Contents PART II. Foreword

Transcription

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2 Contents PART II Foreword v Preface vii 7. Integrals Introduction Integration as an Inverse Process of Differentiation Methods of Integration Integrals of some Particular Functions Integration by Partial Fractions Integration by Parts 7.7 Definite Integral 7.8 Fundamental Theorem of Calculus Evaluation of Definite Integrals by Substitution Some Properties of Definite Integrals 4 8. Application of Integrals Introduction Area under Simple Curves Area between Two Curves Differential Equations Introduction Basic Concepts General and Particular Solutions of a 8 Differential Equation 9.4 Formation of a Differential Equation whose 85 General Solution is given 9.5 Methods of Solving First order, First Degree 9 Differential Equations 0. Vector Algebra Introduction Some Basic Concepts Types of Vectors Addition of Vectors 49

3 iv 0.5 Multiplication of a Vector by a Scalar Product of Two Vectors 44. Three Dimensional Geometry 46. Introduction 46. Direction Cosines and Direction Ratios of a Line 46. Equation of a Line in Space Angle between Two Lines 47.5 Shortest Distance between Two Lines 47.6 Plane Coplanarity of Two Lines Angle between Two Planes Distance of a Point from a Plane Angle between a Line and a Plane 49. Linear Programming 504. Introduction 504. Linear Programming Problem and its Mathematical Formulation 505. Different Types of Linear Programming Problems 54. Probability 5. Introduction 5. Conditional Probability 5. Multiplication Theorem on Probability Independent Events 54.5 Bayes' Theorem Random Variables and its Probability Distributions Bernoulli Trials and Binomial Distribution 57 Answers 588

4 INTEGRALS 87 Chapter 7 INTEGRALS Just as a mountaineer climbs a mountain because it is there, so a good mathematics student studies new material because it is there. JAMES B. BRISTOL 7. Introduction Differential Calculus is centred on the concept of the derivative. The original motivation for the derivative was the problem of defining tangent lines to the graphs of functions and calculating the slope of such lines. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions. If a function f is differentiable in an interval I, i.e., its derivative f eists at each point of I, then a natural question arises that given f at each point of I, can we determine the function? The functions that could possibly have given function as a derivative are called anti derivatives (or primitive) of the function. Further, the formula that gives G.W. Leibnitz (646-76) all these anti derivatives is called the indefinite integral of the function and such process of finding anti derivatives is called integration. Such type of problems arise in many practical situations. For instance, if we know the instantaneous velocity of an object at any instant, then there arises a natural question, i.e., can we determine the position of the object at any instant? There are several such practical and theoretical situations where the process of integration is involved. The development of integral calculus arises out of the efforts of solving the problems of the following types: (a) the problem of finding a function whenever its derivative is given, (b) the problem of finding the area bounded by the graph of a function under certain conditions. These two problems lead to the two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus.

5 88 MATHEMATICS There is a connection, known as the Fundamental Theorem of Calculus, between indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering. The definite integral is also used to solve many interesting problems from various disciplines like economics, finance and probability. In this Chapter, we shall confine ourselves to the study of indefinite and definite integrals and their elementary properties including some techniques of integration. 7. Integration as an Inverse Process of Differentiation Integration is the inverse process of differentiation. Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti differentiation. Let us consider the following eamples: d We know that (sin ) cos... () d d d ( )... () d and ( e ) e... () d We observe that in (), the function cos is the derived function of sin. We say that sin is an anti derivative (or an integral) of cos. Similarly, in () and (), and e are the anti derivatives (or integrals) of and e, respectively. Again, we note that for any real number C, treated as constant function, its derivative is zero and hence, we can write (), () and () as follows : d d d (sin + C) cos, ( +C) and ( e +C) e d d d Thus, anti derivatives (or integrals) of the above cited functions are not unique. Actually, there eist infinitely many anti derivatives of each of these functions which can be obtained by choosing C arbitrarily from the set of real numbers. For this reason C is customarily referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function. d More generally, if there is a function F such that F( ) f ( ), I (interval), d then for any arbitrary real number C, (also called constant of integration) d d [ F( )+C] f (), I

6 INTEGRALS 89 Thus, {F + C, C R} denotes a family of anti derivatives of f. Remark Functions with same derivatives differ by a constant. To show this, let g and h be two functions having the same derivatives on an interval I. Consider the function f g h defined by f () g() h(), I df Then d f g h giving f () g () h () I or f () 0, I by hypothesis, i.e., the rate of change of f with respect to is zero on I and hence f is constant. In view of the above remark, it is justified to infer that the family {F + C, C R} provides all possible anti derivatives of f. We introduce a new symbol, namely, f() d which will represent the entire class of anti derivatives read as the indefinite integral of f with respect to. Symbolically, we write f() df()+c. dy Notation Given that f () d, we write y f () d. For the sake of convenience, we mention below the following symbols/terms/phrases with their meanings as given in the Table (7.). Symbols/Terms/Phrases f() d f () in f() d in f() d Integrate An integral of f Integration Constant of Integration Table 7. Meaning Integral of f with respect to Integrand Variable of integration Find the integral A function F such that F () f () The process of finding the integral Any real number C, considered as constant function

7 90 MATHEMATICS We already know the formulae for the derivatives of many important functions. From these formulae, we can write down immediately the corresponding formulae (referred to as standard formulae) for the integrals of these functions, as listed below which will be used to find integrals of other functions. Derivatives Integrals (Anti derivatives) (i) n + d n ; d n + Particularly, we note that d d d d ( ) (ii) ( sin ) d d (iii) ( cos) d d (iv) ( tan ) sec d d ; d + C n + n d + C, n n + cos ; cos d sin + C sin ; sin d cos + C ; (v) ( cot) cosec d d (vi) ( ) d d ; sec d tan + C cosec d cot + C sec sec tan ; sec tan d sec + C cosec coseccot ; cosec cot d cosec + C (vii) ( ) d (viii) ( sin ) d d (i) ( cos ) d d () ( tan ) d d (i) ( cot ) d ; + ; ; + ; d d sin + C cos + C d tan C + + d cot C + +

8 INTEGRALS 9 d (ii) ( sec ) d d (iii) ( cosec ) d ; ; d sec + C d cosec + C d (iv) ( e ) e ; ed e + C d d d (v) ( log ) a ; d log + C d a (vi) a ; C d log a ad + log a Note In practice, we normally do not mention the interval over which the various functions are defined. However, in any specific problem one has to keep it in mind. 7.. Geometrical interpretation of indefinite integral Let f (). Then f() d + C. For different values of C, we get different integrals. But these integrals are very similar geometrically. Thus, y + C, where C is arbitrary constant, represents a family of integrals. By assigning different values to C, we get different members of the family. These together constitute the indefinite integral. In this case, each integral represents a parabola with its ais along y-ais. Clearly, for C 0, we obtain y, a parabola with its verte on the origin. The curve y + for C is obtained by shifting the parabola y one unit along y-ais in positive direction. For C, y is obtained by shifting the parabola y one unit along y-ais in the negative direction. Thus, for each positive value of C, each parabola of the family has its verte on the positive side of the y-ais and for negative values of C, each has its verte along the negative side of the y-ais. Some of these have been shown in the Fig 7.. Let us consider the intersection of all these parabolas by a line a. In the Fig 7., we have taken a > 0. The same is true when a < 0. If the line a intersects the parabolas y, y +, y +, y, y at P 0, P, P, P, P etc., respectively, then dy at these points equals a. This indicates that the tangents to the d curves at these points are parallel. Thus, d + C F ( ) (say), implies that C

9 9 MATHEMATICS Fig 7. the tangents to all the curves y F C (), C R, at the points of intersection of the curves by the line a, (a R), are parallel. Further, the following equation (statement) f() d F() + C y(say), represents a family of curves. The different values of C will correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself. This is the geometrical interpretation of indefinite integral. 7.. Some properties of indefinite integral In this sub section, we shall derive some properties of indefinite integrals. (I) The process of differentiation and integration are inverses of each other in the sense of the following results : d f () d d f () and f () d f () + C, where C is any arbitrary constant.

10 INTEGRALS 9 Proof Let F be any anti derivative of f, i.e., d F( ) d f () Then f() d F() + C d Therefore f () d Similarly, we note that d ( F( )+C) d d d F( ) f( ) d d f () f () d (II) and hence f () d f () + C where C is arbitrary constant called constant of integration. Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. Proof Let f and g be two functions such that d f () d d d () d g d d or f () d g () d d 0 C, where C is any real number (Why?) Hence f() d g () d or f() d g () d+ C So the families of curves { f() d+ C,C R} and { g () d+ C,C R} are identical. are equivalent. Hence, in this sense, f() dand g() d

11 94 MATHEMATICS Note The equivalence of the families { f() d+c,c } { g () d+c,c } f d R and R is customarily epressed by writing () g() d without mentioning the parameter. (III) [ ] f()+ g () d f() d+ g () d Proof By Property (I), we have, d [ f ( )+ g( )] d d f () + g ()... () On the otherhand, we find that d d f () d+ () f () d+ g() d d d d d g d f () + g ()... () Thus, in view of Property (II), it follows by () and () that ( () + ()) f g d f() d+ g() d (IV) For any real number k, kfd () k fd (). d Proof By the Property (I), kfd () kf () d. d Also k f() d d d k f() d k f() d Therefore, using the Property (II), we have kfd () k fd () (V) Properties (III) and (IV) can be generalised to a finite number of functions f, f,..., f n and the real numbers, k, k,..., k n giving [ () + () + + n n() ] kf k f... k f d k f() d+ k f () d kn fn() d.. To find an anti derivative of a given function, we search intuitively for a function whose derivative is the given function. The search for the requisite function for finding an anti derivative is known as integration by the method of inspection. We illustrate it through some eamples.

12 INTEGRALS 95 Eample Write an anti derivative for each of the following functions using the method of inspection: (i) cos (ii) + 4 (iii), 0 Solution (i) We look for a function whose derivative is cos. Recall that d d sin cos or cos d d (sin ) d sin d (ii) (iii) Therefore, an anti derivative of cos is sin. We look for a function whose derivative is + 4. Note that 4 ( ) d + d + 4. Therefore, an anti derivative of + 4 is + 4. We know that d d (log ), > 0 and [log ( )] ( ),< 0 d d d d Combining above, we get ( ) log, 0 Therefore, d log is one of the anti derivatives of. Eample Find the following integrals: (i) d (ii) ( + ) d (iii) ( + e ) d Solution (i) We have d d d (by Property V)

13 96 MATHEMATICS C + C + + ; C, C are constants of integration + C C + +C C + +C, where C C C is another constant of integration. Note From now onwards, we shall write only one constant of integration in the final answer. (ii) We have ( + ) d d + d (iii) We have C C ( + e ) d d+ e d d + + e log +C e log +C 5 Eample Find the following integrals: (i) (sin + cos ) d (ii) cosec (cosec + cot ) d (iii) sin d cos Solution (i) We have (sin + cos ) d sin d+ cos d cos+ sin+ C

14 INTEGRALS 97 (ii) (iii) We have + (cosec (cosec + cot ) d cosec d cosec cot d We have sin sin d d d cos cos cos cot cosec+ C sec d tan sec d tan sec + C Eample 4 Find the anti derivative F of f defined by f () 4 6, where F (0) Solution One anti derivative of f () is 4 6 since d 4 ( 6 ) 4 6 d Therefore, the anti derivative F is given by F() C, where C is constant. Given that F(0), which gives, C or C Hence, the required anti derivative is the unique function F defined by F() Remarks (i) (ii) We see that if F is an anti derivative of f, then so is F + C, where C is any constant. Thus, if we know one anti derivative F of a function f, we can write down an infinite number of anti derivatives of f by adding any constant to F epressed by F() + C, C R. In applications, it is often necessary to satisfy an additional condition which then determines a specific value of C giving unique anti derivative of the given function. Sometimes, F is not epressible in terms of elementary functions viz., polynomial, logarithmic, eponential, trigonometric functions and their inverses etc. We are therefore blocked for finding f() d. For eample, it is not possible to find e d by inspection since we can not find a function whose derivative is e

15 98 MATHEMATICS (iii) When the variable of integration is denoted by a variable other than, the integral formulae are modified accordingly. For instance y 5 y dy + C y + C Comparison between differentiation and integration. Both are operations on functions.. Both satisfy the property of linearity, i.e., (i) d [ k f() + k f () ] k d f () + k d f () d d d (ii) [ + ] + k f () k f () d k f () d k f () d Here k and k are constants.. We have already seen that all functions are not differentiable. Similarly, all functions are not integrable. We will learn more about nondifferentiable functions and nonintegrable functions in higher classes. 4. The derivative of a function, when it eists, is a unique function. The integral of a function is not so. However, they are unique upto an additive constant, i.e., any two integrals of a function differ by a constant. 5. When a polynomial function P is differentiated, the result is a polynomial whose degree is less than the degree of P. When a polynomial function P is integrated, the result is a polynomial whose degree is more than that of P. 6. We can speak of the derivative at a point. We never speak of the integral at a point, we speak of the integral of a function over an interval on which the integral is defined as will be seen in Section The derivative of a function has a geometrical meaning, namely, the slope of the tangent to the corresponding curve at a point. Similarly, the indefinite integral of a function represents geometrically, a family of curves placed parallel to each other having parallel tangents at the points of intersection of the curves of the family with the lines orthogonal (perpendicular) to the ais representing the variable of integration. 8. The derivative is used for finding some physical quantities like the velocity of a moving particle, when the distance traversed at any time t is known. Similarly, the integral is used in calculating the distance traversed when the velocity at time t is known. 9. Differentiation is a process involving limits. So is integration, as will be seen in Section 7.7.

16 INTEGRALS The process of differentiation and integration are inverses of each other as discussed in Section 7.. (i). EXERCISE 7. Find an anti derivative (or integral) of the following functions by the method of inspection.. sin. cos. e 4. (a + b) 5. sin 4 e Find the following integrals in Eercises 6 to 0: 6. (4 e + ) d 7. ( ) d 8. ( a + b + c) d 9. ( + e ) 0. d d d. + d 4. ( ) d ( + + ) d 6. ( cos + ) ( sin + 5 ) d 8. sec (sec + tan ) sec d sin 0. d. 9. cosec cos Choose the correct answer in Eercises and.. The anti derivative of + equals (A) C + + C (C) C + + C. d If f ( ) 4 4 d such that f () 0. Then f () is (A) (C) e d d

17 00 MATHEMATICS 7. Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions. It was based on inspection, i.e., on the search of a function F whose derivative is f which led us to the integral of f. However, this method, which depends on inspection, is not very suitable for many functions. Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms. Prominent among them are methods based on:. Integration by Substitution. Integration using Partial Fractions. Integration by Parts 7.. Integration by substitution In this section, we consider the method of integration by substitution. The given integral f() d can be transformed into another form by changing the independent variable to t by substituting g (t). Consider I f() d Put g(t) so that d g (t). dt We write d g (t) dt Thus I f( ) d f( g( t)) g ( t) dt This change of variable formula is one of the important tools available to us in the name of integration by substitution. It is often important to guess what will be the useful substitution. Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following eamples. Eample 5 Integrate the following functions w.r.t. : (i) sin m (ii) sin ( + ) (iii) tan 4 sec (iv) sin (tan ) + Solution (i) We know that derivative of m is m. Thus, we make the substitution m t so that md dt. Therefore, sin m d sin t dt m m cos t + C cos m + C m

18 (ii) INTEGRALS 0 Derivative of + is. Thus, we use the substitution + t so that d dt. Therefore, sin( + ) d sin t dt (iii) Derivative of is cos t + C cos ( + ) + C. Thus, we use the substitution t so that d dt giving d t dt. 4 4 tan sec ttan tsec t dt Thus, d t Again, we make another substitution tan t u so that Therefore, 4 4 t t dt u du tan sec 5 4 tan t sec t dt u + C 5 sec t dt du tan 5 C 5 t + (since u tan t) (iv) Hence, 4 tan sec d 5 tan 5 + C (since t ) 5 tan 5 + C Alternatively, make the substitution tan t Derivative of tan. Thus, we use the substitution + d tan t so that dt. + sin (tan ) Therefore, sin d t dt + cos t + C cos(tan ) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique. These will be used later without reference. (i) tan d log sec + C We have tan d sin d cos

19 0 MATHEMATICS Put cos t so that sin d dt Then tan d dt log t + C log cos + C t or tan d log sec + C (ii) cot log sin + C cos We have cot d d sin Put sin t so that cos d dt Then dt cot d log t + C log sin + C t (iii) sec log sec + tan + C We have sec (sec + tan ) sec d d sec + tan Put sec + tan t so that sec (tan + sec ) d dt Therefore, sec d dt log t t + C log sec + tan + C (iv) cosec log cosec cot + C We have cosec (cosec + cot ) cosec d d (cosec + cot ) Put cosec + cot t so that cosec (cosec + cot ) d dt So cosec d dt log t log cosec + cot + C t cosec cot log + C cosec cot log cosec cot + C Eample 6 Find the following integrals: (i) sin sin cos d (ii) d (iii) sin ( + a) d + tan

20 INTEGRALS 0 Solution (i) We have sin cos d sin cos (sin ) d ( cos ) cos (sin ) d Put t cos so that dt sin d Therefore, sin cos (sin ) d ( t ) t dt (ii) Put + a t. Then d dt. Therefore sin sin ( ) d t a dt sin ( + a) sin t 5 4 t t ( t t ) dt + C 5 5 cos + cos + C 5 sin t cos a cos tsin a dt sin t cos a dt sin a cot t dt (cos a) t (sin a) log sin t + C (cos a) ( + a ) (sin a) log sin ( + a) + C + + cos a acos a (sin a) log sin ( a ) C sin a (iii) sin Hence, d sin ( + a) cos a sin a log sin ( + a) + C, where, C C sin a + a cos a, is another arbitrary constant. d cos d + tan cos + sin (cos +sin +cos sin ) d cos + sin

21 04 MATHEMATICS cos sin d + d cos + sin C cos sin + + d cos + sin... () Now, consider I cos sin d cos + sin Put cos + sin t so that (cos sin ) d dt Therefore I dt log t + C log cos + sin + C t Putting it in (), we get d C C + + log cos sin + tan + + C C + log cos + sin + + C C + log cos + sin + C, C + EXERCISE 7. Integrate the functions in Eercises to 7:. +. ( log ) 4. sin sin (cos ) 5. sin ( a + b) cos ( a + b). + log 6. a + b (4+ ) ( ) ( + ) e 7. e + 4, > 0 (log ) m, > 0

22 INTEGRALS e tan + 9. e e + 0. e e e + e. tan ( ). sec (7 4). sin 4. cos sin 6cos + 4sin 7. sin cos cos ( tan ) cos + sin cos cot log sin 0. sin + cos. sin ( + cos ). + cot. tan ( ) 4. tan sin cos ( ) 4 5. ( + log ) 6. ( + ) + log sin tan Choose the correct answer in Eercises 8 and log 0 d e equals (A) C (B) C (C) (0 0 ) + C (D) log (0 + 0 ) + C 9. d equals sin cos (A) tan + cot + C (B) tan cot + C (C) tan cot + C (D) tan cot + C 7.. Integration using trigonometric identities When the integrand involves some trigonometric functions, we use some known identities to find the integral as illustrated through the following eample. Eample 7 Find (i) cos d (ii) sin cos d (iii) sin d

23 06 MATHEMATICS Solution (i) Recall the identity cos cos, which gives cos + cos Therefore, cos d ( + cos ) d cos d + d + sin + C 4 (ii) Recall the identity sin cos y [sin ( + y) + sin ( y)] (Why?) Then sin cosd sin 5 d sin d (iii) cos 5 cos C cos 5+ cos + C 0 From the identity sin sin 4 sin, we find that Therefore, Alternatively, sin sin sin 4 sin d sin sin 4 d d 4 cos + cos+ C 4 sin d sin sin d Put cos t so that sin d dt Therefore, sin d ( t ) dt ( cos ) sin d t dt + t dt t + + C cos + cos + C Remark It can be shown using trigonometric identities that both answers are equivalent.

24 INTEGRALS 07 EXERCISE 7. Find the integrals of the functions in Eercises to :. sin ( + 5). sin cos 4. cos cos 4 cos 6 4. sin ( + ) 5. sin cos 6. sin sin sin 7. sin 4 sin 8 8. cos cos 9. + cos + cos 0. sin 4. cos 4.. cos cos α cos cos α tan sin cos 0. cos sin + sin sin + cos sin cos cos ( cos + sin ) sin + cos 5. tan sec 8. cos + sin cos. sin (cos ). cos ( a) cos ( b) Choose the correct answer in Eercises and 4.. sin cos is equal to d sin cos (A) tan + cot + C (B) tan + cosec + C (C) tan + cot + C (D) tan + sec + C 4. e ( + ) equals d cos ( e) (A) cot (e ) + C (B) tan (e ) + C (C) tan (e ) + C (D) cot (e ) + C 7.4 Integrals of Some Particular Functions In this section, we mention below some important formulae of integrals and apply them for integrating many other related standard integrals: d a () log +C a a + a

25 08 MATHEMATICS d a + () log +C a a a d () +a tan C + a a (4) (5) a d a d log + a +C sin +C a (6) d + a log + + a +C We now prove the above results: () We have a ( a)( + a) ( + a) ( a) a ( a)( + a) a a + a d d d Therefore, a a a + a log ( ) log ( + ) + C a [ a a] a log C a + a + () In view of () above, we have a ( a+ ) + ( a ) a ( a+ )( a ) a + a a+

26 INTEGRALS 09 d Therefore, d d a a + a a+ [ log a log a ] C a a+ log + C a a Note The technique used in () will be eplained in Section 7.5. () Put a tan θ. Then d a sec θ dθ. Therefore, d + a asec θdθ a tan θ + a dθ θ C tan C a + + a a a (4) Let a secθ. Then d a secθ tan θ d θ. Therefore, d a asecθ tanθdθ a sec θ a secθd θ log secθ +tanθ +C log + + C a a log + a log a + C log (5) Let a sinθ. Then d a cosθ dθ. Therefore, d a (6) Let a tanθ. Then d a sec θ dθ. Therefore, d + a + a + C, where C C log a acosθdθ a a sin θ dθ θ +Csin asec θdθ a tan θ + a C a + secθd θlog (sec + tan ) + C θ θ

27 0 MATHEMATICS log C a a log + + a log a + C log a C, where C C log a Applying these standard formulae, we now obtain some more formulae which are useful from applications point of view and can be applied directly to evaluate other integrals. d (7) To find the integral, we write a + b + c a + b + c b c b c b a + + a a a + + a a 4a Now, put b + t so that d dt and writing a c b k a a ±. We find the 4 dt c b integral reduced to the form a depending upon the sign of t ± k a 4a and hence can be evaluated. d (8) To find the integral of the type, proceeding as in (7), we a + b + c obtain the integral using the standard formulae. p + q (9) To find the integral of the type d, where p, q, a, b, c are a + b + c constants, we are to find real numbers A, B such that d p + q A ( a + b + c)+ba( a + b)+b d To determine A and B, we equate from both sides the coefficients of and the constant terms. A and B are thus obtained and hence the integral is reduced to one of the known forms.

28 INTEGRALS ( p+ q) d (0) For the evaluation of the integral of the type a + b + c as in (9) and transform the integral into known standard forms. Let us illustrate the above methods by some eamples. Eample 8 Find the following integrals: d d (i) (ii) 6 Solution, we proceed d d (i) We have log C [by 7.4 ()] (ii) d d ( ) Put t. Then d dt. Therefore, d dt t sin ( t ) + C [by 7.4 (5)] sin ( ) + C Eample 9 Find the following integrals : d d d (i) (ii) 6+ (iii) Solution (i) We have ( ) + 4 So, Let Therefore, d ( ) 6+ + d t. Then d dt d 6+ t + tan + C dt tan t + C [by 7.4 ()]

29 MATHEMATICS (ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand, (completing the square) Thus d d Put + t. Then d dt. 6 Therefore, d dt t 6 7 t log t C [by 7.4 (i)] 7 + log C 6 4 log C log + C + log log C, where C C + log 7

30 INTEGRALS (iii) We have d 5 d 5 5 Put t. Then d dt. 5 Therefore, d 5 d (completing the square) dt 5 t 5 t t log + + C 5 5 log + + C [by 7.4 (4)] Eample 0 Find the following integrals: + + (i) d (ii) d Solution (i) Using the formula 7.4 (9), we epress d + ( ) A B A(4 + 6) + B d Equating the coefficients of and the constant terms from both sides, we get 4A and 6A + B or A 4 and B. Therefore, d d I + I (say)... () 4

31 4 MATHEMATICS In I, put t, so that (4 + 6) d dt dt Therefore, I log t + C t log C... () d and I d + + d Put + t, so that d dt, we get dt I t + tan C t + [by 7.4 ()] tan + + C Using () and () in (), we get ( ) tan + + C... () + d log tan ( + ) + C (ii) C C where, C + 4 This integral is of the form given in 7.4 (0). Let us epress d + A (5 4 )+B A ( 4 ) + B d Equating the coefficients of and the constant terms from both sides, we get A and 4 A + B, i.e., A and B

32 INTEGRALS 5 Therefore, + d 5 4 ( 4 ) d d I + I... () In I, put 5 4 t, so that ( 4 ) d dt. ( ) Therefore, I 4 d dt 5 4 t t + C C... () d d Now consider I ( + ) Put + t, so that d dt. dt t Therefore, I sin + C [by 7.4 (5)] t sin + + C... () Substituting () and () in (), we obtain + + C 5 4 +sin + C, where C C 5 4. EXERCISE 7.4 Integrate the functions in Eercises to ( ) a 9. sec tan + 4

33 6 MATHEMATICS ( )( ) ( a)( b) ( 5)( 4) Choose the correct answer in Eercises 4 and d equals + + (A) tan ( + ) + C (B) tan ( + ) + C (C) ( + ) tan + C (D) tan + C 5. d equals 9 4 (A) 9 8 sin + C 9 8 (B) 8 9 sin + C 9 (C) 9 8 sin + C Integration by Partial Fractions (D) 9 8 sin + C 9 Recall that a rational function is defined as the ratio of two polynomials in the form P( ), where P () and Q() are polynomials in and Q() 0. If the degree of P() Q( ) is less than the degree of Q(), then the rational function is called proper, otherwise, it is called improper. The improper rational functions can be reduced to the proper rational

34 INTEGRALS 7 functions by long division process. Thus, if P( ) Q( ) is improper, then P( ) P( ) T( ) +, Q( ) Q( ) P( ) where T() is a polynomial in and is a proper rational function. As we know Q( ) how to integrate polynomials, the integration of any rational function is reduced to the integration of a proper rational function. The rational functions which we shall consider here for integration purposes will be those whose denominators can be factorised into P( ) linear and quadratic factors. Assume that we want to evaluate d, where P( ) Q( ) Q( ) is proper rational function. It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods. The following Table 7. indicates the types of simpler partial fractions that are to be associated with various kind of rational functions. Table 7. S.No. Form of the rational function Form of the partial fraction. p+ q, a b ( a)( b) A B + a b p+ q. ( a) A B + a a ( ). p + q + r A + B + C ( a)( b)( c) a b c 4. p + q+ r ( a) ( b) A + B + C a ( a) b 5. p + q + r A B + +C ( a)( + b+ c) a + b + c, where + b + c cannot be factorised further In the above table, A, B and C are real numbers to be determined suitably.

35 8 MATHEMATICS Eample Find d ( + )( + ) Solution The integrand is a proper rational function. Therefore, by using the form of partial fraction [Table 7. (i)], we write A + B ( + )( + ) + + where, real numbers A and B are to be determined suitably. This gives A ( + ) + B ( + ). Equating the coefficients of and the constant term, we get A + B 0 and A + B Solving these equations, we get A and B. Thus, the integrand is given by + ( + )( + ) () Therefore, d ( + )( + ) d d + + log + log + + C log + + C + Remark The equation () above is an identity, i.e. a statement true for all (permissible) values of. Some authors use the symbol to indicate that the statement is an identity and use the symbol to indicate that the statement is an equation, i.e., to indicate that the statement is true only for certain values of. Eample Find + d Solution Here the integrand by and find that is not proper rational function, so we divide

36 INTEGRALS ( )( ) 5 5 Let A + B ( )( ) So that 5 5 A ( ) + B ( ) Equating the coefficients of and constant terms on both sides, we get A + B 5 and A + B 5. Solving these equations, we get A 5 and B Thus, d Therefore, d d 5 d log + 0 log + C. Eample Find d ( + ) ( + ) Solution The integrand is of the type as given in Table 7. (4). We write ( + ) ( + ) A + B + C + ( + ) + So that A ( + ) ( + ) + B ( + ) + C ( + ) A ( ) + B ( + ) + C ( + + ) Comparing coefficient of, and constant term on both sides, we get A + C 0, 4A + B + C and A + B + C. Solving these equations, we get 5 A, B and C. Thus the integrand is given by ( + ) ( + ) 4( + ) ( + ) 4( + ) d 5 d d Therefore, ( + ) ( + ) 4 + ( + ) log + + log + + C 4 ( +) log + + C 4 + ( +)

37 0 MATHEMATICS Eample 4 Find d ( + )( + 4) Solution Consider Then ( + )( + 4) ( + )( + 4) and put y. y ( y+ )( y+ 4) Write y A + B ( y+ )( y+ 4) y+ y+ 4 So that y A (y + 4) + B (y + ) Comparing coefficients of y and constant terms on both sides, we get A + B and 4A + B 0, which give 4 A and B Thus, ( + )( + 4) 4 + ( + ) ( + 4) Therefore, d d 4 + ( + )( + 4) d tan + tan + C tan + tan + C In the above eample, the substitution was made only for the partial fraction part and not for the integration part. Now, we consider an eample, where the integration involves a combination of the substitution method and the partial fraction method. Eample 5 Find Solution Let y sinφ Then ( sin ) φ cos φ d φ 5 cos φ 4sinφ dy cosφ dφ

38 INTEGRALS Therefore, ( sin ) φ cos φ φ 5 cos φ 4sinφ ( y ) dy d 5 ( y ) 4y y dy y 4y+ 4 y ( ) y I(say) Now, we write Therefore, y A B + y ( y ) y A (y ) + B ( y ) [by Table 7. ()] Comparing the coefficients of y and constant term, we get A and B A, which gives A and B 4. Therefore, the required integral is given by 4 dy dy I [ + ] dy +4 y ( y ) y ( y ) log y C y 4 log sinφ + + C sinφ 4 log( sin φ ) + +C (since, sinφ is always positive) sin φ Eample 6 Find + + d ( + )( + ) Solution The integrand is a proper rational function. Decompose the rational function into partial fraction [Table.(5)]. Write + + ( + )( + ) A B +C + + ( + ) Therefore, + + A ( + ) + (B + C) ( + )

39 MATHEMATICS Equating the coefficients of, and of constant term of both sides, we get A + B, B + C and A + C. Solving these equations, we get A,B andc Thus, the integrand is given by + + ( + )( + ) ( + ) ( + ) 5 + Therefore, + + d + d + ( +)( + ) d d log + + log + + tan + C EXERCISE 7.5 Integrate the rational functions in Eercises to.. ( + )( + ). 9. ( )( )( ) 4. ( )( )( ) ( ) 7. ( + )( ) 0. ( )(+ ). ( ) ( + ) 8. ( ) ( + ). 5 ( + )( 4) 4. ( + ) n ( + ) [Hint: multiply numerator and denominator by n and put n t ] 7. cos ( sin ) ( sin ) [Hint : Put sin t]

40 INTEGRALS ( + )( + ) ( + )( + 4) ( + )( + ) ( ). [Hint : Put e ( e ) t] Choose the correct answer in each of the Eercises and. d. equals ( )( ). (A) (C) ( ) log + C log d equals ( + ) (A) + C log log ( +) + C (B) ( ) log + C (D) log ( ) ( ) + C (B) log + log ( +) + C log log ( + +) + C (C) log + log ( +) + C (D) 7.6 Integration by Parts In this section, we describe one more method of integration, that is found quite useful in integrating products of functions. If u and v are any two differentiable functions of a single variable (say). Then, by the product rule of differentiation, we have d ( uv ) d Integrating both sides, we get or Let dv du u + v d d dv du uv u d+ v d d d dv du u d uv v d d... () d u f () and dv g(). Then d du d f () and v g () d

41 4 MATHEMATICS Therefore, epression () can be rewritten as f() g() d f () g () d [ g () d ] f () d i.e., f() g () d f () g () d [ f () g () d ] d If we take f as the first function and g as the second function, then this formula may be stated as follows: The integral of the product of two functions (first function) (integral of the second function) Integral of [(differential coefficient of the first function) (integral of the second function)] Eample 7 Find cos d Solution Put f () (first function) and g () cos (second function). Then, integration by parts gives Suppose, we take cos d cos d [ d ( ) cos d] d d sin sin d sin + cos + C f () cos and g(). Then cos d cos d [ d (cos ) d] d d ( cos ) + sin d Thus, it shows that the integral cos d is reduced to the comparatively more complicated integral having more power of. Therefore, the proper choice of the first function and the second function is significant. Remarks (i) It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for sin d. The reason is that there does not eist any function whose derivative is sin. (ii) Observe that while finding the integral of the second function, we did not add any constant of integration. If we write the integral of the second function cos

42 INTEGRALS 5 as sin + k, where k is any constant, then cos d (sin + k ) (sin + k ) d (sin + k) (sin d k d (sin + k) cos k+ C sin + cos + C This shows that adding a constant to the integral of the second function is superfluous so far as the final result is concerned while applying the method of integration by parts. (iii) Usually, if any function is a power of or a polynomial in, then we take it as the first function. However, in cases where other function is inverse trigonometric function or logarithmic function, then we take them as first function. Eample 8 Find log d Solution To start with, we are unable to guess a function whose derivative is log. We take log as the first function and the constant function as the second function. Then, the integral of the second function is. d Hence, (log.) d log d [ (log ) d] d d (log ) d log + C. Eample 9 Find ed Solution Take first function as and second function as e. The integral of the second function is e. Therefore, Eample 0 Find sin d ed e e d e e + C. Solution Let first function be sin and second function be. First we find the integral of the second function, i.e., Put t. Then dt d d.

