www.EngineeringEBooksPdf.com
THE BOOK WAS
DRENCHED
www.EngineeringEBooksPdf.com
www.EngineeringEBooksPdf.com
DIFFERENTIAL
CALCULUS
SIIANTI
NARAYAN
www.EngineeringEBooksPdf.com
www.EngineeringEBooksPdf.com
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
www.EngineeringEBooksPdf.com
(Review published in Mathematical
Gazette,
London, December, 1953
THE
book has reached
can be assumed that
it
Its
meets
5th edition
all
in
demands.
9 years and
Is
it
the revie-
wer's fancy to discern the influence of G. H. Hardy
opening chapter on
clearly dealt with
Or
the
in
numbers, which are well and
real
?
it
this only to
is
be expected from
an author of the race which taught the rest of the world
how
to count
?
*
.-
"i
The course followed
and there
is
a
*
*
comprehensive and thorough,
v
good chapter on curve tracing. The author
is
has a talent for clear
and
exposition,
is
sympathetic to the
difficulties of the beginner.
*
#
*
Answers to examples, of which there are good and ample
selections, are given.
*
Certaianly
excellent
#
Mr.
Narayan's
Our own young
3
command
scientific
of
or
English
is
mathematical
grumbling over French or German or Latin as
additions to their studies, would do well to consider their
specialist,
Indian
confreres,
with
English
to
master
technical education can begin.
www.EngineeringEBooksPdf.com
before
their
DIFFERENTIAL CALCULUS
FOR
B. A.
&
B. Sc.
STUDENTS
By
SHANTI NARAYAN
Principal,
and Head of
Hans Raj
the
Department of Mathematics
College, Delhi University
TENTH REVISED EDITION
1962
S.
CHAND &
CO.
NEW DELHI BOMBAY
JULLUND tTR LUC KNOW
DELHI
www.EngineeringEBooksPdf.com
Books by the same Author
Integral Calculus
Rs.
6'25
Modern Pure Geometry
Rs.
5'00
Analytical Solid
Rs.
5-50
Geometry
A Course of Mathematical Analysis
Rs. 15*00
Theory of Functions of a Complex Variable
A Text Book of Matrices
Rs. 12-50
7'50
Rs.
A Text Book of Vector Algebra
A Text Book of Vector Calculus
A Text Book of Cartesian Tensors
A Text Book of Modern Abstract Algebra
A Text Book of General Topology (Under preparation)
Rs.
6 50
Rs.
7'50
Rs.
6 50
Rs. 14'00
Published by
CHAND &
S.
for
Shyam
Lai Charitable Trust,
16B/4, Asaf
AH
Ram
Road,
New
Delhi.
book are spent on
(All profits from this
S.
CO.
CHAND
charities.)
CO
&
NEW
Nagar
Fountain
DELHI
DELHI
BOMBAY
Lamington Road
Mai Hiran Gate
JULLUNDUR
LUCKNOW
Lai Bagh
First Edition
1942
Tenth Edition 1962
Price
Published by
G S
Shartna. for S.
:
Rs. 7-00
Chand
Printed at Rajendra Printers.
&
Co.,
Ram
Ram
Nagar,
Nagar,
New
www.EngineeringEBooksPdf.com
New
Delhi-1*
Delhi and
Preface to the Tenth Edition
The book has been
revised.
A
few more exercises drawn from
the recent university papers have been given.
SHANTI NARAYAN
30th April, 1962.
PREFACE
This book is meant for students preparing for the B.A. and
Some topics of the Honours
B.Sc. examinations of our universities.
standard have also been included. They are given in the form of
appendices to the relevant chapters. The treatment of the subject
is rigorous but no attempt has been made to state and prove the
theorems in generalised forms and under less restrictive conditions as
It has also been a
is the case with the Modern Theory of Function.
constant endeavour of the author to see that the subject is not preThis is to see that the student
sented just as a body of formulae.
does not form an unfortunate impression that the study of Calculus
consists only in acquiring a skill to manipulate some formulae
through 'constant
drilling'.
The book opens with a brief 'outline of the development of
Real numbers, their expression as infinite decimals and their representation by points 'along a line. This is followed by a discussion
of the graphs of the elementary functions x"> log x, e x sin x, sin- 1 *,
Some of the difficulties attendant upon the notion of inverse
etc,
functions have also been illustrated by precise formulation of
Inverse trigonometrical functions. It is suggested that the teacher
in the class need refer to only a few salient points of this part of the
book. The student would, on his part, go through the same in
complete details to acquire a sound grasp of the basis of the subject.
This part is so presented that a student would have no difficulty in
an independent study of the same.
,
The first part of the book is analytical in character while the
later part deals with the geometrical applications of the subject.
But this order of the subject is by no means suggested to be rigidly
different order may usefully be adopted at
followed in the class.
A
tho discretion of the teacher.
An
analysis of the 'Layman's' concepts has frequently been
to serve as a basis for the precise formulation of the corresponding 'Scientist's' concepts. This specially relates to the two
concepts of Continuity and Curvature.
made
www.EngineeringEBooksPdf.com
Geometrical interpretation of results analytically obtained have
been given to bring them home to the students. A chapter on 'Some
Important Curves' has been given before dealing with geometrical
applications. This will enable the student to get familiar with the
names and shapes of some of the important curves. It is felt that a
student would have better understanding of the properties of a
curve if he knows how the curve looks like.
This chapter will also
serve as a useful introduction to the subject of Double points of a
curve.
Asymptote of a curve has been defined as a line such that the
distance of any point on the curve from this line tends to zero as
the point tends to infinity along the curve. It is believed that, of all
the definitions of an asymptote, this is the one which is most natural.
It embodies the idea to which the concept of asymptotes owes its
importance. Moreover, the definition gives rise to a simple method
for determining the asymptotes.
The various principles and methods have been profusely
by means of a large number of solved examples.
illus-
trated
I
am indebted to
Prof. Sita
Ram
Gupta, M.A., P.E.S., formerly
Government College, Lahore who very kindly went through the
manuscript and made a number of suggestions. My thanks are also
due to my old pupils and friends Professors Jagan Nath M.A.,
Vidya Sagar M A., and Om Parkash M A., for the help which they
of the
rendered
me
in preparing this book.
Suggestions for improvement will be thankfully acknowledged.
January, 1942
SHANTI NARAYAN
www.EngineeringEBooksPdf.com
CHAPTER
1
Real Numbers, Variables, Functions
ARTICLE
ri.
1'2.
1*3.
1*4.
1'5.
1 6.
1*7.
PAGE
Introduction
Rational Numbers
...
...
Real Numbers
Irrational Numbers.
Decimal representation of real numbers
The modulus of a real number
Variables, Functions
Some important types of domains of variation
Some Important
13
14
Classes of Functions and their Graphs
Graphs of y=x*
Monotoiiic Functions. Inverse Functions
n
Graph of y x*i
x
Graph of y=a
Graph of >>^log a x
Graphs of sin x, cos x, tan x, cot x, sec x,
26.
11
...
II
2*2.
2-5.
...
...
2-1.
2-4.
...
...
Graphical representation of functions
CHAPTER
2-3.
...
1
2
5
6
9
=
...
18
...
20
22
24
25
...
...
...
...26
cosec x
2-7.
Graphs of shr^x, cos^x, tan^x, cot~*x see^x,
2*8.
cosec^x
Function of a function
...
2-9.
Classification of functions
...
;
CHAPTER
...
30
34
35
III
Continuity and Limit
3-1.
32.
3*3.
3-4.
35.
361.
3'62.
3-63.
Continuity of a function
Limit
...
...
..
Theorems on limits
Continuity of sum, difference, product and
quotient of two continuous functions. Con...
tinuity of elementary functions
Some important properties of continuous functions
...
Limit of x n as -> oo
...
Limit of x n jn as n -> oo
...
Limit of (1 1//I)" as n ->oo
...
Limit of [(a*~l)/Jt] as x -0
X
...
Limit of [(x
-^)/(x-o)] as x->a
...
as
x->0
Limit of (sin .v/x)
...
Hyperbolic Functions and their graphs
7;.
Inverse Hyperbolic Functions
I
+
-
3-64.
3-65.
3-66.
3-7.
3-8.
www.EngineeringEBooksPdf.com
37
41
52
53
54
57
58
59
63
64
64
66
68
(
VI
)
CHAPTER IV
Differentiation
4-1.
4*11.
4-12.
4-14.
4-15.
4*16.
Indroduction.
Derivability.
Rate of change
Derivative
Derived Function
An Important theorem
Geometrical Interpretation of a derivative
Expressions for velocity and acceleration
4 22.
Derivative of x
4-3.
Spme
4-34.
4*35.
4-36.
4*4.
4*5.
4-61.
4-62.
4-71.
4-8.
4-91.
4*92.
4-93.
72
72
74
74
76
78
TO
...
...
...
...
...
...
a
general theorems on differentiation
Derivative of a function of a function
Differentiation of inverse functions
Differentiation of functions defined
81
86
88
...
...
...
by means
88
89
93
96
97
97
99
100
of a parameter
...
...
Derivatives of Trigonometrical Functions
Derivative of Inverse Trigonometrical Functions
...
Deri vati ve of loga x
...
Derivative of (f
...
Derivatives of Hyperbolic Functions
...
Derivatives of Inverse Hyperbolic Functions
Logarithmic differentiation
Transformation before differentiation
...
Differentiation 'ab
...
initio*
Appendix
...
102
104
108
...
113
...
116
...
118
...
CHAPTER V
Successive Differentiation
5-1.
5*2.
Notation
Calculation of nth derivative.
Some standard
results
5*3.
Determination of the
th derivative of rational
functions
5 -4.
5*5.
5 -6.
Determination of the nth derivative of a product of the powers of sines and cosines
Leibnitz's theorem
Value of the nth derivative for
x0
CHAPTER
General Theorems.
tvl.
6*2.
6*3.
6*4.
...
120
...
121
...
123
VI
Mean Value Theorems
Introduction
...
Rolled theorem
Lagrange's mean value theorem
Some deductions from mean value theorem
Cauchy's mean value theorem
...
www.EngineeringEBooksPdf.com
...
...
...
.
130
130
133
136
137
(
vn
)
development of a function in a finite
form
Lagrange's and Cauchy's forms of remainders
Appendix
6*6,6-7. Taylor's
;
CHAPTER
Maxima and Minima,
7-1.
Greatest and Least values
Definitions
...
A
...
CHAPTER
Evaluation of
...
...
VIII
8-6.
8'7.
Limit of /(x)
.
8-2.
8-4.
8-5.
F'(l
\Y
'
"
148
149
151
157
Indeterminate forms
limits.
Introduction
...
Limit of [/(x)/ir (x)] when/(x)->0 and F(x)->0 ...
Limit of [/(x)/F(x)] when/(;c)->oo and F(xj -*oo ...
Limit of [/(x).F(x)] when/(x)->0 and F(x)->oo ...
Limit of [/(*) F(x)J when/(x)->oo and F(x)->oc
8-1.
140
145
VII
necessary condition for extreme values
7-3,7-4. Criteria for extreme values
7*6.
Application to problems
7*2.
...
...
under various conditions
...
165
166
170
173
174
175
CHAPTER IX
Taylor's Infinite Series
9*
1
.
Definition of convergence
and of the sum of an
...
179
180
181
185
Rigorous proofs of the expansions of
m
cos
...
188
infinite series
...
9-2,9*3. Taylor's and Maclaurin's infinite series
Formal expansions of functions
9-4.
9-5.
Use of
infinite series for evaluating limits
Appendix.
ex
,
sin x,
x, log (l-fx),
...
...
(l+x)
CHAPTER X
FUNCTION OF
VARIABLES
WO
Partial Differentiation
4
10 1.
10 '2.
10*4.
10*5.
10'6.
10*7.
10*71.
...
Introduction
Functions of two variables and their domains
...
of definition
...
variables
a
two
of
of
function
Continuity
...
Limit of a function of two variables
...
Partial Derivatives
Geometrical representation of a function of
...
two variables
Geometrical interpretation of partial derivatives
...
of first order
www.EngineeringEBooksPdf.com
193
193
194
195
196
197
198
(
10*81.
10*82.
Eider's theorem on
10*91.
Theorem on
VIII
)
...
199
...
204
Calculation
...
Differentiation of Composite Functions
...
...
Differentiation of Implicit Functions.
Appendix. Equality of Repeated Derivatives.
Extreme values of functions of two variables.
206
210
213
Homogeneous functions
A new Nota-
Choice of independent variables.
tion
10*93*
10*94.
total Differentials.
Approximate
Lagrange's Multipliers
...
Miscellaneous Exercises I
...
230
232-237
...
238
...
239
CHAPTER XI
Some Important Curves
IT!.
11*3.
11*4.
Catenary
Explicit Cartesian Equations.
Parametric Cartesian Equations. Cycloid,
Hypocycloid. Epicycloid
Branches of a
Implicit Cartesian Equations.
Cissoid, Strophoid, semi-Cubical parabola
Polar Co-ordinates
Polar Equations. Cardioide, Leinniscate.
curve.
11*5.
11-6.
Curves r m =a m cos w0, Spirals,
Curves r~a sin 3$ and r=a sin U0
CHAPTER
...
243
247
...
248
...
XII
Tangents and Normals
and Paramedic Cartesian
12-1;
Explicit, Implicit
12*2.
12*3.
12*4.
12*5.
12-6.
12-7.
12-8.
equations
Angle of intersection of two curves
Cartesian sub-tangent and sub-normal
Pedal Equations. Cartesian equations
Angle between radius vector and tangent
Perpendicular from the pole to a tangent
Polar sub-tangent and sub-normal
Pedal Equations. Polar equations
CHAPTER
..
..
..
..
..
..
..
..
254
262
264
266
267
270
271
272
XIII
Derivative of Arcs
13-3.
13-4.
An axiom
the meaning of lengths of arcs.
a
function
as
an
arc
of
Length
To determine ds/dx for the curve y=f(x)
To determine dsjdt for the curve
13-5.
To determine dsjde
13'1.
13>2.
On
*=/(0, ?=/(')
for the curve
r=/(0)
www.EngineeringEBooksPdf.com
...
.,.
...
...
274
275
275
276
277
CHAPTER XIV
Concavity, Convexity, Inflexion
14-).
14*2.
1 -t-3.
14-4.
281
Definitions
Investigation of the conditions for a curve to be
concave upwards or downwards or to have inflexion at a point
Another criterion for point of inflexion
...
Concavity and convexity
...
282
284
285
...
290
...
291
...
291
w.r. to line
...
...
CHAPTER XV
Curvature, Evolutes
15-1.
15-2.
15vJ.
15-4.
lo'-l'O.
15-47.
15*48.
15-51.
15*5").
Introduction, Definition of Curvature
Curvature of a circle
Radius of Curvature
Radius of Curvature for Cartesian curves.
Explicit Equations.
Implicit Equations.
Parametric
Equations. Newton's formulae for radius of curvature. Generalised
Newtonian Formula.
Radius of Curvature for polar curves
Radius of Curvature for pedal curves
Radius of Curvature for tangential polar curves
Centre Curvature. Evolute. Involute. Circle
of curvature. Chord of curvature
Two
properties of evolutes
...
...
...
...
...
...
292
298
299
300
304
310
CHAPTER XVI
Asymptotes
16-1.
Definition
16'2.
...
Determination of Asj^mptotes
Determination of Asymptotes parallel to co...
ordinate axes
,.:_<
of
rational
Asymptotes
general
Alge^raic'Equations
IO31.
lfi-32.
...
"
1(5-4.
lfi-5.
](>(>.
I(r7.
1C-8.
Asymptotes by inspection
Intersection of a curve and
Asymptotes by expansion
...
its
asymptotes
Position of a curve relative to its asymptotes
Asymptotes in polar co-ordinates
...
...
...
...
313
313
315
317
324
325
328
329
331
CHAPTER XVII
Singular Points, Multiple Points
17*1.
17-2.
17-31.
17*4,
Introduction
...
Definitions
Tangents at the origin
Conditions for multiple points,
...
www.EngineeringEBooksPdf.com
...
,..
335
336
336
339
17-5.
Types of cusps
...
17-6.
Radii of curvature at multiple points
...
CHAPTER
342
345
XVIII
Curve Tracing
18*2.
18-3.
18-4.
18-6.
18-6.
Procedure for tracing curves
2
Equations of the form }> =/(x)
Equations of the form y*+yf(x)+F(x)-^0
Polar Curves
Parametric Equations
...
348
349
357
359
364
...
369
...
...
...
...
CHAPTER XIX
Envelopes
19
f
l.
One parameter family of curves
Definitions.
Envelopes ofy=MX-t-ajm
19*2, 19'3.
19*4.
19*5.
19*6.
...
obtained 'ab initio*
...
Determination of Envelope
Evolute of curve as the Envelope of its normals ...
Geometrical relations between a family of curves
...
and its envelope
Miscellaneous Exercises II
...
Answers
,,.
www.EngineeringEBooksPdf.com
369
270
37:2
372
378-82
383-408
CHAPTER
I
REAL NUMBERS
FUNCTIONS
Introduction.
The subject of Differential Calculus takes its
stand upon the aggregate of numbers and it is with numbers and
with the various operations with them that it primarily concerns
itself.
It specially introduces and deals with what is called Limiting
operation in addition to being concerned with the Algebraic operations of Addition and Multiplication and their inverses, Subtraction
and Division, and is a development of the important notion of Instantaneous rate of change which is itself a limited idea and, as such, it
finds application to all those branches of human knowledge which
deal with the same. Thus it is applied to Geometry, Mechanics and
other branches of Theoretical Physics and also to Social Sciences such
as Economics
It
may
and Psychology.
be noted here that this application
is
essentially based
on the notion of measurement, whereby we employe-numbers to
measure the particular quantity or magnitude which is the object of
investigation in any department of knowledge. In Mechanics, for
instance, we are concerned with the notion of time and, therefore, in
the application of Calculus to Mechanics, the first step is td~ correlate
the two notions of Time and Number, i.e., to measure time in terms
of numbers. Similar is the case with other notions such as Heat,
Intensity of Light, Force, Demand, Intelligence, etc. The formula^
tion of an entity in terms of numbers, i.e., measurement, must, of
course, take note of the properties which we intuitively associate with
the same. This remark will later on be illustrated with reference to
the concepts of Velocity, Acceleration Gwry^fr&e, etc.
The importance of numlTe^l^/Bfe-sradj^bf i^subject in hand
being thus clear, we will in some* .of the; f#ltowig. articles, see how we
were first introduced to the notion of number and how, in course of
time, this notion came to* be subjected to a series of generalisations.
:
It is, however, not intended to give here any logically connected amount of the development of tl>e system of real numbers, also
known as Arithmetic Continuum and only a very brief reference to
some well known salient facts will suffice for our purpose. An
excellent account of the. -Development of numbers is given in
'Fundamentals of Analysis' by Landau.
It may also be mentioned here .that even though it satisfies a
deep philosophical need to base the theory part of Calculus on the
notion *>f number ialohe, to the entire exclusion of every physical
basis, but a rigid insistence on the same is not within the scope of
www.EngineeringEBooksPdf.com
2
this
will
DIFFERENTIAL CALCULUS
book and intuitive geometrical notion of Point, Distance, etc.,
sometimes be appealed to for securing simplicity.
Rational numbers and their representation by points along a
1-1.
straight line.
Positive Integers.
I'll.
It
was to the numbers,
1, 2, 3, 4, etc. r
we were first introduced through the process of counting certain
The totality of these numbers is known as the aggregate of
objects.
that
natural numbers, whole numbers or positive integers.
While the operations of addition and multiplications are^,unrestrictedly possible in relation to the aggregate of positive integers,
this is not the case in respect of the inverse operations of subtraction
and 'division. Thus, for example, the symbols
are meaningless in respect of the aggregate of positive integers.
Fractional numbers.
1*12.
At a later stage, another class of
numbers like p\q (e.g., |, |) where p and q are natural numbers, was
added to the former class. This is known as the class of fractions
and it obviously includes natural numbers as a sub- class q being
;
equal to
in this case.
1
The introduction of Fractional numbers is motivated, from an
abstract point of view, to render Division unrestrictedly possible and,
from concrete point of view, to render numbers serviceable for
measurement also in addition to counting.
1-13.
Rational numbers.
Still later,
the class of numbers was
enlarged by incorporating in it the class of negative fractions including negative integers and zero. The entire aggregate of these numbers
is known as the aggregate of rational numbers.
Every rational
number is expressible as p\q, where p and q are any two integers,
positive and negative and q is not zero.
The introduction of Negative numbers is motivated, from an
abstract point of view/ to render Subtraction always possible and,
from concrete point of view^, to facilitate a unified treatment oi
oppositely directed pairs of entities such as, gain and loss, rise and
fall, etc.
Fundamental operations on rational numbers. An impor1-14.
tant property of the aggregate of rational numbers is that the
operations of addition, multiplication, subtraction and division can
be performed upon any two such numbers, (with one exception which
is considered below in
T15) and the number obtained as the result
of these operations is again a rational number.
This property
rational
numbers
expressed by saying tli^t the ag^egate of
closed with respect to the four fundamental
is
is
operations.
1-15.
Meaningless operation of division by zero. It is important
to note that the only exception to the above property is 'Division
www.EngineeringEBooksPdf.com
REAL NUMBERS
by
zero'
follows
which
3
a meaningless operation.
is
This
may
be seen as
:
To
divide a
by b amounts to determining a number c such that
bc=a,
will be intelligible only, if and only if, the determi-
and the division
nation of c is uniquely possible.
Now, there is no number which when multiplied by zero produces a number other than zero so that a JO is no number when 0^0.
Also any number when multiplied by zero produces zero so that 0/0
may be any number.
On account of this impossibility in one case and indefiniteness in
the other, the operation of division by zero must be always avoided.
A disregard of this exception often leads to absurd results as is
illustrated below in
(i)
(/).
Let
jc-
6.
Then
;c
2
-36:=.x-6,
or
8.
(x--6)(jc-f.6)=Jt
Dividing both sides by jc 6, we get
jc+6=l.
6+6=1,
which
is
12
= 1.
clearly absurd.
Division by jc
absurd conclusion.
(ii)
We may
6,
^lf =
which
remark
also
X
is
zero here,
is
responsible for this
in this connection that
6)
(
*~^~
=x+6,
only when *j66.
...
(1)
hand expression, (je2 36)/(x 6), is meaningwhereas the right hand expression, x-f 6, is equal to 12 so that
For *=6, the
less
i.e.,
left
the equality ceases to hold for Jt=6.
The equality (1) above is proved by dividing the numerator
and denominator of the fraction (x 2 36)/(x 6) by (.x 6) and this
operation of division is possible only when the divisor (jc -6) ^0, i.e*
9
when x^6.
This explains the restricted character of the equality (1).
1.
Ex.
Show that the aggregate of natural numbers is not closed with
respect to the operations of subtraction and division. Also show that the
aggregate of positive fractions is not closed with respect to the operations of
subtraction.
Ex.
2.
Show
that
every rational number
is
expressible as
a terminating
or a recurring decimal.
To decimalise plq, we have first to divide/? by q and then each remainder,
after multiplication with 10, is to be divided by q to obtain the successive
figures in the decimal expression of p/q. The decimal expression will be
terminating if, at some stage, the remainder vanishes. Otherwise, the process
In the latter case, the remainder will always be one of the
will be unending.
finite set of numbers 1,2 .......
tf1 and so must repeat itself at some stage.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
*
From this stage onward, the quotients will also repeat themselves and the
decimal expression will, therefore, be recurring.
The student
some
will understand the argument better if he actually expresses
fractional numbers, say 3/7, 3/13, 31/123, in decimal notation.
Ex.
For what values of x are the following equalities not valid
3.
W --*+.
-!.
(0
:
-
tan -
1*16.
Representation of rational numbers by points along a line
or by segments of a line.
The mode of representing rational numbers
by points along a line or by segments of a line, which may be known
as the number-axis, will now be explained.
We
O
marking an arbitrary point
The
the
origin or zero point.
calling
be represented by the point O.
start with
axis
and
Any
one of these
it
O
The point
-
i
Q
"ft
Fi
and the
On
it
may
o
"
tive
-
divides the
hand
axis into
zero will
two parts or
sides.
be called positive and the other, then negative.
is
number-axis
Usually, the
!
-
left
number
on the number-
number
side of
the positive side,
O
drawn
parallel to the printed
of the page and the right
hand side of O is termed posilines
negative.
we take an
arbitrary length
OA, and
call
the unit length.
We
say, then, that the
number
1 is
represented
by the
point A.
After having fixed an origin, positive sense and a unit length on
the number axis in the manner indicated above, we are in a position
to determine a point representing any given rational number as
explained below
:
Positive integers.
m.
We take
ger,
m.
Firstly,
we
consider
any
positive integer,
a point on the positive side of the line such that its
distance from O is m times the unit length OA. This point will be
reached by measuring successively m steps each equal to OA starting
from O. This point, then, is said to represent the positive inte-.
m, w<
Negative integers. To represent a negative integer,
take a point on the negative side of O such that its distance from
is
times the unit length OA.
m
This point represents the negative integer,
m.
Fractions. Finally, let p\q be any fraction
q being a positive
OB being one
Let OA be divided into q equal parts
integer.
of them. We take $ point on the positive or negative side of O
according as p is positive or negative such that its distance from O is
p times (or, p times jf p is negative) the distance OB.
;
;
The point so obtained represents the
fraction, p\q*
www.EngineeringEBooksPdf.com
REAL NUMBERS
6
If a point P represents a rational number pjq, then the measure
of the length OP is clearly p\q or
p\q according as the number is
positive or negative.
Sometimes we say that the number, p/q,
segment OP.
is
represented
by the
We
Real numbers.
have seen in the
number
can
be
rational
every
represented by a
the line with
see
can
cover
of
a
it
to
that
is
line.
we
Also,
easy,
point
such points as closely as we like. The natural question now arises,
"Is the converse true ?" Is it possible to assign a rational number
to every point of the number-axis ?
simple consideration, as de1-2.
Irrational numbers.
last article that
A
tailed below, will clearly
show that
it is
not
so.
Construct a square each of whose sides is equal to the unit length
P on the nutnber-axis such that OP is equal in
the length to the diagonal of the square.
It will
now be shown +hat the point P
cannot correspond to a rational number
of OP cannot have a
i.e., the length
7i r
rational number as its measure.
OA
and take a point
u
Fig. 2.
measure be a rational number pjq so that,
by Pythagoras's theorem, we have
If possible, let
its
= 2,
i.e.,
p*
(0
2q*.
We may suppose that p and
factors, if any,
Firstly
q have no common factor,
can be cancelled to begin with.
we
for,
such
notice that
so that the square of
an even number
even and that of an odd
is
number is odd.
From thHtion (/), we
Therefore, p itlHpinst be even.
2
that
equal to 2n where n
Let, the
Thus, #
see,
is
p2
an
is
an even number.
integer.
is alsc
Hence p and
#mfcrommo*factor 2 and this conclusion conno common factor. Thus the
the hypothesis
thaPthej^fce
measure -y/2 of OP is not a rationaffiumber. There exists, therefore^
a point on the number-axis not corresponding to any rational number,
traflicts
is
Again, we take a point L on the
any rational multiple say, p/a, of OP.
line
such that the length
www.EngineeringEBooksPdf.com
QL
DIFFERENTIAL CALCULUS
The length OL cannot have a
m\n be the measure of OL.
rational measure.
For, if possi-
ble, let
m
p
mq
/0
or v
-\/2=-\/ 2
v
n
q
p
which states that <\/2 is a rational number,
being equal to mqjnp.
This is a contradiction. Hence L cannot correspond to a
rational number.
=
>
'
Thus we see that there exist an unlimited number of points on
the number-axis which do not
correspond to rational numbers.
If we now require that our aggregate of numbers should be such
that after the choice of unit length on the line,
every point of the
line should have a number
to
it
(or that every length
corresponding
should be capable of measurement), we are forced to extend our system of numbers further by the introduction of what are called irrational
numbers.
We will thus associate an irrational number to every point of
the line which does not correspond to a rational number.
A
method of representing
notation
is
irrational
given in the next article
Def. Real number.
real number.
A
numbers in the decimal
1-3.
number, rational or
irrational, is called
Theaggregate of rational and irrational number
aggregate of real numbers.
is,
a
thus, the
Each real number is represented by some point of the numberaxis and each point of the number-axis has some real number,
rational or irrational, corresponding to it.
Or,
we might
OP
length
say, that each real number is the measure of some
and that the aggregate of real numbers is enough to
measure every length.
1-21.
Number and Point. If any number, say x, is represented by a point P, then we usually say that the point P is x.
Thus the terms, number and point, are generally used in an
indistinguishable manner.
1-22.
Closed and open intervals, set a, b be two given numbers
such that #<6. Then the set of numbers x such thata^x^fe is
called a closed interval denoted by the symbol [a, b].
Also the set
of numbers x such that
a<x<b
open interval denoted by the symbol (a,b).
The number b a is referred to as the length of
<.
is
[a, b]
called
an
as also of
*).
Decimal representation of real numbers. Let P be any
1-3.
given point of the number-axis. We now seek to obtain the decimal
representation of the number associated with the point P.
www.EngineeringEBooksPdf.com
REAL NUMBERS
To start with,
side of O.
we suppose that the
7
point
P
on the positive
lies
Let the points corresponding to integers be marked on the
number-axis so that the whole axis is divided into intervals of length
one each.
Now, if P coincides with a point of division, it corresponds to an
In case P falls between two
integer and we need proceed no further.
l, we sub-divide the interval (a, a+l)
points of division, say a, a
The points
into 10 equal parts so that the length of each part is T1T
+
.
of division, now, are,
#, tf-f TIP
P
coincides with any of these points of division, then it corresIn the alternative case, it falls between
a
rational number.
ponds
two points of division, say
If
to
i.e.,
o.a v
where, a l9
is
We
tf.K+1),
one
of
the
any
integers 0,
1, 2, 3,
,
9.
again sub-divide the interval
r
a,
+ io'
a+
,
,
L*
<Ji+i
10
n
J
into 10 equal parts so that the length of each part
is
1/10
2
.
The points of division, now, are
l
fll
fl
io
_!_"
+io
+ io'
4_
The point P
i
a>
2
9
7
//-u' !
.
.
coincide with one of the above points
corresponds to a rational number) or will
between two points of division say
of division
lie
-
(in
will
cither
which case
it
10*
i.e.,
where a 2
is
one of the integers
0, 1, 2, ...... , 9.
We
again sub-divide this last interval and continue to repeat
the process. After a number of steps, say n, the point P will either
be found to coincide with some point of division (in this case it
corresponds to a rational number) or lie between two points of the
form
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
the distance between which
and smaller as n increases.
is
and which
1/10*
clearly gets smaller
The process can clearly be continued indefinitely.
The successive intervals in which P lies go on shrinking and
clearly close up to the point P.
This point P is then represented by the infinite decimal
Conversely, consider
and construct the
[a,
series
will
any infinite decimal
of intervals
0+1], [a.a& a.a^+l],
[
Each of these intervals lies within the preceding one
their
lengths go on diminishing and by taking n sufficiently large we can
make the length as near to zero as we like. We thus see that these in;
tervals shrink to a point.
This fact is related to the
perceived aspect of the continuity of a straight line.
intuitively
Thus there is one and only one point common to this
and this is the point represented by the decimal
series
of
intervals
Combining the
results of this article with that of Ex. 2, !!,
see that every decimal, finite or infinite, denotes a number which
rational if the decimal is terminating or recurring and irrational in the
p. 3,
is
we
contrary case.
Let, now,
representing it is
P
lie
on the negative
side of O.
Then the number
#-#i#2**#i ......
where
a.a^ ...... a n ......
the number representing the point P' on the positive side oft? such
that PP' is bisected at O.
is
Ex.
1.
Calculate the cube root of 2 to three decimal places.
We have
I=*l<2and23 =8>2.
l<3/2<2.
We consider the numbers
1,M,
1'2, ....... 1-9,2,
which divide the interval [1, 2] into 10 equal parts and find two successive nunv
bers such that the cube of the first is <2 and of the second is >2. We find that
Again consider the numbers
1-2, 1*21, 1-22 ......
which divide the interval
[1*2, 1-3] into
,
1-29, l'3 r
10 equal parts.
www.EngineeringEBooksPdf.com
REAL NUMBERS
We find
that
(l'25)
8
=l-953125<2and(l-26) -2'000376>2.
8
l-25<3/2<l-26.
Again, the numbers
1-25, 1-251, 1-752. ....... ,1*259,1-26
divide the interval [1-25, 1-26] into 10 equal parts
We find
that
8
(1-259)=1-995616979<2 and (1'26) =2'000376>2
!-259<3/2<l-26.
Hence
Thus
to three decimal places,
we have
?/2=l-259.
Ex. 2.
Calculate the cube root of 5 to 2 decimal places.
Note. The method described above in Ex. 1 which is indeed very cumbersome, has only been given to illustrate the basic and elementary nature or
the problem. In actual practice, however, other methods involving infinite
series or other limiting processes are employed.
x,
1*4.
The modulus of a
Def.
the modulus of a real number, x, is meant
according as x is positive, negative, or zero.
x or
real number.
the
By
number
The smybol x is used to denote the modulus ofx.
Thus the modulus of a number means the same thing as ite
numerical or absolute value. For example, we have
Notation*
\
\
|3
=3;| -3
5-7
|
|
|
|
=-(-3)=3;
= 7-5
|
|
[
=0
;
=2.
|
The modulus of the difference between two numbers is the
measure of the distance between the corresponding points on the
number-axis.
We
Some results involving moduli.
now state
useful results involving the moduli of numbers.
a+b |<
1-41.
|
i.e.,
the modulus of the
a
|
|
+
b
|
|
sum of two numbers
sum of their moduli.
The result is almost self-evident.
its truth more clearly, we split it up
,
is less
than or equal to the
To enable the reader
1.
In this
Let
case,
a,
we
I
e.g.,
and
|
|
b have the same
sign.
clearly have
=
b
a +
=
7
+ 3
-7-3 = -7 + -3
a+6
7+3
to see
two cases giving exampfes-
into
of each.
Case
some simple and
.
I
|
|
|
|
|
|
|
|
|
|
|
|
I
|
www.EngineeringEBooksPdf.com
,
.
*
DIFFERENTIAL CALCULUS
Case
Let
II.
In this
case,
we
I
4=
e.g.,
Thus
b have opposite
a,
|
clearly
have
a+b
7-3
<
<
|
|
a
+
+
|
|
7
I
|
\
,
=10.
3
|
|
+
a+b < \a\
\
=
ab
1-42.
b
|
we have
in either case,
I
signs.
|
a
.
b
.
|
|
\b\
|
,
|
|
e.g., the modulus of the product of two numbers
duct of their moduli,
e.g-,
4-3
|
(-4)(-3)
|
If x, a,
1-43.
=12=
=12=
|
/,
|
4
|
3
.
|
-4
-3
.
|.
|
;
|
|
|
equal to the pro-
is
;
|
be three numbers such that
x
<*>
I
I
(^)
then
i.e.,
x
lies
between a
and
I
a-\-l or that
x
belongs to the open
interval
The
inequality (A) implies thai the numerical difference between
less than /, so that the point x (which may He to the
right or to the left of a) can, at the most, be at a distance / from the
point a.
a and x must be
Now, from the
d+t
a,
see that this
Fig 3
may
It
if
x
lies
figure,
I
clearly
and only
and a+l.
possible,
between a
we
if
also be at once seen that
*-!<'
I
is
f
is
closed interval
equivalent to saying that, x, belongs to the
[a-l, a+l].
Ex.
1.
//
1
a-b
|
\
We have
\
b-c
1
a-c
|
\
<m, show
that
</+/w.
= a-b+b
1
c
\
<
1
a-b +
Give the equivalents of the following
2.
notation
a-c
|
</,
\
\
b-c
\
<l+m.
terms of the modulus
in
:
(0
-1<*<3. () 2<x<5.
(m) -3<;t<7.
(iv)
/~s<x</+s.
Give the equivalents of the following by doing away with the modu3.
lus notation :
(/)
4.
If
|
y=
x_2
x
\
\
|
<3.
+
l
1
\x+\\ <2. (7) 0<
(//)
jc
2x,
1
|
,
|
then show that
fora;<0
forO<A:<l.
www.EngineeringEBooksPdf.com
x-\
|
<2.
11
FUNCTIONS
1*5.
Variables, Functions. We give below some examples to
enable the reader to understand and formulate the notion of a variable and a function.
Ex.
"where
Consider two numbers x and y connected by the relation,
1.
we take only
the positive value of the square root.
Before considering this relation, we observe that there is no
is negative and hence, so far as real numbers are concerned, the square root of a negative number does not
real
number whose square
exist.
Now,
1
jc
2
equal to 1. This
the relation
i.e.,
[
,
is
is
x 2 is less than or
any number satisfying
positive or zero so long as
the case
if
and only
if
x
is
when x belongs to the interval [1, I].
If, now, we assign any value to x belonging
1, 1],
to the
interval
then the given equation determines a unique corresponding
value of y.
The symbol x which, in the present case, can take up as its
value any number belonging to the interval [
1, 1], is called the
independent variable and the interval [1, 1] is called its domain o?
variation.
The symbol y which has a value corresponding to each value of
x in the interval [1, 1] is called the dependent variable or & function
1, 1].
of x defined in the interval
["
2.
Consider the two numbers x and y connected by the relation,
Here, the determination of y for x ~2 involves the meaningless
operation of division by zero and, therefore, the relation does not
But for every other
assign any value to y corresponding to jc=2.
value of x the relation does assign a value to y.
Here, x is the independent variable whose domain of variation
consists of the entire aggregate of real numbers excluding the number
2 and y is a function of x defined for this domain of variation of x.
3.
Consider the two
defined by the equations
numbers x and y with
y=^x* when x<0,
y=x when 0<x<l,
y=ljx when x>l.
their
relationship
...(/)
(")
...(iii)
These relations assign a definite value to y corresponding to
every volue of x, although the value of y is not determined by a single
www.EngineeringEBooksPdf.com
DIFFEBENTIAL CALCULUS
12
formula as in Examples 1 and 2. In order to determine a value of
to a given value of x, we have to select one of the
three equations depending upon the value of x in question. For
y corresponding
instance,
forx=
2,
forx=,
for
2
y=(
2)
=4, [Equation
(/)
[Equation
(i7)
[Equation
(in)
j=J
^=|
*=3,
Here again, y
is
v 2<0
v 0<J<1
v
3>1.
a function of x defined for the entire aggregate
9
of real numbers.
This example illustrates an important point that it is not necesthat
sary
only one formula should be used to determine y as a function of x. What is required is simply the existence of a law or laws
which assign a value to y corresponding to each value of x in its
domain of variation.
Let
4.
Here y
is
a function of x defined for the aggregate of positive
integers only.
Let
5.
Here we have a function of x defined
real
for the entire aggregate of
numbers.
It
follows
may
be noticed that the same function can also be defined as
:
6.
ys=
x when
y=
x when x<0.
Let
yt=z\lq f
when x
y=0 when
t
x
is
is
a rational number plq
in its lowest
terms r
irrational.
Hence again y is a function of
of
real numbers.
gate
x defined
for the entire aggre-
1*51.
Independent variable and its domain of variation. The
above examples lead us to the following precise definitions of variable
and function. Ifxis a symbol which does not denote any fixed number
but is capable of assuming as its value any one of a set of numbers,,
then x is called a variable and this set of numbers is said to be its
domain of variation.
1*52. Function and its domain of definition. // to each value of
an independent variable x, belonging to its domain of variation, there
corresponds, by any law, or laws, whatsoever, a value of a symbol, y+
www.EngineeringEBooksPdf.com
FUNCTIONS
13
then y is said to be a function of x defined in the domain of variation of
K, which is then called the domain of definition of the function.
Notation for a function.
1*53.
a
variable
x
is
read as the
is
The
fact that
y is a function of
expressed symbolically as
'/'
of x.
If x, be any particular value of x belonging to the domain of
variation of X then the corresponding value of the function is denoted by/(x). Thus, iff(x) be the function considered in Ex. 3,
1*5,
9
p. 11,
then
If functional symbols be required for two or more functions,
usual to replace the latter /in the symbol f(x) by other
letters such as F, G, etc.
then
it is
Some
important types of domains of variation.
Usually the domain of variation of a variable x is an interval
a, b], i.e., x can assume as its value any number greater than or
equal to a and less than or equal to 6, i.e.,
1*6.
Sometimes
it
becomes necessary to distinguish between closed
and open intervals.
If x can take up
as its value any number greater than a or less
than b but neither a nor b i.e., if a<x<b then we say that its
domain of variation is an open interval denoted by (a, b) to distinguish it from [a, b] which denotes a closed interval where x can take
up the values a and b also.
y
We may similarly
[a, b),
have semi-clos3(l or semi-opened intervals
a<x<6
;
(a, b],
a<x^b
as domains of variation.
We may
bound
also have domains of variation extending
in one or the other directions i.e., the intervals
(00,
b] or
x<6
;
[a, GO
)
or
x>a (00
;
,
oo
)
or
any
without
x.
Here it
00,00 are no
numbers in any sense whatsoever. Yet, in the following pages they
will be used in various ways (but, of .course never as numbers) and
in each case it will be explicitly mentioned as to what they stand for.
Here, for example, the symbol ( 00, b) denotes the domain of variation of a variable which can take up as its value any number less
than or equal to b.
should be noted that the symbols
Similar meanings have been assigned to the symbols
(0, GO
Constants.
called a constant.
A
),
(00
,
oo
).
symbol which denotes a certain fixed number
www.EngineeringEBooksPdf.com
is
14
DIFFERENTIAL CALCULUS
It has
like a, h, c
like JC, y, z
Note.
;
;
become customary to use earlier letters of the alp}ifl,bet r
a, (J, y v as symbols for constants and the latter letters
w, v, w*as symbols for variables.
points about the definition of a function should be
The following
carefully noted
:
A
function need not be necessarily defined by a formula or formulae so
1.
that the value of the function corresponding to any given value of the independent variable is given by substitution. All that is necessary is that some rule or
set of rules be given which prescribe a value of the function for every value of
the independent variable which belongs to the domain of definition of the function. [Refer Ex. 6, page 12.]
It is
2.
not necessary that there should be a single formula or rule for the
[Refer Ex. 3, page 11.]
whole domain of definition of the function.
Ex.
is
Show
1*
that the
domain of definition o f the function
the open interval (I, 2).
For
and
x>2
x=l
and 2 the denominator becomes zero. Also for x<l
the expression (1 -Jt) (x 2) under the radical sign becomes
negative.
is not defined for *<; 1 and x^2.
For x>l
the expression under the radical sign is positive so that a
Hence the function is defined
value of the function is determinable.
Thus the function
and
<2
in^heopen
interval (1, 2).
Show
2.
that the
the closed interval
Show
3.
are
(0, oo)
and
that the
(
domain of definition of the "function
^[(1
x)
(A*
2)]
is
[1, 2].
oo
,
domain of
definition of the functions
0) respectively.
Obtain the domains of definition of the functions
4.
(0 ^(2x+l)
1-7.
(//)
1/U + cos
A-)
(iii)
V(l-f-2 sin x).
Graphical representation of functions.
Let us consider a function
y=f(x)
defined in an interval
...
[a, b].
(/)
represent it graphically, we take two straight lines X'OX
at right angles to each other as in Plane Analytical
fwid Y'OY
Geometry* These are the two co-ordinate axes.
To
We take O
as origin for both the axes and select unit intervals
Also as usual, OX,
lengths).
arq taken as positive directions on the two axes.
on OX,
OF (usually of the same
To any number x
corresponds a point
OY
M on
www.EngineeringEBooksPdf.com
Jf-axis such that
FUNCTIONS
15
To the corresponding number y, avS determined from
on K-axis such that
corresponds a point
N
said
Completing the rectangle OMPN,
to correspond to the pair of
numbers x,
Thus
we obtain a
y*
P
there
which
is
y.
to every number x belonging to the interval [<?, b], there
N
corresponds a number y determined
by the functional equation yc=f(x)
and to
point
(/),
this pair of
corresponds a point
above.
y
o
numbers,
P
x, y
as obtained
Fi g% 4
The
totality of these points, obtained by giving different values,
to x, is said to be the graph of the function /(;c) and y=f(x) is said to"
be the equation of the graph.
1.
page 11,
Examples
The graph of the function considered
in
Ex.
3,
1'5,
is
Fig. 5
2.
The graph
of >>
excluding the point P(l,
=
2
1) is
(jc
the straight line y=x-\-l
2).
o
Fig. 6
3.
The graph of y=*x
consists of a discrete set of points(1, 1), (2, 2), (3, 6), (4, 24), etc.
\
www.EngineeringEBooksPdf.com
DrFFEBENTIAL CALCULUS
The graph of the function
4.
x,
when
1,
when
x when
I
is
as given.
Fig. 7
Draw the graph of
root o/x 2
5.
square
the function
which denotes the positive
.
As
a
*v/x
graph (Fig.
x according
=jc or
8) of
V*2
as
x
is
positive or negative, the
the graph of the function /(x) where,
*s
f x, when
=H
x,
xi
when x<0.
o
Fig. 8
Fig. 9
The student should compare the graph
graph
(Fig. 8) of
2
\/* with the
(Fig. 9) of x.
The reader may
6.
Draw
see that
the graph of
>-
*
We
have
x+lx=l
x+lx=l
2x when x<0,
when 0<x
{ x+x ^l=2x 1 when
Thus we have the graph as
drawn
The graph consists of parts of
:
3 straight lines
^
,,
corresponding to the intervals
^-oo,0].
Fig. 10
www.EngineeringEBooksPdf.com
[0, !],[!,
x).
FUNCTIONS
Dawn
*
7.
17
the graph of
[*],
where
[x] denotes the greatest integer not greater
We
have
y=4
for
1,
1^2,
The value of y
than x.
<*<!,
<x<2,
f 0, for
1
<ix<3 and
for 2
so on.
x can
for negative values of
also be similarly
given.
The right-hand end-point of each segment of the line
point of the graph.
Yk
is
not a
~
O
Fig. 11.
Exercises
1
.
Draw the
<0/(*)
graphs of the following functions
1,
=
when A-<CO
-l,when.v>0
x, when OjcCjc^i
2-*, when^v^i
_
2
f A
A- ,
^
\
2.
x.x
Draw
^v,
when
wften A<TO
x^u
when A>0;
<)/(*)
=
,^
=
,.^ f
(")/(*)
,
( v /)/(A-)
,.^
00 x
(/)
The positive value of the square root
3.
(0
4.
W
where
[*]
is
Draw the graphs of the functions
(//)
|*|.
|*| + |x + l|.
Draw the graphs of the functions
+ [*fl]
()
v
-
(
,
w
when
when
*
{!-*,
,
!
+
,
when x = 4
1
A, when
I/*. when ^<
0, when A =
.
,
1,
^
f
-
the graphs of the following functions
^* 2
:
^
^
;
-l/jc.whe
:
^(x-\)
2
to be taken in each case.
:
(///)
:
MV
M
denotes the greatest integer not greater than
x.
www.EngineeringEBooksPdf.com
2|*-1
|
+3|
:
CHAPTER
II
SOME IMPORTANT CLASSES OF FUNCTIONS AND
THEIR GRAPHS
This chapter will deal with the graphs and some
Introduction.
simple properties of the elementary functions
x n a x loga x
,
,
;
tan x, cot x, sec x, cosec x
1
1
sin~ x, cc5s~ x, tan"~ 3 x, cot~ 1 x, sec^x, cosec- 3*.
The logarithmic function is inverse of the exponential just as the
sin x,
cos x,
;
inverse trigonometric functions are inverses of the corresponding
trigonometric functions. The trigonometric functions being periodic,
the inverse trigonometric are multiple-valued and special care has,
therefore, to be taken to define them so as to introduce them as singlevalued.
Graphical representation of the function
2-1.
y=x*
;
n being any integer, positive or negative.
We have, here, really to discuss a class of functions obtained by
giving different integral values to n.
It will be seen that, from the point of view of graphs, the whole
of this class of functions divides itself into four sub-classes a&
follows
:
(i)
integer
odd
;
when n is a
(Hi) when n
positive even integer
is
;
(//)
a negative even integer
when n is a positive odd
(iv) when n is a negative
integer.
The functions belonging to the same sub -class will be seen to
have graphs similar in general outlines and differing only in details.
Each of these four cases will now be taken up one by one.
Let n be a positive even integer.
2-11.
The following
are,
obviously, the properties of the graph
n
of
whatever positive even integral
n
value,
may have.
y=x
^=0, when x=0
y=l, when x=l
>>=!, when x=
(i)
The
graph,
;
;
1.
therefore,
passes
through the points,
O
(0,0),
A
(1,1), A'
(it,
a positive even integer)
Fig. 12.
(-1,
when x
1).
is posiy
(ii)
positive
tive or negative. Thus no point on the
graph lies in the third or the fourth
is
quadrant.
18
www.EngineeringEBooksPdf.com
19
GRAPHS
We have
(wi)
same value of y corresponds to two equal and opposite
The graph is, therefore, symmatrical about the .y-axis.
so that the
values of x.
The variable y
(iv)
larger numerically.
gets larger
larger, as x gets larger and
made as large as we like by
The graph is, therefore, not
and
Moreover, y can be
taking x sufficiently large numerically.
closed.
for positive even integral value of n
in
as
Fig. 12. page. 18.
general outlines,
given
The graph of y=x*
The following
are the properties
integral value n may have.
y
when
1,
The graph,
of
y=xn
whatever positive
,
^
whenx^O;
x=l
^=0,
(i)
in
Let n be a positive odd integer.
2-12.
odd
is,
9
;
therefore, passes
through the points
0(0,0),
y
(//')
,4 (I, I),
is
according as x
,4'(-l,-l)
positive or negative
is positive or nega-
tive.
lies
Thus, no point on the graph
the second or the fourth
in
(n,
odd integer)
a positive
quadrant.
13.
Fig
The numerical value of y
(///)
increases with an
increase in the
numerical value of x.
Also, the numerical value of y can be
by taking x
2-13.
made
we
as large as
like
sufficiently large numerically.
Let n be a negative even integer, say,
m> (w>0).
Here,
xn
n,
=x
9
(n,
x
a negative even integer)
Fig. 14
~~
xm
-
The following are the
x n whatever
properties of y
negative even integral value,
\U,D
o
1
_______
may have
(i)
Determination
x=Q
of y
involves the meaningless operation of division by
and so y is not defined for
for
www.EngineeringEBooksPdf.com
20
DIFFERENTIAL CALCULUS
y=l when x=l or 1, so that
A (I, l)andX'(-l, 1).
(fl)
the graph passes through
the two points
(itt)
y
Thus no
positive whether x be positive or negative.
lies in the third or in the fourth quadrant.
is
point on the graph
(IF)
We have
so that the same value of y corresponds to the two equal and
opposite values of x. The graph is, therefore, symmetrical about
jr-aaris.
(p)
decreases.
as
As
x, starting
from
Alsojy can be
we like) by
m
increases, x also increases so that y
as small as we like (i.e., as near zero
1,
made
taking x sufficiently large.
m
Again, as x, starting from 1, approaches zero, x decreases so
that y increases. Also y can be made as large as we like by taking x
sufficiently near 0.
The
variation of y for negative values of
down by symmetry.
2-14.
x may
be,
now, put
Let n be a negative odd integer.
The statement of the various
properties for this case
is left
The graph only
reader.
is
to the
shown
here. (Fig. 15)
VJ)
Before considering the
graphs of other functions, we allow
2*2.
*
some digression in this
to facilitate some of the
ourselves
article
later considerations.
2-21.
+. a negat^ odd
integer^
Monotonic
functions.
Increasing or decreasing functions.
Fig. 15.
Det L A function y=*f(x)
is called a monotonically increasing
that
such
interval if
throughout the interval a larger
function in
value of x gives a large value of y, i.e., an increase in the value of x
always causes an increase in the value of y.
an
Def. 2.
it
is
A function is
if His such that an
called a monotonically decreasing function,
increase in the value of x always causes a decrease
in the value of the function.
Thus, if *i and x a axe any two numbers in an interval such that
www.EngineeringEBooksPdf.com
21
GBAPHS
then for a monotonically increasing function in the interval
and
for
a monotonically decreasing function in the interval
A
ing
is
function which
called a
Show
Ex.
is
mono tonic
that for a
either monotonically increasing or
function.
monotonic function /(x)
in
an interval
decreas-
(a, 6),
the
fraction
keeps the same sign for any pair of different numbers, x l9 x, of
Illustration.
xn
Looking
at
the graphs
in
(a, 6).
we
2*1,
notice that
is
(/) monotonically decreasing in the interval (
tonically increasing in the interval [0, oo] when n
integer
00, 0)
is
and mono-
a positive even
;
(//')
when n
monotonically increasing in the intervals
a positive odd. integer.
[
oo, 0], [0, oo ]
is
(in) monotonically increasing in the interval (
00,0) and monotonically decreasing in the interval (0, oo) when n is a negative even
integer.
(iv)
when n
.monotonically decreasing in the intervals
a negative odd integer.
(00,
0), (0, oo)
is
2-22.
Inverse functions.
Let
y=Ax)
...(1)
be a given function of x and suppose that we can solve this equation
for x in terms of y so that we may write x as a function of y, say
x=*(y).
Then
<p(y) is
.-.(2)
called the inverse of f(x).
This process of defining and determining the inverse of a given
function may be accompanied with difficulties and complications as
illustrated
x
below
;
be
(i) The functional equation (1) may not always
as a function of y as, for instance, the case
solvable for
(1) may not always determine a
(ii) The functional equation
unique value of y in terms of x as, for instance, in the case of the
functional equation
j = l+x
which, on solving for x, gives
a
,
,
,
www.EngineeringEBooksPdf.com
...(3)
22
DIFFERENTIAL CALCULUS
so that to each value of y there correspond two values
situation is explained by the fact that in (3) two values
are equal in absolute value but opposite in sign give rise
value of y and so, conversely, to a given value of y there
ally correspond two values of x which gave rise to it.
of x. This
of x which
to the same
must natur-
We cannot, therefore, look upon, x, as a function of, y, in the
usual sense, for, according to our notion of a function there must
correspond just one value of, jc, to a value of, y.
In view of these difficulties, it is worthwhile to have a simple
which may enable us to determine whether a given function
y=f(x) admits of an inverse function or not.
test
If a function y=f(x) is such that it increases monotonically in an
interval [a, b] and takes up every value between its smallest value f(d)
and its greatest value f(b), then it is clear that to each value of y
between f(a) and
/(/>),
of x which gave rise to
there corresponds one and only one value
so that *is a function of y defined in the
it,
interval [f(d)> f(b)}.
Similar conclusion
is
reached
easily
if
y decreases monotoni-
cally.
Illustrations of this general rule will
appears in the following
articles.
2*3.
To draw
the graph of
JL
y=x*
n being any positive or negative
;
integer.
From an examination of the graph
even otherwise, we see that y=x n
(i) is
y~x n
drawn
monotonic and positive in the interval
may have
integral value n
(11)
of
takes
in
(0, oo
)
21, or
whatever
;
up every
positive value.
Thus, from 2*2 or otherwise, we see that to every positive
value of y there corresponds one and only one positive value of x
such that
y=x n
or
xy^
.
Thus
determines
jc
as a function of y in the interval
(0, oo
always positive.
www.EngineeringEBooksPdf.com
)
;
x being
also
OBAPHS
23
The part of the graph of y~x n lying on the
,
also the
graph of
x=y
first
quadrant,
is
n
.
y*
o
when n
is
when n
a positive integer.
is
a negative
integer.
Fig. 17.
Fig. 16.
i
To draw the graph of y~x H when x is independent and y
dependent, we have to change the point (h, k) in
t
i
n
j,
to the point
in the line
(k, h) i.e., find
y=x.
This
is
the reflection of the curve
illustrated in the figure given below.
Cor. x p u means (.x^) 1 /? where the
root is to be taken positively and x is also
positive.
Note. It may be particularly noted that
in order that x lfn may have one and only one
value, we have to restrict x to positive values
n
only and x l l is then to be taken to mean the
positive nth root of x.
It
is
nth root of x
not enough to say that x lln is an
we must say the positive wth root
;
o/the positive number x.
For, if n be even and x be positive, then x
has two nth roots, one positive and the other
negative, so that the nih root is not unique and
if n be even and x be negative, then x has no nth
root so that x lln does not exist.
Thus x and x 1 /" are both positive.
X
when n
a positive integer.
Fig. 18.
is
The exponential function a*. The meaning of a* when a is
and
x is any rational number, is already known to the
positive
2-4.
2*3 above).
student. (It has also been considered in
To give the
rigorous meaning of a* when the index, x is irrational is beyond the
scope of this book. However, to obtain some idea of the meaning of
a? in this case, we proceed as follows
:
www.EngineeringEBooksPdf.com
24
DIFFERENTIAL CALCULUS
We
find points on the graph of y=a* corresponding to the
rational values of x.
then join them so as to determine a continuous curve. The ordinate of the point of this curve corresponding
to any irrational value of x as its abscissa, is then taken to be the
value of a* for that particular value.
We
Note. The definition of a* when x is irrational, as given here, is incomplete in as much as it assumes that what we have done is actually possible and
that also uniquely. Moreover it gives no precise analytical way of representing
a* as a decimal expression.
z
of
y=a
Graph
In order to draw the graph of
.
y*=<f>
we note
its
following properties
2-41.
The function a*
(i)
:
Leta>l.
always positive whether x b3 positive or
is
negative.
(//)
large as
(ill)
(iv)
It increases monotonically as x increases and can be made as
we like by taking x sufficiently large.
tf=l so that the point (0, 1) lies on the graph.
Since a*~\\a~* we see that a* is positively very small when
x is negatively very large and can
be made as near zero as we like
provided we give to x a negative
',
.
value which
is sufficiently
large nu-
merically.
With the help of these facts,
we can draw the graph which is
as shown in (Fig. 19).
Fig. 19.
2-42.
(i)
LetQ<a<\.
The function a*
is
always positive whether x be positive
or negative.
as x increases and can be
(//) It decreases monotonically
as near zero as we like by taking x
sufficiently large.
(ill)
fl=l. Therefore
on the graph.
the
point
(0, 1) lies
(iv)
Since a x ~\\cr x
,
we
see that
positively very large when x is
negatively very large/ Also it can be
cf
is
made
as large as
we like if we
a negative value which
is
give to
x
sufficiently
large numerically.
Thus we have the graph
as
drawn
in (Fig. 20).
www.EngineeringEBooksPdf.com
Fig. 20.
made
GRAPHS
25
Note 1. In the case of the function a* it is the exponent x which is a
variable and the base is a constant. This is the justification for the name
n the base is a variable
"Exponential function" given to it. For the function x
and the exponent is a constant.
,
Note
2.
Ex.
Draw
a* never vanishes and
1
2 /*
(i)
1
2 /*"
</<)
Graph of
2-5.
is
always positive
the graphs of the functions
1
(iiflZ-
WZ-
/*
1
/*'.
the logarithmic function
y-log a x
a,
:
:
;
x being any
We
positive numbers.
know that
y=a*
can be written as
We have seen in 2'4 that y~a* is monotonic and takes up
oo
every positive value as x increases taking 'all values from
to
-f-o>
.
to any positive value of y there corresponds one and only
one value of x.
In the figure of
2*4, we take y as independent variable and
Thus we see that
that
it
to GO
suppose
continuously varies from
x monotonically increases from
x to oo if
(/')
.x
x if
oo
to
from
decreases
(//')
monotonically
x=.0
for
Also,
y~l.
Thus
.
,
,
Considering the functional equation
y=log a x,
we
see that as the independent variable x varies from
oo to
dependent variable y monotonically increases from
and monotonically decreases from
Also, y=Q for x = l.
We
oo
to
oo
if
if
thus have the graphs as drawn.
Fig. 22,
Fig. 21.
Note.
to oo the
oo
The graph of>'=log 3 ;t
is
the reflection
tfy=a*
www.EngineeringEBooksPdf.com
in the line
.y
26
DIFFERENTIAL CALCULUS
Some important properties of lo&x.
We may,
is
now, note the following important properties of
(i) log a x is defined for positive values of
also positive.
(ii)
If
a>l,
x only and the base a
log^x can be made as large as
can be made as small as
sufficiently large and
sufficiently near 0,
we
we
like
like
For #<1, we have similar statements with obvious
log a
(i/O
alterations.
1=0.
When fl>l, we
When 0<1, we
2*6.
by taking x
by taking x
have, log a x>0, if
log a x<0, if
x<l
have,
loga x>0, if
x<l.
x>l, and
log a x<0, if x>l, and
;
Trigonometric functions.
It will be
assumed that the student is already familiar with the
and the nature of variations of the fundamental
functions
sin x, cos x, tan x, cot x, sec x and cosec x.
trigonometrical
The most important property of those functions is their periodic
character, the period for sin x, cos x, sec x, cosec x being 2?r and that
for tan x, cot x being TT.
definitions, properties
We
only produce their graphs and restate some of their
will
important and
2-61.
well
y-sin
known
properties here.
x.
,1\
Fig. 23.
(i)
It is defined for
every value of x.
1 to 1 as x increases from
from
to
1 as x increases
1 to
from
it
decreases
7r/2
7T/2
monotonically
from Tr/2 to 3?r2, and so on.
(ii)
It increases monotonically
;
(Hi)
1
<
sin
x <1,
i.e.,
sin
|
x
[
<1, whatever
have.
www.EngineeringEBooksPdf.com
value x
may
27
GRAPHS
Fig. 24.
It is defined for
(i)
It decreases
(11)
to
TT
;
and
x
)
2-63.
mono ton ically from
1
to
1
as
x
increases from
so on.
cos
(/)
every value of x.
)
y^tan
<1
whatever value x
may
have.
x.
Fig. 25.
(i)
It is defined for all values of
x excepting
+ l)7r/2,
where n is any integer, positive or negative. It fact
be recalled that for each of these angles the base of the rightand so the definition of
angled trangle which defines tan x becomes
tan jc, as being the ratio of height to base, involves division by
which is a meaningless operation.
(2
will
i.e.,
it
www.EngineeringEBooksPdf.com
28
DIFFERENTIAL CALCULUS
(if)
from
x
It increases monotonically
7T/2 to 7T/2
increases from
can
(Hi) It
oo
to
bfe
made
positively as
x<7r/2 and
x>
yr=cot
7T/2
and
large as
oo
to
increases
+00 as
large as well like provided
sufficiently near to it
and can be made negatively as
+00 as x
from
it
we take
2-64.
from
increases monotonically
to
7T/2
37T/2, and so on.
;
we
;
provided we take
like
sufficiently near to
it.
x.
n
Fig. 26.
It
(i)
is
defined for
4?r,
all
STT,
values of x excepting
27r,
TT,
0,
TT, 2?r, STT, 47T,
i.e., nir, where w has any integral value, since for each of these angles
the height of the right-angled triangle which defines cot x is zero and
the definition of cot x being the ratio of base to height involves
division
by
zero.
It decreases monotonically from oo to
to
it decreases monotonically from oo to
TT
/rom
from TT to 2?r and so on.
()
;
<x>
as x increases
as x increases
GO
;
(iii)
It can
be made positively as large as we
like
provided
we take
x>0and
sufficiently near to it
;
and can be made negatively as large as we like provided we take
x<0 and sufficiently near to it.
www.EngineeringEBooksPdf.com
GRAPHS
29
2-65.
Fig.
It is defined for all values of
A:
excepting
+
i.e.,
(2n
l) ?r/2, where n has any integral value, since for each of
these angles, the base of the right angled triangle which defines
sec x is zero and as such the definition of sec x involves division
by
zero.
(11)
x
sec
|
j
^1
whatever value x
may have.
monotonically from
1 to oo when x increases
to Tr/2
it increases
oo to
1 when x
monotonically from
increases from ?r/2 to TT, and so on.
(zv) It can be made positively as large as we like by taking
x<7r/2 and sufficiently near to it
(ill)
It increases
from
;
;
and can be made negatively
x>ir/2 and
2-66.
y=cosec
as large as
we
like
by taking
sufficiently near to
x.
Fig. 28.
www.EngineeringEBooksPdf.com
it.
DIFFBKENTIAL CALCULUS
30
.
It is defined for all values of
(i)
"STr,
-27T,
TT,"
x excepting
0,
TT,
27T, OTT,
when n has any integral value.
cosec x
^ 1 whatever value x may
i.e., nit,
(ii)
(lii)
have.
\
|
oo
from -1 to
decreases monotonically from oo
It decreases monotonically
from
7T/2
creases
from
to
;
it
as~""x
to]*l
increases
as jc in-
to 7T/2.
(iv) It can
be made positively as large as we like by taking
x>0
and
sufficiently near to it
;
and can be made negatively as large as we like by taking
x<0 and sufficiently near to it.
Inverse trigonometrical functions.
2-7.
metrical functions
The inverse
trigono-
sin- 1 *, cos- 1 .*, tan- 1 *, cot~ l jc, sec- 1 *, cosec- 1 x
are generally defined as the inverse of the corresponding trigonometrical functions. For instance, sin" 1 * is defined as the angle whose
sine is x. The definition, as it stands, is incomplete and ambiguous
as will now be seen.
We
consider the functional equation
x== sin y
where y
is
the independent and x the dependent
...
(/)
variable.
Now, to each value of y there corresponds just one value of x
in (i). On the other hand, the same value of x corresponds to an
unlimited number of values of the angle so that to any given value
1 and I there corresponds an unlimited number of
of x between
values of the angle y whose sine is x.
Thus sin" 1 A*, as defined above,
is not unique.
t
The same remark
applies to the remaining
trigonometric
func-
tions also.
We now
proceed to modify the definitions of the inverse
functions
so as to remove the ambiguity referred to
trigonometrical
here.
271.
y-sin^x.
Consider the functional equation
;c=sin>>.
We know that as y increases from Tr/2 to w/2, then x increases
1 and 1, so that to
monotonically, taking up every value between
1 and 1 there
each value of x between
corresponds one and only one
Thus there is one and only
value of y lying between
?r/2 and Tr/2.
and
a given sine.
with
one angle lying between
7T/2
ir/2
www.EngineeringEBooksPdf.com
GRAPHS
sin~ l x
between
5/ne
w
:
the angle lying
is
Tr/2
define sin- 1 * as follows
we
Accordingly
and
Tr/2,
w^ose
x.
To draw the graph of
y= sin- 1 *,
we note that
^
monotoni(i) y increases
to
from
?r/2 as x
cally
Tr/2
1 to 1.
increases from
(if)
sin- 1
(fff)
We,
__
0=0,
sin- 1 *
the interval
_
defined
is
1, 1]
[
in
only.
Fig. 29.
thus, have the graph as drawn.
We
know that if x^siny, then ovaries monotonically taking up
Note 1.
1 and 1 as
y increases from w/2 to 3w/2, 3^/2 to 5w/2, and
every value between
soon. Thus the definition could also have been equally suitably modified by
restricting sin-'a to any of the intervals (w/2, 3*/2), (3*/2, fiw/2) instead of
What we have done here is, however, more usual.
(
7T/2, tf/2).
Note
2*72.
2.
The graph of ^sin-
y=cos""
J
1
.*?
is
the reflection of .y
sin
x in the line
x.
Consider the functional equation
cos y.
increases from
A:
We know
to TT, then x decreases
that as y
1. Thus, there
1 and
value
between
monotonically taking up every
and TT, with a given cosine.
is one and only on? an^lc, lyiiva; botwean
Accordingly we define cos~ x as follows
and TT, whose cosine is x.
cos-*x is the angle lying between
!
:
To draw the graph of
we note that
(/)
from
TT
y
to
decreases
monotonically
as x increases from
1
(/I)
Fig. 30,
COS"
cos-Ml) = 0.
to
1
(
1.
1)=7T,
COS^O^TT^
1
1, 1] only.
defined in the interval [
(MI) cos- ^ is
varies
taking up every
mondtonically,
that
x^cosy
Note. We know
w to 2w, 2* to 3* and so .on. Thus
value between -1 and 1 as y increases from
modified ty restricting
the definition could also have been equally suitably
instead of (On).
1
etc.,
3*]
2w],
[2*,
x
[,
the
intervals,
of
to
cosany
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
32
2-73.
We
y=tair*x.
consider the functional equation
x=tan
We know that
y.
y increases from
7T/2 to 7T/2, then x increases
00 to oo taking up every value.
Thus there is
as
monotonically from
one and only one angle between
7T/2 and 7T/2 with a given tangent.
1
Accordingly we have the following definition of tan- x
l
tari~ x is the angle, lying between
whose tangent
TT/ 2 and Tr/2,
:
isx.
X
Fig. 31
To draw the graph
note that
creases from
y
(Fig. 31)
of
increases monotouically
oo to oo and then tan*" 1
from
?r/2
to Tr/2 as
x
in-
0=0.
1
2-74.
y==cot- x.
Consider the functional equation
x=cot
y.
We know that
to TT, then x decreases
as y increases from
from
oo
to
4-a>
monotonically
taking up every real value. Thus,
and TT with given cotanthere is one and only one angle between
gent.
In view of this, we define cot" 1 x as follows
:
cot~\
x
is
the angle, lying between
and
TT,
whose cotangent
n
o
Fig. 32
www.EngineeringEBooksPdf.com
is x.
GRAPHS
To draw the graph
we note
from
that
oo
(Fig. 32)
of
y=COt~ l
X,
y decreases monotonically from
and cot" 1 0=7r/2.
TT
to
as x increases
to oo
2-75.
y^sec^x.
Consider the functional equation
X=9eC y.
We know
to ?r/2, then x increases
from 7T/2 to TT, then
increases
monotonically from 1 to oo also as y
X increases monotonically from
oo to
1.
that as y increases from
;
is one and only one value of the angle, lying between
1
whose secant is any given number, not lying between
l
x
of
sec
The following is, thus, the precise definition
Thus, there
and
and
1.
TT,
:
and
x is the angle, lying between
To draw the graph of
sec- 1
y=secr*
we note that y increases from
y increases from Tr/2 to TT as x
sec- 1
TT,
whose'$eant
is x.
x,
to ?r/2 as x increases from
1
oo to
increases from
.
1=0, and
sec- 1
(
1
to oo
and
Also
l)=7r.
Ylk
Fig. 33
2-76.
y=cosec-
1
x.
Consider the functional equation
x=cosec
We know
y.
that asj; increases from
7T/2
to 0, then
x decreases
and as y increases from to ?r/2, it
monotonically from
Thus there is one and only one
decreases monotonically from oo to 1
value of the angle lying between
?r/2 and Tr/2 having a given
1
to
oo
.
cosecant.
We thus
1
cosec"
cosecant is x.
x
To draw
is
say that
the
angle,
lying
between
the graph of
www.EngineeringEBooksPdf.com
?r/2
and
*r/2,
whose
DIFFERENTIAL CALCULUS
34
we note that y decreases from
any y decreases from ?r/2 to
J
to
?r/2 as x increases from
as x increases from 1 to oo.
-
oo
to
Fig. 34
1
1
1
Ex. What other definitions of tan- *, cot- *, sec- * and cosec- 1* would
have been equally suitable.
1
cot- 1 *, sec- 1 * and
Note. The graphs of j^s in- 1*, cos- *, tan- 1*,
1
sosec- * are the reflections of y
*, cos *, tan *, cot *, sec * and cosec *
fespectively in the line
yx.
Function of a Function. The notion of & function offunction
will be introduced by means of examples.
2-8.
Ex.
1,
Consider the two equations
>>=sinw.
...(2)
To any given value of *, corresponds a value of u as determined from (1) and to this, value of w, corresponds a value ofy as
determined from (2) so that we see that y is a function of u which isagain a function of x, i.e., y is a function of a function of x.
From a slightly different point of view, we notice that since the
two equations (I) and (2) associate a value of y to any given value
3fx, they determine y as a function of x defined for the entire aggregate of real numbers.
Combining the two equations, we get
2
j>=8in x
which defines y directly as a function of x and not through the intermediate variable w.
Ex.
2.
Consider the two equations
w=sin
x,
.(!)
y=\ogu.
...(2)
The equation
(1) determines a value of u corresponding to every
x.
value
of
The
given
equation (2) then associates a value of y to
But we know that u is positive
this value of u in case it is positive.
if and only if x lies in the open intervals
............... (~47T,
~37T), (-27T, -7T),
(0,
W)
f
(2*, 3*}
www.EngineeringEBooksPdf.com
..,(3)
35
FUNCTION OF A FUNCTION
Thus these two equations
domain of variation of x is the
define y as a function of
set of intervals (3).
x where
'
the.
General consideration.
Let
and
be two functional equations such that/(x)
[a, b]
and
<f>
Let, further, each value of f(x)
x
defined in the interval
is
(u) in [c, d].
lie
in
[c, d\.
Clearly, then, these two equations determine
defined for the interval [a, b}.
Classification of functions.
is either
2-9.
Any
y as a function of
Algebraic and Transcendental.
given function
Algebraic or
(i)
A function
(ii)
transcendental.
said to be algebraic if it arises by performing upon the
constants a finite number of operations
of addition, subtraction, multiplication, division and root extraction.
Thus,
variable
is
x and any number of
y=x+7jc+3, y=(7x 2 +2)/(3;c4 + 5;c 2
y^V(*+ 3 \/* 8 + Vx)
),
are algebraic functions.
A function
is
said to be transcendental if
it is
not algebraic.
The exponential, logarithmic, trigonometric and inverse trigonometric functions are all transcendental functions.
Two
particular cases of algebraic functions. There are
ally important types of algebraic functions, namely,
(ii)
Polynomials
Rational functions.
A
function of the type
(/)
where #
is
two
speci-
;
,
#,, .........
,
a m are constants and
m
is
a positive integer,
called a polynomial in x.
A
function which appears as a quotient of two polynomials
such as
a Qx m +al x m -*+ ...+a m ^x +a
is
called a rational function of x.
A rational function of x is defined for every value of x excludfor which the denominator vanishes.
those
ing
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
36
Exercises
Write down the values of
1.
2.
For what domains of values of the independent variable are the following,
tan-i(-l),
(11)
[Sol.
sin-Nx
(/)
this will
O'O
is
log sin
(iv)
defined only for those values of x for which
and only
if,
open
if,
defined in the interval
x is defined only
sin
in the
2,
log a sinx.
is
be the case
50 that sin-Nx
(///)
:
(//)
I.*.,
cos- 1 (~i).
sin-H,
functions defined
and
sec~i
(/)
[0, 1].
for the values of
x
for
which
x>0
intervals
(0, *), (2n, 3*), ......
or generally in (2nn,
(ill) log sin-i*.
(v)
log
[
2/i-f-l *)
(x-H)/(x-l)
(v/0 (sin x)*.
2
(ix)
log [x+>J(x
~
1)
n being any integer.
;
]
].
Find out the intervals in
3.
cally increasing or decreasing
+*)/(! -x)
(iv)
log
(v/)
log (tan-'x
[
(1
(viii)
(Iog 6 x)^.
(x)
x tan-ix.
]
.
which the following functions are monotoni-
:
(02*.
4.
(//)
(i)*-
(ill)
3
1 ''*'
(iv)
Distinguish between the two functions
1
.
x,
G)
www.EngineeringEBooksPdf.com
4 l/x2
'
(v)
2sinx.
CHAPTER
III
CONTINUITY AND LIMIT
Introduction. The statement that a function of x is defined in a certain
interval means that to each value of x belonging to the interval there corresvalue
ponds a value of the function. The value of the function for any particular
value of x
of x may be quite independent of the value of the function for another
and no relationship in the various values of a function is implied in the defini-
tion of a function as such.
so that
Thus, if *!, x a are any two values of the independent variable
then
the
function
of
)-/(*i)
2
values
/(*
fix),
/(*i) /(*) are the corresponding
may be large even though x a --x,. i? small. It is because the value /(x,)
value /(x^ of the
assigned to the function for x=x 2 is quite independent of the
I
|
I
|
x=x
We now propose
function for
change
|
t.
Xg-Xi
|
relative to the
/(x? )-/(x,)
study the change
of
and
of
discontinuity
notion
the
continuity
by introducing
to
I
|
a function.
In the next article, we first analyse the intuitive notion of continuity that
in the form of
possess and then state its precise analytical meaning
we already
a definition.
31. Continuity of a function at a point. Intuitively, continuous
variation of a variable implies absence of sudden changes while it
varies so that in order to arrive at a suitable definition of continuity,
we have to examine the precise meaning of this implication in its
relation to a function /(x) which is defined in any interval [a, b]. Let,
be any point of this interval.
The function f(x) will vary continuously at x=c if, as x changes
from c to either side of it, the change in the value of the function is
not sudden, i.e., the change in the value of the function is small if
only the change in the value of x is also small.
r,
We
Here,
h,
consider a value c-f-/* of x belonging to the interval [tf, b].
is the change n the independent variable x, may be
which
positive or negative.
f(c), is
Then,/(c+A)
the corresponding change in the depenmay be positive or negative.
dent variable /(x), which, again,
we
For continuity,
numerically small,
can
c _|_/j)_y( c )
small.
sufficiently
|yjr
h
be
if
require
is
c+h)f(c)
that
small.
numerically
as small as
made
we
This
like
be
should
means that
by taking
h
|
\
|
The precise analytical definition of continuity should not involve the US6
and
of the word small, whose meaning is definite, as there exists no definite
absolute standard of smallness. Such a definition would now be given.
To be more
A
precise,
c,
finally say that
c
continuous at
x=
corresponding to any
number S
arbitrarily assigned, there exists a positive
function /(x)
positive number,
such that
we
is
if,
\f(c+h)-f(c)
37
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
38
for
all values of, h, such that
This means that/(c+A)
values of h lying between
8
Alternatively, replacing
is
f(x)
continuous at
around c such
c
that,
for
lies
and
between /(c)
c+h by
x=c,
all values
e
and /(c)+c
for all
8.
we can say that
x,
if there exists an interval (c
ofx in this interval, we have
8,
c-f 8)
being any positive number arbitrarily assigned.
Meaning of continuity explained graphically. Draw the graph of
the function /(x) and consider the point P[c,/(c)] on it.
Draw
the lines
which
on
lie
the point
different sides of
are parallel to
P and
x-axis.
I
Here, e is any arbitrarily
assigned positive number and
measures the degree of closeness of the lines (1) from each
other.
