Mechanical BE (Mathematics-II) by IMTS INSTITUTE

I ns t i t ut eofManage me nt & Te c hni c alSt udi e s

MATHEMATI CSII

500

BACHELORI NMECHANI CALENGI NEERI NG www. i mt s i n s t i t u t e . c o m

IMTS (ISO 9001-2008 Internationally Certified) MATHEMATICS-II

MATHEMATICS-II

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II CONTENTS:

UNIT I

Matrices and Theory of Equations

01-49

Introduction ,Objective,Definition,Cayley – Hamilton,Theorem.,Characteristic roots, characteristic vectors. ,Some basic theorems in Theory of Equations,Relations between the root and coefficient of an equation,Problems Type – I,Problems Type – II,Answers to Self Assessment Questions

UNIT II

Theory of Equations (Continued)

Introduction,Objective,Formation

equations

50-100

with

roots

changed

sign,Formation of equations with roots multiplied by a constant,Formation of equation with roots as squares of the given equation,Formation of equation with roots increased or decreased by a constant,Removal of terms,Transformation in general,Symmetric functions of roots.,Answers to Self assessment problems.

UNIT III

Trigonometry

Introduction,Objective,Expansions

101-157

of Sinn , Cosn and tan n

,Expansions of sin n  and cos n  ,Expansions of sin  , cos and tan ,Hyperbolic functions,Inverse hyperbolic functions,Logarithm of a complex number,Answers to Self Assessment Problems

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

UNIT IV

Differential Calculus

158-224

Introduction,Objective,Definition of Derivative.,Some Standard forms. ,Some general theorems,Some important formula,Successive differentiation ,Leibnitz formula,Meaning of the derivative,Answers to Self assessment Problems.

UNIT V

Partial Differentiation and Radius of Curvature

225-290

Introduction,Objective,Partial Differentiation,Eulerâ&#x20AC;&#x2122;s Theorem,Total differential coefficient,Implicit Functions,Curvature,Radius of curvature in Cartesian Coordinates,Radius of curvature in Polar co-ordinates,Answers to Self assessment Problems .

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

UNIT-1 MATRICES

1.0

Introduction

1.1

Objective.

1.2

Definition

1.3

Cayley – Hamilton Theorem.

1.4

Characteristic roots, characteristic vectors.

1.5

Some basic theorems in Theory of Equations

1.6.

Relations between the root and coefficient of an equation.

1.7.

Problems Type – I

1.8

Problems Type – II

1.9

Answers to Self Assessment Questions

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

1.0 Introduction In this unit you will learn about Cayley- Hamilton theorem. The verification of this theorem is illustrated with suitable examples. The characteristic equation of a square matrix is formed and on solving which you can find the characteristic roots of the given matrix. The characteristic vectors of a given matrix are also calculated. You will also learn some basic theorems regarding the theory of equations and using them you can easily solve a given equation. You will know thw relation between the roots and the coefficients of a given equation.

1.1 Objective. After completing this unit you should be able to 

Form the characteristic equation of a given matrix.



Find the characteristic roots and characteristic vectors of a given matrix.



Verify Cayley-Hamilton theorem for a given matrix and how to apply it to find the inverse of a given matrix.



Solve a given equation under given conditions.



Know the relation between the roots and the coefficients of a given equation and how to apply them to solve the given equation.

1.2. Definition A matrix is a rectangular array of numbers arranged into rows and columns.

 a11 a12 a  21 a22 A   a31 a32    am1 am 2

a13 a23 a33  am 3

 a1n   a2 n   a3n      amn 

The above matrix has m rows and n columns The general element is aij th

It occurs at the intersection of i row and j column. Examples: 1.

1 2 A  3 4

 1 3 5 4 B   1 0 2 7 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

1 2   3. C  5 7   3 0 Square Matrix: If the number of rows is equal to the number of columns, then it is called a square matrix.

Examples 1.

1 2  A  4 6

 2 4 6   2. B  1 2 0    1 2 7

1.3. Cayley – Hamilton Theorem. Every square matrix satisfies its characteristic equation. Problems 1. Verify cayley – Hamilton theorem for

Let

1 2  A  4 5

The characteristic equation is i.e.

1 2  4 5  

1 

5

A  I  o

0

i.e.

(1   )(5   )  4.2  0

i.e.

5    5  2  8  0

i.e.

2  6  3  0  1

This is the characteristic equation By Cayley – Hamilton theorem, put We get

  A in (1)

A2  6 A  3I  0  (2)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Now,

1 2 1 2 A2     4 5 4 5

 9 12    24 33  9 12 1 2 1 0 0 0  (2)     6   3   24 33 4 5 0 1 0 0  9 12  6  24 33   24    9  6  3  24  24  0 

12  3 0  0   30  0 3 0 12  12  0  0  33  30  3  0 0 0  0 0 0   0   

0 0  0 0  0 0 

,   Cayley-Hamilton theorem is verified.

1 2 1    2. Verify Cayley-Hamilton theorem for 2 1 2   1 3 1

Let

1 2 1    A= 2 1 2   1 3 1

The characteristic equation is

A  λI = 0 1 

1 

0

1    1   2  6 221     2  16  1     0 1   2  2  5 2 2   5    0 2  2  5  3  22  5  4  5    0  3  32  8  0

3  32  8  0  (1)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

This is the characteristic equation By Cayley – Hamilton theorem, every square matrix satisfies its characteristic equation. Put

  A in (1)  A3  3 A2  8 A  0  (2)

1 2 1  1 2 1     Now, A  2 1 2 2 1 2    1 3 1 1 3 1 2

6 7 6  A2  6 11 6 8 8 8 6 7 6 1 2 1    Next, A  A  A  6 11 6 2 1 2    8 8 8 1 3 1 3

26 37 26 A  34 41 34 32 48 32 3

(2) 

26 37 26 6 7 6 1 2 1 34 41 34  36 11 6  82 1 2  0       32 48 32 8 8 8 1 3 1 26 37 26   18  21  18    8  16  8  0 0 0 34 41 34    18  33  18    16  8  16 = 0 0 0         32 48 32  24  24  24   8  24  8  0 0 0  26  18  8 37  21  16 26  18  18  0 34  18  16 41  33  8 34  18  16   0    32  24  8 48  24  24 32  24  8  0 0 0 0  0 0 0 0   0    0 0 0  0

0 0 0 0  0 0  0 0 0 0  0 0 

Thus Cayley – Hamilton Theorem is verified.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 2 3 1   1  3 3. Find the inverse of the Matrix 3   5 2  4  2 3 1   1  3 Let A  3   5 2  4 The characteristic equation is

A  I  0 2

3

1 

3

5

4

0

2   1    4     6  33 4     15  16  51     0

2   2  3  2 3 3  27  11  5  0 22  6  4  3  2  9  81  11  5  0  3  2  10  66  0

3  2  10  66  0  (1) This is the characteristic equation. By Cayley – Hamilton theorem, every square matrix satisfies its characteristic equation put

  A in (1) A3  A 2  10 A  66I  0 Multiplying by

A 1 ,

A3  A1  A2  A1  10 A1  66I . A1  0. A1 A2  A  10I  66 A1  0

 66 A1   A2  A  10I  (2)  2  3 1  2  3 1   1  3  3 1  3 Now, A  A  A  3   5 2  4  5 2  4 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 10  7 7  A   24  14 12  16 9 5  2

 10  7 7   2  3 1  1 0 0     2  66 A    24  14 12   3 1  3  100 1 0  16 0 0 1 9 5   5 2  4 1

 7 1 0   10  2  10 7  3  0    24  3  0 14  1  10  12  3  0   16  5  0  9  2  0  5  4  10   2 10  8 1  A   27 3  9  66   11  11 9  1

Self assessment questions I

1 2 0   Vertify Cayley – Hamilton theorem for 0 4 5   3 0 1

1 2 2    Find the inverse of the matrix 2 1 2   2 2 1

1.4 Characteristic roots, characteristic vectors. If A is a square matrix, then its characteristic equation is

A  I  0 . The roots of this

equation are called the characteristic roots or Eigen values. The matrix X satisfying the equation characteristic vector corresponding to

 A  I  X  0 or

AX  X is called the

 . It is also called the eigen vector.

1. Find the characteristic roots of the matrix

2 2  18  4  

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

2 2  A  0  4

Let

The characteristic equation is

A  I  0 2

4

0

2    4     0  0 2  2  8  0

  4  2  0   2,4  The characteristic roots are 2, -4

 8  6 2   7  4 2. Find the characteristic roots and characteristic vectors of  6   2  4 3   8 6 2   7  4 Let A   6   2  4 3  The characteristic equation is

A  I  0 8

6

7

4

2 4 0 3

8   7   3    16  6 63     8  224  27     0

8   2 10  5 66 10  210  2   0

82  80  40  3  102  5  36  60  20  4  0  3  182  45  0

3  182  45  0

 2  182  45  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

   3  15  0

  0, 3, 15  The characteristic roots are 0, 3, 15 To find the characteristic vectors Consider the equation

 A  I X  0 2   x1  0 8    6   6 7    4   x   0   2     2  4 3     x3  0

8   x1  6 x2  2 x3  0    6 x1  7   x2  4 x3  0  ( A) 2 x1  4 x2  3   x3  0  If

  0, 8 x1  6 x2  2 x3  0  (1)

 A ,  6 x1  7 x2  4 x3  0  (2) 2 x1  4 x2  4 x3  0  (3) Solving (1) and (2) by the rule of cross – multiplication,

x1 6

4



6 4

x3 8

6

x x1 x2 = = 3 10 20 20 x x1 x = 2 = 3 1 2 2  x1  1;

x2  2;

x3  2;

1  X1   2   2  If

  3,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

10 5x1  6x 2  2x 3  0  (4)

 A  ,

6x1  4x 2  4x 3  0  (5) 2x1  4x 2  0x 3  0  (6)

Solving (5) and (5) we get

x1 4

4



x3 x2 = 6 4 6 4 2

4

x x2 x1 = = 3  8 16  16 x x1 x = 2 = 3 2 2 1 x2  1; x3  2;

 x1  2;

2    X2  1    2  If

  15,  7 x1  6 x2  2 x3  0  (7)

 A ,  6 x1  8 x2  4 x3  0  (8) 2 x1  4 x2  12 x3  0  (9) Solving (7) and (8), we get

x1 6

8  4



x3 x2 = 7 2 7 6 6 4

6 8

x x1 x2 = = 3 40  40 20 x x1 x2 = = 3 2 2 1  x1  2;

x2  2;

x3  1;

2    X 3  2    1   The characteristic root are 0, 3, 15

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

The corresponding characteristic vectors are

1  X1   2  ;  2 

2 X 2   1  ;  2 

2 X 3   2   1 

Self assessment question II

2  2 3   1. Find the characteristic roots and the characteristic vectors of 1 1 1   1 3  1 2. Find the characteristic roots and the characteristic vectors of

1 3 3 1  

1 2 3   3. Find the characteristic roots and the characteristic vectors of 0 4 5   0 0 6 THEORY OF EQUATIONS 1.5 Some basic theorems in Theory of Equations th

The general form of an n degree equation is

an x n  an1 x n1  an2 x n2  .......  a1 x  a0  0 where a0, a1, ………an are all constants, an # 0 and n  1. th

Theorem 1: Every polynomial equation of n degree has exactly n roots. Proof: Let

f (x)  an x n  an1 x n1  an2 x n2  .......  a1 x  a0  0 be a nth degree equation in

x. Here a0, a1, a2………an are all elements, Let

an  0

and

n  1.

1 be a root of the given equation. Then, x  1  is a factor of f (x) .

 f ( x)  x  1  . f1 x  Where f1 (x) is a polynomial of degree n  1 Now, f1 x   0 has one root, say

2

Then f1 x   x   2 . f 2 x  where f 2 x  is a polynomial of degree (n – 2).

 f ( x)  x  1 x   2 . f 2 x  Now, f 2 x   0 has one root, say

3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 f 2 x   x   3 . f 3 x   f x   x  1 x   2 x   3 . f 3 x  Producing like this, we get

f ( x)  x  1 x   2 ............x   n . f n x  Where f n x  is a polynomial of degree n-n=0.

f x   x  1 x   2 ...........x   n .k ………. where k is a constant an .x n  an1 x n1  ......a1 x  a0  x  1 x   2 ......x   n .k Equating the coefficient of

x n , an  k

 we get f ( x)  an x  1 x   2 ...........x   n   The equation f ( x)  0 has n roots 1 , 2 ......... n . If x is given any value other than Hence

1 , 2 ........... n

then

f ( x)  0 .

f ( x)  0 cannot have more than n roots.

 The equation f ( x)  0 has got exactly n roots. Theorem 2: In an equation with rational coefficients, irrational roots occur in conjugate pairs. Proof: Let

f ( x)  an x n  an1 x n1  ........  a1 x  a0  0 be a nth degree equation in x.

Here Let

a0 , a1 , a2 ...........an

xa b

are all rational numerals

an  0

and

n 1.

be one root of this equation where a and be are rational and

irrational. We will show that

xa b

is also a root of the given equation.

Now,

x  a  b x  a  b   x  a  b x  a  b   x  a   b  (1) When f (x) is divided by x  a   b, let Q(x) be the quotient and Rx  R' be the 2

remainder. Here R and

R' are rational numbers and Q(x) is a polynomial of degree n  2 .

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

2  f (x)   x  a   b  .Q  x   Rx  R1  (2)  

Put

xa b



in (2)

 

   b is a root of f ( x)  0 , we get f a  b   0

f a  b   a  b  a  Since

xa





 b  .Q a  b  R a  b  R ' 

0  R.a  R. b  R ' Equating Real parts,

0  R.a  R '  (3) Equating irrational parts, we get

0  R. b R 0

 R'  0

 (3)  0  0  R1





 (2)  f ( x)  x  a   b .Qx 



 



f ( x)  x  (a  b ) x  a  b .Qx 

x  a  b

is a root of

f ( x)  0 .

 Irrational roots occur in conjugate pairs.

Theorem 3. In an equation with real coefficients, imaginary roots occur in conjugate pairs. Proof: Let the given equation be

a0 , a1 , a2 ......an Let

are real numbers.

f ( x)  an x n  an1 x n1..........  a1 x  a0  0 where

an  0

and

n  1.

x    i be one root of this equation. Where ,  are real numbers.

We will prove,

x    i is also a root.

Now,  x     i   x     i    x     i  x     i





 x  2   2



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II When







f (x) is divided by x   2   2 , let Qx  be the quotient and Rx  R' be the

remainder. Here Qx  is a polynomial of degree n  2 and R , R are real numbers. 2





 f ( x)  ( x   ) 2   2 .Qx  Rx  R1  (1) Put

x    i in (1)





 f   i   a  i      2 .Q  i   R  i   R1  (2) Since

x    i is a root of f ( x)  0 , we get f   i   0

 2 

0  0  R.  iR  R1 Equating real parts,

0  R.  R'

Equating imaginary parts,

0  R.

0  R and R' 0  (1) 





f ( x)  x      2 .Q( x)  0 2

f ( x)  x    i .x    i .Qx

 x    i is root of the equation f ( x)  0  Imaginary roots occur in conjugate pairs. 1.6. Relations between the root and coefficient of an equation. Let the given equation be

an x n  an1 x n1  ..........a1 x  a0  0 where

a0 , a1 ,...........an

are constants,

an  0

and

n  1.

Let 1 ,  2 ........ n be the n roots of this equation then, a n x n  a n 1x n 1  ..........a1x  a 0  a n (x-1 )  x- 2  .........  x- n 

 an [ xn  x n 1 (1   2  ................   n  x n  2 1 2  1 2  ........... n 1. n   x n 3 1 2 3   2 3 4  .......  ......   1 1. 2 ..... n  n

Equating the coefficient of x

n 1

, we get

an 1  an 1   2  ......... n  FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 1  1   2  ........ n   Equating the coefficients of

an 1 an

x n  2 , we get

a n 2  a n  1 2   23  .........   1 2 

a n 2 an

/// ly  1 23  1 23   23 4  ........  

a n 3 an

…………………………………………….. ……………………………………………… Equating the constant terms,

a0  an  1 1. 2 ........... n n

1. 2 ............... n

n   1 .a0 . 

Note: 1 Quadratic Equation Let the equation be Let the roots be

ax 2  bx  c  0

 and  b a

     and  

Then

c a

Note: 2 Cubic Equation Let the equation be Let the roots be Then,



  1

  

ax 3  bx 2  cx  d  0

 ,  and 

     

b a

       

c a

d a

Note 3: Bi-quadratic equation Let the equation be Let the roots be

ax 4  bx 3  cx 2  dx  e  0

 ,  ,  and 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Then,



      

b a

 

             

 

         

  

c a

d a

e a

1.7. Problems Type – I 1. Solve the equation

x 4  2 x3  5x 2  6 x  2  0 given that 1  i is a root.

The given equation is

x 4  2 x3  5x 2  6 x  2  0 Given,

1  i is one root. 1  i is also a root.

Let the other two roots be Then,

 and 

  1  i   1  i        1

    4    4    (1) Product of the roots

1  i 1  i   2 1

1  1.. 4     2  4   2  1  2  4  1  0

 



 4  16  4 2 42 3 2

  2  3 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 The roots are 1  i ,  2  3 Another Method The given equation is

x 4  2 x3  5x 2  6 x  2  0  (1) Given,

1  i is a root 1  i is also a root.

x  1  i x  1  i   x  1  ix  1  i

Now,

 x  1  1 2

 x2  2x  2 Next,

x2  4x  1 x 2  2 x  2 x 4  2 x3  5x 2  6 x  2 x 4  2x3  2x 2 4x3  7 x 2  6x 4 x3  8x 2  8x x2  2x  2 x2  2x  2 0  1 

x





 2x  2 x 2  4x  1  0

x 2  2 x  2  0 (or) x 2  4 x  1  0

x 

x

 4  16  4 2

42 3 2

x  2  3  The roots are 1  i,  2  3 . 2.If 1 – 2i and

are two roots of the equation

x5  x4  8x2  9 x  15  0 , find the other

roots .

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

18 x5  x4  8x2  9 x  15  0

The given equation is, Given,

1 2i and

are two roots

1 2i and  3 are also roots of this equation Let

 be the fifth root,

Sum of the roots,

  1  2i  

3  1  2i   3    

 1 1

2  1   1  The roots are 1  2i,  3,  1 . 3.If

2 5

is one root of the equation

3x5  4 x4  42 x3  56 x2  27 x  36  0 , solve the

equation. The given equation is

3x5  4 x4  42 x3  56 x2  27 x  36  0

2 5

Given,

 Let

is one root.

2 5,  2 5

2 5,

are also roots of this equation.

 be the fifth root.

Sum of the roots

    

 

2 5 

 



2  5   2  5   2  5   

 4 3

4 3 4 3

 The roots are  2  5 , .

4. Solve the equation

x4  11x2  2 x  12  0 , given that

The given equation is Given,

5  1 is one root.

x4  11x2  2 x  12  0

5  1 is one root.

  5  1 is also a root. Let the other two roots be

, 

Sum of the roots,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 

5 1 5 1      0

 2     0

  2 

 (1)

Product of the roots







5  1  5  1 . . 

12 1

 5  1. 2     12

2   2  3

 2  2  3  0 (  3)(  1)  0

  3,1  The roots are 5. Solve the equation

5  1,  5  1, 3,  1 .

x4  2 x2  16 x  77  0 , given that  2  i 7 is a root.

The given equation is Given

2i 7

 2  i 7

x4  2 x2  16 x  77  0

is a root

is also a root.

Let the other two roots be Sum of the roots

, 

  2  i

7  2  i 7     0

 4     0

  4  Product of the roots

 . . . 







77 1

 4     2  i 7  2  i 7  77

4   .4  7  77 2

4   2  7

 2  4  7  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



4  16  28 2



4  i2 3 2

  2i 3  The roots are  2  i 7 , 2  i 3

Self Assessment problems III 1.

Solve the equation

x4  5x3  4 x2  8x  8  0 given that 1 5 is a root.

Solve the equation

x 4  12 x  5  0 , given that  1 2i is a root.

Solve the equation

x4  4 x2  8x  35  0 , given that 2  i 3 is a root.

Solve

x6  4 x5  11x4  40 x3  11x2  4 x  1  0 , given that

2 3

is one root.

1.8 Problems-Type – II 1. Solve the equation

x 3 12 x 2  39 x  28  0 , given the roots are in A.P. x 3 12 x 2  39 x  28  0  (1)

The equation is

Given that the roots are in A.P.

a  d , a, a  d .

Let the roots be Sum of the roots

  a  d  a  a  d  

 12 1

3a = 12 a=4

1 0 1

 12

 28

4  32

8





 x  4 x 2  8x  7  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

x  4x 1x  7  0 x  4,1,7  The roots are 1, 4, 7 Another Method

x 3 12 x 2  39 x  28  0

The equation is

Given, the roots are in A.P.

a  d , a, a  d

Let the roots be Sum of the roots

  a  d  a  a  d  

 12 1

3a  12 a4 Product of the roots

a  d aa  d    28 16  d 2  7  d 2  9 d2 9

d 3  The roots are 4 – 3, 4, 4 + 3 1, 4, 7 2. Solve the equation

4 x 3  24 x 2  23x  18  0

Given that the roots are in A.P The equation is

4 x 3  24 x 2  23x  18  0  (1)

Given that, the roots are in A.P. Let the roots be

a  d , a, a  d

Sum of the roots

  a  d  a  a  a  d  

 24 4

3a  6 a2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 24

 18

8  32

 18

 16  9





 1  x  2 4 x 2  16 x  9  0

x  22 x  12 x  9  0

 x  2 or x  1 2 or x  9 2  The roots are  1 2 , 2, 9 2

3. Find the condition that the roots of the equation The equation is

x 3  px 2  qx  r  0

may be in A.P.

 (1)

Its roots are in A.P. Let the roots be a – d, a, a + d. Sum of the roots

  a  d  a  a  d   p 3a   p a

p 3

x  

p is a root. 3

Substituting in (1), 3

 p  p  p     p    q    r  0  3  3  3

 p 3  3 p 3  9 pq  27r 0 27

2p3  9pq  27r  0 This is the required condition. 4. Solve the equation

x 3 19 x 2  11x  216  0 , given that the roots are in G.P.

The equation is

x 3 19 x 2  11x  216  0  (1)

Given that, the roots are in G.P.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Let the roots be

a , a, ar r

Product of the roots

 216 a .a.ar   r 1 a 3  216

a6

 19

114

 216

 76

216

 13





 1  x  6 x 2  13x  36  0

x  6x  4x  9  0

x  6, 4, 9  The roots are 4, 6, 9

5. Solve the equation

2 x 3  21x 2  42 x 16  0

Given that the roots are in G.P. The equation is

2 x 3  21x 2  42 x 16  0  (1)

Given that, the roots are in G.P. Let the roots be

a , a, ar r

Product of the roots

 16 a .a.ar   r 2 a3  8

a2

2 0 2

 21

 16

4  34

 17

 1 

x  22x 2 17 x  8  0 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

x  22x 1x  8  0 1  x  2, , 8 2  The roots are

1 , 2, 8. 2

6. If the roots of the equation

ax 3  3bx 2  3cx  d  0 are in G.P., show that ac 3  db 3 .

The Given equation is

ax 3  3bx 2  3cx  d  0  (1) Its roots are in G.P. Let the roots be,

A , A, R R

Product of the roots

A d . A. AR   R a  A3   Put

d a

x  A in (1)

a. A3  3b. A2  3cA  d  0

 d a.    3bA2  3cA  d  0  a 3bA2  3cA  0 3 AbA  c   0

bA  C b3 A3  C 3

 d b 3 .    C 3  a  db3  ac 3 7. Solve the equation

6 x 3  11x 2  6 x  1  0 , given that the roots are in H.P.

The given equation is

6 x 3  11x 2  6 x  1  0  (1)

Given that, the roots are in H.P.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Put

x

Then,

1 in (1) y 1 1 1  11. 2  6.  1  0 3 y y y

xy 3 ; 6  11y  6 y 2  y 3  0

y 3  6 y 2  11y  6  0

 (2)

The roots of this equation are in A.P.

a  d , a, a  d

Let the roots be Sum of the roots

  a  d  a  a  d  

 6 1

3a  6 a2

6

8

4

2   y  2y 2  4 y  3  0  y  2 y  1 y  3  0  y  1, 2, 3 The roots of equation (2) are 1, 2, 3

 The roots of the given equation are

1 1 1 , , . 1 2 3 8. Solve the equation

x 3  3x 2  4 x  12  0 given that the sum of two roots is zero.

The given equation is Let the roots be

x 3  3x 2  4 x  12  0

, , 

Sum of the roots

        

 3 1

 3 Product of the roots

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 .  .  12   2 .3  12

2  4

  2  The roots are 2, -2, 3. 9. Solve the equation

x 3  9 x 2  14 x  24  0 given that two roots are in the ratio 3 : 2.

The given equation is

x 3  9 x 2  14 x  24  0  (1)

Given two roots are in the ratio 3 : 2 Let the roots be

3 , 2 , 

Sum of the roots

  3  2    

 9 1

5    9

  9  5

 (2)

Next,

  3.2  2.  2.  14 6 2  5  14  2  5 9  5   14 6 2  45  25 2  14  19 2  45  0 19 2  45  14  0

19  7  2  0   2, When Put

7 19

  2, 2  4

x  4 in (1)

42  9.42  14.4  24  0 64  144  56  24  0 144 – 144 = 0 The equation is satisfied

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 x  4 is one root. Put

 2

in (2)

  9  5  2  1  The roots are 3 × 2, 2 × 2, -1 6, 4, -1 10. Solve the equation

x3  x 2  16 x  20  0 given that the difference between two roots is 7.

The given equation is Let the roots be

x3  x 2  16 x  20  0  (1)

 ,   7, 

Sum of the roots

    7    1   8  2  2

     7  .    7.  16  2  7  2  7   16

 2  7  2  8  2   7 8  2   16  2  7  16  4 2  56  14  16  0  3 2  23  40  0

3 2  23  40  0

  53  8  0   5,  Put

8 3

x  5 in (1)

 53   52  16 5  20  0 –125 + 25 + 80 + 20 = 0 –125 + 125 = 0 The equation is satisfied.

 x = -5 is a root Put

  5

in (2)

  8  10  2  The roots are  5,  5  7, 2

 5, 2, 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

11. Solve the equation

x4  4 x3  2 x2  12 x  9  0 given that it has two pairs of equal roots.

The given equation is Let the roots be

x4  4 x3  2 x2  12 x  9  0  (1)

,, , .

Sum of the roots

          4 2     4

    2   2  

 (2)

  .  .  .  .  .   .  2  2  4   2  2  2  4  2      2   2  2  2  8  4 2  4  4   2  2  0  2  8 2  4 2  4  4   2  2  0  2 2  4  6  0

 2  2  3  0

 1  3  0   1,  3  The roots are 1, 1, –3, –3 12. Solve the equation

x 2  4 x 3  25x 2  8x  5  0 given that the product of two roots is unity.

The given equation is Let the roots be Given

x 2  4 x 3  25x 2  8x  5  0  (1)

,  ,  , 

 .  1  2

          4  (3)               25  (4)           8  (5)   5  (6) Put

  1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 6  1.  5

  5

5  1.  1.  .5   .5  8     5     8  (7) (3)  (7)   4     12     3   3  

2    3     1  3   2  1

 2  3  1  0



3 9 4 2



3 5  (8) 2

Put

  3  

in (7)

    5 3       8

   7   7  We have

  5

 7     5 7   2  5  0

 2  7  5  0



7  49  20 2



7  29  9 2

 The roots are

 3  5 7  29 , 2 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

13. Solve the equation x

 2 x3  4 x 2  6 x  21  0 , given that two of its roots are equal in

magnitude and opposite in sign. The given equation is Let the roots be

x 4  2 x3  4 x 2  6 x  21  0  (1)

, ,  , 

          2     2  (2)

                 4   2    4

 (3)

   ( )           6   2      6

 2 .2  6 2  3

   3  (4) Put

2  3

in (3)

 3   2     4

2   2  7  0

 2  2  7  0 

2  4  28 2

  1  i 6  5  The roots are  3, 1  i 6

14.Solve the equation

x 4  8x3  7 x 2  36 x  36  0 given that the product of two roots is

equal in magnitude but opposite in sign to the product of other two roots. The given equation is Let the roots be

x 4  8x3  7 x 2  36 x  36  0  1

 ,  , ,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

          8  2               7  3

          36  4   36  5

    6

given

5 

-     36

 2  36    6  7    6  8 4 

- 6 - 6   .6   .6  -36 -  -       -6  9

2  9 

2     2

   1   1-

8 

 1 -    6

 - 2  6  0

 2   6  0

 - 3  2  0   3,-2  8 Put

  1   in 2

 1      8    7   7 -

7 

 7 -    6 7 -  2  6  0

 2  7  6  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 -1  6  0   1,6  9  The roots are 3, -2, 1, 6. Self assessment problems IV 1.

Solve the equation

4 x3  36 x 2  59 x  39  0 given that the roots are in A.P

Solve the equation

x3  7 x 2  14 x  8  0 given that the roots are in G.P.

Solve the equation

3x3  22 x 2  48x  32  0 given that the roots are in H.P.

Solve the equation

x 4  2 x3  25x 2  26 x  120  0 given that the product of two of its

roots is 8. 5.

Solve

x 4  8x3  14 x 2  8x  15  0 given that the sum of two roots is equal to the sum

of the other two.

1.9. Answers Self Assessment Questions I

1 2 0   1. Verify Cayley – Hamilton theorem 0 4 5   3 0 1 1 2 0   Let A  0 4 5   3 0 1 The characteristic equation is

A  I  0 1 

4

0 5 0 1 

1   4   1     0  20  3.5 0  0

1   4  5  2  30  0

4  5  2  4  52  3  30  0

 3  62  9  34  0  3  62  9  34  0  (1)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

This is the characteristic equation. By Cayley-Hamilton theorem, every square matrix satisfies its characteristic equation. Put

A

in 1

A3  6 A2  9 A  34I  0  (2)

1 2 0 1 2 0    2 Now, A  0 4 5 0 4 5    3 0 1 3 0 1  1 10 10   15 16 25  6 6 1   1 10 10  1 2 0    3 2 Next, A  A  A  15 16 25 0 4 5     6 6 1  3 0 1 31 42 60   90 94 105  9 36 31 

2 , 31 42 60   1 10 10  1 2 0 1 0 0 90 94 105  615 16 25  90 4 5  340 1 0  0          9 96 31   6 6 1  3 0 1 0 0 1 0 0  0 0 0 31 42 60    6  60  60   9 18 0   34 90 94 105   90  96  150   0 36 45   0  34 0   0 0 0         9 96 31   36  36  6  27 0 9   0 0  34 0 0 0 60  60  0  0  0 0 0  31  6  9  34 42  60  18  0 90  90  0  0 94  96  36  34 105  150  45  0  0 0 0     9  36  27  0 36  36  0  0 31  6  9  34  0 0 0

0 0 0  0 0 0  0 0 0   0 0 0      0 0 0 0 0 0 The Cayley-Hamilton theorem is verified.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

1 2 2    2. Find the inverse of 2 1 2   2 2 1 1 2 2    Let A  2 1 2   2 2 1 The characteristic equation is

A  I  0 1 

1 

2 2 0 1 

1    1   2  4 221     4  24  21     0 1   2  2  3 2 2  2  22  2   0 2  2  3  3  22  3  4  4  4  4  0  3  32  9  5  0

3  32  9  5  0 This is the characteristic equation. By Cayley – Hamilton theorem, every square matrix satisfies its characteristic equation. Put

A

in (1)

 A3  3 A2  9 A  5I  0 Multiplying by A

1

A3 . A1  3 A2 . A1  9 A. A1  5I . A1  0. A1 A2  3 A  9I  5 A1  0 5 A1  A2  3 A  9I  (2)

1 2 2 1 2 2    2 Now, A  A  A  2 1 2 2 1 2    2 2 1 2 2 1 9 8 8  8 9 8 8 8 9 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

2 , 9 8 8 1 2 2 1 0 0 5 A1  8 9 8  32 1 2  90 1 0 8 8 9 2 2 1 0 0 1 86 86  9  3  9    86 939 8  6   8  6 86 9  3  9 2  3 2 1  A 1   2  3 2  5  2 2  3 Self Assessment Questions – II

2  2 3   1 1  1. Find the characteristic roots and characteristic vectors of 1  1 3  1 2  2 3   1 1  Let A  1  1 3  1 The characteristic equation is

A  I  0 2

2

1 

3 1 0 1 

2   1   1     3 21   1 33  1     0

2   2  4 4  2 y  6  3  0

22  8  3  4    2  0  3  22  5  6  0

3  22  5  6  0  (1) This is the characteristic equation Put

 1,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

1–2–5+6=0 The equation is satisfied.

