Maths Extension 2 Zndedltion by
S.K. Patel ------J
L~, ~~-et\1'\'c~ X\" ... -.- --~-- -,
-
··~,..,
;.., ')
Copyright © S.K. Patel and Pascal Press First Edition published by Pascal Press April 1989 Revised Edition published October 1989 Second Edition published January 1991 Reprinted 200 I, 2003, 2004
ISBN I 87531 226 9
Pascal Press PO Box 250 Glebe NSW 2037 (02) 8585 4044 '
., ' .
www.pascalpress.com.~u
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PREFACE As always, my aim in this book has been to impart the knowledge of mathematics to students who will need it in their later academic and professional work. I would like to thank and acknowledge the help and contribution of the following individuals: - Roger Myers of the Banks town Technical College for his many thoughtful suggestions - Judy Faulkner who typed the manuscript in the required time - all my colleagues in the Technical Colleges and many other school teachers for their useful suggestions. I sincerely hope that this new edition will be found to be even more useful than the previous one. Suresh Patel
-
./
CONTENTS CHAPTER 1
CURVE SKETCHING 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10
CHAPTER 2
Basic Curves Method for Sketching the Curves Graphs of Basic Functions Curves with Vertical and Horizontal Asymptotes Trigonometric Graphs Exponential Curves Drawing Graphs by Composition of Ordinates Implicit Differentiation and Sketching Curves Applications Miscellaneous Curves Sketching Exercise: l:A Exercise: l:B
1 2 3 8 11 14 15 24 26 30 40 43
INTEGRATION 2.1 Standard Integrals 2.2 Change of Variable: Substitution Exercise 2A 2.3 Integration by Part Exercise 2B 2.4 Trigonometric Integrals 2.5 Use of t = tan (X/2) Exercise 2C 2.6 Reduction Formulas Exercise 20 2.7 Trigonometric Substitution Exercise 2E 2.8 Integration of Rational Functions Exercise 2F 2.9 Method of Partial Fractions Exercise 2G 2.10 Completing the Square (Integration) Exercise 2H 2.11 Integration: Special Properties Exercise 2I Exercise 2J
49 51 53 54 57 58 61 63 64
66 67 68 68 72
73 77 78 80 81 85 86
CHAPTER 3
VOLUMES 3.1 Formulas for Volumes Exercise 3A 3.2 Volumes: Shell Method 3.3 Volumes: Washer Method Exercise 3B 3.4 Worked Examples (Miscellaneous) Exercise 3C
CHAPTER 4
88 97 99 103
104 105 109
COMPLEX NUMBERS 4.1 Introduction 4.2 Operations with Complex Numbers Exercise 4A 4.3 Complex Plane (Argand Diagram) Exercise 4B 4.4 Multiplication and Division of Complex Numbers Using Trigonometry Exercise 4C 4.5 Powers of Complex Numbers: De Moivre's Theorem Exercise 4D 4.6 Roots of Complex Numbers Exercise 4E 4.7 De Moivre's Theorem and its Applications Exercise 4F 4.8 · Square Roots of a Complex Number Exercise 4G 4.9 Properties of Conjugate Complex Numbers Exercise 4H 4.10 The Complex Roots of Unity Exercise 41 4.11 Miscellaneous: Factorisation over the Complex Field Exercise 4J 4.12 Geometric Representation of Complex Numbers Exercise 4K 4.13 Product and Quotient: Rotation ·.Exercise 4L 4.14 Locus Problems with the Complex Variable z Exercise 4M 4.15 Miscellaneous Locus Problems (Including Inequations) Exercise 4N Exercise 40
112 113 116 118 121 122 124 125 126 127 130 131 134 135 136 137 138 139 141 143 145 146 149 150 152 153 154 155 159 160
CHAPTER 5
POLYNOMIALS 5.1 Introduction Exercise SA 5.2 Zeros of a Polynomial/Multiple Roots Exercise SB 5.3 Polynomial over the Complex Field Exercise SC 5.4 Relation between Roots and Coefficients of P(x) = 0 Exercise 50 5.5 Miscellaneous (Worked Examples) Exercise SE
CHAPTER 6
CONIC SECTIONS 6.1 Introduction 6.2 Ellipse (e < 1): (Focus and Directrix Definition) Exercise 6A 6.3 The Hyperbola Exercise 6B 6.4 Shape of the Conics 6.5 ParaiJ~.etric Equations of the Circle and the Ellipse 6.6 Equations of chord, Tangent and Normal to the Ellipse 6. 7 Parametric Equations of the Hyperbola 6.8 A Special (Rectangular) Hyperbola xy = c2 Exercise 6C 6.9 Miscellaneous Problems on Conics Exercise 60 6.10 Tangents and the Chord of Contact 6.11 Geometric properties of the Ellipse 6.12 Geometric Properties of the Hyperbola 6.13 Properties of the Rectangular Hyperbola 6.14 Geometric Properties of Rectangular Hyperbola xy = c2 Exercise 6E
CHAPTER 7
164 167 168 172 173 177 178 183 185 187
191 192 195 196 198 199 201 203 205 206 208 209 214 222 224 227 228 229 232
ELEMENTARY PARTICLE DYNAMICS 7.1 Introduction 7.2 Laws of Motion - Force Exercise 7A
234 234 239
CHAPTER 8
MOTION PROBLEMS IN TWO DIMENSIONS 8.1 Introduction 8.2 Simple Harmonic Motion (Revision) Exercise 8A 8.3 Motion of a Projectile (Revision) Exercise 8B 8.4 Resisted Motion: Other Laws of Motion Exercise 8C
CHAPTER 9
240 241 244 246 249 252 259
CIRCULAR MOTION 9.1 Introduction 9.2 Angular Velocity: Period 9.3 Circular Motion: Tangential Velocity Exercise 9A 9.4 Acceleration of a Particle Rotating in a Circle 9.5 Uniform Circular Motion Exercise 9B 9.6 Conical Pendulum Exercise 9C 9.7 Banked Tracks Exercise 9D 9.8 Components of Acceleration (Variable Angular Velocity Exercise 9E 9.9 Miscellaneous (Worked Examples on Circular Motion) Exercise 9F
265 266 267 269 270 271 272 273 275 276 278 279 281 282 285
CHAPTER 10 HARDER 3 UNIT TOPICS 10.1 Harder Trigonometry Exercise lOA 10.2 3 Unit Co-ordinate Geometry: Circles (Harder Problems) Exercise lOB 10.3 Plane Geometry: Circles (Harder Problems) Exercise lOC 10.4 Inequalities Exercise lOD 10.5 Method of Mathematical Induction Exercise lOE 10.6 Properties of the Integrals Exercise lOP Appendix 1 Appendix 2
Answers
288 291 292 2% 297 300 305 307 308 310 312 315 316 317 318
Chapter 1: Curve Sketching INTRODUCTION A structure, no matter how complicated, is composed of basic building materials. In mathematics, the study of complicated functions such as f(x) = -1 (ex + e-x), log x e x 2 sin-l ex, etc. can be facilitated by sketching the graphs of these functions. But how can we best accomplish this? Firstly,by identifying them by the basic curves (listed below) and then using a number of systematic steps to produce reasonably good graphs. The objective is to draw a quick and neat sketch of the curve showing all the essential features (see section 1.2). Any thought of drawing the graph by brute force i.e. plotting scores of points must be abandoned, as it would most probably miss the essential features, such as asymptotes and the critical points.
1.1
Basic Curves
The following basic curves are well known to students. (See Chapter 1 for the graphs of the basic functions.) Function 1.
Linear: ax + by + c ·y = mx + b
2.
Quadratic: y
+ bx + c
3
2
= ax
Cubic: y
4.
Quartic: y = ax
6.
+ bx 4
.. bx
k Y=x,
2
x + y (x - a)
2 2
A straight line
= ax 2
3.
5.
=0
+ ex + d 3
+ ex
2
2
A cubic curve
+ dx + e
= r
=0
or
2
7.
Exponential: y = ax , ex
8.
Logarithmic: y
9.
Trigonometric: y = sinx, cosx, tanx
10.
Inverse Tng: y
11.
Type: y
.
= x 1/n
An exponential curve
= loga x , .-1 = sm x,
, n
A quartic curve A rectangular hyperbola
xiO
+ 2gx + 2fy + c
+ (y - b)
A parabola
= 2,
loge x
-1
A Iogarithmrc curve
-1
cos x, tan x 3
A sine curve, cosine curve, tangent curve An inverse sine curve etc.
2
1.2
Method for Sketching the Curves
A sketch of the curve shou1d convey the general shape of the curve, showing the following information. 1.
Intercepts on the axes x
2.
=0
gives the y-intercept, y
=0
gives the x-intercept.
Symmetry The curve is symmetric about the (a) y-axis if f(x) = f(-x) (b) origin 0 if f(-x)
3.
= -f(x)
Asymptotes If y = f(x) is a rational function of the form f(x) = _ (a) vertical asymptotes are g1ven by h(x) = 0
~~x~ x
, where g(x) and h(x) are polynomials, then the:
(b) horizontal asymptotes are given by ;i: I co I f(x) if it exists, say y
= c.
2 Example: The vertical asymptotes of the curve y = 2X + 1 are given by 2 X - 1 2 x - 1 = 0 , i.e. x = .! 1 and the horizontal asymptotes are given by 2 Lim 2/ + 1 = 2 and y = Lim 2x + 1 _ . _ Y =X +co. 2 x~-oo 2 - 2 'I.e. Y- 2• X -· 1 X - I 4.
Critical points
=0
(a) Relative maxima:
f' (x)
(b) Relative minima:
f' (x) = 0 , f"(x)
(c) Vertical tangent at x
=a
, f"(x)
0
if f' (a) is undefined
(d) Point of inflection at x = a, if f"(a) = 0 and f"(x) changes sign as x increases through x = a.
5.
Rising and falling curves (a) f' (x) (b) f' (x)
6.
7.
>0 0
indicates where the graph is concave up
Zones (regions) of exclusion: The domain in which the graph of y = f (x) does not exist (i.e. y is undefmed) is an important consideration, because we don't have to worry about the graph in this interval. ~ Example: y = f(x) = V 9 - / The graph of this function is real if -3 ~ x ~ 3. It does not exist for lx I The curve IS the upper half of the circle x 2 + / = '3.
> 3.
j.
3
1.3 1.
Graphs of Basic Functions The graph of a linear function y
= mx
+
b
a straight line, with gradient m and y-intercept b. x = b represents a vertical line.
js
y =c
2.
X
represents a horizontal line.
The graph of a quadratic function
= ax 2
. a para bo Ia. 1s Its vertex lies on the axis of symmetry
y
x = -
+ bx + c
b
2a
•
If a
>
If a
< 0,
0, y has a minimum at x .
b =-2a
y h as a max1mum at x = -
b 2a
Discriminant /J. = b 2 - 4ac . If A
<
the curve does not intersect
0 ,
the x-axis, If
/J.
If IJ.
0,
>
the curve touches the x-axis.
O, the curve intersects the x-axis b
at x =- 2a
Examples: (a)
(c)
+
Sketch the curves:
= x 2 - 4x + 3 y = x2 + x + 1 y
/:;
2a
(b)
y = -x 2 + 2x
tanS
4 Solution:
y =
(a)
a
X
>
2
- ~X
+
3
0, A = 16 - 12
>
0
The intersection with x-axis is given by y = 0 (x - 3) (x - 1) = 0
:.
A(l, 0), B(3, 0)
=
When x
y
0 ,
v (2,
=; · 3
Vertex at x = 2 , y = -1 (b)
y = -x
a<
2
-1)
y V(l,l)
+ 2x
0
y = 0 gives x (x - 2) = 0 0 (0, 0) and
A (2, 0)
Vertex (1, 1) (c)
y
= X2
a>
+ X + 1 A=l-~<
0,
0
No intersection with x-axis Vertex (- l
3 2,4 )
Some points on the graph A(l, 3),
B(O, 1), C(-1, 1)
X
The equation of the form x = ay
2
+ by + c
represents a parabola with its axis horizontal. Vertex on ax1s of symmetry dt y = For a
>
0 and a
< 0,
b 2a
the shapes
are as shown. If A
<
0, the graph does not intersect X
y-axis.
Example: Sketch the curve Solution: a = l > 0. Vertex: y =
X
= y2
y + 2y
b
-2a
x = 0 gives y (y + 2) = 0 (0, 0)
and
(0, -2)
-22
= -1, V(-1, -1)
i.e.
-1
v
X
5
GRAPHS OF POLYNOMIAL FUNCTIONS
3.
For sketching the graph of the polynomial function we need reasonably sufficient information, such as: The function y = f(x) is an even function if f(x)
a.
= f{-x).
The graph of an even function is symmetric about Y-:axis. b.
= f(x)
The function y
is odd if f{-x)
= -f(-x)
and the graph is
symmetric about the origin. The intersection of the curve y = f(x) with
c.
-
the x-axis is given by y = 0 • This is not always simple, but worth trying.
- the y-axis is given by x = 0. d.
The behaviour of the curve as x -
e.
The nature of the turning points, i.e. a maximum or a
+
co
minimum Concavity and points of inflexion
f.
i.e. f" (x)
=0
•
Example: (I)
Sketch the curves y = x
3
and y = -x
3 y=x
Solution:
f (x) = x 3 and f (-x) = -x 3 The graph of y = x
3
is symmetric about the origin
0 (0, 0) is a point on the curve As x X
~ dx
co,
_,.-oo,
y + co y - - co
3x 2
But x = 0 is not a turning point, since
~ does not change its sign in passing
through 0 Some points: A (1, 1) , B (2, 8) • This is reasonably sufficient information to sketch the curve.
X
3
6 Example: (2) y = (x
-t
Sketch the curve:
2) (x - 1) (x_- 3)
y = 0 gives A(-2, 0), B(l, 0), C(3, 0) = 0 gives D(O, 6) ---,....-1:----~-~lc-----:::---:Jf:::-~x
l'
y -+ ~ as x-+ .., y--a.as x-+-~ Other points x X
Example: (3)
y = (x +
= 2, = -1
y 1
= -4
y :: 8
Sketch the curve:
2) (x - 2)
2
y = 0 gives A(-2, 0), B(2, 0) The double zero x = 2 indicates that the x-axis touches the curve at x = 2 ~2
Y-i'.:!:.a.asx~;t""
x "'
o
X
gives y = 8
Other points; C(-1, 9)1 D(l, 3), E(J, 5)
Example: (4)
y = (x
-t 2) (x
Sketch the curve: + 1) (x - 2) (x - 3)
The sketching is similar to that of Example 2, noting thdt y .... dS
X~-
oo
oo
X=O,y~l2
(I, I 2) is another point
X
7 4.
= x 1/2 dnd y =x 1/3 Draw a sketch of y = .JX
Graphs of y
EXAMPLE: (1)
by analysing the behaviour of the function near x =0 •
SOLUTION: The function Also f(x)
f(x) =
,jx
exists and is continuous for
~ 0 for x ~ 0 and f'(x) = -~ 2
for x
x
~
>0
0 • •
V X
The point x = 0 is a critical point as f' (0) does not exist. The curve y = {x has a vertical tangent at x = 0 •
0
X
Since f' (x) > 0 for x > 0, f(x) is an increasing function and f(O) = 0 is the absolute minimum, but f(x) has no absolute maximum. I -3/2 It may be noted that since f" (x) = - ii x < 0 for x > 0, the curve is concave down. This is sufficient information for drawing a reasonable graph of y Note: y
=
vx
is the upper half of the parabola /
Graph of y
= .(X.
= x.
EXAMPLE: (2) Draw a sketch of y = x 1/3 • SOLUTION: Y =x
f (x)
1/3
~
' dx
= x l/ 3
f(-x) = (-x)
I -2/3
=3 x
fr
' dx2 =
exists for all real x
1/3 .
=-x
1/3
f' (x) does not exist for tangent, at x = 0.
= - f(x),
so the curve has point symmetry about the origin.
x = 0, so the curve has a vertical
y
Also f' (x) > 0 for all x, x I 0, so it is an increasing curve. x = 0 is a critical point. f (x) ~
oo
f (x) -t -
oo
as x
-t oo
as x __,. -
oo
f"(x) > 0 for x > 0 and f"(x) < 0 for x < 0, so the origin is an inflection point. Also f"(x) > 0 for x > 0 tells us that the curve is concave down, while f"(x) < 0 tells us that the curve is concave up for x < 0. A few simple points are 0, A(l,l), B(-I,-1).
-J
= ,fX
8
1.4
Curves with Vertical and Horizontal Asymptotes
Example: (1) 2
Sketch the curve y =
X
y has same sign as x, hence curve in first and third quadrant As
X
As x
-IX>' - -co,
y -
2
a+ (from above)
y -+ 0- (from below)
The x-axis and y-axis are the asymptotes of the
2
curve. The asymptotes are very useful in sketching the curves. This curve is a rectangular hyperbola rectangular because the asymptotes are perpendicular.
Example: (2) I
Sketch the curve y = x + x 1
y=x+x'
xl- 0
Near x = 0 , the term
1
x.
dominates, i.e. x is negligible I
X
compared to x :. The curve y = x + .! behaves X
like hyperbola y = .! near x = 0 X
As x-+ ""• y - x i.e. the curve approaches the line y = x The graph is symmetric about the origin as f( -x) = -f{x) The turning points are:
~ dx -
I
1 ---:z X
0
*
p (1, 2)
Q(-1,-2)
The curve does not intersect either axis Other points: (2, 2.5) , (-2, -2.5) This is reasonable enough information to sketch this curve
9
Example: (3)
y
2 --2X-
=
Sketch the curve y
x
-------"-0-+--i'i
Solution: 2 y = -2-
4
X
X-
x
\ I
The function is undefined at x = 0, i.e. y-axis is the asymptote X '"'
2 ' y = 0 • A (2, 0)
> 2,
When x
y
>
0 ,
x
< 2,
<
y
CRITICAL POINTS
y
0
Near x = 0 , the graph behaves more like As x x -+
·~ dx
y -+ 0 + from above
oo , -oo,
y -+ 0
= P (4,
from below 0 gives x
1
2>
=4
is a maximum turning point.
The first sketch shows the important features, the second shows a complete graph 5
Other points: (:-1, -3), (-2, -1}, (-3, -9), (1, -0 y
Example: (4) X
2
Y = (x - 2) (x + 2) Solution:
The curve is undefined at x = + 2 x = 0, y =
o,
0(0, 0) is a point on
the curve. The Jines x = 2 and x = -2 are the asymptotes. As x-+ ""• y - I+ from above x -
- oo
,
I+ from above
y -
This is so because the numerator is greater than the denominator when x is large. For - 2
0 So, x
''
So, the maximum occurs at x = 0 and
minimum occurs at x
= !:
............. .
1
'
,_
"/
X
, /
(2)
g(x) = g(-x), so g(x) is symmetric about
(3) (4)
g (x) ~ 0 for all x, g(x)-. co as x ... !: co . . (x 2 - 1)2 = x 2 - 1 gives X=.!. 1, + l/'[ =.!. 1.4 . Pomts o f mtersect10n:
(5)
Additional points: (0, 1), (.!.1,0), (!: 2, 1), (.!. 2, 9)
the y-axis.
2 Sketch y = sin x 2 period of g(x) = sin x = .!. (1 - cos 2x) is n. 2 f(x) = sinx. ~ 0 for all x. g'(x) = 2sinxcosx = 0, f'(x) = 0 give sinx = O, cosx = 0
EXAMPLE: ( 2) The Let g(x) f(x)
.
.
= 0, n,
-
n
3n
y
..:n, 2, 2• ··· 2 Observe that 5in x < Isin xI except at the stationary points. The graph of g(x) can now be sketched with the points (Fig. 1). So stationary pomt5 are at x
(0,0),
0
, consider the following:
, then
!y >
0
= loge x
and y
y
I
= log
x
e
for y = logex
I Y = logx
y = logex is a known graph 1
y - - - logex as follows.
(base, e, omitted for convenience) I
I.
The asymptote for y = - - 1ogx
2.
The points of intersections are given by solving:
3.
x =e
and
x = e-l = 0.37
~
2.7
As y = log x increases from 1 1
As x - I X-+
5.
is at x = I
i.e.
y = - - - decreases. 1ogx 4.
--
y = logx
is shown by solid line and can be sketched
+
I , y =-1ogx
1-, y -
-oo.
(both positive)
-""
and
Some useful points (0.1, -0.4) , (0.9, -9.5) , (1.1, 10.4).
/
2
e
3
4
X
22 Example: (2) y =
y
Sketch the curve
8
1
2""--:-
6
x - 4
Solution: 4 is a parabola as shown in the
.1'
2
figure. (broken line) The graphs of both functions
= - 2- I.-
y
X
-
I
4
I
= x2 -
y
I
\
-1
2 and y = x - 4
o.
X
4
are both positive or both negative. These two meet at points given by X
2
= - 2-1-
-4
X
2
x - 4
X=~
=
4
-
+1
~,fj
15'
The graph of y = - - 1 2 X
-
has vertical asymptotes at x = ~ 2 and horizontal asymptotes as
4
k.
the x-axis. Also x = 0 , y = -
This is reasonably sufficient information to sketch the
graph. (SOLID LINE)
I
y Example: (3) Sketch the curves
I
3 y = x + I and y = - 3 - x + 1 3
Draw the graph: y = x
2.
1 Asymptote of y = - - at x = -1 3 X + 1
3.
x
4.
As x -
=0 ,
y
=1
I
is a common point.
""• y 1. y = -3-x + 1
>0
+
As y
0
!y <
0
y
and increases,
Two graphs intersect at x = 0 and X
= -1.2
~/11
'\
I
and decreases. 7.
+
I =-3--
y
X
0 for
and decreases. 6.
3
I
1.
As y
X
I
Solution:
.5.
y :
I
I I I I
I
I
I I I
0
+
1
2
X
Solve: y
2
X X
3
+
=0
=+ 1 , -1.2
23 (j)
VtW
Graph of
by graphing f(x)
By notmg the followmg, the sketch of y I.
g(x)
=
2.
g (x)
~
3.
(a)
{Hx)
< f (x)
If f (x)
> 1,
(b)
fiW = f(x)
if f(x)
=1
(c)
J'iW > f (x)
if 0
flit;)• so f'(x)
=0
4.
Jf(x) exists only for
~ 0.
2 2 e.g. Jx - 4 :::> x
=0
.JX < x
e.g.
< f(x) < I,
e.g.
if x
{X>
~4 ~
x
>,
2 or x
~
-2
>I
if 0
X
f(x)
(i;)
for
~
0. From the graph of y
= f(x),
this
? 2. Remembering that:
{ff;) ~ f (x) < f(x) < I ,
g (x) = 0
for
f (x)
We take the square root of the ordinates of y
? I
= f (x)
in the domain of g (x).
The.graph of y = g(x) is shown by the solid line. Verify that the maximum turning point of f(x) and hence g(x) occurs at x
=
A
= 1.15.
24
1.8
Implicit Differentiation and Sketching Curves y =x
Consider the function For y
=x2
~ dx +
i
X
2
+ y
2
= 4, where
~
we have two ways of finding (a)
2 2 and the relation x + y = 4 •
2x.
~ from x 2
But to find
2
y is not defined explicitly in terms of x,
• 2
We differentiate x + y
(b)
= 4
= 4
as it is, term by term.
Solve for y y =
2
{;7
j 4-x 2
or y = -
-x ~-- dx-~
2x + 2y
or ~ = --,-.c;..x=~
~ dx
dx~
X
~
= 0
X
y
X
=-y
=-y
It is not always possible to solve explicitly for
y, as in the example x
3
+
3
y - 3axy = 0 •
For these types of functions, called the implicit functions, we find the derivative by differentiating each term and then solving for
~
•
Example:
(a)
3 3 x + y - 3xy = 0, find
~
(b)
Differentiate with respect to x 3x
2
+
~
dx (3y dy dx =
3/
2
~-
~
- 3x) = 3y - 3x 2
~
y
3 (x
2
- X
+ y) =
2
0
~
x siny = 2, find
Differentiate with respect to x
.
• • X CO/ 0
for 0 ~ x ~ 1 or x ~ 2 •
This 1s the domam of f(x) 2.
= Vx (x
X
- 1) (x - 2)
Relation (1) shows that the required curve is symmetric dbout the x-ax1s. loop m the intervdl 0
y ... .:!: 3.
••• (1)
y
2
,< x ,<
Hence there is a 1. As x +
oo,
co.
3 2 = x - 3x + 2x
y dv
d,c = 3x
2y
2
- 6x + 2 gives the turning point
at x = 0.42.
4.
~
=2
x
5:
are the vertical tangents.
IntersectiOn with the axes: X
= 0, 1, 2
EXAMPLE:
Sketch y
y = .! (x - 3)
We have
v'x'=J
••• (1)
The curve 1s symmetric about the x-axis.
2.
Intersection with the x-ax1s:
x = I, x = 3
The domain 1s x ~ 1. There is a loop between x = 1 and x = 3, due to symmetry.
4.
As x -+
5.
2Y
~
co,
y ..,. .!
y
2 = (x - J) (x - 3) 2
1.
3.
X
is undefined dt x = 0, 1, 2, so x = 0, x = 1,
co
= 2 (x - I) (x - 3) + (x - 3) 2 = (x - 3) (3x - 5)
5 The possible cnucal points are x = 3 or 3 . x = I is the vertical tangent.
0
X
26
1.9
Applications
By using the graphs of functions, we can:
~~
~I
(a)
Solve an inequality such as cos 2x
or 21 xI - I x - 21
(b)
Find the number of solutions of an equation, such as 2 sin x = x or x = e-x f (x )
and by application of Newton's formula x 2 = x 1 to any required degree of accuracy. (c)
.
1 -l'{i(}> , fmd
a particular root
Solve physical and engineering problems involving equations which are either impossible or extremely difficult to solve.
EXAMPLE: (1)
211 X
= sin 2x
The sketch shows the graph of y 1 are obtained by solving sin 2x = 2
and y
=~ •
The intersections P, Q, R, S
=> 2x = ~6 , 511 , 1311 , ..!.Z2rgiving x _ ..2!. 5n 6
Hence the solution of the inequality sin 2x
>, ~
6
1311 1711 - 12' 12 ' I 2 ' 12
6
is given by
;
2
~
x ~ :; or \ ~11 ~ x ~ \7 11 • 2
EXAMPLE: (2) Solve for x: 2lx I - Ix - 21 ~ 2 To solve this, we draw the graph cf y = 2lxl-lx- 21 as follows: (i)
For x
~
(ii)
For 0
< x < 2,
(iii) For
x~
2, y O, y
1
3
= 2x y
- (x - 2)
= 2x
=x
-- (x - 2)
+
2
= 3x
2 = -2x -- (x-2) = -x-2
- 2
Now draw the line y = 2. This line intersects the graph of y = 2lx I -
Ix
- 21 at P and Q.
At P, y3 = 2 =9
-X
-2 = 2 ==>
X
= -4
-4
3
At Q, y 2 = 2 =9 3x- 2 = 2 ==> x = ~ = 1~ Hence the solution of 2lx I - I x - 21 ~ 2 is given by x
~
- 4 or x
-2 A
~ 1~
Note that the sharp corners are located where
lx I = 0 and
Ix -
21 = 0.
X
27 Find the stationary points of the function y = f (x) = (x + l)(x + 4 > •
EXAMPLE: (3)
X
Sketch the graph and find the domain and range of f(x).
Using the graph:
(a)
Solve the inequality (x + 1)x (x + 4) ~ 10
(b)
Shade the region R between the line y = 10 and the curve y =
(c)
(x + 1) {x + 4)
x
• Find the area of this region R.
Find the volume of revolution when the region R is rotated about the x-axis.
SOLUTION: f(x)=(x+0(x+4)
x
2
X
4 f(x) = x + 5 + -
••• (l)
X
f' (x) = 1 -
y
+5x+4
X
4
...
