131 98 112 MB
Urdu Pages [400] Year 2019
Table of contents :
2019-MATH12E-30102019-904AM-AB-00001
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12
"Education is a matter of life and death for Pakistan. The world is progressing so rapidly that without requisite advance in education, not only shall we be left behind others but may be wiped out altogether." (September 26, 1947, Karachi)
Quaid-e-Azam Muhammad Ali Jinnah.-"'"' Founder of Pakistan
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~~~~~~~~~~~~---
6~;Jl~_;JI~\~ ,,, ,,, ,,, ,,, CALCULUS AND ANALYTIC GEOMETRY
MATHEMATICS
12
l I
PUNJAB CURRICULUM AND TEXTBOOK BOARD, LAHORE
All rights are reserved with Punjab Curriculum & Textbook Board, Lahore Prepared by: Punjab Curriculum & Textbook Board, Lahore Approved by: Federal Ministry of Education Curriculum Wing, Islamabad. Vide Letter No. F. 1-6 / 2003 dated June 06,2003.
CONTENTS 1 2
3 4 5 6 7
1:l
Page
Description
Unit
Functions and Limits Differentiation Integration Introduction to Analytic Geometry Linear Inequalities and Linear programming Conic section Vectors Answers
Authors: • Prof. Muhammad Amin Chaudhary •
• Prof. Muhammad Khalid ·Saleem Editor:
e
1 41 119 179 229 249 325 367
Prof. Muhammad Sharif Ghaury Mazhar Hayat
• Mazhar Hayat
Supervision: •
Mazhar Hayat
Senior Artist: Aisha Waheed Director (Manuscripts):
Mrs. Nisar Qamar
Published by : Altaf Raheem Printer Lahore Printed by : Ahmad Tayyeb Printer Lahore. Date of Printing Edition Impression July 2019 1st 28
Copies 4400
Price 166.00
Unit 1: Functions and Limits
1
I Unit 1 I Functions and Limits 1.1
INTRODUCTION
Functions are important tools by whlch we describe the real world in mathematical terms. They are used to explain the relationship between variable quantities and hence play a central role in the study of calculus.
1.1.1 Concept of Function The term function was recognized by a German Mathematician Leibniz (1646-1716) to describe the dependence of one quantity on another. The following examples illustrates how this term is used: (i) The area "A" of a square depends on one of its sides "x" by the (ii)
formula A = x 2 , so we say that A is a function of x. The volume "V" of a sphere depends on its radius " r" by the formula
V
= _±_7r r 3 , so we say that Vis a function of r. 3
A function is a rule or correspondence, relating to two sets in such a way that each element in the first set corresponds to one and only one element in the second set. Thus in, (i) above, a square of a given side bas only one area. And in, (ii) above, a sphere of a given radius has only one volume. Now we have a formal definition:
1.1.2 Definition (Function-Domain-·Range) A Function/from a set X to a set Y is a rule or a correspondence that assigns to each element x in X a unique element y in Y. The set X is called the domain off The set of corresponding elements yin Y is called the range off · Unless" stated to the contrary, we shall assume hereafter that the sets X and Y consist of real numbers.
1.1.3 Notation and Value of a Function If a variable y depends on a variable x in such a way that each value_of x determines exactly one value of y, then we say that ''y is a function of x". Swiss mathematician Euler (1707-1783) invented a symbolic way to write the statement "y is a function of x" as y = f (x), which is read as "y is equal to f of x".
Calculus and Analytic Geometry
2
Note: Functions are often denoted by the letters such asf, g, h, F, G, Hand so on. A function can be considered as a computing machine f that takes an input x, operates on it in some way, and produces exactly one output f (x). This output
Function
f
Inputx
Outputj{x)
Computing Machine
f(x) is called the value of/atx or image of x underf. The outputf(x) is denoted by a
single letter, say y, and we write y = f (x). The variable x is called the independent variable off, and t 1" , ..triable y is called the dependent variable off. For now onward we shall on Iv con sider the function in which the variables are real numbers and we say that f is a real valued function of real numbers. Example 1: Givenf(x) = x 3 (i) f(O)
-
2x 2 + 4x -1 , find
(ii) f(l)
(iii) f(-2) (iv) f(l + x) Solution: f(x) = x -2x + 4x-1 CD ! (O) = o- o+ o-1 = -1 (i) /(1) = (1) 3 -2(1) 2 +4(1)-1=1-2+4-1=2 3
(v) f(l/ x), x
*0
2
(ii) /(-2)= (-2) 3 -2(-2) 2 + 4(-2)-1 = -8-8-8-1 = -25 (iii) /(1 + x) _= (1 + x) 3 -2(1+x) 2 +4(1 + x)-1 2
= 1+3x + 3x + x 3 - 2 - 4x - 2x 2 + 4 + 4x -1 = x 3 +x 2 +3x+2 3 4 (iv) J(Ilx)=(11x) -2(11x)2+4(11x)-l=-;---;-+ -1, x;t:O x x x Example 2: Let f(x) = x 2 • Find the domain and range off.
Solution:
f(x) is defined for every real number x.
Further for every real number x, f(x) = x 2 is a non-negative real number. So Domain/= Set of all real numbers. Range f = Set of all non-negative real numbers. Example 3:
Let f(x) =
Solution:
At x
x
x 2 -4
. Find the domain and range off
=2 and x =-2,
f (x) =
x is not defined. So x 2 -4 Domain!= Set of all real numbers except -2 and 2 Range f = Set of all real numbers.
Unit 1: Functions and Limits
3
= ~x 2 -9. Find the domain and range off
Example 4:
Let f(x)
Solution:
We see that if xis in the interval -3 < x < 3, a square root of a negative
number is obtained. Hence no real number y
=~x 2 -
9 exists. So
Domainf = {xE R: lxl ~ 3 }=(-00 ,-3]u [3,+oo)} Rangef = set of all positive real numbers= [0,+oo)
1.1.4
Graphs of Algebraic Functions
If f is a real-valued function of real numbers, then the graph off in the xy-plane is defined to be the graph of the equation y = f (x) . The graph of a function! is the set of points {(x, y)I y = f(x), xis in the domain off} in the Cartesian plane for which (x, y) is an ordered pair off The graph provides a visual technique for determining whether the set of points represents a function or not. If a vertical line intersects a graph at more than one point, it is not the graph of a function. Explanation is given in the figure. y
(a) a function
y
(b) a function
(c) not a function
y
(d) not a function
Method to draw the graph: To draw the graph of y = f(x), we give arbitrary values of our choice to x and find the corresponding values of y. In this way we get ordered pairs (xp y 1 ), (x 2 , y 2 ), (x3 , y 3 ) etc. These ordered pairs represent points of the graph in the Cartesian plane. We locate these points and join them together to get the graph of the function. Example 5: Find the doniain and range of the function f(x) = x 2 + 1 and draw its graph. 2
Solution: Here y = f(x) = x +1
Calculus and Analytic Geometry
4
We see that f(x) = x 2 +1 is defined for every real number. Further, for every real number x, y = f(x) = x 2 +1 is a non-negative real number. Hence Domain f and
Range
= set of all real numbers
f =set of all non-negative real numbers except the points 0 ~ y < 1.
For graph of f (x) = x 2 + 1, we assign some values to x from its domain and find the corresponding values in the range fas shown in the table: y
I y= ~(x) I
2
~~ 1- ~ I i
I
~ I ~ I ~ I :o
I
Plotting the points (x,y) and joining them with a smooth curve, we get the graph of the function f(x) = x 2 +I, which is shown in the figure.
1.1.5
_ .............-+-_.__...__.... x
.:.2
2
Graph of Functions Defined Piece-Wise.
When the functionfis defined by two rules, we draw the graphs of two functions as explained in the following example: Example 6:
Find the domain and range of the function defined by: when 0 ~ x ~ 1
x
f(x) = [ x -1
Solution: Here domain following two functions: (i) f(x)=x when
when 1 < x
f
~
Also draw its graph.
2
= [O, 1] u (1, 2] = [O, 2]. This function is composed of the
O~x::;;1
Plotting the points (x, y) and joining them we get two straight lines as shown in the figure. This is the graph of the given function.
(ii) f(x)=x-1,when l 0, a t:. 1
is called Logarithmic Function of x.
Unit 1: Functions and Limits
7
(i)
If a _= 10, then we have log 10 x (written as lg x) which is known aS the common logarithm of x.
(ii)
If a= e, then we have log e x (written as In x) which is known as the natural logarithm of x.
1.2.6 Hyperbolic Functions (i)
sinh x =_!_(e x - e-x) is called hyperbolic sine function. Its domain and
(ii)
cash x =_!_(ex + e-x ) is called hyperbolic cosine function. Its domain is
2 range are the set of all real numbers. 2
the set of all real numbers and the range is the set of all numbers in the interval [1, + oo). (iii)"
The remaining four hyperbolic functions are defined in terms of the hyperbolic sine and the hyperbolic cosine function as follows:
tan h x=
sinhx ex -e-x =--cosh x ex + e-x
sechx=
coshx ex +e-x coth x= =--. sinhx e x -e-x
1 2 =--cosh x ex + e-x 1
2
cschx=--=--sinhx ex - e-x
The hyperbolic functions have same properties that -resemble to those of trigonometric functions.
1.2.7 Inverse Hyperbolic Functions The inverse hyperbolic functions are expressed in terms of natural logarithms and we shall study them in higher classes. (i)
sinh- 1 x = ln(x+Jx 2 +l), for allx (iv) coth-' x
di)
2 cosh-1 x = ln(x+Jx -1), x;:::l
(iii)
x , x +=
x->+= xl/5
1.5.4 Method for Evaluating-the Limits at Infinity In this case we first divide each term of both the numerator and the denominator by the highest power of x that appears in the denominator and then use the above theorem.
Example 2:
5x 4 -10x 2 +1 Evaluate Lim - -3 - -2- x->+= - 3x + 10x + 50
Solution:
Dividing up and down by x 3 , we get 4
. L im x->+= -
Example3:
2
5x -10x +1 3x 3 + 10x 2 + 50
. L im x+=
Example 4:
oo-0+0 -3+0+0
=oo
Since x < 0, so dividing up and down by (-x) 5 = -x 5 , we get 4
3
4x -5x 3x 5 + 2x 2 + 1
2
= = L"im - -4/x+5/x - - -3- - 5 H-- -
3 - 2 I x -1 I x
O+O =O -3-0-0
Evaluate (i)
. L im H-=
Solution:
=
4x 4 -5x 3 3x 5 + 2x 2 + 1
. Evaluate Lim H-=
Solution:
3 L" 5x-IO!x+llx = im x->+~ - 3+10 Ix+ 50 I x 3
(i)
2-3x
--=== 2
(ii)
~3+4x
Here
2-3x . L im --::::=== H+=
[;i = \x\ = -x
as
~3+4x 2
x< 0.
:. Dividing up and down by -x, we get
. Lim .H--
(ii)
2-3x
~3+4x 2 Here
. -2/ x-3 0+3 3 =Lim = = 2 x+~ ~3/ x +4 .Jo+4 2
[;i = \x\ = x
as x > 0.
:. Dividing up and down by x, we get
. Lim
2-3x -
x->+~ ~3
+ 4x 2
2/x-3 = L"im -r==== 2 x->+-
~3/ x + 4
0-3 .Jo+4
-3 -2
Unit 1: Functions and Limits
23
ll)'f =e.
1.5.5 Lim (1+ !_ x-H""
By the Binomial theorem, we have
(1 J" ~ 1 +:
2 It)' +
+ n ( : } n(n ~
n(n -
l~~n
-2)(~ J
+ ...
= 1+1 +_!_(1-_!_)+ _!_(1-_!_)(1-3-)+ ... 2!
when
~ oo ,
n
n 3! n n 1 2 3 - , - , - , . . . all tend to zero. n n n
. ( 1+l)n= 1+1+-+-+-+-+ 1 1 1 1 ... Lim x~~ n 2! 3! 4! 5! = 1+1+0.5 + 0.166667 + 0.0416667 + ... As approximate value of e is=:= 2.718281. :.
: . Lim 1+ -1 ( x ~n
Deduction
J" = e .
Lim (1 + x )1/x =e x~O
we know that Lim (1 + __!_) n =e
n
x~~
Put n = _!
x'
then
when x As
ln = x in (i)
~
0, n
. Lim (1 + _! )n
n
n ~~
(i)
~ oo
=e
I
Lim (I+ x) 7 = e x ~o
1.5.6
Lim
ax -1
x~O
X
=loge a
Put a x -I= y then a "'
=1 + y
(i)
= 2.718281...
Calculus and Analytic Geometry
24 From (i) when x
. a x -l . . L1m - x->O
x
-7
0, y
-7
0
.
y
y->0
log a (1 + y)
= Lim - - - -
1
=Lim----y->o ; log a (1 + y)
. 1 1 = Lim - - - - , - = = loge a y->O loga (1 + y)lfy log a e
Deduction Lim x->0
(
y-?0
(ex - 1 J= loge e= 1. X
J
a "" -1 =log e a We know that Lim x->0
·: Lim(l + y);
(
(1)
X
Put a= e in (1), we have e x -1 Lim - - = loge e = 1 x->0
X
Important Results to Remember (i)
Lim (e x)= oo, x->+
(iii)
(ii)
Lim (e x )= Lim (
x --+-oo
x--+-oo
:x)= 0,
e
Lim (E:_)= 0, where a is any real number.
x->±oo
Example 5: (a)
X
Express each limit in terms of the number 'e'
3 )2n Lim 1+x->+ ( n
.(b)
l
Lim (l+2h)h h->0
Solution:
(a)
Observe the resemblance of the limit with Lim (1 + _!_ x->~
n
Y =e J
= eJ
Unit 1: Functions and Limits
25
(1 + ~)n n =Lim (1 + _]_) m 2
:. Lim x -++-oo
x--Hoo [
put m = n/3 when n _, oo,
6
111
=e
]
6
[
m l
Observe the resemblance of the limit with Lim (1 + x):r
(b)
J
_,co = e,
X---l-0
2
:. Lim (1+2h)"hl h-->0
= Lim [ (1+2h) 2Ih ]
(put m = 2h, when h --7 0, m
--7
0)
h-->0
2
= Lim [c1 + m)f.i ] = e 2 m-+0 1.5. 7 The Sandwitch Theorem Letf, g and h be functions such that f (x) ~ g(x) ~ h(x) for all numbers x in some open interval containing "c", except possibly at c itself. If
Lim f(x) x-+c
=L
and
Lim h(x) x-+c
= L, then
Lim g(x) x-+c
=L
Many limit problems arise that cannot be directly evaluated by algebraic techniques. They require geometric arguments, We evaluate an important theorem.
1.5.8 If() is Measured in Radian, then Lim sin()= 1 e--.o
.()
e
roof: To evaluate this limit, we apply a new technique. Take a positive acute .:entral angle of a circle with radius r =l. As shown in the figure, OAB represents a sector of the circle.
Given
In rt ·
joA/ = /oBj = 1
(radii of unit circle) D
~OCB' sine= ',OB Be,' =JBCJ
(·:I OB I= 1)
= IADI
(-: IOA I=I)
In rt AOAD 'tan e
J,:,'
In terms of (),the areas are expressed as: Produce OB to D so thatAD 1- OA. Draw BC 1- OA. Join AB.
O
r= 1
Calculus and Analytic Geometry
26 (i)
~OAB = _!_ IOAllBCI = _!_ (l)(sin ()) = _!_sin()
Area of
2
2
2
1 () ) = -1() Area of sector OAB =-1 r 2 () = -(1)( (·: r = 1) 2 2 2 1 . 1 1 (iii) Area of ~OAD = = -(l)(tan e) = - tan e 2 2 2 From the figure we see that Area of ~OAB
1
()
1
- sin () < - < - tan () 2 2 2
.
As sin 8 is positive, so on division by ()
i.e. ,
0,
sin() cos()0
elfx -1 I/
Lim (1+3xY
I
2
(x)
2
(vi)
x-+0
e ·' +1
CONTINUOUS AND DISCONTINUOUS FUNCTIONS
1.6.1 One-Sided Limits
f (x), we restricted x to an open interval containing c i.e., we studied the behavior of f on both sides of c. However, in some cases it is necessary to In defining Lim x --+c
investigate one-sided limits i.e. , the left hand limit and the right hand limit.
(i)
The Left Hand Limit Li"!
f
(x) =Li read as the limit of
f
(x) is equal to Las x approaches c from
the left i.e., for all x sufficiently close to c, but less than c, the value of f(x) can be made as close as we please to L. (ii)
The Right Hand Limit Lim f(x) = M is read as the limit of f(x) is equal to M as x approaches c \-+ c"'"
from the right i.e., for all x sufficiently close to c, but greater than c, the value of f(x) can be made as close as we please to M. Note: The rules for calculating the left-hand and the right-hand limits are the same as we studied to calculate limits in the preceding section.
1.6.2 Criterion for Existence of Limit of a Function Lim f (x) = L
if and only if Lim _ f(x) = Lim x~c
.• -+c
Example 1:
Determine whether Lim
f
(x) and Lim
.• -+2
2x+1 if
x~c
Osxs2 f(x) = 7 -xx if 2sxs4 { if 4sxs6
x -+i
f
+
f(x) = L.
(x) exist, when
29
Unit 1: Functions and Limits
Solution: (i)Lim f(x)=Lim (2x+1)=4+ 1 =5
(ii)
x ->2
x->2-
Lim f ( x) = Lim (7 - x) = 7 - 2 = 5
x->2+
x->2
Since Lim f(x) =Lim f(x) = 5 x --+2-
:=:}
Lim f(x)= Lim (7-x)=7-4=3
x ->4-
x ->4
Lim f ( x) = Lim ( x) = 4
x ->4+
x->4
Since Lim f(x)
x --+2+
x -+4-
Lim f (x) exists and is equal to 5.
-:f.
Lim f(x) x -+4 +
Therefore Lim f (x) does not exist. x->4
x ->2
We have seen that sometimes Lim f(x) = f(c) and sometimes it does not and x-+c
also sometimes f (c) is not even defined whereas Lim f (x) exists. x ->c
1.6.3 Continuity of a Function at a Number (a) Continuous Function A function f is said to be continuous at a number "c" if and only if the following three conditions are satisfied: (i) f(c) is defined. (ii) Lim f(x) exists. (iii) Lim f(x) = f (c) x-+c
x-+c
(b) Discontinuous Function If one or more of these three conditions fail to hold at "c", then the functionf is said to be discontinuous at "c". Example 2:
x 2 -1 Consider the function f(x) = - x-1
Solution:
Here f (1) is not defined
:=:}
f(x) is discontinuous at 1. 2
Further Lim f(x) =Lim x -l =Lim ( x + 1 ) = 2 (finite) x-+l x->1 X -1 x->1 Thereforef(x) is continuous at any other number x t:-1
Example 3:
For f(x) = 3x 2 -5x+4, discuss continuity offatx = 1
Solution:
Limf(x) =Lim (3x -5x+4)=3-5+4=2 ..
2
x->l
x ->1
and f(1)=3-5+4=2 :=:}
Lim f(x) x -+1
=f(l)
f (x) is ~ontinuous at
x =1
Calculus and Analytic Geometry
30 Example 4:
Discuss the continuity of the function
f (x) and
2
x - 9 if x :;t: 3 (a) f(x)= x-3 { 6 if x=3
x2 (b)
Solution: (a) Given f(3) = 6 : . the function f is defined at x = 3. 2 9 Now Lim f(x) =Lim x x -->3 x-->3 X - 3 =Lim (x+3)(x-3) x -->3 x-3 =Lim (x+3) = 6
g(x) at x -
= 3.
9 if xt:3.
g(x)=
~~~-x_-_3~~~~~~~~ y
(4, 7) (3, 6) (2. 5)
(0, 3)
x -->3
Lim f(x)
As
x -->3
:. f
=6 = f
(x) is continuous at x
(3) 3
0
=3
Fig (i)
It is noted that there is no break in the graph. (See figure (i)) ~~~~~~~~~~~~~~
y
2
(b)
x -9 g(x) = - x-3
if x t: 3
As g(x) is not defined at x ~
=3
g(x) is discontinuous at x = 3
(see figure (ii))
I I I I
I .I
It is noted that there is a break in the
I
graph at x
=3
0
3
Fig (ii)
Example 5: f(x)=
Discuss continuity off at 3, when 1 ifx 0 ~ x > 0
1J r
This means lg x exists only when x > 0
I
~
Domain of the lg x is +ve real numbers.
6
4 -I
-2
Note: lg x is undefined at x
Graph of
I
-3
=0 .
;
--4
y = /(x) = lgx
8
r
I
I
x
~o
rI
I.
For graph of/(x) =lg x, we find the values of lg x from the common logarithmic table for various values of x > 0 Table of some of the corresponding values of x and f(x) is as under: x
-70
0.1
1
2
4
6
8
10
-7 +oo
y=f(x)=lgx
-7-oo
-1
0
0.30
0.60
0.77
0.90
1
-7 +oo
Plotting the points (x, y) and joining them with a smooth curve we get the graph as shown in the figure.
Calculus and Analytic Geometry
34
Graphs of Natural Logarithmic Function f
1.7.4
(x) = In x:
y
The graph of
f(x)
= lnx
properties as that of the graph of
A
has similar
f (x) = lg x.
. .
-- -
~
e--~
I
By using the table of natural logarithm for various values of x, we can get the graph of
y =In x as shown in the figure.
/
0
x
.I .,_
lJ1'1.ph4
-~
;r- -
}\x)= •/.D...K.
)
I I
~
1.7.5 Graphs of Implicit Functions (a) Graph of the circle of the form x 2 + y 2 = a 2 Example 1: Graph the circle x 2 + y 2 = 4 (1) Solution: The graph of the equation x 2 + y 2 = 4 is a circle of radius 2, centered at the origin and hence there are vertical lines that cut the graph more than once. This can also been seen algebraically by solving (1) for yin terms of x.
y
=±-J4-x
2
The equation does not define y as a function of x. For example, if x = I, then y =±J3. Hence, (l,J3) and (l,-J3) are two points on the circle and vertical line passes through these two points. We can regard the circle as the union of two semi-circles.
y = .J4- x2 and y = --J 4- x2 Each of which defines y as a function of x. y
y
__.---,"c_o........ 2)
(--./3, I}
(I,
y
.../3 l
(-1,
(-./3, I)
x'--~---+----+---+X
0
y'
'13)
0, 2) (!,
(-.../3 ,I) x'-----t----- x (2, 0) t-2,0) 0
(-.../3 ,-1) (I,
(-./3, -I} --./3)
x'
(-2,0)
0
(2, 0)
(-./3' -I}
(--./3. -1)
(0,-2)
y'
Graph ofy = ..../4-x 2
Graph ofy = -..../4-x 2
'13) (-./3 ,I)
y'
Graph of x 2 + y 2 = 4
x
Unit 1: Functions and Limits
35
We observe that if we replace (x, y) in tum by (-x, y), (x,-y) and (-x,-y), there is no change in the given equation. Hence, the graph is symmetric with respect to the y-axis, x-axis and the origin.
=4
=> y
= ±2
x = 1 implies y = 3
=> y
= ± .J3
x
=0
implies y
2
2
x
=2
implies y 2
= 0 =>
y
=0
By assigning values of x, we find the values of y. So we prepare a table for some values of x and y satisfying equation (1).
Plotting the points (x, y)and connecting them with a smooth curve as shown in the figure, we get the graph of a circle.
(b)
x2
The graph of ellipse of the form -
a
2
y
y2 +- 2 = 1
(0, 3)
b
x2 y2 Example 2: Graph - 2 + - 2 = 1 i.e., 9x 2 + 4 y 2 = 36
2
3
Solution: We observe that if we replace (x, y) in tum
x'----------x
2• 0
(2,0)
by (-x, y), (x,-y) and (-x,-y), there is no change in the given equation. Hence, the graph is symmetric with respect to the y-axis, x-axis and the origin. y
x
=0 =0
implies x implies y
2
2
=4 =9
=> x => y
=±2
=± 3
(0,-3)
y' x2
y2
22
32
Graphof-+-=1
Therefore, x-intercepts are 2 and -2 and y-intercepts are 3 and -3.
Calculus and Analytic Geometry
36
By assigning values of x, we find the values of y. So, we prepare a table for some values of x and y satisfying equation (1) .
x
0
1
2
-1
-2
y
±3
±ff
0
±ff
0
Plotting the points (x, y), connecting these points with a smooth curve as shown in the figure, we get the graph of an ellipse.
1.7.6 Graph of Parametric Equations (a)
Graph the curve that has the parametric equations x = t2
-2 ~ t ~ 2
Example 3:
Graph
Solution:
For the choice of t in [-2, 2], we
,
y =t
,
(3) y
prepare a table for some values of x and y satisfying the given equation.
t
-2
-1
0
I
2
x
4
1
0
1
4 Graph of x = t 2, y = t
y
-2
-1
0
1
2
We plot the points (x, y), connecting these points with a smooth curve shown in figure, we obtain the graph of a parabola with equation
y2
= x.
Unit 1: Functions an.d Limits
37
1. 7. 7 Graphs of Discontinuous Functions
=
{:-I
Example 4:
Graph the function defined by y
Solution:
The domain of the function is 0 $ x $ 2
when 0 $ x $ 1 when 1 < x $ 2
For 0 $ x $ 1 , the graph of the function is that of y and for 1 < x $ 2 , the graph of the function is that of y
=x
= x -1
We prepare the table for some values of x and y in 0 $ x $ 2 satisfying the equations y =x and y =x-1
x
0
0.5
0.8
y
0
0.5
0.8
1
1.5
1.8
2
0.5
0.8
1
y
.9 .8
Graph of
.7
f(x) = x
.6
Graph of
.5
f(x) =x-1
.4
.3 .2
.I
0
.I
.2
.3
.4
.5
.6
.7
.8
.9
I
I.I 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9
2
Plot the points (x, y). Connecting these points we get two straight lines, which is the graph of a discontinuous function. Example 5:
x 2 -9 Graph the function defined by y = , x-3
Solution:
The domain of the function consists of all real numbers except 3.
When x
x:;t3
=3, both the numerator and denominator are zero, and __Q_ is undefined. 0
Calculus and Analytic Geometry
38 x 2 -9
. = x + 3 provided x :;; 3 . We prepare a x-3 x-3 table for different values of x and y satisfying the equation y = x + 3 and x * 3 .
Simplifying we get y
=- - =
(x-3)(x+3)
4 7 y
(4, 7) (3, 6)
Plot the points (x, y) and joining these points we get the graph of the function which is a straight line except the point (3, 6), The graph is shown in the figure. This is a broken straight line with a break at the point (3, 6).
(2, 5)
I
I I I
I I I I I I I I I
(0, 3)
0
3
1.7.8 Graphical Solution of the Equations Example 6:
Graph
(i)
cos x
=x ·
We solve the equation cos x exercise for the students. Solution:
=x
(ii)
sinx = x
(iii)
and leave the other two equations as an
To find the solution of the equation cos x = x, we draw the graphs of the two functions y = x and y
=cos x
; -
tan x = x
7r $;
x :-:;; 7r
Scale for graphs
Along x-axis, length of side of small square = ~radian 6 Along y-axis, length of side of small square = 0.1 unit Two points (0, 0) and ( ~ ,1) lie on the line y = x
Unit l: Functions and Limits
39
We prepare a table for some values of x and y in the interval the equation y = cos x . x
-n
-sy; -2113' -ri
y=cosx
-1
-.87
-.5
0
-~ - .5
-%
0
.87
1
7r ~
x
~
n satisfying
% ;{ ii 2;{ s3 .87
.5
0
-.5
-.87
7r
-1
y
x' --+---- f'(x) =
1 dx dy
( and g(y)
=x
=> g'(y)
=:
DERIVATIVE OF A FUNCTION GIVEN IN· THE FORM OF PARAMETRIC EQUATIONS
The equations x =at and y = 2at express x and y as function oft. Here, the variable t is called a parameter and the equations of x and y in terms oft are called the parametric equations. Now, we explain the method of finding derivatives of functions given in. the form of parametric equations by the following examples: 2
Example 1: Find
2
if x
=at2 and y = 2at.
2
Solution:
we use the chain rule to find dy d Here, dt = dt (2at) = 2a .1 = 2a
d
and
dx dt
so,
dy dy dy dt dt dx = dy · dx = dx dt
2
= dt (at) = a(2t)= 2at
Eliminating t, we get x
2a
2a
= 2at = y
= a ( Ja
J=
a.
(·: 2at = y)
J~2
= ~:
=> y2 = 4ax
(i)
Differentiating both sides of (i) w.r.t. 'x' we have
d
2
d
dx (y) = dx (4ax) d 2 dy dy (y). dx
d
= 4a dx (x}
dy
=> -2y dx
= 4a(1)
dy _ 2a dx - y dy Example 2: Find dx if x = 1 - t2 and y
= 3t2 -
Solution. Given that x = I - t 2 (i) Differentiating (i) w.r.t. 'x', we get
and
2f. y = 3t2 -
2t3
(ii)
Unit 2: Differentiation dx
dt
67
d
d
2
di
= dt(l-t) = dt(l)-dt(t) = 0-2t = -2t
Differentiating (ii) w.r.t. 'x', we have dy d 2 ,3 d 2 d 3 dt = dt (3t - 2r ) = dt (3t ) - dt (2t )
= 3(2t)-2(31)=
= 6t(l-t)
2
6t-6t
Applying the formula
dy dx
dy
= dt
dy dt
dt . dx
= dx . dt
= 6t(l-t) =
-3(1 + t)
-2t
dy I - i1 Example3: - if x = - - , dx
-
1+1
0
a·iven th at
= 3(t-1)
2t y- - -
-9
-1+1
(i) x = 1+ Differentiating (i) w.r.t. 't', we get
SoIution: ·
!L (] -
dx
1) = di (
d
2t y=1+l
and
(ii)
2 ) 2 ..2 d 2 (1 - t) (1 + t) - (1 - f) . di (1 + t)
dt- dt l+l
(l+ll
=
(-2t)(l + l)-(1-t2)(2t) (1 + ll Differentiating (ii) w.r.t. 't', we have d
)
2
2t(-J -t2-J + 1) (1 + lf d
=
-4t (1 + lf
2
dy !L ( 2t ) - ( di (2t) (1 + t) -2t Xdi (1 + t) dt = dt T+7 (1 + t )
= dy dx
=
2(1 + t2)-2t(2t) (1 + lf dy dy dt dt dt . dx = dx dt
= =
2 + 21-41 2 -21 2(1-1) (1 + lf = (1 + 1/ = (1 + l/ 2(1-1) (1 + 1f 2(1-1) 1-1 4t = -4t = 2t - (1 + lf
2.7 DIFFERENTIATION OF IMPLICIT RELATIONS Sometimes, the functional relation is not explicitly expressed in the form y = f(x) . d but an equation involving x and y is given. To find from such an equation, we
Jx
Calculus and Analytic Geometry
68
differentiate each term.of the equation and use the chain rule where it is required. The process of finding :
in this way, is called implicit differentiation. We explain the
implicit differentiation in the following examples: Example 1: Find: if x 2 + y 2
=4
Solution: Here, x 2 + y 2 = 4
(i)
Differentiating both sides of (i) w.r.t. 'x', we get 2x+ 2ydy =0 dx
dy dy x or x+y-=0 =>-=-dx
dx
y
Solving (i) for y in terms of x, we have
y =±../4-x2
=> y=~4- x 2
(ii)
or y=~4- x 2
(iii)
dy found above represents the derivative of each of functions defined as in
dx
(ii) and (iii). From (ii) dy dx
= --=1==-X (-2x) = --,=x==2 ../4-x 2
2../4-x
=
x
y
... ) dy F rom (m - = dx
~4-x 2 = y)
(·: 1
-x x x(-2x) =-t====- _ - 2v4-x 2 -../4-x 2 y /
. d dy .f 2 2 E xamp1e 2: Fm dx, 1 y + x - 4x =5. Solution: Given that y2 + x 2 - 4x = 5 Differentiating both sides of (i) w.r.t. 'x' d [ 2 dx y
2
d
+ x - 4x] = dx (5)
(i)
Unit 2: Differentiation
or
69
dy 2y dx + 2x-4 dy 2y dx
=0 dy
= 4-2x
2(2 -x) 2 -x 2y = y
=> dx =
(ii)
Note: Solving (i) for y, we have
l=
2
=>
5+4x-x
= ±'15+4x-x2,
y
2
y = '15 + 4x -x
Thus
(iii)
or y = -'15 + 4x -i Each of these equations (iii) and (iv) defines a function. Let
y
= fi(x) = '15 + 4x -i = fz(x) = -'15 + 4x -x2
(v)
(vi)
and y Differentiation (v) w.r.t. 'x', we get
, !. (x) 1
=
21 (5 + 4x 2
From (v), '15 + 4x -x
Also fz'(x) = A
1
-2
x
2)-1/2
= y,
(
so 2 -1/2
I
= y,
2
From (vi) - \15 + 4x -x
)
x 4 -2x =
(5+4x-x)
(iv)
fi'(x)
=
2- x \15 + 4x -i 2 ;x A
x(4-2x) =
so fz'(x)
/
2 -x Al 2 -\f5+4x-x
= 2 Y-x
Thus (ii) represents the derivative of Ji(x) as well as that offz(x).
. d dx dy 1"f y2 - xy - x 2 + 4 = 0 . ExampIe 3 : Fm Solution: Given that y2 - xy - x + 4 = 0 Differentiating of both sides of (i) w.r.t. 'x' gives 2
d
2
2
dx [y -xy -x + 4]
or
(i)
d = dx (0) = 0
dy - ( 1 . y + x dy) 2y dx dx - 2x + 0
=0
dy (2y -x) dx
= 2y-x
= 2x +. y
dy
=> dx
2x+y
(ii)
fx if y3 - 2xy + iy + 3x =0. 2
Example 4: Find
Solution: Differentiating the both sides of the given equation w.r.t. 'x', we have · d
3
2
2
d
dx [y -2.xy + x y + 3x] = dx (0)
=0
Calculus and Analytic Geometry
70 or
d
d
d
d
2
dx (y1) - dx (2xy2) + dx (x y) + dx (3.x) .= 0
2 ;(y1)-2[1.y2+.x;(y2)] +(2xy+x
fx
Using the chain rule on ; (y 1) and 3y2
or
2
+3=O
:
2 2 dy (3y -4.xy + x) dx = 2y 2-2xy-3
dy
=
dx 2
Example 5: Differentiate x + Solution: Let y
= x2+ ;1
;z
and
!h!_
1
cJx = 2x + (-2) . XJ du
1
2y2 -2.xy -3 3y2 - 4.xy + x 2
x
w.r.t. u
-f.
= x - ~ . Then (
=2X-
1)
XJ
1
4 2 2 2(x -1) _ 2(x - J)(x + 1)
= 2 x +1
XJ
-
XJ
dx = 1-(-1). x2 = 1 + x2 = ~
and
!l:J.. - !lJ!.
dx - 2(x2 - l)(x2 + 1)
Thus, du - dx · du -
Find
(y 2), we have
i- 2 [l + x ( 2y i)] + 2.xy + x
=>
1.
i) +3=0
2
•
x2 = 2(x2 - 1) x 2+ 1 x
= 2(x __xl)
EXERCISE 2.4 by making suitable substitutions in the following functions defined as:
y
(i)
x1
=
#
(ii) y =
'1x + ...JX
(iii)
2
(iv) y= (3x -2x + 7)6
2.
. d dy "f Fm dx1: (i) 3x + 4y + 7 = 0 (ii) xy + y2 = 2 2 2 (iii) x - 4.xy - Sy= 0 (iv) 4x + 2hxy + by2 + 2gx + 2fy + c = 0 (v) x-fl+Y + y~ = 0 (vi) y(x2 -1) = x'\/x2 + 4
3.
Find : (i)
of the following parametric functions: 1
x = 8 +9
and y
= 9+ 1
Unit 2: Differentiation .. ) (n
71
_ a(l + ()
I + ti
x -
, y =
4.
dy Prove that y dx + x = 0
5.
Differentiate (i)
2 1 4 x - 2 w.r.t. x
2bt I + ti
if
I -l x = 1 + t2 ,
2t y = 1 + t2
x2
(ii) (1 + i)" w.r.t.
x
2 ax + b ax + b (iv) ex + d w.r.t. ax2+ d (v)
x2 + I x2 _ 1
i+I x-1 x2-1 w·rt· · x +I
(iii)
3
w.r.t. x.
2.8 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS While finding derivatives of trigonometric functions, we assume that x is
lim sin x measured in radians. The limit theorems x-XJ -x-
= I and
lim 1 - cos x x-XJ x
are used to find the derivative formulas for sin x and cos x. We prove from first principles that
! Let and
y
(sin x) = cos x and
!
(cos x) = -sin x
= sinx. Then y + ~ = sin (x + Dx)
~ = sin (x
+ Dx) -sinx
_:__( _§!l
~
ox 2
_:__( _§!']
ox~ . §1. & l zm & = lim cos ( x + '5x-XJ \· &-xl
2}
[
2
l
_: __ ( §! =
lim
& ---70 2
cos ( x
\·
a
+ ; ) lim
& ---70 2
~ & 2
=0
·
Calculus and Analytic Geometry
72
Thus,
= cos x. I
:
l
lim co' (x +
(
Ox/2-XJ
~)= '°'"
_,_ ( ~\
and
Lim Ox/2-XJ
)
::l.Il & =1 2
Let y =cos x, then y +Sy= cos (x + &) and Sy= cos (x +&)-cos x
= cos x cos & - x sin & - cos x = - sin x sin & - cos x (1 - cos &J Oy sin & (1 -cos&) &= (-sinx). ~ -cosxl
lim &-XJ
~=
&
lim[(-sinx) sin Jx -cosx(I-cosJx Jx Jx
&-+o
at
= dy
or
)J
lim [(-sinx)sin Jx]-um [cosx(I-cosJx Jx x-+o Jx
·l:
&-+o
Thus, dx
11
= (-sin x). 1 -(cos x)(O)
d
= cos x
d dx (sec x) = sec x tan x
~~ si"ax& = 1 and)
lix lix)-- 0
Lim (1 - cos &-XJ
dx (cos x) =-sin x.
