Engineering mathematics II

The topics covered are: Unit I and II—Differential Calculus, Unit III and IV—Integral Calculus and Vector Integration, Unit V and VI—Differential Equations and Unit VII and VIII—Laplace

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This Textbook has been prepared as per the syllabus for the Engineering Mathematics Secondsemester B.E classes of Visveswaraiah Technological University The book contains eight chapters,and each chapter corresponds to one unit of the syllabus The topics covered are: Unit I and II—Differential Calculus, Unit III and IV—Integral Calculus and Vector Integration, Unit V and VI—Differential Equations and Unit VII and VIII—Laplace Transforms

It gives us a great pleasure in presenting this book In this edition, the modifications havebeen dictated by the changes in the VTU syllabus The main consideration in writing the bookwas to present the considerable requirements of the syllabus in as simple manner as possible Thiswill help students gain confidence in problem-solving

Each unit treated in a systematic and logical presentation of solved examples is followed by

an exercise section and includes latest model question papers with answers from an integral part

of the text in which students will get enough questions for practice

The book is designed as self-contained, comprehensive and friendly from students’ point ofview Both theory and problems have been explained by using elegant diagrams wherever necessary

We are grateful to New Age International (P) Limited, Publishers and the editorial departmentfor their commitment and encouragement in bringing out this book within a short span of period

AUTHORS

We whole heartedly thank our Chairman Mr S Narasaraju Garu, Executive Director, Mr.

S Ramesh Raju Garu, Director Prof Basavaraju, Principal Dr T Krishnan, HOD Dr K Mallikarjun,

Dr P.V.K Perumal, Dr M Surekha, The Oxford College of Engineering, Bangalore We would like

to thank the other members of our Department, Prof K Bharathi, Prof G Padhmasudha,

Mr Ravikumar, Mr Sivashankar and other staffs of The Oxford College of Engineering, Bangalorefor the assistance they provided at all levels for bringing out this textbook successfully

We must acknowledge Prof M Govindaiah, Principal, Prof K.V Narayana, Reader,Department of Mathematics, Vivekananda First Grade Degree College, Bangalore are the oneswho truly made a difference in our life and inspired us a lot

We must acknowledge HOD Prof K Rangasamy, Mr C Rangaraju, Dr S Murthy, Department

of Mathematics, Govt Arts College (Men), Krishnagiri

We are also grateful to Dr A.V Satyanarayana, Vice-Principal of R.L Jalappa Institute ofTechnology, Doddaballapur, Prof A.S Hariprasad, Sai Vidya Institute of Technology, Prof V.K.Ravi, Mr T Saravanan, Bangalore college of Engineering and Technology, Prof L Satish, RajaRajeshwari College of Engineering, Prof M.R Ramesh, S.S.E.T., Bangalore

A very special thanks goes out to Mr K.R Venkataraj and Bros., our well wisher friend

Mr N Aswathanarayana Setty, Mr D Srinivas Murthy without whose motivation andencouragement this could not have been completed

We express our sincere gratitude to Managing Director, New Age International (P) Limited,and Bangalore Division Marketing Manager Mr Sudharshan for their suggestions and provisions

of the font materials evaluated in this study

We would also like to thank our friends and students for exchanges of knowledge, skillsduring our course of time writing this book

AUTHORS

Dedicated to

my dear parents,

Shiridi Sai Baba,

my dear loving son Monish Sri Sai G

and my wife and best friend S Mamatha

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QUESTION PAPER LAYOUT Engineering Mathematics-II

O6MAT21

Units-1, 2, 3, 4 Units-5, 6, 7, 8

1 Qn from each unit 1 Qn from each unit

To answer five  full questions choosing at least  two questions from each part

1 DIFFERENTIAL CALCULUS-I

Radius of Curvature: Cartesian curve Parametric curve, Pedal curve, Polar curve and some fundamental theorems.

2 DIFFERENTIAL CALCULUS-II

Taylor’s, Maclaurin’s Maxima and Minima for a function of two variables.

5 DIFFERENTIAL EQUATIONS-I

Linear differential equation with constant coefficients, Solution of homogeneous and non homogene- ous linear D.E., Inverse differential operator and the Particular Integral (P.I.)

Method of undetermined coefficients.

6 DIFFERENTIAL EQUATIONS-II

Method of variation of parameters, Solutions of Cauchy's homogeneous linear equation and Legendre’s linear equation, Solution of initial and Boundary value problems.

7 LAPLACE TRANSFORMS

Periodic function, Unit step function (Heaviside function), Unit impulses function.

8 INVERSE LAPLACE TRANSFORMS

Applications of Laplace transforms.