   1 is a root.

2

5

1

6

1

6

1    12    6  0   1  2  3  0   1, 3, - 2  The Characteristic roots are 1, 3, -2 To find the characteristic vectors Consider the equation  A  I X  0

3   x1   0  2    2      1  1   x2    0   1  1 3  1     x3   0  

2   x1  2 x2  3x3  0   1.x1  1   x2  1.x3  0    A 1.x1  3x2   1   .x3  0  If

  1,

 A  , x1  2 x2  3x3  0 x1  0.x2  x3  0 x1  3x2  2 x3  0 Solving

x1  2 3 0



x3 x2  1 3 1 2 1 1

x x1 x  2  3 2 2 2 x1 x2 x3   1 1 1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 x1  1; x2  1; x3  1  1    X1   1  1   If

 3

 A  ,

 x1  2 x2  3x3  0 x1  2 x2  x3  0 x1  3x2  4 x3  0

Solving,

x1  2 3 2 1



x3 x2  1 3 1  2 1

2

x1 x2 x3   4 4 4 x1 x2 x3   1 1 1

 x1  1; x2  1; x3  1  1    X 2   1  1   If

  2

 A  ,

4 x1  2 x2  3x3  0 x1  3x2  x3  0 x1  3x2  x3  0

Solving,

x1  2 3 3



x3 x2  4 3 4 2 1 1

x x1 x  2  3  11  1 14

 x1  11; x2  1; x3  14

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 11    X 2   1   14     The characteristic roots are 1, 3, -2

  1 1   11     The characteristic roots are  1 , 1,   1   1  1  14      2. Find the characteristic roots and characteristic vectors of

Let

1 3 3 1  

1 3 A  3 1

The characteristic equation is

A  I  0 1 

1 

0

1   2  9  0 2  2  8  0

  2  4  0   2, 4  The characteristic roots are –2, 4 To find the characteristic vectors Consider the equation Consider the equation  A  I X  0

3   x1   0  1           3 1     x2   0 

1   x1  3x2  0    A 3x1  1   x2  0 If

  2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

3x1  3x2  0

 A ,

3x1  3x2  0 x1  x2  0 x2   x1

Put x1  1 Then x2  1

1  X 1      1 If

 4  3x1  3x2  0

 A ,

3x1  3x2  0 x1  x2  0 x2  x1

Put x1  1 Then x2  1

 1  X2     1  The characteristic roots are –2, 4 The characteristic vectors are

 1  1  ,     1 1

1 2 3   3. Find the characteristic roots and characteristic vectors of 0 4 5   0 0 6 1 2 3   Let A  0 4 5   0 0 6 The characteristic equation is

A  I  0 1 

4

6

0

1   4   6     0 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

   1, 4, 6  The characteristic roots are 1, 4, 6. Next, Consider the equation  A  I X  0

1     0  0 

3   x1   0      4 5   x2    0  0 6     x3   0  2

1   x1  2 x2  3x3  0   0.x1  4   x2  5.x3  0    A 0.x1  0.x2  6   .x3  0  If

  1,

 A  , 0.x1  2 x2  3x3  0 0.x1  3.x2  5.x3  0 0.x1  0.x2  5.x3  0 Solving

x1 2 3



3 5



x x2  3 0 3 0 2 0 5

0 3

x1 x2 x3   1 0 0

 x1  1; x2  0; x3  0 1    X1   0  0   If

  4,

 A  ,  3.x1  2 x2  3x3  0 0.x1  0.x2  5.x3  0 0.x1  0.x2  2.x3  0 Solving

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II x1 2 3

 

0 5

x x2  3 3 3 3 2 0

0 0

x1 x2 x3   10 15 0 x1 x2 x3   2 3 0

 x1  2; x2  3; x3  0  2    X2   3  0   If 

6,

 A  ,  5.x1  2 x2  3x3  0 0.x1  2.x2  5.x3  0 0.x1  0.x2  0.x3  0 Solving

x1 2

2 5

 

x3 x2  5 3 5 2 0

2

x1 x2 x3   16 25 10

 x1  16; x2  25; x3  10  16     X 3   25   10    Answers Self Assessment Problems – III 1. Solve the equation

x 4  5x3  4 x 2  8x  8  0 given that 1 5 is a root.

The given equation is

x 4  5x3  4 x 2  8x  8  0  (1)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Given

1 5

is one root.

 1 5 is also a root. Let the other two roots be

, 

Sum of the roots

  1 

5 1 5      5

  3    (2) Product of the roots

1  5 1  5 ..  8

1  5. 3     8





 4 3   2  8 3   2  2

 2  3  2  0

   1   2  0   1, 2  The roots are 1 5 , 1, 2. 2. Solve the equation

x 4  12 x  5  0 , given that  1 2i is a root.

The given equation is Given

x 4  12 x  5  0

 1 2i is a root.

  1 2i is also a root. Let the other two roots be

, 

Sum of the roots

  1  2i  1  2i      0   2  Product of the roots

 1  2i  1  2i ..  5 1  4. 2     5





5 2   2  5 2   2  1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 2  2  1  0



2 44 2

  1 2  The roots are  1 2i , 1 2 3. Solve the equation

x 4  4 x 2  8x  35  0 Given that 2  i 3 is a root.

The given equation is Given,

2i 3

x 4  4 x 2  8x  35  0  (1)

is a root

 2  i 3 is also a root. Let the other two roots be

, 

Sum of the roots

  2  i

3  2  i 3     0

  4    (2) Product of the roots

2  i 3 2  i 3 ..  35

4  3  4     35  4   2  5

 2  4  5  0



 4  16  20 2

  2  i  The roots are 2  i 3 ,  2  i 4. Solve the equation

x6  4 x5  11x 4  40 x3  11x 2  4 x  1  0 given that

2 3

is one

root. The given equation is Given



2 3

x6  4 x5  11x 4  40 x3  11x 2  4 x  1  0

is a root.

2  3,  2  3,  2  3 are also roots.

Let the other two roots be

, 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Sum of the roots

 

2  3  2  3  2  3  2  3     4

  4    1 Product of the roots

( 2  3)( 2  3)( 2  3)( 2  3). .  1  2  32  3. .4     1

4   2  1

 2  4  1  0



4  16  4 2

  2 3  The roots are

2  3, 2  3,  2  3,  2  3, 2  3, 2  3 . Self Assessment Problems – IV 1. Solve the equation

4 x3  56 x 2  59 x  39  0 Given that the roots are in A.P.

The given equation is

4 x3  56 x 2  59 x  39  0  (1)

Given, the roots are in A.P.

a  d , a, a  d

Let the roots be Sum of the roots

  a  d  a  a  d  

 36 4

3a  9 a3

4  36

12  72

 39

4  24  13





 1  x  3 4 x 2  24 x  13  0

x  32x  132x  1  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II  x  3,

13 1 ,2 2

 The roots are  2. Solve the equation

1 13 , 3, 2 2

x3  7 x 2  14 x  8  0 Given that the roots are in G.P.

Let the roots be

a , a, ar r

Product of the roots

a .a.ar  8 r

a3  8

a2

0 1

7

8

2  10

5

 1 

x  2x2  5x  4  0 x  2x  1x  4  0 x  2, 1, 4  The roots are 1, 2, 4. 3. Solve the equation

3x3  22 x 2  48x  32  0 given the roots are in H.P.

The given equation is

3x3  22 x 2  48x  32  0  (1) Its roots are in H.P. Put

x

1 in (1) y

1 1 1  22. 2  48  32  0 3 y y y

3  22y  48y 2  32y3  0 32y3  48y 2  22y  3  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Its roots are in A.P.

a  d , a, a  d

Let the roots be Sum of the roots

  a  d  a  a  d  32 3 2 1 a 2

3a 

1 32  48 22 16  16 2 0 32  32

3 3 0

 1 





1  2  y   32 y  32 y  6  0 2  1   y  4 y  13 y  6  0 2  1 1 3 y , , 2 4 4  The roots of (2) are

1 1 3 , , 4 2 4

 The roots of the given equation are 4, 2, 4. Solve the equation

4 . 3

x 4  2 x3  25x 2  26 x  120  0 given that the product of two roots is 8.

The equation is

x 4  2 x3  25x 2  26 x  120  0  (1)

Let the roots be

,  ,  , 

          2  (2)               25  (3)           26  (4)   120  (5)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II Given

 .  8  (6)

5 , 8. .  120  .  15  7 3 , 8  8   .15  15  26 15     8     26  8 2 8  8 ,  7     42   6   6    9

6 ,  6     8  2  6  8  0

  2  4  0   2, 4

  2,

9 ,   6  2  4 Put

  2,   4

in (2)

2  4      2

  8    10

7 ,   8     15  8   2  15  0

 2  8  15  0

  3  5  0   3,  5   3,

10 ,   8  3  5  The roots are 2, 4, –3, –5. 5. Solve

x 4  8x3  14 x 2  8x 15  0 given that the sum of two roots is equal to the sum of the

other two.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

The given equation is Let the roots be

x 4  8x3  14 x 2  8x 15  0

,  ,  , 

          8  (1)               14  (2)           8  (3)   15  (4) Given

        5

        0  6 (1) + (6)

 , 2     8

    4  7     4  8

3 ,             8  .4   .  8     2  9

  2  

4 ,   2     15 Put

  t

t  2  t   15

 2t  t 2  15  0 t 2  2t  15  0

t  5t  3  0 t  3 or  5

  3 or  5 If

  3,

 4     3 4   2  3  0

 2  4  3  0

  1  3  0 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

  1, 3 If

  5

 4     5 4   2  5  0

 2  4  5  0

  1  5  0   1, 5  The roots are 1, 3, -1, 5

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Unit 2

2.0 Introduction 2.1 Objective 2.2 Formation of equations with roots changed in sign 2.3 Formation of equations with roots multiplied by a constant 2.4 Formation of equation with roots as squares of the given equation 2.5 Formation of equation with roots increased or decreased by a constant 2.6 Removal of terms 2.7 Transformation in general 2.8 Symmetric functions of roots. 2.9 Answers to Self assessment problems.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

2.0

INTRODUCTION In this unit will learn different types of transformation of equations. Whenever you

are given an equation you can transform it into another whose roots are changed in sign. You can transform the given equation one whose roots obtained by multiplying the roots of the given equation by a given number. You will also know how to form an equation whose roots are simply the squares of the equation. You can form new equation whose roots are obtained by increasing or decreasing the roots of the given equation by a number. If you are given an equation

you can transform it into one its second term

removed. Also you will learn many other types of transformations. You can also find the values of symmetric functions of roots.

2.1

OBJECTIVE After completing this unit you should be able to  Form an equation with roots changed in sign.  Form an equation with roots multiplied by a constant.  Form an equation whose roots are squares of the given equation  Form an equation with roots increased or decreased by a given number.  Form an equation with its second term removed.  Form an equation whose roots are changed under different conditions.  Find the values of symmetric functions of roots.

Transformation of Equations: If 1 , 2 ,........n are the roots of the equation f(x)=0, then forming the equation whose roots are  ( 1 ),  ( 2 ) . . . . .  ( n ) is called transformation of the given equation.

2.2 TYPE I: Formation of an equation with roots changed in sign of those of the given equation.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

1. If , , ,  are the roots of the equation ax4+ bx3+cx2+dx+e=0, form an equation whose roots are - , , , and -  . The given equation is ax4+bx3+cx2+dx+e=0  (1) Its roots are , ,  and  . We will form an equation with roots - , , , and -  . Put y=-  y=-x x=-y (1)  a(-y) 4+ b(-y) 3 +c(-y) 2 +d(-y) +e = 0

ie.,

ay 4 –by3 +cy2 -dy +e = 0

This is the required equation. 2. If  ,  ,  are the roots of the equation 2x 3 +3x 2 +x -4 =0, form an equation with roots   , and

.

The given equation is 2x3 +3x2 +x -4 = 0

 (1)

Its roots are  ,  ,  . We will form an equation with roots –  , , Put y = - y = -x x = -y (1) 

2(-y) 3 +3(-y) 2 –y -4 = 0 -2y 3 +3y 2 –y -4 = 0 2y 3 -3y 2 +y +4 = 0

This is the required equation.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Self assessment problems I. 1.

If , ,  are the roots of the equation x3 +px2 +qx +r =0, form an equation with roots –  , ,

2.3

TYPE 2

Multiplication of roots by m.

Let 1 ,  2 . . . . . . . . .  n be the roots of the equation anxn + an-1xn-1 + an-2 xn-2 + . . . . . . . . + a1x + a0 = 0  (1) We will form an equation with roots m  1, m  2, . . . . . . . . . .m  n Put

y =m  1 y =mx x=y

n 1 y  y a n   + an-1  y   . . . . . . . . + a1   + a0 = 0 m m m

a n yn + m.an-1 yn-1 + m2.an-2yn-2 + . . . . + mn-1.a,y + mn.a0 = 0 This is the required equation. 1.If , ,  be the roots of the equation 2x3 + 4x2 + 5x – 1 = 0, form an equation with roots 10  , 10  , 10  . The given equation is 2x3 + 4x2 + 5x – 1 = 0 - - - - - - (1) Its roots are , ,  . We will form an equation with roots 10  ,10  ,10  . Put

10 

10x

y 10

 y  y  y (1)  2   + 4   + 5   -1  10   10   10  3

2y3 + 40y2 + 500y – 1000

= =

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

This is the required equation. 2.If , , ,  are the roots of the equation x4 + 4x3 – 5x2 + 10x – 25 = 0 form an equation with roots 2, 2, 2, 2 . The given equation is x4 + 4x3 – 5x2 + 10x – 25 = 0 -------------- (1) Its roots are  ,  ,  and  . We will form an equation with roots 2, 2, 2, 2 . Let

2

y 2

 y  y  y  y (1)    + 4   - 5   + 10   - 25 2 2 2 2 4

y4 + 8y3 – 20y2 + 80y – 400

X1b;

This is the required equation. 3.If , ,  be the roots of the equation ax3 + bx2 + cx + d = 0 form an equation with roots k  , k  , and k  . The given equation is ax3 + bx2 + cx + d = 0 -------------- 1 Its roots are , ,  . We will form an equation with roots k  , k  , and k  . Put

k

y k



 y  y  y a  + b  + c  + d k k k 3

(1)

ay3 + k.by2 + k2.cy + k3.d

This is the required equation. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Self Assessment problem II 1.If , , ,  are the roots of the equation x4 + x3 + fx2 – 5x + 10 = 0 form an equation with roots 3  , 3  , 3  , and 3  .

Type III: Square of the roots. Let the given equation be. an xn + an-1 xn-1 + . . . . . . . . a1x + a0 = 0---------- 1 Let the roots be  1 ,  2 . . . . . . .  n. Put

(1)

12



 y

+ a n 1

 y

n 1

+ . . . . . . + a1

 y + a

This is the required equation. 1.If , ,  be the roots of the equation x3 + 4x2 + 5x + 11 = 0 form an equation with roots

 2 ,  2 and  2 . Given,  ,  ,  are the roots of the equation x3 + 4x2 + 5x + 11 = 0 ------------ (1) We will form an equation with roots  2 ,  2 and  2 Put

(1)

2



y.x + 4y + 5x + 11 = 0

Squaring,

x(y + 5)

-(4y + 11)

x2(y + 5)2

(4y + 11)2

y(y2 + 10y + 25)

16y2 + 88y + 121

y3 – 6y2 – 63y – 121

This is the required equation. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

2. If , ,  are the roots of the equation ax3 + bx2 + cx + d = 0, form an equation with roots  2 ,  2 and  2 . Given, , ,  are the roots of the equation. ax3 + bx2 + cx + d = 0 -------------- (1) We will form an equation with roots  2 ,  2 and  2 . Put

(1)



Squaring,

2

ay.x + by + cx + d

x(ay + c)

1 (by + d)

x2(ay + c)2

(by + d)2

y(a2y2 + 2acy + c2)

b2y2 + 2bdy + d2

a2y3 + (2ac – b2)y2 + (c2 – 2bd)y – d2 This is the required equation.

Self assessment problem III 1.If , ,  are the roots of the equation x2 + 3px2 + 3qx + r= 0 form an equation with roots  2 ,  2 and  2 .

Type 4: Increasing or decreasing the roots by a given number Let  1 ,  2 . . . . . . .  n be the roots of the equation an xn + an-1 xn-1 + . . . . . . . . a1x + a0 = 0

---------- (1)

We will form an equation with roots

1  n,  2  n, . . . . . . .  n  n, Put

x–n

y+n

(1)



1  n

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

an (y + n)n + an – 1 (y + n)n – 1 + . . . . . a1(y + n) + a0 = 0 This is the required equation. 1.Diminish by 2 the roots of the equation x3 + 2x2 – 3x + 5 = 0 Let , ,  be the roots of the equation x3 + 2x2 – 3x + 5 = 0 ----------------- (1) We will form an equation with roots

  2,   2 and   2 Put

 2

x–2

y+2



(y + 2)3 + 2(y + 2)2 – 3(y + 2) + 5 = 0

y3 + 6y2 + 12y + 8 + 2y2 + 8y + 8 – 3y – 6 + 5

y3 + 8y2 + 17y+ 15

This is the required equation.

Another method The given equation is x3 + 2x2 – 3x + 5 = 0 Diminishing the roots by 2, 2

1 0 1 5

1 2 1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

58 7 0

The required equation is x3 + 8x2 + 17x + 15 = 0 2.

Decrease the roots of the equation x4 – 3x3 + 4x2 - 2x + 1 = 0 by unity The given equation is x4 – 3x3 + 4x2 - 2x + 1 = 0 Decreasing the roots by 1 1

2 2

-2

-3

1 1

-1

-2

The new equation is x4 + x3 + x2 + x + 1 = 0 3.Increase the roots of the equation x4 + 3x3 - 2x2 + x + 4 = 0 by unity The given equation is x4 + 3x3 - 2x2 + x + 4 = 0 Increasing the roots by 1,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

-1

-2

-4

-1

+ 4 5 1

0 0

-1

-2

-1 1

The new equation is x4 - x3 - 5x2 + 10x - 1 = 0 4.Find the equation whose roots are the roots of x4 - x3 - 10x2 + 4x + 24 = 0 increased by 2 and hence solve the equation. The given equation is x4 - x3 - 10x2 + 4x + 24 = 0 Increasing the roots by 2, -2

-1

-2

10 6

2 4 -24

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

-3

-2

-5

-2

-7

-2

2 1

1 4 2 0

1 -9 The new equation is x4 - 9x3 + 20x2 + 0x + 0 = 0 x2 (x2 – 9x + 20) = 0 x2 ( x – 5) (x – 4) = 0 therefore

x = 0, 0, 4, 5

The roots of the given equation are 0 – 2, 0 – 2, 4 – 2, 5 – 2 -2, -2, 2, 3

Self assessment problems IV 1.Find

the

equation

whose

roots

are

those

the

equation

those

the

equation

2x4 - 25x3 + 111x2 – 208x + 140 = 0 diminished by 3. 2.Find

the

equation

whose

roots

are

4x4 + 32x3 + 83x2 + 76x + 21 = 0 increased by 2. Hence solve the given equation.

Type 5 : Removal of Terms Let 1 ,  2 . . . . . . . . .  n be the roots of the equation

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

an xn + an-1 xn-1 + an-2 xn-2 + . . . . . . . a1x  a 0  0  (1) We will decrease the roots by h. Put

1  h

x–h

y+h

(1)  an(y + h)n + an-1(y+h)n-1+an-2(y+h)n-2+ . . . . . . .a1(y+h) + a0 = 0 [anyn+an.n h yn-1+. . . . ] + [an-1 yn-1+an-1(n-1)h yn-2+. . . .] + [an-2yn-2+ . . . .] + . . . . . +a1y+a1h+a0 = 0 anyn+yn-1[an.n.h+an-1] + . . . . +a1y+a1h+a0 = 0 This is the new equation. If the second term is to be removed, Coefficient of yn-1

an.n.h + an-1

a n 1 na n

Coeff of x n 1 n. Coeff of x n

Problems 1. Solve the equation x 4 + 20x 3 +143x 2 +430x + 462 = 0 by removing the second term. The given equation is x 4 + 20x 3 +143x 2 +430x + 462 = 0 We will decrease the roots by h h

Coeff of x n 1 n. Coeff of x n

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



coeff of x 3 4  coeff of x 4



20 4 1

= -5. Increase the roots by -5. 5

2 0

43 -

12 -

90 1

8 -

450

-340

The new equation is

y 4 +0.y 3 -7y 2 +0.y +12 = 0.

y 4 -7y 2 +12 = 0. (y 2 -3) (y 2 -4) = 0. y 2 = 3,4. y =  3 ,  2. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 The roots of the given equation are

3 - 5, - 3 -5, 2-5, -2-5 3 - 5, - 3 -5, -3, -7. 2. Solve the equation x 3 -12x 2 +48x -72 = 0 by removing the second term. The given equation is x 3 -12x 2 +48x -72 = 0---------------- (1). We will decrease the roots by h. h

coeff of x 2 3  coeff of x 3

h

(12) 3 1

h4

Reducing the roots of (1) by 4, we get 4

8 4

-72 -

4 1

-8

6 4

16 0

-4

The new equation is x 3 + 0.x 2 +0.x -8 =0. x 3 -8 = 0.----------(2). x 3 -2 2 = 0.

(a 3 - b 3 ) = (a-b) (a 2 +ab+b 2 )

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

(x -2) (x 2 +2x +2 2 ) = 0. (x -2) = 0 or (x 2 +2x +2 2 ) = 0. x=2

x = -2 

4  4.4 /2

x = -2  i 2 3 /2 x = -1  i 3  The roots of 2 are

2, -1  i 3 The roots of the given equation are 2+4, -1  i 3 +4 6, 3  i 3 . 3. Solve the equation x 4 -12x 3 +48x 2 -72x +35 = 0 by removing the second term. The given equation is x 4 -12x 3 +48x 2 -72x +35 = 0. We will decrease the roots by h. coeff of x 3 h 4  coeff of x 4 h

(12) 4 1

h 3

Decreasing the roots of 1 by 3, we get 3

3 2 1

5 6

3 -72

0 1

27 -

9 -

8 9

18 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

65 6 0

9 -

-3

The new equation is x 4 + 0. x 3 -6x 2 +0.x + 8 = 0. x 4 -6x 2 +8 = 0. (x 2 -2) (x 2 -4) = 0. x 2 = 2 or x 2 = 4 x=   The roots of 2 are 

2 , or  2 2 , 2, -2.

The roots of the given equation are 3

2 , 3+2, 3-2

3

2 , 5, 1.

4. Solve the equation x 3 + 6x 2 +12x-19 = 0 by removing the second term. The given equation is x 3 + 6x 2 +12x-19 = 0--------------(1) We will decrease the roots by h, h

coeff of x 2 3  coeff of x 3

(6) 3 1 h  2 h

Increasing the roots of 1 by 2, we get, 1

-19

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

-2

2 0

-2

1 0

-27

-2

1 0

-8

The new equation is x 3 +0. x 2 +0.x-27 = 0. x 3 -27 = 0.------(2) x 3 -3 3 = 0 (x-3) (x 2 +3x+3 2 ) = 0 x-3 = 0 or x 2 +3x +9 =0. x = 3 or x = -3 x  x= x

3  9  36 2

3  i3 3 2

 The roots of (2) are 3, -3  i3 3 /2

The roots of the given equation are 3-2,

3  i3 3  2 , 2

7  i3 3 2

5. Solve the equation x 4 +4x 3 +5x 2 +2x-6=0 by removing second term The given equation is x 4 +4x 3 +5x 2 +2x-6=0----------(1). We will decrease the roots by h, FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

h

coeff of x 3 4  coeff of x 4

4 4 1 h  1 h

Increasing the roots of (1) by -1, we get -1

1 0

-1

-6

-1

-2 2 -

-6

2 2

-1

3 3

1 0

1 1

-1

The new equation is x 4 +0.x 3 -x 2 +0.x -6 = 0 x 4 - x 2 -6 = 0------(2) (x 2 -3) (x 2 +2) = 0 x 2 =3 or x 2 = -2 x =  3, i 2  The roots of (2) are

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 3, i 2 The roots of the given equation are -1  3 , -1  i 2 . 6. Solve the equation 2x 3 -9x 2 +13x-6 = 0 by removing the second term. The given equation is 2x 3 -9x 2 +13x-6 = 0-------------(1). We will reduce the roots by h,

h

coeff of x 2 3  coeff of x 3

h

(9) 3 2

h

3 2

Decreasing the roots of 1 by 3/2, we get, 3 /2

1 -9

6 -

9 4

-6

9/2

1/2

The new equation is 2x 3 +0.x 2 -1/2x+0 =0. 4x 3 -x = 0. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

x(4x 2 -1) = 0 x(2x-1)(2x+1) =0. x = 0, 1/2., -1/2.  The roots of the given equation are

3/2+0, 3/2+1/2, 3/2-1/2 3/2, 2, 1.

Self assessment problems: V 1. Solve the equation x 3 -6x 2 +11x-6=0. by removing the second term. 2. Solve the equation 4x 4 +32x 3 +83x 2 +76x+21=0. by removing the second term. 3. Solve the equation x 4 -4x 3 -7x 2 +22x+24 = 0. by removing the second term. 4. Solve the equation x 4 +16x 3 +83x 2 +152x+84 = 0. by removing the second term.

Type 6:Transformations in general 1. If  ,  ,  ,  are the roots of the equation ax 4 +bx 3 +cx 2 +dx+e = 0, form an equation whose roots are 1/  , 1/  , 1/  , 1/  . Hence find the value of

1 /  .

The given equation is ax 4 +bx 3 +cx 2 +dx+e = 0------------------(1) Its roots are  ,  ,  ,  . We will form an equation whose roots are 1/  , 1/  , 1/  and 1/  . Put

y = 1/  ,

y = 1/x, x = 1/y Substituting in (1), we get a (1/y) 4 +b (1/y) 3 +c (1/y) 2 +d (1/y)+e = 0

 y 4 ; a +by +cy 2 +dy 3 +ey 4 = 0 

ey 4 +dy 3 +cy 2 +by +a = 0------------(2)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

This is the required equation. Its roots are 1/  , 1/  , 1/  , 1/  . Sum of the roots

1/   d / e. 2. If  ,  ,  are the roots of the equation x 3 -px 2 +qx –r = 0 form an equation whose roots are   1 /  ,   1 / and  1 /  Deduce the value of (  1 /  ).(  1 /  ).(  1 /  ) .

The given equation is x 3 -px 2 +qx –r = 0 Its roots are  ,  ,  .

        p         q   r We will form an equation with roots

  1 /  ,   1 / and  1 /  . y =   1 / 

Put,

y = (  1) /  y = (r+1)/  y = (r+1)/x 

x = (r+1)/y

Substituting in (1), we get 3

 r 1   r 1   r 1  3    p   q r  0  y ;  y   y   y  (r+1) 3 - py (r+1) 2 + qy 2 (r+1) - ry 3 = 0 ry 3 - q(r+1) y 2 + p(r+1) 2 y _ (r+1) 3 = 0----------------------(2) This is the required equation. Its roots are   1 /  ,   1 / and  1 /  .

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II Product of the roots, (  1 /  ).(  1 /  ).(  1 /  ) = + (r+1) 3 /r.

3. If  ,  ,  are the roots of the equation ax 3 +bx 2 +cx +d = 0, form an equation whose roots are  ,  ,  . The given equation is ax 3 +bx 2 +cx +d = 0----------------------1 Its roots are  ,  ,      +   b / a

      c / a   d / a. We will form an equation with roots,

 ,  ,  . Let, y =    /   (d / a) /   (d / a) / x  x = -d/ay

Substituting in (1), we get a (-d/ay) 3 + b (-d/ay) 2 + c (-d/ay) + d = 0 -ad 3 + bd. 2 ay – cd.a 2 y 2 + d.a 3 y 3 = 0 -d 2 + dby – acy 2 + a 2 y 3 = 0 a 2 y 3 – acy 2 + bdy – d 2 = 0 This is the required equation. 4. If  ,  ,  are the roots of the equation x 3 + px 2 + qx + r = 0, form an equation whose roots are     2 ,     2 ,     2 . The given equation is x 3 + px 2 + qx + r = 0-----------------1

Its roots are  ,  ,         p

      q   r. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

We will form an equation with roots,

    2 ,     2 ,     2 . Put

y =     2

y = (      )  3 y = - p -3  y = - p -3x 3x = - p – y  x = - (y+p)/3.

Substituting in (1), we get, [ - (p+y)/3] 3 + p [- (p+y)/3] 2 + q [- (p+y)/3] + r = 0 -(y+p) 3 + 3p (y+p) 2 – 9q (y+p) +27r = 0 (y+p) 3 – 3p (y+p) 2 + 9q (y+p) -27r = 0 y 3 + 3py 2 + 3p 2 y+p 3 -3py 2 -6p 2 y – 3p 3 +9qy+9pq -27r = 0 y 3 + (9q – 3p 2 )y – 2p 3 +9pq – 27r = 0. This is the required equation. 5.If  ,  ,  are the roots of the equation x 3 +qx + r = 0, form an equation whose roots are

 2 + 2 ,  2 + 2 ,  2 +  2 . The given equation is x 3 +qx + r = 0--------------------------1. Its roots are  ,  ,        0

      q   r. We will form an equation with roots,

 2 + 2 ,  2 + 2 ,  2 +  2 . Put

y =  2 +  2 = (    )  2 = (a) 2 

2 

y =  2 + 2r/  y = x 2 + 2r/x FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

xy = x 3 - 2r x 3 - xy +2r = 0 -----------(2) Eliminating x from 1 & 2, We get the required equation. 1 - 2  qx +xy –r = 0 x(q+y) = r

x

r yq 3

 r   r  (1)     q r  0  yq  yq

r 3 + qr (y+q) 2 + r (y+q) 3 = 0 y 3 + 3q 2 y +3qy 2 +q 3 +qy 2 +2q 2 y + q 3 +r 2 = 0 y 3 + 4qy 2 + 5q 2 y +2q 3 +r 2 = 0. This is the required equation. 6. If  ,  ,  are the roots of the equation x 3 +qx + r = 0, form an equation whose roots are (    ) 2 , (    ) 2 and (    ) 2 . The given equation is x 3 +qx + r = 0--------------------------1. Its roots are  ,  ,        0

      q   r. We will form an equation with roots, (    ) 2 , (    ) 2 and (    ) 2 . Put,

y =(    ) 2 = (    ) 2 - 4 

= (-  ) 2 - 4  /  =  2 +4r/  = x 2 + 4r/x xy = x 3 +4r x 3 -xy +4r = 0 --------------------(2) FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Eliminating x from 1 & 2, We get the required equation. 1 - 2  qx +xy –3r = 0 x(q+y) = 3r  x

3r yq 3

 3r   3r  (1)     q r  0  yq  yq

 ( y  q) 3 ;

27r 3 + 3qr(y+q) 2 + r (y+q) 3 = 0

 r;

27r 2 + 3q (y+q) 2 + (y+q) 3 = 0 27r 2 + 3qy 2 + 6q 2 y +3q 3 + y 3 + 3qy 2 + 3q 2 y + q 3 = 0. y 3 + 6qy 2 + 9q 2 y + 4q 3 + 27r 2 = 0.

