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. qp
Prove also that OA = CA when tan 6 =
. qp 23. On a fixed diameter of a circle of radius 6 in., and on opposite sides of the centre 0, points A,B are taken such that AO = 3 in., OB = 2 in. The points A,B are joined to any point P on the circle. Prove that, as P moves round the circle, AP + BP takes minimum values 13 in. and 11 in., and takes a maximum value 5^7 in. twice. 24. Prove that, for real values of x, the function 3 sin a? cannot have a value greater than ^/3 or a value less than —
32
LOGARITHMIC AND EXPONENTIAL FUNCTIONS
25. A particle P falls vertically from a point A, its depth below A after t seconds being at2, where a is constant. B is a fixed point at the same level as A, and at a distance b from A. Prove that the rate of increase of Z.ABP at time t is 2abt
and show that this rate of increase is greatest when LABP = 30°. 26. The angle between the bounding radii of a sector of a circle of radius r is 0. Both r and 0 vary, but the area of the sector remains constant and equal to c2. Prove that the perimeter of the sector is a minimum when r = c and 0 = 2 radians. 27. If a variable rectangle has a diagonal of constant length 10 inches, prove that its maximum area is 50 square inches. 28. A straight line with variable slope passes through the fixed point (a, 6), where a, b are positive, so as to meet the positive part of the o;axis at A and the positive part of the yaxis at B. If 0 is the origin, prove that the minimum area of the triangle OAB is 2ab. Find also the minimum value of the sum of the lengths of OA and OB. 29. Prove that of all isosceles triangles with a given constant perimeter the triangle whose area is greatest is equilateral. 30. A variable line passes through the point (2,1) and meets the positive axes 0X,0Y at A, B respectively. If 0 denotes the angle OAB (0<6< ^77), express the area of the triangle OAB in terms of 9, and prove that the area is a minimum when tan 0 = J. Find also the value of 0 if the hypotenuse of the triangle is a minimum. 31. Find the turning points on the graph of the function 2x stating (with proof) which is a maximum and which is a minimum. Sketch the graph and find the equation of the tangent at the point on the curve where x = J.
REVISION EXAMPLES III
33
32. The slant height of a right circular cone is constant and equal to Z. Prove that the volume of the cone is a maximum when the radius of the base is 33. A lighthouse AB of height c ft. stands on the edge of a vertical cliff OA of height b ft. above sea level. From a small boat at a variable distance x from 0 the angle subtended by AB is 0. Prove that tan 0 = —r . Prove also that, if a is the maximum value of 0, then c Integrate with respect to x: 2x—7
34
®
35.
(i)
^ COB«2*,
36. w(i) 5 = i4* cos * 37
(i)
38.
(i) 3 ^
4x Bx2_2x_v
(ii) v(ii)
'
a;tan 1 a;, (iii) » v/ ( )
7r 2 n a ; + 6>
i
2
(ii) ^sina:,
8
(iii) (a*
39. Integrate with respect to x: cos3 a; sin2 a;' Evaluate
—
^rr^—TT dx
J 2 (2a;3)(2a;+1) to three significant figures, given that log10e = 04343. 40. Interpret by means of a sketch the definite integral tx2)dx9 Jo Jo
and evaluate this integral.
34
LOGARITHMIC AND EXPONENTIAL FUNCTIONS
41. Integrate the following functions with respect to x: (I* 2 )*'
x+1
*
2
f xlog xdx. e
Evaluate
Ji 42* By a suitable change of variable, prove that f»»
dx
dx
r*
Jo l + sin#
Jo
and by means of the substitution t = tan^a;, or otherwise, evaluate one of these integrals. 43. Integrate with respect to x:
44. Integrate with respect to x:
[In
sii Jo sin 3# cos 2x dx.
Evaluate
45. Explain the method of integration by parts, and employ it to integrate ^(1 — x2) with respect to x. Integrate with respect to x sin*"1^ 171 1\> ](lX2)
1 Ti
2T#> (1X2)$
.
.
ex(cosxsmx).
46. Integrate with respect to x: .
5
x2+l
47. Integrate with respect to #: 1 sin2 # + 2 cos2 #'
1 a; 3 +l #
By integration by parts, or otherwise, integrate sin1 x.
REVISION EXAMPLES III
35
48. Integrate with respect to x: 1x' Prove that
(1 — cos2 x)2 sin x,
,
xex.
^ &x — s i n ~ x ^
~r
J3 y(1 — X )
49. By integration by parts, show that roc
Jo is equal to
a — sin a, 3
a 1 fa and also to 777—«rr xBsin(ot — x)dx. 3! 3!J 0 From graphical or other considerations prove that, if 0 < a < 77, then a  3 I xzsin(aL~x)dx< x dx, Jo Jo /•a
and deduce that
I x3 sin (oc — x)dx< Ja 4 , Jo
50. Integrate with respect to x:
Evaluate
I xsinxdx. Jo 51. Integrate with respect to x:
smSxcosSxdx. 0
52. Integrate with respect to x: , Evaluate
Jo
cos3 x sin2 xm
x2x2sir sinxdx.
36
LOGARITHMIC AND EXPONENTIAL FUNCTIONS
/ 3\2 53. Integrate 11 +^l with respect to x, and evaluate s
Jo
54. Evaluate the definite integrals: Jo If
cos2 2a; da;,
^w = I xnsinxdx9 Jo
#
Ji
and%>l,
show that
and evaluate w4. 55. (i) Prove that, if In =
seGnxdx,
(n — 1) In = tan # sec w " 2 a; + (7i — 2) ln_2.
then
Use the formula to evaluate
Jo (ii) Find the positive value of x for which the definite integral rx i
dt
Jo 7(
is greatest, and evaluate the integral for this value of x. 56. Prove that — (sinm+1 x cos71"1 x) ax = (m + n) sin m x cos71 x — (n — 1) sin m x cos71"2 xdx, and deduce a formula connecting [In
Jo
smmXGO&nxdx,
[in
Jo
sin™ x cos71"2 xdx.
Evaluate sin 3 x cos 5 x dx, Jo
sin 3 #cos 5 a;d#. J j?r
KEVISION EXAMPLES III
37
57. Prove the rule for integration by parts and use it in finding eaa; cos cxdx,
I eax sin2 bxdx.
a
If In denotes C (a2 — x2)ndx, prove that, if n > 0, Jo
58.
(i) Find a reduction formula for n (l+x2)n+*dx, Jo and evaluate the integral when n = 2. (ii) By integration by parts, show that, if 0 < m < n, and
Deduce that
r
/ = 0.
xm~x
dn~1 n \x {\  x)n}dx. ax
CHAPTER VIII T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S 1. A series giving sin A;. By repeated application of the method given in Volume i (p. 53), we may establish that, when x is positive, sin a; lies between the following pairs of functions: X
(i)
X3
(ii) (iii)
X—
X X—
X3
(iv)
3!
x3
and
of'
and
3
3!
+
X5
IT
and
x>
x~{ + 5! 71 and
Xs
X5
xs
X5
X3
X5
*3!
H
x> 7!' x"1 a;9 ~7!Hh9!'
A simple inductive step completes the argument. Moreover, the results as stated are equally true when x is negative, though the directions of the inequalities must then be reversed; for example, if x is positive, then 3 X >
OXXX si/ *r •*/
77
9
OI
whereas, if x is negative, x* x< smx<x — —. o!
In either case sin# is between x and x ——. There is, however, a more significant form in which the statements may be cast: \x I3 (i) sin# differs from x by not more than ^A, where  x \ stands for the numerical value of x; (ii) sin a; differs from \x\ by not more than —•j;
x3 ^"""ol
A SERIES GIVING SIN X #
(iii) sin x differs from
\x\ by not more than —~;
3
25
x — —r + —. 61 5!
£3
(iv) sin a; differs from la;!9 by not more than ^—j;
39
/p5
z^7
#^7+£7^ oI
oI
71
and so on. We are therefore led to a series
/^.2tl1
whose nth term is ( — I)711 — —, with the property that sin# v^  1 ) 1
differs from the sum of the first n terms by less than Let us examine this 'difference' term j  ~  — — , writing it in the form
I I I I I I I I
II
(2n{1)! II
~F*~2~*~3~#~4~*"#"272
Suppose that x has some definite value, positive or negative. If we 'watch' n increase a step at a time, there will come a point when 2n+l exceeds \x\. Thereafter, the later factors in the product are less than 1; moreover the factor
tends to zero as n continues to increase. By taking n sufficiently large, we may thus ensure that the value of sin# differs from that of the sum of the first n terms of the series
by as little as we please. In that case, sin x is called the 'sum to infinity' of the series, and we describe the series as an expansion of sin x in ascending powers of x.
40
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S EXAMPLES I
1. Complete the inductive step in the argument to prove that sin a; lies between
and
/y.3
zy.5
^3
/^
/y.2fl.— 1
'si+si—•+(1
2. Obtain the expansion for cos a; as a series of ascending powers of x in the form x2
x2n~2
x* x«
2. A series giving 1/(1 + JC). Consider the sum
Sn= lx + x2xs+... + (l)n1xn\ consisting of n terms. By direct multiplication, we have xSn= so that or
x x2 + x*...
+ (  1 ) "  2 ^ " 1 + (1)* 1 **, 1 + (  I)* 1 **,
(1 + x)Sn = ^nn = ,
+ )L—r
1+x
•
l+x
Hence
( — l)nxn __'y3 4_ j _ / _ _ l \nl L —Us  r . . .  r i )
~nl *"
as is probably familiar. We have therefore obtained a series 1 — x + x2 —
4. i
v
/ ^ !
>
\ +x xs+...,
whose nth term is ( — I)nIxn""19 with the property that 1/(1+a?) differs from the sum of the first n terms in this series by precisely the amount \ xn/(l+x) . Consider, then, the 'difference' term \xn/(l+x) . When x lies between —1,1, so that — 1 < x < 1, this difference can be made as small as we please by taking n sufficiently large, and so we may ensure that, when  x \ < 1, the value of 1/(1 +x) differs from that of the sum of the n terms by as little as we please.
A SEBIES GIVING 1/(1+#)
41 n
On the other hand, when a?>l, the difference \x /(l+x) \ becomes larger and larger with increasing n, and the value of so far from approximating to 1/(1+#), oscillates wildly as the number of terms increases when x is positive, and increases beyond all bounds when x is negative. We have therefore obtained a series which represents the function 1/(1+ x) for a certain range of values of x (namely —1<#<1), but whose sum has no value when  x \ > 1; in contrast to the series x3
x5
x7
which was shown (p. 39) to be an expansion for sin# for all values of x. The intermediate values x= + 1 , x = — 1 require separate consideration: When x = + 1 , the series is whose sum to n terms is oscillating, being 1 when n is odd and 0 when n is even. When x = — 1, the series is 1 + 1 + 1 + 1 + ..., and the sum of the first n terms increases indefinitely as n increases. In neither of these cases can we assign a meaning to the sum 'to infinity'. 3. Expansion in series. The two examples given in §§1,2 illustrate the way in which a function f(x) can be expanded as a series of ascending powers of x in the form possibly for a restricted range of values oix. They suffer, however, by referring to very particular functions, and the treatment
42
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S
given for sin a; and 1/(1+ x) leaves us with no idea of how to proceed in more general cases. In the next paragraph we shall give a formula for the coefficients tt o»ai>a2> i n terms of f(x) and its differential coefficients, and later we proceed to a more detailed discussion. First, however, we must say a few words about the meaning of the 'sum to infinity* in general; for a fuller treatment, a textbook on analysis should be consulted. Suppose that we have a series whose successive terms are, say, uvu2,uz, ...,un,.... It may happen that the sum of the first n terms ~ 5
tends to a limit S as n tends to infinity. I For example, if x=$= 1, xn 1 _/y
so that, for this series, 8n = and, if  x \ < 1,
1 X ~~ X
Xx
1
v>
n
,
JL ""^ X
S = lim Sn = ^—.)
In this case we call S the sum to infinity of the series. We write 8
and say that the series converges to 8. On the other hand, if 8n does not tend to a limit as n tends to infinity, the series has no 'sum to infinity', and the expression
has no arithmetical meaning. If the terms uvu2,... of the series depend on x (as in the particular example just quoted) so also does the sum 8n of the first n terms, and the sum to infinity when there is one. We shall be concerned exclusively with series of the form ...+anxn+..., where the coefficients ao,al9... are constants, and we have seen (§§ 1, 2) that such a series may converge for all values of X or for some values only.
EXPANSION IN SEEIES
43
Note. It is important to realize that the word 'converge' implies definite tending to a limit. A series such as for which 8n = 1 when n is odd and 0 when n is even, has a finite sum for all values of n, but does not converge. The series is said to oscillate boundedly. 4. The coefficients in an infinite series. We first assume that expansion in an infinite series is possible, and seek a formula to determine the coefficients: To prove that, iff(x) is a given function which CAN be expanded in 2 ejorm f(x)==a + + ...+a xn +..., n
an = —r^> n n\ where f(n)(0) is the value offin)(x) when x = 0. We assume without proof that (in normal cases) we can differentiate the sum of an infinite series by differentiating the terms separately and adding the results, as we should for a finite number of terms. Then then
f'(x) = f"(x) = and so on. Putting x = 0, we have successively /(0) /'(0) /"(0) /'"(0) /«v>(0)
= a09 = al9 = 2a2, = 3.2a3, = 4.3.2.a4,
and, generally, /<*>(0) = n(nl)...Z.2an
= n\an
Hence
Note. This work gives no help about whether the function CAN be expanded; for that we must go to the next paragraph. 42
44
TAYLOB'S SERIES
AND A L L I E D
RESULTS
ILLUSTRATION 1. A 'particle falls from rest under gravity in a medium whose resistance is proportional to the speed. To find an expression for the distance fallen in time t. If x is the distance fallen, then the acceleration is x downwards; the forces acting per unit mass are (i) gravity, of magnitude g downwards, (ii) resistance of magnitude kx upwards. Hence
x = g — kx.
The formula just given, when adapted to this notation, is t2 .. X =
t3 ...
XQJT tXQ + jry XQ + — XQ + . . . ,
where xQ, x0, x09... are the values of x9 x, x,... when t = 0. From the initial conditions, x0 = 0, xo = 0, so that
x0 = g — kx0 = g.
By successive differentiation of the equation of motion, we have x o = Jcxo = Jcg, XQ
=
KXQ
= /c g9
and so on. Hence
The series converges for all values of t. EXAMPLES II
Use the formula of § 4 to find expansions for the following functions: 1. sin#.
2. cos#.
3. !/(!+«).
5. Taylor's theorem. We come to a somewhat difficult theorem on which the validity of expansion in series can be based. Suppose that f(x) is a given function of x, possessing as many differential coefficients as are required in the subsequent work.
T A Y L O R ' S THEOREM
45
To prove that, if a, b are two given values of x, then there exists a number £ between a, b such that, for given n9
Write
(ft*)* (a?)
~
where Rn is a number whose properties will be described as required, and k is a positive integer to be specified later. We propose to use Rolle's theorem (Vol. I, p. 60) for F(x) exactly as we did (Vol. I, p. 61) for the mean value theorem, of which this is, indeed, a generalization; we therefore want the relations F(a) = F(b) = 0. By direct substitution of b for x in the expression for F(x), we have
F(b) = 0.
In order to obtain the relation F{a) = 0, we substitute a for x on the righthand side and equate the result to zero; thus
0 = ftb)f(a) (ba)f'(a)^^f"(a)
...
We must therefore give to En the value
RnAb) /(a)  ( 6  a ) / » J±
We have now ensured the relations = F(a) = 0, and so, by Rolle's theorem, there exists a value £ of x between a, 6 at which F'{x) = 0.
46
TAYLOR'S SERIES
AND A L L I E D
RESULTS
The next step is to evaluate F'(x). This will involve a number of terms of which
is typical, and the differential coefficient of this term is
Hence, remembering that Rn is a constant, we have F'(x) =
f[x){{bx)f"{x)f(x)}
(n2)\J ~
(nl)\J
J
{X {X)+
{baf
(X) +
j
(ba)*
"n
Kn>
after cancelling like terms of opposite signs. But there is a number £ between a, b for which F'(g) = 0, and so M(6q)(6fl
Equating this to the value of Rn obtained above, we have
The value of k is still at our disposal. If we put k = n (as we might have done from the start, of course, had we so desired) we have (6q)»
TAYLOR'S THEOREM
47
and the result enunciated at the head of the paragraph follows at once. This gives us the most usual form for Taylor's theorem, and we have adopted it in the formal statement; but there are advantages in keeping k more general, as we see below. The theorem is therefore established. The expression
is called the remainder after n terms. It is, in the first instance, what it says, namely a remainder, the difference between
Kb) lh — n\nl
(h — n\%
and
/
The theorem just proved enables us, however, to express this remainder in the suggestive form (with k = n) >(0 by choosing £ suitably. This expression is known as Lagrange's form of the remainder. By giving k other values, we obtain various forms for the remainder. In particular, when k = 1, we have =
"~
(6q)(6g)ni J (n1)!
( }
*
This may be expressed alternatively by writing f, which lies between a, b, as a + 6(b—a), where 6 lies between 0 and 1. Then
This is called Cauchy's form of the remainder. Though less 'in sequence' than Lagrange's form, it enables us to deal with some series for which Lagrange's form does not work.
48
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S
ALTERNATIVE TREATMENT. There is an alternative treatment of Taylor's theorem, based on integration by parts, which leads to yet another form of the remainder. The method is essentially a continued application of the formula
Cb(bt) JO V^
the term
/(fc) (a +1) vanishing for t = b when & > 0.
This formula may be expressed more concisely by writing
bk
uk = T*f{k) {<*>) +
so that Hence
ux = bf (a) + u2
r6 Moreover,
/ ' (a +1) dt = f(a + 6) —/(a). Jo Equating the two values of uv we obtain the 'Taylor' relation (with our original C6' now replaced by ea + 6')
where
%=
T A Y L O R ' S THEOREM
49
EXAMPLE I I I
1. Prove that, if Rn(%) is the function otx defined by the relation
( + ... +* ~ _ ^V then
i?»

^
^
and deduce that Rn{x) = f*^"^ Jb
f{n\t)dt.
(n—1)1
6. Maclaurin's theorem. If we write 6 = a + h, Taylor's theorem (with the Lagrange remainder) becomes
where f is a certain number between a,a + h. A convenient form is found by putting a = 0 and then renaming h to be the current variable x:
f(x) =
J
^^
where f is a certain number between 0,x. This important result is known as Maclaurin's theorem. Compare p. 43. With the Cauchy form of remainder, the corresponding result is
where ^ is a certain number between 0,1. 7* Maclaurin's series. The remainder Rn in Maclaurin's theorem appears in the form or
50
T A Y L O R ' S S E M E S AND A L L I E D R E S U L T S
where £ lies between 09x and 6 between 0,1. Then f(x) = /(0) + xf (0) + . . . + ~ ^
/<»D (0) +
Rn.
It may be possible to prove that, as n becomes larger and larger (x having a definite value for a particular problem) the remainder Rn tends to the limit zero. When this happens, the sum of the first n terms of the series
tends to the limit /(#), and so the sum to infinity of the series exists, and is f(x). The condition for the remainder to tend to zero may involve x9 so that it is fulfilled for some values of x but not for others. It is on this basis that the possibility of obtaining an expansion rests. The succeeding paragraphs give the details for a number of important functions. 8. The series for sinjc and cos*. Let f(x) ss sin x. Then f'{x) a= cos#, f"(x) = —sinx, /'"(#) = —cosx, so that
...,
Hence the Maclaurin series is
a:3 x5 x* or *__ + ___+.... To see whether the series converges to sin x, we consider Rn, the remainder after n terms, where ryfll
The numerical value / (m) (£)l is certainly not greater than 1, since fln) (£) is a s i n e o r a cosine. Hence
THE SEBIES FOR SIN X AND COS X
But (see p. 30)
Urn ^
51
= 0,
for all values of xt and so the series converges to sin x for all values of x. By similar argument, we obtain the expansion x2 1
x4 x* +
+
convergent for all values of x. 9. The binomial series. Let where p may be positive or negative, and not necessarily an integer.
f'"(x)=P(Pl)(p2)(l+z)P\ and, generally, Hence
f{n)(x) = p(p/(0) = 1,
l)...(pn
f"(0)=p(pl), f"(0)=p(pl)(p2),
The Maclaurin series is thus
, ft(ffl)...(fftt+l)
"1
'
1
X "+"••••
n\ If p is a positive integer, the series terminates, giving the expansion, familiar from any textbook on algebra, )v= l+c1x + where
Cn
.
c2x2+...+cpxpt =
_ ^
52
TAYLOR'S
S E R I E S AND A L L I E D
RESULTS
/ / p is NOT a positive integer, we obtain an infinite series, and the conditions for convergence become important. We prove that the expansion
where p is not a positive integer, is valid for values of x in the interval — 1 <x< 1. The Cauchy form of remainder* gives xn( 1 — 0)n
Now
l  0 < l + 0a;
whether x is positive or negative, since — „ Hence Also, if p > 1,
/10V*1 _ I—^ <1. \1 + Ox) (1 + ftu)»i < (1 +1 x l)^1, '
(l + Ox)1* 1
For any given x in the interval — 1 < x < 1, the product
is therefore less than an ascertainable positive number A. Consider next the product (nl)I which we denote by the symbol %w. Then (nl)l un " =
n\ p—n x. n * The proof which follows may be postponed, if desired.
53
THE BINOMIAL SERIES
Take n so large as to be greater than p. Then n—p \x\. n Write 2/ = ^(l + #), so that 0
Uno+2
Then
u
n
y y,
n +1
^w+i < \ n \y ~~ °
Since uno is a definite ascertainable number, and since \y\< I, it follows that as 7&>oo. But
\Rn\
when A is the positive number already defined. Hence as w>oo, and the validity of the expansion is established. 10. The logarithmic series. Let
Then
and generally,
54
TAYLOR'S S E B I E S AND ALLIED RESULTS
Hence
/(0) = log 1 = 0,
so that the Maclaurin series is
or
...+{l)n1
xiz* + $a?l&+
71
This expansion for log (1 +x) is valid for all values of x in the range When x = 1, we have the result
The case x = — 1 is reflected in the graph (Fig. 62, p. 4) where y> —oo as #>0. EXAMPLE IV
I. Use the method given for the binomial series to prove the validity of the logarithmic series when — 1 < x < 1. I I . The exponential series. Let
fix) =e\ / ' (x) = ex,
Then
f" (x) = e*, and, generally^ Hence
{n)
f
(x) = ex.
/(0) =/'(0) =/"(0) = ... = 1.
We therefore have the Maclaurin series 2
ry& +
rv>Yl +
+
+
It may be proved that this expansion for ex is valid for all values ofx. [See Examples V (1) below.]
THE EXPONENTIAL SERIES
55
COROLLARY. The work of this paragraph enables us to fill a gap in the discussion of the exponential function (pp. 1820) by obtaining an expression for the number e. Putting x = 1, we have
EXAMPLES V
1. Use the method given for the sine series to prove the validity of the exponential series for all values of x* By direct calculation of the differential coefficients and substitution in Maclaurin's formula, obtain expansions for the following functions as series of ascending powers of x: 2. e2x. 5. l/(l+a;). 2x
8. e~ .
3. log (1 + 2x). 2
6. 1/(1a;) . 9. yl{l + 2x).
4. sin2#. 7. cos4x. 10. log (l3z).
12. Approximations. If the successive terms in the expansion
become rapidly smaller, a good approximation to the value of the function f(x) may be found by taking the first few terms. ILLUSTRATION 2. To estimate 7(398).
Writing the expression in the form
we may expand by the binomial series (p. 51) to obtain
= 2{1  0025  K000025) + ^ (000000125)...}, and a good approximation is 2(10025) = 19950. A limit may be set to the error by means of Taylor's theorem, as follows:
56
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S
Since, for a general function f(x),
f{x) =
J
the error cannot exceed the greatest value of
rt
or, here,
2x
(005)2
2!
1
00000625
' 4(1  0050)*— (1  0050)*'
In the most unfavourable case, with 0 = 1 , this gives a value less than 0000063 for the error. EXAMPLES VI
Estimate the values of the following expressions:
1. V( 4 '02).
2
 V( 8 * 97 )
3. ^(802).
4. ^(2698).
5. ^(3197).
6. l/V(903).
13. Newton's approximation to a root of an equation. Suppose that we are given an equation f(x) = 0 and know that there is a root somewhere near the value x = a. Newton's method, which we now describe, shows that, under suitable circumstances, a better approximation to the root is a
7W
If the correct root being sought is  == a + h, then
so that, by Taylor's theorem for n = 2, with Lagrange's form of the remainder,
NEWTON'S APPROXIMATION
57
for a value of 77 between a and a + h. If h is reasonably small, then the term involving h2 may be regarded as negligible for practical purposes. We therefore have f(a) + hf'(a) = 0, or so that the root is approximately
a
"7T«T
This crude statement, however, should be supplemented by more careful analysis, and the following graphical treatment shows the precautions which ought to be taken. We begin with an examination of the curve
y=f(x) near the point £, taking first the case where the gradient is positive near and the concavity 'upwards' near that point, so that f(x)J"(x) are both positive.
Fig. 66.
(i) Let X be the point (f, 0) on the curve, and let A be the point for which x — a, where a > £; draw AM perpendicular to Ox, and let the tangent at A meet Ox in P.
58
Then
TAYLOR'S SBBIES AND ALLIED RESULTS
OM = a, AM=f(a), AM p^ = tan^=/(a),
so that and
PM = {jy\, OP = a
But under our assumptions that f{x)J" (x) are positive near X (so that the gradient is positive and the concavity upwards) the tangent AP lies between AM and the curve, so that P lies between X and M. Hence, under these conditions, OP is a better approximation than OM to OX. That is, a — ^V^ is a better approximation rr /'(a) than a to the root. It is assumed that /'(a) is not zero, and, indeed, that/'(#) is not zero near the required root. (ii) Suppose next that, with the same diagram, B is the point for which x = b, where b < £, so that B lies 'below' Ox; draw BN perpendicular to Ox, and let the tangent at B meet Ox in Q. Then
BN = — f(b)
since f(b) is negative,
QH = tan fo =f'(b), so that
and
OQ^bjj^y
But now we cannot be sure that Q is nearer to X than JV, for Q may be anywhere to the right of X according to the shape of the curve. Hence we cannot be sure whether b — ~L is, or is not, a better approximation than b to the root.
NEWTON'S APPBOXIMATION
59
Similar argument applied to the accompanying diagram (Fig. 67) shows that the results (i), (ii) are also true if (the concavity still being 'upwards') the gradient is negative near X; P is closer than M to X, whereas Q may or may not be closer than iVtoX.
Fig. 67.
We have therefore proved, so far, that, iff" (x) is positive near £, the approximation a — J±JL {S better than a itself when f (a) is positive. It is easy to verify in the same way that, if /"(#) is negative near £, the approximation a — JTT\ is better than a itself when f(a) is negative. (The whole diagram is merely 'turned upside down'.) In other words, we can be sure of a better approximation if f"{x) retains the same sign near x = a, that sign being also the sign off(a). In more complicated cases, it is wise to draw sketches such as we have shown in order to determine how the tangent at A cuts the Xaxis. It is, however, clear from the graphs that, for ordinary functions, the approximation a — jrrk is ALWAYS better than a itself once we come sufficiently close to the correct answer. In other words, once it is ascertained that an approximation is reasonably good, it may confidently be expected that Newton's approximation will make it better still. 52
60
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S
ILLUSTRATION 3. An example where the answer is apparent from the outset may help to make the principle clear. Consider the equation
f so that
It is obvious that x = 0 is one root, but let us attempt to reach that value by approximation from (i) x = 1, (ii) x = — 1. When x = 1,
/(!) = ¥ , /'(I) = H, /"(!) = ¥ , so that /(I), / " ( I ) have the same sign; moreover/"(#) remains positive near x = 1 (in fact, down to x = — f, where/(#) is negative, indicating a point on the other side of the root). Hence we expect Newton's formula to give a better approximation; and since _/0^== /'(I)
1 9 _ 14 33 33'
this is actually the case. On the other hand, when x = — 1,
Here, again,/( —1),/"( 1) have the same sign; but f"(x) changes sign from negative to positive at x = — f, which is near — 1; we are therefore in a doubtful region; and since
the approximation is actually worse. ILLUSTRATION 4. To find an approximation to that root of the equation ^Zx+l =d which lies between 0,1.
We have
f(x) = x3  3x + 1, f'(x) = 3x*B, f"(x) = 6x.
NEWTON'S APPROXIMATION
61
Since f"(x) is positive when x lies in the given interval 0,1, it is advisable to begin with an approximation which makes f(x) also positive. Now & We should therefore like an approximation between \ and \ which keeps f(x) positive, and inspection shows that \ appears very suitable. We then have
so that and the corresponding approximation is H TV = ft ='3472. The correct root is '3472..., so we have already obtained four correct figures. 5. / / 77 is a small positive number, to find an approximation to that root of the equation ILLUSTRATION
sin x = rjx which lies near to x = 7r. (The intersection of the graphs y = sin x, y — rjx shows that there is a root near to TT.) Since sin n = 0 and 77 is small, the approximation x = IT is reasonably good. Write f(x) = sin x — qx, so that
f'(x) = cos x7], f"(x) = —sin a;.
Then
/(TT^T^,
/ ' ( 7 7 H  I  7 7 , / ' » = 0.
Although /"(TT) is actually zero, so that the curve (Fig. 68) y = sinxrjx has an inflexion at x = TT, the concavity is 'downwards' in the interval O,TT and /(IT) is negative; moreover the gradient /'(TT) is also negative. The accompanying sketch shows that the tangent lies between the ordinate x = 77 and the curve, so that Newton's method will improve the approximation.
62
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S
The corresponding solution is
— 7
7
y
z
r
— 7
T
—
i
•
(17,) 1+r, l+7j To the first order in q this is, on expansion of (1 + ij)"
o
Fig. 68.
Note. This solution is less than 77, as we expected from the diagram. 14, Leibniz's theorem. The theorem which follows is useful in calculating the higher differential coeflRcients necessary for a Maclaurin expansion. To prove that, iff{x) is the product of two functions u, v, so that f(x) = uv9
then f(n){x)
where ncp is the binomial coefficient nCp nCp
n\
~~p\(np)V
We use the method of mathematical induction, assuming the result to be true for a certain integer N9 so that
LEIBNIZ'S THEOREM
63 (N+1)
Now differentiate this expression to obtain f (x). differential coefficient of a product such as u{N~p)vip) is M(Np+l) v(p)
__ u(Np)
The
v(P+Dt
As we write down these terms for the series on the right, we put the answer in two lines, the top line consisting of terms such as U(NP+DV(P) a n ( j the lower of ^w^dHD; also we displace the lower line one place to the right, thus:
Now the coefficient of u{N~p+1)v{p)
N\ p\(Np)\
is
N\
Hence
It follows that, if the theorem is true for any particular value N9 then it is true for N + 1 , N + 2, and all subsequent values. But it is easily established when N «= 1, being merely the result
It is therefore true generally. Note. The expression is symmetrical when regarded from the two ends, and will equally well be written in the form ln)
ff
(x) = uv{n) + ... + n
64
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S EXAMPLES VII
Use Leibniz's theorem to find the following coefficients: d4 . !• m (#8sina;).
d4 2. r—= i
(XX
3^
differential
iXjJU
ifix QQg Qx),
4.
6. To ap2% Leibniz's theorem in finding a Maclaurin expansion for
We have f {x)
'
=
on simplification. Hence (ar«+1) {/(*)}• = 1 . Differentiate. Then (o:2+ l).2/'(^)/"(^) + 2 4 / ' ( ^ = 0, or
(x2+l)f"(z) + xf'(x) = 0.
Differentiate n times, using Leibniz's theorem. Write the expansion from (x2+ l)f"{x) on the first line, and from xf'(x) on the second; note that the expansions terminate since (x2 +1)'" = 0, x" = 0. We obtain (x2 + 1 )/<"+»> («) + n . 2a; ./««+!> (a;) + n ^ " j " ^ . 2 ./<»> (»)
n
.1 ./ (w) (x) = 0.
LEIBNIZ'S THEOREM
65
We require values when x = 0, so that = 0,
/
or But and so
n2f{n)(0).
/(0) = log 1 = 0, /"(0) =/ ( i v ) (0) = ... = / (2rl) (0) = 0.
Also
/'(0) = 1, /'"(0) =  1 2 . 1 ,
so that
/(v)(0)= + 3 2 .1, /(vii)(0) = ~ 5 2 . 3 2 . l ,
and so on. Hence the Maclaurin expansion is
°r
X
l x s 1.3 a;5 1.3.5 a;7 2 3 + 2 ^ " 5 ~ 2 X 6 7+~"
converging to log {x + <j(x2 +1)} for such values of x (not considered here) as make the series convergent. EXAMPLES VIII
Use the method of Illustration 6 to obtain the following expansions: 1. f{x)= sin"1 a;. 2. f(x)== tan 1 a;. 3. Prove that, if f{x)== cos (m sin 1 x), then (1#2)/<"+2>(Z)(2TI+1^^
= 0.
4. Prove that, if/(a;) = (sin 1 ^) 2 , then n2f
= 0.
66
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S R E V I S I O N E X A M P L E S IV
'Advanced' Level
1. Differentiate
(i)
^
^
}
If y = eaainbx, and y',y" are the first and second differential coefficients of y with respect to x show that Calculate the values of y', y" when x = 0. 2. Differentiate , ,12:*; tan" 1 — , 2+x
. //la?\ sin"1 / , *J\l+x)'
log (sec x + tan x), Find the values of x for which x2 8 # is a maximum or minimum, distinguishing the maximum from the minimum. 3. If y = ext8Hixt prove that
Prove that y = uexta,nx also satisfies this equation when u is a function of x such that
Verify that u = cot x is such a function. 4. Differentiate 2/ = sec m #tan n a\ Deduce that the kth differential coefficient of secm# can be expressed in the form secma;PA.(tana:), where Pfc(tana;) is a polynomial in tan x of degree k. Evaluate J^(tan x) when both m and k are taken equal to 4.
REVISION EXAMPLES IV
67
5. Differentiate with respect to x:  + 2log
,
x — tan x 4 \ tan 3 x>
simplifying your answers as much as you can. Prove that the first function has a maximum at x = 1 and that the second function (whose differential coefficient vanishes at x = 0) has neither a maximum nor a minimum at x = 0. 6. Find the first four differential coefficients of sin4 x. Show that the function has turning values at x = 0 and at x = JTT, and determine whether they are maxima or minima. 7. Prove that, if y = sin2(a;2), then
Prove that the result of changing the independent variable in this equation from x to  , where £ = x2, is
where a, b are constants to be determined. Prove also that, if A9B are any constants, the function y =  + A cos (2a;2) + B sin (2x2) satisfies the first differential equation. 8. Differentiate /
l\ w
(cos x + sec x)n
with respect to x, and find the nth differential coefficient of
K)' for all positive integral values of n, distinguishing the cases n < 2 and n > 2.
68
T A Y L O R ' S S E R I E S AND A L L I E D
RESULTS
9. Show that the polynomial 4 16 (2ir5)x* + —ATT3)x5
f(x) = x
77
77"
and its differential coefficient/'(a;) have the same values as sina? and its differential coefficient respectively, at the values x — 0, x = ± \n.
J
%
i7T
0
f(x)dx as an approximation for
I sinxdx is less than 0*1 per cent. Jo 10. (i) Given that — = ex} j = since, ax ax and that u = 1, v = 0 when x = 0, show that d2(uv) _ ~dtf~ = when # = 0. (ii) Given that y = axn + bx1"71, prove that x7 + (nl)y = (2nl)axn, ax (in d t J
and form an equation in x,y9~9 =^ which does not contain a or 6. ax ax 11. Differentiate with respect to x:
Prove that
r—
1=
, —r,
and find the nth differential coefficient of x 12. Find the nth derivatives with respect to x of
W (ii)
i
and
sin a; and
x>—v x sin #.
REVISION EXAMPLES IV
69
2
13. (i) Find the derivative of y = sin a; with respect to z = cos x by evaluating the limit of 8y/8z. (ii) Differentiate with respect to x: sin# (iii) Prove that
dy2
dx2l \dxt
14. Find from first principles the differential coefficient of \jxz with respect to x. What is the differential coefficient of ljxB with respect to #2? Differentiate with respect to x: logecosa;,
xj{\x%
(1^2)2.
15. Find the nth differential coefficient of y with respect to x in each of the following cases: (i) y = l/x2, (ii) y = sin2#,
(iii) y = e2*sin2#.
There are three values of k for which the function y = xke~x satisfies the equation
Find these values. 16. Differentiate with respect to x: l2x
3 tan2 X
cin~l
Prove that, when y = at2 + 2bt + c, t = ax2 + 2bx + c, and a,b,c are constants,
17. Differentiate
c tanaJ ,
Show that, if y = sin 6, a; = cos 0, then
70
T A Y L O E ' S S E K I E S AND A L L I E D R E S U L T S
18. A particle moves along the axis of x so that its distance from the origin t seconds after starting is given by the formula x = a cos ht + \dkt, where a, k are positive constants. Find expressions for the velocity and acceleration in terms of t, and prove that the particle is not always moving in the same direction along the axis. Find the positions of the particle at the two times t = TT/6& and t = 13TT/6&; also find the position of the particle at the instant between these times at which it is momentarily at rest, and deduce the total distance travelled between the two times. 19. A particle moves in a plane so that its coordinates at time t are given by x = el cos t9y = e'sintf. Find the magnitudes of the velocity and of the acceleration* at time t and prove that the acceleration is always at right angles to the radius vector. Draw a rough sketch of the path of the particle, from time t = 0 to t = 77, and indicate the direction of motion at times 0,  T T , JTT,
f 77, 77.
20. A particle moves in a plane so that its position at time t is given by x = a cos pt,y — b sin pt. Find expressions for (i) the magnitude v of the velocity at time t; (ii) the magnitude / of the acceleration at time t; and prove that there is no value of t for which / = dvjdt. Prove also that the resultant acceleration makes an angle 6 with the normal to the path, where 2ab tan 0 = (a2 62) sin 2pt. 21. A point moves in a straight line so that its distance at time t from a given point 0 of the line is x, where x= Find its velocity at time t, and prove that the acceleration is then — t2 Bint —2t cost+
2smt.
Determine the times (£>0) at which the acceleration has a turning value, distinguishing between maxima and minima. * If v,f are the velocity and acceleration respectively, then v2 = x2+y2,
BEVISIOST EXAMPLES IV
71
22. A particle moves along the axis of x so that its distance from the origin t seconds after starting is given by the formula x = a cos pt. Prove that the velocity of the particle changes direction once, and only once, between the times t = 0, t = 2TTJP and that the change of direction occurs at the point x = — a. The distance from the origin of a second particle is given by the formula . 1 n . x = a cos pt + \a cos 2pt. Write down expressions for its velocity and acceleration at time t. Show that between t = 0 and t = 2TTJP the velocity of the particle changes direction three times, and find the values of x at which these changes occur. 23. Prove that the equation of the tangent to the curve given by at the point where t = tan a is y — #tana+.tan 3 a + t a n a + l = 0. Show that the curve lies on the positive side of the line x = 1 and is symmetrical about the line y = — 1, and prove that the area bounded by the line x = 4 and the curve is ^ . 24. Prove that there are two distinct tangents to the curve
y = x*x + 3 which pass through the origin. Find their equations, and their points of contact with the curve. Give a rough sketch of the curve. 25. A curve is defined by the parametric equations where a is a positive constant. Prove that (i) the curve passes through the points A,0,B whose coordinates are (0,6a), (0,0), (  3a, 0). (ii) the point t is in the first quadrant when  1 < t < 1 and in the third quadrant when 1 < t < 2. Make a rough sketch showing the part of the curve corresponding to values of t between — 1 and + 2. Find the equation of the tangent to the curve at O, and prove that the area bounded by the arc AB and the chord AB is 27a2/2.
72
TAYLOR'S
S E R I E S AND A L L I E D
RESULTS
26. Prove that the slope of the curve whose equation is is always positive. Show that the curve has a point of inflexion where x = — 1, the slope there being \% Prove also that the tangent at the point (0,1) meets the curve again at the point (— 3, — 2). Sketch the curve, indicating clearly the point of inflexion and the tangents at the points (—1,4) and (0,1). 27. A,0,B are three fixed points in order on a straight line, and AO = p, OB = q. A fixed circle has centre 0 and radius a greater than p or q, and P is a point on this circle. Show that the perimeter of the triangle APB is a maximum when OP bisects the angle APB9 and find the corresponding magnitude of the angle POA. 28. Prove that the maximum and minimum values of the function y = x cos 3x occur when 3 tan 3x — Ijx, and discuss the behaviour of the function when x = 0. By considering the curves y = tan 3x, y = \jx, show that maximum values occur near the values x = § kir, and minimum values near x = J(2&+ 1)TT, the approximation becoming more exact as x becomes larger. 29. Given that a 2 # 4 + 62?/4 = c6, where a, b, c are constants, show that xy has a stationary value c3/>/(2a6). Is this value a maximum or a minimum ? 30. (i) A right circular cylinder is inscribed in a given right circular cone. Prove that its volume is a maximum when its altitude is onethird that of the cone. (ii) A right circular cone of height h stands on a base of radius h tan a. A cylinder of height h — x is inscribed in the cone. Prove that 8, the total surface of the cylinder, is equal to 2TT{X2 (tan
2
a — tan a) + xh tan ex},
and prove that, when tan a > ^, S increases steadily as x increases. 31. P is a variable point in the circumference of a fixed circle of which AB is a fixed diameter and 0 is the centre. Prove that,
REVISION EXAMPLES IV
73
when OP is perpendicular to AB, then the function AP + PB has a maximum value and the function AP3 + PBZ has a minimum value. 32. If/(#) is a function otx and/'(#) is its differential coefficient, show that, when h is small, f(a + h) is approximately equal to f(a) + hf'(a). Without using trigonometrical tables, find to three significant figures (i) the value of cos 31°, and (ii) the positive acute angle whose sine is 0*503. Give your answer to (ii) in degrees and tenths of a degree. 33. Calculate (i) <$/805 to 4 significant figures, (ii) cos 59° to 3 significant figures. [Take TT to be 3142.] 34. Determine to two places of decimals that root of the equation ^ ^— = 32104 x+2 whose value is nearly equal to 8. 35. Determine to 3 places of decimals the value of that root of the equation xs which lies between 15 and 16. 36. Prove, graphically or otherwise, that, if n is a large positive integer, there is a root of the equation xmnx = 1 nearly equal to 2n7T. Show that a better approximation is 2rnr\~ (1/2/iTr). 37. Prove that, if TJ is small, the equation 0 + sin0cos0 = 2rjcos0 has a small root approximately equal to
vh3[You may assume the powerseries expansions for sin 9 and cos 0.] 6
M II
74
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S
38. Prove that the result of differentiating the equation ax n+ 1 times (w > 1) with respect to x is
Hence verify that, if h is a positive integer and
then z is a solution of the equation (1 + a;2) ^  + 2(i + l)a; J +fc(*+ 1) z = 0. 39. By using Maclaurin's theorem, or otherwise, obtain the expansion of log (1 f sin x) in ascending powers of x as far as the term in x*. 40. Prove that, if y = loge cos x, then == a
dx Hence, or otherwise, obtain the Maclaurin expansion of logc cos x as far as the term in #4. Deduce the approximate relation
41. Prove that, if y = etanrc, then
where t = tano;. Prove that the expansion of y as far as the term in xz is
y= l+x + ix* + ±x*. 42. Prove that, if y = sin (log a;), where x > 0, then
EEVISION EXAMPLES IV
75
By induction, or otherwise, prove that
where n is a positive integer. 43. If y = e^loga;, show that
Find the equation obtained by differentiating this equation n times. 44. Prove that, when y = eax sin bx,
and that
p = t/(a + 6 cot bx). CbX 00
Prove that, if then
eaxsmbx
£
= T. —%xn,
c n + 2  2acn+1 + (a2 + b2)cn = 0,
and find the values of cl9 c2, c3. 45. If cos y = cos oc cos #, where #,^, a lie between 0 and \TT dii
du
radians and a is constant, find the values oty,^, ~ when x = 0. Taking a; to be so small that xz and higher powers of x are negligible, use Maclaurin's theorem to show that y as a + ^#2 cot a. Hence calculate y in degrees, correct to 0001°, if a = 45° and a; = 1° 48'. [Take IT = 3142.] 46. By using Taylor's theorem obtain the expansion of tan (x + \IT) in powers of x up to the term in #3. Hence calculate the value of tan 44° 48' correct to four places of decimals. [Take TT = 3142.] 62
76
T A Y L O B ' S S E R I E S AND A L L I E D R E S U L T S
47. Prove that, if y = loge I (i)
:—I,
then
 