43 6 MATHEMATICS Therefore, Hence, d dt t t sin d ( ) (sin ) ( ) sin + + C sin + C Alternatively, this integral can also be worked out by making substitution sin θ and then integrating by parts. Eample Find e sin d Solution Take e as the first function and sin as second function. Then, integrating by parts, we have I e sin d e ( cos ) + e cos d e cos + I (say)... () Taking e and cos as the first and second functions, respectively, in I, we get I e sin e sin d Substituting the value of I in (), we get I e cos + e sin I or I e (sin cos ) e Hence, I e sin d (sin cos ) + C Alternatively, above integral can also be determined by taking sin as the first function and e the second function Integral of the type e [ f( )+ f ( )] d We have I e [ f ( )+ f ( )] d e f() d+ e f () d I + e f ( ) d,wherei e f( ) d... () Taking f () and e as the first function and second function, respectively, in I and integrating it by parts, we have I f () e f () e d+ C Substituting I in (), we get I e f() f () e d+ e f () d+ C e f () + C

44 INTEGRALS 7 Thus, e [ f( ) + f ( )] d e f( ) + C Eample Find (i) Solution (i) We have I ( +) e e (tan + ) d (ii) + d ( +) e (tan + ) d + Consider f () tan, then f () + Thus, the given integrand is of the form e [ f () + f ()]. Therefore, I e (tan + ) d e tan + C + (ii) We have ( +) e ++) I d [ ] ( +) e d ( +) e [ ] d e + ( +) ( +) [ + ] d + (+) Consider f(), then f () + ( + ) Thus, the given integrand is of the form e [f () + f ()]. Therefore, + e d e + C ( + ) + EXERCISE 7.6 Integrate the functions in Eercises to.. sin. sin. e 4. log 5. log 6. log 7. sin 8. tan 9. cos 0. (sin ). cos. tan 4. (log ) 5. ( + ) log. sec

45 8 MATHEMATICS e 6. e (sin + cos) 7. ( + ) 8. e + sin + cos 9. e ( ) e 0. ( ). e sin. sin + Choose the correct answer in Eercises and 4.. e d equals (A) (C) e + C (B) e + C (D) 4. e sec ( + tan ) d equals (A) e cos + C (C) e sin + C 7.6. Integrals of some more types e + C e + C (B) e sec + C (D) e tan + C Here, we discuss some special types of standard integrals based on the technique of integration by parts : (i) a d (ii) + a d (i) Let I a d (iii) a d Taking constant function as the second function and integrating by parts, we have I a d a a d a a + a a d a

46 INTEGRALS 9 or I a a d a a I a a a d d a a d a a or I a d a log + a +C Similarly, integrating other two integrals by parts, taking constant function as the second function, we get (ii) a + a d + a + log + + a +C a (iii) a d a + sin +C a Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric substitution a secθ in (i), a tanθ in (ii) and a sinθ in (iii) respectively. Eample Find Solution Note that d d ( + ) + 4 d Put + y, so that d dy. Then d y + dy 4 y y + 4+ log y+ y C [using 7.6. (ii)] Eample 4 Find Solution Note that ( + ) log C d d 4 ( + ) d

47 0 MATHEMATICS Put + y so that d dy. Thus d 4 y dy 4 y y 4 y + sin + C [using 7.6. (iii)] + ( + ) + sin + C EXERCISE 7.7 Integrate the functions in Eercises to Choose the correct answer in Eercises 0 to d is equal to (A) ( ) (B) (D) + + log C ( + ) + C (C) + + log C d is equal to + 9 ( ) C + + (A) (B) (C) (D) ( 4) log C ( + 4) log C ( 4) log C 9 ( 4) 8+ 7 log C

48 7.7 Definite Integral INTEGRALS In the previous sections, we have studied about the indefinite integrals and discussed few methods of finding them including integrals of some special functions. In this section, we shall study what is called definite integral of a function. The definite integral b has a unique value. A definite integral is denoted by f () d, where a is called the a lower limit of the integral and b is called the upper limit of the integral. The definite integral is introduced either as the limit of a sum or if it has an anti derivative F in the interval [a, b], then its value is the difference between the values of F at the end points, i.e., F(b) F(a). Here, we shall consider these two cases separately as discussed below: 7.7. Definite integral as the limit of a sum Let f be a continuous function defined on close interval [a, b]. Assume that all the values taken by the function are non negative, so the graph of the function is a curve above the -ais. b The definite integral f () d is the area bounded by the curve y f (), the a ordinates a, b and the -ais. To evaluate this area, consider the region PRSQP between this curve, -ais and the ordinates a and b (Fig 7.). Y S Q ( ) y f M C DL X' P A B R O a 0 r- r nb Y' Fig 7. X Divide the interval [a, b] into n equal subintervals denoted by [ 0, ], [, ],..., [ r, r ],..., [ n, n ], where 0 a, a + h, a + h,..., r a + rh and b a n b a + nh or n. We note that as n, h 0. h

49 MATHEMATICS The region PRSQP under consideration is the sum of n subregions, where each subregion is defined on subintervals [ r, r ], r,,,, n. From Fig 7., we have area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle (ABDM)... () Evidently as r r 0, i.e., h 0 all the three areas shown in () become nearly equal to each other. Now we form the following sums. s n h [f( 0 ) + + f ( n - )] n h f( r )... () n and S n hf [ ( ) + f ( ) + + f ( )] h f ( )... () Here, s n and S n denote the sum of areas of all lower rectangles and upper rectangles raised over subintervals [ r, r ] for r,,,, n, respectively. In view of the inequality () for an arbitrary subinterval [ r, r ], we have s n < area of the region PRSQP < S n... (4) As n strips become narrower and narrower, it is assumed that the limiting values of () and () are the same in both cases and the common limiting value is the required area under the curve. Symbolically, we write lim S n lim sn area of the region PRSQP () n n f d... (5) a It follows that this area is also the limiting value of any area which is between that of the rectangles below the curve and that of the rectangles above the curve. For the sake of convenience, we shall take rectangles with height equal to that of the curve at the left hand edge of each subinterval. Thus, we rewrite (5) as b h 0 n r 0 r f ( d ) lim h[ f( a) + f( a+ h) f( a+ ( n ) h] a b or f ( d ) a where h ( b a) lim [ f( a) + f( a+ h) f( a+ ( n ) h]... (6) n n b a 0 as n n The above epression (6) is known as the definition of definite integral as the limit of sum. Remark The value of the definite integral of a function over any particular interval depends on the function and the interval, but not on the variable of integration that we r b

50 INTEGRALS choose to represent the independent variable. If the independent variable is denoted by b t or u instead of, we simply write the integral as f () t dt b or ( ) a f u du instead of a b f ( ) d. Hence, the variable of integration is called a dummy variable. a + d as the limit of a sum. Eample 5 Find ( ) 0 Solution By definition b f ( ) d ( b a) lim [ f( a) + f( a+ h) f( a+ ( n ) h], a n n where, h b a n In this eample, a 0, b, f () +, Therefore, + d 0 ( ) 0 h n n 4 ( n ) lim [ f(0) + f( ) + f( ) f( )] n n n n n 4 ( n ) lim [ + ( + ) + ( + ) ] n + n n n n lim [( ) + ( ( n ) ] n n n nterms - lim [ n+ ( ( n ) ] n n n 4 ( n ) n( n ) lim [ n + ] n n n 6 ( n ) ( n ) lim [ n + ] n n n lim [ + ( ) ( )] n n n [ ]

51 4 MATHEMATICS Eample 6 Evaluate Solution By definition 0 e d as the limit of a sum. e d 0 0 ( 0) lim e + e + e e n n 4 n n n n Using the sum to n terms of a G.P., where a, r e n, we have e d 0 n n e lim [ ] n n e n lim e n n e n ( e ) e n lim n n ( e ) e [using lim ] h 0 h h EXERCISE 7.8 Evaluate the following definite integrals as limit of sums.. 4. b d. a 4 d 5. ( ) 5 ( + ) d. 0 e d Fundamental Theorem of Calculus 7.8. Area function b We have defined f ( ) d as the area of a the region bounded by the curve y f (), the ordinates a and b and -ais. Let d 4 ( + e ) d 0 be a given point in [a, b]. Then f ( ) d a represents the area of the shaded region Fig 7.

52 INTEGRALS 5 in Fig 7. [Here it is assumed that f() > 0 for [a, b], the assertion made below is equally true for other functions as well]. The area of this shaded region depends upon the value of. In other words, the area of this shaded region is a function of. We denote this function of by A(). We call the function A() as Area function and is given by a A() f( ) d... () Based on this definition, the two basic fundamental theorems have been given. However, we only state them as their proofs are beyond the scope of this tet book First fundamental theorem of integral calculus Theorem Let f be a continuous function on the closed interval [a, b] and let A () be the area function. Then A () f (), for all [a, b] Second fundamental theorem of integral calculus We state below an important theorem which enables us to evaluate definite integrals by making use of anti derivative. Theorem Let f be continuous function defined on the closed interval [a, b] and F be b a b an anti derivative of f. Then f( ) d [F( )] F (b) F(a). Remarks b (i) In words, the Theorem tells us that f ( ) d (value of the anti derivative F a of f at the upper limit b value of the same anti derivative at the lower limit a). (ii) This theorem is very useful, because it gives us a method of calculating the definite integral more easily, without calculating the limit of a sum. (iii) The crucial operation in evaluating a definite integral is that of finding a function whose derivative is equal to the integrand. This strengthens the relationship between differentiation and integration. b (iv) In f ( ) d a a, the function f needs to be well defined and continuous in [a, b]. For instance, the consideration of definite integral ( ) d is erroneous since the function f epressed by f() ( ) is not defined in a portion < < of the closed interval [, ].

53 6 MATHEMATICS b Steps for calculating f ( ) d. a (i) Find the indefinite integral f( ) d. Let this be F(). There is no need to keep integration constant C because if we consider F() + C instead of F(), we get b b f ( ) d [F( ) + C] a [F( b ) + C] [F( a ) + C] F( b ) F( a ). a Thus, the arbitrary constant disappears in evaluating the value of the definite integral. (ii) Evaluate F(b) F(a) [F ( )] b b a, which is the value of f ( ) d. a We now consider some eamples Eample 7 Evaluate the following integrals: (i) d (ii) 9 4 (0 ) d (iii) d ( + )( + ) (iv) π 4 0 sin t cos t dt Solution (i) (ii) d F( ), Let I d. Since Therefore, by the second fundamental theorem, we get I F() F() Let 9 I d. We first find the anti derivative of the integrand. 4 (0 ) Put 0 t. Then d dt or d dt dt Thus, d t (0 ) t F( ) (0 )

54 INTEGRALS 7 Therefore, by the second fundamental theorem of calculus, we have I F(9) F(4) (0 ) 4 9 (0 7) (iii) Let I d ( + )( + ) Using partial fraction, we get + ( + )( + ) + + d So log + + log + F( ) ( + )( + ) Therefore, by the second fundamental theorem of calculus, we have I F() F() [ log + log 4] [ log + log ] log + log + log 4 log 7 (iv) Let π 4 0 I sin t cost dt. Consider sin t cos t dt Put sin t u so that cos t dt du or cos t dt du So sin t cos t dt udu 4 4 [ u ] sin t F( t)say 8 8 Therefore, by the second fundamental theorem of integral calculus π 4 π 4 I F( ) F(0) [sin sin 0] 4 8 8

55 8 MATHEMATICS EXERCISE 7.9 Evaluate the definite integrals in Eercises to 0.. ( + ) d. d. ( ) d 4. π 4 sin d 5. 0 π cos d ed 7. 4 π 4 0 tan d 8. π 4 π 6 cosec d 9. d 0. 0 d. 0 + d π. cos d. 0 d d e d π 4 (sec + + ) d π (sin cos ) d d ( sin π e + ) d 0 4 Choose the correct answer in Eercises and.. d equals + (A) π (B) π (C) π 6 (D) π. d equals (A) π 6 (B) π 7.9 Evaluation of Definite Integrals by Substitution In the previous sections, we have discussed several methods for finding the indefinite integral. One of the important methods for finding the indefinite integral is the method of substitution. (C) π 4 (D) π 4

56 INTEGRALS 9 b To evaluate f ( ) d, by substitution, the steps could be as follows: a. Consider the integral without limits and substitute, y f () or g(y) to reduce the given integral to a known form.. Integrate the new integrand with respect to the new variable without mentioning the constant of integration.. Resubstitute for the new variable and write the answer in terms of the original variable. 4. Find the values of answers obtained in () at the given limits of integral and find the difference of the values at the upper and lower limits. Note In order to quicken this method, we can proceed as follows: After performing steps, and, there is no need of step. Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step. Let us illustrate this by eamples. Eample 8 Evaluate d. 5 Solution Put t 5 +, then dt 5 4 d. Therefore, d tdt t ( 5 ) + Hence, d 5 ( + ) ( + ) ( ( ) + ) 0 ( ) 4 Alternatively, first we transform the integral and then evaluate the transformed integral with new limits.

57 40 MATHEMATICS Let t 5 +. Then dt 5 4 d. Note that, when, t 0 and when, t Thus, as varies from to, t varies from 0 to Therefore tdt t 0 ( ) 4 0 tan Eample 9 Evaluate 0 d + Solution Let t tan, then dt + d. The new limits are, when 0, t 0 and π π when, t. Thus, as varies from 0 to, t varies from 0 to. 4 4 Therefore tan 0 d + π 4 0 π 4 t tdt π π 0 6 EXERCISE 7.0 Evaluate the integrals in Eercises to 8 using substitution.. 4. d π 5 sin φ cos φdφ. 0 + (Put + t ) 5. d d Choose the correct answer in Eercises 9 and 0. ( ) π 0 sin sin + cos + d e 9. The value of the integral d is (A) 6 (B) 0 (C) (D) 4 0. If f () t 0 (A) cos + sin (C) cos sin t dt, then f () is (B) sin (D) sin + cos d d

58 INTEGRALS Some Properties of Definite Integrals We list below some important properties of definite integrals. These will be useful in evaluating the definite integrals more easily. b a b a P 0 : f ( d ) ft ( ) dt b a P : f ( d ) f( d ) a b. In particular, ( ) 0 b c b a a c P : f ( d ) f( d ) + f( d ) b a b a P : f ( d ) fa ( + b d ) P 4 : P 5 : P 6 : a f ( d ) fa ( d ) 0 0 a (Note that P 4 is a particular case of P ) a a a f ( d ) f( d ) + f( a d ) a a 0 0 a a f d f ( d ) f( d ),if f( a ) f( ) and 0 if f (a ) f () a a P 7 : (i) f ( d ) f( d ) a, if f is an even function, i.e., if f ( ) f (). 0 a (ii) f ( ) d 0, if f is an odd function, i.e., if f ( ) f (). a We give the proofs of these properties one by one. Proof of P 0 It follows directly by making the substitution t. Proof of P Let F be anti derivative of f. Then, by the second fundamental theorem of b a calculus, we have f ( d ) F( b) F( a) [F( a) F( b)] f( d ) a Here, we observe that, if a b, then f ( ) 0. a Proof of P Let F be anti derivative of f. Then b f ( ) F(b) F(a) a... () c f ( ) F(c) F(a) a... () and b f ( ) F(b) F(c)... () c a b

59 4 MATHEMATICS c b b a c a Adding () and (), we get f ( d ) + f( d ) F( b) F( a) f( d ) This proves the property P. Proof of P Let t a + b. Then dt d. When a, t b and when b, t a. Therefore b f ( ) a ( ) a + b b f ( ) (by P a ) b f ( ) d by P a 0 Proof of P 4 Put t a. Then dt d. When 0, t a and when a, t 0. Now proceed as in P. a a a. Proof of P 5 Using P, we have f ( ) d f ( ) d+ f ( ) d 0 0 a Let t a in the second integral on the right hand side. Then dt d. When a, t a and when a, t 0. Also a t. Therefore, the second integral becomes a 0 f ( ) d f ( a t ) dt a a a f ( a t ) dt a ( ) 0 f a d 0 a a a Hence f ( ) d f ( ) d+ f( a ) d Proof of P 6 Using P 5, we have Now, if and if 0 a f ( ) d 0 Proof of P 7 Using P, we have a a a f ( ) d f( ) d+ f( a ) d... () f (a ) f (), then () becomes a a a a f ( ) d f ( ) d+ f( ) d f( ) d, f (a ) f (), then () becomes a a f d 0 0 ( ) f( ) d 0 Let a f ( ) d a 0 a a 0 f ( ) d+ f( ) d. Then t in the first integral on the right hand side. dt d. When a, t a and when 0, t 0. Also t.

60 INTEGRALS 4 a Therefore f ( ) d a 0 a a 0 f ( t ) dt + f ( ) d (i) a a f ( ) d f( ) d (by P 0 )... () Now, if f is an even function, then f ( ) f () and so () becomes a a a a a f ( d ) f( d ) + f( d ) f( d ) (ii) If f is an odd function, then f ( ) f () and so () becomes a a a a f( ) d f( ) d+ f( ) d Eample 0 Evaluate d Solution We note that 0 on [, 0] and 0 on [0, ] and that 0 on [, ]. So by P we write d ( ) d ( ) d ( ) d ( ) d ( ) d ( ) d ( ) Eample Evaluate π 4 π 4 sin d Solution We observe that sin is an even function. Therefore, by P 7 (i), we get π 4 sin d π 4 π 4 0 sin d

61 44 MATHEMATICS π 4 0 ( cos ) d 4 ( cos ) d 0 sin π sin Eample Evaluate 0 d + cos π 4 0 π π sin π 0 π 4 4 π sin Solution Let I d 0. Then, by P + cos 4, we have π ( π )sin( π ) d I 0 + cos ( π ) π ( π )sin d π sin d 0 π I + cos 0 + cos π sin d or I π 0 + cos π π sin d or I 0 + cos Put cos t so that sin d dt. When 0, t and when π, t. Therefore, (by P ) we get π dt π dt I + t + t π dt 0 (by P + t 7, since t is even function) + Eample Evaluate π π π tan t tan tan 0 π π sin cos d Solution Let I 5 4 sin cos d. Let f() sin 5 cos 4. Then f ( ) sin 5 ( ) cos 4 ( ) sin 5 cos 4 f (), i.e., f is an odd function. Therefore, by P 7 (ii), I 0

62 INTEGRALS 45 Eample 4 Evaluate π 0 sin sin cos d Solution Let I π 4 sin d... () sin + cos Then, by P 4 4 π π sin ( ) I d 0 4 π 4 π sin ( ) + cos ( ) Adding () and (), we get I π Hence I 4 π 4 cos d () cos π 4 4 π π sin cos [ ] d d 0 0 sin + π + cos + sin Eample 5 Evaluate π π d + tan 6 Solution Let I π π d π π 6 6 cos d... () + tan cos + sin Then, by P I π π 6 π π cos + d 6 π π π π cos + + sin π π 6 sin d sin + cos... () Adding () and (), we get π π π 6 6 I d [ ] π π π π. Hence I 6 6 π

63 46 MATHEMATICS π Eample 6 Evaluate log sin d Solution Let I log sin d Then, by P 4 Adding the two values of I, we get 0 π 0 π I π log sin d log cos d 0 0 I ( log sin + log cos ) 0 π d ( log sin cos + log log ) 0 π π d(by adding and subtracting log ) π π d 0 0 log sin log d (Why?) Put t in the first integral. Then d dt, when 0, t 0 and when t π. Therefore I π π log sin tdt log 0 π, π 0 π log sin tdt log [by P 6 as sin (π t) sin t) π 0 π log sin d log (by changing variable t to ) π I log Hence π 0 log sin d π log.

64 INTEGRALS 47 EXERCISE 7. By using the properties of definite integrals, evaluate the integrals in Eercises to 9.. cos d. 0 π π 0 sin d. sin + cos π 0 sin sin d + cos 4. π 5 cos d sin + cos 5 + d d 7. ( ) n d 8. 0 π 4 0 log ( + tan ) d 9. 0 d 0.. π 0 (log sin log sin d ). π d. 0 + sin π sin cos 5. d sin cos 4 8. d 0 a π 7 sin d 4. π π log ( + cos ) d a π π π 0 a sin d 5 cos d + a 9. Show that f ( g ) ( ) d f( ) d 0, if f and g are defined as f() f(a ) 0 and g() + g(a ) 4 Choose the correct answer in Eercises 0 and. 0. The value of π 5 ( + cos+ tan + ) d is π (A) 0 (B) (C) π (D). The value of (A) π 0 4+ sin log d 4 + cos is (B) 4 (C) 0 (D) d

65 48 MATHEMATICS Eample 7 Find cos 6 + sin 6d Miscellaneous Eamples Solution Put t + sin 6, so that dt 6 cos 6 d Therefore Eample 8 Find cos 6 + sin 6d t dt ( ) 5 () t + C (+ sin6 ) + C 6 9 d Solution We have 4 4 ( ) ( ) d d Put t,sothat d dt 4 Therefore 4 4 ( ) d t 4 dt C t + C Eample 9 Find 4 d ( )( + ) Solution We have 4 ( )( + ) ( + ) + + Now epress ( )( + ) ( + ) + ( )( + ) A B + C + ( ) ( + )... ()... ()

66 INTEGRALS 49 So A ( + ) + (B + C) ( ) (A + B) + (C B) + A C Equating coefficients on both sides, we get A + B 0, C B 0 and A C, which give A,B C. Substituting values of A, B and C in (), we get... () ( )( + ) ( ) ( + ) ( + ) Again, substituting () in (), we have 4 ( + ) + ( )( + + ) ( ) ( + ) ( + ) Therefore 4 d + + log log ( + ) tan + C ( )( + + ) 4 Eample 40 Find log (log ) + d (log ) Solution Let I log (log ) + d (log ) log (log ) d + (log ) In the first integral, let us take as the second function. Then integrating it by parts, we get d I log (log ) d+ log (log ) Again, consider d d d log (log ) + log (log )... () d, take as the second function and integrate it by parts, log d we have d log log (log )... ()

67 50 MATHEMATICS Putting () in (), we get d d I log (log ) + log (log ) (log ) Eample 4 Find cot tan + d log log (log ) + C Solution We have I cot + tan d tan ( + cot ) d Put tan t, so that sec d t dt or tdt d 4 + t Then t I t + dt 4 t ( + t ) + dt + dt ( t + ) t t dt 4 t + t + t + t t Put t y, so that + t t dt dy. Then I t dy y t tan C tan + + C y + ( ) Eample 4 Find t tan tan + C tan + C t tan sin cos d 4 9 cos ( ) Solution Let sin cos I d 4 9 cos

68 INTEGRALS 5 Put cos () t so that 4 sin cos d dt Therefore dt t I sin C sin cos C t 4 4 Eample 4 Evaluate sin ( π ) d sin π for Solution Here f () sin π sin π for Therefore sin π d sin π d+ sin πd Integrating both integrals on righthand side, we get sin π d sin π d sin π d cosπ sin π cosπ sin π + + π π π π + π π π π π π d Eample 44 Evaluate 0 a cos + b sin π d π ( π ) d Solution Let I 0 0 a cos + b sin a cos ( π ) + b sin ( π ) (using P 4 ) d d π π π 0 0 a cos + b sin a cos + b sin d I a cos + b sin π π 0 π d Thus I π 0 a cos + b sin

69 5 MATHEMATICS π π d π d 0 0 π or I a cos + b sin a cos + b sin (using P 6 ) π sec d π (dividing numerator and denominator by cos ). 0 a + b tan π Put b tan t, so that b sec d dt. Also, when 0, t 0, and when, t. Therefore, dt t 0 a t 0. + π π π π π I tan 0 b b a a ab ab Miscellaneous Eercise on Chapter 7 Integrate the functions in Eercises to a + + b. a [Hint: Put a t ] 4. 4 ( + ) [Hint: + + 6, put t 6 ] 5 6. ( + )( + 9) 7. sin sin ( a) 8. e e 5log log e e 4log log 9. cos 4 sin 0. sin 8 8 cos sin cos. cos ( + a) cos ( + b) e ( + e ) ( + e ) ( + )( + 4) 5. cos e log sin 6. e log ( 4 + ) 7. f (a + b) [f (a + b)] n 8. sin sin ( +α) 9. sin sin cos + cos, [0, ] sin e + cos. + + ( + ) ( + )

70 INTEGRALS 5 + log( ) log. tan Evaluate the definite integrals in Eercises 5 to. sin 5. π π π e d π + cos 6. 4 sin cos cos d d cos + sin 0 cos + 4 sin 8.. π π 6 sin + cos d 9. sin d π sin tan (sin ) d [ + + ] d Prove the following (Eercises 4 to 9) d 4. ( + ) log cos d 0 7. π 4 tan d log 9. 0 e Evaluate d as a limit of a sum. Choose the correct answers in Eercises 4 to d e + e is equal to (A) tan (e ) + C (B) tan (e ) + C (C) log (e e ) + C (D) log (e + e ) + C 4. cos d (sin + cos ) is equal to (A) C sin + cos + (C) log sin cos + C (D) π π 0 π 0 sin + cos d 9+ 6sin tan d sec + tan ed sin d sin π d 0 (B) log sin + cos + C (sin + cos )

71 54 MATHEMATICS b 4. If f (a + b ) f (), then f ( ) d is equal to a (A) a+ b b a+ b b f ( b ) d (B) f ( b ) d a + a (C) b a b a+ b b f ( ) d (D) f ( ) d a a 44. The value of + tan d 0 is (A) (B) 0 (C) (D) Summary Integration is the inverse process of differentiation. In the differential calculus, we are given a function and we have to find the derivative or differential of this function, but in the integral calculus, we are to find a function whose differential is given. Thus, integration is a process which is the inverse of differentiation. d Let F( ) f( ). Then we write f( ) d F( ) + C d. These integrals are called indefinite integrals or general integrals, C is called constant of integration. All these integrals differ by a constant. From the geometric point of view, an indefinite integral is collection of family of curves, each of which is obtained by translating one of the curves parallel to itself upwards or downwards along the y-ais. Some properties of indefinite integrals are as follows:. [ f ( ) + g ( )] d f ( ) d + g ( ) d. For any real number k, kf( d ) k f( d ) More generally, if f, f, f,..., f n are functions and k, k,...,k n are real numbers. Then [ k f ( ) + k f ( ) k f ( )] d n n n n k f ( ) d+ k f ( ) d k f ( ) d π 4

72 INTEGRALS 55 Some standard integrals (i) n+ n d + C, n. Particularly, d + C n + (ii) cos d sin + C (iii) sin d cos + C (iv) sec d tan + C (vi) sec tan d sec + C (vii) cosec cot d cosec + C (v) (viii) cosec d cot + C d sin + C (i) d cos + C () d tan + C + (i) d cot + C (ii) ed e + C + a d (iii) ad + C (iv) sec C log a + d (v) cosec + C (vi) d log + C Integration by partial fractions Recall that a rational function is ratio of two polynomials of the form P( ) Q( ), where P() and Q () are polynomials in and Q () 0. If degree of the polynomial P () is greater than the degree of the polynomial Q (), then we P( ) P( ) may divide P () by Q () so that T( ) +, where T() is a Q( ) Q( ) polynomial in and degree of P () is less than the degree of Q(). T() P( being polynomial can be easily integrated. ) can be integrated by Q( )

73 56 MATHEMATICS P( ) epressing as the sum of partial fractions of the following type: Q( ) p + q A B. + ( a)( b) a b, a b p+ q. ( a) A B + a ( a). p + q + r ( a)( b)( c) A B C + + a b c 4. p + q+ r ( a) ( b) A + B + C a ( a) b 5. p + q + r ( a)( + b+ c) A B + +C a + b + c where + b + c can not be factorised further. Integration by substitution A change in the variable of integration often reduces an integral to one of the fundamental integrals. The method in which we change the variable to some other variable is called the method of substitution. When the integrand involves some trigonometric functions, we use some well known identities to find the integrals. Using substitution technique, we obtain the following standard integrals. (i) tan d log sec + C (ii) cot d log sin + C (iii) sec d log sec + tan + C (iv) cosecd log cosec cot + C Integrals of some special functions d a (i) log + C a a + a d a + (ii) log + C a a a (iii) d + a tan C + a a

74 INTEGRALS 57 d (iv) log + a + C (v) a d (vi) log + + a + C + a Integration by parts For given functions f and f, we have a d sin + C a d f( ) f( ) d f( ) f( ) d f( ) f( ) d d d, i.e., the integral of the product of two functions first function integral of the second function integral of {differential coefficient of the first function integral of the second function}. Care must be taken in choosing the first function and the second function. Obviously, we must take that function as the second function whose integral is well known to us. e [ f( ) + f ( )] d e f( ) d+ C Some special types of integrals (i) (ii) (iii) a a d a log + a + C a + a d + a + log + + a + C a d a + a sin + C a d d (iv) Integrals of the types or a + b + c a + b + c transformed into standard form by epressing a + b + c can be b c b c b a + + a a a + + a a 4a p + q d p + q d (v) Integrals of the types or a + b + c a + b + c can be

75 58 MATHEMATICS transformed into standard form by epressing d p q a b c a b d determined by comparing coefficients on both sides. + A ( + + ) + B A( + ) + B, where A and B are b We have defined f ( ) d as the area of the region bounded by the curve a y f (), a b, the -ais and the ordinates a and b. Let be a given point in [a, b]. Then f ( ) d represents the Area function A(). a This concept of area function leads to the Fundamental Theorems of Integral Calculus. First fundamental theorem of integral calculus Let the area function be defined by A() f ( ) d for all a, where a the function f is assumed to be continuous on [a, b]. Then A () f () for all [a, b]. Second fundamental theorem of integral calculus Let f be a continuous function of defined on the closed interval [a, b] and d let F be another function such that F( ) f( ) for all in the domain of d b f, then [ ] b f ( ) d F( ) + C F( b ) F( a ). a a This is called the definite integral of f over the range [a, b], where a and b are called the limits of integration, a being the lower limit and b the upper limit.