2X
f"(x) = 8 3
(2)
••. (3)
X
~
f' (x) = 0
x
2
- 4 = O, the
stationary points are where x = .!2. f"(i)
>0
and f"(-2)
,.. 10, draw the line X
y = 10. The line
cuts the graph at P and Q. Since x > 0 for P and Q, (x + 1) (x + 4) >,..lOx 2 :. x - 5x + 4 9 0 (x - I) (x - 4) > 0 ~ x 9 4 or 0 < x ~ 1 • (b)
Area of the shaded region between the line y
2
y
1
= f (x) = x + 5 + '!. is given by b X f4 4 A= a (y -y )dx= [IO-(x+5+x)dx 2 1 1
f
= 10 and the curve 2 4 = [5x-y-4logex] 1
A= 5(4- I)- 1 06-1)- 4(log4 -Jogl) = 15- 7.5- 81oge2 = 7.5- 81oge2 sq.u.
2
(cont.)
28 (c)
Volume
=
4
f
=n
v
f
ab (y 2 - Y 2) dx 1 2
n
1
=n
f
4
x) ]
[ 100 - (x + 5 + 4 2 dx
1
2 16 40 (67 - x - lOx - - - - ) dx
l
X
3
16 =n [67x - Tx - 5x 2 + x - 40 logex] 41 16 =n [67(4- 1)- 31 (64- 1)- 5(16- 1) + 0 i.e. f (x) is increasing For -0.4 < x < I, I < x < 2.4, f' (x) < 0 i.e. f (x) is decreasing Also the points A(-0.4, 0.2), B(2.4, 5.8) are respectively, the maximum and the miminum points. This is sufficient information for the sketch (solid line).
31 Sketch the curve y = 2cosx
EXAMPLE: (2)
+
cos 2x, 0 ~ x {. 2n
SOLUTION:
The period of f(x) = 2cosx is 2n and the period of g(x) = cos2x is n, so the function y = 2cosx + cos 2x is periodic with period 2n. A few key points are necessary to guide us correctly along the curve. These are: (0, 3),
•
4n
(3,-
~ 3 z>
It is unnecessary to check the nature of these points as we have sufficient points to sketch the curve.
y
.. .
.
I I • I
•
n
I •
..
,'
··' I
X
-1
-2
We can also sketch this curve by the addition of ordinates of the curves f (x) = 2 cos x
7n
and g(x) = cos 2x at the selected key points x = 0, ; ,
4 , 2n • This is shown in the figure.
~
3 , : ,
.... '
32 EXAMPLE: (3) Sketch the curve Y = 1 + sin x SOLUTION: (i)
(ii)
Intercepts: No x-intercept. The y-intercept
(0, I)
IS
-t----
-i f(x), -f(x), No symmetry
f(-x)
1
-n
(iii) Asymptotes: Vertical asymptotes are where sin x = -I x (iv)
= 2n n
+
31T 0 + 2 , n = , _1
Critical Points: f' (x) = f' (x) = 0 when
COS X
= 0 , giving
'
f(~)
(vii) y
=
>0
~
>0
0
n
3n
1T
1T
7n
31T
21T
X
2
2 l ln
,-2·2•2•2•···>
sin x I - I
X
= - 1.
3n iT 5n '- 2' 2 ' 2' ..• where we exclude
2 f"(x) = sinx. (l + sinx) + 2~osx (l + sinx) cosx (l + sin x) 2 . x -_ sinx(l + sinx) + 2cos x , now usmg cos 2x f "() (l + sin x) 3 . . x , f"(x) __ .· 2- sinx 1 + Sin an d cance 11 Ing · (l + sin x) 2 ,'. f" (x)
(vi)
(x
=>
-cosx . )2 (l + smx
those points given by sin x v)
, -+ 2 , .••
n
-2
· x ) (l = (l - sin
· x) + sm
for all real x, and the curve is concave up.
(All critical points on the line y
for all x (sin x
= ~)
-i - J) and the curve IS periodic with a period of 2n. ~ < x < 32n. The curve is as shown in
The pnncipal branch is in the interval the figure.
(viii) A few simple points (0, 1) and (n, J) are useful in sketching.
EXAMPLE: (4) Sketch the curve y
= ..,....--.....:....,+ sin x
y
by imtially sketching y = 1 + sin x.
I
I I
SOLUTION:
I
I I
After sketching y = sin x, we shift the curve 1 unit vertically up. The asymptotes are where l + sin x = 0. These points are easily seen to be n 3n x= ... ,-2 ' 2 ' ' " The maxima of y = 1 + sin x are the minima of the reciprocal curve y = are at x = -
+l sin x
I I
I
-Tf\
These
3n
n 5n 2 , 2 , 2 , ...
The graph of y
=1
I
.
+ Sin X
\
-!! 2 ....
··....._.,·
can now be sketched. (Heavy line)
A few simple points such as (0, 1) , (n, 1) are useful.
/ 0
n
:3 ,. X :2_.: 2n ·.. : .··y =sin x . ·-·- - .... -- ·..; ·.. - ..... .
2
'Tl·.
--~-----------------------------------------------
33 EXAMPLE: (5)
=x
Sketch /
y
2
2
(1 - x )
SOLUTION: (i)
Intercepts: y = 0 gives the x-intercepts
(ii)
X
= 0,
y
=-
-1
! I x = 0 gives y = 0 Symmetry: -~
+ X
"1 -
X-
,
-J ~ X~ 1
So the curve is symmetric about the x-axis.
/1 - l
.
Let y = f (x) = x We shall eventually combine the graph of y = f (x) with the reflection in the x-axis, i.e. y
=-
x
~
to obtain the
the required graph of
/ = x2
(1 - x2) • Now f(-x) = - f(x), so y = f(x) is point symmetric about 0. Putting these facts together, the curve is symmetric about both axes as well as the origin.
(i-ii) No asymptotes. (iv)
f'(x) =
~-
x2 (1 - x2rl/2 = 1 - 2x2
/1- x2
The critical points are given by 1 - 2x
2
=0
~
X :/:
x
1, -1
=!
1
12
r---7
VI - x- > O, the sign of f' (x) depends on the sign of the numerator 2 N (x) = 1 - 2x (A parabola) This graph tells us that:
Since
1
(a)
f(
(b)
f(
(c)
The curve is rising for
~2
) is the minimum, f(-
.!.:.. ) v2
- ~, ( -~ v2
0.7)
~ ) is the maximum, f (-1-) = -21 • V'i
v 2
-I ~
1
X
~ V"[ and
1
VL
~
Jz <
X
x
- ~ ,
f"(x)
>0
f"(x)
< 0 for x < - 21
for
x
4
x :/. 0
~ is a point of inflection with the curve concave down for x < - 21 and 1 concave up for x > 2 but x i 0. So x = -
This is sufficient information for a sketch of y = el/x.
36 y
EXAMPLE: (8) Sketch y
= ex cos X
0~
for
X,<
2n y
= e-x cos x
3n
Q
2
n
2'1T X
p
SOLUTION: (i)
Intersections: = 01 y = 1 . = 0 gtves cos x
X
y (ii)
=0
x
~
=2n , 23 n
Symmetry: y
f(x) /. f(-x), -f(x) No symmetry about the y-axis or 0. (iii) No asymptotes. (iv)
Critical points: dv
~
= - e-X cos x
-x . - e sm x
X
= -e-x(cosx + sinx)
dv
~
d
2
= 0,
~
when cosx
= e-X (cosx
+
sinx
.
.
=0
~
tanx
+ smx + smx - cosx)
dx
= -1
~
x
3n 7n = 4, 4
· = 2e-X smx
2
When x
d v =43n , -=-t > 0,
soy has a minimum at x
= 43n ,
so y has a maximum at
= 47n , Q (7n 4 ,
dx
When x
d2 Also ~ dx (v)
2
~
= 47n , =0
dx
2
< 0,
~ sinx
=0
x
~ The points of inflection at x
It is useful to draw the basic graphs y = e-x and y From these two we find that e-x cosx for
TT
P(
>0
for
0
3n ) 4 , -0.07.
= O,
n,
0.003). 2n.
= cosx.
< X< ;
and ex cosx
y
=0 =>
logex
x
X
=I
(logex)2
(ii)
f (x)
=
exists for
X
x
>0
f(x) ? 0 for all x (iii) f(x) (iv)
-j: f(-x), -f(x). No symmetry.
Asymptotes: Vertical asymptote x
=0
As x + .,, y-+ 0 y
=0
is the horizontal asymptote. Verify this by taking x
= 1~g ~
f(e10)
=e 10 ,
0
e (v)
Critical Points: _ x (21ogx) 1/x - (logx) 2 dx 2
21ogx - (logx) 2 2
~
X
~ddx = 0 => :. x
=I
log x e
X
=0
and x
or log x e
2
=e =7.I+
For 0
< x < 1, ~ < 0
for
b
C
51
2.2
Change of Variable: Substitution
We are easily tempted to apply a known method to new situations. 2 x dx
J Now
=;
3
+ C
[sin~x]
d dx
=
y
Consider
= X
e
. 2
Slll X COSX
2 then
J2x ex 2 dx ,
But given
. J
2 sm x dx
, so why not
'
3
= -sin3-
x
?
+ C
. 2 not sm x.
~ dx
2x e
X
2
how do we proceed to find such integrals?
The method of substitution, leading to a change of variable, is used to solve such problems. The symbol I will be used to represent the integral in each question. WORKED EXAMPLES
I.
f
2 dx
2x ex
Solution: Let
U
Solution:
= X2
du dx
Let
= x2
du dx
2x
J2x ex
- 4 2x
2 dx
2 2x and x - 4 in terms of u
Substitute for
feu
du dx
Jeu
du
I
c
2
ru
2
~+ c
e
e Thus
u
u
X
+
2 +
f f(x)dx
In either case
f)u
dx
c
Jg(u)du f g(u)du
by using either
is easy to integrate.
du dx
dx
-1/2 u du +
u = r$(x) or
c
x
9(u)
52
Integrate the following: (3)
.
-1 ~
J~ dx
Solution:
(4)
f.
Sln-1 X
(3)
j
Ju. dudx =
du dx
• dx
=
d X
. -1
Sin
X
1
~
Judu u
2
c
+
2
21 (sin-1 x )2 (4)
l
J X loge X
Let u
dx
1 - x2
(5)
+ C
Jexdx
Let u = e
e2x + 1
du dx
du dx2. dx
J
u
X
e
X
+ 1
J- 2du- u
+ 1
-1 tan u +
c
-1 X tan (e ) + (5)
c
Jxlog dx x
du Let u = 1ogx , dx
e
I
.!.X dx
lo!x
J~ .
+- Watch this: -+ dx X
du
loge u +
c
log (log x) + e e
c
= xl = du
53
Exercise 2A
(change of variable)
Find the indefinite integrals: I.
2.
(a)
J 2x ( /
(c)
J 4~
(e)
Jx
(g)
J x dx ~
(a)
Jcos 2 x
(c)
(e)
+ 1)
3 dx
(b)
f 2x (I - x2) 4 dx
(d)
J 4x
3
dx X + I
v'i-7 dx
(f)
J2x d\ I + X
J
(b)
Jsin 3 x cosx
Jtan 4 x • sec 2 x dx
(d)
fcot 4 x cosec 2 x dx
J -si~x
(f)
sinx dx
dx
(g)
Jsec 5 x
(a)
fx2 e
(c)
J e sm
(a)
J~
(c)
J
tanx dx
(h)
3
dx
IX ./I - rx
Jo
+
cosx)
Jo
+
tanx)
dx
4
2
dx
(b)
J
x dx
(d)
J ex dx
logx dx
(b)
Jl (logx) 2 dx
dx x (iogx) 3
(d)
J
X
. -1
4.
dx (x 4 + 1)2
(h)
COS X
3.
3
~
sinx dx 2 sec x dx
e-X dx I + e2x
~
x
x (I
dx logx) 3
+
54
5.
I. I
(a)
2 sinx
(c)
6.
ecos 2x dx
SlnXCOSX
(b)
I
-sinx ~x I +cos X
(d)
cosx dx
Evaluate:
I~
(a)
(c)
(e)
2.3
I
3
2x (x
2
- 3)
3 dx
(b)
Tl/4
f
e
3 sin x cosx dx
0
x dx
(d)
0~
I~
I
eX e
X
(f)
dx
e
2
Irr/
+ I
dx xlog x
2
cosx dx
0 (I + sinx) 2
Integration by Parts
The reverse process of the product rule
d dx (uv)
dv u dx
du + v dx
is the method of
integration by parts. From the above, dv udx
or,
I I
d dx (l•v)
dv • dx u • dx
uv
u dv
uv
-
I - I
-
du vdx
v du • dx dx
.•. (I)
v du
... (2)
WORKED EXAMPLES
Let
du • . dx Using the formula
I
x ex dx
I
uv' dx = uv -
x ex x ex -
I
I
v u' dx
ex • I • dx
ex + C
dv dx
u = x,
=I
,
v
= e
=
x
Je
X
dx
= eX
55
2.
Jx cosx dx U
u'
=X
dv dx
,
=1 ,
Jlogx dx
3.
= COS X
Jcosx dx
v
and let
sinx
• du • • dx
Jx cosx dx
V
=X
f x logx dx ,
u or v can lead to a disastrous situation.
~~
if you choose u = x ,
not easy to integrate. Again in
Jx cosx dx = uv - Ju'v dx = ;
J
x . cosx dx, if
2
cosx +
u
f
= logx , then v = logx dx is
= cosx
~ Jx2 sinx dx ,
dv , dx
=x
2
, then v
a situation worse than
Jx cosx dx • 4.
Jexcosx dx
Let
u
X.
JX· e smx d x
Jex sinx dx
We integrate
ex (-cosx) -
••• () 1 by parts again
Jex (-cosx) dx
-ex cosx + I
where I
Then from ( 1) X .
e smx 2I
ex (sinx + cosx) ,
Je
=ex
dv , dx
u' =ex , v
Jex cosx dx uv - Jvu' dx e smx -
=I
x logx - x + C
A bad selection of
For example in
= x'
dv , dx
uv -
x sinx + cosx + C
Warning:
= log x
Jv u' dx (log x) x - Jx • ~ dx
=
Jsinx dx
x sinx -
u
:. Jlogx dx
Jvu' dx
uv -
flogx dx as f (logx) • 1 dx
Write
divide by 2
x cos x d x = 1 e x(. sm x + cos x) +
2
C
= cosx = sinx
= 2x
56
It is important to note that the given integral may occur while integrating it by
Note:
parts, but actually this occurrence helps to find the solution, as seen above.
l
or x
Integrals with example to find:
Il
(5)
require repeated use of the 'integration by parts' method. For
I
2 dv x , dx = cosx
u
cosx dx
2 x sinx Again
3
I
u 1 = 2x , v = sin x
2x sinx dx
=
x sinx dx
x (-cosx) -
I
I • (-cosx) dx
u
-x cosx + sin x
=x
, v1
= sin x
u 1 = I , v = -cosx
2 x sinx- 2(-xcosx + sinx) + C 2 x sinx + 2xcosx- 2sinx + C (6)
Find
Write
-1
I sin = =
I
x d)(
1 (sin x) • 1 • dx
uv . -1
X Sin
Now for
I
Let u =
I
v u 1 dx
X -
I
x dx
v 1-x~
.•. (I)
~
2 du Use u = I - x , dx = - 2x
~
+-
-~I~ Iu-
dv x , dx =
I I U=T2,V=X
x dx
-~
. -1
Sin
du or xdx =- T
[It's easier to substitute
112
for x dx , than x itself] du
-Vu
- vr:7 . -1
X Sin
X
+
+
c
[from (I) ]
So it seems there is no end to the number of tricks you may be required to play in Integration. But that is what makes it so fascinating!
57
Exercise 28 Integrate the following:
I I
X C-x dX
(b)
x cos 2x dx
(b)
2 sec x dx
(b)
1.
(a)
2.
(a)
3.
(a)
Jx
4.
(a)
J x 2 logx dx
(b)
5.
(a)
f
cos-1 x dx
(b)
6.
(a)
I
log(x
7.
(a)
Jx
(b)
I
8.
9.
2
I I I
(c)
2 x sjnx dx
(c)
2 x sin x dx
(c)
J x cosec 2 x dx
(c)
J
f rx
logex dx
I I
(b)
- 1) dx
I I
x 2 e X dx
tan-1 x dx
(c)
e-2x cos3x dx
(c)
2 tan x dx
(Hint:
2 tan x
2 x cos x dx
(Hint:
2 2cos x
(a)
J~dx
(Hint:
Write
(b)
J~dx
ex sinx dx 2 x cosx dx
(logx)
I I
2
dx
x tan-l x dx
x (loge x)
2
dx
2 sec x-1)
= 1 + cos 2x)
I2
ll"aL+XL
Integrals involving expressions like substitution.
x=atane,
x = a sinS ,
etc. are simplified by
J
x = a cosa
etc.
WORKED EXAMPLES
1.
J /: 2 1
Find
a
+ x
dx d9
a tanS ,
X
dx
2
X
f
2
a
a sec a d9 a seeS
JseeS
Fig. I
d9
c
loge (sec a + tan 9) + log (
1:2?
+
~)
C
+
.;;:z-;-;! + x ) + C a log (x + .;;;z-:-;;z ) - log a e
log (
~)
log (x + 2.
Find
J ~ 2dx
x -a
Let
I
=
2 ,
x
>a >0 a sec9tan9
a seeS tanS d9
2 1 + tan 9
2
(sec a - 1)
Jsec a d9 log (seeS + tanS) + K
x jx2 - I) a2 log (x /x27) e log ( -
a
+
-
+
[Note: log a
C
K
+
x = asec9
I v~/
+
is a constant]
K
+ +
C
68
Exercise 2E
=a
Use x
F-2 x
1.
4.
a
J
Ia2 - x2 ,
sin 9 for
=a
x
tan9 for
2.
dx
~
5.
J?dx
dx J
10.
I
13.
2
-l
/25- x
2
fi J 2 dx x
dx
X~
I
sec9 for
2 J x dx
2 x dx
~
6.
J/xQ
dx
dx
2
2 J x dx
9.
J~
dx
11.
I
dx x2)3/2
12.
I
X d ~X
15.
14. 4 17.
I{.6)3
2.8
=a
8.
dx
f
x
dx
16.
19.
3.
£ Xi X 2 9-
/25- x J /a
and
to find:
,
7.
/a2 + x2
20.
;:G {4
f
J
+
J)2_ 4 dx
18.
dx x./:7-
J9dx
JJT7. 2 dx X
f~dx
21.
X
f.[;.2:9
x2 + 3 dx
Integration of Rational Functions
TYPE 1: We shall consider the rational function We divide P{x)
a~{:) b where P{x) is a polynomial.
P{x) by ax + b by long division and write
=
Q{x) + ~b where Q{x) is a polynomial and R is a constant. ax +
69 EXAMPLES: Integrate the following: (a)
B= ~ ~
dx
(b)
xx~ 2 dx
J
3
x - 1 '"'X""'7T dx
J
(c)
(d)
SOLUTION: 1
(a)
Divide 2x - 1 by 2x
J
+
2x + 1 f2i(':"l 2x + 1
I
~
~dx 2x + 1
Jo-2x:ldx = x - loge (2x + I) + C
When the denominator is linear, we can divide by inspection. For example, 2x - 1 2x + 1 - 2 2x + I = 2x + 1 (b)
2 J ~dx 2 J[x 2x
1
x
X -
8
+ 4 + x _2
+
3
_ __2_ 2x + 1
X-
J
2
dx
2
x + x + 4x + 8 log (x - 2) + C 3 (c)
3 Observe that x - 1 can be factorised as (x - I) (x 2 + x + I) J
L.:.l_ X - I
J0 >0
f' (x) is odd.
>a >0
,
use a sketch to
86
Exercise 2J
(REVISION)
Use any suitable method. Some integrals can be found in more than one way. Integrate the following:
I.
3.
I-x .;x:-'2
r
e
I e
5.
X
2x dx , using u = e
9.
II.
I J
4.
6.
+
15.
17.
2
+
2 2 b sin x)
25.
2 a co/x
+
Jx 2 1ogx dx
x v'f'"+i( dx
14.
X
2
2
16.
dx
18.
+ 4x
5 20.
+
5
dx
22.
2
3x + 2 4x
2
+
4x
+
dx 5 + 4sinx
2 2 b sin x
dx
2 (Hint: U = x )
dx
I + e; dx I - e
(Hint: t = ex)
dx
f Vt-:7 f 2
X
dx
+X +
I
+
X
dx
f
f
12.
f X~ f J f
f
5 sin(x ) dx
2 (Hint: Divide by cos x) 8.
/X
+ 3x - x
23.
4
x 2 sinx dx
4x
21.
~dx
10.
x)
+
dx I +ex
X
19.
X
2
JtfJ
dx 2(1
r f
X
2
- I
2
13.
I: I
1
sin 2x dx 2 2 b sin x
Ja 2
X
- I
(Hint: t = a 7.
2.
dx
dx
24.
5 26.
_e_ _ dx I + e2x
I f f f
I
xsin- 1x dx dx X
2
4x - 5
+
dx y{x
2
+
4x + 5
3x + 2 /4) + 4x
dx +
5
3sinx + 2cosx dx 3cosx + 2sinx
[Hint: 3sinx + 2cosx A Ocosx + 2sinx) + B(-3sinx + 2cosx)]
87
27.
29.
31.
I/§ I I
2 (cos2x = 1 - 2sin x, 2 divide by cos x)
dx 2 - 3cos2x
30.
/x2- X+ 1 dx
32.
2x + 5 dx 2 X - X- 2
34.
f
= t)
36.
J -X -- 3 X
(ex = t)
38.
J
35.
f (x 2 + 3) (x 2 + 1) ,
37.
f + 3ex + 2e 2x cosx dx f (I + sin x) (2 + sin x)
2x dx dx
(x
2
cos
v'X
40.
2
2 (Hint: X = t )
dx
3 X dx 2 X - 3x + 2 dx
I I
(Hint: divide 2 by cos x)
dx (3cosx + 2sinx)
cosx 5- 3cosx dx
33.
39.
I I I
28.
(Hint: multiply by .ja"+X)
a-x dx
[x - x
dx O+X+X
2 + x3)
3
= x(l
(Factorise)
dx sinx + sin 2x (Multiply by sinx, then t
41. Show that
n/ 2
f
0
n/2
I
n n - I sin a da = - n
sin
n-2
I
42. Apply successively the result of exercise 41 to show that n/2
(n - 1) (n - 3) ••• 4.2 n (n - 2) ... 3.1
sinna da
0
if n is odd n
(n - 1) (n - 3) ••• 3.1 n (n - 2) ... 4.2
and
if n is even.
2
43. Establish the reduction formula (using integration by parts)
f
xnsin bx dx = - x: cosbx + n/4
hence evaluate:
I
5
f
x
n-1
cos bx dx ,
2 x sin 2x dx
0
44. Establish the formula (using integration by parts)
f
1 xn+ 1 . -1 xn sin- x dx = n + Sin x 1
Hence find
f
1
n+T
1
x sin-l x dx 0 45. Establish the reduction formula
I
cot
n
dx
= -
[Hint: cotnx
cotn-l x n - 1
-
Jcot n-2 x dx
= cotn- 2 x (cosec2x - 1)]
f
= cosx)
a da , and hence show that
0
n/ 2 . 4 3n sin ada = 16 0 .
f
- x)(l + x)]
xn + 1
~
dx + C
88
CHAPTER 3 VOLUMES 3.1
Formulas for Volumes y
In elementary calculus, we have the following two formulas for calculating the volumes of revolution. 1.
The volume of revolution generated by the region bounded
0
by the curve y = f (x) , above the x-axis, between x = a and
a
b
l:!.x
X
Fig. I
x = b is given by: y V =
f
b
ny
2
dx
b
(Fig. 1)
d
2.
The volume of revolution generated
l:!.y
X
a
---------
f(y)
by the region bounded by the curve x
= f (y),
between y
and
y = b is given by:
v
f
a
b
2
nx
dy
(Fig. 2)
=a
0
X
Fig. 2
In many cases, it is necessary to calculate volumes whose boundaries are not surf.1ces of revolution and hence the two formulas stated above can not be used. For example we cannot find the volume of a pyramid or a doughnut shaped solid by these two formulas.
Fig. 3: Pyramid
Fig. 4-: Doughnut
89
In what follows, the formulas for volumes would be derived intuitively by means of simple examples.
The volume of a solid of uniform cross-section A and height h is h
given by V =A. h
Fig. 5 If h is very smaJJ, we have an
element of volume given by ll V =
where
h
=
A • llz llz
h
Let us calculate the volume of the solid (pyramid) shown in the figure.
We slice
the \l.ho!t'
pyramid by
n planes para1Jel to the base of pyramid. Let the distance between two successive planes be ll z. We find the area A(z) of one of the cross-sections at a distance z
Fig. 7
from the vertex V. Quite clearly A (z) is a function of
We express
z.
A(z) in terms of the sides
a and b of the base. From the two similar
triangles shown in the diagram, we have: X
a
z
h
Similarly
i.e.
x
ahz] y
bz
h
•••• (l)
90
A (z)
area of the rectangular cross-section at P X •
y zb
za abz
[from relation (I) ]
h
h 2
7
The volume of the pyramid is V = f. !:J. V = f. A (z) !:J.z
As n -+ ""•
f. 6. V -+- V
and
f. A (z)!:J.z =
h
A(z)dz
0
I
v
I
h A (z)dz 0
y
X
v
•••• (I)
Now from solid geometry, we know that the volume of the pyramid is given by V:
I
B.h
3
where B = area of the base = ab The formula V
~ Jb A(z)dz
a statt·ment can be mJ.ck:
used above is quite general and the following general
(see diagram above)
The volume of a solid whose cross-sectional ~rea is a continuous function A (z), is given by: V
Ib
A (z)dz
a
where
z
plane).
The limits of integration are chosen to include the entire volume. To evaluate
is the distance of the cross-section from the pre-determined point (or a
V, we must express A (z) and dz in terms of a single variable.
91 We note that the general formula for V =
the volume
Jab A (z) dz
includes y
the volumes of revolution given by
V
=
b
f
2 ny dx •
a
In the diagram: Z
::
X
A(z)
A(x)
=
area of the disc at a distance x from the origin
n/ v
Hence
t
f
A(z)dz
b
Fig. 8
2
ny dx.
a Note that the volume of revolution given by the last formula is also known as a DISC d
method. An Important Method
Finding lengths from the given diagram is of particular interest in calculating the required volume.
An elegant method is presented below and you will be asked to
derive the same result by using similar triangles. Example: (I)
4
Find the length y in terms of h, from the diagram which shows an isosceles trapezium;
r--- ~~-I~ ~--~y~----~
5
L----~~
Solution:
12
We observe that:
Fig. 9
y
4
when h
0
y
12
when h
5
Since y is always proportional to h, y is a linear function of h
y
=
and
T 5
mh+b
Substituting the given values b =4
4
m =
8
5
l
----rT I
h
f - - - - ' - y - - - \ - - -,'
j_
I I
12
Fig. 10
92
Example: (2) Using the intercept properties of parallel lines, prove the above result. (Fig. I 0)
Examp!e: (3) Find the volume of the block
shown in Fig. 3.
Solution: We consider the
block to be made up
of horizontal laminas of length 1, width w and thickness dh •
l .5 = 1 a). Use the method of (a) cylindrical shells and (b) disc (washer method). 2 The area bounded by the parabola y = 4ax and the line
x
=a
is rotated
about the y-axis. Find the volume by two methods. I 2.
A ring of altitude 2h is generated by revolving about the y-axis the area of the segment bounded 2 2 by the circle x + / = a and the chord of
X
length 2h that is parallel to the y-axis.
