An Analytical Calculus: Volume 2: For School and University

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 cos|0 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

8-2

108

CURVES

But

\axj -r-l = 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 counter-clockwise 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 y-axis ? 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 counter-clockwise rotation through an angle 0, it lies along the radius OP produced. A further counter-clockwise rotation <j> brings it to the tangent to the curve, in the POSITIVE sense; this defines . If we assume, as usual, that 0 is the parameter, then the positive sense along the curve is that in which the length increases with 0. Hence the angle lies between 0 and 77, as the diagrams (Figs. 80, 81) indicate. By definition of ifs (p. 108) we have the relation with possible subtraction of 2TT if desired. It is important to remember when using this formula that 0 is the parameter used in defining ifs. To find expressions for sin<£, cos<£, tan :

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 sin 0 + sin ^ cos 0.

= 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 lies between 0, TT, the value of sin is necessarily positive. We may retain the formula p = rsin

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 = ifi—6 depends on the conventions adopted when 0 is the parameter. If another parameter (for example, x) is used, the sign attached to p may require separate checking. 10. Other coordinate systems. The Cartesian coordinates (x, y) and the polar coordinates (r, 6) are by no means the only coordinates available for defining the position of a point of a curve. Others in use are the intrinsic coordinates (s9ift) and the pedal coordinates (p, r). The passage between Cartesian and polar coordinates is familiar. To pass from Cartesian to intrinsic coordinates, we have the relations dx

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 ± is the angle 'behind' the radius vector for the locus of N,

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 = 7-1sin^1 = p sin cf> and so the pedal coordinates (pl9 r±) of N are (p2/r,p). COROLLARY.

Since t a n ^ 1 = ^ - ^ - , and <£ = l9 we have the

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 semi-axes 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*

(y-bf +

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 counter-clockwise 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?-{x-u)% 2/ where the positive sign is attached to the square root. Hence

-(x-u)

y

{a2-(x-uyf9 •*2

so that , and

„

(x-u)2

-1 2

2

y = {a -(x-u) 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. 9-2

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

x-yt + 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

JM-ftt)

{

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+(l-t2)y

+ (l + t2)a = 0.

5. tx + y-a(tz + 2t) = Q. 6. tzx-ty-c(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 COUNTER-CLOCKWISE 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 counter-clockwise 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 counter-clockwise 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

-^Uxdy-ydx)

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 y-a>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 y-axis. 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 y-axis. 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(l-z2)*

= y and (l + z 2 )/(l-z 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 (x2-2x + 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 (m2-l) . (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+2-tfn+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 4-point 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 4-point 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 /(&)-/(*) = (b-a)f'{a+8(b-a)}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 y-axis. 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 (x-l) 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 right-angled 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 + y2-6y + 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 _ e-kx^ y _. _e-kx a n ( j y _ e-kx 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 + 3x2-2xy + 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)dt-e* f V then

(i) F(0) = 0, (ii) J"(0) = 0,

(iii) I"'{z)-3F'(x) + 2F(x) =/(*). (6) Find the most general function {x) which is such that

I

t(t)dt = x2cf>(x).

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 J-i

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 (x) is any polynomial of degree m, 0

where

aw =

^

^

75. If m, n are positive integers greater than unity, prove that 7mw=

cosmxGO&nxdx

Jo m

m m

m-n

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 x-t2y+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 Z-nab. 98. Find the area of the loop (— 1 < £ < 1) of the curve _ l-t2 X

_t(l-t2) 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 277-2/ 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 87r-2>/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 x2-6x =-5, x2-6x + 9 = - 5 + 9, = 4, 2

(x-3)

= 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

x2-6x

= -7,

2

x -6x + 9 = - 7 + 9, = 2, (x-S)

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 text-books 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 x-3 = -<]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 number-world. 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 left-hand side is 1+t 2 = 0 = right-hand side, so that the solution is verified. We now investigate the elementary properties of our extended number-system, 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. x2-4:X + 3 = 0,

x2-±x+l

= 0,

x2-4x + o = 0 .

