 (x + a)} a x2 ± x2 √(− 4a x − 3a2) y= = ⋅ 2 a (x + a) 2a (x + a) D-308 DIFFERENTIAL CALCULUS Now for small values of x ≠ 0, (− 4ax − 3a2) is − ive. Thus y is imaginary in the neighbourhood of origin. H ence origin is a conjugate point. 1. Write down the equations to the tangents at the origin for the following curves : (i) y2 (a − x) = x2 (a + x), (ii) x4 + 3x3 y + 2 x y − y2 = 0, (iii) (x2 + y2) (2a − x) = b2 x . 2. For the curve y2 (a2 + x2) = x2 (a2 − x2) , show that the origin is a node. 3. Show that the origin is a conjugate point on the curve y2 = 2 x2 y + x4 y − 2 x4. 4. 5. Show that the curve x3 + x2 y = ay2 has a cusp at the origin. Find the position and nature of double points of the following curves : (i) y3 = x3 + a x2. (Meerut 2013B) (ii) y2 + 3a x2 + x3 = 0 (iii) x3 + y3 = 3ax y . (iv) x3 (v) a4 y2 (vi) x4 (vii) y3 = 3x y . = x4 + − x4 2y3 y3 + (2 x2 − 3y2 + 2 x2 (Agra 2006; Rohilkhand 07; Kumaun 08; Meerut 12B, 13) (Meerut 2001, 05; Agra 14) − 3a2) . − 2 x2 + 1 = 0. + 3y2 = 0. (Meerut 2007B) (Bundelkhand 2001; Meerut 07; Avadh 13) y2 6. Show that the curve = bx tan (x ⁄ a) has a node or a conjugate point at the origin according as a and b have like on unlike signs. 7. Prove that the curve ay2 = (x − a) 2 (x − b) has at x = a, a conjugate point if a < b, a node if a > b, and a cusp if a = b . 8. Show that the curve y3 = (x − a) 2 (2x − a) has a single cusp of the first species at the point (a, 0). 9. E xamine the curve x3 + 2 x2 + 2 xy − y2 + 5x − 2y = 0 for singular points and show that it has a cusp of the first kind at the point (− 1, − 2) . Determine the position and character of the double points on : 10. (i) y ( y − 6) = x2 (x − 2) 3 − 9. 1) 2 (ii) y ( y − (iii) x3 (iv) y2 (v) y2 − y2 = (x − − 7x2 − x (x − − x3 2) 2. (Rohilkhand 2008, 09) (Meerut 2000, 02; Gorakhpur 05; Rohilkhand 12) + 4y + 15x − 13 = 0. a) 2 = 0, (a > 0). = 0. (vi) a4 y2 = x4 (a2 − x2) . 11. (vii) y2 = x2 (9 − x2). Find the position and nature of the double points on the curve x2 y2 = (a + y) 2 (b2 − y2) if (i) b > a , (ii) b = a , (iii) b < a . SINGULAR POINTS : CURVE TRACING D-309 12. Discuss the nature of double points of the curve (x + y) 3 − √2 (x − y + 2) 2 = 0. 13. Show that the curve (xy + 1) 2 + (x − 1) 3 (x − 2) = 0 has a single cusp of the first species at the point (1, − 1). 1. 5. (ii) y = 0, y = 2 x , (iii) x = 0. (i) y = ± x , (i) A cusp at the origin. (ii) A conjugate point at the origin. (iii) A node at the origin. (iv) A node at the origin. (v) A conjugate point at the origin. (vi) Nodes at the points (0, − 1), (1, 0) and (− 1, 0) . (vii) A conjugate point at the origin. (i) (0, 3) is a conjugate point and at (2, 3) there is a single cusp of the first species. (ii) (2, 1) is a node. (iii) Node at (3, 2) . (iv) Node at (a, 0). (v) Single cusp of the first kind at (0, 0). (vi) Double cusp of the first species at (0, 0). (vii) Node at (0, 0). (i) When b > a, the point (0, − a) is a node. (ii) When b = a , the point (0, − a) is a single cusp of first kind. (iii) When b < a , the point (0, − a) is a conjugate point. There is a single cusp of the first species at (− 1, 1). 10. 11. 12. 14.16 Curve Tracing (Cartesian Equations) To find the approximate shape of a curve whose cartesian equation is given, we should adopt the following procedure : 1. Symmetry : First we should find if the curve is symmetrical about any line. In this connection the following rules are helpful : (i) If in the equation of a curve the powers of y are all even, the curve is symmetrical about the axis of x i.e., the shape of the curve above and below the axis of x is symmetrical. The obvious reason is that the equation of the curve in this case remains unchanged if we replace y by − y . Thus the parabola y2 = 4ax is symmetrical about the axis of x . (ii) If in the equation of a curve the powers of x are all even, the curve is symmetrical about the axis of y . For example, the parabola x2 = 4by is symmetrical about the axis of y. (iii) If the equation of a curve remains unchanged when x is replaced by − x and y is replaced by − y , then the curve is symmetrical in opposite quadrants. For example, the curve xy = c2 is symmetrical in opposite quadrants. (iv) If the equation of a curve remains unchanged when x and y are interchanged, the curve is symmetrical about the line y = x, (i.e., the straight line passing through the origin and making an angle 45o with the positive direction of the axis of x). For example, the curve x3 + y3 = 3axy is symmetrical about the line y = x . D-310 DIFFERENTIAL CALCULUS 2. Nature of the Origin on the Curve : We should see whether the curve passes through the origin or not. If the point (0, 0) satisfies the equation of the curve, it passes through the origin. In order to know the shape of a curve at any point, we should draw the tangent or tangents to the curve at that point. Therefore if the curve passes through the origin, we should find the equation to the tangents at origin by equating to zero the lowest degree terms in the equation of the curve. If there are two tangents at the origin, then the origin will be a double point on the curve. We should also observe the nature of the double point. 3. Points of intersection of the curve with the co-ordinate axes : We should find the points where the curve cuts the co-ordinate axes. To find the points where the curve cuts the x-axis we should put y = 0 in the equation of the curve and solve the resulting equation for x . Similarly the points of intersection with the y-axis are obtained by putting x = 0 and solving the resulting equation for y . We should also obtain the tangents to the curve at the points where it meets the co-ordinate axes. In order to find the tangent at the point (h, k) , we should shift the origin to (h, k) and then the tangent or tangents at this new origin will be obtained by equating to zero the lowest degree terms. The value of dy ⁄ dx at the point (h, k) can also be used to find the slope of the tangent at that point. 4. We should solve the equation of the curve for y or x whichever is convenient. Suppose we solve for y . Starting from x = 0 , we should see the nature of y as x increases and then tends to + ∞ . Similarly we should see the nature of y as x decreases and then tends to − ∞ . We should pay special attention to those values of x for which y = 0 or → infinity. If we solve the equation of the curve for y and the curve is sym m etrical about y-axis, then we should consider only positive values of x . The curve for negative values of x can be drawn from sym metry and there is no necessity of considering them afresh. H owever, if we solve the equation for y and there is symmetry only about x-axis, then we are to consider both positive as well as negative values of x . If the curve is symmetrical in opposite quadrants, or if there is symmetry about the x-axis, then only positive values of y need be considered. If y → infinity as x → a , then the line x = a will be an asymptote of the curve. Similarly if x → infinity as y → b , then the line y = b will be an asymptote of the curve. 5. Regions where the curve does not exist : We should find out if there is any region of the plane such that no part of the curve lies in it. Such a region is easily obtained on solving the equation for one variable in terms of the other. The curve will not exist for those values of one variable which make the other imaginary. For example, in the curve a2 y2 = x2 (x − a) (2a − x) , we find that for 0 < x < a , y2 is negative, i.e., y is imaginary. Therefore the curve does not exist in the region bounded by the lines x = 0 and x = a . For a < x < 2a , y2 is positive i.e., y is real. Therefore the curve exists in the region bounded by the lines x = a and x = 2a . Thus if y is imaginary when x lies between a and b , the curve does not exist in the region bounded by the lines x = a and x = b . 6. Asymptotes : We should find all the asymptotes of the curve. If an infinte branch of the curve has an asymptote, then ultimately it must be drawn parallel to the asymptote. The asymptotes parallel to the x-axis can be obtained by equating to zero the coefficient of the highest power of x in the equation of the curve. Similarly the asymptotes parallel to the y-axis can be obtained by equating to zero the coefficient of the highest power of y in the equation of the curve. SINGULAR POINTS : CURVE TRACING D-311 7. The sign of dy ⁄ dx : We should calculate the value of dy ⁄ dx from the equation of the curve. Then we shall find the points at which dy ⁄ dx vanishes or becomes infinite. These will give us the points where the tangent is parallel or perpendicular to the x-axis. If in any region a < x < b , dy ⁄ dx remains throughout positive, then in this region y increases continuously as x increases. If in any region a < x < b , dy ⁄ dx remains throughout negative, then in this region y decreases continuously as x increases. 8. Special Points : If necessary, we should find the co-ordinates of a few points on the curve. 9. Points of inflexion : While drawing the curve if it appears that the curve possesses some points of inflexion, then their positions can be accurately located by putting d2 y ⁄ dx2 or d2 x ⁄ dy2 equal to zero and solving the resulting equation. Ta kin g all t h e ab o ve iso la t ed fact s int o considerat ion, we can draw t he approximate shape of the curve. Example 1 (a) : Trace the curve ay2 = x3. (semi-cubical parabola). Solution : We note the following facts about this curve : (i) Since in the equation of the curve the powers of y are all even, therefore the curve is sym m etrical about the axis of x . (ii) The curve passes through the origin. (iii) E quating to zero the lowest degree terms in the equation of the curve, we get the tangents at the origin. Therefore the tangents at origin are ay2 = 0 i.e., y = 0 , y = 0 . Thus the origin is a double point and it may be a cusp since there are two coincident tangents at the origin. (iv) The curve does not intersect the coordinate axes anywhere except the origin. (v) Solving the equation of the curve for y , we get x3 y2 = ⋅ a When x = 0 , y2 = 0 . When x > 0 , y2 is positive i.e., y is real. Therefore the curve exists in the region x > 0. As x increases, y2 also increases and when x → ∞ , y2 → ∞ . When x < 0 , y2 is negative i.e., y is imaginary. Therefore the curve does not exist in the region x < 0. (vi) O bviously the curve has no asymptotes. (vii) The curve exists in the neighbourhood of origin where x > 0 . Also x-axis is a common tangent to the two branches of the curve passing through origin. H ence origin is a cusp. Taking all these facts into consideration, the shape of the curve is as shown in the adjoining figure. Example 1 (b) : Trace the curve y2 = x3. Solution : Proceed as in part (a). (Bundelkhand 2006) Example 2 : Trace the curve y2 (2a − x) = x3. (Cissoid) (Meerut 2001, 11; Agra 06; Rohilkhand 06; Bundelkhand 08; Avadh 12; Kashi 12, 14) D-312 DIFFERENTIAL CALCULUS Solution : We note the following particulars about the curve : (i) It is symmetrical about the axis of x , since the powers of y that occur are all even. (ii) The curve passes through the origin and the tangents at the origin are 2ay2 = 0 i.e., y = 0 , y = 0 are two coincident tangents at the origin. Therefore the origin may be a cusp. (iii) The curve meets the coordinate axes only at the origin. (iv) Solving the equation of the curve for y , we get y2 = x3 ⁄ (2a − x) . When x = 0 , y2 = 0 . When x → 2a , y2 → ∞ . Therefore x = 2a is an asymptote of the curve. When 0 < x < 2a , y2 is positive i.e., y is real. Therefore the curve exists in this region. When x > 2a , y2 is negative i.e., y is imaginary. Therefore the curve does not exist in the region x > 2a . When x < 0 , y2 is negative. Therefore the curve does not exist in the region x < 0 . Since the curve exists in the neighbourhood of origin where x > 0 , therefore there is a single cusp at the origin. (v) Putting y = m and x = 1 in the third degree t e r m s i n t h e e q u a t i o n o f t h e c u r ve , we g e t φ 3 (m) = m 2 + 1 . T h e r o o t s o f t h e e q u a t i o n m 2 + 1 = 0 are imaginary, therefore x = 2a is the only real asymptote of the curve. (vi) For the branch of the curve lying above the x-axis, we have y = x3 ⁄ 2 ⋅ √(2a − x) dy (3a − x) √x , = which vanishes when x = 0 , or 3a . dx (2a − x) 2 ⁄ 3 But x = 3a is outside the range of admissible values of x . Therefore dy ⁄ dx vanishes at no admissible value of x except x = 0 . W h e n 0 < x < 2a , dy ⁄ dx is p o sit ive . T h e r e fo r e in t h is r e gio n y in creases continuously as x increases. Combining all these facts, we see that the shape of the curve is as shown in the adjoining figure. ∴ Example 3 (a) : Trace the curve y2 (a + x) = x2 (a − x) . (Meerut 2000, 13; Kanpur 10; Bundelkhand 11; Avadh 13) Solution : (i) The curve is symmetrical about x-axis. (ii) The curve passes t hrough t he origin. The tangents at origin are a (y2 − x2) = 0 i.e., y = ± x . Since there are two real and distinct tangents at the origin, therefore the origin is a node on the curve. ( i i i ) T h e cu r ve i n t e r se ct s t h e x-a xis wh e r e y = 0 i.e., x2 (a − x) = 0 . T h e r e fo r e t h e cu r ve i n t e r se ct s t h e x-a xi s a t (0, 0) , (a, 0) . The curve intersects the y-axis only at origin. (iv) Tangen t at (a, 0) . Shift in g t he origin t o (a, 0) the equation of the curve becomes SINGULAR POINTS : CURVE TRACING D-313 y2 (2a + x) = (x + a) 2 {a − (x + a)} y2 (2a + x) = − x (x2 + 2ax + a2) . E quating to zero the lowest degree terms, we get x = 0 (i.e., the new y-axis) as the tangent at the new origin. Thus the tangent at (a, 0) is perpendicular to x-axis. (v) Solving the equation of the curve for y , we get or y2 = x2 (a − x) ⁄ (x + a) . When x = 0 , y2 = 0 and when x = a , y2 = 0 . When 0 < x < a , y2 is positive. Therefore the curve exists in this region. When x > a , y2 is negative. Therefore the curve does not exist in the region x> a. When x → − a , y2 → ∞ . Therefore x = − a is an asymptote of the curve. When − a < x < 0 , y2 is positive. Therefore the curve exists in this region. When x < − a , y2 is negative. Therefore the curve does not exist in the region x< − a. (vi) Putting y = m and x = 1 in the highest i.e., third degree terms in the equation o f t he cu rve, we get φ 3 (m) = m 2 + 1 . The root s of the equat ion φ 3 (m) = 0 are imaginary. Therefore x = − a is the only real asymptote of the curve. (vii) For the portion of the curve lying in the first quadrant, we have ⎧ (a − x) ⎫ (1 − x ⁄ a) 1 ⁄ 2 ⎨ ⎬= x ⋅ y= x (1 + x ⁄ a) 1 ⁄ 2 ⎩ (a + x) ⎭ When 0 < x < a , y is less than x . Therefore the curve lies below the line y = x which is tangent at the origin. For the portion of the curve lying in the second quadrant, we have √ (1 − x ⁄ a) 1 ⁄ 2 , x < 0. (1 + x ⁄ a) 1 ⁄ 2 When − a < x < 0 , y is greater than the numerical value of x . Therefore the curve lies above the tangent y = − x . H ence the shape of the curve is as shown in the figure. y= − x Example 3 : (b) Trace the curve y2 (a + x) = x2 (3a − x) . (Purvanchal 2011) Solution : Proceed as in part (a). Example 4 : Trace the curve y2 (x2 + y2) + a2 (x2 − y2) = 0 . Solution : (i) The curve is symmetrical about both the axes. (ii) It passes through the origin and a2 (x2 − y2) = 0 i.e., y = ± x are the two tangents at the origin. Therefore the origin is a node. (iii) The curve intersects the x-axis only at origin. It intersects the y-axis at (0, 0) , (0, a) and (0, − a) . (iv) Shifting the origin to (0, a) , the equation of the curve becomes ( y + a) 2 {x2 + ( y + a) 2} + a2 {x2 − ( y + a) 2} = 0 or ( y2 + 2ay + a2) {x2 + y2 + 2ay + a2} + a2 (x2 − y2 − 2ay − a2) = 0 . E quating to zero the lowest degree terms, we get 2a3 y + 2a3 y − 2a3y = 0 i.e., y = 0 as the tangent at the new origin. Thus the new x-axis is tangent at the new origin. We need not find the tangent at (0, − a) as the curve is symmetrical about x-axis. D-314 (v) DIFFERENTIAL CALCULUS Solving the equation of the curve for x , we get x2 = y2 (a2 − y2) ⁄ (a2 + y2) . When y = 0 , x2 = 0 and when y = a , x2 = 0 . W h e n 0 < y < a , x2 is p o sit ive. Th er efor e t he cu rve e xist s in t he re gion 0 < y< a. When y > a , x2 is negative. Therefore the curve does not exist in the region y> a. We need not consider the negative values of y as the curve is symmetrical about x-axis. (vi) The asymptotes parallel to x-axis are given by a2 + y2 = 0 i.e., y = ± ia . Also φ 4 (m) = m 2 (1 + m 2) . Its roots are m = 0 , 0 , i , − i . The asymptotes corresponding to m = 0 are imaginary. H ence all the four asymptotes are imaginary. (vii) In the positive quadrant, we have x = y (a2 − y2) 1 ⁄ 2 ⁄ (a2 + y2) 1 ⁄ 2, y > 0 or x= y 1⁄2 ⎛ y2 ⎞⎟ ⎜ ⎜1 − ⎟ ⎜ a2 ⎟⎠ ⎝ ⎛ y2 ⎞⎟ 1 ⁄ 2 ⎜ ⎜1 + ⎟ ⎜ a2 ⎟⎠ ⎝ ⋅ W h en 0 < y < a , we se e t h a t x is le ss t h an y . Therefore the curve lies above the line y = x which is tangent at the origin. Combining all these facts, we see that the shape of the curve is as shown in the adjoining figure. Example 5 : Trace the curve x2 (x2 − 4a2) = y2 (x2 − a2) . Solution : (i) Symmetry about both the axes. (ii) The curve passes through the origin and a2 y2 − 4a2 x2 = 0 i.e., y = ± 2x are the tangents at the origin. Therefore origin is a node on the curve. (iii) The curve cuts the x-axis at (0, 0) , (2a, 0) , (− 2a, 0) . It cuts the y-axis only at the origin. (iv) Shifting the origin to (2a, 0) , the equation of the curve becomes (x + 2a) 2 (x2 + 4ax) = y2 (x2 + 4ax + 3a2) . The equation to the tangent at the new origin is 16a3 x = 0 i.e., x = 0 . Thus the new y-axis is tangent at the new origin. x2 (x2 − 4a2) ⋅ (v) Solving the equation of the curve for y , we get y2 = (x2 − a2) When x = 0 , y2 = 0 . When x → a , y2 → ∞ i.e., x = a is an asymptote of the curve. When 0 < x < a , y2 is positive i.e., the curve exists in this region. When x = 2a , y2 = 0 . When a < x < 2a , y2 is negative i.e., the curve does not exist in this region. When x > 2a, y2 is positive i.e., the curve exists in this region. When x → ∞ , y2 → ∞ . We need not consider the negative values of x as the curve is symmetrical about the y-axis. (vi) The asymptotes of the curve parallel to y-axis are given by x2 − a2 = 0 . Thus x = ± a are two asymptotes of the curve. D-315 SINGULAR POINTS : CURVE TRACING Also the equation of the curve can be written as x2 ( y2 − x2) − a2 y2 + 4a2 x2 = 0 . ∴ φ 4 (m) ≡ m 2 − 1 = 0 gives m = ± 1 . Also φ 3 (m) = 0 . For m = ± 1 , c is given by c φ ′ 4 (m) + φ 3 (m) = 0. When m = 1 , c = 0 . Also when m = − 1 , c = 0 . Therefore y = ± x are two oblique asymptotes of the curve. (vii) In the positive quadrant, we have x2 (4a2 − x2) , y2 = 0< x< a (a2 − x2) 1⁄2 / 1⁄2 2⎞ ⎛ x2 ⎞⎟ ⎜1 − x ⎟ . ⎟ ⎜ ⎟ 2 ⎜ ⎟ ⎜ 4a ⎠ a2 ⎟⎠ ⎝ ⎝ When 0 < x < a , y is greater than 2x . Therefore the curve lies above the line y = 2x which is tangent at the origin. or ⎛ y = 2x ⎜⎜ 1 − Combining all these facts we see that the shape of the curve is as shown in the above figure. Example 6 : Trace the curve x3 + y3 = 3axy . (Folium of Descartes) (Meerut 2007, 08, 10B, 13B; Rohilkhand 08; Purvanchal 07) Solution : (i) The curve is symmetrical about the line y = x , since its equation remains unchanged on interchanging x and y . (ii) Th e cu rve p asses th ro ugh t he o rigin an d t he t angen ts at origin are 3axy = 0 i.e., x = 0 , y = 0 . Since there are two real and distinct tangents at the origin, therefore the origin is a node on the curve. (iii) The curve intersects the coordinate axes only at (0, 0) . (iv) From the equation of the curve we see that x and y cannot be both negative because then the left hand side of the equation of the curve becomes negative while the right hand side becomes positive. Therefore the curve does not exist in the third quadrant. (v) The curve meets the line y = x at the point (3a ⁄ 2, 3a ⁄ 2) . From the equation of the curve, we have 3x2 − 3ay dy = − 2 ⋅ dx 3y − 3ax D-316 DIFFERENTIAL CALCULUS ⎛ 3a , 3a ⎞ , dy = − 1 . Therefore the tangent at ⎟ ⎝ 2 2 ⎠ dx At ⎜ this point makes an angle of 135o with the positive direction of x-axis. (vi) Asymptotes : φ 3 (m) = m 3 + 1 . The only real root of the equation φ 3 (m) = 0 i.e., Also m 3 + 1 = 0 , is m = − 1 . φ 2 (m) = − 3am . For m = − 1 , c is given by c (3m 2) − 3am = 0 . ∴ when m = − 1 , c = − a . H ence y = − x − a is the only real asymptote of the curve. Combining all these facts we see that the shape of the curve is as shown in the figure. Example 7 : Trace the curve y3 + x3 = a2 x . (Meerut 2006, 10; Kanpur 08; Kashi 13) Solution : (i) If we change the signs of x and y both, the equation of the curve does not change. Therefore the curve is symmetrical in opposite quadrants. (ii) The curve passes through the origin and the tangent at origin is x = 0 i.e., y-axis. (iii) The curve cuts the x-axis where y = 0 i.e., x (x2 − a2) = 0 . Thus the curve cuts the x-axis at (0, 0) , (a, 0) , (− a, 0) . The curve intersects the y-axis only at the origin. dy a2 − 3x2 = ⋅ (iv) From the equation of the curve, we have dx 3y2 dy = ∞ i.e., the tangent is perpendicular to x-axis. At (a, 0) , dx dy Also at (− a, 0) , = − ∞ i.e., the tangent is perpendicular to x-axis. dx dy a (v) = 0 at x = ± ⋅ Therefore the tangents at these points are parallel to dx √3 the x-axis. (vi) Solving the equation of the curve for y , we get y3 = x (a2 − x2) . When x = 0 , y3 = 0 and when x = a , y3 = 0 . When 0 < x < a , y3 is positive i.e., y is positive in this region. When x > a , y3 is negative i.e., y is negative in this region. When x → ∞ , y3 → − ∞ i.e., y → − ∞ . We need not consider the negative values of x as there is symmetry in opposite quadrants. Asymptotes : φ 3 (m) = m 3 + 1 , φ 2 (m) = 0 . The only real root of m 3 + 1 = 0 is m = − 1 . Also c is given by c (3m 2) + 0 = 0 . When m = − 1 , c = 0 . H ence y = − x is the only real asymptote of the curve. Combining all these facts the shape of the curve is as shown in the figure. SINGULAR POINTS : CURVE TRACING D-317 Trace the following curves : 1. 2. x3 y = x + 1 . x = ( y − 1) ( y − 2) ( y − 3) . x (x2 3. y= 4. y2 5. xy 2 = 4a2 6. x2 y2 a2 − 1) . 7. y (x2 − 1) = 8. y (x2 4a2) 9. y2 = 4ax . = + (1 − 10. a2 y2 11. a2 y2 = (Kanpur 2009; Purvanchal 06) (parabola) (2a − x) . (Witch of Agnesi) (x2 + y2) . x2) x2 (x2 = = (a2 (Gorakhpur 2006) + 1). (Bundelkhand 2001; Kashi 11) 8a3. x2 (1 + − x2). (Agra 2008) x2) . (Meerut 2007B; Bundelkhand 07, 10) x3 12. = (2a − x) . y2 (a2 + x2) = x2 (a2 − x2) . 13. y2 x = a2 (x − a) . 14. 9ay2 = x (x − 3a) 2. 15. y2 (x + a) = (x − a) 3. 16. x2 y2 17. y2 18. a3 y2 = (x − − 1) = x . = (1 + y) 2 (4 − (x2 (Meerut 2004, 06B) y2) . (x + 3a) = x (x − a) (x − 2a) . a) 4 (Meerut 2001, 03, 12; Kumaun 08) (x − b), a > b. 19. y2 20. x (x − 2a) y2 = a2 (x − a) (x − 3a) . 21. y2 = (x − a) (x − b) (x − c), a > b > c. (Meerut 2005B) D-318 DIFFERENTIAL CALCULUS SINGULAR POINTS : CURVE TRACING D-319 D-320 DIFFERENTIAL CALCULUS D-321 SINGULAR POINTS : CURVE TRACING 14.17 Polar Equations : Procedure for Tracing 1. Symmetry : (i) If the equation of the curve does not change by changing the sign of θ , then the curve is symmetrical about the initial line. (ii) If the equation of the curve remains unchanged by changing r into − r , then the curve is symmetrical about the pole and the pole is the centre of the curve. 2. Some Special points on the Curve : The curve will pass through the pole if for some value of θ the value of r comes out to be zero. Also if r = 0 when θ = α , then usually the line θ = α will be a tangent to the curve at the pole. We should find the values of θ for which r = 0 , or r is maximum, or r is minimum, or r → ∞ . 3. Solve the equation of the curve for r and consider how r varies as θ increases from 0 to + ∞ , and also as θ decreases from 0 to − ∞ . We should pay special attention to the values of θ found in the paragraph 2 . We should form a table of corresponding values of r and θ which would give us a number of points on the curve. Plotting these points we shall find the shape of the curve. In the polar equations in which only periodic functions (sin θ , cos θ , tan θ etc.) occur, the values of θ from 0 to 2π (or sometimes some multiple or sub-multiple of 2π) need be considered, as the remaining values of θ do not give any new branch of the curve. 4. Regions where the curve does not exis t : I f r is im a gin a r y wh e n α < θ < β , then the curve does not exist in the region bounded by the lines θ = α and θ = β. 5. Asymptotes : Find the asymptotes if the curve possesses an infinite branch. If r → ∞ as θ → α , we should not assume that θ = α is an asymptote. The asymptote might be parallel to the line θ = α or even might not exist at all. The asymptotes should be found by the method given in the chapter on Asymptotes. 6. Find tan φ i.e., r dθ ⁄ dr which will indicate the direction of the tangent at any point. If for θ = α , φ comes out to be zero, then the line θ = α will be a tangent to the curve at the point θ = α . If for θ = α , φ comes out to be π ⁄ 2 , then at the point θ = α , the tangent will be perpendicular to the radius vector θ = α . 7. Important : It is sometimes convenient to change the equation from the polar form to the cartesian form. R emember that the relations between the cartesian and polar coordinates are x = r cos θ , y = r sin θ . Example 1 : Trace the curve r = a (1 + cos θ) . (Cardioid) (Meerut 2009B; Bundelkhand 05; Rohilkhand 07) Solution : (i) The curve is symmetrical about the initial line since its equation remains unchanged by writing − θ in place of θ . (ii) r = 0 , when cos θ = − 1 i.e., θ = π , r is maximum when cos θ = 1 , i.e., θ = 0 . Then r = 2a . Also r is minimum when cos θ = − 1 i.e., θ = π . Then r = 0 . dr (iii) = − a sin θ . dθ When 0 < θ < π , (dr ⁄ dθ) is throughout negative. D-322 DIFFERENTIAL CALCULUS Therefore r decreases continuously as θ increases from 0 to π . a (1 + cos θ) θ dθ = − = − cot ⋅ (iv) Also tan φ = r a sin θ 2 dr φ = 0 when θ = π . Then r = 0 . Therefore the line θ = π is tangent to the curve at the pole. φ = 90o when θ = 0 . Then r = 2a . Therefore the tangent at θ = 0 is perpendicular to the radius vector θ = 0 . (v) Since r is never greater than 2a , therefore the curve will have no asymptotes. (vi) The following table gives the corresponding values of θ and r . θ 0 π 3 π 2 2π 3 π r 2a 3 a 2 a a 2 0 The portion of the curve lying in the region π < θ < 2π can be drawn by symmetry. Hence the shape of the curve is as shown in the figure. Example 2 : Trace the curve r = a cos 2θ . Solution : (i) The curve is symmetrical about the initial line. (ii) r = 0 , when cos 2θ = 0 , i.e., 2θ = ± π ⁄ 2 i.e., θ = ± π ⁄ 4 . Therefore the lines θ = ± π ⁄ 4 are tangents to the curve at the pole. r is maximum when cos 2θ = 1 . Then θ = 0 and r = a . dθ 1 (iii) tan φ = r = a cos 2θ ⋅ dr − 2a sin 2θ 1 = − cot 2θ . 2 φ = 90o when 2θ = 0 i.e., θ = 0 . Therefore at the point θ = 0 , the tangent is perpendicular to the radius vector θ = 0 . (iv) The following table gives the corresponding values of θ and r : θ 0 π 6 π 4 r a 1 a 2 0 π 3 − 1 a 2 π 2 − a The variation of θ from π to 2π need not be considered because of symmetry about the initial line. H ence the curve is as shown in the figure. The curve consists of four similar loops, all lying within a circle of radius a and centre at the pole. Important : T h e a b o ve c u r ve i s a p a r t i cu l a r ca se o f t h e cu r ve s o f t h e t yp e r = a cos nθ which have n loops when n is odd and 2n loops when n is even. 2π 3 − 1 a 2 3π 4 0 5 π 6 π 1 2 a a D-323 SINGULAR POINTS : CURVE TRACING Example 3 : Trace the curve r = a sin 3θ . (Meerut 2003; Rohilkhand 12) Solution : (i) The curve is not symmetrical about the initial line. (ii) r = 0 when sin 3θ = 0 i.e., 3θ = 0 , π , i.e., θ = 0 , π ⁄ 3 . Therefore the lines θ = 0 and θ = π ⁄ 3 are tangents to the curve at the pole. Also r is maximum when sin 3θ = 1 i.e., 3θ = π ⁄ 2 i.e., θ = π ⁄ 6 . The maximum value of r is a . dθ 1 (iii) tan φ = r = tan 3θ . dr 3 φ = 90o when 3θ = π ⁄ 2 i.e., θ = π ⁄ 6 . Therefore at the point θ = π ⁄ 6 , tangent is perpendicular to the radius vector θ = π ⁄ 6. (iv) The following table gives the corresponding values of θ and r : 3θ 0 π 2 π θ 0 π 6 π 3 3π 2 π 2 2π 5π 2 3π 7π 2 4π 9π 2 5π 11π 2 6π 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π − a 0 0 a 0 − a 0 π, H ere one loop of the curve lies in the region 0 < θ < one loop lies in the region 3 2π π < θ< a n d o n e l o o p l i e s i n t h e r e gi o n 3 3 2π < θ < π . If θ increases beyond π to 2π , the same 3 branches of the curve are repeated and we do not get any new branch. H ence the complete curve is as shown in the adjoining figure. Important Note : T h e a b o ve cu r ve i s a particular case of the curves of the type r = a sin nθ which have n loops when n is odd and 2n loops when n is even. Example 4 : Trace the curve r = a + b cos θ , when a < b . (Limacon) Solution : (i) The curve is symmetrical about the initial line. ⎛ a⎞ (ii) r = 0 when a + b cos θ = 0 i.e., θ = cos− 1 ⎜ − ⎟ ⋅ ⎝ b⎠ a a ⎛ ⎞ Since < 1 , therefore cos− 1 ⎜ − ⎟ is real. b ⎝ b⎠ ⎛ a⎞ Therefore the radius vector θ = cos− 1 ⎜ − ⎟ is tangent to the curve at the pole. ⎝ b⎠ r is maximum when cos θ = 1 , i.e., θ = 0 . Then r = a + b . Also r is minimum when cos θ = − 1 , i.e., θ = π . Then r = a − b , which is negative, ( ... a < b) . r 0 a 0 a dr = − b sin θ . dθ dθ (a + b cos θ) ∴ tan φ = r = − ⋅ dr b sin θ (iii) 0 − a D-324 DIFFERENTIAL CALCULUS φ = 90o when θ = 0 and π . Therefore at the points θ = 0 and θ = π , the tangent is perpendicular to the radius vector. (iv) The following table gives the corresponding values of r and θ . θ 0 π⁄2 ⎛ a⎞ cos− 1 ⎜ − ⎟ ⎝ b⎠ ⎛ a⎞ cos− 1 ⎜ − ⎟ < θ < π ⎝ b⎠ π r a+ b a 0 r is negative a− b The variation of θ from π to 2π need not be considered because of the symmetry about the initial line. Hence the curve is as shown in the adjoining figure. Example 5 : Trace the curve r = ae mθ. (Equiangular Spiral) Solution : (i) The curve is not symmetrical about the initial line. (ii) As θ → ∞ , r → ∞ and as θ → − ∞ , r → 0 . Also r is always positive. When θ = 0, r= a. dr = ame mθ. (iii) dθ dr When − ∞ < θ< ∞ , i s t h r o u gh o u t dθ positive. Therefore r increases continuously as θ increases from − ∞ to ∞ . dθ a e mθ 1 (iv) tan φ = r = = ⋅ dr am e mθ m ⎛ 1⎞ φ = tan− 1 ⎜ ⎟ = constant . ⎝ m⎠ Thus in this curve the angle between the radius vector and the tangent always remains constant. H ence the shape of the curve is as shown in the adjoining diagram. ∴ 1. 2. Trace the following curves : r = 2a cos θ . (Circle) r = a (1 − cos θ) . (Cardioid) (Meerut 2001) 3. r = a + b cos θ, when a > b . (Limacon) (Meerut 2000) 4. r2 5. r2 = a2 cos 2θ . = a2 sin 2θ . (Lemniscate) (Lemniscate of Bernouli) (Meerut 2002, 08; Agra 07) D-325 SINGULAR POINTS : CURVE TRACING 6. r = a sin 2θ . (Four leaved rose) (Three leaved rose) 7. r = a cos 3θ . 8. 2a ⁄ r = 1 + cos θ . (Parabola) 9. r= 10. 1 2 (Bundelkhand 2009) (Avadh 2010; Karshi 13) + cos 2θ . (Meerut 2004B) (i) r = a (sec θ + cos θ) . [Hint. Changing to cartesian form, the equation becomes y2 (x − a) = x2 (2a − x)]. (ii) r cos θ = 2a sin2 θ . (Cissoid. For figure, see Example 2 after article 14.16) D-326 DIFFERENTIAL CALCULUS 14.18 Parametric Equations If the equation to a curve is given in a parametric form, x = f (t) , y = φ (t) , then in some cases the curve can be easily traced by eliminating the parameter. But if it is not convenient to eliminate t , a series of values are given to t and the corresponding values of x, y and (dy ⁄ dx) are found. Then we plot the different points and observe the slopes of the tangents at these points given by the values of (dy ⁄ dx) . Exa mple 1 : T ra c e t h e c u rv e x = a (t + sin t) , y = a (1 − cos t) , when (Cycloid) − π ≤ t≤ π. dy dx = a (1 + cos t) and = a sin t . Solution : H ere dt dt dy dy ⁄ dt a sin t t Therefore = = = tan ⋅ dx dx ⁄ dt a (1 + cos t) 2 (i) y = 0 , when cos t = 1 i.e., t = 0 . When t = 0 , x = 0 , (dy ⁄ dx) = tan 0 = 0 . Therefore the curve passes through the origin and the axis of x is tangent at the origin. (ii) y is maximum when cos t = − 1 , i.e., t = π and − π . When t = π , x = aπ , y = 2a and (dy ⁄ dx) = ∞ . D-327 SINGULAR POINTS : CURVE TRACING Therefore at the point t = π , whose cartesian coordinates are (aπ, 2a) , the t a n g e n t i s p e r p e n d i c u l a r t o t h e x- a xi s . W h e n t = − π , x = − aπ , y = 2a , (dy ⁄ dx) = − ∞ . (iii) In this curve y cannot be negative. Therefore the curve lies entirely above the axis of x . Also no portion of the curve lies in the region y > 2a . (iv) Corresponding values of x, y and (dy ⁄ dx) for different values of t are given in the following table : 0 1 π 2 π − a ( π + 1) 0 a ( π + 1) aπ 2a a 0 a 2a − ∞ − 1 0 1 ∞ t − π x − aπ y dy ⁄ dx 1 π 2 − 1 2 1 2 If we put − t in place of t in the equation of t h e c u r ve , we g e t x = − a (t + sin t) , a n d y = a (1 − cos t) . Thus for every value of y , there are two equal and opposite values of x . Therefore the curve is symmetrical about the y-axis. H ence the shape of the curve is as shown in the diagram. Example 2 : Trace the curve x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3. (Astroid) Solution : T h e p ar amet r ic equ at io n s o f t h e x = a cos3 t , y = a sin3 t . dy dy ⁄ dt 3a sin2 t cos t = = − = − tan t . dx dx ⁄ dt 3a cos2 t sin t Also the equation of the curve can be written as We have ⎛ x2 ⎞ 1 ⁄ 3 ⎜ ⎟ + ⎜ 2⎟ ⎜a ⎟ ⎝ ⎠ ⎛ y2 ⎞ 1 ⁄ 3 ⎜ ⎟ = 1. ⎜ 2⎟ ⎜a ⎟ ⎝ ⎠ We observe the following facts about the curve. (i) The curve is symmetrical about both the axes. It is also symmetrical about the line y = x . (ii) The curve does not pass through the origin. (iii) The curve cuts the x-axis, where y = 0 i.e., ⎛ x2 ⎞ 1 ⁄ 3 ⎜ ⎟ = 1 ⎜ 2⎟ ⎜a ⎟ ⎝ ⎠ i.e., x2 = 1 a2 i.e., x= ± a. Thus the curve cuts the x-axis at (a, 0) and (− a, 0) . Similarly the curve crosses the y-axis at (0, a) and (0, − a) . (iv) At the point (a, 0) , we have x = a . Therefore cos3 t = 1 and thus t = 0 . dy = 0. When t = 0 , dx (Rohilkhand 2013B) c u r ve are D-328 DIFFERENTIAL CALCULUS H ence at the point (a, 0) , the x-axis is tangent to the curve. Again at the point (0, a) , we have y = a . π Therefore sin3 t = 1 and thus t = ⋅ 2 π , dy When t = = − ∞. 2 dx H ence at the point (0, a) , the y-axis is tangent to the curve. ( v) T h e va l u e s o f sin t a n d cos t c a n n o t numerically exceed 1. Therefore in this curve the values of x and y cannot numerically exceed a . Therefore the entire curve lies in the region bounded by the lines x = a , x = − a , y = a and y = − a . H ence the shape of the curve is as shown in the diagram. 1. 2. 3. Trace the following curves : x = a (t + sin t ) , y = a (1 + cos t ) , − π ≤ t ≤ π . (Cycloid) x = a (t − sin t ) , y = a (1 − cos t ) . x = a cos t + 1 2 a log tan2 (t ⁄ 2) , y = a sin t . (Tractrix) (Meerut 2005) D-329 SINGULAR POINTS : CURVE TRACING Fill in the Blanks: 1. Fill in the blanks “……”, so that the following statem ents are com plete and correct. At the point of inflexion d2y 2. d3y = 0 and …… . (Agra 2007) dx 2 dx 3 If the two branches through a double point on a curve are real and have different tangents there, the double point is called a …… . (Kumaun 2008) 3. The double point on the curve x 3 + y 3 = 3axy is …… . 4. 5. The curve y 2 (1 − x 2) = x 2 (1 + x 2) is symmetrical about …… . If the equation of the curve r = f (θ) does not change by changing the sign of θ, then the curve is symmetrical about the …… . 6. The curve x 3 + y 3 = 3axy is symmetrical about …… . (Bundelkhand 2006) Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 7. The tangents at origin to the curve x 3 + y 3 = 3axy are (a) x = 0, y = 0 (b) x = 0, y = 1 (c) x = 1, y = 0 (d) x = 1, y = 1 8. The curve y = x 3 is symmetrical about the (a) x-axis (b) (c) both the axes (d) y-axis opposite quadrants (Bundelkhand 2008) 10. The number of loops in the curve r = a cos 2 θ is (a) 1 (b) 2 (c) 3 (d) 4 The curve r = a sin 3θ is symmetrical about the (a) initial line (b) pole π (c) the line θ = (d) there is no symmetry 2 11. At the point of inflexion of the curve x = f (y), 12. (a) 1 (c) − 1 E quation of Lemniscate is (a) r = a cos θ (c) r 2 = a 2 cos 2θ 9. d3y = 0 and is not equal to dx 2 dx 3 (b) 0 (d) 2 (Bundelkhand 2007) (b) (d) d2y r = a sin θ none (Rohilkhand 2008) True or False: 13. Write ‘T’ for true and ‘F’ for false statement. If the two branches through a double point on a curve are real and have coincident tangents there, then the double point is called a node. D-330 14. 15. 16. 1. 6. 11. 16. DIFFERENTIAL CALCULUS If the equation of the curve r = f (θ) remains unchanged on changing the signs of π (Meerut 2001, 09) r and θ both, the curve is symmetrical about the line θ = ⋅ 2 If the equation of a curve remains unchanged even when x and y are interchanged, the curve is symmetrical about the line y = x. A point of inflexion is a point at which a curve is changing concave upward to concave downward, or vice-versa. ≠ 0. y = x. (b). T. 2. node. 7. (a). 12. (c). 3. (0, 0). 8. (d). 13. F. 4. both the axes. 9. (d). 14. T. 5. initial line. 10. (c). 15. T. SECTION B INTEGRAL CALCULUS C hapters 1. Reduction Formulae (For Trigonometric Functions) 2. Reduction Formulae Continued (For Irrational Algebraic and Transcendental Functions) 1. Beta and Gamma Functions 3. 4. Multiple Integrals (Double and Triple Integrals, Change of Order of Integration) 1. Dirichlet’ ' s and Liouville's Integrals 5. 1. Areas of Curves 6. 7. Rectification (Lengths of Arcs and Intrinsic Equations of Plane Curves) 8. Volumes and Surfaces of Solids of Revolution 1.1 Reduction Formulae A reduction formula is a formula which connects an integral, which cannot otherwise be evaluated, with another integral of the same type but of lower degree. It is generally obtained by applying the rule of integration by parts. 1.2 Reduction Formulae for n being a + ive Integer (a) Let In = ∫ sin n x dx or ∫ sin n x dx and In = ∫ ∫ cos n x dx , sin n − 1 x sin x dx . (Note) Integrating by parts regarding sin x as the 2nd function, we have In = sin n − 1 x .(− cos x) − ∫ (n − 1) sin n − 2 x .cos x .(− cos x) dx = − sin n − 1 x .cos x + (n − 1) ∫ sin n − 2 x .cos2 x dx = − sin n − 1 x .cos x + (n − 1) ∫ sin n − 2 x .(1 − sin2 x) dx (Note) = − sin n − 1 x .cos x + (n − 1) ∫ sin n − 2 x dx − (n − 1) ∫ sin n x dx = − sin n − 1 x .cos x + (n − 1) ∫ sin n − 2 x dx − (n − 1) In . I-4 INTEGRAL CALCULUS Transposing the last term to the left, we have In (1 + n − 1) = − sin n − 1 x .cos x + (n − 1) In − 2 , [... In − 2 = ∫ sin n − 2 x dx] nIn = − sin n − 1 x cos x + (n − 1) In − 2 sin n − 1 x cos x n − 1 In = − + In − 2 . n n or or n− 1 ⌠ 1 ⌠ ⎮ sin n − 2 x dx . ⎮ sin n x dx = − sin n − 1 x . cos x + ⌡ n n ⌡ ∴ (b) In = Let ∫ cos n x dx In = or ∫ (Bundelkhand 2008; Agra 14) cos n − 1 x .cos x dx . Integrating by parts regarding cos x as the 2nd function, we have In = cos n − 1 x . sin x − ∫ (n − 1) cos n − 2 x . (sin x) .sin x dx = cos n − 1 x . sin x + (n − 1) ∫ cos n − 2 x . sin2 x dx = cos n − 1 x . sin x + (n − 1) ∫ cos n − 2 x (1 − cos2 x) dx = cos n − 1 x . sin x + (n − 1) ∫ cos n − 2 x dx − (n − 1) ∫ cos n x dx = cos n − 1 x sin x + (n − 1) In − 2 − (n − 1) In . Transposing the last term to the left, we have In (1 + n − 1) = cos n − 1 x . sin x + (n − 1) In − 2 n In = cos n − 1 x . sin x + (n − 1) In − 2 . or ∴ 1.3 ⌠ cos n − 1 x sin x n − 1 ⌠ ⎮ cos n x dx = ⎮ cos n − 2 x dx . + ⌡ n ⌡ n Walli’s Formula ∫ 0π ⁄ 2 To evaluate sin n x dx and ∫ π ⁄2 cos n x dx . 0 Proceeding as in the previous article, we have sin n − 1 x cos x n − 1 ⌠ ⌠ ⎮ sin n − 2 x dx . ⎮ sin n x dx = − + ⌡ n ⌡ n ∴ π ⁄2 ⌠ ⎮ ⌡0 π ⁄2 ⎡ sin n − 1 x cos x⎤⎥ π ⁄ 2 n− 1⌠ + ⎮ sin n − 2 x dx ⎥⎥ n ⌡ n ⎦0 sin n x dx = − ⎢⎢⎢ ⎣ = 0+ 0 π ⁄2 n− 1⌠ ⎮ n ⌡0 sin n − 2 x dx . Putting (n − 2) in place of n in (1), we have π ⁄2 ⌠ ⎮ ⌡0 π ⁄2 n− 3⌠ ⎮ sin n − 2 x dx = n − 2 ⌡0 Substituting this value in (1), we have sin n − 4 x dx . ...(1) REDUCTION FORMULAE (For Trigonometric Functions) π ⁄2 ⌠ ⎮ ⌡0 sin n x dx = = π ⁄2 n− 1 n− 3 ⌠ ⋅ ⋅ ⎮ n n − 2 ⌡0 I-5 sin n − 4 x dx π ⁄2 n− 1 n− 3 n− 5 ⌠ ⋅ ⋅ ⋅ ⎮ n − 2 n − 4 ⌡0 n sin n − 6 x dx . ...(2) Now two cases arise viz, n is even or odd. Case I : When n is odd : In this case by the repeated application of the reduction formula (1), the last integral of (2) is π ⁄2 ⌠ ⎮ ⌡0 sin x dx = ⎡⎢ − cos x⎤⎥ ⎣ ⎦ π ⁄2 0 = 1. H ence when n is odd, from (2), we have π ⁄2 ⌠ ⎮ ⌡0 sin n x dx = = π ⁄2 n− 1 n− 3 n− 5 2⌠ ⋅ ⋅ ...... ⎮ n n− 2 n − 4 3 ⌡0 sin x dx n− 1 n− 3 n− 5 2 ⋅ ⋅ ⋅ ⋅⋅⋅ ⋅ ⋅ 1 n n− 2 n− 4 3 (n − 1) (n − 3) ...... 