This is the required equation.

7. If  ,  ,  are the roots of the equation x 3 +qx + r = 0, form an equation whose roots are  /    /  ,  /    /  and  /    /  . The given equation is x 3 +qx + r = 0 -------------------------- (1) Its roots are  ,  ,        0

      q   r. We will form an equation with roots,

 /    /  ,  /    /  and  /    /  . Put

y =  /  /  y = (  2 +  2 )/  y = [(    ) 2 -2  ]/ 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

y

() 2 2 

y

 2 . 2 

y

3 2 r

-ry = x 3 + 2r x 3 + ry +2r = 0 --------------------------------(2) Eliminating x from 1 & 2, We get the required equation. (1)-(2) 

qx – ry –r = 0

qx = ry+r

x

r(y  1) q

Substituting in 2, we get r 3 (y+1) 3 /q 3 + ry +2r = 0 r 3 (y+1) 3 + q 3 .ry + q 3 .2r = 0

 r;

r 2 (y+1) 3 +q 3 y + 2q 3 = 0

r 2 y 3 + 3r 2 y 2 + (3r 2 +q 3 )y + 2q 3 + r 2 = 0 This is the required equation. 8. If  ,  ,  are the roots of the equation x 3 -x -1 = 0 form an equation with roots

1  1  1  , , Deduce that 1  1  1 

1 

 1    7

The given equation is x 3 -x -1 = 0 -------------------------- (1) Its roots are  ,  ,        0

      1   1 We will form an equation with roots,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

1  1  1  , , 1  1  1  Put

y

1  1 

y

1 x 1 x

y = 1+  / 1   = 1+x/1-x y – xy = 1+x y – 1 = xy +x y – 1 = x (y+1)

x

y 1 y 1

Substituting in (1), we get. 3

 y 1   y 1      1  0  y 1   y 1 

 ( y  1) 3 ;

(y – 1) 3 - (y -1) (y+1) 2 -(y+1) 3 = 0

y 3 -3y 2 +3y -1 –(y -1) (y 2 +2y+1) – y 3 -3y 2 -3y -1 = 0 -6y 2 -2 –y 3 -2y 2 –y +y 2 +2y +1 = 0 -y 3 -7y 2 +y -1 = 0 y 3 -7y 2 -y +1 = 0 This is the required equation. Its roots are

1  1  1  , , 1  1  1 

Sum of the roots,

1 

 1    7 9. If  ,  ,  are the roots of the equation x 3 +qx + r = 0, form an equation whose roots are  (1 /   1 /  ),  (1 /   1 /  ) , and  (1 /   1 /  ) .

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II The given equation is x 3 +qx + r = 0--------------------------(1) Its roots are  ,  ,        0

      q   r. We will form an equation with roots,

 (1 /   1 /  ),  (1 /   1 /  ) , and  (1 /   1 /  ) . Put ,

y =  (1 /   1 /  ), y  [(  ) /()]

y

 (   ) 

 2 ( ) r 3 y r 3 x y r 3 x  ry y

Substituting in 1, we get ry + qx + r = 0 qx = -r(y+1)  q 3 x 3 = - r 3 ( y  1) 3  q 3 ry = - r 3 (y 3 + 3y 2 + 3y +1)  r  q 3 y + r 3 (y 3 + 3y 2 + 3y +1) = 0

r 2 y 3 + 3r 2 y 2 + (3r 2 + q 3 ) y + r 2 = 0. This is the required equation. 10. If  ,  ,  are the roots of the equation x 3 -3x -1 = 0. show that  2 -2 also a root. The given equation is x 3 -3x -1 = 0. -----------------------(1) FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Given, x =  is a root of this equation. We will form an equation for which  2 -2 is a root. Put,

y =  2 -2 y = x 2 -2 x 2 -y -2 = 0 x 2 = y+2. ------------(2)

Put,

x 2 = y+2. in (1)

x (y+2) – 3x +1 = 0 x (y -1) = -1 Squiring,

x 2 (y -1) 2 = 1

(y+2) (y 2 -2y +1) = 1 y 3 – 2y 2 +y +2y 2 -4y +2-1 = 0 y 3 -3y +1 = 0.------------------------(3). This is same as an equation (1)

 2 -2 also a root of equation (1).

11. If  ,  ,  are the roots of the equation x 3 +qx + r = 0, form an equation whose roots are (  2 -  ) /  , ( 2   ) /  and ( 2   ) /  . The given equation is x 3 +qx + r = 0 -------------------- (1) Its roots are  ,  ,        0

      q   r. We will form an equation with roots, (  2 -  ) /  , ( 2   ) /  and ( 2   ) /  .

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II Put,

y = (  2 -  ) /  y = ( 3   ) /  2

2y 3  r x 2 y  x3  r x 3  x 2 y  r  0. --------(2) Eliminating x between (1) and (2), we get the required equation. (1)-(2);

qx+ x 2 y  0 x(q +xy) =0 q+xy =0 x = -q/y

(1) 

(-q/y) 3 + q(-q/y) + r = 0

 y3 ;

-q 3 –q 2 y 2 +ry 3 = 0 ry 3 – q 2 y 2 – q 3 = 0.

This is the required equation. 12..If  ,  ,  are the roots of the equation x 3 +px + q = 0, form an equation whose roots are (1 /  2 )  (1 /  ) , (1/  2 )  (1 /  ) and (1/  2 )  (1 /  ) . The given equation is x 3 +px + q = 0 --------------------- (1) Its roots are  ,  ,        0

      q   r. We will form an equation with roots,

(1 /  2 )  (1 /  ) , (1/  2 )  (1 /  ) and (1/  2 )  (1 /  ) . Put,

y = (1 /  2 )  (1 /  ) y = ( 1 /  2 )  ( /  )

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

y

1   2 q

y = 1/x 2 + x/q x 2 qy = q + x 3

x3  qyx 2  q  0  (2) Eliminating x between (1) and (2), we get the required equation. (1)-(2);

px + x 2 qy = 0 x (p +qxy) = 0 p + qxy = 0 x = -p/qy.

(1) 

(-p/qy) 3 + p (-p/qy) +q = 0

 q 3 y 3 ; - p 3 – p 2 q 2 y 2 +q.q 3 y = 0 q4 y4 – p2 q2 y2 – p3 = 0 This is the required equation.

13. If  ,  ,  are the roots of the equation x 3 +px 2 + qx +r = 0, form an equation whose roots are    2 ,    2 and    2 . The given equation is x 3 +px 2 + qx +r = 0 ----------------- (1) Its roots are  ,  ,        0

      q   r. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

We will form an equation with roots,

   2 ,    2 and    2 . Put,

y =    2 Y=

  2 

y = (-r/  )   2 y = (-r/x) - x 2 xy = -r -x 3 .

 x 3  xy  r  0 ------------- (2). Eliminating x between (1) and (2), we get the required equation. (1)-(2);

p x 2 + qx –xy = 0 x [ px +q –y] = 0 px +q –y = 0 x = (y-q)/p

(2) 

[(y-q)/p] + [(y-q)/p] y + r = 0 (y – p) 3  p 2 y( y  p)  p 3 r  o

y 3  3qy 2  3q 2 y  q 3  p 2 y 2  p 2 qy  p 3 r 0

y 3  ( p 2  3q) y 2  (3q 2  p 2 q) y  p 3 r  q 3  0 . This is the required equation. 14. If  ,  ,  are the roots of the equation x 3 +2x 2 +3x +3 = 0, form an equation whose roots are

   , ,  1  1  1

The given equation is x 3 +2x 2 +3x +3 = 0 ----------------- (1) Its roots are  ,  ,         2

      3   3 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

We will form an equation with roots,

   , ,  1  1  1 Put,

y =  /(  1)

y = x/(x+1) xy +y = x xy – x = -y x(y-1) = -y  x  y /(1  y). 3

(1) 

 y   y   y     2   3 3 0  1 y   1 y  1 y 

 (1  y) 3 ;

y 3  2 y 2 (1  y)  3 y(1  y) 2  3(1  y) 3  0

y 3  2 y 2  2 y 3  3y  6 y 2  3y 3  3  9 y  9 y 2  3y 3  0  y3  5y 2  6y  3  0 y3  5y 2  6y  3  0

This is the required equation.

15..If  ,  ,  are the roots of the equation x 3 +px 2 +qx +r = 0, form an equation whose roots are  /(    ),  /(   ) and  /(   ) . The given equation is x 3 +px 2 +qx +r = 0 - ----------------- (1). Its roots are  ,  ,        0

      q   r. We will form an equation with roots, FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

   , ,       Put,     (     )  

y = x/-p –x -py –xy = x -py = x +xy x = -py/(y+1) (1) 

[-py/(y 1)]3  p[-py/(y 1)]  q[-py/(y 1)]  r  0

 ( y  1) 3 ;

 p 3 y 3  p 2 y 2 ( y  1)  pqy( y  1) 2  r ( y  1) 3  0

 p 3 y 3  p 3 y 3  p 2 y 2  pqy 3  2 pqy 2  pqy  ry 3  3ry 2  3ry  r  0 y 3 (r  pq)  y( p 2  2 pq  3r )  y(3r  pq)  r  0

This is the required equation. Self assessment problems 1.

If  ,  ,  are the roots of the equation x 3 +qx + r = 0, form an equation whose roots are  (1 /   1 /  ),  (1 /   1 /  ),  (1 /   1 /  ) .

If  ,  ,  are the roots of the equation x 3 +qx + r = 0, form an equation whose roots are  ( 2   2 ),  ( 2   2 ),  ( 2   2 )

If  ,  ,  are the roots of the equation x 3 -3x +1 = 0, form an equation whose roots are (  2) 2 , (  2) 2 , (  2) 2 .

If  ,  ,  are the roots of the equation x 3 +2x 2 +3x -4 = 0, form an equation whose roots are (  3) /(  2), (  3) /(   2), (  3) /(  2) .

If  ,  ,  are the roots of the equation x 3 -px 2 +qx -r = 0, form an equation whose roots are  /(      ),  /(     ),  /(     ) .

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Type 7: Symmetric functions of roots 1.If  ,  ,  are the roots of the equation x 3 +px 2 +qx +r = 0 find the value of

3  3  3.  ,  ,  are the roots of the equation

Given,

x 3 +px 2 +qx +r = 0 -------------- (1)        p

      q   r.

 3   3   3 = (     ) 3  3(   )(   ))   ) = (-p) 3 -3(-p -  )( p   )( p   ) = -p 3 +3 ( p   )( p   )( p   ) = -p 3 3[ p 3  p 2 (     )  p(     ) ] = - p 3  3[ p 3  p 2 ( p)  pq  r ]

 3   3   3 = -p 3 + 3 (pq –r). 2.If  ,  ,  are the roots of the equation x 3 +px 2 +qx +r = 0 find the value of (   )(   )(   ) .

Given,

 ,  ,  are the roots of the equation x 3 +px 2 +qx +r = 0 ------------- (1)        p

      q   r. (   )(   )(   ) = ( p   )( p   )( p   )

= - ( p   )( p   )( p   ) = -[ p 3  p 2 (     )  p(     )   ] = -[ p 3  p 2 ( p)  pq  r ] = r- pq.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

3.If  ,  ,  are the roots of the equation x 3 +px 2 +qx +r = 0 find the value of



2.  ,  ,  are the roots of the equation

Given,

x 3 +px 2 +qx +r = 0 ------------- (1)        p

      q   r.



 2 =  2 2    2   2  2 = (     ) 2  2[ .  .   . ] = (q) 2  2 [     ] = q 2 2(r )( p)



 2  q 2  2 pr.

4.If  ,  ,  are the roots of the equation x 3 +px 2 +qx +r = 0 find the value of

[(

  2 ) /  ] .

 ,  ,  are the roots of the equation

Given,

x 3 +px 2 +qx +r = 0 ------------- (1)        p

      q   r.

[(

  2 ) /  ] =

 ( /    /  )

= ( /    /  )  ( /    /  )  ( /    /  ) =    ) /   (   ) /   (   ) /  =

([     )   ] /   [(     )   ] /   [(     )   ] /  = (  p   ) /   (  p   ) /   ( p   ) /  =  p /  1  p /  1  p /  1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

(

MATHEMATICS-II

86 = -p ( 1 /   1 /   1 /  )  3 = -p[(      ) /  ]  3

 q   p    3  r 

5.If

,  ,

  ( 2   2 ) /  =

pq  3r r

are

the

roots

equation

-3ax+b

prove

that

 (   )(   )  9a.  ,  ,  are the roots of the equation

Given,

x 3 -3ax+b = 0 ------------------- (1)       0       (3a)   b.

 ( )(  )   (       ) = [  (     )  2 ] = [  3a  2 ] =   9a  2  2

= (  2   2   2 )  2(     )  9a = (      ) 2  9a = 0 + 9a. = 9a.

Self assessment problems. VII 1. If  ,  ,  are the roots of the equation x 3 +px 2 +qx +r = 0 find the value of

1/ 

 2.

2. If  ,  ,  are the roots of the equation x 3 +px 2 +qx +r = 0 find the value of

 /  .

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Answers. Self assessment problems. I 1. Given  ,  ,  are the roots of the equation

x 3  px 2  qx  r  0 ------------------- (1) We will form an equation with roots -  , , . Put,

y = -

y = -x x =-y (1)  (-y) 3 + p(-y) 2 +q(-y)+r = 0

 y 3  py 2  qy  r  0 y 3  py 2  qy  r  0 This is the required equation. Self assessment problem II: 1. Given,

 ,  ,  ,  are the roots of the equation x 4  x 3  4 x 2  5x  10 = 0 ------------------- (1) We will form an equation with roots

3 ,3 ,3 and 3 . Put,

y = 3

y = 3x x = y/3. (1) 

( y/3)4  (y/3)3  4(y/3)2 - 5(y/3)  10  0

y 4  3 y 3  36 y 2  135 y  810  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

This is the required equation.

Self assessment problem III: 1. Given,

 ,  ,  are the roots of the equation x 3  3 px 2  3qx  r  0 ------------------ (1) We will form an equation with roots

 2 ,  2 and  2 Put,

y = 2

y = x2 x2 = y (1)  x.y +3py +3qx +r = 0 x (y+3q) = -(3py+r) Squaring,

x 2 ( y  3q) 2  (3 py  r ) 2

y ( y  6qy  9q 2 )  9 p y 2  pry  r 2

y 3  y 2 (6q  9 p 2 )  y(9q 2  6 px)  r 2  0 This is the required equation.

Self assessment problems IV: 1. The given equation is, 2 x 4  25x 3  111x 2  208x  140  0 ------------- (1)

Reducing the roots by 3, we get 3

111

208

1 40

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

4 6

5 1 5

138 -

0 2

62 5

6 -1

The new equation is 2 x 4  x 3  6 x 2  x  2  0 2. The given equation is

4 x 4  32 x 3  83x 2  76 x  21  0 Reducing the roots by 2,

4 0

3 -

-70

3 5

12 6

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

90 8 4

6 8

The new equation is 4 x 4  0 x 3  13x 2  0 x  9  0

4x 4  13x 2  9  0            (2) ( 4 x 2  9)( x 2  1)  0 4 x 2  9  0orx 2  1  0

4x 2  9 or x  1 x=  3 / 2 , or x =  1  The roots of (2) are 3/2, -3/2, 1,-1.

The roots of the given equation are 3/2-2, -3/2-2, (1)-(2), -(1)-(2) -1/2, -7/2, -1, -3. Self assessment problems V 1. The given equation is

x3  6x 2  11x  6  0              (1) We will reduce the roots by h Here h = -Coeff of x 2 /3  Coeff of x 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

h = 6/3 h=2 Reducing the roots of (1) by( 2), we get 2

1 -6

-6 -

8 3

-4

4 -

-2

The new equation is

x3  x  0 x(x 2 -1) = 0 x (x+1) (x-1) = 0 x = 0,1,-1  The roots of the given equation are

2+0, 2+1, 2-1 2, 3, 1. 2. The given equation is 4 x 4  32 x 3  83x 2  76 x  21  0 ------------------ (1)

We will reduce the roots by h. h = -Coeff of x 3 /4  Coeff of x 4 h = -32/16

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

h = -2 Increasing the roots of (1) by (2), we get 2

3 2

-70

4 -

6 -

12 6

The new equation is 4 x 4  13x 2  9  0

( 4 x 4  9)( x 2  1)  0 X = -3/2, 3/2, -1, 1 The roots of the given equation are -2-3/2, -2+3/2, -2-1, -2+1 -7/2, -1/2, -3, -1. 3. The given equation is

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

x 4  4 x 3  7 x 2  22 x  24  0              1

We will diminish the roots by h. Here h = -Coeff of x 3 /4  Coeff of x 4 h = -(-4)/4 h=1 Reducing the roots by (1), we get 1

-4

-3 1

-2

1 -

2 4 1

-10

2 1

3 6

-12

-2

0 1

-3

-10

-7

-1 -13

The new equation is

x 4  13x 2  36  0 ( x 2  4)( x 2  9)  0 (x+2)((x-2)(x+3)(x-3) = 0 x = 2, -2, 3, -3. The roots of the given equation are (1)+(2), (1)-(2), 1+3, 1-3 3, -1, 4, -2. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

4. The given equation is x 4  16 x 3  83x 2  152 x  84  0            1

We will reduce the roots by h. Here h = -Coeff of x 3 /4  Coeff of x 4 h = -16/4 h = -4. Increasing the roots by 4 4

1 6+

-140 3

5 -

3 6

3 -

1 0

4 -13 4

The new equation is

x  13x  36  0

( x 2  4)( x 2  9)  0 (x+2)(x-2)(x+3)(x-3) = 0 x = 2, -2, 3, -3. The roots of the given equation are FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II -4+2, -4-2, -4+3, -4-3 -2, -6, -1, -7.

Self assessment problems VI 1. Given

 ,  ,  are the roots of the equation x 3  qx  r  0 --------------- (1) 

     0       q   r.

We will form an equation with roots

 (1 /   1 /  ),  (1 /   1 /  ),  (1 /   1 /  ) Put,

y =  (1 /   1 /  )

y = [(   ) /  ] y =  . ( ) /  y = 3 /(r) ry = x 3 x 3 = ry (1)  ry + qx +r =0 r (y+1) =-qx 

r 3 ( y  1) 3  q 3 x 3

r 3 ( y 3  3 y 2  3 y  1)  q 3 ry r 2 y 3  3r 2 y 2  y(3r 2  q 3 )  r 2  0 This is the required equation. 2. Given FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 ,  ,  are the roots of the equation x3  qx 2  r  0              (1)      0       q   r. We will form an equation with roots

 ( 2   2 ),  ( 2   2 ),  ( 2   2 ) Put, y=

y =  ( 2   ) 2

 [(   ) 2  2 ]

y = [( ) 2  2 ] y =  3  2 y = x 3  2r x 3 -y+2r = 0

------------------------(2)

Eliminating x between 1 and 2, we get the required equation. (1)-(2) 

qx +y +2r = 0 x = (r-y)/q

2

((r-y)/q) 3 -y+2r = 0

(r  y) 3  q 3 y  2rq 3  0 r 3  3r 2 y  3ry 2  y 3  q 3 y  2rq 3  0 y 3  3ry 2  y(q 3  3r 2 )  r 3  2q 3 r  0. This is the required equation. 3. The given equation is

x3  3x  1  0              (1) Its roots are  ,  ,  We will form an equation with roots

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

(  2) 2 , (  2) 2 , (  2) 2 Put,

y = (   2) 2

y = (x-2) 2

y = x-2 x=

y +2

(1) 

( y +2) 3 -3( y +2)+1=0 y y +6y +12 y +8 -3 y -6+1 = 0 y y +9 y = -6y -3

y (y+9) = -3 (2y+1) Squaring y(y+9) 2 = 9 (2y+1) 2 y(y 2 +18y+81) = 9 (4y 2 +4y+1) y 3 +18y 2 +81y-36y 2 -36y-9 = 0 y 3 -18y 2 +45y-9 = 0 This is the required equation . 4. Given

 ,  ,  are the roots of the equation x 3  2 x 2  3x  4  0 

----------------- (1)

      2       3   4

We will form an equation with roots

3 3  3 , , 2 2  2 Put,

y = (  3) /(  2)

y = (x+3)/(x-2) FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

xy -2y = x+3 x(y-1) = 2y+3 x = (2y+3)/(y-1) 3

 2y  3   2y  3   2y  3  ((   2  3   4 0  y 1   y 1   y 1 

(1) 

(2y+3) 3 + 2 (2y+3) 2 (y-1) + 3(2y+3)(y-1) 2 -4 (y-1) 3 = 0 8y 3 +36y 2 +54y+27+(2y-2)(4y 2 +12y+9)+(6y+9)(y 2 -2y+1)-4(y 3 -3y 2 +3y-1) = 0 8 y 3  36 y 2  54 y  27  8 y 3  24 y 2  18 y  8 y 2  24 y  18  6 y 3  12 y 2  6 y  9 y 2  18 y  9  4 y 3  12 y 2  12 y  4  0

18 y 3  61y 2  24 y  22  0 This is the required equation.

5. Given

 ,  ,  are the roots of the equation x 3  px 2  qx  r  0            1 

     p       q   r

We will form an equation with roots

   , ,           Put, Y=

   

 (    )  2

y =  / p  2 y = x/p-2x py-2xy = x py = 2xy+x FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

py = x(2y+1) x = py/2y+1 3

 py   py   py  (1)     p   q r 0  2y  1   2y  1   2y  1 

 (2 y  1) 3 ;

py  py(2 y  1)  qpy(2 y  1)  r (2 y  1)  0

py  2 py  py  4 pqy  4 pqy  pqy  8ry  12ry  6ry  t  0

y 3 (-p 3 +4pq-8r) +y 2 (-p 3 +4pq-12r) + y(pq-6r) – r = 0 This is the required equation. Self assessment problems VII 1. Given

 ,  ,  are the roots of the equation x 3  px 2  qx  r  0



1/ 

     p       q   r.

2=

1 1 1  2 2 2 2    2 2

= ( 2   2   2 ) /  2  2 2 =

(    ) 2  2(    ) ( )2

(p) 2  2q (r) 2

p 2  2q = r2 2. Given,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

100

 ,  ,  are the roots of the equation x 3  px 2  qx  r  0



     p       q   r.

 /  =

       

= ( 2   2   2 ) /  = ( [     ) 2  2(     )] /( ) 2 = [(-p) 2 -2q]/-r = (p 2 -2q)/-r = 2q-p 2 /r

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

101

UNIT – 3

EXPANSIONS

3.0

Introduction

3.1

Objective

3.2

Expansions of Sinn , Cosn and tan n

3.3

Expansions of sin n  and cos n 

3.4

Expansions of sin  , cos and tan

3.5

Hyperbolic functions

3.6

Inverse hyperbolic functions

3.7

Logarithm of a complex number

3.8

Answers to Self Assessment Problems

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

102

3.0 INTRODUCTION

In this unit you will learn about some interesting topics in Trigonometry. You will know how to expand sin n  , tan n  . The method is explained in detail with many examples. The expansion of sin  , cosn  in terms of function of multiples of  are also explained. Then the expansions of sin  , cos  , and tan  as infinite series in ascending powers of  are give with suitable examples. A new concept known as hyperbolic function is introduced. The relation between the circular and hyperbolic functions is derived and many interesting formulae are found out. Different types of problems have been worked out so that you may understand and appreciate the new concept. Inverse hyperbolic functions have also been introduced. You will also learn the logarithm of a complex number.

3.1 OBJECTIVE After going through this unit you should be able to 

Expand sin n  , tan n  in terms of powers of sin  , cos  , and tan

. 

Expand sinn  and cosn  in terms of functions of multiples of 



Expand sin  , cos  , tan  as infinite series in ascending powers of

 

Know hyperbolic functions and do problems



Know inverse hyperbolic functions and do problems



Find the values of the logarithms of complex numbers.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

103

3.1. Expansions of Sinn , Cosn and tan n By De’moivre’s theorem,

cos n  i sin n   cos  i sin  n

[n-a positive integer]

cos n  i sin n  cos n  nC1. cos n1  .(i sin  )  nC2 . cos n2  .i sin    ........  i sin   2

cos n  i sin n  cos n  inC1 . cos n1  .sin   nC2 . cos n2  .sin 2   inC 3 cos n3 .sin 3   nC4 cos n4  .sin 4   ........  i sin  

Equating real parts,

cos n  cosn   nC2 .cosn 2 .sin 2   nC4 .cosn 4 .sin 4 ................ 1 Equating imaginary parts

sin n  nC1cosn-1.sin   nC3.cosn 3 .sin 3   ................  2 nC1cosn-1.sin   nC3 .cos n 3 .sin 3   .................. (2) sin n   (1) cos n cosn   nC2 .cos n 2 .sin 2   nC4 .cos n 4 .sin 4 ............... In the R.H.S. dividing Nr and Dr by cos n 

tan n 

sin  sin 3   nC3 . 3  ........... cos  cos  sin 2  1  nC2 . 2  ............... cos 

nC1.

 tan n 

nC1.tan   nC3. tan 3   nC5 tan 5   .............. 1  nC2 tan 2   nC4 tan 4   .............

Problems: Type1 1. Prove that cos 4  8 sin 4   8 sin 2   1 We have

cos n

 cos n  nC2 . cos n2  . sin 2   nC4 . cos n4  . sin 4   ................

Put n=4

cos 4

 cos 4  4C2 . cos 2  .sin 2   4C4 . cos 0  .sin 4 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

104  1  sin 2   6(1  sin 2 ).sin 2   1.1.sin 4  2

cos 4

[cos 2   1  sin 2 ] cos 4   (cos 2 )2  (1  sin 2 ) 2 43 6 1 2 nCn  1] 4C2 

cos 4  1  sin 2   sin 4   6 sin 2   6 sin 2   sin 4  cos 4  8sin 4  8 sin 2   1

2. Prove that cos 5  16cos 5  20 cos 3   5 cos We have

cos n  cosn   nC2 .cosn 2 .sin 2   nC4 .cosn 4 .sin 4   .......... Put n = 5

cos 5  cos 5  nC2 . cos 3  .sin 2   5C4 . cos .sin 4  cos 5  cos 5  10 cos 3  (1  cos 2  )  5 cos (1  cos 2  ) 2 [ sin 2   1  cos 2  sin 4   (sin 2 ) 2  (1- cos 2 ) 2 5C2 5C4

5 4  10 1 2  5c1  5] 

cos 5  cos 5  10 cos 3   10 cos 5   5 cos   10 cos 3   5 cos 5  cos5  16cos 5  20 cos 3   5 cos  3. Prove that cos 6  32cos 6  48 cos 4   18 cos 2   1 We have

cos n  cosn   nC2 .cosn 2 .sin 2   nC4 .cosn 4 .sin 4 .......... Put n = 6

cos 6  cos 6   6e2 cos 4 .sin 2   6e4 cos 2 .sin 4   6e6 .cos 0 .sin 6 

cos 6  cos 6  15 cos 4  (1  cos 2  )  15 cos 2  (1  cos 2  ) 2  1.1.(1  cos 2  ) 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

105

[ sin 4   (sin 2 ) 2  (1  coh 2) 2 sin 6   (sin 2 )3  (1  cos 2 )3

6C4

65  15 1 2  6C2  15

6C6

 1]

6C2



cos 6  cos 6   15 cos 4   15 cos 6   15 cos 2   30 cos 4   15 cos 6   1  3 cos 2   3 cos 4   cos 6  cos 6  32 cos 6   48 cos 4   18 cos 2   1 4. Prove that

sin 6  32 cos 5   32 cos 3   6 cos  sin 

We have

sin n  nC1cosn-1.sin   nC3.cosn 3 .sin 3   .............. Put n = 6

sin 6  6C1 cosn 1 .sin   6C3 cos3 .sin3   6C5 cos .sin5  

sin 6  6 cos 5   20 cos 3  .sin 2   6 cos  .sin 4  sin 

[6C1  6;

6C3 

6 5 4  20; 1 2  3

6C5  6C1  6]

sin 6  6 cos 5   20 cos 3  (1  cos 2  )  6 cos  (1  cos 2  ) 2 sin   6cos 5  20 cos 3   20 cos 5   6 cos 5   32cos 5  32 cos 3   6 cos 

5. Prove that

sin 7  7  56 sin 2   112 sin 4   64 sin 6  sin 

We have

sin n  nC1cosn-1.sin   nC3.cosn 3 .sin 3   .............. Put n = 7

sin 7  7C1 cos 6  .sin   7C3 cos 4  .sin 3   7C5 cos 2  .sin 5   7C7 . cos 0  .sin 7  FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

106

sin 7  7 cos 6   35 cos 4  .sin 2   21cos 2  .sin 4   sin 6  sin 

[7C1  7;

7C 3 

7C 5  7C2 

7 65  35; 1 2  3

76  21; 1 2

7C 7  1]

sin 7  7(1  sin 2  ) 3  35(1  sin 2  ) 2 .sin 2   21(1  sin 2  ) sin 4   sin 6  sin 

 7 - 21sin 2  21sin 4   7 sin 6   35 sin 2   70 sin 4   35 sin 6   21sin 4   21sin 6   sin 6  

sin 7  7  56 sin 2   112 sin 4   64 sin 6  sin 

6. Prove that

sin 8  128 cos 7   192 cos 5   80 cos 3   8 cos  sin 

We have

sin n  nC1cosn-1.sin   nC3.cosn 3 .sin 3   .............. Put n = 8

sin 8  8c1 cos 7  .sin   8C3 cos 5  .sin 3   8C5 cos 3  sin 5   8C7 cos .sin 7  

sin 8  8 cos 7   56 cos 5  .sin 2   56 cos 3  .sin 4   8 cos  .sin 6  sin  [8C1  8; 8C5  8C3  56;

8 7  6  56; 1 2  3 8C 7  8C1  8]

8c 3 

 8cos 7  56 cos 5  .(1  cos 2  )  56 cos 3  (1  cos 2  ) 2  8 cos (1  cos 2  ) 3  8 cos 7   56 cos 5  (1  cos 2  )  56 cos 3  (1  2 cos 2   cos 4  )  8 cos  (1  3 cos 4   cos 6 

 8cos 7  56 cos 5   56 cos 7   56 cos 3   112 cos 5   56 cos 7   8 cos   24 cos 2   24 cos 5   8 cos 7 



sin 8  128 cos 7   192 cos 5   80 cos 3   8 cos  sin 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

107

7. Prove that cos8  1  32sin 2   160sin 4   256sin 6   128sin 8  We have

cos n  cosn   nC2 .cosn 2 .sin 2   nC4 .cosn 4 .sin 4 ......... Put n = 8 cos 8  cos 8   8C2 cos 6  .sin 2   8C4 cos 4  .sin 4   8C6 . cos 2  .sin 2   8C8 cos 0  .sin 4 





 (1 - sin 2 ) 4  28 sin 2  1  sin 2   70 sin 4  (1  sin 2  ) 2 3

 28 sin 6  (1  sin 2  )  1.1.sin 8  8 7  28; 1 2 8C 6  8C2  28;

[8C 2 

8 7  5 6  70; 1 2  3  4 8C8  1]

8C 4 

 1 - 4sin 2  6 sin 4   4 sin 6   sin 8   28 sin 2  (1  3 sin 2   3 sin 4   sin 6  )  70 sin 4  (1  2 sin 2   sin 4  )  28 sin 6   28 sin 8   sin 8 

 1 - 4sin 2  6 sin 4   4 sin 6   sin 8   28 sin 2   84 sin 2   84 sin 4   28 sin 8   28 sin 6   28 sin 8   sin 8   70 sin 4   140 sin 6   70 sin 8  cos 8  1  32 sin 2   160 sin 4   256 sin 6   128sin 8 

8. Expand cos 7 in powers of cos  We have

cos n  cosn   nC2 .cosn 2 .sin 2   nC4 .cosn 4 .sin 4 ......... Put n = 7

cos 7  cos 7   7C2 . cos 5  .sin 2   7c4 . cos 3  .sin 4   7C5 . cos .sin 6 

 cos 7  21cos 5  (1  cos 2  )  35 cos 3  (1  cos 2  ) 2  7 cos (1  cos 2  ) 3  cos 7  21cos 5   21cos 7   35 cos 3  (1  2 cos 2   cos 4  )  7 cos  (1  3 cos 2   3 cos 4   cos 6  )

 cos 7  21cos 5   21cos 7   35 cos 3   70 cos 5   35 cos 7   7 cos   21cos 3   21cos 5   7 cos 7  cos 7  64 cos 7   112 cos 5   56 cos 3   7 cos

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

108

9. Prove that cos 9  256 cos 9   576 cos 7   432 cos 5   120 cos 3   9 cos We have

cos n  cosn   nC2 .cosn 2 .sin 2   nC4 .cosn 4 .sin 4 ........ Put n = 9 cos 9  cos 9   9C2 cos 7  .sin 2   9C4 cos 5  .sin 4   9C6 cos 3  .sin 6   9C8 cos  .sin 8 

 cos 9  36 cos 7  (1  cos 2  0  126 cos 5  (1  cos 2  )

 84 cos 3  (1  cos 2  ) 3  9 cos  (1  cos 2  ) 4 cos 9  cos 9   36 cos 7   36 cos 9   126 cos 3  (1  2 cos 2   cos 4  )  84 cos 3  (1  3 cos 2   3 cos 4   cos 6  )  9 cos  (1  4 cos 2   6 cos 4   4 cos 6   cos 8 

 cos 9  36 cos 7   36 cos 9   126 cos 5   252 cos 7   84 cos 9   9 cos   36 cos 3   54 cos 5   9 cos 9   36 cos 7  cos 9  256 cos 9   576 cos 7   432 cos 5   120 cos 3   9 cos

10. Wirte down the expansion of tan 10

tan n 

nC1 tan  nC3 tan 3   ...................... 1  nC2 tan 2   nC4 tan 4   .....................

