Applied Mathematics Class XII PDF - CBSE
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2021
CBSE
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This is an e-publication of Applied Mathematics for class 12 from the Central Board of Secondary Education (CBSE). It is a support material for students. The text covers various mathematical concepts related to applied mathematics.
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Grade Xll CENTRAL BOARD OF SECONDARY EDUCATION Shiksha Sadan, 17, Rouse Avenue, New Delhi-110002 Dedicated to: Sh. Pramod Kumar T.K. (15th April 1971 to 25th May 2021) Joint Secretary (Academics) Central Board of Secondary Edu...
Grade Xll CENTRAL BOARD OF SECONDARY EDUCATION Shiksha Sadan, 17, Rouse Avenue, New Delhi-110002 Dedicated to: Sh. Pramod Kumar T.K. (15th April 1971 to 25th May 2021) Joint Secretary (Academics) Central Board of Secondary Education You will always be in our memories... Class Xll Student Support Material CENTRAL BOARD OF SECONDARY EDUCATION Shiksha Sadan, 17, Rouse Avenue, New Delhi-110002 Applied Mathematics- Class Xll Student Support Material Price: Unpriced e-Publication First Edition: July 2021, CBSE, Delhi “This material or part there of may not be reproduced by any person or agency in any manner.” Published by: Central Board of Secondary Education, Academic Unit, Shiksha Sadan, 17, Rouse Avenue, New Delhi-110002 Design & Layout: Anand Book Binding House-1042/A-2, New Chand Mohalla Raghubarpura No-1, Gandhi Nagar, Delhi-110031 Mbl.: 9891110888 H k k jrd k laf o /k u m í sf ' kd k g e]H k k jrdsyk sx ]H k k j rdks, d lEiw. k Z1 i zH k qRo & l ai U u l ek tok nhiaF k f uji s{ kyk sd r a= k k Redx. k j kT ; cuk usd sfy ,] r F kk m l dsl eL ruk xf jdksa dk s% lk e k ft d ]v k f F k Zd v k Sj j k t uSf rd U ;k; ] f op k j ]v f Hk O; fDr ] f o 'o k l ] / e Z v k Sjm i kl u k dh L ora= k r k ] i zfr"B k vkSjv o l j d h l e rk i zk I rd jkusdsf y , r Fk k mu l ce sa O ; f D r d h x f j e k 2 v k Sjjk "V ªd h , d r kv kSjv[k aMr k lqf u f ' p rd j uso kyhca/qr kc svkSj m l d k i f j{k. k d js_ ( N ) i zk o`Q fr d i ;k Zo j. k d h f t l d sv ar x Zro u ] > h y ] u n h ] vkSj oU ; t h o g Sa] j { kkd jsv kSjml d k lao/ Zud jsr F k ki zk.k he k =k d si zf r n ;k H k k o j [k s_ ( t ) o SK kf u d n`f" V dks.k ] e ku o o k n v k Sj K k uk t Zu r F k k l q/ k j d h H k k o u k d kfo d k l djs_ ( > ) l ko Zt f u d l aif Ù k d k sl qj f{ kr j [k sv k Sj fg al kl snwj j g s_ ( × k ) O ; fD r x r v k Sj l kewf g d x f r f o f / ; k sa l sL kH kh{ k s= kksa esa m Rd"k Zd h vk sj c< +u sd k lrr iz; k l d j sftl l sj k" Vªfu jar jc 0 which verifies the first Hawkins-Simon condition Also main diagonal of I-A is positive. both of the Hawkins-Simon conditions are satisfied. Hence required input to fulfil the demand and is viable The requirement output X = (I-A)-1D X =.D Hawkins-Simon Condition In the above example, |I-A| must be positive Diagonal element of I-A X = must be positive. X = Example 48 Prepare an input-output table for Transport industry (TI) and Food industry (FI). Food industry produces 50 units. Out of these 20 units consumed by FI and 25 units by TI. Whereas Transport industry produces 40 units and out of these 10 units used by FI and 25 units by TI. 2.48 Applied Mathematics Construct input- output matrix. Check the condition of Hawkins-Simon condition and decide whether system is viable. If so find the input to fulfil the demand 100 of FI and 80 of TI. Solution: FI TI Total Food Industry 20 25 50 Transport Industry 10 25 40 Input-output coefficient matrix (A) = Firstly we will test the viability using Hawkins-Simon condition Get I- A = Now |I-A|= Clearly diagonal positions of I-A are positive. System is viable. Here demand matrix is given as = D = So X = (I-A)-1D X =.D X = X = Hence, to fulfil the demand FI must share 70/8 units and TI must share 34/5 units. 2.9.2 CHECK YOUR PROGRESS EXERCISE – E 1. Solve the following problems using Leontief input-output model. Sector 1 Sector 2 Total Sector 1 2 5 10 Sector 2 3 4 20 If the system is viable then discuss the situation for new demand 8 and 12 from sector 1 and sector 2 respectively. 2. Solve the following problems using Leontief input-output model. FI AI Total Food industry 20 10 40 Agricultural industry 30 20 60 Algebra 2.49 If the system is viable then discuss the situation for new demand 80 and 120 from FI and AI respectively. 3. Solve the following problems using Leontief input-output model. Sector 1 Sector 2 Total Sector 1 12 20 40 Sector 2 15 20 30 If the system is viable then discuss the situation for new demand 8 and 8 from sector 1 and sector 2 respectively. 4. Solve the following problems using Leontief input-output model. Sector 1 Sector 2 Total Sector 1 5 7 30 Sector 2 6 14 21 If the system is viable then discuss the situation for new demand 8 and 12 from sector 1 and sector 2 respectively. 2.10 UNIT SUMMARY 1. A rectangular array or table of numbers, symbols, expressions or functions, when arranged in rows and columns, is known as a matrix (plural matrices). Each member of this arrangement is called an element of the matrix. 2. A matrix is expressed by using capital English alphabet, say A = [ ] where is the element at the th row and th column of the matrix and 1 3. For a matrix A = having ‘m’ number of rows and ‘n’ number of columns, the expression m x n is called the order of the given matrix A where 1 The elements where are called elements of the diagonal of the matrix And the elements where are called elements of the non-diagonal of the matrix 4. A matrix in which number of rows is not equal to number of columns is called a Rectangular matrix 5. A matrix in which the number of rows and columns are equal is called a Square matrix 6. A matrix having exactly one row is called a Row matrix. 