4 .2 ⋅ 1. n (n − 2) … 3 .1 Case II : When n is even : In this case the last integral of (2) is = π ⁄2 ⌠ ⎮ ⌡0 π ⁄2 ⌠ sin0 x dx = ⎮ ⌡0 π π ⁄2 dx = ⎡⎢ x ⎤⎥ = ⋅ ⎣ ⎦0 2 H ence when n is even, from (2), we have π ⁄2 ⌠ ⎮ ⌡0 sin n x dx = π ⁄2 ⌠ If we evaluate ⎮ ⌡0 ∴ π ⁄2 ⌠ ⎮ ⌡0 π ⁄2 n− 1 n− 3 n− 5…3 1⌠ ⋅ ⋅ ⋅ ⎮ n n− 2 n− 4 4 2 ⌡0 = n− 1 n− 3 n− 5…3 1 π ⋅ ⋅ ⋅ ⋅ n n− 2 n− 4 4 2 2 = (n − 1) (n − 3) … 3 . 1 π ⋅ ⋅ 2 n (n − 2) … 4 .2 sin0 x dx cos n x dx , we get the same results. π ⁄2 ⌠ sin n x dx = ⎮ ⌡0 cos n x dx . (Note) Note : Walli’s formula is applicable only when the limits are from 0 to Example 1 : Establish a reduction formula for ∫ sin n (2x) dx . Solution : Let In = ∫ sin n (2x) dx or In = ∫ sin n − 1 (2x) sin (2x) dx . Integrating by parts regarding sin 2x as the 2nd function, we have 1 2 π. I-6 INTEGRAL CALCULUS 1 In = sin n − 1 (2x) [− cos 2x] 2 ∫ − 1 {(n − 1) sin n − 2 2x . cos 2x .2}. (− cos 2x) dx 2x . cos 2x + (n − 1) ∫ 2 sin n − 1 = − 1 2 1 2 = − 1 2 sin n − 1 2x . cos 2x + (n − 1) ∫ sin n − 2 2x dx − (n − 1) ∫ sin n 2x dx = − 1 2 sin n − 1 2x . cos 2x + (n − 1) In − 2 − (n − 1) In . = − sin n − 2 2x . cos2 2x dx sin n − 1 2x . cos 2x + (n − 1) ∫ sin n − 2 2x . (1 − sin2 2x) dx Transposing the last term to the left, we have 1 2 n In = − In = − or sin n − 1 2x . cos 2x + (n − 1) In − 2 sin n − 1 2x .cos 2x n − 1 + In − 2 , is the reduction formula. 2n n π ⁄2 (2m) ! π ⋅ ⋅ {2m . m !}2 2 Solution : H ere 2m is even. H ence from article 1.3 (Case II), we get ⌠ *Example 2 : Prove that ⎮ ⌡0 π ⁄2 ⌠ ⎮ ⌡0 = sin2m x dx = sin2m x dx = (2m − 1) (2m − 3) … 3 .1 π ⋅ (2m) (2m − 2) … 4 . 2 2 (Walli’s formula) 2m (2m − 1) (2m − 2) … 3 . 2 . 1 π ⋅ ⋅ 2 {2m (2m − 2) .... 4 . 2}2 [Multiplying Nr. & Dr. by 2m (2m − 2) (2m − 4) … 4 . 2] (2m) ! π (2m) ! π = ⋅ = ⋅ ⋅ {2m . m (m − 1) (m − 2) ...... 2 . 1}2 2 {2m . m ! }2 2 2a x9 ⁄ 2 dx ⌠ ⋅ Example 3 : Evaluate ⎮ ⌡0 √(2a − x) Solution : Put x = 2a sin2 θ , so that dx = 2a .2 sin θ cos θ dθ . Also when x = 0 , sin2 θ = 0 i.e., θ = 0 and when x = 2a , sin2 θ = 1 i.e., θ = π ⁄ 2 . π ⁄2 2a Then x9 ⁄ 2 dx ⌠ ⌠ = ⎮ ⎮ ⌡0 √(2a − x) ⌡0 π ⁄2 ⌠ = ⎮ ⌡0 (2a sin2 θ) 9 ⁄ 2. 4a sin θ cos θ dθ √(2a − 2a sin2 θ) (2a) 9 ⁄ 2. 4a sin10 θ. cos θ dθ (2a) 1 ⁄ 2. cos θ π ⁄2 ⌠ = (2a) 4. 4a ⎮ ⌡0 = 64 a5 ⋅ sin10 θ dθ 9 7 5 3 1 π 63 a5 π ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ 10 8 6 4 2 2 8 REDUCTION FORMULAE (For Trigonometric Functions) ∫ 1.4 Reduction Formula for (a) We have = = = ∫ ∫ ∫ tan n x dx = tan n − 2 ∫ tan n x dx and cot n x dx tan n − 2 x . tan2 x dx x . (sec 2 ∫ (Note) x − 1) dx tan n − 2 x . sec2 x dx − (tan x) n − 2 + 1 − n− 2+ 1 ∫ I-7 ∫ tan n − 2 x dx tan n − 2 x dx tan n − 1 x ⌠ ⌠ ⎮ tan n x dx = − ⎮ tan n − 2 x dx , ⌡ ⌡ n− 1 or which is the required reduction formula. Application : Evaluate ∫ tan4 x dx . Putting n = 4 in the above reduction formula, we have ∫ (b) tan4 x dx = 1 3 tan3 x − ∫ tan2 x dx = 1 3 tan3 x − ∫ (sec 2 x − 1) dx = 1 3 tan3 x − tan x + x . We have = ∫ ∫ cot n x dx = ∫ cot n − 2 x . cot 2 x dx cot n − 2 x . (cosec 2 x − 1) dx ⌠ ⌠ = ⎮ cot n − 2 x . cosec 2 x dx − ⎮ cot n − 2 x dx ⌡ ⌡ = − (cot x) n − 1 ⌠ − ⎮ cot n − 2 x dx ⌡ n− 1 cot n x dx = − or cot n − 1 x ⌠ − ⎮ cot n − 2 x dx , ⌡ n− 1 which is the required reduction formula. Application : Putting n = 5 in the above reduction formula and applying it repeatedly, we have ∫ 1 4 cot4 x − = − 1 4 cot4 x − [− = − 1 4 cot4 x + 1 2 cot2 x + = − 1 4 cot4 x + 1 2 cot2 x + log sin x . 1.5 Reduction Formulae for (a) ∫ cot5 x dx = − We have In = ∫ sec n x dx = cot3 x dx 1 2 cot2 x − ∫ ∫ cot x dx] cot x dx ∫ sec n x dx and ∫ ∫ sec n − 2 x . sec2 x dx. cosec n x dx (Bundelkhand 2011) (Note) I-8 INTEGRAL CALCULUS Integrating by parts regarding sec 2 x as the 2nd function, we have ∫ In = sec n − 2 x tan x − (n − 2) sec n − 3 x sec x tan2 x dx = sec n − 2 x tan x − (n − 2) ∫ sec n − 2 x (sec2 x − 1) dx (Note) = sec n − 2 x tan x − (n − 2) ∫ sec n x dx + (n − 2) ∫ sec n − 2 x dx . Transposing the term containing ∫ sec n x dx to the left, we have (n − 2 + 1) ∫ sec n x dx = sec n − 2 tan x + (n − 2) ∫ sec n − 2 x dx or (n − 1) ∫ sec n x dx = sec n − 2 x tan x + (n − 2) ∫ sec n − 2 x dx . Dividing both sides by (n − 1) , we have sec n − 2 x tan x n − 2 ⌠ ⌠ ⎮ sec n − 2 x dx , ⎮ sec n x dx = + ⌡ n− 1 ⌡ n− 1 which is the required reduction formula. (b) To find the reduction formula for same way as in part (a). Thus, we get ∫ cosec n x dx , proceed exactly in the cosec n − 2 x cot x n − 2 ⌠ ⌠ ⎮ cosec n − 2 x dx , ⎮ cosec n x dx = − + ⌡ n− 1 ⌡ n− 1 as the required reduction formula for 1.6 Reduction Formula for Let Im , n = = = ∫ ∫ ∫ ∫ ∫ cosec n x dx . sin m x cos n x dx (kanpur 2014) sin m x cos n x dx sin m x cos n − 1 x cos x dx cos n − 1 x . (sin m x cos x) dx . Integrating by parts taking sin m x cos x as the second function, we get Im , n = n− 1 ⌠ sin m + 1 x ⎮ sin m + 1 x cos n − 2 x sin x dx cos n − 1 x + m+ 1⌡ m+ 1 = n− 1 ⌠ sin m + 1 x ⎮ sin m x cos n − 2 x sin2 x dx cos n − 1 x + m+ 1⌡ m+ 1 = n− 1 ⌠ sin m + 1 x ⎮ sin m x cos n − 2 x . (1 − cos2 x) dx ⋅ cos n − 1 x + m+ 1⌡ m+ 1 = n− 1 ⌠ n− 1 sin m + 1 x cos n − 1 x + ⎮ sin m x cos n − 2 x dx − I m+ 1⌡ m + 1 m,n. m+ 1 Transposing the last term to the left, we have ⎛ ⎝ Im, n ⎜ 1 + or n − 1⎞ n− 1 sin m + 1 x. cos n − 1 x + I ⎟ = m + 1⎠ m + 1 m, n − 2 m+ 1 n− 1 sin m + 1 x cos n − 1 x ⎛ m + n⎞ + I . ⎟ = m+ 1 m + 1 m, n − 2 ⎝ m + 1⎠ Im, n ⎜ REDUCTION FORMULAE (For Trigonometric Functions) I-9 Thus the required reduction formula is Im, n = sin m + 1 x . cos n − 1 x (n − 1) Im, n − 2 + ⋅ m+ n m+ n Note : If we write Im, n = = ∫ ∫ sin m x cos n x dx sin m − 1 x . (cos n x sin x) dx , then integrating by parts regarding cos n x sin x as the 2nd function, the reduction formula can be obtained as sin m − 1 x . cos n + 1 x m − 1 + I . Im, n = − m + n m − 2, n m+ n Similarly other four reduction formulae for as ∫ sin m x cos n x dx may be obtained sin m + 1 x cos n + 1 x m + n + 2 + Im, n + 2 . n+ 1 n+ 1 [To obtain this reduction formula put (n + 2) in place of n in the reduction formula obtained in article 1.6 and adjust the result accordingly] sin m + 1 x cos n + 1 x m + n + 2 Im, n = + Im + 2, n m+ 1 m+ 1 Im, n = − Im, n = − Im, n = sin m + 1 x cos n + 1 x m − 1 + I n + 1 m − 2, n + 2 n+ 1 n− 1 sin m + 1 x cos n − 1 x + I . m + 1 m + 2, n − 2 m+ 1 [This reduction formula has been obtained in article 1.6 at the stage we applied integration by parts.] ⌠ dθ ⋅ Example 1 : Evaluate ⎮ ⌡ sin4 1 θ 2 θ ⌠ ⌠ dθ = ⎮ cosec 4 dθ = 2 ∫ cosec 4 x dx , putting θ = 2x . Solution : We have ⎮ ⌡ sin4 1 θ ⌡ 2 2 But n− 2⌠ cosec n − 2 x cot x ⌠ + ⎮ cosec n − 2 x dx . ⎮ cosec n x dx = − ⌡ n− 1⌡ n− 1 [Derive this formula here] Putting n = 4 , we get ⌠ cosec 2 x cot x 2 ⌠ ⎮ cosec 4 x dx = − + ⎮ cosec 2 x dx ⌡ 3 3⌡ = − 1 3 cosec 2 x cot x + H ence the given integral 2 3 (− cot x) . I-10 INTEGRAL CALCULUS = 2 ∫ cosec 4 x dx = − = − 2 3 1 2 2 3 1 2 4 3 cosec 2 x cot x − cosec 2 θ cot θ − 4 3 cot x [... x = θ ⁄ 2] 1 2 cot θ . Example 2 : Evaluate ∫ (1 + x2) 3 ⁄ 2 dx . Solution : Put x = tan θ , so that dx = sec2 θ dθ . ∫ Then (1 + x2) 3 ⁄ 2 dx = ∫ sec 2 θ sec 3 θ dθ = Now we shall form a reduction formula for 1.5 (a), we get ∫ ∫ sec 5 θ dθ . sec n θ dθ . Proceeding as in article sec n − 2 θ tan θ n − 2 ⌠ ⌠ ⎮ sec n θ dθ = ⎮ sec n − 2 θ dθ . + ⌡ n− 1 n− 1⌡ ∴ ∫ ∫ sec 5 θ dθ = = 1 4 sec 3 θ tan θ + 3 1 [ 4 2 = 1 4 sec 3 θ tan θ + 3 8 = 1 4 3 3 [(1 + x2) 3 ⁄ 2 . x] + x (1 + x2) 1 ⁄ 2 + log {x + √(1 + x2)}. 1 4 sec 3 θ tan θ + 3 4 sec 3 θ dθ sec θ tan θ + sec θ tan θ + 8 3 8 1 2 ∫ sec θ dθ] log (sec θ + tan θ) 8 Evaluate the following integrals : 1. π ⁄2 ⌠ (iii) ⎮ ⌡0 2. π ⁄2 ⌠ (ii) ⎮ ⌡0 ⌠ (i) ⎮ sin6 x dx. ⌡ π ⁄4 ⌠ (i) ⎮ ⌡0 π ⁄2 ⌠ (iv) ⎮ ⌡0 cos9 x dx. (Kanpur 2005) cos10 x dx. a ⌠ (ii) ⎮ x5 (2 a2 − x2) − 3 dx. ⌡0 tan5 θ dθ. π ⁄4 ⌠ (iv) ⎮ ⌡0 ⌠ (iii) ⎮ sec 3 x dx. ⌡ a 3. sin6 x dx. ⌠ (i) ⎮ (a2 + x2) 5 ⁄ 2 dx. ⌡0 π ⁄4 ⌠ (ii) ⎮ ⌡0 sec 3 x dx. sin2 θ cos4 θ dθ. ⌠ (iii) ⎮ tan6 x dx. ⌡ a 4. 3 a4 π x4 ⌠ dx = ⋅ Show that ⎮ ⌡0 √(a2 − x2) 16 5. ⌠ If In = ⎮ ⌡0 π ⁄4 1 , tan n x dx, show that In + In− 2 = and deduce the value of I5. n− 1 (Kanpur 2005, 12; Avadh 06, 11; Bundelkhand 06; Purvanchal 14) REDUCTION FORMULAE (For Trigonometric Functions) 6. 1. π ⁄4 ⌠ If In = ⎮ ⌡0 (iii) 2. 3. (i) 1 2 tan n x dx, prove that n (In− 1 + In+ 1)= 1. sin5 x cos x − 5 24 sin x cos x + 128 ⋅ 315 1 ]. 2 [log 2 − 1 2 a6 (i) [67 √2 + 15 log tan ( π)]. 48 1 5 log (sec x + tan x). (iv) 3 8 tan5 x − (log 2 − 1 3 5 16 x. (ii) 1 1 [log 2 − ]. 2 2 1 1 √2 + log (√2 2 2 (ii) 1 2 (Kanpur 2005, 12; Avadh 06) 5π ⋅ 32 63π ⋅ 512 (iv) sec x tan x + 1 2 5 16 sin3 x cos x − (iii) (iii) 5. 1 6 (i) − I-11 + 1). π 1 + ⋅ 48 64 (ii) tan3 x + tan x − x. 1 ). 2 1.7 Gamma Function ∞ ⌠ The definite integral ⎮ e − x x n − 1 dx is called the second Eulerian integral and ⌡0 is denoted by the symbol Γ (n) [read as G amma n]. Properties of Gamma function. (Commit to memory) 1 2 Γ (n + 1) = n Γ n ; Γ 1 = 1 ; Γ = √π . Γ (n) = (n − 1) ! provided n is a positive integer. Thus Γ (10) = 9 ! . 9 2 Also Γ = 7 2 Γ = 7 2 7 2 ⋅ 5 2 Γ = 5 2 7 2 ⋅ 5 2 ⋅ 3 2 Γ = 7 2 ⋅ 5 2 3 2 ⋅ 1 2 Γ = 1 2 7 2 ⋅ 5 2 ⋅ 3 2 ⋅ ⋅ 3 2 1 2 √π = 105 16 √π . ⌠π ⁄ 2 1.8 Value of ⎮ sin m x cos n x dx in terms of Γ Function, ⌡0 where m and n are positive integers We have already derived in article 1.6 that ∫ sin m x cos n x dx = n− 1 ⌠ sin m + 1 x cos n − 1 x ⎮ sin m x cos n − 2 x dx . + m+ n m+ n⌡ I-12 INTEGRAL CALCULUS ∴ π ⁄2 ⌠ ⎮ ⌡0 sin m x cos n x dx ⎡ sin = ⎢⎢⎢ π ⁄2 π ⁄2 n− 1 ⌠ x cos n − 1 x⎤⎥ ⎥⎥ ⎮ + sin m x cos n − 2 x dx m+ n m + n ⌡0 ⎦0 m+ 1 ⎣ = 0+ π ⁄2 ⌠ ⎮ ⌡0 i.e., π ⁄2 n− 1 ⌠ ⎮ m + n ⌡0 sin m x cos n − 2 x dx sin m x cos n x dx = π ⁄2 (n − 1) ⌠ ⎮ (m + n) ⌡0 sin m x cos n − 2 x dx ...(1) Now four cases arise according as m and n take different types of values, odd or even. Case I : When m and n are both even : Successively applying the formula (1) till the power of cos x becomes zero, we have π ⁄2 ⌠ ⎮ ⌡0 π ⁄2 (n − 1) (n − 3) (n − 5) … 1 ⌠ ⎮ ⋅ ⋅ (m + n) (m + n − 2) (m + n − 4) (m + 2) ⌡0 = Also sin m x . cos n x dx π ⁄2 ⌠ ⎮ ⌡0 sin m x dx = m− 1 m− 3 m− 5 1 π ⋅ ⋅ … ⋅ ⋅ m m− 2 m− 4 2 2 sin m x dx . [See article 1.3] Therefore, π ⁄2 ⌠ ⎮ ⌡0 sin m x cos n x dx = (n − 1) (n − 3) (n − 5) … 1 × (m + n) (m + n − 2) (m + n − 4) … (m + 2) (m − 1) (m − 3) (m − 5) … 3 .1 π ⋅ 2 m (m − 2) (m − 4) … 4 .2 = 1⎫ ⎧⎛ n − 1 ⎞ ⎛ n − 3 ⎞ ⎛ n − 5 ⎞ ⎨⎜ ⎟ ⎜ ⎟ ⎜ ⎟ … ⎬ 2⎭ 2 2 2 ⎩⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1⎫ ⎧⎛ m − 1 ⎞ ⎛ m − 3 ⎞ ⎨⎜ ⎟ ⎜ ⎟ … ⎬ 2⎭ 2 2 ⎩⎝ ⎠ ⎝ ⎠ 4 2⎫ ⎧⎛ m + n ⎞ ⎛ m + n − 2 ⎞ ⎨⎜ ⎟ ⎜ ⎟ … ⋅ ⎬ 2 2⎭ 2 ⎩⎝ 2 ⎠ ⎝ ⎠ ⎛ m + 1⎞ ⎛ n + 1⎞ ⎟ Γ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⋅ π 2 Γ⎜ = ⎛ m + n + 2⎞ ⎟ 2 ⎝ ⎠ 2Γ⎜ ⋅ Similarly, the cases for other values of m and n may be considered and it may be verified that the result is true in other cases too. Thus for all positive integral values of m and n, we have ⎛ n + 1⎞ ⎛ m + 1⎞ ⎟ Γ⎜ ⎟ .Γ⎜ π ⁄2 ⎝ 2 ⎠ ⎝ 2 ⎠ ⌠ ⎮ sin m x cos n x dx = ⋅ (Remember) ⌡0 ⎛ m + n + 2⎞ 2Γ⎜ ⎟ ⎝ ⎠ 2 REDUCTION FORMULAE (For Trigonometric Functions) π ⁄2 ⌠ Walli’s Formula : [An easy way to evaluate ⎮ ⌡0 ⌠ n are + ive integers]. We have ⎮ ⌡ π ⁄2 I-13 sin m x cos n x dx where m and sin m x cos n x dx 0 where k is 1 2 (m − 1) (m − 3) (m − 5) … (n − 1) (n − 3) (n− 5) … × k, = (m + n) (m + n − 2) (m + n − 4) … π if m and n are both even, otherwise k = 1 . The last factor in each of the three products is either 1 or 2. In case any of m or n is 1, we simply write 1 as the only factor to replace its product. This formula is equally applicable if any of m or n is zero provided we put 1 as the only factor in its product and we regard 0 as even. π ⁄2 ⌠ Example 1 : Evaluate ⎮ ⌡0 sin4 x cos2 x dx . (Rohilkhand 2014) Solution : We know that π ⁄2 ⌠ ⎮ ⌡0 ⎛ m + 1⎞ ⎛ n + 1⎞ ⎟ Γ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ Γ⎜ sin m x cos n x dx = ∴ the given integral = ⎛ m + n + 2⎞ ⎟ ⎝ ⎠ 2 ⎛ 4 + 1⎞ ⎛ 2 + 1⎞ 5 3 Γ⎜ ⎟ ⋅ Γ⎜ ⎟ Γ Γ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 2 2Γ⎜ ⎛ 4 + 2 + 2⎞ 2Γ⎜ ⎟ ⎝ ⎠ 2 = 2Γ4 ⋅ = 3 2 ⋅ 1 2 √π ⋅ 1 2 √π 2 .3 .2. 1 = π ⋅ 32 [ ... Γ (n + 1) = n Γ n and Γ = √π ] 1 2 Alternate Solution : U sing Walli’s formula, the given integral π 3 .1 .1 π = ⋅ = ⋅ 6 .4 .2 2 32 π ⁄2 ⌠ Example 2 : Evaluate ⎮ ⌡0 π ⁄2 ⌠ Solution : We have ⎮ ⌡0 sin6 θ dθ . π ⁄2 ⌠ sin6 θ dθ = ⎮ ⌡0 sin6 θ cos0 θ dθ . (Note) Now m = 6 , n = 0 ; using the G amma function, we have the given integral ⎛ 6 + 1⎞ ⎛ 0 + 1⎞ 7 1 5 3 1 Γ⎜ ⎟ .Γ⎜ ⎟ Γ ⋅ Γ ⋅ ⋅ √π . √π 5π 2 ⎝ ⎠ ⎝ 2 ⎠ 2 2 2 2 2 = = = = ⋅ 2Γ4 2 . (3 .2 . 1) 32 ⎛ 6 + 0 + 2⎞ 2Γ⎜ ⎟ 2 ⎝ ⎠ O therwise, by Walli’s formula, the given integral 5 .3 .1 π 5π ⋅ = ⋅ = 32 6 .4 .2 2 I-14 INTEGRAL CALCULUS π ⁄6 ⌠ Example 3 : Evaluate ⎮ ⌡0 sin2 6θ cos5 3θ dθ . Solution : To bring the given integral into the form of G amma function, put 3θ = x, so that 3 dθ = dx . Also for limits, x = 0 at θ = 0 and x = π ⁄ 2 at θ = π ⁄ 6 . ∴ the given integral π ⁄2 = 1 3 ⌠ ⎮ ⌡0 = 1 3 ⌠ ⎮ ⌡0 = π ⁄2 sin2 2x cos5 x dx 4 3 (2 sin x cos x) 2 cos5 x dx = 3 2 Γ .Γ4 3 2 7 5 ⋅ 2 2 ⌠ ⎮ ⌡ π ⁄2 0 Γ . 3 .2. 1 4 4 = ⋅ 3 2 Γ 11 3 2⋅ 9⋅ 2 2 π ⁄4 ⌠ Example 4 : Evaluate ⎮ ⌡0 ⋅ 3 2 sin2 x cos7 x dx 3 2 ⋅ Γ = 64 ⋅ 945 (cos 2θ) 3 ⁄ 2 cos θ dθ . π ⁄4 ⌠ Solution : The given integral = ⎮ ⌡0 (1 − 2 sin2 θ) 3 ⁄ 2 cos θ dθ . Now put √2 sin θ = sin x , so that √2 cos θ dθ = cos x dx . Also when θ = 0 , sin x = √2 sin 0 = 0 giving x = 0 and when θ= 1 4 π , sin x = √2 sin (π ⁄ 4) = 1 giving x = 1 2 π. H ence the given integral π ⁄2 ⌠ = ⎮ ⌡0 = (1 − sin2 x) 3 ⁄ 2⋅ π ⁄2 1 ⌠ ⎮ √2 ⌡0 1 ⋅ = √2 cos4 x dx = π ⁄2 1 1 ⌠ ⎮ cos x dx = √2 √2 ⌡0 π ⁄2 1 ⌠ ⎮ √2 ⌡0 ⎛ 0 + 1⎞ ⎛ 4 + 1⎞ ⎟ ⋅ Γ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ Γ⎜ ⎛ 0 + 4 + 2⎞ ⎟ 2 ⎝ ⎠ 2Γ⎜ 3 cos3 x . cos x dx sin0 x .cos4 x dx 1 = √2 5 2 Γ ⋅ Γ 1 2 2Γ3 1 3π 1 2 ⋅ 2 √π √π = ⋅ = ⋅ √2 16 √2 2 .2 .1 1 ⌠ Example 5 : Evaluate ⎮ x4 (1 − x2) 5 ⁄ 2 dx . ⌡0 Solution : H ere we put x = sin θ , so that dx = cos θ dθ . And now the new limits are θ = 0 to θ = π ⁄ 2 . π ⁄2 ⌠ Thus the given integral ⎮ ⌡0 π ⁄2 ⌠ = ⎮ ⌡0 sin4 θ (1 − sin2 θ) 5 ⁄ 2 cos θ dθ π ⁄2 ⌠ sin4 θ . cos5 θ cos θ dθ = ⎮ ⌡0 sin4 θ cos6 θ dθ (Note) REDUCTION FORMULAE (For Trigonometric Functions) = 3 .1 .5 .3 .1 π , ⋅ 10 .8 .6 .4 .2 2 = 3π ⋅ 512 I-15 [by Walli’s formula] a π a6 ⌠ ⋅ Example 6 : Show that ⎮ x4 (a2 − x2) 1 ⁄ 2 dx = ⌡0 32 (Bundelkhand 2012) Solution : Put x = a sin θ so that dx = a cos θ dθ . 1 2 Also when x = 0 , θ = 0 and when x = a , θ = ⌠ ∴ the given integral = ⎮ ⌡ π ⁄2 π. a4 sin4 θ .a cos θ .a cos θ dθ 0 π ⁄2 ⌠ = a6 ⎮ ⌡0 = 3 2 6 a 1 2 ⋅ sin4 θ cos2 θ dθ = a6 √π 1 2 √π 2 .3 .2 .1 π ⁄4 ⌠ Example 7 : Evaluate ⎮ ⌡0 5 2 Γ .Γ 3 2 2Γ4 πa6 ⋅ 32 = sin4 x cos2 x dx . Solution : The given integral π ⁄4 ⌠ I= ⎮ ⌡0 π ⁄4 ⌠ = ⎮ ⌡0 = 1 8 (sin2 x cos2 x) sin2 x dx 1 4 π ⁄4 ⌠ ⎮ ⌡0 (4 sin2 x cos2 x) ⋅ 1 2 (2 sin2 x) dx sin2 2x (1 − cos 2x) dx . Put 2 x = t , so that 2 dx = dt . Also when x = 0 , t = 0 and when x = π ⁄ 4 , t = π ⁄ 2 . ∴ I= 1 8 π ⁄2 ⌠ ⎮ ⌡0 sin2 t (1 − cos t) ⋅ π ⁄2 π ⁄2 1 2 dt = 1 ⎡⎢ ⌠ ⎢⎮ 16 ⎢⎣ ⌡0 = 1 ⎡ 1 π 1 .1 ⎤ 1 ⎡ π 1⎤ − ⎢ ⋅ ⎥ = ⎢ − ⎥⋅ 16 ⎣ 2 2 3 .1 ⎦ 16 ⎣ 4 3 ⎦ ⌠ sin2 t dt − ⎮ ⌡0 ⎤ sin2 t cos t dt⎥⎥ ⎥ ⎦ a ⌠ Example 8 : Evaluate ⎮ x2 √(ax − x2) dx . ⌡0 a a ⌠ ⌠ Solution : We have ⎮ x2 √(ax − x2) dx = ⎮ x5 ⁄ 2 √(a − x) dx . ⌡0 ⌡0 I-16 INTEGRAL CALCULUS Now put x = a sin2 θ , so that dx = 2a sin θ cos θ dθ , and the new limits are θ = 0 to θ = π ⁄ 2 . Thus the given integral π ⁄2 ⌠ = ⎮ ⌡0 (a sin2 θ) 5 ⁄ 2 (a cos2 θ) 1 ⁄ 2. 2a sin θ cos θ dθ π ⁄2 ⌠ = 2a4 ⎮ ⌡0 = 2a4 sin6 θ cos2 θ dθ 7 2 Γ .Γ 3 2 2Γ5 ⎡5 3 1 ⎤ 1 ⎢ ⋅ ⋅ √π . √π ⎥ 5πa4 2 2 2 2 ⎢ ⎥ = 2a4 ⎢ ⋅ ⎥ = 2 .4 .3 .2 .1 128 ⎣ ⎦ ∞ ⌠ x4 dx Example 9 : Evaluate ⎮ ⋅ 2 ⌡0 (a + x2) 4 Solution : Put x = a tan θ , so that dx = a sec2 θ dθ and the new limits are θ = 0 to θ = π ⁄ 2 . ∴ the given integral π ⁄2 ⌠ = ⎮ ⌡0 π ⁄2 a4 tan4 θ .a sec2 θ dθ 1 ⌠ = 3⎮ 2 2 2 4 a ⌡0 (a + a tan θ) π ⁄2 1 ⌠ = 3⎮ a ⌡0 sin4 θ cos2 1 θ dθ = 3 a 5 2 tan4 θ dθ sec6 θ Γ ⋅ Γ 3 2 2Γ4 3 1 1 ⋅ √π . √π π 1 2 2 2 = 3 = ⋅ 2 .3 .2 .1 a 32 a3 Evaluate the following integrals : 1. π ⁄2 ⌠ (i) ⎮ ⌡0 sin2 x cos3 x dx. π ⁄2 ⌠ (ii) ⎮ ⌡0 sin4 x cos6 x dx. π ⁄2 ⌠ (iv) ⎮ ⌡0 2. π ⁄8 ⌠ (i) ⎮ ⌡0 π ⁄2 ⌠ (iii) ⎮ ⌡0 cos3 4 x dx. π ⁄2 π ⁄6 cos5 x sin 3 x dx. π ⌠ x x (ii) ⎮ sin6 cos8 dx. ⌡0 2 2 π ⁄2 ⌠ (iv) ⎮ ⌡0 sin3 x cos4 x cos 2 x dx. cos4 3 φ sin3 6 φ dφ . 1 3. sin 5 x cos8 x dx. sin12 x cos18 x dx. ⌠ (iii) ⎮ ⌡0 ⌠ (v) ⎮ ⌡0 (Bundelkhand 2009, 10) ⌠ (i) ⎮ x2 (1 − x2) 3 ⁄ 2 dx. ⌡0 1 (ii) ⌠ 4 ⎮ x (1 − x2) 3 ⁄ 2 dx. ⌡0 REDUCTION FORMULAE (For Trigonometric Functions) 1 a ⌠ (iii) ⎮ x6 (1 − x2) 1 ⁄ 2 dx. ⌡0 ⌠ (iv) ⎮ x2 (a2 − x2) 3 ⁄ 2 dx. ⌡0 1 4. I-17 1 ⌠ (ii) ⎮ x3 ⁄ 2 (1 − x) 3 ⁄ 2 dx. ⌡0 ⌠ (i) ⎮ x m (1 − x) n dx. ⌡0 1 ⌠ (iii) ⎮ x3 ⁄ 2 √(1 − x) dx. ⌡0 2a ⌠ (iv) ⎮ ⌡0 (Kanpur 2007, 12; Bundelkhand 07) 2a 5. x m √(2 ax − x2) dx, m being a positive integer. ⌠ (i) ⎮ ⌡0 a x5 √(2 ax − x2) dx. (ii) a a ⌠ x4 (iv) ⎮ dx. 2 ⌡0 (x + a2) 4 ⌠ (iii) ⎮ x2 (2 ax − x2) 5 ⁄ 2 dx. ⌡0 a 6. ⌠ (i) ⎮ x2 ⌡0 √ ⌠ 3 ⎮ x (2 ax − x2) 3 ⁄ 2 dx. ⌡0 a ⎛ a − x⎞ ⎜ ⎟ dx. ⎝ a + x⎠ ⌠ (ii) ⎮ x ⌡0 √ ⎛ a 2 − x2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ a 2 + x2 ⎟ dx. ⎝ ⎠ b ⌠ (iii) ⎮ (x − a) m (b − x) n dx. ⌡a 1 7. Prove that √π Γ (1 ⁄ n) dx ⌠ ⎮ = ⋅ ⌡0 √(1 − x n ) n Γ { 1 + (1 ⁄ n)} 2 ∞ x4 8. π dx ⌠ = ⋅ Show that ⎮ ⌡0 (1 + x2) 4 32 9. If m, n are positive integers, then prove that 1 1 1 . 2 . 3 . …… (m − 1) ⌠ ⌠ ⎮ x m− 1 (1 − x) n− 1 dx = ⎮ x n− 1 (1 − x) m− 1 dx = ⋅ ⌡0 ⌡0 n (n + 1) …… (n + m − 1) 1. (i) 2 ⋅ 15 (iv) (iii) 3. (ii) 13 2 2 Γ (16) 1 3 ⋅ π ⋅ 32 πa6 (iv) ⋅ 32 (i) 19 2 Γ( )Γ( ) ⋅ 2. (i) 3π ⋅ 512 1 6 ⋅ 2 ⋅ 315 3π (ii) ⋅ 256 Γ (m + 1) Γ (n + 1) 4. (i) ⋅ Γ (m + n + 2) (iv) (iii) 8 ⋅ 1287 5π ⋅ 2048 1 (v) ⋅ 15 5π (iii) ⋅ 256 3π (ii) ⋅ 128 (ii) I-18 INTEGRAL CALCULUS π ⋅ 16 33 7 (i) πa . 16 1 ⎡ π 1⎤ 1 ⎢ − ⎥⋅ (iv) 3 ⋅ 16 ⎣ 4 3 ⎦ a (2 m + 1) (2 m − 1) …… 3 . 1 ⋅ (m + 2) (m + 1) m (m − 1) …… 2 . 1 ⎡ 45π ⎛ 9π 23 ⎞⎟ (ii) a7 ⎜⎝ − ⋅ (iii) a8 ⎢⎣ − ⎠ 32 35 256 (iv) am+ 2 (iii) 5. 1 2 6. (i) a3 ( π − ). 4 (ii) 3 1 4 2 ⎤⎥ ⋅ 7⎦ a2 (π − 2). ⎡ Γ (m (iii) (b − a) m+ n+ 1 ⎢⎣ + 1) Γ (n + 1) ⎤⎥ ⋅ Γ (m + 2 + 2) ⎦ 1.9 Integration of x n sin mx and x n cos mx (a) ∫ x n sin mx dx . To form the reduction formula, integrating by parts regarding sin mx as the 2nd function, we have ⌠ n x n cos m x ⌠ n − 1 cos m x dx ⎮ x sin m x dx = − + ⎮ nx ⌡ ⌡ m m = − ⎤ x n cos m x n ⎡⎢ x n − 1 sin m x n − 1 ⌠ n − 2 ⎮x + − sin m x dx⎥⎥ , ⎢ ⎥ m ⎢⎣ m ⌡ m m ⎦ [again integrating by parts regarding cos m x as the 2nd function] = − n (n − 1) ⌠ n − 2 x n cos m x nx n − 1 ⎮x + sin m x − sin m x dx . 2 ⌡ m m2 m Above is the required reduction formula. Successively applying this formula we ⌠ ⌠ are left with ⎮ x sin mx dx or ⎮ sin mx dx according as n is odd or even. ⌡ ⌡ ∫ x n cos mx dx . (b) have Integrating by parts regarding cos mx as the 2nd function, we ⌠ n x n sin m x ⌠ nx n − 1 sin m x dx ⎮ x cos m x dx = − ⎮ ⌡ ⌡ m m = x n sin m x n ⌠ n− 1 ⎮x − sin m x dx m⌡ m = ⎤ n− 1 ⌠ n− 2 x n sin m x n ⎡⎢ n − 1 ⎛ cos m x⎞ ⎮x + ⋅ ⎜ cos m x dx⎥⎥ , ⎟ − ⎢x ⎥ m ⎢⎣ m ⌡ m ⎝ m ⎠ ⎦ [again integrating by parts regarding sin m x as 2nd function] = xn sin m x nx n − 1 cos m x n (n − 1) ⌠ n − 2 + − ⎮x cos m x dx , ⌡ m m2 m2 which is the required reduction formula. 1.10 Reduction Formulae for ∫ x sin x dx and ∫ x cos x dx . n n (Bundelkhand 2009; Meerut 13) Let In = ∫ x sin n x dx = ∫ (x sin n − 1 x) .sin x dx (Note) REDUCTION FORMULAE (For Trigonometric Functions) I-19 ⌠ = (x sin n − 1 x) . (− cos x) + ⎮ cos x [sin n − 1 x + x (n − 1) sin n − 2 x cos x] dx , ⌡ integrating by parts regarding sin x as 2nd function = − x sin n − 1 x cos x + ∫ sin n − 1 x cos x dx + (n − 1) ∫ x sin n − 2 x cos2 x dx 1 ⌠ = − x sin n − 1 x cos x + sin n x + (n − 1) ⎮ x sin n − 2 x . (1 − sin2 x) dx ⌡ n 1 ⌠ ⌠ = − x sin n − 1 x cos x + sin n x + (n − 1) ⎮ x sin n − 2 x dx − (n − 1) ⎮ x sin n x dx ⌡ ⌡ n 1 = − x sin n − 1 x cos x + sin n x + (n − 1) In − 2 − (n − 1) In . n Transposing the last term to the left, we have 1 In (1 + n − 1) = − x sin n − 1 x cos x + sin n x + (n − 1) In − 2 n 1 n In = − x sin n − 1 x cos x + sin n x + (n − 1) In − 2 n or x sin n x (n − 1) . sin n − 1 x . cos x + + In − 2 , n n n2 which is the required reduction formula. In = − or ⌠ Similarly, reduction formula for ⎮ x cos n x dx is ⌡ ⌠ x cos n − 1 x sin x cos n x (n − 1) + + In − 2 . In = ⎮ x cos n x dx = ⌡ n n n2 1.11 ⌠ ax ⌠ ax n n e sin bx dx and ⎮ e cos bx dx . Reduction Formulae for ⎮ ⌡ ⌡ (a) Let ⌠ In = ⎮ e a x sin n bx dx ⌡ = eax nb ⌠ a x ⎮ e sin n − 1 bx cos bx dx , sin n bx − a ⌡ a ...(1) integrating by parts taking e a x as the 2nd function. Now ⌠ ax ⎮ e sin n − 1 bx cos bx dx ⌡ = eax (sin n − 1 bx cos bx) a ⌠ eax − ⎮ [(n − 1) b sin n − 2 bx cos2 bx − b sin n bx] dx , ⌡ a integrating by parts taking e ax as the 2nd function I-20 INTEGRAL CALCULUS = eax (sin n − 1 bx cos bx) a − b ⌠ ax ⎮ e [(n − 1) sin n − 2 bx (1 − sin2 bx) − sin n bx] dx a⌡ = eax b⌠ (sin n − 1 bx cos bx) − ⎮ e a x [(n − 1) sin n − 2 bx − n sin n bx] dx a⌡ a = eax b⌠ nb (sin n − 1 bx cos bx) − (n − 1) ⎮ e a x sin n − 2 bx dx + I . a a⌡ a n Substituting this value in (1), we get In = eax nb sin n bx − 2 e a x sin n − 1 bx cos bx a a b2 ⌠ b2 + n (n − 1) 2 ⎮ e a x sin n − 2 bx dx − n2 2 In . a ⌡ a Transposing the last term to L.H .S., we get 2 2⎞ ax 2 ⎛ ⎜ 1 + n b ⎟ I = e (a sin bx − nb cos bx) sin n − 1 bx + n (n − 1) b I . ⎜ ⎟ n ⎜ a2 ⎟⎠ a2 a2 n − 2 ⎝ n (n − 1) eax (a sin n bx − nb sin n − 1 bx cos bx) + 2 I , 2 2 2 a + n 2 b2 n − 2 a + n b which is the required reduction formula. In = ∴ ⌠ Similarly, ⎮ e a x cos n bx dx ⌡ n (n − 1) eax (a cos n bx + nb sin bx cos n − 1 bx) + 2 I . 2 2 a + n 2 b2 n − 2 + n b Note : The above formulae should not be applied when n is small. In that case sin n bx and cos n bx are converted in terms of multiples of angles. = 1.12 a2 ⌠ ⌠ n ax n ax Reduction Formulae For ⎮ x e sin bx dx and ⎮ x e cos bx dx . ⌡ ⌡ We know that where Now ∫ e a x sin bx dx = r = √(a2 + b2) and eax sin (bx − φ ) , r φ = tan− 1 (b ⁄ a) . 1 ⎡1 ⎤ 1 ⌠ ax ⎮ e sin (bx − φ ) dx = ⎢ e a x sin {(bx − φ ) − φ }⎥ ⎦ r⎣r r⌡ 1 ⌠ = 2 ⎮ e a x sin (bx − 2φ ) . r ⌡ Similarly 1 1 ⌠ ax ⎮ e sin (bx − 2φ ) dx = 3 e a x sin (bx − 3φ ) , and so on. r r2 ⌡ Now ∫ x n e a x sin bx dx can be easily evaluated by repeatedly integrating by parts REDUCTION FORMULAE (For Trigonometric Functions) I-21 taking function of the type e a x sin bx as the 2nd function. Similarly we can obtain a reduction formula for ∫ 1.13 x n e a x cos bx dx . ⌠ m cos x sin nx dx . Reduction Formulae for ⎮ ⌡ (Meerut 2013B) Let Im, n = ∫ cos m x sin nx dx . Integrating by parts regarding sin nx as the 2nd function, we have Im, n = = cos m x (− cos nx) ⌠ m cos m − 1 x .(− sin x) (− cos nx) dx − ⎮ ⌡ n n − cos m x cos nx m ⌠ ⎮ cos m − 1 x cos nx .sin x dx . − n ⌡ n ...(1) Now sin {(n − 1) x} = sin (nx − x) = sin nx cos x − cos nx sin x . ∴ cos nx sin x = sin nx cos x − sin (n − 1) x . Im, n = − ∴ = − cos m x cos nx m ⌠ − ⎮ cos m − 1 x {sin nx cos x − sin (n − 1) x} dx n ⌡ n cos m x cos nx m ⌠ ⎮ cos m x sin nx dx − n ⌡ n m⌠ ⎮ cos m − 1 x sin (n − 1) x dx . n ⌡ + Transposing the middle term to the left and simplifying, we get Im, n = − cos m x cos nx m ⌠ + ⎮ cos m − 1 x sin (n − 1) x dx , m+ n⌡ m+ n which is the required reduction formula. Deduction : If in the above integral we take the limits of integration as 0 to 1 2 π , we find that π ⁄2 ⌠ ⎮ ⌡0 1.14 cos m x sin nx dx = π ⁄2 1 m ⌠ ⎮ + m + n m + n ⌡0 cos m − 1 x sin (n − 1) x dx . ⌠ m Reduction Formulae for ⎮ cos x cos nx dx ⌡ Let Im, n = ∫ cos m x .cos nx dx ⌠ ⎛ sin nx⎞ ⎛ sin nx⎞ ⎟ − ⎮ m cos m − 1 x . (− sin x) ⎜ ⎟ dx , ⌡ ⎝ n ⎠ ⎝ n ⎠ = cos m x ⋅ ⎜ integrating by parts taking cos nx as the 2nd function I-22 INTEGRAL CALCULUS = cos m x sin nx m ⌠ ⎮ cos m − 1 x . sin nx sin x dx . + n n ⌡ But cos (n − 1) x = cos nx cos x + sin nx sin x . ∴ sin nx sin x = cos (n − 1) x − cos nx cos x . H ence Im, n = = cos m x .sin nx m ⌠ ⎮ cos m − 1 x {cos (n − 1) x − cos nx cos x} dx + n ⌡ n cos m x sin nx m ⌠ m⌠ ⎮ cos m − 1 x cos (n − 1) x dx − ⎮ cos m x cos nx dx + n ⌡ n ⌡ n m cos m x sin nx m + Im − 1, n − 1 − I . n n m, n n Transposing the last term to the left, we have = m⎞ cos m x sin nx m ⎛ + I ⎜1 + ⎟ I m, n = n n m − 1, n − 1 n ⎝ ⎠ cos m x sin nx m + I , m + n m − 1, n − 1 m+ n which is the required reduction formula. Im, n = or π ⁄2 ⌠ Deduction : ⎮ ⌡0 cos m x cos nx dx π ⁄2 ⎡ cos x sin nx⎤⎥ ⎥ = ⎢⎢⎢ m + n ⎥⎦ 0 ⎣ m = 0+ m (m + n) π ⁄2 ⌠ Im, n = ⎮ ⌡0 or π ⁄2 m ⌠ ⎮ (m + n) ⌡0 π ⁄2 cos m − 1 x cos (n − 1) x dx cos m − 1 x cos (n − 1) x dx m I . m + n m − 1, n − 1 cosm x cos nx dx = ⌠ Example 1 : If un = ⎮ ⌡0 show that π ⁄2 ⌠ ⎮ ⌡0 + x n sin x dx and n > 1 , 1 un + n (n − 1) un − 2 = n ( π) n − 1. 2 π ⁄2 ⌠ Hence evaluate ⎮ ⌡0 x5 sin x dx . (Kanpur 2005; Gorakhpur 05; Avadh 07; Meerut 12B) π ⁄2 ⌠ Solution : We have un = ⎮ ⌡0 x n sin x dx π ⁄2 π ⁄2 ⌠ − ⎮ = ⎡⎣ x n . (− cos x) ⎤⎦ ⌡0 0 n . x n − 1 . (− cos x) dx , [Integrating by parts taking sin x as the 2nd function] REDUCTION FORMULAE (For Trigonometric Functions) π ⁄2 ⌠ = 0+ n⎮ ⌡0 I-23 x n − 1 cos x dx π ⁄2 π ⁄2 ⌠ ⎡ − ⎮ = n ⎢ ⎧⎨⎩ x n − 1 . sin x ⎫⎬⎭ 0 ⌡0 ⎣ ⎤ (n − 1) . x n − 2 . sin x dx⎥ , ⎦ again integrating by parts π ⁄2 ⌠ 1 = n . ( π) n − 1 − n (n − 1) ⎮ 2 ⌡0 x n − 2 sin x dx . Thus 1 un = n ( π) n − 1 − n (n − 1) un − 2 . 2 ...(1) ∴ 1 un + n (n − 1) un − 2 = n ( π) n − 1. 2 Proved. π ⁄2 ⌠ Now to evaluate ⎮ ⌡0 Then 1 u5 = 5 ( π) 5 − 1 − 5 (5 − 1) u3 2 = 5⋅ = Now x5 sin x dx , put n = 5 in (1). 5 16 1 ( 2 1 π) 4 − 20 [3 ( π) 3 − 1 − 3 (3 − 1) u1] , putting n = 3 in (1) 2 π 4 − 15π 2 + 120 u1. π ⁄2 ⌠ u1 = ⎮ ⌡0 x sin x dx π ⁄2 π ⁄2 ⌠ = ⎡⎢ x . (− cos x) ⎤⎥ + ⎮ ⌡0 ⎣ ⎦0 cos x dx π π ⁄2 ⎡ ⎤ = ⎡⎢ 0 + sin x⎤⎥ = ⎢ sin − sin 0⎥ = 1 . 2 ⎣ ⎦0 ⎣ ⎦ H ence π ⁄2 ⌠ u5 = ⎮ ⌡0 x5 sin x dx = 5π 4 − 15π 2 + 120 . 16 Example 2(i) : Integrating by parts twice or otherwise, obtain a reduction form ula for ∞ ⌠ Im = ⎮ e− x sin m x dx , where m ≥ 2 ⌡0 in the form (1 + m 2) Im = m (m − 1) Im − 2 and hence evaluate I4 . (Agra 2014) ∞ ⌠ Solution : We have Im = ⎮ e− x sin m x dx ⌡0 ∞ ∞ ⌠ = ⎡⎣ sin m x . (− e− x) ⎤⎦ 0 + ⎮ m sin m − 1 x cos x e− x dx , ⌡0 integrating by parts taking e− x as the second function ∞ ⌠ = 0 + m ⎮ (sin m − 1 x cos x) .e− x dx ⌡0 I-24 INTEGRAL CALCULUS ∞ = m ⎡⎣ sin m − 1 x cos x . (− e− x) ⎤⎦ 0 ∞ ⌠ − m ⎮ [− sin m x + (m − 1) sin m − 2 x cos2 x] . (− e− x) dx ⌡0 ∞ ⌠ = 0 + m ⎮ e− x [− sin m x + (m − 1) sin m − 2 x (1− sin2 x)] dx ⌡0 ∞ ⌠ = m ⎮ e− x [(m − 1) sin m − 2 x − m sin m x] dx ⌡0 = m (m − 1) Im − 2 − m 2 Im or (1 + m 2) Im = m (m − 1) Im − 2 or Im = Proved. m (m − 1) I . 1 + m2 m − 2 ...(1) To evaluate I4 , putting m = 4 in (1), we get 4 (4 − 1) 12 I = I 1 + 16 2 17 2 I4 = = 12 ⎡ 2 (2 − 1) ⎤ 24 I ⎥= I , ⎢ 17 ⎣ 1 + 4 0⎦ 85 0 = 24 ⌠ 24 ⎡ − x⎤ ∞ 24 24 ⌠ ⎮ e− x sin0 x dx = ⎮ e− x dx = − e ⎦0 = ⋅ 85 ⌡0 85 ⎣ 85 85 ⌡0 ∞ [To get I2 , we put m = 2 in (1)] ∞ ∞ ⌠ Im = ⎮ e− ax sin m x dx , (a > 0, m ≥ 0); prove that ⌡0 Example 2(ii) : If ∞ ⌠ (m 2 + a2) Im = m (m − 1) Im − 2 and hence evaluate ⎮ e− x sin 4 x dx. ⌡0 (Kanpur 2006) Solution : Proceed as in E x. 2(i). ∞ ⌠ Example 3 : Evaluate ⎮ x e− 2 x cos x dx . ⌡0 (Rohilkhand 2014) Solution : The given integral ∞ ⌠ I = ⎮ x .(e− 2x cos x) dx . ⌡0 Integrating by parts taking e− 2x cos x as the 2nd function, we have − 2x ⎡ e I = ⎢⎢⎢ x ⎣ r ⎤∞ cos (x − φ ) ⎥⎥⎥ ⎦ ∞ e− 2x ⌠ − ⎮ 1⋅ cos (x − φ ) dx , ⌡0 r 0 where φ = tan− 1 (b ⁄ a) = tan− 1 (− 1 ⁄ 2) and r = √(a2 + b2) = √(4 + 1) = √5 REDUCTION FORMULAE (For Trigonometric Functions) I-25 ∞ ⎡ lim 1 x ⎤ 1⌠ = ⎢ x →∞ cos (x − φ ) − 0⎥ − ⎮ e− 2x cos (x − φ ) dx r e2x r ⌡0 ⎣ ⎦ ∞ 1 ⎡ 1 − 2x 1 ⌠ ⎤∞ ⎮ e− 2x cos (x − φ ) dx = − e cos (x − 2φ ) ⎥ ⎢ ⌡ √5 ⎣ √5 √5 0 ⎦0 = 0− 1 = − (1 ⁄ 5) [0 − e0 cos (− 2φ )] = cos 2φ , 5 1 where φ = tan− 1 (b ⁄ a) = tan− 1 (− ) 2 = 1 5 [{1 − tan2 = 1 5 [(1 − 1 ) 4 φ } ⁄ {1 + ⁄ (1 + 1 )] 4 = tan2 1 5 ⋅ 3 5 φ }] , where tan φ = − = 1 2 3 ⋅ 25 Example 4 : Prove that π ⁄2 ⌠ ⎮ ⌡0 cos m x sin m x dx = ⎡ 22 23 24 2m ⎤ ⎢2 + ⎥⋅ + + + … + 2 3 4 m⎦ 2m + 1 ⎣ 1 Solution : Proceeding as in article 1.13 and taking n = m , we first establish the reduction formula 1 1 + Im − 1, m − 1 . ...(1) Im, m = 2 2m ∴ Im, m = 1 1⎡ 1 1 + ⎢ + I 2m 2 ⎣ 2 (m − 1) 2 m − 2 , m − ⎤ 2⎥⎦ ⋅ ⎡. . ⎤ 1 1 ⎢ . from (1), I m − 1, m − 1 = ⎥ + I ⎣ 2 (m − 1) 2 m − 2 , m − 2⎦ = 1 1 1 + 2 + I 2m 2 (m − 1) 22 m − 2 , m − 2 Finally, 1 1 1 1 + 2 + 3 + 3 Im − 3, m − 3 , and so on. 2m 2 (m − 1) 2 (m − 2) 2 1 1 1 Im, m = + 2 + 3 2m 2 (m − 1) 2 (m − 2) 1 1 1 + …+ m − 2 + + I . 2 . 3 2m − 1 .2 2m − 1 1, 1 But ⌠ I1, 1 = ⎮ ⌡0 ∴ Im, m = = π ⁄2 1 ⎡1 ⎣2 ⎤π ⁄2 1 = ⋅ 2 ⎦0 cos x sin x dx = ⎢ sin2 x⎥ ⎡ 22 23 2m ⎤ ⎢2 + ⎥⋅ + + …+ 2 3 m⎦ 2m + 1 ⎣ Example 5 : Prove that if n be a positive integer, π ⁄2 ⌠ ⎮ ⌡0 cos n x cos nx dx = π ⋅ 2n + 1 (Kanpur 2008) Solution : Proceeding as in article 1.14 and taking m = n , we first establish the reduction formula I-26 INTEGRAL CALCULUS Im, n = n 1 I = I . n + n n − 1, n − 1 2 n − 1, n − 1 ...(1) 1 Putting (n − 1) for n in (1), we have In − 1, n − 1 = In − 2 , n − 2 . 2 1 1 1 ∴ In, n = ⋅ In − 2 , n − 2 = 2 In − 2 , n − 2 . 2 2 2 Thus by repeated application of (1), we get 1 1 In, n = n In − n, n − n = n I0, 0 . 2 2 π ⁄2 But ⌠ I0, 0 = ⎮ ⌡0 ∴ In, n = π ⁄2 ⌠ cos0 x cos 0 x dx = ⎮ ⌡0 π π ⁄2 1 . dx = ⎡⎢ x ⎤⎥ = ⋅ 2 ⎣ ⎦0 π 1 1 ⋅ π= n+ 1 ⋅ n 2 2 2 ⌠ sin nx dx and show Example 6 : Find the reduction form ula for the integral ⎮ ⌡ sin x π ⌠ sin nx that ⎮ dx = π or 0, according as n is odd or even. (Kanpur 2009; Rohilkhand 14) ⌡ sin x 0 Solution : To find the required reduction formula, consider sin nx − sin (n − 2) x = 2 cos (n − 1) x sin x or sin nx sin (n − 2) x − = 2 cos (n − 1) x , dividing both sides by sin x sin x sin x or sin (n − 2) x sin nx = 2 cos (n − 1) x + ⋅ sin x sin x Integrating both the sides, we have 2 sin (n − 1) x ⌠ sin (n − 2) x ⌠ sin nx ⎮ dx = + ⎮ dx , ⌡ ⌡ sin x (n − 1) sin x which is the required reduction formula. Now let π π ⎧ 2 sin (n − 1) x⎫ π ⌠ sin nx ⌠ sin (n − 2) x In = ⎮ dx = ⎨ dx = 0 + In − 2 . ⎬ + ⎮ ⌡0 sin x ⌡0 ⎩ ⎭0 n− 1 sin x H ence, In = In − 2 = In − 4 = In − 6 = … i.e., when n is even π π ⌠ sin 2x ⌠ π dx = 2 ⎮ cos x dx = 2 ⎡⎢ sin x ⎤⎥ = 0 In = I2 = ⎮ ⌡0 sin x ⌡0 ⎣ ⎦0 and when n is odd π π ⌠ sin x ⌠ In = I1 = ⎮ dx = ⎮ dx = π . ⌡0 sin x ⌡ 0 REDUCTION FORMULAE (For Trigonometric Functions) I-27 Evaluate the following integrals : 1. π ⁄2 ⌠ (i) ⎮ ⌡0 π ⌠ (ii) ⎮ x sin2 x cos x dx. ⌡0 x3 sin 3x dx. 1 2. π π ⁄2 3. ⎧ ⎛ x⎞ ⎫ ⌠ (iv) ⎮ √(a2 − x2) ⎨⎩ cos− 1 ⎜⎝ ⎟⎠ ⎬⎭ dx. ⌡0 a (Kanpur 2008) ⌠ (i) ⎮ x sin3 x dx. ⌡0 ⌠ (iii) ⎮ ⌡0 ⌠ (ii) ⎮ x sin4 x dx. ⌡ ⌠ (iv) ⎮ e x (x cos x + sin x) dx. ⌡ x cos3 x dx. 1 ⌠ (ii) ⎮ (sin− 1 x) 4 dx. ⌡0 ⌠ (i) ⎮ x2 e x cos α sin (2 x sin α) dx. ⌡ ∞ ∞ x4 + 1 ⌠ dx. (iii) ⎮ ⌡1 x2 (x2 + 1) 2 4. π ⁄2 ⌠ If un = ⎮ ⌡0 2 a ⌠ (iii) ⎮ x6 sin− 1 x dx. ⌡0 ⌠ (iv) ⎮ ⌡1 x n sin mx dx, prove that un = x2 + 3 dx. x6 (x2 + 1) nπ n− 1 n (n − 1) − un− 2 , 2 n− 1 m2 m .2 if m is of the form 4 r + 1. 5. π ⁄2 ⌠ If In = ⎮ ⌡0 (Kanpur 2011) x n sin (2 p + 1) x dx, prove that n (n − 1) n ⎛ π⎞ n − 1 , In − 2 = (− 1) p ⎜ ⎟ 2 2 ( p + 1) (2p + 1) ⎝ 2 ⎠ where n and p are positive integers. In + π ⁄2 ⌠ H ence deduce that ⎮ ⌡0 6. π ⁄2 ⌠ If un = ⎮ ⌡0 x3 sin 3x dx = π2 2 − ⋅ 27 12 θ sin n θ dθ and n > 1 , prove that (n − 1) 1 un − 2 + 2 ⋅ n n 149 H ence deduce that u5 = ⋅ 225 un = 7. (Gorakhpur 2006; Rohilkhand 07) Prove that if n be a positive integer greater than unity, then π ⁄2 ⌠ ⎮ ⌡0 cos n − 2 x sin nx dx = 1 ⋅ n− 1 (Avadh 2004; Kanpur 10) I-28 8. INTEGRAL CALCULUS π ⁄2 ⌠ If I(m, n) = ⎮ ⌡0 cos m x cos nx dx , prove that ⎧ m (m − 1) ⎫⎪ I(m, n) = ⎪⎨ 2 I . ⎪⎩ m − n 2 ⎬⎪⎭ (m − 2 , n) (Purvanchal 2014) 9. π ⌠ ⎛ sin Prove that ⎮ ⎜ ⎟ dθ = nπ . ⌡0 ⎝ sin θ ⎠ π ⁄2 ⌠ 10. If S n = ⎮ ⌡0 π ⁄2 sin (2n − 1) x ⌠ dx , V n = ⎮ ⌡ sin x 0 π2 2 + ⋅ 12 27 πa2 1 (1 + π 2). (iv) 6 8 1. (i) − 2. (i) 2 3 (iii) (i) (ii) − 4 9 ⋅ (iii) π 16 − ⋅ 14 245 π. (ii) − (iv) ⎛ sin nx⎞ 2 ⎜ ⎟ dx, (n is an integer), show ⎝ sin x ⎠ S n + 1 − S n = 0, V n + 1 − V n = S n + 1 . that 3. nθ ⎞ 2 3 2 3 3 x sin 3 x cos x sin4 x + + x − x sin 2x − cos 2x . 16 16 32 4 16 1 7 [π − ] . 3 3 1 x e [x (cos x + 2 1 2 x 2 sin x) − cos x] . e ax sin (bx − φ ) − 1 2 xe ax sin (bx − 2φ ) + 1 4 e ax sin (bx − 3φ ) , where a = 2 cos α , b = 2 sin α and φ = α . (ii) (iv) 1 4 π − 3π 2 + 24 . 16 1 (58 − 15π) . 30 (iii) 3 2 − 1 4 π. Fill in the Blanks: Fill in the blanks “……” so that the following statem ents are complete and correct. 1. n− 1⌠ ⌠ ⎮ sin n− 2 x dx. ⎮ sin n x dx = …… + ⌡ n ⌡ 2. ⌠ ⌠ ⎮ tan n x dx = …… − ⎮ tan n− 2 x dx. ⌡ ⌡ REDUCTION FORMULAE (For Trigonometric Functions) 3. n− 2⌠ ⌠ ⎮ sec n x dx = …… + ⎮ sec n− 2 x dx. ⌡ n− 1⌡ 4. ⌠ ⎮ ⌡0 5. ⌠ ⎮ ⌡0 6. ⌠ If un = ⎮ ⌡0 π ⁄2 π ⁄2 I-29 sin6 θ dθ = …… . sin4 x cos6 x dx = …… . π ⁄2 θ sin n θ dθ and n > 1, then un = (n − 1) un− 2 + …… . n Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 7. π ⁄4 ⌠ If In = ⎮ ⌡0 tan n x dx, then 1 In + In− 2 = n 1 (b) In + In− 2 = n− 1 1 (c) In + In− 2 = n− 2 n (d) In + In− 2 = ⋅ n− 1 ⌠ If In = ⎮ cot n x dx, then ⌡ (a) 8. (a) cot n− 1 x ⌠ + ⎮ cot n− 2 x dx ⌡ n− 1 In = (b) In = − (c) cot n− 1 x ⌠ + ⎮ cot n− 2 x dx ⌡ n− 1 In = − (d) In = cot n− 1 x ⌠ − ⎮ cot n− 2 x dx ⌡ n− 1 cot n− 1 x ⌠ − ⎮ cot n− 2 x dx ⌡ n− 1 True or False: Write ‘T ’ for true and ‘F ’ for false statem ent. 9. cosec n− 2 x cot x n − 2 ⌠ ⌠ ⎮ cosec n− 2 x dx. ⎮ cosec n x dx = − + ⌡ n− 1⌡ n− 1 π ⁄2 10. ⌠ ⎮ ⌡0 11. ⌠ If un = ⎮ ⌡0 cos6 x dx = π ⁄2 5π ⋅ 16 1 x n sin x dx and n > 1, then un + n (n − 1) un− 2 = n ( π) n− 1. 2 I-30 INTEGRAL CALCULUS π ⁄2 12. ⌠ ⎮ ⌡0 1. − 4. 5π ⋅ 32 5. 7. 10. (b). F. 8. 11. sin2 x cos3 x dx = 1 sin n− 1 x cos x. n 1 ⋅ 15 2. tan n− 1 x ⋅ n− 1 3π ⋅ 512 (c). T. 3. 6. 9. 12. sec n− 2 x tan x ⋅ n− 1 1 ⋅ n2 T. F. ⌠ 2.1 Reduction Formulae for ⎮ xm (a + bx n) p dx ⌡ ⌠ m ⎮ x (a + bx n ) p dx can be connected with any one of the following six integrals: ⌡ (i) ⌠ m− n ⎮x (a + bx n ) p dx , ⌡ (ii) ⌠ m ⎮ x (a + bx n ) p − 1 dx , ⌡ ⌠ (iii) ⎮ x m + n (a + bx n ) p dx , ⌡ ⌠ (iv) ⎮ x m (a + bx n ) p + 1 dx , ⌡ ⌠ (v) ⎮ x m − n (a + bx n ) p + 1 dx , ⌡ ⌠ (vi) ⎮ x m + n (a + bx n ) p − 1 dx . ⌡ Rule for Connection : In order to connect the given integral with any one of the six integrals we use the following rule : Let P = x λ + 1 (a + bx n ) μ + 1, where λ and μ are the smaller of the indices of x and (a + bx n ) in the two expressions whose integrals we want to connect. I-32 INTEGRAL CALCULUS Find (dP ⁄ dx) and rearrange this as a linear combination of the expressions whose integrals are to be connected. Finally integrate both sides and transpose suitably to get the required reduction formula. (i) ⌠ ⌠ To connect ⎮ x m (a + bx n) p dx with ⎮ x m − n (a + bx n) p dx . ⌡ ⌡ H ere λ = m − n and μ = p (choosing smaller indices for λ and μ). Let us take P = x λ + 1 (a + bx n ) μ + 1 = x m − n + 1 (a + bx n ) p + 1. ∴ (dP ⁄ dx) = (m − n + 1) x m − n (a + bx n ) p + 1 + x m − n + 1 ( p + 1) (a + bx n ) p bnx n − 1 = (m − n + 1) x m − n (a + bx n ) p (a + bx n ) + ( p + 1) bn x m (a + bx n ) p = a (m − n + 1) x m − n (a + bx n ) p + b (m − n + 1) x m (a + bx n ) p + ( p + 1) bn x m (a + bx n ) p = a (m − n + 1) x m − n (a + bx n ) p + bx m (a + bx n ) p (m − n + 1 + pn + n) (dP ⁄ dx) = a (m − n + 1) x m − n (a + bx n ) p + b (m + pn + 1) x m (a + bx n ) p . Thus (dP ⁄ dx) is expressed as a linear combination of the two expressions whose integrals are to be connected. Integrating both the sides, we have i.e., ⌠ ⌠ P = a (m − n + 1) ⎮ x m − n (a + bx n ) p dx + b (m + np + 1) ⎮ x m (a + bx n ) p dx . ⌡ ⌡ Now putting the value of P and transposing suitably, we get ⌠ b ( pn + m + 1) ⎮ x m (a + bx n ) p dx ⌡ ⌠ = x m − n + 1 (a + bx n ) p + 1 − a (m − n + 1) ⎮ x m − n (a + bx n ) p dx ⌡ ⌠ m ⎮ x (a + bx n) p dx ⌡ or = x m − n + 1 (a + bx n) p + 1 a (m − n + 1) ⌠ m − n − ⎮x (a + bx n) p dx . b ( pn + m + 1) ⌡ b ( pn + m + 1) ⌠ ⌠ (ii) To connect ⎮ x m (a + bx n) p dx with ⎮ x m (a + bx n) p − 1 dx . ⌡ ⌡ H ere λ = m and μ = p − 1 ; (choosing smaller indices for λ and μ). Now let P = x λ + 1 (a + bx n ) μ + 1 = x m + 1 (a + bx n ) p. ∴ dP = (m + 1) x m (a + bx n ) p + x m + 1. p (a + bx n ) p − 1 bn x n − 1 dx = (m + 1) x m (a + bx n ) p + pn . x m (a + bx n ) p − 1 bx n = (m + 1) xm (a + = ( pn + m + 1) bx n ) p xm (a + + pn bx n ) p .xm (a + − a pn bx n ) p − 1 xm (a + (a + (Note) bx n bx n ) p − 1. − a) REDUCTION FORMULAE (For Irrational Algebraic and Transcendental Functions) I-33 Thus (dP ⁄ dx) has been expressed as a linear combination of the two expressions whose integrals are to be connected. Integrating both the sides, we have ⌠ ⌠ P = ( pn + m + 1) ⎮ x m (a + bx n ) p dx − apn ⎮ x m (a + bx n ) p − 1 dx . ⌡ ⌡ Now putting the value of P , dividing by ( pn + m + 1) and transposing suitably, we get anp x m + 1 (a + bx n) p ⌠ m ⌠ m ⎮ x (a + bx n) p − 1 dx . ⎮ x (a + bx n) p dx = + ⌡ np + m + 1 np + m + 1 ⌡ (iii) ⌠ ⌠ To connect ⎮ x m (a + bx n) p dx with ⎮ x m + n (a + bx n) p dx . ⌡ ⌡ H ere λ = m and μ = p . Now let ∴ P = x λ + 1 (a + bx n ) μ + 1 = x m + 1 (a + bx n ) p + 1. dP = x m + 1 ( p + 1) (a + bx n ) p nb x n − 1 + (m + 1) x m (a + bx n ) p + 1 dx = bn ( p + 1) x m + n (a + bx n ) p + (m + 1) x m (a + bx n ) p (a + bx n ) (Note) = bn ( p + 1) x m + n (a + bx n ) p + (m + 1) ax m (a + bx n ) p + b (m + 1) x m + n (a + bx n ) p xm + n bx n ) p = b (np + n + m + 1) (a + + (m + 1) ax m (a + bx n ) p, which is linear combination of the two expressions whose integrals are to be connected. Integrating both sides, we have ⌠ P = b (np + n + m + 1) ⎮ x m + n (a + bx n ) p dx ⌡ ⌠ + (m + 1) a ⎮ x m (a + bx n ) p dx . ⌡ get Now putting the value of P , dividing by a (m + 1) and transposing suitably, we x m + 1 (a + bx n) p + 1 ⌠ m ⎮ x (a + bx n) p dx = ⌡ a (m + 1) − (iv) b (np + n + m + 1) ⌠ m + 1 ⎮x (a + bx n) p dx . ⌡ a (m + 1) ⌠ ⌠ To connect ⎮ x m (a + bx n) p dx with ⎮ x m (a + bx n) p − 1 dx . ⌡ ⌡ H ere λ = m and μ = p , λ being the lesser index of x , and μ being the lesser index of (a + bx n ) in both the integrals. Now let ∴ P = x λ + 1 (a + bx n ) μ + 1 = x m + 1 (a + bx n ) p + 1. dP = x m + 1 ( p + 1) (a + bx n ) p. nb x n − 1 + (m + 1) x m (a + bx n ) p + 1 dx = n ( p + 1) x m (a + bx n ) p bx n + (m + 1) x m (a + bx n ) p + 1 (Note) I-34 INTEGRAL CALCULUS = n ( p + 1) x m (a + bx n ) p (a + bx n − a) + (m + 1) x m (a + bx n ) p + 1 = n ( p + 1) x m (a + bx n ) p + 1 − an ( p + 1) x m (a + bx n ) p + (m + 1) x m (a + bx n ) p + 1 = (np + n + m + 1) x m (a + bx n ) p + 1 − an ( p + 1) x m (a + bx n ) p i.e., (dP ⁄ dx) is a linear combination of the two expressions whose integrals are to be connected. Integrating both the sides, we have ⌠ P = {n ( p + 1) + m + 1)} ⎮ x m (a + bx n ) p + 1 dx ⌡ ⌠ − an ( p + 1) ⎮ x m (a + bx n ) p + 1 dx . ⌡ Now putting the value of P , dividing by {an ( p + 1)} and transposing suitably, ⌠ m ⎮ x (a + bx n) p dx we get ⌡ = − ⌠ ⌠ To connect ⎮ x m (a + bx n) p dx with ⎮ x m − n (a + bx n) p + 1 dx . ⌡ ⌡ (v) H ere ∴ ∴ x m + 1 (a + bx n) p + 1 (np + n + m + 1) ⌠ m ⎮ x (a + bx n) p + 1 dx . + ⌡ an (n + 1) an ( p + 1) λ = m − n and μ = p , P= xλ+ 1 (a + bx n ) μ + 1 [as also in case (i)]. = xm − n + 1 (a + bx n ) p + 1. dP = x m − n + 1 ( p + 1) (a + bx n ) p. nb x n − 1 dx + (m − n + 1) x m − n (a + bx n ) p + 1 = bn ( p + 1) x m (a + bx n ) p + (m − n + 1) x m − n (a + bx n ) p + 1 ∴ = a linear combination of the two expressions whose integrals are to be connected. integrating both sides, we have ⌠ ⌠ P = bn ( p + 1) ⎮ x m (a + bx n ) p dx + (m − n + 1) ⎮ x m − n (a + bx n ) p + 1 dx ⌡ ⌡ ⌠ ⌠ bn ( p + 1) ⎮ x m (a + bx n ) p dx = P − (m − n + 1) ⎮ x m − n (a + bx n ) p + 1 dx . ⌡ ⌡ or Now putting the value of P and dividing by bn ( p + 1) , we get x m − n + 1 (a + bx n) p + 1 ⌠ m ⎮ x (a + bx n) p dx = ⌡ bn ( p + 1) − (vi) (m − n + 1) ⌠ m − n ⎮x (a + bx n) p + 1 dx . bn ( p + 1) ⌡ ⌠ ⌠ To connect ⎮ x m (a + bx n) p dx with ⎮ x m + n (a + bx n) p − 1 dx . ⌡ ⌡ H ere λ = m and μ = p − 1 . REDUCTION FORMULAE (For Irrational Algebraic and Transcendental Functions) I-35 P = x λ + 1 (a + bx n ) μ + 1 = x m + 1 (a + bx n ) p. ∴ dP = x m + 1. p (a + bx n ) p − 1 bn x n − 1 + (m + 1) x m (a + bx n ) p dx And = bp nx m + n (a + bx n ) p − 1 + (m + 1) x m (a + bx n ) p. Thus (dP ⁄ dx) is expressed as a linear combination of the two expressions whose integrals are to be connected. Integrating both sides, we have ⌠ ⌠ P = b pn ⎮ x m + n (a + bx n ) p − 1 dx + (m + 1) ⎮ x m (a + bx n ) p dx ⌡ ⌡ ⌠ ⌠ (m + 1) ⎮ x m (a + bx n ) p dx = P − bn p ⎮ x m + n (a + bx n ) p − 1 dx . ⌡ ⌡ or Now putting the value of P and dividing by (m + 1) , we get ⌠ m ⎮ x (a + bx n) p dx ⌡ = bnp ⌠ m + n x m + 1 (a + bx n) p ⎮x − (a + bx n) p − 1 dx . (m + 1) ⌡ (m + 1) Example 1 : If In denotes ∫ x p (1 − x q) n dx , where p, q and n are positive, prove 0 that (nq + p + 1) In = nq In − 1. 