76 APPLICATION OF INTEGRALS 59 Chapter 8 APPLICATION OF INTEGRALS One should study Mathematics because it is only through Mathematics that nature can be conceived in harmonious form. BIRKHOFF 8. Introduction In geometry, we have learnt formulae to calculate areas of various geometrical figures including triangles, rectangles, trapezias and circles. Such formulae are fundamental in the applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we shall need some concepts of Integral Calculus. In the previous chapter, we have studied to find the area bounded by the curve y f (), the ordinates a, b and -ais, while calculating definite integral as the limit of a sum. Here, in this chapter, we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses (standard forms only). We shall also deal with finding the area bounded by the above said curves. 8. Area under Simple Curves In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by the curve y f (), -ais and the ordinates a and b. From Fig 8., we can think of area under the curve as composed of large number of very thin vertical strips. Consider an arbitrary strip of height y and width d, then da (area of the elementary strip) yd, where, y f (). A.L. Cauchy ( ) Fig 8.

77 60 MATHEMATICS This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of between a and b. We can think of the total area A of the region between -ais, ordinates a, b and the curve y f () as the result of adding up the elementary areas of thin strips across the region PQRSP. Symbolically, we epress b b b A da yd f ( ) d a a a The area A of the region bounded by the curve g (y), y-ais and the lines y c, y d is given by d A dy g( y) dy c c d Here, we consider horizontal strips as shown in the Fig 8. Fig 8. Remark If the position of the curve under consideration is below the -ais, then since f () < 0 from a to b, as shown in Fig 8., the area bounded by the curve, -ais and the ordinates a, b come out to be negative. But, it is only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we b take its absolute value, i.e., f ( ) d. a Fig 8. Generally, it may happen that some portion of the curve is above -ais and some is below the -ais as shown in the Fig 8.4. Here, A < 0 and A > 0. Therefore, the area A bounded by the curve y f (), -ais and the ordinates a and b is given by A A + A.

78 APPLICATION OF INTEGRALS 6 Eample Find the area enclosed by the circle + y a. Solution From Fig 8.5, the whole area enclosed by the given circle 4 (area of the region AOBA bounded by the curve, -ais and the ordinates 0 and a) [as the circle is symmetrical about both -ais and y-ais] Fig a yd (taking vertical strips) 0 4 a a d 0 Since + y a gives y ± a Fig 8.5 As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get the whole area enclosed by the given circle a 4 a + sin a a 0 a a 4 0+ sin 0 a π 4 π a

79 6 MATHEMATICS Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the region enclosed by circle 4 a dy 0 4 a a y dy 0 a (Why?) y a y 4 a y + sin a 0 a a 4 0+ sin 0 a π 4 πa Fig 8.6 y Eample Find the area enclosed by the ellipse + a b Solution From Fig 8.7, the area of the region ABA B A bounded by the ellipse area of theregion AOBAin the first quadrant bounded 4 bythecurve, ais and theordinates 0, a 0 (as the ellipse is symmetrical about both -ais and y-ais) a 4 yd (taking verticalstrips) y b Now + gives y± a, but as the region AOBA lies in the first a b a quadrant, y is taken as positive. So, the required area is 4 a b a d 0 a 4b a a + sin a a 4b a a 0+ sin 0 a a 0 (Why?) 4ba π π ab a Fig 8.7

80 APPLICATION OF INTEGRALS 6 Alternatively, considering horizontal strips as shown in the Fig 8.8, the area of the ellipse is 4 b dy 0 b 0 a 4 b y dy b (Why?) 4a y b y b y + sin b b 4a b b 0+ sin 0 b b 0 Fig 8.8 4ab π π ab b 8.. The area of the region bounded by a curve and a line In this subsection, we will find the area of the region bounded by a line and a circle, a line and a parabola, a line and an ellipse. Equations of above mentioned curves will be in their standard forms only as the cases in other forms go beyond the scope of this tetbook. Eample Find the area of the region bounded by the curve y and the line y 4. Solution Since the given curve represented by the equation y is a parabola symmetrical about y-ais only, therefore, from Fig 8.9, the required area of the region AOBA is given by 4 dy 0 area of theregion BONBbounded bycurve, y ais and thelines y 0and y 4 Fig ydy 0 y 4 8 (Why?) 0 Here, we have taken horizontal strips as indicated in the Fig

81 64 MATHEMATICS Alternatively, we may consider the vertical strips like PQ as shown in the Fig 8.0 to obtain the area of the region AOBA. To this end, we solve the equations y and y 4 which gives and. Thus, the region AOBA may be stated as the region bounded by the curve y, y 4 and the ordinates and. Therefore, the area of the region AOBA yd [y ( y-coordinate of Q) (y-coordinate of P) 4 ] ( ) 4 d 0 (Why?) Remark From the above eamples, it is inferred that we can consider either vertical strips or horizontal strips for calculating the area of the region. Henceforth, we shall consider either of these two, most preferably vertical strips. Eample 4 Find the area of the region in the first quadrant enclosed by the -ais, the line y, and the circle + y. Y Solution The given equations are y... () y and + y... () B (4, 4) Solving () and (), we find that the line and the circle meet at B(4, 4) in the first quadrant (Fig 8.). Draw perpendicular A BM to the -ais. X' X M O (4, 0) Therefore, the required area area of the region OBMO + area of the region BMAB. Now, the area of the region OBMO 4 4 yd 0 0 d... () Fig 8.0 Y' Fig 8.

82 APPLICATION OF INTEGRALS 65 Again, the area of the region BMAB 4 yd d + sin sin 6 + sin 8 π (8 + 4π) 4π 8... (4) Adding () and (4), we get, the required area 4π. Eample 5 Find the area bounded by the ellipse and ae, where, b a ( e ) and e <. a y + and the ordinates 0 b Solution The required area (Fig 8.) of the region BOB RFSB is enclosed by the ellipse and the lines 0 and ae. Y Note that the area of the region BOB RFSB B S ae ae yd b ae a d 0 a 0 F( ae, o) X X ae O b a a + sin a a 0 R B' b ae a a e a sin e a + Y Fig 8. ab e e + sin e EXERCISE 8.. Find the area of the region bounded by the curve y and the lines, 4 and the -ais.. Find the area of the region bounded by y 9,, 4 and the -ais in the first quadrant.

83 66 MATHEMATICS. Find the area of the region bounded by 4y, y, y 4 and the y-ais in the first quadrant. 4. Find the area of the region bounded by the ellipse y Find the area of the region bounded by the ellipse y Find the area of the region in the first quadrant enclosed by -ais, line y and the circle + y 4. a 7. Find the area of the smaller part of the circle + y a cut off by the line. 8. The area between y and 4 is divided into two equal parts by the line a, find the value of a. 9. Find the area of the region bounded by the parabola y and y. 0. Find the area bounded by the curve 4y and the line 4y.. Find the area of the region bounded by the curve y 4 and the line. Choose the correct answer in the following Eercises and.. Area lying in the first quadrant and bounded by the circle + y 4 and the lines 0 and is π π π (A) π (B) (C) (D) 4. Area of the region bounded by the curve y 4, y-ais and the line y is (A) (B) Area between Two Curves Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by cutting the region into a large number of small strips of elementary area and then adding up these elementary areas. Suppose we are given two curves represented by y f (), y g (), where f () g() in [a, b] as shown in Fig 8.. Here the points of intersection of these two curves are given by a and b obtained by taking common values of y from the given equation of two curves. For setting up a formula for the integral, it is convenient to take elementary area in the form of vertical strips. As indicated in the Fig 8., elementary strip has height (C) 9 (D) 9

84 APPLICATION OF INTEGRALS 67 f() g() and width d so that the elementary area Fig 8. da [f () g()] d, and the total area A can be taken as b A [() f g ()] d a Alternatively, A [area bounded by y f (), -ais and the lines a, b] [area bounded by y g (), -ais and the lines a, b] b a b a b [ ] f ( d ) gd ( ) f ( ) g( ) d, where f () g () in [a, b] a If f () g () in [a, c] and f () g () in [c, b], where a < c < b as shown in the Fig 8.4, then the area of the regions bounded by curves can be written as Total Area Area of the region ACBDA + Area of the region BPRQB c b [ ( ) ( )] + [ ( ) ( )] a f g d g f d c Y A C y f( ) B y g ( ) P R D y g ( ) a c Q y f( ) b X O Y Fig 8.4 X

85 68 MATHEMATICS Eample 6 Find the area of the region bounded by the two parabolas y and y. Solution The point of intersection of these two parabolas are O (0, 0) and A (, ) as shown in the Fig 8.5. Here, we can set y or y f() and y g (), where, f () g () in [0, ]. Therefore, the required area of the shaded region [ ( ) ( )] 0 0 f g d d P (4, 4) Thus, the points of intersection of these two curves are O(0, 0) and P(4,4) above the -ais. From the Fig 8.6, the required area of the region OPQCO included between these two curves above -ais is Y Fig 8.6 (area of the region OCPO) + (area of the region PCQP) 4 8 yd + yd 0 Eample 7 Find the area lying above -ais and included between the circle + y 8 and the parabola y 4. Solution The given equation of the circle + y 8 can be epressed as ( 4) + y 6. Thus, the centre of the Y circle is (4, 0) and radius is 4. Its intersection with the parabola y 4 gives or 4 0 or ( 4) 0 X X or 0, 4 O C (4, 0) d + d 0 (Why?) 4 4 ( 4) Fig 8.5 Q (8, 0)

86 APPLICATION OF INTEGRALS , where, 4 + t dt t 0 0 t t + 4 t 4 sin (Why?) sin + π π 4 (8+ π) Eample 8 In Fig 8.7, AOBA is the part of the ellipse 9 + y 6 in the first quadrant such that OA and OB 6. Find the area between the arc AB and the chord AB. Solution Given equation of the ellipse 9 + y 6 can be epressed as y + or 4 6 y + and hence, its shape is as given in Fig Accordingly, the equation of the chord AB is y ( ) 0 or y ( ) or y + 6 Area of the shaded region as shown in the Fig 8.7. (Why?) 4 d (6 ) d sin Fig sin () π 6 π 6

87 70 MATHEMATICS Eample 9 Using integration find the area of region bounded by the triangle whose vertices are (, 0), (, ) and (, ). Solution Let A(, 0), B(, ) and C (, ) be the vertices of a triangle ABC (Fig 8.8). Area of ΔABC Area of ΔABD + Area of trapezium BDEC Area of ΔAEC Now equation of the sides AB, BC and CA are given by y ( ), y 4, y Hence, area of Δ ABC ( ), respectively. d d d ( ) + (4 ) Eample 0 Find the area of the region enclosed between the two circles: + y 4 and ( ) + y 4. Solution Equations of the given circles are + y 4... () and ( ) + y 4... () Equation () is a circle with centre O at the origin and radius. Equation () is a circle with centre C (, 0) and radius. Solving equations () and (), we have ( ) + y + y or y + y or which gives y ± Thus, the points of intersection of the given circles are A(, ) and A (, ) as shown in the Fig 8.9. Fig 8.8 Fig 8.9

88 APPLICATION OF INTEGRALS 7 Required area of the enclosed region OACA O between circles [area of the region ODCAO] (Why?) [area of the region ODAO + area of the region DCAD] y d + y d 0 4 ( ) d + 4 d 0 (Why?) ( ) 4 ( ) + 4sin sin ( ) 4 ( ) + 4sin sin + 4sin 4sin ( ) + 4sin 4sin 0 π π π π π π + π + π 8 π EXERCISE 8.. Find the area of the circle 4 + 4y 9 which is interior to the parabola 4y.. Find the area bounded by curves ( ) + y and + y.. Find the area of the region bounded by the curves y +, y, 0 and. 4. Using integration find the area of region bounded by the triangle whose vertices are (, 0), (, ) and (, ). 5. Using integration find the area of the triangular region whose sides have the equations y +, y + and 4.

89 7 MATHEMATICS Choose the correct answer in the following eercises 6 and Smaller area enclosed by the circle + y 4 and the line + y is (A) (π ) (B) π (C) π (D) (π + ) 7. Area lying between the curves y 4 and y is (A) (B) (C) 4 Miscellaneous Eamples Eample Find the area of the parabola y 4a bounded by its latus rectum. Solution From Fig 8.0, the verte of the parabola y 4a is at origin (0, 0). The equation of the latus rectum LSL is a. Also, parabola is symmetrical about the -ais. The required area of the region OLL O (area of the region OLSO) a yd 0 a a 0 a 4 0 d a d X O Y (D) 4 L (,0) a S X 4 a a 0 Y Fig 8.0 L' 8 a a 8 a Eample Find the area of the region bounded by the line y +, the -ais and the ordinates and. Solution As shown in the Fig 8., the line y + meets -ais at and its graph lies below -ais for, and above -ais for,. Fig 8.

90 APPLICATION OF INTEGRALS 7 The required area Area of the region ACBA + Area of the region ADEA ( + ) d + ( + ) d Eample Find the area bounded by the curve y cos between 0 and π. Solution From the Fig 8., the required area area of the region OABO + area of the region BCDB + area of the region DEFD. Thus, we have the required area Fig 8. π 0 π cos d + cos d + cos d π π π π π [ sin ] + [ sin ] + [ sin ] π π π Eample Prove that the curves y 4 and 4y divide the area of the square bounded by 0, 4, y 4 and y 0 into three equal parts. Solution Note that the point of intersection of the parabolas y 4 and 4y are (0, 0) and (4, 4) as Fig 8.

91 74 MATHEMATICS shown in the Fig 8.. Now, the area of the region OAQBO bounded by curves y 4 and 4y. 4 d () Again, the area of the region OPQAO bounded by the curves 4y, 0, 4 and -ais 4 4 d () 4 Similarly, the area of the region OBQRO bounded by the curve y 4, y-ais, y 0 and y y 4 dy dy y () From (), () and (), it is concluded that the area of the region OAQBO area of the region OPQAO area of the region OBQRO, i.e., area bounded by parabolas y 4 and 4y divides the area of the square in three equal parts. Eample 4 Find the area of the region {(, y) : 0 y +, 0 y +, 0 } Solution Let us first sketch the region whose area is to be found out. This region is the intersection of the following regions. A {(, y) : 0 y + }, A {(, y) : 0 y + } and A {(, y) : 0 } X' O Y Y' Q(, ) R P (0,) Fig 8.4 The points of intersection of y + and y + are points P(0, ) and Q(, ). From the Fig 8.4, the required region is the shaded region OPQRSTO whose area area of the region OTQPO + area of the region TSRQT T S X ( + ) d + ( + ) d (Why?) 0

92 APPLICATION OF INTEGRALS ( ) Miscellaneous Eercise on Chapter 8. Find the area under the given curves and given lines: (i) y,, and -ais (ii) y 4,, 5 and -ais. Find the area between the curves y and y.. Find the area of the region lying in the first quadrant and bounded by y 4, 0, y and y Sketch the graph of y + and evaluate + d Find the area bounded by the curve y sin between 0 and π. 6. Find the area enclosed between the parabola y 4a and the line y m. 7. Find the area enclosed by the parabola 4y and the line y Find the area of the smaller region bounded by the ellipse y line Find the area of the smaller region bounded by the ellipse y line +. a b 0 y + and the 9 4 a y + and the b 0. Find the area of the region enclosed by the parabola y, the line y + and the -ais.. Using the method of integration find the area bounded by the curve + y. [Hint: The required region is bounded by lines + y, y, + y and y ].

93 76 MATHEMATICS. Find the area bounded by curves {(, y) : y and y }.. Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(, 0), B (4, 5) and C (6, ). 4. Using the method of integration find the area of the region bounded by lines: + y 4, y 6 and y Find the area of the region {(, y) : y 4, 4 + 4y 9} Choose the correct answer in the following Eercises from 6 to Area bounded by the curve y, the -ais and the ordinates and is (A) 9 (B) The area bounded by the curve y, -ais and the ordinates and is given by (A) 0 (B) (C) (C) [Hint : y if > 0 and y if < 0]. 8. The area of the circle + y 6 eterior to the parabola y 6 is 5 4 (D) (D) (A) 4 (4 π ) (B) 4 (4 π+ ) (C) 4 (8 π ) (D) 4 (8 π+ ) π 9. The area bounded by the y-ais, y cos and y sin when 0 is (A) ( ) (B) (C) + (D) Summary The area of the region bounded by the curve y f (), -ais and the lines a and b (b > a) is given by the formula: Area yd f ( ) d b a b a. The area of the region bounded by the curve φ (y), y-ais and the lines y c, y d is given by the formula: Area dy φ( y) dy d c d c.

94 APPLICATION OF INTEGRALS 77 The area of the region enclosed between two curves y f (), y g () and the lines a, b is given by the formula, b a [ ] Area f ( ) g( ) d, where, f () g () in [a, b] If f () g () in [a, c] and f () g () in [c, b], a < c < b, then c b [ f g ] d [ g f ] d. Area ( ) ( ) + ( ) ( ) a c Historical Note The origin of the Integral Calculus goes back to the early period of development of Mathematics and it is related to the method of ehaustion developed by the mathematicians of ancient Greece. This method arose in the solution of problems on calculating areas of plane figures, surface areas and volumes of solid bodies etc. In this sense, the method of ehaustion can be regarded as an early method of integration. The greatest development of method of ehaustion in the early period was obtained in the works of Eudous (440 B.C.) and Archimedes (00 B.C.) Systematic approach to the theory of Calculus began in the 7th century. In 665, Newton began his work on the Calculus described by him as the theory of fluions and used his theory in finding the tangent and radius of curvature at any point on a curve. Newton introduced the basic notion of inverse function called the anti derivative (indefinite integral) or the inverse method of tangents. During , Leibnitz published an article in the Acta Eruditorum which he called Calculas summatorius, since it was connected with the summation of a number of infinitely small areas, whose sum, he indicated by the symbol. In 696, he followed a suggestion made by J. Bernoulli and changed this article to Calculus integrali. This corresponded to Newton s inverse method of tangents. Both Newton and Leibnitz adopted quite independent lines of approach which was radically different. However, respective theories accomplished results that were practically identical. Leibnitz used the notion of definite integral and what is quite certain is that he first clearly appreciated tie up between the antiderivative and the definite integral. Conclusively, the fundamental concepts and theory of Integral Calculus and primarily its relationships with Differential Calculus were developed in the work of P.de Fermat, I. Newton and G. Leibnitz at the end of 7th century.

95 78 MATHEMATICS However, this justification by the concept of limit was only developed in the works of A.L. Cauchy in the early 9th century. Lastly, it is worth mentioning the following quotation by Lie Sophie s: It may be said that the conceptions of differential quotient and integral which in their origin certainly go back to Archimedes were introduced in Science by the investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis... The discovery that differentiation and integration are inverse operations belongs to Newton and Leibnitz.

96 DIFFERENTIAL EQUATIONS 79 Chapter 9 DIFFERENTIAL EQUATIONS He who seeks for methods without having a definite problem in mind seeks for the most part in vain. D. HILBERT 9. Introduction In Class XI and in Chapter 5 of the present book, we discussed how to differentiate a given function f with respect to an independent variable, i.e., how to find f () for a given function f at each in its domain of definition. Further, in the chapter on Integral Calculus, we discussed how to find a function f whose derivative is the function g, which may also be formulated as follows: For a given function g, find a function f such that dy g(), where y f ()... () d An equation of the form () is known as a differential equation. A formal definition will be given later. Henri Poincare (854-9 ) These equations arise in a variety of applications, may it be in Physics, Chemistry, Biology, Anthropology, Geology, Economics etc. Hence, an indepth study of differential equations has assumed prime importance in all modern scientific investigations. In this chapter, we will study some basic concepts related to differential equation, general and particular solutions of a differential equation, formation of differential equations, some methods to solve a first order - first degree differential equation and some applications of differential equations in different areas. 9. Basic Concepts We are already familiar with the equations of the type: () sin + cos 0... () + y 7... ()

97 80 MATHEMATICS Let us consider the equation: dy y d (4) We see that equations (), () and () involve independent and/or dependent variable (variables) only but equation (4) involves variables as well as derivative of the dependent variable y with respect to the independent variable. Such an equation is called a differential equation. In general, an equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation. A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation, e.g., d y dy + 0 is an ordinary differential equation... (5) d d Of course, there are differential equations involving derivatives with respect to more than one independent variables, called partial differential equations but at this stage we shall confine ourselves to the study of ordinary differential equations only. Now onward, we will use the term differential equation for ordinary differential equation. Note. We shall prefer to use the following notations for derivatives: dy d y d y y, y, y d d d. For derivatives of higher order, it will be inconvenient to use so many dashes as supersuffi therefore, we use the notation y n for nth order derivative n d y n d Order of a differential equation Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation. Consider the following differential equations: dy d e... (6)

98 DIFFERENTIAL EQUATIONS 8 d y + y 0... (7) d d y d y (8) d d The equations (6), (7) and (8) involve the highest derivative of first, second and third order respectively. Therefore, the order of these equations are, and respectively. 9.. Degree of a differential equation To study the degree of a differential equation, the key point is that the differential equation must be a polynomial equation in derivatives, i.e., y, y, y etc. Consider the following differential equations: d y d y dy + y + d d d dy dy + sin d d y 0... (9) 0... (0) dy dy + sin 0... () d d We observe that equation (9) is a polynomial equation in y, y and y, equation (0) is a polynomial equation in y (not a polynomial in y though). Degree of such differential equations can be defined. But equation () is not a polynomial equation in y and degree of such a differential equation can not be defined. By the degree of a differential equation, when it is a polynomial equation in derivatives, we mean the highest power (positive integral inde) of the highest order derivative involved in the given differential equation. In view of the above definition, one may observe that differential equations (6), (7), (8) and (9) each are of degree one, equation (0) is of degree two while the degree of differential equation () is not defined. Note Order and degree (if defined) of a differential equation are always positive integers.

99 8 MATHEMATICS Eample Find the order and degree, if defined, of each of the following differential equations: dy (i) cos 0 d (ii) y (iii) y + y + e d y dy dy y y d d d Solution (i) (ii) The highest order derivative present in the differential equation is dy, so its d order is one. It is a polynomial equation in y and the highest power raised to dy d is one, so its degree is one. d y The highest order derivative present in the given differential equation is d, so its order is two. It is a polynomial equation in d y d and dy d and the highest (iii) d y power raised to is one, so its degree is one. d The highest order derivative present in the differential equation is y, so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined. EXERCISE 9. Determine order and degree (if defined) of differential equations given in Eercises to d y + sin( y ) 0 4. y + 5y 0. d 4 ds d s + s 0 dt dt 4. d y dy cos 0 + d d 5. d y cos sin d + 6. ( y ) + (y ) + (y ) 4 + y y + y + y 0

100 DIFFERENTIAL EQUATIONS 8 8. y + y e 9. y + (y ) + y 0 0. y + y + sin y 0. The degree of the differential equation d y dy dy sin is d d d (A) (B) (C) (D) not defined. The order of the differential equation d y dy y d d + is 0 (A) (B) (C) 0 (D) not defined 9.. General and Particular Solutions of a Differential Equation In earlier Classes, we have solved the equations of the type: () sin cos 0... () Solution of equations () and () are numbers, real or comple, that will satisfy the given equation i.e., when that number is substituted for the unknown in the given equation, L.H.S. becomes equal to the R.H.S.. y 0 d y Now consider the differential equation +... () d In contrast to the first two equations, the solution of this differential equation is a function φ that will satisfy it i.e., when the function φ is substituted for the unknown y (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S.. The curve y φ () is called the solution curve (integral curve) of the given differential equation. Consider the function given by y φ () a sin ( + b),... (4) where a, b R. When this function and its derivative are substituted in equation (), L.H.S. R.H.S.. So it is a solution of the differential equation (). π Let a and b be given some particular values say a and b, then we get a 4 π function y φ () sin +... (5) 4 When this function and its derivative are substituted in equation () again L.H.S. R.H.S.. Therefore φ is also a solution of equation ().

101 84 MATHEMATICS Function φ consists of two arbitrary constants (parameters) a, b and it is called general solution of the given differential equation. Whereas function φ contains no arbitrary constants but only the particular values of the parameters a and b and hence is called a particular solution of the given differential equation. The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation. The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation. Eample Verify that the function y e is a solution of the differential equation d y dy 6y 0 d + d Solution Given function is y e. Differentiating both sides of equation with respect to, we get dy d Now, differentiating () with respect to, we have d y 9e d, e... () d y dy Substituting the values of and y in the given differential equation, we get d d L.H.S. 9 e + ( e ) 6.e 9 e 9 e 0 R.H.S.. Therefore, the given function is a solution of the given differential equation. Eample Verify that the function y a cos + b sin, where, a, b R is a solution of the differential equation y 0 d y d + Solution The given function is y a cos + b sin... () Differentiating both sides of equation () with respect to, successively, we get dy a sin + b cos d d y a cos b sin d

102 DIFFERENTIAL EQUATIONS 85 d y Substituting the values of and y in the given differential equation, we get d L.H.S. ( a cos b sin ) + (a cos + b sin ) 0 R.H.S.. Therefore, the given function is a solution of the given differential equation. EXERCISE 9. In each of the Eercises to 0 verify that the given functions (eplicit or implicit) is a solution of the corresponding differential equation:. y e + : y y 0. y + + C : y 0. y cos + C : y + sin 0 y 4. y + : y + 5. y A : y y ( 0) 6. y sin : y y + y ( 0 and > y or < y) y 7. y log y + C : y (y ) y 8. y cos y : (y sin y + cos y + ) y y 9. + y tan y : y y + y y a ( a, a) : + y dy 0 (y 0) d. The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) 0 (B) (C) (D) 4. The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) (B) (C) (D) Formation of a Differential Equation whose General Solution is given We know that the equation + y + 4y () represents a circle having centre at (, ) and radius unit.

103 86 MATHEMATICS Differentiating equation () with respect to, we get dy d + y (y )... () which is a differential equation. You will find later on [See (eample 9 section 9.5..)] that this equation represents the family of circles and one member of the family is the circle given in equation (). Let us consider the equation + y r... () By giving different values to r, we get different members of the family e.g. + y, + y 4, + y 9 etc. (see Fig 9.). Thus, equation () represents a family of concentric circles centered at the origin and having different radii. We are interested in finding a differential equation that is satisfied by each member of the family. The differential equation must be free from r because r is different for different members of the family. This equation is obtained by differentiating equation () with respect to, i.e., + y dy d 0 or + y dy d 0... (4) which represents the family of concentric circles given by equation (). Again, let us consider the equation y m + c... (5) By giving different values to the parameters m and c, we get different members of the family, e.g., y (m, c 0) y (m, c 0) Fig 9. y + (m, c ) y (m, c 0) y (m, c ) etc. ( see Fig 9.). Thus, equation (5) represents the family of straight lines, where m, c are parameters. We are now interested in finding a differential equation that is satisfied by each member of the family. Further, the equation must be free from m and c because m and

104 DIFFERENTIAL EQUATIONS 87 c are different for different members of the family. This is obtained by differentiating equation (5) with respect to, successively we get y Y y+ y dy m d d y, and 0 d... (6) X The equation (6) represents the family of straight lines given by equation (5). Note that equations () and (5) are the general solutions of equations (4) and (6) respectively. y y 9.4. Procedure to form a differential equation that will represent a given family of curves (a) If the given family F of curves depends on only one parameter then it is represented by an equation of the form F (, y, a) 0... () For eample, the family of parabolas y a can be represented by an equation of the form f (, y, a) : y a. Differentiating equation () with respect to, we get an equation involving y, y,, and a, i.e., g (, y, y, a) 0... () The required differential equation is then obtained by eliminating a from equations () and () as F(, y, y ) 0... () (b) If the given family F of curves depends on the parameters a, b (say) then it is represented by an equation of the from F (, y, a, b) 0... (4) Differentiating equation (4) with respect to, we get an equation involving y,, y, a, b, i.e., g (, y, y, a, b) 0... (5) But it is not possible to eliminate two parameters a and b from the two equations and so, we need a third equation. This equation is obtained by differentiating equation (5), with respect to, to obtain a relation of the form h (, y, y, y, a, b) 0... (6) Y O Fig 9. X

105 88 MATHEMATICS The required differential equation is then obtained by eliminating a and b from equations (4), (5) and (6) as F (, y, y, y ) 0... (7) Note The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves. Eample 4 Form the differential equation representing the family of curves y m, where, m is arbitrary constant. Solution We have y m... () Differentiating both sides of equation () with respect to, we get dy d m dy Substituting the value of m in equation () we get y d dy or y 0 d which is free from the parameter m and hence this is the required differential equation. Eample 5 Form the differential equation representing the family of curves y a sin ( + b), where a, b are arbitrary constants. Solution We have y a sin ( + b)... () Differentiating both sides of equation () with respect to, successively we get dy d a cos ( + b)... () d y a sin ( + b)... () d Eliminating a and b from equations (), () and (), we get d y y d (4) which is free from the arbitrary constants a and b and hence this the required differential equation.

106 DIFFERENTIAL EQUATIONS 89 Eample 6 Form the differential equation representing the family of ellipses having foci on -ais and centre at the origin. Solution We know that the equation of said family of ellipses (see Fig 9.) is a y +... () b Fig 9. y dy Differentiating equation () with respect to, we get 0 a + b d y dy or d b a Differentiating both sides of equation () with respect to, we get... () dy y y d y d dy + d d 0 + d y dy dy or y y d d 0... () d which is the required differential equation. Eample 7 Form the differential equation of the family of circles touching the -ais at origin. Solution Let C denote the family of circles touching -ais at origin. Let (0, a) be the coordinates of the centre of any member of the family (see Fig 9.4). Therefore, equation of family C is + (y a) a or + y ay... () where, a is an arbitrary constant. Differentiating both sides of equation () with respect to,we get + y dy a dy d d X Y O Y Fig 9.4 X

107 90 MATHEMATICS or dy + y d dy a d or a dy + y d dy d Substituting the value of a from equation () in equation (), we get... () or dy + y d + y y dy d dy ( y dy + ) y + y d d dy or d y y This is the required differential equation of the given family of circles. Eample 8 Form the differential equation representing the family of parabolas having verte at origin and ais along positive direction of -ais. Solution Let P denote the family of above said parabolas (see Fig 9.5) and let (a, 0) be the focus of a member of the given family, where a is an arbitrary constant. Therefore, equation of family P is y 4a... () Differentiating both sides of equation () with respect to, we get dy y d 4a... () Substituting the value of 4a from equation () in equation (), we get or y dy y y ( ) d dy y 0 d which is the differential equation of the given family of parabolas. Fig 9.5

108 DIFFERENTIAL EQUATIONS 9 EXERCISE 9. In each of the Eercises to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. y. +. y a b a (b ). y a e + b e 4. y e (a + b) 5. y e (a cos + b sin ) 6. Form the differential equation of the family of circles touching the y-ais at origin. 7. Form the differential equation of the family of parabolas having verte at origin and ais along positive y-ais. 8. Form the differential equation of the family of ellipses having foci on y-ais and centre at origin. 9. Form the differential equation of the family of hyperbolas having foci on -ais and centre at origin. 0. Form the differential equation of the family of circles having centre on y-ais and radius units.. Which of the following differential equations has y c e + c e as the general solution? d y (A) y 0 d + d y (B) y 0 d d y (C) 0 d + d y (D) 0 d. Which of the following differential equations has y as one of its particular solution? d y dy (A) y d d + d y dy (B) y d + d + (C) dy y 0 d y d d + (D) dy y 0 d y d + d Methods of Solving First Order, First Degree Differential Equations In this section we shall discuss three methods of solving first order first degree differential equations Differential equations with variables separable A first order-first degree differential equation is of the form dy F(, y)... () d

109 9 MATHEMATICS If F (, y) can be epressed as a product g () h(y), where, g() is a function of and h(y) is a function of y, then the differential equation () is said to be of variable separable type. The differential equation () then has the form dy h (y). g()... () d If h (y) 0, separating the variables, () can be rewritten as dy g() d... () hy ( ) Integrating both sides of (), we get hy ( ) dy g( ) d... (4) Thus, (4) provides the solutions of given differential equation in the form H(y) G() + C Here, H (y) and G () are the anti derivatives of C is the arbitrary constant. hy ( ) Eample 9 Find the general solution of the differential equation Solution We have and g () respectively and dy +, (y ) d y dy d +... () y Separating the variables in equation (), we get ( y) dy ( + ) d... () Integrating both sides of equation (), we get ( ydy ) ( + ) d or y y or + y + 4y + C or + y + 4y + C 0, where C C which is the general solution of equation (). C

110 DIFFERENTIAL EQUATIONS 9 Eample 0 Find the general solution of the differential equation dy + y. d + Solution Since + y 0, therefore separating the variables, the given differential equation can be written as dy d + y + Integrating both sides of equation (), we get dy d + y + or tan y tan + C which is the general solution of equation (). Eample Find the particular solution of the differential equation that y, when () dy 4y given d Solution If y 0, the given differential equation can be written as dy 4 d... () y Integrating both sides of equation (), we get or dy y 4 d + C y or y C Substituting y and 0 in equation (), we get, C.... () Now substituting the value of C in equation (), we get the particular solution of the given differential equation as y +. Eample Find the equation of the curve passing through the point (, ) whose differential equation is dy ( + ) d ( 0).