and (ii) the expansion of y in a series of powers of x as far as the term in #4 is 2x + \xz. Find, correct to four significant figures, the value of y when x = 1° 48', taking TT = 3142. 48. Prove that, if y = (sin"1 #)2, then
The Maclaurin expansion of 2/ in powers of x is taken to be
Given that y = 0 when a; = 0, prove that a0 = 0 and ax = 0. Prove also that aw+2 = w2an when n > 0, and hence show that
49. If y =
w"dM, prove that
3<>•*•>£• Find the values of the first four differential coefficients of y when # = 1, and, by using Taylor's expansion in the form 11 deduce that the value of f ' uudu is approximately 0«1053.
h 50. If y = tana:, prove that
and find the third, fourth, and fifth derivatives of y.
BEVISION EXAMPLES IV
77
Hence find the expansion of tana; in a series of powers of x up to x5. 51. Prove that, if y = sin^ttsin^a;), then
Show that the first two terms in the expansion of the principal value of y in ascending powers of x are mx + Jm( 1 — m2)x3. 52. Find the indefinite integrals: x e~x* dx,

2~ fa* x* e* dx
If u, v are functions of x, and dashes denote differentiation with respect to x, show that (uv"f + u'"v)dx — uv" — u'v' + u"v + constant. 53. Show that
\f(x)dx*= \f(ax)dx. Jo Jo
ta,n2xdx, I : ——)dx= Jo \l+sin2aj/ Jo and evaluate the integral. Deduce that
54. Find the indefinite integrals: f\xlogxdx, i *
C dx f\ x*~.dx. J \——,
Prove that, when the expression e^sina; is integrated n times, the result is 2tn ex s i n (£ _ in
wj
where Pn_ x is a polynomial of degree w — 1 in a;.
78
TAYLOR'S
SERIES
AND A L L I E D
RESULTS
55. Find the indefinite integrals of
56. Find the indefinite integrals of 32cos4#,
27x2(logx)2,
~^j>
a*.
JL — X
57. Integrate with respect to x: 4
Evaluate t h e definite integrals: C\
f*» dx Jo 1 + BOOBV
)*****'
58. Find the indefinite integrals:
59. Find the indefinite integrals: f dx J(l3x)3
r J
2O
_
f_ J
,
1
,
f cos J2~cos 2 o;
60. Find the values of sin5^^ Jo
cos 4 a;^, Jo
ex sin 2 x dx. Jo
61. (i) Find t h e indefinite integrals of x{x+l)2' (ii) By substitution, or otherwise, prove that f1/V2 Jo
irsin1^^ = ,
f1 x*J{x*+ l)dx = Jo
62. Find the indefinite integrals of cos3 3x,
— , 3 —2cosa;
Xs sin x.
REVISION EXAMPLES IV
79
Use the method of integration by parts to integrate
twice with respect to x. 63. By means of the substitution x = oc cos2 0 f j8 sin2 0, or otherwise, prove that Cfi dx Ja^{{xoc){px)}^7Tt Prove also that
[
 a) (j8 
64. P r o v e t h a t  c o s 2 Odd = JTT,
Jo
COS4
Jo
§77.
Find the area bounded by the curve r = a(l + cos0) and determine the position of the centre of gravity of the area. 65. Find the equation of the normal at (the point (f, sin £) to the curve whose equation is y = sin x. Prove that, if £ lies between 0, TT, the normal at P divides the area bounded by the #axis and that arc of the curve for which 0 ^ x ^ 77 in the ratio (2  cos £  cos3 £) : (2 + cos £ + cos3 £).  2 + v2= l a2 b2 is rotated through two right angles about the #axis. Prove that the volume generated is ^Trab2. (i) Prove that, if a and 6 are varied subject to the condition a + b = £, then the greatest volume generated is 2TT/81. (ii) The volume is cut in two by the plane generated by the rotation of the yaxis. Prove that the centre of gravity of either part of the volume is at a distance fa from the plane of separation. 66. The ellipse r
67. The complete curve x2/a2 + y2\b2 = 1 is rotated round the 2/axis through two right angles. Find the volume generated by the area enclosed by the curve.
80
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S
The semiaxes a and b of the curve are each increased by €. Prove that, if e is small, the increase in volume is approximately 68. OA is a straight rod of length a in which the density at a point distant x from 0 is b + ex, where b and c are constants. Find the distance of the centre of gravity of the rod from 0. 69. The gradient at any point (x, y) of a curve is given by dx and the curve passes through the point (2,0). Find its equation and sketch the graph, indicating the turning points. Find the distance from the yaxis of the centre of gravity of a uniform lamina bounded by the curve and the positive halves of the x and y axes. 70. A lamina in the shape of the parabola y2 — 4ax bounded by the chord x = a is rotated (i) about the axis of y, (ii) about the line x = a. Prove that the volumes generated in the two cases are 71. Integrate with respect to x: sin2#, sin3 x9 x2 cos x. The portion of the curve y = sin a; from x = 0 to x = \n revolves round the axis of y. Prove that the volume contained between the surface so formed and the plane y = 1 is JT^TT2 — 8). 72. The coordinates of a point on a curve referred to rectangular axes are {at2,2at), where t is a variable parameter which lies between 0 and 1. Make a rough graph of the curve. Calculate (i) the area enclosed by the curve and the lines x = a, y = 0; and (ii) the area of the surface obtained by revolving this part of the curve about the #axis. 73. Find the area contained between the #axis and that part of the curve x = 2£2 f 1, y = t2 — 2t which corresponds to values of t lying between 0 and 2. Find also the coordinates of the centroid of this area.
REVISION EXAMPLES IV
8J
74. Prove that the tangent to the curve
Jx + Jy = Ja at the point (x,y) makes intercepts ^{ax)^(ay) on the axes. A solid is generated by rotating about the #axis the area whose complete boundary is formed by (i) the arc of this curve joining the points (a, 0), (0,a), and (ii) the straight lines joining the origin to these two points. Prove that, when this solid is of uniform density p, its mass M is x^Trpa3. Prove also that the moment of inertia of the solid about Ox is 75. Find the area bounded by the curve x = 2a(t3 — l),y = Sat2 and the straight lines x = 0, y = 0. Prove that the tangent to the curve at the point t = 1 meets the curve again at the point t = —  , and find the area bounded by the parts of the tangent and of the curve that lie between these points. 76. Prove that the parabola y2 = 2ax divides the area of the ellipse 4#2 + 3y2 = 4a2 into two parts whose areas are in the ratio 77. The portion of the curve y2 = 4ax from (a, 2a) to (4a, 4a) revolves round the tangent at the origin. Prove that the volume bounded by the curved surface so formed and plane ends perpendicular to the axis of revolution is ^Tra3, and find the square of the radius of gyration of this volume about the axis of revolution. 78. Find the coordinates of the centre of gravity of the area enclosed by the loop of the curve whose equation is r = acos20, which lies in the sector bounded by the lines 6 = ± \TT. Find also the volume obtained by rotating this loop about the line 0 — ^77. 79. Find the coordinates of the centre of gravity of the loop of the curve traced out by the point x = 1 — t2, y = t — ts. Find also the volume obtained by rotating this loop about the line x = y. 80. A plane uniform lamina is bounded by the curve y2 = 4a# and the straight lines y = 0, x = a. Find the area and the centre of gravity of the lamina.
82
T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S
The lamina is rotated about the axis Ox to form a (uniform) solid of revolution. Find the centre of gravity of the solid and, assuming the density of the solid to be p, find its moment of inertia about the axis Ox. 81. A lamina in the shape of the parabola y2 = 4=ax, bounded by the chord x = a, is rotated (i) about the axis of y, (ii) about the line x = a. Prove that the two volumes thus generated are in the ratio 3 : 2. 82. Prove by integration that the moment of inertia of a uniform circular disc, of mass m and radius a, about a line through its centre perpendicular to its plane is ma 2 . The mass of a uniform solid right circular cone is M, and the radius of its base is a. Prove that its moment of inertia about its 2 . axis is 83. Evaluate
Jo Jo
cosn0d0 when n = 1,2,3,4.
The area bounded by the axis of #, the line x = a, and the curve x = asin0, y = a(l — cos 0), from 0 = 0 to 0 = \n, revolves round the axis of x. Prove that the volume generated is ^7ra3(10 — 3TT). 84. The portion of the curve x2 = 4:a(a — y) from x = — 2a to x = 2a revolves round the axis of x. Prove that the volume contained by the surface so formed is ffrra3, and find its radius of gyration about the axis of revolution. 85. Sketch the curve r = a(l + cos#), and find the area it encloses and the volume of the surface formed by revolving it about the line 0 = 0. 86. Find the three pairs of consecutive integers (positive, negative, or zero) between which the roots of the equation lie, and evaluate the largest root correct to two places of decimals. 87. Show that the equation x*3x7 = 0 has one real root, and find it correct to three places of decimals.
BEVISION EXAMPLES IV
83
88. Show that the equation has one real root, and find it correct to three places of decimals. 89. For a function / having an Nth differential coefficient, Taylor's theorem expresses f(a + x) as a polynomial of degree JV— 1 in #, together with a remainder. State the form taken by this polynomial, and one form of the remainder. Prove by induction, or otherwise, that dn =— (e* sin 3^3) = dX
Hence find the coefficients an in the Maclaurin series Hanxn of the function e^sina;^. By means of Taylor's theorem, show that when x>0 the JV1
difference between exsinx^j3 and 2 anxU *s n o ^ greater than n  0
(2x)Nex
[A proof showing that the difference is not greater than (2x)Nex is acceptable if the form of remainder which you have quoted leads to the result.] 90. If show that
y (l + x)2^+(l+x)y=
1.
dX
Deduce the first four terms in the Maclaurin series for y in powers of x. 91. Show that y = {x + ^ ( l + x2)}k y'y](l+x2) = ley, satisfies the relations Deduce the expansion
Verify that this agrees with the series derived from the binomial series when h = 1.
CHAPTER IX THE HYPERBOLIC FUNCTIONS 1, The hyperbolic cosine and sine. There are two functions with properties closely analogous to those of cos x and sin x. They are called the hyperbolic cosine of x, and the hyperbolic sine of x, and are written as cosh x and sinh#. We define them in terms of the exponential function as follows: cosh# = \{ex\e~x), sinhz =
\{exe~x).
We establish a succession of properties similar to those of the cosine and sine: (i) To prove that
cosh2 a; — sinh2# = 1.
The lefthand side is
l{{e2x + 2 + e~2x)  (e2x  2 + e~2x)}
= 1. (ii) To prove that cosh (x + y) — cosh x cosh y 4 sinh x sinh y It is easier to start with the righthand side: (e* + e~x) (& + e~y) + {ex  e~x) (e^  e~y)} == i{(ex+y + ex~y + ex+y + e~xy) + (ex+y  ex~y 
= cosh (x + y).
THE HYPERBOLIC COSINE AND SINE
85
(iii) Similarly sinh (x + y) = sinh x cosh y + cosh x sinh y. (iv) As particular cases of (ii), (iii), cosh 2x = cosh2 x + sinh2 x, sinh 2# = 2 sinh a; cosh a\ COROLLARY
(i). cosh 2 x = (cosh 2x + 1), sinh2 a; = ^(cosh2# — 1).
COROLLARY
(ii). Since cosh a;—1 = 2 sinh 2 
and sinh2 ~ is positive, it follows that cosh x is greater than unity for all (real) values ofx, i.e. Note.
cosh x ^ 1. coshO = 1, sinh 0 = 0.
(v) To prove that
cosh a; = cosh (— x).
The righthand side is = cosh a;. (vi) To prove that For
sinh x = — sinh (— x).
sinh (  x) = l{e~x  e<*>) = i(e*  ex) = —sinh a:.
iV'o^e. Sinh a; is positive when a; is positive, and negative when x is negative. For example, if x is positive, then ex is greater than e~x, since e is greater than 1. Hence \(ex — e~x) is positive. (vii) To prove that r~ (cosh a;) = sinh a:, CLX
•j (sinh x) = cosh a;. ax
86
THE HYPERBOLIC FUNCTIONS
We have
= (cosh x) =  = (ex + e~x) dxK
'
2dx
and
= cosh a;, (viii) To prove that coshrfa; = sinha;,
= cosh#. These results follow at once from (vii). (ix) By theformulce of (vii), we have the relations =s (cosh a;) = cosh#, cLx
rg (sinh#) = sinha;. dx Thus cosha:, sinha; both satisfy the relation dx*
y
(x) The following expansions in power series are immediate consequences of Maclaurin's theorem: 1
x2
+
x* xB +
+
The series converge to the functions for all values of x.
OTHER HYPERBOLIC FUNCTIONS
87
2. Other hyperbolic functions. The following functions are defined by analogy with the corresponding functions of elementary trigonometry: sinh x cosh a;' KJCVXIXX U/
cotha;
cosh a? sinh a;'
cosecha;
1 sinh x9
seen a;
1 ~~ cosh a;"
sech 2 # + tanh 2 # = 1,
The relations
are found by dividing the equation cosh2 x — sinh2 x = 1 by cosh2#, sinh2 a; respectively. Note the implications sech2 x
tanh 2 x
coefficients are easily obtained from the
If y = tanh a; = *(sinha;) (cosh a?)1,
then
j — (cosh x) (cosh x)"1 — (sinh a;) (cosh a;)~2 sinh x &x = ltanh2a; = sech2 a;. (ii) if y = cotha; = (cosh a:) (sinha:)"1,
then
J = (sinh x) (sinh x)"1 — (cosh a:) (sinh a:)~2 cosh x CLX
= lcoth 2 a; = — cosech2as.
THE HYPERBOLIC FUNCTIONS
88
1 y =• cosecha; = (sinha;)" , dy _= — (sinh#) 2 cosh# dx
(iii) If then
• — cosecho;cotha?. 1 y = sech a; = (cosh a;)*" ,
(iv) If
dy = — (cosha;)2sinha? dx
then
= — sech x tanh x. ILLUSTRATION 1. A body of mass m falls from rest under gravity in a medium whose resistance to motion is gv2/k2 per unit mass when the speed is v. To prove that the speed after t seconds is k tanh(<#/&).
Let x be the distance dropped in time t. Then the acceleration downward is x and the forces are (i) mg downwards due to gravity, (ii) mgx2/k2 upwards due to the resistance. mx = mg — mgx2jk2.
Hence
(it)
Write x = v. Then or Substitute
— = ggv2jk2, at k2t v = &tanh 8.
(This substitution is possible so long as v is less than k, since (p. 87) tanh 2 0
k2. k sech 2 6^ = gk2(l
tanh 2 6)
= gk2 sech 2 6.
Hence
ft = V
so that
0 =  + C,
where C is an arbitrary constant. It follows that
OTHER HYPERBOLIC FUNCTIONS Now we are given that v = 0 when t = 0, so that
89
0 = tanhC. Hence
C = 0,
and so
v = &tanh(<7£/&).
Note that
v = /ctanh (gt/k) egt/k
__ egt/k
=k 1 _ =
JC
e 2flr
l+e2gl/kt
Now as ^ increases, e2gt/k tends to infinity (see, for example, the diagram (Fig. 64) for ex on p. 18), so that e~2gt/k>0. Hence as t increases. In other words, v tends to a terminal value k as the time of falling increases.
The example which follows deserves close attention. It brings out very clearly the points of similarity between the trigonometric and the hyperbolic functions.* ILLUSTRATION 2. Suppose that a particle P (Fig. 69), of mass m, is free to move on a fixed smooth circular wire, of radius a, whose plane is vertical. A light string, of natural length a and modulus of elasticity 2kmg joins P to the highest point B of the wire. We wish to examine what happens if P receives a slight displacement from the lowest point of the wire. Let AB be the diameter through Fig. 69. B, and 0 the centre of the circle. Denote by 6 the angle LAOP. * It may be postponed or omitted by a reader who finds the mechanics difficult.
90
THE HYPERBOLIC FUNCTIONS
If x denotes the horizontal distance of P to the right of the vertical diameter AB, and if y denotes the depth of P below 0, then x = a sin 0, y = a cos 9. Hence, differentiating with respect to time, x = adcosd, y == — adsind, x = aScos 9 — a02 sin 6,
and
y = — ad sin 0 — ad2 cos 9.
The acceleration/ of P in the direction of the (upward) tangent is thus
/ = x cos 9 — y sin 9 = a§.
[This is a standard formula of applied mathematics.] Now the forces on the particle are (i) the reaction R along PO, which has no component along the tangent; (ii) gravity mg, whose component along the (upward) tangent is — mg sin 9;
(iii) the tension T9 where, by definition of modulus, m _ (modulus) (extension) ~~ natural length 2kmg{2a cos 0 — a] a = 2kmg{2 cos 191); the component of T along the (upward) tangent is thus T sin  0
OTHER HYPERBOLIC FUNCTIONS
91
Equating the component of the acceleration to the component of the forces, we have the equation of motion ad = gsin9 + 2kg(2 cos J 0  1 ) sin \9. Now suppose that 0 is small. By the work given on p. 38, the value of sin0 is nearly 6 itself, while cos0 is nearly equal to 1. Hence the equation is approximately
Hence we reach the equation a8 = g(kl)6, valid during the time while 0 is small. The argument now divides, according as k is less than or greater than unity; that is, according as the string is 'fairly slack' or 'fairly tight'. (i) Suppose that k
0= ^
where
an2 = (1 — k) g.
It may be proved that, when 9 = — n2 9, then 9 MUST be of the form 9 — A cosnt + Bsmnt, where A,B are constants. In the meantime, the reader may easily verify the converse result, that this value of 9 does satisfy the relation. If we suppose that the particle is initially drawn aside so that 9 has the small value ex, then 9 = a when t = 0, so that
If also the particle is released from rest, then 0 = 0 when t = 0, so that 0 =
Hence
0.
0 = a cos nt. 72
92
THE HYPERBOLIC FUNCTIONS
Thus if 0 is initially small, it remains small, and its value oscillates between ± a . The equilibrium is stable at the lowest point. (ii) Suppose that k>\.
Then
*<*zl>l!*
where
ap2 = (k—l)g.
This equation is not satisfied by sines and cosines, but we can express the relation between 6 and t in the form where, again, the reader may verify the converse result that this value of 9 does satisfy the equation. With the same initial conditions as before, we have
so that
A=B
Hence
6 =a = a cosh pt.
As t increases, cosh#£ increases steadily to 'infinity' (since epl does), so that 6 ceases to be small. The equilibrium is unstable at the lowest point. The differential equation, in fact, ceases to be accurate once 6 ceases to be small. It is instructive to consider the same problem under the alternative initial conditions that the particle is projected from the lowest point with speed v. Thus 6 = 0,ad = v when t = 0. We take the two cases in succession: (i) where
6 = Acosnt + Bsinnt, 0 = A, vja = nB.
Hence
6 = (v/an) sin nt.
93
OTHER HYPERBOLIC FUNCTIONS
OAefi
(ii)
+ Be*t
0 = A + B,
where
vja = pA —pB, so that
V
A7?.
*= ~2ap%
2ap 6 = — (ept  (
Hence
= (v jap) sinh pt. The analogy between the pairs of solutions oc cos nt, and
otcoshpt
(vjan) sin nt, (vjap) sinh pt
affords striking confirmation of the analogy between the two classes of functions. EXAMPLES I
Differentiate the following functions: 1. sinh 3a;.
2. cosh2 2a;.
3. artanha?.
4. sinh2 (2a;+1).
5. sinh # cos #.
6. secha:sin2a:.
7. (1 +a;)3 cosh3 3a;.
8. a;2tanh24x.
9. log sinh a:.
x
10. log (sinh x + coshx).
12. xe~tl
11. e^ .
Find the following integrals: 1O
I a! YI y% A.*y* fit*
14.
J
1 ^i »•/•
J
16. j a; sinh a;da;.
17. \ex cosh xdx.
C
C a;sinh2#c?a\
20.
22. \x2 cosh xdx.
23.
19.
I Qi in Ti * v /If
J
18.  sinh 3a; cosh xdx.
•I
cosh3 xdx.
21.
e2*sinh5a;da;.
24. I tanha;sech2a;da;.
J
94
THE HYPERBOLIC FUNCTIONS
Establish the following formulae: 25. sinh A + sinh B = 2 sinh —— cosh ——• 26. sinh A — sinh B = 2 cosh —— sinh ——. 2 2 A +B A —B 27. cosh A + cosh B = 2 cosh —— cosh ——. 2 2 A +B A —B 28. cosh A — cosh B = 2 sinh —— sinh ——. 2 2 29. Prove that, for all values of u, the point (a cosh u% b sinh w) lies on the hyperbola whose equation is
and that the tangent at that point is a
= 1.
o
(But note that that point is restricted to the part of the hyperbola for which x is positive.) 3. The graph y = coshx. The two relations y = cosh a;, dy dx
>
= sinh
give us sufficient information to indicate the general shape of the curve: Since cosh (  x) = cosh (x), the curve is symmetrical about the i/axis; and since 1, the curve lies entirely above the line y = 1. The value of y increases rapidly with x.
95
THE GRAPH
Taking x to be positive (when it is negative the corresponding part of the curve is obtained simply by reflexion in the t/axis) we have sinh# positive, so that the gradient is positive. Moreover, dx*
= cosh a;,
which is positive, and so (Vol. I, p. 54) the concavity is 'upwards'. The general shape of the curve is therefore that indicated in the diagram (Fig. 70). 4. The graph y = sinhx. We have the relations y = sinha;, ¥~ — cosh#. dx din Since ~ is positive, the gradient is dx always positive, and y is an increasing function of x, running from — oo to +00 as x increases from — oo to +00.
O
Fig. 71. At the origin, — = 1, so that the ax curve crosses the #axis there at an angle of TT. Also ffiy_ dx2
which is positive for positive x and negative for negative. Hence the curve lies entirely in the first and third quadrants, with (Vol. i, p. 54) concavity 'upwards* in the first and 'downwards' in the third. At the origin, j~ = 0, so that (Vol. I, p. 55) the curve has an inflexion there. Since
sinh (— x) = — sinh (x),
the curve is symmetrical about the origin. The general shape of the curve is therefore that indicated in the diagram (Fig. 71).
96
THE HYPERBOLIC FUNCTIONS
5. The inverse hyperbolic cosine. The problem arises in practice to determine a function whose hyperbolic cosine has a given value x. If the function is y9 then x = cosh y, and we use the notation y = cosh"1 a; to denote the inverse hyperbolic cosine. The graph (Fig. 72) y = cosh"1 x is found by 'turning the graph y = cosh# through a right angle' and then renaming the axes, as shown in the diagram. The graph exhibits two properties at first glance:
Fig. 72.
(i) / / x < 1, the value of cosh"1 a; does not exist] (ii) If x>l, there are TWO values of cosh"1 a;, equal in magnitude but opposite in sign. We can express cosh"1 a; in terms of logarithms as follows: y = cosh"1 a;,
If then
x = cosh y
so that
e*v2xev+l = 0.
Solving this equation as a quadratic in ev, we have and so, by definition of the exponential function, y = log{x±J(x2!)}. These are the two values of y. Moreover their sum is
= 0.
Hence the two roots are equal and opposite. The positive root is log {x + <](x2 1)} and the negative log {x  <J(x2  1)}.
THE INVERSE HYPERBOLIC COSINE
97
1
To find the diflFerential coefficient of cosh" a;, we differentiate 3oefficient o: the relation cosh y = x with respect to x. Then
dy dx
or
1 sinh y
± ^/(cosh2 y — 1) 1
±
The gradient is positive when y is positive and negative when y is negative. In particular, if we take the POSITIVE value of " 1 ^ , then dy 1 dx We can also obtain the result from the formula
taking the positive value. For if
then
ur Hence
dy dy du T~~T~~ 1
Note the corresponding integral 17~2—v\ ~ c 08 ^" 1 x J <J(x — I)
(positive value)
98
THE HYPERBOLIC FUNCTIONS
6. The inverse hyperbolic sine. On 'turning the graph of sinh a; through a right angle and taking a mirror image', and then renaming the axes, we obtain the graph (Fig. 73) y = sinh""1^/ of the inverse hyperbolic sine of x> that is, of the number whose hyperbolic sine is x. The function sinh"1 a; is a SINGLEVALUED function, uniquely determined for all values of x. To express sinh"1 a; in terms of logarithms, we write sinh""1 a; = y, so that x = sinh y = \{ey — e~y)9
O
giving e2y  2xey  1 = 0. Solving this equation as a quadratic in ey, we have
Fig. 73.
But (p. 19) the exponential function is positive, so that we must take the positive sign for the square root. Hence
or without ambiguity. To find the differential coefficient of sinh"1 x, we differentiate the relation sinh y with respect to x. Then
or
1 dy= dx cosh y 1
1 ± ^(sinh2 y + I)
But, from the graph (Fig. 73), the gradient is always positive, 1 and so dy __
dx~~*J{x2+lY
THE INVERSE HYPERBOLIC SINE
We can also obtain this result from the formula
For if then
x
+ (x2 + l)*
du dx " 1
s/(*2f
n YT2 4 1 \ 4. ^1 dy__ dx
Hence
1
w 1
Note the corresponding integral —z—rr = sinh* 1 x
EXAMPLES II
Find the differential coefficients of the following functions: 1. x cosh"1 x. 4. sech"1^.
2. sinh 1 (l + ^ 2 ). 5. cosech""1^.
3. tanh 1 ^. 6. log (cosh*1 #).
7. ajcosh 1 ^ 2 ^!).
8. (cosh"1 a;)2.
9. l/(sinh1o;)
Find the following integrals: dx
11 r
dx
12 2 r ***
J
JV(2 + 2 + 5 ) '
—4arl«)'
CHAPTER X CURVES 1. Parametric representation. Hitherto we have regarded a curve as defined by an equation of the form y =/(*)•
For many purposes it is more convenient to adopt a parametric representation whereby the coordinates x,y of a point T of the curve are expressed as functions of a parameter t in the form
(Of course, there is no reason why the parameter should not be x itself.) For economy of notation, however, we often write x = x(t)9
y = y(t).
Familiar examples from elementary coordinate geometry are the representations x = at2,
y = 2at
for the parabola y2 = 4ax, and x = ct,
y — c/t
for the rectangular hyperbola xy = c2. We confine ourselves to the simplest case, in which x(t),y(t) are singlevalued functions of t with as many continuous differential coefficients as the argument may require. The 'dot' notation x,y,x,y,... will be used to denote the differential coefficients dx dy d2x d2y The tangent at the point ef of the curve
PABAMETEIO EEPEESENTATION
101
may easily be found; it is the line through that point with gradient dx =
x g'(t)
f'(t)' For the reader familiar with determinants, an alternative form of equation may be given: The equation of any straight line is Ix + my + n = 0. If this line passes through the point x = f(t), y = g(t), then lf(t) + mg(t) + n = 0; if it passes through the point x =f(t + St), y = g(t+St), then lf(t + St) + mg{t + 8t) + n = Q. Subtracting, we have the relation l{f(t + St) /(*)} + m{g(t + St)  g(t)} = 0, or, on division by St,
st
°
For the tangent, we must take the limiting form of this relation as 8 ^ 0 , namely lf(t) + mg>{t) = 0. Hence, on eliminating the ratios I :m :n, we obtain the equation of the tangent in the form x y 1 f{t)
g(t) 1 = 0 .
/'(*) g'(t) o 2. The sense of description of a curve. We regard that sense of description of a curve as positive which is followed by a variable point for increasing values of the defining parameter. For example, if the points A9P,Q (Fig. 74) correspond to the values a,p, q respectively, and if a
102
CTJEVES
For example, the positive quadrant of the circle x2 + y2 = 1 may be expressed with x as parameter in the form
y B T
O
Fig. 74.
X
A
x
Fig. 75.
As x increases from 0 to 1, the arc is described in the sense BA of the diagram (Fig. 75). On the other hand, if the polar angle 6 is taken as parameter, we have x = cos 9, y = sin 9 As 9 increases from 0 to TT, the arc is described in the sense AB. 3, The'length* postulate. If we confine our attention to the simplest case, where the curve has a continuously turning tangent, as in the diagram (Fig. 76), then our instinctive ideas will be satisfied if we ensure that the length of a curved arc PQ is nearly the same as that of the chord PQ when the points P,Q are very close together. For this purpose, we shall base our treatment of length on the postulate chord QP
= 1.
Fig. 76.
Our aim is to let the derivation of all the standard formulae of the geometry of curves rest on this single assumption, together with the normal manipulations of algebra, trigonometry and the calculus.
THE ' L E N G T H ' P O S T U L A T E
103
The warning ought perhaps to be added that in a more advanced treatment it would be necessary to examine whether length' exists at all and to proceed somewhat differently. The present treatment suffices when dxjdt, dyjdt are continuous; this is true for 'ordinary' cases such as we shall be considering. 4. The length of a curve. Suppose that U PtP' are three points on the curve (Fig. 77) P'/ * u x = x(t), y = y(t) given by the values u,p,p + 8p of the parameter. For convenience, we assume that
u
whether Sx, 8y are positive or negative. Now the length of the arc UP is a function of p, which we may call s(p). Thus, since arc PP' = arc UP1  arc UP, we have arc PP' = s(p+ Sp)s(p). But
s(p + 8p) s{p) _ s(p + Sp)  s(p) chord PP' Sp ~~ chord PP' * Sp arc PP' chord
If we proceed to the limit, as Sp > 0 so that P' > P, then lim arc PP' _1 '~ ' chord PP' Sx $P
dx dp
8y sP+oSp
dy dp
104 Hence
CTJBVES s'(p) •
where the POSITIVE square root must be taken since 5 increases with p. Replacing p by the current letter t, we have the relation
Integrating, we obtain the formula
measured from the point U with parameter u. 5. The length of a curve in Cartesian coordinates. If the coordinate x is taken as the parameter t, then the formula of § 4 becomes
so that
*W
measured from the point where x = a, where the positive sense of the curve is determined by x increasing. In terms of the coordinate y, we have similarly
measured from the point where y = 6, where the positive sense of the curve is determined by y increasing. ILLUSTBATION 1. To find the length of the arc of the parabola y2 = Aax from the origin to the point (x, y), where y is taken to be positive.
The parametric representation is x = at2, so that Hence
y = 2at,
x = 2at, y = 2a.
[l2aM2
Jo
LENGTH OF CXJBVB IN CARTESIAN COORDINATES
105
To evaluate the integral, write
=U(t I=U(t*+l)dt on integration by parts. Hence
Hence
/ = *V(*2 + l) + ilog{t + j(t*+ 1)},
so that
s = a* V(Z2 + 1) + a log # + j(t2 + 1)}.
6. The length of a curve in polar coordinates. Let the equation of the curve in polar coordinates be If 8 is taken as the parameter, then the formula of § 4 becomes
Now
x = r cos 8, y — r sin 8,
where r is a function of 8. Hence dx j
dr
n
7/j
cLU
(LU
.
n V^VyiO V
/
n
Oi.ll l/j
dy
dr .
J f\
jh
ctu
du
aX±X
V ~J~ T
COS
Uj
To) + 1 ^ 1 = TZ + r •
It follows that so that
8'(0) = 7 ( ( ^ +A, 5(0) =
measured from the point where 8 = a, where the positive sense of the curve is determined by 8 increasing.
106
CURVES
2. To find the length of the curve (a given by the equation r = a ( i + COS 0). ILLUSTRATION
CARDIOID)
The shape of the curve is shown in the diagram (Fig. 78). We have — = — a sin 0, so that do = 2a2 (1 + cos 0) = 4a2 cos2 J0. Hence the length of the curve is I 2a cos \6dQ = 4a  sin \B1 * = 4a[sin (^77) — sin (— TT)]
= 8a. Note. If we had taken the limits of integration as 0, 2TT, we should apparently have had the result that the length is f2ir
r
\2n
2a cos 100*0 = 4a sin £0 = 4a [sin TT — sin 0] = 0. It is instructive to trace the source of error. This lies in our assumption that ^ c o g 2 id) = 2a cos \0. When 0 lies between — TT, TT, this is true, since cos0 is then positive. But in the interval TT, 2TT, the value of cos  0 is negative, so that
V(4a2 cos2 ¥) =  2a cos id
Hence we must use the argument:
f Jo
Jn
Jo Cn =
P
2acos^0a0 — Jo Jn
r
v
r I2*
= 4a sin 10 — 4a s i n  0 Jo L in 8a.L
THE 'GRADIENT ANGLE* \ft
107
1. The 'gradient angle* 4>, If the tangent at a point P of the curve makes an angle if; with the #axis, then
The diagrams (Fig. 79) represent the four ways (indicated by the arrows) in which a curve may leave' a point P on it, the parameter being such that the positive sense along the curve is that of the arrow.
o (i) FIRST QUADRANT
(ii) SECOND QUADRANT
dx + ; dy + ; cos y> + ; sin xp +. dx —; dy 4: cos y) —; sin ip + .
(iii) THIRD QUADRANT
(iv)
— ;dy — ; cosy — ; siny> —.
FOURTH QUADRANT
; dy —; cosy + ; s i n y ^ .
Fig. 79.
dx Thus = iis positive for (i), (iv) and negative for (ii), (iii); while fit]
~ is positive for (i), (ii) and negative for (iii), (iv). ds
82
108
CURVES
But
\axj rl = cos2*/*; ds) dy\2
( pi
= sin 2 ^.
Hence =, p have the numerical values of cos ip, simp respectively; and, if we define the angle \p to be the angle I whose tangent is pi from the positive direction of the #axis round to the positive direction along the curve, in the usual counterclockwise sense of rotation, then the relations dx , dy . , y = cos w, ~ = smTib ds ds hold IN MAGNITUDE AND IN SIGN for each of the four quadrants, as the diagram implies. We therefore have the relations dx
dy
true for every choice of parameter by correct selection of the angle ip. EXAMPLES I
1. If the parameter is x, then ip lies in the first or fourth quadrant. 2. If the parameter is y, then ip lies in the first or second quadrant. 3. If the parameter is 6, then ip lies between 8 and 8 + IT (reduced by 2n if necessary). 4. What modifications are required in the treatment given in § 7 if the curve is parallel to the yaxis ? Prove that the formulae dy dx
— = cos T«/r, — = sm w r are still true. ds ds
ANGLE FROM RADIUS VECTOR TO TANGENT
109
8. The angle from the radius vector to the tangent. The direction of the tangent to the curve r=f(0) may be described in terms of the angle 'behind' the radius vector. For precision, we define <£ as follows: T
Fig. 80.
Fig. 81.
A radius vector, centred on the point P (Figs. 80, 81) of the curve, starts in the direction (and sense) of the initial line Ox; after counterclockwise rotation through an angle 0, it lies along the radius OP produced. A further counterclockwise rotation <j> brings it to the tangent to the curve, in the POSITIVE sense; this defines
From the relations x
we have
r cos 0,
y = r sin 0,
dx dr d0 — = — cos n a  r  r ds ds ds dy dr — = =
ds
ds
d0 ds
cos 0,
110
CURVES
so that (p. 108), when 6 is the parameter, = cos 6 — r j sin 8 = cos I}J = cos (5 + <j>) = cos <> / cos 9 — sin <£ sin si
= cos
= sini/r = si
r
Solving for sin ^, cos ^, we have the formulae .
.
dd dr
so that
dr
9. The perpendicular on the tangent. Let the line ON (Fig. 82) be drawn from the pole 0 on to the tangent at the point P of the curve r= Then, by elementary trigonometry, .dd ds
This formula may be cast into an alternative useful form:
1 = 1^1 p*
r^Xdd}
'dr
A
I I i ~T7» I
I •
Fig. 82.
If we write so that then
du de
1 _ \de) /^Y
P* ~
THE PERPENDICULAR ON THE TANGENT
111
Note. Since
generally if we allow p to take the sign of r. In any case, the numerical value of p is that of rsin<£. The expression for p in terms of Cartesian coordinates follows at once: for . , p = r sin
= xsimfj where (x9 y) are the coordinates of P . It should be noticed that the step
dy
or the equivalent
To pass from polar to pedal coordinates, we have p*
r*\d0)
r2'
We do not propose to develop the theory of these new coordinate systems any further. The following illustration demonstrates the use of some of the formulae.
112
CURVES
ILLUSTRATION 3. The PEDAL CURVE of a given curve with respect to a given pole. The locus of the foot of the perpendicular from a given point 0 on to the tangent at a variable point P of a given curve is called the pedal curve of 0 with respect to the given curve. Referring to the diagram (Fig. 82) on p. 110, we see that N is the point of the pedal curve which corresponds to P . The polar coordinates of N are (rv d^, where
Let us find the pedal coordinates (pl9 r±) of N in terms of the pedal coordinates (p} r) of P . We have at once
Also, if
ddt dp Now
so that
p —
d0 dp
dS dp'
if rsin<£
dp
rsir
dOdr ~ ~j~ \ dr dp
s i n © . / ~z
.dd> dp ^dp
dr dp
a n ? >fdr S m
W
Y
dp
roosd)d
rCOS
^dp)
t a n cf>j(r sin <j> )=tan^
OTHER COORDINATE SYSTEMS
113
Hence
since each angle Kes between 0, TT and 9 is being used as parameter. It follows that ^ 1 = 71sin^1 = p sin cf> and so the pedal coordinates (pl9 r±) of N are (p2/r,p). COROLLARY.
Since t a n ^ 1 = ^  ^  , and <£ =
relation
7,
true for any curve. 11. Curvature. The instinctive idea of curvature, or bending, might be expressed in some such phrase as 'change of angle with distance', and it is just this conception which we use for our formal definition. DEFINITION. (See Fig. 83.) The CURVATURE K at a point P of a curve is defined by the relation
dif; ds
The value of K may be positive or negative, according as ifs increases or decreases with s. For many purposes, the calculation of
O
Fig. 83.
K is best effected by the direct use of the
definition. It is, however, useful to be able to obtain the formulae in the various systems of coordinates which we have described. (i) CARTESIAN PARAMETRIC FORM.
If x, y are functions x(t),y(t) of a parameter t, the relations . dy . , . dx = = cosTib. — = sinT 0 assume the form ds ds x = s cos ifj, y = s sin if/.
114
CURVES
Differentiating,
x = s cos ift — sijj sin iff 't — s2Ksinif/,
since similarly
y = s sin I/J + S2K COS 0.
Hence
y cos I/J— x simp = S2K, ^~ ™~" ~ ——• o / c «
or
,i
yx — xy
i
so that where
#c = s
with (p. 104)
POSITIVE
y
.o
y
.
square root.
Note. We must choose our parameter to avoid the possibiKty that x = 0, y = 0 simultaneously. This can be done, for example, by identifying t with x, when x = 1. (ii) CARTESIAN FORM.
In the particular case when x is the parameter, we have dx x = — = 1, and x = 0. Hence, by (i), dx dx2
Tx
nwr the denominator is
POSITIVE
since 7 is positive when x is the CuX
parameter. It follows that, with our conventions, the sign of K is the same as d2y the sign of r. Thus (Vol. 1, p. 54) the sign of K is positive when CuX
the concavity of the curve is 'upwards', and negative when the concavity is 'downwards'.
CURVATURE (iii)
115
POLAR FORM.
When 8 is the parameter, we have the relation (p. 109)
sothat
d
K=
Af
ds
Now (p. 110)
ddds
cotd> = ~ ,
so that or
/dry Hence
f& d?r
\de)
the positive sign being taken since 0 is the parameter.
Hence
(iv)
PEDAL FORM.
We use the Corollary (p. 113) tan^ = ^ , giving
•
'
•
; dr
116
CURVES
But = = cos T<£, so that T = sec T 6. Hence dr ds dr dp'
tan<£ = I dp r dr
so that
The sign conventions used in this proof imply that 0 was originally taken as parameter on the curve; otherwise, the sign of K may require independent examination.
12. A parametric form for a curve in terms of s. Suppose that 0 is a given point on a curve (Fig. 84). Choose the tangent at 0 as #axis and the normal at 0 as i/axis. We seek to express x, y in terms of s as parameter, assuming that the conditions are such that a Maclaurin expansion is possible. The Maclaurin formulae (p. 49) are O 9Z .' /A\
I
nitf
Fig. 84.
Now so that
and so that
dx
dsi
&y
•
i
£ = sm^r, d2y
4
ds
Sinj/r  f
PARAMETRIC FORM FOR A CURVE IN TERMS OF S 117 dK
If we write K0, KQ to denote the values of /c, = at the origin, then, CIS
since iff = 0 there, *'((>) = 1;
*"(<)) = 0;
^"(0) =
^ ^
y'(0) = 0;
*/"(0) = * 0 ;
We therefore have the expansions
13. Newton's formula. To prove that the curvature at the origin of a curve passing simply through it, and having y = 0 as tangent there, is o /co= l i m   . X+Q**'
We use a method like that of the preceding paragraph to obtain the expansion, assumed possible, of y as a series of ascending powers of x. We have that, if
y=f(x), then
*
» =dydx = tan?/r,
r
sec;
dx
=sec ds dx =sec \fs.K sec 2
K sSv
Thus
/(0)
so that
y
Hence
no) 0) + xf
2^ ~ = K0 + (terms involving x), x
and so, assuming conditions to be such that the remainder tends to zero with x9 2v
i
118
CUKVES
ILLUSTRATION 4. To find the curvature of an ellipse, of semiaxes a,b, at an end of the minor axis. Take the required end of the minor axis as origin (Fig. 85) and the tangent there as the #axis. The equation of the ellipse is
x*
(ybf +
l
Hence
o Fig. 85.
For values of y near the origin, the negative square root must be taken, giving 6
so that Hence
a;2 ~ a K
= Km
i2*
THE CIRCLE OF CURVATURE
119
14. The circle of curvature. The formulae dy
dx*
show that, if the two curves y=/(*),
both pass through a point P(g9rj), so that and if, further,
/'(£) = gr'(f)
and /"(£) = / ' (  ) , then the two curves have the same gradient and curvature at P. In particular, the circle which passes through P, touches the given curve y = f(x) at P, and has the same curvature as the given curve at P, is called the circle of curvature of the given curve at P. Denote by , „ the values of y,y',y" for the given curve at the point P(xP,yP). Suppose that the equation of the circle of curvature at P is where G (a, jS) is the centre of the circle of curvature and p its radius. By differentiation of this equation with respect to x, we obtain the relations
Since x, y9 y\ y" are the same for the circle of curvature as for the given curve,
120
CURVES ( 1 "I" V P
TT
rt
xience
p = yp + 
)
f^— yp
X
—n 2/p
P
>
Vp
where /cP is the curvature of the given curve at P. We therefore have the formulae a

j
x
VP
I —• i
9 Kp
the sign being selected to make p positive. If the coordinates x,y of a point on the curve are given as functions of a parameter t, then dy
dy Idx dt 1 dt dxd2y
d*y dx2
dyd2x dt dt2 dt idx \ 2 dx'
THE CIRCLE OF CURVATURE
121
so that, if dots denote differentiations with respect to t, dy ^y dx x9 d2y
Hence
OC =
___
XT>
xy — yx
—