13.
By using the method of shells, show that the . . 4nh 3 Fig. 38 volume IS g1ven by - 3 x2 2 The area enclosed by the ellipse + = I is rotated about the line x = 8. 25 By using the method of shells, find the volume generated.
h
14.
The ellipse in exercise (13) is revolved about the vertical line through
15.
The triangle
the vertex A (5,0). Find the volume. ABC
formed by the points
A (a,O), B (-a,O), C (O,a) is revolved
about the line x = 2a. Find the volume generated by disc (washer) method. 16.
The cross sections of a certain solid by planes perpendicular to the x-axis are 2 circles with diameters extending from the curve y = x to the curve 2
y = 8 - x •
The solid lies between the points of intersection of these two
curves. Find the volume of this solid. 2 [Hint: radius r is given by 2r = (8 - x ) 2 where A = ll r ] 17.
The area bounded by the curve y = x about the x-axis.
2
+ I
i
=8
- 2x
2
and V =
f
2 A (x) dx,
-2
and the line y
Find the volume of revolution.
=3 -
x is revolved
(Hint: use the washer
method) 18.
The region bounded by the curves y = 3x -
i
and y = x is rotated about the
y-axis. Find the volume by the shell method. 19.
The triangle with vertices
(a, a), (a, 2a), (2a, 2a) is rotated (a) about the x-axis
20.
The area bounded by the curve /
(b) about the y-axis. Find the volume generated in each case. = 4x
and the line x = I is revolved about
the line x = 2. Find the volume generated.
112
CHAPTER 4 COMPLEX NUMBERS 4.1
Introduction - Necessity is the mother of invention.
Invented Number Systems: Complex Numbers Imagination and art of invention were required several times in extending our number system from the counting numbers.
1.
The first invented system J: the set of all integers as developed from the counting numbers. In this system we can solve equations such as x .:!:. 2 = O, in general x + b = 0, where b is any integer.
2.
The second invented system Q: the set of all rational numbers p/q as developed from the integers. In this system we can solve equations such as 2x - 3 = 0, in general ax + b = O, where a and b are rational numbers.
3.
The th1rd invented system R: the set of all real numbers x as developed from the rational numbers. In this system, we can solve not only the types 2 x + a = 0, ax + b = 0, but in addition all quadratic equations ax + bx + c = 0, 2 a ~ 0 and t:. = b - 4ac ~ 0. The roots are not real if 1:!. < 0. The simple 2 2 quadratic equation x + I = 0 or x + x + I = 0 is impossible to solve with the above mentioned three number systems. There is no real number that 2 satisfies the equation x + I = 0, since x = r-f does not exist· in the real number system R.
A new kind of number has to be invented to handle the roots which are not real. The [ i =
symbol
;-:II
is used with the understanding that
I
i
2
= -I
I
i is called
2 the imaginary number. The roots of x = -1 can now be written as 2 2 x = -1 = i -> x =- .:!:. i Again the roots of x 2 + 2x + 3 = 0 can be given as X= -
2
.:!:_
f8
-1
.:!:_
N
= -1.:!:.
Iii
Now we come to the fourth invented system, the set of all complex numbers of the form
We shall use a single pronumeral z to define a complex number
x + iy, i.e. X+
iy
I
113
x is called the real part and y is called the imaginary part of the complex number z. It is important to note that the imaginary part of a complex number is not imaginary! It is i that is called the imaginary number. The history of imaginary numbers is very fascinating. The earlier mathematicians thought that such numbers had no practical use (hence the term 'imaginary'), yet complex numbers are of great importance in fields such as Electronics. A complex number can also be defined by an ordered pair of numbers (x, y), without ever mentioning the imaginary number i, but in this book we shall only work with the binary form z = x + iy. However, the computer requires the form (x, y) for multiplication of two complex numbers.
4.2
Operations with Complex Numbers
We have the following definitions: a + bi
c + di
if and only if a
=c
1.
Equality:
2..
Addition:
(a + bi) + (c + di)
3.
Multiplication:
(a + bi) (c + di) = (ac - bd) + (ad + be) i
=
and b
= d.
(a +c) + (b + d) i
By using these definitions, we can verify that the complex numbers satisfy all the laws of algebra and hence the complex numbers form a field (denoted by C).
=
3 - 5i, then x
= 3,
If x + iy
EXAMPLE: (2)
Find the sum of 2 + 3i and 3 + 2i
SOLUTION:
(2 + 3 i) + (3 + 2 i) = (2 + 3) i
EXAMPLE: (3)
Find the product of 2 + 5i and 3 + 6i
SOLUTION:
(2 + 5i) (3 + 6i)
=
=
y
= -5
EXAMPLE: (l)
5 + 5i
(2x3- 5x6) + (2x6 + 5x3)i
=-24
+
27i
If you ignore the definition of multiplication and expand (2 + 5 i) (3 + 6i), i.e.
6 + 15i + 12i + 30i
2
and put
i
2
= -1,
then the result is
-24 + 27i, and so you
now discover the secret of strange definitions!
Identity elements: By definition: and
(a + bi) + (0 + Oi)
=
(a + 0) + (b + O)i
=
a + bi
(a + b i) (1 + 0 i) = (a • 1 - b • 0) + (a • 0 + b • 1) i = a + b i
Thus, the complex number 0 + 0. i, written as z = 0, is the additive identity and 1 + 0. i written as z = 1, is the multiplicative identity for the set C of complex numbers.
114
Additive inverse: We define the additive inverse of the complex number a + ib to be a number x + iy such that: (a + bi) + (x + iy) = 0 + 0. i (a + x) + (b + y) i = 0 + 0 • i By definition of two equal complex numbers, we have: a + x = 0
and
b +y =0
giving
x = -a, y = -b
Hence the additive inverse of z = a + bi now find the difference: (subtraction)
-a - bi, i.e. -z. Using this, we can
is
(a+ bi)- (c + id) =(a+ bi) + (-c- di) =(a- c)+ (b- d)i. Multiplicative inverse: The multiplicative inverse of the complex number a + bi f. 0 is defined to be a number x + iy such that (a + bi) (x + i y) = l + 0. i. By definition of equality, and
ax - by = l bx + ay = 0
We solve these equations for x and y, then a b and y = - - - - , hence the multiplicative inverse of z = a + bi 2 2 2 2 a +b a +b a -b _1 is the complex number z-1 = [ ~· ~ • Note that z. z-1 = z I = I, z" 0. a +b a +b We can now divide a complex number z by aflother complex number w, but shall use a method used in ex. (6} .below, i.e. complex conjugates. We can easily verify that the complex numbers obey all the rules of Algebra and hence they form a complex field C. Further, if b = 0, then a + bi reduces to the real number a, hence the set of real numbers R is a subset of the complex numbers C. If a = 0, b f. 0, then a + bi reduces to bi, and we call bi a purely imaginary number. x =
J
·z:
Note that a real number can be written in the form a + bi, for example 2 = 2 + 0. i. Similarly a purely imaginary number such as 2i can also be written as 0 + 2i. Calculations with the complex numbers do not require any special rules; wherever i occurs we replace it by -1. Further, 4 2 4 5 3 2 i = -i, i = i x i = (-1) (-I} = 1, i = i • i = i etc.
2
Complex conjugates: If two complex numbers differ only in the sign of their imaginary parts, each is called the conjugate of the other. Thus a + b i and a - i b are the conjugate complex numbers. Notation is used for the conjugate of z, i.e. = a - ib. Since z + z = (a + bi) + (a - bi) = 2a, the sum of two conjugate complex numbers is a real number.
z
z
EXAMPLE: (4)
Find the sum and difference of
SOLUTION:
(3 + 2 i) + (3 - 2 i) = 6
and
3 + 2 i and
3 - 2i
(3 + 2 i) - (3 - 2 i) = 4 i
115
zz
2 2 Again the product = (a + bi) (a - bi) = a + b , hence the product of two conjugate complex numbers is a non-negative real number. (3 + 2 i) (3 - 2 i) = 9 + 4 = I 3
EXAMPLE: (5) QUOTIENT:
a+ bi . We can simplify the quotient c + di , I.e. divide a + bi by c + di by using a+ bi the following procedure: c + di
(ac + bd) + (be - ad) i c2 + d2
c- di c- di
ac+bd be-ad. -2--2 + -2--2 • I c +d c +d EXAMPLE: (6)
Divide 3 + 4i by 2 + i 3 + 4i
SOLUTION:
3 + 4i
6 + 8i - 3i - 4i 4 + I
(3 + 4 i) (2 - i) = (2 + i) • "(2:1}
2+1
2
10 + 5i --5-
2+1
2 + i
EXAMPLE: (7)
If
SOLUTION:
. . 5 + 6i 5 + 6i 2 - 3i 28 - 3i Wnte x + 1Y = 2 + 3i = 2 + 3i • 2- 3i = --13-
(x + iy) (2 + 3i) = 5 + 6i, find x and y.
28 . I d . . Equatmg rea an 1magmary parts, x = TI
,
y =-
TI3
Alternatively, expanding and equating real and imaginary parts, we find: and
5]
2x - 3y = 3x + 2y = 6 '
EXAMPLE: (8)
28 x=n'
solving these:
Expand (a) (I + i) a+ ib.
4
3 y=-n
and (b) (I - i)
5
and simplify in the form
SOLUTION:
(a)
(I + i)
4
+ 4i + 6i
=
2
+ 4i
3
+ i
4
+ 4i - 6 - 4i +
i.
= -4 and
(b)
(I - i)
5 4 2 3 - 5i + !Oi - !Oi + 5i - i
5
- 5i - 10 + !Di + 5 - i =
It is easier to expand by writing z = I + i 2 z4 = z 2 = (I + i) 2 (I + i) = (2i) (2i) = -4 5 2 2 . . I S1m11ar y z = z • z • z =(-2i}(-2i)0-i) = -4 (I - i) etc.
-4 + 4i
EXAMPLE: (9)
Express (2 - 3if
SOLUTION:
in the form a + i b (b) - -.X- IY
·(a)
24.
=2
+ i,
3z + 4
evaluate: 2 (b) z - 2z + 3
(c)
2z - l 2z + l
(d)
(z - 1) (z
2 + z + 1)
If z = x + iy, express each of the following in the form a + ib: (a)
25.
z
-z
(b)
-z
(c)
Solve the following equations for (a)
(l + i)z
(c)
-
=
2 - i
2 l + i + z =
3
r:l
z + l Z-1
(d)
z; express answers in the form
a + ib:
(b)
2z (I - i) z T+i + 3- 2i =
(d)
Z-1
z + 3
2 - 3i
117
26.
Solve the following equations for (a) z 2 + z + I = 0
27.
28.
(b) z
2
z; express answers in the form a + i b:
- 2z + 4 = 0
(c)
2z
2
- 3z + 2 = 0
(d) z + ..!.
z
Find the quadratic equations with wots given below: (a)
i, -i
(d)
3 + i,
(b)
+ i,
+ 3i
(c)
I - i
2 + 3i, 2- 3i
2 + i, 2 + i
(e)
Solve the following pairs of equations for z and w where z and w are complex numbers. Express answers in the form a + ib. (a)
(c)
z + iw
2 + 3i
z- iw
2 - 3i
(b)
(d)
(2 + i)z + (2 - i)w (2 - i) z + (2 + i) w
2
2z + w
I +i
z-w
I - i
z +(I - i) w
2i
w + (I - i) z
29.
Given z = 2 + i, evaluate the following in the form a+ ib: (f) z 4 (e) z 3 I I (c) (b) z 2 (d) z 2 + 2 (a) 2 z z z 2 z2) 2 4 3 z and z (Hint: z z = z
30.
What is thq•fallacy in the following:
V-3 · V-i2
=
1 . 0),
then
[r(cos9 + isin9)]n = rn(cosn9 + isinn9) For r = I, we obtain De Moivre's theorem:
I
(cose + isin 9)n = cosne + isinne
De Moivre's theorem can also be proved by the method of mathematical induction. (This proof is given in a later section.) It also holds for n = 0 and for a negative integer. 6
EXAMPLE: (21)
Find (I + i)
SOLUTION:
Write z = I + i in the mod-arg form.
z = ../2(cos Tl /4 + isin Tr/4) :. z 6 = [ v'2 (cos Tr/4 + isin Tr/4) ] 6
6 ( Vi.) (cos 6 Tr/4 + isin 6Tr/4) 8 (cos 3 TI/2 + i sin 3TI/2) -8i
EXAMPLE: (22)
Evaluate in the form a + ib: (I + .fji)
SOLUTION:
We write
4
+ (-1
+
/3i)
z = I + vJi = 2(cos60° + isin60°) w = -1 + ni = 2(cos120° + isin120°)
z
2 4 (cos60° + isin60°)
4
w
7
z
4
Hence? w
4
/ (cosl20° + isin120°)
~~ ~os (240°
1
=
16 (cos240° + isin240°)
7
and
128(cos840° + isin840°) 128 (cos 120° + i sin 120°)
- 120°) + i sin (240° - 120°)]
(1 /8) (cos 120° + i sin 120°),
i.e. -1/16 + i( '113/16)
7
126
Exercise 40
Evaluate in the form I.
a + ib, by using De Moivre's theorem:
(cos 12° + i sin 12°)
5
2. 4
3.
(cos nl4 + isin nl4)
5.
[cos(-nl4) + isin(-nl4)] 6
4. 6.
[2 (cos 15° + isin 15°) ] 6 [2 (cos nl3 + isin nl3) ] 6 [2(cos2nl3- isin2nl3)] 4 (Hint: bracketed expression = cis - 21113)
Express the following in mod-arg dnd cartesian form 7.
(I
9.
(I
II.
(-1
13.
(I
(3 + 4if 2 10. ( v'j - i)4
+ i)6 - i)4
8.
12. (2 - 2 {jif 4 14. (2 J3 + 2i) 5
- i)IO +
11'304
Simplify the following in the form rcis9 in the form a+ ib (15 to 20). 15. 17.
(1
- i)4 (I
(I
VJi)3 i)4 -
(I
19.
+
(2cis nl6)
(7 to 14).
+ i)3
and wherever possible, express the answers
16. (I 18.
20.
VJi)3 (I
(2 + 2i) 4 (I
4
(4cis nl3) 3
-
-
v'Ji)2
(3cis n I 12)
5
[3cis (-n I 36)] 3
+ i)4
127
4.6
Roots of Complex Numbers
Finding a desired root of a non-negative real number is a simple matter. example: The two square roots of The cube root of
For
4 are .:!:. 2 113 (8) = 2 and so on.
8 is
But finding the desired roots of complex numbers is not so simple, as we must use De Moivre's theorem to find the: three cube roots, a + ib.
four 4th roots,
five 5th roots and so on of a complex number
EXAMPLE: (23)
Find all the cube roots of
SOLUTION:
We use two methods to find the cube roots.
Method 1:
If z
-1
-I, then
is a cube root of
3 2 z + 1 = 0, hence (z + 1) (z - z + 1) = 0
adding -l
=
2kTT to the arg(-1), COS
(TT + 2kTT)
2 cos 11
-1
We write
Method II:
J3i
1 .:!:_
-1,
The three cube roots are
+ i sinTT
and make the expression general by
hence
••. ( l)
+ isin (TT + 2kTT)
Now let the required cube root be W'cos(TT + 2kTT)
R (cos~ + 1sin~)
+ isin(TT + 2kTT)
= R(cos~
+
isin~)
We take the cube of both sides:
R 3 (cos3~+ isin3~) =
cos(TT + 2kTT)
+ isin(TT + 2kTT)
Equating the real and imaginary parts:
R 3 cos3~
= cos(TT + 2kTT)
1,
From these: ~
i.e., R =
TT/3
sin ( 11 + 2kTT)
and and
3~
TT + 2kTT
+ 2kTT/3
Hence the cube roots are given by: z where
= k
(cos~+ isin~)
= 0,
1, 2.
=
cos(TT/3 + 2kTT/3) + isin(TT/3 + 2kTT/3)
128
For k = 3, the root for
the angle is n /3 + 2n and the corresponding cube-root is the same as k = 0. In a similar manner, k = 4 produces the same root as k = 1 1 3 and so on. The three cube roots of (-1) / are: 0,
zl
cos n/3 + isin n/3
1/2 +
k
I,
z2
cosn + isinn
-I
k
2,
z3
cos 5n/3 + isin 5n/3
1/2
k
-
J3/2 . i
r
fi/2 • i
These three roots are the same as given by Method I. The modulus of each root is I, hence these
three distinct cube-roots lie on a circle of radius I and centre the origin. They are equally spaced, the angular distance between any two roots being 360° + 3 = 120" Definition:
If
n is a positive integer, then R(cos\6 + isin\6) is an nth root of
r(cosa + isina) if and only if Rn(cos\6 + isin\6)n = r(cosa + isina) From this, and using De Moivre's theorem: 11
R (cosn16 + isinn\6) = r(cos a+ isina) rcosa = Rncosn\6
and
rsina = Rnsinn\6
By squaring and adding: Rn = r, hence cosa = cosn\6 gives n\6 = a + 2kn
0=
Hence R = r 1/n ,
a +n 2k 1T '
where k = 0, 1, 2, •• • (n - I)
We conclude that the n nth roots of the complex number r(cosa + isina) have the 1/n . a + 2kn ) modulus r and arguments are g1ven by --n--' k = 0, I, 2, ••• (n - 1 •
EXAMPLE: (24) SOLUTION:
Find the 4th roots of - & + & >/3 i
Write -& + &IJi = 16(cos2n/3 + isin2n/3) = 16(cosl20" + isinl20") = z
.. (2kn + 2n/3)] • [ (2kn + 2n/3) .. The 1!-th roots are z = 2 cos + 1sm , 4 4
k =
o,
4
1, 2, 3
The four roots are:
"3
0,
zl
k
I,
z2
2 (cos 120" + isin 120")
-1 + ,/3i
·k
2,
z3
2 (cos210" + isin210")
-.13-l.i
k
3,
z4
2 (cos300" + isin300")
I -
k
2 (cos30" + i sin 30")
+ 1i X
Fig. II
v'3 • i
The points in the Argand diagram that correspond to these four distinct 4th roots are equally spaced (at an angular distance of 360" .;- 4 on a circle of radius 2 and centre the origin.
90")
129
EXAMPLE: (25)
Find all 5 roots of z diagram.
SOLUTION:
z
5
= 32
5
- 32 = 0 and show them in an Argand
= 32(cos0 + isinO)
32[cos2k11 + isin2k11]
The five fifth roots are given by 2k11 . . 2k11] + I Sin T z = 2 [cos T
where k = 0, 1, 2, 3, 4.
To clearly visualise the roots, we replace 11 by 180", then the five roots of z 5 = 32 are: k
o,
zl
2ciso• = 2
k
!,
z2
2cis72"
k
2,
z3
2cis 144"
k
3,
z4
2cis216"
k
4,
z5
2cis288"
(the only real root)
Note that the argument of successive root increases by 360° "'" 5 i.e. 72". This way you can quickly write down all the roots. Observing that: lz 11 = lz 2 1 ••• = lz 51 = 2, we can show these roots on a circle of radius 2, centre 0. From the diagram:
A(z ) 1
-
z5
z2
z4
X
z3 Fig. 12
EXAMPLE: (26)
Solve z
SOLUTION:
Write
z
5
5
16 .f2 + 16 J2i
1+ i
l'i.cis( 11/4) =
32cis (2k11 + 11/4)
The five 5th roots are given by z = 2cis(rr/20 + 2krr/5) The five roots are:
k =
o,
1, 2, 3, 4.
zl
2cis9"
z2
2cis (9" + 72")
z3
2cis(9° + 144") = 2cisl53"
z4
2cis(9" + 216") = 2cis225"
z5
2cis(9" + 288") = 2cis297"
2cis81"
v'2cis(2k11 + 11 /4)
130
Exercise 4E
1.
Find the two square roots of: (a)
2.
(b)
-16i
4.
(b)
-8
Find the four 4th roots of: (b) -16 (a) -1
-2/3- 2i
(d)
4 + 3i
(c)
27(~ + ~ 0
(d)
-8i
(c)
-8 - 8 J3i
(d)
2-2
(c)
-2 + 2 I
(d)
32i
1
V3.1
Find the five roots of: (a)
5.
(c)
Find all the cube roots of: (a)
3.
2 + 213i
(b)
32
-32
.fj
1.
Find the solution set for each of the following equations. Express your answers in the form a+ bi. (a) (c)
X X
(e)
X
(g)
X
3
4 2 4
-
i
2
+ 1
0
(b) (d)
X X
(f)
X
13.
(h)
X
21
4
6
+ i = 0 1 2 -
2
2
+ (3 - i)x + 16i = 0 64 = 0 + i
= 0
-
3i = 0
131
4. 7
De Moivre's Theorem and Its Applications
We have used De Moivre's theorem in finding the powers and the roots of complex numbers. We now prove it by the method of induction and consider some further applications. De Moivre's theorem is: For any integer n, (cos9 + isin9)n = cosn9 + isinn9 Proof:
For
n
= O,
For n = I,
(cos9 + isin9)
0
= I = cosO
••• (I)
+ isinO
cos9 + isin9 = cos9 + isin9
So the theorem is true for
n
=0
and n
= I.
We assume it true for n
=k, a
(cos9 + isin9)k = cosk9 + isink9
positive integer, so:
••• (2)
Multiply both sides of (2) by cos9 + isin9, then (cos9 + isin9)k+l = (cosk9 + isink9) (cos9 + isin9) = (cosk9cos9- sink9sin9)+ i(sink9 cose + icosk9sin9) = cos(k + I) 9 + isin(k + I) 9
This proves that the theorem is true for n = k + I, if it is true for n = k. Since it is true for n = I, then it is true for n = 2 and so on for all positive integers n. The theorem is also true when n is a negative integer. Let n = -m, where m is a positive integer. (cos e + i sine )n= (cos e + i sine fm = - - - - - - (cose + isin9)m by using the theorem for positive integer
cosm9 + isinm9
m.
I (cosm9- isinm9) cosme + isinm6 = (cosm9 + isinme) kosm9- isinm9)
Now Finally
cosme- isinm9 1
cosm9 - isinm9 = cos(-n9)- isin(-n9) = cosn9 + isinn9 •
Hence the theorem is true for negative integers. DERIVATION OF TRIGONOMETRIC FORMULA
By application of both De Moivre's theorem and binomial expansion, we can find: (a)
cos n e and sin n 9 in terms of powers of sine and cos 9 •
(b)
cosn9 and sin 9 in terms of multiple angles (29, 39, etc.)
11
For (a) and (b) we shall need the following:
132
Let
z z
n
= cos9 + isine , then: -n cosne + isin ne and z
cosne- isinne.
-n +z = 2cos ne n -n Subtracting, z - z 2isin ne Adding,
z
n
1 z
For n = I,
2cos e
Z+-
z
n = 2,
2
1 +2 z
••• (1)
•.• (2)
'
z
2cos2e ,
z
1 z 2
-
2isine
1 2= 2 isin 2e z
'
etc.
Express (a) cos4e in terms of cose (b) sin4e in terms of cose and sine
EXAMPLE: (27)
(c) tan4e in terms of
SOLUTION:
tan e.
We use c = cose and s= sine to simplify our work.
By De Moivre's theorem: cos49 + isin49 = (cos9 + isin9) C
4
= (c + is)
4
4 + 4c3.IS + 6c2.2 2 4c.3 3 .4 4 1 S + 1 S + 1 S
c4- 6c2s2 + s4 + i(4c3s- 4cs3) . 1.4 = 1, etc. ) ( Note t hat 1.2 = - 1, 1.3 = -1, Equating real and imaginary parts: (a)
(b)
4 4 2 2 2 4 cos49 = c - 6c s + s = c - 6c (1 - c 2 ) +(I - c 2) 2 , 2 . . 2e usmg cos e = 1 - sm 4 2 :. cos49 = 8cos e- 8cos e + 1 sin49 = 4c\- 4cs
3
3 = 4cos 9sine- 4sin 3 ecose
We note that sin4e cannot be expressed in terms of sine alone. (c)
t
an 4
e _ sin4e - cos4e
4c\ - 4cs3 c4 _ 6c2 5 2 + 5 4
Divide both the numerator and the denominator by c sin9/cose = tan9, we have:
tan49
3 4tane - 4tan e 2 4 1 - 6tan e + tan e
4
remembering
133
EXAMPLE: (28) (a)
If
z " cos 9 + i sin 9, show that:
(i)
zn
_!_ " 2cos n 9
+
z (b)
zn -
(ii)
n
z
Express the foUowing in terms of cos n 9 (i)
cos 4 9
or sin n 9
4 sin 9
(ii)
2isin n9
n
(iii)
3 sin 9
SOLUTION: (a)
z
cos9 + isin9
-
£ 1 = cos9 - isin9
z
By De Moivre's theorem: zn " cosn9
+
isin n9,
z Adding:
z
n
+
-n
z-n
n
cos n9 - isin n9 ••• (i)
2cos n9
z
Subtracting: z
n z
(b)
(i)
(2cos9) 4 4 16cos 9 4 cos 9
(ii)
•.. (ii)
2isin n9
~) 4
(z +
2cos49
= (z
+ -{)
+
+ 4cos29
4
4 16sin 9
2cos49 - 8cos29
(2 i sin 9 )3
3
-
. 39
1
i
4 (z
6
+
+ 6
+
2
1 + /) + 6
[in (a)(i) n = 4, 2]
3)
expand and rearrange.
'
1 1 3z ) - 3(z - -z>
2isin 39
~)
[using results (a)(i) n
+ z4 ) -
1 3
z
(z
Sill
= (z
+
+ 3)
81 (cos49 - 4cos29 (z - -)
2
z
8cos29 + 6
(z - -)
z
4(z
+
z
1 8 (cos49
1 4
4
4 (2 isin 9 )
4 sin 9 (iii)
n
6isin9
(3sin9 - sin39)
[in (a) (ii) n
= 3,
1]
= 4,
2]
134
Exercise 4F I.
Express (a) cos 3 e in terms of tan e
2.
Express (a) cos 5 e in terms of tan e
3.
Express (a) cos6e (b) sin6e in terms of cose and sine and hence express tan 6 e in terms of tan e
4.
Express
5.
Express (a) cos 5 e (c)
6.
(b) sin 3 e in terms of cos e and sine, hence express tan 3 e (b) sin 5 e in terms of cos e and sine, hence express tan 5 e
3
(a) cos 3 e
5 fcos ede
(b) sin e
5
(b) sin e
in terms of multiples of e and hence integrate
5 (d) fsin ede
6 (a) cos 6 e (b) sin e in terms of multiples of e, and hence TI /2 Tl /2 6 6 (d) sin ede cos ede integrate (c) 0 0 Express
f
f
7.
in terms of multiples of e
Find the constants p, q, r and s if: (a) co/ e = pcos?e + qcos5e + rcos3e + s Tl /2 7 and hence evaluate: (b) cos e de 0
f
8.
7
Find the constants p, q, r, s if: (a) sin e = psin?e + qsin5e + rsin3e + ssine, rr/2 7 and hence evaluate: (b) sin e de 0
f
9.
Show that 1'1 + cose + isine)n = 2ncos n( 2 e) is a positive integer
10.
Prove the following: (a)
cot4e
(b)
cot5e
2 4 - 6 tan e + tan e 3 4tane - 4tan e 4 2 I- 10tan 9 + 5tan 9 3 5 5tan9- 10tan e + tan e
~cos n 2e
· n eu + 1· sm
2
, where n
4 2 cot 9- 6cot e + 1 3 4c'ot 9 - 4cot 9
5 3 cot e - I Ocot e + 5cot e 4 5cot e- IOcoie +·I
135
4.8
Square Roots of a Complex Number
We usually use De Moivre's theorem to find the roots of a complex number; but there is also a special requirement to obtain the square roots in the form a + ib. Two methods of obtaining the square roots are: 2 I. We convert z = a + ib to mod-arg form and use De Moivre's theorem to solve
l 2.