111 94 MATHEMATICS Solution The given differential equation can be epressed as dy* + d* or dy + d Integrating both sides of equation (), we get... () dy + d or y + log + C... () Equation () represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (, ). Therefore substituting, y in equation (), we get C 0. Now substituting the value of C in equation () we get the equation of the required curve as y + log. Eample Find the equation of a curve passing through the point (, ), given that the slope of the tangent to the curve at any point (, y) is y. Solution We know that the slope of the tangent to a curve is given by dy d. dy so, d... () y Separating the variables, equation () can be written as y dy d... () Integrating both sides of equation (), we get or ydy d y + C... () * The notation dy due to Leibnitz is etremely fleible and useful in many calculation and formal d transformations, where, we can deal with symbols dy and d eactly as if they were ordinary numbers. By treating d and dy like separate entities, we can give neater epressions to many calculations. Refer: Introduction to Calculus and Analysis, volume-i page 7, By Richard Courant, Fritz John Spinger Verlog New York.

112 DIFFERENTIAL EQUATIONS 95 Substituting, y in equation (), we get C 5. Substituting the value of C in equation (), we get the equation of the required curve as y + 5 or y ( + 5) Eample 4 In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 000 double itself? Solution Let P be the principal at any time t. According to the given problem, dp dt 5 P 00 dp or P dt 0 separating the variables in equation (), we get dp dt P 0 Integrating both sides of equation (), we get t log P + C 0 0 or P e e t C... ()... () or 0 C P C e (where e C )... () Now P 000, when t 0 Substituting the values of P and t in (), we get C 000. Therefore, equation (), gives P 000 Let t years be the time required to double the principal. Then t t e 0 t e 0 t 0 log e EXERCISE 9.4 For each of the differential equations in Eercises to 0, find the general solution:. dy cos. d + cos dy 4 y ( y ) d < <

113 96 MATHEMATICS dy. y ( y ) d + 4. sec tan y d + sec y tan dy 0 5. (e + e ) dy (e e ) d y log y d dy 0 8. dy ( ) ( y ) d + + dy y d 5 5 dy 9. sin 0. e d tan y d + ( e ) sec y dy 0 For each of the differential equations in Eercises to 4, find a particular solution satisfying the given condition:.. ( ) dy d + ; y when 0 dy ( ) ; y 0 when d dy. cos a (a R); y when 0 d dy 4. ytan d ; y when 0 5. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y e sin. dy 6. For the differential equation y ( + )( y + ), find the solution curve d passing through the point (, ). 7. Find the equation of a curve passing through the point (0, ) given that at any point (, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the coordinate of the point. 8. At any point (, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point ( 4, ). Find the equation of the curve given that it passes through (, ). 9. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is units and after seconds it is 6 units. Find the radius of balloon after t seconds.

114 DIFFERENTIAL EQUATIONS In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 00 double itself in 0 years (log e 0.69).. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 000 is deposited with this bank, how much will it worth after 0 years (e ).. In a culture, the bacteria count is,00,000. The number is increased by 0% in hours. In how many hours will the count reach,00,000, if the rate of growth of bacteria is proportional to the number present?. The general solution of the differential equation (A) e + e y C (C) e + e y C 9.5. Homogeneous differential equations Consider the following functions in and y F (, y) y + y, F (, y) y, (B) e + e y C dy + y e is d (D) e + e y C y F (, y) cos, F (, y) sin + cos y 4 If we replace and y by λ and λy respectively in the above functions, for any nonzero constant λ, we get F (λ, λy) λ (y + y) λ F (, y) F (λ, λy) λ ( y) λ F (, y) λy y F (λ, λy) cos cos λ λ0 F (, y) F 4 (λ, λy) sin λ + cos λy λ n F 4 (, y), for any n N Here, we observe that the functions F, F, F can be written in the form F(λ, λy) λ n F (, y) but F can not be written in this form. This leads to the following 4 definition: A function F(, y) is said to be homogeneous function of degree n if F(λ, λy) λ n F(, y) for any nonzero constant λ. We note that in the above eamples, F, F, F are homogeneous functions of degree,, 0 respectively but F 4 is not a homogeneous function.

115 98 MATHEMATICS We also observe that F (, y) y y y + h or F (, y) y + y h y y F (, y) y y h or F (, y) y y h4 y y F (, y) 0 y cos 0 y h5 F 4 (, y) n y h6, for any n N or F 4 (, y) n y h7 y, for any n N Therefore, a function F (, y) is a homogeneous function of degree n if F(, y) n y n g or yh y A differential equation of the form dy F (, y) is said to be homogenous if d F(, y) is a homogenous function of degree zero. To solve a homogeneous differential equation of the type dy F (, y) d y g... () We make the substitution y v.... () Differentiating equation () with respect to, we get Substituting the value of dy d dy d dv v+... () d from equation () in equation (), we get

116 DIFFERENTIAL EQUATIONS 99 dv v+ g (v) d or dv d g (v) v... (4) Separating the variables in equation (4), we get dv d... (5) g () v v Integrating both sides of equation (5), we get dv d C g () v v +... (6) Equation (6) gives general solution (primitive) of the differential equation () when we replace v by y. d Note If the homogeneous differential equation is in the form F(, y) dy where, F (, y) is homogenous function of degree zero, then we make substitution v i.e., vy and we proceed further to find the general solution as discussed y d above by writing F( y, ) h. dy y Eample 5 Show that the differential equation ( y) dy + y is homogeneous d and solve it. Solution The given differential equation can be epressed as dy d + y... () y Let F(, y) Now F(λ, λy) + y y λ ( + y) 0 λ f ( y, ) λ( y)

117 400 MATHEMATICS Therefore, F(, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation. Alternatively, y dy + y g... () d y y R.H.S. of differential equation () is of the form g and so it is a homogeneous function of degree zero. Therefore, equation () is a homogeneous differential equation. To solve it we make the substitution y v... () Differentiating equation () with respect to, we get Substituting the value of y and dy d or or dy d dv v +... (4) d in equation () we get dv v+ + v d v dv + v v d v dv d v + v+ v v d or dv v + v+ Integrating both sides of equation (5), we get v v + v+ dv v + or dv log + C v + v+ d

118 DIFFERENTIAL EQUATIONS 40 v + or log C dv dv v v v + v+ or log v + v+ log C dv + v + + or or v + log v + v+. tan log + C v + log v + v+ + log tan + C (Why?) Replacing v by y, we get or or + y y y log log tan + C + y y y log + + tan + C or y+ log ( y + y+ ) tan + C y or log ( + y+ y ) tan + C which is the general solution of the differential equation () + y dy y Eample 6 Show that the differential equation cos ycos + d homogeneous and solve it. Solution The given differential equation can be written as dy d y ycos + y cos is... ()

119 40 MATHEMATICS dy It is a differential equation of the form F(, y) d. Here F(, y) Replacing by λ and y by λy, we get y ycos + y cos y λ [ ycos + ] 0 F(λ, λy) λ [F(, y)] y λ cos Thus, F(, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it we make the substitution y v... () Differentiating equation () with respect to, we get Substituting the value of y and dy d or or dy d dv v+... () d in equation (), we get dv v v+ cos v+ d cosv dv v cos v+ v d cos v dv d cosv or Therefore cos v dv d cosvdv d

120 or sin v log + log C or sin v log C Replacing v by y, we get sin y log C which is the general solution of the differential equation (). DIFFERENTIAL EQUATIONS 40 y y Eample 7 Show that the differential equation ye d+ y e dy 0is homogeneous and find its particular solution, given that, 0 when y. Solution The given differential equation can be written as y d dy e y y ye... () Let F(, y) ye e y y y λ e y 0 Then F(λ, λy) [F(, y)] λ y λ ye Thus, F(, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it, we make the substitution vy... () Differentiating equation () with respect to y, we get y d dy v+ y dv dy

121 404 MATHEMATICS Substituting the value of and d dy in equation (), we get dv v+ y ve dy v e v or or dv y ve v dy v e dv y dy v e v or e v dv dy y or or e v dv dy y e v log y + C and replacing v by y, we get y e + log y C... () Substituting 0 and y in equation (), we get e 0 + log C C Substituting the value of C in equation (), we get y e + log y which is the particular solution of the given differential equation. Eample 8 Show that the family of curves for which the slope of the tangent at any point (, y) on it is + y y, is given by y c. Solution We know that the slope of the tangent at any point on a curve is dy d. Therefore, dy d + y y

122 DIFFERENTIAL EQUATIONS 405 or dy d y + y... () Clearly, () is a homogenous differential equation. To solve it we make substitution y v Differentiating y v with respect to, we get dy d dv v+ d or dv v+ d + v v or dv d v v v v dv v or dv v d d v Therefore dv v d or log v log + log C or log (v ) () log C or (v ) ± C Replacing v by y, we get or y ± C (y ) ± C or y C

123 406 MATHEMATICS EXERCISE 9.5 In each of the Eercises to 0, show that the given differential equation is homogeneous and solve each of them.. ( + y) dy ( + y ) d. + y y. ( y) dy ( + y) d 0 4. ( y ) d + y dy 0 5. dy y y d + 6. dy y d + y d 7. y y y y cos ysin y d + ysin cos dy 8. dy y y+ sin 0 d y 9. yd+ log dy dy 0 y y 0. + e d+ e dy 0 y For each of the differential equations in Eercises from to 5, find the particular solution satisfying the given condition:. ( + y) dy + ( y) d 0; y when. dy + (y + y ) d 0; y when. y sin 0; π y d + dy y 4 when dy y y 4. + cosec 0 ; y 0 when d 5. dy y + y 0; y when d 6. A homogeneous differential equation of the from d h dy y can be solved by making the substitution. (A) y v (B) v y (C) vy (D) v

124 DIFFERENTIAL EQUATIONS Which of the following is a homogeneous differential equation? (A) (4 + 6y + 5) dy (y + + 4) d 0 (B) (y) d ( + y ) dy 0 (C) ( + y ) d + y dy 0 (D) y d + ( y y ) dy Linear differential equations A differential equation of the from dy Py d + Q where, P and Q are constants or functions of only, is known as a first order linear differential equation. Some eamples of the first order linear differential equation are dy y d + sin dy + y e d dy y + d log Another form of first order linear differential equation is d P dy + Q where, P and Q are constants or functions of y only. Some eamples of this type of differential equation are d dy + cos y d + y e y dy y To solve the first order linear differential equation of the type dy + Py Q... () d Multiply both sides of the equation by a function of say g () to get g() dy + P. (g()) y Q.g ()... () d

125 408 MATHEMATICS Choose g () in such a way that R.H.S. becomes a derivative of y. g (). i.e. g () dy d + P. g() y d d [y. g ()] or g () dy dy + P. g() y g() d d P.g () g () or P g ( ) g ( ) Integrating both sides with respect to, we get + y g () or Pd g ( ) d g( ) P d log (g ()) or g () P d e On multiplying the equation () by g() P d e, the L.H.S. becomes the derivative P d of some function of and y. This function g() e is called Integrating Factor (I.F.) of the given differential equation. Substituting the value of g () in equation (), we get Pd dy P d e + Pe y Q e d d Pd or ( ye ) d P d P d Qe Integrating both sides with respect to, we get Pd ( Q e ) y e P d or y Pd Pd ( Q ) d e e d+ C which is the general solution of the differential equation.

126 DIFFERENTIAL EQUATIONS 409 Steps involved to solve first order linear differential equation: dy (i) Write the given differential equation in the form Py Q d + where P, Q are constants or functions of only. (ii) Pd Find the Integrating Factor (I.F) e. (iii) Write the solution of the given differential equation as y (I.F) ( Q I.F) d + C d In case, the first order linear differential equation is in the form P Q dy +, P dy where, P and Q are constants or functions of y only. Then I.F e and the solution of the differential equation is given by. (I.F) ( Q I.F) dy + C Eample 9 Find the general solution of the differential equation Solution Given differential equation is of the form Therefore I.F dy Py Q d +, where P and Q cos e e d Multiplying both sides of equation by I.F, we get dy y cos d. dy e e y e cos d dy or ( ye ) e cos d On integrating both sides with respect to, we get ye e cos d+ C... () Let I e cos d e cos ( sin ) ( e ) d

127 40 MATHEMATICS cos e sin e d cos e sin ( e ) cos ( e ) d cos e + sin e cos e d or I e cos + sin e I or I (sin cos ) e (sin cos e ) or I Substituting the value of I in equation (), we get ye sin cos e + C or y sin cos + Ce which is the general solution of the given differential equation. dy Eample 0 Find the general solution of the differential equation y ( 0) d +. Solution The given differential equation is dy y d +... () Dividing both sides of equation () by, we get dy + y d dy which is a linear differential equation of the type Py Q d +, where d So I.F e e log e [ as e f ( )] Therefore, solution of the given equation is given by y. ( )( ) d+ C log d+ C log f ( ) or y + C 4 which is the general solution of the given differential equation. P and Q.

128 DIFFERENTIAL EQUATIONS 4 Eample Find the general solution of the differential equation y d ( + y ) dy 0. Solution The given differential equation can be written as d y dy y d This is a linear differential equation of the type P Q dy +, where P and y dy y log y log( y) Q y. Therefore I.F e e e y Hence, the solution of the given differential equation is or y ( y) dy+ C y y ( dy ) + C or y y + C or y + Cy which is a general solution of the given differential equation. Eample Find the particular solution of the differential equation given that y 0 when d ycot dy + + cot ( 0) π. dy Solution The given equation is a linear differential equation of the type Py Q d +, where P cot and Q + cot. Therefore I.F cot d log sin e e sin Hence, the solution of the differential equation is given by y. sin ( + cot ) sin d + C

129 4 MATHEMATICS or or y sin y sin sin d + cos d + C sin cos d + cos d + C or y sin sin cos d+ cos d+ C or y sin sin + C... () Substituting y 0 and π in equation (), we get 0 π π sin + C π or C 4 Substituting the value of C in equation (), we get y sin or y sin π 4 π (sin 0) 4sin which is the particular solution of the given differential equation. Eample Find the equation of a curve passing through the point (0, ). If the slope of the tangent to the curve at any point (, y) is equal to the sum of the coordinate (abscissa) and the product of the coordinate and y coordinate (ordinate) of that point. Solution We know that the slope of the tangent to the curve is dy d. Therefore, or dy d + y dy y d... () dy This is a linear differential equation of the type Py Q d +, where P and Q. Therefore, I. F d e e

130 Hence, the solution of equation is given by ( ) ye DIFFERENTIAL EQUATIONS 4 ( ) e d+ C... () Let I ( ) e d Let t, then d dt or d dt. Therefore, I edt e e t t Substituting the value of I in equation (), we get y e e +C or y + Ce... () Now () represents the equation of family of curves. But we are interested in finding a particular member of the family passing through (0, ). Substituting 0 and y in equation () we get + C. e 0 or C Substituting the value of C in equation (), we get y + e which is the equation of the required curve. EXERCISE 9.6 For each of the differential equations given in Eercises to, find the general solution: dy. y sin d +. dy π 4. + sec y tan 0 < d dy + y e. d dy 6. y log d ( + ) dy + y d cot d ( 0) 5. cos dy y + d dy y tan d + π 0 < dy log + y log d

131 44 MATHEMATICS dy 9. y ycot 0 ( 0) d ( + y) dy d. y d + ( y ) dy 0. dy ( + y ) y d ( y > 0). For each of the differential equations given in Eercises to 5, find a particular solution satisfying the given condition:. dy π + ytan sin ; y 0 when d 4. ( dy ) y ; y 0 when d + 5. dy π ycot sin ; y when d 6. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (, y) is equal to the sum of the coordinates of the point. 7. Find the equation of a curve passing through the point (0, ) given that the sum of the coordinates of any point on the curve eceeds the magnitude of the slope of the tangent to the curve at that point by dy The Integrating Factor of the differential equation y d is (A) e (B) e y (C) (D) 9. The Integrating Factor of the differential equation + y ay( < y < ) is dy (A) y (B) y ( y ) d (C) y Miscellaneous Eamples (D) y Eample 4 Verify that the function y c e a cos b + c e a sin b, where c, c are arbitrary constants is a solution of the differential equation d y dy ( a a b ) y 0 d d + +

132 DIFFERENTIAL EQUATIONS 45 Solution The given function is y e a [c cosb + c sinb]... () Differentiating both sides of equation () with respect to, we get dy d a [ sin cos ] [ cos sin ] e bc b+ bc b + c b+ c b e a dy or d a e [( bc + ac)cos b+ ( ac bc)sin b]... () Differentiating both sides of equation () with respect to, we get a d y d a + + e [( bc ac ) ( bsin b) ( ac bc ) ( bcos b)] [( bc ac ) cos b ( ac bc ) sin b] e. a a + + e [( a c abc b c )sin b ( a c abc b c )cos b] a Substituting the values of d y, dy d d and y in the given differential equation, we get a L.H.S. e [ a c abc b c )sin b + ( a c + abc b c )cos b] a ae [( bc + ac )cos b + ( ac bc )sin b] a + ( a + b ) e [ c cosb+ c sin b] ( ) a c abc b c a c abc a c b c sinb a e + ( ac+ abc bc abc ac+ ac+ bc)cosb a e [0 sinb+ 0cos b] e a 0 0 R.H.S. Hence, the given function is a solution of the given differential equation. Eample 5 Form the differential equation of the family of circles in the second quadrant and touching the coordinate aes. Solution Let C denote the family of circles in the second quadrant and touching the coordinate aes. Let ( a, a) be the coordinate of the centre of any member of this family (see Fig 9.6).

133 46 MATHEMATICS Equation representing the family C is ( + a) + (y a) a... () or + y + a ay + a 0... () Differentiating equation () with respect to, we get ( aa, ) Y dy dy + y a a d + d 0 X O X dy dy or + y a d d + yy or a y Substituting the value of a in equation (), we get Y Fig yy y + yy + yy + + y y y or [y + + y y ] + [y y y y y ] [ + y y ] or ( + y) y + [ + y] [ + y y ] or ( + y) [(y ) + ] [ + y y ] which is the differential equation representing the given family of circles. Eample 6 Find the particular solution of the differential equation log dy + 4y d given that y 0 when 0. Solution The given differential equation can be written as dy d e( + 4y) dy or d e. e 4y... () Separating the variables, we get dy e d 4 y e Therefore 4 y e dy e d

134 DIFFERENTIAL EQUATIONS 47 4 y e e or + C 4 or 4 e + e 4y + C 0... () Substituting 0 and y 0 in (), we get C 0 or C Substituting the value of C in equation (), we get 4 e + e 4y 7 0, which is a particular solution of the given differential equation. Eample 7 Solve the differential equation ( dy y d) y sin y (y d + dy) cos y. Solution The given differential equation can be written as or y y y y ysin cos dy y cos + y sin d dy d y y ycos + y sin y y y sin cos Dividing numerator and denominator on RHS by, we get dy d y y y y cos + sin y y y sin cos... () Clearly, equation () is a homogeneous differential equation of the form dy g y d. To solve it, we make the substitution y v... () dy or d dv v+ d

135 48 MATHEMATICS or or or Therefore or dv v+ d dv d vsin v cosv dv vcosv vsin v cos v dv vcosv tanvdv dv v vcosv+ v sin v vsin v cosv vcosv vsin v cosv d d d or log secv log v log + log C secv or log v log C (using () and ()) secv or v ± C... () Replacing v by y in equation (), we get y sec y ( ) C where, C ± C or sec y C y which is the general solution of the given differential equation. Eample 8 Solve the differential equation (tan y ) dy ( + y ) d. Solution The given differential equation can be written as d + dy + y tan + y y... ()

136 DIFFERENTIAL EQUATIONS 49 d Now () is a linear differential equation of the form P dy + Q, where, P + y and tan y Q. + y Therefore, I.F dy + y tan y e e Thus, the solution of the given differential equation is e tan y tan y tan y e dy+ C... () + y Let I tan y + y e tan y dy Substituting tan y t so that dy dt + y, we get t I tedt t e t. e t dt t e t e t e t (t ) or I tan y e (tan y ) Substituting the value of I in equation (), we get tan y tan y. e e (tan y ) + C or (tan y ) + C e tan y which is the general solution of the given differential equation. Miscellaneous Eercise on Chapter 9. For each of the differential equations given below, indicate its order and degree (if defined). (i) d y dy + 5 6y log d d (ii) dy dy 4 + 7y sin d d (iii) 4 d y d y sin 0 4 d d

137 40 MATHEMATICS. For each of the eercises given below, verify that the given function (implicit or eplicit) is a solution of the corresponding differential equation. (i) y a e + b e + : (ii) y e (a cos + b sin ) : d y dy y 0 d + d + d y dy y 0 d d + (iii) y sin : d y 9y 6cos 0 d + (iv) y log y : ( + y ) dy y 0 d. Form the differential equation representing the family of curves given by ( a) + y a, where a is an arbitrary constant. 4. Prove that y c ( + y ) is the general solution of differential equation ( y ) d (y y) dy, where c is a parameter. 5. Form the differential equation of the family of circles in the first quadrant which touch the coordinate aes. 6. Find the general solution of the differential equation dy y + 0. d dy y + y + 7. Show that the general solution of the differential equation + 0 is d + + given by ( + y + ) A ( y y), where A is parameter. π 8. Find the equation of the curve passing through the point 0, whose differential 4 equation is sin cos y d + cos sin y dy Find the particular solution of the differential equation ( + e ) dy + ( + y ) e d 0, given that y when y y Solve the differential equation ye d e y + dy( y 0).. Find a particular solution of the differential equation ( y) (d + dy) d dy, given that y, when 0. (Hint: put y t)

138 DIFFERENTIAL EQUATIONS 4. Solve the differential equation e y d ( 0). dy dy. Find a particular solution of the differential equation ycot d + 4 cosec ( 0), given that y 0 when π. 4. Find a particular solution of the differential equation ( + ) dy d e y, given that y 0 when The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 0, 000 in 999 and 5000 in the year 004, what will be the population of the village in 009? yd dy 6. The general solution of the differential equation 0 is y (A) y C (B) Cy (C) y C (D) y C d 7. The general solution of a differential equation of the type P Q dy + is (A) ( ) P dy P dy Q ye e dy+ C (B) ( ) P d P d y. e Qe d+ C (C) ( ) P dy P dy Q e e dy+ C (D) ( ) P d P d Q e e d+ C 8. The general solution of the differential equation e dy + (y e + ) d 0 is (A) e y + C (B) e y + y C (C) y e + C (D) y e y + C

139 4 MATHEMATICS Summary An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation. Order of a differential equation is the order of the highest order derivative occurring in the differential equation. Degree of a differential equation is defined if it is a polynomial equation in its derivatives. Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it. A function which satisfies the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution. To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants. Variable separable method is used to solve such an equation in which variables can be separated completely i.e. terms containing y should remain with dy and terms containing should remain with d. A differential equation which can be epressed in the form dy d f ( y, ) or gy (, ) where, f (, y) and g(, y) are homogenous d dy functions of degree zero is called a homogeneous differential equation. dy A differential equation of the form +Py Q, where P and Q are constants d or functions of only is called a first order linear differential equation. Historical Note One of the principal languages of Science is that of differential equations. Interestingly, the date of birth of differential equations is taken to be November,,675, when Gottfried Wilthelm Freiherr Leibnitz (646-76) first put in black and white the identity ydy y, thereby introducing both the symbols and dy.

140 DIFFERENTIAL EQUATIONS 4 Leibnitz was actually interested in the problem of finding a curve whose tangents were prescribed. This led him to discover the method of separation of variables 69. A year later he formulated the method of solving the homogeneous differential equations of the first order. He went further in a very short time to the discovery of the method of solving a linear differential equation of the first-order. How surprising is it that all these methods came from a single man and that too within 5 years of the birth of differential equations! In the old days, what we now call the solution of a differential equation, was used to be referred to as integral of the differential equation, the word being coined by James Bernoulli ( ) in 690. The word solution was first used by Joseph Louis Lagrange (76-8) in 774, which was almost hundred years since the birth of differential equations. It was Jules Henri Poincare (854-9) who strongly advocated the use of the word solution and thus the word solution has found its deserved place in modern terminology. The name of the method of separation of variables is due to John Bernoulli ( ), a younger brother of James Bernoulli. Application to geometric problems were also considered. It was again John Bernoulli who first brought into light the intricate nature of differential equations. In a letter to Leibnitz, dated May 0, 75, he revealed the solutions of the differential equation y y, which led to three types of curves, viz., parabolas, hyperbolas and a class of cubic curves. This shows how varied the solutions of such innocent looking differential equation can be. From the second half of the twentieth century attention has been drawn to the investigation of this complicated nature of the solutions of differential equations, under the heading qualitative analysis of differential equations. Now-a-days, this has acquired prime importance being absolutely necessary in almost all investigations.

141 44 MATHEMATICS Chapter 0 VECTOR ALGEBRA In most sciences one generation tears down what another has built and what one has established another undoes. In Mathematics alone each generation builds a new story to the old structure. HERMAN HANKEL 0. Introduction In our day to day life, we come across many queries such as What is your height? How should a football player hit the ball to give a pass to another player of his team? Observe that a possible answer to the first query may be.6 meters, a quantity that involves only one value (magnitude) which is a real number. Such quantities are called scalars. However, an answer to the second query is a quantity (called force) which involves muscular strength (magnitude) and direction (in which another player is positioned). Such quantities are called vectors. In mathematics, physics and engineering, we frequently come across with both types of W.R. Hamilton quantities, namely, scalar quantities such as length, mass, ( ) time, distance, speed, area, volume, temperature, work, money, voltage, density, resistance etc. and vector quantities like displacement, velocity, acceleration, force, weight, momentum, electric field intensity etc. In this chapter, we will study some of the basic concepts about vectors, various operations on vectors, and their algebraic and geometric properties. These two type of properties, when considered together give a full realisation to the concept of vectors, and lead to their vital applicability in various areas as mentioned above. 0. Some Basic Concepts Let l be any straight line in plane or three dimensional space. This line can be given two directions by means of arrowheads. A line with one of these directions prescribed is called a directed line (Fig 0. (i), (ii)).

142 VECTOR ALGEBRA 45 Fig 0. Now observe that if we restrict the line l to the line segment AB, then a magnitude is prescribed on the line l with one of the two directions, so that we obtain a directed line segment (Fig 0.(iii)). Thus, a directed line segment has magnitude as well as direction. Definition A quantity that has magnitude as well as direction is called a vector. Notice that a directed line segment is a vector (Fig 0.(iii)), denoted as AB or simply as a, and read as vector AB or vector a. The point A from where the vector AB starts is called its initial point, and the point B where it ends is called its terminal point. The distance between initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted as AB, or a, or a. The arrow indicates the direction of the vector. Note Since the length is never negative, the notation a < 0 has no meaning. Position Vector From Class XI, recall the three dimensional right handed rectangular coordinate system (Fig 0.(i)). Consider a point P in space, having coordinates (, y, z) with respect to the origin O(0, 0, 0). Then, the vector OP having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O. Using distance formula (from Class XI), the magnitude of OP (or r ) is given by OP + y + z In practice, the position vectors of points A, B, C, etc., with respect to the origin O are denoted by a, b, c, etc., respectively (Fig 0. (ii)).

143 46 MATHEMATICS Fig 0. Direction Cosines Consider the position vector OP( or r ) of a point P(, y, z) as in Fig 0.. The angles α, β, γ made by the vector r with the positive directions of, y and z-aes respectively, are called its direction angles. The cosine values of these angles, i.e., cos α, cosβ and cos γ are called direction cosines of the vector r, and usually denoted by l, m and n, respectively. Z C z O r y P(,y,z ) B Y P X A O 90 A Fig 0. X From Fig 0., one may note that the triangle OAP is right angled, and in it, we have cos α ( r stands for r ). Similarly, from the right angled triangles OBP and r OCP, we may write cos β y z and cos r γ r. Thus, the coordinates of the point P may also be epressed as (lr, mr,nr). The numbers lr, mr and nr, proportional to the direction cosines are called as direction ratios of vector r, and denoted as a, b and c, respectively.

144 VECTOR ALGEBRA 47 Note One may note that l + m + n but a + b + c, in general. 0. Types of Vectors Zero Vector A vector whose initial and terminal points coincide, is called a zero vector (or null vector), and denoted as 0. Zero vector can not be assigned a definite direction as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any direction. The vectors AA, BB represent the zero vector, Unit Vector A vector whose magnitude is unity (i.e., unit) is called a unit vector. The unit vector in the direction of a given vector a is denoted by â. Coinitial Vectors Two or more vectors having the same initial point are called coinitial vectors. Collinear Vectors Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions. Equal Vectors Two vectors a and b are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written as a b. Negative of a Vector A vector whose magnitude is the same as that of a given vector (say, AB ), but direction is opposite to that of it, is called negative of the given vector. For eample, vector BA is negative of the vector AB, and written as BA AB. Remark The vectors defined above are such that any of them may be subject to its parallel displacement without changing its magnitude and direction. Such vectors are called free vectors. Throughout this chapter, we will be dealing with free vectors only. Eample Represent graphically a displacement of 40 km, 0 west of south. Solution The vector OP represents the required displacement (Fig 0.4). Eample Classify the following measures as scalars and vectors. (i) 5 seconds (ii) 000 cm Fig 0.4

145 48 MATHEMATICS Solution (iii) 0 Newton (iv) 0 km/hr (v) 0 g/cm (vi) 0 m/s towards north (i) Time-scalar (ii) Volume-scalar (iii) Force-vector (iv) Speed-scalar (v) Density-scalar (vi) Velocity-vector Eample In Fig 0.5, which of the vectors are: (i) Collinear (ii) Equal (iii) Coinitial Solution (i) Collinear vectors : a, c and d. (ii) Equal vectors : a and c. (iii) Coinitial vectors : b, c and d. EXERCISE 0. Fig 0.5. Represent graphically a displacement of 40 km, 0 east of north.. Classify the following measures as scalars and vectors. (i) 0 kg (ii) meters north-west (iii) 40 (iv) 40 watt (v) 0 9 coulomb (vi) 0 m/s. Classify the following as scalar and vector quantities. (i) time period (ii) distance (iii) force (iv) velocity (v) work done 4. In Fig 0.6 (a square), identify the following vectors. (i) Coinitial (ii) Equal (iii) Collinear but not equal 5. Answer the following as true or false. (i) a and a are collinear. (ii) Two collinear vectors are always equal in magnitude. Fig 0.6 (iii) Two vectors having same magnitude are collinear. (iv) Two collinear vectors having the same magnitude are equal.