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Type 2 Evaluation of a double integral by changing the order of integration 119

Type 4 Applications of double and triple integrals 124

3.4.3 Relationship between Beta and Gamma functions 133

5.3 Solution of a Homogeneous Second Order Linear Differential Equation 215

5.7 Solution of Simultaneous Differential equations 264

Additional Problems (From Previous Years VTU Exams.) 268

UNIT VI Differential Equations–I 280—320

6.3 Solution of Initial and Boundary Value Problems 308

7.3.1 Laplace Transforms of Some Standard Functions 322

7.3.2 Laplace Transforms of the form e at f (t) 331

7.5 Laplace Transforms of Unit Step Function and Unit Impulse Function 352

7.5.1 Properties Associated with the Unit Step Function 352

8.2 Inverse Laplace Transforms of Some Standard Functions 369

8.3 Inverse Laplace Transforms using Partial Fractions 376

Solution of Simultaneous Differential Equations 406

Additional Problems (From Previous Years VTU Exams.) 417

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UNIT 1

Differential Calculus—I

1.1 INTRODUCTION

In many practical situations engineers and scientists come across problems which involve quantities

of varying nature Calculus in general, and differential calculus in particular, provide the analyst withseveral mathematical tools and techniques in studying how the functions involved in the problembehave The student may recall at this stage that the derivative, obtained through the basic operation

of calculus, called differentiation, measures the rate of change of the functions (dependent variable)with respect to the independent variable In this chapter we examine how the concept of the derivativecan be adopted in the study of curvedness or bending of curves

1.2 RADIUS OF CURVATURE

Let P be any point on the curve C Draw the tangent at P to the

circle The circle having the same curvature as the curve at P

touching the curve at P, is called the circle of curvature It is also

called the osculating circle The centre of the circle of the

cur-vature is called the centre of curcur-vature The radius of the circle

of curvature is called the radius of curvature and is denoted

by ‘ρ’

Note : 1 If k (> 0) is the curvature of a curve at P, then the radius

of curvature of the curve of ρ is 1k This follows from the definition

of radius of curvature and the result that the curvature of a circle is the

reciprocal of its radius

Note : 2 If for an arc of a curve, ψ decreases as s increases, then d

ds

ψ is negative, i.e., k is negative.

But the radius of a circle is non-negative So to take ρ = 1

2 ENGINEERING MATHEMATICS—II

The sign of d

ds

ψ indicates the convexity and concavity of the curve in the neighbourhood of

the point Many authors take ρ = ds

dψ and discard negative sign if computed value is negative.

∴ Radius of curvature ρ = 1

k ·

1.2.1 Radius of Curvature in Cartesian Form

Suppose the Cartesian equation of the curve C is given by y = f (x) and A be a fixed point on it Let P(x, y) be a given point on C such that arc AP = s.

Then we know that

1 2

+FHG IKJ

L NMM dy O QPP L NMM +FHG IKJ O QPP

dx

dy dx

ρ [By using the (1) and (2)]

= 11

2 3 2

2 2

DIFFERENTIAL CALCULUS—I 3

Equation (3) becomes,

ρ = 1 1

2 3 2

2

+ y

y

o t

This is the Cartesian form of the radius of curvature of the curve y = f (x) at P (x, y) on it.

1.2.2 Radius of Curvature in Parametric Form

Let x = f (t) and y = g (t) be the Parametric equations of a curve C and P (x, y) be a given point

//

RST UVW⋅

=

dx dt

d y dt

dy dt

d x dt dx dt

dx dt

FHG IKJ ⋅

2 2

2 2 2

1

d y dx

2

dx dt

d y dt

dy dt

d x dt dx dt

FHG IKJ

2 2

2 2 3

2

2 3 2

2 2

+

=+FHG IKJ

RS|

T| UV| W|

y y

dy dx

d y dx

o t

=

1

2 3 2

2 2

2 2

dt

d y dt

dy dt

d x dt

dx dt

//

dx dt

dy dt dx

dt

d y dt

dy dt

d x dt

2 2

2 2

(6)

This is the cartesian form of the radius of curvature in parametric form

WORKED OUT EXAMPLES

1 Find the radius of curvature at any point on the curve y = a log sec x

x a

2

3 2

x a

x a

secsec

3

2

FHG IKJ FHG IKJ

3 2

coscos

2

3 2

3 2

+ y

y

d i at (a, 0)

6 ENGINEERING MATHEMATICS—IIHere, y = x3(x – a) = x4 – x3a

3 2

3 2

2

+

RST UVWa a

d i

=

16

6 3 2 2

FHG IKJ

2 3 2

FHG IKJ FHG IKJ

x a

x a

x a

sec FHG IKJ+secFHG IKJ FHG IKJtan

DIFFERENTIAL CALCULUS—I 7

6 Find ρ at x = π

3 on y = 2 log sin

x 2

FHG IKJ

Solution We have ρ = 1 1

2 3 2

y1 = 2 1

2

2

12

⋅

FHG IKJ × FHG IKJ×sin

cos

x

x

= cot x2

FHG IKJ

2

12

+

RST UVW e j

–

= 1 32

4 2

3 2

FHG IKJ on x3 + y3 = 3axy.