i
~Kp
The point C (oc, /?) is called the centre of curvature of the given curve at P, and p its radius of curvature. We are adopting a convention of signs in which the radius of curvature is essentially positive. ILLUSTRATION 5. The curvature of a circle, and the sign of the curvature. Too much emphasis may easily be given to questions about the sign of curvature. Usually common sense and a diagram will settle all that is wanted. We give, however, an exposition for the case when the given curve is itself a circle, so that the reader may, if he wishes, be enabled to examine more elaborate examples. The difficulties about sign arise, with our conventions, as a result of varying choices of the parameter used to determine the curve. We begin with the simplest case, in which the parameter is selected so that the O circle is described completely in Fig. 86. one definite sense as the parameter increases. Let A (u,v) be the centre of the given circle (Fig. 86), and a its radius. Take a variable point P (x, y) of the circle, and denote by t
122
CURVES
the angle which the radius vector AP makes with the positive direction of the #axis. As t (the parameter) increases from 0 to 2TT, the point P describes the circle completely in the counterclockwise sense. Then
so that
x = u + acost,
y = v + asint9
x = — a sin t,
y = a cos t9
x = — acost,
y = — asin£.
x2 + y2 = a29
Hence
xij — yx = a2 (sin21 + cos21) = a2. We therefore have the relations , (a cos t) (a2) a = (^ + a cos £) — ^ —  = w, = v9 a3 Hence for all points on, a given circle, the centre of curvature is at the centre of the circle, and the radius of curvature is equal to the radius of the circle. The curvature at any point is therefore ± I/a. With the present choice of parameter, the formula of p. 114 shows at once that K = + I/a, but other parameters may give different signs. For example, if # is the parameter, the sense of description of the curve is LPM in the upper . part of the diagram (Fig. 87) and LQM in the lower, these being Fig. 87. the directions taken by P9Q respectively as x increases. We know (p. 114) that, when x is the parameter, K is positive when the concavity is 'upwards' and negative where it is 'downwards'. Thus K = — I/a in the arc
o
THE CIBCLE OF CURVATURE
123
LPM, and + I/a in the arc LQM. In fact, if P(x, y) is in the arc LPM, then '{a?{xu)% 2/ where the positive sign is attached to the square root. Hence
(xu)
y
{a2(xuyf9 •*2
so that , and
„
(xu)2
1 2
2
y = {a (xu) f a
2
{a (x\
2
Applying the formula (p. 114) for K, we have K = —
a2 {a2  (a  uffl
{a2  (a? 
a" Similarly we may prove that K = + I/a in the arc LQM, where
15. Envelopes. We have been considering a curve as the path traced out by a point whose coordinates are expressed in terms of a parameter. An analogous (dual) problem is the study of a system of straight lines Ix + my + n = 0 when the coefficients I, m, n, instead of being constants, are given to be functions of a parameter t, say
/=/(*), m = g(t), n = h(t). A familiar example is the system x — yt + at2 = 0
consisting of the tangents to the parabola y2 = 4ax. 92
124
CURVES
If we take a number of values of t, we obtain correspondingly a number of lines, which lie in some such way as that indicated in the diagram (Fig. 88). They look, in fact, as if a curve could be determined to which they are all tangents. More precisely, the diagram assumes that the individual lines are numbered in an
Fig. 88.
order corresponding to increasing values of the parameter, and the points of intersection of consecutive pairs (1,2), (2,3), (3,4),... have been emphasized by dots. These dots appear to lie on a curve, and it is easy to conceive of the lines as becoming tangents to that curve as their number increases indefinitely. In that case, the lines are said to envelop the curve, and we make the following formal definition: DEFINITION.
Given a system of lines Ix + my + n = 0,
v)hose coefficients Z, m, n are functions of a parameter t, the locus of that point on a typical line of the system, which is the limiting position of the intersection of a neighbouring line tending to coincidence with it, is called the envelope of the system. Consider, for example, the system of which a typical line is
xyt + at2 = 0. Another line of the system is # — yu + au2 = 0.
They meet where
x = aut, y = a(u +1).
ENVELOPES
125
That point on the typical line to which the intersection tends is found by putting u = t in the expressions for the coordinates, givillg
x = at2, y = 2at.
This is the 'parametric representation of the envelope, whose equation is therefore 2 We may now find the rule for determining parametrically the envelope of the system of which a typical line is Another line of the system is xf(u) + yg(u) + h(u) = 0. Where these lines meet, it is also true that
4/M /(*)}+vW)  0(0}+{* W  *«} = o, or, on division by u — t, that
JMftt)
{
g(u)g(t) {Mn)h(t) =
a
To emphasize the limiting approach of u to tf write u = t + St. Then the point of intersection of the two lines V , H' also lies on the line
J(t+8t)f(t) U
X
, ^g(t + 8t)g(t) , h(t+8t)h(t) _ + 8t ft " 0>
+y
In the limit, as S£>0, this is the line zf'(t)+yg'(t)+h'(t)
= O.
Hence the envelope is the locus, as t varies, of the point of intersection of the lines
ILLUSTRATION
6. To find the envelope of the system zcoat + ysint + a = 0.
The envelope is the locus of the point of intersection of this line with the line . ± x — x sin t + y cos t = 0.
126
That point is
CURVES
x = — acost,
y = — asintf,
and so the envelope is the circle Note. Envelopes exist for many families of curves as well as for families of straight lines; but we are not in a position to give a treatment of the more general case. EXAMPLES II
Find the envelopes of the following families of straight Knes: 2. xsect — yta,nt — a = 0. 3. #eosh£ — ysinht — a = 0. 4. 2tx+(lt2)y
+ (l + t2)a = 0.
5. tx + ya(tz + 2t) = Q. 6. tzxtyc(t*l) = 0. 16. Evolutes. Particular interest attaches to the envelope of those lines which are the normals of a given curve. Using the notation of § 14, denote by yp> yp> yp"
the values of y, y\ y" (where dashes denote differentiations with respect to x) for the given curve at the point P{xF,yP). The equation of the normal at P is or
x + yP'y = xP + yP'yP.
Taking xP as the parameter, the envelope of this line is the locus of its intersection with the line
so that X = Xp
ye"
EVOLUTES
127
Comparison with the results in § 14 establishes the theorem: The centre of curvature at a point P of a given curve is that point on the normal at P which corresponds to P on the envelope of the normals. The envelope of the normals, which is also the locus of the centres of curvature, is called the evolute of the given curve. DEFINITION.
ILLUSTRATION
7. To find the evolute of the rectangular hyperbola xy = c2.
Parametrically, the hyperbola is x = ct, y = c/t, and the gradient at this point is — 1/t2. Hence the normal is or
tzx — For the envelope, we have
The centre of curvature, being the point of intersection of these two lines, is given by = 3c*4+ c, so that
x = v = —u — l9t
2
The evolute is the curve given parametrically by these two equations. EXAMPLES III
Find the evolutes of the following curves: 1. The parabola (at2, 2at). 2. The ellipse (a cos t, b sin t). 3. The rectangular hyperbola (a sect, atan£).
128
CUBVES
17. The area of a closed curve. Consider the closed curve PUQV shown in the diagram (Fig. 89). For simplicity, we suppose it to be oval in shape, and also, to begin with, to lie entirely in the first quadrant. The coordinates of the points of the curve being expressed in the parametric form x = x(t),
y = y(t),
we suppose that the positive sense, namely that of t increasing, is COUNTERCLOCKWISE round the curve, as implied by Fig. 89. the arrows in the diagram. [If this is not so, replacement of t by —t will reverse the sense.] Thus the curve is described once by the point (x9 y) as t increases from a value t0 to a value tv Moreover, since the curve is closed, the two values t0, t± give rise to the
SAME
point, so that
x(t0) = xfa);
y{t0) = y(tx).
A simple example is the circle x = 5 f 3 cos t9 y = 4 + 3 sin t, described once in the counterclockwise sense as t increases from 0 to 2?r. The values t = 0,1 = 2TT both give the point (8,4). In order to calculate the area, draw the ordinates AP9BQ which just contain the curve, touching it at two points P, Q whose parameters we write as tP,tQ9 respectively. For reference, let U be a point in the lower arc PQ and V in the upper. Then the area enclosed by the curve can be expressed in the form area AP VQ  area APUQ. The area APVQ is given by the formula
area APF# = Xp
integrated over points (x, y) on the arc PVQ. Let us suppose first that the 'junction' point given by t0 or tl9 does not lie on this arc.
THE AREA OF A CLOSED CURVE
129
Then the parameter t varies steadily along the arc, and the area is dx T. On the other hand, the 'junction' point J must then lie on the lower arc PUQ, so we write the formula for the area APUQ in the form CXQ
area APUQ =
ydx
integrated over points (x, y) on the arc PUQ, giving area APUQ =
rj
JxP
ydx +
£
rxQ
JJ
ClQ dx
dx , dt+
ydx
)
where the value tx or tQ is taken for J according to the segment of the arc PUQ over which the respective integral is calculated, t± for PJ and t0 for JQ. In all, the area enclosed by the curve is thus
r_y
dx 
fa
dt+
dx
jtyTt dx
dt+ ft*dt+ dx 7
h L
1
/**» dx
Similarly, if J lies on the arc PVQ, we have the formulae rj
area APVQ =
rxQ
ydx+ Jxp
ydx JJ
dx ,.
area APUQ =
CXQ
Ctq dx .
ydx
JXp
I 'Q dx .
130
CURVES
so that the area enclosed by the curve is
[V^dt Hence, in both cases, the area of the closed curve is ** dx ,. In the same way, if we draw the lines CB,DS parallel to Ox (Fig. 90), just containing the curve, to touch it at B, S, and if L, M are points on the left and right arcs B8 respectively, then the area of the closed curve is
y D
c
y ( V
/ R X
0
area CB3IS  area CBLS. Now
S
Fig. 90.
I xdy area CB31S = JVB
integrated over points {x,y) on the arc BM8. point J is not on this arc, we have
If the 'junction'
For the area CEL8, the 'junction' point J must then lie on the arc BLS, so we have In all, the area of the closed curve is (in brief notation)
£{£•£}
THE AREA OF A CLOSED CURVE
131
Similarly for the 'junction' point on the arc BM8, we have
IMHD
+ +
r r r Jt0 ha JtR T1 dy , ]f
dt
Hence the area of the closed curve may be expressed in either of the
r
y dt t. Tt >
where the closed curve is described completely, in the counterclockwise sense, as t increases from t0 to tv
A useful alternative form is found by taking half from each of these: The area of the closed curve is
1 /V dy ILLUSTRATION
dx\ .
8. To find the area of the ellipse
The ellipse is traced out by the point x = a cos t9 y — bsint
as t moves from 0 to 2TT. Then dx = — a sin tdt,
and
^Uxdyydx)
1 •• r r a b .
\.Vn
dy — b cos tdtf
132 CUEVES The restriction for the curve to lie in the first quadrant is not essential. For, if it does not, a transformation of the type for suitable values of a, b, can always be employed to bring it into the first quadrant of a fresh set of axes. But this shows that the area enclosed is dy
since t0, tx give the SAME point of the curve. Hence the area is dy 18.* Second theorem of Pappus. Suppose that a surface of revolution is obtained by rotating an arc PQ (Fig. 91) about the #axis (assumed not to meet it). We proved (Vol. I, p. 130) that, if PQ is the curve = ^x the area S so generated is given by the formula 8=
rb
2<7ryds. Ja Now it is easy to prove that the ^/coordinate of the centre of gravity of the arc is given by the formula
I yds , Ja
o
A
B
Fig. 91.
I
where I is the length of the arc PQ. (Compare the similar work in Chapter vi.) Hence 8 = 2rrrjl * This paragraph may be postponed, if desired.
133
SECOND THEOREM OF PAPPUS
Thus if a given curve, lying on one side of a given line, is rotated about that line as axis to form a surface of revolution, then the area of the surface so generated is equal to the product of the length of the curve and the distance rotated by its centre of gravity. ILLUSTRATION
9. To find the centre of gravity of a semicircular arc.
Suppose that the circle is of radius a (Fig. 92), and that the centre of gravity, lying on the axis of symmetry, is at a distance 77 from the centre. On rotating the semicircle about its bounding diameter, we obtain the surface of a sphere, whose area is known to be Hence so that
7] =
2a Fig. 92. REVISION EXAMPLES V
'Advanced9 Level 1. Find the equations of the tangent and normal at any point of the cycloid given by the equations x = a(2ip + sin 2\jj), y = a(l — cos 2ip).
Prove that ifs is the angle which the tangent makes with the axis of x; and verify that, if p and q are the lengths of the perpendiculars drawn from the origin to the tangent and normal respectively, then q and dpjdijj are numerically equal. 2. Define the length of an arc of a curve and obtain an expression for it. A curve is given in the form x =
— t, y = cosh£ + t.
Express t in terms of the length s of the arc of the curve measured from the point (1,1). The coordinates of any point of the curve are expressed in terms of a and are then expanded in series of ascending powers of s.
134
CURVES
Prove that the first few terms of the expansions are
3. Find the radius of curvature of the parabola y2 = 4A# at a point for which x = c, and deduce that, if A varies (c remaining constant), this radius of curvature is a minimum when A = \c. 4. Prove that the radius of curvature of the curve y = \x2\\ogx (x>0) 2 2 at the point (x, y) is (l+o; ) /4a;. Find the point at which this curve is parallel to the #axis, and prove that the circle of curvature at this point touches the ya>xm. 5. Prove that, if iff is the inclination to the a;axis of the tangent at a point on the curve y = alogsec(#/a), then the radius of curvature at this point is a sec ip. 6. A particle moves in a plane so that, at time t, its coordinates referred to rectangular axes are given by x = a cos 2t + 2a cos t, y = asin2£ + 2asin£. Find the components of the velocity parallel to the axes and the resultant speed of the particle. Show that />, the radius of curvature at a point of the path, is proportional to the speed of the particle at that point. 7. The coordinates of the points of the curve 4yz = 27#2 are expressed parametrically in the form (223,322). By using this parametric representation, or otherwise, prove that the length of the curve between the origin and the point P with parameter tx is 2(1 +
nt
dx\
BEVISION EXAMPLES V
135
A curve is given parametrically by the equations x = 2a cos £ + a cos 2£, y = 2a sin t — a sin 2t. Prove that ifs = —£. P is a variable point on the curve, and Q is the centre of curvature at P; the point R divides PQ internally so that PR = \PQ. Prove that the locus of R is the circle x2 + y2 = 9a2. Prove also that no point of the curve lies outside this circle. Draw a rough sketch of the curve. 9. The coordinates of a point of a curve are given in terms of a parameter t by the equations x = te1, y = t2eK Find dyjdx in terms of t, and prove that d2y dx2 ""
Prove also that the radii of the circles of curvature at the two points at which the curve is parallel to the #axis are in the ratio e2:l. 10. Express cosh4 9 in the form a0+ax cosh 0+a2 cosh 20 + a3 cosh 3d+a4 cosh 40. 11. The area lying between the curve y = cosh (x/2\), 2
the ordinates x = + A , and the #axis is rotated about this axis. Prove that the volume of the solid of revolution so formed is 7rA(A + sinh A). Show that the volumes given by A = 1 , A = 1 + S differ by approximately TT(2 + e) S when S is small. 12. Prove that the two curves y1 = ^ cosh 2x, y2 = tanh 2x touch at one point and have no other point in common. Sketch the two curves in the same diagram and, with them, the curves , . , yz = cosh a;, j / 4 = smh#. Prove that the four curves intersect in pairs for two values of x and indicate clearly in the diagram the relative positions of the four curves for all values of x.
136
CURVES
13. If s be the length of the arc of the catenary y = a cosh (xja) from the point (0,a) to the point (x, y), show that s2 — y 2 — a 2 .
Find the area of the surface generated when this arc is rotated about the yaxis. 14. Sketch the curve whose coordinates are given parametrically by the relations x = a{t + sint),y = a(l + cost) for values of t between — TT9 TT, and find the length of the curve between these two points. 15. Find the radius of curvature of the parabola y2 = x at the point (2,^/2). 16. Find the radius of curvature and the coordinates of the centre of curvature at the point (0,1) on the curve y = cosh a;, 17. For what value of A does the parabola 2/2_
have the same circle of curvature as the ellipse
(
x~a\2
y2
at the origin ? 18. Find the radius of curvature of the curve y = sin ax2 at the origin. 19. Draw a rough graph of the function cosh a;. The tangent at a point P(x, y) of the curve y = c cosh (x/c) makes an angle I/J with the #axis. Show that y — c sec 0, If the tangent at P cuts the ?/axis at Q, show that PQ = xylc. 20. Find the radius of curvature of the curve ay2 at the point (a, a).
BEVISION EXAMPLES V
137
21. Find the radius of curvature at the point t = \TT on the curve . x n. x = a sin t, y = a cos 2t.
22. Find the radius of curvature of the curve y
at the point (x, y) and find a point on the curve for which the centre of curvature is on the yaxis. 23. Show that the radius of curvature of the epicycloid x = 3cos£ — cos 3^,
y = 3sin£ — sin3£
is given by 3 sin t. 24. Sketch the curve whose equation is r = asin30
(a>0).
Find the area of the loop in the first quadrant and show that the radius of curvature at the point (r, 0) is (5+ 4 cos 60)* * a (7+ 2 cos 60)' 25. Show that the two functions sinh"1 (tan a;), tanh""1 (sin x) have equal derivatives, and hence prove that the functions themselves are equal when — \TT < x < \TT. 26. Find the point of maximum curvature on the curve y = logo;, and the curvature at this point. 27. If y is the function of x given by the relation sinhy = tana), where —\TT<X<\n> prove that cosh y = sec x,
y = log tan (\x + \TT).
Show that y has an inflexion at x = 0, and draw a rough sketch of the function in the given range. IO
Mil
138
CURVES REVISION EXAMPLES VI
'Scholarship* Level 1. Transform the equation d2u
2z du
u
to one in which y is the dependent and x the independent variable, Where
u(lz2)*
= y and (l + z 2 )/(lz 2 ) = x,
obtaining the result in the form
2. If f(x) is a polynomial which increases as x increases, show that, when x > 0, is also a polynomial which increases as x increases. Show that, when x > 0, the expression (x22x + 2)ex 2 x x is an increasing function of x. 3. Prove that
Prove also that the function
satisfies the equation
4. If j/ = sin (msiii"1^), show that
REVISION EXAMPLES VI
139
Hence or otherwise show that the expansion of sinm0 as a power series in sin 9 is . J (m2l) . (m 2 l)(ra 2 3 2 ) . An4 msin0 l^———sm29Q fl + ~sin 0...
5. By induction, or otherwise, prove that, if y = cot"1 a?, then dny —* =
(—l)n(n—l)\sinnysmny.
(AJX
Hence, or otherwise, prove that, if
( then
• » asm ex \ 1+X
COS OLj
— = (— 1 ) n ~ 1 —. — sin n (oc — z) sin™ (a — z). v y dxn smna
6. The function / n (re) is defined by the relation
where y = e^2. Prove t h a t dy
and deduce that and that /w(a;) is a polynomial in # of degree n. Prove that
/ ^ ( a ) = fn'(x) + 2^(a;),
and hence express / n+2 (^) in terms of fn{x), fn'(x) Deduce that
» n.
N, n
* ,,
x
and
fn"(x).
o i ? / \ ^
fn"(x) + 2xfn'(x)2nfn(x) = 0. 7. If ^ = sin a;, prove that 1
xdyX Determine the values of y and its successive derivatives when x = 0, and hence expand y in a series of ascending powers of x.
140
CUBVES 1
8. Prove that, if y = tan" ^, then dny n U^T~ = (n— 1) \cos y cos {ny + l(n— 1)TT} for every positive integer n. Deduce, or prove otherwise, that u satisfies the differential equation
9 K prove the relations
^tl
= 2(n+
10. If prove that
l)x^
+ 2(n+ lfy
yn(x) = J ^ yn =
x
n dx
 ^
dx
Hence show that the polynomial yn satisfies a certain linear differential equation of the second order,
i.e. linear in r^r> r^> yn. I dx2 dx *n \ 11. Show that, if the substitution u = ta+1 ,~ is made in at the equation du ~dt then Deduce that, if a is not a negative integer, then the value of ~ when t = 0 is fan
A
where A is the value of y when t = 0.
EEVISION EXAMPLES VI 2
141
n
12. Prove that the polynomial y = (x — 2x) satisfies the equation
13. A function f(x) is defined for 0 < x
3 . TTX
f(x)—sm
has a stationary value at x = 1, and determine whether it is a maximum or a minimum, 14. Prove that, if y = e~x2/2 and n is a positive integer, then
The functions / w (#) are defined by the formula
Prove that
(i) f^+1 = (n + l)/ w ,
(fi)/•**== */»/»» (Hi) fn+2tfn+1 + (n+ l ) / n = 0, (iv) / w is a polynomial in a: of degree w. 15. Obtain the equations for a curve C. Prove that the equation of the general conic having 4point contact (i.e. 4 'coincident' intersections) with C at the origin 0 is where p = l//c and A is an arbitrary constant. Deduce that the length of the latus rectum of a parabola having 4point contact with a circle of radius a is 2a.
142
CTJEVES
16. Obtain the equations The point Q on the curve is such that P is the middle point of the arc OQ; the tangent at the origin 0 cuts the chord PQ at T. Find the limiting value of the ratio TP : TQ as Q tends to 0. Prove that, correct to the second order in s, the angle between the tangents to the curve at P and Q is KSI l + — s\. 17. Show that, if f(x) has a derivative throughout the interval
a^x^b,
then /(&)/(*) = (ba)f'{a+8(ba)}9
where 0 lies between 0 and 1. Find the value of 6 if/(#) = tan#, a = JTT, 6 = TT. Draw an accurate graph of the function y = tana; between the limits \TT, JTT, taking 1 in. to represent ^rr as the #axis and 1 in. to represent O2 on the yaxis. Illustrate the theorem by means of your graph. 18. Given that
eay = cos a;
and that yn denotes dny/dxn9 prove that and express yn+2 in terms ofyvy2, ...,2 The expansion of y as a power series in x is
Prove that bn is zero when n is odd and that, if a is positive, bn is negative when n is even. 19. Prove that the nth. differential coefficient of eaV3cos# is 2neajV3cos
•K)
Deduce, or obtain otherwise, the expansion of the function as a series of ascending powers of x giving the terms up to x% and the general term. Find the most general function whose nth differential coefficient is exV3cos#.
KEVISION EXAMPLES VI
143 2
20. (i) Prove by induction that, if x = cot 6, y = sin 0, then dny dxn
=
(ii) Prove that
,
K
r
=— ^—
=,
Z^L.*
where fn{x) is a polynomial of degree n in as. Prove further t h a t
21. State, without proof, Rolle's theorem, and deduce that there is a number £ between a, 6 such that explaining what conditions must be satisfied by the function f(x) in order that the theorem may be valid. If f(x)=zsinx, find all the values of f when a = 0,6 = §77. Illustrate the result with reference to the graph of sin a;. 22. If a is small, the equation sin x = OLX has a root nearly equal t o 7T. S h o w t h a t
f1
,
2
/1 2 • i \
TT{1  a + a2  (f772 + 1 )a3} is a better approximation, if a is sufficiently small.
23. A plane curve is such that the tangent at any point P is inclined at an angle (k +1)0 to a fixed line Ox, where k is a positive constant and 0 is the angle xOP. The greatest length of OP is a. Find a polar equation for the curve. Sketch the curve for the cases k = 2, k =  . 24. Prove that d*d*
[d*)
dx)
dy2dy*
\dy*)
\dy)
25. Obtain the expansion of sin re in ascending powers of x. For what values of x is this series convergent. A small arc PQ of a circle of radius 1 is of length x. The arc PQ is bisected at Qx and the arc PQt is bisected at Q2. The chords PQ,PQ1,PQ2 are of lengths c,\cx,\c% respectively. Prove that ^5(c — 20cx + 64c2) differs from x by a quantity of order x1.
144
CURVES
26. Sketch the graph of the function where a, b are both positive. Prove that there are always at least two points of inflexion. Find the abscissae of the points of inflexion when a =  ° , 6 = 5.
dt9 dt9 dt*9
(J11
(JOT
s
*
'
(I
•
&
77
fi
v
0T
ta
temls
o t
Prove that (dxVcPy dy)) dx* + \dx) dy* + dx* dy* 28. The functions (X),I/J(X) are difFerentiable in the interval a^x^bf and if*'(x) >0 for a ^ a ; ^ 6 . Prove that there is at least one number £ between a, b such that
If ^(x) = a:2, if*(x) = a;, find a value of  in terms of a and 6. 29. Prove by differentiation, or otherwise, that for all real x and positive y. When does the sign of equality hold ? 30. (i) Prove that, for positive values of x,
(ii) Find whether e'^'sec 2 ^ has a maximum or a minimum value for x = 0. 31. Show that, if /(0) = 0 and iff'(x) is an increasing function fix) of x, then y = ^  ^ is an increasing function of a; for x > 0. a?
32. Prove that, if a; > 0, (1  a;2 + ^a: 4 ) sin a; > ( #  ia^ + rioar5) cos a;.
BEVISION EXAMPLES VI
145
2
33. If x> 0, prove that (xl) is not less than #( Discuss the general behaviour of the function for positive values of x and with special reference to x = 1. Sketch the graph of the function. 34. Prove that, if x is positive, 2x Prove also that, if a, h are positive, then log (a + Oh) — log a — 0{log (a + h) — log a}, considered as a function of 9, has a maximum for a value of 0 between 0 and  . 35. If the sum of the lengths of the hypotenuse and one other side of a rightangled triangle is given, find the angle between the hypotenuse and that side when the area of the triangle has its maximum value. 36. A pyramid consists of a square base and four equal triangular faces meeting at its vertex. If the total surface area is kept fixed, show that the volume of the pyramid is greatest when each of the angles at its vertex is 36° 52'. 37. Find the least volume of a right circular cone in which a sphere of unit radius can be placed. 38. Find the numerical values of y = sin(o; +J7r) + sin4a; at its stationary values in the range — 7r<#<7r. Distinguish between maxima, minima, and points of inflexion, and give a rough sketch of the curve. 39. Prove that, if 0=00^% n
then
(O<0
d9 j  ^ = (  l)n (n  1)! sin718 sinnd$
when n is any positive integer.
146
CURVES
Show that the absolute value of dn6jdxn never exceeds (n— 1)! if n is odd, or (n — 1)! cosn+11
1 if n is even.
\2n + 2j 40. Find the two nearest points on the curves y*±x = 0, x2 + y26y + 8 = 0, and evaluate their distance.
(i) find the maximum and minimum values of y; (ii) find the points of inflexion of the curve; (iii) sketch a graph showing clearly the points determined in (i), (ii), and also the position of the curve relative to the line y = x. 42. Prove that the maxima of the curve y = e~kxsinpx, where Jc,p are positive constants, all lie on a curve whose equation is y = Ae~kx, and find A in terms of k,p. Draw in the same diagram rough sketches of the curves y _ ekx^ y _. _ekx a n ( j y _ ekx s i n ^ # for positive values of x. 43. The tangent and normal to a curve at a point 0 are taken as axes. P is a point on the curve at an arcual distance s from 0. Prove that the coordinates of P are approximately x = s  i* 2 s 3 , y = ^ 2 + $fcV, where K is the curvature at 0, and AC' the value of dKJds at O. The tangents at 0 and P meet at 2\ A circle is drawn through 0 to touch PT at 2\ Prove that the limiting value of its radius as P tends to 0 is l/(4/c). 44. Obtain the coordinates of the centre of curvature at a point (x, y) of a curve in the form (# — p sin if/, y + p cos 0), where p is the radius of curvature and tan 0 the slope of the tangent at the point. Prove also that, when s is the length of arc between (x, y) and a fixed point on the curve, dx/ds = COSI/J, dyfds — sin^r. The centres of curvature of a certain curve lie on a fixed circle. Prove that p ~ is constant. r da
REVISION EXAMPLES VI
147
45. Obtain the formula p = rdrjdp for the radius of curvature of a curve in terms of the radius vector and the perpendicular from the origin on the tangent. Prove that a curve for which p = p satisfies the equation r2 = p2 + a2, where a is constant, and deduce that the polar equation of the curve is <J(r2 a2) = aO + a cos" 1 (a/r),
where the coordinate system is chosen so that the point (a, 0) is on the curve. 46. (i) Prove that the (p, r) equation of the cardioid r = a(l + cos#) 2ap2 = r3.
is
Hence, or otherwise, prove that the radius of curvature is fa cos \ 0. (ii) Prove that the equation of the circle of curvature at the origin of the curve 3(x2 + y2)
is
47. If y is a function of x, and x = g cos oc — 7] sin a,
y =  sin a + rj cos a,
where a is constant, express ^ and ^f in terms of ~ and ^~ dx 2
dy
dx2 d2t]
dt;
dj;2
Deduce that and interpret this result. 48. A curve is such that its arc length s, measured from a certain point, and ordinate y are related by y2 = 52 + C2, where c is a constant. Show that referred to suitably chosen rectangular axes the curve has the equation y = c cosh (x/c).
148
CTJBVES
If C is the centre of curvature at a point P of the curve and G is the point in CP produced such that CP = PG, prove that the locus of G is a straight line. 49. Find the radius of curvature at the origin of the curve y = 2x + 3x22xy + y2 + 2y*9 and show that the circle of curvature at the origin has equation
50. Find the values of x for which y = x2(x — 2)3 has maximum and minimum values, and evaluate for these values of x the curvature of the curve given by the equation above. 51. Sketch the curve defined by the parametric equations x = a cos31, y = a sin31. Show that the intercept made on a variable tangent by the coordinate axes is of constant length. Find the radius of curvature at the point ct\ 52. Find the points on the curve y(x—l) = x at which the radius of curvature is least. 53. Sketch the curves y = x2 — xz,
y2 = x 2 — # 3 ,
and find their radii of curvature at the origin. 54. Prove that all the curves represented by the equation yn+l
a
_ (_ab_\n
b
for different positive values of n9 touch each other at the point / ab b9
ab \ a + b)'
Prove that the radius of curvature at the point of contact is equal to (
REVISION EXAMPLES VI
149
55. Establish the (p, r) formula for the radius of curvature of a plane curve. For a certain curve it is known that the radius of curvature is anlrn"19 where n 4= — 1 and a is positive, and also that p = a/(n+1) when r = a. Show that it is possible to express the curve by the polar equation rn = (n+ l)anGosn0. 56. Show that the curvature of the curve a/r = coshw0 has a stationary value provided that Sn2 is not less than 1. Determine whether this value is a maximum or a minimum. 57. 0 is the middle point of a straight line AB of length 2a, and a point P moves so that AP. BP — c2. Show that the radius of curvature at P of the locus is 2c2/*3/(3r4 + a4 —c4), where r = OP. 58. Find the integrals:
> jeP*ooabxdx
(a #0,6 4=0), J 
dx
59. Prove that the area enclosed by the curve traced by the foot of the perpendicular from the centre to a variable tangent of the ellipse b2x2 + a2y* = a2ft2 is ±7r( 60. (a) Prove that, if
F(x) = e2x [Xe~2lf{t)dte* f V then
(i) F(0) = 0, (ii) J"(0) = 0,
(iii) I"'{z)3F'(x) + 2F(x) =/(*). (6) Find the most general function
I
t
61. Sketch the curve whose polar equation is r2 = a 2 (l + 3 cos 6), and find the area it encloses. *o
/\ T>
j.ce
x J.'
j.1.
•
62. (I) By differentiating the expression otherwise, find
f xdx
Jsin 6 #' (ii) Find
f
w # c o s # + sina;
^zi
sin
QO
> or
150
CURVES
63. Sketch the general shape of the curve x = atcost,
y = at&mt
for positive values of the parameter t. Find an expression for the length of the arc of the curve measured from the origin to the point t. 64. (i) Prove that, if n is a positive integer, [
ex cos nxdx = —^— {Ae** — 1},
where A has one of the values ± 1 , ±n; and classify the cases. (ii) Find the area bounded by the parabola y2 = ax and the circle 2 # + t/2 = 2a2. 65. Find the indefinite integral log (1 + x2) tan* 1 xdx.
/••
66. Show that the four figures bounded by the circle r = 3a cos 8 and the cardioid r = a(l + cos 6) have areas
67. The polar equation of a curve is r 2  2 r c o s 0 + sin20 = 0. Sketch the curve and prove that the area enclosed by it is TT/^2. 68. Determine the function (x) such that \XX(j)(t)etdt=(l+x)2ex, Jo Prove also that it is possible to find a pair of quadratic functions f(x), g(x) such that
s l + f*g(t)dt, Jo and determine these functions.
REVISION EXAMPLES VI
151
69. A plane curvilinear figure is bounded by the parabola x2 = ^y from the origin to the point (13,5); by the hyperbola x2 — y2 = 144 from the point (13,5) to the point (15, 9); and by the parabola y2 = ^x from the point (15,9) to the origin. Prove that the area of the figure is 33^+ 72 log f. 70. Prove that
Verify the identity  xf = (1 + x2) (4  4x2 + 5#4  4x5 + x6)  4,
and, by using this identity in the second of these integrals, prove that 22 1 22 1 7 630 7 1260' _ _
^
fj
^
_ _
71. Determine constants A,B9C,D such that
)+tf+lHence or otherwise prove that <0 503 ° ' 5 0 2 < Jo I (% 7~2—7u^ ' +1)
72. Determine A9B,C,D such that
x2 _ d_ (Ax5 + Bx* + Cx\ 2 +lf~~ dx ) + and show that 73. If yr(x) satisfies the equation
ym(x)yn(x)dx = 0 (m#=7i).
show that Ji
D
152
CUBVES
74. Polynomials fQ{x)if1(x)>f2(x)f...
fm(x)fn(x)^x
Prove that
are defined by the relation
= ° ( m + n)>
Show that, if
where
aw =
^
^
75. If m, n are positive integers greater than unity, prove that 7mw=
cosmxGO&nxdx
Jo m
m m
mn
n+1
~^
j
m+n
Hence show, if p,q are positive integers, that TT( p
+ a)!
76. Find a reduction formula for (1 — x2)n cosh axdx. (1— a;2)3 cosh a; & .
Evaluate Jo
77. If y = sin p a; cos 2 0:^/(1 — & 2 sin 2 #) and p,q,k
are constants,
ax Hence, by taking suitable values for p, q, express m=
in terms of / m _ 2 , 7 m _ 4 .
Jo VU& 2 sin 2 z)
REVISION EXAMPLES VI
153
78. State and prove the formula for integration by parts, and p show that
Jo
xn( 1  x)mdx =
79. Find the volume and area of the surface of the solid obtained by rotating the portion of the cycloid x = a(0 + sin0), y = a(l + cos#) between two consecutive cusps about the axis of x. [Consider the range — TT ^ 6 < TT.] 80. Prove that the envelope of the system of lines is the curve y2(x + y— 1) = xz. 81. Find the envelope of the line x sin 3t — y cos 3t = 3a sin t,
expressing the coordinates of a point of the envelope in terms of a and t. 82. Through a variable point P (at2,2at) of the parabola y2 = 4ax a line is drawn perpendicular to SP, when S is the focus (a, 0). Prove that the envelope of the line is the curve 21 ay2 = x(x — 9a)2. 83. Sketch the locus (the cycloid) given by x = a(t + aint),
y = a(l + cost)
for values of the parameter t between 0, 4TT. Prove that the normals to this curve all touch an equal cycloid, and draw the second curve in your diagram. 84. Find the envelope, as t varies, of the straight line whose equation is xcoaH + ysinH = a. Sketch this envelope, and find the radius of curvature at one of the points of the curve nearest to the origin. 85. As t varies, the line xt2y+2aP = 0 envelops a curve F. Show that for each value of t, other than t = 0, the line cuts F at a point P distinct from the point at which the line touches F.
154
CUBVES
Find the equation of the normal to F at P, and deduce that the centre of curvature at P is given by x =  20atz  (3a/4J), y = 48a£5  3a*. Discuss the case t = 0. 86. Find the equation of the normal and the centre and radius of curvature of the curve ay2 = x3 at the point (at2, at3). Show that the length of the arc of the evolute between the points corresponding to t = 0, t = 1 is (13*/6)a. 87. Show how to find the envelope of the line y = tx+f(t). Show that the length of an arc of the envelope is given by
Hence obtain a formula for the radius of curvature in terms of t. 88. Prove that the equation of the normal to the curve may be written in the form x sin t — y cos 14 a cos 2t = 0, and find parametrically the envelope of the normal. 89. Find the equations of the tangent and normal at any point of the curve x = 3 sin t — 2 sin31, y = 3 cos t — 2 cos31, Prove that the evolute is = 2*. 90. The coordinates of any point on the curve xs f xy2 = y2 can be expressed in the form x = t2l(l+t2),
y=P/(l
+ t2).
Find the equations of the tangent and normal at any point, and deduce that the coordinates of the corresponding centre of curvature are . 2 1M 4 .
REVISION EXAMPLES VI
155
91. Prove that the complete area of the curve traced out by the ^
(2a cos t + a cos 2t,
2a sin t—a sin 2t)
is 2ira2.
92. Find the area inside the curve given by x = a cos t + c sin let, y = b sin t + d cos Jet (0 < t < 2TT), where & is an integer and a, 6, c, d are positive. [It may be assumed that c,d are so small in comparison with a, b that the curve has no double points.] 93. Sketch the curve x = a sin 2t, y = b cos31, where a > 0, b > 0, and find the area it encloses. 94. Show that the curve x = (£l)e',
y = te
has a loop, and find its area. 95. Trace the curve x = cos 2t, y = sin 3t for real values of t, and find the area of the loop. 96. Show that, for the range — a^x^a, curve
the area between the
and the #axis is f 97. Sketch the curve given by x = 2a(sin31 + cos3 £), y = 26(sin3 £ — cos31), and prove that its area is Znab. 98. Find the area of the loop (— 1 < £ < 1) of the curve _ lt2 X
_t(lt2) y
99. Prove that the area of the curve x = a cos t + b sin t + c, y = a' cos £ + 6' sin t + c' is n(ab' — a'b).
156
CURVES
100. Explain the reasons for using the formula 2772/ ds
to calculate the area of a surface of revolution. Apply the formula to show that the area of the surface obtained by revolving the curve r = a(l + cos0) about the line 0 = 0 is equal to
^TTO,2.
101. The curve traced out by the point x = a log (sec t + tan t) — a sin t> y — a cos t,
as t increases from — \n to + \n, is rotated about the axis of x. Prove that the whole surface generated is equal to the surface of a sphere of radius a, and that the whole volume generated is half the volume of a sphere of radius a. 102. Find the length of the arc of the catenary y = c cosh (xjc) between the points given by x = + a. Find also the area of the curved surface generated by rotating this arc about the #axis. 103. Evaluate the area of the surface generated by the revolution of the cycloid x = a(t — sin t)> y = a(l — cos t) about the line y = 0. 104. Two points A(0, c), P(£, rj) lie on the curve whose equation y = ccosh(#/c), and s is the length of the arc AP. If the curve makes a complete revolution about the #axis, prove that the area 8 of the curved surface, bounded by planes through A and P perpendicular to the #axis, and the corresponding volume V are given by c£ = 2V 
REVISION EXAMPLES VI
157
105. A torus is the figure formed by rotating a circle of radius a about a line in its own plane at a distance h( > a) from its centre. Find the volume and surface area of the torus. 106. AB, CD are two perpendicular diameters of a circle. Find the mean value of the distance of A from points on the semicircle BCD, and also the mean value of the reciprocal of that distance. Prove that the product of these means is 87r2>/2log(l+>/2). 107. Prove that the mean distance of points on a sphere of radius a from a point distant / ( ^ a) from the centre is a+(f/3a). What is the mean distance i f / > a ? 108. Calculate the average value, over the surface of a sphere of centre 0 and radius a, of the function (1/r3), where r is the distance of the point on the surface of the sphere from a fixed point C not on the surface and such that OC = /. Distinguish between the two cases f> a, f
CHAPTER X I COMPLEX NUMBERS 1. Introduction. It may be helpful if we begin this chapter by reminding the reader of the types of elements which he is already accustomed to use in arithmetic. These are (i) the integer (positive, negative or zero); (ii) the rational number, that is, the ratio of two integers; (iii) the irrational number (such as ^2, TT, e) which cannot be expressed as the ratio of integers. It is now necessary to extend our scope and to devise a system of numbers not included in any of these three classes. In order to exhibit the need for the new numbers, we solve in succession a series of quadratic equations, similar to look at, but essentially distinct.