. d Now usmg dx (sinx)
11)J
and
d
dx (cos x)
d and dx (cot x)
=-
sin x, we prove that
= - cosec2 x
d
Proof of dx (sec x) = sec x tan x. Let
y
1 = secx = cosx
(i)
Differentiating (i) w.r.t. 'x', we have
d
dx (y) = =
d [
J [~ (1)]co~x-l. ~ (cosx)~Usi~g J quotient
- -1- =
2
dx cos x
(cos x)
O.cosx-1.(-sinx) cos2 x
= cos1 x
sinx cos x
=
secx tanx
ormula
Unit 2: Differentiation
73
d dx (secx)
= secx tanx
d Proof of dx (cot x)
= -cosec2 x
Thus
cosx sinx Differentiating (i) w.r.t. 'x', we get
y =cot x
Let
d dx (y)
=
(i)
J
( Using [!!:._(cosx)lsinx-cosx!!:._(sinx) dx 'j dx . = - [ - -] = quotzent 2 formula dx sin x (sin x ) d
_
cos x
(-sinx)sinx-cosx(cosx) sin 2 x
- (sin 2 x + cos2 x) = sin2 x Thus
1 = - -.-2= cosec2 x sm x
d
d (cot x) = - cosec2 x
Now we write the derivatives of six trigonometric functions:
d
d
(1)
dx(sinx)
= cosx
(2)
dx(cosx)
= -sinx
(3)
d dx (tan x)
= sec2 x
(4)
d dx (cot x)
= - cosec2 x
(5)
d dx (cosec x)
(6)
d dx (sec x)
= sec x tan x
= - cosec x cot x
Example 1: Find the derivative of tan x from first principles. Solution: Let y = tan x, then y + 8y = tan (x + 8.x) and 8y = y + 8y -y = tan (x + 8.x) - tan x sin x _ sin (x + 8.x) cos x - cos (x + 8.x) sin x sin (x + 8.x) = cos (x + 8.x) cos x cos (x +&)cos x =
sin(x+&-x) _ sin& cos (x + 8.x). cos x - cos (x + 8.x) cos x
8y _ 1 8x - cos (x +&)cos x or
sin
8x
8x
lim 8y lim (: 1 ) lim (sin &--XJ 8x = &--XJ cos (x + 8.x). cos x · &--XJ &
&)
Calculus and Analytic Geometry
74
dv Th us ::L. dx
=
&~cos (x + &) lim
1
2 (cos x) (cos x) · 1 = sec x
~ 2 Thus dx = sec x
Lim sin & ( and&~ ~=1
= cos x
J
d . 2 dx (tan x) = see x.
or
Example 2: Differentiate ab-initio w.r.t. 'x' 2 sin ~ (iii) cot x Solution: (i) Let y =cos 2x, then y + 8y = cos 2(x + &) and 8y = cos (2x + 2&) - cos 2.X 2x + 2& + 2x 2x + 2 & - 2x = - 2 sin sin = -2 sin (2x + &) sin & 2 2 sin & Now.§l'. = -2sin(2x+ &). ~ &
(i)
cos 2x
(ii)
lim Thus~ dx = &-4J
=
[
sin&] -2 sin (2x +~ &) . ~
lim lim sin & -2 &-4J (sin 2x + &) . &-4J ~
· 2x) . 1 =-2 sm · 2x ( = (- 2 sm (ii)
Let
y
and
8y
= sin~ , then y + 8y = sin'1x + & -sin~ = 2 cos ( :Jx + &2
As
so
--s;-
Lim 8x---XJ sin (2x +&)=sin 2x and Lim 8x---XJ sin & = 1 )
= sin '1 x + &
:JX) sm. (:Jx + &. -:JX)
+
2
('./~~-+-& + ~) "a' -x'] ![xsin-' =]+!(a' -x' J" 1
=
=
ffi. 1
X = 1 . sm. -1 a + x.
=
. -1
sm
x 1 1 a+x ~2·a +
1-2
d (
dx
1 2 2 ~a + 2. (a -x ) X )
I
d (
2
. dx \Q - x
1
~I 2
2\Ja -x
2.
(-2x)
a
= Example 2: If
a+ x · ~\JaI a-x a1 - ~\JaI -x sm 2 y = tan Gtanshow that dy = 4 0 + Y ) . -1 X
X
sm
2
2
2
dx
=2
tan
2x ,
-1
dy du
y = tan u =>
and
Thus
: dy dx
=
-
a
1 }) ,
\
Solution: Let u
. -I X
2 -
!(
1
2 tan-
dy
= du
4 + x 2)
then 2 = sec u = 1 + tan 2 u = 1 + y 2
~J= 2 .
(I ! )2 ·
l+ 2
du 4 2 · dx = (1 + Y ) · 4 + x 2
=
! (~ J
4(1 + y2) 4 + x2
2
=
x2 ·
1+4
~ = 4: x2
2)
79
Unit 2: Differentiation
EXERCISE 2.5 1.
Differentiate the following trigonometric functions from the first principles. (i) (v)
2.
sin 2x ~
tan~
x
tan3x
(iii)
(vi)
,Jtan x
(vii) cos/;
(iv)
cosx
Differentiate the following w.r.t. the variable involved. 2 (i) x 2 sec 4x (ii) tan3 () sec () (iv) cos --{x +
(iii) (sin 28- cos 38/
3.
Find~ if
4.
y = x cosy Find the derivative w.r.t. "x" (i)
(i)
5.
sin2x+ cos2x
(ii)
cos~
Differentiate · sznx w.r.t. cot x (I.)
=
'1 sin x
y sin y
(ii)
x
(ii)
sm~
6.
4 · 2 x w.r. t. cos x szn dy If tan y(l + tanx) =1- tanx, show that - = -1
7.
If y
= '\)tan x + '1 tan x + '1 tan x
8.
If x
= a cos3 8,
9.
Find :
10.
Differentiate w .r. t. "x"
(1·1·)
dx
y
= b sin3 8,
dy show that a dx + b tan ()
if x =a(cost+ sin t), y -IX
+ .. · =, prove that (2y -1)
-IX
cos a
(ii)
cot a
(iv)
sin-l '11 - x 2
(v)
sec
1
-1
(x2x2_1 + ])
(iii)
=0
(vi)
. -I a -1 sm -
a
cot
x
-I (
2X ) 1 -x2
G: ::)
11.
dy y . y Show that dx = x if x
12.
If y
= tan (p
tan
-1
x -
y
tan-1 x), show that (J + x 2) Y1-p(1 + y2)
=0.
2
= sec
= a(sin t - t cost)
(i)
(vii) cos-
~~
x.
2
Calculus and Analytic Geometry
80
2.10 DERIVATIVE OF EXPONENTIAL FUNCTIONS A function f defined by
= ax
f(x)
if a > 0, a "# 1 and x is any real number. is called an exponential function. If a = e, then y =a x becomes y =ex. ex is called the natural exponential function. Now we find derivatives of ex and ax from the first principles. 1. Let y = ex, then y + 8y = ex+& and 8y = y + 8y - y = ex+& - ex = ex . e& - ex
= ex (e& -1)
That is, 8y
by Thus J!~ bx
=Lim
&-XJ
f. =
and
ex. (
ex .( e&-l) =ex lim
&
. &-XJ
e&a~ 1)
(e&-1) &
xl
.. Lim x ( . &-XJe = e)
~ dx
2.
x
(
=e
•
lim
eh - I
. I Usmg h-XJ -h-
j
= 1)
d
or
dx (~)=ex
Let
y
= ct, then
y +Dy =ax+& and Dy = ct+& -ax Dividing both sides by &, we have
f
=
= ax. a& -ct = ct (a& -
at";'J
l)
Thus ~ = 1:->IJ ,r(& I) = a'. i:->IJ (a"&I) (:· 1:->IJa' =a') =a or
d
x
x
( lim ah-l . (ln a) Using h-XJ - h -
x
dx (a ) = a . (in a) dy
Example 1: Find dx if: (i) y Solution: (i) Let
u
= e~ +I
= X2 + 1 ,
(ii) y = a..fx then du
(A) and dx
d
j
= logae = ln a)
= dx (x2 + 1) = 2x
Unit 2: Differentiatwn
81
Differentiating both sides of (A) w.r.t. 'x'. we have
.!!:_
.!!:_ u d u du dx (y) = dx (e) = du (e). dx II du = e · dx
Thus (ii)
~
= ex
2 +I.
(
.
l
d
Using dx (ex) = ex)
(h)
u =-[x Then du d 1 and dx = dx (x112) = 2
.
(Usmg the cham rule)
( ·.·
X2 + 1 and
= a"
y
Let
u;
~ = 2xJ
(A)
1
= 2-[x
x-112
Differentiating both. sides of (A) w.r.t. x gives
du .)
dy = .!!:_(a") = _!!:._(au) du dx dx du dx
(·: dy = dy. dx du dx
= (~u Zn a) . ~~ Thus
}x
:Jx (a..[;_) = (a..[;_ In a) . 2 = in a
Example 2: Differentiate y
(-. ·
u
!
(ax)
=-{
1, that is, for each x E (1, oo), y ~ (0, oo), so we discard this value of eY. Thus
e = x + '1x 2 -1
which gives y
= In (x + ~)
, that is,
cosh- 1 x = ln(x+'1x 2 -l).
'" ·: rivative of sinh- 1 x: . . h-1 x y = sm Let Then x = sinh y dx
dy
or
'
.. dy -
dy -
1 dx - coshy
= coshy
1 dx = cosh Y =
dy
x,yE R
'1t
1 + sinh2 y
l_)
. dx - dx
(
dy
(": cosh y > 0)
Calculus and Analytic Geomet1y
88 dy dx
d
.
= dx (smh
1
-I
= -{1J
x)
(XE
R)
Derivative of cosh-1 x: Let
y
Then
x
and
dx dy
or
dy dx
dy Thus dx As
cosh- 1 x
= cos h-1 x = cosh y
= sinh =
1 sinhy
[l, oo), y
XE
dy dx
=>
y
='.f
d
= dx (cosh =
;
[0, oo)
E
1
.. dy - _!_J . dx-dx [ dy
= sinhy
1 cosh 2 y -1
(·: sinhy > 0, asy > 0)
1
-1
x)
(x > 1)
= './x2 _ 1
In (x + './x2 -1) , so
1 (1 + 2x ) 1 1 -d [cosh- 1 x] · ~ +x dx - x + './x2 -1 2'./x2 -1 - x + './x2 -1 './x2 -1 - ~
Derivative of tanh - t x: Let
y=tanh-1 x;
Then x
dy dx
= tanh y
=
XE(-1,1),yER
dx dy
and
1 1 - tanh2 y
d
=
= sech2 y
=>
dy dx
1
lx)
( ',': = dy
= sech2 y
1 1 - x2
1
Thus dx (tanh-1 x) = ~
-1 < x < 1 or lxl < 1
1-r
The following differentiation formulae can be easily proved.
d -1 dx {coth x) =
d
or
1
1
-x2 1
-0r 1
-
1
x2 _ 1
dx(sech- x)=
· - x'./1-x 2 '
O 1
Unit 2: Differentiation Example 1: Find
dx
89
if y = sinh-1 (ax + b)
Solution: Let u = ax+ b, then y
= sinh-1 u
dy
dy du du · dx
dx
=
d [ sm . h-1 (ax + b ) Th us dx Example 2: Find :
if y
=
sec x,
Solution: Let u
dy du 1
vl
· dx. .
1 +(ax+
bf
du d ( ·.. dx = dx (ax+ b)
.a
1
=a)
= cash-I (sec x) then
dy 1 du - . ,.-y--:;
y = cash-I u du
= --J 1 + u2 du
= --J 1 + u2
= _/
1
'JU - 1
d
and dx = dx (sec x) = sec x tan x
dy Thus dx
1
= _/
2
\JSeC X -
or
1
dy du ·
= du . dx = '1 u2 -
du 1 . dx
(secxtanx) 1
1
= tanx
(secxtanx)
= secx
d dx [cash-I (sec x)) = sec x.
EXERCISE 2.6 l.
Find (i)
f' (x) if f(x)
= e-fx- t
1
(ii)
f(x)
= x 3 ex, (x :t; 0)
(iii) f(x)
= ex (I+ Zn x)
(vi) f(x)
= eax +e
eax - e-tU:
x
(vii) f(x)
= e-x + 1 (v) = '1zn(e2x + e-2x)
. d dy Fm dx
1
(iv) f(x)
2.
e
(i)
y
(iV) y (vii) y (x) Y
-tU:
•t
= x2 Zn~ = x2 Zn -x1 = Zn (9 - x2) = X esinx
(ii)
y
= x~
(iii) Y
(v)
y
~In~
(vi)
= e-2x sin 2x = 5e3x-4
(ix) y (xii) y
(viii) y
(XI.) y
y
x = Znx
= Zn (x + '1x2 + i)
= e-x (x3 + zx2 + 1) = (x+ ll
Calculus and Analytic Geometry
90 (xiii)
y
(i)
2.15
(xiv)
Y
=
~(x+l) (x3 + 1)312
I
y = cosh 2x
(iii) y (v)
1
.f
. d dy Flil dx
3.
= (ln x)1r x
y
(iv) y
= sinh 3x = sinh- 1 (x3)
(vi) y
= sinh- 1 ( ~ J
(ii)
= tanh-1 (sinx),
-¥ =>
dy
dx = Differentiating both sides of (ii) w.r.t. 'x', we have
2
2ax+x
y2
(ii)
Calculus and Analytic Geometry
92
d2y
.
dx2
=
!!-__ [ 2ax + (-1) dx y2 2(a + x)
=
x2] _
2
2 (
-
(y2)2
2ax + x-?J 2 l- (2ax + x). 2y x(y2
y4 2
2
(2ax + x ) (2ax + x y
2
2 [ (a+ x) y +
=
)]
y4 2 [(a+ x) y3 + (2ax + x 2)2l
=
y
)'4.
2
=
=
2 [(a+ x).(-3ax - x 3 ) + x2(2a + x) 2] 2 3) s (·: y3 =-ax 3 -x y 2 2 2 2x [-(a+ x) (3a + x) + (4a + x + 4ax)] ys 2
2x [-
=
2x2
=
2
(3a + 4ax + x2) + 4a2 + x2 + 4ax] ys
La2 ] = yS
-
2a2x2
YS
Example 4: If x =a( 8- sin 8), y = a(l +cos 8). Then d2 Show that y2 -{- + a =0 dx Solution: Given that x = a ( 8 +sin 8) y = a ( 1 + cos 8) Differentiating (i) and (ii) w.r.t. '8', we get dx d = a(l +cos 8) 8
and
and
Using
~\
(2a + 2x) y - (2ax + x )2y dx)
dy d - a(- sin 0) 8 dy dx
=
(ii)
(iii) (iv)
dy dy d8 d8 d8. dx = dx' we have d8
-a sin 8
(i)
= a (l +cos 8) =
sin 8
1 +cos()
93
Unit 2: Differentiation . dy That 1s , dx
sin() +cos () 1
=
(v)
Differentiating (v) w.r.t. 'x' 2 dy d ( sin() ) . d ( sin() ) 2 dx = dx 1 + cos (} = - d() 1 + cos (}
l-
l
= _cos 8(1+cos9)- sin 8(- sin 9) (1 + cos
d2y dx2
=
8)2
cos () + cos2 () + sin2 (1 +.COS 8)2
= - (11++cos() cos 8) 2
x
1 2 a · (1 + cos fJ) 1
=
a x
a2
7
-
d()
() .
dx (-.·
a · (~
~ = a(l +cos(}))
J ·
.! _L
(·. 1 + cos ()
= ra)
a
= -y2 d2y
-a
or
d(}
dx
1 a( 1 + cos fJ)
= - .!
d(} x dx
.
y2 dx2 +a = 0.
Example 5: Find the first four derivatives of cos (ax+ b). Solution: Let y = cos (ax+ b), then
d
d
YI= dx [cos(ax+b)] = -sin(ax+b). dx (ax+b)
=Y2
y3
y4
sin (ax+ b) x (a+ 0) =- a sin (ax+ b)
= - a! [sin (ax+ b)] = (-a) [cos(ax + b) x (a+ O)] = - a2 cos (ax+ b) = - a 2 [cos(ax + b)] = (-a2) [- sin(ax + b) x (a+ 0) = a3 sin (a~+ b) = a3 [sin(ax + b)] = a 3 x [cos (ax+ b)] x a = a4 cos (ax+ b)
!
!
d3 Example 6: If y =a-ax , than show that : , + a 3y = 0
, dy = dx d(-ax) d( - ax) = e-ax • (-a) SI o ubon: As y = a -ax, so dx e = e-ax . dx
Calculus and Analytic Geometry
94 dy That is dx = - ay
Now
fx d2y
[:J = !
-ax
= y)
(·: e [-ay] =>
~ =-_a:
= (-a) (-ay)
(-:
: .=- ayJ
= a2y (i) dx Differentiating (i) w.r.t. 'x', we get
or
-2
J !!_ dxz = dx [a y] => 2
!!_ [ d y
dx d3y 3 Thus dx3 + a y = 0. •
x = Sln. -l a ' Y1
3
dY
dx3 = a
2
dy dx
= a 2 (-
= sin- 1 ~,then show thaty2 = x(a 2 -x 2
Example 7: If y Solution: y
2
so
=z =~[sin-I~]= =
1
1
H
. a
= - a3 y
ay)
t2 . 3
kl' m X:fx
a
= -;::::::;:=====-'1 a2 - x2
.
1
_a
= (a2 _ x2)-112
EXERCISE 2.7 1.
Findy2 if (i) y
2.
... ) (lll
= (2x + 5)3/2
(ii) y
-c
y ='JX
y=x2 .e-x
(ii) y
= In (~: ~ J
(ii)
x3 -y3
Find Y2 if (i)
x2 + y2 = a2
(v)
= a cos e,
= a sine x2 + y2 + 2gx + 2.fy + c = 0
(iii) x
4.
3x4 + 4x3 + x - 2
Find Y2 if ( 1")
3.
= 2x5 -
y
(iv) x
= a3
= ar, y = bt4 •
Find Y4 if (i)
y
= sin 3x
(ii)
y = cos3 x
(iii)
in (x
2 -
9)
1
+~
Unit 2: Differentiation
95 show that (1 - x2) y 2 - xy 1+ m2y
5.
If x
= sin 0,
6.
If y
=e
7.
If y
8.
If y
= e sin bx, show that dx2 - 2a dx + (a2 + b2)y = 0 = (cos- 1x)2 , prove that (1 - x2) Y2 - xy1 - 2 =0.
If y
= a cos (In x) + b sin (In x), prove that
9.
y
= sin m0,
x
sin x, show that
d2y dx2 -
dy 2 dx + 2 y ~ 0
d2y
ax
=0
dy
x
2
d 2y dx2
dy + x dx +·y
= 0.
2.16 SERIES EXPANSIONS OF FUNCTIONS 3 4 A series of the form ao + a1x + aix2 + a3x + a4 x + ··· + anX' + ···is called a power series expansion of a functionf(x), where ao, ai, a1, ··.,an,··· are constants and x is a variable. We determine the coefficient ao, ai, a1, · . ., an. · ·· to specify power series by finding successive derivatives .of the power series and evaluating them at x = 0. That IS,
4 2 3 5 f(x) = ao + a1x + a1x + aµ + a¥ + asX + ··· + anX' + ··· .f(O) = ao f'(x) = a1 + 2a2x + 3aµ 2 + 4a¥3 + 5asX4 + ·· · + nanXn-I + .'.· f '(0) =a1 2 2 f" (x) = 2a2 + 6aµ + 12a¥ + 20asX3 + ··· + n(n-J)anX'- + ··· j"(O) = 2a2 J:"(x) = 6a3 + 24a¥ + 60aµ 2 + ······ 4 / J(x) = 24a4 + 120aµ ········· .
So we have ao
=f(O),
,
a1 = f (0), a1
=
["(O)
2! , a3
.
Following the above pattern, we can write an =
=
j"'(O) = 6a3 . 4 / )(0) = 24a4 4 f"'(O) / J(O) 3! , a!I - 4!
(n)
L.JSll . n!
Thus substituting these values in the power series, we have , .L1Ql L1!!l 3 / 4J(O) 4 /nJ(O) f(x) = f(O) + f (0) x + 2! x2 + 3! x + 4! x + ··· + n! X' + ··· This expansion of fl..x) is called the ~aclaurin series expansion. The above expansion is also named as Maclaurin's Theorem and can be stated as: Iff(x) is expanded in ascending powers of x as an infinite series then, 4 . f "(O) f"'(O) / J(O) /nJ(O) 3 f(x)=f(O)+f'(O)x+ 21x2+31x +"41x4+ ···+--;i- X'+ .. :
Calculus and Analytic Geometry
96
Note that a function f can be expanded in the Maclaurin series if the functions is defined in the interval containing 0 and its derivatives exist at x = 0. The expansion is only valid if it is convergent. Example 1: Expand/(x) =
!
1 x in the Maclaurin series. Solution: f is defined at x = 0 that is,f(O) = 1. Now we find successive derivatives of f and their values at x = 0. f'(x) = (-1) (1 +x)-2 and f'(O) = -1 f"(x) = (-1) (-2) (1 + xF3 andf "(O) = (-1) 2 ll_ and f"'(O) = (-1)3 LL and f( 4 )(x) = (-1>4 Li_
f'"(x) = (-1)(-2)(-3) (1 + x)-4
f( 4 )(x)
= (-1)(-2) (-3)(-;4) (1
+ xr5
Following the pattern, we can write f (n)(O) = (-1
t
[_!!__
Now substituting f(O) = l, f'(O) = -1, f"(O) = (-1)2 ll_ f"'(O) = (-1)3 LLJ(4)(0) = (-1/ Li_,··· f(n)(O) = (-l)n [_!!__in the formula. f(x)
,
f
= f(O) + f (0) x +
f
"(0)
"'(O)
/
4
J(O)
x2 + LL x3 + Li_
ll_
x4 + · · ·
J(n)(O)
+
211..
1
3~
2
3
1 ,..
t.!!_
41!..
1 +x =l+(-l)x+(-1) ll_ x +(-1) l1_ r+(-1) Li_ x
= 1 - x + x2 -
x3 + x
+ · · · + (-1
t
'
4
+···+ 4
n
x + · · · we have
l!!.. l!!..
(-lt
~+···
~ + ···
1
.
-
Thus, the Maclaurin series for 1 + x is the geometric series with the first term 1 and common ratio - x. Note: Applying the formula 1-
X
+ X2 -
X3
S =
+ ···
~ r , we have
1
1 1 = 1 -(-x) - -1 -+-X
Example 2: Find the Macfatirin series for sin x Solution: Let f(x) = sinx, thenf(O) = sinO = 0. f'(x) =cos x andf'(O) =cos 0 = l;f"(x) = -sinx andf"(O) = -sinO = O; f"'(x) = - cos x and f"'(O) = - cos 0 =-1; jx,f(x +ox)) be two neighbouring points on the arc AB where x, x+oxeD1 . The line PQ is secant of the curve and it makes LXSQ with the positive direction of the x-axis. (Seethe figure 2.21.1)
0
Drawing the ordinates PM, QN and perpendicularPRtoNQ, we have
FIGURE 2.21.l
1 02
Calculus and Analytic Geometry
RQ =NQ-NR =NQ-MP= f(x+ '&x)-f(x) and PR =MN = ON-OM=x+ '&x-x= '&x Thus tan m LXSQ= tan m LRPQ RQ f(x + '&x)- f(x) = PR = '&x Revolving the secant line PQ towards P, some of its successive positions PQ 1, PQ2 , PQ3, ••• are shown in the figure 2.21.2. Points Qi (i = 1, 2, 3, .... ) are getting closer and closer to the point P and PRi i.e; '&xi (i = 1, 2, 3, ... ) are y approaching zero.
In other words we can say that the revolving
secant
line
approaches the tangent line PT as its limiting position at P while
ox
approaches zero, that is,
FIGURE 2.21.2
tanmLXSQ-~
or
f(x+8x) - f(x) '&x
tanmLXTP when '&x
-~
so
f(x + &) - f(x) . L zm Bx ~o &
or
f'(x) =tan mLXTP
~
tan mLXTP as '&x
= t an
m
0 ~o.
L V'T''P ..11. .L
Thus the slope of the tangent line to the graph of 'f' at (x,f(x)) isf(x).
I I
Example 1: Discuss the tangent line to the graph of the function x at x Solution: Let
=Ix I and, f(O) =I 01 = 0 f(O+fu)=I 0 + '&x I= j 8x I, f(x)
so
/(0+8~)- /(O) =
and
f(O + 8x)- /(0) 8x
Iox I - o lox-I =Bx
= O,
Unit 2: Differentiation
103
I
f' (0) = Lim j & 8x ~o 8x Because j 8x = 8x when 8x > 0
Thus
and
I j 8x I= - 8x
when 8x < 0,
so we consider one - sided limits
Lim j 8xj 8x ~o+ 8x Lim and
8x
j &j
~o- 8x
=
8x . L im -= 1 8x ~o+ 8x
=
-8x Lim =-1 8x ~o- Bx y
The right hand and left hand limits are not equal, therefore, the
I
Lim j x does not exist. & ~o & This means that f' (0), the 0
derivative of f at x = 0 does not exist and there is no tangent line
FIGURE 2.21.3
to the graph of f at x = 0 (see the figure 2.21.3). 2
Example 2: Find the equations of the tangents to the curve x _.:._ y2 - 6y = 0 at the point whose abscissa is 4. 2 Solution: Given that x -y2-6y=O (i) We first find the y-coordinates of the points at which the equations of the tangents are to be found. Putting x = 4 is (i) gives 2 16 - y2 - 6y = 0 ~ y + 6y - 16 = 0
or
Y
=
-6 ± ~36 + 64
-6 ± .JiOo
2
2
y= -6+10 =.i=2
2
2
or
=
y =
-6 ± 10 2
-6-10 2
,
that
.
lS
-16
=2 =-8
Thus the points are (4, 2) and (4, - 8). Differentiating (i) w.r.t. 'x' wehave
2x - 2y dy --:- 6 dy = 0 dx dx
~
2 dy (y + 3) = 2x dx
~
dy dx
=
x y+3
- -~---
__
..-...___..
Calculus·and Analytic Geometry
104
4 Theslope of the tangent to (i) at (4, 2) = - - = i 2+3 s Therefore the equation of the tangent to (i) at (4, 2) is 4 y-2=-(x-4) => Sy-10=4x-I6
s
or Sy= 4x - 6 The slope of the tangent to (i) at (4, - 8)
=
4
-8 + 3
4
=
s
Therefore the equation of the tangent to (i) at (4, - 8) is
4 y - (-8) =-s(x-4)
Sy+ 40
or
=
-4x + 16
4x +Sy+ 24
=0
2.19 INCREASING AND DECREASING FUNCTIONS Let/ be defined on an interval (a, b) and let x 1 , x 2 E (a, b) . Then (i)
f is increasing on the interval (a, b) if f(x 2) > f(x 1) whenever x2 > x 1
(ii)
f is decreasing on the interval (a, b) if f(x 2) < f(x 1) whenever x2 > x 1
a
X1
X2
b
f(x,;) > f(x 1) ifx2 > x 1
a
X1
X2
b
f(x 2) x 1
We see that a differentiable function/ is increasing on (a, b) if tangent lines to its graph at all points (x,f(x)) where x E (a, b) have positive slopes, that is,
f' (x) > . 0 for all x such that a < x < b and
f
is decreasing on (a, b) if tangent lines to its graph at all points (x,f(x))
where x E (a, b) , have negative slopes, that is,
f'(x) < Oforallxsuchthata < x < b Now we state the above observation in the following theorem.
Unit 2: Differentiation
105
Theorem: Let/ be a differentiable function on the open interval (a, b). Then (i)
f
is increasing on (a, b) if f ' (x) > 0 for each x E(a, b)
(ii)
/is decreasing on (a, b)if f'(x) < OforeachxE(a, b)
Let
f(x) = x , then ·
2
f(x2)-f(x 1) = x~ -x~ =(x2 -x 1)(x2 +x 1)
If xuX2 E(-oo,O)andx2 > x 1,then f(x2)-f(x 1) < O(·: x 2- x 1 >0andx2 +x 1 x 1 , then f(x2)- f(x 1) > 0 ~
f(x2) >f(x1)
~ f is increasing on the interval (O,oo).
Here f' (x) = 2x and/' (x) < 0 for all x E(- oo, 0) , therefore,
f
is decreasing on the interval (- oo, 0)
Also f'(x) > 0 for all x E (0, oo ), so/ is increasing on the interval (0, oo). From the above theorem we can conclude that 1.
f' (x 1)
2.
/'(x 1) =O ~ f is· neither increasing nor decreasing at x 1
3.
/'(x 1) > 0 ~ f is increasing at x 1
< 0 ~ f is decreasing at x 1
Now we illustrate the ideas discussed so far considering the function f defined as f(x)=4x-x
2
(I)
To draw the graph off, we form a table of some ordered pairs which belong to f
Ir~(x) I ~; I : I : I : I : I : I ~5
I
Calculus and Analytic Geometry
106 The graph off is show in the figure 2.22.1 y
(5, -5)
(-1, -5)
y'
FIGURE 2.22.1
From the graph off, it is obvious that y rises from 0 to 4 as x increases from 0 to 2 and y falls from 4 to 0 as x increases from 2 to 4. In other words, we can say that the function f defined as in (I) is increasing in the interval 0 < x < 2 and is decreasing in the interval 2 < x < 4. The slope of the tangent to the graph off at any point in the interval 0 < x < 2, in which the function f is increasing is positive because it makes an acute angle with the positive direction ofx-axis. (See the tangent line to the graph off at (1, 3)). But the slope of the tangent line to the graph off at any point in the interval 2 < x < 4 in which the functionfis decreasing is negative as if makes an obtuse angle with the positive direction of x-axis. (See the tangent line to the graph off at (3, 3)). As we know that-the slope of the tangent line to the graph off at (x,f(x)) is f' (x), so the derivative of the function/, i.e.,f'(x) is positive in the interval in, which/is increasing andf'(x) is negative in the interval in which f "is decreasing. The function funder consideration is actually increasing at each x for which f'(x)>O. i.e.,
4-2x>O
:::::} - 2x > -4
Thus it is increasing in the interval (- oo, 2). Similarly we can show that it is decreasing in the interval (2, oo). Now we give an analytical approach to the above discussion. Let f be an increasing function in some interval in which it is differentiable. Let x and x + 8x be two, points in that interval such that x+ ox> x. As the function f, is increasing in the interval, it conveys the fact that[(x +ox) >f (x).
I
Unit 2: Differentiation
107
Consequentlywehave,f(x+8x)-f(x)>Oand(x+Ox)-x > 0, that is, f(x+Sx) - f(x)> 0
and
8x > 0
f(x+8x)- f(x) >
or
0 8x The above difference quotient becomes one-sided limit Lim 8x --)-Q+
f(x+8x)- f(x) 8x
As f is differentiable, so f '(x) exists and one sided limit must equal to f '(x). Thusf'(x)>O 2 Example 1: Determine the values ofx for which f defined as f (x) =x + 2x - 3 is (i) increasing (ii) decreasing. · (iii) find the point where the function is neither increasing nor decreasing.
\,, "
2
Solution: The table of some ordered pairs satisfying f(x) = x + 2x- 3 is given below:
1r
~(x) I ; I ~ I =: I ~ I : I ~ 3
I: I
The graph off is shown in the figure 2.22.2. f'(x)=2x+2
(i)
(ii)
(iii)
Theconditionf'(x) > 0 => 2x+2 > 0 => 2x > -2 x which gives x > - 1, so the function f defined asf(x) = x 2 + 2x - 3 is increasing in the interval (-1, oo). Andtheconditionf'(x) < 0 = 2x+2< 0 FIGURE 2.22.2 => 2x 0 f(x) < 0
(x-l)(x-3)
(x-l)(x-3) < 0 0
ifx>landx 3x(x-2)=0 =>x=O or x=2 Now we consider an interval (-8x, 8x) in the neighbourhood of x = 0. Let 0- Eis a point in the interval (-8x, 0). We see that f'(O-s) =3(-s)(-s-2) (·: f'(x)=3x(x-2)) =3s(s+2)>0 (·:s>O,s+2>0) That isf' (x) is positive for all x E (-8x, 0) Let 0 + E1 is a point in the interval (0, 8x), then we have f' (0 +s 1) = 3(s 1)(s 1 - 2) =3s 1 (2-s 1) < 0 (·: 2-E 1 > 0, s 1 > 0), that is y f'(x) is negative for allx E (0, 8x) We note that f' (x) > 0 beforex= 0, f'(x) = 0 atx=Oandf'(x) < Oafterx = 0. The graph of f shows that it has relative maxima atx = 0. 0 (2,0) Thus we conclude that a function has relative maxima at x = c if f' (x) > 0, before x= c,f'(x)=Oandf'(x) < 0 afterx= c.
Calculus and Analytic Geometry
110
Considering an interval (2 - Ox, 2 + 8x) in the neighbourhood of x = 2 we find · the values of f'(2-s)andf'(2+s)when2-E E (2-bx,2)and2+E E (2, 2+bx) f'(2-s)=3(2-s)(2-s-2) [·: f'(x)=3x(x-2)] =3(2-E)(-E) =-3E(2-E) 0, 2 - E > 0) and f' (2+s)=3(2 +s)(2+s-2) =3s(2+s) > 0 (": E > 0, 2 +s > 0) We see that f'(x) < Obeforex= 2, f' (x) =O atx= 2·and f' (x)>O afterx=2: . It is obvious from the grapli that it has relative minima atx = 2. Thus we conclude that a function has relative minima at x = c if /' (x) < 0 beforex = c, f'(x) = 0 at x = c and/' (x) > 0 after x = c. First Derivative Rule: Let f be differentiable in neighbourhood of c wheref' (c) = 0 . 1.
If f' (x) changes sign from positive to negative as x increases through c, then /(c) is the relative maxima off
2.
Iff' (x) changes sign from negative to positive asx increases through c, then f'(c) is the relative minima off
Note: 1. A stationary point is called a turning point if it is either a maximum point or a minimum point. 2.If f'(x)>Obeforethepointx=a,f'(x)=Oatx = aandf'(x)> Oafterx =a, then f does not have a relative maxima. Y 3 See the graph off(x) = x • In this case, we have
f' (x) = 3x2 , that is,
f' (0-s)= 3(-E)2 = 3E2 > 0 and f' (0+s)=3(E) 2 = 3E2 > 0 The function f is increasing before x = 0 and also it is increasing after x = 0. Such a point of the function is called the point ofinflexion.
y'
Second Derivative Test: We have noticed that the first derivative f '(x) of a function changes its sign from positive to negative at the point where f has relative maxima, that is, f' is a decreasing function in the neighbouring interval containing the point where f has relative maxima.
Unit 2: Differentiation
111
Thus f" (x) is negative at the point wher '/ has a relative maxima.
Butj'(x) of a function/ch,'nges its sien from negative to positive at the point where f has relative minima, that is, f' is an increasing function in the neighbouring. interval containing the point where/ has relative minima. Thus /"(x) is positive at the pointwhue.f has relative minima. Second Derivative Rule
Let f be differentia1function in a n.ci ~hbourhood ofc where f' (c) = 0 . Then 1.
f has relative maxima at c if/"
2.
/hasrelativeminimaat c if f"(c) > 0.
(c) < 0.
Example 1: Examine the function defined as f(x) = x' - 6x' -t 9x for extreme values.
Solution:
f' (x) = J/ -
1ix +9
= 3(x"-4x+3)=3("t-l)(x-3) First Method Ifx= 1 -
i::, where i::
is v 0 ry very small positive number, then
(x- l)(x-3 , - (1--1;- l)(l-c: - i) -=-(-s)(-i:-2)=i::(2+i::) > 0 that is, f' (x) > 0beforcx =1 . Forx-:::· l + s, \Ve h:we (x-J)(x-3)=(1 +i::-1)(1
h::-3)~.=s(-2+i::)=-i::(2-i::) Obefore x= I,f'(x)=Oatx-= 1 andf'(x)< Oafterx= 1.
thus f has relative maxima at x = 1 and f( 1) = 1 - 6 + 9 = 4. Letx = 3 - i::, then (x - l)(x - 3) = (3 -
E-
1 )(3 - z -3) = (2 - E) (-E) = - E(2 - i::) < 0
thatis/'(x) < Obefore x= 3.
Forx=3+i:: (x-l)(x-3)=(3 + £-1)(3+i::-3)=(2+s)(-i::)>O that is, f'(x) > Oafterx=3.
Asf'(x) < 0 before x=3,f'(x) =OaLr=3and/'(x) > Oafterx=3, Thus f hasrelativeminimaatx=3, SecondMethod:
f" (x) = f" (1) =
2
and/(3)=3(3) -12(3) +9=0
3(2x-4)
=
6(x-2)
6(1-2)
=-6 < O,therefore,
Calculus and Analytic Geometry
f
hasrelativemaximaatx= 1 and/(1)=(1)3-6(1)2+9(1). =1-6+9=4
f"
= 6 > 0, therefore f has relative
(3) = 6(3 - 2)
minimaatx= 3 and/(3) = 27 -54 + 27 = 0
Example 2: Examine the function defined asf(x) = 1 + x3 for extreme values Solution: Given that/(x) = 1 + x3 Differentiatingw.r.t. 'x', we get ftx) = 3x 2
2
f'(x)=O
=>
3x =0
=>
f'(x)=6x
and
/"(0)=6(0)=0
x=O
The second derivative does not help in determining the extreme values.
f'(O-s)= 3(0-s}2=3s2 > 2
f'(O+s)=3 (O+s) =3s
2
O
>
O
As the first derivative does not change sign at x = 0, therefore (0, 0) is a point of inflexion. . 1 Example 3 : Discuss the function defined as f (x) = sin x + r;:; cos 2x for extreme
2v2
values in the interval (0, 27r ).