1 Hence evaluate In when n is a positive integer. Solution : H ere we have to connect 1 1 ⌠ ⌠ ⎮ x p (1 − x q) n dx with ⎮ x p (1 − x q) n − 1 dx . ⌡0 ⌡0 ∴ H ere λ = lesser index of x = p ; μ = lesser index of (1 − x q) = n − 1 . ∴ P = x λ + 1 (1 − x q) μ + 1 = x p + 1 (1 − x q) n . H ence dP = ( p + 1) x p (1 − x q) n + x p + 1. n(1 − x q) n − 1. (− qx q − 1) dx = ( p + 1) x p (1 − x q) n + nqx p (1 − x q) n − 1. (− x q) = ( p + 1) x p (1 − x q) n + nqx p (1 − x q) n − 1. {(1 − x q)− 1} = ( p + 1) xp (1 − = ( p + 1+ nq) xp x q) n nqx p + (1 − x q) n − (1 − nqx p x q) n nqx p − (1 − x q) n − 1. (1 − (Note) x q) n − 1 Thus (dP ⁄ dx) is expressed as a linear combination of the two expressions whose integrals are to be connected. Therefore integrating both sides, we have ⌠ ⌠ P = ( p + 1 + nq) ⎮ x p (1 − x q) n dx − nq ⎮ x p(1 − x q) n − 1 dx . ⌡ ⌡ I-36 INTEGRAL CALCULUS 1 ∴ ⌠ ( p + 1 + nq) ⎮ x p (1 − x q) n dx ⌡ 0 1 1 ⌠ = ⎡⎣ x p + 1 (1 − x q) n ⎤⎦ + nq ⎮ x p (1 − x q) n − 1 dx , 0 ⌡0 putting the value of P , transposing and also putting the limits of integration 1 ⌠ = 0 + nq ⎮ x p (1 − x q) n − 1 dx . ⌡0 Thus (qn + p + 1) In = nq In − 1 In = or = ...(1) Proved. nq ⋅ I qn + p + 1 n − 1 nq (n − 1) q ⎡ ⋅ ⎢ I qn + p + 1 ⎣ {(n − 1) q + p + 1} n − ⎤ 2⎥⎦ , putting (n − 1) for n in (1) to get In − 1 in terms of In − 2 . Proceeding similarly by successive reduction, we have finally (n − 1) q nq q … In = ⋅ I . nq + p + 1 (n − 1) q + p + 1 q+ p+ 1 0 1 1 But ⌠ ⌠ I0 = ⎮ x p (1 − x q) 0 dx = ⎮ x p dx = ⌡0 ⌡ 0 ∴ In = ⎡⎢ x p + ⎢⎢⎣ p + 1⎤ 1 ⎥⎥ = 1 ⎥⎦ 0 1 ⋅ p+ 1 (n − 1) q nq q 1 … ⋅ ⋅ ⋅ nq + p + 1 (n − 1) q + p + 1 q+ p+ 1 p+ 1 a ⌠ Example 2 : If In denotes ⎮ (a2 − x2) n dx , and n > 0 , prove that ⌡0 In = 2na2 I . 2n + 1 n − 1 (Avadh 2005; Kanpur 06) a ⌠ Hence evaluate ⎮ (a2 − x2) 3 dx. ⌡0 a ⌠ In = ⎮ (a2 − x2) n .1 dx ⌡0 Solution : We have (Note) a ⌠ a = ⎡⎣ (a2 − x2) n . x⎤⎦ 0 − ⎮ n (a2 − x2) n − 1 (− 2x) . x dx , ⌡0 integrating by parts taking unity as the secnod function a ⌠ = 0 + 2n ⎮ (a2 − x2) n − 1 x2 dx , ⌡0 a [... n > 0] ⌠ = − 2n ⎮ (a2 − x2) n − 1 . {(a2 − x2) − a2} dx , ⌡0 [... x2 = − {(a2 − x2) − a2}] REDUCTION FORMULAE (For Irrational Algebraic and Transcendental Functions) a I-37 a ⌠ ⌠ = − 2n ⎮ (a2 − x2) n dx + 2na2 ⎮ (a2 − x2) n − 1 dx ⌡0 ⌡0 = − 2n In + 2na2 In − 1 . (1 + 2n) In = 2na2 In − 1 ∴ or ∴ In = 2na2 I 2n + 1 n − 1 I3 = 6 7 a2 I2 , putting n = 3 in (1) = 6 7 a2. [ a2 I1] , putting n = 2 in (1) to get I2 in terms of I1 5 = 24 4 a I1 . 35 ...(1) 4 a Thus Proved. a ⌠ 24a4 ⌠ ⎮ (a2 − x2) dx I3 = ⎮ (a2 − x2) 3 dx = ⌡0 35 ⌡0 a = 2.2 24a4 ⎡ 2 x3 ⎤ 24a4 ⎡ 3 a3 ⎤ 16a7 ⎢a x − ⎥ = ⎢a − ⎥ = ⋅ 35 ⎣ 3 ⎦0 35 ⎣ 3⎦ 35 ⌠ dx , where n is Positive Reduction Formulae for ⎮ 2 ⌡ (x + a2) n (Bundelkhand 2010; Meerut 12) 1 ⌠ Let In = ⎮ 2 dx . To form a reduction formula for In , we shall integrate ⌡ (x + a2) n 1 ⌠ by parts ⎮ 2 dx , taking unity as the second function. Thus ⌡ (x + a2) n − 1 − (n − 1) 1 x ⌠ ⌠ ⎮ 2 ⋅ 1 dx = 2 − ⎮ x⋅ 2 ⋅ 2x dx 2 n − 1 ⌡ (x + a2) n − 1 ⌡ (x + a ) (x + a2) n x ⌠ x2 In − 1 = 2 + 2 (n − 1) ⎮ 2 dx 2 n − 1 ⌡ (x + a ) (x + a2) n or = = (x2 x ⌠ (x2 + a2) − a2 + 2 (n − 1) ⎮ dx 2 n − 1 ⌡ (x2 + a2) n + a ) (Note) x ⌠ 1 + 2 (n − 1) ⎮ 2 dx ⌡ (x + a2) n − 1 (x2 + a2) n − 1 ⌠ 1 − 2 (n − 1) a2 ⎮ 2 dx ⌡ (x + a2) n x + 2 (n − 1) In − 1 − 2 (n − 1) a2 In . (x2 + a2) n − 1 x + (2n − 2 − 1) In − 1 2 (n − 1) a2 In = 2 (x + a2) n − 1 = ∴ I-38 INTEGRAL CALCULUS In = or 2n − 3 x + I , 2a2 (n − 1) (x2 + a2) n − 1 2a2 (n − 1) n− 1 which is the required reduction formula. m 2 2.3 Reduction Formulae for ⌠ ⎮ x √(2ax − x ) dx ; m being a Positive Integer ⌡ (Bundelkhand 2010) Let ⌠ ⌠ Im = ⎮ x m √(2ax − x2) dx = ⎮ x m + 1 ⁄ 2 √(2a − x) dx . ⌡ ⌡ Integrating by parts taking √(2a − x) as the 2nd function, we have (2a − x) 3 ⁄ 2 ⌠ ⎛⎜ (2a − x) 3 ⁄ 2 1⎞ Im = x m + 1 ⁄ 2 3 − ⎮ ⎝ m + ⎟⎠ x m − 1 ⁄ 2 3 dx ⌡ 2 ( ) . (− 1) ( ) . (− 1) 2 2 2m + 1 ⌠ m − 1 ⁄ 2 2 ⎮x = − x m − 1 x3 ⁄ 2 (2a − x) 3 ⁄ 2 + (2a − x) √(2a − x) dx ⌡ 3 3 = − 2m + 1 ⌠ 2 m− 1 ⎮ 2ax m − 1 ⁄ 2 √(2a − x) dx x (2ax − x2) 3 ⁄ 2 + ⌡ 3 3 − = − (2m + 1) 2a ⌠ m − 1 1 ⁄ 2 2 m− 1 x (2ax − x2) 3 ⁄ 2 + ⎮x x √(2a − x) 1 ⁄ 2 dx ⌡ 3 3 − = − 2m + 1 ⌠ m − 1 ⁄ 2 ⎮x . x √(2a − x) dx ⌡ 3 2m + 1 ⌠ m − 1 ⁄ 2 1 ⁄ 2 1 ⁄ 2 ⎮x . x x (2a − x) 1 ⁄ 2 dx ⌡ 3 2 (2m + 1) a ⌠ m − 1 2 m− 1 ⎮x x (2ax − x2) 3 ⁄ 2 + √(2ax − x2) dx ⌡ 3 3 − = − 2m + 1 ⌠ m ⎮ x √(2ax − x2) dx ⌡ 3 2 (2m + 1) 2m + 1 2 m− 1 x (2ax − x2) 3 ⁄ 2 + 2Im − 1 − Im . 3 3 3 Transposing the last term to the left, we have 2m + 1 ⎞ 2(2m + 1) a 2 m− 1 ⎛ x (2ax − x2) 3 ⁄ 2 + Im − 1 ⎜1 + ⎟ I = − ⎝ 3 ⎠ m 3 3 or 2 (m + 2) 2 (2m + 1) a 2 Im = − x m − 1 (2ax − x2) 3 ⁄ 2 + Im − 1 3 3 3 or Im = − x m − 1 (2ax − x2) 3 ⁄ 2 (2m + 1) a + Im − 1 , m+ 2 m+ 2 which is the required reduction formula. I-40 INTEGRAL CALCULUS (b) ⌠ x n ⎮ (a ⁄ x ) dx . ⌡ ⌠ ⌠ We have ⎮ (a x ⁄ x n ) dx = ⎮ a x x− n dx . ⌡ ⌡ Now integrate by parts taking x− n as the 2nd function. ⌠ 2.6 Reduction Formulae for ⎮ xm (log x) n dx . ⌡ Integrating by parts regarding x m as the 2nd function, we get ⌠ m xm + 1 n ⌠ m ⎮ x (log x) n dx = (log x) n . ⎮ x (log x) n − 1 dx , − ⌡ m+ 1 m+ 1⌡ which is the required reduction formula. ∞ ⌠ Example 1 : Evaluate ⎮ e− x x n dx , n being a positive integer. ⌡0 (Rohilkhand 2007; Kanpur 10, 12) Solution : Integrating by parts regarding e− x as the 2nd function, we get ∞ ⌠ ⎮ ⌡0 ∞ ∞ ⌠ e− x x n dx = ⎡⎣ − x n e− x⎤⎦ 0 + ⎮ ⌡0 ne− x x n − 1 dx . ∞ lim lim x n Now x → ∞ x n e− x = x → ∞ x , which is of the form ∞ ⋅ e ∴ differentiating the numerator and the denominator separately, we get lim x n lim nx n − 1 lim n (n − 1) … 1 = … = x →∞ = 0. x →∞ x = x →∞ x ex e e H ence ⎡ − x n e− x⎤ ∞ = − lim x n e− x − 0 = 0 − 0 = 0 . ⎣ ⎦0 x →∞ ∞ ⌠ Therefore ⎮ ⌡0 ∞ ⌠ e− x x n dx = n ⎮ e− x x n − 1 dx . ⌡0 ...(1) Now applying the reduction formula (1) repeatedly, we get ∞ ⌠ ⌠ ⎮ e− x x n dx = n (n − 1) (n − 2) … 2 . 1 ⎮ ⌡0 ⌡ ∞ e− x x0 dx 0 ∞ ∞ ⌠ = n ! ⎮ e− x dx = n ! ⎡⎣ − e− x⎤⎦ 0 ⌡0 ⎡ ⎢⎣ = n ! ⎢⎢ − 1 ⎤⎥ ∞ ⎥ = (n !) .1 = n ! . e x⎥⎦ 0 REDUCTION FORMULAE (For Irrational Algebraic and Transcendental Functions) I-41 1 ⌠ Example 2 : Evaluate ⎮ x m (log x) n dx , when m ≥ 0 and n is an integer ≥ 0 . ⌡0 1 ⌠ Solution : Let Im, n = ⎮ x m (log x) n dx . ⌡0 Integrating by parts taking x m as the second function, we have ⎡x Im, n = ⎢⎢⎢ ⎣ = 1 m+ 1 1 (log x) n ⎤⎥ n ⌠ ⎮ x m (log x) n − 1 dx ⎥⎥ − m+ 1 m + 1 ⌡0 ⎦0 1 ⎡ m+ 1 lim ⎤ (log 1) n − x → 0 x m + 1 (log x) n ⎥ ⎢1 m+ 1⎣ ⎦ 1 − n ⌠ ⎮ x m (log x) n − 1 dx . m + 1 ⌡0 But log 1 = 0 . Also lim (log x) n lim m + 1 (log x) n = x → 0 − ( m + 1) , x →0 x x [form ∞ ⁄ ∞ ] lim n (log x) n − 1 . (1 ⁄ x) = x →0 − (m + 1) x− (m + 2) n ⎞ (log x) n − 1 lim ⎛ = x →0 ⎜ − ⋅ ⎟ ⎝ m + 1 ⎠ x − (m + 1) Proceeding in this way, we ultimately have lim m + 1 (log x) n x →0 x 1 lim = (some number) × x → 0 − ( m + 1) x lim = some number × x → 0 x m + 1 = 0 . ∴ n I (m + 1) m, n − 1 n − 1⎞ n ⎛ ⋅ ⎜− , applying (1) = − ⎟ I (m + 1) ⎝ m + 1 ⎠ m, n − 2 Im, n = − n (n − 1) I = …= … (m + 1) 2 m, n − 1 Proceeding similarly by successive application of (1), we have ultimately n (n − 1) (n − 2) … 2 .1 Im, n = (− 1) n Im, 0 . (m + 1) n = (− 1) 2 1 But 1 ⌠ ⌠ Im, 0 = ⎮ x m (log x) 0 dx = ⎮ x m dx ⌡0 ⌡0 1 ⎡ x m + 1 ⎤⎥ 1 ⎥ = = ⎢⎢⎢ ⋅ m+ 1 ⎣ m + 1 ⎥⎦ 0 ...(1) I-42 INTEGRAL CALCULUS ∴ Im, n = (− 1) n ⋅ = (− 1) n n! 1 ⋅ (m + 1) n (m + 1) n! ⋅ (m + 1) n + 1 ⌠ Example 3 : Find the reduction form ula for ⎮ {x m ⁄ (log x) n } dx . ⌡ Solution : We have ⌠ 1⎤ 1 ⌠ xm ⎡ ⎮ dx = ⎮ x m + 1 ⎢ ⋅ ⎥ dx n ⌡ ⌡ (log x) n x⎦ ⎣ (log x) ⌠ 1⎤ ⎡ = ⎮ x m + 1 ⋅ ⎢ (log x) − n ⎥ dx . ⌡ x⎦ ⎣ (Note) Now integrating by parts regarding x m + 1 as the first function, we have (log x) − n + 1 ⌠ (log x) − n + 1 ⌠ x m dx ⎮ = xm + 1 − ⎮ (m + 1) x m dx ⌡ (log x) n ⌡ − n+ 1 − n+ 1 = − m+ 1⌠ xm xm + 1 ⎮ + dx , n − 1 n − 1 ⌡ (log x) n − 1 (n − 1) (log x) which is the required reduction formula. 1. Prove the reduction formula x (a2 + x2) n ⁄ 2 ⌠ na2 ⌠ ⎮ (a2 + x2) n ⁄ 2 dx = + ⎮ (a2 + x2) (n ⁄ 2) − 1 dx . ⌡ (n + 1) (n + 1) ⌡ ⌠ H ence evaluate ⎮ (x2 + a2) 5 ⁄ 2 dx . ⌡ 2. ⌠ Find a reduction formula for ⎮ x m (1 + x2) n ⁄ 2 dx , ⌡ where m and n are positive integers. ⌠ H ence evaluate ⎮ x5 (1 + x2) 7 ⁄ 2 dx . ⌡ ⌠ x m dx , 3. If Im, n = ⎮ prove that ⌡ (1 + x2) n 2 (n − 1) Im, n = − x m − 1 (x2 + 1) 1 − n + (m − 1) Im − 2, n − 1 . x ⌠ x n dx , 4. If φ (n) = ⎮ prove that ⌡0 √(x − 1) (2n + 1) φ (n) = 2 x n √(x − 1) + 2 n φ (n − 1) . (Bundelkhand 2005) REDUCTION FORMULAE (For Irrational Algebraic and Transcendental Functions) 5. I-43 ∞ ⌠ If In denotes ⎮ ⌡0 1 dx, (a2 + x2) n where n is a positive integer ≥ 2 , prove that 2n − 3 I . In= 2 a2 (n − 1) n− 1 ∞ ⌠ H ence or otherwise evaluate ⎮ ⌡0 2a ⌠ 6. If Im = ⎮ ⌡0 1 dx. (a2 + x2) 4 x m √(2ax − x2) dx , prove that 2m m ! . (m + 2) ! Im = am + 2 (2m + 1) ! π . 2a ⌠ H ence or otherwise evaluate ⎮ ⌡0 x3 √(2ax − x2) dx . (Bundelkhand 2007; 10) ⌠ 7. If In = ⎮ x n (a − x) 1 ⁄ 2 dx , prove that ⌡ (2n + 3) In = 2an In − 1 − 2x n (a − x) 3 ⁄ 2 . a ⌠ H ence evaluate ⎮ x2 √(ax − x2) dx . ⌡0 ⌠ 8. If un = ⎮ x n (a2 − x2) 1 ⁄ 2 dx , prove that ⌡ un = − x n − 1 (a2 − x2) 3 ⁄ 2 n − 1 2 + a un − 2 . n+ 2 n+ 2 a ⌠ H ence evaluate ⎮ x4 √(a2 − x2) dx . ⌡0 9. ∞ ⌠ n! Show that ⎮ e− ax x n dx = n + 1 , ⌡0 a where a is a positive quantity and n is a positive integer. 1 10 ⌠ E valuate ⎮ (log x) 4 x m dx . ⌡0 11. If m and n are positive integers, and 1 ⌠ f (m , n) = ⎮ x n − 1 (log x) m dx , ⌡0 prove that f (m , n) = − (m ⁄ n) f (m − 1, n) . Deduce that f (m , n) = (− 1) m . m ! ⁄ nm + 1. ∞ ⌠ x 12. E valuate ⎮ dx . ⌡0 (1 + e x) (Bundelkhand 2011) I-44 1. 2. 5. 8. INTEGRAL CALCULUS x (a2 + x2) 5 ⁄ 2 5a2 5a4 ⎡ x⎤ + x (a2 + x2) 3 ⁄ 2 + ⎢ x √(a 2 + x2) + a 2 sin− 1 ⎥ ⋅ 6 24 16 ⎣ a⎦ ⎡ ⎤ 4 2 8 1 (1 + x2) 9 ⁄ 2 ⎢⎣ x4 − x (1 + x2) + (1 + x2) 2⎥⎦ ⋅ 11 143 9 5π ⋅ 32 a7 πa6 ⋅ 32 6. 10. 7πa5 ⋅ 8 24 ⋅ (m + 1) 5 7. 12. 5πa4 ⋅ 128 π2 ⋅ 12 Euler’s Integrals : Beta and Gamma Functions : 3.1 Beta Function (Meerut 2012; Kashi 13, 14) Definition : The definite integral 1 ⌠ ⎮ x m − 1 (1 − x) n − 1 dx, for m > 0, n > 0 ⌡0 is called the Beta function and is denoted by B (m, n) [read as “Beta m , n” ] . 1 Thus ⌠ B (m, n) = ⎮ x m − 1 (1 − x) n − 1 dx, ⌡0 where m, n are any positive numbers, integral or fractional. Beta function is also called the Eulerian integral of the first kind. I-46 INTEGRAL CALCULUS 3.2 Elementary Properties of Beta Function (Meerut 2012B; Lucknow 11) (i) Symmetry of Beta function i.e., B (m, n) = B (n, m) : 1 ⌠ We have B (m, n) = ⎮ x m − 1 (1 − x) n − 1 dx, by the def. of the Beta function ⌡0 1 ⌠ = ⎮ (1 − x) m − 1 {1 − (1 − x)}n − 1 dx ⌡0 a ⎡ . . ⌠a ⎤ ⎢ . ⎮ f (x) dx = ⌠ ⎮ f (a − x) dx⎥⎥ ⎢ ⌡0 ⌡0 ⎢ ⎥ ⎣ ⎦ 1 1 ⌠ ⌠ = ⎮ (1 − x) m − 1 x n − 1 dx = ⎮ x n − 1 (1 − x) m − 1 dx ⌡0 ⌡0 = B (n, m), by the def. of Beta function. H ence B (m, n) = B (n, m). (ii) If m or n is a positive integer, B (m, n) can be evaluated in an explicit form: Case I : When n is a positive integer : If n = 1, the result is obvious because 1 1 ⌠ ⌠ B (m, 1) = ⎮ x m − 1 (1 − x) 1 − 1 dx = ⎮ x m − 1 dx = ⌡0 ⌡0 ⎡⎢ x m ⎤⎥ 1 1 ⎢⎢ ⎥ = ⋅ m ⎣ m ⎥⎦ 0 So let us take n > 1. We have 1 ⌠ B (m, n) = ⎮ x m − 1 (1 − x) n − 1 dx ⌡0 1 ⌠ = ⎮ (1 − x) n − 1 x m − 1 dx ⌡0 1 1 ⎡ x m ⎤⎥ ⌠ xm ⎥⎥ − ⎮ (n − 1) (1 − x) n − 2 (− 1) ⋅ dx , = ⎢⎢⎢ (1 − x) n − 1 ⋅ m ⎦ 0 ⌡0 m ⎣ integrating by parts taking x m − 1 as the second function 1 = 0+ n− 1 ⌠ ⋅ ⎮ x m (1 − x) n − 2 dx ⌡0 m [ ... n > 1] 1 = n − 1 ⌠ (m + 1) − 1 ⋅ ⎮ x (1 − x) (n − 1) − 1 dx ⌡0 m = n− 1 B (m + 1, n − 1). m By the repeated application of this process, we get B (m, n) = n − 1 n − 2 n − 3 ...... 1 ⋅ ⋅ B (m + n − 1, 1) m+ 1 m+ 2 m+ n− 2 m I-47 BETA AND GAMMA FUNCTIONS 1 = n − 1 n − 2 n − 3 ...... 1 ⌠ ⎮ x m + n − 2 (1 − x) 0 dx ⋅ ⋅ m m+ 1 m+ 2 m + n − 2 ⌡0 = n − 1 n − 2 n − 3 ...... 1 ⌠ ⎮ x m + n − 2 dx ⋅ ⋅ m+ 1 m+ 2 m + n − 2 ⌡0 m = ⎡ x m + n − 1 ⎤⎥ 1 (n − 1) ! ⎥ ⋅ ⋅ ⎢⎢⎢ m (m + 1) (m + 2) ...... (m + n − 2) ⎣ m + n − 1 ⎥⎦ 0 1 ∴ B (m, n) = 1 ⋅ m (m + 1) (m + 2) ..... (m + n − 2) (m + n − 1) Case II : When m is a positive integer : Since the Beta function is symmetrical in m and n i.e., B (m, n) = B (n, m), therefore by case I, we conclude that (m − 1) ! ⋅ B (m, n) = n (n + 1) (n + 2) ...... (n + m − 2) (n + m − 1) (iii) If both m and n are positive integers, then B (m, n) = (m − 1) ! (n − 1) ! ⋅ (m + n − 1) ! From (ii), we have B (m, n) = = (n − 1) ! m (m + 1) (m + 2) ...... (m + n − 2) (m + n − 1) (n − 1) ! (m − 1) ! , (m + n − 1) (m + n − 2) ...... (m + 1) m (m − 1) ! writing the denominator in the reversed order and multiplying the Nr and Dr by (m − 1) ! = (m − 1) ! (n − 1) ! ⋅ (m + n − 1) ! Example 1 : Express the following integrals in term s of Beta function : 1 ⌠ (i) ⎮ x m (1 − x2) n dx, m > − 1, n > − 1 ; ⌡0 1 (Lucknow 2010) x2 ⌠ (ii) ⎮ dx ⌡0 √(1 − x5) (Meerut 2013B) 1 ⌠ (iii) ⎮ x m − 1 (1− x2) n − 1 dx . ⌡0 Solution : (i) 1 (Garhwal 2003) We have 1 ⌠ ⌠ ⎮ x m (1 − x2) n dx = ⎮ x m − 1 (1 − x2) n . x dx ⌡0 ⌡0 [ Note] I-48 INTEGRAL CALCULUS 1 ⌠ dy , = ⎮ y (m − 1) ⁄ 2 (1 − y) n ⋅ putting x2 = y so that 2x dx = dy ⌡0 2 1 = 1⌠ ⎮ y (m − 1) ⁄ 2 (1 − y) n dy 2 ⌡0 = 1⌠ ⎮ y [(m + 1) ⁄ 2] − 1 (1 − y) (n + 1) − 1 dy 2 ⌡0 = 1 2 1 ⎛ ⎞ 1 2 B ⎜⎝ (m + 1), n + 1⎟⎠ ⋅ 1 (ii) 1 x2 ⌠ ⌠ We have ⎮ dx = ⎮ x2 (1 − x5) − 1 ⁄ 2 dx ⌡0 √(1 − x5) ⌡0 1 1 ⌠ = ⎮ x2 ⋅ 4 (1 − x5) − 1 ⁄ 2 . x4 dx ⌡0 x 1 ⌠ = ⎮ x− 2 (1 − x5) − 1 ⁄ 2 x4 dx ⌡0 1 ⌠ 1 = ⎮ y − 2 ⁄ 5 (1 − y) − 1 ⁄ 2 ⋅ dy, putting x5 = y so that 5x4 dx = dy 5 ⌡0 1 = 1⌠ ⎮ y − 2 ⁄ 5 (1 − y) − 1 ⁄ 2 dy 5 ⌡0 = 1⌠ ⎮ y (3 ⁄ 5) − 1 (1 − y) (1 ⁄ 2) − 1 dy 5 ⌡0 = 1 ⎛3 , B⎜ 5 ⎝5 1 1⎞ ⎟ ⋅ 2⎠ (iii) Proceed as in part (i). Example 2 : Prove that a a m+ n− 1 Γ m Γ n ⌠ ⎮ (a − x) m − 1 . x n − 1 dx = a m + n − 1 B (m, n) = ⋅ ⌡0 Γ (m + n) Solution : We have a ⌠ ⎮ (a − x) m − 1 x n − 1 dx ⌡0 1 ⌠ = ⎮ (a − ay) m − 1 (ay) n − 1 a dy, putting x = ay ⌡0 1 ⌠ = ⎮ a(m − 1) + (n − 1) + 1 (1 − y) m − 1 y n − 1 dy ⌡0 I-49 BETA AND GAMMA FUNCTIONS 1 ⌠ = am + n − 1 ⎮ y n − 1 (1 − y) m − 1 dy ⌡0 [ ... B (m, n) = B (n, m)] = am + n − 1 B (n, m) = am + n − 1 B (m, n) = a m+ n− 1 Γ m Γ n ⋅ Γ (m + n) ⎡ ⎢ ⎣ ... B (m, n) = Γ m Γ n ⎤⎥ Γ (m + n) ⎦ Example 3 : Show that if m, n are positive, then b ⌠ ⎮ (x − a) m − 1 (b − x) n − 1 dx = (b − a) m + n − 1 . B (m, n) ⌡a = (b − a) m+ n− 1 ⋅ Γm Γn ⋅ Γ (m + n) (Agra 2003; Avadh 04) Solution : The given integral is b ⌠ ⎮ (x − a) m − 1 (b − x) n − 1 dx . ⌡a Put x = a + (b − a) y so that dx = (b − a) dy. Also when x = a, y = 0 and when x = b, y = 1. b ⌠ ∴ ⎮ (x − a) m − 1 (b − x) n − 1 dx ⌡a 1 ⌠ = ⎮ [(b − a) y]m − 1 [b − a − (b − a) y]n − 1 . (b − a) dy ⌡0 1 ⌠ = ⎮ (b − a) m − 1 . y m − 1 . (b − a) n − 1 . (1 − y) n − 1 . (b − a) dy ⌡0 1 ⌠ = (b − a) m + n − 1 ⎮ y m − 1 (1 − y) n − 1 dy ⌡0 = (b − a) m + n − 1 B (m, n) = (b − a) m+ n− 1 ⋅ Example 4 : Show that Γm Γn ⋅ Γ (m + n) ⎡ ⎢ ⎣ ... B (m, n) = Γ m Γ n ⎤⎥ Γ (m + n) ⎦ 1 ⌠ x m − 1 (1 − x) n − 1 1 ⎮ dx = B (m, n) . ⌡0 (a + b) m . a n (a + bx) m + n Solution : The given integral 1 ⌠ x m − 1 (1 − x) n − 1 I= ⎮ dx ⌡0 (a + bx) m + n 1 1 ⌠ ⎛ x ⎞m− 1 ⎛ 1− x⎞n− 1 ⋅ ⎜ ⋅ dx . = ⎮ ⎜ ⎟ ⎟ ⌡0 ⎝ a + bx⎠ ⎝ a + bx⎠ (a + bx) 2 [ Note] I-50 INTEGRAL CALCULUS Put y (a + bx) . 1 − x . b x dy = so that dx = 2 a + bx a + b a + b (a + bx) 1 dy dx = ⋅ a (a + b) (a + bx) 2 i.e., Further 1− x x (a + b) ⎤ 1− y 1 a − ax 1 ⎡ a + bx − ax − bx⎤ 1⎡ = = ⎢ ⋅ ⎥ = ⎢1 − ⎥ = ⎦ a + bx a a + bx a ⎣ a⎣ a + bx ⎦ a a + bx Also when x = 0, y = 0 and when x = 1, y = 1. ∴ 1 ⌠ ⎛ y I= ⎮ ⎜ ⌡0 ⎝ a + dy ⎞ m − 1 ⎛ 1 − y⎞ n − 1 ⋅ ⎟ ⎜ ⎟ a (a + b) a ⎝ ⎠ b⎠ 1 ⌠ ⎮ y m − 1 (1 − y) n − 1 dy ⌡0 = 1 (a + b) m . a n = B (m, n) ⋅ (a + b) m . an B (m + 1, n) m = ⋅ B (m, n) m+ n Solution : We have B (m + 1, n) = B (n, m + 1) [By the symmetry of Beta function] Example 5 : Prove that 1 ⌠ = ⎮ x n − 1 (1 − x) (m + 1) − 1 dx ⌡0 1 ⌠ = ⎮ (1 − x) m x n − 1 dx ⌡0 1 [ Note] 1 ⎡ x n ⎤⎥ xn ⌠ ⎥⎥ − ⎮ m (1 − x) m − 1 (− 1) ⋅ = ⎢⎢⎢ (1 − x) m ⋅ dx , ⌡ n ⎦0 n ⎣ 0 (integrating by parts) = 0+ m n 1 ⌠ ⎮ x n − 1. x (1 − x) m − 1 dx ⌡0 1 = m n ⌠ ⎮ x n − 1 [1 − (1 − x)] (1 − x) m − 1 dx ⌡0 = m n 1 ⎡⌠1 ⎤ ⎢ ⎮ x n − 1 (1 − x) m − 1 dx − ⌠ ⎮ x n − 1 (1 − x) m dx⎥⎥ ⎢⌡ ⌡0 ⎢ 0 ⎥ ⎣ ⎦ = m ⎡ B (n, m) − B (n, m + 1) ⎤⎥ n ⎢⎣ ⎦ = m m B (m, n) − B (m + 1, n) n n or ⎛ m⎞ m B (m, n) ⎜1 + ⎟ B (m + 1, n) = ⎝ n⎠ n or (n + m) B (m + 1, n) = m B (m, n) [By transposition] I-51 BETA AND GAMMA FUNCTIONS B (m + 1, n) m = ⋅ B (m, n) m+ n or 3.3 Another form of Beta Function ∞ xm − 1 dx , m > 0 , n > 0 . (1 + x) m + n ⌠ B (m, n) = ⎮ ⌡0 Proof : By the definition of Beta function, we have 1 ⌠ B (m, n) = ⎮ x m − 1 (1 − x) n − 1 dx . ⌡0 Put x = 1 1 so that dx = − dy . 1+ y (1 + y) 2 Also when x → 0 , y → ∞ and when x = 1 , y = 0 . 0 1 1 ⎤n − 1 ⎡ 1 ⌠ ⎡ ⎤ ⋅ 1 − ⋅ ⎢− ∴ B (m, n) = ⎮ ⎢ ⎥ ⎥ dy ⌡∞ (1 + y) m − 1 ⎣ 1 + y⎦ ⎣ (1 + y) 2 ⎦ ∞ yn − 1 ⌠ ⋅ dy = ⎮ m − 1 + 2 n − 1 ⌡0 (1 + y) (1 + y) ∞ xn − 1 dx . (1 + x) m + n ⌠ = ⎮ ⌡0 ⌠ = ⎮ ⌡0 ∞ 1 yn − 1 dy (1 + y) m + n [By a property of definite integrals] …(1) Again since Beta function is symmetrical in m and n , we have ∞ ⌠ B (m, n) = B (n, m) = ⎮ ⌡0 Thus ∞ xm − 1 dx , by (1). (1 + x) m + n ∞ xm − 1 ⌠ dx = ⎮ m + n ⌡0 (1 + x) ⌠ B (m, n) = ⎮ ⌡0 ∞ ⌠ Example 1 : Prove that ⎮ ⌡0 xn − 1 dx , m > 0 , n > 0 . (1 + x) m + n xm − 1 − xn − 1 dx = 0 , m > 0 , n > 0 . (1 + x) m + n Solution : The given integral is ∞ ⌠ = ⎮ ⌡0 ∞ xm − 1 ⌠ dx − ⎮ ⌡0 (1 + x) m + n xn − 1 dx (1 + x) m + n = B (m, n) − B (n, m) = B (m, n) − B (m, n) = 0 . ∞ ⌠ Example 2 : Express ⎮ ⌡0 xm − 1 dx in term s of Beta function where (a + bx) m + n m > 0, n > 0, a > 0, b > 0. Solution : In the given integral put bx = ay i.e., x = (a ⁄ b) y so dx = (a ⁄ b) dy . When x = 0 , y = 0 and when x → ∞ , y → ∞ . that I-52 INTEGRAL CALCULUS ∴ ∞ ⌠ ⎮ ⌡0 ∞ xm − 1 ⌠ ⎛ a ⎞m − 1 1 a ⎮ dx = ⋅ ⋅ dy ⎜ y⎟ ⌡0 ⎝ b ⎠ (a + ay) m + n b (a + bx) m + n ∞ ⌠ = ⎮ ⌡0 = am − 1 y m − 1 a 1 dy = n m m − 1 m + n m + n a b .a (1 + y) b b ∞ ⌠ ⎮ ⌡0 1 B (m, n) . a n bm ym − 1 dy (1 + y) m + n [By article 3.3] 3.4 Gamma Function (Gorakhpur 2006; Meerut 12; Kashi 13, 14; Kanpur 14) The definite integral ∞ ⌠ ⎮ e− x x n − 1 dx , for n > 0 ⌡0 is called the Gamma Function and is denoted by Γ (n) [read as “Gamma n”]. Thus ∞ ⌠ Γ (n) = ⎮ e− ∞ x n − 1 dx , for n > 0 . ⌡0 G amma function is also called Eulerian integral of the second kind. 3.5 Elementary Properties of Gamma Function To prove that (i) Γ (n + 1) = n Γ (n) , where n > 0 and(ii) Γ (n) = (n − 1) ! , where n is a positive integer. (Gorakhpur 2006; Kashi 14) Proof : By the definition of gamma function, we have ∞ ∞ ⌠ ⌠ Γ (n + 1) = ⎮ e− x x(n + 1) − 1 dx = ⎮ x n e− x dx ⌡0 ⌡0 ∞ ∞ ⌠ = ⎡⎣ − e− x x n ⎤⎦ 0 + ⎮ e− x. nx n − 1 dx , ⌡0 ...(1) integrating by parts taking e− x as the second function. xn lim lim x n Now x →∞ x = x →∞ e 1 + x + (x2 ⁄ 2 !) + … + (x n ⁄ n !) + … 1 1 lim = x →∞ = ∞ = 0. 1 1 1 x + + …+ + + ...... n ! (n + 1) ! xn xn − 1 ∞ ⌠ ∴from (1), we get Γ (n + 1) = 0 + n ⎮ e− x x n − 1 dx ,[ ... n > 0] ⌡0 = n Γ (n) , which proves the result (i). (ii) We have Γ (n) = Γ [(n − 1) + 1] = (n − 1) Γ (n − 1) . [... Γ (n + 1) = n Γ (n)] Similarly Γ (n − 1) = (n − 2) Γ (n − 2) , … etc . I-53 BETA AND GAMMA FUNCTIONS H ence if n is a + ive integer, then proceeding as above, we get Γ (n) = (n − 1) (n − 2) … 2.1 Γ (1) . ∞ ∞ ∞ ⌠ ⌠ ⌠ Γ (1) = ⎮ e− x x1 − 1 dx = ⎮ e− x x0 dx = ⎮ e− x. 1 dx ⌡0 ⌡0 ⌡0 But ∞ ⎡ e− x ⎤⎥ ⎡ lim 1 ⎤ ⎥⎥ = − ⎢⎢ x → ∞ = ⎢⎢⎢ − e0⎥⎥⎥ = − [0 − 1] = 1 . x ⎢ e ⎣ − 1⎦ 0 ⎣ ⎦ H ence Γ (n) = (n − 1) (n − 2) …2.1.1 = (n − 1) ! if n is a + ive integer. Remember : Γ (n) = (n − 1) Γ (n − 1) , where n > 1 and Γ (1) = 1 . Also it may be remarked that Γ (0) = ∞ and Γ (− n) = ∞ where n is a positive integer. 3.6 Some Transformations of Gamma Function ∞ ⌠ We have Γ (n) = ⎮ x n − 1 e− x dx . ⌡0 (i) ...(1) Put x = ay so that dx = a dy ; when x = 0 , y = 0 and when x → ∞ , y → ∞ . ∞ ∴ ⌠ Γ (n) = ⎮ e− ay an y n − 1 dy . ⌡0 H ence Γ (n) ⌠ ⎮ e− ay y n − 1 dy = ⋅ ⌡0 an (ii) ∞ (Remember) (Kanpur 2006) In (1) if we put x = log (1 ⁄ y) or y = 0 (iii) 1 (Kanpur 2006) In (1) if we put x n = y so that nx n − 1 dx = dy , we get Γ (n) = 3.7 so that dy = − e− x dx , 1⎞ n − 1 1⎞ n − 1 ⌠ ⎛ ⌠ ⎛ Γ (n) = − ⎮ ⎜ log ⎟ dy = ⎮ ⎜ log ⎟ dy . ⌡1 ⎝ ⌡0 ⎝ y⎠ y⎠ then or e− x 1 n ∞ − ( y) 1 ⁄ n ⌠ ⎮ e dy ⌡0 ∞ − ( y) 1 ⁄ n ⌠ dy = n Γ (n) = Γ (n + 1) . ⎮ e ⌡0 Relation Between Beta and Gamma Functions To prove that B (m , n) = Γ (m) Γ (n) , where m > 0 , n > 0 . Γ (m + n) (Kanpur 2009, 12; Kashi 13, 14; Rohilkhand 13) Proof : We have ∞ Γ (m) ⌠ = ⎮ e− z x x m − 1 dx . ⌡0 zm [See article 3.6, part (ii)] I-54 INTEGRAL CALCULUS ∴ ∞ ∞ ⌠ ⌠ Γ (m) = z m ⎮ e− z x. x m − 1 dx = ⎮ z m e− z x x m − 1 dx . ⌡0 ⌡0 Multiplying both sides by e− z zn − 1, we get ∞ ⌠ Γ (m) e− z z n − 1 = ⎮ e− z (1 + x) z m + n − 1 x m − 1 dx . ⌡0 ...(1) Now integrating both sides of (1) with respect to z from 0 to ∞ , we get ∞ ∞ ⎡ ⌠ ⌠ Γ (m) ⎮ e− z z n − 1 dz = ⎮ ⌡0 ⌡0 ∞ ⎡ ∞ ⌠ ⌠ Γ (m) Γ (n) = ⎮ ⌡0 or ∞ ⌠ = ⎮ ⌡0 ⎢⎮ ⎢⌡ ⎢ 0 ⎣ ∞ ⎢ ⌠ ⎢ ⎮ ⎢ ⌡0 ⎣ ⎤ e− z (1 + x) z m + n − 1 x m − 1 dx⎥⎥ dz ⎥ ⎦ ⎤ e− z (1 + x) z m + n − 1 dz⎥⎥ x m − 1 dx ⎥ ⎦ Γ (m + n) m − 1 x dx (1 + x) m + n ∞ ⌠ = Γ (m + n) ⎮ ⌡0 [By article 3.6, part (ii)] xm − 1 dx (1 + x) m + n = Γ (m + n) . B (m, n) , by § 3 . ∴ B (m, n) = Γ (m) Γ (n) ⋅ Γ (m + n) 1 Thus remember that Γ (m) Γ (n) ⌠ ⎮ x m − 1 (1 − x) n − 1 dx = ⋅ ⌡0 Γ (m + n) Corollary : Γ (n) Γ (1 − n) = π , where 0 < n < 1 . sin n π ∞ ⌠ Proof : We know that B (m, n) = ⎮ ⌡0 B (m, n) = and ∴ (Kanpur 2009) (Avadh 2008) xn − 1 dx , [See § 3] (1 + x) m + n Γ (m) Γ (n) , where m > 0 and n > 0 . Γ (m + n) ∞ Γ (m) Γ (n) ⌠ = ⎮ ⌡0 Γ (m + n) xn − 1 dx . (1 + x) m + n Putting m + n = 1 or m = 1 − n in the above relation, we get ∞ Γ (1 − n) Γ (n) ⌠ x n − 1 = ⎮ dx , where 0 < n < 1 . ⌡0 1 + x Γ (1) [Note that m > 0 ⇒ 1 − n > 0 ⇒ n < 1 .] But Γ (1) = 1 . Also ∞ ⌠ ⎮ ⌡0 ∴ π xn − 1 dx = ⋅ 1+ x sin nπ Γ (n) Γ (1 − n) = π , where 0 < n < 1 . sin nπ [ Remember it] I-55 BETA AND GAMMA FUNCTIONS ⎛ ⎞ 3.8 The Value of Γ ⎜⎝ 12 ⎟⎠ 1 2 To prove that Γ ( ) = √π . (Agra 2000; Meerut 12) Γ (m) Γ (n) Proof : We know that B (m, n) = Γ (m + n) If we take m = 1 2 ...(1) , n = 1 , then from (1), we have 2 1 1 1 2 Γ ( ) Γ ( ) [Γ ( )]2 ⎡ ⎛ 1 1⎞ ⎛ 1⎞ ⎤ 2 2 2 ⎢Γ ⎜ ⎟⎥ ⋅ B ⎜⎝ , ⎟⎠ = = = ⎣ ⎝ 2⎠ ⎦ 1 1 2 2 Γ (1) Γ( + ) 2 Thus 1 [ Γ ( )]2 2 = B 1 ( 2 , [... Γ (1) = 1] 2 1 ) 2 1 ⌠ = ⎮ x1 ⁄ 2 − 1 (1 − x) 1 ⁄ 2 − 1 dx , ⌡0 by the definition of Beta function 1 ⌠ = ⎮ x− 1 ⁄ 2 (1 − x) − 1 ⁄ 2 dx . ⌡0 Now put x = sin2 θ so that dx = 2 sin θ cos θ dθ . Also when x = 0 , θ = 0 and when x = 1 , θ = ∴ π ⁄2 ⌠ 1 [Γ ( )]2 = ⎮ 2 ⌡0 π. 1 1 ⋅ ⋅ 2 sin θ cos θ dθ sin θ cos θ π ⁄2 ⌠ = 2⎮ ⌡0 1 2 π ⁄2 dθ = 2 ⎡⎢⎣ θ ⎤⎥⎦ 0 1 2 = 2 [ π − 0] = π . Taking square root of both the sides, we get 1 2 Γ ( ) = √π . (Remember) ∞ 2 √π ⌠ ⋅ Important Deduction : To prove that ⎮ e− x dx = ⌡0 2 Proof : Let ∞ 2 ⌠ I = ⎮ e− x dx . ⌡0 Put x2 = z so that 2x dx = dz or 1 1 1 dz = dz = z− 1 ⁄ 2 dz . 2 √z 2 2x Also when x = 0 , z = 0 and when x → ∞ , z → ∞ . dx = ∴ ∞ ∞ √π 1 1⌠ 1 ⎛ 1⎞ ⌠ ⋅ I = ⎮ e− z z− 1 ⁄ 2 dz = ⎮ e− z z1 ⁄ 2 − 1 dz = Γ ⎜⎝ ⎟⎠ = ⌡0 2 2 ⌡0 2 2 2 I-56 INTEGRAL CALCULUS H ence ∞ 2 √π ⌠ ⎮ e− x dx = ⋅ ⌡0 2 (Remember) ⌠π ⁄ 2 cos m θ sin n θ dθ in terms of Γ, 3.9 Value of ⎮ ⌡0 where m > − 1 , n > − 1 ⌠ ⎮ ⌡0 Proof : ⎛ m + 1 ⎞ ⎛⎜ n + 1 ⎞⎟ ⎟ Γ ⎝ 2 ⎠ ⎜⎜ 2 ⎟⎟ ⎝ ⎠ Γ⎜ π ⁄2 cos m θ sin n θ dθ = ⎛ m + n + 2⎞ ⎟ 2 ⎝ ⎠ Put sin2 θ = x so that 2 sin θ cos θ dθ = dx 2 sin θ. √(1 − sin2 θ) dθ = dx or or ∴ 2 x1 ⁄ 2. √(1 − x) dθ = dx . dx ⋅ 2 x1 ⁄ 2 (1 − x) 1 ⁄ 2 dθ = Also when θ = 0 , x = 0 and when θ = ∴ ⋅ 2Γ ⎜ π ⁄2 1 π,x= 2 π ⁄2 ⌠ ⎮ ⌡0 ⌠ cosm θ sinn θ dθ = ⎮ ⌡0 1. (1 − sin2 θ) m ⁄ 2. sin n θ dθ 1 ⌠ dx = ⎮ (1 − x) m ⁄ 2. x n ⁄ 2. 1 ⁄ 2 ⌡0 2 x (1 − x) 1 ⁄ 2 1 = 1 ⌠ (n − 1) ⁄ 2 ⎮ x (1 − x) (m − 1) ⁄ 2 dx 2 ⌡0 = 1 ⌠ {(n + 1) ⁄ 2} − 1 ⎮ x (1 − x) {(m + 1) ⁄ 2} − 1 dx 2 ⌡0 = 1 ⎛ m + 1 , n + 1⎞ , B⎜ provided m > − 1 and n > − 1 ⎟ 2 ⎝ 2 2 ⎠ 1 1 1 Γ (m + 1) Γ (n + 1) 1 2 2 = 2 Γ 1 (m + 1 + n + 1) 2 = 3.10 1 2 1 2 Γ (m + 1) Γ (n + 1) 1 2 2 Γ (m + n + 2) Γ (m) Γ (n) ⎤ ⎡ .. ⎢ . B (m, n) = ⎥ ⎣ Γ (m + n) ⎦ ⋅ Some Important Transformations of Beta Function Beta function can be transformed into many other forms. A few of them are given below. I-57 BETA AND GAMMA FUNCTIONS (i) ∞ ⌠ ym − 1 We know that ⎮ dy = B (m, n) . ⌡0 (1 + y) m + n ∞ ∞ 1 ym − 1 ⌠ ym − 1 ⌠ ym − 1 ⌠ ⎮ ⎮ ⎮ dy = dy + dy . ⌡0 (1 + y) m + n ⌡1 (1 + y) m + n ⌡0 (1 + y) m + n Now Making the substitution y = 1 ⁄ x in the last integral, we get ∞ 1 ym − 1 ⌠ x n − 1 dx ⌠ ⎮ dy = ⎮ ⋅ ⌡0 (1 + x) m + n ⌡1 (1 + y) m + n ∞ ⌠ B (m, n) = ⎮ ⌡0 ∴ ym − 1 dy (1 + y) m + n 1 1 1 1 ⌠ ym − 1 ⌠ xn − 1 = ⎮ dy + ⎮ dx ⌡0 (1 + y) m + n ⌡0 (1 + x) m + n xm − 1 ⌠ xn − 1 ⌠ dx + ⎮ dx = ⎮ ⌡0 (1 + x) m + n ⌡0 (1 + x) m + n 1 ⌠ xm − 1 + xn − 1 = ⎮ dx . ⌡0 (1 + x) m + n 1 H ence (ii) Γ (m) Γ (n) ⌠ xm − 1 + xn − 1 ⎮ dx = B (m , n) = ⋅ ⌡0 (1 + x) m + n Γ (m + n) ∞ xm − 1 dx = B (m, n) . (1 + x) m + n ⌠ We know that ⎮ ⌡0 a ay , so that dx = dy , we get b b If we put x = ∞ ∞ ∞ ⌠ xm − 1 1 ⎮ dx = m n B (m, n) . ⌡0 (1 + x) m + n a b xm − 1 ⌠ ym − 1 ⌠ ⎮ dx = am bn ⎮ dy . m + n ⌡0 (ay + b) m + n ⌡0 (1 + x) ∴ ⌠ ym − 1 1 ⎮ dy = m n ⌡0 (ay + b) m + n a b H ence ⌠ ⎮ ⌡0 ∞ ∞ Γ (m) Γ (n) ym − 1 dy = m n ⋅ a b Γ (m + n) (ay + b) m + n Again putting y = tan2 θ i.e., dy = 2 tan θ sec 2 θ dθ in the integral just obtained, we get π ⁄2 ⌠ ⎮ ⌡0 sin2 m − 1 θ cos2 n − 1 θ dθ Γ (m) .Γ (n) = ⋅ 2 2 m + n 2 am b n Γ (m + n) (a sin θ + b cos θ) 1 (iii) ⌠ We know that ⎮ x m − 1 (1 − x) n − 1 dx = B (m, n) . ⌡0 Putting x = sin2 θ , so that dx = 2 sin θ cos θ dθ , we have I-58 INTEGRAL CALCULUS π ⁄2 1 ⌠ ⌠ ⎮ x m − 1 (1 − x) n− 1 dx = 2 ⎮ ⌡0 ⌡0 π ⁄2 ⌠ ⎮ ⌡0 ∴ sin2 m − 1 θ cos2 n − 1 θ dθ . B (m , n) Γ (m) Γ (n) sin2 m − 1 θ cos2 n − 1 θ dθ = = ⋅ 2 Γ (m + n) 2 This result may also be written in the form ⎛ p + 1⎞ ⎛q+ Γ⎜ ⎟ ⋅ Γ ⎜⎜ ⎝ 2 ⎠ π ⁄2 ⎜⎝ 2 ⌠ ⎮ sin p θ cos q θ dθ = ⌡0 ⎛ p + q + 2⎞ 2Γ ⎜ ⎟ 2 ⎝ ⎠ by putting 2m − 1 = p and 2n − 1 = q . 1 ⎞⎟ ⎟⎟ ⎠ , 1 (iv) ⌠ We know that ⎮ y m − 1 (1 − y) n − 1 dy = B (m, n) . ⌡0 Putting y = x− b , dx , so that dy = we have a− b a− b 1 a ⌠ ⎛ x − b⎞m − ⌠ ⎮ y m − 1. (1 − y) n − 1 dy = ⎮ ⎜ ⎟ ⌡b ⎝ a − b ⎠ ⌡0 = a − x⎞ n − 1 dx ⋅ ⎜ ⎟ a− b ⎝ a − b⎠ 1⎛ a ⌠ ⎮ (x − b) m − 1 (a − x) n − 1 dx . (a − b) m + n − 1 ⌡b 1 a 1 ⌠ ⌠ ∴ ⎮ (x − b) m − 1 (a − x) n − 1 dx = (a − b) m + n − 1 ⎮ y m − 1 (1 − y) n − 1 dy ⌡b ⌡0 a ⌠ ⎮ (x − b) m − 1 (a − x) n − 1 dx = (a − b) m + n − 1 B ( m, n) ⌡b or Γ (m) Γ (n) = (a − b) m + n − 1 ⋅ Γ (m + n) 3.11 Duplication Formula ⎛ To prove that Γ (m) Γ ⎜⎝ m + √π 1 ⎞⎟ = 2 m − 1 Γ (2 m) , where m > 0 . 2⎠ 2 (Agra 2001, 03; Kanpur 09; Rohilkhand 13, 13B; Avadh 13; Purvanchal 14) Proof : We know that Γ (m) Γ (n) , where m > 0 , n > 0 . Γ (m + n) If we take n = m , then B (m, n) = B (m, m) = [Γ (m)]2 ⋅ Γ (2 m) Again by the definition of Beta function, we have ...(1) I-59 BETA AND GAMMA FUNCTIONS 1 ⌠ B (m, m) = ⎮ x m − 1 (1 − x) m − 1 dx . ⌡0 Let us put x = sin2 θ so that dx = 2 sin θ cos θ dθ . Also when x = 0 , θ = 0 and when x = 1 , θ = Then π ⁄2 ⌠ B (m, m) = ⎮ ⌡0 1 2 π. sin2 (m − 1) θ .cos2 (m − 1) θ .2 sin θ cos θ dθ π ⁄2 ⌠ sin2m − 1 θ .cos2m − 1 θ dθ = 2 ⎮ ⌡0 π ⁄2 π ⁄2 1 ⌠ ⎛ sin 2θ⎞ 2m − 1 dθ = 2m − 2 ⎮ sin2m − 1 2θ dθ ⎜ ⎟ ⌡0 ⎝ 2 ⎠ 2 ⌠ = 2⎮ ⌡0 ⌠ = 2⎮ ⌡0 π ⁄2 (sin θ cos θ) 2m − 1 dθ π = dφ , ⌠ 1 ⎮ sin2m − 1 φ ⋅ putting 2θ = φ so that dθ = dφ 2m − 2 2 ⌡ 2 2 0 = ⌠ 1 ⌠ ⎮ sin2m − 1 φ dφ = 2m − 1 ⋅ 2 ⎮ ⌡0 22m − 1 ⌡0 2 = ⌠ ⎮ 2m − 2 ⌡0 2 = = 1 π 1 π ⁄2 π ⁄2 1 sin2m − 1 φ dφ sin2m − 1 φ .cos0 φ dφ 1 2 1 2 [ Note] 1 Γ (2m − 1 + 1) Γ (0 + 1) 22m − 2 2 Γ (2m − 1 + 0 + 2) 1 2 [ Note] = 1 1 2 1 ) 2 Γ (m) Γ ( ) ⋅ 22m − 1 Γ (m + Γ (m) √π 1 ⋅ ⋅ 22m − 1 Γ (m + 1 ) ...(2) 2 [... Γ ( ) = √π] 1 2 Now equating the two values of B (m, m) obtained in (1) and (2), we get [Γ (m)]2 Γ (m) .√π 1 = 2m − 1 ⋅ 1 Γ (2m) 2 Γ (m + ) 2 Γ (m) Γ (m + or ⎛ 1⎞ ⎟ ⎝ n⎠ Value of Γ ⎜ 3.12 1 ) 2 = ⎛ 2⎞ ⎟ ⎝ n⎠ Γ⎜ √π Γ (2 m) . 22 m − 1 ⎛ 3⎞ ⎟ ⎝ n⎠ Γ⎜ ⎛n − ⎝ n …Γ⎜ (Remember) 1⎞ , ⎟ ⎠ where n is a Positive Integer Let ⎛ 1⎞ ⎛ 2⎞ ⎛ 3⎞ ⎛ n − 1⎞ A = Γ⎜ ⎟ Γ⎜ ⎟ Γ⎜ ⎟ …Γ⎜ ⎟⋅ ⎝ n⎠ ⎝ n⎠ ⎝ n⎠ ⎝ n ⎠ Writing the above expression in the reverse order, we have ...(1) I-60 INTEGRAL CALCULUS ⎛ ⎝ A = Γ ⎜1 − n − 2⎞ ⎛ n − 1⎞ 1⎞ ⎛ 2⎞ ⎛ ⎟ Γ ⎜1 − ⎟ … Γ ⎜1 − ⎟ Γ ⎜1 − ⎟⋅ n⎠ ⎝ n⎠ n ⎠ ⎝ n ⎠ ⎝ ...(2) Multiplying (1) and (2), we get n − 1⎞ 1⎞ ⎛ 2⎞ ⎛ 2⎞ ⎛ 1⎞ ⎛ ⎛ n − 1⎞ ⎛ A 2 = Γ ⎜ ⎟ Γ ⎜ 1 − ⎟ Γ ⎜ ⎟ Γ ⎜ 1 − ⎟ …… Γ ⎜ ⎟ Γ ⎜1 − ⎟ n⎠ ⎝ n⎠ ⎝ n⎠ n ⎠ ⎝ n⎠ ⎝ ⎝ n ⎠ ⎝ πn − 1 . [See corollary of article 3.7] π 2π n− 1 sin sin … sin π n n n To calculate this expression, we factorize 1 − x2n . = ...(3) Now the roots of the equation x2n − 1 = 0 are given by x = (1) 1 ⁄ 2n = (cos 2 r π + i sin 2 r π) 1 ⁄ 2n rπ rπ , + i sin where r = 0 , 1, 2 , … , 2n − 1 . n n H ence, we have 1 − x2n π π⎞ ⎛ π π⎞ ⎛ = (1 − x) (1 + x) ⎜ x − cos − i sin ⎟ ⎜ x − cos + i sin ⎟ … n n⎠ ⎝ n n⎠ ⎝ = cos ⎛ ⎝ … ⎜ x − cos n− 1 n− 1 ⎞ ⎛ n− 1 n− 1 ⎞ π − i sin π ⎟ ⎜ x − cos π + i sin π⎟ ⎠ ⎝ ⎠ n n n n ⎛ ⎝ = (1 − x2) ⎜ 1 − 2x cos π 2π ⎞ ⎞ ⎛ + x2⎟ ⎜ 1 − 2x cos + x2⎟ … ⎠ n n ⎠ ⎝ ⎛ ⎝ … ⎜ 1 − 2x cos ∴ 1− 1− n− 1 ⎞ π + x2⎟ ⋅ ⎠ n π 2π x2n ⎛ ⎞ ⎞ ⎛ = ⎜ 1 − 2x cos + x2⎟ ⎜ 1 − 2x cos + x2⎟ … n n ⎠ ⎝ ⎠ ⎝ x2 ⎛ ⎝ … ⎜ 1 − 2x cos n− 1 ⎞ π + x2⎟ ⋅ n ⎠ Putting x = 1 and x = − 1 respectively, we have in the limit, π ⎛ 2π ⎞ n− 1 ⎞ ⎛ n = ⎛⎜ 2 − 2 cos ⎞⎟ ⎜ 2 − 2 cos ⎟ … ⎜ 2 − 2 cos π⎟ n⎠ ⎝ n⎠ n ⎝ ⎠ ⎝ and or π ⎛ 2π ⎞ n− 1 ⎞ ⎛ n = ⎛⎜ 2 + 2 cos ⎞⎟ ⎜ 2 + 2 cos ⎟ … ⎜ 2 + 2 cos π⎟ ⋅ n⎠ ⎝ n⎠ n ⎝ ⎠ ⎝ Multiplying these, we get π 2π n− 1 n2 = 22n − 2 sin2 ⋅ sin2 … sin2 π n n n π 2π n− 1 π. n = 2n − 1 .sin ⋅ sin … sin n n n H ence, from (3), we get πn − 1 (2π) n − 1 (2π) (n − 1) ⁄ 2 = or A = ⋅ A2 = n n ⁄ 2n − 1 n1 ⁄ 2 I-61 BETA AND GAMMA FUNCTIONS Remark : The value of sin π 2π n− 1 sin … sin π can also be found by using n n n the trigonometrical identity π⎞ 2π ⎞ 3θ⎞ sin nθ ⎛ ⎛ ⎛ = 2n − 1 sin ⎜ θ + ⎟ sin ⎜ θ + ⎟ sin ⎜ θ + ⎟ … n⎠ n⎠ n⎠ sin θ ⎝ ⎝ ⎝ ⎛ ⎝ … sin ⎜ θ + n− 1 ⎞ π⎟ ⋅ n ⎠ From the above identity, we have sin nθ θ π⎞ 2π ⎞ n− 1 ⎞ ⎛ ⎛ ⎛ ⋅ ⋅ n = 2n − 1 sin ⎜ θ + ⎟ sin ⎜ θ + π⎟ ⋅ ⎟ … sin ⎜ θ + ⎝ ⎝ ⎠ nθ sin θ n⎠ n⎠ n ⎝ Taking limit as θ → 0 , we get π 2π 3π n− 1 n = 2n − 1 sin sin sin … sin π. n n n n 3.13 ⌠∞ Values of the Integrals ⎮ e− ax cos bx . x m − 1 dx ⌡0 ⌠ ∞ − ax sin bx . x m − 1 dx and ⎮ e ⌡0 We have ∞ ∞ Γ (m) ⌠ ⌠ ⎮ e− ax e ibx x m − 1 dx = ⎮ e− (a − ib) x x m − 1 dx = ⌡0 ⌡0 (a − ib) m [See article 3.6, part (i)] = (a − ib) − m Γ (m) . ...(1) Let us first separate (a − ib) − m into real and imaginary parts. Put a = k cos α and b = k sin α so that α = tan− 1 (b ⁄ a) and k = √(a2 + b2) . Then (a − ib) − m = [k (cos α − i sin α)]− m = k − m (cos α − i sin α) − m = k − m (cos mα + i sin mα) , by De-Moivre’s theorem. Now from (1), we have ∞ ⌠ ⎮ e− ax e ibx x m − 1 dx = k − m (cos mα + i sin mα) Γ (m) ⌡0 or ∞ Γ (m) ⌠ ⎮ e− ax (cos bx + i sin bx) x m − 1 dx = (cos mα + i sin mα) , ⌡0 km [... e iθ = cos θ + i sin θ , by E uler’s theorem] or ∞ ∞ ⌠ ⌠ ⎮ e− ax cos bx . x m − 1 dx + i ⎮ e− ax sin bx . x m − 1 dx ⌡0 ⌡0 = Γ (m) Γ (m) cos mα + i m sin mα . km k ...(2) I-62 INTEGRAL CALCULUS E quating real and imaginary parts in (2), we get ∞ Γ (m) ⌠ ⎮ e− ax cos bx . x m − 1 dx = cos mα , ⌡0 km ∞ Γ (m) ⌠ ⎮ e− ax sin bx . x m − 1 dx = sin mα , ⌡0 km and where k = √(a2 + b2) and α = tan− 1 (b ⁄ a) . Deductions : (i) If we put a = 0 , then α = π ⁄ 2 and k = b . H ence ∞ Γ (m) mπ ⌠ ⎮ x m − 1 cos bx dx = cos ⌡0 2 bm ∞ Γ (m) mπ ⌠ ⎮ x m − 1 sin bx dx = sin ⋅ ⌡0 2 bm and If we put m = 1 , then (ii) ∞ Γ (1) k cos α ⌠ a ⎮ e− ax cos bx dx = cos α = = 2 ⌡0 k k2 a + b2 ∞ Γ (1) k sin α ⌠ b ⎮ e− ax sin bx dx = sin α = = 2 ⋅ 2 ⌡0 k k a + b2 and Example 1 : Evaluate the following integrals. ∞ x8 (1 − x6) dx , (1 + x) 24 ⌠ (i) ⎮ ⌡0 ∞ (Garhwal 2000) ⌠ (ii) ⎮ ⌡0 x4 (1 + x5) dx , (1 + x) 15 ∞ x dx ⌠ ⋅ (iii) ⎮ ⌡0 1 + x6 Solution : (i) ∞ ⌠ ⎮ ⌡0 We have ∞ x8 (1 − x6) ⌠ dx = ⎮ ⌡0 (1 + x) 24 ∞ ∞ x8 dx ⌠ − ⎮ (1 + x) 24 ⌡0 x14 dx (1 + x) 24 ∞ ⌠ x9 − 1 ⌠ x15 − 1 ⎮ = ⎮ dx − dx ⌡0 (1 + x) 9 + 15 ⌡0 (1 + x) 15 + 9 = B (9, 15) − B (15, 9) , [By article 3.3] = B (9, 15) − B (9, 15) , by symmetry of Beta function = 0. (ii) We have ∞ ∞ ∞ ⌠ x4 (1 + x5) ⌠ x4 dx ⌠ x9 dx ⎮ dx = ⎮ + ⎮ 15 15 ⌡0 (1 + x) ⌡0 (1 + x) ⌡0 (1 + x) 15 ∞ ∞ ⌠ x5 − 1 ⌠ x10 − 1 = ⎮ dx + ⎮ dx 5 + 10 ⌡0 (1 + x) ⌡0 (1 + x) 10 + 5 = B (5, 10) + B (10, 5) = B (5, 10) + B (5, 10) I-63 BETA AND GAMMA FUNCTIONS = 2 B (5, 10) = 2 = 2⋅ (iii) Γ5 Γ10 Γ15 4.3.2.1 1 = ⋅ 14.13.12.11.10 5005 ∞ x dx ⌠ ⋅ Let I = ⎮ ⌡0 1 + x6 1 Put x6 = y or x = y1 ⁄ 6 , so that dx = y− 5 ⁄ 6 dy . 6 ∴ ∞ y1 ⁄ 6. y− 5 ⁄ 6 1⌠ dy = ⎮ 6 ⌡0 1+ y ∞ y(1 ⁄ 3) − 1 1 ⎛ 1 2⎞ dy = B ⎜⎝ , ⎟⎠ , (1 ⁄ 3) + (2 ⁄ 3) 6 3 3 (1 + y) I= 1⌠ ⎮ 6 ⌡0 = 1⌠ ⎮ 6 ⌡0 ∞ 1 2 1 y− 2 ⁄ 3 dy 1+ y [By article 3.3] 1 Γ Γ Γ Γ (1 − ) π 1 1 3 1 3 3 3 = = = ⋅ 1 2 6 Γ( + ) 6 6 sin 1 π Γ1 3 3 3 π ⎤ ⎡. . ⎢ . Γn Γ (1 − n) = ⎥ sin n π ⎦ ⎣ = π π 1 1 2π ⋅ = ⋅ = ⋅ 6 (√3 ⁄ 2) 6 √3 3 √3 1 √π Γ (1 ⁄ n) dx ⌠ = ⋅ Example 2 : Show that ⎮ ⌡0 (1 − x n ) 1 ⁄ 2 n Γ (1 ⁄ n + 1 ⁄ 2) Solution : Let xn = θ i.e., x = θ so that 2 dx = sin(2 ⁄ n) − 1 θ cos θ dθ . n π ⁄2 1 Then (Lucknow 2010) sin2 ⁄ n sin2 dx 2⌠ ⌠ = ⎮ ⎮ ⌡0 √(1 − x n ) n ⌡0 = 2 n π ⁄2 ⌠ ⎮ ⌡0 sin(2 ⁄ n) − 1 θ cos θ dθ cos θ sin(2 ⁄ n) − 1 θ cos0 θ dθ 1 Γ (1 ⁄ n) Γ ( ) 2 2 = ⋅ n 2 Γ (1 ⁄ n + 1 ⁄ 2) = √π Γ (1 ⁄ n) ⋅ ⋅ n Γ (1 ⁄ n + 1 ⁄ 2) 1 ⌠ xm − 1 + xn − 1 Example 3 : Evaluate ⎮ dx . ⌡0 (1 + x) m + n (Kumaun 2002) Solution : We have 1 1 1 ⌠ x m − 1 dx ⌠ x n − 1 dx ⌠ xm − 1 + xn − 1 ⎮ ⎮ ⎮ dx = + ⋅ ⌡0 (1 + x) m + n ⌡0 (1 + x) m + n ⌡0 (1 + x) m + n ...(1) I-64 INTEGRAL CALCULUS Now in the second integral on the R .H .S. of (1), we put x = 1 ⁄ y so that dx = − (1 ⁄ y2) dy ; also when x → 0 , y → ∞ and when x = 1 , y = 1 . 1 ∴ 1 (1 ⁄ y) n − 1 ⎛ 1 ⎞ x n − 1 dx ⌠ ⌠ ⎮ = ⎮ − dy ⌡0 (1 + x) m + n ⌡∞ (1 + 1 ⁄ y) m + n ⎜⎝ y2 ⎟⎠ ∞ 1 y m + n dy ⌠ ⌠ = − ⎮ = ⎮ ⌡∞ (1 + y) m + n . y n − 1. y2 ⌡1 ym − 1 dy (1 + y) m + n ∞ (Note) b ⎡ . . ⌠b ⎤ ⎢ . ⎮ f (x) dx = ⌠ ⎮ f ( y) dy⎥⎥ ⎢ ⌡a ⌡a ⎢ ⎥ ⎣ ⎦ xm − 1 ⌠ = ⎮ dx. ⌡1 (1 + x) m + n Now from (1), we have 1 ∞ 1 ⌠ xm − 1 ⌠ xm − 1 ⌠ xm − 1 + xn − 1 ⎮ dx = ⎮ dx + ⎮ dx m + n m + n ⌡0 (1 + x) ⌡1 (1 + x) m + n ⌡0 (1 + x) ∞ ⌠ xm − 1 = ⎮ dx , by a property of definite integrals ⌡0 (1 + x) m + n = B (m, n) = Γ (m) Γ (n) ⋅ Γ (m + n) π ⁄2 ⌠ Example 4 : Show that ⎮ ⌡0 [R efer article 3.3 and 3.7] (tan x) n dx = π nπ , sec where − 1 < n < 1 . 2 2 Solution : We have π ⁄2 ⌠ ⎮ ⌡0 = π ⁄2 ⌠ tan n x dx = ⎮ ⌡0 1 2 π ⁄2 sin n x ⌠ dx = ⎮ n ⌡0 cos x 1 2 Γ (n + 1) . Γ (− n + 1) 1 2 sin n x cos− n x dx , 2 Γ (n − n + 2) where − n + 1 > 0 i.e., n < 1 and n + 1 > 0 i.e., n > − 1 = 1 2 1 2 1 2 = π 1 , 2 sin 1 (n + 1) π Γ (n + 1) Γ (1 − n) = 1 2 1 2 Γ (n + 1) Γ [1 − 2 1 2 (n + 1)] [ ... Γ (n) Γ (1 − n) = π , sin n π by corollary to article 3.7] = π 1 ⋅ 2 sin ( 1 π + 1 n π) 2 = 2 π 1 ⋅ 2 cos ( 1 n π) 2 π nπ , = sec where − 1 < n < 1 . 2 2 I-65 BETA AND GAMMA FUNCTIONS Example 5 : Prove that ∞ 2 Γ (n) ⌠ ; (i) ⎮ x2 n − 1 e− ax dx = ⌡0 2an ∞ Γ [(m + 1) ⁄ n] n ⌠ ⎮ x m e− a x dx = ; ⌡0 na(m + 1) ⁄ n (ii) 1 dx ⌠ = √π . (iii) ⎮ ⌡0 √(− log x) Solution : (i) Let ∞ ∞ 2 2 ⌠ ⌠ I = ⎮ x2 n − 1 e− ax dx = ⎮ x2 n − 2 e− a x x dx . ⌡0 ⌡0 Put ax2 = z so that 2ax dx = dz . When x = 0 , z = 0 and when x → ∞ , z → ∞ . ∞ 1 ⌠ ⎛ z⎞n − 1 − z 1 e dz = I= ⎮ ⎜ ⎟ ⌡0 ⎝ a ⎠ 2a 2an ∴ = (ii) Let ∞ ⌠ ⎮ e− z z n − 1 dz ⌡0 1 Γ (n) , by definition of G amma function. 2an ∞ ∞ n ⌠ ⌠ I = ⎮ x m e− ax dx = ⎮ ⌡0 ⌡0 xm xn − 1 n e− a x x n − 1 dx [ Note] ∞ n ⌠ = ⎮ x m − n + 1 e− a x x n − 1 dx . ⌡0 Put ax n = t so that na x n − 1 dx = dt . Also when x = 0 , t = 0 and when x →∞ , t →∞ . ∞ ⌠ ⎛ t ⎞ (m − n + 1) ⁄ n − t 1 e ⋅ dt , I= ⎮ ⎜ ⎟ ⌡0 ⎝ a ⎠ na ∴ Let ∞ = ⌠ ⎮ t {(m + 1) ⁄ n} − 1 e− t dt na . a(m − n + 1) ⁄ n ⌡0 = 1 Γ {(m + 1) ⁄ n}, by the definition of G amma function. na(m + 1) ⁄ n 1 1 (iii) ⎡ .. ⎛ t ⎞1⁄n ⎤ ⎢ . a xn = t ⇒x = ⎜ ⎟ ⎥ ⎣ ⎝ a⎠ ⎦ 1 1 dx ⌠ dx ⌠ ⌠ = ⎮ = ⎮ I= ⎮ ⌡0 √(− log x) ⌡0 √{log (1 ⁄ x)} ⌡0 1⎞ − 1 ⁄ 2 ⎛ dx . ⎜ log ⎟ x⎠ ⎝ Put log (1 ⁄ x) = y i.e., 1 ⁄ x = e y i.e., x = e− y so that dx = − e− y dy . Also when x → ∞ , y → ∞ and when x = 1 , y = 0 . ∴ ⌠0 I = − ⎮ y− 1 ⁄ 2 e− y dy ⌡∞ ∞ ⌠ 1 = ⎮ e− y y1 ⁄ 2 − 1 dy = Γ ( ) , 2 ⌡0 = √π . by the def. of G amma function I-66 INTEGRAL CALCULUS Example 6 : Evaluate the integral b ⌠ ⎮ (x − a) p (b − x) q dx , where p and q are positive integers. ⌡a b ⌠ I = ⎮ (x − a) p (b − x) q dx . ⌡a Solution : Let Put x = a cos2 θ + b sin2 θ so that dx = − 2a cos θ sin θ dθ + 2b sin θ cos θ dθ dx = 2 (b − a) cos θ sin θ dθ . i.e., Also x − a = a cos2 θ + b sin2 θ − a = b sin2 θ − a (1 − cos2 θ) = b sin2 θ − a sin2 θ = (b − a) sin2 θ and b − x = b − a cos2 θ − b sin2 θ = b (1 − sin2 θ) − a cos2 θ = (b − a) cos2 θ . To find the limits for θ , when x = a , we have a = a cos2 θ + b sin2 θ i.e., (b − a) sin2 θ = 0 i.e., sin2 θ = 0 as a ≠ b i.e., θ = 0 and when x = b , we have b = a cos2 θ + b sin 2 θ i.e., (a − b) cos2 θ = 0 i.e., cos2 θ = 0 as a ≠ b i.e., θ = π ⁄ 2 . Thus the new limits for θ are 0 to π ⁄ 2 . H ence the given integral π ⁄2 ⌠ I= ⎮ ⌡0 (b − a) p sin2 p θ .(b − a) q cos2 q θ .2 (b − a) cos θ sin θ dθ π ⁄2 ⌠ = 2 (b − a) p + q + 1 ⎮ ⌡0 = 2 (b − a) p + q + 1 sin2p + 1 θ cos2q + 1 θ dθ 1 2 1 2 Γ { (2p + 1 + 1)} Γ { (2q + 1 + 1)} 2 Γ (2p + 1 + 2q + 1 + 2) , provided 2p + 1 > − 1 and 2q + 1 > − 1 i.e., p > − 1 and q > − 1 which is so because p and q are given to be + ive integers Γ ( p + 1) Γ ( q + 1) = (b − a) p + q + 1 Γ ( p + q + 1 + 1) = (b − a) p + q + 1 p!q! , ( p + q + 1) ! because Γ (n + 1) = n ! if n is a positive integer. ⎛ 1⎞ ⎛ 2⎞ ⎛ 3⎞ ⎛ 8⎞ Example 7 : Find the value of Γ ⎜ ⎟ Γ ⎜ ⎟ Γ ⎜ ⎟ … Γ ⎜ ⎟ ⋅ ⎝ 9⎠ ⎝ 9⎠ ⎝ 9⎠ ⎝ 9⎠ (2 π) (n − 1) ⁄ 2 , ⎛ 1⎞ ⎛ 2⎞ ⎛ 3⎞ ⎛ n − 1⎞ Solution : We know that Γ ⎜ ⎟ Γ ⎜ ⎟ Γ ⎜ ⎟ … Γ ⎜ ⎟ = ⎝ n⎠ ⎝ n⎠ ⎝ n⎠ ⎝ n ⎠ n1 ⁄ 2 where n is a positive integer. I-67 BETA AND GAMMA FUNCTIONS Putting n = 9 in the above relation, we get (2 π) (9 − 1) ⁄ 2 (2 π) 4 16 4 ⎛ 1⎞ ⎛ 2⎞ ⎛ 8⎞ = = π . Γ⎜ ⎟ Γ⎜ ⎟ …Γ⎜ ⎟ = ⎝ 9⎠ ⎝ 9⎠ ⎝ 9⎠ 3 3 91 ⁄ 2 Example 8 : Show that (i) (ii) 2n Γ (n + 1 ) 2 = 1.3.5.… (2n − 1) √π , where n is a + ive integer, (Lucknow 2009) ⎛3 ⎞ ⎛3 ⎞ ⎛1 ⎞ Γ ⎜ − x⎟ Γ ⎜ + x⎟ = ⎜ − x2⎟ π sec πx , provided − 1 < 2 x < 1 . ⎝2 ⎠ ⎝2 ⎠ ⎝4 ⎠ Solution : (i) We have 1 ) 2 Γ (n + 1⎞ ⎛ 3⎞ ⎛ 3⎞ ⎟ ⎜n − ⎟ Γ ⎜n − ⎟ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 1⎞ ⎛ 3⎞ ⎛ 5⎞ 3 1 ⎛ ⎛ 1⎞ = ⎜n − ⎟ ⎜n − ⎟ ⎜n − ⎟ … ⋅ ⋅ Γ ⎜ ⎟ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 2 2 ⎝ ⎝ 2⎠ = (n − 1 ) 2 Γ (n − 1 ) 2 ⎛ ⎝ = ⎜n − 2n − 1 2n − 3 2n − 5 3 1 ⋅ ⋅ … ⋅ ⋅ √π 2 2 2 2 2 1 = n (2n − 1) (2n − 3) (2n − 5) … 3.1 .√π . 2 = ∴ (ii) 1 ) 2 2n Γ (n + = 1.3.5. … (2n − 1) √π . ⎛3 ⎞ ⎛3 ⎞ 1 1 1 1 We have Γ ⎜ − x⎟ Γ ⎜ + x⎟ = ( − x) Γ ( − x) . ( + x) Γ ( + x) 2 2 2 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎛1 ⎞ ⎛ 1 − 2 x⎞ ⎛ 1 + 2 x⎞ = ⎜ − x2⎟ Γ ⎜ ⎟ Γ⎜ ⎟ 2 ⎠ ⎝ 2 ⎠ ⎝4 ⎠ ⎝ 1 − 2 x⎞ ⎛1 ⎞ ⎛ 1 − 2 x⎞ ⎛ = ⎜ − x2⎟ Γ ⎜ ⎟ Γ ⎜1 − ⎟ 2 ⎠ ⎝ 2 ⎠ ⎝4 ⎠ ⎝ ⎛1 ⎞ = ⎜ − x2⎟ ⎝4 ⎠ π ⎛1 − 2x ⎞ sin ⎜ π⎟ 2 ⎝ ⎠ ⎛1 ⎞ = ⎜ − x2⎟ ⋅ ⎝4 ⎠ π ⎛1 ⎞ sin ⎜ π − xπ ⎟ ⎝2 ⎠ π ⎛1 ⎞ ⎛1 ⎞ = ⎜ − x2⎟ ⋅ = ⎜ − x2⎟ ⋅ π sec π x . ⎝4 ⎠ cos x π ⎝4 ⎠ Example 9 : With certain restrictions on the values of a, b, m and n , prove that ∞ ∞ 2 2 Γ (m) Γ (n) ⌠ ⌠ ⎮ ⎮ e− (a x + by ) x2 m − 1 y2 n − 1 dx dy = ⋅ ⌡0 ⌡0 4 a m bn Solution : Let us denote the given integral by I . Then ∞ ∞ 2 2 ⌠ ⌠ I = ⎮ e− ax x2 m − 1 dx × ⎮ e− by y2 n − 1 dy = I1 × I2 . ⌡0 ⌡0 To evaluate I1 , put ax2 = t so that 2ax dx = dt . ∴ ∞ ⌠ dt 1 = I1 = ⎮ e− t (t ⁄ a) (2 m − 1) ⁄ 2 ⋅ ⌡0 2 √(at) 2am ∞ Γ (m) , ⌠ ⎮ e− t t m − 1 dt = ⌡0 2am provided a and m are + ive. I-68 INTEGRAL CALCULUS Γ (n) , Similarly, I2 = provided b and n are + ive. 2bn Γm Γn ⋅ H ence I= 4am bn Example 10 : Show that the sum of the series m (m + 1) m (m + 1) (m + 2) 1 1 1 1 + m + ⋅ + ⋅ + … n+ 1 n+ 2 2! n+ 3 3! n+ 4 is Γ (n + 1) Γ (1 − m) , where − 1 < n < 1 . Γ (n − m + 2) Γ (n + 1) Γ (1 − m) = B (n + 1, 1 − m) Solution : We have Γ (n − m + 2) 1 ⌠ = ⎮ x n (1 − x) − m dx ⌡0 1 m (m + 1) 2 m (m + 1) (m + 2) 3 ⌠ ⎡ ⎤ x + x + …⎥ dx = ⎮ xn ⎢1 + mx + ⌡0 2! 3! ⎣ ⎦ 1 ⎤ m (m + 1) n + 2 m (m + 1) (m + 2) n + 3 ⌠ ⎡ x + x + …⎥ dx = ⎮ ⎢ xn + m xn + 1 + ⌡0 ⎣ 2! 3! ⎦ ⎤1 ⎡ xn + 1 m (m + 1) (m + 2) x n + 4 x n + 2 m (m + 1) x n + 3 = ⎢⎢⎢ + m + + + …⎥ 2! 3! n+ 2 n+ 3 n+ 4 ⎦0 ⎣n+ 1 = m (m + 1) m (m + 1) (m + 2) 1 1 1 1 + m. + ⋅ + + … n+ 1 n+ 2 2! n+ 3 3! n+ 4 ∞ π ⌠ sin bz dz = ⋅ Example 11 : Prove that ⎮ ⌡0 z 2 Solution : We have ∞ ∞ ⌠ ⌠ I = ⎮ ⎮ e− x z sin bz dx dz ⌡0 ⌡0 ∞ ⎡ − x z⎤ ∞ e ⌠ = ⎮ ⌡0 ∞ ⌠ = ⎮ ⌡0 ⎢⎢ ⎥ ⎢⎣ − z ⎥⎥⎦ sin bz dz , on first integrating w.r.t. x 0 sin bz dz . z ...(1) Again on first integrating w.r.t. z , we have ∞ ∞ ⌠ ⌠ I = ⎮ ⎮ e− x z sin bz dx dz ⌡0 ⌡0 ∞ ⎡ ∞ ⌠ ⌠ = ⎮ ⌡0 ∞ ⌠ = ⎮ ⌡0 ⎢⎮ ⎢⌡ ⎢ 0 ⎣ ⎤ e− x z sin bz dz⎥⎥ dx b dx , b2 + x2 ⎥ ⎦ [See article 3.13, Deduction (ii)] I-69 BETA AND GAMMA FUNCTIONS x⎤ ∞ ⎡ = ⎢ tan− 1 ⎥ b⎦ 0 ⎣ π ⋅ 2 H ence equating the two values (1) and (2) of I , we have = ∞ ⌠ ⎮ ⌡0 ...