Following the usual 'completing the square' process, we have the solution x26x =5, x26x + 9 =  5 + 9, = 4, 2
(x3)
= 4.
The important point at this stage is the existence of two integers, namely + 2 and —2, whose square is equal to 4. Hence we can find integral values of x to satisfy the equation: #  3 = +2 so that
or z  3 =  2 ,
x = 5 or x — 1.
The equation x2 — 6x f 5 = 0 can therefore be solved by taking x as one or other of the rational numbers (integers, in fact) 5 or 1.
INTRODUCTION 2
II.
159
X 6X+7 = O.
Proceeding as before, we have the solution
x26x
= 7,
2
x 6x + 9 =  7 + 9, = 2, (xS)
2
= 2.
Now comes a break; for there is no rational number whose square is 2, and so we cannot find a rational number x to satisfy the given equation. We must therefore extend our idea of number beyond the elementary realm of integers and rational numbers. This is an advance which the reader absorbed, doubtless unconsciously, many years ago, but it represents a step of fundamental importance. The theory of the irrational numbers will be found in textbooks of analysis; for our purpose, knowledge of its existence is sufficient. In particular, we regard as familiar the concept of the irrational number, written as ^2, whose value is 1*414..., with the property that (yj2)2 = 2. Once the irrational numbers are admitted, the solution of the equation follows: #  3 = +>/2 or x3 = <]2t
so that
x = 3 + sj2 or £=3^/2.
The equation x2 — 6x + 7 = 0 can therefore be solved by taking x as one or other of the irrational numbers 3 + <J2 or 3 — ^2.
III. As before,
X2  6 # + 1 0 = 0 .
=—10,
X2
_6#
X2
6x + 9 =  1 0 + 9,
(£3)2
=_1.
Again comes the break; for there is no number, rational or irrational, whose square is the NEGATIVE number —1. We have therefore two choices, to accept defeat and say that the equation has no solution, or to invent a new set of numbers from which a
160
COMPLEX NUMBERS
value of x can be selected. The second alternative is the purpose of this chapter. To be strictly logical, we should now proceed to define these new numbers and then show how to frame the solution from among them. But their definition is, naturally, somewhat complicated, and it seems better to begin by merely postulating their 'existence'; we can then see what properties they will have to possess, and the reasons for the definition will become more apparent. Just as, in earlier days, we learned to write the symbol '^2' for a number whose square is 2, so now we use the symbol 'J( — 1)' for a 'number* (in some sense of the word) whose square is — 1; but, in practice, it is more convenient to have a single mark instead of the five marks J, (, —, 1,), and so we write
as a convenient abbreviation. We shall subject this symbol i to all the usual laws of algebra, but first endow it further with the property that
This will be its sole distinguishing feature—though, of course, that feature is itself so overwhelming as to introduce us into an entirely new numberworld. It may be noticed at once that the 'number' — i also has — 1 for its square, since, in accordance with the normal rules,
= 1. We round off this introduction by displaying the solution of the above equation: x — 3 = t or x — 3 = — t, so that x = 3 + i or x=3 — i. By selecting x from this extended range of 'numbers', we are therefore able to find two solutions of the equation. It may, perhaps, make this statement clearer if we verify how 3 + i, say, does exactly suffice. By saying that 3 f i is a solution of the equation x2 — 6x + 10 = 0, we mean that +10 = 0.
INTRODUCTION
161
The lefthand side is 1+t 2 = 0 = righthand side, so that the solution is verified. We now investigate the elementary properties of our extended numbersystem, and then, when the ideas are a little more familiar, return to put the basis on a surer foundation. EXAMPLES I
Solve after the manner of the text the following sets of equations: 1. x24:X + 3 = 0,
x2±x+l
= 0,
x24x + o = 0 .
2. x2 + 8x+15 = 0,
#2 + 8a;+ll = 0,
z2 + 8a; + 20 = 0.
3. x22x3
z22z4
x22x+10
=0,
=0,
= 0.
2. Definitions. The numbers of ordinary arithmetic (positive or negative integers, rational numbers and irrational numbers) are called real numbers. If a, b are two real numbers, then numbers of the form a + ib
(i2 =  l )
are called complex numbers. When 6 = 0, such a number is real. When a = 0, the number ib
(b real)
is called a pure imaginary number. Two numbers such as a + ib,
a — ib
(a, b real)
are called conjugate complex numbers. A single symbol, such as c, is often used for a complex number.
162
COMPLEX NUMBERS
then a is called the real part of c and b the imaginary part. The notations a = Me, b = Jc are often used. The conjugate of a complex number c is often denoted by c, so that, if c = a 4 ib, then c = a — ib. Clearly the conjugate of c is c itself. If c is real, then c = c. The magnitude + V( 2 + &2) (a, 6 real) is called the modulus of the complex number c = a + i6; it is often denoted by the notations
\c\, It follows that
\a + ib\.
 c  =  c .
Moreover, we have the relation cc =  c  2 , for
cc = (a + i6) (a — i&)
Finally, since
c = a + ib,
c = a — ib,
we have the relations
Jrc = b =
.(cc).
Note. All the numbers now to be considered are actually complex, of the form a + ib, even when 6 = 0. Correctly speaking we should use the phrase 'real complex number' for a number such as 2, and 'pure imaginary complex number' for 2i. But this becomes tedious once it has been firmly grasped that all numbers are complex anyway.
DEFINITIONS
163
EXAMPLES II
The following examples are designed to make the reader familiar with the use of the symbol i. Use normal algebra, except that i 2 =  l . Express the following complex numbers in the form a + ib, where a,b are both real: 1. (3 + 5i) + (72i).
2.
3. (4 + 5i)(62i).
4.
5. 3 + 2i + i(5 + i).
6.
2
7. (1+i) .
8.
2
2
9. ( 2  i )  ( 3 + 2i) .
10.
Prove the following identities:
n.Ut. 13 * i6 ' 3^4i p + iq
12. ^
« 1(1).
l
3. Addition, subtraction, multiplication. Let c = a + ib9 w~u\iv
(a, b, u, v real)
be two given complex numbers. In virtue of the meaning which we have given to i, we obtain expressions for their sum, difference and product as follows: (i)
SUM
(ii) DIFFERENCE (iii) PRODUCT
C+W
=
C — IV = (a — u) + i(b — v). CW = (a + ib)(u + iv)
= au + ibu + iav + i2bv = (au — bv) + i(bu + av). The awkward one of these is the product. At present it is best not to remember the formula, but to be able to derive it when required.
164
COMPLEX NUMBERS EXAMPLES III
Express the following products in the form a + ib (a, 6 real): 1. (2 + 3i)(4 + 5i). 2. (l + 4i)(26i). 3. ( _ i _ ; ) ( _ 3 + 4i). 4. 5. i(4 + 3i).
6.
7. (3f)(3 + t).
8. (2 + 3i)3.
4. Division. As before, let c = a + ib, w = u + iv Then the
QUOTIENT
(a, b, u, v real).
is, by definition, the fraction c _a + ib
It is customary to express such fractions in a form in which the denominator is real. For this, multiply numerator and denominator by w = u — iv. Then ib)(u — iv) (u + iv) (u — iv)
w =
(au + bv) + i(bu — av) u2 + v*
=
au + bv .bu — av u2 + v2+l u2 + v2'
It is implicit that u, v are not both zero. For example, (2 + i) ~ (3  i) 2+i
9i 2 10
"
9+1
DIVISION
165
EXAMPLES IV
Express the following quotients in the form a + ib (a,b real): 1. —.. li A 2 — 3i
2. —^. 4 + 3^ 3 + 5i # 5 . 1 —6i —..
3.
„. . 1% ,, cos 20 + i sin 29 6." cos^ jr + i s i n ^
5. Equal complex numbers. To verify that, if two complex numbers are equal, then their real parts are equal and their imaginary parts are equal. Let c = a + ib, r==p + iq be two given complex numbers {a,b,p,q all real), such that c = r. Then ., a + %b = p + iq, or a— p =i(q — b). Squaring each side, we have
But a—p,q — b are real, so that their squares are positive or zero; hence this relation cannot hold unless each side is zero. That is, a = p,
b = q,
as required. COROLLABY.
/ / a, 6 are real and a + ib = 0,
then
a = 0, 6 = 0. EXAMPLES V
Find the sum, difference, product and quotient of each of the following pairs of complex numbers: 35i. 3. 4, 2i. 5. 2 + 3i,
2. 4. 3 + 2i,  i .
23i.
6.  3 + 4i,
34^.
166
COMPLEX NUMBERS
Find, in the form a + ib (a, b real) the reciprocal of each of the following complex numbers: 7. 3 + 4i. 9. 6i.
8. 5+12^. 10.  6  8 ^ .
Solve the following quadratic equations, expressing your answers in the form a±ib:
11. #2 + 4#+13 = 0.
12. # 2 
13. x2 + 6x+l0 = 0.
14. 4x2 + 9 = 0.
15. x28x + 25 = 0.
16. x2 + 4x + 5 = 0.
6. The complex number as a numberpair. A complex number consists essentially of the pair of real numbers a, b linked by the symbol i to which we have given a specific property (i2 negative) not enjoyed by any real number. Hence c is of the nature of an ordered pair of real numbers, the ordering being an important feature since, for example, the two complex numbers 4 + 5i and 5 + 4i are quite distinct. The concept of numberpair forms the foundation for the more logical development promised at the end of § 1, for it is precisely this concept which enables us to extend the range of numbers required for the solution of the third quadratic equation (x2 — 6x + 10 = 0). We therefore make a fresh start, with a series of definitions designed to exhibit the properties hitherto obtained by the use of the fictitious 'number' i. We define a complex number to be an ordered numberpair (of real numbers), denoted by the symbol [a, 6], subject to the following rules of operation (chosen, of course, to fit in to the treatment given at the start of this chapter): (i) The symbols [a, 6], [u,v] are equal if, and only if, the two relations a = u,b = v are satisfied; (ii) The sum [ is the numberpair [a + u,b + v];
THE COMPLEX NUMBER AS A NUMBERPAIR
(iii) The product
r
...
r
167
n
is the numberpair [au — bvtbu + av\. Compare the formulae in § 3. A number [a, 0], whose second component is zero, is called a real complex number) a number [0,6], whose first component is zero, is called a pure imaginary complex number. These terms are usually abbreviated to real and pure imaginary numbers. (Compare p. 162.) In order to achieve correlation with the normal notation for real numbers, and with the customary notation based on the letter i for pure imaginaries, we make the following conventions: When we are working in a system of algebra requiring the use of complex numbers, (i) the symbol a will be used as an abbreviation for the numberpair [a, 0], (ii) the symbol ib will be used as an abbreviation for the numberpair [0,6]. It follows that the symbol a + ib may be used for the numberpair [a, 0] + [0,6], or [a, 6]. In particular, we write 1 for the unit [1,0] of real numbers, and i x 1 for the unit [0,1] of pure imaginary numbers; but no confusion arises if we abbreviate i x 1 to the symbol i itself. We do not propose to go into much greater detail, but the reader may easily check that the complex numbers defined in this way by numberpairs have all the properties tentatively proposed for them in the earlier paragraphs of this chapter. We ought, however, to verify explicitly that the square of the numberpair i is the real number — 1. For this, we appeal directly to the definition of multiplication: [a, 6] x [u} v] = [au — bv, bu + av].
Write a = u = 0, 6 = v = 1. Then
or, in terms of the abbreviated symbols, i x i = — 1.
168
COMPLEX NUMBERS
Finally, we remark that, although the ordinary language of real numbers is retained without apparent change, the symbols in complex algebra carry an entirely different meaning. For example, the statement 2x2 = 4 remains true; but what we really mean is that
[2,0] x [2,0] = [2 x 2  0 x 0,0 x 2 + 2 x 0] = [4,0]. From now on, however, we shall revert to the normal symbolism without square brackets, writing a complex number in the form a + ib as before. It must always be remembered that numberpairs are intended. 7. The Argand diagram. Abstractly viewed, a complex number a + ib is a 'numberpair' [a, b] in which the first component of the pair corresponds to the real part and the second component to the imaginary. A numberpair is, however, also capable of a familiar geometrical interpretation, when the two numbers a,b in assigned order are taken to be the Cartesian coordinates of a point in a plane. We therefore seek next to link these two conceptions of numberpair so that we can incorporate the ideas of analytical geometry into the develop•P(x,y) ment of complex algebra. With a change of notation, denote a complex number by the letter z, where •?? We do this to strengthen the implication of the real and imaginary parts x and y Fi g 93« as the Cartesian coordinates of a point P(x,y) (Fig. 93). There is then an exact correspondence between the complex numbers z = x + iy and the points P(x, y) of the plane: (i) If z is given, then its real and imaginary parts are determined and so P is known; (ii) If P is given, then its coordinates are determined, and so z is known.
169 THE ARGAND DIAGRAM We say that P represents the complex number z. The diagram in which the complex numbers are represented is called an Argand diagram, and the plane in which it is drawn is called the complex plane, or, when precision is required, the zplane.
EXAMPLES VI Mark on an Argand diagram the points which represent the following complex numbers: 1. 3 + 4i.
2. 5i.
3. 6i.
4.  3 .
5. 1+i.
6.  2  3 i .
7. cos 0 + isin 0 for 2
8. 2cos 0 + i8iu0
0 = 0°, 30°, 60°,..., 330°, 360°. for
0 = 0,45°, 90°, ...,315°, 360°.
8, Modulus and argument. Given the point P(x, y) representing the complex number
we may describe its position alternatively by means of polar coordinates r, 0 (Fig. 94), where r is the distance OP, ESSENTIALLY POSITIVE and 0 is the angle LxOP which OP makes with the xaxis, measured in the counterclockwise sense from Ox. Thus r is defined uniquely, but 0 is ambiguous by multiples of 2TT. We know from elementary trigonometry that y x = r cos 0,
y — r sin 0,
whatever the quadrant in which P lies. Squaring and adding these relations, we have
^\p
Fig. 94. so that, since r is positive,
Dividing the relations, we have tan 0 = yjx, or
0 = tan%/#).
170
COMPLEX NUMBERS
The choice of quadrant for 9 depends on the signs of BOTH y AND x; if x, y are + , + , the quadrant is the first; if x, y are —, + , the second; if x, y are —, —, the third; if x, y are 4, —, the fourth. r = J{x2 + y2)y
The two numbers
9 = tan 1 ^/*), with appropriate choice of the quadrant for 9, are called the modulus and argument respectively of the complex number z. If r, 9 are given, then z = r(cos 9 + i sin 8). We have already (p. 162) explained the notation \z\ = r, and proved the formula zz =  z j 2 . ILLUSTRATION 1. To find the modulus and argument of the complex number —l+i<j3.
If the modulus is r, then r2=
so that
( _ l ) 2 + ( ^ 3 ) 2 = 1 + 3 = 4,
r = 2. The number is therefore
so that, if 9 is the argument, cos 9 = — J,
sin 0 = ^  .
Hence 0 *= f TT, and so the number is 2{cos TT + i sin f TT}. EXAMPLES VII 1. Plot in an Argand diagram the points which represent the numbers 4 f 3i, — 3 + 4i, — 4 — 3i, 3 — 4i, and verify that the points are at the vertices of a square. 2. Plot in an Argand diagram the points which represent the numbers 7 + 3i, 6i, — 3 — i, 4 — 4i, and verify that the points are at the vertices of a square.
MODULUS AND AKGUMENT
171
3. Find the modulus and argument in degrees and minutes of each of the complex numbers ^3 i,
3 + 4i,
5+12i,
3, 6  8 i ,
2i.
4. Prove that, if the point z in the complex plane lies on the circle whose centre is the (complex) point a and whose radius is the real number Jc, then \z — a\ = k. 5. Find the locus in an Argand diagram of the point representing the complex number z subject to the relation \z + 2\ =
\z5i\.
9. The representation in an Argand diagram of the sum of two numbers. Suppose that
are two complex numbers represented in an Argand diagram by the points
respectively (Fig. 95), Their sum is the number
O represented by the point P(x, y), where
Fig. 95.
By elementary analytical geometry, the lines Px P2 and OP have the same middle point ( ^ ± ^ , ^ ± * ! ? ) f so that OPXPP2 is a \ 2 2 / parallelogram. Hence P (representing zx + z2) is the fourth vertex of the parallelogram of which 0Pv OP2 are adjacent sides. In particular, OP is the modulus \z1 + z2\ of the sum of the two numbers zlt z2.
172
COMPLEX
NUMBERS
10, The representation in an Argand diagram of the difference of two numbers. To find the point representing the difference z==z2zl9 we proceed as follows: The relation is equivalent to so that OP2 (Fig. 96) is the diagonal of the parallelogram of which OPX, OP (where P represents z) are adjacent sides. Hence OP is the line through 0 parallel to P±P2y the sense along the lines being indicated by the arrows, where OP, P±P2 are equal in magnitude. COROLLARY. If P1(x1, yj, P2(x2, y2) represent the two complex numbers z1 = x1 + iy1,z2 = x2 + iy2, then the line PiP2 represents the difference z2 — zly in the sense that the length PiP2 is the modulus z2~""zil ana* the angle which ^ ' ^^ l ]* 1 makes with the xaxis is the argument of z2 — zv This corollary is very important Fig. 96 in geometrical applications. [The reader familiar with the theory of vectors should compare the results in the addition and subtraction of two vectors.] 11. The product of two complex numbers. In dealing with the multiplication of complex numbers, it is often more convenient to express them in modulusargument form. We therefore write zx = rx(cos 9X + i sin 0J, Then
z2 == r2(cos 92 + i sin 92).
zxz2 = rx r2(cos 9± + i sin 0X) (cos 92 + i sin 92) = /•^{(cos 9X cos 92 — sin 9X sin 92) + i(sin 9X cos 92 + cos 0± sin 92)} = ^^{cos {0t + 92) + i sin (91 + 92)}9
which is a complex number of modulus rxr2 and argument 0±+ 92.
THE PRODUCT OF TWO COMPLEX NUMBERS
173
Hence (i) the modulus of a product is the product of the moduli', (ii) the argument of a product is the sum of the arguments.
We may obtain similarly the results for division: (iii) the modulus of a quotient is the quotient of the moduli', (iv) the argument of a quotient is the difference of the arguments.
12. The product in an Argand diagram. In the diagram (Fig. 97), let PVP2,P represent the two complex numbers zvz2 and their product z = zxz2 respectively, and let U be the 'unit' point 2 = 1 . Then (§ 10) LxOP = 91+92, where 0l9 92 are the arguments of zl9 z2, and so »
LxOPLxOP9
= LU0P±. Moreover, if rv r2 are the moduli of zv z2, OP\OP2 = (rxr2)/r2 = rjl = OPJOU.
O
u Fig. 97.
Now in measuring the angle LP20P or LUOPV we proceed, by convention, from the radius vector OP2 or OU, through an angle 9V in the counterclockwise sense, to the vector OP or OPV It may be that 9X itself does not lie between 0, TT, but nevertheless the fact that the angles LP20P, LU0P1 so described are equal ensures that the Euclidean angles LP2OP9 LUOP1 of the triangles AP20P, hU0Px are also equal. Moreover, the sides about these equal angles have been proved proportional, and so the triangles are similar. Hence follows the construction for P: Let Pl9P2 represent the two given numbers zl9z2; let 0 be the origin and U the unit point 2 = 1 . Describe the triangle P0P2 similar to the triangle PxOU, in such a sense that LP20P = LUOPV Then the point P represents the product zxz2.
174
COMPLEX
NUMBERS
COKOLLABY. The effect of multiplying a complex number z by i is to rotate the vector OP representing z through an angle \TT in the counterclockwise sense. This follows at once if we put zx = i in the preceding work, so that Px in the diagram becomes the point (0,1) or, in polar coordinates, (1, £TT). Thus i acts in the Argand diagram like an operator, turning the radius vector through a right angle. EXAMPLES VIII
1. Mark in an Argand diagram the points which represent (a) 3 + 2i, (6) 2 + i, (c) their product. Verify from your diagram the theorem of the text. 2. Repeat Ex. 1 for the points (a) 1,
(6) 35i,
(c) their product;
(a) 2  i ,
(6) 2 + i,
(c) their product.
13. De Moivre's theorem. (i) To prove that, if n is a positive integer, then (cos 6 + i sin 0)n = cos nd + i sin nO. We use a proof by induction. Suppose that the theorem is true for a particular number N, a positive integer, so that (cos 0 + i sin 0)N = cos JV^ + i sin NO. Then (cos 9 f i sin 0)*+* = (cos N9 + i sin N6) (cos 6 + i sin 6) = (cos Nd cos 6 — sin NO sin 6) + i(cos NO sin 0 + sin NO cos 6) Hence if the theorem is true for any particular value N, it is true for N + 1, N + 2,... and so on. But it is clearly true for N = 0, and so the theorem is established. (ii) To prove that, if n is a negative integer, then (cos 0 + i sin 0)n = cos nO + i sin nO. Write
w = w.
DE MOIVRE'S
THEOREM
Then m is a positive integer, and so (cos 9 + i sin 9)n = (cos 9 + i sin 0)~m 1 (cos d + i sin 1
(as above)
cosmd — isinmd (cos m9 + i sin m#) (cos md — i sinmO) cos mO — isinmO cos2m0 —i2sin2ra0 cosra0 —isinrafl = cosrafl —isinmfl = cos (— n0) — i sin (— nO) = cos nd + i sin nd, as required. (iii) To prove that, if n is a rational fraction, then (cos 9 + i sin 9)n = cos n(9 +
2^TT) +
i sin w
i5 an integer, positive or negative. Suppose that n = where p, q are integers (q positive) without a common factor. Consider the expression (cos0 +isinfl) 1 /*, and suppose that it CAN be expressed in the form c o s cf> + i s i n
cos 6 + i sin 9 = (cos
Equating real and imaginary parts, we have cos 9 = cos q
sin 0 = sin q
175
176
COMPLEX NUMBERS
These equalities hold simultaneously if, and only if, where k is any integer, positive or negative. Hence A
n
Q
so that
q
1 2k \ /I 7r+isin(0H (~9\ 9 9/ \9
2k \ TT). ? / Raising each side to the power jp, by (i) or (ii) above, and then writing p\q = n, we obtain the relation (cos 9 + i sin 0)n = cos n(9 + 2&7r) + i sin w(0 + 2kir). The expression on the righthand side takes various values according to the choice of k. That is to say, there are several values for (cos 9 + i sin 9)n when n is a rational fraction; the case n = \ is familiar, leading to two distinct square roots. We see by elementary trigonometry that distinct values are obtained when k assumes in succession the values 0,1,2, ...,g—l, where q is the denominator in the expression of n as a proper fraction in its lowest terms. Thereafter they repeat themselves, any block of q consecutive values of k giving the whole set. SUMMARY.
The formula
(cos 6 + i sin 6)n = cos n(9 + 2&TT) + i sin n(9 + 2krr)
is true for all values of n, where n may be a positive or negative integer or rational fraction. When n is an integer, k may be taken as zero; when n is fractional, distinct values are obtained on the righthand side for k = 0, l,2,...,g— 1, where q is the denominator in the expression of n as a proper fraction in its lowest terms, EXAMPLES IX
1 J3 1. Express ^ + i~ in modulusargument form r(cos 6 + i sin 9) and hence, with the help of your tables, find (i) its square, (ii) its square roots, (iii) its cube roots, (iv) its fourth roots. 2. Repeat Ex. 1 for the number 4 + 3i. 3. Repeat Ex. 1 for the number 12 — 5i.
THE WTH ROOTS OF UNITY
177
14. The nth roots of unity. In virtue of the formula 1 = cos 0 f i sin 0, it follows from De Moivre's theorem that, if n is any integer (assumed positive here) then an nth root of unity, that is
can be expressed in the form COSi (.2kn)
\n
)
+ isin (  . 2kir\ \n )
for & = 0,l,2,...,7&—1. Hence for any given positive integer n, there are n distinct roots of unity, namely cos
2kn . . 2kir h % sin , n n
where k = 0,1,2, ...,w— 1. For example: When n = 2, there are two square roots, namely 2&7T
. .
2klT
cos — + t s i n — , where k = 0,1. When k = 0, this gives cosO, or 1; when k = 1, it gives cos 77, or — 1. When n = 3, there are three cube roots, namely 2kir
. . 2k7T
cos — + i s i n — , where £ = 0,1,2. When k — 0, this gives 1 itself. The values k = 1,2 give 2TT
. .
2TT
,,
cos — +1 sin —==!(6 6 4TT . . 4TT „ cos — 4 * sin — = J ( — 1 — i ^3). o o
These complex cube roots of unity are usually denoted by the symbols o>, o>2, either being the square of the other. They are also connected by the relation = 0.
178
COMPLEX NUMBERS EXAMPLES X
Use De Moivre's theorem to express the following powers in 'modulusargument' form r(cos 8 + isin 6): 1.(1+0*
2<
4. (V3»)*.
5. (l»)*.
3. (l + iV3)6.
(13^
6. (V3 + 0*.
7. Find expressions for (i)
1*,
(ii) 1*
analogous to the roots evaluated in the text.
15. Complex powers. The reader will remember that, when seeking an interpretation for symbols such as a0,
a~n,
a1/n,
a~1/n
for real positive a and positive integral n, he was guided by the principle that the formula of multiplication
should hold for all values of p and q. This led to the interpretations a 0 = 1, a~n = \\an,
a1/n = tya, a~1/n =
We now consider the meanings which should be given to an when a,n are complex numbers, still retaining the validity for the formula of multiplication. (i) When n is real and rational, the treatment is comparatively simple, for we know that any complex number a can be expressed in the form ,»../» a = r(cos O + t&in 6), and so, by De Moivre's theorem, an =
rn{cos n(9 + 2lc7T) + i sin n(9 +2 JCTT)}
where the values h = 0,1,..., q— 1 (q being the denominator in the expression of n as a proper fraction in its lowest terms) give distinct values for an. Thus an is, in general, a gvalued function. (For example, if n =  , there are two square roots.)
COMPLEX POWERS
179
(ii) When n is complex, say n = x + iy,
(x, y real)
n
then we must interpret a in such a way that ax+iy
_
a
We have already dealt with the factor ax for real and rational values of x, so that we are left to consider what meaning may be given to the expression aiy when the power iy is pure imaginary. 16. Pure imaginary powers. In the preceding paragraph we reduced the interpretation of an to the case, which we now consider, when n is pure imaginary. With a slight change of notation, we write n = ix, where x is real. (i) When a is real and positive. We note first that any real positive number a (other than zero) can be expressed in the form
so that, if the laws of indices are to be preserved, an = (el0K«a)n = enloe*a. Writing
n logea = ix logea = ix'
(x' real),
we obtain the expression, as an imaginary power of e, in the form
which is found to be more amenable to treatment. Dropping the dash, then, we consider what interpretation should be given to the number
** (a; real).
We take the hint from the expansion of p. 54, which, if valid for pure imaginary numbers, would yield the relation
/
X2
X*
180
COMPLEX
NUMBERS
Now (p. 50) we have the relations, valid for all real x,
X5
and so we are led to propose for consideration the relation eix = cos x + i since. Before we can accept this as a valid interpretation for eix% however, we must satisfy ourselves that it obeys the index law eix x eiy =
eUx+y)
fa
y
r e a J) #
Under the proposed interpretation, the lefthand side is (cos x + i sin x) x (cos y + i sin y) = (cos x cos y — sin x sin y) + i(sin x cos y f cos x sin y) = cos (x + y) + i sin (x + y), and this is precisely the interpretation to be given to the righthand side also. The interpretation is therefore consistent with the series expansions and with the index law, and so we DEFINE the interpretation of eix (x real) to be given by the relation eix = cos x + i sin x. Note. The mere writing of ix in the expansion in series does not of itself allow us to use the notation eix under the index laws. The step that the product of the expressions proposed for eix, eiv is indeed eiix+v) is vital to the interpretation. (ii) When a is any complex number. We are now in a position to give an interpretation to the expression aix for any complex number a. We have seen that a may be written in the form a == r(cos 9 + i sin 6) (r positive) and, by what we have just done, we have the two relations r==eiog,r
cos 8 4 i sin 0 = eid.
PURE IMAGINARY POWERS loe r
Hence
181
id
a = e < xe = eu x eid,
say, where u} or loger, is a real number, positive or negative. For interpretations consistent with the laws of indices, we must therefore take
aix
=
{eU)ix x
= e
iux
{eid)ix
x e~6x
= e~0x (cos ux + i sin ux)
= e~^{cos (x loge r) + i sin (# log,, r)}. COROLLARY. Any (nonzero) complex number can be expressed in the form eu+iv, where uyv are real.
17. Multiplicity of values. There is one difficulty which we have slurred over, as it should not be overemphasized at the present stage. An example will illustrate the point. Let us consider what interpretation we are to give to the symbol Our first task is to express the number to be raised to the power i (in this case, i itself) in the form reid, and this we do easily by noticing that .= o
for any integer Jc, positive or negative. Hence the interpretation (i
s
fe(2k+i)7riy
s e(2k+*)7TiXi
gives an infinite succession of values, all of which, oddly enough, are real. In a full discussion, we should therefore be on our guard to include manyvalued powers. A safe way of doing this is to include the factor e2kni (which is just cos 2&77 + i sin 2&TT, or 1, for any integer k) in the expression in exponential form of the number whose power we seek. That is to say, in order to evaluate an, we write a in the form and allow k to take integral values.
182
COMPLEX NUMBEES
Ifc should be noticed that we established the interpretation eix
—
ii
under the conditions of De Moivre's theorem, namely that a; is a real number, being a positive or negative integer or rational fraction. When x is not integral, there is really an ambiguity of interpretation, since eix =
( c t)z
__ eUl+2k7r)x = cos (1 + 2kir) x + i sin (1 f 21CTT) X,
where distinct values are obtained for k = 0,1, ...,# — 1 as usual, q being the denominator when the fraction x is expressed in its lowest terms. Our interpretation is therefore subject to the convention k = 0. ILLUSTRATION
2. To find an expression for (l+iV3
We first write 1 + i ^3 in the form
= 2ei^7T+2kn)
(k integral).
We next use the fact that
Now
AISO
(1 + i V3)4
(1 + i V3)** = 2 ^ X = 2* f x e~
Finally, giving
2 = el0«*2, 2** = e*il0«*
MULTIPLICITY OF VALUES
183
so that, in all,
= l6e<*+^> {cos (f77 + 1 log, 2) + i sin (§77 + J log, 2)}, where & is any integer, positive, negative, or zero. EXAMPLES XI
Find expressions for 1. (l+iy.
2. (I*) 2 "*.
4. (liV3)3+2*.
5
3. (V3 + »)H.
 (^
18, The logarithm of a complex number, to the base e. We have seen (p. 181) that any (nonzero) complex number can be expressed in the form a==eu+ivm
We DEFINE the logarithm of a to be the complex number u + iv
(u, v real).
This is consistent with the treatment already given for real numbers (v = 0) and, in virtue of the interpretation already given to indices, it retains the property We have now, of course, released the restriction (p. 5) that a must be positive in order to have a logarithm. It need not even be real. The ambiguous determination of v9 as seen by the equivalent formula (j integral) a == eu+uv+2kn) means that the logarithm is ambiguous to the extent of additive multiples of 2iri. Note. The relation (pp. 170, 180) z = reid, where r is the modulus and 9 the argument of z, gives the formula logz = log I z I + i*argz.
184 NOTATION.
COMPLEX
NUMBERS
The notation Logea
is often used to denote a value of the logarithm when ambiguity may be present. Then the notation denotes that determination of the logarithm whose imaginary part lies between — IT and TT, and logea is called the principal value of Loge a. This distinction, which is often of importance, will not, in fact, be used much in this book, and we shall usually take the principal value without further comment. 19. The sine and cosine. We have made no statements so far about the formula eix — cosa;fisin# except when x is real; naturally, because cos a;, sin a; are otherwise undefined. Our next task is to define the trigonometric functions of a complex variable z = z + iy (x,y real), and this we do by what is in some ways a reversal of the process hitherto adopted. We first observe that, for real x, eix = cos x + i sin x, eix
so that
=
cog x
__ i s
m x^
cos a; sin a; = —,(eix — e~ix).
We now adopt these formulae, and make them the basis of the following more general definitions: / / z is a complex number, then
sin z = —. (eiz  e~iz).
It follows at once that these definitions give the normal functions when z is real, and also that eiz = cos z + i sin z.
THE SINE AND COSINE
We must, however, now make sure that they retain the properties of the trigonometric functions. Firstly, we have
185 NORMAL
4(cos2 z + sin2 z) = (eiz + e~iz)2  (eiz  e~iz)2 = (e2iz + 2 + e~2iz)  (e2iz  2 + e~2iz) = 4, so that
2
2
cos z + sin z = 1.
Again, 4(cos z1 cos z2 — sin z± sin z2) = (e^1 + ei2!i) (eiz* on reduction, so that cos zx cos z2 — sin zx sin z2 = cos (zx 4 z2). Also 4i(sin zx cos z2 4 cos zx sin z2)
so that
sin zx cos z2 + cos zx sin z2 = sin (zx + z2).
These are the formulae on which the theory of real angles is based, and so the structure will stand for complex 'angles' also. Further details are left to the reader. Note. The functions cosz, sinz are not now subject to the restriction of being less than unity. For example, the equation cos z = 2 is satisfied if or so that
eiz + e~iz = 4, 6 2te_4 C <* + i
= 0,
eiz = 2 + ^/3.
Thus
iz = log"(2 + ,/3)
or
z =  i log (2 + ^/3).
186
COMPLEX
NUMBERS
For reference we record the following formulae: (i)
cosh iz = \ (eiz + e~iz) = cos z;
(ii)
sinh iz = \{eiz  e~iz) = isinz;
(iii)
e2k7Ti = cos 21CTT + isin 2TCTT = 1 (k integral);
(iv) e(2A:+1)^ = cos(2fe+l)7r + i s i n ( 2 i + l ) 7 7 =  l ; (V) e(2k+»ni =
c o s (2fc +
1) 77 + i sin (2Jfe +  ) 77 = i.
20. The modulus of e*. If z is complex, so that z=x + iy then
(x, y real),
ez = e x +^ = exeiy
Hence the modulus of ez is ex and its argument is y, the argument being undetermined to the extent of multiples of 2TT. COROLLARY.
is that
An important corollary, found by putting x = 0, , .,
1^1 = 1, when iy is a pure imaginary. 21. The differentiation
and integration
of
complex
numbers. We confine our attention to the only case which we shall use, the complex functions of a REAL variable x. The general theory for a complex variable is much beyond the scope of this book. By a complex function of a real variable x, we shall mean a function w(x) which is either given in the form w(x) = u(x) + iv(x), where u(x), v(x) are real functions of x, or can be reduced to that form. For example, the function eix can be reduced to cos x + % sin x.
DIFFERENTIATION AND INTEGRATION
We then use the following ,..
187
DEFINITIONS:
W
dw _du .dv dx~ = dx' + idx>
(ii)
wdx = udx + i vdx.
Particular interest is attached to the function ecx, where c may be complex of the form a + ib. We have
= — {eax (cos bx + i sin bx)} = 7  [{e
+ i(a e°* sin bx + 6 eax cos 6a:) a e°* (cos 6a; + i sin bx) + i6 e ^ (cos 6a; + i sin 6a;) (a +
ib)eax.eibx
ce™.
Hence the rule
— (eM) = ao;
holds whether c is real or complex. Similarly we may prove that the formula \ecxdx =  e« holds whether c is real or complex. 132
188
COMPLEX
NUMBERS
Because of its importance, we ought perhaps to mention also the differentiation and integration of #c, where x is assumed to be real but where c may be complex. Since xc =
we have
ecio«x9
j (xc) = ec]ogx.c.j (loga;) ax ax
(as above)
in accordance with the similar rule for real functions. It follows that \xcdx = ^jXc+1
(c*l)
whether c is real or complex. ILLUSTRATION
3. To prove the rule
whenf(x),g(x) are complex functions of the real variable x. Write
f(x) = u(x) + iv(x) = u + iv, g(x) =p(x) + iq(x) ==p + iq.
Then
f(x) g(x) = (up — vq) + i(uq + vp),
so that
sOTrt*» = [(u'p  v'q) + (up'  vq')] + i[{u'q + v'p) + (uq' + vp')]
Reversal of this formula leads to the rule for integration by parts. Hence this method is also at our disposal for these functions. The two illustrations which follow show how complex numbers may be used to sum series and to evaluate integrals.
DIFFERENTIATION AND INTEGRATION ILLUSTRATION
189
4. To find the sum of the first n terms of the series
1 + cos 0. cos 0 + cos2 9. cos 2 9 + ... + cos 71 " 1 9. cos (n  1 ) 9, where 9 is a real angle.
Write cos0.sin0 + cos 2 0.sin20+...+ cos71"10. sin (71  1)0.
S= Then
C + iS = 1 + cos 0 eie + cos2 0 e2ie + ... + cos"10 e
C + iS = —
In order to find 'real and imaginary parts', multiply numerator and denominator by 1 — cos 0 e~ie. The new denominator is (1  cos 0 eie) (1  cos 0 e*°) = 1  cos 9(eie + e^) + cos2 0 = 1cos 0(2 cos 0) +cos 2 0 = lcos 2 0 = sin20. Thus C + ttf = J^
{(1  cos* 0 en<*) (1  cos 0 e'*)}
sm20 l Equating real parts, we have C = roTT {lCOS n 0COS7l0COS 2 0 + COS n+1 0COS(n 1)0}
sm 2 0 ^ \ 2 n = . 2 {sin 0  cos 0(cos n9  cos 0 cos (n  1) 0)}
/>
= —  ^ {sin2 0  cos n 0(  sin 0 sin (n  1) 0)} sin (/
_ cos n 0sin(ril)0 "~ sin 0 It is implicit in the working that sin 0 is not zero. That case can easily be given separate treatment.
190
COMPLEX NUMBERS
ILLUSTRATION
5. Tofindthe real integral xexcosxdx. C=\xex cos x dx,
Write
8= xexmxxdx9 C + iS = \x ex eix dx
so that
= (xea+i)xdx. Integrating by parts (p. 188), we have C + iS = —!; eu+o*.x —U 1+^ 1+iJ
[e^*A.dx
Now Hence = [\x — \ix + \i\ ex (cos x + i sin #). Equating real parts, we have C= \ EXAMPLES XII
Use the method of the two illustrations to find: 2. sin# — #sin20+ #2sin30 — o;3sin40+... (to n terms). Find the integrals: 3. zcoszote. 6. I xe%*sin xdx.
4. e^sinzda;. 7. xsin Zxdx.
5. [e^smZxdx. 8. I xer**cos 3xdx.
DIFFERENTIATION AND INTEGRATION
191
Finally, we give an illustration to show how the use of complex numbers can help in dealing with the geometry of a plane curve. We use for the curve the notation of the preceding chapter. 6. Let P(x9 y) be a point of a plane curve, and let s9 if* denote as usual the length of the arc measured from a fixed point and the angle between the tangent at P and the #axis. (It is assumed that x9 y are real.) Write z = x + iy. dz dx ,dy mi Then  =   + 4/ ILLUSTRATION
ds
ds
ds
= cos ip f i sin if*
(p. 108)
= e** (p. 180). 7 = 1 .
In particular,
We also have the relation
ds (p. 113).
This is equivalent to d2z ds2
. ds2
so that, taking moduli on each side,
. A Again,
so that
dx Ay . . . . =— % ~ = cos \p — i sin \p (dx \ds
.dy\(d2x ds) \ds2
.d2y\ ds2)
.
Equating imaginary parts, we obtain the formula _ dxd2y dyd2x 2 ds ds ds ds2'
192
COMPLEX NUMBERS REVISION EXAMPLES VII
'Scholarship' Level 1. Two complex numbers z, z' are connected by the relation z' = (2 + z)/(2 —z). Show that, as the point which represents z on the Argand diagram describes the axis of y, from the negative end to the positive end, the point which represents z' describes completely the circle x2 + y2 = 1 in the counterclockwise sense. 2. Explain what is meant by the principal value of the logarithm of a complex number x + iy as distinct from the general value. Show that, considering only principal values, the real part of
2*log«2e*7rl cos (JTT loge 2).
is
3. Express tan (afit) in the form A + iB, where A,B are real when a,b are real. Show that, if x + iy = tan J d + fq), then i# = ev sin  — e~2ri sin 2  1 e~3^ sin 3£ —..., when 7] is positive; and that there is a like expansion with the sign of 7] changed, valid when TJ is negative. 4. Prove that, if sin (x + iy) = cosec (u + iv), where x, y, u, v are real, then cosh2 v tanh 2 y = cos2w, cosh2 y tanh 2 v = cos2#. 5. Solve the equation i)n = 0, giving the roots as trigonometrical functions of angles between 0 and 77.
6. Prove that, if x + iy = c cot (u + iv), then x sin 2u
y sinh 2v
c cosh 2v — cos 2u'
REVISION EXAMPLES VII
193
and show that, if x, y are coordinates of a point in a plane, then for a given value of v the point lies on the circle x2 + y2 + 2cy coth 2v + c2 = 0. Also verify that, if alta2 denote the radii of the circles for the values vl9 v2 oiv, and d denotes the distance between their centres, then 2 *> 72 *
2
= cosh 2(vt — Vo).
2axa2 7. Obtain an expression for  o,o&{x\iy)  2 in terms of trigonometric and hyperbolic functions of the real variables x, y. Show that  cos (x + ix) \ increases with x for all positive values of x. 8. Express