=a+ ib.
We assume the solution z = x + iy and hence solve two simultaneous equations involving x and y. Using De Moivre's theorem.
METHOD 1:
EXAMPLE: (29)
Let
Find the two square roots of 2 + 2/Ji
z 2 = 2 + 2 _,i = 4(cos n/3 + isin n/3) = 4[cos(2kn + n/3) + isin(2kn + n/3)]
By De Moivre's theorem:
~os (
z = 2
2kn ; n/3) + isin ( 2kn ; n/3)
J,
k = 0, I.
The two roots are:
zl
2(cos n/6 + isin n/6)
z2
2(cos7n/6 + isin7n/6) =
Verify that,
=
tf3 + -13-
i
[!. ( J3 + i) ] 2 = 2 + 2/3i
METHOD U:
EXAMPLE: (30)
Let z = x + iy, then
Find the square roots of 7 + 61/ii
2
SOLUTION: Let (x + iy)
z =a + ib X2 -
y 2 + 2"IXY = a + 1"b
Equating real and imaginary parts. x 2 - y 2 = a and 2xy = b, then: (x2 +/)2 = (x2 -/)2 + 4x2/ = a2 + b2
.22~ 22 + y = '1/a- + b. as x + y
•• x
>0
= a and
for x and y.
i
+ /
=
= 7 + 6 /2i
:. x - / = 7 and 2xy = 6 v'f 2 2 2 2 2 Now (x + (x + 4x /
/>
/l + b2
/>
49 + 72
121 X
2
+y
2
2
and x - y
Finally we solve
i -/
2
2
2
=11
••• (1)
=7
••• (2)
Solving (1) and (2): X=!.3,
y=!.fi
We check xy = 6{2 for the proper combination of x and y. The required roots are:
136
Exercise 4G 1.
Find the square roots of the following in the form x + iy. (a)
3+4i
(b)
3-4i
(c)
(e)
5 + 12i
(f)
8 + 6i
(g)
(d)
5+2J6i
(h)
7-6/ii -8i
2.
Solve the following equations, expressing the answers in the form x + iy. 2 2 2 (b) z = I + {:Ji (c) z = 2i (a) z = -15 + 8i
3.
Use the formula z = (-b :!:. lb equations in x + iy form:
4.
(a)
z
(b)
z
(c)
z
(d)
z
2 2 2
=
- (I - 4 i) z - (5 - i)
- 4ac)/2a to express the roots of the following 0
+ (2 + 4i) z - 11 - 2i
0
=
+ ( 4 + 2 i) z + (3 + 2 i) + (4 - 2i) z + 6
(Hint:
J5:l2i
3 - 2i)
(Hint:
.Js
3+i)
+ 6i
0
=0
Write each of the following in the form a+ ib. (Take (a)
5.
2
2
,fZ to mean the square root whose real part 1 +i ./8+6i
(b)
/5 -
l2i 5+12i
Prove the following: (a)
~
(b)
..ra+l>1
6.
If X+ iy
7.
Simplify:
=
+
h
~
va:bT
~ c + !d ' ..t5:lii v57iTi
(c)
=
i
(./a
2
~ I + z + z 2 where z = v& + 6i
+b
2
+a)
h< E:;}_
prove that (X2 + y 2)2 +
~ {5 - 12i
> 0)
a)
=
2 + b2 2 c + d2 a
137
4.9
Properties of Conjugate Complex Numbers
z= x
- iy
is the conjugate of
If the point
P(x,y) represents
complex plane, then
z
in the x-axis.
z = x + iy in the
is the reflection of
z = 2x
Fig.l3
arg -z = -argz
(a real number)
5.
Further, for two complex numbers z
6.
7.
9.
10.
1
z and
(a) z
1
Let + z
2
z
= a + ib and 1 = a + ib + c + id
+ c) -
i (b
z
2
1
• z
2
= c + id,
then 1
- z
= a + ib - c - id
2
=(a- c)+ i(b- d)
+ d)
z
1
- z
2
= (a - c) - i (b - d) = (a - ib) - (c - id)
= zl
=zl
+ z2
= (ac - bd) - i (be + ad)
= (a - ib) (c - id)
I
z
z
- z2
(d)
We have
zI
(ac + bd) - i (ad - be)
z2
(see Section 4.2, example 5)
c2 + d2
zI Again, z2
• z-2
= z = zz = Izl2
I
I
2
= (a - ib) + (c - id)
= zI
.-~ Q
we prove the following properties:
2
(b) z
= (a + ib) (c + id) = (ac - bd) + i(be + ad) z
''
z
z-I
z
2 = (a
X
(a purely imaginary number)
=(a+ c)+ i(b +d) ~
I I
: -y
- = x 2 + y 2 = lzl2 -- l-zl2 zz
3.
z = 2iy
X
8.
lzl
Proof:
..
We have:
2. z +
y
Note that x + iy is the
x - iy.
I.
4.
P(z)
Q(x,-y) is the reflection of
P in the x-axis and hence z
conjugate of
y
z = x + iy.
(a - ib) (a - ib) (c + id) _ id) = (c _ id) (c + id)
=(c
(ac + bd) + i(ad - be) '
c
2
+
d
2
( :~)
using property (3)
We easily generalise, but shall not prove that: II. z I + z 2 + z 3 + ••• + zn 12. ZIWJ + Z2W2+ In particular, if
w
•••
= z I + z 2 + ••. + zn
+ZnWn
= zl.wl +Z2
is real, then using
13. z 1w"l + z 2w 2 + •••
+ znwn
w = w,
W2+ •••
+Zn.wn
we have
w 1z 1 + w 2z 2 + •••
h
, ence
+ wnzn
We shall use the properties (9), (10) and (11) in the next chapter on Polynomials.
138
EXAMPLE: (31):
=~
If z
+ 3i, express the following in a + ib form:
(a) z
(b) z + z
(c) z - z
(d) zz
(e)
£1
SOLUTION:
(a)
z =~-
(d) zz
3i
=(~
z =~
(b) z +
+ 3i + ~- 3i
z =~
(c) z-
~ - 3i ~ =z1 =zzz =Izlz2 =2 5 = 25
-1 (e) z
+ 3i)(~ - 3i)
=8
+ 3i- (~ - 3i)
=6i
3. . ( ) - 25 1, usmg d •
= 16 + 9 -1 Note that z
= 25
= (X
If x + 1 y
. )(
+ IY
X
+
-z
Using ,
,
=arg
z .
prove t h at ( x 2 + y 2)2
We square both sides and write
SOLUTION:
w
~
= Vr;;-:;Tb ~
.
EXAMPLE: (32)
-1 z in general, even though arg z
.1-
=
1 • z2
= ca
. )
IY
z1z
+ ib
••• (I)
+ id
and
(z1) Z2
2
w = (X-
. )(
IY
. )
=
X- IY
a - ib c - id
••• (2)
= x2
Multiplying (I) and (2) and using (x + iy) (x- iy) (x
2
2 +b 2 c +d
+/,etc.
2
2 2 2 + y ) (x + y )
= a2
, hence the required result.
Exercise 4H I.
Given o> z
(a) z
= 1 + 2i
(ii) z + z
x + iy x2 + /
Oii) z-
3.
If a+ ib
(x + i)2 2x - i
+ iy
~ 1 '
~-
5.
If a
3 - i, find and sketch the following: 1 ov> z • -z lzl 1zl 0 By squaring and adding, r whence 7 9 = 2kn • cos79 =cos2kn, 2kn 9 = T , where k = 0, 1, 2, ••• , 6
The seven roots of unity are given by: z
1
=cos4n/7 z =cos8n/7 5
z
+ isin4n/7
z4
=cos2n/7 =cos6n/7
+ isin8n/7
z
= coslOn/7 + isin lOn/7
z
=cosO + isinO = 1
3
2 6
+ isin2n/7 + i sin 6n /7
z = cos12n/7 + isinl2n/7 7 Now we apply De Moivre's theorem, in reverse order, i.e. cosn9 + isinn9 = (cos9 + isin9) 7 to each of the above complex roots. Let
z
2
= w =cos2n/7
+ isin2n/7
2 = w 3 3 and so on, z =cos6n/7 + isin6n/7 = (cos2n/7 + isin2n/7) = w 4 4 .5 6 hence, z = w , z 6 = w , z 7 =
6.
other complex roots. (a) (b)
(c)
4 2 3 1 + w + w + w + w = 0
Prove that
4 Find the quadratic equations whose roots are a = w + w and 7 6 5 2 3 B = w + w (Hint: use w = 1 to reduce w and w ) Show the roots of
z 5 - 1 = 0 in an Argand diagram.
(d)
7.
Find the area of the pentagon formed by the roots. If w is a complex root of z 6 - 1 = 0 with the smallest positive argument, then
2
3
4
5
show that the other roots are w , w , w , w • Prove that:
2
3
4
5
(a)
1 + w + w + w + w + w
= 0.
(b)
Find all the roots in the form
a + ib
and indicate these roots in an
Argand diagram. Find the area of the hexagon formed by the roots. (c)
Find the two quadratic equations whose roots are 4 (i) w and (ii) and w
(d)
Using part (c), show that 4 2 5 (i) z 6 - 1 = (z- 1) (z + 1) [(z- w) (z- w )] [(z- w ) (z- w )] 2 2 = (z 2 - 1) (z + z + p (z - z + 1)
J
(ii)
The roots of
d
z
4
+ z
2
+ 1
=0
2
4
are w, w , w
5 and w
142
8.
Show that if. w is one complex root of the equation zn - I = 0, then - I) (z - w) (z -
i> ...
(z - wn-l)
zn - I
(b)
Deduce from part (a) that: zn-l + zn- 2 + ••• + z + I = (z - w) (z - w2) ••• (z - wn-l) (1 - w) ( I - w2 ) ••• (1 - wn-l) = n
(c) 9.
= (z
(a)
Prove by mathematical induction that for any real 9, cos ne + isin ne = (cose + isin9)n (a) (b)
Find the 6 sixth roots of I, expressing each in the form a + ib. 4 Using part (a), find the four roots of z + z 2 + I = 0 and show their positions in an Argand diagram.
143
4.11 Miscellaneous: Factorisation over the Complex Field 2 3 3 Consider the factorisation of z - 1 = 0 over C. z - 1 = (z - 1) (z + z + 1) 2 Now z + z + 1 has no real linear factors, but over C we can write z 2 + z + 1 = (z - a ) (z - 8 ) where a = -1 +2 t/3i , 8 = - 1 - 2/3i Thus it appears that we can factorise expressions of the form zn - 1, zn + 1 and (by extension) zn-l + zn- 2 + ••• + 1, into either: (a) real quadratic factors or (b) complex linear factors.
6
Factorise z - 1 into real quadratic and real linear factors, hence 4 2 factorise z + z + 1. 6 We solve the corresponding equation z - 1 = 0
EXAMPLE: (36)
SOLUTION: z 6 = 1 = cos(2kn) + isin(2kn), z z
hence the six sixth roots are given by
= cos(2kn/6) + isin(2kn/6), k = 0, 1, 2, 3, 4, 5.
z 1
2
z
i.e.
= cos 0 + i sin 0 = 1
z
=cos n/3 + isinn /3
z5
=cos4n/3
= cos2n/3 + isin2n/3 ,
z
= cos5n/3 + isin5n/3 = cosn/3- isinn/3
3 We find that z
4 6
= -1 +
isin4nj3
=cos2n/3-
isin21T/3
z 5 = z 3 , hence:
=z and 2 6 z2 + z6 = z2 + z2 = 2cosn/3
and
z3 + z5 = z3 + z3 = 2cos2n/3
Also z z = z z = 1 , z z =z z = 1 2 2 3 5 3 3 2 6 6 Now z - 1 = (z- z ) (z- z ) (z- z ) (z- z ) (z- z ) (z- z ) 4 6 1 2 5 3 = (z- 1) (z + 1) [(z- z ) (z- z )] [(z- z ) (z- z )] 6 3 5 2
= (z
-l)(z + l)[z 2 - (z 2 + z6 )z + z 2z6 ][z 2 - (z 3 + z5)z + z z ] 3 5
2 (z - 1) (z + 1) 2cos(n /3) z + 1] [z - 2cos(2n/3) z + 1] 2 4 2 (z - 1) (z + z + 1), we at once have:
[i -
Since z 6 z
4
+ z
2
6
+ 1
~ =
[z
2
- 2cos(n/3) z + 1] [z
2
- 2cos(2n/3)z + 1]
z - 1 EXAMPLE: (37) Solve z
6
+ 1 = 0. Express the roots in the form a + ib. Show these 6 roots in an Argand diagram. Factorise z + 1 into real quadratic factors. 6 SOLUTION: z = -1 = cos(n + 2kn) + isin(n + 2kn), hence the six sixth roots are given
.
1T + 2k1T
by z = c1s - -, where k = 0, 1, 2, 3, 4, 5. 6 z = cis7n/4 =- 13/2- i/2 z = cisTT/6 = 13/2 + i/2 4 1 z = cisn/2 = 0 + i z = cis3n/2 = 0- i 5 2 z = cis 5n/6 = - /3/2 + i/2 z = cis llTT/2 = .fJ/2 - i/2 6 3 We have z = z l , hence Z l + z 6 = Z l + Zl = V3 , Z l z 6 = 6 and z5 = z2' z2 + z5 = z2 + z2 = 0 z2z5 = 1 and z4 = z3 '
z2
x
z3 + z4 = z3 + z3 = - VJ' z3z4 = 1 Fig.l6
144
6 The factors of z + 1 are: z 6 + 1 = [(z- z ) (z- z )) [(z- z ) (z- z )] [(z- z ) (z- z )] 2 5 3 4 6 1 2 2 2 = [z - (z + z )z + z z ] [z - (z + z )z + z z ] [z - (z + z )z + z z ] 2 5 2 5 6 1 6 4 3 4 1 3 2 2 = (z - .f3 z + 1) (z + 1) 2 y SOLUTION: (a) (a) Re(z) = 3, z = x + iy Re(x + iy) = 3 X = 3 2 The locus is the vertical line x = 3 Im (z) > 2 , z = x + iy y >2 The locus is a half-plane above the line y =2
(b)
EXAMPLE: (49)
Sketch the curve
3
X
WJJ!J/111/IIJ/1_ ____. 0
Fig.33 X
y
lzl
2
SOLUTION:
Izl is the distance of a point from the origin, so the locus of I z I = 2 is a circle of radius 2, centre 0(0,0).
X
The cartesian equation of this circle is X
2
+
y
2
= 4 Fig.34
EXAMPLE: (50)
Iz
Describe the locus
- 2 + 3i I
3
X
SOLUTION:
Write:
lz- (2- 3i)l = 3
... (I)
We know that the equation of the form Iz - w I = r, represents a circle. Hence (I) represents a circle of radius 3 and centre (2,-3) whose cartesian equation is (x - 2)
2
+ (y + 3)
2
= 9
Fig.35
EXAMPLE: (51) X
Sketch the region defined by Iz - 2 + 3 i I ~ 3. SOLUTION: From example (50) this region is the set of points within and on the boundary of the circle, radius 3, centre (2,-3)
Fig.36
156
EXAMPLE: (52)
P(z)
Describe the loci of z if
Iz Iz
- 2 - 2
I Iz I < lz
+ 1- i
+ 1- i
I I I
I I
SOLUTION: (a)
P(z) where z
=x
B
I
+ iy
1A
We have:
Iz
- 21
= Iz -
X
Let A be (2, 0) and
Fig.37
B be(-1,1)
The relation (1) says that PA = PB for all positions of P and from plane geometry we know that the locus of P is then the perpendicular bisector of AB.
(b)
lz-21
<
y
lz-(-l+i)l
represents a set of points closer to A(2,0) than B(-1,1). Hence the locus of P(z) is the region on the right hand side of the perpendicular bisector of AB, excluding the bisector itself.
A
X
Fig.38
EXAMPLE: (53) (may be deferred until after Chapter 6)
y
Describe the locus lz-21 + lz+21 = 6
SOLUTION: Here A(2,0), B(-2,0) we have PA + PB = 6 (given). This is the condition for the locus of z to be an ellipse. The locus of P is an ellipse. The foci are (.:!: 2, 0). The centre is o.
Fig.39
Length of semi-major axis is 3 Length of semi-minor axis is .f5
2 Equation of the ellipse is x 9
X
+
2 ~
157 EXAMPLE: (54) Sketch the region defined by: (a) 0 ~ argz ~}
(b) lzl ~ 2 and 0 ~ argz ~%
SOLUTION:
(a)
argz =
Locus of
y
j
= 60°
P(z) satisfying Fig. 40
~
0 .{. drgz
n/3 is the set of points within the angular region [POX = 60° (shaded) including the boundaries, excluding the origin. z =0
argz is not defined for (b)
Izl
~ 2
and
0 ~ argz ~ n/4
The region shaded is within the circle of radius 2, centre 0, restricted in the sector 0 ~ arg z ~ 45°, excluding the origin 0.
Fig. 41
EXAMPLE: (55) Find the locus of w if w - z z- 1
Izl
given
2
SOLUTION:
We eliminate z from w Solve for
z,
lz I lw -
=z
then zw
II -
z --I =z - I
4
II =
=- I
su z(w - I)
II - wI = Iw - I I
but
wl '
Iz I =
. by usmg
Hence the locus of w is a circle of radius 2 2 I of the circle is (x - I) + Y = 4
2 and z =
and
I r-:-w
lzl
=
2 ,
then
4
and centre (1,0). The cartesian equation
EXAMPLE: (56)
z- 2 Find the locus of z if w = - - , given that w is purely imaginary.
z
SOLUTION:
z- 2 z2z
w w
--z2 1
zz
Z
X
2 X
+y
2
X
+ iy
_ 2 (x - iy)
2
2x
w
=
+ y
2
2iy 2 2
+ X
+y
Now if w is purely imaginary, then Re(w) = 0 I
-
2 2
X
2
2x + y
2
=
x + y - 2x = 0 (x - 1)2 + y2 = I
0
or lz-11=1 Hence locus of z is a circle of radius I, centre (1,0).
158
EXAMPLE: (57)
Describe the locus given by the equation
Iz
+ 21
2lz-2+il
SOLUTION:
Let z = x + i y, then the given equation becomes: lx+iy+21
I.,. 1
w,
--
(c)
I > Iz
3
n/6 ~ arg z ~ 2n/3
(c)
(a)
~
(f)
z - 2 + i z + 2 - i
'
lzl
where z is
160
Exercise 40 (cosa- cos 9 >
into linear
~nd quadratic factors with real coefficients.
5 2 n 2 3n Prove that z + l = (z + I) (z - 2zcos + I) (z - 2zcos 5 5 + l) . 15 3 5 By observmg that z + 1 = (z ) + l, prove that 15 2 6 3 n 6 3 3n z + 1 = (z + 1) (z - z + I) (z - 2z cos + 1) (z - 2z cos 5 + l)
5
=- ~ ,
3 w
28.
Show that if lw I
29.
1 + sinS + icosa . . Prove that 1 +Sin . a -lOS c . 9 = sma + 1COS9 and deduce that
=1
and Re(w)
then
. n . n]5 + 1·[ 1 + Sin n]5 . 5n - !COS . 5 [ 1 + Sin 5 + !COS 5 30.
Express integers.
. 4k + 1 [ 1 + 1tan ~n
J
=1
=
0
m m . t he f orm a+ 1. b , where m and k
are
164
CHAPTER 5 POLYNOMIALS 5.1
Introduction
In 3U Mathematics, we learned many important properties of polynomials in the real variable x. In this chapter we shall study these properties and a few more theorems about polynomials over the complex field C. Let us first revise some of the important work on polynomials. The general nth degree polynomial function cf a real variable x is of the form: P ( x) " a x
n
n
+
a
x
n-1
n- 1
+ a
x
n-2
n- 2
+ •• • +
a
1
x + a
0
,
a
n
.J 0
where the coefficients a , a , ••• are real, and n is a non-negative integer. 0 1 OPERATIONS ON POLYNOMIALS
When two or more polynomials are added, subtracted or multiplied, the result is another polynomial. The division of one polynomial P(x) by another polynomial A(x) may or may not be exact. When P(x) is div1ded by A(x), we can write the identity: P(x) where, P(x) Q(x) A(x) R (x)
" Q(x) . A(x) + R(x) is called the dividend, is called the quotient, is called the divisor, is called the remainder. ..
[deg R (x)
< deg
R (x) " 0, P(x) is exactly divisible by A (x), then Q(x) and factors of P(x).
If
A (x)] A (x) are called the
THE REMAINDER AND FACTOR THEOREMS
In the division of P(x) by A(x), if A(x) is a linear polynomial degree of R (x) must be zero, i.e. R (x) is a constant. We write:
x - a,
tht·n the
P(x) " (x - a) Q(x) + R Substitute x " a,
then
P(a) " 0 . Q(a) + R R " P(a) . So, the remainder theorem is: If P(x) is divided by (x - a), the remainder is P(a). Further, it P(a) " 0, then x - a is a factor of P(x), and conversely, if x - a 1s a factor of P(x) then P(a) " 0. The result is known as the factor theorem. A value of x, such that P(x) " 0, IS called a ZERO of the polynomial P(x) or a root of equation P(x) = 0. For example, x = 2 is a zero of 3 P(x) = x - 8.
165
EXAMPLE (1)
Find the remainder when P(x) = 2x 3 - 3x A(x) = X - 2
SOLUTION:
2x X-
2
+ x - 2
) 2x 3 - 3x 2 - ~x - 6 3 2 2x - 4x 2 x - 4x
2
x
2
- 2x
2
-
~x
- 6 is divided by
If we only want the remainder, we use:
R = P(2) = 16- 12- 8- 6 = -10 The remainder is -10. We can write 2 P(x) = (x - 2) (2x + x - 2) -10
- 2x - 6 - 2x + 4
- 10 EXAMPLE (2)
Show that x + 3 is a factor of x other factors.
SOLUTION:
We have P(x) = x P(-3)
= -27
+
9
+
3
3
- 3x + 18 and hence find the
- 3x + 18, x- a = x + 3, so a= -3 and
18
=0
•
Since P(-3) = 0, x + 3 is a factor of P(x) we divide P(x) by x + 3 to find other factors. 2 2 Observe that the missing term x x - 3x + 6 X +
3
) 3 X
x
3
2
+ 0. X - 3x + 18 +
3/
- 3x - 3x
2 2
is arranged as 0 . / . This is important and rt•duces the chance
- 3x
of t'rrors.
- 9x 6x + 18 6x + 18 0
over the R-field)
P(x) is /
The other factor of - 3x + 6. (Irreducible
EXAMPLE (3)
Find the values of the constant m if the polynomial 3 2 2 P(x) = 4x - m x - 4mx + 64 is divisibk by x + 1, hence find the other fdctors of P(x)
SOLUTION:
P(x) = 4x
3
2 2 - m x - 4mx + 64 is divisible by x + 1, i.e. x - (-1} 2 P(-1} =- 0 gives - 4 - m + 4m + 64 = 0 2 :. m - 4m - 60 = 0, i.e. (m- 10) (m + 6) = 0
••• m = 10 or m
= -6
2 3 2 3 For m =- 10, P(x) = 4x - 100x - 40x + 6~ = 4 (x - 25x - lOx + 16) 3 2 Divide x - 25x - lOx + 16 by x + 1, then 2 P(x) = 4(x +I) (x - 26x + 16), ( / - 26x + 16 is irreducible over R-field) 2 3 3 For m = -6, P(x) = 4x - 36x + 24x + 6~ =- 4(x - 9x + 6x + 16) 3 2 Divide x - 9x + 6x + 16 by x + 1, then P(x) = 4(x +I) (x 2 - lOx+ 16) = 4(x + l)(x- 2) (x- 8)
166
EXAMPLE: (4)
Given that x roots.
=2
is a root of x
3
- 4x
2
+ 14x - 20
= C,
find the other
SOLUTION:
2 3 Divide x -4x +14x-20 by x-2 2 2 (x 3 - 4x + 14x - 20) = (x - 2) (x - 2x + 10) 2 The equation is (x - 2) (x - 2x + 10) = 0
l=3b -
2 + 2 . . x = 2 (given root ) or x - 2x + 10 = 0 , I.e. x -~ - 1 _+ 3.1 Observe that the complex roots 1 + 3i and 1 - 3i are conjugate. We shall later prove that the complex roots of a real polynomial occur in conjugate pairs.
EXAMPLE: (5)
Solve: x 3 - 3x 2 + 4x - 2 = 0.
SOLUTION:
The constant term 2 = 2 x 1. This suggests we try 2 3 Let P(x) = x - 3x + 4x - 2 P(2) = 8 - 12 + 8 - 2
:1
x =
~
2,
~
1 for the roots.
0
P(l) = 1 - 3 + 4 - 2 = 0
x = 1 is a root of P(x) = 0 Divide P(x) by x - 1 :. P(x) = (x- 1) (x 2 - 2x + 2) The roots of
P(x) = 0 are
1,
2~
r-4
2
i.e.
1, 1
~
i
Again observe that the complex roots occur in a pair of conjugates.
167
Exercise SA Perform the following divisions, and check the remainder by using the remainder theorem.
I.
3.
3
- 2i + 3x - 1) 7- (x - 2) 4 3 2 2 (x + 2x + 2x - 2x - 3) 7- (x - 1) (x
In each of the following, decide whether
5. 6. 7. 8.
P(x)=x
4
3
2.
4
2
- 2x + 3x - 2) 7- (x + 2) 3 2 (2x - x + x - 1) + (x - 1)
(x
4.
A(x) is a factor of
2
-2x +x +x-6 3 P(x) = 2x - 3x + 1 2 3 P(x) = x + 2x - x + 6 2 3 P(x) = 5x + 7x + 3x - 1
P(x).
A(x)=x-2 A(x) = x + 1 A(x) = x + 3 A(x) =- x + 3
Find the remainder of the following without actually dividing, i.e. use the remainder theorem. 9.
P(x) = 4x
10. P(x) = 5x
3 4 4
- 3x - 2x
2 3
+ x + 8
A(x) = x + 3
2
11. P(x)=2x -4x +5 2 3 - ax + bx + 2
Find values of k such that
A(x) = 2x + 1
A(x) is a factor of
= x4 -
3 3k x + 3x - 1 2 3 14. P(x) = x .:. kx + 4x - 4 2 3 15. P(x) = x - 3x - 6kx + 8k 4 3 16. P(x) = x + kx + 7x + 21
Use the factor theorem to find the value of and hence find the zeros of P(x). 17. P(x) = 3x
3 3
- 12x
+ 2
A(x)
12. P(x) = x
13. P (x)
=x
- x + 7
2
- llx- k
18. P(x) = 2x - 6 i + kx + 4 2 3 19. P(x) = k x + x - 8x + 6 4 3 2 20. P(x) = x + x + k x + 4x - 24
A(x)
=x
- 2
A(x)
=x =X =x
- 1
P(x).
A(x) A(x)
-
2
- 2
A(x) = x + 3 k
that makes A(x)
A(x) A(x) A(x) A(x)
=X =X =X =X
a
-
5
-
2
-
1
+ 3
factor of
P(x)
168
5.2
Zeros of a Polynomial/Multiple Roots
Rational or Integral Zeros We use the formula x
=-
2 b !. /b - lfac 2a
zeros of a second degree polynomial P(x)
(or factorise by inspection) to find the
= ax 2 + bx
+ c, but it is generally very
difficult to find the zeros of higher degree polynomials such as 3xlf + 2x
3
- x
2
+ 7.