7. A matrix having exactly one column is called a Column matrix. 8. A square matrix in which all the non-diagonal entries are zero, i.e. is called a Diagonal matrix 9. A diagonal matrix having same diagonal elements, i.e. where is called a Scalar matrix. 2.50 Applied Mathematics 10. A scalar matrix in which all the diagonal entries are equal to 1, i.e. is called an Identity matrix (Unit matrix), denoted by English alphabet I 11. A matrix with each of its elements as zero, i.e. is called a zero matrix 12. Two matrices A = [ ] and B = [ ] having same order m n are called Equal matrices when each element of A is equal to the corresponding element of B, i.e. 13. For a matrix A = [ ] of order m × n and k is a scalar quantity, then kA is another matrix obtained by multiplying each element of A by the scalar quantity k, i.e. kA = k [ ] = [ ], 14. If A = [ ] and B = [ ] are two matrices of the same order, say m × n, and k and p are scalars, then (i) k (A +B) = k A + k B (ii) (k + p) A = k A + p A (iii) k (A + B) = kA + kB 15. For a non-zero matrix A, of order m×n, a matrix B of same order is called Negative matrix of matrix A such that A + B = O, where O is the zero matrix of the same order. We denote negative matrix A as – A 16. For two matrices A = [ ] and B = [ ], of same order m × n, the sum of two matrices A and B is defined as a matrix S = A+B = [ ] of order m n such that 17. Addition of two or more matrices is possible only when the given matrices are of same order. The order of resultant matrix is also same Am×n + Bm×n = Cm×n Matrix addition is commutative Am×n + Bm×n = Bm×n + Am×n Matrix addition is associative Am×n + (Bm×n + Cm×n) = (Am×n + Bm×n) + Cm×n Zero matrix is the additive identity Am×n + Om×n = Am×n = Om×n + Am×n Negative of a matrix is the additive inverse in matrix addition Am×n + (-Am×n) = Om×n 18. For two matrices A = [ ] and B = [ ], of same order m × n, the difference of two matrices A and B is defined as a matrix D = A - B = [ ] of order m n such that 19. For two matrices A = [ ] of order m n and B = [ ] of order n × p, the multiplication of the matrices AB is defined as a matrix P = AB = [ ] of order m p such that for finding the we multiple th row of frist matrix A with the th column of second matrix B and calculate the sum of these products i.e. Algebra 2.51 20. i. The multiplication of matrices is associative i.e. for any three matrices A, B and C, (AB) C = A (BC), whenever order of multiplication is defined on both sides ii. Distributive property of multiplication holds true for multiplication of matrices. i.e. for three matrices A, B and C, A (B+C) = AB + AC iii. A (B - C) = AB - AC, whenever order of multiplication is defined on both sides. 21. For a given square matrix A of order m m, there exists a multiplicative identity matrix of same order such that IA = AI = A. 22. For a matrix A = [ ] of order m n, the matrix obtained by interchanging the rows and columns of the matrix is called the transpose of matrix A 23. For a matrix A of order m n, the order of transpose matrix A is n m 24. For given matrices A and B: i. (A’)’ = A ii. (kA)’ = kA’ (where k is any constant) iii. (A + B)’ = A’ + B’ iv. (AB)’ = B’A’ 24. For a given square matrix A = [ ], if A’ = A, , then the matrix A is called a symmetric matrix 25. For a given square matrix A = [ ], if A’ = - A, , then the matrix A is called a skew-symmetric matrix 26. For a square matrix A having real values as elements, i. A + A’ is a symmetric matrix ii. A – A’ is a skew symmetric matrix 27. Let X be a set of square matrices and R be the set of numbers (real or complex) such that a function f is defined as f : M K by f (A) = k, where A X and k R, then f(A) is called the determinant of A 28. In a given determinant, minor of an element is the determinant obtained by deleting its th row and th column in which element lies, and is denoted by. 29. In a square matrix, cofactor of an element , denoted by or Cij is defined by = , where is the minor of 30. The transpose of cofactors matrix of a square matrix A = [ is called Adjoint matrix and is denoted by adj A. 31. For a given square matrix A, of order n, i. A(adj A) = (adj A) A = |A| , where is the identity matrix of order n ii. |adj(A)| = |A|n-1 32. If the determinant of any matrix is zero then that matrix is called Singular Matrix and if |A| 0 then A is called Non-singular matrix. 33. Let A(x1,y1), B(x2,y2) and C(x3,y3) be the vertices of the triangle ABC, then 2.52 Applied Mathematics Area ABC = 34. For a given square matrix A of order n, if there exists another square matrix B of the same order n, such that AB = BA = , then A is said to be invertible and B is called the inverse matrix of A and denoted by A-1 i. Inverse of a matrix, if it exists, is unique ii. For two invertible matrices of same order, say A and B, then (AB)-1 = B-1 A-1 iii. For an invertible matrix A, (A-1)-1 = A iv. For an invertible matrix A, (AT) -1 = (A-1) T 35. A system of equations can be solved by using any of the following methods: i. inverse of coefficient method ii. Cramer’s method iii. Row reduction method 36. Hawkins-Simon Conditions to check for viability of an economy are defined as: i. | I – A | must be positive ii. Diagonal elements of (I – A) must be positive 2.11. ANSWERS TO THE EXERCISES EXERCISE – A 1. i) Row, 1×2 ii) Column, 3×1 iii) Rectangle, 3×2 iv) Row 1×3 v) Square 2×2 vi) Square 3×3 vii) Zero 2×2 viii) Zero 2×3 ix) Identity 3×3 x) Scalar 3×3 i) Scalar 3×3 2. i) – 4 ii) 0 + 2 iii) 4 3. 