Put n=10

tan 10 

10C1 tan  10C3 tan 3   10C5 tan 5   10C7 tan 7   10C9 tan 9  1  10C2 tan 2   10C4 tan 4   10C6 tan 6   10C8 tan 8   10C109 tan10 



10tan  120 tan 3   252 tan 5   120 tan 7   10 tan 9  1  45 tan 2   210 tan 4   210 tan 6   45 tan 6   tan10 

Self assessment problems I 1. Prove that

sin 5  16 sin 4   20 sin 2   5 sin 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

109

2. Prove that cos 6  1  18 sin 2   48 sin 4   32 sin 6 

3.3. Type 2. Expansions of sin n  and cos n  Let x  cos   isin  Then

1  (cos   i sin  ) 1  cos   i sin  x

1  2 cos  x 1 x 2i sin  x 1 x k  k  2 cos k x 1 x k  k  2i sin k x x

1. Prove that 16 cos 5   cos 5  5 cos 3  10 cos  Let x  cos  i sin  Then

1  cos   i sin  x

x

1  2 cos  x

1  (2 cos  ) 5  ( x  ) 5 x 1 1 1 1 1  x 5  5C1 .x 4 .  5C2 .x 3 . 2  5C3 .x 2 . 3  5C4 .x. 4  5 x x x x x  x 5  5x 3  10 x   (x 5 

10 5 1  3 5 x x x

1 1 1 )  5(x 3  3 )  10(x  5 x x x

32cos5   2.cos5  5.2cos3  10.2cos  16cos5   cos5  5cos3  10cos 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

110

2. Prove that 32 cos 6   cos 6  6 cos 4  15 cos 2  10

x  cos  isin 

Let Then, x

1  cos   i sin  x 1  2cos x

 2 cos 

1   x   x 

1 1 1 1 1 1  x 6  6C1.x 5 .  6C 2 .x 4 . 2  6C3 .x 3 . 3  6C 4 .x 2 . 4  6C5..x. 5  6 x x x x x x  x 6  6 x 4  15 x 2  20 

15 6 1  4 6 2 x x x

1  1  1      x 6  6   6 x 4  4   15 x 2  2   20 x   x  x    64 cos 6   2 cos 6  6.2 cos 4  15.2 cos 2  20 32 cos 6   cos 6  6 cos 4  15 cos 2  10

3. Prove that 64sin 7   35sin   21sin 3  7sin 5  sin 7 Let x  cos  i sin then

1  cos   i sin  x

x

1  2 cos  x

x

1  2i sin  x

(2i sin  ) 7



( x  1x ) 7

x 2  7c1x 6 . 1x  7c2 x 5 =

1 x2

 7c3 x 4 . x13  7c4 x 3

1 x4

 7c5 x 2 . x16  7c7 . x17

x 7  7 x 5  21x 3  35x 

35 21 7 1  3 5 3 x x x x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

-i.128sin7 

111

1  7 1   5 1   3 1    x  7   7 x  5   21 x  3   35 x   x x   x  x    

2isin 7  7.2isin 5  21.2isin 3  35.2isin 

35sin   21sin 3  7sin 5  sin 7

 - 2i;

64sin7 

4. Prove that 128 Sin8  = Cos8  - 8cos6  + 28 cos4  - 56cos2  + 35 Let

cos θ +isin θ

1  cos θ - i sin θ x x

1  2 cos θ x

x

1  2i sin θ x

xn 

1  2 cos nθ xn

xn 

1  2i sin nθ xn

1  (2i sin θ )8 =  x   x 

1 1 1 1  x 8  8c1 x 7 .  8c2 x 6 . 2  8c3 x 5 . 3  8c4 x 4 . 5 x x x x 1 1 1 1  8c5 x 3 . 5  8c6 x 2 . 6  8c7 x. 7  8 x x x x 28 sin 8   x 8  8x 6  28x 4  56 x 2  70 

56 28 8 1  4 6 8 2 x x x x

1  1 1    1   x 2  8   8 x 6  6   28 x 4  4   56 x 2 2   70 x   x  x     x  256 sin 8   2 cos 8  8.2 cos 6  28.2 cos 4   56.2 cos 2  70

 2; 128sin8   cos8  8cos 6  28cos 4  56cos 2  35

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

112

5. Prove that 28. cos 9   cos 9  9 cos 7  36 cos 5  84 cos 3  126 cos  Let

x  cos   i sin  1  cos   i sin  x x

1  2 cos  x

x

1  2i sin  x

x n  cos n  i sin n 1  cos n  i sin n xn xn 

1  2 cos n xn

xn 

1  2i sin n xn

2 cos  

1   x   x 

1 1 1 1 1 29 cos 9   x 9  9c1.x 8 .  9c2 .x 7 . 2  9c3 .x 6 . 3  9c4 .x 5 . 4  9c5 .x 4 . 5 x x x x x 1 1 1 1 9c6 .x 3 . 6  9c7 .x 2 . 7  9c8 .x. 8  9 x x x x 29 cos 9   x 9  9 x 7  36 x 5  84 x 3  126 x 

126 84 36 9 1  3 5 7 9 x x x x x

1   1  1 1 1     29 cos 9    x 9  9   9 x 7  7   36 x 5  5   84 x 3  3   126 x   x x   x  x  x      28.cos9   2.cos9  9.2cos 7  36.2cos5  84.2cos3  126.2cos 

 2;

28. cos 9   cos 9  9 cos 7  36 cos 5  84 cos 3  126 cos  6. Prove that 32 sin 4  . cos 2   cos 6  cos 4  cos 2  2 Let x  cos  i sin 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

113

1  cos   i sin  x x

1  2 cos  x

x

1  2i sin  x

xn 

1  2 cos n xn

xn 

1  2i sin n xn 4

1 1 4  2i sin  .  2cos    x    x   x  x  2

1  1  1  2 6.sin 4  . cos 2    x    x    x   x  x  x 

1  1     x    x2  2  x  x  

1  1     x 2  2  2  x 4  2  4  x  x  

 x6  2 x 2 

1 2 2 1  2x4  4  4  x2  2  6 x x x x

26.sin 4  . cos 2   2. cos 6  2.2 cos 4  2 cos 2  4

64.sin .cos2   2.cos6  2.2 cos 4  2 cos 2  4 2

32 sin 4  . cos 2   cos 6  2 cos 4  cos 2  2 7. Prove that cos 4  .sin 3   

1 sin 7  sin 5  3sin 3  3sin   26

Let x  cos  i sin 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

114

1  cos   i sin  x x

1  2 cos  x

x

1  2 cos  x

x

1  2i sin  x

xn 

1  2 cos n xn

xn 

1  2i sin n xn 4

1  1    x4    x   x  x 

2 cos  .2i sin   4

1  1  1    x   x    x   x  x  x  1  1     x   x 2  2  x  x  

1  1 1 1     x   x 6  3x 4  3x 2 4  6  x  x x x   1  3 1     x   x 6  3x 2  2  6  x  x x    x 7  3x 3 

3 1 3 1  5  x 5  3x  3  7 x x x x

1   1  1  1  2 4 cos 4  .23.i 3 .sin 3    x 7  7    x 5  5   3 x 3  3   3 x   x x   x   x     i 27. cos 4  ..sin 3   2i sin 7  2i sin 5  3.2 sin 3  3.2i sin   2i;  26 cos 4  .sin 3   sin 7  sin 5  3sin 3  3sin   cos 4  .sin 3   

1 sin 7  sin 5  3sin 3  3sin   26

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

115

8. Show that 64 sin 5  .cos 2   sin 7  3sin 5  sin 3  5sin . Let x  cos  i sin 1  cos   i sin  x x

1  2 cos  x

x

1  2i sin  x

xn 

1  2 cos n xn

xn 

1  2i sin n xn

2i sin  5 .2 cos  2   x  1  

1   x   x 

1  . x   x  x

1  1  . x    x   x  x 

1  1     x   . x 2  2  x  x  

1 1 1  1 1     x 3  3x 2 .  3x. 2  3  x 4  2 x 2 . 2  4  x x x  x x   3 1  4 1   3  x  3x   3  x  2  4  x x  x   i 5 2 5 sin 5  .2 2 cos 2   x 7  2 x 3 

1 3 6 3 2 1  3x 5  6 x  3  3x 3   5  x  3  7 x x x x x x

i 2 7 sin 5  . cos 2   2i sin 7  3.2i sin 5  5.2i sin 

 2i; 26 sin 5  . cos 2   sin 7  3 sin 5  sin 3  5 sin 

9. Prove that cos 5  . sin 3  

1 6 sin 2   2 sin 4  2 sin 6  sin 8 27

Let x  cos   i sin 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

116

1  cos   i sin  x x

1  2 cos  x

x

1  2i sin  x

xn 

1  2 cos n xn

xn 

1  2i sin n xn

2 cos  5 2i sin  3   x  1  

1  . x   x  x

1  x   x 

1  1  . x    x   x  x 

1  x   x 

1   . x 2  2  x  

1 1  1 1 1     x 3  2x.  2  x 6  3x 4 . 2  3x 2 . 4  6  x x  x x x   1  3 1     x 2  2  2  x 6  3x 2  2  6  x  x x    x 8  3x 4  3 

1 6 2 3 1  2x6  6x 2  2  6  x 4  3  4  8 4 x x x x x

1  1   1   1   25 cos 5  .23.i 3 .sin 3    x 8  8   2 x 6  6   2 x 4  4   6 x 2  2  x   x   x   x    i28 cos 5  .sin 3   2i sin 8  2.2i sin 6  2.2i sin 4  .2i sin 2  2i; 27 cos 5  .sin 3   6si2  2 sin 4  2 sin 6  sin 8 cos 5  .sin 3  

1 6 sin 2  2 sin 4  2 sin 6  sin 8  27

10. Show that cos 6   sin 6  

1 3 cos 4  5 8

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

117

Let x  cos  i sin  1  cos   i sin  x x

1  2 cos  x

x

1  2i sin  x

xn 

1  2 cos n xn

xn 

1  2i sin n xn

 2 cos 

1   x   x 

1 1 1 1 1 1  x 6  6c1 .x 5 .  6c2 .x 4 . 2  6c3 .x 3 . 3  6c4 .x 2 . 4  6c5 .x. 5  6 x x x x x x

= x 6  6 x 4  15 x 2  20 

15 6 1  4 6 2 x x x

1   1  1      x 6  6   6 x 4  4   15 x 2  2   20 x   x  x   

26.cos6   2cos 6  6 /.2cos 4  15.2cos 2  20  (1) Next,

2i sin  6   x  1  

x

1 1 1 1 1 1  x 6  6c1 .x 5 .  6c2 .x 4 . 2  6c3 .x 3 . 3  6c4 .x 2 . 4  6c5 .x. 5  6 x x x x x x

1   1  1    2 6 i 6 sin 6    x 6  6   6 x 4  4   15 x 2  2  - 20 x   x  x   

26 sin 6   2cos 6  6.2cos 4  15.2cos 2  20  (2) (1)  (2) 





26 cos 6   sin 6   24 cos  40

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

118

 8;





8 cos 6   sin 6   3 cos 4  5  cos 6   sin 6  

1 3sin 4  5 8

Self assessment problems II 1. Prove that 16 sin 5   sin 5  5 sin 3  10 sin  2. Prove that 64 cos 7   cos 7  7 cos 5  21cos 3  35 cos 3. Prove that 64 sin 4  . cos 3   cos 7  cos 5  3 cos 3  3 cos





4. Prove that 64 cos 8   sin 8   cos 8  26 cos 4  35 3.4 Type 3. Expansions of sin  , cos and tan If n is an positive integer, we have

cos n  cos n   nc2 . cos n2  .sin 2   nc4 . cos n4  .sin 4  ............. put

n  

 cos   cos n 

or  

 n



 n

nn  1   . cos n2 .sin 2 2 n n

nn  1n  2n  3   . cos n1 .sin 4 ......... 4 n n

n n  1  sin 2  n  cos  n n . cos n 2 . 1 n 2 n n2 n  n  1  n  2  n  3       sin 4  n n  n  n  n ..........   . cos n 4 . 1 4 n n2 n 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

119

 1   1.1   n  2 sin n       n . 2 cos   cos    .cos  .  n 4 n       n  4  1  2  3    11  1  1   n  4 sin n  n  n      n  . 4................   .cos  .  4 n      n  n

Taking the limit as n  

  sin  n  0, cos  1, 1 n n n

 cos   1 

1.1  0 11  01  01  0 .1.1. 2  .1.1. 4 ........... 2 4

 cos   1 

2 4  ................  (1) 2 4

Next,

  n 1 n 3  sin  n    nn  1n  2     n   ............. sin n  .cos  . sin  . cos  . 1 n n 3 n  1      n  3

n    n  n  1  n  2      n1  sin n 3  sin     n   n   n  n  n  . cos  . n   ........    . cos  . 1  n  1  3 n  1        n   n  3  1  2      1 1  1     n1 sin n3  sin   1   n .   n  n  . cos   . n  . 3 ...........  . cos  . 1 n    3 n          n   n 

Taking the limit as n  

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 n

 0, cos

120



sin

 1,



 n 1

sin  1.1.  sin    -

1.1.1 .1.1. 3 .............. 3

  ....................  (2) 3

Next, tan  

sin  cos 

3





2

1

 

5 5

4 4

 ..................  ................

    3 5  2  4       ................ 1     ............. 3 5 4     2 

1

2 3 5 2 4 2 4                       ................1     .......     ........  .............. 3 5 4 4       2   2  

   2  4  4  3 5      ................1     ........... 3 5 2 24 4       2 5 4  3 5      ................ 1   .  ........... 6 120 2 24   

 

3 6



5 5 3 5 5      ........... 24 6 12 120

1 1  1 1  5     3     5   ............. 2 6  24 24 120 

tan    

3 3



2 5   ................  3 15

Note: For small values of  measured in radians FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

cos   1 

2 2

3

sin    

tan    

3 3

121

 



4 4

5 5

 ...........  .......

2 5   ......... 15 Problem

1. Evaluate Lt

x0

sin2x - 2sinx x3

sin 2 x  2 sin x x3 3 5     2x   2x   x3 x5 2x  ......  2 x    ......    3 5 3 5      Lt  3 x 0 x

 Lt

2x 

x 0

8x 3 32x 5 2x 3 2x 5   2x    ........ 6 120 6 120 x3

 3 x5    x   .....  4   Lt  x 0 x3   x3 x 3   1   ..... 4   Lt  3 x 0 x

 1 0 sin 2 x  2 sin x  1 x0 x3

 Lt

2. Evaluate Lt

x0

sinx - tanx x3

sinx - tanx x0 x3 Lt

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

122

 x 3 x5    x3 2 5 x   .........  x   x  ........    3 5 3 15      Lt  3 x0 x

 Lt

x

x0

x3 x5 x3 2   x   x 5  ..... 6 120 3 15 3 x

3  1  16  5  x3    x  ..... 6 120    Lt 3 x0 x  1 1  x 3   x 5  ......... 2 8   Lt  3 x0 x 

1 2

tan 2 x  2 tan x x0 x3

3. Evaluate Lt

tan 2 x  2 tan x x0 x3 Lt

 2 x 3  2 2 x 5  .....  2 x  x 3  2 x 5  ...... 2 x      3 15 3 15      Lt 3 x0 x

8 64 2 4 2 x  x 3  x 5  2 x  x 3  x 5  ......... 3 15 3 15  Lt 3 x0 x

 8  2   64  4  5 x3    x  ........ 3   15    Lt x0 x3





x 3 2  4 x 2  ........ x0 x3

 Lt 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

123

tan x  sin x x0 sin 3 x

4. Evaluate Lt Lt

x0

tan x  sin x sin 3 x

 Lt

    x3 2 5 x3 x5 x   x  .....  x   ........    3 15 3 5       x3 x5 x    .........  3 5  

x 0

 Lt

x

x 0

x3 2 5 x3 x5  x x   ...... 3 15 6 120 3  x2 x4 3 x 1    ...... 6 120  

1  1 1 2 x3     x5     ...... 3 6 5 120     Lt 3 x 0  x2 3 x 1   ..... 6  

1 1  x 3   x 2 .  ...... 2 8   Lt  2 x 0  x  x 3 1   ........ 6  

1 0 2  1  03 

1 2

5. Determine a, b, so that Lt

x 0

a   sin   b cos  1  4 2 Lt

x0

a   sin   b cos 





1 12

   2 4  3 5 a       ......   b1    ........  3 5 2 4     1 Lt 4 x 0 12 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

124

a  b   2  1  b    4  1  

2



x 0

6

b  ...... 24 



1 12

cons tan term  0

and

coeff of 2  0

a  b  0

and

1 

ab

and

b 0 2 b2

 a  2 and b  2

6. If

sin x 863 find approximate value of x  x 864

Given,



2



5045 5406

5405 5406

 1

2





3

 1

sin 

nearly

1 nearly 5046



1 nearly 5046

2 

6 nearly 5046

2 

1 nearly 841



1 radiant nearly 29



1 57 0 28 nearly 29

  1058 nearly 7. If

tan





2524 find the approximate value of  2523

Given,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II tan







2524 2523

3

3  2524  2523

1

3

2

 1

2524 nearly 2523



1 nearly 2523

2 

3 nearly 2523

2 

1 nearly 841



1 radiant nearly 29

8. Evaluate tan 3x  3 tan x x0 x3 Lt

tan 3x  3 tan x x0 x3 Lt

 3x 3  2 3x 5  ......  3 x  x 3  2 x 5  ..... 3 x      3 15 3 15     Lt 3 x0 x

27 x 3 2 5 5 x3 6 5 3x   .3 .x  3x  3.  x  ...... 3 15 3 15  Lt 3 x0 x

8 x 3  31x 5  ........ x0 x3

 Lt

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

125

MATHEMATICS-II

126





x 3 8  32 x 2  ...... x0 x3

 Lt

80 8 9. Evaluate Lt sinx tan x x / 2

Let A  Lt sinx tan x x / 2

log A  Lt tan x. log sin x x / 2

log sin x x / 2 cot x

 Lt

1 .cos x sin x  Lt x  / 2  cos x

 Lt

x  / 2

cos x  sin 2 x sin x

0  A  e0 , A  1  Lt sin x 

tan x

x / 2

1

Self assessment problems III sin 2 x  2 sin x x0 sin 3 x

1. Evaluate Lt

cos ecx  cot x x0 x

2. Evaluate Lt 3. Evaluate Lt

x0

4. If 5. If

sin 

 tan



3sin x  sin 3x x  sin x



5765 find the approximate value of  5766



433 find the approximate value of  432

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

127

3.5 Type 4 Hyperbolic functions

Definition: For all real or complex values of x,

e x  ex cosh x  2

sinh x 

e x  ex 2

tanh x 

sinh x e x  e  x e 2 x  1   cosh x e x  e  x e 2 x  1

Relation between circular and hyperbolic functions We know that ei  cos  i sin  e i  cos  i sin 

ei  ei   cos   (1) 2 ei  ei  sin   (2) 2i

put   ix e i x  e i x 1  cosix  2 2

cos ix 

ex  e x 2

i



 1

cos ix  cosh x

e i x  e i x 2  sin ix  2i 2

e i

x

 ex 2i 2



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



 i e x  ex 2

e i

128



 e x  2

sin ix  i sinh x Next, tan ix  

sin ix cos ix

i sinh x cosh x

 tan ix  i tanh x Some Important Formula 1. cosh 2 x  sinh 2 x  1 Proof: we know that cos 2   sin 2   1

Put   ix

 cos ix   sin ix   1 2

cosh x2  i sinh x2  1 cosh 2 x  i 2 sinh 2 x  1

II. cosh 2 x  sin 2 x  cosh 2x

Proof: We know that cos 2   sin 2   cos 2

put   ix

 cos ix   sin ix   cos i 2 x 2

cosh x2  i sinh x2  cosh 2x FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

129

cosh 2 x  i 2 sinh 2 x  cosh 2x

cosh 2 x  sinh 2 x  cosh 2x

III . sinh2 x  2 sinh x.cosh 2x. Proof: We know that

sin 2  2 sin  .cos Put   ix sin i 2x  2 sin ix. cos ix i sinh 2x  2i sinh x.cosh x

sinh 2x  2 sinh x.cosh x Problems 1. Find the real and imaginary parts of sin(x  iy)

 sin x.cos iy  cos x.sin iy  sin x.cosh y  cos x.(i sinh y)  sin x.cosh y  i cos x.sinh y

 Re alpart

 sin x.cosh y

Imaginarypart  cos x.sinh y

2. Find the real and imaginary parts of cos( x  iy ) cos( x  iy )  cos x. cos iy  sin x.sin iy  cos x. cosh y  sin x(i sinh y )  cos x. cosh y  i sin x.sinh y R.P  cos x. cosh y I .P   sin x.sinh y

3. Separate into real and imaginary parts tan(x  iy )

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

130

sin(x  iy) cos(x  iy) sin(x  iy).cos(x  iy)  cos(x  iy).cos(x  iy) 1 [sin(x  iy  x  iy)  sin(x  iy  x  iy)]  2 1 [cos(x  iy  x  iy)  cos(x  iy  x  iy)] 2 sin 2x  sin i2y  cos 2x  cos i2y sin 2x  i sinh 2y tan(x  iy)  cos 2x  cosh 2y sin 2x sinh 2y  i cos 2x  cosh 2y cos 2x  cosh 2y sin 2x  R.P  cos 2x  cosh 2x sinh 2y I.P  cos 2x  cosh 2y tan(x  iy 

4. Separate into real and imaginary parts cot(x  iy )

cot(x  iy)

 R.P I.P

cos(x  iy) sin(x  iy) 2 cos(x  iy).sin(x  iy)  2sin(x  iy).sin(x  iy) sin(x  iy  x  iy)  sin(x  iy  x  iy)  [cos(x  iy  x  iy)  cos(x  iy  x  iy)] sin 2x  sin i2y   cos 2x  cosh 2y sin 2x  i sinh 2y  cosh 2y  cos 2x sin 2x  cosh 2y  cos 2x  sinh 2y  cosh 2y  cos 2x 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

5. If tan(x  iy )  u  iv , prove that

131

u sin 2 x  v sinh 2 y

Given, tan(x  iy)  u  iv sin(x  iy) cos(x  iy) 2sin(x  iy).cos(x  iy)  2 cos(x  iy).cos(x  iy) sin(x  iy  x  iy)  sin(x  iy  x  iy)  cos(x  iy  x  iy)  cos(x  iy  x  iy) sin 2x  sin i2y  cos2x  cos i2y sin 2x  i sinh 2y u  iv  cos 2x  cosh 2y

 u  iv 

Equating real parts,

u

sin 2 x cos 2 x  cosh 2 y

(1)

Equating imaginary parts,

v

sinh 2 y cos 2 x  cosh 2 y

(2)

(2) (1)

 cos 2x  cosh 2y  u sin 2x   v  cos 2x  cosh 2y  sinh 2y



u sin 2x  v sinh 2y

6. If tan(x  iy)  sin(  i), prove that

tan  sin 2x  tanh  sinh 2y

Given,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

132

tan(x  iy)  sin(  i) sin  x  iy 

cos  x  iy 

 sin .cos i  cos .i sinh 

sin(x  iy  x  iy)  sin(x  iy  x  iy)  sin .cosh   i cos .sinh  cos(x  iy  x  iy)  cos(x  iy  x  iy) sin 2x  i sinh 2y  sin .cosh   cos .sinh  cos 2x  cosh 2y

Equating real parts, sin 2x  sin .cosh  cos 2x  cosh 2

(1)

Equating imaginary parts, sinh 2y  cos .sinh  cos 2x  cosh 2 (1) (2)

(2)

sin 2x sin .cosh    tan .coth  sinh 2y cos .sinh  sin 2x tan    sinh 2y tanh 

7. If cos(  i )  k (cos   i sin  ) , prove that cos 2  cosh 2  2k 2 Given, cos(  i )  k (cos   i sin  )

cos  . cosh   sin  . sin i  k (cos   i sin  ) cos  . cosh   i sin  . sinh   k cos   ik sin  Equating real parts cos  . cosh   k cos 

(1)

Equating imaginary parts  sin .sinh   k sin 

(2)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

133

(1)2  (2)2 cos2 .cosh 2   sin 2 .sinh 2   k 2  cos2   sin 2  

1  cos 2  1  cos 2  . cosh 2    . sinh 2   k 2    2 2   

cosh 2   cos 2 . cosh 2   sinh 2   cos 2 . sin 2   2k 2

(cosh 2   sinh 2  )  cos 2 (cosh 2   sinh 2  )  2k 2 That is,

cosh 2  cos 2  2k 2 Another method, Given, cos(  i )  k (cos   i sin  )

(1)

cos(  i )  k (cos   i sin  )

(2)

(1)  (2)

cos(  i ). cos(  i )  k 2 (cos i sin  )(cos  i sin  ) 1 [cos(  i ) cos(  i )  cos(  i    i )]  k 2 (cos 2   sin 2  ) 2 cos 2  cos i 2  2k 2 cos 2  cosh 2  2k 2

8. If cos(  i )  tan  i sec , prove that cos 2 . cos 2  3 Given,

cos .cosi  sin .sin i  tan   i sec  cos .cosh   i sin .sinh   tan   i sec 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

134

Equating real parts cos  . cosh   tan 

(1)

Equating imaginary parts  sin  . sinh   sec

(2)

(2)2  (1)2 ; sin 2  . sinh 2   cos 2  . cosh 2   sec2   tan2   1  cos 2   1  cos 2  2 2   .sinh     .cosh   1 2 2     2 2 2 sinh   cos 2 . sinh   cosh   cos 2 . cosh 2   2  (cosh 2   sinh 2  )  cos 2 (sinh 2   cosh 2  )  2  1  cos 2 . cosh 2  2  1  cos 2 . cosh 2  3  cos 2 . cosh 2  3 x y  tanh , prove that cos x. cosh y  1 . 2 2 x y Given tan  tanh 2 2 Next, x 1  tan 2 2 cos x  2 x 1  tan 2 y 1  tanh 2 2 cos x  2 y 1  tanh 2

9. If tan

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

y 2 1 2 y cosh 2  2 y sinh 2 1 y cosh 2 2 y cosh 2  sinh 2 2  y cosh 2  sinh 2 2 1  cosh y  cos x. cosh y  1

135

sinh 2

y 2 y 2

   10. If u  log tan   , prove that i) sinh u  tan ii) tanh u  sin  and  4 2 iii) cosh u  sec Given   u  log tan     4 2   e u  tan     4 2   tan  tan 4 2 eu    1  tan .tan 4 2  1  tan 2 eu   1  tan 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

136

i. now eu  e u 2    1  tan 1  tan 1 2 2    2 1  tan  1  tan    2 2

sinh u 

2 2       1  tan   1  tan   1   2  2  sinh u      2    1  tan 2  1  tan 2       

   4 tan   1 2    2 1  tan 2    2  2 tan 2  tan1.    2 1  tan 2 2  sinh u  tan 

ii.

cosh 2 u  1  sinh 2 u  1  tan 2   sec2   cosh u  sec 

iii.

tanh u 

sinh u cosh u



tan  sec



sin  cos   cos  1

 tanh u  sin 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

137

  u  11. If u  log tan(  ) , prove that tanh  tan 4 2 2 2 Given,



   u  log tan    4 2    eu  tan    4 2

eu 

tan

 4

1  tan

eu 

1  tan 1  tan

 tan

 4

 2

. tan



 2



Now,

u e 2  e 2 tanh  u u 2 e 2  e 2 u

eu  1  u e 1 1  tan 

1  tan 1  tan 1  tan

1  tan 

 2 1



2 1



   1  tan  2  2

1  tan



 1  tan



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



2 tan 2

 tanh

138

 2

u   tan 2 2

   12. If cosh u  sec , prove that u  log tan    4 2 Given, cosh u  sec eu  e u  sec 2 1 eu  u  2 sec  e Put, eu  x 1 x   2 sec x 2 x  1  2 x sec x 2  2 x sec  1  0

 2 sec  4 sec 2   4 x 2





2 sec  4 sec2   1 x 2

2 sec  2 tan 2  x 2 x  sec   tan  1 eu   tan  cos    1  tan 2 2 tan 2 2    1  tan 2 1  tan 2 2 2   1  tan 2  2 tan 2 2   1  tan 2 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

139

  1  tan  2       1  tan 1  tan  2  2  2

eu 

1  tan 1  tan



tan

 4

1  tan



 tan

 4

 2

. tan



2    eu  tan    4 2     u  log tan    4 2

3.6 Inverse hyperbolic functions 1. cosh 1  log( x  x 2  1) Proof Let cosh 1 x  y x  cosh y

e y  e y x 2 1 2x  e y  y e Put ey  z 1 z 2 2 xz  z  1 z 2  2 xz  1  0

2x  z 

z  ey 

2x  4x2  4 2

2x  2 x2  1 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

140

e y  x  4x2  4  y  log( x  x 2  1) ie,

cosh 1 x  log( x  x 2  1)

2. sinh 1 x  log( x  x 2  1) Proof: Let  x  sinh y

e y  e y x 2  2x  e y 

1 ey

Put, e y  z 1 2x  z  z 2 xz  z 2  1 z 2  2 xz  1  0

ey 

2x  4x2  4 2

e y  x  x2  1



 y  log x  x 2  1





 sinh 1  log x  x 2  1



1 1 x 3. tanh 1 x  log 2 1 x Proof, Let tanh1 x  y x  tanh y

e y  e y e y  e y 1 ey  y e x 1 y e  y e 2y e 1 x  2y e 1 2y xe  x  e2 y  1 x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