This is because some, or even aJJ the zeros of a polynomial may not be integers, 2 e.g. x + 2 = 0 has no real zeros, let alone integer ones. The following theorem is very useful in finding rational zeros (if any), though we shall mainly be concerned with zeros which are integers. Theorem: Let P (x )
= anx n
coefficients. If x
+ an_ x n-1 + ••• + a x + a
=%,
must be a factor of a
Proof:
Since x n
:. a ( !:.) n s
a
be a polynomial with integer
0
and s must be a factor of a
is a zero of P(x), then P n-1
+
0
where % is in its lowest terms, is a rational zero of P(x),
then r
=%
1
1
r + ••• + a 1 ( s) + ao
( !:.) n- 1 s
0::
n
•
= 0.
0 •
. Multiplying both sides by s n and rearrangmg:
anr
n
an-{
+
Or, r(a r n
n-1
n-1 + a
•s
+ •••
r
n- 1
n-2
+ air. s
n-1
. s + •• • + a s 1
n-1
- aos )
n ••• (1)
The relation ( 1) shows that r is a factor of a sn • But r cannot be factor of 0 s
n
as r and s have no common factor, hence r must be a factor of a • In 0
the same manner, we can prove that s is a factor of an. Further, if an is monic and the zeros of
P(x) are integers as s must be a factor of I.
1, P(x)
169
Find all the zeros of P(x) =
EXAMPLE:
X
3
+
2x
2
- 3x - 6
SOLUTION: Since P(x) is monic, the integral zeros must be the factors of the constant term -6. All the possible factors are .!: I, .:!::2, .!: 3, .!: 6. We find that only x = -2 satisfies P(x) = 0, so -2 is a zero of P(x). P(x) is of degree 3 and therefore has 3 zeros. The other two zeros must be irrational or complex. To find these, we divide P(x) by X + 2. 2 :. P(x) = (x + 2) (x - 3) Hence the zeros of P(x) are: -2, .!: Vf. It is possible that EXAMPLE:
P(x) has no integral zeros, as the following example shows.
Show that
P(x) = x
4
- x
2
- 2x + 6 has no rational zeros.
SOLUTION: All the possible factors of 6 are: .!: 1, .:!:: 2, .!: 3, .!: 6. None of these numbers satisfy P(x) = 0, hence P(x) has no integral zeros. Since a fourth degree polynomial must have 4 zeros (see section 5.3, theorem 3), we conclude that the zeros of P(x) are either irrational or complex. EXAMPLE:
Find all the roots of
5x
3
+ 28x
2
+ I Ox - 3
=0
SOLUTION: 2 3 Let P(x) = 5x + 28x + lOx - 3 = 0. To make our work easier, we transform this equation so that the leading term has a coefficient equal to I. 3 Observing that the coefficient of x is 5, we multiply the equation by 25 and then put y = 5x. (5x) Or y
3
3 +
+ 28 (5x)
28y
2
2
+ 50
(5x) - 7 5
=0
+ 50y - 75 = 0
•·· (1)
By testing the possible factors of 7 5 i.e . .:!:: I, .!: 3, .!: 5, .!: 15, .!: 25, .!: 7 5, we find that y = -3 is a root of (1), hence P(y) = (y + 3)
1,
PS = PM, PS PS
< PM, > PM,
the conic is a parabola the i::'onic is an ellipse d
the conic is a hyperbola.
M
Fig. 3
6.2
Ellipse (e < 1): (Focus and Directrix Definition)
Let P (x, y) be any point on ellipse.
y
S is the focus, ZM is the directrix.
P(x, y)
Draw SZ 1 the directrix , and mark the points A and A' on this line so that SA AZ
SA' A'Z
e,
e A'
0
Fig. 4 By definition, A and A' are on the ellipse. 0 is the mid-point of AA'. rectangular axes at 0, as shown. S (c, 0), then Let AA' = 2a; SA' eA'Z SA e. AZ and SA' and
SA'
+
SA
A'A = 2a
SA
e (A'Z - AZ)
(l)
2ae
(2)
Adding and subtracting: SA = a (I - e) Then
c = OS = OA
SA'
a (1 + e)
SA= a - a(l-e)
ae
Take the
193
and
oz = s z_
Now
OA + AZ
is (c, 0)
2
(ae, 0)
(x - ae) (x - ae) (I -
e 2~
2
e
OD
a --x e
2
(PM )
+ /
2
+ /
2
2
Let
-
(by definition)
2
x
oz
PM
e
=
a e
a + .!. (a - ae) e
is (~ 0) e'
PS PM SP
=
= e
2
(~e
- x) 2
2 = (a - ex) 2 2 + / a (1 - e )
2
a (1-e)
••• (3)
+
Equation (3) is the equation of the ellipse in standard form. The point S is called the focus of the ellipse and the line x = ~ is the directrix. By e symmetry of the equation of ellipse, we infer that it has two foci S(ae, O) and S' (- ae, O) and two directrices (See Fig. 5)
d'
y
d
y
B
d
B
X
Z'
B' B'
d'
Fig. 5
Fig. 6
194
Important Features of the Ellipse:
2 2 The eJlipse x + ~ = 1 has the following properties. (Fig. 5) 2 a b 1.
The curve is symmetric about both axes.
2.
The major axis AA'
2a
3.
The minor axis BB'
2b
2
2
Note that since b
=
a
2
A'(-
A(a, 0)
B'(O,- b)
B(O, b)
(l - e ) , e
< 1,
then b
4.
The centre of the ellipse is at 0 (0, 0)
5.
Foci: S (ae, 0) ,
6.
Equation of two directrices are x
a, 0)
< a.
S' (-ae, 0)
a ± e
lf the foci are at S (0, be) and S (0, -be), the standard form of the equation of the 2 2 2 2 2 ellipse is x + ~ 1, with a = b (l - e ) , a < b 2 a b
and its properties are: (see Fig. 6) l.
The major axis is along the y-axis,
BB' = 2b
2.
The minor axis is along the x-axis,
AA' = 2a
3.
The foci are given by (0, ±. be) with centre 0 (0, 0)
4.
The directrices are y
=
.±.
eb
WORKED EXAMPLES Find the eccentricity, foci, directrices of the following ellipses:
2
(l)
2
~5
+
Solution: a b
~ = 1
(see Fig. 5)
= 5, 2
b 2
=3 2
=a 0 - e )
2
9 = 25 (I - e ) e =
4
3
195 The foci are (! ae, 0) = (± 4, 0) a The directrices are X = ± = .± 425
e
2 (2)
2
x9
Solution:
h
+
=
1
(see Fig. 6)
For this ellipse, the major axis is along the y-axis.
a = 3,
b = 5
a2 = b2 (l - e2) 2 9 = 25 (l - e ) gives e
4
5
The foci are (0, .± be) = (0, .± 4) The directrices are:
y
= ± eb = .± 425 Exercise 6A
Find:
(a) the centre (b) the eccentricity (c) the foci (d) the lengths of axes (e) the equations of directrices for the following ellipses: 2
2
1.
X
3.
X
5.
X
7.
4/ + y
9.
16
+
L
+
L 25
+
L
9 =
2
36
9/
X
9
4.
T6
+
X
2
10 2
4/
=
4
-
36 = 0
2
2.
2
2
25
1
X
8.
4x X
+
L
+
L
16
2
2
2
48
+
+
2
36
2
T2
10.
L
2
6.
2
+
=
9/ = 2/
-
I
1 = 0
Find the equations of the ellipses described as follows.
=4
11.
Foci on the x-axis, centre (0, 0) , a
12.
Foci on the y-axis, centre (0, 0) , a = 2 , b = 5
13.
Foci ( ± 4, 0) , b = 3
14.
Foci (0, .±
15.
Centre (0, 0), the length of major axis = 10 and length of the minor axis = 6
16.
Centre (0, 0), minor axis = 12, focus (0, 4).
17.
Eccentricity e
/5),
, b = 2
a = 3
=~,major
axis= 12
196
6.3
The Hyperbola
Starting with the definition
~~
= e , e
> I for the hyperbola and using the
following diagram, you can easily derive the standard equation of the hyperbola as:
X
2
2 a
where b
'1
2
- Lb2 2
= a
2
(e
2
p
- I) A'
0
X
Fig. 7
2
2
The hyperbola x L a2 - b2
Y.
has the following properties (see Fig. 8)
X
Fig. 8 I.
It is symmetric about both axes.
2.
The transverse axis is the line segment joining the vertices A (a, 0), A'(- a, 0) and AA' - 2a.
3.
The foci are S (ae, 0), A' (-ae, 0),' centre 0 (0, 0)
4.
The conjugate axis is the segment BB', where B (0, b) and B' (0, -b)
5.
The equations of two directrices are x
6.
The asymptotes are:
+ ~
-
e
The rectangle through the vertices A, A', B and B', is very useful in drawing the graph b of the hyperbola. Also draw the asymptotes y = ± X through the corners of this
a
rectangle. The hyperbola fits nicely between the two asymptotes.
197 2 x2 The hyperbola given by L - a2 b2 details.
I is shown in Fig. 9 with all the important
I.
It is symmetric with respect to both axes.
2.
The foci are on the y-axis, S (0, ae), S' (0, - ae).
3.
The transverse axis AA' = 2a, A (0, a), A' (0, - a).
4.
The conjugate axis BB' = 2b, B (b, 0), B' (-b, 0).
5.
The directrices are
6.
The asymptotes are: y = ;!;.
y = .± ~ e
6x
Fig. 9 WORKED EXAMPLES I.
Find the foci, directrices, asymptotes and vertices of the following hyperbolas and hence sketch them. y 2 2 ~-L(a) (See Fig. 10) 16 25 -
(i)
a = 4, b = 5 c2 = a2e2 = a2 +b2
c =
.J4J
or c = -
ll4i. X
The foci are ( ;t v'iii, 0) (ii)
The directrices are
But
X
a -e
= +-
c = ae .Jii1=4.eore=
V4l 4
. . 16 t h e d trectnces are x = ± .~ v41
Fig. 10
198
(iii)
The vertices are A (4, 0), A' (-4, 0),
(iv)
The asymptotes are y = ± (b)
45 x
(See Fig.
a = 3,
b =4
c2 = a2 + b2 = a2e2 5
:. c = 5,
e =3
If we put y = 0, we have x
2
X
-16 and this shows that the foci lie on the y-axis • . ·.
the foci are (0, ± 5)
and vertices (!: 0, 3) The directrices are
y = .:!:.
9
3
The asymptotes are + 3 y = - 7f
X
Fig. II
Exercise 68 1.
For the following hyperbolas, find the foci, directrices and vertices and sketch them: (a)
(b)
X
2
25
; T6
2
2
-
L
16
=
1
(c)
X
-
x2 25
=
I
(d)
4/
4/ = - y2
I
4 2
2. 3.
Find the angle between the asymptotes of the hyperbola x a2 2 2 Draw an accurate sketch of x - y = a 2 Find the angle between the asymptotes. 2 [The curve given by x - /
= a
2
is called a RECTANGULAR HYPERBOLA]
199
6.4
Shape of the Conics
-
We can study the variation of the shapes of conics in general by varying the eccentricity e or changing
e
the ratio b/a.
c
=0 =0
'
Eccentricity
< e < I,
For 0
where c
c =
the conic is an ellipse and
is the focal distance given by
~.
a>b>O.
If we keep a fixed and vary c
0~ c
~
over the interval
e
a, the resulting ellipses vary in shape,
For e = 0, i.e. b = a, the shape is circular.
and when e is nearly I, the ellipse reduces towards a line-segment.
I
S,S'
-
"o·•E ~ 0
as shown in the diagram.
As e increases, the shape becomes flatter
\_
I
e"O·\C e
= 0.97
c
y
0
::>
0
e~l
S'
The orbit of a planet around the sun is elliptical in shape, but the orbits of most of the planets
Planet
s
0
e
Neptune
0.01
Earth
0.02
Jupiter
0.05
Mercury
0.21
are nearly circular, as can be seen from the table of eccentricities.
The varying shape of an ellipse can also be examined by changing the ratio b : a, as shown below. (cont)
200
b
5
a
3
E---+---3> the variation of the shape of conics as e
vanes from zero to a
ver e
>1
hyperbola~
e circles
e
1.1
O 0)
2
~ to ~h; ellipse l 2
lies on the ellipse ~ 5 + ~
= 1.
+ 6/
= 15.
Find the equations of
the tangent and the normal at P. Also find the coordinates of the point in which this tangent intersects the directrix corresponding to the focus S. 13.
Find the equations of the tangent and the normal at P(9,-3) on the hyperbola
~
2
2
4 - fg
= 1.
The normal meets the curve again in Q. Find the coordinates of the
point of intersection of the tangents at P and Q. 14.
2
=2x
-y2
- 4 intersects the curve x =1 at P and Q. Find the 3 coordinates of the point of intersection R of the tangents at P and Q. What The line y
is the angle between the normals at P and Q? 15.
Prove that the line x + y
=5
is a tangent to the el!ipse 9x
2 + 16y 2
144.
Find the coordinates of the point of contact. 16.
Show that the line 3x + 4y = 10 is a normal to the hyperbola 2x 2 - 3/ = 5 and find the point at which the line is normal.
17.
Find the equations of the tangents to the hyperbola 2x parallel to the line x
+
y
= 0.
2 - 3y 2
Also find the points of contact.
=6
which are
209
18.
2
(a)
X
2
+
L
2 hyperbola X 2a
-
ellipse
(b)
19.
=0
Show that the line lx + my + n
2 a
if a212 + b2m2
I '
b2
L
touches the
2 I'
b2
= n2
if a212 - b2m2
= n2
•
x2 2 The tangent at P (xI' y 1) on the hyperbola a - ~ = 1, x .> 0, intersects 2 1 the directrix at
Q.
S
is the focus
(ae,O).
Prove that
PSQ
is a right
angle. 20.
Find the equations of the four tangents common to the hyperbola 2 x - 2/ = 4 and the circle + / = 1. Find the points of contact of these
l
tangents with the circle. [Hint: Let xx
6.9 1.
1
+ yy
1
=1
be a tangent to x
2 + y2
1 at P(xl'yl)]
Miscellaneous Problems on Conics x2 2 Find the equation of the tangent at P(x , y ) to the ellipse a + ~ 1 1 2
Find the point Q(X, 0) where this tangent meets the x-axis, and prove that Xx
= 1
1.
= a 2•
· Find the point R (0, Y) where the tangent meets the y-axis. Show that the locus of a a2 b2 point T (X, Y) is given by the equation 2X + 2y = 1. y
Solution: The equation of the tangent at P (x , y ) is 1
YYt
XXI
-2- + -2- = I. a b
1
(Prove it) X
Substituting for Q (X, O) ,
Xx
1
= a
2
or
.... (l)
210
••• (2)
Similarly
To find the locus of T (X, Y), we eliminate x and y by using: 1 1 2 Xl -2 a
2
yl +
17
If
blf
a + x2a2
Y2b2
Hence the locus of R is given by the equation 1 •
2.
Find the equations of the tangent and normal to the hyperbola 3x 2 - y 2
at P (If, 3{5).
The tangent meets the x-axis in M and the normal meets the y-axis in
N, find the length MN.
Solution: There is no need to remember the eq·Jations of the tangent and normal. 3x
2
6x
- y
2
=
~
2y
-
At P(lf, 3/5),
~ dx
Differentiating
3 or
0
~
_ 3x dx - y
If
v'5
:. Equation of the tangent at P (If, 3/5) is or
r5y
lfx -
••• (I)
=
For M, put y = 0 • :.
M is
M
2 = 4c2xy.
43.
Find the locus of the mid-points of chords of the curve
xy
c
2
drawn
parallel to the line lx + my = 0. 44.
A tangent at P(ct,
~)
to the hyperbola
xy = c
2
intersects the axes in
A
and B and 0
is the origin. Prove that the area of triangle OAB is 2 independent of the position of P on the curve xy = c •
4.5.
The tangents to xy
=c 2
at A(cp,
c
p)
If the chord AB touches the curve xy 2
and B(cq, ~) intersect at R. q
2 4c , show that the locus of
R is
given by 4xy = c 46.
47.
A variable chord PQ of the hyperbola xy = c 2
where P is (x , y ) and Q 1 1 is (x , y ), is such that Ix - x I = 2c. Prove that the locus of the mid-point 1 2 2 2 2 2 of PQ is given by the equation x y = c (x + y). 2 P(cp, ~) , Q(cq, ~) are two points on the conic xy = c • Show that the
p
q
gradient of PQ is
~~
. If PQ subtends a right angle at a third point
R (cr, ~) on the conic, prove that the tangent at R is perpendicular to 48.
PQ.
P(cp, £) and Q(cq, £) are variable points on the conic xy = c 2. Prove that p
q
the tangents at P and Q intersect at T ( ~ , ~ ). Hence prove that if p+q p+q pq
=k ,
origin.
a constant, then the locus of T is a straight line passing through the
221
49.
PQ
is
is ( 4p,
~)
a
variable
chord
and Q is 4q,
*).
of
the
xy
hyperbola
= 16,
where P
Prove the following: (a) (b)
The equation of the chord PQ is x + pqy = 4 (p + q) 2 The equation of the tangent at P is x + p y = 8p
(c)
8 - -) The point of intersection of the tangents at P and Q is T(_!Es_ p+q'p+q
(d)
If the chord PQ passes through the point R (0, 8), show that the locus of
T is a straight line, x 50.
=4. 2
P(x 1' y ) is a point on the rectangular hyperbola x - y 1
the foci. Prove the following: (a)
The eccentricity e
(b)
IPS I= a- .fix 1
(c)
PS • PS'
=OP 2,
,
= Vi
ls'PI
where 0
=a+
.fix 1
is the origin.
2
2
= a • S and S' are
222
6.10 Tangents and the Chord of Contact Many properties of the two central conics, the ellipse and hyperbola, are so common that it is convenient and instructive to treat them as one by writing the equation in the form
2 ~
2
+
1r-
=a 2
where A
= 1
••• (1)
and B
=b2
The tangent at P (x 1' y 1)
. IS
for the ellipse but B xx1
A
=- b2
for the hyperbola.
yy 1
+
---a-
••• (2)
= 1
To find the condition for a line to touch the conic, the line:
=0
lx +my + n
must have the form of the tangent (1). Comparing coefficients: X/A
-,- =
y1/B
---m-
-AI
= -n-
xl
=
- 1
--n
-Bm -n
'
But P (x , y ) lies on the conic (1 ), so that: 1 1 2 2 AI Bm - 2 + -2n n A1
2
Bm
+
1 , which simplifies to:
2
n
2
This is the condition for the line lx + my + n
2
••• (2)
=0
to be a tangent to the conic
2
+~=1.
XA
If the equation of the tangent is expressed in the form
y
= mx
+ b, the condition
becomes: Am
2
+
B = b
2
••• (3)
This is left as an exercise to the student.
Chord of Contact: The chord
PQ,
jommg the points of the contact of tangents drawn to the conic from
an external point T(x , y ), is called the chord of contact. 1 1
223 y
Let P(x ,y ) and Q(x ,y ) be the points of 3 3 2 2 contact of tangents drawn to the conic from T(x l' y ). The tangent at (x , y ) to the conic 1 2 2 x2
A
2
1f-
+
xx 2
yy 2 + B"" '"'
"A"
= 1 is
l and
T (x l' y ) lies on it, so 1 xlx2 Y1Y2 -A-+ - B -
=
••• (1)
l
Similarly, for the tangent at Q(x ,y ), 3 3 xlx3 Y1Y3 -A-+-B-=l (1) and (2) clearly indicate that the points XX
I
T
••• (2)
P and Q lie on the line
yy 1
+
r=
••• (3)
1
which is therefore the equation of the chord of contact of T (x , y ). 1 1 Though the equation of the C.O.C. is the same form as that of a tangent, it must be remembered that T(x l' y ) does not lie on the conic. 1
The method shown above can be used to find the equation of a chord of contact to 2 2 . 1.e. . . Ie x 2 + y 2 = a 2, a h yper bl any come, a para bl o a x = 4 ay, a Clrc o a xy = c etc.
It is left as an exercise for the student to prove the following:
(Chord of
contact = COC) 1.
The equation of the COC, from xx
2.
1
2
2
+y
2
=a
2
is
. P(x I' y ), to a parabola x 1
2
= 4ay is
= 2a(y + y ) 1
2 2 2 The equation of the COC, from P(xl'y ), to a rectangular hyperbola x -y = a is XX l
4.
1
=a
The equation of the COC, from xx
3.
1
+ yy
P(x I' y ), to a circle x 1
- yy} = a
2
1
•
The equation of the COC, from P(x I' y I), to a (special) rectangular hyperbola xy = c
2 IS . xx I + yy = 2c 2• 1
224
6.11 Geometric Properties of the Ellipse The standard equation of the ellipse 2 2 is ; - + a b
7
=
I; Centre 0.
Foci: S(ae,O), S'(-ae, 0)
X
Directrices: x = .! ~ e Length of semi-major axis Length of semi-minor axis Vertices: (.!a, 0), (0,.! b)
=a =b
B'
-a
X=-
X
e
The following properties of the ellipse are proved here: I.
The sum of the focal lengths is a constant, i.e. SP
+
PS We have PM
(Definition)
=e
and
PS' PM'
=e
S' P = 2a
.". PS + PS' = e (PM + PM') = e • MM' But MM' = The distance between two directrices
= 2ea
2 a = 2a , which is a constant e This fact gives us a fast, accurate (and inexpensive)
:. PS + PS' = e .
method of drawing
the ellipse. A thread of length 2a, fastened at S and S' is kept tightly stretched by a pencil at P. As the pencil moves, it traces out an ellipse. (See the diagram above)
2.
(a) The normal at P bisects the angle between SP and S'P. (b) The tangent at P is equally inclined to SP and S'P. These properties are equivalent and we only have to prove one of them.
A
purely algebraic method is too laborious, so we use the following geometric result. (The proof is given in the appendix) If
AM is the internal (external) bisector of LBAC of l1ABC, then
AB BM AC = MC
BM'
( = M'C
) D
........... '
..... ...
..... .......
... M'
a e
225
The normal at P(xl'yl) is
y
2 2 ~ - l2._ = a 2 - b 2 = a 2 e 2 , and meets X1
M
y1
the x-axis, where y
= 0,
2 in G(e x 2
Then, SG = OS - OG = ae - e x SP Now PM
=
e
1
,o)
T
1
1
(definition)
SP = e • PM = e(NM - NP) = e(~- x ) 1
=a
SP
X
= e (a -ex )
- ex
a
X=-
e
(Note: this result is very 1 useful)
L-----------------------
SG =e. SP
=e
Similarly S'G SG S'G
=
• S'P
SP S'P
PG bisects the LSPS' Thus, the normal at P bisects the angle between SP and S'P. (b)
Since the normal PG J. the tangent T'PT, LGPT'
=
(both are 90°)
LGPT
LGPT'
LS'PG
LT'PS'
= LTPS
=
LGPT
-
and LS'PG
LGPS
LGPS
Hence, the tangent at P is equally inclined to SP and S'P. This is the reflecting property of the ellipse.
A ray of light or a sound wave
originating from the focus S, will be reflected through the other focus S'.
3.
The chord of contact from a point on a directrix is a focal chord. Proof: Let T(xl'yl) be a point on the directrix, SO
that
Xl
a =e . y
The equation of the chord of contact is
XXI yy 1 -2- + -2- =I.
a
b
X
For PQ, this bec.omes ea
yy 1 b2 -
·
= 0,
i.e.
+ -- - 1
PQ meets the x-axis where y
x = ae, which is the focus S(ae, 0). Hence the result.
226 II.
That part of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus. Proof: Using the diagram from property 3, we have to prove
l PST = 90° •
7
2 2 · xx · m · Th e tangent at ·p( x ,y 2) IS a 2 + yy = 1 an d t h.IS meets t he d.1rectnx 2
T( ~, k), where e
••• (1)
=x
The gradient of SP is m The gradient of ST is m
mm'
I
=
ke
=
k a -e ae
:. LPST
=90°
ke a(l - e 2 ) aeky
2
a(l - e ) Using (1), we have, mm'
y2 - ae 2
= 2 a (1
2
2 - e ) (x - ae)
=- 1
2
and this proves the result.
The ellipse possesses a wealth of useful and interesting properties.
Some of these
follow simply from the definition and. others can be proved by co-ordinate geometry and plane geometry. The reader who masters the general techniques of proving these properties will have no trouble in proving the same properties when particular values of a and b are used.
227
6.12 Geometric Properties of the Hyperbola y
d'
d
G
0
-a e
X=-
X
= -ea
Many properties of the hyperbola are similar to those of the ellipse, so to avoid repetition, the properties are stated without proof.
lt would be instructive for the
reader to supply the proofs referring, if necessary, to the corresponding results for the ellipse. 1.
The difference of focal distances is constant. 5'P- 5P 5P - 5'P
i.e. 2.
= 2a, = 2a,
if
P is on the branch near 5, and P is on the other branch
= 2a
I5P- 5'PI
The tangent at
if
P
bisects the angle
5P5'
internally and hence the normal at
P bisects L5P5' externally. (Using the equation of the tangent at 5T 5P prove that ,T = ,p • ) 5 5
P, find the co-ordinate of T and hence
This is the reflection property of the hyperbola. towards the focus
5
A ray L P
of light
directed
of a hyperbolic mirror, is reflected towards the other
focus 5'. 3.
The chord of contact from a point on the directrix is a focal chord. In the figure, RP, RQ are tangents from
4.
R on the directrix.
That part of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus ( L P5R = 90°).
X
228
6.13 Properties of the Rectangular Hyperbola y y
=c
2
a
2
=y
2
=a
2
Fig. 2 In this section we investigate the geometric properties of two special hyperbolas 2 2 2 2 and xy =c . x - y =a 2 2 2 2 2 = I reduces to x - y = a and the If b = a, then the equation ~ - L a2 b2 asymptotes are y = .:!: x, each
other.
A
i.e.
hyperbola
x + y = 0 and x - y = 0, which are perpendicular to whose
asymptotes
are
perpendicular
is called
a
RECTANGULAR hyperbola (Fig. 2). The reader is familiar with the equation y =
~
,
i.e.
hyperbola (Fig. I) with the axes as asymptotes.
xy = k, which represents a =c 2 is also a rectangular
Thus xy
hyperbola, but the reader who is not satisfied by this reasoning should refer to two 2 a explanations given in the appendix, where it is shown that the hyperbola xy = is
2
obtained by rotating the hyperbola x . 2 a2 . . wnte c = our work. , to Simplify
2
- y
2
a
2
2
The eccentricity of both hyperbolas is given by b
2
through 45° about the origin. We
= a 2 (e 2
- I) and with b
2
the
same,
2
~- L a2 b2 2 but a few peculiar to xy = c are
this gives e = {2. Obviously most of the properties of hyperbolas 2 2 2 2 x - y = a and xy = c are
= a, I,
given below with proofs. In science and engineering, it is the form xy = k which is 2 2 2 useful rather than x - y = a • The law connecting the pressure and volume of a perfect gas under constant temperature is
pv = k,
and in electricity the law
connecting the current C, the resistance R and the E.M.F. V is given by CR = V.
229
6.14 Geometric Properties of the Rectangular Hyperbola xy = c2 1.
The area of the triangle bounded by a tangent and the asymptotes is a constant. y
Proof: The equation of the tangent at any point 2 P(ct, on the conic xy = c is:
r>
2
x + t y = 2ct
B
This meets the axes where OA = x = 2ct , OB = y = Area of the t.OAB = 1
=2 . 2.
2ct .
2c
t
1
2
t2c 0
X
OA • OB
2
= 2c = a
constant.
The length of the intercept, cut off from a tangent by the asymptotes, equals twice the distance of the point of contact from the intersection of the asymptotes. Proof: In the diagram, we want to prove AB = 20P. Using the previous example: P{ct,
r) ,
2 A(2ct, 0), B(O, tc ) ... (I)
••• (2)
From (I) and (2): AB
2
= 40P
2
which gives AB = 20P.