(2) π sin bz dz = ⋅ z 2 Example 12 : Show that ∞ nπ ⌠ 1 ⎮ cos (bz1 ⁄ n ) dz = n Γ (n + 1) . cos ⋅ ⌡0 2 b (Kanpur 2014; Agra 14) Solution : Put z1 ⁄ n = x i.e., z = x n , so that dz = nx n − 1 dx . ∴ ∞ ∞ ⌠ ⌠ ⎮ cos(bz1 ⁄ n ) dz = ⎮ cos (bx) . nx n − 1 dx ⌡0 ⌡0 ∞ ⌠ = n ⎮ x n − 1 cos (bx) dx ⌡0 ∞ ⌠ = real part of n ⎮ e− ibx x n − 1 dx ⌡0 = real part of n Γ (n) (ib) n = real part of n Γ (n) 1 1 ⋅ (cos π + i sin π) − n 2 2 bn = real part of Γ (n + 1) ⎛ nπ n π⎞ − i sin ⎜ cos ⎟ ⎝ 2 2 ⎠ bn 1 ⎛ n π⎞ ⋅ Γ (n + 1) .cos ⎜ ⎟⋅ ⎝ 2 ⎠ bn = ⎮ Example 13 : Show that ⌠ ⌡ 0 ∞ Γ (c + 1) , xc dx = c > 1. (log c) c + 1 cx (Gorakhpur 2005; Kanpur 07; Lucknow 10; Rohilkhand 13) Solution : We have ∞ ∞ ∞ xc xc xc ⌠ ⌠ ⎮ ⎮ dx = ⌠ dx = e− x log c x c dx . ⌡ ⌡ ⌡ x dx = ⎮ x 0 c 0 e log c 0 e x log c 0 Put x log c = y so that (log c) dx = dy . When x = 0, we have y = 0 and when x → ∞ , y → ∞ because c > 1 ⇒ log c > 0 . ∞ ∞ y ⎞ c dy 1 − y⎛ ⌠ ⎮ ⎮ e− y y(c + 1) − 1 dy ∴ I= ⌠ ⌡ ⌡ e ⎜⎝ log c ⎟⎠ log c = ( log c) c + 1 0 0 ⌠ I= ⎮ ⌡ = ∞ 1 Γ (c + 1) , ( log c) c + 1 provided c + 1 > 0 which is so because c > 1 . I-70 1. INTEGRAL CALCULUS Prove that a π dx 1 ⌠ (i) ⎮ = ⋅ ⋅ ⌡0 (an − x n ) 1 ⁄ n n sin (π ⁄ n) (Kanpur 2010) 2 2π ⌠ (ii) ⎮ (8 − x3) − 1 ⁄ 3 dx = ⋅ ⌡0 3 √3 (Kumaun 2008) 1 2. 3. Show that Γ (m) Γ (n) ⌠ x m− 1 (1 − x) n− 1 ⎮ dx = n ⋅ m+ n ⌡0 a (1 + a) m Γ (m + n) (a + x) Hint. Put x (1 + a) = y. a+ x Show that ⌠ ⎮ ⌡0 π ⁄2 sin p θ cos q θ dθ = 1 ⎛⎜⎜ p + 1 , q + 1 ⎞⎟⎟ , B p > − 1, q > − 1. 2 ⎜⎝ 2 2 ⎟⎠ 2 ⌠ 4 16 ⎛⎜ 5 , 2 ⎞⎟ ⎮ x (8 − x3) − 1 ⁄ 3 dx = B ⋅ ⌡0 3 ⎝ 3 3⎠ Deduce that 4. Prove that B (m, n) = B (m + 1, n) + B (m, n + 1) for m > 0, n > 0. (Kanpur 2005, 11; Gorakhpur 05; Bundelkhand 11; Avadh 06, 11, 14) 5. Prove that ∞ Γ (n) ⌠ ⋅ (i) ⎮ e− a x x n− 1 dx = ⌡0 an 1 1⎞ ⌠ ⎛ (ii) ⎮ ⎜⎝ log ⎟⎠ ⌡0 x 6. n− 1 dx = Γ (n). Show that, if m > − 1, then ∞ ⎛ m + 1⎞ 2 2 ⌠ 1 ⎟⋅ ⎮ x m e− n x dx = Γ⎜ ⌡0 2 nm+ 1 ⎝ 2 ⎠ 7. Prove that 1 ⎞ ⌠ 1 ⎛m + 1 , (i) ⎮ x m (1 − x n ) p dx = B ⎜⎝ p + 1⎟⎠ ⋅ ⌡0 n n 1 ⌠ 1 [Γ (1 ⁄ n)]2 (ii) ⎮ (1 − x n ) 1 ⁄ n dx = ⋅ ⌡0 n 2 Γ (2 ⁄ n) 8. Show that Γ (0⋅ 1) Γ (0⋅ 2) Γ (0⋅ 3) … Γ (0⋅ 9) = (2 π) 9 ⁄ 2 ⋅ √(10) I-71 BETA AND GAMMA FUNCTIONS 9. Show that π ⁄2 ⌠ (i) ⎮ ⌡0 π ⁄2 ⌠ √(sin θ) dθ × ⎮ ⌡0 1 dθ = π. √(sin θ) 1 π ⌠ x2 dx ⌠ dx (ii) ⎮ × ⎮ = ⋅ ⌡0 (1 − x4) 1 ⁄ 2 ⌡0 (1 + x4) 1 ⁄ 2 4 √2 π ⁄2 ⎛ 1⎞ ⎛ 3⎞ ⌠ Γ ⎜⎝ ⎟⎠ Γ ⎜⎝ ⎟⎠ = 2 ⎮ ⌡0 4 4 ∞ ⌠ √(tan θ) dθ = 4 ⎮ ⌡0 x2 dx = π √2 . 1 + x4 10. Show that 11. Show that the perimeter of a loop of the curve r n = an cos nθ is [Γ (1 ⁄ 2 n)]2 a ⋅ 2(1 ⁄ n)− 1 ⋅ ⋅ Γ (1 ⁄ n) n 1 12. Prove that √2 ⌠ dx ⎮ = ⌡0 √(1 − x4) 8 √π 13. Show that ⌠ ⎮ ⌡0 14. Show that (i) π ⁄2 sin p θ dθ = ⎡ ⎛ 1⎞ ⎤ 2 ⎢Γ ⎜ ⎟⎥ ⋅ ⎣ ⎝ 4⎠ ⎦ 1 ⎛⎜ 1 , p + 1 ⎞⎟ B ⋅ 2 ⎠ 2 ⎝2 ∞ (α2 − β2) ⌠ ⎮ xe− α x cos βx dx = ⌡0 (α2 + β2) 2 ∞ 2 αβ ⌠ (ii) ⎮ xe− α x sin βx dx = ⋅ ⌡0 (α2 + β2) 2 ∞ 15. ⎛1 ⎞ ⌠ Prove that ⎮ cos ⎜⎝ π x2⎟⎠ dx = 1. ⌡− ∞ 2 16. Show that ⎛ B (m, m) . B ⎜⎝ m + πm − 1 1, 1⎞ m + ⎟⎠ = 4m− 1 ⋅ 2 2 2 (Rohilkhand 2005) Fill in the Blanks: Fill in the blanks “……” so that the following statem ents are complete and correct. 1 1. ⌠ The definite integral ⎮ x m− 1 (1 − x) n− 1 dx, for m > 0, n > 0 is called the …… . ⌡0 2. ⌠ The definite integral ⎮ e− x x n− 1 dx, for n > 0 is called the …… . ⌡0 3. Γ (m) Γ (n) B (m, n) = ⋅ …… ∞ (Garhwal 2001) I-72 INTEGRAL CALCULUS 4. B (m + 1, n) m = ⋅ B (m, n) …… 5. ⌠ For m > 0, n > 0, ⎮ ⌡0 6. ⌠ For m > 0, n > 0, ⎮ ⌡0 7. For n > 0, Γ (n + 1) = …… Γ (n). 8. If n is a positive integer, then Γ (n) = …… . 9. If 0 < n < 1, then Γ (n) Γ (1 − n) = …… . ⎛ 1 ⎞⎟ 2⎠ ∞ x m− 1 dx = …… . (1 + x) m+ n ∞ x m− 1 − x n− 1 dx = …… . (1 + x) m+ n 10. Γ ⎜⎝ 11. If m > − 1, n > − 1, then = …… . π ⁄2 ⌠ ⎮ ⌡0 cos m θ sin n θ dθ = ⎛ m + 1⎞ ⎛ n + 1⎞ ⎟ Γ⎜ ⎟ 2 ⎠ ⎝ 2 ⎠ Γ ⎜⎝ …… ∞ 12. 2 ⌠ ⎮ e− x dx = …… . ⌡0 13. For m > 0, Γ (m) Γ ⎜⎝ m + 14. Γ (n) ⌠ For a > 0, n > 0, ⎮ e− a x x n− 1 dx = ⋅ ⌡0 …… 15. The value of Γ ( )⋅ Γ ( ) is …… . ⎛ √π 1 ⎞⎟ = Γ (2 m). ⎠ 2 …… ∞ 1 3 2 3 (Agra 2002) Multiple Choice Questions: 16. 17. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). For m > 0, n > 0, (a) B (m, n) = Γ (m) Γ (n) (b) B (m, n) = Γ (m) Γ (n) Γ (m + n) (c) B (m, n) = Γ (n) Γ (m) (d) B (m, n) = Γ (m + n) Γ (m) Γ (n) ∞ ⌠ The value of the integral ⎮ e− x x− 1 ⁄ 2 dx is ⌡0 (a) √π 2 (c) √π (b) π 2 (d) π (Kumaun 2008) I-73 BETA AND GAMMA FUNCTIONS 18. For m > 0, n > 0, 1 ⌠ x m− 1 (b) B (m, n) = ⎮ dx ⌡0 (1 + x) m+ n 1 1 ⌠ (d) B (m, n) = ⎮ ⌡0 ⌠ x m− 1 + x n− 1 dx (a) B (m, n) = ⎮ ⌡0 (1 + x) m+ n ∞ ⌠ x n− 1 (c) B (m, n) = ⎮ dx ⌡0 (1 + x) m+ n 19. xm dx (1 + x) m+ n ∞ ⌠ If a > 0 and n > 0, then the value of the integral ⎮ e− ax x n− 1 dx is ⌡0 (b) a− n Γ (n) (a) an Γ (n) (c) Γ (n) 2 an (d) Γ (n) n2 1 20. 21. ⌠ The value of ⎮ x4 (1 − x) 3 dx is ⌡0 (a) 1 280 (b) 1 180 (c) 1 380 (d) 1 80 π ⁄2 ⌠ The value of the integral ⎮ ⌡0 sin4 x cos2 x dx is π 8 π (d) 32 π 4 π (c) 16 (b) (a) 22. ∞ ⌠ The value of ⎮ ⌡0 (Garhwal 2001) x m − 1 dx is (1 + x) m + n (a) Γ (m) + Γ (n) (b) Γ (m) Γ (n) (c) Γ (m) . Γ (n) (d) Γ (m) Γ (n) Γ (m + n) 1 23. (Garhwal 2001) ⎛ ⌠ 1⎞ If m, n > 0, then the value of ⎮ x n − 1 ⎜⎝ log ⎟⎠ ⌡0 x (a) Γ (m) nm (b) Γ (n) nm (c) Γ (n) mn (d) Γ (m) mm (Garhwal 2002) m− 1 dx is equal to (Garhwal 2003) I-74 INTEGRAL CALCULUS True or False: Write ‘T ’ for true and ‘F ’ for false statem ent. ∞ 24. ⎛ 1⎞ ⌠ ⎮ e− x x1 ⁄ 2 dx = Γ ⎜⎝ ⎟⎠ ⋅ ⌡0 2 25. Γ ⎜⎝ 26. ⌠ ⎮ ⌡0 27. ⌠ ⎮ ⌡0 28. ⌠ For m > 0, n > 0, B (m, n) = ⎮ x m (1 − x) n dx. ⌡0 29. ⌠ For m > 0, n > 0, B (m, n) = ⎮ ⌡0 30. ⌠ For m > 0, n > 0, ⎮ ⌡0 31. Γ (6) = 120. 32. For m > 0, n > 0, B (m, n) = 33. 2 √π ⌠ ⎮ e− x dx = ⋅ ⌡0 4 34. B (m + 1, n) + B (m, n + 1) = B (m + 1, n + 1). ⎛ 1 ⎞⎟ 4⎠ ⎛ Γ ⎜⎝ 3 ⎞⎟ 4⎠ = π √2 . ∞ π x dx = ⋅ 1 + x6 3 √3 ∞ x8 (1 − x6) dx = 1. (1 + x) 24 1 1. ∞ ∞ x n− 1 dx. (1 + x) m+ n ∞ x m− 1 ⌠ dx = ⎮ ⌡0 (1 + x) m+ n x n− 1 dx. (1 + x) m+ n Γ (m) Γ (n) ⋅ 2 Γ (m + n) ∞ Beta function. 4. m + n. 5. B (m, n). (Agra 2003) 2. G amma function. 3. Γ (m + n). 6. 0. 8. (n − 1) !. 7. n. 9. π ⋅ sin nπ 10. √π. ⎛ m + n + 2⎞ ⎟⋅ 11. 2 Γ ⎜⎝ ⎠ 2 13. 22 m− 1. 14. an. 15. 2π ⋅ ⎯⎯3 √ 16. (b). 17. (c). 19. (b). 20. (a). 21. (d). 22. (d). 23. (a). 24. F. 25. T. 26. T. 27. F. 28. F. 29. T. 30. T. 31. T. 32. F. 33. F. 34. F. 12. √π ⋅ 2 18. (a). 4.1 Double Integrals The concept of double integral is an extension of the concept of a definite integral to the case of two arguments (i.e. a two dimensional space). Let a function f ( x, y ) of the independent variables x and y be continuous inside some domain (region) A and on its boundary. Divide the domain A into n subdomains A 1, A 2 , … , A n of areas δA 1, δA 2 , … , δA n . Let ( xr , yr) be any point inside the rth elementary area δA r . Form the sum S n = f ( x1, y1) δA 1 + f ( x2 , y2) δA 2 + … + f ( xr , yr) δA r + … + f ( xn , yn ) δA n n = Σ f (xr , yr) δA r . ...(1) r= 1 Now take the limit of the sum (1) as n→∞ in such a way that the largest of the areas δA r approaches to zero. This limit, if it exists, is called the double integral of the function f ( x, y ) over the domain A . It is denoted by ∫ ∫ f ( x, y ) dA and is read A as “the double integral of f ( x, y ) over A”. I-76 INTEGRAL CALCULUS Suppose the domain (region) A is divided into rectangular partitions by a network of lines parallel to the coordinate axes. Let dx be the length of a sub-rectangle and dy be its width so that dx dy is an element of area in Cartesian coordinates. The integral ∫ ∫ f ( x, y ) d A is written as ∫ ∫ f ( x, y ) dx dy and is called the double integral A of f ( x, y ) over the region A . 4.2 Properties of a Double Integral I. If the region A is partitioned into two parts, say A 1 and A 2 , then ∫∫A f ( x, y ) dx dy = ∫∫A f (x, y ) dx dy + 1 ∫∫A f ( x, y ) dx dy . 2 Similarly for a sub-division of A into three or more parts. II. The double integral of the algebraic sum of a fixed number of functions is equal to the algebraic sum of the double integrals taken for each term. Thus ∫∫A [ f1 ( x, y ) + f2 ( x, y ) + f3 ( x, y ) + …] dx dy = ∫∫A f1 ( x, y ) dx dy + ∫∫A f2 ( x, y ) dx dy + ∫∫A f3 ( x, y ) dx dy + … III. A constant factor may be taken outside the integral sign. Thus ∫ ∫ m f ( x, y ) dx dy = m ∫ ∫ f ( x, y ) dx dy , A A where m is a constant. 4.3 Evaluation of Double Integrals (a) If the region A be given by the inequalities a ≤ x ≤ b , c ≤ y ≤ d , then the double integral ∫∫A f ( x, y ) dx dy = = or ∫∫A f ( x, y ) dx dy = = ∫ ab∫ cd f ( x, y ) dx dy ∫ ab ⎡⎢⎣ ∫ cd ∫ cd∫ ab ⎤ ⎦ f ( x, y ) dy⎥ dx , ...(1) f ( x, y ) dy dx ∫ cd ⎡⎢⎣ ∫ ab ⎤ ⎦ f ( x, y ) dx⎥ dy ...(2) i.e., in this case the order of integration is immaterial, provided the limits of integration are changed accordingly. d Important Note : In formula (1) the definite integral ∫ c f ( x, y ) dy is calculated first. During this integration x is regarded as a constant. While in the formula (2) the definite integral ∫ ab f ( x, y ) dx is calculated first and during this integration y is regarded as a constant. (b) If the region A is bounded by the curves y = f1 (x) , y = f2 (x) , x = a and x = b , then I-77 MULTIPLE INTEGRALS b f (x) ⌠⌠ ⌠ ⌠2 ⎮⎮ f ( x, y ) dx dy = ⎮ ⎮ f ( x, y ) dx dy ⌡⌡A ⌡a ⌡f (x) 1 b ⎡ f2 (x) ⌠ ⌠ = ⎮ ⌡a ⎤ ⎢⎮ f ( x, y ) dy⎥⎥ dx , ⎢⌡ ⎢ f (x) ⎥ ⎣ 1 ⎦ where the integration with respect to y is performed first treating x as a constant. Similarly, if the region A is bounded by the curves x = f1 ( y) , x = f2 ( y) , y = c , y = d , we have d f ( y) ⌠ ⌠2 ⌠⌠ ⎮⎮ f ( x, y ) dx dy = ⎮ ⎮ f ( x, y ) dy dx ⌡c ⌡f ( y) ⌡⌡A 1 d f ( y) ⎤ ⌠ ⎡⌠ 2 = ⎮ ⎢⎢ ⎮ f ( x, y ) dx⎥⎥ dy , ⌡c ⎢ ⌡f ( y) ⎥ ⎣ 1 ⎦ where the integration with respect to x is performed first treating y as a constant. Remember : While evaluating double integrals, first integrate w.r.t. the variable having variable limits (treating the other variable as constant) and then integrate w.r.t. the variable with constant limits. b Remark : In the double integral ∫ a ∫ cd f ( x, y ) dx dy , it is generally understood that the limits of integration c to d are those of y and the limits of integration a to b are those of x . H owever this is not a standard convention. Some authors regard these limits in the reverse order i.e. they regard the limits c to d as those of x and the limits a to b as those of y . So it is better to write this double integral as ∫ xb= a ∫ yd= c f ( x, y ) dx dy so that there in no confusion about the limits. H owever in b f (x) ⌠ ⌠2 the double integral ⎮ ⎮ F ( x, y ) dx dy , there is no confusion about the limits. ⌡a ⌡f (x) 1 O bviously the variable limits are those of y because they are in terms of x and so the constant limits must be those of x . H ere the first integration must be performed with respect to y regarding x as constant. Example 1 : Evaluate the following double integrals : a b 2 x ⌠ ⌠ (i) ⎮ ⎮ ( x2 + y2) dx dy ⌡0 ⌡0 ⌠ ⌠ dx dy (ii) ⎮ ⎮ 2 ⋅ ⌡1 ⌡0 x + y2 Solution : (i) (Kanpur 2006; Purvanchal 14) a b a y3 ⎤ b ⌠ ⌠ ⌠ ⎡ We have ⎮ ⎮ ( x2 + y2) dx dy = ⎮ ⎢ x2 y + ⎥ dx , ⌡0 ⌡0 ⌡0 ⎣ 3 ⎦y= 0 (integrating w.r.t. y treating x as constant) I-78 INTEGRAL CALCULUS a a ⎡ x3 ⌠ ⎡ b3 ⎤ b3 ⎤ ba3 b3a 1 ⎥ dx = ⎢ b = ⎮ ⎢ bx2 + + x⎥ = + = ab (a2 + b2) . 3 ⌡0 ⎣ ⎣ 3 3⎦ 3 ⎦0 3 3 2 (ii) x 2 x ⌠ ⎡⌠ dy ⎤⎥ ⌠ ⌠ dx dy = ⎮ ⎢⎢ ⎮ 2 dx We have ⎮ ⎮ 2 2 ⌡1 ⎢ ⌡0 x + y2 ⎥⎥ ⌡1 ⌡0 x + y ⎣ ⎦ 2 y⎤ x ⌠ ⎡1 dx = ⎮ ⎢ tan− 1 ⎥ ⌡1 ⎣ x x⎦ y = 0 (integrating w.r.t. y treating x as constant) 2 2 π ⌠ dx π ⎡ 2 ⌠ ⎡1 ⎤ = ⎢ log x ⎤⎥ = ⎮ ⎢ (tan− 1 1 − tan− 1 0) ⎥ dx = ⎮ ⌡1 ⎣ x 4 ⌡1 x 4⎣ ⎦1 ⎦ = 1 4 π [log 2 − log 1] = 1 4 π log 2 . Example 2 : Evaluate 3 1 √(1 + x2) ⌠ ⌠ dx dy (ii) ⎮ ⎮ ⋅ ⌡0 ⌡0 1 + x2 + y2 2 ⌠ ⌠ (i) ⎮ ⎮ xy (1 + x + y ) dx dy . ⌡0 ⌡1 (Gorakhpur 2005; Kanpur 12; Avadh 14) 3 Solution : (i) We have 2 ⌠ ⌠ ⎮ ⎮ xy (1 + x + y) dx dy ⌡0 ⌡1 3 y2 y2 y3 ⎤ 2 ⌠ ⎡ + x2 ⋅ + x⋅ ⎥ dx , = ⎮ ⎢x ⋅ ⌡0 ⎣ 2 2 3 ⎦y= 1 (integrating w.r.t. y treating x as constant) 3 ⎤ ⌠ ⎡x x2 x = ⎮ ⎢ (4 − 1) + (4 − 1) + (8 − 1) ⎥ dx ⌡0 ⎣ 2 ⎦ 3 2 3 3 ⎡ 23 x2 3 ⎤ 3 x3 ⎤ ⌠ ⎡ ⎛ 3 7⎞ ⋅ + ⋅ ⎥ = ⎮ ⎢ ⎜ + ⎟ x + x2⎥ dx = ⎢ ⌡0 ⎣ ⎝ 2 3 ⎠ ⎣ 6 2 2 ⎦ 2 3 ⎦0 23 9 27 123 3 = ⋅ + = = 30 ⋅ 6 2 2 4 4 (ii) 1 √(1 + x2) ⌠ ⌠ dx dy ⎮ ⎮ ⌡0 ⌡0 1 + x2 + y2 We have 1 ⌠ 1 = ⎮ ⌡0 √(1 + x2) 2 y ⎡ ⎤ √(1 + x ) − 1 tan dx , ⎢ ⎥ ⎣ √(1 + x2) ⎦ y = 0 (integrating w.r.t. y treating x as constant) 1 1 1 dx ⌠ ⎡ tan− 1 1 − tan− 1 0⎤ dx = π ⌠ ⎮ = ⎮ ⎦ ⌡0 √(1 + x2) ⎣ 4 ⌡0 √(1 + x2) π π 1 = ⎡⎣ log {x + √(1 + x2)}⎤⎦ 0 = log (1 + √2) . 4 4 a √(a2 − y2) ⌠ ⌠ Example 3 : Evaluate ⎮ ⎮ ⌡0 ⌡0 √(a2 − x2 − y2) dy dx . I-79 MULTIPLE INTEGRALS Solution : H ere the variable limits are those of x and so the first integration must be performed w.r.t. x taking y as constant. a √(a2 − y2) ⌠ ⌠ ∴⎮ ⎮ ⌡0 ⌡0 √(a2 − x2 y2) − a ⌠ dy dx = ⎮ ⌡0 ⎡ ⎢ ⎢ ⎢ ⎣ √(a2 − y2) ⌠ ⎮ ⌡0 ⎤ √{(a2 − y2) − x2} dx⎥⎥ dy ⎥ ⎦ 2 2 a ⎤ √(a − y ) ⌠ ⎡ x √(a2 − y2 − x2) (a2 − y2) x ⎥ = ⎮ ⎢⎢ + sin− 1 dy, ⎥ ⌡0 ⎢⎣ 2 2 √(a2 − y2) ⎥⎦ x = 0 (integrating w.r.t. x treating y as constant) a a a2 − y2 π ⎤ π⎡ π⎡ y3 ⎤ a3 ⎤ 1 ⌠ ⎡ ⎥ = ⋅ ⎥ dy = ⎢ a2y − ⎥ = ⎢ a3 − π a 3. = ⎮ ⎢0 + ⌡0 ⎣ 2⎦ 4⎣ 6 2 3 ⎦0 4 ⎣ 3⎦ 2 y⁄ 2 2 ⌠ ⌠ Example 4 : Show that ⎮ ⎮ ⌡1 ⌡0 x⁄2 ⌠ ⌠ y dy dx = ⎮ ⎮ ⌡1 ⌡0 x dx dy . Solution : We have 2 y⁄ 2 ⌠ ⌠ ⎮ ⎮ ⌡1 ⌡0 y⁄ 2 2 ⌠ ⎡ ⌠ y dy dx = ⎮ ⎢⎢ y ⎮ ⌡1 ⎢ ⌡0 ⎣ 2 ⎤ y⁄ 2 ⌠ dx⎥⎥ dy = ⎮ y ⎡⎢ x ⎤⎥ dy , ⌡1 ⎣ ⎦ 0 ⎥ ⎦ (integrating w.r.t. x treating y as a constant) 2 2 2 ⌠ 1⌠ 2 1 ⎡ y3 ⎤ 1 7 ⎡y ⎤ = ⎮ y ⎢ − 0⎥ dy = ⎮ y dy = ⎢ ⎥ = [8 − 1] = ...(1) ⌡1 ⎣ 2 ⎦ 2 ⌡1 2 ⎣ 3 ⎦1 6 6 2 Again x⁄2 ⌠ ⌠ ⎮ ⎮ ⌡1 ⌡0 x⁄2 2 ⎡⌠ ⌠ x dx dy = ⎮ x ⎢⎢ ⎮ ⌡1 ⎢ ⌡0 ⎣ 2 x⁄2 ⎤ ⌠ dy⎥⎥ dx = ⎮ x ⎡⎢⎣ y ⎤⎥⎦ dx , ⌡1 0 ⎥ ⎦ (integrating w.r.t. y treating x as a constant) 2 2 2 ⌠ 1⌠ 1 ⎡ x3 ⎤ 1 7 ⎡x ⎤ = ⎮ x ⎢ − 0⎥ dx = ⎮ x2 dx = ⎢ ⎥ = (8 − 1) = ⌡1 ⎣ 2 2 ⌡1 2 ⎣ 3 ⎦1 6 6 ⎦ From (1) and (2), we see that 2 y⁄ 2 ⌠ ⌠ ⎮ ⎮ ⌡1 ⌡0 2 x⁄2 ⌠ ⌠ y dy dx = ⎮ ⎮ ⌡1 ⌡0 ...(2) x dx dy . Examples on the region of integration (Double Integration) ⌠⌠ Example 5 : Evaluate ⎮⎮ x2 y2 dx dy over the region x2 + y2 ≤ 1 . ⌡⌡ Solution : Let R denote the region x2 + y2 ≤ 1 . Then R is the region in the xy-plane bounded by the circle x2 + y2 = 1 . The limits of integration for this region can be expressed either as − 1 ≤ x ≤ 1 , − √(1 − x2) ≤ y ≤ √(1 − x2) or as − √(1 − y2) ≤ x ≤ √(1 − y2) , − 1 ≤ y ≤ 1 . Because from the equation of the circle x2 + y2 = 1 , we have x2 = 1 − y2 so that x = ± √(1 − y2) . Thus for a fixed value of y, x varies from − √(1 − y2) to √(1 − y2) in the area bounded by the circle x2 + y2 = 1 . Also y varies from − 1 to 1 to cover the I-80 INTEGRAL CALCULUS whole area of the circle x2 + y2 = 1 . Therefore if the first integration is to be performed w.r.t. x regarding y as constant, then √(1 − y2) 1 ⌠ ⌠ ⌠⌠ 2 2 ⎮ ⎮⎮ x y dx dy = ⎮ x2 y2 dx dy ⌡y = − 1 ⌡x = − √(1 − y2) ⌡⌡R 2 ⎡ √(1 − y ) ⎤ x2 dx⎥⎥ dy ⎢⌡ ⎢ x = − √(1 − y2) ⎥ ⎣ ⎦ 1 ⌠ = ⎮ ⌡y = − 1 ⌠ y2 ⎢ ⎮ √(1 − y2) 1 ⎡ ⌠ ⌠ y2 ⎢⎢ 2 ⎮ = ⎮ ⌡− 1 ⎢ ⌡x = 0 ⎣ x2 1 2 1 ⎤ ⎡ x3 ⎤ √(1 − y ) ⌠ 2 ⎥ dx⎥ dy = ⎮ 2y ⎢ ⎥ dy ⌡− 1 ⎣ 3 ⎦0 ⎥ ⎦ 1 ⌠ 2 2 2⌠ = ⎮ y (1 − y2) 3 ⁄ 2 dy = 2 . ⎮ y2 (1 − y2) 3 ⁄ 2 dy . 3⌡ ⌡− 1 3 0 Put y = sin θ so that dy = cos θ dθ ; when y = 0 , θ = 0 and when y = 1 , θ = π ⁄ 2 . ∴ π ⁄2 ⌠⌠ 2 2 4⌠ ⎮⎮ x y dx dy = ⎮ ⌡⌡R 3 ⌡0 = π ⁄2 4⌠ ⎮ 3 ⌡0 sin2 θ (1 − sin2 θ) 3 ⁄ 2. cos θ dθ sin2 θ cos4 θ dθ = π 4 1.3.1 π ⋅ ⋅ = ⋅ 3 6.4.2 2 24 Example 6 : Find by double integration the area of the region bounded by the circle x2 + y2 = a2 . (Agra 2007; Kanpur 09) Solution : The area of a small element situated at any point ( x, y ) is dx dy . To find the area bounded by the circle x2 + y2 = a2 , the region of integration R can be expressed as − a ≤ y ≤ a , − √(a2 − y2) ≤ x ≤ √(a2 − y2) , where the first integration is to be performed w.r.t. x regarding y as constant. ∴ the required area a ⌠ ⌠⌠ = ⎮⎮ dx dy = ⎮ ⌡y = − a ⌡⌡R √(a2 − y2) a ⎡ ⌠ ⌠ ⎢2 ⎮ = ⎮ ⌡− a ⎢⎢ ⌡0 ⎣ √(a2 − y2) ⌠ ⎮ 1 . dx dy ⌡x = − √(a2 − y2) a 2 2 ⎤ ⌠ ⎡ x ⎤ √(a − y ) dy 1. dx⎥⎥ dy = 2 ⎮ ⌡− a ⎢⎣ ⎥⎦ 0 ⎥ ⎦ a a ⌠ ⌠ = 2 ⎮ √(a2 − y2) dy = 2.2 ⎮ √(a2 − y2) dy ⌡− a ⌡0 (Note) a ⎡ y √(a 2 − y2) ⎡ ⎤ a2 y⎤ a2 + sin− 1 ⎥ = 4 ⎢ 0 + sin− 1 1⎥ ⎣ ⎣ ⎦ a⎦ 0 2 2 2 = 4⎢ 1 2 1 2 = 4 . a2 . π = π a2 . ⌠⌠ Example 7 : Evaluate ⎮⎮ ( x2 + y2) dx dy over the region in the positive quadrant ⌡⌡ for which x + y ≤ 1 . (Rohilkhand 2012; Avadh 14) I-81 MULTIPLE INTEGRALS Solution : The region of integration R is the area bounded by the coordinate axes and the straight line x + y = 1 . Therefore the region R is bounded by y = 0 , y = 1 − x and x = 0 , x = 1 . Therefore 1− x 1 ⌠⌠ ⌠ ⌠ ⎮⎮ ( x2 + y2) dx dy = ⎮ ⎮ ( x2 + y2) dx dy , ⌡⌡R ⌡x = 0 ⌡y = 0 the first integration to be performed w.r.t. y regarding x as constant 1 1 y3 ⎤ 1 − x (1 − x) 3 ⎤ ⌠ ⎡ ⌠ ⎡ ⎥ dx = ⎮ ⎢ x2 y + ⎥ dx = ⎮ ⎢ x2 (1 − x) + ⌡0 ⎣ ⌡ ⎦ ⎣ ⎦ 3 0 3 0 ⎡ x3 = ⎢ ⎣3 − 1 x4 (1 − x) 4 ⎤ 1 1⎤ 1 ⎡1 ⎥ = ⎢ − − + ⎥ = ⋅ 4 3 × 4 ⎦ 0 ⎣ 3 4 12⎦ 6 ⌠⌠ Example 8 : Evaluate ⎮⎮ xy ( x + y ) dx dy over the area between y = x2 and ⌡⌡ y = x. (Gorakhpur 2005, 06) Solution : Draw the given curves y = x2 and y = x in the same figure. The two curves intersect at the points whose abscissae are given by x2 = x or x ( x − 1) = 0 i.e., x = 0 or 1. When 0 < x < 1 , we have x > x2 . So the area of integration can be considered as lying between the curves y = x2, y = x , x = 0 and x = 1 . Therefore the required integral x x 1 1 ⎤ ⌠ ⌠ ⌠ ⎡⌠ ⎮ = ⎮ xy ( x + y ) dx dy = ⎮ ⎢⎢ ⎮ ( x2 y + xy2) dy⎥⎥ dx ⌡x = 0 ⌡y = x2 ⌡0 ⎢ ⌡x2 ⎥ ⎣ 1 ⌠ ⎡ x2 y2 x xy3 ⎤ 1 ⌠ ⎡ ⎛ x4 x4 ⎞ ⎦ ⎛ x6 x7 ⎞ ⎤ ⎥ dx = ⎮ ⎢ ⎜ + ⎟ − ⎜ + ⎟ ⎥ dx + = ⎮ ⎢ ⌡0 ⎣ ⎝ 2 ⌡0 ⎣ 2 3 ⎦ x2 3⎠ ⎝ 2 3⎠⎦ 1 1 ⎡ x5 x7 x8 ⎤ ⌠ ⎡ 5x4 x6 x7 ⎤ ⎥ − − ⎥ dx = ⎢ − − = ⎮ ⎢ ⌡0 ⎣ 6 ⎣6 2 3⎦ 14 24⎦ 0 28 − 12 − 7 1 1 1 9 3 = − − = = = ⋅ 6 14 24 168 168 56 Example 9 : Prove by the m ethod of double integration that the area lying between 16 2 a . the parabolas y2 = 4ax and x2 = 4ay is 3 Solution : Draw the two parabolas in the same figure. The two parabolas intersect at the points whose abscissae are given by (x2 ⁄ 4a) 2 = 4ax i.e., x (x3 − 64a3) = 0 i.e., x = 0 and x3 = 64a3 . Thus the two parabolas intersect at the points where x = 0 and x = 4a . Now the area of a small element situated at any point ( x, y ) = dx dy . ∴ the required area 4a √(4 ax) 4a ⌠ ⌠ ⌠ ⎮ = ⎮ dx dy = ⎮ ⌡x = 0 ⌡y = x2 ⁄ 4 a ⌡0 ⎡ y ⎤ √(4 ax) dx ⎢ ⎥ 2 ⎣ ⎦ x ⁄4a I-82 INTEGRAL CALCULUS 4a 4a ⎡ ⎡ 1 2⎤ 2 1 x3 ⎤ . x ⎥ dx = ⎢ 2 √a . x3 ⁄ 2 . − ⋅ ⎥ ⎢ 2 √a. x1 ⁄ 2 − ⎣ ⎣ ⎦ 4a 3 4a 3 ⎦ 0 ⌠ = ⎮ ⌡0 4 1 32 2 16 2 16 2 √a. (4a) 3 ⁄ 2 − . 64a3 = a − a = a . 3 12a 3 3 3 = Evaluate the following double integrals : 2 1. √(4 + x2) ⌠ ⌠ (i) ⎮ ⎮ ⌡0 ⌡0 a dx dy ⋅ 4 + x2 + y2 (Rohilkhand 2005) b ⌠ ⌠ dx dy (ii) ⎮ ⎮ ⋅ ⌡1 ⌡1 xy π ⁄2 π ⌠ ⎮ cos (x + y) dy dx. ⌡π ⁄ 2 ⌠ (iii) ⎮ ⌡0 (Kanpur 2007, 11) x2 1 ⌠ ⌠ (iv) ⎮ ⎮ e y ⁄ x dx dy. ⌡0 ⌡0 2 3y ⌠ ⌠ (v) ⎮ ⎮ y dy dx. ⌡1 ⌡0 √(2 x− x2) 2 ⌠ ⌠ (vi) ⎮ ⎮ ⌡0 ⌡0 1 2. x dx dy. 1 1 ⌠ ⌠ dx dy (i) ⎮ ⎮ ⋅ ⌡0 ⌡0 √{(1 − x2) (1 − y2)} 1 √x 3 y− 1 ⌠ ⌠ (iii) ⎮ ⎮ ⌡0 ⌡x a √(a2− y2) √(a2− x2) ⌠ ⌠ (vi) ⎮ ⎮ ⌡0 ⌡0 (a2 − x2 − y2) dy dx. (x + y) dx dy. 2 Show that 1 4 y dy dx. dy dx ⋅ y ⌠ ⌠ (v) ⎮ ⎮ ⌡0 ⌡0 a √(1− y2) ⌠ ⌠ (ii) ⎮ ⎮ ⌡0 ⌡0 (x2 + y2) dx dy. ⌠ ⌠ (iv) ⎮ ⎮ ⌡2 ⌡0 3. (Kanpur 2008) 4 4 2 ⌠ ⌠ ⌠ ⌠ (i) ⎮ ⎮ (xy + e y) dx dy = ⎮ ⎮ (xy + e y) dy dx. ⌡1 ⌡3 ⌡3 ⌡1 1 1 1 x− y x− y ⌠ ⌠ ⌠ ⌠ dy ≠ ⎮ dy ⎮ dx. (ii) ⎮ dx ⎮ ⌡0 (x + y) 3 ⌡0 ⌡0 (x + y) 3 ⌡0 Find the values of the two integrals. I-83 MULTIPLE INTEGRALS a 4. √(a2− x2) ⌠ ⌠ (i) E valuate the double integral ⎮ ⎮ ⌡0 ⌡0 x2 y dx dy. Mention the region of integration involved in this double integral. ⌠⌠ (ii) E valuate ⎮ ⎮ x2 y3 dx dy over the circle x2 + y2 = a2. ⌡⌡ (Rohilkhand 2013B) 5. ⌠⌠ E valuate ⎮ ⎮ (x + y + a) dx dy over the circular area x2 + y2 ≤ a2. ⌡⌡ 6. ⌠⌠ E valuate ⎮ ⎮ x2 y2 dx dy over the region bounded by x = 0, y = 0 and x2 + y2 = 1. ⌡⌡ (Avadh 2012) 7. ⌠⌠ E valuate ⎮ ⎮ xy dx dy over the region in the positive quadrant for which x + y ≤ 1. ⌡⌡ 8. ⌠⌠ E valuate ⎮ ⎮ e 2 x+ 3y dx dy over the triangle bounded by x = 0, y = 0 and x + y = 1. ⌡⌡ 9. ⌠⌠ xy E valuate ⎮ ⎮ dx dy over the positive quadrant of the circle x2 + y2 = 1. ⌡ ⌡ √(1 − y2) 10. Find the area of the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1, by double integration. 11. ⌠⌠ Compute the value of ⎮⎮ y dx dy, where R is the region in the first quadrant ⌡⌡R 12. bounded by the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1. Find the mass of a plate in the form of a quadrant of an ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 whose density per unit area is given by ρ = k x y. 1. (i) 1 4 π log (1 + √2). (ii) (log b) (log a). (iii) − 2 . 1 2 (v) 7. (vi) π 2. (ii) ⋅ 4 3 (iii) 3 ⁄ 35. (v) πa4 ⁄ 8. (vi) 2 a3 ⁄ 3. (iv) ⋅ 2. (i) 1 4 (iv) 1 − log (3 ⁄ 2). 3. (ii) 4. (i) 7. 10. 1 1 and − ⋅ 2 2 5 a ⁄ 15. The area (ii) 0. 1 ⋅ 24 π ab. 1 2 π. of the circle x2 + y2 = a2 in the positive quadrant. 5. πa3. 8. 1 6 (e − 1) 2 (2 e + 1). 11. ab2 ⁄ 3. 6. π ⁄ 96. 9. 1 ⋅ 6 12. k a2 b2 ⁄ 8. I-84 4.4 INTEGRAL CALCULUS To Express a Double Integral in Terms of Polar Coordinates Let a function f ( r, θ) of the polar coordinates ( r, θ) be continuous inside some region A and on its boundary. Let the region A be bounded by the curves r = f1 (θ) , r = f2 (θ) and the lines θ = θ1 , θ = θ2 . Divide the area A into elements by a series of concentric circular arcs with centre at origin and successive radii differing by equal amounts and a series of straight lines drawn through the origin at equal intervals of angles. Let δr be the distance between two consecutive circles and δθ be the angle between two consecutive lines. There is thus a network of elementary areas (say n in number) of which a typical one is PQRS . If P is the point (r, θ) , the area of the element PQRS situated at the point P is 1 2 (r + δr) 2 δθ − 1 2 r 2 δθ = r δθ δr , by neglecting the term 1 2 (δr) 2 δθ being an infinitesimal of higher order. Now by the definition of the double integral of f (r, θ) over the region A , we have n ⌠⌠ lim n→∞ ∑ f (r , θ ) r δθ δr , ⎮⎮ f (r, θ) d A = δr→0 , δθ→0 k k k ⌡⌡A k= 1 where rk δθδr is the area of the element situated at the point (rk , θk) . U sing the area of integration, this double integral is generally written as θ f (θ) θ f (θ) ⌠ 2 ⌠2 ⌠ 2⌠ 2 ⎮ ⎮ f (r, θ) dθ dr , or ⎮ dθ ⎮ f (r, θ) dr . ⌡θ ⌡f (θ) ⌡θ ⌡f (θ) 1 1 1 1 The first integration is performed with respect to r , keeping θ as a constant. After substituting the limits for r , the second integration with respect to θ is performed. Remark : The area of the typical element PQRS situated at the point P (r, θ) can also be found as below : We have OP = r , OQ = r + δr so that PQ = δr . Also PS is the arc of a circle of radius r subtending an angle δθ at the centre of the circle and so arc PS = r δθ . Therefore the area of the element PQRS is δr. r δθ i.e., r δθ δr . I-85 MULTIPLE INTEGRALS π a (1 + cos θ) ⌠ ⌠ Example 1 : Evaluate ⎮ ⎮ ⌡0 ⌡0 r2 cos θ dθ dr . Solution : We have π a (1 + cos θ) ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 π a (1 + cos θ) ⎡ r3 ⎤ ⌠ r2 cos θ dθ dr = ⎮ cos θ ⎢ ⎥ ⌡0 ⎣ 3⎦0 dθ π = 1⌠ ⎮ cos θ . a3 (1 + cos θ) 3 dθ 3 ⌡0 = a3 ⌠ ⎮ cos θ (1 + 3 cos θ + 3 cos2 θ + cos3 θ) dθ 3 ⌡0 = a3 ⌠ ⎮ [cos θ + 3 cos2 θ + 3 cos3 θ + cos4 θ] dθ 3 ⌡0 π π = 2. π ⁄2 a3 ⌠ ⎮ 3 ⌡0 [3 cos2 θ + cos4 θ] dθ , π ⁄2 ⌠ or 2 ⎮ ⌡0 ⎡ . . ⌠π ⎢ . ⎮ cos n θ dθ = 0 ⎢ ⌡0 ⎢ ⎣ ⎤ cos n θ dθ according as n is odd or even⎥⎥ ⎥ ⎦ 2a3 ⎡ 1 π 3.1 π ⎤ 2a3 3π ⎡ 1⎤ ⋅ + ⋅ ⎥= ⋅ ⎢3 ⋅ ⎢1 + ⎥ 2 2 4.2 2 ⎦ 4 ⎣ 4⎦ 3 ⎣ 3 3 3 3π 5 5πa 2a = ⋅ ⋅ = ⋅ 4 4 3 8 = ⌠⌠ r dθ dr Example 2 : Evaluate ⎮⎮ ⌡⌡ √(a2 + r2) over one loop of the lem niscate r2 = a2 cos 2θ . Solution : In the equation of the lemniscate r2 = a2 cos 2θ , putting r = 0 , we get cos 2θ = 0 i.e., 2θ = ± π ⁄ 2 i.e., θ = ± π ⁄ 4 . Therefore for one loop of the given lemniscate θ varies from − π ⁄ 4 to π ⁄ 4 and r varies from 0 to a √ (cos 2θ) . Therefore the required integral π ⁄4 a √(cos 2θ) ⌠ ⌠ ⎮ = ⎮ ⌡θ = − π ⁄ 4 ⌡ r = 0 π ⁄4 a √(cos 2θ) ⌠ ⌠ ⎮ = ⎮ ⌡− π ⁄ 4 ⌡0 1 2 r dθ dr √(a2 + r2) (a2 + r2) − 1 ⁄ 2 (2 r) dθ dr π ⁄4 ⌠ ⎡⎢ (a 2 + r2) 1 ⁄ 2⎤⎥ a √(cos 2θ) dθ = ⎮ ⎦0 ⌡− π ⁄ 4 ⎣ π ⁄4 ⌠ [a (1 + cos 2θ) 1 ⁄ 2 − a] dθ = ⎮ ⌡− π ⁄ 4 I-86 INTEGRAL CALCULUS π ⁄4 ⌠ = 2a ⎮ ⌡0 π ⁄4 ⌠ [(2 cos2 θ) 1 ⁄ 2 − 1] dθ = 2 a ⎮ ⌡0 (√2 cos θ − 1) dθ π⎤ π 1 a ⎡ = 2a ⎡⎣ √2 sin θ − θ⎤⎦ π ⁄ 4 = 2a ⎢ √2 . − ⎥ = 2a ⎡⎢ 1 − ⎤⎥ = (4 − π) . √2 4 ⎦ 4⎦ 2 ⎣ 0 ⎣ Example 3 : Find by double integration the area lying inside the circle r = a sin θ and outside the cardioid r = a (1 − cos θ) . Solution : The given circle is r = a sin θ and the cardioid is r = a (1 − cos θ) . Note that the given circle passes through the pole and the diameter through the pole makes an angle π ⁄ 2 with the initial line. E liminating r between the two equations, we have a sin θ = a (1 − cos θ) 1 1 or 2 sin θ cos θ θ sin θ 2 2 = = tan 1= 1 2 1 − cos θ 2 cos2 θ or 1 2 2 θ= 1 4 π i.e., θ = π ⁄ 2 . Thus the two curves meet at the point where θ = π ⁄ 2 . Also for both the curves r = 0 when θ = 0 and so the two curves also meet at the pole O where θ = 0 . To cover the required area the limits of integration for r are a (1 − cos θ) to a sin θ and for θ are 0 to π ⁄ 2 . Therefore the required area π ⁄2 ⌠ = ⎮ ⌡0 a sin θ ⌠ ⎮ r dθ dr ⌡a (1 − cos θ) π ⁄ 2 ⎡ 2 ⎤ a sin θ r ⌠ = ⎮ ⌡0 1⌠ π ⁄2 ⎢ ⎥ dθ = ⎮ ⎣ 2 ⎦ a (1 − cos θ) 2 ⌡0 π ⁄2 [a2 sin2 θ − a2 (1 − cos θ) 2] dθ = a2 ⌠ ⎮ 2 ⌡0 = π⎤ 1 π⎤ a2 ⎡ a2 a2 ⎡ 1 π π − + 2.1 − ⋅ ⎥ = (4 − π) . ⎢ ⋅ ⎢2 − ⎥ = 2 ⎣2 2 2 ⎣ 4 2 2 2⎦ 2⎦ [sin2 θ − 1 + 2 cos θ − cos2 θ] dθ 2 √(2 x − x2) ⌠ ⌠ Example 4 : Transform the integral ⎮ ⎮ ⌡0 ⌡0 polar coordinates and hence evaluate it. Solution : From the limits of integration it is obvious that the region of integration is bounded by y= 0, 2 y = √(2x − x ) and x = 0 , x = 2 i.e., the region of integration is the area of the circle x2 + y2 − 2x = 0 between the lines x = 0 , x = 2 and lying above the axis of x i.e., the line y = 0 . x dx dy by changing to √( x2 + y2) (Kumaun 2008) I-87 MULTIPLE INTEGRALS Putting x = r cos θ , y = r sin θ the corresponding polar equation of the circle is r2 (cos2 θ + sin2 θ) − 2 r cos θ = 0 , or r = 2 cos θ . From the figure it is obvious that r varies from 0 to 2 cos θ and θ varies from 0 to π ⁄ 2 . Note that at the point A of the circle, θ = 0 and at the point O , r = 0 and so from r = 2 cos θ , we get θ = π ⁄ 2 at O . The polar equivalent of elementary area dx dy is r dθ dr . ⌠⌠ ⌠⌠ ⎮⎮ f ( x, y ) dx dy = ⎮⎮ f ( r cos θ, r sin θ) r dθ dr , ⌡⌡A ⌡⌡A ∴ where A is the region of integration. H ence transforming to polar coordinates, the given double integral 2 cos θ π ⁄2 ⌠ ⌠ ⎮ = ⎮ ⌡θ = 0 ⌡r = 0 π ⁄2 ⌠ = ⎮ ⌡0 1. π π ⁄2 1 ⌠ cos θ .4 cos2 θ dθ = 2 ⎮ ⌡0 2 a sin θ ⌠ ⌠ (i) E valuate ⎮ ⎮ ⌡0 ⌡0 π ⁄2 ⌠ (ii) E valuate ⎮ ⌡0 π 2 cos θ ⎡ r2 ⎤ cos θ ⎢ ⎥ ⎣ 2⎦0 cos3 θ dθ = 2 . dθ 2 4 = ⋅ 3 3 r dθ dr. (Kashi 2013) a cos θ ⌠ ⎮ ⌡0 r sin θ dθ dr. a (1 + cos θ) ⌠ ⌠ (iii) E valuate ⎮ ⎮ ⌡0 ⌡0 2. π ⁄2 r cos θ ⌠ r dθ dr = ⎮ ⌡0 r r 3 sin θ cos θ dθ dr. ⌠⌠ E valuate ⎮⎮ r 2 dθ dr over the area of the circle r = a cos θ. ⌡⌡ (Agra 2003) (Kanpur 2010) 3. Integrate r sin θ over the area of the cardioid r = a (1 + cos θ), lying above the (Kanpur 2010) initial line. 4. Find the mass of a loop of the lemniscate r2 = a2 sin 2 θ if density ρ = kr 2. 5. Find by double integration the area lying inside the cardioid r = a (1 + cos θ) and outside the circle r = a. 6. Find by double integration the area lying inside the cardioid r = 1 + cos θ and outside the parabola r (1 + cos θ) = 1. Transform the following double integrals to polar coordinates and hence evaluate them : a 7. √(a2− y2) ⌠ ⌠ ⎮ (i) ⎮ ⌡y = 0 ⌡x = 0 (a2 − x2 − y2) dx dy. I-88 INTEGRAL CALCULUS 1 √(2 x− x2) ⌠ ⌠ (ii) ⎮ ⎮ ⌡0 ⌡x a (x2 + y2) dx dy. √(a2 − x2) ⌠ ⌠ (iii) ⎮ ⎮ ⌡0 ⌡0 1 4 1. (i) 2. 4a3 ⋅ 9 5. 7. πa2. (ii) a2 ⋅ 6 16 4 a . 15 π ka4 4. ⋅ 16 (iii) 4a3 ⋅ 3 9π + 16 1 2 a (π + 8). 6. ⋅ 4 12 π ⁄2 a πa4 ⌠ ⌠ ⎮ (a2 − r2) r dθ dr ; ⋅ (i) ⎮ ⌡0 ⌡0 8 3. π ⁄2 ⌠ (iii) ⎮ ⌡0 4.5 y2 √(x2 + y2) dx dy. π ⁄2 2 cos θ ⌠ ⌠ (ii) ⎮ ⎮ ⌡π ⁄ 4 ⌡0 ⎛ 3π ⎞ ⎟ − 1. 8⎠ r 3 dθ dr ; ⎜⎝ a πa5 ⌠ 4 2 ⎮ r sin θ dθ dr ; ⋅ ⌡0 20 Triple Integrals Let the function f ( x, y, z ) of the point P ( x, y, z ) be continuous for all points within a finite region V and on its boundary. Divide the region V into n parts; let δ V 1, δ V 2 , … , δ V n be their volumes. Take a point in each part and form the sum S n = f ( x1, y1, z1) δV 1 + f ( x2 , y2 , z2) δV 2 + … + f ( xn , yn , zn ) δV n n = Σ f ( xr , yr , zr) δV r . ...(1) r= 1 Then the limit to which the sum (1) tends when n tends to infinity and the dimensions of each sub-division tend to zero, is called the triple integral of the function f ( x, y, z ) over the region V . This is denoted by ⌠⌠⌠ ⎮⎮⎮ f ( x, y, z ) dV ⌡⌡⌡V 4.6 ⌠⌠⌠ ⎮⎮⎮ f ( x, y, z ) dx dy dz . ⌡⌡⌡V or Evaluation of Triple Integrals (a) If the region V be specified by the inequalities a ≤ x ≤ b, c ≤ y≤ d, e ≤ z ≤ f, then the triple integral ⌠⌠⌠ ⎮⎮⎮ f ( x, y, z ) dx dy dz = ⌡⌡⌡V = ∫ ab∫ cd∫ ef ∫ ab dx ∫ d c f ( x, y, z ) dx dy dz dy ∫ f e f ( x, y, z ) dz . I-89 MULTIPLE INTEGRALS H ere the order of integration is immaterial and the integration with repsect to any of x, y and z can be performed first. (b) If the limits of z are given as functions of x and y , the limits of y as functions of x while x takes the constant values say from x = a to x = b , then b y (x) z ( x, y ) ⌠ ⌠ 2 ⌠ 2 ⌠⌠⌠ ⎮⎮⎮ f ( x, y, z ) dx dy dz = ⎮ dx ⎮ dy ⎮ f ( x, y, z ) dz . ⌡a ⌡y (x) ⌡z ( x, y ) ⌡⌡⌡V 1 1 The integration with respect to z is performed first regarding x and y as constants, then the integration w.r.t. y is performed regarding x as a constant and in the last we perform the integration w.r.t. x . 3 2 Example 1 : Evaluate ∫ y = 0 ∫ x = 0 Solution : The given integral ∫ y3= 0 ∫ x2= 0 ⎧⎨⎩∫ 01 = 3 2 ⌠ ⌠ ⎮ = ⎮ ⌡y = 0 ⌡x = 0 ∫ z1= 0 ( x + y + z ) dz dx dy . ⎫ ⎭ ( x + y + z ) dz⎬ dx dy 3 1 ⎧⎪ z2 ⎫⎪ ⌠ ⎨⎪ xz + yz + ⎬⎪ dx dy = ⎮ ⌡0 2 ⎭0 ⎩ ⎧⌠ 2 ⎫ ⎪ ⎮ ( x + y + 1) dx⎪ dy ⎨⌡ ⎬ 2 ⎪ 0 ⎪ ⎩ ⎭ 3 3 2 3 ⎡ ⌠ ⎧ x2 x⎫ ⌠ 2y2 ⎤ ⎥ = 18 . = ⎮ ⎪⎨⎪ + xy + ⎪⎬⎪ dy = ⎮ (3 + 2y ) dy = ⎢ 3y + ⌡0 ⎩ 2 ⌡0 ⎣ 2 ⎦0 2 ⎭0 Example 2 : Evaluate the following integrals. 1 1− x 1− x− y ⌠ ⌠ (i) ⎮ ⎮ ⌡0 ⌡0 c ⌠ ⎮ ⌡0 b x yz dx dy dz ; a ⌠ ⌠ ⌠ ( x2 + y2 + z2) dx dy dz . (ii) ⎮ ⎮ ⎮ ⌡− c ⌡− b ⌡− a Solution : (i) We have 1 1− x ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 1− x− y ⌠ ⎮ ⌡0 1 1− x ⌠ ⌠ x yz dx dy dz = ⎮ ⎮ ⌡0 ⌡0 ⎡ z2 ⎤ 1 − x − y ⎥ dx dy , ⎣ 2 ⎦0 xy ⎢ integrating w.r.t. z regarding x and y as constants 1 1− x 1 1− x = 1⌠ ⌠ ⎮ ⎮ 2 ⌡0 ⌡0 = 1⌠ ⌠ ⎮ ⎮ 2 ⌡0 ⌡0 = xy {(1 − x) − y}2 dx dy x [ y (1 − x) 2 − 2 (1 − x) y2 + y3] dx dy 1 ⎡ 1− x (1 − x) 2 y2 2 (1 − x) y3 y4 ⎤ 1⌠ ⎮ x⎢ − + ⎥ dx , 2 ⌡0 ⎣ 2 3 4 ⎦0 integrating w.r.t. y regarding x as constant e 3: I-91 MULTIPLE INTEGRALS 4 4 ⌠ 2 √z ⌠ ⎡x 4z x ⎤ 2 √z = ⎮ ⎡⎣ 0 √(4z − x2) dx⎤⎦ dz = ⎮ ⎢ √(4z − x2) + sin− 1 dz ⎥ ⌡0 ⌡0 ⎣ 2 2 2 √z ⎦ 0 4 4 4 2 √z ⎤ π 4z ⌠ ⌠ ⌠ ⎡ sin− 1 = ⎮ ⎢0 + ⎥ dz = ⎮ 2 z . dz = ⎮ πz dz ⌡0 ⌡0 ⌡0 ⎣ 2 2 √z ⎦ 2 ⎡ z2 ⎤ 4 π ⎥ = [16] = 8π . ⎣ 2 ⎦0 2 = π⎢ Example 4 : Find the volume of the tetrahedron bounded by the coordinate planes and the plane x + y + z = 1 . (Rohilkhand 2013B) Solution : H ere the region of integration V to cover the volume of the tetrahedron can be expressed as 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y . Therefore the required volume of the tetrahedron 1 1− x ⌠ ⌠ ⌠⌠⌠ = ⎮⎮⎮ dx dy dz = ⎮ ⎮ ⌡0 ⌡0 ⌡⌡⌡V 1 1− x ⌠ ⌠ = ⎮ ⎮ ⌡0 ⌡0 1− x− y ⌠ ⎮ ⌡0 1 dx dy dz 1− x ⌠ ⎡ z ⎤ 1 − x − y dx dy = ⌠ ⎮ ⎮ ⎢ ⎥ ⌡0 ⌡0 ⎣ ⎦0 (Note) (1 − x − y ) dx dy 1 1 1− x (1 − x ) 2 ⎤ ⌠ ⎡ y2 ⎤ ⌠ ⎡ ⎥ dx = ⎮ ⎢ (1 − x ) y − ⎥ dx = ⎮ ⎢ (1 − x ) 2 − ⌡0 ⎣ ⌡0 ⎣ ⎦ 2 ⎦0 2 1 1 1 ⎡ (1 − x ) 3 ⎤ ⌠ 1 ⎥ (1 − x ) 2 dx = ⎢ = ⎮ ⌡0 2 2 ⎣ 3 .(− 1) ⎦ 0 1 1 = − [0 − 1] = ⋅ 6 6 ⌠⌠⌠ Example 5 : Evaluate ⎮⎮⎮ ( x + y + z ) dx dy dz over the tetrahedron x = 0 , ⌡⌡⌡ y = 0 , z = 0 and x + y + z = 1 . Solution : The region of integration V for the given tetrahedron can be expressed as 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y. ⌠⌠⌠ H ence the required triple integral = ⎮⎮⎮ ( x + y + z ) dx dy dz ⌡⌡⌡V 1 1− x ⌠ ⌠ = ⎮ ⎮ ⌡0 ⌡0 1− x− y ⌠ ⎮ ⌡0 1 1− x ⎡ 1 1− x ⎡ ⌠ ⌠ = ⎮ ⎮ ⌡0 ⌡0 ⌠ ⌠ = ⎮ ⎮ ⌡0 ⌡0 1 1− x ( x + y + z ) dx dy dz 1− x− y z2 ⎤ ⎢ ( x + y) z + ⎥ dx dy ⎣ 2 ⎦0 ⎢ ( x + y ) (1 − x − y ) + ⎣ (1 − x − y ) 2 ⎤ ⎥ dx dy ⎦ 2 1 − x − y⎞ ⎟ dx dy 2 ⎠ ⌠ ⌠ = ⎮ ⎮ ⌡0 ⌡0 (1 − x − y ) ⎜ x + y + ⌠ ⌠ = ⎮ ⎮ ⌡0 ⌡0 1 (1 − x − y ) (1 + x + y ) dx dy 2 1 1− x ⎛ ⎝ I-92 INTEGRAL CALCULUS 1 1− x 1⌠ ⌠ ⎮ ⎮ 2 ⌡0 ⌡0 = [1 − ( x + y ) 2] dx dy = 1 1− x ( x + y )3⎤ 1⌠ ⎡ ⎥ ⎮ ⎢y − dx ⎦0 3 2 ⌡0 ⎣ 1 1 1 x3 ⎞ 1 ⌠ ⎛2 x3 ⎞ 1⌠ ⎛ = ⎮ ⎜ 1 − x − + ⎟ dx = ⎮ ⎜ − x + ⎟ dx ⌡ ⌡ 3 2 0 ⎝3 3⎠ 3⎠ 2 0 ⎝ (Note) 1 x2 x4 ⎤ 1 ⎡2 1 1⎤ 1 1 1 1 ⎡2 ⎢ x− ⎥ = + + ⋅ = ⋅ ⎢ − ⎥ = 2 4 8 2 3 × 4 ⎦ 0 2 ⎣ 3 2 12⎦ 2 ⎣3 = ⌠⌠⌠ Example 6 : Evaluate ⎮⎮⎮ z2 dx dy dz over the sphere x2 + y2 + z 2 = 1 . ⌡⌡⌡ Solution : H ere the region of integration can be expressed as − 1 ≤ x ≤ 1 , − √(1 − x2) ≤ y ≤ √(1 − x2) , − √(1 − x2 − y2) ≤ z ≤ √(1 − x2 − y2) . ∴ the required triple integral √(1 − x2) 1 √(1 − x2 − y2) ⌠ ⌠ ⌠ ⎮ z2 dx dy dz = ⎮ ⎮ ⌡− 1 ⌡− √(1 − x2) ⌡− √(1 − x2 − y2) 2 2 1 √(1 − x2) ⎡ z3 ⎤ √(1 − x − y ) ⌠ ⌠ ⎢ ⎥ = ⎮ ⎮ dx dy ⌡− 1 ⌡− √(1 − x2) ⎣ 3 ⎦ − √(1 − x2 − y2) 1 √(1 − x2) ⎤ 1 ⌠ ⎡⌠ = ⎮ ⎢⎢ ⎮ 2 (1 − x2 − y2) 3 ⁄ 2 dy⎥⎥ dx ⎥ 3 ⌡− 1 ⎢⎣ ⌡− √(1 − x2) ⎦ ⎤ 2 1 ⎡ π ⁄2 ∫ ⎢∫ [(1 − x2) cos2 θ]3 ⁄ 2. √(1 − x2). cos θ dθ⎥ dx 3 − 1 ⎣ − π ⁄2 ⎦ = [putting y = √(1 − x2) sin θ so that dy = √(1 − x2) cos θ dθ ; also when y = 0 , θ = 0 and when y = √(1 − x2) , θ = π ⁄ 2] 2 1 ⎡ ∫ ⎢ 2 .∫ 0π ⁄ 2 (1 − x2) 2 cos4 θ dθ⎤⎥⎦ dx 3 − 1 ⎣ π 1 3.1 π 4 1 (1 − x2) 2⋅ ⋅ dx = ∫ (1 − = ∫ 4.2 2 4 − 1 3 − 1 1 π π⎡ 2 = . 2 ∫ (1 − 2x2 + x4) dx = ⎢ x − x3 + 0 4 2⎣ 3 = π⎡ π 8 4π 2 1⎤ + ⎥= ⋅ = ⋅ ⎢1 − 2⎣ 3 5⎦ 2 15 15 = Evaluate the following integrals : 1 1. 2 2 ⌠ ⌠ ⌠ (i) ⎮ ⎮ ⎮ x2 yz dx dy dz. ⌡x = 0 ⌡y = 0 ⌡z = 1 1 1 1 ⌠ ⌠ ⌠ (ii) ⎮ ⎮ ⎮ e x + y + z dx dy dz. ⌡0 ⌡0 ⌡0 x2) 2 dx 1 5⎤ 1 x⎥ 5 ⎦0 I-93 MULTIPLE INTEGRALS 1 z x+ z log 2 ⌠ ⎮ ⌡0 ⌠ ⌠ (iii) ⎮ ⎮ ⌡− 1 ⌡0 ⌠ (iv) ⎮ ⌡0 1 2. ⌠ ⎮ (x + y + z) dy dx dz. ⌡x− z x x + log y ⌠ ⎮ ⌡0 1− x 1 ⌠ ⌠ ⌠ (i) ⎮ ⎮ ⎮ ⌡0 ⌡y2 ⌡0 1 1− x ⌠ ⌠ (ii) ⎮ ⎮ ⌡0 ⌡0 3 x dy dx dz. 1− x− y ⌠ ⎮ ⌡0 √(xy) 1 ⌠ ⌠ ⌠ (iii) ⎮ ⎮ ⎮ ⌡1 ⌡1 ⁄ x ⌡0 π ⁄2 ⌠ (iv) ⎮ ⌡0 a 3. x a sin θ x+ y a a− x dx dy dz ⋅ (1 + x + y + z) 3 (Kanpur 2008; Avadh 13) xyz dx dy dz. ⌠ dθ ⎮ ⌡0 ⌠ ⌠ ⌠ (i) ⎮ ⎮ ⎮ ⌡0 ⌡0 ⌡0 ⌠ ⌠ (ii) ⎮ ⎮ ⌡0 ⌡0 e x+ y+ z dx dy dz. (a2− r 2) ⁄ a ⌠ dr ⎮ ⌡0 r dz. e x+ y+ z dx dy dz. a− x− y ⌠ ⎮ ⌡0 x2 dx dy dz. 4. E valuate the triple integral of the function f (x, y, z) = x2 over the region V enclosed by the planes x = 0, y = 0, z = 0 and x + y + z = a. 5. Find the volume of the tetrahedron bounded by the plane x ⁄ a + y ⁄ b + z ⁄ c = 1 and the coordinate planes. (Rohilkhand 2012; Avadh 12) 6. ⌠⌠⌠ dx dy dz (i) E valuate ⎮⎮⎮ over the region x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1. ⌡⌡⌡ (x + y + z + 1) 3 (Avadh 2013) x2 y2 z2 ⌠⌠⌠ (ii) E valuate ⎮⎮⎮ xyz dx dy dz over the ellipsoid 2 + 2 + 2 = 1. ⌡⌡⌡ a b c (Kanpur 2011) ⌠⌠⌠ (iii) E valuate ⎮⎮⎮ (z5 + z) dx dy dz over the sphere x2 + y2 + z2 = 1. ⌡⌡⌡ ⌠⌠⌠ (iv) E valuate ⎮⎮⎮ u2 v2 w du dv dw, where R is the region u2 + v2 ≤ 1, 0 ≤ w ≤ 1. ⌡⌡⌡R 1. (i) 1. 19 8 ⋅ (iv) log 2 − 9 3 (ii) (e − 1) 3. 4 2. (i) ⋅ 35 (iii) 0. (ii) 1 ⎛⎜ log 2 2⎝ − 5 ⎞⎟ ⋅ 8⎠ I-94 INTEGRAL CALCULUS ⎞ 1 ⎛⎜ 26 − log 3⎟⎠ ⋅ 6⎝ 3 a5 ⋅ (ii) 60 (iii) 6. 4.7 (i) 1 2 ⎛ ⎜ ⎝ log 2 − 5 ⎞⎟ 8⎠ (iv) 3. (i) a5 ⋅ 60 4. ⋅ 5a3 π ⋅ 64 5. (ii) 0. 1 8 (e 4 a − 6 e2 a + 8 e a − 3). abc ⋅ 6 (iii) 0. (iv) π ⋅ 48 Change of Order of Integration If in a double integral the limits of integration of both x and y are constant, we can generally integrate ∫ ∫ f ( x, y ) dx dy in either order. But if the limits of y are functions of x , we must first integrate w.r.t. y regarding x as constant and then integrate w.r.t. x . In this case the order of integration can be changed only if we find the new limits of x as functions of y and the new constant limits of y . This is usually best obtained from geometrical considerations as will be clear from the examples that follow. Example 1 : Change the order of integration in the double integral a x ⌠ ⌠ ⎮ ⎮ f ( x, y ) dx dy . ⌡0 ⌡0 (Kashi 2013) Solution : In the given integral the limits of integration are given by the straight lines y = 0 , y = x , x = 0 and x = a . Draw these lines bounding the region of integration in the same figure. We observe that the region of integration is the area ON M . In the given integral, the limits of integration of y being variable, we are required to integrate first w.r.t. y regarding x as constant and then w.r.t. x . To reverse the order of integration, we have to integrate first w.r.t. x regarding y as constant and then w.r.t. y . This is done by dividing the area ON M into strips parallel to the x-axis. Let us take strips parallel to the x-axis starting from the line ON (i.e., y = x ) and terminating on the line MN (i.e., x = a) . Thus for this region ON M , x varies from y to a and y varies from 0 to a . H ence by changing the order of integration, we have a x a a ⌠ ⌠ ⌠ ⌠ ⎮ ⎮ f ( x, y ) dx dy = ⎮ ⎮ f ( x, y ) dy dx . ⌡0 ⌡0 ⌡0 ⌡y b x b b ⌠ ⌠ ⌠ ⌠ Example 2 : Prove that ⎮ dx ⎮ f ( x, y ) dy = ⎮ dy ⎮ f ( x, y ) dx . ⌡a ⌡a ⌡y ⌡a b Solution : Let x ⌠ ⌠ I = ⎮ dx ⎮ f ( x, y ) dy . ⌡a ⌡a I-95 MULTIPLE INTEGRALS We are required to change the order of integration in the integral I . In the integral I the limits of integration of y are given by the straight lines y = a and y = x . Also the limits of integration of x are given by the straight lines x = a and x = b. Draw the straight lines y = a , y = x , x = a and x = b, bounding the region of integration, in the same figure. We observe that the region of integration is the area of the triangle A BC . In the integral I we are required to integrate first w.r.t. y and then w.r.t. x . To reverse the order of integration we have to integrate first w.r.t. x and then w.r.t. y . This is done by dividing the area ABC into strips parallel to the x-axis. Let us take strips parallel to the x-axis starting from the line A C (i.e., y = x ) and terminating on the line BC (i.e., x = b) . Thus for the region ABC , x varies from y to b and y varies from a to b . Hence by changing the order of integration, we have b x b b ⌠ ⌠ ⌠ ⌠ ⎮ dx ⎮ f ( x, y ) dy = ⎮ dy ⎮ f ( x, y ) dx . ⌡a ⌡a ⌡y ⌡a 2a Example 3 : Change the order of integration in √(2ax − x2 ) ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 f ( x, y ) dx dy . (Meerut 2013B) Solution : In the given integral the limits of integration of y are given by y = 0 (i.e., the x-axis) and y = √(2ax − x2) i.e., y2 = 2ax − x2 i.e., ( x − a) 2 + y2 = a2 which is a circle with centre (a, 0) and radius a . Again the limits of integration of x are given by the straight lines x = 0 ( i.e., the y-axis) and x = 2a . Draw the curves ( x − a) 2 + y2 = a2 , y = 0 , x = 0 and x = 2a , bounding the region of integration, in the same figure. From figure we observe that the area of integration is OMN O . In the given integral we are required to integrate first w.r.t. y regarding x as a constant and then w.r.t. x . To reverse the order of integration, divide the area OMN O into strips parallel to the x-axis. These strips will have their extremities on the portions ON and NM of the circle. Solving the equation of circle ( x − a) 2 + y2 = a2 for x , we get ( x − a) 2 = a2 − y2 i.e., x − a = ± √(a2 − y2) x = a ± √(a2 − y2) . i.e., √(a2 So for the region OMN O , x varies from a − − to a + √(a2 − y2) and y varies from 0 to a . Therefore, changing the order of integration, the given double integral a transforms to a + √(a2 − y2) ⌠ ⌠ ⎮ ⎮ f ( x, y ) dy dx . ⌡0 ⌡a − √(a2 − y2) y2) I-96 INTEGRAL CALCULUS Example 4 : Change the order of integration in the double integral ∞ ∞ ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡x e− y dx dy y (Agra 2000, 02; Kumaun 01; Avadh 07; Kashi 14; Purvanchal 14) and hence find its value. Solution : In the given integral the limits of integration are given by the lines y = x , y = ∞ , x = 0 and x = ∞ . Therefore the region of integration is bounded by x = 0 , y = x and, an infinite boundary. In the given integral the limits of integration of y are variable while those of x are constant. Thus we have to first integrate with respect to y regarding x as constant and then we integrate w.r.t. x . This is done by first integrating w.r.t. y along a strip drawn parallel to the y-axis and then integrating w.r.t. x along all such strips so drawn as to cover the whole region of integration. If we want to reverse the order of integration, we have to first integrate w.r.t. x regarding y as constant and then we integrate w.r.t. y . This is done by dividing this area into strips parallel to the x-axis. So we take strips parallel to the x-axis starting from the line x = 0 and terminating on the line y = x . Now the limits for x are 0 to y and the limits for y are 0 to ∞ . H ence by changing the order of integration, we have ∞ ∞ ∞ y ∞ ⌠ ⌠ e− y ⌠ e− y ⎡ ⎤ y ⌠ ⌠ e− y dx dy = ⎮ ⎮ dy dx = ⎮ ⎮ ⎮ ⎢ x ⎥ dy ⌡0 ⌡0 y ⌡0 y ⎣ ⎦ 0 ⌡0 ⌡x y ∞ ∞ ⎡ e− y⎤⎥ ∞ ⌠ ⌠ e− y ⎥ = 1. ⋅ y dy = ⎮ e− y dy = ⎢⎢⎢ = ⎮ ⌡0 ⌡0 y ⎣ − 1 ⎥⎦ 0 Example 5 : Change the order of integration in the integral a √(a2 − x2) ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 f ( x, y ) dx dy . Solution : In the given integral the limits of integration of y are given by the straight line y = 0 (i.e., the x-axis) and the curve y = √(a2 − x2) i.e., y2 = a2 − x2 i.e., x2 + y2 = a2 which is a circle with centre at the origin and radius a . Again the limits of integration of x are given by the lines x = 0 and x = a . We draw the curves y = 0 , x2 + y2 = a2 , x = 0 and x = a , giving the limits of integration, in the same very figure and we observe that the region of integration is the area OAB of the quadrant of the circle x2 + y2 = a2. To change the order of integration in the given integral, we have to first integrate w.r.t. x regarding y as a constant and then we integrate w.r.t. y . This is done by covering the area OA B by strips drawn parallel to the x-axis. These strips start from the line OB (i.e., x = 0) and terminate on the arc AB of the circle x2 + y2 = a2. So on these strips x varies from 0 to √(a2 − y2) . Also to cover the area OA B , y varies from 0 to a . H ence by changing the order of integration, we have the given integral I-97 MULTIPLE INTEGRALS a 2 2 √(a − y ) ⌠ ⌠ = ⎮ ⎮ ⌡0 ⌡0 f ( x, y ) dy dx . Example 6 : Change the order of integration in a x+ 2a ⌠ ⌠ f ( x, y ) dx dy . ⎮ ⎮ ⌡0 ⌡√(a2 − x2) Solution : H ere the area of integration is bounded by the curves y = √(a2 − x2) i.e., x2 + y2 = a2 which is a circle with centre (0, 0) and radius a , y = x + 2a which is a straight line passing through (0, 2a) , x = 0 i.e., the y-axis and the line x = a which is a line parallel to the y-axis at a distance a from the origin. We draw the curves x2 + y2 = a2, y = x + 2a , x = 0 and x = a , giving the limits of integration, in the same figure. We observe that the region of integration is the area ML AN M . To reverse the order of integration, cover this area of integration ML A N M by strips parallel to the x-axis. Draw the lines MC and N B parallel to the x-axis so that the region of integration ML A N M is divided into three portions ML C , N MCB and NAB . For the region ML C , x varies from the arc ML of the circle x2 + y2 = a2 to the line x = a i.e., x varies from √(a2 − y2) to a and y varies from 0 to a . For the region N MCB , x varies from 0 to a and y varies from a to 2a . For the region N BA , x varies from y − 2a to a and y varies from 2a to 3a . Therefore, changing the order of integration, the given integral transforms to a 2a a a 3a a ⌠ ⌠ ⌠ ⌠ ⌠ ⌠ ⎮ ⎮ f ( x, y ) dy dx + ⎮ ⎮ f ( x, y ) dy dx + ⎮ ⎮ f ( x, y ) dy dx . ⌡0 ⌡√(a2 − y2) ⌡0 ⌡0 ⌡2a ⌡y − 2a Change the order of integration in the following integrals. 