Solution We have ρ = 1 1

2 3 2

,

FHG IKJ.Here, x3 + y3 = 3axy

Differentiating with respect to x

2

32

2

2

FHG IKJ−FHG IKJ FHG IKJ − FHG IKJ

3

32

9

94

32

323

3 4

a

Using these

ρ 3 2 3 2

2 3 2

DIFFERENTIAL CALCULUS—I 9

8 Find the radius of curvature of b2x2 + a2y2 = a2b2 at its point of intersection with the y-axis.

Solution We have ρ = 1 1

2 3 2

2

+ y

y

o t at x = 0Here, b2x2 + a2y2 = a2b2

Differentiating again w.r to x

y2 = – b

a

y xy y

2 2

1 2

3 2

a b

a b

10 ENGINEERING MATHEMATICS—II

ρ(0, – b) = 1 0

3 2

a b

Solution The curve is xy = c2

Differentiating w.r to x

xy1 + y = 0

y1 = – y

x Again differentiating w.r.t x

y2 = – xy y

x

1 2

−

R S|

T|

U V|

2+

2 .

10 Show that, for the ellipse x

a

y b

2 2 2 2

perpen-DIFFERENTIAL CALCULUS—I 11

Solution The ellipse is x

a

y b

2 2 2 2

+ = 1

Differentiating w.r.t x

2 1 2

x a

yy b

a

x y

2 2

Again Differentiating w.r to x

y2 = – b

a

y xy y

2 2

1 2

2 2

2 2 2

2

L N

MM MM

O Q

PP PP

= – b

a y

y b

x a

4

2 2 2 2

2 2 2

HG I KJ

b x

a x b

3 2

2 2 2 2

+ = 1 is

x x a

y y b

0 2

0 2

12 ENGINEERING MATHEMATICS—IILength of perpendicular from (0, 0) upon this tangent

0 2

2 0 2

2

x a

y b

FHG IKJ +FHG IKJ

a y b x

2 2 4 0

0 2

0 2

1+

2 3

ρ

FHG IKJ = x

y

y x

b g b g

DIFFERENTIAL CALCULUS—I 13

2 3 2

3 2

2 3

++

RS|

T| UV| W|

−+

F

HG I KJ2 +FHG IKJ2

2 3

ρ

a

FHG IKJ =

22

3 2

2 3

T||

U V||

b g

b g

2 2

T|

U V|

T|

U V|

b g

b g

2 2

2

∴ L.H.S = R.H.S using (2) and (3)

14 ENGINEERING MATHEMATICS—II

12 Find ρ at any point on x = a (θ + sinθ) and y = a (1 – cosθ).

Solution Here x = a ( θ + sinθ), y = a (1 – cosθ)

= θθ

= a

a

sincos

θθ

2

sin coscos

FHG IKJ

= d

d

d dx

θ

tan2

FHG IKJ ×

= sec2 θ

θ2

12

11

2

2

3 2

4

+

RST R UVW S|

T|

U V|

= θθ

= a

a

tantan

2θθ

y1 = tan θ

y2 = d y

dx

d dx

θtan

2θθ

a

2 3 2

+

F

HG I KJ

tansectan

θθθ

16 ENGINEERING MATHEMATICS—II

= sec

3 2

⇒ d dxθ = a (– sin θ + θ cos θ + sin θ) = a θ cos θ

sincos tan

y2 = d

dx

dy dx

d dx

FHG IKJ = b gtanθ

= d

d

d dx

θtan

2 3 2

15 If ρ1 and ρ2 are the radii of curvatures at the extremities of a focal chord of the parabola

y2 = 4ax, then show that ρ1 – 2 3 ρ2

–2 3

2 3

11

2 3

2

2

12

11

–

b g d i

ρ2

2 3 –

2 3

2

2 1

12

11

b g d i

2 3 1 2

1

2

a

t t

b g d i× +

2 3 2

2 3

1 2

t t

a

t t

2 3

2 3

a

b g–

· Hence proved.