Solution: Given that f(x) = sinx +
f' (x) =cos x + =cos x-
~
2v2
(cos 2x) .
lr;:; (-2 sin 2x) =cos x - ~sin 2x 2v2 v2
}i
(2 sinx cos x) =cos x -
J2 sin x cos
(i)
x
=cos x (1- Ji sin x) Now f'(x)= 0
or
1-Jisinx=O
=>
cos x(l-{isin x)
=>
cosx
=0
=>
sinx
= J2
1
=>
=>
=0 1t 37t X=- 2' 2 1t 37t X=- 2' 2
Differentiating (i) w.r.t. 'x', we have
f"(x)
= -sinx -
}i (cos 2x) x 2 =-sin x - J2 cos 2x
Unit 2: Differentiation As
113
/"(7t) =-sin~ -J2cos n = -1- J2 x (-1)= J2 2
and
.
2
3
3
2
2
/" ( n) = - sin n - J2 cos 3n =
Thus f(x) has minimum values for x As and
/" (
~) = - sin :
-1>0 t
- J2 cos
-1 (-1) - J2 ( -1 )= 1 + J2 .
=~- and 2
3 x = n
2
~=- ~-
3 3 3 /" ( :) = - sin : - J2 cos ;
Thus f ( x) has minimum values for x =
=-
>0
J2. 0 = -
~-
~ 0 a_,_ l ~ _,_ 0) - 'A In a · ' ' -r- ' . I\, .....
f cosec x cot x dx = cosec x+c x
x
fe dx=e +c 1 fa dx= -Ina. a x
x
+c,
(a> 0, a* 1)
10.
11.
1 J-dx ax+b
=
-1
(ax+b) dx
x:t:O
= ~ lnlax+bl+c,(ax+b-:t-0)
Jtan (ax + b) dx = 1In Isec(ax+ b)I + c =-l. lnlcos(ax+b)l+c
a
12.
J~dx=lnlxl+c,
ftan x dx = In Isec xi + c =-lnlcos xl+c
Jcot(ax+ b)dx =~lnjsin (ax+ b)I +c
fcot x dx =Zn Isin xi + c
J
Jsec x dx =in jsec x+ tan xi + c
J
Jcosec xdx=ln lcosec x-cotxl+c
13. sec (ax+ b)dx = *lnlsec(ax +b) + tan(ax + b)I +c 14. cosec(ax + b)dx=~Znlcosec(ax + b)-cot(ax+ b)I + c
These formulae can be verified by showing that the derivative of th~ right hand side of each with respect to xis equal to the corresponding integrand.
Calculus and Analytic Geometry
126 Examples with solution:
3
2.
J
} -Gdx=f x 2
3
\JX-
3.
J
- - +I 2
I
--
x 2 2 dx=~l = -1 =- Fx +c x
-+ 2
-2
4 1
1
(2x+3)
4
dx = f (2x + 3)--4 dx =
(2x + 3r
+
=
(2x + 3f3
2(-4+1)
(
-6
1
=-
6(2x+3/
) =---((2x+3) l d _,- ) · :d- ( - · l 3 d>: 6(2x + 3) 6 dx
l -3 = --(-3)(2x+3)
J
(2)
6
4.
Jcos2xdx=~sin 2x+c
6.
Jcosec
2
d(l 1d 1 - -sin 2x ) =--(sin 2x)=-(cos ( · :dx 2 2d'r 2
x dx =-cot x+c
7. Jsec 5x tan 5x dx
= -sec5x -+c 5 .
+c
=
l
~x+~
4
J
2x) x 2=cos 2x )
d 2 2 ) ( ·: dx (-cot x) = -(-cosec x) = cosec x
(sec-5x) .S1(sec 5x tan 5x)
d ( ·: dx -
5
=
x 5 = sec 5x tan 5x
)
. Unit 3: Integration
127
ax+b ax+bdx e 8. Je =--+c - a
f3
3Ar
Ax
9.
10.
( ·:
d ·: -
dx=--+c
A.ln3
1
J
a.x+fJ
d e ax+b) 1 ax+ b ax+b ) dx ( -a- = ;_; (e x a)= e
3 1 ).x ).x -).x- ) = - ( 3 (/n3)A.)=3 )
dx
(
(
A.ln3
A.ln3
1
dx=f(a.x+bf dx=.I_ln(ax+b)+c, a
(a.x+b>O)
1 1 1 ) ·: -d(l-ln(ax+b)=-.--a=-. ( dx a a ax+b ax+b
11.
f
1
~ 2
,-;----;
dx=ln(x+'\/x- +a-)+c
x +a
2
I
d
·:-(ln(x+~))= (
dx
1
0)
flf(x)f1 f' (x) dx =lnf(x)+c,
Thus we can prove that n+l (i) xn dx = .!__ + c, n+l ndx (ax+bf+l (ii) ax+ b) = +c, a(n + 1)
J J(
(iii)
J_!_dx = lnjxj +c x
(iv)
J
1
(n "#-l) (a"# 0, n "# -1)
1
- - dx=- lnj(ax+b)j+c, ax+b a
(a"# 0)
Example 12 : Evaluate (i)
J-2) x+2 .
(iv)
Jdi $' (x>O) x+l- x 3-cos 2x (vii) - - dx, (cos 2x "#-1) f l+cos 2x Solution: J(x+l)(x-3)dx =J (x -2x-3)dx (i) (v)
Jx~x -I dx f $x(~ dx, (x>O) x+I) 2
•
(vi)
3
sm x +cos x f
2
•
cos x sm x
2
(By theorems I and II)
=Ji dx-2Jxdx-3fldx 3
2
x x =--2. --3.x+c 3 2
1 3 2 =-x -x -3x+c
3
n+l
·: Jxn dx=~ +cl and n+l J. Ji~= xo ~=T+c2
(
O+l
Unit 3: Integration (ii)
129
Jx.Jx -1 dx = J(x 2
=
I
2
-1)2 x dx
J[f(x)]
=~
x
(If
~ f' (x) dx
J[f (x)]~ f' (x) dx
f(x)
=
x' -1,
then/' (x) = 2x
~ x = ~ f' (x) J
3
= 1 [/(x)]2 + c = _!.(x2 +1)% =c 2 (iii)
f _!_dx=f x+2-2 dx, x+2 x+2
= (iv)
3 2
3 (x>-2)
J(1- x~ 2 )dx = J
J
1
dx-2 (x+2f .ldx=x-2 ln(x+2)+c
f J;cJ;+1) 1 dx
=f
=
1 _1 dx J;+1 · J; ·
(x>O)
f[f(x)r' . 2 f' (x)dx [·: J' (x)
=
=2f [f(x)r1 f' (x) dx =2/n f(x)+c
= 2/n(J;+l)+c
Rationalizing the denominator, we have
:
rx+i+~ dx f .Jx +dx1 -J; -f - (.Jx + 1 - J;)(.Jx + 1 + ~ =
J.Jx+i +.Jx dx= j[(x+I)t +xt] dx x+I-x
=J
I
f
I
2 (x +1)2 dx + x dx
if f (x) =Fx +I
or
(x > 0)
(v)
1 l
2
=2f'(x)]
Calculus and Analytic Geometry
130 3 -
3 -
(x + 1) 2 x 2 __.:.__--+- + c 3 3
2 .
3
2
= -(x+l)
-2
3
2
3
-2
+- x +c 3
2
3
sznx+cos x dx 2 • cos xsznx
(vi)
J
So1u tion ..
Jszn x +cos •
= J(
dx
3
•
2
cos xsznx
=
. x szn + cos 3 x• ) dx • 2 2 cos xsznx cos xszn x
J(-1 cos2
+cos x)dx sin x
= Jsec2 xdx+ Jcot xdx = tanx+lnlsin xl+c (cos 2x ":t:.- l)
- cos 2x dx, J31+cos 2x
(vii)
= J4-(I+cos
Solution: J3-cos 2x dx 1+cos 2x
2x) dx
1+cos 2x
=
J( +
4
1 cos 2x -
=J2 cos4 2x dx -Jldx =J2 sec
2
-JI dx
x dx
2
1) dx
= 2 tan x-x+c
EXERCISE 3.2 1.
Evaluate the following indefinite integrals (i)
J(3x
2
-2x+1) dx
(v)
Jx(...fx +I) dx, Jc...fx +1) cJx,
(vii)
f3~2
(iii)
(ix)
2
f(JO -1)2 dB To 2x
(xi)
dx,
Je e:e
x
dx
(ii)
J(Fx+ JJtx.
(x>O)
1
(x > 0)
(iv)
(x>O)
(vi)
(x>O)
(viii)
(8 >0)
(x)
J(2x+3)2 dx
J(Fx- JJ
dx, (x>O)
fJY(y +1) dy, y
so-ff>'
O) (x>O)
Unit 3: Integration · 2.
Evaluate
[x+a>OJ
(i)
dx J .Jx+a +.Jx+b' x+b > 0
(iii)
J
(v)
J(l+~.t)3 dx
(vii)
JJi - cos 2x dx,
(ix) (xi)
2
(iv)
Jl-x - - 2 dx, l+x 3 J (a-2x) 2 dx
(vi)
fsin (a+b) x dx
(viii)
J (In x) x .!.. dx, (x > 0) x
J sin x dx
(x)
J
J
(xii)
J cos 3x sin 2x dx
d
Fx,(x>O,a>O) x+a+ x ·
e (1 - cos 2x > O)
2
(xiii)
3.3
131
ax+b dx 2 ax +2bx+c Jcos 2x-1 dx, (1 +cos 2x l+cos 2x
-:t:.
0)
(ii)
n n)
1 dx ( -- 0) 2"'\/at+b
f
Solution:
Thus
Let
f
at+ b = u. Then adt=du
a dt _ 2.Jat+b -
J du
.
-I
=.!._Ju 2 du 2~ 2
=~[ :~~+: ]+c=~[ ~ ]+c
I
=u
2
J
+ c = at + b + c
Calculus and Analytic G~ometry
132 Example 2 : Evaluate
4+x
2 dx.
2
Solution: =>
Ji;l
j
Put 4 +x = /
2x dx
= di
j 4 + x2 x
1
x dx = - dt, therefore
or
2
j 1 ( 1) 1j dx = Ji 2 dt = 2 I
112
-1/2
1 1 di = 2·1I2 + c
2
=Ji +c = ~4+x +c
Example3: Evaluate j x.Jx-a dx, (x >a)
Solution: Thus
Let x - a= I =>
x = a+ t => . dx=dl
Jx.Jx-a dx = J(a+l)Ji di =
r(at~ +ti) 3 2
dt
=aft~ dt +fti dt
5· 2
3
5
2a 2 =a-+-+ c = -1 2 + -1 2 +c 3 5 3 5 t
2
1
2
~ ~t )+c = 2(x-a)i ( ~ + ~(x-a)) +c
=2ti ( + =2(x-a)
% (5a+3(x-a))
2
= -(x-a)
15
2 ~ +c =-(x-a) 2(5a+3x-3a)+c 15
~2
(2a+3x)+c 15 co1Fx Example4: Evaluate .£ dx, (x > 0)
I
Solution: Put Fx then d(.£)
or
= z, = dz =>
1 1
2vx
dx=dz
1
- - dx = 2dz Fx
Thus
1 dx = Jcot z. (2dz) .£ dx = Jcotvxr . .£ I co1Fx
Unit 3: Integration
133
f
f
f
COS Z =2 cot zdz = 2 . - dz= 2 (sin z) -I cos zdz szn z = 21njsin zl +c, (z > 0 as x > 0)
= 2 In lsinFxl + c
fcosec x dx (ii) fsec x dx x (cosec x-cot x) dx (.) fcosec x dx f cosec(cosec x - cot x)
Example 5 : Evaluate Solution:
(i)
=
1
Put or
2
cosec x cot x = t, then (-cosec x cot x + cosec x) dx cosec x(cosec x- cot x) dx = dt
= dt
x(cosec x-cot x) dx = f -1 d = 1n I I+c f cosec(cosec x-cot x) t fcosec x dx = In lcosec x - cot xi + c [·: t = cosec x -cot x]
so
t
Thus
t
x(secx+ tan x) dx (n. )Jsec x dx = Jsec(secx+tanx) Put or
(secx+tanx)
Solution : or
=In It l+c
t
Jsecx dx =In jsecx+tanxl+c
Example6: Evaluate Put
2
(secxtanx+sec x)dx = dt
Jsecx(secx+tanx) dx = J!dt
so Thus
'
sec x + tan x = t, then secx(secx+tanx)dx = dt
(·: t = secx+tanx)
Jcos3 x Jsin x dx,(sinx > 0).
~.cos x] dx
J sin x = t, then dt = [
2 szn x
[·: ~sinx = t
2t dt =cos xdx
Putting Jsinx =t . and cos x dx
= 2t dt in
Jco/ xJsinx cos x dx = J(I-t4).tx 2t dt, =2 J(t -r6) dt = 2 ft 2
/3
the integral, we have, 2
(·:co/ x = 1-sin =1-t 2
dt - 2
4 )
Jr6 dt
/7
=2.-2. -+c 3 7 3 7 3 7 2 2 -2 2 -2 2 -2 2 = -(sinx) --(sinx) +c = -sin x--sin x+c 3
7
3
7
--- -- -- ~- -- -- ~-- -- -- ~~-~- -- -~--- -
Calculus ana Analytic Geometry
134 Example 7 : Evaluate J
(-%
O, a :;t: 1)
= t, then x dx =
Thus a x x dx
2
1 dx = 1 . dt = t -3 dt = - t- + c 3 x t -2
. -
fa
t
1 dt 2
-
2'
x 1 dt I
x2
= .!_ Ja' dt = .!_ ~ + c . = _a_+ c. 2 2 Ina 21na Example IO: Evaluate (i)
J~
1 2
a -x
2
dx, (-an:
Let f(x);::: ln(x + ~i'+1) and g'(x)::;;; 1. Then
J
f'(x) =
2
1
[
1
2
- ~ x 1+-(x +1)
:r+vx- +I
2
.!._1 2
•
2x
]
Ix dx 5
2 -
x -c1
2
Calculus and Analytic Geometry .
140
=
x+~·(I+ ~J 1
=
f;2;1 +x = 2
x
~x
2
x+.Jx +1
+1
1
and
g(x) = x
2
.Jx +1
f
Using the formula f f(x) g' (x) dx = f(x) g(x)- g(x) f' (x) dx, we get
fln(x+.Jx' + 1).1 dx: = [ln(x+.Jx' +I)]. x- fx. ~ ,1
dx:
x +1
I
2
=(ln(x+.Jx +1).
x-~f (x2 +1)-2(2x)dx
~+1)- 2 +1)~ +c, xln(x+vx· 2
=
1 [(x ~
2
l
2
= xln(x+.Jx +1) -.Jx +1 +c, where c=-_!_c
2
Example 6: Evaluate
Solution:
fx
2 •
a eax dx 2
and
= 2x
and
If we put f(x) = x f' (x)
f
Using the formula f(x) g'(x) dx 2
f x .axax dx
But
I
= x 2 eax = x 2 eax
g' (x) = a eax, then g(x)
= eax
= f(x) g(x)-f g(x)f'(x)
-I eax.(2x)dx
-2J x eax dx fxem dx: = x( •: )- f(•: Jx I. dx: J
1 e ax - -1 e ax dx =-x 1 e ax --. 1 (eax = -x -
a
Thus
dx,weget
a
a
a
a
J
+c
1
fx' aem dx: = x' em -2[~ .x eQ - :, eQ +c J 1
= x2ea x- 2-a . x eux + -a2ax + c1 2 e
where c =-2c1
Unit 3: Integration
141
Example 7: Evaluate f eax cos bxdx and
g' (x) = cos bx
f' (x) = a.eax and
sin bx gx ( )=-b
Solution : Let
f(x)
then
= eax
Thus f e ax cos bx dx
J . bx dx sm bx - -a f eax sm
= eax x (sin - bbx) - - (sin - bbx) - x ( ae ax ) dx
= -1 eax
•
b
J
(I)
b
Integrating eax sin bx dx, by parts, we get
ax · b dx f e sm x
= eax x ( -
b bx) -
COS
J(-
b bx) x (ae ax ) dx+c1
COS
1 ax cos bx+af eax cos bx dx +c =--e 1
b
(II)
b
J
Putting the value of e= sin bx dxin(I), we get
ax dx 1 ax smbx-. a[ --e 1 ax cos bx+af e ax cos bx dx +c f e cos bx . =-e 1 b b b b
J
2
lax.b aax.cos bx -a- fax a =-e sm x +-e e cos bx dx --c b b2 b2 b I
( 2J a b
ax
.
or 1Je cos bx dx 2
.
lax. aax a = -e szn bx+-2 e cos bx--c1 b
Jeax cos bx dx= a b+b
.
1.e.,
2
2
2
b
[
e
2
a +b
2
J
lax. aax b a -e szn bx+ 2 e cos bx - 2 2 2 x-c1 b b a +b b
ax
=
b
(b sin bx+ a cos bx]+c, where c=
ab b(a2 + b2)
--=---C I
If we put a = r cos e and b = r sine, 2 2 2 2 2 then a +b = r => r = .Ja +b
b rsin e e => = tan -I -b = tan a rcos e a and a cos bx +b sin bx= rcos ecos bx+ r sine sin bx = r [cos bx cos a+sin bx sin0] = r cos (bx-0) - =
Calculus and Analytic Geometry
142 2 2 1 .Ja +b cos(bx-tan-
=
!)
The answer can be written as:
Je~ cos bx dx ~ ../a'l+b' e~ cos( bx-tan-' !)+c J.Ja
Example 8: Evaluate
J.Ja
Solution:
2
2
2
+ b dx
2
= (.Ja
+x .1 dx
2
2
+x )x-
fx.~(a
2
1
+x )-2 . 2x dx 2
2
=x.Ja2+x2-f
x dx .Ja2 +x2 2
=
2
2
t 2 2 fa +x -a dx xva +x .J 2 2 a +x
=x.Ja2+x2-f.Ja2+x2dx+f.J~
2J .Ja2+x
2
dx
= x.Ja
2
2
+x +a
2 •
J.Ja 1+x 2
2
2dx
a +x 2
dx
= x.Ja 2 +x 2 +a 2 [/n(x+.Ja 2 +x 2 )+c1] (See Example 1Article3 .4) 2
2
/2 2 x/2 2 a /2 2 ac va +x dx =-va +x +-ln(x+va +x )+c, where c=--1 2 2 2 2 2 2 2 Similarly integrals .Ja - x dx and .Jx - a dx can be evaluated.
f
f
J
Solution:·
f sin
4
x dx. 2 2 2 fsin xdx;;; fsin x.sin xdx= sin x(l-co/x)dx
Example 9 : Evaluate
f
4
= f szn. 2x dx - f·2 sm x cos 2x dx
I
2x 2 2 = 1- cos dx - sin x cos x dx 2 .
I
2
IntegratingJ stn xcos2 x dx by parts, we have
Jstn
2
co/ x dx = Jcos x sin x cos x dx 2
(I)
Unit 3: Integration ·
. 143
. 3 x szn =cos x ( - -
3
= -i cos x sin 3 3
J-J.- x x (-sin x) dx szn 3
[·: if f(x)
3
= szn. 2 xcos x.
g , ( x)
I
x + -i sin 4 x dx ... (II) 3
then f' (x)= -sinx and g(x)
J
= cos x and
= -szn. 3-x ] 3
2
Putting the value of sin x co/ x dx in (I), we obtain
Jsin
4
= J(~- co:
x dx
2
x)
dx-[~cos x sin
= .!.J 1 dx _ _!_Jcos or
(1+!) 3
2
4
Jsin x dx
2
2
I
3 szn ·2 x --1 cos = -3 x--
Solution:
i.e.,
f
4
ex (I +sin x) dx. 1+cosx
ex(l+sin x)dx--f f 1+cos x ex (I+ sin x) dx I+cos x
ex
(i
4
3
3
x +c1
.3 x sm x +c
'
J h 3 were c=-c 4 I
x x)
2
2
1 2 -+tan x =J ex ( -sec -x) dx 2
2
2
2
2
+ f e x tan -xdx 2
J(tan~ ).ex dx =(tan~ ).ex - Jex ( sec 1). t dx+ c, i.e., I e x tan =e x tan-2 --21Jx e sec +c 2 2 2
Xdx
Putting the value of
X
2Xdx
(I) (Integrating by parts)
(II)
J,/ tan ~ dx in (I), we get
tl (l +sin x) dx'"" .! Jtlse(i ~ dx+[tl tan ~-.!.Jtl sec 2 :_ dx + 1 + cos x 2 2 2 2 2
f
J
. + 2 szn-cos2 2 dx 2 x . [·:l+cosx=l+2cos --1] . 2 cos 2 -x
= -If e x sec 2X - dx
But
x dx
_!cos x sin 3 x 3
4
Example 10: Evaluate f
4
3
2
16
~ Jsin
3
f sm. x dx = 43[12x- 41.szn 2x- 31 cos x szn. 8
x+
2x dx-!cos x sin x-! Jsin x dx
= _!_ x-_!_(sin 2x)+c 2
3
c] ==tl tan ~+c 2
Calculus and A11alytic Geometry
144 Example 11 :Show that Solution:
fe"'[af(x)+ f'(x)] dx
= e
0 -'
f(x)+c.
Jeax[af(x)+f'(x)]dx .= Jeax af(x)dx+ Je=.f'(x)dx
(i)
In thesecondintegral, let a) f 2x 3,t2 - x - 7 dx 2.x2- 3x- 2 f x (x -2x-1)(xI - 3) dx f (1 + 4+7x x) 2(2 + 3x) dx 3
8.
dx
(x - 3)(x + 2)3
J (x +5x+ 8 3)(2.x- 1) dx J (x-{a-b}x a)(x - b) dx,(a > b)
(x + x + 1)(x2 + 2x + 3) dx
22. 24. 26. 28. 30.
J
3x-8 (.x2 - x + 2)(.x2 + x + 2) dx
Unit 3: Integration
3.7
151
THE DEFINITE INTEGRALS
We have already discussed in section 3.2 about the indefinite integral that is, if '(x) =fix), then
Jf(x) dx = 0. The graph off is shown in the figure. y
We define the function A(x) as the area above the x-axis and under the curve y = j(x) from a to x. Let ox be a small positive number and x + 8x be any number in the interval [a, b] such that a sin- 1x = t)
dt
7C
6 7C
= [~J: = ~[(f)2-(~)2] = t(~ -3~ ) 2
6
= t(
4n;6n2) = ~~2 = 2~
!! 6
fxcos xdx
Example 7: Evaluate
0
Solution:
Applying the formula
ff(x) '(x) dx = f(x) 0.
3
3.9 DIFFERENTIAL EQUATIONS An equation containing at least one derivative of a dependent variable with respect to an independent variable such as dy
y dx
+ 2x
(1·)
=0
2
or
xd y dy dxz + dx - 2x
=0
(ii)
is called a differential equation. Derivatives may be of first or higher orders. A differential equation containing only derivative of first order can be written in terms of differentials. So we can write the equation (i) as y dy + 2x dx = 0 but the equation (ii) cannot be written in terms of differentials. Order: The order of a differential equation is the order of the highest derivative in the equation. As the order of the equation (i) is one so it is called a first order differential equation. But equation (ii) contains the second order derivative and is called a second order differential equation.
3.9.1 Solution of a Differential Equation of First Order Consider the equation
y = Ai+4 where A is a real constant Differentiating (iii) with respect to x gives dy dx = 2Ax
(iii)
(iv)
... ) y-4 . h From ( m A = -y, so puttmg t e value of A in (iv), we get
Unit 3: Integration
xdy dx
169
= 2y-8
dy 2y-xdx
which is free of constant A
=8
(v) y
Substituting the value of y and its derivative in (v), we see that it is satisfied, that is, 2(Ai + 4) - x(2Ax) =2Ai + 8 - 2Ax =8 which shows that (iii) is a solution of (v) 2
Giving a particular value to A, say A =-1, we get y
= -i+4 Y'
We see that (v) is satisfied if we put y =-i + 4
and~ =- 2x, soy= -x
2
+4
is also a solution of (v). · For different values of A, (iii) represents different parabolas with vertex at (0, 4) and the axis along the y-axis. We have drawn two members of the family of parabolas. y = Ax + 4 for A = -1, 1 All solutions obtained from (iii) by putting different values of A, are called particular solutions of (v) while the solution (iii) itself is called the general solution of (v). 2
A solution of differential equation is a relation between the variables (not involving derivatives) which satisfies the differential equation.
Here we shall solve differential equations of first order with variables separable in the forms dy dx
f(x)
= g(y)
or
dy dx
=
g(y) f(x)
Example 1: Solve the differential equation (x- 1) dx + y dy =0 Solution: Variables in the given equation are in separable form, so integrating either terms, we have
, Calculus and Analytic Geometry
170
f (x or
1) dx +
f y dy = c
where c 1 is a constant
1,
(~ -x) + t = c,' x2 -
2x + y2
= 2c
which gives
1
Thus,the required general solution is Example 2: Solve differential equation 2 dy x (2y + 1) dx =1 = 0
x2 + y2- 2x = c,
where c = 2c 1•
Solution: The given differential equation can be written as 2 2 I) dx dy x(y+
=1
Dividing by x , we have (2y + 1) ~ 2
(1·)
=
;1,
(x
* 0)
(ii)
Multiplying both sides of (i) by dx, we get
=
(2y + 1) ( : dx)
or
:2dx
1 = -x2 dx
(2y + 1) dy
Integrating either side gives
f(2y+l)dy
= f ;2dx
or Thus, y2 + y
x1 is the general solution of the given differential equation.
=c -
Example 3: Solve the differential equation
~:
- 2y
=
0 x i- 0, y > O
Solution: Multiplying the both sides of the given equation by ..! dx, gives y
~ (~ dx J - 2x dx = 0
or
~ dy 2
Now integrating either side gives Zn y = x + c 1 or
x2+ c
y = e
Thus, y = cex'
x
2
= 2x dx
( ·: :
where c, is a constant
c
'= e .e' where ec,
dx =dy J
=c
is the required general solution of the given differential equation.
Unit 3: Integration
Example 4: Solve
171 dy y2 +I dx = ~
Solution: Separating the variables, we have 1
1
= e-x
y2 + 1 dy
dx = exdx
Now integrating either side gives
tan or
-I
y
x
y= e + c,
where c is a constant.
= tan(ex + c)
which is the general solution of the given differential equation.
2
Example 5: Solve 2ex tan y dx + (1- e} sec y dy
=0
0 0)
=>
In p kt
= kt+ c 1
c
' =e.e ' c
(where e ' = c)
(i)
Applying the given condition. that is p =JOO when t
100
= ce k=
(': e0
c
= 1)
Putting c = 100. (i) becomes p = 100 l'
(ii)
p will be 200 when t = 2 (hours), so (ii) gives 200 = 100 e2" => or
2k
= ln2
=>
e 2k = 2
k
1
= 2 In 2
1 Substituting k =2 In 2 in (ii), we get
If t
p
= 100 ed ln2)t = lOO efr In 2 = lOOe'n ( 2 t)
p
= 100 (2t)
=4 (hours), then p = 100 (1!) = 100 x 4 =400.
= 0. we have
Calculus and Analytic Geometry
176
Example 11: A ball is thrown vertically upward with a velocity of 1470 cm/sec, Neglecting air resistance, find (i)
velocity of ball at any time t
(ii)
distance traveled in any time t
(iii) maximum height attained by the ball.
Solution. (i)
Let v be the velocity of the ball at any time t, then by Newton's law of motion, we have dv dt = - g or
=>
Jdv = J-g dt
(i)
dv = - g dt (Integrating either side of (i))
v =-gt+ c,
(ii)
Given that v = 1470 (cm/sec) when t = 0, so 1470 = - g(O) + c1 => (ii)
c1 = 1470
Thus (ii) becomes v = - gt+ 1470 = 1470- 980 t Let h be the height of the ball at any time t, then dh dt or
dh h
= 1470 - 980 t = (1470 - 980 t) dt r = 1470 t - 980 2 + c 2
( ·.. v
(taking g = 980)
= ddht)
= 1470 t - 490 t2 +
C2
(iii)
h = 0 when t = 0, so we have
= 1470 X 0- 490(0)2 + c 2 =0 in (iii), we have
0 Putting c 2 (iii)
h = 1470t-490t2 The maximum height will be attained when v =0, that is 1470 3 1470- 980 t= 0 => t = 980 = 2 (sec) Thus the maximum height attained in (ems)
= 2205 -
1102.5
=
1102.5
= 1470 x ( ~ J - 490 x ( ~
J·
Unit 3: Integration
177
EXERCISE 3.8 1.
Check that each of the following equations written against the differential equation is its solution. (i)
dy xdx =l+y
(ii)
dy 2 x (2y + 1) dx - 1
= cx-1
y 2
=0
y +y
= c-~1
dy 2 (iii) y -dx - ex
=1
l=e2x+2x+c
1 dy (iv) ~ dx -2y
=0
y
dy _ dx -
(v)
l±.1
= =
y
e-x
tan (ex + c)
Solve the following differential equations:
2. 4. 6. 8.
dy dx
3. y dx + x dy = 0
= -y
dy dx
=
dy 5. dx
1 -x
y
. dy sin y cosec x dx = 1 ~+]
Y+1
(x, y
> 0)
2 dy 2 2xydx=x-1
12.
(x
13.
sec x tan y dx + sec y tan x dy
14.
dy) ( y -x dx
-
1 dy 9. ~ dx
dy 2 yx ) dx + y2 + xi = 0
2
1
2
= 2 ( y2 + dy) dx
=0
2
= 2 (1 + y)
dy 2.xy 11• dx + 2y + 1
10.
2
(y > 0)
7. x dy + y (x - 1) dx =0
x dy
= Y. dx '
y
= x2 ,
=x
Calculus and Analytic Geometry
178 15.
dy
1 + cos x tan y dx
=0
16.
=0
17.
secx +tan y dy dx
18.
(e
19.
Find the general solution of the equation : - x
x
dy x +e -x) -=e -e -x dx
Also find the particular solution if y
= x/
= 1 when x =0.
20.
dx Solve the differential equation dt = 2x given that x = 4 when t = 0.
21.
Solve the differential equation
!~ + 2st =0. Also find the particular solution if
s = 4e, when t = 0. 22.
In a culture, bacteria increases at the rate proportional to the number of bacteria present. If bacteria are 200 initia11y and are doubled in 2 hours, find the number of bacteria present four hours later.
23.
A ball is thrown vertically upward with a velocity of 2450 cm/sec. Neglecting air resistance, find (i)
velocity of ball at any time t
(ii)
distance traveled in any time t
(iii) maximum height attained by the ball.
Unit 4: Introduction to Analytic Geometry
Unit 4.1
179
·Introduction to Analytic Geometry
INTRODUCTION
Geometry is one of the most ancient branches of mathematics. The Greeks systematically studied it about four centuries B.C. Most of the geometry taught in schools is due to Euclid who expounded thirteen books on the subject (300 B.C.). A French philosopher and mathematician Rene Descartes (1596-1650 A.D.) introduced algebraic methods in geometry which gave birth to analytical geometry (or coordinate geometry). Our aim is to present fundamentals of the subject in this book.
Coordinate System I
Draw in a plane two mutually perpendicular number lines x'x and y y, one horizontal and the other vertical. Let their point of Y intersection be 0 which we call the origin and the real number 0 of both the lines is represented by 0 . The two lines are ca1led the coordinate axes. The horizontal Jine x'Ox is caJled the x -----------x x-axis and the vertical line y'Oy is called the yO axis. As in the case of number line, we fo11ow the convention that all points on the y-axis above y y x'Ox are associated with positive real numbers, those below x'Ox with negative real numbers. p (x, y) Similarly, all points on the x-axis and Jying on the s right of 0 will be positive and those on the left of 0 and lying on the x-axis will be negative. Suppose P is any point in the plane. Then x _ _ _.._..._..._x_.._..._..._:- - - - • _x P can be located by using an ordered pair of real O R numbers. Through P draw lines parallel to the y coordinates axes meeting x-axis at R and y-axis at
S. Let the directed distance OR = x and the directed distance OS = y. The ordered pair(x, y) gives us enough information to locate the point P. Thus, with very point P in the plane, we can associate an ordered pair of real numbers (x, y) and we say that P has coordinates (x, y). It may be noted that x and y are the directed distances of P from the y-axis and the x-axis respectively. The reverse
180
Calculus and Analytic Geometry
of this technique also provides a method for associating exactly one point in the plane with any ordered pair (x, y) of real numbers. This method of pairing off in a one-toone fashion the points in a plane with ordered pairs of real numbers is ca11ed the two dimensional rectangular (or Cartesian) coordinate system. If (x, y) are the coordinates of a point P, then the first number (component) of the ordered pair is called the x-coordinate or abscissa of P and the second member of the ordered pair is called they-coordinate or ordinate of P. Note that abscissa is always first element and the ordinate is second element in an ordered pair.
The coordinate axes divide the plane into four equal parts called quadrants. They are defined as follows:
y
II
Quadrant I: All points (x, y) with x > 0, y > 0 Quadrant II: All points (x, y) with x < 0, y > 0
I
xI
x
Quadrant III:All points (x, y)with x < 0, y < 0
v
I
Quadrant IV:All points (x, y) with x > 0, y < 0 The point P in the plane that corresponds y' to an ordered pair (x, y) is called the graph of (x, y). Thus given a set of ordered pairs of real numbers, the graph of·the set is the aggregate of all points in the plane that correspond to ordered pairs of the set. }'
Challenges!
ffl I\'
I
1-
Write down the coordinates of the points if not mentioned.
~
v
,_
x' ~
(
-
'
."
x
'
/. f I
-1 , -~
1T 11-
Locate (0, -1), (2, 2), (-4, 7) and (-3, -3)
n 1,.
A
K
iii
y'
Unit 4: Introduction to Analytic Geometry
4.1.1
181
The Distance Formula Let A(x,,y.) and B(x2,yJ be two points in the
plane. We can find the distance d =
Note that: AB stands for mAB or IABI
IAB Ifrom the right
triangle AQB by using the Pythagoras theorem, we have d2 = AB2 = AQ2 + QB2 .
(1)
y '
IAQI =IRSI = IRO +OSI =1-0R+osl =lx2 -x,I
R A(J I' Y1
IQBI = ISB - SQI = IOM - ONI =ly2-Y,i
)__
M i--
r
.
i....---~
~
N
-
-n
I R
Iv )
;:.
~
0
x
Therefore, (1) takes the form di =(x2-x1)2+(y2 -y1)2 (2)
which is the formula for the distance d. The distance is always taken to be positive and it is not a directed distance from A to B when A and B do not lie on the same horizontal or vertical line. · If A and B lie on a line parallel to one of the coordinate axes, then by the formula (2), the distance AB is absolute value of the directed distance AB . The formula (2) shows that any of the two points can be taken as first point. Example 1: Show that the points A(-1,2), B(7,5) and C(2,-6)are vertices of a right triangle. y Solution: ~t, a, b and c denote the lengths of the ·~ sides BC, CA and AB respectively. l
i....ol"'J
By the distance formula, we have . 2
c= AB=~(7-(-1)) +(5-2)
=J73
a=BC=~(2-7) 2 +(-6-5) 2 b
~
~,
2
=CA= ~(2-(-1))2 +(-6-2)
=.J146 2
= J73
~...-
'"') _, I-'
I
' ,.
I
\
1
J \
J
l 2
2
Clearly: a = b +c
2 •
Therefore, ABC is a right triangle with right angle at A.
l ' J
I
-11
-x
Calculus and Analytic Geometry
182
Example 2: The point C(-5,3) is the centre of a circle and P(7,-2) lies on the circle. y What is the radius of the circJe? Solution:
The radius of the circle is the distance from Cto P. · C(-5.3)
By the distance formula, we have
Radius= CP=~(7-(-5)) 2 +(-2-3) 2 =
.J144 + 25 = 13
4.1.2 Point Dividing the line segment in a Given Ratio Theorem 1: Let A(xl'y1 ) and B(x2 .y2 ) be the two given points in a plane. The coordinates of the point dividing the line segment AB in the ratio k 1 : k 2 are
l
(
k1x2 + k2x1, k1Y2 + k2Y1 kl +k2 kl +k2 )
Proof: Let P(x, y) be the point that divides AB in the ratio k1 : k 2
From A, B and P draw perpendiculars to the x-axis as shown in the figure. Also draw BC J_AQ. Since LP is paralJel to CA, in the triangle ACB, we have k
AP
CL
QM
x-x
y
1 -1= - = - = - - = - k2 PB LB MR x2 - x
So,
k,
-
x-x, x. -x
= k 2 x-k 2 x 1
or
k 1x 2 -k1x
or
(k 1 + k 2 )x = k1x 2 + k 2x 1
or
k1X2 + k1X1
Q
0
x=---k1
M
R
x
+k2
Similarly, by drawing perpendiculars from A, B and P to the -y-axis and proceeding as before, we can show that
k1Y2
+ k2Y1
y = ---"----"'-----':__:_ kl +k2
Unit 4: Introduction to Analytic Geometry Note: (i) (ii)
183
If the directed distances AP and PB have the same sign, then their ratio is positive and Pis said to divide AB internally. If the directed distances AP and PB have opposite signs i.e., Pis beyond AB, then their ratio is negative and Pis said to divide AB externally. AP k AP k1 - = - ' or -BP k1 PB k2
Proceeding as before, we can show in this case that X
= k1Xz- kzX1 kl -k2
v = k1Y2 - k1Y1 . kl -k1
'
Thus P is said to divide the line segment AB in ratio k : k , internally or externally according as Plies between AB or beyond AB . (iii) If k 1 : k2 =I:1, then P becomes midpoint of AB and coordinates of P
(iv) The above theorem is valid in whichever quadrant A and B lie.
