Download ecc math 1 notes 2k18 internal notes and more Study Guides, Projects, Research Mathematics in PDF only on Docsity!
SYLLABUS
Engineering Mathematics-I
Subject Code: 15MAT11 IA Marks: 20
Hours/Week: 04 Exam. Hours: 03
Total Hours: 50 Exam. Marks: 80
Course Objectives
To enable students to apply knowledge of Mathematics in various
engineering fields by making hem to learn the following:
- nth derivatives of product of two functions and polar curves.
- Partial derivatives.
- Vectors calculus.
- Reduction formulae of integration to solve First order
differential equations
- Solution of system of equations and quadratic forms.
Module – 1
Differential Calculus -1:
Determination of nth order derivatives of Standard functions -
Problems. Leibnitz‟s theorem (without proof) - problems.
Polar Curves - angle between the radius vector and tangent, angle
between two curves, Pedal equation for polar curves. Derivative of
arc length - Cartesian, Parametric and Polar forms (without proof)
- problems. Curvature and Radius of Curvature – Cartesian,
Parametric, Polar and Pedal forms(without proof) and problems.
10hrs
Module – 2
Differential Calculus -
Taylor‟s and Maclaurin‟s theorems for function of o ne
variable(statement only)- problems. Evaluation of Indeterminate
forms.
Partial derivatives – Definition and simple problems, Euler‟s
theorem(without proof) – problems, total derivatives, partial
differentiation of composite functions-problems, Jacobians-
definition and problems. 10hrs
Module – 3
Vector Calculus:
Derivative of vector valued functions, Velocity, Acceleration and
related problems, Scalar and Vector point functions.Definition
Gradient, Divergence, Curl- problems. Solenoidal and Irrotational
vector fields. Vector identities - div ( F A), curl ( F A),curl (grad F
), div (curl A). 10hrs
Module- 4
Integral Calculus:
Reduction formulae ∫ sin
n
x dx ∫cos
n
x dx ∫sin
n
xcos
m
xd x ,, (m and n
are positive integers), evaluation of these integrals with standard
limits (0 to л/2) and problems.
Differential Equations :
Solution of first order and first degree differential equations –
Exact, reducible to exact and Bernoulli‟s differential equations.
Applications - orthogonal trajectories in Cartesian and polar forms.
Simple problems on Newton‟s law of cooling.. 10hrs
Module – 5
Linear Algebra Rank of a matrix by elementary transformations,
solution of system of linear equations - Gauss- elimination method,
Gauss- Jordan method and Gauss-Seidel method. Rayleigh‟s
power method to find the largest Eigen value and the
corresponding Eigen vector. Linear transformation, diagonalisation
of a square matrix, Quadratic forms, reduction to Canonical form
10hrs
MODULE I
DIFFERENTIAL CALCULUS-I
CONTENTS:
Successive differentiation ………………………………………….. 3
nth derivatives of some standard functions…………………...
Leibnitz‟s theorem (without proof)………………………..…
Polar curves
Angle between Polar curves…………………………………….
Pedal equation for Polar curves………………………………...
Derivative of arc length………………………………………….
Radius of Curvature……………………………………………….
Expression for radius of curvature in case of Cartesian Curve …
Expression for radius of curvature in case of Parametric
Curve………………………………………………………………..
Expression for radius of curvature in case of Polar Curve……...
Expression for radius of curvature in case of Pedal Curve……...
SUCCESSIVE DIFFERENTIATION
In this lesson, the idea of differential coefficient of a function and its successive
derivatives will be discussed. Also, the computation of n
th derivatives of some
standard functions is presented through typical worked examples.
1. Introduction:- Differential calculus (DC) deals with problem of calculating rates of
change. When we have a formula for the distance that a moving body covers as a
function of time, DC gives us the formulas for calculating the body‟s velocity and
acceleration at any instant.
Definition of derivative of a function y = f(x):-
Fig.1. Slope of the line PQ is x
f x x f x
The derivative of a function y = f(x) is the function f ( x )whose value at each x is
defined as
dx
dy = f ( x )= Slope of the line PQ (See Fig. 1 )
0
lim x x
f x x f x
0
lim x
(Average rate change)
= Instantaneous rate of change of f at x provided the limit exists.