146 VECTOR ALGEBRA Addition of Vectors A vector AB simply means the displacement from a point A to the point B. Now consider a situation that a girl moves from A to B and then from B to C (Fig 0.7). The net displacement made by the girl from point A to the point C, is given by the vector AC and epressed as AC AB + BC This is known as the triangle law of vector addition. Fig 0.7 In general, if we have two vectors a and b (Fig 0.8 (i)), then to add them, they are positioned so that the initial point of one coincides with the terminal point of the other (Fig 0.8(ii)). b a A a + b a (i) (ii) (iii) C Fig 0.8 For eample, in Fig 0.8 (ii), we have shifted vector b without changing its magnitude and direction, so that it s initial point coincides with the terminal point of a. Then, the vector a+ b, represented by the third side AC of the triangle ABC, gives us the sum (or resultant) of the vectors a and b i.e., in triangle ABC (Fig 0.8 (ii)), we have AB + BC AC Now again, since AC CA, from the above equation, we have AB + BC + CA AA 0 This means that when the sides of a triangle are taken in order, it leads to zero resultant as the initial and terminal points get coincided (Fig 0.8(iii)). C b B A a b a C b B b

147 40 MATHEMATICS Now, construct a vector BC so that its magnitude is same as the vector BC, but the direction opposite to that of it (Fig 0.8 (iii)), i.e., BC BC Then, on applying triangle law from the Fig 0.8 (iii), we have AC AB + BC AB + ( BC) a b The vector AC is said to represent the difference of a and b. Now, consider a boat in a river going from one bank of the river to the other in a direction perpendicular to the flow of the river. Then, it is acted upon by two velocity vectors one is the velocity imparted to the boat by its engine and other one is the velocity of the flow of river water. Under the simultaneous influence of these two velocities, the boat in actual starts travelling with a different velocity. To have a precise idea about the effective speed and direction (i.e., the resultant velocity) of the boat, we have the following law of vector addition. If we have two vectors aand b represented by the two adjacent sides of a parallelogram in magnitude and direction (Fig 0.9), then their sum a + b is represented in magnitude and direction by the diagonal of the parallelogram through their common point. This is known as Fig 0.9 the parallelogram law of vector addition. Note From Fig 0.9, using the triangle law, one may note that OA + AC OC or OA + OB OC (since AC OB ) which is parallelogram law. Thus, we may say that the two laws of vector addition are equivalent to each other. Properties of vector addition Property For any two vectors a a + b and b, b + a (Commutative property)

148 Proof Consider the parallelogram ABCD (Fig 0.0). Let AB aand BC b, then using the triangle law, from triangle ABC, we have AC a + b Now, since the opposite sides of a parallelogram are equal and parallel, from Fig0.0, we have, AD BC b and DC AB a. Again using triangle law, from triangle ADC, we have AC AD + DC b + a Hence a + b b + a Property For any three vectors ab, andc VECTOR ALGEBRA 4 Fig 0.0 ( a + b) + c a+ ( b + c ) (Associative property) Proof Let the vectors ab, andc be represented by PQ, QR and RS, respectively, as shown in Fig 0.(i) and (ii). Fig 0. Then a + b PQ + QR PR and b + c QR + RS QS So ( a + b) + c PR + RS PS

149 4 MATHEMATICS and a + ( b + c ) Hence ( a + b) + c a+ ( b + c ) Remark The associative property of vector addition enables us to write the sum of three vectors a, b, c as a + b + c without using brackets. Note that for any vector a, we have a a a Here, the zero vector 0 is called the additive identity for the vector addition. 0.5 Multiplication of a Vector by a Scalar Let a be a given vector and λ a scalar. Then the product of the vector a by the scalar λ, denoted as λ a, is called the multiplication of vector a by the scalar λ. Note that, λ a is also a vector, collinear to the vector a. The vector λ a has the direction same (or opposite) to that of vector a according as the value of λ is positive (or negative). Also, the magnitude of vector λ a is λ times the magnitude of the vector a, i.e., λa λ a A geometric visualisation of multiplication of a vector by a scalar is given in Fig 0.. a a a a a Fig 0. When λ, then λ a a, which is a vector having magnitude equal to the magnitude of a and direction opposite to that of the direction of a. The vector a is called the negative (or additive inverse) of vector a and we always have a+ ( a) ( a) + a 0 Also, if λ, provided a 0, i.e. a λ a λ a a a a is not a null vector, then

150 VECTOR ALGEBRA 4 So, λ a represents the unit vector in the direction of a. We write it as â a a Note For any scalar k, k Components of a vector Let us take the points A(, 0, 0), B(0,, 0) and C(0, 0, ) on the -ais, y-ais and z-ais, respectively. Then, clearly OA, OB and OC The vectors OA, OB and OC, each having magnitude, are called unit vectors along the aes OX, OY and OZ, respectively, and denoted by iˆ, ˆj and k ˆ, respectively (Fig 0.). Fig 0. Now, consider the position vector OP of a point P(, y, z) as in Fig 0.4. Let P be the foot of the perpendicular from P on the plane XOY. We, thus, see that P P is Fig 0.4 parallel to z-ais. As iˆ, ˆj and k ˆ are the unit vectors along the, y and z-aes, respectively, and by the definition of the coordinates of P, we have PP ˆ OR zk. Similarly, QP ˆ OS yj and OQ iˆ.

151 44 MATHEMATICS Therefore, it follows that OP OQ + QP ˆ ˆ i + yj and OP OP ˆ ˆ ˆ + PP i + yj+ zk Hence, the position vector of P with reference to O is given by OP (or r ) iˆ+ yj ˆ+ zkˆ This form of any vector is called its component form. Here,, y and z are called as the scalar components of r, and iˆ, yj ˆ and zk ˆ are called the vector components of r along the respective aes. Sometimes, y and z are also termed as rectangular components. The length of any vector r iˆ+ yj ˆ+ zkˆ, is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle OQP (Fig 0.4) OP OQ + QP + y, and in the right angle triangle OP P, we have OP OP + PP ( + y ) + z Hence, the length of any vector r iˆ+ yj ˆ+ zkˆ is given by r iˆ+ yj ˆ+ zkˆ + y + z If a and b are any two vectors given in the component form ai ˆ + a ˆ+ j ak ˆ and bi ˆ + b ˆ j+ bk ˆ, respectively, then (i) the sum (or resultant) of the vectors a and b is given by a + b ( a+ b ˆ ˆ ˆ ) i + ( a + b) j+ ( a + b) k (ii) the difference of the vector a and b is given by a b ( a b ˆ ˆ ˆ ) i + ( a b) j+ ( a b) k (iii) the vectors a and b are equal if and only if (iv) a b, a b and a b the multiplication of vector a by any scalar λ is given by λa ( λ a ˆ ˆ ˆ ) i + ( λ a) j+ ( λa) k

152 VECTOR ALGEBRA 45 The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws: Let a and b be any two vectors, and k and m be any scalars. Then (i) ka + ma ( k + m) a (ii) kma ( ) ( kma ) (iii) ka ( + b) ka+ kb Remarks (i) One may observe that whatever be the value of λ, the vector λa is always collinear to the vector a. In fact, two vectors a and b are collinear if and only if there eists a nonzero scalar λ such that b λa. If the vectors a and b are given in the component form, i.e. ˆ ˆ a ai + aj+ aˆ k and b bi ˆ ˆ + bj+ bˆ k, then the two vectors are collinear if and only if bi+ b j+ bk λ ( ai ˆ ˆ + aj+ ak ˆ ) ˆ ˆ ˆ bi ˆ + b ˆ j+ bk ˆ ( λ a ˆ ˆ ˆ ) i + ( λ a) j+ ( λa) k b a λ, b λ a, b λa b a b b λ a a (ii) If a ai ˆ ˆ + aj+ aˆ k, then a, a, a are also called direction ratios of a. (iii) In case if it is given that l, m, n are direction cosines of a vector, then liˆ+ mj ˆ+ nkˆ (cos α ) iˆ+ (cos β ) ˆj+ (cos γ ) kˆ is the unit vector in the direction of that vector, where α, β and γ are the angles which the vector makes with, y and z aes respectively. Eample 4 Find the values of, y and z so that the vectors b ˆ i + yj ˆ+ kˆ are equal. a iˆ+ ˆj+ zkˆ and Solution Note that two vectors are equal if and only if their corresponding components are equal. Thus, the given vectors a and b will be equal if and only if, y, z

153 46 MATHEMATICS Eample 5 Let a iˆ+ ˆj and b iˆ+ ˆj. Is a b? Are the vectors a and b equal? Solution We have a + 5 and b + 5 So, a b. But, the two vectors are not equal since their corresponding components are distinct. Eample 6 Find unit vector in the direction of vector a iˆ+ ˆj+ kˆ Solution The unit vector in the direction of a vector a is given by aˆ a. a Now a Therefore aˆ (iˆ+ ˆj+ kˆ ) iˆ+ ˆj+ kˆ Eample 7 Find a vector in the direction of vector a iˆ ˆj that has magnitude 7 units. Solution The unit vector in the direction of the given vector a is aˆ a ( iˆ ˆj) iˆ ˆj a Therefore, the vector having magnitude equal to 7 and in the direction of a is 7a 7 i j ˆ 4 i 5 5 ˆj Eample 8 Find the unit vector in the direction of the sum of the vectors, a iˆ+ ˆj 5kˆ and b iˆ+ ˆj+ kˆ. Solution The sum of the given vectors is a+ b( c,say)4iˆ+ ˆj kˆ and c ( ) 9

154 Thus, the required unit vector is ˆ ˆ ˆ 4 ˆ ˆ cˆ c (4i + j k) i + j kˆ c VECTOR ALGEBRA 47 Eample 9 Write the direction ratio s of the vector a iˆ+ ˆj kˆ and hence calculate its direction cosines. Solution Note that the direction ratio s a, b, c of a vector r iˆ+ yj ˆ+ zkˆ are just the respective components, y and z of the vector. So, for the given vector, we have a, b and c. Further, if l, m and n are the direction cosines of the given vector, then a b c l, m, n as r 6 r 6 r 6 r 6 Thus, the direction cosines are,, Vector joining two points If P (, y, z ) and P (, y, z ) are any two points, then the vector joining P and P is the vector PP (Fig 0.5). Joining the points P and P with the origin O, and applying triangle law, from the triangle OP P, we have OP+ PP OP. Using the properties of vector addition, the above equation becomes PP OP OP i.e. PP ( iˆ+ y ˆ ˆ ˆ ˆ ˆ j+ zk) ( i + yj+ zk) ( ˆ ˆ ˆ ) i + ( y y) j+ ( z z) k The magnitude of vector PP is given by Fig 0.5 ( ) + ( y y ) + ( z z ) PP

155 48 MATHEMATICS Eample 0 Find the vector joining the points P(,, 0) and Q(,, 4) directed from P to Q. Solution Since the vector is to be directed from P to Q, clearly P is the initial point and Q is the terminal point. So, the required vector joining P and Q is the vector PQ, given by PQ ( ) iˆ+ ( ) ˆj+ ( 4 0) kˆ i.e. PQ iˆ 5ˆj 4 kˆ Section formula Let P and Q be two points represented by the position vectors OP and OQ, respectively, with respect to the origin O. Then the line segment joining the points P and Q may be divided by a third point, say R, in two ways internally (Fig 0.6) and eternally (Fig 0.7). Here, we intend to find the position vector OR for the point R with respect to the origin O. We take the two cases one by one. Case I When R divides PQ internally (Fig 0.6). If R divides PQ such that m RQ n PR Fig 0.6, where m and n are positive scalars, we say that the point R divides PQ internally in the ratio of m : n. Now from triangles ORQ and OPR, we have RQ OQ OR b r and PR OR OP r a, Therefore, we have m( b r) n( r a) (Why?) or r mb + na (on simplification) m+ n Hence, the position vector of the point R which divides P and Q internally in the ratio of m : n is given by mb + na OR m+ n

156 VECTOR ALGEBRA 49 Case II When R divides PQ eternally (Fig 0.7). We leave it to the reader as an eercise to verify that the position vector of the point R which divides the line segment PQ eternally in the ratio PR m m : n i.e. is given by QR n mb na OR Fig 0.7 m n Remark If R is the midpoint of PQ, then m n. And therefore, from Case I, the midpoint R of PQ, will have its position vector as a+ b OR Eample Consider two points P and Q with position vectors OP a b and OQ a+ b. Find the position vector of a point R which divides the line joining P and Q in the ratio :, (i) internally, and (ii) eternally. Solution (i) The position vector of the point R dividing the join of P and Q internally in the ratio : is ( OR a b ) ( a b + + ) 5 a + (ii) The position vector of the point R dividing the join of P and Q eternally in the ratio : is ( a+ b) (a b) OR 4b a Eample Show that the points A( iˆ ˆj+ kˆ),b( iˆ ˆj 5),C( kˆ iˆ 4j 4) kˆ are the vertices of a right angled triangle. Solution We have AB BC and CA ( ) iˆ+ ( + ) ˆj+ ( 5 ) kˆ iˆ ˆj 6kˆ ( ) iˆ+ ( 4+ ) ˆj+ ( 4+ 5) kˆ iˆ ˆj+ kˆ ( ) iˆ+ ( + 4) ˆj+ ( + 4) kˆ iˆ+ ˆj+ 5kˆ

157 440 MATHEMATICS Further, note that AB BC + CA Hence, the triangle is a right angled triangle. EXERCISE 0.. Compute the magnitude of the following vectors: ˆ ˆ ˆ ˆ ˆ ˆ ˆ a i + j+ k; b i 7j k; c i + j kˆ. Write two different vectors having same magnitude.. Write two different vectors having same direction. 4. Find the values of and y so that the vectors iˆ+ ˆj and iˆ+ yj ˆ are equal. 5. Find the scalar and vector components of the vector with initial point (, ) and terminal point ( 5, 7). 6. Find the sum of the vectors a iˆ ˆj+ kˆ, b iˆ+ 4ˆj+ 5kˆ and c iˆ 6 ˆj 7kˆ. 7. Find the unit vector in the direction of the vector a iˆ+ ˆj+ kˆ. 8. Find the unit vector in the direction of vector PQ, where P and Q are the points (,, ) and (4, 5, 6), respectively. 9. For given vectors, a iˆ ˆj+ kˆ and b iˆ+ ˆj kˆ, find the unit vector in the direction of the vector a + b. 0. Find a vector in the direction of vector 5iˆ ˆj+ kˆ which has magnitude 8 units.. Show that the vectors iˆ ˆj+ 4 kˆ and 4iˆ+ 6ˆj 8kˆ are collinear.. Find the direction cosines of the vector iˆ+ ˆj+ kˆ.. Find the direction cosines of the vector joining the points A (,, ) and B(,, ), directed from A to B. 4. Show that the vector iˆ+ ˆj+ kˆ is equally inclined to the aes OX, OY and OZ. 5. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are iˆ+ ˆj kˆ and iˆ+ ˆj+ kˆ respectively, in the ratio : (i) internally (ii) eternally

158 VECTOR ALGEBRA Find the position vector of the mid point of the vector joining the points P(,, 4) and Q(4,, ). 7. Show that the points A, B and C with position vectors, a iˆ 4ˆj 4 kˆ, b ˆ i ˆj+ kˆ and c iˆ ˆj 5kˆ, respectively form the vertices of a right angled triangle. 8. In triangle ABC (Fig 0.8), which of the following is not true: (A) AB + BC + CA 0 (B) AB + BC AC 0 (C) AB + BC CA 0 (D) Fig 0.8 AB CB+ CA 0 9. If a and b are two collinear vectors, then which of the following are incorrect: (A) b λa, for some scalar λ (B) a ± b (C) the respective components of a and b are proportional (D) both the vectors a and b have same direction, but different magnitudes. 0.6 Product of Two Vectors So far we have studied about addition and subtraction of vectors. An other algebraic operation which we intend to discuss regarding vectors is their product. We may recall that product of two numbers is a number, product of two matrices is again a matri. But in case of functions, we may multiply them in two ways, namely, multiplication of two functions pointwise and composition of two functions. Similarly, multiplication of two vectors is also defined in two ways, namely, scalar (or dot) product where the result is a scalar, and vector (or cross) product where the result is a vector. Based upon these two types of products for vectors, they have found various applications in geometry, mechanics and engineering. In this section, we will discuss these two types of products Scalar (or dot) product of two vectors Definition The scalar product of two nonzero vectors a and b, denoted by a b, is

159 44 MATHEMATICS defined as a b a b cos θ, where, θ is the angle between a and b, 0 θ π (Fig 0.9). If either a 0orb 0, then θ is not defined, and in this case, Fig 0.9 we define a b 0 Observations. a b is a real number.. Let a and b be two nonzero vectors, then a b 0 if and only if a and b are perpendicular to each other. i.e. a b 0 a b. If θ 0, then a b a b In particular, a a a, as θ in this case is If θ π, then a b a b In particular, a ( a ) a, as θ in this case is π. 5. In view of the Observations and, for mutually perpendicular unit vectors iˆ, ˆj and k ˆ, we have iˆ iˆ ˆj ˆj k ˆ k ˆ, iˆ ˆj ˆj kˆ k ˆ i ˆ 0 6. The angle between two nonzero vectors a and b is given by a b ab. cos θ, or θ cos a b a b 7. The scalar product is commutative. i.e. ab b a (Why?) Two important properties of scalar product Property (Distributivity of scalar product over addition) Let a, b any three vectors, then a ( b + c) a b + a c and c be

160 VECTOR ALGEBRA 44 Property Let a and b be any two vectors, and λ be any scalar. Then ( λa) b ( λa) b λ( a b) a ( λb) If two vectors a and b are given in component form as ai ˆ+ a ˆj + ak ˆ and ˆ ˆ ˆ bi+ b j+ bk, then their scalar product is given as ab ( aiˆ+ a ˆj + akˆ) ( biˆ+ b ˆj + bkˆ) Thus aiˆ ( biˆ+ b ˆj + bkˆ) + a ˆj ( biˆ+ b ˆj + b kˆ) + akˆ ( bi ˆ ˆ+ b ˆj + bk ) ab( iˆ iˆ) + ab ( iˆ ˆj) + ab ( iˆ kˆ) + a b( ˆj iˆ) + a b ( ˆj ˆj) + a b ( ˆj kˆ) + ab ( kˆ iˆ) + ab ˆ ˆ ˆ ˆ ( k j) + ab ( k k) (Using the above Properties and ) a b + a b + a b (Using Observation 5) a b ab + ab+ ab 0.6. Projection of a vector on a line Suppose a vector AB makes an angle θ with a given directed line l (say), in the anticlockwise direction (Fig 0.0). Then the projection of AB on l is a vector p (say) with magnitude AB cosθ, and the direction of p being the same (or opposite) to that of the line l, depending upon whether cos θ is positive or negative. The vector p A a (i) B θ p C 0 0 (0 < θ < 90 ) l B θ C A p 0 0 (90 < θ < 80 ) a (ii) l C p A θ l θ A p C l a B 0 0 (80 < θ < 70 ) (iii) Fig 0.0 a B 0 0 (70 < θ < 60 ) (iv)

161 444 MATHEMATICS is called the projection vector, and its magnitude p is simply called as the projection of the vector AB on the directed line l. For eample, in each of the following figures (Fig 0.0(i) to (iv)), projection vector of AB along the line l is vector AC. Observations. If ˆp is the unit vector along a line l, then the projection of a vector a on the line l is given by a pˆ.. Projection of a vector a on other vector b, is given by b a b ˆ, or a, or ( a b) b b. If θ 0, then the projection vector of AB will be AB itself and if θ π, then the projection vector of AB will be BA. 4. If π θ or π θ, then the projection vector of AB will be zero vector. Remark If α, β and γ are the direction angles of vector a ai ˆ ˆ + aj+ aˆ k, then its direction cosines may be given as a iˆ a a a cos α, cos, and cos ˆ β γ a i a a a Also, note that a cos α, a cos β and a cosγ are respectively the projections of a along OX, OY and OZ. i.e., the scalar components a, a and a of the vector a, are precisely the projections of a along -ais, y-ais and z-ais, respectively. Further, if a is a unit vector, then it may be epressed in terms of its direction cosines as a cosα iˆ+ cosβ ˆj+ cos γkˆ Eample Find the angle between two vectors a and b with magnitudes and respectively and when a b. Solution Given a b, a and b. We have a b π θ cos cos a b

162 VECTOR ALGEBRA 445 Eample 4 Find angle θ between the vectors a iˆ+ ˆj kˆ and b iˆ ˆj+ kˆ. Solution The angle θ between two vectors a and b is given by a b cosθ a b Now a b ( iˆ+ ˆj kˆ) ( iˆ ˆj+ kˆ). Therefore, we have cosθ hence the required angle is θ cos Eample 5 If a 5iˆ ˆj kˆ and b iˆ+ ˆj 5kˆ, then show that the vectors a+ b and a b are perpendicular. Solution We know that two nonzero vectors are perpendicular if their scalar product is zero. Here a+ b (5iˆ ˆj kˆ) + ( iˆ+ ˆj 5 kˆ) 6iˆ+ ˆj 8kˆ and a b (5iˆ ˆj kˆ) ( iˆ+ ˆj 5 kˆ) 4iˆ 4ˆj+ kˆ So ( a+ b) ( a b) (6iˆ+ ˆj 8 kˆ) (4iˆ 4ˆj+ kˆ) Hence a+ b and a b are perpendicular vectors. Eample 6 Find the projection of the vector a iˆ+ ˆj+ kˆ on the vector b iˆ+ ˆj+ kˆ. Solution The projection of vector a on the vector b is given by ( a b ( + + ) 0 5 ) 6 b () + () + () 6 Eample 7 Find a b, if two vectors a and b are such that a, b and a b 4. Solution We have a b ( a b) ( a b) aa. a b b a+ b b

163 446 MATHEMATICS a ( a b) + b () (4) + () Therefore a b 5 Eample 8 If a is a unit vector and ( a ) ( + a ) 8, then find. Solution Since a is a unit vector, a. Also, ( a ) ( + a ) 8 or + a a a a 8 or 8 i.e. 9 Therefore (as magnitude of a vector is non negative). Eample 9 For any two vectors a and b, we always have a b a b (Cauchy- Schwartz inequality). Solution The inequality holds trivially when either a 0 or b 0. Actually, in such a situation we have a b 0 a b. So, let us assume that a 0 b. Then, we have a b cos θ a b Therefore a b a b Eample 0 For any two vectors a and b, we always have a+ b a + b (triangle inequality). a a + b Solution The inequality holds trivially in case either A B a 0orb 0 (How?). So, let a 0 b. Then, a+ b Fig 0. ( a+ b) ( a+ b) ( a+ b) a a + a b + b a + b b a + a b + b (scalar product is commutative) a + a b + b (since R ) a + a b + b (from Eample 9) ( a + b ) b C

164 VECTOR ALGEBRA 447 Hence a b + a + b Remark If the equality holds in triangle inequality (in the above Eample 0), i.e. a+ b a + b, then AC AB + BC showing that the points A, B and C are collinear. Eample Show that the points A( iˆ+ ˆj+ 5 kˆ), B( iˆ+ ˆj+ kˆ) and C(7 iˆ kˆ ) are collinear. Solution We have AB ( + ) iˆ+ ( ) ˆj+ ( 5) kˆ iˆ ˆj kˆ, BC (7 ) iˆ+ (0 ) ˆj+ ( ) k ˆ 6iˆ ˆj 4k ˆ, AC (7 + ) iˆ+ (0 ) ˆj+ ( 5) k ˆ 9iˆ ˆj 6 k ˆ AB 4, BC 4 and AC 4 Therefore AC AB + BC Hence the points A, B and C are collinear. Note In Eample, one may note that although AB + BC + CA 0 but the points A, B and C do not form the vertices of a triangle. EXERCISE 0.. Find the angle between two vectors a and b with magnitudes and, respectively having a b 6.. Find the angle between the vectors ˆ iˆ ˆj+ k and iˆ ˆj+ kˆ. Find the projection of the vector iˆ ˆj on the vector iˆ+ ˆj. 4. Find the projection of the vector iˆ+ ˆj+ 7kˆ on the vector 7iˆ ˆj+ 8kˆ. 5. Show that each of the given three vectors is a unit vector: ˆ ˆ (iˆ+ ˆj+ 6 k), (iˆ 6 ˆj+ k), (6iˆ+ ˆj kˆ) Also, show that they are mutually perpendicular to each other.

165 448 MATHEMATICS 6. Find a and b, if ( a+ b) ( a b) 8 and a 8 b. 7. Evaluate the product (a 5 b) (a+ 7 b). 8. Find the magnitude of two vectors a and b, having the same magnitude and such that the angle between them is 60 o and their scalar product is. 9. Find, if for a unit vector a, ( a ) ( + a ). 0. If ˆ ˆ ˆ, ˆ ˆ ˆ a i + j+ k b i + j+ k and c iˆ+ ˆ j are such that a+λb is perpendicular to c, then find the value of λ.. Show that a b+ b a is perpendicular to a b b a, for any two nonzero vectors a and b.. If a a 0 and a b 0, then what can be concluded about the vector b?. If abc,, are unit vectors such that a+ b + c 0, find the value of a b + b c + c a. 4. If either vector a 0 or b 0, then a b 0. But the converse need not be true. Justify your answer with an eample. 5. If the vertices A, B, C of a triangle ABC are (,, ), (, 0, 0), (0,, ), respectively, then find ABC. [ ABC is the angle between the vectors BA and BC ]. 6. Show that the points A(,, 7), B(, 6, ) and C(, 0, ) are collinear. 7. Show that the vectors iˆ ˆj+ kˆ, iˆ ˆj 5kˆ and iˆ 4ˆj 4kˆ form the vertices of a right angled triangle. 8. If a is a nonzero vector of magnitude a and λ a nonzero scalar, then λ a is unit vector if (A) λ (B) λ (C) a λ (D) a / λ 0.6. Vector (or cross) product of two vectors In Section 0., we have discussed on the three dimensional right handed rectangular coordinate system. In this system, when the positive -ais is rotated counterclockwise

166 VECTOR ALGEBRA 449 into the positive y-ais, a right handed (standard) screw would advance in the direction of the positive z-ais (Fig 0.(i)). In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z-ais when the fingers are curled in the direction away from the positive -ais toward the positive y-ais (Fig 0.(ii)). Fig 0. (i), (ii) Definition The vector product of two nonzero vectors aand b, is denoted by a b and defined as a b a b sinθ nˆ, where, θ is the angle between aand b, 0 θ π and ˆn is a unit vector perpendicular to both a and b, such that ab, and nˆ form a right handed system (Fig 0.). i.e., the right handed system rotated from a to b moves in the Fig 0. direction of ˆn. If either a 0orb 0, then θ is not defined and in this case, we define a b 0. Observations. a b is a vector.. Let aand b be two nonzero vectors. Then a b 0 if and only if a and b are parallel (or collinear) to each other, i.e., a b 0 a b

167 450 MATHEMATICS In particular, a a 0 and a ( a) 0, since in the first situation, θ 0 and in the second one, θ π, making the value of sin θ to be 0.. π If θ then a b a b. 4. In view of the Observations and, for mutually perpendicular unit vectors iˆ, ˆj and k ˆ (Fig 0.4), we have iˆ iˆ ˆj ˆj kˆ kˆ 0 iˆ ˆj k ˆ, ˆ j k ˆ i ˆ, k ˆ i ˆ ˆ j Fig In terms of vector product, the angle between two vectors a and b may be given as sin θ a b a b 6. It is always true that the vector product is not commutative, as a b b a. Indeed, a b a b sinθnˆ, where ab, and nˆ form a right handed system, i.e., θ is traversed from a to b, Fig 0.5 (i). While, a sin, where b, aandn form a right handed system i.e. θ is traversed from ˆ b to a, Fig 0.5(ii). Fig 0.5 (i), (ii) Thus, if we assume aand b to lie in the plane of the paper, then nˆ and n ˆ both will be perpendicular to the plane of the paper. But, ˆn being directed above the paper while ˆn directed below the paper. i.e. nˆ ˆ n.

168 Hence a b a b sinθnˆ a b sin θnˆ 7. In view of the Observations 4 and 6, we have VECTOR ALGEBRA 45 b a ˆj iˆ kˆ, kˆ ˆj iˆ and iˆ kˆ ˆj. 8. If aand b represent the adjacent sides of a triangle then its area is given as a b. By definition of the area of a triangle, we have from Fig 0.6, Area of triangle ABC AB CD. But AB b (as given), and CD a Fig 0.6 sinθ. Thus, Area of triangle ABC sin b a θ. a b 9. If a and b represent the adjacent sides of a parallelogram, then its area is given by a b. From Fig 0.7, we have Area of parallelogram ABCD AB. DE. But AB b (as given), and DE a sin θ. Thus, Area of parallelogram ABCD b a sinθ Fig a b 0.7. We now state two important properties of vector product. Property (Distributivity of vector product over addition): If a, b and c are any three vectors and λ be a scalar, then (i) a ( b + c) a b + a c (ii) λ ( a b ) ( λ a) b a ( λb)

169 45 Let MATHEMATICS aand b ˆ ˆ ˆ be two vectors given in component form as ai ˆ+ a ˆj + ak ˆ and bi+ b j+ bk, respectively. Then their cross product may be given by a b iˆ ˆj kˆ a a a b b b Eplanation We have a b ( aiˆ+ a ˆj + akˆ) ( biˆ+ b ˆj + bkˆ) ab( iˆ iˆ) + ab ( iˆ ˆj) + ab ( iˆ kˆ ) + a b( ˆj iˆ) ab( ˆj ˆj) + ab( ˆj kˆ ) + + ab ( kˆ iˆ) + ab ˆ ˆ ˆ ˆ ( k j) + ab ( k k) (by Property ) ab ( iˆ ˆj) ab ( kˆ iˆ) a b ( iˆ ˆj) ab( ˆj kˆ) + ab( kˆ iˆ) ab( ˆj kˆ) + (as iˆ iˆ ˆj ˆj kˆ kˆ 0 and iˆ kˆ kˆ iˆ, ˆj iˆ iˆ ˆj and kˆ ˆj ˆj kˆ) abkˆ ab ˆj abkˆ+ abiˆ+ abj ˆ abiˆ (as iˆ ˆj kˆ, ˆj kˆ iˆ and kˆ iˆ ˆj) ( ab ab ) i ˆ ( ab ˆ ˆ ab ) j+ ( ab ab ) k iˆ ˆj kˆ a a a b b b Eample Find a b, if a iˆ+ ˆj+ kˆ and b iˆ+ 5 ˆj kˆ Solution We have a b iˆ ˆj kˆ 5 iˆ( 5) ( 4 9) ˆj+ (0 ) kˆ 7iˆ+ ˆj+ 7kˆ Hence a b ( 7) + () + (7) 507

170 VECTOR ALGEBRA 45 Eample Find a unit vector perpendicular to each of the vectors ( a+ b ) ( a b ), where a iˆ+ ˆj+ kˆ, b iˆ+ ˆj+ kˆ. Solution We have ˆ ˆ 4ˆ a+ b i + j+ k and a b ˆj kˆ A vector which is perpendicular to both a+ b and a b is given by and ( a+ b) ( a b) iˆ ˆj kˆ 4 ˆ 4ˆ ˆ i + j k ( c, say) 0 Now c Therefore, the required unit vector is c i ˆ + ˆ j k ˆ c Note There are two perpendicular directions to any plane. Thus, another unit vector perpendicular to a+ b and a b will be i ˆ ˆ j+ k ˆ. But that will be a consequence of ( a b) ( a+ b). Eample 4 Find the area of a triangle having the points A(,, ), B(,, ) and C(,, ) as its vertices. Solution We have AB ˆj + kˆ and AC iˆ+ ˆj. The area of the given triangle is AB AC. Now, iˆ ˆj kˆ AB AC 0 4iˆ+ ˆj kˆ 0 Therefore AB AC Thus, the required area is

171 454 MATHEMATICS Eample 5 Find the area of a parallelogram whose adjacent sides are given by the vectors a iˆ+ ˆj+ 4kˆ and b iˆ ˆj+ kˆ Solution The area of a parallelogram with aand b as its adjacent sides is given by a b. Now a b Therefore a b and hence, the required area is 4. iˆ ˆj kˆ 4 5iˆ+ ˆj 4kˆ EXERCISE 0.4. Find a b, if a iˆ 7ˆj+ 7kˆ and b iˆ ˆj+ kˆ.. Find a unit vector perpendicular to each of the vector a+ b and a b, where a iˆ+ ˆj+ kˆ and b iˆ+ ˆj kˆ.. If a unit vector a π makes angles with ˆ π i, with ˆj and an acute angle θ with 4 ˆk, then find θ and hence, the components of a. 4. Show that ( a b) ( a+ b) ( a b ) 5. Find λ and μ if (iˆ+ 6 ˆj+ 7 kˆ) ( iˆ+λ ˆj+μ kˆ) Given that a b 0 and a b 0. What can you conclude about the vectors aand b? 7. Let the vectors ab,, c be given as ai ˆ+ a ˆ ˆ ˆj + ak, bi ˆ+ b ˆj + bk, ˆ ˆ ci + cj+ cˆ k. Then show that a ( b + c) a b + a c. 8. If either a 0 or b 0, then a b 0. Is the converse true? Justify your answer with an eample. 9. Find the area of the triangle with vertices A(,, ), B(,, 5) and C(, 5, 5).