4. 5. 81 6. 4 7. a = 14, b = -7, c = 7, d = 1 Algebra 2.53 EXERCISE – B 1. Order of the matrix A B A B AB 2 2 2 2 2 2 2 2 2 3 3 2 Not possible 2 2 3 4 4 1 Not possible 3 3 3 3 3 3 3 3 2 3 2 3 2 3 Not possible 3 3 2 Not possible 1 2 5. i) ii) iii) iv) p=1, q = 2, r = 3, s = 4 v) 6. 7. i. iii. iv. v. 9. In the month of March P = In the month of April Q = Total Sell P+Q = + = 2.54 Applied Mathematics 10. Sell matrix pen notebooks S = Cost Matrix C = Amount Matrix A= S C A= = = Bookseller P generates the amount Rs 249/- and Q generates the amount Rs 180/- EXERCISE – C 1. i) 0, ii) -53, iii)25, iv) 0, v) 18 2. 15 sq. units 3. -5 7. i. ii. iii. EXERCISE – D 1. i) ii) iii) iv) : 2. i) ii) 3. a) x =1, y =2 b) x = 3, y = -2 c) x = 1, y = -1 d) x = 0, y = 5, z = 3 e) x = 1, y = 2, z = 3 Algebra 2.55 4. i) x = 1, y = 2 ii) x = 3, y = -2 iii) x = 1, y = -1) iv) Infinite many solution v) No solution vi) x = 0, y = 5, z = 3 vii) x = 1 , y = 2, z = 3 viii) Infinite many solution ix) No solution EXERCISE – E 1. Coefficient unit matrix = Get I-A = , Check conditions det (I-A) = 16/25 – 3/40 = 109/200 > 0, main diagonal entries positive So X = (I-A)-1D = 2. Coefficient unit matrix = Get I-A = , Check conditions det (I-A) = 1/6 – 5/12 = - 3/12 < 0, First condition of Hawkins-Simon is not true. So condition is not viable 3. Coefficient unit matrix = Get I-A = , Check conditions det (I-A) = 7/30 – 1/4 = (14-15)/60 < 0, And main diagonal entries positive. First condition of Hawkins-Simon is not true. So condition is not viable. 4. Coefficient unit matrix = Get I-A = , Check conditions det (I-A) = 5/18 - 1/15 = (25-6)/90 = 19/90 > 0, main diagonal entries positive So X = (I-A)-1D = 2.56 Applied Mathematics it n U 3a 3.0 LEARNING OUTCOMES After completion of this unit the students will be able to Find derivative of implicit functions, parametric functions Apply logarithmic differentiation in the functions of the type [f(x)]g(x) Find second order derivative Define Cost function and Revenue function Understand derivatives as the rate of change of various quantities Define marginal cost and marginal revenue Understand the gradient of a tangent and normal to a curve at a given point Write the equation of tangent and normal to a curve at a given point Recognize whether a function is increasing or decreasing or none Determine the condition for an increasing or a decreasing function Find the maximum and minimum values of a function at a given point Determine turning points (critical points) of the graph of a function Find the values of local maxima and Local minima at a point Find the absolute maximum and Absolute minimum value of a function on a closed interval Apply the derivatives in real life problems Differentiation and Its Applications 3a.1 3.1 CONCEPT MAP Equation of tangent and Normal 3.2 RECALL SOME STANDARD RESULTS OF DIFFERENTIATION 1. Derivatives of standard functions i. nx n 1 d xn ii. d x a a x log a dx dx d x d 1 iii. dx e ex iv. dx log x x d v. constant 0 dx 2. Basic rules of differentiation d d i. dx k f x k dx f x , where ‘k’ is some real number.. d d d ii. dx f x g x dx f x dx g x . d d d iii. dx f x g x dx f x g x f x dx g x , also called as product rule. d d d f x dx dx f x g x f x g x iv. , also called as quotient rule. dx g x g x 2 3a.2 Applied Mathematics 3.3 DIFFERENTIATION OF IMPLICIT FUNCTIONS 3x 2 The equation 3x2 + xy + y = 0 can be written as y i.e. ‘y’ can be expilictly expressed in 1x terms of ’x’ only i.e. in terms of independnent variable ‘x’. We have been finding derivatives in such cases. As, in above example dy 6x 1 x 3x 2 3x 2 6x 3x x 2 or dx 1 x 2 1 x 2 1 x 2 Let us consider the equation. The graph of this equation is the union of the graphs of the functions y = f1(x), y = f2(x), y = f3(x), as shown below, which are differentiable except at O and A. How do we get the derivative when we can not conveniently find the functions? Here, in this example, ‘y’ can not be explicitly expressed as a function of ‘x’. Such functions are called implicit functions. We treat y as a differentiable function of x and differentiate both sides of the equation with respect to x using the differentiation rules. The process by which we find derivative is called implicit differentiation. The equation above defines three functions and we find their derivatives implicitly without knowing explicit formula to work with. The process by which we find is called implicit differentiation. The equation above defines three functions f1, f2, f3 and we find their derivatives implicitly without knowing explicit formula to work with. FIG. 1 dy The process by which we find is called implicit differentiation. The equation above dx defines three functions f1, f2, f3 and we find their derivatives implicitly without knowing explicit formula to work with. Example 1 If y = f(x) is a real function, then find derivative of the following with respect to ‘x’. x2 i. y2 ii. x3 y5 iii. log (xy2) iv. 1 e xy Differentiation and Its Applications 3a.3 Solution: i. In order to differentiate y2 with respect to ‘x’, we shall have to use chain rule, as y2 depends on ‘y’ and ‘y’ depends on ‘x’. d 2 dy 2 dy dy dx y dy dx 2y dx ii. We shall use product rule to find the derivative of x3y5 with respect to ‘x’ d dx 3 5 dy 5 dy dy dx x3y5 dx y x3 dx 3x 2 y 5 x 3 5y 4 dx 3x 2 y 5 5x 3 y 4 dx iii. Both chain rule and the product rule will be used to diffrentiate log (xy2) d 1 d 1 dy 1 2 dy dx log