2. x2 + 8x+15 = 0,

#2 + 8a;+ll = 0,

z2 + 8a; + 20 = 0.

3. x2-2x-3

z2-2z-4

x2-2x+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 =

-.(c-c).

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) + (7-2i).

2.

3. (4 + 5i)(6-2i).

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.U-t. 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)(2-6i). 3. ( _ i _ ; ) ( _ 3 + 4i). 4. 5. i(4 + 3i).

6.

7. (3-f)(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

9-i 2 10

"

9+1

DIVISION

165

EXAMPLES IV

Express the following quotients in the form a + ib (a,b real): 1. -—-.. l-i 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: 3-5i. 3. 4, 2i. 5. 2 + 3i,

2. 4. 3 + 2i, - i .

2-3i.

6. - 3 + 4i,

-3-4^.

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. x2-8x + 25 = 0.

16. x2 + 4x + 5 = 0.

6. The complex number as a number-pair. 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 number-pair 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 number-pair (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 number-pair [a + u,b + v];

THE COMPLEX NUMBER AS A NUMBER-PAIR

(iii) The product

r

...

r

167

n

is the number-pair [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 number-pair [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 number-pairs 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 number-pair 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 'number-pair' [a, b] in which the first component of the pair corresponds to the real part and the second component to the imaginary. A number-pair 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 number-pair 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 z-plane.

EXAMPLES VI Mark on an Argand diagram the points which represent the following complex numbers: 1. 3 + 4i.

2. 5-i.

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 x-axis, measured in the counter-clockwise 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\ =

\z-5i\.

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==z2-zl9 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 x-axis 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 modulus-argument 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 »

LxOP-LxOP9

= 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 counter-clockwise 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 counter-clockwise 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) 3-5i,

(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 . If so, then

cos 6 + i sin 9 = (cos + i sin )* = cos g<£ + i sin #<£ [by (i)].

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 right-hand 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 + 2k-rr)

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 right-hand 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 modulus-argument 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

2k-n . . 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 'modulus-argument' 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 g-valued 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 left-hand 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 (non-zero) 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 over-emphasized 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 many-valued 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. (l-iV3)3+2*.

5

3. (V3 + »)H.

- (^

18, The logarithm of a complex number, to the base e. We have seen (p. 181) that any (non-zero) 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, e-ix

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 + e-i2!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. 13-2

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 = 1-cos 0(2 cos 0) +cos 2 0 = l-cos 2 0 = sin20. Thus C + ttf = -J^

{(1 - cos* 0 en<*) (1 - cos 0 e-'*)}

sm20 l Equating real parts, we have C = -r-oTT {l-COS n 0COS7l0-COS 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(ri-l)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 counter-clockwise 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 (a-fit) 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# = e-v 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 = x1-i-iy1, 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 (z-l)» = 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: 1-k; (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 = — (Z-f 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) z2-2z = 3, (ii) | z - 2 | = 3, (iii) | z- 11 + | z-2 \ = 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

a-j8

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: z-2 (i) 2* + 2 = 23f (ii) z + 2 2-1

(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 mid-point 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 60-asm 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(2K2-I

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 text-book should be consulted. 1. Polynomials. The first type presents no difficulty. If f(x) is the polynomial f(x) = aoxn + axxn-1 +... + 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(ab-h2). C dx _ r qP^dt^ab-h2) 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 -(2p-l)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 ? Ip-2> • ••> &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'2-q2

(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. 14-2

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, ab-h2> 0,

(ii) J(x'2 -q2)

a> 0, ab-h2< 0,

(iii) <]{r2-x'2)

a<0,ab-h2<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(x2-q2). 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 <](r2-x2).