in the form X + iY, where z1 = x1iiy1, z2 = x2 + iy2. Deduce that the points which represent three complex numbers zl9 z2, z3 in the Argand diagram cannot all lie on the same side of the real axis if 9. Prove that the roots of the equation (zl)» = where n is a positive integer, are (4=0, Deduce, or prove otherwise, that S cot^——— = 0,
2 cosec2^——— = n2.
10. Write down the complex numbers conjugate to x + iy and to r(cos# + isin 6). Prove that, if a quadratic equation with real coefficients has one complex root, the other root is the conjugate complex number. Deduce a similar result for a cubic equation with real coefficients. Given that x = 1 + 3i is one solution of the equation a 4 +16z 2 +100 = 0, find all the solutions.
194
COMPLEX NUMBERS
11. Express each of the complex numbers in the form r(cos 0 + isin 0), where r is positive. Prove that z\ = z2, and find the other cube roots of z2 in the form r(cos 9 f i sin 9). 12. The complex number z = x + iy = r(cos 5 + i sin 0) is represented in the Argand diagram by the point (x,y). Prove that, if three variable points zvz29zz are such that z3 = Az2+(1 — X)z19 where A is a complex constant, then the triangle whose vertices are zvz2>zz is similar to the triangle with vertices at the points 0,1, A. ABC is a triangle. On the sides BC,CA,AB triangles BCA\ CAB\ ABC are described similar to a given triangle DEF. Prove that the median points of the triangles ABC, A'B'C are coincident. 13. The complex numbers a,b, (1 — k)a + kb are represented in the Argand diagram by^the points A}B,C. Prove that the vector AB represents b — a, and that (i) when k is real, C lies on AB and divides AB in the ratio k: 1k; (ii) when k is not real, ABC is a triangle in which the sides AB, AC are in the ratio 1 :  k \ and the angle of turn from AB to AC is 9, where k =  k \ (cos 9 + i sin 9). The vertices of two triangles ABC and XYZ represent the complex numbers a,b,c and x,y,z. Prove that a necessary and sufficient condition for the triangles to be similar and similarly situated is i1 l l
1
1
a
b
c = 0.
X
y
z
14. [In this problem small letters stand for complex numbers and capital letters for the corresponding points in the Argand diagram.] The circumcentre, centroid, and orthocentre of a triangle ABC are S, G, H respectively. Prove that g = £( a f 6 + c),
2$ + h = a + b + c.
REVISION EXAMPLES VII
195
ABC is a triangle with its circumcentre at the origin. The internal bisectors of the angles A,B,C of the triangle meet the circumcircle again at L,M,N respectively. Prove (by pure geometry if you wish) that the incentre P of ABC is the orthocentre of LMN, and deduce that p= Prove also that the excentres Pv P2, P3 of the triangle ABO are given by px = l — m — n,
p% = —l + m — n,
p3 = —l — m + n,
and that the circumcentre K of P1P2P3 is given by k = — (Zf m + n).
Prove, finally, that KP1 is perpendicular to BO. 15. Points A,B in the Argand diagram represent complex numbers a,b respectively; 0 is the origin, and P represents one of the values o{yj(ab). Prove that, if OA = OB = r, then also OP = r, and OP is perpendicular to AB. A,B,0 lie on a circle with centre 0, and represent complex numbers a, b, c respectively. Prove that the point D which represents — bcja also lies on the circle, and that AD is perpendicular to BO. The perpendiculars from B, O to CA,AB meet the circle again at E, F respectively. Prove that OA is perpendicular to EF. 16. Prove by means of an Argand diagram, that Find all the points in the complex plane which represent numbers satisfying the equations (i) z22z = 3, (ii)  z  2  = 3, (iii)  z 11 +  z2 \ = 3. 17. Find the fifth roots of unity, and hence solve the equation ( 2 #  l ) 5 = 32a5. 18. Prove that, if a, /? are the roots of the equation = 0, n
aj8
where
cot 9 = x+ 1.
sinnd
196
COMPLEX NUMBERS
19. A point z = x + iy in the Argand diagram is such that z = 2 , # = l , and y>Q. Determine the point and find the distance between the point and \z2. Show also that the points 2, z, ^z2, z 3 , Jz4, ^z5 are the vertices of a regular hexagon. 20. The points zi,z2,z3 form an equilateral triangle in the Argand diagram, and zx = 4 + 6i, z2 = (1 — i)zv Show that z3 must have one of two values, and determine these values. What are the vertices of the regular hexagon of which zx is the centre, and z2 is one vertex ? 21. A regular pentagon ABODE is inscribed in the circle x2 + y2 = 1, the vertex A being the point (1,0). Obtain the complex numbers x + iy of which the points A,B9C,D,E form a representation in an Argand diagram. 22. Use De Moivre's theorem to find all the roots of the equation (2x I) 5 = ( z  2 ) 5 in the form a + ib, where a, b are real numbers. 23. Mark on an Argand diagram the points ^3 + i, 2 + 2 ^ 3 , and their product. What is the general relation between the positions of the points a + ib and (^/3 + i) (a + ib) ? A triangle ABO has its vertices at the points 0,2 + 2 ^ 3 , — l+i^/3 respectively. A similar triangle A'B'C has its vertices A', B' at the points 0, Si respectively. Find the position of C". 24. Find all values of (5 — 12i)*, (2i — 2)*, expressing the answers in the form a + ib, where a, b are fractions or surds. 25. Prove that the triangle formed by the three points representing the complex numbers f,g,h is similar to the triangle formed by the points representing the numbers u, v, w if fv + gw + hu =fw + gu + hv. 26. By expressing 3 + 4i, l + 2i in the form r(cos 9 i i sin 6), or otherwise, evaluate in the form a + ib, obtaining each of the real numbers a, b correct to two significant figures.
REVISION EXAMPLES VII
197
27. Identify all the points in the z plane which satisfy the following relations: z2 (i) 2* + 2 = 23f (ii) z + 2 21
(iii)
(iv) 
Z+l
Z

