Sixth Form Pure Mathematics. Volume 2

412 Pages·1963·11.879 MB·English

by  C. Plumpton and W. A. Tomkys (Auth.)

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Sixth Form
Pure Mathematics
VOLUME TWO
C. PLUMPTON
Queen Mary College, London
W. Λ. TOMKYS
Bi//c Fue Bofî Grammar School
Bradford
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First edition 1963
Reprinted (with corrections) 1967
Reprinted 1971, 1977, 1979
Library of Congress Catalog Card
Number 61-10010
Printed in Great Britain by Aberdeen University Press
ISBN 0 08 009383 3 flexicover
PREFACE
IN this volume the pupil is introduced to inverse trigonometric functions,
to hyperbolic and inverse hyperbolic functions and to a new range
of mathematical methods including the use of determinants, the mani-
pulation of inequalities, the solution of easy differential equations and
the use of approximate numerical methods. Complex numbers are defined
and the various ways of representing them, and of manipulating them,
are considered.
Polar coordinates, curvature, an elementary study of lengths of curves
and areas of surfaces of revolution, a more mature discussion of two-
dimensional coordinate geometry than was possible in Volume I, and
an elementary introduction to the methods of three dimensional co-
ordinate geometry comprise the geometrical content of the book.
Throughout, the authors have tried to preserve the concentric style
which they used in Volume I and the many worked examples and exer-
cises in each chapter are designed or chosen to provide a continuous
reminder of the work of the preceding chapters.
The aim of the book has been to provide an adequate course for
mathematical pupils at Grammar Schools and a useful introductory
course for Science and Engineering students in their first year at Uni-
versity or Technical College or engaged in private study. Except for Pure
Geometry, the two volumes cover almost all of the syllabuses for
Advanced Pure Mathematics of the nine Examining Boards.
The authors wish to thank the authorities of the University of London,
the Cambridge Syndicate, the Oxford and Cambridge Joint Board and
the Northern Joint Board for permission to include questions (marked L.,
C, O.C., and N. respectively) from papers set by them.
C. PLUMPTON
W. A. TOMKYS
CHAPTER XI
LINEAR EQUATIONS AND DETERMINANTS
11.1 The solution of linear simultaneous equations in two unknowns
If
ax + by + c = 0,
x x x
ax + by + c = 0, (11.1)
2 2 2
where none of a , a, 6 , b, c , c is zero, then
x 2 X 2 x 2
62(043? + ^y + c ) — M^a; + b y + c) = 0.
x 2 2
.·. x(ba — 6^2) = b c — bc.
2 x x 2 2 x
r.x = (bc - bc)j(ba ~ M2) (H·2)
x 2 2 x 2 x
unless a^ — 6 α = 0. Also
χ 2
a(ax + by + c) — 04(0^ + 6 s/ 4- c) = 0.
2 x x 2 2 2
... y = _ ( c - ac)j(ab - a«^) (11.3)
ai 2 2 x x 2
unless a 6 — «2^1 = 0· These results can be shown by substitution to
1 2
satisfy the original equations (11.1). We write the solution in the form
x -y 1
(11.4)
bc — bc ac — ac α^ — ab
x 2 2 x x 2 2 x 2 2 x
and, at this stage, it is best remembered in the form
x —y 1
(11.5)
(X2 C2 a2 62
where the arrows indicate cross-multiplication and each set of coefficients
in the denominator is obtained by "covering-up" the column in the
equations corresponding to x, y and the constant respectively. __
(a) Iïab —ab = 0, eqn(l 1.2) is meaningless unless also b c — bc =0
x 2 2 x x 2 2 x
and eqn (11.3) is meaningless unless also a c — ac = 0. In fact, if
2 x x 2
ab — ab =0 and one or both of (b c — bc), (ca — ca) are
x 2 2 x x 2 2 x x 2 2 x
non-zero, equations (11.1) have no solutions.
2 PURE MATHEMATICS
(b) If (a b — a b) = {bc — &2ci) = (a2ci ~ a\c2) == 0, then
x 2 2 x x 2
«j/ag = bjb = Cj/02 and the original equations (11.1) can assume
2
identical forms if one of them is divided by a numerical factor. In this
case eqns (11.1) have an infinity of pairs of solutions in which x can
take any arbitrary value and y = ( — ax — c^jb-y.
x
(c) If a = 0, b φ 0, a φ 0, then y = — cjby.
x 1 2
.·. a x — 6ci/&i + c2 = 0.
2 2
·'·# = — (&iC2 — &2Cl)/a2^1
and this result agrees with the one we have already obtained.
Similar results are obtained in other cases in which one of a , α , or
x 2
one of b b is zero. If a = a = 0 or if b — b = 0, one equation
lf 2 x 2 x 2
becomes trivial or untrue.
If one of a, a is zero and one of b , b is zero, the solution is evident
x 2 x 2
and in each case it is included in the general solution obtained above.
Geometrical Interpretation. The eqns (11.1) each represent a straight