230
Worked Examples: Geometric Properties EXAMPLE: ( 1) Find the equation of the tangent to the ellipse 5x
j>.
P(2,
2
+ 9y
2
= 45 at the point
Find the coordinates of the foci S and S'. SV and S'V' are the
perpendiculars to the tangent at P. Prove that: 2 2 x + y = 9 and SV • S'V' = 5.
V and V' lie on the circle
y
SOLUTION:
2
T
The equation of the ellipse is
+ 2
+
The equation of the tangent at P(2, 2x
j>
1. is
2l
l ( . XX 1 yy 1 15 = • usmg -2- + -2- = I] a b
9
+
or
2x + 3y
=9
v
••• (1)
2
X
2
We have a = 9, b = 5, then 2 2 2 2 . a e = a - b g1ves ae = 2, so the foci are: S(2, 0) and S' (- 2, 0) SV J. to the tangent (1), so the equation of SV is Solving (l) and (2), we find S'V'
J. to
V(
~;
3x - 2y " 6
••• (2)
, : ~)
the tangent (l) and its equation is 3x - 2y = -6
••• (3)
Solving (l) and (3): V' (0, 3) V' (0, 3) obviously satisfies the equation x Substituting for
2
+ y
2
= 9.
V in the L.H.S. of this equation: 2
2 13
hence both V and Now SV and
2
V' lie on the auxiliary circle x
2
+
( ~; - 2)2 + ( : ~ )2 = ( : ~ )2 + ( : ~ )2 =
2
s•v• 2 SV
2
3 X 13 ----,~-
L.H.S. =
(- 2) 2 + (3)
• S'V'
2
=
325 . 169
2
y
2
= 9 = R.H.S., and
= 9.
~~~
= 13
13
=
25 g1vmg SV • S'V'
=
5
This example is a particular case of the general property of the ellipse
2
x
2 Y
7+~ SV . S'V'
i.e.
= I,
= b
2
.
V and V'
lie on the auxiliary circle x
For this example a
2
= 9, b
2
= 5.
2
+ y
2
= a
2
and
231
EXAMPLE: (2): Prove that the portion of the tangent at P (x 1' y ) to the hyperbola 1 2 2 ~ = 1 intercepted by the
7a - b
asymptotes is bisected at the point of contact. SOLUTION: X
The equation of the tangent at P(x , y ) is 1
XXI yyl -2--2= a b
1
1
... (l)
Let R and Q be the intersections of the tangent with the asymptotes. We find the quadratic equation which gives the ordinates of R and Q. The equation of the asymptotes is
2X
2
2
~
a
... (2)
= 0.
From (I)
or
Substituting into (2): a
2
~
(l
X
=
2 2 y yl
2yyl
+7+~
2 2 2 a YI I Re-arranging: y ( 4 b2 - -b 2)
XI
+
)
2 2 2YY Ia a b 2 2 + -2
XI
0
XI
This equation gives the ordinates of R and Q. Let M(X, Y) be the mid-point of RQ. If the roots are
y
1
and y , then: 2
2 2 2 22 - a b y1 1 - b x1 ~ ( 42 / 22 22 bx ay -bx XI 1 1 1 22 2 2 2 2 Since P(x , y ) lies on the conic, a y - b x -a b 1 1 1 1 :. y = yl 2 Xxl yl Then using (I), since M(X, Y) lies on the tangent, we have - 2 - -y= a b 2 2 2 Xx XI yI XI 1 But - - - - I so - - = - giving X= x 1 2 2 a2 b2 a a 2
yla - -b2 2 -
2
a y
Thus M (X, Y) = (x , y > which proves the result. 1 1
232
Exercise 6E: I.
PROPERTIES OF CONICS
2 2 144 meets the directrices The tangent at P (xI' y ) to the ellipse 9x + 16y 1 at T and T' respectively. S and S' are the foci. Prove that L PST and
LPS'T' are both right angles. 2.
Show that the equation of the normal to the ellipse
point P(x ,y ) is 25xy - 16x 1y 1 1 1 (a)
3.
9x 1y1 •
=e
+
25/
=e
400 at the
. PS and
. PS', where S and S' are the foci.
PS Hence prove that PS'
GS = GS'
. PG btsects the
and that
V and V' are the feet of the perpendiculars from the tangent at P(x I' y ) to the ellipse 4x 1
4.
2
The normal meets the x-axis at G. Prove that GS GS'
(b)
~
16x
2
+
=4
(a)
SV • S'V'
(b)
V and V' lie on the auxiliary circle x
2
S and S'
9/ = 36.
+ y
2
LS'PS. respectively to
Prove that:
= 9.
M is the mid-point of a variable chord PQ of the ellipse
16x
2
+ 25y
2
400,
where P is (x , y ) and Q is (x , y ). Prove that the product of the 1 1 2 2 gradients of PQ and OM is constant. 5.
6.
22 22 22 NP is the ordinate of a point P(x , y ) on the ellipse b x + a y = a b . 1 1 2 The tangent at P meets the x-axis at A. Prove that ON . OA = a , where 0 is the origin. 2 2 P(x , y ) is a point on the ellipse ; 1 1
+
a
·1e ctrc
x
2
+ y
2
=a 2
~ = 1 and Q is the point on the b
h · . P avmg t he same a b sctssa. roveh t at t he tangents at
P
and Q meet on the x-axis. 7.
2 2 P(x I' y ) on the hyperbola 16x - 25y = 400 meets 1 the directrix at T. Show that L PST = 90°, where S is the corresponding
The tangent at a point focus.
8.
2
NP is the ordinate of a point P(x I' y ) on the hyperbola 1 The tangent at P 0
meets the x-axis at L
X
P(xl'y 1> is a point on the hyperbola } If
a
= I.
b
2 Prove that ON . OT = a , where
is the origin. 2
9.
2
7 -~
2 v - ~ = I with the focus at S.
PS is parallel to the asymptote, prove that the directrix, the asymptote
and the tangent at
P are concurrent.
233 PQ is a chord of a hyperbola 9x 2 - 16y 2 = 144 passing through s. The tangents at P and Q intersect at T. Prove that T lies on the directrix corresponding to the focus S. 11. Prove that the point of intersection T of the tangents at P(cp, ~) and 10.
Q(cq' £q ) on the hyperbola xy = c 2 is given by T ( ~ , .1£._) • OT p+q p+q produced meets the chord PQ at R. Prove that PQ is bisected at R. 12.
Prove that the portion of the tangent at P(l, -1) to the hyperbola 2 Jx - 2y 2 = 1 intercepted between the asymptotes is bisected at the point of contact. (Hint: Use the equations of asymptotes Jx 2 - 2i = 0 and the tangent Jx + 2y = 1 to find a quadratic in either x or y.)
234
CHAPTER 7 ELEMENTARY PARTICLE
DYNAMICS 7.1
Introduction
Dynamics is the branch of Mechanics (Physics) that deals with the conditions under which bodies move. The other branch of mechanics is called STATICS, which deals with bodies at rest or under equilibrium under the action of some forces. Two branches of Dynamics are called: 1.
KINETICS: Kinetics is the study of effects produced by forces acting on the bOdies in motion.
2.
KINEMATICS: This deals with the motion of the body without regard to the cause, effect or result of the motion. So far in our work, we have done just that, i.e. the motion of a particle in straight line, the motion of a projectile, the SHM etc. We discussed the motion in terms of the position, velocity, time, acceleration. It did not matter which forces caused this motion.
We shall now introduce "the elements of KINETICS" which relate the forces with the motion of the body. This not only enhances our knowledge of the subject, but we can now solve a wide variety of motion problems, such as the motion in a RESISTING medium, the motion in a circle etc. Remembering that the 4- Unit Mathematics syllabus requires us to study not only the harder new topics, but also 3U-Maths harCier motion problems, we shall first completely summarize the previous work, then revtse with harder 3U problems, and then extend to the required new topics.
7.2
Laws of Motion - Force
In everyday life we use force to pull or push an object. In this chapter we study the cause-effect relation between the observed motion and the system of forces. Newton (1642-1727), one of the most famous and greatest scientists, formulated laws of motion after studying the motion problems which involve application of natural (gravitational) or mechanical (push, pull, friction) forces. Newton's First Law of Motion. (Inertia) A body remains in a state of rest, or of uniform motion in a straiRht line (a constant velocity, no external force) in the absence of a force.
235
A force is an invisible entity, it is recognised only by its effect, so it is a CONCEPT. A heavier object requires a greater force to move than a lighter object, hence we say a heavier object has greater inertia. The First Law introduces us to the idea of a Force and mass (inertia), i.e. the definition of what is a force. Newton's Second Law of Motion This law relates the change in velocity i.e. acceleration with the magnitude of the force that produced the motion. It states that: A force acting on a body produces an acceleration which is proportional to the magnitude of the force and this acceleration is in the direction of the force. The mathematical form of the second law is: Force Mass x Acceleration F = m a The mass m is a measure of the amount of material and hence the inertia of the body. In the Sl units: Mass is in kilograms (kg) Acceleration is in m.s- 2 Thus a force of 1 N acts on an object of mass of 1 kg, the object accelerates by 1 m.s-2•
Mass and Weight of a Body The weight is the force acting on an object of mass m, due to gravity. The value of g, the acceleration due to gravity is 9.8 m.s- 2 (at earth's surface). Weight = m x g W (newtons) = m(kg) x g (m.s- 2) A weight of 1 kg is equal to 9.8 N. Never confuse the weight with the mass. Weight is a force and hence a vector, but mass is a scalar quantity.
236
Newton's Third Law of Motion: This law states that the force exerted by one body on another body i.e. action force, is equal to the force exerted by the second body on the first, the ·reaction force, and they are opposite in direction. N
When you kick a f.oot ball, you are applying a force on the foot ball, and at the same time, the ball's reaction applies an equal force to your foot, it hurts!
mg Fig. I
A body of mass m lying on a horizontal smooth surface is pressing the table with a downward force of its weight, but at the same time the equal and opposite force N acts on the body. This force is the reaction in the direction at right angles to the surface.
WORKED EXAMPLES
EXAMPLE 1: A body of mass 10 kg is suspended by a string from a ceiling. Find the tension in the string.
SOLUTION: You need not show the entire system, only the forces acting on the particle.
T
The two forces acting on the particle are the tension T, upwards and the gravitational force mg , downwards.
We use Newton's second law: mg
The resultant force = mass x acceleration.
Fig. 2
T - mg = m x O, because the system is at rest, hence, acceleration = 0 T = mg
9.8 = 98 kg. m.s-2
m = 10 kg,
= 98 N
(I N
= 10
X
=I
g
= 9.8
kg. m.s- 2 )
m.s-2
237
EXAMPLE 2:
A particle of mass 20 kg is suspended by two strings. Calculate the tensions in the strings, shown in the diagram. (Acceleration due to gravity g = 10 m.s-2). SOLUTION: y
The forces acting on the particle P are the two tensions T and T 1 2 and the gravitational force mg.
Fig. 3
We select the rectangular coordinate system at P and decompose (resolve) the system of forces into the horizontal components Fx and the vertical components, F Y l:Fx = T 1 cos 4.5° - T2 cos 60° lF y = T 1 sin 4.5° + T2 sin 60° - mg The upward components are considered positive and downward components negative. Since there is no acceleration in any direction, we have: l:Fx
=
0 and lFY
=
0
T l cos 4.5° - T2 cos 60° = 0
and
T 1 sin 4.5° + T2 sin 60° - mg = 0 Remembering, cos 4.5° = sin 4.5°, subtract (1) from (2) T (sin 60° + cos 60°) = mg 2 substitute m = 20, g = 10 T2
200 + cos 60°
= sin 60°
=
146 N
From (1):
...
T I cos 4.5° = T2 cos 60° Tl =
146 cos 60° cos 49
= 104 N
•••• (1)
•••• (2)
238
Note: It is very important that: 1.
A free-body diagram is drawn, showing all the forces on the particle.
2.
The proper resolution of each force into two components at right angles to each other is shown. The two perpendicular directions need not always be the HORIZONTAL and the VERTICAL directions,
3.
If there is no acceleration, i.e. the system is at rest, only then is:
IF X = 0 and IF y = 0. Normally write, F x = m.ax and F y = m.ay' and then substitute for ax and a , for each problem. y EXAMPLE 3: A truck of mass 3 tonnes is descending an inclined plane at a speed of 20 m/s. Find the retarding force R, necessary to stop the truck in 30 m. (Angle of the incline is 10°). SOLUTION: The forces on the truck are: X
Normal reaction N, perpendicular to plane. Retardation force R, along the incline. mg, force due to gravity. Angle of incline = 10° Resolving the forces along and at right-angles to the plane:
and
R - mg sin 10°
••• (1)
N- mg cos 10°
••• (2)
Now the net force on the truck is along the incline, given by m.a , where retardation to be calculated using: 2 u + 2ax 2
given u
0
20
a
2 -2 -6 J m.s
IF x
rna = 3000 x
From(l), 20000
+ 30 x 2a
= R-
20
3
X
20 m/s 30m
= 20000 N
3000 x 10 sinl0°, giving R • 25200 N.
a
is the
239
Exercise 7A In the following examples, take g :: 10 m.s 2 1.
2.
A particle of mass 10 kg is suspended by two strings of length 3 m and 4 m attached to two points at the same level S m apart. Find the tensions in the strings. 2 The combined air and road resistance of a car in motion is proportional to v , where v is its speed. When the engine is disengaged the car moves down an incline making an angle sin- I (l/30) with the horizontal, with a velocity of 30 m/s. Find the force required to drive the car up the incline with a steady speed of 24 m/s, g!ven that the mass of the car is 1200 kg.
3.
A truck of mass M is driven up a road inclined at an angle 9 to the horizontal. After its speed has reached u m/s the engine continues to exert a constant force of F newtons. If the resistance R is constant, find the time taken to reach the velocity v m/s.
4.
A car of mass 1SOO kg is moving at 60 km/h; when the brakes are applied with a braking force of 12000 N. Find:
(a) (b) (c)
the acceleration the time taken by the car to stop the distance travelled before coming to rest.
s.
A body of mass m is pulled up a smooth incline making an angle 9 with the horizontal, and has an acceleration f. Find the force F that pulls the body.
6.
A mass of 10 kg is pulled along the horizontal by a chain making an angle of 30• with the horizontal. If the tension in the chain is SON, find the acceleration of the body and the magnitude of the normal reaction.
7.
A smooth block of mass 2 kg slides down an incline making an angle of tan- I (3/4) with the horizontal. Find the acceleration and the magnitude of the normal reaction.
8.
A truck of mass 2000 kg starts to climb an incline of angle given by 1 9 = sin 0/JO). The total resistive force is 2000 N. Find the retardation it experiences.
9.
Find the magnitude of the braking force to stop a car of mass 1200 kg in 20 m when it is travelling at 60 km/h (a) on a horizontal road (b) down an incline of an angle sin-l (1/40).
240
CHAPTER 8 MOTION PROBLEMS IN TWO DIMENSIONS We shall first completely summarise the results of various types of motion studied so far in 3U Mathematics, then revise the harder problems of simple harmonic motion and projectile motion, then embark on 4U Motion.
8.1 Introduction: Motion In a Straight Une
= =
Displacement Velocity
x dx dt :
v
(x, t, v) X
Acceleration a
= ••x
2
..
Fig. 1
2
dx = dlv) 'd;2 dx \ 2 =
v
dv
dx
For constant acceleration ONLY, the equations of motion are: u
initial velocity
a
constant acceleration
I.
v
2.
v
3.
X
2
= u + at 2 = u + 2ax 2 = ut + !2 at
Do not use these formulas for the variable acceleration
Method of Solution (variable acceleration)
I.
Given
X
=
x
2.
Given
x =
I I
f(t), integrate
f(t)dt
=
g(t) + c
g(t)dt + ct
x
f(x),
use
p
0
Time = t
~x (
2 T) = f(x) and integrate.
X
241
3.
. ••x = f (v ), use v dv . G1ven dx = f (v ), and mtegrate
y For VERTICAL MOTION under gravity only replace a, by, g acceleration due to gravity. x may be replaced by y. I.
y
2.
v
3.
v
I
ut u 2
u
2
2 gt
-
gt
-
2gy
2 p
•
mg Fig. 2 0
For downward vertical motion under gravity, assuming the object falls from the rest, we have: (u = 0) 1
2 gt
y
gt
v
v
8.2
2
2
Fig. 3
2gy
f
Simple Harmonic Motion (Revision)
Definition: A particle M on a straight line is said to perform a SHM if its displacement satisfies the differential equation y 2 d x 2 -n x 2 dt
X
X
A'
0
M
A
X
X
Fig. 4 The acceleration is always directed towards, and proportional to the displacement from the centre. 1.
ix
The general solutiOn of 2 dt x = a cos (nt + a), where a is the amplitude of the SHM, n and
a
are constants.
242
2.
If the motion starts at A, where x
substitute in
x
a cos (nt + ex)
a
a cos ex giving
cos ex
X
0.
t
ex = 0
a cos nt
x 3.
+a,
A cos (nt) + B sin (nt) is also a general solution.
Details of the motion: Let x = a cos (nt + ex) be the solution of x (i)
-n
2
x.
The greatest displacement given by the function x = a cos (nt. + ex) is
Ix I
a and is called the amplitude of SHM.
period of SHM is T
(ii)
The
(iii)
dx The velocity v = dt
2n n
-an sin (nt + ex)
The maximum velocity is v = ±an v
+
v (iv)
an
for the particle moving to the right.
an for the particle moving to the left.
We have:
2 X
+
v
2 a
2
v
2
2
Fig. 5
n
or v
2
Since v
n2 (a2 - x2) 2
then
3> 0,
lx I ~ a
2
The graph of v shows that v is greatest at x = 0 and v = 0 at x = ±a. Hence the particle oscillates between two extreme positions why it is also called an oscillatory motion (-a< x gd,
.
where g is the acceleration due to gravity.
If U 2 = 4gd, also prove that the maximum height that can be reached by the 15d . on t h"1s wa 11 IS . g1ven . by 8 Jet .
5.
A missile is fired from 0 with initial velocity U at an angle horizontal. Prove that it describes a parabola of focal length
a
with
the
u 2 cos 2 a 2g 4 2 Also prove that any point P(x,y) within and on the circle x + / = v 1s m 2 2 danger of being hit by the missile. (g m.s- is the acceleration due tog gravity).
6.
A stone is projected upwards at an angle 9 to the horizontal. Find an expression for the velocity v at time t in terms of g, t and the initial velocity U. If the stone at time
velocity, show that t v 7.
= U cot
t
= -u g
is moving in a direction perpendicular to the initial cosec 9 and that the stone's speed is given by
9 •
A body is projected with speed U from a height h, above a horizontal plane, at an angle 9 to the horizontal. Show that the range R on the horizontal is given by 2 2 2 2 gR sec 9 - 2U R tan 9 - 2hU
=0
Further show that the maximum range R
1
is given by
and the corresponding angle of projection is given by tan9
u2
=· gR1
Hence prove that h
= R 1 cot
2 9.
251
8.
A particle is projected from a point 0 with velocity v at an angle 9 to the horizontal. It passes through the point P (x,y) in the vertical plane through 0 where (x,y) are the co-ordinates of P with respect to the rectangular axes at 0. Prove that .y
=. 20
=x
2
2
tan 9 - gx se;: 9 2v
-2
m, g = 10 m.s , v = 20 m/s, find the two values of tan 9 . . 2 9 =- 2t . o f two ranges and usmg, t = tan 9 an d sm prove t h at t he ratto 2 . 5 I + t IS j . If x
9.
= 10
m, y
A particle is projected with velocity v at an angle 9 to the horizontal. Show that by SUitable choice of axes, the equation of the path of the projectile is y =
2
X
2
tan 9 - gx sec B 2 2v
Prove the following: (a)
There are two possible directions of projection given by tan 9 and 1 tan 9 for a given range R. 2
(b)
tan 9 1 + tan 9 2
(c)
Let T
2 2v gR
Tl
sin 9
T2
sin 9
(d)
From (b) prove tha1 9
(e)
R
(f)
and tan 9
1
tan 9
2
=
I.
and T be the times of flights corresponding to angles of 2 1 projections 9 and 9 · 1 2 Prove that
10.
=
=
R
1
1 2
1
• + 9
2
=
90°.
sin 29 , where R is the maximum range. 1 2
T'
then
R'
4
~
5
[Hint: sin29
= ~] I + t
Two particles P and Q are projected from the same point 0 with the same velocity 25 m/s. They both strike the horizontal plane through 0 at the point A, 60 m from 0. P reaches A before Q. Show that:
~
(a)
the angle of projection of P is tan-!
(b)
the time of flight of P is 3 seconds.
(c)
the distance between P and Q at the instant P reaches A is
(d)
and that of Q is tan-!
the ratio of the maximum heights reached by P and Q is
15
9
~
/2
16 ·
•
m.
252
8.4
Resisted Motion: Other Laws of Motion
In our previous studies of motion, we neglected the effect of air-resistance or air-
friction. A body moving in a fluid experiences a resistance which tends to stop the motion. In many cases of motion, the resistance is an important consideration. Cars, planes and boats are streamlined so as to reduce the frictional drag and improve fuel economy. The air or fluid resistance on an object depends on its: (i) shape
(ii) size (exposed area)
(iii) speed.
For example, a sky-diver with an unopened parachute falls quite rapidly, but the descent is slowed when the parachute opens. The parachute encounters greater resistance due to its shape and size. A sky-diver can enjoy a free fall (without an open parachute) by employing a spread-eagle position to increase the air resistance and prolong the time of fall. Air or fluid resistance also depends on the speed of the object. The greater the speed, the greater the air resistance. We shall mostly be concerned with motion for which the resistance is proportional to the speed v or v 2• Contrary to our perception of resistance, it is quite beneficial to us. Actually it is the road friction that makes car driving possible! Sky-diving is pleasant and possible because the air-resistance helps to slow the descent. Though streamlined cars cost more, they are at least pleasing to our eyes, if not to our purse! EXAMPLE : (1)
A particle of mass m moves along the x-axis. It experiences a resistive force R given by R = kv, where k is a constant and v is the velocity at any time t and position x. Discuss the motion. SOLUTION: (x,t,v) p
0
X
The equation of motion at time t is given by dv m. dt
=
-kv
(F
= ma)
We attach a negative sign because the resistance opposes the motion. k • dv • • dt = .v
m
Integrating loge v
I~v
-
f~ .dt
kt A m +
or v = B e-kt/m ' where B is a constant.
Fig. 10
253
If v = v
at t = 0, then B = v
0
0
-kt/m v = voe As t ...
the function e-t ... 0 and hence v ... 0
co
dx -kt/m Further v = dt = v0 e The distance travelled in time t X
- Jt
-
0 or x
jp-
vO
is
v e-kt/m dt
0
[I -e-kt/mJ mvo
Again as t
-+- "'• x-+-
k
i.e. it moves with decreasing velocity towards the
limiting position xt
EXAMPLE : (2)
A particle of mass m falls under gravity from rest in a medium with the resistive force given by R (v) = kv. Discuss the motion. SOLUTION: 0
We take the initial position as the origin and x-axis along the direction of motion. The equation of motion at time t is: dv m dt dv
mg -
mkv p
dt = g - kv
t
or
dt
- kI [log (g- kv)- log (g)]
Fig. 11
- k1 loge ( 1 _kgv)
X
254
We can solve this equation for
v
;kt = 1 _ kv g
f (1
v
- e-kt)
••• (1)
As t-+oo,
Fig. 12 vT
=f
is called the terminal velocity and the particle continues to travel with
constant velocity vr
This happens when the resisting force
mkv
balances the
gravitational force on the particle, i.e. mkv
=
mg
or
-
V
-
gk
[Have you ever wondered how a team of sky-divers frolic (in the sky) with their parachutes not open! From above· you see that a sky-diver should enjoy a free fall until his weight balances the resistive force, thereafter his parachute must open, and with good luck the diver should then enjoy his fall with the reduced terminal velocity. Terminal velocity before the parachute opens is about 200 km/h and it is 40 km/h after the parachute opens.] Further
dx
v
+
X
When t
hence
from (1)
dt
= O, x
x
=0,
c
this gives c
255
EXAMPLE : (3)
A particle of unit mass is thrown vertically upwards with a velocity of U into 2 the air and encounters a resistance kv . Find the greatest height h achieved by the particle and the corresponding time. SOLUTION:
Equation of motion is dv v dx
I
=
-g - kv
X
I
vdv g + kv2
dx
1
p
2
- 2k loge (g + kv ) + c
X
When x
2
= 0,
v
= U,
this gives c 2 g + kU 2 g + kv
1
x = 2k loge
= 21k
2 loge (g + kU )
••• (1)
For the greatest height, x = h, v = 0 1
h = 2k log ( 1
+
k 2 g U )
.•• (2)
0
For the corresponding time, we use:
2 dv dt = - g - kv
t
t
I
-dv g + kv2
Fig. 13 dt dv
or
=
-1 g+kv2
1
=- k
1 .fiJk
r"
Ltan
1
1/fkg . v
]o
U
256
EXAMPLE : (4)
A particle is thrown vertically upwards with speed U in a medium with resistance R = mkv, where m is the mass of the particle and k is a constant. Find the greatest height h reached and the corresponding time. SOLUTION:
Selecting the origin as the point of projection, the equation of motion is dv m.v. dx·.
X
- mg- mkv p
X
_I ~
x
= -~+~
When x
= 0,
v
k1
X
(g + kv) g + kv
= U,
mkv • dv
loge (g + kv) + c hence 0
=-
¥
+
~
log (g + kU) + c
k ••• (1)
(U - v)
At the greatest height, v h
~
= .k!d. -
= 0,
x
=h
&... log ( 1 + kgU )
••• (2)
k2
To find the time, we use dv dt
=
-g - kv
o t--
J
~
U g + kv
or 1
k log
dt .dv
( 1 +
=
~u)
-1 g + kv
••• (3)
0
Fig. 14
257
EXAMPLE : (5) A body is projected vertically upwards from the earth's surface with velocity U. The 2 acceleration of a particle in space is given by kx , towards the centre of the earth, where x is the distance of the body from the centre of the earth. Given that the acceleration is g at the earth's surface, prove that the velocity v at time t is given by v
2
= u
2
2 - 2gR (
~-~
)
where R is the radius of the earth. If u
2
~~
= 2gR, find
in terms of x and show that the body will reach a distance &R
from the earth's surface in 2.72 hours. Also find the velocity of ESCAPE, i.e. the velocity never returns to the earth.
U of projection so that the body
(R = 6400 km, g = 10 m.s-2 ). SOLUTION:
Selecting the axes as shown, the equation of motion is X
dv m.v. dx and
g
mk
(l)
-7 ...
k (given) at x = R ;2 2
k = gR '
hence
dv vdx
R2
(2)
p
l
mk x2
-&y X
x=R Integrating , At
X
l
2
=
= R, v = U,
gR2
+
X
c
u2 so C=y - gR
v 2 = U 2 - 2gR 2(
1
R.
X
)
Fig. 15 ••• (3)
258
If U
2
2gR then from (3), v
= dx dt
v
For
=
x
2 R T
=
i 112 (sign
+ v'22gR.
IR
9R
=
xl/2 d
ffgR
+ as v t from
1
2
= 6.4
6
x 10 m, g
= 10
1
-2 , t
m.s
Finally, the particle never returns if x
2
=
U
2
The body never returns if v
2
~
(27 R/R - R {R)
H.
= 2.72 -+
hours
oo,
(since
- 2gR
R
3
/ZgR
5: .
v
0)
2 [ x3/2] 9R
ffgR .
x
=3
For R
2
112 x dt dx Rffg 8R + R = 9R (from the centre) time taken is
=
t
2
hence from (3)
.!X
-+
0)
0 •
2
u > 2gR U
> /2gR
U must be slightly greater than i.e.