172 VECTOR ALGEBRA Find the area of the parallelogram whose adjacent sides are determined by the vectors a iˆ ˆj+ kˆ and b iˆ 7ˆj+ kˆ.. Let the vectors a and b be such that a and b, then a b is a unit vector, if the angle between a and b is (A) π/6 (B) π/4 (C) π/ (D) π/. Area of a rectangle having vertices A, B, C and D with position vectors ˆ iˆ+ ˆj+ 4 k, iˆ+ ˆj+ 4kˆ, ˆ ˆ ˆ i j+ 4k and iˆ ˆj+ 4kˆ, respectively is (A) (B) (C) (D) 4 Miscellaneous Eamples Eample 6 Write all the unit vectors in XY-plane. Solution Let r i+ y j be a unit vector in XY-plane (Fig 0.8). Then, from the figure, we have cos θ and y sin θ (since r ). So, we may write the vector r as r ( OP ) cosθ iˆ+ sinθ ˆj... () Clearly, r cos θ+ sin θ Fig 0.8 Also, as θ varies from 0 to π, the point P (Fig 0.8) traces the circle + y counterclockwise, and this covers all possible directions. So, () gives every unit vector in the XY-plane.

173 456 MATHEMATICS Eample 7 If iˆ+ ˆj+ kˆ, iˆ+ 5 ˆj, iˆ+ ˆj kˆ and iˆ 6 ˆj k ˆ are the position vectors of points A, B, C and D respectively, then find the angle between AB and CD. Deduce that AB and CD are collinear. Solution Note that if θ is the angle between AB and CD, then θ is also the angle between AB and CD. Now Therefore Similarly Thus AB Position vector of B Position vector of A (iˆ+ 5 ˆj) ( iˆ+ ˆj+ kˆ) iˆ+ 4ˆj kˆ AB () + (4) + ( ) CD iˆ 8 ˆj+ kˆ and CD 6 AB CD cos θ AB CD ( ) + 4( 8) + ( )() 6 ( )(6 ) 6 Since 0 θ π, it follows that θ π. This shows that AB and CD are collinear. Alternatively, AB CD which implies that AB and CD are collinear vectors. Eample 8 Let ab, and c be three vectors such that a, b 4, c 5 and each one of them being perpendicular to the sum of the other two, find a+ b + c. Solution Given a ( b + c ) b ( c + a ) 0, c ( a+ b ) 0. Now a+ b + c ( a+ b + c) ( a+ b + c) ( a+ b + c) a a+ a ( b + c) + b b + b ( a+ c) + c.( a+ b) + c. c a + b + c Therefore a + b + c 5

174 VECTOR ALGEBRA 457 Eample 9 Three vectors a, b and c satisfy the condition a+ b + c 0. Evaluate the quantity μ a b + b c + c a, if a, b 4 and c. Solution Since a+ b+ c 0, we have a ( a + b + c ) 0 or a a + a b+ a c 0 Therefore a b+ a c a... () Again, b ( a+ b + c) 0 or a b+ b c b 6... () Similarly a c + b c () Adding (), () and (), we have ( a b+ b c+ a c) or μ, i.e., μ Eample 0 If with reference to the right handed system of mutually perpendicular unit vectors ˆ ˆ ˆ i, j and k, α iˆ ˆj, β iˆ+ ˆj kˆ, then epress β in the form ββ +β,where β is parallel to α and β is perpendicular to α. Solution Let β λα, λ is a scalar, i.e., β ˆ ˆ λi λj. Now β β β ˆ ˆ ˆ ( λ ) i + ( +λ) j k. Now, since β is to be perpendicular to α, we should have α β 0. i.e., ( λ) ( +λ ) 0 or λ Therefore β ˆ ˆ i j and β ˆ ˆ i + ˆ j k

175 458 MATHEMATICS Miscellaneous Eercise on Chapter 0. Write down a unit vector in XY-plane, making an angle of 0 with the positive direction of -ais.. Find the scalar components and magnitude of the vector joining the points P(, y, z ) and Q (, y, z ).. A girl walks 4 km towards west, then she walks km in a direction 0 east of north and stops. Determine the girl s displacement from her initial point of departure. 4. If a b + c, then is it true that a b + c? Justify your answer. 5. Find the value of for which ( iˆ+ ˆj+ kˆ ) is a unit vector. 6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a iˆ+ ˆj kˆ and b iˆ ˆj+ kˆ. 7. If ˆ ˆ ˆ ˆ ˆ ˆ a i + j+ k, b i j+ k and c iˆ ˆj+ kˆ, find a unit vector parallel to the vector a b + c. 8. Show that the points A (,, 8), B(5, 0, ) and C(,, 7) are collinear, and find the ratio in which B divides AC. 9. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ( a+ b) and ( a b) eternally in the ratio :. Also, show that P is the mid point of the line segment RQ. 0. The two adjacent sides of a parallelogram are iˆ 4 ˆj+ 5 kˆ and iˆ ˆj kˆ. Find the unit vector parallel to its diagonal. Also, find its area.. Show that the direction cosines of a vector equally inclined to the aes OX, OY and OZ are,,.. Let ˆ ˆ ˆ ˆ ˆ ˆ a i + 4j+ k, b i j+ 7 k and c iˆ ˆj+ 4kˆ. Find a vector d which is perpendicular to both a and b, and c d 5.. The scalar product of the vector iˆ+ ˆj+ kˆ with a unit vector along the sum of vectors iˆ+ 4ˆj 5kˆ and λ iˆ+ ˆj+ kˆ is equal to one. Find the value of λ. 4. If a, b, c are mutually perpendicular vectors of equal magnitudes, show that the vector a + b + c is equally inclined to a, b and c.

176 VECTOR ALGEBRA Prove that ( a+ b) ( a+ b) a + b, if and only if a, b are perpendicular, given a 0, b 0. Choose the correct answer in Eercises 6 to If θ is the angle between two vectors a and b, then a b 0 only when π π (A) 0 <θ< (B) 0 θ (C) 0 < θ < π (D) 0 θ π 7. Let a and b be two unit vectors and θ is the angle between them. Then a + b is a unit vector if (A) π θ (B) 4 π θ (C) π θ (D) 8. The value of iˆ.( ˆj kˆ) + ˆj ( iˆ kˆ) + kˆ ( iˆ ˆj) is (A) 0 (B) (C) (D) 9. If θ is the angle between any two vectors a and b, then a b a b when θ is equal to (A) 0 (B) π 4 (C) π (D) π θ π Summary Position vector of a point P(, y, z) is given as OP( r) iˆ+ yj ˆ+ zkˆ, and its magnitude by + y + z. The scalar components of a vector are its direction ratios, and represent its projections along the respective aes. The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of any vector are related as: a b c l, m, n r r r The vector sum of the three sides of a triangle taken in order is 0.

177 460 MATHEMATICS The vector sum of two coinitial vectors is given by the diagonal of the parallelogram whose adjacent sides are the given vectors. The multiplication of a given vector by a scalar λ, changes the magnitude of the vector by the multiple λ, and keeps the direction same (or makes it opposite) according as the value of λ is positive (or negative). For a given vector a a, the vector aˆ gives the unit vector in the direction a of a. The position vector of a point R dividing a line segment joining the points P and Q whose position vectors are aand b respectively, in the ratio m : n na + mb (i) internally, is given by. m+ n mb na (ii) eternally, is given by. m n The scalar product of two given vectors a and b having angle θ between them is defined as a b a b cosθ. Also, when a b is given, the angle θ between the vectors aand b may be determined by a b cosθ a b If θ is the angle between two vectors aand b, then their cross product is given as a b a b sinθnˆ where ˆn is a unit vector perpendicular to the plane containing aand b. Such that abnform,, ˆ right handed system of coordinate aes. If we have two vectors aand b, given in component form as a ai ˆ a ˆ + j+ a k and b bi ˆ ˆ + bj+ bˆ k and λ any scalar, ˆ

178 VECTOR ALGEBRA 46 then a + b and a b ( a + b ) iˆ+ ( a + b ) ˆj+ ( a + b ) kˆ ; λa ( λ a ˆ ˆ ˆ ) i + ( λ a) j+ ( λ a) k; ab. ab + a b + ab ; iˆ ˆj kˆ a b c. a b c Historical Note The word vector has been derived from a Latin word vectus, which means to carry. The germinal ideas of modern vector theory date from around 800 when Caspar Wessel (745-88) and Jean Robert Argand (768-8) described that how a comple number a + ib could be given a geometric interpretation with the help of a directed line segment in a coordinate plane. William Rowen Hamilton ( ) an Irish mathematician was the first to use the term vector for a directed line segment in his book Lectures on Quaternions (85). Hamilton s method of quaternions (an ordered set of four real numbers given as: a+ biˆ+ cj ˆ+ dkˆ, iˆ, ˆj, kˆ following certain algebraic rules) was a solution to the problem of multiplying vectors in three dimensional space. Though, we must mention here that in practice, the idea of vector concept and their addition was known much earlier ever since the time of Aristotle (84- B.C.), a Greek philosopher, and pupil of Plato (47-48 B.C.). That time it was supposed to be known that the combined action of two or more forces could be seen by adding them according to parallelogram law. The correct law for the composition of forces, that forces add vectorially, had been discovered in the case of perpendicular forces by Stevin-Simon (548-60). In 586 A.D., he analysed the principle of geometric addition of forces in his treatise DeBeghinselen der Weeghconst ( Principles of the Art of Weighing ), which caused a major breakthrough in the development of mechanics. But it took another 00 years for the general concept of vectors to form. In the 880, Josaih Willard Gibbs (89-90), an American physicist and mathematician, and Oliver Heaviside (850-95), an English engineer, created what we now know as vector analysis, essentially by separating the real (scalar)

179 46 MATHEMATICS part of quaternion from its imaginary (vector) part. In 88 and 884, Gibbs printed a treatise entitled Element of Vector Analysis. This book gave a systematic and concise account of vectors. However, much of the credit for demonstrating the applications of vectors is due to the D. Heaviside and P.G. Tait (8-90) who contributed significantly to this subject.

180 THREE DIMENSIONAL GEOMETRY 46 Chapter THREE DIMENSIONAL GEOMETRY. Introduction v The moving power of mathematical invention is not reasoning but imagination. A. DEMORGAN v In Class XI, while studying Analytical Geometry in two dimensions, and the introduction to three dimensional geometry, we confined to the Cartesian methods only. In the previous chapter of this book, we have studied some basic concepts of vectors. We will now use vector algebra to three dimensional geometry. The purpose of this approach to -dimensional geometry is that it makes the study simple and elegant*. In this chapter, we shall study the direction cosines and direction ratios of a line joining two points and also discuss about the equations of lines and planes in space under different conditions, angle between two lines, two planes, a line and a plane, shortest distance between two skew lines and distance of a point from a plane. Most of Leonhard Euler (707-78) the above results are obtained in vector form. Nevertheless, we shall also translate these results in the Cartesian form which, at times, presents a more clear geometric and analytic picture of the situation.. Direction Cosines and Direction Ratios of a Line From Chapter 0, recall that if a directed line L passing through the origin makes angles α, β and γ with, y and z-aes, respectively, called direction angles, then cosine of these angles, namely, cos α, cos β and cos γ are called direction cosines of the directed line L. If we reverse the direction of L, then the direction angles are replaced by their supplements, i.e., π α, π β and π γ. Thus, the signs of the direction cosines are reversed. * For various activities in three dimensional geometry, one may refer to the Book A Hand Book for designing Mathematics Laboratory in Schools, NCERT, 005

181 464 MATHEMATICS Fig. Note that a given line in space can be etended in two opposite directions and so it has two sets of direction cosines. In order to have a unique set of direction cosines for a given line in space, we must take the given line as a directed line. These unique direction cosines are denoted by l, m and n. Remark If the given line in space does not pass through the origin, then, in order to find its direction cosines, we draw a line through the origin and parallel to the given line. Now take one of the directed lines from the origin and find its direction cosines as two parallel line have same set of direction cosines. Any three numbers which are proportional to the direction cosines of a line are called the direction ratios of the line. If l, m, n are direction cosines and a, b, c are direction ratios of a line, then a λl, bλm and c λn, for any nonzero λ R. ANote Some authors also call direction ratios as direction numbers. Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines (d.c s) of the line. Then l a m b n k c (say), k being a constant. Therefore l ak, m bk, n ck... () But l + m + n Therefore k (a + b + c ) or k ± a + b + c

182 Hence, from (), the d.c. s of the line are THREE DIMENSIONAL GEOMETRY 465 a b c l±, m±, n± a + b + c a + b + c a + b + c where, depending on the desired sign of k, either a positive or a negative sign is to be taken for l, m and n. For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k 0 is also a set of direction ratios. So, any two sets of direction ratios of a line are also proportional. Also, for any line there are infinitely many sets of direction ratios... Relation between the direction cosines of a line Consider a line RS with direction cosines l, m, n. Through the origin draw a line parallel to the given line and take a point P(, y, z) on this line. From P draw a perpendicular PA on the -ais (Fig..). OA Let OP r. Then cosα. This gives lr. OP r Similarly, y mr and z nr Thus + y + z r (l + m + n ) But + y + z r Hence l + m + n Fig... Direction cosines of a line passing through two points Since one and only one line passes through two given points, we can determine the direction cosines of a line passing through the given points P(, y, z ) and Q(, y, z ) as follows (Fig. (a)). X A R Z O O S P (, y, z) r Y P A Fig.

183 466 MATHEMATICS Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ with the, y and z-ais, respectively. Draw perpendiculars from P and Q to XY-plane to meet at R and S. Draw a perpendicular from P to QS to meet at N. Now, in right angle triangle PNQ, PQN γ (Fig. (b). Therefore, cosγ NQ PQ z z PQ Similarly cos α y and cos y β PQ PQ Hence, the direction cosines of the line segment joining the points P(, y, z ) and Q(, y, z ) are PQ, y y PQ, z z PQ where PQ ( ) ( ) + ( y y ) + z z A Note The direction ratios of the line segment joining P(, y, z ) and Q(, y, z ) may be taken as, y y, z z or, y y, z z Eample If a line makes angle 90, 60 and 0 with the positive direction of, y and z-ais respectively, find its direction cosines. Solution Let the d.c.'s of the lines be l, m, n. Then l cos , m cos 60 0, n cos 0 0. Eample If a line has direction ratios,,, determine its direction cosines. Solution Direction cosines are + ( ) + ( ), + ( ) + ( ), + ( ) + ( ) or,, Eample Find the direction cosines of the line passing through the two points (, 4, 5) and (,, ).

184 THREE DIMENSIONAL GEOMETRY 467 Solution We know the direction cosines of the line passing through two points P(, y, z ) and Q(, y, z ) are given by, y y, z z PQ PQ PQ where PQ ( ) ( + + z z ) ( y y) Here P is (, 4, 5) and Q is (,, ). So PQ ( ( )) + ( 4) + ( ( 5)) 77 Thus, the direction cosines of the line joining two points is 8,, Eample 4 Find the direction cosines of, y and z-ais. Solution The -ais makes angles 0, 90 and 90 respectively with, y and z-ais. Therefore, the direction cosines of -ais are cos 0, cos 90, cos 90 i.e.,,0,0. Similarly, direction cosines of y-ais and z-ais are 0,, 0 and 0, 0, respectively. Eample 5 Show that the points A (,, 4), B (,, ) and C (, 8, ) are collinear. Solution Direction ratios of line joining A and B are,, + 4 i.e.,, 5, 7. The direction ratios of line joining B and C are, 8 +,, i.e.,, 0, 4. It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel to BC. But point B is common to both AB and BC. Therefore, A, B, C are collinear points. EXERCISE.. If a line makes angles 90, 5, 45 with the, y and z-aes respectively, find its direction cosines.. Find the direction cosines of a line which makes equal angles with the coordinate aes.. If a line has the direction ratios 8,, 4, then what are its direction cosines? 4. Show that the points (,, 4), (,, ), (5, 8, 7) are collinear. 5. Find the direction cosines of the sides of the triangle whose vertices are (, 5, 4), (,, ) and ( 5, 5, ).

185 468 MATHEMATICS. Equation of a Line in Space We have studied equation of lines in two dimensions in Class XI, we shall now study the vector and cartesian equations of a line in space. A line is uniquely determined if (i) it passes through a given point and has given direction, or (ii) it passes through two given points...equation of a line through a given point and parallel to a given vector b r Let a r be the position vector of the given point A with respect to the origin O of the rectangular coordinate system. Let l be the line which passes through the point A and is parallel to a given vector b r. Let r be the position vector of an arbitrary point P on the line (Fig.4). uuur r Then AP is parallel to the vector b, i.e., uuur r AP λb, where λ is some real number. Fig.4 uuur uuur uuur But AP OP OA i.e. λb r r r a Conversely, for each value of the parameter λ, this equation gives the position vector of a point P on the line. Hence, the vector equation of the line is given by r r a+ ë b r... () r Remark If b aiˆ+ bj ˆ+ ckˆ, then a, b, c are direction ratios of the line and conversely, r if a, b, c are direction ratios of a line, then b aiˆ+ bj ˆ+ ckˆ will be the parallel to the line. Here, b should not be confused with b. r Derivation of cartesian form from vector form Let the coordinates of the given point A be (, y, z ) and the direction ratios of the line be a, b, c. Consider the coordinates of any point P be (, y, z). Then r iˆ + yj ˆ + zkˆ r ; a iˆ y ˆj z kˆ + + r and b a iˆ + b ˆj + c kˆ Substituting these values in () and equating the coefficients of i ˆ, ˆ j and kˆ, we get + λ a; y y + λ b; z z + λ c... ()

186 THREE DIMENSIONAL GEOMETRY 469 These are parametric equations of the line. Eliminating the parameter λ from (), we get y y z z... () a b c This is the Cartesian equation of the line. ANote If l, m, n are the direction cosines of the line, the equation of the line is y y z z l m n Eample 6 Find the vector and the Cartesian equations of the line through the point (5,, 4) and which is parallel to the vector iˆ + ˆj 8kˆ. Solution We have a r r 5iˆ+ ˆj 4kˆ and b iˆ+ ˆj 8 kˆ Therefore, the vector equation of the line is r 5iˆ + ˆj 4 kˆ + λ ( iˆ + ˆj 8 kˆ) Now, r is the position vector of any point P(, y, z) on the line. Therefore, iˆ+ y ˆj + z kˆ 5iˆ+ ˆj 4 kˆ+ λ ( iˆ + ˆj 8 kˆ) Eliminating λ, we get (5 + λ ) $ i+ ( + λ ) $ j+ ( 4 8 λ) k$ 5 y z+ 4 8 which is the equation of the line in Cartesian form... Equation of a line passing through two given points Let a r and b r be the position vectors of two points A(, y, z ) and B(, y, z ), respectively that are lying on a line (Fig.5). Let r be the position vector of an arbitrary point P(, y, z), then P is a point on uuur r r the line if and only if AP a and uuur r r AB b a are collinear vectors. Therefore, P is on the line if and only if r r r r aλ( b a) Fig.5

187 470 MATHEMATICS or r a r r + λ ( b a r ), λ R.... () This is the vector equation of the line. Derivation of cartesian form from vector form We have r ˆ ˆ ˆ r r i+ y j+ zk, a iˆ ˆ ˆ + y j+ zk and b ˆ ˆ ˆ i + y j+ zk, Substituting these values in (), we get i $ + y $ j+ zk$ $ i+ y $ j+ z k$ +λ[( ) $ i+ ( y y ) $ j+ ( z z ) k$ ] Equating the like coefficients of iˆ, jˆ, kˆ, we get + λ ( ); y y + λ (y y ); z z + λ (z z ) On eliminating λ, we obtain y y z z y y z z which is the equation of the line in Cartesian form. Eample 7 Find the vector equation for the line passing through the points (, 0, ) and (, 4, 6). Solution Let a r and b r be the position vectors of the point A(, 0, ) and B(, 4, 6). r Then a iˆ + kˆ r and b iˆ+ 4ˆj+ 6kˆ r r Therefore b a 4iˆ+ 4ˆj+ 4kˆ Let r be the position vector of any point on the line. Then the vector equation of the line is r iˆ+ kˆ+λ (4iˆ+ 4ˆj+ 4 kˆ) Eample 8 The Cartesian equation of a line is + y 5 z Find the vector equation for the line. Solution Comparing the given equation with the standard form y y z z a b c We observe that, y 5, z 6; a, b 4, c.

188 THREE DIMENSIONAL GEOMETRY 47 Thus, the required line passes through the point (, 5, 6) and is parallel to the vector iˆ+ 4 ˆj+ kˆ. Let r be the position vector of any point on the line, then the vector equation of the line is given by r ( iˆ+ 5ˆj 6 kˆ ) + λ (iˆ+ 4 ˆj+ kˆ ).4 Angle between Two Lines Let L and L be two lines passing through the origin and with direction ratios a, b, c and a, b, c, respectively. Let P be a point on L and Q be a point on L. Consider the directed lines OP and OQ as given in Fig.6. Let θ be the acute angle between OP and OQ. Now recall that the directed line segments OP and OQ are vectors with components a, b, c and a, b, c, respectively. Therefore, the angle θ between them is given by Fig.6 cosθ aa + bb + cc a b c a b c... () The angle between the lines in terms of sin θ is given by sin θ cos θ ( aa + bb + cc ) ( a + b + c )( a + b + c ) ( a + b + c )( a + b + c) ( aa + bb + cc ) ( a + b + c ) ( a + b + c ) + + ( a b a b ) ( b c b c ) ( c a c a ) a b c a b c... () A Note In case the lines L and L do not pass through the origin, we may take lines L andl which are parallel to L and L respectively and pass through the origin.

189 47 MATHEMATICS If instead of direction ratios for the lines L and L, direction cosines, namely, l, m, n for L and l, m, n for L are given, then () and () takes the following form: cos θ l l + m m + n n (as l + m + n and sin θ ( ) l + m + n )... () lm lm ( mn mn) + ( nl nl)... (4) Two lines with direction ratios a, b, c and a, b, c are (i) perpendicular i.e. if θ 90 by () a a + b b + c c 0 (ii) parallel i.e. if θ 0 by () a a b c b c Now, we find the angle between two lines when their equations are given. If θ is acute the angle between the lines r a +λb ur and r r a +µ b then cosθ r r b b r r b b In Cartesian form, if θ is the angle between the lines y y z z a b c... () and a y y z z... () b c where, a, b, c and a, b, c are the direction ratios of the lines () and (), respectively, then cos θ aa + bb + cc a b c a b c Eample 9 Find the angle between the pair of lines given by r iˆ+ ˆj 4 kˆ+λ ( iˆ+ ˆj+ kˆ) and r 5iˆ ˆj+µ (iˆ+ ˆj+ 6 kˆ )

190 THREE DIMENSIONAL GEOMETRY 47 Solution Here b r ˆ ˆ ˆ i + j+ k and b r iˆ+ ˆj+ 6kˆ The angle θ between the two lines is given by r r b b ˆ ˆ ˆ ˆ ˆ ˆ ( i + j+ k) (i + j+ 6 k) cos θ r r b b Hence θ cos Eample 0 Find the angle between the pair of lines and + y z y 4 z 5 Solution The direction ratios of the first line are, 5, 4 and the direction ratios of the second line are,,. If θ is the angle between them, then cos θ Hence, the required angle is cos 5..5 Shortest Distance between Two Lines If two lines in space intersect at a point, then the shortest distance between them is zero. Also, if two lines in space are parallel, then the shortest distance between them will be the perpendicular distance, i.e. the length of the perpendicular drawn from a point on one line onto the other line. Further, in a space, there are lines which are neither intersecting nor parallel. In fact, such pair of lines are non coplanar and are called skew lines. For eample, let us consider a room of size,, units along, y and z-aes respectively Fig.7. Fig.7

191 474 MATHEMATICS The line GE that goes diagonally across the ceiling and the line DB passes through one corner of the ceiling directly above A and goes diagonally down the wall. These lines are skew because they are not parallel and also never meet. By the shortest distance between two lines we mean the join of a point in one line with one point on the other line so that the length of the segment so obtained is the smallest. For skew lines, the line of the shortest distance will be perpendicular to both the lines..5. Distance between two skew lines We now determine the shortest distance between two skew lines in the following way: Let l and l be two skew lines with equations (Fig..8) r r a +λb r... () and r r a +µ b r... () Take any point S on l with position vector a r and T on l, with position vector a r. Then the magnitude of the shortest distance vector T will be equal to that of the projection of ST along the Q direction of the line of shortest distance (See 0.6.). l If PQ uuur is the shortest distance vector between l and l, then it being perpendicular to both b r and l S P b r, the unit vector nˆ along PQ uuur would therefore be r r Fig.8 b b ˆn r r... () b b Then PQ uuur d nˆ where, d is the magnitude of the shortest distance vector. Let θ be the angle between ST uur and PQ uuur. Then PQ ST cos θ uuur uur PQ ST But cos θ uuuur uur PQ ST r r dnˆ ( a a) uur r r (since d ST ST a a) r r r r ( b b) ( a a) r r [From ()] ST b b

192 THREE DIMENSIONAL GEOMETRY 475 Hence, the required shortest distance is d PQ ST cos θ r r r r ( b b ). ( a a) or d r r b b Cartesian form The shortest distance between the lines l : and l : is a a y y z z b c y y z z b c y y z z a b c a b c + + ( bc bc) ( ca ca) ( ab ab).5. Distance between parallel lines If two lines l and l are parallel, then they are coplanar. Let the lines be given by r r a +λb r... () and r r a +µ b r () where, a r is the position vector of a point S on l and a r is the position vector of a point T on l Fig.9. As l, l are coplanar, if the foot of the perpendicular from T on the line l is P, then the distance between the lines l and l TP. Let θ be the angle between the vectors ST uur and b r. Then b r ST uur r uur ( b ST sin θ) nˆ Fig.9... () where nˆ is the unit vector perpendicular to the plane of the lines l and l. But ST uur r r a a

193 476 MATHEMATICS Therefore, from (), we get r r r r b ( a a) b PT nˆ (since PT ST sin θ) i.e., r r r r b ( a a) b PT (as ˆn ) Hence, the distance between the given parallel lines is r uuur r r b ( a a) d PT r b Eample Find the shortest distance between the lines l and l whose vector equations are r iˆ + ˆj + λ ( iˆ ˆj + kˆ )... () and r iˆ + ˆj kˆ+ µ (iˆ 5ˆj + kˆ )... () Solution Comparing () and () with r r a +λb r r r r and a + µ b respectively, we get a r r iˆ + ˆj, b ˆ ˆ ˆ i j + k a r î + ĵ ˆk and b r î 5 ĵ + ˆk r r Therefore a a î kˆ r r and b b ( iˆ ˆj + kˆ ) (iˆ 5 ˆj + kˆ ) iˆ ˆj kˆ iˆ ˆj 7kˆ 5 r r So b b Hence, the shortest distance between the given lines is given by r r r r ( b b ). ( a a ) d r r b b Eample Find the distance between the lines l and l given by r iˆ + ˆj 4 kˆ + λ ( iˆ + ˆj + 6 kˆ ) and r r iˆ + ˆj 5 kˆ + µ ( iˆ + ˆj + 6 kˆ )

194 Solution The two lines are parallel (Why? ) We have THREE DIMENSIONAL GEOMETRY 477 a r iˆ+ ˆj 4kˆ, a r iˆ+ ˆj 5kˆ and b r iˆ+ ˆj+ 6kˆ Therefore, the distance between the lines is given by d r r r b ( a a) r b iˆ ˆj kˆ or 9iˆ+ 4 ˆj 4 kˆ EXERCISE.. Show that the three lines with direction cosines 4 4 4,, ;,, ;,, are mutually perpendicular.. Show that the line through the points (,, ), (, 4, ) is perpendicular to the line through the points (0,, ) and (, 5, 6).. Show that the line through the points (4, 7, 8), (,, 4) is parallel to the line through the points (,, ), (,, 5). 4. Find the equation of the line which passes through the point (,, ) and is parallel to the vector iˆ + ˆj kˆ. 5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector iˆ j + 4kˆ and is in the direction iˆ+ ˆj kˆ. 6. Find the cartesian equation of the line which passes through the point (, 4, 5) + y 4 z+ 8 and parallel to the line given by y+ 4 z 6 The cartesian equation of a line is. Write its vector form Find the vector and the cartesian equations of the lines that passes through the origin and (5,, ).