The continuity of /(x)
x=c,
then,
should be
F
for
requires that we
able to draw two
lines
g. 35.
x=c
8,
x=c+S,
...(2)
which are parallel to >>-axis and which lie around the line x=c, such
that every point of the graph between the two lines (2) lies also
between the two lines (1).
311. Continuity of a function in an interval. In the last
articles,
at
we
dealt with the definition of continuity of a function /(x)
now extend this definition
its interval of definition.
We
a point of
to continuity in an interval and say that
A function /(x)
is
continuous in an interval
[a, &], if it is
continuous
at every point thereof.
Discontinuity. A function /(x) which
said to be discontinuous for
is
not continuous for
x=c
is
xc.
The notion of continuity and discontinuity
ted by means of some simple examples.
Examples
1.
is
Show
continuous at
that
x=l.
www.EngineeringEBooksPdf.com
will
now be
illustra-
39
CONTINUITY AND LIMIT
Here,
-and
We
will
now attempt
to
make the numerical
value of this
difference smaller than
(/)
any preassigned positive number, say '001.
Let x>\, so that 3(# -1) is positive and is, therefore, the
numerical value off(x)
/(I).
Now
|/(*)-/(l)| =3(x-l)<-001,
if
x-KOOl/3
ie., if
x<l+-001/3=*l-0003.
Let JC<1, so that 3(x
value of/(jc)-/3;i) is 3(L-x).
1) is
(11)
-..(0
negative and the numerical
Now
|
/(*)-/(!)
I
=3(l-*)<-001,
if
l-x<
001/3,
'
i.e., if
l~-001/3<x,
or
Combining
(/)
and
|
(//'),
we
/(*)-/(!)
seo that
|
<-001,
for all values of x such that
1
<
-0003
x <1 +-0003.
test of continuity for x~l is thus satisfied for the particular value -001 of e.
may similarly show that the test is true for
the other particular values of e also.
The
We
The complete argument
Let
e
however, as follows
be any positive number. We have
is,
:
Now
3
|x-
if
|*i.e
,
if
l-e/3
<
x
<
l+e/3.
+
Thus, there exists an interval (1 e/3, l e/3) around 1 such
that for every value of x in this interval, the numerical value of the
difference between f(x) and /(I) is less than a preassigned positive
number e. Here S=e/3,
Hence /(x) is continuous
Note*
lit
may be shown
for
x=l.
that, 3;c-M, is continuous for every value of x.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
2.
Show
Let
e
f/Ktf
/(x)=3x
2
+2x
1 is
We have
be any given positive number.
= 3x*+2x-l-15
= x-2
3x+8
|
|
I
|
|
We
suppose that x
between 1 and 3, 3x + 8
Thus when
lies
is
between
positive
and
x = 2.
continuous for
1
less
.
|
and 3. For values of x
than (3*3+8)= 17.
we have
Now
17
if
|
Thus,
we
x-2
<e/17.
|
see that there exists a positive
number e/17 such that
,
when
x-2
<e/17.
Therefore /(x) is continuous for x=2.
Note. It may be shown that 3x*-f2x 1,
|
of
continuous for every value
is
x.
Prove that
3.
;
say
sin
x
continuous for every value of x.
is
shown that
It will be
x
|
sin
x
is
continuous for any given value of
c.
Let
e
be any arbitrarily assigned positive number.
|
/(x)
/(c)
|
=
=
sin
x
x+c
-~ sin x
c
X + C!
xc
.
2 cos
have
\
|
.
We
sin c
1
Now
cos
x+c
;
1
for
every value of x and
xc^lx
.
Also
sm
c.
c
(From Elementary Trigonometry)
Thus we have
sm x sm
cos
c
xc
x+c
XC
Hence
sin
|
x
sin c
I
<e when x
|
c
|
<
www.EngineeringEBooksPdf.com
e.
CONTINUITY AND LIMIT
Thus, there exists an interval
every value of x in this interval
x
sin
(c
e,
sin c
|
every value of x
4.
continuous for
is
c being
;
c such that
for
<s.
xc
|
Hence, sin x
c+e) around
and, therefore,
also for
any number.
sm* x
is continuous for every value ofx.
be any given positive number.
S7z0w
Let
e
We
have
f/?tff
=
\f(x)-f(c)
sin 2 x-sin 2 c
|
|
=
sin
|
<
(x+c)
sin (x
|
c)
x-c
|
|
|
sin (x
c)
|
|
.
when
1
2
Hence, sin x
every value of x
5.
Show
atx^.
;
is
x-c
<e.
continuous for
c being
I
x=c
and, therefore, also for
any number.
that the function f(x), as defined below,
x,
[x,
The argument
will
discontinuous
is
when
when
be better grasped by considering the graph*
of this function.
consists of the point P (,
the
Our intuition
excluding
point A.
there
that
is a disimmediately suggests
continuity at A there being a gap in the
graph at A.
The graph
1)
and the
y
*
*
lines
OA, AC:
*p ^ *
,
r
;
we can see that
round about x=,
/(x) differs from /() which is equal to 1,
by a number greater than so that there
is no question of making the difference
between /(x) and/ () less than any posiAlso, analytically,
for
every value of x,
tive
number
3-2.
for
Fi
X.
^
arbitrarily assigned.
Therefore /(x)
will
C
B
is
discontinuous at
x=:
Limit. The important concept of the limit of a function
introduced.
now be
For the continuity of /(x) for x = c, the values of the function
values of x near c lie near/(c). In general, the two things, v/z.
The value /(c) of the function for x= c, and
(i)
;
(ii)
the values of /(x) for values of
x near
c,
www.EngineeringEBooksPdf.com
42
DIFFBBENTIAL CALCULUS
It may, in fact, sometimes happen that the values
are, independent.
of the function for values of x near c lie near a number / which is not
equal to f(c) or that the values do not lie near any number at all.
Thus, for example, we have seen in Ex. 5, above, that the values of
the function for values of x near
lie near
which is different from
the value, 1, of the function for
In fact this inequality itself
was the cause of discontinuity of this function for x= J,
The above remarks lead us to introduce the notion of limit as
x=.
follows
:
Mm
Def.
f(x;=l.
x -> c
A function f(x)
x
tends to c, if, corany positive number e, arbitrarily assigned, there exists
a positive number, 8, such that for every value of x in the interval
S, c+S), other than c,f(x) differs from I numerically by a number
(c
is
said to tend to a limit,
as
I,
responding to
which
is
less
than
e, i.e.,
\A*)-l\ -<e,
for every value
ofx
other than
t
such that
c,
\
Right handed and
left
handed
lim
xc
/(*)=/.
x->(c+0)
A function f(x)
is
<8.
\
limits.
said to tend to
lim /(*)=/
x->(c-0)
a limit, I, as x tends
above, if, corresponding to any positive number,
there exists a positive number, 8, such that
I/W-M
e,
to,
c,
from
arbitrarily assigned,
<*
whenever
c<x<c+8.
In this case, we write
lim fix)
and say
=
1,
the right handed limit of f(x).
A function f(x) is said to tend to a limit, I, as x tends to c from
below, if corresponding to any positive number, e, arbitrarily assigned,
their exists a positive number, S, such that
that,
/ is
whenever
c8<x<c.
In this case, we write
lim
/(*)=/
x -> (c-0)
and say that, /, is the left handed limit of/(x).
From above it at once follows that
lim
*)
=
/,
www.EngineeringEBooksPdf.com
CONTINUITY AND LIMIT
jf,
and only
if,
Jim
It is
/fx)=/=
lim
important to remember that a limit
f(x)
may not
always exist.
(Refer. Ex. 3, p. 46)
Remarks
In order that a function may tend to a limit it is necessary
and sufficient that corresponding to any positive number, e, a choice of 8 is
possible. The function will not either approach a limit or at any rate will not
have the limit /, if for some e, a corresponding 8 does not exist.
2.
1.
The question of the
x approaches 'c* does not take
The function may not even be defined for
limit of f(x) as
any note of the value of /(or) for x^c.
x=c.
3-21.
Another form of the definition of continuity. Comparing
the definitions of continuity and limit as given in
3*1, 3*2, we see
that
x=c
f(x; is continuous for
if,
and only
if,
lim f(x)=f(c).
X ->C
Thus the limit of a continuous function f(x)
equal to the value f(c) of the function for x=c.
as
y
x tends
to
c,
is
>
We
in
any
continuous for X = c
see that a function /(x) can fail to be
one of the following three ways :
(/)
(11)
is
f(x)
not defined for
x=c
;
f(x) does not tend to a limit as x tends to
lim f(x) does not exist.
r, i.e.,
x ->c
Hm
(Hi)
f(x) exists and/(c)
is
defined but
lim f(x) ^f(c).
Examples
1.
Examine
<w x tends to
1
the limit of the function
.
The function
is
y=
defined for every value of
x2
_
x
1
=*+!, when x^l.
Case 1. Firstly consider the behaviour of the values of
values of x greater than L
Clearly, the variable
than
y
is
greater than 2
1.
www.EngineeringEBooksPdf.com
when x
is
y for
greater
^4
DIFFERENTIAL CALCULUS
If x, while remaining greater than
difference from 1 constantly diminishes,
In
takes
up values whose-
then y, while remaining
takes up values whose difference from 2 constantly
greater than 2,
diminishes also.
like
1,
fact, difference
by taking x
between y and 2 can be made as small as we-
sufficiently
near
1.
For instance, consider the number
"001.
Then
if
x<l-001.
Thus, for every value of x which is greater than 1 and less thai*
1*001, the absolute value of the difference between y and 2 is less than*
the number -001 which we had arbitrarily selected.
Instead of the particular number -001,
number
Then
we now
consider
any
e.
positive
if
x<l +
e.
Thus, there exists an interval (/, 1 -f e), such that the value ofy,
for any value of x in this interval, differs from 2 numerically, by a
number which is smaller than the positive number e, selected arbitrarily.
/, is
Thus the limit of y as x approaches
2 and we have
Case
x
II.
We now
less
than
When x
2,
consider the behaviour of the values of
y for
1.
than
1, }> is less
than
2.
while remaining less than 1, takes up values whose
1 constantly diminishes, then y, while remaining less
takes up values whose difference from 2 constantly diminishes
If,
difference
than
is less
through values greater than
=2.
lim
values of
I
x,
from
also.
Let, now, e be
any
arbitrarily assigned positive
number, however
small.
We
then have
|
if
y-2
|
=2->>=2-(x+l)=l-x<
1
e,
<x,
so that for every value of x less than 1 but >1
value of the difference between y and 2 is less than the
www.EngineeringEBooksPdf.com
s,
the absolute
number
e.
CONTINUITY AND LIMIT
Thus there
Jor any x in the
is
such that the value of y,
s, 1)
2 numerically by a number which
selected positive number e.
exists an interval (I
interval, differs from
smaller than the arbitrarily
Thus the
1, is
45
ofy, as x approaches
limit
1,
values
through
less
than
2 and we write
lim
y=2.
jc~(i-0)
Combining the conclusions arrived at in the last two
any arbitrarily assigned positive
e, 14**) around 1, such that
for every value of x in this interval, other than 1, y differs from 2
numerically by a number which is less than e, i.e., we have
Case HI.
cases, we
number e,
see that corresponding to
there exists an interval (1
y-2
|
or
any
x, other
than
1,
|
<
e
such that
|x-l|<e.
Thus
lim
y=2,
or,
y ->
2 as
x ->
1.
x->l
Examine
2.
of the function
the limit
y=x sin x,
as x approaches 0.
Case
Let x
I.
>
that y
0, so
X
<
is
also
> 0,
if
we suppose
7T/2.
For 0<x<7T/2, we have
sin
BO
r<x,
that
x
Let
e
sin
x<x 2
,
if
0<x<7r/2.
be any arbitrarily assigned positive number.
For values of x which are positive and
less
than
y'e,
we have
2
X <e
we have
x sin x<e.
so that for such values of x,
V
e ) such that for every value
exists an interval (0,
of the difference between
value
numerical
the
interval
of x in this
is less than the arbitrarily assigned positive number
x sin x and
1
so that we have a situation similar to that in case I of the Ex.
Thus there
e,
above.
Thus
lira
Case
II.
Let
x<0
so that
x
sin
^>0,
x=0.
if
we suppose
x>
Tr/2.
The values of the function for two values of a which are equal
Hence, as in case I,
in magnitude but opposite in signs are equal.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
46
we easily see that for any value of x in the interval
the numerical value of the difference between x sin x and
than
\/e, 0)
is^ less
(
e.
lim
x sin x=0.
x -> (0-0)
Case HI. Combining the conclusions arrived at in the last two
cases, we see that corresponding to any positive number e arbitrarily
\/s, \/ e ) around 0, such that
assigned, there exists an interval (
for any x belonging to this interval, the numerical value of the
Hence
difference
between x sin x and
i.e.,
|
Hence
x
<e,
x-~Q
is
sin
x
lira
sin
\
<e.
x=0.
x->0
It will be seen that the inequality
Note.
x
sin
|
x
|
< e is
even for x=0. But it should be carefully noted that no
difference would arise as to the conclusion
lim xsin x=0,
satisfied
even
if
x=0
were an exception.
we
Again,
see that for
#=0,
sin
which
is
for
is
it is
1
,
limit
conti-
l)/(jtl), as considered in Ex. 1, possesses
but does not have any value for x=l, go
discontinuous for
Examine
3.
Thus in this case, the
Thus this function is
x=0.
The function (x2
a limit as x approaches
that
is
0=0
x approaches 0.
the same as its value.
also its limit as
of the function
nuous
the value of the function
the limit
x = l.
of sin
(1/jc)
as x approaches 0.
Let
so that
y
is
a function defined for every value of x, other than
to understand
The graph of this function will enable the student
the argument better.
To draw the graph, we note the following points
(/)
from
1
toO
As x
7T/2
to
:
increases monotonically from 2/7T to oo, 1/x decreases
and therefore sin (I/*) decreases monotonically from
;
as x increases monotonically from 2/3-7T, to 2/7T, Ifx de(//')
greases monotonically from 3?r/2 to Tr/2 and therefore sin 1/x increases
1 to 1
monotonically from
;
x
increases from 2/5?r to 2/37T, l-/jc decreases monotoni(i/i)
to 3?r/2 and therefore sin l]x decreases from I to
from
5?r/2
cally
as
-i;
and so on.
www.EngineeringEBooksPdf.com
CONTINUITY AND LIMIT
Thus the positive values of
number of
jc
47
can be divided in an
infinite-
intervals
r j-
2
-Li
n
r j!_
I*'
* J'
LlT' 5;: J' L57T> 3lTj'
such that the function decreases from 1 to in the first interval on
the right and then oscillates from
1
1 to 1 and from 1 to
on
interval
second
in
the
the
others
from
alternatively
beginning
the right.
It may similarly be seen that the negative values of
divide themselves in an infinite set of intervals.
A
[-
_
x also
_? 1
37T'
57TJ
such that the function decreases from
1 in the first interval
to
1
on the left and then oscillates from
1 to 1 and from 1 to
alternately in the others beginning from the second interval on the^
left.
Hence, we have the graph (Fig. 37) as drawn.
more and more
1
1 and
The function oscillates between
rapidly as x approaches nearer and nearer zero from either side. If
we take any interval enclosing x=0, however small it may be then
for an infinite number of points of this interval the function assumes
the values
1
and
]
.
Fig. 37.
There can, therefore, exist no number which differs from the
function by a number less than an arbitrarily assigned positive number
for values of x near 0.
Hence
lira
x
(sin 1/x)
does not exist.
->0
This example illustrates an important fact that a limit
always
may
not
exist.
Note.
The function considered above
the function defined for
jc=0. Neither
exist when x tends to zero.
is
is
x=0
www.EngineeringEBooksPdf.com
not continuous for
nor does its limit
48
DIFFERENTIAL CALCULUS
Examine
4.
the limit
of
x
sin
x
as x approaches
0.
Obviously the function
zero values of
is
not defined for
x=0.
Now,
for
non-
x
*rini-
=
]*|
sin~L]<|*|
so that
x
xsm
sin
x
j\
when
Thus, we see that
be any positive number, then there
0, such that for every value of x in
vthis interval, with the sole exception of 0, x sin (I/*) differs from
than e.
Jt>y a number less
an interval
exists
if,
e, s)
(
e,
around
Hence
In
fact, as
may
be easily seen, the graph of
y=x
oscillates
x
between
y=x
as
1
.
sin
x tends to
and>>=
x
zero.
This function
is
not continuous for x=0.
and variables tending to
Infinite limits
3*22.
infinity.
Meanings
of
(i)
lim/(x)=oo
lim f(x)=
(//)
oo
;
A function f(x) is said to tend to oo as x tends to c, if,
Def.
corresponding to any positive number G, however large, there exists a
positive number 8 such that for all values ofx other than c lying in the
,
9
interval
9
[c-8, c+S],
/(*)
>
G.
as x tends to c, if,
oo
Also, a function f(x) is said to tend to
Corresponding to any positive number G, however large, there exists a
positive number S such that for all values of x in [c8, c+8] exclud,
ing
c,
AX)
Right handed and
left
<
-G.
handed
limits can be defined as in
p. 42.
We shall now consider
some examples.
www.EngineeringEBooksPdf.com
3*2,
4
INFINITE LIMITS
Examples
Show
I.
that
lim
(
^cA*
The function
is
52
not defined for
]=OQ
.
'
x=0.
We
write
y=l/x
2
.
Case I. Lef ;t>0. If x, while continuing to remain positive,
diminishses, then 1/x 2 increases.
Also, 1/x* can be made as large as we like, if only we take x
sufficiently
and greater than
near
it.
Consider any positive number, however large, say 10 6
.
Then
l/x
2
>10 6
so that for all positive values of
if
X<1/10*,
3
x<l/10 , we have
,
Instead of the particular number
positive number
Then
10 6
we may
consider
any
G.
Thus, there exists an interval [0, lj\/G] such that for every
x belonging to it, y is greater than the arbitrarily assigned
t
value of
positive
number G.
Hence
lim
l*2 = oo.
Here also y is positive. If, x while contiII. Let x<0.
2
the
remain
nuing
negative, increases towards zero, then 1/x also
Case
increases.
Also
if,
G
be any* arbitrarily assigned positive number, then,
x in the interval ( l/y^* 0)> we have
for every value of
so that 1/x 2 tends to oo as
and we write in smybols
x tends to
2
lim
(l/JC )
a->(0-0)
through values
less
than
it
= oo.
Case III. Combining the conclusions arrived at in the last two
cases we see that corresponding to any arbitrarily assigned positive number G, there exists an interval [
II\/G, lj\/G], around 0,
such that for every value of x (other than 0) belonging to this interval,
we have
Thus
lim
lx*z=oo
,
or (l/;c*)-*oo as
www.EngineeringEBooksPdf.com
50
WFFBRENTIAL CALCULUS
2.
Show
that
lim
=POO
--
lim
The function
is
1.1 =
1
,.
lim
,
00,
cfoes wo/ exist.
We
not defined for x=0.
write
y=I/x.
Case I. Let x>0 so that j is positive. If x, while
continuing
to remain positive, diminishes, then
Ijx increases.
G
Also, if
be any positive number taken arbitrarily, then
l/x>G,
if
x<l/G.
Hence
lim (l/x)
Case
Let
= oo
whenx->(0+0).
#<0
so that y=ljx is negative.
If x, while con^
tinuing to remain negative, increases towards zero, then y=ljx
decreases and numerically increases.
II.
G
Also, if
be any positive number taken arbitrarily, then
l/x<-G,
if
x>-l/G.
Hence
lim (l/x)=
Case
when x->(0
0).
Clearly when x->0, lim (1/x) does not exist.
We will now give a number of definitions whose mean-
III.
3-23.
ings the reader
(i)
oo
A
may
easily follow.
function f(x)
is
said to tend to a limit
/,
as
x
tends to
QO ), if, corresponding to
(or
any arbitrarily assigned positive
e, there exists a positive number G, such that
oo
number
'
!/(*)-/
for every value
ofx>G, (x<
(11)
i/,
A function f(x)
corresponding to
a positive number
is
oo
as x
number G, however
,
oo
].
tends to oo (or
oo )
large, there exists
such that
we
r
Symbolically,
[lim/(x)=/ when x->
,
said to tend to
any positive
A
<>
G).
we write
= / when x->oo
Symbolically,
lim /(*)
|
write
lim/(x)=oo when x~oo [lim/(x)=<x> when x-> oo )]
The reader may now similarly define
oo when x-*oo
lim /(*) =
oo when x-*
lim/(x) =
,
;
Note
1.
Instead of
oo,
we may
write +00 to avoid confusion with
oo
oo
.
Note
Rigorous, solution of any .problem on limits requires that the
problem should be examined strictly according to the definitions given above in
terms, of e'$, g'j, G's etc. Bat in practice this proves very difficult except in some
2.
www.EngineeringEBooksPdf.com
THEOREMS ON LIMIfS
We may
very elementary cases.
not,
J5J
therefore, always insist
on a rigorous
solution of such probelms.
Exercises
Show
1.
lim
(2*+3)=5.
lim
(**+3)=4.
(0
(///)
that the following limits exist
(//)
2.
(3*-4)=2.
o;~>3
lim
Show
(l/aO=i
(vO
lim
that
(0
5x+4
is
continuous for z=2.
(11)
z a +2
is
continuous for *=3.
Also show that two functions are continuous
3.
:
lim
(iv)
a?->l
(v)
and have the values as given
lim
Examine the continuity
for every value of x.
for oj=0, of the following functions
:
(0
.
/()=/
(,7)
6,
sin
Prove that cos x
is
~,
whena?=0,
0,
j^
4.
when #=3,
continuous for every value of x.
Show that cos 2 x and 2+x+x* are continuous for every value of x.
6.
Draw the graph of the function V(x) which is equal to
when x
when 0<a;^l and to 2 when s>l and show that it has two points of
5.
to
1
;
discontinuity.
7.
function
Investigate the points of continuity and discontinuity of the following
:
f(x)=(x*la)-a for x<a, /(a)=0,/(aO=a-(aV#), for
8.
Show
that
when
/(z)=cos
is
discontinuous for
9.
Show
g=0.
that
lim
tan a?=oo
[Refer
lim
,
tan a?=
tan x does not exist.
lim
but
10.
x^O and/(0)=l.
2-63, p. 27]
Show
that
lim
logx=
oo.
*->(<HO)
[Refer
11.
2-5, p. 25].
Examine
lim
5^
www.EngineeringEBooksPdf.com
oo
DIFFERENTIAL OALOTTLtTS
[If x tends toO through positive values, then I/* tends to ooand, therefore,
21 /* tends ooo.
i
If x tends to
through negative values, then I/* tends to
l/a
tents toO.
fore, 2
oo
and there-
>
Thus
JL_
-JL,
lim
2* =
lim
2*~=oo,
s-KO+0)
but
2*
lim
does not
exist.]
Show
12.
that
lim
Show
13.
where
that if
denotes the greatest integer not greater than rr, then
lim
lim
lim f(x) does not
/(*) = !,,
/(aO=0,
[a?]
*-Hi -0)
What
a?->(l
the value of the function for
is
exist.
<c->J
+0)
x=\
?
[Refer Fig. 11, p. 17]
Explain giving suitable examples, the distinction between the value of
the function for a?=a and the limit of the same as x tends to #.
14.
15.
but
is,
Give an example of a function which has a definite value at the origin
nevertheless, discountinuous there.
Theorems on
3*3.
Let/(x) and
lim
<p(x)
limits.
be two functions of x such that
/(*) = /,
lim
?(x)=w.
Then
lim
a?-0
(0
i.e.,
the limit of the
limits
lim [f(x)
the limit
of their
limits
(Hi)
i.e.,
of the
equal to the
sum of
their
(iv)
?(x)]=lim
difference
/(x)lim
of two functions
y>(.x)
is
/
m,
equal to the difference
lim [f(x)-9(x)]=limf(x) lim *'(*)
x-*a
x~>a
x >a
.
of the product of two functions
is
= /m,
equal to the product of
;
lim
of the quotient of two functions is equal to the quotient of
provided the limit of the divisor is not zero.
^Ae limit
their limits
=
;
ih* limit
their limits
i.e.,
is
;
(ii)
i.e.,
x-*a
x >a
sum of two functions
www.EngineeringEBooksPdf.com
FEOPERT1ES OF CONTINUOUS ITUNCTIONS
63
These results are of fundamental importance but their formal
proofs are beyond the scope of this book.
These results can easily be extended to the case of any finite
number of
functions.
A
Note. Inversion of operations.
beginner who notices the above statements for the first time may fail to see any meaning in them and he may not be
able to appreciate the significance of the same. In order, therefore, to help
him dp so, some observations seem to be necessary here.
In mathematics we deal with different types of operations and sometimes the mathematical expressions are subjected to two or more operations
and in such a case the order in which the operations are made must be taken
into account. It cannot just be assumed that the order can be changed at will
without altering the final outcome, i.e., the final result may not be independent
of the order of the operations.
The fact is that in each case the question as to whether the inversion of
the order of operations is or is not valid has to be separately examined in
relation to the nature of the operations.
We now examine the following statements
:
(i)
logOnfn) ;*logw4-log
n.
+ B) ^sia A+ sin B.
=0x6 + axc,
(Hi) ax(b+c)
In (/) we find that the two operations
(it)
sin (A
involved,
viz.,
that of addition
and taking of log arc not invertible.
Similar is the case in (//) where we have the operations of adding and
taking of sine.
In (//'/), however,
plication are invertible.
we
two operations of addition and multi-
see that the
The theorems in this article amount to asserting the validity of the inversion of the operations of taking the limits with each of the four operations of
addition, subtraction, multiplication and division.
The reader is advised to think of other similar cases also.
3-4.
Continuity of the sum, difference, product and
quotient of
two continuous functions.
Let /(;c),
x=0,
<p(x)
be any two functions which are continuous for
so that
=
Prom
lim/(x) ^(a),
x -> a
the theorems in 3*3,
lim
x ->
lim 9(x)=?(0).
x -> a
we
see that
a.
=value of
so that/(x)+9(.x)
is
[/(*) +?(*)], for
The continuity
of/(x)
y(x)
and
f(x)<p(x)
proved.
For the quotient, we have
lim
*=0
continuous at x = (t.
/M -?^M
,
'
*here lim
www.EngineeringEBooksPdf.com
may
similarly
be
D1FFB&BKTUL CALCULUS
54
JM
^value of t&.
so that/(x)-r?(x)
for
x-a,
?(*)'
?(a)
also continuous at x=fl, provided
is
when x ->
lim <p(x)=<p(a)^iQ,
a.
Thus, f&e sum, the difference, the product and the quotient of two
continuous functions are also continuous (with one obvious exception
in the case of a quotient).
3*41.
Continuity of elementary functions.
We have seen in Ex. 3, p. 40 and Ex.
GOB x are continuous for all values of x.
Hence from
-
sin
tan
x= cos
3*4, above,
x
x
cot
x
sm x
x=- --
51, that sin
x and
see that
cos
.
,
we
4, p.
,
sec
x=-
1
1
cos
x
,
'
cosec
x=- ------sin
x
all those values of x for which they are defined
points of discontinuity of these four functions arise when the denominators cos X sin x become zero and for such values of x, these functions themselves cease to be defined.
are also continuous for
;
9
Also it may be easily seen that, x, and a constant are continuous functions of x. Thus by repeated applications of the results of
3/4 above, we see that every polynomial
a continuous function for every value of x and that every rational
function
is
<*
Q
x+a x-l +...+a n
l
a continuous function of x for every value of x except those
which the denominator becomes zero.
is
for
1
1
1
1
1
Finally the functions, sin- x, cos- x, tan"" x, cot- x, sec- x,
1
a?
are, continuous for all those values of x for which
cosec- x, log a x,
they are defined. This is geometrically obvious from the graphs
drawn in Chap. II. Analytical proofs are beyond the scope of this
book.
3*5.
Some
important properties of continuous functions.
Letf(x) be continuous for x=c and f(c)^0. Then there
an interval [c S, c+S] around c such that /(x) has the sign of
3-51.
exists
f(c)for all values
ofx
in this interval.
Its truth is obvious if we remember that a continuous function
does not undergo sudden changes so that if /(x) is positive for any
value c of x and also continuous thereat, it cannot suddenly become
negative and must, therefore, remain positive for values of x in a
certain neighbourhood of c.
www.EngineeringEBooksPdf.com
PROPERTIES OF CONTINUOTJS FUNCTIONS
In informal, precise manner,
it
may
be proved as follows
To
every positive number s, however small
correspond a positive number S such that
for
65
it
may
;
be, there
every value of x such that
c
Letf(c)>Q. In this case we take e any positive number less
than/(c) so that/(c)-~ e and/(c)+e both are positive, Thus, /(*) lies
between two positive numbers and
x lies between
and c+S.
c8
is,
therefore, itself positive
when
Hence /(c)e is already negative and /(c) +s will
Ltf(c)<0.
be negative if e< /(c). Thus in this case if we take e any positive
number which is smaller than the positive number /(c), then we see
that /(x) lies between the two negative numbers /(c) e and/(c)+s
and is, therefore, itself negative when x lies in the interval
~
[C-8, C+8].
:_
*
:
;
Hence the theorem.
Note. This simple but important property of continuous functions will
be used in the chapters on Maxima, Minima and Points of inflexion.
3*52.
Let f(x) be continuous in a closed interval [a, b] and let
/(a), f(b) have opposite signs ^ Thenf(x) is zero for at least one value of
*> lying between a and b.
Its truth is intuitively obvious, for, a continuous curve
y=f(x)
going from a point on one side of x-axis to. a point lying on the other
cannot do so without crossing it.
Its formal,
analytical proof
however, beyond the scope of
is,
this book.
3-53.
Letf(x) be continuous in a closed interval [a, b]. Then
there exists points c and d in the interval [a, b], where f(x) assumes its
andm, i.e.,
greatest and least values
M
The proof
is
beyond the scope of
this book.
continuous func_ .The. theorem states that there is a value of a
tion greater than every other of its values and as also a value smaller
than every other value.
.
A
as
discontinuous function
we now
may
not possess -greatest or least values
illustrate.
Consider the function defined as follows-:
I
x,~fbr
\,~-
0<;
ibrJfc=X
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
point C(0 *>' and the line AB excu u point 5. The function
.
the
possesses no greatest value; 1 not being
a value of the function. If we consider
>
any^ value less than 1, however near 1
may be seen that there is a value
of the function greater than that value.
;
it
(0'i)C
This
explained by the fact that
the function is not continuous in the
is
interval (0, 1)
Fig. 38.
;
#=0
being a point of
discontinuity.
Note. This property will be required to prove Rolle's Theorem in Chap.
Example
Show
when, n
is
that
any
integer.
We cannot
write
,.
hm
jc*~ a*
------
=
yr-S
--- ~>
r
for
,.
/
hm(x
In the present case the limit of the numerator
Case
Firstly suppose that n
I.
is
x
A
a)=0.
is
also zero.
a positive integer.
actual
By
division,
x
a
the equality being valid for every value of x other than a.
limit does not depend upon the value for x=a, we can write
-
lim
As the
~
~~
x-*a x a
x -> a
Being a polynomial, the function
is
continuous for every value of x and, as such,
Thus
to its value for x~a.
its limit
when x -> a
must be equal
x
n
-*a+ ...... +d-i=na*-*
-^-=a*-i+a
~
x ->a x a
lim
Case
where
m
is
Now suppose that n is a negative integer, say
a positive integer. We have
II.
www.EngineeringEBooksPdf.com
m,
67
iMPOKTAisri LiMi'fs
-- =
x*
,.
lira
a*
,.
lim
-
1
x m -a m
m
a xm
x a
x m a m ,.
1
hm
m m
employing Case I and the fact that l/a x
In the Case II, we must suppose that
'
-
is
continuous for x=a.
Exercises
1.
Show
2.
If
that
^-2 and/(-2)*t
find
A so that/(x) may be continuous for
Show
3.
*=
2.
that
lim sin(*-f /t)=sin x.
/t~>0
Some important and
3*6.
Limiting value of x" when n tends to infinity through posi
x being any given real number.
3*61.
live integral values
Let
(i)
useful limits.
;
x>\.
x= 1 +h
We
write
By
the Binomial theorem,
where each term
is
so that h
is
positive.
we have
positive.
Hence
n
x=(l+h) >l+nh.
As
(\-\-nh) tends to infinity
with
n,
therefore x n also tends to
infinity.
To be more
rigorous,
we
consider any pre-assigned positive number G.
Then
If
A
be any^positive integer greater than (G l)/t, then
x n >G for H> A so that x*-*oo when w~>oo
.
www.EngineeringEBooksPdf.com
58
DIFFERENTIAL CALCULUS
(ii)
Let
x~l. Here
x*
=l
and therefore
for all values of n
in this case.
(m) Let
We
0<x<l.
write
Here, also, x"
so that A
x=l/(l +h)
is
positive.
a positive number.
is
As before
<z __ __
^
_
1+nh'
Now
ll(l+nh)-*Q as n tends to
Therefore x"-*Q as
infinity.
oo.
To be more
rigorous,
,
If
m be any
we consider any
number
positive
t.
Then
if
integer which
>
is
--
f
then
1
J[A,
;c*<c for ri^m*
Thus we
n^m
so that for
see that x w , which
is
lim x w
(iv)
Let
(v)
Le/
lies
always positive,
between
e
and e
0<jc<l
x=0.
Here
when
.
^=0 for all n so that x->0.
Here x
l<x<0.
>co
negative so that x* is positive
We write .*= a
is
for even values of n and negative for odd values.
so that a is a positive number less than 1.
Absolute value of x n
a
is
n
x"
|
But a*->0.
i.e.,
,
|
=an
.
Hence
1 and 1, therefore
Since x n is alternately
1.
Let x=
i n this case.
neither tends to any finite limit nor to Jt
(vi)
x<
it
n
Here, again, x is alternatively negative and
in this case it takes values numerically greater than
But
positive.
n
any assigned number. Hence x does not tend to a limit.
{vii)
Let
/.
Thus, we see that lim x*, when n ->
oo,
exists finitely
if,
and only
if,
Abo ttisOif-l<x<I
9
9
i.e.,
if\x\ <1 and
Iforx^l.
is
n
3'6i. Limiting value of x jn !, when n tends to infinity
positive integral values and x is any given real number.
Let x be positive and
integers between..which x
lies
let
m>
ao that
m+l
we haye
through
be the two consecutive
.
,
.,
www.EngineeringEBooksPdf.com
.
[
IMPOETAN'T LlMlfs
We
write
s
=
^L
1*2
nl
^L
-JL
'
'
"m
IT'
*
'
*
m+1
p=^.JL
12
Let
so that
'
'
'
'
^n+2'
'7T*
m..
a positive constant.
/? is
^.
m+3
-^~,
Also, each of
m+2
x*
x
f
\*~
,
'
m
is
n
'
"
x
/
p
<-~ ^
8ay
where
A:
is
a positive constant independent of
Since, x/(w
+ l)
is
n.
and <1, therefore when
positive
m being a constant.
Hence
xn
lim -^^=^0,'
when w-^oo
.
!
let
Again,
tive.
We
x be any negative numbed, say
a, so
have
that a
is
posi-
x*
n
n
Ij
Y
Now,
since
j-->0,
Hence
we
-
whatever value x
may
see that
lim
also-0.
x*
n-y*=
have.
Here, of course, n tends to infinity through positive integral
values only.
3 63.
Limiting value of
,
when n tends
to infinity through positive integral values.
www.EngineeringEBooksPdf.com
60
DIFFERENTIAL CALCTJLXJS
Step
1.
we have
By
the Binomial Theorem for a positive integral index,
The expression on the right
Changing n to fl + 1, we get
a
is
sum
of (n +1) positive terms.
__L_ViA_^
A n+i
)
......
V
(w-i-1
Here, the right-hand side consists of the
Also
sum
fi
of
(+2)
n
n+l
\
J
positive
terms.
'
1
i_l_<i__2_
< n + l' !_!
n
so that each term in the expansion of
41
i
is
greater than the corresponding term in the expansion of
C'+r)'Thus, we conclude that
l
/
1
\*
K'+T)
for all positive integral values of w,
cally as n increases.
Step
II.
We
+ l/M)"
1
i.e.,
(l
have
www.EngineeringEBooksPdf.com
increases monotoni-
SOME IMPORTANT LIMITS
<l+l+-2-+
t
2"+. ..+2.2.2. ..(-!)
-g.
=1 +
61
factors
<3) foralln.
...(//)
2
/
Thus we
see that
N*
-1
1
f
-j
steadily increases
J
successively the series of positive integral values
less than 3 for all values of n.
as
/i
takes up
1, 2, 3, etc.,
and
re-
mains
H
(1
\
-----
Note
1,
From
n
J
(i)
approaches a finite
and
(11),
we
for every value of n so that the limit
limit as
n ->
oo
.
see that
is
some number which
lies
between 2 and
3.
The mere
existence of the limit of (1 -f l/w)*has beea shown here and at
this stage nothing more can be said about the actual value of the limit which is
denoted by e, except that it lies between 2 and 3.
The reader will be enabled to appreciate the argument better and also
begin to feel a little more acquainted with, e, if he calculates the value of
(
1+
for various successive positive integral values of n.
<?-,
tfj-2, a 2 =2 25, a 3 -2-37, a4 -2 441,
In Chapter
IX
will
it
be shown how the value of e, correct to any number
of places, can be determined without
Note
much
inconvenience.
The proof for
the existence of the limit has been based on an
intuitively obvious fact that a monotonically increasing function which remains less
than a fixed number tends to a limit. The formal proof of this fact is beyond the
2.
scope of this book.
Cor.
real
I.
Lim
numbers as
its
(1
4-l/x)*=e, when x tends to
infinity taking all the
values.
If x be any positive real number, then there exists a positive
integer n such that
JL > JL
x
or
n+i'
i.j__^],4_ -L>i_|__J_,
x
'
n+1
www.EngineeringEBooksPdf.com
82
DIFFERENTIAL CALCTTLtT&
/
or
1
\
O+T)
number to a
(In case the base is greater than 1, raising a greater
Qf.itie inequality.) __
greater
power, does not alter the. direction
We thus
have
Let x ~>
values.
oo
.
We know
Then n and (w+1) ->
00
through positive integral
that
lim
lim
H-|
= lr=:lim
J
/
x*
1
(l-|
J
M+ ""Xf
/I
J
\ n4
*
=^ = lim (l+^~TTj
1
Therefore
Urn
x-+<
COT. 2.
lim
l-.
-C^-ccV
Let
x=
so that >>->
y
=
+
oo
l
'
Z
lim (l+z)
3.
x
oo
-4-
We have
.
a
lim fl-f
=e
--j-Y
^y-V
-
Cor.
as
when z ->
-+
Let z=l/x so that jc
positive or negative values.
T ^(l+~-
lim
ooV
0.
according as z ->
or
Now
lim
(1+z)
z->(0+0)
a=s
lim
*->
[1+
)
s=f
and
z
lim
(1+2)
-> (0-0)
lim
f
1+ ~)
www.EngineeringEBooksPdf.com
=^.
through
63
SOME IMPORTANT LIMITS
Cor.
4.
x
Hl+
^
-+0\
*.y'*-Iim
a)
X ^.Q
V
Note, tn higher Mathematics the number
logarithms which are then called Natural logarithms.
not mentioned so that log x means log e x.
To show
3-64.
that
_
a*
lim
Let
taken as the base of
is
The base
i
------ =log, a.
x
a?l=y so that y ->
as x ->().
We have
or
x
a=log
log
(1 +>>),
i.e.,
=x
log (l+>0/loga.
Hence
a*
1
y
1
-.
--..
&v
y
,
'logo
--- _
1
,
=loga.
-;
log
hm
..
...
X
^Q
hm
'.
y
r.
^Q]_
a
,
j
log (1+7)
(1+7)
a*-\
--=
x
--------lus M .
.
logo
----1
T7/:K
Hm
lim
log
(1+7)
-^0
3-65.
Iy
To
www.EngineeringEBooksPdf.com
1
J
e
is
generally
DIFFEBENTIAL CALCULUS
Let x=a(l+y) BO that y ->
*X -<* X
= aX
...
Again,
X
[(1+JO
0, as
-1]
x
=
-
a.
x~ 1
fl
we put
(1-BO
X
l=z
=i +2
so that z -+ 0, as
y ->
0.
or
-l)
_ a (X-D
a
(X-D
Fog
_
(1+z)
log
(1+z j
_z
'
z_
>
*
'
log (1+z)
~~~y
.
'
log (1+z)
Hence
..
Urn -
-
..
~
->0 x a
3*66.
lim
hm
z-*0
=?va
---------
-^0 x
=1,
--------------
,,
log
as proved in books
on Elementary Trigono-
metry.
Examples
1.
Show
that
ll
is
continuous for
X
when
x=0.
Now
lim/(x)=lim f
#->0
aothat
lim
a?->OL
f[x)=*
->0
Hence the
hm
,.
'
result.
www.EngineeringEBooksPdf.com
EXAMPLES
2.
Show
65
that
i
/(x)=(e*--l)/(e* +1)
whenx^Q,
/(0)=0
if
x=0.
discontinuous at
When x
through positive values, then
tends to
i
+
therefore e* -
Thus
oo.
tT
..
_1
-
.
.
^
When x
through negative values, then
tends to
JL
And therefore e * ->0.
Thus
hm
Hence we see that
lim
f(x)^
Hm
eo that
does not
f(x)
f(x)
exist.
Hence the
3.
lim
Show
as
-
that
continuous for
Now,
result.
x=l.
may
easily be seen
_
_
lim
;r
^ - 1 =B
oo,
e*"1 =0.
lim
a?->(l
0)
www.EngineeringEBooksPdf.com
l/JC->
<x>
DIFFERENTIAL CALCULUS
Thus
lim
Km
/(x)=0,
*-Hl+0)
x=0=
lim
Hence the
4.
annum.
/(x)-0,
*->(!- 0)
result.
A sum of P rupees is given on interest at the rate of r% per
What mil be the amount after t years, when the interest is
being continuously added ?
Supposing that the interest
year, the
amount
after
t
is
added after every nth part of a
years will be
r
The required amount
We
is its
\tn
limit as n-> c.
have
"
fr/100
as n
which -> Pe
--*>
oo
and
is
thus the required
amoun
^
Exercises
1.
Prove that
sin
a
ax
-- COS
l
..
.
taf^
*
1
(j/i)
2.
lim
-
-
Examine whether or not the following functions are continuous
0,
I
C
(iO
when
:r
(
when
1,
2r
"5
rc=0.
when a; =0.
1
tan 2*
|>
i^O.
when #-0,
sin
A*)-}"
^
a;
wbencc^O.
when a;
0.
whena;=0.
www.EngineeringEBooksPdf.com
HYPERBOLIC FUNCTIONS
e2 ,
(
e
when #=0.
-I/*
2
*)=)
C
when #=0.
1,
e-
f
(v/0 /(*)=
1
/*
whcn x
1+~^T~'
<(
f
vi/0
L
Show
whens^O.
-1
l
3.
-
I/a'
"777
*
e 11
/()--(
^
when # = ().
U,
(
67
when #=0.
1,
that
sin
lim
a?
sin
------
(a?-|-/0
-
=cos
.
A-->0
Carefully state the results you employ.
Show
4.
and
that
(//) (
when w->oo through
Draw
5.
positive integral values.
the graphs of the functions
3-7. Note on Hyperbolic Functions. In analogy with Trigonometric functions, the Hyperbolic functions are defined in the following
manner
:
QX _ p X
03?
.
sinh
.
x=
^
sinh
.
tanh X=s
sech
3-71.
x
=e^
e-^
=
Graph of
The following
.
C0th
;
cosech
5
sinh
x
is
x=
sinh
properties of sinh
x
will
;
-- ass ~
enable us to
continuous for every value of #.
0=-
-^
2
sinh x.
V>__gO'
(tf)
g
X= cosh x =
graphs:
(i)
I
2
12
x=
_
cosh x==
5
=0.,
www.EngineeringEBooksPdf.com
draw the
68
DIFFERENTIAL CALCULUS
-_-L
e x -er*
e*
~
2
and as x increases from
0,
2
monotonically increases and
Thus as x
monotonically decreases.
increases from 0, sinh x monotonically
e*
increases.
(iv)
lim (sinh
x->oo
x)=oo
.
e~*- e*
(v)
sinh
-a
(-*)=
..
=
sinh x.
Thus we have the graph as
shown, in general outlines. (Fig. 39).
Fig. 39
3'72.
(/)
Graph of cosh
x
cosh
x.
continuous for every value of x.
is
(ii)
cosh
0=
(iii)
cosh
x=
and as seen above ( ex e-x)l2 monotonically
increases, as x increases from 0. Thus cosh x
increases
as x
increases
monotonically
from
0.
(iv)
(v)
x=
lim cosh
X->co
cosh
(
oo.
x)=cosh
x.
Hence the graph as shown.
3*73.
(i)
Graph of tanh
x.
tanh x is continuous for every value of x, the denominator
not vanishing for anv value of x.
() tanh 0=0.
(tit)
tanh x
=
e*+e~*
1-
www.EngineeringEBooksPdf.com
69
HYPERBOLIC
BO that tanh
x
increases as
x
increases from
onward.
vi
y^tanh x
Fig. 41
lim tanh
(iv)
(v)
tanh
x
We may note
\-e~**
e*-fi-
lim
lim
e*+e~*
=
x)
(
=
1+e
tanhx.
that
tanh x
j
for every value of x.
Note. The graphs of coth x, sech x and cosech x have not been given.
The reader may, ifhe so desires, draw the same himself.
Some fundamental relations. The following fundamental
3-74.
relations can be at once deduced from the definitions
:
r
<
I
a)
2
cosh x
sinh
2
x
tanh x
2
=1
.
=sech2 x.
(2)
1
(3)
coth2 x-l
(*)
cosh 2x+sinh 2x
i
2 sinh x cosh x
=sinh 2x.
=cosech2 x.
cosh 2x.
Inverse hyperbolic functions. In this section we obtain the
3*8.
logarithmic expressions for the inverse Hyperbolic Functions
:
1
sinh- x, cosh- 1
X tanh- 1
9
1
1
1
x, coth- x, sech"" x, cosech"" x,
which are defined as the inverses of the corresponding hyperbolic
functions. This will also necessitate a slight modification of the definitions so as to
A.
make them
single-valued.
Let
1
y=sinh"" x,
<
so that
y
is
the
number whose
sinh
is
x.
or j
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
It is easy to see that x+
V(* 2 1) is positive and x <\/(x*+l)
negative for every value of x, positive or negative. Also we know
that the logarithm of a
negative number has no meaning in the
2
field of real numbers so that
[x
<\/(*
1)] has to be rejected. Hence
+
is
+
we have
B.
Let
We
will see that there are
a given number.
always two values of y whose cosh
is
Now
y= J
*=xi/(x*
x=cosh
(<# -f-<r")
l)
e 2 *-2,x.^+l=0.
i.e.,
or >>=log [x
v^C*
1
!)]
22
Here we see that both jt +
1 ) are
v/(* -l), an d x-Vf*
positive and real when x>l so that in this case y has two values.
For x>l, we have
,
and
BO that
log
[*4V(* 2 --1)]>0, and
To avoid this ambiguity,
cosh"" x a little and
say that
cosh is x so that we have
cosh-1
log
[x-
usual to modify the definition of
cosh- 1 x is the positive number y whose
it is
x=log
[x
+ vXx*
1)].
Note. The ambiguity referred to here is a consequence of the fact that
in xcosh y two values of .y
give rise to the same value pf x so that to a given
value of x correspond two values of y which
gave rise to it.
(Refer 3*72).
C.
Let
7=tanh~1 x, where
.-.
|
x
\
x=
Thus
l
A "T" A
x== ___ fog
.
It is easy to see that
^0,
if
I
x
www.EngineeringEBooksPdf.com
HYPB11BOLIC FUNCTIONS
We may
D.
similarly
show that
eoth-ix=^^f
log
[<*+!)/(*- 1)]>0,
E.
71
sech-*x=log
X
A ^]
~
if
|
x
|
M.
>1.
jc|
1+ ^fl-Tlg2)-.
[0
<x< 1].
'
F.
x
where the sign of the radical
is
positive or negative according as
x
is
positive or negative.
Ex.
oo
or
oo
oo
Show
that
sinh* tends
to oo
or
oo
according as x tends to
3.
Show that cosh x tends to oo whether x tends to oo or to oo.
1 according
Show that tanh x -> 1 or ->
as x tends to
4.
Show
2.
,lo
1.
.
.
that
sinh (;c+^)==sinh
x cosh .y+cosh x sinh y
cosh (x-f v)=cosh x cosh y i sinh x sinh y.
www.EngineeringEBooksPdf.com
oo
or
CHAPTER
IV
DIFFERENTIATION
Rate of Change. The subject of Differenits origin mainly in the geometrical problem of the determination of a tangent at a point of curve, has
rendered possible the precise formulation of a large, number of physical concepts such as Velocity at a point, Acceleration at a point,
Curvature at a point, Density at a point, Specific heat at any temperature, etc. each of which appears as a Local or instantaneous Rate
of change as against the Average Rate of Change which pertains
to a finite interval of space or time and not to an instant of time
Introduction.
4-1.
tial Calculus
which had
and space.
The fundamental idea of Local or instantaneous Rate of Change
all
pervading
these concepts underlies the analytical definition of
differential co-efficient.
4-11.
defined in
We consider a function f(x)
Let c be any number of this interval
Derivability. Derivative.
any interval
(a, b).
the corresponding value of the function. We take
of this interval which lies to the right or left
of c (i.e., c+/*> or <c) according as h is positive or negative. The
value of the function corresponding to it is f(c -{-/?).
o that /(c)
is
c+h any other number
Now,
h, is
the change in the independent variable x, and
the corresponding change in the dependent variable /(c).
The expression [f(c+h)f(c)]jh which is the ratio of these two
changes, is a function of h and is not defined for A=0 c being a fixed
is
9
;
number.
It is possible that the ratio tends to a limit as h tends to 0.
This limit, if it exists, is called the derivative of f(x) for x==c and
the function, then, is said to be derivable for this value.
Def.
f(x)
is
said to be derivable at
xc
if
^,
/r->0
exists
and
the limit
is
the function f(x) for
called the Derivative or Differential co-efficient
of
x=c.
The limit must be the same whether h tends to zero thcpugh
positive or through negative values. The function will not b&'derivable if these limits are different.
The function /(x)
is
said to be finitely derivable if its derivative
isftflffe.
72
www.EngineeringEBooksPdf.com
DIFFERENTIATION
Ex.
Show that x2
1.
derivable forx=l
is
for *=7.
73r
and obtain
its
derivative
Let
2
/(jc)=x so that/(l)
To find the derivative for
that the function changes from
Change
in the
=l
Ex.
is
1
derivable and
x
that
\
is
\
1
to
1-f-ft
&
to
functional +h)*-l=2h+h
Show
2.
=l.
x=l, we change x from
which approaches 2 as h approaches
Hence /(x)
2
2
.
0.
its
derivative
is
not derivable for
2 for x
1.
x=0.
Let
f(x)=
x
\
\
s>othat/(0)=0.
shown that the
when h tends to 0.
It will be
exist,
limit of [/(04-/J)
f(0)]//J
does not
Now
f(0+h)-f(0)
h
according as h
is
=
-=-f(h)
/r lor-l
positive or negative.
Hence, [f(Q+h) f(0)]/h -~> 1 as h ->
through positive values
1 as h ->
through negative values.
and ->
Thus [f(0+h)f(0)]/h approaches different limits when h
approaches
through positive or negative values so that it does not
*
tend to a limit as h tends to 0.
Hence
Ex.
|
3.
show thatf(x)
for
x
is
\
If
is
not derivable for
f(x)=xsin
continuous for
x=0
x=0.
We
x=0.
but has no differential co-efficient
have
=x sin
-0=xsin
X
www.EngineeringEBooksPdf.com
-
X
.
DIFFERENTIAL OALOTTLUS
-
x
|
Thus
if e
|
|
*-
Bin
|
<
|
*
be any pre-assigned number, we have
!/(*)-/(0)
<e when x-0
[
!
<e.
|
Hence
lim /(x)=0
x-0
so that f(x)
is
continuous for
Again
x=0.
==
x
jc
8n
x
and, as seen in Ex. 3, p. 47 lim sin (I/*) does not exist when x ->
Thus/(x) has no differential co-efficient for JC=0.
>
Ex.
0.
Find the derivatives of
4.
(/)
2x 8 -f 3x-4 for .v=5/2.
(11)
1/jc
for
x-5.
have defined the
Derived function. In
4-11, Ave
of
a
for
derivative
a function /(x)
particular value, c, of the independent variable. Instead of considering a particular number c, we, now,
consider any number x and determine
4*12.
>/v
lim
~ -yv
when h -> A
.
;
;
n
,
where x is kept constant in the process of taking the limit. We
suppose here that the limit exists whatever value x may have provi-
ded
it
belongs to the interval of definition of the function.
\
^his
limit which
a function of x
is
of Derivative off(x) and
is
is
called the Derived function
denoted by/'(jc).
Derivative of function
also
is
called its
Differential co-effi-
cient.
The symbol /'(c) then denotes the derivative of f(x)
Examples
1.
Find the derivative of x 2
.
Let
/(*)=**.
/'<*)-lim
&^^
when h ->
www.EngineeringEBooksPdf.com
for x=*c.
DIFFERENTIATION
X
_
when h ->
^^I^
(
]i ra
ft
=
Thus 2x
is
lim
(
the derivative or the derived function of
x2
.
Puting x a=l, we get
which is the derivative of
Ex. 1, 4-11, p. 73.
2.
Find the
.x
2
for
x=l
and agrees with the
differential co-efficient
result of
of \/x.
Let
f(x)=\(m
^
We start
afresh to find derivative at
when'/* ->
through positive values
tive values of h.
Ex.
Find the
;
--r
,
'
x=0.
when X
We have
being not defined for nega-
<\Jh
of
derivations
(///)
(i/y 1/^Jx.
x8
.
</v)
ax*+bx+c.
Another notation for the Derivative. In this notation the
variables x and y are denoted by the composite symthe
changes in
bols Sx and Sy respectively, so that
4-13.
The
derivative
t.e. 9
lim
(Sj>/Sx),
as 8x ~+ 0,
is
then denoted by
another composite symbol dyjdx.
Thus
<
dx
.
lim
=
when 3x
^ 0.
ox
of y=f(x) for any partiAgain, the value f'(c) of the derivative
cular value,
c,
of x
is
denoted by
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
76
dyldx is a composite symbol denoting lim SylSx and is not to be
regarded as the quotient of dy by dx which have not so far been defined as separate symbols.
Note
1.
Note
2.
The changes gx and Sy
1.
Find
Ex.
2.
If .K=M(x*-i-l), find (dy/dfc)
An
(-! ) x=Q
y are
*- when y
and
Ex.
4-14.
ts
in x,
*f-
also
known as
increments.
=-,
when *=-l-
important theorem.
Every
finitely derivable function
continuous.
x=c
Let/(x) be derivable for
so that the expression
[f(c+h)-f(c)]lh
We
finite limit as h tends to 0.
tends to a
write
lim
J^L^'J^J- x
lim
"
ii
m
(
A--0
Hence
lim fa+h)=f(c),
Therefore /(x)
Cor.
nition,
then
Note.
is
i.e.,
lim /(x)=
x ~> c
continuous at x=c.
If f(x) is derivable for every point of
it is continuous in that interval.
The converse of this theorem
is
its
not necessarily true
interval of defi-
i.e.,
a function
be continuous for a value of x without being derivable for that value.
may
For
example, the function
yis
1*1
continuous but not derivable for x=0.
[Ex. 2, 3
Ex. 1. Show that the function x + x
of x but is not derivable for x=0 and xl.
|
is
|
Ex. 2. Construct a function which
not derivable for three values of x.
Ex. 3. Show that/(x)=x*
and derivable for x=0.
[For
mor examples and
sin (1/x)
|
is
1
|
is
page 73)
continuous for every value
continuous for every value of x but
when x-^0, and/(0)-0,
is
continuous
exercises refer to the appendix to this chapter.)
www.EngineeringEBooksPdf.com
DIFFERENTIATION
To show
Geometrical interpretation of a Derivative.
4*15.
/'(c),
77
that
t.e. 9
is the tangent of the angle which the tangent line to the curve
at the point P[c,f(c)] makes with x-axis.
y=f(x)
We take two points P\c, f(c)] and Q[(c+h), f(c+h)] on the
curve yf(x).
and
Draw the ordinates PL,
draw
PN
JL
MQ. We
QM
have
yi
and
T
L
NQ
Here,
_XRQ
is
M
Fig. 42.
the angle which the chord
makes with the A'-axis.
As h approaches
PQ
of the curve
0, the point Q moving
along the curve
the
the chord PQ approaches the tangent line
P,
point
approaches
TP as its limiting position, and XRQ approaches /. XTP which we
^
denote by $.
On
taking limits, the equation
(1) gives
tan</r=f'(c).
Thus f'(c)
is
the slope of the tangent to the curve
y=f(x) at
P[cJ(c)}.
Cor.
The equation of the tangent at any point P[c, f(c)] of the
curve y =/(*) is
y-f(c)=:f'(c)(x-c).
Note. The student should note that it is not necessary for every curve to
have a tangent line at every paint thereof. The existence of the tangent demands
the existence of the derivative and we have seen in Ex. 2, and 3, 3 4'11, p. 73
that every function
is
not derivable for every value of jc.
For example, we know that x is not derivable at JC=0. The
curve y= x Cannot therefore possess tangent at (0, 0).
This fact may be seen directly from the graph Fig. 8,
16 also.
]
|
|
|
1.
Find the slope of the tangent to the parabola
p.
Ex.
point
thS
(2, 4).
Derivative of x* is 2x and its value for
required slope of the tangent is 4.
angle
y=x 2 at
x=2
is
4.
Ex. 2
Show that the tangent to the hyperbola y^ljx at
with x-axis.
3*r/4
www.EngineeringEBooksPdf.com
Hence the
(1, 1)
makes an
DIFFEBBNTIAL CALCULUS
78
Ex. 3.
points (-1,
Find the equations of the tangents to the parabola y=x* at the
and (2, 4).
1)
4*16*
Expressions for
in a straight line.
and acceleration of a particle
velocity
Every thinking person is aware of
and Acceleration of a moving point. The
moving
the concepts of Velocity
In practice,
difficulty arises in assigning precise measures to them.
velocity at any instant is calculated by measuring the distance
travelled in some short interval of time subsequent to the instant
under consideration. Ihis manner of calculating the velocity cannot
clearly be precise, for different measuring agents may employ different
In fact this is only an approximate value
intervals for the purpose.
of the actual velocity and some approximate value is all that we need
in practice. The smaller the interval, the better is the approximation
to the actual velocity.
In books not employing the method of Differential
Calculus,,
velocity
any instant is defined as the distance travelled in an
Now there exists
infinitesimal interval subsequent to the instant.
no such thing as an infinitesimal interval of time. We can take
intervals of time as small as we like and in fact interval with duration smaller than any other is conceivable. The definition as it stands
A meaning can, however, be attached to the
is thus meaningless.
above definition by supposing tluat the words 'velocity' and 'infinitesimal' in it really stand for approximation to the velocity' and 'some
short interval of time', respectively.
at
The
meaning to the velocity of a moving particle at any
by employing the notion of Derivative.
Expression for velocity. The motion of a particle
precise
instant can only be given
4-161.
along a straight line
equation
is
analytically represented
by a functional
where, 5, represents the distance of the particle measured from some
fixed point O on the line at time /.
Let P be the position of the particle at any given time /. Let,
some short interval 8t, and let PQ=8s.
again, Q be its position after
ratio 8s/S/ is the average velocity over this interval and
know intuitively
is an approximation to the actual velocity at P.
smaller
be
obtained
will
better
considering
by
that
approximations
The
We
values of 8t .
We
are thus led to define the measure of the velocity at
J>eing equal to
f
.
lim
Hence,
if v
8s
T
6t
ds
.
,
denotes the velocity,
i.e.,
fr
dt
we have
www.EngineeringEBooksPdf.com
P as-
DIFFERENTIATION
19
Expression for acceleration. Let v be the velocity at
t, and lot v-\-Sv be the velocity after some short
interval of time St 8v is the change in velocity during time 8t.
4*162.
any given time
:
8t
The ratio SvjSt is the average acceleration during this interval
and is an approximation to the actual acceleration at time t. The
smaller values of St will correspond to better approximations for the
acceleration at time t
We are thus led to define the measure of
acceleration as
.
.
7
Jim
Bv
dv
.
--,
.
i.e.,
dtf
/
Ex.
Find the velocity and acceleration
each of the following cases
1.
(//) initially,
in
(a)
j=i*-f2f+3.
Ex,
2.
function of/
Ex.
A particle
;
3.
(/')
at
the end of 3 seconds,
:
(b)
(c)
moves along a
prove that
Ff ,y-/ 3
s= \l(t -{-!).
its
straight line such that
acceleration remains constant.
5=
V
is
a quadratic'
-2/ a -h3f-4, give the position, velocity and acceleration
the particle at the end of 0,
1,
of
2 seconds.
4-2.
The remaining part of this chapter is devoted to determining the derivatives of functions. Home general theorems on differentiation which are required for the purpose will- also be obtained"
To provide for illustrations of these general theorems,
4*3.
in
we obtain, in this section, derivative of x a where a is any real'
number.
4-21.
Derivative of constant.
Let
V=r,
where,
c, is
a constant.
To every value of x corresponds the same value of y, so that theincrement Sv, corresponding to any increment Sx, is zero.
*y
8x
dy
..? -o
8.x
..
By
...
dc
_
Note. Looking at derivative as the rate of change, this result appearsalmost intuitive, as the rate of change of anything which does not change i
necessarily zero.
The result may also be geometrically inferred from the fact that the sloper
cf the tangent at any point of the curve y=c, which is a straight line parallel to*
x-axis, is 0.
4-22,
Derivative of
xa where a
is
any real constant number.
"*
Let
www.EngineeringEBooksPdf.com
90
DIFFERENTIAL CALCULUS
Let 8y be the increment in y corresponding to the increment 8x
in x.
y+8y=(x+8x)*
,
"
Sx
8x
=
(x+8x)-x
'
lim
(
3-65, p. 63)
Hence
d(x
a
=a x^
Vdx
where a
is
any
real
)
number, rational or
,
1,*
irrational.
Another method. The derivative of xa for rational values of a
can also be obtained without employing the general limit theorem of
3-65,
Case
I.
Let a be any positive rational number, say, p\q*
Here
We write
Then
z+8z=(x+8x)
8y
l lq
or
(z+Sz^^x+Sx.
=
8x
Let 8x ->
so that Sz also -> 0.
www.EngineeringEBooksPdf.com
SOME GENERAL THEOREMS ON DIFFERENTIATION
81
= axoc-1
Case II. Let a be any negative rational number, say
We have
being both positive.
Sv
oy
8x
Writing
Sx
Let
dy
Jx
;
p,
q
v~*Pl4
x
fYJL&v-\~~PlQ
(x-^-bx)
~~
p\q
*
Sx
z=x llq
1
and z+Sz=(x+Sx) ^, we obtain
Sx ->
so that Sz ->
1
""
z*.z p
^
p'
^1
As
also.
'
a-2^"-
before,
w
get
1
9
1
Ex.1.
(/)
()
dx
ar
4*3.
Some
general theorems on differentiation.
Derivative of the
4-31.
of x.
functions
able
derivable
Denoting their sum by
y,
sum
we
or difference.
Let
w, v
be two
write
...(i)
Sw, Sv, 8y be the respective increments in u, v, y, corresan increment Sx in x so that x, u, v, y become x-\-8x,
to
ponding
Let
u-\-8u,
r+Sv,
y+Sy
respectively.
We
have
y+Sy^U+Sll+V+Sv.
Subtracting
(i)
from
(H),
JSj^ ==s
Sx
and dividing by Sx we get
SM
Sv
'
"SjT"^"8jc
www.EngineeringEBooksPdf.com
...(//)
82
DIFFERENTIAL CALCULUS
Let Sx ->
0.
8y
to
,.
=lim
/Sw
8u\
(to+to)du
dy
We may
"
1
Sv
v
,
+lim S*
^
>
dv
dx^dx +
r
Sw
,.
1
ta-
similarly prove that
d(u
~
v)
dx
du
dv
dx
dx
Generalisation.
By a repeated
obtained above, it can be proved that
number of derivable functions and
"
application
if
u l}
t/
2,
,
of the results
u n be buy finite
then
dy
dx
We
du
= ~dx
duz
du 3
du n
~dx
~dx
dx
'
thus have the theorem
:
Algebraic
derivable functions is itself derivable
equal to the sum of their derivatives.
and
sum of any finite number of
sum is
the derivative of the
Ex.1.
_ 11
_
= 1 4.
-TV
/
(0
1
3
_
i '\
x -i- 1 _
2
__
x
-i
J
(io
'
3
Find the derivatives of
i*.
>|X
I
dx
dx
Ex. 2.
,
(l-/o
""""
www.EngineeringEBooksPdf.com
z
_
DIFFERENTIATION
4-32.
Derivative of a product.
Let
y~uv
where
are
u, v f
Let
two derivable functions of
Sw, 8v, 8y,
be the increments in
ponding to the increment 8x in
We
x.
x.
y respectively corres-
u, v,
have
8y=u.8v+v.8u+8u.8v,
or
Let 8x ->
function,
is
Then 8u
0.
Sw
8v
8y
*
also
;
for
8v
w,
which
is
a derivable
continuous.
W
-limf
L
,.
*V
+
:
.
'
fix
Sv \
/
.
=hm(
ljc
H-to.
f
Sx
v.
,
when S* -,
'
,.
8
Su \
/
,
=
dx
.
..
)+hm( v.---)+hm3w.l,m
dv
..
,.
du
dv
dx
.
Thus we have the theorem
functions is itself derivable and its
The product of two derivable
is the sum of the twa
with
each
obtained
derivative of the
products
by multiplying
function
:
derivative
other.
=
The
first function
derivative of the product of two functions
derivative of the second -f second function x derivative of the first.
The
result
(i)
may
.-
y dx
Note.
It
x
also be re-written as
^
u dx
may be noted
dx
v
,
'
that the operations of differentiation and multi-
plication are not invertible.
l
'
d(u
e"
__
"^
v)
"dx
,
du
dx
dv
dx'
'
Cor.
1.
Generalisation.
Directive of the product
number of derivable functions.
We first take
www.EngineeringEBooksPdf.com
of any finite
84
DIFFERENTIAL CALCULUS
dy
(
f
du,
Hi Un
*
"
du
On
by
dividing
I
l
du l
1
dy
dx
dx^U!
~if
By
y=u u2 u^ we
obtain
1
du z
u%
dx
*
repeated application of this result,
l^
dki
1
dy
27
,-a
1
-r~
du
dc
'3x~~
dx
dx
du
du z
u3
dx
,
*
we obtain
du^t
i
J_
^^
where y==w1t/ 2 w3 ...t/ n
Cor. 2. Derivative of cu, where c
vable function of x. Let
dy
1
is
=
TT
wr
.
a constant and u any deri-
t/w
.
Hence
du
d(cu)
v
2.
;
dx
t/x
Find the derivatives of
(0 (x+2)(3+x).
4-33.
a
(//)
(x-l-2) (2x:-3).
Derivative of a quotient.
Let
J>=n/v,
are two derivable functions of
for the value of x under consideration.
w, v
x such that
www.EngineeringEBooksPdf.com
v is not zero
85
DIFtffiBENTIATION
V.SM--M.8V
T(v+Sv)"'
By
Sx
Now,
~
v(v
+Sv)
being a derivable function of x,
v,
is
continuous
Hence
as Sx -> 0.
Sv ->
Thus
dv
*du
<.:* =
d(
,
"~~
dx
v2
dx
so that derivative of the quotient of two functions
= [Derivate of Numer.
(Denomr.)
(Numer.) (Derivative of Denomr .)]
x
,/
d(
...
Ex.
^ dx
(1+^) ax
7
\
,
--
+x--)/
i
\l
Square of Denominator.
-
dx
(lJr x).lx.l
L 1+X*J
-[--i]
Obtain the derivatives of
2.
n\
'
ft:\
,
^x -f~5
v-*
r *)
x*
"f
x
s
Being a derivable function, v is continuous. Also, we have
value of x under consideration. There is, therefore,
an interval around x such that v^O for any point of the interval. (3*51, p. 54.)
Note
supposed
tf
at
1.
v=o Tor the
www.EngineeringEBooksPdf.com
86
DIFFERENTIAL CALCULUS
Thus
if
ding value of
v,
we suppose x+ Sx
/.<?.,
to lie wittvn this interval then, the corresponjustifies division by v+gv in step (/)
This fact
v-j-gv^O.
Note 2. The importance of the results obtained above in 4 3 lies in
the fact that the derivative of any function which is an algebraic combination
[/.*., built up through the operations, of addition subtraction,
multiplication md
division of several others is itself expressible as an algebraic combination of the
derivatives of the latter.
4
34.
Derivative of a function of a function.
11= <l>(x), soy
is
y=f(u) and
//
a function of x,"*then
dy
dy
du
^
dx
i4r
'and<t> being derivable functions
du
dx
*
'
ofu and x
respectively.
Let Sx be any increment in x and Su the corresponding increment in u as determined from u = (f>(x). Again, corresponding to the
increment Su, in u let 8y be the increment in y as determined from
9
y=f(u).
We
write
Sy
Let Sx ~>
Su
Sy
8u
8x~~~
Sx
so that Su -> 0.
n
nm
Sx ->
.
( _ S ^y \\
i
sx
_
.
.
f
o
->
lim
^
'
Su
8>L
->
6//
Su
/ Sy
-rL,
nm
lirv,
-
\ Sx J
Sx
lim
a
Sw
5x
8u
.
S-^
>
Hence
du
jdy ~~ dy
dx
The
functions so that
dy
dx
Ex.
(i)
dx
immediate generalization.
result is capable of
be three derivable
'
du
>>
a function of x,
is
du
dy
~ du
'
dv
'
dx
dv
Find the derivatives of
We
w
write
"
du
^
dy
= l +x 2
,
j=w2.
1
i
-
Hence
www.EngineeringEBooksPdf.com
Thus
if
we have
87
DIFFERENTIATION
or,
without introducing
u,
_
~
dx
(U)
Let
we have
'
</(l+x
2
dx
u=-~, y=u
du
</x~
(I-*)
2
-1)
,
(1-x)
__
J2
'
2
~~(1
x)*
Hence
dy
~~
C/A:
dy
du
du
1
~~
'
dx
/l4-x\
:
or, directly
l+x
Ex.
(i)
Find the derivatives of
(ax+b)
n
(
.
(M
{tv)
Differentiation of inverse functions.
4-35.
function derivable in its interval of definition.
admits of an inverse function
Let y=f(x) be any
suppose that it
We
as explained in 2*22, p. 21.
We
have to find a relation Jbetween/'(x) and
www.EngineeringEBooksPdf.com
f '(y).
88
DIFFERENTIAL CALCULUS
Let 8y be the increment in y corresponding to the increment
8x in x, as determined from y~-f(x). The increment bx in x corresponds to the increment 8y in y as determined from
We have
,
1
.
-
Sx
Sy
Let 8x -
^y
by
d[v/rfx
I
dy
and dx/dy are reciprocal to each
y=x* when x = 2.
Differentiation of functions defined
4*36.
We consider
by means of a para-
two derivable functions
x =ft) y=9(t)
9
of
other.
Verify the theorem for
Ex.
meter.
.
0.
dx
Thus
8x
" A*
:
t>x
9
'*'.
Assuming that x=f(t) admits an inverse .function t=$(x) (2-22)
we
obtain
so that
y
By
(
4*34),
is
a function of x.
the rule for the differentiation of a function of a function
we have
dy
dx
~
dt
dy
*
dt
'
dx
dt
AI
Also
_
.
dx*=
Note,
Ex.
We
'f* is
1.
dt)
-
'
~dt~T\t)
called a parameter.
Finddyjdx, when
x=a/ 2 y=
,
have
www.EngineeringEBooksPdf.com
89
DIFFESENTIATION
2.
Find dyjdx, whea
...
(0
I/'
,.
*- flt
,-6
{
2t
....
.
*=
*=
(ii)
,.
3at
8,
7-
2a/
.,
44. derivatives of Trigonometrical functions. The symbols
Bx and 8y stand for the increments in x and y and will always be
used in this sense without any frequent mention ^of their meanings.
4 41. Derivative of sin x.
Let
>>=sin x.
sin (x+8x)
8y
""
~
8x
sin
_x
S^
_2
""
cos
%(2x+8x)
Bx
sin |8x
sin iSx
gy-
^
=lim
S
cos
(x+^Sx)
UJ\
As
cos
AT
Also
x
is
.
lim
wo
a CDntinuous function,
lim cos
o
when Sx
->
r,
i
0,
sin iSx
- .2
T
hm
r
-
-^$ O^t
,
.
when Bx ->
have,
when Bx
0.
-> 0,
= 1.
,
dy -=
,-
cos.v.
Thus
d
(sin
x)y
v
dx
cos x.
Derivative of cos x.
4-42.
Let*
By
_cos (,V+SA:)~ cos x
'Bx^
SAT
2 sin |(2x
-
_
=
As
sin
A: is
dy
+Sx)
sin
s^
-sin
sin iS
.
$A
a continuous function, \ve have,
lim sin (x+^Sx) = sin x.
= hm
,.
.
i* v, iC- sm x +iSx)] hm
.
/
,
1-
(
a?->0
=
sin
xl
sin x.
www.EngineeringEBooksPdf.com
when Bx ~>
90
DIFFBBBNTIAB CALCULUS
*->(Cos x/)
Tk
Ihus
Ex,
Let
We
smx.
Find the derivatives of
1.
sin 2x.
(/)
(i)
=
j;
= sin
cos 3 x.
(ii)
(Hi)
^(sin
2x.
write
= 2x,
w
dy
so that >>:=sin w.
du
dy
or briefly
~~=
d(sm 2x)
(ii)
Let
We
write
2x\
d(sm
~
}
d(2x)
- cos
2x
2==2 cos
.y
t/=cos
.y=w 3
so that
.
du
rfv
rfy
A:
3 cos 2 JC
)=
.
sin x.
or, briefly
__ </(cos x)
~
dx
dx
=3
Let^^V
(Hi)
We
write
3
(cos x)
rf(cos x)
3
2
x(
~
rf(cos x)
*
~d(Qo* x)
sin x)
=
dx
3 cos 2 x
(sin
v=sin
u=<\/x=x^
u,
j= v/v=v2.
so that
dy
dx
~~
'
dv
\v
""
du
dv
dy
du
*.
*
dx
cos w
.
x~~^
^
COS <\/X
1
VC^ V*)
11
*
V^*
or, briefly
d\/(sm \/x) ~ d\/(sm \/x)
dx
d sin \/x
= | (sin
sss
JL
*
v^*)""""
dsin y/x
cos
cos
*
4
www.EngineeringEBooksPdf.com
V^- I*
.
sin
x
91
DIFFERENTIATION
Find dy\dx for /=7r/2, when
2.
x=2
cos
y=2
cos 2/,
t
sin
f
sin 2t.
We have
sin
^=--2
-^ =2
cos
/
(cos 2/)2
dy dx
dy
dx ~~dTj ~dt
,
Putting
= 7T/2,
/
z=r
=2
~2(sin
sin 2/,
2 cos 2/>
/
cos 2/)
sin 2/)
/
2(cos
cos
/+2
2 sin
/-(-sin 2/)(2)=
/
we obtain
Find the derivatives of
3.
w
(0 sin x,
(//)
sin x
(1/1)
^
m
(/>)
,
cos 8
1
--*,
(v)
cos mx,
(vi)
.f
cos
(v/0
.
w
sin^x. cos x.
<v//7)
4.
Find dy\dx when
y
(/)
(i/y
x=a (cos /-f sin /), y=a
x = 3 cos r~ 2 cos 3 /,r=3
x-a cos 3 /, ^- sin /.
/
sin
/
r
(sin
/
cos
2 sin
3
/).
t.
3
(///)
4-43.
(D.U. 1955)
Derivative of tan x.
Let
>=tan x
^
(x
tan x
Sx
sin
_
+ Sx)
cos
(x+Sx) sin x
(x+Sxj"~cos x
Sx
cos x
~ sin (x+Sx)
Sx cos
cos
(A:
(x 4 S^j
sin (x-4-S.x
x)
8x. cos (x+ox). cos
*
cbs~fx
x
sin
1
1
+8x)
cos :f
+8x)
x
cos
*
Sx
www.EngineeringEBooksPdf.com
sin
DIFFERENTIAL CALCULUS
= cos
J_._L.l
=8ec*x
x cos x
dx
d
Thus
Or,
we
x.
^>=sec*
dx
write
x= sin
,
}>=tan
cos
d
cos x.
,
x
x
,,
(sin x)^
d
.
sin x.
-,
dx
dy
dx~~
,
so that
(cos x)'
j
dx
[
2
cos lc
cos x. cos
x+sin
x. sin
x
cos 2
1
"cos 2
/
4-44.
7
=
proof
Ex
1.
(i)
tan x-f cot x.
.
N
(iv)
cosec 2 x.
Find the derivatives of
tan x
cot
sin x. tan 2x.
(/i)
A
;c
(v^
.
tanx-t-cotjc
4-45.
x.
to the reader.
Its
is left
=sec 2
x
tanx\
A //I
\
VJ
x tan x cot 2x.
(Ill)
/
i.
(
\l + tan x/
.,
(v/)
A
A
//"I
/ i 1i
y V
Derivative of sec x.
Let
>>=sec jc=-
1
COS
X
1
" Sy
cos_(x+8x)
~
cos
x
cos
x
cos
"Sx. cos (Jt-f6x). cos
x
2 sin | (2x+$x) sin
bx. cos (x+S.x) cos
1
Sx
x
sin i
1
OX)
1
COS
X
COS
-=tan x
www.EngineeringEBooksPdf.com
sec x.
8x
D CFFEK ENTI ATIO N
d(sec
~ x)
Thus
Or,
dx
we
.
-=tan x
,
dy
-/-
dx
cos
so that
x
x 01
---= cos
cos
-
,
sm x)
--=tan x
(
s-2
dfcosec x)
'=
-^-,
Its proof
x cosec
cot
,
/
x
sec x.
x.
to the reader.
is left
Find the derivatives of
Ex.