28. Prove that three points z1,z2)zz in the complex plane are collinear if, and only if, the ratio (2 8 2 1 )/(2 2 2 1 )
is real. 29. Find the real and imaginary parts and the modulus for each of the expressions 1 (1)
TT2i'
.... a + ib (U)
~b+Ta9
1 + cosa + isina {m)
30. Find the logarithms of  1, 2  i , 31. The three points in an Argand diagram which correspond to the roots of the equation r = 0 are the vertices of a triangle ABC. Prove that the centroid of the triangle is the point corresponding to p. If the triangle ABC is equilateral, prove that p2 = q. 32. Find the cube roots of 4  3 i . 33. If the vertices A, B, C of an equilateral triangle represent the numbers zl9z29zz respectively in an Argand diagram, state the number represented by the midpoint M of AB, and show that J3 ) ± * ( 2  « )
If
21 = 2 + 2£,
22 = 4
and if G is on the same side of AB as the origin, find the number z3. 34. If 2 = cos 0 + i sin 6, express cos 9, sin 9, cos n09 sin nO in terms of 2, where n is an integer. Hence express sin7 9 in the form
198
COMPLEX NUMBERS
35. Prove that 6 . + sin0 —icos0\ icosey
a)
= 
cos 60asm 60.
&cos0/ 36. Prove that
37. Show that the equation » — 1 — 2i  = 3 represents a circle in an Argand diagram. If 2 lies on  z — 1 — 2i  = 3, what is the locus of the point Find the greatest and least values of  z — 4 — 6i \ if z is subject to the inequality 12 — 1 — 2% \ ^ 3. 38. Find the modulus of Sz + i when  z  = 1. 39. Two complex numbers z, zx are connected by the relation Zl
__ 2 + z "2^Tz
Prove that, if z = iq, where q is real, then the locus of z1 in an Argand diagram is a circle. Describe the variation in the amplitude of zx as q increases from — oo to + oo. 40. If a, b are real and n is an integer, prove that
has n real values, and find those of 41. If x,y are real, separate
sec (x + iy) into its real and imaginary parts.
REVISION EXAMPLES VII 2
199 2
2
2
42. If aib,c,d are real, express each of a + b and c + d as a product of linear factors. Deduce that the product of two factors, each of which is a sum of two squares, is itself a sum of two squares. 43. The intrinsic coordinates of a point on a plane curve are s, xfs, and the cartesian coordinates are x, y. The complex coordinate x f iy is denoted by z and the curvature by K. Prove that dz ds
d2z ds2
.
Hence, or otherwise, prove that K=(z"* + y"*)*=(z'y"x"y'), where dashes denote differentiations with respect to s. Prove further that x'
y'
1
x"
y"
K
x'" y"'
=
K(2K2I
K2
44. Any point P is taken on a given curve. The tangent at P is drawn in the direction of increasing s, and a point Q is taken at a constant distance I along this tangent. In this way Q describes a curve specified by X, F, 8>Z analogous to the specification x,y,s,z for P . [Notation of previous problem.] Show that
(ii) the curvature K of the derived curve at Q is given by the formula where K = dfc/ds.
CHAPTER X I I SYSTEMATIC I N T E G R A T I O N AT first sight the integration of functions seems to depend as much upon luck as upon skill. This is largely because the teacher or author must, in the early stages, select examples which are known to 'come out'. Nor is it easy to be sure, even with years of experience, that any particular integral is capable of evaluation; for example, #sin# can be integrated easily, whereas sinz/a; cannot be integrated at all in finite terms by means of functions studied hitherto. The purpose of this chapter is to explain how to set about the processes of integration in an orderly way. This naturally involves the recognition of a number of 'types', followed by a set of rules for each of them. But first we make two general remarks. (i) The rules will ensure that an integral of given type MUST come out; but it is always wise to examine any particular example carefully to make sure that an easier method (such as substitution) cannot be used instead. (ii) It is probably true to say that more integrals remain unsolved through faulty manipulation of algebra and trigonometry than through difficulties inherent in the integration itself. The reader is urged to acquire facility in the normal technique of these subjects. For details a textbook should be consulted. 1. Polynomials. The first type presents no difficulty. If f(x) is the polynomial f(x) = aoxn + axxn1 +... + an, then
(f(x)dx = ^ ^ J
tb ~T~ J
+
^ * ! + ... +an*. 7h
2. Rational functions. (Compare also p. 11.) A rational function f(x) of the variable x is defined to be the ratio of two polynomials, so that
W)
RATIONAL FUNCTIONS
201
for polynomials P{x),Q(x). If the degree of P(x) is not less than that of Q(x), we can divide out, getting a polynomial (easily integrated) together with a rational fraction in which the degree of the numerator is less than that of the denominator. We therefore confine our attention to the case in which the degree of P(x) is less than that of Q(x). It is assumed, too, that all coefficients are real. It is a theorem of algebra that any (real) polynomial, and, in particular, Q(x), can be expressed as a product of factors, of which typical terms are
where oc, jS, a, h, b are real constants, but where h2
A
l
,
^2
.
•
A
m ]
Bnx + Cn \ a where the first summation extends over all linear factors ocx and the second over all quadratic factors ax2 + 2hx + b. The integrals from the first summation are of the type Adx
which give
202
SYSTEMATIC INTEGRATION
The integrals from the second summation are of the type
Bx + O
f
and require more detailed consideration. Note first that ax + h /<
7
dx
and so, by writing Bx + C = (J5/a) (ax + A) + (a(7  hB)ja, we can reduce our problem to the evaluation of integrals such as
Write
2
h
dx
a(ax + 2hx + b) = (ax + h)2 + ab h2,
and make the substitution (remembering that ab — h2 is positive , . ,, , , 9N by assumption) J r ' ax + h = tj(abh2). C dx _ r qP^dt^abh2) 1
J (ax2 + 2hx + b)v~] {t2 (ab  h2) + (ab  h2)}* aP~x
C dt
Our final problem is therefore to evaluate dt and for this we need a formula of reduction. On integration by parts, we have r ., ( •'•in
/ . n .
i\/«
I / J O .
T
RATIONAL FUNCTIONS
Hence
2plp+1 (2pl)Ip
=
203
^
or, replacing p by p — 1,
By applying this formula successively, we make the evaluation of Ip depend on that of / p _ 1 ? Ip2> • ••> &nd, ultimately, on Iv But
= tan" 1 *, and so the whole integration is effected. 3. Integrals involving \/(ax + b). Rational functions oix and ^(ax + b) may be integrated readily by means of the substitution
s = (J 2 &). a The result is the integration of a rational function of t.
or
Note. The reader is unlikely to remember all the details to be given in § 4 following. The methods should be thoroughly understood, but it may well be found necessary to refer to the book for details in actual examples. If desired, §§ 5, 6 and 7, which will be of more immediate practical value, may be read next. 4. Integrals involving y(ax2 + 2hx + b). We first take steps to simplify the quadratic expression under the square root sign, (i) Suppose that a is positive. Then a(ax2 + 2hx + b) = (ax + h)2 + ab h2. Write then
ax + h = x'; xf2+p2 a(ax2 + 2hx + b) = [x'2q2
(ab>h2) (ab
where p2 = ab — h2, q2 = h2 — ab respectively. (ii) Suppose that a is negative. Then — a is positive, so we write  a(ax2 + 2hx + b) = h2  ab  (ax + h)2. 142
204
SYSTEMATIC INTEGRATION 2
If h — ab is negative, the right side is necessarily negative, so that ax2 + 2hx + b is also negative and (in real algebra) has no square root. We therefore take h2 — ab to be positive, and write thus, if as before, we have
ax + h = x',  a{ax2 + 2hx + b) = r2
x'2.
If, then, we have to evaluate a rational function of x and y](ax2 + 2hx\b), we may first apply the transformation ax + h = xr. The integrand becomes a rational function of x' and of a surd which may assume one or other of the three forms (i) yj{x'2 +p2)
a> 0, abh2> 0,
(ii) J(x'2 q2)
a> 0, abh2< 0,
(iii) <]{r2x'2)
a<0,abh2<0.
We may now drop the dashes and treat x as the variable. (i) The surd <J(x2+p2).
Consider the transformation x =
2pt
The graph (Fig. 98) indicates (what can also be proved algebraically) that all values of x are obtained by allowing t to vary continuously from — 1 to + 1 . We therefore impose the restriction
O
Fig. 98.
on the values of t which we select. We have
(l< 2 ) 2
INTEGRALS INVOLVING J(ax2+2JlX+b)
205
2
Since £ <1, we have, without ambiguity (assuming, as we may, that p is positive), Hence a rational function of x,J(x2+p2) is transformed into a rational function of t, and may be integrated accordingly. (ii) The surd J(x2q2). Consider the transformation t2+l The graph (Fig. 99) indicates (what can also be proved algebraically) that all values of x for which x > q are obtained by allowing t to vary continuously from + 1 to +00. 1
~ ^
1
x
J q "
1 „
H
1
0
J — —
/ 11
Fig. 99.
[We can restrict ourselves to positive values of x, since a range of integration which involved the two signs for x would have to pass through the region —q<x
Now
dx —
4qtdt
206
SYSTEMATIC INTEGRATION
Since t>l, we have, without ambiguity (assuming, as we may, that q is positive), Hence a rational function of x, ^{x2 — q2) is transformed into a rational function of t, and may be integrated accordingly. (iii) The surd <](r2x2).
Consider the transformation t2!
x = t2+l r.
The graph (Fig. 100) indicates (what can also be proved algeFig. 100. braically) that all values of x for 2 2 which x
dx =
4rtdt
r2x2 =
(t2+l)2'
Since t>0f we have, without ambiguity (assuming, as we may, that r is positive), ~ Hence a rational function of x9 ^{r2 — x2) is transformed into a rational function of t, and may be integrated accordingly. Note. The work just completed proves that the integrations are it does not necessarily give the easiest method. For example, the quadratic surds may also be subjected to the following substitutions: POSSIBLE;
(i) For y](x2+p2), let x = #tan 9, or x = p sinh 9; (ii) For sl{x2 — q2), let x = gsec 9, or x = gcosh 9; (iii) For <j(r2x2), let x = r sin 9.
ALTERNATIVE TREATMENT OF INTEGRALS
207
2
5. Integrals involving V(ax +2hx + b): alternative treatment. Integrals of rational functions of #, <J(ax2 + 2hx + b) may also be treated by methods which, by leaving transformation until the last stages, retajn the identity of the surd. We begin by reducing such a rational function to a more amenable form. Since even powers of yj(ax2 + 2hx + b) are polynomials and odd powers are polynomials multiplying the square root, we may express any polynomial in x, ^](ax2 + 2hx + 6) in the form
where P,Q are polynomials in x. Hence any rational function, being by definition the quotient of two polynomials, is
where P,Q}R,S are polynomials in x. Multiply numerator and denominator by R — S yj(ax2 + 2hx + 6), so that the new denominator is the polynomial R2 — S2(ax2 + 2hx + b) and the numerator PR  QS(ax2 + 2hx + b) + (QR  PS) <J(ax2 + 2hx + b), and we obtain the form '
0
where A,B,C are polynomials in x. We already know how to deal with the rational function AjC, so our problem reduces to the integration of 0
'
It is found (surprisingly, perhaps) more convenient to have the surd on the denominator, so we multiply numerator and denominator by yj(ax2 + 2hx + b), obtaining U F or
77—
where F is a rational function of x.
20S
SYSTEMATIC INTEGRATION
By § 2 above, the problems to which F gives rise involve as typical terms J xm (ra^O), 1 Ax + B We have therefore to consider the three types of integral: xmdx
J ... I* JJ (P
U (iii)
x2
+ 2g# + r)n ^(ax2 + 2hx + 6)
(i) The evaluation of the integral C
xmdx
Preparing the ground for an integration by parts, we observe the identity
J =\ Now performing the integration, we have, on the righthand side, X
f 
m+lj Equating the two sides, we have the recurrence relation dm + 2) alm+i + (2m + 3) hlm+1 + (m+l)blm = a;m+1 ^/(aa;2 + 21tx + b).
ALTERNATIVE TREATMENT OF INTEGRALS
209
This enables us, once 70, Ix have been determined, to calculate 72, / 3 , / 4 ,... successively, and so to evaluate Im for any positive integral value of m. For 70, we must reduce the surd to one or other of the forms enumerated earlier in this section, giving
f—
For Il9 we have to consider the integral
1= NOW
J
/ i + hIQ=
J so that
7j =  {yl(ax2 + 2hx + b) 
ex (ii) The evaluation of the integral dx
We can reduce this type to the form of type (i) and evaluate it at once by the substitution.
dx = ~72^« Then
ax2 + 2hx + b = alk + 7) + 2^(& + 7) + b
where
a = aJe2 + 2hJc 46,
/? = ak + h,
y = a.
210
SYSTEMATIC INTEGRATION
The integral is therefore dt
tn t
which is of the first form
k
xmdx
(ra ^ 0).
(iii) The evaluation of the integral (Ax + B)dx 2 px + 2qx + r)n J(ax2 + 2hx + 6)' where n ^ 1, g2 <^r. The general case is very difficult. It is, of course, possible to avoid it by expressing the rational function as a sum of complex partial fractions of the type \j(x — h)ny where h is complex. This reduces the integration to type (ii) ,
7i—77—o—^
rr
I \X — 1c\n' (nor 4 2hor 4 h)
(& complex)
already discussed. The final return to real form is an added point of difficulty. We begin with an algebraic lemma, designed to reduce two quadratic expressions simultaneously to simpler form. LEMMA.
TO
establish the existence of a (real) transformation of
the type which reduces the quadratic expressions ax2 + 2hx + b, px2 + 2qx + r to the forms
ut2 + v u'P + v'
simultaneously.
ALTERNATIVE TREATMENT OF INTEGRALS
Since
2
211
2
(*+l) (ax + 2hx + b)
under the transformation, we see that the coefficient of t vanishes on the righthand side if a, j8 are chosen so that aa/8 + % + j3) + 6 = 0. Similarly the coefficient of t from px2 + 2qx + r vanishes if poLp + q(aL + P) + r = 0.
Also a, j8 necessarily satisfy the equation in 9 On eliminating the ratios a/2: a + /3:1 between these three relations, we obtain the equation i h b p q r o,
i
e e2
which is a quadratic in 9 with (by definition) roots a, /?. Hence a, j3 may be found and the transformation determined. (If, exceptionally, a/h = pjq, the two quadratics differ only by a constant. The substitution ax + h = at reduces the integral In to one or other of the types discussed on p. 213.) Moreover a, p are real. If not, they must be conjugate complex numbers, since all coefficients are real. Suppose that a = A + i/x,
/? = A — ijj,
(A, jix r e a l ) .
Since p«.p + g(a + j8) + r = 0, we have the relation Multiply by _p (which is not zero, by the nature of the problem). Then
or
^2A2 + 2^gA +pr +p22 2
= 0,
2
(pX + q) + (pr  q ) + (p/x)a = 0.
Since p, g, A, JU, are all real, the two squares are positive; and, by hypothesis, pr — q2 > 0. Hence the lefthand side is positive. We are therefore led to a contradiction, and so the supposition that a, j8 are not real is untenable. Moreover, the condition pr — q2 > 0 shows that a, j3 are DIFFERENT numbers.
212
SYSTEMATIC
INTEGRATION
SUMMARY. We can find two distinct real numbers, a,j8, the roots of the quadratic equation
a P 1
h
b = 0,
6
62
which enable us, by means of the transformation ctt + p
x= t+l' to reduce the two quadratic forms ax2 + 2hx + b = 0, px2 + 2qx + r = 0 (q2 < pr) ut2 + v
, ,, , to the forms
n't2 + v' simultaneously. Let us now return to the integral. On substituting for x in the expression Ax + B, we obtain an expression of the form Ct + D t+l ' where C = Aoc + B, D = AfS + B; also
Hence
Ct + D aft dt t+l '(t+l)2 r 2 n (u t + v') (t+l)2n ' t+l __ [(otP)(Ct + D)(t+l)2n2dt
=J!
The numerator, which is a polynomial in t, can be expressed in the form P + Qt, where P,Q are polynomials in t2; and so, writing u' s, or u'
ALTERNATIVE TREATMENT OF INTEGRALS
213
we obtain P and Q as polynomials in s. Hence the numerator can be expressed as a polynomial in s = (u't2 + v'), together with t times another polynomial in (u't2 + v')9 the order of the whole numerator being 2n — 1 in t, which is less than that of the denominator. Hence In is found as a sum of integrals of the form
r k
v)'
") (u't2 + v'
For the latter, put
y2 = ut2 + v,
so that
ydy = utdt; ydy K= K lu' )k o u — (y2 — v) + v' y [u if r 2 dy {u y +{uv'u'v)}k*
hence
which reduces the problem to the integration of a rational function. Finally, consider
uft2 + v' =  ,
Write
2u'tdt = —9dz9 z* so that
t= Ut2 + V =
I—.— I, u — (uv' — u'v) z uz
Then UkJ[^2 /(A)U /(—j4* K
I I
j ^
V'J/ y * KI \ 1 — t i c IX
A / 111
AIL 6
/\
_ __ 1 r ~~2J
/y
yjL^c/ «ty j
i'iltr
] 1/0)1 71
^M/ — yu/U — U/ vj /if
zk~1dz yl[(lv'z){u(uv'u'v)z}Y
But the integral is of the form
whose evaluation we considered in the preceding section. The whole integration may therefore be effected.
214
SYSTEMATIC INTEGRATION
6. Trigonometric functions. Since 1 — tan 2 Ix . 2 tan \x 2 sin a; — cos a; = —•—rf. 1+tan ^ 1 + tan2^' trigonometric integrals may be reduced by means of the substitution M A 1 t = tan \x. Then
cos x = , . ,o,
sin a; =
2dt Hence a rational function of since and cos# is transformed into a rational function of t, and may be integrated by the methods of § 2. For reduction formulae for
I
&ii\mxcosnxdx
and similar integrals, see Vol. I, pp. 10610. The substitution t — tan \x should not be applied blindly. For example, if the integrand has period TT, the alternative t = tana; may be better. Thus, consider the integral rC
Since
dx
{sec (x + TT) + cos (x + TT)}2
(sec x + cos a;)2'
we may use the substitution t = tana;. Now
(sec2 x + 1 ) 2 dt
This is the integral of a rational function, and may be evaluated according to the usual rules. Alternatively, the substitution t = y/2 tan 6 makes the integral trigonometric again, but leads to an easy solution.
EXPONENTIAL TIMES POLYNOMIAL IN X
215
7. Exponential times polynomial in JC. A polynomial is a sum of multiples of powers of x, and so, in order to evaluate the integral ^ \ePxf{x)dx, where f(x) is a polynomial, we need only consider integrals of the type r un = epx xn dx (n ^ 0). Integrating by parts, we have the relation 1 nC un =  evxxn evxxn\ dx n P PJ leading to the recurrence formula 1 n un =  ePxxn — un_x p p which enables us to express un in terms of = \e pxdx = epx.
J
P
8. Exponential times polynomial in sines and cosines of multiples of JC. Consider the integration of a sum of terms of the type epxsinaxxsina2x ... sin amx cos bxxcosb2x... cosb n x, involving m + n factors in sines and cosines. The use of the formulae such as 2 sin ux cos vx = sin (u + v) x + sin (u — v) x, 2sinw#sin vx = cos (u — v)x — cos (u + v)x, and so on, enables us to reduce the number of terms in a typical product to m + n— l,m + n — 2,m + n — 3,... successively, while retaining the same type of expression. Ultimately we reach one or other of the forms C = \epx cos qxdx, S= \epx sin qxdx,
216
SYSTEMATIC INTEGRATION
whose solution should now be familiar. We find
°  ? C 0 S 9*)
p2^2
WARNING. The work of this chapter will enable the reader to evaluate (ultimately) any integral that comes within its scope. But there are many functions which cannot be integrated in terms yet known to him. Some of these are very innocent to look at; for example, „. „ fsm x j C ,, dx, \exdx,
dx 1
MC Thorough familiarity with the FORMS of the integrals that can be evaluated should help in the avoidance of these pitfalls. [For examples on the work of this chapter, see Revision Examples VIII and IX, pp. 226, 227.]
CHAPTER XIII INTEGRALS INVOLVING 'INFINITY' 1. 'Infinite' limits of integration. It is often necessary to evaluate an integral such as !"f(x)dx Ja
under conditions where b tends to infinity; we then speak about 'the integral from a to infinity'
f
f(x)dx.
Ja
Similarly we meet the integral f(x)dx, or, combining both possibilities,
r
f(x)dx.
f
f(x)dx
Our problem is to discussJcc what is meant by such integrals, and to show how to evaluate them. The general theory is difficult, so we confine ourselves to the simplest cases. We also assume that the function f(x) is continuous throughout the range of integration. We define the integral to infinity
by means of the relation
Ja Ja
f(x)dx= lim Ja
f(x)dx.
N>coJa
In the cases with which we shall be concerned, the indefinite integral F(x) off(x) is considered to be known, so that rN
15
f Ja
= F(N)F(a).
218
INTEGRALS INVOLVING 'INFINITY*
We can therefore evaluate the integral to infinity in the form *f(x)dx = Hm
F(N)F(a).
It may be added that it is often possible to evaluate the definite integral
f(x)dx even when the indefinite integral
cannot be found. A wellknown example is
/(#)cfo
dx whose value
is ILLUSTRATION
1. To evaluate rco
xndx.
Consider the integral xndx which, by elementary theory, is
[ n+l or
If n + 1 is that
n+l POSITIVE,
n+l
then Nn+1 increases indefinitely with N, so hm { N^ooin+l
} n+l)
does not exist. If n+l is NEGATIVE, then Nn+1 tends to zero as N increases indefinitely, so that hm Hence
^1
l
1 } } =
n+l)
r°° xndx =
h
1
.
n+l
i = (n<  1). n
+l
' I N F I N I T E ' LIMITS OF INTEGRATION
219
ILLUSTRATION 2. To evaluate /•oo
I xe~xdx. Jo We know that, on integrating by parts, xe~xdx = —xex+\ rN r xe~xdx=
so that
Now
l.e~xdx
IN
(x+l)ex\
WWZ£
^ __+...
and, for positive values of N, the denominator is certainly greater than %N2, so that (N+l)eN+O
Thus
as N increases indefinitely. Hence
r
xe~xdx = Km { (N+ l)e~N+ 1} iV>oo
= 0+1 = 1.
ILLUSTRATION 3. To evaluate
dx Consider the integral
where M, N are large positive numbers. The value of this integral is
or
tan" 1 a; L \M tan* 1 N — tan" 1 (— M).
Fig. 101. 152
220
INTEGRALS INVOLVING 'INFINITY*
Care must be exercised in selecting the correct angles from the manyvalued inverse tangents. The graph y = tan" 1 a;, shown in the diagram (Fig. 101), illustrates the 'parallel' curves along which y must run, and the values selected must be confined to one of them. The most natural choice is to work with the curve through the origin 0. Then for large negative values of x, the value tan~ 1 ( — M) just exceeds — \n\ as x increases, tan" 1 x rises continuously through, say, — \TT, — JTT, — £77,0, JTT, £TT and so on, approaching the value \TT for the limit of tan" 1 N. /•oo
Hence
,7™
f~
= lim {tan1 i f }  lim {tan1 ( 
=
ILLUSTRATION
7T.
4. (Change of variable.) To evaluate
r
dx
h
Jo
N
r Consider the integral
dx
uN=
Make the substitution Then
., oxo. Jo (1+x 2 ) 2 a; = tan 0, dx =4 sec20d0. sec
between appropriate limits. When x = 0, we may conveniently take 0 = 0. As a; increases, 0 also increases, assuming a value very near to \TT for large values of N. In the limit, we obtain the value ^n. Hence
= Jir. ILLUSTRATION
5. (Formula of reduction.) To evaluate /•oo
xmexdx> Jo where m is a positive integer, greater than zero. Jm=
'INFINITE'
Let
LIMITS OF INTEGRATION
221
xme~xdx
Im =
Jo IN [N Tn 1 x r = —xmex\ +m\ x ~ e~ dx. Jo Jo L Now
Nm N™e~N = ~ =
and the denominator, being certainly greater than Nm+1l(m+l)\9 greatly exceeds the numerator for large values of N, whatever m may be, so that xme~x = 0
Also when x = 0, since m > 0. Hence
lim
— a^c*
2V»ooL
and so, as N+00,
= 0,
JO
Jm 
The formula of reduction enables us to make the value of Jm depend on that of Jo; for Jm = mJm_x
= m(ml)J m _ 2 ra(ral)...2.U0 m\J0. Moreover,
Jroo== fI° erxdx Jo Jo = lim  e~x N*ao[_
Jo
= lim \e~* = 1.
222
INTEGRALS INVOLVING
'INFINITY*
/•oo /•oo
Hence
Jo
xm erx dx = ml
EXAMPLES I
Evaluate: 1.
2. f j k
r^. 2
Jl ^ 4.
J2 W * r.
Ji 1 + a: 7.
Jo
Jo
5.
2
/•oo
3. f r.
Jool+sc
2
6.
Jo Jo
e~x8inxdx.
eax, stating for what values of a the integration is possible.
2, 'Infinite' integrand. It may happen that, in evaluating the function \f(x)dx, Ja we find that the integrand f(x) tends to infinity—or, indeed, has some other discontinuity—in the range of integration. Suppose, for example, that f(x) increases without bound near x = c. If c is inside the interval (that is, not equal to a or 6) we 'cut it out' by considering the sum *ce
J
fb
f(x)dx+ a
f(x)dx, Jc+17
where €,77 are small positive constants, over ranges which just miss c on either side. We then define the value of the integral to be rce rb lim f(x)dx+]im f(x)dx, supposing that these limits exist. When c is at a or 6 the modification is obvious; only one limiting value is then required. ILLUSTRATION 6. To determine the values of n for which the integral exists.
' I N F I N I T E ' INTEGRAND
223
Consider the integral
I —n 1 71
As € tends to zero, the term c " also tends to zero if the exponent 1— n is positive; otherwise e1~n increases indefinitely. Hence the integral exists provided that n< 1, and its value is then
{1 — 0} \ — nK
J
1 ln
The case n = 1 requires separate treatment. We then have
j ~^ =  1 O g H
=1
°g 1  1 °g € >
so that (compare p. 4) the integral does not exist. The required condition is therefore n< 1. ILLUSTRATION
7. To evaluate dx
This integral involves two infinities; the upper limit of integration is infinite, and the integrand is infinite when x = 1. The two phenomena must be kept separate. We consider the integral r
dx
= Jl+e
224
INTEGRALS INVOLVING 'INFINITY*
Write
xl = t*, dx = 2tdt. 2tdt
Then 
/
•
between appropriate limits. Now we have cut out the value x = l,t = 0 by making the lower limit 1 + e. Thus t is never zero, so that the factor t may be cancelled from numerator and denominator of the integrand. Hence
between appropriate limits. Consider the range of variation of t as x increases continuously from 1 f e to N. We have
having committed ourselves to the positive square root by putting = t in the integrand during the substitution. Hence t increases continuously from the small value <Je to the large value yl(N— 1); and as it does so, tan~ 1 (^) increases continuously from just above zero to just short of \n. Thus, in the limit, yl(x—l)
/ = ITT0
ILLUSTRATION
8. To evaluate dx
f
Ja The denominator in the integrand becomes zero at x = a and at x = 6, and so we must consider the integral
.
rb*
dx
' I N F I N I T E ' INTEGRAND
225
Make the substitution x = a cos21 + 6 sin21, dx = 2(6 — a) sin t cos £d£. a; — a = {b — a) sin2 £,
Then
6 — x = (b — a) cos2 £. Consider the range of integration in the variable t. When x = a, we have asin2£ = 6sin2£, so that sint = 0; and when x = 6, we have 6 cos2 2 = a cos21, so that cos t = 0. We must select as startingpoint for £ a value for which sin t = 0 (corresponding to # = a), and the obvious value is t = 0. The relation dx
— = 2(6 — a) sin £ cos £ U/t
shows that, at any rate for small values of t, the variables x, t increase together; and this process continues until t = %TT, at which point x has the value 6. The range is therefore 0, \TT. But we have had to exclude these points themselves because of trouble with the integrand, and so the range must run from just above zero to just short of \TT\ say from e' to \TT — rj'. Thus ,
• 2(6 — a) sin t cos tdt V{(6 a)2 sin21 cos21}'
Moreover the POSITIVE value of ^{(6 — a)2 sin21 cos21) in the interval 0, \TT is (6 — a) sin t cos t, so that ^ n*i 2(6  a) sin £ cos tdt ~ Je'
(b — a) sint cost
We have excluded the endpoints, so that sin t and cos t are not zero in the range of integration; we may therefore cancel factors in the numerator and denominator of the integrand, giving =/;
2dt
Proceeding to the limit as e',7?' tend to zero independently, we 77.
226
INTEGRALS INVOLVING
'INFINITY*
REVISION EXAMPLES VIII
'Advanced* Level 1. Find the following integrals: dx
x2dx
r
r
dx
l) (a ._3)' J (a:l)(a;3)' JJ O
T t
(^ 2 +l)
+
J{(xl)(3x) )Y
X
2. Integrate j — . — = j—z—rjr cos" 1 1 (6 < a).
Show that
<J(a2b2)
Jo a + bsmx 3. Evaluate the integral
\a/ v
'
Cb___xdx___ Ja<j{(x<*)(bx)} by means of the substitution x = a cos2 0 + b sin2 0. Prove that the integrals
fV(la;) 8 dz,
fz 8 (l z)7dte
Jo
Jo
are equal, and show that their common value is 7!8! 16! '
4. Integrate a,3 )(X
2)'
5. Integrate Xs
X
X 2
(1 + X )(
2
:)'
<J(x + 4:X + l
Prove that, when a,b are positive,
r a coscosa; +a;^b sin a; 2
2
2
2
2
77
a(a + 6)#
6. Integrate (2x + 3)
1
x
e
x
REVISION EXAMPLES VIII
227
7. Integrate
Jltl
, flog*)2
8. Find a reduction formula for
and evaluate the integral when n = 2. 9. By integration by parts, show that, if 0 < m < n, and
then
/ =  m\ x™1
Deduce that
n_1
{xn( 1  x)n) dx.
7 = 0.
10. Taking
y=
verify that the result of changing the variable from y to x in the integral # J V<s/ 2 — :i) is
Deduce that
f2 rfcc ^ — ^ ^ —  ^
= loge2.
REVISION EXAMPLES IX
'Scholarship9 Level 1. Show that, if
cos nxdx
un
then, provided thattt#1, Hence, or otherwise, show that, if n is a positive integer, «„ =77/(3.2").
228
INTEGRALS INVOLVING
'INFINITY*
2. Prove that
Jo a W J + 6 W g Jo
=
2^6' Jo l 0 g t a n
sin 2 01ogtan0d0 = JTT.
3. (i) Prove that values of a,b can be chosen so that the substitution , , X =
at + b HT
yields the result dx dx _f f J (2x2  6x + 5) V(5z2~ 12a + 8) " J
(t+l)dt
(ii) Find the indefinite integral of t+l
4. Evaluate J^sinx)^,,
J^^)
5. Prove that o
J(cos 2x  cos 4a) da = i J6  J V2 l o g (2 + V3)f1
6. Evaluate
da
r
,
f1
^ 4 fc
 y 
^.
7. Evaluate r^atan 1 ^
C*n
2 2
Jo (1+a )
oos2 9d0
' Jo a*
Find r°°
8. Evaluate
xdx
Jo Jo
Find
r 9. I i
T
i n11 ==
C2ncos(n — l)x — cosnx , aa, Jo 1cosa
show that In is independent of w, where n is a positive integer.
REVISION EXAMPLES IX
229
Hence evaluate In and prove that C2n ($m\nx\2 Jo \sin^/
1
10. Evaluate Ca
xdx
' CLX
I
%j i
** \j\jij
T
~"2 2\ > Q> —X )
^~
%\J