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

r2-x2 =

(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(r2-x2), 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 right-hand 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 4-6,

/? = 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 right-hand 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 left-hand 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 a-ft dt t+l '(t+l)2 r 2 n (u t + v') (t+l)2n ' t+l __ [(ot-P)(Ct + D)(t+l)2n-2dt

=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/ «t-y j

i'iltr

] 1/0)1 71

^M/ — yu/U — U/ vj /if

zk~1dz yl[(l-v'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; = -—•—r-f-. 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. 106-10. 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 discussJ-cc 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 well-known 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 = —xe-x+\ rN r xe~xdx=

so that

Now

l.e~xdx

IN

-(x+l)e-x\

WW-Z£

^ __+...

and, for positive values of N, the denominator is certainly greater than %N2, so that (N+l)e-N-+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. 15-2

220

INTEGRALS INVOLVING 'INFINITY*

Care must be exercised in selecting the correct angles from the many-valued 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 {tan-1 i f } - lim {tan-1 ( -

=

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

xme-xdx> 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 = —xme-x\ +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(m-l)J m _ 2 ra(ra-l)...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.

J-ool+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 *c-e

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 rc-e 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 l-n

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

x-l = 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)

/ = ITT-0

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 starting-point 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 end-points, 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{(x-l)(3-x) )Y

X

2. Integrate j — . — = -j—z—rjr cos" 1 -1 (6 < a).

Show that

<J(a2-b2)

Jo a + bsmx 3. Evaluate the integral

\a/ v

'

Cb___xdx___ Ja<j{(x-<*)(b-x)} by means of the substitution x = a cos2 0 + b sin2 0. Prove that the integrals

fV(l-a;) 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^sin-x)^,,

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 1-cosa

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^ji-cw (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

l4-cosasin# —^— ,

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) w

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 =

xne-axcosbxdx,

Jn=

xne-ax&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

16-2

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 single-valued 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 8z-7-k 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 y-directions respectively. EXAMPLES II

1. Show that the function with the z-axis 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 z-axis measured vertically upwards, is represented by a 'bowl-shaped' 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(r2-y2), ,., and then

8x

r

— = -r—^

r rr = - .

dr J(r2-y2) 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-\-nxn-1\ogx.

18. 2xj(l+x2).

19. a; log a;-a;.

20. ^(loga:)2.

21. log sin x.

22. ia-logx-lx-.

(

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;2-2a;+17). 28. - 12 tan- 1 (x + 3) + log (a;2 + 6a; +10).

29. 9tan- 1 fcp)+ flog{x2-8a; + 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 = = - ++ccotx-o 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 4-1)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(x-l) !)(»-3)' 13. ilog ( « -(z-2) 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(asinx-ncosx)

+

n(n-l)In_2.

3. (a2 + 6 2 ^ 2 )7 n = eax cos71"16a;(a cos bx + nb sin 6z) + n(n — l)b2ln_2. 4. 120-44e.

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' 1-a;

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;), 2-24.

.

21og2-f. 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.

^773-127T + 24.

(i) ^ + | l o g ( l + V 2 ) -

56.. (m + n)

ex(x— 1).

(2) 1,

K4V2-5).

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. 2-005.

2. 2-995.

3. 2-0017.

4. 2-999.

5. 1-9996.

6. 0-3328. 17-2

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;4-l) 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'I-6-T+"""'

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+(m4-W')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. 3x-y = Oat (1,3); 5a; + y =» 0 at (-1,5). 25. a?-2y = 0. 27. ooB

29. Maximum. 33. (i) 2-004. O*J»

X <J*J
rrU.

"~~ "2*1/

43

32. (i) 0-857. (ii) 0-515.

(ii) 30-2.

34. 8-03. 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. 0-06285.

45-028°.

0-9930.

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.

I-JTT.

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 tan-ia; -2a;, \l-a;/

=-^-. loga

57. 4 log a; - 2 log (1 + x)- log (1+ #2) - 2 tan" 1 x9 21og2-l,

itan-H-

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 (2V2-l).

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,

2-88.

87. 2-426.

88. -1-844.

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; (l-a;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°x-XZ

W ^ | *6I 'XT + XZ^\

sauivxa 01

'LI

258

ANSWERS TO EXAMPLES

Examples III:

2. x = - (a2-b2) cos 3 1, y = -ha2-b2)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. 0-55.