1 x (2− x) 1. ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡x 2. ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡1 3 √(4− y) f (x, y) dx dy. (x + y) dy dx. (Meerut 2013) I-98 INTEGRAL CALCULUS √(a2− x2) a cos α 3. ⌠ ⎮ ⌡0 4. ⌠ ⌠ ⎮ ⎮ f (x, y) dx dy. ⌡0 ⌡m x 5. ⌠ ⌠ ⎮ ⎮ f (x, y) dx dy. ⌡0 ⌡x2 ⁄ 4a 6. ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 7. ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡x 8. ⌠ ⌠ ⎮ ⎮ f (x, y) dx dy, where ⌡c ⌡(b ⁄ a) √(a2− x2) 9. ⌠ ⎮ ⌡0 a ⌠ ⎮ ⌡x tan α f (x, y) dx dy. (Kanpur 2005; Avadh 11) lx 2a 3a− x b ⁄ (b+ x) a a2 ⁄ x a a f (x, y) dx dy. f (x, y) dx dy. (Kanpur 2010) b x− (x2 ⁄ a) a⁄2 ⌠ ⎮ ⌡x2 ⁄ a c < a. f (x, y) dx dy. √(2 ax) 2a 10. ⌠ ⌠ ⎮ ⎮ f (x, y) dx dy. ⌡0 ⌡√(2 ax− x2) 11. ⌠ ⎮ ⌡0 12. ⌠ ⎮ ⌡0 13. Change the order of integration in the double integral ab ⁄ √(a2+ b2) π ⁄2 (a ⁄ b) √(b2− y2) ⌠ ⎮ ⌡0 2 a cos θ ⌠ ⎮ ⌡0 a f (x, y) dy dx. f (r, θ) dθ dr. (Kanpur 2009) x φ ′ ( y) dx dy ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 √{(a − x) (x − y)} 1 y 2 4 − x2 1. ⌠ ⌠ ⎮ ⎮ f (x, y) dy dx. ⌡0 ⌡1− √(1− y) 2. ⌠ ⌠ ⎮ ⎮ ⌡1 ⌡0 3. ⌠ ⎮ ⌡0 4. ⌠ ⎮ ⌡0 a sin α am y⁄ m ⌠ ⎮ ⌡y ⁄ l (x + y) dx dy. y cot α ⌠ ⎮ ⌡0 and hence find its value. a √(a2− y2) ⌠ ⌠ ⎮ f (x, y) dy dx + ⎮ ⌡a sin α ⌡0 al a ⌠ ⌠ f (x, y) dy dx + ⎮ ⎮ f (x, y) dy dx. ⌡am ⌡y ⁄ l f (x, y) dy dx. I-99 MULTIPLE INTEGRALS a √(4ay) 3a 5. ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 6. ⌠ ⎮ ⌡0 7. ⌠ ⌠ ⌠ ⌠ ⎮ ⎮ f (x, y) dy dx + ⎮ ⎮ ⌡0 ⌡0 ⌡a ⌡0 8. ⌠ ⎮ ⌡0 9. ⌠ ⎮ ⌡0 b ⁄ (a+ b) a 3a− y ⌠ ⌠ f (x, y) dy dx + ⎮ ⎮ ⌡a ⌡0 a f (x, y) dy dx. b (1− y) ⁄ y 1 ⌠ ⌠ ⌠ ⎮ f (x, y) dy dx + ⎮ ⎮ ⌡0 ⌡b ⁄ (a+ b) ⌡0 ∞ y b √{1− (c2 ⁄ a2)} a2 ⁄ y f (x, y) dy dx. a b a ⌠ ⌠ ⌠ ⎮ ⎮ f (x, y) dy dx. f (x, y) dy dx + ⎮ ⌡a √{1− ( y2 ⁄ b2)} ⌡b √{(1− (c2 ⁄ a2)} ⌡c a⁄4 ⌠ ⎮ f (x, y) dy dx. ⌡1 [a − √(a2− 4ay)] 2 a a − √(a2− y2) √(ay) a 2a 2a 10. ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡y2 ⁄ 2 a 11. ⌠ ⎮ ⌡0 12. ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 13. φ ′ ( y) dy dx ⌠ ⌠ = π [φ (a) − φ (0)]. ⎮ ⎮ ⌡0 ⌡y √{(a − x) (x − y)} ab ⁄ √(a2+ b2) cos− 2a a f (x, y) dy dx. 1 2a ⌠ ⌠ ⌠ ⌠ f (x, y) dy dx + ⎮ ⎮ f (x, y) dy dx + ⎮ ⎮ f (x, y) dy dx. ⌡0 ⌡a + √(a2− y2) ⌡a ⌡y2 ⁄ 2 a ab ⁄ √(a2+ b2) ⌠ ⎮ ⌡0 (r ⁄ 2 a) a (b ⁄ a) √(a2− x2) ⌠ ⌠ ⎮ f (x, y) dx dy + ⎮ ⌡ab ⁄ √(a2+ b2) ⌡0 f (x, y) dx dy. f (r, θ) dr dθ. a To evaluate the integral put x = a sin2 θ + y cos2 θ. 4.8 Change of Variables in a Double Integral Sometimes, the evaluation of a double integral becomes more convenient by a suitable change of variables from one system to another system. ⌠⌠ Let the variables in the double integral ⎮⎮ f ( x, y ) dx dy be changed from x, y ⌡⌡A to u, v where x = φ (u, v ) and y = ψ (u, v ) . Then on substituting for x and y , the double integral is transformed to ⌠⌠ ⎮⎮ F (u, v ) J du dv , where J (u, v ) is the Jacobian of x, y w.r.t. u, v i.e., ⌡⌡A′ ⎪ ∂x ∂y ⎪ ∂ ( x, y ) ⎪⎪ ∂u ∂u ⎪⎪ , J= = ⎪ ⎪ ∂ (u, v ) ⎪ ∂x ∂y ⎪ ⎪ ⎪ ⎪ ∂v ∂v ⎪ and A ′ is the region in the uv-plane corresponding to the region A in the xy-plane. Thus remember that dx dy = J du dv . I-100 INTEGRAL CALCULUS Special case : Change to polar coordinates from the cartesian co-ordinates : To change the variables from cartesian to polar coordinates we put x = r cos θ , y = r sin θ . In this case ⎪ ∂x ∂x ⎪ ∂ ( x, y ) ⎪⎪ ∂r ∂θ⎪⎪ ⎪ cos θ − r sin θ⎪ J= = ⎪ ⎪ = r, ⎪= ⎪ r cos θ ⎪ ⎪ ∂y ∂y ⎪ ∂ (r, θ) ⎪ sin θ ⎪ ⎪ ⎪ ∂r ∂θ⎪ and therefore dx dy = J dθ dr = r dθ dr . This change is specially useful when the region of integration is a circle or a part of a circle. ⌠⌠ Example 1 : Transform ⎮⎮ f ( x, y ) dx dy by the substitution x + y = u , y = uv . ⌡⌡ Solution : We have x + y = u and y = uv . From these, we have x = u − y = u − uv and y = uv . ∂x ∂x ∂y ∂y ∴ = 1 − v, = − u, = v and = u. ∂u ∂v ∂u ∂v ⎪ ∂x ∂x ⎪ ∂ ( x, y ) ⎪⎪ ∂u ∂v ⎪⎪ ⎪ 1 − v − u⎪ ∴ J= = ⎪ ⎪ = u. ⎪= ⎪ u ⎪ ∂ (u, v ) ⎪ ∂y ∂y ⎪ ⎪ v ⎪ ⎪ ⎪ ∂u ∂v ⎪ ∴ dx dy = J du dv = u du dv . H ence the given integral transforms to ⌠⌠ ⎮⎮ F (u, v ) u du dv . ⌡⌡ ...(1) ...(2) ⌠⌠ Example 2 : Transform ⎮⎮ f ( x, y ) dx dy to polar coordinates. ⌡⌡ Solution : We have x = r cos θ , y = ⎪ ∂x ∂x ⎪ ∂ ( x, y ) ⎪⎪ ∂r ∂θ⎪⎪ = ⎪ Now J= ⎪= ⎪ ∂y ∂y ⎪ ∂ (r, θ) ⎪ ⎪ ⎪ ∂r ∂θ⎪ ∴ r sin θ . ⎪ cos θ ⎪ ⎪ sin θ − r sin θ⎪ ⎪ = r. r cos θ ⎪ dx dy = J dθ dr = r dθ dr . ⌠⌠ H ence the given integral transforms to ⎮⎮ F (r, θ) r dθ dr . ⌡⌡ ⌠⌠ Example 3 : Evaluate ⎮⎮ √(a2 − x2 − y2) dx dy over the sem i-circle x2 + y2 = ax ⌡⌡ in the positive quadrant. Solution : H ere the region of integration is a semi-circle. Therefore, for the sake of convenience, changing to polar coordinates by putting x = r cos θ and y = r sin θ in x2 + y2 = ax , we have I-101 MULTIPLE INTEGRALS r2 cos2 θ + r2 sin2 θ = a r cos θ or r2 (sin2 θ + cos2 θ) = ar cos θ or r = a cos θ . The equation r = a cos θ represents a circle passing through the pole and diameter through the pole along the initial line. For the given region r varies from 0 to a cos θ and θ varies from 0 to π ⁄ 2 . π ⁄2 ⌠ ⌠⌠ ⎮⎮ √(a2 − x2 − y2) dx dy = ⎮ ⌡0 ⌡⌡ ∴ π ⁄ 2 ⎡ a cos θ ⌠ ⌠ = ⎮ ⌡0 π ⁄2 ⌠ = ⎮ ⌡0 ⎢⎮ ⎢⌡ ⎢ 0 ⎣ − a cos θ ⌠ ⎮ ⌡0 √(a2 − r2) . r dθ dr , [... x2 + y2 = r2 and dx dy = r dθ dr ] ⎤ 1 2 (a − r2) 1 ⁄ 2 . (− 2 r) dr⎥⎥ dθ ⎥ 2 ⎦ ⎡ 1 2 2 ⎤ a cos θ ⋅ (a − r2) 3 ⁄ 2⎥ dθ ⎢− ⎣ 2 3 ⎦0 π ⁄2 = − 1⌠ ⎮ 3 ⌡0 a a− x (a3 sin3 θ − a3) dθ = − π⎤ a3 ⎡ 2 1 1 2 − ⎥ = a3 ( π − ) . ⎢ 3 2 3 3 ⎣ 3.1 2 ⎦ 1. ⌠ ⌠ Transform ⎮ ⎮ ⌡0 ⌡0 2. By using the transformation x + y = u, y = uv, show that 1 1− x ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 3. 4. (Note) f (x, y) dx dy, by the substitution x + y = u, y = uv. e y ⁄ (x+ y) dx dy = 1 (e − 1). 2 By using the transformation x + y = u, y = uv, prove that ⌠⌠ ⎮⎮ {xy (1 − x − y)}1 ⁄ 2 dx dy ⌡⌡ taken over the area of the triangle bounded by the lines x = 0, y = 0, x + y = 1 is 2 π ⁄ 105. ⌠⌠ E valuate ⎮⎮ (x2 + y2) 7 ⁄ 2 dx dy over the circle x2 + y2 = 1. ⌡⌡ 5. ⌠⌠ E valuate ⎮⎮ xy (x2 + y2) 3 ⁄ 2 dx dy over the positive quadrant of the circle x2 + y2 = 1. ⌡⌡ 6. 2 2 ⌠⌠ E valuate ⎮⎮ e− (x + y ) dx dy over the circle x2 + y2 = a2. ⌡⌡ 1. ⌠ ⌠ ⎮ ⎮ F (u, v) u du dv. ⌡0 ⌡0 4. 2 π ⁄ 9. 5. 1 ⁄ 14 . 2 6. π (1 − e− a ). a 1 I-102 INTEGRAL CALCULUS Fill in the Blanks: Fill in the blanks “……” so that the following statem ents are com plete and correct. 3 2 1. ⌠ ⌠ The value of the double integral ⎮ ⎮ dx dy is …… . ⌡0 ⌡1 2. ⌠ ⌠ The value of the double integral ⎮ ⎮ xy dx dy is …… . ⌡0 ⌡0 3. ⌠ ⌠ The value of the double integral ⎮ ⎮ xy dx dy is …… . ⌡0 ⌡0 4. ⌠ The value of the double integral ⎮ ⌡0 5. ⌠ ⌠ ⌠ The value of the triple integral ⎮ ⎮ ⎮ xyz dx dy dz is …… . ⌡0 ⌡0 ⌡0 6. ⌠ ⌠ ⌠ The value of the triple integral ⎮ ⎮ ⎮ dx dy dz is …… . ⌡1 ⌡1 ⌡1 7. ⌠ ⌠ The value of the double integral ⎮ ⎮ ⌡− a ⌡0 1 1 1 x π ⁄2 ⌠ ⎮ ⌡0 2 2 2 2 2 3 2 a cos θ √(a2− x2) a (Agra 2002) r dθ dr is …… . dx dy is …… . Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 8. 2π a ⌠ ⌠ ⎮ The value of the double integral ⎮ r dθ dr is ⌡θ = 0 ⌡r = 0 πa2 2 (d) 2 πa2. (a) πa2 (b) (c) πa 1 9. 1 1 ⌠ ⌠ ⌠ The value of the triple integral ⎮ ⎮ ⎮ xyz dx dy dz is ⌡0 ⌡0 ⌡0 (a) (c) 1 2 1 4 (b) 1 8 (d) 1. I-103 MULTIPLE INTEGRALS √(a2− y2) a 10. ⌠ ⌠ The value of the double integral ⎮ ⎮ ⌡0 ⌡0 (a) πa2 (c) 11. (b) 2 πa2 πa2 2 (d) π ⁄2 ⌠ The value of the triple integral ⎮ ⌡0 π ⁄2 ⌠ ⎮ ⌡0 cos2 x cos2 y cos2 z dx dy dz is (b) π 64 (c) π3 8 (d) π3 ⋅ 64 1 1 ⌠ ⌠ ⌠ The value of the triple integral ⎮ ⎮ ⎮ e x+ y+ z dx dy dz is ⌡0 ⌡0 ⌡0 e3 4 (d) (e + 1) 3. (b) (c) (e − 1) 3 π ⁄2 ⌠ The value of the triple integral ⎮ ⌡0 π ⁄2 ⌠ ⎮ ⌡0 π ⁄2 ⌠ ⎮ ⌡0 cos x cos y cos z dx dy dz is π 2 3π (d) ⋅ 2 (a) 1 (b) (c) π 14. π ⁄2 ⌠ ⎮ ⌡0 π2 16 (a) e 3 13. πa2 ⋅ 4 (a) 1 12. dy dx is π ⁄2 ⌠ The value of ⎮ ⌡0 π ⁄2 ⌠ (a) ⎮ ⌡0 π ⁄2 ⌠ (c) ⎮ ⌡0 sin θ ⌠ ⎮ ⌡0 (Rohilkhand 2005) r dθ dr is equal to sin θ π r dr 2 sin θ dθ ⌠ (b) ⎮ ⌡0 sin2 θ dθ 2 (d) none of these. (Garhwal 2003) True or False: Write ‘T ’ for true and ‘F ’ for false statem ent. π ⁄2 a 15. πa2 ⌠ ⌠ ⎮ The value of the double integral ⎮ r dθ dr is ⋅ ⌡θ= − π ⁄ 2 ⌡r= 0 2 16. ⌠ ⌠ The value of the double integral ⎮ ⎮ dx dy is πa2. ⌡− a ⌡− √(a2− x2) 17. ⌠ ⌠ The value of the double integral ⎮ ⎮ ⌡− a ⌡0 √(a2− x2) a a √(a2− x2) x dx dy is 0. I-104 INTEGRAL CALCULUS 1 4 ⋅ ⋅ 3. 2. 4. πa2 ⋅ 2 5. 8. 6. 2. 8. (a). 9. (b). 11. (d). 14. (c). 17. T. 12. (c). 15. T. 7. 10. 13. 16. πa2 ⋅ 2 (d). (a). T. 3. 1 8 1. 5.1 Dirichlet’s Theorem for Three Variables If l, m, n are all positive, then the triple integral Γ (l) Γ (m) Γ (n) , ⌠⌠⌠ l − 1 m − 1 n − 1 ⎮⎮⎮ x y z dx dy dz = ⌡⌡⌡ Γ (l + m + n + 1) where the integral is extended to all positive values of the variables x, y and z subject to the condition x + y + z ≤ 1 . (Agra 2001; Kanpur 14; Avadh 14; Kashi 14; Purvanchal 14) Proof : Let us first consider the double integral ⌠⌠ I2 = ⎮⎮ x l − 1 y m − 1 dx dy , ⌡⌡ where the integral is extended to all positive values of the variables x and y subject to the condition x + y ≤ 1 . O bviously the region of integration of I2 , in the 2-dimensional E uclidean space, is bounded by the straight lines x = 0 , y = 0 and x + y = 1 . The limits of integration for this region can be expressed as 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 − x . 1 ∴ 1− x ⌠ ⌠ I2 = ⎮ ⎮ x l − 1 y m − 1 dx dy ⌡x = 0 ⌡y = 0 I-106 INTEGRAL CALCULUS 1− x 1 ⎡ ym ⎤ ⌠ = ⎮ x l − 1 ⎢⎢⎢ ⎥⎥⎥ ⌡0 ⎣ m ⎦0 dx 1 ⌠ 1 l− 1 = ⎮ x (1 − x) m dx ⌡0 m = 1 m 1 ⌠ 1 ⎮ x l − 1 (1 − x) m + 1 − 1 dx = B (l, m + 1) , ⌡0 m by the def. of Beta function 1 Γ (l) Γ (m + 1) 1 Γ (l) . m Γ (m) , = = m Γ (l + m + 1) m Γ (l + m + 1) = Γ (l) Γ (m) ⋅ Γ (l + m + 1) [... Γ (n + 1) = n Γ (n)] (Remember) ...(1) This is Dirichlet’s theorem for two variables. ⌠⌠ Now consider the double integral U2 = ⎮⎮ x l − 1 y m − 1 dx dy , ⌡⌡ where the integral is extended to all positive values of the variables x and y subject to the condition x + y ≤ h . y x We have x + y ≤ h ⇒ + ≤ 1 . h h So putting x ⁄ h = u and y ⁄ h = v so that dx = h du and dy = h dv , the integral U2 becomes ⌠⌠ U2 = ⎮⎮ (hu) l − 1 (hv) m − 1 h2 du dv ⌡⌡ ⌠⌠ = hl + m ⎮⎮ ul − 1 vm − 1 du dv , where u + v ≤ 1 ⌡⌡ Γ (L ) Γ (m) , = hl+ m by (1). Γ (l + m + 1) Now we consider the triple integral ⌠⌠⌠ I3 = ⎮⎮⎮ x l − 1 y m − 1 z n − 1 dx dy dz , ⌡⌡⌡ subject to the condition x + y + z ≤ 1 i.e., y + z ≤ 1 − x and 0 ≤ x ≤ 1 . We have 1 ⌠ I3 = ⎮ ⌡x= 0 ⎡ ⌠⌠ ⎤ ⎢ ⎮⎮ y m − 1 z n − 1 dy dz⎥ x l − 1 dx , where y + z ≤ 1 − x ⎢ ⌡⌡ ⎥ ⎢ ⎥ ⎣ ⎦ 1 Γ (m) Γ (n) ⌠ x l − 1 dx , by using (2) = ⎮ (1 − x) m + n ⌡0 Γ (m + n + 1) 1 = Γ (m) Γ (n) Γ (m + n + 1) = Γ (m) Γ (n) B (l, m + n + 1) Γ (m + n + 1) ⌠ ⎮ x l − 1 (1 − x) m + n + 1 − 1 dx ⌡0 ...(2) I-107 DIRICHLET’S AND LIOUVILLE’S INTEGRALS Γ (m) Γ (n) Γ (l) Γ (m + n + 1) ⋅ Γ (m + n + 1) Γ (l + m + n + 1) Γ (l) Γ (m) Γ (n) , = which proves the required result. Γ (l + m + n + 1) Remark : The triple integral Γ (l) Γ (m) Γ (n) , ⌠⌠⌠ l − 1 m − 1 n − 1 ⎮⎮⎮ x y z dx dy dz = h l + m + n ⌡⌡⌡ Γ (l + m + n + 1) = where the integral is extended to all positive values of the variables x, y and z subject to the condition x + y + z ≤ h . Alternative proof of Dirichlet’s theorem for three variables : ⌠⌠⌠ Let I3 = ⎮⎮⎮ x l − 1 y m − 1 z n − 1 dx dy dz , ⌡⌡⌡ where the integral is extended to all positive values of the variables x, y and z subject to the condition x + y + z ≤ 1 . O bviously the region of integration, in the 3-dimensional E uclidean space, is the volume bounded by the coordinate planes x = 0 , y = 0 , z = 0 and the plane x + y + z = 1 . After a little geometric consideration, we observe that the limits of integration for this region can be expressed as 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y. H ence the triple integral I3 may be written as 1− x 1 ⌠ ⌠ ⎮ I3 = ⎮ ⌡x= 0 ⌡y= 0 1− x 1 ⌠ ⌠ = ⎮ ⎮ ⌡0 ⌡0 1 1− x− y ⌠ ⎮ ⌡z= 0 x l − 1 y m − 1 z n − 1 dx dy dz n 1 − x− y ⎡z ⎤ x l − 1 y m − 1 ⎢⎢⎢ ⎥⎥⎥ ⎣ n ⎦0 1− x = 1 n ⌠ ⌠ ⎮ ⎮ ⌡0 ⌡0 = 1 n ⎡⌠ ⌠ ⎮ x l − 1 ⎢⎢ ⎮ ⌡0 ⎢ ⌡0 dx dy x l − 1 y m − 1 (1 − x − y) n dx dy 1− x 1 ⎣ ⎤ y m − 1 {(1 − x) − y }n dy⎥⎥ dx . ⎥ ⎦ To integrate w.r.t. y , put y = (1 − x) t so that dy = (1 − x) dt ; also when y = 0 , t = 0 and when y = 1 − x , t = 1 . ∴ the required integral I3 = 1 n 1 = n = = 1 n 1 1 ⎡⌠ ⎤ ⌠ ⎮ x l − 1 ⎢⎢ ⎮ (1 − x) m − 1 t m − 1 {(1 − x) n (1 − t) n } (1 − x) dt⎥⎥ dx ⌡0 ⎢ ⌡0 ⎥ ⎣ 1 ⎦ 1 ⌠ ⌠ ⎮ ⎮ x l − 1 (1 − x) m + n t m − 1 (1 − t) n dx dt ⌡0 ⌡0 1 1 ⌠ ⌠ ⎮ x l − 1 (1 − x) m + n dx × ⎮ t m − 1 .(1 − t) n dt ⌡0 ⌡0 1 B (l, m + n + 1) B (m, n + 1) , n (by the definition of Beta function) 5.1, I-109 DIRICHLET’S AND LIOUVILLE’S INTEGRALS ⌠⌠ ⌠ = h l1 + l2 + … + ln ⎮⎮ … ⎮ u1 l1 − 1 u2 l2 − 1 … un ln − 1 du1 du2 … dun ⌡⌡ ⌡ subject to the condition u1 + u2 + … + un ≤ 1 = h l1 + l2 + … + ln Γ (l1) Γ (l2) … Γ (ln ) Γ (1 + l1 + l2 + … + ln ) , ...(3) using the assumed result (2). Now for n + 1 variables the condition is x1 + x2 + … + xn + xn + 1 ≤ 1 i.e., x2 + x3 + … + xn + xn + 1 ≤ 1 − x1, and 0 ≤ x1 ≤ 1 . We then have ⌠⌠ ⌠ l − 1 l − 1 ⎮⎮ … ⎮ x1 1 x2 2 … xn ln − 1 xn + 1 ln+ 1 − 1 dx1 dx2 … dxn dxn + 1, ⌡⌡ ⌡ where x1 + x2 + … + xn + 1 ≤ 1 1 ⎡ ⌠⌠ ⌠ ⌠ x l1 − 1 ⎢⎢ ⎮⎮ … ⎮ x2 l2 − 1 … xn + 1 ln + 1 − 1 dx2 … dxn + = ⎮ ⌡x = 0 1 ⌡ ⎢ ⌡⌡ ⎣ 1 1 ⌠ = ⎮ ⌡x = 0 1 x1 l1 − 1 ⋅ Γ (l2) Γ (l3) … Γ (ln + 1) Γ (1 + l2 + l3 + … + ln + ln + 1) ⎤ ⎥ dx 1 ⎦ 1⎥⎥ ⋅ (1 − x1) l2 + l3 + … + ln + 1 dx1, using (3) = Γ (l2) Γ (l3) … Γ (ln + 1) Γ (1 + l2 + … + ln + ln + 1) ⋅ 1 ⌠ ⎮ x1 l1 − 1 (1 − x1) (1 + l2 + l3 + … + ln + 1)− 1 dx1 ⌡0 = = Γ (l2) Γ (l3) … Γ (ln + 1) Γ (1 + l2 + … + ln + ln + 1) Γ (l1) Γ (l2) … Γ (ln + 1) ⋅ Γ (l1) Γ (1 + l2 + … + ln + 1) Γ (1 + l1 + l2 + … + ln + ln + 1) ⋅ ...(4) Γ (1 + l1 + l2 + … + ln + 1) The result (4) shows that the theorem holds for (n + 1) variables if it holds for n variables. But we have seen that the theorem is true for two variables. Hence by mathematical induction the theorem is true for all values of n . ⌠⌠ Exam ple 1 : Evaluate ⎮⎮ x2 l − 1 y2 m − 1 dx dy for all positive values of x and y such ⌡⌡ that x2 + y2 ≤ c2. Solution : Let us denote the given integral by I . Then we have to find the value of I extended to all positive values of x and y subject to the condition I-110 INTEGRAL CALCULUS ⎛ x ⎞ 2 ⎛ y⎞ 2 ⎜ ⎟ + ⎜ ⎟ ≤ 1. ⎝ c⎠ ⎝ c⎠ Put (x ⁄ c) 2 = u i.e., x = cu1 ⁄ 2, so that dx = ( y ⁄ c) 2 = v i.e., y = cv1 ⁄ 2, so that dy = and 1 cu− 1 ⁄ 2 du , 2 1 − 1⁄2 cv dv . 2 Then the required integral ⌠⌠ 1 1 I = ⎮⎮ (cu1 ⁄ 2) 2 l − 1 (cv1 ⁄ 2) 2 m − 1 . cu− 1 ⁄ 2 ⋅ cv− 1 ⁄ 2 du dv 2 2 ⌡⌡ = 1 2l+ 2m c 4 ⌠⌠ l − 1 m − 1 ⎮⎮ u v du dv , where u, v take all + ive values ⌡⌡ = 1 2l+ 2m c 4 subject to the condition u + v ≤ 1 Γ (l) Γ (m) , ⋅ by Dirichlet’s theorem. Γ (l + m + 1) ⌠⌠ ⌠ Example 2 : Find the value of ⎮⎮ … ⎮ dx1 dx2 … dxn extended to all positive ⌡⌡ ⌡ values of the variables, subject to the condition x12 + x22 + … + xn 2 < R 2. Solution : Let us denote the given integral by I . Then we have to find the value of I extended to all positive values of x1, x2 , … , xn subject to the condition x12 x22 xn 2 + + … + < 1. R2 R2 R2 Put 1 (x1 ⁄ R) 2 = u1 i.e., x1 = Ru11 ⁄ 2, so that dx1 = Ru−1 1 ⁄ 2 du1, 2 1 (x2 ⁄ R) 2 = u2 i.e., x2 = Ru21 ⁄ 2, so that dx2 = Ru−2 1 ⁄ 2 du2 , and so on. 2 Then the required integral ⌠⌠ ⌠ I = ⎮⎮ … ⎮ ⌡⌡ ⌡ ⎛ 1⎞ n n − 1 ⁄ 2 − 1 ⁄ 2 ⎜ ⎟ R u u2 … un − 1 ⁄ 2 du1 du2 … dun 1 ⎝ 2⎠ ⌠ ⎛ R ⎞ n ⌠⌠ = ⎜ ⎟ ⎮⎮ … ⎮ u1(1 ⁄ 2) − 1 u2(1 ⁄ 2) − 1 .... un (1 ⁄ 2) − 1 du1 du2 … dun , ⌡ ⎝ 2 ⎠ ⌡⌡ subject to the condition u1 + u2 + … + un < 1 1 {Γ ( )}n 2 ⎛ R⎞n , by Dirichlet’s theorem = ⎜ ⎟ ⎝ 2 ⎠ Γ (1 + n ⋅ 1 ) 2 ⎛ R⎞n πn⁄2 ⋅ = ⎜ ⎟ ⋅ 1 ⎝ 2⎠ Γ (1 + n) [... Γ ( ) = √π] 1 2 2 Example 3 : Find the volum e of the solid surrounded by the surface (x ⁄ a) 2 ⁄ 3 + ( y ⁄ b) 2 ⁄ 3 + (z ⁄ c) 2 ⁄ 3 = 1 . Solution : Since the equation (x ⁄ a) 2 ⁄ 3 + ( y ⁄ b) 2 ⁄ 3 + (z ⁄ c) 2 ⁄ 3 = 1 does not change by putting − x for x , − y for y and − z for z , therefore the surface represented by this equation is symmetrical in all the eight octants. I-111 DIRICHLET’S AND LIOUVILLE’S INTEGRALS So the volume of the solid surrounded by this surface = 8 × the volume of the portion of this solid lying in the positive octant. Now the volume of a small element situated at any point (x, y, z) = dx dy dz . ∴ the volume of the solid in the positive octant ⌠⌠⌠ = ⎮⎮⎮ dx dy dz , ⌡⌡⌡ where the integral is extended to all positive values of the variables x, y, z subject to the condition (x ⁄ a) 2 ⁄ 3 + ( y ⁄ b) 2 ⁄ 3 + (z ⁄ c) 2 ⁄ 3 ≤ 1 . Now put (x ⁄ a) 2 ⁄ 3 = u , ( y ⁄ b) 2 ⁄ 3 = v , (z ⁄ c) 2 ⁄ 3 = w i.e., x = au3 ⁄ 2, y = bv3 ⁄ 2, z = cw3 ⁄ 2 so that dx = 3 2 au1 ⁄ 2 du , dy = 3 2 bv1 ⁄ 2 dv , dz = 3 2 cw1 ⁄ 2 dw . ∴ the volume in the positive octant ⌠⌠⌠ 27 = ⎮⎮⎮ abc u(3 ⁄ 2) − 1 v(3 ⁄ 2) − 1 w(3 ⁄ 2) − 1 du dv dw , ⌡⌡⌡ 8 where u + v + w ≤ 1 = 1 ( 2 √π) 3 ⋅ [Γ (3 ⁄ 2)]3 27 27 abc = abc ⋅ 9 7 5 3 1 9 8 8 Γ ( + 1) ⋅ ⋅ ⋅ ⋅ √π 2 2 2 2 2 2 π π abc 4 27 32 = abc ⋅ ⋅ = ⋅ ⋅ 8 8 27.35 35 8 H ence the required volume π abc 4 4π abc = 8⋅ ⋅ = ⋅ 35 8 35 5.3 Liouville’s Extension of Dirichlet’s Theorem If the variables x, y, z are all positive such that h 1 ≤ x + y + z ≤ h 2, then the triple integral ⌠⌠⌠ ⎮⎮⎮ f (x + y + z) x l − 1 y m − 1 z n − 1 dx dy dz ⌡⌡⌡ = Γ (l ) Γ (m) Γ (n) Γ (l + m + n) h ⌠ 2 ⎮ f (u) u l + m + n − 1 du . ⌡h 1 (Garhwal 2002; Kashi 14) ⌠⌠⌠ Proof : Let I = ⎮⎮⎮ x l − 1 y m − 1 z n − 1 dx dy dz , integrated over some region. ⌡⌡⌡ Subject to the condition x + y + z ≤ u , we have by Dirichlet’s theorem Γ (l) Γ (m) Γ (n) ⋅ I = ul + m + n Γ (l + m + n + 1) If the condition be x + y + z ≤ u + δu , then ...(1) I-112 INTEGRAL CALCULUS Γ (l) Γ (m) Γ (n) I = (u + δu) l + m + n ⋅ ...(2) Γ (l + m + n + 1) Therefore the value of the integral I extended to all such positive values of the variables as make the sum of the variables lie between u and u + δu is Γ (l) Γ (m) Γ (n) = [(u + δu) l + m + n − ul + m + n ] , Γ (l + m + n + 1) [subtracting (2) from (1)] l + m + n δu ⎞ Γ (l) Γ (m) Γ (n) l + m + n ⎡ ⎛ ⎤ u − 1⎥ = ⎢⎜1 + ⎟ Γ (l + m + n + 1) u⎠ ⎣⎝ ⎦ Γ (l) Γ (m) Γ (n) l + m + n ⎡ δu ⎤ = u + … − 1⎥ ⎢ 1 + (l + m + n) Γ (l + m + n + 1) u ⎣ ⎦ Γ (l) Γ (m) Γ (n) = (l + m + n) ul + m + n − 1 δu , Γ (l + m + n + 1) to the first order of approximation Γ (l) Γ (m) Γ (n) l + m + n − 1 = u δu . Γ ( l + m + n) ⌠⌠⌠ ⎮⎮⎮ f (x + y + z) x l − 1 y m − 1 z n − 1 dx dy dz , ⌡⌡⌡ Now consider the integral subject to the condition h1 ≤ x + y + z ≤ h2 . If x + y + z lies between u and u + δu , the value of f (x + y + z) can only differ from f (u) by a small quantity of the same order as δu . H ence neglecting square of δu , the part of the integral ⌠⌠⌠ ⎮⎮⎮ f (x + y + z) x l − 1 y m − 1 z n − 1 dx dy dz ⌡⌡⌡ which arises from supposing the sum of the variables to lie between u and u + δu is Γ (l) Γ (m) Γ (n) f (u) . ul + m + n − 1 δu . ultimately equal to Γ (l + m + n) Therefore the whole integral ⌠⌠⌠ ⎮⎮⎮ f (x + y + z) x l − 1 y m − 1 z n − 1 dx dy dz , where h1 ≤ x + y + z ≤ h2 , ⌡⌡⌡ is equal to Γ (l) Γ (m) Γ (n) Γ (l + m + n) h ⌠ 2 ⎮ f (u) . ul + m + n − 1 du . ⌡h 1 Remark : The above theorem holds good even if we take the condition as h1 < x + y + z < h2 in place of h1 ≤ x + y + z ≤ h2 . ⌠⌠⌠ Example 1 : Find the value of ⎮⎮⎮ log (x + y + z) dx dy dz , the integral extending ⌡⌡⌡ over all positive values of x, y, z subject to the condition x + y + z < 1 . (Kanpur 2014; Purvanchal 14) Solution : H ere the integral is to be extended for all positive values of x, y and z such that 0 < x + y + z < 1 . I-113 DIRICHLET’S AND LIOUVILLE’S INTEGRALS ∴ the required integral ⌠⌠⌠ = ⎮⎮⎮ log (x + y + z) dx dy dz , where 0 < x + y + z < 1 ⌡⌡⌡ ⌠⌠⌠ = ⎮⎮⎮ log (x + y + z) x1 − 1 y1 − 1 z1 − 1 dx dy dz ⌡⌡⌡ (Note) 1 = Γ (1) Γ (1) Γ (1) Γ (1 + 1 + 1) = 1 Γ (3) = 1 1 ⎡ ⎤ u3 ⎞ ⌠ 1 u3 ⎥ , 1 ⎢⎛ ⎢ ⎜ (log u) . ⎟ − ⎮ ⋅ du⎥ ⎥ ⌡0 u 3 3⎠ 2.1 ⎢⎣ ⎝ ⎦ ⌠ ⎮ (log u) u1 + 1 + 1 − 1 du , ⌡0 by Liouville’s extension of Dirichlet’s theorem 1 ⌠ ⎮ u2 log u du , ⌡0 [... Γ (1) = 1] 0 integrating by parts taking u2 as the second function 1 = ⎤ 1 lim 3 1⌠ 1 ⎡⎢ u log u − ⎮ u2 du⎥⎥ ⎢0 − u → 0 ⎥ 3 3 ⌡0 2 ⎢⎣ ⎦ 1 = − 1 ⎡ u3 ⎤ , ⎢ ⎥ 6 ⎣ 3 ⎦0 = − 1 ⋅ 18 ⎡ . . lim ⎤ ⎢ . u → 0 u 3 log u = 0⎥ ⎣ ⎦ 1⁄u 1 lim log u lim lim Lim = u →0 = u → 0 − u3 = 0. Note : u → 0 u3 log u = u → 0 3 4 3 1⁄u − 3⁄u Example 2 : Prove that π 2a 2 , ⌠⌠⌠ dx dy dz ⎮⎮⎮ = ⌡⌡⌡ √(a2 − x2 − y2 − z2) 8 the integral being extended for all positive values of the variables for which the expression (Garhwal 2002; Avadh 08) is real. Solution : The given expression is real when x2 + y2 + z2 < a2 . Therefore the required integral is to be extended to all positive values of x, y and z such that 0 < x2 + y2 + z2 < a2 i.e., 0 < x2 ⁄ a2 + y2 ⁄ a2 + z2 ⁄ a2 < 1 . Put (x2 ⁄ a2) = u1 , ( y2 ⁄ a2) = u2 and (z2 ⁄ a2) = u3 i.e., x = au11 ⁄ 2 , y = au21 ⁄ 2 and z = au31 ⁄ 2 so that dx = 1 2 1 1 au−1 1 ⁄ 2 du1 , dy = au−2 1 ⁄ 2 du2 and dz = au−3 1 ⁄ 2 du3 . 2 2 With these substitutions the given condition reduces to 0 < u1 + u2 + u3 < 1 and the required integral becomes 1 ( ) 3 . a3 u−1 1 ⁄ 2 u−2 1 ⁄ 2 u−3 1 ⁄ 2 du1 du2 du3 ⌠⌠⌠ 2 = ⎮⎮⎮ ⌡⌡⌡ a √{1 − (u1 + u2 + u3)} I-114 INTEGRAL CALCULUS = 1 ⁄ 2 − 1 u 1 ⁄ 2 − 1 u 1 ⁄ 2 − 1 du du du a2 ⌠⌠⌠ u1 2 3 1 2 3 ⎮⎮⎮ 8 ⌡⌡⌡ √{1 − (u1 + u2 + u3)} = 1 3 a2 [Γ (1 ⁄ 2)] ⌠ 1 ⎮ ⋅ ⋅ u3 ⁄ 2 − 1 ⋅ du , 3 ⌡0 √(1 − u) 8 Γ( ) 2 by Liouville’s extension of Dirichlet’s theorem = a2 8 π ⁄2 [√π]3 ⋅ 1 2 ⋅ √π π ⁄2 π a2 ⌠ ⎮ = 2 ⌡0 ⌠ ⋅ ⎮ ⌡0 sin θ .2 sin θ cos θ dθ , putting u = sin2 θ etc. √(1 − sin2 θ) sin2 θ dθ = π a2 1 π π 2 a2 ⋅ ⋅ = ⋅ 2 8 2 2 Example 3 : Prove that when x and y are positive and x + y < h , π ⌠⌠ ⎮⎮ f ′ (x + y) x l − 1 y− l dx dy = [ f (h) − f (0)] . (Kumaun 2008) ⌡⌡ sin lπ Solution : The given integral ⌠⌠ I = ⎮⎮ f ′ (x + y) x l − 1 y(1 − l ) − 1 dx dy , where 0 < x + y < h ⌡⌡ = Γ (l) Γ (1 − l ) Γ (l + 1 − l ) h ⌠ ⎮ f ′ (u) ul + (1 − l ) − 1 du , ⌡0 by Liouville’s extension of Dirichlet’s theorem h Γ (l ) Γ (1 − l ) ⌠ ⎮ f ′ (u) du ⌡0 Γ (1) π π ⎡ h f (u) ⎤⎥ = [ f (h) − f (0)] . = sin π l sin π l ⎢⎣ ⎦0 = ⌠⌠⌠ Example 4 : Evaluate ⎮⎮⎮ x α y β z γ (1 − x − y − z) λ dx dy dz over the interior of ⌡⌡⌡ the tetrahedron form ed by the coordinate planes and the plane x + y + z = 1 . Solution : H ere the region of integration is bounded by the planes x = 0 , y = 0 , z = 0 and x + y + z = 1 . So the variables x, y, z take all positive values subject to the condition 0 < x + y+ z < 1. H ence the given integral ⌠⌠⌠ = ⎮⎮⎮ x(α + 1) − 1 y( β + 1) − 1 z(γ + 1) − 1 [1 − (x + y + z)] λ dx dy dz ⌡⌡⌡ 1 = Γ (α + 1) Γ ( β + 1) Γ (γ + 1) ⌠ ⋅ ⎮ uα + 1 + β + 1 + γ + 1 − 1 (1 − u) λ du , ⌡0 Γ (α + β + γ + 3) by Liouville’s extension of Dirichlet’s theorem 1 = Γ (α + 1) Γ (β + 1) Γ (γ + 1) ⌠ ⋅ ⎮ u(α + β + γ + 3) − 1 (1 − u) (λ + 1) − 1 du ⌡0 Γ (α + β + γ + 3) = Γ (α + 1) Γ (β + 1) Γ (γ + 1) B (α + β + γ + 3 , λ + 1) Γ (α + β + γ + 3) I-115 DIRICHLET’S AND LIOUVILLE’S INTEGRALS Γ (α + 1) Γ (β + 1) Γ (γ + 1) Γ (α + β + γ + 3) Γ (λ + 1) ⋅ Γ (α + β + γ + 3) Γ (α + β + γ + λ + 4) Γ (α + 1) Γ (β + 1) Γ (γ + 1) Γ (λ + 1) = ⋅ Γ (α + β + γ + λ + 4) = Example 5 : Evaluate ⌠⌠⌠ ⎮⎮⎮ √(a2 b2 c2 − b2 c2 x2 − c2 a2 y2 − a2 b2 z2) dx dy dz ⌡⌡⌡ taken throughout the ellipsoid x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 = 1 . Solution : The given ellipsoid x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 = 1 is symmetrical in all the eight octants. Let us first evaluate the given integral over the region of the ellipsoid which lies in the positive octant i.e., where x, y, z are all positive. Put x2 ⁄ a2 = u , y2 ⁄ b2 = v , z2 ⁄ c2 = w . 1 2 Then x = au1 ⁄ 2, dx = au− 1 ⁄ 2 du etc. Now the given integral extended over the positive octant of the given ellipsoid is ⌠⌠⌠ I = abc ⎮⎮⎮ ⌡⌡⌡ √ 1 − ax 2 ⎛ ⎜ ⎜ ⎜ ⎝ 2 − y2 z2 ⎞ − 2 ⎟⎟ dx dy dz , where 0 < x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 ≤ 1 2 b c ⎟⎠ ⌠⌠⌠ 1 = abc ⎮⎮⎮ √(1 − u − v − w) ⋅ abc u− 1 ⁄ 2 v− 1 ⁄ 2 w− 1 ⁄ 2 du dv dw 8 ⌡⌡⌡ where 0 < u + v + w ≤ 1 = a2 b2 c2 8 ⌠⌠⌠ (1 ⁄ 2) − 1 (1 ⁄ 2) − 1 (1 ⁄ 2) − 1 ⎮⎮⎮ u v w √{1 − (u + v + w)} du dv dw ⌡⌡⌡ 1 [Γ ( )]3 1 ⌠ a2 b2 c2 2 ⋅ ⎮ √(1 − t) . t1 ⁄ 2 + 1 ⁄ 2 + 1 ⁄ 2 − 1 dt , = 8 Γ (3 ⁄ 2) ⌡0 by Liouville’s extension of Dirichlet’s theorem = a2 b2 c2 8 (√π) 3 ⋅ 1 2 1 ⌠ ⎮ (1 − t) (3 ⁄ 2) − 1 t(3 ⁄ 2) − 1 dt . √π ⌡0 Γ (3 ⁄ 2) Γ (3 ⁄ 2) π 2 a2 b2 c2 a2 b2 c2 ⋅ 2π ⋅ = ⋅ = Γ (3) 8 32 H ence if the integration is extended throughout the ellipsoid, the given integral π 2 a2 b2 c2 π 2 a2 b2 c2 = 8I= 8⋅ = ⋅ 32 4 ⌠⌠ Example 6 : Evaluate ⎮⎮ ⌡⌡ √ ⎧ 1 − x2 ⁄ a 2 − y2 ⁄ b2 ⎫ ⎪ ⎪ dx dy ⎨ ⎬ ⎪ 1 + x2 ⁄ a 2 + y2 ⁄ b2 ⎪ ⎩ ⎭ where x2 ⁄ a2 + y2 ⁄ b2 ≤ 1 . Solution : The ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 is symmetrical in all the four quadrants. Let us first evaluate the given integral over the region of the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 which lies in the first quadrant i.e., where x and y are both positive. Put x2 ⁄ a2 = u , y2 ⁄ b2 = v . I-116 INTEGRAL CALCULUS Then x = au1 ⁄ 2, dx = y = bv1 ⁄ 2, dy = 1 au− 1 ⁄ 2 du , 2 1 − 1⁄2 bv dv . 2 ∴ the given integral extended over the region of the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 which lies in the first quadrant is given by ⌠⌠ I = ⎮⎮ ⌡⌡ = √ ab ⌠⌠ ⎮⎮ 4 ⌡⌡ ab ⋅ = 4 ⎛ 1 − u − v⎞ 1 abu− 1 ⁄ 2 v− 1 ⁄ 2 du dv , ⎜ ⎟ ⋅ ⎝ 1 + u + v⎠ 2 √ 0< u+ v≤ 1 ⎧ 1 − (u + v) ⎫ (1 ⁄ 2) − 1 (1 ⁄ 2) − 1 v du dv ⎨ ⎬u ⎩ 1 + (u + v) ⎭ 1 2 1 2 Γ( )Γ( ) Γ (1) 1 ⌠ ⎮ ⌡0 √ ⎛ 1 − t⎞ ⎜ ⎟ ⋅ t1 ⁄ 2 + 1 ⁄ 2 − 1 dt ⎝ 1 + t⎠ π ⁄2 1 = where 1− t π ab ⌠ π ab ⌠ ⎮ ⋅ ⎮ dt = ⌡0 √(1 − t2) 4 ⌡0 4 1 − sin θ cos θ dθ , cos θ putting t = sin θ so that dt = cos θ dθ = π ⁄2 π ab ⌠ ⎮ 4 ⌡0 (1 − sin θ) dθ = π ab ⎡ ⎤π ⁄2 ⎢ θ + cos θ⎥ 4 ⎣ ⎦0 π ab ⎡ ⎛ π π ab ⎛ π ⎞ ⎞ ⎤ ⎢ ⎜ + 0⎟ − ⎛⎜ 0 + 1⎞⎟ ⎥ = ⎜ − 1⎟ ⋅ 4 ⎣⎝ 2 4 ⎝2 ⎝ ⎠⎦ ⎠ ⎠ H ence the given integral extended over the whole region of the ellipse = 1 x2 ⁄ a2 + y2 ⁄ b2 = 1 = 4. I = π ab ( π − 1) . 2 ⌠⌠⌠⌠ Example 7 : Prove that I = ⎮⎮⎮⎮ dx dy dz dw , for all positive values of the ⌡⌡⌡⌡ variables for which x2 + y2 + z2 + w2 is π 2 (b4 − a4) ⁄ 32 . Solution : We have to evaluate I a2 < x2 + y2 + z2 + w2 < not less than a2 and not greater than b2 is subject to the condition b2 . 1 Putting x2 = u1 i.e., x = u11 ⁄ 2, dx = u1− 1 ⁄ 2 du1 etc., we get 2 ⌠⌠⌠⌠ 1 1 1 1 I = ⎮⎮⎮⎮ u−1 1 ⁄ 2 ⋅ u−2 1 ⁄ 2⋅ u−3 1 ⁄ 2 ⋅ u−4 1 ⁄ 2 du1 du2 du3 du4 ⌡⌡⌡⌡ 2 2 2 2 subject to the condition a2 < u1 + u2 + u3 + u4 < b2 or I= 1 ⌠⌠⌠⌠ (1 ⁄ 2) − 1 (1 ⁄ 2) − 1 (1 ⁄ 2) − 1 (1 ⁄ 2) − 1 ⎮⎮⎮⎮ u u2 u3 u4 du1 du2 du3 du4 16 ⌡⌡⌡⌡ 1 1 [Γ ( )]4 1 2 = ⋅ 16 Γ ( 1 + 1 + 1 + 1 ) 2 = (√π) 4 16 Γ (2) 2 b 2 2 b2 ⌠ ⎮ t1 ⁄ 2 + 1 ⁄ 2 + 1 ⁄ 2 + 1 ⁄ 2 − 1 dt , ⌡a2 2 by Liouville’s theorem 2 π 2 ⎡ t2 ⎤ b π2 4 ⌠ ⎮ t dt = ⋅ ⎢ ⎥ = (b − a4) . ⌡a2 16 ⎣ 2 ⎦ a2 32 I-117 DIRICHLET’S AND LIOUVILLE’S INTEGRALS 1. ⌠⌠⌠ (i) Show that the integral ⎮⎮⎮ x l− 1 y m− 1 z n− 1 dx dy dz integrated over the region ⌡⌡⌡ in the first octant below the surface (x ⁄ a) p + ( y ⁄ b) q + (z ⁄ c) r = 1 is, Γ (l ⁄ p) Γ (m ⁄ q) Γ (n ⁄ r) a l bm c n ⋅ ⋅ Γ (l ⁄ p + m ⁄ q + n ⁄ r + 1) pqr (Avadh 2011) (ii) Show that if l, m, n are all positive, al bm c n Γ (l ⁄ 2) Γ (m ⁄ 2) Γ (n ⁄ 2) , ⌠⌠⌠ l− 1 m− 1 n− 1 ⎮⎮⎮ x y z dx dy dz = ⋅ ⌡⌡⌡ Γ (l ⁄ 2 + m ⁄ 2 + n ⁄ 2) 8 where the triple integral is taken throughout the part of the ellipsoid x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 = 1, which lies in the positive octant. 2. Prove that the area in the positive quadrant between the curve x n + y n = an and the coordinate axes is a2 [Γ (1 ⁄ n)]2 ⋅ 2 n Γ (2 ⁄ n) (Kanpur 2009) 3. (i) Find the volume in the positive octant of the ellipsoid x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 = 1. (ii) Find the volume of the ellipsoid x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 = 1. (iii) E valuate 4. ⌠⌠⌠ ⎮⎮⎮ dx dy dz, where x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 ≤ 1. ⌡⌡⌡ ⌠⌠⌠ E valuate ⎮⎮⎮ xyz dx dy dz for all positive values of the variables throughout the ⌡⌡⌡ ellipsoid x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 = 1. 5. 6. 7. (Kanpur 2011) The plane x ⁄ a + y ⁄ b + z ⁄ c = 1 meets the coordinate axes in the points A, B, C. Use Dirichlet’s integral to evaluate the mass of the tetrahedron OA BC , the density at any point (x, y, z) being k xyz. (Garhwal 2003) ⌠⌠⌠ E valuate the integral ⎮⎮⎮ x2 y z dx dy dz over the volume enclosed by the region ⌡⌡⌡ x, y, z, ≥ 0 and x + y + z ≤ 1. (i) Evaluate the double integral ⌠⌠ 1 ⁄ 2 1 ⁄ 2 ⎮⎮ x y (1 − x − y) 2 ⁄ 3 dx dy ⌡⌡D (Agra 2003) over the domain D bounded by the lines x = 0, y = 0, x + y = 1. ⌠⌠ (ii) E valuate ⎮⎮ x1 ⁄ 2 y1 ⁄ 2 (1 − x − y) 3 ⁄ 2 dx dy, where T is the region bounded by ⌡⌡T x ≥ 0, y ≥ 0, x + y ≤ 1. ⌠⌠⌠ (iii) E valuate ⎮⎮⎮ e x+ y+ z dx dy dz taken over the positive octant such that ⌡⌡⌡ x + y + z ≤ 1. (Kanpur 2005, 10) I-118 8. INTEGRAL CALCULUS ⌠⌠⌠ (i) E valuate ⎮⎮⎮ x− 1 ⁄ 2 y− 1 ⁄ 2 z− 1 ⁄ 2 (1 − x − y − z) 1 ⁄ 2 dx dy dz extended to all posi ⌡⌡⌡ tive values of the variables subject to the condition x + y + z < 1. (Kanpur 2007) π2 ⌠⌠⌠ dx dy dz , the integral extended to all positive (ii) Prove that ⎮⎮⎮ = ⌡⌡⌡ √(1 − x2 − y2 − z2) 8 values of the variables for which the expression is real. (iii) If S is a unit sphere with its centre at the origin, then prove that ⌠⌠⌠ dx dy dz ⎮⎮⎮ = π 2. ⌡⌡⌡S √(1 − x2 − y2 − z2) 9. ⌠⌠⌠ E valuate ⎮⎮⎮ (x + y + z + 1) 2 dx dy dz, where R is the region defined by ⌡⌡⌡R x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1. 10. ⌠⌠ ⎮⎮ ⌡⌡ Show that ⎛ 1 − x2 − y2 ⎞ 1 ⁄ 2 π ⎜ ⎟ ⎜ ⎟ dx dy = (π − 2) ⎜ 1 + x2 + y2 ⎟ 8 ⎝ ⎠ over the positive quadrant of the circle x2 + y2 = 1. 11. ⌠⌠⌠ E valuate ⎮⎮⎮ ⌡⌡⌡ √ ⎛ 1 − x2 − y2 − z2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 + x2 + y2 + z2 ⎟ dx dy dz integral being taken over all positive ⎝ ⎠ values of x, y, z such that x2 + y2 + z2 ≤ 1. 12. ⌠⌠ (i) E valuate ⎮⎮ √(x2 + y2) dx dy ⌡⌡R where R is the region in the xy-plane bounded by x2 + y2 = 4 and x2 + y2 = 9. ⌠⌠ (ii) Evaluate ⎮⎮ √(x2 + y2) dx dy where R is the region x2 + y2 ≤ a2. ⌡⌡R 13. E valuate the integral ⌠⌠⌠ ⎮⎮⎮ √(1 − x2 − y2 − z2) dx dy dz ⌡⌡⌡R 14. 15. where R is the region interior to the sphere x2 + y2 + z2 = 1. Find the mass of the region bounded by the ellipsoid x2 ⁄ a2 + y2 ⁄ b2 + z2 ⁄ c2 = 1 if the density varies as the square of the distance from its centre. ⌠⌠⌠ dx dy dz 1 Prove that ⎮⎮⎮ = ⌡⌡⌡ (x + y + z + 1) 3 2 ⎡ ⎢ log 2 ⎣ − 5 ⎤⎥ 8⎦ throughout the volume bounded by the coordinate planes and the plane x + y + z = 1. (Rohilkhand 2013) 3. (i) π abc ⁄ 6. 4. a2b2c2 ⁄ 48. (ii) 4 3 π abc. 5. k a2b2c2 ⁄ 720. 4 3 (iii) π abc. 6. 1 ⁄ 2520. I-119 DIRICHLET’S AND LIOUVILLE’S INTEGRALS 7. (i) 27π ⁄ 1760. 8. (i) π 2 ⁄ 4. 11. 1 8 ⎡ ⎢ ⎣ 13. 2 ‘π ⁄ 4. π B ⎛ ⎜ ⎝ 3 4 (ii) 2 π ⁄ 315. 9. , 1 ⎞⎟ 2⎠ − B ⎛ ⎜ ⎝ 5 4 , 1 ⎞⎟ 2⎠ ⎤ ⎥ ⎦ (iii) (e − 2) ⁄ 2 . 31 ⁄ 60. (ii) 2 πa3 ⁄ 3. ⋅ 12. (i) 38π ⁄ 3. 14. 8π abck (a2 + b2 + c2) ⁄ 30, where k is constant. Fill in the Blanks: Fill in the blanks “……” so that the following statem ents are com plete and correct. 1. If l, m, n are all positive, then the triple integral Γ (l) Γ (m) Γ (n) , ⌠ ⌠ ⌠ l− 1 m− 1 n− 1 ⎮⎮⎮ x y z dx dy dz = ⌡⌡⌡ …… where the integral is extended to all positive values of the variables x, y and z subject to the condition x + y + z ≤ 1. 2. If the variables x, y, z are all positive such that h1 ≤ x + y + z ≤ h2 , then the triple integral ⌠⌠⌠ ⎮ ⎮ ⎮ F (x + y + z) x l− 1 y m− 1 z n− 1 dx dy dz ⌡⌡⌡ = Γ (l) Γ (m) Γ (n) …… h ⌠ 2 ⎮ F (t) t l+ m+ n− 1 dt. ⌡h 1 Multiple Choice Questions: 3. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). If x, y, z ≥ 0 and h1 ≤ x + y + z ≤ h2 , the value of ⌠⌠⌠ ⎮⎮⎮ F (x + y + z) x l − 1 y m − 1 z n − 1 dx dy dz is equivalent to ⌡⌡⌡ (a) (b) Γ (l) Γ (m) Γ (n) Γ (l + m + n) 1 Γ (l) Γ (m) Γ (n) Γ (l + m + n + 1) Γ (l) Γ (m) Γ (n) (c) Γ (l + m + n) (d) h ⌠ 2 ⎮ F (u) u l + m + n − 1 du ⌡h h ⌠ 2 ⎮ F (u) u l + m + n − 1 du ⌡h 1 h2 ⌠ ⎮ F (u) u l + m + n du ⌡h Γ (l) Γ (m) Γ (n) Γ (l + m + n + 1) 1 h ⌠ 2 ⎮ F (u) u l + m + n du ⌡h 1 (Garhwal 2002) I-120 4. INTEGRAL CALCULUS Dirichlet’s theorem can be generalized for n variables, where (a) n ≤ 4 (b) n ≤ 100 (d) none of these ‘(c) n is any positive integer (Kumaun 2008) True or False: Write ‘T ’ for true and ‘F ’ for false statem ent. 1 5. 1⌠ u2 ⌠⌠⌠ dx dy dz = ⎮ du, ⎮⎮⎮ 3 ⌡ ⌡ ⌡ (x + y + z + 1) 3 ⌡0 (u + 1) 3 where the region of integration is the volume bounded by the coordinate planes and the plane x + y + z = 1. 1 6. 7. 1⌠ ⌠⌠⌠ ⎮ ⎮ ⎮ (x + y + z + 1) 2 dx dy dz = ⎮ u2 (u + 1) 2 du, ⌡⌡⌡ 2 ⌡0 where the region of integration is the volume bounded by the coordinate planes and the plane x + y + z = 1. If l and m are both positive, then the double integral Γ (l) Γ (m) , ⌠ ⌠ l− 1 m− 1 ⎮⎮ x y dx dy = ⌡⌡ Γ (l + m) where the integral is extended to all positve values of the variables x and y subject to the condition x + y ≤ 1. 1. 4. 7. Γ (l + m + n + 1). (c). F. 2. Γ (l + m + n). 5. F. 3. (a). 6. T. 6.1 Quadrature The process of finding the area of any bounded portion of a curve is called quadrature. 6.2 Areas of Curves given by Cartesian Equations If f ( x ) is a continuous and single valued function of x , then the area bounded by the curve y = f ( x ) , the axis of x and the ordinates x = a and x = b is b b ⌠ ⌠ ⎮ y dx , or ⎮ f ( x ) dx. ⌡a ⌡a Proof. Let CD be the arc of the curve y = f ( x ) and A C and BD be the two ordinates x = a and x = b . and C o n si d e r P ( x, y ) Q ( x + δx, y + δy ) , the t wo neighbouring points on the curve. Draw PM and QN perpendiculars to the axis of x , then PM = y , QN = y + δy a n d MN = δx . I-122 INTEGRAL CALCULUS Draw PR and QS perpendiculars to N Q and MP produced respectively. The area A MPC depends upon the position of P on the curve. Let A denote the area A MPC and A + δA be the area A NQC . Then the area MN QP = area A N QC − area A MPC = A + δA − A = δA . But clearly this area δA (i.e., the area MN QP) lies in magnitude between the areas of the rectangles MN RP and MN QS . Thus, we have Area of the rectangle MN QS > δA > area of the rectangle MNRP δA or y + δy > > y. i.e., ( y + δy ) δx > δA > y δx δx Now as Q → P , δx → 0 and δy → 0 . Therefore we have dA = y = f ( x ) , or dA = y dx . dx Integrating both sides between the limits x = a and x = b , we have x= b b b ⌠ ⌠ dA = ⎮ y dx ⎮ ⌡a ⌡x = a ⎡ A ⎤x= b = ⌠ ⎮ y dx ⎢ ⎥ ⎣ ⎦ x = a ⌡a or b or ⌠ (Area A when x = b ) − (Area A when x = a) = ⎮ y dx ⌡a or ⌠ Area A BDC − 0 = ⎮ y dx ⌡a or ⌠ ⌠ Area ABDC = ⎮ y dx = ⎮ f ( x ) dx . ⌡a ⌡a b b b Similarly, it can be shown that the area bounded by the curve x = f ( y ) , the axis of y and the abscissae y = a and y = b is b b ⌠ ⌠ ⎮ x dy , or ⎮ f ( y ) dy . ⌡a ⌡a Note 1 : In choosing the limits of integration, the lower limit of integration should be taken as the smaller value of the independent variable while the greater value gives us the upper limit of integration. Note 2 : If the curve is symmetrical about x-axis or y-axis or both, then we shall find the area of one symmetrical part and multiply it by the number of symmetrical parts to get the whole area. Example 1 : Find the area bounded by the ellipse x = c , x = d and the x-axis. Solution : E quation of the ellipse is x2 y2 + 2= 1 2 a b or y2 x2 = 1− 2 2 b a y2 x2 + 2 = 1, the ordinates 2 a b (Meerut 2000) I-123 AREAS OF CURVES b √(a2 − x2) . ...(1) a ∴ the required area = the area ABDC giving y= d d ⌠ ⌠ b = ⎮ y dx = ⎮ √(a2 − x2) dx , ⌡c ⌡c a from (1) ⎡1 ⎛ x⎞⎤d 1 ⎢ x √(a 2 − x2) + a 2 sin− 1 ⎜ ⎟ ⎥ 2 ⎝ a⎠⎦c ⎣2 = b a = b ⎡ d c⎞ ⎤ ⎛ ⎢ d √(a 2 − d2) − c √(a 2 − c2) + a 2 ⎜ sin− 1 − sin− 1 ⎟ ⎥ ⋅ 2a ⎣ a a⎠ ⎦ ⎝ Example 2 : Find the area bounded by the parabola y2 = 4ax and its latus rectum. (Agra 2005; Avadh 05; Bundelkhand 08) Solution : Latus rectum is a line through the focus S (a, 0) and perpendicular to x-axis i.e., its equation is x = a . Also the curve is symmetrical about x-axis. ∴ the required area L OL′ a ⌠ = 2 × area OSL = 2 . ⎮ y dx ⌡0 a ⌠ = 2 ⎮ √(4ax ) dx , ⌡0 [ ... y2 = 4ax , i.e., y = √(4ax )] ⎡ 2 3 ⎤ = 2 √(4a) ⎢⎣ x3 ⁄ 2⎥⎦ a 0 = 8 3 √(a) . a3 ⁄ 2 = 8 3 a 2. Example 3 : Find the area of a loop of the curve xy2 + ( x + a) 2 ( x + 2a) = 0 . T h e cu r ve is sym m e t r ica l Solution : about x-axis. Putting y = 0 , we get x = − a and x = − 2a . The loop is formed between x = − 2a and x = − a . To find the area of the loop, we first shift the origin to the point (− a, 0) . The equation of the curve then becomes ( x − a) y2 + {( x − a) + a}2 ( x − a + 2a) = 0 or y2 ( x − a) + x2 ( x + a) = 0 or y2 = x2 (a + x ) ⋅ a− x ...(1) I-124 INTEGRAL CALCULUS Note that the shifting of the origin only changes the equation of the curve and has no effect on its shape. Now the origin being at the point A , the new limits for the loop are x = − a to x = 0 . ∴ required area of the loop = 2 × area CPA 0 ⌠ = 2 ⎮ y dx , [the value of y to be put from (1)] ⌡− a 0 ⎧ ⌠ = 2⎮ ⌡− a √ ⎪ ⎨− x ⎪ ⎩ ⎫ ⎛ a + x⎞ ⎪⎬ ⎜ ⎟ ⎪ dx , ⎝ a − x⎠ ⎭ [Note that in the equation (1), for the portion CPA , y = − x √{(a + x ) ⁄ (a − x )}] 0 ⌠ − x (a + x ) = 2⎮ dx , multiplying the numerator and the ⌡− a √(a2 − x2) denominator by √(a + x ) 0 − (− a sin θ) (a − a sin θ) ⌠ ⋅ (− a cos θ) dθ , = 2⎮ ⌡π ⁄ 2 a cos θ putting x = − a sin θ and dx = − a cos θ dθ π ⁄2 0 ⌠ ⌠ (sin θ − sin2 θ) dθ = 2a2 ⎮ = − 2a2 ⎮ ⌡π ⁄ 2 ⌡0 = 2a2 [1 − = 2a2 (1 − Example 4 : 1 2 1 4 ⋅ 1 2 (sin θ − sin2 θ) dθ π] , by Walli’s formula π) . Find the whole area of the curve a2 y2 = x3 (2a − x ) . (Meerut 2006B; Bundelkhand 12; Avadh 13; Rohilkhand 14) Solution : The given curve is a2 y2 = x3 (2a − x ) . It is symmetrical about x-axis and it cuts the x-axis at the points (0, 0) and (2a, 0) . The curve does not exist for x > 2a and x < 0 . Thus the curve consists of a loop lying between x = 0 and x = 2a . 2a ⌠ ∴ the required area = 2 × area OBA = 2 ⎮ ⌡0 y dx 2a ⌠ x3 ⁄ 2 √(2a − x ) dx , from (1). = 2⎮ ⌡0 a Now put x = 2a sin2 θ so that dx = 4a sin θ cos θ dθ . When x = 0 , θ = 0 and when x = 2a , θ = ∴ the required area = π ⁄2 ⌠ = 32a2 ⎮ ⌡0 = πa2. 2 a π ⁄2 ⌠ ⎮ ⌡0 1 2 π. (2a) 3 ⁄ 2 sin3 θ .√(2a) .cos θ . 4a sin θ cos θ dθ sin4 θ cos2 θ dθ = 32a2. 3.1.1 π , ⋅ by Walli’s formula. 6.4.2 2 ...(1) I-125 AREAS OF CURVES Example 5 : Find the whole area between the curve x2 y2 = a2 ( y2 − x2) and its asymptotes. Solution : The given curve is symmetrical about both the axes and passes through the o r igin . T h e t a n gen t s at (0, 0) a r e gi ve n b y y2 − x2 = 0 i.e., y = ± x are the tangents at the origin. E qu at in g t o zero th e coefficient of th e highest power of y (i.e., of y2) the asymptotes parallel to y-axis are given by x2 − a2 = 0 i.e., x= ± a. The asymptotes parallel to x-axis are given by y2 + a2 = 0 which gives two imaginary asymptotes. ∴ the required area = 4 × area lying in the first quadrant a a ⌠ ⌠ = 4 ⎮ y dx = 4 ⎮ ⌡0 ⌡0 [ ... √ ⎛ a 2 x2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ a 2 − x2 ⎟ dx , ⎝ ⎠ from the equation of the given curve, y2 = a2 x2 ⁄ (a2 − x2)] a a a ⎡ (a 2 − x2) 1 ⁄ 2 ⎤ ⌠ ax dx ⌠ − 2x dx ⎢⎢ ⎥⎥ = 4⎮ = − 2a ⎮ = − 2a ⎢⎣ ⎥⎦ ⌡0 √(a2 − x2) ⌡0 √(a2 − x2) 1⁄2 0 = − 4a [0 − a] = 4a2. 