195 478 MATHEMATICS 9. Find the vector and the cartesian equations of the line that passes through the points (,, 5), (,, 6). 0. Find the angle between the following pairs of lines: (i) r iˆ 5 ˆj+ kˆ+λ (iˆ+ ˆj+ 6 kˆ) and r 7iˆ 6 kˆ+µ ( iˆ+ ˆj+ kˆ) (ii) r iˆ+ ˆj kˆ+λ( iˆ ˆj kˆ) and r iˆ ˆj 56 kˆ+µ (iˆ 5ˆj 4 kˆ). Find the angle between the following pair of lines: (i) y and + y z (ii) y z 5 and y z 4 8. Find the values of p so that the lines 7 y 4 z p and 7 7 y 5 6 z are at right angles. p 5. 5 Show that the lines y + z y z and are perpendicular to 7 5 each other. 4. Find the shortest distance between the lines r ( iˆ + ˆj + kˆ ) + λ( iˆ ˆj + kˆ ) and r iˆ ˆj kˆ+µ (iˆ+ ˆj + kˆ) 5. Find the shortest distance between the lines + y + z + and y 5 z Find the shortest distance between the lines whose vector equations are r ( iˆ+ ˆj + kˆ ) + λ( iˆ ˆj + kˆ ) r and 4iˆ+ 5ˆj + 6 kˆ+ µ (iˆ + ˆj + kˆ) 7. Find the shortest distance between the lines whose vector equations are r ( t) iˆ+ ( t ) ˆj + ( t) kˆ and r ( s+ ) iˆ+ (s ) ˆj (s+ ) kˆ

196 .6 Plane THREE DIMENSIONAL GEOMETRY 479 A plane is determined uniquely if any one of the following is known: (i) the normal to the plane and its distance from the origin is given, i.e., equation of a plane in normal form. (ii) it passes through a point and is perpendicular to a given direction. (iii) it passes through three given non collinear points. Now we shall find vector and Cartesian equations of the planes..6.equation of a plane in normal form Consider a plane whose perpendicular distance from the origin is d (d 0). Fig.0. If ON uuur is the normal from the origin to the plane, and nˆ is the unit normal vector along ON uuur. Then ON uuur Z d nˆ. Let P be any uuur point on the plane. Therefore, NP is perpendicular to ON uuur. uuur uuur Therefore, NP ON 0... () Let r be the position vector of the point P, r uuur r uuur uuur uuur then NP d nˆ (as ON + NP OP ) d N Y Therefore, () becomes O r ( d n) dn 0 X r Fig.0 or ( d n) n 0 (d 0) r or n d n n 0 r i.e., n d (as n n ) () This is the vector form of the equation of the plane. Cartesian form Equation () gives the vector equation of a plane, where nˆ is the unit vector normal to the plane. Let P(, y, z) be any point on the plane. Then uuur r OP iˆ+ y ˆj + z kˆ Let l, m, n be the direction cosines of nˆ. Then ˆn liˆ+ mˆj + nkˆ P(,y,z)

197 480 MATHEMATICS Therefore, () gives ( iˆ+ y ˆj + zkˆ) ( liˆ+ mj ˆ+ nkˆ) d i.e., l + my + nz d... () This is the cartesian equation of the plane in the normal form. r ANote Equation () shows that if ( a iˆ + b ˆj + c kˆ ) d is the vector equation of a plane, then a + by + cz d is the Cartesian equation of the plane, where a, b and c are the direction ratios of the normal to the plane. 6 Eample Find the vector equation of the plane which is at a distance of 9 from the origin and its normal vector from the origin is iˆ ˆj + 4kˆ. Also find its cartesian form. r Solution Let n iˆ ˆj + 4 kˆ. Then ˆ n r iˆ ˆj + 4 kˆ iˆ ˆj + 4 kˆ n r n Hence, the required equation of the plane is r ˆ ˆ 4 ˆ 6 i + j + k Eample 4 Find the direction cosines of the unit vector perpendicular to the plane r (6 iˆ ˆj kˆ ) + 0 passing through the origin. Solution The given equation can be written as r ( 6 iˆ + ˆj+ k ˆ )... () Now 6iˆ + ˆj + kˆ Therefore, dividing both sides of () by 7, we get r 6 ˆ ˆ ˆ i + j + k r which is the equation of the plane in the form nˆ d. 6 This shows that nˆ iˆ + ˆj + kˆ is a unit vector perpendicular to the plane through the origin. Hence, the direction cosines of nˆ are 6 7,,. 7 7

198 THREE DIMENSIONAL GEOMETRY 48 Eample 5 Find the distance of the plane y + 4z 6 0 from the origin. Solution Since the direction ratios of the normal to the plane are,, 4; the direction cosines of it are 4,, + ( ) ( ) ( ) + 4, i.e., 4,, Hence, dividing the equation y + 4z 6 0 i.e., y + 4z 6 throughout by 9, we get y + z This is of the form l + my + nz d, where d is the distance of the plane from the 6 origin. So, the distance of the plane from the origin is. 9 Eample 6 Find the coordinates of the foot of the perpendicular drawn from the origin to the plane y + 4z 6 0. Solution Let the coordinates of the foot of the perpendicular P from the origin to the plane is (, y, z ) (Fig.). Then, the direction ratios of the line OP are Z, y, z. Writing the equation of the plane in the normal form, we have 4 6 y + z O Y 4 where,,, are the direction X cosines of the OP. Fig. Since d.c. s and direction ratios of a line are proportional, we have 9 y z k P(, y, z ) i.e., k, 9 y k 4k, z 9 9

199 48 MATHEMATICS Substituting these in the equation of the plane, we get k Hence, the foot of the perpendicular is,, ANote If d is the distance from the origin and l, m, n are the direction cosines of the normal to the plane through the origin, then the foot of the perpendicular is (ld, md, nd)..6.equation of a plane perpendicular to a given vector and passing through a given point In the space, there can be many planes that are perpendicular to the given vector, but through a given point P(, y, z ), only one such plane eists (see Fig.). Let a plane pass through a point A with position vector a r and perpendicular to the vector ur Fig. N. Let r be the position vector of any point P(, y, z) in the plane. (Fig.). Then the point P lies in the plane if and only if uuur ur uuur ur AP is perpendicular to N. i.e., AP. N 0. But uuur r r r r r AP a. Therefore, ( a) N 0 () This is the vector equation of the plane. Cartesian form Let the given point A be (, y, z ), P be (, y, z) and direction ratios of ur N are A, B and C. Then, Fig. r ˆ ˆ ˆ r a ˆ ˆ ˆ r i+ y j+ zk, i+ y j+ zk and N Aiˆ+ Bˆj+ Ckˆ r r Now ( a) N0 r ˆ ˆ ˆ ˆ ˆ ˆ i y y j z z k + + (Ai+ B j+ C k) 0 So ( ) ( ) ( ) i.e. A( ) + B (y y ) + C (z z ) 0 Eample 7 Find the vector and cartesian equations of the plane which passes through the point (5,, 4) and perpendicular to the line with direction ratios,,.

200 THREE DIMENSIONAL GEOMETRY 48 r Solution We have the position vector of point (5,, 4) as a 5iˆ+ ˆj 4kˆ normal vector N r r perpendicular to the plane as N iˆ+ ˆj kˆ r r r Therefore, the vector equation of the plane is given by ( a).n 0 or [ r (5iˆ+ ˆj 4 kˆ)] (iˆ+ ˆj kˆ) 0 Transforming () into Cartesian form, we have [( 5) iˆ+ ( y ) ˆj+ ( z+ 4) kˆ] ( iˆ+ ˆj kˆ) 0 or ( 5) + ( y ) ( z+ 4) 0 i.e. + y z 0 which is the cartesian equation of the plane..6. Equation of a plane passing through three non collinear points and the... () Let R, S and T be three non collinear points on the plane with position vectors a r, b r and c r respectively (Fig.4). Z P R (RS X RT) O r a b S c T Y X Fig.4 uuur uuur uuur uuur The vectors RS and RT are in the given plane. Therefore, the vector RS RT is perpendicular to the plane containing points R, S and T. Let r be the position vector of any point P in the plane. Therefore, the equation of the plane passing through R and uuur uuur perpendicular to the vector RS RT is r r uuur uuur ( a) (RS RT) 0 r r r r r r or ( r a) [( b a) ( c a )] 0 ()

201 484 MATHEMATICS This is the equation of the plane in vector form passing through three noncollinear points. Note Why was it necessary to say that the three points A had to be non collinear? If the three points were on the same R line, then there will be many planes that will contain them (Fig.5). S These planes will resemble the pages of a book where the T line containing the points R, S and T are members in the binding of the book. Cartesian form Fig.5 Let (, y, z ), (, y, z ) and (, y, z ) be the coordinates of the points R, S and T respectively. Let (, y, z) be the coordinates of any point P on the plane with position vector r. Then uuur RP ( ) î + (y y ) ĵ + (z z ) ˆk uuur RS ( ) î + (y y ) ĵ + (z z ) ˆk uuur RT ( ) î + (y y ) ĵ + (z z ) ˆk Substituting these values in equation () of the vector form and epressing it in the form of a determinant, we have y y z z y y z z 0 y y z z which is the equation of the plane in Cartesian form passing through three non collinear points (, y, z ), (, y, z ) and (, y, z ). Eample 8 Find the vector equations of the plane passing through the points R(, 5, ), S(,, 5) and T(5,, ). r r Solution Let a iˆ+ 5 ˆj kˆ, b iˆ ˆj+ 5kˆ r, c 5iˆ+ ˆj kˆ Then the vector equation of the plane passing through a r, b r and c r and is given by r r uuur uuur ( a) (RS RT) 0 (Why?) r r r r r r or ( a) [( b a) ( c a)] 0 i.e. [ r (iˆ+ 5ˆj kˆ)] [( 4iˆ 8ˆj+ 8 kˆ) (iˆ ˆj)] 0

202 THREE DIMENSIONAL GEOMETRY Intercept form of the equation of a plane In this section, we shall deduce the equation of a plane in terms of the intercepts made by the plane on the coordinate aes. Let the equation of the plane be A + By + Cz + D 0 (D 0)... () Let the plane make intercepts a, b, c on, y and z aes, respectively (Fig.6). Hence, the plane meets, y and z-aes at (a, 0, 0), (0, b, 0), (0, 0, c), respectively. Therefore Aa + D 0 or A Bb + D 0 or B D a D b D Cc + D 0 or C c Substituting these values in the equation () of the plane and simplifying, we get Fig.6 y z () a b c which is the required equation of the plane in the intercept form. Eample 9 Find the equation of the plane with intercepts, and 4 on the, y and z-ais respectively. Solution Let the equation of the plane be y z () a b c Here a, b, c 4. Substituting the values of a, b and c in (), we get the required equation of the y z plane as + + or 6 + 4y + z Plane passing through the intersection of two given planes Let π and π be two planes with equations r r nˆ d and nˆ d respectively. The position vector of any point on the line of intersection must satisfy both the equations (Fig.7). Fig.7

203 486 MATHEMATICS If t r is the position vector of a point on the line, then r r t nˆ d and t nˆ d Therefore, for all real values of λ, we have r t ( nˆ+λnˆ) d+λd Since t r is arbitrary, it satisfies for any point on the line. r r r Hence, the equation ( n+λ n) d+λdrepresents a plane π which is such that if any vector r satisfies both the equations π and π, it also satisfies the equation π i.e., any plane passing through the intersection of the planes n r r d and n d has the equation ( n + λn ) d + λd... () Cartesian form In Cartesian system, let n r Ai ˆ + B ˆ ˆ j+ Ck n r Ai ˆ + B ˆ ˆ j+ Ck and r iˆ+ y ˆj + zkˆ Then () becomes (A + λa ) + y (B + λb ) + z (C + λc ) d + λd or (A + B y + C z d ) + λ(a + B y + C z d ) 0... () which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ. Eample 0 Find the vector equation of the plane passing through the intersection of the planes ˆ ˆ ˆ r ( i + j+ k) 6and (iˆ+ ˆj+ 4 kˆ) 5, and the point (,, ). r Solution Here, n ˆ iˆ+ ˆj+ k and n r iˆ+ ˆj+ 4 kˆ ; and d 6 and d 5 r r r Hence, using the relation ( n+λ n) d+λd, we get or r [ iˆ+ ˆj+ kˆ+λ (iˆ+ ˆj+ 4 kˆ)] r [( + λ ) iˆ+ (+ λ ) ˆj+ (+ 4 λ) kˆ ] where, λ is some real number. 6 5λ 6 5λ ()

204 Taking r iˆ+ y ˆj+ zkˆ, we get THREE DIMENSIONAL GEOMETRY 487 ( iˆ+ y ˆj + zkˆ) [(+ λ ) iˆ+ (+ λ ) ˆj + (+ 4 λ ) kˆ] 6 5λ or ( + λ ) + ( + λ) y + ( + 4λ) z 6 5λ or ( + y + z 6 ) + λ ( + y + 4 z + 5) 0... () Given that the plane passes through the point (,,), it must satisfy (), i.e. ( + + 6) + λ ( ) 0 or λ 4 Putting the values of λ in (), we get r ˆ 9 ˆ 6 i j k ˆ r 0 or ˆ ˆ ˆ i+ j+ k r or (0iˆ+ ˆj+ 6 kˆ ) 69 which is the required vector equation of the plane..7 Coplanarity of Two Lines Let the given lines be r r a+λb r... () and r r a+µ b r... () The line () passes through the point, say A, with position vector a r and is parallel to b r. The line () passes through the point, say B with position vector a r and is parallel to b r. uuur r r Thus, AB a a uuur r r The given lines are coplanar if and only if AB is perpendicular to b b. uuur r r r r r r i.e. AB.( b b) 0 or ( a a) ( b b) 0 Cartesian form Let (, y, z ) and (, y, z ) be the coordinates of the points A and B respectively.

205 488 MATHEMATICS Let a, b, c and a, b, c be the direction ratios of b r and b r, respectively. Then uuur AB ( ˆ ˆ ˆ ) i+ ( y y) j+ ( z z) k r r b aiˆ+ b ˆ ˆ ˆ ˆ ˆ j+ ck and b ai+ b j+ ck uuur r r The given lines are coplanar if and only if AB ( b b) 0. In the cartesian form, it can be epressed as y y z z a b c a b c Eample Show that the lines 0... (4) + y z 5 + y z 5 and are coplanar. 5 5 Solution Here,, y, z 5, a, b, c 5, y, z 5, a, b, c 5 Now, consider the determinant y y z z 0 a b c 5 0 a b c 5 Therefore, lines are coplanar..8 Angle between Two Planes Definition The angle between two planes is defined as the angle between their normals (Fig.8 (a)). Observe that if θ is an angle between the two planes, then so is 80 θ (Fig.8 (b)). We shall take the acute angle as the angles between two planes. Fig.8

206 THREE DIMENSIONAL GEOMETRY 489 If n r and n r are normals to the planes and θ be the angle between the planes r n r d and r. n r d. Then θ is the angle between the normals to the planes drawn from some common point. r r n n We have, cos θ r r n n A Note The planes are perpendicular to each other if n r. n r 0 and parallel if n r is parallel to n r. Cartesian form Let θ be the angle between the planes, A + B y + C z + D 0 and A + B y + C z + D 0 The direction ratios of the normal to the planes are A, B, C and A, B, C respectively. Therefore, cos θ A A + B B + C C A B C A B C A Note. If the planes are at right angles, then θ 90 o and so cos θ 0. Hence, cos θ A A + B B + C C 0. A B C. If the planes are parallel, then. A B C Eample Find the angle between the two planes + y z 5 and 6y z 7 using vector method. Solution The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are ur ur N iˆ + ˆj kˆ and N ˆ iˆ 6 ˆj k ur ur N N (iˆ + ˆj kˆ) (iˆ 6 ˆj kˆ) 4 Therefore cos θ ur ur N N Hence θ cos

207 490 MATHEMATICS Eample Find the angle between the two planes 6y + z 7 and + y z 5. Solution Comparing the given equations of the planes with the equations A + B y + C z + D 0 and A + B y + C z + D 0 We get A, B 6, C A, B, C cos θ + ( 6) () + () ( ) ( + ( 6) + ( ) ) ( + + ( ) ) Therefore, θ cos -.9 Distance of a Point from a Plane Vector form Consider a point P with position vector a r and a plane π whose equation is r nˆ d (Fig.9). Z Z O a P N d Q N Y N a P d N O Y X (a) Fig.9 X (b) Consider a plane π through P parallel to the plane π. The unit vector normal to r r π is nˆ. Hence, its equation is ( a ) nˆ 0 r r i.e., nˆ a nˆ r Thus, the distance ON of this plane from the origin is a nˆ. Therefore, the distance PQ from the plane π is (Fig.. (a)) r i.e., ON ON d a nˆ

208 THREE DIMENSIONAL GEOMETRY 49 which is the length of the perpendicular from a point to the given plane. We may establish the similar results for (Fig.9 (b)). A Note r ur. If the equation of the plane π is in the form N d, where ur N is normal to the plane, then the perpendicular distance is r ur a N d ur. N ur. The length of the perpendicular from origin O to the plane r N d is d ur N (since a r 0). Cartesian form Let P(, y, z ) be the given point with position vector a r and A + By + Cz D be the Cartesian equation of the given plane. Then a r iˆ + y ˆ ˆj + z k ur N A iˆ+ B ˆj+ Ckˆ Hence, from Note, the perpendicular from P to the plane is ( iˆ + y ˆj + z kˆ ) ( A iˆ + B ˆj + C kˆ ) D A + B + C A + B y + C z D A + B + C Eample 4 Find the distance of a point (, 5, ) from the plane r (6iˆ ˆj + kˆ ) 4 r ur Solution Here, a iˆ + 5 ˆj kˆ, N 6iˆ ˆj + kand ˆ d 4. Therefore, the distance of the point (, 5, ) from the given plane is ( iˆ + 5 ˆj kˆ) (6 iˆ ˆj + kˆ) 4 6iˆ ˆj kˆ

209 49 MATHEMATICS.0 Angle between a Line and a Plane Definition The angle between a line and a plane is the complement of the angle between the line and normal to the plane (Fig.0). Vector form If the equation of the line is r r r a + λ b and the equation of the plane is r n r d. Then the angle θ between the line and the normal to the plane is r r b n cos θ r r b n and so the angle φ between the line and the plane is given by 90 θ, i.e., sin (90 θ) cos θ r r b n b n i.e. sin φ r r or φ sin b n b n Eample 5 Find the angle between the line and the plane 0 + y z. + y z 6 Fig.0 Solution Let θ be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have r ( iˆ + kˆ) + λ (iˆ + ˆj + 6 kˆ ) r and ( 0 iˆ + ˆj kˆ ) Here b r iˆ + ˆj + 6 kˆ r and n 0 iˆ + ˆj kˆ sin φ ( iˆ + ˆj + 6 kˆ) (0 iˆ + ˆj kˆ) or φ 8 sin

210 THREE DIMENSIONAL GEOMETRY 49 EXERCISE.. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a) z (b) + y + z (c) + y z 5 (d) 5y Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector iˆ + 5 ˆj 6 kˆ.. Find the Cartesian equation of the following planes: (a) r ( iˆ + ˆj kˆ ) (b) r (iˆ + ˆj 4 kˆ ) (c) r [( s t) iˆ + ( t) ˆj + ( s + t) kˆ ] 5 4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (a) + y + 4z 0 (b) y + 4z 6 0 (c) + y + z (d) 5y Find the vector and cartesian equations of the planes (a) that passes through the point (, 0, ) and the normal to the plane is iˆ+ ˆ j kˆ. (b) that passes through the point (,4, 6) and the normal vector to the plane is iˆ ˆj+ kˆ. 6. Find the equations of the planes that passes through three points. (a) (,, ), (6, 4, 5), ( 4,, ) (b) (,, 0), (,, ), (,, ) 7. Find the intercepts cut off by the plane + y z Find the equation of the plane with intercept on the y-ais and parallel to ZOX plane. 9. Find the equation of the plane through the intersection of the planes y + z 4 0 and + y + z 0 and the point (,, ). 0. Find the vector equation of the plane passing through the intersection of the r planes.( ˆ ˆ ˆ r i + j k ) 7,.( iˆ + 5 ˆj + kˆ ) 9 and through the point (,, ).. Find the equation of the plane through the line of intersection of the planes + y + z and + y + 4z 5 which is perpendicular to the plane y + z 0.

211 494 MATHEMATICS. Find the angle between the planes whose vector equations are r ( iˆ + ˆj kˆ ) 5 and ( iˆ ˆj + 5 kˆ ).. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. (a) 7 + 5y + 6z and y 0z (b) + y + z 0 and y (c) y + 4z and y + 6z 0 (d) y + z 0 and y + z + 0 (e) 4 + 8y + z 8 0 and y + z In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane (a) (0, 0, 0) 4y + z (b) (,, ) y + z + 0 (c) (,, 5) + y z 9 (d) ( 6, 0, 0) y + 6z 0 Miscellaneous Eamples Eample 6 A line makes angles α, β, γ and δ with the diagonals of a cube, prove that cos α + cos β + cos γ + cos δ 4 Solution A cube is a rectangular parallelopiped having equal length, breadth and height. Let OADBFEGC be the cube with each side of length a units. (Fig.) The four diagonals are OE, AF, BG and CD. Z The direction cosines of the diagonal OE which is the line joining two points O and E are i.e., a 0 a 0 a 0,, a + a + a a + a + a a + a + a ( a, 0, a) C(0, 0, a) F(0, aa, ) A( a,0,0) D( a, a, 0),, X Fig. G O E( aaa,,) Y B(0, a, 0)

212 THREE DIMENSIONAL GEOMETRY 495 Similarly, the direction cosines of AF, BG and CD are,, ;,, and,,, respectively. Let l, m, n be the direction cosines of the given line which makes angles α, β, γ, δ with OE, AF, BG, CD, respectively. Then cosα (l + m+ n); cos β ( l + m + n); cosγ (l m + n); cos δ (l + m n) (Why?) Squaring and adding, we get cos α + cos β + cos γ + cos δ [ (l + m + n ) + ( l + m + n) ] + (l m + n) + (l + m n) ] [ 4 (l + m + n ) ] 4 (as l + m + n ) Eample 7 Find the equation of the plane that contains the point (,, ) and is perpendicular to each of the planes + y z 5 and + y z 8. Solution The equation of the plane containing the given point is A ( ) + B(y + ) + C (z ) 0... () Applying the condition of perpendicularly to the plane given in () with the planes + y z 5 and + y z 8, we have A + B C 0 and A + B C 0 Solving these equations, we find A 5C and B 4C. Hence, the required equation is 5C ( ) + 4 C (y + ) + C(z ) 0 i.e. 5 4y z 7 Eample 8 Find the distance between the point P(6, 5, 9) and the plane determined by the points A (,, ), B (5,, 4) and C(,, 6). Solution Let A, B, C be the three points in the plane. D is the foot of the perpendicular drawn from a point P to the plane. PD is the required distance to be determined, which uuur uuur uuur is the projection of AP on AB AC.

213 496 MATHEMATICS uuur uuur uuur Hence, PD the dot product of AP with the unit vector along AB AC. uuur So AP iˆ + 6 ˆj + 7 kˆ iˆ ˆj kˆ uuur uuur and AB AC iˆ 6ˆj + kˆ uuur uuur iˆ 4 ˆj + kˆ Unit vector along AB AC 4 Hence PD ( i ˆ + 6 ˆj + 7 kˆ ). iˆ 4 ˆj + kˆ Alternatively, find the equation of the plane passing through A, B and C and then compute the distance of the point P from the plane. Eample 9 Show that the lines a + d α δ y a z a d α α + δ and b + c β γ y b z b c β β + γ are coplanar. Solution Here a d b c y a y b z a + d z b + c a α δ a β γ b α b β c α + δ c β + γ Now consider the determinant y y z z a b c a b c b c a+ d b a b + c a d α δ α α+δ β γ β β+γ

214 Adding third column to the first column, we get b a b a b + c a d α α α + δ β β β + γ THREE DIMENSIONAL GEOMETRY 497 Since the first and second columns are identical. Hence, the given two lines are coplanar. Eample 0 Find the coordinates of the point where the line through the points A (, 4, ) and B(5,, 6) crosses the XY-plane. Solution The vector equation of the line through the points A and B is r iˆ + 4 ˆj + kˆ + λ[ (5 ) iˆ + ( 4) ˆj + (6 ) kˆ ] i.e. r iˆ + 4 ˆj + kˆ + λ ( iˆ ˆj + 5 kˆ )... () Let P be the point where the line AB crosses the XY-plane. Then the position vector of the point P is of the form iˆ + y ˆj. This point must satisfy the equation (). (Why?) i.e. iˆ + y ˆj (+ λ ) iˆ + ( 4 λ ) ˆj + (+ 5 λ) kˆ Equating the like coefficients of iˆ, ˆj and k ˆ, we have + λ y 4 λ λ Solving the above equations, we get Hence, the coordinates of the required point are and y 5 5, 5 5 0, 0. Miscellaneous Eercise on Chapter. Show that the line joining the origin to the point (,, ) is perpendicular to the line determined by the points (, 5, ), (4,, ).. If l, m, n and l, m, n are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these m n m n, n l n l, l m l m are

215 498 MATHEMATICS. Find the angle between the lines whose direction ratios are a, b, c and b c, c a, a b. 4. Find the equation of a line parallel to -ais and passing through the origin. 5. If the coordinates of the points A, B, C, D be (,, ), (4, 5, 7), ( 4,, 6) and (, 9, ) respectively, then find the angle between the lines AB and CD If the lines y z and y z are perpendicular, k k 5 find the value of k. 7. Find the vector equation of the line passing through (,, ) and perpendicular to the plane r. ( iˆ + ˆj 5 kˆ) Find the equation of the plane passing through (a, b, c) and parallel to the plane r ( iˆ+ ˆj + kˆ ). r 9. Find the shortest distance between lines 6iˆ + ˆj + kˆ + λ ( iˆ ˆj + kˆ) r and 4 iˆ kˆ +µ (iˆ ˆj kˆ). 0. Find the coordinates of the point where the line through (5,, 6) and (, 4,) crosses the YZ-plane.. Find the coordinates of the point where the line through (5,, 6) and (, 4, ) crosses the ZX-plane.. Find the coordinates of the point where the line through (, 4, 5) and (,, ) crosses the plane + y + z 7.. Find the equation of the plane passing through the point (,, ) and perpendicular to each of the planes + y + z 5 and + y + z If the points (,, p) and (, 0, ) be equidistant from the plane r ( iˆ+ 4 ˆj kˆ ) + 0, then find the value of p. 5. Find the equation of the plane passing through the line of intersection of the planes r ( iˆ+ ˆj + kˆ ) and r ( iˆ+ ˆj kˆ ) and parallel to -ais. 6. If O be the origin and the coordinates of P be (,, ), then find the equation of the plane passing through P and perpendicular to OP. 7. Find the equation of the plane which contains the line of intersection of the planes r ( iˆ+ ˆj + kˆ ) 4 0, r ( iˆ+ ˆj kˆ ) and which is perpendicular to the plane r (5 iˆ+ ˆj 6 kˆ )

216 THREE DIMENSIONAL GEOMETRY Find the distance of the point (, 5, 0) from the point of intersection of the r line iˆ ˆj + kˆ+λ (iˆ+ 4 ˆj+ kˆ) and the plane r ( iˆ ˆj + kˆ ) Find the vector equation of the line passing through (,, ) and parallel to the planes r ( iˆ ˆj + kˆ ) 5 and r ( iˆ+ ˆj + kˆ ) Find the vector equation of the line passing through the point (,, 4) and perpendicular to the two lines: 8 y + 9 z 0 5 y 9 z 5 and Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then + +. a b c p Choose the correct answer in Eercises and.. Distance between the two planes: + y + 4z 4 and 4 + 6y + 8z is (A) units (B) 4 units (C) 8 units (D) 9 units. The planes: y + 4z 5 and 5.5y + 0z 6 are (A) Perpendicular (B) Parallel (C) intersect y-ais Summary (D) passes through 5 0,0, 4 Direction cosines of a line are the cosines of the angles made by the line with the positive directions of the coordinate aes. If l, m, n are the direction cosines of a line, then l + m + n. Direction cosines of a line joining two points P(, y, z ) and Q(, y, z ) are, y y, z z PQ PQ PQ where PQ ( ) ( + + z z ) ( y y) Direction ratios of a line are the numbers which are proportional to the direction cosines of a line. If l, m, n are the direction cosines and a, b, c are the direction ratios of a line

217 500 MATHEMATICS then a b c l ; m ; n a + b + c a + b + c a + b + c Skew lines are lines in space which are neither parallel nor intersecting. They lie in different planes. Angle between skew lines is the angle between two intersecting lines drawn from any point (preferably through the origin) parallel to each of the skew lines. If l, m, n and l, m, n are the direction cosines of two lines; and θ is the acute angle between the two lines; then cosθ l l + m m + n n If a, b, c and a, b, c are the direction ratios of two lines and θ is the acute angle between the two lines; then a a + b b + c c cosθ a + b + c a + b + c Vector equation of a line that passes through the given point whose position vector is a r and parallel to a given vector b r r r is a + λ b r. Equation of a line through a point (, y, z ) and having direction cosines l, m, n is y y z z l m n The vector equation of a line which passes through two points whose position vectors are a r and b r r r r r is a + λ ( b a). Cartesian equation of a line that passes through two points (, y, z ) and (, y, z ) is y y z z. y y z z If θ is the acute angle between r r a +λb r r r and a +λb r, then r r b b cosθ r r b b If y y z z y y z z and l m n l m n are the equations of two lines, then the acute angle between the two lines is given by cos θ l l + m m + n n.

218 THREE DIMENSIONAL GEOMETRY 50 Shortest distance between two skew lines is the line segment perpendicular to both the lines. Shortest distance between r r a + λb r r r and a +µ b r is r r r r ( b b) ( a a) r r b b Shortest distance between the lines: y y z z and a b c y y z z a b c is y y z z a b c a b c + + ( bc bc ) ( ca c a ) ( ab a b) Distance between parallel lines r r a +λb r r r and a +µ b r is r r r b ( a a) r b In the vector form, equation of a plane which is at a distance d from the origin, and nˆ is the unit vector normal to the plane through the origin is r nˆ d. Equation of a plane which is at a distance of d from the origin and the direction cosines of the normal to the plane as l, m, n is l + my + nz d. The equation of a plane through a point whose position vector is ar and perpendicular to the vector ur r r ur N is ( a ).N 0. Equation of a plane perpendicular to a given line with direction ratios A, B, C and passing through a given point (, y, z ) is A ( ) + B (y y ) + C (z z ) 0 Equation of a plane passing through three non collinear points (, y, z ),

219 50 MATHEMATICS (, y, z ) and (, y, z ) is y y y y y y z z z z z z 0 Vector equation of a plane that contains three non collinear points having position vectors a r b r r r r r r r r, and c is ( a ).[( b a ) ( c a ) ] 0 Equation of a plane that cuts the coordinates aes at (a, 0, 0), (0, b, 0) and (0, 0, c) is a + y b + z c Vector equation of a plane that passes through the intersection of r r r r r r r planes n d and n d is ( n +λ n) d +λd, where λ is any nonzero constant. Vector equation of a plane that passes through the intersection of two given planes A + B y + C z + D 0 and A + B y + C z + D 0 is (A + B y + C z + D ) + λ(a + B y + C z + D ) 0. Two planes r r a + λb r r r and a + µ b r are coplanar if r r r r ( a a ) ( b b ) 0 Two planes a + b y + c z + d 0 and a + b y + c z + d 0 are coplanar if a a y y b b z z c c In the vector form, if θ is the angle between the two planes, r r n d and r n r r r n d, then θ cos n r r. n n r The angle φ between the line r a +λr r band the plane nˆ d is r b nˆ sin φ r b nˆ 0.

220 THREE DIMENSIONAL GEOMETRY 50 The angle θ between the planes A + B y + C z + D 0 and A + B y + C z + D 0 is given by A A + B B + C C cos θ A + B + C A + B + C The distance of a point whose position vector is ar from the plane r nˆ d is r d a nˆ The distance from a point (, y, z ) to the plane A + By + Cz + D 0 is A + By + Cz + D. A + B + C v

221 504 MATHEMATICS Chapter LINEAR PROGRAMMING The mathematical eperience of the student is incomplete if he never had the opportunity to solve a problem invented by himself. G. POLYA. Introduction In earlier classes, we have discussed systems of linear equations and their applications in day to day problems. In Class XI, we have studied linear inequalities and systems of linear inequalities in two variables and their solutions by graphical method. Many applications in mathematics involve systems of inequalities/equations. In this chapter, we shall apply the systems of linear inequalities/equations to solve some real life problems of the type as given below: A furniture dealer deals in only two items tables and chairs. He has Rs 50,000 to invest and has storage space of at most 60 pieces. A table costs Rs 500 and a chair Rs 500. He estimates that from the sale of one table, he can make a profit of Rs 50 and that from the sale of one L. Kantorovich chair a profit of Rs 75. He wants to know how many tables and chairs he should buy from the available money so as to maimise his total profit, assuming that he can sell all the items which he buys. Such type of problems which seek to maimise (or, minimise) profit (or, cost) form a general class of problems called optimisation problems. Thus, an optimisation problem may involve finding maimum profit, minimum cost, or minimum use of resources etc. A special but a very important class of optimisation problems is linear programming problem. The above stated optimisation problem is an eample of linear programming problem. Linear programming problems are of much interest because of their wide applicability in industry, commerce, management science etc. In this chapter, we shall study some linear programming problems and their solutions by graphical method only, though there are many other methods also to solve such problems.