141

e2y (x  1)  x  1 1 x e2 y  1 x

1 x 1 x 1 x  y  log 1 x 1 1 x  tanh 1 x  log 2 1 x  2 y  log

Problems 1. Separate into real and imaginary part tan 1 (x  iy) Let, u  iv  tan1 ( x  iy ) (1)

u  iv  tan1 ( x  iy ) (1)  (2)

2u  tan1 x  iy   tan1 x  iy 

(2)

 1 1 1 A  B  tan A  tan B  tan 1  AB 

 x  iy  x  iy  2u  tan 1   1  (x  iy)(x  iy)    2x 2u  tan 1  2 2  1  (x  y ) 

u 

 1 1  2x tan  2 2 2 1  x  y 

(3)

Next,

i 2v  tan1 ( x  iy )  ( x  iy )  1 1 1 A  B  tan A  tan B  tan 1  AB   ( x  iy )  ( x  iy )  i 2v  tan 1   1  ( x  iy )( x  iy )  i2 y tan i 2v  1  x2  y 2 i2 y i tanh 2v  1  x2  y 2   2y  2v  tanh 1  2 2 1  x  y  (1)  (2)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

v 

142

  1 2y  tanh 1  2 2  2 1 x  y 

Real part u 

 1 1  2x  tan  2 2  2 1 x  y 

v

 1 1  2y  tan  2 2  2 1 x  y 

Im part

(4)

Self assessment problems IV 1. If sin( A  iB)  x  iy , prove that

x2 y2   1 and cosh 2 B sinh 2 B x2 y2  1 2. sin 2 A cos 2 B 2. If tan(  i )  x  iy prove that 1.

x 2  y 2  2 x cot 2  1 3. If tan(  i )  cos   i sin  prove that n  1.    and 2 4 1    2.   log tan   2 4 2 4. Separate into real and imaginary parts tanh  x  iy  3.7 Logarithm of a complex number Definition:

if ez  w where Z and W are complex numbers, then Z is called the

logarithm of W and is written as Z  log e W To find the real and imaginary parts of Log a  ib  Let a  ib  r cos  i sin   1



a  rcos

and

b  rsin





a 2  b 2  r 2 cos 2   sin 2   r 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

143

b r sin   tan a r cos   tan 1

b a

Now 1  a  ib  rcos  i sin  a  ib  rcos  2n   i sin  2n 

where n  0,1,2,3........ a  ib  re 

i  2n

 Loga  ib   Logr  Logei  2n 





 Log a 2  b 2  i  2n .Loge e





1  log a 2  b 2  i  2n  2





1 b   Log a  ib   log a 2  b 2  i 2n  tan 1  2 a  Where n  0,1,2,3..................... This is the general value. Put n = 0





1 b loga  ib   log a 2  b 2  i tan 1 2 a

This is the principal value. Problems 1. Find the value of Log 1  i 





1 1  Log 1  i   log 12  2 2  i 2n  tan 1  2 1  1    log 2  i 2n   2 4  Where n  0,1,2,3.......... FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

2. Show that i log L.H .S  i log

144

x i    2 tan 1 x xi

x i xi

 ilogx - i   logx  i   1   1   1  1   i  log x 2  12  i tan 1     log x 2  12  i tan 1    x   2  x   2









 1  1   i - tan -1    i tan 1    x  x  

 -2i 2 tan 1

 -1 1 1  tan x  cot x

1 x

tan x cot

 -2cot -1x

-1

1

x

 2

   2  tan 1 x  2    - 2tan-1 x

 R.H.S

a - ib  2ab  3.Pr ove that tan log  2 2  a  ib  a  b



log a  ib 



1 b log a 2  b2  i tan 1  1 2 a 1  b log a  ib   log a 2  b 2   i tan 1   2  a 





1 b log a  ib   log a 2  b2  i tan1   2 2 a

2  1; b b  log a  ib   log a  ib   i tan1  tan1  a a 

 1 1 1 A  B  tan A  tan B  tan 1  AB 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

145

b b     a  ib 1 a a  log  i  tan b b a  ib  1 .  a a 

  b  2  a  ib a i log  i 2  tan 1   b2 a  ib  1 2  a 

     

 2ab    tan-1  2 2  a b   a  ib  2ab   tan i log   a  ib  a 2  b2   4.If  i  b x iy

 y  prove that  x log   2  2  2 tan 1

Given   i  b x iy log  i   x  iy  log b

1  log   2  i2   i tan 1     x  iy  log b 2  Equating real parts, 1 log   2  2   x log b  (1) 2 Equating imaginary parts,

 tan 1    y log b  (2)   tan -1   

(2) y log b ;  (1) 1 log  2  2 x log b   2   2 tan 1      y 2 log    2 x





FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

5. Prove That i i  e

146

4 n 1

where n is a positive integer

we have

i i  elogi

 x  elog x 

 ei logi

e

1 1  1 i  log 0 2 12  i  2 n  tan 0  2  



1

 ei  2

e



   

Where n is a positive integer



 log1 i 2 n  tan 1   

    i  0  i  2 n    2   

e e



 4 n     2  

  4 n 1.



Where n is a positive integer

Self assessment problems V 1. Prove that log 1  i tan x   logsec x  i  2n  x  2. If tan logx  iy   a  ib, prove that





tan log x 2  y 2 

2a 1  a 2  b2

ANSWERS 3.8 Self Assessment Problems I 1. We know that

sin n  nc1.cos n 1  .sin  nc3.cos n 3 .sin 3   ............ Put n=5

sin 5  5c1. cos 4  . sin   5c3. cos 2  . sin 3   5c5 cos0 . sin 5  

sin 5  5. cos 4   10 cos 2  .sin 2   1.1.sin 4  sin 









 5 1  sin 2   10 cos 2  .sin 2  1  sin 2 gq  sin 4  2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

147

 5  10 sin 2   5sin 4   10 sin 2   10 sin 4   sin 4 

 16 sin 4   20 sin 2   5 2. We know that

cos n  cos n   nc2 .cos n  2 .sin 2   ..........  cos 6  cos 6   6c2 .cos 4  .sin 2   6c4 cos3  .sin 4   6c6 .cos 0  .sin 6   cos 6   15 cos 4  .sin 2   15 cos 2  .sin 4   sin 6 













 1  sin 2   15 sin 2  1  sin 2   15 sin 4  1  sin 2   sin 6  3

 1  3sin 2   3sin 4   sin 6   15 sin 2   30 sin 4  15 sin 6   15 sin 4   15 sin 6   sin 6  cos 6  1  18sin 2   48sin 4   32 sin 6 

Self assessment problems II 1. Let x  cos  i sin 1  cos  i sin  x x

1  2 cos x

x

1  2i sin  x

2i sin  

1   x   x 

1 1 1 1 1 i 5 25 sin 5   x5  5c1x 4 .  5c2 .x3. 2  5c3.x 2 . 3  5c4 .x. 4  5c5 . 5 x x x x x i32 sin 5   x5  5 x3  10 x 

10 5 1   x x3 x5

1  1 1   i32 sin 5    x5    5 x3  3   10 x   x  x  x   i32 sin 5   2i sin 5  5.2i sin 3  10.2i sin 16 sin 5   sin 5  5sin 3  10 sin

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

148

1  2. 2 cos 7   x   x 

1 1 1 27. cos 7   x 7  7c1.x 6 .  7c2 .x5 . 2  7c3 .x 4 . 3 x x x 1 1 1 1 7c4 .x3 . 4  7c5 .x 2 . 5  7c6 .x. 6  7 x x x x  x 7  7 x5  21x3  35 x 

35 21 7 1  3 5 7 x x x x

1  1 1 1      x 7  7   7 x5  5   21 x3    35 x   x   x  x x    27.cos 7   2 cos 7  7.2 cos 5  21.2 cos 3  35.2 cos  64 cos 7   cos 7  7 cos 5  21cos 3  35 cos 4

3. 2i sin   2 cos  4

1  1   x   x   x  x 

1  1  1  i 2 sin  .2 cos    x   x    x   x  x  x  4

1  1  2 .sin  . cos    x   x 2  2  x  x   7

1  1 1 1    x    x 6  3x 4 . 2  3x 2 . 4  6  x  x x x   1  3 1    x   x 6  .3x 2  . 2  6  x  x x    x 7  3x 3 

3 1 3 1   x 5  3x  3  7 x x x x

1  1  1  1    x 7  7    x 5  5   3 x 3  3   3 x   x x   x   x    27.sin 4  .cos 3   2 cos 7  2 cos 5  3.2 cos 3  3.2 cos 64 sin 4  .cos 3   cos 7  cos 5  3cos 3  3cos 1  4. 2 cos 8   x   x 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

149

1 1 1 1 28. cos 8   x 8  8c1 .x 7 .  8c2 .x 6 . 2  8c3 .x 5 . 3  8c4 .x 4 . 4 x x x x 1 1 1 1 8c5 .x 3 . 5  8c6 .x 2 . 6  8c7 .x. 7  8  2 x x x x

2i sin 

1  x  x 

1 1 1 1 1 (1)  (2)  x 8  8C1.x 7 .  8C2 .x 6 . 2  8C3 .x 5 . 3  8C4 .x 4 . 4  8C5 .x 3 . 5 x x x x x 1 1 1  8C6 .x 2 . 6  8C7 .x1. 7  8  (2) x x x





1 1  28 cos 8   sin 8   2 x 8  8c2 x 4  8c4  8c6 . 4  8  x x     1 1   2 x 8  8   28 x 4  4   70 x  x    



2 2.cos 8  282 cos 4  70

 4;







64 cos 8   sin 8   cos 8  28 cos 4  35 Self assessment problem III 1. Lt

xo

sin 2 x  2 sin x sin 3 x

 Lt

3 5      2 x  2 x  x3 x5   ........  2 x    ........ 2 x  3 5 3 5    

xo

 Lt x0

2x 

  x3 x5  ........... x   3 5  

8 x 3 32 x 5 2x3 2x5   2x   ......... 6 120 6 120 3 2 4   x x x 3 1   ........ 3 5  

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II x4 x   ......... 4  Lt 3 xo  x2 x4  3 x 1   ........ 3 5   3

  x2 x 3  1   ........ 4   Lt 3 x0  x2 x4  3 x 1   .......... 3 5  

1 0 1

 1

2. Lt

x0

cos ecx  cot x x

1 cos x   Lt sin x sin x x0 x 1  cos x x0 x sin x

 Lt

  x2 x4 1  1    ...... 2 4    Lt x 0   x3 x5 xx    ........ 3 5   x2 x4   ........ 2 24  Lt x0  x2 x4  x 2 1    ........ 6 120  

 1 x  x2    ...... 2 24   Lt  2 x0  x  x 2 1   ....... 6  

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

150

MATHEMATICS-II

151

1 0 2  1 0 

3. Lt

x0

1 2

3 sin x  sin 3x x  sin x 3 5      x3 x5 3x  3x  3 x    .....  3x    ...... 3 5 3 5      Lt  3 5 x0   x x x   x    ....... 3 5  

3x 

 Lt

x0

 Lt

x0

3 3 3 5 27 3 243 5 x  x  3x  x  x 6 120 6 120 x3 x5 xx   ......... 6 120

4 x 3  2 x 5  ...... x3 x5   .......... 6 120





x 3 4  2 x 2  .... x0  x2 3 1 x    ....  6 120 

 Lt



4 1   6

 24 4. Given,

sin 





3 3

 

 1

5765 5766

5 5

..... 

2 1  1  5766

5765 5766

nearly

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

2 6

1 961

2 

tan 







nearly

3 3

2

nearly

433 432

 .......



1

nearly

1 radians 31



5. Given

6 5766



152

 1



433 432

1 nearly 432



1 nearly 432

2 

3 nearly 432

2 

1 144



1 radians approximately 12

Self assessment Problems IV 1. Given x  iy  sin A  iB 

 sin A. cos iB  cos A. sin iB  sin A. cosh B  cos A.i sinh B Equating Real Parts,

x  sin A.cos B  (1) Equating imaginary Parts, y  cos A.sinh B  (2)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

153

x2 y2 sin 2 A. cosh 2 B cos 2 A. sinh 2 B    cosh 2 B sinh 2 B cosh 2 B sinh 2 B

 sin 2 A  cos 2 A

1

ii )

x2 y2 sin 2 A. cosh 2 B cos 2 A. sinh 2 B    sin 2 A cos 2 A sin 2 A cos 2 A

 cosh 2 B  sinh 2 B

1 2. Given tan  i   x  iy

  i  tan 1  x  iy   (1) lll ly   i  tan 1  x  iy   (2)

(1)  (2);

2  tan 1 x  iy   tan 1 x  iy   x  iy  x  iy  2  tan 1   1  x  iy x  iy 

tan 2 

2x 1  x2  y2





1 2x  cot 2 1  x 2  y 2

1  x 2  y 2  2 x cot 2 1  x 2  y 2  2 x cot 2

3. Given tan  i   cos   i sin 

  i  tan 1 cos   i sin    1 lll ly  - i  tan-1 cos   i sin    2 (1)  (2);

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

154

2  tan 1 cos   i sin    tan 1 cos   i sin    cos   i sin   cos   i sin   2  tan 1   1  cos   i sin  cos   i sin  

  2 cos  2  tan 1   2 2 1  cos   sin  





 2 cos   2  tan 1   1  1 

2  tan 1 



2  n    n

 2

2 



where n  0,1,2,3,......

Next, 1  2;

i 2  tan 1 cos  i sin    tan 1 cos  i sin    cos   i sin   cos   i sin   i 2  tan 1   1  cos   i sin  cos   i sin    tani 2   i tanh 2 

i 2 sin  1  cos 2   sin 2 

i 2 sin  2



2 tan e 4  1 2  4 e  1 1  tan 2  2 4

 

4

e 1  e e 4  1  e 4

   1  tan 2  1 2  2     1 2 tan  1  tan 2  2  2

 

2 tan



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

155

  1  tan  4 2e 2   2 2    1  tan  2  2

e 

2 2

    1  tan   2       1  tan   2  

e 2 

1  tan 1  tan

e 2 

tan

 4

 2



 tan



 2



1  tan . tan 4 2

    tan   4 2     2  log tan   4 2

1      log tan   2  4 2

1 4. tanhx  iy   tan ix  iy  i  i tanix  y 





 tan ix  i tanh x     tanh x  1 tan ix  i  

i sin  ix  y  cos  ix  y 

 2i sin  y  ix . cos y  ix  2 cos y  ix . cos y  ix 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

156



isin  y  ix  y  ix   sin  y  ix  y  ix  cos y  ix  y  ix   cos y  ix  y  ix 



isin 2 y  sin i 2 x cos 2 y  cos 2 x 

isin 2 y  i sinh 2 x cos 2 y  cosh 2 x



i sin 2 y  sinh 2 x cos 2 y  cosh 2 x



sinh 2 x cos 2 y  cosh 2 x

Imaginary Part 

sin 2 y cos 2 y  cosh 2 x

Real part

Self assessment Problems V

 1  tan x  1. log 1  i tan x   log 12  tan 2 x  i 2n  tan 1   2  1  





1  log sec2 x  i2n  x 2 1  log sec x  i2n  x 2

Where n=0,1,2,3,…….. 2. Given, tan logx  iy   a  ib

 log  x  iy   tan 1  a  ib 

 (1)

logx  iy x  iy   tan1 a  ib   tan1 a  ib  lll ly log  x-iy   tan 1  a  ib   (2) (1)  (2)

log( x  iy )  log( x  iy )  tan1 a  ib   tan1 a  ib 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

157

 a  ib  a  ib  log x  iy x  iy   tan 1   1  a  ib a  ib 





2a   tan log x 2  y 2   2 2 1  a  b 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

158 UNIT-4

DIFFERENTIATION

4.0

Introduction

4.1

Objective

4.2.

Definition of Derivative.

4.3

Some Standard forms.

4.4

Some general theorems.

4.5

Some important formula

4.6

Successive differentiation

4.7

Leibnitz formula

4.8

Meaning of the derivative

4.9

Answers to Self assessment Problems

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

159

4.0 Introduction

In this unit you will learn an interesting and important branch of mathematics known as Differential Calculus. The derivative of a function is defined and from the basic principles how to derive the differential coefficient of a function is explained with many examples. List of formulae for many standard functions is given. Suitable examples were given for different types of functions. Next you will learn about successive differentiation. Leibnitz theorem is proved and many examples were given to illustrate the same. Also you will know the meaning of the derivative.

4.1 Objective

After going through this unit, you will be able to

4.2. Definition:



Understand what is derivative of a function



The meaning of the derivative



Find the derivatives of different kinds of functions



Find the successive derivative



Apply Leibnitz theorem and do problems.

Derivative.

Let y be a continuous function of the independent variable x. We can give arbitrary values for x. The values of y depend on those of x. y is called the dependent variable. Let x be any small increment in x. Let y be the corresponding increment in y. i.e.,

Let y  f (x)

Then y + y = f ( x + x )

y  f ( x  x)  f ( x)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

160

y f ( x  x)  f ( x)  x x dy y f ( x  x)  f ( x)  Lt  Lt dx xo x xo x

If the limit on the R.H.S exists, then it is called the differential coefficient of y with respect to x. It is also called the derivative.

4.4 Some Standard forms. 1. To find the derivative of x n . Let y  x n _____________(1) Let x

be an arbitrary increment in x.

Let y

be the corresponding increment in y.

i.e.,

y  y  ( x  x) n ___________(2)

(2)  (1)  y  ( x  x) n  x n dy  dx

y x

( x  x) n  x n x

x o



x o



x  x  x

 n.x 

( x  x) n  x n ( x  x)  x

  xn  an  n.a n1   Lt  x a x  a 

n 1

d n ( x )  n.x n1 dx

for all values of n.

2.To find the derivative of ex: Let y  e x Let x

be an arbitrary increment in x.

Let y

be the corresponding increment in y.

Then

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

161

y  y  e x  x  y  e x  x  e x y

dy  dx

Lt x

x 0



e x  x  e x Lt x x 0 e x  1 x x 0

 e x . Lt

 eh 1   1  Lt  h 0 h 

 e x .1 

d x (e )  e x . dx

3.To find the derivative of log x: Let y  log x ___________(1) Let x

be an arbitrary increment in x.

Let y

be the corresponding increment in y.

Then (2)  (1);

y  y  log  x  x  __________(2) y  log  x  x   log x dy  dx

y

Lt x

x 0



x 0



x 0

log  x  x   log x x

 x  x  log    x  x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

162



x  0

 x  log 1   x    x  x.    x 

x h x As x  0, h  0 put



dy dx

d  log x  dx

log 1  h  1 . Lt x h 0 h 1 1/ h  Lt log e 1  h  x h 0



1 .1 x



1 . x

n    1  Lt 1    e   n   n   1/ h  1  h   e   Lt h o 

4. To find the derivative of sin x: Let y  sin x ____________(1) Let x

be an arbitrary increment in x.

Let y

be the corresponding increment in y.

Then (2)  (1);

y  y  sin  x  x  __________(2) y  sin  x  x   sin x. dy  dx

y

Lt x

x 0

sin  x  x   sin x x x 0  x  x  x   x  x  x  2 cos   .sin   2 2      Lt x x 0 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

163

S 

x 0

x    x  cos  x   .sin   2    2   x     2 

  x   sin      x     2     Lt cos  x   2    Lt  x    x 0  x 0     2    cos  x  0  .1

sin     Lt   1  0 

 cos x 

d  sin x   cos x. dx

5. To find the derivative of cos x: Let y  cos x __________(1) Let x

be an arbitrary increment in x.

Let y

be the corresponding increment in y.

Then, (2)  (1);

y  y  cosx  x  __________(2) y  cosx  x   cos x dy y  Lt dx x0 x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

164

cos x  x   cos x x x  0  x  x  x   x  x  x   2 sin  . sin   2 2      Lt x x  0  x  sin   x    2   Lt  sin  x  . 2   x   x  0    2  

d cos x  dx



  x   sin      x     2    Lt  sin  x   . Lt 2  x 0  x     x 0      2    sin      sin  x  0 .1  Lt   1   0    sin x.

4.4 Some general theorems. 1. The differential coefficient of a constant is zero. Let y = c, constant._____________(1) Let x

be an arbitrary increment in x.

Let y

be the corresponding increment in y.

y  y  c ______________(2)

Then, (2)  (1);

y  0 dy y  Lt dx x0 x

dy 0  Lt dx x0 x  Lt 0 x 0

 0 

d (c ) 0 dx

Where c is any cons tan t.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

165

1. To find the differential coefficient of a product of a constant and a function of x. Let y = c.u _____________(1) Where c is a constant. Where u is a function of x. Let x be an arbitrary increment in x. Let y be the corresponding increment in y. Let u be the corresponding increment in u.

Then, y  y  cu  u  ____________(2) (2)  (1); y  c.u dy y  Lt dx x0 x c.u  Lt x 0 x u  c. Lt x 0 x du  c. dx d du  cu   c. . dx dx 2. To find the derivative of the sum or difference of two functions. Let y = u+v___________(1) Where u and v are two functions of x. Let x be an arbitrary increment in x. Let u, v and y be the corresponding increments in u,v and y respectively.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

166

y  y  u  u   v  v  _________(2)

Then, (2)  (1);

y  u  v dy  dx

y

Lt x

x 0



u  v x x 0



Lt  x  x 

 u

v 

x 0

dy du dv   . dx dx dx Note:

If y = u-v, then dy du dv   . dx dx dx

3. Product Rule: Let y = u.v____________(1) Where u and v are two functions of x. Let x be an arbitrary increment in x. Let u, v and y be the corresponding increment in u,v and y respectively. Then, y  y  u  u v  v  ________(2) (2)  (1);

y  u.v  v.v  u.v dy y  Lt dx x0 x u.v  v.u  u.v  Lt x x 0 u u   v  Lt u.  v.  .v  x x x  x 0 



dy dx

v   u     u    u. Lt   v. Lt   Lt  . Lt v   x0 x   x0 x  x0  x   x0  dv du du  u.  v.  .0 dx dx dx dv du  u.  v. dx dx

4. Quotient Rule: FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Let y =

167 u __________(1) v

Where u and v are functions of x. Let x be an arbitrary increment in x. Let u, v and y be the corresponding increment in u,v and y respectively.

u  u _______(2) v  v u  u u (2)  (1); y   v  v v u  u .v  u v  v   v  v .v

Then,

y  y 

v.u  u.v v  v .v dy y   Lt dx x 0 x v.u  u.v v  v .x  Lt x x  0 u v v.  u. x  Lt x   v   v . v x  0 du dv v.  u. dx dx  v  0.v du dv v.  u. d u     dx 2 dx dx  v  v y 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II 4.5

168

Some Important Formulae: y ex

dy dx

1 x n.x n 1

sin x

cos x

 sin x

log x

tan x

sec 2 x

cot x

 cos ec 2 x

sec x cos ecx

sec x. tan x  cos ecx. cot x

sinh x

cosh x

sinh x

sin 1 x

1 1 x2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

y cos 1 x tan 1 x cot 1 x sec 1 x sinh 1 x cosh 1 x

169 dy dx 1 1 x2 1 1 x2 1 1 x2 1 x x2 1 1 1 x2 1 x2 1 0

c  cons tan t  c.u uv u.v u v ax xx 1 x x

du dx du dv  dx dx dv du u.  v. dx dx du dv v.  u. dx dx v2 a x log a c.

x x 1  log x  1 x2 1 2 x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

u v ax xx 1 x x

170

du dv  u. dx dx 2 v x a log a x x 1  log x  1 x2 1 2 x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

171

Function of a function rule: Let y  f u  Where u  g x  Then, dy dy du  . dx du dx

(or)

y  f g x 

y1  f 1 g x .g 1 x . Differentiate the following functions: 1. 2 x 3  4 log x  5 sin 2 x Let y  2 x 3  4 log x  5 sin 2 x

Differentiating with respect to x

dy 1  2.3x 2  4. 5.cos 2x.2 dx x dy 4  6x 2   10 cos 2x. dx x 2.  2x  1  x 2  1 2

Let y   2x  1  x 2  1 2

2 3 dy 2   2x  1 .3  x 2  1   x 2  1 .2  2x  1 .2 dx

 2  2x  1  x 2  1 3x  2x  1  2  x 2  1  2

2 dy  2  2x  1  x 2  1 8x 2  3x  2  dx

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

172

4. e  x  sin 2x  2 cos x  Let y e  x  sin 2x  2 cos x  dy  e  x  cos 2x.2  2sin x    sin 2x  2 cos x   e  x  dx  e  x  2 cos 2x  2sin x  sin 2x  2 cos x  5. x 3e3x sin 4x Let y  x 3e3x sin 4x dy 3 3x  x e cos 4x.4  x 3 sin 4x e3x .3  e3x sin 4x 3x 2 dx dy  x 2 e3x  4x cos 4x  3x sin 4x  3sin 4x  dx 6. x  sin 1 x 

Let y  x  sin 1 x 

2 dy 1  x.2  sin 1 x  .   sin 1 x  .1 dx 1 x2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

173

7. tan 1 x Let y  tan 1 x dy 1 1  . 2 dx 1   x  2 x   1  2 x 1  x 

 a  b cos x  sin 1   b  a cos x   a  b cos x  Let y  sin 1   b  a cos x  b  a cos x  b sin x   a  b cos x  a sin x  dy 1  . 2 dx b  a cos x 2  a  b cos x  1    b  a cos x  b  a cos x

dy  dx

b  a cos x 2  a  b cos x 2



 b 2 sin x  ab sin x cos x  a 2 sin x  ab sin x cos x b  a cos x 2

sin x a 2  b 2



1 b 2  2ab cos x  a 2 2 cos 2 x  a 2  2ab cos x  b 2 cos 2 x b  a cos x  sin x a 2  b 2 1 . b 2  a 2  a 2  b 2 cos 2 x b  a cos x 

 



  

sin x a 2  b 2

 



1 b  a 1  cos x b  a cos x 

 





sin x a 2  b 2

.



1 b  a . sin x b  a cos x  2

dy  b 2  a 2  . dx b  a cos x 9. x sin x Let y  x sin x Taking log, log y  sin x log s Differenti ating with respect to x, 1 dy 1 .  sin x.  cos x. log x y dx x 

dy  sin x   y  cos x. log x  dx  x 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



174

dy  sin x   y  cos x. log x  dx  x  dy sin x  sin x  x   cos x. log x  dx  x 

10. tan x 

log x

Let y  tan x 

log x

 log y  log x. log tan x  Differenti ating with respect to x 1 dy 1 1 .  log x. . sec 2 x  log tan x . y dx tan x x dy cos x 1 1    y log x. .  . log tan x  2 dx sin x cos x x   dy log x 1  log x   tan x    . log tan x  dx  sin x. cos x x 





 x x 1  11.    2 x  3  2

3 2





 x 2 x3 1  2 Let y     2x  3  Taking log,









3 2 log x  log x 3  1  log 2 x  3 2 Differenti ating with respect to x,

log y 

1 dy 3  2 1 1     3 3x 2  2 y dx 2  x x  1 2x  1  dy 3  2 3x 2 2   y  3   dx 2  x x  1 2 x  1 12.



 3x  4 x 

sin 1 3x  4 x 3 Let y  sin

1

put x  sin 



  sin 1 x.

 y  sin 1 3 sin   4 sin 3 



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

175

y  sin 1 sin 3 y  3 y  3sin 1 x 

13.

dy 3  dx 1 x2

1  x 2  cos 1 x  2 1  x 

1 x2   Let y  cos 1  2  1  x  

Put x  tan

  tan 1 x

 1  tan 2     y  cos 1  2  1  tan   y  cos 1 cos 2  y  2 y 

 2 tan 1 x dy 2  dx 1  x 2 1

14.

 x2  2  tan 1  2 2  a x  1

Let

 x2  2 y  tan 1  2 2 a  x 

Put x  a sin  or

x x  sin  ;   sin 1 a a

 a 2 sin 2   2  y  tan 1  2  2 2  a  a sin   1

 a 2 sin 2   2 y  tan  2  2  a 1  sin   y  tan 1 tan   1





FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

176

y  y  sin 1 

dy  dx

x a 1 1

x a2

1 a

 

a2  x2

1 a

dy 1  2 dx a  x2

15. If x 1  y  y 1  x  0, prove that

dy 1  dx 1  x 2

Given x 1  y  y 1  x  0

x 1 y   y 1 x Squaring ,

x 2 1  y   y 2 1  x  x 2  x 2 y  y 2  xy 2  0

x

 



 y 2  x 2 y  xy 2  0

x  y x  y   xy x  y   0 x  y x  y  xy   0  x  y or x  y  xy  0 y 1  x    x

x 1 x dy 1  x  1  x.1   2 dx 1 x y 





dy 1  dx 1  x 2 16. If y  x

..  x.

find

dy dx

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

177

Given y  x

. ..

y  xy Taking log, log y  y log x

Differentiating with respect to x 1 dy 1 dy .  y.  log x. y dx x dx  y dy  1   log x   dx  y  x dy  1  y log x  y    dx  y  x 

dy y2  dx x1  y log x 

17. If x y  y x show that

dy y y  x log y   dx xx  y log x 

Given x y  y x Taking log, y log x  x log y

Differentiating with respect to x 1 dy 1 dy y.  log x.  x. .  log y.1. x dx y dx 

dy  x y  log x    log y  dx  y x



dy  y log x  x  x log y  y    dx  y x 

dy y  x log y  y   dx x y log x  x  dy y  y  x log y    dx x x  y log x  

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

178

18. If x y  e x y prove that

dy log x  dx 1  log x 2

Given x y  e x y Taking log,

y log x  x  y log e e

log e e  1

y log x  x  y y  y log x  x

y 1  log x   x y

dy  dx

x 1  log x

1  log x .1  x. 1 1  log x 

19. If x  at 2 ; y  2at find

x 

log x 1  log x 2

dy . dx

Given x  at 2

; y  2at

dx dy  2at ;  2a dt dt dy dy / dt   dx dx / dt 2a  2at 1  t 

20. If x  a  sin   ; y  a1  cos  find

dy dx

Given x  a  sin   ; y  a1  cos 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

179

dx dy  a 1  cos   ;  a sin  d d dy dy / d   dx dx / d a sin   a 1  cos  

  2sin cos 2 2   2 cos 2 2   tan . 2 21.

sin 2A  2sin A.cos A    2 1  cos 2A  2 cos A 

If x  a cos 3  ; y  a sin 3  find Given x  a cos 3 

dy dx

; y  a sin 3 

dx  a.3 cos 2   sin   d dy  a.3 sin 2  . cos  d dy dy / d   dx dx / d 3a sin 2  . cos    3a cos 2  . sin  dy    tan dx x

22.

dy a If y    find dx x a Given y    x

Taking log, log y  xlog a  log x

Differentiating with respect to x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

180

1 dy  1 .  x   log a  log x .1 y dx  x  dy a  y 1  log a  log x  Where y    dx  x

23.

 x  1 x2  dy If y  log   find 2 dx  x  1  x   x  1 x2 y  log   x  1  x 2



  





 log x  1  x 2  log x  1  x 2



    1 1 1 1  2x   1 1  2x     2 dy 2 1 x2    2 1 x     2 2 dx x  1 x x  1 x





1 x2  x





1 x . x  1 x 2





x  1 x2



  2





 x

1 x2 x  1 x2

 1 x2  x   

 1 x 

1 x2 x  1 x2 2

 x 



1 x2



dy x 2  1  x 2  2x 1  x 2  x 2  1  x 2  2 x 1  x 2  dx 1  x 2  x 2  1  x 2   dy 2  dx 1  x 2  2x 2  1

1 x2 dy 24. If y  x find 2 dx 1 x 1 x2 Given, y  x 1 x2







1 1 log y  log x  log 1  x 2  log 1  x 2 2 2



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

181

Differentiating with respect to x, 1 dy 1 1.2x 1 1 2x     . 2 y dx x 2 1  x  2 1  x 2 

dy x x  1  y   2 2 dx  x 1  x 1  x 

1  x 4  x 2 1  x 2   x 2 1  x 2     y x 1  x 2 1  x 2    1  x 4  x 2  x 4  x 2  x 4    y x 1  x 4   

  x 4  2x 2  1  dy  y 4 dx  x 1  x   24.