^
aX
'
Jo
11. Evaluate the integrals:
dx 2
1
JopT^^ Jo
2 cos # + 2 cos x sin a; + sin2 x* dx x2 dx z )(l+a; 2 ) 2 #
12. Evaluate
2
00
13. Evaluate where a, 6, c are positive. 14. Prove that f00
rf^
a sin a
2 Jo # + 2a; cos a + 1
(0
Evaluate r°° I
,/0
dx /V«4 _l
"^
r«
O/y«2 pr%Q / v 4— 1
i^ •«•*/
v U i j LX T^ x
I *v
/o
15. Prove that, when b > a > 0, 2ab cos 6) n
C sin 0 cos cc 0 d# 2 2 Jo <J{a + b —
b'
2a
Make it clear at what points of your proofs you use the condition &>a>0. 16. Evaluate the definite integrals:
230
INTEGKALS INVOLVING
'INFINITY'
17. Evaluate
7_f
Ipl^jicw (o<8<w)
when (i) 0 < r < 1, (ii) r > 1. Prove that, S being fixed, / tends to one limit as r> 1 through values less than 1, and to a different limit as r> 1 through values greater than 1. Show, also, that neither limit is equal to the value of I when r = 1. 18. Show that, by proper choice of a new variable, the integration of any rational function of sin x and cos x can be reduced to the integration of a rational algebraic function of that variable. Integrate sin a; 1 in(# —a)' sin a; cos a; + sin a; + cos a;—1# (x2— 1)
19. Integrate Prove that
p_
dx
a. 2
Ji (iE + c o s a ) ^ — 1)
r 0
20. Integrate
sin ex
rroc
l4cosasin# —^— ,
sina 2—^
(0 <
(0 <
OL
OL
<
< TT),
\TT).
(b
Evaluate Cn x sin xdx l\l a s i n x) Prove that I /(sin 2x) sin xdx = y]2\ /(cos 2x) cos xdx. Jo Jo 21. Evaluate the integrals:
REVISION EXAMPLES IX
231
22. Find the integrals:
Evaluate
'
23. Prove that, if a is positive, the value of the integral
r
1—acos0 Jo 1 — 2a cos 0 + a2 Jo is 77 —cot"1 a or —cot"1 a, according as a<\ or a > l , where the 1 value of cot" is taken between 0, ^77. What is the value of the integral when a = 1 ? 24. Show that, if P, Q are polynomials in s, c (where s = sin 0, c = cos 0), of which Q contains only even powers of both s and c, then can be expressed as a sum of integrals of the form 12^(5)
J
l + cos20
'
v y
1 :—p: dd.
J l + sinfl
In case (ii) obtain the integral also by the substitution x = tan^0, and reconcile the results obtained by the two methods. 25. Show that the substitution b( 1\2 d( 1\2 a\ tj c\ t) reduces the integral \F{x9 J(ax + 6), J{cx + d)} dx, where F(x, y, z) is a rational function of x> y, z, to the integral of a rational function of t.
232
INTEGRALS INVOLVING
'INFINITY'
Hence, or otherwise, find
J (x +
,dx.
26. Find a reduction formula for dx J (5 + 4cos#) n
n
in terms of In_l9 I n _ 2 {n ^ 2), and use it to show that
f1** da; Jo ;o 27. Find a reduction formula for dx Evaluate the integral when n = 1 and when w = 2. 28. If
Ip q =
si Jo
show that
(p + q)Iv a =
and evaluate / a _ x 5 where a is any positive real number. 29. If In =
xneaxcosbxdx,
Jn=
xneax&inbx,
Jo Jo where w is a positive integer and a, 6 are positive, prove that
Show that
(a2 +fe2)**^1)In = n! cos (n + 1 )a, (a2 + 6«)*(»+D J n = w! sin (n + l)a,
where tan a = bja and 0 < a < \ir. 30. (i) Prove that, if dx then
(a2 + 62) Mn  2a«n_1 + « n _ 2
REVISION EXAMPLES IX
233
(ii) Prove that, if n is an integer and m > 1, then sinm x cos 2nx dx = m(m — 1) o
sinm~2 x cos 2nx dx.
Jo
31. Find the reduction formula for dx Prove that, if a and ab — h2 are positive and n is a positive integer, then J _oo (ax2 + 2te + 6)w+x "
2n 'ab h2] _« (ax2
2.4... 32. Show that J (aa2 + 2hx + 6)w+*
(A2  ah)**1 J Vi/
ff
g2 = a^)2 + 2hp + 6,
where
Deduce that, when a,b,h + <J(ab) are positive, dX
l
\a* Jo (aa:2 + 2te + 6)*
33. Find a reduction formula for the integral f Prove that
dx
;
^'
234
INTEGRALS INVOLVING 'INFINITY*
34. (i) Integrate
\
(ii) Find a formula of reduction for j
(l+x2)neaxdx.
35. (i) Evaluate f1
dx
I
(Q
(ii) If prove that ?mn — (2w — 1) un_x cos a + (n — 1) wn_2 = 2 sin  a when n > 2, and evaluate w0, i^!. 36. Obtain reduction formulae for the integrals
r»(log^ Ji *2
(Vsin^, Jo
and evaluate the first integral for any positive integer n. pi
prove that (2p  g + 2) 7p + (2p Hence, or otherwise, prove that Jo 38. If
/,
Ja
verify by differentiation that, when n is a positive integer or zero, 2(n +1) k(k 1) In+2  (2n + 1) (2*  1 ) In+1 + 2nln
Show further that I^ can be integrated by the substitution
REVISION EXAMPLES IX
235
Hence, or otherwise, find dx 39. Show that, if I
i
i*°°
n then
2Jn = /n_2,
2
Hence, or otherwise, prove that cos 40. Find a reduction formula for the integral
r**_d^_ Jo eosna;' and evaluate the integral for the cases n = 1,2. 41. Prove that, if dx
then Hence evaluate the integral
42. Find a reduction formula for the integral
4=
xndx
and use it to evaluate
J
xz dx
162
APPENDIX First Steps in Partial Differentiation The functions which we have considered in this volume have always involved a single, variable. The work on functions of several variables belongs to a later stage, but it may be convenient to set down one or two of the most elementary properties—mainly definitions and first applications. Consider, as an illustration, the expression As x, y, z take various values, so also does u. For example, y = 2,
3=1,
if then
2=3,
w= 72;
if
3 =  1,
then
y = 0,
2=3,
w = 0;
if
3 = 2,
then and so on.
2/=  2 , 2 = 1 , t* =  1 2 8 ;
We say that u is then & function of the three independent variables x,y,z.
To denote this functional dependence, we may use the
notation
u(xyz)
The function u no longer has a unique differential coefficient. Each of the variables x,y,z is capable of its own independent variation, and each of these variations produces a differential coefficient of its own. More precisely, we use the notation — = the differential coefficient of u with respect to x only, calculated on the assumption that y, z are constant; = the differential coefficient of u with respect to y only calculated on the assumption that z9x are constant; = the differential coefficient of u with respect to z only calculated on the assumption that x, y are constant.
APPENDIX Thus, if
then
237
2
u = x*y*z ,
f^ = 4xyz,
2
ox
to
3*223xyzt
=
cy
to oz
As another illustration, suppose that u = cos (ax + by2) is a function of the two variables x, y. Then du . . . o. du _. . . , ov — = — a sin (ax + by2), — = — 26^/ s m (aa; + 02/2). mi r. .. Bu du du The functions ^r, ^, 7&? ^ dz
are called the partial differential coefficients of w with respect to x,y,z respectively. These partial differential coefficients are, in their turn, also functions of the three variables x, y, z, and have their own partial differential coefficients. We write d
d (du\_d2u Bz\dz)~ dz2' The 'mixed* coefficients are a little more awkward. We write dx \dy) ~ dxdy du\
9
d2u dydz'
d2u dx \dz) ~~ dxdz9 2 d [du\ 3 (du\ (du\ _ d u dy\fa)~dydx'
3 (du\ __ d2u d (du\ Tz\dx] dzdx' ~j
82u
In practice, however, it may be proved that for 'ordinary' functions (a term which we do not attempt to make more precise) interchange of the order of partial differentiation leaves the result
238
APPENDIX
unaltered. Thus
00 2
du dy dz
dz dy9
d2u _ d2u dzdx ~ dxdz' d2u _ d2u dxdy ~ dydx
For example, returning to our function we have the relations dxdy
ex
dydx
dzdy
dz 7)
dzdx
dz
dxdz
dx EXAMPLES I
2
2
. . du du d u d n d2u d2u Find — 2 , T  ^  , a—^> X T 9   , ^o c^o; dy dx dxdy dydx dy2 functions: 1. x*yz. 4. excosy. 2
7. x* sin y. 10. secx + secy. 13. Prove that, if f(x,y)
_
for e a c h o f t + lie
__ . following&
2. xy2.
3 . a;2 + ^ 2 .
5. Iog(#f2/).
6.
8. e^sinz.
9.
x
2
11. e sin 2y.
12. a^e^.
is any jpolynomial in #,?/, then
a2/ _ a 8 / .
APPENDIX
14. Prove that, if f(x,y,z)
239
is any polynomial in x9y,z, then 2
d f _ d2f dydz " dzdy' ILLUSTRATION 1. The 'homogeneous quadratic form*. Let
Then
u = ax2 + by2 + cz2 + 2/?/z + 2gzx + 2hxy. du ~dx~~
du
These expressions are probably familiar from analytical geometry. To show how the partial differential ~ \x^z) coefficients are linked with the idea of gradient, we use an illustrative example. Let OXtOY be the axes for a system of rectangular coordinates in a horizontal plane. This is illustrated in the QUy.o) diagram (Fig. 102), where the reader may regard himself as looking 'down7 R{x,o,o) upon axes drawn in the usual position. / Fig. 102. The straight line OZ is drawn vertically upwards. Given a point P in space, let the vertical line through it meet the plane XOY in Q; draw QR perpendicular to OX. Denote by x,y,z the lengths OR,RQ,QP respectively; then the triplet x,y,z may be used as coordinates for the point P in space, just as the pair x, y is used for a point in a plane. If P is the point (x, y, z), then Q is the point (x, y, 0) in the horizontal plane; the coordinate z gives the height of P referred to the plane XO Y as zero level. (Of course, P may be below the plane, in which case z is negative.) In particular, if x} y, z are connected by the relation
240
APPENDIX
where we assume f(x, y) to be a singlevalued function defined for each pair of values of x, y, then, as x, y (and consequently z) vary, the point Q moves about the plane XOY, while P describes the surface whose height at any point is equal to the corresponding value of the function. We say that the surface represents the function f(x, y). For instance, it is an easy example on the theorem of Pythagoras to show that the function
is represented by the hemisphere of centre 0 and unit radius lying above the plane XOY. We now assume, for convenience of language, that OX is due east and 0 Y due north. We regard the surface
as a hill, and P as the position of a climber on it.
OR
R'
Fig. 103.
Suppose that the climber is at the point P (Fig. 103) defined by the values x, y of the easterly and northerly coordinates, and that he wishes to climb to the point P' defined by x + h, y + k. The crux of the difference between functions of one variable and functions
APPENDIX
241
of two lies in the fact that, whereas for one variable motion along the CURVE representing the function is defined all the way, for two variables the SURFACE may be traversed by an innumerable choice of paths. Moreover, each way of leaving P will demand a gradient all of its own. The partial differential coefficients are the bases of the mathematical expressions for such gradients corresponding to the various paths. From the mathematical point of view, the obvious way to pass from P to P' is firstly to move the distance h easterly, to B, and then to move the distance k northerly. The climber thus describes in succession the two arcs PB, BP' shown in the diagram. Now suppose that P' is very close to P, so that the arcs PB, BP' are very small. The arc PB may be regarded as almost straight, so that the 'rise' between P and B is proportional to the length h, say Sz(easterly) = ah. Similarly BP' is almost straight, so that the 'rise' between B and P' is proportional to k, say Sz(northerly) = j8&. If 8z is the total 'rise' between P, P', then Sz = Sz(easterly) + Sz(northerly) = ah + fSJc.
If the climber had gone first northerly and then easterly, following the course PD9 DP' in the diagram, then, for distances so small that the paths may be regarded as straight, PBP'D is approximately a parallelogram, and so, once again, Sz = ah + fik
for the SAME values of a, j8. Two simple observations complete the illustration. Geometrically, a, j8 are the gradients of those curves which are the sections of the hill in the easterly and northerly directions respectively. Analytically, we see by putting k = 0 that a is the ratio Sz H h calculated on the assumption that y is constant; thus
242
APPENDIX
evaluated at P. Similarly we see by putting h = 0 that jS is the ratio 8z7k calculated on the assumption that x is constant; thus R=8z P By
calculated at P. Hence the partial differential coefficients —, — are identified as the ex cy gradients of the surface z = f(x9 y) in the x and ydirections respectively. EXAMPLES II
1. Show that the function with the zaxis measured vertically upwards, is represented by a hemisphere. Prove also that the gradients in the x and ^/directions at the point (x,y,z) are in the ratio(# — 1) : y, and that these gradients are equal only for points on a certain vertical diametral plane. 2. Prove that the function with the zaxis measured vertically upwards, is represented by a 'bowlshaped' surface whose horizontal sections are ellipses of eccentricity £^/3. Prove that the gradient in the ^direction at the point (1,2,17) is 2, and that the gradient in the ^direction at the point (3,1,13) is 8. du d^u d^u 3. Find —, ^9 7— for each of the following functions: (i) eax sin (by + cz), (ii) xyze~x\
(iii) (y2 + z2)log(a
4. Prove that, if u = ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy, and if then
d2u d2u d2u ^ "i~2+^~2+"^2" = °> dx2
dy2
dz2
a + b + c = 0.
243
APPENDIX 2
5. Prove that, if then
2
r = yj(x + y + z2 ) . dr
Deduce that
x
d2r
1
j
j
—
X2
2 r
Prove also that
Finally, there is a point of notation which the reader may meet in physical applications. Consider, as an illustration, the transformation . n n x = r cos 9, y = r sin 8 between the Cartesian and the polar coordinates of a point. Four variables are involved, of which two are independent—for example, r and 6, When we form the partial differential coefficient —, we naturally have in mind that 6 is the other independent variable, and the relation x = r cos 6 ., thus gives
8x x Q — = cos 0 =  .
But it is possible to express x in terms of r and y, in the form
x = J(r2y2), ,., and then
8x
r
— = r—^
r rr =  .
dr J(r2y2) x These two values are quite different; they are, indeed, calculated under the quite different hypotheses 6 = constant, y = constant respectively. To make sure what is intended, we often use the notation
to denote the partial differential coefficient of x with respect to r when 6 is the other independent variable (being kept constant
244
APPENDIX
during differentiation). With such notation, the two formulae given above may be expressed in the form
cx The following examples should serve to make the notation clear. EXAMPLES III
Given the relations x = r cos 9,y = r sin 9, establish the following formulae: 1 fk\ dr)0
dx 5.
=?.
2
r
d
~y\ =¥.
' dr)e
r
r
9. ^ ) =  r .
11. «
1.
10.
.2.
«.i.
ANSWERS TO E X A M P L E S CHAPTER VII
Examples I: + l).
2. ilog(2z+l).
4. lx* + logx.
3.  J l o g ( 2  3 s ) .
5.  l o g ~ .
6.
X ~T i.
7. log 2.
8. log 2.
9. Jlogf.
10. JlogS. 13. . 3 +2 16. x + 2x\ogx.
11. Jlogf. 14. 2cosec2x.
12. JlogJ^. 15. —coto;.
17. xn~1\nxn1\ogx.
18. 2xj(l+x2).
19. a; log a;a;.
20. ^(loga:)2.
21. log sin x.
22. ialogxlx.
(
23.
1 ^— o o s ^c\
25. log (x2 + 5a; +12).
^—1.
1 + COS X)
26. log {x2 3x + 7). 27.  tan 1 ^ ^  ^ W log (a;22a;+17). 28.  12 tan 1 (x + 3) + log (a;2 + 6a; +10).
29. 9tan 1 fcp)+ flog{x28a; + 25). 30. ~
Examples II: ydx
3. 2/aa; \*l
1+a;
1—a;*
ydx
«*.11 —2a; 2
a;+ iaAx+
33
4. !2/da; £  ?a; d
1+a; 2 1+a; 1+a;4
246
ANSWERS TO EXAMPLES
^ Idy I 3 1 4x L n Idy 5. ff = =  ++ccotxo t x + +.. 6. 6. // = = , s ydx x l+z 1—z ydx l + x* z d l+ 1 d 4 3 m Idy 7. r = — ydx x l—x
8 cosec 4#.
1 dy ydx
2 sin a; 1 + cos x
1 + 2x
\dy _ y dx
1 , 4 t 1 — X 1 + 2X
9
16
Examples III:
3. l o g ^ ^ .
4. (or 41)2
5. tan 1 x + 1 logf^fr. *g(^)2(^Ty 9.
6. 8
 