xy 2 2 xy dx xy xy 2 2 y 2 2xy dx x y dx But, one can also first simplify log(xy2) = log x + 2log y and then find the derivative. (Try youself !) iv. We shall use quotient rule, chain rule and product rule to find the derivative as follows dy xy 3 dy d x2 2x 1 e xy x 2 e xy y x dx 2 2x e 2x x y x dx dx 1 e xy 2 2 1 e xy 1 e xy Example 2 dy Find , when x3 + y3 = xy. dx Solution. Differentiating with respect to ‘x’, d 3 d 3 d dx x dx y dx xy dy dy 3x 2 3y 2 yx dx dx dy 3y 2 x dx y 3x 2 dy y 3x 2 dx 3y 2 x Example 3 m n mn dy y If x y x y , then show that dx x Solution. We first take log of both sides of the equation mn log x m y n log x y mlog x n log y m n log x y 3a.4 Applied Mathematics Differentiatiing both sides with respect to ‘x’. m n dy m n dy 1 x y dx xy dx dy n m n m n m dx y x y x y x dy n x y m n y m n x m x y dx y x y x yx dy nx my nx my dy y dx y x y x y x dx x 3.4 DIFFERENTIATION OF PARAMETRIC FUNCTIONS It is sometimes convenient to represent the relation between the variables x and y by two equations x = g(t), y = f(t). For example, the equations x = at2, y = 2at, where t R, a is a constant >0, are the parametric equations for the curve (rightward parabola) y2 = 4ax, a > 0. The variable t is a parameter for the curve. We can verify that the parametric equations represent the parabola as y 2at y 2 4a 2 t 2 y 2 4a at 2 y 2 4ax , i.e., the points (at2, 2at) satisfy the equation y2 = 4ax, a > 0, where t R. The equations x = at2, y = 2at define y as a composite function of x and are said to represent the function in parametric form. If represent a function in parametric form, then where is an inverse function with respect to the function. Using chain rule and applying , the derivative of ‘y’ with respect to ‘x’ can be obtained as dy dy dt dx dx dt Example 4. dy Find if x = at2, y = 2at dx 2 dx dy Solution. Differntiating, with repect to ‘t’, x at 2at , and y 2at 2a dt dt dy dy dt 2a 1 dx dx 2at t dt Differentiation and Its Applications 3a.5 Example 5 dy 1 t Find if x , xy = 2t3. dx t 1 1 t Solution. Differntiating with respect to ‘t’, we get 1t dx 11 t 1 t 2 dy x 2 3 2 and y 2t 6t 2 1 t dt 1 t 1 t dt dy dy dt 1 t 2 3t 2 1 t dy 3 2 2 12 2 6t 2 dx dx 2 dx t 1 dt 3.5 LOGARITHMIC DIFFERENTIATION d n d x We know that dx x nx n 1 where ‘n’ is any real number and dx a a x log a , where ‘a’ is any positive real number, other than 1. Both of these formulae can not be used in the differentiation x2 gx 1 x of the functions of the type f x like xx , 1 x etc x x x 1 or d x x x x log x d xx dx dx g x In the functions of the type f x we use logarithm to find the derivative of the function as shown the following examples. Example 6 Differentiate the following with respect to ‘x’ i. y = xx ii. y = xy Solution: i. y = xx, Taking log of both sides, we get log y x log x Differentiating both sides with respect to ‘x’, we get 1 dy 1 dy = 1.logx + x (. ) y log x 1 x x log x 1 y dx x dx ii. y = xy, Taking log of both sides, we get log y y log x Differentiating both sides with respect to ‘x’, we get 1 dy dy y log x y dx dx x dy 1 y log x dx y x dy y y y2 dx x 1 y log x x 1 y log x 3a.6 Applied Mathematics Example 7. dy If xy + yx = ab, then find. dx Solution. Let u = xy, v = yx x y y x a b u v a b , where u = xy ; v = yx Differntiating both sides of the equation u + v = ab with respect to ‘x’, we get du dv 0 — (i) dx dx 1 du dy y du dy y u x y log u y log x log x xy log x —(ii) u dx dx x dx dx x 1 dv x dy dv x dy v y x logv x logy 1 logy y x log y —(iii) v dx y dx dx y dx du dv Substituting and from (ii) and (iii) in (i), we get dx dx dy y x dy xy log x y x log y 0 dx x y dx dy y dx x log x xy x 1 x y 1y y x log y dy x y 1y y x log y y dx x log x xy x 1 3.6 SECOND AND HIGHER ORDER DERIVATIVES Let y = f (x) be a differentiable function of ‘x’, then d dy d2y i. Derivative of f’(x) with respect to ‘x’ = y f x , is the second order dx dx dx2 derivative of ‘y’ or f(x). d d2 y d3 y ii. Derivative of f”(x) with respect to ‘x’ = y f x , is the third order dx dx 2 dx 3 derivative of ‘y’ or f(x). Similarly, we can find the other higher order derivatives of y f x . Note: For derivatives higher than three we do not use primes, instead we write the, n th order derivatives as dn y yn fn x dx n Differentiation and Its Applications 3a.7 Example 8 d2 y Find for the following functions dx 2 i. y = x ii. y = log x iii. y x 2 1 Solution: dy i. y = x, differentiate with respect to ‘x’, 1, dx 2 d dy d y Differentiating again with respect to ‘x’, we get 0 i.e. 0 dx dx dx 2 dy 1 ii. y = log x , Differentiating again with respect to ‘x’, we get dx x d2 y 1 2 dx x2 dy d 1 1 2 12 d 2x x iii. y 2 x 1 dx dx x 1 2 2 2 x 1 dx x2 1 2 x2 1 2 x 1 x 2 1 x2 1 x d y x2 1 x2 1 x 2 1 2 2 3 dx x 1 x 2 1 x2 1 x 2 1 2 Example 9 t2 t If x and y , find y 2. 1t 1 t t2 dx 2t 1 t t 2 2t t 2 Solution. x 1 t dt 1 t 2 1 t 2 t dy 1 1 t t 1 1 y 2 1 t dt 1 t 1 t 2 dy 1 1 t 2 1 y1 dt dx 1 t 2t t 2 2 2t t 2 dt 2 3 d d 1 d 1 dt 2 2t 1 t 2 1 t y2 y1 2 2 2 2 3 3 dx dx 2t t dt 2t t dx 2t t 2t t 2 t t 2 3a.8 Applied Mathematics Exercise 3.1 dy 1. Find from the following dx i. x 3 y 3 3axy ii. e xy axy a 1 1 2 iii. 3x 3 5x 2 y 2xy 2 4y 3 0 iv. x 3 y 3 a 3 v. x ylog xy dy 2. Find from the following parametric equations dx a log t i. x at , y t ii. x t log t , y t iii. x a 1 t2 , y 2bt 1 t 2 1 t2 dy 3. Find from the following equations dx i. x y yx ii. x y exy x xy y xlog x iii. x y e 7 iv. d2 y 4. Find from the following dx 2 (i) y x log x (ii) y x 2e x (iii) y log logx (iv) y 3e2x 2e3x 5. If x 1 y y 1 x 0 , show that 1 x 2 dy dx 1 0. 