The instantaneous velocity and acceleration of a body (moving along a line) at any instant
x is the derivative of its position co-ordinate y = f(x) w.r.t x, i.e.,
Velocity = dx
dy = f ( x ) --------- (2)
And the corresponding acceleration is given by
Acceleration ( ) 2
2
f x dx
d y ---------- (3)
Calculation of n
th derivatives of some standard functions
Below, we present a table of n
th order derivatives of some standard functions for
ready reference.
We proceed to illustrate the proof of some of the above results, as only the
above functions are able to produce a sequential change from one derivative to
the other. Hence, in general we cannot obtain readymade formula for nth
derivative of functions other than the above.
1. Consider
mx e. Let
mx y e. Differentiating w.r.t x , we get
y 1
mx me. Again differentiating w.r.t x , we get
mx y (^) 2 m me =
mx m e
2
Similarly, we get
y 3 =
mx m e
3
y 4 =
mx m e
4
And hence we get
Sl.
No
y = f(x) D y dx
d y y
n n
n
n
(^1) emx mn^ emx
mx
a
n n mx m log a a
3 ax b m i. m m 1 m 2 .... m n 1 a n^ ax b m n for all m.
ii. 0 if m n
iii. n !a
n if m n
iv.
mn x m n
m (^)
if m n
ax b
n n
n
a ax b
n 1 ( )
m ax b
n mn
n
a m ax b
m n
6. log( ax b ) n n
n
a ax b
n
1
7. sin( ax b ) ) 2
sin( a ax b n
n
8. cos( ax b ) ) 2
a cos( ax b n^
n
9. e sin( bx c ) ax
r n^ eax sin( bx c n ),^22 tan 1 ( )
a
r a b b
10. eax^ cos( bx c )
r e cos( bx c n )
n ax , tan ( )
2 2 1 a
r a b b
yn =
n mx me
mx n
n
e dx
d n mx me.
m ax b
let y
m ax b Differentiating w.r.t x ,
y 1 = m ax b a
m 1 . Again differentiating w.r.t x , we get
y 2 = m m 1
(^2 ) ax b a
m
Similarly, we get
y 3 = m m 1 m 2
3 3 ax b a
m
And hence we get
yn = m m 1 m 2 ………. m n 1
m n n ax b a
for all m.
Case (i) If m n (m-positive integer),then the above expression becomes
yn = n n 1 n 2 ……….3.2.1
n n n ax b a
i.e.
n y (^) n n! a
Case (ii) If m<n,(i.e. if n>m) which means if we further differentiate the above
expression, the
right hand site yields zero. Thus D ax b if m n
n m 0
Case (iii) If m>n, then
mn n yn mm m m n ax b a
1 2 ...... 1 becomes
m n n ax b a m n
m m m m n m n
. i.e
mn n n ax b a m n
m y
m ax b
m m
ax b ax b
Let y
Differentiating w.r.t x
y max b a max b a
m 1 m 1 1 1
11 2 2 (^2) y 2 (^) 1 m m 1 ax b a 1 mm 1 ax b a
m m
Similarly, we get
3 3 (^3) y 3 (^) 1 mm 1 m 2 ax b a
m
4 4 (^4) y 4 (^) 1 mm 1 m 2 m 3 ax b a
m
……………………………
n mn (^) n yn mm m m n ax b a
1 1 2 ..... 1
This may be rewritten as
mn n
n
n ax b a m
m n m n m mm y
6
2
log 2
log 10
x
x x
e
A B A
B log log
A B
B
A
log (^) log log
2 6 log( 3 5 ) log( 2 3 ) log( 1 ) 2 log 10
x x x
e