If x m . y n  x  y 

m n

, prove that

dy y  dx x

Given x m . y n  x  y 

m n

Taking log, m log x  n log x  m  nlogx  y 

Differentiating with respect to x,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

182

 1  1 1 dy dy  m.  n. .  m  n  1  1.  x y dx dx  x y  m n dy m  n m  n dy    . x y dx x  y  x  y  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   x  y . y x  y .x dx   dy  nx  ny  my  ny  mx  nx  mx  my  dx  y x  dy  nx  my   nx  my     dx  y   x  

dy y  . dx x

Self assessment problems. I

1. Find the differential coefficient of tan x from first principles. Find the derivatives of 2. e 2 x , sin 3x, cos 4 x.

1 x  3. sinh 1  . 1 x  4. sin x, sin 2 x, sin 3x, sin 4 x.

 sin x  5. tan 1  .  1  cos x  6. If sin y  x sina  y  prove that



7. If x 2  y 2







 a 2 x 2  y 2 find

dy sin 2 a  y   dx sin a dy . dx

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

4.6.

183

Successive differentiation: Let y be a function of x. On differentiating y with respect to x, we get

called the first derivative of y with respect to x. Generally we can differentiate

dy . It is dx

dy is also a function of x. So dx

d  dy  dy with respect to x. We get the second derivative   or dx  dx  dx

d3 y d2y . Similarly the derivative of is called the third derivative. Differentiating y dx 3 dx 2

dn y successively n times, we get dx n

Some Standard results.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

184

1. e ax 2. 3.

a n .e ax

ax  b m

mm  1......m  n  1.a n ax  b 

mn

 1n n!.a n ax  b n1  1n1 n  1!a n ax  b n

1 ax  b

4. log ax  b 

 n  a n . sin   ax  b   2   n  a n cos  ax  b   2 

5. sin ax  b  6. cos ax  b  7.e ax . sin bx  c 

r n .e ax sin n  bx  c 

8. e ax . cos bx  c 

r n .e ax . cos n  bx  c 

Where r 2  a 2  b 2 , tan  

9. x n

Type-I: 1. Find the nth derivative of

Let



x  1x  2 A B  x 1 x  2

x  1x  2 1  Ax  2  Bx  1 x  2;

1  B.1

1  A 1

x  1; 

Let y  y

x  1x  2



B 1  A  1

1 1  x 1 x  2

x  1x  2

1 1  x 1 x  2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

b a

MATHEMATICS-II

185

Differentiating n times successively, we get

yn 

 1 1n n!.1n   1n .n!.1n x  1n1 x  2n1

 1  1 n y n   1 n!  n 1 n 1  x  2   x  1 2. Find the nth derivative of Let y  Let

1 x 2 x  3 2

1 ____________ 1 x 2 x  3 2

1 A B C   2 2x  3 x 2 x  3 x x 2

1  Ax 2 x  3  B2 x  3  Cx 2 x  0; 1  B.3 x

1 3 4 C 9

B

3 9 1  c.  2 4

coeff .of x 2 0  2 A  C c 2 A 2 9  2 / 9 1/ 3 4 / 9 y   2  x 2x  3 x A

Differentiating n times successively, we get

4 1 n  1n n!.1n 4 . 1n .n!.2 n . 1 .n!.1n yn  9 3 9 x n1 x n 2 2 x  3n1 1 4   4 2 n.  n 9  y n   1 .n! 9n1  n3 2   x 2 x  3n1  x   3. Find the nth derivative of sin2x. Let y  sin 2 x FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

186 y

1  cos 2 x 2

Differentiating n time successively, 1  n  y n  0  2 n cos  2 x  2  2   n  y n  2 n1. cos  2 x .  2 

4. Find the nth derivative of cos3x. Let y  cos 3 x. y

1 cos 3x  3 cos x 4

cos 3 A  4 cos 3 A  3 cos A 4 cos 3 A  cos 3 A  3 cos A cos 3 A 

1 cos 3 A  3 cos A 4

Differentiating n times successively,

yn 

1 n  n   n  3 . cos  3x   3. cos  x   4  2   2 

5.Find the nth derivative of sin2x.cosx. Let y  sin 2 x. cos x

1 sin 2 x  x   sin 2 x  x  2 1 y  sin 3x  sin x 2 y

Differentiating n times successively,

1  n   n  y n  3n sin  3x   1n.sin  x  2  2   2  6. Find the nth derivative of ex.cosx Let y  e x cos x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II Then,

187

yn  r n .e x . cosn  x  ______1

Where, a=1; b=1;c=0 r 2  a 2  b 2  12  12 r 2 b a 1  1

tan  

 tan





1  y n  



n  n  2 e x cos  x  4 

 n  y n  2 e x cos  x  4  n 2

7. Find the nth derivative of e3xsin2x

Let y  e3 x sin 2 x y  e3x

1  cos 2x 

2 1 1 y  .e3x  .e3x cos 2x 2 2

1  cos 2A   2 sin A   2

Differentiating n times successively, 1 1 yn  .3n.e 3 x  .r n .e 3 x cosn  2 x  ______ 1 2 2

Where,

a=3; b=2; c=0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

188

r 2  a 2  b 2  32  2 2  13 r  13 tan  

2 3

2 3

  tan 1  

 

n   1 1 2  1  y n  .3n.e 3 x  13 .e 3 x . cos n. tan 1    2 x  2 2 3  

Self assessment problems – II 1. Find the nth derivative of

1 x  5x  6 2

2. Find the nth derivative of cos 2 x 3. Find the nth derivative of cos 3x. cos x

Type-2

d4y  4y  0 1. If y  e cos x, prove that dx 4 x

Given, y  e  x . cos x  y1  e  x  sin x   e  x cos x y1  e  x sin x  cos x 

 y2  e  x cos x  sin x   e  x sin x  cos x  y2  e  x  cos x  sin x  sin x  cos x  y2  2e  x sin x  y3  2e  x. cos x  2e  x sin x

y3  2e  x cos x  sin x 

 y4  2e  x  sin x  cos x   2e  x cos x  sin x  y4  2e  x  sin x  cos x  cos x  sin x  y4  4e  x cos x y4  4 y 

d4y  4y  0 dx 4

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

189

Another method Given y  e  x . cos x Differentiating n times successively,

 n  yn  r n e  x cos  x  _________ 1  2  Where a=-1; b=1; c=0

r 2  a2  b2  11  2 r  2 tan  

b 1   1 a 1

   tan 1  1   . 4

1  yn  



n      2 e  x cos n   x   4  

put n  4 4      y 4   2 e  x cos 4.   x     4    4.e  x cos   x 

cos    cos 

 4e  x cos  x 

 4 y 

d y  4 y  0. dx 4

d2y dy 2. If xy  ae  be , prove that x 2  2  xy  0 dx dx x

Given xy  ae x  be  x Diff, Diff,

dy  y.1  ae x  be  x . dx

d 2 y dy dy x 2  .1   ae x  be  x dx dx dx  xy x

d2y dy  2  xy  0 2 dx dx

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

190





d2y dy  4x  x 2  6 y  0 2 dx dx

3. If y  x 2 cos x, prove that x 2

Given, y  x 2 cos x _______ 1 Diff,

dy  x 2  sin x   2 x cos x _______ 2 dx

Diff,

d2y  x 2  cos x   2 x sin x   2 x sin x   2 cos x dx 2

d2y   x 2 cos x  4 x sin x  2 cos x ______ 3 dx 2 Now, x2





d2y dy  4x  x 2  6 y 2 dx dx  x 2  x 2 cos x  4 x sin x  2 cos x  4 x  x 2 sin x  2 x cos x  x 2  6 .x 2 cos x







 



  x cos x  4 x sin x  2 x cos x  4 x sin x  8 x cos x  x cos x  6 x 2 cos x 4

0

 , prove that 1  x y  xy  m y  0 Given, y  x  1 x    1 y  mx  1  x  1  .2 x Diff, 2 1 x 

4. If y  x  1  x 2

m 1





y1  m x  1  x 2



m 1



1  x 2 y2  y1



 1 x2  x    2  1  x 

1  x 2 y1  m x  1  x 2

Diff,

1 2 1 x2





2 x  m.m x  1  x 2



m1

  1 2 x 1  2  2 1 x 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

191



  1  x  x  m x  1  x  



m 1

 1  x 2 ; 1  x 2 y 2  xy1  m 2 x  1  x 2 2



 m2 y



 1  x 2 y 2  xy1  m 2 y  0





5. If y  sin 1 x , show that 1  x 2 y2  xy1  2  0



Given , y  sin 1 x Diff,



y1  2 sin 1 x



1 1 x2



 1  x 2 y1  2 sin 1 x Diff,

1  x 2 y2  y1



 1

2 1 x



 2 x   2.

 1  x 2 ; 1  x 2 . y2  xy 1  2



1 1 x2



 1  x 2 y2  xy1  2  0 6.

If y  a coslog x   b sinlog x , show that x 2 y 2  xy1  y  0 Given, y  a coslog x   b sinlog x 

Diff,

1 1 y1  a sin log x .  b coslog x . x x

xy1   a sin  log x   bcos  log x  1 1 Diff, xy 2  y1 .1  a coslog x .  b sin log x . x x

x 2 y 2  xy1   a cos  log x   b sin  log x    y  x 2 y 2  xy1  y  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

192

7. If y  ax n1  bx  n , prove that x 2

d2y  nn  1 y. dx 2

Given, y  ax n1  bx  n Diff,

y1  an  1x n  b n.x  n1

Diff,

y2  an  1.n.x n1  b n n  1.x  n2

x 2 y 2  an  n  1 .x n 1  bn  n  1 .x  n  n  n  1 ax n 1  bx  n 

 n  n  1 .y

 x 2 y 2  n  n  1 y 8. If x  at 2 , y  2at find y 2 Given, x  at 2 , y  2at

dx dy  2at ;  2a dt dx dy dy / dt  dx dx / dt 2a  2at dy 1  dx t



Differentiating with respect to x,

d2y 1 dt  2. 2 dx t dx 1 1  2 t 2at 2 d y 1  2  dx 2at 3 9. If x 

d2y 1 t 2t ,y , find 1 t 1 t dx 2

Given, x 

1 t ; 1 t

y

2t 1 t

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Then,

193

dx 1  t  1  1  t .1  dt 1  t 2 

1  t  1  t

1  t  

2

1  t  dy 1  t  .2  2t.1  2 dt 1  t  2

dy 2  dt 1  t 2



dy dy / dt  dx dx / dt 

1  t 

1  t  

2

 1 

d2 y 0 dx 2

10. If x  acos t  t sin t , y  asin t  t cos t 

d2y find dx 2

Given, x  acos t  t sin t 

dx  a sin t  t. cos t  sin t.1 dt dx  at cos t. dt Now,

y  asin t  t cos t 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

194

dy  acos t  t. sin t   cos t.1 dt dy  at sin t dt dy dy / dt   dx dx / dt at sin t  at cos t dy  tan t dx

Differentiating with respect to x,

d2y dt  sec 2 t. 2 dx dx 1  sec 2 t. at cos t 3 sec t  . at

Self assessment problems- III









1. If y  sin m sin 1 x , prove that 1  x 2 y2  xy1  m 2 y  0 2. If y  e a sin

1





prove that 1  x 2 y2  xy1  a 2 y  0





3. If x  sin t , y  sinpt prove that 1  x 2 y2  xy1  p 2 y  0

d2y 4. If x  2 cos  cos 2t , y  2sint - sin2t find dx 2 5. If ax 2  2hxy  by 2  1 , show that

d2 y h 2  ab  dx 2  hx  by 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

195

4.7 Leibnitz formula If u and v two functions of x , then

D n u.v   u n .v  nC1.un1.v1  nC2 .un2 .u2  ............  nCr .unr .ur  ........u.vn Proof: Let u and v be two functions of x We have to prove

D n u.v   u n .v  nC1.un1.v1  nC2 .un2 .u2  ............  nCr .unr .ur  ........u.vn  1 We will prove by the method of induction d u, v  du dv  .v  u. dx dx dx

ie, D ' u, v   u1.v  u.v1 ie, the theorem is true when n  1 Let us assume 1 to be true for n  2,3........m. ie,

D m u.v   um .v  mC1.um1.v1  mC2 .um2 .u2  ............  mCr .umr .ur  ........u.vm  2 We will prove that 1 is true for n  m  1 Now differentiating 2 w.r.t.x, D m1 u.v   um1.v  um. .v1   mC1 um .v1  um1v2   mC2 um1.v2  um2 .v3 ............  mCr umr 1.vr  umr .vr 1   ......u1.vm  u.vm1 

 u m1.v  1  mc1 .u m .v1  mc1  mc2 .u m1.v2  ....  mcr 1  mcr .u m1r .u r  .....  u.vm1   3

Now, 1  mc1  m  1c1 mc1  mc2  m  1c2 ..................... .....................

mcnr  mcr  m  1cr .................... ...................

 3 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

196

Dm1 u.v  um1.v  m  1c1..umv1  m  1c2um1.v2  ......  m  1cr .um1r  .....  u.vm1 ie, The theorem is true for n  m  1  By the method of induction, the theorem is true for all the integral values of n

Problem 1. Find the n th derivative of x 2 .e x Let

y  x 2e x

Using Leibniz theorem, differentiating n times,



yn  Dn x 2e x



u  ex ; v  x 2 

yn  e x .x 2  nc1.e x .2 x   nc2 .e x .2  nn  1    e x  x 2  n.2 x  2  1.2  





yn  e x x 2  2nx  nn  1   Find the n th derivative of x 3e 2 x Let y  x3e2 x Using Leibnitz theorem differentiating n times, yn  Dn  x 3e2x 





u  e2x ; v  x 3



 









yn  2n e2 x .x3  nc1. 2n1 e2 x 3x 2  nc2 2n2.e2 x 6 x  nc3 2n3.e2 x .6 nn  1 n2 nn  1n  2 n3   yn  e 2 x 2n.x 3  n.3.2n1 x 2  .2 .6 x  .2 .6 1 2 1.2.3  





yn  2n3.e2 x 8x3  12nx 2  6nn  1x  nn  1n  2 3. Find the n th derivative of x sin 2 x. Let y  x sin 2 x Using Leibnitz theorem differentiating n times,

yn  Dn x. sin 2 x 

u  sin2x; v  x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

197

      yn  2n. sin n  2 x .x  nc1.2n1 sin n  1  2 x.1 2  2           yn  2n1 2 x sin n  2 x   n. sin n  1  2 x 2  2     4. Find the n th derivative of e x . log x. Let y  e x . log x Using Leibnitz theorem, differentiating n times u  ex ;

v  log x

 1 . n  1 1  1  y n  e D (e ) log x  nc1.e .  nc 2 .e x .  2   .......  e x . x xn x  n 1

 1 . n  1 n nn  1 yn  e log x    ......  x 2x2 xn n1

5. If y  a coslog x  b sinlog x, prove that

i  x 2 y2  xy1  y  0 and ii  x 2 yn2  2n  1xyn1  n2  1yn  0 Given y  a coslog x   b sinlog x  1 1 Diff, y  - a sin log x .  b coslog x . x x  xy 1  -a sinlog x  b coslog x 

1 1 Diff; xy 2  y1.1  a coslog x .  b sin log x . x x

xx; x 2 y2  xy1  a coslog x  b sinlog x -y

 x 2 y2  xy1  y  0  1 Using Leibnitz theorem, differentiating n times

y

n 2



.x 2  nc1. yn1.2 x  nc2 . yn .2  yn1.x  nc1. y1   yn  0

x 2 yn2  n. yn1.2 x 

nn  1 yn .2  x. yn1  n. yn  yn  0 1.2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

198

x 2 yn2  xyn1 2n  1  yn nn  1n  1  0 x 2 yn 2   2n  1 xyn 1  yn n 2  n  n  1  0





x 2 yn2  2n  1xyn1  n2  1 yn  0 6. If y  ea sin

1





prove that i  1 - x 2 y2  xy1  a 2 y  0 and

ii  1 - x 2 yn2  2n  1xyn1  n2  a 2 yn  0 y  ea sin

Given,

1

y1  e a sin x .a.

Diff,

 1  x 2 . y1  a.ea sin

1

1  x2

Diff,

1  x 2 . y2  y1.

1 2 1 x



 2 x   a.ea sin



x 1  x 2 ; 1 - x 2 y2  xy1  a 2 .ea sin

1

.a.

1  x2

 a2 y





 1  x 2 y2  xy1  a 2 y  0  1 Using Leibnitz theorem, differentiating n times

y 1  x  nc .y 2

n2

1  x y 2

 n. yn1

n1

1.2

1  x y  xy 2n  1  y nn 1  n  a   0 1  x y  2n  1xy  y n  n  n  a   0  1  x y  2n  1xy  n  a y  0 y  sin m sin x , prove that 1  x y  2n  1xy  m Given y  sin m sin x  2

n 2

n1

n 2

7. If

n 2

 2x  nc2 . yn  2 yn1.x  nc1. yn .1  a 2 . yn  0 nn  1 . 2 x   . y . 2  xy  ny  a 2 y  0

n1

n 2

n1

1

n 2

n1



 n2 yn  0

1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

199





Diff, y1  cos m sin 1 x .m.

1  x2



1  x 2 y1  m. cos m sin 1 x



Diff,

1  x 2 . y2  y1.

1 2 x 





 m. sin m sin 1 x .

2 1  x2

m 1  x2

x 1  x2 ;

1  x y 2



 xy1  m2 sin m sin 1 x



 - m2 y





 1  x 2 y2  xy1  m2 y  0 Using Leibnitz theorem, differentiating n times,

y 1  x  nc .y 2

n2

1  x y 2

n 2



8. If y  x 

n 2

 xyn1 2n  1  yn nn  1  n  m2  0

n 2

n1



n1





 n2  m2 yn  0



y1  m x  1  x



n 2



y1  m x  1  x 2



Given, y  x  1 x 2 Diff,

n1

1.2

   2n  1xy  y n  m   0 1 x  , prove that 1  x y  2n  1xy

1  x y 1  x y 2

 n. yn1

 2x  nc2 . yn  2x yn1.x  nc1. y1.1  m2 yn  0 nn  1 .2 x  . y  2  xy  ny  m2 y  0 n1



m1



m1

1  x 2 . y1  m x  1  x 2

 12 x   .1  2   2 1 x 

 1  x2  x    2  1  x 



 my Squaring

1  x y 2

 m2 y 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

200

Diff,

1  x 2 y y 2

1 2

 2 x. y1  m2 2 yy1 2

 2y1

1  x y  xy  m y ie, 1  x y  xy  m y  0 2

Differentiating n times using Leibnitz theorem,

y 1  x  nc .y 2

n 2

1  x y 2

n 2

1  x y 1  x y 1  x y 2

n1



.2 x  nc2 . yn .2  yn1.x  nc1. yn .1  m2 yn  0

 n.2 xyn1 

nn  1 . yn .2  x. yn1  n. yn  m2 yn  0 1.2

   y n  n  n  m   0  n  m y  0

n 2

 xyn1 2n  1  yn nn  1  n  m2  0

n 2

 2n  1xyn1

n 2

 2n  1xyn1

9. If y  x n1 log x, prove that i  xy 1  n  1y  x n1 and ii  y n 

n 1 . x

Given, y  x n1 log x Diff,

1 y1  x n1.  n  1x n2 . log x x



xy 1  x n1  n  1x n1. log x xy 1  x n1  n  1y

Differentiating n  1 times using Leibnitz theorem,

yn .x  n  1e1. yn1.1  n  1  n  1 yn1 xyn  n  1yn1  n  1  n  1yn1  xyn  n  1  yn 

n 1 x









10. If y  x 2  1 , prove that x 2  1 yn2  2 xyn1  nn  1yn  0 n

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

201





Given y  x 2  1





y1  n x 2  1

Diff







n1

.2 x



 x 2  1 y1  n x 2  1 .2 x

x



 1 y1  2nxy

diff

x



 1 y2  2 xy1  2nx  y1  y 1

( x 2  1) y2  2 xy1 (1  n)  2 xy  0 differentiating n times,

[ yn2 ( x 2  1)  nC1. yn1.2 x  nC2 . yn .2]  2(1  n)[ yn1.x  nC1. yn .1]  2nyn  0

( x 2  1) yn2  n.2 xy n1 

n(n  1) . yn .2  2(1  n) xy n1  2(1  n).n. yn  2nyn  0 1.2

( x 2  1) yn2  xy n1 (2n  2  2n)  yn [n 2  n  2n  2n 2  2n]  0 ( x 2  1) yn2  2 xy n1  n(n  1) yn  0

Self assessment problems IV 1. Find the n th derivative of x cos 3x 2. Find the n th derivative of x 2 e 5 x 3. If y  e tan

1

prove that

(1  x 2 ) yn2  [2 x(n  1)  1] yn1  n(n  1) yn  0 4. If y  sin 1 x prove that

(1  x 2 ) yn2  (2n  1) xy n1  n 2 yn  0 1

5. If y m  y



1 m

 2 x prove that

( x 2  1) yn2  (2n  1) xy n1  (n 2  m2 ) yn  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

202

4.8 Meaning of the derivative In the previous section, we have calculated the derivative of many function. In this section we will learn about the meaning of the derivative. I. Geometrical Interpretation: Let p( x, y) the any point or the curve y  f (x) . Let Q( x  x, y  y) be a neighboring point on the curve. Let PM  OX , QN  OX and PR  QN . Let PQ be inline at an angle  with OX . In PQR QPR   , PR  x and QR  y

tan 

y x

As Q  P on the curve, the chord PQ will become the tangent at P and tan will be the slope of the tangent at P also x  0 .

y

Lt tan  Lt x

Q P

x0

slope of the tangent at P 

dy . dx

Meaning of the sign of the derivative. Let y  f (x)  (1) Let x be an arbitrary increment in x. Let y be the corresponding increment in y.  y  y  f (x  x)  (2)

(2)  (1);

y  f ( x  x)  f ( x)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



203

dy y  Lt dx x0 x

dy f (x  x)  f (x)  Lt dx x 0 x Put,

x  a and x  h

Then,

dy f ( a  h)  f ( a )  Lt dxatxa h0 h

Let h  0 If

dy  0 then f (a  h)  f (a)  0 ie, f (a  h)  f (a) ie, f (x) is an increasing dx

function at x  a .

dy  0 , then f (a  h)  f (a)  0 ie, f (a  h)  f (a) ie, is a decreasing dx

function at x  a .

Problem 1. Show that 2 x 3  3x 2  12 x  7  0 when x  1. Let f ( x)  2 x 3  3x 2  12 x  7

f 1 ( x)  6 x 2  6 x  12  6(x 2  x  2)  6(x  1)(x  2)

when x  1, x  1  0 and x  2  0

 f 1 ( x)  0 when x  1  f (x) is an increasing function when x  1

At x  1

f ( x)  2  3  12  7  0

 when x  1 , f ( x)  0

2. Find the range of value of x for which x 3  6 x 2  36 x  7 is an increasing function.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

204

Let f ( x)  x 3  6 x 2  36 x  7

f 1 ( x)  3x 2  12 x  36

 3( x 2  4 x  12)  3( x  6)( x  2)

when  2  x  6, f 1 ( x)  0  f (x) is a decreasing function when  2  x  6 for all other values of x ,

ie    x  2 and x  6 f 1 ( x)  0 and f (x) is increasing function. 3. For what values of x is 2 x 3  15x 2  84 x  7 a decreasing function ? Let f ( x)  2 x 3  15x 2  84 x  7

 6( x 2  5x  14)  6( x  2)( x  7)

when  2  x  7 , we get f 1 ( x)  0  f (x) is a decreasing function when  2  x  7

4. Show that x 

x2  log(1  x)  x when x  0 2

x2 Let, f ( x)  x   log(1  x) 2 f ' ( x)  1  x 



1 1 x

1  x 2 1 1 x

 f ' ( x)  

x2 1 x

when x  0 , we get f ' ( x)  0  f (x) is a decreasing function,

At x  0 f ( x)  0  0  log(1  0)  0  when x  0 , we get f ( x)  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

ie, x 

x 

205

x2  log(1  x)  0 when x  0 2

x2  log(1  x) when x  0 2

 (1)

Next let, f ( x)  log(1  x)  x f ' x   

1 1 1 x

1 1  x 1 x

f ' ( x)  

x 1 x

when x  o , we get f ' ( x)  0  f ( x)  0 when x  0

ie,

log(1  x)  x  0 when x  0

ie, log(1  x)  x when x  0

 (2)

(1)  (2) 

x

5. Show that 1 

x2  log(1  x)  x when x  0 . 2

x2  cos x  1 when x  0 2

Let f ( x)  1 

x2  cos x 2

f ' ( x)   x  sin x when x  0 we get f ' ( x)  0

(sin x  x)

 f (x) is a decreasing function At x  0 , f ( x)  1  0  1  0

 f ( x)  0 when x  0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

206

ie, 1 

x2  cos x  0 when x  0 2

ie, 1 

x2  cos x when x  0 2

 (1)

next, let f ( x)  cos x  1

f ' ( x)   sin x

| f ' ( x)  0

when x  0,

 f (x) is a decreasing function.

At x  0

f (0)  1  1  0

 f ( x)  0 when x  0

cos x 1  0 when x  0 cos x  1 when x  0

 (2)

1& 2 

1

x2  cos x  1 When x  0 . 2

Self assessment problems V 1. Prove that 3x 4  8x 3  6 x 2  24 x  19  0 when x  1 2. For what values of x, is 2 x 3  9 x 2  12 x  4 a decreasing function? 3. Show that x  log(1  x) 

x when x  0 1 x

1  4. Prove that x sin x  cos x  cos 2 x is an increasing function in 0, 2 2

Velocity and acceleration The velocity of a moving particle is defined as the rate of displacement w.r.t. time ie, velocity v 

ds when s is the position vector of the particle at time t. Next the dt

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

207

acceleration of a particle is defined as the rate of change of velocity w.r.t.t ie, Acceleration a 

d 2s dv or a  2 dt dt

1. The displacement of a moving particle is given by s  t 3  2t 2  4 find its velocity and acceleration when t  2 sec. Given s  t 3  2t 2  4 Vel

v

Acces a 

ds  3t 2  4t dt dv  6t  4 dt

When t  2 sec Velocity v  3  4  4  2  4units Acceleration a  6  2  4  8units 2. The displacement of a moving particle in t secs is given by s  t 3  9t 2  24t 18. Prove that the velocity is zero for two values of t. Find its velocity when the acceleration is zero and find the acceleration when the velocity is zero. The displacement is t secs is Given by s  t 3  9t 2  24t 18  (1) Velocity v 

ds  3t 2  18t  24  (2) dt

Acceleration a 

dv  6t  18  (3) dt

When the velocity is zero

 0 3t 2  18t  24  0





3 t 2  6t  8  0

t  2t  4  0  t  2or 4

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

208

 When t=2 secs and t=4 secs, velocity  0

when t=2, (3)  a  6  2 18  6units when t=4, (3)  a  6  4 18  6units When the acceleration is Zero

a0 6t  18  0 6t  3  0

t  3 secs When, t=3 2    3  32  18  3  24

  3 units 3. The distance described by a particles in t seconds is given by s  ae t  be t . prove that the acceleration is always equal to the distance passed over. The distance described in t second is

s  ae t  be t Velocity  

d  ae t  be t dt

Acceleration a 

d  ae t  be t dt

as

Self assessment Problems 1. A stone thrown vertically upward risesrmts in t seconds where s  48t  16t 2 . Find its velocity when t= 1 second and find the time at which it is momentarily at rest. 2. The distance described by a particle in t seconds is given by s  10  27t  t 3 . Find its velocity when t=2 seconds and its acceleration when its velocity is zero.

4.8.3. Rate of change of variables If y is a faction of x, then y  f x  Here x is the independent variable and y is the dependent variable FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

209

dy is called the rate of change of y w.r.t.x dx

Using this definition we will do problems.

1. If the rate of increase of x 3  5x 2  5x  8 is twice the rate of increase of x, what are the values of x? Let y  x 3  5x 2  5x  8 Diff

dy  3x 2  10x  5  (1) dx

Given, the rate of change of y w.r.t. x =2. ie.,

dy  2. dx

2  3x 2  10 x  5 3x 2  10 x  3  0

3x 1x  3  0 1  x  or 3 3

2. An inverted cone has a depth of 10cms and a base of radius 5 cms. Water is poured into it at the rate of 1.5 c.c per minute. Find the rate at which the level of water in the cone is rising when the depth is 4 cms. Given, Base radius=5 cms

Depth=10cms When water is pouring into the cone, at any time t, let r be the radius, h be the depth and v be the volume of the water

cone. 1 Then, v  r 2 h  1 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

210

From the figure, we get h r  10 5 r 

5 h h 10 2 2

1 h 1  v  . .  .h 3 2

v

 12

Differentiating w.r.t.t, dv  dh  .3h 2 .  (2) dt 12 dt

Given, h  4cms and  (2)  

dv 3  1.5  dt 2

3  dh   3 2 12 dt

dh 3 12  dt 2  3 16

dh 3  cm / min . dt 8  The rate at which the level of water in the cone is rising is

3 cm/min. 8

3. A balloon which always remains spherical is being inflated by pumping 10 c.c of gas per sec. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm. At any time t, let r be the radius of the balloon. 4 It volume v  rh3 3

Differentiating w.r.t. t dv 4 dr  .3.r 2 . dt 3 dt

 (1)

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II dv  10 (given) dt

when r= 15,  (1) 

211

10  4   225.

dr dt

dr 10  dt 4    225 dr 1  cm / sec . dt 90

Self assessment problems: VII 1.

A ladder of 20mts long has one end on the ground and the other in contact with a

vertical wall. Then the foot of the ladder is 16mts away from the wall, the lower end is slipping at 2mt/sec. find rate at which its upper end is moving downwards. 2.

A hollow right circular cone of height 10 cms and semi vertical angle 30  is full

of water. When water is drawn off such that the height of the water decreases at a uniform rate of 1cm/sec, find the rate at which the volume is decreasing when the height of the water is 5 cms.