U
> 11300
U
> 11.3
m/s
km/s
The escape velocity
12 km/s
h x 10 x 6.4 x 106
259
Exercise SC I.
A particle of. unit mass falls vertically from rest in a medium with resiStive force R = kv, where v is the velocity of the particle at time t. (k is a constant). Find the velocity v and hence show that the terminal velocity is given by
2.
f.
2 A particle of unit mass falls from rest in a medium with resistive force R = kv , where k is a constant. Prove that the distance x fallen when the velocity is v, is given by: 1 = 2k
x
loge (
g 2 ) • g- kv
Find the distance fallen when it reaches half of its terminal velocity. 2 = 0, i.e. v = .fijk).
(Hint: The terminal velocity is given by g - kv 3.
An object is projected vertically upwards with initial velocity U from the earth's surface. The acceleration obeys the law given by 2 d x dt 2
k - x2
where x is the distance of the particle from the centre of the earth whose radius is R.
= R,
Given the acceleration is g when x is given by: v
2
= u
2
-
Hence show that
2gR
2
(
~
-
~
show that velocity v in any position
) •
U = 12 km/s
for the escape velocity. not return to the earth). (R = 6400 km, g = 10 m.s- 2 )
(Hint: x 4.
+ oo
for escape from the earth and v
2
(i.e. the object does
> 0)
A particle moves under gravity in a resistive medium with the resistance = kv, where v is the velocity in any position and k is a constant. The
R
particle is projected vertically upwards with velocity U time the expressions for the velocity v and position -kt
v = U \2e
- l)
and
x =
ku (2
-kt - kt ) . - 2e
Find the greatest height achieved by the particle. (Hint: v = 0 for the greatest height).
x
=f .
Show that at any
are given by:
260
5.
A particle of mass m is projected vertically upwards under gravity with velocity 2 at any time t and position x.
U m/s. The resistive force is R (v) = mkv Show that the expression for x is given by: x = 2k1 loge
[g
J
2
+ kU2
g + kv
Show that the greatest height H is given by:
H= ....!.
2k
6.
log
[1
e
2 kU ] g
+
A body of mass m falls from rest in a medium with resistive force R = kv, where k is the coefficient of air resistance and v is the speed of the object. (k is a constant.) Prove that the distance x fallen when the velocity is v, is given by:
= - mkv -
X
m2 g log k2 e
[I -
kv ] mg
Find the terminal velocity for a falling 70 kg sky-diver, if k = 14 and g 7.
= 10
m.s
2
•
Express your answer in km/h.
A ball of mass
m
is thrown vertically upwards with velocity
resistance is per unit mass f(v) (i)
(ii) (iii)
=
2~
loge
[1
+
~
u
•
2
J
Draw a neat sketch of the downward motion of this ball after it reaches the greatest height H. Show that the distance y fallen when velocity is W, is given by
= 2~
loge [
g
g- kW
2]
Deduce from (iv) that the terminal velocity V is given by that H can also be given by: H =
2~
loge [
g g- kW
W .is the velocity when y (vi)
The air
Show that the greatest height achieved is
Y (v)
U.
, at a distance x when velocity is v.
Draw a neat sketch of the motion, showing the forces acting at a distance x from the point of projection. 2 Show that x = kI loge [ g + kU ] 2 2 g + kv
H
(iv)
= kv 2
2]
=H
From (iii) and (v) deduce that
where for downward motion. I
u2
+
I
7
-
-
I
::::2
w
v2
=
f , and
261
8.
9.
A particle is thrown vertically upwards where the air resistance is given by 2 R = 0.01 mv • If the velocity of projection is 60 m/s, find the (a)
time to reach the highest point
(b)
greatest height H.
2 A particle falling from rest in a vertical line in a medium with resistance kv per unit mass, k is a constant, v is the velocity at any time and position x, prove that it acquires a speed
[/I -
e-2kh
J vr
in falling through a distance h, where VT is the terminal velocity given by
ff
A particle projected upwards in the same medium with initial speed returns to the point of projection with speed VR .
VI
and
Prove that 10.
An object of mass 20 kg experiences a resiStive force, in newtons, of one-fifth of the square of its velocity in m/s, when it moves through the air. This object is projected verticaHy upwards from a point 30 m/s and reaches the highest point H in time T.
. Given g
11.
= 10
0
with velocity of
-2 m.s , find
(a)
the time T
(b)
the distance OH.
A particle of unit mass moves in a straight line against a resistance R given by = v(l + v 2), where v is the velocity of the particle at a distance x metres
R
from the origin. Prove that x = tan-l v
0
- tan-l v,
velocity at the origin. Use the formula
l
-1 -1 -1 [A - B tan A - tan B = tan I + ABj to show that: v
v - tan x 0 1 + v tan x 0
where
v
0
is the initial
262
12.
A particle of unit mass travels in a straight line against the resisting force f(v) = v(l + v 2 ). Its initial velocity is c m/s at the origin. Show that the time t, when velocity is v, is given by: =
t
2] · 1 ' [1 Iog _ 2 2 + v-
1 +
c
2 Find v as a function of t and hence the limiting value of v at t 13.
14.
+ co •
A particle of mass m moves along a straight line under the action of a constant (propelling) force P, and a resistive force mkv, where k is a constant, v is the speed at any time t. Show that if the speed increases from 2 m/s to 4 m/s over a time interval of 5 seconds, = 2km
rL :~
(a)
p
2e e
- I] . - 1
(b)
Find the corresponding distance moved.
(c)
Find the propelling force P for k
= 0.5.
The acceleration due to gravity at a distance
r
from
centre is directed towards the centre and equal to !5. when r 2
the earth's
> R,
equal to
r
kr when r < R and equal to g when r = R. Imagine a narrow tunnel along a diameter XY of the earth and the particle is projected from X with initial velocity U towards Y. (R = radius of earth). If U
2
< 2gR,
prove that the motion is oscillatory and the amplitude is given by
R
15.
A particle is projected vertically upwards from the surface of the earth with initial velocity U. The acceleration due to gravity at a distance x from the centre of the earth is given by k , directed towards the centre. Prove that 2 X
2 the rocket will escape from the earth provided u ? 2gR, where g is the acceleration due to gravity at earth's surface and R is the earth's radius. 2 Further, if u = 2gR, show that the time to achieve the height R above the 2 earth's surface is approximately equal to 0.273 fR , given g = 10 m.s- •
263
16.
A body of mass m is released from a height h above the ground and it experiences a resistive force R given by R = 0.1 mv 2, where v is tile velocity achieved by the body at time t. 2 If the object falls from rest under gravity (g = l 0 m.s- ) Find: (i)
the terminal velocity U
(ii) the height h the ground. 17.
if the velocity is
0.5U
+ m ).
just before the body strikes
A ball of mass, m, is descending vertically in a tank of fluid (under constant gravity). The resistive force is kv per unit mass, where v is the speed and k, a constant. (i)
Draw a motion diagram at time t and write down the equation of motion dv using F = m • dt
(ii)
Write down an expression for time t taken by the ball to acquire the velocity v from rest and hence show that: v
(iii) 18.
(i.e. velocity as t
= ~
(l - e-kt )
Find the terminal velocity vT. If g of v against t.
= 10
and k
= 0.2,
draw the graph
A particle P of unit mass is projected from a point 0 with velocity U at an angle 9 to the horizontal. The particle moves under gravity and each component of its velocity experiences a resistance k times the magnitude of the component. By considering the rectangular axes Ox and Oy and the particle has components v and v of the velocity v at time t: X y
(a)
Draw a diagram of the motion of P
(b)
Write down the equations of motion in the form: and ~ (v ) dt y
(c)
Prove that v
(d)
Prove: x
X
= Ucos9e-kt
u (1 = kcos9
y =
e-kt)
(kUsin9 +g) 2
k
and
v
y
= . ) e-kt - & k1 (g + kUsme k
264
19.
A projectile is fired vertically upwards from the earth's surface with velocity U m/s. The retardation due to gravity is given by the law k X
2
where x is the
distance of the projectile from the centre of the earth, and k is a constant. The acceleration due to gravity on the earth's surface is g. The earth's radius is R. 2
Neglecting the air resistance, show that if u reaches the height R above the earth's surface. for this journey is [
20.
~
+
= gR,
then the projectile Show further that the time
1] A"·
An object of mass mv
resistance R
=k ,
m
is thrown vertically upwards in a medium with a
where v is the velocity of the object and k is a constant.
Given t = 0 = x and the initial velocity is k(c - g), where c is a constant and g is the acceleration due to gravity, x is the distance travelled in time t. (a)
Draw a neat sketch of the motion and the forces acting at a point P, distance x from the origin. Hence write down the equation of the motion.
(b)
Find the time taken by the object to reach the highest point H and find the height of H above the point of projection.
(c)
The object falls to its original position with the same law of resistance. Will the time of descent be the same as that of ascent? Give your reasons.
265
CHAPTER 9 CIRCULAR MOTION 9.1
Introduction
The study of circular motion is of great importance in science and engineering. The orbit of the earth around the sun, or the moon around the earth, can be considered circular for practical calculations. The safe speed on a circular section of a highway or railway track is governed by the laws of circular motion. We shall also study the problems related to circular motion, such as conical pendulum and banked tracks. In solving the motion problems, we often require resolution of the forces in two perpendicular directions OX and OY. These directions need not always be the horizontal and the vertical. Study the following examples of two resolved parts of the force F.
y y
y
0
X
Fig. 1 We shall study the dynamical problems, in which Newton's laws have to be used, namely
=
(1)
F
(2)
Force of action
rv
ma (mx or my)
= Force
of reaction
To solve any dynamical problem, 1.
Resolve all the forces acting on the system in two perpendicular directions.
2.
Then using F = mx and F = my : mx
3.
= .EX
- .EFx and my
= .EY
.EF
0
X
.EX
- .EFy
Substitute known values in (2) and solve for the unknown (velocity, force etc.)
.EF
y
Fig. 2
266
9.2
Angular Velocity: Period
For a point moving in a straight line, its velocity is defined as the rate of change of its displacement.
(x, t) p
0
X
Fig. 3 v
= ~~
in m/s where x
= f(t) 2 d x
-2
Linear acceleration of a particle is given by a = dt 2 in m.s
When a point moves on a curve, we talk about its angular velocity, i.e. the rate of change of its angular displacement, as defined below. ANGULAR VELOCITY of a point about a given point Let 0 be the given point, and OX a line through 0 of fixed direction. Suppose P moves in the plane containing OX and LPOX = 9 at any time t. The anticlockwise rotation is considered positive and the clockwise rotation negative. The angular velocity w (omega) of the moving point P about 0 is defined as the rate of change of 9, i.e. X
Fig. 4
The unit of angular velocity is the radian/s and is abbreviated as rad/s. Angular velocity is a vector and when the direction is not significant we speak of angular speed. It must be remembered that as defined above w is a variable, i.e. a function of t. In most of our applications in circular motion, w is a constant, i.e. a uniform angular velocity. In a later section, we shall talk about the angular acceleration of a point, about a given point.
267 EXAMPLE: (I)
A point P moves on a circle with uniform angular velocity of
j
rad/s. Show the
positions of P for t = 0, 1, 2, 3, 4, 5, 6. What is the time taken by the point P to describe the circle once completely? ll 3 = 60° .
SOLUTION:
t
0
9
0
2
5
3
4
6
ll
4n
5n
3
3
We have: ll
2n
J
3
2n
t=3
t=O t=6
Time to describe the circle once completely is 6 s. T = Angular displacement Angular speed
2n
=
1iT3
= 6 s Fig. 5
9.3
Circular Motion: Tangential Velocity
Suppose a particle moves in a circle of radius r (anticlockwise) and sweeps out angle 9 in time t. Let it describe a small arc
PQ in time 8t.
arc PQ = r x 89 Then v,
the tangential velocity at P is given by:
d d d9 v = dt (arcPQ)=dt (r.89)=rdt asllt~O d9 Now the angular velocity is w = dt , hence: Fig. 6 ••. (l)
The angular velocity w is usually defined in radians per second, but if the radius of the orbit is extremely large as in planetary motion, w may be defined in radians per hour or a day or even a year. It should be remembered that w is a variable in formula (1), but most of our work deals with the constant angular velocity and hence constant tangential velocity. In that case; at time t, 9 1 9
angular velocity x time =
wt 1
... (2)
It is usual to describe w as revolutions per second or minute or hour etc.
268
EXAMPLE: (2)
Convert the angular velocity w radians per second.
SOLUTION:
Since 1 rev
=
w
50 revolutions per second to
2n radians
=
50 rps
=
50 x 2n
lOOn rad/s
EXAMPLE: (J)
A satellite moves in a circular orbit with 20 revolutions per day. Describe w in rad/s.
SOLUTION:
(J)
=
20
X
=
20 rpd
=
3
=
11 36 X 60
rad/s
=
11 2160
rad/s
51! 60
24
211 radians per hour
rad/min
X
EXAMPLE: (4)
A wheel of radius 2 m revolves at 1200 rounds per minute. Find: (a) its angular velocity (b) the tangential velocity of a point on the wheel.
SOLUTION:
(a)
w
(b)
v
=
=
1200 rpm
=2x
rw
1200 X 211 rad/s 60
40n
=
=
SOn
= 40rt rad/s.
251 m/s.
THE PERIOD: The period T of a circular motion with constant angular velocity the time for one revolution. T
EXAMPLE: (.5)
=
w is defined to be
angular displacement in one revolution angular velocity
An artificial satellite travels in a circular orbit of radius 9000 km. If the period is 90 minutes, find the angular and the tangential velocity in km/h.
SOLUTION: T (a)
(J)
(b)
v
=
211
given T
(J)
211
T
=r
x w
211
T.5
=
9000
rad/h X
4.19
=
90 min
= 4.19 rad/h = 37700 km/h
=
1.5 h
269
Exercise 9A 1.
A particle on a disc rotating with a uniform angular speed of 4 revolutions per second is 0.2 m from the centre of the disc. Find: (a) (b)
the tangential speed v of the particle the angle through which it rotates in 0.4 s.
2.
A car travels halfway around a circular track in 12 s. What is the angular speed of the car? If the radius of the track is 100 m, find the velocity of the car in km/h, to the nearest km /h.
3.
A motor shaft has a speed of 240 revolutions per minute. speed in rad/s.
4.
Find the average angular speed in rad/s of the earth's (a) (b)
Find its angular
rotation about ·its axis. (Hint: period is 24 h) revolution about the sun. (Assuming a circular orbit and a period of I year = 365.25 days)
5.
Find the tangential speed of the earth at the equator due to its rotation about its axis, given that the radius of the equatorial circle is 6440 km. [Use exercise 4 (a)]
6.
Find the tangential speed of the earth due to its revolution about the sun, given that the radius of the earth's orbit around the sun is 1.5 x 108 km. [Hint: use exercise 4(b)]
7.
What is the angular speed of the particle in a circular path of radius 5 m and a tangential velocity of 100 km/h?
8.
A belt is wrapped around a pulley that is 0.4 m in diameter. If the pulley rotates at 240 revolutions per minute, what is the linear velocity in m/s of the belt?
9.
An aeroplane propeller revolutions per minute. (a) (b) (c)
whose blades are
1.5 m long is rotating at 2400
Express the angular speed in rad/s Find the angular displacement in 4 s Find the linear speed of a point on the end of a blade.
10.
An artificial satellite travels in a circular orbit of radius 10000 km. If the period is 2 h, find the angular speed in rad/h and the tangential speed in km/h of the satellite.
11.
The average distance of the moon from the centre of the earth is 3.85 x I o km and the period of the moon's revolution about the earth is 27.3 days. Find the angular speed in rad/h and the linear speed of the moon in km/h.
5
270
9.4
Acceleration of a Particle Rotating in a Circle y
The diagram shows a particle P moving anticlockwise on a circle. At time t,
P is (x, y)
e
LPOX w
angular velocity (constant)
v
tangential velocity
w
Cit = a constant, so dt
de
X
dw
0
The horizontal com onent X
The vertical com onent
= rcose
y
=
rsine
The horizontal velocity
The vertical velocity
dx . e de . e dt = -rsm dt = -rw sm
dv dt
The horizontal acceleration
The vertical acceleration
de = -rw 2 cos a -d2x 2- = -rw cose • dt dt 2 =-rw cose
~
d9 = rcose dt
= rwcose
2
dt
:. x
y
. e de 2 . e 2 = -rwsm • dt = -rw sm 2 = -rw sine
The resultant acceleration
a
fx2 + -y2 2
1 = tan I 5°
2
4! {i2
i.e. -1, 2!
1/3.
we have tan I 5° = 2 -
/3.
2
Observing that
291
Exercise 1OA For questions 1 to 5 prove that: 2 2 1. (a) 8sin ecos e = 1 - cos49, hence 2 4 (b) 32cos 9sin e = 2 -cos29- 2cos49 +Cos69 8 8 2 2. (a) cos e- sin e = cos29 (1 - ~ sin 29) (b)
6 6 cos e + sin e =
i
(1 + 3cos 2 29)
jl1 _sine + sine
3.
n e) tan ( 4 + 2
4. 5.
(2cos9 + 1) (2cos9- J) (2cos29- J) (2cos49- J) = 2cos89 + 1 9 cosec 9 + cosec 29 +cosec 49 + cot49 = cot < >
6.
if A + B + C
= seca +tanG
2
(a) (b)
7.
=
= 180°,
prove that:
cos2A + cos2B + cos2C = -1 - 4cosAcosBcosC 2 2 2 cos A + cos B- cos c 1- 2sinAsinBcosC
1f A + B + C
(a)
= 90°,
prove that:
sin2A + sin2B + sin2C . 2B +Sin . 2C . 2A +Sin
(b)
4cosAcosBcosC
=
Sin
1 - 2sinAsinB sinC
8.
Solve the following equations: 0 ~ 9 ,< 2n
9.
Solve the following equations: 0° .$- 9 .$- 360°
(a)
sin4a = cos29
(b)
sin39 = sin29
4 sin 2 a - 3cos 2 9 = 1
(a)
sin59- sin39 + sin9 = 0 = 4cos 3 9 - 3cos9 3 Put x = 2cos9 in the equation x - 3x
(b)
10.
Show that cos39
reduces to cos 3 a (a)
.
= V2
and show that the equation
1
= V2 2
3
Hence solve the equation x - 3x -
.f2 =
0, giving your answer in the
exact form, x = 2cos9 (b)
n +cos (9rr 17n Prove that: cos (TI) TI) + cos(TI)
=
0
(Do not use a calculator)
11.
(a)
Show that sin (A + B) - sin (A - B) = 2cos A sin B
(b)
Show that 2 sin x (cos 2x + cos 4x + cos 6x) = sin 7x - sin x
(c)
6n) Deduce that cos ( 2n ) + cos (411) +cos = l and (x - 5) + (y - 5) = 25
294
3.
Equation of a Circle having AB as Diameter
Let A be (h, k) and B be (h', k'), and P(x, y) be any point on the required circle. We use the fact that the angle APB in the semicircle is a right angle.
m
m'
gradient of AP = gradient of
=
BP =
Since AP J..BP, mm'
= -1
~ X.-
h
v - k' ~
and this gives:
(x - h) (x - h') + (y - k) (y - k') = 0
as the equation of the required circle. 0
EXAMPLE: (4) Find the equation of the circle on AB as diameter where A is (2, 3) and B is (4, 5). SOLUTION: The required equation is obtained by substituting for
A (2, 3) and B ( 4, 5) in the
equation above, i.e. (x - 2) (x - 4) + (y - 3) (y - 5)
or
x
2
+ y
2
- 6x - 8y + 23
=
= 0
0
As a check, we find that the centre (3, 4) is the mid-point of AB.
X
295 T
4.
Tangents to a Circle
We use the fact that the tangent is perpendicular to the radius at the point of contact. Consider the equation of the circle x
2
+ y
2
+ 2 gx + 2 fy + c
= 0.
•• . (I)
The centre is (- g, -f). If P(x , y ) is a point on the circle, then m 1
of CP
1
Y1+f
=-xl+g
= gradient
X
and the gradient of the tangent PT is then m'
The equation of the tangent is y - y 1 Or (y - y ) (y + f) 1 1
= - (x 1 + g)
=-
-J
=-m =-
X}+g --yl+f
XI + g 'Yi"+T ·(x - x 1)
(x - x ) 1
Now P(x I' y ) lies on the circle (1), 1 Hence the equation of the tangent is
The reader is advised not to memorise this equation, but derive the result for each question. It will be instructive to prove the following results: 1.
2.
2 2 2 The equation of the tangent at P(x I' y ) to the circle x + y = a is 1 2 xx + yy = a and at P(acos9, asin9) is xcos9 + ysin9 =a. 1 1 Prove that
~
- ; :
f
from the equation (I) and hence derive the equation
2 of the tangent at P(xl'yl) to the circle x + /
+ 2gx + 2fy + c
= 0.
EXAMPLE: (5) Find the equation of the tangent at A(-1, 3) to the circle x . C ( 4, T -7). SOLUTION: The centre IS
2
+/
- 8x + 7y - 39
= 0.
A is (-1, 3)
. 3 + 7/2 = lO - 13 Th e grad 1ent o f CA = :-y-:t;""
The tangent is perpendicular to the radius, so the slope of the tangent is
10 13
and
hence its equation is y - 3 = : ~ (x + I) , which simplifies to lOx - 13y + 49 = 0. Alternative method: We find the gradient of the tangent by differentiating the given equation: 10 :. 2x + 2y ~- 8 + 7 ~ = 0. At (-1, 3), ~ = etc. dx dx dx 13
296
Exercise 1OB 1.
Find the equation of the circle passing through the points: (a)
2.
(1,0), (0,-1),(0,0)
(b)
(-1,3),(2,2),(1,4)
Find the centre and the radius of the following circles: 2 2 2 2 x + y + 2x - 4y + 1 = 0 (b) 8x + 8y - 12x + 20y
(a)
=
3.
Find the equation of the circle whose centre is on the x-axis and which passes through the points (0, 3) and (4, 1).
4.
Find the equation of the circle whose centre is at the point C (-3, -4) and which is tangential to the line 3x + 4y = 20.
5.
Find the equation of the circle through the point A (-1, 2) and which is tangential to both axes.
6.
Find the equation of the circle centred on the line y = 2x, which passes through the point A (-2, 4) and is tangential to the x-axis.
7.
Show that the chord whose equation is x - 3y + 8 = 0 subtends a right angle at 2 the centre of the circle 9x + 9/ - 18x + 6y = 170. 2 Show that A (1, 1) lies on the circle x + / + 4x + 6y - 12 = 0. Find the co-ordinates of B, if AB is a diameter of the circle.
8. 9.
Find the equation of the circle passing through the origin and making intercepts a and b on the x- and y-axis respectively.
10.
Find the intercept made on the x-axis by the circle which has AB as diameter, where A is (0, -1) and B is (2, 3).
11.
Find the equations of the tangents to the following circles at the points indicated: 2 (a) x + / - 6x - 2y - 3 = O, (5, 4) (b) (c)
(x - 1)
x
2
+ y
2
2
+ (y + 2)
2
- 4x + 2y
= 5, = 20,
= mx
(3, -3)
(5, 3)
= a 2,
13.
2 Find the equations of the two tangents to the circle x + / - 2x - 6y + 6 = 0 which pass through the point P (-1. 2). Use lx + my + n = 0. (Warning: do not use y = mx +b)
14.
Write down the equations of the circles on AB as diameter where:
15.
(a)
A (4, -8), B (3, 5)
(c)
A (a, a), B (-a, -a)
+y
2
Show that the line y 2 b =~a /1 + m •
+ b
is a tangent to the circle x
2
12.
if
(b) A (a, b), B (b, a)
2 2 2 2 Show that the circles x + y = 4 and x + y + 6x - 8y + 16 = 0 touch externally.
297
10.3 Plane Geometry: Clrcles(Harder Problems) A
EXAMPLE: (l) ABC is an equilateral triangle, inscribed in a circle. X is a point on the minor arc BC. Prove that: (i)
t.BDX
(ii)
Ill
XB + XC
t.ACX
=
XA
SOLUTION: Given:
t.ABC is equilateral
Prove:
(i)
u
Ill
(ii)
t.ACX
XB + XC
XA
We have,
Proof: (i)
t.BDX
=z =60°
x =z
(t.ABC is equilateral)
and y = u ( Ls
in the same segments on arcs AB and AC respectively)
Then, in t.BDX and t.ACX: X
= y = 60°
p
=q
L BDX
(ii)
XB XA
( L s in the same segment, arc CX)
= LACX t.BDX Ill t.ACX BD AC
giving
( L sum of a t. being I 80°)
XB • AC = XA • BD
Similarly we can prove that t.CDX XC • AB
= XA
Ill
t.ABX and the corresponding result
• CD
••• (2)
Adding ( 1) and (2): XB . AC
+
XC . AB = XA (BD
+
CD)
Now t.ABC being equilateral, AB length, we have the required result: XB
+
CX = XA.
••• (I)
=
= BC
XA • BC
= CA, so, removing the common
298
EXAMPLE: (2) In the diagram, AD and BE are perpendicular to BC and AC respectively. Prove that: (a)
HDCE is a cyclic quadrilateral on HC as diameter
(b)
AH
= AK
Given: Prove:
ADlBC, BEJ.AC (a) HDCE is a cyclic quadrilateral (b)
=
AH
Proof:
LHEC
(a)
...
LHDC
AK
= 90° = 90°
(BE lAC) (AD .i BC)
E and D are on the circle with HC as diameter.
Hence, HDCE is a cyclic quadrilateral on HC as diameter. y =z
(b)
x
=z X :
(Ext. L of eye. quad. HDCE = Internal opp. L)
(L s in the segment, arc AB) y
t.AHK is isosceles AH
=
AK
EXAMPLE: (3) In the diagram, BC is a fixed chord of a circle, A is a variable point on the major arc on the chord BC. BD J.AC and CE (a)
j_ AB.
Prove that:
BCDE is a cyclic quadrilateral on a circle with BC as diameter.
(b)
as A varies, the segment ED has constant length.
(c)
the locus of the mid-point of ED is a circle whose centre is the mid-point of BC.
299
SOLUTION: Given:
BC is a fixed chord, BD .1 AC, CE .1 AB
Prove:
(a)
BCDE is a cyclic quadrilateral
(b)
ED has constant length
(c)
Locus of mid-point of ED is a circle.
(a)
LBEC and LBDC are both 90°.
Proof:
(BD .i AC, CE .lAB)
:. E and D lie on the circle whose diameter is BC. :. BCDE is a cyclic quadrilateral. (b)
Since chord
BC
is of constant length, it subtends a constant angle,
say a, at the circumference of the given circle. Now m =
L ABO
= 90° - a
(BD J_AD)
Since a is constant, m is also constant. Using the fact that equal chords subtend equal angles at the circumference of a circle, we conclude that for various positions of chord ED
on the fixed circle, (on
BC
as diameter)
ED
must be of
constant length. (c)
Let P and M be the mid-points of ED and BC respectively. Join MP and MD. M is the centre of the circle BCDE. MP J_ED MP:2
=
MD
2
- PD
Now r = MD= MC =
2
= r 2 - s2
1 2 sc = a constant
4
s = PO = ED which is a constant 2 MP is a constant Hence, the locus of P is a circle with the centre at the mid-point of BC and the radius
300
Exercise 10C: 1.
PLANE GEOMETRY: CIRCLES (HARDER PROBLEMS)
In Fig. 1, if PQ is parallel to RS, prove that PQ = RS.
2.
In Fig. 2, P is any point on a diameter AB of a circle; QPR is a chord such that
L APQ = 4.5°, prove that 2 2 2 AB :: 2PQ + 2PR • 3.