2 log( 3 5 ) log( 2 3 ) 6 log( 1 )
2 log 10
y x x x e
Hence,
n n
n n n
n n n
n
e
n x
n
x
n
x
n y ( 1 ) ( 1 )
2 log 10
1 1 1
2. (a)
2 4 2 4 6
x x e (b)cosh 4 x cosh 4 x
2
(c) e x x
x sinh 3 cosh 2
(d)
5 4
x x x
Sol : (a) Let
2 4 2 4 6
x x y e
2 4 2 4 6 6
x x e e
( ) 1296 ( 6 )
4 2 x 2 x y e e
hence ( ) 1296 ( 6 )
4 2 2 x n
x n (^) n dx
dn e dx
dn y e
n x n n x e e
4 2 2 2 12962 (log 6 ) 6
(b) Let y cosh 4 x cosh 4 x
2
4 4 4 4 2
x x x x e e e e
( ) ( ) 2 ( )( ) 4
(^1 4) x 4 x 4 x 2 4 x 2 4 x 4 x e e e e e e
2 4
x x x x y e e e e
hence, 8 ( 8 ) 0 4
n x n x n n n n yn e e e e
(c) Let y e x x
x sinh 3 cosh 2
2 2
3 x 3 x 2 x 2 x x e e e e e
( )( ) 4
3 x 3 x 2 x 2 x
x
e e e e
e (^)
x x x x
x
e e e e
e (^) 5 5
x x x e e e
4 2 6 1 4
x x x y e e e
4 2 6 1 4
Hence,
n x n x n x y (^) n e e e
4 2 6 0 ( 4 ) ( 2 ) ( 6 ) 4
(d) Let
5 4
x x x
y
Hence,
5 4
x dx
dn
dx x
dn
dx x
dn y n n n n
1 4
n n
n n n
n
x
n
x
n
i.e
n n
n n n
n
n x
n
x
n y ( 5 ) 3 !( 5 4 )
1 4
Evaluate
- (i) 6 8
2 x x
(ii) 2 3 1
x x x
(iii) 2 7 6
2
2
x x
x
(iv) 4 12 9
2
x x x
x (v) a
tan^1 x (vi) tan^1 x (vii)
x
x
tan
1
Sol : (i) Let 6 8
2
x x
y. The function can be rewritten as ( 4 )( 2 )
x x
y
This is proper fraction containing two distinct linear factors in the denominator.
So, it can be split into partial fractions as
x
B
x
A
x x
y Where the constant A and B are found
as given below.
x x
Ax Bx
x x
1 A ( x 2 ) B ( x 4 ) -------------(*)
Putting x = 2 in (*), we get the value of B as 2
B ^1
n n
n n n
n n n
n
n x
n
x
n
x
n y ( 1 ) ( 1 )
1 2 1
( 1 ) 2
1 1 xn
n
x x
n
n
n n
n
(iii) Let 2 7 6
2
2
x x
x y (VTU July-05)
This is an improper function. We make it proper fraction by actual division
and later
spilt that into partial fractions.
i.e 2 7 6
2
2
7 2 2
x x
x x x x
x x
x y Resolving this proper fraction into partial fractions,
we get
x
B
x
A
y. Following the above examples for finding A &
B , we get
x x
y
Hence,
n n
n n n
n
n x
n
x
n y ( 1 ) ( 2 )
1 1
i.e
1 1
2
9
n n
n n n x x
y n
(iv) Let ( 1 ) 4 12 9
2
x x
x
x
x y
(i) (ii)
Here (i) is improper & (ii) is proper function. So, by actual division (i)
becomes
x x
x
. Hence, y is given by
2 ( 2 3 )
x x
y [( 2 3 ) 4 12 9
2 2 x x x ]
Resolving the last proper fraction into partial fractions, we get
2 2 ( 2 3 ) ( 2 3 ) ( 2 3 )
x
B
x
A
x
x
. Solving we get
A ^1 and 2
B ^3
2
2
3 2
1
x x x
y
n
n n n
n n n
n
n x n
n
x
n
x
n y ( 2 ) ( 2 3 ) 2
1
(v) a
tan^1 x
Let a
y tan^1 x
(^1222)
x a
a
a a
x
y
2 2
1 1
1 ( ) x a
a y D y D y D
n n n n
Consider ( )( )
2 2 x ai x ai
a
x a
a
( ) ( x ai )
B
x ai
A
, on resolving into partial fractions.
x ai
i
x ai
i
, on solving for A & B.