Example 3: Find the coordinates of the point that divides the join of A(-6,3) and B(5,-2) in the ratio 2 : 3.
internally
(i)
Solution:
(i)
Here k, = 2, k1
(ii) externally
= 3,
x, = -6, x2 = 5 .
By the formula, we have
x=
2x5+3x(-6) 2+3
=
-8 5
Coordinates of the required point are ( -
(ii)
y
and
= 2(-2) + 3(3) = 1 2+3
8 5
,I)
In this case
x= 2x5-3x(-6) =-28 2-3
and y = 2(-2)-3(3) 2-3
=13
Thus the required point has coordinates (-28, 13). Theorem 1:
The centroid of a MBC is a point that divides each median in the ratio 2 : 1. using this show that medians of a triangle are concurrent.
Calculus and Analytic Geometry
184
Proof: Let the vertices of a MBC have coordinates as shown in the figure. Midpoint of BC is D X2+X3 , Y2+Y3J . ( 2 2 Let G(x, y) be the centroid of the ~ Then G divides AD in the ratio 2: 1. Therefore
X2 + X3 + l.x 1 2 x=------= 2+1 2.
. . l SHill1ar y, y
XI
+X2 + X3 3
B "'-----....___ ____.. C
= Y1 + Y2 + Y3
D
(Xi, Yi)
3
(X3, Yi)
In the same way, we can show that coordinate of the point that divides BE and CF x 1 +x +x 3 y 1 + y + y 3 ] )• , each in the ratio 2 : 1 are ( ; ;
Thus (x, y) lies on each median and so the medians of the MBC are concurrent.
Theorem 3: Bisectors of angles of a triangle are concurrent. Proof:
Let the coordinates of the vertices of a triangle be as shown in the figure. Suppose
JscJ = a, JcAJ = b and JABJ = c
Let the bisector of LA meet BC at D. Then D divides BC in the ratio
.
c : b . Therefore coordmates of D are
(ex +bx 3
b+c
2
,
cy 3 +by 2 b+c
J
The bisector of LB meets AC at I and 1 - - - - - - - - - - - - - - - - divides AD in the ratio c:
JBDJ _c or JDCJ-b IBDI-~ JDCJ+JBDJ _b+c JEDI c __!!____ = b +c or IBDI = ~ IBD/ c b+c
Now
or or
JBDI IDCl _b
Unit 4: Introduction to Analytic Geometry
185
Thus I divides AD in the ratio c: ___!!!!___ or in the ratio b + c: a b+c Coordinates of I are
(b by2 + cy3 (b + C) bx2 + cx3 -----=b_+'-c=---_+_ax_1 ' + c) b + c + ay, a+b+c a+b+c [ i.e.,
(
ax, +bx2 +cx3 a+b+c
,
ay, +by2 +cy3 a+b+c
]
)
The symmetry of these coordinates shows that the bisector of LC will also pass through this point. Thus the angle bisectors of a triangle are concurrent.
EXERCISE 4.1 Describe the location in the plane of the point P (x, y) for which
1.
.
(i)
x >0
(ii)
x > 0 and y > 0
(iii)
x =0
(iv)
y=0
(v)
x < 0 and y
(vi)
x=y
(vii)
lxl = -IYI
(viii)
Ix!~ 3
(ix)
x > 2 and y
(x)
x and y have opposite signs.
-··' (i)
the distance between the two given points
(ii)
midpoint of the line segment joining the two points
=2
A( - 8'
3); 8(2, -1) (c)
A(-.JS.-~} B ~ 3,/5,s)
Which of the following points are at a distance of 15 units from the origin? (a)
4.
0
Find in each of the following
,a) A(3, 1); B (-2,-4) (b)
3.
~
(.Ji76,7)
(b)
(10, -10)
(c)
(1, 15)
(d)
1 1 ( ;, ; )
Show that (i) the points A(O, 2), B
(J3,-1)
and C(O, - 2) are vertices of a right triangle.
186
Calcitlus and Analytic Geometry (ii) the points A(3, 1), B (-2,-3) and C(2, 2) are vertices of an isosceles triangle.
(iii) the points A(5, 2), B(-2, 3), C(-3,-4) and D(4, -5) are vertices of a parallelogram. Is the parallelogram a square? 5.
The midpoints of the sides of a triangle are (1, -1), (-4, -3) and (-1, 1). Find coordinates of the vertices of the triangle.
6.
Find h such that the points A(.J3,-1), B(O, 2) and C(h, -2) are vertices of a right triangle with right angle at the vertex A.
7.
Find h such that A(- I, h), B(3, 2) and C(7. 3) are col1inear.
8.
The points A (-5,-2) and B(5,-4) are ends of a diameter of a circle. Find the centre and radius of the circle.
9.
Find Ii such that the points A(h,l), B(2,7) and C(-6,-7) are vertices of a right triangle with right angle at the vertex A.
10.
A quadrilateral has the points A(9, 3), B( - 7 , 7), C ( -3,-7) and D(5, - 5) as
its vertices. Find the midpoints of its sides. Show that the figure formed by joining the midpoints consecutively is a parallelogram. 11.
Find h such that the quadrilateral with vertices A( -3, 0), B(l, -2 ), C(5, O) and D(l, h) is parallelogram. Is it a square?
12.
If two vertices of an equilateral triangle are A(-3, 0) and B(3, 0), find the third vertex. How many of these triangles are possible?
13.
Find the points trisecting the join of A( -:-1, 4) and B(6, 2).
14.
Find the point three-fifth of the way along the line segment from A( - 5, 8) to B(5. 3). _
15.
Find the point Pon the joint bf A(l, 4) and B(5, 6) that is twice as far from A as Bis from A and lies (i) on the same side of A as B does.
16.
(ii) on the opposite side of A as B does.
Find the point which is equidistant from the points A(5, 3), B( - 2, 2) and C (4, 2). What is the radius of the circumcircle of the L\ABC ?
17. The points (4, -2), (-2, 4) and (5, 5) are the vertices of a triangle. Find in-centre of the triangle. 18. Find the points that divide the line segment joining A(x 1 , y 1 ) and B(x , y ) into 2 2 four equal parts.
Unit 4: Introduction to Analytic Geometry
187
4.2 TRANSLATION AND ROTATION OF AXES Translation of Axes Let .xy-coordinate system be given and O' (h, k) be any y P(x, v) or P(X , Y) y ' . point in the plane. Through O' draw two mutually perpendicular r: lines O'X, O'Y such that O'X is ······K· ····· parallel to Ox. The new axes O'X --+-----1--~--•X O'(h , k) M ': and O'Y are called translation of y the Ox- and Oy-axes through the .... . ...'. ·······X···· ···· . . .......... . point Q'. In translation of axes, origin is shifted to another point •--+-0-(0-,--) ____N.__ _ ___.M_ _ _ __._ x 0 in the plane but the axes remain parallel to the old axes. Let P be a point with coordinates (x, y) referred to xy-coordinate system and the axes be translated through the point O' (h, k) and O'X, O'Y be the new axes. If P has coordinates (X, Y) referred to the new axes, then we need to find X, Y in terms of x,y. Draw PM and O' N perpendiculars to Ox.
From the figure, we have OM=x,MP=y, ON =h, NO'= k
Now
X
Similarly,
Y
= =
O'M' M'P
= =
=
NM
MM'
=
OM - ON
MP-MM'=
=
x- h
y-k
Thus the coordinates of P referred to XY-system are (x - h, y - k) i.e.,X
= x-h
Y=y-h
Moreover, x
=
X + h, y
=
Y + k.
Example 1:
The coordinates of a point P are (-6, 9). The axes are translated through the point O' (-3, 2). Find the coordinates of P referred to the new axes.
Solution:
Here
h =-3, k = 2
Coordinates of Preferred to the new axes are (X, Y)given by X= -6-(-3)
Thus
P (X, Y)
=
P (-3, 7).
= -3
and
Y= 9-2
=7
188
Calculus and Analytic Geometry
Example 2:
The xy-coordinate axes are translated through the point O' (4, 6). The coordinates of the point Pare (2, -3) referred to the new axes. Find the coordinates of P referred to the original axes.
Solution: Here X = 2, Y = -3, h = 4, k = 6. We have
x=X+h=4+2=6
y=Y+k=-3+6=3 Thus required coordinates are P (6, 3).
Rotation of Axes Let xy-coordinate system be given. We rotate Ox and Oy about the origin through an angle 8 (0 < e < 90°) so that the new axes are OX and OY as shown in the figure. Let a point P have coordinates (x, y) referred to the xy-system of coordinates. Suppose P has coordinates (X, Y) referred to the XY-coordinate system. We have to find X, Y in terms of the given coordinates x, y. Let a be measure of the angle that OP makes
with Ox.
y
P(x,y)or P(X, Y)
y
x
x
·
From P, draw PM perpendicular to Ox and PM' perpendicular to OX. Let IOPI
= r.
From the right triangle L1 OPM: we have
8)]
OM' = X = rcos (aM'P = Y = rsin(a- 8) Also from the L1 OPM, we have
(1)
x = r cos a, y = r sin a System of equations (i) may be re-written as:
(2)
X Y
= r cos a cos()+ r sin a sin()] = r sin a cos() - rcosa sin()
Substituting from (2) into the above e~uations, we have x = x cos e + y sin y = y cos e - x sin e
8]
i.e., (X, Y)
= (x cos 8 + y sin 8, -
(3)
x sin 8 + y cos fJ)
are the coordinates of Preferred to the new axes OXand OY
---~---~---~~--------------~------..,____ _ _ _
Unit 4: Introduction to Analytic Geometry
189
Example 3: The xy-coord!nate axes are rotated about the origin through an angle of 30°. If the xy-coordinates of a point are (5, 7), find itsXY-coordinates, where OX and OY are the axes obtained after rotation. Solution: Let (X, Y) be the coordinates of Preferred to the XY-axes. Here From equations (3) above, we have X
=
X
=~ 7 2 + 2
an
d
y =
= ( 5j32 + 7 ,
- 5+ 2
113 }
.
i.e., (X, y)
5 cos 30° + 7 sin 30° and Y
=
e = 30°.
-5 sin 30° + 7 cos 30°
?Ji 2
-_J_
2 +
are the required coordinates. Example 4: The xy-axes are rotated about the origin through an angle of arctan
4
3
lying in the first quadrant. The coordinates of a point P referred to the new axes OX and OY are P (-1, - 7). Find the coordinates of Preferred to the xy-coordinate system. Solution: Let P (x, y) be the coordinates of Preferred to the xy-coordinate system. 4 4 3 Angle of rotation is given by arctan 8 = j . Therefore, sin 8 = S , cos 8 = S. From equations (3) above, we have = x cos () + y sin and
x
e
3 4 -1 =- x + - y
y = - x sin e+ y cos e 4
3
- 7 =- - x + - y 5 5 5 5 or 3x + 4y + 5 = 0 => -4x + 3y + 35 = 0 Solving these equations, we have ___L_ .1_ x 125 = - 125 = 25 => x = 5 , y = - 5 or
and
Thus coordinates of Preferred to the xy-system are (5, -5).
EXERCISE 4.2 1.
The two points P and O' are given in xy-coordinate system. Find the XY-coordinates of P referred to the translated axes O'X and O'Y . . (i) P (3, 2);
O' (1, 3)
(iii) P (-6, -8);
O' (-4, -6)
(ii)
P (- 2, 6); O' (-3, 2)
. (32' 25) ;
(IV)
p
I
O
(
1 7)
-2' l
190 2.
Calculus and Analytic Geometry
The xy-coordinate axes are translated thr 0
c and the expression (2) have opposite signs. Thus 0(0, 0) and P(5, - 8) are on the opposite sides of (1).
Note: To discuss whether a point P(x 1 , y 1 ) lies above or below the line ax+by+c=O we make the co-efficient of y as positive by multiplying the equation by (-1) if needed
4.4
TWO AND THREE STRAIGHT LINES For any two distinct and lines 11 , 12 .
11 : a 1x + b1y + c = 0 and 12 : a 2 x + b2 y + c = 0 one and only one of the following holds: (i) l1 11 12 (ii) l 1 ..L 12 (iii) l 1 and l 2 are not related as (i) or (ii). The slopes of 11 and 12 are m1 =-
~1
,
m2
l
(i) 11 II .12 ~ slope of 11 (m 1 ) = slope of 12 (m 2 ).
=- ~2 . 2
Recall that: Two non-parallel lines intersect each other at one and only one point.
Unit 4: Introduction to Analytic Geometry a
b
a2
b2
1 1 -=-
207
a 1b2 -b1a2 =0
(ii) 11 J_ 12 m 1m 2 = -1
(-::I-:}-1
=0
a,a, +b,b,
(iii)If 11 and 12 are not related as in (i) and (ii), then there is no simple relation of the above forms.
4.4.1 The Point of Intersection of Two Straight Lines Let 11 : a1x +b1y + c1 = 0 (1) and 12 : a 2 x+b2 y + c 2 = 0 (2) be two non-parallel lines. Then a 1b2 -b,a 2 ::t 0 Let P (x 1 , y 1 ) be the point of intersection of 11 and 12 • Then (3) a2X1
+ h2Y1 + C2 = 0
(4)
Solving (3) and (4) simultaneously, we have X1
Y1
b1C2 -b2 c 1
a 2 c1 -a1c 2
---=---=
1.e.,
x1 =
1 =----
a1b2 -a 2b1
qc2 -b2c1 d a c - a 1c 2 an y 1 = 2 1 a 1b2 -a2q a 1b2 -a 2b1
is the required point of intersection. ·
INote:
Recall that: Two non-parallel lines intersect each other at one and only one point.
a 1b2
-
a 2 b1 ::t 0, otherwise 11 j j 12 •
I
Examples 1: Find the point of intersection of the lines ....-----------~ 5x + 7 y = 35 (i) Remember that:
3x-7y =21
(ii)
* If the lines are parallel,
then solution does not exist (·: a 1b2 - a 2 b1 = 0) * Before solving equations one should ensure that lines are not Setting this value of x into (1), we find y = 0. parallel. Thus (7, 0) is the point of intersection of the two lines.
Solution: We note that the lines are not parallel and so they must intersect at a point. Adding (i) and (ii), we have or 8x=56 x=7
'--=-----------
208
4.4.2
Calculus and Analytic Geometry
Condition of Concurrency of Three Straight Lines
Theorem 1: Three non-parallel lines 11 : a 1x+b 1y+c 1 =0
(1)
12 : a 2 x+b 2 y+c 2 =0
(2)
a 3 x+b3 y+c3 =0
(3)
/3 :
a,
b,
ct
are concurrent iff a1
b2
C2
a3
b3
C3
Proof: P(x 1 , y 1 )
=0
If the lines are concurrent then they have a common point of intersection say. As 11 .H' 12 , so their point of intersection (x, y) is b1C2 -b2C1
x = - - - - and a1b2 -a2b1 This point also lies on (3), so a3 ( or
c
J (a c-a c J _0
b 1 2 -b2 c1 + b3 2 1 1 2 + C3 a1b2 - a 2 b1 a 1b2 - a 2 b1
a 3 (b1c 2
- b2 c 1 )
-
+ b3 (a 2 c 1 -a 1c 2 ) + c 3 (a 1b2 - a 2 b 1 ) = 0
An easier way to write the above equation is in the following determinant form:
a2
bl b2
c2
a3
b3
C3
al
Cl
=0
This is a necessary and sufficient condition of concurrency of the given three lines. Example 2: Check whether the following lines are concurrent or not. If concurrent, find the point of concurrency. 3x - 4 y - 3 = 0 (1)
5x+l2y+l=O
(2)
32x + 4 y -17 = 0
(3)
Solution: The determinant of the coefficients of the given equations is
3
-4
-3
5 32
12 4
1 -17
=
18
32
0
5 117
12 208
1 , by R1 + 3R 2 o and R3 + 17 R 2
Unit 4: Introduction to Analytic Geometry
=-1
18
32
117 208
209 '
= -(208xl8-117x32) =0
Thus the lines are concurrent. The point of intersection of any two lines is the required point of concurrency. From (I) and (2), we have x y 1 ---= =---4+36 -15-3 36+20 32 4 -18 -9 x =-56 = and y= = 28 1.e., 2' 28
7
. [4 -91
56
is the point of intersection.
4.4.3 The Equation of Lines Through the Point of Intersection of Two Lines We can find a family of lines through the point of intersection of the two nonparallel lines 11 and l 2 • Do you remember? Let l :a x+b y+c =0 1
1
1
(1)
1
(2)
For a non-zero real h , the equation a 1x+b1 y+c 1 +h(a 2 x+b2 y+c 2 ) =0 (3)
An infinite number of lines can pass through a point.
This, being a linear equation, represents a line. For different values of h,(3) represents different lines. Thus (3) is a family of lines. If (xp y 1 )
is any point lying on both (I) and (2), then it is their point of
intersection. Since (x1 , y 1 ) lies on both (I) and (2), we have a 1x+b1 y+c 1 =0
and
a 2 x 2 +b2 y 1 +c 2 =0
From the above two equations, we note that (x1, y 1 ) also lies on (3). Thus (3) is the required family of lines through the point of intersection of (1) and (2). Since h can assume an infinite number of values, (3) represents an infinite number of lines. A particular line of the family (3) can be determined if one more condition is given. Example 3: Find the family of lines through the point of intersection of the lines
3x - 4 y -10 = 0
( 1)
=0
(2)
x + 2 y -10
Find the member of the family which is
210
Calculus and Analytic Geometry (i)
(ii)
-2
parallel to a line with slope -
3 perpendicular to the line l : 3x - 4 y + 1 = 0 .
Solution: (i) A family of lines through the point of intersection of equations (1) and (2) is 3x-4y-10+k(x+2y-10) = 0 or
(3 + k)x +(-4 + 2k)y + (-10-lOk) = 0
. b Slope m of (3) is given y: m
(3)
= - -_-3+k -+--k
4 2 This is the slope of any member of the family (3). . withs . 1ope - -2 then If (3) is parallel to the lme 3 3+k -2 =or 9 + 3k = -8 + 4k i.e., k = 17 -4+2k 3 Substituting k = 17 into (3), equation of the member of the family is 20x+30y-180=0 i.e., 2x+3y-18=0. (ii) Slope of 3x- 4y + 1=0 (4) is i . Since (3) is to be perpendicular to (4 ), we have 3 + k x i = -1 4 -4+ 2k 4 or 9 + 3k = -16 + 8k or k = 5 Inserting this value of k into (3), we get 4x + 3y- 30 = 0 which is required equation of the line. Theorem 2: Altitudes of a triangle are concurrent A (x ,,y ,) Proof: Let the coordinates of the vertices of .1 ABC be as shown in the figure. Then Y2-Y3 ~~ Xz-X3 X2-X3 Therefore slope of the altitude AD = Y2-Y3
Slope of BC =
Equation of the altitude AD is X2 -X3
Y - Y1
= - Y2 -y3 . (x -
xi)
(Point-Slope form)
or x (x2 - X3) + Y (y2 - Y3) - X1 (x2 - X3) - Y1 (y2 - y3) = 0 Equa.tions of the ·altitudes BE and CF are respectively (by symmetry)
(1)
x (x3 - X1) + Y (y3 - Y1) - X2 (x3 - x1) - Y2 (y3 - y 1) = 0 x (x1 - Xz) + Y (yl - Y2) - X3 (x1 - X2) - Y3 (y3 - Y1) = 0 The three lines (1), (2) and (3) are concurrent if and only if
(2)
(3)
Unit 4: Introduction to Analytic Geometry
211
Y2-Y3
.D=
is zero
Y3 -y1 X1 - x2
Y1 - Y2
- X3 (x1 - x2) - Y3 (y1 - Y2)
Adding 2nd and 3rd rows to the 1st row of the determinant, we have
0
0
0
=0
y3-Y1 Y1 - Y2
- X3 (x1 - x2) - Y3 (y1 - Y2) Thus the altitudes of a triangle are concurrent. Theorem 3: Right bisectors of a triangle are concurrent. X1 - X2
Proof. Let A (x 1, y 1), B (x2, y2) and C (x3, y3 ) be the vertices of .1 ABC.
The midpoint D of BC has coordinates X2 + X3 (
2
Y2 + Y3)
'
E
2 c
B
Since the slope of BC is
y2 -y3 X2 -X3
,
the slope of the right bisector DO of BC is -
x2 -x3 Y2 -y3
Equation of the right bisector DO of BC is Y-
or
Y2
x (x2 - X3)
+ Y3 __ _ X2 - X3 2 Y2 -y3
(x _
X2
+ X3) 2
(Point-slope form)
. I 2 2 1 2 2 + Y (y2 - Y3) - 2 (Yz - Y3 ) - 2 (xz - X3 )
=0
(1)
By symmetry, equations of the other two right bisectors EO and FO are respectively:
and The lines (1), (2) and (3) will be concurrent if and only if
(3)
Calculus and Analytic Geometry
212
Y2-Y3 =0
y3-Y1 Y1 -y2
Adding 2nd and 3rd rows to 1st row of the determinant, we have
0
0
0
y3-Y1 Y1 -y2 Thus the right by bisectors of a triangle are concurrent.
Note: If equations of sides of the triangle are given, then intersection of any two lines gives a vertex of the triangle.
4.4.4 Distance of a Point From a Line Theorem 4: The distanced from the points P(x1, y1) to the line l I : ax + by + c = 0
is given by:
d
( 1)
= \a.xi + by1 + c\ ~a2 +b2
Proof: Let l be non-vertical and non-horizontal line. FromP, draw PQR .L Ox and PM .l I. Let the ordinate of Q be y 2 so that coordinates of Q are
(xi' y 2 ). Since Q lies on/, we have a.x 1 + by2 + c = 0 or
-axI -c Y2--
b From the figure it is clear that LMPQ = a= the inclination
of I.
-a
Now tana= slop of l = b
=0
Unit 4: Introduction to Analytic Geometry
Therefore, icosal Thus
J
PM
=
213
lbl
.Ja2 +b2
I=d =I PQ 11 cos a I=I y
I
y 2 cos a
1-
J
I
_ _ - ax, - c I Ib I - Yi b . I z z -va +b
= lhY1 +a.xi +cl·lbl lhi.Ja2+b2
=ax, +by1+c .Ja2+b2
If l is horizontal, its equation is of the form y = _!:._ and the distance from . b , P(x1, y 1) to l is simply the difference of they-values.
:.
d = Y1
-(-~) =
by,b+c
Similarly, if the line is vertical and has equation: x
Note:
=- c a
then d
= a.xi + c a
If the point P(x1 , y1) lies on l, then the distance d is zero, since (x 1, Y1 ) would satisfies the equation i.e., ax1 + by 1 + c =0
4.4.5 Distance Between Two Parallel Lines The distance between two parallel lines is the distance from any point on one of the lines to the other line. Example 4: Find the distance between the parallel lines Challenge! l1 : 2x-5y + 13.= 0
Check the answer by taking (i) any other point on 11
and
12 :2x-5y+6 =0
Solution: First find any point on one of the lines, say !1 • If x =1 lies on 11 , then y =3 and
(ii) any point of 12 and finding its distance from 11
the point (1,3) lies on it. The distanced from (1,3) to 12 is
d
= 12(1)-5(3)+61 =_12-;=.-=15=+=--61 =
J22 + 52
.J4+25
7
.fi9
The distance between the parallel lines is
~.
v29
Calculus and Analytic Geometry
214
y
4.4.6 Area of a Triangular Region Whose Vertices are Given
R
To find the area of a triangular region whose vertices are: P(x1 , y 1 ), Q(x2, y 2) and R(x 3, y3) Draw perpendiculars -P-L, -QN- and RM on x-axis.
Q
p
-~----lLl......"-..:..-....:..~M~---..::......::.J.N-• x
0
Area of the triangular region PQR =
Area of the trapezoidal region PLMR + Area of the trapezoidal region RMNQ -
Area of the trapezoidal region PLNQ.
= ![(Y1 + Y3)(x3 - X1) + (y3 + Y2)(x2 -x3) -(Y1 + Y2 )(x2 - X1 )] 2 1 =1(X3Y1 +X3Y3 -X1Y1-X1Y3 +X2Y3 +X2Y2 -X3Y3 -X3Y2 -X2Y1 -X2Y2+X1Y1 1 = -(X3Y1 - X1 Y3 + X2Y3 -X3Y2 - X2Y1 + 2 Thus required area ti is given by:
X1
+X1Y2)
Y2) Have you observed that?
x,
1
ti=-(x1(Y2 -y3)+x2(Y3-Y1)+x3(Y1 -y2)) 2 Corollary: If the points P, Q and R are collinear, then
Ii=.!_ X1 2 X3
Y1
1
Y2
1
Y3
1
ti=O Note: · In numerical problems, if sign of the area is negative, then it is to be omitted.
Example 5: Find the area of the region bounded by the triangle with vertices (a,b+c),(a,b-c) and (-a,c). Solution: Required area ti is ti =.!_ 2
=
a a
-a
1 a
b+c 1 b-c 1
c
Trapezium: A quadrilateral having two parallel and two non-parallel sides.
1
Area of trapezoidal region:
b+c 1 0 -2c 0 , by R2 -R1 2 -a c 1
21 (sum of 11 sides) (distance between 11 sides)
-~------------~---
Unit 4: Introduction to Analytic Geometry
215
=![-2c(a +a)], expanding by the second row 2
= -2ca Thus ti= 2ac Example 6: By considering the area of the region bounded by the triangle with vertices A(l,4), B(2, -3) and C (3, -10), check whether the three points are collinear or not. Solution: Area ti of the region bounded by the triangle ABC is, 1
1 ti = - 2
2
4
- 3
1
1
3 -10 1.
I
=-
2
1
4
1
- 7
1
0 by R 2 - R1 and R 3 - R1
2 -14 0
= _!_ [1(-14+14)], expanding by third column
2 = 0 Thus the points are collinear.
EXERCISE 4.3 1.
2.
3.
Find the slope and inclination of the line joining the points: (ii) (3, - 2) ; (2, 7) (iii)(4,6); (4,8) (i) ( - 2 ,4) ;(5, 11) Sketch each line in the plane. In the triangle A (8, 6) ,B( - 4 ,2), C (-2,-6) , find the slope of (i) each side of the triangle (ii) each median of the triangle (iii) each altitude of the triangle By means of slopes, show that the following points lie on the same line·: (a) (-1,-3) ; (1, 5); (2, 9) (b) (4,-5); (7, 5); (10,15) (c )
4. 5. 6.
(-4, 6): (3, 8): (10. 10)
(d)
(a. 2b); (c, a+ b); (2c- a. 2a)
Find k so that the line joining A (7, 3); B (k, -6) and the line joining C(-4, 5); D ( - 6 ,4) are (i) parallel (ii) perpendicular. Usin~ slopes, show that the triangle with its vertices A (6,1), B (2,7) and C (-6.-7) is a right triangle. The three points A (7, -1), B(-2, 2) and C (1,4) are consecutive vertices of a parallelogram. Find the fourth vertex.
216 7.
8.
9.
Calculus and Analytic Geometry The points A (-1: 2), B (3, -1) and C (6,3) are consecutive vertices of a rhombus. Find the fourth vertex and show that the diagonals of the rhombus are perpendicular to each other. Two pairs of points are given. Find whether the two lines determined by these points are: (i) parallel (ii) perpendicular - (iii) none (a) (l,-2),(2,4)and(4, 1),(-8,2) (b) (-3,4),(6,2)and(4,5), (-2,-7) Find an equation of (a) the horizontal line through (7, -9) (b)
the vertical line through ( -5, 3)
(c)
the line bisecting the first and third quadrants.
(d)
the line bisecting the second and fourth quadrants.
10. Find an equation of the line (a)
through A( -6, 5) having slope 7
(c)
through (-8, 5) having slope undefined
(d)
through (-5,-3) and (9, -1)
(e) y-intercept-7 and slope -5
(f)
x-intercept :-3 and y-intercept:4
(g) x-intercept:-9 and slope :-4
(b) through (8, - 3 ) having slope 0
11.
Find an equation of the perpendicular bisector of the segment joining the points A(3, 5) and B(9,8)
12.
Find equations of the sides, altitudes and medians of the triangle whose vertices are A( - 3, 2), B(5, 4) and C(3, - 8 ).
13.
Find an equation of the line through (-4, -6) and _perpendicular to a line
-3 . 2
having slope -
.
14.
Find an equation of the line through (11, -5) and_parallel to a line with slope-24.
15.
The points A( -1, 2), B(6, 3) and C(2, -4) are vertices of a triangle. Show that the line joining the midpoint D of AB and the midpoint E of AC is parallel to BC and DE= _!_BC. 2
16.
A milkman can sell 560 litres of milk at Rs.12.50 per litre and 700 litres of milk at Rs. 12.00 per litre. Assuming the graph of the sale price and the milk sold to be a straight line, find the number of litres of milk that the milkman can sell at Rs.12.25 per litre.
Unit 4: Introduction to Analytic Geometry 17.
18.
19.
20.
21.
22.
23.
24.
217
The population of Pakistan to the nearest million was 60 million in 1961 and 95 million in 1981. Using t as the number of years after 1961, find an equation of the line that gives the population in terms oft. Use this equation to find the population in (a) 1947 (b) 1997. A house was purchased for Rs. I million in 1980. It is worth Rs.4 million in 1996. Assuming that the value increased by the same amount each year, find an equation that gives the _value of the house after t years of the date of purchase. What was its value in 1990? Plot the Celsius (C) and Fahrenheit (F) temperature scales on the horizontal axis and the vertical axis respectively. Draw the line joining the freezing point and the boiling point of water. Find an equation giving F temperature in terms ofC. The average entry test score of engineering candidates was 592 in the year 1998 while the score was 564 in 2002. Assuming that the relationship between t\ time and score is linear, find the average score for 2006. Convert each of the following equation into (i) Slope intercept form (ii) two intercepts form (iii) normal form (a) 2x-4y+ll=O (b) 4x+7y-2=0 (c) 15y-8x+3=0 Also find the length of the perpendicular from (0, 0) to each line. In each of the following check whether the two lines are (i) parallel (ii) perpendicular (iii) neither parallel nor perpendicular (a)
2x+ y-3 =0
4x+ 2J1+5 = 0
(b)
3y = 2x+5
3x+2y-8 = 0
(c)
4y+2x-1=0
x-2y-7=0
(d)
4x-y+2=0
12x-3y+l = 0
(e)
12x+35y -7=0
105x-36y + 11=0
Find the distance between the given parallel lines. Sketch the lines. Also find an equation of the parallel line lying midway between them. (a)
3x-4y+3=0
3x-4y+7=0
(b)
12x + 5 y -.- 6 = 0
12x + 5 y + 13 = 0
(c)
x+2y-5=0
2x+4y = 1
Find an equation of the line through ( - 4 , 7) and parallel to the line 2x-7y+4 = 0.
218
Calculus and Analytic Geometry
25.
Find an equation of the line through (5, - 8) and perpendicular to the join of A( -15, -8 ), B(lO, 7).
26.
Find equations of two parallel lines perpendicular to 2x - y + 3 = 0 such that the product of the x- and y-intercepts of each is 3. One vertex of a parallelogram is (1, 4); the diagonal intersect at (2, 1) and the
27.
sides have slopes 1 and
=-!:..
Find the other three vertices. 7 Find whether the given point lies above or below the given line (a) (5,8); 2x-3y+6=0
28.
(b)
29.
(-7,6);4x+3y-9~0
Check whether the given points are on the same or opposite sides of the given line. (a) (0, 0) and (-4, 7); 6x-7y + 70 = 0 (b)
(2, 3) and (-2, 3); 3x-5y+8 = 0
30.
Find the distance from the point P(6,-1) to the line 6x.-4y + 9 = O.
31. 32.
Find the area of the triangular region whose vertices are A(5, 3 ), B(-2, 2), C(4, 2). The coordinates of three points are A(2, 3), B(-1, 1) and C(4, -5). By computing the area bounded by ABC check whether the points are collinear.
4.5. ANGLE BETWEEN TWO LINES Let 11 and l 2 be two intersecting lines, which meet at a point P. At the point P two supplementary angles are formed, by the lines / 1 and /2 •
Unless l 1 1- 12 one of the two angles is acute. The angle from 11 to 12 is the angle () through which 11 is rotated anti-clockwise about the point P so _that it coincides with 12 •
In the figure below 8 is angle of intersection of the two lines and it is measured from l1 to 12 in counterclockwise direction ,l/f is also angle of intersection but it is measured from 12 to f 1 • With this convention for angle of intersection, it is clear that the inclination of a line is the angle measured in the counterclockwise direction from the positive . x-axis to the line and it tallies with the earlier definition of the inclination of a line.
Unit 4: Introduction to Analytic Geometry
219
Theorem 1: Let 11 and 12 be two non-vertical lines such that they are not perpendicular to each axes. If m, and m 2 are the slopes of I, and 12 respectively, then the angle
e from
LI to / 2 is given by;
tan()=
l,
m 2 -m 1
y
I,
I+m 1m 2
Proof: From the figure, we have
a 1 = a1 +8 8=a 2 -a 1
or
Corollary 1.
11
tan () = 0 = m1
11
12 if and only if m 1 = m 2
m 2 -m1 l+m1m 2
= m1
Corollary 2.
11, _l_ 12 iff 1 + m1m2 = 0
These two results have already been stated in 4.3.1 . Example 1: Find the angle from the line with slope -
5
7 to the line with slope ~ . 3 2
-7
Solution: Here m 2 = - , m, = - . If () is measure of the required angle, then 2 3
~-( =21
tanB=~=_3.2_=-l
5(-7) -29
1+- 2 3
Thus () = 135~ Example 2: Find the angles of the triangle whose vertices are A(-5,4),B(-2, -1),C(7, -5)
Solution: Let the slopes of the sides AB, BC and CA be denoted by me, m 0 , mb respectively. Then
y
A1 ~
' i"r-...
c\
!"-. I\
I" b ' 1:n ~
B
!'-....
-x ~
~
" a-..... ,.,_',"-!:::... Ir
Calculus and Analytic Geometry
220
4+1 -5 -5+1 -4 me= -5+2 =},ma= 7+2 =g,mb Now angle A is measured from AB to AC. -3 5
fn
-+-
tan A= ;:
~,:::, = 1+ ( ~:
=
-5-4 7+5
=
-3 4
11 or mLA = 22.2°
= 27
. . The angle B is measured from BC to BA tan B = me - ma -
··
l+m,m.
-5 4 -+-
sr- J
3 l+(3
9
=
4 9
-33 47
or mLB= 144.9°
The angle C is measured from CA to CB. -4 3
-+-
.. tan
4.5.1
c
=
r
:::~.::. = 1+( ~n4_43 :~
or mLC= 12.9°
Equation of a Straight Line in Matrix Form
It is easy t? solve two or three. simultaneous linear equations by elementary methods. If the number of equations and variables become large, the solution of the equations by ordinary methods becomes very difficult. In such a case, given equations are written in matrix form and solved.
One Linear Equation: A linear equation l : ax+ by + c = 0 in two variables x and y has its matrix form as: [ax + by] = [-c]
er [a bt]=[-c] AX=C
or
where A=
[a
b], X
=[:]and C=[-c]
(1)
Unit 4: Introduction to Analytic Geometry
221
A System of two Linear Equations: A system of two linear equations
o}
11 :a1x+b1y+c1 = 12 :a 2 x+b2 y+c 2 = 0
(2)
in two variables x and y can be written in matrix form as
(3)
or or
AX =C .
where A =[a
1
a2
Equations (2) have a solution iff det A :;:. 0 . A system of three equations: A system of three linear equations
1 a1x+b y +c1 = 0} 1:
1
12 :a 2 x+b 2 y+c 2 =0
(5)
13 :a 3 x+b3 y+c 3 =0
in two variables x and y talces the matrix form as
Cl]
c 2 is singular, then the lines are concurrent and so the c system (5) has a unique solution. 3
222
Calculus and Analytic Geometry
Example 3: Express the system
=0}
3x+4y-7 2x-5y+8= 0 x+y-3=0
in matrix form and check whether the three lines are concurrent. Solution: The matrix form of the system is
[~ ~:rn=m -45
Coefficient matrix of the system is 4
A=[~
-5
-87]
1
-3
0 ==:}
det A= 0 1
1
2
:_ 7
14
1
-3
by
R1 -3R3
and
R2 -2R3
and det A= 1(14+14) = 28 :;t: 0 As A is non-singular, so the lines are not concurrent. Example 4: Find a system of equations corresponding to the matrix form
[! ~ lJ[~J=m
(1)
Are the lines represented by a system concurrent? Solution: Multiplying the matrices on the L.H.S. of (1), we have ;x++2;y++51] [
=[~]
(2)
0
4x+7y+6
By using the definition of equality of two matrices, we have from (2),
x+2y+5=0] 3x+5y +l = 0 4x+7y+6=0
as the required system of equations. The coefficient matrix A of the system is such that 1 2 5 1 2 5 det A = 3 5
1
4 7 6
=0
-1 -14 0 -1 -14
Thus the lines of the system are concurrent.
=0
Unit 4: Introduction to Analytic Geometry
223
EXERCISE 4.4 1.