18 ENGINEERING MATHEMATICS—II

EXERCISE 1.1

×FHG IKJ

L N

4 4

+

L N

12 Find the radius of curvature at (–2a, 2a) on x2y = a (x2 + y2) Ans 2a

13 Show that ρ at (a cos3θ, a sin3θ) on x y

2 3 2 3

124+

L N

MM e j O Q PP

15 Find ρ for x = t – sin ht cos ht, y = 2cos ht. Ans. 2 cosh t2 sinht

DIFFERENTIAL CALCULUS—I 19 1.2.3 Radius of Curvature in Pedal Form

Let polar form of the equation of a curve be r = f (θ) and

P(r, θ) be a given point on it Let the tangent to the curve

at P subtend an angle ψ with the initial side If the angle

between the radius vector OP and the tangent at P is φ then

we have ψ = θ + φ (see figure)

Let p denote the length of the perpendicular from the

pole O to the tangent at P Then from the figure,

sin φ = OM

OP

p r

=

d ds

ψ

= d

ds

d ds

d ds

d dr

dr ds

r

d dr

= 1

d dr

This is the Pedal form of the radius of curvature

1.2.4 Radius of Curvature in Polar Form

Let r = f ( θ) be the equation of a curve in the polar form and p(r, θ) be a point on it Then we know

Differentiating w.r.t r, we get

P (r, ) G

r = f ( ) G r

X O

p

M B

Fig 1.2

– –52

d dr

dr d

2 2

dr

dr d

d r d

d dr

dr

d r d

3

2 4 2 2

=

r p

d r d

6 3

d r d

6

2 3 2

2

2 3 2

2

2

2+FHG IKJ

θ ·

2 1 2 3 2 2

1 2 2

DIFFERENTIAL CALCULUS—I 21

WORKED OUT EXAMPLES

1 Find the radius of the curvature of each of the following curves:

(i) r 3 = 2ap 2 (Cardiod) (ii) p 2 = ar

(iii) 1

p

1 a

1 b

dp r

ap r

ap r

⋅ = ⋅4 =

3

43

3 1 2

a r

3 2

ρ = r dr

dp r

ar a

r a

= ⋅ ⋅2 =2

3 2

2 2

dr dp

⋅

1 2 2

3 2

cos cos sin

++

coscos

θθ

ab + cos θg

=

23

2

1 2

aFHG IKJcos θ

ρ = 4

3a cos2θ

Squaring on both sides, we get

DIFFERENTIAL CALCULUS—I 23

ρ2 = 8

9

89

+ 1 1

Solution Here r n = a n cos nθ

Taking logarithms on both sides, we get

n log r = n log a + log cos nθDifferentiating w.r.t θ, we have

n r

dr

dθ = 0 –

sincos

n

θθ

2 1 2 3 2 2

1 2 2

3 2

1

θθ+

n n

+ 1 1

b g – ·

24 ENGINEERING MATHEMATICS—II

4 Find the radii of curvature of the following curves:

(i) r = aeθ cot α (ii) r (1 + cos θ) = a

(iii) θ = r a

a r

θθ

= 111

2

2 2

+

L N

MM sin O Q PP

cos

θθ

L N

MM cos O Q PP

cos

θθ

= 2 2

3 2

a r

a

a r

1 2

r

a r

a r

26 ENGINEERING MATHEMATICS—II

d dr

⋅ = = 2 − 2⋅

EXERCISE 1.2

1 Find the radius curvature at the point ( p, r) on each of the following curves:

a

3 2

+

L NMM b g1 − 1O QPP

++

L N

MM MM

O Q

PP PP

L N

MM MM

O Q

PP PP

+

L NMM b g1 − 1O QPP

(v) r2 cos 2θ = a2 Ans r

a

3 2

(vii) r = a sec 2θ Ans r

p

4 2

3 If ρ1 and ρ2 are the radii of curvature at the extremities of any chord of the cardiode

r = a (1 + cos θ) which passes through the pole Prove that ρ12 ρ

2 2

1 continuous in a closed interval [a, b],

2 differentiable in the open interval (a, b) and

3 f (a) = f (b).

Then there exists at least one value c of x in (a, b) such that f ′ (c) = 0

(No proof)

1.3.2 Lagrange’s Mean Value Theorem

Suppose a function f (x) satisfies the following two conditions.

1 f (x) is continuous in the closed interval [a, b].

2 f (x) is differentiable in the open interval (a, b).

Then there exists at least one value c of x in the open interval (a, b), such that

( ) ––

b g = f ′ (c)

28 ENGINEERING MATHEMATICS—II

Proof Let us define a new function

where k is a constant Since f (x), kx and φ (x) is continuous in [a, b], differentiable in (a, b).

From (1) we have, φ (a) = f (a) – k·a

φ (b) = f (b) – k·b

f (a) – k·a = f (b) = k·b i.e., k (b – a) = f (b) – f (a)

This proves Lagrange’s mean value theorem

1.3.3 Cauchy’s Mean Value Theorem

If two functions f (x) and g (x) are such that

1 f (x) and g (x) are continuous in the closed interval [a, b].

2 f (x) and g (x) are differentiable in the open interval (a, b).

3 g ′ (x) ≠ 0 for all x ∈ (a, b).

Then there exists at least one value c ∈ (a, b) such that

( ) ––