x ai
D
x ai
D
x a
a D
n n i n 2 i
1 2 1
1 1 2 2
1
n
n
n
n
x ai
n
x ai i
n
i ( )
1 1
-----------(*)
We take transformation x r cos a r sinwhere
2 2 r x a ,
x
1 a
tan
i x ai r cos i sin re
i x ai r i re
cos sin
n
in
n n in r
e
x ai r e
n
in
n r
e
x ai
now(*) is
in in n
n
n e e ir
n y
1
n
r
n i n ir
y n
n
n
n
n sin
2 sin 2
1 1
2. e sin bx c .
ax
Let y e sin bx c .....( 1 )
ax
Differentiating using product rule ,we get
y 1
ax ax e cos bx cb sin bx cae
i.e. y 1 e a bx c b bx c
ax sin cos . For computation of higher order
derivatives
it is convenient to express the constants „a‟ and „b‟ in terms of the
constants r and
defined by a r cos& b r sin,so that
2 2 r a b and a
tan^1 b .thus,
y 1 can be rewritten as
y e r bx c r bx c
ax 1 ^ cos sin sin^ cos
or y 1 e r {sin bx c cos cos bx c cos }
ax
i.e. y 1 re sin bx c .......... .( 2 )
ax
Comparing expressions (1) and (2), we write y 2 as
sin 2
2 y 2 (^) re bx c
ax
sin 3
3 y 3 (^) re bx c
ax
Continuing in this way, we get
sin 4
4 y 4 (^) re bx c
ax
sin 5
5 y 5 (^) re bx c
ax
y r e bx c n
n ax n ^ sin
D e sin bx c r e sin bx c n ,
n ax n ax where
2 2 r a b & a
tan^1 b
Solve the following:
- (i) x x
2 3 sin cos (ii) x
3 3 sin cos (iii)cos x cos 2 x cos 3 x
(iv) sin x sin 2 x sin 3 x (v) e x
x cos 2
3 (vi) e x x
2 x 2 3 sin cos
The following formulae are useful in solving some of the above problems.
(i) 2
1 cos 2 ( )cos 2
1 cos 2 sin
2 2 x ii x
x x
(iii)sin 3 x 3 sin x 4 sin x ( iv )cos 3 x 4 cos x 3 cos x
3 3
(v) 2 sin A cos B sin A B sin A B
(vi) 2 cos A sin B sin A B sin A B
(vii) 2 cos A cos B cos A B cos A B
(viii) 2 sin A sin B cos A B cos A B
Sol : (i) Let x x
x y x x cos 3 3 cos 4
1 cos 2 sin cos
2 3
2
3 cos 2
3 cos 3 4
0 2 cos 2 2
y n x n ^ n x n x n^
n
(ii)Let y =
sin 6 3 sin 2
sin 2
sin 2 sin cos
(^33) 3 3 x x x x x x
3 sin 2 x sin 6 x
6 sin 6 2
- 2 sin 2 32
1 n
x
n y x
n n n
(iii) )Let y =cos 3 x cos x cos 2 x
= x x x cos 4 x cos 2 x cos 2 x 2
cos 4 cos 2 cos 2 2
1 cos 4 cos 6 cos 2 2
1 x x x
x
x x 1 cos 4 4
cos 2 cos 6 4
4 cos 4
2 cos 2
6 cos 6 4
n x
n x n y x
n n
n n
(iv) )Let y =sin 3 x sin xsn 2 x
sin 2 x sin 4 x sin 2 x
sin 2 x sin 4 x sin 2 x 2
x x
x sin 2 sin 6 2
1 cos 4
x x
x sin 2 sin 6 4
1 cos 4
6 sin 6 2
2 sin 2 2
4 cos 4 4
1 n
x
n x
n y x
n n n n
(v) Let y e x
x cos 2
3
y re x n
x (^) n cos 2
3
where
2 2 r 3 2 13 &
3
tan
1
1 0
2 2 1
2 x y xy p y --------------- (1) [ 2 y 1 , throughout]
Equation (1) has second order derivative y 2 in it. We differentiate (1), n times,
term wise,
using Leibnitz‟s theorem as follows.