Find the point of intersection of the lines
W
x-2y+1=0
and
2x-y+2=0
(ii)
3x + y + 12 =0
and
x + 2 y -1 =O
(iii)
2.
x + 4 y -12 = 0 and x - 3y + 3 = O Find an equation of the line through (i) the point (2, -9) and the intersection of the lines 2x + 5 y - 8 =0
(ii) (a)
(iii) 3.
and
3x - 4 y - 6 = 0
the intersection of the lines x- y- 4 =0 and
7 x + y + 20
and
parallel (b) perpendicular to the line 6x + y-14 =0 through the intersection of the lines x + 2 y + 3 = 0 ,
3x + 4 y + 7 =0
and making equal intercepts on the axes. Find an equation of the line through the intersection of 16x-l0y-33 = 0 12x+ 14y + 29 = 0 and the intersection of
x-y+4=0
x-7y+2=0
4.
Find the condition that the lines y = m1x+c1 ;y = m 2 x+c 2 and y = m3 x+c3
5.
are concurrent. Determine the value of p such that the lines 2x - 3 y - 1 0 , 3x - y - 5 = 0 and
=
3x + py + 8 = 0 meet at a point. 6. 7.
8.
9.
Show that the lines 4x - 3 y - 8 = 0 , 3x - 4 y - 6 :::: 0 and x - y - 2 = 0 are concurrent and the third-line bisects the angle formed by the first two lines. The vertices of a triangle are A( - 2, 3), B(-4, 1) and C(3, 5). Fmd coordinates of the (iii) circumcentre of the triangle (ii) orthocentre (i) centroid Are these three.points collinear? Check whether the lines x-y-2=0 3x-4y-6 =0; 4x-3y-8=0; are concurrent. If so, find the point where they meet. Find the coordinates of the vertices of the triangle formed by the lines x - 2y - 6 = 0 ; 3x - y + 3 = 0 ; 2x + y - 4 = 0 Also find measures of the angles of the triangle.
224 10.
Calculus and Analytic Geometry
Find the angle measured from the line 11 to the line 12 where (a) /1 : Joining (2,7) and (7, 10)
(b)
12 : Joining (2, 4) and ( -8, 2)
12 : Joining (1, 1) and (-5, 3) (c) 11 : Joining (1, -7) and (6, -4) 12 : Joining (-1, 2) and (-6, -1) 11.
12.
13.
14.
15. 16.
17.
11 : Joining (3, -1) and (5, 7)
(d)
11 : Joining (-9, -1) and (3, -5) 12 : Joining (2, 7) and (-6, -7)
Also find the acute angle in each case. Find the interior angles of the triangle whose vertices are (a) A(-2,11),B(-6,-3),(4,-9) (b) A(6, 1),B(2, 7),C(-6,-7) (c) A (2, -5), B(-4,-3), (-1, 5) (d) A (2, 8), B (-5, 4), C (4,-9) Find the interior angles of the quadrilateral whose vertices are A (5, 2), B(-2,3), C (-3,-4) andD(4,-5) Show that the points A (0, O), B (2, 1), C (3, 3), D ( 1, 2) are the vertices of a rhombus and find its interior angles. Find the area of the region bounded by the triangle whose sides are 7x-y-10=0; lOx+y-41=0; 3x+2y+3=0 The vertices of a triangle are A( - 2, 3), B( - 4, 1) and C(3, 5). Find the centre of the circum centre of the triangle. Express the given system of equations in matrix form. Find in each case whether the lines are concurrent. (a) x+3y-2=0; 2x-y+4=0; x-lly+l4=0 (b) 2x + 3y + 4 = O; x - 2y - 3 = O; 3x + y - 8 = 0 (c) 3x-4y-2 = 0, x+2y-4 = 0, 3x-2y+5 = 0. Find a system of linear equations corresponding to the given matrix form. Check whether the lines represented by the system are concurrent.
(•)(~ ~I -J~1=m
~) (~ : ~~1~1=rn
4.6 HOMOGENEOUS EQUATION OF THE SECOND DEGREE IN TWO VARIABLES We have already seen that if a graph is a straight line, then its equation is a linear equation in the variables x and y. Conversely, the graph of any linear equation in x and y is a straight line.
. 225
Unit 4: Introduction to Analytic Geometry
Suppose we have two straight lines represented by and
a 1x + b 1y + c 1 = 0
(1)
a 2 x+b2 y+c 2 =0
(2)
Multiplying equations (1) and (2), we have (a 1x + b1y + c 1 )(a 2 x + b2 y + c 2 )
=0
(3)
It is a second degree equation in x and y. Equation (3) is called joint equation of the pair of lines (1) and (2). On the other hand, given an equation of the second degree :in x and y, say ax 2 + 2hxy + by 2 + 2gx + 2.fY + c =: 0
(4)
where a"# 0, represents equations of a pair of lines if (4) can be resolved into two Jinear factors. In this section, we shall study special joint equations of pairs of lines which pass through the origin. Let y = m 1x and y = m 2 x be two lines passing through the origin. Their joint equation is: (y- m 1x)(y-m 2 x) = 0
y 2 -(m 1 +m 2 )xy+m1m 2 x 2 =0
or
(5)
Equation (5) is a special type of a second degree homogeneous eq\lation.
4.6.1 Homogeneous Equation Let
f(x, y)
=0
(1)
be any equation in the variables x and y . Equation (1) is called a homogeneous equation of degree n (a positive integer) if f(kx,ky) =
e f(x,y)
1
for some real n1_i.~1 oer k. For e'°.dmple, in equation (5) above if we replace x and y by kx and ky respectively, 'Zve have
k2y2-k2(m 1 +m2 )xy+k2m,m2 x 2 =0 or
k 2 [y2 - (m 1 + m2 )xy + m1m 2 x 2 ] = 0 i.e.,
k2 f (x, y) =()
Thus (5) is a homogeneous equation of degree 2. A general second degree homogeneous equation can be written as:
ax 2 + 2hxy +by
2
=0
provided a, h and b are not simultaneously zero.
-
--~--- ~- ~-- -- -- ~~---~-- ~-~---
226
Calculus and Analytic Geometry
Theorem 1: Every homogenous second degree equation 2
a.x + 2hxy +by
2
=0
(1)
represents a pair of lines through the origin. The lines are (i) real and distinct, if h 2 >ab (ii) real and coincident, if h 2 =ab (iii) imaginary, if h 2 ab.
(ii) real and coincident, if h 2 =ab.
(iii) imaginary, if h 2 II =135'
-5
Acute angle between the lines= 180° - () = 180° -135° = 45°
228
Calculus and Analytic Geometry
Example 3: Find a joint equation of the straight lines through the origin perpendicular to the lines represented by x 2 +xy- 6 y 2 -o (1) Solution: (1) may be written as (x-2y)(x+3y) = 0 Thus the lines represented by ( 1) are x-2y=O ~) and x + 3y = 0 (3) The line through (0, 0) and perpendicular to (2) is y=-2x or y+2x=O (4) Similarly, the line through (0, 0) and perpendicular to (3) is y=3x or y-3x=O (5) Joint equation of the lines (4) and (5) is
(y+2x)(y-3x)=O or
y 2 -xy-6x 2 =0
EXERCISE 4.5 . Find the lines represented by each of the following and also find measure of the angle between them (Problems 1-6):
1.
10x 2 -23.xy-5y 2 = 0
2.
3x 2 + 7 xy + 2 y 2 = 0
3.
9x 2 +24.xy+16y 2 = 0
4.
2x 2 +3.xy-5y 2 = 0
5.
6x 2 -19.xy+15y 2 = 0
6.
x2
-
2xy sec a+ y2 = 0
7. Find a joint equation of the lines through the origin and perpendicular to the lines: x 2 -2xytana-y2 =0 8. Find a joint equation of the lines through the origin and perpendicular to the lines: 2 ax + 2hxy +by 2 = 0 9. Find the area of the region bounded by: 2
10x -xy-21y2=0and
x+y+l=O
Unit 5: Linear inequalities and Linear Programming
229
Linear Inequalities and Lin~ar programming
Unit5
5.1 INTRODUCTION Many real life problems involve linear inequalities. Here we shall consider those problems (relating to trade, industry and agriculture etc.) which involve systems of linear inequalities in two variables. Linear inequalities in such problems are used to prescribe limitations or restrictions on allocation of available resources (material, capital, machine capacities, labour hours, land etc.). In this chapter, our main goal will be to optimize (maximize or minimize) a quantity under consideration subject to certain restrictions. The method under our discussion is called the linear programming method and it involves solutions of certain linear inequalities.
5.2 LINEAR INEQUALITIES Inequalities are expressed by the following four symbols; > (greater than); < (less than); ~ (greater than or equal to); ~ (less than or equal to) For example (i) ax < b (ii) ax + b ~ c (iii) ax + by > c (iv) ax + by ~ c are inequalities. Inequalities (i) and (ii) are in one variable while inequalities (iii) and (iv) are in two variables. The following operations will not affect the order (or sense) of inequality while changing it to simpler equivalent form: (i) Adding or subtracting a constant to each side of it. (ii) Multiplying or dividing each side of it by a positive constant. Note that the order (or sense) of an inequality is changed by multiplying or dividing its each side by a negative constant. Now for revision we consider inequality, x < 2_ 2 All real
nu~bers < 2_ 2
Thus the interval (-
oo,
~)
(A)
are in the solution set of (A).
or -
oo
< x < ~ is the solution set of the inequality (A)
which is shown in the figure 5.21
••
-5 -4 -3 -2 -1 0
3/2 I
1
0
I
2
I
3
4
5
•
Fig. 5.21
Calculus and Analytic Geometry
230
We conclude that the solution set of an inequality consists of all solutions of the inequality. 5.~1
Graphing of A Linear Inequality in Two Variables
Generally a linear inequality in two variables x and y can be one of the following forms: ax+byc;
ax + by $ c;
ax + by '?:. c
where a, b, c are constants and a, b are not both zero. We know that the graph of linear equation of the form ax + by which divides the plane into two disjoint regions as stated below: ( 1) The set of ordered pairs (x, y) such that
ax + by < c
(2) The set of ordered pairs (x, y) such that
ax+ by> c
= c is a line
The regions (1) and (2) are called half planes and the line ax + by called the boundary of each half plane.
=c
is
Note -that a vertical line divides the plane into left and right half planes while a non-vertical line divides the plane into upper and lower half planes. A solution of a linear inequality in x and y is an ordered pair of numbers which satisfies the inequality. For example, the ordered pair (1, 1) is a solution of the inequality x + 2y < 6 because 1 + 2(1) = 3 < 6 which is true. There are infinitely many ordered pairs that satisfy the inequality x + 2y < 6, so its graph will be a half plane. Note that the linear equation ax + by = c is called "associated or corresponding equation" of each of the above mentioned inequalities.
Procedure for Graphing a linear Inequality in two Variables (i)
The corresponding equation of the inequality is first graphed by using 'dashes' if the inequality involves the symbols > or < and a solid line is drawn if the inequality involves the symbols '?:. or $.
(ii)
A test point (not on the graph of .the corresponding equation) is chosen which determines that the half plane is on which side of the boundary line.
Unit 5: Linear inequalities and Linear Programming
Example 1: Graph the inequality
231
x + 2y < 6.
Solution: The associated equation of the inequality x + 2y < 6 (i) is x + 2y = 6 (ii) The line (ii) intersects the x-axis and y-axis at (6, 0) and (0, 3) respectively. As no point of the line (ii) is a solution of the inequality (i), so the graph of the line (ii) is shown by using dashes. We take 0(0, 0) as a test point because it is not on the line (ii).
y
...
·· ~ .,
· .. :.:- ""' 6 •••• (0, 3)
x' 4-.-~
_
·.
·(6, 0) __....__..,..,.;,.;.....;'-x
Fig. 5.22(a)
y'
Substituting x = 0, y = 0 in the expression x + 2y gives 0 - 2(0) = 0 < 6, so the point (0, 0) satisfies the inequality (i). Any other point below the line (ii) satisfies the inequality (i), that is all points in the half plane containing the point (0, 0) satisfy the inequality (i). Thus the graph of the solution set of inequality (i) is the y region on the origin-side of the line (ii), that is, the region below the line (ii). A portion of the open halfplane below the line (ii) is shown as shaded region in figure 5.22(a) x· ---~..........i........,........,.._..._.. x .. All points above the dashed line satisfy the (iii) inequality x + 2y > 6 Fig. 5.22(b) A portion of the open half plane above the line (ii) is shown by shading in figure 5.22(b) y'
...
Note: 1.
The graph of the inequality x + 2y ~ 6 .. (iv) includes the graph of the line (ii), so the open half-plane below the line (ii) including the graph of the line (ii) is the graph of the inequality (iv). A portion of the graph of the inequality (iv) is shown by shading in figure 5.22(c)
y
Note: 2.
All points on the line (ii) and above the line (ii) satisfy the inequality x + 2y ~ 6 .... (v). This means that the solution set of the inequality (v) consists of all points above the line (ii) and all points on the lines (ii). The graph of the inequality (v) is partially shown as shaded region in figure 5.22(d)
y'
Fig. 5.22(d)
Note: 3.
The graph of x + 2y :'S 6 and x + y ~ 6 are closed half planes.
y'
Calculus and Analytic Geometry
232
Example 2: Graph the following linear inequalities in .xy-plane; (i) 2x ~ -3 (ii) y ~ 2 Solution: The inequality (i) in .xy-plane is considered as 2x + Oy ~ -3 and its solution set consists of 3 all point (x, y) such that x, y E 1R.,. and x ~ - 2
The corresponding equation of the inequality (i) (1) is 2x =- 3
y
Fig. 5.23(a)
y'
which a vertical line (parallel to the y-axis) and its graph is drawn in figure 5.23(a) The graph of the inequality 2x > -3 is the open half plane to the right of the line (1 ). Thus the graph of 2x ~ - 3 is the closed half-plane to the right of the line (1). (ii)
The associated equation of the inequality (ii) is y
=2
which is a horizontal line (parallel to the x-axis) and its graph is shown in Figure 5.23(b) Here the solution set of the inequality y < 2 is the open half plane below the boundary line y = 2. Thus the graph of y ~ 2 consists of the boundary line and the open half plane below it.
y
(2)
x' ..........-..,...,.....,.....,....,.,..._,'""""'""...........,...,...?X y2 because 2(0)+0 = 0 -:f 2. Thus the graph of the inequality 2x + y ;;::: 2 is the closed half-plane not on the origin-side of the line 2x + y
= 2.
Thus the closed half-plane is shown partially as a shaded region in figure 5.31(b) The solution region of the given system of inequalities is the intersection of the graphs indicated in figures 5.31(a) and 5.3l(b) and is shown as shaded region in figure 5.31(c). The intersection point (2, -2) can be found by solving the equations x :- 2y = 6 and 2x+ y = 2. Note that the lines x - 2y = 6 and 2x + y
y' y
y'
= 2 divide the xy-plane into four region
bounded by these lines. These four (bounded) regions are displayed in the adjoining figure.
y'
Calculus and Analytic Geometry
234
Example 2: Graph the solution region for the following system of inequalities. 2x + y 2: 2,
x - 2y ~ 6,
x + 2y ~ 10
Solution:
The graph of the inequalities x - 2y ~ 6 and 2x + y ~ 2 have already drawn in figure 5.3l(a) and 5.3l(b) and their intersection is partially shown as a shaded region in figure 5.3l(c) of the example 1 (Art 5.3). Following the procedure of the example 1 of Art (5.3) the graph of the inequality x + 2y ~ 10 is shown partially in the figure 5.32(a). The intersection of graphs is the required solution which is the shaded triangular PQR (including its sides) partially in the figure 5.32(b).
three region region shown
Now we define a comer point of a solution region. Fig. 5.32(b)
y'
DEFINITION: A point of a solution region where two of its boundary lines interest, is called
a corner point or vertex of the solution region. Such points play a useful role while solving linear programming problems. In three comer points are ·obtained by solving the corresponding equations (?f linear inequalities given in the example 2) in pairs. ~xample 2, the following
Corresponding lines of inequalities: x-2y x-2y 2x+y
= 6, = 6, = 2,
2x+y x+2y x+2y
=2 = 10 = 10
Comer Points P(2, -2)
Q(8, 1) R(-2, 6)
Unit 5: Linear inequalities and Linear Programming
235
Example 3: Graph the following systems of inequalities. (i)
2x + y 2: 2 x + 2y :s 10
(ii)
2.x + y 2: 2 x+2y:S 10
(iii) 2.x + y 2: 2 x + 2y :s 10
x2: 0
y2:0
x 2: 0, y 2: 0,
Solution:
(i)
The corresponding equations of the inequalities 2.x+y
~
2
2.x+y
=
2
and
x + 2y
x + 2y
(I) and
=
$
10
10
are
(II)
For the partial graph of 2.x + y ~ 2 see figure 5. 31 (b) of the ex amp le 1 and the graph of the inequality x + 2y $ 10 is partially shown in figure 5.32(a) of the example2 >
The
solution
inequalities 2x + y
~
region
of
2 and x + 2y
$
the 10 is
the intersection .of their individual graphs. The common region of the graphs of inequalities is partially shown as a shaded region in figure 5.33(a). y' y
The graph of y plane
including
~
the
0 is the upper half graph
of
the
corresponding line y = 0 (the x-axis) of the linear inequality y ~ 0. The graph of y 2: 0
x' .,....,..,..........,..,....,...,....,.....O +,-,...,...v =.... O..,......,,........._. x
is partially displayed in figure 5.33(b). Fig. 5.33(b)
.J''
The solution region of the system of inequalities in (i) is the intersection of the graphs shown in figure 5.33(a) and 5.33(b) This solution region is displayed in figure 5.33(c). y'
236
Calculus and Analytic Geometry See figure 5.33(a) for the graphs of the inequalities 2.x + y
(ii)
~
2 and x + 2y 5 10. y
The graph of x
~
half-plane
the
open
to
corresponding line x inequality x
~
=
0 consists of the of
right
the
0 (y-axis) of the
0 and its graph. See figure
5.34(a). Fig. 5.34(a)
Thus the solution region of the inequalities in (ii) is partially shown in figure 5.34(b). This region is the intersection of graphs in figure 5.33(a) and 5.34(a).
y'
(iii)
The graphs of the system of inequalities
in (iii) are drawn in the solution of (i) and (ii). The solution region in this case, is shown as shaded region ABCD in figure 5.34. (c).
EXERCISE 5.1 1.
Graph the solution set of each of the following linear inequality in xy-plane: (i)
2.
2x + y
5 6
(ii) 3x + ?y ~ 21
(iii)
3x - 2y ~ 6
(iv) Sx - 4y 5 20 (v) 2x + 1 ~ 0 (vi) 3y - 4 5 0 Indicate the solution set of the following systems of linear inequalities by shading: (i)
2x-3y 5 6 2x+ 3y 5 12
(ii) x+y
~
5
-y+x 5 1
(iii)
3x+?y
~
x-y 5 2
21
Unit 5: Linear inequalities and Linear Programming (iv)
~
4x - 3y
3.
(v) 3x + ?y 2:: 21
12
3
x 2::
y~4
2
Indicate the solution region of the following systems of linear inequalities by shading: (iii) x + y 2:: 5 (ii) x + y ~ 5 (i) 2x- 3y ~ 6 2x + 3y
~
(iv)
2
x-y2::1
x 2:: 0
y 2:: 0
3x+7y
~
21
(v) 3x + 7y ~ 21
x-y
~
2
x-y
~
(vi)
3x+7y
~
21
2x- y 2:: -3
2
x 2:: 0
y 2:: 0
x2::0
Graph the solution region of the following system of linear inequalities and find the corner points in each case. (i)
2x - 3y
~
6
2x + 3y
~
12
x 2:: 0 (iv) 3x + 2y 2:: 6
x+ 3y
~
6
y 2:: 0
5.
~
y-2.X
12
y 2:: 0
4.
237
(ii)
x +y
~
5
-2x + y
~
2
(iii)
3x + 7y
~
5x+7y
~
35
x-2y
~
4
-3 y 2:: 0
(v) 5x + 7y ~ 35 ~
21
2x - y
y 2:: 0
-x+ 3y
~
(vi)
3
x 2:: 0
x 2'. 0
Graph the solution region of the following system of linear inequalities by shading. (iii) 2x + y ~ 4 (ii) 3x - 4y ~ 12 (i) 3x - 4y ~ 12 3x+ 2y 2:: 3 x+ 2y ~ 9
2x-3y 2:: 12 x+ 2y ~ 6
x+2y ~ 6 x+ y 2:: (v) 2x+3y ~ 18
(vi)
3x - 2y 2:: 3
(iv) 2x + y
~
10
x+y
~
7
2x+y
~
10
x+4y
~
12
-2x+y
~
4
-2x+ y
~
2
3x+y
~
12
5.4 PROBLEM CONSTRAINTS In the beginning we described that linear inequalities prescribe limitations and restrictions on allocation of available sources. While tackling a certain problem from every day life each linear inequality concerning the problem is named as problem constraint. The system of linear inequalities involved in the problem concerned are called problem constraints. The variables used in the system of linear inequalities relating to the problems of every day life are non-negative and are called non-
.____
--- -- ~~---~--~~----- -~~~ -
Calculus and Analytic Geometry
238
negative constraints. These non-negative constraints play an important role for taking decision. So these variables are called decision variables.
5.5 FEASIBLE SOLUTION SET We see that solution region of the inequalities in example 2 of Art 5.3 is not within the first quadrant. If the nonnegative constraints x ;::: 0 and y ;::: 0 are included with the system of inequalities given in the example 2, then the solution region is restricted to the first quadrant. It is the polygonal region ABCDE (including its sides) as shown in the figure 5.51.
y
Fig. 5.51
y'
Such a region (which is restricted to the first quadrant) is referred to as a feasible region for the set of given constraints. Each point of .the feasible region is called a feasible solution of the system of linear inequalities (or for the set of a given constraints). A set consisting of all the feasible solutions of the system of linear inequalities is called a feasible solution set.
Example 1: Graph the feasible region and find the comer points for the following system of inequalities (or subject to the following constraints). x-y :'.S3 x + 2y :'.S 6 ' x 2: 0, y 2: 0 Solution: The associated equations for the inequalities
are
x-y
~
3
(i)
and
x + 2y
6
(ii)
x-y
=3
(1)
and
x + 2y = 6
(2)
~
y
As the point (3, 0) and (0, -3) are on the line (1), so the graph of x - y = 3. is drawn by joining the points (3, 0) and (0, -3) by solid line. , x...-.............
.,............,....,.-.:+-~:,...;;:o.,~:,....--+
Similarly line (2) is graphed by joining he points (6, 0) and (0, 3) by solid line. For x = 0 and y = 0, we have;
0- 0=0~ 3
and
0 + 2(0) = 0 ~ 6,
Fig. 5.52(a)
y'
so, both the closed half-planes are on the origin sides of the lines (1) and (2). The intersection of these closed half-planes is partially displayed as shaded region in figure 5.52(a).
Unit 5: Linear inequalities and Linear Programming
239 y
For the graph ofy 2: 0, see figure 5.33(b) of the example3 ofart5.3. The intersection of graphs shown in figures 5.52(a) and 5.33(b) is partially graphed as shaded region in figure 5.52(b). y' y
The graph of x 2: 0 is drawn in figure 5.34(a). The intersection of the graphs shown in figures 5.52(a) and 5.34(a) is graphed in figure 5.52(c).
x'....,....,..........,...,...,........,...,..
y' y
Finally the graph of the given system oflinear inequalities is displayed in figure 5.52(d) which is the feasible region for the given system of linear
x'....,,........,........,...,...,...,..~l"'l'~,...,...~..-.--x
inequalities. The points (0, 0), (3, 0), (4, 1) and (0, 3) are comer points ofthe feasible region. y'
Example 2: A manufacturer wants to make two types of concrete. Each bag of Agrade concrete contains 8 kilograms of gravel (small pebbles with coarse sand) and 4 kilograms of cement while each bag of B-grade concrete contains 12 kilograms of gravel and two kilograms of cement. If there are 1920 kilograms of gravel and 480 kilograms of cement, then graph the feasible region under the given restrictions and find comer points of the feasible region. Solution: Let x be the number of bags of A-grade concrete produced and y denote the number of bags of B-grade concrete produced, then 8x kilograms of gravel will be used for A-grade concrete and 12y kilograms of gravel will be required for B-grade concretes so 8x + 12y should not exceed 1920, that is,
Calculus and Analytic Geometry
240 8x+ 12y:S 1920
Similarly, the linear constraint for cement will be 4x + 2y :S 480
Now we have to graph the feasible region for the linear constraints 8x + 12y :S 1920 4x + 2y :S 480,
y
x~O, y ~O
Taking the one unit along x-axis and y-axis equal to 40 we draw the graph of the feasible region required. The shaded region of figure 5.53(a) shows the graph of 8x + I2y :S 1920 including the nonnegative constraints x ~ 0 and y ~ 0.
Fig. S.53(a)
y'
y
In the figure
5.53(b), the
graph of
4x+ 2y :S 480 including the non-negative constraints x ~ 0 and y ~ 0 is displayed as shaded region. Fig. S.53(b)
y' y
The intersection of these two graphs is shown as shaded region in figure 5.53(c), which is the feasible region for the given linear constraints. The point (0, 0), (120, 0), (60, 120) and (0, 160) are the comer points of the feasible region. Fig. S.53(c)
y'
Unit 5: Linear inequalities and Linear Programming
241
Example 3: Graph the feasible regions subject to the following constraints.
(a)
2x- 3y :S 6
(b)
2x+y~2
2x- 3y :S 6 2x+y~2
x ~ 0, y ~ 2
x + 2y
:s
' x ~ 0, y ~ 0
8 y
Solution: The graph of 2x -:- 3y
~
6 is the closed half-plane on the origin side of 2x - 3y = 6. The portion of the graph of system 2x - 3y ~ 6, x~O.y~
0
is shown in figure 5.54(a). Fig. 5.54(a)
Y' The graph of 2x + y
~
y
2 is the
closed half-plane not on the origin side of 2x + y
= 2. The portion of the graph
of the system 2x + y ~ 2, x=O
x
;?:
0, y ~ 0
is displayed in figure 5.54(b).
Fig. 5.54(b)
Y' y
The graph of the system
2x - 3y $; 6, 2x + y $; 2,
y=O
is the intersection of the graphs shown in figures 5.54(a) and 5.54(b) and it is partially displayed in figure 5.54(c).
Fig.5.54(c)
Y'
X
Calculus and Analytic Geometry
242
y x=O
The graph of system x+2y ~ 8, x;?: 0, y ;?: 0 is a triangular region indicated in figure 5.45(d). (b)
Thus the graph of the system 2x-3y
~
6
2x+y;?:2 x
+ 2y
~
8'
x ~ 0, y ;?: 0
Fig. 5.54(d)
Y' y
is the intersection of the graphs shown in figures 5.54(c) and 5.54(d). It is the indicated in the figure 5.54(e).
Note: The comer points of feasible region for the set of constraints in (a) are (1, 0) (3, 0) and (0, 2) while the comer points of the feasible region for the set of constraints 36 1 (0, 4) and in (b) are (1, 0), (3, 0), ( , 7 7
°),
(0, 2). Y'
We see that the feasible solution regions in example 3(a) and 3(b) are of different types. The feasible region in example 3(a) is unbounded as it cannot be enclosed in any circle how large it may be while the feasible region in example 3(b) can easily be enclosed within a circle, so it is bounded. If the line segment obtained by joining any two points of a region lies entirely within the region, then the region is called convex. Both the feasible regions of example 3(a) and 3(b) are convex but the ~ regions such as shown in the adjoining figures are not convex.
EXERCISE 5.2 1.
Graph the feasible region of the foil owing system of linear inequalities and find the corner points in each case. (i) 2x - 3y ~ 6 (ii) x + y ~ 5 (iii) x + y ~ 5 2x + 3y ~ 12 -2x + y ~ 2 -2x + y 2 2 x ~ 0, y 2 0
x 2 0, y ~ 0
x 2 0
Unit 5: Linear inequalities and Linear Programming (iv) 3x + 7y ~ 21 x- y ~3 x 2 0, y 2 0
2.
(v)
3x + 2y 2 6 x +y $ 4 x 2 0, y 2 0
243 (vi)
5x + 7y $ 35 x - 2y $ 4 x 2 0, y 2 0
Graph the feasible region of the following system of linear inequalities and find the corner points in each case. (i) 2x + y ~ 10 (ii) 2x + 3y $ 18 (iii) 2x+ 3y $ 18 x + 4y ~ 12 2x + y $ 10 x + 4y $ 12 x + 2y ~ x 2 O,y 2 (iv) x + 2y $ 3x + 4y$ 2x + y $ x 2 O,y 2
10 0
14 36 10 0
x + 4y $ x 20, y 2 (v) x + 3y $ 2x + y $ 4x + 3y $ x 2 0, y 2
12 0 15 12 24 0
(vi)
3x + y $ x 2 0, y 2 2x + y $ 8x +15y $ x +y $ x 2 0, y 2
12 0 20 120 11 0
5.6 LINEAR PROGRAMMING A function which is to be maximized or minimized is called an objective function. Note that there are infinitely many feasible solutions in the feasible region. The feasible solution which maximizes or minimizes the objective function is called the optimal solution. The theorem of linear programming states that the maximum and minimum values of the objective function occur at corner points of the feasible region.
Procedure for determining optimal solution: (i) (ii) (iii)
Graph the solution set of linear inequality constraints to determine feasible region. Find the comer points of the feasible region. Evaluate the objective function at each corner point to find the optimal solution. Example 1: Find the maximum and minimum values of the function defined as: f(x, y)= 2x + 3y subject to the constraints; x- y ~ 2 x +y ~ 4 2x - y ~ 6, x ~ 0 Solution: The graphs of x - y ~ 2 is the closed halfY . plane on the origin side of x - y = 2 and the graph of x + y ~ 4 is the closed half-plane not on the origin side of x + y = 4. The graph of the system x - y ~ 2, x + y ~ 4 including the non-negative constraints x ~ 0 is partially displayed in the figure 5.61. The graph of 2x - y $ 6 consists of the graph of the · line 2x - y = 6 and the half plane on the origin side of the line 2x - y = 6. A portion of the solution of the given system of inequalities is in the figure 5.62.
Y' Fig. 5.61
Calculus and Analytic Geometry
244
y
We see that feasible region is unbounded upwards and its comer points are A(O, 4), B(3, 1) and C(4, 2). Note that the point at which the lines x+y = 4 and 2x - y = 6 intersect is not a comer point of the feasible region. It is obvious that the expression 2x + 3y does not posses a maximum value in the feasible region because its value can be made larger than any number by increasing x and y. We calculate the values off at the comer points to find its minimum value:
y'
Fig. 5.62
f(0,4) = 2(0)+3x4 = 12
/(3, 1) = 2 x 3 + 3 x 1 = 6 + 3 = 9 /(4, 2) = 2 x 4 + 3 x 2 = 8 + 6 = 14 Thus the minimum value of 2x + 3y is 9 at the comer point (3, 1).
Note: If/ (x, y) = 2x + 2y, then/ (O·, 4) = 2 x 0 + 2 x 4 = 8,/(3, 1) = 2 x 3+2xl
= 6+2
= 8 and/(4, 2) = 2 x 4 + 2 x 2 = 8 + 4 = 12. The minimum value of 2x + 2y is the same at two comer points (0, 4) and (3, 1). We observe that the minimum value of 2x + 2y at each point of the line f(x, y) = 2x + 2(4-x)
segment AB is 8 as:
(°: x + y
=4
=}
y =4 - x)
= 2x+8-2x = 8 Example 2: Find the minimum and maximum values of/and
1, then the conic is a hyperbola.
273
Unit 6: Conic Section
The fixed line L is called a directrix and the fixed point F is called a focus of the conic. The number e is called the eccentricity of the conic.
6.4.1 PARABOLA
We have already stated that a conic section is a parabola if e = I
We shall first derive an equation of a parabola in the standard form and study its important properties. If we take the focus of the parabola as F(a,O), a > 0 and its directrix as line L whose equation is x =-a, then its equation becomes very simple. Let P(x. y) be a point on the parabola. So, by definition
jPFj = 1. IPM j Now and
or
jPFj=jPMj
jPMj=x+a
iPFj =~(x-a) 2 +(y-0)2
Substituting into (l), we get
~(x-a) 2 + y2 = x+a y2 = (x+a) 2
or
(x-a) 2 +
or
Iy 2 = (x + a) 2 -
(x
2
- a) = 4ax
or y
2
=4ax I
(2)
which is standard equation of the parabola.
Definitions (i)
The line through the focus and perpendicular to the directrix is called axis of the parabola. In case of (2), the axis is y = 0.
(ii)
The point where the axis meets the parabola is called vertex of the parabola. Clearly the equation (2) has vertex A(0,0). The line through A and perpendicular to the axis of the parabola has equation x = 0 . It meets the parabola at coincident points and so it is a tangent to the curve at A.
(iii)
A Jine joining two distinct points on a parabola is called a chord of the parabola. A chord passing through the focus of a parabola is called a focal chord of the parabola. The focal chord perpendicular to the axis of the parabola ( l) j s called latusrectum of the parabola. It has an equation x = a and it intersects the curve at the points where
Calculus and Analytic Geometry
274
= 4a2
)'2
or
y=±2a
Thus coordinates of the end points L and L' of the latusrectum are
'L(a,2a) and L'(a,-2a). The length of the latusrectum is (iv)
ILL'I =4a.
The point (at 2 , 2at) lies on the parabola x
y2
= 4a.x
for any real t.
= at 2 , y = 2at
are called parametric equation& of the parabola y 2
= 4ax .
6.4.2 General Form of an Equation of a Parabola. Let F(h,k) be the focus and the line lx+my+n =0 be the directrix of a parabola.
An equation of the parabola can be derived by the definition of the
parabola . Let P(x, y) be a point on the parabola. Length of the perpendicular PM from P(x, y) to the directix is given by; Jlx+my+nJ IPMI-- .J12 +m2
.
Le.,
(
X
-/) 2
'l
+ ( Y -k)2 -_ (lx+my+n) 2 2
2
l +m
is an equation of the required parabola. A second degree equation of the form 2
ax + by
2
+ 2 gx + 2 fy + c = 0
with either a= 0 or b = 0 but not both zero, represents a parabola. The equation can be analyzed by completing the square.
6.4.3 Other Standard parabolas There are other choices for the focus and directrix which also give standard equations of parabolas. (i)
If the focus lies on the y-axis with coordinates F(O,a) and directrix of the parabola is y = -a, then equation of the parabola is 2
x = 4ay
The equation can be derived by definition
(3)
Unit 6: Conic Section
275
If the focus is F(O,-a) and directrix is the line y =a, then equation of the parabola is x 2 = -4ay (4) Opening of the parabola is upwards in case of (3) and downwards in case of (4). Both the curves are symmetric with respect to the y-axis.
(ii)
The graphs of (3) and (4) are shown below.
y
y 0 F(O, a) F(O,-a)
(iii)
If the focus of the parabola is F(-a,0), and its directrix is the line x =a, then equation of the parabola is
y2
= -4ax.
The curve is symmetric with respect to the x-axis and lies in the second and third quadrants only. Opening of the parabola is to the left as shown in the figure
6.4.4
Graph the Parabola y 2 = 4ax
(1)
We note that corresponding to each positive value of x there are two equal and opposite values of y. Thus the curve is symmetric with respect to the x-axis. The curve passes through the origin and x =0 is y tangent to the curve at (0,0). If xis negative, then
y2 is negative and so y is imaginary. Thus no portion of the curve lies on the left of the y - axis. As x increases, y also increases numerically so that the curve extends to infinity and lies in the first and fourth quadrants. Opening of the
--~~~~~~~~~~--
Calculus and Analytic Geometry
276 parabola is to the right of y - axis.
Sketching graphs of other standard parabolas is similar and is left as an exercise.
Summary of Standard Parabolas Sr.No. Equation
1 y
2
2
3
4
y =-4ax
x 2 =4ay
x 2 = -4ay
2
= 4ax
Focus
(a, o)
(-a, o)
(o, a)
(o. -a)
Directrix
x=-a
x=a
y =-a
y=a
Vertex
(o,o)
(o,o)
(0,0)
(o,o)
Axis
y=O
y=O
x=O
x=O
Latusrectum
x=a
x=-a
y =a
y =-a
-~)'
Graph
Example 1:
+E·
~
0
0
I
'
;\
'
Analyze the parabola x 2 = -16 y and draw its graph.
Solution: We compare the given equation with x 2 = -4ay Here
x
\:' )
4a =16
or
a=4.
)'
0
The focus of the parabola lies on the y-axis and its opening is downward. Coordinates of the focus= (0, -4). Equation of axis is x
F(0,-4)
=0
Length of the latusrectum is 16 and y = 0 is tangent to the parabola at its vertex. The shape of the curve is as shown in the figure.
Unit 6: Conic Section
277
Example 2: Find an equation of the parabola whose focus is F (-3, 4) and directrix
is 3x-4y+5 =0. Let P (x, y) be a -point on the parabola. Lentgh of the perpendicular
Solution:
/PM/ from P(x,y)tothedirectrix 3x-4y+5=0 is /PM/= /3x-4y+5/
~3 2 +(-4)2
By definition, /PF/ =/PM/
2
/PF / =/PM1
or
= (3x-4v+5) 25
2
2
or
(x+ 3) 2 + ( y- 4 )2
or
25(x 2 +6x+9+ y 2 -8y+16)= 9x 2 +16y 2 +25-24.xy+30x-40y
or
16x 2 +24.xy+9y 2 +120x-160y+600=0
is an equation of the required parabola. Example 3: Analyze the parabola x
2
-
4x - 3 y + 13 = 0
(1)
and sketch its graph Solution: The given equation may be written as
x2
-
4x + 4 = 3 y - 9 2
=3(y -
or
(x - 2)
Let
x-2=X, y-3=Y
(2)
3).