1. Find the area bounded by the axis of x , and the following curves and the given ordinates : (i) y = log x ; x = a , x = b (b > a > 1) . (ii) xy = c2 ; x = a , x = b , (a > b > 0) . 2. (Kashi 2012) x3, (i) Find the area bounded by the curve y = the y-axis and the lines y = 1 and y= 8. (ii) Show that the area cut off a parabola by any double ordinate is two third of the corresponding rectangle contained by that double ordinate and its distance from the vertex. 3. (i) Find the area of the quadrant of an ellipse ( x2 ⁄ a2) + ( y2 ⁄ b2) = 1 . (Bundelkhand 2010; Kanpur 11) (ii) Find the whole area of the ellipse ( x2 ⁄ a2) + ( y2 ⁄ b2) = 1. (Avadh 2010; Rohilkhand 10B) 4. (i) Trace the curve ay2 = x2 (a − x ) and show that the area of its loop is 8a2 ⁄ 15 . (Avadh 2008) I-126 INTEGRAL CALCULUS (ii) Find the area of the loop of the curve 3ay2 = x ( x − a) 2. (iii) Find the area of the loop of the curve y2 = x ( x − 1) 2. 5. Find the area (i) of the loop of the curve x ( x2 + y2) = a ( x2 − y2) or y2 (a + x ) = x2 (a − x ) . (Meerut 2004) (ii) of the portion bounded by the curve and its asymptotes. 6. (i) Trace the curve y2 (2a − x) = x3 and find the entire area between the curve and (Avadh 2011) its asymptotes. (ii) Find the area between the curve y2 (4 − x ) = x2 and its asymptote. (Avadh 2012; Kanpur 14; Bundelkhand 14) (iii) Find the whole area of the curve a2 x2 = y3 (2a − y ) . 7. (i) Find the area bounded by the curve xy2 = 4a2 (2a − x ) and its asymptote. (Rohilkhand 2009B) (ii) Find the area enclosed by the curve (iii) Trace the curve a2 y2 = a2 x2 x4 − 8. (i) Prove that the area of a loop of xy2 = a2 (a − x ) and y-axis. and find the whole area within it. (Rohilkhand 2012; Avadh 12, Bundelkhand 14) the curve a4 y2 = x4 (a2 − x2) is πa2 ⁄ 8 . (ii) Show that the whole area of the curve a4 y2 = x5 (2a − x ) is to that of the circle (Kanpur 2010) whose radius is a, as 5 to 4. 9. (i) Find the area between the curve y2 (a − x ) = x3 (cissoid) and its asymptotes. Also find the ratio in which the ordinate x = a ⁄ 2 divides the area. (ii) Find the area of the loop of the curve y2 (a − x ) = x2 (a + x ) . (Purvanchal 2011) 10. Trace the curve asymptote. y2 (a + x ) = (a − x ) 3. Find the area between the curve and its (Purvanchal 2007) 1. 2. (i) b log (b ⁄ e) − a log (a ⁄ e) 3. (i) πab ⁄ 4. (i) (45) ⁄ 4. 4. (ii) 8a2 ⁄ (15 √3). 1 2 a2 (4 − π). (iii) 8 ⁄ (15). 1 2 (ii) 5. (i) 6. (i) 3πa2. 7. (i) 4πa2. 9. (i) 3πa2 ⁄ 4 ; (3π − 8) : (3π + 8). 10. 3πa2. (ii) c 2 log (a ⁄ b). (ii) πab. a2 (4 + π). (ii) (64) ⁄ 3. (ii) πa2 (iii) πa2. (iii) 4a2 ⁄ 3. (ii) 1 2 a2 (4 − π). I-127 AREAS OF CURVES 6.3 Area Between Two Curves It is a clear from the adjacent figure that the area lyin g bet ween t he cu rves y = f (x), y = g (x) an d t h e ordinates x = a, x = b is = area ABNM − area CDN M b b ⌠ ⌠ = ⎮ f (x) dx − ⎮ g (x) dx ⌡a ⌡a b ⌠ = ⎮ {f (x) dx − g (x)} dx. ⌡a Example 1 : Find the area included between the curves y2 = 4ax and x2 = 4by . (Bundelkhand 2011; Rohilkhand 10) Solution : Solving the equations of the two given curves, we have y4 = 16a2 (4by ) = 64a2 by . y ( y3 − 64 a2b) = 0 , giving y = 0 , 4a2 ⁄ 3 b1 ⁄ 3. ∴ W h e n y = 0 , x = 0 a n d wh e n y = 4a2 ⁄ 3 b1 ⁄ 3, x = 4a1 ⁄ 3 b2 ⁄ 3. H ence, the points of intersection of the given curves are O (0, 0) and A (4a1 ⁄ 3 b2 ⁄ 3, 4b1 ⁄ 3 a2 ⁄ 3) . ∴ the required area (i.e., the shaded area) = area OPA L − area OQA L 4a1 ⁄ 3 b2 ⁄ 3 ⌠ = ⎮ ⌡0 y dx , from the curve y2 = 4ax 4a1 ⁄ 3 b2 ⁄ 3 ⌠ − ⎮ ⌡0 y dx , from the curve x2 = 4by (Note that for the required area x varies from 0 to 4a1 ⁄ 3 b2 ⁄ 3) 4a1 ⁄ 3 b2 ⁄ 3 ⌠ = ⎮ ⌡0 4a1 ⁄ 3 b2 ⁄ 3 ⎛ 2 ⎞ x ⌠ √(4ax ) dx − ⎮ ⌡0 ⎡ 2x3 ⁄ 2 ⎤⎥ 4a ⎥ = 2 √a ⎢⎢⎢ ⎣ 3 ⎥⎦ 0 = 1⁄ 3 2⁄ 3 ⎜ ⎟ dx ⎝ 4b ⎠ 4a1 ⁄ 3 b2 ⁄ 3 b − 1 ⎡ x3 ⎤ ⎢ ⎥ 4b ⎣ 3 ⎦ 0 4 √a 1 32 16 16 [8√(a) . b] − (64 ab2) = ab − ab = ab . 12b 3 3 3 3 Example 2 : Find the area of the segm ent cut off from the parabola y2 = 2x by the straight line y = 4x − 1 . I-128 INTEGRAL CALCULUS Solution : The given curves are y2 = 2x , ...(1) ...(2) and y = 4x − 1 . The two curves have been shown in the figure. Solving (1) and (2) for y we have 1 4 y2 = 2 . ( y + 1) or 2y2 − y − 1 = 0 or ( y − 1) (2y + 1) = 0 . ∴ y= − 1 , 1. 2 Thus the curves (1) and (2) intersect at the points where y= − 1 2 and y= 1. Now the required area of the segment POQ (i.e., the dotted area) = the area bounded by the st. line y = 4x − 1 and the y-axis from y= − 1 2 to y = 1 − the area bounded by the parabola y2 = 2x and the y-axis from y= − 1 1 2 to y = 1 1 ⌠ ⌠ x dy , from (2) − ⎮ x dy , from (1) = ⎮ ⌡− 1 ⁄ 2 ⌡− 1 ⁄ 2 1 ⌠ 1 = ⎮ ( y + 1) dy − ⌡− 1 ⁄ 2 4 1 ⌠ 1 2 ⎮ y dy ⌡− 1 ⁄ 2 2 ⎤1 1 ⎡1 2 1 1 ⎡ 3 ⎛ 1 1⎞ ⎤ 1⎛ 1⎞ 1 − ⎡⎣ y3 ⎤⎦ − 1 ⁄ 2 = ⎢ − ⎜ − ⎟ ⎥ − ⎜ 1 + ⎟ ⎢ y + y⎥ ⎦− 1⁄2 4 ⎣2 6 4 ⎣ 2 ⎝ 8 2⎠ ⎦ 6⎝ 8⎠ 1 ⎛ 3 3⎞ 1 9 1 15 1 9 15 3 9 = ⎜ + ⎟− ⋅ = ⋅ − ⋅ = − = ⋅ 4 ⎝ 2 8⎠ 6 8 4 8 6 8 32 16 32 = Example 3 : If P ( x, y ) be any point on the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 and S be the sectorial area bounded by the curve, the x-axis and the line joining the origin to P, show that x = a cos (2S ⁄ ab) , y = b sin (2S ⁄ ab) . Solution : The given ellipse is shown in the figure. We have S = the sectorial area OA P (i.e., the dotted area) = the area of the Δ OMP + the area PMA a = 1 ⌠ OM . MP + ⎮ y dx , ⌡x 2 for the ellipse a = ⌠ b 1 xy + ⎮ √(a2 − x2) dx ⌡x a 2 [... from the equation of the ellipse, y = (b ⁄ a) √(a2 − x2)] I-129 AREAS OF CURVES 1 b b⎡x 1 x⎤ a x . √(a2 − x2) + ⎢ √(a2 − x2) + a2 sin− 1 ⎥ 2 a a ⎣2 2 a⎦ x π bx b⎡ 1 x 1 x⎤ = √(a2 − x2) + ⎢ 0 + a2. − √(a2 − x2) − a2 sin− 1 ⎥ 2a a⎣ 2 2 2 2 a⎦ bx b 1 ⎛π x⎞ bx = √(a2 − x2) + ⋅ a2 ⎜ − sin− 1 ⎟ − √(a2 − x2) 2a a 2 ⎝2 a⎠ 2a x⎞ ab x ab ⎛ π cos− 1 ⋅ = ⎜ − sin− 1 ⎟ = a⎠ 2 a 2 ⎝2 x ab cos− 1 ⋅ S= a 2 x 2S x 2S 2S cos− 1 = or = cos or x = a cos ⋅ a ab a ab ab b 2S b ⋅ y = √(a2 − x2) = √{a2 − a2 cos2 (2S ⁄ ab)} = b sin a ab a = Thus ∴ Also Example 4 : S h ow th at th e larger of the two areas in to wh ich the circle 16 2 + = 64a2 is divided by the parabola y2 = 12ax is a [8 π − √3] . 3 Solution : x2 + y2 = 64a2 is a circle with centre (0, 0) and radius 8a and y2 = 12ax is a parabola whose vertex is at (0, 0) an d lat u s rectum 12a . Both the curves are symmetrical abou t x-axis. So lvin g t he two equ atio ns, t he co -o r di n a t e s o f t h e co m m o n p o in t P a r e (4a, 4a √3) . D r a w PM p e r p e n dicu la r fr o m P to the y-axis. Now the area of the larger portion of the circle (i.e., the shaded area) = the area PRSTQOP = the area of the semi-circle RST + 2 area OPR x2 y2 = 1 2 . π (8a) 2 + 2 [area OPM + area MPR] = 1 2 ⌠ π (8a) 2 + 2 ⎮ ⌡0 4a √3 4a √3 ⌠ = 32π a2 + 2 ⎮ ⌡0 8a ⌠ x dy , for y2 = 12ax + 2 ⎮ x dy , for ⌡4a √3 x2 + y2 = 64a2 8a y2 ⌠ dy + 2 ⎮ √(64a2 − y2) dy ⌡4a √3 12a 8a ⎡ y3 ⎤ 4a √3 ⎡1 64a2 y⎤ ⎢ ⎥ + 2 ⎢ y √(64a2 − y2) + sin− 1 ⎥ ⎣ 3 ⎦0 ⎣2 2 8a ⎦ 4a √3 = 32 πa2 + 1 6a = 32 πa2 + 1 ⎡ 64 × 3 √3a3 ⎤ ⎢ ⎥ ⎦ 3 6a ⎣ + 2 [{0 − 8a2 √3} + 32a2 {sin− 1 1 − sin− 1 (√3 ⁄ 2)}] = 32 πa2 + 32 √3a2 32 2 − 16a2 √3 + a π 3 3 I-130 INTEGRAL CALCULUS = 128 2 16 2 16 2 a π− a √3 = a (8π − √3) . 3 3 3 Example 5 : Find by double integration the area of the region enclosed by the curves x2 + y2 = a2, x + y = a (in the first quadrant). Solution : The given equations of the circle x2 + y2 = a2 [centre (0, 0) and radius a] and of the st r a i gh t l i n e x + y = a ( wi t h e q u a l intercepts a on both the axes) can be easily traced as shown in the figure. T h e r e quir e d a r e a is t he ar ea bounded by the arc AB an d t he line A B . To find it with the help of double integration take any point P ( x, y ) in this portion and consider an elementary area δx δy at P . The required area can now be cover ed by fir st mo ving y fr o m t h e straight line x + y = a to the arc of the circle x2 + y2 = a2 an d then moving x from 0 to a . √(a2 − x2) a ⌠ ⌠ ⎮ dx dy , the first integration to be ∴ the required area = ⎮ ⌡x = 0 ⌡y = (a − x ) performed w.r.t. y whose limits are variable a a √(a2 − x2) ⌠ ⌠ dx = ⎮ [√(a2 − x2) − (a − x)] dx = ⎮ ⎡⎢ y ⎤⎥ ⌡0 ⌡0 ⎣ ⎦ (a − x ) a ⎡ ⎧1 ⎫ 1 1 ⎤ = ⎢ ⎨ x √(a2 − x2) + a2 sin− 1 ( x ⁄ a) ⎬ − ax + x2⎥ ⎣ ⎩2 = 1 2 2 1 2 a2. ( π) − a2 + 1 2 2 ⎭ a2 = 1 2 1 2 a2 ( π − 1) = 1 4 ⎦0 a2 (π − 2) . Note : The required area can also the covered by first moving x from the st. line x + y = a to the arc of the circle x2 + y2 = a2 and then moving y from 0 to a . 1. Find the common area between the curves y2 = 4ax and x2 = 4ay . (Meerut 2004B, 08; Agra 14) 2. (i) Find the area included between y2 = 4ax and y = m x . (ii) Find the area of the segment cut off from the parabola y2 = 4x by the line y = 8x − 1 . I-131 AREAS OF CURVES 3. (i) Find the area common to the two curves y2 = ax , x2 + y2 = 4ax . (Meerut 2005B, 06, 09B) ( i i) F i n d t h e a r e a lyin g a b o ve x-axis an d included be t wee n t he circle x2 + y2 = 2ax and the parabola y2 = ax . (Bundelkhand 2007) 4. (i) Show that the area included between the parabolas y2 = 4a ( x + a) , y2 = 4b (b − x ) 8 3 is a2 x2 (ii) Show that the area common to the ellipses (ab) − 1 5. 6. (a + b) √(ab) . + b2 y2 = (Rohilkhand 2013) 1 , b2 x2 + a2 y2 = tan− 1 where 0 < a < b , is 4 (a ⁄ b) . I f A is t h e vert ex, O t h e cen t re an d P any point (x, y) on the hyperbola x2 ⁄ a2 − y2 ⁄ b2 = 1 , show that x = a cosh (2S ⁄ ab) , y = b sinh (2S ⁄ ab) , where S is the sectorial area OPA . Prove that the area of a sector of the ellipse of semi-axes a and b between the major axis and a radius vector from the focus is 1 2 ab (θ − e sin θ) , where θ is the eccentric angle of the point to which the radius vector is drawn. 7. Find the area common to the circle x2 + y2 = 4 and the ellipse x2 + 4y2 = 9 . (Purvanchal 2009) 8. Find the area included between the parabola y= 8a3 ⁄ ( x2 + x2 = 4ay and the curve 4a2) . (Rohilkhand 2008B) 9. Find by double integration the area bounded by the curves y (x2 + 2) = 3x and 4y = x2. 10. Find by double integration the area lying between the parabola y = 4x − x2 and the straight line y = x. 1. 16 a2 ⁄ 3. 2. ⎛ ⎜ ⎝ 4 3 7. 4π + 9 ⎧1 sin− 1 ⎨⎩ 3 9. (3 ⁄ 2) log 3 − (2 ⁄ 3). 3. (i) a2 3 √3 + ⎞ ⎟ ⎠ π ⋅ ⎫ √(7 ⁄ 3) ⎬⎭ − (i) 8a2 ⁄ 3m 3. ⎡1 (ii) a2 ⎢⎣ π 4 ⎧1 8 sin− 1 ⎨⎩ ⋅ 2 − (ii) 9 ⁄ (64). 2 ⎤⎥ ⋅ 3⎦ ⎫ √(7 ⁄ 3) ⎬⎭ ⋅ 8. ⎡ ⎢ ⎣ 2π − 4 ⎤⎥ 3⎦ a 2. 6.4 Areas of Curves given by Parametric Equations To find the area of a curve given by parametric equations is explained by the following examples. 1, I-132 INTEGRAL CALCULUS Example 1 : Fin d th e area in cluded between the cycloid x = a (θ − sin θ) , y = a (1 − cos θ) and its base. (Meerut 2007B; Purvanchal 07; Kashi 13) Solution : The parametric equations of the given cycloid are x = a (θ − sin θ) , y = a (1 − cos θ) . We have dx ⁄ dθ = a (1 − cos θ) , dy ⁄ dθ = a sin θ . a sin θ dy dy ⁄ dθ = = ∴ dx dx ⁄ dθ a (1 − cos θ) = 2 sin 2 1 2 1 2 θ cos θ 1 sin2 2 θ 1 2 = cot θ . In this curve y = 0 when a (1 − cos θ) = 0 i.e., cos θ = 1 i.e., θ = 0 . When θ = 0 , x = a (0 − sin 0) = 0 , y = 0 and dy ⁄ dx = cot 0 = ∞ . Thus the curve passes through the point (0, 0) and the axis of y is tangent at this point. I n t h i s cu r ve y i s maximum wh e n cos θ = − 1 i.e., θ = π . W h e n θ = π , 1 2 x = a (π − sin π) = aπ , y = 2a , dy ⁄ dx = cot π = 0 . Thus at the point θ = π , whose cartesian co-ordinates are (aπ, 2a) , the tangent to the curve is parallel to x-axis. This curve does not exist in the region y > 2a . In this curve y cannot be − ive because cos θ cannot be greater than 1. Thus one complete arch of the given cycloid is as shown in the figure. Now this cycloid is symmetrical with respect to the line x = aπ (axis of the cycloid) and its base is the x-axis. Therefore the required area π aπ ⌠ ⌠ y dx = 2 ⎮ = 2⎮ ⌡θ = 0 ⌡x = 0 y dx ⋅ dθ dθ π π ⌠ ⌠ = 2 ⎮ a (1 − cos θ) . a (1 − cos θ) dθ = 2a2 ⎮ (1 − cos θ) 2 dθ ⌡0 ⌡0 π π ⌠ ⌠ 1 1 = 2a2 ⎮ (2 sin2 θ) 2 dθ = 8a2 ⎮ sin4 θ dθ 2 2 ⌡0 ⌡0 π ⁄2 ⌠ = 8a2 ⎮ ⌡0 sin4 φ 2dφ , putting π ⁄2 ⌠ = 16a2 ⎮ ⌡0 sin4 φ dφ = 16a2 ⋅ 1 2 θ = φ so that 1 2 dθ = dφ 3.1 π , ⋅ by Walli’s formula 4.2 2 = 3π a2. Example 2 : Find the whole area of the curve (hypocycloid) given by the equations x = a cos3 t , y = b sin3 t . (Gorakhpur 2005; Rohilkhand 09; Kashi 11) Solution : E liminating t from the given equations the cartesian equation of the curve is obtained as (x ⁄ a) 2 ⁄ 3 + (y ⁄ b) 2 ⁄ 3 = 1 i.e., {(x ⁄ a) 2 }1 ⁄ 3 + {( y ⁄ b) 2}1 ⁄ 3 = 1 . I-133 AREAS OF CURVES Since the powers of x and y are all even, the curve is symmetrical about both the axes. It does not pass through the origin. It cuts the axis of x at the point s (± a, 0) a n d t h e a xis o f y a t t h e p o in t s (0, ± b) . The tangent at the point (a, 0) is x-axis. At th e point B , x = 0 and t = 1 2 π . At the p oint A , x = a and t = 0 . ∴ the required area = 4 × area OAB a 0 dx ⌠ ⌠ = 4⎮ y dx = 4 ⎮ y⋅ ⋅ dt ⌡x = 0 ⌡t = π ⁄ 2 dt 0 ⌠ = 4⎮ b sin3 t .(− 3a cos2 t sin t) dt , ⌡π ⁄ 2 π ⁄2 ⌠ = 12ab ⎮ ⌡0 sin4 t cos2 t dt = 12ab ⋅ 3.1.1 π 3 ⋅ = π ab . 6.4.2 2 8 (putting for y and dx ⁄ dt) (Note) Find the area included between the curve x = a (t + sin t) , y = a (1 − cos t) and (Agra 2005) its base. 2. Find the area of a loop of the curve x = a sin 2t , y = a sin t or 1. a2 x2 = 4y2 (a2 − y2) . 3. Show that the area bounded by the cissoid x = a sin2 t , y = (a sin3 t) ⁄ cos t and its asymptote is 3 πa2 ⁄ 4 . (Purvanchal 2006; Avadh 09; Rohilkhand 11; Purvanchal 14) 4. Find the area of the loop of the curve x = a (1 − t2) , y = at (1 − t2) , where − 1 ≤ t ≤ 1 . 1. 3πa2. 2. 4a2 ⁄ 3 3. 3πa2 ⁄ 4. 4. 8a2 ⁄ (15). 6.5 Areas of Curves given by Polar Equations If r = f (θ) be the equation of a curve in polar coordinates where f (θ) is a single valued continuous function of θ , then the area of the sector enclosed by the curve and the two radii vectors θ = θ1 and θ = θ2 (θ1 < θ2) , is equal to 1 2 θ ⌠ 2 ⎮ r2 dθ . ⌡θ = θ 1 I-134 INTEGRAL CALCULUS Proof : Let OAB be the area of t he cu r ve r = f (θ) b e t we e n t h e r a d ii ve ct o r s θ = θ1 and θ = θ2 . Let P (r, θ) be any point on the curve Take a p o in t b e t we e n A and B . Q (r + δr, θ + δθ) on the curve very near to P a nd dr aw t h e radius vect o r OQ . L e t t h e sectorial areas A OP and A OQ be denoted by A and A + δA respectively. Then the curvilinear area OPQO = A + δA − A = δA . Also we have OP = r ; OQ = r + δr and ∠ POQ= δθ . The area of the circular sector POQ′ = 1 2 (radius × arc) = 1 2 r . rδθ = 1 2 r 2 δθ , and the area of the circular sector P′ OQ = 1 2 (r + δr) .(r + δr) δθ = 1 2 (r + δr) 2 δθ . Now, area POQ′ < area OPQ < area P′ OQ , i.e., i.e., 1 2 1 r δθ < δA < (r + δr) 2 δθ , 2 2 1 2 1 r < δA ⁄ δθ < (r + δr) 2. 2 2 Proceeding to limits as δθ→ 0 , we get dA 1 = r 2 or dθ 2 dA = 1 2 r dθ . 2 θ ⌠ 2 1 2 θ ∴ ⎡⎢ A ⎤⎥ 2 = ⎮ r dθ . ⌡θ 2 ⎣ ⎦θ 1 1 Now the L.H.S. = the value of A for θ equal to θ2 − the value of A for θ equal to θ1 = (the area A OB) − 0 = area A OB . H ence the required area A OB = θ 1⌠ 2 2 ⎮ r dθ . 2 ⌡θ 1 Note : In some cases it is more convenient to find the required area by using double integration. In that case the area is given by θ f (θ) ⌠ ⌠ 2 ⎮ ⎮ r dθ dr , (θ1 < θ2) . ⌡θ = θ ⌡r = 0 1 Remember : The number of loops in r = a cos nθ or r = a sin nθ is n or 2n according as n is odd or even. Example 1 : Find the area of the curve r2 = a2 cos 2θ . (Agra 2006, 07; Rohilkhand 07; Meerut 10B) Solution : The given curve is symmetrical about the initial line θ = 0 and about the pole. Putting r = 0 in the given equation of the curve, we get I-135 AREAS OF CURVES cos 2θ = 0 or 2θ = ± 1 2 π or θ = ± 1 4 π. Thus two consecutive values of θ for which r is zero are − 1 4 1 4 π and π . Therefore for one loop of the curve θ varies from − π ⁄ 4 to π ⁄ 4 . When 1 2 π < 2θ < 3 2 π i.e., 1 4 π < θ< 3 4 π , r2 is negative i.e., r is imaginary. Therefore this curve does not exist in the region 1 4 π < θ< 3 4 π. H ence this curve has only two loops as shown in the figure. ∴ whole area of the curve = 2 × area of one loop π ⁄4 π ⁄4 ⌠ 1 2 ⌠ = 2⎮ r dθ = ⎮ a2 cos 2θ dθ , ⌡− π ⁄ 4 2 ⌡− π ⁄ 4 π ⁄4 ⌠ = 2a2 ⎮ ⌡0 cos 2θ dθ , [... r2 = a2 cos 2θ] [by a property of definite integrals] 2a2 ⎡ sin 2θ⎤ π ⁄ 4 = = a 2. ⎥ ⎣ 2 ⎦0 2 = 2a2 ⎢ Example 2 : Find the area of the cardioid r = a (1 + cos θ). (Meerut 2003, 04B, 10B; Kashi 12) Solution : The given curve is symmetrical about the initial line since its equation remains unaltered when θ is changed into − θ . We have r = 0 , when cos θ = − 1 i.e., θ = π . Therefore the line θ = π is tangent at the pole to the curve. Also r is maximum when cos θ = 1 i.e., θ = 0 and then r = 2a . When θ increases from 0 to π , r decreases from 2a to 0. Thus the curve is as shown in the figure. Now the required area = 2 × area of the upper half of the curve π ⌠ 1 2 = 2⎮ r dθ ⌡0 2 π ⌠ 1 2 a (1 + cos θ) 2 dθ , = 2⎮ ⌡0 2 [... r = a (1 + cos θ)] π ⌠ 1 = a2 ⎮ (2 cos2 θ) 2 dθ 2 ⌡0 π ⌠ 1 = 4a2 ⎮ cos4 θ dθ . ⌡0 2 Now put 1 2 θ = φ so that 1 2 dθ = dφ . Also when θ = 0 , φ = 0 and when θ = π , φ = π ⁄ 2 . π ⁄2 ⌠ ∴ the required area = 8a2 ⎮ ⌡0 = 3π a2 ⁄ 2 . cos4 φ dφ = 8a2 ⋅ 3.1 π , ⋅ by Walli’s formula 4.2 2 I-136 INTEGRAL CALCULUS Example 3 : Find the area of a loop of the curve r = a cos 3θ + b sin 3θ . (Meerut 2000) Solution : In the given equation of the curve put a = k cos α , b = k sin α so that k = √(a2 + b2) and α = tan− 1 (b ⁄ a) . Thus the given equation reduces to r = k cos 3θ cos α + k sin 3θ sin α r = k cos (3θ − α) = k cos 3 (θ − or 1 3 α) . (Note) Now rotating the initial line through an angle α ⁄ 3 , the given equation of the curve becomes r = k cos 3 (θ + 1 3 α− 1 3 α) = k cos 3θ . (Note) It should be noted that the rotation of the initial line changes only the equation of the curve and has no effect on its shape. Therefore the area of a loop of the given curve is the same as the area of a loop of the curve r = k cos 3θ . The curve r = k cos 3θ is symmetrical about the initial line. Putting r = 0 in it we have, cos 3θ = 0 i.e., 3θ = ± π ⁄ 2 i.e., θ = ± π ⁄ 6 . ∴ one loop of this curve lies between θ = − π ⁄ 6 and θ = + π ⁄ 6 an d it is symmetrical about the initial line. ∴ π ⁄6 ⌠ the required area = 2 . ⎮ ⌡0 π ⁄6 ⌠ = ⎮ ⌡0 1 2 r dθ , 2 [By symmetry] k 2 cos2 3θ dθ . Now put 3θ = t , so that 3 dθ = dt . Also when θ = 0 , t = 0 and when θ = π ⁄ 6 , t = π ⁄ 2. ∴ the required area = k2 3 π ⁄2 ⌠ ⎮ ⌡0 cos2 t dt = k2 1 1 k2 ⋅ ⋅ π= π 3 2 2 12 [... k 2 = a2 + b2] = (a2 + b2) π ⁄ 12 . 1. Find the area between the following curves and the given radii vectors : 2. (i) The spiral r θ1 ⁄ 2 = a ; θ = α , θ = β . (ii) The parabola l ⁄ r = 1 + cos θ ; θ = 0 , θ = α . Find the area of the loop of the curve r = a θ cos θ between θ = 0 and θ = π ⁄ 2 . (Kanpur 2009) 3. (i) Find the area of one loop of r = a cos 4θ . 4. (ii) Find the area of a loop of the curve r = a sin 3θ . (i) Find the whole area of the curve r = a sin 2θ . (Rohilkhand 2007) (Bundelkhand 2009) (ii) Find the whole area of the curve r = a cos 2θ . 5. (i) Find the whole area of the curve r2 = a2 cos2 θ + b2 sin2 θ . (ii) Find the area of the cardioid r = a (1 − cos θ) . I-137 AREAS OF CURVES 6. (i) Show that the area of the limacon r = a + b cos θ , (b < a) is equal to π (a2 + 7. 1 2 b ). 2 ( ii) P r o ve t h a t t h e su m o f t h e ar e a s o f t h e t wo lo o p s o f t h e lim a co n r = a + b cos θ , (b > a) is equal to π (2a2 + b2) ⁄ 2 . Calculate the ratio of the area of the larger to the area of the smaller loop of the curve r = 1 2 + cos 2θ . Show that the area of a loop of r = a cos nθ is π a2 ⁄ 4n , n being integral. Also prove that the whole area is π a2 ⁄ 4 or π a2 ⁄ 2 according as n is odd or even. 9. Trace the curve r = √3 cos 3θ + sin 3θ , and find the area of a loop. 8. 1 2 a2 log ( β ⁄ α), (ii) 1 2 l 4 ⎡ ⎢ ⎣ 1 2 tan α + 1 3 1 2 ⎤ tan3 α⎥⎦ ⋅ 2. πa2 2 (π − 6). 96 1. (i) 3. (i) πa2 ⁄ (16). (ii) πa2 ⁄ (12). 4. (i) πa2 ⁄ 2 . (ii) πa2 ⁄ 2 . 5. (i) 1 2 (ii) 3πa2 ⁄ 2 . 4π + 3 √3 ⋅ 2 π − 3 √3 9. π ⁄ 3. 7. π (a2 + b 2). 6.6 Area Bounded by Two Curves (Polar Form) To find the area bounded by two curves given in polar form is explained by the following examples. Example 1 : Find the area comm on to the circles r= a √2 and r = 2a cos θ . (Meerut 2005, 11; Kumaun 08; Rohilkhand 11B; Kanpur 14; Purvanchal 14) Solution : The given equations of circles are r = a √2 and r = 2a cos θ . The first equation represents a circle with centre at pole and radius a √2 . The second equation repre- sents a circle passing through the pole and the diameter through the pole as the initial line. Both these circles are symmetrical about the initial line. E liminating r between the two equations, we have at the points of intersection a √2 = 2a cos θ , i.e., cos θ = 1 ⁄ √2 , i.e., θ = ± π ⁄ 4. T h u s a t P , θ = π ⁄ 4 . F o r t h e cir cle r = 2a cos θ , a t O , r = 0 a n d s o cos θ = 0 i.e., θ = 1 2 π. Now the required area = Area OQA PBO = 2 (area OA PBO) , (by symmetry) I-138 INTEGRAL CALCULUS = 2 [Area OA P + Area OPBO] π ⁄4 ⎡ 1⌠ ⎮ ⎣ 2 ⌡0 = 2⎢ r2 dθ , for the circle r = a √2 + π ⁄4 ⌠ = ⎮ ⌡0 π ⁄2 π ⁄2 1⌠ ⎤ ⎮ r2 dθ , for the circle r = 2a cos θ⎥ 2 ⌡π ⁄ 4 ⎦ ⌠ (a √2) 2 dθ + ⎮ (2a cos θ) 2 dθ ⌡π ⁄ 4 = 2a2 ⎡⎢ θ ⎤⎥ ⎣ ⎦ π ⁄4 0 π ⁄2 ⌠ + 2a2 ⎮ (1 + cos 2θ) dθ ⌡π ⁄ 4 sin 2θ⎤ π ⁄ 2 πa2 π 1⎤ π ⎡ ⎡π = 2a2 ⎛⎜ ⎞⎟ + 2a2 ⎢ θ + = + 2a2 ⎢ − − ⎥ ⎥ 2 2 ⎦π ⁄4 4 2⎦ ⎣ ⎣2 ⎝ 4⎠ = 1 2 1 2 π a2 + π a2 − a2 = π a2 − a2 = a2 (π − 1) . Example 2 : Fin d t h e ra t io o f t h e t w o p a rts in to w h ich t h e p a ra b o la 2a = r (1 + cos θ) divides the area of the cardioid r = 2a (1 + cos θ) . Solution : E liminating r between the given equations of the curves, we get 2a (1 + cos θ) = 2a ⁄ (1 + cos θ) (1 + cos θ) 2 = 1 cos θ (cos θ + 2) = 0 or cos θ = 0 , [ ... cos θ ≠ − 2 ] or θ = ± π ⁄ 2. Thus at the point of intersection P of the two curves, θ = π ⁄ 2 . Now area of the whole cardioid or or = 2× π 1⌠ 2 ⎮ r dθ , 2 ⌡0 (by symmetry) π π ⌠ ⌠ 1 = ⎮ 4a2 (1 + cos θ) 2 dθ = 4a2 ⎮ (2 cos2 θ) 2 dθ 2 ⌡0 ⌡0 π π ⁄2 ⌠ ⌠ 1 = 16a2 ⎮ cos4 θ dθ = 16a2 ⎮ 2 ⌡0 ⌡0 cos4 φ .2 dφ , (putting 1 2 θ = φ so that 12 dθ = dφ ; also when θ = 0 , φ = 0 and when θ = π , φ = = 32a2 Area OA CPO = = 3.1 π ⋅ = 6π a2. 4.2 2 π ⁄2 1⌠ ⎮ 2 ⌡0 π ⁄2 1 ⌠ ⋅ 4a2 ⎮ ⌡0 2 1 2 π) ...(1) r2 dθ , for the parabola r = π ⁄2 dθ a2 ⌠ ⎮ = 2 2 ⌡0 (1 + cos θ) 2a 1 + cos θ sec4 1 θ dθ 2 I-139 AREAS OF CURVES = 1 2 π ⁄2 ⌠ a2 ⎮ ⌡0 ⎡ ⎣ 1 2 1 2 = a2 ⎢ tan θ + Also area OPBO = = 1 2 1 2 (1 + tan2 θ) sec2 θ dθ 1 3 π ⁄2 1 ⎤ 2 ⎦ 0 tan3 θ⎥ = a2 (1 + 1 ) 3 = 4a2 ⋅ 3 ...(2) π ⌠ ⎮ r2 dθ , for the cardioid r = 2a (1 + cos θ) ⌡π ⁄ 2 π 1⌠ ⎮ 4a2 (1 + cos θ) 2 dθ 2 ⌡π ⁄ 2 π ⌠ = 2a2 ⎮ [1 + 2 cos θ + cos2 θ] dθ ⌡π ⁄ 2 π ⌠ 1 1 = 2a2 ⎮ (1 + 2 cos θ + + cos 2θ) dθ 2 2 ⌡π ⁄ 2 ⎡3 ⎣2 = 2a2 ⎢ θ + 2 sin θ + 1 4 ⎤π 3 = π a2 − 4a2. 2 ⎦π ⁄2 sin 2θ⎥ ...(3) Adding (2) and (3) and multiplying by 2, we get the whole area included between the two curves i.e., the area of the smaller portion of the cardioid 16⎤ ⎡4 ⎛3 ⎞⎤ ⎡ = 2 × ⎢ a2 + ⎜ πa2 − 4a2⎟ ⎥ = a2 ⎢ 3π − ⎥ 3⎦ ⎣3 ⎝2 ⎠⎦ ⎣ = 1 3 a2 [9π − 16] . ...(4) Also the shaded area (i.e., the area of the larger portion of the cardioid) = (Area of the whole cardioid) − (unshaded area) i.e., = (1) − (4) = 6 π a2 − 1 3 a2 (9π − 16) = 1 3 a2 (9π + 16) . ...(5) 1 a2 (9π + 16) 9π + 16 Larger area 3 = = ⋅ ∴ R atio of the two parts = Smaller area 1 a2 (9π − 16) 9π − 16 3 Example 3 : Find the area lying between the cardioid r = a (1 − cos θ) and its double tangent. Solution : L e t PQ b e t h e d o u b le t a n ge n t o f t h e ca r d io id . C le ar ly it is perpendicular to OX i.e., it must be inclined at an angle of 90o to the initial line i.e., ψ = 90o at P . Also we know that at any point of a curve, ψ= θ+ φ. Now tan φ = r (dθ ⁄ dr) = r ⁄ (dr ⁄ dθ) = a (1 − cos θ) ⁄ (a sin θ) , ...(1) [... r = a (1 − cos θ)] I-140 INTEGRAL CALCULUS = ∴ φ= 1 2 2 sin2 θ 1 2 2 sin 1 2 1 2 θ cos θ 1 2 = tan θ . θ. Putting the value of φ in (1), we get ψ = θ + 1 2 Since at P , ψ = 60o . 1 2 3 2 π , therefore at P , π = 1 2 θ= 3 2 θ. θ or θ = π ⁄ 3 . ∴ the vectorial angle of the point of contact P of the double tangent is π ⁄ 3 i.e., Substituting this value of θ in the equation of the curve, we get the radius vector OP = a (1 − cos 60o ) = a ⁄ 2 . Thus in the triangle OPM , 1 a , ∠ POM = 60o , ∠ PMO = 90o . 2 1 1 1 1 1 OM = a cos 60o = a ⋅ and PM = a sin 60o = a (√3 ⁄ 2) . 2 2 2 2 2 1 1 1 area of the triangle OPM = OM . PM = ( a) (√3a ⁄ 4) = (1 ⁄ 32) 2 2 4 OP = ∴ ∴ a2 √3 . Also the sectorial area OPO of the cardioid r = a (1 − cos θ) i.e., the dotted area π ⁄3 π ⁄3 = 1⌠ ⎮ 2 ⌡0 = 1 2⌠ a ⎮ 2 ⌡0 = 1 2 1 ⎡3 ⎤π ⁄3 a2 ⎢ θ − 2 sin θ + sin 2θ⎥ ⎣2 ⎦0 4 = 1 2 a2 ( π − √3 + r2 dθ = π ⁄3 1⌠ ⎮ 2 ⌡0 a2 (1 − cos θ) 2 dθ (1 − 2 cos θ + 1 2 1 8 √3) = 1 2 + 1 2 1 2 a 16 cos 2θ) dθ (4π − 7 √3) . H ence the required area (i.e., the area shaded by vertical lines) = 2 [area of Δ OPM − area of sector OPO] ⎡ = 2 ⎢⎣ 1 2 a 32 √3 − 1 2 a 16 ⎤ (4π − 7 √3) ⎥⎦ = 1 2 a 16 (15 √3 − 8π) . Example 4 : Find the area of a loop of the curve r = a sin 3θ outside the circle r = a ⁄ 2 and hence find the whole area of the curve outside the circle r = a ⁄ 2 . Solution : E liminating r between the two given equations, we get I-141 AREAS OF CURVES i.e., (a ⁄ 2) = a sin 3θ sin 3θ = i.e., i.e., 3θ = π ⁄ 6 or 5π ⁄ 6 θ = π ⁄ 18 or 5π ⁄ 18 1 2 θ = 10o or 50o . T h u s t h e l o o p o f t h e c u r ve r = a sin 3θ l yi n g b e t we e n θ = 0 a n d θ = π ⁄ 3 intersects the circle r = a ⁄ 2 at the points B and B′ where θ = 10o at B and θ = 50o at B′ . This loop is symmetrical about OA and θ = π ⁄ 6 at A . Now the required area of a loop of t h e cu rve r = a sin 3θ lying ou tside t he circle r = a ⁄ 2 = the area BAB′ CB (i.e., the shaded area) = 2 × area BA CB , (by symmetry) = 2 × [(area of the curve r = a sin 3θ between the radii vectors OB and OA i.e., θ = π ⁄ 18 and θ = π ⁄ 6) − (area of the circle r = a ⁄ 2 between the radii vectors OB and OC i.e., θ = π ⁄ 18 and θ = π ⁄ 6 )] i.e., π ⁄6 ⎡1 ⌠ = 2 ⎢⎢ ⎮ r2 dθ , for the curve r = a sin 3θ ⎢ 2 ⌡π ⁄ 18 ⎣ − π ⁄6 π ⁄6 π ⁄6 1⌠ a⎤ ⎮ r2 dθ , for the circle r = ⎥ 2 ⌡π ⁄18 2⎦ π ⁄6 ⌠ ⌠ a2 a2 ⌠ a2 ⎡ ⎤ π ⁄ 6 = ⎮ a2 sin2 3θ dθ − ⎮ dθ = ⎮ (1 − cos 6θ) dθ − θ ⌡π ⁄ 18 ⌡π ⁄ 18 4 2 ⌡π ⁄ 18 4 ⎢⎣ ⎥⎦ π ⁄ 18 = sin 6θ⎤ π ⁄ 6 π ⎤ a2 ⎡ ⎧ π sin π ⎫ ⎧ π π ⎫⎤ a2 ⎡ a2 ⎡ π 1 a2 π − − sin ⎬ ⎥ − ⋅ ⎢θ − ⎥ ⎢⎨ − ⎬− ⎨ ⎢ − ⎥= 6 ⎦ π ⁄ 18 6 ⎭ ⎩ 18 6 3 ⎭⎦ 2 ⎣ 4 ⎣ 6 18⎦ 2 ⎣⎩6 4 9 a2 π a2 π 1 √3 ⎫ ⎤ a2 ⎡ π √3 ⎤ a2 a2 ⎡ π ⎧ π − ⋅ = = [2π + 3 √3] . ⎢ − ⎨ ⎬⎥ − ⎢ + ⎥ − 12 ⎦ 36 36 2 ⎣ 6 ⎩ 18 6 2 ⎭ ⎦ 2 ⎣9 72 Again the curve r = a sin 3θ has 3 equal loops ( ... n = 3 which is odd). = ∴ whole area of the curve r = a sin 3θ outside the circle r = a ⁄ 2 = 3 × area BA B′ CB i.e., 3 times the shaded area 1 2 1 2 a [2π + 3 √3] = a [2π + 3 √3] . = 3× 24 72 1. Find the area outside the circle r = 2a cos θ and inside the cardioid r = a (1 + cos θ) . 2. Find the total area inside r = sin θ and outside r = 1 − cos θ . I-142 INTEGRAL CALCULUS Find by double integration the area lying inside the circle r = a sin θ and outside the cardioid r = a (1 − cos θ). 4. Find the area common to the circle r = a and the cardioid r = a (1 + cos θ) . 3. (Meerut 2007) 5. Find the area of the portion included between the cardioids r = a (1 + cos θ) and r = a (1 − cos θ) . 6. Show that the area contained between the circle r = a and the curve r = a cos 5θ is equal to three-fourth of the area of the circle. 7. Find the area between the curve r = a (sec θ + cos θ) and its asymptote. (Purvanchal 2010) r2 a2 8. O is the pole of the lemniscate = cos 2θ and PQ is a common tangent to its two loops. Find the area bounded by the line PQ and the arcs OP and OQ of the curve. 1. πa2 ⁄ 2 . 4. a2 ⎛ ⎜ ⎝ 5 4 3. a2 {1 − (π ⁄ 4)}. 2. 1 − (π ⁄ 4). ⎞ ⎟ ⎠ π − 2 . 5. 2 a2 ⎛ ⎜ ⎝ 3 4 ⎞ ⎟ ⎠ π− 2 ⋅ 7. 5πa2 ⁄ 4. 8. 1 8 a2 (3 √3 − 4). 6.7 Evaluation of Area by Changing the Cartesian Equation to Polar Form Sometime it is convenient to evaluate the area of a given curve by changing its cartesian equation to polar form by using the substitution x = r cos θ and y = r sin θ . Example 1 : Find the area of a loop of the folium x3 + y3 = 3axy . (Meerut 2001) Solution : Changing the equation of the curve x3 + y3 = 3axy into polar form by putting we h a ve x = r cos θ and y = r sin θ , (r cos θ) 3 + (r sin θ) 3 = 3a (r cos θ) .(r sin θ) r = 3a cos θ sin θ ⁄ (cos3 θ + sin3 θ) . F r o m ( 1) , r = 0 wh e n θ = 0 a n d wh e n θ = π ⁄ 2. ∴ the loop lies between θ = 0 and θ = π ⁄ 2 . H ence the required area of the loop or = π ⁄2 1⌠ ⎮ 2 ⌡0 r2 dθ ...(1) I-143 AREAS OF CURVES π ⁄2 = 1⌠ ⎮ 2 ⌡0 = 9a2 ⌠ ⎮ 2 ⌡0 ⎛ 3a cos θ sin θ ⎞ 2 ⎜ ⎟ dθ , putting for r from (1) ⎝ cos3 θ + sin 3 θ⎠ π ⁄2 π ⁄2 cos2 θ sin2 θ 9a2 ⌠ ⎮ dθ = 2 ⌡0 (cos3 θ + sin3 θ) 2 tan2 θ sec2 θ dθ , (1 + tan3 θ) 2 dividing the numerator and the denominator by cos6 θ . Now put 1 + tan3 θ = t so that 3 tan2 θ sec2 θ dθ = dt . Also when θ = 0 , t = 1 and when θ →π ⁄ 2 , t →∞ . ∴ area of the loop = 9a2 2 ∞ ⌠ 1 dt 3a2 ⎡ 1 ⎤ ∞ 3a2 ⎮ 2⋅ = ⋅ ⎢− ⎥ = ⌡1 t 3 2 ⎣ t ⎦1 2 1. Find the area of a loop of the curve x4 + y4 = 4a2xy . 2. Find the area of a loop of the curve (x2 + y2) 2 = 4axy2. 3. Prove that the area of a loop of the curve x6 + y6 = a2 y2 x2 is π a2 ⁄ 12 . 4. Find the area of a loop of the curve x4 + 3 x2 y2 + 2y4 = a2 xy . 5. Prove that the area of a loop of the curve x5 + y5 = 5ax2 y2 is five times the area of one loop of the curve r2 = a2 cos 2θ . (Purvanchal 2014) 1. a2 (π ⁄ 2). 2. πa2 ⁄ 4. 4. 1 4 a2 log 2 . Fill in the Blanks: 1. 2. 3. 4. 5. Fill in the blanks “……”, so that the following statem ents are com plete and correct. The process of finding the area of any bounded portion of a curve is called …… . If f (x) is a continuous and single valued function of x then the area bounded by the curve y = f (x) the axis of x and the ordinates x = a and x = b is …… . The area between the curve r = a e m θ and the given radii vectors θ = α, θ = β is …… . The curve r = a sin 3 θ has …… loops. ⎛ x⎞ The area bounded by the axis of x, and the curve y = c cosh ⎜⎝ ⎟⎠ and the ordinates c x = 0, x = a is …… . I-144 INTEGRAL CALCULUS Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 6. The area bounded by the axis of x, and the curve y = sin2 x and the given ordinates π is x = 0, x = 2 (a) 7. 8. π 4 (b) (c) π 2 (d) π The loop of the curve 3 a y2 = x (x − a) 2 will lie between (a) x = 0, x = a (b) x = − a, x = a (c) x = 0, x = − a (d) y = 0, y = a The area of one loop of the curve r2 = a2 cos2 θ is (a) a2 9. π2 4 (b) a2 2 (c) The whole area of the ellipse (a) π ab 2 3a2 2 (d) a2 4 (d) 2π ab 3 2 x2 y + = 1 is a2 b 2 (c) π 2 ab (b) π ab (Agra 2006; Bundelkhand 06, 08) 10. The area of the curve r = a is (a) π a2 (c) 2 π a2 (b) 2 π a (d) 4 π a2 (Rohilkhand 2006) True or False: Write ‘T’ for true and ‘F’ for false statement. 11. If r = f (θ) be the equation of a curve in polar co-ordinates where f (θ) is a single-valued continuous function of θ, then the area of the sector enclosed by the cu r ve a n d t h e t wo r a d ii ve ct o r s θ = θ1 a n d θ = θ2 (θ1 < θ2), i s e q u a l t o θ= θ 2 1⌠ ⎮ r 2 dθ. ⌡ 2 θ= θ 1 12. 13. The number of loops in r = a cos n θ is n or 2n according as n is even or odd. 3 The area of the astroid x 2 ⁄ 3 + y 2 ⁄ 3 = a2 ⁄ 3 is π a2. 8 b 1. quadrature 4. three 8. 13. (b). T. 9. (b). ⌠ 2. ⎮ f (x) dx ⌡a a 5. c 2 sinh ⋅ c 10. (a). 3. a2 2 m β (e − e 2 m α) 4m 6. (a). 11. T. 7. (a). 12. F. 7.1 Rectification The process of finding the length of an arc of a curve between two given points is called rectification. 7.2 Lengths of Curves (Meerut 2009B) If s denotes the arc length of a curve measured from a fixed point to any point on it, then as proved in Unit I on Differential Calculus, we have ds = ± dx √ ⎧ ⎛ dy⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ , ⎩ ⎝ dx⎠ ⎭ where + ive or –ive sign is to be taken before the radical sign according as x increases or decreases as s increases. H ence if s increases as x increases, we have ds = dx √ ⎧ ⎛ dy⎞ 2 ⎫ ⎨ 1 + ⎜ ⎟ ⎬ or ds = ⎩ ⎝ dx⎠ ⎭ √ ⎧ ⎛ dy⎞ 2 ⎫ ⎨ 1 + ⎜ ⎟ ⎬ dx . ⎩ ⎝ dx⎠ ⎭ I-146 INTEGRAL CALCULUS x ⌠ Integrating, we have s = ⎮ ⌡a √ ⎧ ⎛ dy⎞ 2 ⎫ ⎨ 1 + ⎜ ⎟ ⎬ dx , ⎩ ⎝ dx⎠ ⎭ where a is the abscissa of the fixed point from which s is measured. H ence the arc length of the curve y = f (x) included between two points for which x = a and x = b is equal to b ⌠ ⎮ ⌡a √ 1+ ⎛ dy⎞ 2 ⎫⎬ ⎜ ⎟ ⎭ dx , ⎝ dx⎠ ⎧ ⎨ ⎩ (b > a) . Sometimes it is more convenient to take y as the independent variable. Then the length of the arc of the curve x = f ( y) between y = a and y = b is equal to b ⌠ ⎮ ⌡a √ ⎧ dx 2 ⎫ ⎨ 1 + ⎛⎜ ⎞⎟ ⎬ dy , ⎩ ⎝ dy⎠ ⎭ (b > a) . Remark : Suppose we have to find the length of the arc of a curve (whose cartesian equation is given) lying between the points (x1, y1) and (x2 , y2) . We can use either of the two formulae x ⌠ 2 s = ⎮ √{1 + (dy ⁄ dx) 2} dx and ⌡x 1 y ⌠ 2 s = ⎮ √{1 + (dx ⁄ dy) 2} dy . ⌡y 1 If we feel any difficulty in integration while using one of these two formulae, we must try the other formula also. Example 1 : Show that the length of the curve y = log sec x between the points where x = 0 and x = 1 3 π is log (2 + √3) . (Kanpur 2005; Rohilkhand 14) Solution : The given curve is y = log sec x . Differentiating (1) w.r.t. x , we get 1 dy = sec x tan x = tan x . dx sec x Now ...(1) ⎛ ds ⎞ 2 ⎛ dy⎞ 2 ⎜ ⎟ = 1 + ⎜ ⎟ = 1 + tan2 x = sec 2 x . ⎝ dx⎠ ⎝ dx⎠ ...(2) If the arc length s of the given curve is measured from x = 0 in the direction of x ds increasing, we have = sec x or ds = sec x dx . dx 1 Therefore if s1 denotes the arc length from x = 0 to x = π , then 3 s π ⁄3 ⌠1 ⌠ ⎮ ds = ⎮ ⌡0 ⌡0 or π ⁄3 sec x dx = ⎡⎢ log (sec x + tan x) ⎤⎥ ⎣ 1 3 ⎦0 1 3 s1 = [log (sec π + tan π) − log 1] = log (2 + √3) . Example 2 : Find the length of the arc of the parabola y2 = 4ax extending from the vertex to an extremity of the latus rectum . (Meerut 2009) Solution : The given equation of parabola is y2 = 4ax . ...(1) I-147 RECTIFICATION The point O (0, 0) is the vertex of the parabola and the point L (a ,2a) is an extremity of the latus rectum L SL′. We have to find the length of arc OL . Defferentiating (1) w.r.t. x , we get 2y (dy ⁄ dx) = 4a . or dx ⁄ dy = y ⁄ 2a . ∴ dy ⁄ dx = 2a ⁄ y Now y2 ⎛ ds⎞ 2 ⎛ dx⎞ 2 ⎜ ⎟ = 1+ ⎜ ⎟ = 1+ ⎝ dy⎠ ⎝ dy⎠ 4a2 1 (4a2 + y2) . ...(2) 4a2 I f ‘s’ deno t es the arc length of th e parabola measured from t he vertex O t o a n y p o in t P (x, y) towards the point L , then s increases as y increases. Therefore ds ⁄ dy will be positive. So extracting the square root of (2) and keeping the positive sign, we have ds 1 1 = √(4a2 + y2), or ds = √(4a2 + y2) dy . dy 2a 2a Let s1 denote the arc length OL . Then = s 2a ⌠ 1 ⌠1 ⎮ ds = ⎮ √(4a2 + y2) dy ⌡0 2a ⌡0 2a or ⎤ 1 ⎡y 4a2 ⎢ √(4a 2 + y2) + log { y + √(4a2 + y2)}⎥ ⎦0 2 2a ⎣ 2 1 = [a √(4a2 + 4a2) + 2a2 log {2a + √(8a2)} − 0 − 2a2 log (2a)] 2a 1 = [2 √2a2 + 2a2 log {(2a + 2 √2a) ⁄ 2a}] 2a 2a2 = [√2 + log (1 + √2)] = a [ √2 + log (1 + √2)] . 2a Example 3 : Find the perim eter of the loop of the curve 3ay2 = x2 (a − x) . s1 = (Meerut 2000, 04, 06B, 07B, 11B; Purvanchal 10; Kashi 14) Solution : The given curve is 3ay2 = x2 (a − x) . ...(1) H ere the curve is symmetrical about the x-axis. Putting y = 0, we get x = 0, x = a . So the loop lies between x = 0 and x = a . Differentiating (1) w.r.t. x, we get dy dy x (2a − 3x) 6ay = 2ax − 3x2 or = ⋅ dx dx 6ay ∴ x2 (2a − 3x) 2 x2 (2a − 3x) 2 ⎛ dy⎞ 2 = 1+ ⎟ = 1+ 2 2 ⎝ dx⎠ 36a y 12a x2 (a − x) 1+ ⎜ [Substituting for 3ay2 from (1)] (2a − 3x) 2 12a2 − 12ax + (2a − 3x) 2 (4a − 3x) 2 = = ⋅ 12a (a − x) 12a (a − x) 12a (a − x) the required length of the loop = twice the length of the half loop lying above the x-axis, [By symmetry] = 1+ ∴ a ⌠ = 2⎮ ⌡0 √ a ⎧ ⎛ dy⎞ 2 ⎫ ⌠ ⎨ 1 + ⎜ ⎟ ⎬ dx = 2 ⎮ ⌡0 ⎩ ⎝ dx⎠ ⎭ √ ⎧⎪ (4a − 3x) 2 ⎫⎪ ⎨⎪ ⎬ dx ⎩ 12a (a − x) ⎪⎭ I-148 INTEGRAL CALCULUS a a = 1 √(3a) ⌠ (4a − 3x) 1 ⌠ 3 (a − x) + a ⎮ ⎮ dx = dx ⌡0 √(a − x) √(3a) ⌡0 √(a − x) = 1 √(3a) ⌠ ⎡ 3 (a − x) a ⎤ ⎮ ⎢ + ⎥ dx ⌡0 ⎣ √(a − x) √(a − x) ⎦ = 1 √(3a) ⌠ ⎮ [3 √(a − x) + a (a − x) − 1 ⁄ 2] dx ⌡0 = 1 ⎡ 2 ⎤a (a − x) 3 ⁄ 2 − a . 2 (a − x) 1 ⁄ 2⎥ ⎢− 3 ⋅ √(3a) ⎣ 3 ⎦0 = 1 ⎡ 3⁄2 4a ⎢ 2a + 2a3 ⁄ 2⎤⎥⎦ = ⋅ √(3a) ⎣ √3 a a Example 4 : Find the length of the astroid x 2 ⁄ 3 + y 2 ⁄ 3 = a 2 ⁄ 3. (Meerut 2002, 03, 13B; Agra 05; Purvanchal 08; Kashi 12) astroid is x 2 ⁄ 3 + y 2 ⁄ 3 = a 2 ⁄ 3. ...(1) Solution : The given The curve is symmetrical in all the four quadrants. For the arc of the curve in the first quadrant x varies from 0 to a. Differentiating (1), w.r.t. x, we get 2 − 1 ⁄ 3 2 − 1 ⁄ 3 dy x + y = 0 so that 3 dx 3 dy ⎛ y⎞ 1 ⁄ 3 = − ⎜ ⎟ ⋅ dx ⎝ x⎠ ∴the required whole length of the curve = 4 × length of the curve lying in the Ist quadrant a ⌠ = 4⎮ ⌡0 √ a ⎧ ⎛ dy⎞ 2 ⎫ ⌠ ⎨ 1 + ⎜ ⎟ ⎬ dx = 4 ⎮ ⌡0 ⎩ ⎝ dx⎠ ⎭ a ⎛ ⎜ ⎜ ⎜ ⎝ 2 ⁄ 3⎞ √ 1 + yx ⎟ ⎟ dx ⎠ 2 ⁄ 3⎟ a ⌠ √(x2 ⁄ 3 + y 2 ⁄ 3) ⌠ √(a2 ⁄ 3) ⎮ = 4⎮ dx = 4 dx ⌡0 ⌡0 x1 ⁄ 3 x1 ⁄ 3 a ⌠ ⎡3 ⎤a = 4a1 ⁄ 3 ⎮ x − 1 ⁄ 3 dx = 4a1 ⁄ 3 ⎢ x2 ⁄ 3⎥ = 6a . ⌡0 ⎣2 ⎦0 1. (i) Find the arc length of the curve y = 1 2 x 2 − 1 4 log x from x = 1 to x = 2 . (Meerut 2012B) ex − 1 (ii) Find the length of the curve y = log x from x = 1 to x = 2 . e + 1 (Meerut 2004B; Agra 06; Avadh 08; Kanpur 11; Rohilkhand 13; Kashi 13) I-149 RECTIFICATION (i) Show that in the catenary y = c cosh (x ⁄ c) , the length of arc from the vertex to any point is given by s = c sinh (x ⁄ c) . (ii) If s be the length of the arc of the catenary y = c cosh (x ⁄ c) from the vertex (0, c) to the point (x, y) , show that s2 = y2 − c2 . 3. (i) Find the length of an arc of the parabola y2 = 4ax measured from the vertex. (ii) Find the length of the arc of the parabola y2 = 4ax cut off by its latus rectum. 4. (i) Find the length of the arc of the parabola x2 = 4ay from the vertex to an (Kanpur 2008; Purvanchal 09) extremity of the latus rectum. (ii) Find the length of the arc of the parabola x2 = 8y from the vertex to an extremity of the latus rectum. 5. (i) Find the length of the arc of the parabola y2 = 4ax cut off by the line y = 3x . (ii) Show that the length of the arc of the parabola y2 = 4ax which is intercepted between the points of intersection of the parabola and the straight line 3y = 8x is ⎛ 15 ⎞⎟ ⋅ a ⎜⎝ log 2 + 16 ⎠ (Gorakhpur 2006; Purvanchal 06) 2. 6. (i) Find the perimeter of the curve x2 + y2 = a2. 7. (Avadh 2010; Rohilkhand 13B) (ii) Find the length of the arc of the semi-cubical parabola ay2 = x3 from the vertex to the point (a, a). (Bundelkhand 2010) (i) Show that the length of the arc of the curve x2 = a2 (1 − e y ⁄ a) measured from the origin to the point (x, y) is a log {(a + x) ⁄ (a − x)} − x . (Rohilkhand 2010B) (ii) Prove that the length of the loop of the curve 3ay2 = x (x − a) 2 is 4a / √3. (Meerut 2005B, 08, 09B) 8. (i) Find the perimeter of the loop of the curve 9ay2 = (x − 2a) (x − 5a) 2. (ii) Show that the whole length of the curve x2 (a2 − x2) = 8a2 y2 is π a √2 . (Bundelkhand 2006; Purvanchal 11) 1. 3. 4. 5. 6. 8. (i) 3 2 + 1 4 log 2 . ⎛ 1⎞ (ii) log ⎜⎝ e + ⎟⎠ e ⎡ ⎧ 2 2 ⎫⎤ ⎪ y + √( y + 4a ) ⎪ ⎥ 1 ⎢⎢ ⎨ ⎬⎥ ⋅ 2 2 2 ⎢ y √( y + 4a ) + 4a log ⎪ ⎪⎥ ⎩ ⎭⎦ 4a ⎣ 2a (ii) 2 a [√2 + log (1 + √2)]. (i) a [√2 + log (1 + √2)]. (ii) 2 [√2 + log (1 + √2)]. ⎡ 2 √(13) ⎧ 2 + √(13) ⎫ ⎤ ⎬⎥ ⋅ + log ⎨⎩ (i) a ⎢⎣ ⎭⎦ 9 3 1 (i) 2 aπ. (ii) a [13 √(13) − 8]. 27 (i) 4a √3. (i) I-150 7.3 INTEGRAL CALCULUS Equations of the Curve in Parametric form (Meerut 2009B) If the equations of the curve be given in the parametric form x = f (t ), y = φ (t ) , then s is obviously a function of t . In this case if we measure the arc length s in the direction of t increasing , we have ds = dt √ ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎫ ⎨⎜ ⎟ + ⎜ ⎟ ⎬ ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ or √ ds = ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ ⎨⎜ ⎟ + ⎜ ⎟ ⎩ ⎝ dt ⎠ ⎝ dt ⎠ 2 ⎫ ⎬ dt . ⎭ O n integrating between proper limits, the required length t ⌠2 s= ⎮ ⌡t 1 √ ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎫ ⎨ ⎜ ⎟ + ⎜ ⎟ ⎬ dt . ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ (Meerut 2003) Example 1 : Show that 8a is the length of an arch of the cycloid whose equations are x = a (t − sin t ), y = a (1 − cos t ) . (Meerut 2006; Rohilkhand 08; Kashi 11; Avadh 12; Purvanchal 14) Solution : The given equations of the cycloid are x = a (t − sin t ), y = a (1 − cos t ) . We have dx ⁄ dt = a (1 − cos t ), and dy ⁄ dt = a sin t . 1 ∴ 1 2 sin t cos t a sin t dy dy ⁄ dt 2 2 1 = = = = cot t . 1 2 2 dx dx ⁄ dt a (1 − cos t ) 2 sin t 2 N o w y = 0 wh e n cos t = 1 i.e., t = 0. A t t = 0, x = 0, y = 0 and dy ⁄ dx = ∞ . Thus the curve passes through the point (0, 0) and the tangent there is perpendicular to the x-axis. Again y is maximu m when cos t = − 1 i.e., t = π . When t = π , x = aπ , y = 2a , dy ⁄ dx = 0. Thus at the point (aπ, 2a) the tangent to the curve is parallel to the x-axis. Also in this curve y cannot be negative. Thus an arch OBA of the given cycloid is as shown in the figure. It is symmetrical about the line BM which is the axis of the cycloid. ⎛ ds⎞ 2 ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎟ = ⎜ ⎟ + ⎜ ⎟ ⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠ We have ⎜ = {a (1 − cos t )}2 + (a sin t ) 2 1 2 1 2 1 cos2 2 1 2 = a2 {(2 sin2 t ) 2 + (2 sin t cos t) 2} 1 2 1 2 = 4a2 sin2 t (sin2 t + 1 2 t ) = 4a2 sin2 t . ...