1 1 2 2 2 2 6. If y m y m 2x then prove that (x – 1)y1 m y. 7. 2 2 2 2 If y log x a x , show that a x y 2 xy 1 0. p 8. If y x x 1 2 , prove that x 2 1 y 2 xy 1 p 2 y 0. 3.7 COST AND REVENUE FUNCTION Any manufacturing company has to deal with two types of costs, the one which varies with the cost of raw material, direct labour cost, packaging etc. is the variable cost. The variable cost is dependent on production output. As the production output increases (decreases) the variable cost will also increase (decrease). The other one is the fixed cost, fixed costs are the expenses that remain the same irrespective of production output. Whether a firm makes sales or not, it must pay its fixed costs. Differentiation and Its Applications 3a.9 Cost Function: If V(x) is the variable cost of producing ‘x’ units and ‘k’ the fixed cost then, the total cost C(x) is given by C(x) = V(x) + k Revenue Function: if R is the total revenue a company receives by selling ‘x’ units at price ‘p’ per unit produced by it then the revenue function is given by R(x) = p.x Note: Generally, it is assumed that a company sells the number of units it produces. Example 10 A company produces ‘x’ units in a year and the variable cost is V(x) = x2 – 2x. Also, the company spends a fixed cost of Rs15,000 on commissions and rent, then (i) Find the total cost function C(x) (ii) If ‘p’ the price per unit is given by p = 5–x then find its revenue function. Solution: (i) The variable cost, V(x) = x2 – 2x The total cost function is C(x) = V(x) + 15000 = x2 – 2x + 15000 (ii) The revenue function is given by R = px R 5 x x 5x x2 3.8 DERIVATIVE AS RATE OF CHANGE OF QUANTITIES In science, business and economics there are variables one depending on the other such as distance and time, cost and production, revenue and production, price and demand etc. In all these examples we are interested in the rate at which one variable changes with respect to other to know the micro details of relation between these variables. In class XI we have discussed that if y = f(x) is a real function then, dy Rate (or instantaneous rate) of change of 'y' with respect to 'x' dx Example 11 A boy is blowing air into a spherical balloon and thus the radius r of the balloon is changing, then find the rate of change of surface area of the balloon with respect to the radius r. Also find the rate of change of surface area when r = 2cm. Solution. Let Area of balloon be A at the radius r then, A = 4r2 Rate of change of surface area of balloon with respect to the radius is dA dA 8r and 8r r 2 8 2 16cm 2 /cm. dr dr r2 3a.10 Applied Mathematics 3.8.1 RELATED RATES In related rate problems the quantities change with respect to time, you may be given rate of change of one variable with respect to time and rate of change of other variable has to be found with respect to time. dy dy dt dy dy dx dx Recall: If y f x , then using chain rule, f x dx dx dt dx dt dt dt Example 12 Find the rate of change of volume of a sphere with respect to its surface area when the radius is 5 m. 4 3 Solution: For the radius r, the volume V and the surface area S of the sphere is given by V r and 3 S = 4r2. As V and S both are functions of radius r, we will use chain rule to find derivative of V with respect to S dv 4 dS Since, dr 3 3 r 2 4 r 2 , dr 8 r dV dV dr 4r 2 r dV 5 m 3 /m 2 dS dS 8r 2 dS r 5 2 dr Example 13 A cylindrical vessel of radius 0.5 m is filled with oil at the rate of 0.25πm 3 / min. Find the rate at which the surface of the oil is rising. Solution: The rate at which the surface area rises is the rate of change of height of oil in the vessel with respect to time. Let r be the radius, h be the height and V the volume of the oil at time t. 2 1 Then V r h h as r 0.5 4 2 As we are given rate of change in volume with respect to time t, therefore differentiating V with respect to t, we get dV dh dh 0.25 , FIG. 2 dt 4 dt 4 dt dV as we are given 0.25 m 3 dt dh 1m/ min ute dt Differentiation and Its Applications 3a.11 Example 14 A boy of height 1m is walking towards a lamp post of height 5 meters at the rate of 0.5 m/sec. Then find the rate at which the length of the shadow of the boy is decreasing. Solution: Let RT = y, be the length of the shadow of the boy when he is x m away from the lamp post at time t. dx 0.5m /sec , as the boy is moving towards dt lamp post the distance between the boy and the lamp post is decreasing and hence the rate of change of the distance will be negative. FIG. 3 5 xy 4 y x , differentiating with respect to t 1 y dy dx dy 0.5 1 4 m/sec , hence the shadow of the boy is decreasing at the rate of dt dt dt 4 8 1 m /sec. 8 3.8.2 MARGINAL COST AND MARGINAL REVENUE Marginal cost and marginal revenue are the instantaneous rate of change of cost and revenue with respect to output i.e. rate of change of C(x) (or C), the cost function and R(x) (or R), the revenue function, with respect to production output ‘x’. Therefore, the Marginal cost (MC) and the Marginal revenue (MR) are given by dC MC (Marginal cost) C x dx dR MR (Marginal revenue) R x dx Note: Marginal cost is an important factor in economics theory because a company that needs to maximize its profits will produce up to the point where marginal cost (MC) equals marginal revenue (MR). Beyond that point, the cost of producing an additional unit will exceed the revenue generated Example 15 A toy manufacturing firm assesses its variable cost to be ‘x’ times the sum of 30 and ‘x’, where ‘x’ is the number of toys produced, also the cost incurred on storage is 1500. Find the total cost function and the marginal cost when 20 toys