The equation (2) becomes X 2 = 3Y
(3)
which is a parabola whose focus lies on X = 0 and whose directrix is Y = Thus coordinates of the focus of (3) are
-3 . X=O Y= -
'
i.e.,
x-2= 0
4
and
3 y-3= 4
3 . 4
Calculus and Analytic Geometry
278 15 x = 2, )' = -
or
4
y
-F(2, 15/4)
Thus coordinates of the focus of the parabola (1) are
(2. 1:) =0
Axis of (3) is X is the axis of (I)
or
A (2.3)
x-2
=
y = 914
Z(2,9/4)
0
---0-l----+--x-=~2=-----x
Vertex of (3) has coordinates
x = 0, y = 0 x - 2 = 0, y - 3 = 0
or
i.e., x
= 2, y = 3
are coordinates of the vertex of (1 ).
Equation of the directrix of (3) is
y
= --
3 . 3 1.e., y- 3 = -- or y 4 4
= -9
4
. of (1) . an equation . of the ct•irectnx is · .
Magnitude of the latusrectum of the parabola (3) and also of (1) is 3. The graph of (1) can easily be sketched and is as shown in the above figure. Theorem 1: The point of a parabola which is closest to the focus is the vertex of the parabola y Proof: Let the parabola be x
2
= 4ay ,a > 0
with focus at F (0, a) and P(x, y) be any point on the parabola.
---"'--+--===------- x 0
jPFI = ,jx2 +(y-a)2
A (0,0)
=
~4ay+(y-a)
2
= y+a Since y can take up only non-negative values,
jPFI
is minimum when y = 0.
Thus P coincides with A so that all points on the parabola, its vertex A it is closest to the focus.
Unit 6: Conic Section
279
Example 4: A comet has a parabolic orbit with the sun at the focus . When the comet is 100 million km from the sun, the line joining the sun and the comet makes an angle of 60° with the axis of the parabola. How close will the comet get to the sun? Solution: Let the sun S be the origin. If the vertex A of the parabola ZM has coordinates (-a, 0) then directrix of the parabola is (a >O)
x =-2a,
if the comet is at P(x, y). then
i.e.,
z
IPsj;,,, jPMI
by definition
x 2 +y 2 =(x+2a) 2
or
y
Now
2
·
= 4ax + 4a2 is orbit of the comet
jPsl =~ x2 + y 2 = x+2a = 100,000,000
The comet is closest to the sun when it is at A.
x = PScos60°
Now
= jPsj = x+2a
Ix I or
2
2
x+2a
2
x
1
or
100,000,000 2a
or
a
or
x+2a
~
= 2, (lxl = l-2al =
2a)
=2
=25,000,000
Thus the comet is closet to the sun when it is 25,000,000 km from the sun.
Reflecting Property of the parabola. A frequently used property of a parabola is its reflecting property. If a light source is placed at the focus of a parabolic reflecting surface then a light ray travelling from F to a point Pon the parabola will be reflected in the direction PR parallel to the axes of the parabola.
Calculus and Analytic Geometry
280
The designs of searchlights, reflecting telescopes and microwave antenas are based on reflecting property of the parabola. Another application of the parabola is in a suspension bridge. The main cables are of parabolic shape. The total weight of the bridge is uniformly distributed along its length if the shape of the cables is parabolic. Cables in any other shape will not carry the weight evenly. Example 5: A suspension bridge with weight uniformly distributed along the length has two towers of 100 m height above the road surface and are 400 m apart. The cables are parabolic in shape and are tangent to road surface at the centre of the btidge. Find the height of the cables at a point 100 m from the centre. Solution: The parabola formed by the cables has A(O,O) as vertex and focus on the y-axis. An equation of this parabola is x 2 = 4ay .
p
(~ ' ] ~200, 100) A (0 , 0)
The point Q(200,100) lies on the parabola and so (200) 2 = 4a x 100
or
a =100
Thus an equation of the parabola is x
2
= 400y.
To find the height of the cables when x = 100, we have from (1) (100) 2 = 400y
or
y=25
Thus required height = 25 m
(1)
Unit 6: Conic Section
281
EXERCISE 6.4 1.
Find the focus. vertex and directix of the parabola sketch its graph. 2
(i)
y =8x
(iv)
y
2
= -12x
(vii) (x-1) 2 =8(y+2) (x) 2.
(ii)
x 2 =-16y
(v)
x 2 = 4(y -1) (vi)
(viii) y
= 6x 2 -J
(iii)
(ix)
x 2 =5y
y2 = -8(x-3) x+8-y2+2y=O
x 2 -4x-8y+4=0
Write an equation of the parabola with given elements. (i) Focus ( - 3, l); directrix x
=3
(ii) Focus (2, 5);
directrix y
=1
(iii) Focus ( -3, l); directrix x-2y-3 = 0 (iv) Focus (1, 2); vertex (3, 2) (v) Focus (-1, 0); vertex (-1. 2)
(vi) Directrix x
= -2, Focus (2. 2)
(vii)Directrix y = 3 ; vertex (2, 2) (viii) Directrix y (ix) Axis y
= 1, length of latusrectum is 8. Opens downward.
= 0, through (2,
1) and (11, -2)
(x) Axis parallal to y-axis, the points (0, 3), (3, 4) and (4, 11) lie on the graph.
3.
Find an equation of the parabola having its focus at the origin and directrix parallel to the (i) the x-axis (ii) they - axis. Show that an equation of the parabola with focus at (a cos a, a sin a) and directrix xcosa+ysina + a=Ois 2
(xsina - ycosa) = 4a(xcosa+ysina)
,.
Show that the ordinate at any point P of the parabola is a mean proportional between the length of the latusrectum and the abscissa of P.
6.
A comet has a parabolic orbit with the earth at the focus. When the comet is 150,000 km from the earth, the line joining the comet and the earth makes an angle of 30 with the axis of the parabola. How dose will the comet come to· the earth?
7.
Find an equation of the parabola formed by the cables of a suspension bridge whose span is a m and the vertical height of the supporting towers is b m.
282 8.
Calculus and Analytic Geometry
A parabolic arch has a 100 m base and height 25 m. Find the height of the arch at the point 30 m from the centre of the base. Show that tangent at any point P of a parabola makes equal angles with the line PF and the line through P parallel to the axis of the parabola, F being focus. (These angles are called respectively angle of incidence and angle of reflection).
9.
6.5
ELLIPSE AND ITS ELEMENTS We have already stated that a conic section is an ellipse if e < 1.
Let 0 < e < 1 and F be a fixed point and L be a fixed line not containing F. Let P(x, y) be a point in the plane and IPMI be the perpendicular distance of P from L. The set of all points P such that
is called an ellipse. The number e is eccentricity of the ellipse, Fa focus and La directrix.
6.5.1
Standard Form of an Ellipse Let F(-c,0) be the focus and the line x
=- ~
be the directrix of an ellipse e Let P(x, y) be any point on the ellipse and
with eccentricity e, (0b b" a 2
.r:
'
/\
B'
c
-~ Fig. l
y
B
~A' _/
B
x
Fig.2
c
\_cJ A
Note: In each ellipse Length of major axis = 2a, Length of minor axis= 2b. Length of Latusrectum
~
2b2 - , Foc1- r1e on the major axis. a
-x
B'
Unit 6: Conic Section
287
Example 1: Find an equation of the ellipse having centre at (0,0), focus at (0,-3) and one vertex at (0,4). Sketch its graph. Solution: The second vertex has coordinates (0, -4). Length of the semi-major axis is
,.-------------------. y A'(O, 4)
a=4 c=3
Also From
p2 =
a2
c 2 , we have
-
F'( 0,-3)
2
b =16-9=7
.J7 which is length of
b=
B'( fi,O)
B(-J7, 0)
c
x
the semi-minor axis. Since the foci lie on the y-axis, equation of the el1ipse is
F(O, 3)
y2 x2 -+-=l.
16
A (0,-4)
7
The graph is as shown above
Example 2: , Analyze the equation 4x 2 +9y 2
= 36
(1)
and sketch its graph
Solution:
The given equation may be written as
,.-----------------"\ y
x2 y2 -+-=1
9
B'(O, 2)
4
which is standard form of an ellipse. Semi-major axis Semi-minor axis
=3 b =2 a
F ( .JS,O)
F '( .JS,O)
A~~-3~·~0)~--1-----1-------+---+---+X 11'(3,0)
c (0' 0)
From
b 2 = a 2 -c 2 we have B (0,-2) ' 2 2 2 c =b - a = 9- 4 = 5 '---------~------_...;
or
c
=±JS
Calculus and Analytic Geometry
288 Foci:
F(-J5,0), F'(J5,0);
Vertices:
f5
c . . Eccentnc1ty = - = - -.. a 3
Covertices: B(0,-2),B'(0,2) ;
Directrices:
A( - 3,0), A'(3,0)
c .[5 9 2b 2 4 x = + - 2 = + - = +--; Length of latusrectum = =- . - e a 3 - 519
f5
The graph is as shown above.
Example 3:
Show that the equation 2 9x -18x+4y 2 +Sy-23 =0
(1)
represents an ellipse. Find its elements and sketch its graph. Solution: We complete the squares in ( 1) and it becomes (9x 2 -18x+9)+(4y 2 +8y+4)-36=0 9(x-1) 2 +4(y+l) 2 =36
or
(x-1)2 + (y+l)2 · =l
or
4
(2)
9
If we set x - 1 = X, y + 1 = Y in (2), it becomes
x2 y2 -+-=1 22 32
(3)
which is an ellipse with major axis along X = 0 i.e., along the line x-1=0 (i.e., a line parallel to the y-axis) Semi-major axis= 3,
Semi-minor axis = 2
c = .J9 - 4 = J5, Eccentricity = .!!.._ = J5 .
a Centre of (2) is X = 0, Y = O or
x=l, y=-1 i.e., (1,-l)iscentreof(l)
The foci of (2) are
x
=0, y =±JS
3
Unit 6: Conic Section
289
=0, y + 1 =±.JS
i.e.,
.X-1
i.e.,
(1,-l+.J5) and (1,-1-JS) arefociof(l).
Vertices of (2) are X
=0,
Y
= ±3
i.e.,
x = 1, y
=-1±3
or (1,-4) and (1, 2) y
are the vertices of (1). Covertices of (2) are
x =±2, i.e.,
Y=O
x -1 = ±2, y + 1 = 0 (-1,-1)
or
and
B (-1,-1)
(3,-1)
are the covertices of (1 ). A (1 , -4)
The graph of (1) is as shown
Example 4: An arch in the form of half an ellipse is 40 m wide and 15 m high at the centre. Find the height of the arch at a distance of 10 m from its centre.
Solution: Let the x-axis be along the base of the arch and the y-axis pass through its centre. An equation of the ellipse representing the arch is
__£_+L=1 2 2 20
(1)
15
.Let the height of arch at a distance of 10 m from the centre bey. Then the points (10, y) lies on (1 ). For. x
= 10, we have
15./3
so that y = - 2
. d heig . ht Require
=-152./3- m.
B(O, 15)
P(l 0, 0)
A(-20, 0)
C
A'(20, 0)
Calculus and Analytic Geometry
290
EXERCISE 6.5 1.
Find an equation of the ellipse with given data and sketch its graph: (i)
Foci (±3,0) and minor axis of length 10
(ii)
Foci (0,-1) and (0,-5) and major axis of length 6.
(iii) (iv)
Foci (-3J3,0) and vertices (±6,0) Verti~s (-1, J ), (5, I); foci (4, 1) and (0, 1)
(v)
Foci (±J5,0) and passing through the point (
(vi)
Vertic•s (0, ± 5), eccentricity
'
%,J3)
~. 5
(vii) (viii)
2.
Centri (0,0), focus (0, - 3 ), vertex (0,4) Centre (2, 2), major axis parallel to y-axis and of length 8 units, minor axis parallel to x -axis and of length 6 units. (ix) Centre (0, 0), symmetric with respect to both the axes and passing through the points (2, 3) and (6, 1). (x) Centre (0, 0), major axis horizontal, the points (3, 1), (4, 0) lie on the graph. Find the centre, foci, eccentricity, vertices and directrices of the ellipse whose equation is given: 2 2 2 (i) x +4y 2 =16 (i) 9x + y = 18
'
(iii) (v)
3.
2
25x + 9y 2 = 225 2
(iv)
2
x +16x+4y -16y+76=0
(2x -1) 2 4
+
(y + 2) 2
16
=1
(vi) 25x 2 +4y 2 -250x-16y+541=0
Letabeapositivenumberand O', y·=rsin(B+a)
or
x = rcosBcosa-rsin e sin y =.r sine cos a - r cos e sin a
a}
(2)
Substituting the values of r cos a, r sin a from (1) into (2), we get
BJ
x = X cosB-Y sin y=XsinB+Ycose
as the required equations of transformation for a rotation of axes through an angle Example 3:
Find an equation of 5x 2
6.xy + 5 y 2
-
-
e.
• 8 = 0 with respect to new axes
obtained by rotation of axes about the origin through an angle of 135°. Solution: Here 8 = 135°. Equations of transformation are
-x y -1 x = X cos135° -Y sinl35° = J2 - J2 = J2 (X + Y) •
0
x=Xszn135 -Ycos135
0
x y = J2- J2=
1 J2(X-Y)
Substituting these expressions for x, y into the given equation, we have
s(- xJ{)'- 6(- xJiY. x;/}s(xiiYJ-s~o 5
2
.
5
or
-(X +2XY+Y 2 )+3(X 2 -Y 2 )+-(X 2 -2XY+Y 2 )-8=0
or
8X 2 +2Y 2 -8=0
2
2
4X 2 +Y 2 =4
or
is the required transformed equation. Example 4: Find the angle through which the axes be rotated about the origin so that the product term XY is removed from the transformed equation of 2 5x + 2.fjxy + 7 y2 -16 = 0 . Also find the transformed equation. Solution: Let the axes be rotated through an angle
x=
e. Equations of transformation are
x cos e - y sine ; y = x sine + y cos e
Substituting into the given equation, we get
Unit 6: Conic Section
313
5(X cos e-Y sine) 2 + 2.J3cx cose-Y sine) (X sine+ Y case)
+ 7(X sine+ Y cos e) 2 -16 = 0
(1)
Since this equation is to be free from the product term XY, the coefficient of Xf is zero, i.e., -10 sine case+ 2.J3(cos 2 e- sin 2 e)+14 sine cos e = 0 or
2 sin 2e + 2.J3 cos2e = 0
or
-2.J3 tan2e=--=tan120° 2
e=60°.
or
Thus axes be rotated through an angle of 60° so that XY term is removed from the transformed equation. Setting
e = 60° in (1), the transformed equation is (after simplification) 8X 2 +4Y 2 -16=0 or2X 2 +Y 2 -4=0
EXERCISE·6.8 1.
Find an equation of each of the following with respect to new parallel axes obtained by shifting the origin to the indicated point:
(i)
x 2 +16y-16
(ii)
4x 2 +y2+16x-10y + 37 = 0,
0'(-2, 5)
(iii)
9x 2 +4y 2 +18x-16y-11 =0,
0'(-1, 2)
'iv)
x 2 -y 2 +4x+8y-11
=0,
0'(-2, 4)
(v)
9x 2 -4y 2 +36x+8y-4
=0,
0'(-2, 1)
= 0,
O'(O, 1)
;
2.
Find coordinates of the new origin (axes remaining parallel) so that first degree terms are removed from the transformed equation of each of the following. Also find the transformed equation: (ii) 25x 2 + 9y 2 + 50x-36y-164 = 0
(i)
3x 2 -2y 2 +24x+12y+24=0
(iii)
x 2 -y 2 -6x+2y+7=0
3.
In each of the following, find an equation referred to the new axes obtained by rotation of axes about the origin through the given angle:
(i)
xy = 1, e = 45°
(ii)
7x 2 -8.xy + y2 -9 = 0, e = arctan 2
Calculus and Analytic Geometry
314 2
(iii)
9x2+12.xy +4y2 -x- y = 0,
e =arctan3
(iv)
x -2.xy + y -2.fix-2.fiy + 2 = 0'
4.
Find measure of the angle through which the axes be rotated so that the product term XY is removed from the transformed equation. · Also find the transformed equation:
2
2
/
e = 45
(i)
2x 2 +6xy+10y 2 -11
=0
(ii)
5x 2 -6xy+5y 2 -8
=0
6.9
THE GENERAL EQUATION OF SECOND DEGREE
xy + 4x- 3y -10 = 0
(ii)
Standard equations of conic sections, namely circle parabola, ellipse and hyperbola have already been studied in the previous sections. Now we shall take up the general equation of second degree viz. 2
2
Ax + By + Gx + Fy + C
=0
(1)
The nature of the curve represented by (1) can be determined by examining the coefficients A, B in the above equation. The following cases arise: (i)
If
A = B -:t 0, equation (1) may be written as 2 2 2 2 G F C A (x + y ) + Gx + Fy + C = 0 or x + y + A. x +A y + A
=0
I
. h . 1 'th ( G F) - G2 p2 whic represents a clfc e w1 centre at - 2A , - 2A) and radius -\j A 2 + A 2 4 4 (ii)
-
c
A·
If A -:t B and both are of the .same sign, then we have (Ax2 + Gx) + (By2 + Fy) + C = 0
or A
:;2
(x + ~ x+ ~2J+ B (i + ~ y + J = ~~ + :: - C 2
2
( F) G )2 or A ( x + 2A) + B y + 2B)
2
=
G p2 4A + 4B - C
(2)
' wnte ·x = x + 2A G , Y =y + FB , then (2) can be written as If we 2 2
2
AX + BY
G2
= 4A
p2 + 4B - C
X2
= K (say)
or (
'1KJA) 2
y2 + (1K/B) 2 =1
which is standard equation of an ellipse in XY-coordinate system;
Unit 6: Conic Section
315
If A :t Band both have opposite signs (say A is positive and Bis negative),
. (iii)
we can write (1) as
Ax 2 -B
y 2 + Gx + Fy + C=
0, where B = -B' and'B'is positive.
l
( F F2 G . G2 ) 2 or A (x +Ax+ A2 -B' / - B,y+ 4 B,z) 4
or
+ ~J+
B'(y-
G2
xz or
- 4B' -C
.
=M
. (say)
:S,) = M
G 2 2 or AX -B'Y = M , where X = x + 2A , Y = y i
F2
= 4A
F
B, 2
y2
'1
( MI A )2
-
(
'1M I B')
2
= 1
and this is standard equation of a hyperbola in XY-coordinates system. (iv)
If A = 0 or B = 0 (both cannot be zero since in that case the ·equation (1) reduces to a linear equation). Assume A 0 and B = 0.
*
The equation ( 1) becomes 2 G G ) 2 or A .(x + A x + 4A 2
or
2
Ax + Gx + Fy + C
=0
2
= - Fy -
G C + 4A
A(x+ ~J = -F(y + ; - Z~J
.G C G2 or AX = - FY, where X = x + 2A , Y = Y + F - 4AF · 2
which is standard equation of a parabola in XY-coordinates system. We _summarize these results as under: 2 Let an equation of second degree be of the form Ax + B/ + Gx + Fy + C =O·. It represents: (i) a circle if A = B :t 0 (ii) an ellipse if A B and both are of the same sign (iii) a hyperbola if A B and both are of opposite signs (iv) a parabola if either A= 0 or B = 0.
*
*
6.9.1 Classification of Conics by the Discriminant The most general equation of the second degree
ru:2 + 2hxy + by2 + 2gx + 2fy + c = 0
(1)
Calculus and Analytic Geometry
316
represents a conic. The quantity h 2 - ab is called the discriminant of (1). Nature of the conic can be determined by the discriminant as follows. (1) represents: 2
(i) an ellipse or a circle if
h
(ii) a parabola if
h
2 2
(iii) a hyperbola if
h
-
ab < 0
-
ab
-
ab > 0
=
0
The equation (1) can be transformed to the form 2
2
AX + BY + 2GX + 2Fy + C
=
if the axes are rotated about the 01igin through an angle given by tan 28
=
(2)
0
e, (0 < e < 90°) where e is
2h
a-b
If a= b or a = 0 = b, then the axes are to be rotated t~ough an angle of 45°.
Equations of transformation (as already found) are x
=
x cos e- y sin e ] (3)
y = x sine+ y cos e
Substitution of these values of x, y into (1) will result in an equation of the form (2) in which product term XY will be missing. Nature of the conic (2) has already been discussed in the last article. Solving equations (3) for X, Y we find
x = x cos e+ y sin e J y = - x sin e+ y cos e
(4)
These equations will be useful in numerical problems. Note: Under certain conditions equation (1) may not represent any conic. In such a_ case we say (1) represents a degenerate conic. _One such degenerate conic is a pair of straight lines represented by (I) if
a
h
g
h
b
f
g
f
c
= 0.
The _proofs of the above observations are beyond our scope and are omitted. 2
Example 1: . Discuss the conic 7x
and find its elements.
-
6°'13 xy + 13y2- 16 =O
(1)
317
Unit 6: Conic Section
Solution: In order to remove the term involving xy, the angle through which axes · be rotated is given by
-613 = \J3 _r:;
tan 20 = 7 - 13
e = 30°.
or
Equations of transformation are
x y
= X cos 30° ~ Y sin 30° = j3 ~ = X sin 30° + Y cos 30°
y
= x+13 Y
l
(2)
2 Substituting these expressions into the equation (1), we get
7(:f3;-YJ - 6:}3 (fi;-Yw+t y)+u(X+f YJ = 16 which simplifies to 2
4X + 16Y
2
xz
= 16
-
or
4
yz + 1
=
1
(3)
This is an ellipse. Solving equations (2) for X and Y,(or as already found in (4) of 7.7.1) we have
X=
j3x+y 2
Y=
,
-x+j3y 2
Centre of the ellipse is X = 0, Y = 0
{3 x + y = 0 and - x + {3 y = 0 giving x =0, y =0. Thus centre of (1) is (0, 0) Length of the major axis = 4, length of minor axis = 2 Vertices of (3) are: X = ± 2, Y =0 i.e.,
.
j3x+y
= ±2 and . 2 Solving these equations for x, y, we have
i.e.,
({3 ,
1) , ( -
{3 , -
- -x+j3y 2
= 0
1) , as vertices of (1).
· · are X =0 an d Y =± 1. i.e.; · j3x+y = 0 Ends of the mmor axis 2 - x + -}3 y = ± 1. 2 as ends of the minor axis.
and
So~ving these equations, we get (k,
Equation of the major axis:
Y
= 0,
i.e., - x + {3 y
Equation of the minor axis:
X
= 0,
i.e.,
{3
x +y
=0
=0
-r) (-k, ~) and
Calculus and Analytic Geometry
318
Analyze the conic xy = 4 and write its elements. Example2: Solution: Equation of the conic is
=
xy-4
(1)
0
Here a = 0 = b, so we rotate the axes through an angle of 45°. Equations of transformation are
X-Y ~
x = X cos 45° - Y sin 45° = y = X sin 45° - Y cos 45° =
(2)
X+Y ~
Substituting into (1 ), we have
(X~Y) (XfiY)or or
4=
0
x2_ y2 = 8 x2 y2 g--g- = 1
(3)
which is a hyperbola. Solving equations (2) for X, Y, we have x+y -x+y X=~, Y=~ Centre of the hyperbola (3) is
x = 0, i.e.,
£.±2
or
x=O,
~
=
y=0 0,
and
Y
Equation of the conjugate axis:
Foci of (3): X
=~ = ± 2~
or
x+y ~
or
x+y
and
-x+ y
=0
or
y=O isthecentreof(l)
Equation of the focal axis: Eccentricity
-x+y ~
. ~ ±4,
= 0 i.e., y = x X = 0 i.e., y = - x. Y=O
-x+y
= ~ = ±4~ =
0
=0
x
=0,
y
=0
Unit 6: Conic Section
319
Solving the above equations for x, y, we have the foci of ( 1) as ( 2fi , 2fi) and ( -2
fi , - 2fi)
Vertices of (3) X i.e.,
:{iy
= ± 2 fi,
Y
=0
= ± 2fi,
and -x+ ·y
=0
Solving these equations, we have (2, 2) , (-2, - 2) as vertices of (1). Asymptotes of the hyperbola (3) are given by or
x2_ y2 = o X-Y = 0
X+Y =0
and
i.e., 1.e.,
Example3:
x=0
y = 0 are equations of the asymptotes of (1).
and
By a rotation of axes, eliminate the xy-term in the equation
=0
9x2 + 12xy + 4/ + 2x - 3y
(1)
Identify the conic and find its elements.
Solution:
Here a
=9,
or or
= 4,
2h
tan 28
rotated to given by or
b
2 tan(} 1 - tan2 fJ
= 12 the angle e through which axes be 12 12 -- 9-4 - - 5
12
= -5 5 tan fJ = 6-6 tan2 fJ 6 tan2 fJ + 5 tan fJ - 6 = 0 tan fJ
-5±13 + 144 -2 = -5 ±'125 = = 12 3' 12
Since fJ lies in the first quadrant, tan fJ
tan fJ
2
=3
=>
sin fJ
=- ~ 2
= '113 ,
Equations of transformation become
-3 2
is not admissible.
cos
3
(J
= '113
Calculus and Analytic Geometry
320
~
x
y
.Ju
~XY 2 3 = '113 X + -{13 Y
X cos 8- Y sin 8 =
=
X sin 8 + Y cos 8
l
(2)
Substituting these expressions for x and y into (1 ), we get
9 '( 113 )
2
(3X - 2Y)
2
12
4
+ii
(3X - 2Y) (2X + 3Y) + l3 (2X +3Y)
+ 9 2 2 13 (9X - 12XY +4Y )
or
12
+ -13
-
l~
+
2
(4X + 12XY + 9Y
81 72 (ii + 13
or
16)
+ii
2
X +
'1213
(
3
(3X - 2Y)- -{13 (2X + 3Y)
13X
2
-
ffi Y = 0
= 0
(6X 2 +5XY - 6Y 2 )
)-'113 Y = 0
2
108 60 48) -13 +ii+ ii XY
36 72 36) 2 • ,..,-;; + ( 13 - 13 + 13 y - \J 13 y or
2
X
or
2
=
Ju
=0
Y
which is a parabola.
.
.
Solvmg equation (2) for X, Y, we have X
2y = 3x+ -{13 ,
-2x+ 3y
y
= ffi
Elements of the parabola are: . Focus: X i.e.,
= 0,
3x+ 2y _
Y
ffi - 0
1
= 4-{13 and
-2x + 3y
1
ffi = 4-{13 . these equations, . . SI o vmg we have x =- 1 , y = 3 . 1.e., Focus = (- 1 , 3) 2 26 52 26 5 3x + 2y = 0 and - 2x + 3y = 0
Vertex:X
= 0,
Y=0
x
= 0,
y =0 i.e., (0, 0) i.e., 3x + 2y = 0
i.e.,
Axis: X = 0
i.e.,
~---~--~---~--------~------~~--~~-----
Unit 6: Conic Section
.
x-mtercept =
2 -9,
321
y-intercept =
3 4
.
Example 4: Show that 2x2 - xy + 5x - 2y + 2 =0 represents a pair of lines. Also fmd an equation of each line. 1 5 Solution: Here a= 2, b =0, h = - 2, g = 2 , f = -1, c = 2.
a
h
h
b
f
g
f
c
g
1 2
=
-5
-1 2
2
5 2
2
0
-1
-1
2
= k(-1+~)+1 (-2 3
3
=4-4=
+
%)
0
The given equation represents a degenerate conic which is a pair of lines. The given equation is
2x2 +x(5-y) + (-2y+2)
x
or
=
= =
=
y- 5 ± j(y- 5)
2
-
=0
8 (-2y + 2)
4
y - 5 ±Ji - 10y + 25 + l 6y - 16 4 y-5 ± (y + 3)
4 2y-2 4 '-2
Equations of the lines are 2x - y + 1
=0
and x + 2
= 0.
Tangent Find an equation of the tangent to the conic ax
2
+ 2hxy+by 2 + 2gx+ 2.fy +c =0
at the point (x1. Yi) Differentiating (1) w.r.t. x, we have
(1)
Calculus and Analytic Geometry
322
2ax + 2hy + 2hx dy + 2by dy + 2g + 2f dy dx dx dx
or
dy = dx
or
dy] dx (x,,y, ) =
=0
ax+hy+g hx+by+ f ax 1+ hy 1 + g hx1 +byl + f
= slope of the tangent at (x
1,
Yi)
Equation of the tangent at (x 1 , y 1 ) is ax1 + hy1 + g ( ) y - Y1 = - hx1 + byl + f x - Xi
or
(x-x1)(ax1 + hy 1+ g) +(y- y 1)(hx1+by 1+f)=0
·or
axx1 +hxy1 +gx+hx1y+bY1Y+ fy =ax~ + 2hx 1y 1 + gx1 +by[ + fy 1
Adding gx1 + fy 1 + c to both sides of the above equation and regrouping the terms" we have
ax:x1 + h(xyi + yx1) + hw1+ g(x + x1) + f(y + Y1) + c =ax; + 2hx1y1+by; + 2gx1 + 2fy 1 + c =0
since the point (x 1 , y 1 ) lies on (1). Hence an equation of the tangent to (1) at (x 1 , y 1 ) is
axx1 + h(xy1 + YX1) + hw1+ g(x+ x1) + f(y + Y1) + c = 0 Note: An equation of the tangent to the general equation of the second degree at the point (x 1 , y 1 ) may be obtained by replacing
x2
by
XX1
y2
by
YY1
2xy
by
xyl + YX1
2x
by
x+x1
2y
by
y+yl
in the equation of the conic. ~
Unit 6: Conic Section
323 2
Find an equation of the tangent to the conic x 2 -xy + y -2 = 0 at the
Example 5:
.J2, .
point whose ordinate is Solution:
Putting y = .J2 into the given equation, we have
x 2 -.J2x = 0 x(x-../2) = 0
x
The two points on the conic are (0, Tangent at (0,
= 0, .J2
.J2) and (../2, .J2) .
.J2) is
O.x-__!_(x .../2 + O.y) + ../2y-2 = 0 2 or
x - 2 y + 2../2 =0
Tangent at (.J2, .J2) is
.J2 x-__!_(.J2 x+../2 y)+../2 y-2=0 2 or
.J2 x + .J2 y - 4 = 0
EX-ERCISE 6.9 1.
By a rotation of axes, eliminate the xy-term in each of the following equations. Identify the conic and find its elements:
(i)
4.x2 - 4.xy + y2 - 6 = 0
(ii)
.x2 - 2xy + y2 -
(iii)
x 2 + 2.xy + y 2 + 2.J2x-2../2y + 2 = 0
(iv)
.x2 + xy + y2 -
8x - 8y
4
=0
=0 2
(v)
7x 2 -6.J3xy+13y -16 = 0
(vi)
4x2 - 4.xy + 1y2 + 12x + 6y - 9 = 0
(vii)
xy-4x-2y =0
(viii)
x 2 +4xy-2y 2 -6=0
324
Calculus and Analytic Geometry
(ix)
x 2 -4.xy-2y 2 + lOx+ 4y = 0
2.
Show that
(i)
10 xy + Sx - 15y - 12 = 0
(ii)
6x 2 +xy-y2 -21x-8y +9 = 0
and
each represents a pair of straight lines and find an equation of each line . . 3.
Find an equation of the tangent to each of the given conics at the indicated point.
(i)
3x 2
(ii)
x
(iii)
x 2 +4.xy-3y 2 -5x-9y+6=0, at x=3.
2
-
7 y2 + 2x - y - 48 = 0 , at (4, 1)
+ 5.xy - 4 y2 + 4 =0 , at y =-1
Unit 7: Vectors
325
Junit 11
Vectors
7.1 INTRODUCTION In physics, mathematics and engineering, we encounter with two important quantities, known as "Scalars and Vectors".
A scalar quantity, or simply a scalar, is one that possesses only magnitude. It can be specified by a number alongwith unit. In Physics, the quantities like mass, time, density, temperature, length, volume, speed and work are examples of scalars. A vector quantity, or simply a vector, is one that possesses both magnitude and direction. In Physics, the quantities like displacement, velocity, acceleration, weight, force, momentum, electric and magnetic fields are examples of vectors. In this section, we introduce vectors and their fundamental operations we begin with a geometric interpretation of vector in the plane and in space.
(;,al
flil)
7.1.1 Geometric Interpretation of Vector Geometrically, a vector is represented by a directed line segment AB with A its initial point and B its terminal point. It is often found 1convenient to denote a vector by an arrow and is written either as AB or as a boldface symbol like v or in underlined form.!:'.· (i)
The magnitude or length or norm of a vector AB or .!'.'., is its absolute value and is written as IAB I or simply AB or
(ii)
l.!:'.I ·
A unit vector is defined as a vector whose magnitude is unity. Unit vector of vector.!:'. is written as Q(read as .!:'. hat) and is defined by Q= I~
I
326
Calculus and Analytic Geometry
(iii)
If terminal point B of a vector AB coincides with its initial point A, then
(iv)
magnitude AB = 0 and AB =Q, which is called zero or null vector. Two vectors are said to be negative of each other if they have same magnitude but opposite direction. If AB=!'..
and
then
BA=-AB=-v
IBA I=I- AB I
7.1.2 Multiplication of Vector by a Scalar We use the word scalar to mean a real number. Multiplication of a vector !'. by a scalar 'k ' is a vector whose magnitude is k times that of !'. . It is denoted by k!'. . i)
If k is +ve, then !'. and k!'. are in the same direction.
(ii) (a)
If k is -ve, then !'. and k!'. are in the opposite direction.
Equal vectors
Two vectors AB and CD are said to be equal, if they have the same magnitude and same
I I
direction i.e., AB = j CD j
(b)
Parallel vectors
Two vectors are parallel if and only if they are non-zero scalar multiple of each other. (see figure).
7 .1.3 Addition and Subtraction of Two Vectors Addition of two vectors is explained by the following two laws:
a. (i) Triangle Law of Addition If two vectors !! and }'. are represented by the two sides AB and BC of a triangle such that the terminal point of ~ coincides with the initial point of~. then the third side AC of the triangle gives vector sum ~ + ~, that is
AB+BC=AC ==>
A~
!!+!:=AC
!!..
c
Unit 7: Vectors
327
Parallelogram Law of Addition
(ii)
D
\
If two vectors ~ and .!:'. are represented by two adjacent sides AB and AC of a parallelogram as shown in the figure, then diagonal AD give the
I
I
~ I I
sum or resultant of AB and AC ,
AD=AB+AC
I I
'c
that is A
AD= AB+AC=u+v
I Note: This law was used by Aristotle to describe the combined action of two forces. I b.
Subtraction of two vectors The difference of two vectors AB and AC is defined by
AB-AC= AB+(-AC)
In figure, this difference is interpreted as the main diagonal of the
-
-
parallelogram with sides AB and -AC. We can also interpret the same vector difference as the third side of a triangle with sides AB and AC. In this second interpretation, the vector difference AB - AC = CB points the terminal point of the vector from which we are subtracting the second vector.
7 .1.4 Position Vector
y
p
The vector, whose initial point is the origin 0 and whose terminal point is P; is called the position vector of the point P and is written as OP.
0
x
The position vectors of the points A and B relative to the origin 0 are defined by OA =f! and OB ='2_ respectively. In the figure, by triangle law of addition,
OA+AB=OB
-a+AB=b==>
AB= b-a
B
Calculus and Analytic Geometry
328
7 .1.5 Vectors in a Plane Let R be the set of real numbers. The Cartesian plane is defined to be the R = {(x, y) : x, y E R}. 2
An element (x, y) E R 2 represents a point P(x, y) which is uniquely determined by its coordinate x and y. Given a vector ~ in the plane, there exists a unique point P(x, y) in the plane such that the vector OP is equal to ~ (see figure). So we can use rectangular coordinates (x, y) for P to associate a unique ordered pair [x, y] to vector ~ .
y P(x , y)
0
We define addition and scalar multiplication in R 2 by: (i)
Addition:
For any two vectors ~ = [x, y] and ~ = [x', y'], we have ~ + ~ = [x, y]+ [x', y'] = [x + x', y + y']
(ii)
Scalar Multiplication:
For ~ = [x: y] and a E R , we have
a~ =a[x, y]= [ax,ay]
Definition: The set of all ordered pairs [x, y] of real numbers, together with the rules of addition and scalar multiplication, is called the set of vectors in R 2 • For the vector ~ = [x, y], x and y are called the components of ~ .
Note: The vector [x, y] is an ordered pair of numbers, not a point (x, y) in the plane. (a)
Negative of a Vector In scalar multiplication (ii), if a = -1 and ~ = [x, y], then a~=(-1)
[x, y] = [-x,-y]
which is denoted by-~ and is called the additive inverse of ~ or negative vector of ~.
(b)
Difference of two Vectors We define~-~ as~+(-~) If~= [x,
y] and ~ = [x', y'], then
~-~=~+(-~)
= [x, y]+ [- x',-y'].= [x- x', y- y']
Unit 7: Vectors
(c)
329
Zero Vector Clearly ~ + (-~)= [x, y ]+ [-x,-y ]= [x-x, y- y ]= [0,0]= Q
Q= [0,0] is called the Zero (Null) vector.
(d)
Equal Vectors
Two vectors ~ = [x, y] and .!:'. = [x', y'] of R 1 are said to be equal if and only if they have the same components. That is, [x, y]
and we write
(e)
!:!:
= [x', y']
if and only if x = x' and y
= y'
=.!:'.
Position Vector For any point
P(x, y) in R1 a vector ~ = [x, y] is represented by a directed
line segment OP, whose initial point is at origin. Such vectors are called position vectors because they provide a unique correspondence between the points (positions) and vectors. (f)
Magnitude of a Vector
y
For any vector ~ = [x, y] in R 1 , we define the magnitude or norm or length of the vector as the distance of the point P(x, y) from the origin 0.