AA
10. l o g C x  ^  ^  ^ .
11.
12. ^ t a n  H ^  ^ l o g (a; ) ^2+T 4) ^ + 25(xl) !)(»3)' 13. ilog ( « (z2) 4 14.
4
15. 16. 17. ^  ^ 
247
ANSWERS TO EXAMPLES
19. a:20. 21.
24. ^ r tan"1
Examples IV: 1. 2e 2 x .
3.
2.
x
x
5. e
4. e cos a; — e sin ic.
sinx
7. e 2x — e~2x. 9 9

6. — e~x(l—a:)2.
cos#
8. 2
g^fl + 2x 7 i ^9
^
x) ^
1 0

2 2
(1—a; ) 3x
11. e (3cos4a; — 4 sin 4a;). 2x
13. %e . x
16. \e~ \ 8111
19. e
*.
14.
12.
15. c^.
5x
%e .
18. 6x(a;2
x
17. (a;l)e .
21. etanx
20. e s i n 2 x .
22. 23. 24.
Examples V: 1. aln = a ^ c ^  n / ^ j . 2. (a2 + n 2 )/„ = eaxsin71"1 x(asinxncosx)
+
n(nl)In_2.
3. (a2 + 6 2 ^ 2 )7 n = eax cos71"16a;(a cos bx + nb sin 6z) + n(n — l)b2ln_2. 4. 12044e.
5. A(41c*»24).
6.  f ( l + e»).
248
ANSWERS TO EXAMPLES REVISION EXAMPLES III
1. (i) (l + x^Q
+ nlogll + x)}.
(ii)  + log(l+s). 7b
2. ^ 3_ 4 ^ = dx dx
0.
— 2x
4. 2cot#,
5. 2 sin x — 3 sin 3 x, „
1
.
—2
2ae ax cosaa;,
*
2x
7.  I ,
{]i)secx
)i' 1a;
12/
{iii
' 2x
3\ *'
1 + sin2s #
tana; —cot a;.
1 16. Velocity en9 acceleration —en. 18. Tangent: xt&nt + y — asint = 0. Normal: xco8t — ysint — acos2t = 0.
2a; 2
ANSWERS TO EXAMPLES
aft sin 0
249
3V3
h — a cos 0
2
21. a; = 0, y = 1, minimum;
2 8 . (<Ja + y
x = I,?/ = 2, neither; a; = — 2,2/ = 29, maximum. 30. Area: 2 +  c o t 0 + 2tan0; angle: tan 1 31. ( — 1 ,  1 ) minimum; (1,1) maximum; 24a; —25?/+ 8 = 0. 34.
(i) 2log(2z + 3 )  l o g ( x  1). (ii) Jcos 3 # — cos a;. (iii) a;2 sin a; + 2a; cos a; — 2 sin a;.
35.
(i) £a; (ii) (iii) (i) J sec 3 x — sec x.
36.
(ii) ^(l+aj^tan^la;. (iii) log (4a;  1)  log (a; + 1). 37.
(i) 2 log (2a;  1 )  2 log {x + 2). (ii) J sin 3 a; — J sin 5 a;.
38.
(i) 3 log (x  3 )  f log (3a;  2 ) . (ii) sin x — x cos a;. (iii) fa 4 8 + Ja 4 sin 2 6 + ^a 4 sin 40,
39. — sin a; — coseca;, 40.
3
^
2
where
a; = a sin 0.
x
£e (sina; — cos a;), 224.
.
21og2f. 42. 1. 43. a;2 sin a; + 2a; cos x — 2 sin x, 17
 log (1 + 3a;)  log (1  3a;).
250
ANSWERS TO EXAMPLES
44. (1 +x*)l, ixJ(l+
45. Ja;^(l—rc^ + Jsin 1 ^, 46. 21og(l+^»),
^sin^a;) 2 ,
or
T—
gr,
e*cosa;.
— cos a; + § cos3 x — J cos5 x%
x + 37 log (x  6)  26 log (a;  5).
—1 a: sin" 1 a; + ^(1 — a:2). 48. x — log(l— x)9 tan" 1 a;,
50.
x 51. a; —log (a:+2),
— cosa; + f cos 3 # — cos 5 a;, J tan 3 a; — tan x + #, a; tan x + log cos x,
52. log (2a:2 x  3 ) ,
J sin3 a;  £ sin5 a;,
53. «  ~ ^ ,
Ilog2.
A,
54. JTT, 16 log 2 — ^ , 55.
1. J. 7r~2.
^773127T + 24.
(i) ^ +  l o g ( l + V 2 ) 
56.. (m + n)
ex(x— 1).
(2) 1,
K4V25).
fin
r\n sinm a: cos n a:rfa;= (n— 1) sinm a: cosn~2 x dx, ^ , 0. Jo Jo
57. 1
Ya ^
P ( l +x*)«+*dx = 2n+*+(2n+l) F{1 + z*)n*
58. (i) 67 24 J\
ANSWERS TO EXAMPLES
251
CHAPTER VIII
Examples II:
v.2
3.
/y.4
/y.2n
l
Examples V: o
1
^to)
(^
(3x)n
3
3. 2o;i(2a;)2 + i(2ic) 3 +... I ^ 1
K
'
n
(<)~\n i
(x) I •
(2n+l)l .. + (  1 nxn
"I __ /y» _l 1*2 __ sy3 1
+ (^+l]
[/uXf
(i
2! i
XV.
"~~ OJU — ^yOJUJ
l
3
5
^ _
3
)
—— • ,
Examples VI: 1. 2005.
2. 2995.
3. 20017.
4. 2999.
5. 19996.
6. 03328. 172
252
ANSWERS TO EXAMPLES
Examples
VII:
1. 1680a;4 sin x + 1344a;5 cos x  336a;6 sin x  32a:7 cos x + x8 sin x. 2. a;2sina; — 8a;cosa; — 12sin#. 3. e2*(122cos3a; + 597siii3a;). 4. 34e3*(9a;3 + 54a:2 + 90a; + 40). 5. x3 cos {\UTT + x} + 3na;2 cos {(n — 1) TT + a;} + 3w(7i — 1) a; cos {(TI — 2) \TT + x} + n(n — 1) (TI  2) cos {(w  3) JT7 + a;}. 6. 2n~3e2a;{8a;3+127ia;2 + 6 n ( n  l ) a ; + n ( n  l ) ( n  2 ) } . 7. 2 5 .10.9.8(1  2a;)5 (  132a;2 + 55a;  5 ) . 2 3 6 12' 8. —r.— : (3a;4l) 4 (819a; 2 + 312a; + 28). o!
Examples VIII: 1  a? +
1 a;3 1 3 a:5 1 3 5 a;7 2 ' " 3 + 2  i ' 5 "+ 2'I6T+"""'
2. x\x* + \x*\x'J + .... REVISION EXAMPLES IV
1. (i) ax(\
^ ab,
1
a262.
1
x = 1 maximum, x = — 2 minimum. 4. secmo;tann~1a;{7i+(m4W')tan2a;}. 840 tan 4 a; + 640 tan 2 x + 56.
2(a; 2 q;+2)
ANSWEBS TO EXAMPLES 3
6. 4 s i n x c o s x ,
2
12sin s—16sin s,
24 sin x cos x — 64 sin 3 x cos x, x = 0 minimum, # + 1
253
4
256 sin 4 x — 240 sin 2 x + 24,
x — \TT maximum. nsinx tan 2 o;(cos x + sec a?)*1"1,
11 —  1 ,
li when «> 2, 2 + \ when »  2, 2 2a:—r3 when TI = 1. a; 10. (ii)
2nn\
1 777 9
12
t a n x,
—
—
n\ 1
——

m (ii) sin(a; + n7r),
a; sin (a; + ^ T T ) + TI sin {a;
13. (i)  2 cos a;. .... cos3a; —sin3a;
1— x+2x2
ex — e
(22)
(cosx + sinx) 2 '
3
15.
(i) (  1 ) * 1 " ' V ' .
(ii) 2nsin(2s + 4nw).
(iii) 2 n e2x I sin 2a; + n sin (2s + fa)
k= 0,1,2. ~~ »,
2tanssec 2 a;e t a n t a ; ,
17. sec 2 se tajaa; , ^ 7 
(1+s 2 ) 3
Q
""
254
ANSWEBS TO EXAMPLES
18. Velocity, \ak{\ — 2sinAtf); acceleration, — ak2 cos Jet.
At rest when t = —r9
x = — (5TT —
On)
J.J!
Total distance, \ (n + 6 ^/3). o
19. Velocity, e'^2; acceleration, 2e'. 20.
(i) v2 = j92(a2si (ii) /2 = p4(a 2 co
21. Velocity, £2 cos t — U sin t — 6 cos £; Minimum at J = ( 4 i + 1)TT;
Maximum at t = (4i
22. Velocity, — a p ( s i n ^ + sin2_pf); Acceleration, — ap 2 (cos _p^ + 2 cos 2pt); 3a
a
3a
24. 3xy = Oat (1,3); 5a; + y =» 0 at (1,5). 25. a?2y = 0. 27. ooB
29. Maximum. 33. (i) 2004. O*J»
X <J*J
rrU.
"~~ "2*1/
43
32. (i) 0857. (ii) 0515.
(ii) 302.
34. 803. oc<
•€/
2
^^ 6
^^ 1 2
™T" • • •
— 12*^ •
Xt/^"^" ^ "~~ ^2iIC ~~ 1 ~" 7i^ i2/^"^"A^ j (% —" 1 — 2?2<}  / ^ ) J 77,1/^—1) rrs 0 2
44. cx = 6, c2 = 2a6, 45. y = a,
c 3 = 3a2 6  6 3 .
j / ' = 0, y " = cot a, 2
3
46. l + 2a;+2a; + fa + ..., 47. 006285.
45028°.
09930.
ANSWERS TO EXAMPLES
255
49. y' = 1, y" = 1, y"' = 2, y" = 3. 50. / " = 2 + 8/ + 6/, y* = 16y + 40/ + 24/, 2/v = 16+ 136/+ 240/+ 120/, 52. e* f , 53.
IJTT.
54. Jo;2 log x — Jo;2,
log tan \x,
55. sin a;£ sin3 a;,
lo
x {(log a;)3  3(log a;)2 + 6 log x  6}, 56. 12a; + 8 sin 2a; + sin 4a;,
1 — \ tan" 1 x.
\ log I
j tan" 1 (^5 tan %x).
a;3 {9(log a;)2  6 log x + 2},
log & ( ~ W 2 tania; 2a;, \la;/
=^. loga
57. 4 log a;  2 log (1 + x) log (1+ #2)  2 tan" 1 x9 21og2l,
itanH
x tan x + log cos x9 59.———rs, 6(1 — ox)
% tan 3 x — tan x + x.
^tan2o; — xt
(1 + x2) tan" 1 a; — x,
tan 1 (sina;).
60. A, frr, (e 2 "l). log (a;2 + 2a; + 2) + tan"1 (x + 1).
61. (i)  J  + log (^A, A+ #
\ •»• + xj
62. ^ sin 3a;  ^ sin 3 3a;,
2 y tan" 1 (^/5 tan Jo;),
— a;3 cos x + 3a;2 sin a; + 6a; cos x — 6 sin a;,
256
ANSWERS TO EXAMPLES
64. $ 7ra2,
(fa, 0).
65. x sec g + y  sin £  £ sec £ = 0.
69. 2/ = 2 + 3 x  ^ ; 71. £a; — J sin 2x, 2
72. fa ,
(1,4), (  1 , 0 ) ;
f77a (2V2l).
75. ±M
j ^
2
^ cos x — cos #,
2
2
ft.
3
a; sin a; + 2a; cos a; — 2sina;.
73. V,
.
(¥,§)•
77.
i  >)•• 80. o 2 , ( K f a ) ; 83. / x = 1 ,
(a,0),
/ 2 = Jw,
J3 =  ,
84
85>
 V(^r)
86.  1 and 0,
74 =
0 and 1,
2 and 3,
288.
87. 2426.
88. 1844.
89. ^sin^WTT.
90. a;a; 2 + ^ x 3 
I
CHAPTER IX
Examples I: 1. 3cosh3a\
2. 4 cosh 2a; sinh 2a;. 2
3. tanh x + a; sech x.
4. 4 cosh (2a; + 1) sinh (2a; + 1).
5. cosh x cos a; — sinh x sin a;. 6. 2 sech a; sin a; cos x — sech a; tanh x sin2 a;. 7. 3(1 + a;)2 cosh3 3a; + 9(1 + a;)3 cosh2 3a; sinh 3a;. 8. 2a; tanh 2 4a; + 8a;2 tanh 4a; sech2 4a;. 9. cotha;. 11. cosha;e8inhaj.
10. 1. 12. e" tanha; (la;sech 2 a;).
• H ) 4  fan)*= s^3^ #I
'(!<»)
zfi•0I
'feV
L
i . ^
•o „ .g
.8 •—
*Q
1+
=^
'^
IT+
'f Z
I
*8
'——£ii + a;I_qsoo 'i X
• II
— x '\z XZqso°xXZ
W ^  *6I 'XT + XZ^\
sauivxa 01
'LI
258
ANSWERS TO EXAMPLES
Examples III:
2. x =  (a2b2) cos 3 1, y = ha2b2)sin31. o a 3. x = 2asec3£, 2/ = —2atan 3 £. REVISION EXAMPLES V
1. Tangent, xsinifj — ycosI/J — 2a\\ssini/r = 0; normal, # cos ip + ysimfj— 2ai/j cos i/r — 2a sin if; = 0.
4. ( l , i ) . 6. ^ = — 2asin 2t — 2a sin f,
y = 2a cos 2i f 2a cos ^;
speed, 4a cos ^ (numerical value).
13. 2Tra(a;sinh — a cosh — + a I. \ a a
14. 8a.
15. ¥•
16. 1, (0,2).
17. (6«/2a).
18.
20. 1*5.
21. 1??.
6 22. V2(l + 2a: + 2^2)^, 24. ^ira2.
1. 4
(1,1). 26. 1^/3
REVISION EXAMPLES VI
6. fM(x)
= K(z) + 4fl/;(aj) + 2(1 + 2x2)fn(x).
at
a: =
ANSWERS TO EXAMPLES
259
13. f(x), f'(x)f f'{x) continuous everywhere in the range; f'"(x) has a discontinuity at x = 1; Maximum. 16. £.
17. 055.
19. 2n e xv3 cos I a; —^j + arbitrary polynomial of degree n— 1. 21. 18,
45 (radians).
fc
23. r = Of? M
__
i
flti
flu \H'y*
27 ± ~ 777 ^

'dx
=
dx d2y
d2x dy
rlt
i
fl^ii
1
~dildl'
dx ^
fff^
TobY TdxV
'
\dt) dx dy
d2x dy dx d2y d x dt2 dt dt dt2 dy2 idy 2
dx Idy dtj dt'
28. %{b + J(3a2 + b2)}.
29. Equality when ex~x = y.
30. (ii) Minimum.
35. \IT.
37. §77.
38. x = — YTT,
y = —065, maximum;
x = — IQTT,
y = — 1*19, minimum;
x = — TT,
y = 0, inflexion;
23b7r,
?/ = 119, maximum;
x = J^TT,
y = 0*65, minimum;
x = ^77,
i/ = 073, maximum;
x = iltr,
y = 044, minimum.
x =
182
260
ANSWERS TO EXAMPLES
41. (i) Maximum, y = ^ 2 — 1 ) ;
minimum, y = —1(^2+ 1).
(ii) Inflexions, (1,0), {2 + ^/3, i(3,/3)}, and{2V3,
d2rj
drj
' dx
d£_ dv^ ' 1 — 7^ tan a
2
d y __ dx* /
d^ drj . \ 3 # cos a — ~ sm a
50. a; = 0, maximum, /c = —16; x = f, minimum, * = V^51. 3a cos £ sin t (numerical value of). 52. (0,0), (2,2). 53. l, 2^2. 56. Minimum.
(a cos bx + b sin bx), \x2 — \x <J(x2 — 1) + \ cosh" 1 x. 60. (b) —, A an arbitrary constant. x 61. a 2 (a + 3sina), where cos a = —% (^TT < a < TT). 62. (i) Y5 log sinx — YsX cos #(3 cosec5 x + 4 cosec3 a; + 8 cosec #) — ^o cosec4 X — TS cosec2 x. ... 65 aa
(ii) ^ v 6
63. 4a«
6
/a + bx\ \ x I
64 a5x
63 2a*x2
62 3a 3 x 3
6 4a2a;4
1 5az 5 '
i.NSWEES TO EXAMPLES
64. (i) n = 4k,X=l;n
261
= 4:k+l,X = n;
n= 4&+2.A =  1 ; n = 4& + 3, A = n. (ii) (l + inW. 65. (tan 1 x)2  2x tan' x + x tan 1 x log (1 + xz) + log(l+a; 2 )£{log(l + a;2)}2. 68. <£(*) = (l+x)(3 + x). 71. ^ =  , B = f, C = f, D = f. 72. ^ = ^,J5=iC =
^,i)
76.
3
77. cos3"1 x{psinP1 x(p + q + k2+pk2)sin^+1 x
79. Volume 5?r2a3, area 81. x = a(cos ^ cos 3£ + 3 sin t sin 3^), y = a(cos ^ sin 3£ — 3 sin t cos 3^).
85. P (  8 a £ 3 ,  6 a O ; Normal, 4 ^ + 2/+6a« + 32a^5 = 0; Inflexion a t origin. 86. Normal, 2x + Sty  Sat*  2at2 = 0; Centre of curvature, (— f at* — at2,4at3 + %at); Radius 87. Envelope, x = f'(t), + t2)K P=f"(t)(l
y
=f(t)tf'(t);
88. Envelope, x = a sin 2(3  2 sin 2 1), y — a cos t(S — 2 cos 2 1). 89. Tangent, x sin t — y cos £ + cos 2t = 0; Normal, a; cos £ + y sin t — 2 sin 2£ = 0.
262
ANSWERS TO EXAMPLES
90. Tangent, t(3 + t2)x2yt* = 0; Normal, 2x + t(2 + t2)yt2(2 + t2) = 0. 92. 7r(abcdk).
93.
94. ± .
95.
98. 2\
102. 2csinh; 7rc(2a + csinh — c \ c 2 2 105. Volume, 2>rr a h; area
103. ^?ra2. 4/2 106. *a, 7T
«2 107. / + "
2 — log(l+V2).
O/
TTd
1O9 a
'

CHAPTER XI
Examples I: 1. 1 and 3; 2 + ^/3; 2±i. 2.  5 and ~ 3 ;  4 + ^/5;  4 ± 2 i . 3.  1 and 3; l±^\
1 ± 3i.
Examples II: 1. 10 + 3*. 2.  8 + 8J. 3. 34 + 22£. 4.  1 6  3 i . 6. 0 + 0i.
7. 0 + 2i.
8. 2+ Hi.
9.  2  1 6 i .
5. 2 + 7i. 10.  1 0 + 0i.
Examples III: 1.  7 + 22i.
2.
3. 1L
4.
5.  3 + 4i.
6.
7. 10.
8.  4 6 + 9i
Examples IV: 1. i. 4. iVfK
2.   + 2V. 5. ff + ffi.
3. ffi. 6. oosO + imnO.
ANSWERS TO EXAMPLES
263
Examples V: 1. 52i,

2.  7 + 9i,  l  5 i ,  2  3 4 i , H + i K 3. 4 + 2i, 4  2 i , 8f, 2i.
4. 3 + i, 3 + 3i, 2  3 i ,  2 + 3£.
5. 4, «, 13,  A + ift.
6.  6 , 8i, 25,A —if f.
9. Jf.
10. 
11.  2 ± 3 i .
12.
l±i.
13. 3±i.
14.
±i.
15. 4±3i.
16.
2±t.
Examples VII: 3. 2,30°; 5, 53° 7'; 13,112° 36'; 3,0°; 10,53° 7'; 2,90°. 5. Straight line, 4x+ 10^21 = 0. Examples
VIII:
1. (a) (3,2); (6) (2,1); (c) (4,7). 2. (i) (a) (1,0); (b) ( 3 ,  5 ) ; (c) ( 3 ,  5 ) . (ii) (a) ( 2 ,  1 ) ; (6) (2,1); (c) (5,0). Examples IX: 1.
(i) 05 + 0866i. (ii) ± (0866 + 05i). (iii) 0940+ 0342i,  0766 + 0643i,  0174 0985i. (iv) ± (0966+ 0259i), ± (02590966i).
264
ANSWERS TO EXAMPLES
2.
(i) 7 + 24i.
(ii) ±(2121+ 0707*). (iii) 1671 + 0364i, 1151 + 1265*", 0520l629i. (iv) ± (1476+ 0239i), ± (02391476*). 3.
(i) 119120i. (ii) ± (3535 + 0708*). (iii) 2331+ 0308i, 1432+1865*, 08992173*. (iv) ±(1890 + 0187*), ±(01871890*).
Examples X:
5TT
3. 4. 2(cos^(12fcl) + isin^ (
5.
lo
18
(
6. „
... ,
2TT
. . 2TT
4TT
. . 4TT
7. (I) 1, cos —±*sin—, cos —±*sin—. 5 5 5 5
(ii) ±1, ±i(l±t'V 3 )Examples XI: 1. e(in+2kn) { cog (1 l o g 2 ) + { gin (J log 2)}. 2.  2e(2***ir) {sin (J log 2) + i cos (J log 2)}. 3# e(&n+kn)  c o s (  iOg 2) + i sin (  log 2)}. 4.  8e(4A;7r*;r) {cos (log 4) + i sin (log 4)}. 5. 26 (A:7r ^ ) {cos (£ log 2 (/ r
)
JTT) +
i sin (  log 2 
6.  4e " +** {cos (J log 2) + i sin (J log 2)}.
TT)}.
ANSWERS TO EXAMPLES
265
Examples XII: '
1 — cos 0 + cos 7i0 — cos (n + 1) 9 # 2 ( 1  c o s 0) gin 0 4  f 
1 + 2x cos 6 + x2 3. cosx + x sinx. 4. ^e^sina; — cos a;). 6. ^xe2x(2 sin x — cos #) + ^e 2 ; r (4 cos a: — 3"sin x). 7.  sin 3# — \x cos 3a;. 8. £sxe* x (4: cos 3a:  3 sin 3a;) ^>s e ~* x 0 c o s %x
REVISION EXAMPLES VII
sin a cos a ~~ cos 2 a + sinh 2 6'
— sinh6cosh6 cos 2 a + sinh 2 b *
5. — tan—(4&— 1),fc= 1, ...,n. 4n 7. cos2 a; + sinh 2 y. =
(a:2 + yj) a;2 + (a:2 + yj)x1 (xt + a:2) (4 + 2/?) ( 4 + y\) ~ 2(^x^2 " 2 ) + 1# (a:2 + y\) {x% + y\) 
10. x — iy9r(cos6 — isin6);
2(Xlx2±l±3i.
11. zx = 2(c
16.
(i) (  1,0), (3,0); (ii) circle, centre (2,0), radius 3; (ill) ellipse, foci (1,0), (2,0), eccentricity  .
17. 1, co, co2, o>3, co4 where o) = cosfTr + isinf n; x = J(l + i cot ££77), k = 1,2, 3,4. 19. (1,^/3), distance 2.
2 4sin3
^)«
266
ANSWERS TO EXAMPLES
20. «8 = 7 + 273 + 1(4 + 373) or 7  2 7 3 + i(4373). Vertices (4 + 4 sin \1CTT + 6 cos \kiry 6 + 6 sin \JCTT — 4 cos \1CTT), 21. cc
23. (<j3 + i)(a + ib) and a + ib subtend an angle ^TT at the origin, and the distance of (<j3 + i)(a + ib) from the origin is twice that of a + ib from the origin. C is at ±273 + 2%. 24. ±(3 — 2i); l + i , 72cosYTT + i72sinJ^TT,
72 cos jf 77 + i 72 sin yf TT. 26. 23002100*. 27.
(i) (1,1), ( 1 ,  1 ) , (ii) all points with x^ 0, (iii) interior of circle, centre ( — f, 0), radius f, (iv) interior of ellipse, foci (± 1,0), eccentricity £.
29.
(i)if
I;
i, " ;
1,
coscxcos(a —j3) V
30. tV,
/
___ 1 Q
^
coscxsin^(a —j8)^ >
^_„ 1 O
COS i* ^
>
e
32. l67036i, 052+l63i, 115127*. 33. i
—i (sin 19  7 sin 59 + 21 sin 30  35 sin 9). 37. Circle, centre (5,  1 ) , radius 3; 8, 2. 38
V(10 + 6y) ' " 1062/1*
COS i
ANSWERS TO EXAMPLES
267
39. Amplitude increases from — n to IT. 40. 2*cos7T, — 2*COST7, — 2* cos f jr.
cos x cosh y + i sin a; sinh y cos2 a; + sinh 2 y 42. CHAPTER XIII
Examples I: 1. 1.
2. ^/2.
3. 2.
4.
5. TT.
6. .
7.  I / a ( a < 0 ) .
REVISION EXAMPLES VIII
1.  l o g ^  ; ^r + J l o g i ^  9 ; coSi(2x). *C — x
JO — X
l^""3^
2.
X —
3.
4.  ( x + 6)s +  ( a — 6) (x + 6)*; § cos 3 # — cosx — \ cos 5 x.
+ 4a: + 5)  2 log {# + 2 + V(z2 + 4a; + 5)}; £(a + a;2)  a 2 log (a2 + x2)  ^a 4 /(a 2
2 N/(z 2
a;{(log x)2  2 log x + 2}.
REVISION EXAMPLES IX _
3
.... 1
(
<"> V a I
JTT.
268
ANSWERS TO EXAMPLES
JLi ^2
8
8. ITT;  X / V ( X 2  1 ) ; ^ 4 (2x 3 3x)sin2x^(2a; 2 I)cos2a; 9. 4 = 2w.
11. 1  J T T ;
fw;
12.  (   l o g 3 ) . 10 y
Z
2a(a 2 6 2 )(a 2 c 2 )'
14. J77 sec \a, \n sec a.
16.
fc 2tan1 ( (  ^ ) tan] + S, r< 1,
17. 1
 2 tan1 {£t) tan) + 8, r > 1.
Limits iT + S , r < l ; —7r + S , r > l ; 7 = S when r = 1. 18. a; cos a + sin a log sin (x — a); 1
11 * J. 1 , 2 t + l \log——jtan —j—,
19. lo 20
1  ji=W)
21.
Jx (
Jog
, , . where t = tanx. x
ANSWEKS TO EXAMPLES
269
23. ITT.
2 tan0 — sec#= —  —
^ , differing by a constant.
M
_2=(5
+
27. (7i2)^ n _ 2 ~(72,l)^ n = 0, n>2;
31. If the given integral is un+1, then
33.
u
n2
a 35. 77/^(a — 1); u0 = log (1 + cosec  a ) , % = u0 cos a — 1 + 2 sin Jo 2
36. un = nun_l9 un = n\\ un +
un_2 =
38.   ^ t a n  1 ! ^  ) . 40. (n\ 41.
3(2a; 2 +l) 3
42. Formula: see p. 208. 4)  i log {x + 1 + J(x* + 2x + 2)} +1 (2X*  5x + 7) 4(x* + 2x + 2).
270
ANSWERS TO EXAMPLES APPENDIX
Examples I: 1. 4:X3ys, 3x*y2, \2x2yz,
12xzy2, I2xzy2,
6x*y.
2
2. y , 2xyy 0, 2y, 2y, 2x. 3. 2a:, 2y, 2, 0, 0, 2. 4. excosyf —exsinyy excosy, —exsiny, —exsiny, —excosy. I
l
l
1
1
1
5.
7. 3a;2 sin 2 1/, 2#3 sin i/ cos y, 6xsin 2 2/, 6a:2 sin y cos y, 6a;2 sin y cos i/, 2o;3(cos2i/ —sin2i/). 8. yexysmx + exycosx,xexv8inxf y2 exy sin a; + 2yexy cos a; — exy sin a;, eX2/ sin x + #2/eXI/ sin a; + xexy cos a:, eX2/ sin x + a^e^ sin x + a:eX2/ cos a:, a?2 e ^ sin a:. 9. t a n y, j ^ ,
0, r ^ , y  ^ ,  ( 1 + y a ) s 
10. sec a; tan a;, sec y tan y, sec x tan 2 a; + sec3 xf 0, 0, sec i/ tan 2 y 4 sec31/. 11. exsin22y9 4ex sin 2y cos 2y, exsin22y, 4:exsin2ycos2y, 4e* sin 2y cos 2?/, Sex cos2 2y  8e^ sin 2 2y. 12. c^, a:e^, 0, ey, ey, xey. Examples 3.
II:
(i) aeax sin (by + cz), — b2eax sin (by+ cz), aceax cos (by+ cz); (ii) yze~x%2x2yze~x\
0, ye~x*
INDEX Approximations, 5562 Area of closed curve, 128 Argand diagram, 168, 1714
Formula of reduction, 20016 for infinite integrals, 220 Gradient angle \jjt 107
Cauchy's form of remainder, 47 Centre of gravity of semicircular arc, 133 Complex numbers, 158—91 Argand diagram, 168, 1714 argument, 169 conjugate complex numbers, 161 De Moivre's theorem, 174 differentiation and integration, 186 logarithm, 183 modulus, 162, 169 numberpair, 166 powers, 178 pure imaginary, 161 quotient, 164 roots of unity, 177 sine and cosine, 184 sum, difference, product, 163 Convergence of series, 42 Coordinates, intrinsic, 111 pedal, 111 Cosh a;, 84 Curvature, 113 circle of, 119 Newton's formula, 117 Curves, 10033 angle 'behind' radius vector, 109 gradient angle, 107 length, 103 parametric form, 116 sense of description, 101 De Moivre's theorem, 174 Differentiation, logarithmic, 10 partial, 236 Envelopes, 123 Evolutes, 126 Expansion in series, 413 Exponential function, 1820, 217
Hyperbolic functions, 8499 inverse, 969 Integrals involving 'infinity', 21725 Integration, systematic, 20016 involving surds, 203, 207 polynomials, 200 rational functions, 201 Intrinsic coordinates, 111 Irrational number, 158 Lagrange's form of remainder, 47 Leibniz's theorem, 62 Length of curve, 1026 Logarithm, 118, 27 complex numbers, 183 in differentiation, 10 series for, 53 Maclaurin's series, 49 Modulus, 162, 169 of e\ 186 Newton's approximation, 56 Newton's formula, 117 Pappus, second theorem of, 132 Partial differentiation, 236 Pedal coordinates, 111 Pedal curve, 112 Powers, complex, 178 Rational functions, 11 integration of, 11, 200 number, 158 Remainder Cauchy's form, 47 Lagrange's form, 47 Roots of unity, 177
272 Series, binomial, 51 convergence, 42 exponential, 54 logarithmic, 53 Maclaurin's, 49 oscillating, 43 for since, 38, 50 for sinha;, cosh#, 86 sum to infinity, 42 Taylor's, 3862 for (1 +x)~\ 40
INDEX Sine and cosine of complex numbers, 184 Sinh#, 84 Taylor's series, 3862 Taylor's theorem, 44 Trigonometric functions, integration of, 214