1 0
2 2 1
2 D x y xy p y
n
i.e ( 1 ) 0
2 2 1
2 D x y D xy D p y
n n n ---------- (2)
(a) (b) (c)
Consider the term (a):
2
2 D 1 x y
n . Taking u y 2 and ( 1 )
2 v x and applying Leibnitz‟s theorem
we get
3 3 3
2 2 2
1 1
D uv Duv CD uDv C D Dv CD uDv
n n n n n n n n
i.e
( 1 ) ( ).( 1 ) ( ). ( 1 ) ( ) ( 1 ) ( ) ( 1 )
3 2 2
3 3
2 2 2
2 2
2 2
1 1
2 2
2 2 ^
D y x D y x CD y D x C D y D x CD y D x
n n n n n n n n
2 ( ) 2
(^) n n n yn
nn n y
nn y x ny x
(^) (^) (^) n n n
n D 1 x y 1 x y 2 2 nxy 1 n ( n 1 ) y
2 2
2 ^ ----------- (3)
Consider the term (b):
D xy 1
n
. Taking u y 1 and v x and applying Leibnitz‟s theorem,
we get
2 1
2 1 2
1 1 ^1 1
D y x D y x CD y Dx C D y D x
n n n n n n
( ) 1.^ ( 1 ) 1 ( 2 ) 2
(^) n n yn
nn y x ny
n n
n D xy 1 xy 1 ny ---------- (4)
Consider the term (c):
n
n n D p y p D y p y
2 2 2 ( ) ( ) --------- (5)
Substituting these values (3), (4) and (5) in Eq (2) we get
1 2 ( 1 ) 0
2 2 1 1
2 x yn nxyn nn yn xyn nyn p yn
ie 1 ( 2 1 ) 0
2 2 2 1
2 x yn n xyn n yn nyn nyn p yn
1 ( 2 1 ) 0
2 2 2 1
2 x yn n xyn p n yn as desired.
2. Ifsin 2 log( 1 )
1
y x or y sin 2 log( x 1 ) or
2 y sin log( x 1 ) or
sin log( 2 1 )
2 y x x , show that 1 2 1 1 4 0
2 2 1
2 x yn n x yn n yn
(VTU Jan-03)
Sol: Out of the above four versions, we consider the function as
sin ( ) 2 log( 1 )
1
y x
Differentiating w.r.t x, we get
1 (^2) x
y y
ie
2 ( x 1 ) y 1 2 1 y
Squaring on both sides
2 2 1
2 x y y
Again differentiating w.r.t x,
2 1 2 1
2 x yy y x yy
or 1 2 ( 1 ) 1 4 ( 2 1 )
2 x y x y y y
or 1 2 ( 1 ) 1 4 0
2 x y x y y -----------*
Differentiating * w.r.t x, n-times, using Leibnitz‟s theorem,
1 2 1
2 2
2 1 2
D y D g x nD y D y
nn D y x nD y x
n n n n n n
On simplification, we get
2 2 1
2 x yn n x yn n yn
3. If x tan(log y ), then find the value of
2 1 x yn 2 nx 1 yn n ( n 1 ) yn (VTU July-04)
Sol: Consider x tan(log y )
i.e. tan x log y
1
or
x y e
tan^1
Differentiating w.r.t x,
2 2
tan 1 1 1
1
x
y
x
y e
x
2 1
2 x y y ie x y y -----------*
We differentiate * n-times using Leibnitz‟s theorem,
We get
2 D x y D y
n n
ie.
2 2 1
2 2
2 1
1 1
2 1 ^
D y x CD y D x CD y D x D y
n n n n n n
ie ( 2 ) 0 .... 0 2!
2 1
n ^ n yn yn
nn y x ny x
2 x yn nx yn nn yn
4. If y m^ y m 2 x
1 1
, or
m y x x 1
2
or
m y x x 1
2
Show that 1 ( 2 1 ) 0
2 2 2 1
2 x yn n xyn n m yn (VTU Feb-02)
Sol: Consider x
y
y y x y m
m m m 2
1 1 1
1 2 1 ym^ xym Which is quadratic equation in y m
1