P(x , y)
y
0
..
Magnitude of OP =
JoPJ =l~I =~ x + y 2
2
7.1.6 Properties of Magnitude of a Vector Let .!:'. be a vector in the plane or in space and let c be a real number, then (i)
I.!:'. I ;;:::o, and l.!:'.I =O if and only if v =0
(ii)
ic.!:'.i = lei i.!:'.I
Proof: (i) We write vector .!:'. in component form as .!:'. =
l.!:'.I= ~x 2 + y 2 ;;:::o
[x, y], then
for all x and y.
Further l.!:'.I= ~x 2 + y 2 = 0 if and only if x=O, y =0 In this case _!:'.=[0,0] = Q
-- -- -- ~ --
--~ --- ---
330
Calculus and Analytic Geo"!fletry
(ii)
7.1.7 Another Notation for Representing Vector in Plane We introduce two special vectors
y
·~
i = [l, 0], j = [O, I] in R 2
i
As magnitude of
j (0, I)
= ..J1 2 + 0 2 = 1
I 2 magnitude of l · =-vO + 12 = 1
0
"!.
• x (1 , 0) -
So ! and j are called unit vectors along x-axis, and along y-axis respectively. Using the definition of addition and scalar multiplication, the vector [x, y] can be written as y
= [x,
!:! =[x,y]
O]+[O, y]
= x[l, O] + y[O, I]
= x! + yj in R 2 can be uniquely represented by x! + y j.
[x, y]
Thus each vector
In terms of unit vector !:! = [x,
y]
! and
1 , the sum ~ + !: of two vectors
and !: = [x', y'] is written as
!:
~ + = [x + x', y + y'] = (x +
x') ! + (y + y')j
7.1.8 A Unit Vector in the Direction of Another Given Vector A vector His called a unit vector, if 1~
I= 1
Now we find a unit vector H in the direction of any other given vector.!::'.· We can do by the use of property (ii) of magnitude of vector, as follows: 1
1
l!:I !: = l!:l l!:I =1 the vector
~ = l~I !:
is the required unit vector
Unit 7: Vectors
331
It points in the same direction as~. because it is a positive scalar multiple of~Example 1:
For v=[l,-3] and w=[2,5] (i)
}:'. + w=[l,-3]+[2,5]=[1+2,-3+5]= [3,2]
(ii)
4}:'. + 2w = [4,-12]+ [4,10]= [8,-2]
(iii)
}:'.- w = [1,-3 ]-[2,5 ]= [1-2,-3-5 ]= [-1,-8]
(iv)
}:'.-}:'.=[1-1,-3+3]= [o,O]=Q
(v)
l}:'.l=~(1) 2 +c-3) 2 =./1+9=Jl0
Example 2:
Find the unit vector in the same direction as the vector v = [3, - 4] .
Solution:
}:'. = [3, -4] = 3!-4 j
'}:'.' = ~3 + (-4)
2
Now
~
1
2
=Es.=5
1
=r~f =5 [3, -
4] (!!is unit vector in the direction of v)
~=[+·~] Verification:
lul= = ~16 -+-=1 (-53J + r-4J 5 25 25
Example3:
Find a unit vector in the direction of the vector
2
2
(i) }:'.=2t+6j
Solution:
(ii)
}:'.=[-2,4]
(i) }:'.=2!+6j
l}:'.1=~(2)2 +(6)2 = ,J4+36 = J40 . . h d. . f :. Aurutvector mt e irect10no }:'. (ii) .
}:'.=[-2,4]=-2! + 41
2 . 6 . 1 . 3 . =-,-,vv = -v40 ~!+ ~J= ~!+ ~J -v40 - -vlO -vlO -
'}:'.' =~(-2)2 + (4) = ,J4+ 16 = .JW 2
Calculus and Analytic Geometry
332
. . h d' . f v - 2 . 4 . -1 . 2 . :. Aumtvector mt e irectiono v =-=-=--t+--1=-t+-1 1~1 Ea- En- JS- JS-
Example 4: If ABCD is a parallelogram such that the points A, B and C are respectively (-2, -3), (1, 4) and (0, -5). Find the coordinates of D.
Solution: Suppose the coordinates of D are (x, y)
,,----------------..
A_/_D----=•
·: AB=DC and AB// DC ~
7
•~----YC
As ABCD is a parallelogram
AB=DC
:. (l+ 2)! + (4+ 3)j = (0-x)t +(-5- y)j -
-
~
3! + 7 j =- x! + (-5 - y) j
Equating horizontal and vertical components, we have
and
-x=3
~
x=-3
-5-y=1
~
y=-12
Hence coordinates of Dare (-3, 12).
7.1.9 The Ratio Formula Let A and B be two points whose position vectors (p.v.) are g and p_ respectively. If a point P divides AB in the ratio p: q, then the position vector of P is given by qa+pbr= p+q Proof: Given f! and /2. are position vectors of the points A and B respectively. Let r be the position vector of the point P which divides the line segment AB in the ratio p : q . That is A mAP: mPB = p : q
So,
mAP mPB
p
~=-
q(mAP
q
)= p(mPB)
Unit 7: Vectors
Thus
333
q(AP
)= p(PB)
q(r. - f!) = p('2_- r.).
qr_-qf! =pl]_- P!. pr..+qr.=qf!+ pl]_ r_(p + q) r=
-
= q{!: +pl]_
qa+ pb q+ p
If P is the mid point of AB, then p : q
Corollary:
=1 : 1
.. a+b :. pos1t1ve vector of p = r = =----=
-
2
7!1.10 Vector Geometry Let us now use the concepts of vectors discussed so far in proving Geometrical Theorems. A few examples are being solved here to i11ustrate the method. Example 5: If a and!!. be the p.vs of A and B respectively w.r.t. origin 0 and C be a point on AB such that OC = a+ b , then show that C is the mid-point of AB. 2 -
Now
20C=a+b
~
OC+OC=OA+OB
~
OC-OA=OB-OC
~
OC+AO=OB+CO
~
AO+OC=CO+OB
:.
1
Given OA=a , OB=b_ and OC=-(a+b) 2 - -
Solution:
A
AC=CB
Thus mAC;;;; mCB ~
C is equidistant from A and B, But A, B, C are collinear.
Hence C is the mid point of AB.
334
Calculus and Analytic Geometry
Example 6: other.
Use vectors, to prove that the diagonals of a parallelogram bisect each
Solution: Let the vertices of the parallelogram be A, B, C and D (see figure) Since AC= AB+ AD, the vector from A to the mid point of diagonal AC is }::'.=
1(-) AB+AD
2
Since DB = AB - AD, the vector from A to the mid point of diagonal DB is
~=AD+~(AB-AD) =AD +_!_AB-_!_AD
2
2
=_!_(AB+ AD)
2
=v
Since
:!::'.=~,these
mid points of the diagonals AC and DB are the same.
Thus the diagonals of a parallelogram bisect each other.
EXERCISE 7.1 1.
Write the vector PQ in the form xi+ y j. (i)
2.
Q(6,-2)
y_
= 2! - 7j
(ii)
If Y:. = 2! - 7 j , :!: '. = ! (1.)
4.
=(2,3),
(ii)
p
=(0,5),
Q(-1,-6)
Find the magnitude of the vector Y:. : (i)
3.
p
Y:.+}:'.-w
61
c··)n
y_
= ! + j_
(iii) y__= [3, - 4]
and ~= -! + 1· Find the following vectors:
2Y:.- 3}::'.+ 4 w
Find the sum of the vectors
c·· .) 111
i i i -u+-v+-w
2-
2-
2-
AB and CD, given the four points A(l, -1),
B(2, 0), C(-1, 3) and D(-2, 2).
5.
Find the vector from the point A to the origin where AB= 4! - 2 j and B is the point (-2, 5).
Unit 7: Vectors
6.
• of the vector given below: Find a unit vector in the direction (i)
7.
(ii)
1.
fj .
(iii)
v=-t+-1 - 2- 2 -
If A, Band Care respectively the points (2, -4), (4, 0) and (1, 6). Use vector method to find the coordinates of the point D if:
(i) 8.
335
ABCD is a parallelogram
(ii)
ADBC is a parallelogram
If B, C and Dare respectively (4, 1), (-2, 3) and (-8, 0). Use vector method to find the coordinates of the point:
(i) A if ABCD is a parallelogram.
(ii)
E if AEBD is a parallelogram.
9.
If 0 is the origin and OP= AB , find the point P when A and B are (-3, 7) and (1,0) respectively.
10.
Use vectors, to show that ABCD is a parallelogram, when the points A, B, C and Dare respectively (0, 0), (a, 0), (b, c) and (b - a, c).
11.
If AB =CD. Find the coordinates of the point A when points B, C, D are (I, 2), (-2, 5), (4, 11) respectively.
12.
Find the position vectors of the point of division of the line segments joining the following pair of points, in the given ratio: (i)
Point C with position vector 2! - 3 j and point D with position vector 3{+2j in the ratio 4: 3
(ii)
Point E with position vector Si and point F with position vector 4!+ j_ in ratio 2 : 5 ·
14.
Prove that the line segment joining the mid points of two sides of a triangle is parallel to the third side and half as long.
15.
Prove that the line segments joining the mid points of the sides of a quadrilateral taken in order form a paral1elogram.
7.2
INTRODUCTION OF VECTOR IN SPACE
In space, a rectangular coordinate system is constructed using three mutually orthogonal (perpendicular) axes, which have origin as their common point of intersection. When sketching figures, we follow the convention that the positive x-axis points towards the reader, the positive y-axis to the right and the positive z-axis points upwards.
z /
x'
/ / /
/
' Y ------~--- Y /
,o I I I
X
I I
z'
336
Calculus and Analytic Geometry
These axis are also labeled in accordance with the right hand rule. If fingers of the right hand, pointing in the direction of positive x-axis, are curled toward the positive y-axis, then the thumb will point in the direction of positive z-axis, perpendicular to the xy-plane. The broken lines in the figure represent the negative axes. z A point P in space has three coordinates, one along x-axis, the second along y-axis and the third along z-axis. If the distances along x-axis, y-axis and z -axis respectively are a, b, and c, then the point P is written with a unique triple of real numbers as P = (a,b,c) 0 (see figure).
right hand rule P(a,b,c)
•! !
JC I I I I
, /
! ,./a "
y
____________ y
b
7.2.1 Concept of a Vector in Space The set R 3 = { (x, y, z):x, y, z E R} is called 3-dimensional space. An element (x, y, z) of 3 R represents a point P(x, y, z), which is uniquely determined by its coordinates x, y and z. Given a vector !:!. in space, there exists a unique point P(x, y, z) in space such that the vector OP is equal to
~
H.
! ! I
Jz
I
>-----.-~~-y
I / I
_________ __y,/'X
y
x
(see figure).
P(x, ', z)
z
Now each element (x, y, z) E R 3 is associated with a unique ordered triple [x, y, z], which represents the vector ~=OP= [x, y, z].
W~ define addition and scalar multiplication in R 3 by: (i)
Addition:
For any two vectors ~ =
~ + !'.'. = (ii)
[x, y, z] and
!'.'. =
[x', y', z'], we have
[x, y, z]+ [x', y', z']= [x + x', y + y', z + z']
Scalar Multiplication:
For ~ =
[x, y, z] and a E
R, we have
a~ =a[x,y,z]= [ax,ay,az]
Definition: The set of all ordered triples [x, y, z] of real numbers, together with the rules of addition and scalar multiplication, is called the set of vectors in R 3 • For the vector ~ =
[x, y, z], x, y and z are called the components of ~.
Unit 7: Vectors
337
The definition of vectors in R 3 states that vector addition and scalar multiplication are to be carried out also for vectors in space just as for vectors in the plane. Similarly we define in R 3 :
!i=[x,y,z]
as -!!=(-1)!!=[-x,-y,-z].
a)
Thenegativeofthevector
b)
The difference of two vectors .!: = [x', y', z'] and ~ = [x . ., y ..., z'] as
_!:- w = .!: + {- !!:'.) = [x' - x. ., y' - y ... , z' -
z']
c)
The zero vector as Q= [0,0,0]
d)
Equality of two vectors ~ = [x', y', z'] and ~ = [x . ., y", z'] by .!: = w if and on1y i"f xI = x# , y I = y # an d z I = z I
e)
Position Vector For any point
P(x, y, z)
in R 3 , a vector !! = [x, y, z] is represented by a
directed line segment OP, whose initial point is at origin. Such vectors are called position vectors in R 3 • f)
Magnitude of a vector : We define the magnitude or norm or length of a vector !! in space by the distance of the point P(x, y, z) from the origin 0.
:. loPI = l!!I = ~x2 +
Y2
+ z2
Example 1: For the vectors, .!: = [2,1,3] and !!:'. = [-1,4,0], we have the following: (i)
.!: + w=[2-1,1+4,3 + 0]= [1,5,3]
(ii)
.!: - w = [2 + 1,1- 4, 3 - 0] = [3,-3, 3]
(iii) (iv)
2w=2[-l,4,0]= [-2,8,0]
1_!:-2~ I= ![2+ 2,1-s,3-o]l=l[4,-1,3]1=~(4) 2 + {-1)2 + (3)2 =-J16+ 49+9 =.fi4
7.2.2 Properties of Vectors Vectors, both in the plane and in space, have the following properties: Let
~.!'.
and
!'.!:'.
be vectors in the plane or in space and let a,bE R, then
they have the following properties: (i)
~ + !'. = !'. + ~
(Commutative Property)
Calculus and Analytic Geometry
338 (ii)
~ + ~) + !!'. = ~ + ~ + !!'.)
(Associative Property)
(iii)
~+(-1)~=!:'_-!:'_=Q
(Inverse for vector addition)
(iv)
a{!i +!!'.)=a~+ a!!'.
(Distributive Property)
(v)
a(b!:l_)= (ab)!'.'.
(Scalar Multiplication)
Proof: Each statement is proved by writing the vector/vectors in component form in R 2 I R 3 and using the prop.e rties of real numbers. We give the proofs of properties (i) and (ii) as follows: (i) Since for any two real numbers a and b a + b == b +a ,
it follows, that
for any two vectors ~ = [x, y] and ~ = [x', y'] in R 2 , we have
g + ~ = [x, y] +[x' + y']
= [x+x' , y+ y']
= [x' +x,y' + y]
= [x', y'] +[x, y] =~+~
So addition of vectors in R 2 in commutative. (ii)
Since for any three real numbers a, b, c, (a +b)+c = a+(b+c)
it follows that
for any three vectors, ~ = [x, y], ~ = [x', y'] and w = [x", y"] in R 2 , we have (~ + ~) + w = [x +·x', y + y'] + [x', y"]
= [(x+x')+x",(y+ y')+ y"] = [x+ (x' + x"), y + (y' + y")] = [x, y]+[x' + x', y' + y"] = ~+(~+w)
So addition of vectors in R 2 is associative. The proofs of the other parts are left as an exercise for the students.
Unit 7: Vectors
339
7.2.3 Another Notation for Representing Vectors in Space As in plane, similarly we introduce three special vectors
£= [1,0,0], j = [0,1,0] As magnitude of
and !_ = [0,0,1] in R 3
(0, 0, ll
i. = ,J1 2 + 0 2 + 0 2 =1
magnitude of and magnitude of !_
l=.Jo
2
2
j
(0, I, 0)
2
+1 +0 =1
= ,Joi + 0 2 + 12 =1 So L j
and k are called unit vectors
along x-axis, along y-axis and along z-axis respectively. Using the definition of addition and scalar multiplication, the vector [x, y, z] can be written as ~ = [x, y,z]
=
[x,O,O]+ [O, y,O]+ [o,o, z]
= x[l,0,0] + y[0,1,0] + z[0,0,1] =
Thus each vector
[x, y, z] in
x£ + yj + z!_
R 3 can be uniquely represented by
In terms of unit vector
x£ + y j + z!_.
£ , j and !_ , the sum ~ + .!:'. of two vectors
~=[x,y,z] and
,!:'.=[x',y',z'] is written as
~ + .!:'. = [x + x', y + y', z +
z']
=(x + x')! + (y + y')l + (z + z')!_ 7.2.4 Distance Between Two Points in Space If
Ofi and0P2 are the position vectors of the points
fi (x1, Y1, Z1) and The vector
P2 (x2. Y2, Z2)
fi P2 , is given by
fiP2 = OP2 - Ofi = [x2-X1,Y2 - Yi· Z2-z1] :. Distance between. fi and P2 =
y
j
fiP2
= ~(x2 -
j
X1
)i + (Yi -
Y1
)i + (zi - zi)i
This is called distance formula between tWo points
fi
and P2 in R 3 • --~----------~-
340
Calculus and Analytic Geometry
If ~ = 2! + 31 + k, ~ = 4£ + 61 + 21.i and ~ = -6! -
Example 2:
(a)
91- 31.i, then
Find (ii)
(i) u+2v
Show that g, }:'., and w are parallel to each other.
(b)
Solution: (a)
(i) ~ + 2~ = 2! + 3-j + k + 2( 4i + 6-j + 2k_)
= 2! + 3-j + k + 8! + 12-j + 4k = 10! + 15 j + 51.i (ii) ~ - ~ - ~ = (2! + 31 + k_) - (4! + 61 + 2k_) - (-6! -
91- 31.i)
= (2 - 4 + 6)! + (3 - 6 + 9) j + (1- 2 + 3)1.i = 4! +6j +2k ~
(b)
= 4! + 6 j + 21.i =2(2! + 3 j + k) :.
v = 2u
Y: and }:'. are parallel vectors and have same direction.
=::}
Again
w=
-6! - 9 j -
3k_
= -3(2! + 3 j + k) :. w=-3u =::}
Y: and w are parallel vectors and have opposite direction.
Hence M, .!:'. and w are parallel to each other.
7 .2.S Direction Angles and Direction Cosines of a Vector Let r. = OP= x! + yj + zk:_ be a non-zero vector, let a , fJ and y denote the angles formed between r and the unit coordinate vectors i, l and If respectively,
z
p
such that (i) the angles a , angles.
fJ,
y are called the direction
.......................................J..·············..···
y
Unit 7: Vectors
341
(ii) the numbers cos a' cos fJ and cos y are called direction cosines of the vector r_. z
Important Result: Prove that cos 2 a + cos 2 p + cos 2 'Y
=1
Proof:
-•y
l~~:.__-~B
Let
r = [x, y, z ]= xt + y j + zl.i
(O,y. 0) A
Id= ~x2 + y2 + z2 then
(x, O. 0)
=r
x
l~I =[:,~,-;.]is the unit vector in the direction of the vector r= OP.
It can be visualized that the triangle OAP is a right triangle with LA = 90 . Therefore in right triangle OAP, OA cos a = = OP
.. I 1ar y = -x , sum r
cos fJ =L, cosy=!:._ r r x The numbers cos a = - , cos p r
=y r
and cosy=!:_ are called the direction
r
cosines of 0 P . x2 y2 2 cos 2 a+cos 2 p+cos 2 y=++~ 2 r r2 r
x2 + y2 + z2 r
2
r2 =1 r2
EXERCISE 7.2 1.
Let A= (2, 5), B = (-1, 1) and C = (2, -6). Find (i)AB
2.
(ii) 2AB-CB
(iii) 2 CB -2 CA
Let !i = ! + 2j_ - l.i , !:'. = 3! - 2 j_ + 2/i , w = 5! - j_ + 3/i . Find the indicated vector or number. (ii) _!:'.-3!!:'.
(iii) j 3}::'.+ !!:'.!
342 3.
Calculus and Analytic Geometry
Find the magnitude of the vector l'. and write the direction cosines of .!'. • (i) y_=2£+31+4k
(ii)
y_=t-1-k
(iii)
y_=4t-sL
4.
Find a, so that ja£+(a+1)_L+2kl=3.
5.
Find a unit vector in the direction of
6.
If g_=3{-j-4k, '2_=-2£-4_L-3k and f=!+21-k·
Find a unit vector para1lel to 3g_ 7.
8.
y_ = £+ 2j-k.
21z + 4f
.
Find a vector whose (i)
magnitude is 4 and is parallel to 2! - 3 j + 6k
(ii)
magnitude is 2 and is parallel to - ! + j + k·
If !! = 2! + 3 j + 415:_,
y_ = -! + 3_L - k
and ~ = ! + 6_L +
zk
represent the sides of
a triangle. Find the value of z.
9.
The position vectors of the points A, B, C and D are 2! -
2! + 4 j- 215:_ and -
10.
-! -2j- + k
1+ k,
3£ + l
respectively. Show that AB is parallel to CD.
We say that two vectors l'. and win space are parallel if there is a scalar c such that l'. = c~. The vectors point in the same direction if c > 0 and the vectors point in the opposite direction if c < 0 (a) Find two vectors of length 2 parallel to the vector (b) Find ~
the
constant
a
so
that
the
y_ = 2! -
vectors
4 j + 4k.
!'. = ! - 3 j + 4k
and
=at + 9 j -12k are parallel.
(c) Find a vector of length 5 in the direction opposite that of !'. = ! - 2 j + 3k . (d) Find a and b so that the vectors 3!-1+4k and a!+ b j-2k are parallel.
11.
Find the direction cosines for the given vector: (i) }:'. = 3! - j + 2k
(ii) 6!-2j + k
(iii) PQ, where P = (2, 1, 5) and Q = (1, 3, 1).
12.
•
Which of the fo11owing triples can be the direction angles of a single vector: (i) 45° ,45° ,60°
(ii) 30° ,45° ,60°
(iii) 45° ,60° ,60°.
Unit 7: Vectors
7.3
343
THE SCALAR PRODUCT OF TWO VECTORS
We shall now consider products of two vectors that originated in the study of Physics and Engineering. The concept of angle between two vectors is expressed in terms of a scalar product of two vectors.
Definition 1: Let two non-zero vectors!:!. and}:'., in the plane or in space, have same initial point. The dot product of Y:. and }:'., written as y_.}:'., is defined by ?.!. •~ = ?!.
I II~ Icos e
. where ()in the angle between Y:. and .!:'. and
Definition 2: (a) If~= a,!+b 1 j and ~ = a2 !+b 2 j are two non-zero vectors in the plane. The dot product Y:.·.!:'. is defined by ~-~ = a 1a2 +b1b2 (b) If~= a 1!+b11+c 1k and ~ = a 2 !+b2 1+c2 k are two non-zero vectors in space. The dot product Y:.·.!:'. is defined by ?.!.·~ = a 1a2 +b1b2 +c 1c 2
I Note:
The dot product is also referred to the scalar product or the inner product.
.3.1 Deductions of the Important Results By Applying the definition of dot product to unit vectors (a)
(c)
!-!= l!l l!I 1·1=/11111 k·k=lkl lkl
cos0°=1 cos0°=1 cos0°=1
(b)
!,
!·1= I! 111 1·k=l11 lkl k·! =lkl l!I J
?.!.·~=1?.!.11~1 cos(} =1~11?.!.1 cos(-(}) =1~11?.!.1 cos(} -u.v=v.u - - x
j, j, we have,
=0 cos90° =0 cos90° =0
cos9o·
I
344
Calculus and Analytic Geometry
..
Dot product of two vectors is commutative.
7.3.2 Perpendicular (Orthogonal) Vectors Definition:
Two non-zero vectors if u. v = 0.
~
and
.!:'.
are perpendicular if and only
Since angle between u and v is .!!_ and cos.!!_ = 0 2 2 !i·~ =
7C
l!il 1~1 cos 2
:. ~.~=0
Note: As Q•!z = 0, for every vector Q . So the zero vector is regarded to be perpendicular to every vector.
7.3.3 Properties of Dot Product Let~ • .!:'.
(i) (ii)
and w be vectors and let c be a real number, then u. v = 0 ==> ~ = 0 or ~ =0 u.v=v.u (commutative property)
(iii) (iv)
~· (~+w) = ~·~ + ~·!!'.' (c ~).~ = c (~.~)
(v)
~ ·~ = l!il
(distributive property) (c is scalar)
2
The proofs of the properties are left as an exercise for the students.
7.3.4 Analytical Expression of Dot Product "·!:'.. (Dot product of vectors in their components form) Let be two non-zero vectors. From distributive Law we can write:
·· !i ·~ = ( a1! + b1i_ + c1k).( a2! + b2j_ + c2kJ = a1a2(!.!) + a1b2(!.j_) + a1c2(!.k) + b1a2 (j_ •D + b1b2 (j_ • j_) + b1c2 (j_ • k)
-.·!.!=j_.j_=k.k=l !.j_ = j_.k = k·! = 0
+c1a2(k.D+ C1b2(k.j_)+ C1C2(k.k)
==> !i·~ = a1a2 +b1b2 +c1c2 Hence the dot product of two vectors is the sum of the product of their corresponding components. Equivalence of two definitions of dot product of two vectors has been proved in the following example. Example 1: (i) If ~ = [x 1 , Y1 ] and w = [x 2 , Y2 ] are two vectors in the plane, then ~.w
= X1X2 + Y1Y2
Unit 7: Vectors
345
(ii) If r and w are two non-zero vectors in the plane, then
!'.·~=l~l l~Jcose where () is the angle between !'. and w and 0 5 e 5 7r • Proof: Let r and w determine the sides of a triangle then the third side, opposite to the angle 9, has length I!'. - ~I
(by triangle law of addition of vectors)
By law of cosines, I!'. - w I2 = I!'. I2 +I ~I 2
-
21 !'.I I~I cos e
(1)
~
2
I x 1 -x2 l +I Y 1 - yJ
Ls -
-
()
= [xp y 1 ] and~= [x 2 , y 2 ], then _!!'- w = [x 1 - x 2, Yi - Y2] So equation (1) becomes: if
w
=Ix~+ y~j+j xi+ Yij-21 !'.I I~lease
-2x1x 2 -2y 1 y 2 =-21!'.l I wlcose
=>
x 1x 2 + y 1 y 2 =I!'. I I~lease= !'.·w
Example 2:
If !± = 3! - j - 2k and !'. = ! + 2j - k, then
Example 3:
-
-
!±·!'.
= (-3)(1)+(-1)(2)+(-2)(-1)=3
If !± = 2! - 4 j + 5k and !'. - 4! - 3 j -
-
4k , then
!±. !'. = (2)(4) + (-4)(-3) + (5)(-4) = 0
=>
!± and !'. are perpendicular
7.3.5 Angle Between Two Vectors .. The angle between two vectors !± and !'. is determined from the definition of iot product, that is {a) !±·!'. = 1!±11!'.lcose, where 05e57r u. v e :. cos = 1;11~1
(b)
If!±= a1£+b1 j_+c 1k and
_!:'.=a 2 !+b2 j_+c2 !s_, then
!± • !'. = a1a 2 + b1b2 + c1c 2
I!± I=~ a12 + b12 + C12 and I!'. I=~ ai + bJ + ci
Calculus and Analytic Geometry
346 Corollaries:
= 0 or n , the vectors y_ and .!:'. are collinear.
(i)
If ()
(ii)
If () = - ' cos() = 0 ~ u . v = 0
7r
2
-
~
So the vectors y_ and .!:'. are perpendicular or orthogonal.
Example 4: Find the angle between the vectors
= 2i - j_ + Ii and .!:'. =- i + j_ Solution: Y: . .!:'. = (2i - j_ + /i) . (-i + j_ + 0/i) Y:
= (2) (-1) + (-1)(1) + (1)(0) =-3
1~1=I2!-1 +kl= ~(2)2 + (-1)2 + (1)2 = J6 and Now
() cos
u.v
=I ~1-1~ I -3
cos()=
fj
J6J2 = - 2
. ()=Sn
.. Example 5:
6
Find a scalar a so that the vectors
2! +a j + 5k and
3! + j + ak are perpendicular.
Solution: Let
Y: =2!+a1+5k and
.!:'. =3!+ 1 + ak
It is given that Y: and .!:'. are perpendicular :. Y: . .!:'. = 0 ~
(2! +a 1+5k) . (3! + j + ak_) = 0
~
6+a+5a=O
a =-1
Unit 7: Vectors
347
Example 6: Show that the vectors 2! - 1 + k , ! - 31- 5k and 3! - 41- 4k form the sides
of a right triangle.
c
Solution: AB = 2! - j + k and BC
Let Now
=! -
3 j - 5k_
AB+ BC= (2! - j + k_) + (! - 3 j - 5k_) -
-
=3£-4j-4k_ =AC (third side) ·
AB, BC and AC form a triangle ABC.
Further we prove that MBC is a right triangle AB. BC
=
(2!- j +k).(£-3j-5k_)
= (2)(1) + (-1)(-3) + (1)(-5) = 2 + 3-5
- =O .. AB 1- BC
Hence MBC is a right triangle.
7.3.6 Projection of One Vector Upon Another Vector
~~~~~~~~~~
B
In many physical applications, it is required
to know "how much" of a vector is applied along a given direction. For this purpose we find the projection of one vector along the other vector. Let OA = !:!. and OB
= !:'.
Let () be the angle between them, such that 0 ~ fJ ~ ir . Draw BM
J_ OA.
Now
l.!!.lcosO
Then OM is called the projection of .r along H·
OM = COS () , that IS, . -=OB OM =IOBicosfJ =l_!:'.icosfJ
(1)
Calculus and Analytic Geometry
348
By definition, cos
u.v
e = I;, ,~ I --
From (1) and (2),
OM
(2)
u.v
=I y I· I;II~ I
ll
. . f 1 u.v :. ProJection o }:'.a ong !! = ~
Similarly, projection of ff: along l'. =
l~J
Example 7:
Show that the components of a vector are the projections of that vector along f:. , j and Is. respectively.
Solution:
Let y = a!+ b j +els:_ , then
fif= (a!+ b1+els:_).!= a
Projection of l'. along f:. =
Projection Of E along j_ =
f~i
Projection of}:'. along Is.=
f
= (ai + b j_ + c/£). j_ = b
k1=(a!+b1 + c!s:_).k:_ = c
Hence components a, b and e of vector y = a!+ b j +ck:_ are projections of vector l'. along !, j and Is. respectively
Example 8:
Prove that in any triangle ABC 2
(i)
a = b 2 + c 2 - 2bc cos A
(Cosine Law)
(ii)
a = b cos C + c cos B
(Projection Law)
Solution: Let the vectors ~, "Q and c be along the sides BC, CA and AB of the triangle ABC as shown in the figure. (i)
a+b+c=O
=>
~
Now
~-~ = (Q+f).(Q+f)
=-(Q+f) B~--------c Q.
349
Unit 7: Vectors
=b.b+b.c+c.b+c.c
==:}
a 2 =b 2 +2b.c+c 2 a2
(:. !z.q_=q_.!z)
= b 2 + c 2 + 2bc .cos(n -
A)
A
a' =b' +c' -2bccosA (ii)
a+b+c=O a =-b-c
7C-C B "'--_ _ _ _ ___..___,_ __..
Take dot product with
~
7C-B
c
a.a= -a.b-a.c
= -ab cos( n -
C) - ac cos( n - B)
a ' = abcosC+ac cosB ==:}a= bcosC + c cosB Example 9:
Prove that: cos(a -
fi) = cos a cos P +sin as in p
Solution: Let OA and OB be the unit vectors in the .xy-plane making angles a and p with the positive x-axis. y A
So that LAOB = a - p Now OA= cosa!+sinaj and OB= cosP!+sinP!
:. OA.OB= (cosa!+sinaj). (cosfi!+sinfij)
-
==:}
joAl!osJ
:. cos(a -
-
cos(a-P)=cosacosp+sinasinp
fi) .= cos a cos p +sin a sin P
EXERCISE 7.3 1.
Find the cosine of the angle (} between !:!:. and ~: (i)
l_i
=
3! + j_ - k , Y. = 2! - j_ + k
(ii)
l_i
= ! - 3j_ + 4k,
Y. = 4! - j + 3k
Calculus and Analytic Geometry
350 (iii)
2.
-u=[-3,5], -v=[6,-2]
Calculate the projection of £!along
'2.
and projection of
'2.
~
= [2, 4, I]
along £! when:
(ii)
(i)
3.
!i = [2,-3, l] ,
(iv)
Find a real number a so that the vectors Y: and =2a!+ j-k_,
(i)
!_!_
(ii)
I_!_= a.£+ 2a.j -k
'
~
= ! +aj +4k
~
= ! +aj + 3k_
~
are perpendicular.
4.
Find the number z so that the triangle with vertices A(l,-1,0), B(-2,2,1) and C(0,2, z) is a right triangle with right angle at C.
5.
If~
is a vector for which
~. ! = 0,
6.
(i)
~. j = 0,
~. k = 0, find~·
Show that the vectors 3£ - 2 j + k,
£- 3 j + 5/s_ and 2£ + j - 4/s_ form a -
-
right angle. (ii)
Show that the set of points P = (1,3,2), Q( 4,1,4) and R = (6,5,5) form a right triangle.
7.
Show that mid point of hypotenuse a right triangle is equidistant from its vertices.
8.
Prove that perpendicular bisectors of the sides of a triangle are concurrent.
9.
Prove that the altitudes of a triangle are concurrent.
10.
Prove that the angle in a semi circle is a right angle.
11.
Prove that cos( a.+ /J) =cos a. cos /3- sin a. sin fJ
12.
Prove that in any triangle ABC.
7.4
(i)
b = c cos A+ a cos C
(iii)
b2
=c2 + a 2 -
2cacos B
(ii) (iv)
c = a cos B + b cos A 2
2
c = a +b 2 -2abcosC.
THE CROSS PRODUCT OR VECTOR PRODUCT OF TWO VECTORS
The vector product of two vectors is widely used in Physics, particularly Mechanics and Electricity. It is only defined for vector in space.
351
Unit 7: Vectors
~
Let ~ and and.!:'., written as
~X_!:'.
.!:'.
be two non-zero vectors. ~ x .!:'., is defined by
The cross or vector product of
= (1~11.!:'.1 sinO) fl:
a
where () is the angle between the vectors, such that 0 ~ 0 ~ 7t: and is a unit vector perpendicular to the plane of ~ and .!:'. with direction given by the right hand rule.
Figure (a)
Figure (b)
Right hand rule (i)
If the fingers of the right hand point along the vector
~
and then curl
a
which is u xv. towards the vector .!:'.' then the thumb will give the direction of It is shown in the figure (a). (ii) In figure (b), the right hand rule shows the direction of .!:'. x ~.
7.4.1 Derivation of Useful Results of Cross Products (a)
By applying the definition of cross product to unit vectors i,j and Is:, we have: (a)
z
!X!=l!Jl!lsinD° fl=O _Lx_L=llllllsinD° fl=O
k
kxk=lkJlklsinD° fl=O (b)
!X_L=J!lillsin90°k=k
o~-----•y
j
x
_L xk =I.L 11 k Isin 90° !=!
(c)
kX!=JkJj!Jsin90°_L=_L ~X.!:'. = 1~11.!:'.J sin() fl = J.!:'.11~1 sin(-{}) fl= -1_!:'.11~1 sinO fl
=>
UXV=-VXU
(d)
~X~=1~11~1sin0° fl=O
Calculus and Analytic Geometry
352
Note: The cross product. of !, j and k are written in the cyclic pattern. The given figure is helpful in remembering this pattern.
7.4.2 Properties of Cross Product The cross product possesses the following properties: (i) !!:. X~ = Q if !!:. = Q or ~ = Q (iii)
-uxv - =-vxu - !!:.X(~+ W) = !!:.X~+!!:.XW
(iv)
!!:.X(k~) = (k!!:.)X~
(v)
~X!!:.=Q
(ii)
= k(!!:.X~)
(Distributive property) k is scalar
,
The proofs of these properties are left as an exercise for the students.
7.4.3 Analytical Expression of
~x ~
(Determinant formula for u xv ) Let !f:.=a 1!+b11+c1k_ and· ~=a 2 !+b2 1+c 2 k_, then !!:.X~
= (a 1 ! + b1 1 +c1 k_)x(a 2 ! +b2 1 + c 2 k_) = a 1a 2 (!x!) + a 1b2 (!x D+ a 1c 2 (!xk_)
(by distributive property)
+~~vxo+~~vxD+~~vx~
+ c1a 2 (k_x!) + c1b2 (k_x
'.·fXj=k_=-jXf
-
D+ c1c2 (k_xk_)
-
=~~k-~~1-~~k.+~~!+~~1-~~! !f:.X~=(b 1 c 2 -c1b2 )!-(a 1c 2 -c1a 2 )1+(a1b 2 -b1a 2 )k_
=>
(i)
The expansion of 3x3 determinant
i
j
k
=a 1
b1
c 1 =(b1c 2 -c1b2 )!-(a 1c 2 -c1
a2
b2
C2
a )1+(a b 2
1 2
-b,a 2 )k
The terms on R.H.S of equation (i) are the ·same as the terms in the expansion of the
above determinant .
l
j
k
Hence ~x~ = a 1
b1
c1
az
bz
Cz
(ii)
Unit 7: Vectors
353
which is known as determinant formula for uxv. Note: The expression on R.H.S. of equation (ii) is not an actual detenninant, since its entries are not all scalars. It is simply a way of remembering the complicated expression on R.H.S. of equation (i).
7.4.4 Parallel Vectors If Y: and .!:' are parallel vectors, ( () = 0 :::::} sin 0 = 0 ) 1 then ~ x ~ = 1~11~1 sin 8 fl
!! x ~ =Q And
if
gx~
I!! x ~ I =0
or
= Q, then
I!! I= 0
either sin () = 0 or
or I~ I= 0
(i)
If sin()= 0 :::::} () = 0° or 180°, which shows that the vectors Y: and parallel.
(ii)
If Y: = Q or .!:'. = Q , then since the zero vector has no specific direction, we adopt the convention that the zero vector is parallel to every vector.