(1) If s denotes the arc length of the cycloid measured from the cusp O to any point P towards the vertex B, then s increases as t increases. Therefore ds ⁄ dt will be taken with positive sign. So taking square root of both sides of (1), we have 1 2 ds ⁄ dt = 2a sin t , or 1 2 ds = 2a sin t dt . At the cusp O, t = 0, and at the vertex B, t = π . Now the length of the arch OBA = 2 × length of the arc OB I-151 RECTIFICATION π π π ⎡ ⎡ ⌠ 1 1 ⎤ 1 ⎤ = 2 ⎮ 2a sin t dt = 4a ⎢ − 2 cos t⎥ = − 8a ⎢ cos t⎥ 2 2 ⎦ 2 ⎦ ⌡0 ⎣ ⎣ 0 0 = − 8a [0 − 1] = 8a . Example 2 : Find the length of the loop of the curve x = t2, y = t − Solution : E liminating the parameter t from x = t2 an d y = t y2 = x (1 − 1 3 1 3 t . 3 (Kanpur 2010) 1 − t3, we get 3 x) 2 as the cartesian equation of the curve and hence we observe that the curve is symmetrical about the x-axis. The loop of the curve extends from the point (0, 0) to the point (3, 0). Putting y = 0 in y = t − 1 3 t , we 3 get t = 0 and t = √3. Therefore the arc of the upper half of the loop extends from t = 0 to t = √3. Now the required length of the loop = 2 × length of the half of the loop which lies above x-axis √3 √3 ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎫ ⌠ 1 ⎨ ⎜ ⎟ + ⎜ ⎟ ⎬ dt = 2 ⎮ √{(2t ) 2 + (1 − ⋅ 3t2) 2} dt 3 ⌡0 ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ ⌠ = 2⎮ ⌡0 √ ⌠ = 2⎮ ⌡0 ⌠ √(1 + 2t2 + t4) dt = 2 ⎮ ⌡0 √3 ⎡ = 2 ⎢t + ⎣ √3 (1 + t2) dt √3 t3 ⎤ ⎥ = 2 [√3 + √3] = 4 √3. 3⎦0 Example 3 : Show that the length of an arc of the curve x sin t + y cos t = f ′ (t ), x cos t − y sin t = f ′′ (t ) is given by s = f (t ) + f ′′ (t ), where c is the constant of integration. Solution : The given equations of the curve are x sin t + y cos t = f ′ (t ) and x cos t − y sin t = f ′′ (t ) . Multiplying (1) by sin t and (2) by cos t and adding, we get x (sin2 t + cos2 t ) = sin t . f ′ (t ) + cos t . f ′′ (t ) or x = sin t f ′ (t ) + cos t f ′′ (t ) . Again, multiplying (1) by cos t and (2) by sin t and subtracting, we get y = cos t f ′ (t) − sin t f ′′ (t ) . Now differentiating (3) and (4) w.r.t. t , we get dx ⁄ dt = cos t f ′ (t ) + sin t f ′′ (t ) + cos t f ′′′ (t ) − sin t f ′′ (t ) = [ f ′ (t ) + f ′′′ (t )] cos t and dy ⁄ dt = − [ f ′ (t ) + f ′′′ (t )] sin t . Now if s be the arc length in the direction of t increasing, then ds = dt √ ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎫ ⎨⎜ ⎟ + ⎜ ⎟ ⎬ ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ = √[cos2 t { f ′ (t ) + f ′′′ (t )}2 + sin2 t { f ′ (t ) + f ′′′ (t )}2] = [ f ′ (t ) + f ′′′ (t )] √(cos2 t + sin2 t ) = f ′ (t ) + f ′′′ (t ) . ⌠ Integrating both sides, we have s = ⎮ [ f ′ (t ) + f ′′′ (t )] dt + c ⌡ = f (t ) + f ′′ (t ) + c , where c is the constant of integration. ...(1) ...(2) ...(3) ...(4) I-152 1. INTEGRAL CALCULUS (i) Find the whole length of the curve (astroid) x = a cos3 t , y = a sin3 t . (Rohilkhand 2011) 2. 3. 4. (ii) Find the whole length of the curve (H ypocycloid) x = a cos3 t , y = b sin3 t . R ectify the curve or find the length of an arch of the curve x = a (t + sin t ), y = a (1 − cos t ). (Rohilkhand 2009B) Prove that the length of an arc of the cycloid x = a (t + sin t ), y = a (1 − cos t ) from the vertex to the point (x, y) is √(8ay) . (Bundelkhand 2007; Meerut 12) Find the length of the arc of the curve x = e t sin t , y = e t cos t , from t = 0 to t = 1 2 π. (Kumaun 2008; Kanpur 09) 5. Show that in the epi-cycloid for which x = (a + b) cos θ − b cos {(a + b) ⁄ b} θ, y = (a + b) sin θ − b sin {(a + b) ⁄ b} θ, the length of the arc measured from the point θ = πb ⁄ a is {4b (a + b) ⁄ a} cos {(a ⁄ 2b) θ} . 6. In the ellipse x = a cos φ , y = b sin φ , show that ds = a √(1 − e2 cos2 φ ) dφ , and hence show that the whole length of the ellipse is ⎡ ⎤ ⎛ 1 ⎞ 2 e2 ⎛ 1 . 3 ⎞ 2 e4 ⎛ 1 . 3 . 5 ⎞ 2 e6 − ⎜ − ⎜ − …⎥ , 2 πa ⎢ 1 − ⎜ ⎟ ⋅ ⎟ ⋅ ⎟ ⋅ ⎣ ⎦ 1 3 5 ⎝ 2⎠ ⎝ 2 . 4⎠ ⎝ 2 . 4 . 6⎠ where e is the eccentricity of the ellipse. (Meerut 2005) 1. (i) 6a. (ii) 4 (b 2 + ab + a2) ⁄ (b + a). 2. 8a. 4. √2 [e π ⁄ 2 − 1]. 7.4 Equation of the Curve in Polar Form For the curve r = f (θ), if we measure the arc length s in the direction of θ increasing, we have ds = dθ √ 2 ⎧ ⎪⎨ 2 ⎛ dr ⎞ r + ⎜ ⎟ ⎪⎩ ⎝ dθ⎠ ⎫ ⎪⎬ ⎪⎭ or ds = √ 2 ⎧ ⎪⎨ 2 ⎛ dr ⎞ r + ⎜ ⎟ ⎪⎩ ⎝ dθ ⎠ ⎫ ⎪⎬ ⎪⎭ dθ . O n integrating between proper limits, the required length θ ⌠ 2 s= ⎮ ⌡θ 1 √ 2 ⎧⎪ ⎛ dr ⎞ ⎨⎪ r 2 + ⎜ ⎟ ⎝ dθ⎠ ⎩ ⎫⎪ ⎬⎪ dθ . ⎭ (Meerut 2003) I-153 RECTIFICATION If the equation of the curve be θ = f (r) , then the required length is given by r ⌠2 s= ⎮ ⌡r 1 √ 2 ⎧⎪ ⎛ dθ⎞ ⎨1 + ⎜ r ⎟ ⎪⎩ ⎝ dr ⎠ ⎫⎪ ⎬ dr . ⎪⎭ Example 1 : Find the perim eter of the cardioid r = a (1 − cos θ) . (Meerut 2007; Bundelkhand 11) Solution : The given curve is r = a (1 − cos θ). It is symmetrical about the initial line. We have r = 0 when cos θ = 1 i.e., θ = 0. Also r is maximum when cos θ = − 1 i.e., θ = π and then r = 2a. As θ increases from 0 to π , r increases from 0 to 2a . So the curve is as shown in the figure. By symmetry, the perimeter of the cardioid = 2 × the arc length of the upper half of the cardioid. Now differentiating (1) w.r.t. θ, we have dr ⁄ dθ = a sin θ. We have ...(1) ⎛ ds ⎞ 2 ⎛ dr ⎞ 2 ⎜ ⎟ = r2 + ⎜ ⎟ = a 2 (1 − cos θ) 2 + a 2 sin2 θ ⎝ dθ⎠ ⎝ dθ⎠ 1 2 1 2 1 2 = a2 (2 sin2 θ) 2 + a2 (2 sin θ cos θ) 2 1 2 1 2 1 2 1 2 = 4a2 sin2 θ (sin2 θ + cos2 θ) = 4a2 sin2 θ. ...(2) If s denotes the arc length of the cardioid measured from the cusp O (i.e., the point θ = 0) to any point P (r, θ) in the direction of θ increasing, then s increases as θ increases. Therefore ds ⁄ dθ will be positive. H ence from (2), we have 1 2 ds ⁄ dθ = 2a sin θ, or ds = 2a sin 1 2 θ dθ. ...(3) At the cusp O, θ = 0 and at the vertex A, θ = π . ∴ π ⌠ 1 the length of the arc OPA = ⎮ 2a sin θ dθ 2 ⌡0 θ⎤ π θ⎤ π ⎡ ⎡ = 4a ⎢ − cos ⎥ = − 4a ⎢ cos ⎥ = − 4a (0 − 1) = 4a . 2 2⎦ 0 ⎣ ⎦0 ⎣ ∴ the perimeter of the cardioid = 2 × 4a = 8a . Example 2 : Find the length of the arc of the equiangular spiral r = ae θ cot α between the points for which radii vectors are r1 and r2 . (Kanpur 2007) Solution : The given curve is r = ae θ cot α. Differentiating (1) w.r.t. θ, we get dr ⁄ dθ = ae θ cot α . cot α = r cot α , from (1). ...(1) I-154 INTEGRAL CALCULUS ∴ dθ ⁄ dr = If s denotes the increasing, we have ds = dr 1 ⁄ (r cot α) i.e., (r dθ ⁄ dr) = tan α . ...(2) arc length of the given curve measured in the direction of r √ 1+ r or ⎧ ⎨ ⎩ 2 ⎛⎜ dθ⎞⎟ ⎝ dr ⎠ 2 ⎫ ⎬ ⎭ (Note) = √(1 + tan2 α) = √(sec 2 α) = sec α , from (2) ds = sec α dr . Let s1 denote the required arc length i.e., from r = r1 to r = r2 . s Then (Meerut 2001B) r ⌠2 ⌠1 ⎡ ⎤ r2 ⎮ ds = ⎮ sec α dr = (sec α) ⎢ r ⎥ ⌡r ⌡0 ⎣ ⎦r 1 or s1 = (sec α) (r2 − r1) . 1 Example 3 : Prove that the perim eter of the lim acon r = a + b cos θ, if b ⁄ a be sm all, is approximately 2 πa (1 + 1 2 2 b ⁄a ). 4 Solution : The given curve is ...(1) r = a + b cos θ, (a > b) . Note that b ⁄ a is given to be small so we must have b < a. Th e cu rve (1) is symmetrical about the initial line and for the portion of the curve lying above the initial line θ varies from θ = 0 to θ = π . By symmetry, the perimeter of the limacon = 2 × the arc length of the upper half of the limacon. Now differentiating (1) w.r.t. θ, we have dr ⁄ dθ = − b sin θ. We have 2 ⎛ dr ⎞ ⎛ ds ⎞ 2 ⎜ ⎟ = r 2 + ⎜ ⎟ = (a + b cos θ) 2 + (− b sin θ) 2 ⎝ dθ⎠ ⎝ dθ⎠ = a2 + b2 cos2 θ + 2ab cos θ + b2 sin2 θ = a2 + b2 + 2ab cos θ. If we measure the arc length s in the direction of θ increasing, we have or ds ⁄ dθ = √(a2 + b2 + 2ab cos θ) ds = √(a2 + b2 + 2ab cos θ) dθ . The arc length of the upper half of the limacon π π ⌠ ⌠ ⎛ 2b b2 ⎞ = ⎮ √(a2 + b2 + 2ab cos θ) dθ = a ⎮ ⎜⎜ 1 + cos θ + 2 ⎟⎟ ⌡0 ⌡0 ⎜⎝ a a ⎟⎠ ⎡ 1 1 ( − 1) ⎛ 2 2 ⎞ ⎢ 1 + b cos θ + 1 ⋅ b + 2 2 ⎜ 4 b cos2 θ⎟ ⎜ ⎟ ⎢ 2 2 2 ⎜ a ⎟ a 2! a ⎣ ⎝ ⎠ π⎢ ⌠ = a⎮ ⌡0 1⁄2 dθ ⎤ ⎥ ⎥ dθ ⎥ ⎦ [E xpanding by binomial theorem and neglecting powers of b ⁄ a higher than two because b ⁄ a is small] π⎡ ⌠ = a⎮ ⌡0 2 ⎤ ⎢ 1 + b cos θ + 1 b (1 − cos2 θ) ⎥ dθ ⎢ ⎥ 2 ⎢ ⎥ a 2 a ⎣ ⎦ I-155 RECTIFICATION π⎡ 2 ⎤ ⎢ 1 + b cos θ + 1 b sin2 θ⎥ dθ ⎢ ⎥ 2 ⎢ ⎥ a 2 a ⎣ ⎦ ⌠ = a⎮ ⌡0 π ⁄2 b 1 b2 ⌠ ⎫π sin θ⎬ + 2⎮ ⎭0 a 2 a2 ⌡0 ⎡⎧ = a ⎢⎢ ⎨ θ + ∴ ⎢⎩ ⎣ ⎡ 1 b2 1 π⎤ = a ⎢⎢ π + ⋅ 2 ⋅ ⋅ ⎥⎥ 2 ⎢ 2a 2 2 ⎥⎦ ⎣ 2 ⎡ ⎤ b ⎥ ⋅ = aπ ⎢⎢ 1 + 2 ⎥⎥ ⎢ 4a ⎣ ⎦ ⎤ sin2 θ dθ⎥⎥ ⎥ ⎦ the perimeter of the limacon = 2 × aπ [1 + (b2 ⁄ 4a2)] = 2aπ [1 + (b2 ⁄ 4a2)] . 1 2 Example 4 : If s be the length of the curve r = a tanh θ between the origin and θ = 2π, and Δ be the area under the curve between the sam e two points, prove that Δ = a (s − aπ) . 1 2 Solution : The given curve is r = a tanh θ. ...(1) Differentiating (1) w.r.t. θ, we get dr ⁄ dθ = a ⋅ ⎛ ds ⎞ 2 ⎛ dr ⎞ 2 1 ⎟ = r 2 + ⎜ ⎟ = a 2 tanh2 θ + 2 ⎝ dθ⎠ ⎝ dθ⎠ We have ⎜ = = 1 4 1 4 1 2 1 2 1 4 a2 [4 tanh2 θ + sech4 θ] = 1 1 sech2 θ. 2 2 2 a 1 sech4 θ 2 4 1 2 1 2 a2 [4 (1 − sech2 θ) + sech4 θ] 1 2 a2 [2 − sech2 θ]2. ...(2) If we measure the arc length s in the direction of θ increasing, we have ds ⁄ dθ = 1 2 1 2 a (2 − sech2 θ) [R etaining + ive sign while taking the square root of (2)] or ds = 1 2 1 2 a (2 − sech2 θ) dθ . Now at the origin r = 0 and putting r = 0 in (1), we get θ = 0. ∴ the arc length of the given curve between the origin (θ = 0) and θ = 2π is given by 2π s= = 1 2 1 ⌠ a⎮ 2 ⌡0 1 2 (2 − sech2 θ) dθ = a . 2 ⎡⎢⎣ θ⎤⎥⎦ 2π 0 − 1 2 1 2 2π 1 ⎤ 2 ⎦ 0 ⎡ ⎣ a ⎢ 2 tanh θ⎥ 2π ⌠ a⎮ ⌡0 2dθ − 1 2 2π ⌠ a⎮ ⌡0 = 2aπ − a tanh π . Also the area between the radii vectors θ = 0, θ = 2π and the curve is 2π Δ= 1⌠ ⎮ 2 ⌡0 1 2 ⌠ a2 ⎮ ⌡0 = 2π r 2 dθ = 2π 1 2⌠ a ⎮ 2 ⌡0 1 2 tanh2 (1 − sech2 θ) dθ = 1 2 1 θ dθ 2 ⎡ ⎣ 2π 1 ⎤ 2 ⎦ 0 a2 ⎢ θ − 2 tanh θ⎥ 1 2 sech2 θ dθ ...(3) I-156 INTEGRAL CALCULUS = 1 2 a2 [2π − 2 tanh π] = a2 [π − tanh π] = a [aπ − a tanh π] = a [(2aπ − a tanh π) − aπ] = a (s − aπ), from (3). 7.5 Equation of the Curve in Pedal Form sLet p = f (r ) be the equation of the curve and r1 and r2 be the values of r at two given points of the curve. Then by differential calculus we know that ds r r = or ds = dr , 2 dr √(r 2 − p2) √(r − p2) where s increases as r increases. O n integrating between proper limits, the required length r ⌠ 2 r s= ⎮ dr . ⌡r √(r 2 − p2) 1 The value of p should be put in terms of r from the equation of the curve. Important Remark : If the curve is symmetrical about one or more lines, then find out the length of one symmetrical part and then multiply it by the number of symmetrical parts. ⌠ r dr ⋅ Example 1 : Prove the form ula s = ⎮ ⌡ √(r 2 − p2) Show that the arc of the curve p2 (a4 + r4) = a4 r 2 between the limits r = b, r = c is equal in length to the arc of the hyperbola xy = a2 between the lim its x = b, x = c . Solution : From differential calculus, we know that tan φ = r ∴ Thus dθ ds and = dr dr √ 1+ ⎡ ⎢ ⎣ ⎛ dθ⎞ 2 ⎤ ⎜r ⎟ ⎥ ⋅ ⎝ dr ⎠ ⎦ ds = √(1 + tan2 φ ) = √(sec 2 φ ) = sec φ dr 1 1 1 = = = cos φ √(1 − sin2 φ ) √{1 − ( p2 ⁄ r 2)} r ⋅ = 2 √(r − p2) r dr . ds = 2 √(r − p2) [ ... p = r sin φ ] r ⌠ Integrating between the given limits, we get s = ⎮ dr . ⌡ √(r 2 − p2) Now the given curve is p2 (a4 + r 4) = a4 r 2 or ...(1) p2 = a4 r 2 ⁄ (a4 + r 4) . a4 r 2 r6 = ⋅ ...(2) 4 4 + r ) (a + r 4) Therefore from (1), the arc of the given curve between the limits r = b, r = c is We have r 2 − p2 = r 2 − (a4 I-157 RECTIFICATION c c ⌠ r dr ⌠ r dr = ⎮ = ⎮ ⌡b √(r 2 − p2) ⌡b √{r 6 ⁄ (a4 + r 4)} c [From (2)] c ⌠ √(a4 + r 4) ⌠ r √(a4 + r 4) ⎮ dr = dr . = ⎮ ⌡b ⌡b r3 r2 ...(3) Also, for the hyperbola xy = a2 i.e., y = a2 ⁄ x, dy ⁄ dx = − a2 ⁄ x2. ∴ the arc length of the hyperbola xy = a2 between the limits x = b, x = c c ⌠ = ⎮ ⌡b c √ 2 ⎧⎪ ⎛ dy⎞ ⎨⎪ 1 + ⎜ ⎟ ⎝ dx⎠ ⎩ c ⎫⎪ ⌠ ⎬⎪ dx = ⎮ ⌡ ⎭ √ 1 + ax b c 4 ⌠ √(r + ⌠ √(x4 + a4) = ⎮ dx = ⎮ ⌡b ⌡b x2 r2 ⎧ ⎪ ⎨ ⎪ ⎩ a 4) 4⎫ ⎪ dx ⎭ 4⎬ ⎪ dr [Changing the variable from x to r by a property of definite integrals] c ⌠ √(a4 + r 4) dr . = ⎮ ⌡b r2 From (3) and (4) we observe that the two lengths are equal. ...(4) 1. Find the entire length of the cardioid r = a (1 + cos θ). 2. Find the perimeter of the curve r = a (1 + cos θ) and show that arc of the upper half is bisected by θ = π ⁄ 3. (Gorakhpur 2005; Purvanchal 07) Prove that the line 4 r cos θ = 3a divides the cardioid r = a (1 + cos θ) into two parts such that lengths of the arc on either side of the line are equal. Show that the arc of the upper half of the curve r = a (1 − cos θ) is bisected by θ = 2π ⁄ 3. Find the length of the cardioid r = a (1 − cos θ) lying outside the circle r = a cos θ. (Purvanchal 2007; Rohilkhand 09, 11B) 3. 4. 5. 6. Find the length of the arc of the equiangular spiral r = a e θ cot α, taking s = 0 when θ = 0. 7. 8. Find the length of any arc of the cissoid r = a (sin2 θ ⁄ cos θ) . Show that the whole length of the limacon r = a + b cos θ, (a > b) is equal to that of an ellipse whose semi-axes are equal in length to the maximum and minimum radii vectors of the limacon. 1. 8a. sec α [e θ cot α − 5. 4a √3. 6. a 7. f (θ2) − f (θ1), where f (θ) = a √(sec 2 θ + 3) − a √3 log {cos θ + √(cos2 θ + 1]. 1 )}. 3 I-158 7.6 INTEGRAL CALCULUS Intrinsic Equations Definition : By the intrinsic equation of a curve we mean a relation between s and ψ, where s is the length of the arc A P of the curve measured from a fixed point A on it to a variable point P, and ψ is the angle which the tangent to the curve at P makes with a fixed straight line usually taken as the positive direction of the axis of x . The co-ordinates s and ψ are known as Intrinsic Co-ordinates. (a) To find the intrinsic equation from the cartesian equation : Let the equation of the given curve be y = f (x) . Take A as the fixed point on the curve from which s is measured and take the axis of x as the fixed straight line with reference to which ψ is measured. Let P (x, y) be any point on the curve and PT be the tangent at the point P to the curve. Let arc A P = s and ∠ PTX = ψ . Now, we have tan ψ = dy ⁄ dx = f ′ (x) . ...(1) Let a be the abscissa of the point A from which s is measured. Then x ⌠ s= ⎮ ⌡a √ ⎡ ⎛ dy⎞ 2 ⎤ ⎢ 1 + ⎜ ⎟ ⎥ dx ⎣ ⎝ dx⎠ ⎦ x ⌠ = ⎮ √[1 + { f ′ (x)}2] dx . ⌡a ...(2) E liminating x between (1) and (2), we obtain the required intrinsic equation. Note : To find the intrinsic equation from the parametric equations dy dy ⁄ dt = and then proceed as in case (a). we use dx dx ⁄ dt (b) Intrinsic equation from Polar equation : Let the equation of the given curve be r = f (θ) . Take A as the fixed point on the curve from which s is measured. Let P be any point (r, θ) on the curve. Let arc A P = s and ∠ PTX = ψ , where OX is the initial line. If φ is the angle between the radius vector and the tangent at P, then f (θ) , dθ r = = ...(1) tan φ = r dr dr ⁄ dθ f ′ (θ) and ψ = θ + φ . ...(2) Let α be the vectorial angle of the point A . Then we have θ ⌠ s= ⎮ ⌡α θ √ ⎧ 2 ⎛ dr ⎞ 2 ⎫ ⎨ r + ⎜ ⎟ ⎬ dθ ⎩ ⎝ dθ⎠ ⎭ ⌠ = ⎮ √[{ f (θ)}2 + { f ′ (θ)}2] dθ ⌡α ...(3) I-159 RECTIFICATION E liminating θ and φ between (1), (2) and (3), we get a relation between s and ψ, which is the intrinsic equation of the curve. (c) Intrinsic equation from Pedal Equation : Let the pedal equation of the curve be p = f (r) . ...(1) r Then ⌠ r dr , s= ⎮ ⌡a √(r2 − p2) ...(2) the arc length s being measured from the point r = a . ds dr Also the radius of curvature ρ = = r ⋅ …(3) dψ dp E liminating p and r between (1), (2) and (3), we obtain the required intrinsic equation. Example 1 : Show that the intrinsic equation of the parabola y2 = 4ax is s = a cot ψ cosec ψ + a log (cot ψ + cosec ψ ), ψ being the angle between the x-axis and the tangent at the point whose arcual distance from the vertex is s. Solution : The given parabola is y2 = 4ax . ...(1) Differentiating (1) w.r.t. x , we get 2y (dy ⁄ dx) = 4a . ...(2) ∴ tan ψ = dy ⁄ dx = 4a ⁄ 2y = 2a ⁄ y. If s denotes the arc length of the parabola measured from the vertex (0, 0) in the direction of y increasing, then √ = √ ds = dy ∴ ds = ⎧ ⎛ dx⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ = ⎩ ⎝ dy⎠ ⎭ √ 2 ⎫ ⎧ ⎪1 + y ⎪ ⎨ ⎬ ⎪ 4a2 ⎪⎭ ⎩ y⎤ ⎡ . . dx = ⎢ . ⎥ dy 2a ⎦ ⎣ ⎧ 4a 2 + y2 ⎫ ⎪ ⎪ = 1 √(4a 2 + y2) . ⎨ ⎬ ⎪ 4a 2 ⎪ 2a ⎩ ⎭ 1 √(4a2 + y2) dy . 2a s y ⌠ 1 ⌠ ⎮ √(4a2 + y2) dy Integrating, ⎮ ds = ⌡0 2a ⌡0 or s= 1 2a ⎡1 ⎤y 1 ⎢ y √(4a 2 + y2) + ⋅ 4a 2 log { y + √(4a 2 + y2)}⎥ 2 2 ⎣ ⎦0 1 2 = (1 ⁄ 2a) [ y √(4a2 + y2) + 1 2 ⋅ 4a2 log { y + √(4a2 + y2)} − = y + √(4a2 + y2) ⎤ 1 ⎡ ⎢ y √(4a 2 + y2) + 4a 2 log ⎥⋅ ⎦ 4a ⎣ 2a 1 2 ⋅ 4a2 log 2a] ...(3) Now to obtain the intrinsic equation of the given parabola we eliminate y between (2) and (3). From (2), we have y = 2a cot ψ . Putting this value of y in (3), we get I-160 INTEGRAL CALCULUS s= 1 4a ⎡ ⎢ ⎢ 2a cot ψ √(4a 2 + 4a 2 cot2 ψ) ⎢ ⎣ + 4a2 log 2a cot ψ + √(4a2 + 4a2 cot 2 ψ) ⎤ ⎥ ⎦ 2a 1 [(2a cot ψ) . 2a √(1 + cot2 ψ) + 4a2 log {cot ψ + √(1 + cot 2 ψ)}] 4a = a cot ψ cosec ψ + a log (cot ψ + cosec ψ), which is the required intrinsic equation. = Example 2 : Show that the intrinsic equation of the cycloid x = a (t + sin t ), y = a (1 − cos t ) is s = 4a sin ψ . Hence or otherwise find the length of the com plete cycloid. (Meerut 2001, 06B, 07, 10; Kanpur 04; Avadh 04, 09, 10; Rohilkhand 07B) Solution : The given equations of the cycloid are x = a (t + sin t ), y = a (1 − cos t ) . We have dx ⁄ dt = a (1 + cos t ), and dy ⁄ dt = a sin t . 1 ∴ ...(1) 1 2 sin t cos t a sin t dy dy ⁄ dt 2 2 1 = = = = tan t . 1 2 2 dx dx ⁄ dt a (1 + cos t ) 2 cos t 2 H ence 1 2 tan ψ = dy ⁄ dx = tan t or ψ= 1 2 t. ...(2) If s denotes the arc length of the cycloid measured from the vertex (i.e., the point t = 0) to any point P (i.e., the point ‘t ’) in the direction of t increasing, then t ⌠ s= ⎮ ⌡0 t √ ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎫ ⎨ ⎜ ⎟ + ⎜ ⎟ ⎬ dt ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ ⌠ = ⎮ √{a2 (1 + cos t ) 2 + a2 sin2 t } dt ⌡0 t ⌠ = ⎮ √{2a2 (1 + cos t )} dt ⌡0 t t ⎡ ⌠ 1 1 ⎤ 1 = 2a ⎮ cos t dt = 2a ⎢ 2 sin t⎥ = 4a sin t 2 2 ⎦ 2 ⌡0 ⎣ 0 ...(3) E liminating t from (2) and (3), we get s = 4a sin ψ , ...(4) which is the required intrinsic equation of the cycloid. Second Part : In the intrinsic equation (4) of the cycloid the arc length s has been measured from the vertex i.e., the point ψ = 0. At a cusp, we have t = π and ψ = π ⁄ 2 . If s1 denotes the length of the arc extending from the vertex to a cusp, then 1 from (4), we have s1 = 4a sin π = 4a . 2 ∴ the whole length of an arch of the cycloid = 2 × 4a = 8a . Example 3 : Find the intrinsic equation of the cardioid r = a (1 + cos θ), and hence, or otherwise, prove that s2 + 9ρ2 = 16a2, where ρ is the radius of curvature at any point, and s is the length of the arc intercepted between the vertex and the point. (Meerut 2005B) I-161 RECTIFICATION Solution : The given curve is r = a (1 + cos θ) . Differentiating (1) w.r.t. θ, we have dr ⁄ dθ = − a sin θ. ...(1) 1 2 cos2 θ dθ a (1 + cos θ) r 2 ∴ tan φ = r = = = 1 1 dr dr ⁄ dθ − a sin θ − 2 sin θ cos θ 2 1 2 = − cot θ = tan 1 2 Therefore φ = π+ 1 2 1 2 θ= 1 3 1 2 (ψ − π+ 1 2 2 θ) . θ, so that ψ= θ+ φ = θ+ or 1 ( 2 1 2 1 2 π+ θ= 1 2 π+ 3 2 θ π) . ...(2) If s denotes the arc length of the cardioid measured from the vertex (i.e., θ = 0) to any point P (i.e., θ = θ) in the direction of θ increasing, then θ ⌠ s= ⎮ ⌡0 √ ⎧ 2 ⎛ dr ⎞ 2 ⎫ ⎨ r + ⎜ ⎟ ⎬ dθ ⎩ ⎝ dθ⎠ ⎭ θ ⌠ = 2a ⎮ √{(1 + cos θ) 2 + sin2 θ} dθ ⌡0 θ ⌠ = 2a ⎮ √{1 + 2 cos θ + cos2 θ + sin2 θ} dθ ⌡0 θ ⌠ = 2a ⎮ √{2 (1 + cos θ)} dθ ⌡0 θ ⌠ 1 = 2a ⎮ cos θ dθ 2 ⌡0 ⎡ ⎣ θ 1 ⎤ 2 ⎦ 0 1 2 = 2a ⎢ 2 sin θ⎥ = 4a sin θ. ...(3) 1 3 E liminating θ between (2) and (3), we get s = 4a sin { (ψ − 1 2 π)} , ...(4) which is the required intrinsic equation. 4a ds 1 1 = cos (ψ − π), from (4) Also ρ= 3 2 3 dψ or 1 3 3ρ = 4a cos (ψ − 1 2 π) . ...(5) Squaring and adding (4) and (5), we get 1 3 s2 + 9ρ2 = (4a) 2 {sin2 (ψ − 1 2 1 3 π) + cos2 (ψ − 1 2 π)} = 16a2 . 1 = 16a2. Example 4 : Find the intrinsic equation of the equiangular spiral p = r sin α . (Meerut 2000, 01, 04, 06, 09, 10B) Solution: The given pedal equation of the curve is p = r sin α . s Differentiating (1) w.r.t. r, we have dp ⁄ dr = sin α . ...(1) I-162 INTEGRAL CALCULUS ds dr r r = r = = = r cosec α . ...(2) dψ dp dp ⁄ dr sin α If we measure the arc length s from the point r = 0 in the direction of r increasing, we have r r r r dr ⌠ r dr ⌠ ⌠ ⎮ ⎮ = = sec α dr s= ⎮ ⌡0 √(r2 − p2) ⌡0 √(r2 − r2 sin2 α) ⌡0 ∴ ρ= r ⌠ = sec α ⎮ dr = sec α ⎡⎢⎣ ⌡0 or or r r ⎤⎥⎦ = r sec α . ...(3) 0 E liminating r between (2) and (3), we have (ds ⁄ dψ) cosec α = = cot α [Dividing (2) by (3)] sec α s ds ⁄ s = cot α dψ. Integrating, log s = ψ cot α + log a, where a is constant of integration log (s ⁄ a) = ψ cot α s = a eψ cot α, or which is the required intrinsic equation of the curve. Example 5 : Find the intrinsic equation of the curve for which the length of the arc m easured from the origin varies as the square root of the ordinate. Find also parametric equations of the curve in term s of any param eter. Solution : Let s denote the arc length of the curve measured from the origin to any point P (x, y) such that s increases as y increases. As given s ∝ √y so that s = λ √y, where λ is some constant. Choosing this constant λ = √(8a), we have (Note), s = √(8ay) or or or s2 = 8ay. Now differentiating (1) w.r.t. y, we have 2s (ds ⁄ dy) = 8a ds ⁄ dy = 4a ⁄ s . Now we know that dy ⁄ ds = sin ψ. ∴ sin ψ = dy ⁄ ds = s ⁄ 4a s = 4a sin ψ, which is the required intrinsic equation. 16a2 sin2 ψ s2 = 8a 8a = a (1 − cos 2ψ). Again from (1), we have y = Also dψ ds ds dψ = ⋅ = 4a cos ψ dx dψ dx dx or dψ 1 = 4a cos ψ cos ψ dx or dx = 4a cos2 ψ dψ = 2a (1 + cos 2ψ) dψ. If x = 0 when ψ = 0, then integrating (4), we get ...(1) ...(2) [From (2)] [ ... s = 4a sin ψ] ...(3) ⎡ . . ds ⎤ = 4a cos ψ⎥ ⎢ . dψ ⎣ ⎦ ⎡ . . dx ⎤ = cos ψ⎥ ⎢ . ds ⎣ ⎦ ...(4) I-163 RECTIFICATION ψ x ⌠ ⌠ ⎮ dx = 2a ⎮ (1 + cos 2ψ) dψ ⌡0 ⌡0 ⎡ ⎣ 1 2 ⎤ψ ⎦0 or x = 2a ⎢ ψ + or x = a [2ψ + sin 2ψ] . sin 2ψ⎥ ...(5) So from (3) and (5), the required parametric equations of the curve are x = a (2ψ + sin 2ψ) and y = a (1 − cos 2ψ), which are the parametric equations of a cycloid. 1. Prove that the intrinsic equation of the parabola x2 = 4ay is s = a tan ψ sec ψ + a log (tan ψ + sec ψ) . 2. Find the intrinsic equation of the parabola y2 = 4ax . Hence deduce the length of the arc measured from the vertex to an extremity of the latus rectum. 3. Show that the intrinsic equation of the semi-cubical parabola 3ay2 = 2x3 is 9s = 4a (sec3 ψ − 1) . (Meerut 2005, 09B; Rohilkhand, 08B) 4. Find the intrinsic equation of the catenary y = c cosh (x ⁄ c) . (Rohilkhand 2007; Kumaun 08; Kanpur 14) H ence show that cρ = c2 + s2, where ρ is the radius of curvature. 5. Prove that the intrinsic equation of the curve x = a (1 + sin t), y = a (1 + cos t) is s + aψ = 0. 6. Find the intrinsic equation of the cardioid r = a (1 − cos θ) . (Meerut 2007B; Avadh 05, 12; Rohilkhand 12) 7. Find the intrinsic equation of r = a (a, 0). 8. eθ cot α, where s is measured from the point Find the intrinsic equation of the spiral r = aθ, the arc being measured from the pole. 9. Find the intrinsic equation of the curve p2 = r2 − a2. 10. In the four-cusped astroid x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3, show that (i) s = (ii) s = 3 a cos 2 ψ, s being measured from the vertex; 4 3 a sin2 ψ , s being mesured from the cusp on 2 x-axis; (Purvanchal 2014) (iii) whole length of the curve is 6a . 11. Find the cartesian equation of the curve whose intrinsic equation is s = c tan ψ when it is given that at ψ = 0, x = 0 and y = c . I-164 INTEGRAL CALCULUS 2. 4. s = a cot ψ cosec ψ + a log (cot ψ + cosec ψ), a {√ ⎯⎯2 + log (1 + √ ⎯⎯2 )}. s = c tan ψ. 6. s = 8a sin2 ψ. 7. s = a sec α [e (ψ− α) cot α − 1]. 8. s= 9. s= 11. 1 6 1 2 1 2 a [θ √(1 + θ2) + log {θ + √(1 + θ2)}], where ψ = θ + tan− 1 θ. aψ2. y = c cosh (x ⁄ c). Fill in the Blanks: Fill in the blanks “……”, so that the following statem ents are com plete and correct. 1. The process of finding the length of an arc of a curve between two given points is (Kumaun 2008) called ……. 2. The arc length of the curve y = f (x) included between two points for which x = a and x = b (b > a) is …… . 3. The arc length of the curve y = 4. If r = a e θ cot α, then ds = …… . 5. ds = √ ⎯⎯⎯⎯ …… ⎯ dr 6. 1 2 x 2 − 1 4 log x from x = 1 to x = 2 is …… . T h e le n gt h o f a n a r ch o f t h e x = a (t − sin t ), y = a (1 − cos t ) is …… (Meerut 2001, 03) (Meerut 2001) c yc l o i d wh o s e e q u a t i o n s a r e (Agra 2005) Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 7. If the equations of the curve be given in the parametric form x = f (t), y = φ (t ), and the arc length s is measured in the direction of t increasing, then on integrating between the proper limits, the required length s is given as 2 t ⎛ dy⎞ 2 ⎫ ⎧ ⎛ dx⎞ ⌠2 ⎟ ⎨ ⎜ (a) ⎮ + ⎜ ⎟ ⎬ dt ⌡t ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ 1 t ⌠2 (b) ⎮ ⌡t 1 √ ⎛ ⎞⎫ ⎧⎛ ⎞ √ ⎨⎩ ⎜⎝ dxdt⎟⎠ + ⎜⎝ dydt⎟⎠ ⎬⎭ dt I-165 RECTIFICATION t ⌠2 (c) ⎮ ⌡t 1 t ⌠2 (d) ⎮ ⌡t 1 8. √ 3 ⎛ ⎞3⎫ ⎧ ⎛ dx⎞ ⎨ ⎜ ⎟ + ⎜ dy⎟ ⎬ dt ⎝ dt ⎠ ⎭ ⎩ ⎝ dt ⎠ √ 1⁄2 ⎛ ⎞ 1⁄2 ⎫ ⎧ ⎛ dx⎞ ⎨ ⎜ ⎟ + ⎜ dy⎟ ⎬ dt. ⎩ ⎝ dt ⎠ ⎭ ⎝ dt ⎠ For the curve r = a (1 + cos θ), 1 2 1 cos 2 ds is dθ (a) 2 cos θ (c) a 9. 10. If x = a 1 2 1 cos 2 (b) 2 a cos θ θ cos3 (Meerut 2003) (d) t, y = a sin 3 3 2 a θ ⎛ ds ⎞ 2 t, then ⎜ ⎟ is ⎝ dt ⎠ (a) (a sin t cos t) 2 (b) (sin t cos t) 2 (c) (3 a sin t cos t) 2 (d) 3a sin t cos t The entire length of the cardioid r = a (1 + cos θ) is (a) 8 a (b) 4 a (c) 6 a (d) 2 a (Rohilkhand 2008) True or False: Write ‘T’ for true and ‘F’ for false statement. 11. The length of the arc of the curve x = f ( y) between y = a and y = b, (b > a) is equal to b ⌠ ⎮ ⌡a √ ⎛ ⎞2⎫ ⎧ ⎨ 1 + ⎜ dx⎟ ⎬ dy. ⎩ ⎝ dy⎠ ⎭ 12. The relation between s and ψ for any curve is called its polar equation. 13. If the equation of the curve be θ = f (r ), then the arc length from r = r1 to r = r2 is given by r ⌠2 ⎮ ⌡r 1 14. √ ⎞2⎫ ⎛ ⎧ ⎨ 1 + ⎜ r d θ ⎟ ⎬ d r. ⎩ ⎝ dr⎠ ⎭ If the equation of the curve be r = f (θ), then the arc length from θ = θ1 to θ = θ2 is given by θ ⌠ 2 ⎮ ⌡θ 1 √ ⎛ ⎞2⎤ ⎡ ⎢ r2 + ⎜ dr ⎟ ⎥ dr. ⎣ ⎝ dθ ⎠ ⎦ (Meerut 2003) I-166 INTEGRAL CALCULUS b 1. 4. 7. 10. 13. R ectification. r cosec α d θ . (a). (a). T. 2. ⌠ ⎮ ⌡a √ 5. ⎧ ⎪ ⎨1 ⎪ ⎩ ⎛ ⎜r ⎝ 8. 11. 14. + (b). T. F. ⎛ ⎞2⎫ ⎧ ⎨ 1 + ⎜ dy⎟ ⎬ dx . ⎩ ⎝ dx⎠ ⎭ d θ⎞⎟ d r⎠ 2⎫ ⎪ ⎬⋅ ⎪ ⎭ 3. 3 1 + log 2 . 2 4 6. 8a. 9. 12. (c). F. 8.1 Revolution : Definitions Solid of Revolution : If a plane area is revolved about a fixed line lying in its own plane, then the body so generated by the revolution of the plane area is called a solid of revolution. Surface of Revolution : If a plane curve is revolved about a fixed line lying in its own plane, then the surface generated by the perimeter of the curve is called a surface of revolution. Axis of Revolution : The fixed straight line, say A B, about which the area revolves is called the axis of revolution or axis of rotation. 8.2 Volumes of Solids of Revolution (a) The axis of rotation being x-axis : If a plane area bounded by the curve y = f (x) , the ordinates x = a, x = b and the x-axis revolves about the x-axis then the volume of the solid thus generated is b b ⌠ ⌠ ⎮ π y2 dx = ⎮ π [ f (x)] 2 dx, ⌡a ⌡a where y = f (x) is a finite, continuous and single valued function of x in the interval a ≤ x ≤ b. Or The volum e of the solid generated by the revolution of the area bounded by the curve b ⌠ y = f (x) , x-axis and the ordinates x = a, x = b about the x-axis is ⎮ π y2 dx . ⌡a I-168 INTEGRAL CALCULUS Proof : Let A B be the arc of the curve y = f (x) included between the ordinates x = a and x = b. It is being assumed that the curve does not cut the x-axis and f (x) is a continuous function of x in the interval (a, b) . Let P (x, y) and Q (x + δx, y + δy) be any two neighbouring points on the curve y = f (x). Draw the ordinates PM and QN. Also draw PP′ and QQ′ perpendiculars to these ordinates. L e t V d e n o t e t h e vo lu m e o f t h e so l id generated by the revolution of the area A CMP about the x-axis and let the volume of revolution obtained by revolving the area A CNQ about x-axis be V + δV, so that volume of the solid generated by the revolution of the strip PMNQ about the x-axis is δV. Now PM = y, QN = y + δy and MN = (x + δx) − x = δx . Then the volume of the solid generated by revolving the area PMN P′ = πy2 δx and the volume of the solid generated by revolving the area Q′MN Q = π ( y + δy) 2 δx . Also the volume of the solid generated by the revolution of the area PMNQP (i.e., the volume δV ) lies between the volumes of the right circular cylinders generated b y t h e r e vo l u t i o n o f t h e a r e a s PMN P′P a n d MN QQ′ i.e., δV l ie s be t we e n π y2 δx and π ( y + δy) 2 δx (δV ⁄ δx) lies between π y2 and π ( y + δy) 2 or i.e., π y2 < (δV ⁄ δx) < π ( y + δy) 2. In the limiting position as Q → P, δx → 0 (and therefore δy → 0), we have dV ⁄ dx = π y2 H ence or dV = π y2 dx . b b ⎡ ⎤x= b ⌠ ⌠ ⎮ π y2 dx = ⎮ dV = ⎢⎣ V ⎥⎦ x= a ⌡a ⌡a = (value of V for x = b) − (value of V for x = a) = volume generated by the area A CDB − 0 = volume of the solid generated by the revolution of the given area A CDB about the axis of x . b ∴ ⌠ the required volume = π ⎮ y2 dx . ⌡a (Meerut 2003) (b) The axis of rotation being y-axis : Similarly, it can be shown that the volume of the solid generated by the revolution about y-axis of the area between the curve x = f ( y), the y-axis and the two abscissae b ⌠ y = a and y = b is given by ⎮ π x2 dy. ⌡a Important Remarks : (i) If the given curve is symmetrical about x-axis and we have to find the volume generated by the revolution of the area about x-axis, then in such case we shall revolve only one of the two symmetrical areas and shall not double it as in the case of area or length. O bviously each of the two symmetrical parts will generate the same volume. I-169 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION (ii) If the curve is symmetrical about x-axis and it is required to find the volume generated by the revolution of the area about y-axis, then the volume generated will be twice the volume generated by half of the symmetrical portion of the curve. 4 πa3. 3 (Bundelkhand 2010; Avadh 10) Example 1 : Show that the volum e of a sphere of radius a is Solution : The sphere is generated by the revolution of a se m i-cir cu la r a r e a a b o u t it s bo u n din g diamet er . The equation of the generating circle of radius a and centre as origin is x2 + y2 = a2. Let A A′ be the bounding d i a m e t e r a b o u t wh i c h t h e semi-circle revolves. Take an elementary strip PMNQ where P is the point (x, y) and Q is the point (x + δx, y + δy). We have PM = y and MN = δx . Now volume of the elementary disc formed by revolving the strip PMN Q about the diameter A A′ is = π . PM 2. MN = π y2 δx = π (a2 − x2) δx . Also the semi-circle is symmetrical about the y-axis and for the portion of the curve lying in the first quadrant x varies from 0 to a . ∴ the required volume of the sphere a a ⎡ ⌠ 1 ⎤ 1 4 = 2 ⎮ π (a2 − x2) dx = 2π ⎢ a2 x − x3⎥ = 2π [a3 − a3] = πa3. 3 ⎦ 3 3 ⌡0 ⎣ 0 Example 2 : The curve y2 (a + x) = x2 (3a − x) revolves about the axis of x . Find the volum e generated by the loop. (Meerut 2004; Bundelkhand 05) Solution : The given curve is y2 (a + x) = x2 (3a − x) . ...(1) It is symmetrical about x-axis. Putting y = 0 in (1), we get x = 0 and x = 3a i.e., a loop is formed between (0, 0) and (3a, 0) . The volume generated by the revolution of the whole loop about x-axis is the same as the volume generated by the revolution of the upper half of the loop about x-axis. Take an elementary strip PMNQ where P is the point (x, y) a n d Q i s t h e p o i n t (x + δx, y + δy) . W e h a ve PM = y and MN = δx . Now volume of t he element ary disc formed by revolving the strip PMNQ about the axis of x is = πPM 2 . MN = π y2 δx . ∴ the required volume generated by the loop I-170 INTEGRAL CALCULUS 3a 3a ⌠ x2 (3a − x) πy2 dx = π ⎮ dx , ⌡0 a+ x ⌠ = ⎮ ⌡0 3a ⎧ ⌠ = π⎮ ⌡0 ⎪⎨ − x2 + 4ax − 4a 2 + 4a ⎪⎩ x+ from (1) 3 ⎫ ⎪⎬ dx , dividing the Nr. by the Dr. a ⎪⎭ 3a ⎤ x3 4ax2 + − 4a2x + 4a3 log (x + a) ⎥ ⎦0 3 2 ⎡ = π ⎢− ⎣ = π [− 9a3 + 18a3 − 12a3 + 4a3 (log 4a − log a)] = π [− 3a3 + 4a3 log 4] = πa3 [8 log 2 − 3] . Example 3 : Find the volume of the solid generated by the revolution of the cissoid (2a − x) = x3 about its asymptote. (Meerut 2007; Kanpur 14) Solution : T h e gi ve n c u r ve i s y2 (2a − x) = x3. Its shape is as shown in the figure. E quating to zero t he coefficient of highest power of y, the asymptote parallel to the axis of y is x = 2a . Take an elementary strip PMNQ perpendicular to the asymptote x = 2a where P is the point (x, y) and Q is the point (x + δx, y + δy) . We have PM = 2a − x and MN = δy . N ow vo lume of t h e ele ment ary disc formed by revolving the strip PMN Q about the line x = 2a is y2 = π . PM 2. MN = π (2a − x) 2 δy. The given curve is symmetrical about x-axis and for the portion of the curve above x-axis y varies from 0 to ∞ . ∞ ⌠ ∴ the required volume = 2 ⎮ π (2a − x) 2 dy. ⌡y = 0 ...(1) From the given equation of the curve y2 (2a − x) = x3 we observe that the value of x cannot be easily found in terms of y. H ence for the sake of integration we change the independent variable from y to x . (Note) The curve is y2 = ∴ or 2y x3 ; 2a − x dy (2a − x) . 3x2 − x3 (− 1) 2 (3a − x) x2 = = dx (2a − x) 2 (2a − x) 2 (3a − x) x2 √(2a − x) (3a − x) √x √(2a − x) ⋅ dx = dx . x √x (2a − x) 2 (2a − x) 2 Also when y = 0, x = 0 and when y → ∞ , x → 2a . H ence from (1), the required volume dy = 2a ⎡ (3a − x) √x √(2a − x) ⎤ ⌠ (2a − x) 2 ⎢ = 2π ⎮ ⎥ dx ⌡x = 0 ⎣ ⎦ (2a − x) 2 I-171 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION 2a ⌠ = 2π ⎮ ⌡0 (3a − x) √x √(2a − x) dx . Now put x = 2a sin 2 θ so that dx = 4a sin θ cos θ dθ. When x = 0, θ = 0 and when x = 2a, θ = π ⁄ 2 . Therefore the required volume π ⁄2 ⌠ (3a − 2a sin2 θ) √(2a) sin θ √[2a (1 − sin2 θ)] × 4a sin θ cos θ dθ = 2π ⎮ ⌡0 π ⁄2 ⌠ = 16πa3 ⎮ ⌡0 (3 sin2 θ cos2 θ − 2 sin4 θ cos2 θ) dθ ⎡ 3 3 5 3 ⎤ 2 Γ ( ) Γ ( )⎥ ⎢ 3Γ ( ) Γ ( ) 2 2 2 2 ⎥ − ⎢ ⎥ 2 Γ (4) ⎦ ⎣ 2 Γ (3) = 16πa3 ⎢ = 16πa3 ⎢ ⎡ ⎤ 1 1 3 1 1 2 ⋅ ⋅ ⋅ √π ⋅ ⋅ √π ⎥ ⎢ 3 ⋅ ⋅ √π ⋅ ⋅ √π 2 2 2 2 2 ⎥ − ⎢ ⎥ 2.2.1 2.3.2.1 ⎣ ⎦ π⎤ ⎡ 3π − ⎥ = 2π 2a 3. 16⎦ ⎣ 16 = 16πa3 ⎢ Note : If the given curve is y2 (a − x) = x3, then the required volume can be 1 4 obtained by putting a for 2a in the above E xercise. The volume so obtained is π 2a3. Important Remark : When we are to revolve an area about a line which is neither the x-axis nor the y-axis we m ust take an elem entary strip which is perpendicular to the line of revolution as explained in the above exam ple. Example 4 : The area between a parabola and its latus rectum revolves about the directrix. Find the ratio of the volum e of the ring thus obtained to the volum e of the sphere whose diam eter is the latus rectum . Solution : Let the parabola be y2 = 4ax . Then the directrix is the line x = − a . Let L L ′ be the latus re ct um. The area L OL′SL is revo lved abo u t t h e directrix. The volume of the ring thus obtained = the volume V 1 of the cylinder formed by the revolution of t h e r e ct a n gle L L′R′R a bo u t t h e di r e ct r ix − the volume V 2 of the reel formed by the revolution of the arc L OL′ about the directrix. N o w t h e vo l u m e V 1 o f t h e c yl i n d e r = π r2 h = π (L R) 2 . L L′ = π (2a) 2. 4a = 16πa3. To find the volume V 2 of the reel consider an e l e m e n t a r y s t r i p PMN Q wh e r e P (x, y) a n d Q (x + δx, y + δy) are two neighbouring points on the arc OL and PM, QN are perpendiculars from P and Q on the directrix. We have PM = a + x and MN = δy. ∴the volume V 2 of the reel 2a ⌠ = 2⎮ ⌡0 π (a + x) 2 dy [By symmetry about x-axis] I-172 INTEGRAL CALCULUS 2a ⌠ = 2⎮ ⌡0 2a ⎛ ⌠ π (a2 + 2ax + x2) dy = 2π ⎮ ⌡0 2 4 ⎞ ⎜ a 2 + 2a ⋅ y + y ⎟ dy ⎜ ⎟ ⎜ 4a 16a2 ⎟⎠ ⎝ [ ... x = y2 ⁄ 4a] 2a 1 y3 1 y5 ⎤ + ⋅ ⎥⎥ 2 5 ⎥⎦ 2 3 16a ⎣ 0 4 3 2 3⎤ 56 112 πa3 ⎡ 3 = 2π ⎢ 2a + a + a ⎥ = 2πa3 ⋅ = ⋅ 15 3 5 ⎦ 15 ⎣ ∴ Volume of the ring = volume of the cylinder − volume of the reel 112 3 128 3 = V 1 − V 2 = 16πa3 − πa = πa . 15 15 Volume of the sphere whose diameter is the latus rectum 4a i.e., the radius is 2 a ⎡ = 2π ⎢⎢ a2y + ⎢ = ∴ 4 3 the required ratio = 4 32 π (2a) 3 = 3 3 3 128πa ⁄ 15 4 π r3 = 32πa3 ⁄ 3 = 5 πa3. ⋅ 1. (i) Find the volume of a hemisphere. (ii) Find the volume of a spherical cap of height h cut off from a sphere of radius a. (Kanpur 2010) 2. (i) A segment is cut off from a sphere of radius a by a plane at a distance 1 2 a from the centre. Show that the volume of the segment is 5 ⁄ 32 of the volume of the sphere. 3. 4. 5. 6. (ii) The part of the parabola y2 = 4ax cut off by the latus rectum revolves about the tangent at the vertex. Find the volume of the reel thus generated. Prove that the volume of the solid generated by the revolution of an ellipse round its minor axis is a mean proportional between those generated by the revolution of the ellipse and of the auxiliary circle about the major axis. (Rohilkhand 2010) (i) Find the volume of the solid generated by the revolution of an arc of the catenary y = c cosh (x ⁄ c) about the x-axis. (Meerut 2009B; Purvanchal 11) (ii) Find the volume of the solid generated by the revolution of the curve y = a3 ⁄ (a2 + x2) about its asymptote. (Meerut 2009) If the hyperbola x2 ⁄ a2 − y2 ⁄ b2 = 1 revolves about the x-axis, show that the volume in clu ded bet ween t h e su rface t h u s gen erat ed, t h e co n e gen erat ed by t he asymptotes and two planes perpendicular to the axis of x, at a distance h apart, is equal to that of a circular cylinder of height h and radius b. (i) F in d t he volu me fo rmed by t h e revolut ion o f t he loop of t he curve y2 (a + x) = x2 (a − x) about the axis of x . (Kanpur 2008) (ii) Find the volume of the solid generated by the revolution of the loop of the curve y2 = x2 (a − x) about the axis of x . (Kanpur 2011) I-173 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION 7. Show that the volume of the solid generated by the revolution of the upper half 4 of the loop of the curve y2 = x2 (2 − x) about x-axis is π . (Meerut 2005) 3 8. 9. The area of the curve x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3 lying in the first quadrant revolves about (Agra 2014) x-axis. Find the volume of the solid generated. Find the volume of the solid obtained by revolving the loop of the curve a2 y2 = x2 (2a − x) (x − a) about x-axis. 10. 11. A basin is formed by the revolution of the curve x3 = 64y, ( y > 0) about the axis of y. If the depth of the basin is 8 inches, how many cubic inches of water it will hold ? Show that the volume of the solid generated by the revolution of the curve 1 2 3 π a . 2 (Meerut 2004B, 06B; Kumaun 08; Rohilkhand 12) (a − x) y2 = a2 x , about its asymptote is 12. The figure bounded by a quadrant of a circle of radius a and tangents at its extremities revolves about one of the tangents. Prove that the volume of the solid 5 3 generated is ( − 13. 1 2 π) πa3. The area cut off from the parabola y2 = 4ax by the chord joining the vertex to an end of the latus rectum is rotated through four right angles about the chord. Find the volume of the solid generated. (Rohilkhand 2008; Bundelkhand 09) 2 3 ⎡ πa3. (ii) πh2 ⎢⎣ a − 1. (i) 2. (ii) 6. (i) 2 a3π ⎢⎣ log 2 − 9. 23 3 πa . 60 4 5 πa3. 4. (i) ⎡ 2 ⎤⎥ 3⎦ ⋅ 1 3 ⎤ h⎥⎦ ⋅ πc 2 ⎡⎢ c 2 x⎤ x + sinh ⎥⎦ ⋅ 2 2 ⎣ c 1 πa4. 12 1536 10. π cubic inches. 5 (ii) π 2 a3 ⋅ 2 16 8. πa3. 105 2 13. √5 πa3. 75 (ii) 8.3 Volume of a Solid of Revolution when the Equations of the Generating Curve are given in Parametric Form (i) If the curve is given by the parametric equations, say x = φ (t), y = ψ (t), then the volume of the solid generated by the revolution about x-axis of the area bounded by the curve, the axis of x and the ordinates at the points where t = a and t = b is b b ⌠ dx ⌠ = ⎮ π y2 dt = π ⎮ { ψ (t)} 2 φ ′ (t) dt . ⌡a ⌡a dt (ii) The volume of the solid generated by the revolution about y-axis of the area between the curve x = φ (t ), y = ψ (t ), the y-axis and the abscissae at the points where t = a, t = b is I-174 INTEGRAL CALCULUS b b ⌠ dy ⌠ = ⎮ π x2 dt = π ⎮ { φ (t)} 2 . ψ′ (t) dt . ⌡a ⌡a dt Example 1 : Find the volum e of the solid form ed by revolving the cycloid x = a (θ − sin θ), y = a (1 − cos θ) (i) about its base (ii) about the y-axis. Solution : The given equations of the cycloid are ...