line. The lines intersect at one point [Fig. 101 (i)] if a b — α^φθ.
1 2
The lines are parallel [Fig. 101 (ii^ifa^ — α2^ι ^ 0 and a c —a c 4=0.
x 2 2 x
The lines are coincident [Fig. 101 (iii)] if a b — a b =0 and
x 2 2 x
a c — a c = 0.
1 2 2 1
Examples, (i) Solve the equations
23* - \ly - 5=0,
19* + 2ly + 18 = 0.
FIG. 101 (i).
LINEAR EQUATIONS AND DETERMINANTS 3
FIG. 101 (ii).
FIG. 101 (iii).
y 1
- 17 x 18 + 5 x 21 23 x 18 + 5 x 19 23 x 21 + 19 x 17
,·.χ = -201/806, y = -509/806.
(ii) Solve the equations
(a + l)x - Sy = 4a + 3,
2x — (a + l)t/ = 5a — 1,
and find the values of a for which the lines represented are parallel.
Discuss the case a = — 1.
4 PURE MATHEMATICS
We write the equations as
(a + l)x - 3y - (4a -f 3) = 0,
2x - (a + 1)2/ - (5a - 1) = 0.
Then
x -y
3(5a - 1) - (a + 1) (4a + 3) - (a + 1) (5a + 1) + 2 (4a + 3)
1
-(a + l)« + 6
unless (a -f l)2 = 6, i.e., unless a = — 1 ± ^6.
4a2-8a 4-6 5a 2 - 2a - 5
*'* * a2 + 2 a - 5' y a2 + 2a - 5
unless a = — 1 db ]/6
If a = — 1, the lines are parallel to the x and y axes respectively and
they intersect at ( — 3, — ).
Two important cases of elimination,
(i) If ab2 — ab =+= 0, the equations
x 2 x
ax2 4- bx + Cj = 0,
x x
αα;2 + 6 x + c = 0,
2 2 2
have a common root a (say) if
a(x2 4- ί^Λ -f- c = 0,
t x
α Λ2 4- 6 Λ 4- c = 0,
2 2 2
oc2 —oc 1
i.e., if
bxC2 — &2C1 aiC2 ~~ a2Cl αΐ^2 ~ a2^J
i.e., if x2 *iCa —fcaCj
a 6 — ab '
x 2 2 x
i.e., (c^c, - ac)2 = (Ojfc, - a 6 )(6 c - kjcj.
2 1 a 1 1 s
(ii) If a cos 0 4- 6 sin Θ 4- c = 0,
x X x
a cos 0 4- b sin 0 4- c = 0,
2 2 2
then
cos 0 — sin 0 1
&1 C2 — 62Cl «1C2 — «2 Cl ai ^2 ~~ a2 ^1
LINEAR EQUATIONS AND DETERMINANTS 5
Hence the identity sin20 + cos20 = 1 gives
bxc2 — &2cix2 1.
<hf>\) \«1^2— αΦ\Ι
Examples, (i) Show that the result of eliminating 0 between the
equations
x = (4 - &)cos0 + 2 sinfl, y = 2cos0 + (1 - &)sin0,
in which k is constant, is generally an equation of the second degree
in x and y. Show, however, that there are two values of k for which
the éliminant is of the first degree in x and y, and find the éliminant
in all cases. (N.)
We have
2 sin0 + (4 - k) cosO - x = 0,
(1 - &)sin0 + 2cos0 - y = 0.
sin 0 — cos 0 1
' - (±-k)y + 2x -2y + x(l-k) 4- (4 - k) (1 - k)
if 4 — (4 — k)(l — k) φ 0, i.e., if k2 — 5k φ 0, which is so provided
k 4= 0 and & 4= 5. Hence in general
{-2y + x(l - k)}* + {2* - y(4 - &)}2 = {4 - (4 - *)(1 - k)}*
which reduces to
x2(k2 - 2k + 5) + xy(Sk - 20) + y2(k2 - 8ß + 20) = {&(& -5)}2.
This is the éliminant in the general case and it is an equation of the
second degree in x and y. This solution is not valid in the cases k = 0
and k = 5 which must now be considered separately.
If & = 5, the original equations become
x = — cos 0 + 2 sin 0, y = 2 cos 0—4 sin 0
and the éliminant is y = —2x.
If k — 0, the original equations become
a: = 4 cos 0 + 2 sin Θ, y = 2 cos 0 + sin 0
and the éliminant is a: = 2y.
x2 y2
(ii) As the parameter k varies, the equation 4- -j- = 1 represents
a variable conic 8 . Prove that there is just one value of k for which S
k k
touches the line
x cosa + y sinoc = p.
6 PURE MATHEMATICS
If now this line moves parallel to itself, prove that the locus of the
point of contact is
x2 - 2xy cot 2oc - y2 = a2. (N.)
In this example the word parameter is used to mean the variable
element in the equation of a family of curves. Hitherto we have used it
to mean the variable element in the coordinates of a family of points.
In Vol. I we showed that y = mx -\- c is a tangent to x2\a2 ± y2/b2 = 1
if c2 = a2m2 ± b2. Hence y = — cota; + pcoseeoc is a tangent to
χ2 y2 .,
Λ
—z r + -;— = 1 if
a2 -f k ^ k
p2 cosec2 oc = (a2 + k) cot2 oc + k,
i.e., if
k(l + cot2<%) = p2 cosec2 oc — a2 cot2 oc,
which reduces to
k = p2 — a2cos2<%.
This is a unique value of k since a, p, oc are constants. Hence there is
just one value of k for which S touches the line x cos oc -f y sin oc = p.
k
If the line touches the conic at (χ , y) then, by eqn (3.9) of Vol. I,
λ x
the equation of the line is
x*i , Mi
= χ
a2 + k k
But the equation of the line is
x cos oc + y sin oc = p
and hence by comparing coefficients
xi Vi J(_
= =
(a2 + k) cos oc k sin oc p
But
k = p2 — a2 cos2&.
.·. px = (a2 -f p2 — a2 eos2<x)eos oc = p2 cos oc -\- a2 sin2oc cos oc.
1
.·. p2 cos oc — px + a2 sin2a cos oc = 0.
x
Similarly p2 sin oc — py — a2 cos2& sin a = 0.
x
. P2 ^JP
= =
a2 x cos2 a sin oc + a2y sin2 a cos a — a2 cos3 a sin oc — a sin3 Λ cos a
x 1
_ 1
— y cos oc + χ sin a
x λ