(/)
sec x.
write
v=
A A*
4-46.
93
cosec8 3 x.
(//)
(Hi) secVfc-f&x).
(/v)
>l[sec
(ax+b)].
sec (cosec x).
Darivatives of inverse trigonometrical functions. The precise
4*5.
definitions of inverse trigonometrical functions as given in
2*7, p.
30 will have to be kept in mind to obtain their derivatives.
4-51.
Derivative of
sin"" 1
x.
Let
1
^=sin"" x so that x=sin y.
dx
4y
j^
__
^
1
where tha sign of the radical
By
sin"" 1 x,
the def. of
is
we have
1
A:<'7r/2,
i.e.,
positive.
Hence
4-52.
1
it
the same as that of cos y.
is
7r/2<sinso that cos y
1
_
,,
=
fi
Derivative of cos-
1
^/(TT^p
x.
Let
y^cos- x
1
so that x^=cos y.
dx
www.EngineeringEBooksPdf.com
7r/2<><7r/2
DIFFEBENTIAL CALCULUS
94
1
dy
f\f
-
_.__
_.
-
where the sign of the radical
def. of cos"" 1 x, we have
<
Also
if
y
lies
4-53.
.._
and
dfcos- 1 x)/
same as that of sin
is .the
cos" 1 * <TT
between
Hence
1
,
I
then sin y
TT,
=-
---,-
<
i.e.,
is
<
y
By
y.
the
TT.
necessarily positive.
1
Derivative of tan- 1 x.
Let
"1
dx
x
so that
x=tan
y.
9
or
1
4.54
454
d(cot-
dx
x)_
~
J^
1+x*
Its proof is left to the reader.
4-55.
Derivative of see" 1 x.
Let
y=sec~ 1 x so that #=sec
dx
j =sec v tan v
y.
dy
dy
dx
or
sec
_
We take,
y tan y
L
,
--^-
c*f\f* i^ //afvr>2
2
sec
^ V(8e
s ig n before
+>
S"
=
i
1 \
J'
1)
=
,
1
x)
.
-*>2
'
the radical and write
(Refer note below)
x7(^)
Thus
d(sec~
-_
// 2 x-^ VCx
1)
>
1
1
'
dx
www.EngineeringEBooksPdf.com
95
DIFFERENTIATION
Note. This note is intended to show precisely what sign should be
chosen before the radical. Now it is clear that the sign before the radical is
the same as that of tan y. By the definition of sec~ T x. [Refer 2*75, p. 33],
we have
When x
ween
is
positive so that it lies in the interval
is positive ;
[1,
oo].
then y
[
oo,
1],
lies
bet-
then y
lies
and n j2 and so tan y
When x is negative so that it lies in the
between w/2 and n and so tan y is negative.
Thus the sign of the
Hence
radical
is
interval,
positive or negative according as
x
is
posi-
tive or negative.
:r"=
2
^
dx
x^(x*l)
if
*>
=
and
-------,/-
2
~ir
if
<-
*
jH(x -l)
so that
dy
_._
1
=-:
ctx
i
x
I
Thus
*\
[x~
for every admissible value of x.
1 )
__~
dx
_
2
|x| V(x -!)'
Derivative of cosec- 1 x.
4-56.
Let
x
dx
-j-
or
=
dx
cot
x=cosec
so that
y.
y cosec
cosec
cot
y
y
__
-1
cosec 2
We take,
+,
-1
yl)~-^
x\7(x
a
-1)
sign before the radical and write
"
dfcosec" 1 x)
-
Thua
Note.
By
dx"
The
and
nr/2
when x
ween
7T/2
and
xV(x-iy
sign before the radical
the definition of cosec
When x
ween
1
1
*,
is
the
same
as that of cot y.
we have
is positive so that it lies
and so cot y is positive
in
the interval
[1, oo],
then y
lies bet-
then y
lies bet-
;
is
negative so that it lies in the interval
and so cot y is negative.
www.EngineeringEBooksPdf.com
[oo
,
1],
DIFFERENTIAL CALCULUS
96
fckw
Thus the
sign of the radical
is
positive or negative according as
=^br
and
so that
we
dKcosec^x)^^
Thus
Ex.
(v;
'
posi-
if
x<0
>
for every admissibl e value of x.
1_
^
Find the derivatives of
1.
sin-H*.
(i)
is
can write
l-
(//;
tan-H(l +*)/(! -x)].
(i//)
x
Hence we have
tive or negative.
x
sec
>t
_
tan-Xcos Vx).
(/v)
^
.
/ X-
Jt!~~l\
(v/iFcos-
l4x
,
Ex.
2.
4-61.
.
Find
when
rfy/d*
Devivative of
(, x are both
log,, x.
positive)
Let
J>=log x.
tf
Now
Cor.
lim (
--^
Let
a=e
1+
X
* )
^
=e.
so that
^=log^ x=log
f
_
-.,
rrs"^
dx
x
t
iut&
6e
c
x
www.EngineeringEBooksPdf.com
(
3-63, cor. 3)
D1FFBBENTIATION
JL
*&*_
x
dx
Thu8
4-62.
Derivative of ax
.
Let
a*.
y
~x -f ox
*
a
=
&x~
by
dy
_
2~*
a
x
ax
8x
,
a
,.
8x
a
&x
-,
1
*
*
8x
-l
ii? """&T~
=a*
(3-64, p. 63)
log^a.
Thus
-f-J=a-
Cor.
Let a^=e so that ^=6^.
Thus
-fdx
.
Find the derivatives of
Ex.
97
(/)
log sin x.
(//)
(/v)
log[sin (log x)].
(vi)
log tan (Jx -hj").
:
cos (log
x).
(Ill)
(v)
(vii)
*
sln *
logV(#
-I y.
(V//0 -,-~r'
log X
(x) logioCsin"
(x/v)
('*>
1
^1 ).
w *+e- w;r
(jci")
).
eV^7
(jci//)
(jrv)
v \
V(
fl
log[x+
)/
f
V(jc
a** sin 1*.
a+b *
x
los-a b tan x
Derivatives of hyperbolic functions.
4*71.
-t-;t4-l).
log(sec x-ftan x)
0^X
(xi7) log(e
2
Derivative of sinh x.
Let
e*-e~*
*
www.EngineeringEBooksPdf.com
98
DIFFBBENTIAL CALCULUS
m,
d(sinh x)
Thus
4-72.
.
===coshx *
d
Derivative of cosh x.
Let
x=
.
x
e -e._.__
dy
^
d <co h
*)=sinh
;
dx
Thus
4-73.
!t
x.
Derivative of tanh x.
Let
,
,
>>=tanh
x
x= sinh
-,
x
cosh
^(sinh-- x)
,
cosh x
-
-,
.
d(cosh
'
- x)
,
~smh x
.
-
.
t/x
T
Jx
""
cosh 2 x
cosh x.cosh
"
cosh 2
sinh 2 x
x
_ ----------
_
cosh 2
_.
Thus
A
*A
4-74.
xsinh
x
1
.^ ---
,
dx
-=
^-^dx
,
cosch 2 x.
ft
Its proof is left to the reader.
4*75.
Derivative of sech x.
Let
v=sech
Q
dx
^
""
x
cosh
1
x=~cosh
vx
xO
:
l.sinh
cosh a x
sinh
^^ _- S ech10
cosh 2 x
.
d(tanh
x)/
~
=sech29 x.
d(coth x)
/
x.sinh
cosh 2 x
x
www.EngineeringEBooksPdf.com
x
2
x.
DIFFERENTIATION
d <S
Thus
h
-*>=-tanhxsechx.
dx
An*
4-70.
d(cosech
x
x)<
-_
Its proof
=
..
,
coth x cosech x.
to the reader.
is left
Derivatives of inverse hyperbolic functions.
4-81.
Derivative of sinh- 1 x.
Let
3^= sinh- 1 x so that x=sinh}>.
dx
dy
-A:
or
1
1
where the sign of the radical is the same as that of cosh y which we
know, is always positive, ( 3'72, page 68).
^
Hence
4-82.
1
1
dfsinhv x)
-'
...
dx
Derivative of cosh- 1 x.
Let
x=cosh
1
j^cosh""" x so that
y.
dx
dy'
^_J_
Gin
n
Ol
1111
//~Y*
t*^V
I
v
X
"
1
A /^rr^ftri*
tV ^^ A
1
I v/v/Oll
\/
?
^
i
H /iI
^^
J
I**
*Hs
^^
where the sign of the radical is the same as that of sinh y.
Now, cosh- 1 x i.e., y is always positive so that sinh y
(3-71, page 68).
'
4-83.
^
ll00
^^^
Hence
Derivative of tanh- 1 x.
[
x
\
|
.
<1]
.
Let
j^tanh- 1 x
so that
x=tanh
y.
dx
l^l
Thus
T~- =sech* "
V,
dy
1
1
dy ^.^
dx ~sech 2 >~ 1-tanh2
_r_ _
._
-
--=.- -----
dx
= *1
x^a
>>
1
~~l-x*
.
Its proof is left to the reader.
www.EngineeringEBooksPdf.com
*
1
*
is
\I
positive
100
DIFFERENTIAL CALCULUS
Derivative of sech- 1 x.
4*85.
Let
1
>>=sech~ x so that x=sech
=
sech y
-j
sech y tanh y
--1
where the sign of the radical
is
is
-1
the same as that of tanh y.
But we know that sech- 1
tanh y
y.
1
dy
dx
or
tanh
.
y.
x,
y
i.e.,
always positive, so that
is
always positive.
dteechr 1 x)
--- J _.
_,
Hence
._!.
dx
.
xyXl
Derivative of cosech-
4*86.
1
_
1
x
)
x.
Let
1
>?=cosech~ x so that x=coseoh y.
dx
T
=
.
coseoh y
or
cosech y
_
,
sign of the radical
Now,
as
x
is
y,
.
coth
____
^cosech y
when the
,.
coth y.
.
is
.
________
2
=,
-
,
^(cosech y~+I)
x<\/(x*+l)'
the same as that of coth y.
and therefore coth y
is
positive or negative according
positive or negative.
.
*2?!*^x) -
TH
rhus
x
dx
)
|
ifx<0.
=1
VO
for all values of x.
Find the derivatives of
Ex.
(i)
log (cosh x),
4-91.
(//)
necessary to take
is
slnh2
*,
Logarithmic differentiation.
v
a function of the form u
which
*
known
its
(///)
In order
tan
x
to
.
tanh x.
differentiate
where w, V are both variables, it is
logarithm and then differentiate. This process
,
as logarithmic differentiation
is
www.EngineeringEBooksPdf.com
also useful
when the
101
DIFFERENTIATION
function to be differentiated
The following examples
Ex.1.
Differentiate
log
1
dy
y
dx
,-
.
x
sln x
^ = 16g
we
Differentiating,
x
x sin
=x*
Let
(x
Hence
X
=:COS
-
+ Sin X.
,
.
,
log
&
JC
y=x *
l
n
log
.
+ (sin
Differentiate [x
write
log x.
.
get
tan *
Ex. 2.
*+(sin
x
sin x\
x -----
y
-\
cos x
x)
].
cos
*.
x)
tan
Let
t/=;c
and
v=(sin x)
so that
y
..
*,
(1)
cos
dy _ du
*
-
-
-
From
(1),
we
-
*,
dv-
r
\
obtain, taking logarithm,
w=tan x
log
1
du
u
dx
,
log
.
,
we
tana;
/
f
^ec
log
v=cos x
</v
,
--==
dx
v
sin
dx =(sm x)
3.
(3)
and
.
.
log
x+
tanx
\
-
J-
log sin x.
x
1
.
cos^/
dv
Adding
x
.
obtain, taking logarithm
1
*-e.,
x
.
rfx^^
(2),
;c.
21logx+tan x
=sec 2 x
rfw
j.^.,
From
...(2)
dx^dx
rfx
Ex.
)=sin x
dy = sin * /
~
x
(cos x
TT
We
the product of a nuniber of factors.
is
illustrate this process.
(4),
log
a sin
jc+cosx.
.
^-sin
x
.
log sin
we obtain
Differentiate
www.EngineeringEBooksPdf.com
.
sin
x
COBJC
/ov
...(3)
DIFFERENTIAL CALCULUS
102
Putting
it
equal to y, and taking logarithms,
logx+| log (l-2x)-f
Differentiating, we obtain
log }>=
!
dy
1
2
x
3
log
-2
(2-3*)- J
l-2x~~
4
log (3-4x).
-Z'
3
_
2
we obtain
2-"3x
4
5
3-4x
_ __
9
+5(3-.4x)
9
2JC
Ex.
(/)
4.
Find the
(cos x)
(1
(3-j?)
sin
x
.
e
16
"I
+ 5(3-4x)" J'
:
(in
x
tanA:
-f(cotx)
8/4
x
.
(4-^ 4 ) 4
log x
'
*(v)
(log
;c)4(sin-
1
.
5
xx
.
/C
,
-*)V (2-x)/
;
(vi/j)
differential co-efficients of
.
cot
A
"
x
log
(/vfaanx)
v
^3(l-2;c)
+ 4(2 -
.
Preliminary transformation. In some cases, a preliminary
[transformation of the function to be differentiated facilitates the pfocess of differentiation a good deal, as is illustrated by the following
4*92.
examples.
Ex.
1.
Differentiate
Sin
!+-*
Putting
x=tan
we have
6,
Yr
1
-|-t/an
dy
_
~dx
-!+*'
*'
Ex.
2.
=Bin-i
(sin
Q
1
2^)=2^=2 tan-
2_
Differentiate
Putting x=cos
0,
we have
2
^'
=V2
www.EngineeringEBooksPdf.com
'
cos
x.
DIFFERENTIATION
103
=
e
COS x
.
F
f\
cos
'2 sin
l\
+sin-
_
rt
l-tan
l+tan-2-
~
7T
4
^
=
2
7T
J.
4
2
1
Ex.
F/wJ /A^ differential
3.
tan~ l
,
JL
of
to sin" 1
j
2x
x=tan
.*
J.
-
_
Putting
r^c/
wft/i
*~2
X
co- efficient
0,
we
2
(P.C7. 1954,
1956)
2y
_ sin-i
see that
2
-
.
z
J*
1
-
(tan 20)
3
2= sin~
z=sin- 1
1
( sin
= 2^=2
tan-1 x.
tan~i x.
26)^20=2
2
dy^
dy ,dx
dz~dx dz~
Also otherwise, we have
y***>
dy
j- ==1.
so that
Ex.
4.
//
V( !*)+ V(l -y*)=a(x-y), prove
A
Putting x=sin 6 and y=ain
sin
that
(D.U. Hons. 1949
<f>,
we have
+ *>(lBm*
)=a(sin B
www.EngineeringEBooksPdf.com
sin
;
P.U. 1952)
DIFFERENTIAL CALCULUS
or
cos
0+cos ^=0(sin
sin
<f>)
2 cos |(04-0) cos \
~
2 cos i(0+<) sin~ (V
or
<
^(0~<)= cot- 1 a
.
0^=2 cot"
1
a
"**
sin^
1
Differentiating,
x
sin"
we
get
1
y=2
1
cot" #.
or
<fa
Ex.
5.
.(/)
.....
%
(ltt)
Find
tan- 1
.
.
tan-
the differential co-efficients
(>.t/. 1952)
j^.
Vx
of
(it)
x
..
sin~ l
.
.
.
Ov) tan
/\l*vos x\l/2
y
[Show that
Ex.
6.
this is equal to sin""
1
*
sin
Express in their simplest forms the differential co-efficients with
respect to x of
4 93. Differentiation "ab initio". To differentiate "ab initio"
from first principles, means that the process of differentiation is
to be performed without making any use of the theorems on the
or,
differentiation of sums, products, functions of functions etc., nor is
any use to be made of the differential co-efficients of standard forms.
We have already had numerous illustrations of it.
Ex.
1.
Differentiate sin-*x <ab
Let
^sssin-
1
initio'.
,*.
1
.y+Sy^sin- (x+Sx).
We have
and
so that
x=sin y
x + &t s= sin
(y
+
8>>)
Sx=sin (y+8y)~8my.
www.EngineeringEBooksPdf.com
(D.U. 1955)
DIFFERENTIATION
105
Thus
Sx
sin
(y+Sy^&'my
_
1
Let Sx
(y+ %Sy)
~"
2.
F//irf,
cos
3;
/rom
>0.
~
~~~
*
cos
\
$8y
Vsin ^ Sy )
so that 8y also
-->
</x
Ex.
/
'
cos
y
^(F
~x*y*
first principles, the differential co-efficient
V
of
~sin*:~
Let
y~ Vsin
x.
\/sin x
=
'
Sx
sin
sin
(x+Sx)
= COS (X + iSx)
.
Let Sx->
\/sin
x
1
sin A Sx
TO
'
---- ------
1
0,
^V
=cos x
,
,
COS
^
1
.^
X
~~
Vsinlc
Ex.
Find, from
3.
sin
<'>
^
first
principles, the differential coefficients of
2
()
-
sin
2
x.
(i/O
Vx.
Exercises
Find, from
* 8)
*
*(
4-
tanx 8
first
principles, the differential coefficients of
2.
-
.
5.
>/(tanx).
sec^x.
www.EngineeringEBooksPdf.com
3.
:
x
:-
(D.U. 7955)
106
J>^FFBBBNTIAlj CALCULUS
Find the
differential co-efficients of
6XT^iwo 3T
:
__
-.*.
7.
.
X*
b tan- 1
9.
tt tan
*
11.
*
i
tan- 1
^
~\
** --*
tan-*
12.
1+sin-jc
-I
..
O-fcOjH^*).
14
O
.
16.
.
8
a
.sin- 1^.
17. [l-x")"
22. 10
lo 8 sin
Mj
tan-11
.
e
26.
/7
V
23.
.
Y*
A-
/
-T-J
b + a cos x
IJX*
.
'
21. sin-
x
L./
f
t/ f*O^
V^^Jo
^
18. log [tanh (fcc)].
\/(f^H).
20-
3
(l-4x)
~
j^
x log x
log(log x).
.
f <^Q.
v^-V/O
/
^
V
^V
25. (sin
v
x)y
.
sin (x log x).
27.
^29.
&*
/ eaX
_"~_?_
sin^ 1
f
sin- 1
-;^^-^
,
,
.
31.
*^ v-v/k
X.
t
x
33. 9x4 sin (3 jc- 7) log (1-5.)
34. e aj?cos (^ tan- 1 x).
35. log [l-**i
x
37. xa sinh *.
cot x coth x.
^
.
^^
-
,
Find the
.
+)
differential coefficient of
:
www.EngineeringEBooksPdf.com
^~ w sin
/
107
DIFFERENTIATION
42.
Differentiate
'
(0
x
1
,.,,
00
1
1
,2x+l
t
+ ^T tan
^3-
Find dyjdx when
43.
. .,
(i)
(/f)
(///)
sin 8 /
x =
cos 8
1
'a"-*/ 6
-^(cos2/)
'
x=sin H(cos 2/), y=cos /V(cos 20x=a(cos /-flog tan i/), .y=a sin /.
44.
If x* **<-*, prove that
JI5.
Differentiate sin*
46.
Differentiate
47.
Differentiate tan- 1
dy/dx^log
jc/(l
+log
x)
a
(P.U. 7955, 56)
.
(Differentiate logarithmically)
x
x with respect
sm x
(D.U. 1950)
to (log x)*.
with respect to (sin
(P.U. 1959)
x)*.
1
2
with respect to tan[{V(l -hx )-l)}/x]
x
(P.U. 1956)
when x=e
48.
Find
49.
Differentiate
with respect to cos- 1 x 2
50.
(D.U. Hons. 1949)
.
Differentiate (log x)
tan*
with regard to sin (m cos- 1
x.)
(P.
51.
Differentiate the determinant.
F(x)=
From
first
/,(*)
<fr,(x)
principles
we have
F(x+A)-F(x)
;-j
;
S
/<*)
AW
fi(x+h)-A(x)
*)
f&c+h)
(*+*)
www.EngineeringEBooksPdf.com
U".
1955, 56)
108
DIFFERENTIAL CALCTJLUS
<P,W
Dividing by h and making h tend to zero,
we
get
<Pi
!
^*
^.
The
rule can be easily extended to the case of determinants of any order.
APPENDIX
EXAMPLES
1.
Show
that
/(x)=x
8 sin
(//x)
when
x^O
/(0)=0
is
derivable for every value
but the derivative
ofx
not continuous for
is
x=0.
(D.U. Hons. 1954)
For xr0,
sin
f'(x)=2x
coa(
o-sin
=2x
cos
.V
^ )(_
---
X
For x=0, we have
1
x'sin
_!1
fw-m^
x
x
=x
* _
sin
>
x
as
x
/'(0)=0.
x and
Thus the function possesses a derivative /'(x)
is given by
f'(x)=2x
We
have to show that
cos
X
=2*
sin
X
cos
for every value
when
-
X
f'(x) is not continuous for
sin
X
(2x
\
of
-
sin
*
www.EngineeringEBooksPdf.com
x=0.
Y
cos
-*
/
We write
...(1)
DIFFERENTIATION
109
Here
lim ( 2x sin
J=0.
y ^Q
'
In case
lim (2xsin --- cos
x
^
i.e.,
had
x-0
existed,
would follow from
it
x
)
'
that lim (cos 1/x) would also
(1)
;c->0
But
exist.
this
is
Henca lim
x-0
for
THUS
/'(*) doas not exist.
f'(x)
is
not continuous
x=0.
Examine
2.
(
not the case.
oo
,
oo )
the continuity
the
following function
/or
)
=l
and
derivability
in
the
interval
oo<x<0,
in
n x
in
-|7r)
O<X<|TT,
2
in
JTT<X<
oo.
(Mysore)
The function /(x) is derivable for every value of x except
parhaps for x*=0 and x = 7T/2. Thus we shall now consider x=0 and
X=7T/2.
Firstly
we consider x=0.
Now
lim
/(x)
-0)
=
l
and
lim
/(*)=
x->(0+0)
lim
Hence /(x)
is
/x=l=:
continuous for
x=0.
for
x<0,
Again,
lim
so that
Also for
x>0
"""
1+sinx-^.
x
www.EngineeringEBooksPdf.com
(l+sinx) =
DIFFERENTIAL CALCULUS
80 that
lim
*->(OH-0)
M*~
x-KO+0) *
Thus
*
Hence the function
Now we
not derivable for
is
*
x-KO~0)
x=0.
consider x=irj2.
We have
lim
is
X ^(-|7T
0)
lim
Hence /(jc)
lim
/(x)=
W
lim
/(x)=
continuous for
Again, for
Putting
x=/ we
^TT
see that
t
1
sin
x~"
==
Isin
"" (JTT
_
1
cos
f
/)
_
2 sin 2 \~t
t
so that
lim
x
|TT
lim
iii-u
(}n^
exists
and
is
equal to
0,
www.EngineeringEBooksPdf.com
_ sm
.
sin \
"""t
DIFFBBBNTIATION
111
Exercises
Discuss the existence of/'(x) and/"(x) at the origin for the function
1.
/(x)=x
2
sin -
when
x
(B.U. 1953
Examine the
2.
;
D.U. Hons. 1952)
differentiability of the function
m
/(x)= x
-
sin
x
when
x^r.0,
w>0
when x=0
at the point
x=0.
Determine
m when/'(x)
Determine whether /(x)
3.
is
is
continuous at the origin.
(D.U. Hons. 1952)
continuous and has a derivative at the origin
where
Show
4.
is
that
continuous but not derivable for
5.
x=0
and x=l.
Examine the function
and the existence of its derivative at the
as regards its continuity
origin.
(D.U. Hons. 1951)
6.
Discuss the continuity and
the differentiability of the function f(x)
where
o,
when x
is
W ^ en
irrational or zero
JC==
^
a fraction in
'
its
lowest terms.
(D.U. 1953)
1.
Find from
and/(0)=0
first
at a point
principles the derivative of f(x)
x=0 and show
that the derivative
=~
is
-
when x-^0
continuous at x=0.
(B.U. 1953)
8.
Show
that the function
9
/(x)=x{l-fisin(logx )} for x^0,/(0)=0
but has no differential co-efficient at x=0.
continuous
everywhere
;
is
(B.U. 1952)
9.
If /(*)=*
tan- 1
ous but not derivable for
when x^O and /(0)=0/show
x0.
www.EngineeringEBooksPdf.com
that /(x)
is
continu-
112
DIFFERENTIAL CALCULUS
10.
Is the function
f(x)**(x-a) sin
continuous and differentiate at
x=a
?
_-- for
Give your answer with reasons.
(P. U.)
11.
f(x)
is
Discuss the continuity of f(x) in the neighbourhooi of the origin
defined as follows
(/)
f(x)=*x log sin x for x?*Q, and /(0)=0.
(ii)f(x)=e
12.
5x
A
when
:
function f(x)
4whenO<x^l,
11 *
is
to 4x a
when x^O and/(0)=0.
defined as being equal
3x when
l<x<^2
(D.U. 1955)
2
Jo
discuss the continuity of f(x) and the existence" of f'(x) for
x
when ^0, to
3x+4 when x=2
,
and to
;
x=0,
1
and
2.
(D.U. Hons. 1957)
www.EngineeringEBooksPdf.com
CHAPTER V
SUCCESSIVE DIFFERENTIATION
5*1.
Notation. The derivative f'(x) of a derivable function
f(x) is itself a function of x.
suppose that it also possesses a
derivative, which we denote by/"(;c) and call the second derivative of
The third derivative off(x) which is the derivative of f"(x) is
f(x).
denoted by/"'(jc) and so on.
We
Thus, the successive derivatives of f(x) are represented by the
symbols,
/'(*),/"(*), ........ ,/"(*), ......
where each term
is
the derivative of the preceding one.
dny\dx*
Alternatively, if y =/(*), then
derivative of y. Sometimes
also
denotes the
rtth
are used to denote the successive derivatives of y.
The symbols
denote the value of the nth derivative of y=f(x) for x=fl.
Examples
1,
Ifx=a
0+0
(cos
sin 0),
)>=0
(sin
00 cos 0),find
d*yldx*,
We have
-.
c/v
-V--
M0
rfy
~a (sin
0-j-sin 0-J-0 cos
=a (cos
rfy
~dx=~d~e
rf*v
Q)=a
0+0 sin 0)=0
cos
cos
0sin
0,
0.
_dy \dx ~ tan ^
'dt~dijd7
f
v-Y=sec
^~SeC 3 va
t/(9
^0
-v-
[Note this step]
sec s
1
-
a
--
"~~
"^~ rsz
cos d
fl
ad
113
www.EngineeringEBooksPdf.com
114
DIFFERENTIAL CALCULUS
2.
Ify=sin
(sin x),
prove that
Tx
COS* JC==0
+y
(/>
'
^
*
1953)
We have
dy
^
Making
.
,
=cos (sm
v
x) cos x,
=
sin (sin x) cos
x
=
sin (sin x) cos 2
x
substitution,
we
cos
dy
Change
cos (sin x) sin
cos (sin x) sin x.
see that
^ x j- +}>
-j-^+tan
3.
x
cos*
#=0.
the independent variable to Q in the equation
d 2y
2x
_
y
dy
by means of the transformation
x=tan
We
6.
have
_dy
dt ~de
dd__ dyjdx _dy
dx ~~~de/de '"del
I
dy^
= -2 cos
rt
dd
Oane..
.
.
=-
2 cos 6 sin
=
2 sin 6 cos 3
cos 2
dy
'
,
+">*
dy
^T+ cos2
2>in
cos8
^--|-cos
dy
do
dO
d*y
n
'
'
'
j~
cos 2
dv
0.
^r
Substituting the values of x, dy\dx and
dififerential equation, we see that it becomes
or
U. 1932)
(P.
0^4-2
d ly\dx*
4
sin 0. cos 3
or
www.EngineeringEBooksPdf.com
in the given
^j-|-co8
SUCCESSIVE DIFFERENTIATION
Ifax*+2hxy+by*+2gx+2fy+c=Q, show
4.
that
d*y
dx z
Solving the given equation as a quadratic in y, we get
x [2(h*-ab)x+2(hf-bg)]
Differentiating again,
for
Substituting
b
we get
dy
+h
,-
from
(1),
we
get the "required
result.
Exercises
;>~log (sin x) show that y9
r.
9
Show
^3.
that
y=x f tan x
-
/^/AO
--
-
V
.
.
satisfies
the differential equation
/;
If >~[(fl+te)/(c + <fc)] t then
If
5.
x2
cos /-cos It and
^=2
sin
/-sin
2/, find
the value otd*y!dx*
when =i*.
y\
If
(f.
'
#=a
sin 2^ (l-fcos 20), >>=a cos 29 (I
If ^-(I/A:)*,
show
that
\If p
29),
^ (1) -0.
%
'4s
,
-cos
1
^^ 1 cos 2 9+6 a sin'fl,
prove that
www.EngineeringEBooksPdf.com
prove that
DIFFERENTIAL CALCULUS
116
9.
.J*
1 j.
lfy**x log Kaxy-i+fl- 1 ], prove that
If
*=sin
/,
y=sin />/, prove that
Change the independent variable to
^-4 cot x
~J~
z in the equation
+ 4>> cosec*x=-0
by m^ans of the transformation
z=log tan
12.
of x and
If y is a function
jc<=l/z,
i*.
show
that
dz*'
.
and
Show
that
dx
__
~dy
~dy$x
1
find the value ofd*xfdy* in
.
'
""
tf^^^
dy*
terms of
4vA/Jc, d*yldx*> d*y[dx*.
Also show that
dx*
Calculation of the nth derivative.
5*2.
Let
5-21.
dy*
Some
standard results.
y=(
so that in general
jV=w(m
In case,
m
is
a positive integer, yn can be written as
7--
(m
w)
m
60 that, the /wth derivative of (^x+fr) m is a constant viz.,
\ a
and the (m-fl)th derivative along with the other higher successive;
derivatives are all zero.
m
www.EngineeringEBooksPdf.com
SUCCESSIVE DIFFERENTIATION
Cor.
1.
m=
Putting,
1,
we
,117
get
JV(-lX-2) ...... (-n)a"(
_
~
dx"
Cor.
Let y=\og(ax+b).
2.
a
yi== a
dx*
5-22.
(ax+b)*
Let
y l --=ma mx
log a,
so that, in general
Cor.
Putting *e for
a,
we
~~
dx
5-23.
I
get
7*
m
e
'
Let
jsin (ax+b).
y =a cos (ax-\-b) = a
l
J 3 =a
8
cos
sin (tfx-f 6-f |TT),
(ax+b+i7i)=a*
sin
(ax+6+|7r),
o that, in general
d*
5-24.
sin(ax+b)
_
n
.
r
mr"|
Similarly
r
d*cos (ax+b)
<=a wn cos
j\
dx*
L
.
ax+b+
,
.
5-25.
Let
y=ea * sin
(6^:+c).
ax
y^=ae sin
*=eax
(6jc+c)+e
a*
6 cos
\a sin
www.EngineeringEBooksPdf.com
,
DTTI
^~
2 J'
118
DIFFERENTIAL CALCULUS
In order to put y in a form which will enable us to make the required generalisation, we detarmine two constant numbers r and
f such that
a=r cos
b=r
V,
sin ?
r^i/
Hence, we have
ax
Thus y l
Thus
from
arises
increasing the angle
by
>>
y l =re sin
on multiplying by the constant
the constant
r
and
<f>.
similarly
y 2 ==r*e
ax sin
Hence, in general
d w 1[e * sin (bx-f c)]l
a*
-----^V___r_y ==r e sm
-
,.
fl
where
5-26.
Similarly
ffi!.fll)].
(
m.
+ W)4^
cos
tan-^
(bx+c-fn
1.).
Determination of nth derivative of Algebraic rational funcIn order to determine the nth derivative of
Partial Fractions.
rational
function, we have to decompose it into partial
any algebraic
fractions.
5-3.
tion.
Sometimes it also becomes necessary to
theorem which states that
(cos
where n
is
0/ sin 0)
any
n
=cos
n0i
sin
n()
integer, positive or negative,
apply
Demoivre'a
t
andi=\/(
l).
Examples
1.
Find the nth derivative of
Throwing
it
dr
into partial fractions,
_ __
(x+2)(2x+3)~ 2 L
xa
"I
we obtain
"
1
x+2 "2x+3j.
''
2(2*
8
www.EngineeringEBooksPdf.com
9
SUCCESSIVE DIFFERENTIATION
Find the differential
2.
119
of
co-efficient
x
We
have
x2 + a2
(jc
-~~
2
form,
+ <r/j(x
a.
-+
f
L*
a
J
-
-
^x + a/
i
To render the result free from T and express the same
we determine two numbers r and such that
x=r
-r-rr?
n+1
and
^--
=
cos
9,
a=r
sin 9.
'
nl
"i i v 1
j=
/
.(cos
v
0-\-i
(cos
= ^-ii
^
sin 0)y
sin
t cos
Hence
x
~"
n
(-l) n!
.
cos(K-fl)fl
rfx"
where
r=\/(x
+
1.
Find the
2
a
-f-0
),
0=tan- 1
(ajx).
Exercises
/ith differential co-efficients
of
^.^
2.
in real
Find the tenth and the wth differential co-efficients of
www.EngineeringEBooksPdf.com
x+1
DIFFERENTIAL CALCULUS
120
Find the nth differential co-efficients of
3.
(//)
X
Show
4.
that the wth differential co-efficient of
1
is
K-l)w ("
!)
sin
<l
+ 0[sin (n 4- 1)9 -cos
l
(w
+ (sia
Prove that the value of the nth differential co-efficient of
if /i is even* and
(
!) if n is odd and greater that 1.
5.
for
0-fcos 0)-*- 1 ]. where fl^
x=0 is zero,
Show
6.
that the wth derivative of .y^tan- 1
(-l)n-i (w -i)
x
is
1
w (Jn-^) sin' (iw+^).
s in
f
(P-U.)
(P.U. 1935)
Find the wth differential co-efficients of
7.
_..
N
1
.
If ^=tan-i x,
8.
ton -,
show
If
y=jc(x+
provided that
log
cos a
!
(jc-f I)
cos
3
,
[n,y
+(-!)
]
cos"^.
prove that
/*
If >;=A:
10.
1)
Jf
that
=(n-l)
9.
....
1
A:
log
-~, prove
that
X~j~ I
The nth derivative of the product of the powers of sines ai*4
In order to find out the nib derivative of such a product,
5-4.
cosines.
we have to express it as the sum of the sines and cosines of multiples
of the independent variable. Example 1 below, which has been
solved, will illustrate the process.
Ex.
(i)
i)
.
Find the nth differential co-efficients of
() #*
cos* x.
cos 2
x
sin x,
We know that
1
+cos 2x
www.EngineeringEBooksPdf.com
(/>./. 1951)
SUCCESSIVE DIFFERENTIATION
cos *
121
2x
2x +1(1+008
1+4
-.
)
cos
cos
4x
*+- +4
cos
x=4(l + cos
cos 2* sin
2x+J
cos
2x) sin x
= J sin x+J.2 sin X.cos 2x
x+J (sin 3x
= J sin x+J sin 3x.
==| sin
sin
x
Hence
sn x
=
sin
-(
x)+J
~
sin
sn
e
(e
sn
sin
2.
Find the nih differential co-efficients of
sin 8Ar.
(/)
(//)
2
sin
(v) e
2ay
(i/)
x cos
cos
A:
3
x.
cos x cos 2x cos 3x.
(/v)
e* sin 4 x.
sin 2 2x.
55. Leibnitz's Theorem. The nth derivative of the "product of
two functions. If w, v be two functions possessing derivatives of the
nth order, then
This theorem
Step
I.
By
will
be proved by Mathematical induction.
direct differentiation,
and
we have
(wv),j=w a
t/
Thus the theorem
is
2
v
true for
+
2
C 1t/1 v 1
n=l,
2,
Step II. We assume that the theorem
value of n, say m, so that we have
is
true
for
a particular
mC w
a w-a vJ + ......
t
+
www.EngineeringEBooksPdf.com
......
+ m C muv m
.
122
DIFFERENTIAL CALCULUS
Differentiating both sides,
+i
v+u m Vj+^C,
um
B,
we
get
^+"0, M W _,
va
V^rC^C.,)
V2
!/_!
U^-OK^ V,+
...+
OT
+
CW
,
But we know that
from which we see that if the theorem is true for any value
then it is also true for the next higher value m + l ofn.
Conclusion.
for==2.
In step
Therefore
=3 + l,
i.e., 4,
it
and
I,
m
we have seen that the theorem
for /f=ii + l, i.e., 3 and
must be true
of
is
so
tt,
true
for
so for every value of n.
Examples
1.
Find the nth derivative of x* ex cos
2
To
find the ttth derivative of ;c e
as the first factor and x 2 as the second.
.-.
2
(x e* cos x) n ^(e
x
x
x.
(P-U-)
cos x,
we
x
look upon e cos x
cos x) n x*+ n C l (e* cos x) n .,.2x
-t-"C 2 (e
=
n x
2k e cos
.
(x+n tann
l).x
~ 1)
" 1
e
x
n
)
cos
.
2.
e
x
cos
x+n^l -J-) +w(w-l)
(x+^"2 tan-U).2
cos
//
.V=a cos
show
1
(x+n-lj.tan- l)2x
~2
.2*
+2?.nx cos f
cos x) w _ 2 .2
a
J
(/og
x)+& ^m
(fog x),
that
(D.C7. 1952)
www.EngineeringEBooksPdf.com
12$
SUCCESSIVE DIFFERENTIATION
we
Differentiating,
get
or
xj^ 1
x)
=
x)+b
+
x
a sin (log
Differentiating again,
xy*+yi=*
or
b ??_C!.o 8^t
-
a sin (log
-----
__
Vl
we
cos (log x}.
get
----
^(xya+J'iH
x
a cos (log
x)
-
x
1> cos (log
x)
- b sin(log
*-
x)+6
sin (log *)];=
y
2
* j>3+*yi+:y=o.
or
Differentiating n times
by Leibnitz's theorem, we obtain
* aJ^2+"Ci-2x^,fi+^
or
x'y n
Exercises
Find the nth derivative of
1.
(i)
jc
n
e*.
x
(i/i)
2nax+n(n + l)].
4* [a*x*
(//)
x 3 cos
x.
(iv)
e" log
x
(P.U.1954,56)>
2.
If y=x 2 sin x, prove that
sn
3.
If/(x)=tan
x+-x
cos
x, prove that
[P.J7. 5w/?/>. 7936).
[Write /(x) cos x
= sin x and
apply Leibnitz theorem.]
Differentiate the differential equation
4.
-
dx*
d
times with respect to x.
Find the nth derivative of the
5.
(D.U. 1950)1
differential equation
Determination of the value of the nth derivative of a funcit is possible to obtain the value of the nth
derivative of a function for x =0 directly without finding the general
expression for the nth derivative which cannot, in general, be
obtained in a convenient form. Examples 1 and 2 below which have
5*6.
tion for
x=o. Sometimes
www.EngineeringEBooksPdf.com
124
DIFFBRBNTIAL QALCIJTJS
been solved will make the procedure cl *ar. As will be seen in Ch. IX,
the values of the derivatives for x=0 are required, to expand a
function by Maclaurin's theorem.
Examples
1.
Find the value of the nth derivative of e m 8in
or
"w=*r
Differentiating,
get
^
(1-x 2 2y,y 8 2xy1 -2ifi f
Dividing by 2y t) we obtaain
(l-jffo-xyi^m'y.
)
Differentiating n times
Putting
From
x=0, we
(1), (2)
...
(1)
-
(2)
-
(3)
ox
,
x*)y*=nfy*.
(1
we
x0,
m
msiiTl x
*-
for
"
sin Ax
y=e m
Let
"1 *
by
Leibnitz's
1
.
theorem, we get
get
and
WO)HHm%
(3),
B (0).
we obtain
XOHl^OHm^COHin".
Putting n=l,
2, 3, 4, etc. in (4),
we
get
In general
^
'~
fm2 (2 a +/n 2 )(4 2 +/M 2 )...[(n-2) 2 +m 2], when
(w(lHJ
2
2
)
(3
+m )...[(n-2) +m
3
2
2
],
n
is
when n
is
even,
odd.
(P.U. 1955)
2.
1
If >>= (sin- x)
(1
x
2
,
prove f/wf
x 2 ')^-x^-2:=0
ax*
dx
...
(1)
Differentiate the above equation n times with respect to x,
Also find the value of all the derivatives of y for
x=0.
(P.U, 1955)
Differentiating
y
(sin-
1
2
x)
,
we
get
2 sin" x
1
*=
v (!-*).
www.EngineeringEBooksPdf.com
,
.
9
,^>
SUCCESSIVE DIFFERENTIATION
.
(1
.
- x*) y* = (2
125
= 4y.
sin- 1 x)*
we get
Differentiating again,
2
2(l-x ^2-2x^=4^.
Dividing by 2y l9 we get
2
(l-* )>>2-*yi-2=0
...(3)
)
which
...(4)
is (1).
Differentiating this n times
by Leibnitz's theorem, we obtain
(1-^+2+^+1
n(
(-2x)+
~^y(-2)-xy ni
.
l
-ny n
.
1=0.
l-*
or
Putting
x=0, we
obtain
JWOHa^CO).
From
From
(2),
>>i(0)^0.
(4),
y,(0)
Putting H=l,
In general,
if
= 2.
(7)
3, 5, 7 successively in
w=2,
Again, putting
n
is
>> n (0)
4,
even,
= 2.2
2
,.,(6)
6 ...... in
(6),
we
(6),
we
see that
see that
we obtain
'4 ? .6 2
..
(~-2)
2
,
when w
Note. The result (5), obtained on dividing (4) by 2y t is not a legitimateconclusion when ^t=0, which is the case when x=0. Thus, it is not valid to
derive any conclusion from (5) and (6) for x=0.
,
But these results may be obtained by proceeding to the limit as x->0 instead of putting x=0. This may be shown as follows
:
We can
easily convince ourselves that the derivative of every order of y
x 2 ) in its denominator
as calculated from (2) will contain some power of (1
1 , so
that lim ^t
and will therefore be always continuous except for
and lim ^==^(0), as x-0.
x=
Exercises
1.
If
u~ tan~
x
x, prove that
and hence determine the values of the derivatives of u when
2.
x=0
(M.T,
)
If
1
j>=sin (m sin" x),
show
that
(1 -x')*, + ,=(2n f IkVn-n+V-'"')*,
and
find
yn
3.
(P.U.
(0).
Find y n
(0)
when
j>=*log
[x+
4(\
{ x')].
www.EngineeringEBooksPdf.com
/P5>
126
DIFFERENTIAL CALCULUS
4.
It y=lx+4(I +**)]", find
5.
If
yn (Q).
ys~JM
*how
COS"" 1*
that
(1
*nd find yn (0).
6.
-x^+i-Vn + ^xyn^-W+m^y^O
(/)./. Hons. 1953)
If ^(sinh- 1 *)
Hence
1
x=0
find at
,
prove that
the value of
dnyfdx n
.
(B.U.1952)
Exercises
1.
Show
that if
l
- 36^-0
then
2.
n" 1
If >>n 'denotes
(bja),
the
th
differential co-efficient of e ax sin
bx and
prove that
.
y n =(a sec
n aa! sin
^) ^
Also show that
y=*(ABx)
cos
;c+(C4-/>;c) sin AJC, prove that
3.
If
4.
Find the third differential
:5.
Prove 'that
6.
Show
that
if
co-efficient of
=sin wx+cos nx then
y
w r =^[l-f(~lKsin2x] i
when
Wr denotes tht rth differential co-elficient of u with respect to x.
(Lucknow)
7.
Show
that
__
23
......
__
n
(P.U.)
j '
8.
Prove that
where P and
and
Q stand
for
nx^-Mi-lXfl
f|
2)jt
-t
4-
respectively.
www.EngineeringEBooksPdf.com
EXERCISES
9.
127
Find the value of nth derivative of
jx*^x
2
a
-4)
U
for
x0.
(
10.
Prove that
dn
if
;t=cot
fi
^=(-l)
where n
is
11.
Trinity College
(0<0 <n), then
0,
n- 1
(-l)
sin nO sin*0
!
;
any positive integer.
If
Prove that
IifIn-i+(-l)
!
and hence show that
(D.U. Hons. 1949)
Rewriting this relation as
Jn_
/j
and replacing
by
If
y=
12.
1,
,
d n
U
2
In-i
~(/i-i)
!
____
-l)
n,
,
,
r
3, 2,
show
JL
we
n
get the required result.
that
(x*-l)y n +t+2xy n + l -n(n+ l)^ n
Hence show
13.
If
that
.y
U n denotes
satisfies
0.
the Legendre's equation
the nth derivative of (Lx-t-M)l(x*2Bx+C)., prove
that
Wn+t +
14.
? (-*r*Ji
If y=x*e x , then
z,
15.
P./., D.(/., 7955)
If
prove that
(D.I/. //ow5.,
www.EngineeringEBooksPdf.com
and Pass, 1949 ;
P.U. 1958 Sept.)
DIFFERENTIAL CALCULUS
128
If
16.
(tan"
;'
1
*)*,
then
Deduce
that
(Birmingham}
If
17.
x=^ tan (log y),
prove that
(1
+*)+
(2n X
-l)% +(-!)
m +y __m _
=2x, prove
!>=(>.
(B.U. }
__
18.
y
I
that
(D.U. Hons it 1947
where > w denotes the wth derivative of y.
x
If
19.
cos x,
y=*e*
show
:
P.U. 7959)
that
>W (0) -4n>>2n(0) + (2- l)2n ^ n - (0) =0.
2
y =(l+ x rf
If
20.
2
m
gin (m tan" 1^),
^ n(0)=Oand^ 2n+1(0)-(-l)
show
that
w(m-l)'w~2).
If
22.
If x4-y==l,
.(m-2n).
prove that
(D.U.Hons.l950\M.U.)
/
/
.
y=sin (m cos" Vx) then
21.
23.
.
1
By forming
in
two
different
8
ways the nth derivative of* ", prove
that
1
+ "!-+
is
723
+
i*
.
2 2 73 2
^"f
an
with the
[Equate the nth derivative of x
of x n and x n and put x=l].
24.
"Of!?
th derivative of the product
Prove that
sin
www.EngineeringEBooksPdf.com
129
EXERCISES
25.
If
jf=s~.__ prove that
9=tan ^
Where
26.
show
Ifw=
that
().
www.EngineeringEBooksPdf.com
(7,
CHAPTER VI
GENERAL THEOREMS
MEAN VALUE THEOREMS
Introduction. By now, the student must have leanrt to distinguish
between theorems applicable to a class of functions and those concerning some
particular functions like sin x, log x etc. The theorems applicable to a class of
known
functions are
as general theorems.
Some
general theorems which will play a very important part in the
following chapters will be obtained in this chapter. Of these Rollers theorem is
the most fundamental.
y
M &1.
Merval
'c'
ofx
Rollers
(a, b],
Theorem.
// a
is
function f(x)
and also f(a) ~f(b), then there
exists at
derivable
least
in
an
one value
lying within [a, b] such thatf'(c)=Q.
The function f(x) being derivable
in the interval [a, b] is con4*14, p. 76). By virtue of continuity, it has a greatest
value
and a least value in in the interval, (3-53, p. 55) so that
there are two numbers c and d such that
tinuous,
M
(
Now
M~w,
either
... (i)
M^m.
or
..
.(//)
When
case
(/),
the greatest value coincides with the least value as in
the function reduces to a constant so that the derivative
for every value of x and therefore the theorem is
equal to
f'(x) is
true in this case.
When
M and m are unequal, as in case
(//), at least one of them
from the equal values /(a), f(b). Let
/(c) be
different from them. The number 'c' bsing different from a and 6,
must be
M
different
lies within
the interval
[a, b].
The function /(x) which is derivable
x~c, so that
in the interval [a, b]
is,
in
particular, derivable for
Hra
exists
and
is
As/(c)
the sani3
is
/(c+/t)-/(c)
when h ->
wh0nA
_|.
through positive or negative values/
the greatest value of the function,
we have
f(c+h)<f(c)
whatever positive or negative value h has.
Thus
...(I,
130
www.EngineeringEBooksPdf.com
GENERAL THEOREMS
131
and
f(c+h)-f(c)
Let h -*
>0for h<0.
From
..-(2)
through positive values.
(1),
we
get
...(3)
Let h -+
through negative values.
f'(c)
The
relations (3)
and
From
(2), .we
get
>0.
(4) will
...(4)
both be true
if,
and only
if
/'(c)=0.
The same conclusion would be
Beast value
m which
differs
Hence the theorem
is
similarly reached
if
it
is
the
from /(a) and/(6).
proved.
Geometrical Statement of the theorem.
If a curve has a tangent at every point thereof, and the ordinates
extremities
A, B are equal, then there exists at least one point
of
curve
than A, B, the tangent at which is parallel to x-axis.
the
other
of
P
its
In this geometrical form the theorem is quite self-evident, as
the student may himself realise by drawing some curves satisfying
the conditions of the statement.
B
M
"X
Fig. 43.
Fig. 44.
This intuitive consideration
proof of the theorem.
is
also
sometimes regarded to be
"a
V
Note 1. It will be seen that the point
where the derivative has been
to vanish lies strictly within the interval [a, 6], so that it coincides
neither with a nor with b. In view of this fact, the theorem is generally stated
shown
as follows
:
If a function f(x)
\
.
is
such that
It is
continuous in the closed interval
It is
derivable in the open interval (a, b),
[a, b],
*
2.
3.
/()=/<.
then, there exists at least one point
V of the
open interval
(a, b),
www.EngineeringEBooksPdf.com
such thatf'(c)~Q
132
DIFFERENTIAL CALCULUS
Note 2. The conclusion of Rolle's theorem may not hold good for a
function which does not satisfy any of its conditions.
To illustrate this remark, we consider
the function >>=/(*)= x
the interval
|
[-1,
lt
/(
1)
is
continuous in
are both equal to
Its
[
7
x
1,
1]
and
/(I),
1.
derivative/ (x)
1^
for
1,
m
\
li-
< 0, and
is 1
for
0<z<l
and
does not exist for
jc=0, so that/'(x) nowhere vanishes.
(Ex. 2. p. 73>
~%r
The
^
failure is explained by the fact that
x is not derivable in [ 1, 1], in as much as
the derivative does not exist for jc=0, which is
a point of the interval.
|
|
Fig. 45.
Ex.
1.
Verify Rolle's theorem for
x2
(i)
in
[-7,
1].
'
(ii)
[-3,
0].
(0 Let
2
/(;c)=;c so that /(!)=! =/(~l).
Also,
f'(x)
x1
is
derivable in
[
1, 1].
The conditions of the theorem being satisfied, the derivative
must vanish for at least one value of x lying within [1,1].
Also directly we see that the derivative 2x vanishes for
lies within [1,
Hence the .verification.
1].
which value
x=O
Let
(ii)
We have
/(- 3) =0 =
*nd/(x)
is
derivable in the interval
[3,
0]*
We
have
*
x
Putting/ (A:)=0, we get
This equation
satisfied by x==
Of these two values of
2, 3.
Jis
is zero,
2, belongs to the interval [ '3, 0] under
x, for which /'(x)
consideration.
Hence the
Ex.
verification.
Verify Rolle's theorem in the interval [a, b] for the functions
(0 log[(x+*&)/(<i+&)K].
() (x 0)*(x 6)* m, n being positive integer*.
2.
;
www.EngineeringEBooksPdf.com
GENERAL THEOREMS
Ex.
Verify Rolle's theorem for the functions
3.
sin x/e
(/)
Ex.
jc
log
133
x
in [0, *]
(//)
;
e
x
(sin
x-cos
By considering the function (x2)
4.
x-2 x
is satisfied
x) in [w/4, 5w/4].
log x,
show that the equation
1 and 2.
by at least one value of x lying between
6*2. Lagrange's mean value theorem.
If a function f(x) is derivable in the interval [a, b], then there exists at least one value c of x
lying within [a, b] such that
9
f(b)-f(a)
"
b-a
_
-
1
(C) '
To prove the theorem, we define a new function <p(x) involving
and
f(x)
designed so as to satisfy the conditions of Rolle's theorem.
Let
V(x)=f(x)+Ax,
where
A
is
a constant to be determined such that
Thus
-~
Now, /(x)
constant.
is
derivable in
<?(x), is
Therefore,
Thus,
<p(x)t satisfies all
b-a
[a, b]. Also x is derivable and, A, is a
derivable in [a, b] and its derivative is
the conditions of Rolle's theorem. There
'c' of x, lying within [a, b] such that
therefore, at least one value
is,
f'(c)=0.
Q = <e>(c)=f'(c)+A,
JWz>
=f (c)
i.e.,
A = -f(c),
,
...
.
(0
Another form of the statement of Lagrange's mean value theorem.
is derivable in an interval [a, a +h], there exists at
and 1 such that
least one number '#' lying between
If a function f(x)
f(a
We
Ja, b]
-a
write 6
which
+ h)=f(a)+hf(a+0h).
a = h so that h denotes the length
may now be
written as
a+h],
between a and
The number, c, which lies
a-\-h
by some fraction of A, so that we may write %
where
becomes
or
is
of the interval
[a,
some number between
and
1.
is
greater than
Thus the equation
f(a-fh)=f(a)+hf'(a+0h).
www.EngineeringEBooksPdf.com
(i)
134
DIFFERENTIAL CALCULUS
Geometrical statement of the theorem. If a curve has a tangent
at each of its points, then there exists at least one point P on the
curve such that the tangent at P is parallel to the chord AB joining its
extremities.
01
Fig. 46.
Fig. 47.
We
will now see the equivalence of the analytical
cal statements.
and geometri-
Iff(x) is derivable in [a, b], then the curve y=f(x) has ar
tangent at each point of the curve lying between the extremities
The
slope
r
of the chord
.
Let
Pbe
AB= f(b)-f(a)
L
b a
V
-
-
the point (c,f(c)) on the curve
;
c,
being such that
...(01
^>-M=/'(c).
The
slope of the tangent at P~f'(c).
From
chord
(i7),
we
see that the
AB are equal.
Thus there
is parallel to
exists a point
the chord AB.
slopes of the tangent at
P on
P and
thex
the curve the tangent at which
Note 1. Now [f(b)-f(a)] is the change in the function /(x) as x changes
from a to b so that [f(b)f(a)]l(ab) is the average rate of change of the
function f(x) over the interval [a, b]. Also/'(c) is the actual rate of change of
the function for x=c- Thus the theorem states that the average rate of change of
a function over an interval is also the actual rate of change of the function at
some point of the interval. In particular, for instance, the average velocity
over any interval of time is equal to the actual velocity at some instant belonging to the interval ; velocity being rate of change of distance w.r. to time. This
interpretation of the theorem justifies the name 'Mean Value' for the theorem.
we
Note 2. If we draw some curves satisfying the conditions of the
will realise that the theorem, as stated in the geometrical form,
www.EngineeringEBooksPdf.com
theorem*
almost
is
GENERAL THEOREMS
Ex.l:
135
//
f(x)=(x-l)(x-2)(x-3)
find the value
of
a=0, 6=4,
;
c.
(
D.U. Hons. 1954)
We have
/(6)=/(4)=3.2.1=6,
f(a)=f(Q)
f(b)-f(a)
=
-7,~a
12
4
= -6,
~6
'
Also
Consider
now
the equation
i.e.,
3=3c 2 -12c+ll.
or
3c 2 -12c+8=0
Taking \/3 = 1*732..., we
may
see that both these values of
c'
belong to the interval [0, 4],
Ex.
(i)
(ii7)
Ex.
2.
Verify the
log x in
Find
'c'
value theorem for
.
[1, e].
lx*+mx+nin
3.
mean
(//)
x3
in [a, 6].
[a, b].
of the
mean
value theorem,
/\x)=x(x- l)(x-2)
;
if
a=0, &=*.
(D.U. Hons. 1951)
Ex.
cases
4.
Find
'c'
so
that
/
/
the following
(c)=[/(/))~/(fl)]/(6~a) in
:
6=3.
fl=2,
(x-4)
/(*)=*; a-0, 6=1.
;
Ex.
tions log
5.
x and
Applying Lagrange's mean value theorem, in turn to the funcdeteimine the corresponding values of in terms of a and h.
ex ,
Deduce Jhat
(0 0<[log
Ex.
6.
(1
+*)]-*-*-'<!
;
(//)
< 1 log
~e
"
Explain the failure of the theorem in the interval
www.EngineeringEBooksPdf.com
]
~
- <1.
[1,
1}
when
DIFFERENTIAL CALCULUS
136
important deductions from the mean Value theorem.
consider a function /(x)
the sign of Derivative.
derivable in an interval [a, &]. Let xv x2 be any two points belonging to the interval such that x^>xv Applying the mean value
theorem to the interval [xly xj, we see that there exists a number
Some
6-3.
We
Meaning of
between xl and X2 such that
'
f^-ftoHfv-Xi)*'^
Letf(x)0
6-31.
From
(/)
...
throughout the interval [a, b].
we get
where xl9 x 2 are any two values of x. Thus we see that every two
We
values of the function are equal. Hence f(x) is a constant.
thus prove :
"If the derivative of a function vanishes for
interval, then the function must be a constant.
This
all values
of x
in
an
the converse of the theorem, "Derivative of a constant
is
is
zero/'
Cor. If two functions f(x) and F(x) have the same derivative for
every value ofx in [a, b] then they differ only by a constant.
We
write
Hence, ^(x),
f(x)F(x)
i.e.,
6-32.
Letf(x)>0for
From (i), we get
for,
*!*!
is
a constant.
every value
ofx
in [a, b].
and/' () are both positive.
f(x) is an increasing function of
Hence
proved
x.
We
have thus
:
"A function
interval is
whose derivative is positive for every value ofx in an
a monotonkally increasing function ofx in that interval."
6-33.
From
JOT,
X2
Xi
Letf'(x)
(/),
is
Hence
proved
we
every value
ofx
in [a, b].
get
positive
f(x)
<Qfor
is
and/'()
negative.
a decreasing function of x.
We
have thus
:
whose derivative is negative for every value of x in an
a monotonically decreasing function of x in that interval."
Note. The above conclusions remain valid even if/'(x) vanishes at the
*n4 points a, * of the interval, for the V of the mean value theorem never
"A function
interval is
www.EngineeringEBooksPdf.com
GENERAL THEOREMS
137
6'4.
Cauchy's mean value theorem. If two functions f(x) and
F(x) are derivable in an interval [a, b] and F'(x)-^Qfor any value ofx
in [a, b], then there exists at least one value 'c'
ofx lying within [a, b],
such that
f(b)-f(a)
f'(c)
F(a)^
F'(c)
we note that
Firstly,
[F(b)
F(a)]^0 for if it were 0, then
would
the
of
conditions
the Rolle's theorem and its deriF(x)
satisfy
vative would therefore vanish for at least one value of x and the
would be contradicted.
hypothesis that F (x) is never
:
r
Now, we
define a new function <p(x) involving f(x) and
to satisfy the conditions of Rolle's theorem.
F(x)
and designed so as
Let
where
A
a constant to be determined such that
is
f (*
=
Thus
f(a)
Now,
Therefore,
f(x)
+ AF(a)=f(b)AF[b).
and F(x) are derivable in [a, b]. Also A is
derivable in [a, b] and its derivative is
a constant.
<f(x) is
f(x)+AF'(x).
the conditions of Rolle's theorem. There
therefore, at least one value, c, of :c lying within [a, b] such that
Thus,
-f(x) satisfies
is,
f(c)=0.
or
f'(c)=-AF'(c),
Dividing by F'(c ) which^f-O, we get
_
F'(c)
~F(b)-F(a)
Hence the theorem.
Another form of the statement of Cauchy's mean value theorem.
If two functions f(x) and F(x) are derivable. in an interval [a, a+h] and
and 1,
F'(x)^0, then there exists at least one number 6 between
such that
The equivalence of the two statements can be easily seen as in
the case of Lagrange's mean value theorem.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
138
Note. Taking F(*)=jc, we
only a particular case of Cauchy's.
may
theorem
easily see that Lagrange's
is
Ex. 1. Verify the theorem for the functions x* and x* in the interval
b being positive.
Ex. 2. If, in the Cauchy's mean value theorem, we write for/Yx), F(x)
x
is the arithe- x show that in each case
(i) **, x
(ii) sin *, cos x
(///) e
metic mean between a and b.
[a, b]
;
a,
;
;
Ex.
3.
If,
,
in the Cauchy's
V* and 1/Vx respectively then,
and
we
write l/x 2 and
Ijjc
then,
6*5.
value theorem, we write for f(x), and
the geometric mean between a and b>
the harmonic mean between a and
mean
F(x),
if
;
V
;
c, is
c, is
.
Examples
1.
w
monotonically increasing
in
every interval.
Let
Thus f'(x) >0 for every value of x except 1 where
Hence f(x) is monotonically increasing in every interval.
2.
is
Separate the intervals
it
vanishes.
which the polynomial
in
increasing or decreasing.
Also draw a graph of the function.
Let
so that
for
x<2
;
/'(*)<0for2<x<3
x>3
=
x=2
for
/'(*)
/'(*)>0for
()
is
;
;
and
3.
oo , 2)
positive in the interval (
negative in the interval (2, 3).
and
(3,
oo
)
and
monotonically increasing in
2] [3, oo) and monotoni(
cally decreasing in the interval [2, 3].
To draw the graph of the function, we
note the following additional points
Hence /(x)
the intervals
is
oo
,
:
(iii)
(iv)
/(0)=2,
/(#)->
<
as
(v)
Fig. 48.
www.EngineeringEBooksPdf.com
oo
GENERAL THEOREMS
Show
3.
that
*/(!+*)
We
<
<
log
(1+x)
>
and
xfor x
>
0.
write
1
Thus
/'(x)
Hence f(x)
>
x
for
for
x=0.
monotonically increasing in the interval
is
= 0,
/(*) > /(O) =0
Also
=0
/(0)
Hence /(x)
for
x
>
0.
that
positive for every positive value of x, so
0.
x/(l+x) for x
log (1+*)
is
>
>
Again, we write
BO that
/
F'(jc)==l
Thus,
oo],
>
F(x)
Therefore F(x)
[0,
[0, QO ].
is
for
x
>
2
and
monotonically
for
is
increasing
#=0.
in
the
interval
AlsoF(0)=0.
>
F(x)
Hence
F(x)
is
>
F(0)=0forz
0.
positive for positive values of x, so that
x
>
log (1+Jt) for
x
>
to
x=n/2.
0.
Exercises
1.
(i)
(ii)
(//i)
Show
that
x/sin x increases steadily from
x=0
x/tan x decreases monotonically from x=0 to jc=n/2.
the equation tan x x=6 has one and only one root in
(P*U.%
(
Jw, Jn).
(.{/. 7952)
O'v)
2.
*>-2 ^
tan-
<
tan-"
1
x
<
that x~sin x is an increasing function throughout any
Determine for what values of a axsin x is a steadily
Show
of values of x.
y
ing function.
3.
interval
increas-
(M.J7.)
Determine the
(x
intervals in which the function
4
+ 6jcH 17x a + 32x+ 32)*-*
is
increasing or decreasing.
is
Separate the intervals in which the function
(*+*+!) /(*-- * + l)
increasing or decreasing.
4.
5.
Determine the intervals in which the function (4-x 8 ) 8
Also draw its graph.
or decreasing.
www.EngineeringEBooksPdf.com
is
increasing:
140
DIFFEBENTIAL CALCULUS
6.
Find the greatest and least values of the function
x 8 -9x 2 -f24xin[0,
x-
7.
Show
that
$.
Show
that if
x-
(//)
Show
x
>
6].
decreases as x increases from
(i -f x)
+
<
log
< -log(l-x) <
that e~ x lies
Prove
Show
If
.
*.
d+x) < x-
x
(1
-x)-
*
l\
-
(B.U. 1953)
1
that sin
x
<
x
<
1,
lies
<
x
<
1.
.
between
,
x5
show that
taking * =2/1
+
Hence
for
between
x9
12.
o
0,
1-xand 1-x + Jx 2
11.
to
that
x
10.
log
,-*-< lo, (!+)< -
(/)
9.
1
W
1'
>
rdcduce that
5.
13.
Show
that
x
tan^c
sm
x
14.
Show
ifo<x<
..
jc
that
x-1 > logx>
jc~l
x
and
for
15.
7957)
If /(x) is
where
(jc-l)x-
1
,
> 2xlogx > 4 (x-l)-2 log*
>
(M.T.)
1.
derivable in the interval [a-h, a+h], prove that
<^<
1
;
(a+h)-2f(a)+f(a-h
where < t < 1.
fl
The derivative of a function /(x) is positive for every value of x in an
16
show that /(c)
interval [c-A, c], and negative for every value of x in [c, c+h] ;
b the greatest value of the function in the interval [c-h, c+h].
*
of
Higher mean value theorem or Taylor's development
derivatives
the
function in a finite form.
// a function f(x) possesses
*
to a certain or^er n f r every value
f'(x)>f"( x )'f'"( x )> ..... /"(*)> UP
x in the interval, [a, a+h], then there exists at least one number 0,
between and 7, such that
6-6.
www.EngineeringEBooksPdf.com
TAYLOB'S THEOREMS
f(a +
141
h)=f(a)+hf'(a)+*f f" (a)+
We define a new function 9(x)
x
f'( )>f"(x), ...... fn (x) designed so
,
Rollers theorem.
involving /(x) and its drivatives
as to satisfy the conditions for
Let
......
^
,i
where A
is
*
a constant to be determined such that
Thus we get
A
from the equation
-
Now,
it is
in the interval
given that/(x), f'(x),f"(x) .......
[a,
,
A
fn ~\x) are derivable
a+h].
2
Also, a+h-x, (a+A-x) /2 !, ........ , (a+h-x)"ln I are derivable in [a, a+h]. Therefore <t(x) is derivable in [a, a-f^]- Also
f'(x)=f(x)-f'(x)+(a+h-x)f*(x)-(a+h-x)f'(x)
ether terms cancelling in pairs.
Thus
<p(x)
satisfies all
exists, therefore, at least
the conditions for Rolle's theorem. There
and 1 such that
one number 6 between
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
142
f"(a+6h)=A
;
for
Substituting this value of
(1-
A
in
(i),
we
get
f(a+h)-f(a)+hf\a)+~f''(a)+ ...... +
^
The (+l)th term
Lagrange's form of remainder after n terms in the Taylor's
expansion of/(0-f7z) in ascending integral powers of //.
Note. Taking /?=! we see that Lagrange's mean value theorem is only
a particular case of the Taylor's development obtained here.
Cor. Maclaurin's development. Instead of considering the interval [a, a+h], we now consider the interval [0, x] so that we change
a to and h to x in (//). We get
is called
'-
'
'
-.
..(//I)
which holds when the function f(x) possesses the derivatives
f'(x),f"(x), ......... ,/(*)
Mervai
in the
[0, x].
f
,
The formula
[0, x] in
the
finite*
(HI)
is
known
as Maclaurin's development of /(x) in
form with Lagrange's form of remainder.-
67. Taylor's development of a function with Caiichy's form of
If a function f(x) possesses the derivatives /'(*)> f"( x ) ......
fn (x) up to ft certain order n in the interval [a, a +/*], then there exists
between
and 1 such that
at least one number
remainder.
;7 2
We define
/)-i
+_
a ftew function f(x) as follows
f(x)=/(x)-f (a+h-x)f(x)
where
A
is
a constant to be determined such that
www.EngineeringEBooksPdf.com
1)
r
/-i
(a)
TAYLOR'S THEOREMS
Thus, we get
A
143
from the equation
h n ~i
f
It is easy to see that v(x)
Hence
<p(x) satisfies
there exists at least one
derivable in
is
[a,
a+h].
the conditions for Rolle's theorem so that
between and 1 such that
number
Now,
y'(x)~
other terms cancelling in pairs.
o=?'(a+0/o=
hn
Substituting this value of
.
h*
A
ri-i)'i
f n ( a + 6h )- A
'1
in
(/),
.
,
we
get
,
The (+l)th term
is called Cauchy's form of remainder after n terms in the expansion
of f(a+h) in ascending integral powers of h.
Maclaurin's development.
Cor.
(//),
we
Changing a to
and h
to
in
get
is known as Maclauriris development off(x) in the interval
with
Cauchy's form of remainder after n terms. It holds when
x]
n
which
[0.
jc
f(x) possesses
Ex.
1.
^derivatives /'(*), /"(*), ...... f
,
Show
that
www.EngineeringEBooksPdf.com
(x) in [0, x].
DIFFERENTIAL CALCULUS
144
Substituting these values in (Hi)
Ex.
2.
Show
JC
3.
JC
5
we
obtain the result,
x
~
X*** 1
8
"2~!
Ex.
6*7, p. 142),
that, for every value of
3
JC
(
+
Show
X^
4!
that
Ex. 4. Find, by Maclaurin's theorem, the first four terms and the remainder after n terms of the expression of e ax cos bx in terms of the ascending
powers of x.
www.EngineeringEBooksPdf.com
APPENDIX
Examples
Show that the number 6 which occurs in the Taylor's theorem
1.
with Lagrange's form of remainder after n terms approaches the limit
n+1
(x) is continuous and
ll(n-\-J)ash approaches zero provided that f
different from zero at x=a.
(D.U. Hons. 1950)
Applying Taylor's Theorem with remainders after n terms and
terms successively, we obtain
(rt-f-1)
f(a+h)=f(a)+hf'(a) + ...... +
n-l
fn
\
These give
hn
*
-
hn
/(+*)=-
f n ()+
,
,-
n
fn (a+eh)-f (a)= n
or
(n+l)
,/"(+'*),
h
+l f<*\a+e'h).
Applying Lagrange's Mean Value Theorem to the left-hand
we have
f
1
n
Let h ->
+i
'/" +
0.
lim
2.
Show
0=^.
that
x*>(l+x)
We
.-.
put
side,
f(x)
= x* -
/'(x)=2x-l.
(
1
[log
+ x)]*for x>0.
[log 1 + x)].
[log (l
+x)
(l+x)]-(l+x).
(
2
=2x-[log (l+x)] -2 log (l+x).
The form of /'(*) is such that we cannot immediately
2
as to its sign.
vative.
We,
therefore, proceed to determine the second
145
www.EngineeringEBooksPdf.com
decide
deri-
DIFFERENTIAL CALCULUS
146
/"(x)=2-2
which
is
>0
for
'
/'(*)
Also
log (l+x).l/(l
+ x)-2/(l+x)
x>Q.
i"3
(See Ex. 3, p. 139)
monotonically increasing in the interval [0, oo
).
/'(0)=0.
Therefore
f'(x)>Qforx>0.
,
Hence f(x)
is
monotonically increasing in the interval
[0, oo).
Also
/(0)=0
Therefore
f(x)>0
for
x>0.
Hence
x>(l+x)[log (1+*)]
3.
2
for
x>0.
If <p"(x)>Q for every value of x, then
f [*(*!+*)]< tt*(*i)
for 'every pair of values of x1
Suppose that xa >xt
+ *(*.)],
and x z
.
.
We write
Applying Lagrange's mean value theorem to the function
X 2 ] we see that there
for the intervals [x l9 (x l
2 )/2] and [(x 1
2 )/2,
exist numbers 19 % belonging to the two intervals respectively, such
+x
+x
that
and
= [K*i+^) -*i]^ (i)=l(*i-^'(&)
f
[f K*i+*i)-^(*i)]
Thus from
(1), (2), (3),
-(3)
we obtain
Applying the mean value theorem to the function 9' (x) for the
interval [^, | 2 ], we see that there exists a numer y such that
From
(4)
9'(&)-9'(4i)=(!t-Si)9"(*)and (6), we obtain
www.EngineeringEBooksPdf.com
..(5)
APPENDIX
Since Jta xl9 | 2
required result.
147
an(* ? "(?) are
i
ft ll
we obtain the
positive,
Exercises
Show
1.
that
*0M which
approaches the limit i as
'A'
mean
occurs in the Lagrange's
approaches
0,
7
value theorem)
'
provided that/ (a)
(P.U. 1949
is
[It
2.
-where
should also be assumed that f"(x)
Show that
/(a+/0=/(aH
lies
between
and
1
V W+4* /
/
i
//
is
not zero.
Hons. 1949)
).(/.
;
not continuous.]
(fl+ 0/0,
and prove that
lim 0=J,
H-+Q
specifying the necessary conditions.
3.
where
(C.U. Hons. 1953)
Assuming f"(x) continuous
in [a, b] 9
show
both lie in [a, 6].
and
(Take <v(x)^f(x)+Ax+ Bx* and determine
that
c
Now apply Rollers Theorem
4.
^^*^^
/4,
to the intervals
B
such that
[a, c]
and
[c, 6].]
The second
derivative /"(*) of a function f(x) is continuous for
anc^ at eac^ P^ nt x tlle s *8ns of /(x) and /"(jc) are the same. Prove that
if f(x) vanishes at points c
between c and
and
where a<^c<d<^b, then
d,
it
vanishes everywhere
(B.U. 1952)
d.
/(*) 9(*) an(* ^U) ar e three functions derivable in an interval
such that
that there exists a point
5.
*how
f(b)
Deduce Lagrange's and Cauchy's mean value theorems.
6.
Prove that
lim
if /
7/
(jc)
/
exists.
7.
Discuss the applicability of Rolle's
(/)
/U)tan x and a=0,
Theorem when
b^n.
.
(m)
/W=U-c) 3 -c 3
when a=0 and 6=2c
(iv)
www.EngineeringEBooksPdf.com
(a,
b)
;
CHAPTER
VII
MAXIMA AND MINIMA
GREATEST AND LEAST VALUES
In this chapter we shall be concerned with the application
7'1.
of Calculus for determining the values of a function which are greatest
or least in their immediate neighbourhood technically known as
Maximum and Minimum values. A knowledge of these values of a
function is of great help in drawing its graph and in determining its
greatest and least values in any given finite interval.
assumed that/(x) possesses continuous derivatives of
come in question.
that
order
every
It will be
%
Maximum value of a function. Let, c, be any interior point
of the interval of definition of a function /(x). Then we say that/(c)
is a maximum value of /(*), if it is the greatest of all its values for
values of x lying in some neighbourhood of c. To be more definite
and to avoid the vague words 'Some neighbourhood', we say that
f(c) is a maximum value of the function, if there exists some interval
(c8, c+S) around c such that
for
all values
ofx, other than
c,
lying in this interval.
This, again,
c
_
f(C)
c+8
c
8
is
a
f(c)
for values ofh lying between
small in numerical value,
8
equivalent to saying that
value of /(;c), if
is
maximum
> f(c+h),
and
i.e.,
S, i.e., for
f(c+h)-f(c)
<
values of h sufficiently
Minimum
value of a function, /(c) is said to be a minimum value
the least of all its values for values of x lying in some
neighbourhood of c.
of /(#),
if it is
This
is
equivalent to saying that f(c)
is
a minimum value off(x)
if there exists a positive 8 such that
f(c)
<f(c+h), i.e.J(c+h)-f(c)
for values ofh lying between
small in numerical value.
8
and
8, i.e.,
>
for values of h sufficiently
Note 1. The term extreme value is used both for a maximum as well as
for a minimum value, so that /(c) is an extreme value if f(c+h) f(c) keeps an
invariable sign for values ofh sufficiently small numerically.
148
www.EngineeringEBooksPdf.com
MAXIMA AND MINIMA
Note
149
While ascertaining whether any value /(c)
is an extreme value or
the values of the function for values of x in any
immediate neighbourhood of c, so that the
values of the function outside the neighbourhood do not come into question at all.
Thus, a maximum value may not be the
greatest and a minimum value may not be the
least of all the values of the function in any
finite interval.
In fact a function can have
several maximum and minimum values and a
minimum value can even be greater than a
not,
2.
we compare /(c) with
maximum
value.
A glance
at the adjoining
graph of f(x)
shows that the ordinates of points P 19 P3 P5
are the maximum and the ordinates of the points
P2 P4 are the minimum values of f(x) and that
Fig. 49.
the ordinates of P4 which is minimum is greater than the ordinate of P x whi:h
a maximum.
,
,
is
A
To prove that a
necessary condition for extreme values.
is that
value
extreme
an
condition
be
to
off(x)
for f(c)
necessary
7-2.
Let f(c) be a
c-\-h
maximum
value of f(x)<
There exists an interval (c S, c+8), around c, such that,
any number, belonging to this interval, we have
Here, h
may
/(c+/0</(c),
be positive or negative.
Thus
'^<0iffc>0,
If h tends to
...
through positive values, we obtain from
If A tends to
through negative values,
.-(Hi)
we obtain from (),
...
/'(c)>0.
The
relations (in)
and
(/v)
(0
(/),
/'(c)<0.
only
if,
is
will simultaneously be true,
if
(iv)
and
if
/'(c)=o.
It
can similarly be shown that /'(c)=0,
if /(c)
is
minimum
value of f(x).
Cor. Greatest and least values of a function in any interval. The
either f(a)
greatest and least values off(x) in any interval [a, b] are
x
andf(b), or are given by the values of for w/i/c/i/'(x)=0.
The greatest and least value of a function are also its extreme
values in case they are attained at a point strictly within the interval
so that the derivative must be zero at the corresponding point.
The theorem now
easily follows.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
150
Note
for
1.
P
where the ordinate
point
x-axis."
When
the necessary condition
to a curve at a
Geometrically interpreted,
extreme values obtained above states
a
is
stated in this
maximum
"Tangent
:
or
minimum
is
parallel
to
geometrical form, the theorem appears
[Refer Fig, 48 p. 149.]
The vanishing off (c) is only a necessary but not sufficient condition for f(x) to be an extreme value. To see this, we consider
almost self-evident.
Note
;
2.
the function /(Jc)==x 3 for
x=Q.
For value of x greater than 0,/(x) is positive and is, therefore,
and for values of x less than 0, f(x) ia
greater than/(0) which is
less
negative and is, therefore,
than/(0).
Thus/(0) is not an extreme value even though /'(0)=0.
Note 3. Stationary value. A function f(x) is said to be stationary
;
also
for x^cif the derivative f (x) vanishes for x=c, i e. if /'(c) =
then f(c) is said to be a stationary or a turning value of f(x). The
term stationary arises from the fact that the rate of change /'(#)
of the function f(x) with respect to x is zero for a value of x for which
;
9
f(x)
is
stationary.
noted that a maximum or a minimum value is also a
stationary value but a stationary value may neither be a maximum
nor a minimum value.
It will be
Ex.