are produced. Solution: The total cost function C(x) is given by, C(x) = x (x + 30) + 1500 = x2 + 30x + 1500 The marginal cost MC is given by, 3a.12 Applied Mathematics dC MC 2x 30 dx dC Marginal cost of producing 20 toys is MC 20 2 20 30 70 dx x 20 The marginal cost of producing 20 toys is 70. Example 16 The price per unit of a commodity produced by a company is given by p 30 2x and ‘x’ is the quantity demanded. Find the revenue function R, the marginal revenue when 5 commodities are in demand (or produced). Solution. The revenue function R (or R(x)) is given by, R = px = (30 – 2x) x = 30x – 2x2 dR The marginal revenue MR 30 4x dx The marginal revenue of producing 5 commodities is, dR 30 4 5 10 dx x 5 The marginal revenue when 5 commodities are in demand is 10. Exercise 3.2 1. Find the rate of change of circumference of a circle with respect to the radius r. 2. Find the rate of change of lateral surface area of a cube with respect to side x, when x = 4cm. 3. If the rate of change of volume of a sphere is equal to the rate of change of its radius, then find its radius. Also find its surface area. 4. The volume of a cone changes at the rate 40 cm 3/sec. If height of the cone is always equal to its diameter, then find the rate of change of radius when its circular base area is 1m 2. 5. For what values of x is the rate of increase of total cost function C(x) = x3 – 5x + 5x + 8 is twice the rate of increase of x? 6. The radius of the base of a cone is increasing at the rate of 3cm/minute and the altitude is decreasing at the rate of 4cm/minute. Find the rate of change of lateral surface area when the radius is 7cm and the altitude 24 cm. 7. A ladder 10 meters long rests with one end against a vertical wall, the other on the floor. The lower end moves away from the wall at the rate of 2 meters / minute. Find the rate at which the upper end falls when its base is 6 meters away from the wall. 8. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm 3 /min. When the thickness of ice is 5 cm, find the rate at which the thickness of ice decreases. 9. A stationery company manufactures ‘x’ units of pen in a given time, if the cost of raw material is square of the pens produced, cost of transportation is twice the number of pens produced and the property tax costs 5000. Then, (i) Find the cost function C(x). Differentiation and Its Applications 3a.13 (ii) Find the cost of producing 21st pen. (iii) The marginal cost of producing 50 pens. 10. A firm knows that the price per unit ‘p’ for one of its product is linear. It also knows that it can sell 1400 units when the price is 4 per unit, and it can sell 1800 units at a price of 2 per unit. Find the price per unit if ‘x’ units are sold (or demanded). Also find the revenue function and the marginal revenue function. 3.9 SLOPE (OR GRADIENT) OF TANGENT AND NORMAL Consider the graph of a curve y = f(x) as shown in the Fig 4. Let A (x, f(x)) be a point on the graph and B(x +x, f(x +x)) be a neighbouring point on the curve. Then, The slope of the line (secant) AB = tan f x x f x , x where is the angle of inclination of the line AB. The limiting position of the secant AB, when the point B moves along the curve and tends to coincide with the point A (i.e., when x 0), if the limiting position exists, is the tangent to the curve at the point A. Hence, FIG. 4 The slope of the tangent (non-vertical) to the curve at the point A is given by , where is the angle of inclination of the tangent line at the point A. Please note that in the figure x chosen is positive, whereas x could be negative and x has to approach 0 from both sides. Therefore, we conclude that, dy Slope (or gradient) of a non-vertical tangent at a point A(x0,y0) = dx A x0 ,y0 Normal to a curve at a point: If y = f(x) is a real function then normal to the curve at a point A(x0,y0), i.e., at ((x0, f(x0)) on it is a line perpendicular to the tangent at the given point on the curve. As shown in the adjoining figure 5. FIG. 5 3a.14 Applied Mathematics 1 Slope of a normal line to the curve at a point A x0 ,y0 = , provided the dy dx A x0 ,y0 normal is not perpendicular to the x-axis. Example 17 Find slope of the tangent and normal at a point (2,6) to the curve y = x3 – x 3 dy Solution: y x x 3x 2 1 dx Slope of the tangent at (2,6) is given by dy dx 2,6 3x 2 1 2,6 2 3 2 1 11 1 1 And, slope of normal to the curve = dy 11 dx 2 ,6 3.9.1 EQUATION OF TANGENT AND NORMAL TO A CURVE Recall that equation of a line passing through (x 1 ,y 1) and having slope m is given by (y–y1) = m(x–x1). Using this we can find equation of tangent and normal to a curve at a given point on it. Let y = f(x) be a real function and A(x0,y0) be a point on it then, dy Equation of tangent to the curve at the point A(x0,y0): y y0 x x0 dx x 0 ,y 0 1 Equation of normal to the curve at the point A(x0,y0) : y y0 dy x x0 dx x0 ,y0 Example 18 2 2 Find the equation of the tangent and normal to the curve x 3 y 3 2 at (1,1). Solution. Differentiating with respect to ‘x’, we get 1 2 1 3 2 1 3 dy dy x 3 x y 0 1 3 3 dx dx y 3 Differentiation and Its Applications 3a.15 Equation of the tangent is p dy s hel sure t y 1 x 1 y 1 1 x 1 x y 2 0 en ea ang ally m of dx 1,1 e t Th etric ticity Equation of normal is g eom e elas curve to pric and 1 dem y 1 dy x 1 y 1 1 x 1 y x 0 dx 1,1 Example 19 Find the equation of the tangent and normal to the curve f(x) = ex + x2 + 1 at the point (0,2) on it. Solution. f’(x) = ex + 2x then slope of tangent at the point (0,2) is f’(0) = 1 and slope of normal is 1 1 f 0 Therefore, equation of tangent is (y – 2) = 1(x – 0) x – y + 2 = 0 And equation of normal is (y – 2) = –1(x – 0) x + y – 2 = 0 Example 20 Find the equation of the tangent to the curve x2 + 3y – 3 = 0, which is parallel to the line y = 4x – 5. Solution. Differentiating the given equation we get, dy dy 2x 2x 3 0 dx dx 3 Slope of the tangent to the curve = slope of the line y = 4x – 5 dy 2x 2 33 4 4 x 6 6 3y 3 0 y 11 dx 3 3 Therefore, the point on the curve is (–6, –11) and equation of tangent is (y + 11) = 4(x + 6) 4x – y + 13 = 0 Example 21 Find the equation of the tangent to the curve y = (x3 – 1) (x – 2) at the points where the curve cuts the x-axis. Solution. Putting y = 0 in the equation of the curve to get the points where it cuts the x-axis. (x3 – 1) (x – 2) = 0 = 1, 2 Thus, the points of intersection of curve with x-axis are (1,0) and (2,0) Differentiating the equation with respect to ‘x’, The equations of the tangents are 3a.16 Applied Mathematics dy y 0 x 1 3x y 3 0 dx 1,0 dy y0 x 2 7x y 14 0 dx 2 ,0 Example 22 Find the equation of the tangents to the curve 3x2 – y2 = 8, which passes through the point 4 3 , 0. Solution. Note: The given point does not lie on the curve Let us assume the tangent touches the curve at the point (h,k) 3h 2 k 2 8 —————— (i) Differentiating the equation of the curve with respect to ‘x’, we get dy dy 3x dy 3h 6x 2y 0 dx dx y dx h ,k k 4 3h Equation of the tangent is: y 3 k x , also it passes through (h, k) 4 3h k h 9h 2 3k 2 4k 0 —————— (ii) 3 k Replacing h2 from (i) in (ii), we get, k2 8 2 44 9 3k 4k 0 k 6 h 3 3 4 3 44 The equation of the tangents is y x 33x 3y 4 0 3 6 3 Example 23 n n x y x y Prove that 2 touches the straight line 2 for all n N, at the point (a,b). a b a b n n y Solution. x 2 bn x n a n y n 2a n bn Differentiating with respect to ‘x’, we get a b n n1 n n1 dy dydy dy bn xn 1 b nb x na y = 0 n n1 dx dxdx dx a,b a y a,ba a Equation of the tangent at (a,b) is Differentiation and Its Applications 3a.17 Example 24 Find the points on the curve y = 4x3 – 2x5, at which the tangent passes through the origin. Solution. Let (h,k) be any such point on the curve k 4h 3 2h 5 - (i) Differentiating the given equation with respect to ‘x’, we get dy dy 12x 2 10x 4 12x 2 10x 4 12h 2 10h 4 dx dx h,k h,k The equation of the tangent through origin is, y 0 12h 2 10h 4 x 0 k 0 12h 2 10h 4 h 0 As (h,k) lies on the tangent, k 12h 3 10h 5 (ii) Solving (i) and (ii), 4h 3 2h 5 12h 3 10h 5 8h 5 8h 3 0 8h 3 h 2 1 0 h 0, h 1 The points on the curve are (0,0), (1,2) and (–1, –2). Exercise 3.3 1. Find the slopes of the tangents and normal to the curves at the indicated points. i. y x 3 x at x = 1. ii. y = 3x2 – 6x at x = 2. x 1 2 2 iii. y ,x 2 at x = 10. iv. x 3 y 3 2 at (1,1). x2 2. Find the equations of the tangent and normal to the curves at the indicated points. i. y = x3 – 3x + 5 at the point (2,7). ii. x = at2, y = 2at at t = 2 3. Find the equations of the tangents to the curve at points where the tangents to the curve y = 2x3 – 15x2 + 36x – 21 are parallel to x-axis. 4. Find the equation of the tangents to the curve y = x3 + 2x – 4, which is perpendicular to the line x + 14y + 3 = 0. x7 5. Find the equation of the tangent and the normal to the curve y 2 at the point, where it x 5x 6 cuts x-axis. 6. Find the equation of the normal to the curve x2 = 4y which passes through the point (1,2). 7. For the curve y = x2 + 3x + 4, find all points at which the tangent passes through the origin. x y 8. Show that the line 1 touches the curve y b e x a at the point where it crosses the y-axis. a b 9. Show that the curves xy = a2 and x2 + y2 = 2a2 touch each other. 10. Prove that the curves xy = 4 and x2 + y2 = 8 touch each other. 3a.18 Applied Mathematics 3.10 INCREASING AND DECREASING FUNCTIONS (MONOTONIC FUNCTIONS) Increasing function: A real function y = f (x) is said to be an increasing function (abbreviated as) on an open interval (a,b) if the value of ‘f’ increases as ‘x’ increases or vice – versa. i.e., the graph of f(x) rises from left to right see Figure 6. Here, the open interval (a, b) includes (i) (a, b), where a, b are fixed real numbers such that a < b, (ii) (a, ), where a is a fixed real number, (iii) (-, b), where b is a fixed real number, (iv) (-, ) = R Following is an analytical condition for a function to be increasing on (a, b) x1 x2 f x1 f x2 x1 , x 2 a, b Fig. 6 Or x1 x2 f x1 f x2 x1 , x 2 a, b Decreasing function: A real function y = f(x) is said to be a decreasing function (abbreviated as) on an open interval (a,b) if the value of ‘f’ decreases as ‘x’ increases or vice – versa, i.e. the graph of f(x) falls from left to right see Figure 7 Following is an analytical condition for a function to be decreasing on (a, b) x1 x2 f x1 f x 2 x1 , x 2 a, b Or Fig., 7 x1 x 2 f x1 f x 2 x1 , x 2 a, b Note: 1) A function which is either increasing or decreasing on its domain (an open interval) is termed as a monotonic function 2) You may note that a function may be defined to be increasing/decreasing in any of the following intervals represented as I: (a, b), (a, ), (– , b), (– , ), [a, b], [a, ), (– , b] Not all cases have been included here in this topic. Differentiation and Its Applications 3a.19 Example 25. Show that f(x) = x2 is an rd p wa e s (i) Increasing function on (0, ) u t im term some sing e ea (ii) Decreasing function on (– , 0) Th lso ncr i sa i Solution: pin g an slo for ion d ct (i) Let x1, x2 (0, ) such that x1 < x2 us e f un x1 x 1 x1 x 2 x1 2 x1 x 2 ard w n w im e s Also, x1 x 2 x 2 x 2 x1 x 2 x 2 2 do met g rm so n e te al so creasi Th g is d e From above we conclude that x12 x 22 f(x 1 ) f(x 2 ) lo pin for a t ion s d c us e f un hence, x1 x 2 f(x 1 ) f(x 2 ) i.e., f(x) is an increasing function over (0, ) (ii) Let x1 ,x2 ,0 such that x1 < x2 x 1 x 1 x 1 x 2 x 12 x 1 x 2 x1 , x2 are negative numbers Also, x 1 x 2 x 2 x 2 x 1 x 2 x 2 2 From above we conclude that, x12 x22 f x1 f x2 , hence, x1 x2 f x1 f x2 i.e., f(x) is a decreasing function over (– , 0) Note: f(x) is decreasing on (– ,0) and increasing on (0,) therefore we say that f(x) is neither increasing nor decreasing on (– ,). 