.!:'.
are
Note: Zero vector is both parallel and perpendicular to every vector. This apparent contradiction will cause no trouble, since the angle between two vectors is never applied when one of them is zero vector. Example 1: . Find a vector perpendicular to each of the vectors ~
Solution:
...
= 2! + j + k and q_ =4! + 2 j -
-
k
A vector perpendicular to both the vectors f! and Q is ~ x q_ i
j
axb= 2 -1 4
2
k 1 =-!+6j+8k
-1
Verification: ~.~xq_ = (2!- j + k_).(-! + 6j + 8k_) = (2)(-1) + (-1)(6) + (1)(8) = 0 and
q_.~XQ = (4! + 21-k_).(-! + 61+8/s)
= (4)(-1) + (2)(6) + (-1)(8) =0
Hence ~ x q_ is perpendicular to both the vectors g_ and l2. . Example 2:
If f!
=4! + 3J.. + Ii and q_ =2i - j
+ 2k . Find a unit vector perpendicular
to both g and b_. Also find the sine of the angle between the vectors g and l2. •
354
Calculus and Analytic Geometry
l
k
j
axb= 4 3 1 = 7!-6j-10! 2 -1 2
Solution:
and A unit vector
fl:
Iaxb I
perpendicular to f! and Q = ~ x ~
1
= --(7i-6j-10k)
JIBs- -
Now
1£!1=~(4)2+(3)2+(1)2
-
=J26
IQl=J bsinA = asinB b
a sinA
(ii)
--=--
sinB
similarly by taking cross product of (i) with Q, we have
a c --=-sinA
(iii)
sinC
From (ii) and (iii), we get
a sinA
b
--=--=
sinB
c sinC
7.4.5 Area of Parallelogram
I!! I
If H and .l::'. are two non-zero vectors and 0 is the angle between H and .l::'. , then and represent the lengths of the adjacent sides of a parallelogram, (see figure)
I.!:'. I
We know that:
---------------, u
Area of parallelogram = base x height =(base) (h) =
I!! II.!:'. I sin8
•· ~I ,;••
///
1~1
~
Calculus and Analytic Geometry
356
:. Area of parallelogram= I!! x .!:'.I
7 .4.6 Area of Triangle u
From figure it is clear that
- ...- -- -- - - -- ----1
......
Area of triangle = _!_ (Area of parallelogram) 2 Area of triangle =
/
.........
//
......
/
...... / '------"_ _ _!'. _
~ J!! x :!:'. J
__,
where _!Land :!:'. are vectors along two adjacent sides of the triangle. Example 5: Find the area of the triangle with vertices A(l, -1, 1), 8(2, 1, -1) and C(-1, 1, 2) Also find a unit vector perpendicular to the plane ABC.
i
Solution: AB= (2-1)! + (1+1) + (-1-1) k
::;:
! + 21- 2k
AC=(-l-l)!+(l+l)j+(2-l)k = -2!+2j+k_
Now
ABxAC= _:
2
~ ~i~(2+4)i-(l-4),i+(2+4)!
= 6i+3j+6!
The area of the parallelogram with adjacent sides AB and AC is given by I ABxAcl = J6! +31 +6kJ = "'36+9+36 = Areaofthetriangle=
J8i = 9
~IABxACI= ~ J6!+3j+6k.J= ~
A unit vector l. to the plane ABC=
1
~x~
ABxAC
1
= !(6i +3j +6!)=!(2i + j+ 2!) 9 3 -
Example 6: Find area of the parallelogram whose vertices are P(O, 0, 0), Q(-1, 2, 4), R(2, -1, 4) and S(l, 1, 8).
Solution: Area of parallelogram = I!! x .!:'. \ where !! and :E are two adjacent side of the parallelogram PQ=(-1-0)!+(2-0)1+(4-0)k_ = -!+2j+4k_ and
PR = (2- 0) ! + (-1- 0) l + (4- 0) k = 2! -
- - -- -- - - - - - -
l + 4k_
Unit 7: Vectors .
357 i
Now
k
j
-PQxPR= - -1
2 4 =(8+4)!-(-4-8)j+(l-4)k 2 -1 4 Be careful!: Not all pairs of vertices . . Area of parallelogram= x PR I= j 12! + 12j -3k give a side e.g. PS is =.J144+ 144+9 not a side, it is diagonal = .J297 since PQ +PR = PS
IPQ
I
-
-
I
Example7: (i)
If fl:. = 2! - j + k and y_ = 4! + 2j u xv
(ii)
fl:.Xfl:.
Solution: 1l:. = 2! - j + k
(iii)
k , find by detenninant formula
vxu
y_=4!+2j-k
and
By determinant formula i (i)
(ii)
(iii)
j
uxu= 2 -1
k
2 -1
1
i
k
j
uxv= 2
-1
4
2
i j }!_Xfl:.= 4 . 2 2 -1
( :. Two rows are same)
1 =0
1 =(1-2)!-(-2-4)j+(4+4)k = -!+6j+8k
-1 k
-1 =(2-1)!-(4+2)j+(-4-4){f = !-6j-8'5. 1
EXERCISE 7.4 1.
Compute the cross product f! x Qand ~ x f! . Check your answer by showing that each f! and ~ is perpendicular to f! x ~ and Qx f! .
(i)
g_ = 2! + -j -
(iii)
g_ =3!-2j +k -
2.
Find a unit vector perpendicular to the plane containing f! and ~ . Also find sine of the angle between them.
k ' Q= ! - -j + k ' Q=! + -j
(ii)
g_ = ! + -j ' ~ = ! - -j
(iv)
f! = -4! + j - 2! , ~ = 2! + j + k -
-
358
Calculus and Analytic Geometry
(i)
g_=2!-6j_-3k '!!..=4!+3j_-k
(ii)
g_=-!-j_-k '!!..=2!-3j_+4k
(iii)
g_ = 2!-21_ + 4k , !!.. = -! + j_-2k
(iv)
g_ = ! +
3.
Find the area of the triangle, determined by the point P, Q and R.
(i).
P(O, 0, 0); Q(2, 3, 2); R(-1, 1, 4)
4.
Find the area of parallelogram, whose vertices are:
(i)
A(O, 0, 0) ; B(l, 2, 3) ;C(2, -1, 1); D(3, 1, 4)
(ii)
A(l , 2, -1); 8(4, 2, -3); C(6, -5, 2); D(9, -5, 0)
(iii)
A(-1, 1, 1); B(-1, 2, 2); C(-3, 4, -5); D(-3, 5, -4)
S.
Which vectors, if any, are perpendicular or parallel
(i)
~ = 5!- j_ + k
(1·1·)
~=!+
6.
Prove that:
7.
If. g_ + !!.. + f = 0 , then prove that g_ x !!.. = !!.. x f = f x g_
8.
Prove that: sin( a -
9.
If g_ x !!.. = 0 and g_ · !!.. = 0 , what conclusion can be drawn about g_ or !!.. ?
7.5
SCALAR TRIPLE PRODUCT OF VECTORS
(ii)
j_ , !!.. = !- j_
P(l, -1, -1); Q(2, 0, -1); R(O, 2, 1)
; Y. = j_-5k ; !!' = -15! + 3 j_- 3!:_
• 2 . k
. . k 7r . . 7r k l--; y_=-!+ l+ - ; w=-2!-7rl+2 _ g_ x (!!._ + f) + !!.. x (f + g_) + f x (g_ + !!._) = 0
p) = sin a cos p + cos a sin p .
There are two types of triple product of vectors: (a)
Scalar Triple Product: (~ x y_) · w or !i · (y_ x w)
(b)
Vector Triple product: ~ x (~ x w)
In this section we shall study the scalar triple product only Definition Let
~=a 1 !+b 1 j_+c 1 k, ~=a 2 !+b2 j_+c 2 k and w=a 3 f+b 3 j_+c3 k:_
be three vectors The scalar triple product of vector g, ~ and w is defined by !i_.(~xw) or y_.(wx~) or w.(!i_X~)
The scalar triple product ~.(y_ x w) is written as
Unit 7: Vectors
359
!!·(!:'.X w)
=[!! !:'. w]
7.5.1 Analytical Expression of !!· Volume= 1(8 + 21) -2(-4-3)-1(7 + 2) = 29 + 14 + 5 = 48
362
Calculus and Analytic Geometry
Example 2:
Prove that four points
A(-3, 5, -4), B(-1, 1, 1), C(-1, 2, 2) and D(-3, 4, -5) are coplaner.
Solution: AB = (-1 + 3) ! + (1 - 5) j + (1 + 4) k
AC= (-1 +3)! + (2-5) j + (2+ 4)'5_
= 2! - 4 j + 5'5_ = 2!-3j +6'5_
AD= (-3+3)! + (4-5) _L + (-5 +4)'5_ = 0!- _L-k=- _L-k Volume of the parallelepiped formed by AB, AC and AD is
2 -4
5
[AB AC AD]= 2
-3
6 = 2(3+ 6) +4(-2-0)+5(-2-0)
0
-1
-1
=18-8-10=0 As the volume is zero, so the points A, B, C and D are coplaner.
Example 3:
Find the volume of the tetrahedron whose vertices are A(2, 1, 8), B(3, 2, 9), C(2, 1, 4) and D(3, 3, 0)
Solution: AB = (3 - 2)! + (2 -1) j + (9 - 8) k
AC= (2- 2) ! + (1-1) j + (4-8) '5_
= !+ j +k = 0!-0 j -4'5_
AD= (3-2)! +(3-l)j +(0-8)'5_ = !+2j-8'5_
l(---1
Volume of the tetrahedron= - ABAC AD
.
6
Example 4: Find the value of a , so that a! + j , ! + j + 3'5_ and 2! + j - 2'5_ are coplaner. Solution:
Let
Triple scalar product
~ = a!+ _L + k, y_ = ! + _L + 3'5_ and ~ = 2! + .L - 2'5_
363
Unit 7: Vectors
[u v
a 1 0 w]= 1 1 3 = a(-2-3)-1(-2-6)+0(1-2) 2 1 -2 = -5a+8
The vectors will be coplaner if - 5a + 8 =0 Example 5:
:=:::)
8 a=5
Prove that the points whose position vectors are A(-6! + 3 j + 2k),
B(3!-2j + 4k), C(5! + 7 j + 3k_), D(-13! + 17 j-k_) are coplaner.
-
-
-
Let 0 be the origin.
Solution:
OA=-6!+3j+2k ; OB=3!-2j+4k OC=5!+7j+3k; OD=-13!+17j-k_ -
-
AB= OB-OA = (3!-2j +4k_)-(-6! + 3j + 2k_)
= 9!-5j + 2k · AC= OC-OA = (5! + 7 j +4k)-(-6! +3j + 2k_) -
-
= 11! + 4j +k
AD = OD - OA = (-13! + 17 j -
k) - (-6! + 3-j + 2!~)
=-7!+14j-3k_
9
Now
-5 2 AB.(ACxAD) = 11 4 1 -7 14 -3 = 9(-12-14) +5(-33+ 7) + 2(154+ 28) = -234-130+364 = 0
.. AB, AC, AD are coplaner :=:::)
The points A, B, C and D are coplaner.
Calculus and Analytic Geometry
364
7 .5.4 Application of Vectors in Physics and Engineering (a) Work done. If a constant force F, applied to a body,
F
acts at an angle () to the direction of motion, then the work done by F is defined to be the product of the component of F in the direction of the displacement and the distance that the body moves. In figure, a constant force F acting on a A d body, displaces it from A to B. :. Work done= (component of F along AB) (displacement)
B
= (F cos B)(AB) = F . AB
Example 6: Find the word done by a constant force F = 2! + 41, if its points of application to a body moves it from A(l, 1) to B( 4, 6) .
(Assume that
IFI
Solution:
The constant force F = 2! + 4 j ,
is measured in Newton and
141
The displacement of the body =
in meters.)
4 = AB
= (4-1)! + (6-l)j- = 3! +5j:. work done = F .d
= (2! + 4-j) • (3! + 5-j)
= (2)(3) + (4)(5) = 26 nt. m The constant forces 2! + 51 + 6k and - ! + 21 + k act on a body, which is displaced from position P(4,-3,-2) to Q(6,1,-3). Find the total word done. Solution: Total force= (2! + 51+6!f) + (-!- 2j- !f)
Example 7:
:=:}
F = ! + 3 j + 5!f
The displacement of the body = PQ = (6 - 4)! + (1+3) j + (-3 + 2)k
4 = 2! + 4 j - k Work done= F .4 :=:}
·
= (! +31 +5~J .(2! +41-k) 2+12-5 =9 nt. m
=
Unit 7: Vectors (b)
365
Moment of Force Let a force F (PQ) act at a point P as
shown in the figure, then 0
moment of F about 0. ; product of force F and perpendicular ON • fl.
= (PQ)(ON)(fl.) = (PQ)(OP)sin8 .B_ =OPxPQ
Example 8:
B
F Q
= r_xF
Find the moment about the point M (-2,4, -6) of the force represented
-AB = (3 -1)! + (-4 - 2) j + (2 + 3)! = 2! - 6 j + 5k
by AB, where coordinates of points A and Bare (1, 2, -3) and (3, -4, 2) respectively. ..
Solution:
-
-
MA= (1+ 2)! + (2-4)j + (-3+6)! = 3!-2j +3! -
Moment of AB about (-2, 4, -6) = !.. x F l
=
---+
-
--+
= MA x AB
j
k
3 -2 3
2 -6 5 = (-10+ 18)!-(15-6)j + (-18+ 4)k
=8!-9j-14k Magnitude of the moment= ~(8) 2 +(-9) 2 +(-14) 2 =.J341
EXERCISE 7.5 1.
2.
. Find the volume of the parallelepiped for which the given vectors are three edges. (i)
~=3!
(ii)
~=!-41- !;
(iii)
~=!-
+ 2!;
21 +3!£;
'£.=! + 2j + !;
w= -
y_=! -1-2!;
w= 2!- 3j+k
y_= 2!
-1-1£;
j + 4!
~=1+1£
Verify that f! •
ex £
=
if g_ = 3! -
e..
£ x f! = £. !!: x e.
1+ 5k,
P. =
4! +
31 -
2!£ and f
= 2! + 51 + !£
Calculus and Analytic Geometry
366
• 3.
Prove that the vectors ! -2j + 3k:_, -2! + 3j_- 4k:_ and
!-
3j + 5k:_ are
coplanar. 4.
5.
Find the constant a such that the vectors are coplanar. (i)
!-
(ii)
! - 2 a j - k:_,
(a)
Find the value of:
j + k:_,
6.
7.
! - j + 2k:_ (ii) 3 j.
(i)
(b)
! - 2j-3k:_ and 3! -a j_ +5k:_,
k x!
and
a! -
(iii) [
j +
k!
k (iv)
j]
[!
i
k:_]
Prove that _H.(!'. x w) + .!:'.·~ x .H) + w.(.H x ,!:'.) = 3_M.(}'. x w)
Find volume of the Tetrahedron with the Vertices (i)
(0, 1, 2),
(3, 2, 1),
(1, 21) and
(5, 5, 6)
(ii)
(2, 1, 8),
(3, 2, 9),
(2, 1, 4) and
(3, 3, 10).
Find the work done, if the point at which the constant force F is applied to an object, moves from
Pi (3,1,-2)
= 4! + 3j_ + 5k:_
to P2 (2,4,6).
k , is displacement
8.
A particle, acted by constant forces 4£ + j_ - 3k:_ and 3! - j_ -
9.
from A(l,2,3) to 8(5,4,1). Find the word done. A particle is displaced from the point A(5,-5,-7) to the point 8(6,2,-2) under the action of constant forces defined by 10 ! - j + 1 lk_, 4! + 5 j + 9k:_ and - 2£ + j
-9k .
-
-
Show that the total work done by the forces is 102 units.
10.
A force of magnitude 6 units acting parallel to 2!-2j + k displaces, the point
11.
of application from (1, 2,3) to (5,3,7). Find the work done. A force f_ = 3! + 2j-4k:_ is applied at the point (1,-1, 2). Find the moment of
12.
the force about the point (2, -1, 3). A force E = 4£-3k, passes through the point A(2,-2,5). Find the moment of E about the point 8(1,-3,1).
13.
Give a force
E=
,
2£ + j_-3k:_ acting at a point A(l,-2,1). Find the moment off_
about the point 8(2,0,-2) .
14.
Find the moment about A(l, 1, 1) of each of the concurrent forces !-2j, 3! + 2 j_ - k:_, 5 j_ + 2k_ , where P(2, 0, 1) is their point of concurrency.
15.
A force F = 7£ + 4j_-3k:_ is applied at P(l,-2,3). Find its moment about the point Q(2, 1, 1) .
Answers
367 EXERCISE 1.1
1.
(a)
(i) 6
(b)
(i)
Ji
2
3x + 2
(ii) 0
(iii) x
(ii) 2
(iii) .Jx+3
-
(iv) x 4 +7x 2 +12 (iv)
~x 2 +8
2.
. ) -sm - 2 . ( a+h ) sm. h (i) 6 (ii) l:.....cos(a+!!_)sin!!_ (iii) h 2 +(3a+2)h+3a 2 +4a (1v h 2 2 h 2 2
3.
(a) P = 4.JA.
4.
(i)
4n-. Dom g = Set of real numbers;
(ii)
Domg=R-(-2,2); Rangeg= [O,+oo]
(iii)
Dom g =
(iv)
Domg= (-oo,+oo); Rangeg= (O,+oo)
(v)
Domg= (-oo,+oo); Rangeg= [-oo,+oo)
(vi)
Domg=(-oo,+oo); Rangeg=(-oo,+2)u[7,+oo)
(vii)
Domg= R-{-1}; Rangeg= R-{l}
(viii)
Domg= R-{4}; Rangeg= R-{8}
5. 7.
1 (b) A = - C 2
a= 2, b =-2
[-1, + oo);
(c) V = (A) 312
Range g = Set of real numbers
Range g = [O, + oo]
(i)(a) 30 m (b) 17.5 m (c) 11.1 m (ii) 2sec (i) odd (ii) neither even nor odd (iii) odd (iv) neither even nor odd (v) even (vi) odd 6.
EXERCISE 1.2 1.
(i)
(a) 5+ x x-1
(ii)
(a)
(jii)
(a)
(iv)
2.
3.
(b)
~1 + x2 1
x~x
(c)
(b) ( - x x-l
+2
(a) 8(6\-1) x 8-x
(c) 4x+ 3
2x
(b) _1_ x+l
x 2
2-
(b)
J
2
x~3x
2
9 2
(i)
(a)
(iii)
(a) 9 -x 1/3 (b) 9-(-1) 113
(i) (iii)
2
(b)
~J;+I +1
(d) 3(x -1) 4-x (d) x 4
2 2 4 ( c ) f g f i (d) (x + 2x + 2)
I-~
(c) 3(3x 4 -2x 2 ) 4 -2(3x 4 -2x 2 ) 2
(d)~2../x
-2 3
3
(ii)
-(a) (x-7r 3
(b) (-8r 3
(iv) (a) x+ l (b) 0 x-2 (ii) Dom =R-{-1},Range =R-{4} Dom =[0, oo), Range =[-2, oo) Dom =[0,+oo), Range =[5,+oo) Dom= R-{O} , Range= R-{-3} (iv)
368
Calculus and Analytic Geometry
EXERCISE 1.3 (iv) 0
1.
(i) JO
(ii) 5
(iii) 4
2.
(i) 2
(ii) 4
C°) Ill -12 (iv) 0
3.
I (v) - 2
5
7r:
(i) 7
(ii)-
(x) 0
(xi) -
(iii) 0
180 p2
( VI") -13 4
(v) 0
.. (vi) I (vn)
a
(iv) I
1
.. . (vm)
.fi.
2 2
I
(ix) .!!_ xn-m
m
. (viii)
(vi) I (vii) 2
(v) b
rx
2 x
_!.
(ix) 0
2
( XII.. ) - I 2
q2
4.
(i) e 2 (ii) e 112 (iii)
e- 1
1. 2. 3.
(i) -2, -2, -2
(ii) 0, 0, 0
(i) Continuous
(ii) Discontinuous
(iv) e 111 (v) e 4 (vi) e 6 (vii) e 2 (viii)
e-1
(ix)
e - 1 (x) -J (x) I
EXERCISE 1.4
s.
(iii) 0, 0, 0
=2, Discontinuous at x =-2 (i) m =1, n =3 (ii) m =3
Continuous at x
4.
c=-1
6.
1 k=-
6
EXERCISE 1.5 Draw the graph yourself EXERCISE 2.1 1.
.. I (11) - - -
(i) 4x
2/;
(vi) 2x-3
(xi)
2.
(i)
...
(111)
mxm-I
l
8
(vii)--
(viii)
x5
(X.11.. ) .. ) ( 11
2../x+2
m
(xiii)
---
xm+I
I
---
(iv)
2x3/2
3(x+4) 213
(ix)
-3
(v)
x4
3rx
(x)
2
-100 . )(XIV -
40x 39
XIOI
I
-----
2(x+ a) 312
EXERCISE 2.2 (i) 3a(ax+b) 2 (ii) 10(2x+3) 4 (iii)
-6 3 (3t+2)
(iv)
-Sa (ax+b) 6
(v)
-7a (az-b) 8
EXERCISE 2.3 1. 4x 3 +6x 2 + 2x
2.
-3[-l x4
+-1] xs12
3
2a • (a-x) 2
8
4.--(2x+ 1) 2
l
(x-a) 2
-5x 3/2 2
I'" :'.
11 ·r'rs
_"I\
•
369 x2
x
-I
1 -
\)~
rI
u.
or
-
1
-2.x
J,2+l(x2
14.
-1) 3/2
9
(x2-l)2
x+ 2 (x2+1)Jx2+I
11.
(l-x) 11 2~
2
S. 2x(x +1) (x -3)
2.,Jx
x2
Ill.
2
l-3x 7. ----=,......
6.--
I-~
~or 2 2
x vl-x
.
x+2 (x2 + l)312
or
I
-8x (x 2 -3) 2
12.
r--
-a
va-x(a+x)3 12 2
a -x 2 + ax
15.
~(l+Jl-x 2 )
-{ r;:;:S5) =. 27
~ • tr)
372
Calculus and A11c1/r1ic ""
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Answers
383 y
y
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(ii)
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(v)
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..........
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x
Answers
385 y
y
' (ii)
"
(-
.
...
)
~ .t
'
...... ......_6
"II'\
(0 3
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0
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x
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v..
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l: 0
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(0, 3 1....
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.... (6JU
~
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\
.....
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..... ~
0 l'N
x
(iii)
:t_
I
.....
:. .,.."·~~
•• 1(....
~
i ,. T
y'
y
y'
y
(iv)
(v)
y
y'
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-'
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EXERCISE 5.2 Draw the graph yourself
~ ' 1} (0, 4)
(ii)
(0, 0), (5, 0), (1, 4), (0, 2)
(0,2), (1, 4 ), (0, 5)
(iv)
(0, 0), (3, O'), [ 21
(2; 0), (4, 0), (0, 4), (0, 3)
(vi)
(0, 0), (4, 0), - , 15) - '(0, 5) 17 17
l(i).
(0, 0), (3, 0), [
(iii)
(v)
5' 56 ) ' (0, 3)
(98
ru
,.._,
.. x
Calculus and Analytic Geometry
386 2(i).
(0, 0), (5, 0), (4, 2), (0, 3)
(ii)
(0, 0), (5, 0), (4, 2), (0, 3)
(iii)
(0, 0), (4, 0), (36 - ,24) - , (0, 3) 11 11
(iv)
(0, 0), (5, 0), (2, 6), (0, 7)
(v)
(0, 0), (6, 0), (3, 4), (0, 5)
(vi)
45 7 32) , (0, 8) (0, 0), (10, 0), (9, 2), ( 7,
EXERCISE 5.3 1. 3.
Maximum at the comer point (16, 12) Maximum at the comer point (0, 3)
5.
Maximum at the comer point
l~ ,°)
2. 4.
Maximum at the corner point (0, 5) Minimum at the comer point (0, 3)
2 6. Minimum at the corner point (0, 3) 3 Minimum cost if 3 units of food X and 2 units of food Y are fed to each animal. Maximum profit if the investor purchases 8 fans and 12 sewing machines. Maximum profit if 200 units of product A and 400 units of Product B are produced.
7. 8. 9.
EXERCISE 6.1 2
(a) x +y2-10x+4y+l3=0 (b) x 2 +y 2 -2.fi.x+6.J3y+21=0
1.
7 2 6 (c) x +y2-2x+4y-27=0 2.(a) (-6,5);.J61 (b) (- , 5 5 (d)
(i, -
3 }.J19/2 2
2
2
(3, -2);0
2
3(x + y2 )-l4x-8y- 65 = 0 4.(a)x 2 + y 2 -15x-18y+l7 =0 2
2
(b) (x + 3) + (y + 1) = 4 or x + y + 6x + 2 y + 6 = 0 and ( x + 2
(c)
3.(a)x 2 +y 2 -4x+2y-35=0 (b)x 2 +y 2 -10x-24y=O
(c)x + y2-ax-by =0 (d) 2
}!¥
2
2
~
!J
+( y-
3 13
J
= 4 or
2
169 (x + y )+390x-78y-442=0 (c)x + y +52x-2ly-265 = 0 (d) (x+2)2+(y-5)2=10 2
2
2
or x + y +4x-10y+19 = 0 and (x-14) +(y-13) 2 = 250 or x 2 + y 1 -28x-26y+115 = O 5. x 2 + y 2 -2ax-2ay+a 1 =0 2
9. (x+.fi.-1)2 +(y+3-J2)2 = 4 or
1
x + y +2(J2-1)x+2(.J3-2)y+10-8.J2=0 and (x-.fi.-IJ
+~+3+.fi.J = 4
or
x + y -2(.J2 +1~+2(3+J2~+10+8.J2 = 0 2
2
EXERCISE 6.2 1.(i) 4x+3y=25; 3x-4y=O
21x-33y+89=0
xcos(J+ysin =5; xsinB-ycos =0 (ii) 33x+2ly-103=0;
1 2. 24x-10y+91=0;5x+12y+26=0 at (-4, - }24x+l0y+151=0; 2
Answers
387
Sx-12y-46=0
at(-4,-~ 1 )
(3:, -; J
3.(i) Inside
(ii) Outside 4.
7. (i) x-2y-.JlO=O; x-2y+.JJQ =0
(ii)
. 3x-4y+20=0; (-12 16J ;3x+4y-20=0; (12 8. (1) -,- ,16) 5 s , 5 5 x-2y+5 =0;(-3,1)
(iii)
~
5.2.J13 6.(-4,5);
2x-3y-J26 =0; 2x-3y+.J26 =0
(ii)
2x + y =
o; (o,o)
5x-y+33 =0; (-6, 3) x+5y+l7 = 0;(-2,-3)
9. -4x+l6y+35=0
EXERCISE 6.4 y M
1(i)
Focus: (2,0); Vertex: (0, 0) ; Directrix: x + 2 = 0; Axis: y = 0 F(2, 0)
-+---=+------x
z
0
y
(ii)
Focus: (0,-4);Vertex: (O,O);Directrix: y-4=0;Axis:x=O
z
M
F(0,-4) y
(iii)
Focus: ( 0, ~); Vertex: (0, 0); Directrix: y + ~ = 0 ; Axis : x = 0
--~z-=-----M
y
M
(iv)
Focus: (-3,0); Vertex: (O,O);Directrix: x-3=0 Axis:y=O
388
Calculus and Analytic Geometry
(v)
Focus: (0,2); Vertex: (0,1); Directrix: y = 0 ; Axis: x = 0
(vi)
Focus: (1,0); Vertex: (3,0); Directrix: x = 5 ; Axis : y = 0
(vii)
Focus: (1,0); Vertex: (1,-2); Directrix: y = -4; Axis: x = 1
(viii)
Focus:
(ix)
Focus: (-
(x)
Focus: (2,2); Vertex: (2,0); Directrix: y + 2 = 0 ; Axis : x = 2
2(i)
(y -1 ) = - l 2x or y 2 - 2x + l 2x + 1 = 0
(ii)
(x-2) 2 = 8(y-3) or x 2 -4x-8y+28 = 0
(o.- 2324 l;) 3 } ,1}
25 ; Axis: x = 0 24
Vertex: (0,-1); Directrix: y = -
Vertex: (-9,1); Directrix: x =
-!
7
; Axis: y = 1
2
2
2
2
(iv) (y-2) =-8(x-3) or y 2 -4y+8x-20=0
(iii) 4x +4.xy+ y +36x-22y+41 =0
2
2
(v) (x+ 1) = -8(y-2) or x 2 + 2x+8y-15 = 0
(vi) (y-2) = 8x or y 2 -4y-8x+4 = 0
2
(vii) (x-2) = -4(y-2) or x 2 -4x+4y-4 = 0 (viii) (x-hf = -8(y + 1) or x 2 - 2hx+ 8y +8+ h2 = 0 harbitrary
(ix) 3l=x+l
(x) (x-7 I 5)2 =
3. (i) x 2 - 4h2 ± 4ax = 0, h arbitrary
(ii) y 2
~(y+~ l or 5x2 -14x-3y+9 = O 5
15)
+ 2hx- h 2 = 0, h arbitrary 2
6. 37500(2-.J3) km, 37500(2+.J3) km
7.
2 a x =-y 4b
8. 9m
EXERCISE 6.5 Sketch the graphs yourselves where asked. 2
l.(i)
2
.::.___ + 2'._ = 1 34
25
(ii)
(y+3)2 +~=l 9
5
2
2
y ... ) x +(111 =1 36 9
(iv)
(x-2)2 (y-1)2 + 9 5
(v)
x2 y2 -+-=l 9 4
(vi)
Y2 x2 -+-=1 25 16
(vii)
y2 x2 -+-=l 16 7
(viii)
(y-2)2 (x-2)2 + 16 9
(ix)
x2 y2 -+-=1 40 10
(x)
x2 7y2 -+-=l 16 16
Answers
2(i) (ii)
389
Centre: (0,0); Foci:
~2../3,o); Vertices:
(±4,0);Eccentricity:
.j3; Directrices:
Centre: (0, 0), Foci: { 0,±4); Vertices: (o,±3./2); Eccentricity:
~;
3v2
~
x=±
2
v3
Directrices: y = ± 2. 2
(iii)
Centre: (0, 0), Foci: { 0,±4); Vertices: { 0,± 5) ; Eccentricity: : ; Directrices: y = ± 2_%'
(iv)
Centre: +.-2} Foci:
l
l
+.-2±2../3} Vertices:
l
+.2 ){+.-6); Eccentricity:
~,
. 2 8 . D irectnces: y = - ± .j3 (v)
Centre: (-8, 2),
Foci:
Directrices: x = -8 ± (vi)
(-8 ± ,,/3, 2),
Vertices:
(-10, 2), (-6, 2) Eccentricity:
.j3
~
Centre:(5, 2), Foci: (5, 2±../2i), Vertices: (5,-3), (5, 7) Eccentricity:
m, Directrices: 5
25 y=2±--
5i
4.
3x 2 -2xy+3y 2 -2x-2y-1 =0
6.
~+L=t 8 4
9.
Greatest distance = 404582 km Least distance = 364224 km
2
2
2
7.
2
_x_+_Y_=l 100 2 3711
8.
15m
EXERCISE 6.6 Sketch the Graphs yourselves. 2
1. (i) (iv)
(vii)
2
2
2
~-1'.__=1 4 20
(ii)
~-1'.__=1 9 16
(iii)
y2 x2 ---=l 9 27
(v)
y2 x2 ---=l 36 45
(vi)
y2 9
(x-2)2 =1
914
4 '
(viii)
(y-1)2
4
(x-5)2 =1
5
(x-2) 2 25 (x-2) 2
9
(y+ 7)2 = 1 25
(y-2)2 = 1 27
Calculus and Analytic Geometry
390
..fi; Directrices: x = ±3 I ..fi
2. (i).
Centre: (0, 0), Foci: (±3/2, 0) ; Vertices: (±3, 0) Eccentricity
(ii).
Centre: (0, 0), Vertices: (±2,0); Foci: (±.Jl3, 0) Eccentricity: .J1312; Directrices: x = ±4 I .Jl3
(iii)
Centre: (0, 0), Vertices: (0, ±4); Foci: (0, ±5) Eccentricity:
(iv)
Centre: (0, 0), Vertices:
(0, ±2); Foci:
2_; Directrices: 4
y=
±~ 5
./5 , 2
(0, ±./5) Eccentricity :
Directrices: y = ±4 I ./5 Centre: (1 , 1), Vertices: (1±/2, 1); Foci: (1±.Jil, 1) Eccentricity :
(v)
J11i2;
. . 2 D 1rectnces: x = -1 ± r.-; vl l 5
Centre: (2, -2), Vertices: (2, l); (2, - 5) ; Foci: (2, 3) and (2, - 7) Eccentricity =
(vi)
D"1rectnces: . y= - 2
(vii)
Ceotre;
3
,
9 ±S
U,-
I } Vertim; (
+, -
I } (1, -1) ; Foci; (
~ ± ~ , - I JEceeotridty = ,,/JO ;
. . 2 I D tTectnces: x = - - ± - 3
(viii)
Centre: ( 2, -
3..[10
~} Vertices:
( 2, -
~ }( 2,-~} Foci:
(2,-
~ ±./5)
Eccentricity = ./5;
. . 3 l D1rectnces: y = ± ./5
2
(ix)
Centre: (-4 ,-1) ; Eccentricity Foci : (-4 ±
(x)
5.fi.
=Ji. , Vertices:
, -1); Directrices: x = -4 ±
Centre: (2, - 3) ; Eccentricity =
~
, Foci : (2
Vertices : (3, - 3) ; (1, - 3) , Directrices : x
=
2
(1, -1) ; (-9, -1)
}i ±
JlO ,- 3) ;
± ~ vlO
4.
llx
2
-
2
50xy+l ly +504 = 0
5.
2
2
7x -9y -84x+36y+l53=0
EXERCISE 6. 7 l(i).
yt=x+at 2 ;y=-tx+2at+at 3
(ii)
_::_cos e+l'....sin a
b
=1; ax sec B-by csc = a 2 -b 2
x2
291600
y
2
198400
=l
Answers
391
(iii)
~sec e _ l'._ tan = 1; ax cos e +by cot = a 2 + b 2
2(i).
3x+2y-6=0 at (4, -3) ; 3x-2y+6=0 at (-4,-3)
(ii)
9x+7y-20=0 at (3,-1), 9x-7y+20=0 at (-3, -1)
(iii)
15 304 8 26x-15y-89=0 at (4, l); 13x+-y--=0 at ( 4, - 2 7 7
3(i).
3x-4y-25=0, 4x+3y-25=0
(iii)
x+y+l=0,5x-y-7=0
4.
18x+ 27y-88 =0
6.
5x-2y+8 = 0,5x-2y-8 = 0
7(i).
±3x-4y-45 =0
8(i).
(iv)
a
(
±
b
(99 (601 vii· vii ±
(±~. ±2l
J
x - y + 3 = 0, 3x- y + 1=0
(ii)
5.
x-2y+2.J2 = O,x-2y-2.J2 = 0
(ii)
2x+y+2=0
(ii)
[±J+.±#l
(v)
(0, -4) and ±
(iii)
~8, ±F45)
[~ 6 .341
EXERCISE 6.8 l(i).
X 2 +16Y=0
(ii)
2X 2 +Y 2 -4=0
(iii)
9X 2 +4Y 2 .:_36=0
(iv)
X 2 -Y 2 +1=0
(v)
9X 2 -4Y 2 -36=0
2(i).
(-4, 3) ;3X 2 -2Y 2 -6 = 0
(iii)
(3,l);X 2 -Y 2 -1=0
3(i).
(ii)
(-1, 2);25X 2 +9Y 2 -225 =0
X 2 -Y 2 =2
(ii)
X 2 -9Y 2 +9=0
(iii)
13.Ji3 X 2 -5X-Y=0
(iv)
2Y 2 -2X +1 =0
4(i).
tan
(ii)
e = 45°, x 2 -Y 2 +.J2 x - 1.J2 Y -20 = o
(iii)
8=45°, X 2 +4Y 2 -4=0
e = 3, x 2 + y2 -11=0
EXERCISE 6.9
±.J6
l(i).
Pair of lines : 2x - y =
(ii)
Parabola: Vertex: (0, 0); Focus [l, I] Axis: x- y = 0; Directrix; x+ y + .J2 = 0
392
(iii)
Calculus and Analytic Geometry
Parabola: Vertex: (-
2~, 2~ J;
Focus
(-~,~}Axis:
x+ y = 0;
Directrix ; x - y = 0 Ov)
2 Ellipse: Centre: (0, 0); Foci [ :
2 2 2 ,- Jr)aod [- : , : }
~
Vertices: (2, - 2) ; (-2, 2) ; Major axis : x + y = 0 , Minor axis : x - y = 0 ; Eccentricity =
(v)
Ellipse ; Centre : (0, 0) ; Foci:
Uf ,) 3
and [-
~
3f).
,-
Vertie" : (-Jl, I),
(-.J3, -1); Major axis : x-.J3y .= 0 ; Minor axis : .J3 x+ y = 0 ; Eccentricity= .J3 2 (vi)
Ellipse : Cenrre : (-2, -1) ; Foci : (0, 0) and (-4, - 2) ; Vertiees:
(-fl(vii)
2, -
~ -1 Jmd
Hyperbola: Centre: (2, 4), Foci: (-2, 0) and (6, 8), Vertices: (2+2.J2, 4 +2.J2
x-y+2=0
Focal axis
Conjugate axis
Eccentricity : J2
FJ+ }[-
Hypecbofa:Centre : (0, 0), Foei : (2, I), (-2, -1) , Vertiee•{2f
Focal axis: x-2y = 0 ; Conjugate axis: 2x+ y = 0; Eccentricity: (ix)
2,
~ - I ).Majornxi
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