Find the greatest and
1.
least values
of
in the interval [0, 2J.
Let
Thus
/'(*)=0forx=l, -1,
The value
x=
J.
does not belong to the interval
to
be
considered.
not, therefore,
1
[0, 2]
and
Now
Also
Thus the least value
Ex.
2.
is 1
and the greatest value
Find the greatest and
least values
is
21.
of
in the interval [-2, 5].
Change of Sign. A function is said to change sign from
to
negative as x passes through a number c, if there exists
positive
some left-handed neighbourhood (c ft, c) of c for every point of
which the function is positive, and also there exists some rightfhanded neighbourhood (c, c+h) of c for every point of which the
Def
unction
is
negative.
www.EngineeringEBooksPdf.com
151
MAXIMA AND MINIMA
A similar
the statement
meaning with obvious alterations can be assigned to
function changes sign from negative to positive, as
"A
x
passes through c".
It is clear that if a continuous function /(x) changes sign as
passes through c then we must have/(c)=0.
Ex. 1. Show that the function
x
y
= (x + 2)(x-m2x- l)Cx~3)
<?(x)
changes sign from positive to negative as x passes through 4 and from negative
to positive as x passes through
2 or 3 ; also show that it does not change sign
as x passes through 1.
Ex. 2. Show that the function
changes sign from positive to negative as x passes through
.* and 2.
negative to positive as x passes through
4 and
1
and from
72. Sufficient criteria for extreme values. To prove that f(c) is
an extreme value off(x) if and only iff'(x) changes sign as x passes
through c, and to show that f(c) is a maximum value if the sign
changes from positive to negative and a minimum value in the contrary
case.
Case
Let f(x) change sign frcm positive to negative as x
I.
passes through
c.
In some left-handed neighbourhood of c,f'(x) is positive and so
/(x)is monotonically increasing in his neighbourhood. (6*32, p. 136),
Therefore /(c)
is
the greatest of
all
the values of /(x) in this
handed neighbourhood.
In some right-handed neighbourhood of
so
f(x)
p. 136).
in
is
monotonically increasing
Therefore /(c) is the greatest of
left-
negative and
neighbourhood ( 6*33,
the values off(x) in this
c,
f'(x)
is
this
all
right-handed neighbourhood.
the greatest of all the values of /(x) in a certain
value
complete neighbourhood of c and so, by def., f(c) is a maximum
Hence
f(c)
is
of/(x).
Case
Let f'(x) change sign from negative to positive as
II.
passes through
x
c.
the least of all
the values of /(x), in a certain complete neighbourhood of c and so,
by def., /(c) is a minimum value of /(x).
It can similarly be
shown that
in this case/(c)
is
Case III. If /'(x) does not change sign, i.e., has the same sign
in a certain complete neighbourhood of c> then /(x) is either monothis
tonically increasing or monotonically decreasing throughout
value
extreme
so
of/(x).
neighbourhood
that/(c) is not an
Note. Geometrically interpreted, the theorem states that the tangent to
a curve at every point in a certain left handed neighbourhood of the point P
whose ordinate is a maximum (minimum) makes an acute angle (obtuse angle)
and the tangent at any point in a certain right-handed neighbourhood of P makreJJ
an obtuse angle (acute angle) with x-axis. In case, the tangent on either suit of r
makes an acute angle (or obtuse angle), the ordinate of P is neither a maximum
nor a minimum.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
152
Ex.
I.
Examine
the polynomial
10x6
for
24x 5 +15x 4 ~4Qx 8 +108
maximum and minimum
values.
Let
t
2)
2
=60x*(x +l)(x-2)
Thus, /'(;t)=0 for x-=0 and x=2
values of f(x) for #=0 and 2 only.
so that
we expect extreme
Now,
forx<0,
/'(*)<<>;
forO<x<2
/'(x)<0;
forx>2
/'(*)> 0.
3
Here,/'(*) does not change sign as x passes through
/(O) isjieither a maximum nor a minimum value,
so that
since f'(x) changes sign from negative to positive as
100 is a minimum value.
passes through 2, therefore /(2)
Also,
Ex.
2.
x
=
Find the extreme values of
and distinguish between them.
Let
24
/
so that
/(x)=0
for
x=l,
2, 3.
Therefore, /(x) can have extreme values
for *==-!, 2, 3 only.
Now, for* <1,
1<*<2,
for2<x<3,
for
x
/ (x)<0;
/'W>0
;
forx>3,
pig, 50
therefore /(l)--8
Since /'(x) changes sign from negative
to positive as x passes through 1 and 3,
are the two minimum values.
and/(3)=-8
from positive to negative as
Again since, f'(x) changes sign
is a maximum value.
-7
therefore, .fl2)
passes through 2,
www.EngineeringEBooksPdf.com
MAXIMA AND MINIMA
Ex.
153
Find the extreme values of
3.
5*+18;c 5 +15;c*-10.
Ex.
are
Show
4.
that the
maximum and minimum
values of
-9 and - 1/9 respectively.
Ex.
Show
5.
that
9x5 +30;c4 +35;c 8 -fl5;c 2 +l
is
maximum when x=
Also, find
[-2,
2/3
its
and minimum when x=0.
greatest
Ex.
6.
Show
that
value
-5x+5;c3 -l
when x = l, a minimum
jc
has a
and least values in the intervals [2/3,
0],
and
2].
maximum
8
value
when x=3 and
neither
*=0.
when
(D.U. 1948)
Use of derivatives of second and higher orders. The derivatives
order only have so far been employed for determining and
As shown
distinguishing between the extreme values of a function.
in the present article, the same thing can sometimes be done more
conveniently by employing derivatives of the second and higher
7-4.
of the
first
orders.
All along this discussion it will be assumed that f(x) possesses
continuous derivatives of every order, that come in question, in the
neighbourhood of the point c.
7*41.
Theorem
1.
f(c)
is
a minimum value off(x) if
y
f'(c)=0andf"(c)>0.
Applying Taylor's theorem with remainder after two terms, we
get
f(c+h)^f(c)+hf'(c)+
r(c+0ji)
;
is positive for X
c, there exists an
for every point of which the second derivative is
3-51, p. 54)
(
As the value f"(c) of f(x)
interval around,
c,
positive.
Let
point of
ft
a
/2
!
is
c+h
be any point of this interval.
th''s
interval
Then,
and accordingly /"(C +M)
is
c+6 2 h,
a
Also
is also
positive.
positive.
see that there exists an interval around c for every
of
which, f(c+h)f(c) is positive, i.e.,f(c)<f(c+h).
point/ +A,
.Thus
we
Hence /(c)
is
a
minimum
value of/(x).
www.EngineeringEBooksPdf.com
154
DIEFERENTIAL CALCULUS
Theorem
7-42.
As
in theorem
f(c
i'.*.,
2.
1,
f(c) is a
maximum
we have by
value off(x) 9 if
Taylor's theorem,
+h)-f(c) =
Asf"(c) is negative, there exists an interval around c for every
Thus as in the
point of which the second derivative is negative.
c
for every point,
preceding case, there exists an interval around
c+h of which /(c+/0/(c), is negative, i.e.,f(c)>f(c+h)
;
Hence /(c)
is
a
maximum
General Criteria.
7-43.
value of/(x).
Let
Thenf(c)is
a minimum value off(x), iffn (c)>Q and n
(i)
is
even
;
n
() a maximum value of(x), iff (c)<0 and n is even
neither a maximum nor a minimum value ifn is odd.
(iii)
;
Applying Taylor's theorem with remainder after n terms, we
so that because of the given conditions,
n h).
As/n (c) ?^0,
there exists an interval around c for every point
the nth derivative fn (x) has the same sign, viz., that of
x of which
Thus
for every point, c-f-A, of this interval,
fn (c + 9 n h)
has the
n
sign of f
(c).
is positive, whether, h, be positive
Also when n is even, h n \n
or negative and when nis odd, h n \n !, changes sign with the change
in the sign of h.
!
;
Hence, as in the preceding cases we have the criteria as stated.
Note. The result proved in 7*41 and 7*42 is only a particular case of
the general criteria established
in^
7*43 above.
www.EngineeringEBooksPdf.com
MAXIMA AND MINIMA
155
Examples
maximum and minimum
Find the
1.
Let
values of the polynomial
4
2
5
/(x)-=8x --15x -l-10x
.
3
4
/'(x) = 40x -60x +20x
=20x(2x 3 ~-3x* + l)
2
Hence
-20x(x-l) (2x+l).
/'(*)=0 for x=0, 1,- J.
Again
/"(x)=-160x
3
- 180x 2 +20
9x 2 +l).
Now, /"(
maximum
45 which
)=
negative so that /(
is
I)
=
T-S-
is
a
value.
Again, /"(0)=: 20 which
is
positive so that/(0)
=
is
a
minimum
value.
As/"(l)-0, we have
Now
to
examine /
(1).
2
/'"(*)= 480x -360x.
= l20
f'"(l)
Hence /(I)
is
neither a
which
maximum
Investigate for 'maximum
2.
//;
sin JC+cos
is
nor a
not zero.
minimum
and minimum values
value.
the function
2jc.
Let
x+cos 2x.
x 2 sin 2x
=cos
x 4 sin x cos
%
we
dyldx=Q,
get
>>=sin
...(i)
dy fdx = cos
Putting
cos
We
function
x=0
or sin
consider values of
is
x= J.
x between
and
cos
x~ Ogives
x=
x = siii"
and
J lying
sin
x=|
gives
and
between
Now
d*y/dx
^
J
=
sin
x
For x=7r/2, d y/dx*=:3 which
For x=37r/2, d 2y/dx 2 = 5 which
For x=sin~ 1 J and TT sin* 1 J,
df^/rf^^-sin
and
1
TT
sin* 1 J,
Tr/2.
2
2
is
only, for the given
and
-
j
which
2ir
periodic with period 2?r.
Now
sin*" 1
x.
x-4 (1-2
4 cos 2x.
is
positive
;
is
positive
;.
sin 2 x)
= -15/4
negative.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
156
Therefore y is a maximum for x
minimum for x=7r/2, 37T/2.
= sin~"
1
J4
sin" 1 J
TT
and
is
a
Putting these values of x in (i), we see that |, | are the two
2 are the two minimum values.
values and 0,
maximum
Exercises
maximum and minimum
Investigate the
1.
(/)
(it)
values of
2x 3 -15x 2 + 36x+10.
(P.U.I945)
3x*-4x 3 +5.
Find the values of x for which x e -6ax 3 -f 90 2 x4 -f
(a>0).
'
2.
values.
Determine the values of x
3.
minimum
has
which the function
for
12x 5 -45x<+40x 3 -f6
attains (1) a
maximum value, (2) a minimum value.
maximum value of (x-l)(x-2)(x-3).
(P-U- 1941)
4.
Find the
5.
Find the maxima and minima as well as the greatest and the
values of the function
Find
6.
minimum
for
y=x*
12x 2 +45x in the interval
(P>V. 1939)
least
[0, 7].
what values of x the following expression
is
a
maximum
or
:
2x 3 -21x 2 4- 36x-20.
7.
Find the extreme values of x 3 /(x 4 -f
8.
Show
that (x + !)*/(*+ 3) s has a
9.
Show
that x x
10.
Show
1).
value 2/27 and a
maximum
minimum
value 0.
is
that the
minimum
maximum
maximum
x=e~ l
for
value of
(1
.
l
/*)* is e
l*.
<x<oo
11.
Find the
12.
Find the extreme value of a 3*
13.
Find the maximum and minimum values of x+sin 2x
14.
Find the values of x for which
value of (log x)lx in
1
~as -x.
sin
x
(a>
(P.U. 1955)
.
1).
x cos x
is
a
in
<S x
maximum
^
2*.
or a
minimum.
15.
Find the
maximum and minimum
x
in(
*^jc5^0;
Show
16.
17.
that sin
Discuss the
(/)
values of
sin 2x-f J sin 3x,
x (1-fcos x)
is
a
maximum when x=$n.
maxima and minima
sinx-fisin 2x+ j
in the
interval
[0, n]
of the sums
sin 3x.
cos K^\ cos 2x + $ cos 3x.
Also, obtain their greatest and the least values in the given interval.
(//)
18.
(i)
(Hi)
19.
for
Find the minimum and
sinxcos^x.
(i"0
sinxcos 2x.
Show
maximum
(/v)
that (3
x)e**
4x^-x
values of
flsecx-f&cosecx,
ex cos (x
has no
(0<a<6).
a).
maximum
x=0.
www.EngineeringEBooksPdf.com
or
minimum
value
EXAMPLES
Find the maxima and minima of the radii vectors of the curve
20.
ra
21.
Find the
P
~ _a?__ +
"cos
sin*0
l
(Delhi, Aligarh)
"fl
maximum and minimum
values of x*+y* where
ax*+2hxy+by*=l.
[Taking x=rcos
treme value of r 2 where
0,
y = rs'm
0, the
question reduces to finding the ex-
,
2
-T=fl cos 0-f 2/i cos $ sin
+
sin 2 0]
In the following, we shall apply the theory of
maxima
and minima to solve problems involving the use of the same.
It /will
7*5.
be seen that, in general, we shall not need to find the second derivative and complete decision would be made at the stage of the first
derivative only when we have obtained the stationary values. In this
connection, it will be found useful to determine the limits between
which the independent variable lies. Suppose that these limits are
a, b.
.
If
y
is
for
x=a
xb and positive otherwise and has only
and
owe stationary value, then
maximum and the greatest.
the stationary value
is
necessarily the
If y -> QO as x -> a and as x -> b and has only one stationary
value, then the stationary value is necessarily the minimum and the
least.
In connection with the problems concerning spheres, cones
cylinders the following results would be often needed :
1
Sphere of Radius
.
r.
Volume = ?Ttr*.
Cylinder of height,
2.
Surface
Cone of height,
h,
47ir.
and radius of circular base, r.
Curved surface =2Trrh.
h,
Volume =7rr 2 /?.
The area of each plane
3.
and
face
Trr
2
.
and radius of circular base,
r.
=
tan" 1 (/*//*),
Semi-vertical angle
Slant height- V( r2 +/* 2 ),
Volume = |TT r 2 h,
Curved 8urface=-7Tr\/(r 3 +/j 2).
Here cones and cylinders are always supposed to be right
circular.
Application to Problems.
7-6.
Examples
Show
1.
greatest volume
Let
surface
r
that the height of an open cylinder
equal to the radius of its base.
of given surface and
is
be the radius of the circular base
and K, the volume of the open
;
h,
cylinder.
www.EngineeringEBooksPdf.com
the height
Therefore,
;
S
9
the
158
DIFFERENTIAL CALCULUS
Here, as given, S is a constant and
are variables. Substituting the value of
(')> we get
V in
which gives
terms of one variable
V
is a variable.
Also, h, r
as obtained from (i), in
h,
r.
dV __S
'
dr~~~
2
so waat ur /u/=v only when r =\/(Sl37i)
negative value of
inadmissible. Thus V has only one stationary value.
r
:
As
V must be
positive,
Srnr >
we have
3
Thus
r varies in
Now F=0
for every other
r=V
being
0, i.e.,
>
Sr
the interval
for the points
Trr
(0,
3
V
or r
>
y"
(S/Tr).
(5/7:)).
r=0 and
admissible value of x.
^/ (S/n)
9
Hence K
and
is
is
positive
greatest for
(S/37T).
Substituting this value of r in
(i),
we
get
c
h
=
3?r
2?rr
_2S
~
3
Hence
1
27T
hr for a cylinder of greatest volume
and given
surface.
Show that the radius of the right circular cylinder of greatest
2.
curved surface which can be inscribed in a given cone is half that of the
cone.
Let
r
be the radius
OA
of the base and
h,
OV of the
the height
given cone.
We inscribe in it a cylinder, the radius
of whose base is OP=x as shown in the
figure. We note that x may take up any value
y
between
and
r.
To determine the height PL,
cylinder, we have
PL
PA
= ~'
Fis. 51.
www.EngineeringEBooksPdf.com
PL
of this
r-o:
'
r
159
EXAMPLES
If
S be
the curved surface of the cylinder,
we have
= -?5^?L ^^
S=27T. OP. PL
^J^
dx
(r-2x)
r
=0
(rx-x
for
Thus S has only one stationary value.
Now S is for x=0 and x=r and is
and r.
between
lying
S
is greatest for x=r/2.
Therefore
2
),
x=r/2.
positive
for values of
x
circular cylinder of greatest
surface of the right
a
in
sphere of radius r.
surface which can be inscribed
is the
construct a cylinder as shown in the figure.
radius of the base and CB is the height of this cylinder.
3.
Fi/irf
//*
OA
We
Let Z_AOB==0 so that
between and 7T/2.
t
lies
OA
^
Q
=008
0.
= OB cos
AB
=sm
OB
.-.
cos
6.
0=rsin
6.
0,
AB=OBsm
IfS be the
0=r
surface,
we have
5=27r. OA* +2v.OA.BC.
=27rr 2 (cos2 0+sin
dS
=27r/-
= 27rr
2
(
2
20). ..(1)
2 cos
(2 cos
sin
20
Fig. 52.
0+2 cos
20)
sin 20).
=0
gives
2 cos 20
sin
20=0,
i.e.,
tan 20=2.
...(2)
Let, 0j, be a root of tan 20 = 2.
sin 20 1 = 2/\/5 and cos 20 1 =l/'V/5.
.-.
As tan 20X =2,
From (1) ,we see that
when 0=0, S=27rr 2 when
;
when 0=0,, 5=27rr2
which
is
2
greater than Sirr
Hence
0=7T/2,
(1+^^+
5=0,
sin
20,)
.
wr
is
the required greatest
www.EngineeringEBooksPdf.com
sjurface. (Cor.
p,149),
160
DIFFERENTIAL CALCULUS
Prove that the least perimeter of an isosceles
triangle in which
of radius r can be inscribed is Qr^/3.
(P. u. 1934)
We take one vertex A of the triangle at a distance x from the
centre O. Let AO meet the circle at P.
The two tangents from A
.
and the tangent at P determine an isosceles
4.
a
circle
ABC
triangle
We
Also
B
p
circumscribing the given
circle.
have OL~r.
BP=AP tan
r
denote the perimeter of the triangle,
If* P>
Fig. 53.
/_BAP
we have
p=AB+A C+BG
=2AB+2BP
=2(AL+LB)2BP
=2AL+4BP,
(for,
BL=BP)
-rr
"dx
=2 2{x+r)(x*-r
for
so that dpjdx=0
admissible.
x
z
)-x(x+r}
2r
;
z
negative value,
Now, x may take up any value >r
the interval
From
(r,
(i),
oo
only,
r,
of x being in-
so that
p-
x
Hence p
we
see that
p
also as X->QO
is
varies in
).
->
oo
as x ->
r.
Again, dividing the numerator and denominator by
so that
it
least for
x2 we get
,
.
x=2r.
Putting this value of x in
see that the least value of p is
www.EngineeringEBooksPdf.com
(/),
we
EXAMPLES
A
161
circumscribed to a sphere of radius r ; show that
is a minimum, its altitude is 4r and its
semi-vertical angle sin~i f
(Madras 1953 ; P.U. 1930)
5.
cone
is
when the volume of the cone
We take the vertex A of the cone at a
O of he sphere. .(See Fig. 53. p. 160).
distance
x from the
centre
By drawing tangent lines from A, as shown in the figure, >ve
construct the cone circumscribing the sphere.
Let the semi- vertical angle BAP of the cone be 0.
Now,
if v
be the volume of the cone, we have
V=i7T .BP*. AP,
which
now expressed
be
will
Since sin
=7=r-,= x
OA
A
Again, since
Thus
.'.
,
BP
tan
have
0=
*
,
-=tan
AL
(r+x)
mi
We
in terms of x.
0,
2
r2
v=ir\,- 92----
9
2
.
^
(x+r)=
,
--
Trr
,
'
v
r)
Thus dvjdx is for x=3r.
Here x can take up any value ^r and v -* cc when x
Thus t; is minimum and least for x=3r.
when x -> QO
>> r
and
.
Hence, for minimum volume, the altitude of the cone
and the semi -vertical angle
= sinNormal
6.
is
drawn
at
1
x
~sin- 1 r
= sin~
1
3r
a variable point
P of an
i.
^
ellipse
maximum
distance of the normal from the centre of the ellipse.
(P.U. 1935)
take any point P(a cos 6, b sin 0) on the ellipse
being
the eccentric angle of the point.
find the
We
;
The equation of the tangent at P
at
Therefore, slope
r of the tangent
&
of the normal at
Hence, slope
*
1
-p
is
P~ ----a sin
.
P=tb ---cos
www.EngineeringEBooksPdf.com
-
-
-r
=1.
DIFFERENTIAL CALCULUS
Therefore equation of the normal at
,
"
:
0x
>P> be
,.
:
sin
.
y-b sm
0=^~
0by cos
0=(a
/
<5r
a sin
P
is
,
(x-a
2
2
)
cos 0.
sin
tance from the centre
perpendicular distance
its
(0, 0),
we
obtain
jm
(9
cosj?
*~6+b*^o&
-
,--
-
/
(
CL
^ 2 cos4
ka\\
u
2
---------
(a
Putting
2
------
sin
2
6-}
<
-
ft
-
2
-
-
cos
2
dpld6=Q, we get
Because of the symmetry of the ellipse about the two co-ordinate axes, it is enough to consider only those values of 6 which lie
between and 7r/2 so that we reject the negative value of tan 0.
Now, />=0 when 0=0
between
and
or
7T/2
and p
is
positive
when
lies
Therefore p is maximum when tan 0=v'(^0Substituting this value in (/), we see that the maximum value of
p
is
a
7T/2.
b.
7.
Assuming that the petrol burnt (per hour) in driving a motor
boat varies as the cube of its velocity, show that the most economical
speed when going against a current ofc miles per hour is | c miles per
hour.
Let v miles per hour be the velocity of the boat so that (v -c)
miles per hour is its velocity relative to water when
going against
the current.
Therefore the time required to cover a distance of d miles
d
-
u
hours.
3=
v
c
The
8
petrol burnt per hour=A:v where k
the total amount, y> of petrol burnt is given by
,
,
y=k
'
v*d
vc
y
Thus
Of these r;=0
is in-
t8
,
7
=fo/
VC
.
.
2
dv
Putting dy/dv=0
a constant.
is
(v
we
get
c)
0=0 and |#
admissible.
Also y _>
oc
when
t?
Thus v=^c gives the
-> c and when
_> oo
least value of y.
www.EngineeringEBooksPdf.com
.
183
EXAMPLES
Exercises
1.
Divide a number 15 into two parts such that the square of one
multiplied with the cube of the other is a maximum.
Show
2.
that of all rectangles of given area, the square has the smallest
perimeter.
3.
Find the rectangle of greatest perimeter which can be inscribed in a
of radius a.
4.
If 40 square feet of sheet metal are to be used in the construction of
an open tank with a square base, find the dimensions so that the capacity is
circle
(P-U.)
a semi-circle with a rectangle on its diameter.
Given that perimeter of the figure, is 20 feet, find its dimensions in order that
its area may be a maximum.
(Patna, Allahabad)
greatest possible.
A figure consists of
5.
6.
A, B are fijced points with co-ordinates (0, a) and (0, b) and P is a
variable point (x t 0) referred to rectangular axes ; prove that x*o6 when the
( p -# 1935 )
angle APB is a maximum.
7.
given quantity of metal is to be cast into a half-cylinder, /.*., with
a rectangular base and semi-circular ends. Show that in order the total surface
area may be minimum the ratio of the length of the cylinder to the diameter
of its circular ends is */(*+2).
(Aligarh 1949)
A
8.
The sum of the surfaces of a cube and a sphere is given show that
when the sum of their volumes is least, the diameter of the sphere is equal to
;
the edge of the cube.
9.
The strength of a beam varies as the product of its breadth and the
square of its depth. Find the dimensions of the strongest beam that can be
cut from a circular log of wood of radius a units.
(D.U. 1953)
10.
The amount of fuel consumed per hour by a certain steamer varies as
the cube of its speed. When the speed is 15 miles per hour, the fuel consumed
The other expenses total Rs. 100
is 4 J tons of coal per hour at Rs. 4 per ton.
per hour. Find the most economical speed and the cost of a voyage of 1980
(P.V. 1949)
miles.
11.
Show
maximum volume
12.
Show
that the right circular cylinder of the given surface and maxisuch that its height is equal to the diameter of its -base.
that the semi-vertical angle of the cone of
and of given slant height is tan~N2.
mum volume is
13.
surface
is
that the height of a closed cylinder of given
equal to its diameter.
Show
(D.V. 1952)
volume and
least
14.
Given the total surface of the right circular cone, show that when the
l
volume of the cone is maximum, then the semi-vertical angle will be sin
15.
Show that the right cone of least curved surface and given volume
has an altitude equal to 42 times the radius of its base.
.
16. Show that the curved surface of a right circular cylinder of greatest curved surface which can be inscribed in a sphere is one-half of that of the
sphere.
17.
well as
its
A cone is inscribed in a sphere of radius r
curved surface
is
greatest
when
its
;
altitude
prove that
its
volume as
is 4r/3.
Find the volume of the greatest cylinder that can be inscribed in a
18.
(D.U. 1955)
cone of height ft and semi-vertical angle a.
the length
n
one
times
have
to
is
box
edge
19. A thin closed rectangular
of another edge and the vojume of the box is given to bev. Prove that the
least surface s is given by
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALOVLUS
164
JO. Prove that the area of the triangle formed by the tangent at anj
9
point of the ellipse x*la*+y*lb **l and its aves is aminimuu for the point
21. Find the area 9f the greatest isosceles triangle that can be inscribed
in a given ellipse, the triangle having its vertex coincident with one extremity
ef the major axis.
(Allahabad 1939)
A perpendicular is let fall
from the centre to a tangent to an ellipse.
Find the greatest value of the intercept between the point of contact and the
22.
foot of the perpendicular.
A
show
23.
;
tangent to an ellipse meets the axes in P and
least value of PQ is equal to the sum of the semi-axes of the ellipse
that PQ is divided at the point of contact in the ratio of its semi-axes.
Q
that the
and also
24.
is the foot of the perpendicular drawn from the centre O on to the
2
2
tangent at a variable point P on the ellipse x*la +y lb*=l (a>b). Prove that
2
2
the maximum area of the triangle OPN is (a
i )/4.
N
25.
One corner of the rectangular sheet of the paper, width one foot, is
folded over so as to just reach the opposite edge of the sheet ; find the minimum length of the crease.
a^x^b,
26. If f'(x) exists throughout an interval
prove that the greatleast value of/(x) in the interval are either /(a) and/(fc) or are given by
the values of x for which /'(*)=0.
est
and
A
grocer requires cylindrical vessels of thin metal with lids, each to contain exactly a given volume V. Show that if he wishes to be as economical as
8
possible in metal, the radius r of the base is given by 2wr =V.
If, for other reasons, it is impracticable to use vessels in which the diameter exceeds three-fourths of the height, what should be the radius of the base
of each vessel
A
?
(P.U.)
/, feet
long is in the shape of a frustum of a cone the
radii of its ends being a and b feet (a>b). It is required to cut from it a beam
of uniform square section. Prove that the beam of the greatest volume that
can be cut is alll(a-b) feet long.
(Agra ; P.U.)
27.
tree trunk,
28. Find the volume of the greatest right circular cone that can be described by the revolution about a side of a right-angled triangle of hypotenuse
1 foot.
(P.U. 1940)
A
29.
rectangular sheet of metal has four equal square portions reat the corners, and the sides are then turned up so as to form an open
rectangular box. Show that when volume contained in the box is a maximum,
moved
the depth will be
where
a,
b are the sides of the original rectangle.
(Banaras 1953)
The parcel post regulations restrict parcels to be such that the length
plus the girth must not exceed 6 feet and the length must not exceed 3 feet
Determine the parcels of greatest volume that can be sent up by post if the
from of the parcel be a right circular cylinder. Will the result be affected if the
30.
greatest length permitted were only If feet.
31.
Show
that the
maximum
(Patna)
rectangle inscribable in a circle
(P.C7.
www.EngineeringEBooksPdf.com
is
a square.
Suppl 1944)
CHAPTER VIII
EVALUATION OF LIMITTS
INDETERMINATE FORMS
We know that x ->
8'1.
a,
hmF(x)
F(x)
so that this theorem on limits fails to give any information regarding
the limit of a fraction whose denominator tends to zero as its limit.
Now, suppose, that the denominator F(x) -> as x -> a.
The numerator f(x) may or may not tend to zero. If it does
not tend to zoro, then f(x)/F(x) cannot tend to any finite limit. For,
if possible, let-it
tend to
finite limit,
say
/.
We
write
we have
so that, in this case,
lim/(x)=lim
[fe
.
F(x)
lim
.
lim
]
F(x)=/.0=
Thus we have a contradiction.
Three types of behaviour are possible in this case. The fraction
oo or it 3 limit may not exist. For example,
to + oo or
tend
may
when JC -> 0, (so that the limit of the denominator in each of the
,
,
following cases is 0),
8
(i) lim (I/* )
we
see that
=+oo
(Hi)
zero
lim
(I I x)
a
;
(//)
lim(l/-Jt )=:-ao
;
does not exist.
The case where the limits of the numerator of a fraction is ale
more important and interesting. A general method of deter-
is
mining the limit of such a fraction will be given in this chapter. For
the sake of brevity, we say that a fraction whose numerator and
denominator both tend to zero as x tends to a, assumes the indeterminate form 0/0 for x=a.
may be
of interest to notice that the determination of the
dyjdx is itsolf equivalent to finding the limit
of a fraction 8yjbx which assumes an indeterminate form 0/0.
It
differential co efficient
Other cases of limits which are reducible to this
be considered in this chapter.
<(.x)
form
will
also
In what follows it will always be assumed that /(x), F,x) and
possess continuous derivatives of every order that come in
question in a certain interval enclosing x^=a.
165
www.EngineeringEBooksPdf.com
186
DIFFABBNTUL OALOVLV8
8-2.
The Indeterminate form
lim
-^
To determine
.
[
x-*a
when
lim
/(x)=0
Urn F(x).
As/(x), F(x) are assumed to be continuous for
f(a)= lim
By
=
yfr-0 ;*
x=0, we have
Fx=Q.
lim
Taylor's theorem, with remainder after one term,
we have
Hence
r
i
\f
f
y._
f,\
ttli
This argument fails if F'(a)=0.
~
has already been discussed in
The
u
J \ )
Tflf/**\i
Jj \CL\
case
:f
pt
when F'(a)=Q but
8-1.
Now, lQtf'(a)=F'(a)=0.
Again, by Taylor's theorem with remainder
after
two terms,
we get
Hence
lim
-^-li
=
lim
h^(
The case of failure which
as before*
In general,
arises
when -F"(a)=0 can be examined
let
f(a)=*f'(a)=*f"(a)=*
=
F
Mid
www.EngineeringEBooksPdf.com
INDETERMINATE FORM 3
By
167
Taylor's theorem, with remainder after n terras, we get
An
/iti-i
...+
(a )JL. f( a +6n h)
+
^Lfn-i
n
-l
~
]
h
F^(a)+ ~- F(a+0' nh)
F(a+6' nh)
Hence
Ex.
Determine lim
1.
x
-.
sin
where x ->0.
.,'
x
/(0)=0,
F(0)=0.
-.
F(x)=x
{
sin
jc
-.
;
f'(x)=e*+e-*- r
~,
{ F'(x)=x cos x+sin x.
.
.
/'(0)=0,
'.
F'(0)=0.
...
/" (0 )= 2>
Again
x
)=
_hm ----
sin
JC+2 cos
F"(0)=2.
x,
~
x
_.>
The process may be conveniently exhibited as
follows
:-
*- e-_21og(l + x)
x cos
^o
Ex.
2.
e'-e-a +2/(l+x) 8
-x sin x+2
cos
z
x~
2
_
~
Find the values of a and b in order that
x(l
a cos
f
x) - b sin x
^
fim
nm
3
*
x~>0
+
.
ay be equal
to
'-
.
L
'
www.EngineeringEBooksPdf.com
U. 1944, 1959)
DIFFERENTIAL CALCULUS
The function
of the form (0/0) for
is
all
values of a and
6.
-frsinx
->-i- ----hm x(l+acosx)
x-3'
,.
x ->
=
l+a
,.
cos
x
x -> .
finite
ax sin x
3x 2
JX
b cos x
'
:
The denominator being for x=0, the fraction will tend to a
limit if and only if the numerator is also zero for x=0. This
requires
Again supposing
l+a
we have
this relation satisfied,
cos
xax sin x
b cos x
3x2
6X
2# sin
------
= hm
..
xax cos x+& sin x
-
-
,.
bx
x-*a
30 cos x+tf* sin x+fe cos x
.
A
*}/
As given,
From
Ex.
and
(1)
=1,
A
(2),
6
i.e.,
3a = 6.
we have
Determine the limits of the following
3.
X
C
....
sm x x
x..v
lv
,
.
f
cos x
cosh
A:
cos
_
( V)7
(x->0).
v
log cos x
,- M -,,-v
(vi)
;
Jog(14-6x)
f
A:
-.
(D.C/. //b/i5. 7952)
Ex.
Evaluate the following
4.
Jim
*"
.
^^
Af
:
.(D.U.1952)
(//)
lim
(D. U. Hons. 1951, P.U. 1957)
i
"
'
\
1'
0*1*
.*
*v
^v
/
r\ T
(D.U. 1955)
www.EngineeringEBooksPdf.com
INDETERMINATE FORMS
Ex.
If the limit
5.
169
of
2x+a
sin
x
sin
X9
as
x tends to zero, be
the value of a and the limit.
finite, find
(P.U.)
Preliminary transformation. Sometimes a preliminary
transformation involving the use of known results on limits, such as
8*3.
..
hm
sin
x ->
simplifies the process
x
tan x
..
lim
* =1, x->
a good deal.
x
.
=1
These limits
may
also be used
to
shorten the process at an intermediate stage.
Ex.
x
hm 1+sin
j
*
1.
i?
|.
ITFind
cos
(1x)
x+log
J
x tan 25x
,
~> 0).
(x
v
'
The inconvenience of continuously differentiating the denomiwhich involves tan 2 x as a factor, may be partially avoided as
nator,
We
follows.
1
write
+ sin
x
cos
x+log
x)
(1
x tan^x
1
~~
1
,.
101
+ sin xcos x+log
x
+ sin x
x
/
x)
\tan x/
(1
x)
xcos x+log (1
x)
Ax 3
1+sin
-
-
~
\2
"
cos x+log
x tan 2 x
1-j-sin
(1
3
x-*
xcos~ x+log
-
(1
-
x)
~
.
L
x >0
1
lim
jc^O
tor
+ sin
x
cos x+log(l
-3
x
x)
To evaluate the limit on the B.H.S., we notice that the numeraand denominator both become for x=0.
l+sin x cos x + log (I x)
..
hm
x3
cosx fsui-v-
1
_x- T
-cos x^sin
------
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
Ex,
Evaluate the following
cosh # cos x
..
2.
:
x un x
We
have
xcos x = cosh x cos
^g
x
^
cosh
x
x
*
-
g
n
----- cos x =
cosh x
,.
.....
lira
xsmx
cosh x
cos x
-----x*
,.
lim
cosh
.
rvi
CQ sh
_
x
cos
x
cos
..
.
hm
x
x
and denominator are both zero
Since the numerator
therefore
I*
i
x
sin
x
sinh
__ ,.
Aim
^
for
x=0,
*+sin~~x
OY
cosh
x
lim ____ - x+cos
iim
J
x->0
,.
=1.
Ex.
3.
Determine the
(i)
limits of the following functions
te -
-"J~.
dV) log (1
0).
:
-) cot
x.
(
->
0).
*..(- o,
8-4.
The Indetermine form
~.
Tc?
determine
00
ton
lim
*Let f'(x)IF'(x) -+
also ->
/(x)=
I
oo
asx-+a.
=
//m F(x).
It
mil be shown
that f(x)/F(x)
/.
Suppose, x>a. As f\x)IF'(x) -> /, when x -> a, we make it
arbitrarily near / by taking x sufficiently near a, say between a, and
a+8.
I* Another
proof is given io the following sub-section.]
www.EngineeringEBooksPdf.com
INDETERMINATE FORMS
171
We now
take any two numbers c and x which lie between a
apply Cauchy's mean value theorem f(x) and F (x) for
the interval (c, x). We thus have
and
a+8 and
where
We
F(x)-F(c)
between c and x and, therefore, between a and a
lies
re-write
(/)
as
7"T~T~I
x
'/'(I)
F(x)
a
l-FWlFWJ'F'fc)
c
a+S
c fixed, we make x -> a.
Tlien/(c) and F(c) are fixed
to infinity.
tend
and
our
F(x)
and, by
hypothesis, f(x)
Keeping
1
Therefore,
F(c)
1
as *~* 00
made arbitrarily near / by taking x
as x -> a through values greater than
Similarly,
values less than
'
Thus
^)/^W
Y-f(c{ f(x)
it
sufficiently near
can be
a so that
it
-W
a.
may be shown that/(x)/F(x)
-*
as
/
x -> a through
a.
Hence
^ hm
hm
,.
,.
/'(*)
when
lim
/x=
oo
=s lim
Note. The above investigation rests on the hypothesis that /'(x)/F'(*)
tends to a limit as x -+ a. This part of the hypothesis necessarily implies that
x near a so that f'(x), F'U) both exist
f'(x)IF'(x) has a meaning for values of
and F'OOs^O for such values. This justifies the use of Cauchy's mean value
theorem in the above investigation.
8*41.
A proof of the
above result
lim /(*):=
oo
is
also often given as follows
and lim F(x)=
A*)
F'(x)
3BES
f'(x)
www.EngineeringEBooksPdf.com
oo
,
:
172
DIFFERENTIAL CALCULUS
Let
lim [f(x)IF(x)]
.-,
from
we
(1),
=
where /^O and
I
get
Supposing now 1=0.
^
/+.-U-
.-.
Applying the above
+ l-
result,
F'(x)
/
Finally
By
let
/=*>
= m
^
)
-
...(8)
)
so that
the preceding result,
.
..
0=hm
F(x)
.
-
..
=lnn
F'(x)
r' -<
-
- (4)
i)*'
Hence we
see that always
when lim/JC
,
= 30
^- ^
"m
-
<>
lim JPx.
Note 1. The above second proof is incomplete in the sense that it
assumes that Urn lf(x)IF(x)] exists. The first proof did not assume this
existence.
Note
2.
While evaluating
Urn
-j~r*
when
it is
of the form
.
we must try to change over to the form 0/0 as soon as it may be conveniently
possible, for, otherwise we may go on indefinitely without ever arriving at the
end of the process.
www.EngineeringEBooksPdf.com
INDETERMINATE FORMS
Ex.
1.
Determine lim
y-p^
173
L, as
Here, the numerator and the denominator both tend to
tends to a.
oo
as
x
I
log(x-a)
.-.
lim T --~-JT --L
=
=
x
..
Inn
lim
a
=*
-JT-
e*
Ex.
2.
Determine the
w,.
A
Hint,
(v/)
8-6.
limits of the following functions
^
i
log
tan x
tan 2x =
log (l~.v) cot (xw/2),
U ->
The indeterminate form
log tan 2x
log tan *
:
1
-
,
J
1).
0*oo
.
7b determine
Urn [/(x).
x->a
when
lim /(x)=0,
To determine
this limit,
we
write
so that these new forms are of the type 0/0 and oo /oo respectively
8-4.
limit can, therefore, be obtained by 8*2 or by
and the
In this case, we say that /(*). f(x) assumes the indeterminate
form
O'oo
Ex.
at
1.
x=a.
Determine lim (x logx), as x->0.
We write
x
log
x-
_.
f/i
lim (x log
x)-
lim
lim
^
(*-)
r=
www.EngineeringEBooksPdf.com
DIFFERENTIAL
The reader may
see that writing
*
xlogx=
(I/log x)
which is of the form
(0/0) and employing the corresponding result of
8-2 would not be of
any avail.
'
Note
w e know that
-
haye
1
does nottend to a limit as x -*
lx
lim
=00,
*-(0+0) *
Ag ~ in
0.
In fact
*-(0-0)
x is defined for positive values of x only so that there
* *
question of making x ->
through negative values while determining
log
'
we
is
no
lim (xlogx).
x->0
Thus, here x <+
Ex.
(/)
(///)
8-6.
means x -> (0+ 0) so
really
Determine the
2.
x log tan
x,
that
1
/*
does tend to a limit.
limits of the following functions
(x -*
0).
->
The Indeterminate for
lim
x -> a
x tan (w/2-x).
(//)
(a~-x) tan (nxfta), (x
:
(x
-*
0).
0).
oo
<*>
7b determine
.
[f(x)-F(x)l
when
lim /(x)=<x>
X->fl
We
write
so that the numerator and denominator both tend to
as x tends to
a.
The limit may now be determined with the help of 8*2.
In this case, we say that [/(x)
nate form oo
oo for x=a.
F(x)] assumes the indetermi-
Note. In order to evaluate the limit of a function which assume! the
form, oo -oo , it is necessary to express the same as a function which assumeg
the form 0/0 or oo/oo.
Ex.
1.
Determine
We write
___
_1
x-2
and
see that the
log
1
new form
*.
tog
(X- !)-(*'
(x-1)
is
(x-2)
(x-1)
of the type 0/0 when x -*
_ __L__1
r
log
2.
iog(y~i)-(y~2
www.EngineeringEBooksPdf.com
175
FORMS
for x=2. On
The numerator and the denominator are both
the
that
we
show
of
the
method
required limit is
8-2,
may
using
Ex.
limits of the following functions
Determine the
2.
'
The Indeterminate forms,
8'7.
L
^
1
as
'
,
oo
.
:
To determine
x
J
when
(i)
(fl)
limf(x)=:Q
F(x)=0.
limf(x)=0; lim F(x)=<x>
(Hi) //w/(jc)
We
lim
;
= oo
;
.
lim F(x)=0.
write
so that
In each of the three cases, ^ e see that the right hand side
assumes the indeterminate form 0. oo and its limit may, therefore, be
determined by the method given in 8*5.
Let
lim
x->
[^(*),log /(*)]==/.
a
;\
limlogj;=/,
or
log lim
Hence
lim
or brevity,
forms
,
we say
1,
oo
y=l or
that
j/(x)
lim y=e*.
^1
respectively for
assumes the
x=a.
www.EngineeringEBooksPdf.com
indetermi-
DIFFERENTIAL CALOVLVS
176
Ex.
Determine
I.
x~a
Urn (x
a)
os x ->
a.
form).
(0
Let
lim
log>>= lim
{} ^~^
I
~~
.
Hence
log lim }>=0,
Thus
i.e.,
lim
lim y~e<>
x~a
lim (x
=
=
(xtf)=0.
:
=I when
a)
Note. Here it is understood that x tends to a through values greater
x ~ a would
for otherwise the base (x-a) would be negative and (x a)
have no meaning.
than
a,
Ex.
2.
Determine
lim (cos x)
1/x 2
tfs
1/x
Let
y=(cos
IT
\f\Ct
y
iv'fti
,.
o
,.
,
hm logy= hm
=
log (lim y)
..
lim
=
2
i or
x
"~
x
log cos
---^
tan x
~
s
^JT
lim (cos x)
1
=
JL-
2
y=e
lim
x~>0
Hence
0.
x)
cos
log
^^
,
x ->
'
=
'
T
e
*.
x->0
Ex.
3.
Determine the
->
(/)
x*, (x
ii )
(cot x)
sin
limits of the following functions
0).
(//)
^
,
(x
->> 0).
x
(l
~' x
)~\ (x ->
tan x
-
(/v)
(sin x)
www.EngineeringEBooksPdf.com
,
(x
s
(P.C7. 7923)
1).
->
1.
12).
INDETERMINATE FORMS
177
(v)
(vii)
*
x*
(viii)
(D.C7. 7949)
(x-->0).
,
(x ->1).
,
(P.C/.
7957)
Exercises
Determine the limits of the following functions
e_e- x
12.
o
->0).
v
^ 2 sin A: (x
:
.
3.
1+^cos A:~cosh x
(2jc
tan x
^3
,
'
(A:
v
-> <>.
'
log(l-f x)
(
7.
log
& -AC
>
>-
n sec x),
(A:
->
n/2).
j_
8
log x
.
(cot *)
,
u -> o).
9
a*
.
10.
13.
(cos ax)
,
(x -> 0).
14.
15.
tan
16.
(2-O
17.
(
sin
2
^
-=-
2a
S
Ct
\w
(
',
(*->).
(B.V.1953)
sec 2
^
_
_
20.
cot x
(sec x)
,
(x -> n/2).
21.
(2-x)
www.EngineeringEBooksPdf.com
tan
ff
^
,
(x^-
1).
'
DIFFIRENTIAL CALCULUS
178
u
a t5>*', (,+,).
'*,
..
26.
(fl.tf.
[For solutions of Ex. 25 and Ex. 26 by Infinite series refer
log
cos
Hons. 1959)
9-5, p. 185
]
*
sect*_
sec
ijc
(
28.
*
!+-
,(*->
co)
a
31.
Evaluate
sinxi
lim
32.
If
/(0)=0,
show
that tho derivative of every order of
/(x) vanishes for
/
33,
x=0,
/.<?.,
n
(0)=0,foralln.
Discuss the continuity of f(x) at the origin when
/(*)=* log
sin
x for
x^O and/(0) =0.
(D.U. Hons. 1955)
www.EngineeringEBooksPdf.com
CHAPTER IX
TAYLOR'S INFINITE SERIES
EXPANSIONS OF FUNCTIONS
9-1.
Infinite Series.
Its
convergence and stun.
Let
be an
infinite set of numbers
given according to some law.
Then a
symbol of the form
an
Here each term is followed by another so
no last term.
If we add the first two terms of this infinite series, and then
add the sum so obtained to the third, and thus go on adding each
term to the sum of the previous term, we see that, as there is no
last term of the series, we will never arrive at the end of this process.
In the case of a finite series, however, this process of addition will be
completed at some stage, howsoever large a number of terms the series
is
called
that there
may
infinite series.
is
consist of.
Thus, in the ordinary sense, the expression 'Sum of an infinite
has no meaning. A meaning is assigned to this expression by
employing the notion of limit in the manner we now describe.
series',
Let Sn denote the sum of the first n terms of the series so that
a function of the positive integral variable n. If Sn tends to a
finite limit S, as n tends to infinity, then the series is said to be comergent and S is said to be its sum.
Sn is
In case, Sn does not tend to a
does not converge.
finite limit,
we say that the
The question of the sum of a non-convergent
series
series
does not
arise.
We may find an approximate value of the sum
infinite series
by adding a
sufficiently large
Consider the infinite geometric series
Illustrations.
l+r+r*+r* + ......
We know
of a convergent
number of its terms.
that
#n ,=-
1
JL
r*
*f
Sn =n
when
179
www.EngineeringEBooksPdf.com
DlFFEBENTIAL CALCULUS
180
We have now to
[Refer
examine lim Sn when n ->
oo
.
3-61, p. 57].
r=l, S n =n which tends
For
For
<
1,
lim r"=0,
r>
1,
lim
r
k
|
For
For
r^
1,
r
lim
r
w
<
r
only if
\
9
\
1,
and
the
Sn
does not
.
exist.
geometric series converges if and
infinite
sum of the
infinite
2."
.
Sn =l/(l-r).
therefore, lim S n =oo
and, therefore, lim
Hence, we see that the
oo
so that lim
=oo and,
n
to
infinite series then,
is
We suppose that a
given function
series.
Taylor's
f(x) possesses derivatives of every order in
an interval
Then, however large a positive integer n
and 1, such that
0, lying between
may
[a,
1/(1
r).
a+h].
be, there
exists
a
number
H
f(a+h)=f(a)+hf'(a)+
=^
R
where
*^
n
(Taylor's development with Lagrange's form of remainder.)
We
write
so that
Suppose that R n ->
0,
lim
so that
we
as n ->
ao
.
It
is
then clear that
Sn =f(a+h),
see that the series
f(a)+hf'(a)+
converges and that
*",
its
sum is equal
to /(
Thus we have proved that
(0
[a,
(/*/(*)
possesses derivatives
of every order
a+H] and
(ii)
the remainder
tends to
as n tends to
infinity,
then
www.EngineeringEBooksPdf.com
in
the
interval
EXPANSIONS
This
known
is
as Taylor's theorem
j(a+h) in an infinite
power series in h.
The
as Taylor's series.
infinite series.
series,
for the
we
for a
Putting
(ii)
for
h
in
in
the interval [0, x]
the remainder
tends to
as n tends to
This
known as
is
expansion of/v x) in
i.e.,
and x
see that if
(0 /(*) possesses derivatives of every order
and
x,
known
Maclaurin's
the Taylor's infinite
development of
ascending integral powers of h, i.e.,
series of
series (1) is
9'3.
181
power
The
an
infinity,
then
Maclaurin's theorem for the development or
infinite series of ascending integral powers of
series in x.
series (2) is
known
as Maclaurin's series.
Note. It may be seen that instead of considering Lagrange's form of
remainder, we may as well consider Cauchy's form.
Formal expansion of functions. We have seen that in order
any given function can be expanded as an infinite
Taylor and Maclaurin series it is necessary to examine the behaviour
9-4.
to find out if
of R n as n tends to infinity. To put down R n we require to obtain
the general expression for the nth derivative of the function, so that
we fail to apply Taylor's or Maclaurin's theorem to expand in a
power series a function for which a general expression for the nth
derivative cannot be determined
Other more powerful methods
have, accordingly, been discovered to obtain such expansions whenever they are possible. But to deal with these methods is not within
the scope of this book.
,
Formal expansion of a function as a power series may, however,
be obtained by assuming that it can be so expanded, i.e., R n does
as n tends to infinity.
Thus we have the result
tend to
:
then
Iff(x) can be expanded as an infinite Maclaurin's series,
/(*)=/(0)+*/'(0)+
**,
/*(<))
+ ......
...(1)
Such an investigation will not give any information as to the
range of values of x for which the expansion is valid.
To obtain the expansion of a function, on the assumption that
values of its derivatives
possible, we have only to calculate the
x=0 and substitute them in (1).
it is
for
www.EngineeringEBooksPdf.com
182
DIFFERENTIAL CALCULUS
In the Appendix, we shall obtain the expansions of
e, sin
x, cos x, log
(1+x). (l+x)
m
without assuming the possibility of expansion by actually examining
the behaviour of R n for the functions,
In the following, however, we obtain these and other expansions
the possibility of expansion.
by assuming
Expansion of
9-41.
e*.
Let
Thus
Substituting in the Maclaurin's series,
we obtain
-v
-v-3
which
is
known
Cor.
1.
a
as Exponential Series.
Changing x into x log a in
* log *
,
=e
This result
(1),
we
get
= 1+(x log a) +
may
also
be obtained directly by employing Maclau-
rin's series.
Cor
2.
Putting,
1
for x,
we obtain
.LI
+ 2! +
L
from which we
!
\
~3!
L.4.
+ '"
+ ~4!
JL
obtain the values
may
of,
e,
upto any number of
decimal places.
9-42.
Expansion of sin
x.
Let
/(*):=: sin
x
mr
Thus
/-(0)-sin
1,
so that
we
-*
AOJ^O,/^)^-!,
/-'(0)-0,
n
etc.,
see that the values of / (0) for different values of n
a successively appearing periodic cycle of four values
1,0,
-1,0,
www.EngineeringEBooksPdf.com
form
183
EXPANSIONS
Making these substitutions
ths sine
in the Maclaurin's series,
we obtain
series.
sin
*=*-__
We may
9-43.
similarly obtain the cosine series
x2
cos
Expansion of log
9-44,
x 2n
xl
*=:!-_ +_-_ ...... +
(1
:
(-!)
+x).
Let
Thus
.\/'(0)
= l,/"(0)=-l
Making substitutions
Logarithmic Series
log
;
/'''(0)==2
in the Maclaurin's seriep,
wo
so on.
obtain the
:
(i+x)=.x-^+-^+
9 45.
!,/""(0)= -3 land
Expansion of
(1
+x)
......
+ (-i)"-i.
+
..... .
OT
.
Let
Thus
Making
substitutions,
2f
In case
m
is
we
obtain^the Binomial Seriesj
......
a positive integer,
n\
we obtain
a finite series on the
right.
Note 1. If we examine the behaviour of R n as is done in the Appendix,
can show that (i) the Exponential, sine and cosine series are valid for every
value x, (ii) the Logarithmic Series is valid for
Kx<J, and (in) the Binomial
Series is valid for
Note 2. In the following we shall consider some more cases of expansions of functions, in each case assuming the possibility of expansion. It will
be seen that in some cases we may also obtain the expansion by vsing any of
the series obtained above.
,
we
www.EngineeringEBooksPdf.com
184
DIFFERENTIAL CALCULUS
Examples
1.
Assuming
as the term inx*.
the possibility
of expansion, expand, tan
as far
x,
Lot
/(x)=tan
/.
/'(x)=sec
2
x.
x=l+tan 2 x.
/"(x)=2 tan x
=2
sec 2 x
tan x(l+tan 2 x)
= 2 tan x+2 tan
/'"(x)=2
sec
2
x+6
3
x.
tan x sec 2 *
2
=2 (l+tanx)+6 tan 2 x (l+tan2 x)
= 2+8tan 2x+6 tan x.
/ '"(x) = 16 tan x sec 2 x+24 tan 3 x sec2 *
4
;
= 16 tan x +tan x) -f 24 tan x(
= 16 tan x -f40 tan x+24 tan x.
= 16 sec x + 120 tan x sec x + 120
(
2
1
3
3
/ v (jc)
2
1
+tan 2 x)
5
2
2
tan 4 x sec 2 x.
Thus
Substituting these values in the Maciaurin's series
obtain
anx-x+
x
+
x
9*4,
+
e
Assuming the possibility of expansion, expand /(x)
ascending integral powers of x.
In Ex. 1, 5-6, p. 124 we proved that
2.
in
I
(w 2 (2 2 +w 2 )(4 2 +w 2
2
2
2
2
(m (! +m )(3 +m
[(w-2)
)
2
-fw 2
2
+m
2)
[(
)
Substituting these values in the Maciaurin's series,
Use of known
3.
find the expansion of
m
2
2i
\
2
by
the term in
w(l -fw
o
2
)
8
!
m
2
2
(2
4
]
m sm x
;
2
]
we
+m
By Maciaurin's theorem
series.
sin
upto and including
2
we
;
n even
w odd.
get
2
)
i
!
or otherwise
(e*-l),
x4
.
First four derivatives are neaded to expand the given function
Maciaurin's theorem upto the term required.
The required expansion can also be obtained by employing the
Exponential and the sine series and thus avoiding the process of calculating the derivatives which is often very inconvenient.
www.EngineeringEBooksPdf.com
EXPANSIONS
=z,
Now,
sin (e
x
say.
l)=sinz
z
O
z5
,
O
I
I
v*2
"v4
"vS
+
2T
9*5.
185
3~"T
+ 4l +
Use of infinite series for evaluating the limits of indeterThe following examples will illustrate the procedure.
minate form?.
Examples
1.
find Urn
Using the
?*?
*-*-**
(D.U.1953)
infinite series for e
e" sin
x
1
x) 1
x
.x
x
,
sin
x and
log (I
.v),
we obtain
2
= lim
=lim
= lim
2.
-i-i
Eh
(0
(//)
lira
x->0
L
'
I
-_e
ie v
:
(D.U. Hons. 1954)
.
x
(Cf. Ex. 25,
www.EngineeringEBooksPdf.com
Ex. 26,
p. 178)
186
DIFFERENTIAL CALCULUS
Let
y=(i+ X
log
7=
l lx
)
log(l+x)
o
2
L
(T>
or
o
'6
t
a
i
as
1^
O
3
a
Q
2
-e.e.
24
2
IX
^
.
JL
i
\
./V
e
+-STV-
'
2
/
x
,
e
2
(
""
24
llx
+ "24
5
.
'
Exercises
(/w f/w following ,
1.
//re
possibility
of expansion may also be assumed)
Prove that
w
!
www.EngineeringEBooksPdf.com
187
EXPANSIONS
2.
Prove that
e a * sin
3.
b -~-
3
bx=bx+abx*+-
*+....
3
Obtain the wth term in the expansion of tan" 1 x
in ascending
of*.
powers
(P.U.1951)
4.
Show
5.
Prove that
6.
If >>=log
x=*l-x 2 +$x*-^\x
that cos 2
e x sin
2
*= *
2
3
4
-f * -f
Jx
......
____
2
[x+ >!(!+ x )], prove that
Differentiate this n times and deduce the expansion of y in ascending
in the form
powers of x
yss *~
x3
1
T*
x5
3
1
+ T-' 4
3
x7
5
3
1
'
'
'
~5~~ ~2~' 4
6
7
'+
,
'
'
'
(P.C/.
7.
If >>=sin log
2
(;c
7P57)
-}-2x-M), prove that
Hence or otherwipe expand
8.
'
.v
ascending powers of x as
in
far as
Prove that
.
,
9.
Show
that
2)
Obtain the following expressions
10.
(/)
log tan
(l
(i"fx)=Zx+*x
.
log
,
.
(v)
tan
^
X
__.1
e.^!-
(vi7)
A:
(m
2
8)
-JxH ----
log sec
,
T~ 2T~
=-^ +
7
xM ""
A'
1
47
,
"90"
2
'
2"+fi
30"
.
'
X
'
*
2
y
x4
I
,
^c
sm-^^xl,1
i
(v/)
2
:
x2
1
4
...
-f
m
,
+ sin
1+
(iv)
3
,
-
2!
^8
'
1
""
30
l
X*
3
x5
4
"5
'
3
--f-2^-
2
*=2> +
12
+ *"-
4"!
^+45
x%
.
-+
+"-
www.EngineeringEBooksPdf.com
(D.V.1953}
188
DIFFERENTIAL CALCULUS
log sin (x+/i)=log sin
(v//7)
x+ h
cot
/i
8
+JY
-
tan-Ms+AJ-tan-^+Asinz
(/AT)
2
cosec x
*--yT
-~?
-(A
1
sin z) --
2*
1
^
--
v.sin3z
8
.
.,
x- ----
8
2 cot x cosec
.
4-(Asmz)
...
3
when zcofr^x.
Obtain the expansion of e
11.
sm X
in
powers of x as
far as
x4
.
(D.I/.
12.
Obtain
the
first
three terms of the expansion of log
Powersoft
13.
App]y Maclaurin's theorem
as the term in x 8
14.
to find the expansion
Expand log
sin
x
in
powers of U-3).
2
Expand 3x3 -2x 4-x-4
16.
Show
that
(/)
Uin
in
3
~
S
x
in
.
lim
(//)
^
and A by
x~3
in
sin
x
"-.
x-smx
=1.
^
*3
Obtain by Maclaurin's
3
powers of x-2.
*
*-55=;
^
x->0
X
+
(D.U. 1955)
15.
sion of e
in
(P. U. 1955)
of e x l(e x 1) as far
.
[Write log sin x=log sin (3+ x-3) and replace x by
the expansion of log sin \x+h) in powers of h.]
17.
(1-Manx)
16
theorem, the
first
four terms of the expan-
ascending powers of x. Hence or otherwise show that
xcosx
_
~-A-
lim
x-0
x
7
--
=3.
smx
Appendix
We shall now
e*,
obtain the expansions of
sin x, cos x, log
(1+x),
(l
+ x) m
without making any assumption as to the possibility of expansion,
r
Expansion of e
.
x
n
Let /(*)=*.
Therefore, f (x) = e so that/(x) possesses
vatives of every order for every value of x.
Taking Lagrange's form of remainder, we have
,
..
deri-
'
V!
We know
that
when
->
oo
x//i
whatever value x
may
,
!
->
have.
www.EngineeringEBooksPdf.com
(See
3*62, p. 58)
EXPANSIONS
Rn
->
as n ->oo
189
.
Conditions for Maclaurin's infinite expansion are thus satisfied.
/(0) = e=l
Also
Making
v-2
--l+*+
which
is
we
substitutions in the Maclaurin's series,
+
2!
get
v-n
y3
+
3,
+-;,-!+"
valid for every value of x.
Expansion of sin
Let f(x)= sin x.
x.
.-.
f n (x) = sm
+ ^nn),
(x
so that/(x) possesses derivatives of every order for every value
We
of
jc.
have
Since
*
x"
n\
we
see that
Rn
->
as n -> oo for every value^of x.
Thus the conditions
Maclaurin's infinite
for
Now
satisfied.
expansion are
n*
/"(0)=sin
Making these
substitutions in the Maclaurin's series,
sinx=x
f
which
is
_
-j-
*
,-
+
we
get
....
valid for every value of x.
Expansion of cos
x.
cos x
for
.
As above, it may
2
x4
x6
_i x
2
j-
+-4
?
6
easily
be shown that
+
j
every value of x.
Expansion of log (1+x).
Let
We know that log (1+*)
(l+x)>0,
i.e.,
for
possesses derivatives of every order for
x>~l.
Moreover
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
190
If
Rn
,
denotes Lagrange's form of remainder,
<
Let
we have
x
<;i, so that x/(l+0.x) and, therefore,
whatever value n may
[xl(l
(tx)]
positive and <1,
Since, also, 1/n -^ 0, as n -> oo we see that R n -> 0.
(i)
n
+
is
also,
have.
,
Thus
/? n
Let
(ii)
as H ->
->
1<JC<0.
In this case x/(l+fix)
so thafc
we
when 0^x<J 1.
oo
to
fail
may
draw any
not be numerically less than unity
from Lagrange's form
definite conclusion
.
Taking Cauchy's form of remainder, we have
Here
(1
0)/( !
+ #*).
is
positive
v
and
r
less
than
1,
and
i_
Also
Hence,
x n ->
as n -+
co
.
Rn
as n ->
oo
.
->
(
3'61, p. 57)
the conditions for Maclaurin's theorem are satisfied for
Also
in the Maclaurin's
Making these substitutions
x*
series,
we get
(l+x)=x- -2+-$-- ~T~ + ---.^
log
for
v3
A:
<v2
_
A.
.
./>
.
!<*<!.
m
Expansion of (l+x)
;
(m
is
any
real number).
Let
/w=(i +*r-
Whenm
is
any
real
number, (l+x)
vatives of every order only
when
l
w
+ x>0,
possesses continuous deri/.^.,
when
x>
1.
Now,
m
f(x)=m(m-l)(m-2) ...... (m-n+l)(l+x) ~.
We notice that
if
m is any positive integer, then the derivatives
www.EngineeringEBooksPdf.com
EXPANSIONS
191
otf(x) of order higher than mth vanish identically and thus, for
m
n
w, R n identically vanishes, so that (l+x) is expanded in a
finite series consisting of (m+l) terms.
>
Ifwbe
not a positive integer, then no derivative vanishes
we have
identically so that
If
Rn
to examine this case
still
denotes Cauchy's form of remainder, we
further.
get
Let
1
<
x
<
x
1, i.e.,
|
|
<
1.
Also
0<0<
< 1<(
~
or
t
<
1.
l+Ox,
<
;-}
1,
0<
Let (m
1) be positive.
Now
< l+9x <
ml be
Let, now,
Now, since
negative.
x
x
I
,
I
or
X )-!.
(l + fl*)-l^(l_
we know that
m(m
l)...(w
w+1)
^
_J
""*
when
\
*Also
--
1
/
Thus,
Hn
~>
The conditions
fied.
"ZTiri
as n -> oo
\
^
when
,
|
x
\
<
,
I
x
,
I
<
,
1
-
1.
for Maclaurin's expansion are, therefore,
Now
/(0)=w(w-l) ...... (w-n
Aefer to the foot note on the next page.
www.EngineeringEBooksPdf.com
satis-
DIFFERENTIAL CALCULUS
192
Making substitutions, we get
m(m--l)
,+
xm t
L__/xa
,
,
ft
,
(l+x)*=l+mxH
when
<
1
<
A:
JL
-
l)(m2) x
-v
+ m(m
3
,
'>
-
7
,
3
,
-f ...
1.
Ex.
1.
Justify the Maclaurin's expansions of
1
e ax sin bx e ax cos bx, tan" x.
Ex,
2.
Show
3.
Show
9
that log
x and cot x cannot be expanded
as
Maclaurin's
series.
Ex.
f(x)
that the Maclaurin's
infinite
expansion
is
not valid for
where
e~
when x^O and /(0)=0.
[Refer Ex. 32. p. 178.]
*To prove
that
when
\
x
<
\
1) ____
Changing n
to w-f
1,
we
1,
(w
"^
get
x
or
As
p such
|
x
|
<
we can
1,
<
x
that
Wn
>
k, for
I
1,
and a positive integer
K
*
<
k
I
I
W
I
n
we get
Now
|
up
|
//c^ is
a constant and k* ->
and so also Un ->
'
|
<
Thus
/>.
2
Multiplying
number k
find a positive
!
as n >oo.
n
|
as w
->
oo.
www.EngineeringEBooksPdf.com
as n
->
oo.
CHAPTER X
FUNCTIONS OF
TWO
VARIABLES
PARTIAL DIFFERENTIATION
The notions of
two
10*1.
continuity, limit and differentiation in
variables, will be briefly explained in
few theorems of elementary character will also be
relation to functions of
this chapter.
A
proved.
The sub ject of functions of two variables is capable of extension to functions of n variables, but the treatment of the subject in
this generalised form is not within the scope of this book.
Only a
few examples dealing with functions of three or more variables may
be given.
102. Functions of two variables. As in the case of functions
of a single variable, we introduce the notion of functions of two
variables by considering some examples.
The
(/)
between
relation
x, y, z,
of numbers x,
determines a value of z corresponding to every pair
which are such that x 2 + y 2 ^l.
y,
i
Denoting a pair of numbers x, y, geometrically by a point on
a plane as explained in
1-7, p. 14, we see that the points (#, y) for
which X 2 +J> 2 <1 lie on or within the circle whose centre is at the
origin and radius is 1.
The region determined by the point (.x, y) is called the domain
of the point (x, y).
Now, we say that the relation (1) determines z as a function of
two variables, x 9 y defined for the domain bounded by the unit circle
>=!.
(/)
Consider the relation
a<b c<d.
Now (a x)(x
where
is
;
non-negative
if
b)
is
non-negative
if
a^x^b
and
(cy)(yd)
ct^y^d.
The points (x, y) for which a^x^b, c^yt^d determine a
rectangular domain bounded by the lines
x=a x=b y=c,y=d.
y
;
The
relation (2) determines a value of z corresponding to every
Thus z is a function of x
point (x, y) of this rectangular domain.
and y defined for the domain.
193
www.EngineeringEBooksPdf.com
194
DIFFERENTIAL CALCULUS
The
(ill)
relation
r2
V2
z=e x y
determines z as a function of
x,
y
,
defined for the whole plane.
In general, we say that z is a function of two variables defined
for a certain domain, if it has a value corresponding to every point (x y)
of the domain.
y
The
etc., as in
relation of functionality is expressed by the symbols /, ?
the case of functions of a single variable, so that we may
write z=f(x,y),
Ex.
</(*, y), etc.
Determine the domains of
definition of
Neighbourhood of a point
(x, y) such that
10-3.
Let 8 be any positive
(a, b).
The points
number.
determine a square bounded by the lines
x=a
Its centre
is
S,
at the point
hood of the point
(a, b).
x = a+8
y=^b~8, y
;
This square
(a, b).
For every value of,
is
called
S,
we
a neighboura neigh-
will get
bourhood.
10*4.
Let
Continuity of a function of two variables.
(a, b)
be any point of the domain of definition of
As in tho case of functions of one variable, we say that f(x, y)
continuous at (a, b), if for points (x, y) near (a, b), the value
f( x y) of the function is near f(a, b) i.e., f(x, y) can be made as
near f(a> b) as we like by taking the points (x, y) sufficiently near
is
>
(a. b).
Formally, we say that /(A: y) is continuous at (a, b), if, corresponding to any pre-assigned positive number e, there exists a positive
,
number
8
such that
1
P(a,b)
for
y)-f( a b )
f(*>
I
>
all points (x, y) in the
<e
I
square
Thus for continuity at (a, b), there
exists a square bounded by the lines
x=a
8,
such that,
square,
x=fl+S; y^b
for
any point
between
8,
y=b+8
(x, y)
of this
/(x, y) lies
f(a, fc)-e,
p.
j.
where
e is
any positive number, however
small.
www.EngineeringEBooksPdf.com
PARTIAL DIFFERENTIATION
Continuity in a domain. A j"unction f(x, y) is said to be
if it is continuous at every point of the same.
10 41.
continuous
195
a domain
in
10-42.
Let /(.x, y) be continuous at (a, b) and
Special Case.
be any positive number, however small. Then there exists a
square bounded by the lines
let e
x-=a8, x = a-\-8
;>'=
S, >'
=
/
such that for points (x, y) of the square, the numerical
difference bet ween f(x, y) and f(a, b) is less than e.
value of the
if we consider points of this square lying on the
see that for values of x lying between a
8 and a
8,
the numerical value of the difference between f(x b) and f(a, b) is less
than e. Also, such a choice of S is possible for every positive number
This is equivalent to saying that f(X, b) is a continuous function
.
of a single variable x for x-a.
In particular,
line
+
y=b, we
t
It
of y
for
may
y = b.
be similarly shown that f(a, y)
a continuous function
is
Thus we have shown that a continuous function of two
is also
variables
a continuous function of each variable separately.
10
5.
Limit of a function of two variables.
lim/(x, y)
=
l,
as (x, y) -> (a b).
y
A function f(x, y) said to tend to the limit /, as x tends to a
andy tends to b, i.e , as (x, y) tends to (a, ft), //, corresponding to any
pre-assigned positive number z, there exists a positive number 8 such
that
is
\f( X ,y)-l\
for
<
all points (x, y), other than (a, 6), lying within the square,
This means that corresponding to every positive number e,
there exists a neighbourhood such that for every point (x, y) of this
neighbourhood other than f(a, b),f(x, y) lies between / e and /+e.
,
Limit of a continuous function. Comparing the
10-51.
104, and
10'5,
tions of limit and continuity as given in
y) is continuous at (a, 6), if and only if
defini-
we
see
that/(x,
limf(x, y)=f(a, b) as
(x, y)
->
(a, b),
ie.,
the limit of the function^ actual value
The same thing may
nuity at (a, 6), we have
as
also be expressed
of the function.
by saying that
(0,0).
www.EngineeringEBooksPdf.com
for conti-
196
DIFFEBENTIAL CALCULUS
Partial derivatives.
10-6.
Let
*=/(*>
J>).
Then
if
it
exists, is
said to be the partial derivatives of f(x, y) w.r.t. x at
and is denoted by
(a, b)
3z
f_
V
\
or
8x /(a, b)
x at
fA
.(a,
b).
'
It will be seen that to find the partial derivative of/(x, y) w.r.t.
we put j equal to b and consider the change in th&
(a, b),
function as
x changes from a
to a
-\-h
so that
/(*, b)
is the
x=a.
ordinary derivative off(x, b) w.r. to xfor
Again,
if
it
and
15
exists,
is
is
called the partial derivative off(x, y) w.r. to
y at
(a, b),
denoted by
the ordinary derivative off(a, y) w.r. to
y for y=b.
If f(x, y) possesses a partial derivative w.r. to x at every point
of its domain of definition, then the values of these partial derivatives
themselves define a function of two variables for the same domain.
This function is called the partial derivative of the function w.r. to x
and is denoted by gz/gx or fx (x, y) or simply fx
.
Thus
fix v]-\im
lim
")
Jx\x>
<*
f(X+h
-------'
y)
-f(X
' y)
-----------
as/z-^0
AI -> u,
as
f,
We
where y is kept constant in the process of taking the limit.
can
similarly define the partial derivative of/, w.r. toy which is denoted
by
gz/8^, f*(x, y) or fy .
10 61. Partial derivatives of higher orders. We can form partial
derivatives of 3z/gx and dzjfty just as we formed those of z.
www.EngineeringEBooksPdf.com
PARTIAL DIFFERENTIATION
197
Thus we have
which are called the second order
partial derivatives of z and are
denoted by
oZ
Z
Y>
or
f
9
v,
f
lvx
respectively.
Similarly the second order partial derivatives
aj \dy
-are
respectively denoted by
Thus, there are four second order partial derivatives of z at
*tny point (a, b).
2
Ths
by
y.
is
2
partial d3rivatives 3 z/g>> 3* and 3 z/dx $y are distinguished
the ordzr in which z is successively differantiated w.r. to x and
But, it will be sesn that,
given in the Appendix.
in
general, they are
The proof
equal.
Geometrical representation or function of two variables.
and OY. Through the
perpendicular lines
plane and
point 0, we draw a line Z'OZ perpendicular to the
10
7.
OX
We take a pair of
XY
oall it Z-axis.
Any one
sense.
of the two sides of Z-axis may be assigned a positive
z, will be measured parallel to this axis.
Lengths,
The three co-ordinate axes, taken in pairs
determine three planes, viz
XY, YZ and ZX
which are taken as the co-ordinate planes.
Let z --/(#, y) be a function defined in any
,
domain lying
in the
XY plane,
To each point (x, y) of this domain, there
corresponds a value of z. Through this point,
we draw the line perpendicular to XY plane
qual in length to z, so that we arrive at another
point P denoted as (x, y, z), lying on one or the
other side of the plane according as z in positive
or negative.
Fi g 55
.
Thus to each point of the domain in the XY plane there corresponds a point P. The aggregate of the points P determines a surface
which is said to represent the function geometrically.
www.EngineeringEBooksPdf.com
198
first
DIFFERENTIAL CALCULUS
10 71.
order.
Geometrical interpretation of partial derivatives of the
Let
z=f(x
We have seen
9
y).
...(/>
that the functional equation (/) represents a surface
We now seek the geometrical interpretation of the
geometrically
partial derivatives
.
The point P[a,
b.
f(a, b)]
corresponds to the values
dent variables
If a
A
#,
on the
surface*
b of the indepen-
x, y.
variable
P
from
starting
point,
position on the surface such that
changes its
remains constantly equal 1o b
y
then
it
is
r
y
clear
that the locus of the point is the curve of intersection of the surface and the plane y=b.
Fig. 56
On
this curve
x and
z vary according to the relation
*=/(*,*).
T
--
the ordinary derivative of/(x, b) w.r. to x for
is
)
3* Aa>
(C5
&)
x=a.
Hence, we see that
--
is
(
\OX
)
J(a>
the tangent
of the angle which
6)
the tangent to the curve, in which the plane y = b parallel to the
plane cuts the surface at P[a, bj(a, b)], makes with X-axis.
/
Similarly,
it
may
be seen that
ZX
)T V
is
-
(
)
^ oy /( a
>
the
tangent of the-
ft)
angle which the tangent to the curve of intersection of the surface and
the plane x=a makes with Y-axis.
Homogeneous Functions.
10-8.
a homogeneous function of order
n. if
Thus
in
x and y
is
a homogeneous function of order
is
equal to
n.
Ordinarily, /(x, y) is said to be
the degree of each of its terms
n.
This definition of homogeneity applies to polynomial functions
To enlarge the concept of homogeneity so as to bring even
only.
transcendental functions within its scope, we say that z is a homogeneous function of order or degree n, if it is expressible as
xf(ylx).
www.EngineeringEBooksPdf.com
199
PARTIAL DIFFERENTIATION
The polynomial function
is
(/)
which can be written as
a homogeneous function of order n according to the
The functions
new
definition
also.
are homogeneous according to the second definition only.
n
degree of x sin (yjx) is n. Also
'
L
so that
it is
of degree
Here the
1+
* J
J.
10 81. Euler's theorem on Homogeneous Functions.
7/z be a homogeneous function of.x, y of order
n, then
x
We
have
Thus
Cor.
// z
y
a homogeneous function of x, y of degree n,
Differentiating
/
^"r
^T
?-!
-i,
www.EngineeringEBooksPdf.com
...(i)
200
DIFFERENTIAL CALCULUS
x and y
partially with respect to
separately,
we obtain
z
d z
32
'" (
9*'
'
8z
Multiplying
(2), (3)
by
y respectively and adding, we obtain
x,
8
where we have employed
and
(1)
*
and assumed the equality of
Examples
I-
//
-1
2
^
v
_
2 f
-1
*
yv
^v
'
prove that
2
_9
"
3x3^
Wa
=
x
l~ y
x 2 +>' \2
-
/955; P.f/.)7
(D.t/.
v
have
rtAtan-1l
---- 2v
-1
_
dxdy
*
= 12.
//
www.EngineeringEBooksPdf.com
PARTIAL DIFFERENTIATION
show
that
3
9
-i^
We
201
"-+
_|_
a"^
-
(D.U. 1952
;
P.U. 1949)
have
Similarly or
by symmetry
2
3 w
,
---=:
f
J
Adding, we obtain the result as given.
3.
//
prove that
=sm
.
d*
Here u
is
not a homogeneous function.
z=tan
so that z
is
ft
2u.
dy
u= -
xy
We, however,
l(ylx)
write
'
a homogeneous function of x, y of order
2.
%7
>?
...(i)
dy
But
--*-
'
Substituting in
(1),
we obtain
sec 2 w
-
)=2z=2 tan
d* J
x
3w
+y
ax
,
3w
-<
W~
2 sin w
cos
cos
9
w,
M=
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
202
Exercises
Find the
1.
first
order partial derivatives of
e ax
l
(/)
(Hi)
tanr (x+y).
logU
2
(//)
sin by.
2
).
-j->>
Find the second order partial derivatives of
2.
X
/
V
/
.
r^
.
Verify that
3.
2
a
//
a*a>
when
u
is
(/)
(Hi)
4.
sin- 1
Xy
/;;x
-
y
log (y sin
x+x sin
y
Find the value of
1
a2
when
5.
Verify Euler's theorem for
liii)
(v)
z^sin- 1
xy
y
+ taxri- x
-
(iv)
.
z=jc
log
~.
*
2=
rj7l'
6.
If u=f(yjx),
show
that
*!"_
1.
If
(D.[/. 7950
;
P.U. 1954)
t
dz
If z=-tan
.
dZ
-j-^-r.
(y+ax)+(y-ax)'*
J!^
>.
.
z=xyf(x/y) show that
x~~^
8.
+J 8_ =0
^J!?.
-
=2z.
,
\
p rj
^.L/./
i
find the value of
.
(P.U. 1945)
1
If r=tapr (y/x), verify that
1-^-4-1^ -0,
www.EngineeringEBooksPdf.com
(AC/.)