4.9 Answers Self assessment problems I 1. Let y  tax  1 Let x be an arbitrary increment in x Let y be the corresponding increment in y Then y  y  tanx  x   2

2  1;

y  tanx  x   tan x



sin x  x  sin x  cosx  x  cos x



sin x  x . cos x  cosx  x .sin x cosx  x . cos x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II





212

sin x  x  x  cosx  x . cos x

y sin x  x x. cosx  x . cos x

y sin x 1  Lt . x 0 x x0 x cosx  x . cos x Lt



dy 1  1. dx cosx  0. cos x



dy 1  dx cos 2 x



d tan x   sec 2 x dx

sin     1  Lt x0  

2. Let y  e 2 x .sin 3x.cos 4 x Taking log,

log y  2 x.log ee  log sin 3x  log cos 4 x Diff, 

1 dy 1. cos 3x 1 sin 4 x  .  2.1  .3  .4 y dx sin 3x cos 4 x

dy  y2  3 cot 3x  4 tan 4 x dx

 1 x  3. Let y  sinh 1   1 x  dy  dx



1 1 x  1   1 x 

1  x  1  1  x .1 1  x 2

 1  x  1  x  1  x  .  2 2 2 1  x     1  x   1  x   -2 1  2x  x  1  2 x  x 2 1  x  2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II 

-2

x  1



213



2 x2 1

4. Let y  sin x.sin 2 x.sin 3x.sin 4 x Taking log, log y  log sin x  log sin 2x  log sin 3x  log sin 4 x

Diff,

1 dy cos x 2 cos 2 x 3 cos 3x 4 cos 4 x .     y dx sin x sin 2 x sin 3x sin 4 x dy  ycot x  2 cot 2 x  3 cot 3x  4 cot 4 x dx

 sin x  5. Let y  tan 1    1  cos x  x x  2 sin . cos   2 2 y  tan 1   x  2 cos 2  2  

x  y  tan 1  tan  2  y



x 2

dy 1  dx 2

6. Given sin y  x sina  y 

Diff` 

cos y.

dy dy  x. cosa  y .  sin a  y .1 dx dx

dy cos y  x cosa  y   sina  y  dx

 dy  sin y cos y  . cos(a  y   sin a  y   dx  sin a  y  

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



214

dy  sin a  y  y   sin a  y  dx  sin a  y  

dy sin 2 a  y    dx sin a

x

7. Given

 y2





Let f x, y   x 2  y 2



 a2 x2  y2





 a2 x2  y2



f x  2 x 2  y 2 .2 x  a 2 2 x

f dy  x dx fy

  dy x2x   dx y2x

    y   a   x2x  y   a    y   a  y2x  y   a  

dy  2 x 2 x 2  y 2  a 2  dx 2 y 2 x 2  y 2  a 2 2 2

Self Assessment Problems II 1. Let y 

1 x  5x  6 2

y

1  (1) x  2x  3

y

1 1  x2 x3

Differentiating n times successively,

 1n . n.1n  1n . n.1n yn   x  2n1 x  3n1   1 1 n y n   1 . n   n 1 x  3n1    x  2

2. Let y  cos 2 x y

1  cos 2 x  2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II y

215

1 1  . cos 2 x 2 2

Differentiating n times successively,

1    y n  0  2 n. cos n  2 x  2  2      y n  2 n1. cos n  2 x   2  3. Let y  cos 3x. cos x cos A.CosB  y

1 cos3x  x   cos3x  x  2

y

1 cos 4 x  cos 2 x 2

1 cos A  B   cos A  B  2

1  n   n   y n  4 n. cos  4 x   2 n. cos  2 x  2  4   4 

Self Assessment Problems III



1. y  sin m sin 1 x







y1  cos m sin 1 x .m

1 1 x2



1  x 2 . y1  m. cos m sin 1 x



Diff,

1  x 2 . y 2  y1 .

1 2 x  2 1 x2





 m sin m sin 1 x .m.

1 1 x2

1 x2 ,

1  x y 2



 xy1  m 2 sin m sin 1 x



 m 2 y

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



216



 1  x 2 y2  xy1  m 2 y  0 2. y  e a sin

1

y1  e a sin x .a.

1  x2

1  x 2 y1  a.ea sin

1

Diff,

1  x 2 y2  y1.

1 2 x  2 1 x

1

 a.e a sin x .a.

1 1  x2

 1  x2 ;

1  x y

 xy1  a 2 .ea sin x 1

(1  x 2 )y2  xy1  a 2 y





 1  x  y2  xy1  a 2 y  0 3. x  sin t ;

y  sin pt

dx  cos t; dt

dy  cos pt. p dt

dy dy   dt dx dx dt dy p cos pt  dx cos t dy p 1  sin 2 pt  dx 1  sin 2 t dy p 1  y 2  dx 1  x2



p2 1  y2 y  1  x2 2 1

  1  x y 2

2 1





 p2 1  y2



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

217

Diff,

1  x 2 y y 2

1 2

 y12  2 x   p 2  2 yy1 

 2 y1 ,

1  x y  xy  2 p y  1  x y  xy  p y  0 2

4. x  2 cos t  cos 2t;

y  2 sin t  sin 2t

dx  2 sin t  2 sin 2t; dt  2sin 2t  sin t 

 2.2 cos

3t t . sin 2 2

dy  2 cos t  2 cos 2t dt  22 cos t  cos 2t 

 2.2 sin

3t t . sin 2 2

dy dy   dt dx dx dt 3t t . sin 2 2  3t t 4 cos . sin 2 2 4 sin

dy 3t  tan dx 2

Differentiating w.r.t.x,

d2y 3t 3 dt  sec 2 . . 2 dx 2 2 dx d2y 3 1 1  . . 2 dx 2 cos 2 3t 4 cos 3t . sin t 2 2 2

d2y 3 3t t  . sec2 . cos ec 2 dx 8 2 2 5. Given ax 2  2hxy  by 2  1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

218

dy  dy  Diff; 2ax  2h x.  y.1  2by.  0 dx  dx  

dy hx  by   ax  hy  dx



ax  hy   1 dy  hx  by  dx

dy  dy     hx  by a.1  h   ax  hy h  b  d y dx  dx    Diff;  2 2 dx hx  by  2









 hx  by  ahx  aby  ahx  h 2 y  ax  hy  h 2 x  hby  abx  bhy hx  by 3



hx  by yh 2  ab  ax  hy xh 2  ab hx  by 3



h

 ab hxy  by 2  ax 2  hxy hx  by 3

h 

 ab ax 2  2hxy  by 2 hx  by 3











d2y h 2  ab  2  dx hx  by 3

Self Assessment problems IV 1. Let y  x. cos 3x Differentiating n times using Leibnitz theorem,

u  cos 3x;

vx

yn  D n x cos 3x 

      yn  3n. cos n  3x .x  nC1.3n1. cos n  1  3x.1 2  2    2. Let y  x 2e5 x Differentiating n times using Leibnitz theorem, FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



yn  D n x 2 e 5 x

219



u  e5 x ;

v  x2

yn  5n.e5 x .x 2  nC1 5n1.e5 x .2 x  nC2 .5n2.e5 x .2 nn  1   yn  5n2.e5 x  x 2  n.5.2 x  .2 1.2  





y n  5 n2.e 5 x x 2  10nx  nn  1 3. y  e tan

1

y1  e tan

1

1  x y 1  x y

1 1 x2

 e tan

y

1

Diff,

1  x .y 2

 y1 .2 x  y1

Differentiating n times using Leibinitz formula,

y 1  x   nC .y 2

n 2

1  x y

nn  1 yn .2  2 xy n1  2n. yn  yn1  0 1.2

 n.2 xy n1 

n1

 xyn1 2n  2  yn1  yn nn  1  2n  0

n1

 yn1 2 xn  1  1  yn n 2  n  2n  0

n1

 yn1 2 xn  1  1  nn  1yn   0

1  x y x  1y x  1y 2



.2 x  nC2  y n .2  2y n1 .x  nC1 . yn .1  yn1  0

n 2

n 1





4. y  sin 1 x

y1 

1 1  x2

1  x 2 . y1  1 Diff,

1  x 2 y2  y1

1 2 1  x2

 2 x   0

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

1  x y 2

220

 xy1  0

Differentiating n times using Leibnitz formula,

y 1  x   nC y 2

n 2

1  x y 2

n 2

1  x y 1  x y 2

1 m

y y 1

ym 



1 n1

 n.2 xy n1

 2x  nC2 . yn  2 yn1.x  nC1 yn .1  0 nn  1  . y ,2  xy  ny  0 1.2

n1

n 2

 xyn1 2n  1  yn nn  1  n  0

n 2

 2n  1xyn1  n 2 yn  0

1 m

 2x

1  2x 1 y m 1

m put y  z

z

1  2x z

z 2  2 xz  1  0

2x  4x2  4 z 2 1 m

y  x  x2 1 y  ( x  x 2  1) m



diff, y1  m x  x 2  1



y1  m x  x 2  1





m1



m1

.[1 

1 2 x2 1

.2 x]

 x2 1  x  .  x 2  1  

x 2  1. y1  m x  x 2  1



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

221

x 2  1. y1  m. y Squaring

x



 1 y1  m2 y 2 2

diff

x



 1 2 y1 y2  y1 .2 x  m2 .2 yy1 2

x

 2y1

x 1y 2



 1 y2  xy1  m2 y

 xy1  m2 y  o

Differentiating n times using leibnitz theorem,

( y

n 2



)( x 2  1)  nC1. yn1.2 x  nC2 . yn .2  yn1.x  nC1. yn .1  m2 yn  0





( x 2  1) yn2  xyn1 (2n  1)  yn n(n  1)  n  m2  0  ( x 2  1) yn2  (2n  1) xyn1  (n2  m2 ) yn  0

Self assessment problems V s  48t  16t 2

1.Given

Velocity v 

ds  48  32t dt

When t  1 , velocity  48  32.1  16 units When the stone is at rest, Its velocity  0

 48  32t  0 32t  48 t

48 32

3 t  sec s 2

2. Given

s  10  27t  t 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

Velocity

222 v

Acceleration a 

ds  27  3t 2 dt

dv  6t dt

When t  2, velocity  27  3  4  15units When the velocity is zero 27 - 3t 2  0 3t 2  27

t2  9

t  3 secs. Then, acceleration =-6  3

 -18 units

1. Self assessment Problem VI Let AB be the ladder Given OA=16 and AB = 20 B

Then, OA2  OB2  AB 2

20 B'

16 2  OB 2  20 2 OB2  20 2  16 2

OB  144 20

OB  12

When the ladder is shipping,

At any time t, let A' B' be the position of the ladder. Let OA'  x and OB'  y Then, x 2  y 2  20 2 Diff 2 x.

dx dy  2 y.  0 dt dt

1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

223

Given, x  16, y  12. 1  16.2  12.

dx 2 dt

dy 0 dt

dy  32 dt

dy  32  dt 12 dy 8   mt / sec . dt 3

The upper end is moving downwards at

8 mt / sec . 3

2. Given, Height of the cone = 10 cms

  300

Semi vertical angle

When the water is drawn off, At any given time t, Let h be the height and r be the radius of the water cone. 1 Its volume v  r 2 h 3 2

1  h  v     .h 3  3 

v

 g

tan 30 0 

1 r  3 h r

h 3

r h

h 300

10 cms

Differentiating w.r.t. t,

dv  2 dh  .3h . dt g dt Given, h=5; 1 

 (1)

dh  1 dt

dv    52   1 dt 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II dv 25    c.c/sec. dt 3  The volume is decreasing at

25 c.c/sec . 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

224

MATHEMATICS-II

225

UNIT V

5.0

Introduction

5.1

Objective

5.2

Partial Differentiation

5.3

Eulerâ&#x20AC;&#x2122;s Theorem

5.4

Total differential coefficient

5.5

Implicit Functions

5.6

Curvature

5.7

Radius of curvature in Cartesian Co-ordinates

5.8

Radius of curvature in Polar co-ordinates

5.9

Answers to Self assessment Problems

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

226

5.0 Introduction

In this unit you will learn a new concept known as partial differentiation. Also you will learn Euler’s theorem and how to apply it to solve many problems. You will also know about the total differential coefficient. You will also learn bow to differentiate an inplict function. Next you will know the curvature of a given curve of a given curve both in caritisian and polar co-ordinates.

5.1 Objective

After completing this unit, you will be able to  Find the partial derivatives of a function of many variables.  Apply Euler’s theorem and solve many problems  Find the total differential coefficient  Find the derivative of an implicit function  Calculate the radii of curvature of different curves both in Cartesian and polar coordinates.

5.2 PARTIAL DIFFERENTIATION

In unit 4, we have seen the differentiation of functions of one variable. In this unit, we will consider the differentiation of functions of two or more variables.

Let u be a function of two independent variables x and y. ie, u= f (x,y).

The derivative of u w,r,t,x . when x varies and y remains constant is called the partial derivative of u w, r, t, x. It is denoted by

u . x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

227 f (x  x, y)  f (x, y) u = Lt  (1) x x 0 x

Similarly , when x remains constant and y varies, the partial derivatives of u w,r,t,,y is

f ( x1 y  y)  f ( x1 y ) u = Lt  (2) y 0 y y

Generally

u u and are functions of x and y .  They may be y x

differentiated again partially w,r, t , x or y . Then we get the higher order partial derivatives

 2 u  2u  2 u , , , ........ x 2 xy y 2

Problems 1. If u = log ( x 3  y 3  z 3  3xyz ) , show that

u  u 3 u + + = y  z x yz x Given , u = log ( x3  y 3  z 3  3xyz ) Differentiating partially w,r,t, x

1 u (3x 2  3 yz )  (1) = 3 3 ( x  y  z 3  3xyz ) x III ly Diff u partially w,r,t, y,

u 1 (3x 2  3xz )  (2) = 3 3 3 y ( x  y  z  3xyz ) FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II Diff u partially w,r,t,z

1 u (3x 2  3xy )  (3) = 3 3 3 ( x  y  z  3xyz ) z (1)+(2)+(3);

u  u 1 u [3x 2  3xy  3 y 2  3xz  3z 2  3xy ] + + = 3 3 3 y  z ( x  y  z  3xyz ) x 3( x 2  y 2  z 2  xy  yz  zx) = ( x  y  z )( x 2  y 2  z 2  xy  yz  zx) =

3 x yz

2. If u= log (tan x+tan y+tan z), show that Sin 2x.

u u u +sin2y . +sin2z . =2. y x z

Given u = log (tanx+tany+tanz) Differentiating partially w,r,t,x

1 u = . sec2 x  (1) tanx  tany  tanz x

Similarly

u 1 = . sec2 y  (2) y tanx  tany  tanz

1 u = . sec2 z  (3) tanx  tany  tanz z (1)  sin 2 x  (2)  sin 2 y  (3)  sin 2 z;

sin 2 x.

u u u + sin 2 y . + sin 2 z . y x z =

1 2 [ sin 2 x. . sin 2 x. + sin 2 y . sin 2 y + sin 2 z. sin z ] tan x  tan y  tan z

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

228

MATHEMATICS-II

1 tan x  tan y  tan z

[2 sin x. cos x. =

229

1 1 1  2 sin y. cos y. .  2 sin z. cos z. ] 2 2 cos x cos y cos 2 z

1 2(tan x  tan y  tan z ) tan x  tan y  tan z

=2. 3. If u = ( y  z)( z  x)( x  y), show that Given u=

u  u u + + =0 y  z x

( y  z)(z  x)(x  y)

Differentiating partially w,r,t,x, u = ( y  z )[(z  x).1  ( x  y).(1)] x

u = ( y  z )( z  x)  ( y  z )( x  y )  (1) x

///

u = ( z  x)( x  y)  ( z  x)( y  z )  (2) y

u = ( x  y)( y  z )  ( x  y)(z  x)  (3) z (1)+(2)+(3);

u u u (y  z)(z  x)  (y  z)(x  z)  (z  x)(x  y)  (z  x)(y  z) + + = (x  y)(y  z)  (x  y)(z  x) x y z =0

4. If u =

1 x2  y2  z 2

u  (x2  y2  z 2 )

1

Differentiating partially w,r,t, x,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

230

3 u  1 = ( x 2  y 2  z 2 ) 2 2 x 2 x

Again differentiating partially w,r,t,x, 3 5  2u 2 2 2 2 .1  3 ( x 2  y 2  z 2 ) 2 .2x  ( x  y  z ) = 2 x 2

 2u  ( x 2 = x 2

y z ) 2

3

 3x ( x  y  z ) 2

5

 2u 3 2 5 =  u  3x u  (1) 2 x ///ly

 2u 3 2 5 =  u  3 y u  (2) 2 y

And

 2 u =  u 3  3z 2u 5  (3) z 2

(1)+(2)+(3); 2  2u  u  2u 3 5 2 2 2 + + =  3u  3u ( x  y  z ) 2 2 2 x y z

 3u 3  3u 5 .u 2 )

= 3u

 3u3

[u  ( x 2  y 2  z 2)

1

u 2  ( x 2  y 2  z 2 )1 ]

 2u  x 2

= 0

5. If r  x  y , 2

show that

 2 r  2 r = 1 ( r ) 2  ( r ) 2    y  x 2 y 2 r  x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II Given, r  x  y 2

231 2

Differentiating partially w,r,t,x,

r  2x x r x    (1) x r r y /// ly   (2) y r 2r.

In (1), differentiating partially w,r,t,x,

 r  x 2 2

r.1  x.

r x

r2 

r  x.

x r

 2r r 2  x2   (3) x 2 r3

 2r r 2  y 2   (4) /// y 2 r3 ly

2  2r  r r 2  x2 r 2  y 2 Now, + +  x 2 y 2 r3 r3

2r 2  ( x 2  y 2 ) = r3 2r 2  r 2 = r3 =

r2 r3

1 =  (5) r

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

232

2 2 2 2 1  r   r   1  x   y   Next,       =       r  x   y   r  r   r    

1  x2  y2   =  r  r 2  1 r2  =  2  rr  =

1  (6) r

(5)and(6)  2 2  2 r  2 r 1  r   r   + =       x 2 y 2 r  x   y  

6. If Z= x 2 tan 1

y x  y 2 tan 1 , prove that x y

2z x  y = xy x 2  y 2 2

Given,

Z = x 2 tan 1

x x  y 2 tan 1 y y

Differentiating partially w, r, t, y,

  z 1 1  2. 1 2 x .   y y y 2 x  x2 1 2 1   x y2 



 x    2   2 y. tan 1  y 

  x y  

x3 xy 2 1 x   2 y tan y x2  y2 x2  y2





x x2  y2 1  x   2  2 y tan   x  y2  y FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

z y

233

 x  x  2 y tan 1    y

Differentiating partially w, r, t, x,

2z  1  2 y . xy

1 x2 1 2 y

1 y

2y2 1 2 x  y2 2z x2  y2  2 y2  xy x2  y2

x2  y2  2 x  y2 7. If u  tan( y  ax)  ( y  ax)

show that

2  2u 2  u  a x 2 y 2

Given, u  tan(y  ax)  (y - ax)

Differentiating partially w.r.t.x, 1 u 3  sec2 ( y  ax).a  ( y  ax) 2 , (a) x 2

Differentiating again partially w.r.t.x, 1  2u 3 1 2 2 (a )  a . 2 sec ( y  ax ). sec( y  ax ). tan( y  ax ). a  a . ( y  ax ) 2 2 x 2 1  2u 3 2 2 2 2  (1)  2 a sec ( y  ax ). tan( y  ax a ( y  ax ) 2 4 x

Next, differentiating u partially w.r.t.y,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II 1 u 3  sec2 ( y  ax).1  ( y  ax) 2 .1 y 2

Differentiating again partially w.r.t.y, 1  2u 3 1 2  (1)  2 sec( y  ax ). sec( y  ax ). tan( y  ax )  . ( y  ax ) 2 2 2 y 2  2u 2  u (1) and (2)  a x 2 y 2

 2u  2u  8. If u  x sin y  y sin x, prove that xy yx Given,

u  x sin y  ysin x Differentiating partially w.r.t.x,

u  sin y  y. cos x x Differetiating partially w.r.t.y,

 2u  cos y  cos x  (1) xy Next, differentiating u partially w.r.t.y,

u = x cos y  y. cos x x Differentiating partially w.r.t.x,

 2u  cos y  cos x  (2) xy (1) and (2) 

 2u  2u  yx xy

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

234

MATHEMATICS-II

9. If u 

xz , prove that 2 x  y2

Given, u 

235

2u  x 2  0

xz , x2  y2

Differentiating partially w.r.t.x,

u ( x 2  y 2 ).xz.2 x  x (x2  y2 )

 2u z( y 2  2 )  x (x2  y 2 )2  2u ( x 2  y 2 ) 2 (2 xz )  z ( y 2  x 2 ).2( x 2  y 2) .2 x   x (x2  y 2 )4 

(x 2  y2 )2xz  4xz(y 2  x 2 ) (x 2  y2 )3

2 2 2 2  2 u 2 xz[ x  y  2 y  2 x )  ( x 2  y 2 )3 x 2

 2 u 2xz[x 2  3y2 ]   (1) (x 2  y 2 )3 x 2 Next, differentiating u partially w.r.t.y,

u (1)  xz. 2 .2 y 2 x (x  y 2 )2  2u ( x 2  y 2 ) 2 (2 xz )  2 xyz.2( x 2  y 2) .2 y  y 2 (x2  y 2 )4 

 ( x 2  y 2 )2 xz  8 xy 2 z ( x 2  y 2 )3



2 xz[ x 2  y 2  4 y 2 ] ( x 2  y 2 )3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

236

2 2  2 u 2 xz[ x  3 y ]   (2) ( x 2  y 2 )3 x 2

Next, differentiating u partially w.r.t.z,

u x  2 z x  y 2 Differentiating again partially w.r.t.z,

 2u  0  3 z 2 (1)+(2)+(3)  2 2 2 2 2  2 u  u  2u 2 xz ( x  3 y )  2 xz ( x  3 y ) 0 + + = ( x 2  y 2 )3 x 2 y 2 z 2

=0 10. If u  e ( x cos y  y sin y), show that x

2u 2u  0 x 2 y 2

Given, u  e ( x cos y  y sin y),  (1) x

Differentiating partially w.r.t. x u  ex (cos y)  e x (x cos y  ysin y) 2 x

Differentiating again partially w.r.t.x,

 2u  e x cos y  e x ( x cos y  y sin y)  e x (cos y) 2 x

 2u  e x [2 cos y  x cos y  y sin y ]  (2) 2 x Next, differentiating u partially w.r.t.y,

u  e x [ x sin y  { y. cos y  sin y.1}] y Differentiating again partially w.r.t.y,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

237

 2u  e x [ x cos y  { y sin y  cos y  cos y}] 2 y  2u  e x [ x cos y  y sin y  2 cos y ] 2 y  e x [ x cos y  y sin y  2 cos y] 2  2u  u (1)+(2)  + =0 x 2 y 2

Self assessment problems I 1. If u  log( x  y  z ) find 2

u

 x

 2r  2r  2r 2   2. If r  ( x  a)  ( y  b)  ( z  c) , show that 2  x y 2 z 2 r 2

3. If v  (1  2 xy  y ) 2

1

prove that x

v v y  y 3v 3 x y

4. If u  x ( y  z )  y ( z  x)  z ( x  y), prove that 2

u

 x  0

 2u  2u  5. If u  x , prove that xy yx y

5.3 Euler’s Theorem If f(x, y) is a homogeneous function of degree n, then x

f f y  nf x y

Proof: Let f(x,y) be a homogeneous function of degree n in x and y. Let

f ( x, y)  a0 x n  a1 x n1. y  a2 x n2. y 2  ....... an y n

where

a 0 ,a1 ,a 2 ................a n are constants.

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

238

 y

ie, f ( x, y )  x n. f  x   (1) Differentiating partially w.r.t.x,

x f n 1  y   y  x .F  .  2   n.x n 1.F    (2) x x x   y Differentiating (1) partially w.r.t.y,

f  y 1  x n .F 1  .  (3) y x x (2)  x  (3)  y x

x

f f +y   x n 1. yF 1  y   n.x n .F  y   x n 1. y.F 1  y   x   x   x  x y

f f y + y = n.x n .F    x  x y = n f.

Problems 1. Verify Euler’s theorem for u  x  y  z  3xyz 2

Given, u  x  y  z  3xyz 3

U is a homogeneous function of degree n=3.

u  3x 2  3 yz x u  3 y 2  3xz y

u  3z 2  3xy z FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

x

239

u u u y z  3x 3  3xyz  3 y 3  3xyz  3z 3  3xyz x y z = 3( x 3  y 3  z 3  3xyz ) = 3 u.

Thus Euler’s theorem is verified. 2. Verify Euler’s theorem for u 

u

1 x  xy  y 2 2

1 x 2  xy  y 2

This is a homogeneous function of degee n=-2.

u 0  (2 x  y )  2  (1) x ( x  xy  y 2 ) 2

u 0  (2 y  x)  2  (2) y ( x  xy  y 2 ) 2 x

u u 2x 2  xy  2y 2  xy y  x y (x 2  y2  xy) 2 2(x 2  xy  y 2 ) 1   2. 2 2 2 2 (x  y  xy) x  xy  y 2 =-2.u.

Thus Euler’s theorem is verified. 3. If u  sin

1

x2  y2 u u , show that x  y  tan u. x y x y

 x 2  y2  Given u  sin 1  ,  xy 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

240

x2  y2  sin u  . x y R.H.S is a homogeneous function of degree f= sin u

n= 2-1 =1.

 By Euler’s theorem,

x x

u u y  nf x y

  (sin u )  y (sin u )  n. sin u x y

x. cos u

u u  y cos u.  1. sin u y y

u u sin u y  x y cos u

x

u u y  tan u x y

4. If u  sin

x y u u show that x y 0 x y x y

1

Given, u  sin

 sin u

1

x y x y

R.H.S is a homogeneous function of degree

n 1  1 0 2 2  By Euler’s theorem,

f  sin u

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

241

f f y  nf x y

(sin u ) (sin u ) y  o. f x y

cosu x

u u  cos u. y o x y

:- cosu,

u u y o x y 3

5. If u = tan

1

x y 2 u u   , prove that x  y 0 x y x y 3

 x  y 2 Given, u  tan 1   xy

tan u 

( x  y) ( x  y)

, 2

R.H.S is a homogeneous function of degree n= 3 2 - 3 2 =0 By Euler’s theorem, x

f f y  nf x x

(tan u ) (tan u ) y  o. f x x

sec2 . x

:-sec 2 u ;

f=tan u

u u  sec 2 u. y o x x x

u u y 0 x x

 x3  y 3  u u , prove that x 6. If u= tan  y  sin 2u. x x  x y  1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

242

 x3  y 3   Given, u  tan 1   x y 

tan u 

x3  y 3 x y

R.H.S is a homogeneous function of degree n=3-1=2

By Euler’s theorem, x

f f y  nf x x

(tan u ) (tan u ) y  2 tan u x x

sec2 u. x

u u  sin 2 u. y  2 tan u x y

:- sec2 u. x

f f 2 tan u y  x x sec2 u

sin u cos2 u   2sin u.cos u =2 cos u 1 = sin 2u. 7. If u  sin 1

u u 1 x y  tan u. ,prove that x  x y 2 x y

Given, u  sin 1

Sin u=

x y x y

R.H.S is a homogeneous function of degree

n  1 1  1 2 2.

By Euler’s theorem,

f = sin u

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

243

f f  y  nf x y

(sin u ) (sin u ) 1 y  sin u x y 2

cos u.x

x

u u 1  cos u. y  sin u x y 2

f f 1 sin u y  . x y 2 cos u 1 = tan u . 2

Self assessment problems II  x2  y2  , show that 1. if u  tan 1   x y 

f f 1  y  sin 2u. x y 2

f f  x2  y 2  1 , prove that x  y 2. If u  log x y  x y  x x f f 3. If u  sin 1    tan 1  , show that x  y  0 . x y  y  y

5.4 Total differential coefficient Let u be a function of two variables x and y .Let x and y be functions of another variable t . Ie, u  f ( x, y); x=f 1 (t) and y=f 2 (t). Then , the total differential of u is

du 

u u .dx  .dy x y

Total differential coefficient

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

244

du u dx u dy  .  . . dt x dt y dt

1. If u  2 x 2  3xy  y 2 , x  et , y  e 1 find

du . dt

x  et and y  e t dx dy  et and  e t dt dt

du u dx u dy  .  . . dt x dt y dt = 4 x  6 y.et   3x  2 y.(e1 )



 



= 4et  6et .et   3et  2et .(e1 ) = 4e2t  6  3  2e2t du = 4e2t  2e2t  3 dt

2. If u  xyz and x  et , y  e 1 sin t , z  sin t find

du . dt

u  xyz x  e t ;

y  e t ;

z  sin t

dx dy  e  t ;  e  t .cos e  t sin t; dt dt du u dx u dy u dz  .  .  . dt x dt y dt z dt

dz  cos t dt

 yz(e  t )  xz.[e  t cos t  e  t sin t]  xy.cos t  e t sin t.sin t.e  t  e  t .sin t[e  t cos t  e  t sin t]  e  t .e  t sin t cos t  e2 t sin 2 t  e2t sin t.cos t  e 2t sin 2 t  e 2t .sin t.cos t  2e2t .sin t(cos t  sin t) 3. If u  x 2 y 3 z 4

and x  t; y  t 2 ;

z  t 2 ; find

du dt.