3.10.1 DERIVATIVE CONDITIONS FOR A MONOTONIC (INCREASING OR DECREASING) FUNCTION. As discussed in section 3.4 that the derivative of a function at a given point on it is slope of the tangent to the curve at that point. It is an interesting fact to note that slope of tangents also determines the monotonicity of functions. It can be observed from the following figures. In Fig.8 one can observe that the function is increasing in its domain and the tangents drawn to the curve are making acute angles of inclination at any point within the domain of the curve. You may encounter the graph of a certain increasing function in an open interval, where the angle of inclination of the tangent at a point is 0° or 90°. Please, check the graph of the functions Fig. 8 3a.20 Applied Mathematics DERIVATIVE TEST FOR INCREASING FUNCTIONS A real function y = f(x) is an increasing function on (a,b) if f x 0 , x a , b Please note that the condition mentioned above is sufficient for a function f to be an increasing function in (a, b), but not necessary. The functions are examples of increasing functions over R, but is not defined. In Fig.9, one can observe that the function is decreasing in its domain and the tangents drawn to the curve are making obtuse angles of inclination at any point within the domain of the curve. Fig. 9 But, you may encounter some exceptions similar to what was observed above in case of an increasing function. DERIVATIVE TEST FOR DECREASING FUNCTIONS A real function y = f(x) is a decreasing function on (a,b) if f x 0, x a, b The condition is sufficient for a function f to be a decreasing function in (a, b), but not necessary. CRITICAL POINTS Definition: An interior point of the domain of a function f where f’ is zero or undefined is a critical point of f. Hence, the critical points are essentially interior points of the domain of the function f together with a second condition as mentioned above. The word critical probably has been used because at this point an abrupt change in the behaviour of the graph of the function is noted. i. The curve f(x) = (x – 1)2 + 2, in the Fig.10, is turning at the point ‘A’ on it and tangent at ‘A’ is parallel to x-axis, i.e., slope of tangent at ‘A’ is 0, i.e., f ’(1) = 0. The curve takes a smooth turn at the point (1, 2). Fig. 10 1 is a critical point of the function. Differentiation and Its Applications 3a.21 ii. The graph of the function f (Fig. 11) is turning at the points B and C. f’(-2) = 0, f’(3) = 0. –2 and 3 are the critical points of the function f. The curve takes a smooth turn at the points B and C. iii. f’(0) does not exist. 0 is a critical point. Note that the graph of the function (Fig. 12) takes a sharp turn at (0, 0). There is a corner at the point (0, 0). Fig. 11 Fig. 12 iv.. g’(0) is not defined. 0 is a critical point. Note that there is a vertical tangent at the point (0, 0), which is the y-axis itself and 0 is a point of inflexion. A point where the graph of a function (Fig. 13) has a tangent line and where the concavity changes is called a point of inflexion. Fig. 13 v. h’(0)=0. 0 is a critical point. The tangent line at the point (0, 0) is the x-axis itself. 0 is a point of inflexion. (Fig. 14) Fig. 14 3a.22 Applied Mathematics vi. The following is the graph of the function f(x)=x2/3 f’(0) is not defined. 0 is a critical point. Note that at (0, 0), there is a cusp in the graph. (Fig. 15) Fig. 15 vii. 0 is a point of discontinuity. The derivative of the function at 0 is not defined. 0 is a critical point. (Fig. 16) Fig. 16. A function f(x) is defined in an interval I and c is an interior point of I. If f(x) is continuous at x = c, f’(c) = 0, then c is a critical point. If f(x) is continuous at x = c, f’(c) is not defined, then c is a critical point. If c is a point of discontinuity, then c is a critical point. Here also, f’(c) is not defined. Differentiation and Its Applications 3a.23 Note: The conditions for the graph of a function to have a vertical tangent at a point, a corner at a point, a cusp at a point and a point of inflexion at a point may be explored in higher courses of mathematics. Stationary points: A Stationary point is the point where the derivative of the function is 0. It is essentially the point where the curve is momentarily at rest and then it either takes a smooth turn or becomes a point of inflexion. Stationary points are necessarily the interior points of the domain of the function. Note: 1) All stationary points are critical points, but not every critical point is stationary. In part i above, 1 is a stationary point. In part ii, -2 and 3 are stationary points. In part v, 0 is a stationary point. 2) If f(x) is differentiable in an open interval (a, b), even if the function is defined in the closed interval [a, b], the only critical points are the interior points of the domain of the function where f ‘(x) = 0. These are stationary points too. Example 26 Find the critical point(s) of the following functions 4 1 x4 11 i. f x 12x 3 6x 3 on 1,1 ii. f x 2x 3 x 2 6x 4 2 Solution: 1 2 16x 2 i. f x 16x 3 2 2 x3 x3 1 1 f x