205
EXERCISES
r
10.
If
z(x+y)=x
2
y
2
show
,
V
J
*L
3y
11.
11
12.
t
that
=
1
lfz=3xy-ya -\-(y -2x) i
Tr
If w =log
4
(
\*
-
1
dx
.
A
If
( fl )
(A) If
w-sin- 1
^
w-sia- 1
x
show
,
'
^^"^
(AlhLatcd)
show
=1.
,
that
*
3w
y ^
9x^9^
A:
>>
-f
\
J
,
-^, prove that ^ ^- + 7
i
9Z-
dy
verify that
,
2
13.
-
az-
4
=tan
w.
(P.t/.
1954)
that
^x+->!y
/
14.
^^
3 A:
If
z=f(x
a
a*
15.
V9^
_9 z\
2
If w=/(cix -f
2
2
9y9^y
2/7^+^
2
If
9=t
n
9^
e~ r2/4/
i
r
18.
If
-=/(r)
2
^^^
\9y
*
2
v-^(^
),
V,
x
V .Jr.V
9^
17.
'
2
ltz = (x-{-y) + (x+y)<9(ylx), prove that
/9 2 z
16.
(P.U. 7955)
m
prove that
ay),
ay)-\-<f>(x
-I
.=
9y
9*
2
-f-2/?xy-h^
),
prove that
\
-f
9^ x
f
V
find the value of n
,
2
(M.r.)
which
will
.dY^ao^^ao
/
V
8r
9r
9^
where r= VU 2 -f j/2 ), prove that
&+C;-^m
19.
If
=log
UHy-hz
f a
2
),
If
K=r m where
21 .
If
w = tan- 1
[Refer Ex.
3, p.
r
_
=
2
- y-
=x2 -Fy
,
9
2
w
dxdy'
9zax
3>az
20.
prove that
2
2
+ z 2 show that
,
find
201]
www.EngineeringEBooksPdf.com
make
(M
.
r<)
204
DIFFERENTIAL CALCULUS
Differentiating
(2), w.r.
to
x and
~
2
y,
we
=2
<>
2
*
sec w tan w
?
As
z is a
w
9*
^
2
*
-=2
w
9;>
A
2 sec w tan
get
/0-~
(
2
^sec w
,
-
Y +sec
,
8*
2
;^.
2
}
\9^ /
a^
homogeneous function of jc, y of order
2,
we
have, by Cor.
10*81
P. 199.
Making substitution, we obtain
Using
(3) p.
201,
we
obtain
/ ^ ]=2 tan
22.
-2 sec*
tan
.
sin
2
2u.
If
w~sin~
show
that
10 82.
introduce the
that
notation,
we
x=r cos
a<jid
we
A
new Notation. To
consider a particular case.
Suppose
Choice of independent variables.
new
6,
y=r
sin 6
;
.
.
(0
are required to find 9x/9r.
Before beginning partial differentiation, we have to ask ourselves a question.
"What are the independent variables ?" Hitherto
the notation has always been such as to suggest readily what the
independent variables are and no ambiguity was possible.
connected
In the present case we have four variables x, y, r,
relations so that any one of them may be expressed in terms
by two
www.EngineeringEBooksPdf.com
PARTIAL DIFFERENTIATION
Thus, x
of two of the remaining three.
(a) r, e
;
or (b)
r,
y
may
;
or
205
be expressed in terms of
(c) 0, y.
In case (c), 3.x/9r has no meaning. In cases (a) and (&), where
9x/gr has a meaning there is no reason to suppose that the two values
of (9x/3r) as determined from them, where we regard
and>> conSome modification of the notation is
stants respectively, are equal.
therefore necessary to distinguish between these two values.
For the sake of distinction, these two values are respectively
denoted as
Thus
means
L
r,
0?
Jv
-i.
rax
J0,
Ldr
of x w.
the partial derivative
r.
to r
when
are the independent variables.
From x
To
r
cos
,
find
Ldrjy
From
(/),
we have
.we have
1.
If
*
*!
>'
fl
x
in
terms of
r
and
y.
we obtain
r2
Ex.
to express
rcos
r
8
=x +y
2
z
so that
5, >'=' sin 9, find
*!
L ar Je'
r^-i
L'w
V
r
L
x= \?(r
z
-y*).
the values of
9x
a<-
i
Jfl
1
r^
L'w
(P.t/.
Ex. 2. v is the volume, s the curved surface, h, the height and
radius of the circular base of a right circular cylinder, show that
Ex.
are
its
r
the
Explain the meanings of the partial differential co-efficients
x, y are the rectangular cartesian co-ordinates of a point and
polar co-ordinates. Prove that
3.
and ar/Jx where
r,
1938)
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
206
Ex.
4.
Prove that
)
-&r
to
- ~
0*(logr)
8
/
--
1
-F cos
2
>
\vhere
x=rco5
0,
yr sin
0.
For the following developments,
10-9.
>
10 91.
Theorem on Total
will
it
f( x y) possesses continuous partial derivative
domain of definition of the function.
w.
Differentials.
r.
We
be assumed that
x and y in the
to
consider a
func-
tion
z=f(*, y)..(')
be
two
so
that
Sx 8y are
any
(x, y), (x-\-8x. y+8y)
points
the changes in tlie independent vnriablcs x B,ruly. Lst Sz be the
consequent change in z.
Let
We
t
have
From
(/)
and
z+Zz=f(x \-Sx,y+8y).
we get
(//'),
...(//)
t-ix,y)~f(x
so that
y)]
...(Hi)
we have subtracted and added
Here the change Sz has been cxpresse 1 as the sum of two
which we shall apply Lag range's mean v.ilue
differences, to eaoh of
theorem.
We regard f(x-\-8x, y) as a function of y only
supposed constant, so that by the mean valu? theorem,
;
x \-x being
We write
=
...(/v)
fy(x+&x, y+OW-f^x, ^} e a
so that e 2 depends on $x, Sy, aid because of the assumed continuity
offy (x y) tends to zero as x and Sy both tend to 0.
t
Again we regard f(x, y) as a function of x only, y feeing sup*posed constant, so that by the mean value theorem, we have
f(x+8x, y)-f(x, y)=Sxft (x \-Ojx,
We
y).
write
Ux + OJx.yl-Ux.y),**!
...(v)
s, depends
upon 8x and, because of the assumed ontinuity
to
tends
as 8x tends to 0.
off
y),
tso x (x
that
t
www.EngineeringEBooksPdf.com
EXAMPLES
Prom
(i/7), (iv), (v),
we
get
,
V
207
y)+Byfv
(x,
----A----
---y---
v
;
;
in r consists of two points as marked.
the differential of z and is denoted by dz.
Thus the change Sz
these the
first is called
Of
Thus
9
dz=^~8x +
-S>>.
dx
...(v/)
v
dy
Let
z
=x
so that
dx=dz=l 8x^8x.
.
by taking z=y, we show that
8y=dy.
Similarly,
Thus
(v/)
takes the form
dz=-^dx n+ a/
ax
dy.
ay
It should be carefully noted that the differentials dx and dy of
the independent variables x and y are the actual changes 8x and &y, but
the differential dz of the dependent variable z is not the same as the
change Sz it being the principal part of the increment Sz.
;
Approximate Calculations. From above we see that the
approximate change dz in z corresponding to the small changes 8x and
&y in x, y is
Cor.
dz
8r
dx ^
+
dx
2y
which has been denoted by dz.
</>>,
Examples
in the area of an ellipse when an
per cent, is made in measuring the major and minor axes.
If a, b, A denote semi-major axis, semi-minor axis and area of
an ellipse respectively, we have the relation
1.
error of
Find the percentage error
+
1
-._
=7rV
TT
T
Here
^ =7m
.
dATrbda+Tradb.
Since
we
are given that
j
a
JL
www.EngineeringEBooksPdf.com
208
DIFFERENTIAL CALCULI! S
-Too
dA
+ ioo ""loo
'
A = 2.
Therefore the percentage of error in
2,
The sides of an acute-angled triangle are
that the increment in
due to small increments in a,
A
measured. Prove
b c is given by the
y
equation
be
.
sin
A
S/l
.
=a(cos
CS&-f cos J3&c$a).
Supposing that the limits of error in the length of any side are
M per cent, where p is small, prove that the limits of error in A are
approximately
1
2
5(fjLa
jbc sin
A
)
From elementary Trigonometry, we know
A~
2 cos
Here
A is
.*.
9 sin
-2
(M. T. >
degrees.
that
,
be
a function of three variables,
a, b, c.
A A 2b*c-c(b*+c*-a*)
--->--------- ' db
AA dA=*~
'
.
~ ----------
a*)'
--------
i
2a
,
^c
b'c*
or
-2O
sin A
^
2
JA
<M=
.
2ab cos
.-.
Tho
be sin
A dA~
C
a(cos
.
(7
.
&/i/100.
is
2 c# cos
c/6+cos
B dcdd).
.
triangle
c<*/100,
acute-angled,
therefore cos
dA are
JB,
C
cos
are
Therefore the limits of the error
C+c
cos ff-f a)
0(6 cos
6c sin"^
^
a(
^
100
180
'
100
1
15
A
*
'
(
Cc cosj8a)
Bc"s5tt
u
or
^
6cos
'TOO'
sin
or
,
limits of the errors d&, Jc, da are
As the
positive.
222
2
IT
At
)
.--A
t/c
-,
sin A.
degrees approximately.
www.EngineeringEBooksPdf.com
M
'
100
PABTIAL DIFFERENTIATION
209
Exercises
Find the percentage error in calculating the area of a rectangle when
error of 2 per cent is made in measuring its sides.
1.
an
Show
that the error in calculating the time period of a pendulum at
zero if an error of +/* per cent be made in measuring its length
gravity at the place.
2.
anyplace
and
is
In a triangle ABC, measurements are taken of the side c and angles
is calculated from these measurements.
If Ac, A/4, A/*
are the small errors in these measurements, show that the error Atf in a is
given by
3.
A,
B
and length a
is an acute-angled triangle with fixed base EC.
If Sb, Sc, 8A
small increments in 6, c, A and 5 respectively, the vertex A is given
small displacement S* parallel to EC, prove that
4.
ABC.
and SB are
a
(i)
c8b+b8c+bc
cot
/I
S/l-0.
(//)
cSB-hsin BSjt=0.
(M.T.)
a triangle whose sides are a, b, c is A. Prove that the
error corresponding to errors 8a, 8b, 8c in the sides is approximately given by
5.
The area of
2 A S A = S 8p - s8q - abcSs,
2
2s=a+b + c,
Avhere
6.
The work
distance s in time
/
is
2/?=0
that
2
-f
6
2
+c
2
,
3?
must be done
to propel a ship of displacement
D
for
a
proportional to
rZ>f/ 2
.
Find approximately the percentage increase of work necessary when the displacement is increased by 1%, the time diminished by 1%, and the distance
diminished by 3%.
7. The height // and the semi-vertical angle a of a cone are measured, and
from them A, the total area of the cone including the base, is calculated. If h
and a are in error by small quantities S/z and 8* respectively, find the corresponding error in the area. Show further that if a=n/6, an error of -fl per cent
0'33 degree in a.
in h will be approximately compensated by an error of
(M.T.)
10*92.
Composite
functions.
Let
-=A*.
and
y)
0")
;
let
so that x, y are themselves functions of a third variable t.
The functional equations, (/), (//), (///) are said to define z as a
composite function of t.
Again, let
x=^(ii
f
y^(u
9
...(h)
t>,),
v)
;
,..(*;)
so that x y are functions of the variables u, v. Here the functional
zas a function of w, v, which is called
equations (i), (/v), (v) define
a composite function of u and v.
t
www.EngineeringEBooksPdf.com
210
DIFFERENTIAL CALCULUS
Differentiation of composite functions.
10*93.
Let
***f(x>y)*
possess continuous partial derivatives and
*
let
possess continuous derivatives.
Then
Jz =
dF
'Let
in x, y, z
f-fSf be
f,
dx
gz
dt
ax~
dy
gz
+ 8y~
'
any two values.
Let
consequent to the change St in
'
'
dt
be the change*
Sx, Sy, Sz
t.
We have
y+8y)-f(x, y)
,
+ [f(x,y+*y)-f(x, y )1
As
we apply Lagrange's mean value theorem
in
10'91, p. 207,
to the two differences on the right,
and obtain
Let 8f->0 so that ?x and 8y-*Q.
Because of the continuity of partial derivatives, we have
Urn
(
fJx+Ol
lim
Hence, in the
W
t
fi
limit, (/)
dz
becomes
_
"~
dt
Cor
Bx y+Sy)=fj(x,y)=-&--,
Sx, 8y)->(0, 0)
1.
dx
9z
'
dt
9x
+
8z
dy
9y
dt
'
Let
z=/(^,
possess continuous
first
y),
order partial derivatives w.r. to x, y.
Let
X=>(,
y=$(u,
possess continuous
first
,),
v),
order partial derivatives.
www.EngineeringEBooksPdf.com
'
'
*' '
211
To obtain 3z/gw, we regard v as a constant so that x and y may
be supposed to be functions of u only. Then, by the above theorem,
we have
az
It
may
similarly be
a*
az
a*
.
ay
shown that
_az __ az
~av ~~ax
ax
a_z^
av
ay
*
j?y
*
av"'
Examples
1.
Find dzjdt when
z=xy 2 +x*y, x=at 2 y
,
Verify by
direct substitution.
Now
-.
.
Substituting these values in
(//),
10*93, p. 210,
*L = (yZ+2xy) 2at+(2xy+x*)2a
Again
~
Hence the
2.
z
/
verification.
a function ofx andy.
Prove that
if
then
We look
upon z as a composite function of u,
.-
.
'
8j>
9"
"
3V
www.EngineeringEBooksPdf.com
v.
we
get
212
DIFFERENTIAL CALCULUS
Subtracting,
we
du ~"av
get
^a
---.
9*
3.
9>>
JfHf^yz, zx, xy)
;
prove that,
^
^c
z
Let
u=y
vzx, w=xy,
z,
so that
H=f(u,
We
have expressed
v,
w).
H as a composite
function of x,
>>,
2.
'
'
9*
Similarly
__
'
'
'dy
9M
'
3"
Adding, we get the result.
4.
H
th(it
is
a homogeneous function of x, y, z of ordern
;
prove
v
[This
is
Euler's theorem for a homogeneous
independent variables.]
H=xnf
f
We
,
function of three
have
|j=x/\w,
v)
where yjx-u, z 'x=v.
www.EngineeringEBooksPdf.com
(
IMPLICIT FUNCTIONS
213
But
8w
=
~~ - JL
a*
;c*~
?
>
v
^ _
z
'
~~;c a
a*
Hence
= njt-'/(,
/V
0*
v)'
'
'
~ x* -
8
8
+z -
( v
V '9W
0V /
Again
dH^xnfdf
a^
8", ?/
+ 8v~
8.y
'
\8
?/,
for
aw
?
'
v >
aj/
8B
1
a>>
A;
.
-3J!..o.
ay
Similarly
3-
two
10*94.
Let /(jc, _y)
Implicit functions.
variables.
Ordinarily, we say that, since
f(x,
we can
an
be any function of
on solving the equation
y)=o
...(0
obtain y as a function of x, the equation
(/)
defines
y as
implicit function of x.
difficulty here. Without investigation
that corresponding to each value of A', the equation (/)
must determine one and only one value of y so that the equation (/)
always determines, y, as a function of x and in fact, this is not the
The investigation of the conditions under which
case, in general.
the equation (/) does define y as a function of x is not, however,
within the scope of this book.
There arises a theoretical
we cannot say
Assuming that the conditions under which the equation (/) dey as a derivable function of x are satisfied, we shall now obtain
the values ofdy/dxandd 2yldx 2 in terms of the partial derivatives 3//3X,
2
2
2
3 //ay, 3 //8*
d fld*dy, d*f/dy* of / w.r. to x and y.
fines
'
,
Now, f(x y) is a function of two variables x, y and y is again
a function of x so that we may regard f(x, y) as a composite function
of x. Its derivative with respect to x is
9
dx
3/
'
3x
~dx
df
+ dy~
a/
dy
'
dx
*'
ax
dy
df
+9*
'
dx
'
Also/(x, y) considered as a function of x alone,
equal to 0.
Therefore
derivative w.r. to
its
-
x
is 0.
-
www.EngineeringEBooksPdf.com
is
identically
214
DIFFERENTIAL CALCULUS
__
,_.
/v
dfldy
Differentiating again w.r. tox, regarding 9//9X
composite functions of x, we get
_
'
~dx
_
and 3//8y as
'
dx\dxdydy*
>dy
dy
dy
a.y
2V
dx
8*A
>3
3
Hence
-__
~
dx
ftf
'
and
_._
.^__
Without making use of
follows
10'93,
we may obtain dyjdx
...
also
as
:
Now
f(x>y)=0Let Sx be the increment in x and Sy the consequent increment
in y, so that
f(x+Sx,
.or
or
,
Let 8x ->
y+Sy)
fy (x,
Sx
0.
Example
Prove
/fart
i//
2
3flrx
+x3 =0, then
www.EngineeringEBooksPdf.com
2J6
EXERCISES
f(x, >>)=>>
-3ax2 +Jts =0.
3
Substituting these values in
on
(Hi),
p. 214,
we
get
_
dx*
27>>
- a)(3ajc2 (x
_
y
Thus
j
.
5
<7r, directly, differentiating the
given relation w.r. to x,
(2a2x)y*2y(2axx 2 )(2ax-x 2 )ty
as before.
Exercises
1
1.
Ifii=Jt- y ,JC=2r-35 + 4,
2.
If
3.
(
z=(cos y)/x and
i-
If
-,
-
,
cosv"
y=-r+85-5,
x=w 2 - v, y.=^ v find
sin*
,
v=
-
smy
If
5.
Find dy/dx in the following cases
(/)
x
;
sin (x-^)-(x-|-y)=0.
(j/0 (cos x)f -(sin >^)*=0.
(v)
(tan x)* +y
eoi
z)=0
6.
If F(x, y,
"7.
If 2=xyf(vjx)
,
find v
ar/
/
x=tan (2r-5 2 ), y-eot(r 2j),
4.
Ji=(x+ y)l(l-*y)
find
.
(//)
(i v)
y
x
=sin
(P.U. 1955)
find 0z/ax, 3z/ay.
is
x.
x* ^y*.
x =a.
and z
find
:
a constant, show that
www.EngineeringEBooksPdf.com
216
DIFFERENTIAL CALCULUS
8.
y)=Q,
If/(jc,
<p(y,
z)=0, show that
9
?/
If
9.
.
JH(l-yH^(l-*
If u
and
v
8
1*
?/..
show
)=fl,
(/>./.)>
.
that
a
-
y?=2
dX
10.
4L=
?,.
(P.U. 1935\
-
.
_^
,j
3
-J
are functions of x and
xu + e~
z'
sin w,
>>
defined by
v
yv-\-e~ cos
w,
prove that
d^ = dv
If
11.
,4,
C
B,
(P.U. 1936)
dx
dy
are the angles of a triangle such that
sin
2
/1
2
+ sin B]
sin
2
C=constant,
prove that
dA^
C-lan B
tan
p
,
Ifax 2 -} 2hxy + by* -\-2gx-\-2fytc-Q, prove that
12.
......
"dx*
Show
13.
where
I
//*-!-
that at the point of the surface
x=yz,
r
=
ff~z
"" 1
"I"
X log
tfX
14.
(/)
(Hi)
15.
Find dy/dx 2
in the following cases
x *+y*=laxy.
Jc
6
+^ 5 =5fl 8
If AX,
(P.U.)
Jc.y.
=_
d 2x
/y
'
dy
fx
(//)
x 4 -i y*=
(iv)
jc
5
-!
y
5
=
/(x,
>>)=.xM y
dy
^_S^fv}*-2fa
2
(/
.
Given that
dx 2
17.
:
>^)=0 and/.T ^0, prove that
dx
16.
.
J
L
If
is
3
3ax>>:=0,
show
that
_
'
dy
2
a homogeneous function of the nth degree in
and if
(x, y> z)
where X, Y, Z are the first differential co-efficients of, w, with respect to x, y,
respectively, prove that
.
www.EngineeringEBooksPdf.com
r
1950)
217
PARTIAL DIFFERENTIATION
APPENDIX
EQUALITY OF REPEATED DERIVATIVES
It has been seen that the two1.
Equality of fxy and fyx
are generally equal. They
derivatives
second
order
repeated
partial
ate not, however, always equal as is shown below by considering twoexamples. It is easy to see & priori also whyfyx (a, b) may bo diffe-
A.
rent
.
from/^
We
(a, b).
have
-Iim
A ->
/
i
and
t,
/
/y (0, &)=
i-
Inn
k -*
/K
yv
f(a,b)
b-\-k)'
--
,
k
Jim,
li
>0
A-
=
r
Iim
/7
It
_>
similarly bo
may
A
r
Inn
A ->
<
,
7
say.
/7/c
shown that
->
Iim
->
//fc
/z
Thus wo sec that/ (a, ft) and/^ffl, /;) are repeated limits of the
same expression taken in different orders. Also the two repeated
limits may not be equal, as, for example
rj!/
r
lirn
v
Jim
^
fc
.
j
hk
and
A
Iim
Iim -.----_> o/j-^0 n
+K
=
=
v
Iim
''
A->0
ll
i
1,
_k
Iim
A:->0
7
^
=
Examples
1
.
x,
Prove that fxy ^ fyx at the origin for the funct ion
y are not simultaneously zero andf(Q, 0)=0. (D.U. Hons. 1954)
We
have/_(0, 0)= Iim
A->0
'--
^
'-J v^
L
*
.
www.EngineeringEBooksPdf.com
9
-'
...(1)
218
DIFFERENTIAL CALCULUS
Also
/,(0,
and
/,(
Thus from
Again
0)=
lim
=
lim
MH
/(0.
0+k)-fM
--
... (2 )
= m
li
K
=0..
lim
and
(1), (2)
(3)
/((>,
OH
lim
/,,(0 (
0)-
lim
fl( 0, 0)=
lim
t- =
l.
-
...,4)
Also
**
and
/fin
x (0,
From
(4), (5)
to
/c)=
and
/.>,
Thus
/JO,
==
"
h^-0
OH *->o
I
.
"
=0,
.
.(5)
-
lim
lim
(6),
lim
A->0
*=
K
0)=!^- 1 =/^0, 0),
2.
V
2
1 -/(x, .y)=* tan- x
and
We
is
2
>^
tan- 1
^/
zero elsewhere.
have
www.EngineeringEBooksPdf.com
(B.U. 1953)
219
PARTIAL DIFFERENTIATION
= hm
..
r
\
,
h
fc-oL
=A
.
1
/ta,n- l klh\
'
(
\
.
.
h 1
,
,
k tan- 1
)
,,
J
klh
0=/J, for [tan-
/c
1
*/*]
->
1
as
*
J
-* 0.
A:->0
*
We may
similarly
Hence the
show that
result.
A. 2. Equality of fxy and f^. Theorem. Ifz=f(x, y) possesses
continuous second order partial derivatives d^ld^dy and d*?ldydx, then
Consider the expression
For the sake of brevity, we write
t(x)=f(x,y+k)-f(x,y),
...(1)
so that
<?(h,
By
Lagrange's
k)=$(x+h)-t(x).
...(2)
mean value theorem,
Also
...(5)
ji, y)].
Again applying the mean value theorem to the right side of
(5),
we obtain
)-
0<0 t <l.
Thus
Again considering
proceeding as before,
/r(^)=/(A:+/j, y)
f(x, y) instead of
we may prove that
www.EngineeringEBooksPdf.com
</t(x)
and
DIFFERENTIAL CALCULUS
220
Let h and k -> 0. Then, because of the assumed continuity
of the Partial Derivatives, we obtain
fvx (x,y)=fz V (x,y).
A.
3.
Taylor's theorem for a function of two variables.
V /(*
y) possesses continuous partial derivatives of the nth
any neighbourhood of a point (a, V) and if (a+h, b+k) be
any point of this neighbourhood, then there exists a positive number 9
which is less than J, such that
order
in
f(a+h, b+k)=f(a,
b)
+
+k A/(<7,
h
b)
Lemma.
We
write
-=/(*, y)
;
and
so that z
is
a function of
t.
Now, we have, by
10'93, p. 210,
dz
8z
dx_
*
di'^'Wx
~dt
dz
Now, we agree
dz
dy
'
"di
dy
dz
.
.
d
8z
,
to write
/
dz
in the
+
,
*
form of the operator
hL
operating on the operand
1
+k
,
ax
z.
This equality shows that the operators
d
are equivalent.
www.EngineeringEBooksPdf.com
d
dz
TWO VARIABLES
TAYLOR'S THEOREM FOR
221
Employing the equivalence of these two operators, we obtain
d / dz \
d*z__
dt^ dt \dt )
d
8
d \
'
,
z
8y)
8
lx +k
,
If,
7
8
now, we agree to write
we have
d*z
Continuing in this manner, we see that
dn z (
8
,
8 \w
*
'
where the operator
implies the repeated application of the operator
/i
Thus we have arrived at the following
Jfz=f(x, y) and x=a+ht, y = b-\-kt, where
times.
result
a, b,
:
//,
k, are cons-
tants, then
8
\n
z
'
By)
Proof of Taylor's Theorem.
We
write
f(x,y)=f(a+ht b+kt)=g(t)<
and apply Maclaurin's theorem to the function
9
variable
of the single
g(t)
t.
There exists a positive number 6 between
For
/
= !,
this
becomes
www.EngineeringEBooksPdf.com
and
1
(0
<
such that
6
<
1)
.
.(/)
DIFFERENTIAL CALCULUS
222
Now
)=/te+A, b+k).
f(0)-/te
b).
Also since
Substituting these values in
as stated.
(i),
we have the Taylor's theorem's
Exercises
1.
Show
>
2.
that
(*
By mathematical
indication or otherwise,
3 7
/ h. 9 +
+/t _3_\" -A" V
a>-/
3^
3^"
show
that
"
8 z^^ A-^ a^-^r
+
l
......
on,
n K
-"
4-"c.h
f..t
C,n -rkr
n-r
4-
3.
Show
(i)
that
sin
K
sin
^^ary
T[(**+ 3x^
2
)
cos 0a sin
sin
(W)
e*
sin
ex cos
0,y]
;
<
< K
6^A>+afrx,y
2
u
+|. e [(a*x* -3ab*xy ) sin
where u~a9x, v~b$y.
8
v-|- (3fl
.
www.EngineeringEBooksPdf.com
^ ^-6^) cos
a
],
MAXIMA AND MINIMA FOB TWO VARIABLES
MAXIMA AND MINIMA
A.
Maxima and Minima of a function of two variables.
Maximum value. f(a, b) is a maximum value of the function
4.
Def.
fix, y)> if there exists some neighbourhood of the point (a, b) such
for every point (a-\>h, 6+fc) of this neighbourhood,
Minimum
value,
f(a,b)>f(a+h,b+k).
/(a, b) is minimum value of the function
that
f(x, y) r
some neighbourhood of the point (a, b) such that for
every point (a+A, b+ k) of this neighbourhood
f(a, b)<f(a+h, b+k).
Extreme valuej /(a, b) is said to be an extreme value of f(x, y),
if there
if
it is
a
exists
maximum
A. 41.
or a
minimum
value.
The necessary conditions for f(a, b) to be an extreme
value offix, y) are that
fx (a b)=Q,fv (a,b)=0.
9
If f(a b) is an extreme value of the function /(;c, y) of two
variables x and y then, clearly, it is also an extreme value of the
and as such its derivative
function/(jc, b) of one variable x for
t
x=a
x
f
x (a, b) for
a must necessarily be zero.
Similarly
we may show
that
/,(*,
Note
As
1.
*)=0.
in the case of single variable, the conditions obtained above
not sufficient. For example, if f(x,y)Q when jc=0 or
are necessary and
and/(x, v) = l elsewhere, then
but/(0, 0)
is
Note
2.
jc=<i,
y=b
/*(0,0)-0,/V(0,0)=:0,
not an extreme value.
Stationary value. A function f(x, y) is said to be stationary for
or f(a, b) is said to be a stationary value of /(jc, y) if
Thus every extreme value
be
y=0
is
a stationary value but the converse may not
true.
A. 42.
Sufficient conditions.
To show
that
/.(M)=O //a,&HO,
f
and
f*(a.
b)=AJxy (a,
b)=B,tf(a,
is
a max. value
ifACB*>0 and A<0,
a min. value
ifAC-B*>0
b)=C
then
(i)
(ii)
(iii)
f(a b)
t
f(a.
b) is
f(a> b) is not
an extreme value if
AC-B*<O
(iv) the case
It
may
is
and A>0,
y
doubtful and needs further
be noticed that
A^O
consideration, if
if
www.EngineeringEBooksPdf.com
224
DIFFERENTIAL CALCULUS
Taylor's Theorem with remainder after three terms, we.
By
obtain
fta+h. b+k)=f(a,
or
&)+(
~ +k
h
f(a+h, b+k)-f(a, b)=hfx
(, b)+2hkfacv(a,
(a,
b)+kf,
b)+k%*
(a, b)
(a, b)]
where u=a+0h. o=b+6k.
Now
fx (a,
Also,
where
p is
We
6)=0,/,(a, 6)=0.
we have written
of the third degree in h and
fc.
assume that for sufficiently small values
of
h,
k,
the
sign of
is
the same as that of
Case
1.
Let
ACB >0.
2
In this case neither
A
nor
C can
be zero.
We
write
=-r
Since
is
ACB*
is
positive,
we
[(Ah+Bk)*+(AC-B*)k*].
see that
always positive except when
Ah+Bk=Q,
i.e.,
when
/*=0, fc=0,
Thus we
which
is
when
it is
fc=0.
zero.
see that Ah?-\-2Bhk-\-Ck 2 always retains the
same sign
that of A.
Thus f(a>b)
maximum
or
is
minimum
an extreme value
according as
A
in tbja case
a^J ,will be
or
negative
positive.
t
is
www.EngineeringEBooksPdf.com
a
EXAMPLES
Case
225
,
II.
Let
AC-B*<0.
Firstly,
We
we suppose that A^O.
write
ACB
2
Since
is negative, \ve see that this expression lakes
values with different signs when fc=0 and when Ah-\-Bk= Q;
up
}
Thus in this case/(#, b) is not an extreme
The proof is similar when Cr^O,
In case
A=Q
as well as
value.
C=0, we have
so that the expression does assume values with different signs
accordingly f(a, b) is not an extreme value.
and
Case HI.
Let
AC-B*=0.
Suppose that
We
A^Q.
have
Here the expression becomes
zero,
when
so that the nature of the sign of
.
f(a+h,b+k)-f( a,b)
depends upon
If,
have
now,
the consideration of
A = Q then,
p.
The
case
is,
because of the condition
therefore, doubtful.
AC=B
2
,
we must
5=0.
so that the expression
case is again doubtful.
is
zero
when fc=0 whatever h may
Examples
1.
Find the extreme values of
We
write
xy(axy).
/(*, y)
We now
=xy(a-x-y)=axy-x*y-.\y*.
solve the equations fx
~Q and/y=0.
www.EngineeringEBooksPdf.com
be!
The
DIFFERENTIAL CALCULUS
226
Thus we have
2xy
x2
ay
ax
2
=0,
.y
These are equivalent to
so that
we have to consider the four
pairs of equations, viz.,
Solving these, we obtain the following pairs of values of x and
the function stationary
y which make
J"
:
(0, 0), (0, a), (a, 0), (J*. Jfl).
1
(0, 0),
^1=0,
5= a, C=0
is
Thus/(0, 0)
For
(0, a),
so that AC B* is negative.
not an extreme value of /(*, y).
A2a, B=
Thus/\0,
is
^r)
We may
a,
C=0 so that AC U2 is
negative.
also not
an extreme value of /(x,
>>).
show that
similarly
f(a, 0)
also not
is
an extreme
value of the function.
For
(\a, la),
A\<t, B-
\a,
C=
\a so that
AC
J3* is
positive.
Thus /(|0, i#) is an extreme value and will be a maximum or
a minimum according as, -4, is negative or positive, i.e., according as,
a, is positive or negative.
The extreme value /(^a, Ja)=^T^ 3
2.
JVmf
f/ie
2(x-^)
We
write
/(x,
fy*(x,
We now
.
extreme value of
a
-x -^4
4
(/>.[/.
.
Hons.)
j;)
y)=4-
solve the equations fx -=zQ
t
fy=Q which
y
are
...(1)
...(2)
Adding these, we obtain
-4(x3 +}")=()
or
(x+y)(x*- xy+y*) =0,
www.EngineeringEBooksPdf.com
EXAMPLES
227.
which shows that
x -\-y~Q
either
or
x
We
2
xy+y*=0.
have, thus, to consider the two pairs of equations,
--
and
viz.,
>
x
The equations
(3)
give the following pairs of solutions
(0, 0), (v/2,
The equations
(4)
:
-V2), (-V2, V2).
give only
as the real solution.
(0, 0)
For (0,0),
5= -4, C=4 so that ACB*^Q
A =4,
and
accordingly this case needs further examination.
For(V2,
X=
-V 2
20,
.. /(\/2,
value
as,
A,
)
5= -4, C=-20 so
is
\/2)
negative.
is
We may
that
4C B
an extreme value and
similarly see that/(
is,
a
is positive.
in fact, a
maximum
also a
maximum
as well as ,y=0 can
be disposed
y2,
\/2)
is
value.
Note.
The
case which arises
of by an elementary consideration as
when x=0,
follows:
Now/(0,0)=0.
For points, (x
9
0) along x-axis,
where .y=0, the value of the function
=2x-;c=;t'(2--;c
which
is
which
is
positive for points in the
8
)
neighbourhood of the
origin.
for points along the line, y=*x the value of the function
negative.
Again
2x 4
Thus in every neighbourhood of the point (0, 0) there are points where
function assumes positive values i.e., >/(o, 0) and there are points where the
function assumes negative values i.e., </(0, 0).
Hence /(O,
0) is not
an extreme value.
3.
Find the minimum value of
We
write
when
so that we have to find the minimum
variables x, y z which are connected
t
We
value of a function of three*
by a
single relation, viz.,
re-write this relation in the form
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
228
so that z has been expressed
We now
a function of x and y.
as
obtain
where,
ables
w,
has been expressed as a function of two
x and
We
independent vari-
y.
have
(*>
;>)=?*-
(p- a * - hy)
"*
>
<>b
u(x.
y)=2y-~c t[(p- ax-by).
Equating to zero these two
order
first
partial derivatives,
we
obt&jn
Again, we have
2
2^2
*
x
2
C'
2ab
a
-
t
Since
minimum
4C
jB
2
positive and,
is
for the values of
x and y
A is also positive, therefore, w, is
in question.
The minimum value
}
of, w, therefore, is
Exercises
1.
Examine the following functions
(/) ^+4x^+3**+ x.
//)
for extreme values:
(11 )
x*+xy+y*+ax+by.
(vi)
4).
(v/i)
(ix)
3x 2 ~^ 2 4-x a
2 sin (jc+2y)+3 cos (2x-y).
.
^
(iv)
(vi/i)
(
2x*y+x*~y*+2y.
z
-x-y).
(x) x*y (\
(B.U. 1952)
2.
Find
all
the stationary points of the function
examining whether they are maxima or minima.
www.EngineeringEBooksPdf.com
(D.U. Hens. 1952)
STATIONARY VALUES
3.
A.
229
Find the shortest distance between the lines
Stationary values under subsidiary conditions.
5.
To find
the
stationary values of
u=f(x,y, z,w).
where the four variables x, y,
r,
w
...(i)
are subjected to
the
two subsidiary
conditions
...(2)
...(3)
ing
Now, we can look upon the equations
any two of the four variables x, y, z, w
We may
two.
(2)
and
(3)
as determin-
in terms of the remaining
suppose, for the sake of definiteness, that (2) and (3)
determine z and w as functions of x, y. Thus u is a function of x,
V, z, w where z and w are functions of A: and y so that u is essentially
a function of two independent variables x and y. Therefore for
stationary values of, w, the two partial derivatives of, w, with respect
to x and y obtained after z and w have bo3n replaced by their values
in terms of x and y, are respective^ zero.
Equating to zero the partial derivatives of, w, with respect to
>
x and
y\
we obtain
.-0.
.-(4)
-0.
...(5)
+/.
Again, differentiating
v,
(2)
and
(3) partially
with respect to
and
.v
we obtain
"-=0.
...(7)
-<>.
...(8)
^-0.
'...(9)
9z
--
If, now, 3Z/3X, 3>v/gx, 9z/3j^, 3w/9>> be eliminated out of the six
equations (4) (9), we shall obtain two eliminants, say,
F,(x y z w)=0,
9
9
...(10)
9
...(11)
^t(^, y, z, >v)=0.
Then the four equations (2), (3), (10) and
values of the unknowns x, y, z and w for which u
www.EngineeringEBooksPdf.com
(11) determine the
is
stationary.
230
DIFFERENTIAL CALCULUS
Note. This method for determining stationary "values of a function
under subsidiary conditions outlined above is unsymmetrical in character
regard to its treatment of the variables involved. This defect has been remedied
by Lagrange who has given a symmetrical method based on the introduction of
m
certain undetermined multipliers.
This method
is
explained in the following
section.
Lagrange's method of undetermined multipliers. We
the
multiply
equations (6) and (8) of the preceding section by Aj and
A 2 respectively and add to (4), so that we obtain
A. 51.
*)
Again,
section
by
Xl
we multiply the equations (7) and
and A, and add to (5), so that we
(9)
-1-AalM
We, now, suppose
~=0.
...(12)
ox
of the preceding
~ =0.
...(13)
make
that A t and A 2 are determined so as to
...(14)
...(15)
With
this choice of the values of A x
and
A2
,
the equations (12)
and
(13) give
...(16)
The equations (2) and (3) of the preceding section along with
the equations (14), (15), (16), (17) determine values of A x A 2 and of
x, y, z and w, which render u stationary.
,
,
If,
now, we write
we
see that the equations (14) to (17) are obtained by equating to zero
the four partial derivatives of g, with respect to x, y> z and w so that
In practice,
all the four variables are being treated on a uniform basis.
therefore, the necessary equations are to be put
the auxiliary function g.
down by
setting
up
Note. The method outlined above is applicable to a function of any
number of variables which are subjected to any set of subsidiary conditions.
Examples
Find the lengths of the axes of the section of the
*+z*lc*=l, by the plane lx+my+nz^Q.
Let (x, y> z) be any point of the section. We write
1.
www.EngineeringEBooksPdf.com
ellipsoid
EXAMPLES
231
We
have to find the stationary values of
subjected to the two subsidiary conditions
jfrVa'
We
g(x, y,
x,
>>
r
2
,
where
/x=0.
1,
z are
x, y,
...
()
write
zM
Equating to zero the
and z we obtain
partial
derivatives of g(x, y, z) w.r. to
9
The equations
AI, A lf x, y,
z.
and
(/v)
(11), (///),
These values of x,
y, z,
r2 .
determine the stationary values of
eliminating
^^
x,
y and
z
from
(v)
will
when
determine the values of
substituted in
(/),
will
These operations amount to
(/), (11), (///), (iv)
and
(v).
that,
Wo multiply (///), (iv) and (v) by x, y, z respectively and add so
on making use of (/) and (//), we obtain
With
this value of 7^,
the equations
(//'/),
(iv)
and
(v)
can be re-
written as
2x
/
n
r^~
!-_-'.
1
c2
Multiplying these by
/
/,
2z
A"
Ai
m, n respectively and adding, we obtain
a
a quadratic in
8
stationary values of r
which
~
is
r*
and (Jetermines, as
.
www.EngineeringEBooksPdf.com
its
roots,
the two
232
DIFFERENTIAL CALCULUS
The geometrical considerations, now show that these two
2
stationary values of r are the squares of the semi-axes of the section,
in question.
,
2.
u
If
where
x
show
'
z
y
that the stationary value ofu,
is
given by
We write
g(x, y,
z)=flx'+&V+c'2 a +>
( I
+
y
f
Equating to zero the partial derivatives of
x,
y and
z,
I
-1
)-
g(x, y, z)
w
r,
to
we obtain
y-
,
=0,
-
2A
2
=0.
These give
or
ax=by~cz.
The equation
=1, determine
(/)
x,
.,,
(/)
along with the given subsidiary condition
y and
z.
Exercises
2
1.
Find the minimum value of
2.
xyz=a*.
Find the extreme value of xy when
3.
Find the greatest value of axby when
,v
a
-h^
fz 2 when
(w)
(MI)
(D.U. Hons. 1953)
x*+xy+y*^3k*.
Find the perpendicular distance of the point
4.
(a> b, c)
from the plane
b^the Lagrange's method of undetermined multipliers.
Which point of the sphere 2x*=l is at the maximum distance from
5.
tbepohU{2, 1,3)?
6.
Find the lengths of the axes of the conic
z
ax*+2hxy+by ~l.
7.
In a plane triangle, find the maximum value of
ens A COS B COS C.
'
8.
Find
the?
extttme Va lues of
www.EngineeringEBooksPdf.com
MISCELLANEOUS EXERCISES
233
subject to
Show that the maximum and minimum values of, r 2 where
2
r =a*x*+b 2y 2 + c 2 z 2 jc'+yH z ? =l and /x-f iny+wz=0
9.
,
,
are given by the equation
-.2
72
2
>7_^ + i,*,v + f .l>.-0.
10.
show
will
that
(D.C/./9J5)
two variables x and y are connected by the relation
ax*+by*=ab,
the maximum and the minimum values of the function
If
be the values of
given by the equation
4(0-a)(0 -b)~ah.
(D.I/,
//ww. 7957)
Miscellaneous Exercises
Find the points of continuity and discontinuity of the following
1.
functions
2.
:
Determine the points of continuity and discontinuity of the function
f(x) defined by
fin
where n
is
3.
l-/)-o, if r = i
(/-I, ifi</<l,
(B.U. 1)52)
any integer.
Determine the points of discontinuity of
tan
4.
and
Draw
X
f"
i
L
the graph of
find the points of discontinuity.
5.
Determine the points of discontinuity of
(0 [*] + [-*?.
/ 2
lim
(///)
(
n
/I
6.
Show
~>
oo
(')
tan^/
\
that the function fPU)
to J when * = ; to
three points of discontinuity which
whenO<x<J
7.
:
which
is
equal to
when 4<x<l
%xare
to
you
required
;
when x=-0
and to
find.
1
;
to
i
x
whenx--=l. has
(Patna)
Examine the continuity of the function
/(0)=0.
www.EngineeringEBooksPdf.com
(D.U. Hons. 1947)
234
DIFFERENTIAL CALCULUS
d0W4
Find
8.
a
where
Care
,4, J0,
relation
S 5/it
J/H
where,
/r,
C sin A+sin A
sin
B=h,
constant.
is
9.
C-f sin
the angles of a triangle and satisfy the
Find
V + ^0'(~r
\
in terms of
r,
/)'
when
r a ~a*cos 20.
10.
Itax*+2hxy+by***l, prove that
x
11.
Find
W
^ V i+
12.
show
If
/
1
cos fl+a
2
1
j,
that
*
13.
If ^
when iw=6,
If >>=cos
(w sin" 1 ^:), prove that
can be expanded in a series of ascending powers of
.v,
prove that
(M.T.)
14.
If
prove that
Assuming
that
,y
flo
can be expanded as a series
2
+ QiX+ a z x -\- ----
prove that
00=0, #!=! and for
2-4 6.... (2m)
-
=(^iivm
/
^ cosa
if e
cos (jc sin a) can be expanded
that the co-efficient of x n is cos wa/w !
15.
show
16.
(a)
Give the
first
(b)
powers of or,
(B.U.)
three non-vanishing terms in the expansion of
sin"" 1
Show
in ascending
(
J sin x).
that the expansion upto
x4 of x/sinh x
www.EngineeringEBooksPdf.com
is
MISCELLANEOUS EXERCISES
Use Taylor's theorem
17.
g=
is
satisfying
prove that the only f unction /(x)
to
-/(*) for
/(O)-
all x,
-
1
,
/'(0)= 0,
given by
/W=1 ~^!+4T ......
adinf
'
(D.U. Hons. 1952)
Prove that whatever the functions /and
18.
z**xf(x+y)+yg(x
(I)
g,
\ry)
satisfies the relation
z=xf(ylx)+g(ylx),
(//)
satisfies the relation
Given that
19.
a function of u and
z is
= jc
M
2
t?,
/-2xy,
r
while
= v,
prove that the equation
<*>) g+f*-y)j;-ois
equivalent to 9z/av^0.
20.
21.
If M
If
= (l
ii^sin' 1
,2
22.
t2
2xy F3 r^, prove that
W
,
J_7w
;
then
=
tv
_
4cos 2 w
Prove that
'
if
w
dxdy
is
a
dy*
homogeneous function of the wth degree
in x, y,
then
x .*?+ y
9w
3x+
ay
+,-
. nM
8?l
.
a^
Prove also that
/*
^2
^2
^2
(bF+a7
+
2
\
9* w
9 "
9 2u
i /
"+ 3 J*' ( 3x + 3? + 'a?-
^)(*>-
2
where
13.
show
that
3'/(r)
9 V(r)
3 V(r)
3?
www.EngineeringEBooksPdf.com
,
i
236
DIFFERENTIAL CALCULUS
24
^^ =
If
9*
dV
Ir
2
97
'
has a solution of the form
Ae~ a
prove
sin (\vt~bx).
that
H=Io
25.
(x
3
x*y-xy*), prove that
+y*
'
Show
26.
mean
1
to
that
value theorem
as,
/*,
27.
increases
=0
if
is
and /U)=log
(l
+ x)then
a continuous function of,
from
Show that
(/)
*<log
1
[1 /(I
to oo
A,
'0'
of the Lagrange's
which decreases steadily from
.
-*)!<*/( I
-.Y)
where
0<x<l.
,where-l<x<0.
28.
A
slight error
x is
made
measuring the semi-vertical angle of a
in
cone which circumscribes a sphere of radius R. Find the approximate error in
the calculated volume of the cone and show that, for a given Sac, the least value
of the error is
647TR3
V5 -* a
-
75-
29.
'
If the three sides a, b, c of a triangle are
angle A, due to given small error in the sides,
A
c/A=
,
A
da
--cot C
sin
r~ir-; ~.
sin B
sin
A
C
measured, the error
in
the
is
~,
db
a
.
cot
B
dc
c
-.
when
(i)
31.
a=3, 6=~5,
Findlim
c
=4
;
(ii)
a=3, 6=~4, c=2
;
~~-
^
32.
33.
</)
34.
Obtain
lim
Find
.7.,
lim tan(sin^-sin(tan^)
z
sin x "" x cos ^^ sm x
x >o
35.
.(//)lim
(///)
(D.U. Hans. 1955)
2
Exarr i le where the following functions are continuous for *=0.
www.EngineeringEBooksPdf.com
MISCELLANEOUS EXERCISES
237
^
cot2 x
(/) f(x) = (cos *)
36.
Having
,
(*?*0),/(0)-
sin v
n
+jc +BJC 3 )/x
(tanx-A:e
non-zero limit as x tends to zero, prove that n must be equal
'
maintains a
i
5
finite
a
-
(
37.
38.
from
2
-3;c-f 2)/(x
2
+ 3x+ 2).
Sketch the curve
*0 to
sin
indicating correctly the positions of
any, of the function.
39.
Indian Pol ice 1932)
Find the maximum values of the function
U
if
1/>U?.
given that the fraction
A:
==2rr,
Show
maxima and minima
that the function
4(jc-l)-
1
has a single maximum and a single minimum value and that the function does
not lie between its maximum and minimum. Draw the graph of the function.
40.
Find the maximum and the minimum values of (1 x)"e x
.
Show
to
1
and that
that e x
it
*= + <.
(1
steadily decreases as x increases from
only one minimum for values of x between x=l
+ x)j(l~x)
has one and
Prove that |(35 sin 4 * 40 sin 2*-}
and has also | as a maximum value.
41.
and
.J.
42.
Show that #(*)
the interval
<jc<n/2.
43.
sin
x tan
A:
(P. U. 1938)
in
value
between
ranges
8)
unity
log sec
x
is
positive and increasing in
(M.T.)
Find the maxima and the minima
of,
2 cos
x+
>', where
1 and (adbc)y*Q.
(B.U.)
x*axy* oPy = Q prove that y is maximum where 3xy+ 4o =0 and
minimum where * = 0.
(B.U.)
45.
Draw the curve r=a(2 cos 04-cos 39).
Show that the extreme
44.
a
=i
oo
and
2
If
9
values of the radius vector are 3a and
/3>/3.
The sum of the surfaces of a cube and a sphere is given.
when the sum of their volumes is least, the diameter of the sphere
46.
Show
is
that
equal to
the edge of the cube.
47.
surface
Given the volume of a
is least
right cone
;
required
its
dimensions when the
possible.
48. Prove that the volume of aright circular cylinder of greatest volume
which can be inscribed in a sphere, is \3/3 times that of the sphere.
49. An ellipse is inscribed in an isosceles triangle of height h and base
2k and having one axis lying along the perpendicular from the vertex of the
triangle to the base.
Show
that the
maximum
area of the ellipse
is
>|3nM:/9.
A
sector has to be cut from a circular sheet of metal so that the
remainder can be formed into a conical shaped vessel of maximum capacity.
Find the angle of the sector.
(M.U.)
50.
51. Find the greatest rectangle which can be described so as to have two
of its corners on the latus rectum and the other two on the portion of the curve
cut off by the latus rectum of the parabola.
52.
Find the greatest and
least values of (sin x)
sm *
www.EngineeringEBooksPdf.com
(B.U.)
CHAPTER XI
SOME IMPORTANT CURVES
The following chapters will be devoted to a discussion of
11-1.
such types of properties of curves as are best studied with the help
of Differential Calculus. It will, therefore, be useful if we acquaint
ourselves at this stage with some of the important curves which will
frequently occur.
11-2. Explicit Cartesian equations of Curves.
A few curves
whose equations are of this form have already been traced in Chapter
We now trace another very important curve
II.
,
which
known
is
as Catenary.
The following
trace
it
particulars about the
curve will enable us to
:
Since
(i)
cosh
on changing x to
x,
=s
we
"o-l
see that
y=c cosh
c
=c cosh (.
\
-
\
c J'
so that the two values of x which are equal in magnitude but
opposite in sign give rise to the same value of y.
Hence
(ii)
(111)
Thus, y
is
the curve
When x=0,
~j- =sinh
,
which
is
positive
monotonically increasing in
when x
[0, oo
is
positive.
).
=0, i.e., the slope of the tangent
-Jx \j
a
so that the tangent is parallel to x-axis at A (0, c).
Also,
I
x=0
symmetrical about y-axis.
y=c so that A (0, c) is a point on the curve.
is
\
"
is,
0, for
\
/
/
-^^
Hence if the point P(x, y) on the
curve starts moving from the
position
A (0, c) such that its abscissa increases,
then its ordinate also increases. Also
since y -> GO when x -^ oo the curve is
not closed.
,
The curve being symmetrical about
>>.axis, we have its shape as shown in the
adjoining figure.
Fig. 57
238
www.EngineeringEBooksPdf.com
SOME IMPORTANT CURVES
239
Note. In books on Statics it is shown that the curve in which a uniformally heavy fine chain hangs freely under gravity is a Catenary.
113.
We
Parametric Cartesian Equations of Curves.
two functions of t defined for some interval.
Let/(/), F(t) be
write
To each value of /, there corresponds a pair of numbers x, y as
determined from the equations (/), (i7). To this pair of numbers x, y
there corresponds a point in a plane on which a pair of rectangular
co-ordinate axes has been marked. Thus the two equations associate
The two equations (i) and
to each value of, t, a point in the plane.
(ii) then constitute the parametric equations of the curve determined
by the points (x, y) which arise for different values of the parameter t.
The point P on the curve corresponding to any particular value,
of the parameter is denoted as the point P (t) or simply 'f.
We assume that the reader is familiar with the standard parametric equations of a parabola and the Ellipse so that we may only
restate them here. Afterwards the parametric equations of a Cycloid,
Epicycloid and Hypocyoloid will be obtained from their geometrical
f,
definitions.
11*31.
Parabola.
The parametric equations
x=at a , y=2at
represent the parabola having its axis along x-axis and the tangent
at the vertex along >>-axis, and latus rectum equal to 4a.
The point P(t) describes the part ABO of
y,
the parabola in the direction of the arrow-head
as the parameter, f, continuously increases from
oo to 0.
For the vertex 0, f=0. Again, the point
of the parabola in the
describes the part
direction of the arrow-head as the parameter t
to oo
continuously increases from
OCD
.
11-32.
The
Ellipse.
The parametric equations
x=a
cos 0,
y=b
Fig. 58
sin
represent the ellipse whose axes lies
along x-axis and y-axis and are of
lengths 2#, 26, respectively.
The point P(0) describes the
parts AB, BA', A'B, B'A of the ellipse
in the direction of the arrow-head as
the parameter, 6, continuously increases in the intervals (0,
TT), (I?r, TT)
(TT,
ITT), (ftf, 2ir) respectively.
59.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
240
The Cycloid is the curve traced out by a point
11*33.
Cycloid
the circumference of a circle as it rolls without sliding along
marked on
a fixed straight
line.
Let the rolling circle start from the position in which the
line X'X.
generating point P coincides with some point O of the fixed
O
as origin and
the fixed line as *-axis. When the
Take the point
has rolled on to the
position shown in the figure, the
generating point has moved from O
rolling
/
Fig.
circle
-*X to P(x, y) so that
O/=arc PL
60
Let, 0, be the angle between
Let, a be the radius of the circle.
and C/so that it is the angle through which the radius drawn to
the general point has rotated while the circle rolls from the initial
have
Thus arc PI=a0.
to its present position.
CP
We
x=OL=OI-LI--=OI-PM=aO-a
sin
6)
;
= <j(l-cos 0).
Thus
x=a(0-sin
y=a(l cos
6))
6))
are the parametric equations of the Cycloid
;
being the parameter.
P
Note. While the circle makes one complete revolution, the point
describes one complete arch OVO of the Cycloid, so that
increases from
to
2n as the point P moves from O to O'.
The position K of the generating point which is reached after the circle
revolved through two right angles is shown as the verttx of the Cycloid.
The Cycloid evidently consists of an endless succession of exactly congruent portions each of which represents one complete revolution of the rolling
circle.
11-34.
The curve traced out by
Epicycloid and Hypo cycloid.
a point marked on the circumference of a circle as it rolls without sliding along a fixed circle, is called an Epicycloid or Hypocycloid accord-
jpg
as the rolling circle
We
is
outside or inside the fixed circle.
shall first consider the case of
an Epicycloid.
Let O be the centre and, a, the radius of the fixed circle. Let
the rolling circle start from the position in which the generating
point P coincides with some point O of the fixed circle. We take the
puint O as origin and OA as A^-a is.
www.EngineeringEBooksPdf.com
SOME IMPORTANT CURVES
The generating point has moved on from
the
241
A
P
(x, y)
Fig.
61.
to
when
has rolled on to the
position shown in the figure so that
rolling circle
arc
Let
a0
J/~arc PL
Z_40C=0,
arc ^47
i.e.,
/_ICP=<f>
arc
PIb<t>
(/>=aejb.
Here, 0, is the angle through
which the line joining the centres of
the two circles rotates while the rolling
circle rolls from its initial to its
present
position.
A PP= /. OCP- /_ OPiV
T
/_
Therefore
x=OL
= OP cos + OP sin
|TT)
cos 6
b cos
cos
b cos
sin 6
b cos
sin 6
b sin (0+<)
,
;
y -LP
sin
Thus
ft
sin
(0+^
fl+fc
b
x=(a+b)
cos
b cos
y=(a+b)
sin
b sin
are the parametric equations of the
meter.
Epicycloid
a+b
were within the fixed
obtained by changing
btob
a+b
'
0,
;
being tbe para-
y
b) cos
0+b
if
the rolling
equation can easily be
in the equations of the Epicycloid.
Its
circle.
Thus
x=(a
0,
.
The tracing point would describe an Hypocycloid,
circle
Jir)
cos
= (a-b) sin 0+b sin b
*
b
0,
are the parametric equations of the Hypocycloid.
www.EngineeringEBooksPdf.com
242
DIFFERENTIAL CALCULUS
In making one complete revolution the rolling
cribe
p\q
2bno
circle will
the length of the circumference of the fixed
Let the ratio a/b of the radii of the
terms so that
des-
circle.
be a rational number
circles
in its lowest
a
b
v
= -p
,
i.e.,
q
aq~bp
or 2na.q.~27Tb.p.
This relation shows that in making, /?, complete revolutions,
the rolling circle doscrib3S the circumference of the fixed circle q,
times and then the generating point returns to its original position.
Therefore the path consists of the repetition of the same, p, identical
portions.
In case a\b
is
its original position
irrational, the tracing point will never return to
and so the path will consist of an endless series
of exactly congruent portions.
Some
Particular cases of Epicycloid and Hypocycloid.
For a
epicycloid become
the
b,
(i)
x=2a
cos
y2a sin
Fig.
ing
The shape of
6
a sin 20.
will
original position after the
roll-
circle
the curve
0a cos 20,
In this case the generating point
return to
62.
equations of the
is
its
has made one complete revolution.
shown
in the adjoined figure.
Four cusped hypocycloid. If a~4b, the equations of hypobecome
cycloid
jt=| a cos + tf cos 30,
(//)
^=|
a sin
\a sin 30.
Now,
cos
30=4
cos 8
x=a
3 cos 0, sin 30
cos
3
0,
y=a
=3
sin
3
are the parametric equations of a curve
hypocycloid or astroid.
Here 0//?=4 so that the path consists
of the repetitions of four portions. The
thickly drawn curve ABA'B' gives the
path of the point.
sin
04 sin
0.
0,
known
For the points A, B, A' B' the values
are 0, Tr/2, TT, 3?r/2 respectively.
9
of
3
between the parametric
Eliminating
this
of
we obtain
curve,
equations
www.EngineeringEBooksPdf.com
four
cusped
SOME IMPORTANT CURVES
h
is
cycloid
243
form in which the equation of a four-cusped hyposometimes given.
also the
is
Show
Ex.
becomes a
that hypocycloid
straight line for
114.
Implicit cartesian equations of curves.
function of the two variables x and y, then
a=2b.
If f(x y) be a
f
fa, y)=0.
the implicit equation of the curve determined by the
points whose
co-ordinates satisfy it.
is
many
Curves whose equations are of the form f(x, y)~Q possess
points of interest not offered by curves with explicit equations
of the form y=F(x).
In the following, we consider only rational algebraic functions
f( x y) so that the form of the equation, when arranged according
to ascending powers of x and y, is
>
The same equation may,
in a concise form, be written as
where, u k represents the general homogeneous polynomial of the
degree in x and y.
,
y
The degree of any term means the sum of the indices of x and
term and the degree of the curve means the highest of the
in that
degrees of each term.
Branches of a curve.
of degree w, we
an equation in,
in the rational
If,
algebraic
equation
x by any fixed value, then there will result
whose degree will bo less than or equal to n. On
replace
y,
many values of, y, as is its degree.
as abscissa there will be as many
real roots of the equation.
points on the curve as are the difTeront
As, x, goes on taking different values, each of these points will separately describo what is known as a branch of the curve.
solving, this equation will give
Thus with the given value
as
of, x.
We
now trace a few important curves. In each case we have
to examine how, y, will vary as x starting from some fixed value,
increases or decreases.
}
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
244
11-41.
We
so that,
(x
2
+y2)x-ay 2 =0.
>
(a
0).
Cissoid of Diodes.
write the equation as
we
see, that
to a value of
x correspond two
values" of
y
which are equal in magnitude but opposite in sign. The two values
of y determine the two branches of the Cissoid which are symmetrically situated about x-axis.
We consider
one branch,
y=>
and the form of the other branch can be seen by symmetry.
We have
dx
The following
trace
it
=~
about the curve
particulars
will
enable us to
:
xj(a-x) under the radical is negative when
(/) The expression
x is negative or when x is greater than a and is positive when x lies
and a. Thus for y to be real, x must lie between
between
and a.
Hence the curve
and x=a.
(ii)
is
entirely
When x=0, >>=0
so
situatedjjsetween the lines A'=0
that the ^branch passes through the
origin.
.(ill)
x=0 or %a. The values
of the admissible values of x.
dy/dx=Q, when
the interval
[0, a]
a of x
is
outside
Thus the slope of the tangent at the origin to the branch
so that the branch touches the x-axis at the origin.
(iv)
For values of x between
and
a,
the value
positive so that the ordinate
increases.
Also as x
a,
is
of dy\dx
is
y monotonically
y
Hence we have]" the shape of the
curve as shown, the* two symmetrical
branches lying in the first and the fourth
quadrants.
Note. It is important to notice that origin
a point common to the two branches and the
two branches have a common tangent there.
is
Fig 64.
www.EngineeringEBooksPdf.com
SOME IMPORTANT CURVES
11-42.
We
(x
2
+y 2 )x~a(x 2 -y 2 )=0,
245
Strophoid.
write the given equation as
i.e.,
so that we see that to a value of x there correspond two values of
v (giving rise to two branches) which are equal in magnitude but
opposite in sign. The two branches of the curve are, therefore,
symmetrically situated about X-
We have
dy
__
dx
Some
enable us to trace the curve will
particulars which
now
be obtained.
(/) The expression
x)l(a-i-x) under the radical is positive
(a
a and a. Thus the curve entirely lies
and only if .v lies between
between the lines x~ a and x = fi.
if
(ii)
points
When .x=0
(0, 0)
and
When
(a,
0)
both the values of y are
on both the branches.
or 0,
lie
so
that the
=
I so
that the slopes of the two tandy/dx
(///)
gents at the origin to the two branches are 4-1. Hence y=x and
x are two distinct tangents to the two brandies at the origin.
v
at
.x=0,
For both the branches dyjdx tends to
is
(a, 0) the tangent to either branch
infinity as
x ~> a so that
parallel to j'-axis.
(/v)
dyjdx0
for values of
.v
given
by
a8
i.e.,
a.x
.x^O,
x-=-a(liV 5 )/--
for
The value
0(1
belong to the interval [
-f-
\/5)/2 does not
of the ad-
a, a]
missible values of x.
Thus
for
both the branches,
for
only.
(v)
When x->-a,
v->-foo
for
one branch and
oo
for the
other.
Hence we have the shape of the curve
Note.
as shown.
Both the branches of the curve pass through the origin and have
distinct tangents there.
www.EngineeringEBooksPdf.com
246
DIFFERENTIAL CALCULUS
ay2- x (x-f a)^0,
11-43.
(a
>
0).
Clearly,
are the
two branches of
this curve.
Also
dx
(/) The point (a, 0) lies on both the branches.
negative value of x corresponds a real value of y.
The value of dyjdx
(-*,
is
To no other
not real
for
0).
()
on both the branches.
lies
(0, 0)
Also, dy/dx tends to infinity for either branch
as x ~> 0.
Thus .y-axis is a tangent to the
two branches at the origin.
(Hi)
As x takes up
one value of dy/dx
is
positive values only
always positive and the
Thus the ordinate y
one branch monotonically increases and
other always negative.
for
for the other monotonically decreases,
Fig. 66.
y ->
oo
(iv)
and ->
for one branch
Also when
x ->
oo
dy
= -^
-t
rf*
1
~T~
When x
oo
oo
for the other.
,
P3V*'
/
y'tfL
\2
a
4-i
;
j
1 tends
j.
-V* J
c -x
to inanity.
*
-
This shows that as we proceed to infinity along the curve, the
tangent tends to become parallel to j^-axis. This is possible if and
only if at some point, the curve changes its concavity from down-
wards to upwards.
We have the shape of the curve as in Fig. 66.
Note. The peculiar nature of the point ( a, 0) on
the curve may be
This point lies on either branch, but no point in its immediate
neighbourhood lies on the curve.
carefully noted.
11-44.
x3
ay
2
=0,
semi-cubical para-
The
discussion of this equation which
very simple is left to the student.
Its
shape
is
shown
in
the
y
is
adjoined
Pig. 67.
11 45.
x 3 +y 3 ~3axy=:0. Folium of DesFig. 67.
cartes.
It will be traced in
Chapter XVIII.
www.EngineeringEBooksPdf.com
247
SOME IMPORTANT CURVES
Important Note
we obtain
Putting
y=tx
in
2
the equation y (a-x)=x* of
the
Cissoid,
at 3
at*
which are
its
parametric equations
We may
similarly
show
;
t
being the parameter.
that
*~
3af
'
1+Y
3
^
__
3af
2
'
1+t 3
are the parametric equations of the Strophoid, Semi-cubical Parabola and
Folium
respectively.
This method of determining parametric equations
we have here considered.
not general and
is
applies only to such curves as
115. Polar co-ordinates. Besides the cartesian, there are other
systems also for representing points and curves analytically. Polar
system, which is one of them, will be described here.
line
In this system we start with a fixed line
fixed point on it, called the pole.
OX,
called the initial
and a
P be any given point, the distance
and /_XOP~0, the vectorial angle.
If
vector
OP=r
is
called the radius
re-
The two together are
ferred to as the polar co-ordinates of P.
11 51.
we were
If
Unrestricted variation of polar co-ordinates.
concerned with assigning polar co-ordinates to only individual points
in the plane, then it would clearly be enough to consider the radius
vector to have positive values only and the vectorial angle 9 to lie
between and 2ir. But, while considering points whose co-ordinates
satisfy a given relation between r and 6, it becomes necessary to remove this restriction and consider both r and 6 to be capable of varyx oo ). The necessary conventions for this will
ing in the interval (
be introduced now.
,
The angle, 0, will be regarded as the measure of rotation of a
which starting from OX revolves round it the measure being
positive or negative according as the rotation is counter-clock-wise
line
;
or clock-wise.
To find the point (r,
value, we proceed as follows
0)
where
r is
negative and,
6,
has any
:
OX
revolve though 0. We
Let the revolving line starting from
produce this final position of the revolving line backwards through
r
is the
O. The point P on this produced line such that OP
\
required point
The
(
1,
(r,
positions of the points
97T/4)
\
6).
(1, 97T/4),
(1,
9ir/4),
have been marked in the diagrams below.
www.EngineeringEBooksPdf.com
(1,
-
248
DIFFERENTIAL CALCULUS
'X
Fig. 68.
Fig. 69.
v\
Fig. 70.
Fig. 71.
It will be seen that
according to the conventions introduced
here a point can be
For
represented in an infinite number of ways.
example, the points (-1, ir/4), (I, oir/4), (I, 2/17T 57T/4), (n is any
integer), are identical.
+
11 52.
Transformation of co-ordinates. Take the initial line
as the positive direction of Jf-axis and the
The positive direction of
pole O as origin for the Cartesian system
K-axis is to be such that the l\neOX after
revolving through ?r/2 in
counter-clock-wise direction comes to coincide with it.
OX of the polar system
Let
(x, y)
and
(r,
be the cartesian and polar co-ordinates respectively of any point P in the plane.
9)
From
the
A
OMP, we
OJf/OP=cos o,
AfP/0P = sin 0,
The equations
get
'X x=r cos
i.e.,
(/')
y=r
and
(//)
6
sin d
..
(i)
(//')
determine
cartesian co-ordinates (x, y) of the point
terms of its polar co-ordinates (r, 6) and
Fig. 72
..
the
P
in
vice
versa
6.
Polar Equations of Curves. Any explicit or implicit rebetween, r and 9 will give a curve determined by the points
whose co-ordinates satisfy that relation.
11
Thus the equations
determine curves,
The co-ordinates of two points symmetrically situated about the
or of the form (r, 9) and (r,
9) so that their vectorial
fnitial line
angles differ in sign only.
www.EngineeringEBooksPdf.com
SOME IMPORTANT CURVES
Hence a curve
changing f) to
curve /=#(! +cos
249
symmetrical about the initial line if, on
equation does not change. For instance, the
symmetrical about the initial line, for,
will be
its
0,
j
0)
s
/-=tf(l
+ cos
0)
= a[l+cos(
0)].
be noted that
may
r=a represents
It
a circle with
its
centre at the pole and radius
a and
;
9=b
the
represents a lino through the pole obtained
through the angle b.
by revolving
initial line
A
curves,
To trace polar
will now be treated.
varies.
generally consider the variation in r as
few important curves
we
cos
11-61.
r=a(l
The curve
(/)
(//)
(/'//)
is
Cardioide.
0).
symmetrical about the
When 0-0, r = 0.
When increases
initial line.
from
to 7r/2, cos 9 decreases from 1 to
and, therefore, r increases contito a. When
nuously from
7r/2,
r
a.
=
When 9 increases from
to
cos 9 decreases from
1 and, therefore, r increases from
to 2a.
When
IT, r='2a.
(iv)
Tr/2 to
a
TT,
-~..^
=
Fig. 73
The variation of 9 from TT to 2ir ne^d not bo considered because
of symmetry about the initial line.
Hence the curve is as shown.
Ex.
Trace the curve
11*62.
r
2
^ a-
r
--
cos 20.
--0(1-1-
cos
0).
Lemniscate of Bernoulli!.
It is symmetrical about the initial line and so
the variation in r as 9 varies from
to TT only.
of
r
values
We consider positive
only.
(/)
(//)
When 0=0, r=a.
When increase from
so that cos
a to 0.
2'y
decreases from
1
to
we need
consider
to ir/2
to ir/4, 20 increases from
and, therefore, r decreases from
When = ir/4, r = 0.
When increases from
(in)
y4
'
,
.*''
I
^v
Tr/4
to
to 3-7T/4, cos 20 remains
7T/2 and from ir/2
Thus no
r is not real.
so
and
negative
to these
curve
the
on
corresponds
point
values of 0.
When = 3ir/4,r=0.
When increases from
(iv)
fl
NX
X
;
Pig. 74
TT,
20 increases from 3ir/2
www.EngineeringEBooksPdf.com
to
2-rr
3ir/4 to
to that
250
DIFFERENTIAL CALCULUS
cos 20 increases from
Hence we
to
1
and, therefore,
have the curve as shown.
r
increases
from
to a.
P
It is easy to see that the point
will describe exactly the
if we take, r, to be negative.
The curve consists of two
situated between the lines
same
curve even
loops
To obtain the cartesian equation
the polar equation as
r2
=a
r4
=a
re-write
sin 2 0)
2
2
we
of the lemniscate,
(cos
or
Therefore,
which
2
2
(r
r 2 sin 2 0).
cos 2
we obtain
a well-known form of the equation of the lemniscate and
is
of the fourth degree.
11 63.
rw
=a m
(1)
r=a
^
or
1T^
'
*
r
/.*.,
For
(2)
is
m=
1
,
J.
,
cos
=0rcos
a
circle
(Fig.
and radius
(0/2, 0)
75
2
2,
x 2 +^ 2 =^x,
which
Fig.
cos m0,
m = \, 1, 2,
For w 1, we have
for
is
75)
with
its
centre at
(a/2).
we have
r^sstf" 1 cos
0),
(
# = r cos
or
a~-x
which
a
is
straight
line
(Fig. 76) to the initial line
a,
from
perpendicular
at a distance,
it.
(3)
For
w~2, we
r*=a
which
and
is
2
have
cos 20,
lemniscate.
Fig. 76.
(4)
/w m=
r-
or
2
>X
we have
=a- cos(-20)
a 2 =r 2 cos 20
=r
or
2,
2
2
a 2 =x
2
sin*0)
(cos
y
,
which is known to be a rectangular
hyperbola (Fig. 77). To trace this
curve, write its equation in the form
20
Fig. 77
www.EngineeringEBooksPdf.com
SOME IMPORTANT CURVES
and note the following points about
(0
(//)
When 0=0, r=a.
creases from a to oo
(iii)
(i'v)
When
When
:
initial line.
When
increases from
to
7T/4, r in-
.
varies from Tr/4 to 37T/4, r remains imaginary.
varies from 3?r/4, to
m=
F0r
(5)
it
symmetrical about the
It is
251
TT,
r
decreases from
to a.
oo
we have
J,
=a cos 2
or
or
2r=a(l+cos
which
a Cardioide.
For
(6)
0),
is
w
|,
we have
i.e.,
=r
or
cos 2
i
=r(l+co"s
0)J2
= l+cos0,
or
v/hich
is
To
known
trace
Fig. 78.
to be a parabola (Pig. 78).
this curve (Fig. 78), we re- write its equation in
form
2a
r ~~
1+cos
and note the following points about
(/)
(//)
It
is
symmetrical about the
When 0=0, r~a.
decreases from 2 to
11-64!
(/)
(//)
it
r--a0,
initial line.
When
r
and, therefore,
:
to
increases from
increase from a to oo
TT,
1+cos
.
Spiral of Archimedes.
so that the curve goes through the pole.
(a>0).
When 0=0, r=0,
When increases,
r increases.
Also,
when
when
~>
->
+
oo
oo
;
,
,
,
oc
Thus the curve
starting fron*
the pole goes round it both ways an
infinite number of times. The continuously draw n line corresponds to
and the dotted
positive values of
r
one to negative values of
Fig. 79.
www.EngineeringEBooksPdf.com
0.
252
DIFFERENTIAL CALCULUS
1165.
Here
Hyperbolical Spiral.
r0:=a,(a>0).
r=flf/0.
(i)
(iV)
r is positive or
When
6
|
\
negative according as
increases,
[
is
positive or negative.
decreases.
r
\
Also
when |0
when
0,|r|->x>.
oo
|
(ill)
->
r
,
0.
Now
or
r sin
or
0=(fl sin 0)/0
>>=(a sin 0)/0 which
-;>
a as
-> 0.
The ordinate of every point
on
curve
the
approaches
as drawn.
approaches
a
as
We thus have the curve
0.
The continuously drawn
line
corresponds to positive values of
r=aeJ>0
1166.
(ii)
Also when
(i/i)
(a,
,
b>0). (Equiangular Spiral)
When 0=0, r=a.
When increases,
(/)
r is
GO
,
r
r
also increases.
> GO
;
when
->
-
x
,
/*
->
.
always positive,
We thus
The
->
0,
while the dotted one corresponds to
negative value of 0.
Fig. 80.
have the curve as drawn.
justification
of the
adjective
'Equiangular' will appear in Chapter XII,
Ex,
1. p.
269.
Fig. 81
11-67.
r=asin30,
(a >$}. Three leaved rose.
www.EngineeringEBooksPdf.com
(D.U. 1949)
SOME IMPORTANT CURVES
When 0=0, r=0. As
(i)
from
increases from
to 7T/2 and, therefore, r increases
from
to
to a.
253
7T/6,
30 increases
(ii)
As
increases from 7T/6 to Tr/3, r decreases from a to 0.
(///)
As
increases from 7T/3 to
to a.
remains negative and
r
7T/2,
numerically increases from
(iv)
As
increases from ?r/2 to
27T/3,
r
remains negative and
numerically decreases from a to 0.
It
57T/6
may
increases from 27T/3 to
similarly be shown that as
P
the
to
describes
the second loop
TT,
57T/6
point
(r, 0)
and from
above the
If
ted and
initial line.
increases
we do not
11-68.
r=a
beyond TT, the same loops of the curve are repeaget any new point.
sin 20.
Four leaved
The discussion of this equation
shown in figure 83.
is
only
rose.
is left
to the reader.
Fig. 83.
www.EngineeringEBooksPdf.com
Its
shape
CHAPTER
XII
TANGENTS AND NORMALS
[Introduction. This and the following chapters of the book will be devoted to the applications of Differential Calculus to Geometry. The part of
Geometry thus treated is known as Differential Geometry of plane curves.]
Section I
Cartesian Co-ordinates
12
EQUATIONS OF TANGENT AND NORMAL.
1.
It was shown in Ch.
Explicit Cartesian Equations.
be
the
if
which
the
77
that
^
IV 4-15, p.
angle
tangent at any point
makes
with
then
curve
the
on
x-axis,
y=f(x)
(x, y)
12 11.
tan
*=
=/'(*).
-j
Therefore, the equation of the tangent at
curve y=f(x)
&ny point
(x, y)
on the
is
Y-y-f'()(X-x),
where X,
Y
are
the
...(i)
current co-ordinates of
on the
any point
tangent.
The normal to the curve y=f(x) at any point (x, y) is the straight
which
line
passes through that point and is perpendicular to the
tangent to the curve at that point so that its slopa is, -!//'( jt).
Hence the equation of the normal at (x, y) to the curve y=f(x) is
(X-x)+f(x)(Y-y)=0.
Implicit Cartesian Equations.
12-12.
the curve /(*,
y)=Q
where dfldy*
0,
For any point
(x, y)
on
we have
'
dy
dx
Hence the equations of the tangent and
are
(f,y) on the curve /(x, y)=0,
rt
the
normal
=0.n d(X-x,
at
any point
_(V-n
=0.
respectively,
A
symmetrical form of the equation of the tangent to a
rational Algebraic curve. Let f(x, y) be a rational algebraic function
of x, y of degree n. We make f(x, y) a homogeneous function of
Cor.
254
www.EngineeringEBooksPdf.com
TANGENTS AND NORMALS
255
three variables x, y, z by multiplying each of its terms with a suitable power of z. Then by Euler's theorem, ( 10'81, p. 199) we have
i.e.,
z
"^
'
dz
d'y~~'~
so that the equation
/
V_ Y\ df
,
v
/
yr__
8/
^Q
of the tangent takes the form
or
x.j
9.x
or
X
f
8f
+Y
?
cx
8y
f
+Z?8z
=0,
where, for the sake of symmetry, the co-efficient r of 8//8- has been
replaced by Z.
The symbols
Z
and
both put equal to unity after
r are to be
differentiation.
This elegant form of the equation of the tangent to a rational
algebraic curve proves very convenient in practice.
Parametric Cartesian Equations. At any point *f of
12-13.
the curve x=f(t) y=F(t), where/' (0^0, we have
y
dy _
~ dy
dx
Hence
9
*t
dt
dt
'
dx
_F'(t)
^f
(t)
and
the equations of the tangent
y=F(t) are
'
r
the
normal
at
any point
of the curve *=/(/),
[X -f(t)]F'(t) - [Y- F(t)]f '(t) =0,
and
[X-f(t)]f'(t)+[Y-F(t)]F(t)=0,
respectively.
Examples
1.
(x, y)
Find the equations of the tangent and the normal at any point
of the curves
(/)
y=c
cosh (xjc).