19. 2-n e xv3 cos I a; —-^j + arbitrary polynomial of degree n— 1. 21. 1-8,

4-5 (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 = — Y|TT,

y = —0-65, maximum;

x = — IQ-TT,

y = — 1*19, minimum;

x = — |TT,

y = 0, inflexion;

-23b-7r,

?/ = 1-19, maximum;

x = J^TT,

y = 0*65, minimum;

x = ^77,

i/ = 0-73, maximum;

x = il-tr,

y = 0-44, minimum.

x =

18-2

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{-2-V3,-

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{psinP-1 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)x-2y-t* = 0; Normal, 2x + t(2 + t2)y-t2(2 + t2) = 0. 92. 7r(ab-cdk).

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. 1-L

4.

5. - 3 + 4i.

6.

7. 10.

8. - 4 6 + 9i

Examples IV: 1. i. 4. iV-fK

2. | | + -2V. 5. -ff + ffi.

3. f-fi. 6. oosO + imnO.

ANSWERS TO EXAMPLES

263

Examples V: 1. 5-2i,

-

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) -0-5 + 0-866i. (ii) ± (0-866 + 0-5i). (iii) 0-940+ 0-342i, - 0-766 + 0-643i, - 0-174 -0-985i. (iv) ± (0-966+ 0-259i), ± (0-259-0-966i).

264

ANSWERS TO EXAMPLES

2.

(i) 7 + 24i.

(ii) ±(2-121+ 0-707*). (iii) 1-671 + 0-364i, -1-151 + 1-265*", -0-520-l-629i. (iv) ± (1-476+ 0-239i), ± (0-239-1-476*). 3.

(i) 119-120i. (ii) ± (3-535 + 0-708*). (iii) 2-331+ 0-308i, -1-432+1-865*, -0-899-2-173*. (iv) ±(1-890 + 0-187*), ±(0-187-1-890*).

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(4-373). 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 , 72cosY|TT + i72sinJ^TT,

72 cos jf 77 + i 72 sin yf TT. 26. -2300-2100*. 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,

cos|cxcos|(a —j3) V

30. tV,

/

___ 1 Q

^

cos|cxsin^(a —j8)^ >

^_„ 1 O

COS i* ^

>

e

32. l-67-0-36i, -0-52+l-63i, -1-15-1-27*. 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) ' "| 10-62/1*

COS i

ANSWERS TO EXAMPLES

267

39. Amplitude increases from — n to IT. 40. 2*cos|7T, — 2*COS|T7, — 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 ; coS-i(2-x). *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| 2tan-1 ( ( | ^ ) tan|] + S, r< 1,

17. 1

- 2 tan-1 {£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-—-—j-tan —j—-,

19. lo 20

1 - ji=W)

21.

Jx (

Jog

, , . where t = tan|x. x

ANSWEKS TO EXAMPLES

269

23. ITT.

2 tan0 — sec#= — - —

^ , differing by a constant.

M

_2=-(5

+

27. (7i-2)^ n _ 2 ~(72,-l)^ n = 0, n>2;

31. If the given integral is un+1, then

33.

u

n-2

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, 55-62 Area of closed curve, 128 Argand diagram, 168, 171-4

Formula of reduction, 200-16 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, 171-4 argument, 169 conjugate complex numbers, 161 De Moivre's theorem, 174 differentiation and integration, 186 logarithm, 183 modulus, 162, 169 number-pair, 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, 100-33 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, 41-3 Exponential function, 18-20, 21-7

Hyperbolic functions, 84-99 inverse, 96-9 Integrals involving 'infinity', 21725 Integration, systematic, 200-16 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, 102-6 Logarithm, 1-18, 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, 38-62 for (1 +x)~\ 40

INDEX Sine and cosine of complex numbers, 184 Sinh#, 84 Taylor's series, 38-62 Taylor's theorem, 44 Trigonometric functions, integration of, 214