In Fig. 3, 0
is the centre of the circle,
prove that a + b 4.
=c.
Two lines OAB, OCD cut a circle at A, B and C, D. If OB = BD, prove OC =CA •
.5.
(Fig.2)
ABCD is a quadrilateral inscribed in a circle. BA and CD when produced meet at P. 0
is the centre of the
circle PAC. Prove that BD is perpendicular to OP. (See Fig. 4) (Fig.3) 6.
Two circles ARPB, AQSB intersect at A and B. PAQ and RAS are straight lines. PR and SQ are produced to meet at M. Prove that MPBQ is a cyclic quadrilateral. (See Fig• .5)
7.
AB is a chord of a circle. The tangents at A and B meet at T. AP is drawn perpendicular to AB. TP is drawn perpendicular to TA. Prove that PT is equal to the radius of the circle. (See Fig. 6)
8.
s
Two circles of radii 3 em and 12 em touch each other externally. Find the length of their common tangents.
(Fig.6)
301 9.
In Fig. 7, two tangents from T (to the circle) meet the two tangents from D (to the T
same circle) at A and C, as shown. Prove that AT - CD 10.
= TC
- AD.
The altitudes PM and QN of an acuteangled triangle PQR meet at H. PM produced cuts the circle PQR at A. Prove that HM
= MA.
(Hint: Join AQ)
(See Fig. 8) 11.
AD is an altitude of the triangle ABC, inscribed in a circle. DP is drawn parallel to BA and meets the tangent at A at P. Prove l CPA = 90°. (See Fig. 9) \Hint: Show that x
12.
= y = z)
AB is a diameter of a circle, AC is any chord. M is the mid-point of the arc BC. Prove that AC is perpendicular to the tangent at M.
13.
In Fig. 10, three circles intersect at P. Prove that AB is parallel to DC. (Hint: Join PF and PE)
14.
In Fig. 11, 0
is the centre, TP is the
tangent and TC bisects LOTP, prove that [TCP
= 45°.
(Hint: Join AP)
(Fig.10)
(Fig. II)
302
15.
Prove that the quadrilateral formed by the angle bisectors of a cyclic quadrilateral is also cyclic. (Fig. 11)
16.
If the two non-parallel sides of a trapezium
are equal, prove that the trapezium is cyclic (Fig.l2)
(Fig. 12)
17.
Prove that the angle bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral intersect at right angles. (Fig. 13) (We have to prove L FMP = 90° Hint: Produce FM as shown and prove AFDE
18.
Ill
AFBN
The bisectors of the opposite angles LP and L R of a cyclic quadrilateral meet the circle at A and B as shown. Prove that AB is a diameter of the circle. (Fig. 14)
(Fig.l4) 19.
Two circles intersect at A and B.
T
A straight line PAQ cuts the circles at P and Q. The tangents at P and Q intersect at T. Prove that PBQT is a cyclic quadrilateral. (See Fig. 15) (Hint: Join AB, x = y etc.)
{Fig.l5)
303 20.
AB and CD are parallel tangents to a circle, centre 0. APC is another tangent at P. Prove that
LAOC is a right angle.
(See Fig. 16)
21.
(a)
In Fig. 17 CP is a tangent and CAB is any secant. CP
2
Prove that
=CA. CB
(Hint: Join AP, show that t.PCA,
t. BCP are similar)
(b)
In Fig. 18 PQ is a common tangent to the two circles, and CAB is a common secant. Prove that CP = CQ. [Hint: Use the result (a) ]
(Fig.18) 22.
PQ and RS are the common tangents to two circles intersecting at A and
B. AB produced
(Fig.l9)
either way meets the tangents in M and N as shown. (Fig. 19)
23.
2
Prove:
= MA • MB and RN
(a)
MP
(b)
PM = MQ
2
= NB. NA
(See question 21)
(c)
PQ = RS
(Hint:
t.OEC : t.OFC)
(d)
MA = BN
(Hint:
MP = RN)
P is any point on the circle ABC. PL, PM, PN are the perpendiculars to the sides BC, CA and AB respectively of t.ABC.
Prove that the points
L, M, N are collinear. (Fig. 20)
(Fig.20)
304 p
24.
In the diagram, two fixed circles intersect at A and B. Prove that
L PMQ is of
constant size for all positions of Q. (Fig. 21)
25.
ABCD is a quadrilateral inscribed in a circle.
R (Fig.2l)
X, Y, Z, W are the mid-points of the arcs AB, BC) CD and DA respectively. Prove that XZ
26.
.L YW.
In Fig. 22, two fixed circles intersect at A
M
and B. P is a variable point on one circle. PA and PB when produced meet the other circle at M and N respectively. Prove that MN is of constant length.
N
(Fig.22)
27.
In the diagram (Fig. 23), 6ABC is inscribed in a circle. AD, BE, CF are the altitudes of 6ABC.
H, the point of intersection of
the altitudes,is called the ORTHOCENTRE of the triangle. Prove that: (a) (b)
l BHF = LBAC l BHC + LBAC
Ill Ill
=
180°
(c)
6AEF
(d)
6BDF
6ABC
(e)
AD bisects L FOE
(f)
l EDF = 180° - 2 l BAC
(g)
If BC is fixed and A varies remaining
6EDC
on the major arc on BC, find the locus of H. (h)
[Hint: Use part (b) ]
BFEC is a cyclic quadrilateral lying on the circle on BC as diameter.
(Fig.23)
305
10.4 Inequalities Definition:
>y
x
if and only if x - y
> 0.
On a real number line this means x lies to the right of y.
y
X
Worked Examples
1.
Prove that a ; b
Solution:
b are positive real numbers.
Consider
a+b2 1 2 2 1 2 Now ( 2-) - ab = 7i (a + 2ab + b ) - ab = 7i (a - b) ~ 0
~ -Jib,
a ; b
2.
> 0,
a
From example (1), a+b'l2 1
=i ,
Using (1),
From (2),
3.
, b
Given that x + y = c, prove that
Solution: Put
>0
for a
b
1
=y ,
then:
VxY ,<
~
VxY ,<
c
1
-y1
X
+
!
y
1 4 i1 + y 'l c for x > 0 , y > 0 •
vas
••• (1)
~ _2_
1 -~ VxY
or ~
!
= b.
••• (2)
VXY
and putting x+y = c
2
2
+
X
and equality when a
2
c
4
c
If x, y, z are positive real numbers, prove the following:
2
2
(a)
x
+ y 'l2xy
(d)
x +) +z
'l
X
v
(b) - + L. 'l2 y X
(c)
x
3
+y
3
+z
3
~
3xyz
(xyz)l/3 (cont)
306
Solution: (a)
Consider X
2
X
(b)
X
2
+y
2
+ y
+ y
2
2
2
~ 0
- 2xy
2
(x - y)
=
- 2xy
~ 2xy
Divide the result in part (a) by
xy,
then:
~+x.:V2
y
(c)
X
We have
x
3
From (a), x
+ y x
2
3
=
+ y
2
(x + y) (x ~
- xy
2
+ y
2
- xy)
xy
3 + y 3 >.,- (x + y) xy ~ xyz ( ~ + ~)
z
>,.xyz(~+i),
y3+z3
Similarly
z
Adding these: 2 (x
3
+ y
3
3 + z )
XV
VZ
ZX
~ xyz [ (- + .._)
+ ( L + -)
+ (- + -) ]
>.,.. xyz [ ( ~ + X.)
+ (X. + ~)
+ ( ~ + ~) ]
Z
Z
y
X
X
Z
X
y
y
X
y Z
Now using (b): 2 (x
3
+ y
Hence (d)
3 x
3 + z ) 3
+ y
>, xyz (2 + 2 + 2)
3
+ z
From (c), we have: Put
a
3
=x
x+y+z Hence
,
b
3
3
~
a
=y
3
3xyz + b
, c
3
3
3
~
,
then:
+c
=z
3abc
'-v 3 x 1/3 .y 1/3 .z 1/3
x +
J
+ z
~
(xyz)
113
The general result for a set of
,
equality for
n
Arithmetic Mean
= y = z.
n positive numbers
x 1 + x2 + .•• + xn
1.e.
x
>, Geometric Mean
x , x , ••• , xn 1 2
is
307
-Exercise 100 1.
(a)
Show that, for a
2
+ b
2
+ c
2
>0
a ~
, c
> 0,
ab + be + ca.
Hence show that a 2 + b
(b)
>0
, b
2
+
c 2 >,... 3 (abc) 2/ 3
[Hint: Use worked example 3 (d) ]
2.
Prove that for (Hint:
3.
(a + b + c) 2 >.... 3 (ab + be + ca)
Prove that
(c)
:r
x, y, z
0,
x + y + z
:r
[3 (xy + yz + zx) ] l/
Use question 1c)
Prove that for
a, b, c, d >,... 0, a + b : c + d >,... (abcd)l/ 4
. a+b c+d (Hmt: Let x = - - , y = - - , then use worked example 2 2 4.
A rectangular box of sides a, b, c
formulas S
= 2 (ab
(ii)
x, y
prove that ~ (x 3 + y 3) ~ (x ; y)3
> 0,
2
+
/> (z 2 + i>
>,
(xz + yu)
2
(Hint: Show that L.H.S. - R.H.S. 7.
~ 0)
For any real numbers x, y, z, u, prove that: (x
>,. 0)
For any real numbers x, y, z, a, b, c, prove that: (x
2
+ y
2
2
+ z ) (a
2
+ b
2
2
+ c )
Using the
(volume), prove that:
V is a maximum when the box is a cube.
(Hint: Show that L.H.S. - R.H.S. 6.
= abc
Let x = ab, y = be, z = ca, then use worked example 3d)
(Hint: If
repeatedly)
has a constant surface area S.
and V
+ be + ca)
(i)
5.
2
>, (ax
+ by + cz)
2
308
10.5 Method of Mathematical Induction Worked Examples I.
(i)
Show that for each positive integer n, there are unique positive integers a
(ii)
n
and b
= a n + /3.
such that (1 + Jj)n
-n
b
n
Hence show that a 2 - 3b 2 = 2n • (-l)n n n
Solution: (i)
.{3;
+ .fj. b 1 1 = 1 which are unique integers.
For n = I, L.H.S. = 1 + a1
= I,
b
1 Assume that (1 + o/'J)k
R.H.S. = a
= ak
+
o/1.
bk , where k is a positive integer.
(l + v'j) (1 + o/3)k
Then (1 + v'J)k+l
= (l + Jj) (ak +
.r::3)k+l ::: ( 3b ) r::3 ( b ) (1 + '" ~ + k + v" ak + k
=
VJ.
•.. (1)
bk), using (1).
ak+l +
and Since ak and bk are unique integers, so are ak+l and bk+l . Thus the statement is true for n = k + I. Since true when n Since true when n
= 1, =2,
the statement is also true for n
= 2.
the statement is also true for n :: 3, and so on, the
statement is true for all positive integral values of n. Rather than repeating the foregoing three lines, in the subsequent examples we will say "by the principle of induction, the statement is true for each positive· integer n". (ii)
To prove: a 2 - 3 b 2 n
n
= (- l)n
=a 12 -
• 2n
2
= I - 3 = -2, using part (i). 1 Also R.H.S. = -I • 2 = -2, so the statement is true for n = 1. Assuming it to
For n = I, L.H.S. be true for n
=k,
3b
we have:
a / - 3 b/ = (-l)k • 2k Then,
a~+l
- 3 b~+l
This simplifies to - 2
••• (2)
= (ak + 3bk)
(a~
a~+l-3b~+l =
-
3b~)
2
- 3 (ak + bk)
, using part (i).
and using (2), we have:
-2.(-l)k.2k
Hence the statement is true for
2
n
=k
= 2k+l(-l)k+l + 1 and, by the principle of induction,
the statement is true for each positive integer n.
309 2.
(b)
Hence, by induction prove that E 1
2n + 3
2
(2n + 3)
=
4n
>
2 ( .fi1:;l - 1)
1 .;J
+ ••• +
For n = 1, L.H.S.
=1
and
=
1+
1 ./2
Assume it is true for n
We add
+ 12n + 9 - 4 (n
1::'"7""1 v'il1 > 2 ( -vn + ,
2
+ 3n + 2)
=
1
>0
Vi
+ ••• +
-~
=k, 2(
••• (1)
- 1)
R.H.S. = 2 ( {'1. - 1) ~ 0.8
So, the statement is true for n
ff1
2
> 4 (n + 1) (n + 2) > 2 J(n + 1) (n + 2) +
1 +
I
• r.:vr
We have:
2 (2n + 3) - 4 (n + 1) (n + 2)
S(n)
.J(n + I) (n + 2)
n
(2n + 3)
(b)
>2
Show that for n
Solution: (a)
> 0,
(a)
= I.
where k is a positive integer, then: ~
vk
+ 1 - 1)
on both sides, then
V" + 1
s (k
+ 1)
>
2 v'i('"":;'"1 +
1
v1
Using part (a), S(k + I)
This simplifies to S(k + 1)
- 2 = 2k + 3 - 2
v'i)
('e)
/ X (
z
0
z-
(
z
/
A
(
(q) 9 X
!;""0
(l!)
x,
0
•r
£-
q
'e
0
zq + z-e q-
VN
~
= -edJ\1
·~
327
8 (a)
A
y
(b)
2
y
Max. A (- J , 5.6)
(c)
(d)
-· (e)
-1
9 (a)
0
y
X
~)_/;' _____ I_ 2_ .. I I
12
)X
(d)
-~.=..L
__ _
X
A
0 _j
0
X
I
\j
(!)
v.
~
X
7
r (4) !.
1I
X
,....1,....
X
/
//
I
~I
I
0
329 10 (c)
"T X
X
------~--~-~~-
(a)
- - --2 ·-
See Section 1.5, page 13, Example (c) (c)
(b)
X
(d)
X
(f)
X
2n
ll
3n
4n
f(x) = f (-x) = si~l x , the graph is symmetric about the y-axis. (h)
(i)
y
Y n/2
X
X
-n/2
331 12 (k)
12 (m)
13 (a)
(b)
-1 X
y
(d)
(c)
X
X (e)
y
(f)
(1nl4, 0.003)
i
0
X
I (3m4,- 0.067)
-x y = e COS X
,
0
4
5
~ X~
211
y = 0 for x = 11/2, 311/2 Minimum at (311/4, - 0.067) Maximum at (711/4, 0.003)
2n 6 .
:)
(d)
y
i\__/ A
I X
2
Min. at A (e-1 , -e-1)
e
Min. at
X A (e, e)
(f)
y
Max. at
1 A (e, e)
A
X
0
X
e)
2 2 Max. at A (e ,
y
(d)
-----,--1
~~
X
333
16.
y
0
,
U = x + I
~ Jog (x + ~ + lx 2 + x
+ I)
1)
Sin
- /1 - 2x- x
12.
V3
. -1 (2x + \;:-3-
I o.
2
U = x 2 + 2x + 3
sin 1 (~1 )
-
15.
tan-
1
3- tan-
r--2 3 sin-1 ( 2x-1 ) -ojx-x-+
18.
2
v'3
tar1 1 (~)
sin- 1 (x-l)
6.
+x + 1 +
3) Vtl) '
2
+ 2x + 4) 4. log (U +
+2x+4]
fx 4 - 3x2 +I)
+ 3 log (x + 1 + log
2
8.
v'39
2
v'3
+2x+4-log[x+I+Vx
V39
log (x
2.
2
2 /x +
log (x + 1 + .r;;r-;;x)
2x -
Exercise 21 I.
(a)
0
(e)
3
2 11
2.
(a)
4
3.
(a)
2
- 112
(b)
1/2
(c)
0
(f)
0
(g)
8
(b)
I 840
(b)
-1 .;ab tan
2
(d)
0
(h)
0
(d)
4
3
(c)
Ifa
11
96
11
(e)
1
2
344
Exercise 2J I.
32 (x - 2) 3/2
+ 4 (x - 2)
4. -0.059 7.
1/2 I
5.
2b
1og (a
8.
2 tan
I
-1
tan-! (.JX)
16
2.
+c
X
2
2 . 2
+ b sm x)
2
3.
e
6.
b tanx~ I tan-1 ( a ab
9.
x - log (I + ex)
m
2
- e + log (I + e)
2
10. x - 2Jog (I - ex) 13.
~
(! + x)5/2
-l
II. (I +
X
3.3
)3/2
,;;-:-;
14.
X
3
12.
X
X 3 Jog x- 9
15.
-I sec X
1
-1
1i tan
19.
(x +
I
2)
(x + ~)
25.
-1 [4 2 3 tan
27.
a sm
29. 31.
. -1
~
a
+ 5 ~an x/2
- /i-
x
J
2
V5 tan x- I I --log \1'5 tan x + I 2./5 - X + 5 tan-J (2 tan~)
3
6
33.
3 log (x - 2) - log (x + I)
35.
I 2 log
[~] 3 X
26.
12x TI - TI5 log ( 3 cos x
28.
2x - I -4-
30.
- (4 tan x + 6f 1
32.
2Vx sin
36.
I 2 log [-)
log (I + sin x) - log (2 + sin x)
40.
I 6 log
(I - cos x) +
4 3.
8 - 4
n
I
(I + cos x) -
44.
11
- I 83 log (-2x -2-
+
VP) p ' P =
X
2
-
X
+ I
rx + 2cos rx
2
39.
+
2
34.
+
2.I log
X
.;pP
.
+ 2 sm x)
8
+ 3x + 8 log (x - 2) - Jog (x - I)
~:21 I
38.
2 1og
2 3 log
(I + 2 cos x)
(x + 1) -
1
1i log (x
2
+ 1) +
1
2
-1
tan
x
345
CHAPTER 3 VOLUMES
Exercise 3A 20611
(1)
211
(2)
15
(4)
i
(5)
11
2
(7)
51211
IT
(8)
16
(IO)
4
(II)
T
(13)
15
(16)
(a)
144
(b)
36
(d)
72
(e)
36
(17)
11
-2 e
[e: - 2
+ 2]
51211
2
11
- 211 8
411 ab
11 30
(6)
11 11 (1- 4)
(9)
211 [(log 2) 2 - 21og 2
(12)
12811 -5
2
5611
(14)
(15)
411log2- 1.511
v'3
2 (18)
---r-
(3)
{c)
112
T
1811
40011 -3-
Exercise 38 (I)
108811 -15-
(2)
16211 -5-
(3)
2711
4011
(4)
2
(5)
3
(6)
(7)
811
(8)
6411
5
(9)
(10)
211
(II)
11
5 1611 -3811
3
11 (e 8 +4e 4 +3 )
4
Exercise 3C (2)
3211 (b)
(6)
(3)
3
81Th
15
81!
(c)
3 (8)
311
(4)
TO
6411
3
3211
(5)
4811
3
(9)
3 32na _!_5_
!3.25 m
~
I]
346
10.
2 2 2n a c
11.
3
16.
15. 4113 19.
.5na
(a)
J 16na -.5-
4na
3
---r
(b)
J20n
2
14.
117n 17. -5-
51211
15
3
-r
13.
20.
200n
2
8 ;
18. 112n
15
CHAPTER 4 COMPLEX NUMBERS
Exercise 4A ).
5 - i
2.
2
4.
4i
5.
2
III
6.
II - 23 i
7.
3 + 6i
8.
7 - 24 i
9.
2i
10.
2 - 25
II.
5i
12.
8
14.
19 + 4 i _2_9_
15.
-5
17.
X
18.
X
=3 , y =- 2
20.
X
14 23 = - 25 ' y = - 25
3 .
+ 50 I
13. 16.
- 2 + i
19.
(x
21.
2 + II i
(b)
TO
I
= I,
-
V3. 10 1
19 + 4 i 29
(b)
~
25.
2
(a)
+y I
5 4 , y =2
o
y -= -I) or (x : -1, y ' I)
(c)
X
3. +
2
3 . 2-2 1
13
5
2' y =- 5 2
22.
X
23.
(a)
(d)
I + IIi
(c)
=
X
10 + 3 i
2
+ ~
2
-
(x - 1)
(b)
7 .
9 2+2 1
2
-15 + i
2
3.
-
51 I
25
-2
(b)
2 + 2i
24.
(a)
1 - 2~i 2 + y
(d) (c)
X
X
2
5
-
25 .,fj i 2
-
(a)
2
-
iy
- y 2.
51
2
- 1 + 2xyi
347
~
~i
26.
25.
(d)
(c)
4-
(b)
x
(e)
5i - (12 + 4 i) x + 5
3 + 2
+
./'li
- 2x + 2 = 0
2 x - 4x + 13
(c)
=
30.
1.
2
=0
+ I
28. (a) z = 2, w = 3
0
z
i
3
=8 + 4
5- 5 1
(b)
3 + 4i
(e)
2 + 11 i
(f)
- 7 + 24 i
Va . Vb = ..raE
•w
i
3
=8 - 4
=I
z
(d)
4. 3 25- 25 1
(c )
(a )
- 6, Rule
x
2 x -(4+4i)x+l0i=0
(d)
0
v'Ji
.:!:
(b)
2
27. (a)
(c)
2
v'Ji
+
-
(d)
-4-
(b)
29 •
_!2
(a)
+ i,
w
= -1
78 96. 25 + 25 1
(d)
is not defined for the imaginary numbers.
Exercise 48 5cis 0
I.
2. 5. 8.
5cis n 2 Jicis (n /4)
3.
4.
5 cis (- n/2)
7.
2
I 0.
20 cis ( n/6)
11.
cis (2n/3)
13.
~
14.
2cis( 9) where
15.
2 cis(- n/2)
16.
5i
v"i CIS (3 n/4) cis(-n/4)
18.
-4
21.
..ti(~
24.
5 (I - .fj i)
y
~
+
2
.Ji cis(-
5cis n/2
6.
2 .ffcis (-11/4)
9.
8 cis (3 n/4)
12.
5 cis(- 2 n/3)
3n/4)
tanS=-~ 17.
and 9 in 4th quadrant 10 5
5..;3.
19.
-2i
20.
2--2-'
22.
I
23.
5
25.
2 (-I
+ +
~~
-
~v
.f3i)
Exercise 4C 1.
8i
5.
-2 + 2
9.
- .,fj + i
(c)
2 V'I cis (
(c)
- 8- 8
14. z (c)
1
./Ji ~;)
VJi
15. (a) z (c)
-10
3.
-6
6.
27 i
7.
-1-
10.
~(I
(d)
"; cis (-n/4)
(d)
512-512Vli
= 2cis(n/3), z
1024cis(n/2) 1
+
z2
- (2 +
=
2
- >/3i)
= 2cis(n/6)
V2 (1 (d)
+ v'3) cis (n/4) - 7 - 4
+ i)
.f3i
(a)
'+
32 cis(- n/3)
- 5
8.
2i
t/3 +
Si
1 . 7n (b ) \i'iClS T2
12. (a) cis ( n/2) 1.3. (a) - 4 V3
4.
4i
(b)
2i
(b)
16 cis (2n /3)
(e) cis(- n/6)
(d) cis(n/6)
113) i
.f2(1
2.
V3
(b)
z
1
- z
2
=
Vi ( '113 -
1) cis (3n/4)
348
Exercise 40 J.
2.
64i
3.
-I
5.
6.
-8-8VJi
7.
8cis(-11/2),-8i
8.
I cis9, w here 25
9.
4cis( 11),- 4
12.
2 ~6
tan 9
64
-24 (9 m . 3r d qua drant ) - 7 "--::'7 - 24 1. 625 625
10. 16cis(-211/3),- 8-8 t/3 II. 32cis(11/2), 32i 1 cis(-211/3), (- I - ~i) 13. 16cis(-211/3),- 8- 8 VJi 5 12
V3
14. 512 (-
18. 16cis(-11/3), 8-8 .IJi
17. 2cis 0, 2
i cis (-11/ 3), i
v'3 i)
(I -
V'i cis (-11/4), 8- 8i
15. 8
+ i), 1024 cis(511/6)
16. 32cis0,32 19.
4.
20. 9 cis ( 11/2), 9 i
Exercise 4E J.
(a)
(d)
{2(1- i)
.± 2
3.
4.
5.
:t
-f
.±
cv'J + i)
(c)
2cis(- 511/12)'
/5cis (n + 9/2), where tan 9
.f5cis(9/2),
vr3
2.
(b)
. + ~
2cis (711/12)
= ~ , Cartesian
v2
- i,
(c)
3cis
(a)
_I_ (+ l + .)
(d)
cis(k;- ; ) . k = 0,1,2,3-+ cis (-15•, 75•, 165•, -105°) 1
(a)
z = 2cis
(b )
Z
o
2CJS - -- ; , 5
(c)
z
o.
cisQ;
(d)
2 z = 2cis( ;
(a)
r
(b)
- 0.820 + J.02i, 0.472- l.22i 2 11 - 3, i (c) cis n ~ k
(d)
2cis ( k; -
(f)
.:!: 2, (.:!: 1 .:!: .fS i) (six roots)
2k11 11 - - + 9 3
.J2-
,
II 3
. Cis
11 -
11
VJ - i
, I +
J3 i,
+
i),
.fJ +
.r::
v3- i
i, - 1 - ,fj i
to) , k = 0,1,2,3,4 r7"
.
where r - v 5, t , C(O, 1), r = 2
(e)
X
Locus of z is the perpendicular bisector of OA, A (I, 0), 2x = I
The circle x2 + (y + 2)2 = 9 C(O, -2), r = 3
,,A
(i) I
y
Locus of z is the perpendicular bisector of AB, A(2, 0), B(O, -1), 4x + 2y = 3 (I)
I
l
P(z) /
X
x' Locus of z is the perpendicular bisector of AB, A (-2, 3), B(-2,-1). y = I
2 2 x4 + ~ = I
The locus is a circle Locus, a circle (x + 4/3)2 + (y + 2/3)2 (x + 2)2 + (y + 7/3)2 Locus is the ellipse = 20/9. C(-4/3, -2/3), = 64/9, C(-2, -7/3), Centre (0, 0), Foci (~ I, 0)
r
=
V25/3
r
y
(m)
= 8/3
~y
(n)
Ellipse Centre (0, 0) Foci (0, ."!:I)
P(z)
X
2
2
2.
!:(,)
(a)
( 0
I
(x - 1/2) 4
2
2 _y__ +15/4=
~~ (c)
(b)
y
~
2
Foci (0, O), (1, 0)
X
~ + L 3 4
Ellipse 1 Centre < , 0)
y
)
/3 .. X
Locus is half ray along the line y = VJx excluding the point 0(0, 0), y > 0
y
5n/6
A\-2,0) 0 Locus is the half ray Locus is half ray AP along the line y = -x where A (-2, 0) and excluding 0(0, 0) and y > 0 excluding A y < o. < PAx = 5n/6 < POX = -n/4 x+y J3+2=0
Locus is half-ray from 0(0, 0), y ) 0, X = 0 excluding 0.
355
.t~
(e)
~ox
y
(f)
Locus is half-ray OP, X< 0, y = 0
0
}(z)
)f~~--
Locus is the half-ray AP, A (0, -2), excluding A. Y= JJx-2
X
3. (d)
(a)
•
1
y
~T~h~e~se~t~o~f~p~:~i~:~ts~ ~The:~ ~i":' ~"'ide'
c -~·
The set of points of The set of points within outside the circle of outside the circle the circle, C (2, 0), the circle of radius 2, radius 3, not including centre 0(0,0) excluding radius 3, centre 0(0,0) C(O,J), radius 2, not including the boundary including the boundary the bounCiary. the boundar (e)
0
The set of points inside the circle, C(-2,-3), radius 2, excluding the boundary
The set of points The set of points between two conbetween two concentric circles of centric circles, centre radii 2 and 3, centre C(-2, 1), radii 2 and 0(0, 0), including the 4, including the boundary of the larger boundaries circle, but excluding the boundary of the smaller circle.
The set of points within the angular region