(Catenary)
(ii)
x m la m -\-y m lb m = l.
www.EngineeringEBooksPdf.com
256
DIFFERENTIAL CALCULUS
(i)
For
y~c
we have
cosh (xjc),
dy =
.
sinh
,
ax
x-
.
c
Hence the equations of the tangent and the normal
point
(x, y), i.e., (x, c
y
x
y
respectively
(//)
;
X,
c
Y
any
x
r
c cosh
=(A
(x
and
at
cosh xjc) are
x) sinh
\
sinh
c J
cosh
)
x
^X
+
c
x^O,
being the current co-ordinates.
Let
/(x,
+
y)^*
df
~~
dx
1=0.
tt
my
m -*
a
Hence
Therefore, the equation of the tangent at (x, y)
or
y
m
_
on the given curve.
Also, the equation of the normal at (x,y)
for (x, y) lies
is
or
x-x
The given equation
Another method.
it
homogeneous, we
y-v
is
of degree m.
get
M **)%. +>-*".
jW
yw
so that
www.EngineeringEBooksPdf.com
Making
TANGENTS AND NORMALS
Hence the equation of the tangent
^
Putting
Z.mz-
\-~/ti---
Z=z = l, we
257
is
1
^.
12*12, p. 254)
(cor.
obtain
am
bm
'
'
as the required equation of the tangent.
2.
Find the equations of the tangent and normal
at
6=n/2
to the
Cycloid
x^a(9sin
6),
y=a(lcos
9).
Wejiave
-r-
a(
a0
dy
cos
\
)
= 2a
sin 2
=a
.
sin
n
9
.
rt
2a sin
civ
dy
dx
,
~i
6
cos
~
dy
d9
dd
=
"
dx
dx
dy
;
rftf
.
2i
2i
__
~" C
9
'
2
rf^
Thus
Also, for
7T/2,
we have
jc=0(7T/;2
equation of the tangent at 0=7T/2,
-a =
i.e.,
-^Tr-
and the equation of the normal
1)
and j
a.
Hence the
at the point [0(7T/2
or
-=aTr-a
t
is
y-a=-l[*--<i(i7r--l)] or -Y+r=Jair.
3.
may
Prove that the equation of the normal to the Aitroid
be written
in
the form
x
sin
<f)y cos
(f>
+a
cos
Differentiating
we
get
www.EngineeringEBooksPdf.com
1), a] is
DIFFERENTIAL CALCULUS
268
Therefore the slope of the normal at any point
(x,
y)
=**/A
But the
We
slope of the given line
= tan
<f>.
write
...()
.
Equations
They
(/)
and
have now
(11)
to be solved to find x and y.
give
or
cos 2 ^,
y$=a*
Substituting this value of y in
we
(//),
x*=flrsin
i.e.,
}>=0cos
3
<.
get
<f>,
x~a
i.e.,
sin 3 <.
the slope of the normal at (a sin 3 ^, a cos 3 <).
equation of the normal at the point is
Thus tan
<f>
is
a cos 3 <i=-^-^(x
T
cos
y
The
a sin 3 <i),
Y/
<f>
or
y
cos <f>acos*<f>~x sin
a sin 4 ^,
<f>
i.e.,
x
sin
2
y cos ^+fl(cos
<
sin 2 ^)(cos 2 ^+sin 2 ^)=0 ?
</>
i.e.,
x
which
is
4.
sin
y cos
</)
<^-f
a cos
2^=0,
the given equation.
Find
the condition for the line
x cos
0+y
sin
6=p,
...(/)
x/a+ylb m =l.
...(/I)
to touch the curve
was shown that the equation of the
In Ex.
1, (//) page 255, it
tangent at any point (x, y) of the curve
(11) is
where X, Fare the current co-ordinates.
We re-write
the equation
XGOB
taking X,
Y as
(/)
in the
0+Y
sin
form
0~p=0,
current co-ordinates instead of x,
The equations
(in)
and
(/v)
>>.
represent the same line.
www.EngineeringEBooksPdf.com
TANGENTS AND NORMALS
a m cos 6
b m sin
'
p
or
X
(a* cos
~\
The point
6
\l/(m
>
p~
)
(x, y) lies
1
cos S
_
1)
/6 W sin 0\l/(m
y ~(
p
on the given curve.
\w/(w
1
1)
1)
J
Therefore
1
/6* sin
-r
i
or
'
(0 cos 0)
which
is
+
(&sin0)
^
==P
the required condition.
Show that the length of the perpendicular from the foot of the
5.
ordinate on any tangent to the Catenary
y=c
is
cosh (x/c),
constant.
Equation of the tangent at any point
X
sinh x/c
The
foot
F+(rcosh
xjcx
sinh
_c
is
6.
free of x,
cosh x/r __
~~
cosh x/r
y and
is,
is
1. p.
the point
is
255).
(x, 0).
0) to the tangent
x/c
x sinh x/c)
*
therefore, constant.
SViou' that the length
of the portion of the tangent
intercepted between the co-ordinate axes
is
Differentiating the given equation,
,''-*+
dy
(x, y)
(x,
+ (r cosh
__x sinh x/c
of the Catenary
x/c)=0 (See Ex.
of the ordinate of the point
The length of the perpendicular from
which
(x, y)
to the astroid
constant.
we
get
!'-'y*
Therefore the equation of the tangent at any point
yt
Y-y=- y r (X- X
),
www.EngineeringEBooksPdf.com
(x, y)
is.
DIFFERENTIAL CALCULUS
260
or
=x*
The tangent
with
y*
where y=0. Hence
(/)cuts JV-axis
(0
fl*.
its
intersection
its
intersection
A'-axis is
A(x^
on putting ^=0 in
Similarly,
with F-axis
9
0).
(/),
we
find
that
is
5(0, yl($).
AB=<x$a
Therefore
which
is
x and y and
free of
is,
therefore, a constant.
Exercises
I-
Find the tangent and normal to each of the following curves
(//) x*l<P+y*lb*=l at (*', /).
20).
(/) y*=4ax at (0,
:
xy=c f
at (cp, c/p).
(iv)
2
2
<v) (x +^ )A:~0(jc ->' )=0 for jc=-3a/5.
2
?
<v/) x (jc-^)-f0 (x+>>)=0 at (0, 0).
<iii)
i
(v/i)
x = 20cos^
2
2
2
(x 4-y
2.
a
+/)*-0>>
2
=0
at
x-0/2.
0cos 2^,^=2a
)=^V
a sin 2^ at (9=^/2.
sin ^
at (c/cos 0, c/sin ^).
Find the equation of the tangent to the carve c*U 8 -fy t )=JcV
form x cos 3
3.
U
2
Show
in
the
= c.
3
0-f>> sin
that the tangent to the curve
curveagainat (16/5,
3xy*-2x-y=l
meets the
at (1,1)
1/20).
4.
Prove that ths equation of the tangent at any point (4m 2 8m 3 ) of the
2
3
semicubical parabola x 3 =y is
and show that it meets the curve
m 1 ), where it is a normal if 9m a =2.
(D.C/. Hons., 1957)
again at (m\
,
y~3mx4m
Find where the tangent
5.
F-axis for the following curves
is
parallel to yf-axis
and where
it is
parallel
to
:
(/)
6.
x*+y*=a*.
x 3 4-^ 3 = 30xy.
(Folium)
Find the equations of the tangent and the normal to the curve
at the point where
7.
(//)
Show
it
cuts JT-axis.
(Delhi 1948)
that the line
x cos O+y
sin
/?=0,
touch the curve
if
sin n ^.
(P.I/. 7957)
www.EngineeringEBooksPdf.com
261
TANGENTS AND NORMALS
8.
Tangent
ordinate axes in
co-ordinates is
,4
at
and
any point of the curve Oc/a)l +(ylb$ =1 meets the coShow that the locus of the point with (OA, OB) as
2
Show
9.
2
li
y m =ax m
makes
=
x*/a +y /b
that the tangent at any point x, y)
-l
1
-\-x
on the curve
in
t
intercepts
ax
ay
,
and
\)a+mx
(m
-
m(a-\-x)
on the co-ordinate axes.
10.
Prove that the sum of the intercepts on the co-ordinate axes of any
tangent to
is
constant.
11.
Prove that the portion of the tangent to the curve
a
y
intercepted between the point of contact and A'-axes
12.
Show that the normal at any
x=acos Q+aQ sin 0,
constant.
point of the curve
sin
cos
ya
is
is
QaQ
at a constant distance from the origin.
Show that the length of the portion of the normal to the curve
13.
3
x=a(4 cos
3 cos 0),
y=a
3
(4 sin
09 sin 0)
intercepted between the co-ordinate axes is constant.
Show that the tangent and the normal at every point of the curve
14.
x=ae u
fl
(sin
are equidistant from the
15.
Show
0-cos0), y=ae
fl
u
+ cos
(sin
0)
origin.
that the distance from the origin of the normal at any point
of
the curve
is
x=ae^ (sin i0 1-2 cos
twice the distance of the tangent,
16.
Show
makes angle i
0),
y=ae^
(cos
i0- 2
sin 10)
that the tangent at any point of the curve
x = a(/-f sin t cos /), >>=c(l-f-sin /) 2 ,
(*
+ 2/)
with
A'-axis.
origin to the curve v=sin x.
Tangents are drawn from the
=x 2 -/.
their points of contact lie on
Prove that
(D.U 7955>
17.
xY
18.
If
p~ x
cos
0+y
sin 0,
touch the curve
prove that
p n = (tfCOS
(P.U. 1941
www.EngineeringEBooksPdf.com
;
D.U. 7955)
262
DIFFERENTIAL CALCULUS
The tangent at any point on the curve x3 +y*=2a*
7tm die co-ordinate axis show that
cuts off lengths
p and
;
The tangent
20.
at
any point P(x v
x3
y=x*
nusefcs it
again at
chat its locus is
g
.
of the curve
y^)
Find the co-ordinates of the mid-point of
PQ
and show
y=l-9x -\-2Zx* ~28x*.
(B.U.)
Prove that the distance of the point of contact of any tangent to
2a mxy m ~~i =zy2in a 2m
21.
t
from the
origin as
the segment of the
-Ac-axis
between the origin and the
tangent.
Show
22.
at
Show
23.
at
(
tangent at the two points where
3) is
(0,
12-2.
it
meets the curve again.
that the tangent to the curve
also a normal at
1, 1) is
Def.
that the normal to the curve
two points of the curve.
Angle of intersection of two curves.
The angle of intersection of two curves
to
intersection is the angle between the tangents
the two
a point of
curves at that
at
point.
Examples
Find the angle of intersection of the parabolas
1.
/>
and
at the point other than the origin.
We
have
or
x=0
.
Substituting these values of
x
and
in (#),
we
>=4a^
Therefore
(0, 0)
and (4a*6 f
,
4a*b*)
get
for
x=
are the two
intersection.
www.EngineeringEBooksPdf.com
points of
TANGENTS AND NORMALS
Differentiating
(i),
we get
or dy/dx=2a/y.
2ydyjdx=4a
Therefore, for the curve
(//),
we
n
^L
2x=46
(/),
4(0M, 40M) =^12$.
(dyfdx) at
Differentiating
get
^V
-~-
Therefore, for the curve
Thus,
if
/,
wi
(//),
(4a%$,
(dy/dx)Q,i
x
= !-/
dx 26
t/y
or
t/x
;
263
4a^)
=2a*/A*.
be the slopes of the tangents to the two curves,
we have
The required angle
ft^
2.
JF/w/
//?e
condition that the curves
should intersect orthogonally.
Let (x
r
,
y')
be a point of intersection so that we have
or
Differentiating the
first
equation,
we
.
get
dy
v
=_
dx](x', y')~
by"
www.EngineeringEBooksPdf.com
ax
264
DIFFERENTIAL CALCULUS
Similarly for the second curve
__.
*L1
G
For orthogonal intersection, we have
ax'
by'
*
0ijc'__
~~~~
'
hy'
Substituting the values of
*-.
_r
L.
\
l'
e
'
9
aa ^ x
x', y' in (/),
J
**._..
!'
we obtain
r=0
or
/.*.,
b
bl
a
a^
as the required condition.
Exercises
1.
2.
Find the angle between
2
2
2
(/) x -/=a and x +/=a^2.
(//) ^=ax and ^-f-^^
(P.C7.
7955)
Prove that the curves
interesect at tan" 1 (88/73) at {3a, -2a).
3.
Find the condition for
y=mx
to cut at right angles the conic
ax*+2hxy+by*=l.
Hence
find the directions of the axes of the conic.
12*3.
Lengths of the tangent, normal, sub-tangent and subnormal at any point of a curve.
Let the tangent and the normal at any point (x, y) of the curve
meet the A'-axis at Tand G respectively. Draw the ordinate PM.
Then
sub- tangent
the lines
TM,
MG are
called the
and sub-normal respectively.
The lengths PT, PG are sometimes
referred to as the lengths of the tangent
and the normal respectively.
Clearly
Also
dy
www.EngineeringEBooksPdf.com
TANGEENTS AND NORMALS
From the
/i
we have
Tangent ^TTWMPcosec
(i)
(
figure,
Sub-tangent
)
^=
= TM = A/P cot ^ = y
Normal =GP==MP
(/'//)
265
-=?
.
sec /r=
-V
dv
-
(iv)
.
Exercises
1.
Prove that the sub-norrral at any point of the parabola
y*=4ax
is
constant.
2.
Show
that the sub-normal at any point of the curve
2
y*x =a*(x*-a
2
)
varies inversely as the cube of its abscissa.
3.
Find the length of the tangent, length of the normal, sub-tangent
and sub-normal at the point 0, on the four cusped hypocycloid
x=a cos 3 0, y-=a sin 3 0.
Show
4.
that the sub-tangent at any point of the curve
^m _. Q rn+ n
yti
varies as the abscissa.
5.
Show
(Banaras}
that for the curve
x=a+b log
[
+ V( &'->>)] -V(6 a ->' 2 ),
sum of the sub-normal and sub-tangent
6.
Show
constant.
is
(P.C/.
1938}
that for the curve
the product of the abscissa and the sub-tangent is constant.
Show that the sub-tangent at any point of the exponential curve
7.
is
constant.
8.
For the catenary
y-c cosh U/c),
2
(P-U. 1941}
prove that the length of the normal is v /c.
Prove that the sum of the tangent and sub-tangent at any point of
9.
the curve
*
/<
= *-<!.
varies as the product of the corresponding co-ordinates.
10.
Show
that in the curve
y=a
log
U
2
-0
a
),
the sum of the tangent and the sub-tangent varies as
ordinates of the point. [Compare Ex. 9 above].
www.EngineeringEBooksPdf.com
the product of the co(D.U. 1952}
268
DIFFERENTIAL CALCULUS
Pedal equations or p, r equations.
12*4.
Del A relation between the distance, r, of any point on the
wrvefrom the origin (or pole), and the length of the perpendicular* p,
from the origin (or pole) to the tangent at the point is called pedal equa*
lion of the curve.
To determine the pedal equation of a curve whose Car-
12-41.
tesian equation
is
given.
Let the equation of the curve be
Equation of the tangent at any
point
(X, y) is
Y-y=f'(x)(X-x)
y>
or
If,
cular
p be the length of the perpendi}
from
(0, 0)
to this
tangent,
we
have
Fig. 85.
y-xf'(x)
Also
r*=x*+y 2
Eliminating x
y b3twe3n
t
...(Hi)
.
(/),
required pedal equation of the curve
(il)
and (m), we obtain the
(/).
Example
Find the pedal equation of the parabola
Tangent at
(x, y) is
Y-y^-.(X-X)
or
The
tangent
length,
(/')
is
p
given
9
of the perpendicular from the origin to the
by
<
ISO
www.EngineeringEBooksPdf.com
TANGENTS AND NORMALS
From
which
is
(//)
and
(///),
267
we obtain
the required pedal equation.
Exercises
1.
Show
that the pedal equation of ellipse
;
2.
Show
j 4 1 ;;
-
IP.*. 1955)
.
that the pedal equation of the astroid
x=a
cos 3 0,
y=a
sin 8 0,
is
3.
Show
that the pedal equation of the curve
A'
-a
a cos 29, ;y=2a sin
2a cos Q
sin 20,
43
2
9(/-
4.
Show
-^)=8//-.
that the pedal equation of the curve
x^--ae
+ cos
cos 0),,)'=^ (sin
(.sin
0),
is
r=
5.
Show
V2p.
that the pedal equation of the curve
x=a
(3
0-cos 3 0), y=a
cos
(3 sin
0-sin 8 0),
as
s
2
3/> (7fl
6.
Show
-^)
s
8
=(10fl
-r)
a
.
that the pedal equation of the curve
<*(*'
f >')=**>'.
is
s
l/P*+3/r'=J/r
.
Section II
Po/^fr Co-ordinates
12-5.
Angle between radius vector and tangent.
Let P(r,
d)
be any given point on the curve.
Take any other point Q(r+8r, 6+80) on the curve.
OP to L.
www.EngineeringEBooksPdf.com
Produce
268
CALCULUS
Let
PT
P
be the tangent at
Let
and
let
^LPT=
Z_PQ=a so that
limit of a as
Q ->
P.
We
=<>
^_
is
the
have
Also
the
Applying
"\ we
sine
formula
to
the
get
OQ
OP
Fig. 86
sin
=
sin
or
sin
sin
a)
(TT
a
*
sin (a
r
8r
_
sin (a
80)
80)
sin a
"""
sin (a
r
8^)
sin a
sin (a
sin (a
=2
8
89)
cos
sin
se
2 sin
or
5r
1
Let
Q
80
P so
-*
sin
=
~~ cpsJa-4S0)
sin (a -80)
t
'
r
that 80 -> 0, 8r
^
'
-->
0,
and a
Therefore
1
dr __ cos
~~
r
dti
(^
"
sin
<
or
.
,
tan cj=r
^
r
dr
Cor. Angle of intersection of two curves. If <^,
be angles
2
between the common radius vector and the tangents to the two curves
it a point of intersection, then their angle of intersection is
<
i
Note.
#1-*,
Precise meaning of
<p.
i
For a point P of the curve,
q> is precisely defined to be the angle through which the
of the radius vector (/.<?., the direction of the radius vector
rotate to coincide with the direction of the tangent in which
direction of the tangent in which, 0, increases is taken as the
of the tangent.
www.EngineeringEBooksPdf.com
positive direction
produced) has to
9
increases.
The
positive direction
TANGENT3 AND NORMALS
269
Examples
Show
1.
that the radius vector
inclined, at
is
a constant angle to
the tangent at any point on the equiangular spiral
Differentiating (i) w.r. to
dr
,
hfi
-
*
or
which
<=
<i
,
=
dO
,,'
i
do
tan
we get
6,
b
dr
,
tan" 1
,
b
,
a constant.
This property of equiangular spiral justifies the
il'66, p. 252]
[Refer
Find the angle of intersection of the Cardioides
is
adjective 'Equiangular'.
2.
r=fl(l+eos
Let
P (r
r=6(l
0),
-
cos
6).
OJ be a point of intersection. Let <^ 1
which OP makes with the tangents to the two curves.
For the curve r=a(l +cos 0) we have
t
,
<
,
dr
--
2
be the angles
*
or
dO
__
2Ujos* 0/2
____
"~
2 sin 6]"2.GOs 0/2
_a(i+cos0) ~~
a sin
dr
/
=tan
tan
,1T
-
(
\
/)
\
**
/
\
+
*^
or
Hence
^i=*f + 2
r=6 (I cos
For the curve
dr
_b
'd9~~
0),
we have
.
8m
'
or
tan cA=
Y
t/0
r
t
dr
fe(l
=--.
-
cos0)
b sin
--tan
,
Hence
www.EngineeringEBooksPdf.com
.
2
*
2
DIFFERENTIAL CALCULUS
270
Therefore,
^^
and hence the curves cut each other
-Tr/2
afc
right angles.
Exercises
Find for the curves
1.
(i)
cos
r=0(l
0)-
(Cardioide).
m =a rn cos m 6m
m (cos m$ + sin wifl).
r -a
r
(11)
cos 0. (Parabola)
(/v)
that the two curves
2
a
r =a cos 20 and r=a(l-f cos 0)
(Hi) 2alr=*\ -f
Show
2.
1
intersect at an angle 3 sin- (3/4)*.
m
m
3.
Show that the curves r =a cos mfl,
r
w- w
cut each other
sin
orthogonally.
4.
(i)
Find the angles between the curves
r=aO, rS=a
(Hi) Y
=a
r=ae
(vi)
Show
npp' be drawn
meet in
T, then
6.
cosec (0/2),
r=6
sec (0/2)
re
+ 0),
2
r=-0((l-f
(v) r
2
sin
20-4,
r
a
- 16
;
sin 20
;
=ft.
that in the case of the curve
P and
P'
r=a
(sec 0-J-tan 0), if a radius
and if the* tangents at P, P'
(Af-C/.>
that the tangents to the cardioide
r^=(l+cos 0)
whose vector ial angles are rr/3 and 2^/3 are respectively
perpendicular
)
;
;
cutting the curve in
TP= IT'.
Show
at the points
,
19
r=flfl/(l
2
r=alog0, r-0/log0
(iv)
5.
(//)
;
2
parallel
and
to the initial line.
on the same side of the initial
7. The tangents at two points P, Q lying
show that
0), are perpendicular to each other
r=a(H-cos
cardioide
the
of
line
at the pole.
an
n/3
angle
subtends
the line PQ
chord of the
B Show that the tangents drawn at the extremities of any
the pole are perpendicular to
which
cos
through
passes
0)
cardipide r=a(l-f
;
each other.
r-=<i(i + cos 0) are parallel, show that
an angle 2"/3 at the pole.
subtends
of
contact
the line joining their points
that
10. Show
9.
If
two tangents to the cardioide
(Agra 1952}
from pole to the tangent.
From the pole O draw OY perpendicular to the tangent at any point
P (r e) on the curve r=/(0) Y being its foot.
Let OY^p
126*.
Length of the perpendicular
;
*
From the &OPY, we
p
r
-
=sin
r
get
.
.
<f>
or
p=r
which
is
a very useful expression for p.
We now
ordinates
9.
sin <,
r,
express, p, in terms of the co~
and the derivative
We have
www.EngineeringEBooksPdf.com
of, r, w.r.
to
271
TANGENTS AND NORMALS
d
tan
O
j
cfc=?r -~j-
v
-*7\*
which we can write as
1
2
__r
+(dr/d0)_
11
2
/dr_\
Yet, another form of p will be obtained,
if
we
write
r=l/,
so that
dr
~"~
^
Substituting these values in
1
w2
(//),
^
'
</6i
we
get
12 7. Lengths of polar sub-tangent and polar sub-normal. Let
the line through the pole 0, drawn perpendicular to the radius
vector OP, meet the tangent and normal at P in points T and G
respectively.
Then, OT, OG are respectively called
polar sub-normal at P.
From
the
the polar sub-tangent
&OPT, we get
OT
OP =tan
or
Hence
polar sub-tangent
=r
z
d^
,-
dr
=
d#
where
j
du
.
From the A^PG, we
w=
get
Fig.
88.
= cot ^,
or
OG=OP cot
Hence, polar sub-normals
<A
du
,
~A* where
dtf
www.EngineeringEBooksPdf.com
w=
1
r
and
272
DIFFERENTIAL CALCULUS
Note. The lengths PT and PGara sometimes called the lengths of the
polar tangent and polar normal at P. We have
>
/
2
r=OPseccp=rV[l-ftan q>]=r/\/rH-r
8
PG=OPcosec<p=rV[l+cot <p]=<\/r
12-8.
is
('
2
r -f
(
-fa
)
1.
To obtain the pedal equation of a curve
Pedal equations.
whose polar equation
* VI
a
given.
Let
r~f(6) be the given curve.
We
...(/)
have
1
1
1
-
Eliminating 6 between
equation of the curve.
and
(i)
fdr
we obtain the required pedal
(//),
The pedal equation is sometimes more conveniently obtained
between (/) and
6 and
by eliminating
<f>
dQ
tan <=/
r
,
dr
p=r
'
sin <.
Exercises
Show
1.
that for the curve
the length of polar tangent
is
constant.
Prove that for the curve
2.
the polar sub -tangent
is
constant.
a9, the polar sub-nDrmal
3.
Show
4.
Find the polar sub-tangent of
(/)
that for the spiral
r-a(H-cos0).
r
(Cirdioide)
(i7)
2^/r^=l
+e
2
(Hi)
V
r~a0 1(0-1).
Show
5.
that for the curve
:
*
rae L/)2
.
pol^r sub-normal varies as 9 2
'
polar sub- tangent
.
6
.
Find the polar sub-normal of
(/)
1.
r=aje
(//)
;
r
=a+b cos
Find p for the curve
www.EngineeringEBooksPdf.com
Q.
is
C3nstant.
cos Q.
(Conic)
KXERCISE8
Find the pedal equation ofr m =am cos m0.
8.
Logarithmically differentiating,
dQ
=r
or
f^
Thus
p-r
-.
sin
mm
is
we get
-=~-m
,
.
which
273
tan/n 9
-=-cot w0 = ta
p=r
P
(sin Jfi-}-m0)=r cos
m9.
or
the required pedal equation.
9.
Find the pedal equations of
(iV)
(v)
//r-
1 -j
<?
cos 0-
r-ci(Mcos0).
(v) r(l-sinj0) 2
(ix)
r
m =u m
a.
(iv)
r= o0.
(v/)
r--asin/ii0.
(viii)
sin mO~\ b m cos
r^-a + bcos$
iwfl.
10. Prove that the locus of the extremity of the
polar sub-normal of the
curve r=/(0) is the curve
r=-/'lfl-l").
Hence show
that tlie locus of the extremity of the polar
the equiangular spiral
11.
r
a?
w^
is
sub-normal of
another equiangular spiral.
(P.U. 1940)
Prove that the locus of the extremity of the polar sub-tangent of the
curve
!
-
r
4/(0)--0
it
u
Henoe show
/'(R4
that the locus of
ttie
0).
extremity of the polar sub-tangent of
the curve
r
is
(H
tan 10)/(m f-wtan 10)
a cardioide.
12.
Show
(
that the pedal equation of the spiral
form
www.EngineeringEBooksPdf.com
Allahabad 1943)
r=^a sech
is
of the
CHAPTER
XIII
DERIVATIVE OF ARCS
On the meaning of the lengths of arcs. The intuitive
13*1.
notion of the length of an arc of a curve is based upon the assumption
that it is possible for an inextensible fine string to take the form of
the given curve so that we may then stretch it along the number
axis and find out the number which measures its length.
This assumption, which is not analytical, cannot be the basis
of analytical treatment of the subject of lengths of arcs of curves.
Also to define lengths of arcs analytically is not within the scope of
this book.
Hence we have to base our treatment on an axiom which concerns the numerical measure of lengths of arcs and we now proceed
to give it.
Axiom. If P, Q, be any two points on
a curve such that the arc PQ is throughout its
length concave to the chord PQ, then
Chord
PL
where
PQ<arc
and
QL
PQ<PL+QL
9
are any two lines enclosing
the curve.
Fig. 89
An
important deduction from the axiom.
If P,
Q
be any two
points on a curve, then
PQ
= l.
rfxs
PQ
near P that the
arc
,.
lim
.
Q-+P
we take Q
-
chord
so
To prove it.
concave to the chord PQ.
Let PL be the tangent at P.
arc
PQ
is
Draw
Ai/)*
QLPL.
Let /_LPQ=a.
chord
PQ
<arc
;
we have
PQ<PL+LQ,
^ PQ
chord
or
,
chord
PQ
everywhere
.
cc+sm
a.
274
www.EngineeringEBooksPdf.com
Fig.
90
275
DERIVATIVE OF ARCS
Let
-*
Q
P so
limiting position
From
(/),
that chord
and a ->
we
PQ
tends to the tangent
PL
as its
0.
get
arcPQ
_^ ^ chord
^^*
~~
PQ
'
for
(cos
a+sin
a)
>
as a -> 0.
1
13-2.
Length of arc as a function. Let y =f(x) be the equation
of a curve on which we take a fixed point A.
To any given value of x corresponds a value of y, viz. 9 f(x)
to this pair of number x and/(x) corresponds a point P on the curve,
and this point P has some arcual lengths 's* from A. Thus 's' is a
function of x for the curve y=f(x).
Similarly, we can see that 's* is a function of the parameter '*'
for the curve
;
(Parametric Equation)
x=f(t) y=F(t),
a function of for the curve
9
and
is
r=f(6).
Cartesian Equations.
13*3.
(Polar Equation)
To prove that
for the curve
Let
V denote arcual distance of any
fixed point
We
A on
point P(x 9 y) from some
the curve.
take another point
AQ=s+Ss
Let an arc
Q(x+8x
9
yi-By) on the curve near P.
so that
PQ=Ss.
arc
From the
rt.-
&PQN,
angled
by
we
N
have
or
L
Fig. 91
or
fchord PQ-\*
L~a?cPe J
Let
Q
->
P so
( 8s \
^J
that in the limit
'
or
'
-
()'='+()
www.EngineeringEBooksPdf.com
MX
DIFFERENTIAL CALCULI^
276
We make a
convention that for the curve >> =/(*), 's
ed positively in the direction of, x, increasing so that,
with x. Hence ds\dx is positive.
r
is
measurincreases-
s,
Thus we have
taking positive sign before the radical.
Cor 1. If x=f(y) be the equation of the curve, thenfunction of y and, as above, it can be shown that
Cor.
From the
2.
cos
right-angled
A
Aror,
= &*
L NPQ=
pQ s/
's'
is
a>
PNQ, we h ave
arc
/
.
chor
is the angle which
-> ^ so that
-> //, where
j2
A'-axis.
with
at
jP'nmkes
direction
of
the
positive
tangent
NPQ
Let
,
cos r
d/=
dx
_
.
J^
1=
dx
ds
Hence
003
Similarly,
i
*=
dx
'
ds
we can show that
.
,
sin r
dt=
dy
,
ds
Parametric Equations.
13-4.
To prove
that
for the curve
Let *s' denote the actual distance of any point P(t) on
from a fixed point A of the same.
We
be
take another point
Q(t+St) on the curve.
its co-ordinates.
Let
B,TC.PQ
From
the
A PNQ>
= SS.
(Fig- 91, p. 275),
we get
www.EngineeringEBooksPdf.com
Let
the-
277
DERIVATIVES OF ARCS
'
St
Bt
or
2
/chord PQ\*( &s
1.
arcPQ / ( St
Let
We
that dsjdt
Q -> /* so that
measure
ie
V
/
2
8y \
/
^Uf
in the limit
positively in the direction of,
/,
increasing so
Thus
positive.
ds
:
It
dt
Polar Equations.
13-5.
To prove
)'+(dt )*T
that
for the curve
denote th'3 arcivvl length of any point P(r, J) from some
fixed point A on the curve.
We take another point Q(r+Sr, fl-f 80) on the curve near P.
Lot
s
'5
Let arc
AQ=s+$s
so that arc
PQ^bs.
Q
Fig. 92
Draw PN.OQ.
Now
JP7V/OP=sjn 85 so that
PJVW
sin S^.
Again
ONjOP^cos 89
Hence
so that OJV^==r cos 89.
NQ^OQ-ON
= r+8r~- r cos 80
=:rl
cos 80
+ 8r
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
278
From the
&PNQ
Dividing by
9
2
(S0)
,
we
get
we
get
~~^
PQ^
/chord
_
~~
f
2
Let
Q
We
measure ^' positively in the direction of Q increasing so
that dsjdff
-> P.
is
Therefore
positive.
-2
Thus
-V
Cor. 1. If 0=f(r) be the equation of the curve, then
function of r and, as above, it can be shown that
_
Cor.
.
2.
..,,,
sin
PN
c
s' is
r sin
= T sin 80
arc
80
"
*
~8lT
Us
chord
P so that /_PQN ->
PQ
^ where, <, is the angle between
of
the
the positive directions
tangent and the radius vector.
Let
Q
->
sm ^=r.
,
^
/.
.
,
1.
-j~
.
I
or
Again,
cos
2r sin2
TQ
www.EngineeringEBooksPdf.com
279
EXERCISES
2r sin 2
~
JSj+Sr
'
*
80
sin
arc
80^
S7~
^
'
chord
PQ
PQ
Sr \ 80
80
arc
,
-
8T
G
Let
->
chord
P
f r\
cos r
<i=(V r. 0.
i
i
^r N do .
.1
1+
,
.
)
dQ ) as
or
dr
.
Exercises
Find
1.
(i)
for the curves
t/5/d[x
cosh
y=c
.
jc/c.
^==a log
(//)
3V=^ (~ x).
5
(///)
(v)
(v/i)
8ay-x
4^-f
ind
2.
I
3.
Show
1
2
2
2
(a -jc
2
c/5/c/y
2
that ydsjdy
(v/)
).
logx=x
for the
2
.
curve
is
constant for the curve
2
j^jhi^jz^
_i
~ g
)
tf^^-rr)
>
fl
Find dsldQ for the following curves,
4.
(/)
(//)
(111)
d'v)
x=^a cos
^,
y^b
x=a cos 3
^,
^=sin 3 ^. (Astroid)
x=a(9
sin ^),
x=ae
sin 0,
(v) jc==fl(cos
Find dsjdB
5.
(/)
r=>ae
(iv)
r =00.
(v/i)
(v/ii)
6.
y=a(l-cos
0-f0 sin
0),
cos
0).
.
(Cycloid)
cos
y=a(sin
(Cardioide).
0).
:
r
(//)
a
2
=fl cos 20. (Lemniscate)
(Equiangular spiral)
8
(v)
r=a(0 -l).
w =aw cos nt0. (Cosine spiral).
m m (cos w0+sin w0).
r =a
r
Show
:
0.
for the fo'lowing curves
cot a
being the parameter
sin ^ (Ellipse)
y=ae"
r=a(l+cos0).
(///)
x 3 =ay*.
(iv)
that for the hyperbolical spiral
r0=a,
rfr
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
280
7.
Show
that r ds/dr
8.
Show
that
is
.
^r
constant for the curve
.is
constant for the curve
dr
If /?i and /? 2 be the p^rp^ndicular^ from tlv^ origin on the tangent and
9.
normal respectively at the point (x, y\, and if tan ^ -dy/dx, prove that
p^x sin ty~y cos ^,
and
/i a =.=a:
cos ipi-y sin
hence prove that
10.
Show
that for any pedal curve
ds
dr
~
pf(r],
r
2
V(r -/?
"
2
)
www.EngineeringEBooksPdf.com
^f
;
CHAPTER XIV
CONCAVITY CONVEXITY
;
POINTS OF INFLEXION
14-1.
Definitions.
Draw
P[c,f(c)] thereon.
We suppose
f'(c) is
some
Now
sider
finite
Consider a curve y=f(x) and any point
the tangent at P.
that this tangent
is
not parallel
X-axis so that
number.
there are three
mutually exclusive possibilities to con-
:
Y
V"
+X
Fig. 93
(/)
may
of
be,
Fig. 94
Fig. 95
A
portion of the curve on both sides of P, however small it
lies above the tangent P (i.e., towards the positive direction
X-axis).
In tliis case \vo say that the curve
convex downwards at P. (See Fig 93).
(//)
it
may
A
be,
is
concave upwards or
portion of the curve on both sides of P, however small
lies below tho tangent at P (i.e., towards the
negative
direction of X-axis).
In this ease we say that the curve
convex upwards at P. (See Fig. 94).
(Hi)
is
concave downwards or
The two portions of the curve on the two
different sides of the
at P. In this case we
tangent at
say that P
P,
is
sides of
(See Fig. 95).
Thus, the curve in the adjoined
figure 06 is concave upwards at every
point between P and P2 and concave
Y
1
downwards
and Po.
at every point between
P2 and P 3 are
flexion
P
lie
on
the curve crosses the tangent
a point of inflexion on the curve.
/ e.,
Pa
the points of in-
Q
on the curve.
Fig. 96
281
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
282^
From these
figures, it is easy to see that the curve changes the
of its bending from concavity to convexity or vice versa as a
point, moving along the curve, passing through a point of inflexion.
This property is also sometimes adopted as the definition of a point
of inflexion.
Note. The property of concavity or convexity of a curve at any point
direction
is
not an inherent property of the curve independent of the position of axes.
directions, as also positive and negative directions of
by convention and have no absolute meaning attached to
them. But this is not the case for a point of inflexion. The point where a
curve crosses the tangent is an inflexional point so that its existence in no way
depends upon the choice of axes.
Upward and downward
y-axis, are fixed
14 2. Criteria for concavity, convexity and inflexion. To determine whether a curve y~f(x) is concave upwards, concave downwards,
or has a point of inflexion at P [c,f(c)].
We take a point
the point P[c,/(c)].
Q[c+h, f(c+h)] on the curve y=f(x)
lying near
Q lies to the right or left of the point P according as
or
negative.
positive
Draw
\_X axis meeting the tangent at P in R.
The point
h
is
QM
The equation of the tangent at P is
y-Ac)=f'(c) (x-c),
and therefore the ordinate
abscissa c+h is
MR of this
tangent corresponding to the
f(c)+hf'(c).
www.EngineeringEBooksPdf.com
CONCAVITY, CONVEXITY, INFLEXION
MQ
Also, the ordinate
283*'-
of the curve for the abscissa
c+h
is-
= MQ~MR=f(c+h)-f(c)-hf'(c).
For concavity upwards
MQ >
MR,
at P, (Fig. 97).
i.e.,
RQ,
For concavity downwards
MQ <
For
i.e.,
positive
when Q
lies
on
either side of P.
lies
on
either side of P.
at P, (Fig. 98).
RQ is
when Q
negative
inflexion at P, (Fig. 99).
RQ
Q
MR,
is
is
when Q
positive
on the other side of
lies
lies
on one side of P and negative when?
P.
Thus, we have to examine the behaviour of RQ,
i.e.,
f(c+h)-f(c)-hf'(c)
which are
for values of, h,
theorem, with remainder after two terms,
By Taylor's
Case
sufficiently small numerically.
Let
I.
/"(c)
As the value f"(c) of/"(x)
is
>
we have
0.
for
positive
x=c,
there exists an-
every point x of which the second derivativef(x) ia positive. (3*51, p. 54). Let c+h be any point of thisThen c+yji is also a point of this interval and accordingly
interval.
interval around
/"(-f
2 A) is
c
for
positive.
Also
/*
2
/2
!,
is
positive.
RQ >0,
for sufficiently small positive
Hence
Case
and negative values of
the curve is concave
II.
Let
upwards at
f"(c)
<
P iff"(c) >
the curve is concave
Case HI.
By
Let
downwards at
/"(O^O
0.
0.
In this case we may how, as above, that
small positive and negative values of h.
Hence
h.
RQ <
P iff"
for sufficiently
(c)
<0.
fy///'"(c):0.
Taylor's theorem, with remainder after three terms, we*
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
:284
have
h*
3
!
so that
RQ=
3*
f'"( c
+e
*
h )> f r /"(c)=0.
There exists an interval around c frr every point X of which
the sign of/"'(c). Let c+li be any point on this interval.
f'"(x) has
Then
c-\-Q.Ji is also a point of this interval and accordingly f'"(c-{-Pji)
has the sign of f"'(c)
But A 3 /3
!
changes sign with the change in the sign of h.
changes sign with the change in the sign h.
Thus RQ
Jhe curve has inflexion
Case IV.
at
Hence
P iff"(c)^0 andf"
Generalisation
.
Let
Taylor's theorem, with remainder after n terms,
By
f(c+h)=f(c)+hf'(c) +
**
f"(c) +
+
^7!
/n
we have
~
1(C)
so that
There exists an internal around,
sign
off
n
(c)
c,
such that / n (c-f-0 n /i) has tho
or every point c4-h of this interval.
Also h n fn changes its sign or keeps the same sign while the
sign of, /i, changes, according as n is odd or even.
I
Thus
/i
is
RQ
changes sign ifrcis odd and keeps tho sign of/
n
(c),
if
even.
Hence if n is odd th*. curve hxs inflexion atP\ifn is even,
n
concave upwards or downwards according as f (c) > 0, or <0.
it
is
Another Criterion for points of inflexion. We know that
inflexion at P if it changes from concavity upwards to
concavity downwards, or vice versa, as a point moving along the curve
P such
passes through P, i.e., if there i a complete neighhourhood of
that at every point on one sid^ofP, lying in this neighbourhood, the
curve is concave upwards and at every point on the other side of P
14-3.
a curve has
www.EngineeringEBooksPdf.com
CONCAVITY, CONVEXITY, INFLEXION
the curve
IR
concave downwards. We thus see that the curve, has
iff(x) changes sign as x passes through c.
in-
flexion at P[c,f(c)],
'n
Note. We have already remarked that the position of the point of inflexion on a curve is independent of the choice of axes so that, in particular, the
positions of x and y axes may he interchanged witnout affecting the positions or
the points of inflexion on the curve. Thus the points of inflexion may also be
For pomts \vhere
determined by examining d z xldy 2 just as we examine d zy/dx 2
the tangent is parallel to X-axis, i e., where dy/dx is infinite, it becomes neces.
ary to examine d*x/dy instead
14-4.
Concavity and convexity with respect to a line. Let P be
a given point on a curve and, /, a straight line not passing through
Then the curve is said to be concave or convex at P with respect
P.
to /, according as a sufficiently small arc containing P and extending
to both 'sides of it lies entirely within or without the acute angleformed by, /, and the tangent to the curve at P.
Fig. 100
Thus the curve
Fig. 101
is
convex to
/
at
P
in Fig. 100
and concave
inr
Fig. 101.
In the noxt section, we deduce a test for concavity and convexiwith
ty
respect to the X-axis.
A
14-41.
test
of concavity and convexity with respect to the X-axis.
o
Fig. 103
Fig. 102
103,
From the examination of
we deduce the following
the figures 97, 98 and the figures 102 r
:
(i)
positive)
A
is
curve lying above, the axis of. x (so that the ordinate y is
convex or concave wit h respect to the axis of x according
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
286
as
it
concave upwards or downwards,
is
i.e.,
according a,ad*y/dx*
is
positive or negative.
A curve lying
(ii)
below the axis of x (so that the ordinate y is
convex or concave with the respect to the axis of x accoris concave downwards or upwards i.e., according as d 2yl<tx*
is
negative)
ding as it
is
negative or positive.
Thus, in either case, we see that a curve
.P with respect to the axis of x according as
convex or concave at
is
d*y
dx*
y
ds positive or negative at P.
Examples
I.
.is
Find if the curve
concave or convex upwards throughout.
(P.U. 1932)
Now
dy
dx
_
1
~~~
x
d*y
which
i.e.,
is
1
~
always negative.
Hence the curve is concave downwards,
convex upwards throughout.
Note.
We know that y=Iog x is negative or posi0<x<l or x >1. Thus for 0<x<l,
tive according as
2
yd*yldx is positive and for x>\,yd*y(dx* negative, so that
*k e curve is convex w.r. to x-axis if 0<x<l and concave
Fie
104
rig. ju^
2.
iy.r.
5Aow
//ra/
to a?-axis if
j=x
4 is
#>1.
concave upwards at the origin.
We have
=
dx
~dx*
24,
that
but positive at
Therefore the curve
3.
is
concave upwards at the origin.
Find the ranges of values of x for which
as concave upwards or downwards*
Also, determine
(0, 0).
its points
of itfft&xion.
www.EngineeringEBooksPdf.com
CONCAVITY, CONVEXITY, INFLEXION
287
We have
Now,
-V~>0
forx<l,
forx = l
;
"'=0;
dz v
for
1<*<2,
<0; forx=2,
for
*>2,
>0.
=0,
Thus the curve is concave upwards in the intervals
and is concave downwards in the interval [1. 2].
[
oo, 1),
<(2, GO ]
It has inflexions for
Therefore
the curve.
4.
Show
(1, 19)
x=l
and
(2,
that the curve
and x = 2.
33) are the
2
(a
two points of
+x 2 )^=a 2^
inflexion
has three points of
on
in-
flexion.
'
Here
_
~
dy
dx
~~ a 2
dx*
"
*=Q
for
x= ^3^,
0,
-
2
2
It is easy to see that d y[dx changes sign as, x, passes
the
Hence
curve has inflexions at the
ach of these values.
through
corres-
ponding points.
Thus
(V3a, V3a/4),
(0, 0),
(-
the three points of inflexion of the curve.
5.
Find the points of inflexion on the curve
Here, y,
is
the independent and, x, the deper\dei|t variable,
dx
--3
xl
.
(logy)-,
1
.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALC0LUS
288
d'x
1
3
.y=0 or
log
>
=0
2
if log
y~2,
i.e., if
=e=
e*.
Thus we expoct points of
Again
6
~~
or
inflexion oti the curve for
121oj; """ Qlocry
3
3
3
^~
J
;
=l
and".
3
Therefore
and e 2 give points of
are the two points of inflexion.
Hence
9
(8, e )
y~
I
inflexion,
so that
(0,1)
and'
Exercises
y~e x
everywhere concave upwards.
1.
Show
2.
Examine the curve .y^sin x for concavity and
that
is
convexity in the-
interval (0, 2*).
3.
is
concave upwards or downwards.
4.
is
Also,
which the curve
fin
1
the points of inflexion.
Also
find its points
of inflexion.
Find the intervals in which the curve
y= (cos* i sin xie x
concave upwards or downwards
6.
in
Find the ranges of the values of x in which the curve
concave upwards or downwards.
5.
i*
Find the ranges of values of x
;
x varying
in the interval (0, 2n).
Find the points of inflexion the curves
2
(i)
(Hi)
* V'
y--ax*-\-bx +<QX
4
A*=^-3.y
y ~~a 2
+x
3
4.v -f 5.
2
[xi)
y*^x(x\-\)*.
ny
(n)
(iv)
y=(x*
(yla).
=jc f (fl 2 -x).
(x)
2
*)/(3.x -f !)
x-~(y~\)(y2)(y3)"
a--j-x
xy^a* log
(xii)
.
Vl ' ;
'
(ix)
f
\-d.
2
2
y- x log
(D.U. Rons. 1947)
(D.U. Hons. 7955)
www.EngineeringEBooksPdf.com
'
(
289
EXERCISES
Also, obtain the equations of the inflexional tangents
(7) and
jjto
the curves
(it),
(xi).
Show
1.
that in the curve
the abscissae of the points of inflexion are
Show
8.
that the line joining the
h.
two points of inflexion of the curvs
y*(x-ay x*(x+a)
subtends an angle w/3 at the origin.
Find the values
9.
of, x, for
which the curve
54,y=U+ 5)^-10),
has an inflexion and draw a rough sketch of the curve for
-6<x<3, making
Show
10.
the
(MT.)j
inflexions.
that the curve
ay*=x(xa)(xb)
has two and only two points of inflexion.
Find the points of inflexion on the curves
11.
(/)
(ii)
;
/,
y~a
sin
cos
/
x2a cot Q, y=^2a sin
2
f.
0-
of the points of inflex on on
Show
that the abscissae
satisfy the equation
12.
y*=f(x)
(Hi)
jc=0(2<9-sin 9),;y=fl(2-cos 0).
x--=a tan
[f'(x)?=2f(x)f''(x).
13.
Show
the curve
(PV.)
that the points of inflexion of the curve
2
2
y =(jf-fl) (.v-6)
lie
on the
line
4/>.
www.EngineeringEBooksPdf.com
(Luckjow)
CHAPTER XV
CURVATURE
EVOLUTES
Introduction, Definition of Curvature.
15'1.
In our everyday
language, we make statements which involve the comparison of bendFor instance, at
ing or curvature of a road at two of its points.
times, we say, "The bend of the road is sharper at this place than at
that." Here we depend upon intuitive means of comparing the
curvature at two points provided the difference is fairly marked. But
we are far from intuitively assigning any definite numerical measure
to the curvature at any given point of a curve. In order to make
'Curvature' a subject of Mathematical investigation, we have to
assign, by some suitable definition, a numerical measure to it and
this has to be done in a way which
may be in agreement with our
intuitive notion of curvature.
This we proceed to do.
We
take a point Q on the curve
P.
near
lying
Let A be any fixed point on the
Let
curve.
arc
AP=s,
arc
AQ=s-\- 8s,
arc
PQ=Ss,
so that
Let
i/s
i/'+St/',
be the angles which
the positive directions of the tangents
at P and Q make with some fixed line.
The symbol, 8^ denotes the angle
through which the tangent turns as a
point moves along the curve from P to
a
8s.
distance
According to our intuitive feeling, 8$ will
Q through
be large or small, as compared with 8s, depending on the degree of
sharpness of the bend.
This suggests the following definitions
Fig. 105
:
(i)
the total bending or total curvature of the arc
be the angle 8$
(//)
the average curvature
ratio 8(pj8s
and
(ill)
PQ
is
defined to
;
of the arc
PQ
is
defined to be the
;
the curvature of the curve at
lim
Q->P
'"
Ph
defined to be
ds
290
www.EngineeringEBooksPdf.com
CURVATURE
Thus, by def.,
a
A/r/ds is the curvature
15-2.
Curvature of
circle is constant.
a
291
of the curve at any point
To prove
circle.
Intuitively, we feel that the curvature
throughout its circumference and that the
that the
P
curvature oj
of a circle
is
uniform
larger the radius of the
It will now be shown that
the smaller will be its curvature.
these intuitive conclusions are
consequences of our formal definition
of curvature.
circle,
Consider any circle with radius, r, and centre O.
Q be any points on the
Let P,
circle
and
'
Also let
tangents PT,
We
T
let arc
L
be the point where the
QT at P and
Q
meet.
have
From Elementary Trigonometry
r
Fig. 106
top
!_
""""
Us
Let
Q
->
P
r
so that in the limit,
we have
~~
ds
its
r
Thus the curvature at awjy point of a
radius and is, thus, a constant.
Also,
r increases.
15-3
it is
circle is
clear that the curvature, 1/r,
Radius of curvature.
and
is
decreases as the radius
The
reciprocal of the curvature
curvis called its radius o
generally denoted by, p, so that we have
of a curve at any point, in case
ature at the point
the reciprocal of
it is
=0,
= dThus the radius of curvature of a
circle at
any point
is
equal
t<?
its radius.
only
15-31.
The expression ds}dfy for radius of curvature is suitable
for those curves whose equations are given by means of a
between s and (p. We must, therefore, transform it so that
be applicable to other types of equations such as Cartesian,
relation
it
may
Polar, Pedal, Tangential Polar, etc.
It may be remarked that in the case of Cartesian and Polar
equations, the fi^ed line will be taken as X-axis and initial line res-
For the curve y=f(x), the positive direction of the tanone
in 'which, x, increases, and for
the
gent
r=/(0) the positive*
is the one in which, 0, increases.
direction of the
pectivelyis
tang&j^
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
292;
Ex.
Find the radius of curvature
(j)
any point of the following
(//)
i
(///)
(v)
at
j=c tan (Catenary).
s-4a sin ty (Cardioide).
(/v)
s=4a
s~c
sin
ty
log sec
:
(Cycloid).
$
(Tractrix).
sa log (tan 1^4-sec t//)fa tan ^ sec ^ (Parabola).
Radius of curvature for Cartesian Curves.
15*4.
15 41.
Explicit equations
y=f(x).
:
Now,
tan Y
\b =
Differentiating, w.r. to
s,
dy
/--
.
dx
we
get
dx
But, we have
(13-3, p. 275}*
where the positive sign
to be taken before the radical.
is
Hence
_~
~"d^
.
~
d2y
y2
dx 2
The radius of curvature,
Cor.
is positive or
P,
negative
or
d*yjdx*
positive,
negative ie., according as the
curve is concave upwards or downwards. Also, the equation (A)
shows that curvature is zsro at a point of inflexion.
according as
iSi
is
S'mss the valus of p is indspendent of the choice of X-axis
interchanging x and y we see that, p, is also gfcven by
Tfhis
formula
is
specially useful
when the tangent
is
perpendicular
K^axis in which case
L5;42.
Implicit equations
:
f(x,
y)=0.
^t points where dfjdy^fyjkQ-, we have, (JlO-94/ p. 213)
www.EngineeringEBooksPdf.com
and
to?
CURVATURE
_
"
293
....... "
.....
<***
(/,)*
2
2
Substituting these values of dy\ dx and d yjdx in the formula
A) above, we get
the sign
Note.
"
is
+
or
according as fv
The form of the formula
Parametric equations
At points where
f'(t)^0
9
negative or positive,
(B) shows that the expression for p will
^0 but/^0, The exceptional case
etain the same format points where /
y
^hich arises for the points where fx and fv
onsidered in chapter XVII.
1543.
is
:
become simultaneously
2
be
x=f(t), y=F(t).
we have
*L_^/_>dx~f'(t)'
(l
will
"'
w
/
,
1
yf'(t)F"(t)-F'(t)f"(t)
2
2
Substituting these- values of dyjclx and d y/dx in the formula
above, we get
Lf(t)+F'(t)]*_
vhere the sign
Note.
ame form
is
+
or ~~ according as/'(0
The formula
for the points
is
'
>t<l
positive or negative.
(C) shows that the expression for, p, will retain the
where /'(0=0 but F r U)^0.
If a curve passes
Newtonian Method.
15-44.
at
is
the
x
origin, then
tangent
origin and the axis of
,
rives the radius
;
of curvature at the
as
through
the
x ->
origin.
Here we obtain the values of y l and y%
at the origin.
2
Now, x l'2y, assumes the indeterminate form 0/0
Iim -*-= Urn
= hm
..
^
1
=
as
x ->
0.
(--)
1
www.EngineeringEBooksPdf.com
204
DirFBBBNTlAL CALCULUS
Also, from the formula (A) of
Thus at the
15 41,
we have
at the origin
origin where x-axis is a tangent,
<.=
Um
shown
It can similarly be
that
at the origin where
Y-axis
is
a
tangent,
=
lim
(g-).
These two formulae are due to Newton.
15-45.
Generalised Newtonian Formula.
If a curve
through the origin and X-axis
tangent at the origin, we have
/
X 2 -4-V 2
lim
SX^
=lim
'X
V \
+-~^
2 J
\2y
,~=lim[
2y
M
passes
is the
x
]
-----
2y
= p, at the orgin.
Fig. 107
=OP
2
2
2
is the square of the distance of any point
Here, X +J
n
the
curve
the origin O and, j, is the distance of the
from
y)
P
from
at O.
the
X-axis
point
tangent
,
P( x *
Interpreted in general terms this conclusion can be stated a
follows (see Fig. 108.)
:
of a curve, and
If OT be the tangent at any given point
PAf, the length of the perpendicular drawn from any point P to the
tangent at O, then the radius of curvature
at
O
lim
when the point
OP 2
2PM'
P tends
to
O
as
M
its limit.
Fig. 108
Note. It will be proved later on in 17-31, on p. 332 that the tangent
at the origin lying on a curve is obtained by equating to zero the lowest degree
terms in its equation.
Examples
1.
In the cycloid
x=a(/+sin
t),
j;=
cos
www.EngineeringEBooksPdf.com
t),
CURVATURE
295
prove that
L
&
P=4acos
(P.U.1944)
We have
~a
dy
>
s t),
_
a sin
fc
a(
I
_ 2_8i ?
*
+0087)
1
g
QPO ,
8ec
-
2
1
s
_
*/
__
'~
*
2
A
,..__
'
-_
11
rfx'
<
T
/.
2"ooB772
/
.
2
^sec
sin
.
-
1
--
2a cos 2
_
4a
cos
=4a
__
C08
.
2
"
*
>
COS*
2.
5Aoif
f/iar
Folium x*-\-y*=Zaxy
Differentiating,
/Ae curvature
is
we
of the point (3a/2, 3a/2) on
8^2/30.
get
xl+^^^ + flx.
or
Again, differentiating
x
(/),
we
...(/)
get
+V [*J^.g..
Substituting 3a/2,
1 for x, y,
3fl/2,
Hence the curvature
*
(
V
dyjdx respectively,
"o )
.
8V2
f
www.EngineeringEBooksPdf.com
we
get
DIFFERENTIAL CALCULUS
296
Find the radius of curvature at the origin of the curve
3.
It is easy to see that X-axis is the tangent at the origin.
15*45, p. 294]
[Refer note
Dividing by y
Let x -
y
we
get
so that lim
= 2p.
2
(x jy)
x->0
0.2p+5.2p-8=0,
or
p=4/5.
Exercises
Find the radius of curvature at any point on the curves
c cosh (xlc) (Catenary).
(/) y
1.
(//)
x-=a (cos
sin
t-l-t
(W)x% -Kyf =--fll
(iv) x^(a cos t)lt,
2.
.
/),
y^a(sin
t
t
cos
:
t).
(D.U. 1953)
(Astroid)
y^=(a sin
t)it.
Find the radius of curvature at the origin for
2
(//)
x*y-xy*-\ -2x y
2
(ill)
3.
is
2x*'\-ly*-\-4x
Show
y+xy-y
2
-\-2x- -0.
(P.U. Supp. 1939)
that the radius of curvature of any point of the astroid
x=a cos 8 0, y-^a sin 3
equal to three times the length of the perpendicular from the origin to the
(Andhra 1951)
tangent.
4.
Show
that for the curve
),
the radius of curvature
is
is, a,
Show
.
2c cosh 2
6.
(tic)
Show
sinh
t
.
(//c),
c
where
t
.
t
.
cosh
c
is
,
y^~-2c
,
cosh
-
t
.
c
the parameter.
that the radius of curvature at a point of the curve
x^ae
*
*
fi
fi
(sin 9
cos
0),
y=a?
(sin 0-fcos 0)
twice the distance of the tangent at the point from the oiigio.
7.
Prove that the radius of curvature at the point
(
8.
at
0),
that the radius of the curvature at any point of the curve
x=tc smh
is
cos
which the value of the parameter
TT/4.
5.
is,
(H
>>=flsin
at the point for
Show
2a, 2a)
on the curve
jc
f
l
y=flU
i
+y i )
is
"~2a
that the radius of curvature of the Lemniscate
the point where the tangent
is
parallel to x-axis is
www.EngineeringEBooksPdf.com
CURVATURE
Prove that the radius of curvature
9.
is
f
297
point (2a, 0) on the curve
at the
(65)1 n/(536).
Find the radius of curvature for 4(xla)-4(ylb)=l at the points
10.
where
it
touches the co-ordinate axes.
Show
11.
that the ratio of the radii of curvature at points
on the two
curves
which have the same abscissa varies as the square root of the ratio of the ordinal es.
Show
12.
that the radius of curvature at each point of the curve
#=-a(cos /-flog tan J/), y=a sin f,
the
is inversely proportional to the
length of the normal intercepted between
point on the curve and the X-axis.
13. Show that 3V3/2 is the least value of
for j?=log x.
p
j
|
at which the curvature is maximum,
and sh^w that the tangent at this point forms with the axes of co-ordinates a
triangle whose sides are in the ratio 1
V2 ^3.
(Rajputana 1951)
14.
Find the point on the curve
y=e x
:
15.
is least for
16.
Show
:
that the radius of curvature of the curve given by
the point x^a and its value there
Prove that for the ellipse
is
9a/lO.
=
p, being the perpendicular
from the centre upon the tangent at any point
(x, y).
(P.U. 1944)
17.
Prove that for the ellipse
x*la*-'r y*lb*
where
CD
18.
is
-l,p-C>
3
/a&
(Madras 1953)
the semi-conjugate diameter to CP.
Employ
generalised
Newtonian Formula to show that the radius of
curvature at any point of the ellipse x 2 /a 2 -! j 2 //> 2 -- 1 is equal to
3
(normal)
__
2
(scmi-idtus rectum)
19. The tangents at two points P,
Q on the cycloid
___
;
show
cos 0)
sin0), y=--a(\
;c=--a(0
are at right angles
that if p lf p a be the radii of curvatures at these points,
then
a
Pi
20.
ture
is 7.i
+P,
i
lfa l
.
Find the points on the parabola y a =8x
at
which the radius of curva-
.
l
21.
(a)
Prove that
parabola y*=4ax and S be
if,
its
p
be the radius of curvature
at
2
3
focus, then p varies as (SP)
.
any point
P on
the
(P-C/- 1952)
a focal
(b) Pi Pi''are the radii of curvature at the extremities of
chord of a parabola whose semHatus rectum is / prove that
;
22.
Show
that in the curve
x=- A fl(sinh u cosh u
if the
normal at P(x,
y)
fi/),
meets the axis of x
y=a
in G,
cosh 3 w,
the radius of curvature at
equal to 3 PG.
P
is
(B.U.1954)
www.EngineeringEBooksPdf.com
298
DIFFERENTIAL CALCULUS
23.
If x,
y are given
** -
dylds
p~
also,
show
show that
as functions of the arc s 9
dxjd*
'
d*ylds*
that
jL/^iY
*2y
(
Find p for the catenary
x=c log
24.
Show
LS+ V(c*+s
)], y--
that for the curve s=f(x),
_ [V~ VA /
25.
8
(
f^\_
Prove that for the curve s=ce xlc >
Find p for the curve s*=8ay.
(Patna, 1952)
Radius of curvature for polar curves
15*46.
Here we are to express
with respect to
terms of
ds\dty in
:
r=f(0).
and
r
its
derivatives
0.
From
the figure,
_
we
see that
_>
ds
_
~
ds
dO
ds
di
+ de
dQ
'
ds
IT
Fig. 109
tan
Now,
^= -T
.
-
_T
'
860
>
i.
A.
or
I
-"
,
TT-
s
I
_
,
d<f>
r.
tt
=
ae
dff*
tie
_^V J
I
-T"
77a
r.
\
CII+-"
Also
...(3,
www.EngineeringEBooksPdf.com
CURVATURE
where positive sign
From
(1), (2)
ds
is
to be taken before the radical.
and
we
(3),
obtain]
__
f
do
dp
.
*
f
J
dr
Therefore
where
and
r=
Cor. Since curvature is zero at a point of inflexion, thereforeat a point of inflexion on a polar curve,
ra
15-47.
+2r1 2 -rra =0.
Radius of curvature for pedal curves.
We know that
= 6 +<f>.
//
Differentiating w.r. to
r,
the relation
/?=r sin
we.
^,
have
=sm +r cos
A
Now
sin
^=r
,
-,
cos
^=
.
dr
(
13-5, Cor. p. 278)
-
~~ds
r
www.EngineeringEBooksPdf.com
300
DIFFERENTIAL CALCULUS
AT
.
or
Note. The relation $^0+$ is true whatever be the position of the point
on a curve and the tangent at it relative to the initial line. We can satisfy
ourselves on this point, it we examine a few different curves, keeping in view
the following conventions
:
6 is the angle through which the positive direction of the initial line has
to rotate to coincide with the positive direction ot the radius vector ;
is the
angle through which the positive direction of the radius vector has to rotate to
coincide with the positive direction of the tangent which is that of, 0, increasing;
$ is the angle through which the positive direction of the initial line has to
rotate to coincide with the positive direction of the tangent.
or
For
Fig. 111.
7
or
A
15 48. Radius of curvature for tangential polar equations.
called
of
is
a
for
an'd
between
curve,
relation
holding
every point
p
i/>,
Tangential polar equation.
The perpendicular drawn to the
tangent at any point (x, y) from the origin makes angle,
?r/2, with #-axia.
i//
Also p,
is
the length of this per-
pendicular.
Therefore the equation of the tan-
gent at
Fig. 112
P
is
o
p=sX sini/'
y cos
www.EngineeringEBooksPdf.com
&
:
CURVATURE
,301
X, Y being the current co-ordinates of any point on this tangent.
Since the point P(x, y) lies on this tangent, we have
p=x
which
is
Differentiating
w
(/),
ipy
sin
a relation between
;;,
i//,
to
r.
cos
we
i//,
(p
...(/>>
any point on the curve.
for
y
x,
get
sn
XT
Now
dx
,
a^
dx
-
=-->rfs-
^-f^ sin
-^
= x cos i^+>
(//) H'.r.
i//-f
sn
f
Differentiating
dy
to
ds
dy
-TTP
d*p
ds
d>,-= --}
(p,
dip
=x cos
-
ds
,-,-=? cos
p cos
t//
sin
/>
j//
sin T
<//
sin ^ cos
^.
i//,
we
get
dx
COS
=
=
x sin
x
p=x
i//
COS
cos
-f>'
sin i/^+j cos
sin
y cos
ip
i//
or
For the
First
Example
m
m
curve r =a cos md, prove
Method.
By
logarithmic differentiation,
m
dr
~~~~
'
dO
r
r,1
=
-
that
,.
=
r
dd
m9
m
cos
6
sin
tan md.
dV*
=
rtn sec 2
*s}
rm*'Qd(fi
md
f
tan
dr
rn9.
m0+r tah 2 mtf.
www.EngineeringEBooksPdf.com
we obtain
DIFFBBENTIAL CALCULUS
Hence
gec 2
_______
m 0_ tan m g
2
r2
15-46, p. 298>
(See
r a sec 8
_
r
mQ
2
1
(cos mtt
Second
is
273
tpage
Method.
Its pedal
equation, as obtained in Ex.
8,
*~ r~~dn
Hence
r.a m
Exercises
.r.
1.
Find the radius of curvature of the curve r=a cos H0 as a function of
2
(P.C/.)
Also, show that at a point where r=a its values is a/( 1 + /i ).
Find the radius of curvature
2.
ing curves
at
the point
(r, 0)
on each of the follow-
:
Kl+cos
3.
Find the radius of curvature of the curve r=a sin n$
4.
Prove that for the cardioide r=a(l +cos
at the origin.
(Allahabad)
2
0), p /r is
constant.
(P.U. 1954
5. Find the
following curves
radius of curvature at
the
point (p,
r)
;
/>.(/.
1955)
on each of the
:
6.
Find the radius of curvature for the ellipse
7.
Find the
radii of curvature of the curves
(/)
p=a sin # cos
(//)
pwA'm+1)
^.
-a
/(+)
sin
8. Show that at the points in which the Archimedean spiral r=a$ interthe hyperbolical spiral rQa, their curvatures are in the ratio 3 : 1
www.EngineeringEBooksPdf.com
CURVATURE
9.
If Pi> P2
be the
303
of curvature at the extremities of any chord of
radii
the cardioide
r=a(l-fcos
0),
which passes through the pole, then
10.
Find the radius of curvature to the curve
point where the tangent is parallel to the initial line.
r=a
(1
+cos
0) at the
A
line is drawn
through the origin meeting the cardioid
a*id the normals atP,
cos 0) in the points P,
meet in O; show
are proportional to PC and QC.
that the radii of curvature at P and
11.
Q
r==fl(l
Q
Q
Find the points of
12.
inflexion
(0 r-aJVW'-l)-
Show
13.
(11)
on the curves
r
2
0=a2
(See Cor.
.
that (a, 0), in polar co-ordinates,
is
15'46, p. 299)
a point of inflexion on
thecurver=0e0/(l-{-0).
Show
14.
lies
r=aQ n has
that the curve
and
between
1
fi
15.
Show
16.
Prove that
points of inflexion if and only
if,
n
and they are given by 0=^[/i(/i-f 1)].
that the curve re
for
=a (1+0)
has no point of inflexion.
any curve.
r/p=sin 9[l+<fr/</0],
where
p is
the radius of curvature and tan <p~rd$ldr.
(PU.1951\
If
17.
the equation to a curve
e be given in polars
pola r=/(0) and
prove that the curvature
is
Deduce or otherwise prove
The curve r=ae"
a
c
Pn
_i_
mn
.
cuts
If Pn
log
is
given by
any radius vector in the consecutive
denotes the radius of curvature at P n ,
j*.
pn
constant for all integral values of m and
19.
u*=
sin ' ? -
that the curvature
points PI, Pi, .............. ,
prove that
is
if
given by
($F+ U )
18.
Delhi 1948)
n.
(Delhi Hens. 1947)
Prove that in the curve
r
-a f sin
a
20.
the tangent turns three times as fast as the radius vector and that the curvature
varies as the radius vector.
(Delhi, 1949)
Centre of curvature for any point P of a curve
on the positive direction of the normal at P and
which
lies
point
15-5.
distance,
p,
The
from
is
is
the
at
a
it.
distance,
p,
must be taken with a proper sign so that the
pn the positive or negative direction of the
normal of /curvature
lies
centre according as,
p, is
positive or negative.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
304
thepositive direction of the normal is obtained by rotating
direction of the tangent
of
the
direction
(the
positive
tangant
positive
to y~f(x) is the one in which x increases) through ir/2 in the anti-
The
clockwise direction.
X
o
o
Fig. 114
Fig. 113
Fig. 116
Fig. 115
From an examination of the figures above, we see that thecentre of curvature at any point of a curve lies on the side towards which*
the concavity is turned.
15-51.
To find the
co-ordinates of the centre of curvature for any
point P(x, y) of the curve y =f(x).
Let the positive direction of the tangent make angle, $ 9 with
X-axis so that the positive direction of the normal makes angle
-j-^/2 with X-axis.
\jf
The equation of the normal
X-x
Y-y
_
""
cos ($ +7T/2)
is
'sin
(4r
+7T/2)
or
X-x
sin
\p
COS
wheje X, Y are the current co-ordinates of any potrit 6nthe
mal and, r is the variable distance, of the variable pbftit (X, Y>
:
www.EngineeringEBooksPdf.com
fir
CURVATURE
305
Thus the co-ordinates (X, Y) of the point on the
distance, r from (x, y) are
(x
normal at a
r sin ^, y-\-r cos
^),
For the centre of curvature,
if
Hence,
=y+p
But,
we know
sin ^
we hare
(X, Y) be the centre of curvature,
cos w.
that
1
=
;
X=x
^
cos
^~h ~
1
Another method.
Y=v4-
*
r
be the centre of curvature for P,
If
we
have
'
-
nrn
'
/ft
^
* TD/1f
Also,
=x
p sin
=}>+P
cos
L
Fig, 117
</r.
Substituting the value of sin ^, cos ^ and
and Y.
values of
X
curve
M
T
^LP+POcos^
p,
we can
obtain the
Evolute.
The locus of the centres of curvature of a
15*52.
is called its evolute and a curve is said to be an involute of its
evolute.
15-53.
circle
The
whose centre
P
of curvature of any point
of a curve is the
at the centre of curvature G and whose radiu
circle
is
The circle of curvatura Will Clearly touch the curve at
Curvature will be the same #3 thdfr of the curve.
Pand
its
;
15 54. Chord of curvature drawn in a given direction for any
a curve is the chord 6f fche circle of curvature through tu ~
point of
point drawn in the gaid direction. ^
.
www.EngineeringEBooksPdf.com
306
DIFFERENTIAL CALCULUS
We will now
determine expressions fpr the lengths of someimportant particular cases of the chords of curvature.
Let
(i)
PT
be the
tangent
and C
thfr
centre of curvature for
We draw
PC
any point -Pof a curve.
the circle with C as centre and
as radius.
PI?
to A'-axis
and PQ are chords of this
and Y-axis respectively.
Clearly, /.
RSP= $
==--
/_
SPQ
Thus, the chord of curvature
=2p
PQ=PS cos
Thus
the chord
of curvature
||
\\
sin
t//.
X-axis
i/>.
cos
<Jr.
Y-axis =2/ cos
i/>.
the pole. PT is the chord
of curvature through the
is
(ft")
of the
pole O.
Z_S?# = 2p
sin
parallel
.
PR=PS sm_RPS=2p
Now,
Also,
circle
circle
PS is the chord of the circle of
curvature perpendicular to the radius
vector OP,
Clearly
TQP=<l>=:^SPQ.
sin /_TQP=:2f
PT=PQ
/.
Thus,
f/ie
the pole=:2p sin
Also,
sin
<f>.
chord of curvature through
Fig. 119
<f>.
PS=PQ cos /_SPQ2p cos
<?>
Thus, the chord of curvature J_ wrf/wj vec/or=2 p co^
i//.
Examples
1.
(*>
J')
/
JTwrf /te co-ordinates of the centres
of curvature at any point'
pcrabola y*=*ax. Hence obtain its evolute.
^
Differentiating,
we
get
d*y
Jf
(X Y) be the
y
2a
centre of curvature
.
www.EngineeringEBooksPdf.com
CURVATURE
307
-
(0
4*2
14-
Y=
4a*
UOX}t
To
= ^^~.
-..(>
we have to eliminate x from
find the evolute,
(/)
and
(//).
Thus,
a
27^r2
or
is
the required evolute.
2.
.F/n J //i^ evolute
2
^_^ w
We
of the four cusped hypocycloid
^,
^
v,
,,.
2
^ +j;3=a?.
..^.,
can easily show that
dy
tan
d 2y
1
-.-=-
;
cosec
sec*
9.
-.
sec* g coseo
sin 2
cos g,
----- 4 -
=a sin3 0+3a
To
eliminate
0,
we
-----
sec 4
cosec
cos 2
sin
^
...
(/)
...
(ii)
,
0.
separately add and subtract
There-
(/), (//).
fore
X+ Y=a (cos 0+sin 0)
X-Y=a (cos 0-sin 0)
On
3
3
or
(X+ 7)*=a* (cos
or
(^-7)'=^*
(cos
^-f sin
0).
0-sin
0)
squaring and adding, we obtain
as the required evolute.
3.
y=a log sec
constant
length.
of
In the curve
to Y-axis is
(x/tf),
the chord
www.EngineeringEBooksPdf.com
of cuivature parallel
308
DIFFERENTIAL CALCULUS
TT
d)>
,
Here
tan ^^-r-
AI
^0=
Also,9
dx*
X
x
=tan
-
a
sec*
.'.
.
/.
is
*
= a sec
*
.
x
a
chord of curvature parallel to F-axis
=2p
which
X
a
(14. tan 8 xla)*
'
sec x/a
p=a ------- -29
^=
X
X
cos
cos i/>=2a sec
= 2a,
constant*
Exercises
1
.
Show
tint the evolute of the ellipse
$9
y=b sin
18
2.
(a)
Show
that for the hyperbola
ad the equation of the evolute is
(axfi
(6)
-(byfi -(
Prove that the evolute of the hyperbola
s
3.
Shaw
that the evolute of the tractrix
xa (cos
it
f
-Hog tan
J/),
y=
sin
/
the catenary
>>=0cosh
(xla).
IProve that the centres of curvature at points of a cycloid lie on an
(P.U. Supp. 1944)
equal cycloid.
4.
Show that (210/16, 210 (16) is the centre of curvature for the point
5.
(3a/2, 3n/2) of the Folium x*+y*=laxy.
6. Show that the centre of curvature of the point P(a, a) of
xi +y*=2cPxy divides the line OP in the ratio 6:1; O being the
co-ordinates.
7.
Show
that the parabolas
j^-aHx+l. *=-;K2 -f.y-f
have the same circle of curvature at thQ point
8.
is
ftt
Show
1
(1, 1).
that
the circle of curvature of the curve
the point (a/4, a/4).
www.EngineeringEBooksPdf.com
the curve
origin of
OTJKVATURB
9.
Find the
309
circle of curvature at the origin for the curve
x+ y=ax*+by*+c**.
10.
Show
(Delhi Hons. 1951)
that the circle of curvature at the origin of the parabola
is
mx ).
(D.U. 1955)
2
(a) Show that the circle of curvature, at the point (am 2am) of the
parabola y*=4ax has for its equation
x*+y*-6am2x-4ax+4am*y=3a*m*.
(D.U. Hons. 1957)
11.
,
9
(b)
the ellipse
12.
Find the equation of the
*
2
Show
+ ^=
the point (0, b) of
(P.U. Hons. 1959)
1
that the radii of curvature of the curve
x==ae^ (sin
and
circle of curvature at
cos^),^=a^
(sin
0+cos
0),
evolute at corresponding points are equal.
13.
Find the radius of curvature at any point P of y=c cosh (x]c) and
show that PC=PG where C is the centre of curvature at P and G the point of
intersection of the normal at P with x-axis.
(Allahabad)
its
14.
Show
m
spiral
spiral
that the chord of curvature through the pole of the equiangular
is 2r.
r=ae
15. Show that
r=ae
16.
^
bisected at the pole.
is
If cx
the chord of curvature through the pole of the equiangular
and cv be the chords of curvature
x a
point of the curve
y=ae
l
,
parallel to
prove that
17.
If Cx
and
cv
point of the catenary
18.
is
Show
the axes at any
{P
'
U
'
1948)
be the chords of curvature parallel to the axes at any
cosh (xlci, prove that
y=c
that the chord of curvature through the pole
r=0(l
Jr.
of the cardioide
cos 0).
and c * be the chords of curvature of the cardioide r=o(l 4-cos
through the pole and perpendicular to the radius vector, then
19.
Lf c r
20.
Show
0)
that the chord of curvature through the pole of the curve
r
m_ m C08 m Q
fl
t
is
2r/(ro4-l).
21.
Show
(Guj rat 1952)
that the chord of curvature through the pole for the curve
is
2/(r)//'(r).
(Lucknow)
br the chord of
curvature through the
22. Show that for the curve p~ae
,
pole is of constant length.
For the lemniscate r 2 =fl2 cos 20, show that the length of the tangent
23.
the
from
(B.U.)
origin to the circle of curvature at any point is r^3/ 3.
www.EngineeringEBooksPdf.com
310
DIFFERENTIAL CALCULUS
24. If P is any point on the curve r 2 =a2 cos 29 and Q is the intersection
of the normal at P with the line through O at right angles to the radius vector
OP, prove that the centre of curvature corresponding to P is a point of trisection
ofPQ.
(L.U.)
Pis any point on the curve r=a (1-l-cos 9) and Q is the intersection
of the normal at P with the line through the pole O perpendicular to OP, prove
that the centre of curvature at P is a paint of trisection of PQ remote from P.
25. If
26. 1 he circle of curvature at any point
meets the radius vector OP at A, show that
OP: AP=l
P of the
Lemniscate
r2
=a2 cos
29
2;
:
O being the pole.
27.
p lf p, are the radii of curvature at the corresponding points of a
f
8
cycloid and its evolute ; prove that Pi +p t is a constant.
28. Show that the chord of curvature through the focus of a parabola is
four times the focal distance of the point and the chord of curvature parallel to
the axis has the same length.
(Rajputana 1952)
29. Prove that the distance between the pole and the centre of curvature
n
w
is
corresponding to any point on the curve r =a cos
15-55.
Two
important properties of the evolute.
If (X, Y) be the centre of curvature for
any point P(x, y) on
the given curve, we have
X=x
Differentiating w.r. to
dX
p sin0,
9
.
ds- dx
.
T
-=-
dif>
dY
cos
X we obtain
=1
dx
Y=y+p
dib
y--
ds dx
.
sin
,
rfp
-=-.