(1) x = a (θ − sin θ), y = a (1 − cos θ) . (i) The arc OBA is revolved about the base i.e., the x-axis. For the arc OBA, θ varies from 0 to 2π and at B, θ = π . Take an elementary strip PMN Q where P is the point (x, y) and Q is the point (x + δx, y + δy) . We have PM = y and MN = δx . Now the volume of the elementary disc formed by revolving the strip PMN Q about the base (i.e., the x-axis) is π PM 2 . MN = π y2 δx . Now the cycloid is symmetrical about the line BH . ∴ the required volume ⌠ = 2 ⎮ π y2 dx , the limits of integration being extended from O to B ⌡ π π ⌠ dx ⌠ = 2π ⎮ y2 dθ = 2π ⎮ a2 (1 − cos θ) 2 a (1 − cos θ) dθ [From (1)] ⌡θ = 0 ⌡0 dθ π ⌠ = 2π ⎮ a3 (1 − cos θ) 3 dθ ⌡0 π ⌠ = 2π a3 ⎮ ⎛⎜ 2 sin2 ⌡0 ⎝ π ⁄2 ⌠ = 32πa3 ⎮ ⌡0 = 32π a3 ⋅ 5 6 ⋅ π θ⎞ 3 θ ⌠ ⎟ dθ = 16π a 3 ⎮ sin6 dθ ⌡0 2⎠ 2 sin6 φ dφ , putting 3 4 ⋅ 1 2 ⋅ 1 2 θ = φ so that dθ = 2 dφ 2 π = 5π 2a3. (ii) When the curve revolves about y-axis, the required volume of the solid generated I-175 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION = the volume generated by the revolution of the area OA BDO about y-axis − the volume generated by the revolution of the area OBDO about the y-axis. ...(2) Also at A, θ = 2π ; at B, θ = π and at O, θ = 0. Now the area OA BD is bounded by the arc A B of the cycloid and the axis of y . Therefore volume of the solid generated by the revolution of the area OA BDO about y -axis π π dy ⌠ ⌠ = ⎮ π x2 dy = ⎮ π x2 dθ ⌡θ = 2π ⌡θ = 2π dθ π ⌠ = π ⎮ a2 (θ − sin θ) 2 a sin θ dθ ⌡2π [From (1)] π ⌠ = π ⎮ a2 (θ2 − 2θ sin θ + sin2 θ) a sin θ dθ ⌡2π π ⌠ = πa3 ⎮ (θ2 sin θ − 2θ sin2 θ + sin3 θ) dθ ⌡2π π ⌠ 1 = πa3 ⎮ [θ2 sin θ − θ (1 − cos 2θ) + (3 sin θ − sin 3θ)] dθ 4 ⌡2π ⎡ ⎣ 1 2 θ 2 = πa3 ⎢ θ2 . (− cos θ) − 2θ (− sin θ) + 2 cos θ − − 1 (− 1 4 cos 2θ) − 3 4 (Note) 1 2 + θ ( sin 2θ) cos θ + ⎤π cos 3θ⎥ , ⎦ 2π 1 12 ⌠ ⌠ the values of the integrals ⎮ θ2 sin θ dθ and ⎮ θ cos 2θ dθe ⌡ ⌡ have been written after applying integration by parts = πa3 [(π 2 = πa3 [ 13 2 − 2− π2 − 1 2 π2 + 1 4 + 3 4 − 1 ) 12 − (− 4π 2 + 2 − 2π 2 + 1 4 3 4 − 8 ]⋅ 3 + 1 )] 12 ...(3) Again volume of the solid generated by the revolution of the area OBDO about y-axis π π π ⌠ ⌠ dy ⌠ = ⎮ π x2 dy = ⎮ π x2 dθ= π ⎮ a2 (θ − sin θ) 2 . a sin θ dθ ⌡θ = 0 ⌡θ = 0 ⌡0 dθ π ⌠ = πa3 ⎮ (θ2 − 2θ sin θ + sin2 θ) sin θ dθ ⌡0 π ⌠ = πa3 ⎮ (θ2 sin θ − 2θ sin2 θ + sin3 θ) dθ ⌡0 π ⌠ 1 = πa3 ⎮ [θ2 sin θ − θ (1 − cos 2θ) + (3 sin θ − sin 3θ)] dθ 4 ⌡0 ⎡ ⎣ = πa3 ⎢ θ2 (− cos θ) − 2θ (− sin θ) + 2 cos θ − − 1 (− 1 4 1 2 θ 2 cos 2θ) − 1 2 + θ ( sin 2θ) 3 4 cos θ + 1 12 ⎤π ⎦0 cos 3θ⎥ I-176 INTEGRAL CALCULUS = πa3 [(π 2 − 2 − 1 2 1 2 π2 + 1 4 + 3 4 − 1 ) 12 − (2 + 1 4 − 3 4 + 1 )] 12 8 ). 3 = πa3 ( π 2 − ...(4) ∴ from (2), the required volume = (3) − (4) = πa3 [ 13 2 π2 − 8 ] 3 1 2 − πa3 [ π 2 − 8 ] 3 = π a3 [6π 2] = 6π 3a3. Example 2 : Find the volum e of the solid generated by the revolution of the tractrix x = a cos t + 1 2 a log tan2 (t ⁄ 2), y = a sin t about its asymptote. (Meerut 2000, 05B; Rohilkhand 06; Avadh 09, 11; Kashi 12; Purvanchal 14) Solution : The given curve is x = a cos t + ∴ 1 2 a log tan2 (t ⁄ 2), y = a sin t . ...(1) 1 dx 1 1 = − a sin t + a ⋅ ⋅ 2 tan (t ⁄ 2) sec2 (t ⁄ 2) ⋅ 2 2 2 dt tan (t ⁄ 2) a a = − a sin t + = − a sin t + 2 sin (t ⁄ 2) cos (t ⁄ 2) sin t (1 − sin2 t) cos2 t = a sin t sin t N o w t h e g i ve n c u r ve i s symmetrical about both the axes and the asymptote is the line y = 0 i.e., x-axis. For the portion of the curve lying in the second quadrant y varies from a to 0, t varies from π ⁄ 2 to 0 and x varies from 0 to − ∞ . ∴ the required volume = a ...(2) 0 ⌠ = 2 ⎮ π y2 dx ⌡− ∞ π ⁄2 ⌠ = 2⎮ ⌡0 π y2 π ⁄2 ⌠ = 2π ⎮ ⌡0 a2 sin2 t . π ⁄2 ⌠ = 2πa3 ⎮ ⌡0 1. dx ⋅ dt dt a cos2 t dt sin t cos2 t sin t dt = 2πa3 [From (1) and (2)] 1 2 = πa3. 3.1 3 Find the volume of the solid generated by the revolution of the cycloid x = a (θ + sin θ), y = a (1 − cos θ), − π ≤ θ ≤ π , (i) about the x-axis, (ii) about the base. I-177 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION 2. Show that the volume of the solid generated by the revolution of the cycloid x = a (θ + sin θ), y = a (1 − cos θ), 0 ≤ θ ≤ π , 3 2 about the y-axis is πa3 ( π 2 − 8 ) 3 ⋅ 3. Prove that the volume of the reel formed by the revolution of the cycloid x = a (θ + sin θ), y = a (1 − cos θ) about the tangent at the vertex is π 2a3. 4. Prove that the volume of the solid generated by the revolution about the x-axis 1 3 π. of the loop of the curve x = t2, y = t − t3 is 3 4 5. Find the volume of the spindle shaped solid generated by revolving the astroid 6. x2 ⁄ 3 + y2 ⁄ 3 = a2 ⁄ 3 about the x-axis. Find the volume of the solid generated by the revolution of the cissoid x = 2a sin2 t , y = 2 a sin3 t ⁄ cos t about its asymptote. (Kanpur 2006; Bundelkhand 14) 1. (i) π 2 a3, (ii) 5π 2 a3. 5. 32 πa3. 105 6. 2 π 2 a3. 8.4 Volume of Solid of Revolution when the Equation of the Generating Curve is given in Polar Co-ordinates If the equat ion of the generating curve is given in polar co-ordinates, say r = f (θ), and the curve revolves about the axis of x, the volume generated b β ⌠ ⌠ dx s= π ⎮ y2 dx = π ⎮ y2 dθ, ⌡x = a ⌡θ = α dθ where α and β are the values of θ at the points where x = a and x = b respectively. Now x = r cos θ and y = r sin θ. Therefore the volume β d ⌠ r 2 sin2 θ (r cos θ) dθ, = π⎮ ⌡θ = α dθ in which the value of r in terms of θ must be substituted from the equation of the given curve. A similar procedure can be adopted in case the curve revolves about the axis of y. Alternative method in the case of polar curves : The volum e of the solid generated by the revolution of the area bounded by the curve r = f (θ) and radii vectors θ = θ1, θ = θ2 θ ⌠ 2 2 3 (i) about the initial line θ = 0 (i.e., the x-axis) is ⎮ π r sin θ dθ, ⌡θ 3 θ 1 ⌠ 2 2 3 π r cos θ dθ, (ii) about the line θ = π ⁄ 2 (i.e., the y-axis) is ⎮ ⌡θ 3 1 I-178 INTEGRAL CALCULUS θ ⌠ 2 2 (iii) about any line (θ = γ) is ⎮ π r 3 sin (θ − γ) dθ, ⌡θ 3 1 where in each of the above three formulae the value of r in terms of θ must be substituted from the equation of the given curve. Note : The above results are important and should be committed to memory. 8.5 Volume of the Solid Generated by the Revolution when The Axis of Rotation being any Line If, however, the axis of rotation is neither x-axis nor y-axis, but is any other line CD, then the volum e of the solid generated by the revolution about CD of the area bounded by the curve A B, the axis CD and the perpendiculars A C, BD on the aixs is OD ⌠ ⎮ π (PM) 2 d (OM), ⌡OC where PM is the perpendicular drawn from any point P on the curve to the axis of rotation and O is som e fixed point on the axis of rotation. Example 1 : The cardioid r = a (1 + cos θ) revolves about the initial line. Find the volum e of the solid thus generated. (Meerut 2001, 03, 07B; Agra 06, 07, 08; Rohilkhand 13, 13B) Solution : The given curve is r = a (1 + cos θ) . ...(1) It is symmetrical about the initial line. We have r = 0 when cos θ = − 1 i.e., θ= π. Also r is maximum when cos θ = 1 i.e., θ = 0 and then r = 2a . As θ increases from 0 to π, r decreases from 2a to 0. H ence the sh ape o f the curve is as shown in the figure. For the upper half of the curve, θ varies from 0 to π . ∴ the required volume = π 2⌠ ⎮ π r 3 sin θ dθ 3 ⌡0 π ⌠ ⎮ a3 (1 + cos θ) 3 sin θ dθ ⌡0 [From (1)] π ⌠ 2 = − πa3 ⎮ (1 + cos θ) 3 (− sin θ) dθ 3 ⌡0 = 2π 3 = − 2 3 (Note) ⎡ (1 + cos θ) 4 ⎤ π ⎥ , using power formula ⎣ ⎦0 4 πa3 ⎢ [ f (x)]n+ 1 ⌠ i.e., ⎮ [ f (x)]n f ′ (x) dx = ⌡ n+ 1 = − 1 6 πa3 (0 − 24) = 8 3 πa3. I-179 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION Aliter : (By double integration) Take a small element r δθ δr at any point P (r, θ) lying within the area of the upper half of t he cardioid. Draw PM p e r p e n d icu la r to OX . Th en PM = r sin θ. T h e vo l u m e o f t h e elementary ring formed by revolving the element r δθ δr about OX = 2π (r sin θ) r δθ δr = 2πr 2 sin θ δθ δr . ∴ the required volume formed by revolving the whole cardioid about the initial line π a (1 + cos θ) ⌠ ⌠ ⎮ = ⎮ ⌡θ = 0 ⌡r = 0 2π r 2 sin θ dθ dr π π ⎡ r 3 ⎤⎥ a (1 + cos θ) 2π ⌠ ⌠ ⎮ a3 (1 + cos θ) 3 sin θ dθ ⎥⎥ = ⎮ 2π ⎢⎢⎢ sin θ dθ = ⌡0 3 ⌡0 ⎣ 3 ⎦0 π π = − 2πa3 ⌠ 2πa3 ⎡ (1 + cos θ) 4 ⎤ ⎢ ⎥ ⎮ (1 + cos θ) 3 (− sin θ) dθ = − ⎦0 3 ⌡0 3 ⎣ 4 = − 2πa3 1 2 1 8 ⋅ [0 − 24] = ⋅ πa3 ⋅ ⋅ 16 = πa3. 4 3 4 3 3 Example 2 : Find the volum e of the solid form ed by revolving one loop of the curve r 2 = a2 cos 2θ about the initial line. (Rohilkhand 2007B) Solution : For the upper half of the loop θ varies from 0 to π ⁄ 4 . H ere the curve is revolving about the initial line (i.e., x-axis). ∴ the required volume = π ⁄4 ⌠ ⎮ ⌡0 = 2π 3 = 2πa3 ⌠ ⎮ 3 ⌡0 π ⁄4 2 ⌠ π⎮ 3 ⌡0 r 3 sin θ dθ [ ... r 2 = a2 cos 2θ] {a √(cos 2θ)}3 sin θ dθ π ⁄4 (2 cos2 θ − 1) 3 ⁄ 2 sin θ dθ. (Note) Put √2 cos θ = sec φ so that − √2 sin θ dθ = sec φ tan φ dφ . When θ = 0, φ = π ⁄ 4 and when θ = π ⁄ 4, φ = 0. ∴ the required volume 0 (− sec φ tan φ ) 2πa3 ⌠ ⎮ (sec 2 φ − 1) 3 ⁄ 2 dφ = 3 ⌡π ⁄ 4 √2 π ⁄4 = √2πa3 ⌠ ⎮ 3 ⌡0 = √2πa3 ⌠ ⎮ 3 ⌡0 π ⁄4 tan4 φ sec φ dφ = π ⁄4 √2πa3 ⌠ ⎮ 3 ⌡0 (sec 5 φ − 2 sec 3 φ + sec φ ) dφ . (sec 2 φ − 1) 2 sec φ dφ ...(1) I-180 INTEGRAL CALCULUS Also we know the reduction formula sec n − 2 φ tan φ n− 2⌠ ⌠ ⎮ sec n − 2 φ dφ . ⎮ sec n φ dφ = + ⌡ n− 1⌡ n− 1 [ Establish it here] π ⁄4 ⌠ ⎮ ⌡0 ∴ ⎡ sec 3 φ tan φ ⎥ ⎥⎥ 4 ⎦ sec 5 φ dφ = ⎢⎢⎢ ⎣ 0 + π ⁄4 √2 3 ⎧⎪ ⎡ sec φ tan φ ⎤ π ⁄ 4 1 ⌠ + ⎨⎢ + ⎮ ⎥ 2 4 ⎪⎩ ⎣ 2 ⌡0 2 ⎦0 = π ⁄4 3⌠ ⎮ 4 ⌡0 sec 3 φ dφ ⎫ sec φ dφ ⎪⎬ ⎪ ⎭ π ⁄4 = √2 3 ⎧ √2 1 ⎡ + ⎨ + ⎢ log (sec φ + tan φ ) ⎤⎥ 2 4⎩ 2 2⎣ ⎦0 = √2 3√2 3 7√2 3 + + log (√2 + 1)= + log (√2 + 1), 2 8 8 8 8 π ⁄4 ⌠ ⎮ ⌡0 π ⁄4 ⌠ ⎮ ⌡0 ⎫ ⎬ ⎭ π ⁄4 ⎡ sec φ tan φ ⎤ π ⁄ 4 1⌠ + ⎮ ⎥ ⎣ ⎦0 2 ⌡0 2 sec 3 φ dφ = ⎢ = and ⎤π ⁄4 sec φ dφ √2 1 + log (√2 + 1) 2 2 sec φ dφ = log (√2 + 1) . H ence the required volume from (1) is = √2πa3 ⎡ 7√2 3 1 ⎧ √2 ⎫ ⎤ + log (√2 + 1) − 2 ⎨ + log (√2 + 1) ⎬ + log (√2 + 1) ⎥ ⎢ 8 2 3 ⎣ 8 ⎦ ⎩ 2 ⎭ = √2πa3 ⎡ 3 √2 ⎤ ⎢ log (√2 + 1) − ⎥ 3 ⎣8 8⎦ = πa3 √2 [3 log (√2 + 1) − √2] . 24 Aliter : The equation of the given curve is r 2 = a2 cos 2θ or r 4 = a2 r 2 (cos2 θ − sin 2 θ) . Changing to cartesians, the equation becomes (x2 + y2) 2 = a2 (x2 − y2) or y4 + y2 (2x2 + a2) + x4 − a2 x2 = 0. Solving for y2, we have y2 = [− (2x2 + a2) ± √{(2x2 + a2) 2 − 4 (x4 − a2 x2)}] / 2 . Neglecting the negative sign because y2 cannot be –ive, we have 1 − (2x2 + a2) + 2 √2a √(x2 + a2) − (2x2 + a2) + √(8a2 x2 + a4) 8 = ⋅ y2 = 2 2 Now for one loop of the given curve x varies from 0 to a . a ⌠ ∴the required volume = π ⎮ y2 dx ⌡0 = π 2 a ⌠ 1 ⎮ [− 2x2 − a2 + 2 √2a √(x2 + a2)] dx 8 ⌡0 I-181 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION = π⎡ 2 3 x 1 ⎢− x − a2 x + 2 √2a ⋅ √(x2 + a2) 8 2⎣ 3 2 + 2√2a ⋅ 1 2 a 16 log {x + √(x2 + 1 8 ⎤a ⎦0 a2)}⎥ π⎡ 2 3 a 3a a − a3 + 2 √2a ⋅ ⋅ ⎢− 2⎣ 3 2 2 √2 ⎧ ⎛ a ⎫⎤ 1 3a ⎞ ⎟ − log ⎬⎥ + √2a3 ⎨ log ⎜ a + 2 √2 ⎭ ⎦ 8 2 √2 ⎠ ⎩ ⎝ π⎡ 5 3 1 ⎧ a (2 √2 + 3) 2 √2 ⎫ ⎤ = ⎢ − a3 + a3 + √2a3 log ⎨ ⋅ ⎬⎥ 2⎣ 3 2 √2 a ⎭⎦ 2 8 ⎩ π⎡ 1 1 ⎤ = ⎢ − a3 + √2a3 log (2 √2 + 3) ⎥ ⎦ 2⎣ 6 8 π⎡ 1 3 1 ⎤ = ⎢ − a + √2a3 log (√2 + 1) 2⎥ ⎦ 2⎣ 6 8 = = πa3 ⎡ πa3 ⎡ 1 1 1⎤ 1⎤ √2 log (√2 + 1) − ⎥ = ⎢2 ⋅ ⎢ √2 log (√2 + 1) − ⎥ 2 ⎣ 2 ⎣4 8 6⎦ 6⎦ = πa3 πa3 √2 ⎡ 3√2 log (√2 + 1) − 2⎤ = [3 log (√2 + 1) − √2] . ⎣ ⎦ 24 24 Example 3 : Show that if the area lying within the cardioid r = 2a (1 + cos θ) and without the parabola r (1 + cos θ) = 2a revolves about the initial line, the volum e generated is 18πa3. Solution : The equation of the cardioid is r = 2a (1 + cos θ), ...(1) ...(2) and that of the parabola is r = 2a ⁄ (1 + cos θ) . E quating the values of r from (1) and (2), we get 2a (1 + cos θ) = 2a ⁄ (1 + cos θ) (1 + cos θ) 2 = 1 cos θ (cos θ + 2) = 0. Now cos θ ≠ − 2 . Therefore cos θ = 0 i.e., θ = π ⁄ 2 , − π ⁄ 2 . Thus the curves (1) and (2) intersect where θ = π ⁄ 2 and θ = − π ⁄ 2 . A l s o b o t h t h e c u r ve s a r e symmet rical abo ut th e initial line (i.e., x-axis). The required volume is generated by revolving the upper half of the shaded area about the initial line. ∴ the required volume = Volume generated by the revolution of the area OA BO o f t h e c a r d i o i d − volume generated by the revolution of the area OPBO of the parabola or or I-182 INTEGRAL CALCULUS = 2π 3 π ⁄2 ⌠ ⎮ ⌡0 r 3 sin θ dθ − (for cardioid) 2π 3 π ⁄2 ⌠ ⎮ ⌡0 r 3 sin θ dθ (for parabola) π ⁄2 ⎡ ⌠ ⎤ 8a3 ⎢ 8a 3 (1 + cos θ) 3 − ⎥ sin θ dθ ⎢ ⎥ 3 ⎢ (1 + cos θ) ⎥⎦ ⎣ = 2π ⎮ 3 ⌡0 = − 16πa3 ⌠ ⎮ ⌡0 3 = − 16πa3 ⎡⎢ (1 + cos θ) 4 (1 + cos θ) − − ⎢⎢ 3 4 − 2 ⎣ π ⁄2 [(1 + cos θ) 3 − (1 + cos θ) − 3] (− sin θ) dθ 2⎤ π ⁄ 2 ⎥⎥ ⎥⎦ 0 (Note) , using power formula − 16πa3 ⎡ 1 1⎛ 1⎞ ⎤ ⎢ (1 − 16) + ⎜1 − ⎟ ⎥ ⎣4 2⎝ 4⎠ ⎦ 3 − 16 3 ⎡ 15 3 ⎤ πa ⎢ − + ⎥ = ⎣ 4 8⎦ 3 − 27 ⎛ 16 ⎞ ⎛ ⎞ = ⎜− πa3⎟ ⎜ ⎟ 3 ⎝ ⎠ ⎝ 8 ⎠ = = 18πa3. 1. 2. 3. Find the volume of the solid generated by the revolution of r = 2a cos θ about the initial line. F ind t he volume of the solid generated by t he revolution of the cardioid r = a (1 − cos θ) about the initial line. (Rohilkhand 2010) The arc of the cardioid r = a (1 + cos θ) , specified by − π ⁄ 2 ≤ θ ≤ π ⁄ 2 , is rotated 5 2 about the line θ = 0, prove that the volume generated is πa3. 4. Show that the volume of the solid formed by the revolution of the curve r = a + b cos θ (a > b) about the initial line is 5. πa (a2 + b2) . (Meerut 2008) Find the volume of the solid generated by revolving one loop of the lemniscate r2 = a2 cos 2θ about the line θ = 1. 4 3 4 3 π a 3. 8 3 2. π a3. 4. 4 3 1 2 π. πa (a2 + b 2). (Meerut 2006) 5. (π 2 a3) ⁄ 4 √2 . VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION 8.6 Surfaces of Solids of Revolution I-183 (Agra 2014) (a) Revolution about the axis of x : To prove that the curved surface of the solid generated by the revolution, about x-axis, of the area bounded by the curve y = f (x), the ordinates x = a, x = b and the x-axis is x= b ⌠ ⎮ 2 πy ds , ⌡x = a where s is the length of the arc m easured from x = a to any point (x, y). Or Show that the area of the surface of the solid obtained by revolving about x-axis the arc of the curve intercepted between the points whose abscissae are a and b is b ds ⌠ ⎮ 2πy dx . ⌡a dx Proof : Let AB be the arc of the curve y = f (x) included between the ordinates x = a and x = b. It is being assumed that the curve does not cut x-axis and f (x) is a continuous function of x in the interval (a, b) . Let P (x, y) and Q (x + δx, y + δy) be any two neighbouring points on the curve y = f (x) . Let the length of the arc AP be s and arc A Q = s + δs so that arc PQ = δs. D raw t h e o rdinat es PM a n d QN . L e t S denote the curved surface of the solid generated by the revolution of the area CMPA about the x-axis. Then the curved surface of the solid generated by the revolution of the area MN QP = δS . We shall take it as an axiom that the curved surface of the solid generated by the revolution of the area MN QP about the x-axis lies between the cu rved su rfaces of th e right circular cylinders whose radii are PM and N Q and which are of the same thickness (height) δs. There is no loss in assuming so because ultimately Q is to tend to P . Thus δS lies between 2π yδs and 2π ( y + δy) δs i.e., 2π y δs < δS < 2π ( y + δy) δs or 2π y < (δS ⁄ δs) < 2π ( y + δy) . Now as Q approaches P i.e., δs → 0, δy will also tend to zero. H ence by taking limits as δs → 0, we have dS = 2π y or dS = 2π y ds. ds ∴ x= b x= b x= b ⌠ ⌠ ⎮ 2π y ds = ⎮ dS = ⎡⎢⎣ S ⎤⎥⎦ ⌡x = a ⌡x = a x= a = (the value of S when x = b) − (the value of S when x = a) = surface of the solid generated by the revolution of the area A CDB − 0. x= b ⌠ 2π y ds ∴ the required curved surface = ⎮ ⌡x = a I-184 INTEGRAL CALCULUS x= b √ ⎧ ⎛ dy⎞ 2 ⎫ ⌠ ds ds = ⎮ 2π y dx , where = ⎨1 + ⎜ ⎟ ⎬ ⋅ ⌡x = a ⎩ ⎝ dx⎠ ⎭ dx dx (b) Axis of revolution as y-axis : Sim ilarly the curved surface of the solid generated by the revolution about the y-axis, of the area bounded by the curve x = f ( y), the lines y = a , y = b and the y-axis is y= b b ⌠ ds ds ⌠ x ds or S = 2π ⎮ x dy , where = 2π ⎮ ⌡y = a ⌡y = a dy dy √ ⎧ ⎛ dx⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ ⋅ ⎩ ⎝ dy ⎠ ⎭ Important Remark : If an arc length revolves about x-axis, the basic formula for ⌠ the surface of revolution in all cases is ⎮ 2π y ds , between the suitable limits. If we want ⌡ to integrate w.r.t. x , we shall change ds as (ds ⁄ dx) dx and adjust the limits accordingly. A similar transformation can be made if we want to integrate w.r.t. y or with respect to θ or w.r.t. some parameter, say t . Example 1 : Find the curved surface of a hem isphere of radius a . (Agra 2005; Kanpur 14) Solution : A hemisphere is generated by the revolution of a quadrant of a circle about one of its bounding radii. ...(1) Let the equation of the circle be x2 + y2 = a2. Let the hemisphere be formed by revolving about x-axis the arc of the circle (1) lying in the first quadrant. Differentiating (1), w.r.t. x , we get 2x + 2y (dy ⁄ dx) = 0 or dy ⁄ dx = − x ⁄ y . Therefore √ = √ ds = dx ⎧ ⎛ dy⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ = ⎩ ⎝ dx⎠ ⎭ ⎧ y2 + x2 ⎫ ⎪ ⎪= ⎨ ⎬ ⎪ y2 ⎪ ⎩ ⎭ √ √ 2⎫ ⎧ ⎪1 + x ⎪ ⎨ ⎬ ⎪ y2 ⎪⎭ ⎩ ⎛ a2 ⎞ ⎜ ⎟ ⎜ 2⎟ ⎜y ⎟ ⎝ ⎠ [From (1)] = a ⁄ y. For the arc of the circle (1) lying in the first quadrant x varies from 0 to a . ∴ the required surface a a a a ⌠ ⌠ ds = 2π ⎮ y ds = 2π ⎮ y ⋅ dx ⌡x = 0 ⌡0 dx ⌠ a ⌠ a = 2π ⎮ y ⋅ dx = 2π ⎮ a dx = 2π a ⎡⎢ x ⎤⎥ ⌡0 ⌡0 y ⎣ ⎦0 = 2πa . a = 2πa2. Example 2 : Find the surface generated by the revolution of an arc of the catenary y = c cosh (x ⁄ c) about the axis of x . (Meerut 2000, 04B, 07, 07B, 10; Rohilkhand 14) Solution : The given curve is, y = c cosh (x ⁄ c) . ...(1) Differentiating (1) w.r.t x , we get I-185 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION x 1 x dy = c sinh ⋅ = sinh ⋅ c c c dx ∴ ds = dx √ ⎧ ⎛ dy⎞ 2 ⎫ ⎨1 + ⎜ ⎟ ⎬ = ⎩ ⎝ dx⎠ ⎭ √ 1 + sinh xc ⎧ ⎨ ⎩ 2 ⎫ ⎬ ⎭ = cosh x c ...(2) If the arc be measured from the vertex (x = 0) to any point (x, y), then the required surface formed by the revolution of this arc about x-axis x x ⌠ = ⎮ ⌡x = 0 2π y ds x x ⌠ dx = 2π ⎮ c cosh ⋅ cosh dx , from (1) and (2) ⌡0 dx c c x x ⌠ x ⌠ ⎡ 2x⎤ = π c ⎮ 2 cosh2 dx = π c ⎮ ⎢ 1 + cosh ⎥ dx ⌡0 ⌡0 ⎣ c c⎦ ⎡ ⎣ = π c ⎢x + (Note) c 2x⎤ x c 2x⎤ ⎡ sinh ⎥ = π c ⎢ x + sinh ⎥ 2 c ⎦0 2 c⎦ ⎣ ⎡ ⎣ = π c ⎢ x + c sinh x x⎤ cosh ⎥ ⋅ c c⎦ Example 3 : Prove that the surface of the prolate spheroid form ed by the revolution of the ellipse of eccentricity e about its m ajor axis is equal to 2 × area of the ellipse × [ √(1 − e2) + (1 ⁄ e) sin− 1 e] . Solution : [ Note : Prolate spheroid is generated by the revolution of an ellipse about its major axis] x2 y2 Let the equation of the ellipse be 2 + 2 = 1, a b the x-axis being the major axis so that a > b. The parametric equations of (1) are x = a cos t , y = b sin t . ∴ dx ⁄ dt = − a sin t and dy ⁄ dt = b cos t . We have ds = dt √ ...(1) ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎫ ⎨ ⎜ ⎟ + ⎜ ⎟ ⎬ = √(a 2 sin2 t + b2 cos2 t ) ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ = √{a2 sin2 t + a2 (1 − e2) cos2 t }, [ ... for the ellipse b2 = a2 (1 − e2)] ...(2) = a √(1 − e2 cos2 t ) . Now the ellipse (1) is symmetrical about y-axis and for the arc of the ellipse lying in the first quadrant t varies from 0 to π ⁄ 2 . At the point (a, 0) we have t = 0 and at the point (0, b) we have t = π ⁄ 2 . H ence the required surface S formed by the revolution of the ellipse (1) about the x-axis ⌠ = 2 ⎮ 2πy ds , between the suitable limits ⌡ π ⁄2 ⌠ = 4π ⎮ ⌡0 y π ⁄2 ds ⌠ dt = 4π ⎮ ⌡0 dt b sin t . a √(1 − e2 cos2 t ) dt , [ ... y = b sin t and ds ⁄ dt = a √(1 − e2 cos2 t ), from (2)] I-186 INTEGRAL CALCULUS π ⁄2 ⌠ = 4π ab ⎮ ⌡0 sin t √(1 − e2 cos2 t ) dt . Pu t e cos t = z so t h at − e sin t dt = dz . W h e n t = 0, z = e an d wh en t = 1 2 π, z = 0. 0 e 4πab ⌠ ⌠ 1 ⎮ √(1 − z2) dz √(1 − z2) dz = S = − 4π ab ⎮ ⌡e e e ⌡0 ∴ 4πab ⎡ e 4πab ⎡ z 1 1 ⎤e ⎤ sin− 1 z⎥ = sin− 1 e⎥ ⎢ √(1 − z2) + ⎢ √(1 − e2) + 2 e ⎣2 2 e ⎣2 ⎦0 ⎦ = = 2πab [ √(1 − e2) + (1 ⁄ e) sin− 1 e] = 2 × area of the ellipse × [ √(1 − e2) + (1 ⁄ e) sin− 1 e] . Remark : The solid of revolution formed by revolving an ellipse about its minor axis is called an oblate spheroid. Example 4 : The part of the parabola y2 = 4ax cut off by the latus rectum revolves about the tangent at the vertex. Find the curved surface of the reel thus generated. (Bundelkhand 2011) y2 Solution : The given parabola is = 4ax . Differentiating (1) w.r.t. x , we get dy ⁄ dx = 2a ⁄ y . 2⎫ ⎧ ds ⎧ ⎛ dy⎞ 2 ⎫ ⎪ 1 + 4a ⎪ ∴ = ⎨1 + ⎜ ⎟ ⎬ = ⎨ ⎬ 2 ⎪ ⎩ ⎝ dx⎠ ⎭ dx y ⎪⎭ ⎩ ⎧⎪ 4a2 ⎫⎪ ⎛ x + a⎞ = ⎜ ⎟ ⋅ ⎨⎪ 1 + ⎬⎪ = 4ax ⎝ x ⎠ ⎩ ⎭ The required curved surface is generated by the revolution of the arc L OL′ (L SL′ is the latus rectum), about the tangent at the vertex i.e., y-axis. The curve is symmetrical about x-axis and for the arc OL, x varies from 0 to a . ∴ the required surface √ √ ...(1) √ √ a ds ⌠ 2π x dx = 2⎮ ⌡x = 0 dx a ⌠ ⎛ x + a⎞ = 4π ⎮ x ⎜ ⎟ dx ⌡0 ⎝ x ⎠ a √ ⌠ = 4π ⎮ √(x2 + ax) dx ⌡0 a ⌠ = 4π ⎮ ⌡0 √ ⎧⎛ a⎞ 2 ⎛ a⎞ ⎨⎜ x + ⎟ − ⎜ ⎟ ⎩⎝ ⎝ 2⎠ 2⎠ 2 ⎫ ⎬ dx (Note) ⎭ a ⎡1 ⎛ ⎧⎛ ⎫⎤ a⎞ 1 a2 a⎞ = 4π ⎢ ⎜ x + ⎟ √(x2 + ax) − ⋅ log ⎨ ⎜ x + ⎟ + √(x2 + ax) ⎬ ⎥ ⎣2 ⎝ ⎩⎝ ⎭⎦ 0 2⎠ 2 4 2⎠ ⎤ ⎡ .. ⌠ 1 1 ⎢ . ⎮ √(x2 − a2) dx = x √(x2 − a2) − a2 log {x + √(x2 − a2)}⎥ 2 2 ⌡ ⎦ ⎣ I-187 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION 1 3 1 3 1 a a √2 − a2 log { a + a √2} + a2 2 2 8 2 8 3 2 1 2 3 1 4π [ a √2 − a log {( a + a √2) ⁄ ( a)}] = 4 8 2 2 1 πa2 [3 √2 − log (√2 + 1) 2] 2 πa2 [3 √2 − log (√2 + 1)] . 1 2 = 4π [ ⋅ log ( a)] = πa2 [3 √2 − = = 1 2 log (3 + 2 √2)] (Note) 1. Find the surface of a sphere of radius a . 2. 3. (Kanpur 2006) Show that the surface of the spherical zone contained between two parallel planes is 2πah where a is the radius of the sphere and h the distance between the planes. (Kanpur 2009) parabola y2 = 4ax Find the area of the surface formed by the revolution of the about the x-axis by the arc from the vertex to one end of the latus rectum. (Rohilkhand 2011) 4. 5. 6. 7. F ind t h e sur face gene rat e d by t he re volu t ion of an arc of t he cat enary y = c cosh (x ⁄ c) about the axis. of x , between the planes x = a and x = b. For a catenary y = a cosh (x ⁄ a) , prove that aS = 2V = πa (ax + sy), where s is the length of the arc from the vertex , S and V are respectively the area of the curved surface and volume of the solid generated by the revolution of the arc about x-axis. F in d t h e su rface o f t h e so lid gen erat ed by t h e revo lu t ion o f th e ellip se x2 + 4y2 = 16 about its major axis. (Meerut 2005, 06) Find the surface of the solid formed by the revolution, about the axis of y, of the part of the curve ay2 = x3 from x = 0 to x = 4a which is above the x-axis. 1. 4πa2. 4. πc ⎢⎣ (b − a) + 7. 128 πa2 [125 √(10) + 1]. 1215 3. ⎡ 8.7 c 2b c 2 a ⎤⎥ sinh − sinh ⋅ 2 2 c c ⎦ 8 3 πa2 [2 √2 − 1]. ⎡ 4π ⎤⎥ 6. 8π ⎢⎣ 1 + ⋅ 3 √3 ⎦ Surface Formula for Parametric Equations S u p p o s e t h e e q u a t i o n o f t h e c u r ve i s gi ve n i n p a r a m e t r i c fo r m x = f (t ), y = φ (t ), t being the variable parameter. Then the curved surface of the solid formed by the revolution about the x-axis ⌠ ds = ⎮ 2π y dt , between the suitable limits ⌡ dt where ds = dt √ ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎫ ⎨⎜ ⎟ + ⎜ ⎟ ⎬ ⋅ ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ I-188 x2 ⁄ 3 INTEGRAL CALCULUS Example 1 : Find the surface of the solid generated by revolution of the astroid + y2 ⁄ 3 = a2 ⁄ 3 or x = a cos3 t , y = a sin3 t about the x-axis. (Meerut 2006, 09; Kanpur 05; Rohilkhand 07, 09, 11B; Avadh 11; Kashi 12) Solution : The parametric equations of the the curve are x = a cos3 t , y = a sin3 t . dx = − 3a cos2 t sin t ∴ dt dy and = 3a sin2 t cos t . dt H ence ds = dt √ ⎡ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎤ ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎣ ⎝ dt ⎠ ⎝ dt ⎠ ⎦ = √[9a2 cos4 t sin2 t + 9a2 sin4 t cos2 t ] = √[9a2 sin2 t cos2 t (cos2 t + sin2 t )] = 3a sin t cos t . Also t he given curve (astro id) is symmetrical about both the axes and for the curve in the first quadrant, t varies from 0 to π ⁄ 2 . π ⁄2 ds ⌠ 2π y dt ∴ the required surface = 2 ⎮ ⌡t = 0 dt π ⁄2 ⌠ = 4π ⎮ ⌡0 π ⁄2 ⌠ a sin3 t . 3a sin t cos t dt = 12 π a2 ⎮ ⌡0 sin4 t cos t dt ⎡ sin 5 t ⎤ π ⁄ 2 12πa2 1 ⎥ = 12π a2 [ − 0] = ⋅ 5 ⎣ 5 ⎦0 5 = 12πa2 ⎢ Example 2 : Prove that the surface of the solid generated by the revolution of the tractrix x = a cos t + 1 2 1 2 a log tan2 t , y = a sin t about its asym ptote is equal to the surface of a sphere of radius a . (Gorakhpur 2006; Meerut 09) Solution : The given tractrix is x = a cos t + 1 2 1 2 a log tan2 t , y = a sin t . 1 ∴ sec2 t 1 dx 2 ⎛ ⎞ 1 = − a sin t + a ⋅ = a ⎜ − sin t + ⎟ 1 1 1 ⎟ 2 ⎜ dt 2 sin t cos t ⎟ tan t ⎜ 2 ⎛ ⎝ = a ⎜ − sin t + ⎝ 2 2 ⎠ (− sin2 t + 1) a cos2 t 1 ⎞ = ⎟ = a sin t ⎠ sin t sin t dy = a cos t . dt and H ence ds = dt √ ⎧ ⎛ dx⎞ 2 ⎛ dy⎞ 2 ⎫ ⎨⎜ ⎟ + ⎜ ⎟ ⎬ = ⎩ ⎝ dt ⎠ ⎝ dt ⎠ ⎭ √ ⎧ a 2 cos4 t ⎫ a cos t ⎪ + a2 cos2 t⎪⎬ = ⋅ ⎨ ⎪ sin2 t ⎪ sin t ⎩ ⎭ I-189 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION The given curve is symmetrical about both the axes and the asymptote is the line y = 0 i.e., x-axis. For the arc of the curve lying in the second quadrant t varies from 0 to 1 2 π. π ⁄2 ⌠ ∴ the required surface = 2 . ⎮ ⌡0 π ⁄2 ⌠ = 4π ⎮ ⌡0 a sin t . = 4πa2 ⎡⎢ sin t⎤⎥ ⎣ ⎦ π ⁄2 0 2π y ds dt dt (Note) π ⁄2 a cos t ⌠ dt = 4πa2 ⎮ ⌡0 sin t cos t dt = 4πa2 = the surface of a sphere of radius a . 1. Find the surface area of the solid generated by revolving the cycloid x = a (θ − sin θ), y = a (1 − cos θ) about the x-axis. 2. Find the area of the surface generated by revolving an arch of the cycloid x = a (θ + sin θ), y = a (1 − cos θ) about the tangent at the vertex. 3. The portion between two consecutive cusps of the cycloid x = a (θ + sin θ), y = a (1 + cos θ) is revolved about the x-axis. Prove that the area of the surface so formed is to the area of the cycloid as 64 : 9. 4. Prove that the surface area of the solid generated by the revolution, about the x-axis of the loop of the curve 1 x = t2, y = t − t3 is 3π . 3 5. 1. 8.8 Prove that the surface of the oblate spheroid formed by the revolution of the ellipse of the semi-major axis a and eccentricity e is ⎡ 1 − e2 ⎛ 1 + e⎞ ⎤ . 2πa2 ⎢ 1 + log ⎜ ⎟⎥ ⎣ ⎝ 1 − e⎠ ⎦ 2e 64 2 πa . 3 32 2 πa . 3 2. Surface Formula for Polar Equations Suppose the equation of the curve is given in the polar form r = f (θ) . Then the curved surface generated by the revolution about the initial line, of the arc intercepted between the radii vectors θ = α and θ = β is θ= β ds ds ⎧ 2 ⎛ dr ⎞ 2 ⎫ ⌠ ⎮ 2π (r sin θ) dθ, where = ⎨ r + ⎜ ⎟ ⎬ ⋅ [ ... y = r sin θ] ⌡θ = α dθ dθ ⎩ ⎝ dθ ⎠ ⎭ Note. In some cases we may use the formula ds ds ⎧ ⎛ dθ ⎞ 2 ⎫ ⌠ = S = ⎮ 2π y dr , where ⎨1 + ⎜ r ⎟ ⎬ ⋅ ⌡ ⎩ ⎝ dr ⎠ ⎭ dr dr √ √ I-190 8.9 INTEGRAL CALCULUS Curved Surface Generated by Revolution about any Axis If the given arc A B is revolved about a line CD other than the coordinate axes, then the curved surface thus generated is ⌠ = 2π ⎮ (PM) ds, (between the proper limits of integration) ⌡ where PM is the perpendicular drawn from any point P on the arc A B to the axis of revolution CD and ds is the length of an element of the arc A B at the point P. Example 1 : Find the surface of the solid generated by the revolution of the lemniscate r 2 = a2 cos 2θ about the initial line. (Meerut 2004, 10B, 11; Rohilkhand 08B; Agra 14; Purvanchal 14) or Solution : The given curve is r 2 = a2 cos 2θ. Differentiating (1) w.r.t. θ, we get dr = − 2a2 sin 2θ 2r dθ dr − a2 sin 2θ = ⋅ dθ r ds ⎧ 2 ⎛ dr ⎞ 2 ⎫ ∴ = ⎨r + ⎜ ⎟ ⎬ dθ ⎩ ⎝ dθ⎠ ⎭ √ = √ a cos 2θ + a sinr 2θ 4 ⎧ 2 ⎪ ⎨ ⎪ ⎩ 2 2 ...(1) ⎫ ⎪ ⎬ ⎪ ⎭ 1 √{r 2 . a2 cos 2θ + a4 sin2 2θ} r 1 [ ... r 2 = a2 cos 2θ] = √{a4 cos2 2θ + a4 sin2 2θ} , r = a2 ⁄ r . ...(2) The given curve is symmetrical about the initial line and about the pole. = Putting r = 0 in (1), we get cos 2θ = 0 giving 2θ = ± Therefore one loop of the curve lies between θ = − 1 2 1 4 π i.e., θ = ± π and θ = 1 4 1 4 π. π. There are two loops in the curve and for the upper half of one of these two loops θ varies from 0 to 1 4 π. ∴ the required surface = 2 × the surface generated by the revolution of one loop π ⁄4 ds ⌠ 2π y dθ, where y = r sin θ = 2.⎮ ⌡0 dθ π ⁄4 ⌠ = 4π ⎮ ⌡0 r sin θ ⋅ π ⁄4 ⌠ = 4πa2 ⎮ ⌡0 a2 dθ r sin θ dθ = 4πa2 ⎡⎢ − cos θ⎤⎥ ⎣ ⎦ [From (2)] π ⁄4 0 = 4πa2 [− (1 ⁄ √2) + 1] = 4πa2 [1 − (1 ⁄ √2)] . I-191 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION Example 2 : A circular arc revolves about its chord. Find the area of the surface generated, when 2α is the angle subtended by the arc at the centre. Solution : Let the parametric equations of the circle be x = a cos θ, y = a sin θ, ...(1) θ being the parameter. Take any point P (a cos θ, a sin θ) on the circular arc A BC which is symmetrical about the x-axis and which subtends an angle 2α at the centre O so that ∠ A OB = α . W e h a ve OD = OA cos α = a cos α . Draw PM perpendicular from P to A C , the axis of rotation. Then PM = ON − OD = a cos θ − a cos α . ...(2) For the upper half of the arc to be rotated i.e., for the arc BA, θ varies from 0 to α . Also ds = dθ √ ⎧ ⎛ dx ⎞ 2 ⎛ dy⎞ 2 ⎫ ⎨⎜ ⎟ + ⎜ ⎟ ⎬ ⎩ ⎝ dθ⎠ ⎝ dθ⎠ ⎭ = √{a2 sin2 θ + a2 cos2 θ} = a . ∴ the required surface = 2 × surface generated by the revolution of the arc BA about the chord A C α ds ⌠ dθ = 2 × ⎮ 2π (PM) ⌡0 dθ α ⌠ = 4π ⎮ (a cos θ − a cos α) . a . dθ ⌡0 [From (2)] α = 4πa2 ⎡⎢ sin θ − θ cos α⎤⎥ = 4πa2 [sin α − α cos α] . ⎣ 1. 2. 3. ⎦0 F ind t he area of the surface of revolution formed by revolving the curve r = 2a cos θ about the initial line. F in d t h e su rface o f t h e so lid fo rmed by t h e revo lu t io n o f t h e card io id r = a (1 + cos θ) about the initial line. (Purvanchal 2006, 10; Kashi 11) The arc of the cardioid r = a (1 + cos θ) included between − rotated about the line θ = 1 2 1 2 π≤ θ ≤ 1 2 π is π . Find the area of the surface generated. (Purvanchal 2010) 4. A quadrant of a circle of radius a revolves about its chord. Show that the surface of the spindle generated is 2πa2 √2 (1 − 5. 1 4 π) . The lemniscate r2 = a2 cos 2θ revolves about a tangent at the pole. Show that the surface of the solid generated is 4πa2. (Meerut 1993, 2005B) I-192 1. INTEGRAL CALCULUS 4πa2. 8.10 2. 32 2 πa . 5 3. Theorems of Pappus and Guldin 48 √2 πa2. 5 (Agra 2014) State and prove the theorems of Pappus and Guldin. Theorem 1 : Volume of a Solid of Revolution : If a closed plane curve revolves about a straight line in its plane which does not intersect it, the volum e of the ring thus obtained is equal to the area of the region enclosed by the curve m ultiplied by the length of the path described by the centroid of the region. Proof : Let A P1BP2 A be the closed plane curve and let it rotate about the axis of x . Let A L (x = a) and BN (x = b) be the tangents to the curve parallel to the y-axis (a < b) . Also let any ordinate meet the curve at P1, P2 and let MP1 = y1, MP2 = y2 so that y1, y2 are functions of x . N o w vo lum e o f t h e rin g ge n er at ed by t h e revolution of the closed curve AP1BP2 A about the axis of x = volume generated by the area A L NBP2 A − volume generated by the area A L N BP1 A b b b ⌠ ⌠ ⌠ = π ⎮ y22 dx − π ⎮ y12 dx = π ⎮ ( y22 − y12) dx . ...(1) ⌡a ⌡a ⌡a _ Also if y be the ordinate of the centroid of the area of the closed curve, then b _ y= ⌠ 1 ⎮ ( y + y2) ( y2 − y1) dx ⌡a 2 1 1 2 b ⌠ ⎮ ( y22 − y12) dx ⌡a , = ...(2) A A where A is the area of the closed curve. [See the chapter on centre of gravity] _ _ H ence from (1) and (2), the required volume = 2π A y = A × 2π y _ = area of the closed curve × circumference of the circle of radius y = (area of the curve) × (length of the arc described by the centroid of the region bounded by the closed curve). Theorem 2 : Surface of a solid of revolution : If an arc of a plane curve revolves about a straight line in its plane, which does not intersect it, the surface of the solid thus obtained is equal to the arc m ultiplied by the length of the path described by the centroid of the arc. Proof : Let l be the length of the arc AB and let it revolve about OX . Let the abscissae of the extremities A and B of the arc be a and b. VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION I-193 Then the surface generated by the revolution of the arc A B about x-axis is x= b ⌠ 2π y ds ...(1) = ⎮ ⌡x = a _ Also we know that (see the chapter on centre of gravity) the ordinate y, of the centroid of the arc from x = a to x = b, of length l, is given by b ⌠ ⎮ y ds _ ⌡x = a y= ...(2) l _ _ From (1) and (2), we get the required surface = 2π y l = l × 2π y = (length of the arc) × (length of the path described by the centroid of the arc). Note 1 : The closed curve or arc in the above theorems must not cross the axis of revolution but may be terminated by it. Note 2 : When the volume or surface generated is known, the theorems may be applied to find the position of the centroid of the generating area or arc. Example 1 : Find the volum e and surface-area of the anchor-ring generated by the revolution of a circle of radius a about an axis in its own plane distant b from its centre (b > a) . Solution : H ere the given curve (circle) does not intersect the axis of rotation, so Pappus theorem can be applied. In this case A = area of the region of the closed curve = area of the circle of radius a= πa2 and l = length of the arc of the curve = circumference of the circle = 2πa . As the centroid of the area of a circle and also of _ its circumference lies at the centre, so y = b in both the cases and hence the length of the path described by the C.G. = 2πb . Now by Pappus theorem, the required volume of the anchor-ring = (area of the circle) × (circumference of the circle generated by the centroid) = πa2 . 2 πb = 2 π 2 a2 b. And the surface area of the anchor-ring = (arc length of the circle) × (circumference of the circle generated by the centroid ) = 2πa . 2πb = 4π 2ab. Example 2 : Show that the volum e generated by the revolution of the ellipse x2 ⁄ a2 + y2 ⁄ b2 = 1 about the line x = 2a is 4π 2 a2 b. I-194 INTEGRAL CALCULUS Solution : Area of th e given ellipse is πab. ...(1) T h e C.G . o f t h e ellip se will describe a circle of radius 2a when revolved about the line x = 2a . H ence the length of the arc described by the C.G. = 2π (2a) = 4πa . ∴ by Papp us th eo rem t he required volume = (area of the ellipse)× (length of the arc described by its C.G.) = πab . 4πa = 4π 2 a2 b. Example 3 : The loop of the curve 2ay2 = x (x − a) 2 revolves about the straight line y = a . Find the volume of the solid generated. Solution : The given curve is 2ay2 = x (x − a) 2. The cu rve (1) is symmet rical about the x-axis and the loop lies between x = 0 and x = a . Differentiating (1) w.r.t. x , we get ...(1) 4ay (dy ⁄ dx) = 2x (x − a) + (x − a) 2 = 3x2 − 4ax + a2. Now (dy ⁄ dx) = 0 when 3x2 − 4ax + a2 = 0 or when x = a ⁄ 3 which gives from (1), y = (a √2) ⁄ (3 √3) i.e., < a showing that the loop does not intersect the straight line y = a . By symmetry the C.G . of the loop lies on x-axis i.e., the distance of the C.G . from the axis of revolution ( y = a) is a . When the loop is rotated about y = a, its C.G . will describe a circle of radius a whose perimeter is 2πa . Also the area A of the loop a a (x − a) √x⎤ ⌠ (x − a) √x ⌠ ⎡ .. dx , = 2 ⎮ y dx = 2 ⎮ ⎢ . from (1), y = ⎥ ⌡0 ⌡0 √(2a) √(2a) ⎦ ⎣ √ = √ = a ⎛ 2⎞ ⌠ ⎜ ⎟ ⎮ (x3 ⁄ 2 − ax1 ⁄ 2) dx ⎝ a ⎠ ⌡0 ⎛ 2⎞ ⎜ ⎟ ⎝ a⎠ a ⎡⎢ x5 ⁄ 2 ax3 ⁄ 2 ⎤⎥ 4 ⎢⎢ ⎥⎥ = − √2 a2. 5 ⁄ 2 3 ⁄ 2 15 ⎣ ⎦0 ∴ by Pappus theorem, the required volume 4 8 = 2πa × A = 2πa × √2a2 = √2πa3. 15 15 Example 4 : Prove that the volum e of the solid form ed by the rotation about the line θ = 0 of the area bounded by the curve r = f (θ) and the lines θ = θ1, θ = θ2 is θ 2 ⌠ 2 3 π ⎮ r sin θ dθ. 3 ⌡θ 1 VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION I-195 Solution : L e t OA B b e t h e a r e a bounded by the curve r = f (θ) and the radii vectors θ = θ1 and θ = θ2. We have to find the vo lu me fo rmed by th e revolut io n o f area OA B about the initial line OX . Take an y point (r, θ) in side the area OA B and take a small element of the area r δθ δr at the point P. Drop PM perpendicular from P to the axis of rotation OX . We have PM = OP sin θ = r sin θ. Now the volume of the ring formed by revolving the element of area r δθ δr about OX = 2 πr sin θ . rδθ δr = 2 π r 2 sin θ δθ δr. Therefore the whole volume formed by revolving the area OA B about OX f (θ) θ ⌠ 2 ⌠ = ⎮ ⎮ 2 π r 2 sin θ d θ dr ⌡θ = θ ⌡r = 0 1 f (θ) θ ⎡ 3⎤ ⎢r ⎥ ⌠ 2 2 π sin θ ⎢⎢ ⎥⎥ = ⎮ ⌡θ = θ ⎣ 3 ⎦0 1 = θ dθ θ 2 ⌠ 2 2 ⌠ 2 π⎮ [ f (θ)]3 sin θ d θ = π ⎮ r 3 sin θ d θ, 3 ⌡θ 3 ⌡θ = θ 1 1 where r is to be replaced from the equation of the curve r = f (θ). Note : Proceeding as above we can also show that the volume of the solid formed π by the rot ation of the above ment ioned area about the line θ = is equ al to 2 θ 2 ⌠ 2 3 π ⎮ r cos θ d θ. 3 ⌡θ 1 1. 2. 3. 4. Use Pappus theorem to find : The position of the centroid of a semi-circular area. The volume generated by the revolution of an ellipse having semi-axes a and b about a tangent at the vertex. Find by using Pappus theorem the volume of the ring generated by the revolution of an ellipse of eccentricity 1 ⁄ √2 about a straight line parallel to the minor axis and situated at a distance from the centre equal to three times the major axis. F ind t he volume of t he ring generat ed by the revolut ion of t he cardioid r = a (1 + cos θ) about the line r cos θ + a = 0, given that the centroid of the cardioid is at a distance 5a ⁄ 6 from the origin. I-196 INTEGRAL CALCULUS 5. A semi-circular bend of lead has a mean radius of 8 inches; the initial diameter of 1 2 the pipe is 4 inches and the thickness of the lead is inch. Applying the theorem of Pappus and G uldin find the volume of the lead and its weight, given that 1 cubic inch of lead weighs 0⋅ 4 lb. [Hint. Internal diameter of pipe = 4 inches. Thickness of metal = 1 2 inch ∴ external diameter of the pipe = 4 + 1 = 5 inches. ∴ area of lead = 1 4 π (52 − 42) = 9 4 π. The centroid of this area is at a distance of 8 inches from the axis of rotation. Therefore the length of path traced out by its centroid in describing a semi-circle = 8π inches. ∴ volume of the lead = 8π × 6. 1. 3. 4. 9 4 π = 18π 2 cu. inch. ∴ weight of the pipe = volume × density = 18π 2 × 0⋅ 4 lb. = 71⋅ 1 lb] (Meerut 2008) State the theorems of Pappus and G uldin. 4a ⁄ 3π. 6 √2 π 2 a3, where a is the semi-major axis. 11 2 2 π a . 2 2. 2 π 2 a2 b or 2 π 2 ab 2. Fill in the Blanks: Fill in the blanks “……”, so that the following statem ents are com plete and correct. 1. The volume of the solid generated by the revolution of the area bounded by the curve y = f (x), x-axis and the ordinates x = a, x = b about the x-axis is …… . (Meerut 2003) 2. The volume of the solid generated by the revolution of the area bounded by the curve r = f (θ) and the radii vectors θ = θ1, θ = θ2 about the initial line is …… . (Meerut 2001) 3. If the equation of the curve in the polar form is r = f (θ), then the curved surface generated by the revolution about the initial line of the arc intercepted between ⌠ the radii vectors θ = α and θ = β is ⎮ ⌡ θ= β θ= α 2 π (r sin θ) ds ds d θ, where = …… . dθ dθ VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION 4. I-197 If the equations of the curve in parametric form are x = f (t ), y = φ (t ), t being the variable parameter, then the curved surface of the solid formed by the ds ⌠ d t, between the suitable limits, where revolution about the x-axis is ⎮2 π y ⌡ dt ds = …… . dt Multiple Choice Questions: Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 5. The volume of the solid generated by the revolution of the area bounded by the curve r = f (θ) and the radii vectors θ = θ1, θ = θ2 about any line ( θ = γ) is θ ⌠ 2 2 π r 3 cos(θ − γ) d θ (a) ⎮ ⌡θ 3 1 θ2 ⌠ 2 (b) ⎮ π r 3 sin(θ − γ) d θ ⌡θ 3 1 θ2 ⌠ (c) ⎮ π r 2 sin(θ − γ) d θ ⌡θ 1 θ2 ⌠ (d) ⎮ π r 2 cos(θ − γ) d θ ⌡θ 1 6. The volume of the paraboloid generated by the revolution about the x-axis of the parabola y2 = 4 a x from x = 0 to x = h is (a) 2 π a h 2 (c) 7. 2 3 π a h2 (b) 2 π a h (d) 2 3 πah (Rohilkhand 2005) The curved surface of the solid generated by the revolution about the y-axis of the area bounded by the curve x = f ( y), the lines y = a, y = b and y-axis is b ⌠ (a) ⎮ π x d s ⌡a b ⌠ 2 π x ds (c) ⎮ ⌡a 3 b ⌠ (b) ⎮ 2 π x ds ⌡a b ⌠ (d) ⎮ π 2 x ds ⌡a True or False: 8. Write ‘T’ for true and ‘F’ for false statement. The volume of the solid generated by the revolution of the area bounded by the curve r = f (θ) and the radii vectors θ = θ1, θ = θ2 about the initial line θ = 0 is θ ⌠2 2 ⎮ π r 3 sin θ d θ. ⌡θ 3 1 I-198 9. INTEGRAL CALCULUS If an arc length revolves about x-axis, the basic formula for the surface of ⌠ revolution in all cases is ⎮ 2 π y d s, between the suitable limits. ⌡ b 1. ⌠ ⎮ π y 2 dx . ⌡a ⎞2⎫ ⎛ ⎧ ⎨ r 2 + ⎜ d r⎟ ⎬ ⋅ ⎩ ⎝ d θ⎠ ⎭ θ ⌠ 2 2 2. ⎮ π r 3 sin θ d θ . ⌡θ 3 1 3. √ 5. (b). 6. (a). 7. (b). 8. T. 9. T. 4. √ ⎧⎛ ⎨⎜ ⎩⎝ 2 2 ⎛ dy⎞ ⎫ ⎜ ⎟ dx⎞⎟ ⎜ ⎟ ⎬⋅ + ⎜ ⎟ d t⎠ ⎝ dt ⎠ ⎭