Given u  x 2 y 3 z 4

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II xt

y  t2

;

245 ;

z  t 3;

dx dy dz 1 ;  2t ;  3t 2 dt dt dt du u dx u dy u dz  .  .  . dt x dt x dt x dt  2 xy 3 z 4 .1  x 2 .3 y 2 .z 4 .2t  x 2 y 3 .4 z 3 .3t 2  2t.t 6 t 12  t 2 .t 4 .t 12 .t  12t 2 t 6 .t 9 .t 2 du  20t 19 dt

4. If u  log( x  y  z )

andx  cos t; y  sin 2 t; z  cos 2 t

find

du . dt

Given u  log( x  y  z) x  cos t

;

y  sin 2 t

;

z  cos 2 t

dx dy dz   sin t;  2sin 2 t.cos t ;  2sin t.cos t dt dt dt du u dx u dy u dz  .  .  . dt x dt y dt z dt

1 1 1 .( sin t)  .(2sin t.cos t)  .( 2sin t.cos t) xyz xyz xyz 1 (1sin t)  cos t  sin 2t  cos 2 t du  sin t  dt cos t  1 

Self assessment problems: III 1. If u  x 2  y 3 2. If u 

x y  y x

and x  1  t , y  1  t find

and x  et ; y  t find

du dt

3. If u  xyz and x  t; y  t ; z  e t find

du dt

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II 4. If u  x 2  y 2

246 and x  t 3 , y  t  t 2 find

5. If u  x 2 y 3 and x  1  t , y  1  t

find

du dt du dt

5.5 Implicit Functions

Consider the function f (x  y)  c, constant. Here the values of y is given implicity interms of x . Differentiating (1), we get f f y  . 0 x y x 

ie,

f dy   x f dx y dy fx  dx fy

Note: If u  f (x, y)

and

1. If x  y  3axy 3

find

y  g(x),

then

du u u y   . . dx x y x

dy . dx

Set f ( x, y)  x  y  3axy 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

247

f x  3x 2  3axy f y  3y 2  3ax dy fx  dx fy  3x 2  3ay    2   3y  3ax  dy ay  x 2  dx y 2  ax 2. If x  y  c, y

find

dy dx

Set f ( x, y)  x  y  c. y

f x  y.x y 1  y x .log y f y  x y .log x  x.y x 1 dy fx  dx fy dy (yx y 1  y x log y)  y dx (x log x  xy x 1 )

3. If ax  2hxy  by  2 gx  2 fy  c  0 find 2

dy dx

let f ( x, y)  ax  2hxy  by  2 gx  2 fy  c 2

 2ax  2hy  2 g

 2hx  2by  2 f

dy dx



fx fy

( 2ax  2hy  2 g ) ( 2hx  2by  2 f ) dy (ax  hy  g )   dx (hx  by  f ) 

2 2 dy 4. If x  y  1 , find 2 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

248

2 2 Let f ( x, y )  x  y  1 2 2

2x a2 2y fy  2 b f dy  x dx fy fx 

2x 2 a 2y b2 b2 x  2 a y 5. If ( x  y ) 2 3  ( x  y ) 2 3  a 2 3 let f ( x, y)  ( x  y)

 ( x  y)

f x  2 ( x  y) 3 f y  2 ( x  y) 3 dy fx  dx fy

1

find 3

a

dy . dx

 2 ( x  y) 3  2 ( x  y) 3

1 1

1

2 ( x  y) 3  2 ( x  y) 3 3  3 1 1 2 ( x  y) 3  2 ( x  y) 3 3 3 1

1

dy ( x  y) 3  ( x  y) 3  1 1 dx ( x  y) 3  ( x  y) 3 6.If u  x  y  a 2

Given u  x  y  a 2



where x 3  y 3  a 3 find 2

and

du dx

x3  y 3  a3

du dy dy and 3x 2  3 y 2  2 x  2 y. 0 dx dx dx

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

249

dy x2   2 dx y  x2   2x  2y   2   y  2x 2  2x  y 2xy  2x 2 y 2x(y  x)  y 

7. If u  x 3 y 3 where x 2  xy  y 2  a 2 Given u  x . y 3

du dy  x 3 .2y.  3x 2 y 2 dx dx

and

find

du . dx

x 2  xy  y 2  a 2

dy  dy  and 2x   x  y.1   2y 0 dx  dx  dy (2y  x)  y  2x dx dy y  2x  dx 2y  x

 y  2x  2x 2 y(y  2x)  3x 2 y 2 (2y  x) 2 2  2x y    3x y  2y  x  2y  x  2x 3 y 2  4x 4 y  6x 2 y 3  3x 3 y 2  2y  x 3

 

6x 2 y3  4x 4 y  x 3 y 2 2y  x

du x 2 y(6y 2  4x 2  xy)  dx 2y  x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

250

5.6 CURVATURE Let P be any point on the curve

y  f (x). Let the

y y=f(x)

tangent at P make an angle

 with x axis. Let A be a fixed point on the curve from which arc distances are measured.

s A

Let AP  s Then the rate of change of w,r,t,s is called the curvature =

d ds

 x

The reciprocal of the curvature is called the radius of curvature. It is denoted by P.



ie,

ds . d

Cartesian formula for the radius of curvature Let P be any point on the curve

y  f (x) .

y y=f(x)

Let A be a fixed point on the curve

From which one distances are measured Let AP  s Let the tangent at p make an angle  with

s A

x =axis The

dy  tan  dx

 O

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

251

Self assessment problems IV 1. If ax 2  2hxy  by 2  1

find

2. If (sin x) tan y  (tan y ) sin x 3. If u  sin( x 2  y 2 )

where

find

dy dx

x2 y2  1 a2 b2

find

dy . dx

On the curve form which arcual distances are measured. Set Now,

AP=S dy  tan dx

Differentiating w.r.t..x

d2y d  sec 2  . 2 dx dx d ds  sec 2  . . ds dx d  sec 2  . . sec ds d  sec3  . ds  (sec 2  )

[cos 

dx ] ds

2 3

 (1  tan 2  ) 2 .   dy  2  1     ds   dx     d2y d dx 2

d ds 3

i.e., Radius of curvature at P is (1  y12 )  y2

Note: If

x  f (t ) and y  g (t )

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

252 3

(x1  y1 ) 2   1 11 1 11 x y y x 2

Problems 1. Find the radius of curvature at the point (0,1) on the curve y  e x. Given equation of the curve is

y  ex dy  ex dx d2y  ex dx 2



At the point P (0,1),

dy  e0  1 dx d2y  e0  1 2 dx Radius of curvature at P is

  dy  2  1       dx    d2y dx 2 (1  12 )  1

 2

 x1 . x

 x x)

 2 2 2. Find the radius of curvature at (1,1) on the curve xy=1 The given equation of the curve is xy  1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

253

y 1

x dy 1  2 dx x d2y 2  3 2 dx x

At the point P (1,1),

dy 1    1 dx 1 2 d y 2  2 2 dx 1 Radius of curvature at P(1,1) is

(1  y12 )  y2 [1  (1) 2 ]  2

2 2  2

 2 x y2 3. Show that the radius if curvature at any point (x,y)on the curve y  c cosh is c c The given equation of the curve is

y  c cosh

x c

dy x 1  c sinh . dx c c 2 d y x 1  cosh . 2 dx c c Radius of curvature at an point P (x,y) is

(1  y12 )  y2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

254

(1  sinh 2 x ) c  x 1 cozh . c c

(cosh 2 x ) 2 .c c  x cozh c



c cosh 3 x cosh x

  c cosh 2 x c

 

c 2 cosh 3 x

( y  c cosh x ) c

y c

4. Show that the radius if curvature at any point on the curve y  a log (sec x ) is

a(sec x ) . a The given of the curve is y  a log (sec x ) a

dy 1 x x 1  a. . sec . tan . x dx a a a sec a

dy  tan x a dx d2y 1  sec 2 x . . 2 a a dx The radius of curvature at auy point (x,y) is

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II (1  y12 )  y2

255

(1  tan 2 x ) 2 a  2 x sec . 1 a a 3 (sec 2 x ) 2 .a a  2 x sec a sec3 x  a 2 a sec x a   a sec x a

5. Find the radius of curvature at

(a,2a) on the parabola y 2  4ax .

The given equation of the curve is

y 2  4ax y  2 a. x 1 dy 1  2 a.  a .x 2 dx 2 x

3 d2y 1 ).x  2  a .(  2 dx 2

At the point P(a,2a)

dy 1  a. 1 dx a 3 d2y 1 2   . a . a 2 2 dx 1 1   . a. 3  2 a 2 1  . 2a

The radius of curvature at P(a,2a) is

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

(1  y1 )  y2

256

(1  12 ) 2  ( 1 ) 2a 3

  2  2a 2

[ is

always

 ve]

  4 2a. 6. Find the radius of curvature at (1,0) on the curve y  The equation of the curve is y 

log x x

x. 1  log x.1 x x2 1  log x y1  x2 x 2 ( 1 )  (1  log x).2x x y2  x4 y1 

At the point P (1,0) y1 

1  log 1 1  0  2 12 1

y1  1 12 (1)  (1  log 1).2.1 14  1  (1  0).2 y2  1 y2  3 y1 

Radius of curvature at P (1,0) is

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

(1  y12 )  y2 (1  12 )  3 2 2  3

257

(Radius of curvature is always positive)

7. Find the radius of curvature at x  

on the curve

y  4 sin x  sin 2 x

The equation of the curve is y  4 sin x  sin 2 x

y1  4 cos x  2 cos 2 x y 2  4 sin x  4 sin 2 x At x   2

y1  4 cos

 2 cos 2 y1  4.0  2(1) y1  2 Next y2  4 cos

 4.sin 

y 2  4.1  4.0 y 2  4. Radius of curvature is (1  y12 )  y2

(1  2 2 )  4 5 5  4

8. Show that the radius of curvature at any point on the cycloid x  a(  sin  ), y  a(1  cos  ) is 4a cos  . 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

258

The given equation of the curve is x  a(  sin  )

y  a(1  cos  )

;

x'  a(1  cos  )

;

y '  a sin 

x"  a sin 

;

y"  a cos 

Radius of curvature 3

( x'2  y '2 ) 2  x' y" y ' x" 3

[a 2 (1  cos  ) 2  a 2 sin 2  ] 2  a (1  cos  ).a cos   a sin  .a sin  (a 2 ) 3 [1  2 cos   cos 2   sin 2  ]  a 2 [cos   cos 2   sin 2  ] a 3 [1  2 cos   1]  a 2 [cos   1] a[2(1  cos  )]  [1  cos  ]

 a 2 2 (1  cos  )

[1  cos 2 A  2 cos 2 A] 1

 a.2 2 (2 cos 2  ) 2  a 2 2 . 2 . cos     4a cos  . 2

Another method The equation of the curve is

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II x  a (  sin  )

259 y  a (1  cos  )

;

dx  a(1  cos  ) ; d dy dy  d dx dx d a sin   a(1  cos  ) 

dy  a sin  d

2 sin  . cos  2 2 2 2 cos 2

dy  tan  . 2 dx

Differentiating

w, r, t, x

d2 y 1 d  sec 2  . . 2 2 dx dx 1 1  cos 2  .2 2 a(1  cos ) 2 

cos  .2 2 cos 2  2 2 1  . 4a cos 4  2 2

Radius of curvature   dy 2  1       dx     d2 y dx 2

(1  tan 2  ) 2 2  1 4 4 cos  2. 4a cos 4  2 3

  (sec 2  ) 2  4a cos 4  2 2 2  4    sec  4a cos 2 2    4a cos . 2 3

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

260

 on the curve x  a cos3  , y  a sin 3  .

9. Find the radius of curvature at any point The equation of the curve is

x  a cos3 

y  a sin 3  .

;

dx  a. 3 cos 2  ( sin  ) d d 2x  3a cos 2  . cos   sin   2 cos  . sin   2 d







 3a cos  cos 2   2 sin 2 



dy  a.3 sin 2  . cos  d



d2y  3a sin 2   sin    cos  .2 sin  . cos  2 d



 3a sin   sin 2   2 cos 2 



Radius of curvature

x'  y'   2

2 32

x' y" y ' x"



9a cos  . sin   9a sin  . cos    . sin  .3a sin   sin   2 cos    3a sin  . cos  .3a cos  cos 4

 3a cos 2

  2 sin 2  

9a 2 cos 2 .sin 2   cos 2   sin 2  3 2     2 2  2 2 2 2 9a sin .cos    sin   2cos   cos   2sin 2 

3 a 2



cos 2 .sin 2  

  32 a 2 cos 2 .sin 2    cos 2   sin 2  

  32 a  sin 2  . cos 2  

  3a sin  . cos  10. Show that the radius of curvature at any point on the curve x  acos    sin  ; y  asin    cos   is a

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

261

The given equation of the curve is x  acos    sin  ; y  asin    cos  

dx  a sin    cos   sin   d

d 2x  a  sin    cos   d 2 dy  acos     sin    cos   d

 a sin 

d2y  a . cos   sin   d 2 Radium of curvature 3

( x'2  y '2 ) 2  x' y" y ' x"

a 



cos 2  2  a 2 2 sin 2   a cos  .a cos   sin    a sin  .a  sin   cos   2

a  cos

  sin 2    2 a   cos 2   cos  sin    sin 2   sin  . cos   2



a 3 3 a 2 2 cos 2   sin 2 





  a





11. Find the radius of curvature at at 2 ,2at on the parabola y 2  4ax The equation of the curve is y 2  4ax In parametric co-ordinates x  at 2

;

y  2at

x' 2at

;

y' 2a

x" 2a

;

y" 0

Radius of curvature

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

x'  y'  

2 32

x' y" y ' x"

2at  



 2a  2at.0  2a.2a

2 32

4a t 



1  4a 2

32 3  2a  t 2  1 

 4a 2





8a  t 2  1   4a 2

  2at 2  1

12. Find the radius of curvature at any point  on the curve

   x  a log tan  ; y  a sec .  4 2    x  a log tan  ; y  a sec  4 2 x'  a.

   1 .sec 2   .     4 2 2 tan    4 2 1

   cos   1  4 2 . x'  a.       2 sin    cos 2     4 2  4 2 x' 

a    sin 2    4 2

x'

a   sin     2 

x' 

a cos 

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

262

MATHEMATICS-II

263

x'  a sec x" a sec . tan Next, y'  a sec . tan



y"  a sec .sec 2   tan .sec . tan



y"  a sec sec 2   tan 2 



Radius of curvature



x'  y' 

2 32

x' y" y ' x"

a sec   a sec  . tan   a sec .a sec sec   tan    a sec . tan .a sec . tan a sec  1  tan    a sec   a sec sec   tan    a sec . tan .a sec . tan a sec  1  tan    2



a sec   a sec  . tan   a sec2  . tan 2  2

a 

sec2  .sec2  a 2 sec4 



  a 2 sec4  2 1

  a sec2  13. Show that radius of curvature at a,0 on the curve xy

 a 3  x 3 is 3a 2

The equation of the curve is

xy 2  a3  x3 x.2 y 2 xy

dy  y 2 .1  3x 2 dx

dy  3x 2  y 2 dx





dy  3x 2  y 2   1 dx 2 xy

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

264

At P (a, 0)



dy  3a 2  0  dx 2a.0



dy  dx



dx 1  0 dy 

1 , dx   dy

2 xy 3x 2  y 2





Differentiating w.r.t.y, 2

d x  dy 2

3x

  dx  dx   y 2  2 x  2 y   2 xy 3x  2 y  dy    dy  2 2 2 3x  y







At P (a, 0)





d 2 x 3a 2  0  2a  2 y.0  2a.0  2 dy 2 3a 2  0







 6a 3 9a 4



2 3a

Radius of curvature   dx  2  1       dy    d 2x dy 2



1  03 2  2     3a 

3a . 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

265

14. Find the radius of curvature at (–2, 0) on curve y 2  x3  8 The equation of the curve is

y 2  x3  8 2y

dy  3x 2 dx

dy 3x 2   1 dx 2 y At P(–2, 0)

dy 3 2  dx 2.0

dy  dx

dx 1  0 dy 

1 

dx 2 y  dy 3x 2

Differentiating w.r.t. y 2

d x  dy 2

3x 2 .2  2 y.6 x.

dx dy

9x4

At P (–2, 0) d 2 x 6 2  0  4 dy 2 9 2 2



24 9  16



1 6

The radius of curvature

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II   dx  2  1       dy    d 2x dy 

1  0  

266

1   6

6

15. Show that the radius of curvature at (a, 0) on the curve xy 2  a 2 a  x  is

a 2

The equation of the curve is

xy 2  a 2 a  x  Differentiating w.rt.x x.2 y

2 xy

dy  y 2 .1  a 2 .1 dx

dy   y2  a2 dx





dy y2  a2   1 dx 2 xy At (a, 0)



dy  0  a 2  dx 2 xy



dy  dx



dx 1  0 dy 

1 ,

dx  2 xy  2 dy y  a 2

Differentiating w.r.t. y,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

d x  dy 2

y

267

 dx   a 2  2 x  2 y   2 xy.2 y dy   2 y2  a2







At P (a, 0)





d 2 x 0  a 2  2a  0  0  2 dy 2 o  a2





d 2 x  2a 3  4 dy 2 a d 2x 2  2 dy a The radius of curvature at P (a, 0) is   dx  2  1       dy    d 2x dy 2 



1  03 2 2    a 

a 2

Self Assessment Problems V

a a 1. Find the radius of curvature at  ,  on the curve 4 4

x y a

 3a 3a  2. Find the radius of curvature at  ,  on the curve x 3  y 3  3axy  2 2 3. Show that the radius of curvature at any point  on the curve x  3a cos   a cos 3 , y  3a sin   a sin 3 is 3a sin 

4. Find  at ‘t’ on the curve x  6t 2  3t 4 , y  8t 3 5. Find  at ' ' on the curve x  a logsec  tan , y  a sec

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

268

5.8. Radius of curvature in Polar co-ordinates Let O be the pole and OA be the initial line. Let Pr ,   be any

point on the curve r  f   P(r, 

Let the tangent at P make an angle 

 with the initial line. Let  be the angle between the

tangent and the radius vector at P. The,

  



d d  1  1 d d

 A

We know that, tan   r

d dr

i.e. tan  

r  dr     d 

Differentiating w.r.t.  , we get

d 2r  dr   dr   .   r. 2 d  d   d  d sec 2 .  2 d  dr     d  2

d2r  dr   r   2 1  tan 2  dd   d  2 d  dr     d 

   d 2r   dr      r    2  d  1  r . d   d  2   dr  2  d  dr         d    d   2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

269

 d 2r   dr     r  2  d  d   d    d   2  2  1  dr . dr    dr  2   d        d   2

 d 2r   dr     r  2  d  d   d   2 d  dr  2   r d    2

d 2r  dr    r 2 d 1  d  1   d  2 d  dr  2 r    d  2

d 2r  dr   dr  r     r 2 d d  d   d   2 d  dr  r2     d  2

d 2r  dr  r 2  2  r  d d 2  d     2 2 d dr   r2     d 

Next, we know that 2

 ds   dr  2    r    3  d   d 

Radius of curvature



ds d  ds   d .  d   d

 

  

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

270 2



 dr    d 

  r 2   

 

2 12

  

 dr  r2     d  . 2 d 2r  dr  2 r  2  r 2 d  d 

 2  dr  2    r    d   

d 2r  dr  r 2  2  r  d 2  d 

Note: The pedal equation of a curve is given by 1 1 1  dr   2 4  2 p r r  d 

Radius of curvature in pedal co-ordinates

  r.

dr dp

Problems 1. Show that the radius of curvature at (r,  ) on the cardioid r  a1 cos   is

2 2ar 3

The equation of the Cardioid is r  a1  cos    1

dr  a sin  d

The pedal equation of the curve is 1 1 1  dr   2 4  2 p r r  d 





1 1  4 a 2 sin 2  2 r r



r 2  a 2 sin 2   r4

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

271



a 2 1  cos    a 2 sin   r4



a 2 1  2 cos   cos 2   a 2 sin 2  r4



a2 1  2 cos   1 r4





a2  4 21  cos   r a4 r  4 .2. r a 1 2a  p2 r3

p2 

r3 2a

p

r3 2  2 2a

This is the pedal equation of the given curve differentiating w.r.t. r

dp 1 3 12  . .r dr 2a 2 Radius of Curvature

  r.

dr dp

  r.

2 2. a 3 r



2 2ar 3

Another Method The equation of the cardioid is r  a1 cos  

dr  a sin  d

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

272

d 2r  a cos  d 2 Radius of curvature 32

 2  dr  2    r    d      2 d 2r  dr  2 r  2  r 2 d  d 



a 1  cos    a sin   2

a 2 1  cos    2a 2 sin 2   a1  cos  .a cos  2

a 1  2 cos   cos    a 2

 a 1  2 cos   cos   sin    a 1  3 cos   2cos   sin   3





sin 2  a 2 1  2 cos   cos 2   2 sin 2   cos   cos 2  2





a21  cos   31  cos  

a.23 2  r   .  3 a



a.2 2 r . 3 a 2 3

  . 2ar 2. Find the radius of curvature at any point on the curve r 2  a  cos 2 The equation of the curve is r 2  a 2 cos 2

Diff, 2r

dr  a 2 . sin 2 .2 d

Diff,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

273 2

d 2 r  dr  2    a .2 cos 2 d 2  d 

d 2r  dr  r 2  2r 2    d  d 

Radius of curvature



 2  dr  2    r    d   

d 2r  dr  r  2  r d 2  d  2

 2  dr  2  r      d     2 2  dr   dr  2 2 r  2   2r     d   d  32

 2  dr  2    r    d     2   dr   3r 2      d   

1 a 4 sin 2 2   r 2   3 r2 







1 4 r  a 4 sin 2 2 3r



1 4 a cos 2 2  a 4 sin 2 2 3r



12 a2 cos 2 2  sin 2 2 3r









a2 3r

3. Find the radius of curvature at any point on the parabola

2a  1  cos  . r

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

274

The equation of the parabola is

2a  1  cos  r

Taking log, log 2a  log r  log1  cos 

Diff, 1 dr 1  sin    .  r d 1  cos  

dr r sin   d 1  cos 

The p–r equation of the curve is, 1 1 1  dr   2 4  2 p r r  d 



1 1 r 2 sin 2   . r 2 r 4 1  cos  2

2  1  cos    sin 2   2 r 2 1  cos   21  cos    2 2 r 1  cos  



2 r . r 2 2a

1 1  2 ar p

p 2  ar p  a. r Diff w.r.t.r

dp 1  a. dr 2 r Radius of curvature FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

  r.



275

dr 2 r  r. dp a

2 32 .r a

Self Assessment Problems VI 1. Find the radius of the curvature at any point on the curve r  a cos  2. Find the radius of curvature at any point on the curve r 2  a 2 sin 2

Answers 5.9. Self Assessment Problems – I



1. u  log x 2  y 2  z 2



1 u  2 .2 x x x  y 2  z 2

u 1  2 .2 y y x  y 2  z 2 u 1  2 .2 z z x  y 2  z 2



u u u 2x  y  z     x y z x 2  y 2  z 2

2. r 2  x  a    y  b  z  c  2

Differentiating partially w.r.t. x, 2r. 

r  2x  a  x

r x  a  x r

Differentiating again partially w.r.t. x,

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

 r  x 2 2

276

r.1  x  a 

r x

r2 r  x  a .

x  a  r

 2 x r 2  x  a   x 2 r3

Similarly,

 2r r 2   y  a  y 2 r3

 2 r r 2  z  a   z 2 r3

 2u  2u  2u r 2    x  a   r 2   y  b   r 2   z  c   2 2 2  x y z r3 2



3r   x  a    y  b   z  c   r3 2



3r 2  r 2 r3



2r 2 r3



2 r



3. v  1  2 xy  y 2



1 2



 1

Differentiating partially w.r.t.x, v 1   1  2 xy  y 2 x 2



x

  2 y  3 2

v  xy 1  2 xy  y 2 x





3 2

 2

Differentiating (1) partially w.r.t.y, FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

277

v 1   1  2 xy  y 2 y 2



v   1  2 xy  y 2 y



y

  2 x  2 y 3 2

  x  y  3 2

v   1  2 xy  y 2 y



  xy  y   3 3 2

(2) – (3);

v v  y  1  2 xy  y 2 x y



 xy  xy  y  3 2

 v 3 .y 2  y 2 .v 3 4. u  x 2  y  z   y 2 z  x   z 2 x  y  u  2 x y  z   y 2  z 2 x

u  x 2  2 yz  x   z 2 y u   x 2  y 2  2 x x  y  z



u u u    2x y  z   yz  x   z x  y   y 2  z 2  x 2  z 2  x 2  y 2 x y z

 2xy  xz  yz  xy  zx  yz 

= 2.0 =0 5. u  x y

u  x y . log x y  2u 1  x y .  log x . yx y1 xy x

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

278

 2u  x y1 1  y log x  1 xy Next,

u  y.x y 1 x

 2u  y.x y1. log x  x y1 .1 yx  2u  x y1  y log x  1  2 yx (1) & (2) 

 2u  2u  xy yx

Self Assessment problems III  x2  y 2   1. u  tan 1   x y 

tan u 

x2  y2 x y

R.H.S is a homogeneous function of degree n=2–1=1

f  tan u

By Euler’s theorem,

f f  y  nf x y

 tan u   y  tan u   1. tan u x y

sec2 u.x

u u tan u  sec2 u. y.  x y sec2 u



sin u cos 2 u  cos u 1



2 sin u . cos u 2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

279

1  sin 2u 2  x2  y 2   2. u  log  x y 

eu 

x2  y2 x y

R.H.S is homogeneous function of degree

f  eu

n = 2 – 1= 1 By Euler’s theorem,

f f  y  nf x y

 u  u e y e  1.eu x y

 

e u .x

 

u u u  e . y.  eu x y

 eu , x

f f  y 1 x y

x  y 3. u  sin 1    tan 1   x  y u  v    1

 x  y Where v  sin 1    2 and   tan 1    3 x  y

sin 

x y

R.H.S is a homogeneous function of degree

n  1 1  0

f  sin

By Euler’s theorem,

f f  y  nf x y

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

280

 sin   y  sin   0. f x y

cos .x

v v  cos . y  0  4 x y

Next, 3  tan  

y x

R.H.S is a homogeneous function of degree

f  tan 

n=1–1=0 By Euler’s theorem,

f f  y  nf x y

 tan    y  tan    0. f x y

sec2 .x

   sec2 . y 0 x y

 sec2 ; x

f f y  0  5 x y

(4) + (5); x

v    v    y 0 x y

f f y 0 x y

u  v  

Self Assessment problems III 1. u  x 2 y 3 ;

x  1  t; y  1  t

dx dy  1;  1 dt dt du dx dy  2 xy 3 .  x  .3 y 2 . dt dt dt

 21  t 1  t  .1  31  t  1  t  . 1 3





 1  t 1  t 1  t  .2  31  t 1  t  2

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

   1  t 5t

281

 1  t 2 2  4t  2t 2  3  3t 2 2



 4t  1

 5t 2  4t  1  5t 4  4t 3  t 2  5t 4  4t 3  6t 2  4t  1

2. u 

x y  ; y x

x  et ; y  t

dx dy  et ; 1 dt dt du  1 dx   1  dx     1  dy 1 dy    .  y 2 .    x. 2   .  dt  y dt  x  dt    y  dt x dt 

t t   et 1  1 t   .e  2t .e    2 .1  t .1 e e  t   t

et t et 1    t et t 2 et te 2t  t 3  e 2t  t 2  t 2 et du t 2  te 2t  e 2t  t 3  dt t 2 e 2t 3. u  xyz ;

x  t;

y  t;

z  et

dx dy 1 dz  1;  ;  et dt dt 2 t dt du dx dy dz  yz.  xz  xy dt dt dt dt

 t .et .1  te t .  t .et 

1 2 t

 t t .et

t t e  t t et 2

3   t et   t  2  FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

282

4. u  x 2  y 2 ;

x  t 3; y  t  t 2

dx  3t 2 ; dt

dy  1  2t dt

du dx dy  2 x.  2 y dt dt dt





 2t 3 .3t 2  2 t  t 2 1  2t 



 6t 5  2 t  2t 2  t 2  2t 3



 6t 5  2t  6t 2  4t 3 du  6t 5  4t 3  6t 2  2t dt

5. u  x 2 y 3 ;

x  1 t;

y  1 t

dx 1 dy 1  ;  dt 2 t dt 2 t du dx dy  2 xy 3  x 3 .3 y 2 . dt dt dt







 2 1 t 1 t . 



 



2 2  1   1  t .3 1  t .  2 t 2 t 

1  t 1  t  21  t   31  t 1  t    2 t 2

1  t  2  4

t  2t  3  3t

2 t



du 1  t   5t  4 t  1 dt 2 t





Self Assessment Problems IV 1. ax 2  2hxy  by 2  1 Let f x, y   ax 2  2hxy  by 2  1

f x  2ax  2hy

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

283

f y  2hx  2by f dy  x dx fy

dy 2ax  2hy  dx 2hx  2by

ax  hy  dy  hx  by  dx 2. sin x 

tan y

 tan y 

sin x

Let f x, y   sin x 

tan y

f x  tan y sin x 

tan y 1

f y  sin x 

tan y

 tan y 

sin x

.cos x  tan y 

sin x

.logtan y .cos x

.logsin x .sec 2 y  sin x.tan y 

sin x 1

.sec 2 y

f dy  x dx fy



 

dy cos x tan y.sin x   tan y  . log tan y   tan y sin x 1 2 dx sec y sin x  . log sin x   sin xtan y 



tan y 1





x2 y2  1 a2 b2

3. u  sin x  y ; 2

sin x



du  cos x 2  y 2 dx

 2 x  2 y. dy dx  



  b 2 x   cos x  y 2 x  2 y.  2   a y  











2x . cos x 2  y 2 . a 2  b 2 2 a

2 x 2 y dy  . 0 a  b 2 dx

dy b2 x  2 dx a y



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

284

Self Assessment Problems – V 1. The equation of the curve is

 2 x



x y a

dy 0 2 y dx .

y dy  dx x

d2y  dx 2

 1 dy  1 x . .   y.  2 y dx  2 x   x

a a At P  ,  , 4 4 a dy   4  1 dx a 4

    a 1 a 1  . 1  . 2  a a  2 2  2 2  d2y 2  2 a dx 4 1 1    2 2     a   4



4 a

Radius of curvature

1  y  

2 32 1

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II



285

1  13 2 4   a



2 2a 4



a 2

2. The equation of the curve is x 3  y 3  3axy

3x 2  3 y 2 .

dy  dy   3a  x.  y  dx  dx 





dy 2 y  ax  ay  x 2 dx

dy ay  x 2  dx y 2  ax



d y  dx 2

y





 dy   ax a  2 x   ay  x 2  dx  2 2 y  ax





 2 y dy  a dx 



 3a 3a  At P ,  ,  2 2 3a a2 9 dy 4  1  22 dx a 3a 9  a. 4 2 a.

 9a 2 3a  3a   3a a 2  3a      2.  1  a   a .  a  2 .  a .  9      2 2  2  2 4  2 d y  4    2 2 dx  a 3a   9  a  2  4  3a 2   3a 2    4a      4a  4  4     2  3a 2    4   FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II  6a 3  

286

16 9a 4

32 3a

Radius of curvature

1  y  

2 32 1

 

1  13 2  3a  32 2 2  3a 32



3 2a 16

3. The equation of the curve is x  3a cos   a cos 3 ;

y  3a sin   a sin 3

x'  3a sin   3a sin 3 ;

y '  3a cos   3a cos 3

x"  3a cos   9a cos 3 ;

y"  3a sin   9a sin 3

Radius of curvature

x'  y'  

2 32

x' y" y ' x" 32

 3a sin   3asin3 2   3a cos   3a cos 3 2      3a sin   3a sin 3   3a sin   9a sin 3   3a cos   3a cos 3       3a cos   9a cos 3 



9a sin

  sin 2 3  2 sin  .sin 3  cos 2   cos 2 3  2 cos  . cos 3  9a 2 sin 2   27a 2 sin  .sin 3  9a 2 sin  .sin 3  27a 2 sin 3 3

 9a 2 cos 2   27a 2 cos  cos 3  9a 2 cos  . cos 3  27a 2 cos 3 3



9a 2 3a1  1  2cos 3 cos   sin 3 .sin   9a 2  27a 2  36a 2 cos 3 . cos   sin 3 .sin  



27a 2 2  2 cos3    9a 2 4  4 cos3   

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

287

3a21  cos 2 2  41  cios 2  3



3a 3 2 12 2 .1  cos 2  4



3a 2 sin 2  2



3a 2 sin  2





  3a sin  4. The equation of the curve is x  6t 2  3t 4 ;

y  8t 3

x'  12t  12t 3 ;

y '  24t 2

x"  12  36t 2 ;

y"  48t

Radius of curvature

x'  y'  

2 32

x' y" y ' x"





12t 12t   24t  

32 2 2

3 2

12t  12t .48t  24t .12  36t  3

144t 1  t   4t  2 2



  1  t 

24 12t 2 2 1  t 2  1  3t 2

12 t  24 12t 2  2t 2 2

 1  3t 2



12 12  12  t 3  1  t 2  24  12t 2  1  t 2



 6t 1  t 2









5. The equation of the curve is x  a logsec  tan ;

x'  a.



1 . sec . tan   sec2  sec  tan  

y  a sec



FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

x' 

288

a sec  sec  tan   sec  tan  

x'  a sec x"  a sec tan y'  a sec tan



y"  a sec  . sec2   tan  . sec  . tan 



y"  a sec  sec2   tan 2 





Radius of curvature

x'  y'   2

2 32

x' y" y ' x"

a



sec2   a 2 sec2  . tan2   a sec  .a sec  sec2   tan2   a 2 sec2  . tan2  32

a









sec2   1  tan2   2 4 a sec   a 2 sec2  tan2   a 2 sec2  tan2 

a 

sec2  . sec2  a 2 sec4 



a 3 sec 6   2 4 a sec 

  a sec2 

Self Assessment Problems – VI 1. The equation of the curve is r  a cos  dr  a sin  d

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II

289

d 2r  a cos  d 2 Radius of curvature 32

 2  dr  2  r      d      2 d 2r  dr  2 r  2   r 2 d  d 

a



cos 2   a 2 sin 2   2 2 a cos   2a 2 sin 2   a cos  .a cos  32

a cos   sin    2a cos   sin   2



a3 2a 2



a 2

2. The equation of the curve is r 2  a 2 sin 2

Taking log,

2 log r  2 log a  log sin 2 Diff, 1 dr 1 2. . 0 . cos 2 .2 r d sin 2 dr  r cot 2 d

The p – r equation of the curve is 1 1 1  dr   2 4  2 p r r  d 



1 1 2  .r cot 2 2 r2 r4

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

MATHEMATICS-II 

1 1  2 cot 2 2 2 r r



1  cot 2 2 r2



cos ec 2 2 r2

1 1  2 2 2 p r sin 2

p 2  r 2 sin 2 2 p  r sin 2

 r.

r2 a2

p

r3 a2

Differentiating w.r.t. r. dp 1   .3r 2 dr a

Radius of curvature

  r.

dr dp

  r.

a2 a2   . 3r 2 3r

------------------------------------------THE END---------------------------------------------------

FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

290

MATHEMATI CSII

Publ i s he dby

I ns t i t ut eofManage me nt& Te c hni c alSt udi e s Addr e s s:E4 1 , Se c t o r 3 , No i da( U. P) www. i mt s i ns t i t ut e . c o m| Co nt a c t :+9 1 9 2 1 0 9 8 9 8 9 8