* dx
dy
^x
= dy
ds dy
dip,
,
cos
Jx-d* dsdx-+
+
dp
dx,
rfp
-COS ^
fc
From
(i)
and
...
(//)
(//),
v
Now, dY/dX is the slope of the tangent to the evolute at pf
cot </>, is the slope of the normal PP' in the
and,
original curve at
P. By (//) the slopes of two lines, which have a point P' in
common,
are equal, and therefore they coincide.
www.EngineeringEBooksPdf.com
CURVATURE
311
Thus, the normals to a curve are the tangents
we square
Again,
one
(/)
and
(//)
and add.
Let, S, be the length of the arc of the
fixed point on it upto (X, Y) so that
Here, x
is
to its evolute.
evolute measured
from
a parameter for the evolute.
...(/*)
dx)
dS
dx
dx'~
or
where,
c, is
constant.
M
on the
,
2 be the two points
evolute corresponding to the points L 1?
L2 on the original curve. Let p lf P 2 be
the values of, p, for L l9 L 2 and S 19 S 2 be
Afa
the values of 8 for
19
Let Af
f
M
.
Thus
i.e.,
arc A/,
M = difference
2
the radii of curvatures at
L L9 L 2
between
,
Fig. 120
we have shown
that difference between the radii of
Thus,
curvatures at two points of a curve is equal to the length of the arc of the
evolute between the two corresponding points.
Note. We suppose that 5 is measured positively in the direction of x, increasing so that dSldx is positive. Also, dpfdx is positive or negative according
as, p, increases or decreases as x increases.
Thus
.-= T
ax
according as,
p,
dx
or
,
dx
increases or decreases for the values of, x, under consideration.
easy to see that the conclusion arrived at in this section remains the
same if we consider dSldx^dpfdx instead ofdS'dx^dpldx. It should, however,
be noted that the conclusion holds good only for that part of the curve for which
p , constantly increases or decreases so that d?/dx keeps the same sign.
Tt is
www.EngineeringEBooksPdf.com
312
DIFFERENTIAL CALCULUS
Ex.
which
is
Find the length of the arc of the evolute of the parabola
intercepted between the parabola.
The evolute is
Let L,
M
27a^=4(x-2a)
be
8
(Ex.
.
1,
the points of intersection of the evolute
p.
30)
LAM
and the parabola.
To find the co-ordinates of the points
of intersection L, M, we solve the two
equations simultaneously.
We get
07/7
^
*i I C*
X
A
/ry
^IM^V
--
---- ft/r\*^
A.("Y
-*1
*\^^
AC*y y
*
a are the roots of
x=Sa is the
of
which
equation
on ly admissible value
a being negative.
Now,
8a,
fl,
this cubic
*
Fie 121
/.
;
(84','4\/2a), (8a,
4^20)
are the co-ordinates of L,
If(Xt Y) be the centre of curvature for any point
parabola,/we have
X^3x+2a,
/
4\/2a)
is
3
(x, y)
(Ex.
I4:a*.
the centre of curvature for
f
the centre of curvature for P(2a,
/Thus A(2a, 0)
L(8<t,
Y=y
is
M.
1, p.
arc
Hence the required length Af4L=4a(3 v'3
Ex.
2.
Show
Ex.
3.
Show
that the
that the
ellipse
whole length of the evolute of the astroid
x~
it
1).
whole length of the evolute of the
120.
www.EngineeringEBooksPdf.com
306)
O(0, 0) and
The radius of curvature at/? = 0.4=20.
The radius of curvature at P=PL=(
/;
on the
CHAPTER XVI
ASYMPTOTES
16-1.
an
A
Definition.
straight line
said to be an Asymptote of
recedes to infinity along
from the straight line-
is
P
P
branch of a curve, if, as the point
the branch, the perpendicular distance of
infinite
tends to 0.
Illustration.
The
line
#=a
is
an asymptote of the Cissoid
(Seell-41>
y\a-x)=x*.
P
moves to
It is easy to see that as
(x, y)
line x--a tends to zero.
infinity,
its
distance
from the
Ex.
What
arc the asymptotes of the curves
j>=tan x ; ,y=cot x ; .y=sec x and >>=cosec x.
We know that the equation
16-2. Determination of Asymptotes.
of a line which is not parallel to X-axis is of the form
ymx+c.
The
...(/>
must tend to
abscissa, x,
infinity as the point P(x, y}
recedes to infinity along this line.
We shall now determine, m, and,
an asymptote of the given curve.
c,
so that the line
(i)
may
M
X
Fig. 123
Fig. 122
Ifp=MPbethe perpendicular distance of any point P(x,
on the infinite branch of a given curve from the line (/), we have
v mx c
p ->
as
x ->
lim(>>~ mx
oo
.
c)=0,
which means that, when x ->oo
.
313
www.EngineeringEBooksPdf.com
be
314
DIFFEBENTIAL CALCULUS
Also,
lim (yl*
w)=lim
mx). lim l/x=c.
(y
0=0
or
lim (yly)~m,
-when
x ->
oo.
Hence
m=
lim (y/x),
X ->
c=
lim (y
X ->
oo
mx),
oo
We have thus the following method to determine asymptotes
which are not parallel to j>-axis :
Jet lim (y/x)~m.
(/) Find lim (yjx)
X - oo
x - oo
;
(11)
Find lim (ymx)
X - oo
Then
y=mx+c is an
.
(ymx)^c.
lim
let
;
X ->
oo
asymptote
The values of y will be different according to the different
branches along which -Precedes to infinity, and so we expect several
values of lim (yjx) corresponding to the several values of y and also
Thus a curve may have
several corresponding values of lim (y mx).
more than one asymptote.
This method will determine all the asymptotes except those
parallel to 7-axis. To determine such asymptotes, we
start with the equation x^my+d which can represent every straight
line not parallel to A'-axis and show, that when y -^ oo
whiah are
m^lim
The asymptotes not
(xjy)
and d~]im (xmy).
parallel to
any
axis can be obtained
way.
Ex.
Examine
1.
the Folium
for asymptotes.
The given equation is of the third degree.
To find lim (y/x), divide the equation (/) by X s
xx
x J
Let
Jc
-
oo.
We then get
l-[-m
3
=Oor
(m+l)(/n
2
The roots of w 2 -- m + 1 =0 are not
m+l)~Q.
real.
www.EngineeringEBooksPdf.com
,
so that
either
ASYMPTOTES
To
put
mx) when
find lim (y
so that,
y+xp
Putting
x
p
/?, is
for
y
m=
315
l,
i.e.,
to find lim (y+x),
a variable which -> c when x ->
in the equation
(i),
we
oo
we
.
get
or
which
is
of the second degree in x.
2
Dividing by x
Let # -*
oo
.
,
we
get
We then have
3(c+a)=0
or
c=
a.
Hence
x
>>=
is
the only asymptote of the given curve.
If
is
a or .*+}> 4-0=0,
we
start
with x=my-\-d, we get no new asymptotes.
the only asymptote of the given curve.
Ex. 2. Find the asymptotes of the following curves
Thus
:
(i)
(ill)
16*3.
Working
rules
for
determining
asymptotes.
Shorter
In practice the rules obtained below for determining
-asymptotes are found more convenient than the method which involves direct determination of
Methods.
lim (yjx) and lim
(ymx).
Firstly we shall consider the case of asymptotes parallel to the
co-ordinate axes and then that of oblique asymptotes.
16-31.
Determination of the asymptotes parallel to the co-ordi-
nate axes.
Asymptotes parallel to Y-axis.
Let
x^k
..(t)
be an asymptote of the curve, so that we have to determine k.
Here, y, alone tends to infinity as a point P (x, y) recedes to
infinity along the curve.
www.EngineeringEBooksPdf.com
816
DIFFERENTIAL CALCULUS
The
line
(t) is
PM of
distance
equal to x
any point P(x,
on the curve from
y)
the-
k.
fc)=0 whenj>
oo
lim #=fc when y -
oo
lim (x
,
or
,
which gives fc.
Thus to find the asymptotes parallel
we find
Y-axis,
fc
2,
etc., to
the
which
x=k2
Then x=fc,,
to*
definite value or values fcr
x tends as y tends to oo .
,
etc. are
the required
asymptotes.
We will now obtain a simple rule to
obtain the asymptotes of a rational Algebraic
Fig. 124
curve which are parallel to F-axis.
We
y, so that
.
arrange the equation of the curve in descending powers of
it takes the form
where
are polynomials in x.
Dividing the equation
Let y ->
oo
.
We
(i)
by y
so that,
fc,
Let
is
fe
lf
(ii)
,
get
write
lim
The equation
m we
x=k*
gives
a root of the equation <(x)=0.
fc
2
be the roots of
etc.,
<j)(x)
=0.
Then the asymptotes?
parallel to 7-axis are
x~k
From
l9
we know
x=fc 2
,
etc.
that (x
(x fc 2 ),
fc,),
algebra,
factors of ^(x) which is the co-efficient of the highest
in the given equation.
etc.,
are the
power y
m of
y
The asymptotes parallel to Y-axis are
Hence we have the rule
obtained by equating to zero the real linear factors in the co-efficient of
the highest power of, y, in the equation of the curve.
The curve will have no asymptote parallel to F-axis, if the
co-efficient of the highest power of, y, is a constant or if its linear
factors are all imaginary.
:
Asymptotes parallel to X-axis. As above it can be shown that
the asymptotes, which are parallel to X-axis, are obtained by equating
to zero the real linear factors in the co-efficient of the highest power of>
x, in the equation of the curve.
www.EngineeringEBooksPdf.com
317
ASYMPTOTES
To determine
16-32.
the
asymptotes
the general rational
of
algebraic equation
Ur is a homogeneous expression
We writo Ur in the form
where,
,
where
<f>,(ylx)
is
a polynomial in y\x of degree,
So we write
x
"+-("x
of degree,
(i)
r,
r,
in x, y.
at most,
in the form
)+*""*-' ()+*"<*
(i )+
Dividing by x*, we get
On
taking limits, as x-^oo
,
we obtain the equation
which determines the slopes of the asymptotes.
Let
We
m
l
be one of the roots of this equation so that
write
Substituting this value of yjx in ('),
we
get
Expanding each term by Taylor's theorem, wo get
..(wJ+^-f ..,(111,)+.,
www.EngineeringEBooksPdf.com
.
.
]
DIFFERENTIAL CALCULUS
318
Arranging terms according to descending powers of x, we get
1
Putting ^ n (m,)=0 and then dividing by x"- , we get
-
Let x -
oo
We write lira />!=!.
.
-K
.
.
.
=0
Therefore
or
Therefore
is
the asymptote corresponding to the slope
W
A,
if
^(
Similarly
w
are the asymptotes of the curve corresponding to the slopes
2 m#
which are the roots of ^ n (m)=0, if <f>' n (m 2 ), ^' n (w 3 ), etc., are,
etc.,
not
0.
Exceptional case.
Let
^(m,) =0.
If ^' n (m,)=0 but ^-1(^1)^0, then the equation (vi) does not
determine any value of c l and, therefore, tho/e is no asymptote corresponding to the slope m v
Now
suppose
#'n('i)=-0=^ _ 1 (m 1 ).
In this case, (vi) becomes an identity and we have to
examine the equation (v) which now becomes
fl
On
tion
taking limits, as x ->
oo
,
we
see that
q
is
a root of the equa-
'
which
dejf'ermines
two values of c,, say
c,', c,",
provided that
www.EngineeringEBooksPdf.com
re-
ASYMPTOTES
3 1 fr*
Thus
y=m,x+ c/, yare t he two asymptotes corresponding to the slope
w
l
.
These
are-
clearly parallel.
This
is
known
as the case of parallel asymptotes.
The polynomial J n(m)
Important Note.
x=l andy=m
is
obtained by putting
xn n (ylx), and n -i(tn),
n
^ n - 1 (^/x), x ~^ n ^(yjx), etc., in a'
the highest degree terms
n^1
in
^n- 2 ( m ) etc -> are obtained from x
<f>
<f>
similar way.
Fxunples
Find the asymptotes parallel
1.
curves
2
(x +>>*)* --tfy*-=0.
(//)
(i) The co-efficient of the highest
the asymptote parallel to y axis is
(i)
of
jcy-d*(jt+y)=<).
power y of y is x a.
ther
0.
The co-efficient of the highest power x3 of x is 1 which
Hence there is no asymptote parallel to x -ax is.
(//)
The
the highest power y^ of y
co-efficient of
Hence
1
xa
stant.
to co-ordinate axes,
:
is
x2
a con-
is
a2
.
Also,
rfence
x
are the
two asymptotes
a=Q,
x+a~
parallel to }>-axis.
It may similarly be shown that>> #~0, y 4-^=0 are the twoasymptotes parallel to x-axis.
2. Find the asymptotes of the cubic curve
2x 3 x 2^ +2xy* +y* 4x 2 +8xy 4x + 1 =0.
Putting x=l,
separately,
we
<f>
The
3
y=m in
the third degree and second degree terras
get
2m 2 +m 3
(m)=2m
,
slopes of the asymptotes are given
^3 (w)=w
3
by
2m2
or
111=
Again,
cf
c. is
1
-1,
1,
2
given by
4i+3/ I )+(--4+8m)~0,
www.EngineeringEBooksPdf.com
320
DIFFERENTIAL CALCULUS
m=
1, 1, 2, we get c=2,
Putting
Therefore the asymptotes are
3.
4 respectively.
2,
Find the asymptotes of
x2
4y*
l
m (l
2
m)
The slopes of the asymptotes, given by
To determine c, we have
<
3
m)
(w)=0, are
1.
1, 1,
i.e.,
c(~l-~2m+3m 2)+(2+2m-4w 2 )==0.
For w=s 1, this gives c=l and therefore, j>=
...(0
x+1
the
is
^corresponding asymptote.
For
the equation (/) becomes
In this exceptional case,
w=l,
cally true.
0.
c+0=0
c,
is
which is identidetermined from the
equation
i.e.,
2
(c
For
/w
/2)(-2+6m)+c(2-8m)+l(l+m)=0.
= l, this becomes
2c'~6c?+2=0,
i.e.,
c=(3v^5)/2.
Hence y~x+(3<\/5)l2 are the two
responding to the slope
We have
parallel
cor-
asymptotes
1.
thus obtained
all
the asymptotes of the curve.
Exercises
Find the asymptotes parallel to co-ordinate axes of the following curves
1.
ytx-a^x-a^O.
2.
:
-
x*y~-3x*-5xy+6y+2=Q.
4.
a2 /*2
y=xl(x*-l).
Find the asymptotes of the following curves
3.
+W=i.
:
2
5.
x(y-x) =x(y-x)+2.
6.
x\x-y)*-\-o?(x*-y*)=o?xy.
1.
(x~-y)*(x*+y*)-W(x-y)x*+l2y*+2x+y~0.
x 2y+xy*+xy+y*+3x=Q.
8.
(D.U. 1952)
(P.U.)
(P.U.)
'
10.
(x-y+l)(x-y-2)(x+y)=Sx-l.
y*-x*y+2xy*-y+l=Q.
11.
ylx-y) =x+y.
12.
^V^
9.
2
2
-/) 2
www.EngineeringEBooksPdf.com
(P.U.)
(P.U.)
(B.U.)
ASYMPTOTES
2
321
~* 2 -l.
13.
Xj>-l)
14
xy***(x+y)*.
15.
(x+y)(x-y)(2x-y}-4x(x-2!*)+4x
16.
xy*-x*y-3x*-2xy
17.
2*(>'-3)
18.
(y~a) (x -a*)=x'-\
19.
U-f/r(x
20.
^SyV-x'^-SxM^-ZJOH- 3* 2 4>-f5-0.
2
3
3
(A:-j) (.v--2v)(.v-3r)---2a(A' -v )--2aV-2v)Ul>')^-0,
21.
2
2
y
(L.I7.)
(L
-3X*-l)
2
-t-jty -f-^
}
x~1y-\
U)
(L.U.)
0.
1
(L.U.)
2
2
2
y*
I
0.
(^ra)
.
(Lucknow}
a*.
)-o 2 JC 2
!
a*:y~x).
3
i
(DehiHons. 1950)
22.
yW
Show
2
4
/(fl
-* 2
(P.U. 1955 Supp.)
).
that the following curves have
no asymptotes
23.
jc^f^---^^-^).
24.
25.
aV 2 -.v\2a-JC).
26.
y ---x(xfl)
27.
Find the equation of the tangent to the curve
parallel to its asymptote.
28.
An asymptote
is
:
2
x3
I
2
.
2
3
^ --3ax which
is
sometimes defined as a straight line which cuts the
Criticise this definition and replace it by a
curve in two points at infinity.
(P.U. 1955 Supp.)
correct definition.
16 33.
(/)
Some
deductions from
16 32.
The number of asymptotes of an algebraic curve of the nth
degree cannot exceed
n.
The slopes of the asymptotes which are not parallel to F-axis
are given as the roots of the equation n (m)~ which is of degree n
at the most.
<f>
In casa the curve possesses one or more asymptotes parallel to
is easy to see
that the degree of ^ n (/w) Q will bo
smaller than n by at least the same number.
K-axis, then it
Hence the
result.
an algebraic curve are parallel to the lines
(//) 77?e asymptotes of
obtained by equating to zero the factors of the highest degree terms in its
equation.
y
Let, m, be a root of the equation
^1^=0 is parallel' to an asymptote.
^ n (^)--0,
so that
the line
n (yl*)
elementary algebra, (y/xrn^ is a factor of
n
converse!
t/
Also
a
x
is
of
factor
n
n
y,
i.e.,
hence, yr^x,
(ylx),
we see that if, y m x x, is a factor of U n then m v is a root of
By
<f>
.
<f>
^,(wt)
= 0.
factor, then a
consideration will show that the curve will possess asymptotes
parallel to x=0, i.e., to .y-axis.
In case the highest degree terms contain, x, as a
little
www.EngineeringEBooksPdf.com
DIFFERENTIAL OALOQLUS
322
(iit)
Case of
In this
fore,
case,
m
satisfies
iy
the three equations
Since ^ n (m) and its derivative ^' n (m) vanish for
by elementary 2algebra, m l is a double root of
therefore,
m^
a root of
is
l)th degree terras
m=m
Un _ v
<f>n
_ 1 (m)=0,y
m
L
x
there-
lt
<^ n (m)~0
a factor of the highest degree terms
is
/w^)
(>>
Also, since
(AI
parallel asymptotes.
f/ n
and
.
a factor of the
is
Thus, we see that in the exceptional case of
16*32, a twice
repeated linear factor of U n is also a non-repeated factor of U n _ r
There
be no asymptote with slope m l9
butw>* of ^ n -i(fH)--0, i.e.,
and >> fl^x is not a factor of Un ~i-
will
if
^ n (m)=0, <' n (w)=0
factor of C/ n
w is a root of
(ym^) is a
1
if
2
The
results obtained in the paragraphs (11) and (///) above enable
the process of determining the asymptotes as shown in the
following examples.
Note.
us to shorten
The first step mil always consist in fact or i sing the expression
highest degree terms in the given equation.
formed of
the
Examples
1.
Find the asymptotes of the Folium
The curve has no asymptotes
parallel to co-ordinate axe*.
Factorizing the highest degree terms,
so that,
terms.
y+x,
is
the only real linear factor of the highest degree
Hence the curve has only one
whose slope is
to the line y-\~x~
We have, now
We have
In the limit,
real
asymptote which
when x ->
we have
xa
y
is
parallel
1.
to find, lim (y+x),
y=
is
we get
i.e.,
tha only real asymptote of the folium.
www.EngineeringEBooksPdf.com
<x>
and yjx ->
1.
ASYMPTOTES
323
It is easy to see that we could have eliminated the step
simplified the process by saying that the required asymptote
v 4- x =-- Km
lim
y+X
and
(i),
is
IB
when x ->
oo
and yjx ->
i
1.
Find the asymptotes of
2.
Equating to zero the
we
flO.y
2
-^ f
co-efficient of the highest
1
-0.
power y* of
y,
see that
4xH-10=0,
is
+ 15xy
2
-f 5x
i.e.,
2x+5--=0,
one asymptote.
Factorising the highest degree terms,
we
get
Here 2y +x is a repeated linear factor of highest degree terms,
3rd
There will, therefore, be no asymptote parallel to
i.e.,
degree.
2y+JC~0 if (2y+x) is not a factor of the 2nd degree terms also.
But this is not the case. In fact, the equation is
+5(x +y)(x +2y) - 2y + 1 - 0.
curve has two asymptotes parallel to
x(2y
Therefore, the
2
-f-*)
We have now to find lim (y+\x) when x ->
Let lim (y -f Jx)=--c so that lim (2^+x) 2c.
Dividing by x, the equation becomes
x and
yjx ->
In the limit,
4cM-5.2c(l-i)-2(-i)+0=0
4c 2 +5c + l=-0
or
-
r
C
.
Hence
are the two
It
is
ified
simplified
y=
_
t
5,
...(0
>
i
1.
J^-~J and.y=
Jx
1,
more asymptotes.
easy to see that we could have eliminated the step
that the asymptotes are
the process by saying
sayi
(i)
and
(2y+xy+5(2y+x). lim (l+y/x)+ lim
i.e.,
(ty+x)*+5(2y+x).
J
+1-0
or 2(2>> fx) 2 +5(2>>+*)+2-0,
which gives
=0
and
4>>-f 2x
+ l=0.
www.EngineeringEBooksPdf.com
324
DIFFERENTIAL CALCULUS
Find the asymptotes of
3.
The asymptotes parallel to the two imaginary lines
are imaginary. To obtain the two asymptotes parallel to the lines
x y~Q, we re-write the equation, on dividing it by (x 2 +^ 2 ), an
--
,
1+0
t+--
,
We take the limits when x ->
asymptotes are
i.e..
(x-,y)
and yjx
oo
--> 1
Therefore the
.
8
2=0, x
y
.v
i.e.,
2
/jr)
3>-3-=0.
/7m/ /Ae asymptotes of
4.
(x-y +2)(2jt- 3j +4)(4jc - 5y +6) +5* ~fi v -4-7=0.
The asymptote parallel to x--jH-2=0 is
-
.*
when
.v
->
x and
,
v/x ->
i
,
1
,
f
T
A'--v4-2fhm
5-6^/x-f7/x
---'
I
"1
-v
J
=0.
'
.
,
---,;
,
L^-^/^+^WC
4 "" 5
,
,,,,,.
^-!-^)
x~^+2-0,
or
as the limit
is
zero because of the factor 1/x.
we can show that
Similarly,
3j>+4=0, 4x- -5y -|-6-^0
2x
are also the asymptotes of the curve.
16-4.
Asymptotes by Inspection.
If the equation
of a curve of the
nth degree can be put in the form
Fn _ a is of degree (w
when equated to zero
2) a^ ///^ wort, //^ ev^r.y /me^r /ac/or of
will give an asymptote, provided that no
equating to zero any other linear factor of n is
where
Fn
,
straight line obtained by
parallel to it or coincident with
Let
where
ax+by+c=Q
Fn _ t
ax +by 4-0=0
is
of
F
it.
be a non-repeated factor of
degree
(w
1).
Fn
.
The asymptote
is
www.EngineeringEBooksPdf.com
We
\*rite
parallel
to
32&
ASYMPTOTES
p
ax+by+c+lim /~^0,
fn-i
when x
->
and y/x ->
oo
a\b.
limit (/ rn-a/' rn-i)> w ^ divide the numerator as
well as the denominator by x n ~ l and see that 1/x appears as a factor
To determine the
so that jPn _ 2 /Fn ^
1
-
>
x ->
a*
x
.
Thus
is
on asymptote.
Exercises
Find the asymptotes of the following curves
2.
(x--i)(;c-2)(;c-Kx)+*
2
:
j-x+1 -0.
9
-*M .y2 + X + y- X +\^Q.
3
3.
y
4.
x(y ~lby+2b*)^y*-3bx*+b*.
5.
^ 3 f6.v 2 ^MlA:/--h6^4-3^4-12x^f 1 ly^-ZxfSy f-5-0.
2
6.
.x
2
2 2
2
+ ^-2x 2 )(2>'
7.
(y -HA7^2x
8.
x(y-3)*^.4y(x~\)*.
9.
(a-}-x)
2
2
3 ><
16
5.
(/?
2
5A-y
-}-(y
)
-}A'
2
)--xV
2
-jc)-7/
-l9xy-23x
2
-hx f 2/
|
3-^>.
2
.
3
-h8^^4jc -3/-f 9A7
-6A' 2
Intersection of a curve and
Any asymptote of a curve of the
its
i2^
2x f
-0.
1
asymptotes.
nth degree cuts the curve
in (n
2)
points.
Let y =niK-\-c be an asymptote of the curve
To
find the points of intersection,
we have
to
solve
the two
equations simultaneously.
The
abscissae of the points of intersection are the
roots
of the
equation
xn
<l>
n (m
>
(3>'^,v)MM3vfAr)(xHr)-h9^-|-6A7f9y^6x+9-0.
2
10.
(7\l/.
+clx)
+x
^n-i(m +c/x) +x*-*<f> n - 2 (m +cjx)
+ ...... -0.
by Taylor's theorem
to
according
descending powers of x, we get
Expanding each term
www.EngineeringEBooksPdf.com
...(/)
and arranging
326
DIFFERENTIAL CALCULUS
Aay^mx+c
an asymptote, the
is
co-efficients
of x n and x*~ l
are both zero.
Thus the equation
(11)
reduces to that "of (n 2)th degree
Hence the result.
and,
therefore, determines (ttr-2) values of x.
Cor,
in
1.
n(n2)
The,
/i,
asymptotes of a
curv^e of the
nth degree cut
it
points.
If (he equation of a curve of the nth degree can be put in
where Fn _ 2 is of degree (n 2) at the most and
n consists of n, non-repeated linear factors, then the n(n2) points of
intersection of the curve and its asymptotes lie on the curve
Cor. 2.
the form
F
Fn +Fn _ 2 =0
9
result follows at once from the fact that Fn =0 is the joint
At the points of intersection of
of
the, n, asymptotes.
equation
Fn =0 and
the curve and its asymptotes, the two
equations
Fn n _ 2 ~Q hold simultaneously and therefore at such points we have
The
+F
*-=<>.
Particular cases
3, and therefore the asymptotes cut the
(i) For a cubic, n
curve in 3(3 2) =3 points which lie on a curve of degree
1,
i.e., on a straight line.
32 =
asymptotes cut the
(11) For a quart ic, n=4, and, therefore the
curve in 4(4 2)~8 points which lie on a curve of degree 4 2=2
i.e., on a conic.
Examples
1.
Find the asymptotes of the curve
x*yxjP+xy+yP+x-y=Q
and show
(P.U. 1955)
that they cut the curve again in three points
which
lie
on
the
line
x+y-^0.
The asymptotes of the given curve, as may be
are
The
i.e.,
joint equation of the asymptotes is
x 2y-xy*-\-xy+y*-2y=Q.
The equation of the curve can be written
as
Here
F^tfy
Henee the points of
intersection lie
on the
www.EngineeringEBooksPdf.com
line
(P.U. 1940)
easily
shown
ASYMPTOTES
Show
2.
that the asymptotes
327
of the quartic
2
cut the curve in the eight points which
The asymptotes of the curve
lie
on a
-
1
=0,
circle.
are
x+2y=0,x-2 y+l=0, x-3>'=0,
i
so that their joint equations
is
x
.e.,
The equation of the curve can be written
X
X
5V 2^^
( 2_4^2)( 2_9^2)_|_
Hence the points of intersection
3,
lie
as
on the
circle
Find the equation of the cubic which has the same asymptotes
as the curve
and which passes through
the points (0, 0), (/, 0) and(Q,
I).
(Delhi Hons. 1948)
We
write
Fjssx
8
- Gx-y +
1 1
xv 2 - 6y 3
= (x-y)(x-2y)(x-3y).
Fjsx+y+1.
The equation of the curve can be written in the form /rs +jF1 =0
where F% has non -repeated linear factors. Thus 7^=0 is the joint
equation of the asymptotes of the cubic.
The general equation of the cubic is of the form
or
where ax -\~by-\-c
(0,
is
the general linear expression.
In order that it
we must have
1),
may
pass through the points
c=0,
6+6=0 or
Thus the required cubic
is
www.EngineeringEBooksPdf.com
(0, 0), (1,
0)
and
DIFFERENTIAL CALCULUS
328
Exercises
Show
1.
that the asymptotes of the cubic
2
xy - 2xy -f 2x - y
points which lie on the line
x9
cut the curve again in
0,
3.v-->>---0.
If a right line is drawn through the point (a, 0) parallel to the asymptote of the cubic (Jc a) 3 x 2y=Q, prove that the portion of the line intercepted
2.
by the axes
is
bisected by the curve.
(C.C/.)
2
2
Through any point P on the hyperbola x / ~ 2ax, a straight line is
drawn parallel to the only asymptote of the curve x 3 -f-.y3 =30;t2 meeting the curve
in A and B
show that Pis the mid-point of AB.
3.
;
4.
Show
that y ^x + a
is
the only asymptote of the curve
A straight line parallel to the asymptote
that the mid-point of PQ lies on the hyperbola
meets the curve
in P,
Q
;
show
x(x-y)+ay---0.
5.
Find the equation of the straight
line
on which
lie
three points of
intersection of the cubic
2
8
xV-jn' - 2r
and
its
i
4r 2
I
2xy
v
f
I
asymptotes.
6.
Find the asymptotes of the curve
4.x*
~13*V+V
a
3
f-32xV-42.v -20x
2
i-74>'
-56v-f
4,v f
16
0,
and show that they pass through the intersection of the curve with
(D.U. Hons. 1953)
7.
Find
ail
the asymptotes of the curve
3x 3 -f 2x*y -Ix
Show
on a straight
8.
that the asymptotes
line,
and
meet the curve again
in three
find the equation of this line.
points which lie
(D.U. Hons. 1952)
Find the equation of the cubic which has the same asymptotes as the
curve
x* -6x*y\
\\xy*-6y*i-x+y+\-
0,
and which touches the axis of v at the origin and passes through the point (3, 2).
(Delhi Hons. 1949, 1955)
9.
Find the asymptotes of the curve
4(*H y 1 ) - 17x
2 2
>'
- 4x(4y 2 ~ x*) 4- 2(;c
2
-
2)
-,
and show that they pass through the points of intersection of the curve with the
a
(Delhi Hons. 1951, 1959)
ellipse x + 4/-4.
10.
Find the asymptotes of the curve
9-0
and show that they intersect the curve again
straight line. Obtain the equation of the line.
16-6.
y=mx-\
is
To show
Asymptotes by expansion.
which He on a
(D.U. Hons. 1957)
in three points,
that
c
an asympotate of the curve
(I)
www.EngineeringEBooksPdf.com
ASYMPTOTES
329
Dividing by x we have
9
y\x=m +C/X+A/X* -\-Blx*+C/x* +
so that
when x
*x>
...
,
]im(ylx)=m.
Also from (1)
(y
so that
- mx) = c +A/x + B/;t2 -f Cj 8 + .....
jc
when x->x
,
lini
From
is
(2)
..(2)
we have
and
(ymx)
=
r.
...(3)
we deduce that
(3),
an asymptote of the curve
(1).
Position of a curve with respect to an asymptote.
the position of the curve
16*7.
with respect to
its
To
find
asymptote
y=mx-\
c.
Let A^O. Let > t and >' 2 denote the ordinates of the curve and
the asymptote corresponding to the same abscissa x.
We have
f
By
taking
.v
+....)
we can rnako
sufficiently large,
We
as small as we like.
suppose that x
this expression is numerically less than A.
values of .r, the expression
A
lias
+ B/.v + C/A"
2
|-
D/.v
..(1)
so large numerically that
Thus, for sufficiently large
is
3
+
.
.
.
.
.
.
(2)
the sign of A.
Thus
A
be pjsitive, the expression (1) is positive for suffivalues
of A* so that, from (1) we deduce that when x is
ciently large
lies
positively sufficiently large, then J^
>' 8 is positive, i.e., the curve
above the asymptote and when x is negative but numerically large,
yiy.2 is negative, i.e., the curve lies below the asymptote.
if
Similarly, we may deduce that if A be negative, then the curve
below the asymptote when, jc, is positive but sufficiently large
and lies above the asymptote when, .v, is negative but sufficiently
lies
large numerically.
Let
A=Q, B^O we have
;
As above we can show that
for numerically sufficiently
values of x, the expression
has the sign of B.
www.EngineeringEBooksPdf.com
large
330
DIFFERENTIAL CALCULUS
In this case, the curve lies on the same side of the asympotate
both for positive and negative values of x it will be above or below
the asymptote according as B is positive or negative.
If 5=0 and C^O, we will have a situation similar to that of
;
case
(1).
Ex.
Find the asympotates of the curve
'
y*=x(x~-a)(x-2a)l(x +3a),
and determine on which side of the asymptotes the curve
We have
=
a
i
l
a*
~2x
x ~8&
.....
Y
A
Y
2?-" A
'
a2
a
i
*~
Thus we have two values of y,
y=x3a+-
l
lies.
3fl
i
2^
+ 27aS
8.v
-u
viz.,
a 2 x ......
y=sx+3a*J-a*x ____
Therefore
y=x3a y=
9
are
the asymptote
we
x+3a
two asymptotes.
The difference between the ordinate of the curve and that of
y=x3a being
see that the curve lies
and below
It
it
when x
is
above the asymptote when x
is
positive
negative.
the second
similarly be seen that the curve lies below
is positive and above it when x is negative.
easy to see that
may
asymptote when x
It
is
*=:
3a
To find the position of
we suppose that
x= -3fl +A/y +Bjy* +C/>* + ......
Substituting this value of x in the equation of the
is also
an asymptote of the curve.
the curve
relative to this asymptote,
curve,
have
-5*+
www.EngineeringEBooksPdf.com
+
we
ASYMPTOTES
Equating the
co-efficients
331
of like powers of y, we have
4: = 0.
60a 8
#:=
The
which
is
curve
lies
Ex.
between the abscissae of the curve and the
same value of y, being
difference
x~3tf,
asymptote
etc.
,
for the
negative whether y be positive or negative, we see that ths
towards the negative side of JC-axis.
Find
2.
following curves
the asymptotes and their position with regard to
(Hi)
16*8.
Asymptotes
Lamma.
any
the
:
x*(x-y)i
2
>>
-0.
in polar co-ordinates.
The Polar Equation of a
Line.
The polar equation of
line is
pr cos (6-
a),
where, p, is the length of the perpendicular from the pole to the line
angle which this perpendicular makes with the initial line.
and
a, is the
OY
Let
given line
We
;
be the perpendicular on the
Y being
its foot.
are given that
0y=p;
If
P
(r, 6)
Z_
be any point on the
line,
we
have
Now,
~p= C oS /_YOP.
Fig. 125.
/>/r=cos(0
which
is
a), i.e.,
p=r
cos (0
a),
the required equation of the line.
To determine
the asymptotes of the curve
..(0
r/(0),
we have
to obtain the constants,
p~r
is
/?,
and,
cos(0
,
so that any line
a),
the asymptote of the given curve.
www.EngineeringEBooksPdf.com
..(it)
332
DIFPffiUBNCIAL CALCULUS
Let
(0.
P(r,
be any point on the curve
ff)
Draw OY
Draw PL
J_ the line (n).
and
J_
PM
OY
J_ the line
(ii).
Now.
OY-OL
Fig. 126.
Now
>0 t
=p--r
cos (0-a).
r->oo as the point recedes to infinity along the curve.
when
..(ill)
Let
r->- oo.
We have
Now when
r
oo,
=~
PM
cos (6
->>
a).
Oso that
l
>and
lim cos (0
or
0.
r
r
lim (0
a)=0
a)=7T/2,
or
This gives
a.
Again, p O Y is the polar sub-tangent of the point at which the
asymptote touches the curve, i.e., the point at inanity on the curve. This
may be seen as follows
:
Join the pole O to the point at infinity on the curve i.e., draw
O a line parallel t'> the asymptote. This line is the radius
vector of the point at oo.
through
Draw through O a line perpendicular to the asymptote meetis the polar sub-tangent of the
point
ing it at Y. Then, by def.
at infinity on the curve.
12-7, p. 271).
(
OY
Thus
>
Note
point at
Without employing the notion of the polar sub-tangent and the
value of p, may also be obtained as follows
infinity, the
From
(///)
we
:
have,
when
r->
oo,
/>=lim
[r
cos (0-a)]
lim
[r
cos (e-Oi 4-*/2)l
lim
which
is
of the form (0/0).
www.EngineeringEBooksPdf.com
ASYMPTOTES
..
r
-> p
L
hm
Hence the asymptote
orfei
r*-j
</r
Oi is
the limit of o as r-+
Working
Change
as u ->
^i
,-
L
<fo
i
.
where ii=-
r
J
*~
=-
rcosf
=
r sin
oo re., as
rule for obtaining
r to
r
is
limf-
where
..
=iim
J
33$
6-Oi-f-
^
(Oi0).
->0.
asymptotes -to polar curves.
l\n in the given equation and find out
the limit
of 9
0.
Let
1?
be any one of the several possible limits of
Determine (- dOjdu) and
Let
this limit
its
limit as u -> 9
0.
and 0-+6i-
be p.
Then
p=r
is
sin (0i-0),
the corresponding asymptote.
To draw
the asymptote.
Through the polo O draw a line making angle
on this line take a point Y such that
initial line
(0 L
|TT)
with the
;
OY=\im(-d6ldu).
The
line
drawn through Y perpendicular
to
OY
is
the
asymptote.
Examples
I.
Find the asymptote of the hyperbolic spiral rd~a.
Here
a\r
an so that
~>
as u
> 0.
Here
w = 0/0,
Since
we have
dujde^lla
or
dOjdu^a.
Therefore
/.?.,
is
r
-a=r sin (00)
sin d = a,
r sin 0,
the asymptote.
www.EngineeringEBooksPdf.com
required
334
DIFFERENTIAL CALCULUS
Find the asymptote of the curve
a
__
2.
r
'
~-cos0
Here
= -~(i-COS0).
"=
f
When
w ->
0,
so that cos
cos 0) ->
(
Now,
t/w
1
=
.
or
sin
-rr
d9~
a
,
du
,
=
-
r-
sin0
2a
de
TT
.
and
2a
d0
2a
TT
\
/
_---_r
=r
--
a-r
i.e.,
sin^-g-
and
TT
sin
(
\
\/3
J,
^
4tf
), i.e.,
o
x
/
sm
.
oos
= r(A/3 sin + 3 cos 0),
are the two asymptotes.
Exercises
Find the asymptotes of the following curves
2
'
r
3.
r=o
7.
9.
rn
11.
sec
sinno=? o
w
rologO8
13.
r(l-e )=o.
15.
r
17.
C()s 9
sin
0=o*e
.
=o2 (secl e+cosec 2
4.
rt
6.
2r 2 =tan 20.
8.
10.
.
:
sin 8 e-fccs 8 9'
""e-l"
Q+b tan 6- (P-I/0
r sin 20=a cos 3e.
rO cos 0=0 cos 20. (P.tf.)
5.
"
r=
8).
r sin /io=o.
r=o
tan
12.
rlogO==o.
14.
r(o
16.
r(Tr+0)^o^
2
(P.C/.)
0.
-f 8)-2oO.
6
.
Find the equation of the asymptotes of the curve given by (he
equation
ry-W+r'^/n-itoH
+/t(e)-0
(P.C/.
18.
Show
Hons., 1938)
that all the asymptotes of the curve
r tan n$a,
touch the circle
19.
ro/fi.
Find the asymptotes of the curve
r
cos
2o~o
sin 30.
www.EngineeringEBooksPdf.com
(Delhi
Ho*s
194%)
CHAPTER XVII
SINGULAR POINTS
MULTIPLE POINTS, DOUBLE POINTS
17-1.
Introduction.
Cusps, Nodes and Conjugate points. The
cases of curves considered in
11-4, p. 243 show that curves with
implicit equations of tho form/(x, j>)=0 exhibit some peculiarities
which are not possessed by the curves with
explicit equations of the
form y=F(x). These peculiarities arise from tho fact that the
equation f(x, y)=0 may not
define, >', as a single valued function of*.
In fact, to each value of x
corresponds as many values of y as is tho
degree of the equation in y, and these different values of y give
rise to different branches of the curve.
We
recall to ourselves the
following three curves considered in
11*4.
Fig(i)
Origin
is
127.
a point
Fig. 128 .
common
to the
two branches of the Cisaoid
(Fig. 127)
y*(a-x)=x*,
and the two branches have a common tangent there.
Such a point on a curve is called a cusp.
(ii)
Origin
is
a point
common
to the
two branches of the
phoid (Pig. 128)
and the two branches have
different
Such a point on a curve
is
tangents there.
called a node.
335
www.EngineeringEBooksPdf.com
n 41)
.
(
Stro-
DIFFERENTIAL CALCULUS
336
(Hi)
0,0)
(
is
a point
common
to the
two branches of the
curve (Fig. 129)
ay*-x(x+a)*=^Q
(
and the two branches have imaginary tangents
point in the immediate neighbourhood of the point
lies on the curve.
Here, a, is positive.
There
there.
Such a point on a curve
is
called an
(
isolated
a,
11-43)
is
no
0) which
or conjugate
point.
Fig.
17-2.
129.
Definitions.
Double points. Cusp. Node. Conjugate point. A point through
which there pass two branches of a curve is called a double point.
A
curve has two tangents at a double point,
one for each
The double point will be a node, a cusp or an
according as the two tangents are different and real,
imaginary.
isolated point
coincident or
branch.
A point through which there pass, r, branches
called a multiple point of the rth order so that a curve
tangents at a multiple pDint of the rth order.
Multiple point.
of a curve
has,
r,
is
Thus a double point is a multiple point of the second order.
point of the third order is also called a triple point. A
multiple
^A
multiple point is also, sometimes, called a singular point.
A
17*3.
simple rule for writing down the tangent or tangents
at the origin to rational algebraic curves is obtained in the following
article.
Tangents at the origin. The general equation *>f rational
curve
of the wth degree which passes through the origin 0,
algebraic
17*31.
www.EngineeringEBooksPdf.com
SINGULAR POINTS
when
337
arranged according to ascending powers of x and y,
form
is
of the
...(/),
where the constant term
is
absent.
Let P(x, y) be any point on the curve. The slope of the chord
<OP is y\x. Limiting position of the chord OP, when P -+ O, is the
and y -> 0,
tangent at O so that when x ->
lim
()>/*)
=m,
is the slope of this tangent.
From
On
(/),
we have,
taking limits,
after dividing
when x ->
=0
0,
we
by
x,
get
so that
m=
bjb 2
,
if
Hence
ylx^-bjb*
d.e.,
b,x+b 2 y^Q
is the tangent at the origin.
t;o
zero the lowest degree
If& 2 =0but
6^62=0
(c A x
2
may be written down by equating
degree) terms in the equation (i),
(first
then, considering the slope of OP with
can be shown that the tangent retains the same
6^0,
reference to F-axis,
form.
Let
...(ii)
}
This
it
so that the equation takes the form
+c2 xy +w*) +(4* 3 +djc*y + ......
Dividing by
)
and then taking limits as x ->
x*
C1
+... =0.
0,
we
+c 2m+c 3m 2 =-0,
.
...(wi)
get
...(iv)
a quadratic equation in, m, and determines as its /wo roots
the slopes of the two tangents so that the origin is a double point in
^diich
is
this case.
The equation of
either tangent at the origin
is
y=mx,
m
Tvhen
obtain
is
a root of
(iv).
Eliminating
..(v)
m between
c^+CiXy+Csy^Q,
(iv)
and
(v),
we
...(vi)
as the joint equation of the two tangents at the origin. This can be
written down by equating to zero the lowest degree terms in (111).
The equation (vi) becomes an identity if c l =c2 =c3 =0. In this
case the second degree terms, also, do not appear in the equation of
the curve. It can now be similarly shown that the equation of the
tangents can still be written down by equating to zero the terms of
the lowest degree which is third in this case.
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
338
In general, we see that the equation of the tangent or tangents at
the origin is obtained by equating to zero the terms of the lowest degree
in the equation of the curve.
The
on a curve whose equation
and
contain
the
the first degree terms*
constant
least,
origin will be a multiple point
does not, at
Illustrations.
(i)
The
origin is a
node on the curve
and x=0, y=0 are the two tangents thereat.
(ii) The origin is a cusp on the curve
and
y=Q
( ill)
and
is
the cuspidal tangent.
The
origin
axiby=Q are
(iv)
The
origin
and x=0, x=y,
is
an isolated point on the curve
the two imaginary tangents thereat.
is
a triple point on the curve
xy
are the three tangents thereat.
Exercises
Find the tangerts at the origin to the following curves
1.
.3.
5.
(xH;y
a 2
2
) ==4fl jty.
+y2 )(2a-x)=b2x.
2
2 2
2
2
8
(x +/) =a (x ->> )
(x
2
2
:
-x 2 )=x2 (b-x)^
2.
y*(a
4.
a*(x*-y*)=x*y*.
.
Example
Find the equation of the tangent at
and show
We
that this point is
a
(
2) to the curve
1,
cusp.
will shift the origin to the
point
1,
(
2).
To do
so
we
have to write
Y
where X,
are the current co-ordinates of a point on the curve with
"reference to the new-axes.
The transformed eqation is
or
Equating to zero the lowest degree terms, we get
-y
2
=0,
i.e.,
(r-Jf)
2
==0,
www.EngineeringEBooksPdf.com
339
SINGULAR POINTS
which are two coincident lines, and, therefore, the point is a cusp
and the cuspidal tangent, i.e., the tangent at the cusp with reference
to the
new axes
is
To find the equation of the cuspidal tangent with reference to
the given system of axes, we write
Hence the tangent at
(1,
2) is
Exercises
Find the equations of the tangents to the following curves
:
1.
rV+x
2.
(;c-2)
3.
4
x*-4ax*-2ay* + 40V+3aV--a ==0 a t (a, 0) and (20, a).
Show that the origin is a node a cusp or a conjugate point on the
4.
2
2
2
)=*V--* )at(a,
2
=X>>-l) at(2,
0).
1).
;
curve
2
2
y =ax +ax*
according
as, a, is positive,
Conditions for any point
17*4.
9
zero or negative.
(Delhi Hons. 1950)
(x, y) to
be a multiple point of
the curve
f(x,y)=0.
In
10*94, p.
213,
we have
seen that at a point
(x, y)
of the
curve
/(*.
the slope of the tangent, dy/dx
is
y)=o,
given
by the equation
At a multiple point of a curve, the curve has at least two
tangents and accordingly dy\dx must have at least two values at a
multiple point.
The equation (1), being of the first degree in
by more than one value of dyjdx, if, and only
/.0,/f =0.
satisfied
dyjdx,
can b^
if,
Thus we see that the necessary and sufficient conditions for any
point (x, y) on/(x, }>)=0 to be a multiple point are that
/.(^y)=0,/w (x y)=0.
f
To find
multiple points (x, y),
we have
therefore to find the values.
of(x, y) which simultaneously satisfy the three equations
www.EngineeringEBooksPdf.com
340
DIFFERENTIAL CALCULUS
17*41.
To find
the slopes of the tangents at a double point.
Differentiating (1) w.r. to x,
we have
Ay
Jf*+r
* +J **
dx
d?
<( Jv
f * +J
+f**y
*
+1
dx~) dx
point, where /y =0,/ .=0,
\
so that at the multiple
are the roots of the quadratic equation
a
the values of dyfdx
not all zero andfx =Q=fv the point (x, y)
In
casef^tf^fy*, are
be a double point and will be a node, cusp or conjugate according
as the values ofdyjdxare real and distinct, equator imaginary i.e.,
,
according as
,
will
2
(/*,)
2
2
-/* /,, >0, -0, <0.
2
If/a =/art =y^ =0 the point
order higher than the second.
2
;
,
(x,
will
y)
be multiple point of
Example
Find the multiple points on the curve
Also, find the tangents at the multiple point.
Let
/(*, y)
=x4
f*(x>
J>)=0 gives *=0,
fv( x *
y)=
Hence the two
gives j;=0,
a.
partial derivatives vanish for the points
(0, 0) <0,-fl), (a, 0),
Of these
a.
a,
(fl,-n)
f
(-a,
0),
(-a,-a).
the only points on the curve are
(fl.0),(-fl,0), (0,-a).
Hence these are the only three multiple
To
follows
find the tangents at the
points,
multiple points,
On the curve.
we proceed
as
:
First method.
We have
Since at
(a,
0).
fJ
therefore,
by the equation
(2),
the values of dy/dx at
by
a
-f 8a
z
= 0,
www.EngineeringEBooksPdf.com
(a, 0)
are given
341
SINGULAB POINTS
The two values being both
real,
the point
a node. The
(a, 0) is
tangents at (a, 0) are
It
(0,
may
similarly be
Second method.
we
shown that the tangents at
a, 0)
(
and
a) are
Differentiating the
given equation w.r. to x,
get,
4x 3
&a 2yy l
Qay^
which identically vanishes
for the multiple points.
Differentiating again,
we
12x 2 - 12ayy^
From
this
^=
,
(/) for (a, 0),
for
(ii)
(-fl, 0)^^=1,
(iii)
for(0, -a),
Knowing the
get
6ay*y 2
see that
we
4a*x =0,
^^f,
- Qa yy 2 - 4a
2
6a*y\
/.*.,
2
=0.
J
i>.,^
/.e.,
^-
slopes of the tangents,
we can now put down
their
equations.
To find the tangents at
The transformed equation
Third method.
origin to this point.
(a, 0),
is
we
shift
the
or
The tangents at the new
The tangents
It
origin are
at the multiple point
y=
may similarly
(a, 0),
therefore, are
\/(4/3)(*-).
be shown that
are the tangents at the multiple points
(
0, 0)
and
(0,
a) respec-
tively.
The three mutiple points on the curve are nodes.
Exercises
Find the position and nature of the multiple points on the following
curves
:
1.
x*(x-y)+y*=Q.
2.
y*=*x*+ax*.
x*+y 2x 8 +3>> 2 =0.
3.
www.EngineeringEBooksPdf.com
(D.U.1951}
(D. U. \950\P. U.)
(P. U.)
DIFFERENTIAL CALCULUS
4.
5.
xy
2
>>
2
=(jc-l)(jc--2)
2
.
6.
ay*=(x-a)*(x-b)*.
7.
x*-4ax*+2ay*+4a*x*-3a*y*-a*=Q.
9.
10.
x*+y(y+4a)*+2x*(y-5a)*=5a*x*.
11.
X*+2x*+2xy-y*+5x-2y=Q.
12.
(2 y+x-fl)
13.
(B. U.)
(P. U.)
2
5
-4(l~x) =0.
2
8
(jc+>>) -^2>>-*+2) =0.
(P (70
<
14.
(/-a^-hx^-f
15.
xV 2 =(a+}0 2 (& 2
Find the
following curves
(P.V.)
2
30) ==0.
2
.y )
;
(P. U.)
distinguishing between the cases
equations of the
tangents at
b^a.
(D. U. Hons. 1953)
the multiple points of the
:
16.
17.
18.
19.
(y-2)*=x(x-\)*.
Show that each of the curves
(x cos
cc
y
sin
2
=c(;c sin <x.+y cos a) ,
cusp show also that all the cusps lie
a6)
for all different values of a, has a
3
:
on a
circle.
17 5. Types of cusps. We know that two branches of a
curve have a common tangent at a cusp. There are five different
ways in which the two branches stand in relation to the common
tangent and the common normal as illustrated by the following
figures
:
Fig. 130
Fig. 131
Single cusp of 1st species
Fig. 132
Single cusp of 2nd species
Fig.
Double cusp of
1st species
133
Double cusp of 2nd species
www.EngineeringEBooksPdf.com
SINGULAR POINTS
343
Fig.
Point of oscu-inflexion
mon
In Fig. 130, the two branches lie on the same siue of the comnormal and on the different sides of the tangent.
In Fig. 131, the two branches
mal and on
the
same
lie
on the same
side
of the nor-
side of the tangent.
In Fig. 132, the two branches lie on the different sides of the
different sides of the tangent.
normal and on the
In Fig. 133, the two branches lie on the
normal and on the same side of the tangent.
different
sides
of the
In Fig. 134, the two branches lie on the different sides of the
lie on the same, and on the other on
opposite sides of the common tangent. One branch has inflexion at
the point.
normal but on one side they
It will thus be seen that the cusp is single or double according as
the two branches lie on the same or different sides
of the common nor-
mal. Also it is of the first or second species according as the branches
lie on the different or the same side
of the common tangent.
Examples
Find the nature of the cusps on the following curves
1.
(i)
y*=x\
(ii)
:
^-xi^o.
(0 y=0 is the cuspidal tangent. Since x cannot be negative,
the two branches lie only on the same side of the common normal
eo that the cusp is single. See
Fig. 130.
3
y=xf
so that to each positive value of x
^gain,
correspond
two values of y which are of opposite
signs and hence the two
branches lie on different sides of the common
tangent and the cusp is
of first
species.
(11)
Two
branches of >>*
ndj>+x a =:0 which
y=0
and
the origin
* 4 =0,
are the
two parabolas y
on
x*=0
different sides of the common tangent
extend to both sides of the common normal x=0. Thus
is
lie
a double cusp of
first species.
See Fig. 132.
www.EngineeringEBooksPdf.com
344
D1FFBBENTIAL OALOTTLTJS
(7) Here
where the two signs correspond to the two>
x
\
"^"^
Fig. 135
y~Q
the cuspidal tangent.
Since x cannot take up negative values,
the two branches lie only on the sime sideof the common normal and the cusp, there-
branches
;
is
fore, is single.
Now, one value of .y is always positive and, therefore, the corresponding branch lies above X-axis. Again
JT
QX
^> X
,
II
Thus, for the values of x lying between and 4^, the second
value of y is also positive and, therefore, the corresponding branch
lies above Y-axis in the vicinity of the origin.
Thus the cusp is of
the second species.
Show
2.
that the curve
has a single cusp of the first species at the origin.
(Delhi ffons. 1949}
Equating to zero the lowest degree terms,
origin is a cusp and y=0 is the cuspidal tangent.
We re-write the
we
see
that the
given equation in the form, quadratic in y y
y*-y
and solve
it for y,
so that
_
_.
For positive values of #, we have
*4 (2+x) 2 +4;c 3 >;c 1 (2+;c) 2
,
or
so that to positive values
Thus the
opposite signs.
when x
>
of x correspond two values of y with
two branches lie on opposite sides of x-axis
is positive.
Again,
we have
For values of x which are
is
sufficiently small in
positive, for the same -> 4 when x -> 0.
Thus for negative values of x which are
numerical value,
www.EngineeringEBooksPdf.com
numerical value,
sufficiently
small in
SING o LAB POINTS
345
is
negative so that the values of y are imaginary.
take up negative values.
Hence the curve has a
Thus x cannot
single cusp of first species at the origin.
Exercises
Find the nature of the cusps on the following curves
14
3.
5.
7.
8.
x\x-y)+y*=0.
x *+y*-2ay2 =Q.
2
(y~-x) +x'=0.
x*ax*ya*x*y+a*y 2 =Q.
2.
:
x
4.
a*
6.
x'
Examine the curve
for singularities.
(Delhi Hons. 194SY
Prove that the curve jc s +y 8 =ax2 has a cusp of the first species at the
origin and a point of inflexion where x=a.
(Lucknow Hons. 1950}
10.
Show that the curve
9.
has a single cusp at
(a, 0).
(D.U. 1955)
Radii of curvature at multiple points. The formula for
the radius of curvature at any
point (x y) on the curve f(x, y)~Q,
as obtained in
15-42, page 292, becomes meaningless at a multiple
point where/a. y^~0. At a multiple point we expect as many values
Of course, these values of p may not be all
of, P, as its order.
17-6.
t
distinct.
Tho following examples will illustrate the
ing the values of p at such points.
method of determin-
Examples
1.
Find the radii of curvature at the origin 0* the branches of the
curve
8
Here, 2*y*=*
at the origin so that
To
find,
find Iim(.y 2 /2.x).
p,
,
i.e.,
it is
for
To do
x^Q, )>=(1/V 2 )*
a
are the three tangents
triple point.
the
this,
branch
we
which
touches
jc=0,
wa
write
p l9
.e.,
X=
and
substitute this value of x in the given equation.
Lim
is
the radius of curvature of the
at
branch
the
origin.
corresponding
We get
p^P
or
www.EngineeringEBooksPdf.com
346
D1FFEEENTIAL OALOTLTTS
Let y -*
Thus,
To
we have
so that
branch=
p, for this
find, p, for
1
+a/p=0.
a.
we proceed
the other branches
Suppose that the equation of either branch
J>=/(0)+x/'(0)
YTe have,
+^1/"(0)+
as follows.
is
given by
....
ere
/(0)=0.
Also
we
write /'(0)=p,/"(0)
Making substitution
Equating
=
Thus we have
<7.
in the given equation,
co-efficients of x*
and
x*,
we
get
we get
2ap*=a, p*+2apq=l.
These give
=<
=- 3V2
1
'
for the two branches.
2.
Show
that the pole
is
r=a(2
and
that the radii
a
cos
point on the curve
8+ cos
30),
of curvature of the three branches are
V3a/2,
The radius
triple
vector,
a/2,
V30/2.
vanishes for the values
r,
0+ cos
2 cos
of, 0,
given by
30=0,
!>.,
2 cos
0+4 cos 8
03 cos 0=0,
tpr
cos
2
(4 cos
1)=0,
or
cos
Thus,
r=0 when
0=0,
cos
0=7r/3,
0= J,
?r/2,
cos
0=
J.
27T/3 so that the pole
point.
www.EngineeringEBooksPdf.com
is
a triple
347
SINGULAR POINTS
We now proceed
/
1
=a(2
to find
p.
0-3
sin
We have
2 cos
sin 30), r2 =0(
09 cos 30).
For 0=7r/3,
r^
y^Sa,
r2
=8a
r2
=0
r2
=
;
for 0=7T/2,
r 1==a
;
or 0=27r/3,
r^
Also,
r=0
y3a,
8a.
for each of these branches.
Putting these values in the formula
(r'+r,*)*
we get the required
result.
Exercises
1.
Show
that the radius of curvature at the origin for both the branches
of the curve
is
2.
Find the radius of curvature at the point (1,2) for the curve
3.
Find the radii of curvature at the origin of the two branches of the
curve given by the equations
y=t-t\x=\-t 2
(For the origin /==
values of /).
1,
Find the radii of curvature
4.
x*+y*=*3axy.
5.
-3^y5
jc +ox
6.
7.
JC
at the origin of the following curves
two
:
(Folium).
2
Show
that (a, 0), in polar co-ordinates,
r=a(l4-2sin
and
(P.U.)
.
so that the two branches correspond to these
is
a triple point on the curve
Jo),
find the radii of curvature at the point.
www.EngineeringEBooksPdf.com
CHAPTER XVIII
CURVE TRACING
We have already traced some curves
The general problem of curve tracing,
will be taken up in this chapter.
18*1.
and XII.
aspects,
in
Chapters II
in its elementary
It will be seen that the equations of curves which we shall trace
are generally solvable for y, x or r.
Some equations which are not
solvable for y or x may be rendered solvable for r, on transformation
from Cartesian to Polar system.
18
Procedure for tracing Cartesian Equations.
Find out if the curve is symmetrical about any line. In this
connection, the following rules, whose truth is evident are helpful
2.
I.
:
(/)
A
curve
which occur in
(ii)
A
its
curve
which occur in
its
is symmetrical about A'-axis
equation are all even
if
the powers of
y
the powers of
x
;
symmetrical about
equation are all even
is
7-axis
if
;
(MI) A curve is symmetrical about the line y=x if, on interchanging x and y, its equation does not change.
II.
Find out if the origin lies on the curve. If it does, write
down the tangent or tangents thereat. In case the origin is a multiple point, find out its nature.
Find out the points common to the curve and the co-ordinate
be any. Also obtain the tangents at such points.
there
if
III.
axes
Find out dyjdx and the points where the tangent is parallel
co-ordinate axes.
At such point, the ordinate or abscissa
generally changes its character from increasing to decreasing or vice
IV.
to
the
versa.
V.
Find out such points on the curve whose presence can be easily
detected.
VI.
Find out points of inflexion.
(It
may
not always be neces-
sary).
Find out multiple points, if any, and their nature.
VIII. Find out the asymyplates and the points in which each
asymptote meets the curve.
IX. Find out if there is any region o?the plane such that no part
VII.
of the curve
lies in
it.
Such a region
is
generally obtained on solving the equation
348
www.EngineeringEBooksPdf.com
for
349
CURVE TRACING
one variable in terms of the other, and find out the set of values of
one variable which make the other imaginary.
X. If possible, solve the equation for one variable in terms of
the other and find out, by examining the sign of the derivative, or
otherwise, the ranges of the values of the independent variable for
which the dependent variable monotonically increases.
Important Note. The student must remember that a mere knowledge of
symmetry, asymptotes, double points, etc., will not enable him to trace a curve,
this knowledge being only piece-meal; asymptotes indicate the nature of the
curve in parts of the plane sufficiently far removed from the origin and the
tangent at any point gives an idea of the curve in the neighbourhood of the
point. To draw a curve completely, it may be necessary to solve the given
-equation for one variable, say, y, in terms of another and then to examine how
y varies as x varies continuously. For this purpose th& examination of dyfdx
proves very helpful.
Examples
Equations of the form
18-3.
1.
Trace the curve
We
note the following particulars about this curve
(i)
(11)
It is
symmetrical about both the axes.
through the origin and
It passes
gents thereat.
(in) It
:
Thus the origin
meets A^axis at
dy
r
ax
y=x are the
two tan-
a node.
(a, 0), (0, 0)
The tangents at
axis at (0, 0) only.
jt=- tf respectively.
/M
(V)
is
and (
and
(0, 0)
a, 0)
fl,
(
and meets Yx=a and
0) are
fl
-A(
which becomes
2
fl*-2a x
Thus the only
(v)
(vi)
when
zero,
It has
2
-x4 =0,
i.e.,
real values of
when x2 =(-l
x
for
v'2)fl
s
.
which dyjdx vanishes ar
no asymptotes.
Solving the equation for y,
We see that,
we
re-write
it
as
2
x 2 must be non-negative and
y to be real, a
a
and a. Hence the entire curve lies
therefore, x must lies between
between the lines x=a and x= a.
for
www.EngineeringEBooksPdf.com
350
DIFFERENTIAL CALCULUS
Conclusion. In order to trace the curve we have to connect
the above isolated facts. We proceed to do this now.
The curve
being symmetrical about both the axes,
we firstly consider the part of the curve
lying in first the quadrant where x, y
are positive and
/P
'-* A
o
(a.o)
2
VL
When x=0,
r
then ^=0.
When
increases, starting from 0, then y >
which is positive, also increases and
x
goes on increasing
x=a
t
till
x=v
/
(
l+y'2)^
i.e., where the tangent
Fig. 136.
is parallel to Z-axis. Since ,y=0 when
we see that when x increases from \/(l + \/2) a to fl, y
where dyjdx=0
f
decreases.
/A
<*
Fig. 137
The variable x cannot take up values greater than a, for, then,
y is not real. Thus the part of the curve in the first quadrant is as
shown in Fig. 136. The curve being symmetrical about the two axes,
its complete shape is as shown in Fig. 137.
2.
(/)
Trace the curve
It
is
symmetrical about A'-axis only.
It passes through the origin
are the two imaginary tangents thereat.
lated point.
()
x
;
*axcP
(IV)
ivhich vanishes
Since
Q,
i.e.,
origin
is
an
iso-
meets J-axis at ( a, 0) and (0, 0) and >-axis ,at the
a is the tangent at (a, 0).
(Hi) It
origin only
and yz +xz
Thus the
2
x
when
oxa =0,
2
i.e.,
whenx=J
www.EngineeringEBooksPdf.com
CURVE TRACING
For x=(l
See
(vi)
(v)
(vi)
\S5)aj2,
which
lies
351
a and
between
a,
y
is
not reaL
below],
y=(x+a)
and
see that, for
be
> a and <
three asymptotes.
y= x*^* =x **^^>
Since
we
x=a are its
y
to be real,
jc
2-
a 2 must be non-negative,
i.e.,
must
a.
Hence no part of the curve lies between
a.
and x==
xa
Fig. 138
Conclusion. The curve being symmetrical about the A" axis, we
consider the part of it lying in the first two quadrants where y is
positive so that we have
T
+
where we consider
s ig n
or
according as x
is
positive or
nega-
tive.
When x=
responding point
Let
is
(
y=0.
*
Also jc=
a, is
the tangent at the cor-
a, 0).
vary continuously from
a to
ao
.
Then, since
tfy/flbc
any of these values of x, and y= (x+a) is an asympwe have the part of the curve in the second quadrant as shown
notO
tote,
JC
a,
for
in Fig. 138.
a and fl, y is not real.
For values of x lying between
Since jc=0 is an asymptote and dyjdx^Q for JC=(l+\/*>)
www.EngineeringEBooksPdf.com
fl
/2,
DIFFEBBNTIAL CALCULUS
352
we see that y, starting from oo goes on decreasing as x increases
'from a to (1 4-^5)0/2. As x increases from (l+<\/6)0/2 to oo
y
'
,
,
again goes on increasing and, y=x fa being another asymptote,
have the part of the curve in the first quadrant as shown.
The curve being symmetrical about the
shape is as shown in Fig. 138.
We
A"-
axis,
its
we
complete
thus have the curve as drawn.
Note. The curve appears to have two points of inflexion, apparently
necessitated by the fact that the curve could not be asymptotic to the lines
y= (x+a) unless it changes the direction of its bending after leaving the point
(a, 0) where it touches x= a. In fact, by actual calculation, we see that
- 2a/>|3) are its two points of inflexion.
<-2fl, 2a/>l3) and ( 2a,
3.
Trace the curve
'(/)
It is symmetrical
about both the axes.
does not pass through the origin.
(n) It
a, 0) but it does not meet
(111) It meets A'-axis at (a, 0) and (
'yaxis at any point x=a and ;c = a are the tangents at (a, 0) and
'.{0, 0) respectively.
;
'
_ ~~
dx
JC
2
v (x2
^__
2
}'
which never becomes zero.
(0)
J>=1
Prom
ibut
it
will
(vi)
we
are the
two asymptotes.
x=0 is also an asymptote,
cannot be an asymptote.
16-31. p. 319, it appear*? that
be seen below that
it
Writing the equation in the forms
see that for
x and y
to be real,
we have
<a
a2
>
0, i.e
1-y*
>
0, i.*.,--l
jc
2
,
x
or
x
>
a
and
mo
<y
<1.
As no part of the curve lies between the lines x= fl, and
branch of the curve can possibly be asymptotic to x~0.
Conclusion.
Consider
>which corresponds *to the part of the curve in the
www.EngineeringEBooksPdf.com
first
quadrant.
CURVE TJUGINQ
and a^ y is
values of x lying between
y=Q so that (a, 0) is a point "on thfc curve
seen above, is the tangent there.
As x increases from a onwards, y which is positive
.
If
x=a,
must
t
stantly increase
;
con-
dyjdx being never zero.
y=l is an asymptote, the part of the curve in
quadrant can now be easily shown.
The curve being symmetrical about both the axes, its complete
Since the line
the
first
shape
is
as
shown
in the Fig. 139.
Y*
Fig.
4.
139
Trace the curve
y *=(x-a)(x-b)(x-c).
We
suppose that a, b c are positive and in ascending order of
magnitude so that we have to consider the following four cases
y
:
a<b<c. (//) a = b<c.
Case I. a<b<c.
(in)
(/)
symmetrical about
(/)
It is
(//)
It does not pass
(Hi) It
meets
meet 7-axis
;
x=--a
a<b = c.
a^b^c.
.Y-axis.
through the origin.
A'-axis at (a y 0), (b, 0)
t
(/?)
x=b, x=c
nnd
.
(c, 0),
are the tangents
but
at- (a,
it
docs not
0), (6, 0), (c, 0),
respectively.
(iv)
(v)
It has
no asymptotes.
When x<a,
y*<0,
whena<x<&, .y 2 >0,
when b<x<c, y*<Q
2
whenx>c,
^ <0.
i
;
e.,
is
i.e.,y is
Hence, no part of the curve
between the lines x=6, x = c.
for
y
lies to
not real
;
not real
the
2
(vi) As y vanishes for x=a and x=fe,
some value of x between a and.&.
left
it
;
of the line x = a and
must have a maximum
www.EngineeringEBooksPdf.com
354
DIFFERENTIAL CALCULUS
(viii)
As x
increases
beyond
c,
for
y* also constantly increases,
each of the three factors then increases.
We
Fig. 140
have
S~-2(a+b+c)(\lx)+(ab+bc+ca)(\lx*)
which ->
oo
&s
x ->
ao
.
Thus the curve tends to become parallel to 7-axis as x --> oo
Since the curve, in departing from (c, 0), where the tangent is
parallel to F-axis, must again tend to become parallel to F-axis, we
see that the curve must change its direction of bending at some
and as such must have a point of
point between (c, 0) and oo
.
;
inflexion.
Hence, we have the curve as shown in Fig. 140.
an oval and an infinite branch.
a=b<c. The equation, now, is
It consists of
Case
II.
(a.oi
Fig. 141
www.EngineeringEBooksPdf.com
355
CURVE TRACING
It is easy to show that (a, 0) is an isolated point, as
of case I shrinks to the point (a, 0).
if
the
oval
The curve consists of an isolated point and an infinite branch
and can be easily drawn. (Fig. 141 on the preceding page).
As
in case I,
it
can be shown that the curve has two points of
inflexion.
Case
III.
a<b=c. The
equation, w>w,
is
Fig. 142
It is easily
shown that
(b, 0) is
a node and
are the nodal tangents.
The curve may now be
Case IV.
easily
a=b=c. The
drawn.
equation, now,
(Fig. 142).
is
Fig. 143
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
356
It is easy to see that (a, 0)
is
a cusp and
"j^O
is
the cuspidal
tangent.
The curve may b3
drawn.
easily
(Fig. 143).
Trace the cissoid
5.
y*(a-x)=x*.
This curve has already bsen considered in
11-41, p.
244
also.
(See Pig. 64, p. 244).
We note the
:
symmetrical about x-axis.
It is
(i)
following particulars about the curve
2
through the origin and y =0, i.e., J>=0, jj=0
are the two coincident tangents at the origin.
Thus the origin is a
It passes
(ti)
cusp.
meets the co-ordinate axes at the origin only.
(Hi) It
(iv)
x^a
(v)
Since
is
the only asymptote of the curve.
y*--x*/(ax),
we
see that
and
j
2
is
positive,
y
i.e.,
is
real
only
when x
lies
between
a.
which vanishes when
x=f#
or 0.
fais outside the range of admissible values of x. Thus
vanishes
at no admissible value of x except x=0.
dyjdx
Combining all these facts, we may easily see that the shape of
the curve is as shown in Pig. 64, p. 244.
But X
Note.
1
1
1*4?,
The student would do well
to learn to trace
the curves given in
by the methods of this chapter.
1-43 also
Trace the curve
6.
It is neither
(/)
symmetrical about the coordinate axes nor
about the
(if)
Origin
is
a cusp, and
x=0
is
the cuspidal tangent.
(HI) It meets A"-axis at .(0, 0) and (3a, 0) bjut meets
he origin only x=3a is the tangent at (3a, 0).
F-axis at
;
(10)
y+x=aia
asymptote at
its
only asymptote and the curve meets the
'
(a/3, 20/3).
(t;)
(vi)
x and y cannot both be negative.
Now, y*dyldx=x(%dx) and, therefore,
for x=2a.
y
www.EngineeringEBooksPdf.com
!
OtiBVE TRACING
Solving for y,
357
we obtain
Fig. 144.
If
*=0, then y =0 and
7-axis
is
the tangent there.
<JC<3a, then >>>0, and if x increases from 0, y also inwill go on increasing with x till x =2a where dyfdx=Q.
When x increases beyond 2a, y will constantly be decreasing
y=0
for x=3a and is negative for x>3a.
If
creases ;
y
;
We now consider the
negative values of JC.
If x is negative, y is positive and constantly goes on increasing
as x increases numerically, i.e., as x varies from
to
oo.
Also,
y+x=a is the only asymptote
of the curve.
Taking all the above facts into consideration, we aee that the
complete curve is as shown in Pig. 144.
1
18-4.
Equation of the form y +yf(x) + F(x) = o.
7,
Trace the curve
(i)
It
about the
(ii)
(///)
(tv)
curve at
(v)
it.
is
line
symmetrical neither about the co-ordinate axes, nor
y=x.
a cusp and y =0 is the cuspidal tangent.
It meets the co-ordinate axes at the origin only.
Origin
is
y=x-\-l
(-,
We
We have
is
the only asymptote of the curve.
It
meets the
i).
re- write
the
equation as a quadratic, in y and solve
yx*+x*~ 0,
or
www.EngineeringEBooksPdf.com
358
DIFFERENTIAL CALCULUS
so that
and
y
is
imaginary
if x s (x4.) is
negative
i.e.,
if
x
lies
between
4.
Fig. 145.
Thus, there
is
no part of the curve lying between the
line
x=0
and x=4.
x=4
for both the branches, and ;c=4 is the
y=8 when
the
at
tangent
point.
The following additional remarks about the two branches of
curve will facilitate the process of tracing the curve.
(vi)
y
y=
and
are the
or 4.
two branches. They meet where x 4
Thus the two branches meet
Consider the branch
When
x, starting
4x 8 =0,
at the points (0, 0)
i.e* 9
and
where
#=0
(4, 8).
(i).
from
4, increases,
then y
is
positive
and
also
constantly increases.
*
Also, when x starting from
decreases,
values whose numerical value increases, then
stantly increases.
Now, consider the branch
When x>4,
therefore,
y
x4
4;c
is positive.
3
and takes up negative
y is positive and con-
(//).
positive and <\/(x*4cX*)<\/x*=x* and,
Also, when x, starting from 4, increases, y
is
decreases in the beginning.
When x < 0, x 4
4*3
is
positive
and
\/( x
4
www.EngineeringEBooksPdf.com
4x*)
> -y/x
4
=* x 1
so
CUBVB TRACING
359
that y is negative. Also, when x, starting from 0, decreases, then y
numerically increases. Thus, we have the shape as shown in Fig. 145.
Polar carves.
18-5.
8.
Trace the curve
\
Fig. 146.
We first
consider positive values
only.
dr
2a0
is
have
_
=
^0
which
We
(f-jL02)2>
always positive so that
r,
constantly increases as,
0, in-
creases.
Also,
r=0 when 0=0;
Again we have
-
Thus
s=
-
-
_2
which -> a as
-> oo.
starting from its initial value 0, constantly increases as
-> oo so that the point (r, 0)
and approaches, a, as
and
nearer
nearer
the
circle
whose centre is at the pole
approaches
and radius equal to a.
The circle is shown dotted in the Fig. 146.
The figure shows the part of the curve corresponding to the positive values of
only, and the part of the curve for negative values
r,
increases
is its
reflection in the initial line.
9.
Trace the curve
r~a(sec d+cos
0).
Here
1
+
,
v
\cos
(i)
(ii)
The curve
r
cos
x~
\
cos 9 ]=^a
/
1
+ cos
2
--
cos
symmetrical about the initial
0=a, i.e., x=rtis its asymptote.
is
www.EngineeringEBooksPdf.com
line,
DIFFERENTIAL CALCULUS
360
~
and
so .that
4- is
positive
when
lies
between
ir/2.
Fig, 147.
is
When 0=0, dr/dOQ so that <=Tr/2 and, therefore, the tangent
perpendicular to the initial line at the point (2a, 0)
.
Also, r=2a, when 0=0
Hence we see that when increases from
cally increases that 2a to oo and the point P(r,
of the curve drawn in the first quadrant.
and
r
-> oo as
->
Tr/2.
to TT, or monotoni0) describes the part
When 6 increases from TT to TT, r, remains negative and decreases in numerical value from oo to 2a and so the point P(r 6)
describes the part of the curve as shown in the fourth quadrant.
,
t
As the curve
point
will
is symmetrical about the
be obtained when 6 varies from TT to
initial line,
no new
2ir.
In the case of some curves
it is found convenient to
polar as well as the Cartesian form of their equaare obtained from the Cartesian and the others
from the Polar form.
18*6.
make use of the
tions. Some facts
Curves whose cartesians equations are not solvable for x and
but whose polar equations are solvable for r, are generally dealt
with in this manner.
y,
10.
Trace the curve
(/)
It is symmetrical
about both the axes.
www.EngineeringEBooksPdf.com
CUKVE TRACING
Origin
(//)*
is
a node on
361
the curve and y*
are the nodal
tangents.
and
meets Jf-axis at (0, 0) (a, 0) and ( a, 0), but meets
x=a andx= flare the tangents at (a, 0)
(0, 0) only
It
(Hi)
F-axis at
;
a, 0).
(
(iv)
On
!
no asymptotes.
changing to polar co-ordinates, the equation becomes
It has
I
0* cos 2
,
"cos 4 0+sin 4 0'
(v)
We see
that
cto~~
___
0)~
4
(cos ~0-f sin"*
'
so that drfdQ remains negative as
varies from
to ir/4
fore r decreases from a to
as
increases from Q to ?r/4.
and
there-
2
changes from ir/4 to ir/2, r remains negative and
therefore no point on the curve lies between the lines 0=7r/4 and
0-7T/2.
As
(vi)
Fig. 148.
As the curve is symmetrical about both the axes, we have
(Fig. 148).
shape as shown.
Trace the curve
11.
y*~-x*+xy=Q.
It is
(i)
about the
line
(P.U.)
neither symmetrical about the
It passes through the origin
JC=0,
a
so
that
the
is
thereat
node.
gents
origin
(iv)
On
co-ordinate axes,
nor
y=x.
:
(ii)
(1/1)
its
y~Q
are the two
tan-
It cuts the co-ordinate axes at the origin only.
y=x,
y=*x are its
asymptotes.
transforming to polar co-ordinates, we get
ra
tan 20.
=
(v)
ir/2,
and,
When
increases from
a
therefore, r
to w/4, 20 increases from
from to oo.
increases
monotonically
www.EngineeringEBooksPdf.com
to
362
DIFFERENTIAL CALCULUS
When increases from -ir/4 to ir/2, tan 26 and therefore also
remains negative and, thus, there is no part of the curve lying
between the lines 0=ir/4 and =ir/2.
rf
Y4
Fig. 149.