222 LINEAR PROGRAMMING 505. Linear Programming Problem and its Mathematical Formulation We begin our discussion with the above eample of furniture dealer which will further lead to a mathematical formulation of the problem in two variables. In this eample, we observe (i) The dealer can invest his money in buying tables or chairs or combination thereof. Further he would earn different profits by following different investment strategies. (ii) There are certain overriding conditions or constraints viz., his investment is limited to a maimum of Rs 50,000 and so is his storage space which is for a maimum of 60 pieces. Suppose he decides to buy tables only and no chairs, so he can buy , i.e., 0 tables. His profit in this case will be Rs (50 0), i.e., Rs Suppose he chooses to buy chairs only and no tables. With his capital of Rs 50,000, he can buy , i.e. 00 chairs. But he can store only 60 pieces. Therefore, he is forced to buy only 60 chairs which will give him a total profit of Rs (60 75), i.e., Rs There are many other possibilities, for instance, he may choose to buy 0 tables and 50 chairs, as he can store only 60 pieces. Total profit in this case would be Rs ( ), i.e., Rs 650 and so on. We, thus, find that the dealer can invest his money in different ways and he would earn different profits by following different investment strategies. Now the problem is : How should he invest his money in order to get maimum profit? To answer this question, let us try to formulate the problem mathematically... Mathematical formulation of the problem Let be the number of tables and y be the number of chairs that the dealer buys. Obviously, and y must be non-negative, i.e., 0...() (Non-negative constraints) y 0... () The dealer is constrained by the maimum amount he can invest (Here it is Rs 50,000) and by the maimum number of items he can store (Here it is 60). Stated mathematically, y (investment constraint) or 5 + y () and + y 60 (storage constraint)... (4)

223 506 MATHEMATICS The dealer wants to invest in such a way so as to maimise his profit, say, Z which stated as a function of and y is given by Z y (called objective function)... (5) Mathematically, the given problems now reduces to: Maimise Z y subject to the constraints: 5 + y 00 + y 60 0, y 0 So, we have to maimise the linear function Z subject to certain conditions determined by a set of linear inequalities with variables as non-negative. There are also some other problems where we have to minimise a linear function subject to certain conditions determined by a set of linear inequalities with variables as non-negative. Such problems are called Linear Programming Problems. Thus, a Linear Programming Problem is one that is concerned with finding the optimal value (maimum or minimum value) of a linear function (called objective function) of several variables (say and y), subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). The term linear implies that all the mathematical relations used in the problem are linear relations while the term programming refers to the method of determining a particular programme or plan of action. Before we proceed further, we now formally define some terms (which have been used above) which we shall be using in the linear programming problems: Objective function Linear function Z a + by, where a, b are constants, which has to be maimised or minimized is called a linear objective function. In the above eample, Z y is a linear objective function. Variables and y are called decision variables. Constraints The linear inequalities or equations or restrictions on the variables of a linear programming problem are called constraints. The conditions 0, y 0 are called non-negative restrictions. In the above eample, the set of inequalities () to (4) are constraints. Optimisation problem A problem which seeks to maimise or minimise a linear function (say of two variables and y) subject to certain constraints as determined by a set of linear inequalities is called an optimisation problem. Linear programming problems are special type of optimisation problems. The above problem of investing a

224 LINEAR PROGRAMMING 507 given sum by the dealer in purchasing chairs and tables is an eample of an optimisation problem as well as of a linear programming problem. We will now discuss how to find solutions to a linear programming problem. In this chapter, we will be concerned only with the graphical method... Graphical method of solving linear programming problems In Class XI, we have learnt how to graph a system of linear inequalities involving two variables and y and to find its solutions graphically. Let us refer to the problem of investment in tables and chairs discussed in Section.. We will now solve this problem graphically. Let us graph the constraints stated as linear inequalities: 5 + y () + y () 0... () y 0... (4) The graph of this system (shaded region) consists of the points common to all half planes determined by the inequalities () to (4) (Fig.). Each point in this region represents a feasible choice open to the dealer for investing in tables and chairs. The region, therefore, is called the feasible region for the problem. Every point of this region is called a feasible solution to the problem. Thus, we have, Feasible region The common region determined by all the constraints including non-negative constraints, y 0 of a linear programming problem is called the feasible region (or solution region) for the problem. In Fig., the region OABC (shaded) is the feasible region for the problem. The region other than feasible region is called an infeasible region. Feasible solutions Points within and on the boundary of the feasible region represent feasible solutions of the constraints. In Fig., every point within and on the boundary of the feasible region OABC represents feasible solution to the problem. For eample, the point (0, 50) is a feasible solution of the problem and so are the points (0, 60), (0, 0) etc. Any point outside the feasible region is called an infeasible solution. For eample, the point (5, 40) is an infeasible solution of the problem. Fig.

225 508 MATHEMATICS Optimal (feasible) solution: Any point in the feasible region that gives the optimal value (maimum or minimum) of the objective function is called an optimal solution. Now, we see that every point in the feasible region OABC satisfies all the constraints as given in () to (4), and since there are infinitely many points, it is not evident how we should go about finding a point that gives a maimum value of the objective function Z y. To handle this situation, we use the following theorems which are fundamental in solving linear programming problems. The proofs of these theorems are beyond the scope of the book. Theorem Let R be the feasible region (conve polygon) for a linear programming problem and let Z a + by be the objective function. When Z has an optimal value (maimum or minimum), where the variables and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point* (verte) of the feasible region. Theorem Let R be the feasible region for a linear programming problem, and let Z a + by be the objective function. If R is bounded**, then the objective function Z has both a maimum and a minimum value on R and each of these occurs at a corner point (verte) of R. Remark If R is unbounded, then a maimum or a minimum value of the objective function may not eist. However, if it eists, it must occur at a corner point of R. (By Theorem ). In the above eample, the corner points (vertices) of the bounded (feasible) region are: O, A, B and C and it is easy to find their coordinates as (0, 0), (0, 0), (0, 50) and (0, 60) respectively. Let us now compute the values of Z at these points. We have Verte of the Corresponding value Feasible Region of Z (in Rs) O (0,0) 0 A (0,60) 4500 B (0,50) 650 C (0,0) 5000 Maimum * A corner point of a feasible region is a point in the region which is the intersection of two boundary lines. ** A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle. Otherwise, it is called unbounded. Unbounded means that the feasible region does etend indefinitely in any direction.

226 LINEAR PROGRAMMING 509 We observe that the maimum profit to the dealer results from the investment strategy (0, 50), i.e. buying 0 tables and 50 chairs. This method of solving linear programming problem is referred as Corner Point Method. The method comprises of the following steps:. Find the feasible region of the linear programming problem and determine its corner points (vertices) either by inspection or by solving the two equations of the lines intersecting at that point.. Evaluate the objective function Z a + by at each corner point. Let M and m, respectively denote the largest and smallest values of these points.. (i) When the feasible region is bounded, M and m are the maimum and minimum values of Z. (ii) In case, the feasible region is unbounded, we have: 4. (a) M is the maimum value of Z, if the open half plane determined by a + by > M has no point in common with the feasible region. Otherwise, Z has no maimum value. (b) Similarly, m is the minimum value of Z, if the open half plane determined by a + by < m has no point in common with the feasible region. Otherwise, Z has no minimum value. We will now illustrate these steps of Corner Point Method by considering some eamples: Eample Solve the following linear programming problem graphically: Maimise Z 4 + y... () subject to the constraints: + y () + y () 0, y 0... (4) Solution The shaded region in Fig. is the feasible region determined by the system of constraints () to (4). We observe that the feasible region OABC is bounded. So, we now use Corner Point Method to determine the maimum value of Z. The coordinates of the corner points O, A, B and C are (0, 0), (0, 0), (0, 0) and (0, 50) respectively. Now we evaluate Z at each corner point.

227 50 MATHEMATICS Corner Point Corresponding value of Z (0, 0) 0 (0, 0) 0 Maimum (0, 0) 0 (0, 50) 50 Fig. Hence, maimum value of Z is 0 at the point (0, 0). Eample Solve the following linear programming problem graphically: Minimise Z y... () subject to the constraints: + y 0... () + 4y 4... () 0, y 0... (4) Solution The shaded region in Fig. is the feasible region ABC determined by the system of constraints () to (4), which is bounded. The coordinates of corner points Corner Point Corresponding value of Z (0, 5) 500 (4, ) 00 (0, 6) 000 Minimum Fig.

228 LINEAR PROGRAMMING 5 A, B and C are (0,5), (4,) and (0,6) respectively. Now we evaluate Z y at these points. Hence, minimum value of Z is 00 attained at the point (4, ) Eample Solve the following problem graphically: Minimise and Maimise Z + 9y... () subject to the constraints: + y () + y 0... () y... (4) 0, y 0... (5) Solution First of all, let us graph the feasible region of the system of linear inequalities () to (5). The feasible region ABCD is shown in the Fig.4. Note that the region is bounded. The coordinates of the corner points A, B, C and D are (0, 0), (5, 5), (5,5) and (0, 0) respectively. Corner Corresponding value of Point Z + 9y A (0, 0) 90 B (5, 5) 60 C (5, 5) D (0, 0) 80} Minimum Maimum (Multiple optimal solutions) Fig.4 We now find the minimum and maimum value of Z. From the table, we find that the minimum value of Z is 60 at the point B (5, 5) of the feasible region. The maimum value of Z on the feasible region occurs at the two corner points C (5, 5) and D (0, 0) and it is 80 in each case. Remark Observe that in the above eample, the problem has multiple optimal solutions at the corner points C and D, i.e. the both points produce same maimum value 80. In such cases, you can see that every point on the line segment CD joining the two corner points C and D also give the same maimum value. Same is also true in the case if the two points produce same minimum value.

229 5 MATHEMATICS Eample 4 Determine graphically the minimum value of the objective function Z y... () subject to the constraints: y 5... () + y... () y... (4) 0, y 0... (5) Solution First of all, let us graph the feasible region of the system of inequalities () to (5). The feasible region (shaded) is shown in the Fig.5. Observe that the feasible region is unbounded. We now evaluate Z at the corner points. Corner Point Z y (0, 5) 00 (0, ) 60 (, 0) 50 (6, 0) 00 smallest Fig.5 From this table, we find that 00 is the smallest value of Z at the corner point (6, 0). Can we say that minimum value of Z is 00? Note that if the region would have been bounded, this smallest value of Z is the minimum value of Z (Theorem ). But here we see that the feasible region is unbounded. Therefore, 00 may or may not be the minimum value of Z. To decide this issue, we graph the inequality y < 00 (see Step (ii) of corner Point Method.) i.e., 5 + y < 0 and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then 00 will not be the minimum value of Z. Otherwise, 00 will be the minimum value of Z.

230 LINEAR PROGRAMMING 5 As shown in the Fig.5, it has common points. Therefore, Z y has no minimum value subject to the given constraints. In the above eample, can you say whether z y has the maimum value 00 at (0,5)? For this, check whether the graph of y > 00 has points in common with the feasible region. (Why?) Eample 5 Minimise Z + y subject to the constraints: + y 8... () + 5y 5... () 0, y 0... () Solution Let us graph the inequalities () to () (Fig.6). Is there any feasible region? Why is so? From Fig.6, you can see that there is no point satisfying all the constraints simultaneously. Thus, the problem is having no feasible region and hence no feasible solution. Remarks From the eamples which we have discussed so far, we notice some general features of linear programming problems: (i) The feasible region is always a conve region. Fig.6 (ii) The maimum (or minimum) solution of the objective function occurs at the verte (corner) of the feasible region. If two corner points produce the same maimum (or minimum) value of the objective function, then every point on the line segment joining these points will also give the same maimum (or minimum) value. EXERCISE. Solve the following Linear Programming Problems graphically:. Maimise Z + 4y subject to the constraints : + y 4, 0, y 0.

231 54 MATHEMATICS. Minimise Z + 4 y subject to + y 8, + y, 0, y 0.. Maimise Z 5 + y subject to + 5y 5, 5 + y 0, 0, y Minimise Z + 5y such that + y, + y,, y Maimise Z + y subject to + y 0, + y 5,, y Minimise Z + y subject to + y, + y 6,, y 0. Show that the minimum of Z occurs at more than two points. 7. Minimise and Maimise Z y subject to + y 0, + y 60, y 0,, y Minimise and Maimise Z + y subject to + y 00, y 0, + y 00;, y Maimise Z + y, subject to the constraints:, + y 5, + y 6, y Maimise Z + y, subject to y, + y 0,, y 0.. Different Types of Linear Programming Problems A few important linear programming problems are listed below:. Manufacturing problems In these problems, we determine the number of units of different products which should be produced and sold by a firm when each product requires a fied manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output etc., in order to make maimum profit.. Diet problems In these problems, we determine the amount of different kinds of constituents/nutrients which should be included in a diet so as to minimise the cost of the desired diet such that it contains a certain minimum amount of each constituent/nutrients.. Transportation problems In these problems, we determine a transportation schedule in order to find the cheapest way of transporting a product from plants/factories situated at different locations to different markets.

232 LINEAR PROGRAMMING 55 Let us now solve some of these types of linear programming problems: Eample 6 (Diet problem): A dietician wishes to mi two types of foods in such a way that vitamin contents of the miture contain atleast 8 units of vitamin A and 0 units of vitamin C. Food I contains units/kg of vitamin A and unit/kg of vitamin C. Food II contains unit/kg of vitamin A and units/kg of vitamin C. It costs Rs 50 per kg to purchase Food I and Rs 70 per kg to purchase Food II. Formulate this problem as a linear programming problem to minimise the cost of such a miture. Solution Let the miture contain kg of Food I and y kg of Food II. Clearly, 0, y 0. We make the following table from the given data: Resources Food Requirement I II () (y) Vitamin A 8 (units/kg) Vitamin C 0 (units/kg) Cost (Rs/kg) Since the miture must contain at least 8 units of vitamin A and 0 units of vitamin C, we have the constraints: + y 8 + y 0 Total cost Z of purchasing kg of food I and y kg of Food II is Z y Hence, the mathematical formulation of the problem is: Minimise Z y... () subject to the constraints: + y 8... () + y 0... (), y 0... (4) Let us graph the inequalities () to (4). The feasible region determined by the system is shown in the Fig.7. Here again, observe that the feasible region is unbounded.

233 56 MATHEMATICS Let us evaluate Z at the corner points A(0,8), B(,4) and C(0,0). Corner Point Z y (0,8) 560 (,4) 80 (0,0) 500 Minimum Fig.7 In the table, we find that smallest value of Z is 80 at the point (,4). Can we say that the minimum value of Z is 80? Remember that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality y < 80 i.e., 5 + 7y < 8 to check whether the resulting open half plane has any point common with the feasible region. From the Fig.7, we see that it has no points in common. Thus, the minimum value of Z is 80 attained at the point (, 4). Hence, the optimal miing strategy for the dietician would be to mi kg of Food I and 4 kg of Food II, and with this strategy, the minimum cost of the miture will be Rs 80. Eample 7 (Allocation problem) A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profit from crops X and Y per hectare are estimated as Rs 0,500 and Rs 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 0 litres and 0 litres per hectare. Further, no more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maimise the total profit of the society? Solution Let hectare of land be allocated to crop X and y hectare to crop Y. Obviously, 0, y 0. Profit per hectare on crop X Rs 0500 Profit per hectare on crop Y Rs 9000 Therefore, total profit Rs ( y)

234 LINEAR PROGRAMMING 57 The mathematical formulation of the problem is as follows: Maimise Z y subject to the constraints: + y 50 (constraint related to land)... () 0 + 0y 800 (constraint related to use of herbicide) i.e. + y () 0, y 0 (non negative constraint)... () Let us draw the graph of the system of inequalities () to (). The feasible region OABC is shown (shaded) in the Fig.8. Observe that the feasible region is bounded. The coordinates of the corner points O, A, B and C are (0, 0), (40, 0), (0, 0) and (0, 50) respectively. Let us evaluate the objective function Z y at these vertices to find which one gives the maimum profit. Corner Point Z y O(0, 0) 0 A( 40, 0) B(0, 0) C(0,50) Maimum Fig.8 Hence, the society will get the maimum profit of Rs 4,95,000 by allocating 0 hectares for crop X and 0 hectares for crop Y. Eample 8 (Manufacturing problem) A manufacturing company makes two models A and B of a product. Each piece of Model A requires 9 labour hours for fabricating and labour hour for finishing. Each piece of Model B requires labour hours for fabricating and labour hours for finishing. For fabricating and finishing, the maimum labour hours available are 80 and 0 respectively. The company makes a profit of Rs 8000 on each piece of model A and Rs 000 on each piece of Model B. How many pieces of Model A and Model B should be manufactured per week to realise a maimum profit? What is the maimum profit per week?

235 58 MATHEMATICS Solution Suppose is the number of pieces of Model A and y is the number of pieces of Model B. Then Total profit (in Rs) y Let Z y We now have the following mathematical model for the given problem. Maimise Z y... () subject to the constraints: 9 + y 80 (Fabricating constraint) i.e. + 4y () + y 0 (Finishing constraint)... () 0, y 0 (non-negative constraint)... (4) The feasible region (shaded) OABC determined by the linear inequalities () to (4) is shown in the Fig.9. Note that the feasible region is bounded. Fig.9 Let us evaluate the objective function Z at each corner point as shown below: Corner Point Z y 0 (0, 0) 0 A (0, 0) B (, 6) Maimum C (0, 0) 0000 We find that maimum value of Z is,68,000 at B (, 6). Hence, the company should produce pieces of Model A and 6 pieces of Model B to realise maimum profit and maimum profit then will be Rs,68,000.

236 LINEAR PROGRAMMING 59 EXERCISE.. Reshma wishes to mi two types of food P and Q in such a way that the vitamin contents of the miture contain at least 8 units of vitamin A and units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains units/kg of Vitamin A and 5 units / kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and units/kg of vitamin B. Determine the minimum cost of the miture.. One kind of cake requires 00g of flour and 5g of fat, and another kind of cake requires 00g of flour and 50g of fat. Find the maimum number of cakes which can be made from 5kg of flour and kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.. A factory makes tennis rackets and cricket bats. A tennis racket takes.5 hours of machine time and hours of craftman s time in its making while a cricket bat takes hour of machine time and hour of craftman s time. In a day, the factory has the availability of not more than 4 hours of machine time and 4 hours of craftsman s time. (i) What number of rackets and bats must be made if the factory is to work at full capacity? (ii) If the profit on a racket and on a bat is Rs 0 and Rs 0 respectively, find the maimum profit of the factory when it works at full capacity. 4. A manufacturer produces nuts and bolts. It takes hour of work on machine A and hours on machine B to produce a package of nuts. It takes hours on machine A and hour on machine B to produce a package of bolts. He earns a profit of Rs7.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maimise his profit, if he operates his machines for at the most hours a day? 5. A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 0. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maimise his profit? Determine the maimum profit.

237 50 MATHEMATICS 6. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes hours on grinding/cutting machine and hours on the sprayer to manufacture a pedestal lamp. It takes hour on the grinding/cutting machine and hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 0 hours and the grinding/cutting machine for at the most hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maimise his profit? 7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 0 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are hours 0 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maimise the profit? 8. A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 5000 and Rs respectively. He estimates that the total monthly demand of computers will not eceed 50 units. Determine the number of units of each type of computers which the merchant should stock to get maimum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs A diet is to contain at least 80 units of vitamin A and 00 units of minerals. Two foods F and F are available. Food F costs Rs 4 per unit food and F costs Rs 6 per unit. One unit of food F contains units of vitamin A and 4 units of minerals. One unit of food F contains 6 units of vitamin A and units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of miture of these two foods and also meets the minimal nutritional requirements. 0. There are two types of fertilisers F and F. F consists of 0% nitrogen and 6% phosphoric acid and F consists of 5% nitrogen and 0% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 4 kg of nitrogen and 4 kg of phosphoric acid for her crop. If F costs Rs 6/kg and F costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?. The corner points of the feasible region determined by the following system of linear inequalities: + y 0, + y 5,, y 0 are (0, 0), (5, 0), (, 4) and (0, 5). Let Z p + qy, where p, q > 0. Condition on p and q so that the maimum of Z occurs at both (, 4) and (0, 5) is (A) p q (B) p q (C) p q (D) q p

238 LINEAR PROGRAMMING 5 Miscellaneous Eamples Eample 9 (Diet problem) A dietician has to develop a special diet using two foods P and Q. Each packet (containing 0 g) of food P contains units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains units of calcium, 0 units of iron, 4 units of cholesterol and units of vitamin A. The diet requires atleast 40 units of calcium, atleast 460 units of iron and at most 00 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A? Solution Let and y be the number of packets of food P and Q respectively. Obviously 0, y 0. Mathematical formulation of the given problem is as follows: Minimise Z 6 + y (vitamin A) subject to the constraints + y 40 (constraint on calcium), i.e. 4 + y () 4 + 0y 460 (constraint on iron), i.e. + 5y 5... () 6 + 4y 00 (constraint on cholesterol), i.e. + y () 0, y 0... (4) Let us graph the inequalities () to (4). The feasible region (shaded) determined by the constraints () to (4) is shown in Fig.0 and note that it is bounded. Fig.0

239 5 MATHEMATICS The coordinates of the corner points L, M and N are (, 7), (5, 0) and (40, 5) respectively. Let us evaluate Z at these points: Corner Point Z 6 + y (, 7) 8 (5, 0) 50 Minimum (40, 5) 85 From the table, we find that Z is minimum at the point (5, 0). Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if 5 packets of food P and 0 packets of food Q are used in the special diet. The minimum amount of vitamin A will be 50 units. Eample 0 (Manufacturing problem) A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines. The number of hours required for producing unit of each of M and N on the three machines are given in the following table: Items Number of hours required on machines I II III M N.5 She makes a profit of Rs 600 and Rs 400 on items M and N respectively. How many of each item should she produce so as to maimise her profit assuming that she can sell all the items that she produced? What will be the maimum profit? Solution Let and y be the number of items M and N respectively. Total profit on the production Rs ( y) Mathematical formulation of the given problem is as follows: Maimise Z y subject to the constraints: + y (constraint on Machine I)... () + y (constraint on Machine II)... () y 5 (constraint on Machine III)... () 0, y 0... (4)

240 LINEAR PROGRAMMING 5 Let us draw the graph of constraints () to (4). ABCDE is the feasible region (shaded) as shown in Fig. determined by the constraints () to (4). Observe that the feasible region is bounded, coordinates of the corner points A, B, C, D and E are (5, 0) (6, 0), (4, 4), (0, 6) and (0, 4) respectively. Fig. Let us evaluate Z y at these corner points. Corner point Z y (5, 0) 000 (6, 0) 600 (4, 4) 4000 Maimum (0, 6) 400 (0, 4) 600 We see that the point (4, 4) is giving the maimum value of Z. Hence, the manufacturer has to produce 4 units of each item to get the maimum profit of Rs Eample (Transportation problem) There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of

241 54 MATHEMATICS transportation per unit is given below: From/To Cost (in Rs) A B C P Q How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost? Solution The problem can be eplained diagrammatically as follows (Fig.): Let units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then (8 y) units will be transported to depot at C (Why?) Factory P 8 units Rs 60 y Rs 00 Rs 50 8 y Depot A 5 units B 5 units Rs y Rs 0 Q 6 units Factory Fig. Depot Rs 00 6 [(5 )+(5 y)] C 4 units Depot Hence, we have 0, y 0 and 8 y 0 i.e. 0, y 0 and + y 8 Now, the weekly requirement of the depot at A is 5 units of the commodity. Since units are transported from the factory at P, the remaining (5 ) units need to be transported from the factory at Q. Obviously, 5 0, i.e. 5. Similarly, (5 y) and 6 (5 + 5 y) + y 4 units are to be transported from the factory at Q to the depots at B and C respectively. Thus, 5 y 0, + y 4 0 i.e. y 5, + y 4

242 LINEAR PROGRAMMING 55 Total transportation cost Z is given by Z y + 00 ( 5 ) + 0 (5 y) + 00 ( + y 4) + 50 (8 y) 0 ( 7 y + 90) Therefore, the problem reduces to Minimise Z 0 ( 7y + 90) subject to the constraints: 0, y 0... () + y 8... () 5... () y 5... (4) and + y 4... (5) The shaded region ABCDEF represented by the constraints () to (5) is the feasible region (Fig.). Fig. Observe that the feasible region is bounded. The coordinates of the corner points of the feasible region are (0, 4), (0, 5), (, 5), (5, ), (5, 0) and (4, 0). Let us evaluate Z at these points. Corner Point Z 0 ( 7 y + 90) (0, 4) 60 (0, 5) 550 Minimum (, 5) 580 (5, ) 740 (5, 0) 950 (4, 0) 940 From the table, we see that the minimum value of Z is 550 at the point (0, 5). Hence, the optimal transportation strategy will be to deliver 0, 5 and units from the factory at P and 5, 0 and units from the factory at Q to the depots at A, B and C respectively. Corresponding to this strategy, the transportation cost would be minimum, i.e., Rs 550. Miscellaneous Eercise on Chapter. Refer to Eample 9. How many packets of each food should be used to maimise the amount of vitamin A in the diet? What is the maimum amount of vitamin A in the diet?

243 56 MATHEMATICS. A farmer mies two brands P and Q of cattle feed. Brand P, costing Rs 50 per bag, contains units of nutritional element A,.5 units of element B and units of element C. Brand Q costing Rs 00 per bag contains.5 units of nutritional element A,.5 units of element B, and units of element C. The minimum requirements of nutrients A, B and C are 8 units, 45 units and 4 units respectively. Determine the number of bags of each brand which should be mied in order to produce a miture having a minimum cost per bag? What is the minimum cost of the miture per bag?. A dietician wishes to mi together two kinds of food X and Y in such a way that the miture contains at least 0 units of vitamin A, units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below: Food Vitamin A Vitamin B Vitamin C X Y One kg of food X costs Rs 6 and one kg of food Y costs Rs 0. Find the least cost of the miture which will produce the required diet? 4. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below: Types of Toys Machines I II III A 8 6 B Each machine is available for a maimum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 5 toys of type A and 0 of type B should be manufactured in a day to get maimum profit. 5. An aeroplane can carry a maimum of 00 passengers. A profit of Rs 000 is made on each eecutive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 0 seats for eecutive class. However, at least 4 times as many passengers prefer to travel by economy class than by the eecutive class. Determine how many tickets of each type must be sold in order to maimise the profit for the airline. What is the maimum profit?

244 LINEAR PROGRAMMING Two godowns A and B have grain capacity of 00 quintals and 50 quintals respectively. They supply to ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table: Transportation cost per quintal (in Rs) From/To A B D 6 4 E F.50 How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost? 7. An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 000L and 500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table: Distance in (km.) From / To A B D 7 E 6 4 F Assuming that the transportation cost of 0 litres of oil is Re per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost? 8. A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 40 kg of phosphoric acid, at least 70 kg of potash and at most 0 kg of chlorine. If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

245 58 MATHEMATICS kg per bag Brand P Brand Q Nitrogen.5 Phosphoric acid Potash.5 Chlorine.5 9. Refer to Question 8. If the grower wants to maimise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maimum amount of nitrogen added? 0. A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not eceed 00 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can eceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs and Rs 6 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maimise the profit? Summary A linear programming problem is one that is concerned with finding the optimal value (maimum or minimum) of a linear function of several variables (called objective function) subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). Variables are sometimes called decision variables and are non-negative. A few important linear programming problems are: (i) Diet problems (ii) Manufacturing problems (iii) Transportation problems The common region determined by all the constraints including the non-negative constraints 0, y 0 of a linear programming problem is called the feasible region (or solution region) for the problem. Points within and on the boundary of the feasible region represent feasible solutions of the constraints. Any point outside the feasible region is an infeasible solution.

246 LINEAR PROGRAMMING 59 Any point in the feasible region that gives the optimal value (maimum or minimum) of the objective function is called an optimal solution. The following Theorems are fundamental in solving linear programming problems: Theorem Let R be the feasible region (conve polygon) for a linear programming problem and let Z a + by be the objective function. When Z has an optimal value (maimum or minimum), where the variables and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (verte) of the feasible region. Theorem Let R be the feasible region for a linear programming problem, and let Z a + by be the objective function. If R is bounded, then the objective function Z has both a maimum and a minimum value on R and each of these occurs at a corner point (verte) of R. If the feasible region is unbounded, then a maimum or a minimum may not eist. However, if it eists, it must occur at a corner point of R. Corner point method for solving a linear programming problem. The method comprises of the following steps: (i) Find the feasible region of the linear programming problem and determine its corner points (vertices). (ii) Evaluate the objective function Z a + by at each corner point. Let M and m respectively be the largest and smallest values at these points. (iii) If the feasible region is bounded, M and m respectively are the maimum and minimum values of the objective function. If the feasible region is unbounded, then (i) M is the maimum value of the objective function, if the open half plane determined by a + by > M has no point in common with the feasible region. Otherwise, the objective function has no maimum value. (ii) m is the minimum value of the objective function, if the open half plane determined by a + by < m has no point in common with the feasible region. Otherwise, the objective function has no minimum value. If two corner points of the feasible region are both optimal solutions of the same type, i.e., both produce the same maimum or minimum, then any point on the line segment joining these two points is also an optimal solution of the same type.

247 50 MATHEMATICS Historical Note In the World War II, when the war operations had to be planned to economise ependiture, maimise damage to the enemy, linear programming problems came to the forefront. The first problem in linear programming was formulated in 94 by the Russian mathematician, L. Kantorovich and the American economist, F. L. Hitchcock, both of whom worked at it independently of each other. This was the well known transportation problem. In 945, an English economist, G.Stigler, described yet another linear programming problem that of determining an optimal diet. In 947, the American economist, G. B. Dantzig suggested an efficient method known as the simple method which is an iterative procedure to solve any linear programming problem in a finite number of steps. L. Katorovich and American mathematical economist, T. C. Koopmans were awarded the nobel prize in the year 975 in economics for their pioneering work in linear programming. With the advent of computers and the necessary softwares, it has become possible to apply linear programming model to increasingly comple problems in many areas.

248 PROBABILITY 5 Chapter PROBABILITY The theory of probabilities is simply the Science of logic quantitatively treated. C.S. PEIRCE. Introduction In earlier Classes, we have studied the probability as a measure of uncertainty of events in a random eperiment. We discussed the aiomatic approach formulated by Russian Mathematician, A.N. Kolmogorov (90-987) and treated probability as a function of outcomes of the eperiment. We have also established equivalence between the aiomatic theory and the classical theory of probability in case of equally likely outcomes. On the basis of this relationship, we obtained probabilities of events associated with discrete sample spaces. We have also studied the addition rule of probability. In this chapter, we shall discuss the important concept of conditional probability of an event given that another event has occurred, which will be helpful in understanding the Bayes' theorem, multiplication rule of probability and independence of events. We shall also learn an important concept of random variable and its probability Pierre de Fermat (60-665) distribution and also the mean and variance of a probability distribution. In the last section of the chapter, we shall study an important discrete probability distribution called Binomial distribution. Throughout this chapter, we shall take up the eperiments having equally likely outcomes, unless stated otherwise.. Conditional Probability Uptill now in probability, we have discussed the methods of finding the probability of events. If we have two events from the same sample space, does the information about the occurrence of one of the events affect the probability of the other event? Let us try to answer this question by taking up a random eperiment in which the outcomes are equally likely to occur. Consider the eperiment of tossing three fair coins. The sample space of the eperiment is S {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}