When
$ increases
from
ir/2
to
2
3-7T/4, r
When
there
increases
from
to oo.
2
6 increases from 3?r ,/4 to TT, r remains negative and so
= 3Tr/4, and
no part of the curve lying between the lines
is
0=TT.
We
from
TT
can similarly consider the variations of
r
2
as
$
increases
to 2?r.
Hence we have the curve
12.
as drawn. (Fig. 149)
Trace the Folium of Descartes
y=x
It is symmetrical about the line
point (3a/2, 30/2).
(i)
and meets
it
in
the
x0,
)>=0 are the tangents
(11) It passes through the origin and
there so that the origin is a node on the curve.
(//i)
^
meets the co-ordinate axes at the origin only.
x+y+a=Q is its
only asymptote.
cannot
both
be
00 x, y
negative so that no part of the curve
on the third quadrant.
(iv)
lies
It
On transforming
r
Now, r=0
~
dr^ __
46
for
to polar co-ordinates,
_
~
30 sin
we
get
cos
'
3
cosM~+sin
0=0 and 0=7r/2.
3a(cos
'"
sin
0)(l+sin
81
"(cos ^
+sin
cos
3
0+sin
2
0)
www.EngineeringEBooksPdf.com
2
cos2 0)
363
CURVE TRACING
which vanishes only when
cos
sin
0=0,
For 0=7T/4,
i.e.,
r:-=3a/
tan 0=1.
V2
i.e.,
0=7r/4 or
57T/4.
-
Yt
,
Fig. 150
Thus
from
/*
to
to 3a/\/2, as Q increases
monotonically increases from
and monotonically decreases from 30/ y'2 to 0, as 9
7T/4,
increases from Tr/4 to
Tr/2.
Again as Q increases from ?r/2 to 37T/4, r remains negative and
to oo so that the point (r, 0) describes
numerically increases from
the part of the curve shown in the fourth quadrant. (Fig. 150).
As increases from 37T/4 to TT, r remains positive and decreases
from oo to so that the point describes the part of the curve shown
in the second quadrant.
is easy to see that we do not .get
increases
from TT to 2?r.
when, 0,
It
Find the double point of the curve
13.
and
any new point on the curve
trace
(P.U.I 95 5)
it.
Let
Putting
/JC--0
fv (x,
y)
and
/y =0, we
have
or
(a,
origin
Thus we see that fx and / vanish at the points (0, 0), (a, a)
Of these only (0, 0) is a point of the curve so that the
a).
is
the only double point of the curve.
www.EngineeringEBooksPdf.com
364
DIFFERENTIAL OALODLUS
To
trace the curve
It is
(i)
symmetrical about the line y=*x and meets
Fig.
(it)
It
following particulars aboht
we note the
it
it
at
151
X
passes through the origin and
0,
y0,
are the
two
tangents there.
(Hi)
(tv)
meets the co-ordinate axes at the origin only.
It has no asymptotes.
It
x and y cannot have values with opposite signs, for such
(v)
Thus
values make 4fflxy negative whereas x*+y* is always positive.
the curve lies in the first and fourth quadrants only.
(vi)
On
transforming to polar co-ordinates, we have
r2
dr
r
so that
dr/dO^O
'
if
cos 0/(sin 4 0+cos 4 0).
sin
^ 2a
2
1'
2
4a
4
and only
tan 0=1,
i.e.,
0=7T/4 and
18-6.
Parametric Equations.
trate the process.
x=a(Q+sin
(
co* 2 0)
57T/4.
(Fig. 151).
The
following examples will illus-
Trace the curve
as* 0, varies in the interval
2
if
Thus we have the curve as drawn
14.
4
4
(cos 0-~ sin ~ 0)(cos 0+sift 0-f 4" sin "
0),
y=a(\+cos
0),
TT, TT).
Yt
Fig.
152
www.EngineeringEBooksPdf.com
CURVE TRACING
We
have
dx
dy
~0(l+eos0), -~-=
U0
de
^
dx
365
asin0.
ten-?^
~ VjtlJL
*
2
positive for every value of 0, we see that as, 0,
always increase. Also, dy\dQ is positive when, 0, lies
and negative when, 0, lies in the interval
TT, 0)
(
increases from
TT to
(0, ir).
Therefore}* constantly increases as
and constantly decreases as, 0, increases from to TT,
Since dx/</0
is
increases, .v will
in the interval
The following
arid
dy/dx
table gives the corresponding values of
0,
x,
y,
.
:
Intermediate
TT
an
increases
',
Intermediate
7T
an
increases
I
i
increases
y
dy/dx
;
2a
decreases
oo
i
CO
=
and dy/dxco, we see
Since for
air, y
TT, we have x=
that the point A( an, 0) lies on the curve and the tangent thereat is
increases from --TT to 0, x and y are both
parallel to F-axis when
increasing so that the point P(x, y) moves to the right and upwards
till it reaches the position K(0, 2a) corresponding to
where, dyjdx
being 0, the tangent is parallel to A'-axis.
;
=
When
to TT, x increases but y decreases so
increases from
that the point moves to the right and downwards till it reaches the
position A' (an, 0) corresponding to 0~7r where, dyfdx being infinite,
the tangent is parallel to F-axis.
Thus the point F(x,
from
TT
to
y) describes the curve
AV A'
as
increases
TT.
It is easy to see that the point P(x, y) describes congruent portions to the right and to the left of the portion AVA' as
varies in
the intervals
...... (-57T,
-STT), (-37T, -TT),
(TT, STT),
(3?r, 5?r) ......
Trace the curve
15.
x=^a
sin 20(1
+cos
20),
y=a
cos 2o(l
We have
:
=40
cos 30 cos
d0
-4a cos 30 sin
dx
This shows that
www.EngineeringEBooksPdf.com
0.
cos 20).
DIFFERENTIAL CALCULUS
366
The
following table gives corresponding values of
0,
x,
y and
dyjdx.
7T/2
00
7T
With the help of
shown.
this
X
of values, we have the curve as
table
(Fig. 153).
Since x, y are periodic functions of 0, with TT as their period, the
values of x, y will repeat themselves as
varies in the intervals
(TT,
so that the
same curve
27r), (2;r, 3?r), etc.,
will
be traced.
A
o
6
Fig. 153
The curve has 3 cupps
viz.,
A,
B and C.
www.EngineeringEBooksPdf.com
X
EXERCISES
367
Exercises
Trace the following curves
1.
3.
:
2.
3ay'=x*(x-a).
ay*=x(a
2
*=x 2
2
x
2
4.
).
5.
y
8.
+x2)=a2x.
(B.U.)
2
2
2
2
X (y +x )=a (x*-y 2
9.
13.
y(a
6.
(4-x*).
10.
2
;c=0 8 (a-jc).
15.
yV+x )=0
y x=a(x
19.
/x =x -K
20.
=x
22.
2
2
2
21.
a y
23.
2
25.
2
2
2
-a2
;c
2
).
(D.U. 1955)
2
(2a-x)(x-a).
36;>
^(jc
2
2
-l) (7-;t
8
)
.
Trace the following curves
26.
x-(y-l)(y-2)(.v-3).
28.
r
30.
r
r log
r=a(0 -sin
29.
r
31.
r=alog0.
32.
33.
r=a^
34.
sin 30.
sin 0.
Trace the following curves
35.
x*+y* = 5ax y
37.
x*=ay*-axy*.
39.
x(yx)=(y+x)(y
2
cos 8
2
cos
0=a
0=
cos 20.
2
(L.U.)
sin 30.
0=a.
e).
:
2
36.
.
38.
2
+x*).
Trace the following curves
2
a:
(/*{/.)
:
r-a-t-6cos0.
0=a
^5)
24.
2
27.
cos
(#
18.
).
2
2
flV-x^a 2 -* 1 )-
16.
.
(2;c-l)=x(;t-l).
s
+y2 )=a(x*-y 2
14.
17.
2
>>
2
2
).
;v
x(x
:
41.
40.
.v
42.
<flx?-&lf**\.
43.
44.
axy=x*-a*.
45.
(jt-}-a).
=Q
2
has a cusp at the origin and a
46. Show that the curve x*(x+y)y
rectilinear asymptote x+y=*\ and no part of the curve lies between the lines
4 and that it consists of two infinite branches, one in the second
and
quadrant and the other in the first and fourth. Give a sketch of the curve.
x0
x=
Trace the following curves
47.
**-(* +J0 a
49.
x a (y+l)-y2 (x-4).
51.
2xy=x*+y*.
:
(L.U.)
.
(P.U.)
48.
y(x-y)*=(x+y).
50.
y-x(*
52.
x4
2
-m2>fy
.
www.EngineeringEBooksPdf.com
(B.U.)
(B.C/.
DIFFERENTIAL CALCULUS
368
lies
53.
Xl~*2 )(2-~*) 2 ==*(l--*).
55.
Show
between
54.
-
r(l
that no part of the curve
x=
x=~l
4 and
x0 and x=3
or between
;
give a sketch of the
curve.
(Af.T.)
56.
Show
that the cubic
tl
xy*-x'*y-lx*-2xy+y +x-2y+\=-Q,
has double point at (0,
1)
;
'
asymptotes and sketch the curve.
find its
Trace the following curves
:
57-
\x=a(0~sin0),>'=a(l"-cos0).
58.
60.
0), y^a (1-cos 0).
x=a cos 30, y=b sin 8 0.
x=a cos 8 0, y=a (sin 30+9 sin 0).
61.
x=0(3 cos0
62.
63.
0+0
x/=log (1+cos
64.
x=a
59.
65.
x=a(0+sin
cos 3 0),
sin
x=fl(co$
log
x/a=2
|
sin
0),
y=a
^^=a(sin
0)-~cos
0+tan
(sec
0-log
I
0)
(sec
0,
1
8
(3 sin
,
0-sin
00
cos
y/a=sin
y=a log
0+tan
0)
I
0).
0).
-
i
cosec 0+cot
66.
x=a cos
67.
x=0(sin
68.
*==0(0+sin
;
.y=a(2+sin
+i
0) sin
C03
0/(3+cos
0-log
0).
j=a (cos 0-J cos 30).
6-in 0), y=a(cos 20-cos 0).
sin 30),
cos
.
|
,
ylb=2
2
0)
www.EngineeringEBooksPdf.com
|
(cosec
0+cot
0)
CHAPTER XIX
ENVELOPES
One parameter family of Curves.
19-1.
function of three variables, then the equation
If f(x y, a) be.any
y
f(x,y,*)=o,
determines a curve corresponding to each particular value of
The
a.
of these curves, obtained by assigning different
said to be a one-parameter family of curves.
totality
values to a,
The
is
variable, a,
which
is
different for different curves
is
said to
be the parameter for the family.
Illustration.
(i)
The equation
determines a family of curves which are circles with their centres
on A'-axis and which pass through the origin. Here, a, is the
parameter.
(//)
The equation
y=mx 2am- am 3
,
determines a family of straight lines which are normals to
parabola
Here the parameter
is
m.
the
.
We now
introduce the concept of the envelope of a oneparameter family of curves by means of an example CDnsidered in the
next
article.
'
*
Consider the family of straight lines
19*2.
y^mx+afm,
where, w,
is
the parameter and,
a, is
...(i)
some constant.
The two members of this family corresponding, to, the param l and Wj-f 8m of the parameter, m, are
metric values
~
m
,
i
369
www.EngineeringEBooksPdf.com
..()
370
DIFFERENTIAL CALCULUS
We
4ine
m
fixed and regard. 8m as a variable which
l
so that the line (///') tends to coincide with the
shall
keep
tends .towards
(ii).
The two
lines
As 8w ->
0, this
tion on the
line
(//)
(//), (///)
intersect at (#, y)
which
lies
on
where
point of intersection goes on changing
and, in the limit, tends to the point
1
N
t
its posi-
1
(ii).
This point
is
the limiting position of the point of intersection
latter tends
of the line (ii) with another lino of the family when the
to coincide with the former.
lie a point, similarly, obtained, on every line of the
locus of such points is called the envelope of the given
There will
family.
The
family of lines.
To
the
(//), we notice that
a
on
such
the
line
are
of
*m
y)
point lying
given by
find this locus for the family of lines
oo-ordinates
j
(x,
x=
a
2a
-
m*
,
y= m
Eliminating m, we obtain
as the envelope of the given family of
lines.
The envelope of a one-parameter offamily of
Definition.
19-3.
is the locus of the limiting position of the points of intersection
of any two curves of the family when one of them tends to coincide with
curves
the other which
19*4.
is
kept fixed.
Determination of Envelope.
Let
/(x,y,a)-0,
j^be
any given family of curves.
Consider any two members
and
f(x,y,
a+Sa)=0,
^corresponding to the parametric values a and oc+Sa.
www.EngineeringEBooksPdf.com
..(/)
ENVELOPES
371
The points common to the two curves
(/),
satisfy the
(11)
equation
f(x, y, a +8a)
_ -/(*,
g
Let Sa ->
common
to
(/)
0.
and
JF,
a)
_
.
Therefore the limiting positions of the points,
satisfy the equation which is the limit of (ill)
(11)
viz.,
/a(*,J>,)=0.
..(/v)
Thus the co-ordinates of the points on the envelope
equations (i) and (/v).
Let the elimination of a between
(i)
and
(/v)
lead
satisfy the
to an equa-
tion
This
is,
then, the required envelope.
To
Rule.
obtain the envelope of the family of curves
/(*,J>,a)=0,
eliminate, a, between
f(x, y,
where fax,
is
y, a)
and
are the
then (w)
(v/)
being the parameter.
Illustration.
Tve eliminate, a,
is
To
is
a)=0,
(/)
and
(/v) for x, y,
we obtain
*=#(),
.-(v)
y=*(a),
..(vi)
parametric equations of the envelope
;
a
find the envelope of the family of lines
y <x.x fl/a-=0,
between (v/7) and
obtained by differentiating
The eliminant
which
y,
the partial derivative off(x, y, a) w.r. to a.
on solving the equations
If,
which
a)=0, andfa(x>
(vii) w.r.
-
-(w'O
to a.
is
the envelope of the given family of lines.
This conclusion agrees with the one already arrived at 'ab
9
initio
in
19-2.
www.EngineeringEBooksPdf.com
372
DIFFEKENTIAL CALCULUS
The
Theorem.
19-5.
a curve
evolute of
is
the
env3lope of its
normals.
are the normals and PT,
the tangents at two
of a curve. L is the point of intersection of the tan-
QT
QR
PR,
Q
points P,
/
gents.
T
LPRQ = LTLT'=W,
arc
Applying
PRQ, we
PQ=Ss.
sine-formula to
the
A.
get
sin
/_RQP
P:=sin /_RQP.
^ -~
sin
or
chord
Let
Q
^ P, so that Z./?gP ->
Lim P/?=sin--
.
2
Q-+P
L RPT^ir/2.
~~ 1 = P.
1
.
.
^
Thus the limiting position of R which is the intersection of the
P and Q is the centre of curvature at P.
normals at
Hence the theorem.
19-6.
Geometrical relation between a family of curves and its
envelope.
To prove
19-61.
family
is
a point on
We have to
its
that
any singular point of any curve of a given
envelope.
show that
if for
any point on any member of the
family
we have
/,=/w=0,
then, for such a point,
we
will also
have
/a=0.
From
(i),
we
get
Putting
dfldx=dfldy=Q, we get
9//9a=0.
Hence the
result.
www.EngineeringEBooksPdf.com
ENVELOPES
Thus the
is
373
locus of the singular points of the curves of a
family
a part of its envelope.
1962. To prove that,
curves touches each
in general,
the envelope
of a family of
member of the family.
Let
/(*,?, a)=0,
...(/)
be a given family of curves.
Its envelops is obtained
by eliminating a between
(/)
and
Let
=$
(a)
be the parametric equation of the envelope obtained by solving
and (//) for x and y in terms of a.
The equation
every value of a.
We
(Hi) satisfy
differentiate
(/)
the equation
w.r. to a,
regarding
(i)
jc,
identically,
y
as
i.e.,
(/)
for
functions of a,
we obtain
o that
3/
dx
which, with the help of
dx_
df
'
(//),
dy
'
</<x
^dy
'da
+ 9/_
a<x
becomes
...(IV)
^f()+^*'(a)=0.
The slope of the tangent at an ordinary point
"V of the family is
(x, y)
of a curve
Also the slope of the tangent at the same point to the envelope
<///)iB*'(a)#'(a).
We see from (iv) that these two slopes are the same. Thus
the slopes of the tangents to the curve and the envelope at the common point are equal. This means that the curve and the envelope
have the same tangent at the common point so that they touch.
N ote. If for any point on a curve, 3 ffox and 3 ffoy are both zero, then
the above argument will break down, so that the envelope may not touch a
curve at points which are the singular points on the curve.
7
We
know that a straight lino and a conic cannot have
Cor. 1.
singular point. Hence we can say that the envelope of a family of
straight lines or of conies touches each member of the family at all their
common points without exception.
a
www.EngineeringEBooksPdf.com
DIUFBBBNTIAL CALCULUS
374
Examples
Find the envelope of the family of semi-cubical parabola*
I.
/
Differentiating
w.r. to a,
(/)
*Y
we
.
Eliminating a between
we
(//),
(i)
and
get
y=o,
which
-
the required envelope.
We already know that
is
the locus of singular points
O f i and also it touches each
y=0
i.e.,
-.
155
iff
cusps,
member
of the family.
Find the envelope of the family
2.
where a
is
m
a constant and
we
which
is
a parameter.
to m we
Differentiating w.r.
Eliminating m,
is
get
get
the envelope.
Thus the envelope
consists
of two
ines
x0 and x=a.
If
we
trace the given curves
Fig. 156
x+a
we
will find that j-axis
x=a is tangent
(x=0)
is
the locus of
its
singular points and
to each curve.
Considering the evolute of a curve as the envelope of its
2
2
2
normals, find the evolute of the ellipse x /a +y /6-=l.
3.
The equation of the normal
the ellipse
at any point (a cos
0,
b sin
0)
on
is
ax
^Jby^^tf^ip
cos
Thus, (i)
the parameter.
is
//v
"~~sin 0~~
the equation of the family of normals, where
Differentiating
(/)
ax
partially w.r. to 0,
sin
cos
2
<T +
by GOB
sin
is
we have
__
2
www.EngineeringEBooksPdf.com
,..v
"*
l)
375
ENVELOPES
To obtain the envelope, we have
From (ii), we get
(//).
tanvan v = ---
to eliminate 6 between
(i)
and
.
()*
Substituting these values in
[(<*x)%
(/),
we
get
+(by)$] [(**)* +(*)>)*]* =**-&*,
or
or
which
is
the required evolute.
Find the envelope of the family of ellipses
4.
/z^
two parameter
a, 6, flre
connected by the relation
a+b=c
fomg
;
constant.
(B.U. 1954)*
eliminate one parameter and express the equation of
have
given family in terms of the other.
c,
tz
We will
&
=c
We
a,
so that
ia
the equation of the family involving on parameter
Differentiating (i) partially n\r. to a we have
y
~0
or
ca
-
-
=
which gives
-
---- ^
/(
Substituting these values in
(/),
we
get
or
or
which
is
the required envelope
www.EngineeringEBooksPdf.com
a.
OALOTTIJJ8
f'W>
ewehpt
$) ,:$w4';&?
off he family" of
lines
,(0,
a and b ore connected by the relation
the parameters
Here, if, as in the example abo^e, we eliminate one parameter,
the process of determining the envelope will become rather tedious.
This tedio^sn^ss may be avoided in the following manner.
We
pohsider, b, As a 'functibq. of, a, as determined
Differentiating
(/)
and
(11)
and
(iv),
a,
we
(//).
get
_
b
,
(Hi)
fr6m
.
^
da
'
From
to the parameter,
w.r.
we
y
The equation of the envelope will, now, be obtained by
nating a and b from (/), (11) and (v). Now (v) gives
---xfa+y/b
r =--
1
y\b
n
=:xc w or a
^n
'
eliminate db\da and get
.x
x'/a
-^
N
or
L-
n
[From
L
(/)
v
y
and
elimi-
(//)]
/J
v
=
^^
Substituting these values in
(//),
we
get
or
xn/(n+l)
as the required envelope.
Show that the envelope of a circle whose centre lies on
6.
parabola y*=4ax and which passes through its vertex is the cissoid
Now,
2
(a/
,
the vertex (0, 0)
2at) is
y*(2a+x)+x*=0.
any point on the parabola.
the
(B.U. 7953)
Its distance
from
is
Thus, the equation of the given family of circles
2
2
(x at*)+(y-2at)*=^a t*+4a*t
is
,
or
Differentiating
(i),
Substituting
x*+y*2at*x-4:aty=Q.
t, we get
4a/x 4oy=0, or /= y\x.
this value of t in (/), we get the required
...(vi)
w.r. to
www.EngineeringEBooksPdf.com
envelope.
377
ENVELOPES
7.
Find the envelope of straight
of the Cordioide.
risht angles, to the
drawn at
lines
radii vectors
through their extremities.
Let P be any point on the curve.
then its radius vector OP=^a(l +cos a).
The equation of the
radius vector
line
P at
vectorial
angle,
right angles to the
a)=tf(l
rsin (0
eliminate
(a)
(r
r
+cos
cos
from
we
Qa)
tan
cL
get
,
.
sin a.
a)^
(1)
sin 9 cos a
(4) gives
and
(2),
we
= r sin
Substituting these values in
2
(r cos 0-tf)
2
+a
a
Qa).
0/(r cos
(3),
+r
2
as
.
.(3)
..(4)
0a
r cos
~
a __
C S
y(r
..(2)
them
a=a.
a =0,
re-write
cos a-\-r sin 6 sin
a) sin
(r COR
r sin
__
..(I)-
a).
different for different straight lines.
is
Differentiating (1) w.r. to a,
'
its
OP is
The angle a
Now,
be
drawn through
r cos (6
To
If a
2
'
cos
0'
we obtain
sin 2
0_
"~~
2ar cos 0)
'
or
r 2^02_oar cos
which
is
0a
2
,
i.^.,
r=2a
cos
the required envelope.
Exercises
Find the envelope of the following families of lines
2
2
(i) y^wx+^cFm ^ 6 ), the parameter being m
3
3
the parameter being o
x
cos
sin
0=a,
Q-\-y
(//)
(iii) x sin
y cos G~flO. the parameter being o
n
n
(/v) xcos Oi->>sin 0^=0, the parameter being 6
p
(v) y=mx + am , the parameter being m
^cot o~c.
(v/) xcosec o
2.
Find the envelope of the family of straight lines xla+ylb=-\ where
4, 6 are connected b> the relation
1
2
2
2
(m) a^-c
(/) a+ 6=c.
(//) a +6 -c
c is a const ant.
3.
Find the envelope of the ellipses, having the axes of co-ordinates as
principal axes and the semi-axes a, b connected by tne relation
:
1.
;
;
',
;
;
.
4.
Show
,
that the envelope of the family of the parabolas,
under the condition
(/)
,
(11)
a6=c 2
is
a-f 6==c
a hyperbola having
is
its
asymptotes coinciding with axes.
an astroid.
www.EngineeringEBooksPdf.com
378
DIFFERENTIAL CALCULUS
5*
ag the evolute of a curve as the envelope of
*~A*I>
9on5idcri
find the evolute
of the parabola y*=4ax.
its
normals,,
Prove that the equation of the normal to the curve
6.
may
be written in the form x sin
f-y cos #+ cos 20=0 and find the envelope
of the equation of the normals.
(P.U. 1944}
7. Find the equation of the normal at
any point of the curve
*=0(3 cos
cos8 0, y=a(l sin t-2 sin 8 f)
and also find the equation of its evolute.
[P.U. (Supp.) 19361
8
Fromac y point of the ellipse x2 /a2 -f >2 /6 2 =l, perpendiculars are
A
drawn to the axes and their feet are
joined; show that the straight line thu*
formed always touches the curve
t2
*
,
UAO* +(ylb)% =1
Find the envelope of the curves
9.
x*loP cos Q+b*ly* sin
(B.U. 1952)
6=1.
10.
Circles are described on the double ordinates of the parabola
as diameters
prove that the envelope is the parabola y*=4a(x+a).
2
y =4ax
:
[P.U. (Supp.) J935]
that the envelope of the circles whose centres lie on the
rectangular hyperbola x*-y*=a* and which pass through the origin is the
Show
11.
lemmscate
r2
13.
=4a2
cos
(B.U. 1955)
20.
Find the envelope of the circles described upon the radii vectors of
'*V+y a /6*= las diameter.
Find the envelope of a family of parabolas of given latus rectum and
parallel axes,
when the
locus of their foci
is
a given straight line
y=px+q.
(P.U. 1931)
that the envelope ol the straight line joining the extremities of
2
a pair of
semi-conjugate diameters of the ellipse * 2 /a 2 +rV6 =l is the ellipse
14,
.
Shnw
15.
Find the envelope of straight lines drawn at right angles to the radii
vectors of the following curves
through their extremities
n
n cos n.
(i) r=*a+b cos 6.
(//) r =a
:
Gcota
(///)r=<*
(B.U. 1953)
16.
Find the envelope of the circles described on the radii vectors of the
following curves as diameter
.
:
///= i + e cos 9.
Find the envelope of the curves
(/)
17.
(//)
r
n-. n cos
a
jfQ,
where the parameters a and 6 are connected by the relation
a+
18.
Show
v
v
b =c .
(P.U. 1955)
that the family of circles (x-0) 8 +V 2 =fl 2 has no envelope.
(P.U. 1942)
Miscellaneous Exercises II
1.
Show
that all the curves represented by the equation
*~a~
+ ~~b
VflHhfr/
'
for different values of n. touch each other at the
point x^-y^abl(a
the radius of curvature at this point is
www.EngineeringEBooksPdf.com
f b) and
that
(DV.
1956)
MISCELLANEOUS EXERCISES
its
H
37 9*
2.
If the polar equation of a curve be r**a see* ie, find an expression for
radius of curvature at any point.
3.
Show
that
06
cos
x=a(cos 0+0 sin 8), y=0(sin
is the involute of a circle.
4.
Find the radii of curvature of the curve
nV-a 1* 4 -**,
for the points *=0 and x~a.
5.
If a curve be given by the equations
--'
2
0>,
2=f 2--2,
2 -
-2
of curvature in terms of
to the axis
a curve passes through the origin at an inclination,
of x, show that the diameter of curvature at the origin is the limit of
(x*+y*)l(x sin <t-y cos a)
when x -* 0. Hence show that the radius of curvature at the origin of the curve
y*+ lay 2ax 0,
find the radius
6.
is
t.
If
,
-242a.
7.
If P,, p f be the radii of curvature at the extremities of two conjugate
semi-diameters of an ellipse semi-axes a, b t prove that
1948
that the projections on the X-axis of radii of curvature at the
of >>=log cos x and its evolute are equal.
points
corresponding
m m cos
and Q is the intersec9.
If P is any point on the curve r =a
/up
tion of the normal at P with the .line through O at right angles to the radius
vector OP, prove ihat the centre of curvature corresponding to P divides PQ
the ratio / m.
cot
Prove that if
10.
The equation of the equiangular spiral is r=ae
O be the pole and P any point on the curve, then a straight line drawn through
Oat right angles to OP. intersects the normal at P in C such that PC is.this
(M.u.y
radius of curvature at P.
8.
Show
m
:
.
11.
Show
that the
normal to the Lemniscate
r
2
^?
2
cos 28
at the point whose vectorial angle is */6 is perpendicular to the initial line and
that the centre of curvature at the point at a distance 420/12 below the initial
line.
12.
Show
that a point
*=3
P on
the curve
2
cos 0-cos*0, >'=3 sin 0-sin
and that the centre of
where 0=7r/4, the normal* passes through the origin
(B.U.)
curvature at P divides OP internally in the ratio 4:1.
13. Show that the pole Hes within the circle of curvature at every point
of the cardioide
a(l-f cos 0), and that its power with respect to it is rN3.
14.
Find the co-ordinates of the point in which the circle of curvature of
the parabola y* = 4x at the poini (f*, 2f) meets the curve again.
The circles of curvature of a fixed parabola at the extremities of a focal.
passes through a
chord meet the parabola again at/fandtf; prove that
(Indian Police 1935)
fixed point.
HK
and its evolute at corres15.
p, p' are the radii of curvature of an ellipse
at P make*
ponding points P, P' A is the area of the triangh which the tangent
with the axes ; show that p/p' varies as A*
16.
Prove that the maximum length of the perpendicular from the pole
(B.U.)
on the normal to the curve r=fl(i cos 0) is 4<V3a/9.
17.
For what points on the curve xy*~a*(a-x) is the square of sub-tangent maximum.
www.EngineeringEBooksPdf.com
380
DIFFERENTIAL CALOTTLTT3
18.
Show
19.
Find points
that the tangents at the points of inflexion of the curve
5xH3y-4a=0,
(B.U.)
on
of* inflexion
x(xy-a
z
(M.T.)
)*=b'.
Find the co-ordinates of two real points of inflexion on the curve
2
y =(x-2Y(x-5)
and show that they subtend a right angle at the double point.
21.
If O.is a fixed and Q, a variable point on a given circle whose centre
is C and if QQ is produced to P so that
QP=2. OC, prove that the radius of
20.
curvature of the locus of P at the point
P is
B
22.
If
O
is
the extremity of the polar sub-tangent at a point
P
of the
curve
prove that the locus of
Q
r=atan(0/2).
is.
(B.U.)
r=0(l-fsinO).
23.
Find polar sub-tangent and polar sub-normal at any point of the
-curve
r=*-
-show that the polar sub-tangent constantly increases as
increases and the
polar sub-normal attains its maximum value 3>j3a/8 at the point (a/4, 1/^3).
24.
Show
that the circles of curvature of the parabola
y*=4ax
for the ends of the latus
and
rectum have
for their equations
that they cut the curve again in the point (9a, qp 6n).
25. Show that the centre of curvature at every point of the curve
Avhere
it
r=0(0
meets the positive side .of initial
26.
Show
that
sin 0)
line is the pole.
the co-ordinates of the centre of curvature
in any of the following ways
(P-U.)
may be
given
:
,
dx 1
rf
_1
J
Show
27-
that the area of the triangle
2
j/
its
asymptote and
28.
formed "by a tangent to the cissoid
(2a-x)=A,
A'-axis is greatest for the point given
x 3a/2,
=
by
Find the equations of the tangents to the curve
parallel to the line y=x.
29. Trace the curve
r^=a(l+cos
e)
^ind prove that the greatest distance of the tangent to the curve
point of the axis is >l2a.
30.
Show
of dinate to
;
that the length of the perpendicular
normal at any point of the curve
jt=0(sinh 26-20), y=4a cosh 0,
is constant.
www.EngineeringEBooksPdf.com
from the middle
CP U-)
from the foot of the
MISCELLANEOUS EXERCISES
Show
31.
that for the curve
0)-cos
,
381
II
y/a=sin
0,
0,
the portion of the A'-axis intercepted between the tangent and the norrral at
any point is constant.
32. Find the nature of the double poiut on the curve
and show that
it
has two real points of inflexion.
33.
Show
that the pedal equation of the curve
34,
Show
that the tangential pedal equation of the curve
+(;/*)
<*/*)
=
*
is
1
2
la
sin2
cos 2
$ -f 1 Ib-
on the tangent at each point of curve, a constant length be
measured from the point of contact, prove that the normal to the locus of the
point so formed passes through the corresponding centre of curvature of the
35.
If
given curve.
Show
36.
that there
as the parameter
V
is
a curve for which the circles
cosh a+l=0,
U-a) 2 + y 2 ~2y
varies, are the circles of curvature.
Prove that the distance from a fixed point P (x y e ) to a variable
on the curve f(x,y)^V is stationary when PP is normal to the curve
37.
point
P
,
at P.
The tangents
38.
at any point
x=--a(0
'0'
of
+ sin 0),.v=0(Hcos
0)
normal to the curve at the point where it meets the next span on the right "
cot (0/2)^=-- n/2.
prove that
is
Show
39.
that the locus of the intersection of perpendicular
the Astroid
x=^a cos 3 0,
is
y=a
the curve
2
2(jc*+/)=fl (*
8
- y2
i
2
)
;
and the locus of the intersection of perpendicular normals
to>
is
(D.V. Hans., 1959}
the curve
For the curve
40.
find the
tangents
sin 3
x^=a(2 cos r-f cos 2f), y=a(2 sin /-sin 2f),
equation of the tangent and normal at the point whose parameter
is t
and show that
(0 the tangent at
are -i/ and?r-i/,
P meets
the curve in the points Q,
R whose
parameters
QR is constant,
ihe
tangents at Q and R intersect at right angles on a circle,
(Hi)
(j'v) the normals at P, Q and R are concurrent and intersect on a concentric
(11)
circle.
2
P, Q are any two points on ay =x* such that PQ always passes
9
8
a
or
fixed
on
the
curve
)
show that the locus of the
(at
through
point
lying
point of intersection of the tangents at P, Q is the parabola
41.
,
42,
again at
where
Tangent
Q
;
OX is
show
;
P
of the
at any point
that
cot 2 LXOP+ tan
Folium x*+y*=3axy meets the curve
Jf-axis.
www.EngineeringEBooksPdf.com
DIFFBBENTIAL CALCULUS
43.
Show
that
x=0/4
is
a bitangcnt of the cardioidc
r =a( I
414.
The envelope of the
x cos
-cos
(Birmingham)
0).
straight line
#+y sin
4s the curve whose polar equation
?$=0(cos
is
//(I-*) ~*nl(l-n)
45.
Show
that the radius of curvature of the envelope of the line
XCOS a+ysin
a=/(<x),
that the centre of curvature
is the point
*=-/'(<*) sin a /"(a) cos a
y=/'(a) cos a -/"(a) sin a
Discuss the nature of singularity at the origin of the curves,
a
2
5
2
(i) j>a=*(jc -l).
(//) x -0(x --a>0 ==0.
Show that the curve
|
(M.U.)
|
>46.
-47.
G0.J7.)
bytx* sm2 (xja)
has a cusp at the origin and an infinite series of nodes lying at equal distances
from each other.
Trace the following curves
48.
:
(i)
'(///)
49.
r-fl(2 cos e-f cos 36).
x=0[cos f 4-log taa Jf]-
(11)
y=a sin
>^-2cV+a2^ 2 -0.
/.
The tangent
the parabola
is
to the evolute of a parabola at the point where it meets
also a normal to the evolute at the point where it meets the
evolute again.
50. If p be the radius of curvature of a parabola at a point whose distance
measured along the curve from a fixed point is, s, prove that
51. If p and p' be the radii of curvature at corresponding points of a cusp
evolute, and p, q, r first, second and third differential co-efficients of>>
with respect to x, prove that
.and
its
P
9
2
[Hint. P '=</V<ty -]
52. Show that the radius of curvature at a point of the evolute
-curve
rn
-corresponding to the point
n
53.
(a)
lias a single
~an cos n$,
(r, e) is
/
*
of the
l~l\*
'sec/iO tan n$.
Show that the curve
x*-2x 2 y-xy*-2x 2 -2Ky+y 2 -x+2y+l=*0
1).
cusp of the second kind at the point (0,
Show that the radius of curvature of the curve /(r, 8)=0
(b)
is
rcosecy
the chord of curvature perpendicular to the radius vector
denotes the radius of curvature.
www.EngineeringEBooksPdf.com
is
2p cos ?
where
(D.U. 1956)
ANSWERS
467.]
JPages
Page
4.
Ex.
Page
3.
(//)
a,
(iv) nit
i) 1,
;
n being any integer.
9.
Ex.2.
Page
0.
(/)
1-710.
10.
Ex.2.
Ex.3.
Page
x-1 <2.
(ii)\x-l\ <$.
x-l <e.
x-2 <5.
(iv)
(in)
(i)-l<x<5. (//) -3<x<l. (in) -l<x<3,
(/)
|
|
|
|
\
|
14.
Ex.4.
(J) (-1,00)
The entire aggregate of real numbers excluding the numbers
is any integer.
(2 + l)ir, where
(/') The set of intervals
(')
when n
is
any
36.
Page
1.
(I) 7T/6.
2.
(in)
<v/i)
(*)
[!.
(iv)
(/)
Page
1].
0]
27T/3.
(IV)
[-co
(v)
+ lir),
J
>
,
-1],
[1, oo
].
(vi) [0, oo ].
(viii)
[1, oo
].
(ix) [1, oo
]-
(^"0
The
numbers (H+i)^.
(xiii)
[1,
Increasing in
in
[
and decreasing
!)TT,
oo
[00,00].
,
and
0]
in [0,
oc ].
in
(v)
(ii)
Decreasing in
[00
continuous.
6.
7.
Continuous for every value of
(
11)
x=0
and x = l.
x.
57.
-f
6667.]
(/)
continuous.
(//)
oo
].
[0, oo].
i*v)
Increasing
Increasing in the set of inter-
discontinuous.
(i)
Discontinuous for
2.
,
(2n+^)fr] and decreasing in [(2n+|)7r, (2fl+f)7r].
3.
[Pages
1].
in
51.
2.
].
entire aggregate
[0, QO],
Decreasing
vals [(2n
Page
[-1,
(Ill) 7T/3.
set of intervals [2mr,
t
3.
,
(//)
The
set of intervals (2mr 2n
~" 2
L (^O [~- >
]^ [
<xiv)[-oo,0],
(Hi)
-7T/4.
numbers excluding the
real
[00
(/I)
(i) j(0, 1).
[0, 1).
The
of
integer.
discontinuous,
(ill)
333
www.EngineeringEBooksPdf.com
continuous.
384
DIFFERENTIAL 'tEBptHTMf*
(iv)
(vii)
Page
Page 75,
4.
(vi)
discontinuous.
discontinuous, (vw) continuous.
(/)
13.
() -,V-
4-12.
Ex.1.
Page 76.
Ex.1.
(/)
2,
-2x/(x+3)
2
2
.
2(l-x )/(l+x
(//)
-1/2A
2 2
)
(n/)3x.
Ex.2.
.
(/v)
-1/V2.
78.
Ex.
Page
continuous.
(v)
74.
Ex.
Page
discontinuous,
[Page 74-89
3.
2x+>>+l=:0,
4Ar=^-j-4.
(0)8, 2; 2,
(ft)
79.
Ex.1.
2.
-1/16, 1/32
(c) 1/4, -1/32
1/2, -1/4.
-4, 3, -4; -2, 2, 2; 2, 7,8.
;
-1,
2.
;
Ex.3.
Page
83.
Ex.2,
(i)
(Hi)
Page
(x-l)/2A
(3x*+x-I)l2x%.
(//)
(2x~*+5x~l'*)/l2.
(iv)
(1+7^/6".
85.
Ex.
2.
(i)
2x+5.
(ii)
2(x+2)(3x+6).
2
(Hi)
Page
86.
Ex.2.
(Hi)
Page
(/)
-(4x+6)-.
(tf)
-x(x+2)-.
[(x'-
87.
Ex.2.
(iv)
(v)
Page
(2x+3) (10x-3)/2x^.
(i)an(ax+b)-
1
.
2 2
() -2x/(l +x )
----_. --r
.
,..
.
(w)
.
- ----
89.
www.EngineeringEBooksPdf.com
[Pages 91-97
ANSWERS
91.
Page
3.
(m)
mx m ~i
cos
(v)
(x cos
x
(v/ii)
Ex.
Page
(ii)
...
(in)
{V)
sin-^*
xm
cos x.
.
sin x)/x 2
(ii)
m
(iv)
sin 2x.
(vi)
.
sin x.
2
(ax+b)(ax*+2bx+c)~^ sin V(^
2
(m cos x w sin 2 x) sin- 1 * cos""" 1 *.
4.
(/)
tan
r.
(//)
wx.
sin
cot
+2feo;-f
r.
c).
tan
(ii'i)
r.
'
.4 cosec 2x cot 2x.
(/)
x tan 2x+2 sin x sec 2 2x.
sin 4x
8x sin 2 x
r
2
4 cos x sin 2x
1.
cos
,.
-
.
-;
~
_
.
2
(iv)'
v
.
sm
1
V[(cos 2x)](cos x+sin
(V/)
* 8e
,
2x.
2
x
'
2
x)'
93.
(i)
(i/i)
9 cosec 3 3x cot 3x.
\b sec ^(a-\-bx) tan
(//)
^a^/[seo (ax+b)} tan (ax +6).
a
N/(
+^)/v/ (+*x
-
tan (cosec x) cot x coaec
sec-(cosec x)
(/v)
Page
(/)
92.
Ex.
Page
m
Ex.
(vii)
)-
x.
96.
Ex.
1.
(i)
-
(/V)
1
/X)
Ex.
Page
385
_
Vx
'
1+cosVx2
...
,
__ 2
8l
'
-
2.
1.
97.
n
Ex.
B
,
(/)
,->
cot x.
(ii)
,...,
ain
(in) e
.
x
cos x.
.
cot (log x)
(v >
x
(wi).
sin (log
v
" x);
----
secx.
.
....
(viii)
---ji-f-r,
e 2 ".2x log
x-e-*
,.
.
(w)
www.EngineeringEBooksPdf.com
a
x
.
lo
a
DIFFERENTIAL CALCULUS
386
Page
97.
a 2 cos 2 x
sin 2x).
-!
sin 2
(x/v)
x
100.
Ex.
tanhx.
() sinh 2x
sec 2 x tanh x 4 tan x sech 2 x.
(i)
(///)
Page
2
ft
/V(**)
~
2
log a
.
Page
Pages 97-105]
102.
Ex.
4.
[(log cos x)/x
(i)
(cosx)
(/v)
(tan x)
X
eo
.
tan
x. log x\.
cosoc 2 x log (e cot x)
tau x
(cot x)
(v)
+ lo
-^)-
(logx)*(log>gx
sin
x/
^
i;a
,x)
(co
-x2
(2
1-1
sm
x
S
x)
3
'
-1
log
f"
)
sec 2 x log (e tan x).
.
'
-
.v
r
1
'
9x 2
4.v_
~
2(1
"3(2~Tx
-r)
4
)
+4(3- j
_
(vm)
Page
+ 15jc* + 36)/3(JC
(2.x*
(v/i)
sin
.v
.
c
;r
.
.r
log
.
3
+3)
'
X* fcot .vf
f 4).
(.v
log .v)-'
\-
log
<?-.r|.
104.
Ex.5,
-
(i)
(i/)
,
^
T
(///)
:2
'
Ex.6.
Page
CO
-
1.
(//)
3
105.
Ex.
3.
(/)
2x cos x 2
.
m x cos x.
(ii)
sin 2x.
(///)
l/2y'.v.
'
(iv)
e
s
(v)
eV/Jf /2 v/x.
www.EngineeringEBooksPdf.com
(v/)
]/(l
+x
2
)
ANSWERS
[Pages 105-106
Page
387
105.
-.
*2
2.
sec 2 x\/(cot x)
~
5.
- --1
~
1.
2x e
4.
2jcsec 2 x 2
.
.
3.
'.
(jc
.,
.
cos x
sin x).
t
l
2
Page 106.
COS- J JC
Xy/( I
'
(l-X
^,^
.
)
X
o
/
/'/I
g
"
-^9v
i
9.
"
10.
(tanx-
.
12.
*
A
!
X2 )
"
2 3/ 2
)
log
.
13.
v
(1
~\'"*i
v
/
5
-
(i
1
.'
.(5x)
(
Y
15.
A'
16.
1
"
l
\"
f 7.Y)-/
7
(
I
6.7
'"1
}-7.x
"
V2 3(
3
-1.
14
"
11.
.
l
hotl(f.Y
,
,
i.
y
A
_
*'j_/
7"'""'
-"-Vx)"
4
'.)
v)~4(l
8.v"":V(r
!-3.r)
"f>(l
2
).
"
1
17.
I
-x log
.v-
1
log ex\x
.Y
(xt)
l
.
cosooli
18.
sin-'.v.
-.Y-).
.'5.Yv(l
A
'
.v.
y
I
'
1
.x
v/(l
2a
21
"'
^/
(l
log (^
2
x)
(a-
^
26.
(sin.v)
c
.v
log
-j
-b
1
)
.
los log
.v.
sin
lopRinX
.cot.x.lo^lO.
xf
.
1
log sin
.
x
.
x
cot
.v
^cos-
^.
axl
[cos (x log x)
x
.
log
^v-2ax
sin (x log x)].
2fl
1
27.
r/
win .v)v/(cos 2x)'
.x
cos* jc)-j-4ft cos
1
.cos""
25.
10
.v
^)
loo;
_____
24
22.
-
-
v
23.
(cos
)
sech ax.
28.
a
.
29.
www.EngineeringEBooksPdf.com
2
x
[Pages 106-111
DIFFERENTIAL CALCULUS
388
30
W
*2 ~1
~'
*
1
31
_
32
'
2x cos
Page 106.
9x 4
33.
sin
(3*- 7)
log (1
.
-5x) x
cot S
eax [a(l
34.
+x2
cos (6 tan- 1 *) ~fc sin
)
cosec
x
-
.
cot x,e
x
w/sin
7
oosec 2
37.
a*
sinh x+xfl* sinh x
38.
V3/(sec
cfcth
.
x
^(sec
x)
tan^x)]/^ +x
(fe
2
).
x
_/2/sin
'
/
-
/r1
n
cosech 2
36.
x
.
*~
x
.
w/sinx"
cot x.
log a fxa* cosh
x.
3x).
39.
40
'
(1-x 3 )"^.
41.
Page 107.
42.
(i)
43.
(I)
45.
x
l/(x*+l).
(H)
x/(x^l).
0.
(ii)
-tan
x cos x/log
sin
sin
x
J
_1
x
-,
^o,
~-
47.
i
48.
%
1,
(sin
x+x cos
-__--
(x cos x-f-sin
(sin x)
_/(l-x2
)
(1/1)
tan
t.
x.
.
_
3/.
.
.
x log
.
x
x)
log sin x)
49.
-1.
x(2+2tanlogx+secMogx).
tan * [sec2 xl g (1 S ^)+tan x(xlogx)^]
(logx)
m
*
cog
^
cos""1 x)
Page 111.
1.
/'(O) exists
/x
but/
2.
Differentiable at
the origin for
3.
/(x)
is
5.
/(x)
is
6.
(0)
x=0
m > 3.
does not.
for
m>
2.
/'(x)
is
continuous at
continuous but does not have derivative at the origin.
continuous but not derivable at the origin.
Continuous when x
for other values of x.
is irrational
Differentiable for
or zero
value.
no
7.
www.EngineeringEBooksPdf.com
and discontinuous
ANSWERS
Pages 112-120]
Page
112.
10.
Continuous but not differentiate for x=a.
11.
(i)
12.
Discontinuous at
derivable at
Continuous.
(//)
;
Discontinuous.
derivable at
1
continuous but non
;
2.
Page 115.
5.
Page
116.
Page
119.
-3/2.
^v'r
4
..?.
L (x2)*+
2
10
L(X~I)"
Page
n
!
+
*
i
(x+2)+
1
J'
f
r
t
(-- 1)"
l
^ ( X __ jp,
+
(v
+ i-
^])
1
(jr+o)"* J'
120.
3.
.
r
A
(-!)
_
1
!f
w+1
X
fl)
1
(X + tf)"*
1
2 sin [(n
+ l)
cot~ 1 (
-]
'
(+l)^, where
sin
7.
(j)
(
l)-i( w
(ii)
(
l)
n -J
(M
1)
1)
sin" 6 sin nd, where
!
!
cosec" a sinn
d=cot~ 1 x.
sin n&,
where
[(x
www.EngineeringEBooksPdf.com
coBa)/sin
390
[Pages 121-126
DIFFERENTIAL CALCULUS
Page 121.
+ \n it) - (3/J) sin (3x -f
() J[2 n cos(2x + IWTT) -f/$ cos(4x + JTT) +6"
2.
(i)
<7'0
(iv)
-|
sin.
(x
J/ITT).
cos(6x
+ |/wr)]
.
3 n cos (3.v+|/i7r)--5n co6(5jc+^w)l.
TV t 2 cos (x+imr)
[3e*-4-5^"cos (x+w tan" 1 2)
-f
(v) 4
^
(4x+n tan
17^'cos
4)]/8.
i
(3x+n tan
[25*" cos (jr+n tan-i i)- 13-" cos
|)
Page 123.
1.
e'f ^
(/)
(iii)
.v
8
cos
p
!
(.v
-v
|TT)
-f-
-I
/l2
*
-i-
M
)2
.v
^"-.?
f-3x 2
cos [Ar-j-|(n -l)7r|
3(W- 1)^008
+ J(-2)7T]4
[X
n(/i- ])(/i- 2) cos
(///) 0"+'-.
(iv)
*.
fi
e"[log
x-pfjX
V3
'
-
jr
f-T 3 2
!.
5.
Page
([
2
x-)y n
x ;vh2
x~*~
1
I)""
!-....-}-(
4.
3)ir|.
.
.
(i~l)
"<'
!
.
-v
"|.
^
+(2tt
125.
23-
Page 126.
4.
2
a
^ f n(0)=0;^ ta+1 (0)=/n(L*-./n )(3
V
yan (0) -0 v, -+1 (0) = - 1 )" I
W-
*
-
;
-
.
(
.
.
2
.
)
5
.
.
.
-
(2n
2
1
.
)
56.
^(0) = w-(m= -
2 2 )(m2
-4
a
)
- (2/i - 2)]
.... [w-
2
.... [(2
)
>Wi-=0, ^ J n=(-l)"- 1
2
.
22
.
42
.
-
62.
.
126.
4.
www.EngineeringEBooksPdf.com
.
;
8
.[/
1
.[(2n~2)
.
)
6.
-1)*- m*J.
[(2
-3*)(/i|S-5).
Page
[x+K/i
X2
2
2
)
-(2n-
+/].
4-/n
.(2-2)
2
|.
2
.
I)*].
ANSWERS
Pages 127-156]
Page
391
127.
If n
9.
Page 135.
Ex.3.
Ex.4.
is
even, y n (Q)
-0
n
if
;
odd, y n (0)--~n
is
(3/f- 5)
!
~
.
4
= (6-V2l)/ti.
c
(n)V5.
(/)|.
(//7)log(e
-1).
139.
Page
For a ^1, the function
2.
the function
3.
is
is
for
steadily increasing;
d<[
1,
00
2]
steadily decreasing.
Increasing in
2,
[
1]
and
[0, 1]
decreasing in
;
(
,
1-1,0], [l,oo).
5.
and
-
Increasing in
4.
Decreasing
[2, oo
[
i, 1
in (--oo
]
,
;
decreasing in
and
2]
fO, 2]
(00
;
1],
,
and
increasing in
[1, oo
[
).
2, 0]
).
Page 140.
6.
o.
:it>,
Page 144.
2
A
^
4.
ar
A
cos />.V
l
1
I
X
!
-& 2
--~Jt 2
-^
2t
,
-f
fl(l*-36)^A' 3,
V
:
..+^j
(
fl
24ft)
,
-
o
\
-!
....
!
J/l
1
^cos^ftflx+wtan--
)-
150.
Page
Ex.
2.
i:m.
Ex.
3.
max.
Ex.
5.
-s.
Page 153.
Page
55/27,
8
1
:
;
min.
1109,
21).
-10,
-27.
156.
1.
2.
3.
min. 37.
(/) max. 38
Max. for x -2a and minimum
Max. forx--l, min. for x-^2.
;
(11)
for
min.
x
4.
2^3/9.
5.
max. 64, min. 50 least 0, greatest 70.
max. for x--l, min. for x--=6.
max. for x -=\/3 min. for x-= */3.
6.
7.
= 3</.
;
1
e-
12.
rain,
13.
max. values
.
value
:
miii. values
integer,
.v
;
11.
14.
4.
and
log [(ae
e) log tfj/log a.
'
:
(27T+3v/3)/6
;
(87r+3y/3)/6.
(47T-3V3))6 (10:: 3\/3)/6.
max. for JC=/ITT where n is an odd positive or even negative
and minimum for x~mr where n is an even positive or odd
:
;
negative integer.
www.EngineeringEBooksPdf.com
392
DIFFERENTIAL CALCULUS
[PagOS 156-169
Page 156.
min. values :-J(5ir+3 x/3+2),
15.
max. value
VT
17.
~4
mm. value V3
.
,.v
(i)
,
least value
least value
18.
!
.
J
,
%
(ii)
min.
min.
any
is
0,
y
2/3-v/S
-A'-
max. values
;
;
1-
-V
-V -.
:
max. 1,2/3^6.
max.
;
:
.
1
greatest value
;
min. -1, -2/3v/6
(a* +b%
o
greatest value
:
:
max. values
;
(/)
(in)
where n
;
min. values
(if)
(27r--3 v/3+2).
(-77+3^3-2), ^
:
max.
;
(a$
+b%
0,
2/3V3.
fi
integer.
Page 157.
max. c*l(a+b).
The required values are the roots of the quadratic equation
20.
21.
(a~-r-*)(b
r~*)=h*, in
r.
Page 163.
1.
9,6.
4.
V(40/3),
5.
9.
Sq. whose each Ride =/2ff.
3.
vW3), h/(>/3).
Diameter of the semi-circle = 40/(Tr +4)
Height of the rectangle =-20/(7r+ 4)
Breadth y^ depth \/%a.
;
.
.
15(^
10.
miles per hour, 19800(|-)^ rupees.
18.
Page 164.
21.
3V30&M-
22.
26.
(3v/8ir)*.
28.
Length 2
and girth 4J ft.
30.
ft.,
girth 4
a-ft.
25.
3^3/4.
2V37T/27.
ft.
;
yes, length should
now be
If
Page 168.
Ex.3.
(0-f
(H)**l-2e.
(Hi)
f
(/v)
|
()f
(tfi)
-|.
(iv)
i.
(vi) 2a/6.
Ex.4.
(/)|.
Page. 169.
Ex.
5.
o
2
;
limit is
1.
www.EngineeringEBooksPdf.com
(v) 2.
ft.
Page 170.
Ex.3,
T\.
(i)
1.
(iii)
cofc itx
Change
(v) |.
(0 i
to cot \Ttx.
(v) 4.
173.
Ex.
Page
-J.
(//)
2/7T.
(iv)
Page
393
ANSWERS
Pages 170-188]
2.
(i)
0.
(ii)
3.
(iii)
1.
2.
(i)
0.
(ii)
1.
(iii)
20/7T.
2.
(i)
(ii)
0.
(iii)
3.
(i)
(ii)
e' 1
(iv)
1.
(iv)
2
7r /6.
(iv)
1.
1.
(v)
0.
(vi)
174.
Ex.
Page 175.
Ex.
.
.
Page 176.
Ex.
1.
.
1.
(iii)
Page 177. [8-7].
i
(v)
Page
e.
e.
(vi)
(vii)
e*.
(viii)
J.
6.
e.
177.
1.
i
8
e~\
2.0.
-5.
3.
9.e~*
2A
13.
e-"
17.
e~~
^
-iV
10.
4.
5.
11.
O
O
\4, e
1 2
15.
18.
*.
e"
16.
".
alogrt.
22.
'
<>
.
A /37T/6.
-2.
7.
e~^
12.
2.
2 /7r
21.
-1.
1
""
01 -.
a^ b
4.
.
/
*-'*
20.
19.
2.
23.
tog (e/*) log be.
1.
Page 178.
...aj\
25.
-<?/:>.
26.
29.
aV.
30.
24.
(a,a,<i 3
28.
-iflV.
33.
Continuous at the origin.
lie/24.
27.
f"(a)l'2f'(a).
16.
31.
Page 187.
7.
v
2.v-
*-
g-jc+
*4
-
^5
+
x *2
Page 188.
Y4
Y2
11.
12.
i+x-Hg*
12
-!+ A
.
J
14.
log sin 3
+ (x- 3)
V
1
13.
cot 2
-
^^
.
J-+J
cosec 2 3
.
www.EngineeringEBooksPdf.com
+
C ot
3 cosec 2
3.
1
TB
.
394
DIFFERENTIAL CALCULUS
J
15.
Page
14-f29(x---2)+16(;c-2) 4-3(*-2)
3
.
188.
<ReacU* ->*
Page
LP ft ges 188-215
cos
*
cos x
in place of e-e*
).
202.
a
ae * sin
2.
fofl;r cos
&>>,
--,
eB -' -g*-",
(i)
)
ye^
(ff)
-.
&-*[yx+y-l].
log
'
(i-xy)
4.
3'
(1-xy)*'
(l
0.
8.
]/c*r.
Page 203.
n_3.
2-
Page 205.
Ex.
r
1.
sin
0,
cos
0, r
cos 0, sin Q,
r
cosec 0, cosec 0.
Page 209.
2.
4%.
6.
-
-3J.
Page 215.
1.
3.
-
4.
(M cot
(
1
(oosyxys'my)lx
2.
4x-\-2y.
x cos M
+*)(! +J
2
sin
y+z
sin v sin x)/(cos v sin y).
)(25 f>2)/( 1
-xy)*.
2
5.
()
[>;+x cos (x-y)-\l\x+x* cos (x-^)J.
v ~l
v
(cot x yx.
log y)ylx (y log x log y-4
..
sin
(i)
.
>'(>;
sin
cos jf(sin
(/v)
y
x_+cos x
log cos
log_sin
jc
X^-* log y,lx(x-y
2
1
j(tan
x)*- sec x
""
_
8x
-
~~dFidz
'
s
x
j')
cos
>>)
log x).
cosec*x log y(y) e
8JL
8Fl 8y
8y~~
8F/8z
'
www.EngineeringEBooksPdf.com
"'
*
1).
395
ANS\VKKS
Pages 216-233]
Page 216.
14
,
Page 228.
1.
min. at
(i)
(5,
min. at[J(6
(ill)
(i7)
).
rain, at (0, 0) if
t/
>
.
20), J(0 --26)].
at (2, i).
max. at (2, 1).
(v) min.
at (
2, 0).
1, 0).
(v/i) max.
(vi) min. at (0, 0), (
extreme
value.
(v//7) No
1 + 4m +2/i)|,
(ix) max. at [-&n( I +4w -f!2/i),
min. at [-^-71(74-4^ + 12^), ^7r(l 4 -2m-f -2/?)],
where m and w are integers.
(x) max. at (|,
).
2.
max. value 112 at (4, 0) min. value 108 at (6, 0).
(iv)
M
;
Page 229.
3.
2(x
x 2 )(/;/
l
/2
1
MJh)l\ 2( n h ?L2
a
V
2
1
i)
-
Page 232.
1.
30'-.
(/)
2.
2
fl
3.
2k^(a*
3a*.
(//)
2
(Hi)
3w 2
.
and Ja (max.)
(uiiu.)
ab~\b*).
4.
......
'
x
'
v/14
/14
v'
Squares of the semi-axes are tho roots of the quadratic
6.
71
'
O
""8 O
'2
-
8
'
Miscellaneous Exercises
Page 233.
1.
Zero
2.
w
is
+ i,
the only point of discontinuity.
are the only points of disis any integer,
whore w
continuity.
3.
f2/;H-l)/(l- -2/w), in being
4.
1,
5.
(/)
(11)
(Hi)
any integer and
-1.
All integral values of ,v.
All integral values of x.
x-0.
.
www.EngineeringEBooksPdf.com
x=
I.
DIFFERENTIAL CALCULUS
396
[Pages 233-260
Page 233.
Page
6.
jc=-0,
7.
Discontinuous at the origin.
1.
,
234.
(CA)+ sin
sin
(B- Cf+sin
sin
(cos
_.
-
,4(c~os
A
cos
C)
......
',
_ lf x _
Jv
JA -A-v3
16
3a 4
^
r6
(7--cos 5)'
(-l)n!
J.W.
y
where
tan
^=a
sin
a/(x+a
cos a).
1
Tfg
Page 236.
30.
(i)
31.
|.
33.
(i)
34.
-2.
35.
(i)
!;(//)
-1
-1.
;(ffi)
32.
()
1.
continuous,
i.
(Hi) 0.
2a/b.
(//)
discontinuous, (HI) continuous.
Page 237.
4-3v2
--
max. at 7T/3 and min. at
min.
4/^ max. and
38.
40.
(3db)l(a3c)
43.
min. (max.)
47.
if,
ad
5TT/3.
a max.
is
(min.)
and -(b+d)l(a
[
c) is
Height=2(3i;/7r)*,radius
= (l/V2)(3i;/7r)^
where
v
is
the
volume of the cone.
50.
27r{l
51.
The
-(V2/3)} radians.
vertices of the rectangle are
/
a
('
V
The greatest value
52.
2a \
/
2a\
2a \
/
/
C^'-VS)' (3-
3>
V3')'
ax being the equation of the parabola.
(3'
y* =
is I
least value
and the
is
(l/e)
Page 260.
.
(i)
(ii)
;c+.y+a=-0, x->> -30.
2
7 2
a t/(x-x )=fr 1 .x 0' -/).
v'^-a 6
/
/
6Vx+a
;
(Hi)
.
(v)
(vi)
(vii)
(viii)
(ix)
a
be, is positive (negative).
3lJC8>^+9a=0
;
8x^31^ + 42^-0.
x .y^O.
x+y=--3a x y+a=Q.
x cos 3 5+^ sin 3 0=c x
13x-~16);=2a
x+^^0;
;
;
sin 3
-
8
-v cos ^
www.EngineeringEBooksPdf.com
+ 2c*
cot 20 --Q.
ANSWERS
Pages 260-279]
397
Page 260.
5.
3
(0
(0, a)
(a, 0).
;
(11) (
(^.T|->
>
/2a, y*a)
;
(V4a,*/2a)
(-1-. *
jc2Qy=7, 20* +7 = 140.
6.
Page 262.
"
'
2
2
Page 264.
1.
(i)
ir/4.
(/)
m h+m(a
2
3,
tan" 1
(ii)
/z0.
b)
two
tion in, m, are the slopes of the
V 16
-
The
roots of this quadratic equaaxes.
Page 265.
a sin 2
3.
0,
a tan
sin 2 0, a sin 2
cos 0, a sin 3
tan
0.
Page 270.
(00/2. () 7r/2-fw0.
1
(i) ir/2. (ii) tan- .?.
1.
4.
(v)
(m) Tr/2-0/2. (/v) w/4-fmfl.
1
2 (iv) tan- [2^/(^
(/) ir/2.
2ir/3.
(v/) 7T/2.
Page 272.
6.
(i)
7.
()
0.
0/02.
/>-<*>/(
a^ 3
2fl
cos s |0 cosec
(i) 2fl
4.
g--^ ^
ft
(//)
1+0M2+20
C")
.
tf
_2
sin 0.
2
).
Page 273.
2
9.
(i)
l/2J
1
(iv)
(v/)
(viii)
r
2
.
= l/r 2 +a 2 /r
2
8
=/J [o m +(l
l/p
4
2
=^l/r +l/a
e 2 -!
2
(//)
4
.
2
r 1 .-=(#s
yj^rsina.
2
-a2 +2ar)/>
(v)
J
2
2
)r
J.
(v/7)
3
.
(ix)
Page 279.
'
O
x
cosh
-
c
CO
'
(
,
<
2.
www.EngineeringEBooksPdf.com
;
-
I)]
DIFFERENTIAL CALCULUS
398
[Pages 279-29(>
Page 279.
4.
2
2
2
2
v'(^ sin 0+6 cos 0).
2
2
cos
3 sin
v/(a cos'
(/)
(ii)
(iv)
v/2tf e
(/)
2acos
5.
sin 2 0).
sin J0.
2tf
(ill)
6
a m lr m
(v) 00.
.
0/2.
(//)
ay (sec
2
<V(l+
(iv)
(vii)
2
f/>
(v)
)-
a(0
l
.
2
+l).
m m~
v/2a
(v//7)
20).
cota
a cosec a e G
(111)
.
tf(02+ l)/(02 -1)2.
(v/)
l
/r
.
Page 288.
Concave downwards
Concave upwards in
2.
3.
and upwards in [ir, 2ir].
concave down2, 0] and [2, oo)
Inflexions at (-2, 198), (0, -20),
in [0,
[
TT]
;
wards in ( -oo, -2] and [0,2].
(2, -238).
Concave upwards in ( oo,
concave down4.
1] and [1, GO]
wards in [ 1, 1]. Inflexions at ( I, 2e) and (1, !(>/<?).
Concave upwards in [0, ir/l] and [5?r/4, 2ir] and concave
5.
-
;
downwards
6.
in [ir/4, 5ir/4J.
For x--^blZa.
(/)
(Hi)
(v)
j-
(v/)
(0, 0), (1, 0),
(ii)
(\V- I)a and <r/(2
For x
and itf \/3.
For x(5, 0),
("')
W
For x
(v/7)
-ttf/v'-.
For
(/.v)
x=
2
(-v/)
-
,
ae* }
(/'/')
are
x-\-y-.
(///)
are
x----
9x
Inflexional tangents to (x/) are
0,
5 and
,
(\ \>
:
"f
.x
O
y'*>
---2j'l--
0.
,'{(9.x+<>)>)
3 v /3v
{-
1
0.
Page 289.
...
For*-
2,
(-4
i-
v 'l8).
(.-..-.-l-T).
f].
(0,0).(///)
(
Psge 292.
Ex.
(i) c
(zv)
<
sec
'
1
tan
(/'/)
i//.
(v)
>{i.
4aoos^.
2fl
see 3 !/-.
Page 296.
at.
2.
(i)
-1/2.
(//)
(HI)
3(a.vj)
.
1.
www.EngineeringEBooksPdf.com
(/v)
^
'
Y
/
(0, 0).
(.v/7)
9.
1
3\
~
(\ae
and
*
(1,3)-
Inflexional tangents to
Inflexional tangents to
()).
\ '3).
3
(v//7)
(1,
(0, 2).
-*(
Inl.
V
/:K
ANSWERS
Pages 297-321]
399
Page 297.
10.
W-la.
2tflb,
14.
[-(log2)/2, l/v/2].
20.
(8,
3).
Page 298.
9
23.
25.
(c -+s*)lc.
Page 302.
(r*+aW-n*r*f
,
*
'
*
r*-r*n*+~2a*/i*
/(+ l)r"-i.
(i/i)
3.
o/2.
(/v)
vfy-'-o").
3
5.
(i)
2p
/o
(")
2
(f'O r*//</>.
.
(m) a
6.
7-
(/)
Page 303
10.
--a.
12.
(3<i/2.
(/)
.U
\
i
(\/-.
(//)
1/3).
Page 309.
9.
11.
(a-\-b)(x*+y
2
)
O//)2 _
^
* 2 -fy 2
(0)
>o2
;
Page 315.
Ex.2.
Page 320.
1. * = 0,
y U/ = 0.
.r
(/)
y-v-
A*^ : ui,y^().
5.
.v=(),
y
7.
y=.Y
-2, y---~.v--3.
9.
y-
10.
x,
3,
y=0, y
[
y = x + ].
y ---x+2,
.Y-fy =
= a.
\-y
Lj-</.
3.
r.v--
x
(//)
13.
K&'-2fl-)=<>.
)'
/7
(Hi)
y
x-:2, jc=3, y=-3.
x=a. y-6.
6.
y^xia, x^=a.
8.
.x
I
i
=0, y -0,
().
y ---(),
.Yfy=-h v /2,
.v
-
i:l,y=
:
1
.
Page 321
13.
y = (>,
15.
x+y=2,
16.
y
yA'= 1.
14.
y==x-f -2, y=2jc
*=!.
4.
17.
3=0, JC + 1=0, y-jc
x=0, y=--0, 2,y--,?x =
19.
x-hya^0.
20.
4(y~x) + l=0, 2(y f Jf) = 3, 4(y + 3;c) +9=0.
|
fl.
J
4.
18.
www.EngineeringEBooksPdf.com
/c.
x-\-ajX.
2.
x=l.
12.
2
4.
y = xV2.
11.
y
;c
;
y =-o
400
DIFFERENTIAL CALCULUS
[Pages 321-334
Page 321.
21.
22.
27.
x=y+
x=0.
Page 325.
1.
y^a, x=0, x+y+a=Q.
3.
3(>>-x) 4-2=0.
5.
6.
X4-.X 4-1=0,
7.
;>4-2
8.
10.
x=l, x=2, x 4-^ + 1=0.
2.
3y+
x=
^=2^ + 1, y=*2x+2.
jp=x t
Page 328.
5.
7.
6.
3y+*=l.
y
8(^-x)+7=0. 2(^-3x)+3 =0.
8.
x3
9.
y=2x8 4-l, y=
10.
x 4-^=0, 2x
106^-381x4-105=0.
3(2^+^)4-5=0,
3
x=0.
6x 2 .y4-llx7 3
6j
2*
1,
x+2j=0.
x^=2j,
1=0, 2x->3y4-3-=0, 4x- 6^ 4-9=0.
3^
Page 331.
Ex.
The curve lies above or below the asympx4-y
x is positive or negative.
The curve lies above the asymptote both for
x-f-j; 4-0=0.
2.
(f)
.
tote according as
(//')
positive as well as negative values of x.
(Hi)
The curve
y=zx\-\.
accordiug as x
Page 334.
above or below the asymptote
lies
positive or negative.
is
+ ^2r sin
1.
tf=rsin
2.
a
3.
rcos6b=a.
4.
r cos
5.
2ram0=a,2e=ir.
6.
0=7r/4.
7.
r sin
o
rsmf/ 6
5.
00,
.
/>
\
9.
10.
12.
15.
(1
0).
1
r sin
8a^=0.
cos 0)4-2a=0.
7r(r
-- )=
W TT
>v
n
J
m
7i0=/W7r where
a
n cos
is
rcos0=0.
.
AWTT
any
a=rsin(0-.l).
System of
(04-i7r)=0.
0a=;Q
parallel lines
where mis any
integer.
integer.
11.
0=0.
13.
y+a=0.
y~ae
W7r
where n
14.
is
any integer or
zero.
16.
^4-11=0.
17.
-^T\V
19,
a=2r
f
(ft\
==rr
(cos
sin
(
ei
6 )>
where
sin 0), a4-2r(oos
6 l is
any root of the equa-
0+sin 0)=0.
www.EngineeringEBooksPdf.com
ANSWERS
Pages 338-347
401
Page 338.
I.
4.
x=0, >=0.
>>==*.
2.
5.
Page 339.
1.
*=a.
(y-l) =
2(*-2a) =
;
3.
x=0.
y=x.
2.
V3>'-v'2(*-*)
3.
bx~ay.
(x-2).
V3(y-*).
Page 341.
1.
Cusp at
(0, 0).
3.
Node
at
(0, 0).
Node
Node
at (a, 0).
at (2, 0).
2.
Cusp at
(0, 0).
Page 342.
4.
5.
(a, 0) is a node, cusp or an isolated point according as
than, equal to or greater than 1.
7.
Conjugate point at (a, 0).
6.
b\a
is less
Conjugate point at (2, 3).
9.
Conjugate point at (a,
10.
Cusp at (0, -4a).
llr Cusp at ( 1, -2).
12.
Cusp at (1, -1).
8.
b).
N
13.
No
14-
(0,
multiple point.
0) are triple point
(fa,
:
0) are
cusps
:
node at
(-a,0).
15.
(0,
a) is
point according as
a double point
;
being a node, cusp or conjugate
b>a, b=a or b<a.
-a
16.
17.
18.
Page 345.
Single cusp of
Single cusp of
1.
2.
first species.
first species.
Single cusp of first species.
Single cusp of first species.
Isolated point.
3.
4.
5.
7.
Double cusp of second species.
Oscu- inflexion.
8.
Oscu-inflexion.
6.
Page 347.
2.
2^2.
5.
-VV17,
3.
2v/2 for each.
V2.
7.
www.EngineeringEBooksPdf.com
4.
6.
3fl/2 for each.
a,
-*/2,
-5*
402
DIFFERENTIAL CALCULUS
^
C/O*'
www.EngineeringEBooksPdf.com
[
Page 367
Page 367
]
AXSWEAS
85'
www.EngineeringEBooksPdf.com
403
404
DlFfffiBENTIAL CALOTTLtfS
www.EngineeringEBooksPdf.com
[
Pages
Page 368
405
]
www.EngineeringEBooksPdf.com
DIFFERENTIAL CALCULUS
406
[PagQS 377-382
Page 377.
b*=l.
(iii)
,.
x=a
c*(
(ii)
0+aO sin 0, y=a sin Q-~a
/v
2/(2-w)
/v
2/(2-)
2/(2-n)
/v
'
v
cos
cos
'
,
0.
'
(iv)
(v)
(pX2
(VI)
2.
J
2
=C
2
.
x^+^=ci
(i)
(ii)
x^+^ = ci
(ii)
xyc=0.
(iff)
Page 378.
3.
2xy=c*.
(i)
5.
6.
(jc^j+j^j;)
9.
x
12.
(
13.
The
straight line
= 2a.
7.
p*xpy+(a+ap*+pq)=().
cos
,...,
.
r sin
(///)
16.
(i)
Kae(a
r 2 (e 2
1)
-7T/2)
2/ er cos
cot a
Gr=
*)
cot a
CO
fl
e
2
0+2/ =0.
Page 379.
^
+
17
+ p)
r
Miscellaneous Exercises
II.
Page 379.
2a see 3
2.
4.
(0/2).
4:4/2
a.
;
Page 380.
2
14.
(9/
17.
(Jfl.
,
-6/).
" UlS
1Q
a).
/2256 5
5
448i* \
'
3375V
'
A
(b '
}
*
Page 381.
^
28.
>'=xi*.
1) is an isolated point
(1,
(5, 3) and (5,
5) ate the
two points of inflexion.
x~a tanh a, .y^sech a is the required curve.
36.
382.
Page
x sin \t-\-y cos fat ---a sin /, x cos \ty sin ^/..-3fl cos ?/.
40.
32.
47.
;
(/)
Isolated point.
(//)
Single cusp of second species.
www.EngineeringEBooksPdf.com
INDEX
(Numbers refer
to
pages)
Discontinuity, 38
Double points, 336
Acceleration, 79
Angle of intersection,
of two curves, 262, 268
Approximate calculation, 207
Arcs, 274
, Derivative of, 274
Astroid,212,257
Asymptotes, 313
by expansion, 328
by inspection, 324
parallel to axes, 315
Cardioide, 249
Catenary, 238, 259
Cauchy's Theorem, 137
Cissoid, 244
Fnyelopes, 369
Epicycloid, 240
Equations,
,
,
Concavity, 281
Conjugate point, 336
Parametric, 255, 276
Pedal, 266, 272
Polar, 248, 277, 298
, Tangential polar, 300
Equiangular spiral, 252, 269
Euler's Theorem, 199
Evolutes, 305,310, 372
Expansions, 179
,
,
.
Folium of Descartes, 246, 362
Functions, 11
Algebraic, 35
,
Convexity, 281
Continuity of elementary functions, 54
Continuous functions, 36, 194
properties of, 54
Curvature, 290
Centre of, 303
Chord of, 305
Circle of, 305
,
Composite, 209
Continuous, 37
Exponential, 23, 97
Homogeneous, 199
Hyperbolic, 67, 97
Implicit, 209, 292
,
,
,
,
,
,
,
,
Radius
254
Extreme values, 148
Closed interval, 1*3
Composite functions, 209
,
Implicit, 243,
Intrinsic, 291
of, 291
,
Inverse, 21, 88
-- Inverse Hyperbolic, 70, 99
Inverse Trigonometric, 30, 93
Logarithmic, 25, 96, 100
20
, Monotonic,
of a function, 34, 86
of two variables, 193
Transcendental, 35
Trigonometric, 26, 89
,
,
,
Curve Tracing, 348
,
Cusps, 336
Cycloid, 240, 257
,
,
D
,
,
Derivative, 72
,
,
,
of arcs, 274
Partial, 196
Sign of, 136, 150
Geometric Interpretation,
Differential Coefficient, 72
Differentiation, 72
, Logarithmic, 100
of function of a function, 86
of implicit functions, 209
, of inverse functions, 88
Partial, 196
Repeated, 217
,
77, 131, 197
Graphical representation, 14
Greatest values, 149
Determinant, 107
Differentials, 206
H
Homogeneous
functions, 199
Hypocycloid, 240
,
,
.
,
Successive, 113
292
Implicit Functions, 209,
Indeterminate forms, 165, 185
www.EngineeringEBooksPdf.com
FFEBENTIAL OALOITLUA
408
Infinite limits, 48
179
, series,
Inflexion, 281
Points, Stationary, 150, 223
Polar coordinates, 267, 277, 331, 359
Interval, Closed, 13, 55
, Open, 13
Inversion of operation, 53
Radius of curvature, 291, 345
ve.ctor,
,
Lagrange's Theorem, 133
multipliers. 230
Least values, 149
267
Remainder, 142
Repeated derivatives, 217
Rolle's Theorem, 130
,
Leibnitz's Theorem, 121
Lejnniscate, 249
Limits, 41, 195
Indeterminate, 165
,
Right handed, 42
Left handed, 42
,
,
M
Maximum value, 148,
Minimum value, 148,
Mean value theorem,
223
223
133, 137
Modulus, 9
Monotonic functions, 20
Multipliers, 230
N
Newtonian Method, 293
Node, 336
Normal, 259
Numbers, Irrational, 5
Rational, 2
Real, 6
,
,
Series, Binomial, 183
, Taylor's, 140, 220
Maclaurin's 142
Sign of derivative, 136, 150
Singular points, 335
Smallest values, 149
,
Spirals, 251, 252
Stationary values, 150,223
Strophpid, 245
Subsidiary conditions, 229
Sub-tangent, 264
Sub-normal, 264
Successive differentiation,
,
,
,
,
,
interval, 13
,
Euler's, 199
Lagrange's, 133
Leibnitz's 121
Maclaurin's 142
Mean value, 133, 137
Rolle's, 130
Taylor's, 140, 220
on
limits, 52
Total differentials, 200
,
Parametric equations, 255, 276, 293, 364
Partial differentiation, 196
Pedal equations, 266, 272
Values, Greatest, 149
Maximum, 148, 223
Points, Conjugate, 336
Minimum, 148, 223
Cusp, 336
149
Double, 336
, Smallest,
Stationary, 150, 223
Node, 336
, Multiple, 335
Variables, 11
281
, of inflexion,
Velocity, 78
,
,
,
,
,
,
,
13
Tangent, 254
Tangential Polar equations, 300
Taylor's Theorem, 140, 220
Theorem, Cauchy's, 140
,
Open
1
Singular, 335
www.EngineeringEBooksPdf.com
www.EngineeringEBooksPdf.com
www.EngineeringEBooksPdf.com
www.EngineeringEBooksPdf.com